Kindergarten of Fractional Calculus [1 ed.] 1527544982, 9781527544987

Table of contents :
Dedication
Table of Contents
List of Photographs
List of Figures
List of Tables
Foreword
Preface
Acknowledgements
1 Concept Generalisation
2 Fractional Integration
3 Fractional Derivatives
4 Fractional Differ-integrations’ further Extension
5 Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms
6 Fractional Differential Equations and their Analysis
7 A Formal Method for Solving Fractional Differential and Fractional Integral Equations
8 Fractional Partial Differential Equation
9 Modified Fractional Calculus in conjugation with Classical Calculus
Appendix A: Higher Transcendental and Special Functions
Appendix B: List of Laplace transform pairs of functions related to fractional calculus
Appendix C: Fractional derivatives and integrals of some important functions
Appendix D: Formulas of Modified Fractional Derivative
Appendix E: Branch-points, Branch-Cuts and Riemann Sheets for Multi-valued Functions
Appendix F: Mittag-Leffler Function with Negative Order
Appendix G: Inverse Laplace Transform by Contour Integration
Bibliography

Citation preview

Kindergarten of Fractional Calculus

Kindergarten of Fractional Calculus By

Shantanu Das

Kindergarten of Fractional Calculus By Shantanu Das This book first published 2020 Cambridge Scholars Publishing Lady Stephenson Library, Newcastle upon Tyne, NE6 2PA, UK British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Copyright © 2020 by Shantanu Das All rights for this book reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner. ISBN (10): 1-5275-4498-2 ISBN (13): 978-1-5275-4498-7

Dedicated to my mother, the late Smt.Purabi Das (1934-2018), and my blind father, the late Sri.Soumendra Kumar Das (1926-2009), as well as all of my school and college teachers.

TABLE OF CONTENTS

List of Photographs........................................................................................................................................................ xxi List of Figures............................................................................................................................................................... xxii List of Tables............................................................................................................................................................... xxiv Foreword ...................................................................................................................................................................... xxv Preface ......................................................................................................................................................................... xxvi Acknowledgements .................................................................................................................................................... xxxv Chapter One...................................................................................................................................................................... 1 Concept Generalisation 1.1 Introduction ........................................................................................................................................................... 1 1.2 The thought problem for constructing a ‘half-coin’: an example of concept generalisation................................. 1 1.2.1 The idea of Negative Probability through the generalisation of existing theories........................................ 1 1.2.2 The tossing of coins and Probability Generation Functions (PGFs)............................................................. 2 1.2.3 Generalizing the PGF to get a half-coin construct: a paradox? .................................................................... 2 1.3 The question posed by L’Hospital to Leibniz ....................................................................................................... 3 1.4 A recall for analytic functions............................................................................................................................... 4 1.4.1 Real analytic functions ................................................................................................................................. 4 1.4.2 Complex analytic functions.......................................................................................................................... 4 1.5 Recalling the Cauchy integral formula.................................................................................................................. 5 1.5.1 The derivative is an integration process ....................................................................................................... 5 1.5.2 The Taylor series from Cauchy’s integral formula....................................................................................... 5 1.6 Singularity pole branch point and branch cut........................................................................................................ 6 1.6.1 Isolated singularity ....................................................................................................................................... 6 1.6.2 Pole singularity (removable singularity) ...................................................................................................... 6 1.6.3 Essential singularity ..................................................................................................................................... 6 1.6.4 Multivalued functions, branch points and branch-cuts ................................................................................. 7 1.7 Recalling residue calculus..................................................................................................................................... 8 1.7.1 Finding the residue by power series expansion ............................................................................................ 8 1.7.2 The formula for calculating residue.............................................................................................................. 9 1.8 The basics of analytic continuation..................................................................................................................... 10 1.9 The factorial and its several representations ....................................................................................................... 11 1.9.1 Representation of a falling factorial ........................................................................................................... 11 1.9.2 Double factorial.......................................................................................................................................... 11 1.9.3 Truncated factorial ..................................................................................................................................... 11 1.9.4 Some other historical representations of factorials..................................................................................... 12 1.9.5 Increasing truncated factorials (Pochhammer numbers) ............................................................................ 12 1.9.6 Euler’s integral representation for factorials .............................................................................................. 13 1.10 The gamma function ......................................................................................................................................... 13 1.10.1 The recurring relationship of the gamma function ................................................................................... 14 1.10.2 Factorials related to the gamma function ................................................................................................. 14 1.10.3 Calculation of the factorial of non-integer numbers by gamma function ................................................. 14 1.10.4 Poles of the gamma function .................................................................................................................... 15 1.10.5 Extending the gamma function for negative arguments by analytic continuation.................................... 16 1.10.6 Representations of the gamma function via the contour integral (Hankel’s formula) .............................. 17 1.10.7 Ratio of gamma functions at negative integer points ............................................................................... 18 1.10.8 The duplication formula for the gamma function..................................................................................... 18 1.10.9 Reflection formula of the gamma function............................................................................................... 19 1.10.10 Asymptotic representation of the gamma function................................................................................. 19 1.10.11 Reciprocal gamma function.................................................................................................................... 19 1.10.12 Representing the gamma function as a limit form.................................................................................. 20 1.10.13 Some interesting values of gamma functions that we frequently encountered....................................... 21

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1.11 Ratio of gamma functions ................................................................................................................................. 21 1.12 Generalising binomial coefficients by gamma function.................................................................................... 23 1.13 Incomplete gamma functions ............................................................................................................................ 26 1.13.1 Tricomi’s incomplete gamma function and the complementary incomplete Gamma function ................ 26 1.13.2 Incomplete gamma function ..................................................................................................................... 26 1.13.3 The elementary relationships of Tricomi’s incomplete gamma function ................................................. 26 1.14 Power numbers.................................................................................................................................................. 28 1.15 The beta function and the incomplete beta function ......................................................................................... 28 1.15.1 Integral representations of the beta function ............................................................................................ 29 1.15.2 The derivation for the reflection formula of the gamma function ............................................................ 29 1.15.3 Application of analytic continuation to the reflection formula................................................................. 30 1.15.4 Incomplete beta functions......................................................................................................................... 31 1.16 Psi function ....................................................................................................................................................... 32 1.16.1 The recurring series formula for psi function ........................................................................................... 32 1.16.2 Analytical continuation of a finite harmonic series by using psi function................................................ 32 1.17 A unified notation to represent multiple differentiation and integration........................................................... 33 1.18 Differentiation and integration of series............................................................................................................ 34 1.19 Revising the concepts of mixed differentiation and integration........................................................................ 34 1.20 The lower limit as an important factor in multiple integrals ............................................................................. 38 1.21 Generalising product rule for multiple integration and differentiations............................................................ 40 1.22 Chain rule for multiple derivatives ................................................................................................................... 42 1.23 Differentiation and integration of power functions: unification & generalisation ............................................ 43 1.24 Non-differentiable functions: the motivation for fractional calculus ................................................................ 44 1.24.1 Hurst exponent ......................................................................................................................................... 44 1.24.2 Self-similarity........................................................................................................................................... 45 1.24.3 Nowhere differentiable functions ............................................................................................................. 45 1.25 Short summary .................................................................................................................................................. 46 1.26 References......................................................................................................................................................... 46 Chapter Two ................................................................................................................................................................... 47 Fractional Integration 2.1 Introduction......................................................................................................................................................... 47 2.2 Iterative integration: a review ............................................................................................................................. 47 2.2.1 Double integration...................................................................................................................................... 47 2.2.2 Double integration with a changed order of integration giving the idea of convolution ............................ 48 2.2.3 Generalising the double integration to an n  fold integration .................................................................. 49 2.3 Repeated integration formula by induction......................................................................................................... 50 2.4 Explaining the convolution process in repeated integration numerically............................................................ 52 2.5 Generalising the expression for a repeated integration to get the Riemann-Liouville fractional integration formula.............................................................................................................................................. 54 2.5.1 A generalisation of a repeated integration formula by the gamma function............................................... 54 2.5.2 Applying the Riemann-Liouville formula to get double-integration and semi-integration

x ........................................................................................................................... 55 for the function f ( x ) 2.5.3 Formally defining forward (left) and backward (right) Riemann-Liouville fractional integration............. 56 2.6 Fractional integration of a power function.......................................................................................................... 56 2.6.1 Fractional integration of power functions with the lower limit of integration as zero ............................... 57 2.6.2 Fractional integration of power functions with the lower limit of integration as non-zero ........................ 57 2.6.3 Region of validity of the integer order integration of the power function.................................................. 58 2.7 The extended region of fractional integration of a power function: analytic continuation ................................. 58 2.8 An analytic continuation of a finite harmonic series-derived via a fractional integration concept ..................... 60 2.9 Fractional integral of exponential function by series .......................................................................................... 63 2.9.1 Fractional integration of the exponential function with a start point of integration as zero ....................... 63 2.9.2 Introducing a higher transcendental function in the result of the fractional integration of the exponential function ............................................................................................................................ 63 2.9.3 Fractional integration of an exponential function with the start point of integration as non-zero .............. 64 2.9.4 Fractional integration of a one-parameter Mittag-Leffler function as conjugation to classical integration of the exponential function .......................................................................................................... 65 2.10 Tricks required for solving fractional integrals................................................................................................. 65 2.10.1 Obtaining a useful identity for fractional integration with a lower limit of integration as zero ............... 65 2.10.2 Obtaining a useful identity for a fractional integration with a lower limit of integration as non-zero ..... 67 2.10.3 Application of the obtained identity for acquiring a fractional integration of a power function .............. 68

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2.10.4 Application of the obtained identity for acquiring fractional integration of the exponential function in terms of Tricomi’s incomplete gamma function ......................................................................... 68 2.11 Fractional integration of the analytical function ............................................................................................... 68 2.11.1 Fractional integration related to a series of ordinary whole derivatives of an analytical function ........... 68 2.11.2 Application of the Fractional integration of an analytical function to an exponential function for integer order integration........................................................................................................................... 69 2.11.3 An alternate representation and generalisation of the Riemann-Liouville fractional integration formula of analytical function........................................................................................................................ 70 2.12 A simple hydrostatic problem of water flowing through the weir of a dam with the use of a fractional integration formula........................................................................................................................... 70 2.13 Introducing an initialising process of fractional integration.............................................................................. 72 2.14 Can we change the sign of fractional order in a fractional integration formula to write a formula for the fractional derivative? ............................................................................................................................... 74 2.14.1 Divergence of the Riemann-Liouville integral for a change of sign of order of integration .................... 74 2.14.2 Using Reisz’s logic of analytic continuation to get a meaningful representation of the Riemann-Liouville fractional integration formula with a changed sign of the order of integration .............. 74 2.14.3 Attaching meaning to the Riemann-Liouville fractional integration formula for a positive order elementary approach............................................................................................................... 76 2.15 Fractional integration of the Weyl type............................................................................................................. 77 2.15.1 Weyl transform of the fractional integration of a function on future points to infinity ............................ 77 2.15.2 Weyl fractional integration of the power function.................................................................................... 78 2.15.3 Weyl fractional integration of the exponential function ........................................................................... 79 2.16 Applications using the Weyl fractional integration........................................................................................... 79 2.17 Use of the kernel function in a convolution integral in the fractional integration formula ............................... 82 2.17.1 Defining kernel functions for forward and backward fractional integration ............................................ 82 2.17.2 Additive rule for fractional integrals ........................................................................................................ 83 2.17.3 The choice of a different convolution kernel defines different types of fractional integration operators ...................................................................................................................................... 83 2.17.3.(a) Riesz fractional integration formula .............................................................................................. 84 2.17.3.(b) Further generalisation of the fractional integration formula ......................................................... 84 2.17.3.b(i) Hadamard type fractional integration........................................................................................... 84 2.17.3.b(ii) The modified Erdelyi-Kober (MEK) fractional integration formula .......................................... 85 2.17.4 Generalisation of the fractional integration formula with respect to function .......................................... 85 2.18 Scaling law in a fractional integral.................................................................................................................... 86 2.19 The formula for fractional integration is derived from the generalisation of Cauchy’s integral theorem of complex variables ............................................................................................................................. 86 2.20 Repeated integrals as a limit of a sum and its generalization............................................................................ 90 2.20.1 Limit of a sum formula to have classical repeated n  fold integration................................................... 90 2.20.2 Limit of a sum formula to have fractional integration: the generalisation of classical formula ............... 90 2.20.3 Using a generalised limit of the sum formula derivation of an additive composition rule ....................... 91 2.21 Fractional integration is the area under the shape changing curve.................................................................... 92 2.21.1 Analytical explanation of the area under the shape changing curve for the Riemann-Liouville fractional integration ...................................................................................................... 92 2.21.2 Numerical demonstration of fractional integration as the area under a shape changing curve................. 92 2.21.3 A fractional integration is the case of fading memory while a classical one-whole integration is a case with a constant past memory ........................................................................................................... 94 2.22 The functions which can be fractionally integrated and differentiated ............................................................. 95 2.23 Discrete numerical evaluation of the Riemann-Liouville fractional integration ............................................... 96 2.23.1 Using the average of function values at the end of intervals to have a formula for numerical evaluation of the Riemann-Liouville fractional integration........................................................................... 96 2.23.2 Using the weighted average of the function values at the end of intervals to have a formula for the numerical evaluation of the Riemann-Liouville fractional integration .............................................. 97 2.24 Short summary .................................................................................................................................................. 99 2.25 References......................................................................................................................................................... 99 Chapter Three ............................................................................................................................................................... 100 Fractional Derivatives 3.1 Introduction....................................................................................................................................................... 100 3.2 Iterative differentiation and its generalization .................................................................................................. 100 3.2.1 Euler’s scheme of generalisation for repeated differentiation .................................................................. 100 3.2.2 A seeming paradox in Euler’s scheme of generalisation for repeated differentiation .............................. 101 3.2.3 Euler’s formula applied to find the fractional derivative and fractional integration of power functions.. 101

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3.2.4 That Euler’s formula gives us non-zero as a fractional derivative for a constant function seems to be a paradox............................................................................................................................................. 102 3.2.5 The observation of Euler’s formula: the fractional derivative of a constant becomes zero when the fractional order of the derivative tends to one-whole number in the classical case ............................... 103 3.2.6 Some interesting formulas for the semi-derivative and semi-integration with Euler’s formula ............... 103 3.2.7 Applying Euler’s formula for a function represented as power series expansion..................................... 104 p

3.2.8 Generalising the iterated differentiation for x to get a fractional derivative formula, a different result is obtained by Euler’s scheme .......................................................................................... 105 3.3 A natural extension of normal integer order classical derivative to obtain a fractional derivative: the exponential approach of Leibniz and Liouville ........................................................................................... 106 3.3.1 Obtaining the fractional derivative by exponential approach for x

E

by using the Laplace integral ..... 107

E

3.3.2 Representing x by using the Laplace integral in terms of the exponential function by using the Laplace integral derivation .................................................................................................................... 107 3.3.3 A fractional derivative of the trigonometric and exponential functions using an exponential approach.. 108 3.3.4 An exponential approach postulate of Leibniz and extended by Liouville............................................... 109 3.4 Liouville’s way of looking at a fractional derivative contradicts Euler’s generalization .................................. 109 3.4.1 Liouville and Euler’s fractional derivatives approach is applied to an exponential function ................... 109 3.4.2 Discussing the paradoxes of Euler’s and Leibniz’s by considering formulas to get classical one-whole integration–and a resolution....................................................................................................... 110 3.4.3 The difference between Euler’s and Leibniz’s postulate is due to the limits of integration ..................... 110 3.4.4 The fractional derivative requires the limits of a start-point and end-point for integration...................... 111 a

3.5 Applying Liouville’s logic to get a fractional derivative of x and a cosine function ................................. 111 3.6 Liouville’s approach of a fractional derivative to arrive at formulas as a limit of difference quotients............ 113 3.6.1 Liouville’s formula for fractional integration........................................................................................... 113 3.6.2 Liouville’s formula for a fractional derivative in integral representation................................................. 113 3.6.3 Derivation of Liouville’s fractional derivative formulas .......................................................................... 114 3.6.4 Extension of Liouville’s approach to arrive at formulas of the fractional derivative as a limit of difference quotients ................................................................................................................................. 115 3.7 A repeated integration approach to get the fractional derivative in Riemann-Liouville and Caputo formulations .......................................................................................................................................... 119 3.7.1 Fractional integration is a must to get a fractional derivative................................................................... 119 3.7.2 Different ways to apply a fractional integration formula to obtain a fractional derivative....................... 120 3.8 Fractional derivatives for the Riemann-Liouville (RL): ‘left hand definition’ (LHD)...................................... 123 3.9 Fractional derivatives of the Caputo: ‘right hand definition’ (RHD)................................................................ 124 3.10 Relation between Riemann-Liouville and Caputo derivatives ........................................................................ 125 3.10.1 Riemann-Liouville and Caputo derivatives related by initial value of the function at start point of fractional derivative process.................................................................................................................... 125 3.10.2 A generalisation of a fundamental theorem of integral calculus is the relation between Riemann-Liouville and Caputo fractional derivatives ................................................................................. 126 3.10.3 Further generalisation of the relationship between the Caputo and Riemann-Liouville fractional derivatives ................................................................................................................................... 127 3.10.4 The Caputo and Riemann-Liouville fractional derivatives need an area under the shape changing curve............................................................................................................................................. 128 3.11 Can we change the sign of order of the Riemann-Liouville fractional derivative to write the RL fractional integration formula? .......................................................................................................................... 128 3.12 The Weyl fractional derivative........................................................................................................................ 129 3.13 The most fundamental approach for repeated differentiation ......................................................................... 130 3.14 The unified context differentiation/integration Grunwald-Letnikov formula ................................................. 131 3.14.1 Generalisation by use of the backward shift operator defining the repeated differentiation and integration ............................................................................................................................................. 131 3.14.2 Using the generalised formula obtained by a backward shift operator of fractional integration/differentiation on an exponential function................................................................................. 133 3.14.3 Generalisation by use of the forward shift operator defining repeated differentiation and integration .. 134 3.14.4 Using the generalised formula obtained by a forward shift operator of fractional integration/differentiation on an exponential function................................................................................. 134 3.14.5 The final unified formula for fractional derivatives and integration: the Grunwald-Letnikov formula .................................................................................................................. 135 3.14.6 Reduction of a unified fractional derivative/integration unified formula to a classical derivative and integration ............................................................................................................................ 136 3.15 The whole derivative is local property whereas the fractional derivative is non-local property..................... 137 3.16 Applying the limit of a finite difference formula result to get the answer to L’Hospital’s question............... 137

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3.17 Fractional finite difference (from local finite difference to non-local finite difference)................................. 138 3.18 Dependence on the lower limit ....................................................................................................................... 140 3.19 Translation property........................................................................................................................................ 142 3.20 Scaling property .............................................................................................................................................. 143 3.21 Fractional differentiation-integration behavior near the lower limit ............................................................... 144 3.22 The fractional differentiation-integration behavior far from the lower limit .................................................. 145 3.23 A numerical evaluation of fractional differ-integrals using the Grunwald-Letnikov formula ........................ 146 3.24 The discrete numerical evaluation of the Riemann-Liouville fractional derivative formula........................... 147 3.25 Generalisation of the Riemann-Liouville and Caputo derivatives with the choice of a different kernel, weights and base-function ................................................................................................................................. 149 3.25.1 Generalisation of the left and right fractional derivatives with respect to the base function z ( x ) and weights w( x ) ............................................................................................................... 149 3.25.2 The relationship between the generalised Riemann-Liouville and Caputo fractional derivatives with weight and base function ..................................................................................................................... 150 3.25.3 The composition properties of generalised fractional derivatives .......................................................... 151 3.25.4 The various forms of generalised fractional derivatives......................................................................... 151 3.25.4.(a) A normal Riemann-Liouville/Caputo derivative from a generalised definition .......................... 151 3.25.4.(b) Hadamard type fractional derivative from generalised definitions ............................................. 151 3.25.4.(c) Modified Erdelyi-Kober (MEK) type fractional derivatives from the generalised definition ..... 152 3.26 The Caputo fractional derivative with a non-singular exponential type kernel............................................... 152 3.27 Short summary ................................................................................................................................................ 153 3.28 References....................................................................................................................................................... 153 Chapter Four ................................................................................................................................................................. 155 Fractional Differ-integrations’ further Extension 4.1 Introduction....................................................................................................................................................... 155 4.2 Applying the Grunwald-Letnikov (GL) formula of fractional derivatives to simple functions ........................ 155 4.2.1 Finding the fractional derivative of a constant using the Grunwald-Letnikov (GL) formula .................. 155 4.2.2 Finding a fractional derivative of a linear function using the Grunwald-Letnikov (GL) formula ........... 156 4.2.3 Verifying the GL fractional derivative of a linear function using the Riemann-Liouville (RL) formula of the fractional integration ............................................................................................................ 157 4.2.4 Verifying the GL fractional derivative of a linear function by using the Riemann-Liouville (RL) formula of the fractional derivative ............................................................................................................. 158 4.3 Applying the Riemann-Liouville formula to get the fractional derivative for some simple functions.............. 158 4.3.1- Left fractional derivative of the power function through application of the integral formula of the Riemann-Liouville fractional derivative............................................................................................ 158 4.3.2 Right fractional derivative of the power function through application of the integral formula of the Riemann-Liouville fractional derivative............................................................................................ 160 4.4 Applying the Caputo formula to get a fractional derivative for simple functions............................................. 161 b

4.4.1 Left fractional derivative of the power function x with respect to x through application of the integral formula of the Caputo fractional derivative.......................................................................... 161 X

4.4.2 Left fractional derivative of the power function \ ( x ) with respect to x through application of the integral formula of the generalized Caputo fractional derivative ...................................................... 162 4.5 The fractional differ-integration of the binomial function ................................................................................ 162 4.6 Fractional differ-integration exponential function using the Riemann-Liouville fractional derivative formula ............................................................................................................................................. 164 4.7 Fractional differ-integration for logarithmic functions ..................................................................................... 168 4.8 Fractional differ-integration for some complicated functions described by the power series expansion.......... 171 4.9 Fractional differ-integration of the hyperbolic and trigonometric function using the series expansion method.................................................................................................................................... 173 4.10 Fractional differ-integration of the Bessel function using the series expansion method................................. 174 4.11 Fractional differ-integration of the distribution functions using the Riemann-Liouville formula................... 174 4.11.1 Uniform distribution function and its fractional differ-integration......................................................... 174 4.11.2 The delta distribution function and its fractional differ-integration ....................................................... 175 4.11.3 The relationship between uniform distribution and the delta distribution function................................ 176 4.12 Fractional differ-integration of the saw-tooth function ................................................................................... 176 4.13 Fractional differ-integration of the generalized periodic function .................................................................. 177 4.14 The Eigen-functions for the Riemann-Liouville and Caputo fractional derivative operators ......................... 181 4.14.1 Caputo derivative of the order D for the ‘one-parameter Mittag-Leffler function’: f ( x ) ED (O xD ) . 182

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4.14.2 Riemann-Liouville derivative of the order D for the ‘one-parameter Mittag-Leffler function’: D

ED (O x ) ........................................................................................................................................ 182

f ( x)

4.14.3 Riemann-Liouville derivative of the order D for a function of the ‘two-parameter Mittag-Leffler function’: f ( x )

x

D 1

D

ED ,D (O x ) ...................................................................................... 183

4.15 The fractional derivative of the zero-corrected function and relationship to Caputo derivative ..................... 183 4.15.1 The Riemann-Liouville fractional derivative of the zero-corrected function f ( x ) a  x with a derivative start point at x 0 ........................................................................................................... 183 4.15.2 The Caputo fractional derivative function f ( x ) a  x with a derivative start point at x 0 ........... 184 v

4.15.3 The Riemann-Liouville fractional derivative of the zero-corrected function f (t ) t from the start point as t z 0 ........................................................................................................................ 185 4.15.4 The Riemann-Liouville fractional derivative of the order D for the zero-corrected ‘one-parameter Mittag-Leffler function’: f (t ) ED ( at D ) with a derivative start point at t 0 ........................................ 186 4.15.5 The Caputo fractional derivative of the order D for the ‘one-parameter Mittag-Leffler function’: D

ED ( at ) with a derivative start point at t

0 ........................................................................................... 187 4.15.6 The Riemann-Liouville fractional derivative of the order D of the zero-corrected function D D f (t ) cosD (t ) and f (t ) sinD (t ) with a derivative start point at t 0 ........................................... 187 4.15.7 The Riemann-Liouville fractional derivative of the order E of the zero-corrected ‘one-parameter D

0 and D z E ................ 188 4.15.8 The Riemann-Liouville and Caputo fractional derivatives of the order D for the function Mittag-Leffler function’

f ( x)

x

E 1

D

f ( x)

cosD , E ( x ) and

ED ( x ) with a derivative start point at x

f ( x)

x

E 1

D

sinD , E ( x ) with a derivative start point at x

0 ............... 189 ax

4.15.9 The Riemann-Liouville fractional derivative of the zero-corrected exponential function: f ( x ) e with a derivative start point at x 0 and its relationship to the Caputo derivative.................................... 190 4.15.10 The Riemann-Liouville fractional derivative of the zero-corrected f ( x ) cos( ax ) and f ( x ) sin( ax ) functions with a derivative start point at x 0 and its relationship to the Caputo derivative ................... 191 4.16 The fractional derivative at a non-differentiable point of the continuous function ......................................... 191 4.17 Computation of the fractional derivative and integration-a review and comparison of the various schemes........................................................................................................................................... 194 4.18 Fractional derivatives of the same-order but of different RL-Caputo types: A unified formula ..................... 197 4.19 Generalizing Leibniz’s rule............................................................................................................................. 198 4.20 Chain rule for fractional differ-integration...................................................................................................... 202 4.21 Analytical continuation of the differ-integral operator from integer to real order: Extended fractional calculus ............................................................................................................................. 204 4.22 Analytic continuation of the differ-integral operator from a real to complex order........................................ 208 4.23 Short summary ................................................................................................................................................ 209 4.24 References....................................................................................................................................................... 209 Chapter Five ................................................................................................................................................................. 210 Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms 5.1 Introduction....................................................................................................................................................... 210 5.2 Initialization fractional integration (the Riemann-Liouville approach) ............................................................ 210 5.2.1 The origin of the initialization function (or history function) in the fractional integration process ......... 210 5.2.2 The initialization function is a constant function in the classical integration process .............................. 211 5.2.3 Types of initialization for fractional integration process.......................................................................... 211 5.3 Terminal and side initialization for fractional integration................................................................................. 212 5.3.1 Terminal charging for fractional integration ............................................................................................ 212 5.3.2 Evaluation of initialization function for f (t ) t for semi-integration process starting at t 1 ............ 212 5.3.3 The initialization concept of classical integration in the context of a developed concept of the initialization function for fractional integration .......................................................................................... 214 5.3.4 Side charging for fractional integration.................................................................................................... 214 5.4 Initializing the fractional derivative Riemann-Liouville approach ................................................................... 215 5.4.1 Fractional derivative for a non-local operation requires initialization...................................................... 215 5.4.2 The generalization of the initialization process ........................................................................................ 215 5.5 Terminal initialization for the fractional derivative .......................................................................................... 215 5.5.1 Derivation of terminal charging for the fractional derivative................................................................... 215

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2

5.5.2 Evaluation of the initialized semi-derivative for f (t ) (t  2) starting at t 0 .................................. 216 5.6 Side-initialization of the fractional derivative................................................................................................... 218 5.7 Initializing fractional differ-integrals using the Grunwald-Letnikov approach ................................................ 219 5.8 Criteria for the generalized differ-integration composition............................................................................... 219 5.8.1 Composition rules..................................................................................................................................... 219 5.8.2 Demonstration of the composition of a fractional derivative, and the fractional integration 1

1

and mixed operation that is applied to functions f (t ) t 2 & f (t ) t 2 , and discussion ....................... 221 5.8.3 Composition rules with initialized differ-integrations.............................................................................. 221 5.9 The Relationship between uninitialized Caputo and Riemann-Liouville (RL) fractional derivatives .............. 222 5.9.1 Expression for the RL fractional integration of f (t ) from 0  t as integration of q

q

function f (t  x ) from 0  t ................................................................................................................ 222 5.9.2-Finding an un-initialized Caputo derivative and demonstrating that no singularity term appears at the start point of fractional differentiation ............................................................................................... 223 5.9.3 Finding an un-initialized Riemann-Liouville (RL) derivative and demonstrating the appearance of a singularity term at the start point of a fractional derivative ............................................... 223 5.9.4 Adding singularity function at the start point of a fractional derivative to equate uninitialized RL and Caputo derivatives ................................................................................................................................ 224 5.9.5 Finding out the relationship of the RL-Caputo fractional derivative via Leibniz’s formula .................... 225 5.9.6 Evaluation of the RL and Caputo derivatives from a non-zero start point where the value of the function is zero .................................................................................................................................. 227 5.10 Initialization of Caputo derivatives ................................................................................................................. 227 5.10.1 Initialization with Caputo derivatives and its difficulties ....................................................................... 227 5.10.2 The initialized fractional derivative theory............................................................................................. 228 5.10.3 Demonstration of the initialized function for a fractional derivative (RL and Caputo) as zero when the function value at the start point is zero and the function is zero before the start point ................ 229 5.10.4 Making the initialized Caputo derivative equal to the initialized RL derivative in order to provide an initialization function for the Caputo derivative ........................................................................ 231 5.10.5 The Caputo derivative initialization function: A case of a side initialization process............................ 232 5.11 Generalization of the RL and Caputo formulations and the initialization function......................................... 234 5.12 Observations regarding difficulties in Caputo initialization and demanding physical conditions vis-à-vis RL initialization conditions ................................................................................................................ 236 5.13 The Fractional Derivative of a sinusoidal function with a lower terminal not at minus infinity, and an initialization function ............................................................................................................................. 236 5.14 The Laplace transform of fractional differ-integrals ....................................................................................... 238 5.14.1 The generalization of classical Laplace transform formulas for differentiation and integration ............ 238 5.14.2 Generalization is not possible for a few classical Laplace transform identities ..................................... 240 5.14.3 Evaluating the Laplace transform for function f ( x )

e

 kx



D

0 Dx



ª¬e kx f ( x ) º¼ , D d 0 by

using the Riemann-Liouville integral formula............................................................................................. 241 5.15 Approximate representation of a fractional Laplace variable by rational polynomials in the integer power of the Laplace variable ........................................................................................................................... 241 5.16 Generalized Laplace transform ....................................................................................................................... 242 5.16.1 Laplace transform for the Riemann-Liouville fractional derivative ....................................................... 242 5.16.2 Laplace transform for the Caputo fractional derivative.......................................................................... 243 5.16.3 Requirement of the fractional order of the initial states of the RL derivative and the integer order (classical) initial states for Caputo derivative initializations ....................................................................... 243 5.16.4 Generalized Laplace transforms formula for a fractional derivative with a type parameter................... 244 5.16.5 Generalized Laplace transform demonstrated for use in a fractional differential equation .................... 244 5.17 Generalized stationary conditions ................................................................................................................... 247 5.18 Demonstration of the generalized Laplace transform for solving the initialized fractional differential equation .......................................................................................................................................... 247 5.18.1 Composition of a fractional differential equation for a Riemann-Liouville fractional derivative .......... 247 5.18.2 Composition of the fractional differential equation for the Caputo fractional derivative....................... 248 5.18.3 Applying the concept of the initialization function of fractional derivatives in a fractional differential equation..................................................................................................................................... 249 5.19 Fourier transform of a fractional derivative operator...................................................................................... 249 5.19.1 Fourier transform from a Laplace transform .......................................................................................... 249 5.19.2 Similarity and dissimilarity in Laplace and Fourier transforms ............................................................ 251 5.20 Complex order differ-integrations described via the Laplace transform......................................................... 252 5.21 Short summary ................................................................................................................................................ 253 5.22 References....................................................................................................................................................... 254

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Chapter Six ................................................................................................................................................................... 255 Fractional Differential Equations and their Analysis 6.1 Introduction....................................................................................................................................................... 255 6.2 Tricks in solving some fractional differential equations ................................................................................... 255 6.2.1 A method using the Laplace transform for a fractional differential equation........................................... 255 6.2.2 Alternative method to solve a fractional differential equation ................................................................. 256 6.2.3 A method using the Laplace transform for a fractional integral equation ................................................ 256 6.2.4 An alternative method to solve fractional integral equations ................................................................... 257 6.3 Abel’s fractional integral equation of ‘tautochrone’-a classical problem ......................................................... 257 6.4 Using the power series expansion method to obtain inverse Laplace transforms ............................................. 259 6.5 Using Laplace transform techniques and power series expansion to solve simple fractional differential equations......................................................................................................................... 261 6.6 The contour integration method for obtaining inverse Laplace transforms ...................................................... 262 6.7 Operational calculus of applying Heaviside units to a partial differential equation.......................................... 267 6.8-The response of a fractional differential equation with detailed analysis of power law functions ................... 269 6.8.1 Demonstration of power law functions as a solution to a simple fractional differential equation excited by a constant step function .............................................................................................................. 269 6.8.2 Demonstration of a power series function as a solution to the fractional differential equation excited by a constant step function ............................................................................................................. 270 6.8.3 A demonstration of the power series function as a solution to a fractional differential equation excited by the constant step function through use of the Laplace transform ............................................... 271 6.8.4 Obtaining approximate short-time and long-time responses from a power series solution for a fractional differential equation excited by a constant step function .................................................... 273 6.9 An analytical method to obtain inverse Laplace transforms ‘without contour integration’ the Berberan-Santos method ............................................................................................................................. 274 6.9.1 Development of the Berberan-Santos technique to obtain a distribution function for the decay rates for the relaxation function of time....................................................................................................... 274 6.9.2 Derivation of the Berberan-Santos method .............................................................................................. 278 6.10 A few examples of the inverse Laplace transform of functions available in standard Laplace....................... 280 6.10.1 The inverse Laplace transform of function G ( s )

(s  a)

................................................................. 280

2

6.10.2 The inverse Laplace transform of the function G ( s ) 6.10.3 The inverse Laplace transform of function G ( s )

1

s ( s  1)

e

 O0 s

1

......................................................... 281

...................................................................... 282

D

6.10.4 The inverse Laplace transform of function G ( s ) s ........................................................................ 282 6.10.5 The inverse Laplace transform of function: G ( s ) 1 ........................................................................... 283 1

6.10.6 The inverse Laplace transform of the function G ( s ) s .................................................................... 283 6.11-Examples of using the Berberan-Santos method to obtain the Laplace inversion of a few functions that are not provided in standard Laplace Transform tables.............................................................. 283  s / s0

E

.................................................................. 283

6.11.1 The inverse Laplace transform of function G ( s )

e

6.11.2 The inverse Laplace transform of function G ( s )

1  (1  E )  s s0

6.11.3 The inverse Laplace transform of the function G ( s )

D 1



1   as



 1(1E )

............................................ 284

..................................................... 285

6.11.4 An integral representation of the Mittag-Leffler function using the Berberan-Santos method .............. 285 6.11.5 The inverse Laplace transform of the function G ( s ) 6.11.6 The inverse Laplace transform of function G ( s )

s 1

D 1

D

( s  1)

D

ks ( s  k ) 1

1

1

.................................................. 286

........................................................ 288

b 1

6.11.7 The inverse Laplace transform of function G ( s ) (ln s ) ( s  s a 1 ) ................................................ 290 6.12 The relaxation-response with the Mittag-Leffler function vis-à-vis the power law function as obtained for fractional differential equation analysis ........................................................................................ 291 6.13 Use of several fractional order derivatives in order to obtain a generalized fractional differential equation ......................................................................................................................................... 293 6.14 Short Summary ............................................................................................................................................... 293 6.15 References....................................................................................................................................................... 294 Chapter Seven............................................................................................................................................................... 296 A Formal Method for Solving Fractional Differential and Fractional Integral Equations

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xv

7.1 Introduction....................................................................................................................................................... 296 7.2 A first order linear differential equation and its fractional generalization using the Caputo derivative............ 296 7.2.1 Demonstrating the similarity of a classical differential operator with the Caputo fractional differential operator in a differential equation solution pattern ................................................................... 296 7.2.2 The first order differential equation of a relaxation process with an initial condition.............................. 297 7.2.3 A first order differential equation’s homogeneous and complementary parts.......................................... 297 7.2.4 A fractional order differential equation composed using the Caputo derivative’s homogeneous and complementary parts ............................................................................................................................. 298 7.2.5 Can we just replace the exp( O t ) in the classical case with ED ( O t D ) in order to get the total solution for the fractional differential equation composed using the Caputo derivative?............................ 298 7.2.6 Getting a particular solution to a fractional differential equation composed by the Caputo derivative through a fractional integration operation using a classical convolution process ........................................ 300 7.2.7 A homogeneous fractional differential equation composed using a Caputo derivative is driven by a fractional RL derivative of a Heaviside unit step function................................................................... 301 7.3 The first order linear differential equation and its fractional generalization using the Riemann-Liouville (RL) derivative................................................................................................................... 302 7.3.1 Demonstrating the similarity of a classical differential operator with the Riemann-Liouville fractional differential operator in a differential equation solution ............................................................... 302 7.3.2 A fractional order differential equation composed with the Riemann-Liouville derivative is fundamental and a particular solution is offered by a given fractional order initial state ........................ 303 7.3.3 Verification of a fractional initial state from the obtained solution for a fractional order differential equation composed using the Riemann-Liouville derivative .................................................... 304 7.3.4 Modification in a solution for a fractional differential equation composed using the Riemann-Liouville derivative solution with classical integer order initial states instead of fractional order initial states ......... 305 7.3.5 The requirement of one Green’s function for a fractional order differential equation composed using the Riemann-Liouville derivative and two Green’s functions for the Caputo derivative ............................ 306 7.4 Formal description of a fractional differential and integral equation ................................................................ 306 7.4.1 Concepts of an ordinary differential equation as extended to fractional differential and integral equations .................................................................................................................................. 306 7.4.2 The indicial polynomial corresponding to the fractional differential equation......................................... 307 7.4.3 The order of the fractional differential equation and the number of linearly independent solutions........ 307 7.5 Finding a solution to a homogeneous fractional differential equation .............................................................. 308 7.5.1 An ordinary homogeneous classical differential equation and solution in terms of an exponential function .................................................................................................................................... 308 7.5.2 A fractional differential equation and solution candidate in terms of a higher transcendental function .. 308 7.5.3 A direct approach using the Miller-Ross function as a solution for a fractional differential equation (FDE) ............................................................................................................................................ 309 7.5.4 Getting a solution from the roots of an indicial polynomial corresponding to the FDE using the Miller-Ross function .................................................................................................................... 310 7.6 Motivation for the Laplace transform technique............................................................................................... 311 7.6.1 The indicial polynomial is the same as we get from the Laplace transform of a fractional differential equation..................................................................................................................................... 311 7.6.2 A solution with distinct roots in an indicial polynomial........................................................................... 311 7.6.3 The solution with equal roots for indicial polynomials ............................................................................ 312 7.7 A linearly independent solution of the fractional differential equation (FDE) ................................................. 313 7.8 The explicit solution for a homogeneous fractional differential equation (FDE) ............................................. 315 7.9 The non homogeneous fractional differential equation and its solution ........................................................... 316 7.10 Fractional integral equations and their solution .............................................................................................. 317 7.10.1 Describing a fractional integral equation................................................................................................ 317 7.10.2 Extension of the final value theorem of integral calculus....................................................................... 318 7.10.3 The solution to fractional integral equations .......................................................................................... 319 7.11 Examples of fractional integral equations with explicit solutions................................................................... 319 kD and sequential fractional 7.12 Sequential fractional derivative of the Miller-Ross type aDx differential equations (SFDE) ........................................................................................................................... 321 7.12.1 The sequential fractional differential equation (SFDE).......................................................................... 321 7.12.2 The matrix form representation of SFDE ............................................................................................... 321 7.13 Solution of the ordinary differential equation using state transition matrices -a review................................. 322 7.13.1 Origin of the state transition matrix in a solution to systems of linear differential equations ................ 322 7.13.2 The state transition matrix as Green’s function is a solution to the homogeneous system of differential equations............................................................................................................................... 323 7.13.3 A demonstration of the ways to represent the state transition matrix and its usage in multivariate dynamic systems.................................................................................................................. 324

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7.14 ‘Alpha-exponential functions’ as eigen-functions for the Riemann-Liouville (RL) and Caputo fractional derivative operators .............................................................................................................. 326 7.14.1 The alpha-exponential functions (1 & 2) for fractional derivative operators are similar to the exponential function as an eigen-function for a classical derivative...................................................... 326  (t ) e At as related to the 7.14.2 The alpha-exponential functions )D (t ) eDAt and ) D D Mittag-Leffler function................................................................................................................................ 326 7.14.3 Defining the alpha-exponential functions (1 & 2) via a kernel of convolution as the power law functions and their Laplace transforms ...................................................................................... 327 7.14.4 The alpha-exponential functions (1 & 2) are related via the convolution relationship........................... 329 7.15 Fractional derivatives of the ‘alpha-exponential functions (1 & 2)’ ............................................................... 329  (t  t ) e A( t t0 ) ................................. 329 7.15.1 The Caputo derivative of the alpha-exponential function-2 ) D D 0 A( t t )

7.15.2 The Riemann-Liouville derivative of the alpha-exponential function-1 )D (t  t0 ) eD 0 .............. 330 7.15.3 The backward Riemann-Liouville derivative of the alpha-exponential function-1 )D (T  t ) eDA(T t ) .................................................................................................................. 330

 (t ) .......................................... 331 7.15.4 The relationship between the state transition matrices )D (t ) and ) D 7.16 The general solution to the sequential fractional differential equation using ‘alpha-exponential functions’.. 331 7.16.1 The homogeneous SFDE and its characteristic equation (or indicial polynomial)................................. 331 7.16.2 A generalized Wronskian ....................................................................................................................... 331 7.16.3 Linearly independent solutions of SFDEs .............................................................................................. 332 7.16.4 A demonstration of linearly independent solutions of the SFDE in an equation of motion with fractional order damping...................................................................................................................... 332 7.16.4.(a) SFDE with a Caputo derivative................................................................................................... 333 7.16.4.(b) SFDE with a Riemann-Liouville (RL) derivative ....................................................................... 334 7.17 The solution of a multivariate system of a fractional order differential equation with the RL Caputo derivative using a state transition matrix of the ‘alpha-exponential functions -1 & 2’ ......................... 335 7.17.1 The multivariate system with an RL derivative and its solution with the state transition matrix ........... 335 7.17.2 A Multivariate system with the Caputo derivative and its solution with state transition matrices ......... 335 7.17.3 Formalizing the multivariate problem to get a state trajectory as a solution with the defined state transition matrices........................................................................................................................................ 336 7.17.4 An application to obtain the state trajectory solution for a multivariate fractional differential equation system with a given initial condition and a forcing function ........................................................ 337 7.17.5 An application to obtain the state trajectory solution for a multivariate fractional differential equation with matrix A as a skew-symmetric system with a given initial condition and forcing function ............... 338 7.18 The solution to a fractional differential equation of type V  (b) * DD V E 0H  (E1 ) * DD H with RL or Caputo formulations ....................................................................................................................... 339 7.18.1 Using an RL-derivative formulation for a solution of H with a known V input.................................... 339 7.18.2 Using the Caputo derivative formulation for a solution of H with a known V input .......................... 340 7.18.3 Using the RL-derivative formulation for a solution of V with a known H input ................................... 340 7.18.4 Using the Caputo derivative formulation for a solution of V with a known H input ........................... 341 7.19 A generalization of a fractional differential equation with a sequential fractional derivative......................... 341 7.20 Short Summary ............................................................................................................................................... 343 7.21 References....................................................................................................................................................... 343 Chapter Eight................................................................................................................................................................ 344 Fractional Partial Differential Equation 8.1 Introduction....................................................................................................................................................... 344 8.2 Time fractional diffusion wave equation (TFDWE): Cauchy and the signaling problem................................. 344 8.3 Green’s function................................................................................................................................................ 346 8.3.1 Green’s function for the diffusion equation with auxiliary functions as defined by a similarity variable ..................................................................................................................................... 346 8.3.2 Green’s function for the wave equation with auxiliary functions as defined by the similarity variable... 346 8.4 Solution of TFDWE via the Laplace transform for the Green’s function for Cauchy and signalling problems .................................................................................................................................... 347 8.4.1 The Cauchy problem for a TFDWE solution in the Laplace domain ....................................................... 347 8.4.2 Signalling problem for TFDWE in Laplace domain ................................................................................ 349 8.4.3 The solution of Cauchy and the signaling problems of TFDWE in the time domain by inverting the Laplace domains solution by using auxiliary functions FX and M X .................................................... 349 8.5 Reciprocal relation between Green’s function of Cauchy and signaling problem ............................................ 350

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xvii

8.6 The origin of the auxiliary function from the Green’s function of the Cauchy problem .................................. 351 8.6.1 An inverse Laplace transform of the Green’s function: Gc ( x, t ) in terms of an auxiliary function M X ( z ) with z as a similarity variable ......................................................................................... 351 8.6.2 The description of the Hankel contour from the Bromwich path integral used for an integral representation of the auxiliary function M X ( z ) .......................................................................................... 352 8.7 The integral and series representation of auxiliary functions............................................................................ 354 8.7.1 Getting a series representation of the auxiliary function M X ( z ) from its integral representation ........... 354 8.7.2 The relationship of Wright’s function to M-Wright’s function or the auxiliary function M X ( z ) ........... 355 8.7.3 Proof of the property of the auxiliary function i.e.

³

f

0

M X ( z )dz 1 ....................................................... 356

8.7.4 Moments of the M-Wright function or the auxiliary function M X ( z ) ..................................................... 356 8.8 The auxiliary functions M X ( z ) and FX ( z ) as a fractional generalisation of the Gaussian function.............. 357 8.8.1 The series representation of the Wright function derived from its integral representation formula ......... 357 8.8.2 A further series representation of auxiliary functions FX ( z ) and M X ( z ) ............................................. 358 8.8.3 The relationship between auxiliary functions M X ( z ) and FX ( z ) ........................................................... 359 8.8.4 An auxiliary function M X ( z ) for

X

at

1

2

, 1 3 and 0 ............................................................................. 360

8.9 Laplace and Fourier transforms of the auxiliary functions ............................................................................... 361  sX

giving FX (t X ) and M X (t X ) ....................... 362 8.9.2 The Mittag-Leffler function as an integral representation on Hankel’s path............................................ 362 8.9.3 The Laplace transform of an auxiliary function M X (t ) as the Mittag-Leffler function............................ 363 8.9.1 The inverse Laplace transform of function: X ( s )

e

8.9.4 The Fourier transform of the auxiliary function M X ( x ) ......................................................................... 364 8.10 A graphical representation of the M. Wright or auxiliary function M X ( x ) ................................................... 365 8.11 Auxiliary function in two variables M X ( x, t ) and its Laplace and Fourier transforms.................................. 366 8.11.1 Defining M X ( x, t ) in relation to M X ( xt X ) .......................................................................................... 366 8.11.2 Laplace transform of M X ( x, t ) .............................................................................................................. 367 8.11.3 Fourier transform of M X ( x, t ) ............................................................................................................... 368 8.11.4 Special cases for M X ( x, t ) , for X as 1 2 and one................................................................................. 369 8.12 The use of the two variable auxiliary functions in TFDWE ........................................................................... 369 8.12.1 The time fractional diffusion equation and its solution with M X ( x, t ) .................................................. 369 8.12.2 A time fractional drift equation and its solution with M X ( x, t ) ............................................................ 371 8.13 Reviewing time fractional diffusion equations with several manifestations ................................................... 372 8.13.1 Classical diffusion equation ................................................................................................................... 372 8.13.2 The Green’s function for a classical diffusion equation and its moment................................................ 372 8.13.3 A classical diffusion equation with a stretched time variable in its Green’s function and moment ....... 373 8.13.4 Time fractional diffusion equation ......................................................................................................... 374 8.13.5 The Green’s function for a time fractional diffusion equation and its moments .................................... 374 8.13.6 A time fractional diffusion equation with a stretched time variable....................................................... 375 8.13.7 The Green’s function for a time fractional diffusion equation with a stretched time variable ............... 377 8.14 Short summary ................................................................................................................................................ 377 8.15 References....................................................................................................................................................... 377 Chapter Nine................................................................................................................................................................. 379 Modified Fractional Calculus in conjugation with Classical Calculus 9.1 Introduction....................................................................................................................................................... 379 9.2 Condition of term-by-term differ-integration of a series................................................................................... 379 9.2.1 Linearity and homogeneity of fractional differ-integral operators ........................................................... 379 9.2.2 Convergence of a differ-integration of series ........................................................................................... 380 9.2.3 Discussion on convergence by application of the Riemann-Liouville fractional integration formula from term-by-term to a differ-integrable series.............................................................................. 381

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9.2.4 Discussion on convergence by application of the Euler fractional derivative formula for power functions term-by-term into a differ-integrable series ................................................................ 382 9.3 Composition rule in fractional calculus ............................................................................................................ 384 9.3.1 Condition for the inverse operation f q

Q

9.3.2 Condition D D f N

D q

9.3.3 Condition for D D f

q Q

D

Q

Q

D f to be satisfied ......................................................... 384

f to be satisfied ........................................................................................... 385 q

N

D D f , N as positive integer ..................................................................... 387 q

Q

9.3.4 Generalising the composition D D f to D

q Q

f by initial values of f ............................................. 389

Q

Q

9.4 The reversibility of a differ-integral operator D f g to f D g .......................................................... 390 9.5 An alternate representation for fractional differ-integration for real analytic functions ................................... 392 9.5.1 A representation using the fractional differ-integral formula of the Riemann-Liouville.......................... 392 9.5.2 Representation using the Grunwald-Letnikov formula for differ-integration........................................... 392 9.6 The fractional derivative for non-differentiable functions is defined................................................................ 394 D

9.6.1 Defining a quotient i.e. 'x /  't for a fractional derivative such that the fractional derivative of the constant is zero ................................................................................................................. 394 9.6.2 Corollary to the definition of a fractional derivative ................................................................................ 394 9.7 Utilizing the Mittag-Leffler function to get suitable fractional derivatives as a conjugation to the classical derivative........................................................................................................................................................... 395 9.8 The fractional derivative via fractional difference and its Laplace transform................................................... 396 (D )

via the forward shift operator.................................................... 396 9.8.1 Defining the fractional derivative f 9.8.2 Getting the fractional derivative of the constant as a non-zero using the Laplace transform technique... 397

^ 9.8.4 Getting the Laplace transform $ ^ f

` (t )` ............................................................................................ 398

9.8.3 Getting the Laplace transform $ 'D f (t ) ............................................................................................ 397 (D )

9.8.5 The definition of fractional differentiability is satisfactory for self-similar functions ............................ 399 9.9 The modified fractional derivative for a function with non-zero initial condition-Jumarie type ...................... 399 9.9.1 Defining the fractional derivative via the offset function construction to obtain a fractional derivative of the constant as a zero-Jumarie type ....................................................................................... 399 9.9.2 The Laplace transform of the modified RL fractional Derivative and its conjugation to a classical derivative................................................................................................................................ 400 9.10 Integral representation of modified RL fractional derivatives of Jumarie type............................................... 401 9.11 The application of the modified fractional RL derivatives of Jumarie type to various functions ................... 402 9.11.1 Modified RL-derivative of order D for function f ( x ) ED (O xD ) ..................................................... 402 9.11.2 A modified RL-derivative of order E ! 0 for function f ( x ) 9.11.3 The RL fractional derivative of order

D

ED (O x ) ........................................... 403

D for function f ( x ) xD 1 ED ,D (O xD ) .................................... 404

9.11.4 The modified RL-derivative of order D for function f ( x )

ED (O x ) .................................................. 404 9.11.5 Applying the modified RL fractional derivative to a non-differentiable point of a function: f ( x)

x

1

2

 b ........................................................................................................................................... 404

9.11.6 Defining the critical order for a non-differentiable point by use of the modified RL fractional derivative ..................................................................................................................................... 405 9.12 Fractional Taylor series................................................................................................................................... 405 9.13 The use of fractional Taylor series.................................................................................................................. 407 9.14 Conversion formulas for fractional differentials ............................................................................................. 408 9.15 Integration with respect to fractional differentials .......................................................................................... 409 D

9.15.1 Integration with regards to (d[ ) and its relationship to the Riemann-Liouville fractional integration formula ...................................................................................................................................... 409 D

9.15.2 A demonstration of the integration with regards to (d[ ) for f ( x ) and f ( x )

J

x , f ( x)

1 , f ( x)

G ( x)

D

ED (O x ) .................................................................................................................................. 410 D

9.15.3 Some useful identities for the integration w.r.t. (d[ ) .......................................................................... 410 9.16 Leibniz’s rule for the product of two functions for a modified fractional derivative...................................... 411 9.17 Integration by parts for a fractional differential .............................................................................................. 412 9.18 The chain rule for a modified fractional derivative......................................................................................... 413 9.18.1 The derivation of three formulas for the chain rule with a modified fractional derivative .................... 413

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xix

9.18.2 The application of three formulas for the chain rule with a modified fractional derivative for different cases ....................................................................................................................................... 414 9.19 Coarse grained system .................................................................................................................................... 416 9.19.1 Need for coarse graining ........................................................................................................................ 416 9.19.2 The fractional velocity uD with coarse grained time differential .......................................................... 416 9.19.3 Fractional velocity vD with a coarse grained space differential ........................................................... 416 9.20 The solution for a fractional differential equation with a modified fractional derivative ............................... 417 9.20.1 Fractional differential equations defined with modified fractional derivatives ...................................... 417 9.20.2 Defining the function LnD ( x ) in conjugation with ln x and obtaining the solution to a fractional differential equation with a modified fractional derivative.......................................................................... 418 9.20.3 Proof of the identity

E

D

D

( ax )

 E

D

D





D

( ay ) t ED a ( x  y )

in respect of a modified fractional

derivative definition..................................................................................................................................... 421 9.20.4 The fractional differential equation x

(D )

9.20.5 The fractional differential equation x

(D )

(t )



(2D )

(t ) O  a (t ) x (t )  b(t ) with a modified fractional derivative ..................................................................................................................................... 424

9.20.6 The fractional differential equation a x

Ot

1D

x (t ) with a modified fractional derivative ........... 425



(t )  b x

(D )



(t )  c

0 with a modified fractional

derivative .................................................................................................................................................... 425 9.21 The application to dynamics close to the equilibrium position are subjected to coarse graining.................... 425 9.21.1 The dynamic system at equilibrium........................................................................................................ 425 9.21.2 Dynamic system subjected to coarse graining in time differential dt ................................................... 426 9.21.3 The dynamic system subjected to coarse graining in space differential dx ........................................... 426 9.21.4 The system when both differentials dx and dt subjected to coarse graining ........................................ 427 9.22 Using the Mittag-Leffler function in integral transform formulas .................................................................. 428 9.22.1 Defining the fractional Laplace transform by using the Mittag-Leffler function ................................... 428 9.22.2 Defining a fractional convolution integration process by use of the fractional Laplace transforms....... 428 9.22.3 Demonstrating fractional Laplace transform for the Heaviside unit step function ................................ 429 9.22.4 Defining the fractional delta distribution function G D ( x ) and its fractional Fourier integral representation by use of the Mittag-Leffler function ................................................................................... 429 9.22.5 The inverse fractional Laplace transformation by the Mittag-Leffler function ...................................... 430 9.22.6 The Mittag-Leffler function to define the fractional order gamma function *D ( x ) ............................. 430 9.23 Derivatives with the order as a continuous distributed function ..................................................................... 430 9.23.1 The fractional differential equation generalized from integer order to fractional order ........................ 430 9.23.2 The concept of order distribution ........................................................................................................... 431 9.23.3 From the discrete delta distributed function to a continuous distribution function for describing fractional orders.................................................................................................................... 432 9.23.4 The continuous order distributed differential equation .......................................................................... 432 9.24 Short summary ................................................................................................................................................ 434 9.25 References....................................................................................................................................................... 434 AppendixA ................................................................................................................................................................... 435 Higher Transcendental and Special Functions A.1 Hyper-geometric functions............................................................................................................................... 435 A.2 Mittag-Leffler function ................................................................................................................................... 438 A.2.1 One-parameter Mittag-Leffler function .................................................................................................. 439 A.2.2 The two parameter Mittag-Leffler function ............................................................................................ 440 A.2.3 Graphical representations of Mittag-Leffler function: f ( x ) ED , E ( x ) .................................................. 441 A.2.4 Generalized Hyperbolic and Trigonometric Functions............................................................................ 442 A.3 The error function and its fractional generalization ......................................................................................... 443 A.4 Variants of the Mittag-Leffler function............................................................................................................ 443 A.5 The Laplace Integral and its connection to the Mittag-Leffler function ......................................................... 444 A.6 The Laplace transforms of the Mittag-Leffler function and several other variants.......................................... 446 A.7 Agarwal Function ............................................................................................................................................ 447 A.8 Erdelyi’s Function ........................................................................................................................................... 447 A.9 The Robotnov-Hartley function ....................................................................................................................... 448 A.10 Miller-Ross function ..................................................................................................................................... 449 A.11 The generalized cosine and sine function in the Miller-Ross formulation ..................................................... 450

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Table of contents

A.12 Generalized R function and G function ......................................................................................................... 452 A.13 Bessel Function ............................................................................................................................................. 453 A.14 Wright Function ............................................................................................................................................. 455 A.15 Prabhakar Function ........................................................................................................................................ 455 A-15.1 Three Parameter Mittag-Leffler Function f ( z )

J

ED , E ( z ) .................................................................. 455

A.15.2 Prabhakar Integral ................................................................................................................................. 455 A.15.3 The Prabhakar integral as a series-sum of the Riemann-Liouville fractional integrals ......................... 456 Appendix B................................................................................................................................................................... 457 List of Laplace transform pairs of functions related to fractional calculus Appendix C................................................................................................................................................................... 460 Fractional derivatives and integrals of some important functions Appendix D .................................................................................................................................................................. 464 Formulas of Modified Fractional Derivative Appendix E................................................................................................................................................................... 466 Branch-points, Branch-Cuts and Riemann Sheets for Multi-valued Functions E.1 Branch-points and branches in multi-valued function ...................................................................................... 466 E.2 The branch point of order ( n  1) and branch-cut for n  valued function...................................................... 466 E.3 Multiple Riemann sheets connected along the branch-cut for a multi-valued function.................................... 468 E.4 An example elaborating branches and branch-cut for a two-valued function .................................................. 468 E.5 Choosing the contour with branch-cut for a function having branch-point ...................................................... 470 Appendix F ................................................................................................................................................................... 471 Mittag-Leffler Function with Negative Order F.1 Restriction in Series Representation of the Mittag-Leffler Function ................................................................ 471 F.2 The Integral Representation of the two-parameter Mittag-Leffler function ..................................................... 471 F.3 The formula for a two-parameter Mittag-Leffler function with a negative order ............................................. 472 F.4 The series representation of the two-parameter Mittag-Leffler function with a negative order ....................... 473 F.5 The graphical representation of the Mittag-Leffler function with the negative order....................................... 474 F.6 Comparisons of the functional relationship of ED , E ( x ) and ED , E ( x ) ........................................................... 475 Appendix G .................................................................................................................................................................. 477 Inverse Laplace Transform by Contour Integration G.1 Revising basics of Laplace transforms............................................................................................................. 477 G.2 The inverse Laplace transform via contour integration ................................................................................... 478 G.3 Jordan’s Lemma ............................................................................................................................................... 479 G.4 The application of Residue Calculus to get an Inverse Laplace Transform of

X (s)

2e

2 s

/ ( s 2  4) by contour integration................................................................................................ 480

G.5 The application of residue calculus to get the inverse Laplace transform of X ( s ) sa by contour integration on the branch cut in a complex plane............................................................................ 481 ln s G.6 The application of residue calculus to get the inverse Laplace transform of X ( s ) b a . ( s s )

by contour integration on the branch cut in a complex plane............................................................................ 483 b

a

G.7 The application of residue calculus to get the inverse Laplace transform of X ( s ) s lnss by contour integration on branch cut in the complex plane............................................................................... 486 Bibliography ................................................................................................................................................................. 488

LIST OF PHOTOGRAPHS

Figure-P1: Picture showing magnetic levitation where fractional calculus is used for controls Figure-P2: Picture showing magnetic levitation where classical calculus is used for controls Figure-P3: Picture showing CRO traces for control voltage and ball position for both systems with classical calculus and with fractional calculus Figure-P4: DC motor speed control setup Figure-P5: Armature voltage and current at 1000 RPM DC motor with classical calculus Figure-P6: Armature voltage and current at 1000 RPM DC motor with fractional calculus Figure-P7: Memorizing charging time of applied electric- field in experiments with Laponite for crack formation studies Figure-P8: Charge discharge of a super-capacitor showing the presence of a fractional capacitor

LIST OF FIGURES

Figure 1.1: Hankel’s path for integration Figure1.2: Plot of agamma function Figure1.3: Plot of a reciprocal of the gamma function. Note that values are zero for x integers 1 , 2 , 3 … Figure1.4: Contour for integration with pole and branch point with branch cut Figure 1.5: The sign of d

n

0 and at x equal to negative

n

ª¬( x  a ) p º¼ / >d( x  a )@ for positive and negative integer values of n

Figure 2.1: Geometrically explaining the double integration Figure 2.2: Geometrically explaining the double integration

³ ³

³ ³ x

0

x1

0

dx0 dx1

x

x1

0

x0

dx1 dx0 by interchanging the order of integration

and giving the idea of convolution Figure 2.3: Water flowing through a dam weir Figure 2.4: A function describing the shape of the notch of a dam that depends on flow rate Figure 2.5: Two wires with negative charges attracting a test charge with force Figure 2.6: Contours of integration on the complex plane Figure 2.7: The loop redrawn Figure-2.8: Fractional integration is the area under the shape-changing curve Figure 3.1: Fractional derivative of f ( x ) x for D 0 , 1 4 , 1 3 and 1 2 Figure 3.2: Fractional derivatives of a constant Figure 3.3: The semi-derivative of an exponential function Figure 3.4: Taking the half-derivative of a function by Caputo and the Riemann-Liouville methods Figure 3.5: Fractional differentiation of the left hand definition (LHD) block diagram Figure 3.6: Fractional differentiation of 2.3 times in the LHD Figure 3.7: Block diagram representation of the Caputo RHD Figure 3.8: Differentiation of 2.3 times by the RHD Figure 3.9: The function is divided into a slice Figure-4.1: The sign of the coefficients

* ( P 1) * ( P D 1)

in the differ-integration

dD x P dxD

for various P and D values

Figure-4.2: Different regions of the DDx x P for analytic continuation in D  P plane z w

Figure-4.3: The Zero-region (space) wedges in ( z  w) diagram where D x 0 Figure-5.1: Initialized fractional integration example Figure-5.2: Initialized fractional derivative example Figure-5.3: Adding a singularity at the start point to the Caputo derivative to provide an example of RiemannLiouville (RL) to Caputo conversion Figure-6.1: Abel’s tautochrone curve Figure-6.2: Branch cut and contour in a complex plane for integration Figure-6.3: Plot of function ED , E ( t ) , D 0.25 , E 1 and its derivative Figure-6.4: Plot of function ED , E ( t ) , D

1.75 , E

1 and its derivative

Figure-6.5: Plot of function ED , E ( t ) , D

2.25 , E

1 and its derivative

Figure-6.6: Rate of decay function Figure-6.7: Laplace inversion and Laplace transform Figure-6.8: Plot of ED ( t D ) , 0 d D d 1, t ! 0 in linear scale Figure-6.9: Plot of ED ( t D ) , 0 d D d 1, t ! 0 in log scale Figure-8.1: Showing the Bromwich path and the Hankel path Figure-8.2: Plot of the symmetrical M-Wright function for X = zero to 0.5 Figure- 8.3: Plot of the symmetrical M-Wright function for X = 0.5 to 1.0 Figure-9.1: Integer order delta distributed, fractional order delta distributed and continuously distributed order differential equations

Kindergarten of Fractional Calculus

Figure-A1: Illustrating the general behavior of ED , E ( x ) for 0  D d 1 , and D d E Figure-A2: Illustrating the general behavior of ED , E ( x ) for 0  D d 1 , and D ! E Figure-A3: Illustrating the general behavior of ED , E ( x ) for D ! 1 1

Figure-E1: Branch points for function  ( z  z1 )( z  z 2 )...( z  z n ) 2 Figure-E2: Showing branch-cut at negative real axis and two Riemann-sheets Figure-E3: The Branch line and the contour C is deformed into Cc Figure-F1: General pattern of ED , E ( z ) for 0  D d 1 and for E t D Figure-F2: General pattern of ED , E ( z ) for 0  D d 1 and for E  D Figure-F3: General pattern of ED , E ( z ) for D ! 1 Figure-G1: Contour in the complex-plane to evaluate the inverse Laplace transforms Figure-G2: Contour of integration with branch cut in complex plane

xxiii

LIST OF TABLES

Table 1.1: Some examples of Stirling numbers of the first kind 1

Table 1.2: Some values of binomial coefficients, 2 C j ,

 12

C j and their cumulative sums

Table 1.3: Some values of the Stirling Number of the second kind Table1.4: The sign of d

n

n

ª¬( x  a ) p º¼ / >d( x  a )@ for the positive and negative integers n and p

Table 2.1: Integer order integration of the power function with integer exponents with a fractional integration formula Table 2.2: Integer order integration of the power function with integer exponents Table-4.1: Tabulating the values of the Dxn x m for integer m and integer n Table-6.1: A few examples of Laplace inverted functions using the Berberan-Santos method Table-9.1: Composition rule for differ-integrable units. The requirement for this rule is that fX and D Q fX both are differ-integrable Table-9.2: The composition rule for an arbitrary differ-integrable function f . The requirement for this rule is f and Q

D f both of which are differ-integrable Table-9.3: The composition rule for the differentiable but not essentially the differ-integrable function f . The N

requirement for this rule is f is N times differentiable and D f is differ-integrable Table-9.4: Fractional Laplace transforms identities Table-B-1: Laplace transform pairs of some important functions related to fractional calculus Table-B-2: List of Laplace transforms and inverse Laplace transforms of higher transcendental functions related to fractional calculus Table-C-1: List of semi-differentiation and semi-integration of monomials Table-C2: List of fractional differ-integration of various functions Table-F1: Several functional relationships for a positive and negative order Mittag-Leffler function

FOREWORD

Boundaries between different branches of science seem to be melting away, as knowledge progresses. It is as if we are moving towards a unified interdisciplinary ‘natural science’, held together by a framework of mathematics. Novel materials, as well as familiar, but less understood complex systems, lie on the overlapping areas between the compartments we use for classifying ‘subjects’, such as physics, biology, geoscience, chemical engineering and so on. We need new techniques for understanding and modelling these systems. As we know, Hooke’s law works well for many solid materials within the elastic limit, as does Newton’s law for viscous forces, in the case of many simple fluids. However, what about the other familiar materials we always use – things like sticky pastes, ductile metals, starch gels, and so on? The simple linear laws which work quite well under limiting conditions are not of much use here. An ingenious way of treating such systems, suggested by a group of scientists, was to implement generalized calculus in the governing equations, allowing the order of differentiation to take on non-integer values. This opens up a huge field to play with concerning such peculiar and non-conformist systems. However, of course, a vast amount of intimidating mathematical techniques have to be worked out in order to actually solve problems using this crazy idea. In this kindergarten, or play-school, Shantanu Das develops the ideas behind this new field of mathematics, starting from a most elementary level up to actual applications in different areas of science. Shantanu Das has already published a very successful book on this subject, and has been working in this field for many years having applied generalized calculus in control systems, as well as in the study of visco-elastic materials, ion-conducting polymers and other fields. He has also delivered lectures and courses on the subject at different universities and institutes. I believe the reader will enjoy the experience at this kindergarten and graduate to higher levels in order to make fruitful use of the ideas introduced here.

Sujata Tarafdar Professor of Physics Jadavpur University Kolkata 700032, INDIA [email protected]



PREFACE

This book has been compiled from various ‘hand-written teaching notes’ for classroom lectures and ‘presentations’ that I delivered on the topic of fractional calculus at various universities and institutes, from 2005 onwards. The classroom lectures were termed as a ‘Kindergarten of Fractional Calculus’. I must say that book writing is very tough compared to delivering lectures interactively on a blackboard. The name ‘Kindergarten of Fractional Calculus’ (for the classroom talks) suggests that the treatment is very simple; unconventional, yet serious. It suggests that, without going into ‘formal’ theories ,this topic could be developed following a ‘just do it’ attitude, building on our knowledge of classical calculus and mathematics gained at high school and college. I employ this same, informal style in this book. Therefore, I request purists excuse me for this unconventional and non- formal approach, as it is particularly practical. Why did I write this book in such an unconventional way? It is because my students liked the way I presented this complex subject in class, as it made it easy to grasp and interesting for them, as I first discussed the paradoxes and conditions of the subject, before going on, at a later stage, to discuss its formalism. In this book, I have not dealt with physical interpretations; rather, I took time to write each small step that goes towards every derivation, which is essential in order to develop the subject. This being said, the first chapter and appendix contain a number of relevant definitions and descriptions for terms such as analyticity, analytic continuation, residue calculus, Jordan’s lemma, Laplace transformation, gamma function and its properties, psi function, and Stirling’s number, among others. I also use generalized functions (such as the Mittag-Leffler function and its variants), in other chapters. The first chapter deals with the concept of mixed differentiation and integration, product rules and chain rules, etc., of classical calculus; that I used subsequently for concept generalization. In all other chapters, I carry out detailed derivations in their totality, with each step elaborated and explained. Thus, I am not economical with page space and the descriptive language used. Purists may term this as a verbose treatment, but students nevertheless appreciate this unconventional approach of opening out each step in the derivation process with descriptive explanations, especially those studying Engineering and Applied Sciences. This type of detailed treatment is missing in most of the existing literature as details of derivations are skipped, which students and users usually find difficult to absorb. I wrote this book in this unconventional way mainly because of demand from the student community in my part of the globe. In 2008, the organizers of the ‘International Mathematics Olympiad’ (INMO) at Mumbai asked me to introduce this subject of fractional calculus to a select group of class XI and XII school students. That was a challenge. I started with the idea of tossing several coins. I then arrived at a formula for a Probability Generating Function (PGF) - for one coin, for two coins and then extended for n - coins. Then, I said to the students, “let us put n equal to ½ and see how ‘a half-coin’ should behave”. Mathematically I demonstrated to the students that we have constructed this half-coin, but I stated, “we have our limitations today, because we are presently unable to attach much physical sense (especially in terms of the notion of ‘negative’ probability) to this new construct”. Hence, we are in a paradoxical situation, and I made it clear to the students that “the paradox is because mathematics goes far beyond our physical understanding”. This is how I began the concept generalization. With this example, I stated, “as we have one whole differentiation and two whole differentiations, and n -whole differentiation, we can have ½- differentiations or ½ - integrations”, and proceeded in a very simple way to develop the concept, which was once considered a paradox. Today however, we have a physical and engineering idea of it. This physical and engineering sense I have already dealt with in my previous books and other publications, listed in the Bibliography section. Thus, we can say fractional calculus is concept generalization for the existing classical calculus theory; it is also termed as generalized calculus and sometimes even called on-Newtonian calculus. It is not a paradox anymore. This introduction of the subject to the students gave me confidence that, if I develop the paradoxical, complex-looking mathematics in a different, easy and interesting way, perhaps the students will be hooked on this subject. To some extent I was successful as I delivered detailed classes at Jadavpur University and Calcutta University, and some short classes at Pune University, Mumbai University, IIT- Kharagpur, VNIT- Nagpur, and at several other places. The result is that I see the growth of this subject in this part of globe. Students of Mathematics, Physics, and Engineering have taken up this subject for further research. Some students are presently working, or have previously worked with me in such fields on this subject. Here, in this book I aim to deliver detailed derivations, in a simplified though unconventional way, of the various aspects of the beautiful mathematics of fractional calculus. Recalling the simple classical integration process that is viewed as the area under the curve (that is, the area under the original function being integrated), we take into account all of the values of the function from the present point of interest to the start of the function. We recall that we memorize all the past points (in a causal sense), and sum them up

Kindergarten of Fractional Calculus

xxvii

in order to carry out the classical integration process. The ‘fractional integration’ that we will learn is also the area under the curve, but it is an ‘area under a shape-changing curve’, which keeps on changing as we move ahead. Fractional integration, too, has to account for all the previous values of the function, but here different weights (with decreasing value) are multiplied by each previous and past value of the function, giving a real ‘fading memory’ effect. This is reality as the past memory always fades as we move ahead. Differentiation, as we know classically, is a slope at a particular point in a curve. The ‘fractional differentiation’ that we will learn will be a slope at a particular point of a function, which is the ‘area under the shape changing curve’. Therefore, in order to evaluate fractional differentiation at a desired point, I should consider all the past values of the function. That is the fractional derivative, which has an in-built fractional integration process. We will be calling this the ‘differ-integration’ process. We will also see the embedded ‘memory’ in the concept of fractional differentiation This means that all the past points with decreasing weights are considered in order to evaluate fractional derivatives at the point of interest; unlike classical differentiation, which is a point property or local property. Today, we have several engineered systems based on fractional calculus. These engineered systems, are based on the use of the ‘fractional Laplace variable’, which appears when we conduct Laplace transforms of fractional derivatives and fractional integration operators. We will also learn this here in this book. With the fractional Laplace variable, we have developed analog and digital electronic circuits allowing fractional differentiation and fractional integration. We will also study the use of the Laplace transform method to solve fractional differential equations here. The picture in Figure-P1 below shows a magnetic levitation system, where the metal ball is floating in the air, and is being controlled by current in the coil of a magnet. This current is being governed proportionately by fractional derivatives and fractional integration of the error signal for the ball’s position (Courtesy: BRNS funded joint project of VNIT Nagpur and BARC, which developed a digital fractional order controller for industrial applications. This system also demonstrates that by using fractional calculus in control, we are getting efficient controls, as compared to the classical schemes using classical calculus. This is depicted in Figure-P2. Comparing these two pictures, we see that, in order to do the same job (that is, to position the floating ball and slowly make it follow sinusoidal command), the voltage output of the controller in the case of the second case is fluctuating severely. Many critics will say it is ‘noise’, but why is the ‘so called noisy’ output absent in fractional calculus-based systems? We are using the same electronics and only changing the program of the processor (in this case, the microcontroller); once for classical control governed via classical differentiation and integration, and in the next case for control governed via fractional calculus. The justification for having better efficient control via fractional calculus is that the fractional differentiation and integration operations possess inherent memory. This memory in the system works in order to govern the ball’s position based on its previous or past experience. Therefore, this fractional differentiation and integration process gives us an ideal filtering action, whereas classical differentiation is a point (or local) property that does not therefore have memory, and which acts instantly with no previous experience. Also, classical integration is the summation of all the previous values of function when all are equally weighted; whereas fractional integration is also the summation of previous values, but with decreasing weights. Thus, fractional differentiation and fractional integration give us a memory action where memory fades as we move on. We will study how memory is inherently embedded in fractional calculus.

Figure-P1: Picture showing magnetic levitation where fractional calculus is used for controls

xxviii

Preface

Thus, in the second case, the manoeuvring signal is instantaneously oscillating very dramatically, again giving us the notion that some ‘noise’ is being injected into the system. We also infer in the second case that the controller (with classical calculus) is making a lot of effort to control this ball and keep it afloat, while in the first case the controller’s action (based on fractional calculus) is smooth and effortless. So we can see that the fractional calculus-based system does the control action with less effort than the conventional classical calculus based controllers, and that it is, therefore, better and more efficient. Figure-P3 gives us the experimental records for CRO traces of control voltage (upper trace) and ball position (lower trace) in detail.

Figure-P2: Picture showing magnetic levitation where classical calculus is used for controls

Now we pose a question- what is the implication of ‘lesser effort’ by a control system doing the same job? We have carried out another development where we use classical and fractional calculus to regulate (control) the speed of a DC motor. Figure-P4 gives the picture of the full setup of the DC motor speed control system. Figure-P5 gives us the record of armature voltage and current, while the DC motor is controlled by classical calculus. The multiplication of armature voltage and armature current gives 231.07 Watts as input power to the armature of the DC motor, while regulating the speed at 1000 RPM. Figure-P6 gives the same motor running at the same speed of 1000 RPM but controlled by fractional calculus, which shows an intake power by an armature of 181.61 Watts. Thus, to run a DC motor at a speed of 1000 RPM, the classical calculus-based system takes 17.3% more power than the fractional calculus-based control system. The experiment is done at several speed settings from 500RPM to 1300RPM and at all speeds we observe lesser power intake, when control is based on fractional calculus. This gives us a clue that by using fractional calculus, we are achieving ‘energy/fuel efficiency’. Maybe in future, the industry will adapt this new mathematics to make a fuel/energy efficient control system; at least, I hope so!

Digital CRO traces for control output voltage, and ball position of classical calculus

Digital CRO traces for control output voltage, and ball position of fractional calculus

Figure-P3: Picture showing CRO traces for control voltage and ball position for both systems with classical calculus and with fractional calculus

Kindergarten of Fractional Calculus

xxix

Figure-P4: DC motor speed control setup

Figure-P5: Armature voltage and current at 1000 RPM DC motor with classical calculus

Figure-P6: Armature voltage and current at 1000 RPM DC motor with fractional calculus

Anyway, how is the efficiency achieved? Is it because by using fractional calculus in my control action, I am interacting with the actual plant (i.e. the DC motor or the magnetic levitation system) in a better and more efficient

xxx

Preface

way? Therefore, can I say that fractional calculus is the mathematics with which natural dynamic systems operate better? This thought came to my mind while I was using logarithmic logics and ratio control and doing derivative operations on these logarithmic domains, for nuclear reactor control. The usage of logarithmic logic gave me better and more efficient nuclear reactor control than conventional use of linear logics, and these ‘new’ formulas were implemented in nuclear plants, in 2002. When asked why, I advised that the reason was that the way to govern ‘a natural, exponential system’ (i.e. the nuclear reactor) was using logarithmic power error and its derivative in logarithmic domains which closely match the language of the system. Therefore, maybe this hypothesis is increasing the efficiency of the control action. In the servo-system, the servant, i.e. the plant (the nuclear reactor), is efficiently able to understand the master that is the controller’s command (language, in a logarithmic and exponential domain). Perhaps, in the logarithmic case, ‘we are talking to the plant to be controlled (i.e. the nuclear reactor)’, which is naturally exponential in the language of the process. This delivers an efficient way of achieving better control. The point, which is emphasized here, is that if we communicate in the language of the dynamic system then we will be communicating better. Thus for efficient communication, ‘communicate in French with persons in France’! The above experiments do point out that perhaps fractional calculus is the language that dynamic systems understand better! I had an opportunity to talk on the topic of ‘fuel efficient controls’, in Beijing, China, at the International Conference on Nuclear Engineering (ICONE- 13), in 2005; for the first time worldwide. At that point, it was my conjecture or hypothesis that we should apply non-Newtonian calculus to achieve fuel efficiency. Today, we have energy/fuel efficient controls practically realized, via non-Newtonian calculus. I feel blessed that I could see, within my life span, the original conjecture or hypothesis of mine regarding energy/fuel efficient controls realized practically and become a reality. Therefore, this is one motivation to provide a course on fractional calculus for Science and Engineering students. Here, I may also mention that tuning the controller of a plant to either classical calculus or fractional calculus (say DC motor, magnetic levitation system, nuclear reactor, etc.) uses minimization of the chosen performance index (PI). The PI is a function of the controller’s parameters that we set, and the plant to be controlled is described by plant’s transfer function. By using the minimization technique, we obtain the values of the controller’s parameters. This minimization technique, for minimizing the PI is like minimizing the least square error (LSE), call it ‘ E ’ for the curve fitting the polynomial. Here I give one example that we can get a better fit with a lower value of E if the curve is composed of fractional power monomials. Let us say we have n data points x1 , x2 , x3 ,..., xn and that the corresponding values are y1 , y2 , y3 ,..., yn . Assume by examination of the plotted points; we say linear fitting is good here for this set. Therefore, we need a linear function; E is a function of a and b (i.e. y f ( x ) ax  b is to be fitted so that LSE, that is,

E( a , b )

n

¦ y

i

i 1

2

 ( axi  b ) ) and is minimized. Following simple mathematics: w w a



>E(a, b)@ 2

2

2

¦ 2

n

x i 1 i

 yi  (axi  b)

0



a x1  x2  .......  xn  b  x1  x2  ......  xn w w b

>E(a, b)@

2

n

¦ i 1  yi  (axi  b)

x1 y1  x2 y2  .... xn yn

(P1) 0

n

 

a  x1  x2  ......  xn  b  b  .......b

y1  y2  .... yn

Compactly we use a set of equations, i.e.

ª ¦ n xi « i1 « n x2 ¬¦ i 1 i

º » ªa º « » n ¦ i 1 xi »¼ ¬b ¼ n

ª ¦ n yi º « i1 » « n xy» ¬¦ i 1 i i ¼

(P2)

to get the values of a and b , which minimise LSE, i.e. E( a , b ) . The above expressions ((P1) and (P2)) can be used for the linear fitting which has two degrees of freedom, namely a and b . Now, if E( a, b) z 0 , then we try another D

function ( y ax  b ), and search for the value of, which is close to one (i.e. D  1.00 ) to make the obtained E from the linear fit still lower. This we demonstrate with an example. Take six data points with n function, y ax  b :

6 as follows where we want to fit a linear

Kindergarten of Fractional Calculus

xxxi

i 6

^ xi `i 1 ^0, 0.5,1,1.5, 2, 2.5` i 6 ^ yi `i 1 ^0.43, 0.17, 3.12, 4.79, 4.85, 8.69`

(P3)

Using the formula as indicated above (P2) with n 6 using the above (P3) values of the data points, we get E minimized with a 3.56 and b 0.97 . Therefore, we have a linear function y 3.56 x  0.97 , giving us minimum 6

¦  y  (3.56 x  0.97)

E as E( a , b )

i

i 1

i

2

9.47 . The plot of data fitting by a linear curve y

3.56 x  0.97 gives

minimum E . Now we search for D  1.0 with the new function y

E( a , b, D ) =

¦

6 i 1

 y  (ax i

D

i

 b)



D

3.56 x  0.97 , and see the following results for

2

, which now also depend on D :

¦

E( a , b, D )

6 i 1

 y  (ax i

3.56 ; b

a

D

 b)

i



2

0.97

D

1.05

E

10.93

D

1.00

E

9.47

D

0.95

E

8.56

(P4)

0.90 E 8.14 Thus, we have the possibility of lowering E from the obtained value 9.47 , by choosing the non-integer D . What did the fractional order monomial do? The obvious answer is that it gave me an extra degree of freedom in the parameter D  1.0 . Therefore, in this case, we have three degrees of freedom, namely a , b and D as compared to two degrees of freedom in a linear curve fitting. Now the question comes, how did this curve-fitting example suit our control system example?

D

If we say that the variable x is a Laplace variable called s , then we already know from classical calculus theory, that operator s corresponds to one-whole differentiation, and operator s

1

corresponds to a one-whole integration

D

operation. We will learn in this book that s corresponds to fractional differentiation if D ! 0 , while D  0 corresponds to a fractional integration operation. The polynomial representation of a classical calculus-based controller in the Laplace domain is fint ( s )

k p  ki s

1

 k d s . This controller has three degrees of freedom, namely

k p , ki and k d in order to minimize a control system PI. The latter is a function of controller settings (i.e. k p , ki , k d

) and the plant to be controlled. After evaluating these three parameters in order to minimize PI, we set these values into the control system, while the fractional calculus-based controller structure in the Laplace domain is

ffract ( s )

k p  ki s

O

P

 k d s . This is with five degrees of freedom, namely k p , ki , k d , O

and P ; with

0   O , P  2 ; that is enhancing the degree of freedom with these two extra fractional indices. With the extra

freedom of non-integers O and P , we are enabled to minimize the chosen PI still further as compared to classical case; as we demonstrated in the curve fitting example. This is the essence of fuel/energy efficient controls using fractional calculus.

xxxii

Preface

Figure-P7: Memorizing charging time of applied electric- field in experiments with Laponite for crack formation studies

Figure-P7 presents experimental evidence that a relaxing system (in this case, a Laponite stressed with a DC electric field) has a memory of being connected to a voltage supply during time W and continues as if it is still connected to a non-zero voltage source, even after the voltage source is switched off. The figure shows the variation of discharge voltage Vd (t, W) with time after the electric field is switched off at time t W for different applied voltages (Courtesy of the Condensed Matter Physics Research Center (CMPRC), Dept. of Physics, Jadavpur University Kolkata). The self- discharging curve is a function of time ( W ); therefore in a way the relaxing system is memorizing its history of the charging profile that is described by W . This explanation is only possible if we have a fractional capacitor present in the relaxing system. The impedance of the fractional capacitor is given as Z ( s ) s D CD 1 , which is described in the Laplace variable s . This fractional capacitor has a voltage current relation via fractional calculus, i.e. i(t )

CD

dD dtD

v(t ) . This means that the current i(t )

through a fractional capacitor is a fractional derivative (of the fractional order D with 0  D  1 ) of the voltage v(t ) across it. For D 1 , we get the classical capacitor that we studied in textbooks which has a terminal relationship as

i(t )

C

d dt

v(t ) . We note here that the unit of classical capacitor ( C ) is in Farads while the unit of fractional capacity 1D

CD is in ‘new units’ of Farads / sec . The study of these ‘new types’ of units is also a topic of fundamental research. The memory phenomena are described in the self-discharging curves of Figure- P7 which are different for 15, 30, 45 and 60seconds of charging time W , and which would not be observed if the capacitor behaved as a classical ‘geometric’ capacitor, as described in textbooks. This memory phenomenon is also observed in supercapacitor selfdischarging curves. The presence of a fractional capacitor is identified by the charge-discharge curve of the super-capacitor. See FigureP8(courtesy of the Department of Electrical Engineering, NIT- Raipur) for a joint project with BARC on the identification of fractional orders of super-capacitors. Here we charge a super-capacitor with a constant current source of , 10mA, 20mA and 30mA. These charging currents raise the voltage across the super-capacitor from zero to about 3V, and at this 3V point we discharge the super-capacitor by reversing the constant current to 30mA, until the voltage across it reaches zero.

Kindergarten of Fractional Calculus

xxxiii

Figure-P8: Charge discharge of a super-capacitor showing the presence of a fractional capacitor

If the super-capacitor possessed a textbook type of ideal capacity, then the voltage rise across it would have been a linear function of time (ramp- up i.e. v(t )  t as v(t )

t

³ IdW ) while charging (i.e. proportional to the integration of a 0

constant current; that is, I ). Instead, what we observe is a curved path of voltage rise and fall, which is proportional to D

v(t )  t with 0  D  1 , instead of the ideal which is v(t )  t . These charge-discharge curves are used to measure or find the fractional capacity CD and the value of D present in a super-capacitor. This interesting experiment shows that when the value of D is close to one, the ideal value (but not one) as the charging rate (Coulombs/second) is smaller, and that D is much less than one when charging current is high. Similarly, it is observed that the fractional capacity CD is close to ‘geometric’ capacity ( D  1.00 , i.e. the ideal textbook capacity) for lower charge rates. Not only do super-capacitors show this fractional capacity and memory effect, but large numbers of experiments on insulators also exhibit this. A paper dielectric capacitor which is experimented with is found to have D  0.9 . Thus, D z 1 is a reality; therefore, fractional calculus comes to describe circuits with fractional differential equations, and takes into account fractional capacity, which is a reality. I have been a collaborator and guide of all these engineering and physics experiments that I have described above along with my colleagues (both teachers and students) from Jadavpur University, VNIT- Nagpur, NIT Raipur, and IIT Bombay. Much experimental evidence gives memory-based dynamics like: in the characterization of visco-elastic materials; dielectric relaxation; super-capacitor charging/discharging, wherein the super-capacitor memorizes the ‘time’ that it has been kept in a charging process; and in impedance spectroscopy study of transport phenomena for electrochemical systems, plus several other phenomena, where we have applied fractional calculus. Similarly, the distributed system behaves with fractional order dynamics where we represent it not as a point, but rather having a spread, and thus they are non-local in terms of behavior. As an example, a distributed system behaves like a semi- infinite system, with half-order dynamics. The transport phenomena in porous, rough and fractal media have fractional order dynamics; we have also observed fractional order dynamics in the cooling of liquids. These facts do motivate us to learn fractional calculus and apply the concepts. These above mentioned physics and engineering aspects will not be discussed in this book. One interesting fact is that the points in the graph that are non-differentiable have finite value in their fractional derivative. We use this feature to characterize the points of an ECG graph for differential diagnostic purposes. Thus, fractional calculus is a tool to describe non-differentiable systems, and we will study this aspect in this book. This shows the reality of having a course in fractional calculus, at least at the post-graduate level, for the benefit of further research in mathematics, science and engineering. Calculus is about evaluating rate, i.e. changing the dependent-quantity (say ' f ) per unit of differential independent quantity (say 'x ). In classical calculus, we take this differential independent quantity (variable) towards zero, i.e. in

xxxiv

Preface

the limit 'x p 0 , which gives rise to the notion of a classical derivative ( f (1) ( x ) classical integration ,i.e. f ( 1) ( x )

d dx

f ( x ) ) and (anti-derivative)

³  f ( x) dx . Both classical derivative and classical integration are conducted with

respect to the differential quantity, i.e. 'x , which is made zero. The question is if we have a finite spread of a differential element, by which we mean a classical differential element to the power of a fractional number (less than D

one); i.e. ( 'x ) , with 0 < D < 1 ; then we will have derivative and integration with respect to this finite spread of the D

D

element i.e. (dx ) , with ( 'x ) ! 'x , while we take the limit of 'x p 0 . Thus, we will be dealing with a fractional D

derivative and fractional integration, taking rate (derivative) and integration with respect to ( 'x ) calculus.

into fractional

This argument shows that, when we are observing a system with coarse-grained phenomena, we will be using fractional calculus to describe its dynamics. Why is this coarse graining required? The reason is that this method gives us a way to view the system, taking into account the homogeneities and the roughness, etc. present in real physical systems. On the other hand, the conventional fine graining approach gives us a view of macroscopic behavior. Thus, we move away from a point quantity (which is ideal) to a non- local quantity, which is reality, and thus we use fractional calculus. We will also discuss the development of fractional calculus in conjugation with classical calculus for non- differentiable functions. We may use fractional derivatives to deal with differentiable functions, but we must ask questions about whether this is essential. As far as a function is differentiable and a system is Markovian (i.e. the system’s present state is not affected by its past states), then its dynamic equation is then quite well-defined. If its dynamics are differentiable, strictly speaking, we should not need fractional derivatives in order to analyze its behavior. We may suggest that a (1)

given system, which is defined by a differential equation (such as the classical one, i.e. x (t )

f (t ) ) will show

(D )

dynamics with a fractional derivative i.e. x (t ) g (t ) and will therefore contribute extra information. Many system dynamics show memory-based behaviour where a non- Markovian approach is required; there the dynamics (D )

x (t ) g (t ) , with fractional derivative have been employed. Those ones that we described in Figures P7 and P8, are dynamics with memory. However, if the function in consideration is not differentiable, fractional derivatives should be of use. This concept of using a fractional derivative suitably defining non-differential functions in conjugation with classical calculus is a developing subject. As I mentioned, this book is mainly comprised of all my previous hand-written classroom lecture notes that I have used for more than a decade for teaching in several classes on the subject of fractional calculus, and the treatment here is unconventionally simple, without rigorous formalism, and with elaboration of every step of the discussion. I can guarantee that a high school/college student who has studied differentiation and integration will be hooked on this book, and he/she may develop a liking for this subject, and perhaps even later work towards the advancement of fractional calculus and its applications. I cannot claim that each aspect of fractional calculus will be addressed in full in this book, but I can definitely claim that it will generate immense interest amongst new researchers in order to further investigate and use, develop and apply this subject. I have learnt this subject from various pioneering studies that I have listed in detail in the Bibliography section; and I provide a few notable works at the end of each chapter. I humbly submit that my primary motivation was to make and then engineer electronics circuits and systems, and use fractional calculus in automatic controls for ‘robust and efficient systems’. I started on this subject on my own, more than a decade and a half ago without any guide, and picked up the mathematics part, and simplified the same all by myself; primarily for my own sake. I enjoyed taking this course and enjoyed every bit of interaction with all my students and professors, learning a lot from them. At this time, the subject did not exist for students at most universities and institutes in India, although it is gradually growing in prominence in this part of the world. I am still trying to develop this subject, and its applications in science and engineering, and I am still learning, too. I thank all my colleagues who have taken up this subject passionately with me. I am grateful to my colleagues (both students and professors) who have allowed me to propagate this “new subject” and deliberate with them, even though I am from an engineering background and not qualified with any higher academic degrees. Even today, this subject is not widely accepted, and still I feel no harm in pursuing it. I have several light-years to travel before I fully know what this subject is, and, if our nature truly follows the mathematics of fractional calculus. I wish you pleasant reading.

Shantanu Das

ACKNOWLEDGEMENTS

I have learnt this subject from the pioneering works of several scientists and mathematicians who have enriched this particular subject all over the globe. I convey my gratitude and acknowledgement to all of them. All of their pioneering work has inspired me and provided me with knowledge of this subject, which I taught for a decade and a half at various universities and institutes. I have listed all of their work in the Bibliography section. They are must-read publications. I am grateful to Professor Michelle Caputo (who founded the Caputo fractional derivative in 1967) who read my first edition of Functional Fractional Calculus (2007) with great interest. Professor Caputo gave his appreciation on the modern treatment therein and encouraged directions for further research on this subject. My sincere thanks go to Professor Caputo for having also gifted me with reprints of his pioneering work, which has been a source of inspiration. I took inspiration and learned the subject from several pioneering scholars, including Professor Kenneth S. Miller; Professor Bertram Ross; Professor Keith B. Oldham; Professor Jerome Spanier; Professor Guy Jumarie; Professor Om Prakash Agarwal; Professor J.A.T. Machado; Professor Juan J. Trujillo; Professor Virginia Kriyakova; Professor Alain Oustaloup; Professor Francesco Mainardi; Professor R. Gorenflo; Professor Stefan G. Samko; Professor Katsuyuki Nishimoto; Professor Igor Podlubny; Professor R. Hilfer; Professor Kiran M. Kolwankar; Professor Anil D. Gangal; Professor Varsha Gejji; Professor Carl F. Lorenzo; Professor Tom T. Hartley; Professor John W.Hannekan; and Professor B. N. NarahariAchar. I learnt lot from their pioneering works, and I have duly placed them in the Bibliography section. I consider these scientists and mathematicians as founders of modern fractional calculus of the twenty-first century and salute them. I also acknowledge the encouragement received from Professor Sujata Tarafdar, and Professor Tapas Ranjan Middya of Jadavpur University Kolkata for instituting scholarship for MSc students of pure and applied sciences, and introducing this subject as a special course, at Jadavpur University. I thank them also for inducting this subject into their condensed matter research, and applying it in various physics experiments. I also express my gratitude to Professor Mohan Aware, Professor Ashwin Dhabale, and Professor Anjali Junghare of the Department of Electrical Engineering of Visvesvarya National Institute of Technology (VNIT), Nagpur. They have taken on a challenge in developing analogue and digital circuits with me for fractional order controls, and have set up a laboratory that is the first of its kind in the world (the Fractional Calculus Engineering Laboratory) to carry out further studies and to make circuits and systems that have energy/fuel efficient and robust control systems. I thank Professor Susmita Sarkar and Professor Uttam Ghosh of the Department of Applied Mathematics of Calcutta University, for inducting my lecture notes into the academic curriculum for MSc, M- Phil, and PhD in applied mathematics, and for giving me the opportunity to teach the subject at detailed classroom sessions for postgraduate and PhD students at Calcutta University and Jadavpur University. I acknowledge Professor Tapati Dutta of the Physics Department of St. Xavier’s University Kolkata for introducing this subject and my lecture notes to her undergraduate students and their experiments on physics. I acknowledge the encouragement received from Professor Siddharta Sen and Professor Karabi Biswas of Indian the Institute of Technology (IIT), Kharagpur, Professor Munmun Khanra of N.I.T. Silchar, and Professor Jaylexmi Nair of Mumbai University for introducing my work to their students. I also acknowledge Professor Amitava Gupta of Jadavpur University, Professor Vivek Agrawal of the Indian Institute of Technology Mumbai, Professor Subhojit Ghosh of N.I.T. Raipur, and Dr N. C. Pramanik of CMET Thrissur for accepting the concept of fractional calculus, and for their super-capacitor studies and experiments on system identification and controls. I acknowledge Professor S. Saha Ray of the Department of Mathematics N.I.T. Rourkela, who worked with me on continuously variable fractional order systems. I also thank Professor Suchana Mishra at Dayananda Sagar College of Engineering Bengaluru for having been sufficiently motivated by my teaching/lecture notes and publications to apply the concept of fractional cross product and fractional curl in obtaining radiation patterns for antenna systems . I thank the team of students named as follows: Saptarishi Das, Moutushi Dutta Choudhury, Tania Basu, Somasri Hazra, Simantini Majumdar, Tajkera Khatun, Indranil Pan, Suman Saha, Anish Acharya, Anindya Pakhira, Rutuja Dive, Swapnil Khubalkar, Amit Chopade, Ruchi Jain, Mano Ranjan Kumar, Sumit Mukherjee, Basudev Majumder, Tridip Sardar, Sudeshna Sircar, Joydip Banerjee, Srijan Sengupta, Santanu Raut ,Jitesh Khanna, Vamsi, A Rajashekhar, Ramjan Ali, Adreja Mondol, Rivu Gupta, Archishna Bhattacharya, Geethi Krishnan, and Tapas Das who have worked (and are continuing to work) with me on fractional calculus in various fields of Engineering, Mathematics and Physics.

xxxvi

Acknowledgements

I also thank Nader Maleki Moghaddam of Teheran Polytechnic, Dept. of Energy Engineering & Physics, Iran, who has taken up the concept of fractional divergence and applied the same thinking to nuclear reactor flux mapping problems. Thanks are also due to Professor Maamar Battayeb of the College of Engineering at the University of Sharjah who appreciated my previous books and lecture notes and presentations, and used them to guide his students, and Professor Hammad Khalil of the University of Education Lahore, Pakistan, who appreciated my earlier publications on this subject and used them as teaching matter. I thank Saurav Kumar Giun (FCD), BARC, Alok Kumar (RED), BARC, and Dharmendra Singh Azad Hindustani (SIRD), BARC, for using my lecture notes of mine to learn this subject and to explore applications of fractional calculus in their respective fields of research and development. I acknowledge my senior colleague Dr. .Zafer Ahmed (NPD), BARC, who encouraged me to deliver this topic to school students at the ‘International Mathematics Olympiad’ (INMO) in 2008, and who encouraged me to simplify the topic for the general student, especially with detailed step-wise derivations. I sincerely acknowledge the encouragement of Dr A. P Tiwari (Head HRDD BARC), Sri B. B. Biswas (Ex- Head Reactor Control Division, BARC), and Sri G. P Srivastava (Ex- Director Electronics & Instrumentation Group BARC) who gave me the opportunity to work on, and expand, this subject. My special gratitude and acknowledgement also go to Dr. Srikumar Banerjee (Ex- Director BARC and Ex- Chairman DAE), who encouraged me to carry on with development of ‘curiositydriven science’ and on the subject of fractional calculus.

Shantanu Das Scientist Reactor Control Division, E & I Group, Bhabha Atomic Research Centre (BARC) Mumbai- 400085 (INDIA) Honorary Senior Research Professor, Dept. of Physics Jadavpur University, Kolkata (INDIA) and Ex- Adjunct - Professor DIAT Pune (INDIA), Ex- UGC- Visiting Fellow, Dept. of Applied Mathematics, Calcutta University Kolkata (INDIA). [email protected], [email protected], [email protected] HUhttp://scholar.google.co.uk/citations?user=9ix9YS8AAAAJ&hl=enUH, H www.shantanudaslecture.comU https://www.researchgate.net/profile/Shantanu_Das9

CHAPTER ONE CONCEPT GENERALISATION

1.1 Introduction In this chapter, we will do a ‘concept generalisation’ of what we already know about

d dx

f ( x ) or

x

³  f ( y ) dy i.e. the 0

usual classical derivative and classical integration operation in order to get expressions for n fold differentiation and integration. The further generalisation trick is to have n as non-integers; that we will derive in subsequent chapters. The concept generalisation is like that from natural numbers i.e. n 1, 2, 3, ... we generalise the number line to the negative side i.e. n 1, 2, 3, ... and call them integer numbers. Then we have, in between these numbers:

r 12 , r 13 , r 34 , r 32 ,... or even irrational ones like S , 2 ,  3 etc. and we call them the real line. Similarly, we extend one whole differentiation i.e.

d> f ( x ) @ dx

generalise to get the expression for n folds, that is x

³  f ( y ) dy 0

to

d 2 > f ( x )@

to have twice the differentiation i.e.

n folds integration, that is

x

d n > f ( x )@ dx n

, then thrice i.e.

d3 > f ( x )@ dx3

and

. Similarly we generalise one whole integration i.e.

³  f ( y ) (dy ) 0

dx 2

n

, or

³

x

0

dxn 1

³

xn1

0

dxn  2 ...

³

x2

0

dx1

x1

³  f ( y ) dy . The 0

extension of our classical calculus to fractional calculus is a generalisation of the theory of calculus. This extension in terms of mathematics is an ‘analytic-continuation’ of the operation i.e. n - fold differentiation and integration from the

n integer to entire complex plane z . So we can have the derivatives (or integrals) in between like we have operation

dD dxD

d

1

dx

2

1

2

f ( x ) . That is,

f ( x ) , where D is arbitrary, a ‘real number’, a ‘complex number’, even a ‘function’ as a

continuous distribution. A positive D will signify a differentiation process and a negative D will give us integration (an anti-differentiation process). This is a generalised theory of calculus, a subject as old as Leibniz’s or Newtonian calculus, and is called fractional calculus. Like further generalisation, can we have D as a complex number? The answer is why not. Similarly, can we have D as a continuous distribution function? The answer again is why not. We can also have a system where D is a variable. Therefore, there is fun in learning the subject of fractional calculus. In this chapter, we will be dealing with concept generalisation, with examples, and describing various important functions that are the outcome of the process of generalisation. In addition, we will apply this generalisation to obtain interesting relations in classical calculus theory, for n  fold integration and n  fold differentiation, like chain rule, Leibniz’s rule etc. This chapter thus demonstrates the concept generalisation process.

1.2 The thought problem for constructing a ‘half-coin’: an example of concept generalisation 1.2.1 The idea of Negative Probability through the generalisation of existing theories The idea of negative probability was considered by Richard P. Feynman and M. S. Bartlett as a generalisation of existing theories. Bartlett wrote in ‘Mathematical Proceedings of Cambridge Philosophical Society’ (1945), “It has been shown that orthodox probability theory may consistently be extended to include probability numbers outside the conventional range, and in particular negative probabilities. Random variables are correspondingly ‘generalised’ to include extraordinary random variables; these have been defined in general, however, only through their characteristic functions. This ‘generalised theory’ implies redundancy, and its use is a matter of convenience. Negative probabilities must always be combined with positive ones to give an ordinary probability before a physical interpretation is admissible.” Therefore, the notion of negative probability is the process of concept generalisation. In the following example, we elucidate the possibility of having negative probability, by constructing a hypothetical ‘half-coin’. This notion of a half-coin construct was first introduced in 2005 by Gabor. J. Szekely, and further worked on by J. A. T. Machado, for possible applications in physical systems and in relation to fractional calculus. This concept of negative probability, though it sounds absurd, has recently been getting attention regarding its possible usages in physical systems.

2

Chapter One

1.2.2 The tossing of coins and Probability Generation Functions (PGFs) We know about the full process of coin tossing, and we construct this process in order to realize a Probability Generating Function (PGF). A fair coin has two sides, Heads (H) and Tails (T), which can be represented as ‘states’ say X 0 and X 1 with equal probabilities that are 12 . We write PGF as G1- X ( x ) , indicating the PGF of ‘onewhole coin’ as having two states X

0 and X 1 as in the following expression: 0 1 G1- X ( x ) 12 (1  x ) 12 x  12 x

 P(X : 0) x In (1.1) we call P(X : 0)

1 2

0

1

 P(X : 1) x

(1.1)



the probability of getting a ‘Heads’, i.e. for the first state corresponding to the variable

0

x . Then we have P(X : 1)

1 2

as the probability of getting a ‘Tails’, i.e. for the second state, corresponding to the

1

1 2

variable x . In the PGF of a fair coin, we have a function that is

1 2

(1  x ) . If we write the PGF as

(1  x ) , then we

will have P(X : 1)  12 , assigning negative probability for state two (i.e. for X = 1 ). This looks absurd but mathematically we can have it. Up until now, we have been unable to attach any useful physical meaning to a probability which can be negative or which can have a value of probability greater than one. One may argue that probability as a positive number signifies the occurrence of an “event”, so a negative probability may signify an “antievent”! With two such fair coins, we have the PGF as the following expression, represented as G 2- X ( x )

 G1-X ( x)  G1-X ( x) 2  12 (1  x)

G 2- X ( x )

0

1 4

TH). The probability P(X : 2)

1 4

corresponding to the state-variable x that is state one, and is for the 1

1 2

event HH. The probability P(X : 1)

2 0

1 4

In (1.2) we have the probability P(X : 0)

1

x  12 x  14 x

(1.2)

corresponds to the state variable x that is state two and is for event HT (or

corresponds to the state variable x

2

that is state three and is for event TT. We

extend this PGF to three coins and write the PGF i.e. G 3- X ( x ) as follows

 12 (1  x)

G 3-X ( x )

1 8

We have four states with probabilities

1 8

,

3 8

,

3 8

and

1 8

0

3

1

2

x  83 x  83 x  18 x

(1.3)

3

respectively, for the events HHH, TTH, HTT, HHT, HTH, and

TTT. We extend this (1.3) to the n -coin system; and the ( n  1) states with their corresponding probabilities are expressed in the following steps

 12 (1  x)

G nX ( x)

1 2

n

n

1  nx  1 2n

0

x 

n ( n 1) 2!

n 2n

1

x 

2

x 

n ( n 1)( n  2) 3!

n ( n 1) 2n 2!

2

x 

3

x  .... x

n ( n 1)( n  2) 2n 3!

n



3

x  .... 21n x

(1.4) n

1.2.3 Generalizing the PGF to get a half-coin construct: a paradox? We ask ourselves what happens if n

1 2

, that is in effect, a ‘half-coin’! We place n

1 2

in the above formula (1.4)

and get the PGF of the half-coin denoted as G 1 - X ( x ) , following a series of infinite ‘events’ and ‘anti-events’ (we use 2

the term ‘anti-events’ for state-variables with negative probabilities).

Concept Generalisation

 12 (1  x)

G 1 -X ( x) 2

1 §

1

2©

2

¨1 

1 2

1 2

 12   12 x 2   12   12   32 x3  ..... ·

x 1

0

x 

2!

1

2 2

3

x 

1 8 2

1

2

x 

(1.5)

¸ ¹

3!

3

16 2

x  ...

Now with two half coins we get a PGF of one-whole coin as demonstrated in the following steps:

G1- X ( x )

G

1 -X 2

( x)

 G

1 -X 2 1

( x)



 12 (1  x)  12 (1  x) 2

1

2

(1.6)

1 2

(1  x ) That is (1.6) the PGF we have for a single fair coin (1.1). But what about an infinite number of ‘anti-events’ appearing in the ‘half-coin’ construct above in (1.6)? Those are cancelled as we demonstrate in the following steps: G1-X ( x )

1

 12 (1  x)  12 (1  x) 2

 12 (1  x)

1

2





1

2

2



2

1  1 x  1 x 2  ... 2 2 2 8 2 1  1 x  1 x 2  ... 2 2u2 2u8 1 1 x 2  ....  2u2 x  2u 4 2  21 u8 x  ... 1  1 x  0 x 2  0 x 3  ... 2 2 It is interesting to consider what the case of an n -coins system will be if we determine that n

(1.7)

1 2

gives us a construct

of a ‘half-coin’ with the existence of ‘negative probabilities’. We are in a paradoxical situation, unable to attach physical meaning to this notion of negative-probability or anti-event. A similar query was posed by L’Hospital to Leibniz in respect to

dn dx n

f ( x ) ; questioning what it would become if

n is determined to be half? This brief

demonstration of a half coin construct indicates the method of generalisation of our existing formulations of processes. It would be interesting to use this half-coin to generate, say, a Brownian motion.

1.3 The question posed by L’Hospital to Leibniz In a letter to L’Hospital on September 30th 1695, Leibniz raised the possibility of generalizing the operation of differentiation to non-integer orders, that is finding differentiating the function f ( x )

x ; that is:

d1 2 dx1 2

dn dx n

> x@ .

f ( x ) , and L’Hospital asked what would be the result of half-

Leibniz replied: “It leads to a paradox, from which one day

useful consequences will be drawn”. The paradoxical aspects arise because there are several different ways of generalizing the differentiation operator into non-integer orders, leading to inequivalent results. This aspect we will discuss in subsequent chapters. We can say with accuracy that this query was dated the 30th September, 1695 and that it gave birth to “Fractional Calculus”; therefore this subject of fractional calculus with half derivatives and half integrals is as old as conventional Newtonian or Leibniz’s calculus. Classical calculus is also termed Newtonian calculus. However, this subject of fractional-calculus (some even term this as non-Newtonian calculus) was dormant until the beginning of the century, and only now has it started finding applications in science and engineering. In due course, we will be replying to the question posed by L’Hospital. However, in the case of negative probability as described in Section 1.2, the difficultly in attaching physical meaning was noted, but in the case of

dD dxD

f ( x ) with D

as a non-integer, we do have physical meaning. Today we have systems based on these fractional differential equations. Here we will be doing detailed mathematical derivations without diverging much towards physics or engineering; rather, we will be generating fun in doing and discussing various aspects of

dD dxD

f ( x ) where D is

4

Chapter One

arbitrary and not restricted only to integer numbers, as well as discussing paradoxes and difficulties. Our classical calculus knowledge is enough and we thus generalise in the same way in the next chapters and bring out the essence of fractional calculus, which is a generalisation of our classical integer order and Newtonian or Leibniz’s calculus. In this book, there will be various aspects of classical calculus that we will be using in a fractional calculus context. Therefore, recapitulation of some of these classical concepts is essential.

1.4 A recall for analytic functions (1)

If the function f ( z ) is an analytic at point z then derivative f ( z ) is continuous at z . This implies that if f ( z ) has continuous derivatives of all orders at the point z , the function is analytic. A function is called an analytic function if it is locally given by a convergent power series. There are ‘real analytic’ and ‘complex analytic’ function categories that are similar in some ways, but different in other aspects. Functions of each analytic type are infinitely differentiable, but complex analytic functions exhibit properties that do not hold generally for real analytic functions. We will revise this concept.

1.4.1 Real analytic functions A function is analytic if, and only if, its Taylor series about a point, say x0 , converges to the function in some neighborhood for every x0 in its domain. Formally, a function f is a real analytic function on an open set A on the real line \ , if for any x0 in A one can write the following series expansion: f

f ( x)

¦ a (x  x ) n

0

n

2

a0  a1 ( x  x0 )  a2 ( x  x0 )  ........

(1.8)

n 0

In (1.8), the terms a0 , a1 , a2 , etc., are real numbers and the series is convergent to f ( x ) for x in the neighborhood of x0 . Alternatively, an analytic function is infinitely differentiable such that the Taylor series at any point x0 in its domain is the following f

T ( x)

¦

f

(n)

n 0

( x0 )

n!

( x  x0 )

n

f

(n)

n

( x0 )

d f ( x) dx

(1.9)

n x x0

which converges to f ( x ) for all x in the neighborhood of x0 (locally and uniformly). A function is complex analytic if and only if it is ‘holomorphic’ – that is, it is a ‘complex differentiable’. Examples of analytic functions are polynomials, exponentials, trigonometric functions, logarithmic functions, power functions, special functions like Mittag-Leffler functions, Miller-Ross functions, hyper-geometric functions, Bessel’s functions (refer to Appendix A), or Gamma functions. Special functions are analytic at least in some ranges of the complex plane. Non-analytic functions like f ( x )

x b,

f ( x)

D

x  c ; 0  D  1, x t 0 are not everywhere analytic, as they

are not differentiable at zero, i.e. at x 0 . The functions that are defined at different zones with different functions are not analytic where the zones meet. We have discussed the point that analytic functions are infinitely differentiable, indicating that they are smooth, but all smooth functions need not be analytic.

1.4.2 Complex analytic functions Now we shall look at a complex differentiable concept and a complex analytic condition. In complex variables a function f ( z ) where z x  iy is said to be analytic in region ‘ A ’ of a complex plane ^ if f ( z ) has a derivative at each point of A and if f ( z ) is said to be single valued. In other words, f ( z ) is said to be analytic at a point z , if z is an interior point of some region where f ( z ) is analytic. Hence, the concept of an analytic function at a point implies that the function is analytic in some circles with the centre at that point. The condition for a complex function f ( z ) u ( x, y )  iv ( x, y ) , with z x  iy and conjugate z x  iy to be analytic is something we will discuss. Since x

 z 2 z and y  z2iz ,

we write f ( z , z )

u ( x, y )  iy ( x, y ) . A necessary condition for f ( z , z ) to be

analytic is: w w z

f ( z, z )

0

(1.10)

Concept Generalisation

Therefore, a necessary condition for f u  iv to be analytic is that parts u , v of f , the above (1.10) condition is equivalent to:

w u

5

f depends on z . In terms of real and imaginary

w v

w u

w v

(1.11) w x w y w y w x This expression (1.11) is also called a Cauchy-Riemann equation. The necessary and sufficient condition therefore is that the four partial derivatives satisfy the Cauchy-Riemann equation, and that these four partial derivatives are continuous. From the above expression (1.11) we can also write the following expressions, called Laplace equations: 2

w u



2

2

w u

2

w v

w v

 0  0 (1.12) 2 2 2 2 w x w y w x w y The real and imaginary parts of an analytic function are ‘harmonic conjugate functions’. That is to say that the complex analytic function f u  iv is a solution to the above (1.12) Laplace equations, and satisfies the CauchyRiemann equations (1.11).

1.5 Recalling the Cauchy integral formula In the complex-variables review, we have studied contour integration. We revise those parts here, as in subsequent chapters we will be using this method; especially in order to find inverse Laplace transformations (Appendix G). We studied the Cauchy-Goursat theorem (or simply Cauchy’s theorem); which says that if f ( z ) is analytic in and on closed contour C , then the contour integral

³  f ( z ) dz

0 . We recall that the positive sense of travel on the closed

C

contour is in the anti-clockwise direction. A consequence of this is that

b

³  f ( z ) dz

does not depend on a path.

a

Cauchy’s integral formula states that if f ( z ) is analytic within and on closed contour C , then f ( z0 )

1

f ( z)

2S i ³

( z  z0 )

C

(1.13)

dz

Taking one-whole derivative of Cauchy’s integral formula in (1.13), with respect to f

(1)

( z0 )

1 2S i

f ( z )dz

³

2 C ( z  z0 )

z0

we write

; and by continuing this differentiation n times, we get the following important relationship f

(n)

( z0 )

n!

2S i ³

f ( z)

C

( z  z0 )

n 1

(1.14)

dz

1.5.1 The derivative is an integration process An d dx

interesting

> f ( x )@

1 2S i

observation

³

 f ( y ) dy

2 C ( y x)

and

from dn dx n

(1.14)

> f ( x )@

n! 2S i

is

³

that  f ( y ) dy

n 1 C ( y x)

differentiation

is

an

integration

process,

that

is

.

It is a very important fact that the derivative is actually an integration process that will be helpful in later chapters, where we generalise our earlier classical calculus theories. In the formula (1.14) i.e. f ( n ) ( z0 )

 2nS!i ³C ( z f z( z)) 0

n 1

dz ;

the integration path on the closed contour C is anti-clockwise (say), and a small circle encircling the point z0 ; for n as a zero or positive integer. If n is any real number, then the contour is differently drawn. This concept we will be using in subsequent chapters in order to evaluate the following f

(D )

f

( z ) i.e. D  order derivative is:

(D )

( z)

f (Z )

D!

2S i ³

C

D 1

(Z  z )

dZ

(1.15)

In (1.15) D is a non-integer number.

1.5.2 The Taylor series from Cauchy’s integral formula With Cauchy’s integral formula, we can now derive the Taylor series. As we stated, an analytic function has a convergent expansion about any point z within its domain of analyticity. We write Cauchy’s integral formula (1.13) as f ( z )

1 2S i

³

C

following steps

 f ( z1 ) dz1 ( z1  z )

, where C is circle centered at z0 ; and z is within C so

z  z0 z1  z0

 1 . With this, we write the

6

Chapter One

1

1



z1  z

z  z0 z1  z0

( z1  z0 ) 1  1



  (z  z ) 1

0

f

( z  z0 )

¦ (z n 0

We have used the expansion (1  x ) 1

¦

f n 0

x

n

z  z0 z1  z0

1

1

 z0 )

1

f

1

( z1

¦ z ) n 0

0

z  z0 n z1  z0



(1.16)

n

n 1

when x  1 , that is simply a summation of a geometric series in

(1.16). Therefore, Cauchy’s integral formula becomes

 f ( z1 ) dz1

1

2S i ³

f ( z)

( z1  z )

C

§ f ( z  z0 ) n · f ( z1 )  dz1 ¨¦ n 1 ¸  2S i ³C © n 0 ( z1  z0 ) ¹ § f  f ( z1 ) ( z  z0 ) n · 1 ¨¦ ¸dz1 n 1 2S i ³C © n 0 ( z1  z0 ) ¹ 1

f

1

¦ 2S i ³ n 0

1 2S i

Using (1.14) i.e.

 f ( z1 ) dz1

³

n 1 C ( z1  z0 )

f ( n ) ( z0 ) n!

C

 f ( z1 ) dz1 ( z1  z0 )

n 1

(1.17)

( z  z ) n

0

(this is coming from Cauchy’s integral formula), in the above derived

expression (1.17) we get the following:

§ f ( n ) ( z0 ) · n ¦ ¨ ¸( z  z 0 ) n! ¹ n 0© f

f ( z)

(1.18)

In (1.18) we have achieved the Taylor series for an analytic function f ( z ) .

1.6 Singularity pole branch point and branch cut We shall now revise the concept of singularities of analytic functions, that are points at which f ( z ) is not analytic and which are called singular points or singularities. That is, the function blows up at those points.

1.6.1 Isolated singularity The ‘isolated singularity’ of f ( z ) which is analytic everywhere throughout some neighborhoods of a point z inside a circle

za

R (with radius R and centered at z

a ); except at the point z

a , say

a . Then the point z

a is

called an isolated singularity of f ( z ) . Thus, the function f ( z ) cannot be bounded near an isolated singularity.

1.6.2 Pole singularity (removable singularity) The next type of singularity is called a ‘pole’. If f ( z ) has an isolated singular point at z n

exists an integer i.e. n such that ( z  a ) f ( z ) is analytic at z n

note here that because ( z  a ) f ( z ) is analytic at z example, f ( z )

z

2

have poles of order 2 at z

a , and if in addition there

a , then f ( z ) has a pole of order n at z

a . We

a , such a singularity is termed as a ‘removable singularity’. For

0

1.6.3 Essential singularity ‘Essential singularity’ is an isolated singular point, which is not a pole (that is not a removable singularity); for example, f ( z )

sin

 1z

has essential singularity at z

0.

Concept Generalisation

7

We ought to clarify the above discussions regarding singularity. Suppose f ( z ) has isolated singularity at z0 but is analytic in the neighborhood of z0 , one can write a Laurent series as the following

f ( z)

§ b1 · § f b2 n·   ......... ¨ ¸  ¨ ¦ an ( z  z 0 ) ¸ 2 ¹ © ( z  z0 ) ( z  z0 ) ¹ ©n 0

(1.19)

We classify the isolated singularity as following on from (1.19): Simple pole: bn

0 , n ! 1 , then f ( z )

Pole of order N : bn

b1 ( z  z0 )

0 , n ! N , then f ( z )



¦

f n 0

an ( z  z 0 )

n

bN b1 b2 ( z  z0 )  ( z  z0 )2  ....  ( z  z0 ) N 

Essential singularity: Infinite terms of ( z  z0 ) 1 then f ( z )

¦

f

bm

m 1 ( z  z0 )m

¦

f

a (z n 0 n

 z0 )

n

f

 ¦ n 0 an ( z  z 0 ) n

1.6.4 Multivalued functions, branch points and branch-cuts Branch points and branch-cuts are the methods used to tackle multi-valued functions. There are three types of multivalued functions: a) the logarithmic type, b) the algebraic type and c) the trigonometric type, and they have respective branch cuts and branch points. Knowing this, we can discuss multi-valued functions, for example f ( z )

ln z , which blow out at z

iT

0 . If we write i (T  2S n )

with z re i.e. in polar form, then we have ln z ln r  iT . Thus for apparently the same point i.e. z re n as integers (also termed winding numbers) and for say T  0 ; the function ln r  2S in can have different values as we go around the circle around point z 0 (depending on the winding number, i.e. n ). Thus this function blows out iT

at z re when r tends towards 0 , at infinite T values say T 0, 2S , 3S , 4S ... spread on a primary and infinite number of secondary complex planes, one beneath the other; called Riemann sheets (Appendix E). Therefore the point z 0 , is not a simple essential singularity or pole, rather it is a ‘branch-point’. In this case, an infinite number of values are possible and ln z has an infinite number of ‘branches’. This requires us to introduce ‘branch-cuts’, which are barriers through which z cannot go (an example being the line integral), and thus f ( z ) ln z remains a single value. The logarithm has a jump discontinuity of 2S i when crossing the branch cut. The logarithm can be made continuous by gluing together many copies, called sheets, of the complex plane along the branch cut. On each sheet, the value of the log differs from its principal value by a multiple of 2S i . These surfaces are glued to each other along the branch cut in a unique way to make the logarithm continuous. Each time the variable goes around the origin, the logarithm moves to a different branch (Appendix E). Alternatively one can think of Riemann sheets whereby crossing a branch-cut one moves onto a different Riemann sheet of the function. The number of branches is equal to the number of Riemann sheets. This allows closed contours to be formed by going around branch cuts as many times as required to get back to the original (or primary) Riemann sheet. In the case of ln z ln r  2S in , the primary Riemann-sheet is for n 0 . This example is of a ‘logarithmic branch point’. In the function f ( z ) ln z we can argue the same at the point z f by making a substitution i.e.

z { 1z and then say that the point z f is also a branch point. The branch cut therefore joins these two points z 0 and z f , and makes the barrier that one should not cross, and the function stays single valued (with its ‘principal D

value’). Similarly the function f ( z ) z for D  \ , is multi-valued and has a logarithmic branch point at z will not be discussing the trigonometric multi-valued function. Another type of branch point is the ‘algebraic branch point’. For example point zero ( z

0 . We

0 ) is a branch point of the 1

square root function i.e. f ( z ) z this we explain here ., , Take the function f ( z ) z 2 and take the point z that starts at z 4 and moves along a circle with a radius equalling 4 in the complex plane centered at z 0 . When z has made one full circle ( 0 to 2S ), going from z 4 back to z 4 again, f ( z ) will have made one half-circle ( 0 to S ), going from the positive square root of z 4 i.e., from f ( z ) 2 to the negative square root of z 4 i.e. f ( z ) 2 . Roughly speaking, the branch points are the points where the various sheets of a multiple valued function come together. The branches of the function are the various sheets of the function. For example, the squre root function

8

Chapter One 1

has two branches: one where the square root comes in with a plus sign, and the other with a minus sign and these two branches meet at z 0 that is the branch point. A ‘branch cut’ is a curve in the complex plane from which it is possible to define a single analytic branch of a multi-valued function on the plane minus that curve. Branch cuts are usually, but not always, taken between pairs of branch points. Branch cuts allow one to work with a collection of single-valued functions, ‘glued’ together along the branch cut instead of a multi-valued function. For example, to f ( z)

z

2

>0,1@ on the real axis, connecting the two

make the function single-valued, one makes a branch cut along the interval

1

branch points of the function. The same idea can be applied to the function f ( z ) z 2 ; but in that case one has to perceive that the ‘point at infinity’ is the appropriate 'other' branch point to connect to from z 0 , for example along the whole negative real axis. We will use these concepts of branch cuts, branch points and Riemann-sheets in obtaining an inverse Laplace transform of some special functions. In this chapter too we will deal with branch cuts and branch points depicted in Figure 1.4, while we do contour integrations in the section of Beta functions (Section 1.15). The standard branch cut that we will use in complex planes is a line in the negative real axis from a point of interest (i.e. z 0 or some other branch point, say z Re z0 ) to point the z Re z f . Refer to Appendix E for more details and discussion on branch cuts and Appendix G for an elucidation on obtaining an inverse Laplace transform.

1.7 Recalling residue calculus We will be using the concept of residue calculus in subsequent chapters. We recall that the residue calculus is used for complex differentiation, complex integration, and power series expansion (PSE), and provides three approaches to holomorphic (complex analytic) functions. The Cauchy integral formula provides the relationship between complex differentiation and complex integration. The residue serves to formulate the relationship between complex integration and PSE. Let us recall that the residue of a function f can be holomorphic everywhere within and on closed curve C except possibly at point z0 in the interior of C where f may have isolated singularity. Then the residue of f at z0 is def

Residue z0 f

1 2S i

³  f ( z ) dz C

(1.20) If z0 is a non-singular point then the residue at z0 is zero, else it is non-zero. The use of the definition of residue is to calculate the contour integral

³  f ( z ) dz . If the closed contour encloses say n poles at e.g. C

z1 , z 2 ,.... z n then we

z

write the following rule:



³  f ( z ) dz

2S i Residue z1 f  Residue z2 f  ...Residue zn f

C



n

2S i

¦ Residue i 1

zi

f

Therefore, calculations of residue are very important for the evaluation of

(1.21)

³  f ( z ) dz . C

1.7.1 Finding the residue by power series expansion Consider the Laurent series expansion of f at z0 i.e. f ( z )

¦

n f n f

n

an  z  z0 , where an

Here residue is nothing but a coefficient of ( z  z0 ) 1 , in the expansion. For example say f ( z )

z

1 2

of order n

1 z (1  2 z )

2

The coefficient of term ( z  z0 ) 1 here in (1.22) is exp

 has essential singularity at z 1 z2

1 z (1 2 z )2

³

f ([ )

n 1 C ([ [0 )

d[ .

, has a pole at

2 . We write: f ( z)

f ( z)

1 2S i

§ 1 ·§ 1 · § 1 ·§ 1 ·   ¸¨ ¸ ¨ ¸ ¨¨ 2 ¸ ¨ © 8 ¹ ©  z  12 ¸¹ © 4 ¹ ©  z  12 ¹  14 and that is residue of f at z

(1.22) 1 2

. Similarly, the function

0 , and if we write the series as the following:

Concept Generalisation

f ( z)

We used the series expansion of exp( z )



1 z 

1 2!

5 at z

sin z z6

0 . For f ( z )



2

1 1

(1.23)

0 in this case is zero.

z  ... , with this one potentially verifying that for



pole of order n

1

1

2

exp  1z , the residue at essential singularity point z

f ( z)

1 z2

 .... 4 z 2! z 0 exists; hence the residue of f at z

In (1.23) we find that no term with ( z  z0 ) 1 for z0

exp

9

sin z z6

0 is 1 . We see that for f ( z )

by using an expansion formula i.e. sin z

z

z3 3!



z5 5!

which has the

 .... , we get the

following power series representation:

f ( z)

§ z3 z5 · z   .... ¸ 6 ¨ z © 3! 5! ¹

sin z z

1

6

¦

In (1.24) the function M ( z ) is an analytic part, i.e. M ( z ) residue is

1 120

at point z

f

1 z



5

1 6z

3



1 120 z

(1.24)

 M ( z)

m

m 0

am z . From the above series (1.24) we observe the

0.

1.7.2 The formula for calculating residue Now we shall derive a formula for the calculations of residues in the case of poles of a function f ( z )

h( z ) ( z  z0 )n

, with

function h ( z ) as holomorphic in the entire complex plane ^ . Thus, we have h ( z ) ( z  z0 ) n f ( z ) . From the definition of the residue and then by utilizing the Cauchy formula, we write the following steps: h ([ ) 1 Residue z0 f d[ n 2S i C ([  z )

³

§ d n 1h( z ) · ¨ n 1 ¸ ( n  1)! © dz ¹ § 1 d n 1 · n lim z o z0 ¨ ( z  z0 ) f ( z ) ¸ n 1 © ( n  1)! dz ¹ 1

1 2S i

We used

³

h ( z1 )

m C ( z1  z )

dz1

h( m ) ( z ) m!

1 2

of the order n

n ( z  z0 ) f ( z ) , in the above

i.e. the Cauchy formula (1.14) and then used h ( z )

1 z (1 2 z )2

(1.25) steps of derivation. Let us apply this to our earlier example (1.22) i.e. f ( z )

z

(1.25)

where we have a pole at

2 . We obtain the residue by formula (1.25) as demonstrated in the following steps: Residue z

1 2

f

d

 ( z  dz

lim z o 1

2

lim z o 1

2

1 2 2



f ( z)



d

1 §1 z · ¨ ¸  dz © 4 ¹ 4

(1.26)

That is what we got from the Laurent series expansion (1.22) also. We have calculated in an earlier example of 1 f ( z ) sin6 z (1.24) the residue at z 0 (five-order pole) as z 120 . Now we can use this above described z

formulation (1.25) to calculate the contour integral on a closed curve C i.e. a circular or radial one, call it : centered at the point z 0 , and write that as the following:

³

:

The above integral of (1.27) i.e. radius.

³

:

sin z z

6

dz

2S i  Residue z

0

f



§ 1 · ¸ © 120 ¹

2S i ¨

Si 60

(1.27)

f ( z )dz is valid for any circle centered at origin with any radius, even an infinite

We shall conclude this part by writing a few integrals obtained via residue calculus. Assume the contour C is a circle centered at origin, with a radius large enough so that it is enclosing all the singularities of functions f ( z ) taken as 5 z 2 z ( z 1)

, exp

  1z and

exp S z / 4 z 2 1

, which are presented in the following expressions:

10

Chapter One

5z  2

³

z ( z  1)

C

2S i  Residue z

dz

  1z dz

2S i  Residue z

³

exp

³

dz exp  z S4 2 z 1

C

C

0

f  Residue z 1 f 0

f

2S i ( 1)

2S i  2  3 10S i 2S i

2S i  Residue z i f  Residue z

i

f

(1.28)

§ exp  i  S4 exp  i  S4 ·  ¸¸ ¨ 2i ( 2i ) © ¹

2S i ¨

2S i sin  S4

iS 2

Appendix G gives us the use of residue calculus in order to obtain inverse Laplace transforms; for complex functions with poles and for multi-valued functions, where we employ branch-cuts in the primary Riemann sheet.

1.8 The basics of analytic continuation We will be using the concept of analytic continuation for several concept generalisation processes in this chapter and x

in subsequent chapters. We start with a simple example of the function e , the exponential function originally defined 2 on the real axis \ , the analytical continuation to this is f ( z ) 1  z  z2!  ... a unique function f ( z ) , where z is in x

the complex domain ^ that is equal to e on the real line. This is what is meant by an analytical continuation in this simple example. From the real domain we continued to a complex domain in this example of an exponential function. The intersection of two domains (in the region of the complex plane in our case); say ^1 and ^ 2 denoted as

^1  ^ 2 is a set of all points common to both ^1 and ^ 2 . The union of two domains ^1 and ^ 2 denoted by ^1 * ^ 2 is the set of all points in either ^1 or ^ 2 . Now suppose we have two domains ^1 and ^ 2 , and a function f1 that is an analytic in domain ^1 , if there exists a function f 2 that is analytic in domain ^ 2 and such that f1

f2

on the intersection ^1  ^ 2 , then we say that f 2 is an analytic continuation of f1 into domain ^ 2 . Now, whenever the analytic continuation exists it is unique. This comes from a theory of complex variables: ‘A function is analytic in domain ^ and is uniquely determined over ^ by its values over the domain or along an arc interior to ^ ’. The function F ( z ) analytic over union ^1 * ^ 2 can be defined as:

­ f1 ( z ) when z is in ^1 ® f ( z ) when z is in ^ ¯ 2 2

F ( z)

(1.29)

In other words F is given by f1 over ^1 and f 2 over ^ 2 , and since f1

f 2 over the intersection of ^1 and ^ 2 this is a well-defined holomorphic function. By the definition at the beginning of this section, since F is analytic in ^1 * ^ 2 it is uniquely determined by f1 on ^1 (for that matter it is also uniquely determined by f 2 on ^ 2 ). In other words there is one possible holomorphic function on ^1 * ^ 2 that matches f1 on ^1 . In this case F ( z ) is said to be an analytic continuation over ^1 * ^ 2 of either f1 or f 2 . We therefore can immediately use the result to extend the originally defined function on the real axis to the complex plane example (as we stated at the beginning) with e

z

1 z 

z2 2!

x

 ... being a unique function f ( z ) that is equal to e on the real line.

Let us consider f1 ( z )

¦

f n 0

n

z ; this power series converges when z  1 to

1 1 z

and is not defined for z t 1 . One

understands this series as a geometric series and we can calculate this geometric series so long as we are in the region of convergence. On the other hand, function f 2 ( z ) 11 z is analytic everywhere except at point z 1 . Since f1 f 2 on the disk z  1 ; we can view f 2 as an analytic continuation of f1 to the rest of the complex plane (minus the point z 1 ).

Concept Generalisation

11

³

Consider the function g1 represented as an integral in the equation g1 ( z )

>@

when Re z ! 0 ; and for such a condition this has a value of gives the result

 e zx z

x f x 0

; which equals

1 z

1 z

f

0

e

 zx

dx ; and this integral exists only

. We can check by performing this integration that

only for z ! 0 . Since the function

1 z

>@

matches g1 on Re z ! 0 we say

 1z is an analytic continuation of g1 to non-zero complex numbers. While we are at it, we can define n f i ¦  z i i . This series converges for z  i  1 ; and so g 2 is defined only within a unit-disk centered on n 0 f n 1 i . Using the substitution  z i i p one can show that g 2 ( z ) 1z , by using the fact that ¦ p 1 p n 0

the function g2 ( z)

point z

z i i

which is a summation of the geometric series when p  1 . That gives

i

1 ). Then i

¦

f n 0

disk, we can see that

p

n

1 z

i 1 p

and substituting

p

 z i i

i

f

z i n i

¦  n 0

 1 ; this implies z  i  1 , (note that z i n i

¦  . Since g matches on a i ¦  to non-zero complex numbers. In

gives i

n 0

f

is an analytic continuation of g 2 ( z )

addition, we can view g 2 ( z )

f

n 0

1 z

2

1 z

z i n i

as an analytic continuation of g1 ( z )

³

f

0

e

 zx

dx to the disk z  i  1 .

The above are examples for describing the concept of analytic continuity. However, there are exceptions that may be noted, in particular the fact that not all functions can be continued indefinitely. The functions may have natural barriers of singularities through which one cannot pass. It may happen that the function obtained through this continuation is a multi-valued one. For example, we may carry out analytical continuation along some path that returns to the region of convergence. Then if the path has crossed a branch-cut the values of analytically continued functions will not agree with the original value. The principle of analytic continuation may be involved in the context of integrals, generalising the recurrence relations which we will be seeing in due course.

1.9 The factorial and its several representations The generalisation of factorial n ! to a non-integer number was first done by Bernoulli and Goldback (1720), then later on by Euler (1729) who gave the formulation using integral representation, series representation and then by Gamma function. The concept of a factorial was introduced by French mathematician Louis Franosis Antoine Arbogast. We know the factorial as n ! n u ( n  1) u ( n  2) u ....2 u 1 , this is called a falling factorial, or decreasing factorial.

1.9.1 Representation of a falling factorial In a compact way we write n !

–

n k 1

k , or explicitly we state n ! as the following: n!

^

n 0 1 ( n  1)!( n ) n ! 0

(1.30)

The number n in the above definition is a positive integer.

1.9.2 Double factorial From here the double factorial is defined as the following: ­ n u ( n  2) u ... u 5 u 3 u 1

n!0

n : odd

° n !! ®n u ( n  2) u ... u 6 u 4 u 2 n!0 n : even °¯ 1 n 1, 0

A few values of the double factorial are 0!! 1 , 1!! 1 , 2!! 2 , 3!! 3 , 4!! 8 , 5!! 15 , 6!! 8!! 384 , and 9!! 945 . We have a relationship of the double factorial expressed as: n ! n !!( n  1)!! We are familiar with the above definition of n ! for n as a positive integer.

1.9.3 Truncated factorial k

In addition, a truncated factorial is represented as n  and is defined as:

(1.31)

48 , 7!! 105 , (1.32)

12

Chapter One k

n  We have the following relationship from (1.33) and (1.30):

(1.33)

n u ( n  1) u ( n  2) u ...( n  k  1) k

n 

n ( n  1)( n  2)....( n  k  1)

k!

k ( k  1)( k  2).....(1)

(1.34)

The above expression (1.34) we reckon as binomial coefficients i.e. n Ck , which appears in the binomial expansion of n

(1  x ) . We write the following expression for the binomial coefficients in different forms: k

n

nk

n  n! §n· Ck { ¨ ¸ © k ¹ k !( n  k )! k !

n 

§ n · n ¨ n  k ¸ { Cn  k ( n  k )! © ¹

(1.35)

1.9.4 Some other historical representations of factorials Initially Bernoulli and Goldback wrote in an attempt to generalise the factorial as the following expression: n n § 2 n · §  32 · §  43 · n! ¨ ¸ u ¨ n ¸ u ¨ n ¸ u ..... © 1  n ¹ ¨© 1  2 ¸¹ ¨© 1  3 ¸¹

n

1  1k – n k 1 1  k f

(1.36)

Euler gave the formula originally as the following expression:

§ n z ( n !) · ¸ ( z  1)! lim nnf ¨ ¨ –f (z  k) ¸ © k 0 ¹

(1.37)

In expression (1.37) we have used n as a symbol indicating ‘tending to high’ i.e. f , in the rest of the chapters we also use this symbol; p we will use to indicate ‘tending to low i.e. 0 . For any other value other than 0 or f we will use the usual symbol b i.e. o b , saying ‘tending to b ’. From the basic definition of a factorial that is n ! 1 u 2 u .... u ( n  1) u ( n ) we also say that log( n !)

¦

n x 1

log x . On

this Ramanujan worked and came out with the following representation i.e. approximately representing the factorial in two ways:

log( n !) | n log n  n 



log n 1  4 n (1  2 n ) 6

n log n  n 

log



 log S 2

1 2n



6

1 8 n2

 log(2n)  log S 2

(1.38)

2

Stirling gave the approximation of a factorial as the following:

§n· n ! | 2S n ¨ ¸ ©e¹ where in (1.39) e is y

x

e with x

n

(1.39)

1.

These above expressions are approximation formulas to find the factorial even if n is a non-negative number not essentially an integer.

1.9.5 Increasing truncated factorials (Pochhammer numbers) Here we’ll talk about increasing the truncated factorial which is defined as: n ( n  1)( n  2)...( n  k  1) k ! 0 ( n) k 1 k 0

^

(1.40)

The above expression (1.40) of ( n ) k is also called the Pochhammer number (or polynomial). In terms of factorial notation, we can write an increasing factorial for k ! 0 , as: ( n) k

( n  k  1)! ( n  1)!

(1.41)

Concept Generalisation k

k

The truncated factorial n  defined above in (1.33) i.e. n 

13

n ( n  1)( n  2)...( n  k  1) is also a Pochhammer

number; we may represent this decreasing truncated factorial as ( n) k depicted in the following representation:

( n ) k

n ( n  1)( n  2)...( n  ( k  1))

(1.42)

n ( n  1)( n  2).....( n  k  1)

This is in order to distinguish it from the increasing truncated factorial i.e. ( n ) k

n( n  1)( n  2)...( n  k  1) .

1.9.6 Euler’s integral representation for factorials In terms of integral representation Euler proposed n ! as the following formula: 1

n

³  ln dx 1 x

n!

(1.43)

0

The above expression (1.43) i.e. n !

  1x dx

1

³ ( ln x)

n

0

dx can be put in other forms. Put (  ln x )

t , so we get x

e

t

and

dt . The limits of integration in the variable x , i.e. from 0 to 1 , become the variable t from f to 0 . So,

with these changes of variables we write n ! of the factorial as:

0

n

³ t   xdt ³ f

0

f

n t

t e ( dt ) . This gives another integral representation

n!

³

f

0

t n

(1.44)

e t dt

This integral is also called ‘Euler’s integral of a second kind’, and we will see its usage, shortly as a generalisation of factorial to non-integer numbers. The most popular generalisation of factorials is by use of the Gamma-function, which we will describe in the next section (Section 1.10).

1.10 The gamma function One of the basic functions of fractional calculus is Euler’s gamma function. This function generalises the factorial n ! , and allows n to take non-integer values. The definition of the gamma function in integral form is the following: f

*( z )

³e

 t z 1

t

(1.45)

dt

0

This above (1.45) improper integral defined in complex plane z  ^ converges in the right half of the complex plane

>@

i.e. Re > z @ ! 0 . This convergent improper integral for Re z ! 0 was derived by Daniel Bernaulli. We note here the expression (1.45) is Mellin transform of the function f (t ) z 1

e

t

we represent as * ( z )

^ `

% e t . Also in the above

( z 1)ln t

(1.45) expression we have t and ln t  \ . Considering - z to be a real number this statement implies e that the gamma function is defined continuously for positive real values of z . Compare the above integral (1.45) i.e.

*( z )

³

f

 t z 1

e t

0

³

i.e. n !

f

0

dt with Euler’s integral of a second kind obtained for factorial representation in Section 1.9, (1.44),

t n

e t dt and observe the similarity.

The other representations of gamma functions are the following: f

³

*( z )

2 y

2 z 1  y 2

e

1

dy

*( z )

0

We get the above expression (1.46) by using t

*( z )

³

f

0

t

y

2

and e

³  ln 1 y

z 1

dy

(1.46)

0

t

y respectively in the original integral (1.45) i.e.

z 1  t

e dt . For small iterations of z , between 0 and 1 , we have following approximation: 1 *( z )

|z

(1.47)

14

Chapter One

1.10.1 The recurring relationship of the gamma function The basic properties of the gamma function is the recurrence relation which is as follows: * ( z  1) z* ( z )

³ ( I )( II )dt I ³ ( II )dt  ³ 

The proof is attained through integration by parts formula that is I

z

t and II

e

t

(1.48) d dt



( I ) ³ ( II )dt dt with

demonstrated in the following steps f

f

³e

* ( z  1)

 t ( z 1) 1

³e

dt

t

0

t z

t dt

0

t

z

³

f

0

t

³

e dt  t f

ª¬e  t t z º¼ t

0

f

0

§ d ª z º t · ¨ ¬t ¼ ³ e dt ¸dt © dt ¹

(1.49)

f

³

 t z 1

z e t

dt

0

z  *( z )

This above equality (1.48) i.e. * ( z  1) the integral in (1.49) steps i.e.

³

f

0

>@

z  * ( z ) is true only at the right half plane (RHP), i.e. Re z ! 0 , otherwise

 t z 1

e t

dt

t z

* ( z ) is f , i.e. diverging. In the above steps (1.49) ª¬e t

With this initial recurrence relationship * ( z  1)

t f

º¼ t

0

0 .

z  * ( z ) , we will be able to ‘analytically continue’ the gamma

>@

* ( z 1) z

function for Re z ! 1 (excluding the origin), by writing * ( z )

. This process we will describe in a short

while in this section. Obviously we have * (1)

1 , as * (1)

³

f

0

t

e dt

e

relation of the gamma function, we obtain values for z

t f 0

1 and using the above property (1.48) of the recurrence

1, 2, 3, ... * (2)

1 u * (1)

1!

* (3)

2 u * (2)

2!

* (4)

3 u * (3)

3!

(1.50)

........................... * ( n  1)

n u *( n)

n ( n  1)!

n!

The above property (1.50) is valid for positive values of z .

1.10.2 Factorials related to the gamma function With this observation, we write the decreasing factorial as: ( z )!

(1.51)

* ( z  1)

Increasing the truncated factorial for k ! 0 is done as follows: ( z )k

*( z  k )

(1.52)

*( z )

1.10.3 Calculation of the factorial of non-integer numbers by gamma function We have a recurrence of the relationship for the Gamma-function that is * ( z  1) which is equivalent to stating

  12 ! , as follows:

( z )  * ( z ) . We calculate *  12

Concept Generalisation f

*

1 2

³e

 t  1 2

t

f

³

dt

0

e

t

t

0

f

³

e

³

2 e

³

f

f

I

2

e

 x2

³

dx

f

f

e

dx

 x2

2

dt

2udu

³e

du

u2

du

S

f

S in the above (1.53) derivation. This integral i.e.

dx

was obtained by Pierre-Simon Laplace by considering the square of the integral i.e.

S  x2

f

e

u

(1.53) f

u2

0

f

t

2udu

u f

³

dt

u2

0

We have used a known integral that is

15



2

. Accordingly, we do the following manipulations: I

2

³

f

f

e

 x2

³ ³

dx f

f f

f



e e

2

 x2

dx

 x2

dx

 ³  ³

f

f f

f

e e

 x2

 y2

dx dy



(1.54) f

³ ³

f

e

f f

 ( x2  y 2 )

dxdy

We recognise  dx  dy as an elemental area in Cartesian ( x, y ) coordinates with the domain of integration as on the entire X-Y plane. We have polar representation in ( r , T ) coordinates with transformations as x y 2

 dx  dy

r sin T . In the polar coordinates ( r , T ) the elemental area is transformed as 2

r cos T and

r  dr  dT with

2

r x  y . The entire X-Y plane is now represented by r varying from zero to infinity, and T varying from zero to 2S . With this change in the coordinate system from ( x, y ) to ( r , T ) we write the following steps, with substitution r

2

u and 2 rdr

:

du

I

2

f

³ ³

f

f f

e

 ( x2  y 2 )

dxdy

f

2S

0

0

³ ³ ³

f

e

0

³

f

0

r2

e

e

r2

rdr

u

rdrdT

³

2S

0

³

dT

f

0

e

r2

rdr



f

u

 12 du  2S S ³0

³

The above derivation (1.55) is done by Laplace which gives us I

f

f

we demonstrate the factorial of a non-integer getting defined, i.e.   12 !

e

 x2



(1.55)

 2S

e du

S

S ; a very useful result. Thus, here

dx

*  12

S .

1.10.4 Poles of the gamma function Another important property of the gamma function is that it has simple poles at z explained as splitting the function into two intervals as indicated below 1

*( z )

³e

0, 1, 2, 3, ... . The proof is

f  t z 1

0

t

³

 t z 1

dt  e t

(1.56)

dt

1

The first integral of (1.56) can be evaluated by using a series expansion for the exponential function i.e. e

t

¦

f k 0

(  t )k k!

>@

. With Re z

we write the first term, that is

1

³e 0

x ! 0 , then we have Re >( z  k )@  t z 1

t

dt , in series form as follows:

x  k ! 0 and thus t ( z  k )

t 0

0 . Therefore,

16

Chapter One 1

³e

1 f

 t z 1

t

dt

³¦

( t ) k!

0 k 0

0

f

¦

k

( 1)

k 0 f

t

z 1

dt

k 1

k!

³

( 1)

t

§ ( 1) k dt ¦ ¨ k 0© k! f

k  z 1

0

· §¨ t k  z ¸ kz ¹ ¨©

t 1

t

· ¸ ¸ 0¹

(1.57)

k

¦ k !(k  z ) k 0

The second term of (1.56) i.e.

³

f

1

 t z 1

e t

dt may be represented as an “entire-function”, call it M ( z ) , which permits us

to write the split integral for * ( z ) as follows: f

*( z )

§ ( 1) k

· ¸  M ( z) 0 © k ! (k  z ) ¹ 1

¦¨ k

(1.58)

1 2 § ( 1) 0 1 · ( 1) 1 ( 1) 1    ..... ¸  M ( z ) ¨ 0! 0  z 1! 1  z 2! 2  z © ¹

Clearly this derivation (1.58) is indicating simple poles at z 0, 1, 2, 3, ... . This means that at negative integer points the gamma function asymptotically approaches infinity and is discontinuous at those negative integer values. So

³

by the above (1.45) integral representation of the gamma function * ( z )

f

0

 t z 1

e t

dt defined originally for positive

z , we extended our argument to point zero, and all negative integer points, where the gamma function blows up. What does that not say about the values of the gamma function at other negative non-integer points?

1.10.5 Extending the gamma function for negative arguments by analytic continuation What happens to the Gamma function at say

* ( z  1)

( z )  * ( z ) which we re-write as * ( x  1)

z *( x) ( x 1)

 12 ,

z

 32

etc.? The recurrence relation (1.48)

, also serves as an analytical continuation, extending the

³

definition of the gamma function to the negative arguments to which the definition (1.45) i.e. * ( x )

f

0

y

x 1  y

e dy ,

x ! 0 is inapplicable. This extension shows * (0) to be infinite as with * ( 1) and values of the gamma function at all negative integers (Section-1.10.4). The integral

³

f

0

t

z 1  t

e dt does not make sense in the left half plane (LHP). However if we define * ( z )

³

LHP (and this is not equal to the integral on the LHP); this is consistent with the fact that * ( z )

f

0

t

* ( z 1) z

on the

z 1  t

e dt on the

RHP and provides us with a possible extension on the LHP, except the points of negative integers. Further, this has to be analytic continuation since * ( z ) * ( zz1) is analytic everywhere except at zero and negative integers and matches with the integral on the RHP. Hence it is the uniqueness of analytic continuation and this is the only possible extension.

>@

We can verify this by taking a strip on the LHP i.e. ( 1, 0) such that Re z  ( 1, 0) and define * ( z )

>@

strip. First, note that * ( z  1) is analytic on the strip Re z  ( 1, 0) , since ( z  1)  RHP , and

>@

1 z

* ( z 1) z

on this

is analytic on the

>@

* ( z 1) z

is analytic on the strip Re z  ( 1, 0) . In this way, we have extended

* ( z ) to the region Re z ! 1 (except at z

0 ). Now repeat the same reasoning for the strip Re > z @  ( 2, 1) , since

strip Re z  ( 1, 0) . Hence * ( z )

>@

>@

now * ( z ) is analytic on region Re z ! 1 , and so on. This way the gamma function is extended to the entire complex plane except at points at zero, and all negative integers.

Concept Generalisation

17

1.10.6 Representations of the gamma function via the contour integral (Hankel’s formula) The gamma function is also represented by the contour integral, is also called Hankel’s formula, and is valid for all z z 0, r1, r2, ..... as in the following:

1

2i sin S z ³

*( z )

C

( t )

z 1  t

e dt

(1.59) 

We will derive this (1.59) subsequently. The contour C starts slightly above the real axis at f , runs down to t 0 , where it goes counter-clockwise in a small circle, say of radius  , and returns to f just below the real axis. This integral on C we can represent by

³

(0  )

f

z 1  t

e dt . The complex plane t -plane is cut from 0 to f ; the branch cut

( t )

>@

is the positive real axis (Figure 1.1). To prove the above result (1.59), consider the following integral with Re z ! 0 :

³

I

where ( t )

z 1

e

( z 1)ln(  t )

C

( t )

z 1  t

(1.60)

e dt

and ln( t )  \ . That is, we are defining the logarithm such that ln( t ) is real on the

>

negative t -axis. Thus, on the negative real t -axis we have ‘ ( t )

@

0 or arg( t )

0 , for defining ln( t )  \ . 0

Thus from our convention of positive angles, if negative real axis corresponds to an angle of 0 , then the positive real 0

axis should have an angle of r180 , as depicted in Figure 1.1a. On the contour, i.e. C ; as described above (i.e. the contour enclosing the positive real t -axis), we have S  arg( t )  S ; such that arg( t ) 0 on the negative real t axis. Therefore just above the positive real t -axis we have arg( t ) S , whereas just below we have arg( t ) S ; the angle being measured counter-clockwise (Figure 1.1a). Thus, we write for the part of a contour, which is just above the positive real axis as ( t )

z 1

t

z 1  iS ( z 1)

e

in polar form with its magnitude and its argument. For the part of

the contour just below the positive real axis, we write ( t ) circle enclosing t  dt

0 we have ( t )

 e

iT

z 1

t

z 1  iS ( z 1)

e

. For the part of the contour i.e. the small

where in limit p 0 and T change from S to S ; this makes

 ieiT dT . Thus we write the contour integral as the sum of (1) the integral on the line just above the positive

real axis, (2) the integral on the small circle encircling the origin, and (3) the integral on the line just below the positive real axis, as follows:

³

I

C

( t )

³



f

z 1  t

e dt

t

z 1  iS ( z 1)  t

e



e dt

 ³  S

 e

S



³

f





iT

z 1 (cosT  i sin T )

e

    e

z 1  iS ( z 1)  t

t

e

e dt

(a)-Contour: ( f) : A o 0  o B : (Circle) o D : ( f ) Im t

³

{³

C

(0  )

f

ln( t )  \ ‘  t

00

00

0

0

Re >t @  0 B

1800 1800 Re >t @ ! 0

A f D

(b)-Contour: ( f) : P o 0  o Q : (Circle) o R : ( f )

R 180

Im > x @

Q

0

f

0 P 180

Figure 1.1: Hankel’s path for integration

0

0

³

C

{³

(0  )

f

Re > x @ ! 0

00

iT

idT



(1.61)

18

Chapter One (cosT  i sin T )

In the second integral (1.61) we have written e

for e

t

 eiT

with ( t )

  cos T  i sin T . The z

second integral of (1.61) goes to zero as in the limit p 0 ; i.e. the integral over the small circle vanishes, as does  , which is demonstrated in the following steps:

limp0

 ³  e

iT z 1 (cosT  i sin T )

S

limp0

 ³  S

e

S

S

iT z 1

 e

³ S dT 

limp0

 e

e

S

z



iT

iT

idT

;

(cosT  i sin T )

limp0 e

1



idT

iT z (cosT  i sin T )

ie e

(1.62)



0

So we have the following steps for evaluation I as defined in (1.61)

³

I

C

( t )

z 1  t

e dt

³ ³



f f



e

t

z 1  iS ( z 1)  t

t

e

z 1  iS ( z 1)  t

e

e

 iS ( z 1)

e

We used the known identity sin T sin( S  x )

1 2i

e

iT

e

 sin x and recognising limp0

 iT

³



f



t

z 1  iS ( z 1)  t

e



f

f

³



t

e

e dt

(1.63)

z 1  t



f

e dt

z 1  iS ( z 1)  t



 iS ( z 1)

2i sin S ( z  1)

f

³ t dt  ³ t ³ t

e dt 

e dt

z 1  t

e dt

in the above steps of (1.63). Using the trigonometric identity z 1  t

e dt



* ( z ) from the definition of the gamma function in

(1.45), we write the following: limp0  I

³

C

( t )

z 1  t

e dt

2i  sin(S z )  * ( z )

(1.64)

Thus, we get Hankel’s formula valid for all z z 0, r1, r2, ..... for representation of the gamma function as *( z )

 2i sin(1 S z )

³

C

( t )

z 1  t

e dt that is described in (1.59).

1.10.7 Ratio of gamma functions at negative integer points Even though the values of the gamma function at negative integer points blow up the ratios of gamma functions of negative integers, they are however, finite; thus if N and n are positive integers then *(  n) N n N ! (1.65) (  N )(  N  1)....(  n  2)(  n  1) ( 1) *( N ) n! The proof of (1.65) is left out. The other properties of the gamma function are listed in the following lines (though we are not going to be proving all of them).

1.10.8 The duplication formula for the gamma function The duplication formula is described as follows: x

* (2 x )

4 * ( x )*  x 

1 2



2 S The above (1.66) one is from the Gauss-multiplication formula, which is set out as follows: * ( nx )

§ nx · ¨ ¸ n © 2S ¹

2S

n

n 1

§

k·

– ¨© x  n ¸¹ k 0

(1.66)

(1.67)

Concept Generalisation

19

1.10.9 Reflection formula of the gamma function The reflection formula is as follows (detailed proof is given in Section 1.15, pertaining to the beta function): S S csc(S x ) * ( x )* (1  x ) *( x) sin S x * ( x  1)

(1.68)

1.10.10 Asymptotic representation of the gamma function For x ! 0 , the asymptotic representation of the gamma function is as follows: x x

* ( x  1)  x e

2S x Note earlier that we wrote Stirling’s approximation for factorials (1.39) which is the same as above.

(1.69)

1.10.11 Reciprocal gamma function The reciprocal gamma function is defined as follows: 1

f

§ z· ze – ¨ 1  ¸ k¹ k 1© Jz

 z k

(1.70) *( z ) In (1.70) J 0.577215666.... ; it is Euler’s constant. This reciprocal gamma function is continuous everywhere for the real values of z . The value of the reciprocal gamma function is zero at all the negative integer points and at point zero (this is an important point), whereas the gamma function has discontinuities at all the negative integer points and at point zero. The reciprocal of the gamma function * (1x ) is single valued and finite for all x . The asymptotic representation, for a large approach of x is the following (with x tending to f ):

x

1

 12  x

 *( x) 2S Figure 1.2 gives the plot of a gamma function and Figure 1.3 the plot of a reciprocal gamma function:

(1.71)

*( x)

x Figure 1.2: Plot of a gamma function

1 *( x)

10

5

5

x

Figure 1.3: Plot of a reciprocal of the gamma function. Note that values are zero for x 1 , 2 , 3 …

10

0 and at x equal to negative integers

20

Chapter One 1 2 i sin(S z )

We have derived Hankel’s formula (1.59) for the gamma function, i.e. * ( z )

C

( t )

z 1  t

e dt . We use the

S  csc(S z ) and write the following steps:

reflection formula * ( z )* (1  z )

*( z )

1

2i sin(S z ) ³

C

( t )

z 1  t

e dt (1.72)

  * ( z )* (1  z )

³

2iS

From the above (1.72) steps we write the following useful expression: 1

C

1

( t )

z 1  t

e dt

z 1  t

(1.73) e dt * (1  z ) Z we obtain Hankel’s integral representation for the reciprocal gamma function with the following 2S i ³

( t )

C

Writing (1  z ) expression:

³

1

i

2S ³

* (Z )

Here we used the symbol

³

(0  )

(0  )

f

representing the contour integral on C , i.e.

f

( t )

³

C

Z  t

e dt

(1.74)

, meaning thereby a path starts at

infinity on the positive real axis, encircles 0 in a positive sense (counter-clockwise), and returns to infinity along the positive real axis, respecting the branch-cut along the positive real axis, as we have described previously too (Figure 1.1a). We will be using this expression (1.74) in the subsequent chapter. In the above expression (1.74) we make the substitution, t x then the above integral (1.74) is re-written as the following: 1 i (0  ) Z x (1.75) ( x ) e dx * (Z ) 2S f

³

Here we have used the symbol

³

(0  )

f

representing the contour integral on C , i.e.

³

C

, meaning thereby a path starts

at minus infinity ( f ) on the negative real axis, encircling 0 in a positive sense (counter-clockwise), and returns back to minus infinity, again on the negative real axis, respecting the branch cut along the negative real axis (Figure 1.1b). We write Hankel’s integral for the reciprocal gamma function as the following:

1

1

³

e

t

dt D * (D ) 2S i Ha t Where, for the above expression (1.76), Hankel’s path i.e. Ha is defined as in Figure 1.1b.

(1.76)

1.10.12 Representing the gamma function as a limit form The gamma function can also be represented as a limit, which is from the Euler formula for factorial ( z  1)! that is expressed in the following expression: z § · ( n !) n * ( z ) lim nnf ¨ ¸ © z ( z  1)...( z  n) ¹

(1.77)

Here we initially assume the right half plane Re > z @ ! 0 , in case of real number positive values. Let us introduce an auxiliary function to prove this part, f n ( z ) parts repeatedly, we get the following steps:

n

t n z-1 t dt n

³ 1  0

. Substitute W

t n

, and then performing integration by

Concept Generalisation n

21

t n z -1 t dt n

³ 1 

fn ( z )

0

1

n

z

n

³ (1  W ) W

z 1

dW

0

z

n ª(1  W )

n

¬

n

 º¼ z



n 1 n ª(1  W ) ¬

0

1

n

z

³  n(1  W )  W dW n 1

z

z

0

1

z

z nz z

1

Wz

³

n (1  W )

n 1 z

W dW

0

 W z 1

z 1

1

1

º  ¼0

 n ³  (n  1)(1  W )  dW nz z

n2

W z 1

z 1

0

§ nz ·1 n  2 z 1 (  1) n n ¨ z ( z  1) ¸ ³ (1  W ) W dW © ¹0 ................................................... 1

z

n n!

W z ( z  1)...( z  n  1) ³

z  n 1

dW

0

z

n n!

(1.78)

z ( z  1)...( z  n  1)( z  n ) Taking into account the well known identity i.e. lim nnf 1 

t n n

t



e , we expect the following:

§ n § t · n z 1 · lim nnf ¨ ³ ¨ 1  ¸ t dt ¸ ¨0© n¹ ¸ © ¹

lim nnf  f n ( z ) f

³e

 t z 1

(1.79)

dt

t

0

*( z )

Therefore, we have * ( z )

lim nnf



z

n !n z ( z 1)...( z  n )

.

1.10.13 Some interesting values of gamma functions that we frequently encountered As we have seen, the gamma function of a positive integer n is itself a positive integer, while the gamma function * (  n ) of a negative integer is infinite. The gamma functions *  12  n and *  12  n turn out to be multiples of *  12

S , thus we can write another formula as:

*  12

S,

(2 n )! S

*  12  n

n

4 n!

n

,

( 4) n ! S

*  12  n

(1.80)

(2 n )!

We list the few frequently encountered values of gamma functions as the following:

*   32 * (1) 1,

4 3

S, *  32

* ( 1) rf , 1 2

S,

*   12 2 S ,

* (2) 1,

*  52

3 4

* (0) rf ,

S,

*  12

S

(1.81)

* (3) 2

1.11 Ratio of gamma functions We will be using in subsequent chapters the expression

* ( j q ) * (  q ) * ( j 1)

frequently, where j is a non-negative integer and

q may take any value. For small numerical values of j this expression is readily simplified to a polynomial in q by

22

Chapter One

x  * ( x ) and * ( n  1) n ! . This procedure generalises to give Stirling’s approximation. Stirling’s approximation for the ratio of Gamma functions is given by Stirling’s number, shown by the following expression: application of * ( x  1)

*( j  q)

( 1)

* (  q )* ( j  1)

j

j

¦S

j!

m 0

( 1)

j



(1) Sj q

(2) 2 Sj q

( j) j  ...S j q



(m)

( m 1)

(m)



is Stirling’s number of the first kind with a recurring relation given as S j 1 { S j

(0)

1 , except S 0(0)

Sj

calculate

(0) Sj

(m)

Where S j (m)

(1.82)



j!

S0

(m) m j q

*( j q ) * (  q ) * ( j 1)

j j j

0 0 0

q q q

( 1)

1 2

j

j

¦

j! ( 1)

1 2

j

( 1)

3

q

¦

j

¦S

( 1)

1

 12

S

 12

(0) 1

1!

0

S0

0

1 (1)

 S1

 12 1



j

2

j

(0) 2

 12

0

(1)

 S2

1

 12

(2)

 S2

 12

2





1

(1.83)

8

(m) m j q

m 0 3

S

3!

( 0) 3

 12

0

1

(1)

 12

 S3

(2)

 S3

 12

2

(3)

 S3

 12

1

(1  x ) 2 as:

§ 1  12   12 x 2   12   12   32 x3  ..... · 1 x   ¨ ¸ 2! 3! © 2 ¹ x 

*  j  12

1

1

x 

2

( 1)

*   12 * ( j 1)

coefficients of the power series expansion which are ( 1)

j





16

If we observe the following power series expansion of f ( x )

0

3

1



1

2

(m) m j q

S

¦S

j!

(1  x ) 2

1

m 0

( 1)

0,1, 2, 3... in

(0)

( 1)

(m) m

Sj q

j

0

0!

m 0

j!

1 2

( 1)

(m) m

Sj q

j

j!

1 2

as follows:

m 0

2! j

1 2

0,1, 2, 3... and for q

( 1)

j

with

1 . A few Stirling numbers of the first kind are listed in Table 1.1. We use this table to

for a few terms for j

the terms for various

 jS j

1

2

j 1 j!

1 8

2

x 

¦

j

1 16

x  ...

(m)

m 0

(1.84)

3

Sj

 12

m

are pointing towards binomial



C j . This observation we will be used to generalise the

binomial coefficients. Representation (1.82), i.e. * ( j q ) * (  q ) * ( j 1)

* ( j q ) * (  q ) * ( j 1)

( 1)

j 1 j!

¦

j m 0

(m) m

Sj q

as the polynomial in q , establishes that the expression

is finite and single valued for all finite q and j . This relation with the Stirling number provides an

expression for the gamma function quotient

* ( j q ) *(q)

. The other quotient

* ( j q ) * ( j 1)

appearing in expression (1.82) is also

of interest. This quotient has asymptotic expansion for large j , i.e. in limit j n f , represented as the following expression:

*( j  q) * ( j  1)

 j

1 q

§ q ( q  1) 2 · ¨1  2 j  O( j ) ¸ © ¹

(1.85)

Concept Generalisation

j

m

0 1 2 3 4 5

0

m 1 0 1 -1 2 -6 24

1 0 0 0 0 0

m

23

2

m

0 0 1 -3 11 -50

3

m

0 0 0 1 -6 35

4

0 0 0 0 1 -10

m

5 0 0 0 0 0 1

Table 1.1: Some examples of Stirling numbers of the first kind The approximation, of the ratio of the gamma function as a polynomial in

q establishes the ratio of the gamma

function as finite and a single value function for all q and j . Two examples of the Stirling approximations for x n f are: *( x  a) a b 1 x 1  2 (x ) *( x  b) (1.86) *( x  q) q ( q  1)  (1 q ) § 2 · x  2 (x )¸ ¨1  * ( x  1) 2x © ¹



The result of (1.86) i.e. j n f then the limit of

* ( j q ) * ( j 1)

j q 1



 j

1 q

* ( j q ) * ( j 1)

1 

q ( q 1) 2j

2

 O( j )



is a representation that establishes itself as in the limit

is unity. Thus, one may write the following expression:

­f, § c  q 1 * ( j  q ) · § c q *( j  q) · ° lim ¨ j ¸ lim ¨ j ¸ ® 1, j nf * ( j  1) ¹ j nf © *( j ) ¹ ° © ¯ 0,

0 and writing j

From the above expression (1.87) we get the following, by setting c

§

lim N nf ¨ N

D

c!0 c 0 c0

N:

*( N  D ) · *( N )

©

(1.87)

(1.88)

¸ 1 ¹

The result that may be generalised to the following relationship:

­f, § c  q 1 * ( j  k  q ) · ° lim ¨ j ¸ ® 1, j nf * ( j  k  1) ¹ ° © ¯ 0,

c!0 c 0 c0

(1.89)

1.12 Generalising binomial coefficients by gamma function The expression that we discussed in (1.82), i.e.

* ( j q ) * (  q ) * ( j 1)

, may be regarded as a generalised binomial coefficient;

given by the following relationship: *( j  q) * (  q )* ( j  1)

Equivalently we write

* ( j q ) * (  q ) * ( j 1)



( j  q 1)

Cj



( 1) n

from the definition of a binomial coefficient i.e. Cr

j



q

§ j  q  1· j §q· ¨ j ¸ ( 1) ¨ j ¸ © ¹ © ¹

(1.90)



C j . The above (1.90) equivalents are readily established

n! r !( n  r )! ,

using what we learnt about factorials and the gamma

function and use of a reflection formula (1.68) i.e. * (  x )* ( x  1) steps:

*( j  q)

*  ( q  j )

* (  q )* ( j  1)

* (  q )* ( j  1)

S csc(S x ) . This is described in the following

§ · §  sin S q  * ( q  1) · § 1 · S ¨ ¸¨ ¸¨ ¸ ( S ) ¹ © * ( j  1) ¹ ©  sin S ( q  j )  * ( q  j  1) ¹ © * ( q  1) § sin S q ·§ · ¨ sin S ( q  j ) ¸¨ * ( q  j  1)* ( j  1) ¸ © ¹© ¹

(1.91)

24

Chapter One

The ratio

sin S q sin S ( q  j )

sin S q sin S ( q  2)

sin S q sin(S q  2S )

in (1.91) for sin S q sin S ( q 3)

1 ,

j

sin S q sin S q

0,1, 2, 3, 4... takes the values as

sin S q sin(S q 3S )

sin S q sin S ( q 1)

1,

sin S q sin S ( q  j )

1 …. and so on. Therefore, we can write

sin S q sin(S q S )

1 ,

j

( 1) . From

this we can write the following result:

*( j  q)

* ( q  1) § sin S q · § · ¨ ¸ ¨ ¸ * (  q )* ( j  1) © sin S ( q  j ) ¹ © * ( q  j  1)* ( j  1) ¹ ( 1)

j

( 1)

j

* ( q  1)

( 1)

* ( q  j  1)* ( j  1)

q!

j

§q· ¨ j¸ © ¹

We can write the following expression from the above derived relationship in (1.92), by putting j *( j  q) *(q)

( 1)

j

* ( q  j  1)

*(n  z )

,

( 1)

*( z )

n and q

* (  z  1)

n

z : (1.93)

* (  z  n  1)

z  * ( z ) , we observe from the above expression (1.93), that if n is an

Besides the obvious relationship * ( z  1) integer, then  * ( z  n )  * (  z  n  1)

* ( q  1)

(1.92)

( q  j )! j !

( 1)

n

 *( z )  *(1  z ) , for all ( z)

positive integer, then we may generalise the binomial coefficients

§ z · ¨u¸ © ¹

n . Using D ! * (D  1) , when D is not a

Cu as the following:

(  z )!

* (1  z )

u !(  z  u )!

* (u  1)* (1  z  u )

(1.94)

In particular if u is a non-negative integer, say n , then using the above (1.94) identity, we have following the relationship: * (1  z ) § z · n *( z  n) n § z  n  1· ( 1) ( 1) ¨ ¨n¸ ¸ n !*( z ) © ¹ n ! * (1  z  n) © n ¹ (1.95) Of particular use is the expression we wrote earlier (1.92), i.e.

* ( j D ) * ( D ) * ( j 1)

j D 1

Cj

( 1)

j



D



C j , resulting in the

following (here we have changed our q to D ): *( j  D )

§ j  D  1· j §D · ( 1) ¨ ¸ ¨ ¸ j © ¹ © j¹

* ( D )* ( j  1) n

We simplify (1.96) by following with Ck

n! k !( n  k )!

: *( j  D )

j

§D ·

( 1) ¨ ¸ © j¹

* ( D )* ( j  1)

( 1) *( j  D )

( 1)

D!

j

j !(D  j )!

* (D  1)

j

* ( j  1)* (D  j  1)

( 1)

n

¦ §¨© j 0

¦

n j 0

j D 1

Cj

n D

j  D  1· j

(1.97)

* (D  1)

j

* ( D )* ( j  1) * ( j  1)* (D  j  1) We can then write a useful relationship from a derivation of (1.97) that is the following: * ( p  1) j *( j  p) ( 1) * ( p  j  1) *( p) A useful identity from a standard Handbook of Mathematical Functions is listed below:

(which symbolically also is

(1.96)

§n D · ¸ ¨ n ¸ ¹ © ¹

(1.98)

(1.99)

Cn ), deriving the following relationship: N 1

*( j  D )

¦ *(D )*( j  1)

*( N  D )

(1.100) * (1  D )* ( N ) This is followed by setting N n  1 and expressing the binomial coefficients as their equivalent gamma function combination. Similarly after multiplication by D , it is readily shown as the following expression: j 0

Concept Generalisation

25

n

j  D  1· j  1 ¸¹

¦ §¨© j 1

¦

This is then straight forward from the above expression that is

n

j D 1

j 0

§n D · ¨ n 1 ¸ © ¹ n D

Cj

(1.101)

Cn . On redefining D , j and n

we are led to the following identity: N 1

D  * ( N  D )

*( j  D )

¦ *(D )*( j )

(1.102)

* (2  D )* ( N  1) Likewise, the well-known formula, from the Handbook of Mathematical Functions that is: j 0

j

§ q ·§ Q  k · § q  Q · ¨ j ¸ © ¹ 0

¦ ¨© k ¸¨ ¸ ¹© j  k ¹ k

(1.103)

becomes the following identity: j

* ( q  1)* ( p  1)* ( j  1)

* ( p  j  1)

¦ *(q  k  1)*(k  1)*( p  q  k  1)*( j  k  1)

* ( p  q  j  1) on setting in (1.103) Q { ( p  q  j ) and multiplication by * ( j  1) . Yet again, the binomial relationship is:

(1.104)

k 0

n

¦ §¨© j 0

¦

which is obtainable from our earlier noted relationship (1.99)

n

j  q  1· j  m ¸¹

j D 1

j 0

§nq· ¨n  m¸ © ¹

n D

Cj

(1.105)

Cn , which leads easily to the

following identity: n

¦ §¨© j 0

Table 1.2 gives some values for

1

2

Cn ,

n

¦  j 0

1

2



Cn ,

 12

j  q  1·§ j · ¸¨ m ¸ j ¹© ¹ n

¦ 

Cn and

 12

j 0

§ m  q  1·§ n  q · ¨ m ¸¨ n  m ¸ © ¹© ¹

(1.106)



Cn ; and also values for a very large n . 1

We have used values from column one of Table 1.2 in the expansion of (1  x ) 2 , in the half-coin problem of Section1.2. We then have the following expansion:

( x  1)

n

n

0

n

1

0

1

x  nx  ( x  1)

1

1

2

n

2

n

C0 x  C1 x  C2 x  .........  Cn x

2

n ( n1) 2 n ( n1)( n2) 3 n x  .... x 2! x  3!

1

0

n

1

1

1

2

C0 x  2 C1 x  2 C2 x  ......  2 Cn x

0

x 

1 2

1

x 

1 8

2

x 

1

3

16

x 

5 128

4

x 

(1.107)

n

7

5

256

x .......

that was used earlier in Section 1.2:

n

1

2

n

¦ 

Cn

1

j 0

2

Cj



 12

Cn

n

¦ 

 12

j 0

0 1

1

1

1

1

1 2

3 2

 12

1 2

2

 18

11 8

3 8

7 8

3

1 16

23 16

 165

9 16

4

5  128

179 128

35 128

107 128

5

7 256

365 256

63  256

151 256

6

21  1024

1439 1024

231 1024

835 1024

f

 ( 1)n

2

( 1)n

1 2

2 S (n)

3

2

Table 1.2: Some values of binomial coefficients,

1

2

Cj ,

 12

S (n)

C j and their cumulative sums

Cn



26

Chapter One

1.13 Incomplete gamma functions 1.13.1 Tricomi’s incomplete gamma function and the complementary incomplete gamma function Tricomi’s incomplete gamma function is defined in the integral form:

x

*

J ( a, x) f

³

where we have defined Euler’s gamma function, that is * ( a )

0

gamma function is defined as the following integral:

a

*(a)

x

³e

 t a 1

dt

t

(1.108)

0

 t a 1

dt , whereas a complementary incomplete

e t

f

³e

*( a, x)

 t a 1

t

dt

x t 0,

f  a  f

(1.109)

x

The asymptotic expansion for large x of the complementary incomplete gamma function is the following in limit x n f: *( a, x) # x

a 1  x

e

§ a  1 ( a  1)( a  2) ·   .... ¸ ¨1  2 x x © ¹

(1.110)

1.13.2- Incomplete gamma function The incomplete gamma function is defined as follows: x

³e

J ( a, x)

 t a 1

t

dt

Re > a @ ! 0

(1.111)

*(a)

(1.112)

0

We thus have:

J ( a, x )  *( a, x)

³

where we have defined Euler’s gamma function, that is * ( a )

f

0

 t a 1

dt .

e t

1.13.3 The elementary relationships of Tricomi’s incomplete gamma function Tricomi’s incomplete gamma function is defined in series form as depicted below: *

J (v, z )

e

z

f

z

k

¦ *(v  k  1)

1

*

and

J (v, 0)

def

It is also represented by Kummer’s function, that is M ( a , b, z )

e

*

J (a, x)

1

a z b 1!



a ( a 1) z 2 b ( b 1) 2!

 ... as follows:

x

* ( a  1)

(1.113)

* (v  1)

k 0

M (1, a  1; x )

M ( a , a  1;  x )

(1.114)

* ( a  1)

We have the following identities: a § · x 1  ¨ J * ( a, x) ¸ *( a ) © ¹ *( a, x) · * a § J ( a, x ) x ¨1  ¸ *(a) ¹ ©

*( a, x)

*

J ( a, x)

x

(1.115)

a

*(a)

J ( a, x)

If p is a non-negative integer, the following properties are deduced from a series definition as: *

J ( p, ax )

e

 ax

f

¦ k p

The second expression above in (1.116) is proven as follows:

( ax )

k p

k!

and

*

J (  p, ax )

( ax )

p

(1.116)

Concept Generalisation

*

J (  p, ax )

e

f

 ax

( ax )

¦

 ax

( ax )

f

p

( ax )

¦

 ax

e

 ax

( ax )

p

( ax )

p

0 1 2 ª ( ax )  p ( ax )  p 1 º ( ax ) ( ax ) ( ax )       ......» ..... « (  p )! (  p  1)! 0! 1! 2! ¬ ¼ f

¦

( ax )

 ax

( ax ) In the above steps of (1.117), the

1 p!

p ax

( ax ) e p

 ax

*

k

0 for all p  1  0 . The special values of this version of Tricomi’s

1 * ( p 1)

1 e ax

incomplete gamma functions are J (1, ax )

(1.117)

k!

k 0

e

k

k!

k p

e

k p

k!

k p

e

27

*

*

, J (0, ax )

1 and J ( 1, ax )

ax . Tricomi’s incomplete gamma

>@

function is an entire function of both z and v ; and its integral representation is for Re v ! 0 is from (1.108) is: *

J (v, t )

def

1 t *(v) v

t

³u 0

v *

v 1  u

e du and lim t nf t J

1.

Some elementary relationships that exist amongst Tricomi’s incomplete gamma functions are as follows: *

*

J (v  1, t )  tJ (v, t )

e

t

f

t

k

t

f

t

k 1

e

¦ *(v  k )  e ¦ *(v  k  1) k 0

*

e at *(v)

*

(1.118)

* (v )

k 0

Replacing in (1.118) the variable t , with at , we get J (v  1, at )  ( at )J (v, at )

t

. Iterating this ( p  1) times

we arrive at the following identity: *

p *

J (v, at )

( at ) J (v  p, at )  e

 at

f

( at )

k

¦ *(v  k  1)

(1.119)

k 0

Using the series definition (1.113) of Tricomi’s incomplete gamma function on the above expression (1.119) we get the following identity: f

v at *

( at ) e J (v, at )

( at )

k v

¦ *(v  k  1)

(1.120)

k 0

and taking the p -th derivative of (1.120), leads to the following identity:

d

p

f

( at )

k

ªt v e at J * (v, at ) º¼ t v  p ¦ p dt ¬ k 0 * (v  k  1  p )

(1.121)

v  p at *

t e J (v  p , at ) From this (1.121) and the earlier identity in (1.119) we get the following expression:

d

p

v at *

p





v at *

v p

p 1

¦

( at )

k

ªt e J (v, at ) º¼ a t e J (v, at )  t (1.122) p dt ¬ k 0 * (v  k  1  p ) In this form (1.122) we see that Tricomi’s incomplete gamma function on both sides of the equation have the same arguments. From this with p 1 , we find the integral form to be the following: t v au *

³u e

J (v, au )du

t

v 1 at *

e J (v  1, at ) ;

Re >v @ ! 1 (1.123)

0

Tricomi’s incomplete Gamma function is used in obtaining fractional differentiation and fractional integration of periodic functions, used as the sinusoidal response studies of fractional operators; we will use this in subsequent chapters.

28

Chapter One

1.14 Power numbers Power numbers may be expressed in terms of complete gamma function as follows:

q

j

j

¦ (1)

m

m 0

> m@

The coefficients S j

(1.124)

appearing in this expression are Stirling numbers of a second kind. Examples of these numbers, > m@

> m 1@ > m@ > m@ Sj  mS j , S 0

which obey recurrence S j

m

j 0 1 2 3 4 5

> m@ * ( m  q ) Sj *(q)

0

m 1 0 1 1 1 1 1

1 0 0 0 0 0

> 0@ 0 except S 0

> 0@ Sj

m

2

1 , are listed in Table 1.3.

m

0 0 1 3 7 15

3

m

0 0 0 1 6 25

4

m

0 0 0 0 1 10

5 0 0 0 0 0 1

Table 1.3: Some values of the Stirling Number of the second kind > j@

Note that S j

> m@ 1 for all j and that S j

0 for j

0,1, 2, ..., ( m  1) . The formula m

¦ (1)

l m

l 0

§m· k ¨ l ¸l © ¹

> m@ m ! Sk

(1.125)

may be regarded as defining Stirling numbers of the second kind. This is used in an expansion of the analytic function M of the argument of ( x  jy ) in a rather special way. We give the gist of this here, firstly by relating M ( x  jy ) to the values M ( x ) , M ( x  y ) , M ( x  2 y ) ,…, M ( x  jy ) , by the formula: j

§ j· m l m § m · M ( x  jy ) ¦ ¨ ¸¦ ( 1) ¨ ¸ M ( x  ly ) m ©l¹ m 0© ¹l 0

(1.126)

The inner summation in (1.126) is symbolized as Gm (M , x, y ) and is expanded in the Taylor series as follows: m

¦ (1)

Gm (M , x, y )

l m

l 0

f

y

k

¦ k !M

(k )

k

§ m · f (ly ) ( k ) M ( x) ¨ l ¸¦ © ¹ k 0 k! m

( x)

k 0

It requires now an application of summation (1.125) i.e.

¦

m l 0

( 1)

l m

¦ (1) l 0

 C l m

j

¦G

m (M , x ,

m 0

In (1.128) Gm (M , x, y )

f

¦  S > k 0

m! k!

m@ k ( k ) ( x) . k y M

Since

j

§m· k ¨ l ¸l © ¹

(1.127)

> m@ m ! S k to give the final result:

k

l

M ( x  jy )

l m

§ j· y) ¨ ¸ ©m¹

(1,128)

Cm vanishes when m exceeds integer j , the upper

summation limit in the above expression (1.128) may be replaced by f . Similarly, (as this is a consequence of > m@ Sj

0 , for

is Gm (M , x, y )

j

0,1, ..., ( m  1) ), the k f

¦  S > k 0

m! k!

m@ k ( k ) ( x) , k y M

0 that is the lower summation limit of the expression that

may be replaced by k

m.

1.15 The beta function and the incomplete beta function The derived functions from the complete gamma function are called beta functions, defined as: 1

%( p , q )

³u 0

p 1

(1  u )

q 1

du

* ( p )* ( q ) *( p  q)

(1.129)

Concept Generalisation

For the positive values of p ! 0 and q ! 0 the definite integral

29

1

³u

p 1

0

(1  u )

q 1

du is called the beta integral, also

known as Euler’s integral of a second kind. If either p or q is non-positive the integral diverges and then the beta that %( p , q )

*( p )*(q ) *( pq)

B( p , q )

function is defined by the relationship

p and q . We also note

, valid for all

B( q , p ) . The proofs we are omitting.

1.15.1 Integral representations of the beta function With the above definition (1.129) we can form interesting and useful integral representations. Say, we use u thus we get du

1 ( z 1)2

dz ; this gives limits as for u

0 we get z

0 and for u 1 we get z  1

z z 1

and

z which means

z n f . With these substitutions we get the following steps: 1

³u

%( p , q )

p 1

(1  u )

q 1

du

0

f

§ z · ³0 ¨© z  1 ¸¹

f

z

p 1

f

z

p 1

p 1

z · § ¨1  ¸ © z 1¹

q 1

dz ( z  1)

2

(1.130)

1 § ·§ 1 · ³0 ( z  1) p1 ¨© ( z  1)q1 ¸¹ ¨© ( z  1)2 ¸¹ dz

³ ( z  1)

pq

dz

0

This gives the formula as:

³

f

u

0

If say we have p  q

p 1

(u  1)

 p q

du

* ( p )* ( q )

B( p , q )

(1.131)

*( p  q)

r , then we get another useful formula from (1.131):

³

f

0

u

p 1

r

(u  1) du

* ( p )* ( r  p )

B( p , r  p )

(1.132)

*(r )

From these derived expressions, we can write another expression:

³

f

0

From the formula of (1.132) i.e.

³

f

0

u

p 1

r

(u  1) du

u

X 1

(u  k )

p

du

k

 ( p X )

(1.133)

B( p  X , X )

B( p , r  p )

*( p )*(r  p ) *(r )

B( x,1  x )

* ( x )* (1  x )

we can write the useful integral as

³

f

0

u

x 1

1

(u  1) du

(1.134)

1.15.2 The derivation for the reflection formula of the gamma function We will use expression (1.134) to prove the important formula for gamma functions i.e. * ( z )* (1  z ) S csc S z for z z 0, r1, r2, ... as well using contour integration, first by restricting the real values of x z with the restriction 0  x  1. For this we consider the contour integration i.e. function f (Z )

Z

x 1

1

(1  Z ) . Clearly Z

³

C

Z

x 1

1

(1  Z ) dZ ; with 0  x  1 , in a complex plane Z . Take the

1 is a singularity (pole of order one). For x

singularity (pole of order one); but when 0  x  1 the point Z x 1

0 , we call Z

0 as a

0 is a logarithmic branch point. Writing Z  eiT , x 1

i ( x 1)T

D

D

2S i ( x 1)

; showing as  e i.e. from  to  e and going about point zero in a circle gives values for Z discontinuity at every rotation, as was the case for the complex logarithm function ( ln Z ), described in Section 1.6.4. Thus at Z 0 we draw a branch cut on the negative real axis shown in Figure 1.4. We take contour C in complex plane Z with the branch point at Z 0 as depicted in Figure 1.4 (with a standard branch cut). We notice that the contour encloses the pole at Z 1 , and bypasses the branch point at Z 0 . Therefore, the only residue required is

30

Chapter One

for Z = 1 . The contour is broken as C

C1 * C2 * C3 * C4 , which is depicted in Figure 1.4. We use residue calculus

as follows:

³

Z

x 1

dZ

1Z

C

³

©

x 1

· ¸ 2S i 1Z ¹ Z

(1.135)

iT

x 1

dZ

lim

1Z

C1

2S i ¨ limZ o1 Z  1

Re we write for 0  x  1 the following:

On the C1 large circle with radius R ; Z

Z

§

2S i  ResidueZ 1 f

R

S

³



x 1 i ( x 1)T

e

d Re

1  Re

R nf S



iT

iT

x ixT iT R e dT

S

³

lim

1  Re

R nf S

³

# lim

S

R nf S

iT R

x 1 iT ( x 1)

e

dT

iT

(1.136)

0

iT

On the small circle with radius  ; Z  e we write for 0  x  1 the following:

Z

³

C3

x 1

dZ

1Z

S

³

lim

x 1



³

x

Z

³ On the line C4 from   to  R we write Z

³

iS

dZ

x 1 dZ

³

R

³

 iS

U

# lim

iT

³

p 0 S

U



x 1 i ( x 1)S

e

R

Ue

(1.137) S

x ixT

iT  e dT

0



dU

and we get:

x 1

U



iT

ixT

1  e

1Z

C2

Z

Ue

U



d e

iT

iT  e dT

S

p 0 S

On the line C2 from  R to   , we write Z

i ( x 1)T

1  e

p 0 S

lim

e



d Ue

iS

³



U

R

1 U

x 1 ixS

e

1 U

(1.138)

and we get the following:

x 1  i ( x 1)S

e



d Ue

 iS



³

R

U

x 1  ixS

e

dU

(1.139)   1Z 1 U 1 U The total contour integration on C is the sum of all four segments and we know that is equal to 2S i (calculated from residues). Thus, in the limit R n f and p 0 we have: C4

³

Z

x 1

dZ

³

1Z From above (1.140) we can write the following steps: C

e

2S i

³

f

0

U

 iS x

e

iS x

x 1

dU

1 U

B( x,1  x )

³

f

0

2S i e

iS x

e

U

0

U

e

dU

1 U

f

x 1

dU

1 U

;

S  iS x

x 1 iS x

sin S x

³

f

x 1  iS x

iay

e

dU

1 U e

2S i

(1.140)

B( x,1  x )

(1.141)

 iay

2S i

³

f

0

u

x 1

1

(1  u ) du

S  csc S x

Recognizing the LHS in (1.141) as B( x,1  x ) reflection formula as in (1.68):

* ( x )* (1  x )

³

f

0

Z

x 1

1

(Z  1) dZ , we have proved the following

* ( x )* (1  x )

here for 0  x  1 real z

U

0

e

sin( ay ) ;



S csc S x

(1.142)

x.

1.15.3 Application of analytic continuation to the reflection formula The above obtained result of (1.142) is for 0  x  1 real z x . The full result follows from the concept of analytic continuation. Alternatively, the result is obtained by the following argument. If the identity * ( z )* (1  z ) S csc S z holds for real values of z for 0  z  1 , then it holds for all complex z with 0  Re z  1 by analyticity. Then it also holds for Re z 0 with z z 0 by continuity. Finally, the full result follows for z shifted by integers using

* ( z  1)

z  * ( z ) and sin( z  S )

 sin z . Note that * ( z )* (1  z )

with z z 0, 1, 2, ... . Instead of * ( z )* (1  z )

S csc S z holds for all complex values of z , S csc S z we may write:

Concept Generalisation

31

1

sin S z

* ( z )* (1  z )

S

(1.143)

which holds for all z in complex plane ^ . Im Z

dZ ³C Z (1  Z ) 1 x

C1

C2

C3

Re Z

x

C4

Z 1

Branch point: Z 0

C C1 * C2 * C3 * C4

Figure 1.4: Contour for integration with pole and branch point with branch cut

1.15.4 Incomplete beta functions In addition, there is an incomplete beta function (analogous to the gamma function and its incomplete counterpart) which is defined as: def x

³t

% x (u , v )

u 1

(1  t )

v 1

dt ;

0  x 1

(1.144)

0

If Re >v @ ! 0 and k is a non-negative integer then we have the following definition: x

³

u

v 1

k

( x  u ) du

x

k v

%( k  1, v )

0

( k !) x

k v

* (v )

* (v  k  1)

(1.145)

Hence, we may write the same as the following: x

1 v

k ! x * (v )

³

u

v 1

x

k

( x  u ) du

k

* (v  k  1)

0

(1.146)

Using the series definition of an incomplete gamma function (Section 1.13) and applying the integration by parts formula i.e.

³ uvdx u ³ vdx  ³ u  ³ vdx dx , we obtain: (1)

x ª x v 1 1 v § f ( x u ) j · º « u d u  ³ u ¨ ¦ j ! ¸ du » v x * (v ) ¬« ³0 v0 ©j 0 ¹ ¼»

1

x *

e J (v, x )

1

x

u x * (v ) ³ v

v 1 ( x u )

e

(1.147)

du

0

Thus, we see that the integral representation of incomplete gamma functions is described earlier (Section 1.13). Another useful representation of the incomplete Beta function comes from the hypergeometric function (refer to Appendix A), as follows:

B x (c,1  b  c )

c

1 b  c

x (1  x )

§ f j *( j  1  b) · ¨¦ x ¸ * (1  b ) © j 0 * ( j  1  c ) ¹ * (c )

(1.148)

32

Chapter One

1.16 Psi function The psi function is also called the ‘digamma function’. The derivative of the complete gamma function is a ‘psi’ function \ ( x ) and is defined as following, for x z 1, 2, 3, ... :

\ ( x)

1

d >* ( x )@

*( x)

dx

or

\ ( x)

d >ln * ( x )@

* ( x)

dx

*( x)

(1)

(1.149)

One property ‘psi’ function, that is:

* (c )

¦ * (b )

\ (c )  \ (c  b)

>@

f k 1

§ * (b  k ) · ¨ k * (c  k ) ¸ © ¹

(1.150)

>@

for Re c ! Re b (we are not proving this property), gets derived via use of the ‘hypergeometric function’.

1.16.1 The recurring series formula for psi function However, by the following steps and manipulating the gamma function we get a recurring series representation for  \ ( x  1)  \ ( x  1  m) , where x ! ( m  1) with m  ] . f

* (c )

* (b  k )

¦ k * (c  k ) * (b )

\ (c )  \ (c  b)

k 1

f

* (c )

(b  k 1)(b  k  2).....(b )(b 1)(b  2)....(2)(1) ¦ k * (c k ) (b  1)(b  2)..........(2)(1)

(1.151)

k 1

f

* (c )

¦

(b  k  1)(b  k  2)....b k * (c  k )

k 1

k

( 1) ( b)( b  1)....( b  k  1) . Here we substitute in the above expression

and (b  k  1)(b  k  2)....(b  1)(b) (1.151) c

 m , so x ! ( m  1) where m is a positive integer, so that the condition Re >c @ ! Re >b@ is

( x  1) , b

also maintained, to get the following expression:

\ ( x  1)  \ ( x  1  m)

* ( x  1) f

k

f

( 1) m( m  1)...( m  k  1)

k 1

k *( x  1  k )

¦

k ( 1) m ! * ( x  1)

¦ k (m  k )!*( x  1  k )

(1.152)

x ! ( m  1)

k 1

The

psi

\ (1)

J

1 2

obeys

(1)

0.5772156666... which is called Euler’s constant. Euler’s constant is also noted as

* (1)



3

J



10  1 , also J

recursion

lim nnf

\ (1D ) the expression limD o n ª * (1D ) º ¬ ¼

\ ( x  1)  \ ( x )

1 x

are

\  12 \ (1)  ln 4 \ ( z  1)

J 

\  z \  z 

1 2

the

J  ln 4 .

¦  f

z k 1 k ( zk )

 2 ln 2

.

n

relation

1 x

function

\ ( x  1)  \ ( x )

and

it

has

a

special

value

of

¦   ln n . For D  \ a real number which tends to a natural number n  ` ( 1) * ( n ) is true. In addition, we observe that \ ( n  1)  \ (1) ¦  and 1 k

k 1

n

n

j 1

properties If

z

Similarly

of

the

‘psi’

function.

The

other

interesting

1 j

properties

are

is not a negative integer then we write the relationship as there

is

the

relationship

\ (1  z )  \ ( z )

S cot(S z )

2\ (2 z ) .

1.16.2 Analytical continuation of a finite harmonic series by using psi function We can make an interesting application of ‘analytical continuation’ of the finite harmonic series h ( n ) :

and

Concept Generalisation n

1

¦k

h( n)

33

1

1

2

k 1

1



3

 ..... 

1 n

\ (1  n )  J

(1.153)

ln n  J  O  1n The above derivation (1.153) is from number theory. Here the use of the psi function with Euler’s constant is n

demonstrated. Now we observe that the finite harmonic series h ( n ) appears as the coefficients of x terms when we

³ uvdx u ³ vdx  ³ u  ³ vdx dx ), as follows: (1)

repeatedly integrate ln x (using integration by parts formula i.e.

³

x

0

y x

 ln y  dy

ln y

³

(1)dy

 y 0

 ln x ( x)  ³ That we write as

x

³  ln y dy 0



³

d dy



³

y x

ln y (1)dy dy

  ydy 1 y

y 0

(1.154)

y x

x  ln x  1

y 0

x (ln x  1) . Integrating it once more, we get the following:

³

x

0

ln y (dy )

2

³

³

ln y ydy

 ln x  x

³

ln y (dy )

2

x

0

x2 2

y 0

y x y 0

³



d dy

ln y

 ³  dy y 2

2

2

y x

( y ln y  y )dy

x

 ln x  32

2

2

 ³ ydy dy

y x y 0

y x



y 0

 x2 2

  x ln x  x2 2



2

1 2

3 4

x

2

(1.155)

 ln x  1  12

Continuing n times, we arrive at the following expression: x

³

ln y (dy )

x

n

 ln x  1  n!

n

0

1 2

 13  ..... 

³

Therefore, by analogy to the concept of generating functions, we can take

x

0

1 n



x

n

n!

ln y (dy )

n

 ln x  h(n)

(1.156)

as the generating integral of

the harmonic series h ( n ) . Thus by analytically continuing for a non-integer n , (here D ), we get the following: x

³

x

D

ln y  dy

D

* (1  D )

0

 ln x  h(D )

(1.157)

We will take up this integral (1.157) in a subsequent chapter.

1.17 A unified notation to represent multiple differentiation and integration We know the notation

dn f ( x ) dx n

represents the n  th derivative of a function f ( x ) with respect to x , when n is a non-

negative integer. Because integration and differentiation are inverse operations (with exceptions that we will revise later), it is logical to associate the symbol

d 1 f ( x ) dx 1

with the indefinite integration of f ( x ) with respect to x . However,

it is mandatory to define a lower limit of integration so that the indefinite integration is completely specific, that is

d 1 f ( x ) dx

1

x

{ ³ f ( y )dy . Multiple integrations with a zero-lower terminal limit are symbolized as follows: 0

x

³  f ( y ) (dy )

2

f ( x) 2

x

0

2

> dx @

0

³  f ( y ) (dy )

d

n

d

n

f ( x) n

> dx @

x

x1

³ ³  f ( x ) dx

{ dx1 0

0

x

³

{ dxn 1 0

0

0

xn1

³ dx 0

x2 n2

x1

... ³ dx1 ³  f ( x0 ) dx0 0

0

(1.158)

34

Chapter One x

xa

³a  f ( y ) dy ³0  f ( y  a) dy , we extend the above symbolism (1.158) to a non-zero lower

Using the identity limit as follows:

x

1

d f ( x) 1

>d( x  a)@ d

n

f ( x) n

>d( x  a)@ The notation f

(n)

{

³  f ( y ) dy a

xn1

x

³

{ dxn 1 a

x2

³ dx

n2

a

... ³ dx1 ³  f ( x0 ) dx0 a

a

n

d f ( x)

( x ) will find frequent usage as the short-form of

(1.159)

x1

dx n

. We will also use f

(n)

( x ) to represent

n  fold integration with respect to x , where the lower limits are not specified. That is xn1

x

f

(n)

( x) {

³ dx ³ d x n 1

an

where a1 , a2 , ....an are completely arbitrary. However, for

f

x2

(n)

n2

an1

( x)  f

(n)

x1

... ³ dx1 ³  f ( x0 ) dx0 a2

(1.160)

a1

( a ) we attach the same lower limits

a1 , a2 , ....an to each integral. We have introduced simple symbols in the subsequent chapters; we will be using these and many other symbols that are used historically and for convenience.

1.18 Differentiation and integration of series This idea of classical calculus is very important as the technique we will be using in finding fractional differentiation and fractional integration for functions expanded in a power series. A great many functions are traditionally described by infinite series expansions. It is of very much importance to understand the conditions that permit term-by-term differentiation or integration of such functions. Suppose that f0 ( x ) , f1 ( x ) are functions defined and are continuous on the interval a d x d b , then: d

1

ª f f ( x) º 1 ¬ ¦ j 0 j ¼ >d( x  a )@ provided the series

¦f

j

1

ª¬ f j ( x )º¼ ,

f

d

j 0

>d( x  a )@

¦

1

ad xdb

(1.161)

converges uniformly in the interval a d x d b .

The condition required for a derivative to be distributed through the terms of an infinite series are somewhat different. For this, we need each f j to have a continuous derivative on interval a d x d b then, we have

d ª dx ¬ provided that

¦f

j

converges point-wise and

¦

d ª¬ f j º¼

¦

f j 0

fjº

¼

f

d ª¬ f j º¼

j 0

dx

¦

ad xdb

(1.162)

converges uniformly on the interval a d x d b .

dx

Therefore we note that a uniform convergent series of continuous function (which itself defines a continuous function), may be integrated term by term, and that a continuously differentiable function may be differentiated term by term provided that the derived series is uniformly convergent.

1.19 Revising the concepts of mixed differentiation and integration In classical calculus the identity:

§ d N f ( x) · d n N f ( x) n ¨ N ¸ n N >dx@ ¨© >dx@ ¸¹ >dx@ d

and the identity:

n

d

N

> dx @

N

§ d n f ( x) · ¨ ¸ ¨ >dx@n ¸ © ¹

(1.163)

Concept Generalisation

35

§ d  N f ( x) n ¨ N >d( x  a)@ ¨© >d( x  a)@ d

· ¸ ¸ ¹

n

d

d

 n N

f ( x)  n N

>d( x  a)@

(1.164)

§ d  n f ( x) · ¨ ¸ ¨ >d( x  a )@ n ¸ © ¹

N N

>d( x  a)@

hold for n and N non-negative integers. These (1.163) and (1.164) identities are basics to multiple differentiations and integration. Let us try mixed differentiation and integration. For example, take integrations three times. That is, first, we have x

³ e ³ 0 x

2y

2

 y

0

e2 y 4



y2 2

d dx

ª e8  ¬

x3 6

2x

1 2

 

dy

e2 x  x 2  x  1 . 4 2 2 4

y 2



1 4

dy

x2 4



x 4

 81 º

e2 x 8

¼



e2 x 4

x3 6



 x2 2

x2 4



Using

 4x  81 . x 2



1 4

³

x

0

e

2x

 1 and a

(e 2 y  1)dy

e2 x  x  2

we

one

more

once

the

this, Now

f ( x)

do

differentiating

1 2

0.

First, we do

; doing it once more, we have integration last

and

expression,

write we

get

. Therefore, following our notation (of Section 1.17) we write the result as

below: 3 2x d § d ª¬e  1º¼ · ¨ ¸ >dx@ ¨ >dx@3 ¸

©

d

2

ª¬e 2 x  1º¼

1

2

> dx @

¹

4

e

2x



1

2

x 

2

1 2

1

x

(1.165)

4

2x

Following a similar procedure, we first differentiate the function f ( x ) e  1 once, and then we integrate three folds (from a 0 ). Then we write in the following form the result in our described notation (of Section 1.17) to get the following expression:

§ d ª¬e 2 x  1º¼ · 1 2 x 1 2 1 1 d 2 ª¬e 2 x `1º¼ ¨ ¸ e  x  x z 3 2 2 2 4 >dx@ ¨© dx ¸¹ 4 > dx @ d

3

(1.166)

We note that the above two methods (1.164) and (1.165) give different results.

dN >d( x  a )@N



d n f ( x )

>d( x  a )@ n





d n

Here we will deal with mixing differentiation and integration, that is



dN f ( x)

>d( x  a )@ n >d( x  a )@N

d N n f ( x )

,

>d( x  a )@N n

and

; and then derive similar identities when any two are unequal. If f ( x ) is a function that is

N  fold differentiable with N t 1 then by, the identity (1.163) i.e.

dn >dx @n





dN f ( x)

> dx @

N

d n N f ( x )

dN > dx @ N

n N

> dx @

using a fundamental theorem of calculus we write the following expression (with understanding f 1

(N )

d f

( x) 1

>d( x  a)@ Let us see how (1.167) gets formed. For that take N

f

( N 1)

( x)  f

1 , the LHS of (1.167) reads

x

³ f

( N 1)

(1)

a

(0)

 dn f ( x )

>dx @n

and by

( x ) { f ( x ) ):

(a)

(1.167)



( y ) dy equalling the RHS

of (1.67) i.e. f ( x )  f ( a ) . It is an integration of a once-differentiated function and returns function minus the value of function at the start-point. That we write it again as the following: 1

d f

(1)

( x) 1

>d( x  a)@ This also states, that operation d and d following:

1

x

³ f

(1)



( y ) dy f ( x )  f ( a )

(1.168)

0 . For N

0 we write the

a

are not inverse in operation, unless f ( a ) 1

d f ( x) 1

>d( x  a)@

x

³  f ( y ) dy a

f

( 1)

( x)  f

( 1)

(a)

(1.169)

36

Chapter One

where in (1.169) f

( 1)

(a) {

³  f ( x ) dx

. Integrating once more and using a fundamental theorem again we get

x a

the following expression:

d

2

f ( x)

>d( x  a)@ where in (1.170) f

( 1)

(a) {

³  f ( x ) dx

and f

x a

( 2)

( 2)

f

2

³ ³  f ( x ) dx

(a) {

( 2)

( x)  f

(a)  ( x  a) f

( 1)

d



d  N > f ( x )@

>d( x  a )@n >d( x  a )@ N



d  n N > f ( x )@

>d( x  a )@ n N

When N t 2 , using (1.163) i.e. the following for n

d



N



d  n > f ( x )@

>d( x  a )@ N >d( x  a )@ n



dn

>dx @n

dN f ( x)

> dx @ N



. The LHS of (1.164) is

d n N f ( x )

dN

>dx @n N

> dx @ N

 dn f ( x )

>dx @n

(1.170)

.

x a

The above identity (1.170) in a way is generalised for function differentiated N  folds. Next we integrate a second time making use of identity n

(a)

d 2 f ( N ) ( x )

>d( x  a )@2

, for n

(1.164)

i.e.

2.

, and using the fundamental theorem we have

2: d

2

f

(N )

( x)

f

( N  2)

f

( N  2)

2

>d( x  a)@

( x)  f

( N  2)

(2 1)

( x) 

¦

(a)   x  a f

( x  a)

(a) (1.171)

k

f

k!

k 0

( N 1)

( N 2 k )

(a)

From the above expression (1.171) the following generalisation follows:

d

n

f

(N )

( x)

f

n

>d( x  a)@ In the above expression (1.172) by setting N

( x) 

¦

( x  a)

k

( N n k )

(a)

(1.172)

§ ( x  a)k · (k ) ¸ f (a) 0¨ © k! ¹

(1.173)

k!

k 0

f

n , we find the following identity: d

n

f

(n)

( x)

f ( x) 

n

>d( x  a)@ n

n 1

( N n)

¦

n 1 k

(1)

n

(2)

( n 1)

That is, d and d are not inverse of each other unless all derivatives f ( x ) , f ( x ) …. f ( x ) are zero at x a , along with f ( a ) 0 . After setting N n and rearrangement of (1.173), the formula becomes: n 1

f ( x)

¦ k 0

This (1.174) is a Taylor expansion where the remainder is Rn

d n f ( n ) ( x )

>d( x  a )@ n

( x  a)

k

k!

f

(k )

( a )  Rn

(1.174)

, which gets expressed as n  fold integral of

n  fold derivative of f ( x ) . Now we use (1.172) i.e.

d n f ( N ) ( x )

f

n

>d( x  a )@

( N n)

( x) 

¦

n 1 ( x  a )k k!

k 0

f

( N n k )

( a ) and put N

0 and then write the

following: d

n

f ( x) n

>d( x  a )@

f

(n)

n 1

( x) 

¦ k 0

( x  a)

k

k!

f

( k n)

(a)

2 § (n) · ( x  a) (1 n ) (2  n ) ( )  (  ) ( )  (a)  ¸ f a x a f a f ¨ 2! (n) ¸ ( x)  ¨ f n 1 ¨ ¸ ( x  a) ( 1) ¨ .............. ( n  1)! f ( a ) ¸ © ¹

Differentiating once the above expression (1.175), we write the following:

(1.175)

Concept Generalisation

37

§ 0  f (1 n ) ( a )  ( x  a ) f (2 n ) ( a )  · § d  n f ( x ) · (1 n ) ¨ ¸ n2 f ( x)  ¨ ( x  a) n ¸ ( 1) ¨ ¸ ¨ ¸ d( x  a ) > @ © >d( x  a)@ ¹ ¨ ..............( n  1) ( n  1)! f ( a ) ¸ © ¹ (1 n ) (2  n ) §f (a)  ( x  a) f (a)  · ¨ ¸ (1 n ) n2 f ( x)  ¨ .............. ( x  a ) f ( 1) ( a ) ¸ ¨ ¸ ( n  2)! © ¹ d

f

(1 n )

n 1

( x) 

¦

( x  a)

f

( k  1)!

k 1

We used

k 1

( k n)

(1.176)

(a)

m! m

( m  1)! and then redefined the series in (1.176) in compact form. Again, if we repeat the process on the above-obtained relationship (1.176), i.e. once again differentiating (1.176), we will get the following:

§ d  n f ( x) · 2 ¨ n ¸ >d( x  a)@ ¨© >d( x  a)@ ¸¹ d

2

(2  n )

f

n 1

( x) 

¦ k 2

( x  a)

k 2

( k  2)!

f

( k n)

(a)

(1.177)

f

( k n)

(a)

(1.178)

Therefore repeating N times, we obtain the following:

d

N

>d( x  a)@ Let us see for the case N

N

§ d  n f ( x) · ¨ ¸ ¨ >d( x  a )@ n ¸ © ¹

f

( N n)

n 1

kN

( k  N )!

k N

§ d 1 f ( x ) · 1 ¸ 2 ¨ >d( x  a)@ ¨© >d( x  a)@ ¸¹ 2

(1)

f

0

( x) 

¦

( x  a)

k 2

f

2 and n

( x  a)

2 , and n 1 , then write from (1.178) the following: d

For N

¦

( x) 

(1)

k 2

f

( k  2)!

( k 1)

(a)

(1.179)

( x)

2:

§ d 2 f ( x ) · 2 ¨ 2 ¸ >d( x  a)@ ¨© >d( x  a)@ ¸¹ d

2

1

f ( x) 

¦

( x  a)

k 2

k 2

( k  2)!

f

( k  2)

(a)

(1.180)

f ( x) Note that the summation is zero for N t n . Therefore for N t n we write the following identity:

d

N

>d( x  a)@

N

§ d  n f ( x) · ¨ ¸ ¨ >d( x  a )@ n ¸ © ¹

As choice of the lower terminal does not affect (integer order) differentiation, we have

f

dN f ( x)

>d( x  a )@

( N n)

( x)

dN f ( x)

N

>d( x )@N

(1.181)

, with N as a

positive integer. We have earlier noted (1.173) that

d n f ( x )

f

n

>d( x  a )@

(n)

( x) 

¦

n 1 ( x  a )k k! k 0

f

( k n)

( a ) ; thus for N  n with this formula

we write the following:

d

N n

f ( x)

>d( x  a)@ Earlier we obtained in (1.178) d N n f ( x )

dN >d( x  a )@N

( N n)



¦

d n f ( x ) n

>d( x  a )@



n  N 1 ( x  a )k

f

( N n)

N n

( x) 

( k  N n)

f

( N n)

n  N 1

( x) 

¦

( x  a)

k 0

¦

n 1 ( x  a )k  N k N ( k  N )!

f

( k n)

k!

k

f

( k  N n)

(a)

(1.182)

( a ) , and just above in (1.182)

( x)  ( a ) . The summation expression for both (1.178) and f f k! k 0 >d( x  a )@N n (1.182) is the same so we have to re-adjust the summation index. That makes the RHS of both of these expressions the same as those which imply their LHSs are also the same. This discussion demonstrated the following rule:

we obtained

38

Chapter One

d

N

>d( x  a)@

N

N n § d  n f ( x) · d f ( x) ¨ ¸ n ¨ >d( x  a )@ ¸ >d( x  a )@N  n © ¹ n § d N f ( x ) · n 1 ( x  a ) k ( k  N  n ) d  ¦ f (a) n ¨ N ¸ >d( x  a)@ © d( x  a) ¹ k n N k !

(1.183)

Therefore, we summarise that identity as follows:

§ d M f ( x) m ¨ M >d( x  a)@ ¨© >d( x  a)@ d

m

· ¸ ¸ ¹

d

m M

f ( x)

(1.184)

m M

>d( x  a)@

When in (1.184) m and M are integers necessarily holds ‘unless’ M is positive and m is negative, or in other words ‘unless’ the function f ( x ) is first differentiated and then integrated. The expression:

§ d N f ( x) ¨ n N >d( x  a)@ ¨© >d( x  a)@ d

n

d

· ¸ ¸ ¹

N

>d( x  a)@

N

d

N n

>d( x  a)@

0 , and all derivatives of f ( x ) through ( N  1) -th, are also zero at x

example, that is f ( x )

2x

 1 with a

N n

(1.185)

§ d  n f ( x) · ¨ ¸ ¨ >d( x  a )@ n ¸ © ¹

holds only when f ( a )

e

f ( x)

a . We revert to our

0 and write the following:

§ d 3 ª¬e 2 x  1º¼ · d 2 ¬ªe 2 x  1¼º 1 2 x 1 2 1 1 ¨ ¸ e  x  x 3 2 4 2 2 4 >dx@ ¨© >dx@ ¸¹ > dx @ 3 § d ªe 2 x  1º · d ¼ ¸ 1 e2 x  1 x 2  1 x  1 ¨ ¬ 3 2 2 4 >dx@ ¨© dx ¸¹ 4 d

2

The difference between them is x . From our formula (1.183) we have for N with f ( x )

e

2x

d

¨ ¸ >dx@ ¨© >dx@3 ¸¹

¦

2

3 and a

0 the following,

1: 3 d § d f ( x) ·

Note that

1, n

(1.186)

( x )k k 2 k!

f

( k  2)

(0)

x2 2

f (0)

x2 2

2

3

k

§ df ( x ) · 2 ( x ) ( k  2) f (0) ¸¦ 3 ¨ > d x @ © dx ¹ k 2 k !

f ( x)

d

2

> dx @

(1.187)

2

x which is the difference between the two operations.

(2)

In subsequent chapters, we will be carrying on this generalisation to any arbitrary number n , N (not integers) say D , E i.e. as non-integer numbers.

1.20 The lower limit as an important factor in multiple integrals We will see how multiple integrals behave in a complicated way when their lower limit is changed. In an earlier section (1.175), we derived the following:

d

n

f ( x)

f

n

>d( x  a)@

(n)

n 1

( x) 

¦ k 0

( x  a)

k

f

k!

( k n)

(a)

(1.188)

Say the lower terminal of the n - fold integration gets changed to b , with a  b  x then using the above formula (1.188) we write the difference:

d

n

f ( x) n

>d( x  a)@



d

n

f ( x) n

>d( x  b)@

n 1

1

¦ k !  ( x  b) k 0

(1.189) k

f

( k n)

k

(b )  ( x  a ) f

( k n)

(a)



Concept Generalisation

39

For n 1 : 1

1

d f ( x) 1

>d( x  a)@



0

d f ( x) 1

>d( x  b)@

1

¦ k !  ( x  b)

k

f

( k 1)

k

(b )  ( x  a ) f

( k 1)

(a)

k 0

f

( 1)

( x)

x b

 f

( 1)

( x)

x a

x

b

a

b

a

(b )  f

³  f ( y ) dy  ³  f ( y ) d y ³  f ( x ) d x ,

This is an obvious case that is f

x

( 1)

( 1)



(1.190)

(a)

obtained in the same way as

. The obtained difference expression (1.189), for n other than one; depends on both the

lower terminals of multiple integration, that is a and b ; and on the variable from the obtained expression above (1.189):

d

n

f ( x)

d



n

>d( x  a)@

n

n 1

f ( x)



n

>d( x  b)@ n 1



¦

( x  a)

k 0 n 1



f

( k n)

k

f

k!

k 0

k

k!

¦

( x  b)

x . We do the following manipulation

( k n)

(b )

(a)

1

¦ k !  ( x  b )  (b  a )

k

f

( k n)

(a)

(1.191)

k 0 n 1

k

· ( k n) (a) ¸f k 0 j 0 © ¹ k j j k n 1 § ( x  b ) (b  a ) · ( k n) ¦ f (a) ¨ ¦ ¸ j !( k  j )! k 0 ©j 0 ¹ 

1 §

¦ k!¨¦

k! j !( k  j )! ( x

j

 b ) (b  a )

k j

In the above derivation (1.191) we have used a binomial expansion that is ( x  y ) n

¦

that

¦

n 1 k 0

n 1 ( x b ) K K! K 0

f

( k n)

¦

(a)

¦

k

n 1 ( b  a ) J  K J K ( J  K )!

j 0

f

( x b ) j ( b  a )k  j j !( k  j )!

( J n)

( a ) or

¦

n

n! r 0 r !( n  r )!

r

x y

nr

. The sum

appears above (1.191) may be expressed in various ways, like

¦

n 1 ( x b ) K K! K 0

¦

n  K 1 ( b  a ) j j! j 0

f

( j  K n)

( a ) . We identify the index K with

the k of (1.191) and write the following expression:

d

n

f ( x) n

>d( x  a)@



d

n

f ( x) n

>d( x  b)@

j n  K 1 § ( K n) · (b  a ) ( K  j n)  f ( b ) f (a) ¸ ¨ ¦ ¦ K! © j! K 0 j 0 ¹ nk j n k 1 § (k ) · ( x  b) (b  a ) ( j k )  f ( b ) f (a) ¸ ¨ ¦ ¦ j! k 1 ( n  k )! © j 0 ¹ n 1

using

d n f ( x ) n

>d( x  a )@

becomes f

(k )

f

(n)

(b ) 

¦

( x) 

¦

k 1 ( b  a ) j j! j 0

n 1 ( x  a )k k! k 0

f

( j k )

f

( x  b)

( k n)

K

( a ) i.e. (1.173) for the bracketed expression in (1.192) so that it

d k f (b )

(a)

(1.192)

>d( b  a )@ k

. Substituting this in the above derivation (1.192), we get the

following expression:

d

n

f ( x) n

>d( x  a)@



d

n

f ( x) n

>d( x  b)@

n

¦ k 1

( x  b)

nk

( n  k )!

d

k

f (b ) k

>d(b  a)@

(1.193)

Therefore, we have the expression above (1.193) showing explicit dependence on b  a , shifting the lower terminal of multiple integration from a to b . We will extend this and generalise when n is a non-integer number in later chapters.

40

Chapter One

1.21 Generalising product rule for multiple integration and differentiations dn dx n

Leibniz’s rule for

> f ( x) g ( x)@ is obtained as follows. Writing

f and g ( x )

f ( x)

d dx

> f ( x ) g ( x )@

f

g , we know the formula:

(1)

gg

(1)

(1.194)

f

Applying this again, we get: d

2

> @

fg f 2 dx Repeating this process we get the n -th order generalisation represented below:

d

n

(2)

n

§ n · ( nk ) ( k ) g ¦ ¨k ¸f k 0© ¹

> fg @

g2f

(1)

g

(1)

 fg

(2)

(1.195)

nk k §n· d f d g ¦ ¨ k ¸ nk k dx k 0 © ¹ dx n

(1.196) n dx Now we will see Leibniz’s formula for  n which is the product rule for multiple integrations. Starting with a wellx

x

³  g ( y )  dv( y )  g ( x)  v( x)   g (a)  v(a)  ³  v( y )  dg ( y ) we Let v ( y ) ³  f ( z ) dz , then we write the integration by parts of the formula as

known formula for integration by parts i.e.

a

a

y

will derive some interesting results.

a

follows:

x

³  g ( y)  f ( y) dy

g ( x)

a

1

x

x

y

a

a

a

f and g ( x )

With our new symbolism (Section 1.17) and writing f ( x ) (1.197) as:

(1.197)

g , we shall re-write the above expression

1 § d 1 f · ª (1) d 1 f º d g¨ ¸ «g » ¨ >d( x  a )@1 ¸ >d( x  a )@1 « >d( x  a )@1 » © ¹ ¬ ¼

> fg @ 1 >d( x  a)@ d

§ dg ( y ) · ³  f ( z ) dz  ³  ³  f ( z ) dz ¨© dy ¸¹ dy

(1.198)

We apply this again to the product in the brackets of the second term in the above expression (1.198) and write the following: 1

§

· (1) d 2 f ¸ g 2 ¨ >d( x  a )@1 ¸ >d( x  a)@ © ¹

> fg @ 1 >d( x  a)@ d

1

d f

g¨

(1.199)

ª (2) d 2 f º  g 1 « 2 » >d( x  a)@ «¬ >d( x  a)@ »¼ d

1

Doing this process an infinite number of times, we get the following expression: 1

> fg @ f (1) j g ( j ) d 1 j f ¦ 1 1 j >d( x  a)@ j 0 >d( x  a)@ d

The

alternate

change

j

0,1, 2, 3... .

( 1)! j !( 1 j )! ;

1

Cj

(1  x )

For

sign

1

¦

( 1)! j 1 j (1) ( x ) j 0 j !( 1 j )!

( 1)! (1)!( 2)!

taking ( 1)( 2)! (1)( 2)!

j

j

( 1)

0,

1 x

we

j

0,1, 2, 3...

is

like

for

This

f

verification,

( 1)! j !( 1 j )!

by

1

x

2

can an

be

cast

infinite

by

a

binomial

series

(1.200) expression,

expansion

i.e.

 ... .

obtain

1 ; after that we take j

( 1)! j !( 1 j )!

( 1)! 0!( 1)!

2 and we get

1,

then

( 1)! j !( 1 j )!

taking

( 1)! 2!( 3)!

j 1,

( 1)( 2)( 3)! 2( 3)!

we

obtain

1 and so on.

With this observation, we can write the following formula: 1

> fg @ f § 1·g ( j ) d 1 j f ¦¨ ¸ 1 1 j >d( x  a)@ j 0 © j ¹ >d( x  a)@ d

(1.201)

We perform integration again to the above formula (1.201) with the identity we noted earlier, i.e. those from (1.163) and (1.164) which are written again below:

Concept Generalisation

§ d N f · d n N f n ¨ N ¸ n N >dx@ ¨© >dx@ ¸¹ >dx@ n § dN f d n ¨ N >d( x  a)@ ¨© >d( x  a)@ d

n

41

d

> dx @ · ¸ ¸ ¹

§ dn f · ¨ ¸ ¨ >dx@n ¸ © ¹

N N

d

 n N

f

(1.202)

 n N

>d( x  a )@ d

N N

>d( x  a )@

§ dn f · ¨ ¸ ¨ >d( x  a )@ n ¸ © ¹

Equation (1.202) holds true for non-negative integers N and n . That is demonstrated below: 1 1 j ª f § 1· ( j ) º > fg @ f d d g ¸ 2 1 « ¦ ¨ 1 j » >d( x  a)@ >d( x  a)@ «¬ j 0 © j ¹ >d( x  a)@ »¼

d

2

1 1 j ª ( j) º· f d d § 1· ¨§ ¸ g « » ¦ ¨ j ¸¨ 1 1 j ¸ ¹ © >d( x  a )@ ¬«  d( ) x a j 0© > @ ¼» ¹ f

1 j ª ( j) º d f g « 1 1 j » >d( x  a)@ ¬« >d( x  a)@ ¼»

d

1

f

1

¦ §¨© k ·¸¹g

d

( j k )

 j k 2

(1.203)

f

 j k 2

>d( x  a)@

k 0

We now proceed with the following steps: 2

> fg @ f § 1· f § 1·g ( j  k ) d  j k 2 f ¦ ¨ ¸¦ ¨ ¸ 2  j k 2 >d( x  a)@ j 0 © j ¹ k 0 © k ¹ >d( x  a)@ d

f

f

2  m

f

m

2  m

f § 1·§ 1 · ( m ) d ¦ ¦ ¨ j ¸¨ m  j ¸ g 2  m ¹© ¹ j 0 m j© >d( x  a)@

(1.204)

f § 1·§ 1 · ( m ) d ¦¦ ¨ j ¸¨ m  j ¸ g 2  m ¹© ¹ m 0 j 0© >d( x  a)@ 2  m

f

f § 2 · ( m ) d ¦ ¨ m ¸g 2  m ¹ m 0© >d( x  a)@ In the above derivation (1.204) we have used a permutation that is

f

k

k 0

j 0

¦ ¦

permutation is analogous to the permutation of variables in double integration that is (1.204) we used the following formula of binomial coefficients: j

³

f

0

f

f

j 0

k j

¦ ¦ dy

³

y

0

dx

³ ³

§ n ·§ N · § n  N · ¨ j ¸ © ¹ 0

¦ ¨© k ¸¨ ¸ ¹© j  k ¹ k

f

.

0

dx

f

x

This

dy . In

(1.205)

The iteration of the above procedure (1.204) further leads to the following formula: n

> fg @ f § n ·g ( j ) d  n j f ¦¨ ¸ n  n j >d( x  a)@ j 0 © j ¹ >d( x  a)@ d

n

1, 2, 3,...

(1.206)

A similarity between product rule for multiple differentiations and the expression for product rule for multiple integrations can be seen. Writing both of them below (for n ! 0 ) gives us: d

n

> fg @

dx

n

n

n j j §n· d f d g ¦ ¨ j ¸ n j j dx j 0 © ¹ dx n

> fg @ f § n · d  n j f ¦ ¨ ¸  n j n >d( x  a )@ j 0 © j ¹ dx d

(1.207)

j

d g dx

j

We will use this developed formula for further generalisation when n is a non-integer in later chapters.

42

Chapter One

1.22 Chain rule for multiple derivatives We write what we call chain rule as: d

d

g  f ( x)

d

(1)

(1)

(1.208) dx du dx This enables us to differentiate g (u ) with respect of x , where u f ( x ) , if the derivative of g (u ) with respect to u and the derivative of u f ( x ) are known. By application of this chain rule, and Leibniz’s rule for differentiating a product of two functions we use the following steps: d

2

dx

2

g  f ( x)

d ªd

dx «¬ dx

g  f ( x)

g (u )

f ( x)

g

f

º d ª d d º »¼ dx ¬ du g (u ) dx f ( x ) ¼







§ d d > g (u )@ ·§ d > f ( x )@ · § d > g (u )@ · § d 2 > f ( x )@ · ¸ ¨ ¸¨ ¸¨ ¸¨ 2 © dx du ¹© dx ¹ © du ¹ © dx ¹ § du d 2 > g (u )@ · § d > f ( x )@ · § d > g (u )@ · § d 2 > f ( x )@ · ¨ ¸¨ ¸ ¸¨ ¸¨ 2 2 © dx d x ¹ © dx ¹ © d u ¹ © d x ¹ 2 2 § d > f ( x ) @ d > g (u ) @ · § d > f ( x ) @ · § d > g (u ) @ · § d > f ( x ) @ · ¸ ¨ ¸¨ ¸¨ ¸¨ 2 2 dx ¹ © dx ¹ © dx ¹ © du ¹ © dx

(1.209)

2

§ d > f ( x ) @ · § d 2 > g (u ) @ · § d > g (u ) @ · § d 2 > f ( x ) @ · ¸¨ ¸ ¨ ¸ ¨ ¸¨ 2 2 © dx ¹ © d x ¹ © du ¹ © dx ¹ g

(1)

f

(2)

g

(2)

(1) 2

f

We repeat it and then write the following expressions:

d

3

ª¬ g  f ( x ) º¼ dx

d

4

3

ª¬ g  f ( x ) º¼ dx

4

g

(1)

f

(3)

 3g

g

(1)

f

(4)

 4g

 6g d

5

ª¬ g  f ( x ) º¼ dx

5

 30 g

g

(1)

(3)

f

(3)

(5)

f

(1)

(2)

f

(1)

f

(2)

g

(2)

f

(1)

f

(3)

 6g

f

(2)

(1) 2

f

 5g

(2)

(2) 2

f

f

(1)

f

 10 g

g

(4)

(1) 3

f

(3)

(4)

(1) 4

f

 10 g

(1) 3

f

f

The continuation of the above steps (1.210) to n would give the generalised formula

(1.210) (1) 2

(3)

f

(4)

(2) 2

f

(2)

(2)

dn dx n

g

f

(5)

(3)

(1) 5

f

ª¬ g  f ( x ) º¼ . However, it is

tough and requires a special formula, such as Faa’ di Bruno’s formula (we are not proving this). Faa’ di Bruno’s formula is too complicated to be of any utility for large n . We write that, however, as the following:

d

n

dx

n

n

g  f ( x)

n!

¦g

(m)

m 1

§ f (k ) · ¦– ¨ ¸ k 1 Pk ! © k ! ¹ n

1

Pk

(1.211)

Where in (1.211) the summation ¦ extends over all combinations of non-negative integer values of P1 , P2 , ..., Pn such that

¦

When g

n k 1

n and

kPk

¦

n

P k 1 k

m . However, this complicated formula reveals its utility in some specific cases.

2

u , then we have:

d

n

dx When f ( x )

n

g  f ( x)

d

n

dx

f ( x) n 

2

2f

(n)

n 1

f 

§n·

¦ ¨© k ¸¹ f

(k )

f

( nk )

(1.212)

k 1

x

e , then we have:

d

n

dx > m@ where S n is a Stirling number of a second kind. Again if g

n

g  f ( x) 2

d

n

dx

n



g e

u , then we find:

x

e

nx

n

¦ S> m 1

m@ ( m ) n g

(1.213)

Concept Generalisation

d

n

dx

n

2



> f ( x )@

2 f

(n)

43

( x)

 f ( x) (1.214)

n 1

§ n · (k ) ( nk )  ¦ ¨ ¸  f ( x )  f ( x) k k 1© ¹ a result that may also be obtained from Leibniz’s theorem for multiple integrals.

1.23 Differentiation and integration of power functions: unification & generalisation In Section 1.18, we have dealt with differentiation and the integration of infinite series of functions. We will be p

encountering in the rest of the chapters, the series whose general term is a simple power function i.e. of type ( x  a ) , where p is considered as an integer. We collect here elementary formulae to express

dn >d( x  a )@n

ª¬ ( x  a ) p º¼ for positive

and negative values of n . For differentiation we have the following:

n

0 p d ª¬( x  a ) º¼

0

dx

0

p

1

n

n

p d ª¬( x  a ) º¼

( x  a)

1

1

dx

p 2 d ª¬( x  a ) º¼

2

dx

2

p( x  a)

p 1

p ( p  1)( x  a )

(1.215) p 2

From the above steps of (1.215) we write the following: d

n

ª¬( x  a ) p º¼ dx

n

p ( p  1).......( p  n  1)( x  a )

(1.216)

p! ( p  n )!

We note in (1.216) the term

p! ( p  n )!

p n

( x  a)

p ( p 1)( p  2)....( p  n 1)( p  n )( p  n 1)....(2)(1) ( p  n )( p  n 1)....(2)(1)

p n

n

0,1, 2, ...

p ( p  1)....( p  n  1) . For the integration

we have the following steps: x

n

1

1

p

³ (x

 a ) dx0

x

x1

0

( p  1)

a

n

³ dx ³ ( x

2

1

a

0

p

 a ) dx0

a

( x  a)

p 1

(1.217) 1

( p  1)( p  2)

( x  a)

p2

From the above steps of (1.217), we write the following: d

n

xn1 x1 ª¬( x  a ) p º¼ x p { ³ dxn 1 ³ dxn  2 ........³ ( x0  a ) dx0 n

>d( x  a)@

a

a

a

pn

­ ( x  a) p! pn ( x  a) p ! 1 ° ® ( p  1)( p  2)...( p  n) ( p  n)! °¯ p d 1 f

(1.218)

n 1, 2, 3,...

We note in the above (1.218) that we have

p! ( p  n )!

p ( p 1)( p  2)......(2)(1) ( p  n )( p  n 1)......( p  2)( p 1)( p )( p 1).....(2)(1)

1 ( p  n )( p  n 1)...( p  2)( p 1)

.

From the above formulas (1.216) and (1.218) for differentiation and integration we write the following table, where positive n is differentiation and negative n is integration, and instead of the values we indicate positive and negative signs for the result. Thus, Table 1.4 unifies the above two formulas, for integer p and integer n .

44

Chapter One

3

2

1

0

1

2

3

  

  



0

0 0

0 0 0

1 2

f f

f f

f f

    

 

0

  

 

 

 

pp

no 2 1

n

Table 1.4: The sign of d n ª¬( x  a ) p º¼ / >d( x  a )@ for the positive and negative integers n and p By invoking the properties of the gamma function for the factorials both formulas (1.217) and (1.218) can be generalised for any p by writing

d

n

p! ( p  n )!

as

ª¬( x  a ) p º¼ n

>d( x  a)@ The coefficient

* ( p 1) * ( p  n 1)

* ( p 1) * ( p  n 1)

for positive and negative n .

­ * ( p  1) pn ( x  a) ° ® * ( p  n  1) °¯ f

^

n

0,1, 2,.....; For all p

n

1, 2,....;

p ! 1

n

1, 2,....;

p d 1

(1.219)

may be positive, negative or zero according to the values of p and n . Figure 1.5 displays

these values graphically. We will study the above generalisation in subsequent chapters, for n as a non-integer.

p

Figure 1.5: The sign of d

n

 

 

 

 

 

 

 

3

  

  

  

  

  

  

  

2

 

 

 

 

 

 

0 

  

  

  

  

  





1

0 

0 

  

  

  

  







0

0 

0 

0 

1

 f f

 f f

 f f

  

  

  

  

2

f f f

f f f

f f f

  

  

  

  

3

2

1

0

1

2

3

n

ª¬( x  a ) p º¼ / >d( x  a )@

n

for positive and negative integer values of n

1.24 Non-differentiable functions: the motivation for fractional calculus 1.24.1 Hurst exponent The main aim or at least the motive in investigating fractional differential manifolds is the fact that fractional calculus seems quite relevant to investigate some of the problems, which occur in fractal space-time. The first elementary point of view is the mathematics framework of fractal physics that deals with differentials satisfying the condition:

d  f ( x ) v  dx

H

(1.220)

for the function f ( x ) , where H ! 0 (need not be an integer). This H is referred to as the ‘Hurst exponent’. The above differential condition means that the classical equality d  f ( x )

f

(1)



( x ) dx

no longer holds; in other

words, the function f ( x ) is not differentiable, and that instead we should be describing it in the following form:

d  f ( x)

 g ( x)  dx

H

(1.221)

Concept Generalisation

45

This expression (1.221) leads to fractional calculus. Observe that we have here differential quantity as fractional differential quantity which leads to a derivative and integration with respect to fractional differential elements, and that D

we call it (dx ) , where D is a non-integer, and generally 0  D  1 .

1.24.2 Self-similarity Qualitatively speaking a self-similar function is a function which exhibits similar patterns when one changes the scale of observation. For example, the pattern generated by f ( x ) and f (O x ) with O ! 0 looks the same. Formally, f ( x ) with x  \ is of a similar order to H if one has equality: H

(1.222) f (O x ) O f ( x ) O!0 H !0 which means that the landscape in the vicinity of x , and that of O x look alike. Such a function satisfies the condition, H O f (1) and then by

i.e. f (0) 0 . It also provides the following relation via substitution of x m 1 making f (O ) substituting O m x giving the following expression: H

f ( x)

(1.223)

x f (1)

1.24.3 Nowhere differentiable functions As a special case when 0  H  1 the function f ( x ) is non-differentiable but is fractionally differentiable for the fractional order of H . Like f ( x )

1

x 2;

x t 0 is non-differentiable at x

0 , but is fractionally differentiable by

order 1 2 , as we will learn in subsequent chapters. Then how can we form a Taylor series for a non-differentiable function? This is only possible if we have non-integer (fractional) derivatives; this we will learn subsequently too. 1

We shall soon be dealing with nowhere differentiable functions. The function f ( x ) x 2 is not differentiable at x 0 , but surely has derivatives at points other than zero. However, what is a function that is nowhere differentiable?.

> @

Let us construct one such case. Let us take interval 0,1 on which we will define the nowhere differentiable function. In this interval we have a set of discrete points x1 , x2 ,.... x N such that x1 We construct a function ffract ( x )

¦

N k 1

( x  xk )

1

2

for x t xk and

0 and x N ffract ( x )

1 with x1  x2  x3  ...  x N .

0 for x  xk , with k

We observe that at points xk the function value is 0 but is not differentiable, for any k

1, 2, 3..., N .

1, 2, 3..., N . Thus, the

function ffract ( x ) is continuous at all points xk but not differentiable at any point xk . Therefore, we will write for

> @

infinite points N and the function in the interval 0,1 as: N 1 ­°lim ( x  xk ) 2 N nf ¦ k 1 ffract ( x ) ® 0 °¯

; x t xk

(1.224)

; x  xk

It is continuous everywhere but nowhere differentiable. These types of functions are called ‘fractal functions’.

> @

The arrangement of points in the interval is via some ‘fractal’ arrangement, say as a Cantor set. A line segment 0,1

> @ > 13 , 32 @ and > 23 ,1@ . Delete the middle part i.e. the segment > 13 , 23 @ from the interval >0,1@ leaving the two parts, i.e. >0, 13 @ and > 23 ,1@ . In the next step the open middle third of each segment is deleted leaving the segments: >0, 19 @ , > 92 , 13 @ , > 23 , 79 @ , > 89 ,1@ . This process is continued ad infinitum. Thus, the Cantor set contains all points in the interval >0,1@ that are not deleted at any steps in this infinite process. This idea gives a is cut into three parts. Those are 0, 13 ,

fractal function continuous but nowhere differentiable. In classical calculus, everything happens when an elemental point (the differential element, say 'x ) is made infinitesimally small i.e. made zero. Thereby we take the rate with respect to this element i.e. 'x , and classically we term this rate as a usual derivative i.e. d  f ( x ) / dx in the limit as 'x p 0 ; that is normal differentiation. In a coarsegrained system (or phenomena), everything happens as if this elemental point or the differential element 'x has some D

spread (or thickness), and is non-zero. This can be viewed with fractional differential quantity  'x

with 0  D  1 ;

46

so

Chapter One D

 'x

! 'x as we take the limit 'x p 0 .

This gives rate as per unit of this fractional differential i.e.

D

d  f ( x ) /(dx ) ; suggesting the need of a fractional derivative. These are some motivations to learn from fractional calculus that we will discuss and develop in subsequent chapters.

1.25 Short summary Here in this chapter we have learnt about several generalisation examples. We have come across various functions that generalise our classical concepts like factorial, binomial formulas etc. We have also seen the generalisation of some concepts of classical calculus in terms of multiple whole numbers of differentiation and integration and derived some important formulas and then inferences. We will use all these developed concepts to further generalise classical calculus when the whole number of integration and differentiation that is n or N is an arbitrary non-integer value. The methods of residue calculus, contour integration, analytical continuation etc that we revised in this chapter will be of help in understanding further concepts in subsequent chapters. In subsequent chapters, we shall be trying to answer the question posed by L’Hospital to Leibniz, and discuss whether or not it is a paradox.

1.26 References This chapter is based on pioneering literature. Those which provided motivation are detailed in the bibliography section below. A. D. Polyanin, A. V. Manzhirov, “Handbook of Integral Equations” (2008) Abramowitz. M., Stegun. I.A. (eds.), “Handbook of Mathematical Functions” (1964) Bartlett M. S., “Negative Probability” (1945) Batman H. “Higher transcendental function” (1954) Churchill Ruel V., Brown James W and Verhey Roger F., “Complex Variables & Applications” (1974) Churchill. R. V., “Operational Mathematics” (1948) Davis, Harold T., “The Theory of Linear Operators” (1936) Erdelyi A., “Asymptotic Expansions” (1954); “Tables of Integral Transforms” (1954); “On some functional transformations” (1950) Feynman R. P., “Negative Probability in Quantum Implications: Essays in honor of David Bohm” (1987) Gel’fand. I. M., Shilov G. E., “Generalised functions” (1964) Gorenflo R, Yu. Luchko, Rogosin S., “Mittag-Leffler Type Functions: Notes on Growth Properties and Distribution of Zeros” (1997) Gradshteyn I. S., Ryzhik I. M., “Tables of Integral Series & Products” (1992) J. A. Tenreiro Machado, “Fractional Coins and Fractional Derivatives” (2013); “Entropy Analysis of Systems Exhibiting Negative Probabilities” (2016) LePage W. R., “Complex variables and the Laplace transform for Engineers” (1961) Oldham and Spanier, “Fractional calculus” (1974) Szekely G. J., “Half of a coin: Negative Probabilities” (2005) Scafetta N., Griffin. L, West B. J., “Holder Exponent Spectrum for Human Gait” (2003). Jumarie G., “From self similarity to fractional derivative of non-differentiable functions via Mittag-Leffler function” (2008) Michelle Caputo, “The Riemann Sheet Solutions of Anelasticity” (1994)

CHAPTER TWO FRACTIONAL INTEGRATION

2.1 Introduction In this chapter, we will discuss the process and formulation of fractional integration, which is a concept generalisation of what we know in classical calculus as repeated integration or n -fold integration with n as a positive integer 1, 2, 3 … and so on. We will learn how to formulate the fractional integration when n is a non-integer say

1 2

,

3 2

etc. We

shall start with various notations that are used to represent fractional integration, followed by a review of classical iterative integration, or n -fold repeated integration. Thereafter we will use generalisations of factorials by use of the gamma function to get a fractional integration formula; and apply the obtained formula to get expressions of fractionally integrating various useful and simple functions. We will introduce a few types of higher transcendental functions. In this chapter, we will discuss the notion that if we change the sign of fractional integration processes we get some indication of the formulation of fractional differentiation. In doing so we will discover that there is an indication that the fractional derivative process is similar to the integration process. Here we will give the RiemannLiouville formulation of fractional integration, and discuss Weyle’s definition of fractional integration, with an application to some simple problems. We will also try to give a geometric interpretation to fractional integration as an area under a curve whose shape changes. In this chapter while discussing the various methods, we will also learn many more aspects about fractional calculus. Various notations have been used for the last three hundred years for representing fractional integration operators. Leibniz, Lagrange, and Liouville used

³

D

, and Grunwald wrote

³

D

x x

ª¬...dxD º¼ x

a

, where function is integrated by the D

fractional order D ! 0 and D   , from lower terminal a ! f . Riemann used the symbol w x fractional integration. Several authors use the symbols

D

d dx  D

D

D

D

, fD , I , I x , I a  ,

D

a Ix

,

D

d d( x  a ) D

,

D a Dx

to represent and

D x Db

.

We will be using some of these notations.

2.2 Iterative integration: a review What we mean by iterative integration is n  fold (or many times repeated) integration of a function. Let us symbolise n

n

the n  fold integration as 0 I x , with n as a positive integer or as an anti-derivative 0 Dx , meaning that this operator will do n  fold integration from the lower limit zero to the upper limit x . That we symbolise as the following: n

>

0 I x f ( x)

x

x1

xn1

@ ³0 ³0 .....³0 f ( x0 )dx0dx1...dxn2dxn1   

(2.1)

n

Here in (2.1) we have n t 0 ; therefore, for n

2 , for a function f ( x ) we have the following operation:

§ x1 · f ( x ) d x  ¨ ³0 ¨ ³0 0 0 ¸¸dx1 © ¹ x

2 0 Ix

> f ( x )@

2 0 Dx

> f ( x )@

2.2.1 Double integration Take f ( x )

x , and apply a double integration operation of (2.2) as follows:

(2.2)

48

Chapter Two

§ x1 ª xº ³¨ ³ ¬ ¼ ¨ 0© 0 x

2 0 Ix

2 0 Dx

ª xº ¬ ¼ x

³ ª¬ 23 x02 º¼ 0 dx1 For f ( x )

x

5

1

¹

5

 23 ª¬ 52 x1

0

2

x

º ¼0

(2.3)

2

1 , we explain the above procedure with the aid of Figure 2.1. The two variables x0 and x1 are shown on

³ dx

the X-axis and Y-axis respectively. The first integration

x1

0

2 32 3 x1 dx1

³

0

4 15

0

x

x1

3

·

 x dx ¸¸dx

gives a straight line passing through the origin

0

x0 . The second integration is with respect to x1 for this straight line; where dx1 is segment DG , placed at

position D with AD having the length x1 . The area of the rectangle DEFG is dA , which is almost x1dx1 , that is also

³ x1

0

that is

³

x

0

dx0 dx1

³

dA . The area of AED is

dA which is

x2 2

x1

0

x12 2

dA

. The total area of the triangle ABC , is

³ ³ x

0

x1

0

dx0 dx1 ,

.

2.2.2 Double integration with a changed order of integration giving the idea of convolution Now we interchange the order of integration and derive the following: x

2

0 Ix

2

ª xº ¬ ¼

0 Dx

§x · d x x ¨ ³0 ¨ x³ 0 1 ¸¸dx0 © 0 ¹

ª xº ¬ ¼

x

x § · d x x ¨ ³0 ¨ 0 x³ 1 ¸¸dx0 0 © ¹ x

x0  x  x0 dx0

³

(2.4)

0 x

x

³x

³

x0 dx0  x0 x0 dx0

0

0

x

3

5

x ª 23 x0 2 º  ª 52 x0 2 º

¬

4 15

x

¼0 ¬

5

x

¼0

2

We note that with this change of order of integration we are getting the same result as obtained earlier (2.3) i.e.

4 15

5

x 2.

x

Look at the third step of the above derivation (2.4) i.e. us to write the following general step, with f ( x0 )

2 ª xº 0 Ix

¬

¼

2 ª xº 0 Dx

¬

¼

³

x0  x  x0 dx0 , which allows

0

x0 x

2

0 Dx

> f ( x)@ ³  f ( x0 )  ( x  x0 ) dx0 ;

x0

y

0

x

(2.5)

³  f ( y )  ( x  y ) dy 0

Here we mention that the integral

x

³  f ( y )  g ( x  y ) dy  f ( x) *  g ( x) is a convolution integral with g ( x) 0

x.

Therefore, we say that the double integration operation is a convolution operation i.e. given as 2 0 Ix

> f ( x )@  f ( x ) *  x .

We note that this convolution operation became visible while doing the double

integration by interchanging the order of integration, in steps (2.4).

Fractional Integration

49

2.2.3 Generalising the double integration to an n  fold integration We may guess here that an n  fold integration in convolution form will look like the following: 1 n  f ( x) * x n1 0 I x f ( x) ( n  1)!

>



@

 f ( x ) * x @  0! f ( x)  f ( x ) * x 2  f ( x) *  x , that 0 I x > f ( x )@ (1)!

1



(2.6)

0

>

For n 1 we have 0 I x f ( x ) For n

2 we have

suggests that we are likely to have the following formula: 1 n n f ( x) 0 I x f ( x) 0 Dx ( n  1)!

>

@

>

we just obtained earlier in (2.5). This discussion

 f ( x) *  x n 1

@

1

(2.7)

x



 f ( y) ( x  y) ( n  1)! 0 for generalised n  fold integration. We will further discuss this convolution aspect: x

x1

§ ³0 ¨¨ ©

x1

³ dx

0

0

dy

· dx1 ¸ ¸ ¹

C

x

³

n 1

B Area DEFG dA

x1dx1 x1

F

G dx1

D

§ · ¨ ³ dx0 ¸ ¸ dx1 ¨ 0 © ¹

E

x1 A 0

x12 2

Area ABC

x2 2

x0

x1

Figure 2.1: Geometrically explaining the double integration

For f ( x )

Area AED

³ ³ x

0



x1

dx0 dx1

0

1 , we explain the above procedure with the aid of Figure 2.2. The two variables x0 and x1 are shown on

the X-axis and Y-axis respectively. The first integration

³ dx

1

gives a straight line passing through the origin x0

x1 .

The second integration is with respect to x0 for this straight line; where dx0 is segment UV , placed at position V with RV having the length x0 . The area of a rectangle STUV is dA , which is almost SV u UV , that equals

dA

( x  x0 )dx0 , which is also

giving xx0 

³ x

x0

dx1 dx0

dA . The area of PSVR is half of the product of RV with ( PR  SV ) ,

 . The total area of the triangle PQR , is ³ dA ³  ³ dx dx x02 2

x

0

x

0

x

x0

1

0

, that is

³

x

0

dA which is

x2 2

.

50

Chapter Two

x

§

x

³ ¨¨ ³ dx

1

0

x1

x

©

x0

dx0

· ¸dx0 ¸ ¹

Q

R

V

U

Area STUV dA

( x  x0 )

x0

T

S

W

( x  x0 )dx0

§ x · ¨ ³ dx1 ¸ dx0 ¨x ¸ © 0 ¹ Area PSVR

xx0  x2 2

AreaPQR

P

0

x02 2

x0

x0

Figure 2.2: Geometrically explaining the double integration

³ ³ x

x1

0

x0



dx1 dx0 by interchanging the order of integration and giving

the idea of convolution

2.3 Repeated integration formula by induction Let us write: x

³ f ( x )dx 0

F ( x )  F (0)

0

(2.8)

0

For F (0)

0 we then have F ( x )

we have the following:

³

x

0

1 0 Dx

f ( x0 )dx0

> f ( x )@ .

Further if F (0) x

n

>

0 I x f ( x)

@

n

0 Dx

>

f ( x)

@ ³

F ( F (0))...

f ( x0 )( x  x0 )

0

( n  1)!

0 , then for n t 0

n 1

dx0

(2.9)

If the above expression (2.9) is true for n , then replacing n with ( n  1) gives us the following steps: ( n 1) 0 Ix

> f ( x )@

ª x f ( x0 )( x  x0 ) n 1 º dx0 » > f ( x )@ «³ ( n  1)! ¬0 ¼ x x § f ( x0 )( x1  x0 ) n 1 · ³0 ¨© ³0 (n  1)! dx0 ¸¹dx1

 ( n 1) 0 Dx

1 0 Dx

x

§ x f ( x )( x  x ) n 1 · ³0 ¨¨ x³ 0 (n 1 1)!0 dx1 ¸¸dx0 © 0 ¹ x§ x · f ( x0 ) n 1 ¨ ³0 ¨ (n  1)! x³ ( x1  x0 ) dx1 ¸¸dx0 0 © ¹ x

(2.10)

x

ª ( x1  x0 ) n º ³0 (n  1)! «¬ n »¼ dx0 x0 x

³ 0

f ( x0 )

f ( x0 )( x  x0 ) n!

n

dx0

which is also true for n  1 . In the above derivation (2.10), i.e. demonstration with induction method, we used the concept of obtaining double integration by changing the order of integration, as we described in Figure 2.2, that is:

Fractional Integration

n 0 Ix

1

> f ( x )@

( n  1)! 1

51

 f ( x) *  x n1 (2.11)

x



n 1



 f ( y ) ( x  y ) dy ( n  1)! 0 Now, we formally write for the iterated integrals, for a locally integrable real valued function f :  o  , when

³

> @

domain definition is  { a , b Ž  , that is an interval with f d a  b d f : xn 1

x x1 n

>

a I x f ( x)

@

³ ³ .....

³

a a

f ( xn )dxn dxn 1 ...dx2 dx1

a

(2.12)

x

1

³  ( x  y )  f ( y ) dy n 1

( n  1)! a

where a  x  b , and n is a natural number, (2.12) is also called Cauchy’s formula for repeated iterated integration. Thus, we have n fold integration reducing to a single ‘convolution’ integral. Note that if we take g ( x ) the integral above in (2.12) i.e.

n 1

³ ( x  y )  f ( y ) dy , is a convolution integral as

Generally, the limit of integration of a convolution integral is from y

f to y

n 1

, then

³  g ( x  y )  f ( y ) dy .

g* f

x . As an example, we wish to

O x

perform the convolution of two functions, say f ( x ) e with O as a constant and g ( x ) function for x t 0 and g ( x ) 0 for x  0 . The following steps demonstrate this process:

g* f

x

O K as a constant

x

³  g ( x  y )  f ( y ) dy f

y x

x

³ O K e

O y

0



K e

O x

ª eO y º d y K O « O » ¬ ¼y





1

K 1 e

(2.13) 0



O x

After this comes:

f *g

x

³  f ( x  y )  g ( y ) dy O  O K dy ³e f

x

 ( x y )

0

O Ke

O x

0



K 1 e The above operations (2.13) and (2.14) show that g * f

³

x

Oy

e dy

O x

Ke

(2.14) O x

e

Ox



1



f * g . The convolution is defined as: x

³  f ( x  y )  g ( y ) dy

f *g

(2.15)

f

If the integration is from say point x0 instead of f , we write the convolution as:

f*

x0

x

³  f ( x  y )  g ( y ) dy

g

(2.16)

x0

Here we also state that this convolution is used in finding a particular solution to non-homogeneous differential equations, which we will use in later chapters. Analogous to the above (2.12) explanation for left integration, or forward integration backward integration or right integration as

n x Ib

> f ( x )@ ,

n

>

@

> f ( x)@ , we define the

and write it as follows:

>

x Ib f ( x)

n

n a Ix

@

1

b

³  ( y  x)  f ( y ) dy

( n  1)! x

This above expression (2.16) of x I b f ( x ) we will discuss subsequently.

n 1

(2.17)

52

Chapter Two

2.4 Explaining the convolution process in repeated integration numerically Say we have a function f ( x ) , the values of which are f0 at x x

3h , f 4 at x

4 h and f5 at x

0,

x0

h , f 2 at x

f1 at x

5h . With these six values of function, the integration

³

x

x0

2 h , f3 at

f ( y )dy can be

approximated by the sum of these values, as:

h  f0  f1  f 2  f3  f 4  f5

S

(2.18)

The above expression (2.17) is nothing but a sum of an area of rectangles. With h as a common term, we can say first that the sum is: (2.19) S1 f0  f1  f 2  f3  f 4  f5 The growing sum as x grows is S1 (at x

2 h ) the sum is S1

f0 (at x

0 ), for the next point (at x

h ) it is S1

f0  f1 , for the next point

f0  f1  f 2 and so on, as depicted in the following steps:

x x0

f ( x) f0

S1 f0

x0  h x0  2 h

f1 f2

f0  f1 f0  f1  f 2

x0  3h x0  4 h x0  5h

f3 f4 f5

f0  f1  f 2  f3 f0  f1  f 2  f3  f 4 f0  f1  f 2  f3  f 4  f5

(2.20)

Now the next step is to derive a second sum (equivalent to double integration), that is taking S1 as a function, and the point wise summing to get S 2 . The growing sum as x grows is S 2 (at x

h ) is S 2

f0  ( f0  f1 )

further at x 2 h and write S 2 following steps:

f0 for the first point and then for the next point

2 f0  f1 . We note here that the value of S is S1 f0  ( f0  f1 )  ( f0  f1  f 2 )

f0  f1 at x

3 f0  2 f1  f 2 , and so on, as depicted in the

x x0

f ( x) f0

S2 f0

x0  h x0  2 h x0  3h x0  4 h

f1 f2 f3

2 f0  f1 3 f0  2 f1  f 2 4 f0  3 f1  2 f 2  f3 5 f0  4 f1  3 f 2  2 f3  f 4 6 f0  5 f1  4 f 2  3 f3  2 f 4  f5

x0  5h

h . Now we go

f4 f5

(2.21)

We write the second sum for the six points as follows:

S2

6 f0  5 f1  4 f 2  3 f3  2 f 4  f5

(2.22)

Now we generate the third sum S3 (equivalent to a triple integration) from the obtained S 2 as x grows. The first point is S3

f0 . At the second point at x

go at x 2 h , the sum is S3 following steps:

The six point third sum at x

h , the triple sum is S3

f0  (2 f0  f1 )  (3 f0  2 f1  f 2 )

f0  (2 f0  f1 )

3 f0  f1 , furthermore, as we

6 f0  3 f1  f 2 ; and so on; as depicted in the

x x0

f ( x) f0

S3 f0

x0  h x0  2 h x0  3h x0  4 h x0  5h

f1 f2 f3 f4

3 f0  f1 6 f0  3 f1  f 2 10 f0  6 f1  3 f 2  f3 15 f0  10 f1  6 f 2  3 f3  f 4 21 f0  15 f1  10 f 2  6 f3  3 f 4  f5

f5

(2.23)

5h is the following:

S3 21 f0  15 f1  10 f 2  6 f3  3 f 4  f5 For the six points, we write the first second and third sums as:

(2.24)

Fractional Integration

53

S1

f0  f1  f 2  f3  f 4  f5

S2

(6  0) f0  (6  1) f1  (6  2) f 2  (6  3) f3  (6  4) f 4  (6  5) f5 (6  1)(6  0)

S3

2 

f0 

(6  0)(6  1)

(6  2)(6  3) 2

f1 

2 f3 

(6  1)(6  2) 2

(6  3)(6  4)

f4 

2

(2.25)

f2

(6  4)(6  5) 2

f5

Look at S1 ; multiplying by h and then as h is made very close to zero so we say S1h is indicative of Now we look at S 2 and for small h we may say that S 2 h is indicative of get

x

x

x0

x0

³ ³

x

x

x0

x0

³ ³

³

x

x0

f (u )du .

f (u )du . Let us see, then, how we

f (u )du .

The first term of S 2 in (2.25) is (6  0) f0 ; by taking 6 as x and 0 as u we can write f0 as f (u ) at u Therefore (6  0) f0 is ( x  u ) f (u ) , is the first term at u other terms of S 2 . Therefore, with h derived in an earlier section, that is

x0 .

0

x0 . We use the same argument for (6  1) f1 and all

0

du , we write S 2 h as x

x1

x

x0

x0

x0

³ ³  f ( y )dy dy ³

³

x

x0

( x  u )  f (u ) du . This is the same as what we

( x  u ) f (u )du , which is a double integration, as a

convolution i.e.  x *  f ( x ) in Sections 2.2 and 2.3. Now we take the sum S3 and with h being very small we analyse this triple sum. The first term is Take x

6 , with u

0 and f0 as f (u ) at u

0 ; the term approximately represented as

( x u ) 2

2

(6 1)(6  0) 2

f0 .

f (u ) . This

approximation is valid when h is close to zero. This argument can be extended to all the other terms of S3 and we can see that S3 h is 1 2

 x *  f ( x) 2

or

x ( x u )2 2 x

³

f (u )du ; this is a formula for triple integration

0

1 2

x

³  ( x  u )  f ( u ) du . 2

x0

x

x2

x1

x0

x0

x0

³ ³ ³  f ( y )dy dy dy , that is

Therefore, we have the following generalisation for n  fold

integration with the n  th sum S n i.e. x x1

Sn h

³ ³ ..... a a

xn 1

³

f ( xn )dxn dxn 1 ...dx2 dx1

a

1

(2.26)

x n 1

³ ( x  y )  f ( y ) dy

( n  1)! a

This is another way we could explain the process of repeated integration – and a convolution process. Now, with this background, we can subsequently derive a formula for arbitrary n . We have discussed a very basic way the concept of iterative integration and the concept of convolution. This is necessary as this process that we discussed in this section and previous sections is very important in the generalisation of the concept of n  fold integration to fractional integration (and then fractional differentiation) when n  is not a natural number, i.e. for n  as a real number. We also stress that the convolution process is the basis of fractional integration as well as fractional differentiation.

54

Chapter Two

2.5 Generalising the expression for a repeated integration to get the Riemann-Liouville fractional integration formula We write the following from expression (2.12): x 1 0 Dx

> f ( x ) @ ³  f ( y ) dy 0

x 2

0 Dx

> f ( x)@ ³  f ( y ) ( x  y )dy 0

3 0 Dx

x

1

> f ( x )@

2

 f ( y ) ( x  y ) dy 2³

(2.27)

0

4

0 Dx

>

f ( x)

x

1

@

3

 f ( y ) ( x  y ) dy (2)(3) ³ 0

n

x

1

> f ( x )@

0 Dx

³  f ( y) ( x  y)

n 1

( n  1)! 0

dy

2.5.1 A generalisation of a repeated integration formula by the gamma function Now we have to generalise n ! for non-integer n . This Euler gamma function has the integral representation def

*( n)

³

f

0

x

n 1  x

e dx ; Re > n @ ! 0 (see Section 1.10). Comparing the integral obtained with the definition of the

gamma function, we have the following observation: f

f n t

³t e

n!

dt

* ( n  1)

( n  1)!

0

³t

n 1  t

e dt

*( n)

(2.28)

0

The above integral representation (2.28) is a generalised factorial function known as a gamma function, for any positive n , real or complex, which need not be a natural number. The following is the proof of this gamma function, as a generalisation of factorial function. Let I( n ) be the integral representation of the gamma function, and to that apply the formula for integration by parts, that is: f

I( n )

³e

t n

t dt

0

f

t n f

º¼  n ³ e  t t n 1dt 0

 ª¬e t

(2.29)

0

n  I( n  1)

In addition, we have I(0)

³

f

0

t

e dt

1 . Therefore, we write I( n )

* ( n  1) . Now, consider continuing the

n!

I( n ) of (2.29) to a complex plane, i.e. where n is replaced by a complex number, z  ^ . We require Re > z @ ! 1 , so that I( z )

³

f

0

def

t z

e t dt converges (Refer Section-1.10.1) Thus, we have the definition * ( z  1) z  * ( z ) . This * ( z )

Re > z @ ! 1 , which satisfies * ( z  1)

* ( z 1) z

³

f

0

t z

e t dt , with

we have discussed previously in Section 1.10

as an analytic continuation of the gamma function to the left half plane (LHP), though the integral representation of the gamma function, * ( z )

³

f

0

 t z 1

e t

>@

dt , is defined for the RHP, Re z ! 0 .

We have for the gamma function, *

 12

S and * ( n  1)

n  * ( n ) . Therefore, for n a positive non-integer

number from the above generalisation (2.28), we get an iterated integral formula as follows: n

>

0 I x f ( x)

@

n

0 Dx

>

f ( x)

@

1

x

 f ( y) ( x  y) *( n) ³ 0

n 1

dy

(2.30)

Fractional Integration

55

This expression (2.30) was developed by Liouville (1832) and Riemann (1876), who developed logical definitions of this fractional integration operation when in the above formula, n is a non-integer, in general i.e. arbitrary positive or negative real number. The above definite integral (2.30) i.e.

x

³  f ( y) ( x  y)

n 1

0

dy is a Riemann-Liouville integral.

2.5.2 Applying the Riemann-Liouville formula to get double-integration and semi-integration for the function f ( x) n

>

Let us apply the formula 0 I x f ( x )

@

1 *(n)

x

³  f ( y) ( x  y)

n 1

0

x x , for n

dy for f ( x )

2 0 Dx

x

1

> f ( x )@

2.

³

* (2)

y ( x  y )dy

0

x

³  xy

1

2

y

3

2

0

3

dy

(2.31) x

5

ª 23 xy 2  52 y 2 º ¬ ¼0 4 15

x

5

2

The result of (2.31) we got in (2.3) and (2.4) earlier. Now let us obtain semi-integration of f ( x ) n 0 Ix

> f ( x )@

1 *(n)

x

³  f ( y) ( x  y) 0

n 1

1 2

x i.e. with n

dy obtained for repeated integration and written in the following manner: 1

0 Ix

2

ª xº ¬ ¼

 12 0 Dx

y

* ³

xy  y

0

x2 4

0

1 2

y

x we have T

S

2

x2 4

0

 2x (cos T )dT

dy

(2.32)

x2 4

 2x

y

* ³

dy

2

y 

x

1

 12 1

dy y

³

*  12

 x  x sin T , then we have

2

y ( x  y)

0

x

1

1 2

* ³ 1 2

x

1 2

x

1

ª xº ¬ ¼

1

Now in (2.23) put y

by the above generalised expression i.e.

y 

x 2 2

 y 2

dy

dy



and at y

0 , we have T

 S2 , and at

. With these values, we get the following steps:  12 0 Dx

ª xº ¬ ¼

S

1

³

*  12 S S

1

2



x2 4

2

sin T

 2x  cos T dT

³  12  x  x sin T dT 2

1



2 *

T

1 2



Sx



2 *  12 2

x2 4

2

*  12 S

S

 12

2

x



> xT  x cos T @T

S

2

 S2

(2.33)

56

Chapter Two

We used the value *

 12

S , and with great effort (and with skill in doing the integration) we applied the

generalised expression (2.30) of

n 0 Dx

> f ( x)@ to get the value of semi-integration of

1

x as

0 Ix

2

ª xº ¬ ¼

S 2

x.

2.5.3 Formally defining forward (left) and backward (right) Riemann-Liouville fractional integration Let f d a  x d b d f , then the Riemann-Liouville fractional integration of order D ! 0 is defined for the locally

> @

integrable function f ( x ) , as f : a , b o  , which are followed by: D

a Ix

x

1

> f ( x )@

D 1

( x  y) * (D ) ³

f ( y )dy

a

D

b Ix

1

> f ( x )@

(2.34)

b

D 1

³ ( y  x)

* (D )

f ( y )dy

x

The first one is integration in the left sense, and the second one is integration in the right sense. We may call the first expression a causal one and the second expression an anti-causal one. From above, for x  b and for D 0 , we write the following identity: 0 a Ix

0 x Ib

> f ( x )@

> f ( x )@

f ( x)

(2.35)

2.6 Fractional integration of a power function Next, we deal with the expression of a generalisation of the repeated integration: D

c Dx

x

1

> f ( x )@

1

D ( x  y )  f ( y ) dy ; ³ * (D )

D !0

(2.36)

c

Having taken note that in (2.36) the start point of the fractional integration above is x c , which is the lower limit of integration, earlier in our derivation (2.30) we maintained the lower limit of integration was zero i.e. x 0 . We now try to achieve the fractional integration of the power function: P

(2.37) f ( x) x P ! 1 The condition P ! 1 in (2.37) will be clear as we derive this fractional integration expression. In the expression of the fractional integration (2.36), we put y x (1  u ) and re-wrote the expression with this change of variable as follows with  x  du

dy and the limits of integration from the lower limit ( x  c ) / x to zero. D c Dx

> f ( x )@

x

D

F 1

D u  f ( x  xu ) du ; ³ * (D )

xc

F

x

0

Here x ! c t 0 . For f ( x )

P

x , we have to put f ( x  xu )

P

x (1  u )

D c Dx

P

, and write the following:

ªxP º ¬

x

P D F

x

P D

u * (D ) ³

¼

def

above

³t 0

x 1

P

(1  u ) du

³

F

0

t

x 1

(2.39)

B F (D , P  1)

(1  t )

y 1

dt and using this we write the

B F (D , P  1) for the definite integral. For this beta integral both the arguments D and P  1 need to be

positive; this also gives us the condition that P ! 1 . When c 1

D 1

0

* (D ) We have defined the incomplete beta function (1.144) as B F ( x, y )

(2.38)

(1  t )

y 1

0 , we have F

1 , therefore the integral is

dt ; which we recognise as a complete beta function (1.129) i.e. B( x, y ) which is defined as

With this we write

B(D , P  1)

* (D ) * ( P 1) * (D  P 1)

.

*( x)*( y ) *( x y )

.

Fractional Integration

57

2.6.1 Fractional integration of power functions with the lower limit of integration as zero Therefore in (2.39) by use of the beta function as a ratio of the gamma function (as defined in Section 1.15), we write the fractional integration of

f ( x) D

0 Ix

x

P

with a lower limit of integration c

ª¬ x P º¼

D 0 Dx

x

ª¬ x P º¼

* (D )

B D , P  1 (2.40)

* ( P  1) * ( P  D  1) The beta function B(D , P  1)

* (D ) * ( P 1) * (D  P 1)

0 as follows:

P D

x

P D

D ! 0 , P ! 1

diverges for P d 1 thus we safely write: D

0 Ix

ª¬ x P º¼

D 0 Dx

ª¬ x P º¼ f

P ! 1

D !0

(2.41)

2.6.2 Fractional integration of power functions with the lower limit of integration as non-zero P

Let us evaluate the fractional integration of f ( x ) ( x  a ) from the lower terminal x c where x ! c ! a and c t 0 . Thus, we have from the formula the following expression: x 1 P D 1 P D f ( y) ( y  a) ( x  y )  f ( y ) dy , c Dx  x  a * (D ) c (2.42) x 1 D 1 P ( x  y) ( y  a ) dy * (D ) c

³



³

We substitute in (2.42) y  a

( x  a )(1  u ) so the lower limit y



c becomes u

( x c ) ( xa )

and the upper limit y

x

is changed to u 0 . We then have x  y u ( x  a ) and dy ( x  a )( du ) . With these substitutions, we can write the following expression, on similar lines as obtained for the above derivations (2.39) and (2.40) with the use of an incomplete beta function: x 1 P D 1 P D ª º ( x  y) ( y  a ) dy c Dx ¬ x  a ¼ c * (D )



³

1

0



1

D P  u ( x  a )  ( x  a )(1  u ) ( x  a )  du ³ * (D ) ( x c ) ( x a )

D P

( x  a)

* (D )

x c x a

³u

D 1

P

(1  u ) du ;

(2.43)

xc

F

xa

0

D P

( x  a)

* (D )

B F (D , P  1)

Therefore, we have the formula: D

c Dx

ª x  a P º ¬ ¼

D P

( x  a)

xc

D ! 0 (2.44) * (D ) xa For the start point or the lower limit of integration as non-zero i.e. c z 0 , can we have the following result? * ( P  1) D P D ª P º c DxD ª¬ x P º¼ ; P ! 1 ( x  c) D ! 0 (2.45) c I x ¬x ¼ * ( P  D  1) This expression (2.45) we get by manipulating the derivation of

a

0 , and making F

B( x, y )

1

³u 0

x 1

(1  u )

y 1

x c x

1 , giving x

du

*( x)*( y ) *( x y )

B F (D , P  1) ,

D c Dx

( x  a)

P

F

as done in (2.43) above by setting

( x  c ) and using the definition of a complete beta function i.e.

, (1.129) that is described in the following steps:

58

Chapter Two

D c Dx

x

1

ª¬ x P º¼

1

D ( x  y )  f ( y ) dy ; ³ * (D )

P

f ( y)

x ,

y

x (1  u )

c

x

D P

x c x

u * (D ) ³

D 1

P

(1  u ) du

xc

F

D P 1

( x  c)

³u

* (D )

D 1

1, x

x

0

( x  c)

P

(2.46)

(1  u ) du

0

D P

( x  c)

B(D , P  1)

* (D ) D P

( x  c)

* (D )

* ( P  1) § * (D )* ( P  1) · P D ¨ * (D  P  1) ¸ * (D  P  1) ( x  c ) © ¹ x c xa

The above (2.46) derivation shows that for it to be valid we have taken F above case (2.46) to be valid we thus have c

x

0 . Thus, we recover

constant function

f ( x)

D 0 Dx

P

ª¬ x º¼



1 , thus here we have c

a . For the

0 , i.e. the starting point of the fractional integration process is zero

( x)

P D

>1@

D 0 Dx

* ( P 1) * (D  P 1)

. For P

0 and c

0 , we have fractional integration of a

1 as follows: D

0 Ix

>1@

* (1) * (D  1)

x

x

D

D

1

* (D  1)

D!

x

D

(2.47)

2.6.3 Region of validity of the integer order integration of the power function Let us take the formula for fractional integration i. e.

ª¬ x P º¼

D

0 Ix

* ( P 1) * ( P D 1)

x

P D

, and observe the following simple

calculations for cases where D and P are integers. When both are integers, the formula is: D

0 Ix

ª¬ x P º¼

* ( P  1) * ( P  D  1)

x

P D

P! ( P  D )!

x

P D

,

D

1, 2, 3,.. (2.48)

1 ­ P D P ! 1 x ° . ® ( P  1)( P  2)...( P  D ) °¯ P d 1 f

This we have discussed in Section 1.23. With this, we can compose Table 2.1, showing the region of validity. This is a P

safe bet, to exclude the region for f ( x ) x , when P d 1 (refer to Table 2.1) for doing fractional integration. Nevertheless, can we extend our derived formula for this region? We discuss this issue subsequently.

Pp

Do

3

2

1

0

2

2! 5!

x

5

2! 4!

x

4

2! 3!

x

3

x

1

1 4!

x

4

1 3!

x

3

1 2!

2

x

0

1 3!

x

3

1 2!

x

2

x x

2

1

1

f

f

f

x

1

2

f

f

f

x

2

3

f

f

f

x

3

Table 2.1: Integer order integration of the power function with integer exponents with a fractional integration formula

2.7 The extended region of fractional integration of a power function: analytic continuation

Fractional Integration

59

This is called ‘extended fractional calculus’. In this concept the ratio of the gamma function that appears as a coefficient in fractional integration, the formula of the power-function (2.45) i.e. limp0

* (1 P ) * (1 P D )

is taken as a

fundamental defining expression. Unlike the Riemann-Liouville integral which we saw is defined by the region

P ! 1 for function f ( x )

* (1 P )

P

x (2.45), the limit limp0 * (1 P D ) is well defined in the plane D , P  \ ; except  along the line interval P  ] , D  \ \ ] . Here, ‘ \ ’is a “mod” operation between set \ and set ] . The “mod” operation example is: A

^1, 2, 3, 4, 5, 6`

B

^1, 3, 4` then A \ B ^2, 5, 6` .

Here we will discuss the analytic continuation to these intervals and show it to be: D

0 Ix

* (1  P  ) x

D P

ª¬ x P º¼ limp0  ln x  \ (D  P  1)  J (2.49) * ()* (D  P  1)

where in (2.49) \ is the ‘psi’ function and J is Euler’s constant (we have discussed this function in the previous chapter, in Section 1.16). The demonstration in an earlier discussion (Section 2.6.3) shows that when P  0 , the formula (2.45) i.e. D

0 Ix

* ( P 1) * ( P D 1)

ª¬ x P º¼

x

P D

becomes uncomfortable when P is a negative integer 1, 2, 3, .... , because the value of

the gamma function at point zero or at points of negative integer is infinity. Hence, safely we specify the region of validity P ! 1 , for this formula to be valid, also due to the divergence of the beta-integral in the derivation of this formula (2.45). However, we see that for P d 1 , at the integer points P normal classical integration as demonstrated below:

1, 2, 3,... the following are some relations after doing a

³  dx ln x ³  dx  x ³  dx  x ³  (dx) ³ ln x(dx) ³  (dx) ³ ³   dxdx ³ 1 x

D

1

P

1

D

1

P

2

D

1

P

3

D

2

P

1

D

3

P

3

1

1 x2

1 x3

2

1 2

(2.50)

2

1 x

x2 2

3

1 x3

1 2! x dx

1 2! ln

x

P We construct the following table (Table 2.2), showing the integration of x , with P as an integer in this case of D

0 Ix

ª¬ x P º¼ , also with D as a positive integer. This table has finite entries whereas the entries are f in the

corresponding table (Table 2.1).

Pp

Do

3

2

1

0

2

2! 5!

x

5

2! 4!

x

4

2! 3!

x

3

x

1

1 4!

x

4

1 3!

x

3

1 2!

2

x

0

1 3!

x

3

1 2!

x

2

x x

1 2

3

2

³ ln x(dx)  ³ ln x (dx ) 1 2! ln

x

 ln x 1 2!

x

1

ln x

³ ln x(dx)

x

1

1

1  2! x

2

2

x

1

x

2

x

3

Table 2.2: Integer order integration of the power function with integer exponents With this observation of the breakdown of our obtained formula i. e. and with

D

0 Ix

ª¬ x P º¼

* ( P 1) * ( P D 1)

x

P D

, for P as an integer

P  0 , we note that ,in Table 2.2, a log-region is formed with a pattern. Let us take

60 1 0 Ix

Chapter Two * ( 2 1) * ( 2 11)

ª¬ x 2 º¼

x

2 1

³

integration once, we get

* ( 1) * (0)

1 dx x2

1

x ; this is getting troublesome with formula (2.45). Nevertheless, doing normal

x

1

. We use the following trick, by writing the formula in a limit for P  0 : D

0 Ix

D

Which gives us

* ( 1)

* () (1)

0 Ix

ª¬ x P º¼ limp0

* ( P 1) * ( P D 1)

and using this we get limp0

x

* (1) * ()

P D

x

1

* ( P  1 )

P D ª¬ x P º¼ lim x p 0 * ( P  D  1 )

limp0

limp0

* (1) * ()

1

* ( x 1) x

*( x)

x . Now use

x 1 (1)

(2.51) to write

1

 x . Therefore, we get the same answer as

when doing normal integration; thus we may say the following: 1 0 Ix

One more example we take is again limp0

* (2) * (1)

x

2

1 0 Ix

ª * ( P  1 ) P D º ª¬ x 2 º¼ limp0 « x » ¬ * ( P  D  1 ) ¼P * ( 2) * ( 1)

3

ª¬ x º¼

, and use * ( 2)

x

* (1) (2)

2

* ( 2) * ( 1)

which has ratios of

* (2) * (1)

, we will get limp0

x

x

(2.52)

2,D 1

( 1) 2

1

2 1 1! 2!

1  2! . Well if we write

2

1  2! x . Therefore, we write the

following: 1 0 Ix

§ * ( P  1) P D · ª¬ x 3 º¼ limp0 ¨ x ¸ © * ( P  D  1) ¹P

 3,D 1

1 2!

x

2

(2.53)

Perhaps a better way to write the formula (except for the log-region) in Table 2.1 is as follows: D 0 Ix

§ * ( P  1 ) P D · ª¬ x P º¼ limp0 ¨ x ¸ © * ( P  D  1 ) ¹

(2.54)

Take an example from Table 2.1, and write the few entries of the log-region as follows: 3 0 Ix

) 31 ª¬ x 1 º¼ limp0 * (1*(1 ) ³ ln x (dx )

3 0 Ix

) 3 2 ª¬ x 2 º¼ limp0 * (1*(2 ) ³ ln x (dx )

3 0 Ix

) 3 3 ª¬ x 3 º¼ limp0 * (1*(3 ) ³ ln x (dx )

* ()

2

³ ln x(dx)  ³ ln x (dx ) 1 2! ln

x

limp0 * () limp0

* (1) * ()

* (2) * ()

1 2!

limp0

1 1

Looking at the log-region in Table 2.2 and analyzing it, we get a formula as in the following expression: * (1  P  ) D D P ª P º limp0 ln x (dx ) 0 I x ¬x ¼ * ()

³

limp0

* (1  P  )

D P

(2.55)

(2.56)

> @

ln x 0 Ix * () We will use these interesting results (2.56) when we formally do analytical continuation of this integral operator. What is the fractional integration of ln x , that is formula in the log-region as follows: D

0 Ix

D P

³ ln x(dx)

D P

or 0 I x

* (1  P  ) x

>ln x@ ?

If we write the fractional integration

D P

ª¬ x P º¼ limp0  ln x  \ (D  P  1)  J * ()* (D  P  1)

(2.57)

with the concept of the analytic continuation of the harmonic series represented above in terms of \ ( z ) a psi-function and J as Euler’s constant, we will not be inaccurate. The discussion of this is in the next section, though we have introduced the concept in an earlier chapter (Section 1.16).

2.8 An analytic continuation of a finite harmonic series-derived via a fractional integration concept We have seen in an earlier chapter (Section 1.16) the multiple n  fold integration of ln x and we write that as follows:

Fractional Integration

n

x

> @ ³0

0 I x ln x

ln y (dy )

x

61

n

 ln x  1  n!

n

x n

¦  1 k

n 1

³

is the harmonic series. We say that

 13  ... 

1 n

(2.58)

n

 ln x  h(n)

n! In (2.57), h ( n )

1 2

x

0

ln y (dy )

n

is the ‘generating-integral’ of the

harmonic series. When we change n to any real number, say X , and use the concept of analytic continuity for writing X ! * (X  1) , we write the following as an analytic continuation of the ‘generating integral’ for the harmonic series: X

0 Ix 0 Ix

X

ª¬ x P º¼





We use the formula i.e. writing ln x

limp0

1 

x

* ( P 1) * ( P X 1)

x

ln x

x

0

X

x

>ln x@ ³0 ln y (dy )

P X

³

 1 , in the integral

x

X

* (X  1)

X

ln y (dy )

ln x  ln 1

as shown below: x

1

³

y dy

1 x

³

y

1

limp0 limp0

1 

x



1

dy



x 

 1



³

0

We equate the above (2.61), that is expression,

³

x

0

X

ln y (dy )

xX * (X 1)

³

x

0

x

X

limp0

1 

³y





y

(2.60)

y 1

³

 1 . The analytic continuation of the integral

ln y (dy )

 y x

1 

limp0

x

0

evaluated as follows, with this substitution: x

(2.59)

in the derivation of the analytic continuation of h ( n ) and by

limp0

Therefore, we have ln x

 ln x  h(X )

ln y (dy )

n

thus can be

X

 1 (dy )

0

X

ln y (dy )

limp0

1 

³

limp0

1 



0 Ix

limp0

1 



* (1) * (1X )

limp0

 

limp0

  xX 

x

0

xX 

X



y (dy )  X

³

x

0



X

1(dy )



ª¬ x º¼  0 I Xx >1@ X

x

* (1) * (1X )

* (1) * (1X )



x 

 

* (1) * (1X )

x 

1 * (1X )

(2.61) X

x

1 * (1X )





, with our earlier obtained (2.59)

 ln x  h(X ) , and arrive at the following after simple manipulations i.e. a relation n

¦  :

for h (X ) as the analytic continuation of the harmonic series h ( n )

n 1

1 k

1·  x  ¸ ¹ ©  * (1  X ) The following steps are important to further simplify the above expression (2.62): h(X )

§ * (1 )* (1  X )

ln x  limp0 ¨

(2.62)

62

Chapter Two

h (X )

ln x  lim p 0

1 § * (1 )* (1  X )

¨  © * (1  X )



·

x  1¸

¹

§ * (1 )* (1  X )  * (1 )* (1  X ) · ¨ * (1  X ) x  * (1  X )  ¸ 1 ¸ ln x  lim ¨ p 0  ¨ * (1 ) * (1  X ) ¸ ¨ * (1  X )  1 ¸ © ¹ 1  § * (1 )* (1  X ) · ln x  lim  x  1 ¨ ¸ p 0  © * (1  X ) ¹ 1 § * (1 )* (1  X ) ·  lim ¨1  ¸ p 0  * (1  X ) ¹ © ln x  lim p 0

Taking lim

1 § * (1 )* (1  X ) · x  1  lim ¨1   ¸   * (1  X ) 

p 0

* (1 )* (1  X )

©

§ 1 § * (1 )* (1  X ) · · ¨1  * (1  X ) ¸ ¸ ¹¹ ©©

lim ¨

p 0

¹

1

* (1  X )

p 0

h (X )

1

use

ln x

1 x 1 lim 



p 0





d ª * (1) * (1X ) º «1 » d ¬ * (1X ) ¼ lim d p 0 >@ d (1) (1) § * (1  X )* (1 )  * (1 )* (1  X ) · lim ¨ * (1  X ) ¸ 2 p 0 ¨ ¸ * (1    X )  © ¹

(2.63)

(1)

(1)

* (1) 

* (1  X )

J  \ (1  X ) * (1  X ) We have applied L’Hospital’s rule in (2.63) to calculate the limit in the above derivation (2.63), since we are getting 0 for the following expression: 0 d ª

«1 § 1 § * (1 )* (1  X ) · · d ¬ lim ¨ ¨ 1  lim ¸¸ p 0  * (1  X ) ¹ ¹ p0 © ©

as limp0

* (1) * (1X ) * (1X )

* (1) * (1X ) º » * (1X ) ¼ d >@ d

(2.64)

1 . In addition, we have used a derivative of the quotient formula that is: f ( x)

g ( x)

f

h( x)

(1)

( x)

h( x) g

(1)

(1)

( x)  g ( x)h ( x)

 h( x)

2

(2.65)

in the above steps (2.63) during derivation. Using the definition of the ‘psi’ function (or the di-gamma function),

*(1) ( z ) *( z )

\ ( z ) , and the Euler constant

(1)

* (1) , in the above derivation (2.63) we get analytical continuation of the harmonic series as: (2.66) h (X ) \ (1  X )  J The above (2.66) result is standard formulation proved in ‘number theory’; but here it is also derived from analytic

J

X

continuation the of integration operation, that is, operator 0 I x where X is a non-integer. With this, we can also write fractional integration of ln x as: 0 I x > ln x @

X

x

X

X

x

(2.67)  ln x  \ (X )  J * (X  1) This above expression (2.67) we have used to extend the formula of fractional integration to the log-region (2.57) for D P the formula 0 I x ª¬ x º¼

* ( P 1) * ( P D 1)

x

D P

³

0

ln y (dy )

, where we had difficulty for cases P  0 and D d P .

Fractional Integration

63

2.9 Fractional integral of exponential function by series * ( P 1) * ( P X 1)

X P X P The earlier derived formula (2.45) for fractional integration 0 I x ª¬ x º¼ that is 0 Dx ª¬ x º¼

x

P D

for P ! 1 ,

X ! 0 and X , P   , can be used as a term by term formula for the series expanded function e ax , as indicated below: e

ax

( ax )

1  ( ax ) 

2

( ax )



2!

3

3!

f

¦

 ...

( ax )

k 0

k

f

( ax )

k

¦ *(k  1)

k!

(2.68)

k 0

2.9.1 Fractional integration of the exponential function with a start point of integration as zero We use (2.45) and then write the following steps: X

ª 0 I x ¬e

ax

X

º¼

>@

X

>

@

X

0 I x 1  0 I x ( ax )  0 I x

X 1

X

x

* (X  1) X

x

f



2

a* (2) x

1! * (X  2)

( ax )

ª ( ax ) 2 º « 2! »  ... ¬ ¼



X 2

a * (3) x

2! * (X  3)

3



X 3

a * (4) x

3! * (X  4)

 ... (2.69)

k

¦ *(k  X  1) k 0

X

ª 0 I x ¬e

ax

º¼

X

0 Dx

f

( ax )

k

ª¬e ax º¼ xX ¦ * (X  k  1) k 0

2.9.2 Introducing a higher transcendental function in the result of the fractional integration of the exponential function We have a two-parameter Mittag-Leffler function i.e. ED , E ( x ) defined as follows in the series representation which is also detailed in Appendix A: def f

x

k

¦ *(D k  E )

ED , E ( x )

D, E ! 0

D, E  \

(2.70)

k 0

In Appendix F, we extend the definition of ED , E ( x ) for negative order D . However, we will be considering

ED , E ( x ) positive order D , in our discussions. Thus in terms of the above (2.69) 0 I Xx ª¬e ax º¼

X

x

¦

( ax )k * ( k 0 X  k 1)

f

and

by defining the Mittag-Leffler function (2.70) i.e. ED , E ( x ) , we obtain the fractional integration of the exponential function as follows: X

0 Ix

ª¬e ax º¼

X

x E1,(X 1) ( ax )

(2.71)

We note here that function ED , E ( ax ) is one of several higher transcendental functions. The series that appeared in the above derivation (2.69), that is, xX

¦

( ax )k k 0 * (X  k 1)

f

, is called the Miller-Ross function, represented as E x (X , a ) ,

which is another higher exponential function. Its relation to the two-parameter Mittag-Leffler function is the following (noted in Appendix A): def

E x (X , a )

X

x

f

( ax )

k

¦ *(X  k  1)

X

x E1,X 1 ( ax )

(2.72)

k 0

Thus in (2.72) we have X

0 E x (0, a )

e

ax

and ET (0, ia )

e

i ( aT )

cos( aT )  i sin( aT ) .

What we note here is that while fractionally integrating the exponential function (a transcendental function) we are getting a higher exponential (higher transcendental) function that is: X

0 Ix

ª¬e ax º¼

X 0 Dx

ª¬e ax º¼ E x (X , a ) xX E1,X 1 ( ax )

The complex argument Miller-Ross function is derived from the above definition (2.72) as follows:

(2.73)

64

Chapter Two k

k 1

k f § f ( 1) 2 ( ax ) k ( 1) 2 ( ax ) · i ¦ E x (X , ia ) ( x ) ¨ ¦ ¨ k even * (X  k  1) k odd * (X  k  1) ¸¸ © ¹

X

(2.74)

C x (X , a )  iS x (X , a ) From this we obtain (in Miller-Ross notation) the higher cosine and sine functions as: def

C x (X , a )

k

f

X

¦

x

( 1) 2 ( ax )

k even def

S x (X , a )

f

X

* (X  k  1)

( 1)

¦

x

k odd

k

k 1 2

( ax )

(2.75)

k

* (X  k  1)

0 , we have:

When X

k

f

( 1) 2 ( ax )

¦

C x (0, a ) S x (0, a )

( 1)

¦

1

* ( k  1)

k even f

k

k 1 2

( ax )

( ax )

2



2!

k

( ax )

( ax )

4

3

cos( ax )

 ...

4!

( ax )

(2.76)

5

  ... sin( ax ) ( ax )  * ( k  1) 3! 5! Like we have demonstrated by taking the series expansion of the exponential function and then using the obtained fractional integration formula for a power function, a higher exponential function is obtained; similarly we can derive the series expansion of the cosine and sine function and obtain the following with the method described in (2.69): k odd

X 0 DT

>cos(aT )@

CT (X , a )

X 0 DT

,

>sin(aT )@

ST (X , a )

(2.77)

2.9.3 Fractional integration of an exponential function with the start point of integration as non-zero Now we do the fractional integration of the exponential function with a non-zero start point of integration, i.e. we evaluate

D

a Ix

ª¬ eO ( x  a ) º¼ by using a I Dx ª¬ ( x  a ) E º¼

the series for e

O ( xa )

* ( E 1) * (D  E 1)

D E

( x  a)

, which we derived earlier in (2.46). We write

as:

e

O ( xa )

O ( x  a)

1

2



O ( x  a)

2

3



O ( x  a)

3

1! 2! 3! Now we apply a term-by-term fractional integration formula to (2.78) to get the following: D ª O ( x  a ) º¼ a I x ¬e

D 1

D

* (1)( x  D ) * (D  1)



* (2)O ( x  a )

1! * (D  2)

2



(2.78)

 ... D 2

* (3)O ( x  a )

2! * (D  3)

 ...

(2.79)

D

( x  a ) E1,D 1  O ( x  a ) We have used the series definition of the two-parameter Mittag-Leffler function ED , E ( x ) (refer to Appendix A). D Ox Having derived the above expression, we now get the expression for the fractional integration of a I x ª¬ e º¼ . We

write e

Ox

e

O x O a  O a

Oa O ( xa )

e e

.

Using the above derived expression (2.79) for

D

a Ix

ª¬ eO ( x  a ) º¼ , we get the

following: D

a Ix

ª¬eO x º¼ eO a ( x  a )D  E1,D 1  O ( x  a )

(2.80)

The one parameter Mittag-Leffler function (refer to Appendix A) is defined in series form as follows: D

def

ED ( ax )

D ! 0; We note that ED ,1 ( x )

ED ( x ) , i.e. ED , E ( x )

1

ax

* (1  D )

D \;

ED ( x ) for E

2 2D

D



a x

* (1  2D )

3 3D



a x

* (1  3D )

 ...;

(2.81)

a^

1 (refer to Appendix A).

We integrate the above Mittag-Leffler functions series (2.81) from 0 to x , by fractional order D ! 0 and write the following steps:

Fractional Integration

D D 0 Ix ª ¬ ED ( ax )º¼

65

2 2D ª axD º º D ª a x  I 0 x « » » ¬ * (1  D ) ¼ ¬ * (1  2D ) ¼ 3 3D º D ª a x I 0 x « »  ... ¬ * (1  3D ) ¼

0 I x >1@  0 I x «

D

D

1

* (1  D )

D

* (1  D )

x 

* (1  D )* (1  2D )

* (1  2D ) * (1  2D )* (1  3D )

2 3D

a x

ax

2D



 ...

1 § 1 2 2D D ax  a x ¨ * (1  2D ) 1 * (1  D ) ¨ 1 a¨ 3 3D ¨ * (1  3D ) a x  ... ©

·

¸

¸ ¸ ¸ ¹

D 2 2D §§ · · ax a x  ¨ ¨1  ¸ ¸ * (1  D ) * (1  2D ) 1 ¨¨ ¸  1¸ 3 3D ¸ ¸ a ¨¨ a x  ... ¨¨  ¸ ¸ ¹ ¹ © © * (1  3D )

1

 E a

D

D

(2.82)



( ax )  1

2.9.4 Fractional integration of a one-parameter Mittag-Leffler function as conjugation to classical integration of the exponential function D

Therefore, from (2.82) we have

0 Ix

ª¬ ED ( axD ) º¼

1 a

D

ED ( ax ) 

1 a

D

. Note that the value of ED ( ax ) at x

0 is 1 .

The integration obtained here for the one-parameter Mittag-Leffler function is a conjugation to the exponential x

 1a . This is a very important observation. Like in classical calculus, we get the integration of exponential as exponential; we can make a similar observation for the one-parameter Mittag-Leffler function too. Therefore, this Mittag-Leffler function may play an important role like the exponential function plays, for solving a fractional differential equation. We note this observation for future usage. function that is

³

0

ay

1 a

e dy

e

ax

2.10 Tricks required for solving fractional integrals The integral of the type that has appeared in the fractional integral formula requires proficiency to be solved. The integrals are of the type

³

x

c

D 1

( x  y)

the fractional integral of f ( x ) D 1

integral is of the type ( x  y ) give us useful identities.

f ( y )dy , that is, convolution integral, require some tricks, as indicated in finding

P

x . However, there is no standard universal method. Because the kernel of this , there is some possibility to have some tricks developed. The tricks that we mention

2.10.1 Obtaining a useful identity for fractional integration with a lower limit of integration as zero Let us write the formulation of fractional integral of the function x  f ( x ) as follows D

0 Dx ª ¬ x  f ( x ) º¼

Make the substitution y  f ( y )

1

x

D ( x  y )  y  f ( y ) dy ³ * (D ) 1

(2.83)

0

 x  ( x  y )  f ( y ) in the above (2.83) expression (we have added and subtracted

x ), and with several simple steps of expansion, we get the following:

66

Chapter Two

D 0 Dx

x

1

1

D ( x  y )  x  ( x  y )  f ( y ) dy ³ * (D )

ª¬ x  f ( x ) º¼

0

x

§ 1 © * (D )

·

D 1

³

x¨

 f ( y ) dy ¸ 

( x  y)

(2.84)

¹

0

x

§ 1 · D  x y f y y ( ) ( ) d  ¨ ¸ ³ © * (D ) 0 ¹ D

The first integral term in (2.84) is recognised as 0 Dx the second term in (2.84) by D and use n  * ( n ) D



1 * (D )

³

x

D 1

( x  y)

0

 f ( y ) dy .

We multiply and divide

* ( n  1) , to write the following:

0 Dx ª ¬ x  f ( x ) º¼

x

> f ( x )@



x

D 0 Dx

Now the second integral in (2.85) is recognised as

D

0 Dx

> f ( x ) @ 

> f ( x ) @ 

 (D 1) 0 Dx

D  * (D )

D ³ ( x  y )  f ( y ) dy 0

(2.85)

x

D

D

* (D  1)

> f ( x )@ ,

x

D

³ ( x  y )  f ( y ) dy 0

and we get one useful identity that is represented

below: D 0 Dx

ª¬ x  f ( x ) º¼ x  0 DxD > f ( x )@  D  0 Dx (D 1) > f ( x )@

D

0 Ix ª ¬ x  f ( x ) º¼

x

 I > f ( x ) @  D  D

0 x

(D 1)

0 Ix

> f ( x ) @

(2.86)

From the above obtained expression we can write the fractional integration for some functions with the use of the Miller-Ross function (a higher transcendental function [see Appendix A]) as follows: X 0 Dx

ª¬ xe ax º¼ x





ª¬e ax º¼  X

X 0 Dx

 (X 1) 0 Dx

ª¬e ax º¼



xE x (X , a )  X E x (X  1, a ) X 0 Dx

> x cos(ax)@

x



>cos(ax)@  X  0 Dx(X 1) >cos(ax)@

X 0 Dx

(2.87)

xC x (X , a )  X C x (X  1, a ) X 0 Dx

> x sin(ax)@

x



X 0 Dx

>sin(ax)@  X  0 Dx(X 1) >sin(ax)@

xS x (X , a )  X S x (X  1, a ) X 0 Dx

We can generalise the expression

ª¬ x  f ( x ) º¼

x



X 0 Dx

> f ( x)@  X  0 Dx(X 1) > f ( x)@

p

for function x f ( x )

by the following steps: X

0 Dx

We make y

p

> x  ( x  y )@

p

x

1

X ( x  y)  y ³ * (X )

ª¬ x p f ( x ) º¼

1

p



f ( y ) dy

(2.88)

§ p · pk k ¨ k ¸ x ( x  y) © ¹

(2.89)

0

and expand as follows:

y

p

 x  ( x  y)

p

p

¦ (1) k 0

k

Substituting this expansion (2.89) in the fractional integration expression of (2.88) we write the following: X

0 Dx

Write

³

x

0

X  k 1

( x  y)

 f ( y ) dy

fractional integration to give:

ª¬ x p f ( x )º¼

1 * (X )

x

p

¦ k 0

( 1)

k

§ p · pk X  k 1  f ( y ) dy ¨ k ¸x ³ ( x  y ) © ¹ 0

(2.90)

appearing in the above expression (2.90), using the formula of the Riemann-Liouville

Fractional Integration x

X  k 1

³ ( x  y) 0

67

x § 1 · (X  k ) 1 ( x  y)  f ( y ) dy *(X  k ) ¨  f ( y ) dy ¸ ³ © * (X  k ) 0 ¹



* (X  k )

 (X  k ) 0 Dx

(2.91)

> f ( x ) @

Using the above obtained expressions (2.90) and (2.91) we get the useful expression:

X

0 Ix

p

1

X ª p º 0 Dx ¬ x f ( x ) ¼

¦ (1) * (X )

§ p·  (X  k ) > f ( x ) @ ¨ k ¸ * (X  k )  0 Dx © ¹

k

k 0

p

1

§ p· (X  k ) ( 1) ¨ ¸ * (X  k )  0 I x > f ( x ) @ ¦ * (X ) k 0 ©k¹

p

ª¬ x f ( x ) º¼

(2.92)

k

By use of the identity of the generalisation of binomial coefficients, which is the following: * (1  z ) § z · n *( z  n) ( 1) ¨n¸ n !*( z ) © ¹ n ! * (1  z  n) X p we express the obtained formula (2.92) for 0 Dx ª¬ x f ( x ) º¼ in another form as follows: X

0 Dx

p

1

ª¬ x p f ( x )º¼

¦ (1) * (X )

k

k 0

p

¦ (1) k 0 p

* (X  k )

k !( p  k )!

* (v )

X

¦ §¨© k ·¸¹ ª¬ D

k p xx

k 0

§ z  n  1· ¸ © n ¹

(2.93)

§ p·  (X  k ) > f ( x)@ ¨ k ¸ * (X  k )  0 Dx © ¹

p!

k

n

( 1) ¨



pk

x

 (X  k ) 0 Dx

> f ( x)@

(2.94)

º¼ ª¬ 0 Dx (X  k ) f ( x )º¼

We have used the following identities in the above derivation (2.94):

( 1)

k

* (X  k ) k ! * (X )

§ X · ¨k ¸ © ¹

k

D x

d

p

k

dx

x

k

p

p! ( p  k )!

x

pk

(2.95)

p

That is the k  th derivative of x , and an expression of generalised binomial coefficients, in (2.95).

2.10.2 Obtaining a useful identity for a fractional integration with a lower limit of integration as non-zero Now we try to find a general expression for fractional integration for f ( x ) with a non-zero lower limit: X

c Dx

Put y

x (1  z ) giving dy

upper limit for y

> f ( x )@

1 * (X )

 xdz . Then ( x  y )

x becomes z

x 1

X ³ ( x  y )  f ( y ) dy

X !0

0dcdx

(2.96)

c

xz and for y

c the new lower limit is z

( x c ) x

, and the new

0 . The above integral (2.96) with changed variables substituted is: X

c Dx

> f ( x )@

0

1

X 1

³

* (X ) ( x c ) / x X

( x c ) / x

* (X )

³

x

( xz )

 f ( x  xz )   xdz (2.97)

z

X 1

 f ( x  xz ) dz

0

Therefore, we get an identity that is as follows: X c Dx

X

c Ix

> f ( x )@

> f ( x )@

X

( x c ) / x

* (X )

³

x

X 1

 f ( x  xz ) dz

0

X

( x c ) / x

* (X )

³

x

z

0

(2.98)

z

X 1

 f ( x  xz ) dz

68

Chapter Two

2.10.3 Application of the obtained identity for acquiring a fractional integration of a power function We apply (2.98) for c

0 and for function f ( x )

x

X 0 Dx

P

with P ! 1 , from the above to get: 1

X

x

X P P z  x (1  z ) dz ³ * (X )

ª¬ x P º¼

1

0

x

P X 1

* (X )

³

z

X 1

x

P

(1  z ) dz

* (X )

0

* ( P  1)

P X

B(X , P  1)

(2.99)

P X

x * ( P  X  1) The same expression we obtained earlier too with the beta function, in Section 2.6.

2.10.4 Application of the obtained identity for acquiring fractional integration of the exponential function in terms of Tricomi’s incomplete gamma function at

Let us try to find a fractional integral of order v ! 0 and of exponential function f (t ) integration operation as: v 0 Dt

Make the substitution x

t

1

ª¬e at º¼

e . We write the fractional

(t  u ) * (v ) ³

v 1 au

e du

(2.100)

0

t  u , then the integral expression is changed as follows: v

0 Dt

e

ª¬e at º¼

t

at

x * (v ) ³

v 1  ax

e

dx

(2.101)

0

Clearly, this integral is not an elementary function, but is related to Tricomi’s incomplete gamma function and thus the fractional integral of the exponential function is expressed as follows: v 0 Dt

*

where Tricomi’s incomplete gamma function is J ( a , x )

ª¬e at º¼ t v e at J * (v, at )

x a *(a)

³

x

e

0

Section1.13. We will use this result in detail in the next chapters.

y

y

(2.102)

a 1

dy , which we discussed earlier in

2.11 Fractional integration of the analytical function The Riemann-Liouville fractional integral is expressed in the following convolution integral: D

0 Dx

1

> f ( x )@

x 1

D ( x  u )  f (u ) du ³ * (D ) 0

1 * (D ) This expression (2.103) is achieved by the simple substitution ( x  u )

x

(2.103)

1

D ³ y  f ( x  y ) dy 0

y.

An analytic function is infinitely

differentiable such that the Taylor series at any point of x0 in its domain is the following (Section 1.4): f

T ( x)

¦ n 0

f

(n)

( x0 )

n!

n

( x  x0 ) ,

f

(n)

n

( x0 )

d f ( x) dx

n

D f ( x)

n x x0

The series T ( x ) represents the function f ( x ) in the neighborhood of x0 such that x

x x0

(2.104)

x0  'x .

2.11.1 Fractional integration related to a series of ordinary whole derivatives of an analytical function We write the following steps (with n neighborhood of x0 such that x

0,1, 2, 3... i.e. positive integers), representing the function f ( x ) in the

x0  'x , by using (2.104):

Fractional Integration f

¦

f ( x0  'x )

f

(n)

( x0 )

n!

n 0

f ( x0 )  f f

...

f

....

(1)

f

(2)

( x0 )

2!

2

( x  x0 ) 

n

( x  x0 )  ... f

( x0 )  'x 

( x0 )

 'x

(2)

(2.105)

( x0 )

2!

 'x

2



n

 ... n!  y we can re-write the above (2.105) series in the following way:

x and 'x



f ( x  y)

( x0 )

(n)

n

( x0 )( x  x0 ) 

n!

f ( x0 )  f

With a change in variables x0

( x  x0 )

(1)

(n)

69

(1)

f ( x)  f



f



( x) ( y ) 

( x)

2

( y )  ...( 1)

2! 2

D f ( x)



1

(2)

f ( x)  D f ( x) y 

2

f

n

(n)

( x)

n!

(2.106)

n

D f ( x)

n

n

y  ...

n

y  ...( 1) y  ... 2! n! In the above derivation (2.106), assuming f ( x ) is analytic we thus write a power series expansion for f ( x  y ) as follows: f

f ( x  y)

¦

> f ( x )@

1 * (D )

³

x

0

y

D 1

 f ( x  y ) dy D

0 Dx

( 1)

k

D f ( x)

k

k

y (2.107) k! k 1 We substitute the above series expansion of f ( x  y ) (2.107) into the fractional integration formula (2.103), i.e. D 0 Dx

f ( x) 

which yields the following steps: x

k f § D f ( x) k ·  f ( x ) y ¸ dy ¦ ¨ * (D ) ³0 k! k 1 © ¹ x x k f 1 1 D f ( x) · D 1 D 1 k § y f ( x )dy  y ¨¦ ¸ dy * (D ) ³0 * (D ) ³0 © k 1 k! ¹

1

> f ( x )@

y

D f ( x) ª y º

D 1

x

k

f

1

D f ( x) ª y

»  * (D ) ¦ k 1

«

* (D ) ¬ D ¼ 0

§ ¨ * (D ) © 1 § ¨ * (D ) © 1

f ( x)

D

D

x 

f

¦

D

x 

f

(1)

k! ( x)

D k

D 1



f

(2)

( x)

x

D 2

·

 .... ¸

2!(D  2) ¹ In the above steps (2.108) we have interchanged the order of summation and integration, assuming uniform convergence of the infinite series in the interval of interest. Thus, we have an interesting expression relating to the fractional integration of the function to a series of ordinary whole derivatives of an analytical function, i.e. D

0 Dx

> f ( x )@

D

1 § f ( x)

¨ * (D ) © D

D 1

x

(2.108)

· ¸ D k ¹

D f ( x) x

k 1

f ( x)

º «D  k » ¬ ¼0

k!

k

x

D k

D

x 

f

(1)

( x)

D 1

x

D 1



f

(2)

( x)

2!(D  2)

x

D 2

·

 ... ¸

(2.109)

¹

2.11.2 Application of the Fractional integration of an analytical function to an exponential function for integer order integration Let us put D

1 and see what happens for f ( x )

x , with f (1) ( x ) following steps:

f

(2)

( x)

...

x

x

x

e . This exponential function ( e ) is an analytical function for all

e . Therefore, using the above expansion (2.109) for

D 0 Dx

> f ( x )@

we get the

70

Chapter Two (1) (2) ª º f ( x) 2 f ( x) 3 x  x  ....»  f ( x) x  > f ( x )@ « * (1) ¬ 2 2!(3) ¼ 2 3 ª º x x 1 ª x º « xe x  e x  e x  ...» 0 Dx ¬ e ¼ 2 3! ¬ ¼ 2 3 º x x x ª   ...» e «x  2 3! ¬ ¼ 2 3 º x x x ª   ...» e «1  1  x  2 3! ¬ ¼ 2 3 § ·º x x x ª   ... ¸» e «1  ¨ 1  x  2 3! ¹¼ ¬ © x x e ª¬1  (e ) º¼

1

1 0 Dx

(2.110)

x

e 1 1

Indeed the above is true: 0 Dx

ª¬e x º¼

³

x

0

y

x

e dy

e  1 is the classical integration of an exponential function.

2.11.3 An alternate representation and generalisation of the Riemann-Liouville fractional integration formula of analytical function From the discussion and observations of Sections 2.11.1 and 2.11.2 we can now generalise and proceed to discuss the alternate representation of

D

a Dx

>M ( x)@ , where

M ( x ) is a real analytic function, i.e. this function has a convergent

power series expansion in the interval a d y d x of interest. While D  0 this permits us to use the RiemannLiouville (RL) definition, which is as follows: D a Dx

>M ( x)@ >M ( x)@ D >d( x  a)@ d

D

x

1

M ( y )dy

D * ( D ) ³ ( x  y )

xa

1

1

³

* ( D )

a

M ( x  X )dX

0

X

D 1

(2.111)

We have used X { ( x  y ) . Upon the Taylor expansion of M ( x  X ) about the point x , as we did earlier in (2.106), we write the following: (1)

M ( x  X ) M ( x )  XM ( x ) 

X

2

(2)

k

f

¦

( x )  ....

k

( 1) X M

(k )

( x)

(2.112) k! 2! k 0 The above representation (2.112) involves no remainder since we have assumed M ( x  X ) to have convergent power series. We insert this expansion (2.112) into the RL formula (2.111) and write the following:

M

>M ( x)@ a Dx >M ( x )@ D >d( x  a)@ D

d

D

1 * ( D ) f

¦

xa f

³¦

* ( D ) k

³ k

( 1) ( x  a )

X k D

D 1

M

M ( x  X )dX X

0

( 1) X M

0 k 0

k

xa

1

(k )

( x)

k!

(k )

D 1

dX

(2.113)

( x)

* ( D )( k  D ) k ! k 0 We note here that the analyticity argument in D may be used here to establish the above obtained formula (2.113) for all D positive or negative (differentiation or integration) even though this was derived for D  0 ; (integration) which we will formally elaborate upon subsequently, in a later chapter.

2.12 A simple hydrostatic problem of water flowing through the weir of a dam with the use of a fractional integration formula Here we take a simple problem, where we determine the shape of a notch or a weir of a dam as a function depending on the water flow rate through it. Figure 2.3 gives a side view of the dam and water flow (on the X-Z plane) and a front view showing the shape of the notch (on the Y-Z plane). Figure 2.4 gives the details of the function i.e.

Fractional Integration

y

71

f ( z ) that describes the shape of the notch. In Figure 2.3 the water element I is at position ( x0 , y0 , z0 ) at height

z0 , whereas the water level height is z

h , which keeps element-I under hydrostatic pressure from the water height

above it ( h  z0 ) , apart from the atmospheric pressure ( Patm ) that is always acting. After some time element-I is at position II, that is (0, y0 , z0 ) , which means element-II of the water mass has no pressure from the water head above it and is free from the atmosphere; it is just on the verge of falling off. Side View

z

Front View

z

Dam Wall

x

y

z

z h I

II

z0

y x0

0

0

x

y0

Figure 2.3: Water flowing through a dam weir

Consider element I and element II as unit volumes with density U flowing with respective velocities as shown in Figure 2.3. Here we write pressure heads at I and II from Bernoulli’s law of hydrostatics, given in the following equation: 1 1 2 2 (2.114) PI  U gz0  UVI PII  U gz0  UVII 2 2 2

In the above equation (2.114) the expression g is the gravitational constant (acceleration | 10m / s ); VI is the velocity of element-I, which is far from the opening; and VII is the velocity of element-II gushing out of the notch. Obviously, this gushing out of the element’s velocity will be much greater than the velocity of element-I, which is far from the opening. Thus, we can assume VI  VII , and make an approximation as VI | 0 . The pressure is

PI v ( h  z0 )  Patm and PII v Patm . Therefore, we have PI  PII

( U g )( h  z0 ) . Using all these arguments and

substituting these in the obtained hydrostatic equation i.e. PI  U gz0  12 UVI

2

2

PII  U gz0  12 UVII ; we naturally get

the following: VII ( z0 ) VII ( z )

2 g ( h  z0 ) 2 g (h  z )

1

Shape of the notch a function depending on flow rate of water

z

y

f ( z)

h

dz

y

0

y

Figure 2.4: A function describing the shape of the notch of a dam that depends on flow rate

2

(2.115)

72

Chapter Two

Let the profile of the notch on the Y-Z plane as depicted in Figure 2.4 be represented by the function y f ( z ) . Looking at Figure 2.4, we get the elemental area of a section of the notch as follows: (2.116) dA 2 ydz 2  f ( z ) dz From this area element, the elemental flow rate is velocity at II times are dA , that is: dQ

VII  2 f ( z ) dz

VII dA

2 2 g (h  z )

1

2

(2.117)

 f ( z ) dz

The flow Q as a function of h is: h

Q

h

³ dQ

³

2 2 g (h  z )

0

Let us write the fractional integral of the order

3 2

1

2

 f ( z ) dz

(2.118)

0

by the Riemann-Liouville formula, that is:

 32 0 Dx

> f ( x )@

x

1

³

*  32

( x  y)

 32 1

 f ( y ) dy

0

x

1

1

* 1 

1 2

( x  y )  f ( y ) dy ³ 0

(2.119)

x

1

1

*  12

1 2

2

*  12

³ ( x  y )  f ( y ) dy 2

S

0

x

2

1

( x  y )  f ( y ) dy S ³ 2

0

Comparing the obtained expression for Q in (2.118) i.e. Q ( h )  32 0 Dx

> f ( x )@

2 S

³

x

0

( x  y)

1

2

2 2g

³

h

0

(h  z )

1

2

 f ( z ) dz

with the expression of

 f ( y ) dy , in (2.119) we conclude the following integral (rather than the fractional

integral) equation relating to the flow rate Q ( z ) and the notch-profile f ( z ) :  32

Q ( h)

> f ( h )@

2 gS 0 Dh

For example if the notch profile is a linear function say y

Q( z )

 32

2 gS 0 Dz

> f ( z )@

(2.120)

K1 z , then flow rate as a function of z is

f ( z)

5

2

Q ( z ) k1 z 2 . If the notch profile is curved as in y k 2 z , the flow rate is Q ( z ) k 2 z . Thus, we have used the Riemann-Liouville fractional integration formula to relate the flow rate to the shape of the notch.

2.13 Introducing an initialising process of fractional integration Consider a function f ( x ) such as it starts from x

0 for points x d a . Consider two processes of

a that is f ( x )

> @

fractional integration of f ( x ) , the first is in the interval a , x

> @

and the second is in the interval c, x such that

c ! a . These two integrals are represented as follows: D a Dx

> f ( x )@

1

x 1

D ( x  y )  f ( y ) dy ³ * (D ) a

D c Dx

> f ( x )@

1 * (D )

x 1

D ³ ( x  y )  f ( y ) dy c

In order to have the continuation of the integration process from c ! a to D

integration process c Dx

(2.121)

x ! c ! a , we have to add e.g. \ to the

D

such that it equals the process a Dx . Therefore, for x ! c ! a , we write the following: D c Dx

Thus, we have \ as follows:

> f ( x )@  \

D a Dx

> f ( x )@

(2.122)

Fractional Integration

73

> f ( x)@  c DxD > f ( x)@

D a Dx

\

c

1 * (D )

D 1

³

( x  y)

(2.123)

 f ( y ) dy

a

What we obtained, \ as \ ( x ) , is a function of the independent variable x , and not a constant of integration. Here \ is not a ‘psi’ function, as we have defined and used earlier (Section 1.16). This symbol \ was used by Riemann, and thus we are retaining this symbol for the initialisation function. Therefore, unlike the addition of the constant of integration in the ordinary process of integer order integration, here in the case of the fractional order of integration one has to use the initialisation function. For D 1 , we have \ as:

\

1

c

c

1

D 1

( x  y )  f ( y ) dy * (D ) ³

11

( x  y )  f ( y ) dy * (1) ³

a

a

c

³  f ( y ) dy

(2.124)

K

a

a constant of integration in the case of an integer order for which the whole integration is recovered. Let us take an example with the function f ( x ) x with f ( x ) 0 for x  0 . The semi-integration of this function from x 0 to x 1 is, with the substitution of ( x  y ) z following on in order to get the initialisation function:

\ ( x)

 12 0 D1

> x@

( x  y) * ³ 1 2

x 1

1

1

1

S

³

ydy

0

( x  z )( dz ) z

x

1

 12 1

1

2

x 1

ª 23 z 2  2 xz 2 º ¼x S ¬ 3

1

1

(2.125)

ª 2 ( x  1) 2  2 x ( x  1) 2  23 x 2  2 xx 2 º ¼ S ¬3 3

2

3

1

1

ª 2 x 2  (2 x  1)( x  1) 2 º ¼ 3 S ¬3 The semi-integration of this function from x

3

1

0 to x is, with the substitution of ( x  y )  12

0 Dx

> x@

x

1

( x  y) * ³ 1 2

 12 1

z is the following:

ydy

0

0

1

* ³ 1 2

( x  z )( dz ) x

x

1

1

(2.126)

2

4

0

ª 2 z 2  2 xz 2 º ¼x * 1 ¬3 3

1

3 S

2

Now we do the same semi-integration of f ( x )

x from x  12 1 Dx

x

3

2

1 , as below:

> x@

x

1

( x  y) * ³ 1 2

 12 1

ydy

1

1

0

* ³ 1 2

1 * 2

1 2

( x  z )( dx ) x

x 1

1

2

(2.127) 0

ª 23 z 2  2 xz 2 º ¼ ( x 1) ¬ 3

1

1

ª( x  1) 2 (2 x  1) º ¼ 3 S ¬

74

Chapter Two

Thus, we get the initialisation function as follows:

\ ( x)

 12 0 Dx

1

> x@  1 Dx > x@

2

2

(2.128)

3

ª2 x 2  (2 x  1) ( x  1) º ¼ 3 S ¬

We have obtained the same initialisation function in the beginning of this section. We stop here and we will take up this discussion further in Chapter 5, where we will use this to state that fractional derivatives too require an initialisation function.

2.14 Can we change the sign of fractional order in a fractional integration formula to write a formula for the fractional derivative? In

the

D 0 Dx

Riemann-Liouville

> f ( x )@

1 * (D )

³

x

0

D 1

( x  y)

formula

for

fractional

 f ( y ) dy , we try to replace D D

0 Dx

integration

D

we

reproduce

as

with D , and write the following: x

1

> f ( x )@

* ( D )

³ ( x  y)

 f ( y ) dy

(2.129)

u substitution, the integral in (2.129)

x

1

> f ( x )@

D 1

0

The, * ( D ) is a finite for a non-integer with negative value. With the x  y becomes: 0 Dx

that

* ( D )

³u

D 1

 f ( x  u ) du

(2.130)

0

2.14.1 Divergence of the Riemann-Liouville integral for a change of sign of order of integration The integral

³

x

0

u

D 1

 f ( x  u ) du

will diverge near u

0 no matter how smooth the function f is.

For

generality’s sake take the upper limit of integration and make it f . Take a small number  so the integral is approximated near to the point u 0 by writing f ( x  u ) # f ( x ) in the following way: f

f

³u

D 1

0

 f ( x  u ) du # ³ u D 1  f ( x) du 

f

ª f ( x ) D º «¬ D u »¼ 

(2.131)

f ( x) D

D

As p 0 the above expression (2.131) diverges, thus making it uncomfortable with a change of sign of the order i.e. D. In equation (2.129),

D

a Dx

> f ( x )@

1 * ( D )

³

x

a

( x  y)

D 1

 f ( y ) dy

for D  0 , in the Riemann-Liouville fractional

integration formula worked on by Reisz in 1949, D was regarded as a complex variable; this integral converges for

Re >D @  0 and defines for a fixed f ( x ) an analytic function of D on the left half D -plane.

2.14.2 Using Reisz’s logic of analytic continuation to geta meaningful representation of the Riemann-Liouville fractional integration formula with a changed sign of the order of integration

> @

Although the integral i.e. RL formula diverges for Re D t 0 (2.131) a meaning may be attached to the RL operator D

a Dx

> @

defined above in (1.129), by proper analytic continuation across the line Re D

0 provided the function

f ( x ) is sufficiently differentiable in a d y d x . In fact if the function is n times differentiable the formula D

a Dx

n 1

> f ( x )@ ¦ k 0

( x  a)

D  k

f

(k )

(a)

* ( D  k  1)

> @

D n

 a Dx

ª¬ f ( n ) ( x ) º¼

(2.132)

defines an analytic function of D for Re D  n and thus provides a valid analytic continuation. This above formula (2.132) will be clear shortly by following explanations and steps.

Fractional Integration

75

For say D  1 , considering D as a real number, then we write the above (2.132) expression as:

Well, we know that if D  0 , then 1 *   (D 1)

we have (D  1)  0 , then

D

a Dx

³

x

a

> f ( x )@

( x  y)

1 * ( D )

 (D 1) 1

( x  a)

> f ( x )@

D

a Dx

x

³

a

D

f (a)

* ( D  1)

( x  y)

D 1

 f ( y ) dy

D 1

 f ( y ) dy

a Dx

D 1

 a Dx

ª¬ f (1) ( x ) º¼

(2.133)

is valid. For say D ! 0 but D  1 ,

f ( x ) is also valid; we recognise this (1  D )

as an order of fractional integration. We write the following steps carefully by applying integration by parts i.e.

³ (u )(v)dx

³

(u ) (v )dx 

³  ³ (u )dx v

(1)

dx : D 1

a Dx

1

x

*  (D  1) ³

f ( x)

a

1

* (1  D ) ³

x

a

( x  y)

( x  y)

D

 (D 1) 1

 f ( y ) dy

 f ( y ) dy

§ f ( y ) x ( x  y )  D dy · ³a 1 ¨ ¸ x ¨ ¸ dy D (1) * (1  D ) ¨  ³ ³ ( x  y ) dy  f ( y ) ¸ © a ¹ y x D 1 § · ( x  y) ¨  f ( y) ¸ ( D  1) y a ¸ ¨ 1 ¸ ¨ D 1 * (1  D ) · (1) ¸ ¨ x § ( x  y) ¨  ³a ¨ ( D  1) ¸  f ( y ) dy ¸ © ¹ © ¹





§ f (a) · D 1 (2.134) ¨ ( D  1) ( x  a ) ¸ 1 ¨ ¸ ¸ * (1  D ) ¨ x § ( x  y ) D 1 · (1) f ( y ) dy ¸  ¨  ³a ¨ ¸ © ( D  1) ¹ © ¹ 1 d Now we do one whole differentiation of (2.134) with respect to x , i.e. we operate Dx { dx and write the following steps:

1

Dx



D 1

a Dx

f ( x)



§ f (a) · D 1 ¨ ( D  1) ( x  a ) ¸ 1 d ¨ ¸ ¸ * (1  D ) dx ¨ x § ( x  y ) D 1 · (1) f ( y ) dy ¸  ¨  ³a ¨ ¸ © ( D  1) ¹ © ¹ § § f (a) · · D ( D  1)( x  a )  ¨¨ ¸ ¸ 1 ¨ © ( D  1) ¹ ¸ D 1 ¨ ¸ * (1  D ) · (1) d x § ( x  y) ( ) d f y y ¨¨  ³ ¨ ¸¸  ¸ a © dx © ( D  1) ¹ ¹ f (a) * (1  D ) 

( x  a)

1 * (1  D )

f (a) * (1  D )

D

(2.135)

x

1

a

( D  1)

³

( x  a)

D



d( x  y ) 1

dx

* (1  D ) ³

x

a

D 1

f

( x  y)

(1)

D



( y ) dy

f

(1)



( y ) dy

76

Chapter Two

We recognise the second term of the last line in the above steps (2.135) as the fractional integration of order (1  D ) for function f

(1)

( x ) that is:  (1D ) a Dx

1

ª¬ f (1) ( x )º¼

x

* (1  D ) ³

a

( x  y)

D

f

(1)



(2.136)

( y ) dy , 0  D  1

This RL formula (2.136) is valid as (D  1)  0 . In this way we get a clue, that if D ! 0 but is less than one, i.e. D  1 , we can write the following: D

a Dx



1

> f ( x )@

Dx

D 1

a Dx

f (a)



( x  a)

* (1  D ) ( x  a)

f ( x)

D

D

1

* (1  D ) ³



f (a)

x

( x  y)

a

D 1

D

f

(1)



( y ) dy

(2.137)

(1)

D !0 D 1  a Dx ª¬ f ( x ) º¼ * (1  D ) For D ! 0 and D  2 , by repeating the above procedures i.e. (2.135) and (2.136), we will get the following expression:



2

> f ( x )@

D

a Dx

Dx

( x  a)

D

D 2

a Dx

f (a)



* ( D  1) Continuing this for D  n we get the following expression: D

a Dx

n

> f ( x )@

Dx



( x  a)

D n

f (a)

n 1

¦

( x  a)

D 1

f

(1)

(a)

* ( D  2)

* ( D  1) .... 



( x  a)

a Dx

D

f ( x)



( x  a)

D 1

f

f

(1)

(a)

* ( D  2) f

( n 1)

(a)

* ( D  n )

( x  a)

 a Dx

(2.138)

ª¬ f (2) ( x )º¼

> f ( x ) @

D  ( n 1)

D  k

D 2

(k )

(a)



D n

 a Dx

(2.139)

ª¬ f ( n ) ( x ) º¼

D n

 a Dx

ª¬ f ( n ) ( x ) º¼

* ( D  k  1) k 0 This analytic continuation is very analogous to the way one extends the definition of the gamma function on the complex plane (Section 1.10). This is not surprising since the gamma function plays such a fundamental role in defining the fractional integrals (and the derivatives too). We shall make frequent use of this analyticity in D to

simplify the discussions on the properties of

D

a Dx

, the defined RL operator for general order D , a real number ( \ ),

positive as well as negative.

2.14.3 Attaching meaning to the Riemann-Liouville fractional integration formula for a positive order elementary approach We prefer to take a different and more elementary approach attaching a meaning to the RL operator for D t 0 . The formula

D

a Dx

> f ( x )@

1 * ( D )

fractional integration.

³

x

a

( x  y)

D 1

It is extended to D t 0 , by insisting that

 f ( y ) dy

n a Dx



D

a Dx

for D  0 will be retained only for D  0 , indicating

> f ( x ) @

n D a Dx

> f ( x )@

is satisfied for all positive integers n

and that all D also be satisfied by the RL integral. That is, we shall require that: D

a Dx n

in (2.140) where Dx {

dn dx n

> f ( x)@ { Dxn  a DDx n > f ( x)@

(2.140)

is n -th derivative (ordinary whole derivative operator). Choosing the positive integer i.e.

n so large that we have D  n  0 ; the relationship

D

a Dx

> f ( x)@ { Dxn  a DDx n > f ( x)@ is for all D .

we choose D equal to a negative integer say  n , and then we have generates n -fold integration. If we chose n 1 and D

n a Dx

> f ( x )@

0 , we obtain the following:

1 *(n)

³

x

a

( x  y)

n 1

We see that if

 f ( y ) dy , this

Fractional Integration 0 a Dx

77 1

> f ( x )@

Dx



1 a Dx

> f ( x ) @

(2.141)

f ( x)

Courant and Hilbert in 1962 define semi-derivatives by use of the following expression: 1

2

1

This (2.142) is the same as we described for

1 d § d 2 > f ( x )@ ·

> f ( x)@ 1 d x f ( y )dy ³ S dx a x  y >d( x  a)@ D n D n > f ( x ) @ a Dx > f ( x ) @ { Dx  a Dx d

¨ ©

 12

dx ¨ >d( x  a )@

2

¸ ¸ ¹

(2.142)

i.e. the RL definition (2.140). A

careful integration by part for (2.142) shows the following: 1

> f ( x )@ >d( x  a)@ d

2

1

1

2

d

x

dx ³

S

f ( y )dy x y

a

1

f (a)

S

xa

x

1

f

³ S



(1)

( y )dy

(2.143)

x y

a

1

§ d  2 ª¬ f (1) ( x ) º¼ · ¸ ¨ 1 x  a ¨ >d( x  a )@ 2 ¸ © ¹

f (a) *  12 Equation (2.143) agrees with the earlier expression (2.139) to give D

a Dx

n 1

> f ( x )@ ¦

( x  a)

D  k

f

(k )

(a)

* ( D  k  1)

k 0

D n

 a Dx

ª¬ f ( n ) ( x ) º¼

as it must, for D 12 and n 1 . The semi-derivative defined in this way requires the knowledge of f will discuss in Chapter 3.

(1)

(2.144)

( x ) . This we

2.15 Fractional integration of the Weyl type

> @

We have developed the concept of the Riemann-Liouville fractional integration for a function f : a , b o  , a  x  b where we are doing integration from a to x , from the left side and proceeding towards the right. That is also called a left integration, if the integration axis is time we say that we are evolving the integration from past to present, also called a forward integration. This is a causal way of thinking here in that the past states contribute to the present states. When a f , we have the following representation: D

f I x

> f ( x )@

D f Dx

x

1

> f ( x )@

1

D ³ ( x  y )  f ( y ) dy

* (D ) f

(2.145)

2.15.1 Weyl transform of the fractional integration of a function on future points to infinity We have backward integration that is from the present point to all future points, that is from x to f of a function.

> @

This fractional integration is defined as the Weyl (1917) fractional integration as follows for Re D ! 0 : x I f > f ( x )@

D

D

xWf

> f ( x )@

def

f

1

1

D ( y  x )  f ( y ) dy ³ * (D )

(2.146)

x

The above description is also called the Weyl transform. D

The RL fractional integration of order D ! 0 is f Dx Here we put u

> f ( x )@

1 * (D )

v to get the following: D

f Dx

> f ( x )@

1

³

x

f

D 1

( x  u)

 f ( u ) du .

x 1

D ( x  v )  f (  v ) (  dv ) ³ * (D ) f

( 1) Now we put in the above x

 y and write the following:

1 * (D )

x

D 1

³ ( x  v )  f (  v ) dv

f

(2.147)

78

Chapter Two

D f D y

y

( 1)

> f (  y )@

1

D ³ ( v  y )  f (  v ) dv

* (D ) f

(2.148)

f

1

D 1

³ (v  y )

* (D )

f ( v )dv

y

Call f (  y ) g ( y ) , then the above formula (2.148) with an obvious change in notation becomes Weyl fractional integration. The RHS as described in the abovementioned definition (2.148) is a Weyl transform. D f D y

f

1

> g ( y )@

1

D (v  y )  g (v ) dv ³ * (D ) y

D yWf

The above expression (2.149) of

D yWf

1

> g ( y )@

(2.149)

f

D 1

³ ( v  y )  g ( v ) dv

* (D )

y

> g ( y )@ is a Weyl fractional integration of the right sense. This is a non-causal

type of integration, as it requires the point ahead (or future points), whereas the classical RL integration is of a causal type, which is evaluated on the past points. Put in the Weyl fractional integration formula (2.149) v  y will now be z 0 to z f , and we thus get:

z , so dv

D yWf

dz ; the limit of integration v

1

D (v  y )  g (v ) dv ³ * (D ) y

D 1

* (D )

³ z  g ( z  y ) dz 0

The general Weyl integration can be seen in the following steps with substitution y

> f ( x )@

(2.150)

f

1

D xWc

f

f

1

> g ( y )@

y to v

x ( z  1) and dy

xdz

c

1

1

D ( y  x )  f ( y ) dy ³ * (D ) x

D xWc

> f ( x )@

1

D 1

³

* (D )

( xz )

 f ( xz  x)  xdz

(2.151)

0

D

(c x) / x

* (D )

³

x With c

(c x) / x

z

D 1

 f ( x  xz ) dz

0

f , we get:

> f ( x )@

D

xWf

x

f

D

1

D z  f ( x  xz ) dz ³ * (D )

(2.152)

0

2.15.2 Weyl fractional integration of the power function We take f ( x )

x

p

, with p ! 0 , and find the Weyl fractional integral as follows in these steps: D

xWf

ª¬ x  p º¼

x

f

D

* (D ) x

D

³z

D 1

( x  xz )

p

dz

0

f

z * (D ) ³

D 1  p

x

(1  z )

p

dz

(2.153)

0

x

 p D f

z * (D ) ³

D 1

(1  z )

p

dz

0

We recognise that

³

f

0

z

D 1

(1  z )

p

dz

B(D , p  D )

* (D ) * ( p D ) *( p)

, we derived this in an earlier chapter describing the

beta integral (Section 1.15). By using this formula for the beta function, we get the following:

Fractional Integration

79  p D

§ * (D )* ( p  D ) · ¨ ¸ * (D ) © *( p) ¹

x

D ª  p º¼ xWf ¬ x

*( p  D )

x

*( p) Now we put D

(2.154)

 p D

1 , so we get: 1 xWf

* ( p  1)

ª¬ x  p º¼

*( p) 1

* ( p  1)

 p 1

x

( p  1)* ( p  1)

x

 p 1

(2.155)

 p 1

x p 1 The above expression (2.155) is a classical one-whole integration, i.e.: f 1 xWf

p

ª¬ x º¼

³

y

p

y

dy

(  p  1)

x

1 p 1

x

y f

 p 1

 y x

x

 p 1

(  p  1)

;

p!0 (2.156)

 p 1

2.15.3 Weyl fractional integration of the exponential function We apply this Weyl integration to find the fractional integration of e D xWf

Substituting (u  x )

y p

, du

dy p

ª¬e

 px

 px

f

1

º¼

D 1  pu

(u  x ) * (D ) ³

f

1

ª¬e  px º¼

D

y

 px

D

1 p We have used the definition of the gamma function, * (D )

³

0

e

1

e

(2.157)

dy

 ( y  px )

p

0

p * (D )

f

D 1

D * (D ) ³ p

e

 px

du

, we write: xWf

do one-whole integration of e

e

x

D

e

f

³y

D 1  y

e dy

(2.158)

0

 px

 y D 1

y

dy , in the above derivation (2.158). Now we

in Weyl’s way, giving the following: f 1  px xWf e

³ x

e

 py

dy

e

 py f

( p)

x

e

 px

p

(2.159)

We note that the Weyl fractional integration of exponential function (2.158) returns an exponential function like that in the case of classical integration. Whereas we observed that the Riemann-Liouville fractional integration returns higher transcendental functions on fractionally integrating exponential functions as derived in (2.71), (2.73), (2.80), and (2.102).

2.16 Applications using the Weyl fractional integration With reference to Figure 2.5, we need to find the potential function M ( r ) for the two perpendicular long wires with 

negative charges, such that the force on a test charge Q kept at x wire AB; the condition is thus FCD

2 FAB .

L from the wire CD is twice the force from the

80

Chapter Two

C ds r

ds

s

T

A

Q+

B

0

L

L

y

X

s

Ls

D

Figure 2.5: Two wires with negative charges attracting a test charge with force

Now take an infinitesimal small segment on AB i.e. ds . The infinitesimal force due to this element on the test charge Q is dFAB M ( L  s ) ds , thus FAB , which is directed in X-direction is: f

³  M ( L  s ) ds

FAB

(2.160)

0

From Figure 2.5, we have y L  s , thus dy have the following expression:

0 implies y

ds , and s

L . With these changes in variables we

f

³  M ( y ) dy

(2.161)

c Similarly an element in segment CD has a force directed in r direction meaning dFCD

M (r ) ds , the X-component

FAB

L

of this force is dFCD

M ( r )  cos T (ds ) . The total force due to wire CD on the test charge in X-direction is therefore

as follows: f

³ M (r )  cos T ds

FCD

(2.162)

f L r

From the geometry of Figure 2.5, we write cos T

; therefore (2.162) should be changed to the following: f

FCD

L

³

M (r ) r

f

Again from the geometry of Figure 2.5 we have r

2

2

2

s  L therefore ds

f

ds

2L

³

M (r )

0

r

ds

(2.163)

 rs dr , this substitution gives the limit of

integration of FCD in the variable r , from L to f for the integral describing FCD , i.e. 2 L

f

M ³  ds as in the 0

(r ) r

above steps (2.163), when the limit of integration of s is from 0 to f . Therefore, we get the following expression: f

FCD

2L

³ L

In the above expression (2.164) we’ve now put r

2

k , so 2 rdr

dk , or dr

M (r ) 2

2

dr

(2.164)

r L 1 2 k

2

dk ; and L

expression of integration describing FCD with these new substitutions to get the following steps:

x , and written the

Fractional Integration

81

f

FCD

³

2 x

M ( k ) §

1 · ¨ ¸ dk kx ©2 k ¹

x

M ( k ) (k  x)

f

³

x

 12

k

x f

³  g k k  x

x

(2.165)

dk

 12

dk

x

In (2.165) we call k

M ( k )  g ( k )

1 k

f ( k ) , this is a function of k and we do the integration from k

x to

x to f and obtain the following steps:

f instead of from

f

FCD

x

³  f (k ) (k  x)

 12

dk

x



x *

1 2

ª 1 f º 1 « 1 ³  f ( k ) ( k  x )  2 dk » ¬* 2 x ¼

W

> f ( x ) @

 12 x f

x S

In the above derivation (2.166), we used the Weyl fractional integration definition (given in 2.149), with D

*  12

1 2

, and

S to give:  12

xWf 1 k

With the substitution f ( k )

FAB

(2.166)

M ( k ) , or M ( y) k  f (k ) 2

f

1

> f ( x )@

³

*  12

(k  x)

 12 1

 f ( k ) dk

(2.167)

x

we write the obtained force from AB, which is

f

³ M ( y ) dy , as: L

f

³ k  f (k ) dk 2

FAB

(2.168)

x

Put in the above expression (2.168), [

2

k that gives the following expression:

1

f

1

f ([ )d[ 2³

FAB

2

x

Our condition is FCD

 W > f ( x ) @ 1 x f

(2.169)

2 FAB , and that gives the following integral equation:

Sx

W

> f ( x ) @

 12 x f

1 xWf

> f ( x )@

(2.170)

Now physically we know that an attractive force between charge Q and the AB and CD wires decreases as r increases. Therefore, we can write f ( x ) in the following form:

f ( x)

a1 1 q

x



a2 x

2 q

f

¦a x



a3 x

3 q

 ...

q!0 (2.171)

(nq )

n

n 1

D

Use xWf

ª¬ x  p º¼



* ( p D ) *( p)

x

 p D

from (2.154) which is the Weyl fractional integral of x

f ( x ) written in the series form as above in the integral equation, and write the following steps:

p

, and substitute the

82

Chapter Two

S ( x)

1

2

W

 12 x f

f

¦a x

f ( x)

> f ( x ) @

1 xWf

> f ( x )@

(nq )

(2.172)

n

n 1

S ( x)

1

f

2

¦

an

* n  q 

n 1

1 2



1

*(n  q)

x

f

1

¦ n  q 1 x

n  q  12

n 1

1 n  q 1

Equating the like powers on both sides of (2.172) and along with the simplification by using x  * ( x ) to the following steps:

*  12 *  n  q 

1 2



1

*(n  q) *

1 2

n  q 1

* n  q  1 2

1

( n  q  1)* ( n  q  1) *  12 *  n  q  We get the above equality if n  q

3 2

, that is *  12 * (1)

* ( x  1) leads

1 2



(2.173)

n  q 1

* ( n  q  1)

*  12 . Therefore, we write:

M ( x )

a

f ( x)

x2 x We get the potential function as follows the expression, i.e. an inverse square law: a M( x) M (r ) x This shows an application of the Weyl fractional integration formula.



(2.174)

3



a r

(2.175)

2

2.17 Use of the kernel function in a convolution integral in the fractional integration formula 2.17.1 Defining kernel functions for forward and backward fractional integration From the Weyl representation, we write the forward and backward fractional order integration formula as follows: D

f I x

D

x If

> f ( x )@

1

D ³ ( x  y )  f ( y ) dy

* (D ) f

(2.176)

f

1

D 1

* (D )

The kernel in the above integration (2.176) is a power function, namely  kD

x

1

> f ( x )@

³ ( y  x )  f ( y ) dy x

r kD

( x ) i.e.

( x )D 1 * (D )

, and

(  x )D 1 * (D )

respectively for

( x ) and  kD ( x ) . With this, we can write the left and right integrals as D

>

I r f ( x)

@

def f r kD ( x ) * f ( x )

³

r kD

( x  y )  f ( y ) dy

(2.177)

f

The  is for a left fractional integration operator i.e. D

x If

D

f I x

, and  is for the right fractional integration operator i.e.

. The kernel we define as: D 1

r kD ( x )

where u ( x ) is the Heaviside unit step function defined as u ( x ) x

x

D

D ln x

(r x)

* (D )

u (r x)

1 for x ! 0 and u ( x )

(2.178)

0 for x d 0 . We have

and the conventions that function ln x are real for x ! 0 . For D 0 , the kernel is the delta function at 0 , that is  k0 ( x )  k0 ( x ) G ( x ) . This is the usual kernel called the ‘power-law’ kernel as described in (2.178). e

Fractional Integration

83

2.17.2 Additive rule for fractional integrals We write the following convolution of kernels as: x

  kD ( x) *   k E ( x) ³   kD ( x  y )   kE ( y ) dy 0

x

³

D 1

( x  y)

y

* (D )

0

(2.179)

E 1

*( E )

dy

y x

z gives us dy xdz and new limits of integration for variable z from 0 to 1 . Then, substituting these changes in variables for the above steps (2.179), we get the following: Putting

1

1

  kD ( x) *   kE ( x)

x * (D )* ( E ) ³

D 1

D 1

(1  z )

( xz )

E 1

( xdz )

0

x

D 1

x

E 1 1

D E 1

x x

* (D )* ( E ) x 1

³ (1  x) 0

p 1 q 1

x

dx

z

xdz

0

(2.180)

1

D 1 E 1

³ (1  z )

dz

z

0

D  E 1

* (D )* ( E ) The integral

D 1 E 1

³ (1  z )

* (D ) * ( E )

B(D , E ) *( p )*(q ) *( pq)

B( p , q ) is called the beta function, and B( p , q )

; using this property (as

in Section 1.15), we get the following:

x

  kD ( x) *   k E ( x)

D  E 1

* (D  E )

 kD  E

(2.181)

( x)

With this derivation (2.180) and using (2.181) we imply that the semi-group additive law of exponents in a fractional integration operation is

 I  I I  D  D D

E

a x

a x

a

D x

a

D E

a x

E x

,

D, E ! 0

(2.182)

 (D  E ) a Dx

The above is an index law which holds true for fractional integration. This we will use further and formalise in Chapter-9, for fractional differentiation and integration composition laws.

2.17.3 The choice of a different convolution kernel defines different types of fractional integration operators We saw that fractional integration is defined via a convolution integral and the choice of kernel defines the type of fractional integration. We write in a generalised form the following convolution integrals: t a !t

D

> f (t ) @

³  kD (t  W )  f (W ) dW a

(2.183)

b

D

t

!b

> f (t )@ ³  kD (W  t )  f (W ) dW ,

D !0

t

Obviously, the kernels are kD (t  W )

1 * (D )

D 1

(t  W )

for the first one and kD (W  t )

1 * (D )

D 1

(W  t )

for the second

one, giving us our derived fractional integration formulas. We are deleting the plus and minus signs for the symbol of a kernel as used in previous discussion in (2.178). With the choice of a different kernel, we can have different types of fractional integration formulas.

Fractional Integration

D

a ! x ,> z , w@

D

1

x

D  z ( x )  z (t ) ³  w( x)  *(D )

f ( x)

a ! x ,> ln t ,1@

85

1

a

H D a Ix

f ( x)

1

f ( x)

x

* (D )

ln x D 1 ln t



(1)

w( t ) z ( t )

(2.190)

³   dt , a

 f ( t ) dt

f (t ) t

D !0

2.17.3.b(ii) The modified Erdelyi-Kober (MEK) fractional integration formula By choosing z (t ) operator:

V

t , w( t )

t

VK

and z

D

a ! x ,> z , w@

(1)

(t )

Vt

V 1

we obtain a modified Erdelyi-Kober (MEK) integration

1

x

D  z ( x )  z (t ) ³  w( x)  *(D )

f ( x)

1

a



w( t ) z

(1)

(t )

 f ( t ) dt (2.191)

D !0 MEK D a Ix

D

a ! x , ªtV ,tVK º f ( x ) ¬

¼

Vt

VK

x

D 1 V (1K ) 1

V V x t  ³ * (D )

f ( x)

t

a

 f ( t ) dt

With these different types of fractional integral operators, we will define several types of fractional derivative operators in the next chapter.

2.17.4 Generalisation of the fractional integration formula with respect to function Let us take f ( x ) On

the D

a ! x ,> z , w@

f ( x)

 z ( x)  z (a)

application 1

 w ( x )  * (D )

of

E 1

with E ! 0 and w( x )

the

x

generalised D 1

³  z ( x )  z (t ) a

D

a ! x ,> z ,1@



1.

formula

(2.187),

D !0

for

we

write

 f (t ) dt which gives us the following steps: 1  z ( x )  z ( t )  z ( t )  z ( t )  z ( a ) dt * (D ) ³ (1)

w( t ) z ( t ) x

f ( x)

i.e.

D 1

E 1

(1)

a

1

x

D  z ( x )  z (t ) ³ * (D ) a

1

z ( x)

1

E 1 § dz ( t ) · ¨ ¸  z ( t )  z ( a )  dz ( t ) © dt ¹ 1

(2.192)

1

D E  z ( x )  z ( t )  z ( t )  z ( a )  dz ( t ) ³ * (D ) z(a)

In the last step of the above derivation (2.192), we have integration with respect to z (t ) and hence we re-wrote the z ( a ) and upper limit as z (t ) z ( x ) . For the last integral in (2.192) we see that it is similar to an

lower limit as z (t )

RL fractional integration formula i.e.

D

a Ix

f ( x)

1 * (D )

³

x

a

D 1

(x  t)

 f ( t ) dt ,

 z ( x)  z (a)

with f ( x )

E 1

and

integration with respect to z ( x ) in this case. Therefore, we write the following formula: D

a ! x ,> z ,1@

ª z ( x )  z ( a ) E 1 º ¬ ¼ 1

z ( x)

1

1

D E  z ( x )  z ( t )  z ( t )  z ( a )  dz ( t ) ³ * (D ) z(a)

D

a I z ( x)

D P Using the result obtained in Section 2.6, i.e. c I x ª¬ x º¼ follows in this expression:

D

a I z ( x)

(2.193)

ª z ( x )  z ( a ) E 1 º ¬ ¼ * ( P 1) * ( P D 1)

( x  c)

ª z ( x )  z ( a ) E 1 º ¬ ¼

P D

, P ! 1, D ! 0 , we write the result as

*( E ) *( E  D )

 z ( x)  z (a)

E 1D

(2.194)

86

Chapter Two

2.18 Scaling law in a fractional integral We write the following law called the scaling law: D

a Ix

> f (O x )@

D

O



D Oa I x

> f ( x ) @

D

We apply the RL definition of I to f (O x ) then do changes in the variable i.e. O y u O x . This is demonstrated in the following steps, for the derivation: D

a Ix

> f (O x )@

(2.195)

O!0

z and later do the substitution

x

1

1

D ( x  y )  f ( O y ) dy ³ * (D ) a

Ox

D dz  x  O  f ( z ) §¨ ·¸ ³ * (D ) O ©O¹

1

1

z

a

Ox

1

* (D ) O³

D 1

(O x  z )

O

a

 f ( z ) §¨

1·

¸ dz ©O¹

D 1

(2.196)

Ox

1

1

D ³ (O x  z )  f ( z ) dz

D

O * (D ) O a

In the above derivation (2.196), for a D

0 Ix

> f (O x )@

O

D



D

0 Ix

> f ( x ) @

O

D

0 we have



D

0 IOx

D

0 Ix

> f (O x )@

O O

D

D

> f (O x )@

0 Ix

> f ( O x ) @ .

§ 1 u · D 1 u z f z z  ( ) ( ) d  ¨ ¸ ³ © *(D ) O a ¹

D



D Oa I x

> f ( x ) @ O

D



D

0 Ix

> f ( x ) @ ;

this is also the same as

The following are the steps to prove this result: x

1

f (O y )

D * (D ) ³ ( x  y )

dy ;

 1

z

Oy

a

Ox

1

f ( z)

§ dz · ¨ ¸ ©O¹

D * (D ) ³  O O

x  z  1

0

O

D O x

f ( z)

D * (D ) ³ (O x  z )

 1

dz ;

z

(2.197)

Oy

0

D O x

f (O y )  d(O y ) D ³ * (D ) (O x  O y )

O

 1

0

O Thus, D

0 Ix

we

> f (O x )@

obtained

O

D



D

0 IOx

an

> f (O x)@

important

D



D

0 IOx

> f ( O x ) @ D

identity,

namely a I x

> f (O x )@

O

D



D Oa I x

> f ( x)@

and

which is known as the scaling law of fractional integration.

2.19 The formula for fractional integration is derived from the generalisation of Cauchy’s integral theorem of complex variables Interestingly the derivative is in someway related to the integration process. For that we revised Cauchy’s integral theorem (in Section 1.5), and here we further generalise the same. Let f ( z ) be an analytic single valued function on a complex plane i.e. z , Z   . Cauchy’s integral theorem (1825) statement is the following:

f ( z)

1

2S i ³

C

f (Z ) (Z  z )

dZ

(2.198)

Fractional Integration

87

Im Z C

x z x

Re Z

a

$

Im Z C

u

x z x

Re Z

b

Figure 2.6: Contours of integration on the complex plane  The n  th differentiation ( n  ] being a positive integer) of f ( z ) from Cauchy’s integral formula is:

n

>

Dz f ( z ) We call Z

n!

@

f (Z )

2S i ³

C

(Z  z )

n 1

dZ

(2.199)

z a pole of order ( n  1) .

In a way Cauchy was first to state that differentiation of a function is related to the integration process in complex variable theory. For n as a positive integer, x z is a pole with C as a closed contour (as in Figure 2.6a). For n as a  positive real number (say D  \ ) we call x z a branch point (not a pole), having a branch cut (depicted in Figure 2.6b), and $ a loop (refer to Appendix E). Then Cauchy’s integral formula is generalised as the following expression, by using * ( n  1) n ! :

D

Dz

* (D  1)

> f ( z )@

f (Z )

³

D 1

C (Z  z ) 2S i We re-draw Figure 2.6b as a simplified loop, shown in Figure 2.7. We write the following:

D

c Dx

> f ( x )@

def

* (D  1) 2S i * (D  1)

dZ

(2.200)

( x)

³

(Z  x )

D 1

 f (Z ) dZ (2.201)

c

D 1

(Z  x )  f (Z ) dZ $ 2S i The above expression (2.201) is in a way defining fractional derivatives of the order D in terms of the contour integral in Cauchy’s formulation. We will do the contour integration now:

³

Im Z

J

A cx

L1

B

D

L2

F

xT z x

G Re Z

Figure 2.7: The loop redrawn

The loop ( $ ) is composed of the circle FGB ( J ),

$

segments DF ( L2 ) and segment AB ( L1 ), making

L1 * L2 * J . In addition, the complex quantity (Z  x ) we write in polar form as follows:

(Z  x ) We do the following arithmetic steps:

(Z  x ) e

iT

(2.202)

88

Chapter Two

(Z  x )

D 1

(Z  x )

( D 1) (  D 1)

e

iT ( D 1)

e

ln (Z  x )

e

( D 1)ln (Z  x ) iT ( D 1)

e

iT ( D 1)

(2.203)

e

ª( D 1)ln (Z  x )  iT ( D 1) ¼º

e¬ e

We note that Z  x

( x  Z ) refers to Figure 2.7, the point z

 D 1  ln (Z  x )  iT

x is always at the right side of any other variable Z ,

near or on the real axis, while also on segments AB or DF, thus Z d x , on L1 & L2 .

S , so, we do the following arithmetic steps:

For L1 we have T

(Z  x )

D 1

e

( D 1)ln( x Z )  (D 1) iS

e

e

 iS (D 1)

(x  Z)

( D 1)



(2.204)

S , therefore, we get the following relationship:

For L2 we have T

(Z  x )

D 1

e

( D 1)ln( x Z ) iS (D 1)

e

e

iS (D 1)

(x  Z)

( D 1)



(2.205)

We write the loop integral as in the following steps: ( x)

³

(Z  x )

D 1

c

 f (Z ) dZ ³L

(Z  x )

D 1

2

 f (Z ) dZ  1

D ³J (Z  x)  f (Z ) dZ D  ³ (Z  x )  f (Z ) dZ



 1

L1

(2.206)

( xr )

e

i (D 1)S

³

(x  Z)

D 1

 f (Z ) dZ

c



³J (Z  x)

D 1

 f (Z ) dZ

c

e

 iS (D 1)S

³ (x  Z)

D 1

 f (Z ) dZ

xr

Figure 2.7 shows the integration path as counter clockwise. The radius of the circle is r , so the path L2 is from point

c to point ( x  r ) ; whereas the path L1 is from ( x  r ) to point c . The circle FGB of Figure 2.7 has a very small radius, i.e. r p 0 . The paths considered in the segments AB and DF therefore almost overlap with each other and merge on the real axis (X-axis) and the Z -plane of Figure 2.7; but are on different Riemann sheets of the Riemann



surface for Z  x



D 1

Returning to expression FGB, set (Z  x )

(refer to Appendix E). Figure 2.7 shows only the primary Riemann sheet. * (D 1) 2S i

³$ (Z  x)

D 1

 f (Z ) dZ , we analyse it for

iT

re , for T change from S to S , in addition Z

D  0 . For integration on the circle iT

iT

x  re , and dZ

( riT )e dT . Now as

r p 0 , we write the following integral: S

³J (Z  x)

D 1

³S r

f (Z )dZ

D 1  i (D 1)T

e

iT

iT

f ( x  re )(ire )dT

(2.207)



and:

³J (Z  x)

D 1

f (Z )dZ

d r D ³

S

S

iT

f ( x  re ) dT ,

D 0

(2.208)

Notice that for a small circle r p 0 , and for D  0 the above integral (2.208) on the circle FGB tends to 0 , i.e.

lim r p0

³J (Z  x)

D 1

f (Z )dZ

0

(2.209)

Fractional Integration

89

Therefore, as r p 0 we are left with integration on L1 and L2 only, and thus we write the following steps: ( x)

³

(Z  x )

D 1

 f (Z ) dZ

c

§ i (D 1)S ( x  r ) · D 1 (x  Z)  f (Z ) dZ  ¸ ¨e ³ ¨ ¸ c lim ¨ ¸ c r p0 ¨ e  iS (D 1)S ( x  Z ) D 1  f (Z ) dZ ¸ ³ ¨ ¸ xr © ¹ x

e

i (D 1)S

³ (x  Z)

D 1

 f (Z ) dZ

c

c

e

 iS (D 1)S

³ (x  Z)

D 1

 f (Z ) dZ

(2.210)

x

e

x

i (D 1)S

 i (D 1)S

e

³ (x  Z)

D 1

 f (Z ) dZ

c

Use





sin( pS ) 1 eipS eipS 2i

( x)

³

x

(Z  x )

D 1

 f (Z ) dZ

2i sin S (D  1) ( x  Z )

³

c

³

( x)

c

(Z  x )

integral D 1

formula,

 f (Z ) dZ

 f (Z ) dZ

c

> f ( x )@

D

Therefore from definition (2.201) i.e. Cauchy’s

D 1

c Dx

and

by

using

2i sin S (D  1) D

c Dx

* (D 1) 2S i

³

x

c

the

(x  Z)

> f ( x )@

def

D 1

D 1

above

 f (Z ) dZ ,

derived

which is the generalised

expression

(2.210)

i.e.

 f (Z ) dZ we get the following expression for D  0 :

* (D  1) 2S i

³$ (Z  x)

( x)

³

(Z  x )

D 1

 f (Z ) dZ

c

 *(D  1)  sin S (D  1) x ( x  Z ) D 1

³

S

(2.211)

 f (Z ) dZ

c

Using one of the reflection properties of the gamma function from Chapter-1 (1.68), * (  x )* ( x  1) S csc(S x ) , we get the following simplification:  *(D  1) sin S (D  1)  *(D  1) sin(S  DS )

S

S

 *(D  1)   sin(DS )

(2.212)

S 

* (D  1)

1

S csc(DS )

* ( D )

Now we write the final formula as follows: D

c Dx

> f ( x )@

1

x

(x  Z) * ( D ) ³

D 1

 f (Z ) dZ ,

D 0

(2.213)

c

This is what is obtained by changing the sign of a fractional order in the RL fractional integration formula, as discussed previously, in Section 2.14.

90

Chapter Two

2.20 Repeated integrals as a limit of a sum and its generalisation 2.20.1 Limit of a sum formula to have classical repeated n  fold integration The following notation: 1

d f ( x)

{

1

>d( x  a)@

³

x

a

f ( y )dy

represents integration as an anti-derivative of a function from the start point of integration x

(2.214)

a . This notation we

discussed in Section 1.17. Let the interval of integration from a to x be divided into N slices such that hN are the slice sizes on an infinitesimal width.

xa N

Then we write: x

1

d f ( x) 1

>d( x  a)@

{

³

§ hN f ( x )  hN f ( x  hN ) 

· ¸ hN p 0 h f ( x  2 h )  .... f ( a  h ) © N N N ¹

f ( y )dy

lim ¨

a

(2.215)

lim ¬ªhN  f ( x )  f ( x  hN )  f ( x  2 hN )  .... f ( a  hN ) ¼º

hN p 0

The application of the above definitions (2.214) and (2.215); to double integration gives us the following:

d

2

f ( x) 2

>d( x  a)@

x1

x

³ ³ f ( x )dx

{ dx1

0

a

0

a

ª

§ f ( x )  2 f ( x  hN )  3 f ( x  2hN )  ·º ¸» © ....Nf ( a  hN ) ¹¼

2

lim «hN ¨ h p0 N

¬

§

2

lim ¨ hN

hN p 0

©

N 1

¦ ( j  1) f ( x  jh

N

j 0

We do one more step as follows, writing the triple integration:

d

3

x2

x

f ( x) 3

>d( x  a)@

(2.216)

·

)¸

¹

x1

³ ³ ³ f ( x )dx

{ dx2 dx1 a

a

0

0

a

§ 3 N 1 ( j  1)( j  2) · lim ¨ hN ¦ f ( x  jhN ) ¸ hN p 0 2 © j0 ¹

(2.217)

We notice that coefficients of summation are building up with the order of integration n as described below, and all the signs of the coefficients are positive:

§ j  n  1· ¨ j ¸ © ¹

( j  n  1)!

j !n ! With this, we may write the n -fold repeated integration as a limit of a sum as follows in this formula:

d

n

f ( x) n

>d( x  a)@

§

n

lim ¨ hN

hN p 0

©

(2.218)

N 1

· § j  n  1· f ( x  jhN ) ¸ ¸ j ¹ 0 ¹

¦ ¨© j

N 1 § · n § j  n  1· lim ¨  xN a ¦ ¨ f  x  j  xN a ¸ ¸ N nf j ¹ j 0© © ¹

(2.219)

The expression (2.219) is the formula for repeated integration in classical calculus, taking the limit of a sum.

2.20.2 Limit of a sum formula to have fractional integration: the generalisation of classical formula We take formula (2.219) as derived above using a generalisation formula for binomial coefficients by the gamma function (Chapter-1), that is

§ j  q  1· ¨ j ¸ © ¹

*( j  q) * ( q )* ( j  1)

(2.220)

Fractional Integration

91

We write the fractional integration of order q as follows:

d

q

§  xN a q lim ¨ N nf ¨ * ( q ) ©

f ( x) q

>d( x  a)@

N 1

*( j  q)

¦ *( j  1)

·



¸

f x  j  xN a ¸

j 0

(2.221)

¹

The above obtained formula (2.221) is well valid for any q , say we replace  q by D , then we write the following expression:

§  xN a D lim ¨ N nf ¨ * ( D ) ©

D

d f ( x) D

>d( x  a)@

·

N 1

*( j  D ) ¦ *( j  1) f  x  j  ¸¸ xa N

j 0

(2.222)

¹

We will discuss (2.222) in Chapters 3 and 4. Notice also that the above expression (2.222) involves only an evaluation of the function itself; no explicit use is made of the derivatives or the integrals of f ( x ) .

2.20.3 Using a generalised limit of the sum formula derivation of an additive composition rule n

We would like to establish Dx



D

a Dx

> f ( x ) @

n D a Dx

> f ( x)@ (called the additive composition rule) for all positive xa N

integers n and for all D . To establish this composition, let hN following:

> f ( x )@ D >d( x  a)@ d

D

§ hND

N 1

©

j 0

lim ¨

N nf

, and we will then write from (2.222) the

·

*( j  D ) f  x  jh ¸ ¦ * ( j  1) * ( D ) N

(2.223)

¹

We subdivide the interval that is a d y d x  hN into only N  1 equally spaced subintervals resulting in the following steps:



¼

D d ª f x  hN º

¬

§ hND

lim ¨

D

>d( x  a)@

N nf

©

§ hND

N 2

·

*( j  D ) f x  h ¦ * ( j  1) * ( D )

N

j 0

N 2

 jhN ¸

¹

·

*( j  D ) f  x  ( j  1) h ¸ ¦ * ( j  1) * ( D )

lim ¨

N nf

(2.224)

N

j 0 © ¹ D N 1 § h · * ( j  D  1) f  x  jhN ¸ lim ¨ N ¦ N nf * ( D ) *( j ) j 0 © ¹

On the differentiation of (2.223) and by defining

d dx

> g ( x )@

lim h

N p0

1

hN

 g ( x)  g ( x  hN )

we write the following

by making use of the above derived expressions, (2.223) and (2.224): D d § d > f ( x )@ ·

¨ D ¸ dx ¨ >d( x  a )@ ¸ © ¹

§

1

§ dD > f ( x )@ dD ª¬ f ( x  hN )º¼ · · ¸¸  D ¨ >d( x  a )@D ¸¸ d( ) x  a > @ © ¹¹

lim ¨ hN ¨

N nf

¨ ©

(2.225)

N 1 § h D 1 § § * ( j  D ) * ( j  D  1) · · · lim ¨ N ( ) f ( x ) D *   ¨ ¦ ¨ * ( j  1)  * ( j ) ¸ ¸ ¸¸ N nf ¨ * ( D ) j 0 © ¹¹¹ © ©

Then by making use of recurrence properties of the gamma function * ( x  1) x* ( x ) (as seen in Chapter-1), we obtain the following: * ( j  D ) * ( j  D  1) * ( D )* ( j  D  1) (2.226)  * ( j  1) *( j ) * ( D  1)* ( j  1) Using (2.226) in (2.225) gives us:

92

Chapter Two

§ dD > f ( x ) @ ¨ D dx ¨ >d( x  a )@ © d

· ¸ ¸ ¹

§ hND 1 § N 1 * ( j  D  1) ·· f ( x  jhN ) ¸ ¸ ¨ ¦ ¸ N nf ¨ * ( D  1) © j 0 * ( j  1) ¹¹ © lim ¨

d

D 1

(2.227)

> f ( x )@ D 1

>d( x  a)@ 1

So we have Dx



D

a Dx

> f ( x ) @

1D a Dx

> f ( x)@ , and thus for all D and the positive integer n

we have from induction

the following expression: n

Dx



D

a Dx

> f ( x ) @

n D a Dx

> f ( x )@

(2.228)

2.21 Fractional integration is the area under the shape changing curve 2.21.1 Analytical explanation of the area under the shape changing curve for the Riemann-Liouville fractional integration The Riemann-Liouville formula is to W

tx

1

D

D

0 It

  D1 x

so we have dW

> f (t ) @

1D

D

1 * (D )

t

1

D ³ (t  W )  f (W ) dW .

In this formula we change the variable

0

dx and substitute in this expression to get the following steps: D 0 Dt

> f (t ) @

0

1

D 1 D

x  D x * (D ) ³ t

1

1D

D

dx

D

0

1

 f (t  x ) dx D  * (D ) ³ 1

(2.229)

D

t

D

tD

1

 f (t  x ) dx * (D  1) ³ 1

D

0

1 2

Let us take D

f ( x)

( x  2)

2

, and see what is the area; i.e.

t

³  f (t  x ) dx 2

0

2

indicates for the function

x  4  4 x , when the upper limit of integration keeps changing.

2

 (t  x )  2

2

2

f (t  x ) that is f (t  x )

2

integration is thus 2

³

0

2

.

2

f (0.001  x )dx

2

³



2

2

4

2

2



2

2

0.001

0

(0.001  x  2) # (4  4 x  x ) from 0 to f (1  x ) that is (1  x )  2

say

The changed function is,

First we take a small interval of the integration variable t that is from 0 to 0.001 . Here we have t 0.001

f

2

2

(0.001  x  2) dx .

This is the area under the curve

0.001 . Next is the calculation we do at t 4

or 9  6x  x . Therefore, at point t

0.001 and the

1 , here the function is

1 , we calculate the area under a new curve,

4

that is f (1  x ) 9  6 x  x from 0 to 1 . Thus the fractional integration of f (t ) from 0 to t indicates from this example that the area under the shape changing curve known as t , grows. An interesting observation is contrary to the classical integration where the shape of the curve is fixed.

2.21.2 Numerical demonstration of fractional integration as the area under a shape changing curve We noted in a previous section that the fractional integration is the area under the curve whose shape keeps on changing as we advance. For a demonstration, let us take a constant function i.e. f ( x ) C . From the discussion of the previous section (Section 2.20), we write the integration process of a classical type of one whole order as follows, with the lower limit as zero, i.e. a 0 as: 1 0 Ix

> f ( x )@

lim ª¬hN  f ( x )  f ( x  hN )  f ( x  2 hN )  .... f ( hN ) º¼ (2.230)

hN p 0

Fractional Integration

We are not going to show that for f ( x )

93

C the curve f ( x ) remains the same while we do the process of integration

hN , x1 2 hN , x3 3hN and so on i.e. by increasing the variable x , which is obvious from the above formula, and from our classical calculus theory.

at x0

Now let us write the formula for fractional integration that we discussed in sections (2.221) and (2.223), and do the following manipulation:

> f ( x )@ D >d( x  a)@ d

D

D

0 Ix

§ hDN N 1 * ( j  D ) · lim ¨ f  x  jhN ¸ , a 0 ¦ N nf * (D ) j 0 * ( j  1) © ¹ §

> f ( x )@

D

lim ¨ hN

N nf

·

*( j  D ) ¦ *(D )  j ! f  x  jh ¸ N

j 0

¹

N 1

§

D 1

hN

©

N 1

·

*( j  D ) ¦  *(D )  j ! f  x  jh ¸

lim ¨ hN

N nf

N

© j0 § N 1 · D 1 hN lim ¨ hN ¦ c j f  x  jhN ¸ N nf © j0 ¹

(2.231)

¹

§ hN  c0 f ( x )  c1 f ( x  hN )  c2 f ( x  2hN  · ¸ © .......c N 1 f ( x  ( N  1) hN ¹

K¨

Where in (2.231) we have taken the proportionality constant as K

D 1

hN ; and the above manipulation (2.231) shows

that the fractional integration is: D

0 Ix

§ hN  c0 f ( x )  c1 f ( x  hN )  c2 f ( x  2hN ) · ¸ © .....c N 1 f ( x  ( N  1) hN ¹

> f ( x )@ v ¨

(2.232)

Comparing the above expression (2.232) with the classical one of order integration (2.230), we find that the function * ( j D )

values are modified by coefficients c j We take D

c5

63 256

* (1  D )

1 2

, and f ( x )

 * (D )  j !

.

C . For this we have coefficients such as c0

, for the five points of the function, derived for

D  * (D ) . Thus at point x

cj

1 2

1 , c1

*  j  12

 * 12  j !

j

;

, c2

3 8

, c4

0,1, 2, 3, 4, 5

35 128

and

by using

x5 , we write the following expression: 1

0 I x5

2

§ c0 f ( x5 )  c1 f ( x4 )  c2 f ( x3 )  c3 f ( x2 ) · ¸ ©  c4 f ( x1 )  c5 f ( x0 ) ¹ 5 f (x )· § f ( x5 )  12 f ( x4 )  83 f ( x3 )  16 2 ¸ hN ¨ ¨  35 f ( x1 )  63 f ( x0 ) ¸ © 128 ¹ 256

> f ( x)@ v hN ¨

(2.233)

5  35  63 hN C 1  12  83  16 2.7069 hN C 128 256 The above expression (2.233) is the area under Curve-5 as depicted in Figure-2.8; clearly this is not the area under the original curve i.e. f ( x ) C . In Figure-2.8 the value of the slice on the X-axis is hN x1  x0 x2  x1 .... i.e. the





slices are equally spaced; we have drawn the piece-wise linear curves for demonstration. When examining Figure-2.8 and writing the fractional integration at point x 1

0 I x0

here the fractional integration, i.e.

1

0 I x0

2

2

x0 , which is:

> f ( x)@ v hN f ( x0 )

hN C

(2.234)

> f ( x)@ , is proportional to the area under the curve

curve shown by Curve-0 in Figure-2.8. We proceed and at point x 1

0 I x1

2

f ( x)

C , the original

1.5hN C

(2.235)

x1 we write the following:

> f ( x)@ v hN  f ( x1 )  12 f ( x0 )

hN C 1  12





Fractional Integration

95

happenings, while fractional integration is a process with decreasing/fading memory of all of the past happenings, we would not be inaccurate. We all have fading memory in reality, like fractional integration!

2.22 The functions which can be fractionally integrated and differentiated We have discussed the arbitrary order D of the integral transform equation in the Riemann-Liouville way. There is a class of functions that we can fractionally integrate (or differentiate). In this section, we shall confine ourselves to classically defined functions rather than distributions. These are sometimes referred to as symbolic or generalised functions like delta distribution etc. We take from our classical calculus the definition of such a function that is

> @

defined on a closed interval i.e. a , x and is better behaved than ( x  a )

1

at the lower limit x

a.

We define the class of a differ-integrable series as f ( x)

p

a0 ( x  a )  a1 ( x  a ) ( x  a)

p

p   1n

 a2 ( x  a )

f

¦ a ( x  a) j

 jn

p   n2

 a3 ( x  a )

a0 z 0

p   n3

 ......

(2.240)

p ! 1

j 0

1 The above series (2.240) is a product of the power of ( x  a ) and an analytic function of ( x  a ) n , where n is a positive integer. However, it is likely that the arbitrary powers of ( x  a ) could be treated in the infinite series factor of f ( x ) . This inclusion will complicate the discussion; thus we restrict our attention to the simple series as presented above.

Notice that this p has been chosen to ensure that the leading coefficient is non-zero in the above series. The above series satisfies the following condition

lim  ( x  a ) f ( x )

xoa

§ a ( x  a ) p 1  a ( x  a ) p 1 1n  · 0 1 ¸ lim ¨ p 1  n2 p 1  n3 xoa ¨  a3 ( x  a )  ... ¸¹ © a2 ( x  a ) 0

The expression lim x o a  ( x  a ) f ( x )

f ( x)

( x  a)

p

¦

f j 0

 n j

a j ( x  a)

for

p ! 1

0 , in (2.241) means that the area under the curve, described via the function

, with a0 z 0 , and p ! 1 just near the start point x

we mean when we say that f ( x ) is better behaved at the lower terminal i.e. x The functions f ( x )

( x  a)

 12

, f ( x)

(2.241)

( x  a)

54

sin





x  a , f ( x)

1 2

a , is zero. This is what

a than ( x  a ) 1 . sin(S ( x  a ))  1 , f ( x )

ln( x  a ) are

such a class of functions. The important consequence of this representation of f ( x ) is that the function f ( x ) may be further decomposed as a finite sum of n differ-integrable “units” fX ( x ) each of which is a product of a power (greater than 1 ) of ( x  a ) and a function analytic in ( x  a ) , that is represented below.

f ( x)

f f § · § · p  1n j1 j p ( x  a ) a ( x  a )  ( x  a ) a j2 ( x  a ) 2 ¸  ... ¨ ¸ ¨ ¦ ¦ j1 j1 0 j2 0 © ¹ © ¹ f § · p 1 1n j a jn ( x  a ) n ¸ .......  ¨ ( x  a ) ¦ ¨ ¸ jn 0 © ¹ f f np 1 § · § · j j p ¨ ( x  a ) ¦ a j1 ( x  a ) 1 ¸  ¨ ( x  a ) n ¦ a j2 ( x  a ) 2 ¸  j1 0 j2 0 © ¹ © ¹ f np  n 1 § · j ..  ¨ ( x  a ) n ¦ a jn ( x  a ) n ¸ ¨ ¸ jn 0 © ¹

The functions that cannot be differ-integrated are f ( x )

( x  a)

2

and f ( x )

9 4

(2. 242)

2

 ( x  a ) . Many functions are not

expansible as the differ-integrable series are nevertheless differ-integrable; for example, logarithmic and Heaviside’s

96

Chapter Two

unit step function and even the trivial function f ( x ) { 0 . We used the nomenclature differ-integrable function to include all the above and in fact any function whose differ-integrals can be obtained.

2.23 Discrete numerical evaluation of the Riemann-Liouville fractional integration dD dxD

In this section, we will approximate

f ( x ) for arbitrary D  0 , with known values of f ( x ) at N  1 and evenly

spaced points in the range 0 to x . We designate symbols f N { f (0) , f N 1 { f

 Nx , … f j { f  x  j Nx , and …

f0 { f ( x ) . The Riemann-Liouville (RL) fractional integration formula for D  0 we write as alternative representations that are obtained from the original formula by a change of variables see (2.129) and (2.130), we rewrite as follows: D

d f ( x)

1

* ( D ) ³

f ( y)

x

1

* ( D ) ³

D

d f ( x) dx

1

* ( D ) ³

D

x

D 1

f ( x  y) y

0

1 * ( D )

³

x/ N

D 1

 f ( x  y ) dy y

0

D 1



N 1

1

¦³ * ( D ) j 0

³

dy

f ( x  y)

x

D 0 (2.243) dy D 1 0 ( x  y) 0 dx y Re-writing the above integration (2.243) from 0 to x by splitting the integration into small intervals and then by summing all small integrations in all the discrete intervals we get the following: D

dy

2x/ N

 f ( x  y ) dy y

x/ N

( j 1) x / N

D 1

 ..

f ( x  y) y

jx / N

D 1

x

³

 f ( x  y ) dy yD 1

( N 1) x / N



(2.244)

dy

2.23.1 Using the average of function values at the end of intervals to have a formula for numerical evaluation of the Riemann-Liouville fractional integration Now we use approximation for

in the interval of integration i.e.

f ( x  y)

f  x  j 'x  f  x  ( j 1) 'x

f ( x  y) | , averaging the values at y 2 intervals. We then use them in the following steps:

³

( j 1) x / N jx / N

 f ( x  y ) dy | y



f x

jx N

j 'x and y

 f x 

D 1

( j 1) x N



³

2



f x

jx N

 f x 

x N



jx N

f j  f j 1 § § ( j  1) x · 2D

¨¨ ¨ ©©

2D In the above steps (2.245) for the evaluation of f  x  j 'x  f  x  ( j 1) 'x 2

j 'x and y

 ( j  1) yD 1

jx / N

y

³

D 1

( j 1) x / N

dy y

jx / N

D 1

D § jx · · ¨ ¸ ¸ © N ¹ ¸¹ D

( j 1) x / N f ( x  y )

j

dy

D

(2.245)



the approximated

f ( x  y)

i.e.

comes out of the integration process as this is now not a variable in y . The point should also be

noted that while we are writing an approximation as f ( x  y ) |

y

D

¨ ¸ ©N¹

³

¸ ¹

N

f j  f j 1 § x ·

D

dy

jx / N

2

as

( j  1) 'x i.e. at the end-points of

( j 1) x / N



> j'x, ( j  1)'x@

f  x  j 'x  f  x  ( j 1) 'x 2

, we have used the end points of

( j  1) 'x ; these are a-priori defined as discrete points.

Using the above obtained expression (2.245) i.e. the expression  D  * ( D )

³

( j 1) x / N f ( x  y ) jx / N

yD 1

dy

* ( D  1) we get the following formula:

f j  f j 1 2D

D

 Nx  ( j  1)D  j D and by using

Fractional Integration

D

d f ( x) dx

D

1

§x· ¨ ¸ * (1  D ) © N ¹

 'x

D N 1

* (1  D ) Here 'x

¦

D N 1

¦

f j  f j 1 2

j 0

f j  f j 1 2

j 0

97

 ( j  1)

D

 j

D

;

D 0 (2.246)

 ( j  1)

D

 j

D



 Nx is a discrete step size. From the above derivation, we write a final formula for the discrete numerical

evaluation of an RL fractional integration of order D ! 0 as follows: D

D

0 Ix

> f ( x )@

 'x

N 1

¦ * (1  D )

f j  f j 1

j 0

2

 ( j  1)

D

j

D

;

D !0

(2.247)

Use of the other approximation will lead to a different formula.

2.23.2 Using the weighted average of the function values at the end of intervals to have a formula for the numerical evaluation of the Riemann-Liouville fractional integration We use the approximation for integration i.e. the one based on a linear interpolation between f j 1 and f j . We will use the weighted sum of f j 1 and f j , to have an interpolation in this case. The interval of integration (and our interpolation) is from y

j 'x to y

( j  1) 'x . Say we take any point y between j 'x and ( j  1) 'x . The

distance between j 'x to this point y is ( y  j 'x ) and the distance between y and ( j  1) 'x is  ( j  1) 'x  y . The relative distances from y to these two end-points, in the discrete step size 'x are

( y  j 'x ) 'x

and

 ( j 1) 'x  y 'x

respectively. With this, we write interpolation (or approximation) as

§  ( j  1) 'x  y · ¸ fj 'x © ¹

f ( y) | ¨

§ ( y  j 'x ) · ¸ f j 1 ; © 'x ¹

¨

 ( j  1)  y  f   y   j f N x

j

N x

'x

N x

(2.248)

j 1

The weighted average technique used above for interpolation, gives a larger weight to f j when the point y is close to

j 'x and gives more weight to f j 1 when the point is close to ( j  1) 'x . When the point y is at the mid-point of the interval i.e. at y

 j  12 'x , the two weights are equal to half. This central point interpolation is what we

followed in the previous derivation (2.247), where we approximated it as interpolation formula we now use in the following derivation:

f j  f j 1 2

. This weighted sum (2.248) is the

98

Chapter Two

§ 1  j  Nyx f  x  Njx  · ¨ ¸ ( j 1) x Ny ¨ ¸ ( j 1) x / N  f ( x  y ) dy ( j 1) x / N  x  j f  x  N ¹ © | dy D 1 D 1 ³ ³ y

jx / N

³

y

jx / N

( j 1) x / n jx / N

³

§ (1  j ) f j  ¨¨ ©

Ny x

Ny fj  x D 1

y

f j 1  j f j 1 · ¸dy

¸ ¹

§ (1  j ) f j  j f j 1 · ¨¨ ¸¸dy D 1 jx / N y © ¹ § N · ( j 1) x / N § f j 1  f j · ¨ ¸³ ¸dy ¨ D © x ¹ jx / N © y ¹ ( j 1) x / N

( j 1) x / N

D (1  j ) f j  j f j 1 ª« yD º» ¬ ¼ jx / N

§x· ¨ ¸ ©N¹

1

 ( j 1) x / N

ª D 1 º  f j 1  f j « 1y  D » ¬ ¼ jx / N D

D D j · § x · § ( j  1) j f j f (1 )      ¸ ¨ j j 1 ¨ ¸ D D ¹ ©N¹ © 1

D 1

§ ( j  1) D 1 j D 1 · ¨ 1D  1D ¸ © ¹ D D §  (1  j ) f j  j f j 1  ( j  1)  j · ¸ D ¨ §x· ¨ D ¸ ¨ ¸ ¨ D 1 D 1 ¸ N © ¹  f  f ( j  1)  j ¸ ¨  j 1 j © ¹ 1D §x· §x·   f j 1  f j ¨ ¸ ¨ ¸ ©N¹ ©N¹

(2. 249)

The above steps in derivation (2.249) lead to a more complicated formula described as follows for D  0 : D

d f ( x) dx

D

§  ( j  1) f j  j f j  ( j  1) D  j D · ¨ ¸ D 1 § x · N 1 ¨ D ¸ ¨ ¸ ¦ ¸ * ( D ) © N ¹ j 0 ¨  f  f  ( j  1)1D  j1D ¨  j 1 j ¸ © ¹ 1D §  ( j  1) f j  j f j  ( j  1) D  j D · D N 1 ¨ ¸  'x D ¨ ¸ ¦ ¸ * ( D ) j 0 ¨  f  f  ( j  1)1D  j1D ¨¨  j 1 j ¸¸ 1D © ¹

(2.250)

From the above obtained expression (2.250) we write the fractional integration formula for D ! 0 as a discrete numerical evaluation for RL fractional integration as follows:

D

0 Ix

> f ( x )@

§  ( j  1) f j  j f j  ( j  1)D  jD · D ¨ ¸  'x N 1 ¨ D ¸ ¦ * (D ) j 0 ¨  f  f  ( j  1)1D  j1D ¸ ¨¨  j 1 j ¸¸ 1D © ¹

(2.251)

We demonstrated the use of the numerical integration method to approximately evaluate the Riemann-Liouville fractional integration formula.

Fractional Integration

99

2.24 Short summary In this chapter, we have discussed the various concepts of the classical integration process and concluded that there is an extension of what we know about classical integration. The discussion on fractional integration has given us a thought provoking ‘question’ as to how to unify this concept and then generalise to a differentiation process. This aspect we will take up subsequently. We have also hinted that fractional differentiation is also an integration process; on this issue, we shall build the concept of fractional differentiation further. We took the topic of fractional integration as a start point as we will note subsequently that this formulation is the core process of fractional differentiation. We noted here that the fractional calculus calls for various other types of functions called higher-transcendental functions, which we will be using subsequently too and several of them are described in the Appendices, which are used in the context of fractional calculus. Now we move now to a fractional differentiation process.

2.25 References This chapter is motivated by the following texts, many of which are pioneering pieces of research: Samko S G, Kilbas A A, Maritcheva. O. I, “On Fractional Integrals and Derivatives: Theory and Applications” (1993) Manabe. S, “The Non-Integer Integral and its Application to Control Systems” (1961) Miller K S and Ross B, “An Introduction to Fractional Calculus and Fractional Differential Equations” (1993) Oldham and Spanier, “Fractional Calculus” (1974) Almeida R, Malinowska A. B, Torres D F M, “A fractional calculus of variations for multiple integrals with application to vibrating string” (2010) Agrawal, O. P, “Fractional variational calculus in terms of Riesz fractional derivatives” (2007) and “Generalised variation problems and Euler-Lagrange equations” (2010) Agrawal. O. P, Muslih. S. I, Baleanu. D, “Generalised variational calculus in terms of multi-parameters derivatives” (2011) Batman H, “Higher transcendental functions” (1954) Churchill Ruel V, Brown James W and Verhey Roger F, “Complex Variables & Applications” (1974) Heymans and Podlubny et al., “Physical interpretation of initial conditions for fractional differential equations with Riemann-Liouville fractional derivatives” (2005) Igor Podlubny, “Fractional differential equations” (1999) Kirayakova. V, “Generalised Fractional Calculus and Applications” (1994) Kostadin Trenceveski et al., “On the fractional derivative of some functions of exponential type” (2002) Liouville J, “Memorie sur I integration de I equations a’l’aide des differentielles a’ indices quelconques” (1832) Prabhakar T R. “A singular integral equation with a generalised Mittag-Leffer function in the kernel” (1971) Riemann B, “Versuch einer Auffassung der Integration and Differentiation” (1876) Machado J, Kriyakova V, Mainardi F, “Recent History of Fractional Calculus” (2011) Mauro Bologna, “Short Introduction to Fractional Calculus” (2005) Moshrefi-Torbati. M & Hammond J K, “Physical and geometrical interpretation of fractional operators” (1998) Northover Francis H, “The Foundations of Fractional Calculus” (1977) Ortigueria M D, Machado J A T, “What is a fractional derivative?” (2015) Ross Bertram, “A Brief History and Exposition of the Fundamental Theory of Fractional Calculus” (1975) Sabatier J, Agrawal O. P, Machado J. A (Eds.), “Advances in Fractional Calculus” (2007) Valerio D, Trujillo J J, Rivero M, Machado J A T, Baleanu D, “Fractional Calculus: A survey of useful formulas” (2013) The details of the above works are listed in the bibliography section, in alphabetical order.

CHAPTER THREE FRACTIONAL DERIVATIVES

3.1 Introduction We have stated while discussing concept generalisation in Chapter-1 about Leibniz stating to L’Hospital that putting 1 in the formula of the n -th derivative of a function will lead to a paradox. Why? The paradoxes arise because there 2 are several ways of generalising the differentiation operations to non-integer orders leading to equivalents but not always with the same results. Here we will deal with the most fundamental approaches of using a fractional derivative operator. We will also see that fractional differentiation needs be termed as a differ-integration process, and reason out why the fractional derivative is a non-local property, requiring memory and history. In short we will try to resolve the ‘paradoxes’ that seemed to exist three hundred years ago. We will try to find an answer to L’Hospital’s question and will carry on the generalising of classical integer order calculus. There are several symbols used to represent a fractional derivative operator of order D ! 0 and with D   . Leibniz dD dxD

D

and Euler used the symbol d , while Riemann used wDx . Liouville’s notation is D

d and upper limit in the representation of a fractional derivative operator, that is ª

f

¬ dx

D

, while Grunwald gave a lower x x

º ¼x

a

, for the function f ( x ) from

D

a ! f to x , while x ! a . Marchaud used the notation Da . Hardy-Littlewood used f

modern day use I

D

, I xD , a DDx ,

D

d dxD

,

D

d d( x  a )D

(D )

. Many authors in the

and DaD , for D ! 0 . We will be using some of these symbols.

3.2 Iterative differentiation and its generalisation Like in Chapter-2, where we talked about iterative integration, here we start with a repeated n  fold differentiation or iterative differentiation.

3.2.1 Euler’s scheme of generalisation for repeated differentiation This is Euler’s scheme of a repeated n  fold differentiation (with n as a positive integer), of a power function

f ( x)

p

k

x , p ! 0 which is reproduced below (with the understanding that D { 0 p D ª¬ x º¼

p

x ,

1 p D ª¬ x º¼

px

p 1

,

dk dx k

2 p D ª¬ x º¼

, k

0,1, 2, ....n ):

p ( p  1) x

p2

(3.1)

For n  times differentiation, we obtain the following: n p D ª¬ x º¼

p ( p  1)( p  2).....( p  n  1) x

pn

p ( p  1)...( p  n  1)  ( p  n )( p  n  1)...1

 ( p  n)( p  n  1)...1 p!

x

p n

(3.2)

p n

x ( p  n )! Equation (3.2) is called Euler’s formula. Recall that we did this in Chapter-2. Now we generalize (3.2) for a positive real number D and p ! 0 , by using the gamma function as a generalisation for a factorial; as we did in the case of integration (Chapter-2) to get the following expression: * ( p  1) D p p D (3.3) D ª¬ x º¼ x * ( p  D  1) p

The above expression (3.3) we may term as a fractional differentiation of f ( x ) x . The expression is valid for p ! 1 , the simple reason being that otherwise * ( p  1) will blow up at p 1 . In addition to what we have p

discussed in the last chapter (Section 2.22) that notes; the function (say x ) needs to be better behaved than x

1

at

Fractional Derivatives

101

the start point, (which in this case is x 0 ), for it to be fractionally differ-integrated. When D  0 say D X with X ! 0 , we have a formula for fractional integration as follows, which we have derived in an earlier chapter (Section

X , X ! 0 as D

2.6), that we get to by placing D

* ( p 1) * ( p X 1)

ª¬ x p º¼

X

x

p X

p X I ª¬ x º¼ .

This is an indication that there is a concept unifying fractional integration and fractional differentiation into a single expression. In the previous chapter (Sections 2.14 and 2.19) we hinted at that. We note this point here too, and write the unified formula for the fractional-differ-integration of a power function (discussed in Section 1.23) as for the integer order n case, that is the following:

d

n

^

­ * ( p  1) pn ( x  a) ° ® * ( p  n  1) °¯ f

ª¬( x  a ) p º¼ n

>d( x  a)@

n n

0,1, 2, .....; For all p 1, 2, ....; p ! 1 (3.4)

n

1, 2, ....;

p d 1

We may now generalize (3.4) for a non-integer order D and write the following:

d

D

­ * ( p  1) p D ( x  a) ° D * ( p   1) ® °¯ f

ª¬( x  a ) p º¼ D

>d( x  a)@

p ! 1

For all D

p d 1

For all D

(3.5)

3.2.2 A seeming paradox in Euler’s scheme of generalisation for repeated differentiation  12 , and D

We should now give thought to Euler’s formula (3.5) putting p

1 2

, with a

0 and applying the above

formula (3.5):

D

1

*  12  1

1

ªx 2 º ¬ ¼

2

*     1 1 2

1 2

*  32 * (0) As * (0)

x

1

x

 12  12

(3.6)

0

f we get the above result (3.6).

This seems absurd presently, as we know that a constant function should give zero while doing derivative operations. We will deliberate this observation of getting a zero for a half derivative for non-constant functions, later.

3.2.3 Euler’s formula applied to find the fractional derivative and fractional integration of power functions Applying this formula to find a half derivative and a half integral of f ( x ) *  52

3 S 4

x , using *  12

S , *  32

S

2

, and

we get the following:

D D

1

2

* (2)

> x@

 12

1 12

* 1   1 1 2

> x@

x

* (2) * 1  12  1

1 *

1 12

x

3 2



1 *  52

x x

3 2

2

x

S 4

3 S

(3.7)

x

3 2

The above expression (3.7) shows the applicability of Euler’s formula to get a fractional derivative and a fractional p

integral of functions like x , with p ! 1 . Here we mention that the start point of the process of fractional differentiation and the fractional integration itself is at x Figure-3.1 shows various fractional orders of

D

0 Dx

0 , so the fractional derivative operator is written as 0 DDx .

> x@ , for D

0, 1 4 , 13 , 1 2 . We observe that in the case of Figure-

3.1, all the fractional derivatives pass through the origin and its value is zero at the start point of the fractional derivative process i.e.

D

0 Dx

> f ( x )@ x 0

0 for f ( x )

x . In addition, we observe that as D increases from zero

towards one the fractional derivatives tend to be curved towards y i.e. f

(1)

( x)

1.

1 which is one whole derivative of f ( x )

x;

102

Chapter Three

0,

Fractional derivatives of f ( x ) = x for orders D

1

4

,

1

3

,

1

2

D 0, f ( x) x

D 3

D

2

y

4

DxD > x @

0

2.5

1

1.5

D

1

1

3

2

1 f (1) ( x) 1

0.5

0

0

1.5

1

0.5

2.5

2

x

3

Start point of fractional derivative process Observe how derivatives approach curve y = 1, i.e. one whole derivative of f ( x ) = x but each fractional derivative passes through the origin .

Figure-3.1: Fractional derivative of f ( x )

x for D

0,

1

4

,

1

3

and

1

2

3.2.4 That Euler’s formula gives us non-zero as a fractional derivative for a constant function seems to be a paradox With this Euler’s formula, we write the fractional derivative and fractional integral of the constant f ( x ) follows: 1 * (1) 1 1 1 0 1 D 2 1 x 2 x 2 1 1 *  0  2  1 *2 Sx

1 as

>@

D

 12

* (1)

>1@

*  0  12  1

x

1

0  12

*  23

x

1 2

2

x

(3.8)

S

Here we again come across a paradox that is a constant function, which is not returning a zero after the fractional derivative operation. Figure-3.2 gives the curves of fractional derivatives f ( x)

x

D

0 Dx

> f ( x )@

for D

0, 1 4 , 13 , 1 2 of a constant function i.e.

1 , with a start point of the fractional derivative operation which is at x

0 the values of fractional derivatives blow up, i.e.

D

0 Dx

> f ( x )@ x 0

0 . We observe in Figure-3.2 that at

f , for f ( x )

1.

This implies that for a constant function we get a singularity at the start point of the fractional derivative process. In addition, we observe that all the fractional derivatives tend towards a value of ‘zero’ at x n f , i.e. lim xnf



D

0 Dx

> f ( x ) @

0 , for f ( x )

(1)

1 which incidentally is one-whole derivative i.e. 0 Dx

>1@

f

(1)

( x)

0.

We observe the singularity at the start point of a fractional differentiation process in Figure-3.2, while there is no singularity at the start point x 0 in Figure-3.1. This is a very important observation, and we say that in Figure-3.1 the start value of the function at the start of a fractional differentiation process ( x 0 ) is zero i.e. f ( x ) x 0 0 whereas in Figure-3.2 the start value of the function is non-zero i.e. f ( x ) x of in subsequent chapters.

0

z 0 . This observation we will make use

Fractional Derivatives

103

Fractional derivatives of f ( x ) = 1 for orders D 0, 4

1

4

,

1

3

,

1

2

DxD >1@

0

3

y

D 0, f ( x) 1

2

D

1

1

4

D

0 1

3 D

2

Start point of fractional derivative process

1

1

x

3

2

f (1) ( x) 0

Observe that the fractional derivatives converge to y = 0 for large x; except for an asymptote at x = 0 where there is singularity i.e. 0 DxD >1@ f x 0

Figure-3.2: Fractional derivatives of a constant

Note that all the fractional derivatives of a constant go to zero at a far point; while at the start point of fractional differentiation (that is in this case x 0 ), the value of a fractional derivative is infinity (i.e. we get singularity at the start point).

3.2.5 The observation of Euler’s formula: the fractional derivative of a constant becomes zero when the fractional order of the derivative tends to one-whole number in the classical case C , we get a fractional derivative like D

Well for a function that is constant such as f ( x )

D

>C @

1 * (1D )

Cx

D

, by

applying Euler’s formula. This is counterintuitive as we are used to having zero as a derivative of the constant. However, here in the fractional derivative case, we are getting a non-zero. Rather we are getting a decaying power function which tends to zero at large values of x (see Figure-3.2) In the expression D

D

>C @

1 * (1D )

Cx

D

for the

fractional derivative of a constant, when D tends to 1 , the denominator, i.e. * (1  D ) , tends to * (0) which tends towards f ; thus, one-whole derivative of a constant tends to zero. The fractional derivative of the constant obtained is D

D

>C @

Cx  D * (1D )

, and tends to zero only when D is a natural number. This comes from the property of the gamma

function that is * (  z )

S z* ( z )sin S z

. For z a Natural number ` , sin S z

0 giving * (  z )

f.

This non-zero value for a fractional derivative of a constant function, gives irritation to many scientists, thus they tend not to use fractional calculus. Well, let us carry on for the time being, with this observation seemingly as a paradox.

3.2.6 Some interesting formulas for the semi-derivative and semi-integration with Euler’s formula We obtain an interesting semi-derivative and semi-integration with the Euler generalisation 1

n D 2 ªx 2 º 1

¬

We use the formula *  12  n

(2 n )! S 4n n !

, * ( n  1)

*  n  12  1

¼

* ( n  1) n ! , ( n  1)!

thereby achieve interesting results for the semi-derivative:

x

n

n

0,1, 2, 3,...

(3.9)

( n  1)( n !) and write the following steps and

104

Chapter Three

D

1

2

*  12  ( n  1)

1

ª x n 2 º ¬ ¼

x

n!

 2(n  1) ! 4

( n 1)

n

S

x

( n  1)! n !

(2 n  2)! S

§x· ¨ ¸ 4( n  1)! n ! © 4 ¹

n

(2 n  2)(2 n  1)! S 4( n  1)( n !) (2 n  1)! S 2( n !) The result of semi-integration is as follows:

D

 12

* n  1 

1

ª x n 2 º ¬ ¼

1 2



* n   1  1 2

1 2

*  12  ( n  1)

 2(n  1) ! ( n 1)



x

* ( n  2)

4

2

§x· ¨ ¸ ©4¹

2

x

n 1

0,1, 2, 3,...

n

*  12  ( n  1)

n 1

( n  1)!

S

(2 n  2)! S § x ·

(3.10)

n

n

x

( n  1)!( n  1)!

 (n  1)!

§x· ¨ ¸ ©4¹

n

x

n 1

(3.11)

n 1

n 1

¨ ¸ ©4¹

2

A further result on the semi-derivative is as follows:

D

1

2

ª¬ x n º¼

* ( n  1) * n  1 

1 2



x

n  12

n

( n !)4 ( n !) (2 n )! S

n

0,1, 2, 3,... (3.12)

2 n ( n !) (4 x )

n  12

x x

(2 n )! S x

The following is the result of semi-integration:

D

 12

ª¬ x n º¼

* ( n  1) * ( n  1)  12



( n !)(4)

( n 1)



x

n  12

n

( n  1)!

 2(n  1) !

S

0,1, 2, 3,...

1

n

x x2

n

n

1

2

n

1

1

( n !)( n  1)( n !)(4) (4) x x 2

( n !) (4 x ) (4) 2 x 2

(2 n  2)(2 n  1)! S

(2 n  1)! S

2

( n !) (4 x )

(3.13)

n  12

(2 n  1)! S

3.2.7 Applying Euler’s formula for a function represented as power series expansion Let a function be represented as the series expansion f ( x )

¦

f n 0

n

an x . Then we can take the help of Euler’s

formula and apply it term by term to get the following: D

D

f

> f ( x)@ ¦ an DD ª¬ x n º¼ n 0

f

¦a n 0

* ( n  1) n

* ( n  D  1)

(3.14) x

n D

Fractional Derivatives

Let us evaluate a f ( x)

¦

f n 0

1

2

derivative of f ( x )

n

an x , which is, e x

1 x 

x

by applying Euler’s formula term by term for expansion of

x3 3!

 ... . To this series, if we apply Euler’s formula we get a semi-

e

x2 2!



105

derivative as follows D

1

2

ª¬e x º¼

1 §

4 2 8 3 16 4 · x  ..... ¸ ¨1  2 x  x  x  3 15 105 Sx © ¹

x

(3.15)

x

Figure-3.3 plots this semi-derivative of e and the function e . Fractional Derivative of Exponential Function is not Exponential

f ( x)

1

ex ;

0

ªe x ¼ º z ex Dx 2 ¬

½ Derivative of

y 1

0

Dx 2 ª¬ e x º¼ x

lim xnf



0

f

1

0

Dx 2 ª¬ e x º¼



ex

x Figure-3.3: The semi-derivative of an exponential function

Figure-3.3 elicits that the fractional derivative of an exponential function is not an exponential function, from Euler’s formula. Due to this perhaps, Leibniz thought it a paradox. This is an important observation, and we will discuss it further, subsequently. Also, note that in Figure-3.3, there is a singularity at the start point of the fractional

0 and the curve converges towards e x for large x , with the knowledge that e x is a

differentiation process i.e. at x classical derivative of f ( x )



x

e . This we write as lim xnf D

1

2

ª¬e x º¼



x

e .

3.2.8 Generalising the iterated differentiation for x  p to get a fractional derivative formula, a different result is obtained by Euler’s scheme Now we do the iterated differentiation of the function f ( x ) 1

x

p

, with p ! 0 , as follows:

º¼  px  p 1 2 p  p 2 D ª¬ x º¼  p (  p  1) x D ª¬ x

p

2

( 1) p ( p  1) x 3 p D ª¬ x º¼

 ( p  2)

(  p )(  p  1)(  p  2) x

(3.16)

 p 3

3

( 1) p ( p  1)( p  2) x

 ( p  3)

Continuing this for n  times, we have the following expression: n p D ª¬ x º¼

n

( 1) p ( p  1)( p  2)....( p  n  1) x

We will manipulate the above obtained expression (3.17), as depicted in the following steps:

( pn)

(3.17)

106

Chapter Three

n p D ª¬ x º¼

§ ( p  n  1)( p  n  2)( p  n  3)... ·  ( p  n ) ¸x © ( p  3)( p  2)( p  1) p ¹ ( p  n  1)( p  n  2)...( p  1) p ( p  1)( p  2)...1   ( pn) n ( 1) x ( p  1)( p  2)...1  n

( 1) ¨

n *( p  n)  ( p  n ) x ( p  1)! *( p) For the positive real numbers D and p , we generalise the above obtained expression (3.18) by using the gamma function instead of the factorial and get the following: D p D * ( p  D )  ( p D ) (3.19) D ª¬ x º¼ ( 1) x *( p) We mention here that if we take D  0 say D X , we write the above expression (3.19) as ( 1)

D

X

n

( p  n  1)!

(3.18)

x

( pn)

( 1)

ª¬ x  p º¼ ( 1) X  **( p( pX) ) x  p X .

Note that we have obtained the Weyl fractional integration of x

p

* ( p X ) *( p)

X p as xWf ª¬ x º¼

which is similar to the above derivation except (3.19) with regards to the term ( 1)

X

x

 p X

. In addition, we observe the D

above expression (3.19) returns a complex quantity for the fractional derivative, since ( 1) D

that is ( 1)

e

iSD

is a complex quantity,

cos(SD )  i sin(SD ) . Therefore, we write the following expression: D p D ª¬ x º¼

Here if we place D

(in Section 2.15.2),

1 2

and p

1 2

e

*( p  D )

iDS

*( p)

x

 ( p D )

(3.20)

, we obtain the following result: 1

1

 D 2 ªx 2 º

¬

e

¼

*  12 

i S2

*  12

i

* (1) *  12 i

S p D We had used earlier in (3.5) and (3.6) the formula D ª¬ x º¼

* ( p 1) * ( p D 1)

x

some contradictions. Whereas in this derivation (3.21) we observe D

p D 1

2

1 2

x

x

x   1 2

1 2

1

(3.21)

1

, to find that D 1

ªx 2 º ¬ ¼

1

S

1

2

1

ª x  2 º 0 , revealing ¬ ¼

1

ix , a complex (imaginary)

quantity. Therefore, Euler partially resolved Leibniz’s paradox to state that: D E

d x

* ( E  1)

x

E D

(3.22) dx * ( E  D  1) which is valid for the integer and non-integer D and E , and we can possibly generalisethe above Euler formula (3.21) to all functions that are expanded into a power series, which might seem a natural step. D

3.3 A natural extension of normal integer order classical derivative to obtain a fractional derivative: the exponential approach of Leibniz and Liouville Leibniz implicitly assumed that for any real number D , the fractional order derivative of f ( x ) D Ox

d e

e

Ox

, was:

D Ox

(3.23) O e D dx This was a natural way to extend the concept of classical calculus. For an integer n we are used to formulae i.e. dn dx n

e

Ox

n Ox O e ; in a classical sense. This approach (3.23) is also termed as an exponential approach. Therefore, the

above postulate (3.23) could equally take a natural start point and define the fractional derivative, for functions that are O x expanded in series with exponentials, such as f ( x ) | ck e k , which we can write as:

¦

k

Fractional Derivatives

107 D

d f ( x) dx

kx

¦ c O D eO k k

D

(3.24)

k

3.3.1 Obtaining the fractional derivative by exponential approach for x  E by using the Laplace integral We shall now use the above postulate (3.23) to find the fractional derivative of function f ( x ) representation of x

E

x

E

by using the

by the Laplace integral as in the following expression:

x

f

1

E

*( E )

³e

 yx E 1

dy

y

(3.25)

0

The above expression (3.25) has an integral representation of x  E in terms of the exponential function (the derivation of this integral representation of x write the following:

E

we will describe shortly). Now, applying the above postulate of Leibniz, we can D E

d x dx

f

1

D

E 1

³y

*( E )

0

§ dD  yx · ¨ D ( e ) ¸ dy © dx ¹

f

1

E D y  (  y ) e dy ³ *( E ) 1

 yx

(3.26)

0

D f

( 1)

y *( E ) ³

D  E 1  yx

e

dy

0

Take the integral in the above expression (3.26) i.e. I

³

0

the integral i.e. I with this variable change as I 1

I

D E

x

³

f

0

f

e

 yx D  E 1

dy , put z

y

³

f

0

e

D  E 1

z

 xz

yx , that gives dy

 1x dz ,

dz x

. We write

which after simplification is

 z (D  E ) 1

e z

dz .

Here we recognise the integral

³

f

0

 z (D  E ) 1

e z

dz

* (D  E ) , which comes from the integral definition of the gamma

function. Therefore using all these steps, we obtain the following: D E

d x

D

( 1) * (D  E )

D

(3.27)

D E

*( E ) dx x If we generalise the above to  E (presently disregarding the existence of the integral), we get: D E

d x dx This is not similar to

dD x E dxD

* ( E 1) * ( E D 1)

x

E D

D

D

( 1) * (  E  D ) *( E )

x

E D

(3.28) p

, as obtained by generalising the repeated differentiation of x earlier in

(3.5). The difference becomes obvious for D

D

1

2

1

ª x 2 º ¬ ¼

i

S

1

1 2

, and E

x ; and from the formula

dD x E dxD

 12 , as we get from the formula * ( E 1) * ( E D 1)

x

E D

, we get D

1

2

dD x E dxD

( 1)D * (  E D ) *(E )

x

E D

,

1

ª x  2 º 0 . This is revealing an ¬ ¼

inconsistency again.

3.3.2 Representing x  E by using the Laplace integral in terms of the exponential function by using the Laplace integral derivation E

Now we discuss the Laplace integral representing x that we used in the above discussion regarding (3.25). From the table of the Laplace transforms we have a property called the ‘frequency integration’, while the parameter t is represented as the Laplace complex-frequency (say s ), for a function f (t ) whose Laplace Transform is F ( s ) . The

108

Chapter Three

property is $

^ `{³ f (t ) t

f

s

 F (V ) dV ,

and is ‘frequency integration’. def

Laplace transform, which is $ ^ f (t )`

³

F (s)

f

0

as follows;

e

 st

 f (t ) dt , to the LHS of the property of frequency integration f

³

e

 st

f (t ) t

0

Here f (t )

E

^ `

F (s)

* ( E 1) s E 1

³

t , so we have from the LHS of (3.29) I

Transforms we have $ t

* ( E 1)

E

s E 1

.

From here we apply the definition of the

f

0

e

 st ( t E ) t

f

³  F (V ) dV

dt

(3.29)

s

³

dt

f

e

0

 st E 1

t

dt . From the Tables of Laplace

Use this in the above property of the frequency integration with

, that is at the RHS of (3.29) to get the following: f

f

* ( E  1)

³  F (V ) dV ³ s

V

s

E 1

dV f

f

ª V  ( E 1) 1 º * ( E  1) « » ¬ ( E  1)  1 ¼ s s

E

E In the above steps (3.30) we used * ( z  1)

s

* ( E  1)

E

ªV  E º * ( E  1) « » ¬ E ¼ s

(3.30)

 *( E )

z  * ( z ) as the property of the gamma function. Therefore, from this

derivation we have the following expression: f

s

E

 *( E )

³e

 st E 1

t

dt

(3.31)

0

Change the variables s to x and t to y in the above expression to get the following expressions: f

x

E

 *( E ) ³ e

 yx E 1

y

dy

x

,

f

1

E

e *( E ) ³

0

The above expression (3.32) is the Laplace integral representation of x dD x  E dxD

( 1)D * ( E D ) *(E )

x

 E D

0 E

 yx E 1

y

dy

(3.32)

, which we had used to derive

, via an ‘exponential approach’.

Thus (3.32) is an integral representation of the power law function f ( x )

x

D

, x

D

useful expression.

1 * (D )

³

f

0

O

D 1  O x

e

dO and is a

3.3.3 A fractional derivative of the trigonometric and exponential functions using an exponential approach From our classical calculus, we write the following for repeated differentiation of f ( x ) 0

>

D sin x

@

sin x ,

1

>

@

0

S

D sin x

Every differentiation operation ( D ) on sin x gives a 90 or

cos x ,

D

2

>sin x@

sin x :

 sin x

(3.33)

phase lead. So for n  fold differentiation of sin x ,

2

we write:

D

n

>sin x@

sin  x 

nS 2



(3.34)

For n as a real number, say D , we may write the following:

D

D

>sin x@

sin  x  DS 2 ,

D

D

>cos x@

cos  x  DS 2

(3.35)

110

Chapter Three

d

ª f xk º » D «¦ dx ¬ k 0 k ! ¼

D

d

ªe x º D ¬ ¼ dx

D

f

x

(3.42)

k D

¦ *(k  D  1) z e

x

k 0

These two inconsistencies show that Euler’s rule is not consistent with Leibniz’s postulate. Liouville also found this dD f

inconsistency while calculating the fractional derivative with the rule of Euler, i.e. represented by f ( x ) |

¦

c e k k

Ok x

¦

dxD

D Ok x

c O k k k

e

for functions

.

3.4.2 Discussing the paradoxes of Euler’s and Leibniz’s by considering formulas to get classical one-whole integration–and a resolution

dD eO x dxD

D Ox

O e

, are compared for D

* ( E 1) * ( E D 1)

dD x E dxD

A resolution of Leibniz’s paradox emerges when the formulas of Euler,

x

E D

, and of Leibniz,

1 . That is, when they are interpreted as classical one-whole integrations.

Actually, Leibniz suggested this interpretation as the following: x

1 x

d e dx

1

e

x

y

³ e dy

f x

(3.43)

1

ª f xk º z ³ e dy e  1 » 1 « ¦ dx ¬ k 0 k ! ¼ 0 y

d

x

This suggests that Euler’s fractional derivative on the RHS of (3.43) differs from Leibniz’s/Liouville’s fractional derivative at the LHS of (3.43). Similarly,

dD x E dxD

* ( E 1) * ( E D 1)

x

E D

with D

1 , Euler’s generalisation, corresponds to the following: 1 E

d x dx On the other hand Leibniz’s postulate i.e.

( 1)D * ( E D ) *(E )

dD x  E dxD

x

 ( E D )

1

E 1

1  E



1 E

x

0

E

y dy

(3.44)

1 corresponds to the following:

1 E

x

1

³

E 1

with D

d x dx

x

³

f

x

y

E

dy

³

x

f

y

E

dy

(3.45)

Note how (3.44) and (3.45) differ in the limit of integration.

3.4.3 The difference between Euler’s and Leibniz’s postulate is due to the limits of integration The argument discussed suggests that Euler’s and Liouville’s definitions differ with the limit of integration in (3.44) and (3.45). Let us write the following integration: x 1 ª ax º b Dx ¬ e ¼

³e

ax

dx

b

1 a

e

1 ax

ax

a e

(3.46)

For what value of b this is (3.46) true? Such a question we address as follows: x

³e

ax

dx

b

1 ax

The above integration (3.47) is equal to a e 1 f Dx

ax

1 ax

(3.46),

for ab

1 a

e

ax

f .



1 a

e

ab

So for a ! 0 and b

(3.47) f , we have

ª¬e º¼ a e with a ! 0 . Therefore, the Liouville postulate should be valid if we write D ª ax º aD e ax . f Dx ¬e ¼

Fractional Derivatives

We have discussed regarding Figure-3.3, for x n f we get lim xnf

x

0 , the start point of fractional differentiation.

differentiation is pushed to x formula i.e.

D

f Dx

f , we will get



111

ª¬e x º¼

D

0 Dx



x

e , and observe the singularity at

If the singularity point i.e. the start point of a fractional

lim cpf



D

c Dx

ª¬e x º¼



e

x

; that is what we discussed for the

ª¬e ax º¼ aD e ax .

3.4.4 The fractional derivative requires the limits of a start-point and end-point for integration p

Now consider the integration for x as depicted below: x 1

b Dx

ª¬ x p º¼

³

x

p

x dx

p 1

b

x

p 1



b

p 1

p 1

(3.48)

p 1

for b 0 p 1 The limit of integration plays an important role in the fractional differentiation or fractional integration. This we would like to state here. Therefore, we cannot rule out Liouville’s postulate as a contradiction, but it is dependent on a lower limit which in this case is f . These arguments also point to a question: ‘does fractional differentiation require a lower and higher limit as in integration?’ Indeed, it is true that unlike a normal classical integer order derivative, the fractional derivative does have lower and upper limits like with integration.

3.5 Applying Liouville’s logic to get a fractional derivative of x  a and a cosine function This approach is an exponential one which concerns what we state as Liouville’s postulate. Thus if we write, in terms of Liouville’s logic: rX f Dx

ª¬e ax º¼ a rX e ax

(3.49)

Then this (3.49) becomes a natural way to extend the derivative of any arbitrary order that is X (a natural number, integer, real number or complex number). We will see how Liouville’s logic is applied for finding a fractional derivative of x

a

, with a ! 0 . For that, consider the following integral: f

³u

I

a 1  xu

e

du

a!0

x!0

(3.50)

0

Put xu

t , in the above integral (3.50) to get: f

I

x

a

³t

a 1  t

a

e dt

x *(a)

(3.51)

0

Note that the definition * ( a )

³

f

0

t

a 1  t

e dt is used in (3.51). We thus get the following:

x We will use Liouville’s logic, obtained for x

a

X

f Dx

1

a

*(a)

 I

;

I

ª¬e ax º¼ aX e ax , in the following steps to get

in terms of the integral defined as I in (3.52).

It is depicted in the following steps:

f

x

a

³t

a 1  t

e dt

(3.52)

0

X

f Dx

ª¬ x  a º¼ , by using an expression

112

Chapter Three

X

f Dx

ª f a 1  xu º e du » f Dx « ³ u *(a) ¬0 ¼

ª 1 º  I » f Dx « ¬ *(a) ¼

a

1

X

ª¬ x º¼

f 1 §

X

· X a 1 ª  xu º¼ du ¸ ¨³u f Dx ¬ e *(a) © 0 ¹ f · 1 § a 1 X  xu ¨ ³ u (  u ) e du ¸ *(a) © 0 ¹





(3.53)

f

1

u *(a) ³

a 1

X X  xu

( 1) u e

du

0

X f

( 1)

u *(a) ³

a X 1  xu

e

du

0

We will do some tricks here. In our defined integral I , i.e. I

³

f

0

u

a 1  xu

e

a

du

x * ( a ) we will replace a with

( a  X ) , and get the following expression: f

³u

a X 1  xu

e

du

x

 ( a X )

 *(a  X )

(3.54)

0

Using

this

X f Dx

a

ª¬ x º¼

expression ( 1)X *(a)

³

f

0

u

in

the

a X 1  xu

e

last

line

of

our

derivation

in

the

above

steps

of

(3.53),

i.e.

du , we can write the following expression: X

ª f Dx ¬ x Therefore, we have used Liouville’s logic to obtain

a

 *(a  X ) x  a X

*(a)

( 1)X * ( a X ) *(a)

ª¬ x  a º¼

X

f Dx

X

( 1)

º¼

x

 a X

(3.55)

– a difficult way to derive it.

However, we have derived this via Laplace integration which earlier (3.28) we wrote as

( 1)X * (X D ) * (X )

dX x  a dxX

x

 (X D )

.

Now we apply this Liouville logic to get a fractional derivative of the cosine function, with an exponential approach rX f Dx

ª¬e ax º¼ a rX e ax as depicted in the following steps: X

f Dx

§ eix  e  ix · f Dx ¨ ¸ © 2 ¹

>cos( x)@

X

X ix

X  ix

(i ) e  (  i ) e 2 (e

iXS / 2

ix

)e  (e

X

 iXS / 2

e

 ri

;

)e

 ix

e

r iS / 2 X

e

§ XS · i¨ x  ¸ 2 ¹ ©

e

2 cos  x 



(3.56)

§ XS · i¨ x  ¸ 2 ¹ ©

2

XS



X

>cos( x)@

2

r iSX / 2

We have a nice result, i.e. f Dx

cos  x  XS2

(3.57)

Thus, the generalised differential operator simply shifts the phase of the cosine function (and likewise the sine 0

function) by X u 90 that is in proportion to the order of the differentiation. For differentiation the process advances the phase, the integration makes the phase lagged. After all for X X 1 we have:

d dx

cos( x )

cos  x  S2

,

d

1

dx

1

cos( x )

cos  x  S2

1 and, (3.58)

Fractional Derivatives

113

Needless to say that, this approach can be applied to the exponential, i.e. the 1

f Dx

With this, we do indeed find that the X  th derivative of e approach. Now we write the fractional derivative as

D

ax

1

derivative of e

2

ax

ª¬e ax º¼ ( a 2 )e ax

2

X ax

is simply a e

(3.59)

consistent with the purely exponential

f and a

following on with a lower terminal as a

a Dx

should be:

1

0 in the

following expressions: D

f Dx

D

0 Dx

ª¬e x º¼ e x * ( p  1)

ª¬ x p º¼

* ( p  D  1) f

D ª xº 0 D x ¬e ¼ D

f Dx

D

a Dx

x

x

p D

n D

¦ *(n  D  1)

(3.60)

n 0

ª¬e ax º¼ aD e ax

ª¬ ( x  c ) p º¼

* ( p  1) * ( p  D  1)

( x  c)

p D

only if

a

c

3.6 Liouville’s approach of a fractional derivative to arrive at formulas as a limit of difference quotients 3.6.1 Liouville’s formula for fractional integration dD f ( x )

It has been mentioned that Liouville defined a fractional derivative as represented as the sum of the exponentials f ( x ) |

¦

c e k k

Ok x

¦

dxD

D Ok x

c O k k k

e

for functions

. Liouville seems not to have recognised the necessity

of a limit of integration. Thus from this postulate Liouville derives several integrals and series representations, particularly, he finds the fractional integral of order D ! 0 , as follows:

³

D

f

D

 f ( x) (dx)

³  f ( x  y) y

D 1

dy

0

(3.61)

f

D

1

D D ³  f ( x) (dx) ³  f ( x  y) y

dy

0

The first one in (3.61) has the condition f ( f )

0 , and the second one, also in (3.61) has the condition f (f )

0.

3.6.2 Liouville’s formula for a fractional derivative in integral representation Then Liouville gave the formula of a fractional derivative of the order D ! 0 , as: D

d f ( x) dx

D

( 1)

n D

d f ( x) D

*(n  D ) ³ f

1 ( 1)

n

d f ( x  y) dx

0

D

dx

f

1

n D

*(n  D )

³

n

y

0

dx

n

dy (3.62)

n

d f ( x  y)

n D 1

y

n D 1

dy

where ( n  1)  D  n , with n as a positive integer. Liouville restricts the discussion to functions represented as f ( x ) |

¦

c e k k

Ok x

, where Ok ! 0 so that f ( f )

Let us stop for a moment and study these Liouville formulas. We will derive Liouville’s formulas of (3.62).

0.

114

Chapter Three

3.6.3 Derivation of Liouville’s fractional derivative formulas dD f ( x )

We will apply the Liouville postulates i.e. of exponentials i.e.

D ! 0 , that is,

f ( x)

d  D eOk x dx  D

¦

c e k k

D Ok x

Ok x

¦

dxD

D Ok x

c O k k k

¦

. Put f ( x )

for functions f ( x ) that are represented as the sum

e

c e k k

Ok x

and using the fractional integration of the order

¦

, we write Liouville’s fractional integral formula for f ( x )

Ok e

c e k k

Ok x

. This

integral is finite (bounded) for limits 0 and f , for Ok ! 0 . We write the following: D

f

D ³  f ( x )  dx

¦

Now we substitute f ( x )

c e k k

Ok x

ª ¦ ck (OkD )eOk x º ¬ k ¼0

(3.63)

, in the RHS of (3.61), i.e. Liouville’s fractional integration formula, and get the

following expression: f

³

f

 f ( x  y) y

D 1

³ ¦c e

dy

k

0

Ok ( x  y ) D 1

dy

y

k

0

(3.64)

f

³ ¦c e k

e

dy

y

k

0

We substitute in the above expression (3.64), Ok y

Ok x Ok y D 1

 Odz . In it the y limits are from 0 to f ; so, the

 z , dy

k

f

D ³  f ( x  y) y

changed variable z will have limits from 0 to f and we express the above (3.64) integral

0

as in the following steps: D 1 ³  f ( x  y ) y dy

k

kx

O ³ ¦c e k

e

Ok y D 1

e

z

D 1

  z

k

k

Ok x

Ok

 ³e

D

( 1) * (D ) D

( 1) * (D )

 z D 1

z

dz

(3.65)

0

ck e

¦

Ok x

D

Ok

k

get:

f

( 1)D

OkD

k

0

dz

Ok

¦c e

f

dy

y

k

y 0

z f

z 0

kx

O ³ ¦c e

0

³

dy

y f

f

We have used * (D )

1

D

D ³  f ( x)  dx

 z D 1

e z

dz , i.e. the definition of the gamma function in the above derivation (3.65) and we

D

³  f ( x )  dx The above expression (3.66) is for Ok ! 0 and f ( f )

f

1

D

D

( 1) * (D )

³  f ( x  y) y

D 1

dy

(3.66)

0

0 . Say we write the above fractional integration of the order

(1  D ) ! 0 , the formula will look as follows:

³

(1D )

 f ( x) (dx)

1D

( 1) 1 1D

Differentiating one whole time (3.67), we get by using D I D

d f ( x) dx

D

f

1

(1D )

1

DD

* (1  D )

 (1D )

1 1D

( 1)

* (1  D )

f

³ 0

D

d

³  f ( x  y) y

0 D

D

dy

(3.67)

the following:

 f ( x  y ) y  D dy dx

(3.68)

Therefore, for ( n  D ) ! 0 order integration and then followed by n whole differentiation we get the fractional order derivative formula of Liouville, for ( n  1)  D  n as:

Fractional Derivatives D

d f ( x) dx n n D

n

 ( n D )

The use of D I D D process is from the zero state. Put f ( x )

¦

c e k k

Ok x

D

D

115 f

1

D

( 1)

n D

*(n  D ) ³

d

n

 f ( x  y ) y nD 1dy dx

0

is valid as an initial condition, f ( f )

(3.69)

n

0 and the differentiation integration

d  D eOk x dx  D

, using the fractional integration of the order D ! 0 , which is

D Ok x

Ok e

, and write

Liouville’s fractional integral formula. This integral is finite for limits 0 and f , for Ok  0 . Thus we write: D

f

D

ª ¦ ck (OkD )eOk x º ¬ k ¼0

³  f ( x)  dx ¦

Now we substitute f ( x )

c e k k

Ok x

(3.70)

, in the RHS of (3.61) i.e. Liouville’s fractional integration formula and get the

following: f

f

O ( x  y ) D 1 D 1 y dy ³  f ( x  y ) y dy ³ ¦k ck e k

0

0

(3.71)

f

³¦c e k

dz Ok

z , dy

e

y

dy

k

0

We put in the above expression (3.71) Ok y

Ok x  Ok y D 1

; we get the following steps:

f

f

D 1

O x z D 1 ³  f ( x  y ) y dy ³ ¦k ck e e  Oz k

0

 dz

Ok

k

0

f

¦c e k

Ok x

OkD

k

* (D )

 ³e 1

 z D 1

dz

z

(3.72)

0

kx

¦ (O ) D c e O 

k

k

k

* (D ) From the above steps (3.72) we get the following expression:

The above expression (3.73) is for Ok  0 and f (f )

f

1

D

D ³  f ( x) (dx)

D

D ³  f ( x) (dx)

* (D )

³  f ( x  y) y

D 1

dy

(3.73)

0

0 . Therefore for the ( n  D ) ! 0 integration and then

followed by the n whole differentiation we get the fractional order derivative formula of Liouville, for ( n  1)  D  n : D

d f ( x) dx n n D

n

 ( n D )

The use of D I D D process is from the zero state.

D

D

D

1

f

*(n  D ) ³ 0

d

n

 f ( x  y ) y nD 1dy dx

is valid as an initial condition, f (f )

(3.74)

n

0 , and the differentiation integration

3.6.4 Extension of Liouville’s approach to arrive at formulas of the fractional derivative as a limit of difference quotients In Liouville’s method, we have

dD f ( x ) dxD

¦

c k k

(OkD )eOk x , for the function f ( x )

of exponentials. What are the values of Ok , in the exponential expansion of f ( x ) D

( Ok )

for Ok ! 0 and Ok  0 are the following:

kx

¦ c eO ¦ c eO k k

k k

kx

expanded as the sum ? The expression of

116

Chapter Three

D

1

lim

Ok

D

 hOk

h

hp0

1

D

( 1) lim

Ok

D

1  e

D

hp0

Ok ! 0

D

h

(3.75)

D

1  e hOk

Ok  0

We will find out how in (3.75) the expressions are obtained. Take the function g1 ( x ) (1) g1 ( x ) x 0

ª¬ Ok e

Ok x

º¼ x

0

e

Ok x

with Ok ! 0 .

Differentiating this function i.e. g1 ( x )

Ok . The derivative of g1 ( x ) at x

e

Ok x

at x

0 , we get

0 , with h as an infinitesimal increment in x , we

write the following: (1)

g1 (0)

Placing g1 ( x )

e

Ok x

, and g1(1) (0)

lim

g1 (0)  g1 (0  h )

Ok in (3.76), we have:

lim

Ok

g1 (0)  g1 (  h )

lim

1 e

 Ok x

, Ok ! 0 , and g 2(1) ( x )

h

hp0

(3.77) D

e

 eOk (  h )

 Ok h

§ 1  e  Ok h · lim ¨ ¸ hp0 © h ¹

Ok

Ok (0)

h

hp0

D

e

lim

h

hp0

Ok

Now we write g 2 ( x )

(3.76)

h

hp0

 Ok e

 Ok x

lim hp0

. At x

1

D

1  e  Ok h

D

h

Ok ! 0 (1)  g 2 (0) . As done

0 therefore we have Ok

in the above case for g1(1) (0) , we write the following steps: (1)

g 2 (0)

lim

g 2 (0)  g 2 (0  h )

lim

h

hp0

1 e

lim

g 2 (0)  g 2 (  h ) h

hp0

 Ok (  h )

(3.78)

h

hp0

1 e

lim

Ok h

 Ok

h

hp0

Therefore, we have:

 lim

Ok

1 e

Ok h

(3.79)

h

hp0

From the above expression (3.79) we write: D

Ok In Liouville’s method, we have

dD f ( x ) dxD

¦

c k k

D

( 1) lim hp0

1

Ok h D

1  e

D

h

(OkD )eOk x for the function f ( x )



exponentials. To use the value of OkD , we need the expansion for 1  e

 hOk

D



¦

Ok  0

c e k k



or 1  e

Ok x

hOk

(3.80)

expanded as a sum of

D



. For that, we use the

power series expansion expressed in the following formula: D

(1  x )



We use the above (3.81) to expand 1  e

 hOk

D



D (D  1) 2!

2

(3.81)

x  ...

, for Ok ! 0 and get the following: D

1  e  hOk

Then come the following steps:

1D x 

1 De

 hOk



D (D  1) 2!

e

2 hOk

 ...

(3.82)

Fractional Derivatives D

d f ( x) dx

D

kx

¦ c O D eO k k

117

Ok ! 0

k

lim hp0

lim hp0

lim hp0

lim hp0

¦c



k

k

¦

ck eOk x

¦

ck eOk x

D

h

k

k

¦ ª¬ k

D

ª 1 1  e  hOk «¬ hD

hD 1 hD



º e Ox x »¼

D

1  e  hOk

ª1  D e  Ok h  D (D 1) e 2 Ok h  ...º 2! ¬ ¼

ck e

Ok x



1 hD

D ck e

Ok ( x  h )



1 hD



D (D 1) 2!

c e k

Ok ( x  2 h )

 ...º

¼

ª h1D ¦ ck eOk x  h1D D ¦ ck eOk ( x  h ) º « k » k lim « » O ( x2h) hp0 1 D (D 1)  ... » ck e k ¦ 2! «  hD k ¬ ¼ D (D 1) f ( x  2 h )  ...º lim ª 1D f ( x )  1D D f ( x  h )  1D 2! h h ¼ hp0 ¬ h f º 1 ª m §D · lim D « ¦ ( 1) ¨ ¸ f ( x  mh ) » hp0 h ©m¹ ¬m 0 ¼









(3.83)

We have used the binomial coefficients as appearing in the expansion above (3.83) which are 1 , D ,

D (D 1) 2!

…,

generalised for D as a non-integer, as follows: n!

§n· ¨r¸ © ¹ D

For Ok  0 , we have Ok

r !( n  r )! D 1 hD

lim( 1) hp0

Ok h D

1  e

§D · ¨m¸ © ¹

D!

* (D  1)

m !(D  m)!

m ! * (D  m  1)!

(3.84) D

, we use the power series expansion for (1  x )

as mentioned

above (3.81) to write the following expression D

1  e hOk

From this, we derive the following steps:

1 De

hOk



D (D  1) 2!

e

2 hOk

 ...

(3.85)

118

Chapter Three D

d f ( x) dx

kx

¦ c O D eO k k

D

Ok ! 0

k

lim hp0

lim hp0

lim hp0

¦c

k

k

ª ( 1)D 1  e hOk «¬ hD



¦ (1)D

ck eOk x

¦ (1)D

ck eOk x

D

h

k

º e Ox x »¼ D

1  e hOk

ª¬1  D eOk h  D (D2!1) e 2 Ok h  ...º¼

hD

k

D



ª h1D ck eOk x  h1D D ck eOk ( x  h ) º » lim( 1) ¦ « Ok ( x  2 h ) 1 D (D 1) hp0 »  ... c e k « D k 2! ¬ h ¼ D





ª h1D ¦ ck eOk x  h1D D ¦ ck eOk ( x  h ) º » k k D « lim( 1) « » O ( x2h) hp0 1 D (D 1) ck e k  ... » ¦ 2! «  hD k ¬ ¼



D lim( 1) ª

1

¬ hD

hp0

D

f ( x) 



1 hD

D f ( x  h) 

f

( 1) ª



1 hD

D (D 1) 2!

f ( x  2h)  ...º¼

º §D · ¨ m ¸ f ( x  mh) » © ¹ ¼

(3.86) h m 0 Therefore from Liouville’s postulate for functions that are represented as a series with a sum of exponentials, that is

lim

D

hp0

f ( x)

¦

c e k k

Ok x

« ¦ ( 1) ¬

, the fractional derivative is represented as

m

dD f ( x ) dxD

¦

D Ok x

c O k k k

e

. We arrive at a formula that

contains the representation of the integer order derivatives as limits of difference quotients, those are: D

d f ( x) dx

D

D

d f ( x) dx

D

f

ª1

lim « hp0

D

¬h

¦ (1)

m 0

ª1 lim( 1) « D hp0 ¬h D

m

f

º §D · ¨ m ¸ f ( x  mh) » © ¹ ¼

Ok ! 0

º §D · ( 1) ¨ ¸ f ( x  mh ) » ¦ ©m¹ m 0 ¼ m

(3.87)

Ok  0

where the generalised binomial coefficient (Section 1.12) is the following: * (D  1) §D · ¨m¸ © ¹ * ( m  1)* (D  m  1)

(3.88)

This idea (3.87) was taken up by Grunwald who defined fractional derivatives as limits of the generalised difference quotients; we call this Grunwald’s formula. Grunwald wanted to free the definition of fractional derivatives from their special form of function. Liouville’s postulate was on functions, which are only expanded as an exponential series. Grunwald emphasised that fractional derivatives are ‘interoderivatives’. From calculus based on the limits of the difference quotients, the Grunwald formula is thus the following: N xa k § D · f ( x  kh ) (3.89) F > a, x, D , h@ f ( 1) ¨ ¸ ; N D h ©k¹ h k 0 Grunwald then calls (3.89) the following expression:

¦

D

D

x x

> f ( x )@x

that is the D -th, differential quotient taken over a straight line from a to x .

a

lim F > a , x, D , h@ f hp0

(3.90)

Fractional Derivatives

119

3.7 A repeated integration approach to get the fractional derivative in Riemann-Liouville and Caputo formulations We have seen in Chapter-2 the iterated integrals, and they take the following form:

d

dx d

x u1

2 2

³ ³  f (u1) du du ³ ( x  u )  f (u ) du

f ( x)

2

1

0 0

3

0

x u1 u2

3

dx

x

1

³ ³ ³  f (u1) du du du

f ( x)

3

2

(3.91)

x 2

( x  u )  f ( u ) du 2! ³

1

0 0 0

0

and so on, like in the following:

d

n

dx

n

x u1 u2

un1

0 0 0

0

³ ³ ³ ............. ³  f (u1) du du

f ( x)

n

n 1 .....du1

 

(3.92)

n

x

1

n 1

( x  u )  f ( u ) du ( n  1)! ³ 0

3.7.1 Fractional integration is a must to get a fractional derivative The common point between the ways of obtaining a fractional derivative is fractional integration. This (3.92) we can generalize using the gamma function instead of factorials, i.e. ( n  1)! * ( n ) , for fractional order, as a positive real number n call it X . This we have discussed in Chapter-2, and we express Riemann fractional integration as below:

d

X

dx

X

X

f ( x)

0 Dx

x

1

> f ( x )@

1

X ( x  u )  f ( u ) du ³ * (X )

(3.93)

0

The convergence properties of this formula are best when X has a value between 0 and 1. Let us evaluate the double

> @

integral, in the interval 0, x

of

x , taking X 2 0 Dx

2 , and apply formula (3.93): d

ª¬ x º¼

2

ª¬ x º¼

dx

x

1

2

* (2)

³ ( x  u)

x

u x

³  x u  u du

ª 2 x u3  2 u5 º 5 ¬3 ¼u

3

0

4

x

15

> @

Now let us perform a semi-integration in the interval 0, x of

u du

0

0

(3.94)

5

1 2

x taking X

and applying it to the Riemann

formula (3.93):  12 0 Dx

ª¬ x º¼

d

 12

ª¬ x º¼

dx 1

*  12

1 2



0

1

du

u ( x  u)

0

1

1 2

udu

* ³ 1 2

u

*  ³ ( x  u) 

 12 x

x

1

x

³ 0

u 2

xu  u 1

du

*  12

x

* ³ 1 2

1

0

x

³ 0

x2 4

udu 2

x 4

 u 

(3.95)

udu

x 2 2



2

u 

x2 4

 2 x  u2

120

Chapter Three

 12  x  x sin T

We put u have T

S

2

 2x  cos T dT

so du

and as limits for u

 S2 and for u

0 we have T

x we

. With this change we write the following and proceed:

1

1

2 0 Dx [ x ]

S / 2

³

 12  x  x sin T

*  12 S / 2

x2 4



x2 4

2

sin T

 2x (cos T )dT

S / 2

 12  x  x sin T x (cos T )dT  *  12 S³/ 2  x 1  sin 2 T 2 2 1

S / 2

1

  x  x sin T dT *  S³ 1 2

1 2

 /2

1 *  12

1

T S / 2

*  12

 x2T  x cos2 T T

(3.96)

S / 2

 S2x

 2S x The same we will get via the Euler formula as shown below by putting m X

0 Dx [ x

m

1/ 2 0 Dx

* ( m  1)

]

* ( m  X  1) 1/ 2

ª¬ x º¼

x

1 2

, and X

 12 :

m X

*  12  1 *  12 

1 2

 1

x

11 2 2

*  32 * (2)

(3.97)



S 2 x

*  32 ( x )

x

The above formulas (3.96) and (3.97) demonstrate that the Riemann-fractional integration is the same as we get from Euler’s generalisation.

3.7.2 Different ways to apply a fractional integration formula to obtain a fractional derivative There are two different ways, in which this formula (3.93) of fractional integration might be applied. For example, if we wish to find the

7 3

-rd (say order P

7 3 2 3

7

) derivative of a function that is DxP f ( x ) or Dx 3 f ( x ) then we could begin

3 ). Here we are doing fractional integration first and following that process by a whole differentiation. This is called the Riemann-Liouville or RL (1872) method of fractional differentiation, or left hand definition (LHD). by applying the above formula with X

The X 0 Dx

equation

> f ( x )@

as 1 * (X )

written

³

x

0

X 1

( x  u)

above

and thereafter differentiate the resulting function three whole times ( m

(3.93)

giving

the

Riemann

fractional

integration

formula

i.e.

 f (u ) du , demonstrates the non-local character of fractional operations, because it

explicitly involves an integral, which we have stipulated to range from 0 to x . For any whole number of differentiations we don’t need to invoke this integral, but for a non-integer number of differentiations we must include the effect of this integral, which implies that the result depends not just on the values of functions at x but over the stipulated range from 0 to x . To illustrate the use of the equation for the Riemann fractional integration expression to obtain a fractional derivative we will (again) determine the half-derivative of f ( x ) x . Using the left hand definition, we first apply a half integration to this function using the equation of the Riemann integration with X

1 2

, giving the following:

Fractional Derivatives

 12 0 Dx

x

1

> x@

( x  u) * ³ 1 2

 12

(u )du

put

z , du

xu

 dz

0

0

1 *  12 1 * 4

121

x

0 § 0 1/ 2 · 1/ 2  d d z z xz z ¨ ¸ ³x *  12 © ³x ¹

( x  z)

1

z

x

1

4 x * 3

ª 2 x 2  2 x 2 º ¼ ¬ 3 3

1 2 3

³

( dz )

3

3

(3.98)

2

1 2

2

3 S Then we apply one whole differentiation to give the net result of a half-derivative as: 1

3

§ 4x 2 · x ¨ ¸ 2 S dx © 3 S ¹ d

>@

2 0 Dx x

(3.99)

Formula (3.99) is in agreement with the derived semi-derivative from Euler’s generalisation. In the operator sense we have for Riemann-Liouville (RL) the following fractional derivative: 1

0 Dx

X

0 Dx

2

[ f ( x )]

> f ( x )@

1

D

D

ª D 1 12 f ( x )º ¬0 x ¼

m

ª¬

( m  1)  X  m ,

 ( m X ) 0 Dx

m'

here

f ( x ) º¼



1

1

m

D {

d dx (3.100)

X !0 d

m

D {

m m

dx Alternatively, we could begin by differentiating the function three whole times (taking the nearest integer, say m is just greater than P ; that is, m 3 ), and then apply the above formula (3.93) of the Riemann fractional integration with X

2 3

to “deduct” two thirds i.e.

X

 3  73

(m  P )

2 3

of a thrice-differentiation. This process of first

taking a whole derivative (in this case three) and then applying (the remaining) fractional integration is Caputo’s (1967) method of obtaining fractional differentiation. This process is opposite to that we described for an RL process (3.100). This is also termed the right hand definition (RHD). x , to get a semi-derivative.

In this case, the right hand definition (Caputo) gives the same result as (3.99), for f ( x ) Choose m 1 then differentiate the function f ( x ) this function f

(1)

( x)

x once to have f

(1)

1 , then apply the semi-integration on

( x)

1 that is described in the following expression by using (3.93):  12

0 Dx

x

1

>1@

( x  u) * ³ 1 2

 12

(1)du

0

1

(3.101) 1

ª

º

u x

2( x  u ) ¼u S ¬ 2

We will be getting the same result as (3.101) by using Euler’s formula that is p

0 and m

1

 1 2 . We get the result i.e. 0 Dx 2 [1]

* (1) *  32

x

1

2

2

x S

0

2

x

S

m p 0 Dx [ x ]

* ( p 1) *  p 1 m

x

p m

, placing

, same as obtained in (3.101).

In this process, we first differentiate the function m times, then follow up by a remainder fraction and integrate it by that fraction. In the operator sense the Caputo derivative is as follows (we write C here to distinguish from the Riemann-Liouville fractional derivative): d - 12 C 12 1 ª 1 º here m 1 D { 0 Dx [ f ( x )] 0 Dx ¬ D f ( x ) ¼ dx C X 0 Dx

> f ( x )@

0 Dx

( m  1)  X  m ,

 ( m X )

ª¬ D m f ( x )º¼

m'



m

D {

X !0 d

m

dx

m

(3.102)

122

Chapter Three

The two processes for a semi-derivative of f ( x ) x is shown in Figure-3.4. These two alternatives for fractional derivatives are called the right hand definition (Caputo) and the left hand definition (Riemann-Liouville). Although, these two definitions give the same result in many circumstances especially when the start point of the process is at f or the value of the function at the start point is zero, however, they are not equivalent. This is because (for example) the half-derivative of a constant is zero by the right hand definition. (In Caputo’s process we are first differentiating a constant a whole number of times resulting zero). Whereas the left hand definition gives for the halfderivative of a constant the result given previously as obtained by Euler’s generalisation that is a decaying function. These two are depicted as D

1

2

>K @

K

1

x 2 *  12

K

 and

C

1 Sx

D

1

>K @

2

0 with pre-fix C denoting Caputo.

In general, the left hand definition (RL) is more uniformly consistent with the previous methods, but the right hand definition (Caputo) has also found some applications.

f ( x)

 1 D  2  D (1) f ( x )

1 f  2 ( x)

Caputo Method

1 f  2 ( x)

x

S

f ( x)

0 1. Do one whole derivative of

x

2

f

(1)

x

( x)

1

1

1 2

2. Do half-integration of f (1) ( x)

Riemann-Liouville Method

f

( 1

2)

( x)

4 3 S



f  2 ( x) 1

x3

D (1) D 

f ( x)

1

2

f ( x)

x



f  2 ( x) 1

0

 12

2. Do one whole derivation of f

2

S

1

1. Do half integration for f ( x)

x

(  12 )

( x)



4

3 S



1

x

2

x3

Figure-3.4: Taking the half-derivative of a function by Caputo and the Riemann-Liouville methods

We note that by both methods we are getting a half derivative of f ( x ) it is true because f (0)

x as

2

S

x i.e. the same result. In this case

0 i.e. the start point value of the function and the start point of the half derivative is also 1

from x

0 i.e. we are taking 0 Dx 2 for f ( x ) x . We will elaborate upon this point later. Here we just point out that in all cases the Caputo and Riemann-Liouville derivative need not give the same results.

In the integral form, we write the D ! 0 order fractional derivative in RL form, for a function f ( x ) in the interval

>a, x@ as follows: D

a Dx

> f ( x )@

x ª 1 º m D 1 ( x  u)  f ( u ) du » m « ³ dx ¬ * ( m  D ) a ¼

d

m

(3.103)

In the integral form, we write the D ! 0 order fractional derivative in Caputo form, for a function f ( x ) in the

> @

interval a , x as: C D a Dx

> f ( x )@

1

x

( x  u) *(m  D ) ³

m D 1

f

(m)



( u ) du

a

x

m m D 1 § d f (u ) · x  u ( ) ¨ du m ¸ du * ( m  D ) ³a © ¹

1

(3.104)

Fractional Derivatives

123

3.8 Fractional derivatives for the Riemann-Liouville (RL): ‘left hand definition’ (LHD) The formulation of this definition is done according to the following: Select an integer m greater than the fractional number D ! 0 (i) Integrate the function ( m  D ) folds by the RL integration method. (ii) Differentiate the above result by m -folds The expression is given as: D

a Dt

> f (t ) @

t ª 1 º f (W ) dW » m « D 1 m ³ dt ¬ * ( m  D ) a (t  W ) ¼

d

m

(3.105)

Figure-3.5 gives the process block diagram & Figure-3.6 gives the process of differentiation 2.3 times for a function.

Figure-3.5: Fractional differentiation of the left hand definition (LHD) block diagram

Figure-3.6: Fractional differentiation of 2.3 times in the LHD

In this LHD the limit of integration is from 0 to t , we denote the derivative by notation 0 DtD f (t ) . In fractional calculus we find the limit of the derivative i.e., derivatives are taken in an interval. We call this a ‘forward derivative’. D

Now if the limits of integration are changed to ( t to 0 ) the derivative is denoted as t D0

> f (t ) @

the ‘backward

derivative’. The backward derivative is related to the forward derivative by: t

D0

A fractional derivative of a purely imaginary order i.e. D Riemann-Liouvelli notation (with m 1 ): iT a Dx

> f (t ) @

m

d

m

m D I m t 0

(3.106) > f (t ) @ dt Therefore, in order to obtain a fractional derivative of a function at a point (say 0) we should have the values of these two derivatives be the same: the forward derivative should equal the backward derivative. This implies not only that one should know the function from past to the point of interest (say 0) but also the function should be known into the future – in order to have a fractional derivative at a point! D

( 1)

iT , ( T z 0 ) with a real part as ‘zero’ is expressed in the

> f ( x )@

1

d

x

f (u )

du T * (1  iT ) dx ³ ( x  u ) i

a

and the associated integral of a purely imaginary order in Riemann-Liouvelli definition is the following:

(3.107)

124

Chapter Three

 iT a Dx

> f ( x )@

iT a Ix

> f ( x )@

1

d

* (1  iT ) dx

d

1 iT a Ix

dx

f ( x)

(3.108)

x iT

³ ( x  u )  f ( u ) du a

3.9 Fractional derivatives of the Caputo: ‘right hand definition’ (RHD) The formulation is exactly opposite to the LHD. Select an integer m greater than the fractional number D ! 0 (i) Differentiate the function m times. (ii) Integrate the above result ( m  D ) folded by the RL integration method. In the LHD and RHD the integer selection is made such that ( m  1)  D  m . For example for differentiation of the function by the order D 3.14 we will select m 4 . The formulation of the Caputo RHD is as follows: C D a Dt

> f (t ) @

1

t

d m f (W ) dW m D 1 m

* ( m  D ) ³ (t  W )

t

1

(m)

f

(W )

D * ( m  D ) ³ (t  W )

dW

0

1 m

dW

(3.109)

0

Figure-3.7 gives the block diagram representation of the RHD process and Figure-3.8 represents graphically where the RHD is used for a fractionally differentiating function 2.3 times.

Figure-3.7: Block diagram representation of the Caputo RHD

Figure-3.8: Differentiation of 2.3 times by the RHD

The definitions of Riemann-Liouville fractional differentiation played an important role in the development of fractional calculus. However, the demands of modern science and engineering require a certain revision of the wellestablished pure mathematical approaches. Applied problems require definitions of fractional derivatives allowing the utilisation of physically interpretable “initial conditions” which contain f ( a ), f initial point; and not fractional quantities (presently unthinkable!). The RL definitions require values of fractional initial states like

D 1

a Dt

> f (t ) @ t

a

,

(1)

( a ), f

D 2

a Dt

(2)

( a )... ; where a is the

> f (t ) @ t

a

,…. In spite of

the fact that initial value problems with such conditions can be successfully solved mathematically, their solutions are practically useless, because there is no known physical interpretation for such initial conditions, presently. It is tough to interpret. Caputo requirements are usual initial states i. e. f ( a ), f

(1)

( a ), f

(2)

D

( a )... . For the LHD D C z 0

D C * (1D ) t

, the

derivative of constant C is not zero. This fact led to using the RL or LHD approach with the lower limit of differentiation as f . In the physical world this poses a problem.

Fractional Derivatives

125

The physical meaning of this lower limit extending towards minus infinity is starting the physical process at time immemorial! In such cases, transient effects cannot then be studied. However making start point a as f is a necessary abstraction for consideration of the steady state process, for example for study of sinusoidal analysis for a steady state fractional order system. While today we are familiar with the interpretation of the physical world with integer order differential equations, we do not (currently) have a practical understanding of the world with fractional order differential equations. Our mathematical tools go beyond the practical limitation of our understanding. Therefore, the process is still on to ‘generalise’ the concepts for use in the practical world.

3.10 Relation between Riemann-Liouville and Caputo derivatives 3.10.1 Riemann-Liouvelli and Caputo derivatives related by initial value of the function at start point of fractional derivative process The RL derivative of a function y (t ) from 0 to t of fractional order n such that 0  n  1 is expressed as the derivative of the fractional integral as follows: n

d y (t ) dt We will evaluate

d dt

1

t

d

n

(t  W )  y (W ) dW * (1  n ) dt ³

n

(3.110)

0

ª t (t  W )  n  y (W ) dW º by first doing the evaluation of ¬« ³0 ¼»

t

n

³ (t  W )  y (W ) dW ,

by using the

³  g (W ) dW dW

(3.111)

0

integration by parts formula W t

t

ª f (W ) g (W )dW º ³ ¬ ¼W

³0  f (W )  g (W ) dW

³  t



0

0

³

(t  W )

n

W t

(1)

(W )

t

0

(1)

This is given by assuming in the interval 0 to t that the first derivative of represented by the following expression: t

f

y (t ) exists, call it y (t ) , and that it is t





ª y (W ) (t  W )  n dW º   y (1) (W ) (t  W )  n dW dW ³ ³ ¬ ¼W 0 ³

 y (W ) dW

0

0





t

W t



ª y (W )  (t W )(  n1) º   y (1) (W ) (  n 1) «¬ »¼W 0 ³ 0 t

y (0) (  n  1)

t

 n 1



³ 

(1)

y (W )

0





( 1)( t W ) n1 (  n 1)

( 1)( t W ) n1 (  n 1)

dW

(3.112)

dW

Now we take one-whole derivative of (3.112) with respect to t and write the following:

d ª dt ¬«

t

³  y (W ) (t  W ) 0

n

dW º

y (0)

¼»

d ª (t ) n1 º dt ¬  n1 ¼ t



³y

(1)

0

t

y (0)

§© ddt ª¬ (1)((tnW1))

(W ) ¨

 n 1

º ·¸ dW ¼¹

(1)

y (W )

(3.113)

 n 11 ª¬(  n  1)t  n 11 º¼  ³ (  n  1)(t  W )  dW (  n  1) (  n  1) 0

y (0) t

n

d

* (1  n ) dt



³y

(1)



n

(W ) (t  W ) dW

0

After the above result (3.113), we multiply both sides by

1

t

1 * (1 n )

and write the following expression:

t n

³ (t  W )  y (W ) dW 0

(3 .114)

t

§ y (0) ·  n (t  W )  y (W ) dW  ¨ ¸t , ³ * (1  n) 0 © * (1  n) ¹ 1

n

(1)

0  n 1

The LHS of the above expression (3.114) is the RL derivative of the fractional order 0  n  1 and the first term of the RHS of the expression is the Caputo derivative. Therefore, we write the RL fractional derivative of order 0  n  1 as

126

Chapter Three

the Caputo derivative plus a decaying term based on the initial value of function y (0) at the start of the fractional differentiation: n

d y (t ) dt

1

d

t n

(t  W )  y (W ) dW * (1  n ) dt ³

n

0

(3.115)

t

1

§ y (0) ·  n (t  W )  y (W ) dW  ¨ ¸t ; ³ * (1  n ) 0 © * (1  n) ¹ (1)

n

0  n 1

Therefore, for 0  D  1 , we write the relationship between the RL and Caputo derivative of a function f ( x ) as follows: D

a Dx

For D 1

0 Dx

2

1 2

, we get

> f ( x )@

C 12 0 Dx

1

>

2 f ( x) a Dx

C 12 a Dx

@

> f ( x)@  f S(0)x .

C D a Dx

> f ( x )@

> f ( x)@  f ( a*)(1xa )

1

2

1 2

> f ( x )@  f ( a )

We note that f ( a ) is constant, and f ( a )

of the constant f ( a ) , with start point a , that is

D

a Dx

> f ( a )@

f (a)

D

D

( xa ) * (1D )

(3.116)

* (1  D )

0 and using *  12

With a

.

( x  a)

( x  a ) D * (1D )

S ,

we write

is the RL fractional derivative

. Therefore, we can write the following

steps: D

a Dx

C D a Dx

> f ( x )@

> f ( x )@  f ( a )

C D a Dx

( x  a)

D

* (1  D )

(3.117)

> f ( x ) @  a Dx > f ( a ) @ D

Form (3.117) we have the following relation: C D a Dx

> f ( x )@

D

a Dx

> f ( x )  f ( a )@

(3.118)

That is for the Caputo derivative for a fractional order 0  D  1 the function f ( x ) one needs to offset the function by subtracting the constant value at the start point f ( a ) , and take the Riemann-Liouville (RL) derivative of the offset function i.e. f ( x )  f ( a ) .

3.10.2 A generalisation of a fundamental theorem of integral calculus is the relation between Riemann-Liouville and Caputo fractional derivatives Look at the expression we wrote that is: D

a Dx

1 a Dx

> f ( x )@ ª¬

C D a Dx

 (1D ) a Dx

> f ( x )@  f ( a )

f ( x ) º¼

 (1D ) a Dx

( x  a)

ª¬

D

* (1  D ) 1 Dx

f ( x ) º¼  f ( a )

( x  a)

D

(3.119)

* (1  D )



In this (3.119) we take the limit D p 0 ; and write the following: 1

ª¬ a Dx1 f ( x ) º¼ a Dx1 ª¬ D1 f ( x ) º¼  f ( a ) x ªx § d d · º f ( u )d u « ³ ¨ f ( u ) ¸ du »  f ( a ) ³ dt a ¹ ¼ ¬ a © du

D

(3.120)

The above expression (3.120) is a fundamental theorem of integral calculus, where the differentiation of integration is a constant (i.e. the value of the function at the start point of integration) plus the integration of a differentiated function. Well if the start point of the function is zero the process of differentiation of integration commutes with the process of integration of differentiation; that is,that they are equal. Therefore, we can state that the relationship between the RL and Caputo derivatives is the generalisation of this fundamental theorem of integral calculus.

Fractional Derivatives

127

3.10.3 Further generalisation of the relationship between the Caputo and Riemann-Liouville fractional derivatives We generalize further, for any D need not be between zero and one, and we write it as: D

a Dx

m 1

C D a Dx

> f ( x )@

> f ( x )@  ¦ k 0

C D a Dx

> f ( x )@  f ( a )  ... f

( m 1)

f

(k )

( x  a)

* ( k  D  1)

( x  a)

D

* (1  D )

(a)

(a)

( x  a)

(1)

 f

k D

1D

(a)

( x  a)

(3 .121)

* (2  D )

m 1D

*(m  D ) Here in (3.121), we have ( m  1)  D  m , where m is a positive integer number just greater than the real number for the fractional order D . The above (3.121) formula is applicable, only if the ( m  1) integer order derivatives exist at the start point of the function. The above (3.121) relationship between the RL and Caputo derivatives is the correction by an offset factor at the start point of the function. If the start point of the function is zero that is f ( a ) 0 , then, the RL and Caputo derivatives are the same, or if the start point a is at f , and f ( f ) 0 at which all the function values are zero, then also both are the same. C D a Dx

> f ( x )@ D f Dx > f ( x )@

> f ( x )@ C D f Dx > f ( x )@

D

a Dx

f (a)

0

f ( f )

(3.122)

0

We take the following: D

a Dx

C D a Dx

> f ( x )@

> f ( x )@  f ( a )

 ... f C D a Dx

( m 1)

(a)

(2)

(a)

f

(1)

(a) ,

f

(2)

(a) ,

f

(3)

( a ).......... f kD

derivatives that is, by using a Dx

( m 1)

D

f (a)

> f ( x )@  f

D

a Dx

 f

(1)

1D

(a)

( x  a)

* (2  D )

( x  a)

D

* (1  D )

(a)

(a)

( x  a)

 (D 1)

* 1  (D  1)

(3 .123)

 (D  2)

( x  a)



 (D 1 m )

* 1  (D  1  m)

a i.e. at the start point, of fractional derivative, namely

( a ) are constants. We can write the above (3.123) in terms of fractional

> f ( a )@ a Dx

(1)

m 1D

* 1  (D  2)

The values of integer order whole derivatives of f ( x ) , at x

 f

*(m  D )

( x  a)

( m1)

...  f

D

* (1  D )

( x  a)

> f ( x )@  f ( a )

 f

( x  a)

(2)

> f ( x )@ D 2

 a Dx

( x  a ) kD * (1 kD )

C D a Dx

(a)

, as shown below:

> f ( x )@  f ( a )

( x  a)

D

* (1  D )

 (D  2)

* 1  (D  2)

C D a Dx

( x  a)

 ..  f

( m 1)

 f

(a)

(1)

(a)

( x  a)

 (D 1)

* 1  (D  1)

( x  a)

 (D 1 m )

* 1  (D  1  m)

(3 .124)

> f ( x)@  a DDx > f (a)@  a DDx 1 ª¬ f (1) (a)º¼

ª¬ f (2) ( a )º¼  ....  a DDx  m 1 ª¬ f ( m 1) ( a ) º¼

For ( m  1)  D  m we have a useful relationship between the Caputo & RL derivative i.e. C D a Dx

> f ( x )@

§ a DDx > f ( a )@  a DDx 1 ª¬ f (1) ( a )º¼ · ¨ ¸ a Dx > f ( x )@  ¨ ... a DDx  m 1 ª f ( m 1) ( a )º ¸ ¬ ¼ © ¹ D

(3.125)

128

Chapter Three

3.10.4 The Caputo and Riemann-Liouville fractional derivatives need an area under the shape changing curve Concluding this section, we mention that the RL or Caputo derivatives both have a fractional integration process, therefore we are dealing with the area under the shape changing curve, as we discussed in the previous chapter (Section-2.21). The RL derivative is a process first of fractional integration, then followed by an integer-order whole differentiation. Thus we are required to find the area under a ‘shape-changing’ curve ( f ( x0 ) , f1 ( x1 ) , f 2 ( x2 ) …) as we move along the points x0 , x1 , x2 ,… and after that at each point of interest we have to find the integer-order whole derivative. Thus, the process is of finding a slope of area under the shape changing curve! The Caputo process is the reverse of this; we have to find an integer order whole derivative first at the point of interest, and then find the area under the shape changing curve ( f ( n ) ( x0 ) , f1( n ) ( x1 ) , f 2 ( n ) ( x2 ) …) as we move along the points

x0 , x1 , x2 ,…. and so on. An interesting point is that both these processes require values of the function or values of the integer whole-derivative of the function from the start point to the point of interest; indicating that the RL or Caputo fractional derivative processes have memory.

3.11 Can we change the sign of order of the Riemann-Liouville fractional derivative to write the RL fractional integration formula? In the previous chapter (Section-2.14), we noted that it was difficult to change the sign of a fractional order for a fractional integration formula; to directly get the formula of a fractional derivative. Then we explained via the concept of an analytic continuation that we sought to obtain the validity of this process of sign change. Here we see the reverse operation. The expression given for the RL fractional derivative of the order D ! 0 such that ( m  1) d D  m where m 1, 2, 3, ... is the following:

1

> f ( x )@

D 0 Dx

d

x

m

* ( m  D ) dx

f ( y)

³ ( x  y )D

m

1 m

dy

(3.126)

0

The formula for the RL fractional derivative for order D so that 0 d D  1 is the following: 1 d x f ( y )dy D (3.127) ; 0 dD 1 0 Dx > f ( x )@ D * (1  D ) dx 0 ( x  y ) In the above formula (3.127) we reverse the sign and put D instead of D and write the following steps to see that this change of sign for the RL fractional derivative formula leads to a fractional integration formula: 1 d x  f ( y ) dy D 0 Dx > f ( x )@ D * (1  D ) dx 0 ( x  y )

³

³

1

d

x

* (1  D ) ³

ª¬ ( x  y )D º¼  f ( y ) dy dx

0

1

x

* (1  D ) ³

0

1

D  * (D ) ³

D 1

D ( x  y) x

0

1

* (D ) ³

x

0

D

0 Ix

 f ( y ) dy ,

D 1

D ( x  y) D 1

( x  y)

* (D 1) D  * (D )

(3.128)

 f ( y ) dy

 f ( y ) dy

> f ( x )@

The above (3.128) steps indicate that if we change the sign of the order in the RL-fractional derivative we get the formula of the RL fractional integration as represented below for 0 d D  1 : D

0 Dx

D

0 Ix

> f ( x )@

> f ( x )@

1

d

* (1  D ) dx ³ 1

d

* (1  D ) dx

³

x

 f ( y ) dy D

0

( x  y)

x

 f ( y ) dy

0

( x  y)

D

(3.129)

Fractional Derivatives

129

However, the converse was not quite true; that we have seen in the last chapter. Let us take the case of the RL fractional derivative formula for 1 d D  2 , i.e.

1

> f ( x )@

D

0 Dx

d

2

x

³

 f ( y ) dy ;

1dD  2 D 1 * (2  D ) dx 0 ( x  y ) Now we do the sign change as we did in the earlier case (3.128), demonstrated in the following steps: D

0 Dx

1

> f ( x )@

d

1

d

0

dx

* (2  D ) ³

x

0

D 0 Dx

* (2  D )

> f ( x )@

2

x

³

* (2  D )

Using

2

* (2  D ) dx 1

2

2

 f ( y ) dy

x

³

2

0

( x  y)

D 1

(D  1)(D )( x  y )

x

0

x

0

 f ( y ) dy

(3 .131)

D (D  1)* (D )

D (D  1)* (D ) ³

* (D ) ³

D 1

ª¬ ( x  y )D 1 º¼  f ( y ) dy

1

1

(3.130)

D 1

( x  y)

D 1

(D  1)(D )( x  y )

> f ( x )@

D

 f ( y ) dy

 f ( y ) dy

0 Ix

The above steps (3.131) indicate that if we change the sign of the order in the RL-fractional derivative we get the formula for the RL fractional integration represented below for 1 d D  2 : D

0 Dx

D

0 Ix

> f ( x )@

1

d

2

* (2  D ) dx 1

> f ( x )@

d

2

³

2 2

³

 f ( y ) dy

x

D 1

0

( x  y)

x

 f ( y ) dy

(3.132)

D 1

* (2  D ) dx 0 ( x  y ) We can carry on for other intervals ( m  1) d D  m and write a different form of a fractional integration formula by changing of sign of fractional order D from the RL fractional derivative formula; as indicated below: D

0 Dx

D

0 Ix

> f ( x )@

> f ( x )@

1

d

m

* ( m  D ) dx 1

d

m

m

* ( m  D ) dx

m

x

f ( y)

³ ( x  y )D

1 m

dy

0

x

f ( y)

³ ( x  y) D

 1 m

(3.133)

dy

0

The above (3.133) is a useful observation that we will apply in a unifying fractional differentiation and in a fractional integration formula for discrete numerical evaluation. Thus, fractional integration of the order ( m  1) d D  m is the process of first fractionally integrating ( m  D ) folds by the RL fractional integration formula, and is then followed by m  whole differentiation. Thus, we have: I

D

> f ( x )@

D

m

I

( m D )

> f ( x)@ ;

(3.134)

( m  1) d D  m

If we require the finding of a fractional integration of order D 2.3 , we choose m 3 then we do the fractional integration m  D 5.3 folds, followed by thrice differentiating the result, to get 2.3 fold fractional integration of the function. To get f

( 2.3)

i.e. fractional integration of order 2.3 , we refer Figure-3.6. In this figure we first take the function f

by process of fractional integration to the point in number line towards left side at f

( 5.3)

(0)

and then we perform three

whole differentiations, i.e. moving on the number line of Figure-3.6 towards right to the point f

( 2.3)

.

3.12 The Weyl fractional derivative We have seen in the previous chapter (Section-2.15) that the fractional integration of the Weyl type is given as the following expression:

130

Chapter Three

X xWf

f

1

> f ( x )@

1

* (X )

X ³ ( y  x )  f ( y ) dy

X !0

(3.135)

x

Like Riemann-Liouville, the definition of Weyl’s fractional derivative of the function is defined for an integer just greater than the fractional index of the derivative. For m  1 d D  m , where m  ` , the Weyl’s fractional derivative definition is:

m D

In (3.136), xWf

> f ( x )@

d

 W > f ( x ) @

(3.136)

is the Weyl fractional integration of the order ( m  D ) .

In the integral transform

> f ( x )@

( 1)

m

m

m D x f

D

xWf

dx

m

representation, we have the following: D

xWf

> f ( x )@

m

§

1

( 1) ¨

d

m

© * ( m  D ) dx

m

³

f

x

( y  x)

m D 1

 f ( y ) dy

·¸¹

(3.137)

m 1 d D  m Let f ( x )

e

 px

, with p ! 0 , and under the substitution u  x

gamma function defined as * (D )

³

f

0

( y / p ) , and with the integral representation of the

 t D 1

e t

dt , we get the following expression for the Weyl fractional integration:

ª¬e  px º¼

D xWf

1

p

ª¬e  px º¼

p

 (1X )  px

e

e

du

x

p

 (1X ) xWf

D 1  pu

(u  x ) * (D ) ³ e

Therefore in (3.138) for D

f

D

f

 px

y * (D ) ³

D 1  y

e dy

(3.138)

0

D  px

e

(1  X ) , where 0  X  1 , we have the Weyl fractional integration of e

 px

as

. Now we calculate the X -th order Weyl fractional derivative, from (3.137), where

0  X  1 . The nearest integer for 0  X  1 is m 1 , this gives the Weyl fractional derivative as: d X  (1X ) ª  px º¼ ( 1) ª¬e  px º¼ xWf ¬e xWf dx d ª p  (1X ) e  px º¼ ( 1) p  (1X ) (  p )e  px ( 1) dx ¬









(3.139)

X  px

p e

We observe that the Weyl fractional derivative of e integral.

 px

could be obtained by a sign change in the Weyl fractional

3.13 The most fundamental approach for repeated differentiation This approach is also called the limit of finite difference. In classical calculus, we define the derivative as a limit of finite difference as follows: d f ( x )  f ( x  h) (3.140) f ( x ) lim hp0 dx h Similarly:

d

2

dx d

2

f ( x)

lim

f ( x )  2 f ( x  h)  f ( x  2h) h

hp0

3

2

f ( x )  3 f ( x  h )  3 f ( x  2 h )  f ( x  3h )

(3.140)

f ( x ) lim 3 3 hp0 dx h Repeating this n times for any positive integer we get the following expression:

d

n

dx

n

f ( x)

lim hp0

1 h

n

n

¦ (1) j 0

j

§n· ¨ j ¸  f ( x  jh) © ¹

(3.142)

Fractional Derivatives

131

If the n -th derivative of f ( x ) exists, this equation (3.142) does indeed define

dn f ( x ) dx n

as an unrestricted limit, i.e. as

the limit h tends to zero through values that are totally unrestricted. In order to unify this formula (3.142) with one that defines an integral as the limit of a sum, it is desirable to define derivatives in terms of a restricted limit, namely, as a limit in that h tends to zero through discrete values only. To do this, choose hN { xN a , where a is the start point of the function, and N 1, 2, 3, ... . The number a  x plays the role of a lower limit. If the unrestricted limit exists so does the restricted limit and they are equal; the n -th derivative may be defined as follows:

d

n

dx

f ( x)

n

n

1

lim

¦ (1)

j

hN p 0 h n N j 0

§n· ¨ j ¸  f  x  jhN © ¹

(3.143)

Now since:

§n· ¨ j¸ 0 © ¹

for

(3.144)

j!n

when n is an integer, the above expression (3.143) may be re-written as:

d

n

dx

n

f ( x)

N 1

1

lim

¦ (1)

hN p 0 h n N j 0

j

§n· ¨ j ¸  f  x  jhN © ¹

N 1

(3.145)

§n· xa lim n ¦ ( 1) ¨ ¸  f  x  j  N N nf  xNa j © ¹ j 0 1

The above expression (3.145) is defining

dn dx n

f ( x ) with the understanding that the limit exists in the usual

unrestricted sense. We verify the above expression for n

d

j

2

ª x 4 º¼ 2 ¬ dx

lim

2 and f ( x ) 2

1

¦ (1)

j

hN p 0 h 2 N j 0

1

lim

hN p 0 h 2 N

4

x in the following steps:

4 §n· ¨ j ¸  x  jhN © ¹

ª x 4  2  x  hN 4   x  2hN 4 º ¬ ¼

(3.146)

2 2 lim ª¬12 x  24 xhN  14 hN º¼

hN p 0

12 x

2

> @

We can generalise the summation in (3.145), i.e. ¦ N 1 ... by generalising the binomial coefficients upper summation limit that is n , with the following formula using with the gamma function (Chapter-1): ( 1)

j

§ n · § j  n  1· ¨ j¸ ¨ j ¸ © ¹ © ¹

*( j  n)

(3.147)

* (  n )* ( j  1) We write the n -th derivative to any arbitrary order q by the following formula: q

d f ( x) q

>d( x  a)@

§  xN a  q lim ¨ N nf ¨ * (  q ) ©

N 1

·

*( j  q) ¦ *( j  1)  f  x  j  ¸¸ xa N

j 0

(3.148)

¹

Note that here q is an arbitrary positive or negative. A negative q implies fractional integration, while a positive q implies fractional differentiation. Even for q a non-negative integer say n , so that * (  n ) is infinite, but the ratio * ( j n) *(n)

is finite. Note that the above expression (3.148) involves only the evaluation of the function itself while no

explicit use is made of the derivatives or integrals of the function f ( x ) .

3.14 The unified context differentiation/integration Grunwald-Letnikov formula 3.14.1 Generalisation by use of the backward shift operator defining the repeated differentiation and integration To clarify the situation, as we stated in the previous section, let us start from scratch for the operation of differentiation and integration in a unified context. Let us consider the arbitrary smooth function f ( x ) and divide it with slices of h , as depicted in Figure-3.9.

132

Chapter Three

f ( x) f ( x  5h )

f ( x) f ( x  4h)

f ( x  h) f ( x  3h) f ( x  2h)

x x  4h

x  5h

xh

x  3h x  2h

x

Figure-3.9: The function is divided into a slice

Here we should take note that h we regard as hN , as discussed previously in Section-3.13, without restriction on the slice. We define a backward shift operator E h operating on f ( x ) at x giving the value of the function at x  h that is f ( x  h ) .

E h > f ( x )@

def

f ( x  h)

(3.149)

f ( x  Nh )

(3.150)

The shift operator E hN denotes N back shifts as: N

Eh

> f ( x )@

We apply this defined back shift operator as follows: d

f ( x)

dx

f ( x )  f ( x  h)

lim

h

hp0

lim

f ( x )  E h > f ( x )@ h

hp0

(3.151)

§ 1  Eh · ¸ > f ( x )@ hp0 © h ¹

lim ¨

Therefore, with this shift operator E h we can write the one whole differentiation and one-whole integration. The one whole differentiation is: 1

1

D

> f ( x )@

§ 1  Eh · f ( x )  f ( x  h) lim ¨ f ( x ) lim ¸ hp0 hp0 h © h ¹

(3.152)

The one whole integration is: 1

D

1

> f ( x )@

§ 1  Eh · lim ¨ ¸ f ( x) hp0 © h ¹



2



lim ªh 1  E h  E h  .... f ( x ) º hp0

¬

(3.153)

¼

lim ª¬h  f ( x )  f ( x  h )  f ( x  2 h )  ....  f (0) º¼ hp0

We have used the infinite series expansion (1  x)

1

2

3

1  x  x  x  ... in the above derivation (3.153). However, 1

instead of taking the infinite series expansion of 1  E h , we truncate at f (0) . As the limit, h p 0 the operation 1

D is simply differentiation, and operation D

1

is simply integration. We do n times and write: n

D

n

> f ( x )@

§ 1  Eh · lim ¨ ¸ f ( x) hp0 © h ¹

(3.154)

Fractional Derivatives

133

This reproduces ordinary whole derivatives. For example: 2

D

2

1  2E h  E h > f ( x)@ § 1  Eh · lim ¨ f ( x ) lim ¸ 2 hp0 hp0 h © h ¹

> f ( x )@

2





2

f ( x )  2 E h > f ( x )@  E h > f ( x )@

lim

h

hp0

(3.155)

2

f ( x )  2  f ( x  h)  f ( x  2h)

lim

2

h In the limit h p 0 , which is illustrated above (3.155), how we recover the binomial coefficients for any whole number hp0

of differentiation D

n

> f ( x )@ : D

n

> f ( x )@

However, strictly speaking for (3.155) i.e. the expansion of D

lim hp0

2

n

1 h

¦ (1)

n

j

j 0

> f ( x )@ ,

§n· ¨ j ¸  f ( x  jh) © ¹

(3.156)

this context makes it clear that we should

actually write the second derivative as: 2

§ 1  Eh · D > f ( x )@ lim ¨ ¸ f ( x) hp0 © h ¹ (1) f ( x )  2 f ( x  h )  (1) f ( x  2 h ) (0) f ( x 3h ) .... (0) f (0) 2

lim

h2

hp0

As in this case n as for the integer in D n

n! j !( n  j )!

is C j

n

for j d n , and C j

n

(3 .157)

> f ( x)@ , all the binomial coefficients after (n  1) are identical to zero.

0 for j ! n . Thus, we truncate the series at j

That

n for n as a positive integer.

Note here that if n is a negative or non-integer then the binomial coefficients do not truncate. xa h

Consequently, the upper summation limit in the formula (3.156) i.e. n is n range of the evaluation of the function is a . We often choose a we write:

d

dx where

0 , by convention, but it is actually arbitrary, and

n n

f ( x)

, where the lower bound on the

lim hp0

1 h

x a ¬« h ¼»

¦ (1)

n

j 0

j

§n· ¨ j ¸  f ( x  jh) © ¹

xa ¬« h ¼» is the floor operator returning the integer number just lower than computed n

xa h

(3.158)

. For any arbitrary

n say X  \ , we write a generalised formula as: X

a Dx

> f ( x )@

X

d f ( x)

>d( x  a)@ lim hp0

For a

X lim hp0

1 X

h

1 X

h

«¬ xh a »¼

¦ (1)

j

j 0

x a ¬« h ¼»

¦ (1)

j

j 0

§X · ¨ j ¸  f ( x  jh) © ¹

* (X  1) j ! * (X  1  j )

(3.159)

 f ( x  jh)

f we thus have the following formula: X

>

f Dx f ( x )

@

X

d f ( x) X

>d( x  f)@

lim hp0

1 X

h

f

¦ (1) j 0

j

* (X  1) j ! * (X  1  j )

 f ( x  jh)

(3.160)

3.14.2 Using the generalised formula obtained by a backward shift operator of fractional integration/differentiation on an exponential function For f ( x ) steps:

ax

e , we do fractional differentiation from f to x by the above formula (3.160) as in the following

134

Chapter Three

ax X f Dx ª ¬e º¼

f

1

lim

¦ (1)

X

h

hp0

j 0

e

ax

e

ax

§X · a ( x  jh ) ¨ j¸e © ¹

§ f · j § X ·  ajh ( 1) ¨ ¸ e ¸ X ¨¦ h ©j 0 © j¹ ¹

lim hp0

lim

1  Xe

X

h

hp0

X

Comparing the power series expansion formula that is

j

(1  x)

 ah

1  Xx 



X (X 1)

X (X 1) 2!

2!

e

2 ah

(3.161)



 ....

2 x  ... with the above obtained

expression (3.161) in the brackets i.e. 1  X e  ah  X (X2!1) e 2 ah  ... , we can write the following: X

ª f Dx ¬e

ax

ax

e

X

º¼ lim X 1  e  ah hp0 h

X

§ 1  e  ah · lim e ¨ ¸ hp0 © h ¹ ax

(3.162)

ax X

e a In the above steps we have used lim hp0 gives the final result expansion for e

 ah

X

f Dx



1 e ah h



a ; which we have derived earlier in this chapter (Section-3.6), that

ª¬e ax º¼ aX e ax . This lim hp0





1 e ah h

a can be simply obtained too by using a series

and then simplifying it.

3.14.3 Generalisation by use of the forward shift operator defining repeated differentiation and integration We have used a backward shift operator in the above derivation. We can define a forward shift operator, and from the following definition of one whole derivative, we can build the generalisation. We proceed with the following steps: f ( x  h)  f ( x) 1 D f ( x ) lim hp0 h (3.163) f ( x  2h)  f ( x  h)  f ( x) 2 D f ( x ) lim 2 hp0 h Continuing n times, we obtain the following useful expression:

D Say we have a function with the end-point x

n

>

@

>

@

> f ( x )@

b , then n

n

1

lim

h

hp0

¦ (1)

n

j

j 0

§n· ¨ j ¸  f  x  ( n  j ) h © ¹

(3.164)

«¬ b h x »¼ , and we generalise this derivative with a forward

shift operator for any non-integer X as: X

x Db

For b

> f ( x )@

lim hp0

«¬ bh x »¼

1 X

h

¦ (1) j 0

j

§X · ¨ j ¸  f  x  (X  j ) h © ¹

(3.165)

f we get: x Df > f ( x )@

X

lim hp0

1 X

h

f

¦ (1) j 0

j

§X · ¨ j ¸  f  x  (X  j ) h © ¹

(3.166)

3.14.4 Using the generalised formula obtained by a forward shift operator of fractional integration/differentiation on an exponential function Now, if we want to obtain the D -derivative of the function f ( x ) steps:

e

ax

from point x to f ; we write the following

Fractional Derivatives

ax D x Df ª ¬e º¼

h e

lim

h e

lim

ax

D

h

hp0

§ D · ax ah D  k ¨ k ¸ e e © ¹

k

k 0

ax

§ D · a  x  (D  k ) h ¨ k ¸e © ¹

k

k 0

¦ (1)

D

hp0

h

f

D

hp0

¦ (1)

D

hp0

1

lim

f

1

lim

f

k

¦

§D ·



e

ah D

lim hp0



e

ax

D

h

ah D  k



( 1) ¨ ¸ e ©k¹ 0

k

135

hp0

ah D 1



D e

e

ah

lim

( 1) 

e

ax

f

D ¦ §¨© k ·¸¹  e

D

h

ah D  k

( 1)

k

k 0

D (D 1) 2!

ah D  2

e



(3.167)

expansion

formula

2 ( 1)  ....

D



1

D

§ e ah  1 · lim e ¨ ¸ hp0 © h ¹ ax

ax D

In

the

(x  y)

n

above n

x  nx

D ax result x Df ª¬e º¼

e a (3.167) we

derivation n 1

y

n ( n 1) 2!

x

have

n2 2

y  ... , with n

D, x

g

(1)

( x)

1, x 0

e

the

ah

power

series

 ah

e 1 h

1 . The lim hp0

and y

a , gives us the

D ax

a e .

This is proved via similar lines as done for lim hp0 g (0)

used

and

g

lim hp0



(1)

a.

(0)

g ( h )  g (0) f

g

The

(1)



1 e ah h

( x)

, which is equal to g

(1)



a . That is take g ( x )

lim hp0 ( x)



g ( x h) g ( x ) h

x 0

lim hp0

,

 eah 1 h

and

a.

e

ax

then g

x

at

0,

(1)

ax

( x)

ae ,

we

have

These results are otherwise

ax

obtained via the series expansion of e .

3.14.5 The final unified formula for fractional derivatives and integration: the Grunwald-Letnikov formula In this section, what we observed is the fractional differentiation which is the like weighted sum of all the values of the function from a to x or from x to b ; just like integration. In an earlier chapter (Section-2.20) we have generalised the repeated integration and obtained the formula for fractional integration of the order D ! 0 as a limit of the sum, and we write that as:

d

D

f ( x) D

>d( x  a )@

§  xN a D lim ¨ N nf ¨ * (D ) ©

N 1

·

*( j  D ) ¦ *( j  1)  f  x  j  ¸¸ xa N

j 0

(3.168)

¹

We write the E -th derivative to any arbitrary E ! 0 by the following formula, which we derived previously (Section3.13), that is: E

d f ( x) E

>d( x  a )@

§  xN a  E lim ¨ N nf ¨ * (  E ) ©

N 1

*( j  E )

·

¦ *( j  1)  f  x  j  ¸¸ xa N

j 0

(3.169)

¹

Therefore, in light of these deductions we write a unified formula for fractional differ-integration (differentiation and integration) of any arbitrary order D as: D

d f ( x) D

>d( x  a)@

§  xN a D lim ¨ N nf ¨ * ( D ) ©

N 1

*( j  D )

·

¦ *( j  1)  f  x  j  ¸¸ xa N

j 0

D N 1

§§ N · lim ¨ ¨ ¸ N nf ¨ © x  a ¹ ©

¹

*( j  D )

¦ *(D )*( j  1)  f  j 0

Nx  jx  ja N

(3.170)



· ¸¸ ¹

136

Chapter Three 1 2

With D

, and a

0 we write: d

1

2 0 Dx > f ( x ) @

With D



1 2

and a

1

> f ( x )@

2

dx

1

§ § N · N 1 § (2 j )! · f x  ¸¦¨ j 2 ¸ N nf ¨ x j j  (2 1)(2 !) © ¹ j 0 ¹ ©©

jx N

§ § x · N 1 § (2 j )! · ¸¦¨ j 2 ¸ f x  N nf ¨ © © N ¹ j 0 © (2 j !) ¹

¸¸



lim ¨ ¨

2

·

¸¸ ¹

0 we write:  12 0 Dx

d

> f ( x )@

 12

> f ( x )@

dx (2 j )! S

We have used the formula *  12  j

j

4 j!

1

lim ¨ ¨

2

and *  12

S , to simplify

*  j  12 *  12 * ( j 1)

jx N

·

(3.172)

¹

as described, for the case

 12 in the following expression:

when D

* j 

1 2



§ 1 · § (2 j )! S · § 1 · (2 j )! ¨ ¸¨ 2 j ¸¨ ¸ 2 j *  * ( j  1) © S ¹ © (2 ) j ! ¹ © j ! ¹ (2 j !) 1 2

For D

(3.171)

1 2

, we do the following steps, with *   12

with *  12  k

(2 k )! S 4k k !

2 S and using (2 j  2)!

(2 j )! (2 j )(2 j 1)

(3.173)

and also ( j  1)!

j! j

: * j 

1 2



*  ( j  1) 

1 2



 2 S ( j !)

*   12 * ( j  1)

§ 1 ¨ © 2 S

· §  2( j  1) ! S ¸ ¨ j 1 ¹ © 4 ( j  1)!

·§ 1 · ¸¨ ¸ ¹© j!¹

(3.174)

(2 j )! (2 j )! § 1 · §¨ (2 j )(2 j 1) ·¸ § 1 ·  ¨ ¸ ¨ j 1 j ! ¸ ¨ ¸  2 j (2 j  1)(2 j !) © 2 ¹© 4 4  j ¹© j!¹

When D is negative, the formula gives us a fractional integration and for positive D the formula is a fractional differentiation. These above formulae are useful to compute the fractional differ-integration of a function f ( x ) from a to x .

3.14.6 Reduction of a unified fractional derivative/integration unified formula to a classical derivative and integration The above Grunwald-Letnikov (GL) formula also correctly reproduces the standard classical definition when D is the integer of either sign. Thus when D 1 or D 1 , the GL formula reduces to the Riemann sum limit as follows: 1

> f ( x )@ 1 >d( x  a)@ d

§ § x  a · N 1 · f  x  j  xN a ¸  ¦ ¸ N nf ©© N ¹ j 0 ¹ lim ¨ ¨

(3.175)

or the backward difference quotient limit as follows: 1

> f ( x )@ 1 >d( x  a)@ d

§ § x  a ·1 1 · j lim ¨ ¨ ( 1) f  x  j  xN a ¸ ¦ ¸ ¸ N nf ¨ © N ¹ j 0 © ¹

However, notice that the Riemann sum would converge more rapidly to the integral

³

x

a

f ( x )dx {

d 1 x d( >  a )@1

(3.176)

> f ( x )@

as

N n f if we write the following expression: 1

> f ( x )@ 1 >d( x  a)@ d

§ § x  a · N 1 · f  x   j  12  xN a ¸ ¦ ¸ N nf © N ¹ j 0 © ¹ lim ¨ ¨

(3.177)

Similarly, the difference quotient would converge more rapidly to the true derivative if we write the following expression: 1

> f ( x )@ 1 >d( x  a)@ d

§ § x  a ·1 1 · j lim ¨ ¨ ( 1) f  x   j  12  xN a ¸ ¦ ¸ ¸ N nf ¨ © N ¹ j 0 © ¹

(3.178)

This type of modification gives us the modified Grunwald-Letnikov formula for any arbitrary order D , which is expressed below:

Fractional Derivatives

> f ( x )@ D >d( x  a)@ d

D

§  xN a D lim ¨ N nf ¨ * ( D ) ©

137

N 1

·

*( j  D ) ¦ *( j  1) f  x   j  D  ¸¸ 1 2

xa N

j 0

(3.179)

¹

This modified GL formula we will use in the subsequent section of formulating numerical algorithms to evaluate fractional differ-integrals.

3.15 The whole derivative is local property whereas the fractional derivative is non-local property For n as a non-integer say X ; we write the generalised expression as: X

d

X

dx

f ( x)

lim hp0

1 X

h

«¬ xh a »¼

¦ (1)

* (X  1)

j

j 0

j ! * (X  1  j )

f ( x  jh )

(3.180)

If X is a positive integer the vanishing of binomial coefficients for all j ! n implies we do not really need to carry the summation in the above formula beyond j n , and in the limit h p 0 , the n values of f ( x  jh ) with non-zero coefficients all converge on point x . Therefore, the n whole derivative is the local property of the function. This process of classical Newtonian calculus, where whole derivatives are local, or the point property is referred to as a Markovian process. However, in general, the binomial expansion behaves infinitely and may have non-zero coefficients so the result depends on the value of the function at x plus all the way down to a ; that is, the start point of the function. In that case, we are evaluating the derivative (note this is not the same as slope at point) for the entire interval a to x . The processes where the value of the function at all the previous points is considered to arrive at the derivative are called non-Markovian processes. Therefore the fractional derivative (generalisation of the whole order derivative), is not a local property but a non-local operation and suited for dynamics with non-Markovian systems. Thus, we note that the generalised derivative depends on the value of the function f ( x ) over the entire range a to x .

0,1, 2,.... ¬« x h a ¼» . Meaning that, the argument of the

In the formula the term f ( x  jh ) , has j which ranges from j

function f ranges from point x , down to the start point of the function. It just so happens that this non-locality disappears for positive whole derivatives, when X is a positive integer, say n . However, for n X as a real number, the property is non-local, thus requiring all the points of the function to construct the generalised derivative of the function. This non-locality of the fractional derivative says that the fractional derivative of the function at any point depends on the history in the sense if we consider E h a backward shift operator. While constructing the derivative by forward shift operator, we consider forward history. That is, the future points from x to b are required to compute the derivative.

3.16 Applying the limit of a finite difference formula result to get the answer to L’Hospital’s question We chose the start point a

0 and function f ( x )

x and in the following formula, we chose X

X

d

lim

1

¦ (1)

* (X  1)

j

.

f ( x  jh ) dx h j 0 j ! * (X  1  j ) Using the binomial expansion in the above formula (3.181) we write: 1 ( x  3h ) § 1( x )  12 ( x  h)  18 ( x  2h)  16 · 1 2 ¨ ¸ d 1 1 1 lim 1 > x@ ¨ ¸ § · § · 2 2 5 hp0 h dx 2 ¨  128 ( x  4h)  ...  ¨ « x »  2 ¸ (2h)  ¨ « x »  1¸ ( h)  0 ¸ ©¬h¼ ¹ ©¬h¼ ¹ © ¹ X

f ( x)

«¬ hx »¼

1 2

hp0

X

(3.181)

(3.182)

In this expression (3.182), the binomial coefficient symbol is understood to denote the generalised function, with the factorials expressed in terms of the gamma function. As explained earlier (Section-3.14), the coefficients in the above 1

expressions are just the coefficients in the binomial expansion of 1  E h 2 . The RHS of the above obtained series (3.182) can be evaluated with various values of x , and taking a certain slice h ; and we then see that approximately the expression converges to 2

x S

. Obviously, this numerical evaluation of the series representation is approximate.

The point is that the convergence of the above to the earlier obtained semi-derivative of f ( x ) formula is:

x , via the Euler

138

Chapter Three

1

0 Dx

We used *  32

S 2

* ( m  1)

ª¬ x m º¼

X

0 Dx

2

* ( m  X  1) 1!

> x@

 12

ª2 ¬

x

º S ¼

2

* 1  12  1

S

 12

3 2 *2

S

1

ª¬ x 2 º¼

0 Dx

1!

x

mX x

1 12

. From the Euler formula, we do the anti-derivative of 2 0 Dx

x

2

x

x S

2

*  32

x

(3.183)

S

, that is:

*  12  1

S *  12  12  1

1

x2

  12 (3.184)

x

3.17 Fractional finite difference (from local finite difference to non-local finite difference) In this section, we will find what we have developed in an earlier part of the chapter, the formula for fractional differentiation and fractional integration via the most fundamental definition, applied to the concept of finite differences. Thereby we analytically continue the concept of finite differences to the non-integer order. We here too 1 see the non-local effect and the memory effect described earlier. In a finite difference if we choose the matrix ª¬ D º¼ as:

ª1 «0 «0 1 ª¬ Dx º¼ « «.. «0 «0 ¬ Then ª¬ Dx2 º¼

ª¬ D1x º¼ ª¬ D1x º¼

1 0 0 1 1 0 0 1 1 .. .. .. 0 0 0 0 0 0

0 0º 0 0» » 0 0 »  'x 1 .. .. » 1 1» 0 1 »¼

(3.185)

2

ª¬ D1x º¼ is the following:

ª1 2 1 «0 1 2 «.. .. .. ª¬ Dx2 º¼ « «0 0 0 «0 0 0 «0 0 0 ¬

0 0 1 0 .. .. 1 2 0 0

1 0

0º 0» » .. »  'x 2 1»

(3.186)

2» 1 »¼

1

1 1 For integration ª¬ D º¼ is an inverse of ª¬ D º¼ , that is ª¬ Dx1 º¼

ª¬ D1x º¼ ; we then write the following: ª1 1 1 1 .. 1º «0 1 1 1 .. 1» «0 0 1 1 .. 1» »  'x ª¬ Dx1 º¼ « «.. .. .. .. .. ..» «0 0 0 0 1 1» «0 0 0 0 0 1» ¬ ¼

(3.187)

m For ª¬ Dx º¼ , where m is a positive integer we write the following:

ª¬ Dxm º¼ ª¬

1 m Dx

º¼

m  times

    1 1 ª¬ Dx º¼ ª¬ Dx º¼ .......................... ª¬ D1x º¼

m For ª¬ Dx º¼ , where m is a positive integer we write the following:

(3.188)

Fractional Derivatives

ª¬ Dx m º¼ ª¬ Dx1 º¼

m

139

ª¬ Dx1 º¼ ª¬ Dx1 º¼ ..................... ª¬ Dx1 º¼   

(3.189)

m  times

With the function f ( x ) having values at x1 , x2 , x3 … xn 1 and xn , k

that is with xk

x1  ( k  1) 'x , for

1, 2, ...n the matrix representation of the finite difference as per the above definitions are the following:

ª1 «0 «0 ª¬ D1x º¼ f ( x ) « «.. «0 «0 ¬

1 1

0 1

0

0

0

1

0 0 1 0

..

..

..

0

0

0

0

0

0

0 º ª f ( xn ) º 0 » « f ( xn 1 ) » » « f ( x )» 0 n2

»«

.. » « .. 1 1» « f ( x2 ) « 0 1 »¼ ¬ f ( x1 ) ..

»  'x 1 » » » ¼ (3.190)

ª D1x f ( xn ) º « D1 f ( x ) » « 1x n 1 » « D x f ( xn  2 ) » « » .. « » 1 « D x f ( x2 ) » « f ( x )  'x 1 » 1 ¬ ¼ For integration we have the following:

ª1 «0 « ª Dx1 º f ( x ) «0 ¬ ¼ «.. «0 «0 ¬

1 1 1 1 1º ª f ( xn ) º 1 1 1 1 1» « f ( xn 1 ) » » »« 0 1 1 1 1 f ( xn  2 ) »  'x »« .. .. .. .. ..» « .. » 0 0 0 1 1» « f ( x2 ) » « » 0 0 0 0 1»¼ ¬ f ( x1 ) ¼

ª D 1 f ( x ) xn º x1 » « x x n 1 » « 1 Dx f ( x ) « x1 » xn2 » « 1 « Dx f ( x ) x1 » « » .. « » « Dx1 f ( x ) x2 » x1 » « «¬ f ( x1 )  'x »¼

(3.191)

Now with a fractional generalisation of the m integer to D ! 0 , and D  0 , a real number, we have the analytic continuation to write the following: .. wD , n º ª wD ,1 wD ,2 wD ,3 ..

« « D ª¬ Dx º¼ « « « « «¬

D !0

0

wD ,1

wD ,2

wD ,3

..

wD ,( n 1) »

0

0

wD ,1

wD ,2

..

wD ,( n  2) »

0 0

0 0

0 0

.. 0

.. wD ,1

0

0

0

0

0

» »  'x

.. » wD ,2 » wD ,1 »¼

D

(3.192)

140

Chapter Three

wD , n º .. .. ª wD ,1 wD ,2 wD ,3 « 0 .. wD ,1 wD ,2 wD ,3 wD ,( n 1) » « » wD ,1 wD ,2 wD ,( n  2) » 0 0 .. « D ª¬ Dx º¼ « 0 » 0 0 .. .. .. « » 0 0 0 wD ,1 wD ,2 » « 0 0 0 0 0 wD ,1 »¼ «¬ 0

1

 'x

D

(3.193)

D 0 D

which satisfies the condition ª¬ Dx

D º¼ ª¬ DDx 1 º¼ ª¬ DDx 2 º¼ ...... ª Dx p º ; for D D1  D 2  D 3  ....  D p . The weights are: ¬ ¼

wD , k

( 1)

( k 1)

*(k )

* (1  D )

(3.194)

* (2  D  k ) D

which is incidentally the k  th term coefficient of binomial expansion of 1  ( 1) . Note that we have derived the weight, i.e. wD ,k , for the fractional differentiation and integration formula. These weights ( wD ,k ) are the same as those obtained in our earlier section where we wrote

( 1) j * (X 1) j ! * (X 1 j )

by k  1 and X with D , we obtain our coefficient wD ,k , that is

for X -th order differ-integration. If we replace j

( 1)( k 1) * (D 1) ( k 1)! * (D 1 k 1)

( 1)( k 1) * (D 1) * ( k ) * (D  k  2)

wD , k . We will

learn in the next chapter (Chapter-4) to write in limit form to have a more general representation of the coefficients wD ,k as in the following form:

wD , k m

( 1)

( k 1)

*(k )

limp0

* (1  D  )

(3.195)

* (2  D  k  )

D

The matrix representation D is sparse, while D is generally dense; all the elements in the upper tri-diagonal blocks are non-zero. In finite difference, a sparse matrix entails taking the difference among the neighboring set of points, while a dense matrix entails taking the difference among almost every point in the domain; this is exactly the non-local effect; also called the memory effect. From the above matrix representations, we say: D

D f ( xm )

D lim nnf ª¬ D º¼ > f ( x )@ m

lim nnf

¦

x xm

wD , k  f ( xm  k 1 )  'x

D

(3.196) ,

1  m d n

k 1

We have derived the above (3.196) in an earlier section too. If f ( x ) is an integer power series, then DD f ( xm ) on the LHS in general is a non-integer power series; for example as we have demonstrated (in an earlier section) the half1

x

derivative of e represented as an integer order power-series, when operated by D 2 gives a non-integer order powerseries. The corresponding matrix representation on the RHS above is a sum of ordinary power series. The non-local effect can be then seen to arise from approximating the non-integer power series by sum of an ordinary power series.

3.18 Dependence on the lower limit What is the effect of fractional differentiation and integration if the lower limit changes from say x obviously there will be a different result. In other words, we want to evaluate ' {

D a Dx

> f ( x )@ 

D b Dx

a to x

b,

> f ( x)@ . Here

we derive a formula that precisely demonstrates these phenomena. We consider for study an analytic function M ( x ) , in the interval a to x and assume that a  b  x .

dD M ( x )

dD M ( x )

, one might expect that the >d( x  a )@ >d( x b )@D difference ' is expressible in terms of integrals of M ( x ) in the interval a d x d b ; the formula is

¦

f

dD  k >1@

d kM (b )

Writing '

D



; analogous to the case that we obtained in the earlier chapter (Section-1.20) for a >d( b  a )@ k repeated n -fold integration. Assume D  0 , and we have the Riemann-Liouville formula and state the following:

'

k 1 > d( x b )@D  k

Fractional Derivatives D

>

@

D

>

' { a Dx M ( x )  b Dx M ( x ) D

@

D

d M ( x) D

>d( x  a)@ 1

x



d M ( x) D

>d( x  b)@

M ( y )dy

D * ( D ) ³ ( x  y )

1



a

1

141

b

x

1

M ( y )dy

D * ( D ) ³ ( x  y )

1

b

(3.197)

M ( y )dy

D * ( D ) ³ ( x  y )

1

a

1

b

M ( y )dy

D * ( D ) ³ ( x  b  b  y )

1

a

b

§ f § 1  D · · j 1D  j   ( ) ( ) x b b y ¨ ¸ M ( y ) dy ¦ ¨ ¸ * ( D ) ³a © j 0 © j ¹ ¹ 1

1D

. We have the following In the above steps (3.197) we have used a binomial expansion for  ( x  b )  (b  y ) property of the binomial coefficients and its generalisation by the gamma function i.e. *( j  q) § j  q  1· j §q· (3.198) ¨ j ¸ ( 1) ¨ j ¸ © ¹ © ¹ * (  q )* ( j  1) Replacing in the above formula (3.198) j with r and q with D  r we get: * ( D )

§ 1  D · r §D  r · ¨ r ¸ ( 1) ¨ r ¸ © ¹ © ¹

(3.199)

* ( D  r )* ( r  1) and then using the above relationship from (3.199) in (3.197) we write the following steps:

' { a Dx >M ( x )@  b Dx >M ( x )@ D

D

b f

³¦ a k 0

( x  b)

1D  k

(b  y )

k

* ( D  k ) * ( k  1)

M ( y ) dy

>1@ b M ( y ) dy ¦ D  k 1 ³ k * ( k  1) (b  y ) k 0 > d( x  b ) @ a  D  k 1 f d >1@ d  k 1 >M (b)@ ¦  k 1 D  k 1 k 0 > d( x  b ) @ >d(b  a )@ D k f d >1@ d  k >M (b)@ f DD  k 1 D  k M (b) ¦  b x > @  a b > @ ¦ k D k k 0 > d( x  b ) @ >d(b  a )@ k 0 D  r 1 In the above derivation (3.200) we used b Dx >1@ ( x*(b)D r ) which comes from a DDx >C @ f

d

D  k 1

 ( D  r 1)

where C

C * (1D )

( x  a)

D

,

1. x

>M ( x)@ ³a  * ( rM1)( y ()x dyy ) b  ( r 1) b , the expression a Db >M (b)@ ³a  * ( rM1)( y ()b dyy )

In addition, we used x

(3.200)

 ( r 1) a Dx

r

b

a

 M ( y ) dy in (3.199); from this, we write, by using  * ( r 1) ( b  y ) r

d  ( r 1) >M ( b )@ r

>d( b  a )@ ( r 1)

. We state that

>1@ d  k >M (b)@ f DD  k 1 D  k M (b) ¦ ¦  b x > @  a b > @ D k k k 1 > d( x  b ) @ >d(b  a)@ k 1 f

'

³

d

D k

(3.201)

The expression (3.201) is valid for D  0 . Therefore, we have the above formula (3.201), and argue that, since both sides of the equation are analytic in D by the identity theorem for analytic functions, this above formula gets valid for all D .

142

For D

Chapter Three

n

0,1, 2, 3, ... , we have the following:

>1@ d  k >M (b)@ D n 0,1, 2, 3... ¦ D k k k 1 >d( x  b )@ >d(b  a)@ nk k f d >1@ d >M (b )@ ¦ nk k k 1 >d( x  b )@ >d(b  a)@ n 1 1 n2 2 d >1@ d >M (b )@ d >1@ d >M (b )@  n 1 1 n2 2 >d( x  b)@ >d(b  a)@ >d( x  b)@ >d(b  a)@ n 3 3 d >1@ d >M (b )@   ... n 3 3 >d( x  b)@ >d(b  a)@ f

'

d

D k

(3.202)

0 n 1, 2, 3... Therefore, when D are zero or a positive integer, then all integer order derivatives of unity are zero so we obtain ' 0 for D n 0,1, 2, 3... . . For D 1 , we obtain the classical result, as we demonstrate below:

>1@ d  k >M (b)@ D 1 ¦ D k k k 1 >d( x  b )@ >d(b  a)@ 11 d >1@ d 1 >M (b)@  d 1 2 >1@ d 2 >M (b)@ 11 1 1 2 2 >d( x  b)@ >d(b  a)@ >d( x  b)@ >d(b  a)@ 1 3 3 d 1@ d >M (b )@ >   ... 1 3 3 >d( x  b)@ >d(b  a)@ 1 1 0 d >1@ d >M (b )@ d >M (b )@ 0  1 1 0 >d( x  b)@ >d(b  a)@ >d(b  a)@ f

'

d

D k

(3.203)

b

³ M ( y )dy

1

M ( x)

a

1

x b

 M ( x)

1

1

M (b )  M ( a )

x a

This result we have obtained in an earlier chapter (Section-1.20). For all other values of D , the difference ' is non-zero and its value depends not only on a , b but also on x . For example, for a negative integer D  n , we find: k n

>1@ d  kM (b) , ¦ k n k k 1 > d( x  b ) @ >d(b  a)@ n

'

d

n

( x  b)

nk

¦ *(1  k  n) k 1

Note, for D

 n the sum terminates at k

D

k

d M (b ) k

>d(b  a)@

n , as at k

n n

¦

1, 2, 3,.... ( x  b)

k 1

nk

( k  n )!

n  1 and beyond

(3.204)

k

d M (b ) k

>d(b  a)@

d k n >1@

>d( x b )@k n

0 ; nevertheless for a non-

integer D , the sum extends until infinity.

3.19 Translation property By translation property, we mean f ( x ) is replaced by f (C  x ) , where C is a constant, which we take as positive. We assume that

dD > f ( x )@

>d( x  a )@D apply the RL formula:

is defined and of course f ( x ) is defined between min( a , a  C ) and max( x, x  C ) . We

Fractional Derivatives

> f ( x  C )@ D >d( x  a)@

d

D

x

1

* ( D ) ³

143

 f ( y  C ) dy x C

1

put

D 1

( x  y)

a

yC

[

(3.205)

 f ([ ) d[

³

D 1

* ( D ) a  C ( x  C  [ )

It is evident that translation by a distance C is equivalent to a shift in the upper limit from x to x  C , and shift in lower limit from a to a  C . Representing the effect of this shift in the independent variable, by the symbol ' as we derived in the earlier section, we write the following:

> f ( x  C )@ D >d( x  C  a)@ d

'

D



> f ( x  C )@ D >d( x  a)@

d

D

(3.206)

From here, we write:

> f ( x  C ) @ dD > f ( x  C ) @  ' D D >d( x  a)@ >d( x  C  a)@ D D k k f d > f ( x  C )@ d d > f ( a  C )@ >1@  ¦ D D k k >d( x  C  a)@ k 1 >d( x  C  a)@ >d(a  C  a)@

d

D

(3.207)

Though the use of the RL definition used in calculating ' requires D  0 ; (3.205) the usual analytical argument based on the identity theorem serves to remove this restriction. The above-obtained formula involves an infinite sum that is not, in general, represented by a closed form expression. Thus, this translation represents a process, which is difficult to handle in fractional differentiation and integration. For D 12 , and a 0 then with Euler’s formula we write

d

1/ 2  k

>1@ >d( x  C )@1/ 2 k

1 * 1 12  k

(x  C) d

1

2

1/ 2  k

> f ( x  C )@ >d( x)@ 1

For D

 12 and a

0 we write d

; with this we have:

d

> f ( x  C )@  f ( x  C )   k ¦ >d( x  C )@ k 1 *  12  k

d

>1@

 1/ 2  k

>d( x  C )@

 12

> f ( x  C )@  >d( x)@ 1

1 2

2

1

2

1/ 2  k

1

1 * 1 12  k

2

k

(x  C)

1/ 2  k

(3.208)

k

after which we write: 1 2

 12

1

> f (C )@

dC

2

> f ( x  C )@  f ( x  C )   k ¦  3 k 1 * 2  k >d( x  C )@

d

d

2

d

k

> f (C )@

dC

(3.209)

k

3.20 Scaling property By the scale change of a function f ( x ) with respect of the start point of differentiation (or integration) that is the lower limit x a , it means that the function is replaced by f (O x  O a  a ) , where O is a constant, called a scalefactor. To simplify this take a

0 , then the scale change converts f ( x ) to f (O x ) . Here we will obtain

RL operation on f (O x  O a  a ) . We assume

[

dD f ( x ) D

>d( x  a )@

O y  O a  a , then the following steps are carried out:

is known. Call the variable F

x

dD

>d( x  a )@D

an

 a OaO , and call

144

Chapter Three

> f (OF )@ D >d( x  a)@ d

D

d

> f (O x  O a  a )@ D >d( x  a )@

D

x

1

f (O y  O a  a )

* ( D ) ³

D 1

( x  y)

a

OF

1

* ( D ) ³ a

O

f ([ )



OF

D

dy

§ d[ · ¨ ¸ ©O ¹

OF [ D 1 O



(3.210)

 f ([ ) d[

D * ( D ) ³ (OF  [ )

1

a

O

> f (OF )@ D >d(OF  a)@ d

D

D

When a 0 then we have F x and the scale change of the property is simply a multiplication of the independent variable by the constant O ; the above formula is then:

d

D

> f ( O x )@ D >dx@

O

> f ( O x )@ , D >d(O x)@

d

D

D

D

0 Dx

> f ( O x )@

O



D

D

0 DO x

> f (O x)@

(3.211)

In the previous chapter (Section-2.18), we did a similar derivation for the fractional integration process, that is: D 0 Dx

> f ( O x )@

O

D



D 0 Dx

> f ( x)@

O

D



> f (O x)@

D 0 DO x

(3.212)

Here we mention that the above obtained derivations (3.210) and (3.211) are for a fractional derivative and fractional integration respectively. Say we have a 0 , for D 12 , and D  12 we write the following:

d

1

> f (O x )@ > dx @

2

1

and with O

O

2

1

> f (O x )@ >d(O x)@

d

d

2

1

2

,

 12

> f (O x )@  > dx @ 1

2

O

 12

> f (O x )@  >d(O x)@

1 d

1

2

(3.213)

1 we write the following: d

1

> f (  x )@ >dx@

2

1

2

i

1

> f (  x )@ >d( x)@

d

d

2

1

2

,

 12

> f (  x )@  > dx @ 1

2

i

 12

> f (  x )@  >d( x)@

d

1

2

(3.214)

3.21 Fractional differentiation-integration behavior near the lower limit When D is a positive integer or zero, the operation

dD > f ( x )@

>d( x  a )@D

D

a Dx

or

point property, this we have discussed earlier; and the behavior of other values, i.e. non-integer values of D we will demonstrate that infinity as will show that

x approaches a . We write f ( x ) dD > fX ( x )@

>d( x  a )@D

fX ( x )

( x  a)

p

> f ( x )@

for D

0,1, 2, 3... is ‘local’ and is a

> f ( x)@ near x a is exceptional. For all D a Dx > f ( x ) @ usually approaches either zero or D

a Dx

¦

f j 0

j

a j ( x  a ) ; with p ! 1 and a0 z 0 . We

normally approaches either zero or infinity as x approaches a :

> fX ( x)@ f a j  *( p  j  1) ( x  a) p  j D ¦ D >d( x  a)@ j 0 *( p  j  D  1) d

We know that

1 * ( p  j D 1)

D

(3.215)

is always finite while * ( p  j  1) is finite since p ! 1 (from the properties of the gamma

function). Therefore, the RHS of the above (3.215) is dominated by its first term ( j

0 ) for small x  a , that is:

Fractional Derivatives

145

· § ( x  a ) p D a0  * ( p  1) · ¸ lim ¨ ¸ ¸ xoa © * ( p  D  1) ¹ ¹

§ dD f ( x ) X D xoa ¨ © >d( x  a )@ lim ¨

(3.216)

0 for p  D ! 0 ­ ° ®a0  * ( p  1) for p  D 0 °¯ for p  D  0 f since a0 z 0 .

3.22 The fractional differentiation-integration behavior far from the lower limit We here demonstrate restricting the discussion to an analytical function M ( x ) and for x !! a we write a few terms of the binomial expansion of ( x  a )

k D

as the following steps:

( x  a)

k D

x

k D

§ a· ¨1  ¸ © x¹

x

k D

x

k D

k D

1  (k  D ) (D  k ) ax



a x

O

 a2 x2

(3.217)

k

D 1

x The above approximation (3.217) we substitute in the formula which we obtained previously in Chapter-2 (Section2.11.3) which was: f ( 1) k ( x  a ) k D M ( k ) ( x ) M ( x )@  > ¦ a Dx >M ( x )@ D >d( x  a)@ k 0  *(D ) (k  D )k !

d

D

D

f

¦



( 1)

k



x

k D



(D  k ) ax k xD 1

M

(k )

( x)



(3.218)

 *(D ) (k  D )k !

k 0

In (3.218) we have assumed M ( x ) as an analytic function. Use the reflection formula of the gamma function,

* ( D )



S csc(DS ) * (D 1)

(Section-1.10), to get the following expression for x !! a

>M ( x)@ D >d( x  a)@ d

D

a Dx >M ( x )@ 

D

f

¦

( 1)

k

S D

d M ( x) dx

D



f

¨¦ ©

k D



M

(D  k ) ax k xD 1

(k )

( x)



 *(D ) (k  D )  k !

k 0

sin(DS )* (D  1) §

x 

( 1)k x k D M ( k ) ( x )

k 0



(D  k ) k !

 Da1 x

f

¦



( 1)k x k M ( k ) ( x )

 k !

k 0

·

(3.219)

¸ ¹

a M (0)  sin(DS ) * (D  1)

Sx

D 1

The following expression for the second term is used in the above derivation (3.219) i.e. f

¦ k 0

k

( 1) x

k

M k!

(k )

( x)



1  xM

(1)

 which is the Taylor expansion of M ( x ) at the origin x

( x)  x

3!

0 ; that is M ( x ) x

We also used for the first term in (3.219) the following expression:

3

M

0

x

2

2!

(3)

M

(2)

( x)

( x )  ... M ( x ) x

M (0) .

(3.220) 0

146

Chapter Three k

f

sin(DS )* (D  1)

¦

S

( 1) x

M

k D

(k )

( x)



¦ * ( D )

(D  k )  k !

k 0

k

f

1

( 1) x

k D

M

(k )

( x)



(D  k )  k !

k 0

( x) ¦  *(D ) (k  D )  k ! f

k

( 1) x

k D

M

(k )

(3.221)

k 0

D

0 Dx

Here we used the expression i.e.

>M ( x)@

>

D

D

d M ( x)

a Dx M ( x )

dx



( 1)k ( x  a )k D M ( k ) ( x )

f

@ ¦k

D



with a

 * ( D ) ( k D ) k !

0

0 (Section-2.11.3), in derivation

(3.221). We observe that if D is a positive integer say D 1, 2, 3, ... , the second term of the RHS of the above expression vanishes, indicating again that the integer order derivatives are local (or point) property.

3.23 A numerical evaluation of fractional differ-integrals using the Grunwald-Letnikov formula dD dxD

In this section, we will approximate

f ( x ) for arbitrary D , with known values of f ( x ) at N  1 and evenly

spaced points in the range 0 to x . We designate symbols f N { f (0) , f N 1 { f

 Nx .

f0 { f ( x) ; and discrete steps as 'x

 Nx , … f j { f  x  j Nx ,

…

We write the simplest definition as we saw earlier from the Grunwald-

Letnikov (GL) formula; that is:

> f ( x )@ D >d( x  a)@ d

D

§  xN a D lim ¨ N nf ¨ * ( D ) ©

We omit the limit N n f and make the start point a

d

D

xa N

j 0

(3.222)

¹

0 and write the following:

x D N 1 *( N

> f ( x )@ | 

dx

·

N 1

*( j  D ) ¦ *( j  1) f  x  j  ¸¸

j D) f x  j ¦ * ( j `1) * ( D )

D

x N

j 0

 Nx

D

* ( D ) This is an approximate algorithm to evaluate the

dD dxD

N 1

 'x

*( j  D )

¦ *( j  1) f

j

* ( D )

j 0

> f ( x )@ .

If we call

D N 1

GL

(3.223)

*( j  D )

¦ *( j  1) f

* ( j D ) * ( D ) * ( j 1)

wj

j

j 0

,

GL

w j 1

* ( j 1D ) * ( D ) * ( j )

as

Grunwald’s coefficients then we write: d

D

> f ( x )@ dx

We have a recursion relationship that is

D

* ( j D ) * ( j 1)

D

N 1

*( j  D )

D

N 1

GL f j  'x ¦  w j f j  'x ¦ * ( D )* ( j  1) j 0

j 1D j



(3.224)

j 0

* ( j 1D ) *( j )

. With this, we write a recursive expression for the

Grunwald Coefficient as: GL

wj

§ j 1D · ¨ ¸ j © ¹



GL

w( j 1)



(3.225)

This algorithm may be implemented by the following multiplication-addition, multiplication-addition, multiplicationaddition scheme D

d f ( x) dx

D

D

§ x · ª ªªª ª º N D  2 N D 3 1D º D ¨ ¸ « ¬«¬¬.. ¬ f N 1  N 1  f N 2 º¼  N 2  f N 3 º¼ ..º¼  2  f1 ¼»  1  f0 » ¬ ¼ N © ¹

The above formula (3.226) avoids explicit usage of the gamma function. We have a modified GL definition which has rapid convergence. We write with a

0 the following:

(3.226)

Fractional Derivatives

d

> f ( x )@

D

dx

§ x D N D

N 1

©

j 0

lim ¨

D

N nf

·

*( j  D ) f  x   j  D  ¸ ¦ * ( D ) * ( j  1) x N

1 2

D N 1

§  'x lim ¨ N nf ¨ * ( D ) © We notice that unless D

147

¦

¹

(3.227)

· f  x  j 'x  D '2x ¸ ¸ * ( j  1) ¹

*( j  D )

j 0

0, r2, r4, ... this modified formula requires an evaluation of f ( x ) at other points than as



f j values. We therefore approximate f x  D

x 2N

j



x N

based upon Lagrange’s three point interpolation formula,

that is, upon:

f x D

x 2N

§D D2 · § D2 · x x x  j |¨  ¸ f  x  N  j N  ¨1  ¸ f  x  j N 4 8 4 © ¹ © ¹ 2 §D D · ¨  ¸ f  x  Nx  j Nx 8 4¹ © x N

(3.228)

Thus, our modified GL formula becomes with the above substitution (3.228) the following: 1 D x D N 1 · d > f ( x )@ * ( j  D ) § f j  4 D f j 1  f j 1 N ¨ ¸ | D dx * ( D ) j 0 * ( j  1) ¨  1 D 2 f j 1  2 f j  f j 1 ¸ 8





¦





©

(3.229)

¹

3.24 The discrete numerical evaluation of the Riemann-Liouville fractional derivative formula We continue from the previous section and try to get a numerical scheme to evaluate the Riemann-Liouville (RL) fractional derivative formula; with known values of f ( x ) at N  1 evenly spaced points in the range 0 to x . We

 Nx , … f j { f  x  j Nx ,

designate the symbols f N { f (0) , f N 1 { f

… f0 { f ( x ) . We continue the method of

numerical integration that we applied to the RL fractional integration formula in the previous chapter (Section-2.23). The RL fractional derivative formula for 0 d D  1 ; for a function with a lower terminal set as a 0 is: D

d f ( x) dx

D

x

D

f (0)

* (1  D )



1

* (1  D ) ³

x

0

( x  y)

D

f

(1)



( y ) dy

(3.230) (1)

This above expression (3.230) is a relationship between the RL and Caputo derivative for 0 d D  1 with f the onewhole derivative of function f ( x ) assumed to exist and with an initial value f (0) as finite. We re-write the above (3.230) in the following form, with a change of variables of integration and writing N discrete integration per interval 'x and then summing them up, as was demonstrated in the previous chapter (Section-2.23): D

d f ( x) dx

D

x

D

f (0)

* (1  D )



§ d f ( y ) · § dy · ¸¨ D ¸ 0 ¨ © dy ¹ © ( x  y ) ¹

1

* (1  D ) ³

x

§ f (0) N 1 ( j 1) x / N § d f ( x  y ) · § dy · · ¨ D  ¦ ³ jx / N ¨ ¸ ¨ yD ¸ ¸ * (1  D ) © x dy j 0 © ¹© ¹¹

(3.231)

1

We use an approximation

³

d f ( x y ) dy

( j 1) x / N jx / N

|

f  x  j 'x  f  x  ( j 1) 'x 'x

and use this in the following derivation:

§ d f ( x  y ) · dy § f  x  ¨ ¸ D | ¨¨ dy © ¹y © §x· ¨ ¸ 1D © N ¹ 1

jx N

 f x 

Using the above expression (3.232) we get the following:

x

 x N

D

f

( j 1) N

j

 f j 1

 ( j  1)

·¸

( j 1) x / N

¸ ³ jx / N ¹

1D

 j

1D

dy y



D

(3.232)

148

Chapter Three D

d f ( x) dx

D

§ f (0) N 1 ( j 1) x / N § d f ( x  y ) · § dy · · ¨ D  ¦ ³ jx / N ¨ ¸¨ D ¸¸; * (1  D ) © x dy j 0 © ¹© y ¹¹ 1

f (0)

fN

D § D 1 § x · N 1 1D 1D ¨¨ f N x  ¨ ¸ ¦  f j  f j 1  ( j  1)  j * (1  D ) © (1  D ) © N ¹ j 0

1



x D N



N 1

·

¸¸ ¹

(3.233)

§ · D 1D 1D ¨ (1  D ) N f N  ¦  f j  f j 1  ( j  1)  j ¸ (1  D )* (1  D ) © j 0 ¹

 Nx





D

§ (1  D ) f N N 1 · 1D 1D  ¦  f j  f j 1  ( j  1)  j ¸ ¨ D * (2  D ) © N j 0 ¹





We have the following final formula:

 'x

dx

D

N 1 § (1  D ) · 1D 1D ¨ D f N  ¦  ( j  1)  j  f j  f j 1 ¸ * (2  D ) © N j 0 ¹

D

d f ( x) D





(3.234)

0 dD 1 We take a case for 1 d D  2 and write the RL derivative formula as: D

d f ( x) dx

x

D

D

1D

f (0)



* (1  D )

x

f

(1)

(0)

* (2  D )



1 * (2  D )

³

x

f

(2)



( y ) dy D 1

( x  y)

0

(1)

§ (1  D ) f (0) f (0) ·  D 1 ¨ ¸ D x x 1 ¨ ¸ N 1 ( j 1) x / N f (2) ( x  y ) dy ¸ ¨ * (2  D )  ¸ ¨¨  ¦ ³ jx / N D 1 ¸ y © j0 ¹

(3.235)

The above expression (3.235) is an RL–Caputo derivative relation for 1 d D  2 with the assumption that a second derivative of the function i.e. approximations f

(1)

(0) |

f

f  'x  f (0) 'x

(2)

( x ) exists and the initial values f (0) and f

(2)

( x  y) |

f  x  ( j 1) 'x  2 f  x  j 'x  f  x  ( j 1) 'x

( j 1) x / N jx / N

f

(2)

(1)

(0) |

§ ydy

( x  y) ¨

©

§ f x  |¨ ¨ ©

jx N



 Nx

f

 Nx  f (0)  Nx

§N· ¨ ¸  f N 1  f N ©x¹

(3.236)

i.e. the second derivative and its use as follows:

 'x 2

³

(0) are finite. We make

i.e. the first derivative at the initial point: f

and f

(1)

x N

D 1

· ¸ ¹

 2 f x   f x  jx N

 Nx

jx N



x N

2

·¸

¸³ ¹

( j 1) x / N

dy y

jx / N

(3 .237)

D 1

D

f

j 1

 2 f j  f j 1

( j  1)

2 D

 j

2 D



2 D Using the above approximations of (3.236) and (3.237) we arrive at the following final result (by a similar procedure as done in the previous (3.234) derivation thus not showing all the steps): § (1  D )(2  D ) · fN D ¨ D ¸ D N d f ( x )  'x dx

D

¨

* (3  D ) ¨ (2  D ) ¨  N D 1 ©

N 1

¦ f j 0

j 1

 2 f j  f j 1

( j  1)

1dD  2 Similar steps may be followed for the RL derivative formulas for 2 d D  3 and more.

2 D

j

2 D

¸ ¸¸ ¹

(3.238)

Fractional Derivatives

149

3.25 Generalisation of the Riemann-Liouville and Caputo derivatives with the choice of a different kernel, weights and base-function In the last chapter (Section-2.17) we have seen that the formula of the Riemann-Liouville fractional integration yields several other kinds of formulas by choice of the kernel of integration, the weight function and the base-function; here we apply the same concept for fractional derivatives. We have discussed that the Riemann-Liuoville (RL) fractional derivative is the first fractional integration followed by a whole number of classical differentiation, and the Caputo is just the opposite. Therefore choosing the different kind of integral operators based on a different kernel, weight function and base function and performing either the RL or Caputo process will yield various different kinds of fractional derivative operators. Here we will define the left/forward causal, and right/backward non-causal fractional integrals and fractional derivatives of a function f ( x ) with respect to another function z ( x ) , call it the base function and weight scale w( x ) ; a sufficiently good function. We assume that the base function z ( x ) is positive and a monotonically increasing (1)

function of x in the interval a  x d b ; having a continuous derivative z ( x ) on the interval a  x  b .

3.25.1 Generalisation of the left and right fractional derivatives with respect to the base function z ( x) and weights w( x) A generalised left/forward fractional derivative of the RL sense, with respect to z ( x ) and with the weight w( x ) , is defined with m  ` in the following relationships, just greater than D  \ , i.e. m  1  D  m ; as follows: D

a $ x ,> z ( x ), w ( x )@

f ( x)

 w( x )

1

§ 1 · Dx ¸ ¨ (1) ¨  z ( x) ¸ © ¹

m

 w( x)  a ! xm,>z D( x ),w( x )@ f ( x)

(3.239)

D !0 Where we had defined in an earlier chapter (Section 2.17)

D

a ! x ;> z ( x ), w ( x )@

as a left/forward causal generalised integral,

as in the following expression: D

a ! x ,> z ( x ), w ( x )@

f ( x)

1

x

D  z ( x )  z (t ) ³  w( x) *(D )

1

a



(1)

w( t ) z ( t )

 f ( t ) dt

(3.240)

D !0 a generalised right/backward non-causal fractional derivative of the RL sense with respect to z ( x ) and with weight w( x ) , is defined as follows:

§ 1 · D  w( x) ¨¨ (1) Dx ¸¸ x $ b ,> z ( x ), w ( x )@ f ( x ) ©  z ( x) ¹

m

 w( x )

1



m D x ! b ,> z ( x ), w ( x )@

f ( x)



(3.241)

D !0 We had defined in an earlier chapter (Section-2.17)

D

x ! b ;> z ( x ), w ( x )@

as a right/backward non-causal generalised integral

as follows: D

x ! b ,> z ( x ), w ( x )@

f ( x)

w( x )

b

1

1

D  z ( t )  z ( x )  w( t )  z ³ * (D ) x

(1)

(t )

 f ( t ) dt

(3.242)

D !0 d dx

Where Dx { , is a normal integer order derivative operator. We further define the left and right integer order derivative operator as follows:

150

Chapter Three

§ 1 · D x ¸  w( x ) f ( x ) ¨ (1) ¸ w( x ) ¨  z ( x ) © ¹ º 1 ª§ d ·  w( x ) f ( x ) » ¨ ¸ « w( x ) ¬© d[ z ( x )] ¹ ¼ 1

Dx  ;> z , w@; L f ( x )

(3.243)

§ 1 ·§ 1 · Dx ;> z , w@; R f ( x ) w( x ) ¨ (1) Dx ¸ ¨ f ( x) ¸ ¨  z ( x ) ¸ © w( x ) ¹ © ¹ d

·§ 1 · f ( x) ¸ ¸ ¨ © d[ z ( x )] ¹ © w( x ) ¹ §

w( x ) ¨ 

D

D

f ( x ) (3.239) and x $b ,> z ( x ), w ( x )@ f ( x ) (3.241) which is the original statement written at the start of this section. These formulas (3.243) are like the left/forward causal and the right/backward non-causal operators, with D and  D , except they contain the functions w( x ) and z ( x ) . The above formulas (3.243) are new integer order operators. Therefore, in terms of these new integer

Note that the above expressions (3.243) appear in the formula of

D

order operators, we re-write for the RL type

a $ x ,> z ( x ), w ( x )@

a $ x ,> z ( x ), w ( x )@

D

f ( x ) and

x $ b ,> z ( x ), w ( x ) @

f ( x ) operators in the following

construct: D



m D a ! x ;[ z , w ]

m



x ! b ;> z , w@ f ( x )

Dx ;> z , w@; L

x $ b ,> z , w@ f ( x )

Dx ;[ z , w]; R

D

Similarly, one can construct the Caputo type

m

f ( x)

a $ x ,> z , w@

C D a $ x ,> z ( x ), w ( x )@

m D

C D x $ b ,> z ( x ), w ( x )@

and

f ( x)





(3.244)

left or right generalised fractional

derivative as follows:

Explicitly for w( x )

D

m x ,[ z , w ]



m

C D a $ x ,> z , w@

f ( x)

m D a ! x ,> z , w@

C D x $b ,> z , w@

f ( x)

x ! b ,> z , w@ Dx ,> z , w@ f ( x )

m D

f ( x)





(3.245)

1 , the generalised Caputo derivative is expressed as follows:

C D a $ x ,> z ,1@

f ( x) 1

x

z *(m  D ) ³ a

(1)

(t )

 z ( x )  z (t )

m D 1

§ 1 d· ¨ (1) ¸ ¨  z (t ) dt ¸ © ¹

m

(3.246)

 f ( t ) dt

3.25.2 The relationship between the generalised Riemann-Liouville and Caputo fractional derivatives with weight and base function The generalised RL and Caputo fractional derivatives are related by the following formula, which is for left/forward derivatives: D

a $ x ,> z , w@

f ( x)





§ Dx( k,[)z , w] f ( x ) ¦ ¨ *(k  1  D )x w( x ) k 0 ¨ ©

w( a )

m 1

a

· ¸  z ( x )  z ( a ) k D  C $ D a x ,> z , w@ f ( x ) ¸ ¹

(3.247)

One could similarly write the expression for a right/backward derivative’s relation. The above relationship (3.247) shows a similarity to the expression for the relationship between a normal Caputo and a normal RL derivative. From the above expression, we write the following: a $ x ;> z , w@

D

f ( x)

C D a $ x ;> z , w@

f ( x)

iff

f

(k )

(a)

0; k

0,1,..., m  1

D

f ( x)

C D x $b ;> z , w@

f ( x)

iff

f

(k )

(b )

0; k

0,1,..., m  1

x $ b ;> z , w@

(3.248)

Fractional Derivatives

151

3.25.3 The composition properties of generalised fractional derivatives We now list the properties of these generalised fractional derivatives, and generalised fractional integral operators. The D

operators a ! x ;> z , w@ and

D

x ! b ;> z , w@

  The operator

D

a $ x ;> z , w@

satisfy the semi-group property: D

 @ 

E

@

a ! x ;> z , w@

a ! x ;> z , w@

D

x ! b ;> z , w

x ! b ;> z , w

and the operator

E

D

x $ b ;> z , w@

a

 ! > @  ! > @  ! > @  ! > @

! xD;>z ,Ew@

E

a

D E

D

a

x; z ,w

E

x ! b ;> z , w@

x

x; z ,w

(3.249)

D

x

b; z , w

b; z , w

D

are left inverse of the respective * ! **;> z , w@ operators resulting in

the following rules: D

a $ x ;> z , w@



D

a ! x ;> z , w@

f ( x)



D

f ( x)

x $b ;> z , w@



D

x ! b ;> z , w@

f ( x)



f ( x)

(3.250)

We will prove the composition of these rules as described above in Chapter-9.

3.25.4 The various forms of generalised fractional derivatives The choice of the base function z ( x ) and weight function w( x ) reduces the described generalised definitions to the various types of fractional derivative operators.

3.25.4.(a) A normal Riemann-Liouville/Caputo derivative from a generalised definition The generalized definition of fractional derivative (3.244) and (3.245) is reduced to a normal RL or Caputo type (as (1)

we discussed) and by choosing z (t ) t , w(t ) 1 and z (t ) 1 , we have the generalised fractional integration as the Riemann fractional integration which is recovered as follows: x 1 D 1 D (1) z ( x )  z ( t ) w ( t ) z ( t )  f ( t ) dt  a ! x ,> z , w@ f ( x ) a w( x )* (D )



³



(3.251)

D !0 D

a ! x ,>t ,1@

D

f ( x)

a Ix

1

* (D ) ³

f ( x)

x

D 1

(x  t)

a

 f ( t ) dt

The Riemann-Liouville fractional derivative is recovered as follows: D

a $ x ,> z , w@

 w( x )

f ( x)

1

§ 1 · Dx ¸ ¨ (1) ¨  z ( x) ¸ © ¹

m

 w( x)  a ! xm,[zD( x ),w( x )] f ( x)

D ! 0, m 1  D  m D

a $ x ,>t ,1@ f ( x )

D

a Dx f ( x )

m

Dx



m D

a ! x ,>t ,1@ f ( x )

(3.252)



m

x 1 § · m D 1 (x  t)  f (t ) dt ¸ ¨ ³ a dx © * ( m  D ) ¹

d

m

These are left/forward causal RL fractional integrals and derivatives. In a similar way, we have the other reductions in the RL or Caputo type as follows: D

D

x ! b ,>t ,1@

f ( x)

x Ib

f ( x );

C D a $ x ,>t ,1@

f ( x)

C D a Dx

f ( x) ;

D

x $ b ,>t ,1@

f ( x)

C D x $ b ,>t ,1@

f ( x)

D

x Db

f ( x)

C D x Db

f ( x)

(3.253)

3.25.4.(b) Hadamard type fractional derivative from generalised definitions By choosing z (t )

ln t , w(t )

1 and z

(1)

(t )

 1t we get Hadamard type fractional operators. Following this we

write a Hadamard type left/forward causal RL type integral and derivative operators:

152

Chapter Three

D

a ! x ,> z , w@

1

x

D  z ( x )  z (t ) ³ w( x )* (D )

f ( x)

1

a



w( t ) z

(1)

(t )

 f ( t ) dt (3.254)

D !0 D

a ! x ,> ln t ,1@

D

a $ x ,> z , w@

 w( x )

f ( x)

a $ x ,> ln t ,1@

1

1

x

1

D   f ( t ) dt  ³ * (D )

f ( x)

ln x ln t

a

§ 1 · Dx ¸ ¨ (1) ¨  z ( x) ¸ © ¹

1 t

m

 w( x)  a ! xm,>z D( x ),w( x )@ f ( x)

m 1  D  m

D !0 D

H D a Ix

f ( x)

H D a Dx

f ( x)



m Dx

f ( x)

1

§ d· ¨x ¸ * ( m  D ) © dx ¹

m D a ! x ,> ln t ,1@

m

³  x

a

f ( x)

(3.255)



ln t m D 1 ln x



 dt f (t ) t

The other Hadamard right/backward operators, Caputo left and right, are thus: x ! b ,> ln t ,1@

D

f ( x)

H D x Ib

C D a $ x ,> ln t ,1@

f ( x)

H C D a Dx

D

f ( x );

x $ b ,> ln t ,1@

C D x $ b ,> ln t ,1@

f ( x) ;

H D x Db

f ( x)

f ( x)

H C D x Db

f ( x)

f ( x)

(3.256)

3.25.4.(c) Modified Erdelyi-Kober (MEK) type fractional derivatives from the generalised definition V

(1)

VK

V 1

By choosing z (t ) t w(t ) t and z (t ) V t , we obtain the modified Erdelyi-Kober (MEK) operator. The following is a left causal integral and derivative: x 1 D 1 D  z ( x)  z (t ) w(t ) z (1) (t ) f (t )dt D ! 0 a ! x ,> z , w@ f ( x ) a w( x )* (D ) (3.257) VK x Vt D V V D 1 V (1K ) 1 MEK D x t t  f ( t ) dt a ! x , ªtV ,tVK º f ( x ) a I x f ( x) ¬ ¼ * (D ) a



³

³

D

a $ x ,> z , w@

a $ x , ªtV ,tVK º ¬ ¼

§ 1 · Dx ¸ ¨ (1) ¨  z ( x) ¸ © ¹



m

 w( x)  a ! xm,>z D( x ),w( x )@ f ( x)

m 1  D  m

D !0 D

 w( x )

f ( x)

1



f ( x)

MEK D a Dx

x

m

f ( x)

VK

§ ©

(3 .258)

· ¹

m D

Dx ¨ a ! ª V VK º f ( x ) ¸ x , t ,t ¬

¼

m § VK § 1 d · VK ¨ V t x ¨ V 1 ¸ ¨ * (D ) © V x dx ¹ ©

V (1K ) 1

x

t

a

x

³

V

t

f (t )

V 1D



·

dt ¸

¸ ¹

The other MEK operators are right/backward, Caputo left and right, are thus the following: x ! b , ªtV ,tVK º ¬ ¼

f ( x)

MEK D x Ib

C D a $ x , ªtV ,tVK º ¬ ¼

f ( x)

MEK C D a Dx

D

f ( x );

D

x $ b , ªtV ,tVK º ¬ ¼

f ( x) ;

f ( x)

C D x $ b , ªtV ,tVK º ¬ ¼

MEK D x Db

f ( x)

f ( x)

MEK C D x Db

f ( x)

(3.259)

However, the types of operators described in this section we are not going to use this in the rest of the chapters. In subsequent chapters we will use the regular Caputo and Riemann-Liouville fractional derivatives.

3.26 The Caputo fractional derivative with a non-singular exponential type kernel From (3.109) we write the formulation of the Caputo derivative, as

C D a Dt

f (t )

1 * (1D )

³

t f (1) (W )dW

D a ( t W ) (1)

, for a fractional

order 0  D  1 . This is applicable if the function f (t ) has one-whole derivative f (t ) in the interval under consideration. As per our discussions, the Caputo derivative is a fractional integration of order 1  D for the function

Fractional Derivatives

f

(1)

C D a Dt

(t ) i.e.

t=0 C D a Dt

C

i.e.

f (t )

1 * (1D )

1D a It

f (t )

f

kD (t )  t

D





C

(1)

(t ) . Here the kernel of integration is a ‘power function’ and is a singular function at

.

In

kD (t ) * f

153

(1)

(t )

terms

of

the

convolution

. The kernel if considered as

CF

the

integral

C D a Dt

kD (t )  exp   1DD t , instead of the usual

‘power function’ then we have the Caputo-Fabrizo (CF) definition given as: M (D ) t (1) CF D exp   1DD (t  W ) f (W ) d W ; 0  D  1 a Dt f (t ) 1D a Where M (D ) is a normalization constant M (0) M (1) 1 .

(3.260)

³

Now if we have a kernel as a Mittag-Leffler ‘higher transcendental function’, i.e., ED ( z )

¦

f

zk k 0 * (D k 1)

is

f (t )

ABC



kD (t )  ED  1DD t

D

,

(Appendix-A) then we have the Atangana-Baleanu-Caputo (ABC) definition given as: ABC D a Dt

B (D )

t

 ED   1D ³

f (t )

a

where B (D ) is a normalisation constant B (0)

B (1)

D

D

1D

(t  W )

 f

(1)



(W ) dW , 0  D  1

(3.261)

1.

We note that our classical definition of a fractional derivative (RL or Caputo) is connected to the Riemann-Liouville fractional integration, with the kernel in convolution integration as a ‘power function’. We observe that in those D

which is singular at t 0 ; whereas definitions classical definitions, the kernel of fractional integration is  t (3.260) and (3.261) have a kernel of the convolution integration as a non-singular function– described by the exponential and Mittag-Leffler functions respectively.



One may verify in (3.260) by expanding exp  1DD (t  W )



in the series and then using Cauchy’s formula for

repeated integration, that this expression is a weighted series and the sum of various orders of integer integrals.



D

Similarly by expanding ED  1DD (t  W )

in (3.261) one may verify that this actually is a weighted series sum of

various orders of fractional integrals of the RL type. We have mentioned these new developments, and it is still a developing subject, i.e. using non-singular kernels, in these new definitions of fractional derivatives of the Caputo types. The applicability of these two new definitions is yet a matter of discussions and developments in the fractional order dynamics of the physical systems. We will use only the classical definitions (Riemann-Liouville, Caputo, Grunwald-Letnikov) that are described via a singular kernel, in the rest of the discussions in further chapters.

3.27 Short summary In this chapter, we have unified the context of the integration and differentiation and arrived at several useful definitions and properties of the fractional derivative. Importantly we have seen that unlike the classical integer order derivative, the fractional derivative is a non-local operator requiring the values of the function from the start – that is, this operation requires a lower-limit and upper limit like the integration process. We have also discussed that there is a memory effect due to this non-local effect for fractional differentiation. We have discussed several paradoxes that appeared in the due course of the development of this fractional calculus over the last three centuries. We now proceed with the knowledge developed here and apply it in the next chapters. In the next chapters we will be using the Riemann-Liouville (RL), Caputo, and Grunwald-Letnikov fractional derivatives for an application to the fractional differential equations and discussions on dynamic systems. In the last chapter, we will modify the RL fractional derivative to give a conjugation to classical integer order derivatives, as well as for non-differentiable functions, and that will give a formulation of a ‘local’ fractional derivative operator.

3.28 References This chapter took influence from the following works: Michelle Caputo, “Linear Models of Dissipation whose Q is almost Frequency Independent” (1967) Westerlund, “Dead Matter has Memory” (1991) Scott Blair G W, “Psychoreology: links between the past and the present” (1974) Samko S G, Kilbas A A, Maritcheva. O. I, “On Fractional Integrals and derivatives: Theory and Applications” (1993) Robotnov Y N, “Elements of Hereditary Solid Mechanics”(1980)

154

Chapter Three

Riew. F, “Non-conservative Lagrangian and Hamiltonian mechanics” (1996) and “Mechanics with fractional derivatives” (1997) Manabe. S, “The Non-Integer Integral and its application to Control Systems” (1961) Machado J A T, “I say to myself: What is Fractional World?” (2011) Lubomir Dorcak, “Numerical Models for simulation of fractional order control system” (1994) Kolwankar and Gangal, “Fractional differentiability of nowhere differentiable functions and dimensions” (1996) Balakrishnan V, “Anomalous diffusion in one-dimension” (1985) Miller K S and Ross B, “An Introduction to Fractional Calculus and Fractional Differential Equations” (1993) Oldham and Spanier, “Fractional calculus” (1974) Almeida R, Malinowska A. B, Torres D F M, “A fractional calculus of variations for multiple integrals with application to vibrating string” (2010) Agrawal O. P, “Fractional variational calculus in terms of Riesz fractional derivatives”, (2007) and “Generalised variation problems and Euler-Lagrange equations”, (2010) Agrawal O. P, Muslih S. I, Baleanu D; “Generalised variational calculus in terms of multi-parameters derivatives” (2011) Batman H, “Higher transcendental function”, (1954) Heymans and Podlubny et al., “Physical interpretation of initial conditions for fractional differential equation with Riemann-Liouville fractional derivative” (2005) Igor Podlubny, “Fractional differential equations” (1999), “Fractional derivatives: A new stage in process modeling and control” (1993) “Matrix approach to discrete fractional calculus” (2000); Kirayakova. V, “Generalised Fractional Calculus and Applications” (1994) Kostadin Trenceveski et al. “On fractional derivative of some function of exponential type” (2002) Riemann B, “Versuch einer Auffassung der Integration and Differentiation” (1876) Liouville J, “Memorie sur I integration de I equations a ’l’ aide des differentielles a’ indices quell conques” (1832) Machado J, Kriyakova V, Mainardi F, “Recent History of Fractional Calculus” (2011) Mauro Bologna, “Short Introduction to fractional calculus” (2005) Moshrefi-Torbati. M & Hammond J K, “Physical and geometrical interpretation of fractional operators” (1998) Northover Francis H, “The Foundations of Fractional Calculus” (1977) Ortigueria M D, Machado J A T, “What is fractional derivative?” (2015) Ross Bertram, “A Brief History and Exposition of the Fundamental Theory of Fractional Calculus” (1975) Sabatier J, Agrawal O. P, Machado J. A (Eds.), “Advances in Fractional Calculus” (2007) Valerio D, Trujillo J J, Rivero M, Machado J A T, Baleanu D, “Fractional Calculus: A survey of useful formulas” (2013) Atangana A, Baleanu D. “New fractional derivatives with nonlocal and non-singular kernel”: (2016) Caputo M, Fabrizio M. “A new definition of fractional derivative without singular kernel” (2015) Khaled Saad, Abdon Atangana, Dumitru Baleanu, “New fractional derivatives with non-singular kernel applied to the Burgers equation” (2018) Ortigueira M D, Tenreiro Machado J. “A critical analysis of the Caputo-Fabrizio operator” (2018) Prabhakar T R. “A singular integral equation with a generalised Mittag-Leffer function in the kernel” (1971). The details of the above works, many of which are pioneering, are listed in the bibliography section in alphabetical order.

CHAPTER FOUR FRACTIONAL DIFFER-INTEGRATIONS’ FURTHER EXTENSION

4.1 Introduction The previous chapters highlighted the various ways in which mathematicians described and, by various methods, developed the notion of fractional calculus – namely, fractional integrations and fractional derivatives. We have come across a number of the approaches that can be used to accomplish this. A natural and widely used approach starts from repeated integrations and then extends to fractional integration. Fractional derivatives are then defined either by continuation of the fractional integrals into negative orders (following Leibniz’s idea) or by taking the integer order derivatives of the fractional integrals, as we saw in the Riemann-Liouville integral. We have also discussed why the fractional derivative operation is called ‘differ-integration’. In this chapter, we will be using the formulations developed earlier in the book, and will apply them in order to derive various fractional differ-integrations of some important functions. We will show that fractional derivatives of a non-differentiable point (of a function) exist, and derive their value, in order to provide some examples. We will also extend our general approach to formulate Leibniz’ rule for fractional differ-integrations of the product of two functions and the ‘chain-rule’ for fractional differintegration, and discuss the limitations thereof.

4.2 Applying the Grunwald-Letnikov (GL) formula of fractional derivatives to simple functions 4.2.1 Finding the fractional derivative of a constant using the Grunwald-Letnikov (GL) formula We find the differ-integration of the unit step function, which is f ( x ) 1 for x t a to order D , and f ( x ) 0 for f d x  a . This is generally denoted by f ( x ) u ( x  a ) , and is termed as the Heaviside step function. We have a GL formula as follows (Section-3.23): D

a Dx

> f ( x )@

D

d f ( x) D

>d( x  a)@

§  xN a D lim ¨ N nf ¨ * ( D ) ©

N 1

¦ j 0

· f  x  j  xN a ¸ ¸ * ( j  1) ¹

(4.1)

*( j  D )

Straight forward application of the above formula is: D

a Dx

d

>1@

D

>1@ D

>d( x  a)@

§ § N ·D N 1 * ( j  D ) · lim ¨ ¨ ¸¸ ¸ ¦ N nf ¨ © x  a ¹ *  D *  j ( ) ( 1) 0 j © ¹

We first apply the property of the binomial coefficient, that is

¦

N 1

* ( j D ) j 0 * ( D ) * ( j 1)

* ( N D ) * (1D ) * ( N )

(4.2) (Section-1.12), and

write the following expression: D

a Dx

>1@

§ § N ·D * ( N  D ) · lim ¨ ¨ ¸ ¸ N nf ¨ © x  a ¹ * (1  D ) * ( N ) ¸ © ¹ 1 § D *( N  D ) · lim ¨ N ¸ D *( N ) ¹ ( x  a ) * (1  D ) N nf ©

(4.3)

From the property of the Gamma function (Section-1.11) described below:

§

lim ¨ j j nf

©

c D 1

*( j  D ) ·

¸

* ( j  1) ¹

§

lim ¨ j j nf

©

c D

­f ° ®1 ¸ *( j ) ¹ ° ¯0

*( j  D ) ·

c!0 c 0 c0

(4.4)

156

Chapter Four



We recognize that lim N nf N

D * ( N D ) *( N )



N and c

1 ; i.e. it is obtained by setting j

0 in the above (4.4) formula.

Therefore, we obtain: d

a Dx >1@

D

>1@

D

( x  a) D

>d( x  a )@

D

(4.5)

* (1  D )

This (4.5) we have noted earlier in Section-3.2, but this demonstration is intended to bring out the same result, while noting that here special properties of the Gamma function are used. From the above (4.5), we get for f ( x ) C any constant including zero that is the following result:

>C @ D >d( x  a )@ d

a Dx >C @

D

Since D

a Dx

D

a Dx

> 0@

>1@

dD >1@ D

>d( x  a )@

dD > 0@

>d( x  a )@D

( x  a ) D * (1D )

D

C

D

(4.6)

* (1  D )

x ! a , we conclude that by setting

is never infinite for

C

0

0 for all D . This may sound trivial, and appears obvious. As an example of its importance, dD > f ( x )@

however, it provides a powerful counter example to the belief that if f ( x ) gives zero on

( x  a)

D

>d( x  a )@

g ( x ) then

d  D > g ( x )@

f ( x ) , and if

>d( x  a )@ D

D -th order differentiation, then f ( x ) cannot be restored by D  th order integration.

4.2.2 Finding a fractional derivative of a linear function using the Grunwald-Letnikov (GL) formula Now we take f ( x )

x  a and apply the GL formula in the following steps: a Dx > x  a @

D

§

lim ¨

N nf

D

© § D lim ¨  xN a N nf © § D lim ¨  xN a N nf © § D 1 lim ¨ N D ( xa ) N nf ©



§

 ©

lim ¨

N nf

N 1

·

*( j  D )

 xNa ¦ *(D )*( j  1)  ( x  a)  j  NNa ¸ ¹

j 0

N 1

*( j  D )

¦ *(D )*( j  1) 

Nx  jx  ja N

j 0

N 1

*( j  D )

¦ *(D )*( j  1) 



·

a ¸

Nx  jx  ja  Na N

j 0



¹ · ¸ ¹

N 1

¦ *(*D( )j*(Dj ) 1)  N ( x  a)  j( x  a) ·¸¹ j 0

ND ( x  a )D 1

N 1

¦ *(*D( )j*(Dj ) 1) ·¸¹ j 0

§

 ©

 lim ¨ N nf

N 1

¦ j *(*D( )j*(Dj ) 1) ·¸¹

N D 1 ( x  a )D

j 0

§ · § D N 1 * ( j  D ) · N ¦ ¨ Nlim ¸ ¨ ¸ nf * ( D )* ( j  1) ¹ j 0 ¨ ¸ © 1D ( x  a) ¨ ¸ N 1 ¨  lim § N D 1 j * ( j  D ) · ¸ ¸¸ ¦ ¨ N nf ¨ j 0 * ( D ) * ( j  1) ¹ ¹ © © We first apply the property of the Gamma function, that is * ( j  1) binomial coefficient, that is *( N )

¦

N 1

* ( j D ) j 0 * ( D ) * ( j 1)

* ( N D ) * (1D ) * ( N )

( N  1)  * ( N  1) to write the following steps:

and

¦

(4.7)

j  * ( j ) , and then use the property of the N 1

* ( j D ) j 0 * ( D ) * ( j )

D* ( N D ) * (2 D ) * ( N 1)

, before using

Fractional Differ-integrations’ further Extension

D

a Dx

157

§ § D N 1 * ( j  D ) · · ¨ Nlim ¨N ¦ ¸¸ nf * ( D )* ( j  1) ¹ ¸ 0 j ¨ © 1D ( x  a) ¨ ¸ N 1 *( j  D ) · ¸ ¨  lim § N D 1 ¸¸ ¦ ¨ N nf ¨ j 0 * ( D ) * ( j ) ¹ ¹ © © § · § D *( N  D ) · N ¨ Nlim ¸ ¨ ¸ nf * (1  D )* ( N ) ¹ © 1D ¨ ¸ ( x  a) ¨ § D 1 D* ( N  D ) · ¸ ¨  lim ¨ N ¸¸ * (2  D )* ( N  1) ¹ ¹ © N nf ©

> x  a@

§ 1 § D *( N  D ) · · ¨ *  D Nlim ¨N ¸ ¸ nf *( N ) ¹ (1 ) © 1D ¨ ¸ ( x  a) ¨ D § D 1 * ( N  D ) · ¸ lim ¨ N ¨ ¸¸ * ( N  1) ¹ ¹ © * (2  D ) N nf © § 1 · § D *( N  D ) · lim ¨ N ¨ ¸ ¸ * (1  D ) N nf © *( N ) ¹ 1D ¨ ¸ ( x  a) ¨ D *( N  D ) · ¸ § D lim ¨ N ( N  1) ¨ ¸¸ *( N ) ¹ ¹ © * (2  D ) N nf ©



We recognize that lim N nf N

D * ( N D ) *( N )



1 ; i.e. it is obtained by setting

N and c

j

(4 .8)

0 in the above (4.4)

formula as used previously (Section-1.11). We consequently get the following: D

a Dx

We use * ( x  1)

x  * ( x ) to write * (2  D )

> x  a@

1D

( x  a)

D § 1 · ¨ * (1  D )  * (2  D ) ¸ © ¹

(1  D )  * (1  D ) in order to simplify



1 * (1D )

(4.9)



 * (2DD ) as * (21D ) ,

and arrive at the final result, which is: 1D

( x  a)

a Dx > x  a @

D

(4.10)

* (2  D )

4.2.3 Verifying the GL fractional derivative of a linear function using the Riemann-Liouville (RL) formula of the fractional integration We verify this from the Riemann-Liouville (RL) D

a Dx

> f ( x )@

1 * ( D )

³

x

f ( y)

D 1 a ( x y )

dy . Applying this to f ( x ) a Dx > x  a @

D

formula for fractional integration for

D 0

is

x  a , we get the following: x

1

( y  a)

D * ( D ) ³ ( x  y )

1

dy

put

w

x y

a

xa

1 * ( D )

³

( x  a  w) D 1

w

0

dw

xa § xa ( x  a) 1 d w  ¨ ³ D 1 D ³ w * ( D ) © 0 w 0 1D 1D 1 § ( x  a) ( x  a)  ¨ 1D * ( D ) © D

1

1D

( x  a)

( D )(1  D )* ( D )

·

(4.11)

dw ¸

¹

· ¸ ¹

1D

( x  a)

* (2  D )

D 0

158

Chapter Four

We obtain the denominator in (4.11), the last step of which is ( D )(1  D )  * ( D ) recurrence formula, i.e. x  * ( x )

* (2  D ) , by using the

* ( x  1) . As such, we achieve the earlier obtained result (4.10).

4.2.4 Verifying the GL fractional derivative of a linear function by usingthe Riemann-Liouville (RL) formula of the fractional derivative We can now use the RL-derivative formula

D

a Dx

dn dx n

> x  a@



dD n > x  a @ D n

>d( x  a )@



. Here, we remove the restrictive condition

on D . Thus, for any arbitrary D , one may select an integer n so large that D  n  0 , and the above obtained result (4.11) will give: D n

> x  a@ D n >d( x  a)@ d

1D  n

( x  a)

(4.12)

* (2  D  n )

We use the above expression (4.12) in the following form: n § dD  n > x  a @ · > x  a@ d n ¨ D D n ¸ >d( x  a)@ >d( x  a)@ ¨© >d( x  a)@ ¸¹

d

D

ª ( x  a )1D  n º n « » >d( x  a)@ ¬ *(2  D  n) ¼ d

n

(4.13)

1 § · § * (2  D  n) · 1D ¨ * (2  D  n) ¸ ¨ * (2  D ) ¸ ( x  a ) © ¹© ¹ 1D

( x  a)

* (2  D )

The above (4.13) follows from dn d( x >  a )@n

ª¬ ( x  a ) p º¼

* ( p 1) * ( p  n 1)

( x  a)

We note, as expected, the formula

D

a Dx

> x  a@

( x  a )1D * (2 D )

p

>d( x  a )@n pn

D

a Dx

ª¬ ( x  a ) º¼

p ( p  1)...( p  n  1)( x  a )

p n

,

which is equal to

.

> x  a@

1 ; reduces to ( x  a ) when D

when D that

d

n

( x  a )1D * (2 D )

which reduces to zero when D n 1

( xa ) ( n 1)!

0 ; and reduces to

when compared to

D

a Dx

>1@

( x  a ) D * (1D )

when D

n

2, 3, 4... ; reduces to unity

1, 2, 3,... . We also notice

, and we observe that the D  th differ-integral of the

( x  a ) equals the (D  1) -th differ-integral of the unity function i.e. D

a Dx

D 1

> x  a@

a Dx

>1@

1D

( x  a)

* (2  D )

(4.14)

4.3 Applying the Riemann-Liouville formula to get the fractional derivative for some simple functions 4.3.1 Left fractional derivative of the power function through application of the integral formula of the Riemann-Liouville fractional derivative From the RL fractional derivative formula applied to a function f ( x ) D

0 Dx

Put u

xy ; du

xdy , ( x  u )

b

ª¬ x º¼

b

x , we have the following:

x ª 1 º n D 1 b x  u u u ( ) d « » n dx ¬ * ( n  D ) ³0 ¼

d

n

x (1  y ) in the above expression and obtain the following steps:

(4.15)

Fractional Differ-integrations’ further Extension

D

0 Dx

159

ª 1 n D 1 º n D 1 b b x x y ( xdy ) » (1  y ) n «³ * ( n  D ) dx ¬ 0 ¼ 1

b

ª¬ x º¼

d

n

ª n D b 1 º n D 1 b x y dy » (1  y ) n « ³ * ( n  D ) dx ¬ 0 ¼ 1

d

1

³ (1  y )

n

n D 1 b

y dy § d n ª x n D  b º · ¬ ¼

0

¨ ¨ ©

*(n  D ) B( n  D , b  1) § d

¨ ¨ ©

*(n  D ) n D ! 0

n

dx

¸ ¸ ¹

n

(4.16)

ª¬ x n D b º¼ · ¸ n ¸ dx ¹

b 1 ! 0

Here we used the Beta-integral, that is B( p , q )

1

³u

p 1

0

*( p )*(q ) *( pq)

(1  u ) q 1 du

, with p ! 0 and q ! 0 , giving us

the condition n ! D , and b ! 1 ; so that the Beta-integral does not diverge. Using the Gamma-function *( p )*(q ) *( pq)

representation of the Beta-function, that is B( p , q )

, while also using

* ( p 1) * ( p 1 n )

dn x p dx n

x

n p

, where n is

integer (as discussed in Section-1.15), we write the following: n n D  b

* (b  1)* ( n  D ) § d x

1

D ª bº 0 Dx ¬ x ¼

¨ * ( n  D ) * (b  1  n  D ) © * (b  1)

dx

n

* ( n  D  b  1)

* (b  1  n  D ) * ( n  D  b  n  1) * (b  1) * (b  D  1)

x

b D

The above expression (4.17) we also obtained from Euler’s formula. For b

· ¸ ¹ x

b D

(4.17)

b ! 1

0 , we get 0 DDx >1@

xD * (1D )

, that is the

fractional derivative of the constant function as non-zero in the RL sense. J

Using the formula for RL fractional integration for f (t ) D a Dt

t  a  [ (t  a )

1

J

ª¬(t  a ) º¼

* (D )

t

D 1

³ (t  W )

J

(4.18)

(W  a ) dW

a

a  [ (t  a ) , as we have for W a , [ 0 and for W [ (t  a ) , we get the following:

Using the substitution in (4.18) W

(t  W )

(t  a ) , we see the fractional integral for order D ! 0 as:

t ,[

1 ; dW

(t  a )d[ ,

(t  a )(1  [ ) ; (W  a ) D

a Dt

ª¬(t  a )J º¼

t

1

D 1

* (D ) 1

³ (t  W )

J

(W  a ) dW

a

1

D 1

(t  a ) * (D ) ³

D 1 J

(1  [ )

J

[ (t  a ) (t  a )d[

0

(t  a )

J D 1

³[

* (D ) (t  a )

B(D , J  1)

* (J  1  D )

³[ 0

J

D 1

(1  [ )

d[

(4.19)

J D

* (J  1) 1

D 1

(1  [ )

0

* (D )

We used the Beta-integral B(D , J  1)

J

(t  a )

J D

d[ , and B(D , J  1)

D !0

J 1 ! 0

J ! 1 * (D ) * ( J 1) * (D J 1)

obtained result (4.19) to the fractional integral of the order (1  X ) , with 0 d X  1 , we get:

.

Applying the above

Fractional Differ-integrations’ further Extension

1

( 1) D

D t

 (1X ) A

ª¬ ( A  t )J º¼

( 1)



* (J  1) * (J  2  X )

* (J  1)

d

( 1)

ª

dt ¬



* ( J 1) * ( J  2 X )

(J  1  X )( A  t )

( A  t)

J X t

161

( A  t)

J X

J 1X

º ¼

( 1)

(4.25)

X J D A ª¬ ( A  t ) º¼

* (J  2  X ) We get the following formula for a fractional derivative in the right sense as: * (J  1) X J J X ª º (4.26) ( A  t) J ! 1 t DA ¬( A  t ) ¼ * (J  2  X ) We prefer to take a different and more elementary approach attaching a meaning to the RL operator for D t 0 . D

a Dx

The formula

> f ( x )@

1 * ( D )

³

x

a

( x  y)

D 1

 f ( y ) dy

for D  0 will be retained for only D  0 , indicating

fractional integration; it is extended to D t 0 , by insisting that

n a Dx



D

a Dx

> f ( x ) @

n D a Dx

> f ( x)@ is satisfied for all

positive integers n and for all D ; that it will also be satisfied by the RL integral. That is, we shall require D

a Dx

> f ( x)@ { Dxn  a DDx n > f ( x)@

where Dxn {

dn dx n

is n -th derivative (the ordinary derivative operator), choosing D

the positive integer so large that D  n  0 . The relationship

a Dx

> f ( x)@ { Dxn  a DDx n > f ( x)@ holds for all D .

We

described this earlier in the book (Section-2.14), while performing an extension of the order D , where D  0 , and D

a Dx

> f ( x )@

1 * ( D )

x

³

a

( x  y)

D 1

 f ( y ) dy ; to all D .

We will use this concept subsequently.

4.4 Applying the Caputo formula to get a fractional derivative for simple functions 4.4.1 Left fractional derivative of the power function x b with respect to x through application of the integral formula of the Caputo fractional derivative Let us take the function f ( x )

x

b

x , with b ! 1 , and we will analyse the Caputo derivative from the starting point

0 , as described below: C D 0 Dx

b

ª¬ x º¼

x

1 *(n  D )

³ ( x  u)

n D 1

0

§ d nu b · ¨ du n ¸ du © ¹

x

1

( x  u) *(n  D ) ³

n D 1

0

* (b  1) §

§ * (b  1) b  n · ¨ * ( b  1  n ) u ¸ du © ¹ x

1

( x  u) ¨ * (b  1  n ) © * ( n  D ) ³0 * (b  1) * (b  1  n )



n D 0 Ix

ª¬ x b  n º¼

n D 1 b  n

u

·

du ¸

¹

(4.27)



* (b  1) §

* (b  n  1) b  n  n D · x ¨ ¸ * (b  1  n ) © * (b  n  1  n  D ¹ * (b  1) * (b  1  D ) We note that

1 * ( n D )

³

x

0

( x  u)

n D 1 b  n

u

du

n D 0 Ix

x

b D

ª¬ x b  n º¼ , in the above steps, (4.27) is an expression for the RL

fractional integration of the order n  D , for function p 0 Ix

q

ª¬ x º¼

* ( q 1) * ( q 1 p )

x

q p

f ( x)

x

bn

, and we used the derived formula

(Section-2.6). Note that this (4.27) is the same result as that obtained for the fractional

derivative using the RL formula (4.17). The reason for this is that the value of the function at the starting point of the

162

Chapter Four

differentiation is zero. In a similar way, one can find the Caputo derivative for all other functions described in the previous section. X

4.4.2 Left fractional derivative of the power function \ ( x) with respect to x through application of the integral formula of the generalized Caputo fractional derivative We

have D

a ! x ,> z , w@

demonstrated

f ( x)

1

 w ( x )  * (D )

the

use

of

x

D ³  z ( x )  z (t )

the 1

a

generalized



w( t ) z

(1)

(t )

RL

fractional

integration

formula,

that

is

 f (t ) dt , in an earlier chapter (Sections 2.17 and 3.25).

Now, we demonstrate the use of the generalized fractional derivative formula that we noted earlier with weight function w( x ) 1 , and n  1  D  n , i.e. C D a $ x ,> z ,1@

Let us take f ( x )

1

f ( x)

*(n  D )

³

x

a

z

(1)

(t )  z ( x )  z (t )

X

\ ( x)  \ (a ) ; X E  1 , z ( x) E 1 C D ª º a $ x ,>\ ,1@ ¬\ ( x )  \ ( a ) ¼ 1

n D 1

§ 1 d· ¨ z (1) (t ) dt ¸ © ¹

n

 f (t ) dt

(4.28)

\ ( x ) with E ! 0 and write the following steps:

n

E 1 § 1 d· \ (t ) \ ( x )  \ (t ) \ ( t )  \ ( a ) dt  ¨ ¸ (1) ³ *(n  D ) a © \ (t ) dt ¹ E 1 n · (1) x 1 n D 1 § d \ (t )  \ ( a ) ( ) ( ) x t \ \  ¨ ¸ \ (t ) dt  n ¨ ¸ * ( n  D ) ³a t d ( ) \ > @ © ¹ x 1 n D 1 § * ( E ) E 1 n · (1) \ ( x )  \ (t ) \ (t )  \ ( a ) \ ( t ) dt    ¨ ¸ ³ a *(n  D ) © *( E  n) ¹ \ ( x) 1 *( E ) § n D 1 E 1 n · \ ( x)  \ (t ) \ (t )  \ (a) d \ (t ) ¸ ¨ ³ ( ) a \ *( E  n) © *( n  D ) ¹ x

*( E ) *( E  n)

n D 1

(1)



( n D ) \ ( a ) I\ ( x )

ª\ ( x )  \ ( a ) E 1 n º ¬ ¼

* ( E ) § * ( E  1  n  1)

\ ( x)  \ (a) ¨ *( E  n) © *( E  n  n  D ) *( E ) *( E  D )



\ ( x)  \ (a) d \ ( t ) dt



We note that we have used \ (1) (t ) dt recognize

\ ( x) 1 * ( n D ) \ ( a )

³ \ ( x)  \ (t )

order ( n  D ) of the function f ( x )

n D 1



E 1 n  n D

E 1D

dt

d \ (t ) in the above steps (4.29). We also note that we

\ (t )  \ (a)

\ ( x)  \ (a )

(4.29)

E 1 n

E 1 n

d \ (t ) as a formula for fractional integration of the

, with respect to \ ( x ) from the lower terminal limit \ ( a ) ( n D )

to the higher terminal limit \ ( x ) i.e. compactly expressed as \ ( a ) I\ ( x ) We use

X

a Ix

ª¬ ( x  a )b º¼

* ( b 1) * ( b 1X )

( x  a)

b X

· ¸ ¹

ª\ ( x )  \ ( a ) E 1 n º . ¬ ¼

and obtained the result reported in the above step of the derivation

(4.29).

4.5 The fractional differ-integration of the binomial function We developed the fractional differ-integration of the simple function. The most important application of these techniques is using the fractional differ-integration of functions like representation is presented in the following way:

f ( x)

¦

f j

a ( x  a) 0 j

p

 . j n

A series

Fractional Differ-integrations’ further Extension

f ( x)

p

a0 ( x  a )  a1 ( x  a ) ...an 1 ( x  a ) ( x  a)

p

p   1n

p   nn1

 a2 ( x  a )

 an ( x  a )

f

¦

163

a j ( x  a)

p 1

p   n2



...

(4.30)

 j n

j 0

We refer to the above represented f ( x ) (4.30) as a differ-integrable function with the condition p ! 1 , n as its positive integer, and a0 z 0 as its non-zero constant. This type of function with p ! 1 is called a Riemann-Class function. We conduct term-by-term differ-integration using the arbitrary number D and write the following

ª f a ( x  a ) p  nj º D a x «¦ j 0 j »¼ ¬ D

f

*

¦* j 0





pn  j  n n



pn D n  j  n n



( x  a)

p D 

 j n

(4.31)

The above expression (4.31) is obtained using the earlier derived formula detailed in Section-4.3, the differ-integration p

formula for function ( x  a ) . For the binomial function i.e. f ( x ) theorem, i.e. ( x  y )

m

¦

p

(C  cx ) , we can write C  cx

f

m j j m! x y j 0 j !( m  j )!

(C  cx )

p

C  ac  c ( x  a ) , and then via the binomial

, we write the power series expansion as expressed below:

 (C  ac)  c( x  a ) f

p

p!

¦ j !( p  j )!  (C  ac)   c( x  a) p j

j 0 f

* ( p  1)

¦ *( j  1)*( p  j  1) (c)

j

(C  ac )

p j

j



( x  a)

(4.32) j

j 0

c ( xa ) C  ca

provided 1   1 . Note that the expansion (4.32) is valid even if p is a positive integer, but that in this event the sum is automatically finite, while all terms for which j exceeds p have finite denominators. Using the identity * ( p 1) * ( p  j 1)

( 1)

j *( j p) *( p)

, we write the D -th differ-integral as follows: D

a Dx

ª¬(C  cx ) p º¼ ªf º * ( p  1) j p j j ( c ) (C  ac ) ( x  a ) » «¦ ¬ j 0 * ( j  1)* ( p  j  1) ¼ f § (C  ac ) p ( 1) j c j * ( p  1) j · D ¦ ¨ (C  ac ) j * ( j  1)* ( p  j  1) a Dx ª¬( x  a ) º¼ ¸ j 0© ¹ D

a Dx

(4.33)

§ (C  ac ) p ( 1) j c j * ( p  1) § * ( j  1) · j D · ¦ ¨ (C  ac ) j * ( j  1)* ( p  j  1) ¨ * ( j  1  D ) ¸ ( x  a ) ¸ © ¹ j 0© ¹ f

p

(C  ac ) ( x  a )

D

f

*( j  p)

§ c( x  a) · ¦ ¨ ¸ j 0 * ( j  D  1) © C  ac ¹

j

*( p) We try to obtain the closed form of the above derivation using the identity of the hyper-geometric function and the

Beta function (Appendix-A), i.e. B y (c,1  b  c ) summation in the expression (4.33). Taking y following steps as a final result:

c ( xa ) ( C  ca )

y c (1 y )1bc * ( c ) * (1 b )

¦

f j 0

y

j * ( j 1 b ) * ( j 1 c )

and taking parameters b

, in order to simplify the

( p  1) and c

D , we obtain the

164

Chapter Four p D d ª¬(C  cx ) º¼

D

>d( x  a)@

(C  ac ) D

( x  a)

j

§ cx  ac · *  j  (  p ) ¦ ¨ ¸ D ( x  a ) * (  p ) j 0 © C  ac ¹ *  j  1  ( D ) (C  ac )

p

f

¦y *( p)

j

p

f

*  j  1  ( ( p  1)) *  j  1  ( D )

j 0

(4.34)

§ (C  ac )D  p · § (cx  ac )D ¨ D Dp ¸¨ D ( x  a ) * (  p ) © (C  cx ) ¹ © (C  ac ) (C  ac )

p

D

c (C  cx )

For a

0 and C

c

· *( p) ¸ * ( D ) B y  D , D  p ¹

p D

B y  D , D  p * ( D ) 1 , we write the following simple result:

d

D

(1  x )

ª(1  x ) p º¼ D ¬ dx

p D

B x ( D , D  p )

* ( D )

(4.35)

4.6 Fractional differ-integration exponential function using the Riemann-Liouville fractional derivative formula Let us try to find the fractional integral of the order v ! 0 of the exponential function f (t ) fractional integration operation as follows:

Once the substitution x

t

1

v ª at º 0 Dt ¬e ¼

at

e . We write the

(t  u ) * (v ) ³

v 1 au

(4.36)

e du

0

t  u is made, the above integral expression (4.36) is changed as follows: v 0 Dt

e

at

ª¬e º¼

t

at

* (v )

³x

v 1  ax

(4.37)

dx

e

0

Clearly, this integral is not an elementary function, but is closely related to Tricomi’s incomplete gamma function (discussed in Section-1.13), and thus the fractional integral of the exponential function is expressed as follows: v

0 Dt

ª¬e at º¼

e

at

* (v )

t

³x

v 1  ax

e

dx

0

v at

t e

t

t

v

x * (v ) ³

v 1  ax

e

dx

put

ax

0

v at

t e

v at

t e

t

 v at

* (v ) ( at )

³a

³

at

0

y

v 1  y

e dy

v 1

 v at

³y

* (v )

e

 y § dy ·

¨ ¸ ©a¹

v 1  y

e dy

0

x v *(v)

*

( at ) v *(v)

(4.38)

v 1

0

In the above, Tricomi’s incomplete gamma function is J (v, x ) write

y

y

³

x

0

 t v 1

e t

dt . Therefore, in this derivation we

*

J (v, at ) , and with that, we use the following expression v 0 Dt

ª¬e at º¼ t v e at J * (v, at ) (4.39) We will use this formula (4.39) in the following section. When we take v D , a r1 , we get the following expression as an important formula with a change in variable:

ª rx º 0 Dx ¬ e ¼ D

d

D

ª¬e r x º¼ D

> dx @

e

rx

x

D

*

J ( D , r x )

With C and c as arbitrary constants, using the power series expansion of the f ( x ) expansion at x a in the following steps:

e

( C  cx )

(4.40) , we write the Taylor

Fractional Differ-integrations’ further Extension

f ( x)

(1)

f (a)  ( x  a) f e

C  ca

( x  a)

(a) 

2!

 ( x  a )( c )e

C  ca



2

(2)

f

165

( a )  .....

( x  a)

2

2 C  ca

c e

2!

( x  a)



3

e 

1 c 2 ( x  a ) 2  1 c 3 ( x  a )3  ... 1  c ( x  a )  2! 3!

C  ca

e

C  ca

f

¦

 c ( x  a )

C  cx

e

C  ca

¦

 c ( xa ) j

f

* ( j 1)

j 0

 .... (4.41)



j

* ( j  1)

j 0

Thus, we write e

3 C  ca

( c ) e

3!

, a power series expansion of e

C  cx

at x

a , which is valid for all

x  a . We differ-integrate term by term by order D with respect to cx and with the starting point ca , with the formula

D

ca Dcx

* ( j 1) * ( j 1D )

ª¬ (cx  ca ) j º¼ D

ca Dcx

ª¬ e

C  cx

(cx  ca )

j D

in the above series, and write the following steps

ª C ca f  (cx  ca ) j º º¼ » ¦ D D «e * ( j  1) » j 0 >d(cx  ca)@ >d(cx  ca)@ «¬ ¼ j ª  ( 1) j (cx  ca ) º D f d C  ca « » e ¦ D * ( j  1) » j 0 > d( cx  ca ) @ « ¬ ¼ j f ( 1) § * ( j  1) C  ca j D · e (cx  ca ) ¦ ¨ ¸ j 0 * ( j  1) © * ( j  1  D ) ¹ C  cx D º¼ d ª¬e

e

C  ca

d

D

f

 (1) (cx  ca) (cx  ca)

j 0

* ( j  D  1)

j

¦

 c( x  a)

D

e

j

f

C  ca

(4.42)

D

 c ( x  a )

j

¦ *( j  D  1) j 0

The sum in (4.42) may be expressed as Tricomi’s incomplete gamma function, which is the following: *

J (D , x ) We take J

*

 D , c( x  a )

e

c ( xa )

D

¦

ca Dcx

ª¬ e

f

 c ( xa ) j

j 0 * ( j D 1) C  cx

º¼

D

x c

*( x)

³

y

x 1  y

e dy

e

x

f

x

j

¦ *( j  D  1)

(4.43)

j 0

0

, and use this in the following derivation

C  cx D º¼ d ª¬e

e

D

D

>d(cx  ca)@ e

C  ca

D

c ( x  a)

C  ca  cx  cx

D

D

c ( x  a) e

C  cx  c ( x  a )

D

D

c ( x  a)

f

 c ( x  a )

f

¦ *( j  D  1) j 0

 c ( x  a )

C  cx *

j

¦ *( j  D  1) j 0 f

 c ( x  a )

j

¦ *( j  D  1)

(4.44)

j 0

§ c ( x  a ) f  c ( x  a ) j ¨e ¦ ¨ j 0 * ( j  D  1) © e

j

· ¸ ¸ ¹

 D , c( x  a )

J

D

D

c ( x  a) The scaling property that we evaluated earlier (in Section-3.20) gives the following formula:

O

D

> f (O x )@ D >d(O x)@

d

D

d

D

> f (O x )@ D >dx@

(4.45)

166

Chapter Four

Using these above identities (4.45), we write the following steps:

d

D

ª¬eC cx º¼

c

D

>d( x  a)@

d

D

ª¬eC cx º¼

D

D

>d(cx  ca)@

j f § c ( x  a )  D C  ca c ¨  c( x  a) e ¦ ¨ j 0 * ( j  D  1) ©

D

e

C  ca  cx  cx

 c ( x  a )

f

j

¦ *( j  D  1)

D

( x  a)

(4.46)

j 0

j f § c ( x  a )  cx  ca ¨e ¦ D ( x  a) ¨ j 0 * ( j  D  1) © e

C  cx

e

C  cx

J

D

( x  a)

· ¸ ¸ ¹

*

· ¸ ¸ ¹

 D , c( x  a )

Finally, we have derived the following formula: D

a Dx

Take D

n ,C

a

0 , and c

d

ª¬eC cx º¼

ª¬eC cx º¼

D

e

D

C  cx

J

D

>d( x  a )@

( x  a)

*

 D , c( x  a )

(4.47)

B1 : D

0 Dx

*

d

rx

ª¬e º¼

D

ª¬e r x º¼

e

D

>d( x)@

rx

x

J

D

*

 D , r x

(4.48)

n

If n is a non-negative integer, then J (  n, x ) x , and the above obtained result (4.48) reduces to a formula of the multiple differentiation of an exponential function for the integer order case below: n

ª 0 Dx ¬ e

rx

d

º¼

n

ª¬e r x º¼

>d( x)@ e

rx

x

(r x)

n

n rx

( r1) e

x

e

rx

x

n

J

*

  n, r x

n

(4.49) n!0

The incomplete gamma function has an asymptotic expansion for large x and is given by the following expression: *

J ( D , x )

 1

e

x

§ D 1 2 ·  O( x ) ¸ ¨1  * ( D ) © x ¹

(4.50) x x The above formula (4.50) does not come directly, but results from the asymptotic expansion formula of the complementary incomplete gamma function * ( a , x ) , which is defined for x t 0 and f  a  f , as an integral, D

that is * ( a , x )

³

f

x

D 1

 t a 1

e t

dt . Note that the lower limit is other than zero, as it was in the case of the Euler gamma

function * ( a ) . The asymptotic expansion for large x is given by the following relationship:

*( a, x) # x *

We use the identity J ( a , x ) following:

x

a

1 

*(a,x) *(a)

,

a 1  x

e

§ a  1 ( a  1)( a  2) ·   ... ¸ ¨1  2 x x © ¹

(4.51)

and use the asymptotic expansion of the * ( a , x ) (4.51) to write the

Fractional Differ-integrations’ further Extension

*

J ( a, x)

Putting a J * ( D , x ) xD

D , we get J * ( D , x )

 1

e x xD 1* ( D )

1 

D 1 x

x

D

x 

 O( x

2

a

167

§ e  x x a 1 · ( a 1)( a 2) 1  ax1   ... ¸ 2 ¨1  x *(a) © ¹



x

a

x

a

e x * ( D )





 x 1



 1 x

§ a  1 ( a  1)( a  2) ·   ... ¸ ¨1  2 *(a) © x x ¹

e x

(4.52)

x

§ 1 a  1 ( a  1)( a  2) ·  ... ¸ ¨  2  3 *(a) © x x x ¹ e

( D 1)





x2



( D 1)( D  2)

 ... . From here, we write the expression

x3



) , for large x i.e. x n f . We will use this asymptotic expansion formula in

the following derivation. We have the formula below: D

a Dx

ª¬e

We write the following steps, by putting C D

ª f Dx ¬e

cx

C  cx

d

º¼

>d( x  a )@

d

cx

ª¬e º¼

cx

( x  f)

J

*

§ J * ( D , cy ) · ¸ D y nf y © ¹ y nf

Substituting D

f Dx ª ¬e

cx

cy

J ( D , cy ) ·

D

(4.53)

D cx

y

§ J * ( D , z ) · ¸ D z nf © z ¹

c e lim ¨

¸ ¹

D

(cy )

©

 D , c( x  a )

xf

*

§

cx

*

 D , c( x  f)

e lim ¨

e lim ¨ c

J

c in the above expression (4.53): D

>d( x  f)@ cx

( x  a)

e

D

C  cx

D

f and c

D

e

D

0, a

º¼

ª¬eC cx º¼

D

(4.54)

z

z § e 2 · D cx º¼ cD e cx ¨1  D 1 1  D z1  O ( z ) ¸ c e © z * ( D ) ¹





This simple result for differ-integration with a lower terminal limit is a f , as discussed earlier in Sections 3.4, 3.5 and 3.6, which is provided by Liouville, with postulates about fractional differ-integration of the exponential function. Functions other than exponentials, however, seldom yield finite results for the differ-integrations when the lower limit is pushed towards f . Here, the function e integration process.

C  cx

is analytic in ( x  a ) , and was subject to a term-by-term differ-

Let us evaluate the D -th differ-integral of the f ( x ) D

0 Dx

1

ª xD 1e  x º ¬ ¼

steps:

1 * ( D )

³

x

0

( x  y)

D 1 D 1  1 y

y

e

D

D 1  1 x

1

x

1

y

e

, we get the following

dy

1

0

0

x D 1 xz 1

 * ( D ) ³ f

 1x

f

* ( D )

³z

e x Using the definition of the gamma function * ( n )

x xz 1

D 1  1 y

D * ( D ) ³ ( x  y )

1

following formula:

D  0 using the RL definition we write

. For

e

d y and, followed by the substitution y

ª e x º D « 1D » dx ¬ x ¼ d

x

D 1

³

f

0

z

e x

D 1

 z  1x

 xzxz1

2 x xz 1



dz

(4.55)

e dz

0

e dz , we write

D



D 1  z

n 1  z

d

D 1

1

ª e x º D « x1D » ¼ dx ¬

e x

* ( D )

³

f

0

z

D 1  z

e dz , then the

 1x

D 1

(4.56)

168

Chapter Four

4.7 Fractional differ-integration for logarithmic functions We conducted fractional integration as an analytic continuation of the repeated n -integer order integration of the function i.e. ln x earlier in the book (in Section-2.8). Here, we apply the Riemann-Liouville formula to achieve the same result that we obtained in that section. When f ( x ) ln x , we first find the Riemann-Liouville fractional

x (1  v ) . As such, we write the following

0  D  1 with a change of the variable .i.e. u

integration of the order steps:

D 0 Dx

>ln x@

D 0 Ix

x

1

>ln x@

1

D ( x  u )  ln(u ) du , ³ * (D )

u

x (1  v )

0

1

0

1

D  xv  ln  x(1  v)   xdv ³ * (D ) 1

1

1 * (D ) (ln x ) x

³x

D 1

0

1

³v

as

D 1

0

process:

³v

D 1

dv 

0

x

1

D

1

D v  ln(1  v ) dv ³ * (D ) 0

 D v dv and write the second integral in the above obtained expression (4.57), i.e.

We write here d ª¬(1  v ) º¼ 1

 ln x  ln(1  v) dv

D 1

D

1

v

(4.57)

0

* (D )

D ³ v  ln(1  v) dv

D D 1

D 0 Ix

x

>ln x@ x x

D

0

D

* (D  1) D

1

D ³ ln(1  v)  d ª¬(1  v )º¼ ,

 D1

ln(1  v )dv

ln x 

x

and substitute it to write the following

1

D

D  * (D )

D ³ ln(1  v)  d ª¬(1  v )º¼ 0

 

§ ln x  (1  vD ) ln(1  v ) v 1  1 (1  vD ) ¨ ³0 v 0 * (D  1) © § ln x  (1  vD ) ln(1  v ) v 1  1 (1  vD ) ¨ ³0 v 0 * (D  1) ©

>

d ln(1 v )@ dv

 ( 1) (1 v )

dv ·¸¹



· dv ¸ ¹

(4.58)

1 D 1 § · 1 v D dv ¸ ¨ ln x   (1  v ) ln(1  v ) 0  ³ 1 v * (D  1) © 0 ¹

x

D

By doing integration in parts, we obtained the last steps of (4.58).

Noting

1

v x v y 0 1 v

³

dv \ ( y  1)  \ ( x  1) , with Re > x @ , Re ª¬ y º¼ ! 1 , with a ‘psi’ function defined as the derivative of 1 d* ( x ) * ( x ) dx

the complete gamma function that is \ ( x )

\ ( x  1)  \ ( x )

( x)

1

and \ (1)

J

*(1) ( x ) *( x)

d> ln * ( x )@ dx

, or \ ( x )

, we find that it obeys recursion

0.5772156666... . is Euler’s constant (Section-1.16). As such, we write:

D 0 Dx

>ln x@

D

0 Ix

>ln x@

x

D

* (D  1)

 ln x  \ (D  1)  \ (1)

(4.59)

We previously derived the same result differently in Section-2.8. Thus for m  1 d Re >D @ d m , and using the above D

obtained expression (4.59) for 0 Dx

>ln x@ , D

0 Dx

we write the following:

>ln x@

m Dx



m Dx

Letting m  D

X in the above (4.60) and using

the following expression:

X 0 Dx

 ( m D ) 0 Dx



X 0 Dx

>ln x@

>ln x@

d

m

ª¬ 0 Dx ( m D ) >ln x @º¼ dx

m

(4.60)

>ln x@ xX * (X 1)

 ln x  \ (X  1)  \ (1) , from (4.59) we have

Fractional Differ-integrations’ further Extension

D 0 Dx

>ln x@

169

ª xX º D «  ln x  \ (X  1)  \ (1) » ¬ * (X  1) ¼ m

(4.61)

We introduce < { \ (1)  \ (X  1) which is a constant and independent of the value of x . With this, we write D

0 Dx

>ln x@

1 * (X 1)

m X D ª¬ x  ln x  < º¼ . Now by using Leibniz’s rule regarding the m -th derivative of the product

of the two functions, i.e.

D

m

m

> f ( x) g ( x)@ ¦ k 0  mCk  Dxmk > f ( x)@  Dxk > g ( x)@ ,

we write the following

expression: D

> @

0 Dx ln x

1

m

§m·

¦ ¨© k ¸¹  D * (X  1)

mk x

k 0

Now using D

* (X 1) * (X  m  k 1)

mk X

x

X m k

x

D

; for k 0



ª¬ xX º¼  Dxk >ln x  < @

(4.62)

0,1, ...., m and the integer derivatives of the ln x  < as follows:

>ln x  < @

ln x  < , D

k

>ln x  < @

( 1)

k 1

x

( k  1)! k

(4.63)

k 1, 2, 3..., m With the above expressions ((4.62) and (4.63)), we write the following steps:

D

D

X  m 1 § * (X  1) xX  m m* (X  1) x §1· ·  ln x  <  ¨ ¨ ¸  ... ¸ * (X  m  2) © x ¹ ¨ * (X  m  1) ¸ X mk k 1 ¨ § ( 1) ( k  1)! · ¸ m ! * (X  1) x 1 ¨ ...  ¨ ¸ ¸ k * (X  1) ¨ x k !( m  k )! * (X  m  k  1) © ¹ ¸ ¨ ¸ m 1 ¨ ...  xX § ( 1) ( m  1)! · ¸ ¨ ¸ m ¨ ¸ x © ¹ © ¹

>ln x@

X m § xX  m · mx ln  <  x  ......  ¨ ¸ * (X  m  2) ¨ * (X  m  1) ¸ X m m 1 ¨ m ! xX  m  ( 1) k 1 ( k  1)! x (1) (m  1)! ¸¸ ¨¨   ...  ¸ * (X  1) © k !( m  k )! * (X  m  k  1) ¹ X m X m § x  ln x  < · x ( 1) m ! * (X  m  1)   ...... ¸ ¨ * (X  m  1)  ( m  1)! * (X  m  2) ¨ * (X  m  1) ¸ ¨ ¸ X m k x m ! ( 1) * (X  m  1) ¨ ¸ ¨ * (X  m  1)  k ( m  k )! * (X  m  k  1) ¸ ¨ ¸ X m m x m m ( 1) ( 1)! ( 1) X   *   ¨ ¸  ¨ ....  ¸ * (X  m  1)* (X  1) © ¹ X m

k m § ( 1) m ! * (X  m  1) · ¨ ln x  <  ¦ ¸ * (X  m  1) © k 1 k ( m  k )! * (X  m  k  1) ¹

x

X m

k m § ( 1) m ! * (X  m  1) · \ \ X x     ln (1) ( 1) ¦ ¨ ¸ * (X  m  1) © k 1 k ( m  k )! * (X  m  k  1) ¹

x

Since X

(4.64)

m  D , we have, as found above in (4.64), the following expression: D

D

>ln x@

k m § ( 1) m ! * (1  D ) · ln x \ (1) \ ( m D 1)      ¦ ¨ ¸ * (1  D ) © k 1 k ( m  k )! * ( k  1  D ) ¹

x

D

We use the expression \ ( x  1)  \ ( x  1  m) book (Section-1.16), to write the following:

¦

( 1)k m!* ( x 1) k 1 k ( m  k )!* ( x 1 k )

f

(4.65)

with x ! ( m  1) , as discussed earlier in the

170

Chapter Four

D

D

x

> @ ln x

D

 ln x  \ (1)  \ (m  1  D )  \ (D  1)  \ (D  1  m

* (1  D ) x

D

* (1  D ) x

 ln x  \ (1)  \ (1  D )

D

* (1  D )

(4.66)

 ln x  J  \ (1  D ) ;

 \ (1)

J

n  ` ), the fractional derivative for the RHS of the above expression (4.66) \ (1D ) n should be interpreted as the limit for D o n . In fact, in limD o n * (1D ) ( 1) * ( n ) , the rule When D is a natural number (i.e. D d n > ln x @ dx n

  ( n  1)!   x

n

is fulfilled. That is

D

n

>ln x@

limD o n x

\ (1) \ (1  D ) · § ln x ¨ * (1  D )  * (1  D )  * (1  D ) ¸ © ¹

D

§

limD o n ¨  x

D

\ (1  D ) ·

n n ¸  x ( 1) * ( n)

©

  ( n  1)! (  x )

n

The first two terms in the bracket of the first line of the steps above (4.67) reach zero as limD o n

n

x

>ln x@

1 * (1D )

n

* (1  n )

 ln x  \ (1)  \ (1  n) (4.68)

n

n § 1 · ln x   ¦ k ¸ ¨ n! © k 1 ¹

x

¦

This (4.68) is obtained by using \ ( n  1)  \ (1)

n

1 k 1k

0 , as the

 n; n  ` there holds

reciprocal of the gamma function at points zero and negative integers is zero. However, for D the classical result, as in the following:

D

(4.67)

* (1  D ) ¹

i.e. the properties of the ‘psi’ function (Section-1.16).

Now we do the same thing for f ( x ) ln x with RL-formulation with D replaced by D . We first state that D  0 , before providing the reasoning behind the analytic continuation. D

d ln( x ) dx

x

1

ln( y )dy

D * ( D ) ³ ( x  y )

D

D 0

1

put

x y

X

x

0

x

D

ln( x )

* ( D ) x

D

1

dX

³ XD

1

0

ln( x ) §



x

D

1

* ( D ) ³ 0

x

1·

D

1

¨ ¸  * ( D ) © D ¹ * ( D ) ³0

ln(1  X )dX

X

(4.69)

D 1

ln(1  X )dX

X

D 1

Taking the second definite integral in (4.69) and doing integration by parts, we write the following using

J

\ (1)

0.5772157... , and

³



1 X x X y

0

1X



dX 1

³ 0

\ ( y  1)  \ ( x  1) , as shown earlier in Section-1.16. We get

ln(1  X )dX

X

D 1

1

1

ln(1  X ) ª¬d(1  X D³

D

0

1  X ln(1  X ) D

D 0

J  \ (1  D ) D

) º¼

1

 0



J D

1

D 

1

³ 0

1X

D

1X

dX

(4.70)

\ (1  D ) D

We have used the ‘psi’ function of the argument 1  D in the above (4.70). Combining the above result (4.70) together with a recurrence of the relationship of the gamma function i.e. (  x )  * (  x )

* (1  x ) , we write

Fractional Differ-integrations’ further Extension D

d ln( x ) dx

x

D

D

* (1  D )

171

 ln( x)  J  \ (1  D )

(4.71)

Though the result (4.71) is derived for D  0 (as in Section-2.8), the usual argument of the analytic-continuation says it is valid for all D .

4.8 Fractional differ-integration for some complicated functions described by the power series expansion Let us take the function f ( x )

xD 1 x

and differ-integrate by order D from the starting point or lower limit a

D 1 order to get 0 Dx ª 1x x º . We use the binomial expansion formula (i.e. (1  x )

2

D

¬

1  x  x  ...

¼

¦

f j 0

0 , in

j

x ) here and

write the following

ª xD º » D « dx ¬1  x ¼ d

D

f

¦

d

D

ª¬ x j D º¼ dx

j 0

(4.72)

D

We subject the above expansion (4.72) to the condition that the magnitude of x is less than unity, that is 0  x  1 . If the condition D ! 1 is maintained, then we have the following. Using Euler’s formula for fractional differentiation (as explored in Section-3.2.3), we write:

ª xD º » D « dx ¬ 1  x ¼ d

D

f

¦

* ( j  D  1)

j

x (4.73) * ( j  1) j 0 We have obtained the following from the property of the gamma function (in Sections 1.11 and 1.12) in terms of the binomial coefficients in q and then put q D  1 , i.e. *( j  q)

j §q· ( 1) ¨ ¸ © j¹

* (  q )* ( j  1) * ( j  D  1) * ( j D 1) * ( j 1)

using * (D  1)



* ( j D 1) * (D 1) * ( j 1)

D  1

(4.74)

§ D  1· ( 1) ¨ ¸ © j ¹ j

* (D  1)* ( j  1)

We manipulate

with q

, to write in terms of the binomial coefficients in (4.73) as

follows:

* ( j  D  1)

( 1)

j

§ D  1· ¨ j ¸ * (D  1) © ¹

(4.75)

* ( j  1) By using the above obtained expression (4.75), we thus write the following expression:

ª xD º » D « dx ¬1  x ¼ d

D

f

* ( j  D  1)

j 0

* ( j  1)

¦

x

j

(4.76)

f

§ D  1· j * (D  1)¦ ¨ ¸(  x ) j ¹ j 0© Using the sum in (4.76) as in the following representation f § D  1· j ¨ j ¸(  x ) ¹ j 0©

¦

1

( D  1)!

x

1!( D  2)! (1  x )

( D  1)! 2!( D  3)!

2

x  ...

(4.77)

D 1

we write the simple result as in the following form

ª xD º » D « dx ¬1  x ¼ d

In the next function, we take f ( x ) write the following expression

xE 1 x

D

* (D  1)

(4.78)

D 1

(1  x )

, with a similar procedure to that used for derivation

D

0 Dx

D

ª 1x x º (4.78). We ¬ ¼

172

Chapter Four

ª xE º » D « dx ¬ 1  x ¼ d

E

D ª 1x x º 0 Dx

¬

¼

D

x

f

E D

(4.79)

* ( j  E  1)

¦ *( j  E  D  1) x

j

j 0

with the restrictions 0  x  1 and E ! 1 . To further manipulate the above (4.79) and get a closed form is difficult in this case as it requires a hyper-geometric function (refer to Appendix-A) and its relationship to the incomplete beta* (1 b ) B x ( c ,1 b  c ) * ( c ) x c (1 x )1bc

function, that is ª¬ x bc º¼

. From this we write c

B x (c,1  b  c ) 1 b  c

(1  x )

Putting c

p  q and b

c x * (c ) §

· j *( j  1  b) ¨¦ x ¸ * (1  b ) © j 0 *( j  1  c) ¹

x * (c ) § ª b º · ¨ x ¸ * (1  b ) © «¬ c »¼ ¹

f

(4.80)

p , we obtain B x ( p  q ,1  q )

x

p q

1 q

(1  x )

*( p  q) § ª

* (1  p ) x

p q

*( p  q) §

dD ª¬ x E (1 x )1 º¼

x

dx E

(4.81)

*( j  1  p) ·

j

¨¦ x ¸ © j 0 *( j  1  p  q) ¹

* (1  p ) We use the above identity (4.81) and apply it in

f

p º· ¨ «x »¸ ©¬ p  q¼¹

E D

¦

f j 0

x

j

 

* j  E 1 * j  E D 1



(4.79) to write the

following steps:

d

D

ª¬ x E (1  x ) 1 º¼ dx

x

D

E D

f

*  j  E  1

j

¦ x *  j  E  D  1 j 0

x

E D

§ B x  E  D ,1  D * 1  E · ¨ ¸ 1D E D x * E  D ¹ © (1  x )

(4.82)

 *(1  E )  B x  E  D ,1  D  *( E  D ) (1  x)1D If

we D

0 Dx

apply

the

operator

D 0 Dx

ª¬ x /(1  x ) º¼ * (D  1)  (1  x ) D

on

 (D 1)

both

sides

of

the

above

obtained

and then letting the composition rule i.e.

D 0 Dx

expression



D

0 Dx

f ( x)

(4.78)



i.e.

f ( x ) , we

write the following:

x

D

1 x

* (D  1)

d

D

dx

D

ª¬(1  x ) D 1 º¼

(4.83)

D

0

The above obtained relationship (4.83) for D  1 follows from the fact that the function x /(1  x ) is bound at x D

and the condition D ! 1 , was imposed while deriving the fractional differ-integration of the function x /(1  x ) . Rearrangement and reversal of the sign of the D in (4.83) gives us the following

d

D

ª(1  x )D 1 º¼ D ¬ dx

x

D

(1  x )* (1  D )

(4.84)

with the condition D  1 . This above result (4.84) may be considered as a special case for the following

d

D

ª(1  x ) E º¼ D dx ¬

(1  x )

E D

B x  D , D  E * ( D ) which we have derived in an earlier section (Section-4.5) for the binomial function. We

will

B x (c,1  b  c )

use 1 b  c

(1  x )

the

x

c

*(c) * (1 b )

identity

¦

f j 0

x

j * ( j 1 b ) * ( j 1 c )

of

,

the

incomplete

beta-function

(4.85)

i.e.

which comes from the hyper-geometric function

Fractional Differ-integrations’ further Extension

D  1, c

(Appendix-A), in the above expression (4.85), by putting E

¦

f j 0

x

j

(1  x )

1

173

D , 1  b  c 1 , b

c

D , and using

to write the following: d

D

(1  x )

ª(1  x )D 1 º¼ D ¬ dx

1

* ( D )

B x  D ,1

1

§ x D (1  x )* ( D ) · f j ¨ ¸¦ x * ( D ) © * (1  D ) ¹j 0

(1  x )

(4.86)

1

§ x D (1  x )* ( D ) · 1 1  x ¨ ¸ * ( D ) © * (1  D ) ¹

(1  x )

x

D

(1  x )* (1  D ) That is what we obtained earlier (4.84) in this section too.

4.9 Fractional differ-integration of the hyperbolic and trigonometric function using the series expansion method Here, we follow the series expansion method. First, take f ( x ) 1

f ( x)

sinh

x2

 x

sinh

x

3 2

 x and write its series expansion as 5

f

x2

x

¦

j  12

(4.87)

   .... * (2) * (4) * (6) j 0 * (2 j  2) Using this series, we conduct term-by-term differ-integration and write the following: D ª 0 Dx sinh

¬

 ¼ x º

d

D

dx

D

sinh

 x 1

ª x j 2 º » ¦ D « j 0 dx « ¬ * (2 j  2) »¼ f

d

D

* j 

f

3 2



¦ *(2 j  2)*  j  j 0

We use the simplification



* j  32



S

* (2 j  2)

22 j 1 * ( j 1)

(4.88)

D

3 2

x

j  12 D

, which is a consequence of the duplication property of the Gamma

function (as discussed in Section-1.10.8). Using this, we write

d

D

dx

D

 x ¦ *(2 j  2)*  j  f

sinh

* j

j 0

S  12 D 2

x

3 2

3 2

D

x

j

 14 x ¦ 3 j 0 *  j  D  2 * ( j  1) f

1 D 1 S 2 x 2 I 1 D 2 2



where I 1 D 2

 x denotes the 

1 2

j  12 D



 x



 D -th order hyperbolic Bessel function of the argument

generalization of Rayleigh’s formula. The differ-integration of the f ( x )

sin

(4.89)

x . This comes from a

 x is expressed in a similar way, as

in

ª 0 Dx ¬sin D

where J 1 D 2

 x denotes the 

1 2



x º ¼



d

D

ªsin D dx ¬

 x º¼

1 D 2

  J  x

1 S 2 x 2

1 D 2

 D -th order Bessel function of the first kind of the argument

comes from a generalization of Rayleigh’s formula.

(4.90)

x . This also

174

Chapter Four

4.10 Fractional differ-integration of the Bessel function using the series expansion method We will first conduct differ-integration of the f ( x )

x

X 2



JX 2 x . The

X -th order Bessel function of the

argument 2 x is represented via the series as follows



JX 2 x

x

X 2

j

f

( 1) ( x )

j

¦ *( j  1)*( j  X  1)

(4.91)

j 0

Therefore, we have from (4.91) the following expression:

f ( x)

x

X 2

j

f

( 1) x

j X

 ¦ *( j  1)*( j  X  1)

JX 2 x

(4.92)

j 0

(4.92) represents a differ-integrable function with condition X ! 1 . Below, we conduct term-by-term differintegration and write the following expression

d

D

j

f

( 1) * ( j  X  1) x

j X D

 ¦ *( j  1)*( j  X  1)*( j  X  1  D )

ª xX 2 J X 2 x º D ¬ ¼ dx

j 0

j

f

( 1) x

j X D

¦ *( j  1)*( j  X  D  1)

(4.93)

j 0

x for f ( x )

X



 X 2D



JX D 2 x

 is modified by the Bessel function, and is represented as the series

x 2 IX 2 x , where IX 2 x



IX 2 x

x

X 2

f

( x)

j

¦ *( j  1)*( j  X  1)

(4.94)

j 0

The difference between IX and JX is that, in the modified Bessel function, the alternate sign change in infinite summation does not exist. Carrying out the steps of the term-by-term differ-integration, we obtain the following

d

D

j X D

f

 ¦ *( j  1)**(( jj XX 1)1)*x( j  X  1  D )

ª xX 2 IX 2 x º D ¬ ¼ dx

j 0

f

x

j X D

¦ *( j  1)*( j  X  D  1)

(4.95)

j 0

x Setting y

2 x , or x

2

y 4

; giving dx

1 2

 X 2D



IX D 2 x

 y dy , we get the following result d

D D

> ydy @

ª¬ yX JX ( y ) º¼

y

X D

JX D ( y )

(4.96)

4.11 Fractional differ-integration of the distribution functions using the Riemann-Liouville formula 4.11.1 Uniform distribution function and its fractional differ-integration An example of the uniform distribution function is the Heaviside unit step function. The Heaviside unit step function is described as

u ( x  x0 )

­0 x  x0 ®1 x ! x ¯ 0

(4.97)

The differ-integration of the above is piecewise defined as the function for D  0 , and with a  x0  x via the Riemann-Liouville formula it is

Fractional Differ-integrations’ further Extension

D

>

a Dx u ( x  x0 )

D d ª¬u ( x  x0 ) º¼

@

* ( D ) ³

D

>d( x  a)@ x0

1

(0)dy

1



a

0

D

1

D 0

D 1

( x  y) x

(1)dy

D * ( D ) ³ ( x  y )

1

x0

>@ 1

(4.98)

x ! x0

D

>d( x  x0 )@

( x  x0 )

 u ( y  x0 ) dy ,

a

D * ( D ) ³ ( x  y )

d

x

1

175

D

x ! x0 * (1  D ) Here the results of the earlier discussed differ-integration of the constant function of the unity, i.e. dD >1@ ( x  a ) D D D a Dx 1 * (1D ) (as in Section-4.2), are used. The above expression (4.98) applies for all D . One use of >d( x  a )@ the Heaviside unit step function is to delimit the range of the definition of the function so that we have the product:

>@

­ 0 x  x0 ® f ( x) x ! x ¯ 0

f ( x )u ( x  x0 )

(4.99)

By analogy of the above derivation, we can write: D

a Dx

> f ( x)u ( x  x0 )@ D >d( x  a)@ D d > f ( x )@ u ( x  x0 ) D , a  x0  x >d( x  x0 )@ d

> f ( x)u ( x  x0 )@

D

(4.100)

4.11.2 The delta distribution function and its fractional differ-integration The delta function (distribution) is also termed as the Dirac delta function, and is often used in mathematical physics to state or describe a point source (a monopole). The integer order derivative of a delta function returns two delta functions in opposite directions placed at the same point. This is a dipole placed at that point, so the delta function is a point source or a monopole that moves to a dipole placed at that point, after doing one whole differentiation. We have a property of the delta function as follows d

d

³  G ( x  c ) dx

1

b

³ G ( x  c)  f ( x) dx

f (c )

(4.101)

b

bdcdd We will use this property for a function f ( x ) defined in the interval >b, d @ while integrating that function in the interval after multiplying it by a delta function residing at x f (c ) .

c , returning the value of the function at that point,

We write the Riemann-Liouville formulation, in order to get a fractional derivative of the order D of the delta function, as follows: D

>

f Dx G ( x  c )

@

1

d

n

* ( n  D ) dx

n

x

³ G ( y  c ) ( x  y )

n D 1

(4.102)

dy

f

We have chosen a positive integer n just greater than the real number D , before conducting fractional integration of the order ( n  D ) followed up by whole differentiation of the integer order n . This is what is contained in the above formulation. For 0  D  1 , we have n 1 ; the above formulation becomes D

>G ( x  c)@

1

d

x D

(4.103)  G ( y  c ) ( x  y ) dy * (1  D ) dx ³ Now we use the property of the delta function (4.101) i.e. ³ G ( y  c )  ( x  y ) D dy ( x  c ) D , and write the f Dx

f



following steps



176

Chapter Four

D

f Dx

1

>G ( x  c)@

d

* (1  D ) dx

( x  c)

( D )( x  c )

D

D 1

(4.104)

* (1  D ) ( x  c)

D 1

* ( D ) We have used the property of the gamma function, which is * ( D  1)

write the semi-derivative of the delta function as follows with the value *   12 1

>

2 f Dx G ( x )

x

@

3

* 





and c

0 , we

2 S

1

2

1 2

1 2

D  * ( D ) . For D

x

2 S

 32

(4.105)

We may state that by conducting a semi-derivative of the delta function, we are getting a ‘fractional pole’ which is in between a monopole and a dipole. Now we try an approach similar to that which we used for the Heaviside unit step function in order to obtain the fractional differ-integral of the delta function, f ( x ) G ( x  x0 ) , by means of the property we reproduce again, that is x

³ G ( y  x )  f ( y ) dy 0

a

f ( x0 ) where a  x0  x , for any function f ( x ) . After selecting f ( x )

( x  y)

D 1

,

we write the following relationship: x

G ( y  x0 ) dy

³

( x  y)

a

Dividing the above by

1 * ( D )

( x  x0 )

D 1

D 1

D 0

(4.106)

, we recognize that the LHS of the above (4.106) is an D -th differ-integral of the Dirac

delta function, that is

d

The formula (4.107) can be extended for all D

>G ( x  x0 )@ D >d( x  a)@

D

x

1 * ( D )

³

G ( y  x0 ) dy

a

( x  x0 )

D 1

( x  y)

D 1

(4.107)

* ( D )

.

4.11.3 The relationship between uniform distribution and the delta distribution function Now comparing the earlier derived result, i.e.

dD >u ( x  x0 )@

( x  x0 ) D * (1D )

D

>d( x  a )@

dD >G ( x  x0 )@

(4.98), with

D

>d( x  a )@

( x  x0 ) D 1 * ( D )

(4.107), we

notice the following useful relationships

d

D 1

ª¬u ( x  x0 )º¼ D 1

>d( x  a )@ a Taking D

0

0 , we get the well-known relationship for a

d

D 1 D 1

>u ( x  x0 )@

dx 0 , that is

d

>G ( x  x0 )@ D >d( x  a )@

D

d

D

dx

D

(4.108)

>G ( x  x0 )@

d

>u ( x  x0 )@ G ( x  x0 ) dx The one whole derivative of the Heaviside unit step function is the Dirac delta function.

(4.109)

4.12 Fractional differ-integration of the saw-tooth function First, we take the general piece-wise function and find its D -th differ-integration. Consider the following function

f ( x)

­ f1 ( x ) a d x  x1 ® f ( x) x  x ¯ 2 1

(4.110)

Fractional Differ-integrations’ further Extension

177

With the use of the Heaviside unit step function, we write f ( x ) f1 ( x )u ( x1  x )  f 2 ( x )u ( x  x1 )

 f1 ( x)  f1 ( x)u ( x  x1 )  f2 ( x)u ( x  x1 ) f1 ( x )   f 2 ( x )  f1 ( x ) u ( x  x1 ) since u ( x  x1 )  u ( x1  x ) function (4.100) i.e.

(4.111)

1 for x z x1 . After applying the linearity and the result of the Heaviside unit step

dD > f ( x ) u ( x  x0 )@

u ( x  x0 )

D

>d( x  a )@



dD > f ( x )@

>d( x  x0 )@D

> f ( x)@ dD > f1 ( x)@ D D >d( x  a )@ >d( x  a )@ d



, we establish the following:

D

 u ( x  x1 )

d

D

> f2 ( x)  f1 ( x)@ D >d( x  x1 )@

(4.112)

x z x1

Using the above logic (4.112), we find several piece-wise sections and define a function f ( x) fk ( x) xk 1  x  xk k 1, 2, 3,..., N ; k 1, x0 { a

f1 ( x )  u ( x  x1 )  f 2 ( x )  f1 ( x )  .....

(4 .113)

 u ( x  x N 1 )  f N ( x )  f N 1 ( x )

x z xk

where as in (4.113), x0 { a . The general differ-integral result is the following for the above function:

> f ( x)@ dD > f1 ( x)@  N 1 u ( x  x ) §¨ dD > fk 1 ( x)  fk ( x)@ ·¸ ¦ k D D ¨ >d( x  x )@D ¸ >d( x  a)@ >d( x  a )@ k 1 k © ¹ d

D

(4.114)

We express the saw-tooth function as:

fSAW ( x )

k

( 1) (2 k  x  2) N 1

x

¦ (1)

k

(4.115)

(2 x  4 k  2)u ( x  2 k  1)

k 1

k The application of the formula (4.114), i.e. as a

1, 2, 3,..., N , (2 k 3) x(2 k 1) D

> f ( x )@ >d( x  a )@D d

dD > f1 ( x )@ D

>d( x  a )@



¦

N 1 k

u ( x  xk ) 1

dD > f k 1 ( x )  f k ( x )@

>d( x  xk )@D

, with a lower limit

0 , yields the following expression: d

D

> fSAW ( x)@ dx

D

> x@  N 1 u ( x  2k  1) §¨ dD ª¬(1)k (2 x  4k  2)º¼ ·¸ ¦ D D ¨ ¸ dx k 1 >d( x  2k  1)@ © ¹

d

D

1D

(4.116)

1D § · k ( x  2 k  1)  2¦ u ( x  2 k  1) ¨ ( 1) ¸ * (2  D ) * (2  D ) ¹ k 1 © N 1

x

4.13 Fractional differ-integration of the generalized periodic function The function f PER (t ) , which can be expressed as the sum of the complex conjugated exponential functions is _ _ f f described as f PER (t ) Ak exp i 2STkt  Ak exp i 2STkt , where Ak and Ak are complex-conjugate

¦

k 1



¦k 1





constants and T is the period of the function. Using the scaling property, that is differ-integration of the exponential function, i.e.

dD > exp( cx )@ D

>d( cx )@

 cx

D



*

dD > f ( cx )@ D

dx

c

D D d > f ( cx )@

>d( cx )@D

, and the



exp(cx ) J ( D , cx ) , both discussed earlier (in

Section-3.20), we write the differ-integration expression for the exponential as follows with c

r 2STik and x { t

178

Chapter Four

d

D

D D 2S kt § 2S k · d ª¬exp  ri T º¼ ¨ ri ¸ D © T ¹ ª¬d  r 2STikt º¼

ª¬exp  ri 2STkt º¼ dt

D

D

§ 2S ik · § 2S kt · ¨ ri ¸ ¨ ri ¸ T ¹ © T ¹ © t

D

exp  r i

D

2S kt T

exp  ri

2S kt T

J *  D , ri 2STkt

(4.117)

J *  D , ri 2STkt

Putting this (4.117) into the expression of the f PER (t ) , that is f

f PER (t )

§

¦ ¨© A

k

k 1

_ · exp  i 2STkt  Ak exp  i 2STkt ¸

(4.118)

¹

gives us a complete description of the fractional differ-integration of a periodic function which is

d

D

> fPER (t )@ dt

f

¦ A t D exp i 

k

D

2S kt T

J *  D , i 2STkt

k 1 f



¦A t

D

k

exp  i

(4.119) 2S kt T

*

J  D , i

2S kt T



k 1

For small values of x , and by using the series expansion of Tricomi’s incomplete gamma function, that is *

lim xp0 J ( D , x )

exp(  x )

¦

f

xj j 0 * ( j D 1)

, we write the following steps: d

D

>exp(r x)@ dx

x

D

D r x *

e J ( D , r x ) x

D r x

x

D

e

j § Bx f · rx ¨e ¦ ¸ j 0 * ( j  D  1) ¹ ©

f

rx

(4.119)

j

¦ *( j  D  1) j 0

d »exp  dt D D

Therefore, for small t , we write lim t p0

r i 2 ST kt

¼º

t

D

¦

f

 r i 2STkt

j

j 0 * ( j D 1)

, and elaborating upon this we have the

following form:

lim t p0

d

D

dt

D

ª¬exp  ri

2S kt T

º¼ t

D

t

D

§ f  ri 2S kt j · T ¨¦ ¸ ¨ j 0 * ( j  D  1) ¸ © ¹

2 § 1 · ri 2STkt  ri 2STkt  ¨    ... ¸ ¨ * (1  D ) * (2  D ) * (3  D ) ¸ © ¹

1

t

D

ri 2TS k 1D  ri 2TS k  t  

(4.120)

2

t

2 D

 ... * (1  D ) * (2  D ) * (3  D ) Therefore, in the limit of the small t , i.e. t p 0 , the fractional differ-integration of the periodic function is given in the following steps

Fractional Differ-integrations’ further Extension

lim t p0

d

> fPER (t )@

D

dt

f

D

D

k



¦A t

D

k

k 1

* (1D )

f

¦

 Ak  Ak

k 1

Notice that the coefficients

 Ak  Ak

* (1D )

t

D

f



k 1

 i 2STkt

D

f

A A

¦  i 2TS k *(2k Dk) t1D  ¦  i 2TS k

2

f

i A  A ¦  2TS k *(2D ) t1D  ¦  2TS k k

k

k 1

k 1

2

 Ak  Ak * (3D )

k 1

f



j 0

 Ak  Ak * (3D )

t

t

dD dtD

¦

ª¬ f PER (t ) t 0 º¼

2 D

2 D

(4.121)

 ...

 ...

and i  Ak  Ak in (4.121) are real numbers. The leading term of the above

expression (4.121) is the differ-integral of the initial value of the periodic function f PER (t ) t that is, we say

j

2 § 1 · i 2STkt  i 2STkt  ¨    ... ¸ ¨ * (1  D ) * (2  D ) * (3  D ) ¸ © ¹

k 1

t

f

2 § 1 · i 2STkt  i 2STkt     ... ¸ ¨ ¨ * (1  D ) * (2  D ) * (3  D ) ¸ © ¹

f

 Ak  Ak

D

k

j 0

k 1

k 1

f

k

k 1

¦A t

¦

j

¦ A t ¦ *( j  D  1)  ¦ A t ¦ *( j  D  1)

D

f

f

i 2STkt

f

179

f k 1

t

D



Ak  Ak * (1D )

f

0

¦ A k 1

k

 Ak ,

.

The asymptotic formulation for large x for Tricomi’s incomplete gamma function gives the expression *

lim xnf J ( D , x )

D

x 

e x * ( D )



1 x



lim t nf

( D 1) x2

d

D



( D 1)( D  2) x3

ª¬exp  ri 2STkt º¼ dt

D



 ... , which we use in the following derivation:



lim t nf t

D

e

ri 2 STkt

J

*

 D , ri 2STkt

§  ri 2S kt D · T ¨ ¸ D r i 2 S kt t e T ¨ e B i 2 STkt §  r i 2S kt 1  ( D  1)  r i 2S kt 2  ... · ¸ T T ¨ ¨ ¸¸ ¨ * ( D ) ¨  ( D  1)( D  2)  ri 2S kt 3  ... ¸ ¸ T © ¹¹ © §  ri 2S kt D · T ¨ ¸ ( D 1) ¸ D r i 2 ST kt ¨ § · 2S kt 1 2S kt 2 1 t e ¨ e B i 2STkt ¨ * ( D )  ri T  * ( D )  ri T  ... ¸ ¸ ¨ 3 ¨ ( D 1)( D 2) ¸¸ ¨ ¸¸ ri 2STkt  ... ¨  D ( ) *  © ¹¹ © § §  ri 2S kt 1 t D  ri 2S kt 2 t D · · T T ¨¨  ¸ ¸ 2 kt S D ( ) ( *  * D  1) ¸ D ¨¨ ¸ ri T 2S k ¨  ... ¸  ri T e 3  ¨ ¸ D  2S kt ¨ ¨   ri T t ¸ ¸ ¸ ¸ ¨¨ * ( D  2) ¹ ¹ ©© D

 ri 2TS k

exp  ri

f

2S kt T

j 0

In the above (4.122), we have used ( n  1)  * ( n  1) complex number exp  ri SD 2 to yield the following:

1 j

 ¦ t D §¨  * ( D  j )

* ( n ) with n

r i 2 ST kt

©

· ¸ ¹

(4.122)

D

D . Now, we use the fact that ( ri )

is a

180

Chapter Four

d

D

ª e 2STikt º D ¼» dt ¬«

2S ik D T





D

e

iSD

 2TS k

e2e

D

 2TS k

e

D

 2TS k

e

 2STikt §

Tt

¨

·

T t



2 2 2 2

© 2S ikt * ( D ) 4S i k t * ( D  1)

 2STikt §

Tt

¨

2 D

D

 2S i ktT  D4

i i

Tt

¹

2 D

D

2S kt * ( D ) Tt

¹

 ... ¸

2 2 2

© 2S ikt * ( D ) 4S k t * ( D  1)

 2S i ktT  D4

 ... ¸

·

T t





T t 2 2 2

4S k t * ( D  1)

1D

(4 .123)

 ...

2 2 D



2S k * ( D )

3 3D

i

2 D

D

T t 2 2

4S k * ( D  1)

4 4 D

T t 3 3

8S k * ( D  2)



T t 4 4

16S k * ( D  3)

...

Similarly, we write the following steps: d

D

ª e  2STikt º D « ¼» dt ¬



 2STik

D

 2TS k

D

 2TS k Using

f PER (t )

the

¦

f k 1

e

e

above

two



¦k 1

Ak exp i 2STkt 

f

 iSD 2

D

e

e

  2STikt §

Tt

¨

2 D

D

·

T t



2 2 2 2

© ( 2S ikt )*( D ) 4S i k t *( D  1)

  2STikt §

Tt

¨

2 D

D

Tt

i





¹

2 D

D

T t



2 2 2

2S kt * ( D ) 4S k t * ( D  1) expressions ((4.123) and

_ Ak exp i 2STkt

(4.124)

 ... ¸

2 2 2

© ( 2S ikt )*( D ) 4S k t *( D  1)

 2S i ktT  D4

¹

·

T t



 ... ¸

 ...

(4.124)),

and

putting

into the definition of the periodic function for large t , we

write the following: d

D

f

D

> fPER (t )@ ¦  2TS k t nf dt D

lim

k 1

¦ t f



k 1



Ak e

1D

 2S i ktT  D4



i  Ak  Ak T 2 S k * ( D )

 Ak e



t

 2S i ktT  D4

2 D





 Ak  Ak T 2 4S 2 k 2 * ( D 1)

(4.125)

 ...

In the above expression (4.125), the terms grouped inside the first summation are periodic. They show that the effect of the differ-integration of the order D is to change the amplitude of each component by phase by an angle D

 S2 .

D

 2TS k

and to change the

Within the second summation, the terms are aperiodic. Provided D ! 1 , as the condition

for standard fractional differentiability, the second grouped summation terms of (4.125) represent transients, which die down, and become insignificant at certain (large) times. Let f PER (t )

cos(t )

¦

f k 1

Ak e

 i 2STkt



¦

f k 1

_   i 2S kt Ak e T

1 2

it

e  12 e

 it

. Note that the generalized periodic function

cos(t ) has only a term for k 1 , with all the others being zero. With 2TS 1 and A1 and then applying the fractional derivative formula as obtained above (4.125), we write the following:

here when f PER (t )

A1

1 2

,

Fractional Differ-integrations’ further Extension

lim t nf

d

D

dt

f

D  S ¦ S  A e

>cos(t )@

D

kt  D T 4

2 i

2 k T

k



 Ak e

181

 2S i ktT  D4

k 1







§ t 1D i Ak  Ak T · 2 S k * ( D ) ¨ ¸ ¨ ¸ 2  Ak  Ak T 2 D ¨ ¸ t  f 4S 2 k 2 * ( D 1) ¨ ¸ ¦ i  Ak  Ak T 3 ¨ ¸ 3 D   k 1 t ¨ ¸ 8S 3 k 3* ( D  2) ¨ ¸ 4 ¨ t 4D  A4k 4 Ak T ¸ ... 16S k * ( D 3) © ¹



  



1 2

e

i  t  DS 2

1 2

 e

cos  t  DS 2 

 i  t  DS 2

t





D

dx d

cos( x )@ D >

cos  x  DS 2 

D

2 D

* ( D  1)

2 D

* ( D  1) One can use this argument and can write the following for large x : d

t



x

t



* ( D  3) 2 D

1D

t

4 D

* ( D  3)

 ...

4 D

* ( D  1) x

(4.126)

 x

 ...

x

4 D

* ( D  3)

 ...

3D

(4.127)

sin( x )@ sin  x  DS   ... 2  D > dx * ( D ) * ( D  2) This above described method (4.126) is one of the procedures used to obtain the fractional differ-integrals of the periodic function. When the fractional order D is the positive or negative integer D r1, r2,... , the reciprocal of the gamma function at a negative integer point is zero, the transient term vanishes, and we get well-known classical formulas of the derivatives and integrals of the sine and cosine functions. However, other methods are available, as fractional derivatives of the trigonometric functions may be represented by higher transcendental functions or transcendental trigonometric functions in terms of the Mittag-Leffler or Miller-Ross functions, among others (some of which were discussed in Chapters 2 and 3). For fractional integration of the cosine and sine functions, we reverse the sign of D in the formula obtained for fractional differentiation, and write the following set of expressions, with 2S T

Z and for large t . 1D

D

I

D

D

I

D

D

D

 sin(Zt )  sin(Zt )  cos(Zt )  cos(Zt )

3D

Z t Z t Z sin Zt  2   3  .... Z  * ( D ) Z  * ( D  2) 1D 3D Zt Zt   D SD Z sin Zt  2   3  .... Z  * (D ) Z  * (D  2) 2 D 4 D Z t Z t D Z cos Zt  SD2  2  4  ... Z  * ( D  1) Z  * ( D  3) 2 D 4 D Z t Z t D  4  ... Z cos Zt  SD2  2 Z  * (D  1) Z  * (D  3) D

SD

(4.128)

4.14 The Eigen-functions for the Riemann-Liouville and Caputo fractional derivative operators This is a very important section. We use f ( x ) 1

Ox

e

Ox

as the eigen-function for the classical derivative operator as

O f ( x ) for f ( x ) e . Similarly, we will also deal with the Mittag-Leffler function (described in Appendix-A), which is a higher transcendental function (similar to the exponential function), and derive the eigenfunctions for the Caputo and Riemann-Liouville fractional derivative operators. Dx f ( x )

182

Chapter Four

4.14.1 Caputo derivative of the order D for the ‘one-parameter Mittag-Leffler function’: f ( x ) ED (O xD ) D ED (O x ) (the one parameter Mittag-Leffler function) as follows:

Let us find the Caputo derivative of f ( x )

D

f

D

ED (O x )

¦O

ED ,1 (O x ) 1

Noting that

C D 0 Dx

>1@

0 and

C D 0 Dx

C D 0 Dx

* ( nD 1) * ( nD 1D )

ª¬ x nD º¼

C D 0 Dt

ª¬ ED (O xD )º¼

x

* (1) * (1  D )

§ ©

* (D  1)



O x

* (2D  1)

 ....

, we write the following

2 2D 3 3D D ª§ ·º Ox O x O x 1     ... ¸ » «¨ ¬© *(1  D ) *(1  2D ) * (1  3D ) ¹ ¼

* (1  D )

O ¨1 

(4.129)

2 2D

D

2 2D 3 3D ª O xD º O x O x « *(1  D )  *(1  2D )  *(1  3D )  ...» ¬ ¼

C D 0 Dt

O

nD D

kH

* ( kD  1)

k 0

Ox

x

k

Ox

O

* (1  2D ) x

2

* (1  2D )* (1  D ) 2 2D

D

* (1  D )



D

O

3 3D

O x

* (1  2D )



* (1  3D ) x

3

2D

* (1  3D )* (1  2D )

 ...

(4.130)

·

O x

* (1  3D )

 ... ¸

¹

D

O ED (O x ) D ED (O x ) as an eigen-function.

Therefore, for the Caputo derivative operator we have f ( x )

4.14.2 Riemann-Liouville derivative of the order D for the ‘one-parameter Mittag-Leffler function’: f ( x ) ED (O xD ) We conduct the Riemann-Liouville fractional derivative on

noting that the RL derivative of the constant is non-zero, i.e. D

0 Dx

>1@

D ED (O x ) as depicted in the following steps,

f ( x) D

0 Dx

>1@ z 0 ,

but, as discussed earlier, is rather

D

x * (1D )

D D ª º 0 Dx ¬ ED ( O x ) ¼

x

D

ª§

0 Dx Ǭ 1 

©

D

O

* (1  D )

Ox

2 2D

D

O x



3 3D



O x

* (1  D ) * (1  2D ) * (1  3D )

* (1  D ) * (1)* (1  D )

O

* (1  2D ) x

2

·º

 ................. ¸»

¹¼

D

* (1  2D )* (1  D )

O

3

* (1  3D ) x

2D

* (1  3D )* (1  2D )

2 2D 3 3D D § · O x Ox O x   .............. ¸  O ¨1   * (1  D ) © *(1  D ) *(1  2D ) *(1  3D ) ¹

x

D

x

D

 .. (4.131)

D

* (1  D ) x

 O ED (O x )

D

C

D

D

 0 Dx ª¬ ED (O x ) º¼ * (1  D ) Thus, we note that in the RL case RL operator, i.e. for C D 0 Dx

> f ( x )@

D

0 Dx

D

0 Dx

D

0 Dx

ª¬ ED (O xD ) º¼ z ED (O xD ) so

f ( x) z O f ( x) .

f ( x)

D ED (O x ) is not an eigen-function for the

In the above steps of the derivation (4.131), we also observe that

> f ( x)  f (0)@ ; for 0  D  1 ;

and here f (0)

1 and DDx f (0)

xD * (1D )

.

Fractional Differ-integrations’ further Extension

183

4.14.3 Riemann-Liouville derivative of the order D for a function of the ‘two-parameter Mittag-Leffler function’: f ( x) xD 1 ED ,D (O xD ) D 1

Let us take now f ( x )

x

function, i.e. ED , E (O xD )

¦

D

D

ED ,D (O x ) , where the function ED , E (O x ) is the two-parameter Mittag-Leffler

( O xD ) k * k 0 ( E  kD )

f

. With E

D ! 0 , we write the following series:

f ( x)

x

D 1

D

ED ,D (O x )

x

D 1

x

D 2 2D § 1 · Ox O x ¨ * (D )  * (2D )  * (3D )  .... ¸ © ¹

D 1

* (D )

We note that f (0)



Ox

2D 1

* (2D )

(4.132)

2 3D 1

O x



* (3D )

 ....

f for this case. Now we explore the term-by-term fractional derivative of the RL type: D

0 Dx

> f ( x )@

D

0 Dx

D

0 Dx

ª¬ xD 1 ED ,D (O xD ) º¼

§ xD 1 O x 2D 1 O 2 x 3D 1 ·    .... ¸ ¨ © * (D ) * (2D ) * (3D ) ¹

* (D ) x

1

* (D  D )* (D ) x

1

* (0)

O

§ xD 1

O¨

© * (D )

In the above derivation (4.133)

1 * (0)

x

D 1

* (D ) 

Ox

* (2D ) x

O O

D 1

* (2D  D )* (2D ) 2

* (3D ) x

2D 1

* (3D  D )* (3D )

(4.133)

 ...

 ....

2 3D 1



2

2D 1

* (2D )

2D 1

* (2D )

x

O

O x

* (3D )

·

 ... ¸

¹



O x

D 1

D

ED ,D (O x )



0 is used. Thus, we obtained the eigen-function as f ( x )

x

D 1

E

D ,D

D

(O x )



for the Riemann-Liouville fractional derivative operator.

4.15 The fractional derivative of the zero-corrected function and relationship to Caputo derivative Consider the function f ( x )

x 

non-zero value, which is f (0)

1 2

1 2



in the interval 0 to x . At the starting point ,i.e. x

0 , the function has a

. If we construct a function that is offset by this starting point value (calling that

new function

f0 ( x )

value (say x

a ), where the function has a value f ( a ) ; and in that case the zero-corrected function will be

f0 ( x )

f ( x )  f (0) ), then at the starting point

f0 (0)

0 . The starting point may be any general

f ( x )  f ( a ) . This is what we mean by the zero-corrected function. In addition, the RL fractional derivative

of f0 ( x ) when function f ( x ) is a constant will be zero too. Regarding some functions where the starting point value is undefined (say rf ), then we take for our calculations 

f ( a ) , making f0 ( x ) f ( x )  f ( a  ) finite. These are practical considerations for calculations to take the finite part of the zero corrected function.

4.15.1 The Riemann-Liouville fractional derivative of the zero-corrected function f ( x) a  x with a derivative start point at x 0 In the interval 0 to x , where the function is f ( x )

x 

with 0  D  1 to the offset function, that is f0 ( x ) D

0 Dx

> f0 ( x)@ for the zero corrected function

1 2

, we apply an RL fractional derivative of the order D

f ( x )  f (0)

f ( x )  12 . To obtain a fractional derivative

f0 ( x ) , we must follow the below steps:

184

Chapter Four

D

0 Dx

1

> f 0 ( x )@

x

d

D ( x  y )  f ( y ) dy ³ * (1  D ) dx 

0

0

x 

f0 ( x ) D

0 Dx

 12

1 2

0 D 1

1

x

d

D ( x  y )    y   dy * (1  D ) dx ³

> f 0 ( x )@



1 2

1 2

0xd

1 2

(4 .134)

0

1

x

d

D ( x  y )   y dy ³ * (1  D ) dx 

0

Now we use

³

f1 f 2 dx

f1

³

f 2 dx 

³  f  ³ f dx dx , for (1) 1

³

I

2

x

0

( x  y)

D

  y dy

in (4.134) and write the

following steps: x

x

D 1 ª º § · ( x  y) D I ³ ( x  y )   y dy « (  y ) ( 1)  ³ ¨ ³ ( x  y ) dy ¸ ( 1)dy » D  1 0 © ¹ ¬« ¼» 0 D

x

x

ª y ( x  y )1D ( x  y )1D º ª y ( x  y )1D ( x  y ) 2D º  (  1)d  y « » « » ³ 1D (1  D )(2  D ) ¼ 0 ¬ 1D ¼0 ¬ 1  D 

x

(4.135)

(1  D )(2  D )

0 Dx > f 0 ( x ) @

D

2 D

1D º 1 x  u  D (2 ) » *(1  D ) (1  D )(2  D ) * (1  D ) dx ¬ (1  D )(2  D ) ¼

1

d ª

x

«

2 D

1D



1D

x

(1  D )* (1  D )



x

* (2  D )

Therefore, we write from (4.135) D

0 Dx ª ¬  x 

D

With the fractional derivative of the constant as 0 Dx

> 12 @

1 2 * (1D )

D

0 Dx

x

 12 º¼

1D



x

* (2  D )

(4.136)

, we write 1D

x

x

D

 ª¬  x  12 º¼  * (2  D ) 2  * (1  D )

4.15.2 The Caputo fractional derivative function f ( x)

(4.137)

a  x with a derivative start point at x

Now we derive the Caputo derivative for the original function f ( x ) described in the following steps.

D

1 2

x 

1 2

0

, i.e. without taking zero correction, as

Fractional Differ-integrations’ further Extension

C D 0 Dx

185

§ d f ( y) · ¨ dy ¸ d y ; 0  D  1 © ¹ x · 1 D § d ( x  y ) ¨  ( y  12 ¸ dy ³ * (1  D ) 0 © dy ¹ 1

* (1  D ) ³

f ( x)

x

0

( x  y)

1

x

* (1  D ) ³

0

D

( x  y)

D

 1 dy (4.138)

x § 1 · D ¨ ( x  y ) (1)dy ¸ ³ 0 © * (1  D ) ¹





(1D ) 0 Ix

>1@

1

* (1)



1D

* (1  1  D )

x

1D

x * (2  D ) The result is the same as it was for the RL fractional derivative of the zero-corrected function (4.136), i.e. 

C D x0 Dx

> f ( x )@

D

x0

Dx

> f ( x)  f ( x0 )@ .

This is the relationship between the RL and Caputo derivatives for 0  D  1 . 1 * (1D )

In the above derivation (4.138), we have used order 1  D , for a constant function f ( x )

³

x

0

( x  y)

D

(1)dy as the RL fractional integration formula of the

1.

4.15.3 The Riemann-Liouville fractional derivative of the zero-corrected function f (t ) t v from the start point as t z 0 J

J

J J t  a . Now we

t , with the starting point t a , f ( a ) a , thus f0 (t ) f (t )  f ( a ) apply the definition of the RL fractional derivative as indicated in the following steps. Let us now take f (t )

D

a Dt

> f 0 (t )@

f

(D )

1

t

ª¬t J  aJ º¼ a d

t

J J D (t  [ ) [  a d[ ; ³ * (1  D ) dt 

0 D 1

a

1

d

* (1  D ) dt

t

³

(t  [ )

D

a

 [

J

d[ 

1

d

* (1  D ) dt

t

J D ³ ( t  [ )  a d[ 

(4.139)

a

 (1D ) 1 ª¬t J º¼  a Dt (1D ) ª¬aJ º¼ º D ª a Dt

¬

¼

D

a Dt

ª¬t J º¼  a DtD ª¬aJ º¼

The above expression (4.139) shows that this derivative of the zero-corrected function is composed of the two RL derivatives, which are

D

a Dt

ª¬t J º¼ minus the RL derivative of a constant a DtD ª¬a J º¼ . Going by a similar method to

that used for 0 DtD ª¬t J º¼ , we get first the fractional integral in terms of the Beta function as  (1D ) a Dt

J 1D

(t a ) / t

* (1  D )

³

t

J

ª¬t º¼ t

J

J

(1  z ) dz

0

BK (1  D , J  1)

The fractional derivative of the order D with 0  D  1 of the function t the above expression. As such, we get the following:

The RL-fractional derivative of the a (constant) is:

D

(4.140)

J 1D

* (1  D )

D J a Dt ª ¬t º¼

z

J 1D d ª t

J

K

ta t

is found by taking one whole derivative of

º BK (1  D , J  1) » « dt ¬ * (1  D ) ¼

K

ta t

(4.141)

186

Chapter Four

a

D

ª Jº a Dt ¬ a ¼

J

(t  a )

* (1  D )

D

(4.142)

We write the following further steps: D

a Dt

> f 0 (t )@

(D )

f

1

t

d

t

D J J (t  [ ) [  a d[ ³ * (1  D ) dt

ª¬t J  aJ º¼ a



a

 (1D ) 1 ª¬t J º¼  a Dt (1D ) ª¬aJ º¼ º¼ ; 0D 1 D ª a Dt ¬

(4.143)

D D ªJº ª Jº a Dt ¬t ¼  a Dt ¬ a ¼ J 1D d ª t

«

dt ¬ * (1  D )

J

º

a

¼

* (1  D )

BK (1  D , J  1) » 

(t  a )

D

Therefore, we have: J 1D d ª t

D ªJº a Dt ¬t ¼

For a

º a J ( t  a ) D

BK (1  D , J  1) » 

«

dt ¬ * (1  D )

(4.144)

* (1  D )

¼

0 , we have: D

0 Dt

1

> f 0 (t ) @

t

d

* (1  D ) dt

³ (t  [ )

D

0

ª¬[ J  0º¼ d[ (4.145)

 (1D ) J 1 D ª¬ 0 Dt t º¼

* (J  1)

D J

0 Dt

t

* (J  1  D )

t

J D

4.15.4 The Riemann-Liouville fractional derivative of the order D for the zero-corrected ‘one-parameter Mittag-Leffler function’: f (t ) ED ( at D ) with a derivative start point at t 0 The one parameter Mittag-Leffler ED ( at D ) is defined by an infinite series as: def

D

ED ( at ) By taking f (t )

* (1  D )

* (1  2D )



a t

t

nD

* ( nD 1) D ( n 1) *[D ( n 1) 1] t

D

, is

D

0 Dt

, for n

ª§

at

2 2D

D



3 3D

a t



at

0.

D , with 0 d D  1 and

1, 2, 3, ... ; we will use this

* (1  D ) * (1)* (1  D )

§

a ¨1 

©

at

a

* (1  2D ) * (1  2D )* (1  D ) 2 2D

D

* (1  D )



a t

* (1  2D ) D

aED ( at ) ; we write the following:

2 D

a t 

3 3D



at

* (1  3D )

·

º

¹

¼

 ................. ¸  1»

¬© *(1  D ) *(1  2D ) * (1  3D ) 2 2D 3 3D D º a t at D ª at    .................» 0 Dt « ¬ *(1  D ) *(1  2D ) * (1  3D ) ¼

t D * (1D )

(4.146)

ª¬ ED ( at D )  1º¼ 0 Dt «¨ 1 

Using 0 DtD >1@

 ................

* (1  3D )

ED ( at )  1 . The fractional derivative of the order

0 for f (t )

D 0 Dt



3 3D

a t

D

ª¬t nD º¼ expression in the following derivation for 0 DtD ª¬ ED ( at D )  1º¼

with its starting point as a

2 2D

D

1 . We note that for ED ( at D ) , the first derivative does not exist at t

D ED ( at ) , we have f (0)

The zero-corrected function is f0 (t )

1

at

* (1  3D ) * (1  3D )* (1  2D )

·

 .............. ¸

¹

(4.147) 3 2D

at

 ..

Fractional Differ-integrations’ further Extension

187

1

D

D ª¬ ED ( at D ) º¼ aED ( at D )  t * (1  D )

0 Dt

(4.148)

4.15.5 The Caputo fractional derivative of the order D for the ‘one-parameter Mittag-Leffler function’: ED ( at D ) with a derivative start point at t 0 D

Now we explore the Caputo derivative of the function f (t ) C D 0 Dt

ED ( at ) , depicted in the following steps

ª¬ ED ( atD )º¼ 2 2D 3 3D D ·º at a t a t C D ª§    ... ¸» ; 0 Dt «¨ 1  ¹¼ ¬© * (1  D ) * (1  2D ) * (1  3D ) C D 0 Dt

>1@

0

2 2D 3 3D ª at D º a t a t   « * (1  D ) * (1  2D ) * (1  3D )  .................» ¬ ¼

* (1  D ) * (1)* (1  D ) 

* (1  2D )

a

3 2D

* (1  3D )* (1  2D ) at

2 D

* (1  2D )* (1  D )

* (1  3D )

§

a ¨1 

©

C D 0 Dt

a t

* (1  D )



(4. 149)

 ..........

2 2D

D

a t

3 3D

a t

* (1  2D )

·

a t



 .............. ¸

* (1  3D )

¹

D

aED ( at ) The Caputo fractional derivative of the constant is zero, and, along with

C D 0 Dt

* ( nD 1) D ( n 1) *[D ( n 1) 1] t

ª¬t nD º¼

is used in the

above steps. Therefore, we get the following expression from (4.147) and (4.149): C D 0 Dt

ª¬ ED ( at D )º¼

D

ª¬ ED ( at D )  1º¼ aED ( at D )

0 Dt

(4.150)

4.15.6 The Riemann-Liouville fractional derivative of the order D of the zero-corrected function f (t ) cosD (t D ) and f (t ) sinD (t D ) with a derivative start point at t 0 We defined the complex Mittag-Leffler function in the following form (see Appendix-A for more detail): D

def

ED (it )

D

D

cosD (t )  i sinD (t ) D

D

ED (it )  ED ( it )

D

cosD (t )

2 D

¦

D

2

( 1)

k 1

2 kD

f

¦

(2kD )!

f

¦

t

k

( 1)

k 1

ED (it )  ED ( it )

D

sinD (t )

f

t

k

( 1) f

(2kD  D )!

¦

2 kD

(4.151)

* (2kD  1)

k 1

(2 k 1)D

t

k

( 1)

t

k

k 1

(2 k 1)D

* (2kD  D  1)

D

The series presentation of cosD (t ) is: D

cosD (t ) Take f (t ) find

D

0 Dt

function):

D cosD (t ) , where f (0)

> f 0 (t ) @

D

0 Dt

1

t

2D

* (1  2D )



t

4D

* (1  4D )



t

6D

* (1  6D )

1 , and we have the zero-corrected function f0 (t )

 .....

(4.152)

D cosD (t )  1 , and we will

ª¬cosD (t D )  1º¼ , for 0  D  1 , in the following steps (as we did for the Mittag-Leffler

188

Chapter Four D

0 Dt

ª¬cosD (t D )  1º¼ D

ª§

0 Dt Ǭ 1 

©

t

D

t



* (1  2D )

0 Dt



2D

4D

* (1  4D )

2D

t



6D

* (1  6D )

4D

·

º

¹

¼

 ..... ¸  1»

6D

ª º t t t « * (1  2D )  * (1  4D )  * (1  6D )  ....» ¬ ¼ * (1  2D )

* (1  2D )* (1  D )

t

2D D

* (1  6D )



* (1  4D )



* (1  4D )* (1  3D ) t

* (1  6D )* (1  5D )

6D D

t

(4.153)

4D D

 .....

3D § tD · t D ¨   ....... ¸  sinD (t ) © * (1  D ) * (1  3D ) ¹

Thus, we have: D

ª¬cosD (t D )  1º¼  sinD (t D )

0 Dt

With the Caputo derivative, we can show

C D 0 Dt

(4.154)

ª¬cosD (t D ) º¼  sinD (t D ) . From here we write: D

0 Dt

1

D ª¬cosD (t D ) º¼  sinD (t D )  t * (1  D )

(4.155)

The series representation of sinD (t D ) is the following:

t

D

sin D (t )

D



* (1  D ) Taking a term-by-term fractional derivative, we get: D D 0 Dt ª ¬sinD (t )º¼

D

ª tD

0 Dt «

t

* (1  3D )

t





t

5D

* (1  5D )

3D

t





5D



t

7D

t

* (1  D )* (1  D  D ) 

D D

* (1  5D ) *(1  5D )* (1  5D  D ) t

2D



t

4D



t



t

7D

* (1  3D ) * (1  3D )* (1  3D  D )

5D D

t



 .....

* (1  7D )

(4.156)

º

¬ *(1  D ) *(1  3D ) *(1  5D ) *(1  7D ) * (1  D )

1

3D

t

 .....»

¼

3D D

* (1  7D ) * (1  7D )* (1  7D  D )

t

7D D

 .....

(4 .157)

6D

*(1  2D ) * (1  4D ) * (1  6D )

 .....

D

cosD (t ) Therefore, we have, using f (t )

D sinD (t ) with f (0)

0 , the following useful identity: 0 Dt

D

ª¬sinD (t D ) º¼ cosD (t D )

(4.158)

C D 0 Dt

ª¬sinD (t D ) º¼ cosD (t D )

(4.159)

In addition, we have in a similar way, the following:

4.15.7 The Riemann-Liouville fractional derivative of the order E of the zero-corrected ‘oneparameter Mittag-Leffler function’ f ( x) ED ( xD ) with a derivative start point at x 0 and D z E Let f ( x )

D ED ( x ) , which has f (0)

1 . Now we evaluate D

E

for the function f0 ( x )

D ED ( x )  1 , and using the

fractional derivative of the order D , with 0 d D  1 and with the start point as a D

0 Dx

kD

ª¬ x º¼

* ( kD 1) * D ( k 1) 1

x

D ( k 1)

, for k

1, 2, 3, ... , we write the following:

0 for f ( x )

x

kD

,

Fractional Differ-integrations’ further Extension

E

0 Dx

ª¬ ED ( xD )  1º¼

E

0 Dx

E

0 Dx

x

D E

x x

x

2D 3D D ª§ · º x x x    .... ¸  1» «¨1  ¹ ¼ ¬© * (1  D ) * (1  2D ) * (1  3D ) 2D 3D ª xD º x x   « * (1  D ) * (1  2D ) * (1  3D )  ....» ¬ ¼

D E



* (1  D  E )

D E

189

x

2D  E

* (1  2D  E )

x



3D  E

* (1  3D  E )

 ...

(4.160)

2D

D

§ · 1 x x    ... ¸ ¨ © * (1  D  E ) * (1  2D  E ) * (1  3D  E ) ¹ D k

f

(x )

¦ *(1  D  E  kD ) k 0

x

D E

ED ,D  E 1 ( x

D

)

Where the ‘two-parameter Mittag-Leffler function ‘is placed in the above derivation (4.160), we get the following (see Appendix-A for more details) D k

f

D

ED , P ( x )

(x )

¦ * ( P  kD )

*( P )

k 0

where P

D 2

D

1

( x)



* (D  P )



D 3

(x )

* (2D  P )



(x )

* (3D  P )

x E * (1 E )

D  E  1 in our derivation as above (4.160). Noting that 0 DxE >1@

 ...

(4.161)

, we write the following useful

expression: E

0 Dx

x

E

ª¬ ED ( xD )º¼ xD  E ED ,D  E 1 ( xD )  * (1  E )

(4.162)

In addition, we may write the Caputo derivative as: C E 0 Dx

ª¬ ED ( xD ) º¼

x

D E

D

(4.163)

ED ,D  E 1 ( x )

4.15.8 The Riemann-Liouville and Caputo fractional derivatives of the order D for the function f ( x ) x E 1 cosD , E ( xD ) and f ( x ) x E 1 sinD , E ( xD ) with a derivative start point at x 0 Here we have, as above (4.161), a two-parameter Mittag-Leffler function. Similar to the earlier cosine and sine D

D

functions ( cosD ( xD ) and sinD ( xD ) ), we define the two parameters as cosD , E ( x ) and sinD , E ( x ) , as depicted below: D

cosD ,D ( x )

def f

¦

( 1)

sinD , E ( x )

def f

2 kD

1

* ( E  2D k )

k 0

D

x

k

x

k



*( E )

(2 k 1)D

¦ (1) *  E  (1  2k )D k 0

x

2D

* (2D  E ) x



D

* (D  E )



x

4D

* (4D  E ) x

 ....

3D

* (3D  E )



x

(4.164)

5D

* (5D  E )

 ...

Now placing this (4.164) and the definition of the two-parameter Mittag-Leffler function (4.161) in an imaginary argument, we get the following useful identity: D

D k

f

ED , E (ix )

(ix )

¦ * ( E  kD ) k 0

1 *( E )

D 2

D



(ix )

* (D  E )



(ix )

* (2D  E )

D 3



(ix )

* (3D  E )

2D 4D § 1 · x x ¨ * ( E )  * (2D  E )  * (4D  E )  ... ¸ © ¹ 3D 5D D § x · x x i¨    ... ¸ © * (D  E ) * (3D  E ) * (5D  E ) ¹

D

D

cosD , E ( x )  i sinD , E ( x )

 ... (4 .165)

190

Chapter Four

Now for E ! D , we take a fractional derivative of the order D on the function f ( x ) * (1X ) * (1X D )

0 ) as depicted in the following steps, using the formula 0 DDx ª¬ xX º¼

f (0)

D

0 Dx

X D

x

x

E 1

D

cosD , E ( x ) (here

.

ª¬ x E 1 cosD , E ( xD )º¼ D

ª

0 Dx « x

2D 4D § 1 ·º x x   ¨ * ( E ) * (2D  E ) * (4D  E )  ... ¸» © ¹¼

E 1

¬

E  2D 1 E  4D 1 ª x E 1 º x x    ...» 0 Dx « ¬ * ( E ) * (2D  E ) * (4D  E ) ¼ E D 1 E D 1 x E 3D 1 x x D

*( E  D ) x

E D 1



*( E  D )



(4.166)

 ...

* ( E  3D )

E  3D 1 2D § · x x 1   ¨ * ( E  D ) * ( E  D  2D ) * ( E  D  4D )  .... ¸ © ¹

x

E D 1

D

cosD , E D ( x )

Thus, we get the very useful relationship:

ª¬ x E 1 cosD , E ( xD ) º¼

D

0 Dx

Here, the Caputo derivative will also be the same as above because f (0) C D 0 Dx

ª¬ x

E 1

x

E D 1

D

(4.167)

cosD , E D ( x )

(4.168)

cosD , E D ( x )

0 , that is:

D

cosD , E ( x ) º¼

x

E D 1

D

Similarly, it can be shown that:

ª¬ x E 1 sinD , E ( xD ) º¼

D

0 Dx

Here, the Caputo derivative will also be the same as above because f (0) C D 0 Dx

x

E D 1

D

(4.169)

sinD , E D ( x )

(4.170)

sinD , E D ( x )

0 , that is:

ª¬ x E 1 sinD , E ( xD ) º¼

x

E D 1

D

4.15.9 The Riemann-Liouville fractional derivative of the zero-corrected exponential function: f ( x) e ax with a derivative start point at x 0 and its relationship to the Caputo derivative Now we calculate the fractional order derivative of f ( x ) D

0 Dx

* (1X ) * (1X D )

ª¬ xX º¼

it as e

ax

1

ax * (2)



X D

x

a2 x2 * (3)



and DD >1@ a 3 x3 * (4)

xD * (1D )

e

ax

as we did for ED ( xD ) using the formula ax

e , and f (0) 1 . In series form, we write

. The function is f ( x )

 ... . Now we evaluate the fractional derivative of f0 ( x ) D ax 0 Dx ª ¬e  1º¼

D

ª§

ax

©

* (2)

0 Dx Ǭ 1 

2 2



a x

* (3)

3 3



a x

* (4)

2 2

D

0 Dx 1D

* (2  D )

1D

ED , E (O z )

¹

¼

E

a x

* (3  D )

1,2 D

1:

 ... ¸  1»

3 3

3 3D



a x

* (4  D )

 ...

2 2 § 1 · ax a x   ¨ *(2  D ) *(3  D ) *(4  D )  ... ¸ © ¹

1D

ax ax



º

ax

ª ax a x a x º « * (2)  * (3)  * (4)  ...» ¬ ¼

2 2 D

ax

·

e

( ax )



1  O z  O 2 z 2  ... * ( E ) * (D  E ) * (2D  E )

(4.171)

Fractional Differ-integrations’ further Extension

191

We write the following: D

ª 0 Dx ¬ e

ax

x

D

º¼ ax1D  E1,2D ( ax )  * (1  D )

(4.172)

In a similar way, we write: C D 0 Dx

ª¬e ax º¼ ax1D  E1,2D ( ax )

(4.173)

4.15.10 The Riemann-Liouville fractional derivative of the zero-corrected f ( x) cos( ax) and f ( x) sin( ax) functions with a derivative start point at x and its relationship to the Caputo derivative

cos( ax ) in the same way as we did for cosD ( xD ) , by using the

We find the fractional order derivative of f ( x ) formula

f0 ( x )

D

D

* (1X ) * (1X D )

ª¬ xX º¼

X D

xD * (1D )

and DD >1@

x

cos( x )  1 , with cos( ax ) 1 

2 2

a x * (3)

4 4

a x * (5)



6 6

a x * (7)



0 Dx >cos( ax )  1@

D

D

.

We

a x

* (3  D ) 1D

 ax

calculate

2 2

ª§

a x

©

* (3)

4 4

fractional

6 6

a x



4 4 D



the

derivative

of

the

 .. ; as in the following steps:

0 Dx Ǭ 1 

2 2 D



0

a x



* (5)

* (7)

·

º

¹

¼

 .. ¸  1»

6 6 D

a x

* (5  D )



a x

(4.174)

 ...

* (7  D )

sin1,2 D ( ax )

We obtain: D

0 Dx

1D

>cos(ax)@

 ax

sin1,2 D ( ax ) 

x

D

(4.175)

* (1  D )

Similarly, we can derive the Caputo derivative as: C D 0 Dx

Similarly, the fractional order derivative of f ( x )

1D

>cos(ax)@

 ax

(4.176)

sin1,2 D ( ax )

sin( x ) is the following: D

0 Dx

Here the singular term will not be added as sin( ax )

0x

0

1D

>sin(ax)@

ax

(4.177)

cos1,2 D ( ax )

so in this case f (0)

0 and thus f0 ( x )

f ( x ) . Here too

one may verify that: C D 0 Dx

>sin(ax)@

1D

ax

(4.178)

cos1,2 D ( ax )

4.16 The fractional derivative at a non-differentiable point of the continuous function Consider the function f ( x )

x

1 2

in the interval 0 d x d 1 . Clearly f

(1)

does not exist at x

1 2

,

though at this

point the function is continuous. Now we obtain the fractional derivative RL type on the zero-corrected function for the interval 0 d x d 12 and 12 d x  1 . The value f (0) 12 is still used to offset both regions. Thus, we conduct the

f ( x )  f (0) in two regions that are first 0 to

integration of the f0 ( x )

1 2

, and then in the interval

1 2

to x (as we

show in the following steps), and then take one whole derivative to find the fractional derivative at x 1

2

integral I1 is I1

³

function is f0 ( x )

f ( x )  12

0

( x  y)

 f0 ( y ) dy . In the interval

to x d

1 2

, f ( x)

x 

1 2



so the zero-corrected

2

D ³ ( x  y )    y   dy 

0

1

I1

0

. The

 x . With this, we calculate I1 , which is described as follows: 1

I1

D

1 2

1 2

f0 ( x )

x 

1 2

 12 ;

0 xd

1 2

(4.179)

2

D ³ ( x  y )   y dy 

0

1 2

192

Chapter Four

Writing  y

x  y  x , we get the following steps: 1

2

D ³ ( x  y )  ( x  y )  x dy 

I1

0

1

2

³ ( x  y)

 x( x  y )

2 D

1D

1D

D

dy

0

§ ( x  y)

¨

1 2 D 2

x  

1D

x

2 D



2 D

I 2 is I 2 interval

1 2

³

x 1

( x  y)

2

D

 f 0 ( y ) dy

in the region x t

to x , the zero-corrected function is f0 ( x )

1 2

· ¸ ¹y

x( x  y )



2 D

©

y

xx 

(4.180)

1 2

0 1 1D 2



1D

x

1D 1 2

f0 ( x )

x  1 . With this, I 2 is calculated as follows:

x

D ³ ( x  y )  y  1 dy 

I2

1

Writing y  1

x  12 . Therefore, in the

f ( x )  , and f ( x )

x t 12

 x  1 ;

f0 ( x )

(4.181)

2

( x  1)  ( x  y ) , we obtain the following steps: x

D ³ ( x  y )  ( x  1)  ( x  y ) dy 

I2

1

2

x

³  ( x  1)( x  y )

1

I2

D

1D

 ( x  y)

dy

2

1D

§

( x  y)

©

1D

 ¨ ( x  1)

( x  1)  x 



( x  y)

2 D

2 D

1 1D 2



 x  12 

1D

(4.182)

y x

· ¸ ¹y

1 2

2 D

2 D

After having performed the integrations in the two regions as above, we take one whole derivative then multiply it by

D

1 * (1D )

to get the fractional derivative 0 Dx D

0 Dx

> f0 ( x )@ , 1

> f 0 ( x )@

of I1  I 2 and

as follows for 0  D  1 .

d

* (1  D ) dx 1

d dx

>I1  I 2 @



1 1D 2

1D



(4.183)

2x   x * (2  D ) Therefore, we have the following derivative as above (4.183), and find the fractional derivative at the nondifferentiable point, i.e. x 12 D

0 Dx

0  D  1,

> f0 ( x )@

1

d

* (1  D ) dx

x

³  f ( y )  f (0) ( x  y )

1 * (2  D )

D

dy

0



2 x  12

1D



1D

x



(4.184)

1D

(D )

x

1 2



;

(1)

( x) 1 f x 2 * (2  D ) This example suggests that there exists a finite value of the fractional derivative for points where the integer order’s f

( x)

 12

f

whole derivative is undefined. From the above expression, it is clear that although f

(1)

 12

does not exist, the

Fractional Differ-integrations’ further Extension

fractional D

0 Dx

derivative

> f ( x)  f (0)@x

get at x

1 2

,

D

0 Dx

1 2

D

0 Dx

> f ( x)  f (0)@x

 * (21D )

1D

 12

> f ( x)  f (0)@x

1 2

. For D



2

S

exists

1 2

1 2

1 2

x

at

193

and 1

 *13  12 2 . By putting

, we get the value

2

, whereas f

(1)

 12

is *  32

equal S

2

to , we

f . We solve this example by following the

consolidated steps again D

0 Dx

> f ( x)  f (0)@

1

x

d

D ( x  [ )  f ([ )  f (0) d[ ³ * (1  D ) dx 

(4.185)

0

0 D 1 The above (4.185) provides the definition of the RL fractional derivative of an order less than one. In the above expression, the function for 0 d x d 1 2 is f ( x )  x  12 . D

0 Dx

> f ( x)  f (0)@

1

x

d

D ( x  [ )  f ([ )  f (0) d[ ³ * (1  D ) dx 

0

1

x

d

1

D ³ ( x  [ )  [  

* (1  D ) dx

1 2



1 2

d[

0

d

x

D ( x  [ )  ( x  [ )  x d[ ³ * (1  D ) dx 

0

1

d

* (1  D ) dx

x

³ ( x  [ )

1D

 x( x  [ )

D

(4.186)

d[

0

x

2 D 1D d ª (x  [ ) (x  [ ) º x   « » * (1  D ) dx ¬ 2 D 1  D ¼0

1

2 D

1D º x  » * (2  D ) * (1  D ) dx ¬ ( 2  D )(1  D ) ¼

1

d ª

«

x

d x d 1 , then f ( x ) x  12 . The fractional derivatives from zero to point half and beyond half in two separate segments are depicted below Again when

1

2

194

Chapter Four

D

0 Dx

ª§ 1/ 2 ·º D «¨ ³  ( x  [ )  f ([ )  f (0) d[ ¸ » 1 d «¨ 0 ¸» ¸» * (1  D ) dx «¨ x «¨   ( x  [ ) D  f ([ )  f (0) d[ ¸ » ¸ «¬¨© 1/³2 ¹ »¼

> f ( x)  f (0)@

ª1/ 2 º D 1 1 « ³ ( x  [ )  [  2  2 d[ » 1 d «0 » » * (1  D ) dx « x «  ³ ( x  [ ) D [  1  1 d[ » 2 2 ¬« 1/ 2 ¼» 1D 1/ 2 º ª§ ( x  [ ) 2D · (x  [ ) «¨  x ¸ » D 1 D   2 1 d «© ¹0 » » * (1  D ) dx « x «  ( x  [ ) D [  1 d[ » « 1/³2 » ¬ ¼ 1D 1/ 2 ª§ ( x  [ ) 2D º · (x  [ ) «¨  » x ¸ 2 D 1D 1 d «© ¹0 » » * (1  D ) dx « x «  ( x  [ ) D  ( x  1)  ( x  [ ) d[ » « 1/³2 » ¬ ¼ 1D 1/ 2 ª§ ( x  [ ) 2D º · (x  [ ) «¨  » x ¸ 2 D 1D » ¹0 1 d «© « » x 1D * (1  D ) dx · » (x  [ ) « § ( x  [ ) 2D «  ¨  2  D  ( x  1) 1  D ¸ » ¹1/ 2 ¼ ¬ © 1D ª  x  1 2D  x 2D  x  12  x1D 2 « x 1 d « 2 D 1D 1 1D * (1  D ) dx «  x  1 2 D x  2 2 «  ( x  1) ¬ 2 D 1D

2x 

1 1D 2



(4.187)

º » » » » ¼

1D

x

* (2  D )

Therefore, we obtain the following result:

D

0 Dx

By putting *  32

S

2

­ x1D ° * (2  D ) , ° ® 1D 1 1D °2x  2  x , °¯ * (2  D )

> f ( x)  f (0)@

, we get at x

1 2

> f ( x)  f (0)@x

D

, 0 Dx

1 2



2

S

0 d x d 12 (4.188)

1 d x d1 2 whereas f

(1)

 12

f.

4.17 Computation of the fractional derivative and integration-a review and comparison of the various schemes From the most fundamental approach, we write the derivative of the function as: d f ( x) dx

lim

'x p 0

f ( x )  f ( x  'x ) 'x

(4.189)

The double derivative is: 2

d f ( x) dx The three-whole derivative is:

2

lim

'x p 0

f ( x )  2  f ( x  'x )  f ( x  2 'x ) ( 'x )

2

(4.190)

Fractional Differ-integrations’ further Extension

f ( x )  3  f ( x  'x )  3  f ( x  2 'x )  f ( x  3'x )

3

d f ( x)

lim

3

'x p 0 dx For the n  th whole derivative, we obtain the following: N 1

( 'x )

¦ (1)

n

d f ( x)

195

§n· ¨ j ¸ f ( x  j 'x ) © ¹

j

j 0

(4.191)

3

xa

(4.192) dx N ( 'x ) Now for any real order when n is changed by order D , we can generalize the above (4.192) with j changed to

lim

n

'x

n

'x p 0 N nf

index k , as in: D

Dx f ( x )

lim

'x p 0

f

1

¦ w  f ( x  k 'x)

( 'x )

( xa ) N

d

f ( x) 1

>d( x  a )@

(4.193)

* ( D )* ( k  1)

k 0

We write the anti-derivative, one-whole integration as follows, with 'x 1

*(k  D )

wk

k

D

:

x

{

³ f ( y )dy a

lim ( 'x )  f ( x )  f ( x  'x )  f ( x  2 'x )  ...  f ( a  'x )

(4.194)

'x p 0

§

lim ¨ ( 'x )

'x p 0

©

N 1

·

j 0

¹

¦ f ( x  j'x) ¸

Two-whole integration is expressed as in the following expression:

d

2

2

>d( x  a)@

x1

x

f ( x)

³ ³

{ dx1 dx0 f ( x0 ) a

a

(4.195)

§ f ( x )  2  f ( x  'x )  3  f ( x  2'x )  · 2 lim ( 'x ) ¨ ¸¸ ¨ ....  N  f ( a  'x ) 'x p 0 © ¹ Therefore, for n  th fold integration we write: d

n

§

f ( x)

{ lim ¨ ( 'x ) 'x p 0

n

>d( x  a)@

n

©

N 1

¦ §¨©

· ¸ f ( x  j 'x ) ¸ ¹ ¹

j  n  1· j

j 0

(4.196)

We use the following property of the binomial expansions (as detailed in Section-1.12) and generalize as follows:

( 1)

j

*( j  n) § n · § j  n  1· §  n · ¨ j¸ ¨ j ¸ ¨ j ¸ © ¹ © ¹ © ¹ * (  n)* ( j  1)

(4.197)

Using this generalization for the fractional differ-integration of the D order, we get; D

§

d f ( x)

lim ¨ ( 'x )

D

>d( x  a)@

'x p 0

©

D

N 1

· ¦ w  f ( x  k 'x) ¸¹ k

k 0

*(k  D )

(4.198)

xa

'x , * ( D )* ( k  1) N This (4.198) for D ! 0 is a fractional differentiation; and for D  0 it is a fractional integration. wk

In Chapter-3, we used the Grunwald-Letnikov (GL) formula to determine the numerical evaluation technique that is reproduced again. This is the approximate algorithm to evaluate coefficients i.e.

GL

wj

* ( j D ) * ( D ) * ( j 1)

,

GL

w j 1

* ( j 1D ) * ( D ) * ( j )

dD dxD

> f ( x )@ ,

with weights called Grunwald’s

. Note that wk , which we used above, is the same as

that we described in the previous chapter. Furthermore, in Section-3.23, we wrote the following (with a D 0 Dx

N 1

D > f ( x)@  'x ¦  GL w j f j ;

GL

wj

*( j  D )

,

'x

x

GL

wj ;

0)

(4.199) * ( D )* ( j  1) N In all of these numerical schemes discussed in previous chapters (specifically Chapters 2 and 3), we have, using known values of f ( x ) at N  1 , evenly spaced points in the range 0 to x . We designate symbols f N { f (0) , j 0

196

Chapter Four

f N 1 { f

 Nx , … f j { f  x  j Nx , … f0 { f ( x) ; and the discrete step size as

'x

 Nx .

Although this GL

scheme was derived for D ! 0 , we proved in the beginning of this section that this formula is also valid for D  0 . The above mentioned GL formulation (i.e.

D

0 Dx

utilize f N (the function value) at the start point ( f N

f ( x)

ln x , f ( x )

1 x

N 1 GL wj 0

D



 'x ¦ j

f ( x)

f

j

) is unique as this scheme does not

f (0) ). This permits this formula to be used for functions like

, where the function is singular at the start point, meaning lim xp0 f ( x )

f.

In the GL formula, we need not restrict the function where the value of the function at the start point needs to be merely finite. Thus, the GL formulations for numerical evaluations are important as they span all of the values of D , including positives and negatives and various functions which have singularity at the start point; and can be implemented by convenient multiplication-addition schemes. That is we reproduce them again as: D º º § D · º º §1D · (4.200) d f ( x) D ª ª ª ª ª º§ N D  3· § N D  2 · dx

D

 'x

« « « «.. « f N 1 ¨ ¸  f N 2 » ¨ ¸  f N 3 » ..» ¨ ¸  f1 » ¨ ¸  f0 » © N 1 ¹ ¼© N  2 ¹ ¼ ¼© 2 ¹ «¬ ¬ ¬ ¬ ¬ »¼ ¼© 1 ¹

The modified GL method is as follows, and here we require the value of the function at the start point, i.e. f N

f (0) ,

which needs to be finite. We reproduce this as follows:

d

> f ( x)@ |  'x D N 1 *( j  D )

D

dx

¦ *( j  1)  f * ( D )

D

j 0

N 1

 'x ¦  D

GL

wj

j 0

 f





2

f



1 2 8D

 14 D f j 1  f j 1  81 D

j

f

j 1

 2 f j  f j 1

(4.201)

j



1 4D

j 1

 f j 1 

f

j 1

 2 f j  f j 1



We discussed the first formula in Chapter-2 as an RL-integration formula; i.e. the formula for discrete numerical evaluation is obtained from the RL fractional integration formula, as below x 1 f ( y) D (4.202) dy ; D !0 0 I x > f ( x )@ D 1 0 ( x  y) * (D ) The formula for discrete numerical evaluation is the following:

³

D

D

0 Ix

> f ( x )@

 'x

N 1

f j  f j 1

j 0

2

¦ * (1  D )

 ( j  1)

D

 j

D

;

The above formula (4.203) requires the values at the start point of the function f N

(4.203)

D !0

f (0) which needs to be finite. It

is valid only for fractional integration and not for fractional differentiation. The second formula in Chapter-2 that we obtained for RL fractional integration is:

D

0 Ix

> f ( x )@

§  ( j  1) f j  j f j  ( j  1)D  jD · D ¨ ¸  'x N 1 ¨ D ¸ ¦ * (D ) j 0 ¨  f  f  ( j  1)1D  j1D ¸ ¨  j 1 j ¸ © ¹ 1D

The above formula (4.204) also requires the values at the starting point of the function f N

(4.204)

f (0) , which needs to be

finite. In Chapter-3, for 0 d D  1 , we obtained, a fractional differentiation formula for discrete numerical evaluation, from the RL fractional derivative as below: D

0 Dx

 'x

D

N 1 § (1  D ) · 1D 1D ¨ D f N  ¦  ( j  1)  j  f j  f j 1 ¸ * (2  D ) © N j 0 ¹

> f ( x )@





(4.205)

0 dD 1 We note here that the above formula also requires the values at the start point of the function f N

f (0) , which needs

to be finite. The above formula is derived from the basic formula of the RL fractional derivative i.e. D

0 Dx

>

f ( x)

@

1

d

m

* ( m  D ) dx

m

x

f ( y)

³ ( x  y)D 0

1 m

dy

0 d D  1; m 1

(4.206)

Fractional Differ-integrations’ further Extension

197

We noted in the previous chapter that we get the fractional integration formula, for the fractional order ( m  1) d D  m , by changing the sign of the fractional order D from an RL fractional derivative formula, as indicated below:

> f ( x )@

D

0 Dx

D

0 Ix

> f ( x )@

1

d

m

* ( m  D ) dx 1

d

m

m

* ( m  D ) dx

m

x

f ( y)

³ ( x  y )D

1 m

dy

0

x

f ( y)

³ ( x  y) D

 1 m

(4.207)

dy

0

Thus, with a change of the sign we get the numerical integration formula as follows: D

> f ( x )@

D

0 Ix

 'x

N 1 § (1  D ) · 1D 1D ¨ D f N  ¦  ( j  1)  j  f j  f j 1 ¸ * (2  D ) © N j 0 ¹





(4.208)

0 dD 1 Similarly, we saw the discrete numerical evaluation formula for the RL fractional derivative for 1 d D  2 that is:

§ (1  D )(2  D ) · fN  D ¨ ¸ N  'x ¨ ¸ N 1 * (3  D ) ¨ (2  D ) 2 D 2 D ¸  f j 1  2 f j  f j 1  ( j  1)  j ¸ ¨ N D 1 ¦ j 0 © ¹ D

D

0 Dx

> f ( x )@

(4.209)

The fractional integration formula for discrete numerical evaluation for 1 d D  2 is:

§ (1  D )(2  D ) · fN  D ¨ ¸  'x ¨ N ¸ N 1 * (3  D ) ¨ (2  D ) 2 D 2 D ¸  f j 1  2 f j  f j 1  ( j  1)  j ¸ ¨ N D 1 ¦ j 0 © ¹ D

> f ( x )@

D

0 Ix

(4.210)

In this section, we discussed various formulae in order to provide discrete numerical evaluation for fractional derivatives and fractional integration. The GL method has been shown to be more universal and unified, though several RL methods have also been developed and are useful.

4.18 Fractional derivatives of the same-order but of different RL-Caputo types: A unified formula We have seen two types of fractional derivatives and discussed them. The following formula (4.211) provides fractional derivative definition using a combination of the RL and Caputo types, with E as the type defining parameter as 0 d E d 1 . When E

1 , the definition is the Caputo derivative with the symbol

C

D Dt f (t ) ; while E

0

gives the RL fractional derivative with the symbol DtD f (t ) . The generalized definition is as follows: E D 0 Dt

> f (t ) @

E (1D )

0 It

ª d ª (1 E )(1D ) > f (t )@º¼ º» «¬ dt ¬ 0 I t ¼

(4.211)

The fractional order D is the order of the derivative; here in this definition it is 0  D  1 and E is the type of derivative with 0 d E d 1 . Therefore, the nearest integer is m 1 in the above generalization. Let us see how it works- by using E 1 , we get: E

When E

E (1D )

0 Dt

D

f (t )

0 It

1 D 0 Dt

f (t )

(1D ) 0 It

0 D 1 0 , we get the following:

d

ª¬ 0 I t(1 E )(1D ) f (t )º¼ dt

d

ª 0 I t(11)(1D ) f (t )º¼ dt ¬

E (1D ) 0 It

1

d > f (t )@ dt

C D 0 Dt

f (t )

(4.212)

198

Chapter Four

0 D 0 Dt

0(1D ) 0 It

f (t )

d

d

(1 0)(1D )

ª

0 It dt ¬

(1D ) 0 It

f (t ) º¼

(4.213)

RL D 0 Dt

ª f (t ) º¼ f (t ) , 0 D 1 dt ¬ In a way, we could unify the two formulae for a fractional derivative. This is a different kind of generalization; say with E

1 2

, we get: 1

D 2 0 Dt

f (t )

1 (1D ) 2

0 It

d

1

ª 0 I t2 (1D ) f (t ) º ¼ dt ¬

(4.214)

This (4.214) shows that the fractional derivative operator with the order 0  D  1 is a half Caputo and half RL type. The physical meaning of such a type has yet to be developed.

4.19 Generalizing Leibniz’s rule In this section, let us write f ( x ) f and g ( x ) g . We have a rule for the differentiation of the product of the two functions as discussed earlier in the book (Section-1.21), which is: d

n

> fg @

n j j §n· d f d g ¦ ¨ j ¸ n j j dx j 0 © ¹ dx n

(4.215) n dx In Section-1.21, we derived, via the use of integration by parts, a product rule for multiple-integration, which is: n

j > fg @ f § n · d  n j f d g ¦¨ ¸ n  n j j >d( x  a)@ j 0 © j ¹ >d( x  a)@ >d( x  a)@

d

In the differentiation formula, the sum terminates at j

n , because the binomial coefficients

(4.216) n! j !( n  j )!

are zero when

j ! n for n positive integer; this sum could very well be extended to infinity (making the coefficients zero for the terms j ! n ). We noted this argument while discussing the local properties of the integer order for differentiation. With this, we expect the generalized product rule for any n D real number to be: j > fg @ f § D · dD  j f d g ¦¨ ¸ D D j j >d( x  a)@ j 0 © j ¹ >d( x  a)@ >d( x  a)@

d

D

(4.217)

We have discussed in Section-2.11 that, if M ( x ) is an analytic function, then we have fractional differ-integration of the analytic function in the following expression:

>M ( x)@ f (1) k ( x  a) k D M ( k ) ( x) ¦ D >d( x  a)@ k 0  *(D )  (k  D )k ! k k D ( k ) f ( 1)  * (1  k  D ) ( x  a ) M ( x ) ¦  *(1  k  D )*(D )  (k  D )k ! k 0 d

D

Noting that the fractional derivative of the Heaviside unit step is the function

dD k >1@ D k

>d( x  a )@

(4.218)

( x  a ) D * (1 k D )

, we can re-write,

with the above formula (4.218), the fractional differ-integration of the analytical function as follows

>M ( x)@ f (1) k *(1  k  D ) §¨ dD k >1@ ·¸ M ( k ) ( x) ¦  *(D ) (k  D )k ! ¨ d( x  a) D k ¸ D >d( x  a)@ k 0 @ ¹ ©> d

D

(4.219)

Using the properties of the gamma functions (namely those discussed in Sections 1.10 and 1.12)

* ( x  1)

x  *( x) ,

*( j  q)

* ( n  1)

n  *( n)

§ j  q  1· j §q· ¨ j ¸ ( 1) ¨ j ¸ * (  q )* ( j  1) © ¹ © ¹ we do the following manipulation

n ( n  1)* ( n  1)

...

n! (4.220)

Fractional Differ-integrations’ further Extension

( 1)

* (1  k  D )

k

( 1)

 *(D )  (k  D )k ! ( 1)

*(k  D )

k

199

( k  D )* ( k  D )

k

( k  D )* ( D )* ( k  1)

§ §D · k §D · · ( 1) ¨ ( 1) ¨ ¸ ¸ ¨ ¸ © k ¹¹ © k ¹ ©

(4.221)

k

* ( D )* ( k  1) Thus, we can re-write compactly the fractional differ-integration of the analytical function as D k k f >M @ d M § D · d >1@ ¦¨ ¸ D D k k >d( x  a)@ k 0 © k ¹ >d( x  a)@ >d( x  a)@

d

D

The above (4.222) is similar to Leibniz’s rule with one function f 1 and g formula and, substituting M with the product I\ , we obtain the following

(4.222)

M . Now we start with the above

>I\ @ f § D · dD k >1@ I\ ( k ) ¦¨ ¸ D D k  >d( x  a)@ k 0 © k ¹ >d( x  a)@ D k k f § D · d >1@ § k · (k  j) ( j) \ ¦ ¨k¸ D  k ¦ ¨ ¸I j k 0 © ¹ >d( x  a )@ j 0© ¹ We have made use of the formula (4.215) of d > fg @ as noted at the beginning of this section for an dx d

D

n

n

integer. Note that since j is an integer, then the repeated derivative \

x  a . We use the permutation

f

k

k 0

j 0

¦ ¦

f

f

j 0

k j

¦ ¦

to the permutation of the variables in a double integral that is following:

( j)

We set k  j identity:

n positive

with respect to x equals that with respect to

. This identity is easily established, as it is analogous

³

f

0

dy

³

y

0

dx

³

f

0

dx

³

f

x

dy ; with these we write the

>I\ @ f \ ( j ) f § D · § k · dD k >1@ I ( k  j ) ¦ ¦¨ ¸¨ ¸ D D k >d( x  a)@ j 0 k j © k ¹ © j ¹ >d( x  a)@ d

(4.223)

D

m the second summation of (4.224).Thus, it will be from m

(4.224)

0 to f , and we use the following

§ D ·§ m  j · § D ·§ D  j · ¨ m  j ¸¨ j ¸ ¨ j ¸¨ m ¸ © ¹© ¹ © ¹© ¹

(4.225)

This (4.225) can be easily proven using a factorial notation (or gamma function) for the binomial coefficients with simple manipulation, as demonstrated below:

D! D! · § ( m  j )! · ·§m  j· § j ¹¸ ©¨ ( m  j )!(D  m  j )! ¹¸ ©¨ j ! m ! ¹¸ m ! j !(D  m  j )! D! § D · § D  j · § D ! · § (D  j )! · ¨ j ¸¨ m ¸ ¨ ¸ ¨ ¸ © ¹© ¹ © j !(D  j )! ¹ © m !(D  j  m)! ¹ m ! j !(D  j  m)!

§ D ¨ ©m 

j ¹¸ ©¨

(4.226)

Using (4.224) and (4.226), we re-write the above expression as follows:

>I\ @ f \ ( j ) f § D · § m  j · dD  j m >1@ I ( m ) ¦ ¦¨ ¸¨ ¸ D D  j m >d( x  a)@ j 0 m 0 © m  j ¹ © j ¹ >d( x  a)@ D  j m f f >1@ ( m ) §D · ( j) §D  j · d ¦ ¨ j ¸\ ¦ ¨ m ¸ D  j m I ¹ >d( x  a )@ j 0© ¹ m 0© d

D

(4.227)

D j

f

d I §D · ( j) ¦ ¨ j ¸\ D j j 0© ¹ >d( x  a)@ In the above steps (4.227), we have used the following relationship D k

d M § D · d >1@ ¦ ¨k¸ D k k k 0 © ¹ > d( x  a ) @ >d( x  a)@ f

k

>M @ D >d( x  a)@ d

D

with the symbols changed from D to D  j and from k to m . Therefore, we obtained our expression as:

(4.228)

200

Chapter Four j >I\ @ f § D · dD  jI d\ ¦ ¨ ¸ j D D j >d( x  a)@ j 0 © j ¹ >d( x  a)@ >d( x  a)@

d

D

(4.229)

In all of the above discussions, we assumed that I and \ are analytic functions. A somewhat different argument may be required when one of the functions is a polynomial. Consider the product x  f ( x ) . For D  0 , we write the Riemann-Liouville formula as:

d

D

ª¬ x  f ( x ) º¼

x

1

y  f ( y ) dy

* ( D ) ³

D

>d( x  a)@

* ( D ) ³

D 1

( x  y)

a

x

x

 f ( y ) dy  1

a x

* ( D )

We have used ( D )  * ( D )

³ ( x  y)D

1

a

D

* ( D  1) and *  (1D 1)

D 1

( x  y)

³

D

* ( D ) ³

( x  y )  f ( y ) dy D 1

( x  y) x

( D ) ( D )* ( D )

³

 f ( y ) dy

(4.230)

D

( x  y)

a

D 1

> f ( x )@ D 1 >d( x  a)@ d

x  f ( y ) dy

a

x

1

a

 f ( y ) dy 

> f ( x )@ x D >d( x  a)@ d

 x  ( x  y  f ( y ) dy

a

D * ( D ) ³ ( x  y )

x

x

1

D

( x y )

dD 1 > f ( x )@

>d( x  a )@D 1

. The above technique (4.230) has also

been discussed in Section-2.11. The extension of this result for D ! 0 is easy, as, if ( n  1) d D  n , with n 1, 2, 3... , then:

d

D

ª dD  n ª¬ x  f ( x ) º¼ º D n » n « dx « >d( x  a )@ ¬ ¼»

ª¬ x  f ( x ) º¼

d

D

>d( x  a)@

n

D n D  n 1 f ( x )@ d > > f ( x )@ º d ª d D x ( n )   « n D  D  n 1 » n dx « >d( x  a )@ >d( x  a)@ ¼» ¬

(4.231)

n

We have to use Leibniz’s rule for the first term

d

n

> fg @

dx

dn dx n

ªx «¬



dD n > f ( x )@

>d( x  a )@D n

º»¼

with the following formula

§ dn f · § d n 1 f ·§ d g · § d n  2 f ·§ d 2 g · ( 1)    g n n n ¨ n ¸  ¨ n 1 ¸¨ ¸ ¨ n  2 ¸¨ 2 ¸  ...... © dx ¹ © dx ¹© dx ¹ © dx ¹© dx ¹ n 1 § dn g · § df · § d g · ......  n ¨ ¸¨ n 1 ¸   f ¨ n ¸ © dx ¹ © dx ¹ © dx ¹

n

(4.232)

d> f @

With f

x , we see that all of the derivatives greater than one are zero only terms with dx , and that f exists. So within the above formula (4.232), only the last two terms exist with all other terms vanishing. Therefore, we write dn dx n

ªx «¬



dD n > f ( x )@ D n

>d( x  a )@

º»¼

with f

x and g d

n

ªx n dx ¬«





dD n > f ( x )@

>d( x  a )@D n

dD n > f ( x )@ D n

>d( x  a )@





using the above formula as follows:

n 1 º n§ d x ·§ d ª > @ ¨ ¸¨ ¼» © dx ¹ © dx n 1 ¬«



dD n > f ( x )@ D n

>d( x  a )@

º¼» ·¸¹

§ d n ª dD n > f ( x )@ º ·   x ¨ n D n d( x  a )@ »¼ ¸¹ © dx «¬ > § dD 1 > f ( x )@ · § dD > f ( x )@ n¨ ¸  x¨ ¨ >d( x  a )@D 1 ¸ ¨ >d( x  a )@D © ¹ ©



Also noting that

dn dx n

ª «¬



dD n1 > f ( x )@ D n1

>d( x  a )@

º»¼

dD 1 > f ( x )@

>d( x  a )@D 1



(4.233)

· ¸ ¸ ¹

, we complete the formula as in the following steps:

Fractional Differ-integrations’ further Extension

d

D

D  n 1 n ª dD  n > f ( x ) @ º > f ( x )@ º d ª d   D x n ( ) D n » D  n 1 » n « n « dx « >d( x  a )@ » dx « >d( x  a )@ »¼ ¬ ¼ ¬ D D 1 § dD 1 > f ( x )@ d > f ( x )@ · d > f ( x )@    x n D ( ) ¨n ¸ D 1 ¨ >d( x  a )@D 1 >d( x  a )@D ¸ >d( x  a)@ © ¹

ª¬ x  f ( x ) º¼

d

D

>d( x  a)@

n

> f ( x )@ x D >d( x  a)@ d

For D

1 2

and D

201

 12 with a

(4.234)

D 1

> f ( x )@ D 1 >d( x  a)@

D

d

D

0 , we write the following results: 1

d 2 ª¬ x  f ( x ) º¼

1

1

> f ( x )@  1 d  > f ( x )@ x  >d( x)@ 2 >d( x)@ >d( x)@    d ª¬ x  f ( x ) º¼ d > f ( x )@ 1 d > f ( x )@  x    2 >d( x )@ >d( x)@ >d( x)@ 1

1

2

1

2

1

2

1

dD > f ( x )@

Notice that we have used the relationship

d

D

>d( x  a )@

1

2

3

2

1

2

2

(4.235)

2

3

2

2

ª dD n > f ( x )@ º , i.e. the definition of the Riemann-Liouville «¬ >d( x  a )@D n »¼

dn dx n

{

2

fractional derivative. The inductive argument establishes the following formula: j > fg @ f § D · dD  j f d g ¦¨ ¸ D D j j >d( x  a)@ j 0 © j ¹ >d( x  a)@ >d( x  a)@

d

D

(4.236)

k

x for any non-negative integer k and for any f ( x ) , and thus for g ( x ) any polynomial and any arbitrary f ( x ) . When g ( x ) is a polynomial in ( x  a ) , the RHS of the above formula (4.236) is finite. We may with g ( x )

write for analytic functions I ( x ) and \ ( x ) , with a

0 for the following expression:

>I\ @ f § D · dD  jI d j\ ¦¨ ¸ D D j j >d( x)@ j 0 © j ¹ >d( x)@ >d( x)@

d

For D

1 2

and D

D

(4.237)

 12 , we write the following results:

>I\ @ f § 12 · d  jI \ ( j ) ¦¨ ¸ >d( x)@ j 0 © j ¹ >d( x)@  j

d

1

1 2

2

1

1 2

2

>I\ @ f §  12 · d   jI \ ( j ) ¦ ¨ j ¸  ¹ >d( x )@   j j 0© >d( x)@

d

1 2

 12

1

However, g ( x)

1 1 x

sometimes

the

2

formula

¦

1  x  x  .....

f k 0

has

convergence

(4.238)

1 2

2

issues.

For

example,

take

f ( x) 1

and

k x ; for 1  x  1 . This is certainly a differ-integrable series; in fact, it is

analytic. As such, we write with term-by-term differ-integration the following d

D

> fg @

dx

D

d

D

> 11x @

dx

D

The above (4.239) converges for 0  x  1 . following steps

D

ª f kº x » D «¦ dx ¬ k 0 ¼ d

f

* ( k  1)

¦ *(k  D  1) x

k D

(4.239)

k 0

By using the previously discussed Leibniz’s rule, we obtain the

202

Chapter Four

d

D

> fg @

dx

d

D

1 j D j § D · d >1@ d ª¬(1  x ) º¼ ¦ ¨ j ¸ D j j dx j 0 © ¹ dx

ª¬1  11x º¼

D

dx

f

D

f

x

* (D  1)

j D

¦ *( j  1)*(D  j  1) *( j  D  1)  *( j  1) 1  x

1 j

j 0

* (D  1) 1 x

x

D

f

j

x

1

¦ *(D  j  1)*( j  D  1) (1  x) j 0

* (D  1) 1 x

x

D

f

x

j

¦ (1  x) j 0

* (D  1) 1 x

x

D

j

1 j

(4.240)

( j  D )* ( j  D )* (D  j  1)

f

x

1

j

¦ ( j  D )*  (D  j ) *  (D  j )  1 (1  x) j 0

* (D  1) 1 x

x

D

f

x

( 1)

j

¦ ( j  D )S csc  (D  j )S (1  x) j 0

* (D  1) 1 x

x

D

f

( 1)

j 1

x

j

j

j

¦ ( j  D )S csc(SD ) (1  x) j 0

j

x  * ( x ) and  * (  x )  * ( x  1) S csc(S x ) in the above derivation (4.240). The last line of the above formula is elaborated upon in the following steps: f (1) We have used the identity * ( x  1)

¦ (D  j )S csc  (D  j )S j 0

1 (D )S csc(DS )



1 (D  1)S csc(DS  S )

1 (D )S csc(DS ) 1 (D )S csc(DS )

 

1 ( D )S csc(DS ) f

( 1)



1 (D  2)S csc(DS  2S )

1 (D  1)S csc(S  DS ) 1 (D  1)S csc(DS ) 



1 (1  D )S csc(DS )



1 (D  2)S csc(2S  DS ) 1

(D  2)S csc(DS ) 

 ...  ... (4.241)

 ...

1 (2  D )S csc(DS )

 ...

j 1

¦ ( j  D )S csc(DS ) j 0

The

above

step

(4.241)

uses

the

trigonometric

identities

sin( T )

 sin T ,

sin(S  T )

sin T

and

sin(2S  T )  sin T . The result obtained above is very different to what we obtained when conducting term-by-term differ-integration. Therefore, if we extend Leibniz’s rule to the case with any arbitrary f ( x ) and analytic g ( x ) , the convergence may be a problem.

4.20 Chain rule for fractional differ-integration As we know, the chain rule

d dx

g  f ( x)

d d> f ( x ) @

ª¬ g  f ( x ) º¼ ddx > f ( x )@ is a rule for differentiation. Unfortunately,

there is no analogous rule for integration in integral calculus. If there were an analogous rule for integration, then the process of integration would pose no greater difficulty than differentiation does. We discussed the generalization process for multiple differentiations for the chain rule in Section-1.22; and we emphasise its difficulty in the following expressions:

Fractional Differ-integrations’ further Extension

d

2

dx d

3

2

g  f ( x)

ª¬ g  f ( x ) º¼ dx

d

4

g

3

ª¬ g  f ( x ) º¼ dx

4

(1)

f

5

ª¬ g  f ( x ) º¼ dx

f

(3)

 3g

g

(1)

f

(4)

 4g

(1)

f

dn dx n

(2)

f

(1)

f

(2)

g

(2)

f

(1)

f

(3)

 6g

(1) 2

f

(3)

 30 g The continuation of the above would give

(1) 2

f

(2)

(1)

g

5

g

g

 6g d

(2)

(5)

(3)

 5g

f

(1)

203

f

(2)

(2)

f

(1)

(2) 2

f

g f

(4)

(4)

(3)

(2)

(2) 2

f

(4.242)

(1) 4

f

 10 g

 10 g

(1) 3

f

(4)

(3)

(1) 2

f

(1) 3

f

f

(2)

( 3)

f

g

(5)

(1) 5

f

ª¬ f  g ( x ) º¼ , n ! 0 and an integer but we previously pointed out the

difficulties in Section-1.22. However, there is no generalization for n  0 that can be use for integration in integral dD d( x >  a )@D

calculus. Since the general formula for

ª¬ g  f ( x ) º¼ must encompass integration as a special case, we expect

difficulty in the further generalization for the chain rule for arbitrary D . However, we will develop a chain rule. We start with a formula for analytic function I ( x ) that we used in (4.222), that is:

>I ( x)@ f § D · dD  j >1@ ¦¨ ¸ D D j >d( x  a)@ j 0 © j ¹ >d( x  a)@ d

dD >1@

D

j

d >I ( x )@ dx

j

(4.243)

( x  a ) D * (1D )

; and using this in the above (4.243), we write the following >d( x  a )@ 0 from the rest of the equation;

In Section-4.2, we derived the formula by separating the term for j

D

>I ( x)@ D >d( x  a)@ d

D

( x  a)

j j D d >I ( x )@ §D · ( x  a) ¦ ¨ j¸ j dx j 1 © ¹ * (1  j  D )

D

* (1  D )

f

I ( x) 

(4.244)

Now consider I as function of f ( x ) , that is I { I  f ( x ) ; with this we discuss the difficulties for obtaining dj dx j

ª¬I  f ( x ) º¼ and apply Faa de Bruno’s formula as we did in Section-1.22 of the book. D d ª¬I  f ( x ) º¼ ( x  a ) D D

>d( x  a)@

* (1  D ) f



I  f ( x)

§ D · ( x  a)

j D

j

(m)

j

(k ) · 1 § f

Pk

(4.245)

¦ ¨© j ¸¹ *( j  D  1) j !¦ I ¦– P ! ¨ j 1

m 1

k 1

k

¸ © k! ¹

We provided a description of the third summation in (4.245) and Pk in the same section. We describe the complicacy of the above general result (4.245) by using D

1 as a ‘chain-rule’ for the case of simple integration:

x

³ I  f ( y ) dy a

( x  a )I  f ( x ) 

f

¦ (1) j 1

j

( x  a)

j 1

j 1

j

(m)

j

1

¦ I ¦– P m 1

k 1

k

§f · ¨ k! ¸ © ¹

This gives an infinite series, and there exists a difficulty in expressing the closed form formula. As in Section-1.22, we write the formula when f ( x )

x

e as follows:

(k )

Pk

(4.246)

204

Chapter Four

d

D D

>d( x  a)@

ªI  e x º ¬ ¼

( x  a)

D

j D

f

(4.247)

j

§D · ( x  a) > m@ ( m ) ( jx ) I e  ¦ ¨ ¸ e ¦Sj I j * (1  D ) j 1 © ¹ * ( j  D  1) m 1

> m@

Where S j

x

is Stirling number of the second kind.

4.21 Analytical continuation of the differ-integral operator from integer to real order: Extended fractional calculus The operator Dxn for integer n is well defined as it corresponds to

dn dx n

for n t 1 and

³ (dx)

n

for n d 1 . We have

also on many occasions referred to analytic-continuation and applied this concept. We have studied the Euler formula and we write the following formula for a power function f ( x )

0 for 0  m d n ­ ° ® m ! x m  n for m ! n ! 0 °¯ ( m  n)!

n m

d x dx

³x

n

m

m

x , that is Dxn x m where m, n  ] (i. e. integers):

n

m!

n

­ * ( m  1) m  n x ° ® * ( m  n  1) °¯ f

ª¬ x m º¼ dx

x

mn

for m ! 0 , n  0 ( m  n )! We reproduce this point here and write the unified formula for fractional-differ-integration of a power function (as discussed in Sections 1.23, 2.6, 2.7, and 4.3) as the following expression: d

(dx )

(4 .248)

n

^

n n

0,1, 2, .....; For all m 1, 2, ....; m ! 1

n

1, 2, ....;

(4.249)

m d 1

We may now generalize the above (4.249) for a non-integer order D and write the following for the differ-integration of the power function f ( x )

P

x , P  \ , and D  \ d

D

ª¬ x P º¼

dx

D

­ * ( P  1) P D P ! 1 x ° ® * ( P  D  1) °¯ f P d 1

For all D

(4.250)

For all D

The breakdown of the above formula for P d 1 is associated with the pole of the order unity which occurs at x

a

P

for the function f ( x ) ( x  a ) ; P d 1 (in the above formula (4.250), it is a 0 ). The functions for which such a pole occurs anywhere in the open interval a to x lead to similar difficulties; and for such reasons we have until now purposely excluded these functions from the differ-integrable series. Figure-4.1 here shows the sign of the ratio of the gamma-function in the coefficient of the unified differ-integral formula. Note that the figure stops short of P 1 .

Fractional Differ-integrations’ further Extension

P 3

2

1

0

205













































































 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



0 





















































































0

0 

0  

0

0 

























0





0





0























0





0





0























0





0





0





0  

0   0 





1

2 3

2

* ( P 1) * ( P D 1)

Figure-4.1: The sign of the coefficients

1

0

1

dD x P dxD

in the differ-integration

2

3

P

for various

and

D

D values

P

x . However, if we tabulate Dxn x m , (for integers n and m ), we observe emerging pattern (see Table-4.1). Recall that we previously discussed the integration side of this table in Chapter-2. Now can we extend the generalized formula to the region P d 1 for f ( x )

2

3

mpno 2

2! 5!

x

5

1

1 4!

x

4

0

1 3!

x

3

1 2

3

x

4

1 3!

x

3

1 2!

x

2

2! 4!

2

1

³ ln x(dx)  ³ ln xdx

³ ln xdx

1 2! ln

1 2!

2! 3! 1 2!

x

x

2!x

2!

0

x

x

2

x

1

0

0

1

0

0

0

x

1

1 2!

x



3

3

ln x

1

2

x

x

 ln x

1

0

2

2

x

1

x

x

2

2! x

x

3

3! 4!

2

x

3

4

2! x

3

3! x

4

3! x

4

4! x

5

4! 2!

5

x

 5! 2! x

6

Table-4.1: Tabulating the values of the Dxn x m for integer m and integer n We have started with an n -fold integral

x

³  f ( y ) (dy )

n

a

x 1 ( n 1)! a ( x

³

 y)

n 1

 f ( y ) dy

as a fundamental defining

expression, for an n integer. Then we obtained the Riemann-Liouville fractional integral which is an analytic continuation of this, giving us

D

a Dx

x

D ³a  f ( y ) (dy )

f ( x)

1 * ( D )

³

x

a

( x  y)

D 1

 f ( y ) dy ,

for D  0 and

D , a  \ . After this, we stated that the fractional derivative of the Riemann-Liouville type is an ordinary m differentiation with m ! D , that is DDx f ( x )

m

Dx

 Dx

 ( m D )





f ( x ) for D ! 0 , and m  ] .

However, DDx x P the RL integral is well defined for D  \ and P ! 1 , as we have seen in the previous chapter (Section-3.2). Can we extend this for P d 1 , and conduct the analytical continuation? Let us discuss the Euler formula that we obtained for fractional differ-integration as a generalization, that is

D

0 Dx

ª¬ x P º¼

* ( P 1) * ( P 1D )

x

P D

, for

any arbitrary D . Extended fractional calculus takes the ratio of the gamma-function limp0 expression. Unlike RL integration, giving

D

0 Dx

ª¬ x P º¼

* ( P 1) * ( P 1D )

x

P D

* (1 P ) * (1 P D )

as a fundamental defining

P

with P ! 1 , this limit i.e.

where f ( x )

x

206

limp0

Chapter Four * (1 P ) * (1 P D )



, is well-defined in the plane D , P  \ , apart from along the line intervals P  ] , D  \ \ ] . The

operation denoted by symbol ‘ \ ’ is a “mod” operation in sets, and it is akin to a remainder operation. For example, we have A

^1, 2, 3, 4, 5, 6` , B ^1, 3, 4` then, A “mod” B (symbolically A \ B ) is A \ B ^2, 5, 6` . 

The analytical continuation along the line intervals P  ] , D  \ \ ] will be derived subsequently. The log-region belongs to the integration operation, and, as we discussed in Section-2.7, the analytic continuation of the fractional 

differ-integration to this log zone (in the line intervals P  ] , D  \ \ ] ) is found using the following formula D P

Dx x

limp0

* (1  P  ) * ()

D P

Dx

ln x

· § * (1  P  ) · § x P D limp0 ¨ ¨ ¸  ln x  \ (1  P  D )  J ¸ © * () ¹ © * (1  P  D ) ¹

(4.251)

We documented this in Sections 2.6 and 2.7. D

C

4

P P D line

3

Upper-region 2

Zero-region 1

B

A 2

3

4

:

:

G

1

'

:

Log-region

2

3

'

'

'

'

'

'

'

'

'

1

1

Lower-region

:

:

'

' 2

:

'

'

' 3

E

D

4

½ ° ° ° ° ¾ ° ° ° ° ¿

: horz

F

 

4

: diag

Figure-4.2: Different regions of the DDx x P for analytic continuation in

D P

plane

We draw a D  P diagram, (Figure-4.2) based on our discussion about DDx x P , and its Euler’s generalization, plus the difficulties encountered in integration, with an observation from Table-4.1, showing four different regions; namely the Upper-region (ABCD), Lower-region (EFGB), Log-region (AEB), and Zero-region (BGC). We define them as follows in zones in the D  P plane Upper - region : P t D , Lower - region : P  D , Log - region : P t D Zero - region : P  D

Pt0 P0 Differentiation Right - Half - Plane D ! 0 P0 D 0 Integration Left - Half - Plane Pt0 The ‘ ' points’ are ‘grid-points’ in the Lower-region, and : denotes a ‘grid-point’ in the Log-region, in Figure-4.2. In the Upper-region (region ABCD in Figure-4.2), the ratio

* (1 P ) * (1 P D )

is finite for P t D and P t 0 . Therefore, in this

Upper-region the formula is as follows: D P

0 Dx

x

* (1  P ) * (1  P  D )

x

P D

(4.252)

In the Lower-region (region EFGB in Figure-4.2) and Log-region (region AEB-in Figure-4.1), a region of : P 0 is defined as a union of the Lower and Log-regions, such that P  0 and D  \ (in the lower half plane region AEFG-in Figure-4.2). In this region, the AEFG defines : horz as a set of horizontal lines at negative integer points as with

Fractional Differ-integrations’ further Extension

P

 n , with n

1, 2, 3,... , i.e. n  ]  (integers). In this region, we define : grid as a set of integer points in the

D  ] , P  ]  . Also in Figure-4.2, : diag are a set of right-sloping diagonal lines in

: P  0 region AEFG, such that

the lower region (EFGB), such that P * (1 P ) * (1 P D )

The ratio

207

(D  n )  0 , n  ]  , D , P  \ .

is finite everywhere in the region : P 0 except along the : horz horizontal lines “mod” and the

: grid grid points, (that is in

: horz \ : grid ). The ratio is zero along the diagonals in the lower region : diag “mod”

grid points : grid (that is : diag \ : grid ).

However, the

limp0

* (1 P ) * (1 P D )

evaluated at the points in : grid is

convergent. Thus we write in the region : P 0 “mod” ( : horz “mod” : grid ) or : P 0 \ (: horz \ : grid ) . D P

0 Dx

From Table-4.1, the natural analytic continuation of

limp0

x

D

0 Dx

x

P

* (1  P  ) * (1  P   D )

x

P D

(4.253)

in part of the : horz \ : grid lying within the Log-region

is the following:

* (1  P  )

³ ln x(dx)

P D

* (1  P  )



D P



(4.254) ln x 0 Dx * () * () This is derived from the chapter on fractional integration (Sections 2.6 and 2.7), where the sign of D is shown to be the opposite of the above (4.254). We have also derived the relationship highlighted in Section-2.8, i.e.

limp0

X

³ ln x(dx)

X

Dx ln x

xX * (1X )

{ limp0

 ln x  \ (1  X )  J . With this relationship and the above (4.254) correlation, we

can analytically continue from : horz \ : grid lying in the Log-region into that lying in the lower region. Thus in the 

: horz \ : grid , we have P  ] , D  P  ] and analytic continuation as follows: D P

Dx x

limp0

* (1  P  ) * ()

(D  P )

Dx

ln x

* (1  P  )

x

(4.255)

( P D )





ln x  \ (1  P  D )  J * () * (1  P  D ) Now we deal with the Zero-region (in Figure-4.1, the region BGC). Here we have two possible choices, from two views. The first view is that, with a postulate, the fractional derivative of a constant can be non-zero. As such, the Riemann-Liouville integral should be taken as a fundamental defining relationship and then used to derive a fractional derivative in that Zero-region. Then we write * (1  P ) D P P D (4.256) Dx x x * (1  P  D ) in the Zero-region. limp0

In the second postulate, we say as the ordinary whole integer order derivative of a constant is zero, the fractional derivative shall inherit this property, i.e. the fractional derivative of a constant will be zero. In addition, here we also postulate that the fractional derivatives commute. For a constant function f ( x ) c , we write in this case 1

Dx c As DDx xD

* (1  D ) , for

0

D

Dx c

(D 1)

Dx

D t 0 by continuity DDx x P

 D c 1 x

0 for

(D 1)

Dx

> 0@

0,

for D ! 1

(4.257)

D ! P , P t 0 implies that D

Dx x

P

0

(4.258)

in the Zero-region, and commutability is preserved in this region. Therefore, we have a mutual trade-off between these two postulates, with the options: 1. We choose analyticity and lose commutability in the Zero-region. 2. We choose to preserve and carry commutability and lose analyticity at the edges of the Zero-region. The first type is recognized as the Riemann-Liouville type fractional derivative and the second type is recognized as being similar to the Caputo type fractional derivative. With the above points, we write the analytic continuation for the

208

Chapter Four

fractional derivative formula of the Euler formula for the entire D  P plane. As such, we write the following for the two types of derivatives. a. Riemann-Liouville type: That is ) x ­°limp0 * (1*(P ) * (1 P D )  ln x  \ (1  P  D )  J in : horz \ : grid ® * (1 P ) P D limp0 * (1 P D ) x elsewhere °¯ P D

D P 0 Dx x

(4.259)

b. Caputo Type: That is D P 0 Dx x

­ 0 in Zero - region °° * (1 P ) x P D ®limp0 * () * (1 P D )  ln x  \ (1  P  D )  J in : horz \ :grid ° * (1 P ) P D limp0 * (1 P D ) x elsewhere °¯

(4.260)

4.22 Analytic continuation of the differ-integral operator from a real to complex order To analytically continue D

D

z

to D where z

V  it is a complex number, we just generalize Dxz x w , with complex

z and w . For the Caputo type fractional derivative, we assume that commutability is preserved. That means the differential operator

d dx

commutes with itself and its inverse, which is the integral operator

This assumption makes the complex differ-integration easy, and we discuss this type.

³ dx ;

z

and so does D .

z

With this assumption, the commutative operator D splits into two. Each of these products acts on the function independently, as follows: z

Consider

z w Dx x

with z

V  it

V

it

it

V

D D D D D D u  iv . In the Zero-region of the (V  u ) -plane we have:

V  it and w

V u

Dx x

0

it

implies



V u

Dx Dx x



z u

Dx x

0

(4.261) (4.262)

Similarly in the Zero-region of the (it  iv ) -plane: it iv

Dx x

0

implies

V



it iv

Dx Dx x



z iv

Dx x

0

(4.263)

it

We can think of the above (4.263) as D expanding into the triangular Zero-region of the (V  u ) -plane into a wedge shaped volume of infinite length along t -direction, and similarly expanding that (it  iv ) plane into another along the

V -direction, in order to define D z x w

0 . This is depicted in Figure-4.3. Wedges of Zero-region (space)

t

t

v

t

V

V

u v u

V

D z xw 0 z V  it w u  iv z w

Figure-4.3: The Zero-region (space) wedges in the ( z  w) diagram where D x

0

it

For the line interval in : horz ”mod” : grid , the operator D extends these intervals in the (V  u ) -plane, along the V

t  direction to planer sections, as is the case for D . By now, we have completed the task of defining the action and

Fractional Differ-integrations’ further Extension

209

z

have obtained a complete picture of the operator D . In essence, what we have done is an analytic continuation of an operator to act on a function complex z  times. Let us stop here, and we will find out more about the operator D via Laplace transforms in Chapter-5.

z

4.23 Short summary In this chapter, we have further extended the concept of fractional differ-integration. We have seen a number of interesting results of various fractionally differ-integrating functions. In addition, we have noted the very interesting fact of the existence of the finite value of fractional differ-integration for a non-differentiable point. Here, we have further generalized Leibniz’s product rule and saw the difficulties in generalizing the chain rule, and dealt with the analytic-continuation of fractional differ-integration to complex orders.

4.24 References This chapter is motivated by a number of pioneering works. They are: Jumarie G, “From self similarity to fractional derivative of the non-differentiable functions via Mittag-Leffler function” (2008) Kolwankar and Gangal, “Fractional differentiability of the nowhere differentiable functions and dimensions” (1996) Abramowitz. M, Stegun. I. A. (eds.), “Handbook of the Mathematical Functions” (1964) Michelle Caputo, “Distributed Order Differential Equation Modeling Dielectric Induction and Diffusion” (2001) Michelle Caputo, “Linear Models of the Dissipation whose Q is almost Frequency Independent” (1967) Matignon D, “Generalized Fractional Differential and Difference Equation: Stability Properties and Modeling Issues” (1998) Mandelbrot B. B, “The Fractal Geometry of the Nature” (1977) Miller K S and Ross B, “An Introduction to Fractional Calculus and Fractional Differential Equations” (1993) Oldham and Spanier, “Fractional calculus” (1974) Batman H, “Higher transcendental function” (1954) Igor Podlubny, “Fractional differential equations” (1999) Kirayakova. V, “Generalized Fractional Calculus and Applications” (1994) Kostadin Trenceveski et al, “On fractional derivative of the some function of the exponential type” (2002) Mauro Bologna, “Short Introduction to fractional calculus” (2005) Ross Bertram, “A Brief History and Exposition of the Fundamental Theory of the Fractional Calculus” (1975) Valerio D, Trujillo J J, Rivero M, Machado J A T, Baleanu D, “Fractional Calculus: A survey of the useful formulas” (2013). Further details of these works are listed in the bibliography section,in alphabetical order

CHAPTER FIVE GENERALIZED INITIALIZATION OF FRACTIONAL INTEGRALS/ DERIVATIVES AND GENERALIZED LAPLACE TRANSFORMS

5.1 Introduction In this chapter, we will see how the initialization process of fractional calculus is different from that of classical calculus theory, and the impact this difference has on the Riemann-Liouville fractional derivative and the Caputo fractional derivative. We will discuss why fractional derivatives should require initialization, unlike in the case of classical calculus. We will see that, for fractional integration, the initialization value is not a constant, as it is for classical integration, but is, in fact, a function. We will elaborate upon all these concepts using examples, and adapting Lorenzo and Hartley’s approach. Then we will explore the formulation of Laplace and Fourier transforms in the case of fractional differ-integration operators. Here, we will also discuss the concept of generalized initial conditions and introduce the complex order differ-integrals. Mathematicians have used several different types of notations since the birth of fractional calculus, as mentioned earlier in the book. Here, we attempt to standardize these notations as differ-integral (differentiation integral) operators, with respect to uninitialized and initialized parts. The same operator is used as an integrator when the index is negative and as a differentiator when the index is positive, as we have discussed in previous chapters. Separate notation will be used to indicate the initialized differ-integral operator and the un-initialized operator. q

> f (t )@ represents the ‘initialized’ q  th order differ-integration of f (t ) from the start point c to t . q The notation c d t > f (t )@ represents an ‘un-initialized’ generalized (or fractional) q  th order differ-integral. This is d f (t ) { c d tq > f (t )@ i.e. shifting the origin of the function at the point from where the differalso the same as >d( t  c )@

The symbol c Dt

q

q

q

integration process starts. This un-initialized operator can also be presented in short form as d f (t ) . The initialization function, which is not a constant in the case of fractional calculus, is represented as \ ( f , q , a , c, t ) meaning that this is function \ of independent variable t , and is the differ-integral operator of order q , for function

f born at t a ; before that the function is zero; i.e. f (t ) 0 for t  a , and the differ-integral process starts at t c , where c ! a . This initialization function can be presented in short form as \ (t ) or \ ( f , q , t ) . Therefore, the q

expression between an initialized differ-integral and an un-initialized one is c Dt

> f (t )@

q c dt

> f (t )@  \ ( f , q , a , c , t )

for the function i.e. f (t ) where the function f (t ) is defined for t t a and f (t ) 0 for t  a . The fractional differentiation or fractional integration starts at point t c , where c ! a . This is a causal or left fractional differintegration process. In practical applications, it is usually the case that the problem to be solved is in some way isolated from the past. That is, it should not be necessary to retreat to f in time to start the analysis. Usually, the analyst desires to start the analysis at some time t (1)

a

t0 , with the knowledge (or assumption) of all (2)

values of the function and its derivatives, specifically f (t0 ) , f (t0 ) , and f (t0 ) , and so on, in the case of integer order classical calculus. In modern parlance, this collection is called the ‘state’ and contains the effect of all history. Note here the symbol \ stands for the initialization function. This symbol \ was used by Riemann; and this should not be confused with the ‘psi’ function that we used in earlier chapters.

5.2 Initialization fractional integration (the Riemann-Liouville approach) 5.2.1 The origin of the initialization function (or history function) in the fractional integration process This non-constant initialization function \ (t ) , which shall be elucidated clearly and brings out the past history, also allows to define in the fractional integral the effect of the past; namely, the effect of fractionally integrating the

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

211

function from its birth or start point, before which the function was zero. This added effect will also influence the process after time t c , the start of the integration process. Consider the fractional order q , with q ! 0 integration of f (t ) , the first starting at t t c ! a as depicted in the following expressions: q

t

1

> f (t )@

a dt

a , and the second starting at

q 1

(t  W )  f (W ) dW *(q) ³ a

q c dt

Assume that the function was born at t

1

> f (t )@

a , that is f (t )

*(q)

q 1

³ (t  W )  f (W ) dW c

0 for t  a . As such, the region between a

0 i.e. f (t )

q

and c may be considered as ‘history’. The assumption is that the integral c d t should work as a continuation of the integral starting at t Therefore an initialization must be added to

q c dt

(5.1)

t

> f (t )@ is properly initialized so that it

a.

> f (t )@ , so that the fractional integral starting at

t

c should be

identical to the result starting at t a for t ! c . We call this what Riemann proposed a complementary function to the initialization function \ . We have for the above argument the following statement q c dt

> f (t ) @  \

\

q a dt

q a dt

> f (t ) @

(5.2)

> f (t )@  c dt q > f (t )@

(5.3)

when t ! c . Then for t ! c . Therefore,

\

1 *(q)

c q 1

³ (t  W )  f (W ) dW {

q a dc

f (t ) . t ! c

(5.4)

a

5.2.2 The initialization function is a constant function in the classical integration process Here in (5.4), \ is a function of the independent variable t , and is thus not a constant. For integer order integration, we put q 1 into (5.4) and see that c

\

³  f (W ) dW

K

(5.5)

a

i.e. we get K as a constant. Thus for the integer order integration \ , the initialization function is a constant. Because of the increased complexity of the initialization relative to the integer order calculus, it is important to formalize the initialization process in the case of fractional calculus. This formalization will include the initialization term within the definition of these fundamental fractional order calculus operators.

5.2.3 Types of initialization for fractional integration process Two types of initialization are considered here. The first is the terminal initialization, where it is assumed that the differ-integral operator can be initialized by effectively differ-integrating prior to the start point, i.e. say time t c . The second is the side-initialization where a fully arbitrary initialization may be applied to the differ-integral operator at time t c . In contemporary terms, these may be stated as ‘terminal charging’ and ‘side charging’, as used by Lorenzo and Hartley. First, we restrict it to the RL type of differ-integrals for formalizing these definitions. This initialization function \ has the effect of allowing the function f (t ) and its derivatives to start at a value other than zero, namely q a Dc

> f (t )@ , and continues to contribute to the differ-integral response after

added to the uninitialized integral, (not just a constant at t

t

c . That is, a function of time is

c as in the integer order calculus case).

212

Chapter Five

5.3 Terminal and side initialization for fractional integration 5.3.1 Terminal charging for fractional integration The process of terminal initialization is also termed as terminal charging. The standard contemporary definition of a fractional integral (RL) is accepted if and only if the differ-integrand function f (t ) 0 for all t d a . The initialization region is a d t d c . The fractional integration takes place for t ! c t a . Furthermore, the fractional integration starts at t c (i.e. the point of initialization). q

a Dt

>

t

1

@

q 1

(t  W )  f (W ) dW , q t 0 t ! a *(q) ³

f (t ) {

(5.6)

a

subject to f (t ) t !cta)

0 for all t d a . The following definition of fractional integration will apply generally (at any

q c Dt

with q t 0 , t ! a , c t a , and f (t ) and will be chosen such that

§ 1 t · q 1 (t  W )  f (W ) dW ¸  \ ( f ,  q , a , c, t ) > f (t ) @ { ¨ ³ © *(q) c ¹

(5.7)

0 for t d a . The function \ ( f ,  q , a , c, t ) is called the initialization function q a Dt

q c Dt

> f (t ) @

> f (t ) @

t!c

(5.8)

This above (5.8) condition gives us the following expression 1 *(q)

Since,

t

³

(t  W )

q 1

 f (W ) dW

a

t

c

t

a

a

c

³  g (W ) dW ³  g (W ) dW  ³  g (W ) dW

§ 1 t · q 1 (t  W )  f (W ) dW ¸  \ ( f ,  q, a, c, t ) ¨ ³ © *(q) c ¹

(5.9)

we write the following

\ ( f ,  q, a, c, t )

q

a Dc

>

f (t )

1

@

c q 1

(t  W )  f (W ) dW *(q) ³

(5.10)

a

where we have t ! c and q ! 0 . This expression (5.10) for \ (t ) gives us ‘terminal initialization’, and brings out in the definition of the fractional integral the effect of ‘history’. This history is the effect of fractionally integrating the function f (t ) from a to c . This effect is also called ‘terminal initialization’.

t for semi-integration process starting at t 1

5.3.2 Evaluation of initialization function for f (t ) As an example, let us take function f (t )

t for t t 0 and f (t )

0 for t  0 ; thus we say the start point of a

function is a 0 . We need to find the initialized fractional integration of this function for order q 1 2 , from the start point c 1 z a . The semi-integral process of initialization is demonstrated below ((5.11) to (5.13)) from the start point of the integration, that is at c t 1 . By applying RL formulations to fractional integration, we obtain the following fractional integration from 0 to t .  12 0 Dt

>t @

t

1

³ (t  W ) * 1 2

1 1 2

(W )dW

0

0

1 *  12 4

t

³ t

3

(t  x ) x

1

2

(  dx )

t

3

2

 23  12 *  12

(5.11)

2

3 S Now we conduct the fractional integration process in the interval 1 to t , i.e. for t t 1 , as in the following steps, with the initialization function, i.e.:

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms  12 1 Dt

 12 1 dt

>t @

213

>t @  \  t ,  12 , 0,1, t 1

t

³ (t  W ) * 1 2

 1 2 1

W dW \  t ,  12 , 0,1, t

1

§ 1 t ·  12 ¨¨ 1 ³  t  W W (dW ) ¸¸  \  t ,  12 , 0,1, t , © * 2 1 ¹ § 1 0 (t  x ) · ¨¨ 1 ³ 12 ( dx ) ¸¸  \  t ,  12 , 0,1, t , © *  2 t 1 x ¹ 2

t  W

 12

*

(5.12)

x

S

1

ª (t  1) 2 (2t  1) º  \  t ,  12 , 0,1, t ¼ 3 S ¬ Therefore, the initialization function for semi-integration is derived by subtracting the values obtained in the above process as follows:  12 1 Dt

 12 1 dt

>t @

>t @  \  t ,  12 , 0,1, t  12 0 Dt

\  t ,  12 , 0,1, t \  t ,  12 , 0,1, t

 12

0 Dt

4

>t @

t

3 S 1

3

2

(5.13)

>t @  1 d t >t @

2

2

3

1

ª 2t 2  (2t  1)(t  1) 2 º ¼ 3 S ¬

The nature of the initialization function should be noted, as it is a semi-integration of the function from 0 to 1 , that is

\  t ,  12 , 0,1, t

1

1

*

(t  W )  ³ 1 2

 12

W dW ;

0

*

 12

S , and is a function of variable t . This we calculate in the

following steps:  12 0 D1

\  t ,  12 , 0,1, t 1

>t @

1

³ (t  W ) S

 12

W dW ;

0

1

³ S

t

t 1

x

1

 12

 t  2t S

1

 2t S

2

 2t 3 S

3

2

3

2

1

2

(t  W ) 1

( t  x )  dx

 2(t  1)

1

2

x , dW

S



2 3

³ t

t

3

2

t

t 1

 dx x

 12

 23 (t  1)

3

1

 2t (t  1) 2  23 t 2  23 (t  1) 1

 (t  1) 2 (2t  1)

dx 

3

2

3

2

³

t

1

t 1

x 2 dx

(5.14)







In the above steps for fractionally integrating the function (5.14), there is no requirement for having an initialization function. We observe in the present example that the initialization function \ (t ) has the component and a second part 2( t 1)

1

2 (2 t 1)

3 S

4 3 S

t

3

2

in (5.14). The un-initialized fractional integral, i.e.  12

. With the addition of these two, i.e. 1 d t  12

0,

fractional integration.

1 dt

>t @  \ (t ) , we get the function

2

>t @

4 3 S

1

S

t

³ (t  W ) 1

1

1

2 (2 t 1)

S

W dW is

2

3

t 2 , which is an initialized

>t @ (5.13). This is a continuation of the fractional integration process from the start of the  i.e. 0 Dt >t @ (5.11). Figure-5.1 shows the above described process of initialization of

fractional integration 1 Dt function, from t

1

 2(t 1)3

1

2

214

Chapter Five 3

f (t )

­t ® ¯0

f (t )

tt0 t0

3

\ (t )

0

D1

1

2

0, t  0

f (t )

t

0

Dt

2

2

4t 2 3 S

> f (t )@

1

d t

1

¦

1

2(t  1) 2 (2t  1) 3 S

> f (t )@

2

t, t t 0

0

1

1

4t 2 2(t  1) 2 (2t  1)  3 S 3 S

> f (t )@

1

f (t )

 1 Dt

t 1

t 4

> f (t )@

3 S 0 1

Dt D

1

2

 12

> f (t )@ > f (t )@

1 1

D

d

1

 12 t

t

3

2

> f (t )@ > f (t )@ \ (t )

2

Figure-5.1: Initialized fractional integration example

5.3.3 The initialization concept of classical integration in the context of a developed concept of the initialization function for fractional integration For the integer q

1 , we write the following for a function f (t )



lim q o1 \  t ,  q , 0,1, t



t for t t 0 and f (t )

lim q o1



q 0 D1

0 for t  0

>t @

§ 1 1 · 1 q 1 lim q o1 ¨ (t  W ) W dW ¸ ³ © *(q) 0 ¹ 2 We achieve initialization as constant, i.e.

1

1 2

³ W dW 0

(5.15)

.

The following steps are an initialized integration in classical calculus although proven in a fractional calculus context.



With

lim q o1 \  t ,  q , 0,1, t

lim q o1



q 1 Dt

>t @

 d >t @  lim \ t , q, 0,1, t

lim q o1

lim q o1

q 1 t



q

1 Dt

>t @



q o1

1 2

§ 1

lim q o1 ¨

© * q

§ 1

t

³

t

1

(t  W )

³ (t  W ) © * q

lim q o1 ¨

lim q o1

0



q 0 Dt

 q 1

 q 1

· 1

W dW ¸ 

¹ 2

§ t2 1 · 1 ¨  ¸ ¨ 2 2¸ 2 © ¹

t

2

2

(5.16)

·

W dW ¸

¹

>t @

5.3.4 Side charging for fractional integration When the initial function \ is taken as an arbitrary function, where the terminal initialization equation (5.10) is not valid, then the effect is called ‘side-initialization’. For side initialization, the initialization function \ (t ) could be anything, such as Dirac’s delta function or Heaviside’s step function or the operation of obtaining the initialization function focusing on any other function than the function that is fractionally integrated i.e. f (t ) , call it fi (t ) z f (t ) . This process is also called side-charging.

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

215

5.4 Initializing the fractional derivative Riemann-Liouville approach 5.4.1 Fractional derivative for a non-local operation requires initialization This is contrary to the integer order derivative, which is a point-property and a local quantity (or property) where initialization is not called for. However, the definitions of fractional derivatives do contain fractional integration, and thus the fractional derivative of a function is not a point quantity. Therefore, the process of initialization for a fractional derivative is also necessary. The fractional derivative is a non-local operator, and therefore has history and memory. In a solution of differential equations, the initialization constants, which set the initial values of the derivatives, have the effect of accounting for the integration of the derivative from f to the starting time of the integration of the dynamic system given by the differential equation.

5.4.2 The generalization of the initialization process Extending the generalization concept, the integer order derivative also calls for initialization in a ‘fractional context’. Thus, a generalized integer order differentiation is defined (as with initialization) as the following m

m

>

@

d f (t )

(5.17)  \ ( f , m, a , c , t ) , t ! c m dt Here in (5.17), m is a positive integer and \ ( f , m, a , c, t ) is an initialization function. Now a ‘bare’ or ‘un-initialized’ fractional derivative is defined as the following with q m  p c Dt

f (t ) {

q a Dt

where q t 0 , t ! a and f (t ) by amount p .

> f (t )@ { a Dtm  a Dt p > f (t )@

(5.18)

0 for t d a . This means that m is the integer just greater than the fractional order q ,

The function starts or is born at t a and before that the value is zero. The differentiation starts at t ! c . Now as in the fractional integration case the function \ i.e. \ ( f ,  p , a , a , t ) 0 . Furthermore, consider the function

h (t )

p a Dt

> f (t )@ i.e. the fractional integral of a function starting at

a with an initialized term \ ( h, m, a , a , t )

0;

the initialized fractional derivative looks like, and is defined as, q t 0 and t ! c t a in the following expression: q c Dt

> f (t )@ { c Dtm  c Dt p > f (t )@

(5.19)

5.5 Terminal initialization for the fractional derivative 5.5.1 Derivation of terminal charging for the fractional derivative The definition and concept is similar to that obtained as a terminal initialization for fractional integrals. The requirement is also the same as in the case for fractional integrals, that is q c Dt

> f (t ) @

q a Dt

> f (t ) @

(5.20)

for all t ! c t a . Specifically this requires the compatibility of the derivatives starting at t t ! c . Therefore, it follows that m c Dt



p c Dt

> f (t ) @

m a Dt



p a Dt

a and at t

> f (t ) @

c for all (5.21)

Expanding the fractional integrals in the above expression (5.21) with initialization, we obtain the following m

c Dt

ªt º p 1 « ³ (t  W )  f (W ) dW  \ ( f ,  p, a, c, t ) » ¬c ¼ ª 1 t º m p 1 D (t  W )  f (W ) dW  \ ( f ,  p , a , a , t ) » a t « ³ ¬ *( p) a ¼

(5.22)

for t ! c , while we also have \ ( f ,  p , a , a , t ) 0 , in the above expansion (5.22). Using, the definition of a generalized integer order derivative (5.7), and as defined above in Section-5.4.2, we get the following

216

Chapter Five

ª 1 t º p 1 (t  W )  f (W ) dW  \ ( f ,  p , a , c, t ) »  \ ( h1 , m, a , c, t ) m « ³ dt ¬ * ( p ) c ¼ t m º d ª 1 p 1 (t  W )  f (W ) dW »  \ ( h2 , m, a , a , t ) m « ³ dt ¬ * ( p ) a ¼ d

m

p c Dt

where we now call the new functions h1

> f (t )@ and

p a Dt

h2

> f (t ) @ .

(5.23)

In the RHS of the above expression

(5.23), the integer order derivative is initialized at t

a , thus \ ( h2 , m, a , a , t ) 0 . This means that there is no initialization for the integer order derivative. After rearranging the integrals from the above expression, we get:

ª 1 c º dm p 1 t  W f W W m « ³ ( )  ( ) d »¼  dt m >\ ( f ,  p, a, c, t )@ dt ¬ * ( p ) a d

\ ( h1 , m, a, c, t )

m

(5.24)

ª 1 c º p 1  W W W  \  t f f p a c t ( ) ( ) d ( , , , , )  « » m ³ dt ¬ * ( p ) a ¼ d

m

This formula (5.24) is the requirement for the initialization of the derivative in general. The initialization function of fractional integration under the condition of the terminal charging of the fractional integral

\ ( f ,  p, a, c, t )

1 *( p)

³

c

a

(t  W )

p 1

 f (W ) dW

is as defined and derived earlier (5.10) i.e.

\ ( f ,  p, a, c, t )

p

a Dc

>

f (t )

@

c

1

p 1

(t  W )  f (W ) dW *( p) ³

(5.25)

a

Therefore, in the derived expression for

\ ( f ,  p, a, c, t )

p a Dc

> f (t ) @

1 *( p)

³

c

a

(t  W )

\ ( h1 , m, a, c, t ) p 1

(in (5.24)), we place the expression for

 f (W ) dW and get:

ª 1 c º p 1 (t  W )  f (W ) dW  \ ( f ,  p, a, c, t ) » m « ³ dt ¬ * ( p ) a ¼ c c m º d ª 1 1 p 1 p 1 W W W   t f ( ) ( ) d (t  W )  f (W ) dW »  m « ³ ³ *( p) a dt ¬ * ( p ) a ¼ d

\ ( h1 , m, a, c, t )

m

(5.26)

0 Thus, we have \ ( h1 , m, a , c, t ) 0 , and a very important result is seen, that is ‘integer order differentiation’ which cannot be initialized through the terminal initialization. This is what classical calculus says, that there is no initialization for classical derivatives.

5.5.2 Evaluation of the initialized semi-derivative for f (t ) Let us take the function f (t ) with f ( 2) 0 .

2

(t  2) for t t 2 and f (t )

(t  2) 2 starting at t

0

0 for t  0 ; thus, this function starts at point a

We need to evaluate the initialized fractional derivative of order q

1

2

, from the start point c

2 ,

0 for t ! c , and

c ! a . Thus, we have the start point of the function as t 2 . For q 1 2 , we have m 1 , which means p 1 2 . The initialized semi-derivative from c t 0 is the sum of the uninitialized semi-derivative plus the initialization function, as expressed below 1

2 ª 0 Dt ¬ (t  2)

2

¼º

ª 1 t º  12 1 2 « 1 ³ (t  W ) (W  2) dW » dt «¬ *  2 0 »¼ 0 º d ª 1  12 1 2 W W W  « t   ( ) ( 2) d » dt ¬« *  12 ³2 ¼» d

1

0 dt

2

2 2 1 ¬ª (t  2) ¼º  \  (t  2) , 2 , 2, 0, t

(5 .27)

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

The uninitialized fractional integration of order p , i.e. semi-integration from the start point c

f (t )

217

t

0 for

2

(t  2) , is as follows:  12 0 dt

2

ª¬ (t  2) º¼

t

1

³ * 1 2

(t  W )

 12 1

2

(W  2) dW

0

(5.28)

1

ª16 t *  12 ¬15

5

2

 16 3 t

3

2

1

 8t 2 º

¼

We will discuss this expression (5.28), along with others from Figure-5.2, in detail in the subsequent section in this chapter. Differentiating by one whole integer order the above expression, we obtain the RL un-initialized semi-derivative of the function as expressed in the following: 1

0 dt

2

2

ª¬ (t  2) º¼

ª 1 t º  12 1 2 « 1 ³ (t  W ) (W  2) dW » dt ¬« *  2 0 ¼» d

ª 1 16 5 2 16 3 2 1 « 1 15 t  3 t  8t 2 dt «¬ *  2 d



1 *

1 2

3

1

º

»»

(5.29)

¼

1

ª 83 t 2  8t 2  4t  2 º ¼ ¬

We evaluate as total initialization the terminal initialization case for the semi-derivative operation as in the following steps: 2

0 · 1  1 1 2 ¨¨ 1 ³ (t  W ) 2 (W  2) dW ¸¸ dt © *  2 2 ¹ 3 3 1 1 1 ª8     (t  2) 2  83 t 2  8t 2  4t 2 º 3 1 ¬ « ¼» *

\ ((t  2) , 1 2 , 2, 0, t )

d §

(5.30)

2

Thus, we get an initialized semi-derivative as follows: 1

0 Dt

2

ª¬ (t  2) 2 º¼

1

0 dt

2

ª¬ (t  2) 2 º¼  \  (t  2) 2 , 1 2 , 2, 0, t 8



3 *  12



(t  2)

3

(5.31)

2

2

This (5.31) is the same as the semi-derivative of the function f (t ) (t  2) ; t t 0 from the start point of the function, i.e. a t 2 , given as: 8 3 1 2 2 ª º (5.32) (t  2) 2 2 Dt ¬ (t  2) ¼ 1 3 * 2





Figure-5.2 gives us a diagrammatic representation of the initialization of a fractional derivative for this example.

218

Chapter Five ­(t  2) 2 ® ¯ 0

f (t )

3

t t 2 t  2

f (t )

8(t  2) 2 3  *  12

1

0

Dt 2 > f (t ) @

3

8 º 1 ª 3 (t  2) 2 « 3 » 1 *  2 «¬  83 t 2  8t 12  4t  12 »¼

\ (t ) Dt1  2 d t1/ 2 > f (t )@

f (t )

f (t ) (t  2) t t 2

0, t  2

t

0

2

2

¦

t

1

0

t 8 3 (t  2) 2 3  *  12

1

2

Dt 2 > f (t ) @

2

Dt1 1

0



> f (t )@ 0 Dt1  0 Dt > f (t )@ > f (t )@ 0 dt > f (t )@ \  f (t ), 1 2 , 2, 0, t 2

Dt

1

1

2

1

Dt 2

3

ª 83 t 2  º « 1 » 1 1  2 2 *  2 «¬8t  4t »¼

1 d t 2 > f (t ) @

2

2

Figure-5.2: Initialized fractional derivative example

5.6 Side-initialization of the fractional derivative Refer to expression (5.24) that we have obtained i.e.:

ª 1 c º dm p 1 ( ) ( ) d \ ( h1 , m, a, c, t ) t W f W W   « »  m >\ ( f ,  p, a, c, t ) @ m ³ dt ¬ * ( p ) a ¼ dt d

m

(5.33)

This (5.33) is the requirement for the initialization for a general derivative as obtained in the terminal initialization case (as discussed in Section-5.5). If the side initialization is employed, the function \ ( f ,  p , a , c, t ) is arbitrary. Thus, we can infer from the requirement equation (5.33), the following:

\ ( h1 , m, a , c, t )

m

Dt

ª 1 c º p 1 m (t  W )  f (W ) dW »  Dt >\ ( f ,  p , a , c, t ) @ « ³ ( ) * p a ¬ ¼

(5.34)

which is that either \ ( f ,  p , a , c, t ) or \ ( h1 , m, a , c, t ) can be arbitrary, but not both together. In addition, should this then satisfy the requirement expression derived above (5.34), the generalized expression for the side charging case for t ! c can be stated as follows: q c Dt

> f (t ) @

m is a positive integer and m ! q with q q c Dt

where h (t )

p a Dt

> f (t ) @ .

> f (t )@

m c Dt

§ 1 t · p 1  t  W f W W \ f  p a c t ( ) ( )d ( , , , , )  ¨ ¸ ³ © *( p) c ¹

(5.35)

m  p . From the above expression (5.35), we write the following

d

m

1

m

*( p)

dt

t

³ (t  W ) c

p 1

§ dm · \ ( f ,  p, a, c, t ) @ ¸ ¨ m >  f (W ) dW  ¨ dt ¸¸ ¨ \ ( h, m, a, c, t ) © ¹

(5 .36)

Here in (5.36), both the initialization terms are arbitrary and thus may be considered as a

single (arbitrary) term, namely:

\ ( f , q, a, c, t ) {

d

m

\ ( f ,  p , a , c , t )  \ ( h, m, a , c , t ) m dt In the case of the terminal initialization of the fractional integral, the initialization part is as follows: \ ( f ,  p, a, c, t )

1 *( p)

(5.37)

c p 1

³ (t  W )  f (W ) dW ; t ! c a

(5.38)

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

219

5.7 Initializing fractional differ-integrals using the Grunwald-Letnikov approach Here in this approach too, we must take the function’s starting point as a , and the differ-integration process starting point as t c . An initialization (the notation for which is the same as for the RL-approach) is introduced to account for past-history and goes back to t a , with f (t ) 0 at all points before t a . Then the differ-integration with an arbitrary order q is

with t ! a and f (t ) as:

§  t Na  q { lim ¨ N nf ¨ * (  q ) ©

q

> f (t ) @ > f (t ) @ q >d(t  a)@ d

q a Dt

N 1

¦ j 0

* ( j q ) * ( j 1)



f t j

·

 t Na ¸¸

(5.39)

¹

0 for t d a . The Grunwald-Letnikov (GL) definition for differ-integrals will generally apply q

c Dt

Here again \ ( f , q , a , c, t ) is selected such that

q c Dt

>

@

f (t ) {

> f (t ) @

q

> f (t ) @  \ ( f , q , a , c , t ) q > d(t  c)@ d

(5.40)

will produce the same result as

q a Dt

> f (t ) @

for t ! c .

Expressed as the following: q c Dt

q c dt

> f (t ) @

\ ( f , q, a, c, t ) it

is

self-explanatory,

\ ( f , q, a, c, t )

q a Dt

> f (t ) @ 

for q c dt

t !c,

all

> f (t ) @

> f (t ) @  \ ( f , q , a , c , t ) q q a Dt > f (t ) @  c d t > f (t ) @ and

f (t )

q a Dt

or identifying

> f (t ) @

0 for

q a Dt

all

> f (t ) @ t d a.

(5.41) Therefore,

with

q a dt

as

> f (t )@ i.e. an un-initialized differq q a d t > f (t ) @  c d t > f (t ) @ . In this, with

integral, as per the standard notation, we write the same as \ ( f , q , a , c, t ) (5.41) substituting the GL series we obtain the following:

§ t  a  q N1 1 N1 * ( j q ) t a \ ( f , q, a, c, t ) lim ¨ ¦ * ( j 1) f t  j N1 N1 nf ¨ * (  q ) j 0 ¨ © § t c  q N2 1 N2 * ( j q )  lim ¨ ¦ * ( j 1) f t  j N 2 nf ¨ * (  q ) j 0 ¨ ©



 



For all t ! c and f (t )

N2

 ttac N1 ,

· ¸ ¸¸ ¹

  t c N2

· ¸ ¸¸ ¹

(5.42)

0 for t  a . After considerable manipulation by adjusting the delay element as equal, that is

and adjusting with 'T

 and t a N1

N3

 ct aa N1

we write the following expression for the

initialization function:

§ 'T  q N3 1 * ( N1  1  q  j ) · f  t  ( N1  1  j ) 'T ¸ \ ( f , q, a, c, t ) lim ¨ ¦ N1 nf * (  q ) * ( N1  j ) j 0 © ¹

(5.43)

5.8 Criteria for the generalized differ-integration composition 5.8.1 Composition rules One of the fundamental problems of fractional calculus is the requirement that the function and its derivatives be identically equal to zero at the start of initialization (at the start of the differ-integration process) at time t c . This was needed to ensure that the composition or index law holds, implying that: v c Dt



u c Dt

> f (t ) @

u c Dt



v c Dt

> f (t ) @

u v c Dt

> f (t ) @

(5.44)

It is difficult in engineering and physical sciences, to always require that the functions and their derivatives be at zero (or at rest) at initialization instants. This fundamentally implies that ‘there can be no initialization or that composition is lost’. Thus, it is not in general true that:

f (t ) 

d

q

d f

q

dt

dt

q

q

0

(5.45)

220

Chapter Five

Therefore, while solving a fractional differential equation of the form in the following: q

d f dt

(5.46)

F

q

additional terms must be added, such as:

f (t ) 

d

q

dt

q

f

dt

q

d

q

C1t

q 1

 C2 t

q2

 ...Cm t

qm

(5.47)

to achieve the most general solution, which is: q

d F

q 1

q2

qm

(5.48) dt These issues describe the reality that the index law or composition law is inadequate. A minimal set of criteria have been thought fit to be applied for fractional (or generalized) calculus. They are listed as follows and are called the Ross criteria (1974):

f (t )

 C1t

q

 C2 t

 ...  Cm t

v c Dz

(i) If f ( z ) is an analytic function of the complex variable z ; the differ-integral

> f ( z )@

is an analytic

function of z and v ; v

>

@

(ii) The operator c Dx f ( x ) must produce the same result for the differentiation, when v is a positive integer; (iii) If v is a negative integer (say v integration of the function f ( x )

n c Dx

 n ) then

and

n c Dx

> f ( x)@ must produce the same result of the

> f ( x)@ must vanish along with

( n  1) derivatives at the start point that is x

f

(1)

,f

(2)

,... f

( n 1)

n  th fold

, and all the

c; 0

>

(iv) ‘Zero’ operation leaves the function unchanged, that is c Dx f ( x )

@

f ( x) ;

(v) The linearity of the fractional (generalized) differ-integral operator is: q c Dx

>af ( x)  bg ( x)@

a



q c Dx

> f ( x)@  b  c Dx q > g ( x)@

(5.49)

u v c Dx

(5.50)

(vi) The law of exponents for the arbitrary order holds, meaning: u c Dx



v c Dx

> f ( x )@

 v u c Dx

> f ( x )@

> f ( x )@

The above notations are used by Ross. It should be noted that there is a minor conflict contained in these criteria. Also, a clearly noted explanation can be q

>

@

given as c Dx f ( x ) in the above criteria, which is un-initialized by differ-integrals. It is correct as the function itself starts at t

q

c , c Dx > f ( x ) @

c and before that is zero. So at t

q c dx

> f ( x )@ .

The criteria (ii) & (iii) call for backward compatibility and criterion (vi) calls for the index law to hold vis-à-vis integer order calculus. The fundamental theorem of an integer order calculus violates this “zero law” as: m m

0

d d f ( x) z d f ( x) for all f ( x ) the integer is m . The fundamental theorem states that (take m 1 ): x

³ f

(1)



( y ) dy

f ( x )  f (c ) ,

1

c

dx

(5.51)

f ( x)

1 c dx

f ( x)

f ( x )  f (c ) (5.52)

c

and it can thus be observed that the reversal of differentiation and integration differs from f ( x ) by f (c ) , that is by the initialization which is a constant in the classical integer order calculus. This failure in backward compatibility and index law is handled in the integer order calculus by a constant of integration and by a complementary function for the solution of the differential equations (in an ad-hoc manner). We summarize the composition law of the index with a simplified example for the following: 1. D

P

 D > f (t ) @ v

D

( P v )

> f (t ) @

D

v

 D > f (t ) @ P

for . P , v ! 0 : this implies that commutation and a

computation (additive) are valid for fractional integration. 2. The integer derivative of the fractional integration of a function is not equal to the fractional integration of the

 > f (t ) @

integer derivative of the same function. They are related by D D

v

D

v

 D > f (t ) @ 

t v1 *(v)

f (0) ; that

is the initial condition. Here, the order is 0  v  1 . This follows from the fundamental theorem of calculus-a

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

221

generalization of the same. Thus, the Riemann-Liouville and Caputo derivatives are not the same unless the initial conditions are zero (or static). 3. Fractional integration of order (v  1) of the integer order derivative of a function is not the same as the fractional integration of order 0  v  1 , but again is related by an initial condition as shown in D

 ( v 1)

 D > f (t ) @

D

v

v

> f (t )@  * (tv 1) f (0) .

4. The arbitrary composition is invalid, that is D

u

 D > f (t )@ z D > f (t )@ , for any u, v  \ . This composition v

u v

and commutation is only valid for fractional integrations.

5.8.2 Demonstration of the composition of a fractional derivative, and the fractional integration 1 1 and mixed operation that is applied to functions f (t ) t 2 & f (t ) t  2 , and discussion Let us take f (t ) Similarly, we get D D

u

 D > f (t ) @ v

u v

D

u

v

v

1

2

 D > f (t ) @ u

1

ª t º D 2 ªt 2 º ¬ ¼ ¬ ¼



1 4 t3

 D > f (t ) @ z D > f (t ) @ . v

D

u v

3

2

as

3 2

and v

> f (t )@ , that is D v ª¬

0 that is D

Now calculate D D

1 2

t with u

3

u

> f (t ) @

as D

u

1

1

S

ª t º D 2 ªt 2 º ¬ ¼ ¬ ¼

2

.

1

2 ª 2 º ¼ D ¬t ¼ 0 . Therefore, in this example we have

tº

ª¬ t º¼ D

D

. First, we calculate

v



D

0. u

ª tº ¬ ¼



D

3

2

D

1

2

1

ªt 2 º ¬ ¼



3

D2ª

¬

u



1 4 t3

. Then we calculate

 D > f (t )@ z D  D > f (t )@ and  D > f (t )@ and D  D > f (t )@ both exist but are not

. This calculation demonstrates that This demonstrates that D

º ¼

S 2

D

u

v

v

v

v

u

u

equal. Things could be worse, specifically in that one of them may not exist. Say we take f (t )

1 , then D

v t

³ (t  W ) 0

v

 12  3 2

W

 D > f (t ) @ u

0 but D v > f (t )@  12 t 

3

2

and D

u

 D > f (t ) @ v

t

 12

and let u

 12 and

do not exist since the integral

dW diverges.

The discussion point of all these differ-integrations is limited to the real domain. Under the condition of the terminal initialization of the u  th and v  th differ-integrations u c Dt



v c Dt

> f (t ) @

v c Dt



u c Dt

> f (t ) @

u v c Dt

> f (t ) @

(5.53)

for t ! 0 holds under the following condition: a) u  0, v  0 for continuous f (t ) m a Dt

b) u ! 0, v ! 0 for f (t ) are m times the whole-differentiable i.e.

> f (t ) @

exists and is a non-zero

>@ >@

continuous function of t for t ! a , where m is an integer larger than an integer that is part of u or v . c) u  0, v ! 0 is the same as (b).

5.8.3 Composition rules with initialized differ-integrations Under the conditions of the terminal initialization case, the above-discussed properties and criteria hold, and this in turn provides credibility to the initialized fractional (generalized) calculus. Some conditions are, however, imposed on the linearity of fractional integrals (for t ! c ): v c Dt

>bf (t )  kg (t )@

b



v c Dt

> f (t )@  k  c Dt v > g (t )@

(5.54)

holds only if the initialization functions satisfy the following condition (5.55) \ (bf  kg , v, a, c, t ) b\ ( f , v, a, c, t )  k\ ( g , v, a, c, t ) Relative to the criteria of backward compatibility with the integer order calculus, the addition of the initialized function is clearly a generalization relative to integer order calculus. In a strict sense, \ (t ) z 0 violates criterion 2. However, we are looking for the generalization of integer order calculus and it is clear that this generalization (the addition of the initialization function) will be very useful in many applications.

222

Chapter Five

i. Relative to the criteria of zero order property as holds for terminal initialization; ii. Relative to the linearity as holds for the terminal initialization subject to the above rule; iii. Relative to the composition rule the above should follow. (k )

It is noted that the condition, f (c ) 0 for all k no longer exists. This constraint has effectively been contained (shifted) to the requirement f (t ) 0 for all t d a . This allows for the initialization of fractional differential equations. In summary, the terminal initialization case is backwardly compatible with the integer order calculus, which satisfies the applicable criteria established by Ross. The case for side charging is less definitive. The criteria for backward compatibility are the same as those for the case for terminal charging. Relative to zero property, the condition:

\ ( f ,  p, a, c, t )

c

1

p 1

(t  W )  f (W ) dW *( p) ³

0 \ ( h, m, a , c, t ) (5.56)

a

is required for side initialization since \ is arbitrary. When these conditions are not met, the zero order operation on

f (t ) will return to f (t )  g (t ) , i.e. the original function with an extra time function g (t ) , showing the effect of initialization. Relative to the linearity of the side initialization, it demands additional requirements for initialization. These are not so much of an issue as they initially appear for practical applications. In the solution of fractional differential equations, \ (t ) will be chosen in much the same manner as initializations are currently chosen for ordinary differential equations in classical calculus. This will imply the nature of f (t ) from a to c . The new aspect is that to achieve a particular initialization for a given composition, attention must now be paid to the initialization of the composing elements.

5.9 The Relationship between uninitialized Caputo and Riemann-Liouville (RL) fractional derivatives 5.9.1 Expression for the RL fractional integration of f (t ) from 0  t as q

q

integration of function f (t  x ) from 0  t

We derived the relationship between the RL and Caputo formulations in Section-3.10; here we will derive the same in a different way. The generalized integration (RL) of any real number q -folds is as follows: q

> f (t )@

0 dt

In the above (5.57) expression, let us put W

t

1

³ (t  W )

q 1

*(q) 0

 f (W ) dW ,

q!0

(5.57)

1q

1

 1q x q dx , substituting this change of variable in

t  x q , giving dW

above, to get (5.58). The generalized (fractional) order integration relationship is: 1 t q 1 q ³ (t  W ) f (W )dW 0 d t > f (t )@ 0 *(q) 0

1

t  t  x *(q) ³

1

tq

1

0

³ x

q  *(q) t q 1

tq

1

q

q





q 1

 f (t  x * ( q  1) ³ 0

q 1

x

1

q



 1q x

1q q

1q q



1



f ( t  x q ) dx (5.58)

 f ( t  x ) dx 1

q



) dx

This above expression (5.58) is also discussed in the chapter on fractional integration, and its geometric interpretation is highlighted in Section-2.21. We will stop here and will use the above expression (5.58) later on in this chapter.

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

223

5.9.2 Finding an un-initialized Caputo derivative and demonstrating that no singularity term appears at the start point of fractional differentiation To understand the basic issues of the differences between Caputo and RL derivatives, let us consider a simple function

f (t )

2

(t  2) for t t 2 and f (t )

0 for t  2 . This means that the system f (t ) is at rest before time t 1 2

Our aim is to differentiate from t t 0 , by order D , we take D

f (0)

4,

2 .

as a demonstration, where the function value is

using the forward (left) differentiation method. Therefore, the highest integer order in this case is

1 2

m 1! D , as required by the Caputo and RL definitions. The rule of a fractional differentiable is maintained as per the existence criteria for this function, that is, the candidate function is defined and bounded in the interval, and at the lower terminal a

1

2 , is ‘better’ behaved than (t  a ) .

Using the Caputo definition, choosing the integer order derivative as unity, and then obtaining a semi-integration of that derivative of the function from c 0 , we get the following steps: C 12 0 dt

ª¬(t  2) 2 º¼

0

 12

dt

§d 2· ¨ (t  2) ¸ t d © ¹

  12

0 dt

>(2t )@  0 dt  >(4)@ 1

(5.59)

2

t § t  1 1  1 1 · 2 ³ (t  W ) 2 W dW  4 ³ (t  W ) 2 dW ¸ ¨ 1 *2 © 0 0 ¹

1

With the change of variable (t  W ) x , we get dW changed limits, we get the following: C 12 0 dt

ª¬(t  2) 2 º¼

t  x . By substituting this and integrating it with

dx and W

t § t  12 1  12 1 · 2 ( ) d 4 ( ) dW ¸    t W W W t W ¨ ³0 *  12 © ³0 ¹ 0 0 · 1 § 1 1 2 ³ x 2 (t  x )( dx )  4 ³ x 2 ( dx ) ¸ ¨ *  12 © t t ¹ t t t 3 1 1 ªx 2 º ªx 2 º · 1 § ªx 2 º ¨ 2t « »  2 « »  4 « » ¸ 3 1 *  12 ¨ ¬ 1 2 ¼ ¬ 2 ¼ 0 ¬ 2 ¼ 0 ¸¹ 0 ©

1

(5.60)

From the above (5.60), after simplification, we write the following: C 12 0 dt

ª¬(t  2) 2 º¼

1 *

1 2

ª 8 t 3 2  8t 12 º ¼ ¬3

(5.61)

Note that, in the above expression (5.61), there is no singularity term at the start point, i.e. at t 0 . Note also that the Caputo definition of a fractional derivative is the ‘fractional integration of a derivative of a function’.

5.9.3 Finding an un-initialized Riemann-Liouville (RL) derivative and demonstrating the appearance of a singularity term at the start point of a fractional derivative The un-initialized RL derivative is as follows: 1

0 dt

2

d

 dt

ª¬(t  2) 2 º¼

 12 0 dt

ª¬(t  2) 2 º¼



(5.62)

The un-initialized semi-integration is:  12 0 dt

Changing the variable as (t  W )

ª¬(t  2) 2 º¼

x gives us the following:

1

t

* ³ 1 2

0

(t  W )

 12 1

2

(W  2) dW

(5.63)

224

Chapter Five

 12 0 dt

0

1

2

ª¬(t  2) º¼

* ³ 1 2

x

  12

t

0

1

x * ³ 1 2

*  12 *

  12 ª

(t  2)  x ¬

t

0

1 1

2

(t  x  2) ( dx )

³ ª¬(t  2)

2

x

 12

º (  dx ) ¼

(5.64)

1 2

 2(t  2) x  x

3 2

t

ºdx ¼

ª2(t  2) 2 t 2  43 (t  2)t 2  52 t 2 º ¼ ¬ 3

1

1 2

2

5

Simplifying the above (5.64) yields the following result:  12 0 dt

1

ª16 t 2  16 t 2  8t 2 º 3 ¼ 1 ¬15 *

ª¬(t  2) 2 º¼

5

3

1

(5.65)

2

Differentiating by a one-whole integer order, we obtain the RL un-initialized semi-derivative of the function as expressed below: 1

0 dt

2

ª 1 16 5 2 16 3 2 º 1 2 « 1 15 t  3 t  8t » dt ¬ *  2 ¼ d

ª¬(t  2) 2 º¼



1 * C 12 0 dt



ª 83 t 2  8t 2  4t  2 º ¼ ¬ 3

1 2

1

4t

1

(5.66)

 12

ª¬(t  2) 2 º¼  1 * 2

Note that the above expression (5.66) has a singularity at the starting point of differentiation, i.e. t

0.

5.9.4 Adding singularity function at the start point of a fractional derivative to equate uninitialized RL and Caputo derivatives In (5.66), note that f (0)

4 , and

1

0 Dt

2

> f (0)@

4 , i.e. a semi-derivative of a constant such as f (0) * 12

4 . From

the above example, we have the two following semi-derivatives: C 12 0 dt

ª¬(t  2) 2 º¼

1

2

2 0 dt ª ¬(t  2) º¼

1 *  12

ª 83 t 2  8t 2 º ¬ ¼ 3

1

ª 8 t 3 2  8t 12  4t   12 º *  12 ¬ 3 ¼

(5.67)

1

Thus we find that the relationship between the un-initialized RL derivative and the Caputo derivative is 1   12 2 C 12 2 4 2 , which we write as the following: 0 dt ª ¬(t  2) º¼ 0 d t ª¬(t  2) º¼  * 1 t

2

1

0 dt

1

0 dt

2

2

> f (t ) @ > f (t ) @

C 12 0 dt 1

C 2 0 dt

> f (t ) @ 

f (0) * 1  1

1 2



t

 1 2

(5.68)

> f (t )@  0 Dt > f (0)@ 2

Noting f (0) 4 , the above expressions also equate the two definitions (RL and Caputo) through a singularity at the starting point of the differentiation. We highlighted this relationship between RL and Caputo in Section-3.10, and further explore this process in Figure-5.3.

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

a

­(t  2) 2 ® ¯ 0

f (t )

2

225

t t 2 t  2

Caputo Derivative Process 1

C 0

1 3 1 ª 8 t 2  8t 2 º ¼ *  12 ¬ 3

f (t )

dt 2

Riemann-Liouvelli Derivative Process 1

1

3

1

1

ª 8 t 2  8t 2  4t  2 º ¼ * ¬3

d 2 f (t )

Adding Singularity at start point

¦

1 2

0 t

Start point of Derivative process f (t )

4

f (0)

RL Derivative of constant function 1

0

2

c

0

Dt 2

1

0

4 1 t 2 *  12

f (0) 4

t

3

2

Figure-5.3: Adding a singularity at the start point to the Caputo derivative to provide an example of Riemann-Liouville (RL) to Caputo conversion

We can therefore write a general expression for an D  1 order RL and the Caputo derivative as follows for the function f (t ) : D

> f (t ) @

C D a dt

> f (t ) @ 

f (a)

D

(5.69) * (1  D ) These two derivatives are equal only if at the starting point of fractional differentiation the function is at rest i.e. f ( a ) 0 , or if the start point of differentiation coincides with the start of the function or the process (as demonstrated below). a dt

(t  a )

We say the starting point of the function is t t0 . For t d t0 , we have f (t ) 0 . Generally, if the starting point of differentiation differs from the start of the function (or process), these two derivatives are not equal. The Caputo derivative is added by a function, which is singular at the start point, of the differentiation process (see Figure-5.3), to get the Riemann-Liouville fractional derivative. However, at a steady state (i.e. for t tending to f we have

lim t nf (t  a )

D

0 ; D ! 0 ), these two fractional derivatives converge, as is expressed below: lim t nf



D

a dt

> f ( t ) @

lim t nf



C D a dt

Similarly, if we push the start point of the fractional differentiation process, i.e. t derivatives converge; that is: lim a pf



D

a dt

> f ( t ) @

lim a pf

> f ( t ) @

(5.70)

a , to f , the two fractional



C D a dt

> f ( t ) @

(5.71)

5.9.5 Finding out the relationship of the RL-Caputo fractional derivative via Leibniz’s formula For the generalization proof of

D

a dt

> f (t ) @

C D a dt

> f (t )@  * (1f (aD) ) (t  a) D , let us consider

a

0 , and observe the

effect of the differentiation of the integral. From the definition of the fractional integral as discussed above in (5.58), we express the equivalence obtained as follows: q 0 dt

> f (t ) @

1

t q 1

(t  W )  f (W ) dW , *(q) ³

q!0

0

1 * ( q  1)

tq

³  f (t  x

1

q

0

(5.72)



) dx

To differentiate the above, we use Leibniz’s rule, of differentiation of an integration (of several variables), namely:

d

t

 G ( x , t ) dx dt ³ 0

t

G (t , t ) 

§ wG ·

³ ¨© wt ¸¹dx 0

(5.73)

226

Chapter Five

The rule (5.73) has a function in ( x, t ) . When x

G (t , x )

1

f (t  x q )

1

q

y we obtain x

y

q

 qy dy q 1

and dx

in order to obtain

f (t  y ) . Then the fractional integral expression becomes the following: q

0 dt

>

f (t )

t

1

@

q 1

(t  W )  f (W ) dW *(q) ³ 0

tq

1

 f (t  x * ( q  1) ³

1



q

) dx

0

(5.74)

t

1

³  f (t  y )  qy dy q 1

* ( q  1)

0 t

1 * ( q  1)

³ G (t , y )dy 0

With the differentiation of the above expression (5.74) and using Leibniz’s rule and G (t , x ) obtains the following: 1

dt

1

dt

 

q

0 dt

q 0 dt

> f ( t ) @ > f ( t ) @

putting t  x

1

q

q 1

f (t  x ) , one

§ 1 t · G (t , x )dx ¸ ¨ ³ dt © * ( q  1) 0 ¹ d

1 * ( q  1)

tq

1

(5.75)

§

w

¨ ª f (t  x * ( q  1) ³ © wt ¬

G (t , t ) 

1

q

0

The first term in RHS of the above (5.75) is obtained by placing x thus we obtain G (t , t )

qx

· ¼¹

) º ¸dx

t in the expression G (t , x )

  f (t  x ) ,

q x

q 1

  f (0) . The second in RHS of the above (5.75) expression term, is obtained by

q t

q 1

[ ; we get x

 1q x

q

(t  [ ) , d[ 1

 1dx 1

q

as depicted in the following steps:

tq

0

1

1 §wª · q ¨ ¬ f (t  x ) º¼ ¸dx ³ * ( q  1) 0 © wt ¹

§

(d[ ) ¨ * ( q  1) ³ © t

t

1 q  *(q)

³

§

d[ ¨

q § 1q · ª( t [ )q º ¨© q ¸¹

¨ ©¬

0

¼

t

1

·d q f ([ ) x 1q 1 ¸¹ dt

(t  [ ) *(q) ³

d

q 1

dt

0

·d ¸ f ([ ) ¸ dt ¹

(5.76)

f ([ )d[

 d > f ( t ) @ 1 t

q 0 dt

Substituting these two terms, and simplifying what we obtain, the derivative of the (fractional) integral operation is as follows: 1 dt



q 0 dt

1

> f (t )@

* ( q  1) 1 q  *(q)

tq

1

§w

¨ ª f (t  x * ( q  1) ³ © wt ¬

G (t , t ) 

1

q

0

f (0) qt

q 1



1

t

(t  [ ) *(q) ³

q 1

dt

0

t

q 1

*(q)

q

f (0)  0 d t

d

· ¼¹

) º ¸dx

f ([ )d[

(5.77)

 d > f (t )@ 1 t

Rearranging the above, we obtain the useful expression: 1

dt



q

0 dt

>

f (t )

@

q

0 dt

 > @ 1

d t f (t ) 

t

q 1

f (0) (5.78) *(q) Putting q 1  D , and applying the definition of the RL and Caputo derivatives we find the relation between RL and Caputo, which is:

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

D

0 dt

C D 0 dt

> f (t ) @

> f (t ) @ 

t

227

D

* (1  D )

f (0)

(5.79)

5.9.6 Evaluation of the RL and Caputo derivatives from a non-zero start point where the value of the function is zero Let us now evaluate the fractional Caputo and RL derivatives from the start point of the function itself, that is say at

a 2 , for the function f (t ) derivative is derived as follows: C 12 2 d t

(t  2)

ª¬(t  2) 2 º¼

2

where the value of the function is zero, at f ( 2)  12

2 d t

ªd 1 1 2º (t  2) d t 2 >(2t )@  2 d t 2 >(4)@  2 «¬ dt »¼ t t 1 ª  12 1  12 1 º W W W W t t    2 ( ) d 4 ( ) dW » « ³2 *  12 ¬ ³2 ¼ 0 0 º 1 ª  1  1 2 ³ x 2 (t  x )( dx )  4 ³ x 2 ( dx ) » « *  12 ¬ t  2 t 2 ¼ 1 *

1 2

* 8



3 *



1 2

(5.80)

ª 4t (t  2)  4 (t  2)  8(t  2) º 3 »¼ ¬« 3 2

1 2

1

1 2

0 . The Caputo



(t  2)

(t  2)

 12

1 2

ª¬4t  43 (t  2)  8º¼

 32

The RL derivative is derived as per the below steps:

d

 dt

1

2 2 ª º 2 d t ¬ (t  2) ¼

1

d

0

x *  dt ³ 1 2

1

  12

ª¬ (t  2) 2 º¼

1



t

d

(t  W ) *  dt ³ 1 2

 12 1

2

(W  2) dW

2

2

(t  x  2) ( dx )

t 2

d

0

*  dt ³ 1 2

 12

2 d t

x

t 2

ª¬(t  2) 2  2(t  2) x  x 2 º¼ ( dx )

0

0

0

(5.81)

  º 2  (t  2) x dx  ³ 2(t  2) x dx  ³ x dx » « ³ *  12 dt ¬ t  2 t 2 t 2 ¼ 1 d ª 1 3 5 2 2(t  2) (t  2) 2  43 (t  2)(t  2) 2  52 (t  2) 2 º ¼» *  1 dt ¬« 1

d ª

  12

 12

1 2

3 2

2

1 *  12

d ª 16 5 (t  2) 2 º 15 ¼» dt ¬«

8



3 *  12



(t  2)

 32

The above examples ((5.80) and (5.81)) state that the Caputo derivative and the RL derivative are equal when the differentiation process starts at the point where the value of the function is zero, and also before the start point of the function.

5.10 Initialization of Caputo derivatives 5.10.1 Initialization of Caputo derivatives and the associated difficulties In real applications, it is usually the case that the problem to be solved is in some way isolated from the past. That is, it should not be necessary to retreat to f in time; that is, time immemorial, to start the analysis. Usually, the analyst desires to start the analysis at some time with t t0 , with the knowledge (or assumption) of all values of the function

228

Chapter Five (1)

(2)

(n)

and its derivatives. Specifically, f (t0 ), f (t0 ), f (t0 )..... f (t0 ) , in the case of integer order calculus. In modern parlance, this collection is called a ‘state’, and contains the effect of all that came before it. However, in classical calculus these ‘initial-conditions' are constants. A non-constant initialization or function appears as history in the initializing differentiation (anti-differentiation), as is required for fractional calculus whilst also being well-documented for Riemann-Liouville (RL) and GrunwaldLetnikov (GL) formulations, as described in the earlier sections of this chapter. However, for fractional initialization, (1 )  1 ( n D ) conditions of type, f (t0 ) f 2 (t0 ) f 2 (t0 ) or f (t0 ) where D  \ and n  ` are a natural number, are required. This seems to become physically unrealizable for those formulations, however, great an effort may be required to relate to them physically. The Caputo Derivative definition (1967) was made in order to realise integer order initializations and the making of the fractional derivative of a constant as zero. However, although these may be mathematically pure, as the requirement in the Caputo formulation is for integer order initial conditions, the mathematical purity makes them difficult to realise physically. A popular belief is that the Caputo derivative formulation for a fractional derivative properly accounts for the initialization effect, which is not generally true when applied to fractional differential equations. Constant initialization of the past is insufficiently general, and the widely used contemporary equations for the Laplace transform for differ-integrals based on that assumption lack generality. Therefore, the generalized form requires an initialization function. In a solution for the fractional differential equations with an assumed history, the set of initializing constants representing the values of fractional differ-integrals at t 0 or at start points are ineffective, thus what is required is a ‘non-constant initialization’ for the generalized concept of integration and differentiation, as explained in the earlier sections of this chapter. The requirement of this initialization function for initializing generalized derivatives in RL formulations and the difficulty encountered in Caputo formulation to find an equivalent history function ate demonstrated in this chapter through the use of a few examples. An attempt is made to clarify that initialization function, or assumed history function, which is required for the Caputo formulation, and can give a drastic condition tracing to f , which is much further away from the actual functional behavior that is under consideration. In RL formulation with terminal initialization, the continuity and the nature of the function are maintained throughout the interval. In the next section, an attempt is made through simple examples to point out the difficulties in initializing the Caputo formulation, against the physical reality.

5.10.2 The initialized fractional derivative theory Let \ (t ) be the time dependent initialization function, taking into account the history of the system, function f (t ) , which starts at point t a , and the differentiation process, which starts at time t c , where a  c  t , and f (t ) 0 , for t d a . For an un-initialized fractional integration for t ! a of order q with q ! 0 , the RL formulation q

is a d t

> f (t ) @

1 *(q)

t

q 1

³ (t  W )  f (W ) dW a

q

and for t ! c is c d t

> f (t ) @

1 *(q)

t

q 1

³ (t  W )  f (W ) dW . c

q

The region between a and c is considered to be the history of the fractional integral c d t

> f (t )@ , an-initialization

function. Thus the expression, for the entire integration should be, for t t c ,

\ (t )

q a dt

> f (t ) @ 

q c dt

q c dt

> f (t ) @  \ (t )

q a dt

> f (t )@ with \ (t )

as

> f (t ) @ .

The initialization function in the sense of terminal initialization is about fractionally integrating the function f (t ) from time a to c , i.e. \ (t )

1 *(q)

initialized prior to the start time, t express the initialization as

q c Dt

³

c

a

(t  W )

q 1

 f (W ) dW .

For this terminal initialization, the integral can be

c , of the differ-integration process. Using the standard symbols, one may thus

> f (t )@ { c dt q > f (t )@  \ ( f , q, a, c, t ) .

The meaning of \ ( f ,  q , a , c, t ) is

fractionally integrating (by order of q ! 0 ) the function f (t ) from point a to c , to provide a time varying function \ , for the purpose of initializing the fractional integration process for t ! c .

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

229

We can, therefore, generalize the fractional derivative (as has been done for the fractional integral) for the RL definition as

q c Dt

> f (t )@ { c Dtm  c Dt p > f (t )@ , where

mq,

m is an integer such that ( m  1)  q  m with p

q t 0 and t ! c t a . This expression allows for an initialized derivative and for the converting of the same by expansion with the un-initialized derivative and the initialization function (as done for the fractional integration case). p m m m dm f (t ) , as c Dt h (t ) We get this by taking h (t ) a d t c d t h (t )  \ ( h, m, a , c , t ) , where c d t { dt m , is

>

q

expanded as c Dt

m

> f (t ) @

d dt m



p c dt

@

>

@

>

@

> f (t ) @  \ ( f ,  p , a , c , t )  \ ( h, m, a , c , t ) .

For the terminal initialization for an integer ( m) order derivative, the initialization is zero, that is \ ( h, m, a, c, t ) For terminal initialization, the \ (t ) is the operation on the function itself from time a to c . We calculate the initialized D order derivative 0  D  1 , for a function f (t ) where for t d a  0 , f (t ) is, a function is born at t a  0 . We calculate the fractional initialized derivative from t 0 ! a . D

0 Dt

1 0 Dt

> f (t ) @ d

 dt



 (1D ) 0 Dt

0 . That

> f (t )@

 (1D ) 0 Dt

> f (t )@  \ (h,1, a, 0, t )

 dt

 (1D ) 0 dt

> f (t )@  \  f , (1  D ), a, 0, t  0

d §

1

d

0.

(5 .82)

t

· D (t  W )  f (W ) dW ¸ ¨ ³ dt © * (1  D ) 0 ¹ 0 · d § 1 D t f ( ) ( ) d   ¨ W W W  ¸ dt © * (1  D ) ³a ¹ For the side initialization, the initialization function \ (t ) could be anything, such as Dirac’s delta function or Heaviside’s step function, or the operation of obtaining the initialization of any function other than f (t ) , call it

f i (t ) z f (t ) .

5.10.3 Demonstration of the initialized function for a fractional derivative (RL and Caputo) as zero when the function value at the start point is zero and the function is zero before the start point For the considered example f (t )

2

(t  2) , take m

1 2

1, D

, c

0 and a

2 . We will use the above formulation

(5.82) for an initialized fractional semi-derivative for the RL formulation, for the terminal initialization case for our function f (t )

2

(t  2) , and we write the following: 1

2 0 Dt ª ¬(t  2)

2

º¼

ª 1 t º  12 1 2 « 1 ³ (t  W ) (W  2) dW » dt ¬ *  2 0 ¼ d

(5 .83)

0 º d ª 1  12 1 2 t W W W  «   ( ) ( 2) d » dt ¬ *  12 ³2 ¼

For the first term in RHS of the above (5.83) we have already calculated it as an un-initialized semi integration (5.28) t  1 1 1 1  5 2 16  3 2 1 2 2 2 (t  W ) 2 (W  2) dW , which is 0 d t 2 ª¬(t  2) º¼ * 11 16 i.e. 0 d t 2 ª¬(t  2) º¼ * 11  3t  8t 2 . We 15 t

2

³

0

2



use this derived result in the following expression for the first term of RHS of (5.83), i.e.



230

Chapter Five

1

0 dt

1

0 dt

2

2

d ª 1 1 dt ¬« * 2

ª¬(t  2) 2 º¼

t

³ (t  W )



« dt ¬ *  12

  1

*

1 2

8 3

t

16 15

 32

t

 52

 8t

(W  2) dW º 2

¼»

0

d ª 1

ª¬(t  2) 2 º¼

 12 1

 163 t

 12

 32

 4t

 8t

  12

 12 º

»¼

(5.84)



For the second term of RHS (5.83), we evaluate it as a total initialization, for a terminal initialization case as follows:



\ (t 

2 2) , 1

2 , 2, 0, t

§ 1 0 ·  12 1 2 ¨ 1 ³ (t  W ) (W  2) dW ¸ dt © *  2 2 ¹ d

t

d dt

1

³ * x

  12

1 2

t 2

2

(t  x  2) (  dx )

§ t   12 · 2 ¨ ³ x  (t  2)  x ( dx ) ¸ 1 *  2 dt © t  2 ¹ t · 1 d §  1 2 2 x 2  (t  2)  2(t  2) x  x (  dx ) ¸ ¨ ³ 1 *  2 dt © t  2 ¹ 1

d

t t 1 § t · 3 1 2  2 2 ¨  ³ (t  2) x dx  2 ³ (t  2) x dx  ³ x 2 dx ¸ 1 *  2 dt © t  2 t 2 t 2 ¹

1

d

t t t 5 1 ª§ ·º  32 2 2 4 2 2 2( t 2) x ( t 2) x x      « 3 5 ¨ ¸» 1 *  2 dt ¬ © t 2 t 2 t 2 ¹¼

1

d

1

ª 8 (t  2) 32  8 t  32  8t  12  4t   12 º 3 3 »¼ *  12 «¬

(5.85)

From the above derivation (5.85), what we get is an initialized RL semi-derivative shown as: 1

0 Dt

ª¬(t  2) 2 º¼

2

1

2 0 Dt ª ¬(t  2)

1

This above expression (i.e. 0 Dt 2 ª¬(t  2)

2

8 3 *  12

º¼



a derivative un-initialized starting point at a

f ( 2) C 12 2 d t



2

º¼

1

0 dt

ª¬(t  2) 2 º¼  \  (t  2) 2 , 12 , 2, 0, t 8  32

2



3 *  12

(t  2)

 32



) is the same as the RL and Caputo derivatives in being

2 where the value of the function f (t )

(t  2)

0 as derived earlier in the chapter (see equations (5.80) and (5.81)). This means that it is

ª¬(t  2) 2 º¼ , that are equal to

8 3*  12

(5.86)

(t  2)

2

is zero, that is 1

2 d t

2

ª¬(t  2) 2 º¼ or

3

(t  2) 2 , that is: 1

2 Dt

2

ª¬(t  2) 2 º¼

1

2 d t

2

C 12 2 d t

ª¬(t  2) 2 º¼ 8



3 *  12



ª¬(t  2) 2 º¼

(t  2)

3

2

(5.87)

As it is obvious that in this case \ (t ) that the required initialization will be zero, then: d

ª\  (t  2) 2 ,  12 , 2, 2, t º ¼ dt ¬

d

(5.88) ª¬\  f , (1  D ), a, a, t º¼ 0 dt This is at the starting point of the function where the function value is zero, and prior to that too, where the initialization is also zero and the two definitions (the RL and the Caputo derivatives) are equal.

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

231

5.10.4 Making the initialized Caputo derivative equal to the initialized RL derivative in order to provide an initialization function for the Caputo derivative We have seen the relationship between un-initialized Caputo and RL derivatives as follows (5.89), where, at the start point of the derivative, the function value is non-zero. In our example, this is shown by c 0 with f (0) 4 : D

c dt

D

0 dt

> f (t ) @

C D c dt

> f (t ) @

C D 0 dt

> f (t ) @  f ( c )

(t  c )

D

* (1  D ) t

> f (t )@  f (0)

(5.89)

D

* (1  D ) For case 0  D  1 , to this we add on both sides the expression of the initialization function of the fractional derivative and obtain the following: d D ª¬\  f , (1  D ), a, 0, t º¼ 0 d t > f (t )@  dt C D 0 dt

D

0 Dt

C D 0 dt

> f (t ) @

t

> f (t )@  f (0)

D

0 dt



* (1  D )

> f (t )@  f (0)

where the expression in LHS of (5.90), i.e.

D

t

d dt

ª¬\  f , (1  D ), a, 0, t º¼

D

* (1  D )

d



dt

ª¬\  f , (1  D ), a, 0, t º¼

> f (t )@  d ¬\  f , (1dt D ),a ,0,t ¼ { 0 DtD > f (t )@ , ª

(5.90)

º

is the initialized RL

derivative. The Caputo derivative will be equal to the initialized RL derivative if we make for the above (5.90) obtained expression the following:

f (0) d dt

t

D

* (1  D )

d

* (1  D ) dt d 1 * (1D ) dt

steps. The term

 f (0) D * (1D ) t

³

0

a

(t  W )

D

D

f (W )dW

a D0

D

is recognized as 0 Dt

d dt

ª¬\  f , (1  D ), a, 0, t º¼ 0

\  f , (1  D ), a, 0, t 1

where we have



 f (0)

t

D

0

³

(t  W )

D

 f (W ) dW

 f (0)

a

f (t ) which is equated to

(5.91)

* (1  D )



1 * (1D )

t

D

* (1  D )

t

D

f (0) via above (5.91)

> f (0)@ , that is a RL-fractional derivative of a constant; in this case

the constant is  f (0) . In our example, considered here as f (0)

4 , we can thus write this in the form as expressed below:

d dt

t

 12

ª¬\  f ,  2 , 2, 0, t º¼ 4 * 1  12 1

t

 12

(5.92) 1

2 0 Dt > 4@ *  12 In this case, the initialization is equal to the fractionally differentiating constant function by order of half i.e.

4

f i (t )

1

> @

4 , i.e. 0 Dt 2 4 , so that:

d dt

1

2

1

2

>4@

0d

1

2

>4@



4t

 12

(5.93)

*  12

In this chosen example, one can choose the history function or the initialization function, fi (t )

4 as a constant to

allow the Caputo derivative to be the same as the RL initialized derivative. That function i.e.

f i (t )

differentiated i.e. the Caputo.

1

0 Dt

2

4 is semi-

> fi (t )@ and then subtracted from the Caputo to make an RL initialized derivative the same as

232

Chapter Five

4

However, choosing this constant as 

fi (0 )



4 z f (0 )

will make the function

discontinuous at

f (t )

t

0 , as

4 . However, the argument suggests, that in order to make the RL initialized derivative and the

Caputo process the same, one has to use a constant function (at least in this chosen example) and the concept of terminal initialization will not work, since the choice here is arbitrary. These calls for side-initialization have an initialization function \ (t ) derived from any function other than the one being fractionally differentiated as f (t ) , referred to here as fi (t ) z f (t ) . In the above case, fi (t )

2

4 z f (t ) , where f (t )

(t  2) .

5.10.5 The Caputo derivative initialization function: A case of a side initialization process The initialization function is formed for the Caputo derivative by a history function, which is analytical at t and is represented by the Taylor series, for t d 0 as in the following: f

fi

¦

f i (t )

(n)

n!

n 0



n

t ,

t0

0,

(5.94)



 f (0 ) as described in the previous section. For 0  D  1 , we have

For the Caputo initialization, we have, fi (0 )

m



(0 )

c

1 , and we replace f (t ) by fi (t ) and write the following steps: d dt

\  fi , (1  D ), a, 0, t

t



fi (0 )

D

* (1  D )

§ 1 0 · D (t  W ) fi (W )dW ¸ ¨ ³ dt © * (1  D ) a ¹ d

t



fi (0 )

D

t



 f (0 )

* (1  D )

D

(5.95)

* (1  D )

D ª f fi ( n ) (0  ) 1 0 º t D n  t f   ( ) d (0 ) W W W «¦ » n! dt ¬ n 0 * (1  D ) ³a * (1  D ) ¼

d

D

The LHS of (5.95) is a D0

> fi (t )@ i.e. D

is the order derivative of fi (t ) from a to c  d ª fi (0 )

«

dt ¬ * (1  D ) 

fi (0 ) 

d dt

0

³



D

* (1  D )

D

dW

D

t

D

t

D



t

D



f (0 )

(5.96)

a



lim a pf ¨ fi (0 )

©

t



 f (0 )

¼

³ (t  W )





º

dW »

0 we get:

0

Now, as a tends to f , we get lim a pf fi (0 ) (t  a )

§

D

a

fi (0 ) (t  a )



(t  W )

0 . For n

D

d



0

(t  W ) dt ³ a

D

t t

D

D

·

dW ¸

¹



f (0 ) 

fi (0 ) . With this, we write the following: t

D



fi (0 )

t

D



(5.97)

f (0 ) 

For this above identity (5.97) to be valid, as the initial point a tends to f , it must give us fi (0 )



f (0 ) for

f  t  0 . In our example, therefore, to have the Caputo and RL derivatives initialized and to get them equated, we have to take the function f (t ) at value fi (t )

2

(t  2) and abruptly take a constant value that is the value of the function at t

f (0)

4 , from the start point t

f , until t

0 and a constant

0.

Therefore, the history function for the Caputo derivative in this example is a constant in Heaviside’s step function, multiplied by the constant of value 4 (that is the value of function f (t ) and the point t 0 , i.e. the start point of fractional differentiation), starting at t f . This is, however, in no way a backward continuation of the function f (t ) being considered for the fractional derivative process. It is different from the original form of the function f (t ) from t f to the start point of the differentiation process i.e. t 0 . This condition seems to be physically demanding. Say the function represents a velocity that started at some point in the past, for instance at point t 2 , and we are obtaining fractional acceleration through this process of the use of a

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

233

half derivative, in order to obtain forces that are acting. The situation is such that if we want to initialize this with the Caputo formulation then we have to consider that the system has a constant velocity from time immemorial ( t f ) up to point t 0 , and then suddenly this takes on a defined trajectory of the function. This seems to be unrealistic physically, at least in this case. 

In terms of RL, the initialization lets us take fi (t ) f (0 ) 4 , as a historical function from 2 d t d 0 . In the case of the terminal initialization as demonstrated above, the history function for the RL process was taken as f i (t )

f (t )

2

(t  2) for 2 d t d 0 . We evaluate the initialization term with a revised history function fi (t )

4,

and in so doing we obtain the following:

d

ª¬\  4,  , 2, 0, t º¼ dt 1 2

0 d ª

º 1  12 1 (4)dW » « ³ 1 (t  W ) dt ¬ 2 *  2 ¼ t º 4 d ª   12 x x  ( d ) « » *  12 dt ¬ t ³2 ¼

(5.98)

d ª 8  12 8  12 º « 1 t  1 (t  2) » *2 dt ¬ *  2 ¼ 1 1 4  2 4 2

t  (t  2) *  12 *  12 The un-initialized RL derivative as obtained earlier is reproduced below: 

1

8t

2

0 dt ª ¬(t  2) º¼ 2

 32



3 *  12 C  0 dt

1

2





8t

 12

*  12



4t

*  12 4t



 12

(5.99)

 12

ª¬(t  2) 2 º¼  1 * 2

4 in the region t

Therefore the initialized RL derivative with a history function taken as fi (t ) is different from f (t )

2 to t

0 which

2

(t  2) is: 1

0 Dt

2

ª¬(t  2) 2 º¼

1

0 dt

2

d

ª¬(t  2) 2 º¼  ª¬\  4, 12 , 2, 0, t º¼ dt 8t

3

2



3 *  12





8t

1

2

*  12



4(t  2)

(5.100)

 12

*  12

The observation from the above expression (5.100) is that the singularity is shifted at t 2 from t 0 , by taking this initialization function as a constant of ( fi (t ) 4 ), from the start point of the function ( t 2 ) to that of the fractional derivative ( t 0 ) process. However, pushing the start point towards minus infinity, i.e. taking a makes the initialized RL derivative equal to the Caputo Derivative, exactly as follows: 1 1 d 2 2 1 2 2 0 Dt ª ¬(t  2) º¼ 0 d t ª¬(t  2) º¼  dt \  4, 2 , f, 0, t

8t

3

2



3 *  12 C 12 0 dt





8t

1

f ,

2

(5.101)

*  12

ª¬(t  2) 2 º¼

The exactness in the RL and Caputo derivatives is thus achieved by pushing the singularity to f , at least for this example. We have set the initialization function as \ i (t ) , which is derived from a history function in our case, which is

f i (t )

4 for t

f to t

0 , given in the following expression:

\ i (t )

d dt

ª¬\  fi (t ) º¼

t   f, 0@

fi (t )

4

(5.102)

234

Chapter Five

We calculate the value of the initialization function in this case in the following steps: d d \ i (t ) ª¬\  4,  12 , f, 0, t ) º¼ lim ª¬\  4,  12 , a, 0, t ) º¼ a pf dt dt 0 d ª

1

(t  W ) « a pf dt ³ *  1 ¬a 2 lim

 12 1

º

(4)dW »

¼

t

d ª

º   12  x x ( d ) « » ³ a pf *  1 dt 2 ¬t a ¼ d ª 8  12 8 1 º  t  lim (t  a ) 2 » « 1 a pf dt *  12 ¬ *2 ¼ 4

lim

§

4

lim ¨ 

a pf



4 *

1 2



1 © *2

t

 12

t

4



*  12



 12



f (0 ) * 1  D

t

(t  a )

 12

(5.103)

· ¸ ¹

D

This is much the same as using a constant function from time immemorial ( f ) and putting the formulation of the terminal initialization of the RL derivative for obtaining an initialization function to equate to the Caputo formulation, and to cancel the singularity function which appears as a difference to the un-initialized RL and Caputo derivatives. However, it is observed that this history function makes the first derivative at t 0 a discontinuous one -that is, with this type of history function, the integer order differentiability is lost. This restriction may not be physically viable for physical processes, even though mathematically it sounds correct!

5.11 Generalization of the RL and Caputo formulations and the initialization function The expressions obtained for 0  D  1 in the previous sections are: f (a) C D D D f (t ) f (t )  (t  a ) a dt a dt * (1  D )

>

D

@

>

C D 0 dt

> f (t ) @

@

t

> f (t )@  f (0)

D

(5.104)

d

 ª¬\  f , (1  D ), a, 0, t º¼ * (1  D ) dt This (5.104) may be extended by the above example for cases 1  D  2 ; the greatest integer for the RL and Caputo formulation is m 2 . As such, the initialization expression is as follows: 0 Dt

D

0 Dt

> f (t ) @

C D 0 dt

> f (t ) @  d

(t  0)

D

f (0)

* (1  D )

1D



(t  0)

f

(1)

(0)

* (2  D )

(5.105)

2

\  f , (2  D ), a, 0, t t!0 2 dt For the RL derivative initialized case to be equal to the Caputo derivative initialization, it needs to follow the below expression: 

(t  0)

D

f (0)

* (1  D ) Note that

d 2 ¬ª\  f ,  (2 D ), a ,0,t ¼º dt 2

1D



(t  0)

f

(1)

(0)



* (2  D )

d

2

dt

2

\  f , (2  D ), a, 0, t

0 (5.106)

> f (t )@ , with 1  D  2 . This implies the following: 2 d ª¬\  f , (2  D ), a , 0, t º¼ D a D0 > f (t )@ 2 D

a D0

dt

1

d

(5.107)

2 0

* (2  D ) dt

2

³ (t  W )

2 D 1

 f (W ) dW

a

Again considering the history function fi (t ) as an analytical for t d 0 (as in the Taylor series).it is as follows: f

fi (t )

¦ n 0

(n)

fi



(0 )

n!

t

n

(5.108)

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

235

We obtain the condition for the initialization as described below with fi (t ) z f (t )

1

d

2 0

* (2  D ) dt

2

³

(t  W )

2 D 1

0 and n

t





For the case of f (t ) derivative as D



(0 )

n!

n

W dW

D

t



(5.109)

1D

f (0 ) , and fi (1) (0  )



f (0 )  t f

¦

(n)

(1)



f (0 )  f (0 ) * (1  D ) * (2  D ) 1 in this case provides the history function that will make the RL and Caputo derivatives 

equal, and which will have the condition fi (0 ) to be fi (t )

fi

n 0

a



Equating the n

f

(1)



f

(1)



(0 ) . This causes the history function



(0 ) for f d t d 0 . 2

(t  2) and considering the fractional derivative starting point as t

1.5 , the history function will be fi (t )

0 and the fractional order

4t  4 , for t d 0 . Clearly, this history function will have a

second discontinuous derivative at t 0 . Again, this poses a physical restriction; as we look at the function from time immemorial as being very different from the original function (process) itself. Therefore, we can generalize the expressions obtained earlier as follows: f (a) D D C D (t  a ) , 0 D 1 a d t > f (t )@ a d t > f (t )@  * (1  D ) D

0 Dt

d dt d

> f (t ) @

C D 0 dt

> f (t )@  f (0)

t

D



* (1  D ) t

d dt

ª¬\  f , (1  D ), a, 0, t º¼ (5.110)

D

; ª¬\  f , (1  D ), a, 0, t º¼  f (0 ) * 1  D 

0

1

D

dt * (1  D )

³ (t  W )  f (W ) dW

t



 f (0 )

i

a

0 D 1 D

* (1  D )

;

0 D 1

Thus, by induction, for any integer m , those fractional RL and Caputo derivatives are initialized, for ( m  1)  D  m , as follows: D

0 Dt

C D 0 dt

> f (t ) @

> f (t ) @ 

m 1

t

n D

¦ *(n  D  1) f

(n)



(0 )

n 0

d

(5.111)

m

ª\ f , ( m  D ), a, 0, t º¼ m ¬  dt To make sure the RL and Caputo derivatives are the same, for t ! 0 we have the following rule: 

d

m

dt d

\  f , ( m  D ), a, 0, t m

m

dt

m

m 1



t

n D

¦ *(n  D  1) f

*(m  D )

³ (t  W )

m D 1



(0 )

n 0

0

1

(n)

m 1

 fi (W ) dW



t

n D

¦ *(n  D  1) f

(5.112) (n)



(0 )

n 0

a

Using the Taylor expression for fi (t ) and the history function for f d t d 0 , the expression is as follows: d

m

dt

m

0

1

³

* ( m  D ) f

(t  W )

m 1



t

m D 1

§ f fi( n ) (0  ) n · W ¸ dW ¨¦ n ! n 0 © ¹

n D

¦ *(n  D  1) f

n 0 (n) 

(n)

(n)

(5.113)



(0 )



The above formula (5.113) will thus be valid for fi (0 ) f (0 ) , and for n 0,1, 2, ...( m  1) . Therefore, for the generalized history function to have initialized the RL and Caputo derivatives as being the same as the fractional derivative, which is a polynomial of order ( m  1) for f d t d 0 , it must be represented as the following: ( m 1)



m 1

( m  2)



m2

(5.114)  ... This initialization law for the Caputo derivative that we developed provides a discontinuity for derivatives higher than ( m  1) for the function with a history at t 0 . f i (t )

f

(0 )t

 f

(0 )t

236

Chapter Five

5.12 Observations regarding difficulties in Caputo initialization and demanding physical conditions vis-à-vis RL initialization conditions The discussion in the earlier sections suggests that a Caputo inferred history has discontinuous integer order derivatives at the starting point of the fractional differentiation process. Meanwhile, the RL initialization is a smooth continuation of the function that is being fractionally differentiated, for the terminal initialization. There is a difference in the two RL initialized derivatives compared with the Caputo initialized ones, and it is profound at certain times (that is large). However, the functional forms of the two derivatives are identical in the case of a power function with time for times significantly larger than the start point of fractional differentiation. The RL un-initialized fractional derivative of the function, which begins at the start point of the function itself, is a smooth backward continuation of the initialized fractional RL derivative beginning somewhere other than the start point of the function, for terminal initialization. It is also noted that the RL fractional derivative, using the Caputo inferred history for initialization for f d t d c 0 , is the same as the Caputo fractional derivative. The two definitions of derivatives are the same if and only if the initialization part is zero or the derivative process for both the (1)

RL and Caputo derivatives starts at the birth of the function in our example, as we have f (t ) f (t ) 0 at t d a 2 . Otherwise, the two derivatives will be equated by a polynomial that is singular at the start point of differentiation ( t c 0 ). Say we have a differential equation

y (t )

C D 0 dt

> x (t ) @

(5.115)

Using the Caputo inferred history may be acceptable if: (a) it is found that the history is acceptable to the physics of the problem; (b) it is acceptable to have a discontinuity of the integer order derivative for x (t ) of the order m  1 , where ( m  1)  D  m . Suppose we consider a slightly complicated fractional order differential equation such as

y (t )

C 32 0 dt

1

> x(t )@  C0 dt > x(t )@  x(t )

(5.116)

2

and suppose we assume (for simplicity’s sake) that both the fractional derivative terms have the same history, i.e. xi (t ) , for t d 0 . For this example, the orders of the terms lay between different integer orders. For the first term of the fractional derivative, the order is 1  D 32  2 , and for the second fractional derivative, the order is between 0  D 12  1 . From our discussion on the generalization of the initialization, the history for the first term is xi (t ) and for the second term the history will be xi (t ) fractional terms is xi (t )

(1)



x (0) for f d t d 0 . Clearly the only history that will satisfy both (1)

x (0) , for f d t d 0 , with x (0)

derivative will be discontinuous at t



x (0)  t x (0)

0 (which is not the case, as with this history the first

0 ).

If on the other hand each has a separate history, each of the fractional differential terms in the equation (5.116) will be disconnected from one other, and will be acted upon by their own individual history, in negative time. Then at time (1)

zero, the entire individual position x (t ) and velocity x (t ) will have the same value, and under this condition and scenario the initialization of each term reverts to the case, as discussed for the case y (t )

C D 0 dt

> x(t )@ , (5.115).

This

elaborate discussion is as per theory proposed by Lorenzo and Hartley.

5.13 The Fractional Derivative of a sinusoidal function with a lower terminal not at minus infinity, and an initialization function As described in Section 4-13, the fractional derivative of sin O t from t q f Dt

f to t is:

>sin(Ot )@

q



O sin O t 

qS 2



(5.117)

meaning that, if the derivative (or anti-derivative) process begins from time immemorial, then we will get a steady state condition., and the periodic function will be phase shifted, either leading, or lagging by q  S2 . However, the situation is different if the lower limit of differentiation (integration) is shifted from time immemorial ( t

f ) to

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

237

some other time (say time t c ). In that case, we write the initialized differ-integral of a function as an uninitialized part, plus the initialization function as the following: q c Dt

q a dt

> f (t ) @

> f (t ) @  \ ( f , q , a , c , t )

(5.118)

where the function f (t ) 0 for t  a , and the initialization function is the result of differ-integrating the function from time a to c by order q , for the terminal-initialization case. The initialization function can be chosen as arbitrary as per the requirement for the side-initialization case, whereas the uninitialized differ-integral is q a dt

q

> f (t )@ { d[d t f a(t]) . q

The differ-integration of the sinusoidal function can be represented as follows (as discussed in Section-4.13 of this book):

d

q

>sin(t )@

t  a 

1 q

t  a 

3 q

sin  t  q  ..... (5.119) q d[t  a ] *(q) * (  q  2) This expression suggests that fractional differ-integration of the sinusoidal periodic function provides a phase change in the periodic part; and there are transients associated with the non-periodic part which die down at large times, becoming insignificant. Setting a 0 , we get: q

>

0 d t sin(t )

If we put q

1 2

S 2

@

t

1 q

*(q)



t

3 q

* (  q  2)

 ...

(5.120)

, and a z 0 , the semi-differentiation is:

32

t  a >sin(t )@ sin t  S4  *   12

 12

a dt

Taking a

sin  t  q S2 

7 2

t  a  *   52

f , the polynomial part in the above formula (5.121) becomes zero. Thus, we get:  12 sin(t ) sin  t  S4 f d t

>

Putting the value q



1 2

@

(5.121)

(5.122)

and a z 0 , we have semi-integration set as:  12

 12 a dt

In this case we are also putting a yielding:

>sin(t )@

sin  t 

S 4

Using scaling law (as described in Section-3.20), that is q

0 d t >sin( O t )@

1 , and O

t  a  *  12

52

t  a  *   32

 ...

(5.123)

f , which gives higher order terms in the remainder polynomial as zero,  12 f d t

Putting the value q

 ..

d

q

> f ( O x )@ >dx @q

O

>sin(t )@

O sin  t  q S2  O

q

t

1 q

*(q) 1 in the above expression (5.125), we get: 1 0 dt

>sin(t )@

(5.124)

q

> f ( O x )@ , we get: >d( O x )@q

q d

q

sin  t  S4

sin  t 

S

2



O

t

t

q

* (  q  2)

2

* ( 1)

3 q



t

d

(5.125)

4

* ( 3)

sin  t  S2  0

 ...

 ....

(5.126)

>sin(t )@ dt Here (5.126), we have used the properties of the reciprocal Gamma function, which are zero at all negative integer points (Section-1.10.11). cos(t )

Putting q

q

>

1 in the expression 0 d t sin(t )

@

sin  t  q S2 

t 1q *(q)



t 3q * (  q  2)

 ... , we get:

238

Chapter Five

1

0 dt

>

sin(t )

@

t

sin  t  S2 

0

* (1)



t

2

* ( 1)

 ....

sin  t  S2  1

(5.127) t

³ sin(t )dt

 cos(t )  1

0

The above formula shows that the expression for fractional differ-integration of sin(t ) is correct for one complete integer order of differentiation and one-complete integration of integer order. In the above derivation (5.125), we have assumed that the arbitrary initialization function is zero, i.e. \  sin(t ), q , a , c, t 0 , and the f (t ) sin t 0 , for

ta

0.

5.14 The Laplace transform of fractional differ-integrals 5.14.1 The generalization of classical Laplace transform formulas for differentiation and integration Here we form Laplace transforms of all d

D

> f ( x ) @ / dx D

for all D and differ-integrable functions f ( x ) , which are: f

­ dD f ( x ) ½ $® D ¾ ¯ dx ¿

^

³

e

D

d f ( x)

 sx

dx

0

D

dx

(5.128)

`

We wish to relate the above (5.128) with $ f ( x ) , which is a Laplace transform of f ( x ) (see Appendix-G for more details) defined as:

$ ^ f ( x )`

def f

 sx

³ e  f ( x ) dx

(5.129)

0

We note that $ ^ f ( x )` is a function in complex frequency s ; it is also expressed as $ ^ f ( x )`

F ( s ) . In terms of

engineering science, the variable x is the time variable t . For an inverse Laplace transform, we have a contour integration (see Appendix-G for further information) noted as:

$

1

^$ ^ f ( x)``

$

1

^F ( s )`

def

J  if

1

³  F (s) e

2S i J if

sx

ds

(5.130)

The expression (5.130) is derived from the use of the Fourier integral (see Appendix-G). From a classical Laplace transform of integer order calculus, we write the following for multiple derivative operations as:

­ d n f ( x) ½ n n 1 n2 d f ( x) s $ ^ f ( x )`  s f (0)  s n ¾ dx x ¯ dx ¿

$®

 ...s

0

d

0

n 1

dx

n 1

f ( x)

n

s $ ^ f ( x )` 

; n 1,2,3,..

(5.131)

x 0 n 1

¦

s

n 1 k

k

d f ( x) dx

k 0

.

k x 0

and for multiple iterated integrals as:

­ d n f ( x) ½ n s $ ^ f ( x )` n ¾ ¯ dx ¿

$®

n

0, 1, 2,...

(5.132)

We write the terms in summation of an expression in reverse of the above formula (5.131) and re-index it to give the following:

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms n 1 n2 ­ d n f ( x) ½ n d f ( x) 0 d s $® s $ ^ f ( x )`  s f ( x) n ¾ n 1 n2 dx dx ¯ dx ¿ x 0 x

....  s

d f ( x)

n2

dx

s

n 1

239

0

(5.133)

f (0)

x 0 n 1

n

¦s

s $ ^ f ( x )` 

k 0

k

d

n 1 k

dx

f ( x)

n 1 k x 0

We note that both of the above formulas ((5.131) and (5.133)) are similar and can be transformed into a common formula, (for n 0, r1, r2, .. ) which is as follows: n 1 k n 1 ­ d n f ( x) ½ n f ( x) k d $® s $ ^ f ( x )`  ¦ s n ¾ n 1 k dx k 0 ¯ dx ¿

(5.134) x 0

Here in the above formulation (5.134), the upper summation limit is written as n  1 , and may be replaced by any integer larger than n  1 and even by f . In (5.134), the effect is to add terms containing

d 1 f ( x ) dx 1

x 0

,

d 2 f ( x ) dx 2

x 0

etc.

for n 0, 1, 2,... ; such terms are necessarily zero for any function f ( x ) whose Laplace transform exists. If n is a non-integer, say D , we generalize the above formula for all D and write the following D 1 k n 1 ­ dD f ( x ) ½ D f ( x) k d $  s f ( x ) s ^ ` ¾ ¦ D D 1 k dx k 0 ¯ dx ¿

(5.135)

$®

x 0

where, n is the largest integer such that ( n  1)  D d n . Notice that the sum in (5.135) is zero when D d 0 . In satisfying the above (5.135) generalization, first consider D  0 , i.e. it is a fractional integration, for which the RL formula is: D

d f ( x) dx

x

1

D

1

* ( D )

D ³ ( x  y )  f ( y ) dy

(5.136)

D 0

0

We apply a convolution theorem of Laplace transforms that results in:

$

^³ 

$

^ f1 ( x) * f2 ( x)`  $ ^ f1 ( x)`  $ ^ f2 ( x)`

x

0

`

 $ ^ f1 ( x)`  $ ^ f2 ( x)`

f1 ( x  y )  f 2 ( y ) dy

(5.137)

to the formula given in (5.136), and write for D  0 as follows :

­ dD f ( x ) ½ 1 1D $ ^x ` $ ^ f ( x)` sD $ ^ f ( x)` D ¾ D *  d ( ) x ¯ ¿

(5.138)

$®

^ `

We have used known Laplace transforms, i.e. $ x

* ( m 1)

m

s m1

. We see that for D  0 , negative integers from the

generalized formula described above in (5.138) remain unchanged. For the positive D , i.e. fractional differentiation, we have the RL composition as:

d

D

> f ( x )@ dx

D

Here, n is a positive integer such that ( n  1)  D  n . using $

^ ` dn f ( x ) dx

n

n

s $ ^ f ( x )` 

n 1

k

¦s ¦ k 0

n 1 k

s 0

k d n1k f ( x ) dx n1k

x 0

ª dD  n > f ( x ) @ º » n « D n dx ¬ d x ¼ d

n

(5.139)

Now on the application of the formula as obtained for n

0, r1, r2,... , (5.134), we find that:

­ d n ª dD  n f ( x ) º ½ ­ dD f ( x ) ½ $ ® n« D ¾ D n » ¾ ¯ dx ¿ ¼¿ ¯ dx ¬ dx ­ dD  n f ( x ) ½ n 1 k d n 1 k ª dD  n f ( x ) º n s $® s  n 1 k « D n ¾ ¦ D n » ¯ dx ¿ k 0 dx ¬ dx ¼x

$®

The difference D  n  0 is a negative. The first term of the RHS of

$

^

dD n f ( x ) dxD n

`

s

D n

(5.140) 0

(5.140) is evaluated using

$ ^ f ( x )` as obtained above in (5.138) for D  n  0 . Since D  n  0 , the composition rule may

be applied to the second term of (5.140), within the summation sign, and we therefore write:

240

Chapter Five D 1 k n 1 ­ dD f ( x ) ½ D f ( x) · k §d $® s $ ^ f ( x )`  ¦ s ¨ D ¾ D 1 k ¸ k 0 ¯ dx ¿ © dx ¹x

(5.141)

0

0  D z 1, 2, 3,... Therefore, we have proved our generalization process. The Laplace transform formula in (5.141) is a very simple generalization of the Laplace transform formula of classical integral calculus; that is, of the Laplace transform of the derivative or the integral of a function f ( x ) .

5.14.2 Generalization is not possible for a few classical Laplace transform identities A point may be noted, that no generalization exists for the following classical formulas of integration and differentiation of the Laplace transformed function $ ^ f ( x )` F ( s ) : s

^ ` f ( x)

$ 

1

³

x



^

d

`

^

ds

f

$  x  f ( x)

ds d

`

n

§ d 1$ ^ f ( x )` · ¸ 1 ds © ¹s

d $ ^ f ( x )`

$ ^ f ( x )` ds

¨

1

f

$ ^ f ( x )`

(5.142)

n

$ ^ f ( x )` n 1, 2, 3, ... n ds To prove this statement, that the generalization in (5.142) is not possible for fractional differ-integration of the Laplace $ ( x) f ( x)

$ ^ f ( x )`

transform function; let us take for example F ( s )

^ `

Laplace transform identity i.e. $ x

m

* ( m 1) s m1

s

E

with 1  E  0 . Using the expression of the

, we recognize that for F ( s )

Now we carry out fractional differ-integration for both sides of $ ^ f ( x )`

d

D

D

ds

D

$ ^ f ( x )`

E

d ª¬ s º¼ ds

* ( E  1) * ( E  D  1)

* ( E  1)

D

* ( E  D  1)

­ x

s

s

s

E D

^ `

^

^ csc  (D  E )S x  E 1 *(E )

* ( m  1) s

 ( m 1)

½ * ( E  1) D  E 1 $ ^x x ` ¾ ¯ * (D  E ) ¿ * ( E  D  1)* (D  E )

* ( E  D  1)* (D  E )

f ( x)

m

$®

* ( E  1)* (  E )

 *(X )  *(X  1)

.

as in the following steps:

, $ x

`

 E 1 $ x *x(  E ) ; * ( E  D  1)* (D  E )

We used the formula

x  E 1 *(E )

we have f ( x )

D  E 1

* ( E  1)* (  E )

csc(  ES )

E

E

D

^

f ( x)

x

 E 1

(5 .143)

*( E )

`

D

$ x f ( x)

D

`

$ x f ( x)

^ `

S csc(SX ) and the Laplace transform identity $ x

m

* ( m 1) s m1

with

as depicted in the above steps (5.143).

^

D

`

dD ¬ª$ ^ f ( x )`¼º

dD dsD

ª¬$ ^ f ( x )`º¼

This derivation shows that the fractional differ-integration of $ x f ( x ) z an integer. This we proved via the example of f ( x ) does not hold for any function f ( x ) .

x  E 1 *(E )

. Therefore

dsD

, similar to cases with D as



csc(  ES ) csc (D  E )S

$ ^x

D

`

f ( x)

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

e  kx

5.14.3 Evaluating the Laplace transform for function f ( x)



D

0 Dx

241



ª¬e kx f ( x) º¼ , D d 0 by using the

Riemann-Liouville integral formula Let us evaluate the Laplace transform below:

­

$ ®e

 kx

¯

§ dD kx ·½ D ¨ dxD ª¬e f ( x )º¼ ¸¾ ( s  k ) $ ^ f ( x )` ; © ¹¿

0 , the result is a trivial identity, i.e. at k

For D

0 , we have $

^

dD > f ( x )@ dxD

`

D d0

(5.144)

D

s $ ^ f ( x )` . For D  0 , we use the

RL definition of a fractional integral and write the following:

­° e  kx x e ky f ( y ) ½° ­  kx dD kx ½ ªe f ( x )º¼ ¾ $ ® dy ¾ $ ®e D D 1 ³ dx ¬ °¯ * ( D ) 0 ( x  y ) ¯ ¿ ¿° x k ( x y ) · 1 °­ § e °½ f ( y ) dy ¾ $ ®³ ¨ D 1 ¸  * ( D ) ¯ ° 0 © ( x  y) ¹ ¿° 1 * ( D )

$ ^e

 kx

/x

(5.145)

`  $ ^ f ( x)`

D 1

D

( s  k ) $ ^ f ( x )` In (5.145), we have used the definition of the Laplace transform of a convolution integral of two functions, that is

$

^³  x

0

`

f1 ( x  y )  f 2 ( y ) dy

tables, we have used $

^ ` e kx xD 1

 $ ^ f1 ( x)`  $ ^ f2 ( x)` .

Also, having drawn upon known Laplace transform

D

* ( D )( s  k ) .

5.15 Approximate representation of a fractional Laplace variable by rational polynomials in the integer power of the Laplace variable In the fractional order Laplace transform, we will encounter the fractional power of Laplace variable s , like s

s

D

.

For example,

s represents a semi-differentiation, whereas

fractional Laplace variables, s as

P(s) Q(s)

D

1 s

D

or

is a semi-integration operation. These

for D  \ , are irrational functions of s ; which we shall express first approximately

, where P ( s ) and Q ( s ) are polynomials in an s -variable; and second by simple polynomial expansion. This

is called a rational approximation for irrational functions. For the first representation, we have used a Continued Fraction Expansion (CFE) method as one way to represent the irrational function. The CFE formula is as follows: def 1 D (5.146) (1  x ) x 1D x 1   12 (D  1) x 1   16 (D  1) x 1   16 (D  2) x 1   101 (D  2) 1  .... The other way to write (5.146) is: CFE 1 D x (1  D ) x (1  D ) x (2  D ) x (2  D ) x D (1  x ) (5.147) 1 1 2  3 2  5  ............ CFE converges in a finite complex plane along a negative real axis for x f to x 1 . The formula given in (5.146) is an infinite expansion, although we may terminate it into a few terms, in order to approximate the LHS. For approximating say

s , set x

s  1 and set D

1 2

; and by use of these two terms we get:

242

Chapter Five

s|

3s  1

(5.148)

s3

By the use of four terms of the CFE formula, we get: 2

s| For six terms, we have

s|

7 s 3  35 s 2  21s 1 s 3  21s 2  35 s  7

; for eight terms,

5 s  10 s  1

(5.149)

2

s  10 s  5 s|

11s 5 165 s 4  462 s 3  330 s 2  55 s 1 s 5  55 s 4  330 s 3  462 s 2 165 s 11

approximate the fractional Laplace operator as a ratio of polynomials. For semi-integration

.

In this way, we

1 s

, the ratio will be

s (5.148), (5.149).

‘reciprocal’ for that which we obtained for

The second means of approximation is via the Power Series Expansion (PSE) method, as depicted by the following formula: D (D  1) 2 D (D  1)(D  2) 3 D (1  x ) 1D x  x  x  ... 2! 3! (5.150) f (D ) n n x n! n 0

¦

n

Where D 

D (D  1)...(D  n  1)

(D ) n is a (truncated) falling factorial, such as a Pochhammer number (which

we discussed in Section-1.9). Again, when approximating

s , set x

s  1 and D

1 2

; and with PSE we get two

terms, set as the following: 1 1 1  2 s | 1  12 ( s  1)  2 22! ( s  1)

(5.151)

2

 18 ( s  6 s  9)

5.16 Generalized Laplace transform We have seen two types of fractional derivatives, and previously discussed them in Section-4.18. The following formula gives us a fractional derivative definition with an in-between RL-Caputo type, with E as a type-defining parameter 0 d E d 1 . When E

E

1 , the definition is a Caputo derivative with representation

0 gives an RL fractional derivative with representation

D Dt

C

D

Dt

> f (t )@ , while

> f (t )@ . The generalized definition (as shown in

Section-4.18) is as follows: E

D

0 Dt

> f (t ) @

E (1D )

0 It

ªd (1 E )(1D ) > f (t )@ º» «¬ dt  0 I t ¼

(5.152)

The fractional order D is the order of a derivative; here in this definition it is 0  D  1 and E is the type of derivative with 0 d E d 1 . Therefore, the nearest integer is n 1 in the above generalization (5.152). The two types of fractional derivatives are the Caputo type and the Riemann-Liouville type. We previously generalized these two types in Section-4.18 using the type parameter E . The generalized Laplace transform in integer order theory (5.131), for n  fold differentiation, is:

­ dn

$®

¯ dx

Here, we use the symbol f

( n 1 k )

(0)

d n1k dx n1k

f ( x)

n

x 0

½

f ( x)¾

¿

n

s F (s) 

n 1

¦s

k

f

( n 1 k )

(0)

(5.153)

k 0

, where n  ` is a natural number.

5.16.1 Laplace transform for the Riemann-Liouville fractional derivative Following this, we have generalized the above (5.153) for fractional order D , with ( n  1)  D  n for the RiemannLiouville derivative as follows:

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

$

^

D

0 Dx

243

­ d n n D ½ $ ® n 0 I x f ( x)¾ ¯ dx ¿

`

f ( x)

n

s $

^

n D 0 Ix

n 1

` ¦s

f ( x) 

d

k

dx

k 0

n  ( n D )

s s

$ ^ f ( x )` 

n 1 k

n 1

n 1 k

¦ s D k

ª¬ 0 I xn D f ( x )º¼ x

ª¬ 0 Dx

n 1 k x

 ( n D )

k 0

n 1

D

s F (s) 

¦s

D 1 k

k

0 Dx

0

f ( x ) º¼

(5.154)

x 0

f (0)

k 0

D

s F (s)  f

$

^

D 1 k

The symbol 0 Dx

D

`

n 1

D

0 Dx f ( x )

(D 1)

¦s

s F (s) 

k

(D  2)

(0)  s f (D 1 k )

f

(0)  .......  s

n 1

f

(D  n )

(0)

(0)

k 0

D 1 k

f (0) is represented as 0 Dx

dD 1k dxD 1k

f (0) {

f ( x)

x 0

{ f

(D 1 k )

(0) .

5.16.2 Laplace transform for the Caputo fractional derivative For the Caputo derivative, the Laplace transform can be derived using the above method (5.154), shown as:

$

^

­

`

C D 0 Dx

f ( x)

n

d f ( x) ½

n D

$ ®0 Ix

dx

¯

s

 ( n D )

s

 ( n D )

$

^

n 0 Dx

¾ ¿

n

`

f ( x)

(5.155)

k n 1 ª n º n 1 k d f ( x ) « s $ ^ f ( x )`  ¦ s » k dx k 0 x 0» ¬« ¼ n 1

D

s F (s) 

¦

s

D 1 k

k

d f ( x) dx

k 0

k x 0

Thus, for the Caputo derivative, we have the Laplace transform expression as follows:

$

^

C D 0 Dx

`

f ( x)

D

n 1

s F (s) 

¦ sD

1 k

f

(k )

(0)

(5.156)

k 0

D

s F (s)  s

D 1

f (0)  s

D 2

f

(1)

(0)  ..........  s

D n

f

( n 1)

(0)

5.16.3 Requirement of the fractional order of the initial states of the RL derivative and the integer order (classical) initial states for Caputo derivative initializations D 1 k

The essential difference between the two is that the RL derivative requires 0 Dx

f (0) , k

0,1, 2...( n  1) known as

(k )

fractional initial states, whereas the Caputo requires f (0) , k 0,1, 2,...( n  1) , known as (usual) integer order initial states. For example, in the case where 0  D  1 , the Laplace transforms for RL and Caputo are expressed as follows:

^ $^

$

D

0 Dx

C D 0 Dx

` f ( x )`

D

f ( x)

s F (s)  f D

s F (s)  s

(D 1)

D 1

(0)

(5.157)

f (0)

One may generalize the above two fractional derivatives ((5.154) and (5.157)) as a solution g ( x ) of the integral equation: D

0 Ix

> g ( x )@

1

x 1

D ( x  u )  g ( u ) du ³ * (D )

f ( x)

D !0

(5.158)

0

in cases where it exists. Let us denote the solution by adding equivalent for n  1 d D d n :

* D 0 Dx

f ( x ) , meaning that the following assertions are

244

Chapter Five D

G ( s ) , where $ ^g ( x )`

a) s F ( s ) b) g ( x ) c) f

* D 0 Dx

(D 1 k )

G ( s ) and $ ^ f ( x )`

F (s)

f ( x)

( x ) are equal to ZERO at x

0,1, 2,...( n  1) ,

0 for k

together with

D

0 Dx

f ( x)

g ( x ) , the

Riemann-Liouville derivative. (k )

d) f ( x ) are equal to ZERO at x derivative.

0,1, 2, 3,...( n  1) , together with

0 for k

C D 0 Dx

f ( x)

g ( x ) , the Caputo

5.16.4 Generalized Laplace transforms formula for a fractional derivative with a type parameter With type parameter 0 d E d 1 , we have a generalized definition in the Laplace transform, for a fractional order derivative of order 0  D  1 :

$ Where

(1 E )(D 1) 0 Dt

^

E D a Dt

> f ( x)@`

D

s F (s)  s

E (D 1)



(1 E )(D 1) 0 Dt

> f ( x)@ x 0

(5.159)

0  D  1 , and for

f ( x ) is the Riemann-Liouville fractional derivative for the fractional order

any type described by E we define the fractional derivative as: E

D

a Dx

d

 dx

E (1D )

> f ( x )@

a Ix

(1 E )(1D ) a Ix

> f ( x ) @

(5.160)

5.16.5 Generalized Laplace transform demonstrated for use in a fractional differential equation As an example, consider the differential equation of order 0  D  1 , with type variable 0 d E d 1 , and with K as a constant. The given initial condition is: (1 E )(D 1) 0 Dt

For RL case we have E

(D 1) 0 Dt

0 , the initial condition is

f (t )

f (t )

(1 E )(1D ) 0 It

t 0

f (0)

f0

t 0

(5.161)

f0 that is fractional order initial state (5.161).

t 0

This implies in RL case we write 1  D order fractional integration of function f (t ) at the initial point is a constant

f0 . Therefore, the initialization function at t

D  1 of the constant function f0 ; that is f (t ) t For Caputo case we get initial condition as

0 0 Dt

0 for total solution, is obtained by fractional integration by order D 1 f0 *t (D ) .

0

f (t )

f (t ) t

t 0

1 in (5.161) we write usual

f0 by putting E

0

integer order initial state. We will use these concepts and demonstrate in the use in fractional differential equation. We have a fractional differential equation expressed as the following: E

D

0 Dt

> f (t ) @

(5.162)

K

Using a generalized Laplace transform identity as obtained above (5.159), we write the Laplace transformed equation, (1 E )(1D )

recognizing that 0 I t

{ 0 Dt(1 E )(D 1) , as the following: $

^

E

D

0 Dt

> f (t )@`

D

s F (s)  s D

s F (s)  s

E (D 1)

E (D 1)

$ ^K `



t!0

(1 E )(D 1) 0 Dt

> f (0)@

K s

(5.163)

K

f0

s

giving: K

f0

K

f0

(5.164) s s s s Doing an inverse Laplace transform of the above obtained F ( s ) (5.164), using a known identity, i.e. F (s)

$

1

^

* (D 1) sD 1

`

D

t , we get the following solution for t t 0 :

D 1



D  E (1D )

D 1



 (D  E DE 1) 1

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

K

f (t ) For an RL type, i.e. with E

f0

D

t 

* (D  1)

*  (1  E )(D  1)  1 K * (D  1)

D

D

>K @

giving

f un-init (t )

K

tD * (1D )

f0

D

t 

We observe that in (5.166) the first term is uninitialized solution, got from 0 Dt 0 It

(1 E )(D 1)

(5.165)

0 , we have the solution for t t 0 :

f (t )

i.e.

t

* (D )

> f (t ) @

(D 1)

t

(5.166)

K by fractional integration,

. The second term of (5.166) is initialization function; we get from (D 1)

fractional initial state given by

(D 1) 0 Dt

We note that (5.166) is f (t )

f un-init (t )  finit (t ) . We also observe that for 0  D  1 the RL case has f (t ) t

f (t )

t 0

f0 ; that is by finding 0 I t

f0 , that we write as finit (t )

We verify by obtaining RL fractional derivative of order D for (5.166), that is K  f0 t D

original fractional differential equation with RL derivative i.e. 0 Dt For a Caputo type, i.e. with E

> f (t ) @

1 * (D ) * (0)

tD 1 * (D )

f0 .

0

f.

K . We obtain the

K.

1 , in (5.165) we have the solution:

K

f (t ) In this case the uninitialized function is f un-init (t ) case, and the finit (t )

245

f0 . The (5.167) is f (t )

tD

K

* (1D )

D

* (D  1)

(5.167)

t  f0

, that we get by

D

0 It

> K @ ; as we discussed in the RL

f un-init (t )  finit (t ) . Taking Caputo fractional derivative of order D

for (5.167); recognizing Caputo fractional derivative of constant is zero, we get the original fractional differential equation, that is

C D 0 Dt

> f (t ) @

K . In this case we observe f (t ) t

f0 .

0

Consider another example as a generalized fractional differential equation: E

D

0 Dt

with the given initial condition as

ª¬ 0 I t(1 E )(1D ) f (t ) º¼ t

 K  f (t )

f (t )

(5.168)

f0 with K as a constant, with 0  D  1 and 0 d E d 1 .

0

Using the generalized Laplace transform identity as in the above example we get the following steps: E

D

0 Dt

^

$

 K  f (t )

f (t )

E

D

0 Dt

> f (t )@`

D

s F (s)  s

E (D 1)

$ ^ K  f (t ) `



(1 E )(D 1) 0 Dt

D

s F (s)  K  F (s)

s

D



 K F (s)

s

s

E (D 1)

E (D 1)

t!0

> f (0)@

K

(5.169)

f0

f0

Therefore, we have the following expression in the Laplace domain:

F (s)

s

E (D 1)

f0

s

D D  E (1D )

D

D

s K

s K s

D J

f0 (5.170)

f0 D s K where J D  E (1  D ) . To get an inverse Laplace transform, we re-write the above expression (5.170) as depicted in the following steps:

246

Chapter Five

s

F (s)

D J

Ks

f0

D

s

§ 1 · ¨ D ¸ f 0 © Ks  1 ¹

J

   



J

1  Ks

f0 s

J

1  Ks



f0 s

J

f



 (K )s

¦ ( K )

f0

D 1

f0 s

k

s

D

D 2

  Ks



2 2D J

D  J

(5.171)

 ...

 (K ) s

 ...

D k  J

k 0

We used (1  x)

1

2

3

1  x  x  x  ... with x

Ks

D

. Then by using $

^ ` tD 1 * (D )

1

s

D

, and applying this term

by term for the above expression, we obtain the response as: f

­

1

f (t )

$ ® f0

¯

¦ ( K ) ¦

( K ) t

* (D k  J )

k 0

J 1

½ ¾ ¿

k D k J 1

f

t

s

k 0

f0 f (t )

k D k  J

(5.172)

D k

f

(  Kt )

¦ *(D k  J ) k 0



f0 t

(1 E )(D 1)

D k

·

E

D ,D  E (1D ) (  Kt

D

)

From this, we write:

§

f (t )

f0 ¨ t

J 1

©

f

(  Kt )

¦ *(D k  J ) ¸ ¹

k 0

§

f0 ¨ t

D  E (1D ) 1

©

§

f0 ¨ t



(1 E )(D 1)

·

(  Kt )

¦ * D k  D  E (1  D ) ¸

(D 1)(1 E )

(5.173)

¹

k 0

©

f0 t

D k

f

D k

f

·

(  Kt )

¦ * D k  D  E (1  D ) ¸ ¹

k 0

E

D ,D  E (1D ) (  Kt

D

)

Here, we have used the definition of a two parameter Mittag-Leffler function (refer to Appendix-A), that is: f

Ea ,b ( x )

x

k

¦ *(ak  b)

a!0

b!0

k 0

1 * (b )



x *( a  b)



x

(5.174)

2

* (2 a  b )

 ...

for all a ! 0 and b  ^ . This is an entire function and completely monotone for 0  a d 1 and b t a . Also, note that Ea ,b (0) * (1b ) (as described in Appendix-A). 1) For K

0 , the result is f (t )

f0 (1 E )(D 1) t *  (1 E )(D 1) 1

the RL case, then the solution is f (t ) Take

D

0 Dt

ª¬t D 1 º¼ , which is

solution is f (t ) 2) For E

* (D ) 1 * (0) t

, because Ea ,b (0)

1 * (b )

. In this case, if E

0 , that is

f0 D 1 . * (D ) t

and is zero, since  * (0)

1

0 . For E

1 in the Caputo case, the

f0 , revealing that the Caputo fractional derivative of a constant is zero.

1 in the Caputo derivative type, f (t )



D



f0 ED (  Kt ) , where ED ( x )

one parameter Mittag-Leffler function. For the RL case with E

ED ,1 ( x ) , and it denotes the

0 , the solution is f (t )

f0t

D 1

D

ED ,D (  Kt ) .

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

247

These examples make us think about the concept of ‘stationarity’ in a fresh way, within the domain of fractional (1 E )(D 1) 0 Dt

calculus, leading us to initial conditions, such as

f (t )

t 0

. In addition, we saw that, depending on the type

of fractional derivative which composes a fractional differential equation, the solution varies.

5.17 Generalized stationary conditions Consider a fractional ‘stationary’ differential equation with order 0  D  1 and type 0 d E d 1 as the initial condition shown as

0D

(1 E )(D 1)

E

D

0 Dt

f (t )

0 , with

f0 , which is a fractional Riemann-Liouville derivative at the initial

f (0)

condition. We can re-write this initial condition in order to state it as a fractional integral initial value given as a constant, that is as

1 0 Dt f

E

(1 E )(1D ) 0 It

0 , i.e. f (0)

(t )

f0 . For D

f (0)

1 , the conventional definition of the stationary condition is recovered

f0 is a constant, for any type E . The solution of this fractional differential equation is

as follows:

f0

f (t )

*  (1  E )(D  1)  1

t

(1 E )(D 1)

(5.175)

This solution was stated in the previous section. This may also be seen by inserting the above expression of f (t ) into the generalized definition of a fractional derivative as follows:

ªd (1 E )(1D ) > f (t )@ º» «¬ dt  a I t ¼ D O for 0  D  1 and using the fractional integral of the power function expression a I t ª¬ (t  a ) º¼ E

D

a Dt

For the RL case with E

f (t )

E (1D )

> f (t ) @

f0 D 1 , * (D ) t

0 , the function f (t )

(5.176)

a It

and for the Caputo case with E

* ( O 1) * (D  O 1)

D O

(t  a )

.

1 , the function is

f0 , a constant.

Note that the integral

(1 E )(1D ) 0 It

varies. In particular, lim t p0 f (t )

f0 remains conserved and constant for all t , while the function f (t ) itself

f (t )

f and lim t nf f (t )

0 . For E

1 and D

1 , one recovers f (t )

f0 as usual.

The fractional derivative of the constant for the Caputo type is zero, and the integer order derivative of the constant of any type is zero. The new type of stationary states (in the initial condition) that a fractional integral (rather than the function itself) is a constant that is arrived at. This seems to rather lack knowledge about the fractional ‘stationary-conditions’ and is particularly responsible for the difficulty of deciding which type of fractional derivative should be used when generalizing a traditional equation of the motion of thermodynamics and electrodynamics control systems, among others. Nevertheless, let us carry on with this dichotomy.

5.18 Demonstration of the generalized Laplace transform for solving the initialized fractional differential equation 5.18.1 Composition of a fractional differential equation for a Riemann-Liouville fractional derivative Say we need to find the solution of

1

0 Dt

2

> f (t )@  bf (t )

0 for t ! 0 . We must apply the Laplace transform rule for

an RL fractional derivative, which is:

$

^

D

`

n 1

D

0 Dt f (t )

s F (s) 

¦s  k

D 1 k

0 Dt

f (0)

k 0



(5.177)

( n  1)  D  n

With is $

^

1 2

D 1

0 Dt

2

,

we

> f (t )@`

have 1

n 1,  12

s 2 F ( s )  0 Dt

so

we

get

$

^

1

0 Dt

2

> f (t )@`

1

s 2 F (s) 

¦

0 k 0

s

k

D 1 k

0 Dt

f (0) ,

that

f (0) . Using this expression in the given fractional differential equation, we get

the Laplace transformed equation as follows:

248

Chapter Five

s

1

2



 12

F ( s )  0 Dt

0

(5.178)

> f (t )@ at t

0 ; meaning the value of the

f (0)  bF ( s )  12 0 Dt

Here we require a fractional initialization state, that is, the value of

 12

semi-integral of a function at the initial point. Say that the value is given as a constant, namely 0 Dt

f (0)

C , with

this we write: 1

s 2 F ( s )  C  bF ( s )

From the above (5.178), we get F ( s )

C s b

(5.179)

0

; the inverse Laplace transform of this (tables of Laplace transforms of

the Mittag-Leffler function, refer to Appendix-A) gives the following result:

f (t )

$

1

^ ` C t C s b

 12

E1

 b t

1 2, 2

(5.180)

The solution is formed with a two-parameter Mittag-Leffler function. The two-parameter function is defined as

ED , E ( at )

¦

f

ak t k k E ) ( D * k 0

; D , E ! 0 (refer to Appendix-A). The Laplace transform is given by the following

relationship (again, see Appendix-A):

^

D k  E 1

^

D 1

$ t $ t

^

$ t

1 2

(k )

D

s

`

ED , E ( at )

ED ,D ( at )

k  12

(k ) 1 2, 2

E1

k!

D

(s  a) 1

`

D

D E

D

s a

 a t `

k 1

;

E, k

D k!

( s  a)

1

D

; Re( s ) ! a

D

k 1

(5.181)

0

E

In the above expressions k 0,1, 2... , indicating an integer order whole derivatives of the Mittag-Leffler function. From the Laplace transform relationships, we set k 0 and a b for our case to find the solution. In addition, we

$ ^FD ( b, t )` , where FD ( k , t ) is the Robotnov-

1 sD  b

note from the Laplace transform tables (Appendix-A) that

Hartley function. Thus, the other form of solution of the above is as follows:

f (t ) We observe the FD  b, t

t

D 1

$

1

^ ` C F C s b

1

 b, t

2

(5.182)

D

ED ,D ( bt ) relationship of the Robotnov-Hartley function and the two parameter

Mittag-Leffler function; we write the series form which is given as FD ( b, t ) This can also be cast with a generalized R function i.e. R 1

2 ,0

t

D 1

 b, 0, t , where

D

ED ,D ( bt )

RD ,v [ a , c, t ]

t

D 1

¦

¦

f n 0

(  b )n tD n * n 0 (D n D )

f

.

( a )n ( t  c )( n1) D 1v *  ( n 1)D  v

(refer to Appendix-A).

5.18.2 Composition of the fractional differential equation for the Caputo fractional derivative Now we compose the same differential equation with the Caputo derivative as Laplace i.e. $

^

C D 0 Dx

transform

`

f ( x)



1

rule

s F (s)  s

D 1

 12



D

equation as s 2 F ( s )  s

of

f (0)  s

f (0)  bF ( s )

D 2

the f

(1)

(0)...  s

Caputo D n

f

( n 1)

C 12 0 Dt

> f (t )@  bf (t )

derivative

(0) , we get for D

0 . With the initial condition f (0) F (s)

K

s s

 12

1

2

0 . By using the

(as 1

2,

in

5.156);

the Laplace transformed

1 a constant at t

0 , we write:

K

(5.183)

b

^  `

Using the Laplace transform identity of one parameter of the Mittag-Leffler function (i.e. $ ED O t

D

sD 1 sD  O

), we

write the solution as:

f (t )

E1

2

 b t

(5.184)

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

249

5.18.3 Applying the concept of the initialization function of fractional derivatives in a fractional differential equation Now we apply the concept of an initialized fractional derivative with an initialization function that we have evolved as \ (t ) . We write: 1

0 Dt

The given fractional differential equation (i.e.

1

0 Dt

2

2

1

> f (t ) @

> f (t )@  bf (t )

0 dt

2

> f (t ) @  \ (t )

(5.185)

0 ) is, therefore, as follows in terms of the

initialization function:



1

0 dt

2

> f (t ) @  \ (t )  b  f (t )

(5.186)

0

1

The term 0 d t 2 > f (t )@ is an uninitialized semi-derivative with an initial condition of zero, and its Laplace transform is thus

$

^d

1

0 t

2

> f (t )@`

1

s 2 F ( s ) . The Laplace transform of the fractional differential equation is, with

$ ^\ (t )` \ ( s ) , as follows: 1

s 2 F ( s )  \ ( s )  bF ( s ) \ (s)

(5.187)

0

From here, we write F ( s )



initialization we give \ (t )

CG (t ) , that is, the initialization function is a delta function at t

\ (s)

$ ^\ (t )`

s b

$ ^CG (t )`

; the inverse Laplace transform of this will give f (t ) . Specifically, for a side-

C .

Using this initialization, we get F ( s )

transform will give the same answer as we got above (namely, f (t )

Ct

 12

E1

For any other \ (t ) , we will have a different result. We note that F ( s )

$

1

^ ` 1 s b

R1

2 ,0

C s b

 b t

1 2, 2



,

0 , which gives us

and the inverse Laplace

CF1 ( b, t ) ) in (5.182). 2

 \ (s) . In addition, we note that 1 s b

 b, 0, t , in terms of the ‘Generalized - R ’function defined below with its Laplace transform: n

f

RD ,v [ a, c, t ]

¦

( a ) (t  c )

*  ( n  1)D  v

n 0

^

`

^

`

s

$ RD ,v ( a, 0, t )

( n 1)D 1 v

v

D

s a e

{ RD ,v [ a, t  c ]

; Re >(D  v )@ ! 0 ; Re > s @ ! 0

(5.188)

 cs v

s

; c t 0 Re >(D  v )@ ! 0; Re > s @ ! 0 s a Note that F ( s ) is a product of two functions in the Laplace domain (complex frequency s -domain) and that is a convolution integral in the time domain. Therefore, for a general solution we write the convolution as follows:

$ RD ,v ( a, c, t )

f (t )

D



 R1

2 ,0

 b, 0, t * \ (t ) (5.189)

t

³ ,0  b, 0, (t  W ) \ (W ) dW

 R1

2

0

This demonstrates the use of generalized Laplace transforms and initialized fractional derivatives in solving fractional differential equations. We will devote the next few chapters solely to various types of fractional differential equations. As in the earlier case, the f (t ) can be casted via the use of a two parameter Mittag-Leffler function, and also via the use of a Robotnov-Hartley function, meaning there are several functions (refer to Appendix-A), which are used for fractional differential equations.

5.19 Fourier transform of a fractional derivative operator 5.19.1 Fourier transform from a Laplace transform The Laplace transform of a fractional derivative-integral of order D operation is:

$

^

D

0 Dx

> f ( x)@`

D

s $ ^ f ( x )` 

n 1

¦s  k 0

k

D 1 k

0 Dx

f ( x)



(5.190) x 0

250

Chapter Five

Where the Laplace transform is defined as: f  sx

$ ^ f ( x )`

³ e  f ( x ) dx

(5.191)

0



In the Laplace transform definition above, the order of differ-integration D  \ ; and the positive integer n  ] is such that ( n  1)  D d n . In this expression, when D  0 (that is for fractional integration), the term involving summation in the above expression becomes zero for any function, f ( x ) with the available Laplace transform, i.e.

$ ^ f ( x )` . In addition, one can have something similar to the Laplace transform of fractional differ-integrals of f ( x ) ; a Fourier transform of a fractional differ-integral operation. For a function f ( x ) , which is “well-behaved” at x f , we can have:

^



> f ( x)@`

D

f Dx

D

(iZ ) 

^ f ( x)`

(5.192)

and therefore we have a fractional derivative/integral operation as an inverse Fourier transformed one, which is: D

f Dx

> f ( x )@



1

^(iZ )  ^ f ( x)`` D

(5.193)

Where the Fourier and an inverse Fourier transform are depicted as the following formulae: def f

^ f ( x)`



³ dx  e

 (Z )

 iZ x

f ( x)



f

f ( x)

1

^F (Z )`

def

1

(5.194)

f

³ 



iZ x

dZ e F (Z ) 2S f In some cases (especially for steady state systems with a lower terminal of differ-integration a f ), the Fourier transform method is another way to find a fractional derivative/fractional integration of the function f ( x ) . That is: (i) Obtain the Fourier transform of f ( x ) as F (Z ) ;



D

(ii) This is then transformed into F (Z ) in the frequency Z domain, which we multiply by (iZ ) ; D

(iii) With the resulting function (iZ ) F (Z ) , we conduct an inverse Fourier transform, to get We check the result of the Fourier transform of the derivative of a function, that is, 

^f

(1)

D

f Dx

`

( x)

> f ( x )@ .

(iZ )

^ f ( x)` .

With our definition, we write:

^f



f

` ³e

(1)

( x)

 iZ x

f

(1)

(5.195)

( x )dx

f

Now we write u parts (i.e.

³

f

f

udv

e

 iZ x

, then du f

uv f 

³

f

f

iZ e

 iZ x

dx and dv

f

(1)

dx , v

f ( x ) . Applying the formula for integration by

vdu ), we get the following steps:



^

f

(1)

f

` ³e

( x)

 iZ x

f

(1)

( x )dx

f

e

 iZ x

f ( x)

x f x f

Since we take the function f ( x ) as absolutely integrable, thus we have, e



^

(5.196)

f



 iZ x

³ (iZe

 iZ x

) f ( x )dx

f

x f

f ( x) x

f

0 , so we write:

f

f

(1)

` ³ iZ (e

( x)

 iZ x

)  f ( x ) dx

f

(5.197)

f

(iZ )

 iZ x

³ e  f ( x ) dx

(iZ )

^ f ( x)`

f

Repeating the above n -times, we can write 

^ ` dn f ( x ) dx n

n

(iZ ) 

^ f ( x)` .

We can also verify the same for an

inverse Fourier transform formula and then differentiate the same as described below:

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms f

1

f ( x)

251

i x

2S

Z ³ e  F (Z ) dZ

f

f

F (Z )

^ f ( x)` ³ e iZ x  f ( x) dx



f

(1)

f

§ 1 ¨ dx © 2S d

( x)



^f

(n)

`

( x)

^f

(1)

f

i x

2S



³e

(5.198)

·  F (Z ) dZ ¸ ¹

iZ x

f

1

From the steps in (5.198), we write

f

`

Z ³ e  (iZ ) F (Z ) dZ

1

^(iZ ) F (Z )`

f

(iZ ) F (Z ) ,

( x)



repeating

n

-times so that we get

n

(iZ ) F (Z ) . For n non-integer, say D , we have:



^f

(D )

`

( x)

D

(5.199)

(iZ ) F (Z )

5.19.2 Similarity and dissimilarity in Laplace and Fourier transforms The two integral expressions are rewritten as the following:

$ ^ f ( x )`

³

f

0

e

 sx

 f ( x ) dx

f

^ f ( x)` ³f eiZ x  f ( x) dx



(5.200)

Refer to Appendix-G where the derivation of an inverse Laplace transform is shown using Fourier transforms. They are very similar where we assume the representations of f ( x ) in both cases of (5.200) are integrable, except within the limits of integration. Similarly, we see



^

s

D

f Dx

> f ( x)@`

D

(iZ ) 

^ f ( x)` ;

$

where

^

D

0 Dx

s

> f ( x)@`

D

s $ ^ f ( x )` with all the initial values as zero with

iZ .

is replaced by

Re > s @  i Im > s @ as a complex number, expressed as s

We term the Laplace variable

Re > s @  iZ ; Z

Im > s @

( s is called a complex

frequency), whereas the Fourier variable Z is a real frequency, and the quantity iZ is an imaginary frequency. We

>@

can term the Fourier frequency as the Laplace frequency with Re s

0.

The limits of integration in the Fourier transform expression are from f to f , meaning that we consider all the initial terms as zero; whereas the Laplace transform integration has a limit from 0 to f , and has the effect of initial terms. Physically we mean that the Fourier transform is a steady state representation, while the Laplace transform pictures transient and steady state behavior. However, we can have the definition of a double-sided Laplace transform by extending the lower limit of integration from zero to minus infinity. Therefore, for a given Laplace transform of a function, can we change the Laplace variable s to iZ and get the Fourier transform? For example, if $ ^u ( x )` i.e. u ( x )

1 s

then should we have  ^u ( x )`

0 for x  0 and u ( x )

Let us construct a function y ( x ) and

d dx

y ( x)

1 iZ

? Where u ( x ) is the Heaviside unit step function;

1 for x ! 0 . We also know from  12 for x  0 and y ( x )

d dx

G ( x ) , that it is a Dirac-delta function.

u ( x)

 12 for x ! 0 . Clearly, we can write y ( x )

u ( x) 

1 2

G ( x ) . In addition, we write here from the Fourier transform tables that  ^G ( x )` 1 . In the previous

section, we have derived the expression  (Z )



^ f ( x)`

 ^u ( x )`

iZ 1





^ ddx f ( x)` , and using this we write the following:

^ ` ^ ` d

dx d

u ( x)

1 iZ 1

 ^G ( x )`

1 iZ 1

(5.201)

y ( x)   ^G ( x )` iZ iZ iZ dx We are therefore having the same Fourier transform for two different functions. The functions are different in terms of a constant that is 12 . This is obviously caused by the fact that the derivatives of the constant are zero. Now we discuss where the difference should exist.



^ y ( x)`

1

1 iZ

252

Chapter Five

Take

the

similarly y ( x )

average X X

1 2X

³

X

X

value

as

y ( x) d x

1 2X

u ( x) 0

X

1 2X

X

³

X

X X

³   dx  ³  dx 1 2

X

1 2X

0

1 2

³

X

0

1 2

(1)dx

and

0 . In a physical sense, this average value 1 2

signifies that the function u ( x ) has an average value (dc component) as component. The dc-component corresponds to the frequency Z

1 2X

u ( x )dx

and the function y ( x ) has no such dc-

0 in the Fourier transformed function. Therefore,

we need to add this dc-component in our  ^u ( x )` ; as follows:

 ^u ( x )`



^ y ( x)`   ^ 12 ` (5.202)

1

 SG (Z )

iZ

We used 

^e ` iZ0 x

0 to get  ^1`

2S G (Z  Z0 ) by setting Z0

2S G (Z ) , and used in the above derivation

(5.202). Therefore, we see similarity through relations like the Laplace and Fourier transforms of derivatives of a function, yet the two are different.

5.20 Complex order differ-integrations described via the Laplace transform D

We have discussed the differ-integral operation g (t )

D

0 Dt

> f (t ) @

with D as an arbitrary real number. Say if

u  iv , with u and v as real numbers. Then we have a complex order differ-integration as g (t )

du iv dt u iv

> f (t ) @ ,

which is an analytical continuation of the differ-integral operator, which we discussed in an earlier chapter (Section 4.22).

0 at t

Assuming that our initial values are zero i.e. f (t ) formula to write the following steps:

0 ; then we can use our generalized Laplace transform

$ ^g (t )`

G(s)

s

u  iv

F (s)

u iv

s s F (s) u iv ln s

s e s We write the invert of

F (s)

s

 ( u  iv )

g (t )

( u  iv ) 0 Dt

as

f (t )

u

u ln s iv

s e

F (s)

(5.203)

F (s)

 cos(v ln s )  i sin(v ln s ) F ( s )  ( u  iv ) g (t ) 0 Dt

f (t )

and its Laplace transform as

G(s) .

Using a known Laplace transform (i.e. $

1

^s `

tD 1 * (D )

D

1 or g (t )

and taking G ( s )

G (t ) ), we get the impulse

response f (t ) , which is as follows, by taking an inverse Laplace transform as:

f (t )

$

1

^F ( s )` $

1

t

^s

$

 ( u  iv )

u  iv 1

* (u  iv ) t

u 1

^s `

1

 ( u  iv )

t

`

G(s)

u 1

* (u  iv )

e

iv ln t

(5.204)

 cos(v ln t )  i sin(v ln t )

* (u  iv ) The derivation of f (t ) in (5.204) says we get an imaginary response component along with the real response component, in a time domain. Now we use conjugated differ-integrals where D and D are complex conjugates, as follows g (t )

D

0 Dt

> f (t )@  0 DtD > f (t )@ u  iv > f (t )@  0 Dtu iv > f (t )@ 0 Dt

The Laplace transform of the above conjugated differ-integrals is as follows

(5.205)

Generalized Initialization of Fractional Integrals/Derivatives and Generalized Laplace Transforms

s

$ ^g (t )` G ( s )

s s

u iv

2s We write f (t )

D 0 Dt

which is $ ^ f (t )`

> g (t ) @ 

F (s)

s

D 0 Dt

s

u

 iv

 ( u  iv ) 0 Dt

> g (t ) @

s

iv

u

u  iv

s

u iv

u  iv

s s

253

F (s)

F (s)

s

u

e

iv ln s

e

 iv ln s

F (s)

(5.206)

 cos(v ln s ) F ( s )

> g (t )@  0 Dt (u iv ) > g (t )@ , and take the Laplace transform,

G ( s) . With G ( s)

1 , that is g (t )

^ `

G (t ) , and using $ 1 s D

tD 1 * (D )

,

the impulse response of a conjugated differ-integral is as follows:

f (t )

$

1

^s t

 ( u  iv )

s

u  iv 1

* (u  iv ) From 1 * ( u iv )

the

property

of

the

gamma

function

we

 ( u iv )



t

(5.207)

u iv 1

* (u  iv )

1 * ( u  iv )

have

` Re ª¬ * ( u1iv ) º¼  i Im ª¬ * ( u1iv ) º¼

and

Re ª¬ * ( u1iv ) º¼  i Im ª¬ * ( u1iv ) º¼ , using these we write the following (impulse response for complex conjugated

differ-integration) as:

f (t )

$

1

^s t

 ( u  iv )

s

u  iv 1

* (u  iv ) t

 ( u iv )



t

`

$

1

^2s

u

`

cos(v ln s )

u iv 1

* (u  iv )

§  iv iv ª 1 º ª 1 º·  iv iv ¨ (t  t ) Re « * (u  iv ) »  i (t  t ) Im « * (u  iv ) » ¸ ¬ ¼ ¬ ¼¹ © ª 1 º ª 1 º· u 1 § 2t ¨  cos(v ln t ) Re «   sin(v ln t ) Im « » »¸ ¬ * (u  iv ) ¼ ¬ * (u  iv ) ¼ ¹ ©

u 1

(5.208)

We are getting a pure and real response in time, for these conjugated differ-integrals.



Note that we have ( s

u  iv

)( s

u iv



) F (s)

2u

s F ( s ) and

 F (s) s u iv su iv

s

2 iv

F ( s ) . This means that any real order differ-

integral can be composed of the product of two complex conjugated differ-integral Laplace operators. A pure, imaginary order differ-integral can be composed of the ratio of two complex conjugated differ-integral Laplace operators. This gives a ‘complex’ order calculus; which is still a developing subject.

5.21 Short summary The primary focus of this chapter was the concept of initialization for fractional derivatives and fractional integration. We showed here that the fractional integrations and fractional derivatives require an initialization function, unlike classical integer order calculus. We discussed the initializing processes that require demanding issues for the Riemann-Liouville (RL) and Caputo fractional derivatives. We also stated that to equate the two initializations for RL and Caputo, is a difficult task. We then discussed Laplace transforms for fractional derivatives and fractional integral operations and how initial states matter in both Caputo and RL formulations. We touched briefly upon the Fourier transform of fractional differ-integrations and pointed out the differences between Laplace and Fourier transforms. We also described the approximation method to represent a fractional Laplace operator via continued fraction expansion or the power series expansion method; this is useful for circuit realization in electronic engineering. Then we state the interesting concept of generalized stationary conditions; and we have seen the generalization of the fractional derivative of Caputo and RL types. We also introduced the usage of generalized initialized fractional derivatives and the use of the generalized Laplace transform to solve a fractional differential equation. We also have briefly introduced the composition of complex order differ-integrals via complex Laplace variables, and obtained interesting responses. We will now devote the next few chapters to various aspects of solving fractional differential and fractional integral equations, especially using generalized Laplace transform techniques.

254

Chapter Five

5.22 References The pioneering works which provided inspiration for this chapter are as follows: Carl. F. Lorenzo, Tom T. Hartley, “Initialized Fractional Calculus” (2000) Axtell M, Bise. M E, “Fractional calculus applications in control systems” (1990) Bohannan G. “Analog Realization of fractional order control element revisited” (2002) Bohennan Gary et al, “Electrical components with fractional order impedances” (2006) Carlson G. E, “Simulation of the fractional derivative operator 1/ n

Carlson et al, “Approximation of fractional capacitors 1 / s

s and fractional integral operator 1/ s ” (1961) by regular Newton process” (1964)

Charef. A, Sun H H, Tsao Y Y, Onaral. B, “Fractal systems as represented by singularity function” (1992) Chen, Vinagre, “A new type of IIR digital fractional order differentiator” (2003) Chien-Cheng Tseng, Soo-Chang-Pei, Shih-Cheng Hsia, “Computation of fractional derivatives using Fourier transforms and digital FIR differentiator”(2000) 1/ 2

Halijak C A, “An RC impedance approximation to (1 / s ) ” (1964); Igor Podlubny et al, “Using continued fraction expansion to discretize fractional order derivative”(2003) O

P

Igor Podlubny, Dorcak, Kostial. J, “On fractional derivatives, fractional order dynamic systems and PI D controllers” (2004) Igor Podlubny, Ivo Petras, et al., “Realization of fractional order controllers” (2003) Igor Podlubny, “The Laplace Transform Method for Linear Differential Equations of the Fractional Order” (1994) Jumarie G, “Laplace Transform of Fractional Order via Mittag-Leffler function and Modified Riemann-Liouville derivative” (2009) Oberhettinger. F and Badii L,“Tables of Laplace transforms” (1973) Oldham and Spanier,“Fractional calculus”(1974); “The replacement of Fick’s law by formulation involving semi differentiation” (1970); Oustaloup. A, et al., “Frequency band complex non-integer differentiator characterization and synthesis”(2000) Oustaloup A, “From fractality to non-integer order derivation through recursivity, a property common to these two concepts: a fundamental idea for a new process control strategy” (1988) Sabatier J, Agrawal O. P, Machado J. A (Eds.), “Advances in Fractional Calculus” (2007) Sugi. M et al, “ Z -Variations of AC admittance in the inhomogeneous distributed RC lines” (2000) Sugi. M, Hirano Y, Miura Y. F, Saito. K,. “Simulation of fractal immitance by analog circuits: an approach to optimized circuits” (1999) Tom Hartley and Carl F Lorenzo,“Chaos in fractional Chua’s circuit” (1995); “Fractional order system identification by continuous order distribution”(2003) Vinagre, Ivo Petras, “Two direct Tustin discretization methods for fractional order differentiation / integration” (2003) Wang Jifeng, Yunkai et al,“Frequency domain analysis and application for fractional order control system”(2005) Wang J C, “Realization of generalized Warburg impedance with RC ladder networks and transmission lines” (1987), Jordan Hristov, “Electrical circuits of non-integer order: Introduction to an emerging interdisciplinary area with examples” (Springer, 2018) The above pioneering works are furnished with further detail and listed in alphabetical order in the Bibliography section.

CHAPTER SIX FRACTIONAL DIFFERENTIAL EQUATIONS AND THEIR ANALYSIS

6.1 Introduction Fractional differential equations appear in several physical systems. In systems involving coarse-grained phenomena everything happens as if the elemental point is not infinitesimally zero; rather, it has some thickness (or spread), that D

could be pictured using, say, (dx )

D

with 0  D  1 instead of dx with (dx ) ! dx as dx p 0 . In other words, we D

D

D

would be led to consider the rate of variation as dy / dx , (or d y / dx ), and therefore, in a way, we are going towards a fractional derivative. This finite thickness (or spread) of the fractional differential quantity allows us to view scales where the in-homogeneity and roughness of the systems are captured on micro and meso scales. This is in contrast to fine-grained phenomena where the normal differential element tends towards zero, which make the system uniform. This is one of the reasons for having fractional differential equations, where the differential quantity is D

(dx ) , different from dx . This dx is a standard differential quantity for ‘finely-grained’ phenomena, where normal derivatives and integrals appear. The solution to these fractional differential equations is no more rigorous than their integer order counterparts. The Laplace transform technique is very popular, although several other analytical approaches do exist. However, in this chapter, it is mainly the Laplace transform technique that is considered, as this technique is the most easily understood and the most popular amongst engineers and scientists. We will study methods to obtain an inverse Laplace transform via contour integration and via a power series expansion, and we will discuss the new method of obtaining an inverse Laplace transform via the Berberan-Santos method where the contour integration is not required. We will study the response of a fractional differential equation, obtain response functions, and evolve their properties. We will study how we can determine decay rates for complex decay functions. In this chapter, the Mittag-Leffler function will be used and studied. In addition, various simple types of fractional differential equations will be taken, and their respective solutions will be obtained.

6.2 Tricks in solving some fractional differential equations 6.2.1 A method using the Laplace transform for a fractional differential equation



Let us take the fractional differential equation (FDE), i.e. t D dD y

y t

dtD

, with

1

D

2.

With D

1

2

1

2

> y ( t ) @  y ( t )

0 . This equation is of the type

as a Riemann-Liouville (RL) semi-derivative operator fractional, the initial  12

condition is required to solve this FDE; which is given as Dt

y (t )

 12

Dt

t 0

y (0)

0 . We could also solve this FDE

with the tricks of the Laplace transform and via an alternative method.

^

`

^

$ t D

^

$ D

1

2

1

2

> y (t )@ `

> y (t )@`

^

1

 Ds ª$ D

¬



1

s 2 y(s)  D

1

 12

2

> y (t )@`º¼

> y ( t ) @ t

0

1

1

 12

 Ds ª s 2 y ( s )  Dt

¬

, with

1

 dds ª¬$ ^ f (t )`º¼

The known Laplace transform is $ t  f (t )

y(s)

 Ds ª¬$ ^ f (t )`º¼ .

y (0) º .

Using this, we write

Here

¼

we

have

used

$ ^ y (t )` . Thus, we write the Laplace transform of the

given fractional differential equation as follows:



t D  12

Using the given initial condition, i.e. Dt





1 2

s

 12

1

1



y ( s )  s 2 Ds > y ( s )@  y ( s )

y (0)

1

2

> y (t )@  y (t )

0,

1

1

 12

 Ds ª s 2 y ( s )  Dt

¬

y (0) º  y ( s )

0 , and expanding (6.1), noting that Ds1 {

¼

d ds

0

(6.1)

operation, we get

0 . Rearranging this, we get the first order ordinary differential equation in the

s -variable that is expressed as follows:

256

Chapter Six

d y(s)

§ 1 1  1 ·  ¨ s  s 2 ¸ y(s) 0 , ds ©2 ¹ Integrating both sides of the second equation of (6.2), we get the following: ln  y ( s )

1

 12 ln( s )  2 s

2

d y(s) y(s)

or y ( s )

$

1

^

Ks

1 s

e

 1 2 2 s

a s

e

`

.

1

2 s

(6.3)

 ln K

ln K and simplify (6.3) to write ln y



ln Ks

 1 2 2 s

e



By using the inverse Laplace transform of this expression from standard tables,

e

St

(6.2)

k

 12 ln( s )  ln e With k as an integration constant in (6.3) we can write k

§ 1  12 · ¨   s ¸ ds © 2s ¹

2 a

4t

, we get y (t )

K1t

 1 2  1t

e

, for t ! 0 , with constant K1 including

S .

6.2.2 Alternative method to solve a fractional differential equation



1

We now provide an alternative solution for our example (i.e. t Dt 2 > y (t )@ we D

D

Dx ª¬ x  f ( x ) º¼ equation, get t



¬

1

1

Dt 2 ªt Dt 2 > y (t )@

get

we

1 Dt y (t )

xDx get



1 2

º¼

1

Dt 2 > y (t )@ .

> f ( x)@  D DxD 1 > f ( x)@ ),

a



¼

1

 12





1

1 2

1

¬

Integrating both sides of the second equation of (6.4) we get

1

1

Dt 2 y (t )

y (t ) t

, and substituting this in the

y (t )

,

t

ln y

t

1

d y (t ) y (t )

 1 2  1t

e

§1 1· ¨ 2  ¸ dt © t 2t ¹

(6.4)

 12 ln t  k , where k is a different

ln K and that gives us a further simplification, as ln y Kt

is,

1

§ d y (t ) · 1 ¸  y (t ) © dt ¹ 2

this, we write a solution, that is, y (t ) 6.2.1.

(that

Dt 2 y (t ) ), we get a first order differential equation as follows:

y (t )

t¨

integration constant. We write k

rule

ª Dt 2 y (t ) º Dt 2 y (t ) . Simplifying this, we ¬ ¼

2

obtained equation (i.e. t Dt y (t ) 

Leibniz’s

as described in Section-4.19 of this book, to the LHS of this

1

Dt y (t ) . Using our original equation, that is,

y (t )

y (t ) ). Taking a half derivative of this,

applying

equation t Dt 2 ª Dt 2 y (t ) º  12 Dt

new 1

Then





ln Kt

 1 2  1t

e

. From

, which we also obtained via the Laplace transform tricks in Section-

6.2.3 A method using the Laplace transform for a fractional integral equation

 $ ^D

 12

Now we take a fractional integral equation (i.e. ty (t )  S Dt

^

`

1

$ t  f (t )

(i.e.

 Ds ¬ª$ ^ f (t )`¼º and 1

`

 12 y (t ) t

fractional integral equation, that is,  Ds y ( s )  S s

 12

y(s)

s

> y ( t ) @

 12

0 ). Using the Laplace transform identity

y ( s ) ), we write the Laplace transform of our

0 , which is a first order differential equation in Laplace

variable s . We can rewrite this as follows:

d y(s) Following

ln y ( s ) (i.e. $

1

the

steps

ln K  2 S s

^e ` a s

1

the



ln K  ln e

2

a 2 St

of

3

e

a

2 4t

2 S s

previous

, or ln y(s)

), we can write t ! 0 , y (t )

ds example

Ke

2 S s

Kt

 12

in

Section-6.2.2,

y(s)

0

(6.5) we

can

write

. Using the standard Laplace transform tables

 3 2 S t

e

 Ss

.

Fractional Differential Equations and their Analysis

257

6.2.4 An alternative method to solve fractional integral equations



 12

Here, we carry out alternative methods for obtaining the solution of ty (t )  S Dt 1

derivative of the equation and write Dt 2 >ty (t )@ D

D

Dx ª¬ x  f ( x ) º¼ 1

tDt 2 y (t ) 

1 2

xDx

D

 12 y (t ) t

1

y (t )



0 . Now, take the half-

1

 S Dt 2 ª¬ Dt 2 y (t ) º¼ , and apply Leibniz’s rule to the LHS (that is

> f ( x)@  D DxD 1 > f ( x)@ )

described



 12

S y (t ) . Here, substitute Dt

in t

y (t )

S



 12

y (t )

and

write;

y (t ) and write the following expression:



1

t Dt 2 y (t )  Taking an ordinary whole derivative of ty (t )  S Dt

Section-4.19

t

y (t )

2 S

S y (t )

(6.6)

0 , we get another equation, expressed in the following

step:

 t  D y (t )  y (t )

1

1

t Dt y (t )  y (t )





1

With these t Dt 2 y (t )  1

2

t 2 S

1

Dt 2 y (t ) ) and get t Dt y (t )   32 t  S y (t ) 3

1 t

S y (t ) and

y (t )

0 . Therefore, we have

S Dt 2 y (t ) 1

S Dt 2 y (t ) ,

d y (t ) y (t )



S

t2

(6.7)

we eliminate the term (i.e.



 23t dt , and upon integrating and

S

rearranging it, we get y (t ) Kt 2 e t (i.e. by following the steps as we did for earlier examples). Here we have demonstrated the various tricks used to obtain a solution for a fractional differential equation.

6.3 Abel’s fractional integral equation of ‘tautochrone’-a classical problem Abel’s problem is to find a curve where the time of descent is the same irrespective of the position of release of the ball in a frictionless system. In Figure-6.1, S is the arc length measured along curve C from point O (the origin) to an arbitrary point Q on C (Figure-6.1). The gain in kinetic energy while the ball is descending represents the loss in potential energy as given by the following law: 1

§ ds · m¨ ¸ 2 © dt ¹

2

mg ( y  K )

ds

dt 2 g ( y  K )

(6.8)

y P ( x, y )

C

Q([ ,K )

s O(0, 0)

x Figure-6.1: Abel’s tautochrone curve

The negative square root indicates that the distance (i.e. arc-length) decreases as the time increases. The equation to be solved is, therefore, the following: 1 dt  ds (6.9) 2g( y K) The time of descent from point ‘ P ’ to ‘ O ’ is T , which is constant and is represented as follows: T



1

O

³ 2g P

1 y K

ds

(6.10)

258

Chapter Six

 h (K ) , where h depends on the shape of curve ‘ C ’. We can write it in differential form to show the curve as ds  h (K ) dK . Therefore, this substitution in (6.10) gives us Now take the arc length as a function s

h (K ) or

(1)

ds dK

(1)

the following expression: T

0

1

³ ( y  K )  h 2g



 12

(1)



(K ) dK

(6.11)

³ ( y  K ) (h K ) dK

(6.12)

y

Re-writing the above expression (6.11), we obtain the following: y



 12

2g T

(1)

0

The expression (6.12) means that the integral, that is, the RHS of the previous equation, when ‘a constant’, will be the solution by which we get a constant time of descent ( T ). In the above expression, with k as a constant, the integral of the RHS is as follows: x

³

k { (x  t)

 12

 f ( t ) dt

(6.13)

0

This (6.13) is the standard Abel integral equation. Thus, when the half derivative of a constant k is computed, the function d

1

2

dx

>k @ 1

f ( x ) is obtained. Therefore,

S  f ( x ) . We relate the above Abel’s integral equation (6.13) to the RL-fractional integration formula by

2

taking *  12 The solution to



(1)

S .



³

2g T

y

0

( y K)

 12

h

(1)



(1)

(K ) dK (6.12) can be obtained where h (K )

ds dK

, and through letting



f ( y ) { h ( y ) Subsequently, we will divide both sides of the expression by *  12 and apply the Riemann-

Liouville definition of a fractional integration, through which we get the following: D

 12

x

1

> f ( x )@

 12

( x  t )  f ( t ) dt * ³ 1 2

0

(6.14)

x

SD

 12

 12

> f ( x ) @ ³ ( x  t )  f ( t ) dt 0

Thereby inverting the operator D

 12

in (6.14), we get the following: 2g *

1 2



T

D

 12

> f ( y )@

D

putting the half derivative with regard to y for a constant (i.e. T as

1

2

ª 2g º T» « ¬ S ¼

1

D y 2 >T @

T

Sy

(6.15)

f ( y)

) to get f ( y )

2g T y

S

. By

carrying out algebraic manipulation, the following expressions are obtained:



(1)

f ( y) { h ( y) dx

 f ( y)

dy

§y x ¨³ ¨ ©0



2 gT 2 2

S K

2



2

ds

dx  dy

dy

dy

2

§ dx · 1 ¨ ¸ © dy ¹

2

1

(6.16)



· ¸ ¹

0

y

 1 dK ¸  c

We continue with the following assertions: y

c

0,

x

x

³ 0



2 gT 2

S 2K



 1 dK

(6.17)

Fractional Differential Equations and their Analysis

gT 2

Let a

E

sin

S 1

2

259

2

2 a sin [ , which gives

, and with the change of variables K

x

4a

E

³  cos [ d[ , 2

0

with

 . The parametric equation of the cycloid is thus obtained as follows: y 2a

x

2 a  E  12 sin 2 E

x

a T  sin T

2

y

2 a sin E

2E

T

a

gT

S

2

2

(6.18)

a 1  cos T

y

Cycloid is the tautochrone. It should be mentioned that the cycloid curve shape C - is too curved for the brachistochrone problem as solved by Bernoulli (that is, the determination of the shape of the curve giving us a ‘minimum’ time of descent). Therefore, the constant time T is also the minimum time of descent in a cycloid.

6.4 Using the power series expansion method to obtain inverse Laplace transforms The easiest and most conventional and popular method to solve a differential equation is to form a polynomial equation in the Laplace variable s from the given differential equation and then take the inverse Laplace transform to get the solution. In fractional differential equations, one may follow a similar method. However, to get an inverse Laplace transform, one requires certain skills. In a fractional differential equation, say we need to (for example) obtain an inverse Laplace transform of k with k ! 0 to arrive at y (t ) . The Laplace transformed expression corresponds to the fractional y(s) s ( sD  k ) differential equation: D

1 d y (t )

with y (t ) t 0.

 y (t ) x (t ) D k dt 0 for t  0 ; and x (t ) as Heaviside’s unit step at t 0 , that is x (t )

The Laplace transform of the equation

y(s)

k sD  k

have y ( s )

x ( s ) . In addition, with k s ( sD  k )

x( s)

1 dD k dt D 1 s

y (t )  y (t )

is 1k s

x (t )

D

(6.19) 1 for t t 0 and x (t )

 y(s)  y(s)

0 for

x ( s ) , which gives

(that is, the Laplace transform of the unit step function), we

.

There could be a direct method used here by employing the integral definition of an inverse Laplace transform formula, namely y (t )

1 2S i

J  if

³J

st k e ds D  if s ( s  k )

(see Appendix-G for further details). Here, performing the numerical

integration is tough, and we will elucidate upon that later. We will first resort to a power series expansion by using (1  x )

1

¦

f n 0

n n

( 1) x with x

k sD

, x  1 and for D ! 0 we have the following derivation:

y(s)

k

k

D

s(s  k )

ss

D

1  k sD

2 3 § · k k k 1     .... ¸ 2D 3D D 1 ¨ D s s s © s ¹

k

k s We use a known Laplace transform pair (that is, $ expression in order to write the following:

1

D 1

^



k s

* ( n 1) s n1

2

2D 1

`

 n

k s

3

3D 1



k s

(6.20)

4

4D 1

 ......

t ) by applying it term by term to the above

260

Chapter Six

kt

y (t )

2 2D

D

* (1  D )



* (1  2D ) kt

11

§ ©

k t

kt

 ......

* (1  3D ) 2 2D

D

* (1  D )

1  ¨1 

3 3D

k t



* (1  2D )

k t



2 2D

D



* (1  D )

 ......

* (1  3D ) 3 3D

k t

* (1  2D )



(6.21)

·

k t

* (1  3D )

 ...... ¸

¹

D m

 kt

f

1

3 3D

k t





¦ *  mD  1

1  ED  kt

m 0

D



In the above (6.21) we have used ED ( x ) as a one-parameter Mittag-Leffler function (i.e. ED ( x ) The series expansion formula we use is (1  x ) k sD

1

¦

f

¦

( x )n (1 *  nD ) n 0

f

).

n n

n 0

( 1) x , and this is valid for x  1 ; meaning in our case

 1 . This means our series expansion (for the case of D ! 0 ) is fine for a higher s (frequency) or is true for early

times t . However, as shown by the tables of the Laplace transforms, for the Mittag-Leffler function we have the Laplace

^

transform pair $ t

D p  E 1

( p)

`

D

ED , E ( at )

( p)

D E

p ! sD

s a

D

0,1, 2, 3, ... is the p -th integer order

. The function ED , E ( at ) , p

derivative of the two-parameter Mittag-Leffler function, ED , E ( x ) . The two-parameter Mittag-Leffler function is

¦

defined as ED , E ( x ) and E

$

1

^

f

xm m 0 * (D m  E )

(see Appendix-A). In this, the Laplace transforms the identity by setting p

^

`

1

D D D  1 , meaning that we have $ t ED ,D 1 ( at )

k s ( sD  k )

` k t  E D

D ,D 1 (  kt

D

)

s sD  a

.

Using this, we write for our function:

. Therefore, we have the two equivalent results as depicted below: y (t ) 1  E (  kt ) kt  E (  kt ) D

0

D

D

D ,D 1

D

(6.22)

For the case D  0 , let us represent that by considering D  E with E ! 0 , and for the same function we conduct the following derivation along the same lines as the power series expansion and by using a term-by-term inverse Laplace transform: k k y(s) D E s(s  k ) s(s  k )

1 1  E

s (1  k s 1 s 1 s

1  k 

1  E

s

1 ks

)

E 1



Now conducting a term by term inverse Laplace transform with $

(6.23) 2 2 E

3 3 E

k s 1

2 2 E 1

k s 1 1

^ s`



k s 1

3 3 E 1

k s

 ....



 ...

1 and $ 1

^ ` 1 s n1

tn * ( n 1)

, we achieve

the following: y (t )

1

t

E

k  * ( E  1) f

¦ n 0



t k

2

2E

 *(2 E  1)



t k

3

3E

 *(3E  1)

 ...

(6.24)



n

1  E

 1 , with E ! 0 . This is valid for a large s (frequency)

E  tk * (1  nE )

For the case D  0 , our series expansion requires k s



E EE  tk

and thus an early time for t . In the above cases, we have obtained the solution in the form of a Mittag-Leffler function, for the chosen Laplace domain (a complex frequency domain) expression. This type of Laplace domain

Fractional Differential Equations and their Analysis

261

expression does appear in several physical systems like the low pass filter, the high pass filter, the charging/discharging of fractional order capacitors, etc.

6.5 Using Laplace transform techniques and power series expansion to solve simple fractional differential equations Let us apply the Laplace transform to the FDE, that is: C 13 0D

With x (0)

> x (t ) @  x (t )

y (t )

C 13 0D

0 for t  0 , and the fractional derivative operator

^

1 tt0 0 t0

y (t )

(6.25)

is of Caputo’s type. The source term (i.e. y (t ) ) 1 s

is Heaviside’s unit step function, whose Laplace transform is Y ( s )

. With a generalized Laplace transform, as we

developed in Section-5.16.4, assuming that the fractional derivative operator in the above dynamic equation is of

^

Caputo’s formulation, then $ d

rD

`

f (t ) { s

rD

F ( s ) , when f (0)

0 , that is, with zero initial conditions. Taking the

Laplace transform of the above equation (6.25), we get the following: 1

s 3 X (s)  X (s)

Therefore, rearranging the above and using (1  x )

1

¦

f

Y (s)

(6.26)

1

, and x  1 , we write the

n n

( 1) x , with x

n 0

s

3

following steps:

X (s)

1



s s

1

1



4

1

s

1  (s)

 13

3

s s

1

 13

s

¦ (1)

3

 13

3

f

1 4

4

1  s 1  s

3

k

s

k

¦

n

( 1) s

3

(6.27)

f

k

¦ (1) s

3

k 0

f

4

n3

f

( 1)

n 4

s

¦

n

n

3

In (6.27) we have used a series expansion approximation with the assumption s

^` n

( n )!

* ( n 1)

s n1

s n1

 ( k 3 4 )

k 0

n 4

transform pair (i.e. $ t

1

 13

 1 . Now using the Laplace

), we write the solution as: f

n 1

n

( 1) t 3

¦ *

x (t )

n 4

(6.28)

n 3

Here we are unable to place any closed form result like the one we obtained in Section-6.4. Now consider a ball falling freely under gravity (another case of fractionally damped motion) in a viscous fluid, with a constituent equation as follows:

d v (t )

D

d v (t )

(6.29)  v (t ) 1 0 D 1 D dt dt with the initial condition (velocity) as v (0) 0 . With a generalized Laplace transform (as we developed in Section5.16), assuming that the fractional derivative operator in the above dynamic equation is of Caputo’s formulation, then

^

$ d

rD

`

f (t ) { s

rD



F ( s ) , with zero initial condition (i.e. f (0)

the above equation (6.29) to get sV ( s )  sD V ( s )  V ( s )

$ ^1`

1 s

1 s

0 ). We apply the Laplace transform to both sides of

; where we used the Laplace transform identity, i.e.

. After some algebraic manipulation, the following expression is obtained:

V (s)

1  ( s

1 D

s (1  s  s )

1

s

s 2

D 1

)



1

1  ( s

1

s s

2

D 1

)



1

(6.30)

262

Chapter Six

Expanding the numerator as a binomial series with (1  x )

1

¦

f

n n

n 0

( 1) x , we write x

(s

1

s

D 1

) and with

x  1 the following:

1  ( s

V (s)



1

s

D 1 n



s s

Now we use the binomial expansion formula ( a  b ) coefficients, and expand s

1

n

D 1

)



1

1

2

s

¦

n

n

r 0

2

f

¦ (1)  s n

1

s

D 1 n



n 0

n nr

n

with

Cr a b

(6.31)

n! r !( n  r )!

Cr

as binomial

in the above expression to write the following: V (s)

1 s

2

f

¦

( 1)

n

n 0

r

n ¦ §¨© r ·¸¹  s  sD 1 n

1 n  r

r 0

(6.32)

n

1 §n· ( 1) ¦ ¨ ¸ 2  (2 D ) n  (1D ) r ¦ r n 0 r 0© ¹s f

We use the Laplace transform identity (i.e. $

^

* ( n 1) s n1

`

n

n

t ) to write the solution as follows: f

v (t )

n

t

n

1 (2 D ) n  (1D ) r

¦ (1) ¦ §¨© r ·¸¹ *  2  (2  D )n  (1  D )r n

n 0

(6.33)

r 0

Here too, we are unable to obtain a closed form solution, as we had achieved earlier in the book (Section-6.4). In the above examples, we wrote that the fractional derivative was Caputo’s formulation. In both cases, our initial conditions are zero no matter whether the fractional derivative operator is of Riemann-Liouville or Caputo type.

6.6 The contour integration method for obtaining inverse Laplace transforms We aim to obtain y (t ) Laplace expression y ( s )

$

1

^

k s ( sD  k )

k s ( sD  k )

` , by use of the formula y(t )

, with k ! 0 . Here s

J  if

1 2S i

³J

st k e ds D  if s ( s  k )

, for our example of the

Re > s @  i Im > s @ is a complex quantity or a complex

frequency. It may also be expressed as s V  iZ . The inverse Laplace transform is an integration that is on an imaginary axis (refer to Figure-6.2), which is on the line FA . This is a Bromwich path (written in short form as Br ) on which the Laplace inversion integration is done in a line on the complex plane, from s J  if to s J  if (see Appendix-G). Take the function F ( s ) positions are at s

D

k s ( sD  k )

st

e ; this has a pole at s

0 and is infinite as many numbers of pole

 k . Pole implies that the function F ( s ) goes to infinity at those pole points. Writing  k as

i ( S  2S m )

D

i ( S  2S m )

, with m as integers 0, r1, r2, ... we get s . Therefore, there are multiple poles, in several ke ke Riemann sheets (see Appendix-E and Appendix-G for further details). The real solution y (t ) is obtained from performing integration in a primary Riemann sheet using a complex plane, as shown in Figure-6.2. The primary

>@

Riemann sheet has the region S d arg s d S , with a branch cut at location s

0 and s

f , which is a standard

branch-cut for a complex plane, for a multi-valued function (refer to Appendix-E).

Looking at F ( s ) , for m

0 we write s1

D

ke

i  DS

. For m 1 , we write s2c

>@

fall into the primary Riemann sheet, i.e. S d arg s d S . For m

>@

D

ke

1 , we write s2

is in the primary Riemann sheet, i.e. S d arg s d S . So for the case m

i  S D2 S

D

ke

D i

ke

S 2 S D



i  3DS

D

; this does not ke

 i  DS

, which

r1 , which says the corresponding poles

should be in the secondary Riemann sheet. However, for m 1 , we see that the pole in the neighboring Riemann sheet influences the response in the primary Riemann sheet of integration:

Fractional Differential Equations and their Analysis

263

Inverse Laplace transform of a function by integration in complex plane y (t ) $ 1 Pole at s

^

s( s k )

`

and

s

k

0

D

J  if

ke st 1 ds 2S i J ³if s ( sD  k ) D

ke

i

F ( s)

ke st s ( sD  k )

k !0

m 0, r1, r2,...

 S D2S m

D The power function of non-integer order that is s

k ; a multi-valued and we have standard branch cut

The standard branch cut in complex plane is (f,0] J  iT

y (t )

Im( s )

1 lim e st y ( s )ds 2S i T nf J ³iT

A

B

C

E

D

Re( s )

J

Integration contour is Ao BoC o Do E o F o A F

Figure-6.2: Branch cut and contour in a complex plane for integration

Let us take first the case with 0  D  1 ; say we can choose D

arg > s1 @ s2

0.7

10S 7

ke

0.7

! S . For m 1 , we have s2c

S  2S i  0.7 0.7

> @

ke

2 S i  S 0.7

0.7 . For m

> @

; in this case, arg s2c

0 , we have s1 30S 7

ke

S i  0.7

; here

! S . For the case m

 107S  S . Therefore, in this case, D 0.7 and

; in this case, arg s2

0.7

m

1 ,

0,1, 1 , the poles s1

>@

and s2 are not in the sheet of integration, that is, the primary Riemann sheet (i.e. S d arg s d S ). Therefore, only pole s 0 gives us the result of integration. Only pole s 0 exists and this will not provide any influence in the integration since the contour of Figure-6.2 encircles that and leaves this pole out of the region. Therefore, for the case 0  D  1 , there is no influence of any other poles as they do not exist within this region of integration. Let us now consider 1  D  2 , say with D the case m 1 , we have s2c s2

1.4

ke

i

 10 S 14



> @

; here arg s2

and s2 corresponding to m

1.4

ke

i

30 S 14

1.4 . For m



0 , we have s1

> @

; in this case, arg s2c

30 14 S

1.4

ke

i  10 S 14

> @

; here arg s1

! S . For the case m

10 14 S

 S . For

1 , we have

 10 14 S ! S . Therefore, for the case of 1  D  2 , the poles in primary sheets ( s1

0 and m

1 ) are in the sheet of integration. Since these two poles are within the

region of contour integration, their influence (by their residues) is placed within the contour integration and thus the solution. The pole s 0 is still outside the region. We shall solve these two cases separately. The closed contour integration from the ‘Residue Calculus’ of a complex variable states the following, with s j poles in the region of integration, i.e. the primary Riemann sheet for a branch-cut system (see Appendix-G for more information)

³

C

F ( s )ds B

³ A

C

³

D

³

E

A

³

³

F ( s )ds  F ( s )ds  F ( s )ds  F ( s )ds  F ( s )ds B

2S i

C

¦ Residues

D sj

F

F

(6.34)

264

Chapter Six

The contour is a closed path as shown in Figure-6.2; that is, C { A, B, C , D, E , F , A . Therefore, we get from (6.34), the following: C D §B · F ( s )d s  F ( s )d s  F ( s )ds ¸ ¨ ³ ³ ³ A 1 1 ¨A B C ¸  Residues F F ( s )ds  ¦ sj ³ E F ¨ ¸ 2S i F 2S i ¨  ³ F ( s )ds  ³ F ( s )ds ¸ ¨ ¸ E © D ¹

1 2S i

We note that

³

A

F

F ( s )ds

1 2S i

J  if

³J

 if

F ( s )ds , which is $

1

^ y ( s )`

(6.35)

y (t ) so we have to evaluate the following in

order to get an inverse Laplace transform of y ( s ) to find y (t ) ; that is:

y (t )

J  if

1

³

st

ke D

2S i J if s ( s  k )

J  if

1

ds

³  F ( s ) ds

2S i J if

C §B · ( )d ( )d F s s F s s   ¨³ ¸ ³B 1 ¨A ¸  Residues F  sj D E F ¨ ¸ ¦ 2S i ¨ ³ F ( s )ds  ³ F ( s )ds  ³ F ( s )ds ¸ ¨ ¸ D E ©C ¹

The case of D  1 has no effect on poles as only pole s

(6.36)

0 remains outside the contour of integration, and there are

¦ Residues

no other poles present in that region of Figure-6.2. Therefore, in that case the quantity

sj F

0 for

0  D  1 . We notice that for the case 1  D  2 we have two poles, namely s1 and s2 , which are in the region of

¦ Residues

integration. Thus, in this case we have a non-zero value of

sj

F , that is the sum given as

Residue s1 F  Residue s2 F These residue values are required for calculations.

Now, first, we must calculate the integral obtaining a total solution. y(s)

k s ( sD  k )

³

B

F ( s )ds 

A

k s ( sD  k )

With F ( s )

e

³

C

B

st

F ( s )ds 

³

D

C

F ( s )ds 

³

E

D

F ( s )ds 

³

F

E

st

y ( s )e , we see that if we put s

F ( s )ds , required for iT

re , the function

tends to zero as r n f . Therefore, we write the following as r n f and the integrations on the large

arcs AB and EF are zero; i.e.

Now, we try to find

³

D

C

³

B

A

F ( s )ds 

³

F

E

F ( s )ds

0 ; this is as per Jordan’s Lemma (see Appendix-G).

F ( s )ds . The region CD is a circle encircling the point s

0 ; Figure-6.2 shows it to be iT

iT

enlarged; in fact, it is a very small circle with its overall radius tending to zero. For s re , we write ds ire dT , the angle T varies from S to S as we move from point C to D . Therefore, with this substitution we write the following for D ! 0 : D

³

S

F ( s )ds

C

D

³ F ( s)ds

C

st

S

iT

ke (ir )e dT

st

ke idT

³ (re T )  (re T )D  k S³  rD e TD  k § S k  e idT · lim ¨ ³ D TD ¸ ¨ S r e  k ¸ © ¹ i

i

i

S



r p 0, s p 0

st

(6.37)

i

§ S · kidT lim r p0 ¨ ³ D iTD ¸ ¨ S r e  k ¸ © ¹ st

S

³S

idT

2S i

When the circle CD is tending towards s 0 from all sides, we may argue that e tends towards unity. Now with this argument and making the circle very small, by which we mean making r p 0 , we obtain

³

D

C

F ( s )ds

S

³S

i dT

2S i . That is derived in the above steps (6.37).

Fractional Differential Equations and their Analysis

Now we take the segments BC and DE in Figure-6.2. For BC , we write s write s

xe

 iS

and ds

(dx )e

 iS

265

xe

iS

so ds

(dx )e

iS

and for DE we

. Note that in Figure-6.2, the segments BC and DE are very close to the negative r iS

real axis, so for both the cases we may write s  x xe . Therefore, from B to C , the variable s varies from f to zero, or x  s varies from f to zero. On D to E , the variable s varies from zero to f , or x  s varies from zero to f . With these substitutions and observations, we write the following: C

E

³ F ( s)ds  ³ F ( s)ds B

D

0

ke

 xt

(dx )e

f

iS

³ ( xe S )( xD e SD  k ) ³ ( xe i



i

f

§

e

 xt

³ © x ( xD e

k ¨

dx

 iSD

0

e

r iSD

we

simplify

the

expression

(6.38)

 xt

 iS

within

the

 k)



e

 xt

 iS

D  iSD

(6.38)

 k)

· ¸  k) ¹

dx

D iSD

x( x e

bracket

(dx )e

)( x e

0

f

Now

ke

as

in

the

following

steps,

using

cos(SD ) r i sin(SD ) , and we get: e

 xt

D  iSD

x( x e

 k) e

e

 xt

D iSD

x( x e

 k)

 xt

1 1 § ·  D iSD ¨ D  iSD ¸ k x e k¹ x ©x e 1 § · D D  xt ¨  x cos(SD )  k  ix sin(SD ) ¸ e ¨ ¸ ¸ 1 x ¨ ¨¨  D ¸¸ D ©  x cos(SD )  k  ix sin(SD ) ¹ · D  xt § 2ix sin(SD ) e ¨ ¸ 2 2 x ¨  xD cos(SD )  k   xD sin(SD ) ¸ © ¹ e

Therefore, we write the following: C

³



 xt

D § · 2ix sin(SD ) ¨ 2D 2 D 2 ¸ x © x cos (SD )  2 x k cos(SD )  k ¹ E

f

³

F ( s )ds  F ( s )ds

B

(6.39)

k

³

e

 xt

D § · 2ix sin(SD )dx ¨ D 2D 2 ¸ x © x  2 x k cos(SD )  k ¹

0

D f

2ik sin(SD )

e

 xt

(6.40)

dx

³ x D  x D  2 xD k cos(SD )  k 1

2

2

0

Now we write the expression for y (t ) as:

y (t )



C E §D · F s s  F s s  F ( s )ds ¸  ¦ Residues s F ( )d ( )d ¨³ ³ ³ j 2S i © C B D ¹ f  xt · e dx 1 §  ¨ 2S i  2ik sin(SD ) ³ 1D 2D ¸ D 2 2S i ¨ x  x  2 kx cos(SD )  k ¸ 0 © ¹

1

 1

¦ Residues

k sin(SD )

S

sj

F

f

e

 xt

dx

³ x D  x D  2kxD cos(SD )  k  ¦ Residues 1

0

(6.41)

2

2

sj

F

266

Chapter Six

The above expression (6.41) is a solution for y (t )

$

1

^

k D

s( s k )

J  if

`

³J

ke st ds D  if s ( s  k )

, for the case of positive D ;

especially with 0  D  2 , the residues in the above solution matter. As we have seen that for 0  D  1 there are no residues, so the value of residue in that case is zero; but for the case 1  D  2 , there are two poles. As such, the residues need be calculated. We will calculate them shortly. First, we must say what happens if D  0 , that is, is negative? We represent this by considering D , where D ! 0 k s ( sD  k )

and perform the calculations for y ( s ) has no pole at s

ke st s ( sD  k )

with k ! 0 . We see now that F ( s )

0 . Therefore, the integral is

³

D

C

ksD 1e st (1 ksD )

, and that it

0 . The integral on the arcs AB and FE will be zero for

F ( s )ds

ksD 1 (1 ksD )

a large radius tending to infinity as the value of function y ( s )

is zero (one can apply L’Hospital’s rule to

y ( s ) , and see the limit as a radius which tends to infinity). Therefore we have to evaluate now only the integrals on

lines BC y (t )

 21S i

³

and DE , that is, the integrals given as

³

C

B

F ( s )ds 

³

E

D



F ( s )ds 

¦ Residues

C

B

³

F ( s )ds and

E

D

F ( s )ds . Therefore, we write

F.

sj

Let us evaluate now the integrals first as depicted below, with the same substitution as done earlier, (i.e. with s  x and ds dx , and with the same description for the limit for the integrations for s  x i.e. B to C and D to E , that is, f to zero, and zero to f for x , respectively).

³

C

F ( s )ds 

B

0

³

e

³

E

 xt

kx

³

e

kx e

ke

x

e

D iSD

kx e ke

 xt

³x

1D

0

(  dx )

1

 xt D 1 iS (D 1)

0

f

D 1 iS (D 1) D iSD

f f

F ( s )ds

D



³

e

 xt

kx

f



³

ke

D 1  iS (D 1)

e

D  iSD

kx e

0

dx

1

f

 xt D 1  iS (D 1)

x

e

D  iSD

kx e

0

(  dx )

1

dx

1 (6.42)

 iS (D 1) § eiS (D 1) · e ¨ kxD eiSD  1  kxD e  iSD  1 ¸ dx © ¹

f

2i sin S (D  1) · § ke  xt · § ³0 ¨© x1D ¸¹ ¨© k 2 x 2D  2kxD cos(SD )  1 ¸¹dx f

2 ki sin S (D  1)

³ x D k 1

0

e 2 2D

x

 xt

dx



D

 2 kx cos(SD )  1

We write the solution y (t ) in this case as follows:

y (t )



E §C · F s s F ( s )ds ¸  ¦ Residues s F ( )d  ¨³ ³ j 2S i © B D ¹

1



k sin S (D  1)

k sin SD

S

³ x D k 1

S 

f 2 2D

0

¦ Residues

f

³ x D k 1

0

e

sj

x

dx



D

 2 kx cos(SD )  1

(6.43)

F

e 2 2D

x

 xt

 xt

dx D

Residues  ¦

 2 kx cos(SD )  1

sj

F

Now we calculate the residues of poles at s j , for F ( s ) , which exist for the case 1  D  2 as s1 and s2 . Let us discuss this for 1  D  2 , so that we note that m

0 , and that we have s1

D

ke

i  DS

. This we also represent as

Fractional Differential Equations and their Analysis

D

s1

k cos  DS  iD k sin  DS . For m

complex conjugate of s1 i.e. s2

s1

V  iZ0 and s2 2

2

V  Z0

267

D

1 , the pole in the primary sheet is s2

s1 , meaning that s2

V  iZ0 , noting that

D

s1

Re > s @

V

ke

, which turns out to be a

k cos  DS  iD k sin  DS . For simplicity, we represent

Im > s @

D k cos  DS and Z0

2

D

k D . It may also be noted that s1 and s2 are roots of the equation s  k D

i  S D2 S

D

k sin  DS , with

0 ; which defines the poles of

( s  s1 )( s  s2 ) . With this, let us compute the residues of poles s1

function F ( s ) . As such, we can write s  k

and s2 in the following steps (from the theory of residue calculus, as explained in Section-1.7)

¦

2 j 1

Residues s j F lim

s o s1

lim

s o s1

( s  s1 ) ke

lim

( s  s1 ) ke st

 lim

ke

s o s2

s1t

s1 ( s1  s2 )



ke

 lim

s o s2

ke

( s  s2 ) ke

st

D

s(s  k )

s o s2

st

s ( s  s1 )( s  s2 ) s ( s  s2 )

 lim

D

s(s  k )

s o s1

ke

st

( s  s2 ) ke

st

s ( s  s1 )( s  s2 )

st

s ( s  s1 )

s2t

s2 ( s2  s1 )

§ e s1t e s2t ·  ¨ ¸ s1  s2 © s1 s2 ¹ V t  iZ0t V t iZ0t · e k § e  ¨ ¸ 2iZ0 © V  iZ0 V  iZ0 ¹ V t  iZ0t V t iZ0t ·  ( V  iZ0 )e k § ( V  iZ0 )e ¨ ¸ 2 2 2iZ0 © V  Z0 ¹ k

2

2

Using the above V  Z0

2

k D , sin x

1 2i

e

ix

e

 ix

,

V

k

1

D

cos  DS

and Z0

k

(6.44)

1

D

sin  DS ,

and then

simplifying further, we get the following:

¦

2 j

ke

Residues s j F 1 ke

2

2

V ( e

iZ0t

e

 iZ0t

)  iZ0 ( e

iZ0t

D

e

 iZ0t

)



 2iV sin Z0t  2iZ0 cos Z0t

V sin Z0t  Z0 cos Z0t

D

V t

Z0 k k

D

V t

Z0 k  ke

2iZ0 k

2

V t

2iZ0 k  ke

V t

2

 DD1

D

e

Z0

 k

1

D

cos  DS sin Z0 t  k

(6.45) 1

D

sin  DS cos Z0 t



V t

sin Z0t  DS

One may also use a similar method to calculate the residues of poles for D  0 , in the case of 2  D  1 . The above elaborate method says that it is difficult (but possible) to have an inverse Laplace transform using contour integration for functions with a ‘fractional’ Laplace variable.

6.7 Operational calculus of applying Heaviside units to a partial differential equation Oliver Heaviside (1871) developed the early theory for ‘semi-infinite’ systems described by partial differential equations. The one dimensional diffusion equation is:

268

Chapter Six 2

w u w x 2

c U

a

2

w u

2

(6.46)

w t

2

p with the parameters a and a RC for a semi-infinite heat conductor and semi-infinite lossy transmission k lines respectively. The equation is a standard Fick’s diffusion equation in one dimension with a diffusing quantity shown as u ( x, t ) , a variable of position and time. The initial condition is u ( x, 0) 0 for x ! 0 , and the boundary

u (0, t ) u0 and u (f, t ) 0 . Let the operator s { wwt { d t (as specified by Oliver Heaviside to solve differential equations). By putting this operator in the PDE (6.46), we obtain a differential equation in the position variable as follows, an ordinary second order differential equation in x : condition is

2

d u dx The roots of the above linear differential equations (6.47) are m

u (0, t ) e

by

u0

¦

f m 0

A ; giving the solution as u ( x, s ) m

( by ) m!

0

(6.47)

r a s , giving the standard solution as:



u ( x, s ) Setting the condition for x n f , that is, u ( x, t )

2

 a su

2





A exp  a sx  B exp  a sx



(6.48)

0 . Setting the boundary condition, we get x p 0 ;

0 , we get B





u0 exp  a sx . We expand this exponential as a power series

to get the following:



u ( x, s )

u0 exp  a sx f

u0  u0

¦



 ax s n

f

¦

(6.49)

n!

n 1

u0 

n

(  ax ) ( s )

n

2

u0 n! Segregating odd and even terms in (6.49) and with re-arrangement, we get the following: n 1

f

u ( x, s )

u0 

( ax )

2 m 1

¦ (2m  1)!

s

m

m 0

2n

f

 s u  ¦ ((2axn))! s u 1

2

0

n

(6.50)

0

n 0

1

The operators s 2 u0 and s n u0 in (6.50) are to be analysed further using a half (time) derivative of u0 , i.e. a constant 1

as d t 2 >u0 @

u0

  1

*

1 2

t

 12

u0

St

. Recognizing the fact that the n -th integer derivative of constant u0 is zero and n

thereby putting the values of identities with s { n

> @

n

1

dn dt n

where n is a positive-integer, and s

1

2

{

d1/2 dt1/2

, we

1

have d t u0 o s u0 { 0 and d t 2 >u0 @ o s 2 u0 { u0 / S t . Taking the above formula (6.50), we will change the variable from s to t . We obtain the solution by making the 1

above substitutions for s n u0 { 0 and s 2 u0 { u0 / S t in the obtained solution u ( x, s ) , and thereby get u ( x, t ) as follows:

§ dm ¦ ¨ m m 0 (2 m  1)! © dt f

u ( x, t )

u0 

u0 

( ax )

u0

2 m 1

f

( ax )

2 m 1

u0 

§ dm

¦ (2m  1)! ¨ dt S ©

m 0

u0 

ª u0  1 2 º · «¬ S t »¼ ¸ ¹

u0

S u0

f

( ax )

2 m 1

m

¦ (2m  1)! *   ( ax )

2 m 1

1 2

 m  1

*  12

¦ (2m  1)! *  S m 0

(6.51)

*   12  1

m 0 f

· 1 ªt  2 º ¸ ¬ ¼ ¹

1 2

t

1

 m t m  2 1

 12  m

Fractional Differential Equations and their Analysis

Using the property of the gamma function as *   m 

1 2

( 4)m m! S (2 m )!



u ( x, t )

u0

u0 

S

269

, we simplify the solution as follows:

f

( 1)

¦

m

( ax )

2 m 1

(6.52)

m ! (2 m  1)2 2 m t m  2 1

m 0

writing:

( ax )

(2 m  1)2 and using

¦

f m 0

( 1)m m!

2 m

ax 2 t

y

2 m 1 1 2 m m  2

³

{2

t

y

2m

(6.53)

dy

0

2

( x ) { exp(  x ) , we get the exact solution as follows:

u ( x, t )

u0 

2u0

S

ax 2 t

³ exp( y

2



)dy

u0 1  erf

0

 ax 2 t

(6.54)

This is the work of Heaviside and his operational calculus on a partial differential equation. Oliver Heaviside used the operator symbol as p for dtd ; since we knew that the Laplace operator replaces dtd with s , we have used that symbol. In a way, we have used the Laplace transform technique here and the concept of a half derivative too. We shall deal with a fractional partial differential equation with this Laplace and Fourier technique in Chapter-8.

6.8 The response of a fractional differential equation with detailed analysis of power law functions 6.8.1 Demonstration of power law functions as a solution to a simple fractional differential equation excited by a constant step function Here, we will see the solutions of a fractional differential equation to the standard inputs of step type power-law (or power series) functions, offering either a decaying response or a rising response. Consider the following fractional differential equation of fractional order 0  D  1 as:

V (t )

EW

D

d H (t )

D

or

H (t )

1

D

d V (t )

(6.55) D D D dt EW dt For the functions V (t ) and H (t ) , we consider them to be 0 for t  0 , and consider E and W as constants. The symbol V is a stress function, and the symbol H is for a strain function; in (6.55) they represent the dynamic equation of visco-elastic systems. We have generalized the differ-integral of order q

d dx q

p

ª¬ x º¼

q 0 Dx

p

ª¬ x º¼

* ( p 1) * ( p  q 1)

x

pq

q

for a monomial function

f ( x)

x

p

formula as

for p ! 1 and q  \ , from the Euler formula. The positive order is a

fractional derivative and the negative order is a fractional integration. The above fractional differ-integrals of order D for V (t ) and H (t ) are not equivalent for a general function; however, they are equivalent for functions which are expandable in terms of differ-integrable units, particularly for functions D

which can be expressed as a series of powers of t , and which are bounded for t the fractional integration of the constant V (t ) integration formula (6.55) with q

t t 0 governed by V (t ) equation V

EW

D

EW

V 0 ; ( V 0t ; p

D and p

D

D d H (t ) dt D

0 , we get H (t )

when V (t )

 , is like a power law H  dD H dt D

p

t

0 . The variable H (t ) is found by

0 ) to order D . So, applying it in the above differV0

1 E * (1D )

D

 Wt

. This is a time response of H (t ) for

V 0 is a constant. The time response of H (t ) as obtained from the D

W

  tW

.

270

Chapter Six

6.8.2 Demonstration of a power series function as a solution to the fractional differential equation excited by a constant step function We modify the above equation (6.55) as follows with E , E M and W as constants:

EW

 V0

WDE

>1@  EV

D 0 Dt

D

d H dt

V

D

EW

D

D

d V

E M dt

(6.56)

D

V 0 for t t 0 ) is obtained by operating 0 DtD on both sides of

The time response of H (t ) to a constant (that is, V (t ) the above equation (i.e. H (t )

D

). Now applying the Euler formula, and after rearrangement, we

0

M

get the following: D

§t·  ¨ ¸ EM E * (1  D ) © W ¹ V0

H

1

V0

(6.57)

The above expression (6.57) is a time response function H (t ) for t t 0 ; for constant input V (t ) for EW

D dD H dt D

V

EW D dD V E M dt D

0

. The solution is slightly difficult, as a time response function V (t ) to a step input

0 , for EW D

H (t ) H 0 applied at t

V 0 applied at t

dD H dtD

V

EW D dD V E M dt D

.

In the equation (6.56), we consider a different fractional order derivative in time ( E ) for the variable H (t ) , and rewrite the equation without a loss of generality so that we can have E = E M and write following:

EW We

V (t )

H (t ) H 0

set G

and D

 Wt ¦ ak  Wt

Using the formula



0 Dx

assume

E

d H

E

dt that

V

E

the

EW

D

D

d V

E M dt

time

,

D

response

; G ! 1 , for t t 0 , and put these in E M W

EMW

E

EMW

E

q

E

d H (t ) dt d

E

dt

ª¬ x p º¼



E

V (t )  W

E

>H 0 @ * ( p 1) * ( p 1 q )

pq

x

t

E dE H dt E

dt be

V W

V W

E

a

D

D

d V dt

power

D dD V

(6.58)

D

series,

shown

as

to get the following:

dtD

D

dt

D Dk

§t· ¦ ak ¨© W ¸¹

;

· D dD ª§ t ·G ¸¸  W D «¨ ¸ dt ¬«© W ¹ ¹

Dk §t· º ¦ ak ¨© W ¸¹ » »¼

(6.59)

p ! 1 in the above equation, we obtain the following:

§ § t ·G E M H 0W ¨¨ ¸ * (1  E ) ¨© © W ¹ E

will

E

d H

d V (t )

D

§ § t ·G ¨¨ ¨ ¸ ©©W ¹

EMW

E

E

Dk

¦

· ¸ © W ¹ ¸¹

§t· ak ¨ ¸

(6.60)

§ * (D k  G  1) · § 1 · D k G D D  W ¦ ak ¨ ¸ ¨ D k G ¸ t ¹ © * (D k  G  1  D ) ¹ © W





which we re-write as follows: EM H0

§t· ¨ ¸ * (1  E ) © W ¹ bk

We enter x

E

* (D k  G  1) * (D k  G  1  D )

 W in the above expression to get the following: t

D k G

§t· ¦ ak ¨© W ¸¹

D k G D

§t·  ¦ ak bk ¨ ¸ ©W ¹

(6.61)

Fractional Differential Equations and their Analysis

EM H0 * (1  E )

x

E

¦ a xD

k G

k

G

( a0 x  a1 x  ( a0 b0 x



¦ a b xD

k G D

2D G

 ...)

k k

D G

G D

271

 a2 x

G

 a1b1 x  a2 b2 x

G D

G

(6.62) D G

 a3b3 x

2D G

 ...)

D G

 ( a0  a1b1 ) x  ( a1  a2 b2 ) x  ( a2  a3b3 ) x a0 b0 x Comparing the coefficient of the RHS and LHS for (6.62), we can write the following: EM H0 G  D  E ; G D  E ; a0b0 ; ak  ak 1bk 1 0 * (1  E )

* D ( k  1)  E  1

bk

* D k  1  E b0

b1

2D G

* (D  E  1) * (1  E )

; a0

EM H0

EM H0

* (1  E )b0

* (D  E  1)

(6.63)

* (2D  E ) * (D  E  1) a1



a0 b1



EM H0

* (D  E  1)

* (D  E  1) * (2D  E )



EM H0 * (2D  E )

In general, therefore, we have: k

( 1) E M H 0

ak

(6.64)

* (D k  D  E  1)

H0

Inserting these values in (6.63) and (6.64), we obtain the function V (t ) , for constant H (t ) change in variable of x

as follows, with a

 Wt : V (t )

D E

E M H 0 ( x)

f

( 1)

k

¦ *(D k  D  E  1) ( x)D

k

t

x

This above treatment gives us the idea that the function V (t ) , for constant H (t )

(6.65)

W

k 0

H 0 for t t 0 , is provided in terms of

the power-series of time of the order ( k  1)D  E .

6.8.3 A demonstration of the power series function as a solution to a fractional differential equation excited by the constant step function through use of the Laplace transform This (6.65) can also be obtained from the Laplace transform technique. The Laplace transform of the equation is as described as follows:

­

$ ®E M W

E

E

d H (t ) ½

¯

E E

dt

E M W s Hˆ ( s ) where Vˆ ( s )

$ ^V (t )` and Hˆ ( s )

E

¾ ¿

­

$ ^V (t )`  $ ®W

¯

D

D

d V (t ) ½ dt

D

¾ ¿

(6.66)

D D

Vˆ ( s )  W s Vˆ ( s )

$ ^H (t )` are Laplace transforms, with the initial conditions V (t )

0 for t 0 and for t  0 . For a step function for H (t ) , say H 0 at t 0 , we have Hˆ ( s ) value in the above Laplace transformed equation, and re-arranging terms, we get the following:

H (t )

E E

Vˆ ( s )

EMW s H0

E M H 0W

(W s )

H0 s

0 and

; by entering this

E 1

(6.67) D D 1  (W s ) s 1  (W s ) For this function, no tabulated Laplace transform exists, and therefore we must determine another way to arrive at the solution. We can use the term-wise inverse Laplace transform if we break the above as a power series of s that is, by



power series expansion methods, as we discussed earlier in Section-6.4. The function 1  (W s )

D 1



expanded in a series with decreasing powers, and it will have a radius of convergence different from zero.

can also be

272

Chapter Six

Vˆ ( s )

(W s )

E M H 0W

E 1 D

1  (W s )

E M H 0W

E M H 0W

(W s )

1  (W s)

D

D

(W s ) (W s )

E 1 D

(W s )

E M H 0W (W s ) E M H 0W (W s )

E 1

D 1

1  (W s)

D 1

1  (W s)

E 1D

1  (W s)

E 1D

(6.68)

D

 (W s )

2D

 (W s )

3D

 ...



We can rewrite (6.68) in a compact form, as follows:

§

Vˆ ( s ) We have used the expansion formula (1  x) For the series Vˆ ( s )

¦

EM H0 W

f

E M H 0 ¨W

©

1

¦ (1)

k

(W s )

E 1D D k

k 0

2

· ¸; E D 1  0 ¹

(6.69)

3

1  x  x  x  ... in the above derivation (6.69).

k

k 0

f

( 1) (W s )

, we use a term-wise inverse Laplace transform for

E 1D D k

 t W ! 0 , using a known Laplace transform pair (that is,

$

^ ` s1q

tq (  q )!

, x ! 0:

x

1

1

tq * (1 q )

), so that we invert the above

(6.69) to get the result as described below: f

( 1)

D k D  E

k

§t· V (t ) E M H 0 ¦ ¨ ¸ k 0 * (D k  D  E  1) © W ¹ D E

E M H 0 ( x)

f

( 1)

k

t

W

; D  E ! 1 (6.70)

Dk

¦ *(D k  D  E  1) ( x) k 0

This (6.70) was obtained by a different method in (6.65):

V 0 applied at point t 0 to the following fractional differential equation (given with its Laplace transformed equation), the response H (t ) can be obtained as follows: Similarly for a step input V (t )

­

$ ®E M W

d

E

E

dt

¯

E E

E M W s Hˆ ( s ) For a step input function V (t )

V 0 applied at t

½

E

­

$ ^V (t )`  $ ®W

H (t ) ¾

¿

D

¯

D

d V½ dt

D

¾ ¿

(6.71)

D D

Vˆ ( s )  W s Vˆ ( s )

0 , the Laplace transform is Vˆ ( s )

V0 s

. Putting this in the above

form (6.71), we get the following:

1  (W s) V D

Hˆ ( s )

E E

EMW s

V0 EM

V0 EM

Conducting a term-wise inverse Laplace transform for

$

1

^ ` 1

s1q

q

t (  q )!

q

t * (1 q )

0

s

1  (W s) W D

(W s )

W (W s)

 tW ! 0 ,

(6.72)

E 1

 (1 E )

D  E 1

 W (W s )



using a known Laplace transform pair (that is,

) we carry out an inverse Laplace transform of the above expression (6.72) to get the result

below:

H (t )

V0 §

E

E D

·  ¨ ¸ E M ¨ * (1  E ) * ( E  D  1) ¸ © ¹

 tW

 tW

(6.73)

Fractional Differential Equations and their Analysis

For a case when D

273

E , we have:

V0 §

E

·  1¸ ¨ E M ¨ * (1  E ) ¸ © ¹

H (t )

 tW

(6.74)

This (6.74), we derived earlier too in (6.57).

6.8.4 Obtaining approximate short-time and long-time responses from a power series solution for a fractional differential equation excited by a constant step function In (6.74), we obtained a value for H (t ) that was easy to interpret, and show that the function is solely a power law type. It is easily seen that for the whole D and E parameter range, the function H (t ) is strongly increasing if and only if E  D t 0 is fulfilled. Now, we take both fractional orders to be the same in order to have a fractional differential equation as follows:

EW

E

d H

E

E

dt

EMW

E

V

EW

D

D

d V

E M dt

E

d H

D

,

E

V (t ) =

¦a x k

; x

D

E, (6.75)

E

d V

E

V W E E dt dt With the function V (t ) for a constant step input function, H (t ) H 0 , Ek

EM ,

we have the power law function as

t W , E k ! 1 . We then repeat the earlier steps to get the following: EM H0 E Ek E k E * (1  E )

¦a x

x



k

¦a b x k k

E

( a0  a1 x  a2 x  ( a0 b0 x

E

( a1  a0 b0 ) x

2E

 ...) E

 a1b1  a2 b2 x  a3b3 x E

(6.76)

 ( a1  a2 b2 ) x  ( a2  a3b3 ) x

bk

* ( E k  1) *( E k  1  E )

j



 a j 1b j 1 ( x ) 1

, b0

 ...)

 ( a0  a1b1 ) E

¦ a

2E

* (1  E )

,

2E

 .....

jE

b j 1

*  ( j  1) E  1 *  j E  1

We have: EM H0 * (1  E )

x

E

¦ a

j



 a j 1b j 1 ( x )

jE

,

*  ( j  1) E  1

b j 1

*  j E  1

(6.77)

From the above (6.77), comparing the RHS and LHS, we write the following for all j z 1 : a j  a j 1b j 1

By comparing the j

0,

a j 1

 a j / b j 1 ,

a j 1

a j

* 1  j E * 1  ( j  1) E

(6.78)

1 terms, we get the following condition:

a1  a0 b0 a1  a0

EM H0 * (1  E ) 1

EM H0

* (1  E )

* (1  E )

a0  a1* (1  E ) For a short time response, we can choose a1

(6.79)

EM H0

0 (actually it is so) and have a0

E M H 0 (i.e. at x

 t W | 0 ), with:

274

Chapter Six

a1 a2

 a0  a1

* (1)



* (1  E )

EM H0 * (1  E )

§ E H · * (1  E ) ¨ M 0 ¸ © * (1  E ) ¹ * (1  2 E )

* (1  E ) * (1  2 E )

(6.80)

EM H0 * (1  2 E )

We write from the above (6.80), the following:

aj Therefore, the function V (t ) is obtained for the constant H (t )

( 1)

EM H0

j

(6.81)

* (1  j E )

H 0 , which is: f

k

§t· V (t ) E M H 0 ¦ ¨ ¸ k 0 * (1  k E ) © W ¹ The earlier obtained result (6.70) says that V (t )

EM H0

¦

( 1)

( 1)k k 0 * (D k D  E 1)

f

the above obtained result (6.82). The instantaneous response V (t ) at t

D k D  E

 Wt

Ek

(6.82)

; here we set D = E , to get

0 , for a constant H (t )

H 0 is nevertheless

equal to E M H 0 , and, at short times, the response function V (t ) is ‘some type of exponential’ decay function. For viewing for longer periods of time the function V (t ) , we choose to make a0

0 (it is expected that a function at longer periods of time will have a very small value of the initial value of a function). From the j 1 term of the comparison from (6.81), we get: EM H0 EM H0 j 1 (6.83) a1 a j ( 1) , * (1  E ) * (1  j E ) The expansion for longer periods of time thus contains negative powers essentially limited to j E  1 : we have set the 1

condition above that j E ! 1 , since the fractional derivative is not defined for the larger power of x . Thus, for longer periods of time (i.e. t  W or x  1 ), the response is the power law of exponent E , as follows:

V (t ) | a1  Wt

E

EM H0 § t ·

E

(6.84)

¨ ¸

* (1  E ) © W ¹

6.9 An analytical method to obtain inverse Laplace transforms ‘without contour integration’ - the Berberan-Santos method 6.9.1 Development of the Berberan-Santos technique to obtain a distribution function for the decay rates for the relaxation function of time We offer an alternative method to obtain the inverse Laplace transform, rather than using residue calculus and the described contour integration method. This method was developed by Berberan-Santos, primarily for obtaining a relaxation rate distribution of a time decay function. We know that a simple decay function is of the type e

 k0 t

, with

k0 ! 0 as a relaxation-rate of decay. This simple decaying (or say relaxing) function has only one rate, i.e. k k0 . The function can have a complex way of decaying, meaning that it will have a distribution of several relaxation-rates. In this section, we carry out an inverse Laplace transform of the time domain response of decay of the function to call it f (t ) , in order to obtain the ‘distribution of relaxation rates’, for a complex decay process that is governed by several rates ( k ' s) of an exponential decay. A generally complex decay in time is depicted in Figure-6.3 and Figure-6.4 denoted by the Mittag-Leffler functions ED , E ( t ) . The physical decay function, however, cannot be like that shown in Figure-6.5, which is an unbounded function. Figures 6.3, 6.4 and 6.5 depict the Mittag-Leffler function ED , E ( t ) with E (1)

ED , E ( t )

d dt

1 and its first derivative (i.e.

ª¬ ED , E ( t ) º¼ ) for 0  D  1 , for 1  D  2 and 2  D  3 respectively. We see that for cases of

0  D  1 , the function ED ,1 ( t ) is monotonically decreasing, and is, therefore, a decaying function. For the cases of

Fractional Differential Equations and their Analysis

275

1  D  2 , the function ED ,1 ( t ) has a decaying nature with oscillations. On the other hand, for cases of 2  D  3 , the function ED ,1 ( t ) unboundedly grows with oscillation. The technique discussed here is developed to use an inverse Laplace transform to depict distribution rates or histogram functions (call them H ( k ) ) or time functions (call them f (t ) ), i.e. H ( k )

$

1

^ f (t )` . However, this technique can

be used to provide an inverse Laplace transform of a frequency domain function called G ( s ) to get g (t ) as 1

g (t ) $ ^G ( s )` . Here the rigor of contour integration is eased, and what we get most of the time is an integral representation of an inverse Laplace transform function. First, we will build the essential formulations for Laplace inversion using the Berberan-Santos method, by extracting a histogram function H ( k ) from f (t ) , and give some applications of an inverse Laplace transform. The complex decay may be expressed as follows with several rate constants k1 , k 2 , k3 ,... with weights a1 , a2 , a3 ,...

For a normalized case, we can have f (0) follows:

¦a

n

 k1t

 k2 t

f (t )

a1e

f (0)

a1  a2  a3  ...

 a2 e

 a3e

 k3t

¦a

¦a e

 ...

n

 kn t

(6.85)

n

1 . In the continuum limit, we may write the above (6.85) as f

f (t )

³  H (k ) e

 kt

dk

f (0)

1

(6.86)

0

where H ( k ) is a distribution function (the histogram function) of the rate of the relaxation of the decay process, f (t ) depicted in Figure-6.6. While the discrete rates are shown here as a series of Dirac delta functions, the rate of the distribution function is the following: H ( k ) a1G ( k  k1 )  a2G ( k  k 2 )  a3G ( k  k3 )  ... (6.87) a n G ( k  k n )

¦

ED , E ( t ) 0  D  1,

E

1

1.0

E 14 ,1 (t )

0.75 0.5 0.25 0 0.25 0.5 0.75 1.0

Figure-6.3: Plot of function ED , E ( t ) , D

2

4

6

8

E 1(1)4 ,1 (t ) 0.25 , E

1 and its derivative

10

t

276

Chapter Six

1.0

E7(1) ( t ) 4 ,1

E7 4 ,1 ( t )

0.75 0.5 0.25

t

0

10

30

20

40

50

0.25

ED , E ( t )

0.5

1  D  2, Figure-6.4: Plot of function ED , E ( t ) , D

1.75 , E

1

E

1 and its derivative

The Laplace transform F ( s ) of a function in a time domain f (t )

F (s)

f

³  f (t ) e 0

 st

is defined as the Laplace integral (i.e.

dt ). This is the Laplace integral and is a standard integral transform from a time domain to a

complex frequency domain s

Re > s @  i Im > s @ ; where the real part is significant in a transient response and the

imaginary part of the frequency corresponds to a ‘steady-state’ response. Here, f (t ) is the inverse Laplace transform of F ( s ) (i.e. f (t )

$

1

^F ( s )` ). E9 4 ,1 ( t )

E9(1) ( t ) 4 ,1 2 1

t

0

20

40

60

100

80

1

ED , E ( t ) 2  D  3, Figure-6.5: Plot of function ED , E ( t ) , D

2.25 , E

1

E

1 and its derivative

Compare the Laplace integral with expression (6.86) (i.e. f (t )

f

³  H (k ) e

f

f (t )

 tk ³  H ( k ) e dk 0

 kt

0

dk ) as follows:

f

F (s)

³  f (t ) e

 st

dt

(6.88)

0

Both of the above expressions are Laplace transform expressions, with the first one transforming the function H ( k ) from k domain to t domain; while the second one is transforming f (t ) from t domain to s domain. However, both are Laplace transforms with only a change of variable and symbol. Therefore, we can say that the function H ( k ) is an inverse Laplace transform of f (t ) in the first expression (6.88) and the function f (t ) is an inverse Laplace transform of F ( s ) in the second expression (6.88). Therefore, in order to obtain the rate of the distribution-function H ( k ) from the decay curve, say f (t ) , we need to have an inverse Laplace transform of the time function f (t ) like the conventional definition of the inverse Laplace transform.

Fractional Differential Equations and their Analysis

277

f

¦a e

f (t )

n

f (t )

 kn t

³ H ( k )e

 kt

dk

0

The decay function

t

a3 a1

a2

The rates of decay

xxxx

k1 k2

k3

k4

k5

kn

Distribution function for rates

H (k )

k

Figure-6.6: Rate of decay function

$ 1 H (k ) Rate Domain

k

$

Time Domain

f (t )

F ( s)

Complex-frequency Domain s

$

$ 1

H (k )

$ 1 ^ f (t )`

1 c  if f (t )etk dt 2S i ³c if

f (t )

$ ^ H (k )`

³

f

0

f (t ) $ 1 ^ F ( s )`

H (k )e  tk dk

F ( s ) $ ^ f (t )`

1 c  if F ( s )e st ds 2S i ³c if

³

f

0

f (t )e st dt

Figure-6.7: Laplace inversion and Laplace transform

We get the following expressions for inverse Laplace transforms: f (t )

1

c  if

³

2S i c if

st

 F ( s ) e ds

1

H (k )

c  if

³  f (t ) e

2S i c if

tk

dt

(6.89)

In Section-6.6, we explored out how to carry out inverse Laplace transforms with contour integration. Figure-6.7 describes this process of inversion of the Laplace transform. In the above expression, c is a real number larger than

>@

c0 , where c0 is such that f (t ) has some form of singularity on the real line Re t

c0 but is analytic in the

>@

complex plane to the right of that line, i.e. for Re t ! c0 . The inverse Laplace transform is usually carried out by contour integration (see Appendix-G for more detail). We will develop the formula, where we carry out an inverse Laplace transform from complex s

Re > s @  i Im > s @ V  iZ for a function in s domain called G ( s ) to get the t

domain function g (t ) such that the Laplace transform expression is G ( s ) transform expression is g (t )

V +if 1 2S i V -if

³

 G ( s ) e st ds .

f

³  g (t ) e 0

 st

dt and the inverse Laplace

278

Chapter Six

6.9.2 Derivation of the Berberan-Santos method We will not find H ( k )

$

1

^ f (t )` , but we need to find

g (t )

$

1

^G ( s )` .

However, the methods remain the same

for finding H ( k ) , i.e. the relaxation rate distribution function from a relaxation function f (t ) , where the time variable is considered a complex quantity. This was originally formulated by Berberan-Santos, and we shall be adapting this technique. We develop in this section an analytical Laplace inversion technique in order to obtain the inverse Laplace transform given as g (t ) integral

$

V +if 1 2S i V -if

³

1

V +if 1 2S i V -if

^G ( s )`

³

 G ( s ) e st ds )

 G ( s ) e st ds .

Usually the solution of this integral (i.e. the inverse Laplace

is obtained via the contour integration that we discussed in Section-6.6 (see

Appendix-G for more information). Here, we develop an analytical technique to find the inverse Laplace transform without using the usual contour integration method. Here we describe the Berberan-Santos method of evaluation of the Laplace inversion without going for contour integration. Here is the real part, i.e. V is constant as a vertical line (a Bromwich-path), with V V 0 as a constant.

>@

The variable V 0 is a real number being such that G ( s ) has some form of singularity on the line Re s

V 0 , but an

>@

analytic in the complex plane to the right of that line, i.e. for Re s ! V 0 . Performing the variable change on a Laplace integral to s

V 0  iZ , we get the following steps: g (t )

V 0  if

1

G(s) e 2S i V ³

st

ds ;

s { V 0  iZ

0  if

1

V 0  if

2S i ³V 1 2S i

0  if

V 0  if

³V

0  if

 G (V 0  iZ )  e(V iZ )t  d(V 0  iZ ) ; 0

 G (V 0  iZ )  e

V 0t f

e

2S Writing e

itZ

³ G (V

tV 0 itZ

e

dV 0

0 (6.90)

idZ

iZt

0

 iZ ) e dZ

f

cos Zt  i sin Zt , we get the following form: V 0t

g (t )

f § f · ¨ ³  G (V 0  iZ ) cos(Zt )dZ  i ³  G (V 0  iZ ) sin(Zt )dZ ¸ 2S © f f ¹

e

(6.91)

writing:

G (V 0  iZ )

Re >G (V 0  iZ )@  i Im >G (V 0  iZ )@

(6.92)

and place (6.92) in the preceding expression (6.91) to get the following expression: V 0t

g (t )

§ f · ¨ ³  Re >G (V 0  iZ )@  cos(Zt )   Im >G (V 0  iZ )@  sin(Zt ) dZ ¸ 2S © f ¹ f V t · e 0 § i ¨ ³  Im >G (V 0  iZ )@  cos(Zt )   Re >G (V 0  iZ )@  sin(Zt ) dZ ¸ 2S © f ¹

e

(6.93)

Given that g (t ) is a real function, we get the following (i.e. by equating the imaginary part to zero):

§ f · ¨ ³  Im >G (V 0  iZ )@  cos(Zt )  Re >G (V 0  iZ )@  sin(Zt ) dZ ¸ 0 © f ¹

(6.94)

Thus, the above expression (6.93) for g (t ) reduces to the following (i.e. considering only the real part) and we get: V 0t

g (t )

§ f · ¨ ³  Re >G (V 0  iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ ¸ 2S © f ¹

e

(6.95)

Fractional Differential Equations and their Analysis f

³  g (t ) e

However, we have from a Laplace integral G ( s )

 st

0

following:

279

dt ; and by putting here s

V 0  iZ , we get the

f

³  g (t ) e

G (V 0  iZ )

 t (V 0  iZ )

dt

0

f

(6.96)

f V 0 t

V 0 t

³ e  g (t ) cos(Zt )dt  i ³ e  g (t ) sin(Zt )dt 0

0

This gives the below: f

Re >G @

0t

V ³ e  g (t ) cos(Zt )dt 

0

(6.97)

f

Im >G @

³

 e

V 0 t

 g (t ) sin(Zt )dt

0

> @

We find in (6.97) that function Re G is an even-function in the variable Z (call it Feven (Z ) ) and the function

Im >G @ is an odd-function in variable Z (call it Fodd (Z ) ). We get:

 Re >G@ cos Zt  Im >G@ sin Zt

(6.98)

 Feven (Z ) cos Zt   Fodd (Z ) sin Zt

as an ‘even-function’. That is the even function Feven (Z ) multiplied by another even function (i.e. cos Zt ), which gives a further even function; and an odd-function Fodd (Z ) multiplied by an odd function (i.e. sin Zt ) gives an even function. Therefore for an overall even-function integrand, we have the following integral: V 0t

e

g (t )

2S

f

³  Re >G@ cos Zt  Im >G@ sin Zt dZ f

V 0t

e

S

(6.99)

f

³  Re >G@ cos Zt  Im >G@ sin Zt dZ 0

0 for t

Further simplification of (6.99) can be extended by considering ‘causality’ in g (t ) i.e. g (t ) g (t ) z 0 for t t 0 . For a case V 0

2

f

³  Re >G@ cos Zt dZ

 S2

f

³  Im >G@ sin Zt dZ

0

and

for

tt0

. We are not proving this simplification. One can do this from (6.99) by putting positive t and writing g (t )

0 for

0 we get g (t )

S

0

0

negative t , then adding the two cases (or subtracting the two cases) Write (6.92) in polar form as described below

G (V 0  iZ )

U (Z )

U (Z )e

iT (Z )

G (V 0  iZ )



U (Z ) cos T (Z )  i sin T (Z ) T (Z )

‘G (V 0  iZ )



(6.100)

to get alternate formulas from (6.99) as follows: V 0t f

g (t )

e

S

³  Re >G (V

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

V 0t f

e

S

³  U (Z )  cos T (Z ) cos(Zt )  U (Z ) sin T (Z ) sin(Zt ) dZ

(6.101)

0

V 0t f

e

S

³  U (Z )  cos Zt  T (Z ) dZ 0

We will demonstrate the use of the formulae of (6.99) and (6.101) to obtain the inverse Laplace transforms of various functions and write the equivalent integral representations. The integrals obtained are, however, difficult to solve in closed form, but can be used to get presentation that is useful for numerical integration.

280

Chapter Six

6.10 A few examples of the inverse Laplace transform of functions available in standard Laplace transform tables obtained using the Berberan-Santos method ( s  a ) 1

6.10.1 The inverse Laplace transform of function: G ( s ) Putting s

1

V 0  iZ with V 0 ! a in G ( s )

( s  a ) , we write the following:

V0  a

1

G (V 0  iZ )

2

(V 0  a )  iZ

(V 0  a )  Z

Z

i

2

2

(V 0  a )  Z

2

V0  a

Re >G (V 0  iZ )@

2

(V 0  a )  Z

Im >G (V 0  iZ )@



(6.102)

2

Z 2

(V 0  a )  Z

2

Applying (6.99), we get the following: V 0t f

g (t )

e

³  Re >G (V

S

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

V 0t

e

³

f

(V 0  a ) cos(Z t)dZ 2

(V 0  a )  Z

0

S V 0t

e

(V 0  a )

³

2

V 0t

e

³

f

From the Laplace tables we know that ( s  a )

Z sin(Zt )dZ

f

2

(V 0  a )  Z

0

V 0t

(V 0  a )  Z

2

e



³

2

(V 0  a )  Z

^ `.

1

$ e

at

2

2

(6.103)

Z sin(Zt )dZ

f

2

(V 0  a )  Z

0

S

(V 0  a ) cos(Zt )  Z sin(Zt )

0

S

³

S

cos(Zt )dZ

f

0

S

2

V 0t

e



2

dZ

Therefore, we have an integral representation of e

at

as

follows: V 0t

e

³

b , b ! 0 , we have $

1

^( s  b) ` 1

e

(V 0  a ) cos(Zt )  Z sin(Zt ) 2

(V 0  a )  Z

0

S For a

f

 bt

2

e

dZ

at

(6.104)

0 ! b , in the formulae of (6.99) and (6.104)

. We choose V 0

to write the following:

1

S³

f

b cos(Zt )  Z sin(Zt ) 2

b Z

0

2

dZ

e

 bt

(6.105)

In polar form, we obtain the following using the formula (6.101):

G (V 0  iZ )

V0  a 2

(V 0  a )  Z

T (Z )

‘G (V 0  iZ )

U (Z )

G (V 0  iZ )

tan

i

2

1



Z 2

(V 0  a )  Z Z

(V 0  a )

2

,

(6.106)

1 (V 0  a ) 2  Z 2

We apply (6.101) to get the below: V 0t f

g (t )

e

³  U (Z )  cos Zt  T (Z ) dZ

S

0

V 0t

e

³

0

S Thus, we have another formulation for e

at

S



1 (V 0  a ) 2  Z 2

(6.107)



cos Zt  tan

1



Z

(V 0  a )

dZ

as follows:

V 0t

e

f

³

f

0



1 (V 0  a ) 2  Z 2



cos Zt  tan

1



Z

(V 0  a )

dZ

e

at

(6.108)

Fractional Differential Equations and their Analysis

For the case of G ( s )

1

( s  1) , it has singularity at s 1

S ³

1 . We choose V 0

cos Zt  Z sin Zt

f

1 Z

0

1

³



f

1

(1Z 2

0

S

2

dZ

e

281

0 ! 1 , and we write the following:

t

(6.109)



cos Zt  tan

1

 Z dZ

e

s ( s 2  1) 1

6.10.2 The inverse Laplace transform of the function: G ( s ) For a change of variable s

1 ! 0 , we have the following with s

V 0  iZ with V 0

s

G(s)

2

s 1

,

1  iZ : 1  iZ

G (1  iZ )

>

Re G (1  iZ )

@

2Z

2

4Z

4

(6.110)

2

(1  iZ )  1

ri and at line Re > s @ 0 . We choose V 0

The above expression has singularity at s

at

1 , and, in this case, we get:

>

Im G (1  iZ )

@



Z

2

4Z

4

(6.111)

Using (6.99), we write the following: V 0t f

g (t )

e

S

³  Re >G (V

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

e

t f

S

³  Re >G (1  iZ )@ cos(Zt )  Im >G (1  iZ )@ sin(Zt ) dZ 0

e

t f

e

t

2 § 2  Z2 · Z t t cos( )  sin( ) Z Z ¨ ¸dZ 4 4 4Z S ³0 © 4  Z ¹

S e 1

2

(2  Z ) cos(Zt )dZ 4Z

0

t

S The known Laplace transform is $

³

f

³

f

2

4

e

t

S

³

2

Z sin(Zt )dZ 4Z

4

2

2

4Z

`

f

0

(2  Z ) cos Zt  Z sin Zt

0

^s / ( s



(6.112)

4

dZ

cos t , so we have an integration representation of cos t as

 1)

follows:

e

t

S In polar form, we have G ( s )

2

³

f

2

4Z

0

2(Z 4 Z 2  2)

1

s ( s  1) , U (Z )

2

(2  Z ) cos Zt  Z sin Zt

4

Z 4

; T (Z )

4

tan

1



dZ

Z2 Z2 2

cos t

(6.113)

. Using (6.101), we write the

following: V 0t f

g (t )

e

³  U (Z )  cos Zt  T (Z ) dZ

S

0

e

t f

³

S

0

2(Z 4 Z 2  2)

Z4 4

 

(6.114)

cos Zt  tan

1



Z2 Z2 2

dZ

We have another representation of cos t as follows: t f

 S ³

e

0

2(Z 4 Z 2  2)

Z4 4

 

cos Zt  tan

1



Z2 Z2 2

dZ

cos t

(6.115)

282

Chapter Six

6.10.3 The inverse Laplace transform of function G ( s ) The function G ( s )

e

 O0 s

has no singularity at s ! 0 , so V 0

variable the function as G (V 0  iZ ) as

Re >G (V 0  iZ )@

g (t )

$

1

G (iZ )

e

 iZO0

e

 O0 s

0 must be chosen. Thus, we write in complex

cos(ZO0 )  i sin(ZO0 ) . Thus, we have real and imaginary parts

cos(ZO0 ) ; Im >G (V 0  iZ )@

 sin(ZO0 ) .

Applying

the

formula

(6.99),

we

get

^e ` as follows:  O0 s

V 0t f

g (t )

e

S

³  Re >G (V



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

0

1

³  cos(ZO ) cos(Zt )  sin(ZO ) sin(Zt ) dZ 0

S

(6.116)

0

0

1

S

f

³

f

0

cos Z (t  O0 ) dZ

From the standard Laplace tables, we have $

1

^e `  sO0

G (t  O0 ) ; thus we get the integral representation as

follows:

1

0

S In polar form, we have U (Z )

1, T (Z )

³

f

cos Z (t  O0 ) dZ

G (t  O0 )

(6.117)

 ZO0 . Using (6.119) we write: V 0t f

e

g (t )

³  U (Z )  cos Zt  T (Z ) dZ

S

0

1

(6.118)

f

³



cos Zt  ZO0 dZ S 0 This is the same that we obtained in (6.117) using the formula given in (6.99).

s D

6.10.4 The inverse Laplace transform of function G ( s ) The function G ( s )

s

D

does not have a singularity at s ! 0 . Choosing V 0

G (V 0  iZ ) Applying the formula (6.99), we get g (t )

$

1

e

S

0 , we obtain:

cos  DS2  iZ

D

sin  DS2

(6.119)

D

³  Re >G (V

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

1

f

³ Z

D

S

cos  DS 2 cos(Z t )  Z

1

f

D

cos Zt  DS 2 dZ

0

1

D



sin  DS 2 sin(Z t ) dZ

(6.120)

0

Z S ³ D With the known Laplace transform (i.e. $ ^s ` 

D 1 1 ), * (D ) t

1

S ³

f

0

Putting (D  1)

Z

D

^s ` as the following:

V 0t f

g (t )

(iZ )

D

Z

 n , we write the integral representation of t

we write following:

D

n

* (1  n )

S

1

cos Zt  DS 2 dZ

* (D )

t

D 1

(6.121)

as follows:

³

f

0

Z

n 1



cos Zt 

(1 n )S 2

With use of the polar form formula (6.101), we will get the same result as in (6.120):

dZ

t

n

(6.122)

Fractional Differential Equations and their Analysis

283

6.10.5 The inverse Laplace transform of function: G ( s ) 1 Taking G ( s )

g (t )

$

1

1 and choosing V 0

0 , we have for s

V 0  iZ and G (V 0  iZ ) 1  i (0) . Using (6.99), we write

^1` as follows: V 0t f

g (t )

e

³  Re >G (V

S

1

S 1

S Knowing $

1

^1`

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

f

³  (1) cos(Zt )  (0) sin(Zt ) dZ

(6.123)

0

³

f

0

cos(Zt )dZ

G (t ) , we write the integral representation of G (t ) as: 1

S³

f

0

cos(Zt )dZ

G (t )

(6.124)

6.10.6 The inverse Laplace transform of the function G ( s ) We take G ( s ) to write g (t )

s

$

1

1

and, with V 0

0 , we write G (V 0  iZ )

Z

1

 0  i (1) .

We use the formula in (6.99)

1

g (t )

e

³  Re >G (V

S 1

1



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

f

S

³ Z

1

f

1

(0) cos(Zt )  (1)Z

1



sin(Zt ) dZ

(6.125)

0

0

^s `

0

0

S ³ 1

1

^s ` as follows: V 0t f

We know that $

(iZ )

s 1

sin Zt

Z

dZ

u (t ) (i.e. Heaviside unit step function at t

0 ), so we will write the integral representation

of u (t ) as: 1

S ³

f

sin Zt

0

Z f 1 sin Zt

S

³

0

Z

dZ

u (t )

1, t t 0

(6.126) dZ

u (t )

0, t  0

6.11 Examples of using the Berberan-Santos method to obtain the Laplace inversion of a few functions that are not provided in standard Laplace Transform tables The Berberan-Santos method allows the provision of integral representation of the inverse Laplace transforms of some transfer functions that are useful in a fractional calculus context. Many such functions are however, not listed in the standard Laplace transform tables; we will utilize either the contour integration technique or the Berberan-Santos method to arrive at an expression for inverse Laplace transforms.

6.11.1 The inverse Laplace transform of function G ( s )  s / s

E

0 We obtain G ( s ) e in a complex variable by putting s representation in polar form:

e

 s / s0

V 0  iZ with V 0

E

0,

as in the following

284

Chapter Six

G (iZ )

e

 iZ / s0

e e

E

e

E

Z / s0 ( i ) E

E Z / s0 ª«cos ¬

ª « ¬

 Z

E

s0

cos

 ES i sin  ES º»¼ 2

 ES »¼º 2

2

e

ª «i ¬

E

(6.127)

 sin  »¼º Z

ES

s0

2

 U (Z )  eiT (Z ) whereas in this case we have in polar form a representation for the use of (6.101):

G (V 0  iZ )

U (Z )

‘G (V 0  iZ )

^

Therefore, we write an inverse Laplace transform $ 1 e

 s / s0

T (Z )

`

E

e

ª « ¬



 Z

E

s0

 Z

cos

 ES º»¼ 2

(6.128)

E

sin

s0

 ES 2

as g (t ) in the following expression:

V 0t f

g (t )

e

³  U (Z ) cos Zt  T (Z ) dZ

S

0

f ª « ¬

1

e S ³

 Z

E

s0

cos

(6.129)

 ES º»¼



2

cos Zt 

0

By changing the variable u

 Z

s0

S

f ª E  u cos ¬«

³

e

 ES º¼»



2



1  (1  E )

 s s0

transfer function by putting s

 1(1E )

ES 2

ES

0

(6.130)

2

1  (1  E )  s s0

 1(1E )

. We use the following steps to find an inverse Laplace transform of this

V 0  iZ with V 0

0:

1

1  (1  E )  iZ s0

G (iZ )

 dZ

 du

E

cos ts0 u  u sin

6.11.2 The inverse Laplace transform of function G ( s )

G (iZ )

sin

s0

Z / s0 , we obtain: g (t )

We take G ( s )

E

U (Z )



1



1 1 E

(1 E )Z s0

1 2  2 (1E )



; ‘G (iZ ) T (Z )

Therefore, we write the inverse Laplace transform (i.e. $

1

^

1  (1  E )

tan 

 s s0

1





(1 E )Z s0



(6.131)

1 E 1

(1 E )

`

g (t ) ) as follows:

V 0t f

g (t )

e

S

³  U (Z ) cos Zt  T (Z ) dZ ;

1

S With a change of variable u

(1 E )Z s0

g (t )

V0

0

0

f

ª

³ ¬«1  

(1 E )Z s0

0

2

º¼»

 2 (11E )

(6.132) tan 1  (1sE ) Z · § 0 cos ¨ Zt  ¸ dZ 1 E © ¹

, we get the following expression:

s0

f

1 2  2 (1E )

1  u S (1  E ) ³ 0

cos



tW 0u  tan 1 u 1 E

du ;

u

(1  E )Z

s0

(6.133)

Fractional Differential Equations and their Analysis

285



We conduct an inverse Laplace transform of a simple power law in s domain (i.e. G ( s ) function g (t )

$

1

^1   s D a

1

`

G (iZ )

. Putting s

; D  1 ) to get the

1

iZ D a



Z D

D



 a cos  DS2  i sin  DS2

1   a (i ) 1

1

1 D 1  s / a



0  iZ , we write the following steps:

1 1 

D 1

1   as

6.11.3 The inverse Laplace transform of the function G ( s )

1

 a

Z D

e

i DS 2

1

Z D

(6.134)

1

1  

Z D a





cos  DS 2 i

Z D a



sin  DS 2



From the above (6.134), we get the following: D

Re >G (iZ )@

 Za

2D

cos  DS 2 1 D

 2  Za cos  DS 2 1 Z D

(6.135)

 a sin  2 D   2  Za cos  DS2  1 1 1 $ ^G ( s )` ) with G ( s ) , as follows, using the formula 1  s / a

Im >G (iZ )@ We carry out an inverse Laplace transform (i.e. g (t )

 Za 

DS

Z 2D a

D

(6.99): V 0t f

e

g (t )

S

³  Re >G (V

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

D §§  Za cos  DS2  1 ·¸ cos(Zt ) ·¸ ¨¨ 2D Z D DS ¸ f ¨ Z ¸ 1 ¨ ©  a  2  a cos  2  1 ¹ ¨ ¸dZ S ³0 ¨ § Z D DS · ¸  a sin  2 t sin( Z ) ¸ ¨  ¨ Z 2D ¸ ¨ ¨   2  Z D cos  DS  1 ¸ ¸ a a 2 © ¹ © ¹ Z D cos DS 1 cos Z t  Z D sin DS sin Z t  a   2 · Z 1 f §  a  2 dZ ; u D 2D ¨ ¸ ³ Z Z DS  a  2 a cos 2 1 a S 0 © ¹ a f §  uD cos DS2 1 cos aut   uD sin  DS2 sin aut · ¨ ¸du 1 u 2 D  2 uD cos  DS 2 S ³0 © ¹

(6.136)

6.11.4 An integral representation of the Mittag-Leffler function using the Berberan-Santos method The listed inverse Laplace transform of a two-parameter Mittag-Leffler function (see Appendix-A) is defined as ED , E ( z )

that is $

¦ 1

^

( z )m m 0 * (D m  E )

f

1 D 1  s / a

`

a

S

we have g (t )

³

f

0

$

1

^

1 D 1  s / a

`

D 1

a  at

 E  (at ) . From the obtained expression D

D ,D

§  uD cos DS2 1 cos aut  uD sin  DS2 sin aut ·du (6.136), we write following expressions: ¨ ¸ 1 u 2 D  2 uD cos  DS 2 © ¹

286

Chapter Six

 E  (at )

D 1

D

a ( at )

D ,D

u

f

a

S ³



u

0



at

u

f 1D

( at )

S

D

2D

 2u cos  D



DS

1

2

Z

du ; u

a



D

ED ,D ( at ) 1



DS cos  DS 2  1  cos( atu )  u sin  2 sin aut

D

³





(6.137)



DS cos  DS 2  1  cos( atu )  u sin  2 sin aut

D

u

0

2D

D

du

 2u cos  DS 2 1 D

z



ED ,D  z

f

1

D

z

S

1D

³

u

D





D

DS



1

DS cos  DS 2  1  cos( zu )  u sin  2 sin zu

u

0

2D

 2u cos  D

2

du

We obtained an integral representation of the Mittag-Leffler function via the Berberan-Santos method, i.e.



ED ,D  x With D

f

1

D

1D

x

S

1 and a 1 , we get g (t )

D



1

y

^11s ` 1

 D

S

t

2D

D

³

u

D

 2 y cos  D

DS



1

2

(6.138)

dy

t

E1,1  t f

1D



DS cos  DS 2  1  cos( xy )  y sin  2 sin xy

0

$

ED ,D t

³

y

e . The integral representation is as follows:



D

DS

1





DS cos  DS 2  1  cos(tu )  u sin  2 sin tu

u

0

2D

 2u cos  D

2

du (6.139)

1

D

E1,1  t

1

f

S ³ 0

cos(tu )  u sin tu 2

u 1

Earlier in the expression (6.110), we obtained $

1

^11s `

du 1 S

³

e f

0

received from the Mittag-Leffler function in (6.139).

t

cos Zt Z sin Zt dZ 1Z 2

t

e , which is the same as we

Also, using the expression (6.136), we can write the following useful identity of the Laplace transform pair: k 1

$

^ ` ks D

D

k

0

S u

³

f







§  uD cos  DS2  1 cos ut D k   uD sin  DS2 sin ut D k ¨ 2D D ¨ u  2u cos  DS 2 1 ©

·¸du ¸ ¹

(6.140)

Z D

k

6.11.5 The inverse Laplace transform of the function G ( s )

sD 1 ( sD  1) 1

We will apply the Berberan-Santos method to a known Laplace pair of the Mittag-Leffler function

^

D

`

$ ED ( t )

s

D 1

D

( s  1)

1

(see Appendix-A). For the function G ( s )

representation of the inverse Laplace transformed result, we write s

s

D 1

D

1

( s  1) , in order to get an integral

V 0  iZ with V 0

0 . This is because we do

>@

not expect singularity in the right half plane of complex frequency s (i.e. in region Re s ! 0 ) for function G ( s ) for its well-meaning behavior. With this substitution, we get the following steps:

Fractional Differential Equations and their Analysis

s

G(s)

D 1

1 s

0  iZ

s

D D 1

(iZ )

G (iZ )

Z

D

Z

D 1

1  Z Z

D 1

D 1

(i )

D

1  (iZ )

1  (iZ ) (i )

 cos  DS2  i sin  DS2

1



DS cos  DS 2  i Z sin  2

D

D

 sin   i cos  1  Z

1  Z

D

D

D

DS 2

1  2Z cos  D

Z

(6.141)

DS cos  DS 2  i Z sin  2

2

D 1



DS sin  DS 2  i cos  2

D 1

DS

Z

287

sin  DS 2

D 1





DS cos  DS 2  i Z sin  2 DS 2

Z

D



2D

Z  cos  DS2 D

1  2Z cos  DS 2 Z D

 iZ

2D

D 1

1  2Z cos  DS 2 Z D

2D

From the above (6.141), we write the following:

Z

Re >G (iZ )@

D 1

sin  DS 2

1  2Z cos  DS 2 Z D

Im >G (iZ )@



Z

2D 1

Z

D 1

2D

(6.142)

cos  DS 2 2D

1  2Z cos  DS 2 Z D

We write an inverse Laplace transform using the formula (6.99), as follows: D

ED ( t )

g (t )

$

1

^G (iZ )`

V 0t f

e

g (t )

³  Re >G (V

S



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

0

D 1 §§ · · Z sin  DS2 cos(Zt ) ¸ ¨ ¨ ¸ D D 2 DS f 1 ¨ © 1  2Z cos  2  Z ¹ ¸ ¨ ¸dZ ³ S 0 § Z 2D 1  Z D 1 cos  DS · 2 ¨ sin(Zt ) ¸ ¨ ¨ 1  2Z D cos  DS  Z 2D ¸ ¸ 2 ¹ © © ¹

1

S

³

f

Z

D 1

sin  DS 2 cos(Z t )  Z

2D 1

sin(Zt )  Z

1  2Z cos  DS 2 Z D

0

1

S ³

f

Z

D 1

sin Zt  DS 2 Z 1  2Z cos  D

0

DS 2

2D 1

Z

D 1

1

S

³

f

Z

D 1

sin(Zt )

2D





sin Z D z  DS 2 Z

2D 1

D

dZ

dZ

D

1  2Z cos  DS 2 Z

0

cos  DS 2 sin(Z t )

2D

From here, we write an integral representation of ED (  z ) as follows (i.e. placing t ED (  z )

(6.143)

z ): sin(Z D z )

2D

dZ

(6.144)

Therefore, we can write from the above derivation the following expression: g (t )

$

1

^

sD s (1 sD )

`

1

f

S

³

0

Z

D 1





sin Z D z  DS 2 Z

2D 1

1  2Z cos  DS 2 Z D

sin(Z D z )

2D

(6.145) dZ ,

t

D

z

288

Chapter Six

ks 1 ( sD  k ) 1

6.11.6 The inverse Laplace transform of function G ( s ) Here, we note that we get functions like

sD s (1 sD )

in the Laplace transform of a fractional differential equation (FDE).

We need to carry out an inverse Laplace transform of the same to obtain a solution to the main differential equation. Using a contour integration as described earlier (in Section-6.6) or using this method of Laplace inversion through the Berberan-Santos formulas gives us an integral representation of the solution, which can be numerically solved via numerical integration. It should be noted that the FDE does not always result in a known functional form in solution. We take our earlier example, where we conducted contour integration to get g (t )

$

1

^

k s ( sD  k )

` , which came from

the FDE as follows (given the FDE with its Laplace transformed equation); with a zero initial condition and x (t ) as the Heaviside unit step excitation at D 1 d g (t ) k dt D

 g (t )

G(s)

k s ( sD  k )

We take G ( s )

0 , $ ^x (t )`

t

X (s)

1 s

.

1 k

s

x (t ) . Carrying out the Laplace transform, we get

The fractional differential equation is D



1 s

G(s)  G(s)

.

From this, we get

. 1

D

1

ks ( s  k ) . Putting s G (iZ )

iZ , we have the following steps: k

k

iZ  (iZ )

D

k





D D 1

Z Z i

 ik



k

 cos 

Z Z

D

(D 1)S 2

 iZ sin  D

(D 1)S 2

 ik

k



   k Z cos   i Z sin   k cos   Z sin   2Z k sin   k cos   i Z sin   k Z  2Z k sin  k D

Z Z cos



2D



D 1

k Z

(D 1)S 2

2

D

(6.146)

k

(D 1)S 2

(D 1)S 2

2

(D 1)S 2

D

2

(D 1)S 2

D 1

(D 1)S 2

D

(D 1)S 2

sin

D

2D

(D 1)S 2

2D

i Z

(D 1)S 2

D

Z Z



(D 1)S 2

2

We get from (6.146) the following:

Re >G (iZ )@

Im >G (iZ )@

kZ

Z 

2D

 2Z kZ

Z

2D

 k sin  sin 

D 1 D

D 1

 2Z

k k k sin  k

cos

(D 1)S 2

(D 1)S 2

(D 1)S 2

D

(D 1)S 2

2

(6.147)

2

2

Now using the formula (6.99) and the Berberan-Santos method, we get the integral representation of

g (t )

$

1

^

k s ( sD  k )

` as follows:

Fractional Differential Equations and their Analysis

g (t )

$

1

^G (iZ )`

V 0t f

g (t )

289

e

³  Re >G (V

S

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0

(D 1)S D 1 §§ · · kZ cos  2 ¨ ¨ 2D ¸ ¸ t cos( ) Z (D 1)S D 2 ¸ f ¨¨ Z 2 sin k k Z   ¸  2 1 © ¹ ¨ ¸dZ ³ ( 1) D S  D 1  S 0¨ § · ¸ Z sin  2  k ¸ sin(Zt ) ¸ ¨  ¨ 2D D S ( 1)  D 2 ¨ ¨ Z  2Z k sin  2  k ¸ ¸ ¹ © © ¹

cos(Zt )   kZ sin   k sin(Zt )dZ S³ Z  2Z k sin  k Z cos Zt    k sin(Zt )dZ k S³ Z  2Z k sin  k 1

f

 kZ

D 1

cos



(D 1)S 2

2D

0

f

0

(D 1)S 2

D 1

(D 1)S 2

D

2

2

(D 1)S 2

D 1

2D

(D 1)S 2

D

2

(6.148)

S. No. The transfer function as a function of complex frequency

1

In integral representation of function in time domain by an inverse Laplace transform using theBerberan-Santos method

1 sa

2

1

3

e

et

Z 2  a2

0

³

S

1 S

4 Z 4

³

f

0

1

1

S

5

1 s

6

7 8 9 10

11

s

1

S

1  (1  E )  s a

e

 ( s / a )E

k k  sD

 (11E )

³

0

³

³

e

f k S 0

D

³

³

 u E cos ES /2

cos(Zt )dZ f

sin Zt

Z

0

dZ

f cos Zt  DS 2

ZD

0

³ 1  u

f a S 0

f

1 2  2(1E )

f a S (1 E ) 0

dZ

cos Z (t  Td ) dZ

1 S

D

dZ

f (2 Z 2 )cos Zt Z 2 sin Zt

0

 sTd

4

³

S

s s 2 1

f a cos(Zt ) Z sin(Z t)

cos



dZ

du ; u sin  du ;

aut  tan 1 u 1 E

cos atu  u

E

(1 E )Z a

ES 2

§ uD cos DS2 1 cosut D k uD sin DS2 sinut D k ·du ; u ¨ ¸ u 2D  2 uD cos DS2 1 © ¹

sD s (1 sD )

S

k s ( sD  k )

k

1

S

³

f ZD 1 sin Zt  DS Z 2 D 1 sin(Zt ) 2 Z 2 D 1 2ZD cos  DS 2

0

³

 k sin(Zt ) dZ k sin   k

 

f ZD 1 cos Zt 

0

Z

2D

 2Z

D

dZ

( D 1) S 2

( D 1) S 2

2

Table-6.1: A few examples of Laplace inverted functions using the Berberan-Santos method

Z a

u

Z k

D

290

Chapter Six

6.11.7 The inverse Laplace transform of function G ( s )

(ln s ) 1 ( s b 1  s a 1 )

This type of function appears in fractional differential equations where the fractional order is of a ‘continuous distribution function’. We will explain this type of differential equation in Chapter-9. We take G ( s )

 ln s

1

s

b 1

s

a 1

,

b ! a ! 0 . We may think that there is singularity at s

b  a ; therefore s

L’Hospital rule on G ( s ) , we find lim s o1 G ( s ) lim s nf G ( s )

sb  s a s ln s

0 and G ( s )

, as having singularity at s

1 . While performing the

1 is not a singular point. We also see

0 . Therefore, we put s

V 0  iZ with V 0

0.

This provides the following steps: G (iZ )

(iZ )

b 1

 (iZ )

a 1

; ln(iZ ) ln Z  ln i ; i

ln(iZ )

Z

b 1

cos 

( b 1)S 2

 i sin 

( b 1)S 2

Z

cos



( b 1)S 2

Z

a 1

cos



a 1

( a 1)S 2

S 2

b 1

cos



( b 1)S 2

Z

a 1

cos



( a 1)S 2

 i Z S 2

b 1

cos



( b 1)S 2

Z

a 1

Z

b 1

cos



( b 1)S 2

 i sin 

( a 1)S 2



b 1

sin



( b 1)S 2

Z

a 1

sin



( a 1)S 2



b 1

sin



( b 1)S 2

Z

a 1

sin



( a 1)S 2



(6.149)

 ln Z  i  S2

 (ln Z )  cos   ln Z  Z sin   Z sin    (ln Z )   Z cos    Z sin   Z sin  ln Z  (ln Z )  S2 4

( a 1)S 2

( b 1)S 2

b 1

2

i

( a 1)S 2



 i Z 2

Z

; ln(iZ ) ln Z  i  S2



ln Z  i 

Z

iS /2

 Z  cos 

ln Z  i  b 1

e

( a 1)S 2

a 1

( a 1)S 2

a 1

S 2

S2 4

( b 1)S 2

b 1

S 2

( a 1)S 2

a 1

S2 4

2

This (6.149) gives real and imaginary parts as:

Z  Z

 Z cos   ln Z sin   Z sin   Re ª¬G  iZ º¼  (ln Z )  Z cos   Z cos    Z sin   Z sin  ln Z Im ª¬G  iZ º¼   (ln Z )  b 1

cos



( b 1)S 2

( b 1)S 2

b 1

( a 1)S 2

a 1

( a 1)S 2

a 1

S

2

S2

2

4

( b 1)S 2

b 1

( b 1)S 2

b 1

( a 1)S 2

a 1

(6.150)

S

2

( a 1)S 2

a 1

S2

2

4

We apply (6.99) to get the following:

g (t )

$

1

^ ` sb  s a s ln s

V 0t f

g (t )

e

³  Re >G (V

S

0



 iZ )@ cos(Zt )  Im >G (V 0  iZ )@ sin(Zt ) dZ

0





§ Z b 1 cos  (b 21)S  Z a 1 cos  ( a 21)S  ln Z · ¨ ¸ ¨  Z b 1 sin (b 1)S  Z a 1 sin ( a 1)S  S  2  2 2 cos(Zt ) ¸¸ ¨ 2 2 ¸ f¨ (ln Z )  S4 1 ¨ ¸dZ ¸ S ³0 ¨ Z b 1 cos  ( b 1)S  Z a 1 cos  ( a 1)S  S 2 2 2 ¨ ¸ ¨  Z b 1 sin (b 1)S  Z a 1 sin ( a 1)S ln Z  2  2 sin(Zt ) ¸¸ ¨ 2 2 ¨ ¸ (ln Z )  S4 © ¹



















(6.151)



For this type of function to be obtained, the inverse Laplace transform, the classical contour integration method is demonstrated in Appendix-G. Table-6.1 lists the various integral representations of the inverse Laplace transformed functions obtained by Berberran-Santos method.

Fractional Differential Equations and their Analysis

291

6.12 The relaxation-response with the Mittag-Leffler function vis-à-vis the power law function as obtained for fractional differential equation analysis We have obtained a relaxation function V (t ) for a fractional differential equation of the following form (Section-6.8) E

EW

dt

EMW when the input function is a constant step H (t )

V (t )

EM H0

¦

k

f

( 1) k 0 * (D k D  E 1)

D k D  E

 Wt

Putting D

1 2

E

EW

V (t ) 

E

d H (t )

E

E

D

EM

E

dt for t t 0 .

H0

V (t )  W

D

D

d V (t ) dt

D

(6.152)

D

d V (t ) D

dt This response is the power series

, which we found in Section-6.8 of this book. We rewrite this as follows: D E

V (t )

E

d H (t )

§t· ¸ ©W ¹

EM H0 ¨

Dk k § f ( 1) §t· · ¨¨ ¦ k 0 ¨ ¸ ¸ * (D k  D  E  1) © W ¹ ¸¹ ©

(6.153)

gives:

 

E M H 0  exp( x ) erfc x

V ( x)

1

x

2

t

(6.154)

W

Where erfc is the ‘complementary error function’ (refer to Appendix-A for further information). Putting D E 1 , we obtain a relaxation response of the ordinary integer order differential equation, as follows: dH EW dV dH dV EW EMW V V W (6.155) dt E M dt dt dt

EM H 0  exp(  x ) , with x

as V ( x )

t W

.

The sum contained in the relaxation function (6.153) is the generalized Mittag-Leffler function (GML), or the two-

¦

parameter Mittag-Leffler function (see Appendix-A), which reads as ED , E ( x )

D , E  \ , further ED ,1 ( x )

( x )k * ( D k E ) k 0

f

with D , E ! 0 and

ED ( x ) . Note that the notation of the Mittag-Leffler function uses E(*),(**) , whereas

E M and E which denotes constants (not to be confused with the GML notation): while one is in italics, the other is not. For negative x , we have the expressions for the GML function which are ED , E (  x ) D

ED , E (  x )

¦

( 1)k ( x )D k k 0 * (D k  E )

f

E0,1 (  x )

E0 (  x )

1 1 x

( 1)k x k * k 0 (D k  E )

f

and

.

0 , we have

Interestingly, for D

¦

 ¦ 1 *(E )

E0, E (  x )

f

k k

k 0

1  * ( E ) (1 x ) . With

( 1) x

E

1 , we have

(see Appendix-A). Figures 6.3, 6.4, and 6.5 show the plots of ED , E (  x ) for E

1 with

D 0.25 D 1.75 , D 2.25 . Thus, for a decaying relaxation function, we restrict D  2 ; as we observe from these figures the function is growing and not decaying for case D ! 2 . D E

§t· V (t ) E M H 0 ¨ ¸ ©W ¹

D E

E M H 0 ( x)

D E

E M H 0 ( x)

¦ ¦

f k

f k 0

( 1)

Dk

k

§t· ¨ ¸ 0 * (D k  D  E  1) © W ¹ ( 1)

k

* (D k  J )

Dk

( x)

J

D  E 1

§t· ¨ ¸ x ©W ¹ (6.156)

D

ED ,J (  x )

The asymptotic expansion for the Mittag-Leffler function for a negative argument at x n f is the following (see Appendix-A): D 1 2 1 ED ,D (  x )  x , D z1 ED ,J (  x )  x , J zD * (1  D ) * (J  D ) (6.157) D 1 D D 2D D ED ,D (  x )  x , D z1 ED ,J (  x )  x , J zD * (1  D ) * (J  D )

292

Chapter Six

With these approximations, we express the asymptotic behavior of the relaxation function for short and long periods of time. The relaxation function is V ( x )

EM H0 x

D E

D

ED ,J (  x ) with J

D  E  1 . For a case E

1 , (i.e. an order of

differentiation of the function H (t ) is unity and 0  D d 1 ), we have:

V ( x) = E M H 0 x

D 1

D 1 ­ x E H as x p 0 M 0 °° * (D ) ® °E M H 0 D x  (D 1) as x n f * (1  D ) °¯

ED ,D

(6.158)

For the other case, we have the following:

V ( x) = E M H 0 x

D E

ED ,J

D E ­ x E H as x p 0 M 0 °° * (J ) J ® °E M H 0 1 x  E as x n f * (1  E ) °¯

D  E 1

(6.159)

Figures 6.8 and 6.9 show the plot of ED ( t D ) , 0 d D d 1, t ! 0 in a linear scale and log scale respectively. For

D 1 , the curve is a pure exponential curve. For other cases, for large t we get a power law behavior visible from Figure-6.9; after about t 1 , we get a constant slope. 1

ED ( t D ) ,

0 d D d 1,

t !0

linear scale plot

D

1 4

1 2

D

D

D

3 4

1

0.5

0

0

1

3

2

4

5

t D

Figure-6.8: Plot of ED ( t ) ,

0 d D d 1, t ! 0 in linear scale

A very practical observation from Figure-6.8 is inferred that is useful in studies of the relaxation dynamic. The classical relaxation function is proportional to the exponential decay function (i.e. e  t E1 ( t ) ). When the order of the Mittag-Leffler function is less than one (i.e. ED ( t D ) ; 0  D  1 ), we find that for about the 0  t  1 range (i.e. we call it early times) the decay is steeper (with a faster rate) than the classical exponential decay function. For late times for about t ! 1 , the ED ( t D ) , 0  D  1 falls (decays) at a slower rate compared to the classical exponential function. This means that we observe a ‘long-tailed’ decaying response in cases with ED ( t D ) , 0  D  1 ; as t D e . We point out that the function ED ( t ) ; 0  D  1 is not differentiable at t

compared to E1 ( t ) D

0 , and at

t  0 the variable t becomes imaginary at some values of D . We will discuss the non-differentiability of this function and extend it to negative values in Chapter-9.

Fractional Differential Equations and their Analysis

293

1

1 4

D 1 2

D 1 10

3 4

D D

ED (t ) , 0 d D d 1, t ! 0

D 1

log-log plot 1 100

1 100

10

1

1 10

100

t D

Figure-6.9: Plot of ED ( t ) ,

0 d D d 1, t ! 0 in log scale

Figure-6.9 says that for late times for about t ! 1 , the log-log plot shows an almost linear function (call it log y

 m log t ), which implies that

yt

m

which is called the power law decay.

6.13 Use of several fractional order derivatives in order to obtain a generalized fractional differential equation Another way to generalize the integer order linear differential equation as stated above is via the use of several fractional derivatives operating on an input function (say H (t ) ) and output functions (say V (t ) ) as described below:

V

(D n )

p

(t ) 

¦

p

D

¦ b D E H (t )

ak D k V (t )

k

k 1

ak , bk , D k , E k  \ D

k

(6.160)

k 0

k

0,1, 2, ..., p

E

The operators in the expression (6.160) (i.e. D ; D ) represent an appropriate fractional derivative; such that the frequency of the transformed equation or Laplace transformed equation for the above fractional differential equation model (6.160) is the following:

§ Dn p Dk ¨ s  ¦ ak s k 1 © The steady state frequency domain equation is achieved by using s

· § p Ek ˆ V s ( ) ¸ ¨ ¦ bk s ¹ ©k 1

· ¸ Hˆ ( s ) ¹

(6.161)

iZ , through which we get:

p

§ Dn Dk ¨ (iZ )  ¦ ak (iZ ) k 1 ©

· § p Ek ˆ V i Z ( ) ¸ ¨ ¦ bk (iZ ) ¹ ©k 0

· ¸ Hˆ (iZ ) ¹

(6.162)

From the above expressions (6.161) and (6.162) we have reduced the parameter of the fractional differential equation, now comprising four parameters as depicted below:

1  bD V (t )  E D

0

 E1 D

D

H (t )

(6.163) 0  D  1; E 0 t 0 ; E1 ! 0 ; b t 0 ; E1 t bE 0 It should be noted here that the fractional derivative operator, although it is considered as an RL fractional derivative with a sole consideration, is as follows: D D (6.164)  D f (t ) (iZ ) fˆ (iZ ) fˆ (iZ )  ^ f (t )`

^

`

With this sole consideration there are several definitions of fractional derivatives which can be of usage, like the Grunwald-Letnikov

G D f Dt , 

or the Liouville type forward derivative, the Caputo-Liouville type forward derivatives

over an infinite interval ( f, t ] ), or the RL type

RL D 0 Dt , 

f if f (t )

0 for t  0 . We will continue further in the next

chapters to give formal treatment to these fractional differential equations.

294

Chapter Six

6.14 Short summary In this chapter, we have discussed various tricks, especially the Laplace transform techniques that can be used to solve some fractional differential and integral equations. We observe that these methods are as rigorous as their counterparts in classical calculus. We also showed various ways of obtaining an inverse Laplace transform that we must use to provide a solution to fractional differential equation, and used a method of power series expansion, a contourintegration and a new technique, that is the Berberan-Santos method, to obtain an inverse Laplace transform. We have discussed the power-law function which appears in complex decay for some fractional differential equations and which has applied the Berberan-Santos method to get decay rate distributions of complex decay. In addition, we noted that this method gives us interesting definite integrals representing several functions. We have studied the MittagLeffler function as arising out of a complex decay function, and obtained asymptotic behavior and its integral representation. From the integral representation of the Mittag-Leffler function, it is easy to ascertain the values of the function, as compared to its series representation with gamma-functions, which are difficult to get for large values. The numerical integration is a rather simple way of computing the Mittag-Leffler function. In this chapter, we have informally discussed the various aspects of fractional differential equations and their solutions. We will now carry forward our study to further fractional differential equations.

6.15 References This chapter is based on the following pioneering works: Miller K S and Ross B, “An Introduction to Fractional Calculus and Fractional Differential Equations” (1993) Berberan-Santos Mario. N, “Analytical inversion of Laplace transforms without contour integration: application to luminescence decay laws and other relaxation functions” (2005); “Properties of Mittag-Leffler relaxation function” (2005) Abramowitz. M, Stegun. I.A. (eds.), “Handbook of Mathematical Functions” (1964) Bagley R L and Torvik P J, “Fractional calculus in transient analysis of viscoelastically damped structures” (1985) Bagley R L, “Power law and fractional calculus model of viscoelasticity” (1989) Bagley. R.L, Calico R.A, “Fractional order state equations for control of viscoelastic structures” (1991) Bagley. R.L, “The thermorheological complex materials” (1991) Batman H, “Higher transcendental function” (1954); Churchill Ruel V, Brown James W and Verhey Roger F, “Complex Variables & Applications” (1974) Churchill. R.V, “Operational Mathematics” (1948) Davis, Harold T, “The Theory of Linear Operators” (1936) Diethelm K, “An algorithm for the numerical solution of differential equations of fractional order” (1997) Diethelm, Kup and Ford, “Analysis of fractional differential equations” (2003) Erdelyi A, “Asymptotic expansions” (1954); “Tables of Integral Transforms”(1954); “On some functional transform” (1950) Fracncois Dubois Stephanie Mengue, “Mixed Collocation for fractional differential equations” (2005) Gel’fand, I.M, Shilov,G E ,“Generalized functions”(1964) Glockle W G ,Nonnenmacher. T F, “Fractional Integral Operators and Fox Functions in the Theory of Viscoelasticity” (1991) Gorenflo R, Mainardi F, “Fractional Calculus: Integral & Differential Equations of Fractional Order, CISM Lecture Notes” (2008); “Fractional oscillators and Mittag-Leffler function” (1996); “Fractional calculus: integral and differential equations of fractional order” (1997) Gorenflo R, Yu. Luchko and Rogosin.S, “Mittag-Leffler Type Functions: Notes on Growth Properties and Distribution of Zeros” (1997) Gradshteyn I.S, Ryzhik I M, “Tables of Integral Series & Products” (1992) Heaviside. O, “Electromagnetic Theory” (1971) Hilfer R, “Applications of fractional calculus in Physics” (2000) Igor Podlubny, “Fractional differential equations” (1999); “The Laplace Transform Method for Linear Differential Equations of the Fractional Order”(1994) John W Hanneken, B N. Narahari Achar, Raymon Puzio, “Properties of the Mittag-Leffler function for negative alpha” (2009) Jumarie G, “Laplace Transform of Fractional Order via Mittag-Leffler function and Modified Riemann-Liouville derivative” (2009) Kilbas A A, Srivastava H M, Trujillo J J, “Theory & Application of Fractional Differential Equations” (2006) LePage, W R, “Complex variables and the Laplace transform for Engineers” (1961) Manuel Duarte Ortigueira, “An Introduction to the Fractional Continuous Time Linear Systems, the 21st Century Systems” (2008) Matignon D, “Generalized Fractional Differential and Difference Equation: Stability Properties and Modeling Issues” (1998) Oberhettinger F and Badii L, “Tables of Laplace transforms” (1973)

Fractional Differential Equations and their Analysis

295

Oldham and Spanier, “Fractional calculus” (1974) Rayes-Melo M E, Martinez-Vega J .J, Guerrero-Salazar C A, Ortiz-Mendez U, “Modeling of relaxation phenomena in organic dielectric materials, applications of differential and integral operator of fractional order” (2004) Saxena, R.K, “Fractional calculus and fractional differential equations: Lecture notes (2006) Saxena, R.K, Mathai, A.M, Haubold, H J, “On Generalized Fractional Kinetic Equations” (2004) Shawagfeh N.T, “The Decomposition Method for Fractional Differential Equations” (1999) Shih-Tong, Srivastava, H.M, et al., “A certain family of fractional differential equations” (2000) Srivastava, H.M, Owa S, “Some characterization and distortion theorems involving fractional calculus, generalized hyper geometric functions, Hadamard products, Linear operators and certain subclasses of analytical functions” (1987) Srivastava H M, “On extension of Mittag-Leffler function” (1968) Tarasov, E. Vasily, “Electromagnetic field of fractal distribution of charged particles” (2005) Westerlund S , Eskam L., “Capacitor theory” (1994) Caputo, Michelle, “The Riemann Sheet Solutions of Anelasticity” (1994) The above works are furnished with further detail and listed in alphabetical order in the Bibliography section.

CHAPTER SEVEN A FORMAL METHOD FOR SOLVING FRACTIONAL DIFFERENTIAL AND FRACTIONAL INTEGRAL EQUATIONS

7.1 Introduction In this chapter, we try to formulate a formal development regarding a fractional differential equation (FDE) and a fractional integral equation. We will be using the Laplace transformation technique to find these solutions. We will show the conjugation of the formal method with classical calculus. Here, several higher transcendental functions (like those of Mittag-Leffler, Miller-Ross, and Robotnov-Hartley, etc.) will be used to study various fractional differential and integral equations. We will start our discussion with a first order classical differential equation, and view its counterpart as a fractional differential equation composed by the Caputo derivative. We will then attempt to discern if we can find any similarity and conjugation between the two. We also will discuss the classical methods applied to solve a fractional differential equation composed via the Riemann-Liouville derivative in the Miller-Ross formulation. We will use the concept of a sequential fractional derivative, and will show how the solutions change when the fractional derivatives are either Riemann-Liouville or Caputo type. Here, we will also study a system of fractional differential equations, by introducing a state transition matrix, and explore its usage in the system of fractional differential equations.

7.2 A first order linear differential equation and its fractional generalization using the Caputo derivative 7.2.1 Demonstrating the similarity of a classical differential operator with the Caputo fractional differential operator in a differential equation solution pattern We take simple examples of fractional differential equations in this section using the Caputo derivative, before looking at the Riemann-Liouville fractional derivative in the next section. These two parts form the basis for further sections of this chapter where we will formalize the observations of these two sections. The simple dynamic equations with the Caputo derivative offer the solution

f (t )

D

f0  0 I t

>k @ for

C 0

D

Dt

t t 0 , where f (t ) t

> f (t ) @

k with k as a constant, and 0  D  1 will

f0 is a constant of initialization, and

0

D

0 It

is a

fractional integration operation. We can check this using the property of the Laplace transform (i.e.

$

^

C D 0 Dt

> f (t )@`

D

s F (s)  s

D 1

f (0) , 0  D  1 (as shown in Section-5.16)) in the following steps: D

s F (s)  s

D 1

s

$ ^ f (t )`

;

1 for t t 0 and u (t )

(7.1)

0 for t  0 . The Laplace transform of

1

s . By conducting an inverse Laplace transform of (7.1), we get the following result: f (t )

D

f 0  u (t )  0 I t

1 , we get a classical solution as f (t )

D

* (1  D )

t ;

tt0

(7.2)

f0  kt .

The simple dynamic equations using the Caputo derivative

0  D  1 will have the solution f (t )

>k @

k

f0  For D

F (s)

§ 1 ·§ k · D  ¨ D ¸ ¨ ¸ $ ^ f0  u (t ) `  $ ^I t ª¬ k  u (t ) º¼` s © s ¹© s ¹

Where u (t ) is a Heaviside unit step function, u (t )

^ `

k

f0

F (s)

u (t ) is $ u (t )

f0

D

f0  0 I t

C 0

D

Dt

> f (t ) @

> x(t )@ for t t 0 , where

f (t ) t

We can check this with the Laplace transformation, as we will get the following:

x (t ) with x (t ) as a function, and 0

f0 is a constant of initialization.

A Formal Method for Solving Fractional Differential and Fractional Integral Equations D

s F (s)  s F (s)

f0

D 1

297

$ ^x (t )`

f0

(7.3)

§ 1 · D $ x (t )` $ ^ f0  u (t ) `  $ ^ 0 I t > x (t )@` D ¸ ^ ©s ¹

¨

s An inverse Laplace transformation of the above (7.3) gives the following expression: f (t )

D

f 0  u (t )  0 I t f0 

1 , we get a classical solution as f (t )

For D

f0 

1 * (D )

> x (t ) @

³

t

0

D 1

(t  W )

 x(W ) dW ;

(7.4)

tt0

t

³  x(W ) dW . These two examples show a definite similarity and 0

conjugation of a differential equation of classical calculus with a fractional differential equation composed of the Caputo derivative. We will explore this further now. Why do we specifically use the Caputo derivative operator? It is because, as shown in Section-4.14, fractional



D integration and the fractional derivative (Caputo type) of order D for the Mittag-Leffler function ED O x

are the

following:

ª¬ ED  O xD º¼

D

0 Ix

1

O

 E O x  1 D

D

C D 0 Dx

1 Ox There is a similarity with the classical exponential function as 0 I x ª¬e º¼

D

that ED (O x ) C D 0 Dx

(i.e.

>C @

e

Ox

for D

ª¬ ED  O xD º¼ O ED  O xD (7.5) Ox 1 Ox Ox 1  1 and Dx ª¬e º¼ O e . We note O e

1 . Moreover, like the classical case, the Caputo derivative of a constant function is zero

0 ). With these observations we may state there may exist a conjugation and similarity in the

fractional generalization with a Caputo derivative in order to construct a fractional differential equation system with classical differential equations. To use the Caputo derivative in the fractional differential equation, we will assume that the function is differentiable.

7.2.2 The first order differential equation of a relaxation process with an initial condition Let us say that a process defined by a function T (t ) is a differentiable function of the time variable t , which has a rate relationship given by T

(1)

(t )

O T (t )  Ta , where O ! 0 is a constant. Noting that T (t ) is an absolute

instantaneous temperature, say for hot water, this law corresponds to the classical Newton’s law of cooling, where Ta is room temperature or ambient temperature. Therefore, the rate of change of temperature of hot water is proportional to the difference in temperature. As such, we have the following first order differential equation:

d T (t )

We also set an initial condition as T (0) above differential equation as T

(1)

(7.6) O T (t )  Ta 0dtf dt T0 . Recalling our classical calculus, we put the homogeneous part of the

( t )  OT ( t )

0 . This is given by the solution T1 (t ) and a complementary part

defined by constant OTa , which gives us a particular integral component T2 (t ) . Thus the total solution T (t ) is the sum of these two solutions T1 (t ) and T2 (t ) . Adding a function, which is also called the forcing function to the RHS of the homogeneous differential equation, gives a non-homogeneous differential equation. In the above case (7.6), the forcing function, say x (t ) , is a constant (i.e. x (t ) OTa ).

7.2.3 A first order differential equation’s homogeneous and complementary parts We know that through the use of the property of the exponential function, we get a solution for the homogeneous part (i.e.

G (t )

T

(1)

e

 Ot

(t )

OT (t ) ), which represents the fundamental solution (also called Green’s function).

; and we use this to set the homogeneous solution as T1 (t )

A  G (t )

Ae

 Ot

, with the constant A ,

which is determined via a given initial condition. The initial condition for our case is, at t

A

T0 ; therefore T1 (t )

T0 e

 Ot

That is,

0 , T (t )

. Now to get a particular integral, we use the fundamental solution:

T0 gives

298

Chapter Seven

G (t )

x (t ) * G (t )

process) starts at t

 Ot

(7.7)

OTa , we use the convolution integration (i.e.

In order to have particular solution, for the forcing function x (t )

T2 (t )

e

t

³  x(t  W ) G (W ) dW ). We note that the evolution of the given process (i.e. the cooling f

0 . Thus, we conduct the following convolution integration to get the solution below: T2 (t )  x (t )u (t ) *  G (t ) ; x (t ) OTa t

t

0

0

OW ³  x(t  W ) G (W ) dW ³  OT  e dW O T 1  e 

(7.8)

a

 t

a

This above convolution is indicated as u (t )

T (t )

1 for

t t 0 and u (t )

T1 (t )  T2 (t )

T0 e

 Ot

u (t ) is a Heaviside step function at t

0 (i.e.

otherwise). From these steps in (7.8), we get the total solution as

0



 x(t )u (t )  G (t ) , where

 Ta 1  e

 Ot

, that is: Ta  T0  Ta e

T (t )

 Ot

(7.9)

7.2.4 A fractional order differential equation composed using the Caputo derivative’s homogeneous and complementary parts Now we write the fractional differential equation of the fractional rate of change (using the Caputo fractional derivative), as proportional to the difference in temperature, as: C D 0 Dt

>T (t )@

O T (t )  Ta ; C D 0 Dt

say that the fundamental solution pertaining to the homogeneous part (i.e.



0 T1 (t )

C D 0 Dt

>T (t )@  OT (t )

0 ) is given by: (7.11)

ED ( O t )

D

T0 ED ( O t ) .

This comes from an initial condition, as

t

AE

ª¬ ED ( O t D ) º¼ O ED ( O t D ) . As such, we D

GD (t )

and the homogeneous solution is T1 (t )

(7.10)

T0 . With the property of the Mittag-Leffler

with the same initial conditions as in the integer order case, i.e. T (0) function and the Caputo derivative, we have seen earlier in (7.5) that

0 D 1

T1 (t )

D

D

( O t )





A 1

O tD * (D 1)



O 2t 2D * (2D 1)



 .... . Thus we have at

A T0 . The particular integral, if we write it in the same way via a convolution integral, then, is:

t 0

T2 (t )

t

 x(t )u (t ) GD (t ) ³0 OTa  ED (OW D ) dW



Therefore, we ask ourselves, can we write the total solution as T

D



T0 ED ( O t )  OTa

(7.12)

t

D ³  ED (OW ) dW ? Can we 0

D

just replace the exp( O t ) obtained in the classical case with that obtained by ED ( O t ) to get the total solution?

7.2.5 Can we just replace the exp( O t ) in the classical case with ED ( O t D ) in order to get the total solution for the fractional differential equation composed using the Caputo derivative? We will now verify the above logic of convolution (7.12), by using the Laplace transform. For the classical case, the Laplace transform of a given equation (i.e. T

(1)

( t )  OT ( t )

OTa for 0 d t  f ) is:

 sT ( s )  T (0)  OT ( s ) The RHS of T

(1)

( t )  OT ( t )

OTa s

$ ^T (t )` T ( s )

(7.13)

OTa is transformed to OTa  1s in (7.13) due to the use of the Heaviside step function,

as we are considering the process from t

0 . Before that time, the process is zero. With T (0)

T0 , we obtain the

following:

T (s)

T0 sO



OTa s(s  O )

(7.14)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

 s 1O corresponds to a homogeneous solution, that is,

We note from the above expression (7.14) that the first term T0 an inverse Laplace transform of $

  ) OTa

1 s O

s

$

1

1 s

a

1

to

^T  ` , which gives 1 s O

0

a

convolution

^ OT   `  OT u (t )  e . The  Ot

1 s O

^OT  a

1 s ( s O )

Laplace

^

T0 e

integral;

 Ot

, while the second term of (7.14) (i.e.

giving

a

particular

u (t ) is a Heaviside step function and is u (t )

a

`



Ta 1  e

integral

as

1 for t t 0 ; and

.

 Ot

C D 0 Dt

Now for the fractional differential equation

$

T1

0 , for t  0 . Carrying out an inverse Laplace transform of this second term of (7.14), we get

u (t )

$

corresponds

1

299

transform

C D 0 Dt

> f (t )@`

for

D

s F (s)  s

D 1

the

>T (t )@  OT (t )

Caputo

OTa the Laplace transformed equation using the

derivative

(as

in

Section-5.16)

i.e.

f (0) , 0  D  1 )) is the following:

 s T (s)  s D



D 1

T (0)  OT ( s )

OTa

(7.15)

s

T0 , we obtain the following:

By using the initial condition T (0)

s

T (s)

^

D

`

T0

D

s O

sD 1 sD  a

For the first term in (7.16), we have $ ED ( at )

D 1

OTa



D

s(s  O )

;

0 D 1

(7.16)

, i.e. from the Laplace transform of the Mittag-Leffler

function (see Appendix-A).Thus the homogeneous solution is T1 (t )



T0 ED ( O t

D

, that is, exactly

T1 (t )

T0 GD (t ) ,

D

where we have GD (t ) ED ( O t ) . The second term of (7.16) pertains to a particular integral, and we have derived from Section-6.4 of this book (via the concept of series expansion), the inverse Laplace transform, which is

$

1



k s ( sD  k )





D

1  ED (  kt ) ; it gives T2 (t )



D

TD 1  ED ( O t ) . This gives us the total solution as follows: D

Ta  T0  Ta ED ( O t )

T (t ) The particular integral in this case is T2 (t )

$

1

^OT   ` ³ 1 s

a

1 sD  O

f

f

 OTa u (t )  FD (O , W dW .

have mentioned earlier (7.12)! Here, the Robotnov-Hartley function FD ( a , t ) transform as $ ^FD ( a , t )` Thus,

T2 (t ) z

the

particular

1 sD  a

is

not

the

convolution integral

³  x(t  W ) GD (W ) dW . Let us now manipulate T ( s)

OTa

2

T2 (t )

1

D 1

¦

f

a n tD n n 0 * (D n D )

with the Laplace

D

t

f

t

This is not as we

ED ( O t ) .

appears instead of GD (t )

integral

(7.17)



that

1 s ( sD  O )

we

wrote

earlier

­ ¯

§ 1 · § 1 ·½ ¸¾ D © s ¹ © s  O ¹¿

­

D 1 · § 1 ·½ § 1 ·§ s ¸ ¨ D 1 ¸¾ D © s ¹ © s  O ¹ © s ¹¿

$ ®OTa ¨ ¸ ¨ 1

¯ § 1 1 ­ $ ®OTa ¨ D ¯ ©s

^ `

function G (t ) , as $ G (t ) and, with s for:

D

s

D

 OTa  ss O D 1

D

i.e.

in the following way

$ ®OTa ¨ ¸ ¨

We have manipulated T2 ( s )

(7.12),

s

D

D 1 ·½ ·§ s ¸¨ D ¸¾ ¹ © s  O ¹¿

D

0 It

(7.18)

ª¬ OTa  ED ( O t D º¼

 $ ^OT G (t )` $ ^E (Ot )` . We note that we used the D

a

D

1 . In the Laplace (frequency) domain multiplication is the time domain convolution,

representing a fractional integration operation, we can write the following inverse Laplace transform

300

Chapter Seven

T2 ( s )

s

 OTa  ss O D 1

D

s

D

D

 $ ^OT G (t )` $ ^E (Ot )` D

a

(7.19)

D

as follows:

T2 (t )

D

0 It

ª¬ OTaG (t ) *  ED ( O t D º¼

1

t

D D (t  W )   OT G (W ) *  ED ( OW dW ³ * (D ) 1

1

t

D 1

(t  W ) * (D ) ³ 0

³

W

f



³

f

OTa





D

0 It

> f (t ) @

1 * (D )

t

D 1

³ (t  W ) 0

f (W )dW ).

f

³ G ( x  x) f ( x)dx f ( x ) ), we write the integral in the above G (W  W )  ED (OW D ) dW OT  ED (OW D ) . Therefore, the particular integral in the above

Then, with the property of the delta function (i.e. expression as

D

OTa G (W  W 1 ) ED ( OW 1 ) dW 1 dW

In the above expression (7.20), we have used RL fractional integration (i.e.

W

(7.20)

a

0

1

1

0

f

1

0

a

case is as follows:

T2 (t )

1

* (D ) ³

t

0

D 1

(t  W )

1

³

W

f

t



D





OTa G (W  W 1 ) ED ( OW 1 ) dW 1 dW

D D (t  W )  OT ED ( OW ) dW ³ * (D ) 1

D

(7.21)

a

0

ª¬ OTa ED ( O t D ) º¼

0 It

We note that the fractional integration operator appearing above is due to the operator s D 0 It

D

ª¬ ED ( at )º¼ a

1

E

D

D

and through recognizing



D

( at )  1 , we get the particular integral as:

ª¬ OTa  ED ( O t D ) º¼ § ED ( O t D )  1 · D O T Ta 1  ED ( O t )  a ¨ ¸ ( O ) © ¹ D

T2 (t )

0 It

(7.22)

This is exactly what we have obtained earlier via the Laplace transform method (7.16). Thus, for a differential equation using the Caputo derivative: C D 0 Dt

>T (t )@  OT (t )

We have a fundamental solution as GD (t ) the particular integral known as T2

D

0 It

OTa ;

0  D  1;

T (0)

T0

D

ED ( O t ) , i.e. the homogeneous solution known as T1 (t )

ª OTa  ED ( OW D ) º . The total solution is as follows: ¬ ¼ D D D T (t ) T0  ED ( O t )  0 I t ª OTa  ED ( O t ) º ¬ ¼



D

Ta  T0  Ta ED ( O t )



T0 GD (t ) , and

(7.23)

7.2.6 Getting a particular solution to a fractional differential equation composed by the Caputo derivative through a fractional integration operation using a classical convolution process Thus, in the case of a fractional differential equation composed via the Caputo fractional derivative, our classical convolution integral for obtaining a particular solution is changed to a fractional integral, and we write: D

 f ( t ) g ( t ) (D ) 1

f I t

³

t

ª¬ f (t ) * g (t ) º¼ D 1

(t  W )

³

W



 f (W  W1 )  g (W1 ) dW1 dW f

(7.24)

* (D ) f This is a fractional order convolution, the concept of which is still developing; it is based on the above discussion, and the fractional integration of a classical convolution operation is considered to be a fractional convolution. We will not use this concept in the subsequent sections of this chapter.

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

301

7.2.7 A homogeneous fractional differential equation composed using a Caputo derivative is driven by a fractional RL derivative of a Heaviside unit step function The fundamental solution G (t )

D

(1)

>T (t )@  OT (t )

e

 Ot

is obtained by an indicial polynomial corresponding to the differential equation

0 , which is P ( x )





0 a homogeneous system, where x corresponds

x  O . We call P ( x ) T

1

to operator Dt . The inverse Laplace transform of this indicial polynomial is $

1

^ P(s) ` 1

$

1

^ s 1O `

e

 Ot

. We

can say that this is a solution for a fundamental equation, which is as follows:

d T (t )

 OT ( t )

dt T (t )

$

1

G (t ) ;

T (s)

^( s  O ) ` 1

e

1 sO

(7.25)

 Ot

^ `

That is the homogeneous differential equation, but it is driven by forcing the function G (t ) , noting that $ G (t ) Now for

C D 0 Dt

>T (t )@  OT (t )

Laplace transform of this

D

0 , we have an indicial polynomial such as

 P(s ) D

1

, we get $

1

^ ` 1 P ( sD )

$

1

FD ( O , t ) . Here we have the Robotnov-

sD  O

D

Hartley function (i.e. FD ( O , t ) ) and not the Mittag-Leffler function (i.e. GD (t ) C D 0 Dt

discussion above. Therefore, we cannot use the equation

D

x  O . Performing an inverse

P( x )

^1`

1.

>T (t )@  OT (t )

ED ( O t ) ), which we used in our

G (t ) (i.e. with the Delta function as a 1 sD  O

forcing function to obtain a fundamental solution), which will give T ( s )

and will result in the Robotnov-

Hartley function. Say we write the following homogeneous equation driven by a power function called G D (t ) C D 0 Dt

>T (t )@  OT (t )

C D 0 Dt

>T (t )@  OT (t )

D

The Laplace transform of the above equation (7.26) is s T ( s )  OT ( s ) with

n

D and using ( D )! *  D  1 . We get here T ( s )

which GD (t )

s

D 1

s

D

:

G D (t ) t

D

* (1  D )

D 1

s

D 1 * (1D ) t

;

0 D 1

(7.26)

^ `  n ! s

, obtained by using $ t

O



1

n

 n 1

, an inverse Laplace transform of

D

ED ( O t ) . As such, we obtained a fundamental solution through the modification of using, for

example, G D (t )

D 1 * (1D ) t

- a new fractional order delta- function!

In Section-4.11 of this book, we showed that a fractional derivative (the Riemann-Liouville – RL type) of the Heaviside step function is

D

0 Dt

> u (t ) @

D 1 * (1D ) t

. In addition, we also found that one whole derivative of the 1

>

Heaviside unit step function is a Dirac Delta function D u (t )

@

G (t ) . Therefore, we have conceptualized the

generalized rule to find the fundamental solution of the following homogeneous equation composed via the Caputo derivative: C D 0 Dt

>T (t )@  OT (t )

D

0 Dt

>u (t ) @

G D (t )

1

Dt >T (t )@  OT (t )

T (0)

T0

(7.27)

1

D

0 D 1

1

Dt >u (t )@

G (t )

The solution to the homogeneous differential equation driven by the delta function is also called Green’s function. Therefore, after obtaining Green’s function, with the help of the initial condition, one gets a solution to the homogeneous equation, and, with Green’s function, through carrying out a convolution one gets a particular integral. The sum of these two solutions is then the total solution. What we have observed is that a convolution integral is replaced by fractional integration, and the delta function, which is a derivative of the Heaviside step function, is replaced by a fractional derivative of the Heaviside step function, which we called a fractional delta function. This is also the case for a fractional differential equation using the Caputo derivative. The Riemann-Liouville case is different, however, and we will discuss this in subsequent sections.

302

Chapter Seven

7.3 The first order linear differential equation and its fractional generalization using the Riemann-Liouville (RL) derivative 7.3.1 Demonstrating the similarity of a classical differential operator with the Riemann-Liouville fractional differential operator in a differential equation solution Continuing on from the previous section, we now change the fractional differential equation with the RiemannLiouville fractional derivative. The simple dynamic equations with the Riemann-Liouville derivative D

0 Dt

> f (t ) @

; where f

k with k as a constant, and 0  D  1 will have the solution f (t )

(D 1)

(t )

t 0



C

>G (t )@  0 I tD >k @ for t t 0

D

0 It

C is a constant of initialization, that is, the value of fractional integration of f (t ) , of order

1  D , at the initial point. We can check this with the Laplace transform, as we will see via the use of Laplace transform property for the RL fractional derivative (i.e. $

^

D

0 Dt

> f (t )@`

D

s F (s)  f

(D 1)

(t )

t 0

, 0  D  1 (as shown in Section-5.16)), which

yields the following steps:

k

D

s F (s)  C

§ 1 ·§ k · D ¸¨ ¸ © s ¹© s ¹

C

F (s)

s

s (7.28)

¨

D

$

^

ª¬C G (t ) º¼`  $ ^ 0 I tD ª¬k  u (t ) º¼`

D

0 It

Where the function u (t ) is the Heaviside unit step function (i.e. u (t ) 1 for t t 0 and u (t ) the Dirac-delta function. Performing an inverse Laplace transform on the above (7.28), we get: f (t )

C



C

We have used the formula

X

f Dx

>G ( x  c)@

1 * ( X )

>G (t )@  0 I tD >k @ ;

D

0 It

( x  c)

t

D 1

* (D )

X 1



k

C

t ;

(t )

t 0

tt0

D and c

by setting X

(D 1)

(7.29)

D

* (1  D )

f

0 for t  0 ), G (t ) is

0 , which were described

in Section-4.11. In order to verify the initial condition, we take a fractional integration of order 1  D of the obtained function, i.e.

f

(D 1)

(t )

1D

0 It

§ t D 1 k D · ¨ C * (D )  * (1  D ) t ¸ © ¹

 D 





* (D 11) D 11D t  * (1kD ) * (D 1D )

C *( )





* (D 1) D 1D t * (D 11D )

(7.30)

C  kt From the above (7.30), we find that f

(D 1)

(t )

C ; that is, we have regained our initial condition. For the classical

t 0

case, we have the solution as described in the previous section (i.e. f (t )

f0

1

f0  u (t )  0 I t > k @ for

t t 0 and

f (0) ). t

³ G (W ) dW

We may also write that through recognizing u (t )

f

(i.e. the Heaviside step function) as an integration of

the Delta function, we obtain the following modified expression:

f (t )





1

1

f0 I t >G (t )@  0 I t > k @ 1

f0  0 I t > k @

tt0

f0  kt

f0

f (0)

(7.31)

tt0

For the Caputo case, we have the solution as:

f (t )

D

f 0  u (t )  0 I t



1

>k @ ;



f0

f0 I t >G (t )@  0 I t

D

>k @

f (0) f0 

k * (D  1)

t

D

tt0

(7.32)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

303

For the RL case, we have the solution as: f (t )

C



>G (t )@  0 I tD >k @ ;

D

0 It

t

C

D 1

k



C

f

(D 1)

(t )

t 0

(7.33)

D

t ;

tt0

* (D ) * (1  D ) We have seen similarities in the solutions of fractional differential equations composed via the Caputo derivative, the RL derivative and the classical one-order whole derivative, for simple cases. Note that the initial condition in the case

of RL is C

f

(D 1)

(t )

t 0

, and for the Caputo and classical cases, it is f0

The simple dynamic equations with the RL derivative have the solution

f (t )

C



D

0 It

> f (t ) @

D

0 Dt

>G (t )@  0 I tD > x(t )@

f (t ) t

0

.

x (t ) with x (t ) as a function, and 0  D  1 will

for t t 0 ; where

(D 1)

f

(t )

C is a constant of

t 0

initialization. We can check this with the Laplace transform, as we will get: D

s F (s)  f C

F (s)

s

D

(D 1)

(t )

$ ^x (t )`

t 0

§ 1 · $ x (t )` D ¸ ^ ©s ¹

(7.34)

¨

^ I >CG (t )@`  $ ^ I > x(t )@` D

$

D

0 t

0 t

An inverse Laplace transform of the above expression (7.34) gives the following: f (t )

C



>G (t )@  0 I tD > x(t )@

D

0 It

C

t

D 1

1



³

t

* (D ) * (D ) 0 We find a similarity between the classical case and the Caputo cases here too.

D 1

(t  W )

 x(W ) dW ;

(7.35) tt0

7.3.2 A fractional order differential equation composed with the Riemann-Liouville derivative is fundamental and a particular solution is offered by a given fractional order initial state We have taken the first order classical system in the previous section as follows: 1

Dt >T (t )@

O T (t )  Ta

0dtf

§ d T (t ) · ¨ dt  T0G (t ) ¸  OT (t ) OTa © ¹ s T ( s )  T0  OT ( s ) With T (t ) t

0

T0G (t ) , we have solved the system. The case T (t ) t

Green’s function (i.e. G (t ) i.e.:

e

 Ot

0

(7.36)

OTa s G (t ) gives us a fundamental solution or

). Now we rewrite the equation above this time using the RL fractional derivative, D

0 Dt

>T (t )@

O T (t )  Ta ,

The Laplace transform of the RL derivative operator is $

^

D

0 Dt

0 d t  f, 0  D 1

`

D

f ( x)

s F (s)  f

(D 1)

(t )

t 0

(7.37)

(as shown in Section-

5.16). This gives the following:

 s T (s)  T D

We assume that the fractional initial state is fractional integration of function T (t ) at t

T

(D 1)

(t )

t 0



(D 1)

1 * (1D )

(t )

³

t

f

t 0

 OT ( s ) D 1

(t  W )

0 is a constant, namely C0 . With C0

OTa

T (W )dW

(7.38)

s



t 0

C0 ;

that is the

1 , we attain the homogeneous

part as:

 s T ( s)  1  OT ( s) D

0

(7.39)

304

Chapter Seven

This means that we have the following homogeneous differential equation:

§ d D T (t ) · ¨ dt D  G (t ) ¸  OT (t ) 0 © ¹ From the above (7.40), we have $ function,

which

FD ( O , t )

t

D 1

is

similar

1

to

D

> f (t ) @

1 sD  O

the

(7.40)

GD (t ) as the fundamental solution in this case or using Green’s

Robotnov-Hartley

function

FD ( O , t ) .

We

also

know

that

D

ED ,D ( O t ) , that is, in terms of a ‘two-parameter’ Mittag-Leffler function (see Appendix-A). The

verification is that the function f (t ) for 0 Dt

^ `

§ d D T (t ) · ¨ dt D ¸  OT (t ) G (t ) © ¹

t

D 1

D

ED ,D ( O t ) is the eigen-function for an RL fractional derivative operator

O f (t ) that we described in an earlier chapter (Section-4.14).

We note that for D

1 we have f (t )

t

D 1

D

ED ,D ( O t )

e

 Ot

. Therefore, in this case, the Green’s function is a

solution of a homogeneous system that is driven by the delta function. The full solution comes from the following steps: OTa D (D 1)  OT ( s ) s T (s)  T (t ) t 0 s





OTa

 s T ( s )  C  OT ( s ) D

0

(7.41)

s

§ 1 · § 1 ·§ 1 · ¸  OTa ¨ ¸ ¨ D ¸ D ©s O¹ © s ¹© s  O ¹

T (s)

C0 ¨

While writing an inverse Laplace transform of the above expression, we get: C0  FD ( O , t )   OTa u (t )  FD ( O , t )

T (t )

We have noted in the previous section that

FD ( O , t )

t

D 1

$

1

^OT

a

(7.42)

f

C0  FD ( O , t ) 

³  OT u (t )  FD (O ,W dW   O ` T 1  ED (Ot D ) a

f

1 s

1 sD 

a

and using

D

ED ,D ( O t ) , we write the following expression: T (t )



C0 t

D 1

D





D

ED ,D ( O t )  Ta 1  ED ( O t )



(7.43)

7.3.3 Verification of a fractional initial state from the obtained solution for a fractional order differential equation composed using the Riemann-Liouville derivative For the system

D

0 Dt

>T (t )@  OT (t )

OTa , we have a homogeneous solution as T1 (t )

(D 1)

(t )

t 0

D 1

D



ED ,D ( O t ) , which,

AGD (t ) where A is a constant depending on the fractional initial state, given

in terms of Green’s function, is T1 (t ) as T



C0 t

C0 . Let us verify this: T1 (t )

AFD ( O , t )

§

A¨ t

©

From above expression, we take T

(D 1)

D 1

At

D 1

D

ED ,D ( O t ) k kD

· § t D 1 O t 2D 1 O 2t 3D 1 · A    ... ¸ ¸ ¦ ¨ k 0 *  ( k  1)D ¹ © * (D ) * (2D ) * (3D ) ¹ f

( O ) t

D 1

(t ) by operating Dt

term by term as in the following steps:

(7.44)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

(D 1)

T1

D 1

(t )

0 Dt

D 1

0 Dt

305

§ D 1 f ( O ) k t kD · ¨ At ¦ ¸ k 0 *  ( k  1)D ¹ ©

§ At D 1 O At 2D 1 O 2 At 3D 1 ·    ... ¸ ¨¨ ¸ * (3D ) © * (D ) * (2D ) ¹

A* (D )t

D 1D 1

* (D )* (D  D  1)



AO* (2D )t

2D 1D 1

2



* (2D )* (2D  D  1)

AO * (3D )t

3D 1D 1

 ...

* (3D )* (3D  D  1)

(7.45)

D 2 2D § f (  O ) k t kD · § · Ot O t   .... ¸ A ¨ ¦ A ¨1  ¸ © * (D  1) * (2D  1) ¹ © k 0 * ( kD  1) ¹ D

AED ( O t ) Here we see that T

(D 1)

(t )

t 0

A ; thus, we get A

C0 .

7.3.4 Modification in a solution for a fractional differential equation composed using the Riemann-Liouville derivative solution with classical integer order initial states instead of fractional order initial states What is the value of A when, instead of a fractional order initial state (i.e. T as in the case of a Caputo derivative, or a classical case, is given (i.e. T (t ) t

(D 1)

0

t

D 1

D

ED ,D ( O t ) 1

t 0

1D

 GD (t )  GD1 (t )

(i.e. GD (t ) t

t

D 1

1

T0 GD (t )



t 0

T0 . We note that

0

.

F (t ) represents the inverse function of GD (t ) . Therefore, we have an identity condition

t

GD (t )

AGD (t ) t

0



f . Thus, we have difficulty in writing A

Let us say that GD (t )



1

C0 ), the normal condition

t 0

T0 ).

0

In this case, with the RL fractional derivative, we should have the condition T1 (t ) t

GD (t ) t

(t )

1D

F (t )



1 ).

This

gives

us

the

following

steps,

knowing

that

D

ED ,D ( O t ) :

 GD (t )  GD1 (t )

t

D 1

D

ED ,D ( O t )

t

1D

F (t )



2D 1 3D 1 § t D 1 · 1D t 2 t O O    ...... ¸  t F (t ) ¨ * (D ) * (2D ) * (3D ) © ¹

F (t ) * (D ) From the above (7.46), we write that, for lim t p0

 O F (t )

t

D

* (2D )

  G (t )  G (t ) 1

D

1

we get the inverse of GD (t ) in limit t p 0 as GD (t )

t 0

1D

 O F (t )

2D

* (3D )

 ....

1 , we should have lim t p0 F (t )

D

t

t

2

(7.46)



* (D ) . Therefore, we get A



T0 t

1D

* (D ) . Thus;



* (D ) .

The homogeneous solution is: T1 (t )

AGD (t )

T t

t T  * (D )  E 0

1D

* (D )

0

D 1

D

ED ,D ( O t ) D

D ,D

( Ot )



(7.47)



In series form, we obtain the homogeneous solution as: T1 (t )



D

T0  * (D ) ED ,D ( O t )

§

* (D )

©

* (2D )

T0 ¨ 1  O

D



t O

2

* (D ) * (3D )

t

2D

·

 ..... ¸

¹

(7.48)

306

Chapter Seven

The particular integral in this case is a convolution for forcing the function with Green’s function, i.e.

 OTa u (t )  FD (O , t )  OTa u (t )  GD (t ) .

T2

Thus, we write a fractional differential equation with an RL

derivative and with a fractional order initial state (i.e. T D

0 Dt

>T (t )@  OT (t )

and its solution as depicted (7.49) is GD (t )

(D 1)

x (t )

C0 ) as follows:

(0)

0 d t  f, 0  D 1

FD ( O , t )

t

D 1

T

(D 1)

(t )

t 0

C0

(7.49)

D

ED ,D ( O t )

T (t ) C0  GD (t )   x (t )u (t )  GD (t ) (7.50) We write a fractional differential equation with an RL fractional derivative with an initial condition as an integer order initial state (i.e. T (0) T0 ) using the following system: D

0 Dt

>T (t )@  OT (t )

x (t ) ,

0 d t  f, 0  D 1

T (t ) t

0

T0

(7.51)

The solution of (7.51) is:



T (t )

T0 t

1D

* (D )

 G (t )   x (t )u (t )  G (t ) D

(7.52)

D

7.3.5 The requirement of one Green’s function for a fractional order differential equation composed using the Riemann-Liouville derivative and two Green’s functions for the Caputo derivative We see that for the Caputo derivative, it is based on a non-homogeneous fractional differential equation, say C D 0 Dt

> y (t ) @  O y (t )

x (t ) , with the initial condition y (0)

C D 0 Dt

(or Green’s function) for its homogeneous part driven by C

is GD (t ) The

y0 , and, therefore, we obtain the fundamental solution

> u (t ) @

G D (t )

D 1 * (1D ) t

. Here the Green’s function

D

ED ( O t ) .

homogeneous

solution

is

y1 (t )

A



C

GD (t )



D

y0 ED ( O t ) ,

and

the

particular

integral

is

y 2 (t )

x (t )  GD (t ) , where GD (t ) is the Green’s function in the case of the RL-fractional derivative (i.e.

GD (t )

FD ( O , t )

C D 0 Dt

t

> y (t ) @  O y (t )

D 1

D

ED ,D ( O t ) , the Robotnov-Hartley function). This can be verified as a Laplace transform of





x (t ) , which is sD y ( s )  sD 1 y (0)  O y ( s ) y(s)

s

x ( s ) , and gives: D 1

§ 1 · ¸ D ©s O¹

y (0)  x ( s ) ¨

D

s O The inverse Laplace transform of this above expression (7.53) is following: y (t ) C

y0

C



GD (t )   x (t )  GD (t ) D

GD (t )

GD (t )



(7.54)

ED ( O t ) t

D 1

(7.53)

D

ED ,D ( O t )

FD ( O , t )

The solution requires two Green’s functions, when the fractional differential equation is formed via the Caputo fractional derivative. On the other hand, the fractional differential equation formed via the Riemann-Liouville derivative requires only one Green’s function. This we will discuss in detail in subsequent sections, and we refer to these two separate fundamental solutions (Green’s functions) as ‘Alpha Exponential Function’ type-1 and type-2.  Ot  Ot for the Caputo type operator. They are represented symbolically as eD for the RL-type operator, and eD Obviously for D

1 , we get a classical exponential function (i.e. e  Ot ).

7.4 Formal description of a fractional differential and integral equation 7.4.1 Concepts of an ordinary differential equation as extended to fractional differential and integral equations Consider the ordinary linear differential equation (ODE) as homogeneous with constant coefficients (i.e. 2

D y (t )  aDy (t )  by (t ) P( x)

2

0 ). Then if D

and E

are distinct zeros (roots) of an indicial polynomial (i.e.

x  ax  b ), we know that exp(D t ) and exp( E t ) are linearly independent solutions of the ODE. If D

E,

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

307

then exp(D t ) and t exp(D t ) are the linearly independent solutions of the ODE. As an attempt to define a fractional differential equation (FDE), let rm , rm 1 ,....r1 , r0 be a strictly decreasing sequence of non-negative numbers. Then for the a1 , a2 ,....am constants, we write the following:

D

rm

r

r

 a1 D m1  ....  am D 0

> y (t ) @

(7.55)

x (t )

as a candidate to represent the general fractional differential equations. This equation is also complex looking, so we try to simplify it, through the addition of an imposed condition that rj be rational numbers. Thus, with q as an LCM of the denominators of the nonzero rj ’s, we then formally define the FDE as

D t t 0 , where v

1 q

nv

 a1 D

( n 1) v

 ....  an D

0

> y (t ) @

(7.56)

x (t )

, and n is a positive integer.

 . Then the fractional integral equation is defined as follows: (7.57)  D  a D  ....  a D > y (t )@ x(t )  ....  a D > y (t )@ x (t ) ), the coefficients a , a …, a are remaining 1 q

Let m and q be positive integers and let v

0

v

mv

1

In equation (7.56) (i.e.

D

nv

 a1 D

( n 1) v

m

0

1

n

2

n

constants and so the equation is a linear fractional differential equation. We can also have a non-linear fractional differential equation, such as:

 p (t ) D 0

With



t D

1

pi (t ) 2

as a function of t , say defined as

> y ( t ) @  y ( t )



0

0 or tD  S D

 12

> y (t ) @

nv

 p1 (t ) D

pi (t )

ai t

( n 1) v

 ....  pn (t ) D

( n i ) v

, i

0

> y (t ) @

x (t )

(7.58)

0,1, ...n . As examples, we take

0 . This we have solved in an earlier chapter (Section-6.2) with

the Laplace transform tricks and using an alternate analytical method.

7.4.2 The indicial polynomial corresponding to the fractional differential equation For convenience, we introduce the indicial polynomial P ( x ) corresponding to FDE represented as (7.56), i.e.

D

nv

 a1 D

( n 1) v

 ....  an D

0

> y (t ) @

x (t ) , as follows: n

P( x)

x  a1 x

n 1

(7.59)

 ...  an

 is a fractional differential operator as shown in:

Then, P D

v

 { D a D With this (7.60), the FDE is then compactly described as P  D y (t ) x (t ) P D

v

nv

( n 1) v

1

 ...  an D

0

(7.60)

v

7.4.3 The order of the fractional differential equation and the number of linearly independent solutions The order of this fractional differential equation (7.56) is of ( n, q ) , and is of constant coefficients. If q 1 , then v 1 , and the equation is simply an ordinary differential equation of order n with constant coefficients. The definition in (7.56) is for a non-homogeneous fractional differential equation with constant coefficients. With x (t ) 0 , the homogeneous FDE with constant coefficients vanishes as the source term of the equation. First, this homogeneous FDE has to be solved by obtaining Green’s function, and then for the source term, the standard convolution of this Green’s function using a source function will yield a solution. Consider a simple FDE of order (4, 3) as D

4

3

> y (t ) @

0 . Then if C1 and C2 are arbitrary constants:

y (t ) is a solution. The first solution is y1 (t ) obtained (noting that lim xp0 * ( x )

f ):

1

C1t

1

3

 C2 t

23 4

(7.61)

C1t 3 . Putting this y1 (t ) in the FDE, i.e. D 3 ª¬ y1 (t ) º¼ , the following is

308

Chapter Seven

D

This (7.62) satisfies one solution of D

4

3

4

> y (t ) @

3

> y1 (t )@

D ªC1t

1

¬

14 3

3

*  13  1 t 3

º C1 ¼ *  13  34  1

C1

*  13 * (0)

t

1

(7.62)

0

0 . A derivative of the first solution, i.e. y1 (t ) , shall be the second

that follows from an ODE. This is given as follows:

D > y1 (t )@ which is the solution of D

4

3

> y (t ) @

1

 3 º ¬ ¼ C2 t

D ªt

2

(7.63)

3

0 . The combination of these two gives us a solution for FDE as mentioned

above (7.61). 1

One may draw an inference parallel to the ODE. Consider D constant. Take D

D > y 2 (t ) @

2

> y (t ) @

1.333

(i.e. D

> y (t ) @

0 ; it has one solution, namely y (t )

0 as a second example; it has two solutions, namely y1 (t )

y1 (t ) and the composite solution is y (t )

combined solution is y (t )

> y (t ) @

C1t  C2 . Similarly, D

3

> y (t ) @

C1t and y2 (t )

C, a

C2 . Here

0 has three solutions, and the

2

C1t  C2 t  C3 . A general n order ODE has n linearly independent solutions. The FDE 4 3

0 ) has two solutions and not 1.333.. solutions. Here nv

integer number of solutions, N t nv in this FDE which is two (7.61), i.e. N

2!

1.333.. must logically have an 4 3

.

7.5 Finding a solution to a homogeneous fractional differential equation 7.5.1 An ordinary homogeneous classical differential equation and solution in terms of an exponential function An ordinary homogeneous differential equation, with constant coefficients of order n is:



0D

n

 ( a1 ) 0 D

( n 1)

 ...( an ) 0 D

(0)

> y (t ) @

(7.64)

0

We write the indicial polynomial for this (7.64) as: P( x)

We write the differential operator as P  D {



0D

n

 P ( D ) > y (t ) @

write the differential equation as

 ( a1 ) 0 D

n 1

x  a1 x

n

n 1

0

.

e

ct

 ...( an ) 0 D

0 . Let us try

y (t )

(7.65)

 ...  an

Therefore, in a compact way, we as a solution; placing this in the

differential equation, we obtain:

 P( D) ª¬ect º¼ If c is the root of an indicial polynomial P ( x ) This is because

n 0 Dt

ct

n

x  a1 x

n 1

c

n

 a1c

 ...an , that is, P (c )

n 1



 ...an e

ct

0 , then y (t )

 P(c) ect e

ct

(7.66)

is one solution.

n ct

ª¬e º¼ c e for a n positive integer returns the exponential function itself, without altering its

form.

7.5.2 A fractional differential equation and solution candidate in terms of a higher transcendental function We have seen that the fractional order derivative (say order u ) of an exponential function is not an exponential function. That is

u 0 Dt

ª¬e ct º¼ z c u e ct . Therefore, this exponential function is not a choice candidate function for the

solution of a fractional differential equation. We have seen in an earlier chapter (Section-2.9) that u 0 Dt

ª¬e ct º¼

Et  u , c , where Et

representation (that is Et  u , c

t

u

represents the Miller-Ross function (refer to Appendix-A) in a series

¦

( ct )k k 0 * ( k u 1)

f

this Miller-Ross function also as Et  u , c

). This is a variant of the Mittag-Leffler function. We represent

u

t E1,1u (ct ) , in the form of a two-parameter Mittag-Leffler function

(see Appendix-A). In addition, we have Et  0, c

e

ct



and Et 0, ic



cos(ct )  i sin(ct ) .

A Formal Method for Solving Fractional Differential and Fractional Integral Equations u 0 Dt

Interestingly, the fractional derivative of the Miller-Ross function is

> Et ( w, c)@

309

Et  w  u , c , i.e. retaining its

form. Therefore this Miller-Ross function (a variant of the Mittag-Leffler function), is one possible choice as a candidate function to be a solution to the fractional differential equation. We write the following identities for a Miller-Ross function and the exponential function (see Appendix-A): u 0 Dt

> Et ( w, c)@ Et  w  u, c Dt ª¬ect º¼ cect u 0 Dt ª ¬t  Et ( w, c º¼ t  Et ( w  u , c )  u  Et ( w  u  1, c

 ¼

ct Dt ªt e º

¬

ct

cte  e

ct

ct

We wrote that Et  0, c

(7.67)

u 0 Dt

e , from the above (7.67), and we write



u 0 Dt

> Et (0, c)@

ª¬e ct º¼ Et  u , c (see



Appendix-A). Therefore, we can possibly try functions of type Et kv, c , where k is an integer, as the candidate solution for a fractional differential equation (FDE).

7.5.3 A direct approach using the Miller-Ross function as a solution for a fractional differential equation (FDE) Let us now go for a ‘direct approach’ to an FDE as represented in the following expression:

 In this case (7.68), we have n

2, q

2 and v

 1 q

1 2

1 0 Dt

> y (t ) @

(7.68)

0

, so we can change the FDE to our ( n, q ) notation and

D

write with our standard definition as noted earlier (i.e.

1

 ( a ) 0 Dt 2  b

nv

 a1D

( n 1) v

 ....  an D

0

> y (t ) @

0 ) the following

equations:



0 Dt

1

 ( a ) 0 Dt 2  b

In (7.69) we have ( n, q ) { (2, 2) . In this case, is P ( x )

> y (t )@

 D   ( a ) D   b > y (t )@

0

2 t

0

 { P D

P D

1

v

2

1 2

1 t

0

D  aD

1

2

1 2

(7.69)

0

0

 bD . The indicial polynomial

2

x  ax  b .

With our hypothesis of using Et ( kv, c ) as a candidate, it seems reasonable to construct linear combinations of

Et  0, c , Et   12 , c , and Et  12 , c as a solution of our given FDE. The expression of the Miller-Ross function relates to Et

 12 , c



and Et  12 , c



as Et   12 , c

cEt  12 , c 

t

1

2

S

(see Appendix-A). This renders the term

Et  12 , c as an extra term. We will use this conjecture now. We note here that for an ordinary differential equation (ODE) of order

(2,1) with v

polynomial is the same as for our FDE example of order (2, 2) . For ODE, we have v solutions, namely y1 (t )

 1 q

v

1

0 Dt

2

1 2

,

we

e

try

ct

in

v

and y2 (t )

Dt > y1 (t )@

a

way,

similar

(1)

Dt y1 (t )

> y1 (t )@ , that is e

ct

Et (0, c ) ,

y 2 (t )

and

  1 q

1,

1 q

1 , then the two

the indicial

ct

ce . For the FDE case with

y 2 (t )

v 0 Dt

> y1 (t )@ ,

that

is,

ª¬e ct º¼ Et   12 , c . With this similarity, we try one solution and call it: \ 1 (t )

 {D

With P D following:

1

2

1

 aD

1

2

0

AEt  0, c  Et   12 , c

(7.70)

 bD , we now operate this differential operator on (7.70) (i.e. \ 1 (t ) ), and write the

310

Chapter Seven

 P  D >\ (t )@  D  aD 1

1

2

1

1



2

> AE (0, c)@  bD ª¬ E   , c º¼

 bD

1

 D  aD

1

0

t

0

2

1 2

t

 AcEt  0, c  aAEt   12 , c  bAEt  0, c   Et   32 , c  aEt  1, c  bEt   12 , c  cA  ac  bA Et  0, c   c  aA  b Et  

1 2

t

, c 

(7.71)

 32

*   12 In the above steps (7.71), a property of the derivative of the Miller-Ross function (see Appendix-A) is used, along with other properties (i.e.

X

cEt (X  1, c )  *tX 1 ),

cEt (0, c ) and Et (X , c )

Et ( 1, c )

 32 ; and with a

with X

simplification using these, we arrive at the result (7.71). Now, let A 2

P O

O  aO  b

 P  D >\ (t )@ 2



O P (O ) Et 0, O

1

 P (O ) E  

2

3

2

t 2 *   12

1



, meaning that

0 Dt

1 2

t

If O is the root of the indicial polynomial P ( x ) , then we have P (O )

 P  D >\ (t )@

x  ax  b , we can write

c  aA  b . Then the above result becomes: 1

1

2

2 O . Then from the indicial polynomial, which is P ( x )

O and c

,O

2

t

 32

 * 

(7.72)

1 2

0 . From the above (7.72), we find that



1

 ( a ) 0 Dt 2  b >\ 1 (t )@ z 0 , and implying that \ 1 (t ) which we

constructed via the Miller-Ross functions is not a solution. So, what is the solution?

7.5.4 Getting a solution from the roots of an indicial polynomial corresponding to the FDE using the Miller-Ross function 2

Let us indicate that D and E are the roots of the indicial polynomial P ( x ) formed for these two roots as described below:



\ 1 (t )

D Et 0, D

2

 E  t

1

2

,D

3

  >\ (t )@

These two functions will follow P D

1 2

t 2 *   12

1

2

,

x  ax  b . Let the two functions be



\ 2 (t )

E Et 0, E

2

 E  t

1 2

,E

2



(7.73)

 P  D >\ (t )@ . This derivation we have described in 1

2

1

a previous section (Section-7.5.3). Obviously it states that neither \ 1 (t ) nor \ 2 (t ) is a solution. Thus, let

\ (t ) \ 1 (t )  \ 2 (t ) , meaning that \ (t ) is a solution of the given FDE, which is:



\ (t ) Here we observe that \ (0)

D Et 0, D

1

D  E , 0 Dt 2\ (t )

2

 E E  0, E  E   2

t

1

t 0

1 2

t

0 0 Dt 2\ (t )

f and

t 0

,D

Dt

seen for the ODE solution with e

2

, E  t

1 2

,D

2

and t  E  t

1 2

,D

f . As we have just 1

 ( a ) 0 Dt 2  b

@

of the two distinct solutions); of P ( D ) y (t )

and te



 

0 Dt

2

2

, E  t

(7.74)

2

0 are not equal. When they are equal, say, D

discussion, it appears that a linear combination of terms of the form Et 0, D

t Et  12 , D



>

Dt

2

t

1 0 Dt \ (t ) t 0

seen, \ (t ) given by the above expression (7.74), is a solution of the given FDE (i.e. if the roots D and E of the indicial equation P ( x )

 E  1 , E

2

> y (t ) @

0 ),

E (as we have

0 . So referring to this 1 2

,D

2

, t  E  0, D , 2

t

is a likely candidate for the solution when roots are equal.

If we make this assumption, then proceeding as above, we may write:

\ (t )

1  2D t E  0, D  D E  2

2

t

t

1 2

,D

2

 2D t  E   t

1 2

,D

2



(7.75)

We note here that if the ODE of order (2,1) has the root O for an indicial polynomial P ( x ) , then the FDE of order (2, 2) , which has the same indicial polynomial P ( x ) and has the roots O

1

2

or

2

O . We can thus write the indicial

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

311

q

v

polynomial of FDE of order ( n, q ) , which will have the roots O or O , where q 1 / v . The solution to FDE with a direct method that we described yields a similarity with the ODE. The basis functions of course in the case of the FDE is of higher transcendental functions.

7.6 Motivation for the Laplace transform technique 7.6.1 The indicial polynomial is the same as we get from the Laplace transform of a fractional differential equation



Let us have the same homogeneous FDE (7.68), that is,

1

0 Dt

0

 ( a ) 0 Dt 2  (b ) 0 D

> y (t ) @

0 . Note that this FDE is

composed via an RL-fractional derivative of order half. Applying a generalized Laplace transform formula (as in Section-5.14), to the FDE, we write the following:

 sY ( s )  y (0)  a  s  12

y (0)  0 Dt

Y (s)

have

 12 y (0) 0 Dt

used  12

{ 0 Dt

$ ^ 0 Dt y (t )` y (t )

t 0

y (0)  0 Dt

sY ( s )  y (0)

^ `

and $ y (t )

2

 12

C We

s  as

1

and

1

2

 12

Y ( s )  0 Dt

y (0)

(7.76)



b

^

0

C P s

y (0)

$



y (0)  bY ( s )

1

2

2

P( x) 1

0 Dt

2

`

x  ax  b

1

y (t )

 12

s 2Y ( s )  0 Dt

y (0) ,

noting

that

Y (s) .

Here we discuss two issues. The first one is how we know that

 12

C

y (0)  0 Dt

y (0) is finite. If C is not finite

then the problem is a serious one, and this approach becomes meaningless. Therefore, we must assume that C is finite. The second issue is how to find an inverse Laplace transform that is $

1

^  ` 1

P

s

. We will resolve this second

issue.

7.6.2 A solution with distinct roots in an indicial polynomial Expand

1 P( x)

as partial fractions, assuming two distinct roots D and E for the indicial polynomial P ( x ) , and write it

as: 1

1

P( x)

Put x

s

1

2

1

2

x  ax  b

1 · § 1  ¨ ¸ D  E © x D x  E ¹

(7.77)

in the above (7.77) to get the following: 1

1



P s

We therefore need to find now that $ function, we find that $

1

^ ` 1 sX  c

defined in series form as FX (c, t )

1

^ ` 1

s

1

2 c

1

s  as

2

1

§ 1 1 ·  1 1 ¨ ¸ 2 2 D  E © s D s  E ¹ 1

2

b

(7.78)

. From the tables of the Laplace transform of a higher-transcendental

FX (c, t ) , where FX (c, t ) is the Robotnov-Hartley function (see Appendix-A),

¦

f

c n t ( n1)X 1 n 0 *  ( n 1)X

. Therefore, we write the solution in terms of the Robotnov-Hartley

function (another variant of the Mittag-Leffler function), as:

y (t )

$

1

^Y ( s )`

C

D E

F

1

2

(D , t )  F1 ( E , t ) 2



(7.79)

However, with a direct approach we have found the solution in terms of the Miller-Ross function, but is it the same solution? Let us find out. We manipulate 1 , which appears in partial fractions (7.78), as follows: s c

312

Chapter Seven

1 s

1

2

1

s

c

2

sc

c

s

1

2

2

1  cs  12

sc

2

1  12

2

1

§ 1  cs  2 · ¸ 2 ¨ 1 sc © s 2 ¹ 1

(7.80)

c



2

s (s  c ) s  c From tables of the Laplace transform for the Miller-Ross function (see Appendix-A), we write:

$

1

^ ` 1

s

1

2

$

c

1 ­

1 c ½  ®  12 2 ¾ 2 ¯ s (s  c ) (s  c ) ¿



Et  12 , c

2

(7.81)

 cE  0, c 2

t

With the above derivation, we write:

y (t )

$

1

^Y ( s )`

1

­° C

$ ®



°¯ P s

1

2

½° ¾ °¿

2 2 C § D Et  0, D  E Et  0, E · ¨ ¸ D  E ¨  Et   1 , D 2  Et   1 , E 2 ¸ 2 2 © ¹

The solution in (7.82) is a constant

C

D E

(7.82)

multiplied by an obtained \ (t ) using the direct method.

7.6.3 The solution with equal roots for indicial polynomials When the roots are equal ( D



E ), we write Y ( s )

C s

s

1

2

D



2

D



2

, and carry out the following manipulations:

D





2

§ · D 1  ¨  12 ¸ ¨ s s  D 2 s  D 2 ¸ © ¹

1 1

2

C

Y (s)

s

1

2

1 s

1

2 2



D

2

(7.83)

2 2 2

s  D s  D

D

2 s

 12

2 2

s  D

We use the Miller-Ross function and its Laplace transform identity (see Appendix-A), which is: 1

and write the following inverse Laplace transform:

1

­

½ ¾ t  Et X , a  X Et X  1, a ¯ s (s  a) ¿

$ ®

X

2

(7.84)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

1

­ °

313

½ ­ ½ ­ ½ 1 ° ° 1 ° D 2 ° 1 ° $ ® $ ® ¾ ¾ ¾ 2 2 2 1 1 2 2 2 s s  D ¿ s D ¿ s 2 D ° ° ° ° ° °  ¯ ¯ ¯ ¿ ­ ½ D ° 1 °  2$ ® 1 ¾ 2  2 2 ° ¯ s s  D ° ¿

$ ®

1





(7.85)

t  E  1, D  E  1  1, D  D t E 0, D  D 0 E 0  1, D   2D t  E   , D  2D  E   , D t  E  1, D  E 0, D  D E 0, D   2D t  E   , D  D E   , D 2

2

t

2 0

t

1 2

2

1 2



2

2

t

2

1 2

t

2

2

t

t

2

1 2

t



2

t

2

t

We then use the identity Et 1, a

2

t

t

1 2

2

aEt  0, a (see Appendix-A) to get the following:

§ 1  2D 2t Et  0, D 2  D Et  12 , D 2 · ¸ y (t ) C ¨ ¨ 2D t Et   12 , D 2 ¸ © ¹



(7.86)



We obtained the above solution in terms of the Miller-Ross function. Here, we can write a solution in terms of the Robotnov-Hartley function as a convolution (represented as * ): y (t )

1

­ °

½ ° 2¾ 1 2 °¯ s 2  D °¿ 1 1 § 1 · § 1 · *¨$ C ¨$ 1 1 ¸ 2 2 ¸ 2 2 s D ¹ © s D ¹ ©

$ ®

C





^

`



^

`

C F1 D , t * F1 D , t 2

2

(7.87)



t

C

³  F D , t  W  F D ,W dW 1

1

2

2

0

The starting point of the convolution integral is from 0 , as our FDE has a fractional differentiation operator with zero X

as its starting point, i.e. 0 Dt .

7.7 A linearly independent solution of the fractional differential equation (FDE) Ordinary differential equations (ODEs) have a set of linearly independent solutions, from which Green’s function is obtained via a Wronskian determinant. This Green’s function is a solution of the homogeneous ODE and is utilized to 2

obtain a solution of a non-homogeneous ODE. The ODE is of the form D y (t )  aDy (t )  by (t ) polynomial as P ( x )

2

x  ax  b . We have simplified the symbol from

n 0 Dt

> y (t )@ to

0 with an indicial

n

D y (t ) . Dt

Dt

Let a zero (root) of this indicial polynomial P ( x ) be denoted as D . Then y1 (t ) e and Dy1 (t ) D e will be solutions, as will other higher derivatives. However, they are not linearly-independent. If E is the other zero (root) of the indicial polynomial, then y2 (t )

e

Et

is the other solution along with its higher derivatives. Again, they are not

linearly independent. The combination y (t ) Dt

Et

y1 (t )  y2 (t )

Dt

e

e

Et

is the solution.

are obviously solutions and are too linearly independent .Therefore, the Here, y (t ) and Dy (t ) D e  E e fundamental solution corresponding to the roots is as follows: E y (t )  Dy (t ) Dy (t )  D y (t ) (7.88) y1 (t ) y 2 (t ) E D E D

314

For D

Chapter Seven

 . Dt

E , we have y2 (t )

t e

Let us take an example of the homogeneous FDE as:

D Herein (7.89), n

3

 2D  D

2

1

2

 2D

> y (t ) @

0

(7.89)

0

1 2

3 and v

so q 2 . This is a fractional differential equation (FDE) of the order (3, 2) . Taking the Laplace transform of the FDE (7.89), we get the following:

s

3

2

1

Y ( s )  D 2 y (0)  sD

 12





y (0)  2  sY ( s )  y (0)

1

 s 2Y ( s )  D

 12

(7.90)



y (0)  2 Y ( s )

0

giving the relationships (7.91) : Y (s) C

C s

3

2

 2s  s

1

2

A  Bs

2

We can write the indicial polynomial as



have P s

1

2

s

3

2

 2s  s

1

2

1

A

1

C

D y (0)  2 y (0)  D 2

3

P( x)

 12

D 2 y (0)  sD

 12

2

y (0)  2 y (0)  D

y (0)

B

 12

D

y (0)

 12

2

( x  2)( x  1) with

x  2x  x  2

(7.91)

y (0)

x

1

s 2 , and we

 2 . Therefore, we write:

Y (s)

C s

3

2

A

 2s  s

1

2

Bs



AY1 ( s )  BY2 ( s ) (7.92)

 P s

2

P s

1

2

1

2

The solution is an inverse Laplace transform of the above (7.92), obtained through Y ( s ) , that is:

y (t )

A$

1

^Y1 ( s )`  B$ 1 ^Y2 ( s )` A$

1

­° 1 ® 1 2 ¯° P s



½° ­ s 1 °  B $ ¾ ® 1 2 ¿° ¯° P s



½° ¾ ¿° (7.93)

Ay1 (t )  By2 (t )

^

`

Using the identity $ Dy1 (t ) one-whole derivative y2 (t )

1

y1 (t )

$

y 2 (t )

$

1

1

­°

$ ®

^Y2 ( s )`

sY1 ( s )  y (0) , we get $

½° ¾ 2 °¯ P s °¿ ­ s ½° 1 ° $ ® ¾ 1 2 °¯ P s ¿° 1

^Y1 ( s )`

1

 1



^sY1 ( s )`

Dy1 (t )  y (0)G (t ) . Therefore, y1 (t ) and its

Dy1 (t )  y1 (0)G (t ) are linearly independent solutions, and their linear combination

provides the total solution. Looking at P ( x ) fraction is:

3

2

x  2 x  x  2 , as a cubic equation, it has the roots x 1 P( x)

As demonstrated earlier, we write x

s

1

2

3

1

1

2

3( x  2)

x  2x  x  2



2 , x 1 and x 1

6( x  1)



1 2( x  1)

and re-write the following: 1 1 1 1   1 1 1 1 P s 2 3 s 2 2 6 s 2 1 2 s 2 1

 







1 . The partial (7.94)

(7.95)

We will carry out the inverse Laplace transform using identities of the Miller-Ross function as we demonstrated in an earlier part, to write:

y1 (t )

1 3

  Et  12 ,1  4 Et  12 , 4  2 Et  0,1  2 Et (0, 4)

(7.96)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

We note that y1 (0)

0 ; and from the above (7.96), we carry out one whole derivative and write y2 (t ) : y 2 (t )

1 3

Dy1 (t )

§  Et  12 ,1  16 Et  12 , 4 · t  12 ¨¨ ¸¸  1 E E   2 0,1 8 (0, 4)  t t © ¹ *2

f ; although it is a solution of the given FDE. We have

Note that y2 (0)

315

independent. Therefore, with this, we write y (t )

(7.97)

y1 (t ) and y2 (t ) , which are linearly

Ay1 (t )  By2 (t ) as a solution of the given FDE.

7.8 The explicit solution for a homogeneous fractional differential equation (FDE) From the observation of the previous sections, we now write the following statement for the FDE

D

nv

 a1 D

( n 1) v

 .....  an D

0

> y (t ) @

1 v

0 of order ( n, q ) , with q

. Note that mathematicians write the RHS as

zero, while physicists and engineers will write it as G (t ) (i.e. the Delta-function). The fundamental solution to this homogeneous system is also termed as an ‘impulse response function’. The indicial polynomial is P( x)

n

x  a1 x

n 1

 a2 x

n2

 ...  an , meaning that the fundamental solution is: 1

y1 (t )

­ 1 ½ 1 1 v $ ^P ( s )` v ¾ ¯ P( s ) ¿

(7.98)

$ ®

The integer N is greater than or equal to nv , that is N  1  nv d N , then N numbers of a linearly independent m

solution, such as y1 (t ), y2 (t ), ......, y j (t ), .... y N (t ) , where ym 1 (t ) 1 v

As an example, let us take the FDE of order (2, q ) with q

D 2 , we write P 1 ( s v )

1

$

^P

1

v

`

( s ) . For D1 z D 2 , we write:

­ 1 1 ½  v ® v ¾ D1  D 2 ¯ s  D1 s  D 2 ¿

v

( s  D1 )( s  D 2 ) 2

ei (t ) q 1

¦D

2 q  k 1 Et (  kv, D i ) i

1

A  e1 (t )  e2 (t ) for D1 z D 2 , and y1 (t )

convolution. In addition, we have y1 (0)

y2 (0)

­

½ ¾; v ¯ s  Di ¿ 1

$ ®

k 0

when N ! nv then y N (0)

(7.100)

( s  D1 ) . Let us write the Robotnov-Hartley function and the summation of

the Miller-Ross function (see Appendix-A) as: Fv (D i , t )

Then, y1 (t )

(7.99)

0

1

v

v

> y (t ) @

and it can be factorized and re-written as

1

v

P (s ) and for D1

v

 a1 D  a2

x  a1 x  a2 ,

( x  D1 )( x  D 2 ) . Therefore, we have y1 (t ) 1

2v

2

In addition, the indicial polynomial is P ( x )

0,1, 2, ........, ( N  1) .

as:

D P( x)

D y1 (t ) , with m

A  e1 (t ) * e1 (t ) for D1 0 and y N (0)

........ y N 1 (0)

i

1, 2

(7.101)

D 2 , where the symbol ( * ) means 1 or a constant, with N

nv and

f . We elaborate on this statement with a few simpler examples. 1 v

Let us take an FDE of order (3, q ) , with q

, as:

D The indicial polynomial is P ( x )

3

2

x  a1 x  a2 x  a3

3v

 a1 D

2v

v

 a2 D  a3 D

( x  D1 )( x  D 2 )

2

0

> y (t ) @

0

(7.102)

obtained in its factored form. Therefore,

we write the following partial fraction: B1

1 v

P(s )

With B1

1 (D 2 D1 )2

, B2



1 (D 2 D1 )2

and C2

1 (D 2 D1 )

taking an inverse Laplace transform of (7.103) is y1 (t )

v

( s  D1 )



B2 v

(s  D 2 )



C1 v

(s  D 2 )

2

(7.103)

. The fundamental solution from the above equation when

B1e1 (t )  B2 e2 (t )  C1  e2 (t ) * e2 (t ) , i.e. using:

316

Chapter Seven

$

1

^(s

v

 Di )

`

1

Fv (D i , t )

¦

ei (t )

$

1

^(s

v

 Di )

q 1 q 1

¦¦

2

`

q 1 k 0

ei (t ) * ei (t )

2q  j k 2

Di

j 0k 0

Di

q  k 1

2

Et (  kv, D i )

Fv (D i , t ) * Fv (D i , t )

 

q

t Et ( j  k )v, D i



 

q

 ( j  k )v Et 1  ( j  k )v, D i

(7.104)



We have given the explicit solution as (7.104) above. The other linearly independent solution is obtained via differentiation of this y1 (t ) .

7.9 The non homogeneous fractional differential equation and its solution

D

A type

nv

 a1 D m

conditions P( x)

n

D y (0)

x  a1 x

with x (t )

( n 1) v

n 1

 a2 x

 a2 D

( n  2) v

 .....  an D

for

m

0 n2

0

> y (t ) @

x (t ) ,

represents a non-homogeneous FDE, with

0,1, 2, ...........( N  1) .

The

indicial

polynomial

is

 ....  an . The fundamental solution to the corresponding homogeneous FDE (that is,

0 ) gives Green’s function shown by k (t )

$

1

^P

1

`

v

( s ) , which we earlier termed y1 (t ) . With this

Green’s function, we find the solution of a non-homogeneous FDE as the following in the convolution integral (it is also termed as a ‘Particular-Integral’): t

y (t )

³  k (t  [ )  x([ ) d[

k (t ) * x (t )

(7.105)

0

For example, we have:

D 1 4

a non-homogeneous FDE of order ( n, q ) { (2, 4) , v

2v

, q

 aD

v

> y (t ) @

(7.106)

x (t )

4 ; with the condition given as y (0)

0 . The indicial

2

polynomial is P ( x ) x  ax x ( x  a ) . The fundamental solution or Green’s function is (as obtained from the Miller-Ross function; see Appendix-A): 1

k (t )

1

v

1

1

v

1

$ {P ( s )} $ {s ( s  a ) } 3

(7.107)

¦ a E  ( j  2)v, a j

4

t

j 0

Therefore, we write the solution to the non-homogeneous system as follows: t

y (t )

³  k (t  W )  x(W ) dW 0

t

§ 3 j 4 · a Et W  ( j  2)v, a ¸  x (W ) dW ³0 ¨© ¦ j 0 ¹



We extend further the equation (7.106), i.e. D transform of this is X ( s )

b ( s 2 b2 )

2v

 aD

v

> y (t ) @

x (t ) , with y (0)

0 , and x (t )

(7.108)

sin bt . The Laplace

$ ^sin bt` . Taking the Laplace transform of the entire equation (7.106), after

rearranging, we get the following:

Y (s)

X (s) v

P(s )

(7.109)

b v

v

2

2

s ( s  a )( s  b ) After partial fractions and applying an inverse Laplace transform of the Miller-Ross function and its corresponding cosine functions (see Appendix-A), we get: y (t )

4

1 8

a b

2

¦a k 1

k 1

§ bEt  ( k  1)v  1, a 4  bCt  ( k  1)v  1, b · ¨ ¸ ¨  a 4 S  ( k  1)v  1, b ¸ t © ¹

(7.110)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

317

With Et as the Miller-Ross function, Ct and St are higher trigonometric cosine and sine functions corresponding to the Miller-Ross function, as detailed in Appendix-A. Let us take another example:

D Here, v

X (s)

1 6 55 A s2

n

> y (t ) @

x (t )

x ( x  1) . With

x (t )

6v

D

v

5

6 , and the indicial polynomial is P ( x )

(7.111)



A 55t 

11v 36 * (5 v ) t

, we have

1  . With these, we write: 1 s5 v

X (s)

Y (s)

55 A( s

v

P(s ) s With an inverse Laplace transformation of (7.112), we obtain a solution as shown: 55 A 2 y (t ) t 2

5v

 1)

55 A

5v  2

s

(7.112)

3

(7.113)

7.10 Fractional integral equations and their solution 7.10.1 Describing a fractional integral equation

D

The Laplace transform of an equation of the type

0

 b1 D

v

 b2 D

2 v

 ...........  bm D

 mv

> y (t ) @

x (t ) , with

m ! 0 and q ! 0 , and with a forcing function at the RHS of an exponential order, is:

1  b s

v

 b2 s

1

2 v

 ...........  bm s

 mv

Y (s)

X (s)



s

(7.114)

We can re-write the above (7.114) as:

s

mv

 b1s

( m 1) v

 ......................  bm Y ( s )

 , with R( x) solution of the fractional integral equation is y (t ) $ ^s  ` . Thus, from this (7.115), we have Y ( s )

s

mv

X (s)

m

x  b1 x

R ( sv )

1

m 1

 b2 x

mv

m2

(7.115)

X (s)

 .......  bm .

Then the

X (s)

mv

R ( sv )

v

If we wish to be more explicit and express y (t ) in terms of x (t ) , we should make further assumptions. With R ( D ) as a fractional differential operator of order ( m, q ) and if k (t ) is the Green’s function associated with the v

>

homogeneous FDE (that is, R ( D ) y (t )

@

0 ), that is, a fundamental solution, then k (t )

$

1

^R

1

v

`

(s ) .

Now, let M be an integer just greater than mv (i.e. ( M  1)  mv d M ). The Laplace transform of a fractional

^

`

D

D

derivative of the RL type is $ D x (t ) information). With D (i.e. D

n  M D

x (t )

t 0

s X (s) 

¦

M 1 M  n 1 n 0

s

D

n  M D

x (t )

t 0

(see Section-5.14 for further

mv , and with this rearrangement, we get the following, with an understanding of symbolism

{D

n  M D

x (0) ): s

mv

^

X (s)



In our earlier description (7.115), we obtained Y ( s )

mv

M 1

` ¦s

$ D x (t ) 

M  n 1

n 0

s mv X ( s ) R ( sv )

D

n  M  mv

x (0)



(7.116)

. By placing this in the above expression (7.116), we

get the following: Y (s)

s

mv

X (s) v

R(s )

^

mv

`

$ D x (t ) v

R( s )

M 1

¦ n 0

D

n  M  mv

x (0)

§ s M  n 1 · ¨ R(s v ) ¸ © ¹

(7.117)

318

Chapter Seven

7.10.2 Extension of the final value theorem of integral calculus Let us now revise the final value theorem of integral calculus, before slightly extending it. Let M  1  mv d M , that is, let M be an integer just greater than mv . Also, let us take p as an integer from 0 to ( M  1) . We should also p

assume that D k (t ) is piecewise continuous for values upwards of p

^

`

p

0,1, 2, ...( M  1) . We write p 1

p

s $ ^k (t )` 

$ D k (t )

¦s

p 1 j

j 0

 D k (0) j

(7.118)

p

s k (s) j

The above expression is true for D k (0)

0 with j

0,1, ........, ( M  2) . This is true since y1 (t ) v

0 and since the other linearly independent

fundamental solution of the homogeneous differential equation, R ( D ) j

solutions are the derivatives of y1 (t ) , namely y j 1

y1 (0) p

y2 (0)

....

y M  2 (0)

k (t ) is a

D y1 (t ) We noted in an earlier section that we have

^

`

p

p

s $ ^k (t )` for

0 . With this observation, we therefore have $ D k (t )

0,1, 2, .........., ( M  2) . This will be proven by the extension of a final value theorem.



Extension of the final value theorem states that if lim s nf s



0 , so we have for lim s nf s$ ^ f (t )`

setting v



v 1

$ ^ f (t )`





v

0 , lim t p0 D f (0)

0 , implying lim t p0  f (0)



0 . We test this by

0 -this is the original final value

theorem. We have $ ^k (t )`

K (s)

1 R ( sv )

1 s mv  b1s ( m1) v  b2 s ( m2 ) v ....bm



p

lim s nf s K ( s ) s where, j

( M  2)  mv .

p

j

^

p

`

p

Y (s)

^

mv

v

R(s )

^

0 .Using



`

mv

v

¦ D

s M n1 R ( sv )

n  M  mv

x (0)

n 0

`

$ D x (t ) R(s )

M 1

 $ ^D

M 1

¦ D

M  n 1

p

lim t p0 D k (t )



0 . So, for

`

k (t ) , we write the following (that is,

0,1, 2, ............., ( M  2) ):

s K ( s ) for p

$ D x (t )

(7.119)

§ · 1 1 0 ¨ mv ¸ ( m 1) v ( m  2) v  b2 s  ..  bm ¹ f © s  b1s

p

s $ ^k (t )`

Y (s)



The above derivation (7.119) implies that

0,1, 2, ......, ( M  2) , we state that D k (0)

because $ D k (t )

p

, and we write the following:

( n  M  mv )

 s

M  n 1

v

©

^

x (0) $ D

n 0

§¨ R(1s ) ·¸

M  n 1

¹

`

k (t )

M 1 § 1 · mv ( ) $  D x t ^ `  D(nM mv ) x(0) $ ^D M n1k (t )` ¦ ¨ R(s v ) ¸ © ¹ n 0





(7.120)

 $ ^k (t )`  $ ^D mv x(t )` M 1



¦ D

( n  M  mv )

^

x (0) $ D

n 0

M  n 1

`

k (t )

Getting an inverse Laplace transform of (7.120), we write: t

y (t )

³  k (t  W )  D

mv



x (W ) dW

0

M 1



¦ D n 0

(7.121) n  M  mv

x (0)

 D

M  n 1

k (t )



A Formal Method for Solving Fractional Differential and Fractional Integral Equations t

³  k (t  W )  D

y (t )

We will simplify (7.121) (i.e.



mv

x (W ) dW 

0

M 1

¦ D

n  M  mv

n 0

x (0)

 D

319

M  n 1



k (t ) ) with

0 and M ! mv , the term D  M  mv x (t ) is a fractional integral. This integral, if written in the RL

M t mv . For n

D

formula (that is, a Dt

> f (t ) @

1 * (D )

t

1

D ³ (t  W )  f (W ) dW ) becomes zero, that is: D  x(W ) dW > x(0)@ ³ (t  W ) a

 ( M  mv ) 0

0

0

1 * ( M  mv )

( M  mv ) 1

0

0

(7.122)

7.10.3 The solution to fractional integral equations Thus, for M ! mv , when rewriting the indexes of the summation in the expression (7.121) of y (t ) , we get:

For n

0 and M

t

M 2

0

j 0

mv mv 1 j j ³  k (t  W )  D x(W ) dW  ¦  D x(0)  D k (t )

y (t )

mv , we write that D  M  mv x (0)

0

x (0) . Again, for M

D x (0) indexes of summation in the expression (7.121) of y (t ) we get: t

M 1

0

j 0

(7.123)

mv and rearranging the

M j j ³  k (t  W )  D x(W ) dW  ¦  D x(0)  D k (t )

y (t )

(7.124)

In short, we get the following solutions: t

M

1

M ! mv,

³  k (t  W )  D

y (t )

mv



x (W ) dW

0

(7.125)

t

M

1

M

mv,

³  k (t  W )  Dx(W ) dW  x(0)k (t )

y (t )

0

7.11 Examples of fractional integral equations with explicit solutions We now take explicit examples of the fractional integral equation for a solution. As an example, we take the following equation:

D 

0

Taking the Laplace transform of (7.126), we obtain 1  s

s

2v



 b Y (s)

 bD

2 v

2 v

> y (t ) @



b Y (s)

(7.126)

x (t )

X ( s ) or, after a rearrangement, we write

2v

s X ( s ) . This gives us the following: 2v

s X (s)

Y (s) The equation above shows R ( x )

2

x  b , with k q (t )

$

1

s

2v

b

(7.127)



^ ` . Note here that we have added a sub-script in the 1 s2 v b

 > y (t ) @

earlier notation of k (t ) , which is the fundamental solution of R D

v

0 , the homogeneous FDE.

Drawing from the discussions in the previous section (7.125) we write, for M

y (t )

t

³  k (t  W )  D 0

q

mv



x (W ) dW . For M

1 mv , the solution is as below: t

³  k (t  W )  Dx(W ) dW  x(0)k (t )  D  bD > y (t )@ x(t ) . With say, q

y (t ) With this (7.128), we give an explicit solution for

m

2, M

mv 1 , we write, using $

1

1 and M ! mv , that the solution is

^ s 1b `

e

 bt

0

0

q

2 v

2, v

1 q

1 2

and

Et (0, b ) , the following solutions:

k 2 (t )

$

1

^ s 1b `

Et (0, b )

e

 bt

t

y (t )

(7.128)

q

³e 0

 b ( t W )

 Dx(W ) dW  x(0)ebt

(7.129)

320

Chapter Seven

For q 3 , v 13 , m 2 , and M 1 with M ! mv , we write the following solutions; with the use of the Miller-Ross fractional sine and cosine functions (see Appendix-A)using the following inverse Laplace transform: q q 1 1 1 $ Ct v, b 2  b 2 St 2v, b 2 2v s b (7.130)

^ `









1

 b 2 St 0, b

k 3 (t )

$

1

Ct (v, a )

^ ` 1

2



Ct  13 , b

s b 3

Ct (v,  a )



Ct  13 , b

k 3 (t )

3

2

³ C W  

y (t )

t

0

D

0

 bD

 12

1 3

, m

> y (t ) @

2,M

1 3

3

2

b

St (v, a )

b

t

4, v

2



 1v the solution is:

For (7.130), where q

For q

q



,b

3

2

 12



St  23 , b

b

 12

3



 12

St  23 , b

3

2

 b S  0, b 1

3

2

2

t

 St (v,  a ) (7.131)

 b S  0, b 1

2



St W  23 , b

3

2

3

2

t

b

1

2

2



St W 0, b

3

2

 D D

1 with M ! mv , we write the solution of

0

2

3



x (W ) dW

 bD

2 v

> y (t ) @

x (t ) , that is,

x (t ) :

k 4 (t )

$

1

^ ` 1

s

1

2



Et  12 , b

b

y (t )

³  E W  t

,b

 bE  0, b 2

t

(7.132)

t

1 2

2

2

0

 bE  0, b  D 2

t W

1

2



x (W ) dW

With all these discussions we note that the solution should be obtained by taking the inverse Laplace transform of Y ( s ) . For example, when the forcing function of the fractional integral equation

x (t )

O

t , with the Laplace transform set at X ( s )

* ( O 1)

D

0

 bD

2 v

> y (t ) @

x (t ) , is

, with O ! 1 , we get the following:

s O 1

Y (s)

s s

2v

2v

b

X (s) (7.133)

* (O  1) s For q

2 and v

have Y ( s )

1 2

, we have Y ( s )

 *(O  1)  s

O

s ( s 2  b3 )

* ( O 1) sO ( s b )

O 1 2 v

 b) q

3 and v

1 3

, we



 sO v ( sb2 b3 )  sO v ( sb2 b3 ) as an inverse Laplace transform of this solution i.e.





§ Ct O , b 3 2  b 12 St O  v, b 3 2 y (t )  * (O  1) ¨ 1 ¨ b 2 S O  v , b 3 2 t © § · The solution is: 1 , we have Y ( s )  * (O  1) ¨ O 1 2  O  1 b ¸ 4 s ( s b ) 2 s 2 ( s b ) ¹ ©



4 and v

2v

 *(O  1) Et (O , b) . For

, which gives y (t )



For q

(s

y (t )



·¸

 *(O  1)  Et  O , b 2  bEt  O  12 , b 2

For the inverse Laplace transform and the use of the Miller-Ross function, refer to Appendix-A.

¸ ¹

(7.134)

(7.135)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

321

7.12 Sequential fractional derivative of the Miller-Ross type aDx kD and sequential fractional differential equations (SFDE) 7.12.1 The sequential fractional differential equation (SFDE) A sequential linear fractional differential equation of the order nD and n  ` is represented as: b0 ( x ) y ( x )  b1 ( x )

 D > y( x)@  b ( x)  D > y( x)@  ... ....  b ( x )  D > y ( x)@ f ( x) 2D

D

a

2

x

a

x

(7.136)

nD

n

Where aDx kD

> y ( x )@

is aDx

D

an

RL

or



*

D

above D



*

D

E

(7.137)

f ( x) z

Caputo

ª¬ aDx ( k 1)D > y ( x )@º¼ ; k aDx

The

a

x

is a fractional sequential derivative operator, of commensurate order D , aDx

* D a Dx

where aDx

kD

comes *

D

D E

2D

from

2, 3, ... . For k * D a Dx

> y ( x )@ the

f ( x ) , that is,

operator

fact

* D a Dx

(denoted

2 and 0  D 

1 2

via

* D a Dx

> y ( x )@

D

*).

In

for

addition,

, we have:

ª¬ a* DDx > y ( x )@º¼ z a* Dx2D > y ( x )@

that

> y ( x )@ ,

D ! 0, E ! 0 ,

(7.137)

generally

the

composition

ª¬ a* DDx > y ( x )@º¼ z a* Dx2D > y ( x )@ . We will look at the detailed proof of

this in Chapter-9, while discussing the composition rules, and those of fractional derivatives and integrals.

7.12.2 The matrix form representation of SFDE When representing the SFDE in matrix form, we write it as:

D

nD

( n 1)

> y ( x)@  ¦ ak ( x)D kD > y ( x)@

f ( x) ; n  `

(7.138)

k 0

(7.138) reduces to: *

by changing the variables with y1 ( x )

y ( x) ;

1 § 0 ¨ 0 0 ¨ A( x ) ¨ ... ... ¨ ... ... ¨ a a © 0 1

*

D

D Y( x )

D

D y j ( x)

0 ... 1 ...

(7.139)

A( x )Y( x )  B( x )

y j 1 ( x ) ;

j

1, 2,..., ( n  1) , where, in (7.139), we have:

0 · 0 ¸

§ y1 ( x ) · ¨ y ( x) ¸ ¨ 2 ¸ ¨ ... ¸ ¨ ... ¸ ¨ y ( x) ¸ © n ¹

§ 0 · ¨ 0 ¸ ¸ ¨ ¸ ... ... ... , B( x ) ¸ ¨ ... ¸ ; Y( x ) ... ... ... ¸ ¨ ... ¸ ¨ f ( x) ¸ ¸ ... ...  an 1 ¹ © ¹

As an example, we take the following fractional differential equation:

y cc(t )  3 Setting y cc(t )

0

Dt

4(D )

> y (t )@ with D

1 2



3

0 Dt

2

> y ( t ) @  y ( t )

(7.140)

(7.141)

f (t )

gives SFDE as the following: 0Dt

4D

y (t )  3

D 0

3D

t



y (t )  y (t )

1 2

f (t ) ; D

(7.142)

which is reduced to: * D 0 Dt Y(t )

(7.143)

AY(t )  B(t )

with:

§0 ¨0 A ¨ ¨¨ 0 © 1

1 0

0· 0¸ ¸ ; B(t ) 1 ¸ 0 0 3 ¸¹

0 1 0 0

§ 0 · ¨ 0 ¸ ¨ ¸ ; Y(t ) ¨¨ 0 ¸¸ © f (t ) ¹

§ y1 (t ) · ¨ y (t ) ¸ ¨ 2 ¸ ¨¨ y3 (t ) ¸¸ © y 4 (t ) ¹

(7.144)

The matrix notation then becomes the following:

§ y1 (t ) · ¨ ¸  0*DtD ¨ yy23 ((tt )) ¸ ¨¨ ¸¸ © y 4 (t ) ¹

§0 ¨0 ¨ ¨¨ 0 © 1

0 · § y1 (t ) · § 0 ¸ ¨ y 2 (t ) ¸ ¨ ¸¨ ¸¨ 1 ¨ y3 (t ) ¸ ¸¸ ¨ ¨ 0 0 3 ¹ © y4 (t ) ¸¹ ¨©

1 0

0 1 0 0

0 · 0 ¸ ¸ 0 ¸ f (t ) ¸¹

(7.145)

322

Chapter Seven

From the above (7.145), we get the systems of equations as the following sets: * 0

DtD y1 (t )

* D 0 Dt

y2 ,

y 2 (t )

y3 ,

* D 0 Dt y3 (t )

y4 ,

* D 0 Dt

y4 (t )  y1 (t )  3 y4 (t )  f (t ) Starting from the last one, and replacing the first three, we obtain the following:

> y1 (t )@  3  0* DtD 0* DtD 0* DtD > y1 (t )@  y1 (t )

* D * D * D * D 0 Dt 0 Dt 0 Dt 0 Dt

By writing y1 (t ) as y (t ) , and with a defined sequential fractional derivative aDx that is: 4D

0Dt

y (t )  3

D 0

3D

t



y (t )  y (t )

kD

(7.146)

(7.147)

f (t )

, we attain our original system,

f (t ) ; D

1 2

(7.148)

The solution of this SFDE is similar to what we discussed in earlier sections of the book; we will also use this formulation in later sections.

7.13 Solution of the ordinary differential equation using state transition matrices -a review 7.13.1 Origin of the state transition matrix in a solution to systems of linear differential equations This is also a study of systems of linear differential equations. Let us take a general (classical) differential equation: d (7.149) > x(t )@  a(t ) x(t ) bu (t ) dt Here, we point out that u (t ) is not a Heaviside step function, but a general forcing function. We also set an initial condition as x (t0 )

x0 . By multiplying both sides of (7.149) by e ³ e³

a ( t )dt

d dt

> x (t ) @  e ³

d ª ³ a ( t )dt º e x (t ) »¼ dt «¬ and then integrate (7.150) as demonstrated below: t

a ( t )dt

a ( t )dt

, we get the following:

 a (t ) x (t )

e³

a ( t )dt

a ( t )dt

bu (t )

(7.150)

a ( t )dt e³ bu (t )

t

d

ª§ ³ a (W )dW · º ¸ x (W ) » dW ³t dW «¬¨© e ¹ ¼ 0 Using the notation ) (t )

e³

§ ³ a (W )dW · b u (W ) dW ¸  ¹ t0

³ ¨© e

(7.151)

, we write the following steps: t

t

d

³ dW >)(W ) x(W )@ dW ³ )(W )b u (W ) dW

t0

t0

t

³ )(W )b u (W ) dW

) (t ) x (t )  ) ( t0 ) x ( t0 )

(7.152)

t0

t

x (t )

1

[) (t )] ) (t0 ) x0  [) (t )]

1

³ )(W )b u (W ) dW

t0

For a constant a (t )

a , we have ) (t )

e

at

and ) (t0 )

e

at0

. As such, we have: t

x (t )

e

 at at0

e

x0  e

 at

a

W ³ e b u (W ) dW

t0

t

e

a ( t0  t )

³

x0  e

a (W t )

b  u (W ) u (W )dW

t0

t

) (t0  t ) x0  ) (W  t )b  u (W ) dW

³

t0

t

) (t , t0 ) x0  ) (t , W )b  u (W ) dW

³

t0

(7.153)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

323

7.13.2 The state transition matrix as Green’s function is a solution to the homogeneous system of differential equations The integer order system-solution and the classical multivariate system are defined as : below :: ( Dt x )(t ) A(t ) x (t )  B(t )u (t ) y (t ) C(t ) x (t )  D(t )u (t )

x (t )  \

nu1

C\

; u (t )  \

qu n

; D\

pu1

;

y (t )  \

qu1

; A\

nu n

; B\

nu p

;

(7.154)

qu q

whose entries are a continuous function of time. The operator is Dt

{

d dt

. The solution to (7.154) is as follows:

t

³

x (t )

) (t , t0 ) x0  ) (t , W )B(W )u (W )dW t0

t § · ) (t , t0 ) ¨ x0  ³ ) (t0 , W )B(W )u (W )dW ¸ ¨ ¸ t0 © ¹ t § · y (t ) C(t )) (t , t0 ) ¨ x0  ³ ) (t0 , W )B(W )u (W )dW ¸  D(t )u (t ) ¨ ¸ t0 © ¹

(7.155)

t

Where ) (t0 , t )

( Dt x )(t )

e

³t0 A(W )dW

is the state transition matrix (Green’s function) of the homogeneous system, that is,

A(t ) x (t ) :

For a linear time invariant (LTI) system ) (t0 , t ) by G (t  t0 )

) (t  t 0 )

e

A( t t0 )

) (t  t 0 )

e

A( t t0 )

G (t  t0 ) , the homogeneous solution is given

. This particular solution is given as a convolution integral (the starting point of the

convolution integral is t0 ) depicted as follows: t

x p (t )

t0

t

³ G (t  W )Bu (W )dW ³ e

(G * Bu )(t )

t0

A( t W )

Bu (W )dW

(7.156)

t0

We get a state trajectory and output trajectory as solutions to : , as depicted in the following expressions: t At

x (t )

³

e x0  e

A( t W )

t § ·  AW e ¨ x0  ³ e Bu (W )dW ¸ 0 © ¹ At

Bu (W )dW

0

t § ·  AW y (t ) Ce ¨ x0  ³ e Bu (W )dW ¸  Du (t ) 0 © ¹

(7.157)

At

For a constant a (t )

a , we have ) (t )

e

at

and ) (t0 )

e

at0

, and thus we have:

t

x (t )

e

 at at0

e

x0  e

 at

³

t aW

e bu (W )dW

e

a ( t0  t )

³

x0  e

t0

a (W t )

bu (W )dW

t0 t

³

) (t0  t ) x0  ) (W  t )bu (W )dW

(7.158)

t0

t

³

) (t , t0 ) x0  ) (t , W )bu (W )dW t0

If a (t ) is a matrix A(t )  \ using examples.

nun

, there are several ways to represent e ³

A( t )dt

, which will be elucidated in this section

324

Chapter Seven

7.13.3 A demonstration of the ways to represent the state transition matrix and its usage in multivariate dynamic systems t

³t0 a ([ )d[

Here, we define a state transition matrix as ) (t0  t ) e ) (t , t0 ) which is also a Green’s function of the homogeneous part of the system of a differential equation. One way of representing the matrix exponential, as it is for constant A , is as follows:

e³

A( t )dt

e

At

2 2

# I  At 

A t 2

3 3

A t



(7.159)

 ...

3!

The other way is via an inverse Laplace transform, which is: e

At

$

1

^( sI  A) ` 1

(7.160)

where I is the identity matrix. For example, for illustration’s sake, take the following linear time variant (LTV) system, represented as the following in its state space form:

§ x1 (t ) · § 1 e  t · § x1 (t ) · § 0 · ¨ x (t ) ¸ ¨ 0 1 ¸ ¨ x (t ) ¸  ¨ 1 ¸ u (t ) © 2 ¹ © ¹© 2 ¹ © ¹ t §1 e · §0· §0· § 1· A ¨ , B ¨ ¸ , x (0) x0 ¨ ¸ , x (1) x1 ¨ ¸ ¸ ©1¹ ©0¹ © 1¹ © 0 1 ¹ 2 t

(7.161)

t

Given an excitation such as u (t ) 5.8384e  0.3026e , we need to find x (t ) . Calculation of the state transition matrix ) for this linear time variant (LTV) system is demonstrated as follows from the two sets of coupled differential equations, for which the states are: d d t (7.162) x1 x1  e x2 x2  x2  u ( t ) dt dt (1)

Let us take the second state of (7.162) and solve the ‘homogeneous’ system, that is x2 (1)

take a general differential equation, x (t )  a (t ) x (t ) manipulation as in (7.150), we get the form as follows:

 x2 . To solve this, let us

y (t ) . Multiplying both sides by e ³

a ( t )dt

, and with a

d

In our case, that is x

(1)

x

0 , we have a (t )

1 and y

ª e ³ a (t )dt x º e ³ a (t )dt y (7.163) »¼ dt «¬ ³ a ( t )dt et . We write the above equation, and 0 , so e

integrate it from t0 to t (for our case for x2 in place of x in (7.163)), as follows:

d

t

t

ªe x2 º¼ dt ¬ t

e y t

d

T ³t dT ª¬e x2 (T )º¼ dT 0

T

³e

y (T )dT

(7.164)

t0 t

t0

t

e x2 (t )  e x2 (t0 )

T

³e

y (T )dT

t0

This gives us the following: t

x2 ( t )

t t

e e 0 x2 ( t 0 )  e

t

T

³e

y (T )dT

e

 ( t  t0 )

x2 ( t 0 )

(7.165)

t0

y (T ) 0 Putting the solution (7.165) of the second state’s homogeneous equation into the first state equation in (7.162), we obtain the following: (1)

x1

(1)

 t  ( t  t0 )

x1  e e

x1  x1

e

2 t

e

t0

x2 ( t 0 )

x2 ( t 0 )



x1  e

2 t

e

t0

x2 ( t 0 )



(7.166)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

Comparing (7.166) to x

(1)

 a (t ) x

³ 1 , e

y , we get a (t )

a ( t )dt

e

t

and y (t )

e

e

2 t

t0

325



x2 (t0 ) . Using the same

procedure as state x2 , we write the following similar expressions for state x1 :

d dt t

ª¬e  t x1 º¼ e  t y t

d

T ³t dT ª¬e x1 (T ) º¼ dT 0

³e

T

y (T )dT

(7.167)

t0 t

t

e x1 (t )  e

 t0

³e

x1 (t0 )

T

y (T )dT

t0

From (7.167), we get the following: t

x1 (t )

e

t  t0

x1 (t0 )  e

t

T ³ e e 



2T t0

e x 2 ( t 0 ) dT

t0

e

( t  t0 )

x1 (t0 ) 



1 3

e

( t  2 t0 )

 13 e

(7.168)

( t0  2 t )

x (t ) 2

0

In matrix form, the solution of a homogeneous system, like we obtained for the two state variables, is expressed as follows:

§ x1 (t ) · ¨ x (t ) ¸ © 2 ¹



§ e ( t  t0 ) ¨ ¨ 0 ©

1 3

e

( t  2 t0 )

e

 13 e

( t0  2 t )

·¸ § x (t ) · 1

0

¸ ¨© x2 (t0 ) ¸¹ ¹

 ( t  t0 )

(7.169)

§ x10 · ¸ © x20 ¹

) (t , t 0 ) ¨

Thus, from the homogeneous system’s solution we obtain the state transition matrix as:

§ et W ¨ ¨ 0 ©

) (t , W )

1 3

e

t  2W

e

The state trajectory due to control input u (t )

§ et x (t ) ¨ ¨0 ©

1 3

e

t

t

e

2 t W

·¸ ,

) (0, W )

¸ ¹

 t W

5.8384e



2 t

§ e W ¨ ¨ 0 ©

1 3

e

2W

W

e

W

e

·¸

(7.170)

¸ ¹

t

 0.3026e is as follows:



 e 2t · ª t § 13 e 2W  eW ¸ «³ ¨ W t ¸«0 ¨ e e ¹¬ © 2 t

·¸  5.8384e ¸ ¹

§ 0.3856e  0.1513  1.9966e  1.4596e ¨ t t 2 t 0.1513e  5.9897 e  5.8384e ©

2 t

2 t

t



º

 0.3026e dW »

»¼

(7.171)

· ¸ ¹

The above example elucidates in detail the calculation of the state transition matrix for an LTV system. When the system is a linear time invariant with matrix A (a constant matrix), the simpler method is invoked via an inverse Laplace transform to get to ) (t ) . The equation x ) (t )

e

At

$

1

(1)

(1)

Ax  Bu has a fundamental solution for the homogeneous system x (t )

Ax (t ) as

^( sI  A) ` ; where s is a complex frequency (the Laplace variable) and I is the ‘identity matrix’. 1

Let us elaborate upon this with the following example, where an LTI system matrix is A

§ 3 2 · ¨ 1 0 ¸. © ¹

326

Chapter Seven

­ ª s 0º ª3 2º · 1 °½ °§ ) (t ) $ ^( sI  A) ` $ ®¨ « »« »¸ ¾ °¯© ¬0 s ¼ ¬ 1 0 ¼ ¹ °¿ 1 ­ ½ 1 °§ s  3 2 · ° $ ®¨ ¾ ¸ °¯© 1 s ¹ ¿° 1 § s 2 ·º 1 ª $ « ¨ ¸» ¬ s ( s  3)  2 © 1 s  3 ¹¼ 1

1

1

(7.172)

1 2 2 · § 2  ¨ s  2  s 1 s  2 s 1 ¸ $ ¨ ¸ 1 1 1 2 ¨ ¸    © s  2 s 1 s  2 s 1¹ § 2e 2t  e  t 2e 2t  2e  t · ¨ e 2t  e  t e 2t  2e  t ¸ © ¹ 1

This method will be used in the following section to find the state transition matrix for the fractional order differential equation system as )D (t )

$

1

^( s

D

I  A)

` and 0  D  1 .

1

7.14 ‘Alpha-exponential functions’ as eigen-functions for the Riemann-Liouville (RL) and Caputo fractional derivative operators 7.14.1 The alpha-exponential functions (1 & 2) for fractional derivative operators are similar to the exponential function as an eigen-function for a classical derivative In integer order calculus, the function e de O t dt

systems, as it satisfies C D 0 Dt

ª¬ ED (O t D ) º¼

D

ED (O t ) .

Ot

plays an important role in the solution of ordinary differential equation LTI

Ot

Oe . Thus, t

D 1

We have seen that D

D

0 Dt

ª¬t D 1 ED ,D (O t D ) º¼ O t D 1 ED ,D (O t D ) and

D

ED ,D (O t ) and ED (O t ) could be the candidate functions for fractional

differential equations. We call them alpha-exponential functions 1 and 2. This alpha exponential function-1 (i.e. O (t a )

x (t )

eD

x (t )

O (t a ) eD

) satisfies

D

a Dt

x(t ) O x (t ) , with an RL derivative.

) satisfies the Caputo derivative,

C D a Dt x (t )

The alpha exponential function-2 (i.e.

O x (t ) . These are similar to an exponential function that

1

satisfies Dt x (t ) O x (t ) ; that is, they are eigen-functions for the two types of fractional derivative operators. We have seen this concept in Section-4.14 of the book. We will further develop it in this part. At

We used the notation ) as a ‘state transition matrix’ (an associated Green’s function) which is ) (t ) e for an LTI system of an integer order differential equation system. For a fractional order system, we can (similarly) define the At  (t ) e At . In this case, the notation, e At is an alpha-exponential state transition matrix as )D (t ) eD and ) D D D At function-1, and eD is an alpha-exponential function-2. These are also Green’s functions and eigenvectors for the RL and Caputo derivatives (respectively) based on homogeneous linear differential equations.

 (t ) 7.14.2 The alpha-exponential functions )D (t ) eDAt and ) D related to the Mittag-Leffler function

eDAt as

The alpha-exponential functions follow on from the higher transcendental functions of basic types of the MittagLeffler function. The two parameters of the Mittag-Leffler function are defined by the series as

ED , E ( z )

¦

f

zk k 0 * ( kD  E )

, with D ! 0 , E ! 0

and

z^

(see Appendix-A). For the matrix A  \

nun

,

the ED , E ( z ) is extended as follows: D

ED , E (At )

f

¦ k 0

A

k

t

kD

* ( kD  E )

(7.173)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

By adding E

327

D to the above (7.173), we can define the alpha-exponential function-1 as: At

eD

(t

D 1

D

) ED ,D (At )

(t

D 1

f

)

¦

A

k 0

f

k

t

t

k

kD

*  ( k  1)D

(7.174)

( k 1)D 1

¦ A *  (k  1)D k 0

At e1

1 , we have E1 (At )

For D

the initial point t0

e

At

) (t ) . This is a state transition matrix of an integer order LTI system with

0 , as described in the previous section. By adding E

1 to the two-parameter Mittag-Leffler D

D

function above, we get a single parameter Mittag-Leffler function, as ED ,1 (At )

ED (At )

¦

f k 0

A

k

t kD * ( kD 1)

.

We define the second alpha-exponential function-2 as: At eD

f

D

ED ,1 (At )

¦

A

k 0

At

t

k

kD

(7.175)

* ( kD  1)

is useful for solving the sequential fractional differential equations (SFDEs)

The alpha-exponential function-1; eD

At with the Riemann-Liouville (RL) fractional derivative, while the alpha-exponential-2; eD is a useful solution for a

SFDE with a Caputo derivative. At

Since eD

t

D 1

D

D

ED ,D (At ) and each Mittag-Leffler function ED ,D (O z ) , O ! 0 z  ^ is an entire function on the

complex plane (see Appendix-A). As such, we can have a uniquely determined function f (t ) At

eD f (t )

D

ED ,D (At ) F (t )

At

I for t z 0 and lim t p0 eD f (t ) D

ED ,D (At ) F (t )

t

1D

F (t ) , such that

I , which is depicted as: I,

k kD § f · A t ¨¦ ¸ F (t ) I © k 0 *  ( k  1)D ¹

D § 1 · t   ... ¸ F (t ) I A ¨ * (D ) * (2D ) © ¹

Therefore, we have the following limit:

D § 1 · t lim eD f (t ) lim t p0 ¨ A  ... ¸ F (t ) I t p0 * (2D ) © * (D ) ¹

(7.176)

At

F (t ) * (D )

We have F (t )

* (D ) or f (t )

condition (i.e.

lim t p0 eD f (t )

At

t

1D

(7.177)

I

 *(D ) . Therefore for this particular case

1D

 *(D ) , so that the identity 1D I ) is satisfied. In this way, we can describe the inverse function f (t ) t  * (D ) f (t )

t

At

for eD as t p 0 . We have already documented this discussion in Section-7.3.4.

7.14.3 Defining the alpha-exponential functions (1 & 2) via a kernel of convolution as the power law functions and their Laplace transforms Now let us define a convolution kernel as a power law function and its Laplace transform depicted as: def

kD (t )

D 1

t

* (D )

D !0

,

$ ^kD (t )`

1 s

D

Re > s @ ! 0

(7.178)

1

When D o 1 , then kD (t ) o u (t ) is the Heaviside unit step function; as $ ^kD (t )` o s ; and, when D p 0 , then

^

`

kD (t ) o G (t ) is the Dirac’s delta function, as $ kD (t ) o 1 . The alpha-exponential functions can then, via this kernel, be restructured as:

328

Chapter Seven

Ot

eD

(t

D 1

D 1

D

) ED ,D (O t )

t

D n

f

f

(O t )

¦ *  (1  n)D ¦ O  k n 0

k(1 n )D (t )

t

n

(1 n )D

(t )

n 0



 ( n 1)D 1

(7.179)

*  ( n  1)D

^ ` s Ot

$ eD

D

O



1

The proof of the Laplace transform as described in (7.179) is described next. The alpha-exponential function-1 above is the same as the Robotnov-Hartley function (i.e. FD (O , t ) ). The proof of the Ot

¦

Laplace transform of the alpha-exponential function-1 is the expansion of eD

f n 0

O

n

k

(1 n )D



(t ) , through

which we get the following (it should be noted that we are dropping t  , and writing t as a variable for the defined kernel in the following expansion): Ot

2

eD

kD (t )  O k 2D (t )  O k3D (t )  ... t

D 1

* (D ) Using the known identity (i.e. $ ^kD (t )`

$

^ ` Ot

$ eD

^ ` D 1

t * (D )

1 sD

Ot



2D 1

(7.180)

2 3D 1

* (2D )

O t



 ...

* (3D )

), we get the following:

^

`

2

$ ^kD (t )`  $ ^O k 2D (t )`  $ O k3D (t )  ... $

^ `  O $ ^ tD 1 * (D )

2D

t 2 D 1 * (2D )

2 3D

s

D

 Os

s

D

s

D

1  O s  O s 1  O s s

`  O  $ ^ `  ...

 .....

O s

2 2D

D

t 3D 1 * (3D )

2

 ...



(7.181)

1

D 1

D

1  O s D

1 D

s O 2

3

1

(1  x ) . For the

We have used in the above derivation (7.181) the series expansion formula 1  x  x  x  ...... alpha-exponential function-2, we define the following relationship: Ot eD

D n

f

(O t )

f

¦ *(1  nD ) ¦ O  k

D

ED (O t )

n 0

n

(1 nD ) (t )

n 0



( nD 1) 1

t

k(1 nD ) (t )

^ ` Ot

$ eD

(7.182)

* ( nD  1) s

D 1

s

D

O



1

We derive the Laplace transform shown in (7.182) as we did for an earlier case (7.181). Here we use k1 (t ) { u (t ) , with u (t ) as a Heaviside unit step function with $ ^u (t )`

1

s . The steps are shown below:

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

^ ` Ot

$ eD

^

329

`

2

$ ^k1 (t )`  $ ^O k1D (t )`  $ O k1 2D (t )  ...

 ^

tD * (1D )

$ ^u (t )`  O $  (1D )

`  O  $ ^ 2

s

1

 Os

2  (1 2D )

s

1

s

1

1  O s  O s  ... 1  O s s 1  1O s 

O s

t 2D * (1 2D )

 .....

2 2D

D

(7.183)

1

D 1

D

1 1D

s

s

s

D

`  ...

O



ss

D

s

D

O



D 1

D

s O

7.14.4 The alpha-exponential functions (1 & 2) are related via the convolution relationship The alpha-exponential function-2 is a one-parameter Mittag-Leffler function. We use the symbol ( * ) for the convolution operation, and find it to be an interesting convolution link between these two alpha-exponential functions: Ot eD

k

D

ED (O t )

(1D ) (t )

k The two alpha-exponential functions are the same for D Ot e1

*  (t ) E (t ) *  e D 1

D

D ,D

(O t )



(7.184)

Ot D

(1D )

1 , which is an exponential function:

 e *  k (t ) Ot

E1 (O t )

1

0

Ot

e1

Ot

e ;

k 0 (t )

(7.185)

G (t )

7.15 Fractional derivatives of the ‘alpha-exponential functions (1 & 2)’  (t  t ) 7.15.1 The Caputo derivative of the alpha-exponential function-2 ) D 0 A( t t ) We find the Caputo derivative of eD 0

A( t t ) eD 0

D

ED (A(t  t0 ) ) as follows using the Euler expression of the Caputo

derivative (and the RL derivative) of the power function, denoted as the following (it should be noted that the Caputo derivative of the constant function is zero, but is not for the RL derivative): C D t0  Dt

(t  t 0 )

E

­ * ( E  1) E D (t  t 0 ) ° ® * ( E  D  1) °¯ 0

E z0 E

(7.186)

0

We apply the above (7.186) to the series-expression and get: C D t0  Dt

ª¬ eDA(t t0 ) º¼

C D t0  Dt

ª¬ ED (A(t  t0 )D )º¼ f

¦

A

k 1

k

(t  t 0 )

C D t0  Dt

ª f k (t  t0 )D k º «¦ A » ¬ k 0 * ( kD  1) ¼

( k 1)D

*  ( k  1)D  1

D

AED (A(t  t0 ) )

(7.187)

A( t t0 )

AeD

Therefore, we have a useful relationship, from the calculations (i.e.

C D t0  Dt

ª¬ eDA(t t0 ) º¼ AeDA(t t0 ) ). This is similar to

an exponential function in an integer order differential equation where we have the relationship

de O t dt

Ot

O e . This

At alpha-exponential eD function-2, therefore, is useful in solving a fractional differential equation with the Caputo

derivative formulation. It follows that function) for the Caputo system, i.e.

C D 0  Dt

C D 0  Dt

[eDOt ] O eDOt (that is y (t ) eDOt ) is a fundamental solution (the eigen-

> y (t ) @

O y (t ) .

330

Chapter Seven

7.15.2 The Riemann-Liouville derivative of the alpha-exponential function-1 )D (t  t0 ) A( t t0 )

The RL derivative of eD

eDA( t t0 )

is evaluated by applying the Euler formula term by term, and using limD p0

1 * (D )

0,

for the series expression. As such, we get the following steps:

ª¬ eDA( t t0 ) º¼

D

t0  Dt

ª (t  t0 )D 1 ED ,D  A(t  t0 )D º ¬ ¼

D

t0  Dt

kD f ª º D 1 k (t  t 0 ) ¦ A * (k  1)D » t0  Dt « (t  t0 ) k 0 ¬ ¼ ( 1) 1 D k   ª f k (t  t 0 ) º D D A « » ¦ t0  t *  ( k  1)D ¼ ¬k 0 D 1 2D 1 § º· (t  t0 ) D ª (t  t 0 ) A  ...» ¸ ¨ t0  Dt «I * (2D ) ¬ * (D ) ¼¹ ©

D

I

* (D ) * (D )* (0)

(t  t 0 )

f

¦A

1

A

(t  t 0 )

k

* (2D )

kD 1

f

¦A

* ( kD )

k 1

D 1

* (2D )* (D )

(t  t 0 )

( m 1)

(t  t 0 )

 ... ( m 1)D 1

*  ( m  1)D

m 0

§ f m (t  t0 )( m 1)D 1 · A¨¦A ¸ ¨m 0 *  ( m  1)D ¸¹ © A( t t0 )

(7.188)

AeD

7.15.3 The backward Riemann-Liouville derivative of the alpha-exponential function-1 )D (T  t ) eDA(T t ) Using derivative formulas such as

D

t

DT  ª¬ (T  t )

J



º¼

(T  t )

* ( J 1) * ( J 1D )

J D

which we described in Section-4.3, we

write the following steps: t

*( E )

D E 1 DT  ª¬ (T  t ) º¼

*( E  D )

(T  t )

E D 1

(7.189)

§ E ª (T  t )D 1 º · * (D )(T  t )D 1 E lim ¨ t DT  « »¸ E oD ¬ * (D ) ¼ ¹ * ( E  D )* (D ) ©

0

We get a useful eigenvector for the Riemann-Liouville operator as depicted below: D ª A(T t ) ¼º t DT  ¬ eD

ª§ (T  t )D 1

D

t DT  Ǭ I

* (D )

©

D 1

A

(T  t )

A

* (D ) D 1

(T  t )

f

A

2

A

(T  t )

(T  t )

* (2D )

·º

 ... ¸ »

¹¼

2D 1

 ...............

* (2D ) k

2D 1

(T  t )

(7.190)

kD

¦ A *  (k  1)D k 0

A(T t )

A ª¬ eD Thus, we have the useful property

D

t0  Dt

º¼

ª¬ eDA(t t0 ) º¼ AeDA(t t0 ) , which states that a fractional differential equation At

involving an RL derivative has a solution in terms of the alpha-exponential function-1, that is, eD . This alphaAt

exponential eD

function-1, therefore, is useful in solving the fractional differential equation with the Riemann-

A Formal Method for Solving Fractional Differential and Fractional Integral Equations Ot

D

Ot

Liouville derivative formulation. It follows that D [eD ]

O[eD ] or

fundamental solution (eigen-function) for the RL system, i.e.

0  Dt

D

D

0  Dt

Ot

[eD ]

Ot O eD (that is,

331 Ot

eD ) is a

y (t )

O y (t ) .

y (t )

 (t ) 7.15.4 The relationship between the state transition matrices )D (t ) and ) D Below, we define an interesting relationship between the two alpha-exponential functions: t

t

³

A( t W )

AeD

dW

f

³¦

A

k 1

(t  W )

*  ( k  1)D

t0 k 0

t0

f

¦

A

k

k 1

A( t t0 )

eD

( k 1)D 1

(t  t 0 )

dW

kD

* ( kD  1)



D

ED A(t  t0 )

I

(7.191)

I

 (t  t ) and ) (t  W ) is: The relationship from (7.191) for the two state transition matrices ) D 0 D t A( t t0 )

eD

A( t W )

³

I  AeD

dW

t0

(7.192)

t

 (t  t ) ) D 0

I  A  )D (t  W ) dW

³

t0

At  (t ) e At as state transition matrices or as the Green’s function for homogeneous Denoting )D (t ) eD and ) D D fractional multivariate dynamics with the Riemann-Liouville and Caputo derivative formulations respectively, we have the useful expression as described above (7.192).

7.16 The general solution to a sequential fractional differential equation using ‘alpha-exponential functions’ 7.16.1 The homogeneous SFDE and its characteristic equation (or indicial polynomial) We now write the sequential fractional order differential equation which we described in Section-7.12 as a homogeneous SFDE, as follows:

D a

nD

t

aDt

 kD

¦

n 1

a (t ) aDt k 0 k

aDt

D

( aDt

kD

> x (t ) @

( k 1)D

); k

a\

0

ak (t )  C [ a , b ]

(7.193)

2, 3, .....n

For the constant coefficients ak (t ) { ak , we get the LTI system, and the above system (7.193) exhibits an indicial polynomial (or characteristic polynomial) as P (O )

n

O 

¦

n 1 k 0

k

ak O . In addition, we reiterate that Dt

D

a RL fractional derivative Dt , or a Caputo type fractional derivative

C

D

can be of

D

Dt . This is one important fact about SFDEs,

which shows that the indicial polynomials are integer order polynomials, just as we get indicial polynomials for an integer order differential equation (linear, quadratic, cubic, etc.). We assume here that x1 (t ), x2 (t ),... xn (t ) are n functions, defined in [ a , b ] , as they are seen to be linearly dependent in [ a , b ] if the constants c1 , c2 ,..., cn exist and are

not

zero

simultaneously,

such

that

c1 x1 (t )  c2 x2 (t )  ...  cn xn (t )

0,

for

a dt db.

Otherwise,

x1 (t ), x2 (t ),... xn (t ) will be defined as linearly independent in [ a , b ] .

7.16.2 A generalized Wronskian To check the linear dependence in a fractional calculus context, we use a generalized Wronskian, which is defined as follows:

332

Chapter Seven

WD (t )

WD ( x1 , x2 , ... xn ) x1 (t ) § ¨ DD > x (t ) @ 1 a t det ¨ ... ¨ ¨ D ( n 1)D > x (t )@ 1 ©a t

x2 ( t ) D

a Dt

...

> x2 ( t ) @ ...

( n 1)D a Dt

xn ( t )

· ¸ ... a Dt > xn (t )@ ¸ ... ... ¸ ( n 1)D ... a Dt ( ) x t > n @ ¸¹ D

> x2 ( t ) @

(7.194)

7.16.3 Linearly independent solutions of SFDEs Let us assume that the fractional derivatives in SFDEs are of the RL type. If x1 (t ), x2 (t ),... xn (t ) were a solution to our SFDE, then we would have: D

a Dt

> WD (t )@  (an1 )WD (t )

0

adtdb

(7.195)

The general solution of (7.195) is thus:  an1 ( t  a )

WD (t )

D 1

ceD

c (t  a )



D

ED ,D  an 1 (t  a )



(7.196)

with c as a constant. We have obtained this from the previous sections, which have explained the eigen-value of the RL fractional derivative operator. Therefore, the solutions x1 (t ), x2 (t ),... xn (t ) of the SFDE are linearly dependent in [ a , b ] , if and only if there is a t0  [ a , b ] , for which WD (t0 )

If the indicial polynomial p (O )

n

O 

¦

n 1 k 0

ak O

k

0.

has n different O1 , O2 , ...On simple roots, then the corresponding O (t a )

SFDE with the RL formulation will have x1 (t ) solution to the SFDE is given as:

eD1

O (t a )

x2 ( t )

eD 2

O (t a )

x (t )

......... xn (t )

O (t a )

c1eD1

 c2 eD 2

O (t a )

eD n

and the general

O (t a )

(7.197)

 .....  cn eD n

The c1 , c2 , ...cn are arbitrary constants in (7.197). n

If the indicial polynomial P (O ) O (t a )

eD

O (t a )

, wwO ª¬ eD

O 

¦

2

n 1 k 0

ak O

k

has repeated the roots with multiplicity l , (1  l d n ) , then

l 1

º¼ , wwO 2 ª¬ eDO ( t  a ) º¼ , ..........., wwO l 1 ª¬ eDO ( t  a ) º¼ are l linearly independent solutions of SFDE.

Let O1 , O2 ,..., Ok be different roots with the multiplicities l1 , l2 ,...., lk respectively, with l1  l2  ...  ln

n , meaning

that the general solution of the SFDE is the linear combination of the following fundamental solutions:

w

O ( t D ) eD1 , O (t a )

eD 2

wO1 w

,

wO2

...

O ( t D ) eD1 , O (t a )

eD 2

,

w

2

O ( t D ) e1 , 2 D

wO1 w

2

wO2

O (t a )

2

... w

O (t a )

eD k

O (t a )

w

...

eD 2

, ...

...

...

O (t a )

2

eD k

...

O1 ( t D )

e l 1 D wO1 1 l2 1 w

O2 ( t  a )

e l 1 D

wO2 2

(7.198)

... w

lk 1

O (t a )

e k l 1 D wOk wOk wOk k If the SFDE is formed with the Caputo derivative then we get similar solutions to the above cases with the alpha O (t a ) exponential function-2 (that is, eD

eD k

2

l 1

w1

D

ED ,1 (O (t  a ) ) ).

7.16.4 A demonstration of linearly independent solutions of the SFDE in an equation of motion with fractional order damping To illustrate this, let us take an example of the equation of motion with a fractional damping term as follows: D  or x (t )  P 0 Dt > x (t )@  x (t ) 0 P !0 D 12 D 32





(7.199)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

333

Clearly, this equation (7.199) can be cast into the SFDE as follows:

> x(t )@  P  0Dt D > x(t )@  x(t ) 4E > x(t )@  P  0Dt 3E > x(t )@  x(t ) 0Dt 0Dt

4D

The indicial polynomial reads as P (O ) case for k

1 , then P (O )

4

0

k

O  PO  1 with k

1 2

D

0

E

1 for D

1

1 2

D

3

and k

2

(7.200)

3 2

D

3

3 for D

2.

If we take the

4

O  PO  1 , which has four roots (real or complex, distinct or repeated) depending on

0

the value of P . For a specific case of value P , we have repeated the real roots (say O1

O2 ) and a pair of the

complex-conjugate (which we call O3,4 ). For this combination, we write the general solution as:

x (t )

(7.201)

c1 x1 (t )  c2 x2 (t )  c3 x3 (t )  c3 x3 (t ) 1

with c1 , c2  \ and c3  ^ as arbitrary constants; for the case of SFDE (7.200) with D

2

.

7.16.4.(a) SFDE with a Caputo derivative With the Caputo derivative D

D

C

D

{ D ;D

1

Ot

we take xi (t )

2,

Ot e 11 2

x1 (t )

e 1i ; i 2

f

O t

k 0

k 2

1

1

¦ *

O1t

1

2

1, 3 , that is:

k

 1 1

(7.202) 2 O1 t

2

*  32



3 3 O1 t 2 * 52



* (2)





4 2 O1 t

*  3

 ...

Then we have the following: x2 ( t )

x3 (t )

x4 ( t )

w

ª e 1 º wO1 ¬ 2 ¼ O1t

e 1 2 x3 (t )

f

2

*  32

O t 1

3

¦ * k 0

1

1

t

O1t

k 2

O3t

1

2

*  32

* (2)



k

 1 2



2

2O1t

1

 O3 

2

* (2)

O3t

3O1 t

1

3 2

2

*  52



4O1 t

* (3)

2

2

*  32 t

3



O3 t * (2) 3

3

 O3 t  *  52

2

3



O3 t

 ... 3

4 2

2



*  52

 O3 

O3 t

* (3)

(7.203)

 ...

4 2

t

* (3)



To find the general solution x (t ) for the SFDE of the above formulated system (7.200), it is required that we study the asymptotic behavior of

x (t ) as t p 0 . For the above systems, it holds that for practical cases (first and second (1)

derivatives of a position function) i.e. x (0)  f and x

(2)

(0)  f , depicts the magnitude of velocity and

acceleration respectively. That is a magnitude of velocity and acceleration, which should be bound at the start point of time, t 0 . Equivalently, the terms involving t that we describe below:

1

2

3

and t 2 , etc., in (7.202) and (7.203) vanish in the solution x (t ) ;

The x (t ) is therefore:

§ O t 1 2 O 2 t O 3t 3 2 O 4 t 2 ·  1  1  1  ... ¸ x (t ) c1 ¨1  1 ¨ *  32 * (2) *  5 * (3) ¸ 2 © ¹ 2 3 3 2 § t 12 · 2O t 3O t 2 4O1 t  c2 ¨  1  1   ... ¸ 3 5 © *  2 * (2) *  2 * (3) ¹ 1 2 3 32 4 2 2 § Ot · O t O t O t  c3 ¨1  3  3  3  3  ... ¸ 3 5 © *  2 * (2) *  2 * (3) ¹ 3 1 2 3 § O t 2 (O3 ) t (O3 ) t 2 (O3 ) 4 t 2  c3 ¨1  3    3 5 * (3) * * * (2)   2 2 ©

(7.204)

·

¸

¹

334

Chapter Seven (1)

For the condition x (0)  f to be bounded at the start, t (1)

x (t )  t

x

(2)

1

will blow up at

2

0 , we need

1

t

in the above x (t ) to vanish, or else

2

t p 0 . This means we have a condition c1O1  c2  c3O3  c3O3

(0)  f to be bounded, we need t

3

(2)

terms in x (t ) to vanish, or otherwise x

2

3

2

3

3

t p 0 . That gives us the condition c1O1  3c2 O1  c3O3  c3O3

(t )  t

 12

0 . Similarly, for

terms will blow up at

0 ; both are summarized below:

c1O1  c2  c3O3  c3O3 3 c1O1

2  3c2 O1



3 c3O3



0

3 c3O3

(7.205)

0

With such a complex number c3 , the arbitrary real constants c1 , c2 in the general solution x (t ) of the SFDE equation (1)

x0 ; x (0)

are determined by an initial condition of x (0)

v0 and x

(2)

a0 .

(0)

7.16.4.(b) SFDE with a Riemann-Liouville (RL) derivative For case of SFDE with D

1 2

and with the RL derivative, D O1t

x1 (t )

e 1 2 w

x2 ( t )

t

1

1

¦ *

1

e11

O3t

e 1 2

x4 ( t )

O t

f

1 2

k 0

Ot

wO1

x3 (t )

 1 2 1

x3 (t )

t

 12

¦ * 2

*  12



3

1 2

k

t

2



* (2)

k

t

* (1)

4O1 t

(O3 ) t

1

*  23

3

 12



2

O1



* (1)

O1 t

1

e1 i , i 2

3

O1 t

2

*  23

1, 3 . Then:



* (2)

 ...

2

 ...

*  52

*  12 2



3



Ot

, we take xi (t )



*  12

3O1 t

( k  1)

O3

1 2

 12

2

2

*  32 1

k 0

t

1

O t

f

1

2

D

{ D ;D

( k  1)

2O1t



* (1)

2

D

2

O3 * (1)



O3 t

1

3

O3 t

2

*  23



* ( 2)

 ...

(7.206)

3

2



(O3 ) t * (2)

 ...

Thus, x (t ) is: 2 1 3 § t  12 · O1 O1 t 2 O1 t x (t ) c1 ¨     ... ¸ 3 1 © *  2 *(1) *  2 * (1) ¹

§ 1 2O1t 12 3O12t 4O13t 3 2 ·  c2 ¨     ... ¸ 3 5 © * (1) *  2 * (2) *  2 ¹  12 2 12 3 § t · O O t O t  c3 ¨  3  3  3  ... ¸ 3 1 © *  2 *(1) *  2 * (2) ¹ 1  12 2 3 § t · (O ) t 2 (O3 ) t O  c3 ¨  3  3   ... ¸ 3 1 * (2) © *  2 * (1) *  2 ¹

(7.207)

To find the general solution x (t ) for the SFDE of the system in (7.200), we must study the asymptotic behavior of x (t ) as t p 0 . In order that

initial point of time ( t as t

1

2

and t

1

2

(1)

x (0)  f and x (0)  f , those are magnitudes of the position and the velocity at the

0 ) to be bounded. Therefore, two terms appear in x (t ) (7.207), which has multiplying terms,

are (c1  c3  c3 )

  1

t *

2

1 2



2

2

2

and c1O1  2c2 O1  c3O3  c3O3



1

t 2 *  32



, respectively, should be zero. The

reasoning is similar to that which we discussed above for the Caputo case. Therefore, we have: c1  c3  c3 0 2

2

2

c1O1  2c2 O1  c3O3  c3O3

(7.208)

0

With such a complex number as c3 , the arbitrary real constants c1 , c2 in the general solution x (t ) of the above system SFDE equation (7.200) are determined by the initial condition x (0)

(1)

x0 and x (0)

v0 .

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

335

7.17 The solution of a multivariate system of a fractional order differential equation with the RL Caputo derivative using a state transition matrix of the ‘alpha-exponential functions -1 & 2’ 7.17.1 The multivariate system with an RL derivative and its solution with the state transition matrix We take a fractional differential equation with an RL fractional derivative as follows: D

0 D d1

( D Y)(t )

AY(t )  B(t )

Y0

Y(t0 )

(7.209)

We write the general solution of (7.209) as follows: t A( t t0 )

Y(t )

eD

A( t W )

³

 B(W ) dW

Y0  eD

(7.210)

t0

) D (t  t 0 ) 

³

t

t0

)D (t  W )  B(W ) dW

Where the fundamental solution or the Green’s function (RL-Case) is as follows: GD (t  W )

) D (t  W )

A( t W )

(7.211)

eD

We write the particular integral as the following convolution (with a start point of integration as t0 ): t t

Yp (t )

³ GD (t  W )  B(W ) dW

(GD * 0 B)(t )

(7.212)

t0

7.17.2 A Multivariate system with the Caputo derivative and its solution with state transition matrices For a fractional differential equation with the Caputo derivative for 0  D d 1 , the homogeneous one is:

 While Y(t0 )

C D t0  Dt Y

(t )

AY(t )

A( t W )



b ¨ I  eD

(7.213)

b has a general solution as follows: t

Y(t )

³

b

eD

t0



The RL-Caputo relationship for 0  D d 1 is

C D t0 Dt



D

f (t )

t0

Dt

t

§

Ab dW

³

¨ ©

A( t W )

·

AdW ¸

¸ ¹

t0

(7.214)

> f (t )  f (t0 )@ (Section-3.10). Using this, we get

the following:



C D t0 Dt Y

(t )

By putting Y(t ) Z(t )  b in (7.215), we get Z(t0 ) based on a fractional differential equation is:

0



t0

Dt

>Y(t )  b@

AY(t )

(7.215)

Z0 . Thus, the equivalent equation in the RL derivative



D

t0

D

AY(t )

A  Z(t )  b

Dt Z (t )

AZ(t )  Ab

(7.216)

The solution is as follows:

Z(t )

A( t t0 )

eD

Z0 

t

t

A( t W ) ³ (eD Ab)dW

³ (eD

t0

A( t W )

Y(t )  b

(7.217)

§ t A( t W ) · b ¨ I  ³ eD AdW ¸ ¨ t ¸ © 0 ¹

(7.218)

Ab )dW

t0

We have thus the below result: t

Y(t )

b

A( t W )

³ (eD

Ab )dW

t0

Similarly for a system as described, it is as follows:

 Setting Y(t0 )

b as a constant, we write the solution below:

C D t0 Dt Y

(t )

AY(t )  B(t )

(7.219)

336

Chapter Seven t

³

Y(t )

A( t W )

b  eD

 B(W )  Ab dW

t0

§ t A( t W ) · t A( t W ) b ¨ I  ³ eD AdW ¸  ³ eD B(W )dW ¨ t ¸ t © 0 ¹ 0

which is also:

 (t )  b ) D

Y(t )

t

³  )D (t  W )  B(W ) dW t0

(7.220)

(7.221)

At  (t ) e At . The solution of SFDE with the Caputo’s where state transition matrices are )D (t ) eD and ) D D formulation requires thus two Green’s functions (or state transition matrices).

C D 0  Dt

> x (t ) @

u (t )

0  D  1; x (0)

a  \ ; x (T )

b\; T ! 0

(7.222)

In terms of a system matrix equation, we have: C

in this case, A

D

D

> x (t ) @

(7.223)

0 ; B 1 . Here, u (t ) is a general input function and not a Heaviside step, and: At

) D (t )

eD

t

D 1

t t

 (t ) 0 , we have ) D

I

t

AW

³ eD 0

AdW

I

k kD § f · A t ¨¦ ¸ © k 0 *  ( k  1)D ¹

D 1

D § 1 · t   ... ¸ A ¨ * (D ) * (2D ) © ¹

(7.224)

D 1

;A

* (D ) For A

Ax (t )  Bu (t )

0

1 . Therefore, the state trajectory of the system for the solution is: T

T

 (t )  x (0)) D

x (t ) 0

³  )D (t  W ) u (W ) dW 0

a

1

(7.225)

T 1

D (T  t )  u (t ) dt ³ * (D ) 0

7.17.3 Formalizing the multivariate problem to get a state trajectory as a solution with the defined state transition matrices Now we formalize what we conducted in the above discussions, and use the results, with alpha-exponential functions. Consider a linear time invariant system as denoted by ¦ of the fractional commensurate order D , where 0  D d 1 . C D 0  Dt

¦:

> x (t ) @

Ax (t )  Bu (t )

y (t )

Cx ( t )

(7.226)

Considering what a commensurate order means, for each component the same fractional order of D is used. For a

> @

function x : 0, T o \

nu1

:

C D 0  Dt

Where, x (t )  \

nu1

u (t )  \

used here. The control is observation) is y (t )  \

pu1

mu1

matrix A  \

u (t )  \ .

mu1

> x (t ) @

nun

§ x1 (t ) · ¨ . ¸ C D D ¸ 0 t ¨ ¨¨ . ¸¸ © xn ( t ) ¹

,B\

num

,C\

pu n

§ 0C DtD x1 (t ) · ¨ ¸ . ¨ ¸ . ¨ ¸ ¨ C DD x (t ) ¸ © 0 t n ¹

(7.227)

. The Caputo fractional derivative

(also called forcing input); the state is x (t )  \

nu1

C

D

Dt is

, while the output (or

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

0 and evaluated at t t 0 is the initial value problem of

The forward trajectory of the system ¦ above starting at t0 C D 0  Dt

a fractional differential equation

> x (t ) @

337

Ax (t )  Bu (t ) , with x (0)

a , where a  \

nu1

The solution x (t ) of (7.226) is expressed as a state trajectory as follows: t § t · x (t ) ¨ I  ³ )D (W )AdW ¸ a  ³  )D (t  W ) B  u (W ) dW © 0 ¹ 0

(7.228)

At

eD . Another way to write this expression is as below:

In the above expression (7.228), we have )D (t ) t

 (t )  ) (t  W )B  u (W ) dW a ) D D

³

x (t )

0

t

D

 (t ) ) D

ED (At )

³

I  )D (W )AdW

(7.229)

0

§ t · t y (t ) Cx (t ) Ca ¨ I  ³ )D (W )AdW ¸  C ³ )D (t  W )B  u (W ) dW © 0 ¹ 0 7.17.4 An application to obtain the state trajectory solution for a multivariate fractional differential equation system with a given initial condition and a forcing function Let system ¦ 1 be defined as: 1

­° C D02 > x1 (t )@ x2 (t ) ® C 12 ¯° D0 ª¬ x2 (t ) º¼ u (t )

¦1 : T

1 0

Take the initial point a

A

, and the final point as origin b

) D (t )

$

1

^

s

1

2

>I@  >A@

§0 1· ¨0 0¸ © ¹

­°§ ª s $ ®¨ « °¯© ¬ 0

`

1

T

0 0

1

B

§0· ¨1¸ © ¹

. 1 º ª0 1º · ½° ¸ ¾ » s ¼ «¬0 0»¼ ¹ ° ¿

0

­§ s 1 ·1 ½ ° ° $ ®¨ ¸ ¾ s¹ ° °¯© 0 ¿ 1

1

­1 § s

$ ® ¨

§ 1s ·½ 1 ¨ $ ¸¾ ¨ 0 s ¹¿ ©

1

¯s © 0

Using $

1

^s ` D

tD 1 * (D )

;

$

1

^s `

) D (t )

 12

$ ^1`

1

St

§ 1S t ¨ ¨ 0 ©

(7.231)

1 s 1 s

· ¸ ¸ ¹

, t ! 0 , we write:

1 ·

¸ ¸ St ¹

1

t

 (t ) ) D

1 s

(7.230)

³

I  )D (W )AdW 0

1 t § 1 0 · § SW ¨ ¨0 1¸  ³ ¨ © ¹ 0© 0

·§0 1· ¸ dW 1 ¸ ¨© 0 0 ¸¹ SW ¹ 1

(7.232)

§ 1 2 St · ¨ ¸ ©0 1 ¹ Using the formulas as obtained above in (7.228) and (7.229), we get the following state trajectory: 1 · 1 t § §1 2 t · t W ) S ( ¨ ¸B  u (W ) dW S x (t ) ¨ a 1 ¨ 0 1 ¸¸ ¨ ¸ 0 © ¹ 0 ¨© S (t W ) ¸¹

³

(7.233)

338

Chapter Seven

T

1 0

Taking u (t ) { 1 , then for a given a

T

 0 1

; B

(7.230), we get x (t )



(1  t )



2 St

T



as derived

in the following steps:

1 t § §1 2 t · S t W ) ( S ¸a  ¨ x (t ) ¨ ³ ¨0 1 ¸ ¨ 0 0¨ © ¹ ©

§1 ¨¨ ©0

· ¸B  u (W ) dW 1 ¸ S (t W ) ¸¹ 1 · 1 t § 2 t· 1 S (t W ) § · ¨ ¸ § 0 · dW S ¸¨ ¸  ³ 1 ¸¸ ©¨ 1 ¹¸ 0 1 ¸¹ © 0 ¹ 0 ¨¨ S  W t ( ) © ¹ §1· t § 1 · §1· § t · ¨ 0 ¸  ³0 ¨¨ 1 ¸¸dW ¨ 0 ¸  ¨ 2 t ¸ © ¹ © ¹ © S¹ © S (t W ) ¹ § (1  t ) · ¨2 t ¸ © S ¹ 1

(7.234)

7.17.5 An application to obtain the state trajectory solution for a multivariate fractional differential equation with matrix A as a skew-symmetric system with a given initial condition and forcing function 3

For 0  D  1 , a system in \ , that is, ¦ 2 , is described as: ¦2:

A

The system matrix A is skew symmetric; hence A for k 3, 7,11, ... .

0

k

I

I for k

­° C D0D > x1 (t )@ x2 (t ) ®C D °¯ D0 > x2 (t )@  x1 (t )  u (t ) § 0 1· 2 A ¨ A I ¸ © 1 0 ¹ 0, 2, 4, 8, .. , A

k

A for k

(7.235)

1, 5, 9, ... and A

k

A

We write (7.235) as follows: C

D0  > x1 (t )@ D

C

D0  > x2 (t )@

C

D0  ¨

C

D0  > x (t )@

D

D

x2 ( t )

 x1 (t )  u (t )

§ x1 (t ) · § 0 1 · § x1 (t ) · § 0 · ¸¨ ¸ ¨ ¸  ¨ ¸ u (t ) © x2 (t ) ¹ © 1 0 ¹ © x2 (t ) ¹ © 1 ¹

D

A

§ 0 1· §0· ¨ 1 0 ¸ , B ¨ 1 ¸ © ¹ © ¹

(7.236)

A > x (t )@  Bu (t )

We have for )D (t ) the following series:

) D (t )

t

D 1

D 2D 3D § 0 1 · t 2 t 3 t A  A + A  A  ... ¸ ¨ * (2D ) * (3D ) * (4D ) © * (D ) ¹

(7.237)

Now with the property of this skew symmetric matrix A , we get the following:

D 2D 3D § 1 · t t t ) D (t ) t ¨ I A I A  ... ¸ * (2D ) * (3D ) * (4D ) © * (D ) ¹ D 1 3D 1 2D 1 4D 1 §t · § t · t t I¨   .. ¸  A ¨   ... ¸ © * (D ) * (3D ) ¹ © * (2D ) * (4D ) ¹

(7.238)

§ t 2D 1 t 4D 1 · ¨ * (2D )  * (4D )  ... ¸ ; © ¹

(7.239)

D 1

Use the following notation, to simplify the above expression (7.237):

sD (t )

cD (t )

§ t D 1 t 3D 1 · ¨ * (D )  * (3D )  ... ¸ © ¹

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

339

we get the state transition matrix )D (t ) in this case as follows:

) D (t )

§ cD (t ) sD (t ) · ¨  s (t ) c (t ) ¸ © D ¹ D

I  cD (t )  A  sD (t )

(7.240)

t

I  )D (W ) AdW

 (t ) ) D

³ 0

§ 1 0 · t § cD (W ) sD (W ) · § 0 1 · ¨ 0 1 ¸  ³0 ¨  s (W ) c (W ) ¸ ¨ 1 0 ¸ dW © ¹ ¹ © D ¹© D (7.241)

§ 1 0 · t § sD (W ) cD (W ) · ¨ 0 1 ¸  ³0 ¨ s (W ) c (W ) ¸dW © ¹ © D ¹ D t § 1  t s (W )dW c (W )dW · D ³ ³ 0 0 D ¨ ¸ t ¨ t ¸ ¨ ³ sD (W )dW 1  ³ cD (W )dW ¸ 0 © 0 ¹

Using (7.240) and (7.241), we use the formula x (t )

³ )D (t  W )B u (W ) dW 0

in (7.228) and (7.229) to

T

 x1 (t )

attain the state x (t )

t

 (t )  a ) D

x2 (t ) with the trajectory for any given a acting as the initial point, and the forcing

input u (t ) .

7.18 The solution to a fractional differential equation of type V  b  *DD V

E 0H  E1

 D H * D

with RL or Caputo formulations Let us take the following system of visco-elasticity, as we described and analyzed in Chapter-6, that is following

V (t )  b



*



D

D V (t )

E 0H (t )  E1

 D H (t ) *

D

(7.242)

0  D d 1, b t 0 , E 0 t 0 , E1 ! 0 , b d (E1 / E 0 )

7.18.1 Using an RL-derivative formulation for a solution of H with a known V input We will obtain a solution for H (t ) , for a known input function V (t ) , and then a solution for V (t ) , for a known H (t ) . The solution for a known input function V (t ) is:

 With z (t )

 V (t )  H (t ) , O b E1

*

D

E0 E1

, X (t )

AV (t ) and A

*



D

D z (t )  O z (t )

bE 0 1 E12

(7.243)

X (t )

. The general solution of the above (7.243) with

D

the RL derivative D { a Dt ,  and with V (t ) is continuous and integrable in > a , t @ , and is:

z (t )

 O (t a )

CeD



O (t a )

 eD

 X (t ) a

t

 O (t a )

CeD



³  eD

 O ( t [ )

a

 X ([ ) d[

(7.244)

t  O (t a )

CeD

³

 O ( t [ )

 A eD

V ([ ) d[

a

where C is an arbitrary real constant. Therefore, for the general solution to the above model in (7.242) and (7.243) (with a change of variables from the original one), we have:

H (t )

b E1



E0

V (t )  CeD E1

(t a )

t



 ³ bE 0 1 E12

a



E0

eD E1

( t [ )

V ([ ) d[

(7.245)

340

A t

Chapter Seven b E1

a , we have H ( a )

)

we obtain C

lim t p0 f (t ))D (t )

1

D

(0)



 Ot

V ( a )  C )D (0) , where )D (t )

b E1

eD

t

D 1

D

ED ,D ( O t ) and )D (0)

f . From here,



V ( a )  H ( a ) . However, as per our earlier argument (in Section-7.3 and 7.14),

1 gives us an inverse function for )D (t ) at origin, that is, f (t ) 1

Therefore, we have lim t p0 )D (t )

t

1D

 *(D ) .

 F (t ) with

* (D ) .

F (t )  Ot

eD , we write

a:



1

lim )D (t  a )

t oa

1D

Using this concept of an inverse function of )D (t )

in the limiting case the following value of C at t

C

t





b E1

V (t )  H (t )

1D

lim (t  a )

t oa



 *(D )  Eb V (t )  H (t )

(7.246)

1

 *(D )  Eb limt oa (t  a )1D V (t )  limt oa H (t ) 1

1D

Thus, for the case noted where the initial conditions are specified as lim t o a  (t  a ) 1D

lim t o a  (t  a )

H (t )

k1 , and

k 2 , the solution is:

 b E1

H (t )



C Moreover, if C

V (t )

0 means that z ( a )

b E1



E0

V (t )  CeD E1

t

(t a )



 ³ bE 0 1 E12



E0

eD E1

( t [ )

V ([ ) d[

(7.247)

a

 V (a) , in particular H (a)

k1  k 2 * (D ) b E1

0 , i.e. H ( a )

 V (t )   b E1

H (t )

bE 0 1 E12

t

³

0 , then we write:

V (a) 

E0

eD E1

( t [ )

V ([ ) d[

(7.248)

a

7.18.2 Using the Caputo derivative formulation for a solution of H with a known V input In addition, we can write the general solution of the equation *

D

C



*

D



X (t ) , with the Caputo derivative

D z (t )  O z (t )

D

D z { a Dt ,  z , from the derived cases as in the previous sections, giving: t  O ( t [ )

³

z (t )

k  eD

 AV ([ )  O k d[

z (a)

k

(7.249)

a

The corresponding solution to the following system with a Caputo derivative:

V (t )  b



C

D



D V (t )

E 0H (t )  E1



C

D



D H (t )

E 0  D d 1, b t 0 , E 0 t 0 , E1 ! 0 , b d E1

(7.250)

0

is:

H (t )

k

b E1

t

³

 O ( t [ )

V (t )  eD a

§ bE 0  1 · ¨ 2 V ([ )  O k ¸ d[ © E1 ¹

(7.251)

7.18.3 Using the RL-derivative formulation for a solution of V with a known H input The solution for a known function H (t ) is:



*

D



D y (t )  J y (t ) y (t )

E1 b H (t )  V (t ) ; J

If H (t ) is continuous and integrable in *

D

D

D { a Dt ,  is:

g (t )

> a, t @

1 b ; g (t )

BH (t ) ; B

E1 E02 b2

(7.252)

then the general solution of the above with an RL derivative

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

341

t J ( t  a )

y (t )

 J ( t [ )

³

CeD

H ([ ) d[

 B eD

(7.253)

a

Therefore, the general solution to equation (7.242) with all RL derivatives is:





D



V (t )  b D V (t )



D

E 0H (t )  E1 D H (t )

(7.254)

E 0  D d 1, b t 0 , E 0 t 0 , E1 ! 0 , b d E1

0

and then the following:

 E1 b

V (t )

 1 (t a )

H (t )  CeD b 1D

 E1 b

C For C

0 at t

a , i.e. when V ( a )



E1 b

H (a)

V (t )

b2

 b1 ( t [ )

D

H ([ ) d[

(7.255)

a

While the initial conditions are given as lim t o a  (t  a )

V (t )

t

 ³e E1  E 02



1D

k1 and lim t o a  (t  a )

V (t )  1 (t a )

H (t )  CeD b



t

 ³e E1  E 02 b

2

 b1 ( t [ )

D

H (t )

k 2 , the solution is:

H ([ )d[ (7.256)

a



k 2  k1 * ( a ) 0 , we have:

 H (t )   E1 b

E1  E 02 b

2

t

³e

 b1 ( t [ )

D

H ([ ) d[

(7.257)

a

7.18.4 Using the Caputo derivative formulation for a solution of V with a known H input For the case of using the Caputo derivative for



*



D

C

D

D

D y { a Dt ,  y , we write the

g (t ) with

D y (t )  J y (t )

solution as: t  J ( t [ )

³

y (t )

k  eD

 BH ([ )  J k d[

y(a)

k

(7.258)

a

The corresponding solution to the equation with the Caputo formulation below:

V (t )  b



C

D



D V (t )

E 0H (t )  E1



C

D



D H (t ) E

(7.259)

0  D d 1, b t 0 , E 0 t 0 , E1 ! 0 , b d E1

0

is: t

V (t )

k

 H (t )  ³ e E1 b

 J ( t [ )

D

a

 H [ J [ E1  E 02 b2

( ) k d

(7.260)

7.19 A generalization of a fractional differential equation with a sequential fractional derivative Case-I: We take the following fractional differential equation:

 D z (t )  O z (t ) D

X (t )

(7.261)

EV (t )  H (t )

(7.262)

We generalize this as the following SFDE:

D

nD



z (t ) 

with parameters D , K , E , and O j  \ ( j

D

nD



z (t ) 

with parameters D , E , J and K j  \ , ( j

n 1

¦ O D

jD

j

j 0



z (t )

K V (t )

z (t )

1, 2, 3...n  1) , or as the following SFDE: n 1

¦K D j

j 0

jD



z (t )

EH (t )

z (t )

JH (t )  V (t )

(7.263)

1, 2, 3...n  1) . The model shown in (7.262) in (7.263) with a sequential

fractional derivative can be expressed as follows in a multivariate fractional differential equation formulation:

342

Chapter Seven

 *

*



D

D Z (t )

(7.264)

AZ(t )  B(t )

D

Where D is an RL derivative or Caputo derivative, and with the following representations for A , B(t ) and Z(t ) in matrix form: 1 0 . . . 0 0 · § 0

¨ 0 0 1 ¨ . . ¨ ... ... . . ¨ A ¨ ... . . ¨ ... . . ¨ 0 0 ¨ 0 ¨ O O O © 0 1 2



a Dt

( j 1)D



z (t ) , j

1, 2, ...n or

0 . .

. . . . . .

. .

¸ ¸ ¸ ¸ . ¸ ¸ . ¸ 1 ¸ On 1 ¸¹ 0 . .

. . . 0 . . .  On  2

(7.265)

§ z1 (t ) · ¨ ... ¸ ¨ ¸ ¨ ... ¸ ¨ ... ¸ Z(t ) ¨ ... ¸ ¨ ... ¸ ¸ ¨ ¨ ... ¸ ¨ z (t ) ¸ © n ¹

§ 0 · ¨ 0 ¸ ¨ ¸ ¨ 0 ¸ ¨ 0 ¸ B(t ) K ¨ ... ¸ ¨ 0 ¸ ¨ ¸ ¨ 0 ¸ ¨ V (t ) ¸ © ¹ In (7.265), we have z j (t )

. . . . . . . . .



z j (t )

C ( j 1)D z a Dt

(t ) , j

(1, 2, ..., n ) . The general

solution is the following: t A( t  a )

Z(t )

CeD

³

A( t [ )

 eD

 B([ ) d[

(7.266)

a

where z1 (t )

z (t )

EV (t )  H (t ) , E  \ , and C is an arbitrary constant vector.

Case-II: Take the following multivariate system of a fractional differential equation:



D

a Dt ,  Z

(t )

(7.267)

AZ(t )  B(t )

where:

A

a

§ V 1 (t ) · ¨ ... ¸ ; B(t ) ¨ ¸ ; Z (t ) 1,...n ¨¨ ... ¸¸ © V n (t ) ¹

z j (t )



ij i , j

a Dt

( j 1)D



z (t ) ;

j

§ z1 (t ) · ¨ ... ¸ ¨ ¸ ¨¨ ... ¸¸ © z 2 (t ) ¹

(7.268)

(1, 2,..., n  1)

The solution is: t

Z(t )

A( t  a )

CeD

 ³ eDA( t [ ) B([ )d[

(7.269)

a

Case-III: Take the following multivariate system of a fractional differential equation:



C D a Dt ,  Z

(t )

AZ(t )  B(t )

(7.270)

with the same parameters as used in Case-II. The solution is: t

Z(t )

³

A( t [ )

Z( a )  eD

 B([ )  AZ(a ) d[

(7.271)

a

In particular, if Z( a )

b in (7.271), then we have: t

Z(t )

³

A( t [ )

b  eD a

 B([ )  Ab d[

(7.272)

A Formal Method for Solving Fractional Differential and Fractional Integral Equations

343

Here we saw a generalization method that could be used to provide a response to a fractional order model comprising the fractional differential operators of the Caputo or RL type, using an SFDE.

7.20 Short summary This chapter offered insights into how fractional differential equations and fractional integral equations are formalized. It showed the rigor of solving fractional differential and integral equations, and their conjugation with classical calculus. The topic is perhaps as rigorous as its counterpart in classical calculus. We saw how the solution matters if the formulation of a fractional differential equation is carried out via a Riemann-Liouville fractional derivative or by a Caputo derivative. The use of the Laplace transform technique we have used vigorously is also shown in this chapter, as is the concept of obtaining linearly independent solutions. We may ask ourselves which type of fractional derivative (the RL or the Caputo) is the one that best follows the fractional dynamics of a physical system. This is a good topic for further investigation.

7.21 References This chapter is based on the following pioneering works: Miller K S and Ross B, “An Introduction to Fractional Calculus and Fractional Differential Equations” (1993) Abramowitz. M, Stegun. I.A. (eds.), “Handbook of Mathematical Functions” (1964); Batman H, “Higher transcendental function” (1954) DiethelmKuo and Ford, “Analysis of fractional differential equations” (2003) Francois Dubois and Stephanie Mengue, “Mixed Collocation for fractional differential equations” (2005); Gel’fand I.M, Shilov.G E, “Generalized functions” (1964) Glockle W G and Nonnenmacher. T F, “Fractional Integral Operators and Fox Functions in the Theory of Viscoelasticity” (1991) Gorenflo R, Mainardi F, “Fractional Calculus: Integral & Differential Equations of Fractional Order, CISM Lecture Notes” (2008); “Fractional oscillators and Mittag-Leffler function” (1996); “Fractional calculus: integral and differential equations of fractional order” (1997) Igor Podlubny, “Fractional differential equations” (1999); “The Laplace Transform Method for Linear Differential Equations of the Fractional Order” (1994) Kilbas A .A, Srivastava H .M ,& Trujillo J. J, “Theory & Application of Fractional Differential Equations” (2006) Matignon, D, “Generalized Fractional Differential and Difference Equation: Stability Properties and Modeling Issues (1998) Oberhettinger.F and Badii L, “Tables of Laplace transforms” (1973); Oldham and Spanier, “Fractional calculus” (1974) Shih-Tong, Srivastava, H.M, et al., “A certain family of fractional differential equations” (2000); Srivastava H M, “On extension of Mittag-Leffler function” (1968) Robotnov Y N, “Elements of Hereditary Solid Mechanics” (1980) The above works are furnished with further detail and listed in alphabetical order in the Bibliography section.

CHAPTER EIGHT FRACTIONAL PARTIAL DIFFERENTIAL EQUATION

8.1 Introduction This chapter is a continuation of the application of the Laplace and Fourier Transform tricks used to solve the fractional partial differential equation. Here we have taken the classical partial differential equation of wave propagation and diffusion, and then have developed their fractional counterparts. We have evolved different sets of functions that are similar to what was discussed in earlier chapters like Mittag-Leffler, Miller-Ross, and the RobotnovHartley functions in which we used them. Similarly, we generalised the Wright function and called it an M-Wright function with several auxiliary functions of this type. This M- stands for Prof. Mairandi; this is developed by him and Prof. M. Caputo. We will be using their development and elaborating upon all the derivations, step-wise. We see that these new sets of functions are a generalisation of the Gaussian function that appears in the solution of the classical partial differential equations. We have taken the ‘time fractional diffusion wave equation’ (TFDWE) as developed in this chapter, similar to the integer order counterparts that are studied in details in classical calculus books.

8.2 Time fractional diffusion wave equation (TFDWE): Cauchy and the signalling problem The standard partial differential equation (PDE) governing the known phenomena of diffusion and the wave propagation are governed through the Fourier Diffusion Equation (popularly called Fick’s law) and D’Alembert’s wave equation:

w u ( x ,t ) wt



w 2 u ( x ,t ) wx

2

and

w 2 u ( x ,t ) wt

2

c

2 w 2 u ( x ,t ) w x2

respectively, where u ( x, t ) was denoting a response

variable, with  as a diffusivity constant and c as a characteristic velocity. These above equations are studied in various books, as examples of the standard PDE. We denote the time fractional diffusion wave equation (TFDWE) as follows: w

E

w t

E

u ( x, t )

a

w

2

w x

2

u ( x, t )

The fractional order E is the order of the fractional derivative of the Caputo type w.r.t. time. We define these derivatives for 0  E d 1 and 1  E d 2 as follows:

(8.1)

0E d2 C E 0 Dt

, i.e. a fractional derivative

­ 1 t ª wu ( x, W ) º dW 1 E 1 0  E 1  ° 0 It 0 Dt >u ( x , t )@ E E ³ « » w u ° * (1  E ) 0 ¬ wW ¼ (t  W ) ® E wt wu ° E 1 °¯ wt ­ 1 t ª w 2u ( x, W ) º dW 2 E 2 1 E  2  ° 0 It 0 Dt >u ( x , t )@ « E 1 2 » E ³ w u ° * (2  E ) 0 ¬ w W ¼ (t  W ) ® E 2 wt wu ° E 2 2 °¯ wt

(8.2)

(8.3)

For 1  E  2 , we integrate the Caputo fractional derivative, by order E and get the following: E

0 It



C E 0 Dt 2 0 It

(1)



ut ( x , 0 )

u ( x, t ) >u ( x, t )@ u ( x, t )  u ( x, 0 )  t u

>u ( x, t )@



2 0 Dt

E

0 It



2 E 0 It



2 0 Dt



w u ( x, t ) w t

t 0

For 0  E  1 , we integrate the Caputo fractional derivative, by order E and get the following:

(1)  t ( x, 0 )



(8.4)

Fractional Partial Differential Equation



E

0 It

C E 0 Dt u ( x , t )



345



1 E 0 It

1

1 t

E

0 It

 D u ( x, t )  D >u ( x, t )@ u ( x, t )  u ( x, 0

0 It

1 t

(8.5)



)

Integrating by order E the LHS and RHS of TFDWE (8.1) and by using the above (8.5) expression of fractional integration for the Caputo fractional derivative, for 0  E  1 , we write this by using a definition of fractional integration to RHS of (8.1), given by the following fractional integral equation: E

ª wE º u ( x, t ) E » ¬w t ¼

0 I t «a

0  E 1

t

§ w 2u · § 1 ¨ 2 ¸¨ 1 E ³ * ( E ) 0 © w x ¹ © (t  W ) a



u ( x, t )  u ( x, 0 )

u ( x, t )

ª w 2u º 2» ¬ w x ¼

E

0 It «

t

§ w 2u · § 1 ¨ 2 ¸¨ 1 E ³ * ( E ) 0 © w x ¹ © (t  W ) a



u ( x, 0 ) 

· ¸ dW ¹

(8.6)

· ¸ dW ¹

Integrating by order E the LHS and RHS of TFDWE (8.1) and by using the above (8.4) expression of fractional integration of the Caputo fractional derivative, for 1  E  2 , we write by applying a definition for the fractional integration to RHS of (8.1), the following fractional integral equation: E

ª wE º u ( x, t ) E » ¬w t ¼

E

0 It «









u ( x, t )  u ( x, 0 )  t

u ( x, t )

ª w 2u º 2» ¬ w x ¼

0 I t «a

(1)  ut ( x , 0 )

(1)



1 E  2 t



§ w 2u · § 1 ¨ 2 ¸¨ 1 E ³ * ( E ) 0 © w x ¹ © (t  W )



§ w 2u · § 1 ¨ 2 ¸¨ 1 E ³ * ( E ) 0 © w x ¹ © (t  W )

a

t

a

u ( x , 0 )  t ut ( x , 0 ) 

· ¸ dW ¹

(8.7)

· ¸ dW ¹

Now we set E 2X , to write TFDWE (8.1) as the following (8.8) with the constant a replaced by  as a diffusivity constant or being proportional to the wave propagation constant (velocity): 2X

w u w t 2

2X

2



w u w x

0 X d1

2

(8.8)

!0

2X

The dimension of  is L / T { cm 2 /sec2X , and the u ( x, t ) is a ‘field’ variable which is assumed to be a causal function of time, i.e. u ( x, t ) 0 for t  0 . The above (8.8) is a one-dimensional time fractional diffusion wave 1 2

equation (TFDWE). For X

, the classical integer order diffusion equation is recovered, that is

X 1 , the classical integer order classical wave equation is recovered i.e.

2

w u w t2



2

w u w x2

w u w t



w 2u w x2

. For

. With the fractional order X ,

as in the above (8.8), the meaning of  is different-i.e. having fractional units. The ‘Cauchy problem’ is the Initial Value Problem (IVP) where the initial values are given as: 

u ( x, 0 )

g ( x)

for

 f  x  f;

u ( B f, t ) 0 for t!0 The ‘Signalling problem’ is a Boundary Value Problem (BVP) where the boundary conditions are given as: 

u ( x, 0 )

0

for

x ! 0;



u (0 , t )

h (t )

for

(8.9)

t!0

(8.10) u ( f, t ) 0 for t!0 However, to ensure the continuous dependence of our solution on the ‘fractional parameter’ X , (8.8) where 0  X  1 is also on the transition for X

 12



to X

 12



(1)



, we assume ut ( x, 0 )

wu w t t 0

0 . The reason is that for

X  12 , ut(1) ( x, 0  ) a condition is not required, and that is zero, and that it is continued for X ! 12 , in order to have continuity vis-à-vis X at

1 2

.

346

Chapter Eight

8.3 Green’s function **

The two Green’s functions are represented by the symbol G* ( x, t ) for the Cauchy and signalling problem. The first is Gc ( x, t ) for the Cauchy problem with g ( x ) h (t )

G ( x ) and the second is Gs ( x, t ) for the signalling problem with

G (t ) . The solutions to the above (8.8) problems in terms of the respective Green’s function is therefore: f

³ G ([ , t )  g ( x  [ ) d[

u ( x, t )

c

f

(8.11)

t

u ( x, t )

³ G ( x,W )  h(t  W ) dW s

0

The above (8.11) expressions are a convolution of the Green’s function with forcing functions. We also point out that

Gc ( x, t )





Gc x , t , since the Green’s function turns out to be even a function of x .

8.3.1 Green’s function for the diffusion equation with auxiliary functions as defined by a similarity variable 1 2

The Green’s Function for a Diffusion Equation we now discuss. For X

we get the integer order diffusion

equation and its Green’s function is as follows, (given with the corresponding Fourier transform and Laplace transform respectively): 2 1 x tk 2 d d Gc ( x, t ) e 4t  Gc ( x, t ) e 2 S t (8.12) 2 x  x s  4xt d d $ Gs ( x , t ) e Gs ( x , t ) e 3 2 S t We define the similarity variable as z

x



d

d

d

t G

³

f

0

d

M ( z )dz

^

`

d s

( x, t )



z

d

F ( z)

2

d

(8.13)

M ( z)

1 e  z 2 /4 . We call M d ( z ) the auxiliary function for the diffusion equation. S

It provides the fundamental solution satisfying the condition as

It is easy to see that

`

for x ! 0 , and we get the following expression:

t

x Gc ( x, t )

We define M ( z ) in (8.13) as M ( z )

^

1 . By putting

known definite integral (refer Chapter-1) i.e.

³

f

f

 2z e

 y2

³

y , dz

dy

0

d

M ( z )dz

1.

(2dy ) , we have the following derivation, by using a

S:

f

³

f

f d

M ( z )dz

0

³ 0

1

S

e



1 2 S

 dz z2

1

4

f

³e

y

2 S 2

(2dy )

f

³

e



 dz z2

4

f

(8.14)

1

f

In the above (8.14) the superscript d denotes the diffusion process.

8.3.2 Green’s function for the wave equation with auxiliary functions as defined by the similarity variable Green’s function for the Wave Equation we describe here. For X mentioned above. That is

w 2u w t2

2 2 c w u2 w x

making c

1 in (8.8), we obtain an ordinary wave equation as

 , then for the Cauchy problem with g ( x )

the following along with the Fourier transform expression:

G ( x ) , we obtain

Fractional Partial Differential Equation

1

w

Gc ( x, t )

^

2



1





G xt   G xt  1

`

w

347

 Gc ( x, t )

2

e

2

 itk 



1 2

e

f  x  f

A signalling problem with h (t ) transform expression:

 itk 

c



w

G t x/ 



^

`

w

$ Gs ( x , t )

e



(8.15)



G (t ) , for x ! 0 we obtain the following with c

Gs ( x , t )



 along with the Laplace

 s x

x!0



(8.16)

From the above (8.16) explicit relationship we recognise the reciprocal relationship between the two Green functions (wave and signalling) for x ! 0 and t ! 0 :



w

2 x Gc ( x, t ) w

M ( z)

t G



w

x

zM ( z ) (8.17)

x t

z t

w

F ( z)

G (1  z )

1 , the similarity variable z

However, in the case where X

w s ( x, t )

is not defined, but for a general case of 0  X  1 ,

we can have:

z

x t

x 2X

t

X



; 0 X 1

(8.18)

Then for the generalised case of TFDWE for x ! 0 a certain auxiliary function in M X ( z ) may exist such that the following condition holds: t z x Gc ,X ( x, t ) Gs ,X ( x, t ) MX ( z ) ; 0  X  1 2X 2 (8.19) f



³ MX ( z )dz



1

0

This auxiliary function i.e. M X ( z ) in (8.19) we will use in further study in this chapter:

8.4 Solution of TFDWE via the Laplace transform for the Green’s function for Cauchy and signalling problems We will get G c ( x, s ) and G s ( x, s ) for 0  X  1 by solving an ordinary differential equation of the second order in x and then by an inverse Laplace transformation we will get Gc ( x, t ) and Gs ( x, t ) .

8.4.1 The Cauchy problem for a TFDWE solution in the Laplace domain The Cauchy problem exists with a condition at the initial time i.e. t u ( x, t )

$

^

w u w t 2X

Gc ( x, t ) . We have

transformed C D 0 Dx

2X

equations,

`

f ( x)

D

by

s F (s)  s

D 1



using

2

w u w x2

a

and we use u ( x, s ) Laplace

transform

G c ( x, s )

of

the

0 we have g ( x )

u ( x, 0) then

$ ^u ( x, t )` . The following is the Laplace Caputo

derivative

(Section-5.16)

i.e.

f (0) :  2X 2X 1 s u ( x, s )  s u ( x, 0 )

2X

s u ( x, s )  s For g ( x )

G ( x)

2X 1

g ( x)





2 d u ( x, s )

dx

2

2 d u ( x, s )

dx

(8.20)

2

G ( x ) we write: 2X 2X 1 G ( x) s G c ( x, s )  s



2 d G c ( x, s )

dx

2

(8.21)

348

Chapter Eight

There is a singular term G ( x ) at x 0 in (8.21). We therefore need to consider a solution to the above (8.21) for x ! 0 and x  0 while imposing a boundary condition at x rf , and matching the condition of continuity at

x

r

0 . Look at the following, which is obtained via a rearrangement of the above (8.21) expression i.e. 2 2X 2X 1 d G c ( x, s ) § s ·  s   G ( x) G ( x , s ) ¨ ¸ c 2  dx ©  ¹

(8.22)

The indicial polynomial for the above (8.22) second order differential equation with roots is the following: 2

p 

s

2X

X

0



,

p

r

s

(8.23)



We expect the solution as follows: X

X

s s ­ x x     c2 ( s ) e  , x ! 0 ° c1 ( s )e G c ( x, s ) ® sX sX x x   °   e c ( s ) e  c ( s ) , x0 4 ¯3

Since at x

rf , G c ( x, s )

0 , we have c2 ( s )

(8.24)

0 , ensuring the solution vanishes at x n f . The following

c1 ( s )

is a solution to the Green’s function in the Laplace ( s -frequency) domain: X

s ­ x  ° c1 ( s )e  , x ! 0 G c ( x, s ) ® sX x  °  ( ) , x0 c s e ¯4

Continuity 

, G c (0 , s )

at c1 ( s )

x

0



and



G c (0 , s )

x

0



states

c4 ( s ) , making c1 ( s )

dG c ( x, s ) dx

x 0

From the above (8.26) we write the following, as c1 ( s )

dG c ( x, s ) dx



2X

2X 1

G ( x) 

dx 2

0

³

0 , is as follows:

§ sX · ¸ © ¹

dG c ( x, s ) dx

³

2X 1

G ( x)

0 



(8.27)

z0

0 we get the following: 0

(dx )s

X

s



0 to x 0

(dx )s G c ( x, s ) 

0

(8.26)

  c1 ( s )  c4 ( s ) ¨

x 0



, from x

2X

is

c4 ( s ) :

x 0

d 2G c ( x , s )

That

X

2  c1 ( s ) Integrating s G c ( x, s )  s

0.

r

c4 ( s ) . The first derivative at x

  c1 ( s ) ¨

x 0

  G c (0 , s )  G c (0 , s )

that:

§ s ·  ( sX /  ) x § sX · e c s  ( )  1 ¨ ¸ ¸ © ¹ © ¹ § sX ·  ( sX /  ) x § sX ·   c4 ( s ) ¨  e c ( s )  4 ¸ ¨ ¸ © ¹ © ¹

dG c ( x, s ) dx

(8.25)

³

0

§ d 2G c ( x, s ) · ¸ 2 dx © ¹

(dx ) ¨ 



0 §0 · 2X 1  ( x /  ) sX  ( x /  ) sX s ¨ ³  c1 ( s ) e dx  ³  c4 ( s ) e dx ¸  s ¨  ¸ 0 ©0 ¹ § dG ( x, s ) · dG c ( x, s ) ¨ c  ¸ dx x 0 ¹ © dx x 0 2X

(8.28)

The first term for the LHS of the above (8.28) expression is zero since the function G c ( x, s ) is continuous at x 

0,



i.e. the infinitesimal area under this from 0 to 0 equals zero. The second term in (8.28) is the integral of the delta 2X 1

function which is unity, thus we get the s (a constant function w.r.t. the variable x ). The RHS of (8.28) is expressing a discontinuity in the first derivative of G c ( x, s ) w.r.t. x . Therefore, we have the following:

Fractional Partial Differential Equation

s



2X 1

dG c ( x, s )

dG c ( x, s )



dx



349

x 0

c4 ( s ) , the following:

Substituting the value of RHS as obtained above (8.27) in (8.29) we get, with c1 ( s )



s

2X 1

dG c ( x, s ) dx



(8.29)

dx

x 0

dG c ( x, s )



dx

x 0

x 0

X

s

  c1 ( s )  c4 ( s )

2  c1 ( s )



X

s



(8.30)

X 1

s

c1 ( s ) We have c1 ( s )

sX 1 2 

c4 ( s )

1 2 s1X 

2 

. We therefore write the following: X

s ­ 1 x  ° 1X e  , x ! 0 ° 2s  G c ( x, s ) ® sX  x ° 1  e , x0 ° 2 s1X  ¯

1 1X

2s

e





sX x 

, f  x  f

(8.31)

8.4.2 Signalling problem for TFDWE in Laplace domain For a signalling problem we apply the Laplace Transform to

w 2X u ( x , t )



w t 2X

w 2 u ( x ,t ) w x2

, with 0  X  1 , and  ! 0 , to get

the following expression:  2X 2X 1 s u ( x, s )  s u ( x, 0 ) 

w x



0 , for x ! 0 and

For a signalling problem, the condition is u ( x, 0 )

2 w u ( x, s )



u (0 , t )

(8.32)

2

G (t ) with u ( f, t )

h (t )

0 for

t ! 0 . We therefore write the solution of the above (8.32) problem as Green’s function G s ( x, s ) , and get the following:  2

The corresponding indicial polynomial is p 

s 2X 

2 d G s ( x, s )

Green’s function in the Laplace domain as G s ( x, s )

0 , we have

 u (0 , s )

c1 ( s )e



 s

r

sX 

X

x



(8.33)

0

 c2 ( s ) e



. The solution is:

 sX x



(8.34)

0 , to ensure that the solution vanishes as x n f . Therefore for x ! 0 we have the

Clearly, we must put c2 ( s )

x

2X  s G s ( x, s )

2

0 and gives roots such as p G s ( x, s )



dx



G s (0 , s )

X

x

. At an initial position that is near the origin



c1 ( s ) . Also, we have the condition

$ ^G (t )` 1 G s (0 , s ) 

 s  c1 ( s ) e 

c1 ( s ) . This gives a value of c1 ( s )

Laplace domain for a signalling problem for x ! 0 is G s ( x, s )

e



 s



u (0 , t )

G (t ) , therefore we get

1 . Therefore the Green’s function in the

X

x



.

8.4.3 The solution of Cauchy and the signalling problems of TFDWE in the time domain by inverting the Laplace domains solution by using auxiliary functions FX and M X The transformed solutions for Cauchy and signalling are the following: X  x /  sX 1  ( x , s ) e  x /  s (8.35) ; G c ( x, s ) e G s 1X 2 s The above (8.35) expressions are transformed and the solution must be inverted to provide for the Green’s function in space-time’s domain. For this we require some transformed pair of Wright type function that we write as follows for 0  X  1:

350

Chapter Eight

1

X

FX

1

 1 tX



1 t

X

MX

^ F  ` $ ^ $ ^ F  ` $ ^

 1 tX

X

1

$



1 tX

X

X

1 tX

MX

 ` 1 tX

e

 sX

1X

s

 `

(8.36) X

s

X 1 1 FX 1X M X 1X M X 1X e t X tX X 1 t t tX 1 t t t We note that FX  x used here is not the Robotnov-Hartley function, which is represented as FX ( a , x ) referred to as Appendix-A. The above (8.36) formulas can be used to invert the transformed Green’s function. Then introducing the x similarity variable for x ! 0 and t ! 0 , that is z X ! 0 then by using the rules of scale change of the Laplace t

transform pair i.e. for $ ^ f (t )`

f ( s ) , we have $

^ f (at )`

 as and $ ^ f  `

f

1 a

t a

1 a

f ( as ) where a ! 0 , and

after some manipulations we obtain the Green’s function in variable x and t , with space-time as the following: 1 1 Gc ( x, t ) FX ( z ) MX ( z ) X 2X x 2 t

Xx

1

Gs ( x , t )

FX ( z ) t We get the reciprocal relation of the Green’s function as described below: 2X x  Gc ( x, t ) t  Gs ( x, t )

t

1X

(8.37)

MX ( z )

X z  MX ( z )

FX ( z )

(8.38)

Where the FX ( z ) and M X ( z ) are auxiliary functions for a general case 0  X d 1 . This generalisation those for standard diffusions given for X

1 2

by the following:



d

x Gc ( x, t )

t G 1

d

M ( z) and for a standard wave equation for X



w

 ;

t G

w

M ( z) In fact for X



z2

4

1

d

F ( z)

2



d

z M ( z)

(8.39)

x !0 1 t 2

z

1 given as: 2 x Gc ( x, t )

1 2

( x, t )

e

S



d s

w s ( x, t )



w

F ( z) x

G (1  z ) ; z

t



w

z M ( z)

(8.40)

!0

1 we recover the expressions of the auxiliary functions of classical cases (8.13), (8.17) as

and X

follows: d

M ( z) w

M ( z)

1

S

e



 ; z2

4

G (1  z ) ; z

z x t

x !0 1 t 2

for

!0

X

for

X

1 2

(8.41)

1

8.5 Reciprocal relation between Green’s function of Cauchy and signalling problem We got the transformed Green’s function (in the Laplace domain) for the Cauchy problem for f  x  f as 1

G c ( x, s )

2 s1X 

G s ( x, s )

e



e







 x /  sX

 x /  sX

.

For a signalling problem the transformed Green’s function for

x ! 0 is

. Taking the derivative w.r.t. s for these expressions we get the following:

dG s ( x, s ) ds

§ x ·  x  sX ¨ ¸e © ¹  x sX · 1 § 2X x ¨ 1X e  ¸  © 2s ¹

X 1

Xs

2X x  G c ( x, s )

(8.42)

Fractional Partial Differential Equation

351

Therefore the relationship of the two expressions in the Laplace (frequency) domain is: 1 dG s ( x, s ) x  G c ( x, s )  2X ds With the inverse Laplace transform applied to the above (8.43) we obtain the following relationship:

§ t · ¨ ¸ Gs ( x , t ) © 2X ¹

x  Gc ( x, t )

^

1

^ ` df ( s ) ds

( t ) f (t ) , where $

1

^ f ( s)`

`

n

$ t f (t )

We have used in (8.43) the Laplace transform relationship i.e.

$

(8.43)

(8.44)

( 1)

n



d n f ( s ) ds n



that gives

f (t ) , and finally we obtain (8.44).

8.6 The origin of the auxiliary function from the Green’s function of the Cauchy problem 8.6.1 An inverse Laplace transform of the Green’s function: Gc ( x, t ) in terms of an auxiliary function M X ( z ) with z as a similarity variable We have the transformed Green’s function for the Cauchy problem (8.35) and we then modify it as follows: 1

G c ( x, s )

e

1X



 s , x

X

x

x G c ( x, s )



1X

2s 2s   Apply the inverse Laplace transform to the above (8.45) and write the following steps: 1 x G ( x, t ) $ x G ( x, s )

^

c

1

­



 s x

X



(8.45)

`

c

x

$ ®

e

1X

e

½ ¾ ¿



 x /  sX

 ¯ 2s X x ª 1 ds º st  x /  s e e « 1X » ³ s ¼ 2  ¬ 2S i Br x ª 1 st  x /  sX ds º e « 1X » ³ s ¼ 2  ¬ 2S i Br

(8.46)

The Br Bromwich path on which the Laplace inversion integration is done is a line on the complex plane from s J  if to s J  if , we have used this in Chapter-6 (Appendix-G). Substitute in the above (8.46) V st and ds

(dV ) t

to get the following steps:

x Gc ( x, t )

ª 1 V   x /  V X / tX 1 § dV e « ³ 1X 1X ¨ 2  « 2S i Br V / t © t ¬ x

§ ª V ¨¨ « 1 e ©( ³ 2  « 2S i Br ¬

x

· X x ¸V  ) tX ¸¹

· X x ¸V  ) tX ¸¹

§ ª V ¨¨ 1 ( « e © X ³ i 2 S 2(  )t « Br ¬

x

Using a similarity variable i.e. z

x

  tX

§ 1 · § dV ¨ X ¸ ¨ 1X © t ¹© V § dV ¨ 1X ©V

º ·» ¸ ¹» ¼

·º ¸» ¹ »¼ (8.47)

º ·» ¸ ¹» ¼

and substituting in the above (8.47) expression we get the following

formula:

§z· ¨ ¸ MX ( z ) ©2¹

x Gc ( x, t )

(8.48)

With the function M X ( z ) defined in the following way: MX ( z )

1

2S i ³

Br

V  zV X

e

§ 1 ¨ 1X ©V

· ¸ dV ¹

(8.49)

352

Chapter Eight

for z ! 0 , and 0  X  1 . Therefore M X ( z ) is an auxiliary function that has the following integral representation: 1

V  zV X

e 2S i ³

dV

(8.50) , z ! 0, 0  X  1 1X V Br The representation Br is the Bromwich path on which the Laplace inversion integration is done is a line on the complex plane ( ^ ) from V J  if to V J  if , (Appendix-G). This above (8.50) integral representation of MX ( z )

M X ( z ) (Bromwich representation) can be analytically continued for all z in complex-plane i.e. ^ , by adopting a

suitable integral and a ‘series-representations’ valid in all of ^ i.e. the entire complex-plane For this purpose, let us deform the Bromwich path Br into the Hankel path Ha . Refer to Appendix-G for detailing inverse Laplace transforms.

8.6.2 The description of the Hankel contour from the Bromwich path integral used for an integral representation of the auxiliary function M X ( z ) The Hankel contour that begins at V f  ia , a ! 0 , encircles the branch cut that lies along the negative real axis, and ends up at V f  ib , b ! 0 with a p 0 and b p 0 which is equivalent to the original path of Bromwich. This is shown in Figure-8.1, the complex plane V : Bromwich - Path A o B

Hankel - Path F o E o D o C

B

C

D

C

D

F

E

F

E

A

³

F (V )dV

Bromwich

F (V )dV

³

Hankel

Figure-8.1: Showing the Bromwich path and the Hankel path

The inverse Laplace transform is carried out as an integral on the Bromwich path which is A to B as indicated in Figure-8.1. Notionally we represent it as the following ( F (V ) is inclusive of the reciprocal of 2S i and the

³

exponential kernel in an inverse Laplace transform formula) i.e. f (t )

Ao B

F (V )dV . We make a closed contour

across the branch cut line (i.e. the negative real axis) and call the closed contour : as A, B, C , D, E , F , A in an anticlockwise direction. Assuming the contour made does not include any poles, meaning the Residue is zero, and then we have the following:

³

:

³

³

F (V )dV

Ao B o C o D o E o F o A

F (V )dV 

Ao B

³

F (V )dV 

B oC

³

DoE

F (V )dV 

³

F (V )dV

0

(8.51)

F (V )dV 

C oD

³

EoF

F (V )dV 

³

F (V )dV

0

FoA

As the radius of the arcs BC and FA grows to infinity, the function F (V ) p 0 , for a well behaved function to have an inverse Laplace transform, this condition is satisfied (as per Jordan Lemma Appendix-G). Therefore, the integrals on these arcs vanish. Therefore, we are left with the following paths on which we do the integration:

Fractional Partial Differential Equation

³

F (V )dV 

Ao B

³

³

³

F (V )dV 

C oD

³

F (V )dV 

DoE

³

F (V )dV

Bromwich

353

F (V )dV

0

EoF

F (V )dV

Ao B

§

¨

³

³

F (V )dV 

© C oD

Do E

³

F (V )dV 

³

F (V )dV 

D oC

³

F (V )dV 

³

F (V )dV 

(8.52)

F (V )dV ¸

¹

EoF

EoD

F oE

·

³

F (V )dV 

³

F (V )dV

³

F (V )dV

F oE

EoD

³

F (V )dV

Hankel

D oC

In addition, if the contour ABCDEFA encloses poles, we write:

³

Bromwich

³

F (V )dV

F (V )dV  2S i

Hankel

¦ Re sidues of poles

(8.53)

Recall that we have applied this technique to find the inverse Laplace transform in an earlier chapter (Section-6.6), but here we elaborated upon the Bromwich and Hankel paths. We have described Hankel’s formula with a contour defined by Hankel for an integral representation of the Gamma function, and the reciprocal Gamma function, in Chapter-1 (Section 1.10.6, 1.10.11), that is: 1 1 i (0  )  z t z 1  t (8.54) *( z ) (  t ) e dt (  t ) e dt f C 2i sin(S z ) * ( z ) 2S As defined during the derivation (Section 1.10.6 and 1.10.11) of the above (8.54) formulas the integral is on the contour that is a path starting at f just above the positive real axis, encircling 0 in a positive (anti-clockwise) sense, and returning to f just below the positive real axis, respecting the branch cut along the positive real axis. This was the original contour chosen by Hankel for the derivation of the above (8.54) formula, which is the opposite of what is described in Figure-8.1. We take the formula of the reciprocal Gamma function, and do the substitution t W in order to write the following steps: 1 i (0  )  z t (  t ) e dt f * ( z ) 2S

³

³

³

i

2S ³

(0  )

f

1

2S i ³

(0  )

f

Here we have achieved the contour integration as

³

(0  )

f

z W

W e

i

 dW

2S ³

(0  )

f

z W

W e dW

(8.55)

z W

W e dW

meaning this contour starts at f , just below on the

negative real axis encircling 0 in a positive sense (anti-clockwise) and returning to f just above the negative real axis, respecting the branch cut along the negative real axis. This is as per the Hankel path of Figure-8.1. Thus 1 *( z )

1 2S i

³

Ha

z W

W e dW is a Hankel representation of the reciprocal Gamma function.

We will now demonstrate by use of the inverse Laplace transform formula for the reciprocal of the Gamma function as the Hankel representation obtained (8.55) above. We take the basic definition of the Gamma function in an integral representation, i.e. * ( z ) gives the following steps:

³

f

0

t

z 1  t

e dt (Section-1.10). Substituting t

*( z )

³

f

0

t

s  du into the above formula

su i.e. dt

z 1  t

e dt

³

f

0

s

z

( su )

³

f

0

u

z 1  su

e

z 1  su

e

 s du

du

(8.56)

354

Chapter Eight

Define

*( z )

F (s)

and

sz

F ( s ) $ ^ f (u )`

³

f

0

f (u ) e

use  su

the

definition

of

the

du and use the following to write, with f (u )

^ ` ³

$ u We

f (u )

write

$

1

the

^F ( s)`

inverse

1 2S i

³

C

Laplace

Laplace

z 1

f

0

transform

u

z 1  su

e

Transform u

z 1

(Appendix-G)

:

*( z )

du

(8.57)

F (s) z s (Appendix-G)

equation

i.e.

knowing

that

su

F ( s )e ds , as the following expression: u

z 1

$

1

1

^F ( s )`

su

2S i ³

C

(8.58)

F ( s ) e ds

The path C is any deformed Bromwich contour such that C winds around the negative real-axis in an anti-clockwise sense as depicted in Figure-8.1, i.e. the Hankel path: 1 z 1 1 su $ ^F ( s )` u F ( s ) e ds C 2S i (8.59) *( z ) 1 § * ( z ) · su -z su s e ds ¨ z ¸ e ds 2S i C © s ¹ 2S i C

³

³

Put su

W in (8.59), we get ds

³

 u1 dW , and rearranging the above (8.59) we get the following: u

z 1

*( z )

1

2S i ³

C

-z su

s e ds ;

su

W

z

z 1 § W · W § dW · u z W W e dW ¨ ¸ e ¨ ¸ ³ ³ C C 2S i © u ¹ © u ¹ 2S i

1

(8.60)

From the above (8.60) we get the expression: 1

1

W 2S i ³

*( z )

z W

C

(8.61)

e dW

z

has a branch cut on a negative real axis ( f, 0) (Appendix-E) and is analytic everywhere. Hence the contour C can be chosen for an Inverse Laplace transform as with the Hankel path ( Ha ) and as per the discussion above. With this description, we can represent the formula (8.50) i.e. The function W

MX ( z )

1 2S i

³

Br

V  zV X

e

 dV , 1 V 1X

z ! 0 , 0  X  1 with the Hankel path integral as: MX ( z )

1

2S i ³

Ha

V  zV X

e

§ 1 · ¨ 1X ¸ dV ; ©V ¹

0 X 1

(8.62)

8.7 The integral and series representation of auxiliary functions 8.7.1 Getting a series representation of the auxiliary function M X ( z ) from its integral representation Therefore, this equivalence of both the Bromwich path and Hankel path redefines the auxiliary function on the Hankel path as M X ( z )

1 2S i

³

Ha

V  zV X

e

 dV , 0  X  1 (8.62). This representation is an integral representation of the 1 V 1X

auxiliary function. This auxiliary function as originated earlier is also called M-Wright’s function, as it is very similar to Wright’s function (M- for Mainardi). We study this auxiliary function M X ( z ) , and get its series representation as described in the following steps:

Fractional Partial Differential Equation

1

dV

V  zV X

e 2S i ³

MX ( z )

V

Ha

1 2S i

1X



V

³

355

e

Ha

e

dV

V

1X

 zV X

f

¦

n

X n

( 1) ( zV ) n!

n 0

n n

f

e

use

· dV ¸ 1X © n 0 n! ¹V Ha n n f º ( 1) z ª 1 V X n X 1 e V dV » « ¦ ³ n ! ¬ 2S i Ha n 0 ¼ 1

V

2S i ³

1

Use

*( z )

:

§

 zV X

e ¨

1

¦

( 1) z

x z

2S i

³e x

V

Xn

(8.63)

dx

Ha

n n

f

( 1) z

¦ n !*  X n  (1  X )

MX ( z )

n 0

We have used, in the above (8.63) expansion, the Hankel representation of the reciprocal Gamma function i.e. 1 *( z )

1 2S i

³

Ha

x z

e x dx

(8.61). Therefore, we have a series representation of this auxiliary function M X ( z ) as

mentioned below: n

f

( 1) z

n

¦ n !*  X n  (1  X )

MX ( z )

0 X 1

(8.64)

n 0

Using the reflection formula

1 *( z )

1

S

* (1  z )  sin S z , (Section-1.10.9) we write the auxiliary function in the

following form: n n

f

MX ( z )

( 1) z

¦ n !*  X n  (1  X ) n 0

1

S

f

¦

( 1)

n

 * X (n  1) sin SX (n  1) zn !

n

n 0

1

S

f

¦

( z )

n

n!

n 0

 * X (n  1) sin SX (n  1)

f

1

(8.65)

( z )

n 1

¦ (n  1)!  *(X n) sin(SX n)

S

n 1

8.7.2 The relationship of Wright’s function to M-Wright’s function or the auxiliary function M X ( z ) The Wright’s function (Appendix-A) is defined as follows: def f

W ( z; O , P )

z

n

¦ n ! * (O n  P )

WO , P ( z )

n 0

1 2S i

V  zV  O

³e

dV

V

Ha

P

(8.66)

; O ! 1, P ! 0

Therefore the auxiliary function in terms of the Wright function is: MX ( z ) W (  z ; X ,1  X ) WX ,1X (  z ) For X

1 2

0 X 1

(8.67)

, we get an integer order diffusion problem, and earlier we wrote, for this case, the following function: M 1 ( z) 2

d

M ( z)

1

S

e



 z2

4

(8.68)

356

Chapter Eight



and also wrote the reciprocity relationship as x Gcd ( x, t ) that

³

f

d

M ( z )dz

0

t G

d s

( x, t )



z 2

d

M ( z ) . In addition, we showed

1.

³

8.7.3 Proof of the property of the auxiliary function i.e. Here we try to show that we get

³

f

0

M X ( z )dz

f

M X ( z )dz 1

0

1 via the following derivation: f

³

f

MX ( z )dz

0

1

§

dV

V V X z

³ 2S i ³ ¨© e

V

Ha

0

1

V

³

2S i

e

Ha

§ f V X z · dV ¨ ³ e dz ¸ 1X ©0 ¹V X

§ V z 1 V e e ³ ¨¨ V X 2S i Ha ©

³

2S i

Ha

e

V

 V1 dV ³f

iS

 r ; dV

V

 V1 dV ³0

³

V

re

³

e

F oE

D oC

0

V

e

e

r

 1r dr .

 iS



e

r

V

1X

eV

V

. The first leg is section FE 

the

second

dr and with r varying from 0 f

dV

dr . The r varies from f to 0 so the integration on FE is

 r , dV

Similarly

0

(8.69)

dV

Now we need to evaluate the integral on the Hankel path for the function F (V ) re

z f

· ¸¸ ¹z

V

1

(Figure-8.1). Here we write V

· ¸ dz ¹

1X



leg

of

the

Hankel

path

is

DC

with

to f , yields the integration on path DC (Figure-8.1) as

 1r dr . Summing these two above, the integrations on the legs of the Hankel path FE and

DC gives zero.

iT

We are left with a small circle on the Hankel path as ED , where we write V  e , with  a small constant such iT

that p 0 and T varying from S to S . With this we have dV  ie dT , and the integration on this small circle is: T S

V

³

ED

e

V

dV

lim

p 0

³ T S 

exp( iT )

e

e

iT

 ie dT

iT

S

lim

p 0

exp( iT )

³S e

idT



Therefore, the total integration on the Hankel path gives the following result: f 1 V M X ( z )dz e 0 2S i Ha

³

³

(8.70)

S

i

³S dT

2S i



 V1 dV

1

(8.71)

8.7.4 Moments of the M-Wright function or the auxiliary function M X ( z ) A very important deduction is as follows that is the absolute moment of this M-Wright function that is derived for an 

D order moment with z in \ , for M X ( z ) with D ! 1 and 0 d X  1 :

Fractional Partial Differential Equation f

³

f

D

0

0

f

Use

³e

§ 1 V  zV X dV · e dz ¨ 1X ¸ ³ V © 2S i Ha ¹ f · dV 1  zV X D V § e ¨ ³ e z dz ¸ 1X ³ 2S i Ha © 0 ¹V

³z

z M X ( z )dz

D

* (D  1)

 zV X D

z dz

X D 1

(V )

0

f

³z

D

M X ( z )dz

Use

³z

D

³ V XD

2S i

V

* (XD  1)

2S i

0

dV

e

1

M X ( z )dz

1

Ha

1

f

(8.72)

V

e

* (D  1)

0

357

³ V XD

1

dV

Ha

* (D  1)

D ! 1 ; 0 d X  1

* (XD  1)

Therefore, we have a generalised moment expression for the auxiliary function as follows: f

³x

D

M X ( x )dx

0

* (D  1)



x  \ ; D ! 1, 0 d X  1 (8.73)

* (XD  1)

8.8 The auxiliary functions M X ( z ) and FX ( z ) as a fractional generalisation of the Gaussian function 8.8.1 The series representation of the Wright function derived from its integral representation formula The Wright function we denote by WO , P ( z ) , the function is defined by the series representation (Appendix-A) and is convergent in the entire z  ^ complex plane:

WO , P ( z )

¦

z

f

n

(8.74) n ! * (O n  P ) With O ! 1 and P  ^ . The integral representation of the Wright’s function reads, with O ! 1 and P  ^ as: WO , P ( z )

n 0

1

2S i ³

V  zV  O

Ha

e

§ 1 ¨ P ©V

· ¸dV ¹

(8.75)

The origin of the above (8.75) can be related to as being similar to that which was obtained in the previous section for M X ( z ) . Using the Hankel representation of the reciprocal of the Gamma function that is

1 *( z )

z  ^ , (8.61) we obtain the following series representation: WO , P ( z )

1

V

Ha

2S i

³

V

§

e ¨

f

z

n

0

n z § 1

On

· dV ¸ P ¹V

· V OnP dV ¸ e V ¨ ¦ ³ n 0 n ! © 2S i Ha ¹ f

f

z

n

¦ n ! * (O n  P ) n 0

For O

0 , we get W0, P ( z )

ez *(P )

provided that P z 0, 1, 2, ... .

u z

e u du with

P

¦ n! V

©n

Ha

Ha

dV

V  zV  O

e 2S i ³ 1

³

(8.76)

358

Chapter Eight

The auxiliary functions we introduced as in the previous sections are for 0  X  1 are M X ( z ) FX (X )

WX ,1X (  z ) and

X z  M X ( z ) . As a matter of fact FX ( z )

WX ,0 (  z ) . These two auxiliary functions are related by FX ( z )

and M X ( z ) are particular cases of the Wright function WO , P ( z ) by setting O

X for both P

0 and P

1

respectively.

8.8.2 A further series representation of auxiliary functions FX ( z ) and M X ( z ) Using the identities of the Gamma function z* ( z )

* ( z  1) (Section-1.10) we get the series expression of FX ( z ) , by

using the obtained series expression of M X ( z ) (8.64) as follows: m m

f

FX ( z )

X zM X ( z ) X z

¦ m !*  X m  (1  X )

X

m 0

m m 1

f

X

X

m 0 f

( 1)

( 1) z

¦ m !*  X m  (1  X )

m 0

m m 1

f

( 1) z

¦ m !*  X m  X  1

m m 1

f

( 1) z

( 1) z

¦ m ! (X )(m  1)  *  X m  X

m 0

m 1 m 1

f

z

( 1)

z

¦ m !(m  1)*  X (m  1) ¦ (m  1)!*  X (m  1) ,

m 0 f

(8.77)

m 1 m 1

m 1

n

m 0

( z )

n

¦ n !*(X n) n 1

Using * ( z )* (1  z )

S

sin(S z )



1 * ( X n )

(Section-1.10.9), we get

sin( SX n )

S

* 1  (X n) and placing this in (8.77) we

get following formulas for FX ( z ) : f

FX ( z )

( z )

n

1

§ sin( SX n) · ¨ ¸ * 1  ( X n) * ( X n ) © S ¹

¦ n !*(X n) , n 1

n

f

( z ) § 1

· ¨  * (1  X n)  sin( SX n) ¸ n! © S ¹

¦ n 1

(8.78)

n

f

( z ) § 1

·  ¦ ¨  * (1  X n)  sin(SX n) ¸ n! © S ¹ n 1 

1

S

f

¦

( z )

sin S z

1 * (  z 1) n

f

 *(X n  1) sin(SX n)

n!

n 1

Using the series formula for M X ( z ) (8.64) and using MX ( z )

n

( 1) z

S

n

* ( z ) with z f

n

( 1) z

X ( n  1) we write the following: n

¦ n !*  X n  (1  X ) ¦ n !*  X (n  1)  1 n 0

n 0

f

¦ (1)

§ z · sin SX ( n  1) * X ( n  1) ¨ ¸ S © n! ¹ n

n

n 0

f

1

( z )

n





(8.79)



* X ( n  1) sin SX ( n  1) S n 0 n! Continuing from (8.79) by changing the index of summation from n to n  1 we write the following steps:

MX ( z )

1

S

¦

f

¦

( z )

n 0

1

S

f

n

n! ( z )

 * X (n  1) sin SX (n  1) n 1

¦ (n  1)!  *(X n) sin(SX n) n 1

(8.80)

Fractional Partial Differential Equation

359

8.8.3 The relationship between auxiliary functions M X ( z ) and FX ( z ) Furthermore FX (0)

1 * (1X )

0 and M X (0)

(we use the formula * ( z  1)

, so the relationship between FX ( z ) and M X ( z ) is expressed as follows

z  * ( z ) in the following derivation): f

( z )

n

¦ n !*(X n)

FX ( z )



n 1



1

S

f

¦

( z )

n 1

f

1

¦ S

X

Xz

S

f

S

f

¦

( z )

n

 *(X n  1)  sin(SX n)

n!

n 1

n

(X n )  * (X n )  sin(SX n )

n!

(  z )(  z )

n 1

n  * (X n )  sin(SX n )

n!

n 1

1

1

( z )

(8.81)

n 1

¦ (n  1)!  *(X n) sin(SX n) n 1

X z  MX ( z ) Therefore, we get the relationship:

X z  MX ( z )

FX ( z )

(8.82)

However, we have used (8.82) in deriving (8.77). The

MX ( z )

integral 1 2S i

³

Ha

representation V  zV X

e

for

the

auxiliary

functions

1 2S i

FX ( z )

is

V  zV X

³

Ha

e

dV

and

 dV . The interrelationship between them can be derived from this integral representation 1 V 1X

of M X ( z ) (8.62) as depicted below: 1

V  zV X

e 2S i ³

MX ( z )

dV

V

Ha

1X X

§ e  zV · e ³ ¨¨ V 1X ¸¸ dV 2S i Ha © ¹ X § 1 de  zV 1 V ³ e ¨¨  X z dV 2S i Ha © 1

V

(8.83)

· ¸¸ dV ¹

The obtained result (8.82) is due to the following, which we have used to get the above (8.83), expression:

de

 zV X

dV e Now

we

do

³  f (u )  g (u ) du

integration

³

by

f (u ) g (u )du 

the following derivation:

parts

³ f

(1)

for

(u )

 z (XV

 zV X

V

1X

the



1 de

X 1

)e

 zV X

 zX

e

 zV X

V

1X

(8.84)

 zV X

zX dV expression (8.83)

that

is

by

formula

i.e.

 ³ g (u)du du for the integral expression of M ( z) as depicted in X

360

Chapter Eight X

§ 1 de  zV · MX ( z ) e ³ ¨¨  X z dV ¸¸ dV 2S i Ha © ¹  zV X · 1 § 1 · V § de ¸¸ dV ¨ ¸ ³ e ¨¨  2S i © X z ¹ Ha V d © ¹ X  zV · ­° § de  zV X 1 § 1 · ª V § de ¸¸ dV  ³ ® ³ ¨¨  ¨ ¸ «e ³ ¨¨  2S i © X z ¹ « Ha V d Ha ¯ ° © dV © ¹ ¬ 1

V

· ½°§ d V · º ¸¸ dV ¾ ¨ e ¸ dV » ¹ ¿° © dV ¹ »¼

(8.85)

·· § 1 ·§ 1 §  zV X V ¨ ¸ ¨¨ ¨ ³ (e )(e )dV ¸ ¸¸ © X z ¹ © 2S i © Ha ¹¹ § 1 § V  zV X · · dV ¸ ¸ ¨ ¨³e ¸ X z ¨© 2S i © Ha ¹¹ 1

1

Xz MX ( z )

Which gives

³

Ha



from

 zV X

 dedV

V d

FX ( z )

 X1z FX ( z )

or

X z  MX ( z ) .

FX ( z )

In the above (8.85) derivation we used

0 as this integral along the closed contour is not enclosing any singularity (as the Hankel path X

 zV º f to the origin encircling it and going again to f ). We may write this as ª e

¬

1 2

limit p 0 . The other steps are obvious. For special values say for X

 n

the notation m

m( m  1)( m  2)..( m  n  1) with  m 0

( z )

¼V

fi

 0 with

, we obtain the following identity by using

1.

8.8.4 An auxiliary function M X ( z ) for X at f

V f i

1

2

, 1 3 and 0

n

¦ n !* 

M 1 ( z) 2

1 n 2

n 0

1 *  12 1

S



z *  0

0

 z

z

2

2! *   12



2

2!( 2) S

0

z

3

3! *  1 z



z

4

4! *   23

 ...

4

 ...

4! 43 S

2 4 · § 1 · ( 1) z § 3 · ( 1) z    ... ¸ 1 ¨ ¸ ¨ ¸ ¨ S © © 2 ¹ (2!) © 4 ¹ 4! ¹ 1 2u1 2 2u2 · 1 § § 1 · ( 1) ( z ) § 1 · § 1 · ( 1) ( z )  ... ¸ 1 1    ¨ ¸¨ ¸ ¨ ¨ 2 ¸ (2 u 1)! S © © ¹ © 2 ¹ © 2 ¹ (2 u 2)! ¹

(8.86)

1 §

1

S We have a rising truncated factorial as

 m n

f

¦ n 0

( 1)

n



1 2 n

( z)

2n

(2 n )!

m( m  1)( m  2)..( m  n  1) ;

called a Pochhammer number (Section-1.9.5) which is also ( m) n

*( mn) *(m)

:

 m 0

1 . The notation ( m) n is also

Fractional Partial Differential Equation

361

2 4 · § 1 · ( 1) z § 3 · ( 1) z ¨ ¸  ... ¸ ¨1  ¨ ¸ S © © 2 ¹ (2!) © 4 ¹ 4! ¹

1 §

M 1 ( z) 2

2

1 §

( z )

S ©

2 u 2 u1

¨1 



2 2

3

4 (4 u 3 u 2 u 1)

§  z4 1 ¨ 1 1! S ¨

   2

z2 4

§

f

¦

S ¨n

S

e

z2 4

n

n!

 ... ¸

¹

·

 ... ¸

(8.87)

¸ ¹

· ¸ ¸ ¹

 z2

1



0

©

2

2!

©

1 ¨

·

( z )

4

Therefore, we write:

1

M 1 ( z) 2

For X

1 3

¦

S

f n

( 1) 0

n

§ z 2n ·

1

©

S

 12 n ¨ (2n)! ¸ ¹

  z / 4 2

e

(8.88)

, we have the following formula: f

M 1 ( z) 3

( z )

¦ n !*  2n 3

n 0

f

1 *  23 *  13

z

for X

3n

(8.89)

n 0

¦ n 0

In the above derivations (8.86) and (8.89) the formula M X ( z ) e

z

¦  13 n (3n)! f

1

representation, we get M 0 ( z )

n

2 3 n

¦

z

3 n 1

(3n  1)!

( 1)n z n * n !  X n  (1X ) n 0

f

(8.63) is used. From this series

0 , also by use of FX ( z ) X z  M X ( z ) (8.82) we get F0 ( z ) { 0 .

We know that the Mittag-Leffler function is a (fractional) generalisation of the exponential function; similarly, we may call this M X ( z ) a fractional generalisation of the Gaussian function, which is the solution of an integer order diffusion equation.

8.9 Laplace and Fourier transforms of the auxiliary functions We now state some of the relevant properties of the M X ( z ) and FX ( z ) in view of their role in the TFDWE processes. The Laplace transform pair for 0  X  1 is as follows: 1 X FX 1X M X 1X $ 1t FX X 1 t t t t



1

X

FX

 1 tX

^  ` $ ^



1 t

X

MX

 1 tX

$

1 tX

^ F  ` $ ^ 1

X

X

1 tX

X

tX 1

1 tX

MX

MX

 ` 1 tX

 ` 1 tX

e e

 sX

 sX

1X

s

(8.90)

362

Chapter Eight X

e  s giving FX (t X ) and M X (t X )

8.9.1 The inverse Laplace transform of function X ( s )

We calculate by following steps for an inverse Laplace transform of X ( s )

$

1

1

^e `  sX

(ds )e 2S i ³

st

Br

1

e 2S i ³

st  sX

e

 sX

:

e  sX

ds

put

dV

st ;

V

t

Ha

ds ; s

V t

X V  Vt §1· 1 e dV ¨ ¸ ³ © t ¹ 2S i Ha

§1· ¨ ¸ FX ©t¹ We used FX ( z )

X

 1 tX

t

X 1

(8.91)

 1 tX

MX

 . The above (8.91) result is also obtained via expansion in the power1 tX

X z  M X ( z ) , with z {

series, and the Laplace transformed expression, and through inverting it term by term as follows:

$

1

^ ` e

 sX

$

1 ­

f

®¦ ¯n 0

f

¦ n 0

1

( 1)

n Xn

( 1) s n!

n

t

^

Furthermore, we have $ t

X

M X (t $

1

X

`

)

^e `

^

1

` MX  X

$ X FX (t ) X

 sX

t

X 1

X n 1

X



t

X 1

X 1  sX

s

e

¦ 1

n ! * ( X n )

FX t1X t

f

½ ¾ ¿ t

( 1)

n

$

n 0

n!

f

( 1) t

1

^s ` Xn

n X n

¦ n !*(X n)

(8.92)

n 0



M X t1X

, as shown in the following derivation:

1 tX

X

­ M (t X ) ½ e  s $ ® XX 1 ¾ ¯ t ¿ X

X

s ­ M (t X ) ½ d §e $ ®t XX 1 ¾ ( 1) ¨ ds ©¨ X ¯ t ¿

X

· § e s X 1 X s    ( 1)( 1 ) ¸¸ ¨¨ © X ¹

· ¸¸ ¹

(8.93)

­ M X (t X ) ½ X  sX ¾ s e X ¯ t ¿

$®

8.9.2 The Mittag-Leffler function as an integral representation on Hankel’s path The Mittag-Leffler function (Appendix-A) is defined for any z in the complex ^ -plane, for any X t 0 , by the integral representation EX ( z )

¦

f

zn n 0 * (X n 1)

1 2S i

transform pair of a Mittag-Leffler function which is transform integral as:

xX 1e x dx , X Ha x  z X $ EX ( at )

³

^

X ! 0 and z  ^ . This comes from the Laplace

`

sX 1 sX  a

. From this pair we write an inverse Laplace

Fractional Partial Differential Equation X 1

1

X

s

e X 2S i ³ s  a

EX ( at )

st

ds

put

st

363

dx

x , ds

t

Br

 e ³ 2S i   a

x

, s

t

X 1

x tX 1

1

x

xX tX

Ha

dx t

§ xX 1 · § tX · § e x · ³ ¨© tX 1 ¸¹ ¨© xX  atX ¸¹ ¨© t ¸¹ dx 2S i Ha 1

X 1 x

1

x

e

dx X X 2S i ³ x  at

put

at

X

(8.94)

z

Ha

X 1 x

1

2S i ³

EX ( z )

Ha

x

e

X

x z

dx

^

Similarly following the above steps, and with the use of

E 1

$ t

D

`

ED , E ( at ) 1 2S i

representation of a two-parameter Mittag-Leffler function i.e. ED , E ( z )

sD E sD  a

xD E x e dx D Ha x  z

³

,

we get the following

(Appendix-F).

8.9.3 The Laplace transform of an auxiliary function M X (t ) as the Mittag-Leffler function We now write a Laplace transform pair for the M-Wright function, that is $ ^M X (t )` we derive from the following steps:

EX (  s ) for 0  X  1 , which

f

$ ^M X (t )`

³e

 st

M X (t )dt

0

f

³

e

§ 1 V tV X dV · dt e ¨ 1X ¸ ³ V ¹ © 2S i Ha

 st

0

e

2S i ³ V

1X

Ha

1 2S i

2S i

³e V

X 1

Ha

1

V

e V 2S i ³

X 1

Ha

1

V

0

 st

§ V tV X dV · dt ¨³e 1X ¸ V ¹ © Ha

§ t ( s V X ) · dt ¸ dV ¨³e ©0 ¹ X

V

³

e

f

V

1

f

1

t f

ª e t ( s V ) º « X » ¬« s  V ¼» t

dV

(8.95)

0

§ 1 · dV ¨ X ¸ © s V ¹

X 1 V

e

dV X 2S i ³ V  (  s ) Ha

EX (  s ) In the above (8.95) derivation we used the interchangeability of integrals in view of the analyticity property of the involved functions, and we used the derived integral representation of the one-parameter Mittag-Leffler function. In the second approach we expand the series for an exponential kernel e are obtained earlier, which are

³

f

0

D

x M X ( x )dx

parameter Mittag-Leffler function as follows:

* (D 1) * (XD 1)

 st

and use the absolute moments of M X ( z ) that

(8.73), and then use the series definition of the one-

364

Chapter Eight f

$ ^M X (t )`

³e

 st

M X (t )dt

0

f

§ f (  st ) n · ¸M X (t )dt ³0 ¨© ¦ n! ¹ n 0 f

¦ n 0 f

¦ n 0 f

¦ n 0 f

( s)

nf

³ t MX (t )dt

n! ( s)

f n

n

³ t MX (t )dt

use

0

n

0

* ( n  1) * (X n  1)

(8.96)

* ( n  1)

n ! * (X n  1) ( s)

n

n!

use

n ! * (X n  1) ( s)

* ( n  1)

n!

n

for n  ` def f

¦ *(X n  1)

by u sin g

EX (  s )

EX ( z )

n 0

n

¦ *(Xzn1) n 0

The other way of getting the above (8.96) result is to use the series definition of M X ( z ) , and the Laplace transform expression,

³

f

0

e

 st

n

(  t ) dt

^

$ ( t )

n

`

^ `

n

( 1) $ (t )

n

( 1)

n

 as indicated below: n! s n1

f

$ ^M X (t )`

³e

 st

M X (t )dt

0

f

f

 st

( t )

n

³ ¦ n !*  X n  (1  X ) dt e

0

n 0 f

³e

f

 st

n

(  t ) dt

0

¦ n !*  X n  (1  X )

(8.97)

n 0 f

1

§ ( 1) n n ! · n 1 ¸ © s ¹

¦ n !*  X n  (1  X ) ¨ n 0 f

( 1)

n

§ 1 · n 1 ¸ ¹

¦ *  X n  (1  X ) ¨© s n 0

put

n 1

m

1 · § 1 · § s  ... ¸  EX (  s ) ¦ ¨ m¸ ¨ m 1 * (1  X m ) © s ¹ © * (1  X ) ¹ f

The expression

s 1 * (1X )

( 1)

m 1

is an asymptotic expansion of the Mittag-Leffler function for large s , i.e. for limit





s n f therefore we have s 1 / * (1  X ) as EX (  s ) .

8.9.4 The Fourier Transform of the auxiliary function M X ( x) Now we write the Fourier transform of the symmetric M-Wright function, where it should be extended to the negative real axis as an even function. The Fourier transform is then related to the Mittag-Leffler function via a pair given for

^

`

0  X  1 ,  MX  x

2

2 E2X (  k ) . We now do the one-sided cosine Fourier integration as demonstrated below,

where we used the formula for generalised absolute moments for M X ( z ) (8.63) and made use of the series definition of EX ( z ) .

Fractional Partial Differential Equation f

cos kx

¦

( 1)

n

( kx )

2n

(2 n )!

n 0

f

³

365

f

2n § f · n ( kx ) ( 1)  ¦ ¨ ¸  M X ( x ) dx , ³0 © n 0 (2 n )! ¹

 cos(kx)  MX ( x) dx

0

f

¦

( 1)

n

³x

Use

2n

2n

³ x  MX ( x) dx (8.98)

* (2 n  1)

M X ( x )dx

* (2X n  1)

0

f

2n f

(2 n )! 0

n 0

f

k

f

k

2n

§

(2 n )!

·

n ³ cos(kx)  MX ( x) dx ¦ (1) (2n)! ¨ *(2nX  1) ¸

©

n 0

0

f

¦

( 1)

2 n

(k )

n

*  (2X ) n  1

n 0

* (2 n  1)

¹

(2 n )!

2

EX (  k )

From this we get the two-sided Fourier integral and the Fourier pair for M X

x

and write the following useful

identity: f

^  ` 2³  cos(kx)  M

 MX x

X ( x)

dx

2

2 E2X ( k )

(8.99)

0

8.10 A graphical representation of the M-Wright or auxiliary function M X ( x) We describe the graphs M X ( x ) by schematic diagrams. First, we take the case for 0  X  in Figure-8.2. For X

1 2

we have seen M 1 ( x ) 2

1 2

, depicted schematically

2

1 e  x . This is the standard Gaussian bell shaped curve centred at S

x

0 (when extended to a negative axis by reflecting about x 0 line, shown by a dotted extension in Figure-8.2). x The other extreme is at X 0 where M 0 ( x ) e . Shown in the Figure-8.2 which is a curve for X 14 . The observation points while X is much less than

1 2

, say near zero, indicate that M X ( x ) starts from x

fall as x increases, causing the curve to go to zero. While X is nearing

1 2

0 and has a steep

, the curve becomes Gaussian in nature,

with a slow initial fall followed by a sudden fall and then eventually heading to zero. For the case of X , which is varying in nature from 0 to 12 , we observe the maximum value of M X ( x ) is at x 0 and accordingly, a value of X as the curve of M X ( x ) decays and goes to zero at say x

Now we take the case of M X ( x ) for X varying from value of X

1 2

1 2

5 i.e. for a large x .

to 1 , which is depicted in the schematic of Figure-8.3. At the

, the curve M 1 ( x ) is a Gaussian curve. As X increases from the value of 2

1 2

the peak value of the

curve shifts from x 0 to x ! 0 , and after attaining the peak value the curve falls towards zero. This is as shown for the case X 85 , in Figure-8.3. Still increasing the value of X towards 1 , it makes the peak of the curve shift towards

x 1 , making it steeper and sharper. At the value X 1 , it is expected to take the form of the delta function, as M 1 ( x ) G ( x  1) . For the negative x the function M X ( x ) is plotted by reflection of theses curves about the line x

0 as M X ( x ) is a symmetric function.

366

Chapter Eight

0.8

(1) n x n ¦ n 0 n !*  X n  (1  X ) f

M X ( x)

0.6

1 2

M 0 ( x) e  x

X 0

1 4

X

0 X 

1 2

X

1

f

n

¦ (1)  S

M 12 ( x)

1 2 n

n 0

 e 1

( x) 2 n (2n)!

 x2

S

x

0

3

Figure-8.2: Plot of the symmetrical M-Wright function for X = zero to 0.5

1 2

X

; M 12 ( x)

0.6

1

f

n

( x) 2 n (2n)!

¦ (1)  S

1 2 n

n 0

 e 1

 x2

S

X 1, M 1 ( x) G ( x  1)

0.4

M X ( x)

(1) n x n

f

¦ n!*  X n  (1  X ) n 0

5 8

X

x

0

1

2

3

4

Figure-8.3: Plot of the symmetrical M-Wright function for X = 0.5 to 1.0

Since the function M X ( x ) is a symmetrical function, for X

 e 1

 x2

S

for X

for X

1 2



0 , the function M 0 x is e

, as depicted in the Figure-8.2. In the Figure-8.3 the

 e 1

 x2

S

x

and is transformed into

gets transformed into G ( x r 1)

1.

8.11 Auxiliary function in two variables M X ( x, t ) and its Laplace & Fourier transforms 8.11.1 Defining M X ( x, t ) in relation to M X ( xt X ) 

For 0  X  1 where x, t  \ M-Wright functions in two variables M X ( x, t ) introduced as follows in (8.100): M X ( x, t )

t

X

M X ( xt

X

)

(8.100)

This above (8.100) function defines the probability density function (pdf) in space x , evolving with time t , with exponent X . This exponent is also called self-similarity exponent or Hurst parameter, or the Hurst index H X (Section 1.24). Obviously we should consider a symmetrical version, obtained from the above (8.100) definition and 1 and replacing x by x for x  \ . by multiplying by 2

Fractional Partial Differential Equation

367

8.11.2 Laplace Transform of M X ( x, t )

^

1

$ X FX (t

From the Laplace transform pair i.e.

`

X

^

)

$ (t

X

) M X (t

X

`

X 1  sX

)

s

e

we derive the Laplace



transform of M X ( x, t ) w.r.t. t  \ as:

^

X

`

X

X 1  xsX

$ ^M X ( x, t ); t o s` $ t M X ( xt ); t o s

s

e

(8.101)

The derivation is as follows:

$

1

^e

 xsX

;s o t

`

¦ n 0

n

f

¦

n!

n 0

FX ( z )

^s ` Xn

X n 1

t

(8.102)

n X n

( 1) x t

FX ( xt

X

¦ t

X zM X ( z ) ; z

t

xt

( 1) ( xt

MX

X 1

X n

)

n ! * ( X n )

n 0

Xx

)

n

f

1

n ! * ( X n )

n 0

t

1

* ( X n ) n

f

¦ t

1

n

( 1) x

· ¸ ¹

n!

0

$

n!

n Xn

( 1) x s

¦

n

( 1) x

n

f

©n

n

f

1

§

1

$ ¨

 ; 1 tX

X

We have the following:

$ $

1

^e `  xsX

1

^

t

1

t

M X ( xt

X 1

FX ( xt

X

`

X

1

)

Xx

)

t e

Xx

M X ( xt

X 1

X

)

(8.103)

 xsX

We get now the following steps:

^

X

$ t M X ( xt

X

­§ 1 X · ½ M X ( 1xt ) ¸¾ X 1 ¹¿ ¯ ©t

`

); t o s

$ ®t ¨ ( 1)

d § 1

¨ e ds © X x

( 1)

Xx

 xsX

X

(  xX s )e

· ¸ ¹

(8.104)

 xsX

X 1  xsX

s

Again from the Laplace transform pair i.e.

M X ( x, t )

t

X

M X ( xt

X

e

$

x o s , we derive the Laplace

) l EX (  s ) ;



transform of M X ( x, t ) w.r.t. x  \ as:

^

X

`

X

$ ^M X ( x, t ); x o s` = $ t M X (t x ); x o s f

³e

 sx

(t

X



) M X (t

X



x ) dx

0

f

§ f (  sx ) n · X X ¸ ( t )  M X ( t x ) dx ³0 ¨© ¦ n! ¹ n 0 f

¦ n 0

Now put xt

X

y , dx

X

t dy , x

X

( s) n!

f

n

(t

X

t y and we get the following steps:

X ³  MX ( xt ) dx

) x 0

n



(8.105)

368

Chapter Eight f

¦

$ ^M X ( x, t ); x o s`

( s)

f

n

(t

n!

n 0

f

0

y

n

¦

n!

X n

(  st )

X n

( 1) ( st )

n 0

f

n! * (X n  1) f

X

¦ *(X n  1)

(8.106)

* (X n  1) n

f

0

n!

 MX ( y ) dy

$ ^M X ( x, t ); x o s`

n

³ y  MX ( y ) dy

n!

n 0

³

 MX ( y ) tX dy

X n f

n

( 1) ( s ) (t )

¦ Use

X n

³

) ( yt ) 0

n

f

X

EX (  st )

by using

z

n

¦ *(X n  1)

EX ( z )

n 0

n 0

Therefore, we write:

^

$ ^M X ( x, t ); x o s`

X

$ t M X ( xt

X

`

X

); x o s

EX (  st )

(8.107)

8.11.3 Fourier transform of M X ( x, t ) Note the variable x of representing a quantity position in M X ( x, t ) , then the frequency s , of spatial frequency generally represented as a pure complex number ik (signifying the steady state response). In that case we may write

^

$ ^M X ( x, t ); x o ik`

^  `

 MX x

X

$ t M X ( xt

X

); x o ik

`

X

EX ( ikt ) ,

2 2 E2X (  k ) , we write Fourier of M X

with

x\



and

k \.

From

 x , t as:

^ 



 MX x , t ; x o k

`

2 2X

)

X



2 E2X (  k t

(8.108)

We derive this by the following steps: f

^ 

 MX x , t ; x o k

` 2³ cos(kx) t

X

M X ( xt

) dx

0

f

§

f

³©¦

2 ¨ 0

f

2

¦ n 0

Put xt

X

y , dx

X

t dy to get the following steps:

n

( 1) ( kx )

2n

(2 n )!

n 0

n

( 1) ( k ) (2 n )!

2n f

³ (x 0

· X X ¸  t M X ( xt ) dx ¹ 2n

)(t

X

) M X ( xt

X

)dx

(8.109)

Fractional Partial Differential Equation f

^ 

 MX x , t ; x o k

` 2¦

n

( 1) ( k )

¦

( 1) ( k )

2

( 1) ( k )

³

f

y

0

^ 

m



³ ( yt

f

( 1) ( kt )



X

) M X ( xt

X 2n

) (t

X 2n

(2 n )!

)

X

)  M X ( y ) (t )dy

 M X ( y ) dy

2n

³ ( y )  M X ( y ) dy

(8.110)

0

§ ( 1) n ( k 2t 2X ) n · § (2n)! · ¦ ¨ ¸¨ * (2X n  1) ¸ (2 n )! ¹ n 0© ¹© f

`

2

)

8.11.4 Special cases for M X ( x, t ) for X as

1

and one

2

, we get: 1 2

M1

2

 x ,t

1 2

t

 12

M1

1

1

2 t

S

 xt  12

2

e



 xt 1 / 2

1



2

/4

use

M 1 ( z) 2

1

S

e

z

2 4

(8.111)

2

e

2 St

For limiting the case X

X

* (X m  1)

2 2X

1 2

) dx

m!

2 E2X (  k t

As a special case for X



X

0

X 2n f

n

 M X ( y ) dy

 MX x , t ; x o k

2n f

(2 n )!

n 0

Use

³ ( yt

n 0

¦

)(t

0

n

f

¦

2n f

(2 n )!

n 0

2n

0

n

f

2

³ (x

(2 n )!

n 0

2

2n f

369

 x4 t

1 we obtain:

 12 M1 ( x, t )  12 G ( x  t )   12 G ( x  t )

(8.112)

8.12 The use of the two variable auxiliary functions in TFDWE 8.12.1 Time fractional diffusion equation and its solution with M X ( x, t ) Let us have a time fractional diffusion equation (with the Caputo fractional derivative of order E ): w

2

E

u ( x, t )@ E >

a

w u ( x, t )

0  E 1 2 w t w x Integrating fractionally by order E on both sides of (8.113), we obtain the following steps:

(8.113)

370

Chapter Eight

ª wE º « w t E u ( x, t )» ¬ ¼

E

0 It

0  E 1

t

§ w 2u ( x, W ) · dW ¨ ¸ (t  W )1 E 2 * ( E ) ³0 © w x ¹ a



u ( x, t )  u ( x, 0 )

u ( x, t )

ª w 2u ( x, t ) º «a w x 2 » ¬ ¼

E

0 It

t

§ w 2u ( x, W ) · dW ¨ ¸ (t  W )1 E 2 * ( E ) ³0 © w x ¹ a



u ( x, 0 ) 

(8.114)

t

§ w 2u ( x, W ) · dW  u ( x, t ) u ( x, 0 )  ¨ ¸ (t  W )1 E 2 * ( E ) ³0 © w x ¹ a



With u ( x, 0 )

G ( x)

E

1

> f (t )@

t

(t  W ) *( E ) ³

Use

0 It

u ( x, t )

u ( x, 0 )  0 I t

E 1

f (W )dW ;

0

ª w 2u ( x, t ) º « w t2 » ¬ ¼

E



1

a



GE ( x, 0 ) therefore we obtain the fundamental solution u ( x, t )

GE ( x, t ) and that we write

as: w

E

GE ( x , t )

G ( x)  0 It

2

(8.115)

GE ( x , t )

2

w x

and  ^G ( x )` 1 then

w w x

We take first Fourier transform of the above (8.115) equation, by using ik for operator write the following transformed equation: Gˆ E ( k , t )

1  0 I t ª¬(  k )Gˆ E ( x, t ) º¼ 2

E

Now we apply the Laplace transform to the above (8.116) by using s

E

E

for

(8.116)

(Section-5.14) and, $ ^1`

0 It

1 s

, and

write the following steps:

 1s  k s Gˆ

 Gˆ E ( k , s )

2 E



E 1

 Gˆ E ( k , s )

s

^ `

a

(8.117) 2 E s k We can have two methods, for inverting this (8.117). First, we do an inverse Fourier transform then we do the inverse Laplace transform. That is recalling the Fourier transform pair (where a , b ! 0 ) as follows:

 We set a

s

E 1

and b

1

E (k , s) ;

­ 2 b ½ bt e , ® 2 2¾ ¯b  Z ¿



1

a

bk

2

2 b

e

x b

(8.118)

E

s , and apply the above (8.118) transform pair to get from (8.117) to the following: s

G E ( x, s )

E 1

2 s

^

Now we do an inverse Laplace transform by using $ t

X

e

E

1

 x sE

2

`

1 2

E /2 ( E / 2) 1  x s

e

X 1  xsX

M X ( x, t )

GE ( x , t )

s

s t

 ( E / 2)

(8.119)

e

, here we set X

ME



2

xt

E 2

E 2

to get:



(8.120)

By second method first we do an inverse Laplace transform of Gˆ E ( k , s ) and then apply the inverse Fourier transform. E 1 That is the first recalling Laplace transform pair $ 1 sE s c 2 E 2 setting c k , we get Gˆ ( k , t ) E (  k t ) . Use  M

^

E

 Fourier transformation Gˆ E ( k , s )

E 1

E EE (  ct ) , with c ! 0 (Appendix-A) and then by

^  x , t ; x o k`

E

s

`

2 2X

2 E2X (  k t

X

E

2 1

(s  k )

(8.117), with 2X GE ( x , t )

This is the same as we obtained in (8.120).

1 2

E 2

E; X

M E ( x, t ) 2

1 2

t



) to have an inverse

, to get the following:

E 2

ME

 x

2

t

E

2

(8.121)

Fractional Partial Differential Equation

371

8.12.2 A time fractional drift equation and its solution with M X ( x, t ) We shall now demonstrate the use of the M-Wright function as a solution to time fractional drift equation that is as follows: w

E E

w t

u ( x, t )



w w x

u ( x, t ) ;



u ( x, 0 ) with Applying the Laplace transform on (8.122) we get the following:

f  x  f, t t 0

(8.122)

G ( x)

­ wE ½ ­w ½ u ( x, t )¾ $ ® E u ( x, t ) ¾ $ ® ¯w x ¿ ¯w t ¿



E E 1 s u ( x, s )  s u ( x, t ) t

Now use u ( x, t ) t

0



(8.123)

w



u ( x, s )

w x

G ( x ) to get the following:

0

 E E 1 s u ( x, s )  s u ( x, 0 ) 

u ( x, 0 )

G ( x)



then 

w w x

E ˆ E 1 s G E ( k , s )  s (1)

s

 Gˆ E ( k , s )

wx

u ( x, s )

u ( x, t )

E E 1 s G E ( x, s )  s G ( x )

Now we do the Fourier transform of (8.124), with  ^G ( x )` 1 , and for

w

GE ( x , t )

(8.124)

w  GE ( x , s ) wx

we write ik , to have the following:  ikGˆ E ( k , s )

(8.125)

E 1

E

s  ik We first do the inverse Fourier transform of the above (8.125) by recalling:



1

­ 1 ½ bt ® ¾ e ; ¯ b  iZ ¿

a, b ! 0 ;

We set a

s

E 1

and b

1



^ ` a

b  ik

s , so that we obtain the following:

GE ( x , t )

s

E 1  xs E

(8.127)

e

M E ( x, t )

t

e

E

M E ( xt



^ `

EE ( ct

E

) for c ! 0 . Set c

X

X 1  xs

s

, getting the following E

s E 1 s E  ik

We apply the second method of first using an inverse Laplace transform of Gˆ E ( k , s ) s E 1 sE c

(8.126)

x!0

Now we get an inverse Laplace transform by recalling $ ^M X ( x, t ); t o s` expression:

1

 bx

E

Gˆ E ( x, s )

transform pair $

ae

)

(8.128)

, by using the Laplace

ik , to get:

Gˆ E ( k , t )

E

EE ( ikt )

(8.129)

Now we transform with ik o x to get from the obtained Laplace transform pair (8.107) that is:

$ ^M X ( x, t ); x o ik`

^

X

$ t M X ( xt

X

); x o ik

`

X

EX ( ikt )

(8.130)



x\ , k \

The space x is transformed into a complex spatial frequency ik which is the same as in the Fourier case. Therefore, above (8.130) the Laplace transform pair we use as Fourier transform pair, and we write the following expression from (8.129):

GE ( x , t )

M E ( x, t )

t

E

ME

 x tE

(8.131)

372

Chapter Eight

8.13 Reviewing time fractional diffusion equations with several manifestations 8.13.1 Classical diffusion equation We consider a variety of diffusion equations including the standard integer order diffusion equation whose fundamental solution, the Green’s function, is expressed in terms of the M-Wright function depending on space or time variables. The two variables however are related through a self-similarity condition (Hurst exponent H  (0,1) ) which we discussed in Chapter-1 (Section 1.24). The standard integer order diffusion equation for the field variable u ( x, t ) with initial condition u ( x, 0)

u0 ( x ) is

2

w u

K1

w t

w u w x

 f  x  f; t ! 0

2

(8.132)

2

Where K1 is a suitable diffusion coefficient with dimensions of cm / sec . This is an initial value problem, and we perform integration of both sides to write the solution in terms of Volterra-Integral Equation i.e. t

w u

³wW

t

K1

dW

0

t

2

³

w u ( x, W ) w x

0

2

u ( x, t )

dW

u0 ( x )  K1

³

2

w u ( x, W ) w x

0

2

dW

(8.133)

It is well known that the fundamental solution is also called the Green’s function corresponding to u0 ( x ) G ( x ) and is the Gaussian probability density evolving in time with the mean square displacement proportional to time (that is a 1

variance or second moment). We write the Green’s function as G1 ( x, t ) 2

³

V 1 (t )

f

f

2

x G1 ( x, t )dx

e

2 S K1t

 x 2 /  4 K1t

, having a variance as

2 K1t .

8.13.2 The Green’s function for a classical diffusion equation and its moment

§ ¨ ©

1 2

The Green’s function in terms of the M-Wright function is G1 ( x, t )

1 K1 t

1

2

·M ¸ 12 ¹

 . Let us find the moment x

K1t

of this Green’s function G1 ( x, t ) that is: f

V

2

³x

f

Put

x K1t

y;

x

2

2

y K1t ; dx 2

M1

2 K1t

 x

2

K1t

dx

(8.134)

dy K1t , in the above (8.134) expression to get the following: f

V

1

2

2

y K1t M 1 ( y) 2 f 2 K1t



³

K1t 2



K1t  dy

f

³ y M 2

1

2



(8.135)

( y ) dy

f

f f · § K1t · § 2 2 ¨ ¸ ¨ 2 ³ y M 12 ( y ) dy ¸ K1t ³ y M 12 ( y ) dy © 2 ¹© 0 0 ¹



We use an already derived expression about

³

f

0

D

x M X ( x )dx

* (D 1) * (XD 1)



³

0

x

2

M



D -moments of the M-Wright function (8.63) that is

1

2

1 2

2, X

. In this general expression put D f





( x ) dx

, so we have the following:

* (2  1)

* (3)

2!

*  u 2  1

* (2)

1!

1 2

2

(8.136)

Use of the (8.136) above derived expression gives the variance as proportional to time: f

V

2

³ M

K1t y

2

0

From the self-similarity of the Green’s function, we write the following:

1

2



( y ) dy

2 K1t

(8.137)

Fractional Partial Differential Equation

G1 ( x, t )

1 2 S K1 t

1

2

e

 x 2 /  4 K1t

1

G1 ( x, t )

1

2 K1 t

373

M1

2

2

§ ¨ ©

x K1 t

1

2

· ¸ ¹

(8.138)

Furthermore, we can write: 1

G1 ( x, t )

where H

1 2

K1 t

H

G1 ( z )

x

z

K1 t

(8.139)

H

is the self-similarity (or Hurst) exponent (this Hurst exponent is a self-similarity we have described in x

Chapter-1, Section 1.24). The variable z

acts as the similarity variable, with G1 ( z ) as one variable from the

K1 t H

‘reduced’ Green’s function. Note that the above derived variance (8.137) law v t characterises normal diffusion as it emerges from a Brownian motion approach. With H z with a persistent motion H 

1 2

, H  (0,1) the case is of a fractional Brownian motion

 12 ,1 and anti-persistent with

H   0, 12 , giving us the case for anomalous diffusion;

we will not detail this aspect here.

8.13.3 A classical diffusion equation with a stretched time variable in its Green’s function and moment Now we stretch the time variable in the integer order diffusion equation by replacing t with t

D

where D   0, 2 , and

we write the modified equation as: 2

w u

w u

KD

D

w (t )

w x 2

 f  x  f ; t t 0

2

D

d dt

We call KD a stretched diffusion coefficient having the dimension cm / sec . We will use

(8.140)

ª¬t D º¼ D (t D 1 ) in the

following discussion. We recognise easily that the above diffusion equation is akin to the standard diffusion equation,

 K

but with a diffusion coefficient depending on time K1 (t )

w u w t

D t



D t

w u K1 (t )

, which follows from the following derivation:

w u

KD

w t

w x

D 1

D

 dt

D

use

2

2

d(t )

D t

D 1

2

 K wwxu D t K

D t

w t

D 1

D

2

w u

D

D 1

K1 (t )

2

w u wx

(8.141)

2

D 1

D

The Volterra-integral corresponding to the time stretched case is obtained by taking the integration of both the LHS and RHS of the stretched time diffusion equation

D t

w u w t

D 1

KD



w 2u w x2

(8.141) to get the following: t

u 0 ( x )  D  KD W

³

u ( x, t )

D 1

2

w u ( x, W ) w x

0

2

dW

(8.142)

We write the corresponding fundamental solution that is a stretched-time Gaussian function:

1

GD ( x, t ) The above (8.143) comes from G1 ( x, t )

2 S KD t 1

2 K1 t

1

2

M1

2

§ ¨ ©

D

2

e



 x 2 / 4 KD tD

1

1 2 KD t

x K1 t



2

D

2

M1

2

§ ¨ ©

x KD t

D

2

· ¸ ¹

(8.143)

· , by placing t D instead of t and placing K instead ¸ D ¹

of K1 . We can derive the corresponding variance (via the method described earlier) as: f 2

VD

2

³ x GD ( x, t ) dx

2 KD t

D

(8.144)

f

This comes from V

2

2K1t by placing t

D

instead t and KD in place of K1 . This is characteristic of a general

process of anomalous diffusion, precisely slow diffusion for 0  D  1 , and fast diffusion for 1  D  2 .

374

Chapter Eight

8.13.4 Time fractional diffusion equation Now we take a time fractional diffusion equation and write that for E  (0,1) as the following equivalent fractional differential equations (with the fractional Caputo derivative):

w u w t

§

1 E

K E ¨ 0 Dt

2 w u ·

E  (0,1)

¸ w x ¹ 2

©

apply

§ E 1 § w u · · § E § 1 E w 2u · · D K ¨0 t ¨ E ¨ 0 Dt ¨ 0 Dt ¸¸ 2 ¸¸ w x ¹¹ © w t ¹¹ © © © § E 1 w u · § w 2u · 1 0 Dt ¨ 0 Dt 0 Dt w t ¸ K E ¨ w x 2 ¸ © ¹ © ¹

E 1

0 Dt

(8.145) 1 0 It

2

E

0 Dt u

w u

KE

2

w x Therefore, we have the following equivalent time fractional diffusion equations:

§

w u

2

E

cm / sec . Integrating LHS and RHS of

wu wt

2 w u ·

2

E

;

0 Dt u

KE

w u

(8.146) 2 ¸ 2 w x ¹ w x is a sort of fractional diffusion coefficient having a dimension of

w t In this time fractional diffusion equation K E

1 E

K E ¨ 0 Dt

©

KE



1 E w 2u 0 Dt w x2

(8.146) we obtain the following:

2 § 1 1 E w u · u ( x, t )  u0 ( x ) K E ¨ 0 I t D 0 Dt 2 ¸ w x ¹ ©

§

2 w u ·

E

K E ¨ 0 It

2

©

t

KE

³ (t  W )

*( E )

E 0 Dt

¸

w x ¹ E 1

E

0 It

(8.147)

2

w u ( x, W ) w x

0

dW

2

In (8.147) the last expression is obtained by applying the RL Fractional integration formula for

E

0 It

f (t ) operator

(Chapter-2). Like the diffusion equation of an integer order we consider the equivalent Volterra integral equation corresponding to the fractional diffusion equation of the above (8.147) as by taking the E order fractional integration for both the sides of the LHS and RHS of 0 DtE u

KE

w 2u w x2

(8.146) as demonstrated below: 2

E

0 Dt u

KE



0 Dt u

E

0 It

E

w u w x



2

KE

u ( x , t )  u0 ( x )

§ E w 2u · ¨ 0 It 2 ¸ w x ¹ ©

KE

(8.148)

§ E w 2u · ¨ 0 It 2 ¸ w x ¹ © E

Finally, we write the following by applying the RL fractional integration formula 0 I t f (t ) :

u ( x, t )

u0 ( x ) 

KE *( E )

t

³ 0

(t  W )

E 1

2

w u ( x, W ) w x

2

dW

(8.149)

8.13.5 The Green’s function for a time fractional diffusion equation and its moments The Green’s function, GE ( x, t ) for this time a fractional diffusion equation can be expressed in terms of the MWright function as:

Fractional Partial Differential Equation

§

1

GE ( x , t )

375

2 KE t

E

· ¸ ¹

x

ME ¨ 2

2

KE t

©

E

2

(8.150)

The above (8.150) we have derived in the previous section. Let us find the moments of the above (8.150) Green’s function: f 2

³x

VE

2 KE t

f x

Put

KE t E

y; x

2

2

E

2

E

2

x KE t

E

2

·· ¸ ¸ dx ¹¹

(8.151)

E

y K E t ; dx

V

§ § ¨ ME2 ¨ © ©

1

2

dy K1t , in the above (8.150) expression to get the following: f

2

y KE t

E

KE t

E

³2

f

KE t 2

M

E

2

E f

³y

M

2

f

§ KE t E ¨¨ © 2

( y)



E

KE t

E

dy



(8.152)

( y ) dy

2

f ·§ f 2 · 2 y M y y K t M E ( y ) dy 2 ( ) d E ¸¸ ¨ ³ ¸ E ³y 2 2 0 ¹ ¹© 0









We use an already derived expression about the D moments of the M-Wright function (8.63) that is

³

f

0

x

D

 M X ( x ) dx

* (D 1) * (XD 1)

³

f

0

E 2

2, X

. In this general expression put D

, so we have the following:

* (2  1)

2

x M E ( x )dx 2

*



E 2



u 2 1

* (3)

2

* ( E  1)

* ( E  1)

(8.153)

Use of the above (8.153) derived expression gives the variance as proportional to time as follows: f 2

KE t

VE

E

³y

2

0

M

E

2

2KE t



( y ) dy

E

(8.154)

* ( E  1)

The corresponding variance we write as we obtained from the absolute moments of the M-Wright function as: f

³ x  G E ( x , t ) dx

2

2

2

V E (t )

f

* ( E  1)

KE t

E

(8.155)

Consequently, for 0  E  1 the variance is consistent with a process of slow diffusion with a similarity exponent H

E 2

.

8.13.6 A time fractional diffusion equation with a stretched time variable Now we make one more step as advancement, that is, a stretched one with the time-variable for a fractional diffusion equation, by stretching the time through replacing t with t following (with the Caputo fractional derivative):

wu w (t

D /E

)

D

E

, where

KDE

0  D  2 and 0  E d 1 , so we get the

§ 1 E w 2u · ¨ 0 DtD / E 2 ¸ w x ¹ © 2

D

Where KDE is a sort of stretched diffusion coefficient of dimension cm / sec K E for D

that reduces to KD for E

(8.156) 1 and to

E . From the above (8.156) we get the following equivalent fractional diffusion equation:

w u w t The steps to get the above (8.157) expression are as follows:

§ 1 E w 2u · § D ·  DE 1 KDE ¨ 0 DtD / E ¨ E ¸t 2 ¸ w x ¹ © ¹ ©

(8.157)

376

Chapter Eight

w u w (t

§ 1 E w 2u · KDE ¨ 0 DtD / E 2 ¸ w x ¹ ©

w u

D /E

 t

)

D E

D E

use

d(t

D /E

1

w t D E

)

t

 DE 1dt

(8.158)

§ 1 E w 2u · § D ·  DE 1 t K DE ¨ 0 DtD / E ¨E ¸ 2 ¸ w x ¹ © ¹ ©

w u w t

Now we integrate both sides of the LHS and RHS of the above (8.157) to obtain the following expression: W t

§D ·  DE 1 § D1 E w 2u ( x, W ) · dW u ( x, t )  u0 ( x ) ¨ ¸ KDE ³ W ¨ 0 WD / E ¸ 2 w x ©E ¹ © ¹ W 0 Let

W

D /E

(D / E )W

[, W

(D / E ) 1

dW

[

E /D

for

0,

W

0 and

[

for

t

W

we have

D /E

[

t

E /D

) ·§

(8.159)

. Furthermore, we have

d[ . Substituting all these, we have the following steps: W t

§D ·  DE 1 § D1 E w 2u ( x, W ) · dW K W ¨ 0 WD / E ¸ ¨ E ¸ DE ³ 2 w x © ¹ © ¹ W 0 [ tD / E

§D · ¨ E ¸ KDE © ¹

³ [

[ tD / E

³

KDE

[ 0

§

W

 DE 1 §

2

w u ( x, [

1 E

¨ 0 D[

w x

©

0

· d[ ¸ ¨¨ D (D / E ) 1 ¸¸ ¹ ©  E W ¹

2

§ 1 E w 2u ( x, [ E / D ) · ¨ 0 D[ ¸ d[ 2 w x © ¹

1

1 E

KDE ¨ 0 I[ 0 D[

2

w u ( x, [ w x

©

E /D

)·

¸ ¹

2

§ 1 1  E w 2u ( x, [ E / D ) · KDE ¨ 0 I[ 0 D[ 0 D[ ¸ 2 w x © ¹ KDE E

Using

0 D[

§  E w 2u ( x, [ E / D ) · ¨ 0 D[ ¸ 2 w x © ¹ E 0 I[

f ([ )

9 [

1

([  9 ) *( E ) 9 ³

f ([ )

E 1

f (9 )d9

0

W t

2 D 1 § §D ·  1 E w u ( x , W ) · E ¨ 0 DW D / E ¸ dW ¨ E ¸ KDE ³ W 2 w x © ¹ © ¹ W 0

KDE

9 [

1

([  9 ) *( E ) 9 ³

E 1

w x

0

We have W expression:

D /E

[ , d[

(D / E )W

KDE

(D / E ) 1

dW . Take W

1

9 [

*( E ) 9 ³

([  9 )

E 1

1

9 [

D /E

KDE

Put [

t

D /E

, so we have 9 { [

W

D /E

, thus for 9

2

w x

*( E ) 9

E /D

2

t

D /E

)

2

)

d9

9 {[ d9 , to have the following

d9

D / E E 1

³ [  W 0

E /D

(D / E ) 1 9 and (D / E )W dW

w u ( x, 9

0

(8.160)

2

w u ( x, 9

(8.161)

2

w u ( x, W ) w x

2

§ D  DE 1 · dW ¨ W ¸ ©E ¹

, we have the upper limit W

t:

Fractional Partial Differential Equation

1

9 [

([  W *( E ) 9 ³

KDE

D / E E 1

)

2

w u ( x, W ) w x

0

§D · 1 KDE ¨ ¸ © E ¹ *( E )

9 tD / E

³ 9

W

§D

dW ¨

W

©E

D 1 D /E E

(t

W

· ¸ ¹

(D / E ) 1

D / E E 1

)

2

w u ( x, W ) w x

0

2

dW ; 9 { [

W

D /E

(8.162)

D

W t

KDE

2

377

2 1 §D · 1 D /E D / E E 1 w u ( x , W ) E W W ( t  ) dW ¨ E ¸ *(E ) ³ 2 w x © ¹ W 0

Therefore, our integral equation becomes:

u ( x, t )

u0 ( x )  KDE

t

§ D ·  DE 1 D / E D / E E 1 w 2u ( x, W ) t  W w x 2 dW ¨ ¸ W * ( E ) © E ¹ ³0 1

(8.163)

8.13.7 The Green’s function for a time fractional diffusion equation with a stretched time variable The equation (8.156) has the following Green’s function:

GDE ( x, t )

1 2

1 KDE t

D

2

ME

2



x KDE t ( D / 2 )



(8.164)

which comes from the expression: GE ( x , t )

by placing t

D /E

§

1 2 KE t

E

2

ME ¨ 2

©

x KE t

E

2

· ¸ ¹

(8.165)

in place of t and changing K E to KDE . The variance is: f 2

V DE

³ x GDE ( x, t ) dx 2

f

2

This comes from V E (t )

³

f

f

x

2

 G E ( x , t ) dx

2 * ( E 1)

2 * ( E  1)

KDE t

D

(8.166)

E

K E t , by placing t D / E in place of t and KDE in place of

K E . Consequently, the resulting diffusion process turns out to be self-similar having Hurst exponent i.e. H

D 2

(Section-1.24) and the variance law consistent both with slow diffusion (0  D  1) and fast diffusion (1  D  2) . We note that the parameter E explicitly enters the variance expression only to modify the proportionality constant. It is straight forward to note that the evolution equations of this process reduce to those for fractional-time diffusion if D E  1 , for stretched diffusion if D z 1 and E 1 , and finally to the integer order diffusion equation if D E 1.

8.14 Short summary This chapter extends the lessons of earlier chapters of fractional calculus. Here we have seen how a time fractional partial differential equation, especially a diffusion-wave equation evolves, and after the application of the Laplace Fourier transform tricks we get the solution. The evolution of the M-Wright function and other auxiliary functions is a new idea of Mainardi, as we saw these are a generalisation of a classical Gaussian function, which helps to obtain a solution to a time fractional partial differential equation. We have several derivations regarding these new sets of functions. We conclude with the observation that the treatment is as rigorous as in a classical partial differential equation.

8.15 References This chapter is based on pioneering work on fractional partial differential equations, those arethe following: Wyss, “The fractional diffusion equation” (1986) Michelle Caputo and Mainardi F, “A new dissipation model based on memory mechanism” (1971) Michelle Caputo, “Distributed Order Differential Equation Modeling Dielectric Induction and Diffusion” (2001); “Linear Models of Dissipation whose Q is almost Frequency Independent” (1967) Mainardi F, “Fractional relaxation-oscillation and fractional diffusion wave phenomena” (1996)

378

Chapter Eight

Almeida R, Malinowska A. B, Torres D F M, “A fractional calculus of variations for multiple integrals with application to vibrating string” (2010) Agrawal O P, “Solution for fractional diffusion wave equation defined in bounded domain” (2002) Balakrishnan V. “Anomalous diffusion in one-dimension (1985) Chaves A S, “A fractional diffusion equation to describe Levy flights” (1998) El Sayeed A.M.A,“Fractional order diffusion wave equation” (1996) Enrico Scalas, Rudolf Gorenflo, Francesco Mainardi, Marco Raberto, “Revisiting the derivation of the fractional diffusion equation”(2002) Hilfer R, “Fractional Diffusion based on Riemann-Liouvelli Fractional Derivative” (2000) Juan Bisquert, Albert Compte, “Theory of electrochemical impedance of anomalous diffusion” (2001) Kaya D, “A Numerical Simulation of Solitary-Wave Solution of the Generalized Regularized Long-Wave Equations” (2004) Klafter J, Blumen A, Shlesinger M F, “Stochastic pathways to anomalous diffusion” (1987) Lynch V E, et al, “Numerical methods of solution of partial differential equations of fractional order” (2003) Metzler, R and Klafter J, “The Random Walk’s Guide to Anomalous Diffusion: A fractional Dynamics Approach” (2000) Rina Schumer, David A Benson, Mark M Merscaert, Stephen. W. Wheatcraft, “Eulerian derivation of fractional advection dispersion equation” (2001) Rina Schummer, David A Benson,“Multiscaling fractional advection dispersion equation and their solution” (2003) Shequin Zhang, “Positive solutions for boundary value problems on non-linear fractional differential equations” (2006) Saha Ray S, “Exact solution for time fractional diffusion wave equation by decomposition method” (2007) The details of the above literature are found in the bibliography section, placed in alphabetical order.

CHAPTER NINE MODIFIED FRACTIONAL CALCULUS IN CONJUGATION WITH CLASSICAL CALCULUS

9.1 Introduction We will start the chapter by generalising the concept of term-by-term differentiation as obtained for classical calculus, for the fractional differ-integration context, and then we will formalise the composition rules for a generalised differintegral operator. Here we try to simplify operators of fractional calculus and try for modifications of the classical Riemann-Liouville (RL) fractional integral, and of the RL and Caputo’s fractional derivatives. We merge the two concepts and developed that which is the RL derivative and the Caputo derivative, and call the modified RiemannLiouville derivative, of Jumarie Type, defined for continuous but non-differentiable functions. We will make the fractional derivative of a constant function zero, by this new modified RL fractional derivative. Thereby, we derive a fractional Taylor series, and define the modified RL fractional derivative for a continuous but non-differentiable function. We have seen in the previous chapters how complicated and difficult the Chain rule, Leibniz’s product rule, and Integration by parts formulas are. Here we shall see how this modified RL fractional derivative can make these formulas as a parallel to the classical formulas, by using approximations of fractional differential increments obtained by a fractional Taylor series-for non-differentiable (but D -differentiable) functions. In addition, we shall introduce the notion that by using the (‘fractal’) Mittag-Leffler function, one can construct fractional integral transforms like the fractional Laplace transforms and the fractional order Gamma function and many others. This new development gives a conjugation to classical calculus for a non-differentiable function or ‘fractal functions’, developed by G. Jumarie; that we will elaborate upon. We will also further generalise the notion of fractional differ-integration when the order instead of the non-integer number (that we discussed throughout) is a ‘continuous distribution function’; and we will try to give a solution to such a differential equation. These are very modern developments and are still under review and while controversial presently, are not necessarily wrong.

9.2 Condition of term-by-term differ-integration of a series Here we will generalise the notion of term-by-term differ-integration as we have used in earlier chapters (in Chapters2, 3, 4, 6). We generalise the condition of its applicability or convergence for classical integer order integration and D

differentiation, and extend this to a fractional calculus context. We will use short form symbol that is D f instead of D

a Dx

> f ( x )@

in this discussion.

9.2.1 Linearity and homogeneity of fractional differ-integral operators We note that the fractional differ-integration is a linear operator meaning that we have following relationship: q a Dx

In short, we write D

q

> f1  f2 @

q

> f1 ( x)  f2 ( x)@

q a Dx

> f1 ( x)@  a Dxq > f2 ( x)@

(9.1)

q

D f1  D f 2 . This comes from the direct application of the formulas for fractional

differentiation and integration. In addition, we re-write that the fractional differ-integration operation has the property of homogeneity. That is: q a Dx

>Cf ( x)@

C



q a Dx

> f ( x ) @

(9.2)

Here C is a constant. With these two (9.1), (9.2) obvious properties we write the following: q a Dx

ª¦ n f j º ¬ j0 ¼

¦  n

j 0

q a Dx

ª¬ f j º¼



(9.3)

This is called term-by-term differ-integration, which we have used several times in previous chapters (in Chapters-2, 3, 4, 6). We want to discuss the conditions that permit the term-by-term differ-integration of an infinite series of functions. Our major objective is to establish the term-by-term differ-integrability of the differ-integrable series that we noted earlier, which is f ( y )

( y  a)

p

¦

f j 0

j

a j ( y  a ) n , with p ! 1 and a0 z 0 (Section 2.22).

380

Chapter Nine

We shall make use of classical results on differentiation and integration of an infinite series term-by-term, that we have discussed in an earlier chapter (Section 1.18). In order to use the results of classical differentiation and integration, we must ensure that the terms f j of the series (9.3) are either continuous or continuously differentiable functions. If we restrict our attention to the summands f j that are from differ-integrable series, the form of such a series that is of f ( y )

( y  a)

p

¦

f j 0

j

a j ( y  a ) n , with a0 z 0 and p ! 1 . Its term-by-term derivative shows that

the requisite continuity assumptions are valid away from the lower terminal (i.e. x differ-integration

a ) that is also a start point for

Henceforth we shall consider the infinite sums of differ-integrable series and establish results on the term-by-term differ-integrability of such sums, which in general are valid on open intervals such as a  x  ( a  X ) , (where X is the radius of convergence of the differ-integrable series). We note here that the classical results are stated for closed intervals rather than open ones; thus, in this case of generalisation, we will use term convergence in an open interval (meaning convergence in every closed subinterval of the open interval).

9.2.2 Convergence of a differ-integration of series We now study some of the facts about convergence. Consider an ordinary power series I aj

I( j ) (a) j!

¦

f j 0

j

a j ( x  a ) , with

; it is convergent for 0 d x  a  X . From here, can we say anything about the series that is formed q a Dx

term-by-term to the

§ I ( j ) ( a ) · * ( j  1) j q ( x  a) ¦ ¨ ¸ j ! * ( j  1  q ) j 0© ¹

(9.4)

through term-by-term differ-integration of the stated power series? We apply the operator series of I to get

q a Dx I ( x )

as the following: f

q

¦a

a Dx I ( x )

* ( j  1) j

*( j  1  q)

j 0

( x  a)

j q

f

( x  a)

q

f

I

( j)

(a)

¦ *( j  1  q) ( x  a)

j

j 0

We know that the series I ( x )

¦

f j 0

j

a j ( x  a ) converges for x  a  X , whereby through the ratio test we have

the following condition: X { lim j nf

aj a j 1

lim

( j  1)I

j nf

I

( j 1)

( j)

(a)

(9.5)

(a)

While differ-integrable series will converge for:

I x  a  lim I j nf

§

© j nf

I( j ) (a) I ( j 1) ( a )



( a ) * ( j  q  2)

( j 1)

t ¨ lim where A { lim j nf

( j)



( a ) * ( j  q  1)

( j  1)I

I

( j 1)

( j)

(a)

(a)

lim

( j  1  q )I

j nf

I

( j 1)

( j)

(a)

(a)

(9.6)

( j)

· § I (a) · ¸  q ¨ lim ¸ X  qA ( j 1) (a) ¹ ¹ © j nf I

. We may examine and write that the possibility is that A is invariably negligible in

comparison with X , so the differ-integrated series converges in an open interval 0  x  a  X . Furthermore, the same result is valid for the differ-integrable series whose series has the same radius of convergence as an analytical part i.e.

j -th term is a j ( x  a )

¦ a ( x  a) j

j

j p

; since this

and thus is valid for a general differ-

integrable series. That is if the differ-integrable series f , which is a finite sum of functions, that is each represented as:

Modified Fractional Calculus in conjugation with Classical Calculus

( x  a)

p

f

¦

j1 0

...  ( x  a )

j

a j1 ( x  a ) 1  ( x  a ) f

( np  n 1) n

¦a jn

jn

( np 1) n

381

f

¦a j2 0

j2

( x  a)

j2

 (9.7)

jn

( x  a)

converges for x  a  X , then so does the series obtained by differ-integrating each unit term-by-term, (except possibly at the start point at x

a ). This is a very important point, which we will follow.

9.2.3 Discussion on convergence by application of the Riemann-Liouville fractional integration formula from term-by-term to a differ-integrable series Let f be any differ-integrable series, since f may be decomposed as a finite sum of a differ-integrable series units such as:

fX

( x  a)

p

f

¦ a ( x  a)

j

p ! 1

j

a0 z 0

(9.8)

j 0

The term-by-term differ-integrability of f will follow from that of fX . We will establish the following for all q i.e. q ª p a Dx ( x  a )

¬

¦

f

f

¦a 

j a ( x  a) º 0 j

j

¼

j

q a Dx

j 0

ª¬ ( x  a ) p  j º¼



(9.9)

We will prove the validity of the above (9.9) expression inside the region (interval) of convergence for the differintegrable series

¦ a ( x  a)

p j

j

.

For q d 0 , we extended the result of the integration in classical calculus for term-by-term integration. Suppose the

¦f

infinite series of differ-integrable functions

i

converges uniformly in 0  x  a  X , then: f

q ª a Dx

¬¦ j

fi º

¼

0

f

¦

q a Dx

> fi @

j 0

qd0

(9.10)

and the RHS series (9.10) also converges uniformly in 0  x  a  X . Let us demonstrate this by taking f {

¦

f j 0

f j and let S N {

¦

N

f j . Since q  0 , and the Riemann-Liouville

j 0

formula which is following: q a Dx

f ( x)

x

1

f ( y )dy

*(q) ³ ( x  y )

q 1

,

x

1

q a Dx f j ( x )

f j ( y )dy

*(q) ³ ( x  y )

a

(9.11)

q 1

a

is valid, we can write it as the following expression: q a Dx

f 

1

q a Dx S N

*(q)

x

³

 f ( y )  S N ( y ) dy ( x  y)

a

(9.12)

q 1

The assumption of uniform convergence means that, given H ! 0 , there is an integer N

f ( y )  S N ( y )  H for n ! N and for all following: q a Dx

N (H ) such that

y in the interval a d y d x with x  a  X . Then we write the

f 

f

¦ 

q D f j 0 a x j

d

1

x

*(q) ³ a



x

1 *(q)

³

 f ( y )  S N ( y ) dy

a

f ( y )  S N ( y ) dy ( x  y)

q 1

H ( x  a)

( x  y) 

H

q 1 x

( x  y) *(q) ³ a

q

q  *(q)

 q 1

dy

(9.13)

382

Chapter Nine

which can be made to be small independently of x in the interval converges uniformly to q a Dx

q a Dx

¦

q a Dx

fj

f in the interval 0  x  a  X . The above (9.13) result shows that the equation

ª( x  a) p ¦ f a j ( x  a) j º j 0 ¬ ¼

integrable series the operator

0  x  a  X . This shows that

q a Dx

¦

f j 0



aj

q a Dx



ª¬ ( x  a ) p  j º¼ is valid for

q d 0 and thus if any f is a differ-

may be distributed through the several infinite series that define f as long as

q d 0.

We apply the formula

q a Dx

* ( p 1) * ( p 1 q )

ª¬ ( x  a ) p º¼

q

( x  a)

p q

f

* ( p  j  1)

¦a

a Dx fX

j

j 0

, with p ! 1 and write the following:

* ( p  q  j  1)

( x  a)

p j q

qd0

(9.14)

The equation q a Dx

ª( x  a) p ¦ f a j ( x  a) j º j 0 ¬ ¼

¦

f j 0

aj



q a Dx

ª¬( x  a ) p  j º¼



and the above (9.14)are also valid for q ! 0 ,

which we will discuss now.

9.2.4 Discussion on convergence by application of the Euler fractional derivative formula for power functions term-by-term into differ-integrable series We decompose the series for fX into two parts that are: f

fX

p j

¦ a ( x  a)

¦ a ( x  a)

j

p j

j

j 0



j J1

¦ a ( x  a)

p j

(9.15)

j

j J2

Where J1 is a set of non-negative integers j for which * ( p  q  j  1)

f and J 2 is a set of all non-negative

integers not in J1 , here * ( p  q  j  1)  f means finite. For a fixed, q the properties of the Gamma function ensure that the set J1 has only a finite number of elements, (the Gamma function is infinite for zero and has negative integer points). With this observation, we write: q a Dx fX

q a Dx

¦

ª¦ ¬ j j J1

aj

p j º º  a Dxq ª ¦ a ( x  a) ¼ ¬ j J2 j ¼ q p j ª º¼  a Dxq ª ¦ j J a j ( x  a ) p  j º a Dx ¬ ( x  a ) 2 ¬ ¼

J1



a j ( x  a)

p j



(9.16)

In the above (9.16) expression we have made use of the linearity of the operator a Dxq . The first term of (9.16) will be zero since in this set i.e. J 2 , we have the gamma function expression * ( p  q  j  1) denominator. Now we give proof for the validity of

q a Dx fX

¦

f j 0

aj

* ( p  j 1) * ( p  q  j 1)

( x  a)

p j q

¦

aj

f , which will appear in the

in the case of q ! 0 , by using the

following steps, for q ! 0 : q a Dx

ª¦ ¬ j

J2

a j ( x  a)

¦

j J2

p j

aj

º ¼

j J2

* ( p  j  1) * ( p  q  j  1)



q a Dx

ª¬ ( x  a ) p  j º¼

( x  a)

p j q

(9.17)

Here we assume the series for fX converges uniformly in 0  x  a  X , as will the series on the RHS of the above (9.17) expression, as we have proved at the start. Thus, the operator a Dx1 may be distributed through the terms of this series to give the following steps:

Modified Fractional Calculus in conjugation with Classical Calculus

1 a Dx

ª «¦ j ¬

J2

* ( p  j  1)

aj

* ( p  q  j  1)

¦

( x  a)

p  j q

º » ¼

* ( p  j  1) ª p  j q º « a j * ( p  q  j  1) ( x  a ) » ¬ ¼

1 a Dx

j J2

* ( p  j  1)* ( p  q  j  1)

¦a

j

¦a

j

* ( p  q  j  1)* ( p  q  j  2)

j J2

* ( p  j  1) * ( p  q  j  2)

j J2

¦a  j

383

q 1 a Dx

j J2

( x  a)

ª¬ ( x  a ) p  j º¼

( x  a)

p  j  q 1

(9.18)

p  j  q 1



The last series in the above (9.18) steps converge uniformly in 0  x  a  X , as does the series obtained from it by differentiating each term. We apply a classical result on a term-by-term differentiation as we had already discussed in

¦a 

d ª

¦ 



ª¬ ( x  a ) p  j º¼ that gives: º q 1 ª¬ ( x  a ) p  j º¼ » ¦ a j a Dxq ª¬ ( x  a ) p  j º¼ a Dx ¼ j J2

the previous chapter (Section 1.18), to the series

j

q 1 0 Dx



« aj dx ¬ j J 2 Similarly, we can have, for every positive integer, n the following: n

ª « ¦ aj dx ¬ j J 2 d

n

Again choosing n ! q permits us to use



q n a Dx

D

a Dx

º ª¬ ( x  a ) p  j º¼ » ¼

¦a 

¦

fX

j

j 0

aj

j J2

* ( p  j 1) * ( p D  j 1)

( x  a)

ª¬ ( x  a ) p  j º¼

p  j D

(9.19)



(9.20)

, for D d 0 as discussed in

qn:

(Section 2.14), to give the following result, with D

¦a 

f

q a Dx

j



q n a Dx

j J2

ª¬ ( x  a ) p  j º¼



qn a Dx

ª º p j « ¦ a j ( x  a) » ¬ j J2 ¼

(9.21)

We differentiate both sides of (9.21) n times and we write: d

n

dx

Utilising

dn dx n

¦

j J2

aj



n

¦a  j

q n a Dx

j J2



q n a Dx

ª¬ ( x  a ) p  j º¼





q a Dx

ª º p j « ¦ a j ( x  a) » ¬ j J2 ¼

(9.22)



ª¬ ( x  a ) p  j º¼ ¦ j J a j a Dxq ª¬ ( x  a ) p  j º¼ we finally write: 2 q ª p j º q p j a D ª ( x  a ) º¼ q!0 a Dx ¦ j J a j ( x  a ) 2 ¬ ¼ ¦ j J2 j a x ¬





(9.23)

Therefore the expression that is described earlier (9.17): q a Dx

ª¦ ¬ j

J2

a j ( x  a)

¦

j J2

aj

p j

º ¼

¦

j J2

aj

* ( p  j  1) * ( p  q  j  1)



q a Dx

( x  a)

ª¬ ( x  a ) p  j º¼ p  j q

(9.24)

is valid for q ! 0 hence for any arbitrary order q . We have thus proved the term-by-term fractional differ-integration of the arbitrary differ-integrable series. In the above discussions while proving this we established a generalisation (of a Dxq for any q d 0 ) of the classical calculus result on a term-by-term integration valid for the uniform convergent series. Can this classical result of integration be generalised for q ! 0 ? The answer is yes. If the infinite series for q a Dx

ª ¦ f fX º ¬ j0 ¼

f

¦f

¦  j 0

j

q a Dx

as well as the series

> fX @

¦

q a Dx

ª¬ f j º¼ converge uniformly in 0  x  a  X , then

for q ! 0 and for 0  x  a  X is valid. We note here that, this is a very

384

Chapter Nine

natural extension of the classical result on a term-by-term differentiation, which also requires an addition to the convergence of

¦f

j

¦

which requires the uniform convergence of

d dx

ª¬ f j º¼ .

We use this derivation for a differ-integration of the analytical function M ( x ) represented by series say

M

¦

M( j ) (a) (x j 0 * ( j 1)

f

j

 a ) as: f

a Dx >M @

D

¦M

( j)

(a)

j 0

* ( j  1) * ( j  D  1)

( x  a)

j D

(9.25)

The series above converges uniformly in 0  x  a  X , with X as the radius of convergence.

9.3 Composition rule in fractional calculus We will discuss the validity between the operations



q a Dx

Q a Dx

> f ( x )@

discussions, we assume that f ( x ) is a differ-integrable function along with D

D

a Dx

short form of symbol that is D f instead of

and

q Q a Dx

Q a Dx

> f ( x )@

> f ( x )@ .

For meaningful

as well. We will use the

> f ( x)@ in this discussion. We restrict our discussion to the differ-

integrable series. We have written that the most general non-zero differ-integrable series is a finite sum of differ-integrable units each having a form:

fX ( x )

( x  a)

p

f

¦ a ( x  a)

j

p ! 1

j

a0 z 0

(9.26)

j 0

We will find out that the composition rule may be valid for some units of f ( x ) but possibly not for others.

D  Q D Q f to be satisfied

9.3.1 Condition for the inverse operation f q

It follows from the linearity of differ-integral operators that

Q

D D f q

Q

unit fX of f . First, we see the validity of the composition D D fX

but D fX

q Q

f if D q D Q fX

D

q Q

fX , for every

fX , for a differ-integrable series unit

0 for every Q , and we write D D

composition is trivially satisfied for the differ-integrable function fX Q

D

q Q

q

0 then D Q fX

function fX . Obviously if fX

D

Q

> 0@

D

q Q

> 0@

0 . While this

0 we will see that the possibility when fX z 0

q Q 0 which is exactly the condition that will negate the composition D D fX

D

q Q

fX . Now take fX z 0

and evaluate D Q fX as follows: f

Q

f

¦ a j D ª¬( x  a) p j º¼

D fX

Q

¦a

j 0

j 0

* ( p  j  1) j

*( p  j  1  Q)

( x  a)

p  j Q

(9.27)

Here we note that since p ! 1 we have p  j ! 1 . This makes * ( p  j  1) always finite and non-zero. The Q D fX will vanish only when the coefficients a j are zero or the denominator

individual terms of the series

f is infinite.

comprising of * ( p  j  1  Q )

Therefore, we write a necessary and sufficient condition for

Q D fX z 0 when for j from 0 to f , i.e. we have * ( p  j  1  Q )  f , that is finite for which a j z 0 . We apply

D

Q

to the obtained D Q fX (9.27) and write:

D

Q

Q

D fX

f

¦a

* ( p  j  1) j

j 0

*( p  j  1  Q)

f

¦a f

¦ a ( x  a) j

Q

ª¬ ( x  a ) p  j Q º¼

§ * ( p  j  Q  1) · p  j Q  Q ( x  a) ¨ ¸ *( p  j  1  Q) © *( p  j  Q  1  Q) ¹ * ( p  j  1)

j

j 0

j 0

D

p j

fX

(9.28)

Modified Fractional Calculus in conjugation with Classical Calculus

385

The cancellation of * ( p  j  1) and * ( p  j  1  Q ) is due to these being finite. This condition may be shown to be equivalent to: Q Q (9.29) fX  D D fX 0 meaning that the condition of the differ-integrable units fX be re-generated upon this inverse operation, that is first doing D

Q

and then following it with D

Q

.

9.3.2 Condition D q D Q f Assume that we can apply the operator D

q

D q Q f to be satisfied

on the obtained D Q fX that is the above series (9.27).We will get the

following: q

f

Q

¦a

D D fX

* ( p  j  1)* ( p  j  1  Q ) j

j 0

* ( p  j  1  Q )* ( p  j  1  Q  q ) q a Dx

The above (9.30) is obtained by application of Q

D fX

( x  a)

p  j Q  q

f

> fX @ ¦ j 0 a j **( p( p q j j1)1) ( x  a) p  j q ,

is assumed to be differ-integrable series.

With the condition

fX  D

Q

(9.30)

q d 0 is valid since

Q

D fX

0 , that means

* ( p  j  1  Q )  f (finite) in effect we may cancel the * ( p  j  Q  1) and re-write (9.30) as: q

f

Q

¦a

D D fX

* ( p  j  1) j

j 0

*( p  j  1  Q  q)

( x  a)

p  j Q  q

(9.31)

On the other hand, the same steps show that:

D

q Q

f

¦a D

fX

q Q

j

fX

j 0

(9.32)

f

* ( p  j 1) p  j Q  q a j * ( p  j 1Q q ) ( x  a )

¦ j 0

Thus the composition of D q D Q fX

D

q Q

fX is satisfied as long as

fX  D

Q

q

q

Q

D D fX 0 is satisfied or in other terms

D fX

* ( p  j  1  Q )  f i.e. finite. This condition implies that they are invariably satisfied for Q  0 . Take Q

 M for any M ! 0 , and as p ! 1 is

our condition for differ-integrable units fX . Then p  j  1  M ! 1 and the * ( p  j  1  M )  f will be finite. However, when the condition fX  D  Q D Q fX is not necessary that the case D q  Q fX For example for fX

x

 12

, a

¬

Q

Q

q

Q

D

fX

D

Q

0

q Q

fX  D q

Q

q

1

D D fX z D D

0 . On the other hand it

x

2

1

2

q Q

q Q

D D fX

* (0)

 12

> 0@

 12 , then we observe the following: *  12

¼

D fX

D D fX q Q

, and q 1

fX  D

D

1 2

1  D 2 ªx 2 º

Q

D fX

q Q 0 so that D D fX

0.

0, Q

q

Q 0 is violated, D fX

D

x

1

 12

0 1

1

1  12  1 2 ª *  2 º 1 1 x 2 z0 ¼ x  D ¬« * 0 ¼» x

 D 2 ªx 2 º

¬

0 fX

1

 0 D ªx 2 º

¬

1

 2 ¼ x z0

fX

ª¬ fX  D Q D Q fX º¼ D  D

(9.33)

q Q

fX  D

q Q

1

1 2 2

1

1

1

1

ª x 2 º  D 2 2 ª x 2 º 0 ¬ ¼ ¬ ¼

ª¬ fX  D Q D Q fX º¼

386

Chapter Nine

So the condition fX  D  Q D Q fX and D q D Q fX q

0 while D

Q

D D fX and D

q Q

q Q

Q

0 is certainly violated in the above (9.33) example. Therefore, we have D fX

fX z 0 .

0

Generalising this observation, we easily see the relationship between

fX in the case fX  D

Q

Q

D fX z 0 , to be:

0

q

Q

D D fX

D

q Q

fX  D

q Q

ª¬ fX  D  Q D Q fX º¼

(9.34)

The condition of composition regarding differ-integrable units fX is listed in Table 9.1 Condition Q

D fX

fX

fX  D

0

0 Q

fX z 0 Q

D fX

0 q

Q

q Q

Q

Q

D fX z 0

Q

D D fX D

D fX 0 Not possible

Q

q

0

D D fX D fX z 0

fX  D

q Q

fX  D

q Q

fX  D q

Q

ª¬ fX  D  Q D Q fX º¼ Q

D fX

Q

D D fX

D

q Q

0 fX

Table 9.1: Composition rule for differ-integrable units. The requirement for this rule is that fX and D Q fX are both differ-integrable While the above (9.34) expression is trivial identity for differ-integrable units fX , we shall see its usefulness in q

Q

general differ-integrable series. As the equation D D f q

Q

only if D D fX

D

q Q

D

q Q

f is valid for a general differ-integrable series if and

fX is valid for every differ-integrable unit fX of f . It is straightforward to apply the theory

just discussed for units fX , to write the composition rule for general f . The only difference is that; while the condition fX z 0 and fX  D  Q D Q fX

Q 0 for units guaranteed that D fX z 0 , this is no longer true for arbitrary

case. The reason of course is that some units of f may satisfy the condition fX  D  Q D Q fX q

0 for fX z 0 while

Q

other units may not. This will make it possible to violate the composition rule that is D D f

D

q Q

f even though

Q

f z 0 and D f z 0 . The condition f  D guarantee q

Q

D D f

Q

D f

Q

q Q

f (as was the case for differ-integrable units fX ) is valid at least when Q  0 and even when Q  1

D

q Q

0 for a general differ-integrable series, f is however still necessary and sufficient to

q

D D f D

Q

for a function bounded at x

f . We mention here that for a general differ-integrable

f

the condition is

a . For a general differ-integrable series, f we summarise this as in Table 9.2

Modified Fractional Calculus in conjugation with Classical Calculus

Condition Q

D f

f f D

0

0 Q

f z0 Q

D f

Q

D f z0

q

Q

f D

0 q

387

0

Q

D D f

D

Q

D D f

q Q

D

f

q Q

q Q

D f 0 Not possible

Q

D f z0

Q

ª¬ f  D  Q D Q f º¼

Q

If f  D D f then q

Q

D D f

D

0

q Q

Q

Q

f D

q Q

f

If f  D D f z 0 then q

Q

D D f

D

q Q

ª¬ f  D  Q D Q f º¼

Table9.2: The composition rule for an arbitrary differ-integrable function f . The requirement for this rule is Q

f and D f both of which are differ-integrable We q

have Q

D D f

noted D

q Q

just

f D q

q Q

earlier

that

Q

where

the q

Q

composition

Q

ª¬ f  D D f º¼ relates D D f to D

q Q

rule

is

violated

the

equation

f . The utility of this expression is there for

Q

calculating the D D f i.e. the LHS in marginal general cases.

9.3.3 Condition for D N D q f

D q D N f , N as positive integer

Let us take one marginal general case with Q N as a positive integer. Indeed, we may then use the following equations that we have developed for a classical case in the earlier chapter (Section 1.18): N a Dx

ª¬ a Dx n f º¼

N n a Dx

n a Dx

f

ª¬ a DxN f º¼ n 1



( x  a)

¦

k!

k n N q a Dx

* ( p  1)

ª¬( x  a ) p º¼

* ( p  q  1)

( x  a)

In the above (9.35), with the change of index numbers as k  N  n N a Dx

ª¬

n a Dx

f º¼

N n a Dx

p q

f

( k  N n)

(9.35)

(a)

p ! 1

m we write the following:

f n

a Dx

Put n

k

N 1

( x  a)

(9.36)

m N n

(m) ª¬ a DxN f º¼  ¦ f (a) ( )! m  N  n m 0

 q and we get: N a Dx

ª¬ a Dxq f º¼

N q a Dx q

a Dx

Writing ( m  N  q )!

N 1

( x  a)

m N q

(m) ª¬ a DxN f º¼  ¦ f (a) m 0 ( m  N  q )!

(9.37)

* ( m  q  N  1) and rearranging the above (9.37) expression we get the following: q ª N º a Dx ¬ a Dx f ¼

With the above (9.38) expression and with expression as follows:

f

q

N

D D f

D

N q

a Dx

N 1

f 

( x  a)

m N q

¦ (m  N  q)! f

(m)

(a)

(9.38)

m 0 q N

f D

q N

ª¬ f  D  N D N f º¼ we write the desired

388

Chapter Nine q

N

D D f

D

q N

D

f D q N

q N

ª¬ f  D  N D N f º¼

N 1

f 

¦

( x  a)

For q

and q



1 2

with a

0 , for N

1

ªdN f º 1 « N » dx 2 ¬ dx ¼ d

d

2

 12

dx

1 , f (0) z f with *   12

(9.39)

(a)

1, 2, 3, ... and f (0) z f we have the following:

d

With N

(m)

f

* ( m  q  N  1)

m 0 1 2

mq  N

dx

ªdN f º « dx N » ¬ ¼

 12

N  12

d



N  12

N  12

dx

2 S , and *  12

N 1

f

x

m  N  12

f

(m)

1 2

m 0

f

N  12

(0)

¦ * m  N  N 1



x

m  N  12

(m)

f

(9.40)

(0)

¦ * m  N  3 2

m 0

S we have the following compositions:

ª d > f @ º d 2 f x  2 f (0) d 2 f x  2 f (0)   1 « 3 3 » *   12 2 S dx 2 ¬ dx ¼ dx 2 dx 2 1 1 1 1 1 d 2 ª d > f @ º d 2 f x 2 f (0) d 2 f x 2 f (0)   1 1 » 1 « *  12 S dx 2 ¬ dx ¼ dx 2 dx 2 d

1

3

2

3

3

3

(9.41)

The above (9.40), (9.41) derived equation can be established under the relaxed assumption of differentiability. For p

instance, functions such as x , with p d 1 are differentiable. On differentiation, we get px fractionally differ-integrable. Thus, we can have f be N fold differentiable and that f

0,1, 2,...( N  1) . If all these f ( a ) , f

m

q

then D D

N

D

q N

(1)

(a) , f

(2)

(a) , … f

( N 1)

p 1

(m)

; but this is not

( a ) be finite, for

( a ) are zero, that is the values at the start point,

are valid.

The Table 9.3 gives a summary of the above discussion, the composition rule for the functions that are, N fold differentiable, and its N th derivative are differ-integrable. Condition N

D f

0

f f D

0 N

f z0 N

D f

0 q

N

D f z0

q

N

f D 0

N

D D f

N

D D f

D

q N

N 1

f 

¦

m 0

( x  a ) k q  N f ( m ) ( a ) * ( m  q  N 1)

q N

N

D f z0

D f 0 Not possible

N

N

If f  D D f then q

N

D D f

D

q N

N

N

f 

¦

0 f

If f  D D f z 0 then q

N

D D f

D

q N

N 1

m 0

( x  a ) mq  N f ( m ) ( a ) * ( m  q  N 1)

Table 9.3: The composition rule for the differentiable but not essentially the differ-integrable function f . The N

requirement for this rule is f is N times differentiable and D f is differ-integrable. The general rule was just discussed regarding composing D Q

D and D

Q

Q

q

with D , which makes it clear that while the operators

are ‘usually’ inverse to each other, this is not always the case.

Let us choose fX

x , then D 2 fX

2 2 0 and D D fX

2

2

0 . So certainly in this case D and D are not the inverse of

each other. This difficulty is due to the condition fX  D  Q D Q fX

0 which is violated for the differ-integrable unit

Modified Fractional Calculus in conjugation with Classical Calculus

i.e. fX

fX

389 3 2

x . Nor is this problem restricted to integer q and Q . In fact by choosing Q 1

3

Q

x 2 , we see that

D fX

However, we know that D

q Q

D 2x

1

q

0

fX

Q

0 . So that D D fX

2

D fX

1

fX

q a Dx

x 2 . We used

in this case, too, we chose fX z 0 but D Q fX

D

3

3

D 2x

2

1

0 , and

2

fX  D

* ( p 1) * ( p  q 1)

ª¬ ( x  a ) p º¼

 32 , a

, q

( x  a)

Q

Q

D fX

p q

0 and x

1

2

z 0.

, p ! 1 . Here

Q Q 0 , which guarantees the violation of fX  D D fX

0 . The

Q

composition rule fails even if f z 0 and D f z 0 . For example, choose f ( x ) 1

1 2

and D f

x

1

2

x

, D 1 D1 f

1

2

 1, N

1, a

1

1

0 and arbitrary q . Here f is the sum of two differ-integrable units

1

x 2, f D D f

1 yet: q

*  12

1

D D f

2*   q 

1 2

x



 q  12

(9.42)

and:

D

q 1

*  32

f

*  q  *

1 2

1

1

f D D f

Q

Q

1 2

x





2*   q  The condition f  D

1 2



x

 q  12

 q  12

1



1



x

*(q)

x

*(q)

 q 1

 q 1

N

D

q N

N 1

¦

f 

( x  a)

1

q

D D f

correctly relates

D

q 1

f 

q

1

D D f f (0) *(q)

x

and

 q 1

D

q 1

f .

. Putting q

ª d > f @º 1 « » dx 2 ¬ dx ¼ d

1

2

Realising 1 2

1

f 

f

(m)

(a)

*(q)

x

(9.44)

 q 1

1 , we may write from the above (9.44)

f (0)

, we have the following:

d

3

2

>f@

dx d

3

2

3

2

>f@

dx Now take q

D

q 1

mq  N

* ( m  q  N  1)

m 0

D D f

x

1 and thus the equation: q

1

*(q)

 q 1

0 is violated in this example as we demonstrated above that is we get in this case

D f

D D f

q

D D f 

(9.43)

1

1

q

3

2

f (0) *  f (0)

2 S

1 2

x

x





 1 2 1

*   12

2 S (9.45)

3

2



 12 , and we get the following

d

 12

dx

 12

ª d > f @º « dx » ¬ ¼

d

1

2

>f@

dx d

1

2

1

2

>f@

f (0) *

1 2



x

  1 2 1

*  12

S (9.46)

f (0)

1

Sx dx 2 The LHS of the above (9.46) expression is the Caputo’s fractional derivative and we have earlier seen the relationship between the RL and Caputo derivative (Section 3.10, 5.13). When the order of the Caputo derivative is half we get 1

from the above (9.46) by properly re-writing the expression 0 Dx 2 > f ( x )@

C 12 0 Dx

> f ( x)@  f S(0)x .

9.3.4 Generalising the composition D q D Q f to D q Q f by initial values of f We generalise the composition of (9.44) by the constants

f (m) (a) * ( m  q  N 1) q

N

D D f

as Cm and get the following: D

q N

N 1

f 

¦C

m 0

m (x

 a)

mq  N

(9.47)

390

Chapter Nine

Say N is chosen as an arbitrary non-integer i.e. Q , with N  1  Q  N , where N is a positive integer, then we have the following expressions: q

Q

D D f

D

q Q

N 1

f 

¦

( x  a)

Q

D D f

D

q Q

N 1

f 

f

(m)

(a)

* ( m  q  N  1)

m 0 q

mq  N

¦C

m (x

 a)

(9.48)

mq  N

m 0

Say q

1 2

Q

, then we have N

1 , so we get the composition: 1

1

D 2D 2 f

D

1

2

 12

11

f 

¦C

m (x

 a)

0  12 1

1

D f  C0 ( x  a )

 32

(9.49)

m 0

For q

Q

D and D  1 , we have N

2D

1 , then the sequential derivative D D

2D

D

f

D

D D f

D

2D

f 

is:

( x  a)

D 1

f (a)

(9.50)

*  D

This above (9.50) argument we used in an earlier chapter (Section 7.12).In addition, we note that if f ( a ) D

function value at the start point of differ-integration in the above (9.50) case, then D D

9.4 The reversibility of a differ-integral operator D Q f

D

D

2D

g to f

0 that is a

holds.

D Q g

The discussion of the above sections reveals the danger when one asks about the commutability of differ integral q

Q

Q

q

Q

Q

operators D D D D or the irreversibility of differ-integration that is if D f g then f D g . Of course, this reversibility comes into play while solving fractional differential equations. The above results of various discussions may be used to derive conditions under which commutability holds, and as always, the fractional differintegral operators are applied to a suitable class of functions. Q

Consider the simplest fractional differential equation D f F , where Q is arbitrary. The function F is a known one, and f is an unknown function. Here we choose the lower limit of differ-integration as zero that is a 0 . We Q

apply inversion and write the following as f D F . Nevertheless, we have used this in previous chapters (Section 5.8), to get the solutions of fractional differential equations. With this we get perhaps a solution, but unfortunately this is not the most general solution. In fact, referring to the previous section regarding the composition we say that it is precisely the condition f  D

Q

Q

0 , which guarantees the adherence to the compositional rule for a general

D f

differ-integrable series. In general, the difference f  D

Q

Q

D f will not be zero, but will contain of those parts of the Q

differ-integrable series units fX in f that are sent to zero under the operation D . For example in the case of f

1

0 Cx  x 2 , with Q

1

1

1

operation D , but appeared in f  D D f write:

N

0

1 , the differ-integrable unit that is Cx was sent to zero by

0

Cx , in the example that was shown earlier (9.33). In general, we can f D

N

N

D f

C1 x

N 1

 C2 x

N 2

 ....Cm x

N m

(9.51)

N

with ( m  1) terms which goes to zero upon operation D . We decompose f into differ-integrable series units fX ,i where:

fX ,i

x

p1

f

¦a x

j

pi ! 1

ij

j 0

ai 0 z 0

We now investigate the condition on fX ,i that is required to give f  D earlier section i.e. fX  D

Q

Q

D fX

* ( pi  Q  1)

*( p  j  1  Q)  f

0 . This tells us that f  D

i.e. finite for each Q

j

Q

i

1, 2,...n

(9.52)

Q

D f z 0 . We have conditions in the

for which a j z 0

was equivalent to

Q

D f z 0 obtains i in the range of 1 d i d n , for which the term Q

f , that is not finite (which in turns makes D fX ,i

0 ). This condition can occur only when

Modified Fractional Calculus in conjugation with Classical Calculus

391

pi (Q  1), (Q  2), ..., (Q  m) that is the Gamma function values at zero, and negative integer points. Putting these facts together, we say the most general case is: Q Q Q 1 Q2 Qm (9.53) f  D D f C1 x  C2 x  ...  Cm x Where C1 , ...., Cm are arbitrary constants and 0  Q d m  Q  1 ; and for Q d 0 the RHS members of the above Q

F ):

(9.53) equation is zero. Thus, we have the following (remembering that D f f  C1 x

Q 1

 C2 x

Therefore, we write the most general solution as: f

D

Q2

Q

 .....  Cm x

F  C1 x

Q 1

Qm

 C2 x

D

Q2

Q

Q

D f

D

 ....  Cm x

Q

(9.54)

F

Qm

(9.55)

We note the presence in this general solution (9.55) of m arbitrary constants where 0  Q d m  Q  1 or m Q d 0 . For Q

1 , we have f

calculus result. With Q

1

0 for

0

D F  C1 x , where constant C1 is the integration constant, we recover our classical 2 0 D F  C1 x  C2 x , the same as in classical calculus.

2 , we obtain the solution as f 3

5

Let us take the example of a simple equation D 2 f

f

D

 32

x , whose general solution as per (9.55) is: 1

ª¬ x 5 º¼  C1 x 2  C2 x 

1

* (6)

2

*

15 2



x

13

1

 C1 x 2  C2 x

2



Q

containing two arbitrary constants namely C1 and C2 . Now consider D f  b D

Q 1

arbitrary and b is a known constant, and F is a known function. We apply the operator D given equation, which thereafter yields the following steps:

D

 ( Q 1)

Q

D f  bD



1

D f  b f  B1 x 1Q

1

D f  bf d f ( x)

D

 ( Q 1)

( Q 1) 1

D

 B2 x

F  C1 x 1Q

d

 bf ( x )

( Q 1)

Q2

f

D

( Q 1)  2

 C2 x

F ( x)

 ( Q 1)



(9.56)

F ( x ) , where Q is

 ( Q 1)

1Q

that is D

to this

F

 ....Bm x

Q 3

f

 12

( Q 1)  m

 .....  Cm x

Q2



1Q

D

F (9.57)

Q  m 1

Q 3

Q  m 1

 C1 x  C2 x  ...  Cm x dx dx We got the first order ODE equation, and we can now invoke the standard methods of classical calculus. 1Q

The above two (9.56), (9.57) examples are quite general. The solution of an even slightly more general equation say q

Q

F will encounter great difficulty with this approach, unless the difference q  Q is an integer or half-

D f D f integer.

Q

Q 1

In the above (9.57) example we have demonstrated that when q  Q n , that is D f  aD f F , q { Q and Q { Q  1 which leads to an n -th order ODE. In our example (9.57), we have n 1 . We extend our above (9.57) logic to the following FDE: 1

D 2f  f

Applying in (9.58) D 1

1

D D f 2

2

1

2

1

1

1

to both sides we get D 2 D 2 f  D 2 f

1

D f  C0 ( x  a )

 32

Now apply the composition rule that is

, which we have derived in an earlier i.e. (9.49).Here as per (9.49) we have a 1

1

call C0 of (9.49) as C1 to write D 2 D 2 f 1

0.

(9.58)

0

subtract the original one D 2 f  f

1

D f  C1 x

 32

1

. We now get D f  C1x

 32

1

D 2f

0 and

0 . From this, we

0 , and write the ODE as: 1

D f  f

C1 x

The above equation (9.59) has a standard solution i.e. f ( x )

d f ( x)

 32

dx x

Ae  e

x

³

x

0

y

 f ( x)

e C1 y

 32

C1 x

dy .

 32

(9.59)

392

Chapter Nine

9.5 An alternate representation for fractional differ-integration for real analytic functions 9.5.1 A representation using the fractional differ-integral formula of the Riemann-Liouville To some extent we have discussed this concept in an earlier chapter (Section 2.11). The purpose here is to discuss alternate representations for a Dxq ; for real analytical functions M that have a convergent power series expansion in q a Dx

the interval a d y d x . We restrict here the RL definition with q  0 that is Therefore, we write (with X

1 *(q)

f

³

x

a

( x  y)

 q 1

f ( y )dy .

x  y ) the following: q a Dx M

x

1

M ( y )dy

*(q) ³ ( x  y )

q 1



f ( x)  y f



y2 2!

³

*(q)

a

Using the Taylor expansion that is , f ( x  y )

xa

1

f

(1)

( x) 

(2)

( x )  ......

(2)

M ( x  X )dX X

0

q 1

(9.60)



( x )  .... we write the expansion for

M ( x  X ) as follows: X

(1)

M ( x  X ) M  XM ( x ) 

2

f

¦

k

k

( 1) X M

(k )

( x)

(9.61) 2! k! k 0 The above (9.61) expansion has no remainder term since we have assumed M to have a convergent power series

M

q a Dx M

expansion and since such an expansion is unique. We insert this in the expression of

as in the RL definition

and then write after doing term-by-term integration as shown in the following steps: xa

1

q

a Dx M

*(q)

³

M ( x  X )dX X

0

1 *(q) k

f

q

q 1

¦

( 1) M

k q

M

k

³X

k

( 1) X M

(k )

k!

0 k 0

( k ) xa

k!

k 0

( 1) ( x  a )

³¦

*(q) k

f

xa f

1

1

X

q 1

dX

k  q 1

dX

(9.62)

0

(k )

¦  *(q) (k  q)k !

a Dx M

k 0

Analyticity in the q argument that we have noted in earlier chapters too (Section 2.11, 2.14), may be used to establish the formula

q a Dx M

¦

( 1)k ( x  a )k q M ( k ) k  0  * (  q ) ( k  q ) k !

f

for all q , though it was derived for q  0 , that is the RL fractional

integration formula.

9.5.2 Representation using the Grunwald-Letnikov formula for differ-integration We start another formulation of this

q a Dx M

by starting with the following derived definition for any arbitrary q

(Section 3.14): q

d f q

>d( x  a)@ Then

write

n

j  q 1

¦  j 0

the

Cj



following j

Cm



m  q 1

by

Cm



using nq

§  xN a  q lim ¨ N nf ¨ * (  q ) ©

M ( x  jy )

¦

N 1

*( j  q)

¦ *( j  1) 

f x  j  xN a

j 0

j m 0

Gm (M , x, y )





j



Cm and

· ¸¸ ¹ *( j q ) * (  q ) * ( j 1)

(9.63) j  q 1

Cj ,

Cn  m . These we have discussed in an earlier chapter (Section 1.14) while

discussing Power numbers, (and Stirling numbers), that we again use below:

Modified Fractional Calculus in conjugation with Classical Calculus

q

a Dx M

( x  a)

q

§

lim ¨ N

N nf

( x  a)

q

N 1

·

*( j  q) ¦ *(q)*( j  1) M  x  j  ¸

q

a x N

©

¹

j 0

§

lim ¨ N

N nf

393

q

©

N 1

· § j  q  1· f § j · a x ¦ ¨ j ¸ ¦ ¨ m ¸ Gm M , x, N ¸ ¹m 0© ¹ j 0© ¹

(9.64)

N 1 § q f § j  q  1· § j · · q ( x  a ) lim ¨ N ¦ Gm M , x, aN x ¦ ¨ ¸¨m¸¸ N nf j ¹© ¹¹ j 0© © m0 § q f § m  q  1 ·§ N  q  1 · · q ( x  a ) lim ¨ N ¦ ¨ Gm M , x, aN x ¸ ¸¨ ¸ N nf © m 0 © m ¹© N  m  1 ¹ ¹

Now we use the expression Gm (M , x, y )

¦

f

m! > m @ k ( k ) S y M ( x) k 0 k! k

as discussed earlier (Section 1.14, while discussing

Power numbers, Stirling numbers), to write the following: q a Dx M

§ q f § m  q  1·§ N  q  1 · f § a  x ·k >m@ M ( k ) ( x) · ( x  a ) lim ¨ N ¦ m !¨ ¸ ¸¨ N  m  1¸ ¦ ¨ ¸ Sk N nf ¨ m ¹© k ! ¹¸ ¹k m© N ¹ © m0 © k q f *(m  q) f § qk § N  q  1 · · · > m@ ( k ) § k ( x  a) (  1) S k M ( x ) ¨ lim ¨ N ¨ ¦ ¦ ¸¸¸ k! © N  m  1¹ ¹ ¹ m 0 *(q) k m © N nf © q

(9.65)

The summation on k involves only one non-zero term by virtue of the following: § q 1 § N  q  1 · · °­ * ( m1q 1) , k m lim ¨ N ¨ ¸¸ ® N nf © N  m  1¹ ¹ °¯ k!m 0, ©

(9.66)

Which follows from the following identity described previously (Section 1.11 and 1.12):

­f, § c  q 1 * ( j  k  q ) · ° lim ¨ j ¸ ® 1, j nf *( j  k  1 ¹ ° © ¯ 0,

This results in the following form:

m

f

q a Dx M

¦

* ( m  q ) ( 1) ( x  a ) *(q)

m 0

( 1) ( x  a )

mq

M

(m)

> m@

Sm

m!

m

f

mq

c!0 c 0 c0

M

(m)

(9.67)

( x)

* ( m  q  1)

(9.68)

( x)

¦  *(q)  (m  q)m !

m 0

> j@

which we have obtained by use of S j (9.68) expression is equivalent to

1 for all j and the recurrence relationship i.e. * ( x  1) x* ( x ) . The above

q a Dx M

¦

( 1)k ( x  a )k q M ( k ) k 0 * (  q )( k  q ) k !

f

(9.62). This alternate derivation is to show how is it

possible to handle the definition that is: q a Dx

§  xN a  q lim ¨ N nf ¨ * (  q ) ©

> f ( x )@

N 1

·

*( j  q) ¦ *( j  1) f  x  j  ¸¸ xa N

j 0

(9.69)

¹

In general, however, as with this example, the manipulations required us to use the above (9.62) formulation for an approach that employs the RL formula that is

q a Dx

1 *(q)

f

³

x

a

( x  y)

 q 1

 f ( y ) dy

and which is then followed by

an analytic continuation argument (which we have described in an earlier chapter, Section 2.14). However, the above (9.69) formula we have written in a more concise form i.e. following in Section 2.14: f * (1  k  q ) q k qk ( 1) >1@ M ( k ) a Dx M a Dx  *(q) (k  q)k ! k 0 (9.70) f §q· q qk k ¨ k ¸ a Dx >1@ a Dx M a Dx M k 0© ¹



¦ ¦









394

Chapter Nine

9.6 The fractional derivative for non-differentiable functions is defined D

9.6.1 Defining a quotient i.e. 'x /  't for a fractional derivative such that the fractional derivative of the constant is zero From now on we will consider the order of the fractional derivative D , between 0 and 1 , and develop the modified fractional derivative; that is local in nature as it is parallel to the classical calculus. We will consider all functions primarily to be non-differentiable until Section-9.22 called ‘fractal’ functions ffract ( x ) (Section 1.24). These nowhere differentiable functions are undoubtedly impossible to draw. We cannot graphically draw a ‘fractal Mittag-Leffler function’ which we will consider is nowhere differentiable, by defining it in a suitable way. Here we develop the concept of a local fractional derivative for D -differentiable functions in conjugation with classical calculus. In the earlier chapters, we dealt with non-local fractional derivatives. This is a still evolving concept and is controversial too. However, we give the development of the very modern methods that we discuss now onwards until Section 9.22. A function x (t ) is said to have a fractional derivative x

(D )

(t ) of the order D with 0  D  1 whenever the limit: x

exists and is finite and where, 'x differentials dx

d  x (t )

(D )

(t ) v lim

't p 0

'x (t )

(9.71)

D

 't

x (t )  x (0) . Shortly afterwards, this amounts to us writing the equality in terms of

D

b (dt ) . As a direct result of this (9.71) definition, when x (t ) is a constant function one gets

0 . Thus, the fractional derivative of a constant function should be zero.

x

In a likewise manner, when x (t ) is differentiable we have dx

(1)



(t ) dt and as a result the fractional derivative is

zero i.e.

x

(D )

(t ) v lim

't p 0

'x D

 't

x lim

't p 0

(1)

(t )

 't

 't

't p 0

Thus for a differentiable function we have the fractional derivative x (9.71) deals with non-differentiable functions.

(D )



(1)

lim x (t )

D

(t )

1D

 't

0

(9.72)

0 from (9.72). Clearly, the definition

9.6.2 Corollary to the definition of a fractional derivative The above (9.71) definition indicates that the function x (t ) is D  th differentiable, with 0  D  1 . Corollary for a fractional derivative of a non-differentiable function are as follows: (1) Assume that x (t ) g (t )  f (t ) where f (t ) is differentiable and g (t ) is D  th differentiable, then one has 'x

x

(D )

D

b0  't  b1  't .

(t ) v lim 't p0

Therefore,

 . 'x ( t )

one

can

still

take

the

limit

condition

of

definition

It follows that, exactly how a derivative defines a function up to an arbitrary

'tD

additive constant, a function of which only the fractional derivative is known, is defined up to an additive differentiable function. This is exemplified in (9.73) and (9.74), below. (2) The function x (t ) is self-similar with the Hurst exponent (Section 1.24) D ! 0 whenever there is a positive D

D

constant O such that we have x (O t ) v O x (t ) as a result, these yields x (t ) v t x (1) . This we get by doing D

D

changes 1 m t and t m O in x (O t ) v O x (t ) (Section 1.24). This expression x (t ) v t x (1) suggests, as x (t ) is self-similar, we have x (t ) (i) According to dx

0 . We write two remarks.

D

b (dt ) , that is from the definition (9.71) considering the limit exists and is equal b , for D

small t one has x (t )  x (0) # bt . In addition, we write x (O t )  x (0) # O D  x (t )  x (0) . In other words, if

x (t ) is D  differentiable, then x (t )  x (0) is self-similar with the Hurst exponent as D , or again when x (t ) is locally self-similar at t 0 (ii) The converse of this is if x (t ) is self-similar, then x (0) D

(dx ) v x (1)(dt ) one has x (0)

0 , and according to x (t ) v t D x (1) one has

in such a manner that x (t ) is locally D  th differentiable. In the definition, x  O t v O D x (t )

0 but x (0) may have any value.

Modified Fractional Calculus in conjugation with Classical Calculus

395

D

(3) For (9.71), for the limit existing, say it equals b , then we have dx b (dt ) . Accordingly, fractional the derivative of the constant is zero. (4) For the differentiable case we have following: x (t ) f (t )  C

'x

b1't  0

lim

't p 0

(5) For a fractionally differentiable case we write the following: x (t ) g (t )  f (t ) D

'x

lim

(9.74)

1D

lim b0  lim b1  't

D

 't D E b0  't  b1  't ,

't p 0

More generally, if one has equality for 'x

(1)

x (t )

b1

't

b0  't  b1  't , 0  D  1

'x

differentiable at t

'x

(9.73)

't p 0

b0

't p 0

x

(D )

(t )

D  E then the function x (t ) is D  th

0.

9.7 Utilising the Mittag-Leffler function to get suitable fractional derivatives as a conjugation to the classical derivative The Mittag-Leffler function ED ( y ) , y

t

D

t

1

function e is a solution to the differential equation D D

fractional derivative D D

equation i.e. D ED (t )

y (t ) , y (0)

1 we would like to have a ‘suitable’

ED (t ) as a solution (eigenfunction) to the fractional differential

D

ED (t ) . This way we would be in a position to expand the fractional calculus in

conjugation or parallel with classical calculus, by substituting ED (O t D ) for e D

Ot

almost everywhere. We caution here

D

as a complex-quantity. The negative

for the function ED (t ) , we say t t 0 , as for a negative t , we may have t value for t we will see later is valid for D as On substituting y (t )

k

yk

. Exactly like the exponential 0 D k !

D

which should have

D

 y (t )

f

¦

is defined by a series ED ( y )

D

ED (t )

¦

f

tD k k 0 D k !

1 3

, 15 , 17 ,... not for all D . D

into, D y (t ) D

y (t ) we get:

D

D

D ED (t ) D

D

1 

tD D!

ED (t )  D

D

t 2D (2D )!



t 3D (3D )!



 ...

2D

D

1  D

If in (9.75) we let DD 1

0,D

 tD

D!

D

1 and D



D

definition for a suitable D , that gives DD ED (t D )

t 2D (2D )!

(9.75)

1  DD  Dt !  DD  (2t D )!  DD  (3t D )!  ... tD

D!

3D

tD D!



t 2D (2D )!



t 3D (3D )!



 ...

… then it is fine, and we are at our goal of the

D ED (t ) . Therefore, we arrive at the following ‘so called’

D

definition D for a suitable fractional derivative of order D ! 0 i.e. D

D

C

0

D

D x

Dk

D k ! D k  D !

x

D k D

Where C is a constant. Obviously this fractional derivative provides a customary derivative when D modification looks like conjugation to a classical derivative. Using (9.76), we apply to y (t )

D

ED (O t )

¦

f

O D tD k k 0 D k !

; t t 0 and see DD y (t )

D

O ED (O t ) .

(9.76)

1 . This

396

Chapter Nine D

This gives D y (t )

(1)

D

O , D y (t )

t 0

D O ED (O a ) . This is in conjugation with y (t )

t a

Ot

(1)

(1)

e , with values y (0)

calculus with y (t )

O , y (a)

Oe

Oa

Oe

Ot

as in the classical

.

We define a suitable higher transcendental function, by use of the Mittag-Leffler function and find the conjugation with a classical calculus of this new way to have a fractional derivative.

ED  x

coshD  x  sinhD  x def

 ED ( x)  ED ( x) ED  ix cosD  x  i sinD  x coshD  x

cosD  x

def



D

D ED O x D

D cosD x D

1 2

1 2

sinhD  x

 ED (ix)  ED (ix)

sinD  x

def

def

1 2

1 2i

 ED ( x)  ED ( x)

 ED (ix)  ED (ix)

(9.77)



D

O ED O x

D

 sinD x

D

D

D

D sinD x

D

D

D

cosD x

D

D

D

D

D coshD x  sinhD x D sinhD x coshD x We find an interesting parallel to the usual exponential, hyperbolic, and trigonometric functions. Only with a modified derivative as we described in (9.76), with DD  C

D Dk

0 and D x

D k ! D k D . Now we have to formally define D k D ! x

D

D in conjugation with classical calculus.

9.8 The fractional derivative via fractional difference and its Laplace Transform 9.8.1 Defining the fractional derivative f (D ) via the Forward Shift Operator Let f : \ o \ , t o f (t ) , a continuous (but not necessarily differentiable) function. Let h ! 0 represent a constant

>

discretisation element. We call a forward shift operator E Fh defined as E Fh f (t )

@

f (t  h ) . Then one can write the

finite difference 'f (t ) of f (t ) in the form:

'f (t )

 E Fh  1 f (t )

f (t  h )  f (t )

(9.78)

In quite a natural way, this (9.78) suggests that we define the fractional difference of order D , 0  D  1 of f (t ) by the following expression: f def D k §D · D (9.79) ' f (t )  E Fh  1 f (t ) ( 1) ¨ ¸ f  t  (D  k ) h ©k¹ k 0

¦

D

with the notation E FPh f (t )

f (t  P h ) . This (9.79) is by using the expansion in the series (x  1)

as done in Section

3.13, Section 3.14. In Chapter-3 (Section 3.13, 3.14) we have used the backward shift operator i.e. E h , when operated

>

on f (t ) , to give E h f (t )

@

f (t  h ) . Thus, we have E h D

D

we

D

' (  ) f (t )

D

have D E Fh

' (  ) f (t )

D

1  E f (t ) 1  E f (t ) . As done in Section 3.13, 3.14 expanding on ¦ (1)  C f (t  kh) . We also observe from these discussions

the fractional difference as ' (  ) f (t ) this

1 E Fh and with this backward shift operator E h we write

f

k 0

1 Fh

h

k

D

k

ª¬' f (t ) º¼ , that it is a relationship between the backward fractional differential ( 'D(  ) f ) and the D

D

forward fractional differential ( ' f ). We will consider (9.79) in our further discussions. With these above notations and definitions in a quite natural way, we define the fractional derivative of order D as the limit: (D )

D

' f (t )

(9.80) D h According to the definition (that we described in the previous section of this chapter for a non-differentiable function)

f

i.e. f

(D )

(t ) v lim 't p0

'f ( t ) 'tD

(t )

lim hp0

(9.71) and via the previous argument (9.75) of the eigenfunction i.e. y (t )

D ED (t ) be the

Modified Fractional Calculus in conjugation with Classical Calculus D

solution of D y (t )

397

y (t ) we have seen the fractional derivative of a constant should be zero. However, we have to

examine that if f (D ) (t ) definition (9.71) i.e. f

lim hp0 (D )

'D f ( t ) hD

, does definition (9.80) give a zero for the constant function? While we saw, the

(t ) v lim 't p0

'f ( t )

gives zero for a constant function.

'tD

9.8.2 Getting the fractional derivative of the constant as a non-zero by the Laplace transform technique According

^

$ f

(D )

to

`

the

(9.80)

definition

of

(D )

f

i.e.

f

(D )

D

s F ( s ) . Accordingly, the constant function f (t )

(t )

^

(D )

(t )

^

(D )

(t )

$ f $ f 1

^s

D 1

(t )

s F (s) ,

`

s $ ^C`

D

f (t )

D

`

Laplace

transform

is

D

C

§ C · D 1 ¸ s C ©s¹

s ¨

D

(9.81)

t

(D )

D

C f (t ) Cz0 ( D )! ( D )! That is the fractional derivative of the constant function f (t ) C , which is not zero. However, what if we wrote it as

$

^

$ lim hp0

'D f ( t ) hD

`

C

the

hD

C gives following steps:

`

t

'D f ( t )

lim hp0

D

s F ( s ) ? The proof is in following section.

^

`

9.8.3 Getting the Laplace Transform $ 'D f (t ) $ ^x (t )`

We use the definition of the Laplace transform i.e. X ( s )

³

³

f

0

³

f

h

e

sh

e

e

³

 st

 f ( t  h ) dt

 s (W  h ) f

h

e

 sW

 st

 x ( t ) dt

th

W ; dt

0

following steps:

$ ^ f (t  h)`

f

e

 f (W ) dW

e F (s)  e

e

sh

³

f

0

e

 sW

sh

³

h

0

e

 sW

 f (W ) dW sh

³

h

0

e

 sW

(9.82)

 f (W ) dW

 f (W ) dW

Examine the integral of the second term in the above (9.82) obtained expression i.e. and near W

dW

 f (W ) dW

e sh

(Appendix-G) and write the

0 we can approximately write this expression as

³

considering dW # h . This approximation is the area under the curve e

h

0  sW

e

 sW

³

h

0



e

f (W )dW # e

 sW

 f (W ) dW . For small

 s (0)



f (0) h

h

h  f (0) , by

0.

f (W ) at W

With this derivation, we have $ ^ f (t  h )` # e sh F ( s )  he sh f (0) . With the above (9.82) derived Laplace transform identity we apply to our fractional difference expression (9.79) i.e. 'D f (t )

^

D

`

¦

f k 0

and try to write the expression for $ ' f (t ) , i.e. as described in the following steps:

( 1)

k



D

Ck

f t  (D  k )h ,

398

Chapter Nine

^

­f k $ ®¦ ( 1) ¯k 0

`

D

$ ' f (t )

 C f t  (D  k )h ½¾ ; D

k

¿

^

`

§D · ¨k¸ © ¹

Ck

`

D (D 1) f  t  (D  2) h  .... 2

$ f  t  D kh  D f  t  (D  1) h 

^

D

^

`

$ f  t  D kh)  $ D f  t  (D  1) h $

e

D hs





^D (D21) f t  (D  2)h `  .... 

D hs

F ( s )  D he

f (0)  D e

(D 1) hs

(9.83)

F ( s )  D (D  1) he

D hs

(D 1) hs

D hs

D (D 1) (D  2) hs e 2

(D 1) hs

f (0) 

f (0)



F ( s )  ...

D (D 1) (D  2) hs D (D 1)(D  2) h (D  2) hs F (s)  e e 2 2

 e F (s)  D e F (s)   D he f (0)  D (D  1) he

(D 1) hs

f (0)  ....

D (D 1)(D  2) h (D  2) hs e 2

f (0)  ...



Take the first term in bracket of the last line of (9.83) and we do the following manipulation and approximation for small h as depicted (9.84):

e

D hs

F (s)  D e

(D 1) hs

D hs

e D hs

e

F (s) 





hs

2

F (s) 1  D e

F (s) 1  e

e

D (D 1)

D



1

 hs

 hs

e

(D  2) hs



D

D (D 1) 2

D

1  hs  1

e

2 hs

 e 1  e hs

F (s) ;

e

F (s)

hs

 ... F (s)

F ( s )  ...

 hs

D

(9.84)

# 1  hs

D D

h s F (s)

Therefore, we have the following relationship:

^

D

`

$ ' f (t )

§ D heD hs f (0)  D (D  1) he(D 1) hs f (0) · h s F ( s )  ¨ D (D 1)(D  2) h ¸¸ (9.85) (D  2) hs ¨ e f (0)  ... 2 © ¹ D D

^

`

9.8.4 Getting the Laplace Transform $ f (D ) (t ) D

Divide (9.85) by h to get the following expression:

­ 'D f (t ) ½ D (D 1)(D  2) 2 hs 1D D hs  hs ¾ s F ( s )  D h e f (0) 1  (D  1)e  2 e  ... D ¯ h ¿



$®



(9.86)

Take the limit h p 0 and we get the following:

^

$ f With this fractional difference, we saw that f

(D )

(D )

(t ) z 0 , for f (t )

D

­

`

(t )

$ ®lim

¯

hp0

' f (t ) ½

D ¾ s F (s) ¿

D

h

^

C as we get $ f

(D )

`

(t )

s

(9.87) D 1

C z 0 . In other

words, with this definition (9.80), (9.87) the fractional derivative of a constant would not be zero as that contradicts our assumption.

^

We remark that the classical Laplace transform formula for the first derivative is $ f

^

it is evident it is not equivalent to $ f

(D )

`

(t )

^

$ lim hp0



'D f ( t ) hD

`

D

(1)

s F ( s ) when D

`

(t )

1.

sF ( s )  f (0) and that

Modified Fractional Calculus in conjugation with Classical Calculus

399

^

This gives rise to a query as to whether there is some mistake. First, in two expressions i.e. $ f

^

$ f

(D )

` the consistency is complete when

(t )

1 . This f (0)

0 , for D

f (0)

(1)

`

(t )

and

0 is a condition satisfied by self-

similar functions (Section 1.24).

9.8.5 The definition of fractional differentiability is satisfactory for self-similar functions Therefore, the definition (9.80) f (D ) (t )

lim hp0

'D f ( t )

is quite satisfactory for self-similar functions. According to

hD

our earlier discussion (Section 1.24), i.e. when we restrict ourselves to the class of functions that satisfies the condition i.e. f (0) 0 then, we can claim that: (1) If the function is a fractional differentiable, then it is self-similar. (2) Conversely, if it is self-similar then it is a fractional differentiable. D

C if we assert that D C

To cope with the problem due to a constant term f (0)

0 , then we will have the required

D

modification for the RL fractional derivative (that we represented D ).This modification is called the Jumarie D

fractional derivative (that we represent by D ).

9.9 The modified fractional derivative for a function with non-zero initial condition-Jumarie type 9.9.1 Defining the fractional derivative via the offset function construction to have a fractional derivative of the constant as a zero-Jumarie type Let f : \ o \ , t o f (t ) denote a continuous (but not necessarily differentiable) function, and define f (t ) f (t )  f (0) i.e. the offset function. Let h ! 0 denote a constant discretisation element (span). The fractional derivative of order D of f (t ) is defined as the limit:

f

(D )

def

(t )

'D f (t ) lim hp0 hD

(9.88)

Alternatively assume g (t ) is self-similar with the Hurst index H with 0  H  1 (Section 1.24). Then the fractional derivative of order H is defined by the expression:

g

(H )

(t )

' H g (t ) hp0 hH

(9.89)

lim

According to this (9.88) definition, the fractional derivative of the constant is zero. For f (t ) C we have the offset function as zero; f (t ) 0 . This (9.88) is similar to the Caputo fractional derivative but not the same. They are the same when f (t ) is differentiable. We call it a modified R-L fractional derivative-Jumarie type. Note that the function f (t ) need not be differentiable, in this modified R-L fractional derivative definition (9.88). The definition (9.88) i.e. the definition of the fractional derivative via the forward fractional difference, is the right fractional derivative. This definition is in conjugation with the classical derivative f

(1)

(t ) . We say that f

(D )

D

or D f was defined as in (9.88) and is the ‘local fractional D

derivative’, unlike the classical Riemann-Liouville fractional derivative D f (t ) , which we have dealt with in all previous chapters which are ‘non-local’. By using the (D )

f (  ) (t )

lim hp0



' (  ) f ( t ) D

hD



fractional

, with f (t )

difference,

(1)

(t )

lim hp0

a classical derivative f

(1)

 1h 'f (t )

D

' (  ) f (t )

f (t )  f (0) . We place D

f (t )  f (0) , and '  f (t )  f (0)

Where f (t ) Therefore f

backward

lim hp0

we

have

left the

1 in (9.88) to get f

fractional (1)

(t )

derivative

as

 1h '



lim hp0

f (t ) .

f (t ) .

 1h 'f (t ) is the classical result.

We may infer that in the definition of

(1) , which is a local quantity, f (t ) is inbuilt in the definition f (t )

lim hp0

 , which 'f ( t ) h

400

Chapter Nine

gives f

(1)

(t )

lim hp0



'f ( t ) h

.

That is if we subtract the initial value of the function f (0) from f (t ) and then do

one-whole differentiation, it is the same if we do not subtract the initial value. This brings us to the point that (9.88) is a definition of the local fractional derivative in conjugation with a classical derivative.

9.9.2 The Laplace Transform of the Modified RL fractional Derivative and its conjugation to a classical derivative We find the Laplace transform of the modified R-L derivative and its conjugation with ordinary classical derivatives as demonstrated below: D def ' f (t ) (D ) f (t ) lim D hp0 h

^

$ f

(D )

­ 'D  f (t )  f (0) ½ $ ®lim ¾ D h ¯ hp0 ¿ D D ­ ' f (t ) ' f (0) ½ $ ®lim  lim ¾ D D hp0 h ¯ hp0 h ¿

`

(t )

D

D

D

s $ ^ f (t )`  s $ ^ f (0)` D

D

D 1

s F (s)  s

^

In (9.90) we used (9.87) i.e. $ lim

^

$ f

(1)

D

' f (t ) hD

hp0

`

D

s F ( s )  s $ ^ f (0)`

§ f (0) · ¨ ¸ © s ¹

D

s F (s)  s

(9.90)

f (0)

D

s $ ^ f (t )` . Now for the case in (9.90) with D

1 we get

` s  F ( s)  f (0) here the classical result is recovered. We have found a similarity of (9.90) with the

(t )

classical derivative in the Laplace transform identity. Now we define the integral form of this modified RL fractional derivative of Jumarie type (Section 9.10) However, D ED (O t ) ; t t 0 , to obtain f

before that we make the result of (9.90) for f (t )

(D )

(t ) .

With this formula (9.90) which is from the formula (9.88) we demonstrate that the local value of the fractional derivative can be obtained for D  differentiable function f (t ) at t 0 . For the demonstration of obtaining a local fractional derivative at the point by this modified RL method (Jumarie Type), we take the function f (t ) D

Let us apply the formula (9.90), to f (t )

^

` in (9.90), we obtain $ ^ f as F ( s )

get f (D ) (t )

f (t )

e

D

$ ED (O t )

Ot

giving f

the function is f

(t )

(D )

ED (O t ) ; t t 0 i.e. the Mittag-Leffler function with its Laplace transform

having a non-zero initial condition i.e. f (0)

`

D

(t )

s F (s)  s

D O ED (O t ) , and we have f (1)

(t )

We also have DD ED (O t D ) (D )

sD 1 sD  O

f

(D )

f (t )

Oe

Ot

and f

(1)

(D )

(0)

D 1

f (0)

(0)

O

 . sD 1 sD  O

O . This is in conjugation with classical calculus where we have

O.

D

D



(0) f (t ) similar to f

1 (Appendix-A). Putting these values

Now writing the inverse Laplace transform, we

O ED (O t ) , that is in conjugation with ED (O t ) ,

D

ED (O t ) .

f

(1)

(t )

O f (t ) when f (t )

Ot

e . Therefore, if

we have the expression of conjugation to classical calculus i.e. (1)

(t )

f

(1)



Ot

e , at t

(0) f (t ) , for f (t )

0.

Therefore our fractional differential equation (with modified RL fractional derivative-Jumarie Type) is f

(D )

f

(1)

(t )

(t )

f f

(D )

(1)



(0) f (t )



which has the solution

(0) f (t ) where f (t )

Ot

f (t )

D ED (O t ) , in conjugation with classical calculus

e . As we can write f (t )

e

f

(1)

for

(0) t

f

(1)

(t )

f

(1)



(0) f (t ) , for t t 0

Modified Fractional Calculus in conjugation with Classical Calculus

in the same way we are tempted to write f (t )

ED

 f

(D )

as a solution of

(0) t

D

f

401 (D )

f

(t )

(D )



(0) f (t ) for

t t0.

9.10 Integral representation of modified RL fractional derivative of Jumarie type Assume that f (t ) is a constant function i.e. f (t )

C , then fractional derivative of order D is C

D

Dt C

* (1  D )

t

D

D d0

(9.91)

0 D !0 Note that first expression of (9.91) is fractional integration D d 0 of constant and second one D ! 0 is fractional derivative.

f (0)   f (t )  f (0) and fractional derivative is defined by

When f (t ) is not a constant then we can write f (t ) the expression f

(D )

D

(t )

D

D , is just simply the RL fractional integration i.e. f (D ) (t )

^

(D )

From the derived expression (9.90) i.e. $ f

^

$ f

(D )

`

(t )

ss

D 1

Laplace operator s

F ( s )  ss

D 1

D

 f (t )  f (0) . For D ! 0 we have

Dt f (0)  Dt



D 1 1

s f (0)

s s

`

 f (t )  f (0)

D

(t )

D 1

D

Dt

s F (s)  s

D 1

0 (9.91). In (9.91) the negative

Dt f (0) 1 *  D

t

³ (t  W )

D 1

0

f (W )dW ; D d 0 .

f (0) ; D ! 0 ; we modify the RHS to write it as

 F ( s)  $ ^ f (0)` , where $ ^ f (0)`

f (0) s

. We recognise the

as a fractional integration of the order 1  D i.e. 0 I t1D (Section 5.14) and the Laplace operator d dt

s as a one-order derivative, i.e.

.

Thus by using the definition of fractional integration of the order 1  D , for positive D , i.e. D ! 0 we have thus modified the RL fractional derivative as

f

(D )

(t )

D

Dt

 f (t )  f (0) = $ 1 ^s  sD 1  F ( s )  $ ^ f (0)` ` .

This we write as f

(D )

(t )

d dt



1D 0 It

 f (t )  f (0)

 f

(D 1)

(t )



(1)

where f (t )

f (t )  f (0) . Thus from these

arguments, we write the following: f

(D )

(t )

 f

(D 1)

(t )



1

(1)

d

However f

(1)

(t )

we d dt

have

got

f

 I  f (t )  f (0) 0 0 t

(D )

d dt

(t )



1D 0 It

(D )

(1) f (t ) , with f (t )

(9.92)

D

³ (t  W )  f (W )  f (0) dW 0

D

(t ) or

 f (t )  f (0)

0 D 1 t

* (1  D ) dt

Thus, the above (9.92) is an integral representation of f Jumarie type for 0  D  1 .

,

D f (t ) of the modified RL derivative of the

where

if

we

place

D

1

we

obtain

f (t )  f (0) . Thus, we again infer that the offsetting of a

function is built in for a classical derivative too, (this we discussed in Section 9.9.1). Therefore, the integral formula (9.92) is an analytical way to find the local fractional derivative for order 0  D  1 , and then obtain f can be used to obtain values at the points of interest.

(D )

(t ) which

When n  1 d D  n , and D ! 1 , with n as positive integer, we generalise the above (9.92) as follows:

f

(D )

(t )

 f

(D  n )

(t ) 1



(n)

, d

n 1 d D  n; n

* ( n  D ) dt

n

³

t

0

(t  W )

( n D ) 1

n !1 (9.93)

 f (W )  f (0) dW

402

Chapter Nine

The logic is to have an offset function i.e. f (t )

f (t )  f (0) and then to take the usual RL fractional derivative of

D f (t ) . We clarify the notation while D stands for a modified RL fractional derivative of the Jumarie type i.e. we also

represent this with f

(D )

, while DD stands for the usual RL fractional derivative. We note that when f (0)

D

and

D

0 ; D ! 0 .In addition, we clarify that the fractional integration operation in a modified sense and

D f D C

D D f are the same. While for a constant function,

conventional RL sense is the same, i.e. D D f (t ) Ot

Applying this logic to f (t )

e ,

D

D

D Ot

0 Dt

e

t

(D )

(t )

D Ot

D e

t

D

e

we have

C

D

(t )

0 Dt

> f (t )  f (0)@ ;

D

D C z 0 whereas

D ! 0 where 0 DtD is the

*

D O t *

e J ( D , O t ) , where J ( D , O t ) is called Tricomi’s

Incomplete Gamma function (Section-4.6), and also we have (D )

f (t )

f (t ) , for D ! 0 where f (t ) is unchanged.

1 in order to get f

f (0)

usual RL fractional derivative operator. We have

f

0 , then

D

0 Dt

>1@

D 1 * (1D ) t

, using these results we write



Ot *

J ( D , O t )  * (11D ) , that is a modified fractional derivative of the Jumarie type.

While taking the modified fractional derivative of order D ! 0 we have to take the conventional RL fractional derivative on a changed function called the offset function i.e. f (t ) f (t )  f (0) . The start point t 0 could be any other point say t a then the lower limit of fractional integration will be non-zero. Then the offset function will read as f (t ) f (t )  f ( a ) . Now if at t 0 or at start point, the function is undefined say f (0) rf , then we consider 

the offset function as f (t )

f (t )  f (0 ) or we consider the finite part of the offset function. We also note that

C

i.e. the Caputo derivative for 0  D  1 , but here we are describing it as a modified RL

D

Dt f (t )

D

Dt

 f (t )  f (0) D

fractional derivative, i.e. D . We stress here that the modified RL derivative of the Jumarie type does not demand that the function be a differentiable, whereas the Caputo definition demands the condition of differentiability. These two fractional derivatives i.e. the modified RL (Jumarie) and the Caputo are the same if and only if the function being considered is differentiable. In the earlier chapter (Section 4.15, 4.16) we had obtained the RL fractional derivatives for the offset function for some examples. In the above (9.92) modified definition we have used the left/causal fractional derivative. Similarly a right modified

> @

fractional derivative of a function f defined on the interval say t  0, b is:

f (b )   f (b )  f (t ) ;

f (t ) D

D

> f (t ) @

f

(D )

D

(t )

t

( 1)

d

Db

* (1  D ) dt ³

b

t

f

(D )

(t )

f

(D  n )

(t )



( 1)

(n)

n

f (t )

f (b )  f (t )

 f (t )

 f (b )  f (t )

(W  t )

D

d

n

* ( n  D ) dt

n

³

b

t



,

0 D 1

 f (b)  f (W ) dW

n 1 d D  n;

,

(D 1) (1)

(W  t )

( n D ) 1

(9.94)

n !1

 f (b)  f (W ) dW

9.11 The application of the modified fractional RL derivative of the Jumarie type to various functions ED (O xD )

9.11.1 Modified RL derivative of order D for function f ( x) D

We apply this concept of the modified RL fractional derivative to f ( x ) the Mittag-Leffler function:

f ( x)

D

ED (O x ) 1 

ED (O x ) , D 2

D

(O x )

D!

x t 0 . Take the series form of



(O x )

(2D )!

D 3



(O x )

(3D )!

 ...

(9.95)

Modified Fractional Calculus in conjugation with Classical Calculus

For a modified RL derivative, f

(D )

( x ) , we take an offset function f ( x ) and we get: f ( x ) f ( x )  f (0)

ED (O x )  1 Then take RL fractional derivative D

D

(D !)

D 3

(O x )





(2D )!

(O x )

(9.96)

 ...

(3D )!

of the offset function f ( x ) as depicted in the following steps:

D

D f ( x) D

D 2

D

(O x )

D

D

403

E

(D )

f

D

(O x )

D

( x)

D

E D

(D !)

D



D

 f ( x)  f (0)

(O x )

D

O

D

x

D



D

 ED (O xD )  1

(D )

O



D

2

(2D )!

D



D

x

2D



O

3

D

(3D )!

2

D

 x  ... 3D

(9.97)

3

O § (D !) · O § (2D )! · D O § (3D )! · 2D x  ¨ ¸ ¨ ¸ ¨ ¸ x  ... (D !) © 0! ¹ (2D )! © (D !) ¹ (3D )! © (2D )! ¹ 2 2D

D

§

Ox

©

(D !)

O ¨1 

·

O x



(2D )!

D

O ED (O x )

 ... ¸

¹

D

Therefore, we verified that with the modified RL definition y ED (O xD ) it is a solution of D y fractional integration, we are not taking an offset, as we demonstrate in the following steps: D

D

 f ( x) D

D

D

1 

1

x 

D (O x

O (D !)

D

D



D

) ; D t0

 x

D



O § (D !) ·

D

(D )!

E

D

O y . For a

¸x

¨

2D

(D !) © (2D )! ¹

O

2

D

(2D )! 

D

 x

2D

O



3

(3D )!

D

D

 x  ... 3D

§ (2D )! · 3D O 3 § (3D )! · 4D ¨ ¸ x  (3D )! ¨ (4D )! ¸ x  ... (2D )! © (3D )! ¹ © ¹ O

2

2 3 4 § 1 · § O D O 2D O 3D O 4D · x x x    x  ... ¸ ¨ ¸¨ (2D )! (3D )! (4D )! © O ¹ © (D )! ¹

(9.98)

2 3 4 § 1 · § § O D O 2D O 3D O 4D · · x  x  x  x  ... ¸  1¸ ¨ ¸ ¨1  ¨ (2D )! (3D )! (4D )! © O ¹ © © (D )! ¹ ¹

O

1

E

D (O x

D



) 1

D

This shows that y ED (O x ) is exactly similar to the exponential function y derivative is with conjugation to classical calculus.

e

Ox

and this modified RL fractional

9.11.2 A modified RL derivative of order E ! 0 for function f ( x) Now take the function y ( x ) y

(E )

( x)

ED (O xD )

D ED ( x ) , and calculate the modified fractional derivative of order E ! 0 that is

E D y ( x ) as follows:

y

(E )

( x)

E

D

D

(x ) D

0 x

E



(E )

E

= D y ( x)

E

E

D

D

(x )



D 2D 3D § · x x x    1 ¨ * (1  D ) * (1  2D ) * (1  3D )  .. ¸ © ¹

x

D E

* (1  D  E )

D E

D



x

2D  E

* (1  2D  E )



x

(9.99)

3D  E

* (1  3D  E )

 ...

D

ED ,D  E 1 ( x )

Where, we have a two-parameter Mittag-Leffler function defined as E p , q ( x )

¦

f

xk k 0 * ( q  kp )

(Appendix-A).

404

Chapter Nine

xD 1 ED ,D (O xD )

9.11.3 The RL fractional derivative of order D for function f ( x) RL

If the fractional derivative is a classical RL derivative then we have Ox eD

as follows by letting

(x

D 1

D

D C z 0 , then the solution for

RL D 0 Dx

y

O y is

D

) ED ,D (O x ) we write: Ox

eD

(x

D 1

(x

D

) ED ,D (O x )

D 1

f

)

x

k

kD

f

k 0

RL D 0 Dx

k

x

( k 1)D 1

¦ O *  (k  1)D ¦ O *  (k  1)D

ª¬ eDO ( x ) º¼

§ k ¨¦O ©k 0

RL D 0 Dx

f

¦

k 0

f

O

( x)

k

( x )( k 1) D 1 *  ( k 1)D

· ¸ ¹

(9.100)

kD 1

* ( kD )

k 1

Ox O eD

Thus we have y for

RL D 0 Dx

y

D 1

D 1

RL D 0 Dx

D

ED ,D (O x ) as a solution to FDE with a classical RL derivative

O y is y

we discussed that

x

x

x

C D 0 Dx

D 1

y

y

O y or eigenfunction

D

ED ,D (O x ) . This we have discussed earlier (Section 4.14, 7.3, 7.14, 7.18). In addition,

O y has a solution as y

D

ED ,1 (O x ) , (Section 4.14, 7.2, 7.14, 7.18). Two functions

D

D

ED ,D (O x ) and ED ,1 (O x ) we termed then as the alpha-exponential functions 1 & 2, we had used them in an

earlier chapter (Section 7.14).

9.11.4 The modified RL derivative of order D for function f ( x)

ED (O x)

D

With a modified RL derivative D , we will find DD ED (O x ) knowing the fact that ED (O xD ) is a solution of D

D

D ED (O x ) , i.e. we write it in the following form: D

d

D

D ED (O x ) Let y

D

x then dy

Dx

D 1

dx and

D

 dy

D 1 D

D

 D x  dx

D

dx .

E

D

D

D

(O x )



D

(9.101)

O ED (O x )

With this substitution in (9.101) we write the

following steps:

d

D

 ED (O y ) d y

d

D

D

D D OD D

Therefore, we have an interesting relationship i.e. D ED (O x )

 ED (O y ) D D x(D 1)  dx D

1

§ d D E ( O xD ) · D ¨ ¸ ¨  dx D ¸ © ¹

1

y

D

D 1D

y

D 1D

OD

d

D



D

y

D

 ED (O y ) D  dy D 1

O E

D

D

(O x )

(9.102)



ED (O y )

D 1D

x

ED (O x ) .

9.11.5 Applying the modified RL fractional derivative to a non-differentiable point of a function: f ( x) We now see what is the half derivative at the point x this is non-differentiable function at x

x

1

2

b

0 for function i.e. f ( x )

0 . We use this to calculate the following:

x

1

2

 b , where f

(1)

(0)

f . So

Modified Fractional Calculus in conjugation with Classical Calculus

f

 12

( x)

D

1

2

> f ( x)  f (0)@ *  12  1

1

1 2 D 2 ªx º

¬

We note that for f ( x )

x

1

 b we have f

2

derivative of the Jumarie type and with

 2 1

( x)

1

D 2b

D

1

2

x

1

2

b

0 we have f

405



 12

¼

D

1

2

x

* 1 1

0

(9.103)

 12 !

1

2

 D 2 b . With D

 12 !

( x)

x

1

2

as a modified RL

We observe that the modified RL

D

derivative Jumarie type i.e. D is acting as a local operator like that of classical calculus.

9.11.6 Defining the critical order for a non-differentiable point by use of the modified RL fractional derivative Now we write the fractional derivative D with 0  D  1 of f ( x ) and write via the above (9.103) steps: f

We observe that f ( ) ( x ) D

x 0

(D )

1 2

has values of zero for 0 d D 

( x)

x

1

2

 b by use of the modified RL derivative

 12 !  12  D !

x

 1 2 D

(9.104) 1 2

, a constant value for D

and infinity for

1 2

D d1

, which we summarise below: f

1 2

Therefore, we call D

(D )

as the critical order for f ( x )

( x)

x

1

2

­ 0 , 0 d D  12 ° 1 1 ® 2 ! , D 2 °¯ f , 12  D d 1

x 0

(9.105)

b.

> @

>

Now we take a function f ( x ) that is defined on the interval 0,1 as a set of points x1 , x2 ,... x N

0 and x N

x1

N . Let the function be described as f ( x )

for x  xk , for all xk , k

x

f

x1 , x2 ,.., x N

 12

( x)

lim N nf

N

¦ x  x k

k 1

1

2

@

with N n f ,

 b , for x t xk and f ( x )

b

1, 2, 3..., N , (Section 1.24). We note that it is a function that is not differentiable at

in the interval

>0,1@ ,

but f

 12

( x ) exists at all those non-differentiable points with value

 12 ! . This we will derive in Section- 9.18.2. Thus, a continuous but nowhere differentiable

x x1 , x2 ,... xN

function can have fractional derivatives at all those non-differentiable points; in this case, the critical order of this is nowhere differentiable function is half.

9.12 Fractional Taylor series We started this concept from Section 9.6 of the development of a modified RL fractional derivative Jumarie type for the non-differentiable functions. Until now, we have seen f ( x ) ED (O xD ) ; x t 0 in the context of a modified RL fractional derivative Jumarie type is in conjugation with the function e f

(D )

(0)

f ( x)

D

D

D ED (O x )

x 0

Ox

of classical calculus. We have seen that

O while the classical derivative is not defined as x

D ED (O x ) . Thus, the function

f ( x)

D ED (O x ) is not differentiable at x

0 i.e. f (1) (0)

f , for

0 but certainly differentiable at

other points i.e. at x ! 0 . We want functions that are nowhere differentiable but continuous ones. With f ( x )

D ED (O x ) we make an everywhere non-differentiable function by considering the function f ( x ) defined

in the interval say f ( x)

>0,1@

D ED (O x ) if x

composed of discrete points

x0 , x

x1 , ......... x

x N else f ( x )

> x0 , x1 , x2 ,.........xN @ 0.

with

x0

0 and

xN

1 as

When we have N n f , for the fractal-set

406

Chapter Nine

> x0 , x1 , x2 ,.........xN @

we get a function f ( x ) which is continuous but nowhere differentiable. These types of

functions are called ‘fractal-functions’. This function satisfies DD ED (O xD )

D O ED (O x ) and is D  differentiable at

all points with values at x0 , x1 , x2 ,......... x N where classically it is not differentiable at x0 , x1 , x2 ,......... x N i.e. nowhere differentiable. Similarly, we can have a function f ( x )

lim N nf

¦

N

D

ED (O ( x  xk ) ) for x t xk ; and f ( x )

k 0

1 for x  xk . This is

not differentiable at points x0 , x1 , x2 ,......... x N . We will show later in the Section-9.18.2 that this function is also (D )

having the property f ( x ) O f ( x ) , thus is D  differentiable at points x0 , x1 , x2 ,......... x N . In order to simplify the notations we will imply that the functions are considered that are nowhere differentiable. That is still what we D D represent by normal symbols like x (t ), f ( x ), y ( x ), ED (O x ), cosD ( x ) etc., with an understanding that all these functions are nowhere differentiable functions-i.e. they are ‘fractal functions’ ( ffract ( x ) ) as described in Section-1.24. D

We started defining the fractional derivative (9.71) via defining a quotient i.e. 'x /  't . Then through further



via fractional difference of the zero-corrected function

D

D lim ' x (t ) /  't

definition (9.88) we defined it as

't p 0

described as x (t ) x (t )  x (0) . What is the relationship between these two definitions (9.71), (9.88)? This will make our development and this approach complete. With the above definitions of fractional derivatives it is easy to show a solution of y

(D )

O y (t ) , y (0)

(t )

y0 is

y (t ) y0 ED (O t ) . In addition, we have an operational equality as D h E Fh Dt E Fh , where E Fh  x (t ) x (t  h ) , is a forward shift operator. Now look at the following Laplace identity that we have derived earlier in this chapter (9.82), (9.87): D

D

^ $ ^D

D

` e f (t  h )` # s  e

$ D h f (t  h ) # s D t

D

^

D

st

D

hs

st

F ( s )  te f (0) hs

F ( s )  he f (0)

`

D

^

operational identity i.e. D h E Fh D

Dh y ( h)

(9.106)

`

D $ D h f (t  h) . Therefore with the

Here we see for h p 0 and t p 0 that both are equal i.e. $ Dt f (t  h ) D



D

Dt E Fh which may be considered as a fractional differential equation of type D

K  y ( h ) and taking Dt

Mittag-Leffler function, i.e. y ( h )

E Fh ( h ) . We write the solution in terms of the

K as a constant and y ( h )



D

ED Kh



D

modified RL derivative D we have a solution to DDx y We obtained the operation identity as E Fh



or E Fh

D

O y as y



D

ED h Dt

D

D

ED h Dt



.

Here we have used the fact that with a D

ED (O x ) .

and with this, we operate on x (t ) to write

 E Fh x(t )  ED  hD DDt x(t ) . After expanding the Mittag-Leffler function in series form (see Appendix A), and noting  E Fh x (t ) x (t  h ) , we get the following: x (t  h )

 E  h D

D

Writing DDt x (t )

D Dt

x

(D )

§ hD D h 2D 2D · Dt  Dt  ... ¸ x (t ) (9.107) x (t ) ¨ 1  (2D )! © D! ¹ 2D

(t ) , Dt x (t )

x

(2D )

(t ) we get the fractional Taylor series as: D

x (t  h )

x (t ) 

h

(D )!

x

(D )

f

x (t ) 

(t )  Dk

h

h

2D

(2D )! ( k)

¦ (D k )!x D k 1

Making 0 m t , and t m h gives us the fractional Maclaurent series as follows

x

(t )

(2D )

(t )  ...

(9.108)

Modified Fractional Calculus in conjugation with Classical Calculus f

x (t )

x (0) 

t

Dk

( k)

¦ (D k )!x D

407

(0)

(9.109)

k 1

We note in (9.109) that we have the following notation:

x

(D k )

D

D

(t )

D

D



D

kD

....D x (t ) z D x (t )

 

k

2D

2D

D D

D

D

but D that is D z D D D , that we have discussed earlier in this chapter as the rules of composition (Section 9.3). For a small h we have the following approximation with the first two terms of the fractional Taylor series as follows: D

h

x (t  h )  x (t )  'x (t )   (D !)

(D )!

1

x

(D )

x

(D )

(t )

(9.110) D

(t )  't

D !dx  x

(D )

D

( t )  dt

We remark that the developed fractional Taylor series is suitable for a modified RL derivative of the Jumarie type, and is not valid for a conventional Riemann-Liouville fractional derivative. This fractional Taylor series is for nondifferentiable fractal functions.

9.13 The use of fractional Taylor series Here we will use the fractional Taylor series to expand a non-differential function at a non-differentiable point. In addition, by the fractional Taylor series, we will have conversion formulas to represent fractional differentials to D

D

classical differentials i.e. relating d f to the differential d f and (d f ) . Our fractional Maclaurent series is f

f ( x)

f (0) 

x

Dk

( k)

¦ (D k )! f D

(0)

(9.111)

k 1

The Mittag-Leffler function f ( x ) ED ( xD ) with 0  D  1 is not-differentiable at x (the non-differentiable point) is as follows:

f ( x)

x

f (0)  D

ED ( x )

D

f

D!

ED (0)  1

1

1

x

D

D!

D

D

x





We have obtained the series definition of f ( x )

x

E

f

D

D (x )

x 0



2D

 2D ! x

2D

(2D )!

D

D

D!

(0) 

ED ( x )

D

D! x

x

D

D! x

(D )

x



(2D )

 2D ! D

ED ( x )

x 0

0

(0)  ....



x 0

2D

0 , and its expansion at x

D

D

2D

x

E

 D  E ( x )

D

 2D ! D (x

D

D

D



D

D

x 0

x 0

 .....

 .....

(9.112)

 ....

2D

 2D !

 ...

D ED ( x ) . Thus, a series that defines the Mittag-Leffler function is

D

nothing but a fractional Taylor series at point zero-where the function is not differentiable. We note that D is a modified RL fractional derivative and we have used the above the relationship DDx ED ( xD ) ED ( xD ) which we discussed earlier. We note that ED ( xD )

x 0

ED (0)

1 i.e. used in (9.112).

There is in conjugation with the classical calculus, as for the function such as e x

f ( x)

1 x 

x2 2

f ( x)

x

e we obtain a series representation

 ... i.e. for the exponential function via the classical Taylor series expansion at x

0 for

x

e via the classical Taylor series. 1

Let f ( x ) x 2  b , which is not-differentiable at x Taylor series):

0 , its expansion at x

0 is the following (via a fractional

408

Chapter Nine

f ( x)

f ( x)

x

f (0) 

b

x

1

D

f

D!

2

 ! 1 2

D

x

b

1

1

x

2

2

 ! 1 2

(D )

x

(0)  1

2

2D

(2D )!

b



x 0

 12 !  xD 0!

1

2



(2D )

f

x

1 2

 12 !

!

D

bx

1

1

D  x 1

2

2

1

2

b



x 0

 .....

(9.113)

2

2

We apply the normal Maclaurent series for a polynomial f ( x )

f ( x)

2 12

2

1 2

(0)  ....; D

x  x  c and write the following:



f (0)  x f

x



(1)

2

f 2

(0) 

(2)



(0)  ...

2

 2 x  1  x2  2  0

cx

(9.114)

x 0

x 0

2

cxx We observe a parallel to classical calculus for the fractional Taylor series at a non-differentiable point, for a nondifferentiable function vis-à-vis the classical Taylor series for differentiable functions. Intuitively, if one considers that in fractional calculus the Mittag-Leffler function plays a similar role as the exponential function in classical calculus, one is led to extend the Taylor series f ( x  h )

 exp(hD x )  f ( x) in the form of

f ( x  h)

 E  h D  f ( x) . D

Here we clarify that exp( hD x ) is an operator that operates on f ( x ) . We can write this as f ( x  h )



D

that operates on

D

It is the same as the operator ED h D x

D x

D

 e  f ( x) . hD x

f ( x ) . Of course, despite the simplicity of the formula this

is not sufficient and needs to be supported by a sound derivation. Actually, the problem that appears is how to suitably define the concept of a fractional derivative (that we have tried to do in previous sections) in order that it is fully consistent with the Mittag-Leffler function which is considered as the solution of the fractional differential equation, i.e. x

(D )

(t )

O x (t ) .

9.14 Conversion formulas for fractional differentials From the Fractional Taylor series, we have a conversion formula (approximate) that is:



D !dx  x

(D )

(t )

D

 dt

(9.115)

From this, we can have: D

D !dx 

d x

D

D

 dt

D

 dt

d x

(9.116)

and then the relationship: D

(9.117)

d x  D !dx

From the formula of the fractional derivative, we have: D

d x dx D

that is the Euler formula D x

* (2) * (2 D )

1D

x

D

1

1D

1  D !

(9.118)

x

which gives us another formula i.e. D

d x

1  D ! D

1

1D

x

D

 dx

(9.119) D

These are approximate conversion formulas of differentials relating d x with dx and (dx ) . In the above expressions we have used factorial notations instead of Gamma functions.

9.15 Integration with respect to fractional differentials 9.15.1 Integration w.r.t. (d[ )D and its relationship to the Riemann-Liouville fractional integration formula

Modified Fractional Calculus in conjugation with Classical Calculus (1)

In classical calculus the equation x (t )

g (t ) i.e. the following: dx

g (t ) ;

dt

³

is equivalent to x (t )

t

0

409

x (0)

(9.120)

0

g (W )dW . Here the RHS of (9.120) is considered as an anti-derivative, in the sense that it is

defined by the following equality: d

³

t

³



t

(9.121) g (W )dW g (t )dt 0 dt 0 On applying the same point of view to the modified RL fractional derivative, we can write the following fractional differential quantity for the equation y

(D )

g (W )dW

g (t )

,

d

g (t ) via the definition (9.71) as:

(t )

D

g ( t )  dt ;

dy

³

This (9.122) is equivalent to the integral y (t )

t

0

y (0)

0

(9.122)

0 D 1

D

g (W )  dW . Here the RHS is a fractional anti-derivative of order

D and is described (defined) by the following equality: d

³

t

0



D

g (W )  dW

d

D

g (t )  dt

dt

³

D

t

D

g (W )  dW

0

g (t )

(9.123)

We have a formula (we will prove it subsequently) defined as:

³

x

D

f ([ )  d[

0

D

 dx

relating the integral with respect to the element

def

D

³

x

0

D 1

(x  [ )

 f ([ ) d[ ,

0 D 1

(9.124)

, i.e. LHS of (9.124) and the Riemann-Liouville fractional

integral RHS of (9.124) (with the exception of the Gamma function). D

The integral w.r.t.  dx is defined as a solution to the fractional differential equation, that is: D

f ( x )  dx ,

dy

x t 0,

y (0)

(9.125)

0

that is:

³

y ( x)

x

0

D

f ([ )  d[

(9.126)

D

f ( x )  dx by D ! and we get the following:

Multiply both sides of above (9.125) expression i.e. dy

D

(D !) f ( x )  dx

(D !)dy

(9.127)

D

The fractional Taylor series gives D !dy  y (D ) ( x )  dx (9.115), using this expression in the above (9.127) obtained expression we get the following formulas with a modified RL fractional derivative: D (D ) (9.128) D  y ( x) D ! f ( x) y ( x) D ! f ( x) This provides the following important relationship: y ( x)

D !D

D

§ 1

§

D !¨

1

³ © (D  1)!

D !¨ D

³

x

0

³

x

0

D 1

D

x

0

(x  [ )

f ([ )  d[

D

We have proved our definition (9.124).

x

³ © * (D )

f ( x)

³

x

0

0

D 1

(x  [ )

D 1

(x  [ )

f ([ )d[ ; D 1

(x  [ )

·

f ([ )d[ ¸

¹

·

f ([ )d[ ¸

(9.129)

¹

y ( x)

³

x

0

f ([ )d[

9.15.2 A demonstration of the integration w.r.t. (d[ )D for f ( x ) f ( x) 1 , f ( x) G ( x) and f ( x) On making f ( x )

J

x we get the following from (9.124):

D

f ([ )  d[

ED (O xD )

xJ ,

410

Chapter Nine x

³

0

J

[

D

 d[

D

³

x

D 1 J

(x  [ )

0

[ d[

§ 1

D  * (D ) ¨

³ © * (D )

* D  1 * (D  1)



0

D 1 J

(x  [ )

·

[ d[ ¸

¹



D J 0 Dx x

(9.130)

* (J  1) * (J  1  D )

D ! J ! xJ D , D  J ! On making f ( x )

x

x

J D

0 D 1

1 we get (9.124) the following: x

D ³  d[

D

x , 0  D  1; x t 0

0

(9.131)

In the physical or geometrical systems, the coordinates are drawn in a normal case via the process of integration of the differential element dx i.e. we write as

x

³

0

d[

x on this coordinate we write/draw the function y x

D ³  d[

coordinate x . Thus with the fractional differential the drawn coordinate will be D

0

x

f ( x ) w.r.t. this

D

with a function

D

mapped as f ( x ) w.r.t. the transformed coordinate x , for x t 0 , and 0  D  1 . On making f ( x )

³  f ( x) G ( x) dx

G ( x ) and by using the property of the delta-function i.e.

following relationship by using (9.124):

³

x

0

On making f ( x )

D

G ([ )  d[

D

x

D ³  ( x  [ ) G ([ ) d[ 1

Dx

0

D 1

f (0) we obtain the

, 0 D 1

(9.132)

D ED (O x ) we get from (9.124):

³

x

0

D

D



ED (O[ )  d[

D !O

D

D

D  * (D ) D ED (O x ) 1

E

D



(9.133)



D

(O x )  1

9.15.3 Some useful identities for the integration w.r.t. (d[ )D From our discussions, we have y ( x )

y ( x)

³

x

0

D !D

D

f ( x ) as a result of one-whole integration of dy

D

f ( x )  dx

giving

D

f ([ )  d[ that is: D

D

1

f ( x)

x

D!³

f ( x)

D

f ([ )  d[

0

1 d

D

D ! dx

D

³

x

0

0 D 1 D

f ([ )  d[

(9.134)



We write another useful relationship i.e.

d

D

dx As a direct consequence of the definition, i.e. following for 0  D  E  1 :

³

x

0

D

³

x

0

D

f ([ )  d[ D

f ([ )  d[

D

³

x

0



D ! f ( x) D 1

(x  [ )

(9.135)

f ([ )d[ (9.124), we derive the

Modified Fractional Calculus in conjugation with Classical Calculus

³

x

0

D E

f ([ )  d[

(D  E )

³

(D  E ) From the definition i.e.

³

x

0

D

f ([ )  d[

D

³

x

0

D 1

(x  [ )

x

0

³

D  E 1

(x  [ )

411

f ([ )d[ (9.136)

x

D 1

E

(x  [ ) (x  [ )

0

f ([ )d[

f ([ )d[ , we can write: D

f ([ )  d[

D 1

D (x  [ )

f ([ )d[

(9.137)

Using the above (9.136) relationship we get the following identity: x D E D §D  E · x E f ([ )  d[ ¨ ¸ ( x  [ ) f ([ )  d[

³

³ © D ¹ 0

0

(9.138)

E §D  E · x D ¨ ¸ ³0 ( x  [ ) f ([ )  d[ , © D ¹

0 D  E 1

9.16 Leibniz’s rule for the product of two functions for a modified fractional derivative D

From the fractional Taylor series we have (9.115) D !dy y (D )  dx . Let us write y ( x ) product of two functions. We do the following manipulations: (D )

D !d  uv

 u ( x)  v( x)

that is the

D

uv  dx (D ) D D ! u ( x)dv  v( x)du uv  dx (D ) D uv  dx u ( x) D !dv  v( x) D !du (D ) D uv  dx u ( x)  v (D ) (dx)D  v( x) u (D ) (dx)D

(9.139)

D

Dividing both sides by  dx we obtain the following formula:

 u ( x )v ( x ) For D

(D )

u ( x )v

(D )

( x )  v ( x )u

(D )

( x)

(9.140)

1 we recover the classical formula i.e.

 u ( x )v ( x ) Let u ( x )

G ( x ) and v ( x )

(1)

(1)

(1)

u ( x ) v ( x )  v ( x )u ( x )

(9.141)

f ( x ) with application of above (9.140), we have: (D )

In the above (9.142) derivation we have used

G ( x) f ( x) G ( x) f (D ) ( x)  f ( x)G (D ) ( x) (D ) (D ) G ( x) f ( x)  f (0) 0 (D ) (D ) G  x f ( x ) G ( x ) f ( x ) (D ) G ( x ) f ( x ) f (0) and then DD f (0)  f (0)

modified fractional RL derivative of the constant i.e. f (0) differential, we obtain:

³G

(D )

D

³

 G ([ ) f D 1

³

D ( x  [ )

We have

 u ( x )v ( x )

(D )

u ( x )v

(D )

( x )  v ( x )u

(D )

0;

that is a

C is zero. Now integrating (9.142) with the fractional

([ ) f ([ )  d[

D x

(9.142)

D 1 (D )

f

(D )

G ([ ) f

D

([ )  d[ , 0  D  1

(D )

([ )d[

(9.143)

(0)

( x ) i.e. the product rule. First let u ( x ) D

0  a , b  1 . Then using the product rule we get the following for D

x x : a b

a

x and v ( x )

b

x , with

412

Chapter Nine D

D

x x a b

a

x  x D x b

D

x D a

x



b

a

D



b! x b D  xb a ! x a D (b D )! ( a D )!



(9.144)

b ! · a  b D § a! ¨ ( a  D )!  (b  D )! ¸ x © ¹ D m

m! ( m D )!

While using a direct formula i.e. D x

x

m D

we write the following:

x x a b

D

D



D

( a  b )!



a b

Dx x

( a  b  D )!

a  b D

They (9.144) (9.145) are not the same yet they are of the form Kx

x

a  b D

(9.145)

, with K as a constant. Is there a

a b

a b

contradiction? Do we observe that these two formulas are dealing with x x and x respectively in (9.144) and (9.145)? We must conclude from the point of a fractional derivative that they are not the same in the sense that their a (D )

fractional differentials or differential increments are not of the same value. The ( x ) required in the case-1 i.e. in (9.144) (i.e. using product rule) while ( x (9.145) (i.e. direct calculation). We write these two cases as follows: D

D d

D

x x x x

d

a b

D



a b

are

are required in the case-2 i.e. in

(dx )

D

a

b (D )

D

b

a (D )

a b

(x

)

D

(dx )



(9.146)

a b

D

x x

D

(dx )

a b

dx

x

b (D )

x x ;

D

D

)

D

 ( x ) (dx)  x  ( x d x dx ; d  x x

a b

D

a  b (D )

D

(dx ) , ( x )

D

a  b (D )

)

D

(dx )

9.17 Integration by parts for a fractional differential We have the formula (9.140) i.e.  uv

(D )

D

 dx



u ( x) v

(D )

D

(dx )

 v( x) u

(D )

D

(dx )

.

Now integrating this with

D

respect of fractional differential (dx ) , we obtain the following: ( )

D D D D D D ³ uv  dx ³ u ( x)  v (dx)  ³ v( x) u (dx)

In addition, we have D ! dy

y

(D )

D

 dx

( )

y( x)

0

³

x

0

y

(D )

D

 dx

(9.147)

, (9.115) integrating this on both sides we write:

³ Which gives us D ! y ( x )

( )

D ! dy

³

x

0

y

(D )

D

 dx

(9.148)

. Using this obtained relationship, we write: ( )

D D ³  u ( x ) v ( x )  dx

D ! u ( x )v ( x )

(9.149)

With the use of a concept of fractional differentials, we write the following:

³

b

a b

u

(D )

³  a

b

D

( x ) v ( x )  dx

D

v d u

D ! u ( x )v ( x )  a

b

D ! vu  a

b

³

b

a

u ( x)v

³

a b

u

(1)

³ 

( x ) v ( x )  dx

³ v  du a

b

D

( x )  dx

a

(vu ) a 

(9.150)

D

u d v

It is like how in classical calculus we have the following, which is recovered when we place D b

(D )

b

1

b

 u ( x)v( x) a  ³a u ( x)v (1) ( x)  dx b

³ u  dv a

(9.151)

Modified Fractional Calculus in conjugation with Classical Calculus

413

9.18 The chain rule for a modified fractional derivative 9.18.1 The derivation of three formulas for the chain rule with a modified fractional derivative Assume that f (u ) is D  this differentiable with respect to u and u ( x ) is differentiable with respect to x , then we have the following:

 f u ( x)

§ 'D  f  u ( x ) · lim ¨ ¸ D 'x p 0 ¨ ¸ ' x  © ¹ D § '  f  u ( x ) ·§  'u ( x ) D lim ¨ ¸¨ D 'x p 0 ¨ ¸¨  'x D ' u ( x )  © ¹©

(D )

· ¸ ¸ ¹

(9.152)

D

§ d D f ( u ) · § du · ¸ ¨ D ¸¨ © du ¹ © dx ¹ We write from the above (9.152), the following:

 f u ( x)

(D )

 f u u (D ) u

D (1) x ( x)



(9.153)

Now we assume that f (u ) is differentiable with respect to u and u ( x ) is D  th differentiable w.r.t. x . We have a fractional derivative of f (u ) u as: D

d u du

1

D

u

* 2  D

u

1D

1D

(9.154)

1  D !

From the above (9.154) expression, we get a conversion formula such as:

1  D !

D

d u

1

D

1D

 du

1D

d f

u

(9.155)

Likewise (9.155), we also write: D

d f

1  D !

1

f

D

(9.156)

This being the case one gets the following steps: D

D

d f dx

D

d f d u

D

D

d u dx

D D

1  D ! f 1D  d f 1  D ! u1D  du D

dx

(9.157)

1D

D

d u

§f· ¨ ¸ ©u¹

D

(1) D fu



(D ) u x ( x)

For this case, we write the chain rule as the following expression:

 f u ( x)

1D

§ f (u ( x )) · ¨ u ( x) ¸ © ¹

(D )

f

D (1) u (u ( x ))



u

(D )

( x)

(9.158)

Now we write the chain rule when both are D  th differentiable, then we have the following: D

d f dx D

Use d u

1  D !

1

u

1D

D

 du

D

D

D

d f d u D

d u dx

(9.159)

D

for the denominator on RHS of (9.159) and we get the following expression: D

d f dx

D

D

D

D

d f d u D

d u dx

D

D

d f

1  D !

1

u

d u 1D

D

 du

dx

D

§ dD f · § dD u ¨ D ¸ ¸ ¨  dx D d u  © ¹©

1  D ! uD 1 ¨¨

· ¸ ¸ ¹

1  D ! uD 1  fu(D ) (u ) u x(D ) ( x) Therefore, in this case we have a chain rule expressed as follows:

(9.160)

414

Chapter Nine

 f u ( x)

(D )

1  D ! uD 1  fu(D ) (u ) u x(D ) ( x)

(9.161)

 u ( x)  *(2  D ) uD 1  fu(D ) (u )  u x(D ) ( x) 1 D As a corollary, the D  th derivative of x provides a relationship for dD x  (1  D )! x1D  dx , or D D  (1  D )! xD 1  dD x  dx which allows us to write successively dD f f x(D )  dx . From this we get D D 1 D (D ) (9.162) f x ( x )  (1  D )! x d x d f (D )

fx

Compare this (9.162) with the above (9.161) chain rule. Thus, we have three arrangements for chain rule formulas as listed below:

 f u ( x)

(i)

(D )

 f u ( x)

(ii)

 f u u (D ) u

D (1) x ( x)



1D

§ f (u ( x )) · ¨ u ( x) ¸ © ¹

(D )

f

D (1) u (u ( x ))



(D )

u x ( x)

(9.163)

(D )

1  D ! uD 1  fu(D ) (u ) u x(D ) ( x) (1) 1 all the above formulas obtain the form  f  u ( x )  fu(1) (u ) u x(1) ( x) . The classical chain  f u ( x)

(iii)

Note as we put D

rule is therefore recovered.

9.18.2 The application of three formulas for the chain rule with a modified fractional derivative for different cases The application of these three formulas must be handled with care. The first one (i) of (9.163) is valid when u ( x ) is differentiable and f (u ) is D  th differentiable. The second (ii) of (9.163) is applied while f (u ) is differentiable and

u ( x ) is D  th differentiable. The third formula (iii) of (9.163) applies, when the functions f (u ) and u ( x ) are both D  th differentiable. Let us have f  u ( x )



E (u ) with u  f u ( x)  f (u ) u we have: D

ED (O x )

1

D

D

D

O D x , and see the above chain rules. With a

D

(1) D x

(D ) u

D

formula of (9.163), i.e. D

1

D

ED (O D x )

E

D

D (O x )

E

D

D (u )

D

O 1

D



O ED O x

D

D

f (u )

E

D

D

D

(D )

ED (u ) fu

( O ( x  xk ) )

E

D

D

ED (u ) .

(u ) D

(u )

D

O 1

D

With



the D

O ED O ( x  xk )

formula

lim N nf

N

k 0

(1)

O D ( x  xk ) to get u x of

(9.163),

we

O

1

D

write

. We construct a fractal function (Section 1.24) as

¦ ED (O ( x  x )D ) . Using this derived expression, N N D D D D  lim N nf ¦ k 0 ED (O ( x  xk ) ) O  lim N nf ¦ k 0 ED (O ( x  xk ) ) g ( x)

(9.164)

1

The above (9.164) is a known result. We take ED (O ( x  xk )D ) ; x t xk . Take u and

D

k

we

thus

have

the

following,

(D )

That is, we get g ( x ) O g ( x ) . This example we had in Section 9.12 is a fractal function of a continuous but nowhere differentiable, but everywhere D  differentiable. D

With the formula for (9.163), i.e. D

1D

D

 f u ( x)  uf  fu(1) (u )

(D )

u x ( x ) we write:

1D

D

D



D

ED (O x )



§ ED (O xD ) · 1 ¨ ¸ D © O x ¹

 

D

Du ED (u )

D



§ O · D 1D D Du  ED (u ) ¨ (1  D )! ¸  ED (O x ) © ¹



1

§ O D x1D · ¨ ¸ © (1  D )! ¹ D



(9.165)

Modified Fractional Calculus in conjugation with Classical Calculus

415

We note that the function ED (uD ) is not differentiable at u

0 w.r.t. u so it fails. (Write the function in series form, do normal differentiation term-by-term and observe that it blows up at u 0 ). D

With the formula for (9.163) i.e. D

 f u ( x)

E

D

D

D

(1  D )! u D

(O x )

f

D 1

(D ) (u ) u





1

(1  D )! O x D

u

D 1



(D ) x ( x)

we write 1

§ D 1D ·  ED (O x ) ¨ O(1 xD )! ¸ © ¹ D

(9.166)

D

O ED (O x ) 1

D ED (u ) is not differentiable and u

Again, we get the known result. We note that f (u )

O D x is differentiable as

well as D  th differentiable. Similarly this chain rule (9.163) can be applied for a function which is nowhere differentiable as defined in Section 9.11.6 i.e. f ( x )

lim N nf

x t xk , in the interval (D )

fu f

(u )

 12

( x)

 !   D u  12 D ! 1 2

f u

1 2

1

2



(u )

Let f  u ( x ) x ab formula, we write:

N

¦ x  x k 1

1

2

 b , with f ( x )

b for x  xi and

>0,1@ { > x1 , x2 ,...xN @ . Take u  12 ! , for D 12 . Thus, we

u (1) x

a b

(x )

k

1

2

f

 12

( x)

 12 !.

x x1 , x2 ,... xN

k 1

( x  xk ) , this gives u x(1) get from

1,

f (u )

1

k

u

1

2

2

 b for

giving

formula (i) of (9.163) the result i.e.

Note here that f

 12

b

0.

a

b u , with 0  a , b  1 . Take u

x . According to the application of a standard RL

§ ( ab)! · ab D ¨ ( ab  D )! ¸ x © ¹

ab

D

Dx ( x ) D

N

¦ x  x

f ( x)

(9.167)

D

Since the f (0) 0 for this function D is the same as D the function is self-similar at x 0 . With the first chain rule formula of (9.163) we write: b! b D a 1 D D D  f (u ( x )) u ax b  D !



b!

b  D ! b !a

x

a ( b D ) D aD D

x

ab D

D

b  D !



a x

(9.168)

With second chain rule formula of (9.163) we get: 1D

D

D

 f (u ( x))

§ ub · ¨¨ ¸¸ © u ¹

§

bu ¨  a a D! ! x © b 1 D

1D

x ab  a

a

b ( a !)

 a  D !

x

D

b

x

a ( b 1)

a D

· ¸ ¹

§

¨  a a D! ! x © D

a D

· ¸ ¹

(9.169)

ab D

With the third chain rule formula of (9.163) we write: D

D

 f (u ( x))

(1  D )! x

b!

a (D 1)

b  D !

(1  D )! a ! b ! ( a  D )!(b  D )!

x

( x a )b D

ab D

a!

 a  D !

x

a D

(9.170)

416

Chapter Nine a

b

We note that x and u are both non-differentiable, hence the chain rule formula first and second are disqualified. Therefore, the fractional differentiation with the RL formula (Euler rule) and the third chain rule formula of (9.163) are applicable. The results in (9.167) and (9.170) are different since different differential increments are involved in these two functions, though they are the same.

9.19 Coarse grained system 9.19.1 Need for coarse graining In classical calculus, everything happens when an elemental point (the differential element say 't ) is made infinitesimally small i.e. made zero. Thereby we take the rate for a function x (t ) with respect to this element, and classically the term for this rate as 'x / 't , in the limit as 't p 0 ; that is a normal differentiation dx / dt . In a coarse grained system (or phenomena), everything happens as if this elemental point or the differential element 't is having D

some spread (or thickness), and is non-zero. This can be viewed as a fractional differential quantity ( 't )

with

D

0  D  1 ; so ( 't ) ! 't as we take the limit 't p 0 . This gives us the rate as per the unit of this fractional D

D

differential i.e. 'x /( 't ) , suggesting that the usage of a fractional derivative i.e. dx /(dt ) . D

We have coarse graining in a space variable x and or a time variable t , to have a fractional differential as (dx )

or

D

(dt ) , for 0  D  1 .

9.19.2 The fractional velocity uD with coarse grained time differential With a coarse-grained time (CGT) we write velocity as the following relationship: dx uD D  dt According to the fractional Taylor series approximation we have the following conversion expressions:

dx

1

D!

x

(D )

D

(dt )

,

D !

dx

1

D

d x D

 dt

D

 dt

(9.171)

(9.172)

Therefore, for CGT we get the following:

dx

uD (t )

D

 dt

D !

1

D

d x

(9.173)

D

 dt 1 D !  x (D ) (t )

0  D  1, dt ! 0

9.19.3 Fractional velocity vD with a coarse grained space differential For a coarse grained space (CGS) variable, we write: D

vD (t )

Given a function y following condition

f ( x ) and its inverse x

dx ! 0 , 0  D  1 (9.174) dt g ( y ) , their fractional derivative of order D , with 0  D  1 satisfy the y

The proof is given in the following steps:

 dx

(D )

( x) x

(D )

( y)

2

1D

 (1  D )!  xy

(9.175)

Modified Fractional Calculus in conjugation with Classical Calculus

y

(D )

( x) x

(D )

417

§ dD y · § dD x · ( y) ¨ D ¸ ¨ D ¸ © dx ¹ © dy ¹ § dD y · § dD x · ¨ D ¸¨ D ¸ © dy ¹ © dx ¹

(9.176)

1 § 1 1D · § 1D · ¨ (1  D )! y ¸ ¨ (1  D )! x ¸ © ¹© ¹ 1D

 xy

 (1  D )!

2

Therefore, for CGS we have: D

vD (t )

D

 dx

D ! dx

dt

D ! dt D

 dx D!



,

D

d t

t

(D )

( x)

D

d t ( x)



dx

(9.177)

D

D!

t

(D )

( x)



The above (9.177) steps express the quantity i.e. vD (t ) , in terms of the fractional derivative of the time variable i.e.

t

(D )

( x ) w.r.t. space variable i.e. t

Using y (D ) ( x ) x (D ) ( y )

2

1D

 (1  D )!  xy

vD (t )

(D )

D

d t ( x)

( x)

dx

(9.178)

D

(9.175) and manipulating the above (9.177) steps we get the following:



t

x D !  x (t )

(D )

(D )

(t )



(D )

D! x

D!

(D )

( x)

(t ) t

( x)



(D ) 2

D !  (1  D )!

1D

 (1  D )!  xt D 1 (D ) kD  xt  x (t )

kD

2

D 1

( xt )

D ! (1  D )!

x

(D )

(t )



(9.179)

2

We have fractional velocities described as: 1

x

uD (t )

D !

vD (t )

kD  xt

(D )

D 1



(t ) , 0  D  1, dt ! 0

x

(D )



(t ) , kD

D ! (1  D )!

(9.180)

2

for CGT and CGS cases respectively.

9.20 The solution for a fractional differential equation with a modified fractional derivative 9.20.1 Fractional differential equation defined with modified fractional derivative We take the fractional differential equation: x

(D )

(t )

b (t ) ,

x (0)

(9.181)

x0

That is (9.181) with a modified RL fractional derivative: D

D

Dt x (t )

d x dt

D

D

b (t )

d x

D

b ( t )  dt

(9.182)

One successively has the following:

³

x (t )

x0

D

d [

x (t )  x0

t

D

³ b(W )  dW 0

(9.183)

418

Chapter Nine

Therefore we write from the definition of integration with respect to a fractional differential (9.124) i.e. t

D

³ b(W )  dW

D

0

t

D 1

³ (t  W ) 0

b (W )dW , the following solution: x (t )

t

D 1

³ (t  W )

x0  D

b (W )dW

0

(9.184)

We take the fractional differential equation with a modified RL fractional derivative as: (D ) x (t ) a (t ) x (t )  b (t ) x (0) We will proceed as we do in the case of a classical calculus in following section.

(9.185)

x0

9.20.2 Defining the function LnD ( x ) in conjugation to ln x and obtaining the solution to a fractional differential equation with a modified fractional derivative The solution to the homogeneous equation in classical calculus

x (t )

C exp

³  t

0

x

(1)

 a (t )  x(t )

(t )

is given as



a (W ) dW . This comes from: (1)

x (t )

³

x (t )

x0

§ d[ (t ) · ¨ [ (t ) ¸ © ¹

ln x (t ) x (t )

dx (t )

a (t ) x (t )

a (t ) x (t )

t

³  a(W ) dW 0

ln x0  x0 exp

dt

(9.186)

t

³  a(W ) dW 0

t

³  a(W ) dW 0

D

In conjugation with an exponential function, exp( x ) we have seen that we have the Mittag-Leffler function ED ( x ) in fractional calculus. Similarly, the fractional logarithm function LnD x may be defined as follows: D

³

d x x

§x· ¨ ¸ ©C¹

LnD

§x· ¨ ¸!0 ©C¹

ED  LnD x

x

(9.187)

We note that LnD x should not be confused as a usual logarithm to the base D . With the modified RL derivative we have the property DD LnD x

D

D

 (1  D )!

2

x

D

, for 0  D  1 (we prove this in Section 9.20.3). With the above

(9.187) definition of LnD x we write the solution for x following:

x

³

(D )

(D )

D

(t )

x (t )

x0

LnD

a (t ) x (t ) as done to the above changes to the

(t )

§ dD [ (t ) · ¨ ¸ © [ (t ) ¹

§ x (t ) · ¨ ¸ © x0 ¹

t

D

0

x (t )

³ ³

x0 ED

D

0

t

x (t ) x0

x (t )

a (t ) x (t )

³ a(W )  dW

  ED

D

(dt )

³ a(W )  dW

ED LnD

x0

d x (t )

a (t ) x (t )

t

0

ED

³

t

D

t

0

D

a (W )  dW

0

a (W )  dW

(9.188)



D

a (W )  dW



Some properties of a fractional logarithm function LnD x are produced in Appendix D, where we can see by making

D

1 we get the standard properties of a natural logarithm function. We note that as we said for ED ( x ) that is a

Modified Fractional Calculus in conjugation with Classical Calculus

419

fractal function, the same is LnD x the non-differentiable type, where the modified RL fractional derivative Jumarie type applies. The solution of the homogeneous equation x

(D )

a (t ) x (t ) therefore could be as follows:

(t )

We now do an D derivative of

ED

³

t

0

D

a (W )  dW



³

§ ©

x (t )

C ¨ ED

t

D

a (W )  dW

0

·¸¹

(9.189)

by using the chain rule (9.163) number three (iii) that is as

following:

 f  u (t ) §E ¨ D ©

We note that (D )

 ED (O x) x

³

t

0

D

a (W )  dW

OD

D 1D

x



t

1  D ! uD 1  fu(D ) (u ) ut(D ) (t )

a (W )  dW



· ¸ ¹

1  D ! §¨©

³

t

³

(D )

(D )

t

0

D

dD dtD

(D )

(9.190) D

a (W )  dW

0



D 1

· ¸ ¹

(D )



t

D

 ED (u ) u ³0 a (W )  dW



(D ) t

ª t a (W )  dW D º D ! a (t ) and with the formula (9.102) obtained earlier i.e. ¬« ³0 ¼»

ED  O x we proceed as follows:

§ x (t ) ¨ CED © (D )

³

t

D

0

a (W )  dW

§ ©

³

C 1  D !¨

§ ©

C 1  D !¨

t

0

³

t

0



· ¸ ¹

(D )

D

a (W )  dW

D

a (W )  dW

C (1  D )!D

D

D 1



· ¸ ¹

D 1

·¸¹

D ! a (t ) ED

(D )



t

D

 ED (u ) u ³0 a(W )  dW 1D





§ D t D ¨ D ³0 a (W )  dW ¨ ¨ E t a (W )  dW D ¨ D ³0 ©





³

t

0

D

a (W )  dW



(D ) t

· ¸ ¸ D ! a (t ) ¸ ¸ ¹

(9.191)



(1  D )!D D ! a(t )CE  ³ a(W ) dW C a (t ) x (t ) C  (1  D )!D D ! t

D

D

D

0

D

1

The above (9.191) steps also indicate that x (t )

1

§ ©

C ¨ ED

³

t

0

D

a (W )  dW

·¸¹

is a possible solution to the equation

(D )

x (t ) a (t ) x (t ) (with adjustments of constants). This being the case let us look for a special solution of the complete equation in the form i.e. x (t ) Use formula (9.140) i.e.  u ( x )v ( x )

(D )

u

(D )

( x )v ( x )  u ( x )v

(D )

C (t ) ED

³

t

0

D

a (W )  dW

( x ) to write the following:



(9.192)

420

Chapter Nine

x

(D )

C

(t )

(D )



(t ) ED

³

t

§ dD

  CD (t ) ¨

© dt

C

(D )



D

a (W )  dW

0

t

D

³

¹

a (W )  dW



  CD (t ) a (t ) ED

³



(t ) ED

³

·

a (W )  dW ¸ ED 0

D t

D

0

³

t

0



D

a (W )  dW

(9.193) t



D

0

a (W )  dW



· ¸ ¹

b (t )  a (t ) x (t ) We write CD

(1  D )!D

D

D ! C (t ) while doing §¨ C (t ) ED

©

³

t

D

a (W )  dW

0

(D )

in the above (9.193) steps by using

the chain rule. Use the above (9.193) expression in the given equation i.e. following:

C

(D )

C

(D )

C

(D )



(t ) ED (t )

(t )

³

t

0

x

(D )



D

a (W )  dW

a (t ) x (t )  b(t ) and write the

(t )

b (t )

b (t )



³

ED

t

a (W )  dW

0

(9.194)



D

b(t ) §¨© E  ³ a(W )  dW t

1

D

D

0

·¸¹

We have used the following representation: 1

1

ED ( x )

Integrating (9.194) by recognizing C

(D )

(t )

dD C ( t ) (dt )D

(9.195)

ED ( x )

we have the following:

³ b(u ) ¨© ED  ³ a(W )(dW ) t

C (t )

§

t

1

0

D

0

·¸¹ 

D

du

(9.196)

The general solution we get is:

x (t )



t

C (t )

(9.197)

³ b(u ) ED  ³ a(W )(dW ) du t

t

1

0

§ x  t bE 1 ¨ 0 ³0 D ©



D

D

0

A fractional differential equation with a constant coefficient we write it as: (D ) x (t ) ax (t )  b Using the above (9.197) result, we get:

x (t )



D

 x0  C (t ) ED ³0 a(W )  dW

³

t

0

x0  b

³

x (0) D

a  dW

t

0

1

ED



D

· ¹

du ¸ ED



D

 at  du D

(9.198)

x0

t

³

t

0



D

a  dW



ED at

D

(9.199) D

  bE  at ³ E  at  du

x0 ED at We simplify the second term of (9.199) as follows:

D

D

D

0

1

D

D

Modified Fractional Calculus in conjugation with Classical Calculus

t

D

 ³ E  aW  dW

bED at

D

0

1

D

t

D

·

D

D

0

D

0

D

D

D

0

t

D

D

W t

D

D

D

W 0

D

In the above (9.200) we are

(9.200)

D

D

D

t

D

 ³  E  a(W b ³ E  a ( t  W )  dW b b  ª E  a (t  W ) º  E  at  1 ¬ ¼ a a assuming that  E (ax )  E (ay ) E  a( x  y) bED at

1

§

 ³ ¨¨ E 1aW ¸¸  dW ©  ¹ )  dW

bED at

D

421

D

D

D

D

D

and its inverse as

D ED ( a (  x ) ) . The proofs of these assumptions are mentioned in Section 9.20.3. Now combining them

ED ( ax )

 , we write it in compact form:

with the term x0 ED at

D

 x0  ba ED  atD   ba

x (t )









(9.201)



9.20.3 Proof of the identity ED ( axD ) ED ( ayD ) t ED a ( x  y )D in respect of a modified fractional derivative definition We know from our classical calculus, that a differentiable solution of the functional equation f ( x  y )

f ( x ) f ( y ) is

x

the exponential function i.e. f ( x ) { e , and is a unique one. We have developed in this chapter fractional calculus via the fractional difference of non-differentiable functions. Therefore, the solution of the functional equation f ( x  y ) f ( x ) f ( y ) is exactly defined as the solution for a linear fractional differential equation. Consider the functional equation f ( x  y ) f ( x ) f ( y ) , for x ! 0 and y ! 0 where f is real valued function. If one assumes f ( x ) is not differentiable but is D  differentiable where 0  D  1 for every x , then it is the solution of f

(D )

( x)

f

(D )



(0) f ( x ) for which the solution is f ( x )

ED

 f

(D )

.

(0) x

D

This we have described in Section

9.9.2. Assuming f is D  differentiable, we apply the chain rule formula (9.163) (iii) to the functional equation f ( x  y ) f ( x ) f ( y ) , with x  y u . We perform a first fractional differentiation w.r.t x , and then w.r.t y for this functional equation. We write the following steps:

From the above (9.202) we have f

(D )

( x) f ( y )

f ( x) f

f

(D )

(u )(1)

f

(D )

(u )(1)

(D )

(D )

D

f

D

f ( x) f

( x) f ( y ) ; u

(D )

( x)

f ( x) From the above (9.203) we obtain a fractional differential equation ED

f

(D )

(0) x

D

. We note that x

( y ): u

(9.202)

x y

( y ) , and we have the following ratio as a constant O :

f

f ( x)

(D )

x y

f

(D )

( y)

f ( y) f

(D )

(9.203)

O

( x)

O f ( x) ,

which is exactly

0 is a non-differentiable point in f ( x ) . Thus, we have the function f as

the Mittag-Leffler function i.e. f { ED . Therefore, we write f ( x ) conjugation



to D

ED a ( x  y )



a

D

ED (O x ) , f ( y )

classical D

exponential

D

ED (O y ) thus f ( x  y )

functions,

for

the



D

ED O ( x  y )

, and our identity in the

non-differentiable

function

ED

as

D

ED ( ax ) ED ( ay ) .

We may now prove that f the following:

(D )

( x)

O x satisfies f ( x  y )

f ( x ) f ( y ) . The equation f

(D )

( x)

O x provides us with

422

Chapter Nine (2D )

f

( x)

f

(D )

f

f

(D )

 O f ( x)

(D )



( x) ,

f

Of

(D )

(D )

( x)

O f ( x) (9.204)

( x)

2

O f ( x) ( kD )

k

From (9.204) we have f ( x ) O f ( x ) ; k 0,1, 2, 3, ... . The fractional Taylor series (9.108) for nondifferentiable functions yields the following by making h m y in the expansion f ( x  h ) , considering the nondifferentiability in x : Dk

f

f ( x  h)

h

( k)

( x) ,

ymh

( k)

( x) ;

f

¦ (D k )! f D

f ( x) 

k 1 f

f ( x  y)

y

Dk

¦ (D k )! f D

f ( x) 

(D k )

k

( x)

O f ( x)

k 1

f

¦

f ( x) 

k Dk

O y

(D k )!

k 1

§

f ( x) ¨ Ina similar way with (9.205), we write

k Dk

f

0

f ( x  y)

D ED (O z ) or the functional equation f ( x  y )



D

ED O ( x  y )

D

D

ED (O y )

k 0

¦

¦

f

k Dk

O x k 0 (D k )!

f

O k xD k k 0 (D k )!



f ( y ) f ( x ) by using the fractional Taylor D

ED (O x ) .

This (9.205) says that f ( z ) is D ED (O y ) and

D

ED (O x ) , f ( y )

. 



k Dk

O y

¦ (D k )!

f ( y)

f ( x ) f ( y ) . Thus we have f ( x )



D

Now we verify the identity of ED a ( x  y ) ED a ( x  y )

f ( x) f ( y ) ,

f ( y)

series f ( x  h ) with x m h and then using f ( x )

f ( x  y)

f

· ¸ (D k )! ¹

O y

¦

©k

(9.205)

f ( x)

D

D

ED ( ax ) ED ( ay ) . We apply the fractional Taylor series to

with y m h as follows: f

f ( x  h)

f ( x) 

Dk

h

¦ (D k )!DD

k

f ( x) , y m h

k 1



D

ED a ( x  y )



f

D

y

ED ( ax )  D

§

ED ( ax ) ¨

k Dk

¦



D

f

¦

©k



Dk

k

D

D

D ED ( ax ) a ED ( ax )

0

(9.206)

D

a y ED ( ax ) (D k )!

k 1

In our identity ED a ( x  y )

k

k 1

f

D

D

Dk

¦ (D k )!DD ED (axD )

ED ( ax ) 

k Dk

f

· D D ¸ ED (ax ) ED (ay ) , (D k )! ¹

a y

k Dk

a y

¦ (D k )!

D

ED ( ay )

k 0

D

ED ( ax ) ED ( ay ) , we have considered positive values for x and y .

However x and y may have negative values with some conditions; as for general negative values x be

complex



D

ED a ( x  y )

D

1 2 n 1

numbers



D

depending

on

D

e.g.

for

1 2

D

.

In

to

ED ( ax ) ED ( ay ) for negative values it is better that we restrict D to



D

with n as integers. Thus for these values, we can write ED (  x )

result, we get:

order

D



D

D

have 1

ED (  x ) for D

3

and y this

D

may result

, 1 5 , 1 7 , 1 9 ,... i.e. 1

2 n 1 ;

and as a

Modified Fractional Calculus in conjugation with Classical Calculus

  ( x  x)  ( x  x) D

ED ( x ) ED ( y ) ;

D

D

ED (0)

D

ED

ED ( x  y ) ED ED

D

y

423

x

1

 x  ( x) D

(9.207)

D

1

D

ED ( x ) ED ((  x ) ) ; D D

2n  1

D

ED ( x ) ED (  x ) D

D

From (9.207) we obtain ED ( x ) ED (  x )

1 that gives:

1

D

ED (  x )





D

We note here that f ( x  y )

D

1

1

D

ED ( x ) ; D

D

ED ( x )

(9.208)

2n  1

D

D f ( x ) f ( y ) applies to non-differentiable functions only, (with f (u )

D

ED (u ) )

or to functions defined on a ‘fractal set’. Otherwise the continuous set of definitions we would have is



D

f ( x  y)

d f (x

D

D

) f (y ).

As with classical calculus we have an exponential function i.e. y

f ( x)

e

x

corresponding to x

similarly for non-differentiable functions with the modified RL fractional derivative we have y the

y

(D )

corresponding D

inverse 1

( x) x ( y )

 (1D )! 2

function

1D

( xy )

is

 LnD y

x

1

g ( y) .

D

D

. Using this and y

From

 LnD y

f ( x ) ; and x

ED ( x )

derivation 1

ln y ;

g ( y) D

ED ( x )

(9.176),

f ( x ) and

we

have

g ( y ) , we may write the

D

following:

f

(D )

g

(D )

( x) g

(D )

1

( y)

 (1  D )!

1D

1

( y)

1D

( xy )

2

( xy )

 (1  D )!

2

f

(D )

( x)

;

f

1D

( xy )

1

 (1  D )!

2

 (1  D )!

2

y

 (1  D )!

y

y

(1D )

D

( x)

1

 LnD y

1

(D )

ED ( x )

y

1D 1D

x

2

 LnD y

, x

y

(D )

fx

u ( x)

 f f u

(1) u (u )

g

D



(D )

(D )

u x , by taking f (u ) D

( y) D

 LnD y

D

1D

1

§u D · ¨ ¸ © u ¹ 1 Equating (9.209) and (9.210), we get:

D

D

u

 D1 1

1

u D, u 1D

§ f (u ) · ¨ ¸ © u ¹

1 D

f

D (1) u (u )



1

(D )

uy ,

D

D

D

g ( y)

f (u ) u D , u 1D

D

  

1

D

 LnD y

1

D

i.e.

LnD y we write the following expression

1  D1 1 § 1  D1 1 · D ¨ u ¸  D LnD y D u D ©D ¹

 D Ln y

D

D

On the other hand, by use of the chain rule formula (ii) of (9.163) on 1D

(9.209)

1

 Ln y D  D

1 D

1

 D Ln y D

D

u

 D1 1

LnD y D

D LnD y



(9.210)

424

Chapter Nine

1

D

D

 LnD y

 D1 1

D

D

D LnD y The functional equation f ( x  y ) y ! 0 . In classical calculus g (u )



D

D LnD y

D

 (1  D )!

y

2



D

 LnD y

1

 (1  D )!

;

2

y

(1D )

D

(9.211)

0 D 1

f ( x ) f ( y ) has an inverse equation as g ( xy ) g ( x )  g ( y ) , for x ! 0 and ln u is the solution of this inverse functional equation g ( xy ) g ( x )  g ( y ) .

Now we assume g is a non-differentiable but is D  differentiable with 0  D  1 ; then g (u ) Let us take an D  derivative of g ( xy ) w.r.t. x we get g

(D )

g

D

obtain

(D )



( x) y

(u ) y

D

g

g (D )

(D )



D

g

(D )

(u ) .

(D )

(u ) x

D

g

(D )

g

This gives us the equation

(D )

D



comparing it with this we get g



( x) x

1

D

c , g ( x)

1

u

( y ) . From these two expressions, we

where c is a constant. We have modified the RL derivative of LnD x as D LnD x (D )

LnD u .

g ( x )  g ( y ) . With the chain rule formula (i) of (9.163), making xy

( x ) and w.r.t y we get g

( y) x

D



( x) x D

D

g

D

(D )

 (1  D )!

2



( y) y

x

D

D

c,

(9.211),

LnD x .

1

We note here that  g ( xy ) D  g ( x ) D   g ( y ) D applies to non-differentiable functions only, (with g (u ) LnD u ) or to the functions defined on the ‘fractal set’. Otherwise, through a continuous set of definitions we would have 1

1

1

 g ( xy ) t  g ( x)   g ( y ) .  LnD ( xy )  LnD ( x)   LnD ( y ) D

D

1

1

D

We

D

D

1

D

have

expression

for

g ( x)

LnD x

as

(Appendix D).

9.20.4 The fractional differential equation x

(D )

O  a (t ) x(t )  b(t ) with a

(t )

modified fractional derivative For a fractional differential equation: x

With x (0)

(D )

O  a (t ) x (t )  b (t )

(t )

(9.212)

x0 from the solution as obtained above (9.197) we write the following: t

t

1

³ b(u ) ED

C (t ) In this case with b (t )

 ³

D

 x0  C (t ) ED O ³0 a(W )  dW

x (t )

0

O

t

0

D

a (W )(dW )



D

(9.213)

du

0 and we have for: x

(D )

§ x  t b(u )E 1 O t a (W )  dW D ¨ 0 ³0 D ³0 ©



O  a (t ) x (t )

(t )

(9.214)

The solution is as:

x (t )



³

x0 ED O

t

0

D

a (W )  dW



D

· ¹

³

du ¸ ED O

t

0

D

a (W )  dW



(9.215)

Modified Fractional Calculus in conjugation with Classical Calculus

9.20.5 The fractional differential equation x (D ) (t ) As a special case for (9.212) when a (t )

t

1D

425

O t1D x(t ) with a modified fractional derivative

we have the following: t

t

0

0

D D D ³ a(W )  dW ³ W  dW 1

D

t

D 1 1D

³ t  W

§ 1

t

D  *(D ) ¨

W

0

D 1 1D

³ (t  W ) © * (D )

W

0

dW

·

dW ¸

¹

D  *(D )  0 DtD ª¬t1D º¼ § * (1  D  1)

D  *(D ) ¨

t

© * (1  D  1  D )

1D D

· ¸ ¹

(9.216)

D  *(D )  *(2  D ) t  *(D  1)  *(2  D ) t D !  (1  D )! t x0 ED  OD !(1  D )! t is a solution to x

Therefore x (t )

(D )

(t )

Ot

1D

x (t ) .







9.20.6 The fractional differential equation: a x (2D ) (t )  b x (D ) (t )  c

0 with

a modified fractional derivative We take the fractional differential equation:



a x

(2D )



(t )  b x

(D )



(t )  c

(9.217)

0

This (9.217) can be written as:



D

D



D



a D D x (t )  b D x (t )  c

D

D

0

 r1

 D

D

 r2



(9.218)

0 2

Where r1 and r2 are the roots (or zeros) of the characteristic polynomial (indicial polynomial) i.e. ar  br  c

0.

We write the solutions. For r1 z r2 , we have: x (t )

and for equal roots that is the r1

r2

  C E r t

C1 ED r1t

D

D

2 D

2

(9.219)

rˆ solution is:

C

x (t )

1

 C2 t

D

E  rtˆ D

D

(9.220)

with constants C1 and C2 determined via given initial conditions.

9.21 The application to dynamics close to the equilibrium position are subjected to coarse graining 9.21.1 The dynamic system at equilibrium Consider the dynamics given by the differentials dx and dt related as dx (1)

equation i.e. x (t )

f ( x ) ; with the equilibrium point as x0 , that is f ( x0 )

variation equation i.e. d  'x dy1

f ( x )dt , or the we say via a differential

f

(1) x ( x0 )

 'x dt ,

(1) f x ( x0 ) y1dt , which has a solution as y1 (t )

where

(1)

f x ( x0 )



(1)

d dx

0 . Close to this equilibrium we have a f ( x)

x x0

.

With y1 { 'x

we have



y1 (0) exp f x ( x0 )t . This problem is the standard calculus (1)

problem, where fine-grain differentials are used. Now what happens when x (t ) graining in time (CGT) and the coarse graining in space (CGS)?

f ( x ) is subjected to the coarse

426

Chapter Nine

9.21.2 Dynamic system subjected to coarse graining in time differential dt (1)

As we have seen for a continuously differentiable deviation in the dynamic (1)

y1 (t )

f

(1) x ( x0 )

y (t ) .

(1)

This being the case we assume x (t )

1

x (t )

f ( x)

one has

f ( x ) is subjected to CGT, and we have the

following:

dx

uD (t )

D

 dt

D !

D

d x

1

(9.221)

D

 dt 1 D ! x (D ) (t )

Using the above (9.221) in dx (9.221) relationship we have:

0  D  1, dt ! 0 D

f ( x )dt we have for the CGT case as dx x

(D )

(t )

f ( x )(dt ) ; and by using the above (9.222)

D ! f ( x)

Therefore the deviation equation for the CGT case is: D

D

D

d x0

D ! f ( x0  'x )  f ( x0 )

Dt ( x0  'x )  Dt x0 d ( x0  'x ) dt

D

D



D

dt

0

D

D !¨

D

D

d y2 dt

D ! f ( x0  'x )  f ( x0 ) ,

D

§ f ( x0  'x )  f ( x0 ) · (1) ¸  'x D ! f x ( x0 )  'x ' x © ¹

d ( 'x ) dt

dt

D

d ( x0 )



(1)

D ! f x ( x0 ) y2 (t )

D

y2

'x

(9.223)



That gives: (D )

(1)

y 2 (t )

The solution is y2 (t )



y2 (0) ED D !

and which is a solution of x (D ) (t )

(1) D f x ( x0 )t

.

(9.224)

D ! f x ( x0 ) y2 (t )

We have used the earlier obtained result that is x (t )

D

x0 ED ( at )

a  x (t ) .

9.21.3 The dynamic system subjected to coarse graining in space differential dx Assume now dx

 f ( x ) dt

subjected to CGS. For CGS we have the following from (9.179): D

vD (t )

 dx

D 1

kD  xt

dt

x

(D )

(t )

D

Using the above (9.225) derivation and substituting in (dx )

D ! (1  D )!

kD

2

(9.225)

f ( x )dt we obtain for CGS the following: 1D

x

(D )

 xt

(t )

kD

f ( x)

(9.226)

The deviation equation is done as in the following steps: 1D

D

D

Dt ( x0  'x )  Dt x0

 ( x0  'x)t kD 1D

D

Dt

 'x

(D )

y3

 ( x0  'x)t kD kD

t

f ( x0

kD

f ( x0 )

1D

f ( x0

 x0t  'x ) 

1D 1D

( x0  y3 )

1D

 x0t  'x ) 

kD

f ( x0 )

1D 1D

f ( x0  y3 ) 

x0 t

kD

f ( x0 )

'x

y3

(9.227)

Modified Fractional Calculus in conjugation with Classical Calculus

Write f ( x0  y3 )

(1)

f ( x0 )  f x ( x0 ) y3 as it comes from the fact that f x ( x )

lim 'xp0

427



f ( x 'x )  f ( x ) 'x

,

to get the

following expression: (D )

y3 Use the power series expansion i.e. (D )

y3

§ t1D · 1D (1) 1D ¨ ¸ ( x0  y3 )  f ( x0 )  f x ( x0 ) y3  x0 f ( x0 ) © kD ¹



1D

( x0  y3 )

§ t1D ¨ © kD

1D

# x0



(9.228)

D

 (1  D ) x0 y3  ..... and write:

· § x10D f ( x0 )  (1  D ) x0D f ( x0 ) y3  x10D f x(1) ( x0 ) y3 · ¸ ¸ ¨¨ D (1) 2 1D  (1  D ) x0 f x ( x0 ) y3  .......  x0 f ( x0 ) ¸¹ ¹©

(9.229)

Now put y32 | 0 in the above (9.229) expression to get the final form as follows: (D )

 kD

y3

The above (9.230) expression suggests that x0

1 1D

t

 (1  D ) x

1D

D 0

f ( x0 )  x0



(1)

(9.230)

f x ( x0 ) y3

0 is a special case in the CGS case. The solution to the CGS

deviation equation is (for x0 z 0 ) the following:

y3 (t )

y3 (0) ED

(1  D )! (1  D ) x 1

1D

D 0

f ( x0 )  x0



(1)

f x ( x0 ) (t )

as a solution of x

(t )

Ot

1D

(9.231)



x0 ED OD ! (1  D )! t

From the section of the fractional differential equation in Section- 9.20.5, we have used x (t ) (D )





x (t ) as obtained therein; in (9.231).

9.21.4 The system when both differentials dx and dt are subjected to coarse graining Now we discuss when both the variables x and t are subjected to a coarse-graining process for the same dynamic P

J

system dx f ( x )dt . The differentials (dx, dt ) are changed to (dx , dt ) with 0  P  1 and 0  J  1 . We may write the system as:

 dx

P

J

f ( x )  dt

(9.232)

Depending on whether P ! J or P  J we shall be getting the following:

 dx  dx

f P

J

1

P

f

( x )  dt 1

J

J

P

P !J

(9.233)

( x )  dt

P J The above is related to the CGT and CGS systems discussed earlier, represented as follows: 1 J D P D g1 ( x )  f ( x )  dx g1 ( x)  dt P

 dx

E

g 2 ( x )  dt

E

P J

g 2 ( x)

 f ( x)

1

(9.234)

J

We may use a fractional derivative to deal with differentiable functions, but we ask the question whether this is essential, as far as function is differentiable. As far as the function is differentiable and the system is Markovian (i.e. system state is not affected by past), then its dynamic equation is quite well defined. If the dynamics are differentiable, strictly speaking, we should not need fractional derivative to analyse its behavior. We may ask, given a system which (1)

is defined by a differential equation, x (t ) contribute extra information?

f (t ) will a dynamic with a fractional derivative x

(D )

(t )

g (t ) ,

Many system dynamics show memory based behavior where a non-Markovian approach is required, there the (D )

dynamics x (t ) g (t ) , with a classical fractional derivative of the RL or Caputo type have been employed as we described up to Chapter 8. However, if the function in the consideration is not differentiable, then this modified RL fractional derivative of the Jumarie type should be of use. The basic problem is then to find suitable techniques for modeling such functions, given that they are continuous everywhere but nowhere differentiable. One way is to have the function defined on a ‘fractal set’ and another developing approach (under development and research) is to randomise the problem and describe the non-differentiability by using Gaussian white-noise. We will however not deal with this aspect.

428

Chapter Nine

9.22 Using the Mittag-Leffler function in Integral Transform formulas If the Mittag-Leffler function i.e. ED ( xD ) has a conjugate role in fractional calculus can we use this function instead of the exponential function exp( x ) in integral transform equations like Laplace Transforms and Fourier Transforms and call them D integral transforms. Then this type of new transforms will be applicable to non-differentiable dynamics, with fractional differential equations and functions, which are fractionally differentiable; and is a controversial but evolving topic.

9.22.1 Defining the fractional Laplace transform by using the Mittag-Leffler function The classical Laplace Transform is one such integral transform given by:

$ ^ f (t )`

def

F (s)

lim

³

T

T nf 0

e

 st

 f ( t ) dt

(9.235)

In conjugation with this (9.235), we have a fractional Laplace Transform defined as the following:

$D ^ f (t )`

def

FD ( s )

lim

T

D D D ³  ED ( s t ) f (t )  dt

(9.236)

T nf 0

We will have the following Table 9.4 depicting the operational identity for the fractional Laplace transform, along with listed standard classical Laplace identities. We are not proving them, as one can use all the above techniques to derive these. Fractional Laplace Transform

$D ^ f (t )`

Classical Laplace Transform

$ ^ f (t )`

FD ( s )

f

D D D ³  ED ( s t ) f (t )  dt D D $D ^t f (t )`  D  FD ( s )

³

0

0

^

D

 1a

FD

D D

$D

^ E

D D

D

`

(a t

f (t )dt (1)

 Ds

 F (s)

 1a F  as

$ ^ f (t  W d )`



 exp( sW ) F ( s)

(  s W d ) FD ( s )

D

 st

$ ^ f ( at )`

 as

$D ^ f (t  W d )`

E

e

$ t  f (t )

s

$D ^ f ( at )`

f

F (s)

f (t )`

d

$

FD ( s  a )

^ exp(at f (t )` F (s  a)

$D

^³

t

0

D !

^

$D f

(D )

`

(t )

D

f ( u )  du 1

s

D

`

$

^³

t

0

`

f ( u )  du

s

D

F (s)

FD ( s )

^

D

s FD ( s )  D ! f (0)

$ f

(1)

` s  F ( s)  f (0)

(t )

Table 9.4: Fractional Laplace transform identities The above table is derived by Jumarie, and the use of this is still developing.

9.22.2 Defining a fractional convolution integration process by use of the fractional Laplace Transforms As a fractional order convolution then we represent it as follows: t

D

 f (t ) * g (t ) D ³0 f (t  W ) g (W )  dW $D ^ f (t ) * g (t ) D `  FD ( s )  GD ( s ) While

$

the

classical

one

is

^ f (t ) * g (t ) `  F ( s) G ( s) .

t

 f (t ) * g (t ) ³0 f (t  W ) g (W )  dW

having

the

(9.237)

Laplace

transform

as

Modified Fractional Calculus in conjugation with Classical Calculus

429

9.22.3 Demonstrating fractional Laplace Transform for the Heaviside unit step function We now give a demonstration for finding the fractional Laplace transform for the Heaviside unit step function defined as u (t ) 1 for t t 0 and u (t ) 0 for t  0 . The standard classical Laplace transform is in the following steps: U (s)

$ ^u (t )`

³

f

0

³

e

f

 st

e

0

u (t )dt

 st

e

dt

 st f

( s)

0

(9.238)

1 s

We get a known result. Now we calculate the fractional Laplace transform as follows:

UD ( s )

$D ^u (t )` f

f



D D

ED  s t

0

D

u (t ) dt

D

dt D ³  t  [ E   s [  d[ ³

0



³

D D

ED  s t

f

D 1

D D

D

0

§ 1

f

D t  [ ³ © * (D )

D  * (D ) ¨

1

0

D  * (D ) §¨ Dt

D

 E   s t D D

D

© § E   sD t D f · D ¸ D  * (D ) ¨ ¨   sD ¸ 0 ¹ © Therefore, we have a pair $D ^u (t )`

D! sD

1

. Can we also then write $D



D D

ED  s [ f 0

¹

s

D! sD

(9.239)

· ¸ ¹

 * D  1

^ `

d[ ¸·

D

D! s

D

1 ? Logically, yes. Now what

u (t )

about the inverse Laplace transforms in an integral representation? This representation is a slightly difficult one, which we will discuss shortly.

9.22.4 Defining the fractional delta distribution function GD ( x) and its fractional Fourier integral representation by use of the Mittag-Leffler function First, let us discuss the Dirac Delta distribution of the fractional order. A normal classical Delta function G ( x ) or delta distribution has the property i.e. following property:

³

\

f ( x )G ( x )dx

f (0) . Similarly, we define a fractional delta function with the D

f ( x )G D ( x )  dx

³

\

Define the function in the following way:

G D ( x, H ) Then one has the limit i.e. lim G D ( x, H ) H p0

D  f (0)

­ 0, x  >0, H @ ® D ¯H , 0  x d H

(9.240)

(9.241)

G D ( x ) . A relationship with G D ( x ) and the complex valued Mittag-Leffler

is suitable to form fractional Fourier transform that is in conjugation with a standard Fourier integral that is 1 2S

³

f

f

e

i ( Z x )

dZ

G ( x ) , is following:

D

TD ³ D

f

f



D

ED i ( Z x )



D

D

 dZ D

Here TD is the ‘period’ of a complex valued Mittag-Leffler function ED i (TD ) i.e. T

2S for exp  i T

e

2S i

1.



GD ( x)

(9.242)

1 it is like the classical period

430

Chapter Nine

9.22.5 The inverse fractional Laplace transformation by the Mittag-Leffler Function As we have an inverse Laplace transform defined as: 1

 if

 exp( st )  F ( s ) ds 2S i ³

f (t )

(9.243)

 if

We attempt to write in conjugation with the above (9.243) inverse fractional Laplace transform using the following expression with theMittag-Leffler function:  if 1 D D D 1 f (t ) $D ^FD ( s )` (9.244)  FD ( s )  ds D  if ED s t i TD

³  



In a similar way, one can use the Mittag-Leffler function ED ( xD ) to have the fractional Fourier transform, and other integral transforms of the fractional order. As such, no such close form tables pertain to fractional Laplace and fractional Fourier transform exist.

9.22.6 The Mittag-Leffler function to define the fractional order Gamma function *D ( x ) As we know, the integral representation of the classical Gamma function is: f

³  exp(t ) t

*( x)

( x 1)

0

 dt

(9.245)

We can have a definition of the fractional order Gamma function as the following form: f

1

D ! ³0

*D ( x )

E

D

D

( t )

t

( x 1)D

D

 dt

(9.246)

Some interesting conjugation with the classical Gamma or Beta functions is following:

*D ( x  1)

D !  x  *D ( x)

*D ( n  1)

D !  n !

n

1

³ (1  t )

BD ( x, y )

( x 1)D ( y 1)D

t

0

D

 dt

(9.247)

*D ( x ) *D ( y )

BD ( x, y )

*D ( x  y ) These are very new developments, and further detailed research in this direction is progressing. Here we have utilised the concept developed for integration with respect of fractional differentials and used the Mittag-Leffler function of x

the form ED ( xD ) instead of the exponential function e to get several new types of fractional integral representations of transforms and functions for non-differentiable systems.

9.23 Derivatives with the order as a continuous distributed function 9.23.1 The fractional differential equation generalised from integer order to fractional order Now we generalise the concept of fractional derivatives that we discussed until Chapter-8 the classical RL and Caputo types. Here we discuss considering the fractional order of the derivative as a continuous function (or a distribution function). The first order classical linear differential equation is following on with k1 and k0 as non-zero constants: k1

dx (t ) dt

 k 0 x (t )

(9.248)

f (t )

The fractional differential equation is: D

k0

d 0 x (t ) dt

D0

D

 k1

d 1 x (t ) dt

0 d D n d 1; n The (9.249) becomes (9.248) if D 0 the following:

0 , D1

1 and k 2

k3

D1

 .....k m 1

d

D m1

dt

x (t )

D m1

f (t )

(9.249)

0,1,...( m  1) ...

k m 1

0 , k0 z 0 , k1 z 0 . We write (9.249) as

m 1

d n x (t )

¦ n 0

D

kn

dt

Dn

f (t )

(9.250)

Modified Fractional Calculus in conjugation with Classical Calculus

431

From (9.250) we write the Laplace transformed equation as (9.251) with assuming all the initial values of x (t ) as zero, with X ( s )

$ ^x (t )` and F ( s )

$ ^ f (t )` ; and by using $

^

`

dD dtD

D

s X (s) :

x (t )

§ m 1 D n · ¨ ¦ s kn ¸ X ( s) F ( s) ©n 0 ¹

,

F (s)

X (s)

¦

m 1 n

k s 0 n

Dn

(9.251)

The equation (9.251) is a general fractional differential equation in the Laplace domain, its solution is obtained via the inverse Laplace transform technique and we write: x (t )

The denominator term i.e.

¦

m 1 D s n kn n 0

X (s)

$

1

^

F (s) /

¦

m 1 n 0

kn s

Dn

`

(9.252)

F ( s ) in (9.251) has orders of fractional numbers D n , we need not

be always positive. Thus, we can still generalise this (9.252) by including negative orders as following, with the coefficients k n in (9.251) replaced by an and bn :

¦

m 1 D s n an n 0



¦

m 1  E s n bn n 0

X (s)

F (s) ; D n , E n ! 0

(9.253)

The summation on n from 0 to m  1 in (9.253) we can change in general terms from 0 to f , and write the following:

¦

f

D

n 0

s n an 

¦

f n 0

s

 En



bn X ( s )

F (s) ; D n , E n t 0

(9.254)

This (9.254) gives a much more general expression for a generalised differential equation. The above (9.254) generalised equation says that the fractional orders are distributed in real line from f to f , with orders specified for each fractional differ-integral operator by a fractional number D n for the positive order, and E n for the negative order. The coefficients an and bn are specifying respective weights of fractional orders D n and E n that participate in the generalised fractional differential equation of (9.254).

9.23.2 The concept of order distribution We take the example of a mass spring damper system as represented below: 2

d x (t )

dx ( t )

 kx (t ) f (t ) dt dt By use of the Laplace transforms in (9.255), assuming initial conditions as zero gives the following: m

2

 b0

2

( ms  b0 s  k ) X ( s )

³  g (D ) G (D  D

Using the property of delta function i.e. following expression: 2

( ms  b0 s  k ) X ( s )

0 )dD

(9.255) (9.256)

F (s)

g (D 0 ) we write for (9.256) with g (D ) { s

D

the

F (s)

 m³ s G (D  2)dD  b ³ s G (s  1)dD  k ³ s G (s  0)dD X (s) D

D

D

0

F (s)

(9.257)

§f · D ¨ ³  mG (D  2)  b0G (D  1)  kG (D ) s dD ¸ X ( s ) F ( s ) ©0 ¹ In (9.257) we have only integer orders, those are 2 , 1 and 0 with weights m , b0 and k respectively. This is depicted as the delta distributed order distribution in Figure-9.1a. We can generalise (9.257) by adding a few more terms called the fractional damping terms. Say we add extra fractional damping terms at fractional orders of 3 2 , 1 2 , with weights b1 and b2 respectively in addition to the existing integer orders; then we get the representation as follows (9.258). This is depicted in Figure-9.1b: 2

( ms  b1s

3

2

1

 b0 s  b2 s 2  k ) X ( s )

F (s)

§ f § mG D  2  b1G D  32  b0G D  1 · D · ¨ ³ ¨¨ ¸¸ s dD ¸¸ X ( s ) F ( s ) ¨ 0 b2G D  1  kG D 2 ¹ © © ¹

(9.258)

The corresponding differential equation for (9.258) is as follows, obtained by performing an inverse Laplace transform on (9.258):

432

Chapter Nine 3

2

m

d x (t ) dt

2

 b1

d 2 x (t ) dt

3

 b0

2

dx ( t ) dt

1

 b2

d 2 x (t ) dt

1

2

 kx (t )

(9.259)

f (t )

9.23.3 From the discrete delta distributed function to a continuous distribution function for describing fractional orders We generalise (9.258) with several more delta functions with respective weights as follows:

§f§ N · D · ¨ ³ ¨ ¦ k nG D  D n ¸ s dD ¸ X ( s ) F ( s ) ¹ © f © n 0 ¹

Dn  \

(9.260)

When the adjacent delta functions at D n become close and dense we can write a continuous distribution function

k (D ) . This k (D ) is a distribution function for the order D , for (9.260).For the discrete delta distributed order, we write k (D )

¦

N

k G n 0 n

D  D n . We write the continuous order system as the following (9.261), and the continuous

distributed order is depicted in Figure-9.1c:

§f · D ¨ ³  k (D ) s dD ¸ X ( s ) F ( s ) © f ¹

F (s)

X (s)

,

³

f

f

s

D

 k (D ) dD

(9.261)

In order to get solution x (t ) we need to perform the following inverse Laplace Transform:

x (t )

$

1

^



b

 F ( s ) / ³a sD  k (D ) dD

`

(9.262)

Where k (D ) is an order distribution function, which is a continuous function defined in the range a d D d b . That

> @

derivative interval is D  a , b :

Integer order delta-distributed differential equation

b0

k

m

0 §f · D ¨ ³  mG (D  2)  b0G (D  1)  kG (D ) s dD ¸ X ( s) F ( s) ©0 ¹

Fractional order delta-distributed differential equation

1

k b2

b0

D

a

2

b1

m 1 2

0

1

3 2

b

2

D

§f · D 3 1 ¨ ³  mG D  2  b1G D  2  b0G D  1  b2G D  2  kG D s dD ¸ X ( s) F ( s) ©0 ¹

Continuously distributed order differential equation

§f · D ¨ ³  k (D ) s dD ¸ X ( s ) © f ¹

c k (D )

D F ( s)

Figure 9.1: Integer order delta distributed, fractional order delta distributed and continuously distributed order differential equations

9.23.4 The continuous order distributed differential equation From (9.261) with k (D ) defined in the range a d D d b , generalised differential equation as follows:

³  b

a

D



k (D ) s dD X ( s )

F ( s ) we can write the following

Modified Fractional Calculus in conjugation with Classical Calculus D § d x (t ) · ³a ¨© k (D ) dtD ¸¹dD b

433

f (t )

(9.263)

Substituting the integral with the function k (D ) distributed and limited in the derivative interval a d D d b is the operation of a weighted averaging of fractional orders in a continuous interval. We consider 0 d a  b d 1 , and k (D ) to have a uniform distribution between a d D d b as: Consider in (9.263) f (t )

³

b

a

 k (D ) dD dD x ( t ) dt D

(9.264) k (D ) p ; a d D d b ; 0 d a  b d 1 G (t ) i.e. the driving function as a unit delta function at t 0 . Then we have

G (t ) , as our homogeneous system. For f (t ) G (t ) we have F ( s ) 1 , with this we have X ( s )

in the Laplace domain as in the following:

­ ½ 1 1 ° ° X (s) ® b ¾ D °¯ ³a ps dD °¿ ) ( s )

³

)(s)

b

a

D

ps dD

p

³

b

a

D ln s

e

D b

ª eD ln s º p« » ¬ ln s ¼D

dD ;

b

p a

s s

D ln s

e

s

D

(9.265)

a

ln s

The ) ( s ) in the above (9.265) is similar to the indicial polynomial or the characteristic function for differential equation. Thus x (t )

$

1

^ ` is the solution of a homogeneous equation of (9.265). To get the function x(t ) we 1 )(s)

can resort to the Berberan-Santos technique as we discussed in Section 6.11.7, instead of going in for the contour integration technique. The function lim s o1

s

1 p (ba )

ln s psb  ps a

X (s)

seems to have a singularity at s

. In addition, we note that lim s nf

ln s psb  ps a

1 . Applying L’Hospital’s rule we get

0 . We chose s

V 0  iZ with V 0

0 . To the right of

0 i.e. Re > s @ ! 0 we do not expect to have a singularity for X ( s ) . With this, we have following for X (iZ ) :





X (iZ )

>



§ ln Z Z b cos b2S  Z a cos a2S  S Z b sin b2S  Z a sin a2S 2 ¨ ¨ i ln Z Z b sin bS  Z a sin aS  S Z b cos bS  Z a cos aS 2 2 2 2 2 ©



p Z

@

>

2b

Z

2a

Z

a b

cos



( b  a )S 2





· ¸ ¸ ¹

(9.266)

@

This (9.266) gives Re X (iZ ) and Im X (iZ ) for use in the Berberan-Santos formula, giving the following:

x (t )

$

1

^ X ( s )`

$

1

V 0t f

x (t )

e

³  Re > X (V

S

0

^

ln s ps b  ps a

`,

V0

0



 iZ )@ cos(Zt )  Im > X (V 0  iZ )@ sin(Zt ) dZ

0

f

§ ln Z Z b cos b2S Z a cos a2S  S2 Z b sin b2S Z a sin a2S · cos(Zt ) ¸ dZ ( ba ) S a b 2b 2a ¨ ³ Z Z Z   cos p  2  S 0© ¹ 1



(9.267)

f

§ ln Z Z b sin b2S Z a sin a2S  S2 Z b cos b2S Z a cos a2S · sin(Zt ) ¸ dZ ( ba ) S 2b 2a a b ¨ ³ p Z Z Z cos  2 S 0© ¹ 1

The continuous order distributed differential equations do have significance, but this type of system has not been developed. This is a developing field.

434

Chapter Nine

9.24 Short summary We may use the fractional derivative to deal with differentiable functions, but we ask the question as to whether or not this is essential. As far as the function is differentiable and the system is Markovian (i.e. the present state of the system is not affected by past states), then it is a dynamic equation and is quite well defined and classical Newtonian calculus is sufficient. If the system is differentiable strictly speaking, we should not need a fractional derivative to analyse its behavior. We may ask if a given dynamic system, which is defined by a classical differential equation, then -if the same dynamics be described with the fractional differential equation should contribute any extra information? .Many system dynamics show memory based behavior where a non-Markovian approach is required; there the dynamics, with a classical fractional derivative of the RL or Caputo type have been employed. We started this chapter by formally deriving important rules for fractional differ-integration and its compositions. Though we had used these facts in earlier chapters, we derived them systematically from the theory of classical calculus by its generalisation. After that, we gave detailed treatment to modify the fractional calculus theory that we obtained in earlier chapters, to get parallel to the classical calculus, and called that the modified Riemann-Liouville fractional derivative of the Jumarie type; and obtained various formulas that are in conjugation with classical calculus, for non-differentiable functions. The fractional Taylor series, for non-differentiable functions that was developed using this modified concept gave several conversion formulas for fractional differentials that were used to develop and simplify the Leibniz product rule, integration by parts, and chain rule formulas. We showed how we can use this new developed concept and apply it to study the non-differentiable function, the ease with which we can solve the fractional differential equation, and formulate and solve problems with fractional differentials arising due to the coarse graining process. The new integral transform formulas (of fractional Laplace transforms, and fractional Fourier transforms) and thereby new functions (fractional Gamma, fractional Beta, fractional Delta functions) that are due to this modified fractional calculus by use of the fractal Mittag-Leffler function instead of the classical exponential function, give an immense similarity to classical calculus techniques. However, these concepts, especially with the modified fractional calculus of (Jumarie type) are a yet developing subject (and controversial too), and have immense potential in the research in mathematical sciences. Therefore, if the function in consideration is not differentiable, then this modified RL fractional derivative of the Jumarie type should be of use. The basic problem is then to find suitable techniques for modeling such functions, given that they are continuous everywhere but nowhere differentiable. One way is to have functions defined on the ‘fractal set’ and another developing approach is to randomise the problem and describe the non-differentiability by using Gaussian white noise; that is still under development. We also introduced the concept of generalising the differential equations composing derivatives where the order distribution is of continuous function type instead of discrete fractional numbers. This concept is a developing field. Nonetheless, the acceptability of this subject is still very little, and generally, most established scientists and engineers still turn a blind eye here. However, in this course we have at least learnt about the existence of this vast subject parallel to our classical calculus, and tried to appreciate this so called ‘paradox’!

9.25 References The motivation of this chapter is derived from the pioneering works which are: Michelle Caputo, “Linear Models of Dissipation whose Q is almost Frequency Independent” (1967) Jumarie G, “Laplace Transform of Fractional Order via Mittag-Leffler function and Modified Riemann-Liouville derivative” (2009); “Oscillations of non-linear systems close to equilibrium position in the presence of coarse graining time and space” (2009); “Tables of some basic fractional calculus formulae from modified RiemannLiouville derivative for non-differentiable functions” (2009); “Cauchy Integral Formula via Modified RiemannLiouville derivative for analytic functions of fractional order” (2010); “Modified Riemann-Liouville Derivative and Fractional Taylor Series of non-differentiable functions further results” (2006); “From self similarity to fractional derivative of non-differentiable functions via Mittag-Leffler function” (2008); “On the fractional solution of the equation f ( x  y ) f ( x ) f ( y ) and application to fractional Laplace transform” (2012) Abhay Parvate, Anil.D. Gangal, “Calculus on Fractal Subset on Real line-I: Formulation” (2009) Batman H. “Higher transcendental function” (1954) Igor Podlubny, “The Laplace Transform Method for Linear Differential Equations of the Fractional Order” (1994) Mandelbrot B.B, “Self-Affine and Fractal Dimension” (1985) Miller K S and Ross B, “An Introduction to Fractional Calculus and Fractional Differential Equations” (1993) Oldham and Spanier, “Fractional calculus” (1974) Munkhammar .J.D, “Fractional calculus and Taylor-Riemann series” (2005) Massopust P, Zayed A, “On the validity of Fourier Series Expansion of Fractional Order,” (2015) The details of the above pioneering literature and work are listed in the bibliography section, in alphabetical order.

APPENDIX A HIGHER TRANSCENDENTAL AND SPECIAL FUNCTIONS

This appendix presents a number of functions that are found to be useful in the solution of the problems of fractional calculus. The base function is the gamma function, which generalizes the factorial expression, used in multiple differentiations and repeated integrations, in integer order calculus. We have discussed this and its properties in Chapter One. The Mittag-Leffler function is the basis function of fractional calculus; as an exponential function it is an integer order (classical) calculus. Several modifications of the Mittag-Leffler functions, along with other variants, have been introduced and further developed since 1903, for the study of fractional calculus. These functions are called higher transcendental functions, and their use in solving fractional differential equations is similar to the use of transcendental functions for solving integer order differential equations. The use of some of these functions has been demonstrated in earlier chapters (Chapter Two to Chapter Nine).

A.1 Hyper-geometric functions We have not used these hyper-geometric functions much in the book, but we will list them here, as these functions are the basis for all other functions that we describe. The hyper-geometric function and its generalization encompass an extensive class of analytical functions. Definition and symbolism are as follows: f

ª b1 , b2 , ....., bK º «x » ¬ c1 , c2 , ....., cL ¼

j

 ( j  b1 )!  ( j  b2 )! ...  ( j  bK )!

¦ x  ( j  c )!  ( j  c )! ...  ( j  c )! j 0

1

L

2

K f

¦ j 0

(A1)

– *( j  1  b ) k

j x k L1

– *( j  1  c ) l

l 1

We write from the above (A1), the expanded form below:

ª b1 , b2 , ....., bK º «x » ¬ c1 , c2 , ....., cL ¼

* (1  b1 )* (1  b2 )...* (1  bK ) * (1  c1 )* (1  c2 )...* (1  cL ) x x

* (2  b1 )* (2  b2 )...* (2  bK ) * (2  c1 )* (2  c2 )...* (2  cL ) 2

* (3  b1 )* (3  b2 )...* (3  bK ) * (3  c1 )* (3  c2 )...* (3  cL )

n

(A2)

 ....

* ( n  1  b1 )* ( n  1  b2 )...* ( n  1  bK )

 ... * ( n  1  c1 )* ( n  1  c2 )...* ( n  1  cL ) We assume that parameter values and the values of the argument are such that they ensure convergence of the series; this will often imply restrictions x  1 . x

Say for b1 formula:

c1

1 , and b2

b3

....

bK

c2

.....

ª 1º «¬ x 1 »¼ We shall call this as

K L

cL f

¦x

0 in (A1) and (A2) then, we have the following known j

2

1  x  x  ....

(1  x )

1

(A3)

j 0

hyper-geometric. Note that the leading term (A1) within this summation will be zero if one

term of the denominator parameter is a negative integer; if cl

0

1

2

 n ; terms in x , x , x , ....., x

n 1

are absent.

Conversely, the leading terms (A1) will be infinite if one of the numerators, say b1 , is a negative integer, i.e.  n , due to the value of the gamma function at negative integer points being set at infinity. We consequently encounter this

436

Appendix A

1 * (1 n )

possibility only as the quotient, by representing that case as

,.....,bK ª x cn,,cb2,....., º ¬ 1 2 cL ¼ . While the gamma function at the

negative integer point is infinity, the reciprocal and the ratio of the gamma function at negative integer points are finite, i.e.

*(n) *( N )

  N   N  1 .....  n  2  n  1  1

N n

 Nn!! . * ( n 1) *(n)

On doing one whole derivative of the above (A2) series term by term, and by using n

1 and * (0)

0 , we get the

following steps:

d dx

bK * (2  b1 ) * (2  b2 )...* (2  bK ) ª x bc1 ,,bc2 ,....., º ¬ 1 2 ,.....,cL ¼ 0  * (2 c1 ) * (2 c2 )...* (2 cL )

 2x

* (3 b1 ) * (3 b2 )...* (3 bK ) * (3 c1 ) * (3 c2 )...* (3 cL )

...  nx



n 1 * ( n 1 b1 ) * ( n 1 b2 )...* ( n 1 bK ) * ( n 1 c1 ) * ( n 1 c2 )...* ( n 1 cL )

 ........

* (2  bK ) * (3 b ) * (3 b )...* (3 b ) § 0  **(2(2bc1 ))**(2(2bc2 )...  2 * (3 c1 ) * (3 c2 )...* (3 cK )  · L 1 2 )...* (2  cL ) 1 2 ¸ x ¨ * ( n 1 bK ) ¨ ...  nx n **((nn11bc1 ))**((nn11bc2 )... ¸  ........ 1 2 )...* ( n 1 cL ) © ¹ 1





* (1 bK ) § **(1(11)0) **(1(1bc1 ))**(1(1bc2 )... ·  1 2 )...* (1 cL ) ¨ ¸ ¨ x * (2 0) * (2b1 ) * (2b2 )...* (2bK )  ¸ * (2 1) * (2  c1 ) * (2  c2 )...* (2  cL ) 1 ¨ ¸ x ¨ x 2 * (3 0) * (3b1 ) * (3b2 )...* (3bK )  ......... ¸ * (31) * (3 c1 ) * (3 c2 )...* (3 cL ) ¨ ¸ ¨¨ ¸ * ( n 1 b1 ) * ( n 1 b2 )...* ( n 1 bK ) * ( n 1) n ......  x * ( n 11) * ( n 1 c ) * ( n 1 c )...* ( n 1 c )  .......... ¸ 1 2 L © ¹ 1 ª 0, b1 , b2 , ....., bK º x «x » ¬ 1, c1 , c2 , ...., cL ¼











d

m

bK ª x bc1 ,,bc2 ,....., º ,.....,cL ¼ dx ¬ 1 2 m

x

m



(A4)

ª 0, b1 , b2 , ...., bK º « x  m, c , c , ..., c » ; m 0, r1, r2 , .... ¬ 1 2 L¼

We can derive from similar lines as done above (A4) and write the following formula, with a restriction on m removed. Then from the above (A4) is a fractional differ-integration of order D ; for a generalized hyper-geometric function, given as: d

D

bK ª x bc1 ,,bc2 ,....., º 1 2 ,....., cL ¼ ¬ dx

D

d

D

dx

Representation of

D

ª ¬

E b ,b ,.....,b x c1 ,c2 ,.....,cK 1 2 L

x

D

ª 0, b1 , b2 , ....., bK º « x D , c , c , ....., c » ¬ 1 2 L¼

º x E D ¼

(A5)

ª E , b1 , b2 , ....., bK º « x ( E  D ), c , c , ....., c » ¬ 1 2 L¼

E ! 1

bK ª x bc1 ,,bc2 ,....., º ¬ 1 2 ,.....,cL ¼ in terms of the Gauss function F  , ; ; x is the following: K

ª b1 , b2 , ....., bK º « x c , c , ....., c » ¬ 1 2 L ¼

k § 1,1  b1 ,1  b2 , .....,1  bk ; · – k 1 FL ¨ ¸ L © 1  c1 , .....,1  cL ; x ¹ * (1  c )

* (1  b )

K 1

–

(A6)

l

l 1

The Gauss function or Gauss series is the following (as defined in the Handbook of Mathematical Functions):

F  a , b; c; x

2 F1

 a , b; c ; x

F  b , a ; c; x

f

¦ n 0

* (c ) * ( a ) * (b ) for x  1 .

f

¦ n 0

( a ) n (b ) n x (c ) n

n!

* ( a  n ) * (b  n ) x * (c  n)

n

n

n!

(A7)

Higher Transcendental and Special Functions

We have used ( a ) n

F  a , b; c;1

( a  n 1)! ( a 1)!

* ( c ) * ( c  a b ) * ( c  a ) * ( c b )

*(an) *(a)

437

for n ! 0 Pochhammer number or the rising factorial (Section-1.9.5) and

>

@

with c z 0, 1, 2, ... and Re c  a  b ! 0 , in (A7)

We write the following useful identities for various hyper-geometric functions: 1 ª 1º ª 1º x ª 1º «¬ x 1 »¼ 1  x «¬ x 0 »¼ e «¬ x c »¼

ª bº «¬ x 0 »¼ ª bº «¬ x c »¼

x *

e J (c, x )

* (1  b ) 1 b

(1  x )

* (1  b ) B x (c,1  b  c ) c

* (c )

(A8)

1 b  c

x (1  x )

ª b º * (1  b) «¬ x 0, c »¼ * (1  c ) M (1  b,1  c, x ) ª b1 , b2 º * (1  b1 )* (1  b2 ) F (1  b1 ,1  b2 ;1  c; x ) «¬ x 0, c »¼ * (1  c ) Where J

*

 , x

, B x  ,  M  , , x and F  , ; ; x denote Tricomi’s incomplete gamma, incomplete beta, * (1 b ) B x ( c ,1 b  c ) * ( c ) xc (1 x )1bc

ª¬ x bc º¼

Kummer and Gauss functions respectively. Of special interest, let us take

. From this we

write: c

B x (c,1  b  c )

x * (c )

§ª bº· ¨ x ¸ * (1  b ) © «¬ c »¼ ¹

1 b  c

(1  x )

(A9)

c

§ f j *( j  1  b) · ¨¦ x ¸ * (1  b ) © j 0 *( j  1  c) ¹ x * (c )

Putting c

p  q and b

p , we obtain the following: B x ( p  q ,1  q )

x

p q

1 q

(1  x )

*( p  q) § ª

p º· ¨ «x »¸ ©¬ p  q¼¹

* (1  p ) x

p q

*( p  q) §

· ¨¦ x ¸ © j 0 *( j  1  p  q) ¹

* (1  p )

f

j

(A10)

*( j  1  p)

The above identity (A10) is very useful in order to compactly write the summation in terms of the incomplete betafunction, i.e.: f

¦x j 0

j

*( j  1  b) *( j  1  c)

* (1  b ) 1 b  c

c

x (1  x )

* (c )

B x (c,1  b  c )

(A11)

The above identity is used to reduce the summation term as in the LHS of the above (A10). The generalized hyper-geometric series is defined as the following: p Fq

 a , ..., a 1

p , b1 , ..., bq ; z

* (b1 )...* (bq )

* ( a1  k )...* ( a p  k ) z k

f

* ( a1 )...* ( a p )

¦ *(b  k )...*(b k 0

q

1

(A12)

 k) k !

provided that the arguments bi are not non positive integers. The series converges for all z if p d q , converges for

z  1 if p

q  1 , and diverges for all nonzero z if p ! q  1 . If p 2 F1  a , b, c; z

* (c ) * ( a ) * (b )

2 , and q f

¦ k 0

1 then:

* ( a  k ) * (b  k ) z * (c  k )

k

k!

and all its analytical continuations are called hyper-geometric functions. The series converges for all

>@

>@

Re c ! Re a ! 0 , the integral representation is:

(A13)

z with z  1 . If

438

Appendix A

2 F1

1

* (c )

 a , b , c; z

t * ( a )* (c  a ) ³

a 1

(1  t )

c  a 1

b

(1  tz ) dt

(A14)

0

In particular, if Re >c @ ! Re >( a  b )@ and c is not a non-positive integer, then: 2 F1

* (c )* (c  a  b)

 a , b , c; z

(A15)

* (c  a )* (c  b)

The following are useful identities:

 a , b , c; z 2 F1  a , b, c; z 2 F1

In particular, if p

1

(1  z )

c  a b

(1  z )

a

2 F1 ( c

2 F1

 a , c  b , c; z )

 a, c  b, c; z z1

(A16)

q , then:

* (c )

 a , c; z

f

¦

*(a  k ) z

k

(A17) * ( a ) k 0 * (c  k ) k ! is a confluent hyper-geometric function. It converges for all z provided that c z 0, 1, 2, ... . The name comes from 1 F1

the fact that it may be defined by the limit as 1 F1  a , c; z





lim bnf 2 F1 a , b, c; bz .

If Re >c @ ! Re > a @ ! 0 , the integral representation is: 1

* (c )

1 F1  a , c; z

t * ( a )* (c  a ) ³

a 1

(1  t )

c  a 1 zt

e dt

(A18)

0

The following are interesting identities:

 a, c; z e z  1 F1 (c  a, c; z ) D ª¬ 1 F1  a , c; z º¼  ac  1 F1 ( a  1, c  1; z )

1 F1

(A19)

The relationship of elementary functions to hyper-geometric functions is, for example described in the following expressions:

(1  z )

a

 a; z ;

1 F0

§1 x · ¸ 2x ©1 x ¹

ln ¨

 F 2 1

1

*

J (v, z )

* (v  1)

z 1 1 2

,1, 23 ; x

2

;

0 d x 1

 1 F1  v, v  1; z 1

* (v  1)

e

z

(A20)

 1 F1 (1, v  1; z )

1 x

 2 F1 ( x,1  y, x  1;W ) 1 x y x W (1  W )  2 F1 ( x  y ,1, x  1; W )

%W ( x, y )

x W

A.2 Mittag-Leffler function In the integer order (classical) calculus for the Ordinary Differential Equations (ODE), the exponential function exp( z ) plays an important role. Similarly, in the fractional order calculus of the Mittag-Leffler function, it plays an important part. For this function (i.e. Eq  az , q ! 0 ), Mittag-Leffler (1903) considered that the parameter a be a complex number, as a

a exp(iI ) . As Mittag-Leffler studied this function it became apparent that this function is

either stable (decays to zero), or unstable (goes to infinity) as z increases depending upon how the parameters a and



q are chosen. The result was that the function remained bounded for increasing z if I t q S2 . We will describe

the one parameter and two parameter Mittag-Leffler functions below.

Higher Transcendental and Special Functions

439

A.2.1 One parameter Mittag-Leffler function This function ED ( z ) is defined as the following series form: f

z

k

¦ *(D k  1)

ED ( z )

(A21)

D !0

D \

k 0

We will discuss the case with D  0 separately. The expanded form is the infinite series and looks as follows:

z

z

2

z

3

(A22)  ... * (D  1) * (2D  1) * (3D  1) This function was introduced by Mittag-Leffler in 1903. The Mittag-Leffler function plays an important role in the study of fractional order systems, and is used to express several physical processes. With D 0 and D 1 , we write an interesting relationship as follows:

ED ( z )

1

E0 ( z )



z

1

* (1)





2

z



* (1)

z

3

* (1)

 ...

2

1  z  z  .... 1 1 z 1

E0 (  z )

(A23)

;

z 1

;

1 z

z 1 

The convergence of the above series is valid for z  1 , when for ED ( z ) , we make the limit D p 0 . Therefore the series definition restricts the order D ! 0 , where ED ( z ) is an entire function (analytic everywhere in the complex plane ^ ). With D

1 we write the following:

E1 ( z )

1

z * (2)



1 z  E1 (  z )

Therefore with D

e

z

2

* (3) z

2

2!



z

 z

3

* (4)

 ...

3

f

3!

z

k

¦ k!

 ....

e

z

(A24)

k 0

z

0 , we get E0 (  z ) as the hyperbolic decay function, while with D

1 , we get E1 (  z ) as the

z

pure exponential decay function e . This function incorporates the initially a ‘stretched exponentials’ for small z ; D

(i.e. in limit z p 0 ) which is expressed as ED (  z )  exp



zD * (1D )

and for large z (i.e. in limit z n f ) which is

the asymptotic approximation of an ‘inverse-power-law’; that is expressed as ED (  z D ) 

z D * (1D )

. For D

1 , the

function coincides with the ordinary exponential function. The following are some asymptotic expressions of the Mittag-Leffler function which are useful in system identification:

< ( x)



ED  x

D



# 1

x

D

* (D  1)

# exp



xD * (D 1)

;

(A25)

xp0

for small x , that is 0 d x  1 ; and:

< ( x)

§ sin DS ·  # ¨ S ¸ §¨© *x(D ) ·¸¹ ;

ED  x

D

©

¹

0  D  1;

D

xnf

(A26)

for large x ; it is a power-law asymptote. The asymptotic representation of the derivative of the Mittag-Leffler function is as follows:

\ ( x)



d

x

 (1D )

ª ED   xD º # ¼ * (D ) , dx ¬

xp0

(A27)

xnf

(A28)

for small x , it is 0 d x  1 ; and for large x , it is following:

\ ( x)



§ sin DS · § * (D  1) · ª ED   xD º # ¨ ¸ ¨ D 1 ¸ , ¼ dx ¬ ¹ © S ¹© x d

0 D 1

440

Appendix A

We have derived an earlier integral representation of the Mittag-Leffler function, via the contour integral in Chapter-6 (where Ha represents Hankel’s contour) as follows:

1

x

2S i ³

ED ( z )

D 1

D

x

x z

Ha

e dx

(A29)

The Hankel loop that we have described starts at f and encircles the circular disk x d z

1

D

in a positive sense (i.e.

counter-clockwise) such that S d arg x d S . Some interesting asymptotic expansions of the series for ED ( z ) are listed below for z n f ; and 0  D  2 : ED ( z ) 

1



exp z

D f

ED ( z ) 

z

1

D

f



z

k

¦ *(1  D k ) ;

z n f;

arg z 

DS 2

k 1

k

¦ *(1  D k ) ;

DS

z n f;

2

k 1



(A30)



 arg z  2S  D S2

For the case of D t 2 , we have the following expression:

ED ( z )  Where m takes all integer values such that

¦ exp  z

1

D

1

D

e

 2S imD

m



f



z

k

¦ *(1  D k ) ;

(A31)

z nf

k 1

 DS2  arg z  2S m  DS2 , and

arg z can assume any values between

S and S inclusively.

A.2.2 The two parameter Mittag-Leffler function The ‘two parameter’ Mittag-Leffler functions plays a very important role in fractional calculus. This function type was introduced by R P Agrawal and Erdelyi, in 1953-54. The two-parameter Mittag-Leffler function is defined in series form by use of the following expression: f

ED , E ( z )

z

k

¦ *(D k  E ) ;

D ! 0,

E !0

D, E  \

(A32)

k 0

1 * E

We get the useful recursive relationship as ED , E ( z )



 z ED ,D  E ( z )



from the above series definition as

demonstrated below: f

ED , E ( z )

z

k

¦ *(D k  E ) k 0

1 * E

z

 §

1 * D  E f

¦ ©

1

 z¨ * E

m 0

2

3

 *Dz E  * 2Dz  E  *3Dz  E  ...

1 * E

2

 *D  (zD  E )  * 2D z(D  E )  ...

zm *  mD  (D  E )

· ¸ ¹

1 * E





 z ED ,(D  E ) ( z )

(A33)



The function ED , E ( z ) has a similar integral representation on Hankel’s contour as ED ( z ) , which is:

ED , E ( z )

1

2S i ³

x

D E

x

e dx

(A34) x z (A34) we will derive later in the Appendix, and will use it for extending the series definition for negative arguments of the Mittag-Leffler function, i.e. D  0 for ED , E ( z ) . For E

1 , we have ED ,1 ( z )

¦

f

zk k 0 * (D k 1)

recursion expression (A33) i.e. ED , E ( z )

1 * E

Ha

D

{ ED ( z ) , i.e. the one parameter Mittag-Leffler function. From the





 z ED ,D  E ( z ) , we set E

1 , and write ED ( z )

The following identities were established from the definition of the Mittag-Leffler function:





1  z ED ,D 1 ( z ) .

Higher Transcendental and Special Functions f

z

k

f

z

k

¦ *(k  1) ¦ k !

E1,1 ( z )

k 0 f

z

e

k 0

z

k

f

z

k

k 0 f

z

k

f

z

k

k 0

k 0

2



1 ez  z m1

The above expressions (A35) have the general form E1, m ( z )

k 1

z

e 1

f

z

k 2

z

e 1 z

¦ (k  2)!

z

k 0

¦

m2 zk k 0 k!

(A35)

z

k 0

1 z

z

¦ (k  1)!

z

k 0

¦ *(k  3) ¦ (k  2)!

E1,3 ( z )

f

1

¦ *(k  2) ¦ (k  1)!

E1,2 ( z )

441

2

.

The trigonometric and hyperbolic ones are also manifestations of the two-parameter Mittag-Leffler function, which is indicated below: f

2

E2,1 ( z )

z

2k

f

z

2k

¦ *(2k  1) ¦ (2k )! k 0 f

2

z

2k

1

¦ *(2k  2)

E2,2 ( z )

cosh( z ),

k 0

z

k 0

f

z

2 k 1

¦ (2k  1)!

(A36)

sinh( z ) z

k 0

In addition, we note the following relationships: 2

ED (  x )

2

E2D ( x )  xE2D ,1D ( x ) ED ( x )  ED (  x )

2

E2D ( x ) ED ( ix )

2

(A37)



2

2

E2D (  x )  ix E2D ,1D (  x )

>

Re ED ( ix )

@



2

E2D (  x )

A.2.3 Graphical representations of Mittag-Leffler function: f ( x)

ED , E ( x)

The general graphical representation of ED , E ( x ) is depicted in Figures A1 to A3 for different conditions. The fact that

ED , E ( x ) is a complete monotonic function for 0  D d 1 with E t D is shown in Figure-A1, plotted for E 1 The general behavior of ED , E ( x ) with 0  D d 1 but with E  D is depicted in Figure-A2, plotted for E 1

1 2, 8

3 2, 2

( x) .

( x ) . It is

a smooth monotonic curve, as shown in Figure-A1 for z t 0 , but shows an inflection point in the region of x  0 . For D ! 1 , the graph of ED , E ( x ) is a smooth curve as shown in Figure-A1, for region x t 0 ; but for x  0 , the graph exhibits indefinite oscillations which continue in the region x  0 . This is depicted in Figure-A3 plotted for E2,2 ( x ) . However, depending on the value of E ; the magnitudes of oscillations decrease (or increase). This is shown in FigureA3, in the negative x -direction.

E 12, 3 2

 x

2

E 12, 3 2  x 1

0 2

1

0

Figure-A1: Illustrating the general behavior of ED , E ( x ) for 0  D d 1 , and D d E

1

x

442

Appendix A

All the graphs depicted in Figures A1 to A-3, when rotated about an ordinate axis, or rotate right, will provide figures for ED , E (  x ) which resemble the relaxation responses of process dynamics.

E 1 2 , 18 ( x ) 0.10

1 *  18

0.05

E 1 2 , 18 ( x ) 0.00 0.05 0.10 10

x

5

0

Figure-A2: Illustrating the general behavior of ED , E ( x ) for 0  D d 1 , and D ! E

1.0

E2,2 ( x)

0.5

E2,2 ( x) 0.0 0.2 30

15

0

x

Figure-A3: Illustrating the general behavior of ED , E ( x ) for D ! 1

A.2.4 Generalized Hyperbolic and Trigonometric Functions The generalized hyperbolic function of order n , is represented below: f

hr ( z , n )

z

nk  r 1

¦ (nk  r  1)!

z

r 1

n

En , r ( z ),

r

1, 2, 3, ..., n

(A38)

r

(A39)

k 0

The generalized trigonometric function of order n , is represented below: f

kr ( z , n)

m nm  r 1

( 1) z

¦ (nm  r  1)!

m 0

z

r 1

n

En , r (  z ),

1, 2, 3, ..., n

Higher Transcendental and Special Functions

443

A.3 The error function and its fractional generalization The error function defined as

erf ( z )

z

2

³e S

t 2

dt

(A40)

0

Note that erf (f )

1 . The error function is represented by the series as follows: erf ( z )

n 2 n 1

f

2

S

§ z3 z5 z7 z9 ·  ... ¸ ¨z    3 10 42 216 S © ¹

( 1) z

2

¦ (2n  1)n ! n 0

The complementary error function is defined as:

2

erfc( z ) 1  erf ( z ) 1 

S

z

³

e

t 2

f

2

dt

³

S

0

e

t 2

(A41)

dt

(A42)

z

The series asymptotic expansion of the complementary error function and its relationship with the Mittag-Leffler function is the following:

erfc( z )

 z2

f § n 1.3.5...(2 n  1) · 1  ( 1) ¦ ¨ ¸ 2 n (2 z ) z S © n 1 ¹

e

 z2

f § n (2 n )! · 1 ( 1)  ¦ ¨ 2n ¸ n !(2 z ) ¹ z S © n 1

e f

E 1 ,1 ( z ) 2

¦ * k 0

z k 2

k

 1

(A43)

z2

e erfc(  z )

The error function in terms of the hyper-geometric function is expressed as the following:  1  1  x2 2 2 3 erf ( x ) 2S 2 xe 2S 2 x 1 F1 12 , 32 ;  x 1 F1 1, 2 ; x

 



 



(A44)

Error function and Tricomi’s incomplete-gamma function are related as follows:

J

*

 12 , at

( at )

  12

ªerf ¬

 at º¼

z ªJ

erf ( z )

*

¬



1 2

,z

2

º¼

(A45)

This error function integral (A40) has no algebraic formula, so it is numerically evaluated and is tabulated in mathematical handbooks. The fractional generalization of the error function is serfD ( z ) . Where we define an D -stable error function serfD similar to the error function, call it a generalized error function; that is twice the integral of D ’s stable density for argument 0 to z defined as: z

serfD ( z )

³

2 sD ( x )dx

(A46)

0

Where sD ( x )



exp  x

D

is the standard symmetric D stable density. The error function is thus a special case when 2 , that is:

the probability density function is normal (Gaussian) with D

erf ( z )

(A47)

serf 2 (2 z )

A.4 Variants of the Mittag-Leffler function One variant of the Mittag-Leffler function is described as [ t and is the following:

[ t (Q , a )

Q

t

f

( at )

k

¦ *(Q  k  1)

Q

(A48)

t E1,Q 1 ( at )

k 0

This function (A48) is important for solving fractional differential equations. Another variant D is the following:

D ( E , t )

t

D

f

k k (D 1)

E t

¦ *  (k  1)(D  1) k 0

D



t ED 1,D 1 E t

D 1



(A49)

444

Appendix A

This function above (A49) is called the Robotnov function and one is a special variant too as ScD n (2 D ) n 1

f

( 1) z



¦ *  (2  D )n  2

ScD ( z )

zE2 D ,2  z

n 0

2 D



(A50)

The above function (A50) is a fractional sine function form-I. The other is a special variant known as CsD which is: n (2 D ) n

f

( 1) z



¦ *  (2  D )n  1

CsD ( z )

E2 D ,1  z

n 0

2 D



(A51)

The above function is a fractional cosine function of the form-I. The fractional sine and cosine function form-II are depicted below: k

f

2k

( 1) z



¦ *  2P k  P  O  1

cos O , P ( z )

E2 P , P O 1  z

k 0

k

f

( 1) z

(A52)

2 k 1

¦ *  2P k  2P  O  1

sin O , P ( z )

2



zE2 P ,2 P O 1  z

k 0

2

Generalization of the Mittag-Leffler function to two variables was suggested and was further extended by Srivastava to the following type of symmetric form. f

Q ,V [D , E , O , P

f

x

m  E (Q nD1)1

y

n  P ( V mO1)1

 x, y ¦¦ m 0 n 0 *  mD  (Q n  1) E *  nO  (V m  1) P

(A53)

Several manifestations including several variable representations of the Mittag-Leffler have been made for multidimensional studies on the fractional calculus context.

A.5 The Laplace integral and its connection to the Mittag-Leffler function The Mittag-Leffler function is connected to the Laplace integral by use of the following formula: f f 1 1 D D u  u E 1 e ED (u z )du e u ED , E (u z )du , 0 0 1 z 1 z

³

³

D

¦

The proof is as follows. Here we used ED (u z )

( uD z ) k k 0 * D k 1

f

³

representation of the gamma function (Section-1.10), i.e. * (X )

³

f

0

u

D

³

e ED (u z )du

³

f

0

f

0

e

u

* D  1 ³ 2

*  2D  1 ³



³



f

0

1 * D 1

2



1 z

u D

e u du

 u 2D

e u

du  ...

 u (D 1) 1

du

³

 u (2D 1) 1

f

0

e u 2

f

¦z k 0

1

f

* D  1  z 

1  z  z  ....



e u

1 *  2D 1

1 * D 1

2

f

0

and the integral

 *u 2Dz1  .... du

0

z

( uD z ) k k 0 * D k  E

f

du .

2D 2

uD z * D 1



z 1 z

1 

¦

 u X 1

e u

0

z

u

e du 

1 z

f

D

ED , E (u z )

,

(A54)

k

(A55)

du  ...

1 *  2D 1

*  2D  1  ...

Higher Transcendental and Special Functions

³

f

0

 u E 1

e u

445

D

ED , E (u z )du

³

f

0

 u E 1

e u

  

1

* E ³

f

0



z

2

f

0

f

³

0

z 1 * E

z

2



du  z



³

1 *  2D 1

f

 u E 1 D

e u

0

u du

 u E 1 2D

u du  ..

  ³

 u E 1

1 *  2D 1

* D  E ³

e u

e u

* E  z 2

z

du 

*  2D  E ³

1 * E

 *u2D z E  .... du

 u E 1

e u



2D 2

uD z * D  E

1 * E

f

1 * D E

f

 u (D  E ) 1

du

e u

0

 u (2D  E ) 1

du  ...

e u

0

 * D  E *  2D  E  ... 1 * D  E

f

2

¦z

1  z  z  ....

k

k 0

(A56)

1 1 z D

The above integrals are the fundamental formula for the evaluation of the Laplace transforms of ED (  at ) and D

D

ED , E (  at ) , with D , E ! 0 and a  ^ . Put u z

D

 at , and u

st , with t t 0 , and a  ^ in the above (A54)

Laplace integral relationship to get the following steps by juxtaposition of the variable: f 1 u D e ED (u z )du 0 1 z

³ ³

f

0

s

s

³

e

f

0

³

f

0

³

f

0

e

 st

e

 st

e

 st

 st

Using the definition of the Laplace transform, i.e. $ ^ f (t )` (A57) the following equation:

^

1

D

ED (  at )  s dt



1

D

ED (  at )dt

1 a s

D

ED (  at )dt

0

D

e

 st

`

$ ED (  at )

1 sD

(A57)

D

D

s

ED (  at )dt

³





s a

D

f

D

1   a ut D

D 1

D

s a

f (t )dt , we write from the above derivation

s

D 1

D

s a Similarly as done in (A57) from relationship (A54), we carry out the following steps:

(A58)

446

Appendix A

³

f

0

³

f

0

s

E

s

E

 u E 1

e u e

³

 st

f

0

³

f

0

³

f

0

e

( st )

E 1

e

 st E 1

e

 st E 1

 st

t

t

t

1

D

ED , E (u z )du

E 1

1 z

^

E 1



1 1 a s

D

ED (  at )dt

D

s

D

D E

D

s a

s

`

1 sD

D

ED (  at ) dt

ED , E (  at )

(A59)



s a



D



D

1   a ut D

D

ED , E (  at )dt

We get the following useful Laplace identity:

$ t

1

D

ED , E (  at )  s dt

D E

(A60)

D

s a

A.6 The Laplace transforms of the Mittag-Leffler function and several other variants The following expressions give some identities for Laplace transforms pairs of Mittag-Leffler functions:

^

$ t

D k  E 1 ( k ) ED , E

^

$ t

 12

^

D 1

$ t dk dt k

k  12

^

$ t

Here in (A61), ED( k, E) ( at D )

1 2

E1

D

s

`

( at )

D E

D

(s  a)

E 1 , 1 a t 2 2

1 2, 2

D

`

ED ,D ( at )



1

 a t `

s a 1 D

s a

k 1

; Re( s ) ! a

k!

 `

(k )

k!

;

s a



;

k 1

D D

;

D

E E, k

1 2

(1/D )

E

, k

1 2

(A61)

0

0

D

ED , E ( at ) means the k  th integer order derivative or integration. For k ! 0 , the

operation is the differentiation of the Mittag-Leffler function, and for k  0 the operation is the integration of the Mittag-Leffler function. The following lists some Laplace transform identities:

Higher Transcendental and Special Functions

k

1,

E

^

s

$ ED ( at )

^

^

D

D 1

D

s

`

D

D

D

s a

$ ED ( O t )

^

D

ED , E ( at ) { ED ,1 ( at )

0,

`

D

(k )

447

s

D

s(s  O )

D 1

s

D

s

`

; D !0

D

D

s 1

$ ED (O t )

D

D

s O s

`

$ ED ( t )

D 1

s ( s  1)

D 1

s

D

D

D

s(s  O )

s O

D

s ­d D ½ D ED ( O t )¾ s $ ^ED ( O t )` D s O ¯dt ¿ D  k 1 ­ dk s k D ½ D $ ® k ED ( t ) ¾ s $ ^ED ( t )` D s 1 ¯ dt ¿ D 1 D 2 ­ d 1 s 1 s 1 D ½ D $ ® 1 ED ( t ) ¾ s $ ^ED ( t )` D D s s 1 s 1 ¯ dt ¿ D 1 D  k 1 ­ dk s 1 s k D ½ D $ ®  k ED ( t ) ¾ s $ ^ED ( t )` k D D s s 1 s 1 ¯ dt ¿



$®





d

­ ¯

$ ®G (t ) 

^

$ t

E 1

dt











½ ¿

D

ED ( O t )¾ 1  D

s

`

ED , E (t )

s

D

(A62)

O

D

D

s O

s O

D E

D, E ! 0

D

s 1

A.7 Agrawal Function The Mittag-Leffler function ED ( z ) was further generalized by R.P Agrawal in 1953. This function is particularly interesting to the fractional order system theory due to its Laplace transform as given by Agrawal. The function is defined as follows:

t

E 1

f

D

t

D k  E 1

¦ *(D k  E )

ED , E (t )

D, E ! 0

k 0

^

$ t

E 1

1

E

D

s

`

ED , E (t )

D E

D

s 1

D

ED ,1 (t )

f

D

t

Dk

(A63)

¦ *(D k  1) ;

ED (t )

k 0

^

D

`

$ ED , E (t )

s

D 1

D

s 1

A.8 Erdelyi’s Function Erdelyi (1954) studied the generalization of the Mittag-Leffler function as follows: f

EĮȕ (t )

t

k

¦ *(ĮN  ȕ ) ;

D, E ! 0

k 0

^

f

` ¦s

$ ED , E (t )

k 0

k! k 1

* (D k  E )

(A64)

448

Appendix A

Where the powers of t are integers, k is an integer number.

A.9 The Robotnov-Hartley function To affect the direct solution of the fundamental linear fractional order differential equations, the following function was introduced by Robotnov and Hartley in 1998: Fq (  a , t )

t

q 1

n nq

f

(a) t

^

¦ *(nq  q) ; n 0

Fq (  a , t )

t

q 1

q

Eq , q (  at )

t

`

q ! 0 $ Fq (  a , t )

q 1

1 q

s a

n qn

f

(a) t

¦ *(qn  q) n 0

(A65) This function is the ‘impulse-response’ of the fundamental fractional differential equation, and is used by control system analysis in order to obtain the forced or the initialized system reaction.

A.10 Miller-Ross function Miller and Ross in 1993 introduced a function as the basis of the solution of a fractional order initial value problem. It is defined as the v -th integral of the exponential function, that is:

Et (v, a )

d

v

dt

v

at

ª¬e º¼ t

v

f

( at )

k

s

$ ^Et (v, a )`

¦ *(v  k  1) ;

v

(A66)

sa

k 0

In (A66) we note Re >v @ ! 1 The Miller-Ross function in terms of a hyper-geometric function is:

t

Et (v, a )

* (v  1)

¦

The one-parameter Mittag-Leffler function for w t 0 is Ew (t )

v

f

tn n 0 * (1 nw )

1 F1 (1, v

(A67)

 1; at ) w

. Expanding this Ew (ct ) series and

with the grouping we get the expression as demonstrated in the following steps: w

Ew (ct )

f

w n

(ct )

¦ *(1  nw) n 0

w n

f

n 0, q ,2 q ,.......

* (1  nw)

(ct )

¦



n 1, q 1,2 q 1,...

* (1  nw)

 ....

w n

f

(ct )

¦

... 

w n

f

(ct )

¦

n q 1,2 q 1,3 q 1,... q

(A68)

* (1  nw) 2

q

q

Et (0, c )  cEt ( w, c )  c Et (2 w, c )  ... .....  c

q 1



Et ( q  1) w, c

q



Therefore: w

Ew (ct )

q 1

k

¦ c E (kw, c t

q

)

(A69)

k 0

is the relationship between the one-parameter Mittag-Leffler and Miller-Ross functions. In terms of Tricomi’s incomplete gamma function, we define the Miller-Ross function as: v az *

E z (v, a ) z e J (v, az ) We have the following identity for a generalized cosine and sine function with the Miller-Ross formulation.  Ez (v, ia )  Ez (v, ia ) C z (v, a ) 2 S z (v, a )

 Ez (v, ia )  Ez (v, ia )

(A70)

(A71)

2i where C z (v, a ) and S z (v, a ) are generalized cosine and sine functions respectively, with the Miller-Ross formulation.

These identities of Miller-Ross expressions are similar to the exponential function and its polar representation is to get the cosine and sine trigonometric functions. These functions play a crucial role in fractional differential equation

Higher Transcendental and Special Functions

449

solutions, and are similar to the normal transcendental and trigonometric functions. In the above definition (A66) of the Miller-Ross function let a be replaced by a purely imaginary number as ia . The series form we can re-write by organizing even and odd terms together as the following: ( k 1) § f ( 1) k 2 ( at ) k k f ( 1) 2 ( at ) Et (v, ia ) t ¨ ¦ i ¦ ¨ k even * (v  k  1) k odd * (v  k  1) © v

Et (v, ia )

Ct (v, a )  iSt (v, a )

Et (0, it )

e

· ¸ ¸ ¹

it

(A72)

it

e cos t  i sin t The above (A72) identities indicate the similarity of the Miller-Ross function with an exponential function and trigonometric functions. Integration of the Miller-Ross function is the following: t

³ E (v, a)du

Re >v @ ! 1

Et (v  1, a )

u

0 t

³u

w

Et u (v, a )du

* ( w  1) Et (v  w  1, a )

(A73)

0

Re > w@ ! 1 D

p

Re >v @ ! 1

> Et (v, a)@

Et (v  p , a )

p

Re >v @ ! 1

1, 2, 3, ...

Differentiation of the Miller Ross Function is the following:

D > Et (v, a )@ D p

p

Et (v  1, a )

> Et (v, a)@

k vk  p p 1 a t p a Et (v, a )  ¦ k 0 * (v  k  1  p )

Et (v  p, a )

0,1, 2, 3...

D >tEt (v, a )@

(A74)

tEt (v  1, a )  Et (v, a )

P

D ª¬t Et (v, a ) º¼ P p D ª¬t Et (v, a ) º¼

P

t Et (v  1, a )  P t p

P 1

Et (v, a )

§ p · * ( P  1)

¦ ¨© k ¸¹ *( P  k  1) t P

k

Et (v  k  p , a )

k 0

p 0,1, 2, .. Special values of the Miller-Ross function are the following: at

Et (0, a )

e

E0 ( v , a )

0

Re >v @

Et ( 1, a )

aEt (0, a )

Et (  p , a )

a Et (0, a )

Et (1, a )

p

p

0,1, 2, 3,...

Et (0, a )  1

Et  12 , a Et   12 , a Et (v, 0)

0

a a

 1 2  at

e

 erf at

1 aEt  12 , a  1 t 2 St

t

v

* (v  1) Recursive relationships of the Miller-Ross functions are listed as follows:

(A75)

450

Appendix A

t

Et (v, a )

aEt (v  1, a ) 

Et (v, a )

a Et (v  p , a ) 

v

* (v  1) p 1

p

t

vk

¦ *(v  k  1)

0,1, 2, 3,...

p

k 0

Et (v, a )  Et (v, b )

aEt (v  1, a )  bEt (v  1, b)

Et (v, a )  Et (v, b )

a Et (v  p, a )  b Et (v  p, b)

p

p

p 1



(A76)

¦

k

k

( a  b )t

vk

p

* (v  k  1) Integral representation of the Miller-Ross function is the following:

0,1, 2...

k 1

Et (v, a )

t

1

³u

* (v )

v 1 a ( t u )

e

>@

du

Re v ! 0

(A77)

0

This Miller-Ross function Et (v, a ) is the solution of the ordinary differential equation:

dy dt

t

 ay

v 1

* (v )

(A78)

v!0

A.11 The generalized cosine and sine function in the Miller-Ross formulation In the Miller-Ross modification above (A72), the generalized cosine function is defined as the following:

Ct (v, a )

t

v

k

f

¦

( 1) 2 ( at )

k

Ct (0, a ) cos( at ) * (v  k  1) The Laplace transform of the generalized cosine function is as below: s $ ^Ct (v, a )` Re v ! 1 2 v 2 s (s  a ) In the Miller-Ross modification above (A72), the generalized sine function is defined as follows:

(A79)

k even

>@

St (v, a )

v

f

¦

( 1)

( k 1) 2

( at )

k

St (0, a ) sin( at ) * (v  k  1) The Laplace transform of the generalized sine function is the following: a $ ^St (v, a )` Re v ! 2 2 v 2 s (s  a ) Integral representations of the generalized cosine and sine functions are the following: t

(A80)

(A81)

k odd

>@

Ct (v, a ) St (v, a )

1 * (v ) 1 * (v )

Special values of the generalized cosine and sine functions are:

(A82)

t

³u

v 1

cos  a (t  u ) du

0

(A83)

t

³u 0

v 1

sin  a (t  u ) du

Re >v @ ! 0

Higher Transcendental and Special Functions

Ct (0, a )

cos( at )

C0 ( v , a )

0

Ct ( 1, a )

S0 (v, a ) p

St ( 1, a )

p

( 1) 2 a cos( at )

Ct (  p , a )

( 1)

St (  p , a )

( 1)

Ct (1, a )

1 a

Ct (v, 0)

( p 1)

a cos( at ) p

( 1) 2 a sin( at )

( p 1)

p

a sin( at )

2

a cos( at )

p

2 a

St (1, a )

1  aS  1 , a t 2 St

v 1 * ( v 1) t

St (v, 0)

(A84)

1, 3, 5, ...

p 2

 12 at St   12 , a

sin

aCt  12 , a

0

Ct  12 , a

2 a  (cos at )C ( z )  (sin at ) S ( z )

 12 , a

2 a  (sin at )C ( z )  (cos at ) S ( z )

St

p

St (  p , a )

2

sin( at )

Ct   12 , a

S

>@

Re v ! 1

0, 2, 4, ...

p

2at

sin( at )

0

 a sin( at )

Ct (  p , a )

Where z

St (0, a )

451

and C ( z ) , S ( z ) are used in Fresnel integrals (A84), they are defined as follows: z

³ cos 

C( z)

1 2

2

z



S t dt

S ( z)

0

³ sin 

1 2

2



S t dt

(A85)

0

The recursive relationship of the generalized cosine and sine functions are the following:

Ct (v  1, a )

 aSt (v, a ) 

t

v 1

St (v  1, a )

* (v )

2

Ct (v  1, a )  a Ct (v  1, a ) 2

St (v  1, a )  a St (v  1, a )

t

aCt (v, a )

v 1

(A86)

* (v ) at

v

* (v  1) Integration of the generalized cosine and sine functions are the following: t

³ C (v, a)du u

>@

Ct (v  1, a ) , Re v ! 1

0

(A87)

t

³S

u ( v , a )du

>@

St (v  1, a ) ; Re v ! 2

0

The differentiation of the generalized cosine and sine functions are the following: D Ct (v, a ) Ct (v  1, a ) D St (v, a )

>

D m

m

@

>Ct (v, a)@

>

Ct (v  m, a )

D

m

@

St (v  1, a )

> St (v, a )@

S t ( v  m, a )

0,1, 2,...

> @ D >tSt (v, a )@ D tCt (v, a )

tCt (v  1, a )  Ct (v, a ) tSt (v  1, a )  St (v, a ) P

P 1

P

P 1

P

t Ct (v  1, a )  P t

P

t St (v  1, a )  P t

D ª¬t Ct (v, a ) º¼ D ª¬t St (v, a ) º¼

(A88)

Ct (v, a )

St (v, a )

This generalized cosine and sine function St (v, a ) or aCt (v  1, a ) is the solution to the ordinary differential Equation:

452

Appendix A 2

d y dt

2

2

a y

t

a

v 1

(A89)

v!0

* (v )

This detailed description of Miller-Ross Et and the generalized cosine Ct and sine St functions is given in order to draw attention towards their similarity with exponential, cosine and sine functions. These similarities in property encourage us to select these functions as a solution to fractional differential equations. Similarly, one can construct generalized hyperbolic cosine and generalized hyperbolic sine functions with similar properties as described above for generalized cosine and sine functions. Generalized hyperbolic cosine and sine functions are constructed as follows: t

1

HCt (v, a )

³u

* (v )

cosh  a (t  u ) du

0

(A90)

t

1

HSt (v, a )

v 1

³u

* (v )

v 1

>@

sinh  a (t  u ) du , Re v ! 1

0

A.12 Generalized R function and G function It is of significant usefulness to develop a generalized function, which, when fractionally differentiated or integrated (differ-integrated) by any order, returns itself. Like exponential, trigonometric, hyperbolic functions of integer order calculus, the definitions of such generalized Mittag-Leffler functions are important in fractional calculus. In earlier sections, some variants of Mittag-Leffler were noted. Here a more generalized R function and G function is introduced as: n

f

Rq ,v  a , c, t

¦

( a ) (t  c )

( n 1) q 1 v

{ Rq ,v  a , t  c

*  ( n  1) q  v

n 0

(A91)

In (A91), t is an independent variable where c is the lower limit of fractional differ-integration. Our interest in this function will be normal for the range t ! c . The Laplace transforms of the R function are the following:

^

s

`

$ Rq ,v ( a, 0, t )

^

v

q

s a e

`

$ Rq ,v ( a, c, t )

>

@

>@

; Re ( q  v ) ! 0 ; Re s ! 0

(A92)

 cs v

s

>

@

>@

; c t 0 Re ( q  v ) ! 0; Re s ! 0 q s a The relationship of the R function to elementary functions are following:

R1,0 ( a , 0, t )

e

at

§ a 2t 3 a 4t 5 · aR2,0 (  a , 0, t ) a ¨ t    ... ¸ sin( at ) 3! 5! © ¹ § a 2t 2 a 4t 4 · 2 R2,1 (  a , 0, t ) ¨ 1    ... ¸ cos( at ) 2! 4! © ¹ 2

2

aR2,0 ( a , 0, t )

(A93)

sinh( at )

2

R2,1 ( a , 0, t )

cosh( at )

R1,0 ( a , 0, x )

e

ax

The relationship of the R function to other generalized functions: The Mittag-Leffler function

^ 

q

^

q

$ Eq  at

`

1§

`

s

$ Eq (  at ) q 1

c

Dt

q

· ¨ ¸ q s©s a¹ q

s

q 1

s a

s q

q 1

s a

,

^

q!0

`

$ Rq ,q 1 (  a, 0, t ) ;

ª¬ Rq ,0 (  a, c, t )º¼ Eq   a (t  c ) q

(A94)

Higher Transcendental and Special Functions

453

The Agrawal function

^

s

`

q

$ Eq , p (t )

q p

^

`

$ Rq ,q  p (1, 0, t )

q

s 1

(A95)

q

Rq , q  p (1, 0, t )

Eq , p (t )

Erdelyi’s function

t

1 E

q

Eq , E (t )

Rq , q  E (1, 0, t )

t

1 E

f

t

nq

¦ *(nq  1)

(A96)

n 0

The Robotnov and Hartley function

^

1

`

$ Fq (  a, t )

^

`

$ Rq ,0 (  a, 0, t ) ;

q

s a

n ( n 1) q 1

f

(a) t

¦ *  (n  1)q

Rq ,0 (  a , 0, t )

Fq (  a , t )

n 0

^

1

`

$ Fq ( a, t )

^ 

$ Eq  at

q

q

s a

` 1s  s $ ^F (a, t )` q

q

q 1 0 Dt

q

ª¬ Fq ( a, t ) º¼ 1 ½ 1 ­ $ ® q ¾ ¯ s(s  a) ¿ 1

q!0

,

­ sq ½ ¾ q ¯s  a¿

Eq ( at )

§1 · q ¨  u (t )  Eq (  at ) ¸ ©a ¹

q 0 Dt

$ ®

1

­ ¯

$ ®1  Where u (t ) is the Heaviside unit step function at t The Miller and Ross function:

1 0 Dt

ª¬ Fq (  a, t ) º¼

q 0 Dt

ª¬ Eq ( at q ) º¼

ª¬ Eq (  at q ) º¼

a

½ ¾ G (t )  aFq (  a, t ) s  a¿ q

(A97)

0 and G (t ) is the Dirac delta function in (A97).

s

$ ^Et (v, a )`

v

sa

`

(A98)

n nv

f

R1,  v ( a , 0, t )

^

$ R1, v ( a, 0, t ) ,

(a) t

¦ *(n  v  1)

Et (v, a )

n 0

The further generalized function ( G function)

 (r )(1  r )...(1  j  r ) (a ) j t ( r  j ) q v 1 ¦ * (1  j )*  ( r  j ) q  v j 0 f

Gq ,v , r ( a , t )

^

`

$ Gq ,v ,r ( a, t )

s q

(A99)

v

(s  a)

>

@

>@

; Re ( qr  v ) ! 0 Re s ! 0

a sq

!0

A.13-Bessel Function Perhaps amongst all higher transcendental functions, the Bessel functions are the most ubiquitous; they appear very frequently in theoretical physics and engineering. The Bessel function, J v ( z ) of the first kind and order v is defined as an infinite series as the following:

Jv ( z)

v

f

( 1)

k

 12 z

2k

 z ¦ k ! * (v  k  1) 1 2

k 0

(A100)

454

Appendix A

The, J v ( z ) is an entire function of v . This is the solution of the Bessel equation, which is the following: 2

z D

2

> w@  zD > w@  ( z 2  v 2 ) w

0

(A101)

The Bessel function Yv ( z ) of the second kind and order v is also the solution of the Bessel equation, linearly independent of J v ( z ) . One of the integral representations of the Bessel function is Poisson’s formula, as follows:

Jv ( z)

2

§z· 1 1 ¨ ¸ *  2 * v  2 © 2 ¹

v 1

³ (1  t

2 v  1 2

)

0

cos( zt )dt ; Re v !  12

>@

(A102)

Another representation is Sonin’s formula, as follows:

J P  v 1 ( z )

> @

z

 12 S

v 1



J P  z sin T sin

³

v

2 * (v  1)

P 1

T

 cos



(A103)

cos( z )

(A104)

2 v 1

T dT

0

>@

Re P , Re v ! 1 Some special relationships of the elementary function to the Bessel functions are: 2

J 1 ( z)

Sz

2

sin( z )

,

2

J  1 ( z)

Sz

2

The modified Bessel function I v ( z ) of the first kind of order v is defined as the infinite series: v f

 12 z ¦ k !*(v 1 k  1)  12 z

Iv ( z)

2k

(A105)

k 0

(A105) is the solution of the modified Bessel equation, that is: 2

z D

2

> w@  zD > w@  ( z 2  v 2 ) w

0

(A106)

The modified Bessel function K v ( z ) of the second kind is also the solution of the modified Bessel equation and is linearly independent of I v ( z ) . For Re >v @ !  12 and Re > z @ ! 0 , the integral representation is: Kv ( z)

S

 12 1 2z * v  12

v

 sinh T  ³ f

2 v  z cosh T

e

dT

0

(A107)

f

³  cosh(vT )  e

 z cosh T

dT

0

Particularly for a non-negative integer v , we may write the infinite series for K v ( z ) . For example, we write the series for v

0 as:

K0 ( z)





 ln  12 z I 0 ( z ) 

f

¦ n 0

\ ( n  1) 1 2 2z ( n !)



2n

(A108)

Here \ ( z ) is the ‘psi’ function which is a derivative of the logarithm of the Gamma function that is

\ ( z)

D ln  * ( z )

D * ( z ) z

, which we have discussed in Chapter-1.

Some relationships between special values in relation to the elementary functions are the following:

I 1 ( z) 2

K 1 ( z) 2

2

sinh z ,

Sz

K 1 ( z) 2

I 1 ( z) 2

S 2z

e

z

2

Sz

cosh z (A109)

Higher Transcendental and Special Functions

455

A.14 Wright Function The Wright function, which we denote by WO , P ( z ) , is named after E. Maitland Wright (a British mathematician), who introduced this function in 1933. The function is defined by the series representation convergent in the entire z  complex plane: f

z

n

¦ n ! * (O n  P ) ,

WO , P ( z )

(A110)

O ! 1, P  ^

n 0

The integral representation of the Wright’s function reads as follows: 1 V  zV  O dV WO , P ( z ) e P 2S i Ha V

³

;

(A111)

O ! 1, P  ^

The origin of the above (A111) can be related in a similar way to those obtained for auxiliary function M X ( z ) , in Chapter-8. Using the Hankel representation of the reciprocal of the gamma function, that is

1 *( z )

³

Ha

u z

e u du

for z  ^ , we obtain the following series representation for WO , P ( z ) as:

WO , P ( z )

1

dV

V  zV  O

e 2S i ³

V

Ha

1 2S i

³

P

§ f z n  O n · dV ¨¦ V ¸ P © n 0 n! ¹V

V

e

Ha

(A112)

n

§ 1 · V OnP e V dV ¸ ¨ ¦ ³ n 0 n ! © 2S i Ha ¹ f

z

f

z

n

¦ n ! * (O n  P ) n 0

For O

ez *(P )

0 , we get W0, P ( z )

provided P z 0, 1, 2, ...

A.15 Prabhakar Function A-15.1 The three parameter Mittag-Leffler function f ( z )

EDJ , E ( z )

This function is a modification of the two parameter Mittag-Leffler function carried out by T.R. Prabhakar in 1971, defined in series form as the following: J ED , E

f

(J ) k

z

k

¦ *(D k  E ) k !

( z)

(A113)

k 0

> @

with z  ^ , D , E , J  ^ and Re D ! 0 . In (A113) we have (J ) k as the Pochhammer Number, the rising factorial * (J  k ) * (J )

(Section-1.9.5), which is (J ) k

. This (A113) is the entire function.

The Laplace transformation of the Prabhakar function is the following:

^

$ t

E 1 J ED , E

`

D

( Zt )

1

From (A113) and (A115), we note that ED , E ( z )

ED , E ( z ) and ED ( z )

s

DJ  E

D

(s  Z )

J

; Re > s @ ! Z

1

D

(A114)

1

ED ,1 ( z ) .

A.15.2 Prabhakar Integral J

E 1 J ED , E

Using the function ND , E (t )

t

integral as a EJ(D , E ,Z );t > f (t )@

N

J D ,E

D

(Zt ) as a kernel of the convolution integral, we can define the Prabhakar



(t ) *  f (t ) , described as the following expression: J

a E (D , E ,Z );t

t

> f (t )@ ³a  (t  W ) E 1 EDJ ,E (Z (t  W )D )  f (W ) dW

(A115)

456

Appendix A

We note that the Prabhakar integral (A115) has a kernel of integration different from the power law function (  t that we used as the kernel in the Riemann-Liouville fractional integration formula.

X 1

)

A.15.3 The Prabhakar integral as a series-sum of the Riemann-Liouville fractional integrals X

We know the Riemann-Liouville fractional integral as

a It

1 * (X )

f (t )

inserting (A113) as in the following steps: J

a E (D , E ,Z );t

X 1

a

f (W )dW . Let us expand (A115) by

t

> f (t )@ ³a  (t  W ) E 1 EDJ ,E (Z (t  W )D )  f (W ) dW f

J

t

³ (t  W )

(J ) k

z

k

¦ *(D k  E ) k !

ED , E ( z )

k 0

J

a E (D , E ,Z );t

t

> f (t )@ ³a  (t  W ) E 1 ¦ k 1

X

a I t f (t )

* (X )

³

t

a

f

X 1

(t  W ) f

J

a E (D , E ,Z );t

> f (t )@ ¦ k 0

f

¦

 (J )k Z k





( J )k Z k k!



Dk E

f (t )

a It

a E (D , E ,Z );t > f (t ) @

J

* (X  k )

J

* (X )

a E (D , E ,Z );t

> f (t ) @

f

¦

k

k!

D k  E 1

a

¦

(J ) k Z

E

a It

Z k * (J  k ) k 0 k !* ( J )

f



aI

Dk E



f (t )

f (W )dW Dk E

a It



f (t )





X* (X )

f (t )  JZ

J (J  1)Z

2

2!



E

t

³ (t  W )

; * (X  1)



In (A117) putting J

(A116)

1 * (D k  E )

k 0

(J ) k

 f (W ) dW

f (W )dW

k! Using (A116) we can write the following in expanded form: k 0

( J )k (Z ( t W )D )k k! 0 * (D k  E )



a It



2D  E a It

D E

J (J  1)(J  2)Z

3

3!

f (t )





f (t )

(A117)



3D  E a It



f (t )  ...

1 , we have the following: 1 a E (D ,1,Z );t

t

> f (t )@ ³a  ED (Z (t  W )D )  f (W ) dW 1 a It f

I I

(t )  Z Z

3

D 1

a t

3D 1 a t

 f (t )  ...

f (t )  Z

1

2

We note that the Prabhakar integral defined via the non-singular kernel ND ,1 (t )

2D 1 f a It

(t )

D



(A118)

ED (Zt ) in (A118) is the sum of

the classical Riemann-Liouville fractional integration using the power law (singular) kernel.

APPENDIX B LIST OF LAPLACE TRANSFORM PAIRS OF FUNCTIONS RELATED TO FRACTIONAL CALCULUS

Function

$ ^ f ( x )`

Laplace Transform F ( s )

f ( x) sD 1 sD B O

Re > s @ ! O

,

k !sD E ( sD B O )k 1

Re > s @ ! O

,

k! ( s B O )k 1

Re s ! O

 log  

ae k s s a s



e

1 s

e

1 s

e

e

s2  a2 s2

kt0

,

kt0

,

kt0

,

erfc

2



1 s s

e

1 s

e

1 s

e e

k

e

1 1

Sx

k s

 4k x

  2 kx

J v 1

Sx

k s

2

e

2

sinh 2 kx

1

s

k 2 x

I v 1 2 kx

Sx k

 4k x

2

Sx

k s

e



2 S x3

1

s



erfc a x 

2

k

( v 1)

sin 2 kx

cosh 2 kx cos 2 kx



J 0 2 kx

coth  Ss 2k



k 2 x

erfc

( v 1)

k 2 x

k 2 x

 k erfc 

Sx

 kx

e

ak a 2 x

1

v!0

ax

e )

erfc a x 

kt0

1 s s

bx

e

,

k s2  k 2

(e

2 x  ( k /4 x )

S

k s

1 s

1 x

2 x

 kx

e

(1  cos ax )

k

v!0

1 sv

D 1

2 x

 e

,

e

x

(1  cosh ax )

ak a 2 x

e e

k /s

1 sv



2 x

kt0

,

D

(rO x )

 rO x

(k ) 1 2, 2

x 2 E1



,

k s

k s

k 1

(k )

ED , E

 1x sin(kx)

kt0

k s

k s

2

 ks

,





D k  E 1

 ssba

log e k s s a s

x

1 * (D )

s2 a2 s2

log

^F ( s )` D

D

arctan

1

ED ,1 ( r O x )

 1D

>@

, s

1 s s

 1D

$

sin kx





458

Appendix B 1

1 s

Sx

1 s s

1 sn s

,

n



n 1

1, 2,...

2 2n x 1.3.5...(2 n 1) S

s ( s a )

3



1 s s  a2



 



a2 x

1

a2 x

1

2a

Sx

S



ax

s b

e



2

a x



y2

0

a x

 e ³ e dy ªb  a  erf  a x º  be erfc  b x ¬ ¼ e erfc  a x e erf  b  a x ª erf  a x  1º  e erfc  b x ¬ ¼  a2 x

y2

0

b2 x

a2 x



b2  a 2 s s a2 s b

a x

 a2 x

a2 x

2 a S

1 s b  s  a

 ax

1 ba

e



(1 s )n s

1 Sx

1 a

2

1 s a



e

Sx

1 s s a2

2

bx

  ae erfc  a x   ae erf  a x    e ³ e dy e erf  a x

s s  a2

b a

e

1

s s a2

 s  a 

e (1  2 ax )

2 S x3

1 s a

2

ax

1

Sx

2

sa  sb

s

x S

2

n! (2 n )! S x

 2

n 1

H 2n

a2 x

b2 x

b a

 x ,

H n ( x) e



x2

dn dx n

2

ªe x º ¬ ¼



H n :Hermite Polynomial (1 s ) s

n

n!  (2 n 1)!

 2

n 3

s2a  s s

ae

 ax

H 2 n 1 ( x )

Sx

n

> I1 (ax)  I 0 (ax)@ ,

I n ( x ) i J n (ix )

J n Bessel function of first kind 1 s  a s b *(k ) ( s  a )( s  b )

e

, k !0

1 ( sa)

1

S xe

3 2 ( s b ) 2

 a x b

 12 ( a  b ) x

k  1 2

 12 ( a  b ) x

( a b )k

,

1 x

1 s sa



sa  s



2v

1 s2  a2

k !0 ,

k !0

k x

 12 ( a  b ) x

I k  1 2

 a 2b x

ª¬ I 0  a 2 b x  I1  a 2 b x º¼

s2a  s s2a  s

( s  a  s  b )2 k

e

I 0  a 2 b x

e

e

 ax

I1 ( ax )

 12 ( a  b ) x

1 av

e

 12 ax

Ik

 a 2 b x

I v  a2 x

J 0 ( ax )

List of Laplace Transform Pairs of Runctions related to Fractional Calculus 1

I 0 ( ax ) , Modified Bessel function of the first kind zero order.

s2 a2



s2  a2  s



1





2

s a

2



2

2

s

s2 a2

v

v

,

k

a J v ( ax )

, v ! 1

s2  a2

k !0

S *(k )

k !0

,

v ! 1

,

k !0

S *(k )



N

1

s2  a2 s  s2  a2

s a

s

 x

1 J ( ax ) a 1



1 J ( ax ) aN N

1 2

I k  1 ( ax ) 2

N J ( ax ) aN x N

1

2

k  1 2

1 ax J1 ( ax )

1

s  s2  a2

 2xa

erf

s  s2  a2



J k  1 ( ax ) 2

a I v ( ax )

1 s s1



k  1 2

v

v

s2 a2 1 ( s 2  a 2 )k

,

 2xa

kak J ( ax ) x k

k

s a s



459

s2  a2



N

Table-B-1: Laplace transform pairs of some important functions related to fractional calculus Function

Higher Transcendental Function f ( x ) for x t 0

Mittag-Leffler Agarwal Erdelyi RobotnovHartley Miller-Ross Generalized Cosine

q

¦

Eq ( ax )

x

E 1

D

ED , E ( x )

Laplace Transform F ( s ) sq s ( sq a )

f

a n x nq n 0 * ( nq 1)

¦

f

sD  E sD 1

xD m E 1 m 0 * D m  E f

ED , E ( x )

¦

f

Fq ( a, x )

¦

f

a n x( n1) q1 n 0 *  ( n 1) q

1 sq a

E x (v, a )

¦

f

s v s a

C x (v, a ) x

v

¦

xm m 0 * D m  E

¦

a k x k v k 0 * ( v  k 1) k

( 1) 2 ( ax )k k even * ( v  k 1)

f

$ ^ f ( x )`

, Re >v @ ! 1

m! m 0 s m1* D m  E

s sv ( s 2  a2 )

(Miller-Ross) Generalized Sine (Miller-Ross) Generalized R Generalized G

S x (v, a ) x

v

¦

Rq ,v ( a, x ) Gq ,v ,r ( a, x )

¦

( k 1)

( 1) 2 ( ax )k k odd * ( v  k 1)

f

f j

>@

; Re v ! 2

a sv ( s 2  a2 )

a n x( n1) q1v n 0 *  ( n 1) q  v

sv sq a

 (  r )( 1 r )...(1 j  r ) (  a ) j x( r  j ) qv1 * (1 j ) * ( r  j ) q  v 0

sv ( s q  a )r

¦

f

Table-B-2: List of Laplace transforms and inverse Laplace transforms of higher transcendental functions related to fractional calculus

APPENDIX C FRACTIONAL DERIVATIVES AND INTEGRALS OF SOME IMPORTANT FUNCTIONS

In Table-C1, we tabulate some semi-derivatives and semi-integrals of a few monomials

f

1

0 Dx

p

x ; p ! 1

x

* ( p 1) *  p  12

0

C , any constant

C

,D

0.79195...

x

0.22119

x

x n

x , n

x

n   12

, n

2

0,1, 2, 3...

0,1, 2, 3, ...

3 8

3

 S x

2

5

8x 2 3 S

16 x 2 15 S

( n!)2 (4 x )n

( n!)2 (4 x )

(2 n )! S x

2( n!)

 S x 3

 S x

2

E   12

4x 2 3 S

S

(2 n 1)! S

S

x E 1 E 1

E

x

2 3 4

x 1 2

S x

x

2 1 2

x

 32

S

 E  12 x E 

E   12

x

 12 D D  12

1

1 2

x S

x

Sx

x

p   12

x

2C

0

0

>f@

0

D  12

x {1 E

* ( p 1) *  p  32

Sx

1 x

x , E

p  12

x

0

D

 12 0 Dx

>f@

2

 4x



n 1 2

2

 (2 n 1)! S n

(2 n  2)! S

 ( n 1)!

2

 4x

n 1

Table-C-1: List of semi-differentiation and semi-integration of monomials Here in this Table-C2, examples of fractional integrals and derivatives are tabulated. These are useful functions for physics and engineering. For simplicity, we assume that all variables are real ( \ ) and that x is positive and x ! 0 . The fractional order v is assumed to be arbitrary (positive, negative or zero) unless stated explicitly. The constants a , c , O and P are assumed to be unrestricted unless otherwise indicated. The section demonstrates the use of higher transcendental functions * , %W , 2 F1 , Et , \ ( x ) , and others described earlier. The derivative in the table is the Riemann-Liouville type.

Fractional Derivatives and Integrals of some Important Functions

461

Fractional derivative/ integral expression

Function f ( x )

rv

c

( x  a)

O

Dx ¬ª f ( x ) ¼º

x!c!a

( x  a)

O

(c a )O v x c * (1v ) ( x  c ) u 2 F1  O ,1,1  v;  c  a

v O ª º c D x ¬( x  a ) ¼

v c Dx

ct0

ct0

x!c!a

( x  a)

O

v c Dx

O

v 0 Dx

O

* ( O 1) * ( O  v 1)

v c Dx

f ( x) 1 (a  x)

>1@

v 0 Dx v c Dx

( x  a)

O v

; O ! 1

at0

ª¬ x O º¼

f ( x) 1

* ( O 1) * ( O  v 1)

ª¬( x  a )O º¼

x!c x

( x a )O v * (v ) % F (v, O  1) , v ! 0

ª¬( x  a )O º¼

x

O v

1 * (1 v )

>1@

( x  c)

1 * (1 v )

( a  c )O * (1 v )

ª¬( a  x )O º¼

; O ! 1;

x

v

v

( x  c)

x!c

v

a

0

x!ct0

;

;

x!c

0

u 2 F1  O ,1,1  v; ax cc

a!x!ct0 (a  x)

O

v c Dx

( a  x )O v *(v)

ª¬( a  x )O º¼

a ! x ! c t 0, (a  x) (a  x)

 1 2

 12 0 Dx

 1 2

1

0 Dx

1

v 0 Dx

O

x exp ax

 12 0 Dx v 0 Dx

1

S

ª( a  x )  1 2 º ¬ ¼

exp( ax )

exp( ax )

v!0

ª( a  x )  ( 2 ) º ¬ ¼ 2

% F  v, (O  v )

ln

a x a x

a

1

S x a x

ª¬e ax º¼

,

1

* ( O 1) * ( O  v 1)

x

O v

a!x a!x

E x ( v, a )

ª¬e ax º¼ a  2 e ax erf ª¬ ax º¼ ,

ª¬ x O e ax º¼

,

a!0

u 1 F1 (O  1, O  v  1; ax )

O ! 1 x exp ax

v 0 Dx v 0 Dx

Ex (P , a) O

x Ex (P , a)

v 0 Dx

ª¬ xe ax º¼

xE x (  v, a )  vE x (1  v, a )

> E x ( P , a )@

E x ( P  v, a ) ,

P ! 1

ª¬ x O E x ( P , a )º¼

* ( O  P 1) xO  P v * ( P 1) * ( O  P  v 1) 2 F2 ( O

 P  1,1, P  1, O  P  v  1; ax )

O  P ! 1 xE x ( P , a )

v 0 Dx

> xEx ( P , a)@

xE x ( P  v, a )  vE x ( P  v  1, a )

P ! 2 cos ax

v 0 Dx

>cos ax@

C x ( v, a )

462

Appendix C

cos ax

 12 0 Dx

C( z)

³

z

0

>cos ax@

2

cos 12 S y dy and S ( z )

2

³

z

0

2

x cos ax

v 0 Dx

> x cos ax@

Cx (P , a)

v 0 Dx

>C x ( P , a ) @

v 0 Dx

xC x ( P , a )

a!0

sin 12 S y dy are Fresnel integrals x v 2 * (1 v )

ª¬cos 2 ax º¼

v 0 Dx

cos ax

C ( z ) cos ax  S ( z ) sin ax ;

2 a

> xC x ( P , a)@

 12 C x ( v, 2 a )

xC x ( v, a )  vC x (1  v, a ) C x ( P  v, a ) ,

P ! 1

xC x ( P  v, a )  vC x ( P  v  1, a )

P ! 2 x x

 12

 12

cos x

ª 0 Dx x

cosh x

ª 0 Dx x

cos x º

S 2 x

 12

cosh x º

S 2x

¬

v

¬

¼

v 0 Dx

 12 0 Dx

C( z)

³

z

0

>sin ax@

2

cos 12 S y dy , S ( z )

³

x sin ax

v 0 Dx

v 0 Dx

2

 1 2  v

 x

I  1  v 2

 x

S x ( v, a )

 C ( z ) sin ax  S ( z ) cos ax ,

a!0

2

ª¬sin 2 ax º¼ aS x (1  v, 2a )

> x sin ax@

v 0 Dx

S x (P , a)

>sin ax@

1

J  1  v 2

sin 12 S y dy

v 0 Dx

sin ax

xS x ( P , a )

z

0

2

2 a

 1 2  v



¼

sin ax

sin ax



 12

v

xS x ( v, a )  vS x (1  v, a )

> S x ( P , a )@

> xS x ( P , a)@

S x ( P  v, a ) , P ! 2

xS x ( P  v, a )  vS x ( P  v  1, a )

P ! 3 sin x

v

ª º 0 Dx sin x ¬

sinh x O

x ln x

v 0 Dx v 0 Dx

x

1

2

ln x

ln x

¼

ªsinh x º ¬ ¼

1 2

1

2

 12 v



S 2x

1

2

 12 v

J  1 v 2

 x

I  1 v 2

 x

ª¬ x O ln x º¼

* ( O 1) * ( O  v 1)

ln x



S 2x

1 2

x

O v

v 0 Dx

 ln x  \ (O  1)  \ (O  v  1) ,

>ln x@

 12 0 Dx  12 0 Dx

x v * (1 v )

 ln x  J  \ (1  v)

1

ª x  2 ln x º ¬ ¼

>ln x@

2

S ln  14 x x

S

 ln 4 x  2

O ! 1

Fractional Derivatives and Integrals of some Important Functions

x O  ln x

2

v 0 Dx

ª x O  ln x 2 º ¬ ¼

* ( O 1) * ( O  v 1)

x

O v

§  ln x  \ (O  1)  \ (O  v  1) 2 · ¨ ¸, ¨  D >\ (O  1)@  D >\ (O  v  1)@ ¸ © ¹

O

 x

v 0 Dx

ª xO 2 J O ¬

 x º¼

2

v

x

O

 x

v 0 Dx

ª xO 2 IO ¬

 x º¼

2

v

x

x 2 JO x 2 IO

463

Table-C2: List of fractional differ-integration of various functions

( O v )

( O v )

O ! 1

2

J O v

 x ,

O ! 1

2

I O v

 x ,

O ! 1

APPENDIX D FORMULAS OF MODIFIED FRACTIONAL DERIVATIVE

The formulas of the modified fractional derivative of the Jumarie type are listed as follows for 0  D  1 : D

D

D

D ED (O x )

O ED (O x )

D

D ED (O x )



D

OD

D

D ED u ( x )

³

x

D 1D

x

 E u D

D

D

(dx )

D

D

ED (O[ )  d[

0

ED (O x )

O

1 D

1

 u c( x ) D !  E (O x

D

( x)

D

D

(D1)

D

(D !) d ( x ) D

1D

D !1  D ! dx

 dx 1 D  (1  D )! ³ x1D  dx

D



) 1

x

³d x D D D ³ f ( x )  dx ³ d f ( )

³

D ! dx

D !x

D ! f ( x)

The formulas regarding the Mittag-Leffler function and the application of the modified fractional derivative of the Jumarie type are listed as follows for 0  D  1 : ED  ix

cosD x  i sinD x

D

D cosD x ED ( x )

D

D





D

D

D

D sinD x

D

cosh D x

D

D



D

ED ( x  y ) 1

(( LnD u ) ( LnD u ) D

D

1

D

D

D

D ED (O x )

 LnD u

D

D

 ( LnD v ) D )  ( LnD v )

1

D

(D2)

D

D

cosh D x

D

D

we

ED ( x ) ED ( y ) 1

D

ED ( x ) ED ( y ) for the modified RL fractional derivative Jumarie type for fractal functions

u , which gives x



cosD x

D sinh D x

ED , which are continuous but non- differentiable. The inverse of y set ED ( x )

D

coshD x  sinhD x

D sinh D x

We have ED ( x  y )D

 sinD x

1

D

v

and



ED (u ) we define as u

D

ED ( y ) gives  LnD v

get ED (( LnD u )

1

D

1

D

 ( LnD v ) D )

1

y . Using the derived expression,

D



uv .

From

LnD (uv ) , which gives us the relationship of the

 LnD (uv) D

1

D

in conjugation to ln uv

LnD y . With this, we

here

we

write

D  logarithm function as

ln u  ln v of classical calculus. Like we have

O ED (O x ) ; we have the expression DD LnD x D D  (1  D )!

2

x

D

. The formulas regarding the

fractional logarithm function and the Mittag-Leffler functions are listed as follows for 0  D  1 :

Formulas of Modified Fractional Derivative D

d x

³

 Cx

LnD

x y

y LnD x D

ED ( xy )



 LnD uv

1

 LnD u

D

D

D!

D

1

1

x

(y )

x

E

(x )



y

D

;

D

D

D 1

D

D

1

  LnD v

D

§ d[ · ¨ ¸ , ³ (1  D )! 0 © [ ¹

 dx

x

ED



x 1 (1D )! 0

D

[ ³ [ d



1

D  1  (1  D )!

x

\

D

D

x

D

³

1

(1  D )! x

LnD x

³

E

D D

ED ( x y )

d LnD x dx

§x· ¨ ¸!0 ©C¹

ED  LnD x

x

D

LnD ( x )



465



ED 

D

 x2 2V 2



D

D

dx

 2V

3

§ ¨ ¨ ©

D ! ·¸ S  2D ! ¸¹

D

2

(D3)

Fractional order gamma and beta function are listed as the following for 0  D  1 : 1

f

D

D ! ³0  ED ( yD ) y ( x 1)D  dy *D ( x  1) D !  x  *D ( x ) *D ( x )

n

D !  n !

*D ( n  1)

1

³ (1  t )

BD ( x, y )





( x 1)D ( y 1)D

t

0

D

 dt

*D ( x ) *D ( y )

BD ( x, y ) The formulas derived by using ED ( x  y )D

(D4)

*D ( x  y )

D

D

ED ( x ) ED ( y ) , 0  D  1 are the following:

ED (1)

¦

f

2D n n 0 (D n )!

 ¦

f

1 n 0 (D n )!



1

 E (1) E ( x ) E  ( x  m) E (1) E ( x ) E ( n ) E  (0  n ) E (1) E (0) E  ( mx )  E ( x ) ¦ Cu ¦ C u ; C ( x  y) x ¦ C x y D

D

ED (1  x )

D

D

m

D

D

D

D

D

D

n

D

D

D

D

D

m

D

D

(D5)

D

D

f

f

k

Dn

k

k 0

Dn

Dn

f

k 0

Dk

Dn

k 0

Dk

D k D k

Dn

Dk

n

r

n! r !( n  r )!

APPENDIX E BRANCH-POINTS, BRANCH-CUTS AND RIEMANN SHEETS FOR MULTI-VALUED FUNCTIONS

E.1 Branch-points and branches in multi-valued function A function of a complex variable Z f ( z ) is viewed as a mapping of points in the z  plane to points in the Z  plane. If to each value of the independent variable, z there is one and only one image point Z , the mapping is said to be single valued. In contrast, let us examine a multi-valued function. Consider a small circular path about a point z0 . The circular path is z

z0  re

iT

where r ! 0 is a small constant and T varies in a counter clockwise direction

about the point z0 . If we have a function Z

f ( z ) such that Z

iT

f ( z0  re ) takes on different values as T

increases by 2S , (i.e. one full rotation about z0 ); then the point z0 is called the ‘branch point’ of the function and the different values of Z are called ‘branches’ of the function. By definition, a multi-valued function occurs if for each value of z here is more than one value for the dependent variable Z . Then several values of Z are said to be branches of the complex valued function. If it is possible to solve an equation of the form F ( z , Z ) 0 , connecting the complex variable z x  iy and Z u  iv to obtain a single valued function Z1

f1 ( z ) , Z2

f 2 ( z ) , Z3

f3 ( z ) , … then these functions are called branches of the

function Z .

E.2 The branch point of order (n  1) and branch-cut for n  valued function A point, say z0 , satisfying the property that there is no neighborhood z  z0  in which the function Z single valued, then the point

f ( z ) is

z0 is called a branch point of f ( z ) . A branch point is said to be of order ( n  1)

whenever a function Z f ( z ) is a n  valued function in the neighbourhood z  z0  . A line that connects two and only two branch points is called a branch cut (or branch line). Consider a two-valued function as follows:

Z1

f ( z)

1 ( z  z1 )( z  z 2 )....( z  z n )

(E1)

This (E1) has singularities at the points z1 , z 2 , ....., z n in the z  plane. Let z denote a variable point in the complex

z  plane and construct straight lines from the point z to each point z1 , z 2 ,...., z n and denote the length of these lines by r1 , r2 ,..., rn . These straight lines make angles T1 , T 2 , ..., T n respectively with horizontal lines through each of the points z1 , z 2 ,..., z n , as depicted in Figure-E1.

Branch-points, Branch-Cuts and Riemann Sheets for Multi-valued Functions

467

y

y z

x

x

zk 1

rk

x z1

zk

x

zk x

rm

Tk x

x z2

Tm zm

z

zm 1 x

zm x

x

x

x

x zn x zn 1

z-plane Branch points of the function Z



1 ( z  z1 )( z  z2 )...( z  zn )

f ( z)

Figure-E1: Branch points for function ( z  z1 )( z  z 2 )...( z  z n )



 12

The variable point z can then be represented in terms of the modulus r1 , r2 ,..., rn and the arguments T1 , T 2 , ..., T n by writing z

z1  r1e

iT1

, z

z 2  r2 e

iT 2

,…., z

z n  rn e

Z1

iT n

, and we get the following:

1

f ( z)

( z  z1 )( z  z 2 )....( z  z n )

z  z1

r1e

iT1

, z  z2

r2 e

iT 2

..., z  z n

rn e

iT n

(E2)

1

Z1

r1r2 .....rn e

i (T1 T 2 ...T n )

After moving the point z in a small circle about the point z m , m fixed, the values ri , i

1, 2, ..., n return to their

original values. The angle T m increases to the value T m  2S and the other values of T i , i z m return to their original

Z1e

values. The given equation becomes Z2

 iS

Z1 . Observe that if we move the point z in a small circle about

any one of the points z1 , z 2 ,..., z n , the same thing happens, and changes Z1 to Z2

1 z*

f , we make the substitution z

In order to examine the behavior of the function Z1 at the point z *

Z1 . and examine

*

*

the behavior of Z1 for z near the origin 0 of the z  plane, and write:

Z1

1



1 z*



1 z1*



1 z*



1 z2*

.... 

1 z*



1 zn*



(E3)

* * * z1 z 2 .... z n

 *

*

For point z near the origin 0 let z *

*

*

re

iT *

* z1

z



* z2

z

and as r p 0 , one obtains Z2

*

the origin 0 the angle T changes to T  2S and Z1 changes to Z2 same value and if n is odd, then Z1 becomes Z2 This shows the function of the form:

*

Z1 .

Z1e

inS

*

....  ( re

z *

* zn

i (T * )

z n

*



n

2

*

) 2 . Also as z moves about

. Therefore if n is even, Z1 keeps the

468

Appendix E

1

Z

( z  z1 )( z  z 2 ) 1

Z

(E4)

( z  z1 )( z  z 2 )( z  z3 )( z  z 4 ) 1

Z

( z  z1 )....( z  z 2 m )

The equations have respectively 2, 4, ...., 2m branch points but no branch point at infinity. In contrast the functions with the form:

1

Z

( z  z1 )( z  z 2 )( z  z3 ) 1

Z

(E5)

( z  z1 )( z  z 2 )( z  z3 )( z  z 4 )( z  z5 ) 1

Z

( z  z1 )( z  z 2 )....( z  z 2 m )( z  z 2 m 1 )

have respectively 3, 5,..., (2 m  1) branch points with each function having a branch point at infinity. In case of an even number of branch points, the branch points are connected in groups of any two pairs, where the connecting cuts do not cross one another. In the case of an odd number of branch points the cuts are made in groups of any two pairs, where connecting cuts do not cross one another. The remaining points are joined to infinity by a cut line, which does not cross the other cut lines.

E.3 Multiple Riemann sheets connected along the branch-cut for a multi-valued function To avoid the problem that the same value of z corresponds to two or more values of Z for a multi-valued function Z f ( z ) the z  plane is split into n parallel z  planes called sheets of Riemann surface (or simply Riemannsheets); where n corresponds to the multiplicity of the function. These n sheets are separated by an infinitesimal distance (gap) and are connected along a branch-cut or along each of the branch cuts if more than one branch cut exists. In this way as z moves around the first sheet the image of Z is that of the first branch Z1 f1 ( z ) . Then on, as z moves around the second sheet; the image of Z is that of a second branch Z2

f 2 ( z ) . In general, the value of

z on the i  th sheet, for i 1, 2, ..., n , produces a single valued function Zi fi ( z ) . As z moves around a sheet and crosses a branch-cut then there occurs a change in the branch of the function. All the sheets are connected along the branch-cut(s) and are to be regarded as a continuous surface, called the Riemann surface (or sheets).

E.4 An example elaborating branches and branch-cut for a two-valued function We consider the function Z branches Z1

Z

2

i.e.

z

Z1

when k

re

f1 ( z )

i (T  2 kS )

f1 ( z )

2

z , which has a branch point at z

 z and Z2

f2 ( z )

. From here we write Z

 z

 re

iT /2

 z . Let z z

and for k

1

2

1

r 2e

re

0 and is a two-valued function. It has got two

i (T  2 kS )

i T  2 kS /2

, with k

0,1 . We obtain for k

for k

1 the second branch Z2

3, 5, 7, ... we are on the first branch Z1 and when k

0,1 . Write Z

f2 ( z )

 z

 re

2

z in the form

0 , the first branch, i S  T 2

. Note that

4, 6, 8, ... we are back onto the second branch Z2 .

We desire to define a domain where these branches of the function are single-valued and analytic at each point of the domain. The derivative of the function, i.e. Z1 partial derivatives, i.e.

w u w r

f1 ( z )

r

1

2

cos  T2  ir

1

2

sin  T2

, ww Tu , ww vr , ww Tv , as described in the following expressions:

u ( r , T )  iv ( r , T ) is obtained by

Branch-points, Branch-Cuts and Riemann Sheets for Multi-valued Functions

w u

1

w r

2

w u wT

The total complex derivative is:

dZ1 dz dZ2



r

 12

1

singular point z 1

changes to r 2 e

r 2 sin  T2

(1)

f1 ( z ) (1) f2 ( z )

r 2 e other.

i T  2S /2

d

,

z

dz d

dz 0 and z

1 2

 z

z

w v

1

w r

2

r

1

wT

2

1 2

1

r

sin  T2

(E6)

w v

 12

 12

1

r 2 sin  T2

 1 2  iT /2

e

(E7)

 1 2  i S  T 2

r e 2 f . The points z

f are singular

0 and z

1

z 2 . Let us examine the behavior of the function Z1 as we move around the

0 . If we hold r to be constant and let T vary from T to T  2S , we find that Z1

i T  2S /2

1

iS

e r 2e

iT /2

iS

1

e r 2 e

iT /2

1

r 2e

iT /2

Z2 .

Similarly, for the function Z2 as we move around z 1

,

1

2

dz The functions Z1 and Z2 fail to be analytic at point z points associated with the function Z

cos  T2

469

0 ; we have Z2

1

r 2 e

i T /2

, which changes to

Z1 . This shows that as T increases by 2S the functions Z1 and Z2 change into each

If we construct a branch cut from 0 to f along the negative real axis in the z  plane; and require that z not to be allowed to cross the branch-cut, then the functions Z1 and Z2 become single-valued and analytic when defined as follows: 1

Z1

f1 ( z ) r 2 eiT /2

Z2

f 2 ( z ) r 2 eiT  2S /2

r!0

1

S T d S ; T

r !0

S T d S ; T

arg > z @ ; arg >Z1 @ arg > z @ ; arg >Z2 @

T 2

(E8) T 2

S

Note that at each point on the branch cut, there occurs a discontinuity in the functions Z1 and Z2 . The branch cut is a way of preventing these discontinuities from occurring and keeping the function (in this case the square-root function) single-valued. Figure-E2 depicts the branch cut, and two Riemann-sheets.

>@

0 with S  arg z  S , in the complex z  plane we

For Sheet-1 in Figure-E2, as we move in a circle near z

> @ S  arg > z @  S , in the Complex z  plane, we have its image z moves as S 2  arg >Z1 @  3S 2 , in the complex Z  plane.

have its image moving in Branch-1 i.e. Z1 where  S 2  arg Z1  S 2 , in the complex Z  plane. In Sheet-2 of Figure-E2, as we go around point z

0 , with

moving to Branch-2, i.e. Z2 , where the image of

v x

sheet (1)

u

y

branch-cut x

Z1 for sheet (1)

sheet (2) v

y

x

u

branch-cut

sheet (2) y

sheet (1) z-plane

branch-cut

Figure-E2: Showing branch-cut at negative real axis and two Riemann-sheets

Z2 for sheet (2)

Z  plane

470

Appendix E

E.5 Choosing the contour with branch-cut for a function having branch-point For the evaluation of contour integration

³  (Z  z ) f (Z )dZ ;  n 1

C

where C describes a closed contour surrounding

the point z in the complex Z -plane; and enclosing a region of analyticity of the function i.e. f (Z ) . positive integer as in the integrand (i.e. n ) is replaced by a non-integer say D ! 0 ; then (Z  z ) pole at Z z but a branch point.

D 1

When the

no longer has a

Then one is no longer free to deform (or squeeze) the contour C surrounding the point z at will; since the integral D 1

will depend on the location of the point at which C crosses the branch line (branch-cut) for (Z  z ) . In Figure-E3, this point is chosen to be 0 (the origin of Z -plane). The branch cut (branch-line) then is to be a straight line joining Z

0 and Z

z , and continuing thereafter indefinitely in the third quadrant i.e. Re >Z @ d 0 and Im >Z @ d 0 ; as

depicted in Figure-E3. Then for when D is not a negative integer (i.e. D ! 0 ) we have the integral as contour C begins and ends at Z

D ³  (Z  z ) f (Z )dZ , where the  1

C

0 enclosing z once in a counter clockwise direction (positive sense). To uniquely D 1

specify the denominator in the above expression of the integrand, we define (Z  z ) ln(Z  z ) is real when (Z  z ) ! 0 .

e

(D 1)ln(Z  z )

where

C

z

Im Z Cc

0 0

Re Z

Figure-E3: The Branch line and the contour C is deformed into Cc

We now deform the contour C into contour Cc lying on both sides of the branch line. We write ( z )

D ³  (Z  z ) f (Z )dZ ; also represented in this case as ³  (Z  z ) f (Z )dZ . We have used this concept in  1

Cc

D 1

0

earlier chapters (Chapters-1, 2, and 6), where point z was on a real axis, i.e. z

x , and point 0 was at some other ( x)

D ³  (Z  z ) f (Z )dZ .

point than zero at c and we evaluated the integral by the contour integration method i.e. That is, we employed this method of complex analysis in the evaluation of the integrals

 1

c

³

x

c

( y  z)

D 1

f ( y )d y .

APPENDIX F MITTAG-LEFFLER FUNCTION WITH NEGATIVE ORDER

F.1 Restriction in Series Representation of the Mittag-Leffler Function We recall the series definition of the ‘two parameter’ Mittag-Leffler function, that is f

ED , E ( z )

z

k

¦ *(D k  E ) ;

D, E  \;

(F1)

D, E ! 0

k 0

While ED , E ( z ) is an entire function (e.g. analytic everywhere in complex plane ^ ), with order 1 / D ; here we will consider the function on real z . Furthermore the domain of ED , E ( z ) can be expanded to include a complex D and

E ; with Re >D @ ! 0 and Re > E @ ! 0 . Here we will discuss the Mittag-Leffler ED , E ( z ) function for case D d 0 . Let 

us examine why the domain of D excludes both D 0 and D  0 . In the limit as D p 0 , from the series definition as described above (F1), we find it reduces to a geometric series, that is: f

z

k

1

¦ *( E )  *( E ) 1  z ;

E0, E ( z )

z 1

(F2)

k 0

Keeping the condition E ! 0 ; the above (F2) expression for E0, E ( z )

¦

f k 0

z

k

1 1 z

converges; provided the geometric series

converges, i.e. for z  1 . Consequently, since z must be restricted in the region z  1 , the

analyticity of ED , E ( z ) in the entire complex plane is lost when D

0 . Thus, the restriction that D z 0 is written in

the series representation (F1); so that ED , E ( z ) remains the entire function. In addition, we note that ED ,1 ( z ) for E

ED ( z ) ;

1 , that is the one-parameter Mittag-Leffler function with a positive order D .

The infinite series as defined in the series definition (F1) diverges in the domain of the negative D . We demonstrate this by setting E

0 , thus getting ED ,0 ( z )

¦

f

zk k 0 * D k

. Placing D as a negative alpha by D , we write the ratio

(say) r for ( k  1) and the k -th term in the following expression:

r

z

*  D k *  D ( k  1)

D

z ( 1)

D (k  1) ! D k !

We used the relationship of the ratio of the gamma function with negative arguments, i.e.

*(n) *( N )

(F3)

( 1)

N n N ! n! ;

with n

and N positive numbers; described earlier (in Chapter-1). From the above ratio (F3), it is seen that r ! 1 for z ! 1 . For z  1 , we can have a value of k , beyond which the

r ! 1 and thus the series diverges. Therefore, the Mittag-

Leffler function is defined by the series that is restricted to D ! 0 . We will construct the case for ED , E ( z ) , with the notation D set to always be positive, so D represents a negative order.

F.2 The Integral Representation of the two-parameter Mittag-Leffler function Let us derive the integral representation of ED , E ( z ) with a known Laplace transform identity of the Agrawal function,

^

`

i.e. $ t E 1 ED , E ( at D )

Mittag-Leffler function:

sD E sD  a

. Now we write the following steps as we have learnt previously for a one-parameter

472

Appendix F

t

E 1

D

ED , E ( at )

1

s

D E

e D 2S i ³ s  a

st

ds

Br

Put

t

E 1

st

dx

x , ds

D

ED , E ( at )

t

x

, s

t

 xD E tD E

1

e 2S i ³   a

x

xD tD

Ha

dx t

§ xD  E · § t D · § e x · ³ ¨© tD E ¸¹ ¨© xD  atD ¸¹ ¨© t ¸¹ dx 2S i Ha 1

1

t 2S i ³ E 1

D

ED , E ( at )

ED , E ( z )

D E x

e

D

x  at

Ha

t

x

E 1

D

dx

E 1

§ xD  E e x ³ ¨© xD  atD 2S i Ha

t

· ¸dx ¹

put

at

X

z

§ xD  E x · e ¸dx ¨ D 2S i ³Ha © x  z ¹ 1

(F4)

Therefore, we have an integral representation as follows:

ED , E ( z )

§ t D  E et ¨ D 2S i ³Ha © t  z 1

· ¸dt ¹

(F5)

Where the path of integration Ha is Hankel’s contour (i.e. a loop starting and ending at f ) and encircling the 1

circular disk t d z D , in a positive (counter clockwise) sense. We write the above integral representation (F5) with a modification as follows:

ED , E ( z )

t § e ¨ E E D 2S i ³Ha © t  zt

1

· ¸dt ¹

(F6)

We expand the part of the integrand in the above expression in a partial fraction as follows: 1 1 1  E E E D E 1 D  E t  zt t t z t Using the above (F7), we write expression for ED , E ( z ) as:

ED , E ( z )

§ et ¨ E 2S i ³Ha © t 1

(F7)

t · § 1 e  d t ¸ ¨ E 1 D  E 2S i ³Ha © t  z t ¹

· ¸dt ¹

(F8)

From our derivation in Chapter-1, we have a relationship for the reciprocal gamma function in an integral form as 1 * X

1 2S i

³

Ha

X x

( x ) e dx . Using this formula, we say that the first term of the above expression is *1E .

F.3 The formula for a two-parameter Mittag-Leffler function with a negative order The second term of (F8), i.e. i.e.

ED , E ( z )

1 2S i

³

Ha

E

1 2S i

(t  zt

E

³

Ha

1 D  E 1 t

(t  z t

) e dt we compare with the obtained expression (F6) for ED , E ( z ) ,

E D 1 t

) e dt . We note that the second expression in (F8) is the same as the integral

representation with z changed to z

1

and D changed to D , so we have: 1

E D , E ( z ) With these observations we arrive at the following expression:

ED , E ( z )

t § e ¨ E 1 E D 2S i ³Ha © t  z t

1

1 *( E )

1

 E D , E ( z )

· ¸dt ¹

(F9)

(F10)

Mittag-Leffler Function with Negative Order

Replacing z by z

1

473

, we get the desired expression for ED , E ( z ) , that is:

1

E D , E ( z ) 1 *(E )

Applying the recursion formula (Appendix-A), i.e. ED , E ( z )

*( E )

 ED , E



(F11)

1 z

 zED ,D  E ( z ) , to the above (F11) we obtain:

E D , E ( z )



1 z

ED ,D  E

 1z

(F12)

The above-obtained (F12) formulas are for ED , E ( z ) . We note here that for E

1 , we write ED ,1 ( z )

ED ( z ) , the one-parameter Mittag-Leffler function with the

negative order.

F.4 The series representation of the two-parameter Mittag-Leffler function with a negative order In the formula (F11), i.e. ED , E ( z )

ED , E  z

¦

f

k

z k 0 * D k  E

1 *(E )

 ED , E

 1z ,

using the series definition of ED , E  z (F1), i.e.

, we write the following useful series formula for ED , E ( z ) :

E D , E ( z )

§ f  1 z k ·  ED , E  ¨¦ ¸ * E *  E ¨ k 0 * D k  E ¸ © ¹ k f § 1  1z · 1 ¨ ¦ ¸ *  E ¨ *  E k 1 * D k  E ¸ © ¹ 1

1

1 z

(F13)

k

 1z ¦ k 1 * D k  E f

The series representation of ED , E ( z ) therefore is: k

 1z ; ¦ k 1 * D k  E f

E D , E ( z )





Using the above series formula for ED , E ( z ) we observe that lim z nf ED , E ( z ) positive z -axis, and lim z p0 ED , E ( z )

k

 1z ¦ k 1 * D k  1 f

E D ( z )

(F14)

0 approaches from below the

f . Similarly we observe lim z pf ED , E ( z )

0 approaching from the

above negative z -axis. Now substitute D

0 in ED , E ( z )

1 * E

 ED , E

 1z ,

E0, E ( z )

yielding: 1 * E

 E0, E

1 z



k

 1z ¦ * E k 0 * E 1

f

(F15)

Summing the result in the above (F15) geometric series, and assuming z  1 we have the following:

E0, E ( z )

1 * E



§ 1 · ¨ ¸ *  E © 1   1z ¹ 1

(F16)

1

 *( E ) 1  z We find from the above (F16), that the result is the same as what we obtained for limD p0 ED , E ( z )

1

 * ( E ) 1 z

, i.e.

for positive D . This confirms that the Mittag-Leffler functions defined for positive and negative order ( D ) are equivalent for D

0 and z  1 .

474

Appendix F

F.5 The graphical representation of the Mittag-Leffler function with the negative order Now we see the graphical representation for ED , E ( x ) , depicted in Figures F1 to F3; for various ranges of D and various conditions for E .

2

E 1 2 , 3 2 ( x)

1 *  32

1 0

1 E 12, 3 2 ( x)

2 E 1 2 , 3 2 ( x) 3

4 5 10

4

2

0

2

4

10

x

Figure-F1: General pattern of ED , E ( x ) for 0  D d 1 and for E t D

0.2

E 1 2 , 18 ( x )

(0.77, 0.22) 1 *  18

0.1 0 E 12 , 18 ( x)

0.3

0.6 10

E 1 2 , 18 ( x )

4

2

0

2

Figure-F2: General pattern of ED , E ( x ) for 0  D d 1 and for E  D

4

10

x

Mittag-Leffler Function with Negative Order

475

E2,2 ( x)

1 0 E2,2 ( x)

E2,2 ( x)

2.5

5

0.4 0.2

1

0

0.2

0.4

x

1

Figure-F3: General pattern of ED , E ( x ) for D ! 1

As the value of x n 0



1 * E

we observe from Figure-F1 that lim xn0 ED , E ( x )

, where E 1

3 2, 2

( x ) is pictured, for a

general pattern with 0  D d 1 and E t D . For comparison in Figure-F1, the inset picture shows the graph of

E1

3 2, 2

E 1

E1

( x ) . Figure-F2 shows the general pattern of ED , E ( x ) , with 0  D d 1 and E  D ; the plotted function is

1 2, 8

1 2, 8

( x ) Here a maximum occurs at the negative x -axis point other than x

0 . For comparison the graph of

( x ) is given as an inset in Figure-F2. Figure-F3 shows the case of ED , E ( x ) for D ! 1 . The plot in Figure-F3 is

for E2,2 ( x ) ; and the expanded portion as marked by arrows, and shows very rapid oscillations close to x

0 at a

negative x -axis. The inset in Figure-F3 also depicts the nature of ED , E ( x ) for the case D ! 1 , the plotted function is

E2,2 ( x ) . This Mittag-Leffler function with a negative order function was studied extensively by John W Hannekan and Narahari Achar; some results of their study is presented here.

F.6 Comparisons of the functional relationship of ED , E ( x) and ED , E ( x) Now we list the various functional relationships for ED , E ( x ) and ED , E ( x ) for comparison, in Table-F1. Functional relations for ED , E ( x ) f

ED , E ( x )

¦

ED , E ( x )

1 * E

1 k 0 * D k  E

x

k

E D , E ( x )

 xED ,D  E ( x )

ª x E 1 ED , E ( xD ) º¼ x E  2 ED , E 1 ( xD ) dx ¬ E ED , E 1 ( x )  ax

³

x

0

y

E 1

 1x ED ,D  E ( x )

E ED , E 1 ( x )

 ax D

x ED , E 1 ( x )

³

x

0

y

 E 1

d dx

^

$ x

D

`

ED , E ( r ax )

s

D E

D

s Ba

^

$ x

E 1

ª¬ ED , E 1 ( x ) º¼ D

ED , E ( y )dy x

E 1

k

1

E D , E ( x )

E

1 x  * D  E

1 k 1 * D k  E

ª E11 ED ,E ( xD )º¼  E ED ,E ( xD ) x dx ¬ x

ª ED , E 1 ( x )º¼ dx ¬

D

 1x

¦

d

d

ED , E ( y )dy

f



E D , E ( x )

d

ED , E ( x )

Functional relations for ED , E ( x )

E

 `

ED , E r ax1D

D

ED , E 1 ( x )

Ba E

D

s (s B a)

476

Appendix F

d dx

d

1

> ED ( x)@

D

d

ED ,D ( x )

dx

n

ª E ( x n ) º¼ En ( x n ) ; n  ]  n ¬ n dx 2

ED , E ( x )  ED , E (  x ) n

1 n

EnD , E ( x ) n]

ED (  x )

n 1 k 0



ED , E xe

i 2S k / n



 n

x ED , E D n ( x )

¦

n

x k k 1 *  E D k

2

; n]

2

E2D ( x )  xE2D ,D 1 ( x )

1

1



* D  1 D

ED ,D ( x )

ª¬ E n ( x  n )º¼ E n ( x  n ) ; n  ]  dx n

2

ED , E ( x )  ED , E (  x ) 2 E2D , E ( x ) n

E nD , E ( x )

1 n

¦

n 1

E k 0 D , E

 xe

 i 2S k / n





E D , E ( x ) 



n

n]

ED , E ( x ) 

¦

2 E2D , E ( x )

d

> E D ( x ) @



n

x ED , E D n ( x )

¦

n 1

xk k 1 *  E D k

2

; n]



2

ED (  x ) E2D ( x )  xE2D ,D 1 ( x )

Table-F1: Several functional relationships for a positive and negative order Mittag-Leffler function

APPENDIX-G INVERSE LAPLACE TRANSFORM BY CONTOUR INTEGRATION

G.1 Revising basics of Laplace transforms For a function x (t ) , that is zero for t  0 , and is defined for t t 0 ; the Laplace transform is defined by the following (G1) integral. This is also termed as the Laplace integral; which we have used many times:

³

X (s) Here we have a complex variable s

f

0

e

 st

(G1)

x (t )dt

V  iZ ; with this, we have the following step:

³

X (s)

f

e

0

e

 iZt

V t



x ( t ) dt

(G2)

The inverse problem is stated as how we can get x (t ) from X ( s ) . The Fourier transform of x (t ) is xˆ(Z ) defined as xˆ(Z )

³

f

f

given by the following integral:

e

 iZt

x (t )dt , and the inverse Fourier transform is 1

2S ³

x (t )

³

In the Laplace transform expression (i.e. X ( s )

f

0

following substitution:

e

 iZt

f

e

f

iZt

xˆ (Z )dZ

(G3)

e V x(t ) dt ), as obtained above (G2), let us make the  t

I (t )

e

V t

x (t ) ; t t 0

(G4)

0 t0 where V is a constant in (G4). Taking the Fourier transform of I (t ) we write the following:

Iˆ(Z )

³

f

0

e

 iZt f

³ ³

0 f

0

I (t )dt

e

V t

e

 iZt

e

 (V  iZ ) t



x ( t ) dt

x (t )dt

(G5)

X (V  iZ )

Now we take the inverse Fourier transform of X (V  iZ ) as follows:

I (t ) Therefore, in the above (G6) we obtain x (t )

e

V t

f

e

iZt

X (V  iZ )dZ

(G6)

Vt

1

Z f

2S ³Z

Vt

f

1

V  iZ , we have ds

Z f

2S ³Z

e I (t ) and we write the following expression: x (t )

Letting s

1

x (t )

³

e

e

Z f

iZt



X (V  iZ ) dZ (G7)

(V  iZ ) t

e X (V  iZ )dZ 2S Z f idZ , and get the following from the above (G7):

x (t )

1 2S i

s V  iZ

s

³ e V Z i

0; t  0

st

X ( s )ds ; t t 0

(G8)

478

Appendix G

G.2-The inverse Laplace transform via contour integration The formula that we derived (G8) that is as follows, called the inverse Laplace transform:

x (t )

s V  iZ

1 2S i

s

³ e V Z

st

X ( s )ds

(G9)

i

We ask the following questions on this expression (G9) 1). How do we choose the real part of s , i.e. V ? 2). How do we calculate/evaluate the above integral (G9) in the complex domain? We already know that x (t ) 0 for t  0 ; that will help to answer (1). Im > s @

Im > s @

B

iR

B

iR

R

C1

R

V

iR

Re > s @

C2

V

Re > s @

iR

A

A

a

b

Figure-G1: Contour in the complex-plane to evaluate the inverse Laplace transforms

Consider Figure-G1a, for the closed contour is A o B o C1 . We write the contour integration as per the residue theorem of complex analysis (Section-1.7) as follows:

³

Ao B oC1

st

e X ( s )ds

³

st

e X ( s )ds 

Ao B

³e

st

X ( s )ds

C1

V  iR

³V

iR

st

e X ( s )ds  2S i

¦

At poles

³

C1

st

e X ( s )ds

(G10)

st

Residues ª¬ e X ( s ) º¼

We stress that residues are at the poles inside the closed contour A o B o C1 of Figure-G1a. Now as R n f the V  iR

³V

integral

iR

st

e X ( s )ds is the integral of interest that we require to evaluate the inverse Laplace transform

formula (G9). We note that the integral on the line A o B is the Bromwich integral, and is used for finding the inverse Laplace transform. We will use Jordan’s lemma (described shortly) which says that for t ! 0 , lim Rnf

³

C1

st

0.

e X ( s )ds

Therefore for t ! 0 , we write the following, as R n f 1 V if st st e X ( s )ds Residues ª¬ e X ( s ) º¼ V  i f 2S i Poles at Left of V

³

¦

(G11)

Consider Figure-G1b, the closed contour is A o B o C2 ; we write the contour integration as the following:

Inverse Laplace Transform by Contour Integration

³

st

³

e X ( s )ds

Ao B oC2

479

st

³e

e X ( s )ds 

Ao B

st

X ( s )ds

C2

V  iR

³V

iR

st

e X ( s )ds  2S i

¦

st

³

C2

e X ( s )ds

(G12)

st

At Poles

Residues ª¬ e X ( s ) º¼

We stress that residues are at the poles inside the closed contour A o B o C2 of Figure-G1b. The negative sign in (G12) indicates that the contour is taken in a clock-wise direction. We will use Jordan lemma’s which says for t  0 , lim R nf

st

³

C2

e X ( s )ds

0 . Thus we write for t  0 the following:

1

V  if

2S i ³V

 if

st

e X ( s )ds

st

¦



Poles at Right of V

Residues ª¬ e X ( s ) º¼

We know that this above integral (G13) must be zero, since for t  0 , we have x (t )

x (t )

¦

1 2S i

V  if

³V

 if

st

e X ( s )ds

0; t  0 . st

Poles at Right of V

Residues ª¬e X ( s ) º¼

Thus

the

LHS

of

(G13)

0 . This implies that we have

(G13)

is

zero,

making

0 . This says that there are no poles at the right side of the line Re > s @ V

(Figure-G1b).

>@

Therefore, the Re s

V , the line AB (Figure-G1), or Bromwich line, must be chosen such that the contour of st

Figure-1b, i.e. A o B o C2 does not contain any poles of e X ( s ) as R n f . Thus the contour of Figure-G1a, i.e. st

A o B o C1 must have all poles of e X ( s ) . This gives the answer to point (1) above. In addition, we note that since e

st

st

is analytic everywhere (i.e. it has no poles in the entire Complex- s plane), the poles of e X ( s ) are the same as

that of X ( s ) . This gives a reply to point (2) as posed above. Thus, we apply residue calculus of the complex analysis and write the following: 1 V if st st Residues ª¬ e X ( s ) º¼ (G14) x (t ) e X ( s )ds V  i f 2S i all poles of X ( s ) This is how we must evaluate the integral for obtaining an inverse Laplace transform.

¦

³

G.3 Jordan’s Lemma While we discussed the inverse Laplace transform using contour integration and residue calculus, we also stated that we need a condition that is an integral on arc C1 in Figure-G1a; as follows: lim

³

st

R nf C1

e X ( s )ds

0,

t!0

(G15)

While for t ! 0 , the points on this arc C1 are given as: s

iT

V  Re ;

S

dT d

3S

2 2 Examining the standard Laplace transform tables we observe most functions which satisfy lim s nf X ( s )

example, X ( s )

1 s

, X (s)

will take a case of X ( s )

1 sa

, X (s)

a s2  a2

(G16) 0 . For

etc. Therefore, in those cases where R n f we have X ( s ) p 0 . We

s  a shortly where lim s nf X ( s )

f , and still the inverse Laplace transform is

obtained. This condition (i.e. lim s nf X ( s ) X (s)

iT

0 ) means that for any M R ! 0 a radius R can be found such that

X (V  Re )  M R . By using the inequality (i.e.

following expressions:

³

C

f ( s )ds d

³

C

f ( s )ds ) for this R we have the

480

Appendix G

st

st

³

e X ( s )ds d

³

e X ( s )ds d M R

C1

C1

³

e X ( s ) ds

C1

(G17)

st

iT

On the arc C1 for t ! 0 , s

S

V  Re ;

e

st

e

st

³

e ds

C1

2

dT d

(V  R eiT ) t

e

3S 2

iT

iRe dT . With these we write the following steps:

and ds

(V  R cosT  iR sin T ) t

e

(V  R cosT ) t iRt sin T

,

e

e

(V  R cosT ) t

, e

e

i ( Rt sin T )

(V t  Rt cosT )

1

(G18)

!0

V t Rt cosT

e e

Further, we write the steps from (G17) and (G18), with the observation that e following: MR

3S

st

³

C1

e ds

³

MR

S

V t Rt cosT

2

e e

is an ‘even-function’ of T as the

iT

iRe dT

2

Vt

d M R Re

By changing the variable T

Rt cosT

3S

³

S

2

e

Rt cosT

Vt

2 M R Re

dT

2

³

S

st

³

C1

Vt

e ds d 2 M R Re

S

2

³

0

e

Vt

2 M R Re

sin [ t

2 S

e

Rt cosT

dT

2

[  S2 , we obtain from (G19) the following expression: MR

Plotting the graphs of y

(G19)

S

2

sin [ and a straight line (i.e. y

S

Rt cos[  S2 S

³

d[ (G20)

2

0

e

 Rt sin [

d[

[ ), it is observed that in the region 0 d [ d

S

2

, we see

[ . With this observation we write the following from (G20): MR

³

C1

st

Vt

e ds d 2 M R Re

0

Vt

d 2 M R Re

S

³

S

³

2

2

e

 Rt  S2 [

d[ (G21)

Vt

M RS e



 Rt



1 e t Therefore, for any t ! 0 as R n f , we have M R p 0 , meaning the above quantity in (G21) tends to zero. This proves our case (i.e. lim

³

R nf C1

st

e X ( s )ds

0

e

 Rt sin [

d[

0 , for t ! 0 ).

G.4 The application of Residue Calculus to get an Inverse Laplace Transform of X ( s ) 2e 2 s / ( s 2  4) by contour integration We would like to evaluate the inverse Laplace transform of X ( s )

2 e2 s s2  4

. The Bromwich path integral for the inverse

Laplace transform is the following: x (t )

1 2S i

V  if

³V

 if

e

st

1 2S i

We observe that the given function i.e. X ( s )

Re > s @

0 . In Figure-G1a, we choose V

2 e2 s s2  4

has two simple poles at s

2e

2 s

2

s 4 V  if

³V

 if

ds

2e

(G22)

s ( t  2)

2

s 4

2i and s

ds 2i with both having

0 . Rather we take an arbitrary (positive) small V close to zero. We can

distinguish the two cases (i) t  2 and (ii) t ! 2 .

>@

For case (i), the exponent s (t  2) has a negative real part if Re s ! 0 .

Inverse Laplace Transform by Contour Integration

We note that e

s ( t  2)

e

 Re> s @i Im> s @  t  2

e

the function’s behavior at infinity, since e

( t  2) Re> s @ i ( t  2) Im> s @

. Therefore, it is the part e

e

i ( t  2) Im> s @

481 ( t  2) Re> s @

1 . Thus, as Re > s @ n f , the function e

>@

2

the same time, the denominator s  4 diverges as Re s n f . This means the term

1 s2  4

along the path C2 (Figure-G1b), for R n f , and reaches zero. Therefore, we have lim Rnf

which determines

s ( t  2)

tends to zero. At

that multiplies e 2 e2 s st e ds 2 C2 s  4

³

s ( t  2)

0 , i.e. an

integral on the curve C2 as R n f ; as zero. We can calculate the Bromwich path integral by considering

A o B o C2 , but since this closed contour does not have any poles we say that residues are zero, resulting in x (t ) as zero. For case (ii), the function (i.e. e

s ( t  2)

>@

) tends to zero as Re s tends to f . That means the lim Rnf

³

C1

2 e s ( t 2 ) ds s2  4

0

along the curve C1 (Figure G1 a). For the residue theorem, this integral on the closed path A o B o C1 is given by 2 e s ( t 2 ) s2  4

the sum of residues of the function e st X ( s ) x (t )

1 2S i

2 s

if

2e

 if

s 4

³

2

st

e ds

at all the poles namely r2i ; and we write following steps:

§ 2e 2 s st · e ¸ 2i ¨ 2 ©s 4 ¹ § 2e 2 s st ·  Residue at s 2i ¨ 2 e ¸ ©s 4 ¹ s ( t  2) § · 2e lim ¨ ( s  2i ) 2 ¸ s o2i s 4 ¹ © s ( t  2) § · 2e  lim ¨ ( s  2i ) 2 ¸ s o2 i s 4 ¹ © Residue at s

lim

s o2i

2e

s ( t  2)

s  2i e

2e

 lim 

2i

e

s ( t  2)

s  2i

s o2 i

2 i ( t  2)

(G23)

2 i ( t  2)

2i

sin  2(t  2)

Thus, we write the following result:

x (t )

$

1

^ ` 2 e2 s s2  4

­sin  2(t  2) t ! 2 ® 0 t2 ¯

(G24)

G.5 The application of residue calculus to get the inverse Laplace transform of X ( s ) by contour integration on the branch cut in a complex plane This is for a multi-valued function; we take a function X ( s ) the following:

sa

s  a , with a  \ . The inverse Laplace transform is x (t )

1

V  if

2S i ³V

 if

e

st

s  a ds

(G25)

st

The function e s  a has no poles but the function z (taking z s  a ) is a multi-valued function in the complex plane. Therefore we see a branch point at z 0 , namely at s a . This is the only singularity of our st

function X ( s )e and therefore to evaluate the Bromwich path integral, we have to take V larger than a . Thus, the integral will be the following:

482

Appendix G

x (t )

$

1

^

sa

1

`

1

O  if

2S i ³O e

at

V  if

2S i ³V  if

O  if

2S i ³O

 if

 if

e

( a  z )t

e

zt

e

st

s  ads ; s  a

z dz ; O

z

V a

(G26)

z dz

>@

In this case, the branch point is zero. Therefore, the Re z

O can be arbitrarily small (but is always larger than

zero). Since z 0 is a branch point of the function to integrate, we have to introduce a branch cut to evaluate the integral. Although what we have taken so far is the positive real axis as a branch cut, we have said that this choice is arbitrary and to make the function z into a single value, it is enough that closed curves are not allowed to enclose the origin. We can therefore take a branch cut as the negative real axis. In Figure-G2, we indicate the contour used to integrate the given function. Since the closed contour A o B o C o D o E o F o G o A does not enclose any singularity, its integral is zero. To evaluate the Bromwich path integral, (namely along A o B ), we have to calculate the integral along the arcs C , G along the straight lines D, F and along the circumference on the small circle E (Figure-G2). Im > s @ B

C

D

E

Re > s @

F

G

Bromwich - Path A

Figure-G2: Contour of integration with branch cut in complex plane

The function e e

zt

zt

>@

z goes to zero, for Re z tends towards f (the term

z cannot match the exponential decay of

for t ! 0 as always). Thus, the integral along the arcs C , G is zero and disappears as the radius of these arcs

grows. To evaluate the integral along a small circle E we take z  e

iT

>

in the interval S , S

@ and we take the limit

p 0 . With this we have the following:

³

E

e

zt

z dz

S

³S

teiT

e

e

i  T2

iT

(G27)

i  e dT

The above expression (G27) integrating the function clearly tends to zero for a limit p 0 . Hence, there is no contribution from the integration over the circumference of E . Along the straight lines D, F , we can assume that the arguments of the complex variables lying on them are S (along

D ) and S (along F ) and that their imaginary parts are close to zero. Therefore we have z  iS

iS

 iS

re iS

iS

(for D ) and  iS

(for F ). Consequently, we have dz e dr for D and dz e dr for F . We note e z re e 1 . The parameter r runs between f and 0 for line D and between 0 and f for line F . These integrals are given in the following expressions:

Inverse Laplace Transform by Contour Integration

zt

³

³

ze dz

D

0

re

f 0

³ zt

³

³

ze dz

F

i  S2 treiS iS

f

re

0 f

³

e

e dr

r (i )(e

f

 rt

)( 1)dr

i   S2 tre  iS

e

r (  i )(e

0

483

e

 rt

 iS

i

f

³

re

0

 rt

dr (G28)

dr

)( 1)dr

i

³

f

re

0

 rt

dr

From the above calculations of contour integrations on C , D, E , F , G and with the residue theorem, we get the following (with an observation that the closed contour Figure-G2 A, B, C , D, E , F , G , A does not enclose any poles of X ( s ) , so the residue is zero):

x (t )

$

1

^ X ( s )` $

1

^

sa

`

1

2S i ³

Ao B

 In order to evaluate

³

f

0

re

 rt

2

dr , put rt

W and dr

e

e

at

st

f

³

re

0

S

2W dW t

s  a ds  rt

e

W 2

2W e

W 2

at

(G29)

zt

2S i ³

ze dz

D E

dr

f

re

0

d dW

e

, then write the following:

³ We observe that



 rt

1

dr

t

3

2

³

f

0

We

W 2

2W dW

(G30)

, with this we carry out integration by parts for the above RHS and write the

following:

³

f

re

0

 rt

dr



1

§ ªW e W 2 º f  f e W 2 dW · ¨ ¸ ¼ 0 ³0 ¹ t 2 ©¬ 3

(G31)

S 2t

We have used the known result (i.e. following:

³

f

0

e

W

2

S

dW

2

1

^

sa

`

2

) in the above derivation. Now compactly we write the result as

$ We can write the above derivation $

3

eat 2 S t3

1

^

sa

`



e

at

2 St

(G32)

3

, and thus we have the following useful Laplace transform

identity:

$

1

^

sa  sb

`

bt

at

2 St

3

e e

(G33)

G.6 The application of residue calculus to get the inverse Laplace transform of X ( s )

ln s . ( sb  s a )

by contour integration on the branch cut in a complex plane Let us find x (t )

$

1

^ ln s /  . (s

b

a

`

 s ) ; b ! a ! 0 . The behavior of the function X ( s )

ln s . ( sb  s a )

in limit as

s n f we note by the L’Hospital rule in the following step:

lim s nf X ( s )

lim s nf

K(bs

1 s b 1

 as

a 1

)

0

(G34)

484

We may feel s

Appendix G

1 is a singular point of X ( s ) , but by L’Hospital rule, we see it is not, as follows:

lim s o1 X ( s )

lim s o1

K(bs

1 s b 1

 as

a 1

)

(G35)

1 So we find s

K(b  a ) is a singularity in the X (s) . 0

1 is not a singular point. The point s

Due to the multi-valued nature of the function ln s we call the singular point a branch point and not a pole. The contour shown in Figure-G2 excludes the branch point (zero), and the branch-cut is taken as a standard branch-cut i.e. the negative real axis. The inverse Laplace transform integral is the following: 1 V  if st x (t ) e X ( s )ds 2S i V if (G36) 1 st e X ( s )ds 2S i Bromwich

³

³

>@

The above integration is the Bromwich path integration on the line AB as in Figure-G2. The Re s

V is the line,

to right of which we do not have any singularity, or the singularity is contained in the left side of the line AB (FigureG2). So we select V following:



0 close to the imaginary axis as shown in Figure-G2. By use of Residue Calculus, we get the st

³

The discussion about X ( s )

e X ( s )ds

Ao B o C o D o E o F o G o A ln( s ) indicates that . ( sb  s a )

2S i

¦ Residues of ª¬e

st

X ( s ) º¼

(G37)

there are no singularities in the path enclosed by the contour of

integration A, B, C , D, E , F , G , A . Thus, the residues are zero, and we write the following: st

³

e X ( s )ds

0

Ao B o C o D o E o F o G o A

³

st

Ao B

³ ³ ³

e X ( s )ds



st

C

E

³ X ( s )ds  ³

e X ( s )ds  e

st

D

F

st

e X ( s )ds st

e X ( s )ds

(G38)

st

G

e X ( s )ds

That is we need to evaluate the integral on big arcs C and G ; and on lines just above and the below the negative real axis D and F , and also on a small circle at origin, i.e. E . We take X ( s )

ln s s 2 1

, that is b

2, a

0, K iT

Let the big arcs be represented by C : Re ;

³e

st

S 2

X ( s )ds

C

1 , in our function X ( s )

ln( s ) . ( sb  s a )

and do the calculations.

iT

d T d S and G : Re ; S d T d  S2 , then the integrals are:

³

lim e

R nf

ReiT t

iT

iT

X ( Re )e dR ; s

Re

iT

§ ln( ReiT ) · iT ¨ 2 i 2T ¸ e dR R nf ³ © R e 1¹ R eiT t § ln( R )  iT · iT lim e ¨ 2 i 2T ¸ e dR R nf ³ © R e 1¹ lim e

It may be noted that the function X ( s )

ln s s 2 1

ReiT t

(G39)

vanishes at large values of s ; hence the integral on the large arcs C

and G are zero (we discussed this in the section on Jordan’s Lemma):

³e

C

st

X ( s )ds

³e

G

st

X ( s )ds

0

(G40)

Inverse Laplace Transform by Contour Integration

Now we do an integration on the small circle E with s  e

>S , S @ for T

iT

485

encircling the origin, with p 0 in the interval

that is the integration in a positive sense (anti-clockwise). Therefore, we have the following:

³

st

e X ( s )ds

E

³

lim

S

p 0 S

 lim p 0

S

³S

§ ln  iT · iT ¨ 2 i 2T ¸ i  e dT © e 1¹

eiT t

e

§ ln  iT · iT ¸ i  e dT © 1 ¹

(1) ¨ S

(G41)

T ³  M  iT i  e dT

lim

i

0

p 0, M nf S

2

teiT

iT

In the above steps (G41), we took for  e  1  1; e

 1 and ln   M with M n f as we have for the

limit, i.e. p 0 . The expression (G41) when multiplied by  gives zero, for p 0 ; thus we have an integration on the small circle which is zero. We note that lim s p0 X ( s ) s

0 with limp0

st

³

s eiT

e X ( s )ds

ze

 iS

, ds

e

f , but the integration on a small circle encircling

iS

e dz with z varying from f to zero; and

0.

Thus we are left with integration on D where we take s on line F with s

ln s s 2 1

 iS

iS

ze ; ds

dz and with z varying from zero to f We note that we are on a negative real

axis in both cases where z is positive and e

iS

e

 iS

1 . Thus the integration on line D is:

§ ln  zeiS · iS ³ e X ( s)ds ³f e ¨¨ z 2ei (2S )  1 ¸¸e dz D © ¹ 0  zt § ln z  iS · ³f e ¨© z 2  1 ¸¹ (dz ) 0

st

zeiS t

(G42) f

³

0

e

 zt

Similarly, we write the integration on F as follows:

³e

st

X ( s )ds

F

³

f

0

e

ze iS t

³

f

0

§ ln z  iS · ¨ 2 ¸ dz © z 1 ¹

§ ln  ze  iS ·  iS ¨ 2 i ( 2S ) ¸e dz ¨z e 1¸ © ¹ e

(G43)

 zt § ln z  iS ·

¨ 2 ¸ (  dz ) © z 1 ¹

Adding the two (G42) and (G43), we get the following expression:

³

D F

st

e X ( s )ds

³

f

0

e

 zt

³

f

0

§ ln z  iS ¨ 2 © z 1

e

 zt

f ·  zt § ln z  iS ¸ dz  ³0 e ¨ 2 ¹ © z 1

· ¸ dz ¹

(G44)

2iS

§ · ¨ 2 ¸ dz z  1 © ¹

The inverse Laplace transform integral on the Bromwich path is thus:

x (t )

$

1

^ X ( s )`

1 2S i 

st

³

e X ( s )ds

Bromwich

1 2S i

st

³

e X ( s )ds

D F

§ 2S i · ¨ 2 ¸ dz 0 © z 1¹ f 1 ·  zt § ³ e ¨ 2 ¸ dz 0 © z 1¹ 

1

2S i ³

³

f

0

f

e

e

 zt

 zt

§ 1 · dz ¨ 2 ¸ ©1 z ¹

(G45)

486

Appendix G

Therefore, we have obtained the following: x (t ) ln s sb  s a

The observation is for X ( s )

1

$

^ ` ³ ln s s 2 1

f

0

e

 zt

§ 1 · dz ¨ 2 ¸ ©1 z ¹

(G46)

; b ! a t 0 ; that if a z 0 then the integration on the small circle E blows up,

and the method fails. However, we can resort to Berberan-Santos method that we discussed in Chapter-6.

G.7 The application of residue calculus to get the inverse Laplace transform of b a X ( s ) s lnss by contour integration on branch cut in the complex plane sb  s a ln s

Here we need to do an inverse Laplace transform of X ( s ) b

a

s s s ln s

Thus, if we modify X ( s ) as X ( s )

, b ! a . We note here that lim s nf X ( s )

, then we get lim s nf X ( s )

f .

0 , and we can comfortably apply the formulas

for inverse Laplace transforms. Let us take b

0 , then via the contour integration method as we did in earlier Section-G.6, we see-an

2 and a

s 2 1 s ln s

integral on the big arcs C and G are zero, as X ( s ) circle s  e

iT

>

iT

goes to zero for large R with s

Re . For the small

@

in the interval S , S for T (i.e. E ) (Figure-G2) enclosing the origin we have the following:

³

E

st

e X ( s )ds

limp0

S

³S

limp0 S

³S

S

³S

³

D F

st

³

0

f

e

 zt

st

2

s 1 s ln s

S

³S

ds s eiT 2

 e

eiT

e

e

(1) lim M nf lim M nf

iT

i 2T

1

iT

 ln  iT

( 1)

 ie dT (G47)

idT

  M  iT 1

1 § T 2 2· M ¨ M T ¸ tan © ¹



iS

Thus we are left with integrals on the lines D with s ds  dz . So we have the following:

e X ( s )ds

e



idT

ze , f ! z ! 0 and F with s

0 ze

 iS

, 0  z  f and

2

z 1 (  z )  ln z  iS 

³

f

0

e

 zt

  dz 2

z 1 (  z )  ln z  iS

  dz (G48)

§ z2 1 · § 1 1 · e ³0 ¨© ( z ) ¸¹ ¨© ln z + iS  ln z  iS ¸¹ dz 2 f 2iS ·  zt § z  1 · § e ³0 ¨© z ¸¹ ¨© (ln z )2  S 2 ¸¹ dz f

 zt

As we have seen, the inverse Laplace transform is an integration on the Bromwich path we have from the residue calculus, and contour integration the following, noting that there are no poles enclosed in the contour of FigureG2; 1 1 st x (t ) $ ^ X ( s )` e X ( s )ds 2S i Ao B

³



1 2S i

³

f

0

e

³

st

D F

 zt

e X ( s )ds

2 § · 1 z ¨ ¸dz ¨ z  (ln z ) 2  S 2 ¸ © ¹

(G49)

Inverse Laplace Transform by Contour Integration

487

We started with X ( s ) conditioned it to X ( s ) by dividing by s , and then obtained the inverse Laplace transform. To get x (t )

$

1

^ X ( s )`

obtained x (t ) , i.e. x (t )

in this example of X ( s ) d dt

sb  s a ln s

, b!a

we have to perform a derivative of the

x (t ) . This, we write as follows:

d § f (1 z 2 ) e zt · (G50) ¨ 0 z  (ln z )2 S 2 dz ¸ dt © ¹ We come across these types of ‘characteristic functions’ (transfer functions) discussed in Sections-G.6 and G.7 for a “continuous order distributed differential equation” discussed in Chapter-9, where we used the BerberanSantos method to get an inverse Laplace transform to work out the time responses of these systems. x (t )

$

1

^ ` s 2 1 ln s

³

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A New Fractional Fourier Transform Based design of a Band Pass FIR filter for Power Feed Back in Nuclear Reactor under noisy environment, ICETECT IEEE Conference Kanyakumari-India (2011). 188. Saptarshi Das, Indranil Pan, Shantanu Das, Effect of random parameter switching on commensurate fractional order chaotic systems, Chaos, Soliton and Fractals 91 pp. 157-173, (2016). 189. Saptarshi Das, Indranil Pan, Shantanu Das, Performance comparison of optimal fractional order hybrid fuzzy PID controllers for handling oscillatory fractional order processes with dead time, ISA Transactions 52 pp. 550-556, (2013). 190. Sardar T, Saha Ray S, Bera R K, Biswas B.B. The solution of the multi-term fractionally damped Van-derPol Equation, Physica Scripta IOP Publishing Phys. Scr. 80(2009)025003 (66pp) pp. 1-5, (2009). 191. Sardar Tridip, Saha Ray S, Bera R K, Biswas B.B, Shantanu Das. The Solution of Coupled Fractional Neutron Diffusion Equations with Delayed Neutron, Int. J. of Nuclear Energy Science & Technology,Vol.5, No.2, pp. 105-113, Inderscience Publishers (2010) 192. Saxena. R.K. Fractional calculus and fractional differential equations Lecture Note (chapter-3) pp. 1-39, Jai Narayan Vyas University Rajasthan (2006). 193. Saxena.R.K, Mathai.A.M, Haubold H J. On Generalized Fractional Kinetic Equations. Int J of Math. Phys. 0406046 v1 22-06-2004.(2004). 194. Scafetta. N, Griffin. L, West. B. J. Holder Exponent Spectrum for Human Gait, Physica. A 328 pp. 561-583, Elsevier (2003). 195. Schneider W.R , Wyss W. Fractional Diffusion and Wave Equations, Math. Phys. 30, pp. 134-144. (1989). 196. Shantanu Das, B B Biswas. Fractional divergence for Neutron Flux profile in Nuclear Reactor, Inderscience Publishers Europe Int. J. of Nuclear Energy Science & Technology, Vol. 3, No. 2, pp. 139-159, (2007) 197. Shantanu Das, Aware MV, Jhungare AS and Khubalkar SW. Energy/Fuel Efficient and Enhanced Robust Systems Demonstrated with Developed Fractional Order PID Controller, doi: 10.4172/2576-1463.1000182; Innovative Energy & Research Volume-7, Issue-1 1000182 (2018). 198. Shantanu Das. Convergence of Riemann-Liouville and Caputo Derivative for Practical Solution of Fractional Order Differential Equations, International Journal of Applied Mathematics & Statistics, Vol. 23, Issue No. D11, pp. 64-73, (2011). 199. Shantanu Das Fractional Stochastic Modeling for Random Dynamic Delays in Computer Control System, International Journal of Applied Mathematics & Statistics, Vol. 21, No. J11, pp. 131-140 (2011). 200. Shantanu Das. Generalized Dynamic Systems Solution by Decomposed Physical Reactions, International Journal of Applied Mathematics & Statistics, J10 (special- issue) CESER publications, pp. 44-75, (2010). 201. Shantanu Das, Shiben Bhattacharya, R.T.Keswani, S. Sunder Rajan. Circuit Theory Approach with Fractional Calculus to describe half space Geophysical Analysis for Transient Electromagnetic Method, Geophysical Journal No. 2. T-31-2009, pp. 147-159 (2009). 202. Shantanu Das. Solution of Extraordinary Differential Equation with Physical Reasoning by Obtaining Modal Reaction Series, Modeling and Simulation in Engineering, Hindawi Publishing Corp. ID-739675, pp. 119 (2010). 203. Shantanu Das. Functional Fractional Calculus for System Identification and Controls, Springer-Verlag, Germany, (2007). 204. Shantanu Das. Functional Fractional Calculus II-Edition, Springer-Verlag, Germany, (2011).

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205. Shantanu Das, N C Pramanik. Micro-structural Roughness of Electrodes Manifesting as Temporal Fractional Order Differential Equation in Super-Capacitor Transfer Characteristics, International Journal of Mathematics & Computation Vol. 20, Issue No.3, pp. 94-113, (2013). 206. Shantanu Das. Gramian for Control of Fractional Order Multivariate Dynamic System, International Journal of Mathematics & Computations, Vol. 21, Issue 4, pp. 23-50, (2013) 207. Shantanu Das. Generalization of Fractional Calculus Operators with Applications: Developments, International Journal of Applied Mathematics & Statistics IJAMAS, Vol. 37. Issue No. 7,; pp. 71-96, (2013). 208. Shantanu Das. Geometrically Deriving Fractional Cross Product and Fractional Curl, International Journal of Mathematics and Computation IJMC, Vol. 20, Issue-3, pp. 1-29, (2013). 209. Shantanu Das. Mechanism of Wave Dissipation via Memory Integral vis-a-vis Fractional Derivative. International Journal of Mathematics & Computation IJMC, Vol. 19, Issue No.2, pp. 72-83, (2013). 210. Shantanu Das. Fractional order boundary controller enhancing stability of partial differential wave equation systems with delayed feedback, International Journal of Mathematics & Computation IJMC Vol. 19, Issue No. 2, pp. 42-59, (2013). 211. Shantanu Das. Formation of Fractional Derivative in Time due to Propagation of Free Greens Function in Spatial Stochastic Disorder Field for Transport Phenomena, International Journal of Mathematics and Computation Vol. 17, Issue-4, pp. 68-92, (2012). 212. Shantanu Das. Evolution of Temporal Fractional Derivative due to Spatial Stochastic Disorder in Transport Phenomena, International Journal of Mathematics and Computation (IJMC), Vol. 17, Issue-4, pp. 120, (2012). 213. Shantanu Das. Multiple Riemann Sheet Solution for Dynamic Systems with Fractional Differential Equations, International Journal of Applied Mathematics & Statistics. Vol. 28, Issue-4, pp. 83-89, (2012). 214. Shantanu Das. A new look at formulation of charge storage in capacitors and application to classical capacitor and fractional capacitor theory, doi: 10.9734/AJR2P/2018/43738; Asian Journal of Research and Review in Physics; 1(3): pp. 1-18, (2018) 215. Shantanu Das. Frequency and Time Domain Solution for Dynamic Systems having Differential Equations of Continuous Order, International Journal of Applied Mathematics & Statistics, Vol. 29, Issue No. 5, pp. 616, (2012). 216. Shantanu Das. Revisiting the Curie-von Schweidler law for dielectric relaxation and derivation of distribution function for relaxation rates as Zipf’s power law and manifestation of fractional differential equation for capacitor, Journal of Modern Physics Vol.8, No. 13, pp. 1988-2012; (2017). 217. Shawagfeh N T. The Decomposition Method for Fractional Differential Equations, J. Frac. Calc 16 pp. 2733. (1999) 218. Shequin Zhang. Positive solutions for boundary value problems on non-linear fractional differential equations. Electronic J. of differential equations No.36 pp. 1-12, (2006) 219. Shih-Tong, Yu-Tan Huang, Chun Chen, Srivastava H M. A certain family of fractional differential equations, Taiwanese Journal of Mathematics -, Vol. 4, No .3, pp. 417-426. (2000). 220. Sierociuk Dominik. Fractional Kalman Filter algorithm for states, parameters and order of fractional system estimation, XV Krajowa Konferencja Automatyki, Warsaw, Poland pp. 1-14, (2005). 221. Simantini Majumdar, Somasri Hazra, Moutushi Dutta Choudhury, Suparna Dutta Sinha, Shantanu Das, Tapas Ranjan Middya, Sujata Tarafdar, Tapati Dutta. A study of the rheological properties of visco-elastic materials using fractional calculus, Colloids and Surfaces A: Physicochemical and Engineering Aspects, 516 pp. 181189; 2016, (2017). 222. Somasri Hazra, Sudeshna Sircar, Tajkera Khatun, Moutushi Dutta Choudhury, Abhra Giri, Sanat Karmakar, Tapati Dutta, Shantanu Das, Sujata Tarafdar. Unstable crack propagation in LAPONITER gel: selection of sinusoidal mode in an electric field, RSC-Adv., 6, 64297-Royal Society of Chemistry (2016). 223. Somasri Hazra, Tapati Dutta, Shantanu Das, Sujata Tarafdar. Memory of Electric Field in Laponite & How it Affects Crack Formation: Modeling through Generalized Calculus, American Chemical Society, Langmuir doi: 10.1021/acs.langmuir.7b02034, (2017). 224. Srivastava H M, Owa S. Some characterization and distortion theorems involving fractional calculus, generalized hyper geometric functions, Hadamard products, Linear Operators and certain subclasses of Analytical functions, Nagoya Math J, Vol. 106, pp. 1-128, (1987). 225. Srivastava H M. On extension of Mittag-Leffler function, Yokohama Math. J., Vol. 16, No. 2 pp. 77-88 (1968) 226. Subrata Chandra, Soma Nag, Sujata Tarafdar, Shantanu Das. Analysis of the Spreading of non-Newtonian fluid using Fractional Calculus, National Symposium on Mathematical Modeling of Natural Phenomena NMMNP-2010, , Bose Institute Calcutta-India (2010) 227. Sugi.M, Hirano Y, Miura YF, Saito K, Z -Variations of AC admittance in the inhomogeneous distributed RC lines. Japan. J. Appl. Phys. Vol. 39, no 9A, , pp. 5367-5368 (2000) 228. Sugi.M, Hirano Y, Miura Y.F, Saito.K. Simulation of fractal immitance by analog circuits: an approach to optimized circuits, IEICE Trans. Fundamentals, Vol. E82-A, no. 8, , pp. 1627-1635 (1999).

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229. Suman Saha, Saptarishi Das, R. Ghosh, B. Goswami, Amitava Gupta, R.Balasubramanium, A. K. Chandra, Shantanu Das. Design of Fractional Order Phase Shaper for Iso-Damped Control of PHWR Under SetBack, IEEE Transactions on Nuclear Science, Vol. 57, No. 3, pp. 1602-1612,(2010) 230. Suman Saha, Saptarishi Das, R. Ghosh, B. Goswami, Amitava Gupta, R.Balasubramanium, A.K.Chandra, Shantanu Das. Fractional Order Phase Shaper Design with Bode’s Integral for Iso-Damped Control System, ISA-Transactions 49Elsevier pp. 196-206, (2010) 231. Suman Saha, Saptarishi Das, R. Ghosh, B. Goswami, Amitava Gupta, R.Balasubramanium, A.K.Chandra, Shantanu Das. Fractional Order Phase-Shaper Design with Routh’s Criterion for Iso-Damped Control System, IEEE-India Council Gujarat-Section (Paper ID 740) INDICON, (2009). 232. Sun H H, Abelwahab, A. A, Onaral.B. Linear Approximation of Transfer functions with a pole of fractional power. IEEE Trans on Automatic Control vol. AC-29, No.5, pp. 441-444, (1984). 233. 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…I shall always remain a student of this beautiful subject… There is a lot more to learn & We must develop it further and further in order to know which derivative nature functions best …Thus, it is a small beginning