New Numerical Scheme with Newton Polynomial: Theory, Methods, and Applications [1 ed.] 0323854486, 9780323854481

Table of contents :
Front Cover
New Numerical Scheme With Newton Polynomial
Copyright
Contents
Preface
Acknowledgments
List of symbols
1 Polynomial interpolation
1.1 Some interpolation polynomials
1.1.1 Bernstein polynomial
1.1.2 The Newton polynomial interpolation
1.1.3 Hermite interpolation
1.1.4 Cubic polynomial
1.1.5 B-spline polynomial
1.1.6 Legendre polynomial
1.1.7 Chebyshev polynomial
1.1.8 Lagrange–Sylvester interpolation
2 Two-steps Lagrange polynomial interpolation: numerical scheme
2.1 Classical differential equation
2.1.1 Numerical illustrations
2.2 Fractal differential equation
2.2.1 Numerical illustrations
2.3 Differential equation with the Caputo–Fabrizio operator
2.3.1 Error analysis with exponential kernel
2.3.2 Numerical illustrations
2.4 Differential equation with the Caputo fractional operator
2.4.1 Error analysis with power-law kernel
2.4.2 Numerical illustrations
2.5 Differential equation with the Atangana–Baleanu operator
2.5.1 Error analysis with the Mittag-Leffler kernel
2.5.2 Numerical illustrations
2.6 Differential equation with fractal–fractional with power-law kernel
2.6.1 Error analysis with the Caputo fractal–fractional derivative
2.6.2 Numerical illustrations
2.7 Differential equation with fractal–fractional derivative with exponential decay kernel
2.7.1 Error analysis with the Caputo–Fabrizio fractal–fractional derivative
2.7.2 Numerical illustrations
2.8 Differential equation with fractal–fractional derivative with the Mittag-Leffler kernel
2.8.1 Error analysis with the Atangana–Baleanu fractal–fractional derivative
2.8.2 Numerical illustrations
2.9 Differential equation with fractal–fractional with variable order with exponential decay kernel
2.9.1 Error analysis with fractal–fractional derivative with variable order with exponential decay kernel
2.9.2 Numerical illustrations
2.10 Differential equation with fractal–fractional derivative with variable order with the Mittag-Leffler kernel
2.10.1 Error analysis with fractal–fractional derivative with variable order with Mittag-Leffler kernel
2.10.2 Numerical illustrations
2.11 Differential equation with fractal–fractional derivative with variable order with power-law kernel
2.11.1 Error analysis with fractal–fractional derivative with variable order with power-law kernel
2.11.2 Numerical illustrations
3 Newton interpolation: introduction of the scheme for classical calculus
3.1 Error analysis with classical derivative
3.2 Numerical illustrations
4 Numerical method for fractal differential equations
4.1 Error analysis with fractal derivative
4.2 Numerical illustrations
5 Numerical method for a fractional differential equation with Caputo–Fabrizio derivative
5.1 Error analysis with Caputo–Fabrizio fractional derivative
5.2 Numerical illustrations
6 Numerical method for a fractional differential equation with power-law kernel
6.1 Error analysis with Caputo fractional derivative
6.2 Numerical illustrations
7 Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel
7.1 Error analysis with the Atangana–Baleanu fractional derivative
7.2 Numerical illustrations
8 Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel
8.1 Predictor–corrector method for fractal–fractional derivative with the exponential decay kernel
8.2 Error analysis with the Caputo–Fabrizio fractal–fractional derivative
8.3 Numerical illustrations
9 Numerical method for a fractal–fractional ordinary differential equation with power law kernel
9.1 Predictor–corrector method for fractal–fractional derivative with power law kernel
9.2 Error analysis with Caputo fractal–fractional derivative
9.3 Numerical illustrations
10 Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel
10.1 Predictor–corrector method for fractal–fractional derivative with the generalized Mittag-Leffler kernel
10.2 Error analysis with the Atangana–Baleanu fractal–fractional derivative
10.3 Numerical illustrations
11 Numerical method for a fractal–fractional ordinary differential equation with variable order with exponential decay kernel
11.1 Numerical illustrations
12 Numerical method for a fractal–fractional ordinary differential equation with variable order with power-law kernel
12.1 Numerical illustrations
13 Numerical method for a fractal–fractional ordinary differential equation with variable order with the generalized Mittag-Leffler kernel
13.1 Numerical illustrations
14 Numerical scheme for partial differential equations with integer and non-integer order
14.1 Numerical scheme with classical derivative
14.1.1 Numerical illustrations
14.2 Numerical scheme with fractal derivative
14.2.1 Numerical illustrations
14.3 Numerical scheme with the Atangana–Baleanu fractional operator
14.3.1 Numerical illustrations
14.4 Numerical scheme with the Caputo fractional operator
14.4.1 Numerical illustrations
14.5 Numerical scheme with the Caputo–Fabrizio fractional operator
14.5.1 Numerical illustration
14.6 Numerical scheme with the Atangana–Baleanu fractal–fractional operator
14.7 Numerical scheme with the Caputo fractal–fractional operator
14.8 Numerical scheme for Caputo–Fabrizio fractal–fractional operator
14.9 New scheme with fractal–fractional with variable order with exponential decay kernel
14.10 New scheme with fractal–fractional with variable order with the Mittag-Leffler kernel
14.11 New scheme with fractal–fractional with variable order with power-law kernel
15 Application to linear ordinary differential equations
15.1 Linear ordinary differential equations with integer and non-integer orders
15.1.1 A non-homogeneous linear differential equation
15.1.2 Non-homogeneous linear differential equation with the Atangana–Baleanu derivative
15.1.3 Non-homogeneous linear differential equation with the Caputo derivative
15.1.4 Non-homogeneous linear differential equation with fractal–fractional with the exponential law
15.1.5 Non-homogeneous linear differential equation with fractal–fractional derivative with the Mittag-Leffler kernel
16 Application to non-linear ordinary differential equations
16.1 Non-linear ordinary differential equations with integer and non-integer orders
16.1.1 Non-homogeneous nonlinear differential equation with classical derivative
16.1.2 Non-homogeneous non-linear differential equation with Caputo–Fabrizio derivative
16.1.3 Non-homogeneous non-linear differential equation with fractal derivative
16.1.4 Non-homogeneous non-linear differential equation with the fractal–fractional derivative with exponential law
16.1.5 Non-homogeneous non-linear differential equation with fractal–fractional with variable order with power law
17 Application to linear partial differential equations
17.1 Linear partial differential equations with integer and non-integer orders
17.1.1 Linear partial differential equation with the classical derivative
17.1.2 Linear partial differential equation with the fractal derivative
17.1.3 Linear partial differential equation with the Caputo fractional derivative
17.1.4 Linear partial differential equation with the fractal–fractional derivative with the exponential law
17.1.5 Linear partial differential equation with the fractal–fractional with variable order with the Mittag-Leffler kernel
18 Application to non-linear partial differential equations
18.1 Non-linear partial differential equations with integer and non-integer orders
18.1.1 Non-linear partial differential equation with the classical derivative
18.1.2 Non-linear partial differential equation with the fractal derivative
18.1.3 Non-linear partial differential equation with the Caputo fractional derivative
18.1.4 Non-linear partial differential equation with the fractal–fractional with exponential law
18.1.5 Non-linear partial differential equation with the fractal–fractional with variable order with the Mittag-Leffler kernel
19 Application to a system of ordinary differential equations
19.1 System of ordinary differential equations with integer and non-integer orders
19.1.1 A hybrid attractor with the classical derivative
19.1.2 Shaw oscillator with the Caputo fractional derivative
19.1.3 Dengue model with the Atangana–Baleanu fractional derivative
19.1.4 HIV model with fractal–fractional derivative with power law
19.1.5 Ebola model with fractal–fractional derivative with variable order with the exponential law
20 Application to system of non-linear partial differential equations
20.1 System of non-linear partial differential equations
20.1.1 System of non-linear partial differential equations with the classical derivative
20.1.2 System of non-linear partial differential equations with the Atangana–Baleanu derivative
20.1.3 System of non-linear partial differential equations with the Caputo fractional derivative
20.1.4 System of non-linear partial differential equations with the fractal–fractional with the Mittag-Leffler kernel
20.1.5 System of non-linear partial differential equations with fractal–fractional with the power law
A Appendix
AS_Method_for_Chaotic_with_AB_Fractal-Fractional.m
AS_Method_for_Chaotic_with_AB_Fractal-Fractional_with_Variable_Order.m
AS_Method_for_Chaotic_with_AB_Fractional.m
AS_Method_for_Chaotic_with_Caputo_Fractal-Fractional_with_Variable_Order.m
AS_Method_for_Chaotic_with_Caputo_Fractional.m
AS_Method_for_Chaotic_with_CF_Fractal-Fractional_with_Variable_Order.m
AS_Method_for_Chaotic_with_CF_Fractional.m
AS_Method_for_Differential_Equation_with_AB_Fractal-Fractional.m
AS_Method_for_Differential_Equation_with_AB_Fractal-Fractional_with_Variable_Order.m
AS_Method_for_Differential_Equation_with_AB_Fractional.m
AS_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional.m
AS_Method_for_Differential_Equation_with_Caputo_Fractional.m
AS_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional_with_Variable_Order.m
AS_Method_for_Differential_Equation_with_CF_Fractal-Fractional.m
AS_Method_for_Differential_Equation_with_CF_Fractional.m
AS_Method_for_Differential_Equation_with_CF_Fractal-Fractional_with_Variable_Order.m
AS_Method_for_Differential_Equation_with_Classical.m
AS_Method_for_Differential_Equation_with_Fractal.m
AT_Method_for_Chaotic_with_AB_Fractal-Fractional.m
AT_Method_for_Chaotic_with_AB_Fractal-Fractional_with_Variable_Order.m
AT_Method_for_Chaotic_with_AB_Fractional.m
AT_Method_for_Chaotic_with_Caputo_Fractal-Fractional_with_Variable_Order.m
AT_Method_for_Chaotic_with_Caputo_Fractal-Fractional.m
AT_Method_for_Chaotic_with_Caputo_Fractional.m
AT_Method_for_Chaotic_with_CF_Fractal-Fractional.m
AT_Method_for_Chaotic_with_CF_Fractal-Fractional_with_Variable_Order.m
AT_Method_for_Chaotic_with_CF_Fractional.m
AT_Method_for_Differential_Equation_with_AB_Fractal-Fractional.m
AT_Method_for_Differential_Equation_with_AB_Fractal-Fractional_with_Variable_Order.m
AT_Method_for_Differential_Equation_with_AB_Fractional.m
AT_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional_with_Variable_Order.m
AT_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional.m
AT_Method_for_Differential_Equation_with_Caputo_Fractional.m
AT_Method_for_Differential_Equation_with_CF_Fractal-Fractional.m
AT_Method_for_Differential_Equation_with_CF_Fractal-Fractional_with_Variable_Order.m
AT_Method_for_Differential_Equation_with_CF_Fractional.m
AT_Method_for_Differential_Equation_with_Classical.m
AT_Method_for_Differential_Equation_with_Fractal.m
A.1 Supplementary material
References
Index
Back Cover

Citation preview

New Numerical Scheme With Newton Polynomial Theory, Methods, and Applications

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New Numerical Scheme With Newton Polynomial Theory, Methods, and Applications

Abdon Atangana University of the Free State Bloemfontein, South Africa

˘ Seda I˙gret Araz Department of Mathematics Siirt University Siirt, Turkey

Academic Press is an imprint of Elsevier 125 London Wall, London EC2Y 5AS, United Kingdom 525 B Street, Suite 1650, San Diego, CA 92101, United States 50 Hampshire Street, 5th Floor, Cambridge, MA 02139, United States The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, United Kingdom Copyright © 2021 Elsevier Inc. All rights reserved. MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library ISBN: 978-0-323-85448-1 For information on all Academic Press publications visit our website at https://www.elsevier.com/books-and-journals Publisher: Mara Conner Editorial Project Manager: Aleksandra Packowska Production Project Manager: Bharatwaj Varatharajan Designer: Matthew Limbert Typeset by VTeX

Contents

Preface Acknowledgments List of symbols 1

Polynomial interpolation 1.1 Some interpolation polynomials 1.1.1 Bernstein polynomial 1.1.2 The Newton polynomial interpolation 1.1.3 Hermite interpolation 1.1.4 Cubic polynomial 1.1.5 B-spline polynomial 1.1.6 Legendre polynomial 1.1.7 Chebyshev polynomial 1.1.8 Lagrange–Sylvester interpolation

2

Two-steps Lagrange polynomial interpolation: numerical scheme 2.1 Classical differential equation 2.1.1 Numerical illustrations 2.2 Fractal differential equation 2.2.1 Numerical illustrations 2.3 Differential equation with the Caputo–Fabrizio operator 2.3.1 Error analysis with exponential kernel 2.3.2 Numerical illustrations 2.4 Differential equation with the Caputo fractional operator 2.4.1 Error analysis with power-law kernel 2.4.2 Numerical illustrations 2.5 Differential equation with the Atangana–Baleanu operator 2.5.1 Error analysis with the Mittag-Leffler kernel 2.5.2 Numerical illustrations 2.6 Differential equation with fractal–fractional with power-law kernel 2.6.1 Error analysis with the Caputo fractal–fractional derivative 2.6.2 Numerical illustrations 2.7 Differential equation with fractal–fractional derivative with exponential decay kernel 2.7.1 Error analysis with the Caputo–Fabrizio fractal–fractional derivative

xi xiii xv 1 2 4 4 6 7 7 8 8 9 11 11 12 16 18 22 24 25 29 32 33 38 41 43 48 51 52 57 60

vi

Contents

2.7.2 Numerical illustrations Differential equation with fractal–fractional derivative with the Mittag-Leffler kernel 2.8.1 Error analysis with the Atangana–Baleanu fractal–fractional derivative 2.8.2 Numerical illustrations 2.9 Differential equation with fractal–fractional with variable order with exponential decay kernel 2.9.1 Error analysis with fractal–fractional derivative with variable order with exponential decay kernel 2.9.2 Numerical illustrations 2.10 Differential equation with fractal–fractional derivative with variable order with the Mittag-Leffler kernel 2.10.1 Error analysis with fractal–fractional derivative with variable order with Mittag-Leffler kernel 2.10.2 Numerical illustrations 2.11 Differential equation with fractal–fractional derivative with variable order with power-law kernel 2.11.1 Error analysis with fractal–fractional derivative with variable order with power-law kernel 2.11.2 Numerical illustrations

62

104 106

Newton interpolation: introduction of the scheme for classical calculus 3.1 Error analysis with classical derivative 3.2 Numerical illustrations

113 115 116

4

Numerical method for fractal differential equations 4.1 Error analysis with fractal derivative 4.2 Numerical illustrations

121 123 125

5

Numerical method for a fractional differential equation with Caputo–Fabrizio derivative 5.1 Error analysis with Caputo–Fabrizio fractional derivative 5.2 Numerical illustrations

131 133 135

Numerical method for a fractional differential equation with power-law kernel 6.1 Error analysis with Caputo fractional derivative 6.2 Numerical illustrations

141 143 146

Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel 7.1 Error analysis with the Atangana–Baleanu fractional derivative 7.2 Numerical illustrations

155 157 159

2.8

3

6

7

66 70 73 79 82 84 89 93 95 101

Contents

8

9

10

11

12

13

14

Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel 8.1 Predictor–corrector method for fractal–fractional derivative with the exponential decay kernel 8.2 Error analysis with the Caputo–Fabrizio fractal–fractional derivative 8.3 Numerical illustrations Numerical method for a fractal–fractional ordinary differential equation with power law kernel 9.1 Predictor–corrector method for fractal–fractional derivative with power law kernel 9.2 Error analysis with Caputo fractal–fractional derivative 9.3 Numerical illustrations

vii

169 171 173 175

181 184 187 190

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel 10.1 Predictor–corrector method for fractal–fractional derivative with the generalized Mittag-Leffler kernel 10.2 Error analysis with the Atangana–Baleanu fractal–fractional derivative 10.3 Numerical illustrations

204 206

Numerical method for a fractal–fractional ordinary differential equation with variable order with exponential decay kernel 11.1 Numerical illustrations

215 218

Numerical method for a fractal–fractional ordinary differential equation with variable order with power-law kernel 12.1 Numerical illustrations

225 228

Numerical method for a fractal–fractional ordinary differential equation with variable order with the generalized Mittag-Leffler kernel 13.1 Numerical illustrations

239 243

Numerical scheme for partial differential equations with integer and non-integer order 14.1 Numerical scheme with classical derivative 14.1.1 Numerical illustrations 14.2 Numerical scheme with fractal derivative 14.2.1 Numerical illustrations 14.3 Numerical scheme with the Atangana–Baleanu fractional operator 14.3.1 Numerical illustrations 14.4 Numerical scheme with the Caputo fractional operator

253 253 255 256 258 259 263 264

199 202

viii

Contents

14.4.1 Numerical illustrations 14.5 Numerical scheme with the Caputo–Fabrizio fractional operator 14.5.1 Numerical illustration 14.6 Numerical scheme with the Atangana–Baleanu fractal–fractional operator 14.7 Numerical scheme with the Caputo fractal–fractional operator 14.8 Numerical scheme for Caputo–Fabrizio fractal–fractional operator 14.9 New scheme with fractal–fractional with variable order with exponential decay kernel 14.10 New scheme with fractal–fractional with variable order with the Mittag-Leffler kernel 14.11 New scheme with fractal–fractional with variable order with power-law kernel 15

16

17

Application to linear ordinary differential equations 15.1 Linear ordinary differential equations with integer and non-integer orders 15.1.1 A non-homogeneous linear differential equation 15.1.2 Non-homogeneous linear differential equation with the Atangana–Baleanu derivative 15.1.3 Non-homogeneous linear differential equation with the Caputo derivative 15.1.4 Non-homogeneous linear differential equation with fractal–fractional with the exponential law 15.1.5 Non-homogeneous linear differential equation with fractal–fractional derivative with the Mittag-Leffler kernel Application to non-linear ordinary differential equations 16.1 Non-linear ordinary differential equations with integer and non-integer orders 16.1.1 Non-homogeneous nonlinear differential equation with classical derivative 16.1.2 Non-homogeneous non-linear differential equation with Caputo–Fabrizio derivative 16.1.3 Non-homogeneous non-linear differential equation with fractal derivative 16.1.4 Non-homogeneous non-linear differential equation with the fractal–fractional derivative with exponential law 16.1.5 Non-homogeneous non-linear differential equation with fractal–fractional with variable order with power law Application to linear partial differential equations 17.1 Linear partial differential equations with integer and non-integer orders

267 268 270 270 274 277 278 280 284 287 287 287 289 291 293 296 301 301 301 303 305 307 310 315 315

Contents

ix

17.1.1 Linear partial differential equation with the classical derivative 17.1.2 Linear partial differential equation with the fractal derivative 17.1.3 Linear partial differential equation with the Caputo fractional derivative 17.1.4 Linear partial differential equation with the fractal–fractional derivative with the exponential law 17.1.5 Linear partial differential equation with the fractal–fractional with variable order with the Mittag-Leffler kernel 18

19

20

Application to non-linear partial differential equations 18.1 Non-linear partial differential equations with integer and non-integer orders 18.1.1 Non-linear partial differential equation with the classical derivative 18.1.2 Non-linear partial differential equation with the fractal derivative 18.1.3 Non-linear partial differential equation with the Caputo fractional derivative 18.1.4 Non-linear partial differential equation with the fractal–fractional with exponential law 18.1.5 Non-linear partial differential equation with the fractal–fractional with variable order with the Mittag-Leffler kernel Application to a system of ordinary differential equations 19.1 System of ordinary differential equations with integer and non-integer orders 19.1.1 A hybrid attractor with the classical derivative 19.1.2 Shaw oscillator with the Caputo fractional derivative 19.1.3 Dengue model with the Atangana–Baleanu fractional derivative 19.1.4 HIV model with fractal–fractional derivative with power law 19.1.5 Ebola model with fractal–fractional derivative with variable order with the exponential law Application to system of non-linear partial differential equations 20.1 System of non-linear partial differential equations 20.1.1 System of non-linear partial differential equations with the classical derivative 20.1.2 System of non-linear partial differential equations with the Atangana–Baleanu derivative

315 317 319 322

323 327 327 327 329 330 333

334 339 339 339 342 348 358 365 375 375 375 378

x

Contents

20.1.3 System of non-linear partial differential equations with the Caputo fractional derivative 20.1.4 System of non-linear partial differential equations with the fractal–fractional with the Mittag-Leffler kernel 20.1.5 System of non-linear partial differential equations with fractal–fractional with the power law A

Appendix

384 389 395 401

References

435

Index

439

Preface

... and the Lord God granted mankind power to control the world within which they live. In particular he granted humans the knowledge to understand using mathematics the nature within which they live. The assignment can only be well achieved if humans follow three important steps including, a correct observation of physical problems, a deeper analysis of these observations and finally an accurate prediction in space and time of such observations. The first step thus requires an adequate, efficient and accurate methodology of data collection and elimination of possible uncertainties. The second, which is fundamental, is managing the data-free-uncertainties of a complex observed fact in order to gain a better understanding of it. The genesis of this second step can be traced back to before Aristotle (385–322 B.C). Many forefathers of modern science endorsed this step as a format concept, for instance Alhazen, Galieo Galilei, and Descartes; also it has been credited to Newton who used it as a real-world technique of physical discovery. The last step requires first a systematic conversion of real world problems into a mathematical formulation, solving the mathematical models and fitting the obtained solution with the data-free uncertainties. If the solution of the mathematical model is in good agreement with data-free uncertainties, an accurate and reliable prediction could be performed. We should mention that these mathematical models are constructed using the concept of differentiation and integration. Due to the complexity of nature, the conversion from observed facts to mathematical models usually ends up with highly nonlinear ordinary and partial differential equations that could not be solved using known analytical methods; therefore numerical schemes are needed to at least provide an approximate solution of the problem. The literature counts several suggested numerical schemes, some very efficient and user friendly, with of course limitations and disadvantages besides the advantages. Some of these numerical schemes are based on polynomial interpolation, for instance, the associated Adams–Basforth approach is based on the Lagrange method, and we have the spline polynomial, and many others that will not be listed here. Very recently Atangana and Seda suggested an alternative numerical scheme based on the Newton polynomial. The introduction of the method met a warm welcome and was revealed to be an efficient and accurate numerical scheme able to solve fractional ordinary differential equations and fractional systems of ordinary with different types of fractional integral operators. This book is therefore devoted to the detailed discussions underpinning the theory, methods and applications of this alternative numerical scheme.

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Acknowledgments

The fear of the LORD is the beginning of wisdom; all who follow His precepts have a rich understanding. His praise endures forever! We could not have written this book without God King of heaven, who made it possible for us to write this book, through his mighty grace, divine love, and unending supply of his divine protection, love and knowledge. We thank him for his grace and everyday breath. We thank our respective families for their support, and for their patience for allowing us to finish this book.

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List of symbols

Bm : lk (x): M (α): AB (α): α 0 Dt : α 0 Jt : CH D α : t 0 CH J α : t 0 CF D α : t 0 CF J α : t 0 C Dα : 0 t CJ α: 0 t ABR D α : t 0 ABR J α : t 0 ABC D α : t 0 ABC J α : t 0

F F E D α,β : t 0 F F E J α,β : t 0 F F P D α,β : t 0 F F P J α,β : t 0 F F M D α,β : t 0 F F M J α,β : t 0 F F E D α,β(t) : t 0 F F E J α,β(t) : t 0 F F P D α,β(t) : t 0 F F P J α,β(t) : t 0 F F M D α,β(t) : t 0 F F M J α,β(t) : t 0 Eτα (·): W21 (0, l):

The vector space of polynomials of degree at most m Lagrange basis polynomials Normalization function Normalization function Fractional derivative Fractional integral Fractal derivative in Chen’s sense Fractional integral in Chen’s sense Caputo–Fabrizio fractional derivative Caputo–Fabrizio fractional integral Caputo fractional derivative Caputo fractional integral Atangana–Baleanu fractional derivative in Riemann–Liouville’s sense Atangana–Baleanu fractional integral in Riemann-Liouville’s sense Atangana–Baleanu fractional derivative in Caputo’s sense Atangana–Baleanu fractional integral in Caputo’s sense Caputo–Fabrizio fractal–fractional derivative Caputo–Fabrizio fractal–fractional integral Caputo fractal–fractional derivative Caputo fractal–fractional integral Atangana–Baleanu fractal–fractional derivative Atangana–Baleanu fractal–fractional integral Fractal–fractional derivative with variable order with exponential decay kernel Fractal–fractional integral with variable order with exponential decay kernel Fractal–fractional derivative with variable order with power-law kernel Fractal–fractional integral with variable order with power-law kernel Fractal–fractional derivative with variable order with Mittag-Leffler kernel Fractal–fractional integral with variable order with Mittag-Leffler kernel Error function Hilbert space consisting of the elements L2 (0, l) having generalized derivative of first order

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Polynomial interpolation

1

One of the greatest achievements in the framework of numerical analysis is perhaps the introduction of approximating a giving nonlinear function using a polynomial interpolation. In this framework, a polynomial interpretation can be viewed as the interpolation of a given collected data set by the polynomial of lowest possible degree, which is able to fit the points of the data [1–10]. Such polynomials can be used to approximate complex curves that could be difficult to be evaluated, a clear example has been supplied in the literature, the shapes of the letters in typography given a few points. We have to point out the wider applicability of this technique [1,2,10]. One of the relevant applications is the evaluation of the natural logarithm and trigonometric functions, however, many other important applications can be found in many already published, books, papers and even in Wikipedia. The process of implementing this algorithm follows these steps: Select a few given data points, create a lookup table, then interpolate between those data points [1,2,10]. It is important to recognize the efficiency and the accuracy of these results, as they are significantly faster computations. The polynomial interpolation have been documented to also constitute the basis for algorithms in numerical quadrature and numerical ordinary differential equations, partial differential equations and secure multi party computation and secret sharing schemes. The polynomial interpolation is essential to engage sub-quadratic multiplication and squares like Toom–Cook multiplication and Karatsuba multiplication, where an interpolation via points on a polynomial like expression defines the product yields the product itself [1–10]. Definition 1.1. If we assume a set of given data m + 1 points, where no two xi are the same, one is looking for a polynomial P which is suitable with degree c at most n with the property P (xi ) = γi,

i = 0, ..., m.

(1.1)

The well-established unisolvence theorem stipulates that such a polynomial P exists and more importantly is unique. The proof can be achieved thanks to the Vandermonde matrix. A detailed proof can be found in many text books and also some already published papers, however, to make it easy for readers, we present the proof here [3,4]. The m + 1 interpolation nodes (xi ) polynomial interpolation defines a linear bijection m : K m+1 −→ Bm

(1.2)

where Bm is the vector space of polynomials of degree at most n. Let P be the interpolation polynomial that is of the form P (x) = bm x m + bm−1 x m−1 + ... + b2 x 2 + b1 x + b0 . New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00007-X Copyright © 2021 Elsevier Inc. All rights reserved.

(1.3)

2

New Numerical Scheme With Newton Polynomial

The argument that P interpolates the data point implies P (xi ) = γi ,

for all i ∈ {0, ..., m} .

(1.4)

If m includes Eq. (1.4) in the above polynomial, the substitution yields a system of linear equations with coefficients bl and the following system can be written [3,4]: ⎡ ⎤⎡ ⎤ ⎡ ⎤ x0m x0m−1 x0m−2 · · · x0 1 γ0 b0 ⎢ xm ⎥ ⎢ ⎥ b1 ⎥ x1m−1 x1m−2 · · · x1 1 ⎥⎢ ⎢ 1 ⎥ ⎢ γ1 ⎥ ⎢ . ⎥⎢ ⎢ ⎢ ⎥ . . . . . . . ⎢ . .. ⎥ ⎢ .. ⎥ .. .. .. .. .. ⎥ ⎥ ⎢ . ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎢ . .. ⎥ = ⎢ .. ⎥ . (1.5) .. .. .. .. ⎥ .. ⎢ . ⎥⎢ ⎢ ⎢ ⎥ . ⎢ . . . . . ⎥⎢ . ⎥ ⎢ . ⎥ ⎥ ⎢ m ⎥ m−1 m−2 ⎣ xm−1 xm−1 xm−1 · · · xm−1 1 ⎦ ⎣ bm−1 ⎦ ⎣ γm−1 ⎦ m m−1 x m−2 · · · bm γm xm xm xm 1 m The coefficient bl can be found to establish the existence and the uniqueness of the polynomial [3]. The uniqueness of the interpolating polynomial can be proven in two ways: The first will be to consider another polynomial P1 (x), then to consider g (x) = P (x) − P1 (x) .

(1.6)

Here, g (x) is a polynomial that by definition is an addition of two polynomials. In particular, g (x) has degree at most m due to the degrees of P (x) and P1 (x). However, at the point m + 1, the polynomial g (xi ) = P (xi ) − P1 (xi ) = γi − γi = 0

(1.7)

of course shows that g (x) has m + 1 roots. This implies g (x) = λ (x − x0 ) (x − x1 ) ... (x − xm ) .

(1.8)

Here, λ is a real number and g (x) is a polynomial of degree β < n, which means it has one real too many, with A = 0 or g (x) = 0.g (x) = 0 =⇒ P (x) − P1 (x) =⇒ P (x) = P1 (x). The second approach is to consider the Vandermonde matrix by setting wb = y.

(1.9)

Then we show that w is non-singular.

1.1

Some interpolation polynomials

The literature contains a few different types of polynomial interpolation, each of them having specific advantages and disadvantages [1–10]. We list the Lagrange polynomial

Polynomial interpolation

3

[1–10] , the Newton polynomial, the Bernstein polynomial, the Chebyshev polynomial, the Lagrange–Sylvester interpolation, the Hermite interpolation, the bi-cubic interpolation, the tricubic interpolation, the multivariate interpolation, the spline interpolation, the discrete spline interpolation, and the monotone cubic interpolation. In this section, we show some of the above listed interpolations to help the reader who is not acquainted with the subject [1–10] . For researchers not working within the field of numerical analysis and application, we recall that within the framework of numerical analysis, the Lagrange polynomials are mostly used for polynomial interpolation. Given a set of j + 1 data points, say  (a0 , b0 ) , ..., (ak , bk ) , ..., aj , bj , where indeed no two aj are the similar, the interpolation polynomial in the Lagrange form is a linear combination defined as L (x) =

j

(1.10)

bk lk (x)

k=0

where in general the Lagrange basis polynomials are given by   (x − an ) (x − ak−1 ) x − aj ... ...  lk (x) = (ak − a0 ) (ak − ak−1 ) ak − aj

(1.11)

0≤n≤j n=k

where 0 ≤ k ≤ j . From the given assumption, ai = aj , ∀ (i, j ) ∈ N such that i = j , it is clear that ai = aj =⇒ ai − aj = 0. In this case, ai = aj with bi = bj is possible, so that no interpolation function L as bk = L (ak ) will be possible. Also if bi = bj , then naturally ai = aj .∀i = j , lk (ai ) =

 (ai − an ) (ai − a0 ) (ai − ai ) (ai − am ) = ... ... = 0. (ak − an ) (a1 − a0 ) (ak − a0 ) (ak − am )

(1.12)

 (ak − an ) = 1. (ak − an )

(1.13)

n=j

But lk (ak ) =

n=k

The remainder in the Lagrange interpolation formula. For the process of interpolating a function f by a polynomial of degree c at the nodes a0 , ..., ak , we put L (x) = f (x) − L (x)

(1.14)

where L (x) = f (a0 , ..., ak , x) l (x) = l (x)

f k+1 (ξ ) , (k + 1)!

a0 < ξ < ak ,

(1.15)

and |L (x)| ≤

(ak − a0 )k+1 max f (k + 1)! a0 ≤ξ ≤ak

k+1

(ξ ) .

(1.16)

4

New Numerical Scheme With Newton Polynomial

1.1.1 Bernstein polynomial The m + 1 Bernstein polynomials of degree m are given by

 m λ bλ,m (x) = (1.17) x (1 − x)m−λ , λ = 0, ..., m, λ  where m λ is the well-known binomial coefficient [1–3]. The Bernstein polynomial is therefore a linear combination of Bernstein basis polynomials Bm (x) =

m

(1.18)

γλ bλ,m (x) .

λ=0

Some important properties are here presented. 1. bλ,n (x) = 0, if λ < 0 or λ > n. 2. bλ,n (x) ≥ 0 for x ∈ [0, 1]. 3. bλ,n (0) = δλ,0 .  0, if λ = n 4. bλ,n (1) = δλ,n = . 1, if λ = n 5. bλ,n (1 − x) = bn−λ,n (x). n+1   1 bλ,n (x) dx = n+1 bj,n+1 (x). 6. j =λ+1

n

7.

n 

λbλ,n (x) = nx.

λ=0

1.1.2 The Newton polynomial interpolation In contrast to the Lagrange polynomial, the Newton polynomial uses more data, in addition, a basis polynomial and the corresponding coefficients can be obtained, while all existing basis polynomial and their coefficients stay the same. The basis polynomials of the Newton interpolation [2] are determined as m0 (y) = 1,

mn (y) =

n−1 

(yk − y0 )

(k = 1, .., n)

(1.19)

k=0

and the Newton interpolating polynomial is given as ⎞ ⎛ n n i−1



  Mn (y) = y − yj ⎠ . ai mi (y) = a0 + ai ⎝ i=0

i=1

(1.20)

j =0

We must note that the last data point (xn , yn ) cannot be used for any basis polynomials, nevertheless, it can be used for computing the last coefficient an . For these chosen nth degree polynomials Mn (y) to go to all n + 1 points (xi , yi ), (i = 0, ..., n), one needs

Polynomial interpolation

5

to satisfy the following n + 1 equations: ⎧ y0 = Mn (x0 ) = a0 ⎪ ⎪ ⎪ ⎪ y1 = Mn (x1 ) = a0 + a1 (x1 − x0 ) ⎪ ⎪ ⎪ ⎨ y2 = Mn (x2 ) = a0 + a1 (x1 − x0 ) + a2 (x2 − x0 ) (x2 − x1 ) .. . ⎪ . ⎪ ⎪ ⎪ n−1 ⎪  ⎪ ⎪ (xn − xi ) ⎩ yn = Mn (xn ) = a0 + a1 (x1 − x0 ) + ... + an

(1.21)

i=0

In matrix form, one can rewrite the above ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1 1 1 1 .. . 1

0 (x1 − x0 ) (x2 − x0 ) (x3 − x0 ) .. . (xn − x0 )

⎡ ⎢ ⎢ ⎢ ⎢ ×⎢ ⎢ ⎢ ⎢ ⎣

a0 a1 a2 .. . .. . an

⎤

0 0 (x2 − x0 ) (x2 − x1 ) (x3 − x0 ) (x3 − x1 ) .. . (xn − x0 ) (xn − x1 )

⎡

⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎦ ⎣

y0 y1 y2 .. . .. . yn

⎤ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎦

0 0 0 (x3 − x0 ) (x3 − x1 ) (x3 − x2 ) .. . (xn − x0 ) (xn − x1 ) (xn − x2 )

··· ··· ··· ··· ··· ···

⎤

0 0 0 0 0 n−1 

(xn − xi )

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

i=0

(1.22)

The above leads us to have a0 = y0 = f (x0 ) = f [x0 ] , y1 − y0 a1 = = f [x0 , x1 ] , x1 − x0 y2 −y0 y1 −y0 x −x − x1 −x0 = f [x0 , x1 , x2 ] a2 = 2 0 x2 − x1 y1 y2 y0 + + , (1.23) = (x0 − x1 ) (x0 − x2 ) (x1 − x0 ) (x1 − x2 ) (x2 − x0 ) (x2 − x1 ) y1 y0 a3 = + (x0 − x1 ) (x0 − x2 ) (x0 − x3 ) (x1 − x0 ) (x1 − x2 ) (x1 − x3 ) y3 y2 + + (x2 − x0 ) (x2 − x1 ) (x2 − x3 ) (x3 − x0 ) (x3 − x1 ) (x3 − x2 ) = f [x0 , x1 , x2 , x3 ] .

6

New Numerical Scheme With Newton Polynomial

Thus, in general, we have  f xj  , (xl − xi )



 aj = f x0 , ..., xj = j

i=0

j = 0..., n.

(1.24)

i=0,l+i

In this case, the Newton polynomial can be written as Mn (x) =

n

ai mi (x) =

i=0

n

f [x0 , ..., xi ] mi (x) .

(1.25)

i=0

The Burger error can be calculated as n (x) = f (x) − Mn (x) =

f

(n+1) (ξ )

(n + 1)!

w (x) ,

(1.26)

but we have f (x) = Mn (x) + f [x0 , ..., xn , x] w (x) .

(1.27)

One can conclude that the error can be expressed as n (x) = f [x0 , ..., xn , x] w (x) .

(1.28)

1.1.3 Hermite interpolation This technique is achieved when the function first n derivatives are given above the function value at each of the n + 1 node points x0 , ..., xn . That is to say, one will have (n + 1) (m + 1) values of yji = f (i) xj (j = 0, ..., n, i = 0, ..., m). Thus, the nonlinear function [5] can be interpolated as H(n+1)(m+1)−1 (x) =

(n+1)(m+1)−1

ak x k .

(1.29)

k=0

The coefficient can be obtained by deriving the solution of the following linear equation: ⎛ ⎞ (n+1)(m+1)−1

 di di ⎝ k⎠ (1.30) |x=xj = f (i) xj H (x) |x=xj = i ck x dx i dx k=0

where j = 0, ..., n, i = 0, ..., m. Nevertheless, we have to note that  the method is not practical because of the high computational complexity O (mn)3 .

Polynomial interpolation

7

1.1.4 Cubic polynomial It is well known that in numerical analysis a cubic polynomial is a polynomial of the form ax 3 + bx 2 cx + d = 0.

(1.31)

Thus, given equation that involves a cubic polynomial [6] is known as a cubic equation. It is well known that a closed form solution called a cubic formula exists for the solutions of an arbitrary cubic equation.

1.1.5 B-spline polynomial Another important and very useful polynomial in numerical analysis is perhaps the well-known B-spline [2], also called basis spline function, having a minimal sup-point corresponding to a given degree smoothness and domain partition. It is worth noting that the word B-spline was initiated by Schoenberg. It is noted that a spline function of order n is a piecewise polynomial function of degree n − 1 in the variable t. These values t where the pieces of polynomial intersect are called knots, they can be denoted as λ0 , λ1 , λ2 , λ3 , λ4 , and they are classified into non-decreasing order. With the above arrangement as soon as the knots are distinct, the first derivatives of the polynomial pieces are known to be continuous over the entire knots. Therefore, for given components, there is, up to a scaling factor, a unique spline Bi,n (t) meeting the following condition:  0, if t ≤ λi or t > λi+n , (1.32) Bi,n (t) = nonzero, otherwise . Nonetheless, we may add the condition that

Bi,n (t) = 1, ∀t ∈ [ti , ti+1 ] .

(1.33)

i

Thus, the scaling factor of Bi,n (t) becomes fixed. This leads directly to the name of a B-spline. Another way is to construct B-splines by using the Lox–de Boor recursion formula with a given knot sequence λ0 , λ1 , λ2 , λ3 , ..., then the B-splines of order 1 are defined by  1, if λi ≤ t ≤ λi+1 , (1.34) Bi,n (t) = 0, otherwise.  Again, if i Bi,1 (t) = 1, ∀t we see the fact that for any t exactly for one Bi,1 (t) = 1, but for the others it leads to zero. It is noted that the higher order B-splines are given by Bi,k+1 (t) =

t − λi λi+k+1 − t Bi,k (t) + . λi+k − λi λi+k+1 − λi+1

(1.35)

8

New Numerical Scheme With Newton Polynomial

We also have the derivative expressions of a B-spline of degree n, which is a function of B-splines.

1.1.6 Legendre polynomial Here, the polynomial Pi (t) is a polynomial of degree i, with the following relation: 

1 −1

Pj (t) Pi (t) dt = 0,

if i = j.

(1.36)

The definition obtained from the generating function [8] is provided by ∞

1 = Pi (x) t i √ 2 1 − 2xt + t i

(1.37)

where the coefficient of t i is a polynomial in x of degree i, where P0 (x) = 1, P1 (x) = x. The Pi polynomial defined above can be differentiated with respect to t on both sides to obtain 

 x −t = 1 − 2xt + t 2 iPi (x) t i−1 . √ 1 − 2xt + t 2 i=0 ∞

(1.38)

Rearranging, we obtain (i + 1) Pi+1 (x) = (2i + 1) xPi (x) − iPi−1 (x) .

(1.39)

Here we need P0 and P1 to start.

1.1.7 Chebyshev polynomial The Chebyshev polynomial [7] was initiated from the following Chebyshev ordinary differential equation: 

1 − x2

 dy 2 dx 2

−x

dy + n2 y = 0 dx

(1.40)

where n = 0, 1, 2, 3, 4, .... In this case, when letting x = cos t the initial equation is reduced to the following: dy 2 + n2 y (t) = 0 dt 2

(1.41)

whose general exact solution can be obtained: y (t) = A sin nt + B cos nt.

(1.42)

Polynomial interpolation

9

Here, replacing t by its values, we obtain     y (x) = A sin n cos−1 x + B cos n cos−1 x

(1.43)

where |x| < 1. Thus, we have the solution y (x) = An (x) + Bn (x)

(1.44)

and  n n (x) + n (x) = ent = x + x 2 − 1 ,  n n (x) − n (x) = e−nt = l x − x 2 − 1 .

(1.45)

Then, again 1 ! ent + e−nt = x+ 2 2 1 ! ent − e−nt = x+ n (x) = 2 2

n (x) =

x2 − 1 x2 − 1

n n

 + x−

x2 − 1

 − x−

x2 − 1

n " ,

(1.46)

n " .

The Chebyshev polynomials of the first kind of degree n can be given in general by n (x) =

n−1/2 dn  (−2)n n! 1 − x2 n 1 − x2 dx (2n)!

(1.47)

where n = 0, 1, 2, 3, 4, .... It is well known that the generating function can be written ∞

1 − xt = n (x) t n . 1 − 2xt + t 2

(1.48)

n=0

The following properties can be established: ⎧  π m = n, ⎨ 0, π , m = n = 0, cos (mβ) cos (nβ) dβ = 2 ⎩ 0 π, m = n = 0.

(1.49)

1.1.8 Lagrange–Sylvester interpolation The matrix theory called Lagrange–Sylvester interpolation [10] can be presented in terms of an analytical function λ (B) of a matrix B as a polynomial in B, in the form of the eigenvalues and eigenvectors of B. The formula is defined as λ (B) =

n

 λ γj Bj . j =1

(1.50)

10

New Numerical Scheme With Newton Polynomial

Here γj are eigenvalues of B and the matrices Bj =

 j =1 j =i

1  B − γj I γj − γi

(1.51)

are known as the corresponding Frobenius covariants of the matrix B corresponding to the projection matrix Lagrange of B. It is well known that the formula of Sylvester is only valid for diagonalizable matrices, and this can be extended because of the matrix B. However, using the Hermite interpolating polynomial covers the most extended case, ⎡ ⎤ nj −1 n

⎢

j   n ⎥ 1 (j )   ⎢ γ B − γ B − γj I j ⎥ I λ (B) =

j j ⎣ ⎦ j!

(1.52)

λ (l)

i (l) =   n . l − γj j

(1.53)

j =1

j =0

j =1 j =i

where

j =i

Schwerdtfeger provides the clear form λ (B) =

n

j =1



λ(j ) γj  j B − γj I . Bj j! nj −1 j =0

(1.54)

Two-steps Lagrange polynomial interpolation: numerical scheme

2

Although many researches have investigated classical calculus, like the existence and uniqueness of solutions of differential equations, and numerical and analytical methods to obtain solutions such equations, there are many topics that will open new doors in fractional calculus [11–21]. One of the most important topics is to construct an efficient numerical scheme for solving differential and integral equations with integer and non-integer orders. In the literature, one can find many numerical schemes which are obtained using various polynomials [22–26]. Recently, a new numerical scheme based on the Lagrange polynomial was introduced by Atangana and Toufik [26]. Fractional calculus is a branch of mathematical analysis that has appeared as an extension of derivatives and integrals to non-integer orders. Fractional calculus arises in many fields, such as geology, physic, hydrology, medicine, geohydrology, chemistry, biology, biotechnology and genetics [27–31].

2.1

Classical differential equation

Classical differential equations appear in the mathematical modeling of real world problems in many fields, especially in science and engineering. In this section, we present an efficient numerical scheme that will be useful for solving linear and nonlinear problems. To do this, we deal with a general Cauchy problem with classical derivative, du (t) = h (t, u (t)) dt

(2.1)

with the initial condition u (0) = u0 . Here the function h (t, u (t)) is taken to be a non-linear function. Integrating the above equation, we have the following equation:  t h (τ, u (τ )) dτ. (2.2) u (t) − u (0) = 0

At the point tk+1 = (k + 1) t, Eq. (2.3) is written as follows:  tk+1 h (τ, u (τ )) dτ u (tk+1 ) − u (0) =

(2.3)

0

and also at the point tk = kt  tk h (τ, u (τ )) dτ. u (tk ) − u (0) = 0 New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00008-1 Copyright © 2021 Elsevier Inc. All rights reserved.

(2.4)

12

New Numerical Scheme With Newton Polynomial

If we subtract Eq. (2.5) from (2.4), we obtain the following:  tk+1 u (tk+1 ) − u (tk ) = h (τ, u (τ )) dτ.

(2.5)

tk

As the approximation of the function h (t, u (t)), we can use the Lagrange polynomial which is represented by Pk (τ ) =

h (tk , u (tk )) h (tk−1 , u (tk−1 )) (τ − tk−1 ) − (τ − tk ) . t t

(2.6)

When replacing the Lagrange polynomial into Eq. (2.5), we write the following equality:  tk+1 h (tk , u (tk )) (2.7) uk+1 − uk = (τ − tk−1 ) dτ t tk  tk+1 h (tk−1 , u (tk−1 )) − (τ − tk ) dτ. t tk If we arrange the above equation, we can write the following:  h (tk , u (tk )) tk+1 uk+1 − uk = (τ − tk−1 ) dτ t tk  h (tk−1 , u (tk−1 )) tk+1 − (τ − tk ) dτ. t tk

(2.8)

We can calculate integrals located on the right hand side of Eq. (2.9) as follows:  tk+1 3 (2.9) (τ − tk−1 ) dτ = (t)2 , 2 tk  tk+1 1 (τ − tk−2 ) (τ − tk−1 ) dτ = (t)2 . 2 tk After replacing these calculations into Eq. (2.9), we get the following numerical scheme:   1  3  (2.10) uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t. 2 2

2.1.1 Numerical illustrations In this section, we give illustrations for the numerical solution of the some equations to show the efficiency of the considered numerical scheme. Example 2.1. We consider the following Cauchy problem with classical derivative: u (t) = 2t − 11,

(2.11)

Two-steps Lagrange polynomial interpolation: numerical scheme

13

Figure 2.1 Numerical simulation with classical derivative.

u (0) = 0.1, where the considered equation is linear. For such an equation, the numerical scheme is given by   1  3  (2.12) uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t 2 2     where h tk , uk = 2tk − 11 and h tk−1 , uk−1 = 2tk−1 − 11. We present the numerical simulation for our equation with classical derivative in Fig. 2.1. Example 2.2. We now consider the following Cauchy problem with classical derivative: u (t) = u2 (t) exp (−t) ,

(2.13)

u (0) = −0.5, where the considered equation is non-linear. For such an equation, we have the following scheme:   1  3  uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t 2 2

(2.14)

   2   2  where h tk , uk = uk exp (−tk ) and h tk−1 , uk−1 = uk−1 exp (−tk−1 ). For the classical case, the numerical simulation is given in Fig. 2.2. Example 2.3. We deal with the following problem with integer order: u (t) = t cos (4t) , u (0) = 1.

(2.15)

14

New Numerical Scheme With Newton Polynomial

Figure 2.2 Numerical simulation with classical derivative.

Figure 2.3 Numerical simulation with classical derivative.

Applying the suggested numerical scheme to this example yields   1  3  uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t (2.16) 2 2     where h tk , uk = tk cos (4tk ) and h tk−1 , uk−1 = tk−1 cos (4tk−1 ). For the above problem, the numerical simulation is depicted in Fig. 2.3. Example 2.4. We consider the following problem: u (t) = −2u (t) + 8, u (0) = 1.

(2.17)

The above equation can be solved numerically by the following scheme:   1  3  (2.18) uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t 2 2     where h tk , uk = −2uk + 8 and h tk−1 , uk−1 = −2uk−1 + 8. For the classical derivative, the numerical simulation is given in Fig. 2.4.

Two-steps Lagrange polynomial interpolation: numerical scheme

15

Figure 2.4 Numerical simulation for the classical derivative.

Example 2.5. The Moore–Spiegel system describes a fluid element oscillating vertically in a temperature gradient with a linear restoring force [32]. Associated with this, we consider this model with classical derivative Q1  (t) = Q2 , Q2  (t) = Q3 ,



Q3  (t) = −Q3 − T − R

+ RQ21



(2.19) Q2 − T Q1 ,

with the initial conditions Q1 (0) = 0.3,

Q2 (0) = 0.1,

Q3 (0) = 0.2.

(2.20)

Here, T is similar to the Prandtl number times the Taylor number and R is similar to the Prandtl number times the Rayleigh number, respectively. Applying the suggested numerical scheme to this example yields  3  = Qk1 + h1 tk , Qk1 , Qk2 , Qk3 t − Qk+1 1 2  3  k+1 k Q2 = Q2 + h2 tk , Qk1 , Qk2 , Qk3 t − 2

 1  k−1 k−1 h1 tk−1 , Qk−1 t, 1 , Q2 , Q3 2  1  k−1 k−1 h2 tk−1 , Qk−1 t, 1 , Q2 , Q3 2 (2.21)     1 3 k−1 k−1 k−1 k k k k h h t t − t t, = Q + , Q , Q , Q , Q , Q , Q Qk+1 3 k 3 k−1 3 1 2 3 3 1 2 3 2 2

where

  h1 tk , Qk1 , Qk2 , Qk3 = Qk2 ,   h2 tk , Qk1 , Qk2 , Qk3 = Qk3 ,     2  Qk2 − T Qk1 , h3 tk , Qk1 , Qk2 , Qk3 = −Qk3 − T − R + R Qk1

(2.22)

16

New Numerical Scheme With Newton Polynomial

Figure 2.5 Numerical simulation Moore–Spiegel model with classical derivative.

and

  k−1 k−1 = Qk−1 h1 tk−1 , Qk−1 1 , Q2 , Q3 2 ,   k−1 k−1 = Qk−1 h2 tk−1 , Qk−1 1 , Q2 , Q3 3 ,   k−1 k−1 h3 tk−1 , Qk−1 1 , Q2 , Q3  2   k−1 Qk−1 − T − R + R Q − T Qk−1 = −Qk−1 3 1 2 1 .

(2.23)

With the parameters T = 6, R = 20, we present the numerical simulation for this model in Fig. 2.5.

2.2

Fractal differential equation

Some media like turbulence, porous media, and turbulence usually exhibit fractal properties. The fractal derivative of a function u (t) with respect to a fractal measure

Two-steps Lagrange polynomial interpolation: numerical scheme

17

t β is defined by [33] CH β lim 0 Dt u (t) = t→t

u (t) − u (t1 ) β

t β − t1

1

.

Also, the fractal integral in the sense of Chen has the form of  t CH β I u = β τ β−1 u (τ ) dτ. (t) t 0

(2.24)

(2.25)

0

In this section, to show the validity of the proposed method, we offer a numerical scheme for a general Cauchy problem for which the differential operator is fractal derivative in the sense of Chen. Now, we consider the following problem with fractal derivative: CH β 0 Dt u (t) = h (t, u (t)) .

(2.26)

Applying the fractal integral to the above equation, we have the following equality:  t τ β−1 h (τ, u (τ )) dτ. (2.27) u (t) − u (0) = β 0

At the point tk+1 = (k + 1) t, the above equation becomes  tk+1 u (tk+1 ) − u (0) = β τ β−1 h (τ, u (τ )) dτ

(2.28)

0

and at the point tk = kt, we have  tk τ β−1 h (τ, u (τ )) dτ. u (tk ) − u (0) = β

(2.29)

0

From Eqs. (2.28) and (2.29), we obtain the following:  tk+1 u (tk+1 ) − u (tk ) = β τ β−1 h (τ, u (τ )) dτ.

(2.30)

tk

For the simplification of the above integral, let us use the Lagrange polynomial which is of the form Pk (τ ) =

h (tk , u (tk )) h (tk−1 , u (tk−1 )) (τ − tk−1 ) − (τ − tk ) . t t

(2.31)

Putting this polynomial into Eq. (2.30) and taking g (τ, u (τ )) = βτ β−1 h (τ, u (τ )) for convenience, we have the following:  tk+1 g (tk , u (tk )) k+1 k u −u = (2.32) (τ − tk−1 ) dτ t tk  tk+1 g (tk−1 , u (tk−1 )) − (τ − tk ) dτ, t tk

18

New Numerical Scheme With Newton Polynomial

and we write k+1

u

  t k+1 g tk , uk −u = (τ − tk−1 ) dτ t tk    tk+1 g tk−1 , uk−1 − (τ − tk ) dτ. t tk k

For the above integrals, we have the following calculations:  tk+1 3 (τ − tk−1 ) dτ = (t)2 , 2 tk  tk+1 1 (τ − tk ) dτ = (t)2 . 2 tk

(2.33)

(2.34)

If these calculations are substituted in Eq. (2.33), the following scheme can be written:  3 1  uk+1 = uk + g tk , uk t − g tk−1 , uk−1 t. 2 2

(2.35)

Then, replacing the function g (t, u (t)) by its value, we have the following scheme:   1 β−1  3 β−1  uk+1 = uk + βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t. 2 2

(2.36)

2.2.1 Numerical illustrations In this section, we present numerical illustrations of some differential equations including a fractal derivative. From these examples, we can see that the considered method is accurate. Example 2.6. We deal with the following Cauchy problem: CH β 0 Dt u (t) = 12u (t) + 1,

(2.37)

u (0) = 0.18, where the derivative is fractal. The above equation can be solved numerically by the following scheme:   1 β−1  3 β−1  uk+1 = uk + βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t 2 2     k k k−1 where h tk , u = 12u + 1 and h tk−1 , u = 12uk−1 + 1. For the fractal case, the numerical simulation is depicted in Fig. 2.6.

(2.38)

Example 2.7. We next consider the following problem involving fractal derivative: CH β 0 Dt u (t) = exp (−7u (t)) ,

(2.39)

Two-steps Lagrange polynomial interpolation: numerical scheme

19

Figure 2.6 Numerical simulation with fractal derivative for β = 0.92.

Figure 2.7 Numerical simulation with fractal derivative for β = 0.88.

u (0) = 0.3. Applying the suggested numerical scheme to this example yields   1 β−1  3 β−1  (2.40) uk+1 = uk + βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t 2 2         where h tk , uk = exp −7uk and h tk−1 , uk−1 = exp −7uk−1 . We present the numerical simulation for the above problem with fractal order in Fig. 2.7. Example 2.8. We consider the following test problem having fractal derivative: CH β 0 Dt u (t) = − sin (3 + t) ,

(2.41)

u (0) = 1. The following scheme for the above problem is presented:   1 β−1  3 β−1  uk+1 = uk + βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t 2 2

(2.42)

20

New Numerical Scheme With Newton Polynomial

Figure 2.8 Numerical simulation with fractal derivative for β = 0.9.

Figure 2.9 Numerical simulation with fractal derivative for β = 0.38.

    where h tk , uk = − sin (3 + tk ) and h tk−1 , uk−1 = − sin (3 + tk−1 ). For the above problem, the numerical simulation is presented in Fig. 2.8. Example 2.9. We now consider the following problem with fractal derivative: CH β 2 0 Dt u (t) = 9.5t ln (u (t)) ,

(2.43)

u (0) = 0.2. We obtain the following scheme for the fractal case:   1 β−1  3 β−1  (2.44) uk+1 = uk + βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t 2 2         2 ln uk−1 . The numerical where h tk , uk = 9.5tk2 ln uk and h tk−1 , uk−1 = 9.5tk−1 simulation for the above problem is depicted in Fig. 2.9. Example 2.10. For this example, we deal with the following ACT attractor with fractal derivative: CH β 0 Dt x (t) = α (x

− y) ,

Two-steps Lagrange polynomial interpolation: numerical scheme CH β 3 0 Dt y (t) = −4αy + xz + μx , CH β 2 0 Dt z (t) = −δαz + xy + βz ,

21

(2.45)

with the initial conditions x (0) = 1, y (0) = 1, z (0) = 1.

(2.46)

Applying the suggested numerical scheme to this example yields  3 β−1  x k+1 = x k + βtk h1 tk , x k , y k , zk t 2  1 β−1  − βtk−1 h1 tk−1 , x k−1 , y k−1 , zk−1 t, 2  3 β−1  y k+1 = y k + βtk h2 tk , x k , y k , zk t 2  1 β−1  − βtk−1 h2 tk−1 , x k−1 , y k−1 , zk−1 t, 2  3 β−1  k+1 z = zk + βtk h3 tk , x k , y k , zk t 2  1 β−1  − βtk−1 h3 tk−1 , x k−1 , y k−1 , zk−1 t, 2

(2.47)

where     h1 tk , x k , y k , zk = α x k − y k ,    3 h2 tk , x k , y k , zk = −4αy k + x k zk + μ x k ,    2 h3 tk , x k , y k , zk = −δαzk + x k y k + β zk ,

(2.48)

and     h1 tk−1 , x k−1 , y k−1 , zk−1 = α x k−1 − y k−1 ,   3  h2 tk−1 , x k−1 , y k−1 , zk−1 = −4αy k−1 + x k−1 zk−1 + μ x k−1 ,   2  h3 tk−1 , x k−1 , y k−1 , zk−1 = −δαzk−1 + x k−1 y k−1 + β zk−1 .

(2.49)

With the parameters α = 1.8, β = −0.07, μ = 0.02, δ = 1.5, we depict the numerical simulation for the considered model in Fig. 2.10.

22

New Numerical Scheme With Newton Polynomial

Figure 2.10 Numerical simulation ACT model with fractal derivative for β = 0.95.

2.3

Differential equation with the Caputo–Fabrizio operator

Caputo and Fabrizio introduced a new derivative which does not have singular kernel [34]. Although non-locality cannot be identified for this derivative, the Caputo– Fabrizio fractional derivative established with the exponential function is useful for some physical problems. In this section, we will obtain a new numerical scheme for the following Cauchy problem with exponential decay kernel. Firstly, we recall the definition of the Caputo–Fabrizio fractional derivative. The Caputo–Fabrizio fractional derivative of the function u (t) ∈ W21 (0, l), α ∈ [0, 1] is defined by CF α 0 Dt u (t) =

M (α) 1−α

 0

t

d α u (τ ) exp − (t − τ ) dτ dτ 1−α

(2.50)

where M (α) is a normalization function such that M (0) = 1, M (1) = 1. The fractional integral with exponential decay kernel of the function u (t) is given by CF α 0 It (u (t)) =

α 1−α u (t) + M (α) M (α)



t

u (τ ) dτ. 0

(2.51)

Two-steps Lagrange polynomial interpolation: numerical scheme

23

We now present the following Cauchy problem with new fractional derivative: CF α 0 Dt u (t) = h (t, u (t)) .

(2.52)

From the definition of the Caputo–Fabrizio integral, we can reformulate the above equation as  t α 1−α h (t, u (t)) + u (t) − u (0) = h (τ, u (τ )) dτ. (2.53) M (α) M (α) 0 We write Eq. (2.53) at the point tk+1 = (k + 1) t u (tk+1 ) − u (0) =

1−α α h (tk , u (tk )) + M (α) M (α)



tk+1

h (τ, u (τ )) dτ

(2.54)

h (τ, u (τ )) dτ.

(2.55)

0

and at the point tk = kt 1−α α h (tk−1 , u (tk−1 )) + u (tk ) − u (0) = M (α) M (α)



tk

0

Taking the difference of these equations, we can write the following: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)  tk+1 α h (τ, u (τ )) dτ. + M (α) tk

u (tk+1 ) − u (tk ) =

(2.56)

Putting its Lagrange polynomial into the above equation, we can get the following: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)  tk+1  h(tk ,u(tk )) α t  (τ− tk−1 )  dτ, + h t ,u t M (α) tk − k−1 t k−1 (τ − tk )

uk+1 − uk =

(2.57)

and we get the following: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)  h(t ,u(t )) tk+1 k k − t dτ (τ ) k−1 α t t k   + . ,u tk−1 tk+1 M (α) − h tk−1 t (τ − tk ) dτ tk

uk+1 − uk =

(2.58)

The integrals on the right hand side of the above equation can be calculated as  tk+1 3 (2.59) (τ − tk−1 ) dτ = (t)2 , 2 tk  tk+1 1 (τ − tk ) dτ = (t)2 . 2 tk

24

New Numerical Scheme With Newton Polynomial

Thus, we have the following numerical scheme: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)    3 1  α h tk , uk t − h tk−1 , uk−1 t . + M (α) 2 2

uk+1 = uk +

(2.60)

2.3.1 Error analysis with exponential kernel In this section, we present the error analysis to show the accuracy and efficiency of the considered scheme. To do this, we consider a non-linear fractional differential equation with exponential kernel,  CF α 0 Dt u (t) = h (t, u (t)) , (2.61) u (0) = u0 . Assuming that the function has a bounded second derivative, we shall estimate the error. We can write Eq. (2.61) as follows:  tk+1 1−α α h (tk , u (tk )) + h (τ, u (τ )) dτ (2.62) u (tk+1 ) − u (0) = M (α) M (α) 0  tk+1 α 1−α h (tk , u (tk )) + h (τ, u (τ )) dτ = M (α) M (α) tk  tk+1 1−α α h (tk , u (tk )) + = Pk (τ ) dτ M (α) M (α) tk  tk+1 α (τ − tk ) (τ − tk−1 ) + M (α) tk 2! ∂2 [h (τ, u (τ ))]τ =γk dτ ∂τ 2  1−α  = h tk , uk + Ekα M (α) ⎫ ⎧   h tk ,uk tk+1 ⎬ ⎨ α tk  (τ − tk−1 ) dτ  t + , k−1

tk+1 M (α) ⎩ − h tk−1 ,u (τ − tk ) dτ ⎭ tk t ×

where Ekα =

α M (α)



tk+1 tk

(τ − tk ) (τ − tk−1 ) ∂ 2 [h (τ, u (τ ))]τ =γk dτ. 2! ∂τ 2

We know that there exists γk ∈ [tk , tk+1 ] such that  2  ∂  α   sup h u Ekα = (τ, (τ ))   2   M (α) τ ∈ 0,tk+1 ∂τ τ =γk

(2.63)

(2.64)

Two-steps Lagrange polynomial interpolation: numerical scheme



(τ − tk ) (τ − tk−1 ) dτ 2! tk  2  ∂  α  = sup   2 h (τ, u (τ )) M (α)  ∂τ

×

25

tk+1

τ ∈ 0,tk+1

τ =γk

5 (t)3 . 12

Taking the norm on both sides of the above equation, the error can be estimated as  α E  ≤ k

 2  ∂  5 α  sup h (τ, u (τ )) (t)3 . M (α) τ ∈0,tk+1   ∂τ 2 12

(2.65)

2.3.2 Numerical illustrations In this section, numerical simulation will be presented to solve some equations with new differential operator based on the exponential decay kernel. Example 2.11. We present the following problem: CF α 0 Dt u (t) = 10t

+ 6.7u (t) ,

(2.66)

u (0) = 0.01. Applying the suggested numerical scheme to this example yields 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)    3 1  α h tk , uk t − h tk−1 , uk−1 t + M (α) 2 2

uk+1 = uk +

(2.67)

where h (tk , u (tk )) = 10tk + 6.7uk and h (tk−1 , u (tk−1 )) = 10tk−1 + 6.7uk−1 . For the Caputo–Fabrizio case, we provide the numerical simulation in Fig. 2.11.

Figure 2.11 Numerical simulation with the Caputo–Fabrizio derivative for α = 0.79.

26

New Numerical Scheme With Newton Polynomial

Figure 2.12 Numerical simulation with the Caputo–Fabrizio derivative for α = 0.57.

Example 2.12. We next consider the following nonlinear problem: CF α 0 Dt u (t) = 4t sin (u (t)) ,

(2.68)

u (0) = 0.3. We can get the following scheme for the Caputo–Fabrizio case: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)    3 1  α h tk , uk t − h tk−1 , uk−1 t (2.69) + M (α) 2 2     where h (tk , u (tk )) = 4tk sin uk and h (tk−1 , u (tk−1 )) = 4tk−1 sin uk−1 . The numerical simulation with the Caputo–Fabrizio case is presented in Fig. 2.12. uk+1 = uk +

Example 2.13. We consider the following problem with Caputo–Fabrizio derivative: CF α 0 Dt u (t) = − exp (−2t) u (t) ,

(2.70)

u (0) = 1. The following scheme can be written for the above equation: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)   3 1   α h tk , uk + t − h tk−1 , uk−1 t M (α) 2 2

uk+1 = uk +

(2.71)

where h (tk , u (tk )) = − exp (−2tk ) uk and h (tk−1 , u (tk−1 )) = − exp (−2tk−1 ) uk−1 . For this derivative, the numerical simulation is depicted in Fig. 2.13. Example 2.14. We deal with the problem with the Caputo–Fabrizio fractional operator CF α 0 Dt u (t) = −u (t) + 1.8,

(2.72)

Two-steps Lagrange polynomial interpolation: numerical scheme

27

Figure 2.13 Numerical simulation with the Caputo–Fabrizio derivative for α = 0.9.

Figure 2.14 Numerical simulation with the Caputo–Fabrizio derivative for α = 0.74.

u (0) = 0.03. The scheme for the above equation is as follows: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)    3 1  α h tk , uk t − h tk−1 , uk−1 t + M (α) 2 2

uk+1 = uk +

(2.73) (2.74)

where h (tk , u (tk )) = −uk + 1.8 and h (tk−1 , u (tk−1 )) = −uk−1 + 1.8. We give the numerical simulation in Fig. 2.14. Example 2.15. Many studies have addressed CF α 0 Dt u (t) = au − vw, CF α 0 Dt v (t) = bv + uw,

uv CF α , 0 Dt w (t) = cw + 3

(2.75)

28

New Numerical Scheme With Newton Polynomial

with the initial conditions u (0) = 0.01, v (0) = 0.01, w (0) = 0.01. Here a = 5, b = −10, c = −0.38. To examine the dynamic behavior of the Chen–Lee system based on the Euler equations for the motion of a rigid body [35], we apply the suggested numerical scheme to this example:    1−α   h1 tk , uk , v k , w k − h1 tk−1 , uk−1 , v k−1 , w k−1 M (α)   3  1  α h1 tk , uk , v k , w k t − h1 tk−1 , uk−1 , v k−1 , w k−1 t , + M (α) 2 2 (2.76)

uk+1 = uk +

   1−α   h2 tk , uk , v k , w k − h2 tk−1 , uk−1 , v k−1 , w k−1 M (α)   3  1  α h2 tk , uk , v k , w k + t − h2 tk−1 , uk−1 , v k−1 , w k−1 t , M (α) 2 2

v k+1 = v k +

   1−α   h3 tk , uk , v k , w k − h3 tk−1 , uk−1 , v k−1 , w k−1 M (α)       α k k k 3 k−1 k−1 k−1 1 h3 tk , u , v , w + t − h3 tk−1 , u , v , w t , M (α) 2 2

w k+1 = w k +

where   h1 tk , uk , v k , w k = auk − v k w k ,   h2 tk , uk , v k , w k = bv k + uk w k ,

(2.77)

  uk v k , h3 tk , uk , v k , w k = cw k + 3 and   h1 tk−1 , uk−1 , v k−1 , w k−1 = auk−1 − v k−1 w k−1 ,   h2 tk−1 , uk−1 , v k−1 , w k−1 = bv k−1 + uk−1 w k−1 ,

(2.78)

  uk−1 v k−1 h3 tk−1 , uk−1 , v k−1 , w k−1 = cw k−1 + . 3 We offer the numerical simulations for the considered model with the Caputo–Fabrizio fractional derivative in Fig. 2.15 and Fig. 2.16.

Two-steps Lagrange polynomial interpolation: numerical scheme

29

Figure 2.15 Numerical simulation for Chen–Lee system with the Caputo–Fabrizio derivative for α = 0.99.

2.4

Differential equation with the Caputo fractional operator

The Caputo fractional derivative with singular and non-local properties is used to modeling phenomena taking into account past interactions of daily problems. In this section, before giving the numerical scheme for the general Cauchy problem, we shall recall the definition of the Caputo fractional derivative and integral. The Caputo fractional derivative of order of the function u (t) is defined by C α 0 Dt

1 u (t) = (n − α)



t

(t − τ )n−α−1 u(n) (τ ) dτ, t > 0,

(2.79)

0

where u: R+ → R and α ∈ (n − 1, n), n ∈ N. The fractional integral with power-law kernel is given by the following formula: C α 0 It (u (t)) =

1 (α)

 0

t

u (τ ) (t − τ )α−1 dτ.

(2.80)

30

New Numerical Scheme With Newton Polynomial

Figure 2.16 Numerical simulation for Chen–Lee system with the Caputo–Fabrizio derivative for α = 1.

Now let us consider the following problem with power-law kernel  C α 0 Dt u (t) = h (t, u (t)) , u (0) = u0 .

(2.81)

We convert Eq. (2.81) into u (t) − u (0) =

1 (α)



t

h (τ, u (τ )) (t − τ )α−1 dτ.

(2.82)

0

At the point tk+1 = (k + 1) t, k = 0, 1, 2, ..., the equation can thus be formulated as follows:  tk+1 1 u (tk+1 ) − u (0) = h (τ, u (τ )) (tk+1 − τ )α−1 dτ, (2.83) (α) 0 and we have u (tk+1 ) = u (0) +

k  1  tm+1 h (τ, u (τ )) (tk+1 − τ )α−1 dτ. (α) tm m=0

(2.84)

Two-steps Lagrange polynomial interpolation: numerical scheme

31

To simplify the above integral on the right hand side of Eq. (2.84), we replace its Lagrange polynomial in the above equation,  k  h(tm ,um ) (τ − tm−1 ) 1  tm+1 k+1 t  u = u0 + (2.85) h t ,um−1 (α) − m−1 (τ − tm ) tm × (tk+1 − τ )

m=1 α−1

t

dτ.

Thus, we can reorganize the above equation as follows:  tm+1 h(t ,um ) k m (τ − tm−1 ) (tk+1 − τ )α−1 dτ 1  tm t k+1   u = u0 + , (2.86)

t h t ,um−1 (α) − tmm+1 m−1t (τ − tm ) (tk+1 − τ )α−1 dτ m=0 and we have uk+1 = u0 +

 k 1  h (tm , um ) tm+1 (τ − tm−1 ) (α) t tm

(2.87)

m=0

× (tk+1 − τ )α−1 dτ  t  k m+1 1  h tm−1 , um−1 − (τ − tm ) (α) t tm m=0

× (tk+1 − τ )α−1 dτ. We have the following calculations for the integrals in Eq. (2.87):  tm+1 (τ − tm−1 ) (tk+1 − τ )α−1 dτ tm

(t)α+1 (k − m + 1)α (k − m + 2 + α) = α α (α + 1) − (k − m) (k − m + 2 + 2α)  tm+1 (τ − tm ) (tk+1 − τ )α−1 dτ

(2.88)

tm

=

 (t)α+1  (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) . α (α + 1)

If we replace these calculations into Eq. (2.87), we can numerically solve as follows: k+1

u



k  (k − m + 1)α (k − m + 2 + α) (t)α   m = u0 + h tm , u − (k − m)α (k − m + 2 + 2α) (α + 2) m=0

α

k    h tm−1 , um−1

(t) (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) . −

(2.89)

32

New Numerical Scheme With Newton Polynomial

2.4.1 Error analysis with power-law kernel In this section, we present an error analysis for the considered numerical scheme. For the error analysis, we consider the non-linear fractional differential equation  C α 0 Dt u (t) = h (t, u (t)) , (2.90) u (0) = u0 . To calculate the error, we suppose that the function h (t, u (t)) has a bounded second derivative. Then, we can reformulate Eq. (2.90) as follows:  tk+1 1 h (τ, u (τ )) (tk+1 − τ )α−1 dτ u (tk+1 ) − u (0) = (α) 0 k  1  tm+1 = h (τ, u (τ )) (tk+1 − τ )α−1 dτ (2.91) (α) t m=0 m  k  P (τ ) 1  tm+1  k  = (τ −tm ) τ −tm−1 ∂ 2 (α) + [h (τ, u (τ ))]τ =γm 2! ∂τ 2 m=0 tm × (tk+1 − τ )α−1 dτ ⎧ h(t ,um ) tm+1 m (τ − tm−1 ) ⎪ tm t ⎪ k ⎪ ⎨  τ )α−1 dτ × (tk+1 − 1   = m−1 h t ,u tm+1 ⎪ − m−1 (α) (τ − tm ) tm m=0 ⎪ t ⎪ ⎩ × (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

+ Ekα ,

where Ekα

k  1  tm+1 (τ − tm ) (τ − tm−1 ) ∂ 2 = [h (τ, u (τ ))]τ =γm (α) 2! ∂τ 2 tm

(2.92)

m=0

× (tk+1 − τ )α−1 dτ. Since τ → (τ − tm−1 ) (tk+1 − τ )α−1 is positive within the interval [tm , tm+1 ], one can find γm ∈ [tm , tm+1 ] such that k 1  ∂2 (γm − tm ) [h (τ, u (τ ))]τ =γm 2 (α) 2 ∂τ m=0  tm+1 × (τ − tm−1 ) (tk+1 − τ )α−1 dτ

Ekα =

tm k 1  ∂2 (γm − tm ) [h (τ, u (τ ))]τ =γm (t)α+1 (α) 2 ∂τ 2 m=0

1 (k − m + 1)α (k − m + 2 + α) . × α α (α + 1) − (k − m) (k − m + 2 + 2α)

=

(2.93)

Two-steps Lagrange polynomial interpolation: numerical scheme

33

If we take the norm on both sides of the above equation, we can write the following inequality:  2  α+2 ∂  α   E  ≤ (t)  sup h u (τ, (τ )) k   2 (α + 2) τ ∈0,tk+1  ∂τ 2

(2.94)

 k    (k − m + 1)α (k − m + 2 + α)    ×  − (k − m)α (k − m + 2 + 2α)  . m=0

The sum at the right hand side of the above inequality can be written

 k    (k − m + 1)α (k − m + 2 + α)    α α    − (k − m)α (k − m + 2 + 2α)  = (k + 1) − αk

(2.95)

m=0

×

k (k + 4 + 2α) . 2

Thus, the error can be obtained:  2  α+2 ∂   α    (k + 1)α − αk α E  ≤ (t) sup h u (τ, (τ )) k   2   2 (α + 2) τ ∈ 0,tk+1 ∂τ ×

(2.96)

k (k + 4 + 2α) . 2

2.4.2 Numerical illustrations In this section, we give numerical simulations of the numerical solution of some differential equations with the Caputo fractional operator. Example 2.16. We deal with the following test problem with the Caputo fractional derivative: C α 0 Dt u (t) = log (3t

+ 2) ,

(2.97)

u (0) = 0.1. For the above equation, we obtain the following scheme: uk+1 = u0 +

k  (k − m + 1)α (k − m + 2 + α) (t)α   h tm , um − (k − m)α (k − m + 2 + 2α) (α + 2) m=0

α

k    h tm−1 , um−1

(t) (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) , −

(2.98)

34

New Numerical Scheme With Newton Polynomial

Figure 2.17 Numerical simulation with the Caputo derivative for α = 0.97.

  where h (tm , um ) = log (3tm + 2) and h tm−1 , um−1 = log (3tm−1 + 2). The numerical simulation for the above problem is depicted in Fig. 2.17. Example 2.17. We consider the following problem: C α 4 0 Dt u (t) = t exp (2u (t) + 1.9) ,

(2.99)

u (0) = 0.2, where the differential operator is the Caputo fractional derivative. We have the following scheme for the Caputo case: uk+1 = u0 +

k  (k − m + 1)α (k − m + 2 + α) (t)α   h tm , um − (k − m)α (k − m + 2 + 2α) (α + 2) m=0

α

k    h tm−1 , um−1

(t) (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

(2.100)

  m−1   4 exp (2um + 1.9) and h t m−1 = t 4 where h (tm , um ) = tm + 1.9 . m−1 , u m−1 exp 2u For the Caputo derivative, the numerical simulation is given in Fig. 2.18. Example 2.18. We consider the following problem with power-law kernel C α 2 0 Dt u (t) = −3u

+ 1, u (0) = −0.03.

(2.101)

For the above equation, the following scheme can be written: uk+1 = u0 +

k  (k − m + 1)α (k − m + 2 + α) (t)α   h tm , um − (k − m)α (k − m + 2 + 2α) (α + 2) m=0

Two-steps Lagrange polynomial interpolation: numerical scheme

35

Figure 2.18 Numerical simulation with the Caputo derivative for α = 0.85.

Figure 2.19 Numerical simulation with the Caputo derivative for α = 0.89. k  (t)α   h tm−1 , um−1 (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

(2.102)

  2  where h (tm , um ) = −3 (um )2 + 1 and h tm−1 , um−1 = −3 um−1 + 1. For numerical simulation of the above problem with the Caputo derivative, we refer to Fig. 2.19. Example 2.19. We consider the following initial value problem with the Caputo fractional derivative: C α 0 Dt u (t) = −2t

+ 17

(2.103)

with the initial condition u (0) = 0.1.

(2.104)

36

New Numerical Scheme With Newton Polynomial

Figure 2.20 Numerical simulation with the Caputo derivative for α = 0.76.

The above equation can be solved numerically as follows: uk+1 = u0 +

k  (k − m + 1)α (k − m + 2 + α) (t)α   h tm , um − (k − m)α (k − m + 2 + 2α) (α + 2) m=0

  (t) h tm−1 , um−1 (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

α

k 

(2.105)

  where h (tm , um ) = −2tm + 17 and h tm−1 , um−1 = −2tm−1 + 17. We give the numerical simulation for the above test problem with power-law kernel in Fig. 2.20. Example 2.20. The Dadras–Momeni system is known to be a chaotic system with various parameters and it shows many complex dynamic behaviors. For example, this chaotic model generates a two-wing attractor with the following parameters [37]. The Dadras–Momeni attractor with the Caputo derivative is described by the following system; C α 0 Dt x (t) = y − ρx + σyz, C α 0 Dt y (t) = ry − xz + z, C α 0 Dt z (t) = ζ xy − εz,

(2.106)

with the initial conditions x (0) = −0.5, y (0) = −0.5, z (0) = −0.5.

(2.107)

Applying the suggested numerical scheme to this example yields x k+1 = x0 +

k  (t)α   h1 tm , x m , y m , zm (α + 2) m=0

(2.108)

Two-steps Lagrange polynomial interpolation: numerical scheme

×

(k − m + 1)α (k − m + 2 + α) − (k − m)α (k − m + 2 + 2α)

37



k  (t)α   h1 tm−1 , x m−1 , y m−1 , zm−1 (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

y

k+1

k  (t)α   = y0 + h2 tm , x m , y m , zm (α + 2) m=0

(k − m + 1)α (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α) k  (t)α   − h2 tm−1 , x m−1 , y m−1 , zm−1 (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

k  (t)α   h3 tm , x m , y m , zm (α + 2) m=0

(k − m + 1)α (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

zk+1 = z0 +

k  (t)α   h3 tm−1 , x m−1 , y m−1 , zm−1 (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

where   h1 tm , x m , y m , zm = y m − ρx m + σy m zm ,   h2 tm , x m , y m , zm = ry m − x m zm + zm ,   h3 tm , x m , y m , zm = ζ x m y m − εzm ,

(2.109)

and   h1 tm−1 , x m−1 , y m−1 , zm−1 = y m−1 − ρx m−1 + σy m−1 zm−1 ,   h2 tm−1 , x m−1 , y m−1 , zm−1 = ry m−1 − x m−1 zm−1 + zm−1   h3 tm−1 , x m−1 , y m−1 , zm−1 = ζ x m−1 y m−1 − εzm−1 .

(2.110)

38

New Numerical Scheme With Newton Polynomial

Figure 2.21 Numerical simulation for the Dadras–Momeni attractor with the Caputo derivative for α = 1.

For the parameter set ρ = 3, σ = 2.7, r = 1.7, ζ = 2, ε = 9, the numerical simulations for this system are shown for α = 1 and α = 0.98 in Fig. 2.21 and Fig. 2.22.

2.5

Differential equation with the Atangana–Baleanu operator

Being naturally needed, some powerful mathematical tools called differential and integral operators have been discovered that are successful in modeling problems arising in real life. Very recently, Atangana and Baleanu introduced a new differential and integral operator which contains the generalized Mittag-Leffler function [28]. These operators are highly contributive in modeling real world problems due to its nonsingularity and non-locality. In this section, we deal with the numerical solution for the general Cauchy problem with the Atangana Baleanu fractional derivative in the sense of Caputo. The Atangana Baleanu fractional derivative in the sense of Caputo of the function u (t) is defined by the following formula:

 α AB (α) t d ABC α α − dτ (2.111) u E D u = − τ (τ ) (t) (t ) α t 0 1 − α 0 dτ 1−α

Two-steps Lagrange polynomial interpolation: numerical scheme

39

Figure 2.22 Numerical simulation for Dadras–Momeni attractor with the Caputo derivative for α = 0.98.

and the Atangana Baleanu fractional derivative in the sense of Riemann–Liouville of the function u (t) is as follows:

 t α AB (α) d ABR α α (2.112) Dt u (t) = u (τ ) Eα − (t − τ ) dτ. 0 1 − α dt 0 1−α α Here, u (t) ∈ W21 (0, l), α ∈ [0, 1] and AB (α) = 1 − α + (α) . The fractional integral with the Mittag-Leffler kernel is represented by the following equality;  t α 1−α ABC α u + I u (τ ) (t − τ )α−1 dτ. (2.113) = (t) (u (t)) t 0 AB (α) AB (α) (α) 0

Let us consider the Cauchy problem with the Mittag-Leffler kernel  ABC α Dt u (t) = h (t, u (t)) , 0 u (0) = u0 .

(2.114)

We transform Eq. (2.114) into u (t) − u (0) =

α 1−α u (t) + AB (α) AB (α) (α)



t 0

h (τ, u (τ )) (t − τ )α−1 dτ. (2.115)

40

New Numerical Scheme With Newton Polynomial

At the point tk+1 = (k + 1) t, we have the following: 1−α h (t, u (t)) AB (α)  tk+1 α + h (τ, u (τ )) (tk+1 − τ )α−1 dτ, AB (α) (α) 0

u (tk+1 ) − u (0) =

(2.116)

and we write  1−α  h tk , uk AB (α) k  tm+1  α h (τ, u (τ )) (tk+1 − τ )α−1 dτ. + AB (α) (α) tm

u (tk+1 ) = u (0) +

(2.117)

m=0

After putting its Lagrange polynomial into Eq. (2.117) as the approximation of the function h (τ, u (τ )), the above equation can be written as follows:  α 1−α  h tk , uk + AB (α) AB (α) (α)  k  tm+1 h(tm ,um )  (τ − tm−1 )  t

uk+1 = u0 + ×

m=0 tm

−

h tm−1 ,um−1 t

(τ − tm )

(2.118) (tk+1 − τ )α−1 dτ.

Thus, we get  1−α  h tk , uk AB (α) ⎧ tm+1 h(t ,um ) m (τ − tm−1 ) ⎪ tm t ⎪ ⎪ k ⎨  − τ × (tk+1 )α−1 dτ α +

tm+1 htm−1 ,um−1  ⎪ − AB (α) (α) (τ − tm ) tm m=0 ⎪ t ⎪ ⎩ α−1 dτ × (tk+1 − τ )

uk+1 = u0 +

(2.119) ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

and uk+1 = u0 +  ×

k   h (tm , um ) α 1−α  h tk , uk + AB (α) AB (α) (α) t m=0

tm+1

tm

(τ − tm−1 ) (tk+1 − τ )α−1 dτ

  k  h tm−1 , um−1 α − AB (α) (α) t m=0  tm+1 × (τ − tm ) (tk+1 − τ )α−1 dτ. tm

(2.120)

Two-steps Lagrange polynomial interpolation: numerical scheme

41

We can calculate the above integrals in Eq. (2.120) as follows: 

tm+1

(τ − tm−1 ) (tk+1 − τ )α−1 dτ

tm

(t)α+1 (k − m + 1)α (k − m + 2 + α) = α α (α + 1) − (k − m) (k − m + 2 + 2α)  tm+1 (τ − tm ) (tk+1 − τ )α−1 dτ

(2.121)

tm

=

 (t)α+1  (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) . α (α + 1)

Putting these calculations in Eq. (2.120), we get the following numerical approximation: k     α (t)α 1−α  h tk , uk + h tm , um AB (α) AB (α) (α + 2) m=0

α (k − m + 1) (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

uk+1 = u0 +

(2.122)

k    α (t)α h tm−1 , um−1 AB (α) (α + 2) m=0   α+1 × (k − m + 1) − (k − m)α (k − m + 1 + α) .

−

2.5.1 Error analysis with the Mittag-Leffler kernel Now, we give the error analysis to demonstrate the validity of the presented scheme. Provided that the function h (t, u (t)) has a second order derivative, we recall our problem given by 

ABC D α u t (t) = h (t, u (t)) , 0 u (0) = u0 ,

(2.123)

where the operator has non-local and non-singular kernel. Provided that the function has a bounded second derivative, Eq. (2.123) can be converted as follows: 1−α α h (tk , u (tk )) + AB (α) AB (α) (α)  tk+1 × h (τ, u (τ )) (tk+1 − τ )α−1 dτ

u (tk+1 ) − u (0) =

0

1−α α h (tk , u (tk )) + = AB (α) AB (α) (α)

(2.124)

42

New Numerical Scheme With Newton Polynomial

×

k  

tm+1

h (τ, u (τ )) (tk+1 − τ )α−1 dτ

m=0 tm

= ×

1−α α h (tk , u (tk )) + AB (α) AB (α) (α)   k t  m+1 P (τ )   m m=0 tm

+

(τ −tm ) τ −tm−1 ∂ 2 2! ∂τ 2



[h (τ, u (τ ))]τ =γm

× (tk+1 − τ )α−1 dτ, α 1−α h (tk , u (tk )) + Ekα + = AB (α) AB (α) (α)  h(t ,um ) tm+1 k α−1 m  dτ tm (τ − tm−1 ) (tk+1 − τ ) t  × , h t ,um−1 tm+1 − m−1t (τ − tm ) (tk+1 − τ )α−1 dτ tm m=0 where Ekα

k  tm+1  α (τ − tm ) (τ − tm−1 ) = AB (α) (α) 2! tm

(2.125)

m=0

×

∂2 [h (τ, u (τ ))]τ =γm (tk+1 − τ )α−1 dτ. ∂τ 2

One can find a point γm ∈ [tm , tm+1 ] such that k  ∂2 α (γm − tm ) [h (τ, u (τ ))]τ =γm 2 AB (α) (α) 2 ∂τ m=0  tm+1 × (τ − tm−1 ) (tk+1 − τ )α−1 dτ

Ekα =

tm k  α ∂2 (γm − tm ) [h (τ, u (τ ))]τ =γm (t)α+1 2 AB (α) (α) 2 ∂τ m=0

1 (k − m + 1)α (k − m + 2 + α) . × α α (α + 1) − (k − m) (k − m + 2 + 2α)

=

(2.126)

Taking the norm on both sides of Eq. (2.126), we obtain the following inequality:  α E  ≤ k

  2  ∂ α (t)α+2  sup   2 h (τ, u (τ ))  2AB (α) (α + 2) τ ∈ 0,tk+1 ∂τ

 k    (k − m + 1)α (k − m + 2 + α)    ×  − (k − m)α (k − m + 2 + 2α)  . m=0

(2.127)

Two-steps Lagrange polynomial interpolation: numerical scheme

43

Thus, we can calculate the error as follows:  2  ∂   α α (t)α+2  E  ≤ sup   2 h (τ, u (τ )) k  2AB (α) (α + 2) τ ∈ 0,tk+1 ∂τ

(2.128)

  k (k + 4 + 2α) × (k + 1)α − αk α . 2

2.5.2 Numerical illustrations In this section, in order to see the effect of the new differential and integral operators for modeling real world problems, we first present the simulation for some equations. Example 2.21. We consider a classical problem with the Atangana–Baleanu fractional derivative ABC α Dt u (t) = 10t 0

+ 9,

(2.129)

u (0) = 0.1. For such a problem, we have the following scheme: k     α (t)α 1−α  h tk , uk + h tm , um AB (α) AB (α) (α + 2) m=0

α (k − m + 1) (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

uk+1 = u0 +

(2.130)

k    α (t)α h tm−1 , um−1 AB (α) (α + 2) m=0   α+1 × (k − m + 1) − (k − m)α (k − m + 1 + α) ,

−

  where h (tm , um ) = 10tm + 9 and h tm−1 , um−1 = 10tm−1 + 9. The numerical simulation for the above problem with the Atangana–Baleanu derivative is presented in Fig. 2.23. Example 2.22. We consider the following problem: ABC α Dt u (t) = 2t exp (−2u (t) + 2) , 0

(2.131)

u (0) = 0.2, where the Atangana–Baleanu fractional derivative is used. For the above equation, the following scheme is obtained: uk+1 = u0 +

k     α (t)α 1−α  h tk , uk + h tm , um AB (α) AB (α) (α + 2) m=0

(2.132)

44

New Numerical Scheme With Newton Polynomial

Figure 2.23 Numerical simulation with the Atangana–Baleanu derivative for α = 0.38.

Figure 2.24 Numerical simulation with the Atangana–Baleanu derivative for α = 0.97.

×

(k − m + 1)α (k − m + 2 + α) − (k − m)α (k − m + 2 + 2α)



k    α (t)α h tm−1 , um−1 AB (α) (α + 2) m=0   α+1 × (k − m + 1) − (k − m)α (k − m + 1 + α) ,

−

  where h (tm , um ) = 2tm exp (−2um + 2) and h tm−1 , um−1 = 2tm−1 exp(−2um−1 + 2). For the above problem, the numerical simulation is depicted in Fig. 2.24. Example 2.23. We consider the following linear problem with the Atangana–Baleanu derivative: ABC α Dt u (t) = t 4 0

u (0) = 1.

+ 3t 3 − 1,

(2.133)

Two-steps Lagrange polynomial interpolation: numerical scheme

45

Figure 2.25 Numerical simulation with the Atangana–Baleanu derivative for α = 0.67.

For the above equation, the following scheme is obtained: k     α (t)α 1−α  h tk , uk + h tm , um AB (α) AB (α) (α + 2) m=0

(k − m + 1)α (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

uk+1 = u0 +

(2.134)

k    α (t)α − h tm−1 , um−1 AB (α) (α + 2) m=0   α+1 × (k − m + 1) − (k − m)α (k − m + 1 + α) ,

  3 4 + 3t 3 − 1 and h t m−1 = t 4 where h (tm , um ) = tm m−1 , u m m−1 + 3tm−1 − 1. The numerical simulation for the above problem is given in Fig. 2.25. Example 2.24. Let us consider the following test problem having a Mittag-Leffler kernel: ABC α Dt u (t) = 7.2u (t) + 2, 0

(2.135)

u (0) = 1. The above equation can be solved by the following scheme: k     α (t)α 1−α  h tk , uk + h tm , um AB (α) AB (α) (α + 2) m=0

α (k − m + 1) (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

uk+1 = u0 +

−

k    α (t)α h tm−1 , um−1 AB (α) (α + 2) m=0

(2.136)

46

New Numerical Scheme With Newton Polynomial

Figure 2.26 Numerical simulation with the Atangana–Baleanu derivative for α = 0.91.

  × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,   where h (tm , um ) = 7.2um + 2 and h tm−1 , um−1 = 7.2um−1 + 2. The numerical simulation is given in Fig. 2.26. Example 2.25. We consider the following Hadley attractor problem including the Atangana–Baleanu fractional derivative: ABC α Dt Q1 (t) = −Q22 − Q23 − aQ1 + aζ, 0 ABC α Dt Q2 (t) = Q1 Q2 − bQ1 Q3 − Q2 + δ, 0 ABC α Dt Q3 (t) = bQ1 Q2 + Q1 Q3 − Q3 , 0

(2.137)

where the initial conditions are as follows: Q1 (0) = −0.1, Q2 (0) = −0.1, Q3 (0) = −0.1.

(2.138)

The parameters are chosen as a = 0.2, b = 4, ζ = 8, δ = 1. For the above equation, the following scheme is obtained:   1−α 0 k k k h t (2.139) = Q + , Q , Q , Q Qk+1 1 k 1 1 2 3 1 AB (α) k    α (t)α m m h1 tm , Qm 1 , Q2 , Q3 AB (α) (α + 2) m=0

(k − m + 1)α (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

+

k    α (t)α m−1 m−1 h1 tm−1 , Qm−1 , Q , Q 1 2 3 AB (α) (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

Two-steps Lagrange polynomial interpolation: numerical scheme

Qk+1 = Q02 + 2

47

  1−α h2 tk , Qk1 , Qk2 , Qk3 AB (α)

k    α (t)α m m h2 tm , Qm 1 , Q2 , Q3 AB (α) (α + 2) m=0

(k − m + 1)α (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

+

k    α (t)α m−1 m−1 h2 tm−1 , Qm−1 , Q , Q 1 2 3 AB (α) (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

Qk+1 = Q03 + 3

  1−α h3 tk , Qk1 , Qk2 , Qk3 AB (α)

k    α (t)α m m + h3 tm , Qm 1 , Q2 , Q3 AB (α) (α + 2) m=0

(k − m + 1)α (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α) k    α (t)α − h3 tm−1 , Qm−1 , Qm−1 , Qm−1 1 2 3 AB (α) (α + 2) m=0   α+1 × (k − m + 1) − (k − m)α (k − m + 1 + α) ,

where    m 2  m 2 m m − Q3 − aQm h1 tm , Qm 1 , Q2 , Q3 = − Q2 1 + aζ,   m m m m m m m m h2 tm , Q1 , Q2 , Q3 = Q1 Q2 − bQ1 Q3 − Q2 + δ,   m m m m m m m h3 tm , Qm 1 , Q2 , Q3 = bQ1 Q2 + Q1 Q3 − Q3 ,

(2.140)

and    2  2 = − Qm−1 h1 tm−1 , Qm−1 , Qm−1 , Qm−1 − Qm−1 − aQ1m−1 + aζ, 1 2 3 2 3   m−1 m−1 = Qm−1 h2 tm−1 , Qm−1 , Q , Q Qm−1 − bQm−1 Qm−1 − Qm−1 + δ, 1 2 3 1 2 1 3 2 



(2.141)

h3 tm−1 , Qm−1 = bQm−1 , Qm−1 , Qm−1 Qm−1 + Qm−1 Qm−1 − Qm−1 . 1 2 3 1 2 1 3 3 To test the effectiveness of the proposed scheme, the numerical simulations are provided in Fig. 2.27 and Fig. 2.28.

48

New Numerical Scheme With Newton Polynomial

Figure 2.27 Numerical simulation with the Atangana–Baleanu derivative for α = 0.99.

2.6

Differential equation with fractal–fractional with power-law kernel

The concept of fractal–fractional differentiation has appeared as a combination of two mathematical concepts which are fractal differentiation and fractional differentiation [14]. Besides, these operators are helpful for modeling more complex problems, also they allow us to understand many physical problems which have fractal properties. We assume that u (t) ∈ W21 (0, l), α ∈ [0, 1]. If the function u (t) is continuous and this function is fractal differentiable on the interval (a, b) with order β, then the fractal–fractional derivative of the function u (t) is defined by the following formula;  t d 1 F F P α,β D (2.142) = (u (t)) (t − τ )n−α−1 u (τ ) dτ t 0 (n − α) dt β 0 where the integral has a power-law type kernel and n − 1 < α, β ≤ n ∈ N. If the function u (t) is continuous on the interval (a, b), then the fractal–fractional integral of u (t) is of the form of  t β F F P α,β J τ β−1 u (τ ) (t − τ )α−1 dτ. (2.143) = (u (t)) t 0 (α) 0

Two-steps Lagrange polynomial interpolation: numerical scheme

49

Figure 2.28 Numerical simulation with the Atangana–Baleanu derivative for α = 1.

Here, the integral has power-law kernel. In this section, to present numerical scheme, we consider for a general Cauchy problem with the Caputo fractal–fractional derivative  F F P D α,β u (t) = h (t, u (t)) , t 0 (2.144) u (0) = u0 . We transform Eq. (2.144) into  t β τ β−1 h (τ, u (τ )) (t − τ )α−1 dτ. u (t) − u (0) = (α) 0 At the point tk+1 = (k + 1) t, we have the following:  tk+1 1 u (tk+1 ) − u (0) = g (τ, u (τ )) (tk+1 − τ )α−1 dτ (α) 0

(2.145)

(2.146)

where g (t, u (t)) = βt β−1 h (t, u (t)). Then, we write u (tk+1 ) = u (0) +

k  1  tm+1 g (τ, u (τ )) (tk+1 − τ )α−1 dτ. (α) tm m=0

(2.147)

50

New Numerical Scheme With Newton Polynomial

After putting the Lagrange polynomial into Eq. (2.147), the above equation can be written as follows:  g(t ,um ) k  m (τ − tm−1 ) 1  tm+1 k+1 t  = u0 + u (tk+1 − τ )α−1 dτ. (2.148) g tm−1 ,um−1 (α) − − t t (τ ) m m=0 m t Thus, we have

uk+1 = u0 +

⎧ ⎪ ⎪ ⎪ ⎨

tm+1

g(tm ,um ) t

(τ − tm−1 ) − τ )α−1 dτ × (tk+1  

tm

⎫ ⎪ ⎪ ⎪ ⎬

k 1  m−1

⎪ − tm+1 g tm−1 ,u (α) (τ − tm ) ⎪ ⎪ tm m=0 ⎪ ⎪ ⎪ t ⎭ ⎩ α−1 dτ × (tk+1 − τ )

(2.149)

and we get  k 1  g (tm , um ) tm+1 (τ − tm−1 ) (tk+1 − τ )α−1 dτ (2.150) (α) t t m m=0  t  k m−1 m+1 1  g tm−1 , u + (τ − tm ) (tk+1 − τ )α−1 dτ. (α) t tm

uk+1 = u0 +

m=0

We know that the following calculations for the above integrals in Eq. (2.150) are valid:  tm+1 (τ − tm−1 ) (tk+1 − τ )α−1 dτ tm

= 

(t)α+1 α (α + 1) tm+1



(k − m + 1)α (k − m + 2 + α) − (k − m)α (k − m + 2 + 2α)

(τ − tm ) (tk+1 − τ )α−1 dτ =

tm



(t)α+1 α (α + 1)

(2.151)

 × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) . 

With these calculations, we have the following: uk+1 = u0 + ×

k  (t)α   g tm , um (α + 2) m=0

(k − m + 1)α (k − m + 2 + α) − (k − m)α (k − m + 2 + 2α)

(2.152)

k  (t)α   g tm−1 , um−1 (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) .

+

Two-steps Lagrange polynomial interpolation: numerical scheme

51

Replacing the function g (t, u (t)) by its value, then we can write the following numerical scheme: k  β (t)α  β−1  tm h tm , um (α + 2) m=0

α (k − m + 1) (k − m + 2 + α) × − (k − m)α (k − m + 2 + 2α)

uk+1 = u0 +

(2.153)

k  β (t)α  β−1  tm−1 h tm−1 , um−1 (α + 2) m=0   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) .

+

2.6.1 Error analysis with the Caputo fractal–fractional derivative Let us address the following general Cauchy problem with the fractal–fractional differential operator: F F P α,β Dt u (t) = h (t, u (t)) . 0

(2.154)

At the point t = tk+1 , after applying the Caputo fractal–fractional integral operator, we have  tk+1 β u (tk+1 ) − u (0) = τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ (2.155) (α) 0 k  β  tm+1 β−1 = τ (tk+1 − τ )α−1 h (τ, u (τ )) dτ. (α) tm m=0

We can approximate the function h (τ, u (τ )) within the interval [tm , tm+1 ] using the Lagrange polynomial as follows: h (τ, u (τ )) = Pm (τ ) + E (τ ) =

h (tm , u (tm )) (τ − tm−1 ) t

(2.156)

h (tm−1 , u (tm−1 )) (τ − tm ) t (τ − tm ) (τ − tm−1 ) ∂ 2 + [h (τ, u (τ ))]τ =γm . 2! ∂τ 2 −

Therefore, the error can be evaluated as Eτα (ετ ) =

k  β  tm+1 β−1 (τ − tm ) (τ − tm−1 ) τ (α) 2! tm m=0

×

∂2 [h (τ, u (τ ))]τ =γm (tk+1 − τ )α−1 dτ ∂τ 2

(2.157)

52

New Numerical Scheme With Newton Polynomial

and we have   α   E (ετ ) =  τ 

β (α)

k m=0

  β−1 ∂ 2 [h (τ, u (τ ))] τ  τ =γ 2 m tm ∂τ   × (tk+1 − τ )α−1 dτ

tm+1

k β  (τ − tm ) (τ − tm−1 ) sup   (α) 2! m=0 τ ∈ tm ,tm+1  2  t ∂  m+1 β−1 ×  2 [h (τ, u (τ ))]τ =γm  τ (tk+1 − τ )α−1 dτ ∂τ tm

≤

(2.158)

k β  (τ − tm ) (τ − tm−1 ) sup   (α) 2! m=0 τ ∈ tm ,tm+1  2  t ∂  m+1 β−1 ×  2 [h (τ, u (τ ))]τ =γm  τ (tk+1 − τ )α−1 dτ. ∂τ tm

≤

We know that the following inequality for the integral on the right hand side of Eq. (2.158) is valid:  t   m+1 β−1  α+β−1 α−1  τ dτ  < 2tk+1 B (α, β) (2.159) (tk+1 − τ )  tm

< 2 ((k + 1) t)α+β−1 B (α, β) . Therefore, we present the following error:  α  E (ετ ) < τ

  k  β  (τ − tm ) (τ − tm−1 )  ∂ 2  u [h (τ, (τ ))] τ =γm   2 (α) 2! ∂τ m=0

× 2 ((k + 1) t)α+β−1 B (α, β) 2β ((k + 1) t) (α)

α+β−1


0, β ≤ n ∈ N and M (0) = M (1) = 1. The fractal–fractional integral of u (t) with order α having exponentially decaying type kernel is given by F F E α,β Jt 0

(u (t)) =

αβ M (α)

 0

t

τ β−1 u (τ ) dτ +

β (1 − α) t β−1 u (t) . M (α)

(2.178)

We now present the following Cauchy problem with new exponential decay kernel: F F E α,β Dt u (t) = h (t, u (t)) . 0

(2.179)

Two-steps Lagrange polynomial interpolation: numerical scheme

59

Figure 2.34 Numerical simulation of the Halvorsen attractor having the Caputo fractal–fractional derivative for α = 0.98, β = 0.98.

Using the definition of the Caputo–Fabrizio fractal–fractional integral, we can convert the above equation into  t 1 − α β−1 αβ βt u (t) − u (0) = h (t, u (t)) + τ β−1 h (τ, u (τ )) dτ. (2.180) M (α) M (α) 0 Here we shall take as g (τ, u (τ )) = βτ β−1 h (τ, u (τ )) and write Eq. (2.180) at the point tk+1 = (k + 1) t  tk+1 1−α α g (tk , u (tk )) + u (tk+1 ) − u (0) = g (τ, u (τ )) dτ (2.181) M (α) M (α) 0 and at the point tk = kt u (tk ) − u (0) =

1−α α g (tk−1 , u (tk−1 )) + M (α) M (α)



tk

g (τ, u (τ )) dτ.

(2.182)

0

Subtracting Eq. (2.182) from Eq. (2.181), we have the following: u (tk+1 ) − u (tk ) =

1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)

(2.183)

60

New Numerical Scheme With Newton Polynomial

+

α M (α)



tk+1

g (τ, u (τ )) dτ. tk

If we replace two step Lagrange polynomial into Eq. (2.183), we can get the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)  tk+1  g(tk ,u(tk )) − t (τ ) α k−1   t  dτ, + g t ,u t M (α) tk − k−1 t k−1 (τ − tk )

uk+1 − uk =

(2.184)

and write 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)  g(t ,u(t )) tk+1 k k α t  tk (τ − tk−1 ) dτ  . +

M (α) − g tk−1 ,u tk−1 ttk+1 (τ − tk ) dτ t

uk+1 − uk =

(2.185)

k

The integrals on the right hand side of the above equation can be calculated as  tk+1 3 (2.186) (τ − tk−1 ) dτ = (t)2 , 2 tk  tk+1 1 (τ − tk ) dτ = (t)2 . 2 tk Putting these calculations into Eq. (2.185), we have the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)    3 1  α g tk , uk t − g tk−1 , uk−1 t . + M (α) 2 2

uk+1 = uk +

(2.187)

If we replace g (t, u (t)) by its value, we get the following numerical algorithm:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) uk+1 = uk + (2.188) M (α)   3 1   α β−1 β−1 βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t . + M (α) 2 2

2.7.1 Error analysis with the Caputo–Fabrizio fractal–fractional derivative Let us consider a general Cauchy problem with the Caputo–Fabrizio fractal–fractional differential operator F F E α,β Dt u (t) = h (t, u (t)) . 0

(2.189)

Two-steps Lagrange polynomial interpolation: numerical scheme

61

At the point t = tk+1 , if we integrate with the associated fractional integral operator, we can get the following: 1 − α β−1 βt h (tk , u (tk )) (2.190) M (α) k  tk+1 αβ τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ + M (α) 0 1 − α β−1 βt = h (tk , u (tk )) M (α) k k  αβ  tm+1 β−1 + τ (tk+1 − τ )α−1 h (τ, u (τ )) dτ. M (α) tm

u (tk+1 ) − u (0) =

m=0

With the Lagrange polynomial, we can approximate the function h (τ, u (τ )) within the interval [tm , tm+1 ] as follows: h (τ, u (τ )) = Pm (τ ) + E (τ ) =

h (tm , u (tm )) (τ − tm−1 ) t

h (tm−1 , u (tm−1 )) (τ − tm ) t (τ − tm ) (τ − tm−1 ) ∂ 2 + [h (τ, u (τ ))]τ =γm . 2! ∂τ 2 −

(2.191)

Therefore, we have the following equality: Eτα (ετ ) =

k  αβ  tm+1 β−1 (τ − tm ) (τ − tm−1 ) ∂ 2 τ [h (τ, u (τ ))]τ =γm dτ. M (α) 2! ∂τ 2 tm m=0

(2.192) By applying the norm to both sides of the above equation, we get the following inequality;   k  tm+1  2  α   αβ  ∂  β−1 E (ετ ) =  τ [h (τ, u (τ ))]τ =γm dτ  τ 2  M (α)  ∂τ tm m=0

k αβ  (τ − tm ) (τ − tm−1 ) sup   M (α) 2! m=0 τ ∈ tm ,tm+1  2  t ∂  m+1 β−1 ×  2 [h (τ, u (τ ))]τ =γm  τ dτ ∂τ tm

≤

≤

k αβ  (τ − tm ) (τ − tm−1 ) sup   M (α) 2! τ ∈ tm ,tm+1 m=0

(2.193)

62

New Numerical Scheme With Newton Polynomial

 t  2  m+1 β−1 ∂ ×  2 [h (τ, u (τ ))]τ =γm  τ dτ. ∂τ tm We know that the following inequality is valid:  t   m+1 β−1  1    τ dτ  ≤ (m + 1)β − mβ  β tm

(2.194)

since (t)β < 1. Thus, the error can be obtained in the form of   k    α  (τ − tm ) (τ − tm−1 )  ∂ 2  E (ετ ) < αβ u [h (τ, (τ ))] τ =γm  τ  2 M (α) 2! ∂τ

(2.195)

m=0

×

 1 (m + 1)β − mβ β

2α ((k + 1) t)α+β−1 B (α, β)  (τ − tm ) (τ − tm−1 ) M (α) (α) 2! m=0  2  ∂   × sup  2 h (τ, u (τ )) |τ =γm  (m + 1)β − mβ . 0≤τ ≤tk+1 ∂τ k


0, then a fractional derivative of u (t) with order α and fractal variable dimension β (t) is given by F F E α,β(t) Dt 0

M (α) d u (t) = 1 − α dt β(t)

 0

t

  α u (τ ) exp − (t − τ ) dτ 1−α

(2.243)

where γ (t) − γ (τ ) dγ (τ ) = lim β(t) . t→τ t dt β(t) − τ β(τ )

(2.244)

80

New Numerical Scheme With Newton Polynomial

The new fractional integral with exponential decay kernel is defined by

β (t) 1 − α β(t) β (t) ln (t) + t u (t) M (α) t

 t β (τ ) β(τ ) α + τ u (τ ) β (τ ) ln (τ ) + dτ. M (α) 0 τ

F F E α,β(t) It u (t) = 0

(2.245)

In this section, to derive the numerical scheme, we deal with the following new class of Cauchy problems: F F E α,β(t) Dt u (t) = h (t, u (t)) , 0

(2.246)

u (0) = u0 . When taking the new fractional integral with exponential decay kernel, we can rewrite the above equation as

β (t) 1 − α β(t) β (t) ln (t) + t h (t, u (t)) M (α) t

 t α β (τ ) β(τ ) + h (τ, u (τ )) β (τ ) ln (τ ) + dτ. τ M (α) 0 τ

u (t) =

(2.247)

At the point tk+1 = (k + 1) t, we can get the following:

 β (tk )  1 − α β(tk ) β (tk+1 ) − β (tk ) h tk , uk tk ln tk + u (tk+1 ) = M (α) t tk

 tk+1 β (τ ) β(τ ) α + τ h (τ, u (τ )) β (τ ) ln (τ ) + dτ. M (α) 0 τ

(2.248)

For simplicity, we can take

β (τ ) β(τ ) τ g (τ, u (τ )) = h (τ, u (τ )) β (τ ) ln (τ ) + τ

(2.249)

and taking difference of difference of equations at the points tk+1 = (k + 1) t and tk = kt, we have

 β (tk )  1 − α β(tk ) β (tk+1 ) − β (tk ) h tk , uk tk ln tk + u (tk+1 ) = u (tk ) + M (α) t tk (2.250)

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) t − ln tk−1 + M (α) k−1 t tk−1  tk+1   α g (τ, u (τ )) dτ. × h tk−1 , uk−1 + M (α) 0

Two-steps Lagrange polynomial interpolation: numerical scheme

81

Also, we have the following:

 β (tk )  1 − α β(tk ) β (tk+1 ) − β (tk ) h tk , uk tk ln tk + M (α) t tk (2.251)

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) t ln tk−1 + − M (α) k−1 t tk−1  tk+1   α g (τ, u (τ )) dτ. × h tk−1 , uk−1 + M (α) tk

u (tk+1 ) = u (tk ) +

After putting the Lagrange polynomial into Eq. (2.251), the above equation can be reformulated as follows:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  uk+1 = uk + h tk , uk tk ln tk + (2.252) M (α) t tk

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + − M (α) t tk−1    k  g t ,u   k tk+1 α − t (τ ) k−1 k−1   t × h tk−1 , u + dτ, g t ,uk−1 M (α) tk − k−1t (τ − tk ) and we organize the above equation as follows:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk tk ln tk + uk+1 = uk + M (α) t tk

  1 − α β tk−1 β (tk ) − β (tk−1 ) β (tk−1 ) t ln tk−1 + − M (α) k−1 t tk−1    tk+1 k   α g tk , u × h tk−1 , uk−1 + (τ − tk−1 ) dτ M (α) t tk   t k+1 α g tk−1 , uk−1 − (τ − tk ) dτ. M (α) t tk

(2.253)

By calculating the above integrals in Eq. (2.253), we get 

tk+1

tk  tk+1 tk

(τ − tk−1 ) dτ = (τ − tk ) dτ =

3 (t)2 , 2

(2.254)

(t)2 . 2

If we put them into Eq. (2.253), we obtain the following approximation:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  uk+1 = uk + h tk , uk tk ln tk + M (α) t tk

(2.255)

82

New Numerical Scheme With Newton Polynomial

    1 − α β tk−1 β (tk ) − β (tk−1 ) β (tk−1 ) h tk−1 , uk−1 tk−1 ln tk−1 + M (α) t tk−1    3t  t α α g tk , uk g tk−1 , uk−1 + − , M (α) 2 M (α) 2

−

Thus, we have

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk tk (2.256) ln tk + uk+1 = uk + M (α) t tk

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + − M (α) t tk−1

  αβ β(tk ) β (tk+1 ) − β (tk ) β (tk ) k−1 t ln tk + + × h tk−1 , u M (α) k t tk     3t αβ β tk−1 × h tk , uk − t 2 M (α) k−1

 t β (tk ) − β (tk−1 ) β (tk−1 )  h tk−1 , uk−1 ln tk−1 + . × t tk−1 2

2.9.1 Error analysis with fractal–fractional derivative with variable order with exponential decay kernel   and h (t, u (t)) have a bounded Supposing that the functions t β(t) β (t) ln (t) + β(t) t second and first derivatives, let us consider the nonlinear differential equation with exponential decay kernel F F E α,β(t) Dt u (t) = h (t, u (t)) . 0

(2.257)

At the point t = tk+1 , after applying the associated fractional integral operator, we have

β (t) 1 − α β(t) β (t) ln (t) + t h (t, u (t)) (2.258) u (tk+1 ) − u (0) = M (α) t

 t β (τ ) α + h (τ, u (τ )) β (τ ) ln (τ ) + M (α) 0 τ × τ β(τ ) (tk+1 − τ )α−1 dτ. For simplicity, g (t, u (t)) = t

β(t)



β (t) β (t) ln (t) + h (t, u (t)) . t

(2.259)

Then, we have u (tk+1 ) − u (0) =

1−α g (t, u (t)) M (α)

(2.260)

Two-steps Lagrange polynomial interpolation: numerical scheme

+

α M (α)



83

t

g (τ, u (τ )) dτ. 0

We can approximate the function g (τ, u (τ ))within the interval [tk , tk+1 ] using the Lagrange polynomial as follows: g (τ, u (τ )) = Pk (τ ) + E (τ ) =

g (tk , u (tk )) (τ − tk−1 ) t

(2.261)

g (tk−1 , u (tk−1 )) (τ − tk ) t (τ − tk ) (τ − tk−1 ) ∂ 2 + [g (τ, u (τ ))]τ =γk . 2! ∂τ 2 −

Therefore, the error can be evaluated as  tk+1 α (τ − tk ) (τ − tk−1 ) ∂ 2 α Eτ (ετ ) = [g (τ, u (τ ))]τ =γk dτ, M (α) tk 2! ∂τ 2

(2.262)

and we have



tk+1 (τ −tk )τ −tk−1   α    α  tk 2! E (ετ ) =  M(α)  τ  × ∂ 2 [g (τ, u (τ ))]  τ =γk dτ ∂τ 2  2  ∂  α  ≤ sup   2 g (τ, u (τ ))  M (α) τ ∈ 0,tk+1 ∂τ τ =γk  tk+1 (τ − tk ) (τ − tk−1 ) dτ × 2! tk  2  ∂  α 5  ≤ sup g (τ, u (τ )) (t)3 . M (α) τ ∈0,tk+1   ∂τ 2 12 τ =γk

(2.263)

If we replace g (t, u (t)) by its value, the considered numerical scheme satisfies the following error:  α  E (ετ ) ≤ τ

α 5 (t)3 M (α) 12 ⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ×⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(2.264)

    ∂2 β(τ )  ∂τ  h u τ (τ, (τ )) 2       supτ ∈ 0,tk+1  β(τ )   × β (τ ) ln (τ ) + τ    ∂   h (τ, u (τ ))   ∂τ    + supτ ∈0,tk+1    ) ∂ τ β(τ ) β (τ ) ln (τ ) + β(τ   × ∂τ τ  2     ∂ β(τ ) β(τ )  β (τ ) ln (τ ) + τ τ   + supτ ∈0,tk+1   ∂τ 2    ×h u (τ, (τ ))       ∂  β(τ ) β(τ ) τ β (τ ) ln (τ ) + τ   + supτ ∈0,tk+1   ∂τ  ∂   × h (τ, u (τ )) ∂τ

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

84

New Numerical Scheme With Newton Polynomial

Figure 2.47 Numerical simulation with exponential decay kernel for α = 0.68 fractional order and fractal dimension β = exp (−t).

2.9.2 Numerical illustrations In this section, we provide some examples of some differential equations with new fractal–fractional differential operator with the exponential decay kernel. Example 2.41. Let us consider the following problem: F F E α,β(t) Dt u (t) = 9t 0

− 1, u (0) = 0.01,

(2.265)

with fractal–fractional differential operator with exponential decay kernel. For the above problem, the following scheme can be obtained:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk tk ln tk + (2.266) uk+1 = uk + M (α) t tk

  1 − α β tk−1 β (tk ) − β (tk−1 ) β (tk−1 ) tk−1 ln tk−1 + − M (α) t tk−1

  β αβ β (tk ) − β (t ) (t k+1 k) β(tk ) k−1 t ln tk + + × h tk−1 , u M (α) k t tk     3t αβ β t k−1 × h tk , uk − t 2 M (α) k−1

 t β (tk ) − β (tk−1 ) β (tk−1 )  h tk−1 , uk−1 ln tk−1 + , × t tk−1 2     where h tk , uk = 9tk − 1 and h tk−1 , uk−1 = 9tk−1 − 1. For such a problem, the numerical simulation is given in Fig. 2.47. Example 2.42. In this example, we deal with the following problem: F F E α,β(t) Dt u (t) = 1 + exp (−t) , 0

(2.267)

Two-steps Lagrange polynomial interpolation: numerical scheme

85

Figure 2.48 Numerical simulation with exponential decay kernel for α = 0.95 fractional order and fractal dimension β = cos (t).

u (0) = 0, which has a constant fractional order and variable fractal dimension. Thus, the following scheme is given:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk tk ln tk + (2.268) uk+1 = uk + M (α) t tk

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + − M (α) t tk−1

  β αβ − β β (tk ) (t ) (t k+1 k) β(t ) tk k ln tk + × h tk−1 , uk−1 + M (α) t tk     3t αβ β t k−1 × h tk , uk − t 2 M (α) k−1

 t β (tk ) − β (tk−1 ) β (tk−1 )  h tk−1 , uk−1 ln tk−1 + , × t tk−1 2     where h tk , uk = 1 + exp (−tk ) and h tk−1 , uk−1 = 1 + exp (−tk−1 ). We now present the numerical simulation in Fig. 2.48. Example 2.43. We deal with the following test problem with variable fractal order: F F E α,β(t) Dt u (t) = t 3 u (t) , 0

(2.269)

u (0) = 1. Thus, we can present the following scheme:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk tk ln tk + uk+1 = uk + M (α) t tk

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + − M (α) t tk−1

(2.270)

86

New Numerical Scheme With Newton Polynomial

Figure 2.49 Numerical simulation with exponential decay kernel for α = 0.9 fractional order and fractal dimension β = t + 1.



  αβ β(tk ) β (tk+1 ) − β (tk ) β (tk ) tk ln tk + × h tk−1 , uk−1 + M (α) t tk     3t αβ β t k−1 × h tk , uk − t 2 M (α) k−1

 t β (tk ) − β (tk−1 ) β (tk−1 )  h tk−1 , uk−1 ln tk−1 + , × t tk−1 2     3 uk−1 . The numerical simulation for where h tk , uk = tk3 uk and h tk−1 , uk−1 = tk−1 such a problem is presented in Fig. 2.49. Example 2.44. We consider the fractal–fractional differential equation given by F F E α,β(t) Dt u (t) = sin (2t) , 0

(2.271)

u (0) = 0.02. Thus, the following scheme is obtained:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  uk+1 = uk + h tk , uk tk ln tk + (2.272) M (α) t tk

  1 − α β tk−1 β (tk ) − β (tk−1 ) β (tk−1 ) tk−1 − ln tk−1 + M (α) t tk−1

  β αβ β (tk ) − β (t ) (t k+1 k) β(tk ) k−1 t ln tk + × h tk−1 , u + M (α) k t tk     3t αβ β t k−1 t × h tk , uk − 2 M (α) k−1

 t β (tk ) − β (tk−1 ) β (tk−1 )  h tk−1 , uk−1 ln tk−1 + , × t tk−1 2     where h tk , uk = sin (2tk ) and h tk−1 , uk−1 = sin (2tk−1 ). For this equation, the numerical simulation is provided in Fig. 2.50.

Two-steps Lagrange polynomial interpolation: numerical scheme

87

Figure 2.50 Numerical simulation with exponential decay kernel for α = 0.87 fractional order and fractal dimension β = 0.1t.

Example 2.45. Since piecewise-linear nonlinearity is replaced by an appropriate cubic nonlinearity which displays a very similar behavior, the Chua–Hartley system is different from the usual Chua system [41]. We consider the following Chua–Hartley system which contains a new fractal–fractional differential operator with the exponential decay kernel   u − 2u3 F F E α,β(t) , D u = a v + (t) t 0 7 F F E α,β(t) Dt v (t) = u − v + w, 0 α,β(t) FFE Dt w (t) = −bv + sin v, 0

(2.273)

with the initial conditions u (0) = −0.01, v (0) = −0.02, w (0) = −0.03.

(2.274)

Applying the suggested numerical scheme to this example yields

  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) uk+1 = uk + h1 tk , uk , v k , w k tk ln tk + M (α) t tk (2.275)

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) t ln tk−1 + − M (α) k−1 t tk−1   k−1 k−1 k−1 × h1 tk−1 , u , v , w

β (tk ) αβ β(tk ) β (tk+1 ) − β (tk ) t ln tk + + M (α) k t tk     3t αβ β tk−1 t × h1 tk , uk , v k , w k − 2 M (α) k−1

  t β (tk ) − β (tk−1 ) β (tk−1 ) h1 tk−1 , uk−1 , v k−1 , w k−1 ln tk−1 + , × t tk−1 2

88

New Numerical Scheme With Newton Polynomial



  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h2 tk , uk , v k , w k tk ln tk + M (α) t tk

  1 − α β tk−1 β (tk ) − β (tk−1 ) β (tk−1 ) tk−1 ln tk−1 + − M (α) t tk−1   k−1 k−1 k−1 × h2 tk−1 , u , v , w

β (tk ) αβ β(tk ) β (tk+1 ) − β (tk ) t + ln tk + M (α) k t tk     3t αβ β tk−1 − t × h2 tk , uk , v k , w k 2 M (α) k−1

  t β (tk ) − β (tk−1 ) β (tk−1 ) h2 tk−1 , uk−1 , v k−1 , w k−1 × ln tk−1 + , t tk−1 2

vk+1 = vk +



  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) tk ln tk + h3 tk , uk , v k , w k M (α) t tk

  β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + − M (α) t tk−1   k−1 k−1 k−1 × h3 tk−1 , u , v , w

β (tk ) αβ β(tk ) β (tk+1 ) − β (tk ) t ln tk + + M (α) k t tk     3t αβ β tk−1 × h3 tk , uk , v k , w k − t 2 M (α) k−1

  t β (tk ) − β (tk−1 ) β (tk−1 ) ln tk−1 + , × h3 tk−1 , uk−1 , v k−1 , w k−1 t tk−1 2

wk+1 = wk +

where ,  3   uk − 2 uk k k k k h1 tk , u , v , w = a v + , 7   h2 tk , uk , v k , w k = uk − v k + w k ,   h3 tk , uk , v k , w k = −bv k + sin v k ,

(2.276)

and , 3    uk−1 − 2 uk−1 k−1 k−1 k−1 k−1 h1 tk−1 , u , v , w =a v + , 7   h2 tk−1 , uk−1 , v k−1 , w k−1 = uk−1 − v k−1 + w k−1 ,   h3 tk−1 , uk−1 , v k−1 , w k−1 = −bv k−1 + sin v k−1 .

(2.277)

Two-steps Lagrange polynomial interpolation: numerical scheme

89

Figure 2.51 Numerical simulation with exponential decay kernel for α = 0.98 fractional order and fractal dimension β = 0.07 + 0.01t.

With the parameters a = 12.75, b = 100/7, the numerical simulations are given in Fig. 2.51 and Fig. 2.52.

2.10

Differential equation with fractal–fractional derivative with variable order with the Mittag-Leffler kernel

The generalized Mittag-Leffler function has shown to be the fundamental solution of the evolution equation with power-law kernel. Thus, to obtain a new class of differential and integral operators with the crossover property, the fractal–fractional differential operator with generalized Mittag-Leffler function was suggested as the fractal dimension was replaced by a variable dimension [21]. Let u be a differential function. Let α be a constant fractional order such that 0 < α ≤ 1. Let β (t) be a continuous function β (t) > 0, then a fractional derivative of u with order α and fractal variable dimension β (t) is given by    t α AB (α) d F F M α,β(t) α − dτ (2.278) D u = u E − τ (t) (τ ) (t ) α t 0 1 − α dt β(t) 0 1−α

90

New Numerical Scheme With Newton Polynomial

Figure 2.52 Numerical simulation with exponential decay kernel for α = 0.96 fractional order and fractal dimension β = 0.96.

where du (τ ) u (t) − u (τ ) = lim β(t) . β(t) t→τ dt t − τ β(τ ) The new fractional integral with the Mittag-Leffler kernel is defined by

β (t) 1 − α β(t) F F M α,β(t) β ln + t u (t) I u = (t) (t) (t) t 0 AB (α) t  t α + u (τ ) (t − τ )α−1 AB (α) (α) 0

β (τ ) β(τ ) τ × β (τ ) ln (τ ) + dτ. τ

(2.279)

(2.280)

In this section, to present the numerical scheme, and we derive a numerical solution of the Cauchy problem with the new differential operator with the Mittag-Leffler kernel which is given by F F M α,β(t) Dt u (t) = h (t, u (t)) , 0

u (0) = u0 .

(2.281)

Two-steps Lagrange polynomial interpolation: numerical scheme

Integrating the above equation, we can reformulate this equation as

1 − α β(t) β (t) u (t) = β (t) ln (t) + t h (t, u (t)) AB (α) t  t α + h (τ, u (τ )) (t − τ )α−1 AB (α) (α) 0

β (τ ) β(τ ) τ × β (τ ) ln (τ ) + dτ. τ At the point tk+1 = (k + 1) t, we have the following:

β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) u (tk+1 ) − u (0) = t ln tk + AB (α) k t tk    t k+1 α × h tk , uk + h (τ, u (τ )) AB (α) (α) 0

β (τ ) β(τ ) τ × (tk+1 − τ )α−1 β (τ ) ln (τ ) + dτ. τ

91

(2.282)

(2.283)

For convenience, we put

β (τ ) β(τ ) g (τ, u (τ )) = h (τ, u (τ )) β (τ ) ln (τ ) + τ τ

(2.284)

and then we get

1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) u (tk+1 ) = u0 + t ln tk + AB (α) k t tk  tk+1   α × h tk , uk + g (τ, u (τ )) AB (α) (α) 0

(2.285)

× (tk+1 − τ )α−1 dτ. Also, we write

1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) t ln tk + u (tk+1 ) = u0 + AB (α) k t tk  k   tm+1  α × h tk , uk + g (τ, u (τ )) AB (α) (α) tm

(2.286)

m=0

× (tk+1 − τ )α−1 dτ. After putting the Lagrange polynomial into Eq. (2.286), the above equation can be revised as follows:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk (2.287) tk ln tk + uk+1 = u0 + AB (α) t tk

92

New Numerical Scheme With Newton Polynomial k  tm+1  α + AB (α) (α) tm



m=0

× (tk+1 − τ )

α−1

g(tm ,um ) (τ − tm−1 )  t g tm−1 ,um−1 − (τ − tm ) t



dτ.

So, we can write

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk (2.288) tk ln tk + uk+1 = u0 + AB (α) t tk ⎫ ⎧ tm+1 g(tm ,um ) (τ − tm−1 ) ⎪ ⎪ tm ⎪ ⎪ t ⎪ ⎪ k ⎨ ⎬  − τ )α−1 dτ × (tk+1 α   , +

tm+1 g tm−1 ,um−1 ⎪ − AB (α) (α) (τ − tm ) ⎪ ⎪ t m=0 ⎪ ⎪ ⎪ t m ⎭ ⎩ × (tk+1 − τ )α−1 dτ and we can have



 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  uk+1 = u0 + h tk , uk (2.289) tk ln tk + AB (α) t tk  k t m  g (tm , u ) m+1 α + (τ − tm−1 ) AB (α) (α) t tm m=0   k  g tm−1 , um−1 α α−1 × (tk+1 − τ ) dτ − AB (α) (α) t m=0  tm+1 × (τ − tm ) (tk+1 − τ )α−1 dτ. tm

We use 

tm+1

(τ − tm−1 ) (tk+1 − τ )α−1 dτ

tm

(t)α+1 (k − m + 1)α (k − m + 2 + α) = α α (α + 1) − (k − m) (k − m + 2 + 2α)  tm+1 (t)α+1 (τ − tm ) (tk+1 − τ )α−1 dτ = α (α + 1) tm   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) .

(2.290)

We can rewrite the above calculations in equation (2.289), thus, we have

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  uk+1 = u0 + h tk , uk (2.291) tk ln tk + AB (α) t tk

k    (k − m + 1)α (k − m + 2 + α) α (t)α + g tm , um − (k − m)α (k − m + 2 + 2α) AB (α) (α + 2) m=0

Two-steps Lagrange polynomial interpolation: numerical scheme

93

k    α (t)α g tm−1 , um−1 AB (α) (α + 2) m=0   α+1 × (k − m + 1) − (k − m)α (k − m + 1 + α) .

−

We have the following numerical scheme by replacing the function g (t, u (t)) by its value:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  uk+1 = u0 + h tk , uk (2.292) tk ln tk + AB (α) t tk *   + k −β(t β t ) α m m+1  αβ (t) ln t β(tm ) m t + tm AB (α) (α + 2) + β(ttmm ) m=0

 (k − m + 1)α (k − m + 2 + α)  × h tm , um − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k    αβ (t)α ln t β tm−1 m−1 t   ⎦ − tm−1 ⎣ β tm−1 AB (α) (α + 2) + m=0 tm−1    m−1 α+1 × h tm−1 , u − (k − m)α (k − m + 1 + α) . (k − m + 1)

2.10.1 Error analysis with fractal–fractional derivative with variable order with Mittag-Leffler kernel In this section, we give the error analysis to prove the accuracy of the consid  β(t) β (t) ln (t) + β(t) and ered scheme. Here we shall assume that the functions t t h (t, u (t)) have bounded second and first derivatives. Let us consider the following problem:  F F M D α,β(t) u (t) = h (t, u (t)) , t 0 (2.293) u (0) = u0 , where the differential operator has a non-local and non-singular kernel. We have

β (t) 1 − α β(t) u (tk+1 ) − u (0) = β (t) ln (t) + t h (t, u (t)) (2.294) AB (α) t  t α + h (τ, u (τ )) (tk+1 − τ )α−1 AB (α) (α) 0

β (τ ) β(τ ) τ × β (τ ) ln (τ ) + dτ. τ For simplicity, let us define

β (t) h (t, u (t)) . g (t, u (t)) = t β(t) β (t) ln (t) + t

(2.295)

94

New Numerical Scheme With Newton Polynomial

Then, Eq. (2.294) can be reformulated as follows: 1−α α g (tk , u (tk )) + AB (α) AB (α) (α)  tk+1 × g (τ, u (τ )) (tk+1 − τ )α−1 dτ

u (tk+1 ) − u (0) =

(2.296)

0

1−α α g (tk , u (tk )) + = AB (α) AB (α) (α) k  tm+1  × g (τ, u (τ )) (tk+1 − τ )α−1 dτ m=0 tm

= ×

1−α α g (tk , u (tk )) + AB (α) AB (α) (α)   k t  m+1 P (τ )   m m=0 tm

+

(τ −tm ) τ −tm−1 ∂ 2 2! ∂τ 2



[g (τ, u (τ ))]τ =γm

× (tk+1 − τ )α−1 dτ, 1−α α g (tk , u (tk )) + = AB (α) AB (α) (α)  g(t ,um ) tm+1 k α−1 m  dτ tm (τ − tm−1 ) (tk+1 − τ ) t  × g t ,um−1 tm+1 − m−1t (τ − tm ) (tk+1 − τ )α−1 dτ tm m=0 + Ekα , where Ekα =

k  tm+1  α (τ − tm ) (τ − tm−1 ) AB (α) (α) 2! tm

(2.297)

m=0

×

∂2 [g (τ, u (τ ))]τ =γm (tk+1 − τ )α−1 dτ. ∂τ 2

We can find a point γm ∈ [tm , tm+1 ] such that k  ∂2 α (γm − tm ) [g (τ, u (τ ))]τ =γm 2 AB (α) (α) 2 ∂τ m=0  tm+1 × (τ − tm−1 ) (tk+1 − τ )α−1 dτ

Ekα =

tm k  α ∂2 (γm − tm ) [g (τ, u (τ ))]τ =γm (t)α+1 AB (α) (α) 2 ∂τ 2 m=0

1 (k − m + 1)α (k − m + 2 + α) . × α α (α + 1) − (k − m) (k − m + 2 + 2α)

=

(2.298)

Two-steps Lagrange polynomial interpolation: numerical scheme

95

If we take the norm on both sides of Eq. (2.298), we have the following inequality:  α E  ≤ k

 2  ∂  α (t)α+2   sup g u (τ, (τ ))   2AB (α) (α + 2) τ ∈0,tk+1  ∂τ 2

(2.299)

 k    (k − m + 1)α (k − m + 2 + α)    ×  − (k − m)α (k − m + 2 + 2α)  . m=0

Thus, one can evaluate the error as follows:  2   α ∂  α (t)α+2 E  ≤  sup   2 g (τ, u (τ )) k  2AB (α) (α + 2) τ ∈ 0,tk+1 ∂τ

(2.300)

  k (k + 4 + 2α) × (k + 1)α − αk α . 2 If we replace g (t, u (t)) by its value, the error can be obtained;  α  E (ετ ) ≤ τ

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ×⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

  k (k + 4 + 2α) α (t)α+2 (k + 1)α − αk α 2AB (α) (α + 2) 2    ∂2  β(τ )  ∂τ 2 [h (τ, u (τ ))]τ =γj τ     supτ ∈0,tk+1   β(τ )   × β (τ ) ln (τ ) + τ  

  ∂ β ln (τ ) (τ ) ∂ β(τ )     + supτ ∈ 0,tk+1  ∂τ h (τ, u (τ )) ∂τ τ β(τ )  + τ    2   ∂ β(τ ) β(τ )  β (τ ) ln (τ ) + τ τ   + supτ ∈0,tk+1   ∂τ 2    ×h (τ, u (τ ))       ∂ β(τ )  ∂τ τ β(τ ) β (τ ) ln (τ ) + τ h (τ, u (τ ))   + supτ ∈ 0,tk+1   × ∂ h (τ, u (τ )) ∂τ

(2.301) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥  ⎥  ⎥  ⎦  

2.10.2 Numerical illustrations In this section, we present numerical illustrations of some differential equations with the newly introduced differential operator which has the Mittag-Leffler kernel. Example 2.46. We deal with the following problem: F F M α,β(t) Dt u (t) = 3t 0

− 2,

(2.302)

u (0) = 0.2. Thus, the following scheme is obtained:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  uk+1 = u0 + h tk , uk (2.303) tk ln tk + AB (α) t tk

96

New Numerical Scheme With Newton Polynomial

Figure 2.53 Numerical simulation with the Mittag-Leffler kernel for α = 0.98 fractional order and fractal dimension β = 1 − t.

*   + k β tm+1 −β(tm )  αβ (t)α ln tm β(tm ) t + tm AB (α) (α + 2) + β(ttmm ) m=0

 (k − m + 1)α (k − m + 2 + α)  m × h tm , u − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   α  αβ (t) ln t β tm−1 t   m−1 ⎦ − tm−1 ⎣ β tm−1 AB (α) (α + 2) + tm−1 m=0    × h tm−1 , um−1 (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,   where h (tm , um ) = 3tm − 2 and h tm−1 , um−1 = 3tm−1 − 2. For this equation, the numerical simulation is depicted in Fig. 2.53. Example 2.47. Let us consider the following test problem: F F M α,β(t) Dt u (t) = sin (t) , 0

(2.304)

u (0) = 0, which has a constant fractional order and a variable fractal dimension. Applying the suggested numerical scheme to this example yields

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk (2.305) tk ln tk + AB (α) t tk *   + k −β(t β t m) m+1  αβ (t)α ln t β(tm ) m t + tm AB (α) (α + 2) + β(ttmm ) m=0

 (k − m + 1)α (k − m + 2 + α)  × h tm , um − (k − m)α (k − m + 2 + 2α)

uk+1 = u0 +

Two-steps Lagrange polynomial interpolation: numerical scheme

97

Figure 2.54 Numerical simulation with the Mittag-Leffler kernel for α = 0.85 fractional order and fractal dimension β = sin t.

α

k 





⎡

  β(tm )−β tm−1 ln t t   m−1 β tm−1 + tm−1

⎤

αβ (t) β tm−1 ⎦ tm−1 ⎣ AB (α) (α + 2) m=0    × h tm−1 , um−1 (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

−

  where h (tm , um ) = sin (tm ) and h tm−1 , um−1 = sin (tm−1 ). The numerical simulation is given for such an equation in Fig. 2.54. Example 2.48. We consider the following differential equation with the constant fractional order and variable fractal dimension F F M α,β(t) Dt u (t) = exp (t) , 0

(2.306)

u (0) = 0.1. Thus, we get the following scheme for such a problem:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk (2.307) tk ln tk + uk+1 = u0 + AB (α) t tk *   + k β tm+1 −β(tm )  αβ (t)α ln t β(tm ) m t + tm β(tm ) AB (α) (α + 2) + tm m=0

α   − m + 2 + α) − m + 1) (k (k m × h tm , u − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k    αβ (t)α ln t β tm−1 t   m−1 ⎦ tm−1 ⎣ − β tm−1 AB (α) (α + 2) + m=0 tm−1    × h tm−1 , um−1 (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

98

New Numerical Scheme With Newton Polynomial

Figure 2.55 Numerical simulation with the Mittag-Leffler kernel for α = 0.91 fractional order and fractal dimension β = t.

  where h (tm , um ) = exp (tm ) and h tm−1 , um−1 = exp (tm−1 ). The numerical simulation is presented in Fig. 2.55. Example 2.49. We deal with the following problem: F F M α,β(t) Dt u (t) = −u (t) + 6, 0

(2.308)

u (0) = 0. The above problem can be solved numerically by the following scheme:

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  h tk , uk (2.309) tk ln tk + uk+1 = u0 + AB (α) t tk *   + k β tm+1 −β(tm )  αβ (t)α ln t β(tm ) m t + tm β(tm ) AB (α) (α + 2) + t m=0 m

α   − m + 2 + α) − m + 1) (k (k m × h tm , u − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k    αβ (t)α ln t β tm−1 t   m−1 ⎦ − tm−1 ⎣ β tm−1 AB (α) (α + 2) + m=0 tm−1    × h tm−1 , um−1 (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,   where h (tm , um ) = −um + 6 and h tm−1 , um−1 = −um−1 + 6. For this derivative, we give the numerical simulation in Fig. 2.56. Example 2.50. We consider the newly introduced chaotic problem with two circles of equilibrium points given by F F M α,β(t) Dt x (t) = z + sin z, 0

Two-steps Lagrange polynomial interpolation: numerical scheme

99

Figure 2.56 Numerical simulation with the Mittag-Leffler kernel for α = 0.91 fractional order and fractal dimension β = exp (0.09t).

F F M α,β(t) Dt y (t) = z 0



F F M α,β(t) Dt z (t) = x 2 0

 −ay − by 2 − cxz ,

(2.310)

+ y 2 − |x| ,

with the initial conditions x (0) = 0.01, y (0) = 0.02, z (0) = 0.01.

(2.311)

Applying the suggested numerical scheme to this example yields x k+1 = x0 +



  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) tk ln tk + h1 tk , x k , y k , zk AB (α) t tk *

+

(2.312)

+

 β tm+1 −β(tm ) t + β(ttmm ) α

y k+1 = y0 +



  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h2 tk , x k , y k , zk tk ln tk + AB (α) t tk



k  αβ (t)α ln tm β(tm ) tm AB (α) (α + 2) m=0

  (k − m + 1) (k − m + 2 + α) × h1 tm , x m , y m , zm − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k    αβ (t)α ln t β tm−1 m−1 t   ⎦ − tm−1 ⎣ β tm−1 AB (α) (α + 2) + m=0 tm−1   m−1 m−1 m−1 × h1 tm−1 , x ,y ,z   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

100

New Numerical Scheme With Newton Polynomial

*   + k β tm+1 −β(tm )  αβ (t)α ln t β(tm ) m t + tm AB (α) (α + 2) + β(ttmm ) m=0

  (k − m + 1)α (k − m + 2 + α) × h2 tm , x m , y m , zm − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k    αβ (t)α ln t β tm−1 m−1 t   ⎦ tm−1 ⎣ − β tm−1 AB (α) (α + 2) + m=0 tm−1   m−1 m−1 m−1 × h2 tm−1 , x ,y ,z   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α) ,

z

k+1



  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h3 tk , x k , y k , zk tk ln tk + = z0 + AB (α) t tk  *  + k −β(t β t ) α m m+1  αβ (t) ln t β(tm ) m t + tm AB (α) (α + 2) + β(ttmm ) m=0

  (k − m + 1)α (k − m + 2 + α) m m m × h3 tm , x , y , z − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   α  αβ (t) ln t β tm−1 t   m−1 ⎦ − tm−1 ⎣ β tm−1 AB (α) (α + 2) + tm−1 m=0   × h3 tm−1 , x m−1 , y m−1 , zm−1   × (k − m + 1)α+1 − (k − m)α (k − m + 1 + α)

where   h1 tm , x m , y m , zm = zm + sin zm ,      2 h2 tm , x m , y m , zm = zm −ay m − b y m − cx m zm ,    2  2   h3 tm , x m , y m , zm = x m + y m − x m  ,

(2.313)

and   h1 tm−1 , x m−1 , y m−1 , zm−1 = zm−1 + sin zm−1 ,     2  m−1 m−1 m−1 m−1 m−1 m−1 m−1 m−1 h2 tm−1 , x −ay , =z ,y ,z −b y − cx z     2  2    h3 tm−1 , x m−1 , y m−1 , zm−1 = x m−1 + y m−1 − x m−1  .

(2.314)

Two-steps Lagrange polynomial interpolation: numerical scheme

101

Figure 2.57 Numerical simulation with the Mittag-Leffler kernel for α = 1 fractional order and fractal dimension β = 1 − exp (−t).

The numerical simulation is given for this model with the constant parameters a = 4, b = 4.5, c = 1 in Fig. 2.57 and Fig. 2.58 [42].

2.11

Differential equation with fractal–fractional derivative with variable order with power-law kernel

To present the derivation of the numerical solution of a general problem where the fractal–fractional differential operator with constant order and variable fractal dimension occurs, we deal with the general Cauchy problem where the kernel is the power-law kernel. This differential operator is the extension of the fractal–fractional based on the well-known Caputo differential operator [21]. Let u be a differential function. Let α be a constant fractional order such that 0 < α ≤ 1. Let β (t) be continuous function β (t) > 0, then a fractional derivative of u with order α and fractal variable dimension β (t) is given by  t d 1 F F P α,β(t) D u = (2.315) (t) (t − τ )−α u (τ ) dτ t 0 (1 − α) dt β(t) 0

102

New Numerical Scheme With Newton Polynomial

Figure 2.58 Numerical simulation with the Mittag-Leffler kernel for α = 1 fractional order and fractal dimension β = 0.96.

where dγ (τ ) γ (t) − γ (τ ) = lim β(t) . t→τ t dt β(t) − τ β(τ )

(2.316)

The new fractional integral with power-law kernel is given by

F F P α,β(t) It u (t) = 0

1 (α)



t

u (τ ) (t − τ )

0

α−1

β (τ ) β(τ ) τ β (τ ) ln (τ ) + dτ. τ (2.317)

In this section, we give the derivation of a numerical solution of the following problem: F F P α,β(t) Dt u (t) = h (t, u (t)) , 0

u (0) = u0 .

(2.318)

Two-steps Lagrange polynomial interpolation: numerical scheme

103

Applying the new fractional integral with power kernel, we can rewrite the above equation as 1 u (t) = (α)



t

h (τ, u (τ )) (t − τ )

α−1

0

β (τ ) β(τ ) β (τ ) ln (τ ) + τ dτ. (2.319) τ

At the point tk+1 = (k + 1) t, we can have the following: u (tk+1 ) − u (0) =

1 (α)



tk+1

g (τ, u (τ )) (tk+1 − τ )α−1 dτ

(2.320)

0

where

β (τ ) β(τ ) g (τ, u (τ )) = h (τ, u (τ )) β (τ ) ln (τ ) + . τ τ

(2.321)

Then, we have u (tk+1 ) = u0 +

1 (α)



tk+1

g (τ, u (τ )) (tk+1 − τ )α−1 dτ

(2.322)

0

and we write u (tk+1 ) = u0 +

k  1  tm+1 g (τ, u (τ )) (tk+1 − τ )α−1 dτ. (α) tm

(2.323)

m=0

Using the Lagrange polynomial, the above equation can be revised, k  1  tm+1 uk+1 = u0 + (α) tm m=0



g(tm ,um ) (τ − tm−1 )  t g tm−1 ,um−1 − (τ − tm ) t

(tk+1 − τ )α−1 dτ. (2.324)

Thus, we have k 1  uk+1 = u0 + (α)

m=0



tm+1

g(tm ,um ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ tm t

tm+1 g tm−1 ,um−1  − tm (τ − tm ) (tk+1 − τ )α−1 dτ t

(2.325)

and  k 1  g (tm , um ) tm+1 (τ − tm−1 ) (tk+1 − τ )α−1 dτ (2.326) (α) t t m m=0   t k m−1 m+1 1  g tm−1 , u − (τ − tm ) (tk+1 − τ )α−1 dτ. (α) t tm

uk+1 = u0 +

m=0

104

New Numerical Scheme With Newton Polynomial

We put the calculations for the above integrals into Eq. (2.326), and we obtain the following approximation: uk+1 = u0 +

k  (k − m + 1)α (k − m + 2 + α) (t)α   g tm , um − (k − m)α (k − m + 2 + 2α) (α + 2) m=0

(2.327) −

k α 

α (t) (α + 2)

 g tm−1 , um−1



m=0

(k − m + 1)α+1 − (k − m)α (k − m + 1 + α)

.

Thus, if we replace the function g (t, u (t)) by its value, we have the following:

k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) uk+1 = u0 + (2.328) tm ln tm + (α + 2) t tm m=0

  (k − m + 1)α (k − m + 2 + α) m × h tm , u − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   α  β (t) ln t β tm−1 t   m−1 ⎦ − tm−1 ⎣ β tm−1 (α + 2) + tm−1 m=0

  (k − m + 1)α+1 m−1 . × h tm−1 , u − (k − m)α (k − m + 1 + α)

2.11.1 Error analysis with fractal–fractional derivative with variable order with power-law kernel In this section, to calculate the error of the presented scheme, we consider the following problem 

F F P D α,β(t) u (t) = h (t, u (t)) , t 0 u (0) = u0 ,

(2.329)

  where we assume that the functions t β(t) β (t) ln (t) + β(t) and h (t, u (t)) have t bounded second and first derivatives. We have

 t β (τ ) 1 (2.330) u (tk+1 ) − u (0) = h (τ, u (τ )) β (τ ) ln (τ ) + (α) 0 τ × τ β(τ ) (tk+1 − τ )α−1 dτ. For brevity, we take

β (t) h (t, u (t)) . g (t, u (t)) = t β(t) β (t) ln (t) + t

(2.331)

Two-steps Lagrange polynomial interpolation: numerical scheme

105

Then, Eq. (2.330) can be written as follows: u (tk+1 ) − u (0) = =

1 (α) 1 (α)



tk+1

g (τ, u (τ )) (tk+1 − τ )α−1 dτ

0 k  tm+1  m=0 tm

k  1  tm+1 = (α) tm m=0

(2.332)

g (τ, u (τ )) (tk+1 − τ )α−1 dτ 

P (τ )   m (τ −tm ) τ −tm−1 ∂ 2 + [g (τ, u (τ ))]τ =γm 2! ∂τ 2

× (tk+1 − τ )α−1 dτ ⎧ g(tm ,um ) tm+1 (τ − tm−1 ) ⎪ tm ⎪ t ⎪ k ⎨ α−1  − τ dτ × (t 1 k+1  )  = m−1 g t ,u t m−1 m+1 ⎪ − (α) (τ − tm ) tm m=0 ⎪ ⎪ t ⎩ × (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭



+ Ekα ,

where Ekα

k  1  tm+1 (τ − tm ) (τ − tm−1 ) ∂ 2 = [g (τ, u (τ ))]τ =γT (α) 2! ∂τ 2 tm

(2.333)

m=0

× (tk+1 − τ )α−1 dτ. We can choose a point γm ∈ [tm , tm+1 ] such that Ekα

k 1  ∂2 (γm − tm ) = [g (τ, u (τ ))]τ =γm (α) 2 ∂τ 2 m=0  tm+1 × (τ − tm−1 ) (tk+1 − τ )α−1 dτ

(2.334)

tm k 1  ∂2 (γm − tm ) [g (τ, u (τ ))]τ =γm (t)α+1 2 (α) 2 ∂τ m=0

1 (k − m + 1)α (k − m + 2 + α) . × α α (α + 1) − (k − m) (k − m + 2 + 2α)

=

Taking the norm on both sides of Eq. (2.334), we can write the following inequality:  2  k 

 α+2 ∂    (k − m + 1)α (k − m + 2 + α)   α     E  ≤ (t) sup g u (τ, (τ )) k  − (k − m)α (k − m + 2 + 2α)  .  2 (α + 2) 0,tk+1   ∂τ 2 m=0

(2.335)

106

New Numerical Scheme With Newton Polynomial

Thus, we can evaluate the error as follows:  2  α+2  α ∂   k (k + 4 + 2α) E  ≤ (t)  sup   2 g (τ, u (τ )) (k + 1)α − αk α . k  2 (α + 2) 0,tk+1 ∂τ 2 (2.336) When we replace g (t, u (t)) by its value, we get the following error: α+2    α  k (k + 4 + 2α) E (ετ ) ≤ (t) (k + 1)α − αk α τ 2 (α + 2) 2   ⎡  ∂2   ∂τ 2 [h (τ, u (τ ))]τ =γj τ β(τ )      ⎢ sup 0≤τ ≤tk+1  β(τ )  ⎢  × β (τ ) ln (τ ) + τ  ⎢ ⎢ 

 ⎢  ∂ β ln (τ ) (τ ) ⎢ + sup   h (τ, u (τ )) ∂ τ β(τ ) β(τ ) 0≤τ ≤tk+1  ∂τ ⎢  ∂τ + ⎢ τ    2  ×⎢  ∂ β(τ ) β(τ )  ⎢ β (τ ) ln (τ ) + τ τ   ⎢ + sup0≤τ ≤tk+1  ∂τ 2  ⎢   ×h u (τ, (τ )) ⎢      ⎢   ∂ β(τ ) β(τ ) β (τ ) ln (τ ) + τ ⎣   ∂τ τ + sup0≤τ ≤tk+1     ×h (τ, u (τ ))

(2.337) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

2.11.2 Numerical illustrations In this section, we present numerical illustrations of some differential equations with the new differential operator having power-law kernel. Example 2.51. We consider the following problem: F F P α,β(t) Dt u (t) = 13 − t 2 , 0

(2.338)

u (0) = 5. Thus, the following scheme is given:

k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) ln tm + (2.339) uk+1 = u0 + tm (α + 2) t tm m=0

  (k − m + 1)α (k − m + 2 + α) × h tm , um − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   α  β (t) ln t β tm−1 t   m−1 ⎦ tm−1 ⎣ − β tm−1 (α + 2) + tm−1 m=0

  (k − m + 1)α+1 × h tm−1 , um−1 , − (k − m)α (k − m + 1 + α)

Two-steps Lagrange polynomial interpolation: numerical scheme

107

Figure 2.59 Numerical simulation with power-law kernel for α = 0.9 fractional order and fractal dimension β = 1 − cos (t).

  2 and h t m−1 = 13 − t 2 where h (tm , um ) = 13 − tm m−1 , u m−1 . The numerical simulation is depicted in Fig. 2.59. Example 2.52. We consider the following problem with a constant fractional order and variable fractal dimension: F F P α,β(t) Dt u (t) = log (t) + (u (t))2 , 0

(2.340)

u (0) = 0.002. Thus, the above problem can be solved with the following scheme:

k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) (2.341) ln tm + tm (α + 2) t tm m=0

  (k − m + 1)α (k − m + 2 + α) × h tm , um − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   β (t)α  β tm−1 ⎣ ln t m−1 t   ⎦ − tm−1 β tm−1 (α + 2) + m=0 tm−1

  (k − m + 1)α+1 , × h tm−1 , um−1 − (k − m)α (k − m + 1 + α)

uk+1 = u0 +

 2   where h (tm , um ) = log (tm ) + (um )2 and h tm−1 , um−1 = log (tm−1 ) + um−1 . For such an equation, the numerical simulation is given in Fig. 2.60. Example 2.53. We consider the following test problem F F P α,β(t) Dt u (t) = 1/ (u (t))2 , 0

u (0) = 3.

(2.342)

108

New Numerical Scheme With Newton Polynomial

Figure 2.60 Numerical simulation with power-law kernel for α = 0.85 fractional order and fractal dimension β = exp (−t).

Figure 2.61 Numerical simulation with power-law kernel for α = 0.77 fractional order and fractal dimension β = t 2 .

Thus, the following scheme is given:

k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) (2.343) ln tm + tm (α + 2) t tm m=0

  (k − m + 1)α (k − m + 2 + α) × h tm , um − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   β (t)α  β tm−1 ⎣ ln t m−1 t   ⎦ − tm−1 β tm−1 (α + 2) + m=0 tm−1

  (k − m + 1)α+1 , × h tm−1 , um−1 − (k − m)α (k − m + 1 + α)

uk+1 = u0 +

   2 where h (tm , um ) = 1/ (um )2 and h tm−1 , um−1 = 1/ um−1 . The numerical simulation is given in Fig. 2.61.

Two-steps Lagrange polynomial interpolation: numerical scheme

109

Figure 2.62 Numerical simulation with power-law kernel for α = 0.97 fractional order and fractal dimension β = 1 + t.

Example 2.54. We consider the following differential equation: F F P α,β(t) Dt u (t) = 2 sin (3u (t)) + t 2 , 0

(2.344)

u (0) = 2. For the above problem, we can get the following scheme:

k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) (2.345) ln tm + uk+1 = u0 + tm (α + 2) t tm m=0

  (k − m + 1)α (k − m + 2 + α) m × h tm , u − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   α  β (t) ln t β tm−1 t   m−1 ⎦ tm−1 ⎣ − β tm−1 (α + 2) + tm−1 m=0

  (k − m + 1)α+1 m−1 , × h tm−1 , u − (k − m)α (k − m + 1 + α)     2 and h t m−1 = 2 sin 3um−1 + t 2 where h (tm , um ) = 2 sin (3um ) + tm m−1 , u m−1 . We present the numerical simulation in Fig. 2.62. Example 2.55. The Rucklidge model defines the convection problem of Boussinesq fluid in its horizontal layer with lateral constants and describes the convection at the point where its chaotic solutions appear. We consider the following Rucklidge system: F F P α,β(t) Dt Q1 (t) = −kQ1 + aQ2 0 F F P α,β(t) Dt Q2 (t) = Q1 , 0 α,β(t) FFP Dt Q3 (t) = −Q3 + Q22 , 0

− Q2 Q3 , (2.346)

110

New Numerical Scheme With Newton Polynomial

where the fractional derivative is constant and the fractal is variable. The initial conditions are given by Q1 (0) = 0.01, Q2 (0) = 0.02, Q3 (0) = 0.01.

(2.347)

Applying the suggested numerical scheme to this example yields Qk+1 1

= Q01



k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) ln tm + + tm (α + 2) t tm m=0

  m m × h1 tm , Qm 1 , Q2 , Q3





(k − m + 1) (k − m + 2 + α) − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   β (t)α  β tm−1 ⎣ ln t t   m−1 ⎦ tm−1 − β tm−1 (α + 2) + m=0 tm−1

  (k − m + 1)α+1 m m m , × h1 tm , Q1 , Q2 , Q3 − (k − m)α (k − m + 1 + α) α



k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) ln tm + tm (α + 2) t tm m=0

  (k − m + 1)α (k − m + 2 + α) m m × h2 tm , Qm , Q , Q 1 2 3 − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   β (t)α  β tm−1 ⎣ ln t m−1 t   ⎦ − tm−1 β tm−1 (α + 2) + m=0 tm−1

  (k − m + 1)α+1 m m × h2 tm , Qm , Q , Q 1 2 3 − (k − m)α (k − m + 1 + α)

Qk+1 = Q02 + 2



k β (t)α  β(tm ) β (tm+1 ) − β (tm ) β (tm ) ln tm + tm (α + 2) t tm m=0

  (k − m + 1)α (k − m + 2 + α) m m × h3 tm , Qm , Q , Q 1 2 3 − (k − m)α (k − m + 2 + 2α) ⎡ ⎤   β(tm )−β tm−1 k   β (t)α  β tm−1 ⎣ ln t m−1 t   ⎦ tm−1 − β tm−1 (α + 2) + m=0 tm−1

  (k − m + 1)α+1 m m × h3 tm , Qm , Q , Q 1 2 3 − (k − m)α (k − m + 1 + α)

Qk+1 = Q03 + 3

where

(2.348)

Two-steps Lagrange polynomial interpolation: numerical scheme

111

Figure 2.63 Numerical simulation for the Rucklidge model with power-law kernel for α = 0.99 fractional order and fractal dimension β = 0.08 + 0.01t.

  m m m m m m h1 tm , Qm 1 , Q2 , Q3 = −kQ1 + aQ2 − Q2 Q3 ,   m m m h2 tm , Qm 1 , Q2 , Q3 = Q1 ,    m 2 m m m h3 tm , Qm , 1 , Q2 , Q3 = −Q3 + Q2

(2.349)

and   m−1 m−1 = −kQm−1 h1 tm−1 , Qm−1 , Q , Q + aQm−1 − Qm−1 Qm−1 , 1 2 3 1 2 2 3   h2 tm−1 , Qm−1 = Qm−1 , Qm−1 , Qm−1 , (2.350) 1 2 3 1   2  h3 tm−1 , Qm−1 = −Qm−1 , Qm−1 , Qm−1 + Qm−1 . 1 2 3 3 2 We provide the numerical simulations with the parameters a = 6.7, k = 2 in Fig. 2.63 and Fig. 2.64 [43].

112

New Numerical Scheme With Newton Polynomial

Figure 2.64 Numerical simulation for the Rucklidge model with power-law kernel for α = 0.97 fractional order and fractal dimension β = 0.95.

Newton interpolation: introduction of the scheme for classical calculus

3

Since models may be non-linear, solutions of those models sometimes cannot be obtained by using analytical methods. In cases where analytical methods are insufficient, we need to solve such models numerically. So, we aim to obtain a new numerical scheme that will be useful for solving linear and non-linear differential and integral equations. One can find much research claiming that a Newton polynomial is more accurate than a Lagrange polynomial. That is why the proposed numerical scheme is established with a Newton polynomial, which no one has used until recently [22]. To construct a numerical scheme based on a Newton polynomial for the general Cauchy problem, we start with the classical derivative du (t) = h (t, u (t)) dt

(3.1)

with the condition u (0) = u0 .

(3.2)

For Eq. (3.1), the function h (t, u (t)) is non-linear. We integrate the above equation to find 

t

u (t) − u (0) =

h (τ, u (τ )) dτ.

(3.3)

0

At the point tk+1 = (k + 1) t 

tk+1

u (tk+1 ) − u (0) =

h (τ, u (τ )) dτ

(3.4)

0

and at the point tk = kt, we get  u (tk ) − u (0) =

tk

h (τ, u (τ )) dτ.

(3.5)

0

If we take the difference of Eqs. (3.4) and (3.5), we obtain the following:  u (tk+1 ) − u (tk ) =

tk+1

h (τ, u (τ )) dτ. tk

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00009-3 Copyright © 2021 Elsevier Inc. All rights reserved.

(3.6)

114

New Numerical Scheme With Newton Polynomial

As the approximation of the function h (t, u (t)), we can now use the Newton polynomial which is given by Pk (τ ) = h (tk−2 , u (tk−2 )) +

h (tk−1 , u (tk−1 )) − h (tk−2 , u (tk−2 )) t

× (τ − tk−2 ) h (tk , u (tk )) − 2h (tk−1 , u (tk−1 )) + h (tk−2 , u (tk−2 )) + 2 (t)2 × (τ − tk−2 ) (τ − tk−1 ) .

(3.7)

If we put the Newton polynomial into Eq. (3.7), we have the following:  k+1

u

− u = h (tk−2 , u (tk−2 )) t + k

tk+1

tk

h (tk−1 , u (tk−1 )) − h (tk−2 , u (tk−2 )) t (3.8)

× (τ − tk−2 ) dτ  tk+1 h (tk , u (tk )) − 2h (tk−1 , u (tk−1 )) + h (tk−2 , u (tk−2 )) + 2 (t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ. If we order the above equation, we can write

k+1

u

      h tk−1 , uk−1 − h tk−2 , uk−2 k−2 t + − u = h tk−2 , u t  tk+1 × (τ − tk−2 ) dτ k

(3.9)

tk

      h tk , uk − 2h tk−1 , uk−1 + h tk−2 , uk−2

+

 ×

2 (t)2 tk+1

(τ − tk−2 ) (τ − tk−1 ) dτ.

tk

For the above integrals, we have the following calculations;  

tk+1

(τ − tk−2 ) dτ =

tk tk+1 tk

(τ − tk−2 ) (τ − tk−1 ) dτ =

5 (t)2 , 2 23 (t)3 . 6

(3.10)

Newton interpolation: introduction of the scheme for classical calculus

115

Replacing them into Eq. (3.10), we have the following:       5  uk+1 = uk + h tk−2 , uk−2 t + h tk−1 , uk−1 − h tk−2 , uk−2 t 2 (3.11)      23   t + h tk , uk − 2h tk−1 , uk−1 + h tk−2 , uk−2 12 and we can reorder uk+1 = uk +

3.1

   4  5  23  h tk , uk t − h tk−1 , uk−1 t + h tk−2 , uk−2 t. (3.12) 12 3 12

Error analysis with classical derivative

Let the following be a general Cauchy problem: u (t) = h (t, u (t)) .

(3.13)

At the point t = tk+1 , after applying the associated fractional integral operator, we have  tk+1 h (τ, u (τ )) dτ (3.14) u (tk+1 ) − u (0) = 0

and  u (tk ) − u (0) =

tk

(3.15)

h (τ, u (τ )) dτ. 0

We approximate the function h (τ, u (τ )) within the interval [tk , tk+1 ] using the Newton polynomial as follows: h (τ, u (τ )) = Pk (τ ) + R (τ ) = h (tk−2 , u (tk−2 )) h (tk−1 , u (tk−1 )) − h (tk−2 , u (tk−2 )) + (τ − tk−2 ) t h (tk , u (tk )) − 2h (tk−1 , u (tk−1 )) + h (tk−2 , u (tk−2 )) + 2 (t)2 × (τ − tk−2 ) (τ − tk−1 ) +

(3.16)

k (τ − tm ) (k+1) h (ετ , u (ετ )) . (k + 1)!

m=0

Therefore, the error can be evaluated as  Rτα (ετ ) =

tk+1

tk

k 1 (τ − tm ) h(k+1) (ετ , u (ετ )) dτ (k + 1)! m=0

(3.17)

116

New Numerical Scheme With Newton Polynomial

and we have



k



α

tk+1 1

(k+1)

R (ετ ) =

, u − t h dτ (ε (τ ) (ε ))

m τ τ τ

tk

(k + 1)! m=0



 tk+1 k





1





≤ (3.18) (τ − tm ) h(k+1) (ετ , u (ετ )) dτ



(k + 1)!

tk m=0

k

 tm+1



(τ − t )

m

(k+1)



h sup < sup , u (ε (ε ))



τ τ dτ

 

+ 1)! (k tm τ ∈ tm ,tm+1 m=0 τ ∈ tm ,tm+1

k

(τ − t )



 tk+1 m



(k+1)

sup <  sup

h , u dτ. (ε (ε ))

 τ τ





+ 1)! (m tk τ ∈ t ,t τ ∈ t ,t m m+1

m=0

m m+1

We have interest in the following integral: 

tk+1

dτ = t.

(3.19)

tk

Therefore, finally

m

(τ − t )





(k+1)

m

sup h sup , u (ε (ε ))



τ τ t

 

+ 1)! (k τ ∈ tm ,tm+1 k=0 τ ∈ tm ,tm+1

m



(τ − t )

m

(k+1)

< t h sup

.

∞ τ ∈t ,t (k + 1)!

m m+1

α

R (ετ ) < τ

(3.20)

k=0

From the inequality (3.20), we obtain the following:  m 



α

t k+1 1 (k+1)

R (ετ ) < t .

h

τ ∞ k 4 (k + 1)

(3.21)

k=0

3.2

Numerical illustrations

In this section, we give numerical illustrations of some differential equations with integer order. Example 3.1. We consider the following problem with classical derivative: u (t) = 3t − 7, u (0) = 1.

(3.22)

Newton interpolation: introduction of the scheme for classical calculus

117

Figure 3.1 Numerical simulation for the classical derivative.

Thus, the following scheme is given:    4  5  23  uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t + h tk−2 , uk−2 t (3.23) 12 3 12       where h tk , uk = 3tk − 7, h tk−1 , uk−1 = 3tk−1 − 7 and h tk−2 , uk−2 = 3tk−2 − 7. For the classical derivative, the numerical simulation is depicted in Fig. 3.1. Example 3.2. We consider the following test problem with classical derivative: u (t) = 2t 2 exp (−u (t)) , u (0) = 1.

(3.24)

Applying the suggested numerical scheme on this example yields    4  5  23  uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t + h tk−2 , uk−2 t (3.25) 12 3 12         2 exp −uk−1 where h tk−1 , uk−1 = 2tk2 exp −uk , h tk−1 , uk−1 = 2tk−1 and     2 exp −uk−2 . The numerical simulation is presented for this h tk−2 , uk−2 = 2tk−2 problem in Fig. 3.2. Example 3.3. We deal with the ordinary differential equation u (t) = − cos (2t) u (t) , u (0) = 1.

(3.26)

For the classical case, the following scheme is obtained:    4  5  23  uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t + h tk−2 , uk−2 t (3.27) 12 3 12     where h tk , uk = − cos (2tk ) uk , h tk−1 , uk−1 = − cos (2tk−1 ) uk−1 and   h tk−2 , uk−2 = − cos (2tk−2 ) uk−2 . We present the numerical simulation in Fig. 3.3.

118

New Numerical Scheme With Newton Polynomial

Figure 3.2 Numerical simulation for the classical derivative.

Figure 3.3 Numerical simulation for the classical derivative.

Example 3.4. Let us consider the following problem with classical derivative: u (t) = −u + 2, u (0) = 1.

(3.28)

Then we obtain the following scheme:    4  5  23  uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t + h tk−2 , uk−2 t (3.29) 12 3 12       where h tk , uk = −uk + 2, h tk−1 , uk−1 = −uk−1 + 2 and h tk−2 , uk−2 = −uk−2 + 2. For this equation, the numerical simulation is given in Fig. 3.4. Example 3.5. The new chaotic problem which is given in [44] exhibits two 1-scroll chaotic attractors. The three-dimensional quadratic autonomous system is chaotic in a wide parameter range and displays many interesting complex dynamical behaviors. We consider the chaotic problem with classical derivative x  (t) = −

ab x − yz + c, a+b

Newton interpolation: introduction of the scheme for classical calculus

119

Figure 3.4 Numerical simulation for the classical derivative.

y  (t) = xz + ay, z (t) = bz + xy + sin (x) ,

(3.30)

with the initial conditions x (0) = 1, y (0) = 1, z (0) = 1. Here, we take a = −10, b = −4, c = 18.1. Applying the suggested numerical scheme to this example yields   4  23  h1 tk , x k , y k , zk t − h1 tk−1 , x k−1 , y k−1 , zk−1 t 12 3 (3.31)  5  + h1 tk−2 , x k−2 , y k−2 , zk−2 t, 12

x k+1 = x k +

  4  23  h2 tk , x k , y k , zk t − h2 tk−1 , x k−1 , y k−1 , zk−1 t 12 3  5  + h2 tk−2 , x k−2 , y k−2 , zk−2 t, 12

y k+1 = y k +

  4  23  h3 tk , x k , y k , zk t − h3 tk−1 , x k−1 , y k−1 , zk−1 t 12 3  5  k−2 k−2 k−2 + h3 tk−2 , x , y , z t, 12

zk+1 = zk +

where   ab k h1 tk , x k , y k , zk = − x − y k zk + c, a+b   h2 tk , x k , y k , zk = x k zk + ay k ,

(3.32)

120

New Numerical Scheme With Newton Polynomial

    h3 tk , x k , y k , zk = bzk + x k y k + sin x k ,   ab k−1 h1 tk−1 , x k−1 , y k−1 , zk−1 = − x − y k−1 zk−1 + c, a+b   h2 tk−1 , x k−1 , y k−1 , zk−1 = x k−1 zk−1 + ay k−1 ,     h3 tk−1 , x k−1 , y k−1 , zk−1 = bzk−1 + x k−1 y k−1 + sin x k−1 ,

(3.33)

and   ab k−2 h1 tk−2 , x k−2 , y k−2 , zk−2 = − x − y k−2 zk−2 + c a+b   h2 tk−2 , x k−2 , y k−2 , zk−2 = x k−2 zk−2 + ay k−2     h3 tk−2 , x k−2 , y k−2 , zk−2 = bzk−2 + x k−2 y k−2 + sin x k−2 . The numerical simulation for the considered problem is given in Fig. 3.5.

Figure 3.5 Numerical simulation for the chaotic problem with the classical derivative.

(3.34)

Numerical method for fractal differential equations

4

In this section, we consider the following Cauchy problem for which the differential operator is a fractal derivative in the sense of Chen given as CH β lim 0 Dt u (t) = t→t

u (t) − u (t1 ) β

t β − t1

1

(4.1)

.

However to obtain Eq. (4.1), we assume that the function u is classically differentiable mean lim

t→t1

u(t) − u(t1 ) = u (t) t − t1

then lim

t→t1

u(t) − u(t1 ) tβ

β − t1

= lim

t→t1

u(t) − u(t1 ) t − t1 1 = u (t) β−1 β β t − t1 βt t −t 1

then we set the solution of the following equation CH β 0 Dt u(t) 

= f (t)

u (t) = βt β−1 f (t)  t t  u (τ )dτ = β τ β−1 f (τ )dτ.

 0

Thus

0

 u(t) − u(0) = β

t

τ β−1 f (τ )dτ.

0

Then indeed the associated integral is given by  t CH β I u(t) = β τ β−1 u(τ )dτ. t 0 0

Also the fractal integral in the sense of Chen is of the form  t CH β I u = β τ β−1 u (τ ) dτ. (t) t 0

(4.2)

0

Now, we present the following problem with fractal derivative: CH β 0 Dt u (t) = h (t, u (t)) . New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00010-X Copyright © 2021 Elsevier Inc. All rights reserved.

(4.3)

122

New Numerical Scheme With Newton Polynomial

since u(t) is differentiable then u (t) = βt β−1 h(t, u(t)). Here the function t β−1 h(t, u(t)) satisfies the conditions stated in the Cauchy– Kowalewski theorem. Applying the integral on both sides yields 

t

u(t) − u(0) = β

τ β−1 h(τ, u(τ ))dτ.

0

Integrating the above equation, we obtain the following: 

t

u (t) − u (0) = β

τ β−1 h (τ, u (τ )) dτ.

(4.4)

0

At the point tk+1 = (k + 1) t 

tk+1

u (tk+1 ) − u (0) = β

τ β−1 h (τ, u (τ )) dτ

(4.5)

0

and at the point tk = kt, we have 

tk

u (tk ) − u (0) = β

τ β−1 h (τ, u (τ )) dτ.

(4.6)

0

From Eqs. (4.5) and (4.6), we obtain the following:  u (tk+1 ) − u (tk ) = β

tk+1

τ β−1 h (τ, u (τ )) dτ.

(4.7)

tk

We use the Newton polynomial which is of the form Pk (τ ) = h (tk−2 , u (tk−2 )) +

h (tk−1 , u (tk−1 )) − h (tk−2 , u (tk−2 )) t

× (τ − tk−2 ) h (tk , u (tk )) − 2h (tk−1 , u (tk−1 )) + h (tk−2 , u (tk−2 )) + 2 (t)2 × (τ − tk−2 ) (τ − tk−1 ) .

(4.8)

If we put Newton polynomial into Eq. (4.7) and we take g (t, u (t)) = βt β−1 h (t, u (t)), we have the following: u(tk+1 ) − u(tk ) = g (tk−2 , u (tk−2 )) t  tk+1 g (tk−1 , u (tk−1 )) − g (tk−2 , u (tk−2 )) + (τ − tk−2 ) dτ t tk

Numerical method for fractal differential equations

 +

tk+1

123

g (tk , u (tk )) − 2g (tk−1 , u (tk−1 )) + g (tk−2 , u (tk−2 )) 2 (t)2

tk

(4.9) (τ − tk−2 ) (τ − tk−1 ) dτ. If we order the above equation, we write 



    g tk−1 , uk−1 − g tk−2 , uk−2 t + t

u(tk+1 ) − u(tk ) = g tk−2 , u  tk+1 × (τ − tk−2 ) dτ tk       g tk , uk − 2g tk−1 , uk−1 + g tk−2 , uk−2 + 2 (t)2  tk+1 × (τ − tk−2 ) (τ − tk−1 ) dτ. k−2

(4.10)

tk

For the above integrals, we have the following calculations:  tk+1 5 (τ − tk−2 ) dτ = (t)2 , 2 tk  tk+1 23 (τ − tk−2 ) (τ − tk−1 ) dτ = (t)3 . 6 tk

(4.11)

If these calculations are substituted in Eq. (4.10), the following scheme is written:   uk+1 = uk + g tk−2 , uk−2 t    5   t (4.12) + g tk−1 , uk−1 − g tk−2 , uk−2 2        23 t + g tk , uk − 2g tk−1 , uk−1 + g tk−2 , uk−2 12 and it can be reorganized as   4 β−1  23 β−1  βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t 12 3  5 β−1  k−2 + βtk−2 h tk−2 , u t. 12

uk+1 = uk +

4.1

(4.13)

Error analysis with fractal derivative

Let the following be a general Cauchy problem: CH α 0 Dt u (t) = h (t, u (t)) .

(4.14)

124

New Numerical Scheme With Newton Polynomial

At the point t = tk+1 , after applying the associated fractional integral operator, we have  tk+1 τ β−1 h (τ, u (τ )) dτ (4.15) u (tk+1 ) − u (0) = β =β

0 k  tm+1

τ β−1 h (τ, u (τ )) dτ.

m=0 tm

We approximate the function h (τ, u (τ )) within the interval [tk , tk+1 ] using the Newton polynomial h(τ, u(τ )) = Pk (τ ) + R(τ ).

(4.16)

Replacing h(τ, u(τ )) by the above yields R(β, k) =

k 

tm+1

τ β−1 Rm (τ )dτ

(4.17)

m=0 tm

=

k 

tm+1

τ β−1

m=0 tm

1  h (τ, u(τ ))|τ =εm (τ − tm−1 )(τ − tm )dτ 2!

where 0 < εm < tk , which is the a real number obtained from the Taylor’s theorem. Then we write k  tm+1 1 ∂2 |R(β, k)| ≤ h(τ, u(τ ))|τ =εm (τ − tm−1 )(τ − tm )τ β−1 dτ 2! ∂τ 2 tm m=0

(4.18) and

|R(β, k)| ≤ max

0≤τ ≤tk

×

k 

∂2 h(τ, u(τ ))|τ =ετ ∂τ 2 tm+1

m=0 tm



τ β−1 (τ − tm−1 )(τ − tm ) dτ 2!

whereas τ β−1 (τ − tm−1 )(τ − tm ) = τ β+2 − (tm − tm−1 )τ β+1 + tm tm−1 τ β . Thus the integral produces β+2

β+3 β+3 β+2 tm−1 tm−1 tm tm − − (tm + tm−1 ) − β +3 β +3 β +2 β +2 β+1

β+1 tm−1 tm − +tm tm−1 β +1 β +1

(4.19)

Numerical method for fractal differential equations

125

Then we have

 ∂2 |R(β, k)| ≤ max h(t, u(t))|t=εt 0≤t≤tk ∂t 2 ⎧ ⎡ β+3 mβ+3 ⎪ − (m−1) ⎪ β+3 β+3 k   ⎨ (t)β+3 ⎢ (m−1)β+2 mβ+2 ⎢ − − × β+2 β+2 ⎣  ⎪ 2! β+1 m=0 ⎪ ⎩ mβ+1 − (m−1) +m(m − 1)

≤ max

0≤t≤tk

∂2 h(t, u(t))|t=εt ∂t 2



  k m=0

β+1

(t)β+3 2!

 ∂2 (t)β+3 ≤ max h(t, u(t))| t=εt 0≤t≤tk ∂t 2 2! 

 k m3 m4 m4 + + × β +3 β +2 β +1



(4.20) ⎤⎫ ⎪ ⎪ ⎥⎬ ⎥ ⎦⎪ ⎪ ⎭

β+1

(m−1)β+2 β+2 β+1 m(m − 1) mβ+1

mβ+3 β+3

+



for 0 < β ≤ 1

m=0

whereby

 k 1 1 5 1 4 1 3 m4 1 = k + k + k − k2 , β +3 β +3 5 2 3 12

m=0

 k 1 1 4 1 3 1 2 m3 = k + k + k . β +2 β +2 4 2 4

m=0

Thus we have

 ∂2 (t)β+3 |R(β, k)| ≤ max h(t, u(t))| t=ε t 0≤t≤tk ∂t 2 2! ⎧   1 1 1 5 1 4 1 3 k ⎨ β+3 + β+1  5 k + 2 k + 3 k − × 1 1 4 1 3 1 2 ⎩ + β+2 m=0 4k + 2k + 4k

4.2

1 2 12 k

 ⎫ ⎬ ⎭

.

(4.21)

Numerical illustrations

In this section, we use an efficient numerical scheme in order to obtain a numerical solution for some differential equations and a chaotic system. Example 4.1. We consider the following problem with fractal derivative: CH β 0 Dt u (t) = t

sin (u (t) − 1) ,

u (0) = −0.1.

(4.22)

126

New Numerical Scheme With Newton Polynomial

Figure 4.1 Numerical simulation for the fractal derivative for β = 1.

Then, the above problem can be solved numerically by the following scheme:   4 β−1  23 β−1  βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t (4.23) 12 3   5 β−1 + βtk−2 h tk−2 , uk−2 t 12         where h tk , uk = tk sin uk − 1 , h tk−1 , uk−1 = tk−1 sin uk−1 − 1 and     h tk−2 , uk−2 = tk−2 sin uk−2 − 1 . The numerical simulation is given in Fig. 4.1 for β = 1. uk+1 = uk +

Example 4.2. We present the following test problem with fractal derivative: CH β 0 Dt u (t) = exp(−u (t)),

(4.24)

u (0) = 0.1. Thus the following scheme can be obtained:   4 β−1  23 β−1  βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t (4.25) 12 3  5 β−1  + βtk−2 h tk−2 , uk−2 t 12       where h tk , uk = exp(−uk ), h tk−1 , uk−1 = exp(−uk−1 ) and h tk−2 , uk−2 = exp(−uk−2 ). The numerical simulation is given in Fig. 4.2 for β = 0.55. uk+1 = uk +

Example 4.3. We deal with the following problem with fractal derivative: CH β 0 Dt u (t) = − ln (t

u (0) = 1.

+ 1) u (t) ,

(4.26)

Numerical method for fractal differential equations

127

Figure 4.2 Numerical simulation for the fractal derivative for β = 0.55.

Figure 4.3 Numerical simulation for the fractal derivative for β = 0.72.

Applying the suggested numerical scheme to this example yields   4 β−1  23 β−1  βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t 12 3  5 β−1  k−2 + βtk−2 h tk−2 , u t 12

uk+1 = uk +

(4.27)

    where h tk , uk = − ln (tk + 1) uk , h tk−1 , uk−1 = − ln (tk−1 + 1) uk−1 and   h tk−2 , uk−2 = − ln (tk−2 + 1) uk−2 . The numerical simulation is depicted in Fig. 4.3 for β = 0.72. Example 4.4. We consider the following fractal differential equation: CH β 0 Dt u (t) = cos (u (t)) ,

u (0) = 0.3.

(4.28)

128

New Numerical Scheme With Newton Polynomial

Figure 4.4 Numerical simulation for the fractal derivative for β = 0.52.

For such a problem, we get the following scheme:   4 β−1  23 β−1  βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t (4.29) 12 3  5 β−1  + βtk−2 h tk−2 , uk−2 t 12           where h tk , uk = cos uk , h tk−1 , uk−1 = cos uk−1 , and h tk−2 , uk−2 =   cos uk−2 . The numerical simulation is presented in Fig. 4.4 for β = 0.52. uk+1 = uk +

Example 4.5. We next consider the King Cobra chaotic system introduced by Muthukumar et al. in [45] with fractal derivative, CH β 2 0 Dt x (t) = a (y − x) + byz , CH β 2 0 Dt y (t) = cx + dxz , CH β 0 Dt z (t) = ez + f x,

(4.30)

with the initial conditions x (0) = −1, y (0) = −2, z (0) = −1.

(4.31)

Here a = 7.5, b = 1, c = 1, d = −0.5, e = −2, f = −1. Applying the suggested numerical scheme to this example yields  23 β−1  βtk h1 tk , x k , y k , zk t 12  4 β−1  − βtk−1 h1 tk−1 , x k−1 , y k−1 , zk−1 t 3  5 β−1  + βtk−2 h1 tk−2 , x k−2 , y k−2 , zk−2 t, 12

x k+1 = x k +

(4.32)

Numerical method for fractal differential equations

129

Figure 4.5 Numerical simulation for the King–Cobra attractor with the fractal derivative for β = 1.

 23 β−1  βtk h2 tk , x k , y k , zk t 12  4 β−1  − βtk−1 h2 tk−1 , x k−1 , y k−1 , zk−1 t 3  5 β−1  + βtk−2 h2 tk−2 , x k−2 , y k−2 , zk−2 t, 12

y k+1 = y k +

 23 β−1  βtk h3 tk , x k , y k , zk t 12  4 β−1  − βtk−1 h3 tk−1 , x k−1 , y k−1 , zk−1 t 3  5 β−1  + βtk−2 h3 tk−2 , x k−2 , y k−2 , zk−2 t, 12

zk+1 = zk +

where      2 h1 tk , x k , y k , zk = a y k − x k + by k zk ,    2 h2 tk , x k , y k , zk = cx k + dx k zk ,

(4.33)

130

New Numerical Scheme With Newton Polynomial

  h3 tk , x k , y k , zk = ezk + f x k ,      2 h1 tk−1 , x k−1 , y k−1 , zk−1 = a y k−1 − x k−1 + by k−1 zk−1 ,    2 h2 tk−1 , x k−1 , y k−1 , zk−1 = cx k−1 + dx k−1 zk−1 ,   h3 tk−1 , x k−1 , y k−1 , zk−1 = ezk−1 + f x k−1 ,

(4.34)

and      2 h1 tk−2 , x k−2 , y k−2 , zk−2 = a y k−2 − x k−2 + by k−2 zk−2 ,    2 h2 tk−2 , x k−2 , y k−2 , zk−2 = cx k−2 + dx k−2 zk−2 ,   h3 tk−2 , x k−2 , y k−2 , zk−2 = ezk−2 + f x k−2 . The numerical simulation is given in Fig. 4.5 for β = 1.

(4.35)

Numerical method for a fractional differential equation with Caputo–Fabrizio derivative

5

In this section, we will obtain a new numerical scheme for the following Cauchy problem with exponential decay kernel. Firstly, we recall that the definition of the Caputo–Fabrizio fractional derivative. The Caputo–Fabrizio fractional derivative of the function u (t) ∈ W21 (0, l), α ∈ ]0, 1] is defined by    α M (α) t d CF α u exp − dτ (5.1) D u = − τ (τ ) (t) (t ) a t 1 − α a dτ 1−α where M (α) is a normalization function such that M (0) = M (1) = 1. The space W21 (0, l) is the space of continuous differentiable functions.

M(α) 1−α

CF α 0 Dt u(t) = h(t, u(t)),



t 0

  α (t − τ ) dτ = h(t, u(t)). u(τ ) exp − 1−α

By applying the Laplace transform on both sides yields M(α) 1 (p u(p) ˜ − u(0)) α L(h(t, u(t)))(p). 1−α p + 1−α Rearranging, we get u(p) ˜ =

1 α u(0) 1 − α + L(h(t, u(t))) + L(h(t, u(t))). p M(α) p M(α)

Taking the inverse Laplace yields u(t) = u(0) +

1−α α h(t, u(t)) + M(α) M(α)



t

h(τ, u(τ ))dτ. 0

In the definition of Caputo–Fabrizio, please give the following explanation    M(α) t α CF α (t − τ ) dτ. y(τ ˙ ) exp − 0 Dt y(t) = lim α→1 1 − α 0 1−α We let ξ = 1 − α, thus    M(1 − ξ ) t 1−ξ lim (t − τ ) dτ y(τ ˙ ) exp − ξ →0 ξ ξ 0 New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00011-1 Copyright © 2021 Elsevier Inc. All rights reserved.

132

New Numerical Scheme With Newton Polynomial

lim M(1 − ξ ) = M(1) = 1, then

ξ →0

  1 1 lim exp − (t − τ ) = δ(t − τ ). ξ →0 ξ ξ Indeed M(1 − ξ ) ξ →0 ξ



t

lim

0

   t 1−ξ (t − τ ) dτ = y(τ ˙ ) exp − y(τ ˙ )δ(t − τ )dτ = y(τ ˙ ). ξ 0

We recover the first derivative. The fractional integral with exponential decay kernel of the function u (t) is as follows:  t α 1−α CF α u + I u (τ ) dτ. (5.2) = (t) t (u (t)) 0 M (α) M (α) 0 We now consider the following Cauchy problem with the new fractional derivative: CF α 0 Dt u (t) = h (t, u (t)) .

(5.3)

From the definition of the Caputo–Fabrizio integral, we can reformulate the above equation as 1−α α u (t) − u (0) = h (t, u (t)) + M (α) M (α)



t

h (τ, u (τ )) dτ.

(5.4)

0

We write Eq. (5.4) at the point tk+1 = (k + 1) t 1−α α h (tk , u (tk )) + u (tk+1 ) − u (0) = M (α) M (α)



tk+1

h (τ, u (τ )) dτ

(5.5)

h (τ, u (τ )) dτ.

(5.6)

0

and at the point tk = kt, we have 1−α α h (tk−1 , u (tk−1 )) + u (tk ) − u (0) = M (α) M (α)



tk 0

Taking the difference of these equations, we can write the following: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)  tk+1 α h (τ, u (τ )) dτ. + M (α) tk

u (tk+1 ) − u (tk ) =

(5.7)

Putting the Newton polynomial into the above equation, we can get the following: uk+1 − uk =

1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)

(5.8)

Numerical method for a fractional differential equation with Caputo–Fabrizio derivative

⎧

h tk−2 , uk−2

⎪ ⎪ ⎪  tk+1 ⎪ ⎨ + h tk−1 ,uk−1 −h tk−2 ,uk−2 (τ − t ) α k−2

t

+ h tk ,uk −2h tk−1 ,uk−1 +h tk−2 ,uk−2 M (α) tk ⎪ + ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

133

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

dτ

and write it as follows: uk+1 − uk =

+

1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)

⎧ h t , uk−2 t ⎪

k−2 k−2

⎪ k−1 ⎪  tk+1 ⎪ h tk−1 ,u −h tk−2 ,u (τ − tk−2 ) dτ α ⎨ + tk t

⎫ ⎪ ⎪ ⎪ ⎪ ⎬

M (α) ⎪ ⎪ ⎪ ⎪ ⎩

⎪ ⎪ ⎪ ⎪ ⎭





h tk ,uk −2h tk−1 ,uk−1 +h tk−2 ,uk−2 2 2(t) t × tkk+1 (τ − tk−2 ) (τ − tk−1 ) dτ

+

(5.9)

The integrals on the right hand side of the above equation can be calculated:  tk+1 5 (τ − tk−2 ) dτ = (t)2 , 2 tk  tk+1 23 (τ − tk−2 ) (τ − tk−1 ) dτ = (t)3 . 6 tk

.

(5.10)

Thus, we have the following numerical scheme: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)

⎧ h tk−2 t , uk−2 ⎪ ⎪



  α ⎨ + h tk−1 , uk−1 − h tk−2 , uk−2 52 t



  + h tk , uk − 2h tk−1 , uk−1 M (α) ⎪ 23 ⎪ ⎩ + 12 t +h tk−2 , uk−2

uk+1 = uk +

(5.11) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

and we can rearrange it as 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)



 23  k α , uk−1 t − 43 h tk−1 12 h tk , u t

. + 5 h tk−2 , uk−2 t + 12 M (α)

uk+1 = uk +

5.1

(5.12)

Error analysis with Caputo–Fabrizio fractional derivative

Let us consider a general Cauchy problem: CF α 0 Dt u (t) = h (t, u (t))

(5.13)

134

New Numerical Scheme With Newton Polynomial

where the differential operator is the Caputo–Fabrizio differential operator. At the point t = tk+1 , after applying the associated fractional integral operator, we have  tk+1 α 1−α h (t, u (t)) + h (τ, u (τ )) dτ u (tk+1 ) − u (0) = M (α) M (α) 0  tk+1 α 1−α h (t, u (t)) + h (τ, u (τ )) dτ. = M (α) M (α) tk

(5.14)

We can write the function h (τ, u (τ )) approximately using the Newton polynomial within the interval [tk , tk+1 ] as follows: h(τ, u(τ )) = Pk (τ ) + R(τ ).

(5.15)

Replacing h(τ, u(τ )) by the above yields k  α  tm+1 R(β, k) = Rm (τ )dτ M(α) tm

(5.16)

m=0

k  α  tm+1 1  h (τ, u(τ ))|τ =εm (τ − tm−1 )(τ − tm )dτ = M(α) 2! tm m=0

k  α  tm+1 1 ∂ 2 |R(β, k)| ≤ h(τ, u(τ ))|τ =εm (τ − tm−1 )(τ − tm )dτ M(α) 2! ∂τ 2 tm m=0



|R(β, k)| ≤

∂2



α max h(τ, u(τ ))|τ =ετ M(α) 0≤τ ≤tk ∂τ 2 k  tm+1  (τ − tm−1 )(τ − tm ) × dτ 2! tm

(5.17)

(5.18)

m=0

whereas (τ − tm−1 )(τ − tm ) = τ 2 − (tm − tm−1 )τ + tm tm−1 . Thus the integral produces 

   3 2 3 2 tm−1 tm−1 tm tm − − (tm + tm−1 ) − + tm tm−1 (tm − tm−1 ) . 3 3 2 2

Then we have |R(β, k)| ≤

 2  ∂ α max h(t, u(t))| t=εt M(α) 0≤t≤tk ∂t 2

(5.19)

Numerical method for a fractional differential equation with Caputo–Fabrizio derivative

135

⎧ ⎡ ⎤⎫ 3 3 ⎪ − m3 − (m−1) k ⎪ ⎬ ⎨ (t)3 3   2 ⎢ ⎥ 2 (m−1) m × ⎣ − 2 − 2 ⎦ ⎪ ⎪ 2! ⎭ m=0 ⎩ +m(m − 1)    2  k  α ∂ (t)3 5 2 ≤ max 2m h(t, u(t))| − 3m + t=εt M(α) 0≤t≤tk ∂t 2 2! 6 n=0

whereby k   m=0

5 2m − 3m + 6



2

  1 2k 3 k 2 k 5 = − − + . 6 3 2 3 6

Thus we have |R(β, k)| ≤

 2  α ∂ (t)3 max h(t, u(t))| t=εt 2 M(α) 0≤t≤tk ∂t 12   k  2k 3 k 2 k 5 − − + × 3 2 3 6

(5.20)

m=0

5.2

Numerical illustrations

In this section, we give some numerical illustrations by considering differential equations with the Caputo–Fabrizio fractional derivative. There is no doubt that this new method is influential and applicable for solving linear and non-linear problems. Example 5.1. We consider the following problem with the Caputo–Fabrizio fractional derivative: CF α 0 Dt u (t) = 2tu (t) ,

(5.21)

u (0) = 1. Applying the suggested numerical scheme to this example yields 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)



 23  α h tk , uk t − 43 h tk−1 , uk−1 t 12

+ 5 h tk−2 , uk−2 t + 12 M (α)

uk+1 = uk +

(5.22)





where h tk , uk = 2tk uk , h tk−1 , uk−1 = 2tk−1 uk−1 and h tk−2 , uk−2 = 2tk−2 uk−2 . The numerical simulation is given in Fig. 5.1 for α = 0.9.

136

New Numerical Scheme With Newton Polynomial

Figure 5.1 Numerical simulation for the Caputo–Fabrizio fractional derivative for α = 0.9.

Figure 5.2 Numerical simulation for the Caputo–Fabrizio fractional derivative for α = 1.

Example 5.2. We consider the following problem with the Caputo–Fabrizio fractional derivative: CF α 0 Dt u (t) = sin (5u (t)) ,

(5.23)

u (0) = 1. Thus, the scheme for the above problem is as follows: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)



 23  k α − 43 h tk−1 , uk−1 t 12 h tk , u t

+ 5 h tk−2 , uk−2 t + 12 M (α)

uk+1 = uk +

(5.24)









where h tk , uk = sin 5uk , h tk−1 , uk−1 = sin 5uk−1 and h tk−2 , uk−2 =

sin 5uk−2 . The numerical simulation is depicted in Fig. 5.2 for α = 1.

Numerical method for a fractional differential equation with Caputo–Fabrizio derivative

137

Figure 5.3 Numerical simulation for the Caputo–Fabrizio fractional derivative for α = 0.78.

Example 5.3. We consider the following problem with Caputo–Fabrizio fractional derivative: CF α 0 Dt u (t) = u (t) + t,

(5.25)

u (0) = 1. Thus, the following scheme can be obtained: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)



 23  k α , uk−1 t − 43 h tk−1 12 h tk , u t

+ 5 h tk−2 , uk−2 t + 12 M (α)

uk+1 = uk +

(5.26)





where h tk , uk = uk + tk , h tk−1 , uk−1 = uk−1 + tk−1 and h tk−2 , uk−2 = uk−2 + tk−2 . The numerical simulation is depicted in Fig. 5.3 for α = 0.78. Example 5.4. We consider the following problem with the Caputo–Fabrizio fractional derivative: CF α 0 Dt u (t) = u − 5t,

(5.27)

u (0) = 0.11. Thus the following scheme is given: 1−α (5.28) [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)



 23  4 k k−1 t α tk−1 12 h tk , u t

,u − 3 h k−2 + 5 t + 12 h tk−2 , u M (α)



where h tk , uk = uk − 5tk , h tk−1 , uk−1 = uk−1 − 5tk−1 and h tk−2 , uk−2 = uk−2 − 5tk−2 . The numerical simulation is depicted in Fig. 5.4 for α = 0.82. uk+1 = uk +

138

New Numerical Scheme With Newton Polynomial

Figure 5.4 Numerical simulation for the Caputo–Fabrizio fractional derivative for α = 0.82.

Example 5.5. The Labyrinth system which displays different dynamical behaviors with various parameters describes the mathematical modeling of a particle which moves in a labyrinth under the influence of some external source of energy [46]. We consider the following Labyrinth attractor with the Caputo–Fabrizio fractional derivative CF α 0 Dt x (t) = − sin y (at) − bx (t) , CF α 0 Dt y (t) = − sin z (at) − by (t) , CF α 0 Dt z (t) = − sin x (at) − bz (t) ,

(5.29)

with the initial conditions x (0) = −1, y (0) = 1, z (0) = 1.

(5.30)

Applying the suggested numerical scheme to this example yields    ! 1−α x k+1 = x k + h1 tk , x k , y k , zk − h1 tk−1 , x k−1 , y k−1 , zk−1 (5.31) M (α)



 23  4 k k k k−1 , y k−1 , zk−1 t α 12 h1 tk , x , y , z t − 3 h1 tk−1 , x

+ , 5 h1 tk−2 , x k−2 , y k−2 , zk−2 t + 12 M (α)    ! 1−α h2 tk , x k , y k , zk − h2 tk−1 , x k−1 , y k−1 , zk−1 M (α)



 23  α h2 tk , x k , y k , zk t − 43 h2 tk−1 , x k−1 , y k−1 , zk−1 t 12 , + 5 h2 tk−2 , x k−2 , y k−2 , zk−2 t + 12 M (α)

y k+1 = y k +

   ! 1−α h3 tk , x k , y k , zk − h3 tk−1 , x k−1 , y k−1 , zk−1 M (α)



 23  4 k k k k−1 , y k−1 , zk−1 t α 12 h3 tk , x , y , z t − 3 h3 tk−1 , x

+ , 5 + 12 h3 tk−2 , x k−2 , y k−2 , zk−2 t M (α)

zk+1 = zk +

Numerical method for a fractional differential equation with Caputo–Fabrizio derivative

139

Figure 5.5 Dynamical behavior of the Labyrinth attractor with the Caputo–Fabrizio fractional derivative for α = 0.98.

where

  h1 tk , x k , y k , zk = − sin y (atk ) − bx (tk ) ,   h2 tk , x k , y k , zk = − sin z (atk ) − by (tk ) ,   h3 tk , x k , y k , zk = − sin x (atk ) − bz (tk ) ,   h1 tk−1 , x k−1 , y k−1 , zk−1 = − sin y (atk−1 ) − bx (tk−1 ) ,   h2 tk−1 , x k−1 , y k−1 , zk−1 = − sin y (atk−1 ) − bx (tk−1 ) ,   h3 tk−1 , x k−1 , y k−1 , zk−1 = − sin y (atk−1 ) − bx (tk−1 ) ,

and

  h1 tk−2 , x k−2 , y k−2 , zk−2 = − sin y (atk−2 ) − bx (tk−2 ) ,

(5.32)

(5.33)

140

New Numerical Scheme With Newton Polynomial

Figure 5.6 Dynamical behavior of the Labyrinth attractor with the Caputo–Fabrizio fractional derivative for α = 0.95.

  h2 tk−2 , x k−2 , y k−2 , zk−2 = − sin y (atk−2 ) − bx (tk−2 ) ,   h3 tk−2 , x k−2 , y k−2 , zk−2 = − sin y (atk−2 ) − bx (tk−2 ) .

(5.34)

With the parameters a = 1, b = 0.1 and a = 1, b = 0, the numerical simulations are given in Fig. 5.5 and in Fig. 5.6.

Numerical method for a fractional differential equation with power-law kernel

6

In this section, before giving the numerical scheme for the general Cauchy problem, we recall the definition of the Caputo fractional derivative and integral. The left Caputo fractional derivative of order of the function u (t) is given by  t 1 C α D u = (6.1) (t) (t − τ )−α u (τ ) dτ, t > 0, 0 t  (1 − α) 0 where u ∈]0, 1]. The solution is 2-time differentiable.

1 (1 − α)

C α 0 Dt u(t) = h(t, u(t)),



t

(t − τ )−α u (τ )dτ = h(t, u(t)).

0

Applying the Laplace transform on both sides yields ˜ − p α−1 u(0) = L(h(t, u(t)))(p), p α u(p) 1 1 u(p) ˜ = u(0) + α L(h(t, u(t)))(p). p p Using the convolution theorem, we obtain  t 1 u(t) − u(0) = (t − τ )α−1 h(τ, u(τ ))dτ. (α) 0 The fractional integral with power-law kernel is of the form  t 1 C α u (τ ) (t − τ )α−1 dτ. 0 It (u (t)) =  (α) 0 Now, let us consider the following problem with power-law kernel:  C α 0 Dt u (t) = h (t, u (t)) , u (0) = u0 .

(6.2)

(6.3)

We convert Eq. (6.3) into u (t) − u (0) =

1  (α)



t

h (τ, u (τ )) (t − τ )α−1 dτ.

0

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00012-3 Copyright © 2021 Elsevier Inc. All rights reserved.

(6.4)

142

New Numerical Scheme With Newton Polynomial

At the point tk+1 = (k + 1) t, we can write the following: u (tk+1 ) − u (0) =

1  (α)



tk+1

h (τ, u (τ )) (tk+1 − τ )α−1 dτ.

(6.5)

0

Thus, we have u (tk+1 ) = u (0) +

k  1  tm+1 h (τ, u (τ )) (tk+1 − τ )α−1 dτ.  (α) tm

(6.6)

m=2

Replacing the Newton polynomial into the above equation, we have ⎧

h tm−2 , um−2

⎪ ⎪ ⎪ ⎪ h t ,um−1 −h tm−2 ,um−2 k  (τ − tm−2 ) 1  tm+1 ⎨ + m−1 k+1 0 t

u =u + h(tm ,um )−2h tm−1 ,um−1 +h tm−2 ,um−2 ⎪  (α) + ⎪ m=2 tm ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ (6.7)

× (tk+1 − τ )

α−1

dτ.

Thus, we can rearrange the above equation as follows:  tm+1

⎧ h tm−2 , um−2 (tk+1 − τ )α−1 dτ ⎪ tm ⎪

⎪  tm+1 h tm−1 ,um−1 −h tm−2 ,um−2

⎪ ⎪ ⎪ + (τ − tm−2 ) k ⎨ tm t 1  uk+1 = u0 + × (tk+1 − τ )α−1 dτ ⎪  (α)  tm+1 h(tm ,um )−2h tm−1 ,um−1 +h tm−2 ,um−2

m=2 ⎪ ⎪ ⎪ + ⎪ tm 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(6.8)

and we have k   tm+1 1   h tm−2 , um−2 (tk+1 − τ )α−1 dτ  (α) tm m=2



k 1  h tm−1 , um−1 − h tm−2 , um−2 +  (α) t m=2  tm+1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

uk+1 = u0 +

tm



k 1  h (tm , um ) − 2h tm−1 , um−1 + h tm−2 , um−2 +  (α) 2 (t)2 m=2  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ. tm

(6.9)

Numerical method for a fractional differential equation with power-law kernel

143

We can calculate the above integrals in Eq. (6.9) as follows:  

tm+1 tm tm+1

(tk+1 − τ )α−1 dτ =

(τ − tm−2 ) (tk+1 − τ )α−1 dτ

tm

(t)α+1 = α (α + 1)



tm+1 tm

 (t)α  (k − m + 1)α − (k − m)α , α



(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α)

 ,

(t)α+2 α (α + 1) (α + 2)   ⎤ 2 2 (k − m) + (3α + 10) (k − m) α (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥. 2 ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ = ⎡

⎢ +⎢ ⎣

(6.10)

If we replace these calculations into Eq. (6.9), we have the following scheme: uk+1 = u0 +

k   (t)α   h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

    (t) h tm−1 , um−1 − h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) +

α

k  

(6.11)

k    

(t)α   h tm , um − 2h tm−1 , um−1 + h tm−2 , um−2 2 (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+

6.1

Error analysis with Caputo fractional derivative

To address the accuracy of the suggested scheme, we consider a general Cauchy problem, which has the Caputo derivative 

C Dα u 0 t (t) = f

(t, u (t)) , u (0) = u0 .

(6.12)

144

New Numerical Scheme With Newton Polynomial

Following the methodology presented to derive a numerical solution using the Newton polynomial, we have k  1  tm+1 u (tk+1 ) − u (0) = (tk+1 − τ ) [h (τ, u (τ )) − R (τ )] dτ. (6.13)  (α) tm m=0

Thus, we write the error    k  1  tm+1 α−1 h = (tk+1 − τ ) (τ, u(τ ))|τ =εm (τ − tm−1 )(τ − tm ) dτ (α) 2! tm

Rτα

m=0

(6.14) and  k       tm+1  (τ, u(τ ))| 1 h   τ =ε m |Rτα | = dτ  (tk+1 − τ )α−1  ×(τ − t )(τ − t )  (α)  m−1 m tm

(6.15)

m=0

k  1  tm+1 ≤ (tk+1 − τ )α−1 |h (τ, u(τ ))|τ =εm (τ − tm−1 )(τ − tm )dτ 2!(α) tm m=0

1 max |h (τ, u(τ ))|τ =ετ ≤ 2(α) 0≤τ ≤tk k  tm+1  × (tk+1 − τ )α−1 (τ − tm−1 )(τ − tm )dτ m=0 tm

where ετ is a constant obtained from the Taylor series approximation for the error. We shall calculate 

tm+1 tm

(tk+1 − τ )α−1 (τ − tm−1 )(τ − tm )dτ



=

tm+1

(tk+1 − τ )α−1 (τ 2 − (tm−1 + tm )τ + tm−1 tm )dτ.

tm



tm+1 tm

(tk+1 − τ )α−1 τ 2 dτ −



+



tm+1

(6.16)

(tk+1 − τ )α−1 (tm−1 + tm )τ dτ

tm tm+1

(tk+1 − τ )α−1 tm−1 tm dτ.

(6.17)

tm

whereas 

tm+1 tm

 (tk+1 − τ )α−1 τ 2 dτ ≤ 0

tm+1

(tk+1 − τ )α−1 τ 2 dτ

(6.18)

Numerical method for a fractional differential equation with power-law kernel

therefore  tm+1

 (tk+1 − τ )

τ dτ ≤

α−1 2

tm+1

0

tm

145

2 (tk+1 − τ )α−1 tk+1 dτ

(6.19)

since tk+1 > τ . Then    tm+1 (tk+1 )α (tk+1 − tm+1 )α α−1 2 2 . (tk+1 − τ ) τ dτ ≤ tk+1 − α α tm

(6.20)

Thus 

tm+1

 (tk+1 − τ )α−1 τ 2 dτ ≤ (t)α+2

tm

 (k + 1)α (k − m)α − . α α

(6.21)

Then 

tm+1

 (tk+1 − τ )

α−1

τ dτ ≤ (t)

α+1

tm

(k + 1)α (k − m)α − α α

 (6.22)

and 

tm+1

 (tk+1 − τ )α−1 dτ ≤ (t)α

tm

 (k + 1)α (k − m)α − . α α

(6.23)

Putting all together, we have k  

tm+1

m=0 tm

(tk+1 − τ )α−1 (τ − tm−1 )(τ − tm )dτ

⎧   ⎫ α α ⎪ ⎪ (t)α+2 (k+1) − (k−m) ⎪ ⎪ α α ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ α+2 ⎪ ⎪ −(2m − 1)(t) ⎪ ⎪ k ⎨ ⎬    α α (k+1) (k−m) ≤ × − α α ⎪ ⎪ ⎪ α+2 m=0 ⎪ ⎪ ⎪ +m(m − 1)(t) ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ α α ⎪ ⎪ (k+1) (k−m) ⎩ ⎭ × − α α    α α k  − (k−m) (t)α+2 (k+1) α α ≤

× 1 + 2m − 1 + m2 − m m=0 ! "  k   (k + 1)α (k − m)α (t)α+2 (m2 + m) ≤ − α α m=0 ! " α k(k + 1) k(k + 1)(2k + 1) (k + 1) + ≤ (t)α+2 α 2 6 ! " α+1 2 2k + k (k + 1) k+ ≤ (t)α+2 12α 3

(6.24)

146

New Numerical Scheme With Newton Polynomial

!

≤ (t)α+2

(k + 1)α+1 12α

2k 2 + 4k 3

≤ (t)α+2

(k + 1)α+1 2 (k + 2k). 18α

"

Thus we have the following |Rτα | ≤

6.2

(t)α+2 max |h (τ, u(τ ))|τ =ετ (k + 1)α+1 (k 2 + 2k). 36(α) 0≤t≤tk

(6.25)

Numerical illustrations

In this section, we present numerical solution for some differential equations with a new differential operator called the Caputo fractional derivative. We can see that the suggested method is accurate, efficient and practical with these examples. Example 6.1. We consider the following problem with the Caputo fractional derivative: C α 0 Dt u (t) = −2t cos (2u (t)) ,

(6.26)

u (0) = 0.3. Thus, the following scheme can be written: uk+1 = u0 +

k   (t)α   h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

α

k       h tm−1 , um−1 − h tm−2 , um−2

(t)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) +

(6.27)

k    

(t)α   + h tm , um − 2h tm−1 , um−1 + h tm−2 , um−2 2 (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α 2 +6α + 18α + 12



where h (tm , um ) = −2tm cos (2um ), h tm−1 , um−1 = −2tm−1 cos 2um−1 and



h tm−2 , um−2 = −2tm−2 cos 2um−2 . The numerical simulation is given in Fig. 6.1 for α = 0.67.

Numerical method for a fractional differential equation with power-law kernel

147

Figure 6.1 Numerical simulation for the Caputo fractional derivative for α = 0.67.

Example 6.2. We consider the following problem with the Caputo fractional derivative: C α 3 0 Dt u (t) = t ,

(6.28)

u (0) = −0.3. Applying the suggested numerical scheme to this example yields uk+1 = u0 +

k   (t)α   h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

    (t) h tm−1 , um−1 − h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) +

α

k  

(6.29)

k    

(t)α   h tm , um − 2h tm−1 , um−1 + h tm−2 , um−2 2 (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12



3, h t m−1 = t 3 m−2 = t 3 where h (tm , um ) = tm m−1 , u m−1 and h tm−2 , u m−2 . The numerical simulation is depicted in Fig. 6.2 for α = 0.49.

+

Example 6.3. We deal with the following problem with the Caputo fractional derivative: 1 C α 0 Dt u (t) = − t

+ u (t) ,

(6.30)

148

New Numerical Scheme With Newton Polynomial

Figure 6.2 Numerical simulation for the Caputo fractional derivative for α = 0.49.

u (0) = −0.01. The above problem can be solved numerically by the following scheme: uk+1 = u0 +

k   (t)α   h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

    (t) h tm−1 , um−1 − h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) +

α

k  

(6.31)

k    

(t)α   h tm , um − 2h tm−1 , um−1 + h tm−2 , um−2 2 (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+





1 + um−1 and h tm−2 , um−2 = where h (tm , um ) = − t1m + um , h tm−1 , um−1 = − tm−1 1 + um−2 . The numerical simulation is presented in Fig. 6.3 for α = 0.54. − tm−2

Example 6.4. We consider the differential equation with the Caputo fractional derivative C α 0 Dt u (t) = u (t) t,

u (0) = 0.2.

(6.32)

Numerical method for a fractional differential equation with power-law kernel

149

Figure 6.3 Numerical simulation for the Caputo fractional derivative for α = 0.54.

Thus, the following scheme is given: uk+1 = u0 +

k   (t)α   h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

    (t) h tm−1 , um−1 − h tm−2 , um−2  (α + 2) m=2   − m + 1)α (k − m + 3 + 2α) (k × − (k − m)α (k − m + 3 + 3α) +

α

k  

(6.33)

k    

(t)α   h tm , um − 2h tm−1 , um−1 + h tm−2 , um−2 2 (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+



where h (tm , um ) = um tm , h tm−1 , um−1 = um−1 tm−1 , and h tm−2 , um−2 = um−2 tm−2 . We give the numerical simulation in Fig. 6.4 for α = 0.69. Example 6.5. The Chua attractor describes the asymptotic attractor of solutions of the system of differential equations modeling the dynamics of the Chua circuit, which is the simplest electronic circuit exhibiting chaos. We deal with the following Chua attractor with the Caputo fractional derivative: ⎧ ⎨ ⎩

C Dα P (t) = k (R (t) − P (t)) , 0 t C Dα R = P − P (t) S (t) + mR (t) + sin (P (t) (t) t 0 C Dα S = P (t) R (t) − lS (t) , (t) t 0

(t)) ,

(6.34)

150

New Numerical Scheme With Newton Polynomial

Figure 6.4 Numerical simulation for the Caputo fractional derivative for α = 0.69.

with the initial conditions P (0) = 0.1, R (0) = 0.1, S (0) = 0.1. For this model, we take k = 40, m = 28 and l = 3. Applying the suggested numerical scheme to this example yields k  (t)α   h1 tm−2 , P m−2 , R m−2 , S m−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

 k  (t)α  h1 tm−1 , P m−1 , R m−1 , S m−1

+ −h1 tm−2 , P m−2 , R m−2 , S m−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ m, Rm, S m) k h1 (tm , Pm−1

(t)α  ⎣ , R m−1 , S m−1 ⎦ −2h1 tm−1 , P + 2 (α + 3) m−2 +h1 tm−2 , P , R m−2 , S m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α 2 +6α + 18α + 12

P k+1 = P 0 +

k  (t)α   h2 tm−2 , P m−2 , R m−2 , S m−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

R k+1 = R 0 +

(6.35)

Numerical method for a fractional differential equation with power-law kernel

151



 k  (t)α  h2 tm−1 , P m−1 , R m−1 , S m−1

−h2 tm−2 , P m−2 , R m−2 , S m−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ m, Rm, S m) k h2 (tm , Pm−1

(t)α  ⎣ , R m−1 , S m−1 ⎦ −2h2 tm−1 , P + 2 (α + 3) m−2 +h2 tm−2 , P , R m−2 , S m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

+

k  (t)α   h3 tm−2 , P m−2 , R m−2 , S m−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

 k  (t)α  h3 tm−1 , P m−1 , R m−1 , S m−1

+ −h3 tm−2 , P m−2 , R m−2 , S m−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k h3 (tm , P m , R m , S m ) α 

(t) ⎣ −2h3 tm−1 , P m−1 , R m−1 , S m−1 ⎦ +

2 (α + 3) +h3 tm−2 , P m−2 , R m−2 , S m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

S k+1 = S 0 +

where

h1 tm , P m , R m , S m = k (R (tm ) − P (tm )) ,

h2 tm , P m , R m , S m = P (tm ) − P (tm ) S (tm ) + mR (tm ) + sin (P (tm )) ,

h3 tm , P m , R m , S

m

= P (tm ) R (tm ) − lS (tm ) ,

  h1 tm−1 , P m−1 , R m−1 , S m−1 = k (R (tm−1 ) − P (tm−1 )) ,

(6.36)

152

New Numerical Scheme With Newton Polynomial

Figure 6.5 Numerical simulation for the Chua model with the Caputo fractional derivative for α = 1.

  h2 tm−1 , P m−1 , R m−1 , S m−1 = −P (tm−1 ) − P (tm−1 ) S (tm−1 ) + mR (tm−1 ) + sin (P (tm−1 )) ,

(6.37)

  h3 tm−1 , P m−1 , R m−1 , S m−1 = P (tm−1 ) R (tm−1 ) − lS (tm−1 ) , and

  h1 tm−2 , P m−2 , R m−2 , S m−2 = k (R (tm−2 ) − P (tm−2 )) ,   h2 tm−2 , P m−2 , R m−2 , S m−2 = P (tm−2 ) − P (tm−2 ) S (tm−2 ) + mR (tm−2 ) + sin (P (tm−2 )) ,   h3 tm−2 , P m−2 , R m−2 , S m−2 = P (tm−2 ) R (tm−2 ) − lS (tm−2 ) .

The numerical simulations are given in Fig. 6.5 and Fig. 6.6.

(6.38)

Numerical method for a fractional differential equation with power-law kernel

153

Figure 6.6 Numerical simulation for the Chua model with the Caputo fractional derivative for α = 0.96.

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Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel

7

In this section, we deal with numerical solution for the general Cauchy problem with the Atangana–Baleanu fractional derivative in the sense of Caputo. The Atangana Baleanu fractional derivative in the sense of Caputo of the function u (t) is defined as    α AB (α) t d ABC α α − dτ (7.1) u E D u = − τ (τ ) (t) (t ) α a t 1 − α a dτ 1−α and the Atangana–Baleanu fractional derivative in the sense of Riemann–Liouville of the function u (t) is ABR α Dt u (t) = a

AB (α) d 1 − α dt





t

u (τ ) Eα a

 α α − (t − τ ) dτ. 1−α

Here, u (t) ∈ W21 (0, l), α ∈ [0, 1] and AB (α) = 1 − α + with Mittag-Leffler kernel is given by ABC α It (u (t)) = 0

α 1−α u (t) + AB (α) AB (α)  (α)



t

α (α) .

(7.2)

The fractional integral

u (τ ) (t − τ )α−1 dτ.

(7.3)

0

Now, we consider the Cauchy problem with the Mittag-Leffler kernel 

ABC D α u t (t) = h (t, u (t)) , 0 u (0) = u0 .

(7.4)

We transform equation (7.4) into α 1−α h (t, u (t))+ u (t)−u (0) = AB (α) AB (α)  (α)



t

h (τ, u (τ )) (t − τ )α−1 dτ. (7.5)

0

At the point tk+1 = (k + 1) t, we have the following: 1−α h (t, u (t)) AB (α)  tk+1 α + h (τ, u (τ )) (tk+1 − τ )α−1 dτ AB (α)  (α) 0

u (tk+1 ) − u (0) =

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00013-5 Copyright © 2021 Elsevier Inc. All rights reserved.

(7.6)

156

New Numerical Scheme With Newton Polynomial

and we write  1−α  h tk , uk AB (α) k  tm+1  α + h (τ, u (τ )) (tk+1 − τ )α−1 dτ. AB (α)  (α) tm

u (tk+1 ) = u (0) +

(7.7)

m=2

After putting the Newton polynomial into equation (7.7), the above equation can be written as follows: uk+1 = u0 +

+

 1−α  h tk , uk AB (α)

α AB (α)  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  k tm+1 ⎨ 

m=2 tm

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

 h tm−2, um−2



h tm−1 ,um−1 −h tm−2 ,um−2 t × (τ − tm−2 )  h(t ,um )−2h tm−1 ,um−1 +h tm−2 ,um−2 + m 2(t)2 × (τ − tm−2 ) (τ − tm−1 )

+

(7.8) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we have  1−α  h tk , uk AB (α)  ⎧  tm+1 h tm−2 , um−2 (tk+1 − τ )α−1 dτ ⎪ tm ⎪

⎪  tm+1 h tm−1 ,um−1 −h tm−2 ,um−2  ⎪ ⎪ ⎪ + k ⎨ tm t  α + × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

 ⎪ t AB (α)  (α) m m−1 +h t m−2 m−2 ,u m+1 h(tm ,u )−2h tm−1 ,u m=2 ⎪ ⎪ ⎪ + t ⎪ 2 m 2(t) ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

uk+1 = u0 +

(7.9) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

and we have uk+1 = u0 +  ×

k     α 1−α  h tk , uk + h tm−2 , um−2 AB (α) AB (α)  (α) m=2

tm+1

(tk+1 − τ )α−1 dτ

tm





k  h tm−1 , um−1 − h tm−2 , um−2 α + AB (α)  (α) t m=2  tm+1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ tm

(7.10)

Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel

157





k  h (tm , um ) − 2h tm−1 , um−1 + h tm−2 , um−2 α + AB (α)  (α) 2 (t)2 m=2  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ. tm

Performing the calculations for the above integrals in equation (7.10), we obtain the following approximation: uk+1 = u0 + +

 1−α  h tk , uk AB (α)

k     α (t)α h tm−2 , um−2 (k − m + 1)α − (k − m)α AB (α)  (α + 1) m=2

k       α (t)α h tm−1 , um−1 − h tm−2 , um−2 AB (α)  (α + 2) m=2   α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)  

k   α (t)α h (tm , um ) − 2h tm−1 ,um−1 + +h tm−2 , um−2 2AB (α)  (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

+

7.1

(7.11)

Error analysis with the Atangana–Baleanu fractional derivative

We deal with a general Cauchy problem, which has the Atangana–Baleanu fractional derivative  AB α 0 Dt u (t) = h (t, u (t)) , (7.12) u (0) = u0 . Integrating the above equation, we have u (tk+1 ) − u (0) = +

1−α u (tk ) AB (α)

(7.13)

k  tm+1  α (tk+1 − τ )α−1 h (τ, u (τ )) dτ AB (α)  (α) tm m=0

158

New Numerical Scheme With Newton Polynomial

where h(τ, u(τ )) = Pm (τ ) + R(τ ) = h(tm−2 , u(tm−2 )) (7.14) h(tm−1 , u(tm−1 )) − h(tm−2 , u(tm−2 )) (τ − tm−2 ) + t h(tm , u(tm )) − 2h(tm−1 , u(tm−1 )) + h(tm−2 , u(tm−2 )) + 2(t)2 (τ − tm−1 )(τ − tm ) ∂ 2 × (τ − tm−2 )(τ − tm−1 ) + h(τ, u(τ )). 2! ∂τ 2 Then, we write the error k  tm+1  α α Rτ = (tk+1 − τ )α−1 AB(α)(α) tm m=0



∂2 h(τ, u(τ ))|τ =γm ∂τ 2 × (τ −tm−12!)(τ −tm )

 dτ (7.15)

and |Rτα (γτ )| ≤

 2  k  tm+1    α α−1  ∂  (tk+1 − τ )  ∂τ 2 h(τ, u(τ ))|τ =γm  AB(α)(α) tm m=0

(7.16) (τ − tm−1 )(τ − tm ) dτ 2!  2  k  tm+1  ∂  α ≤ (tk+1 − τ )α−1 sup  2 h(τ, u(τ )) AB(α)(α) tm τ ∈[tm ,τ ] ∂τ τ =γm ×

m=0

(τ − tm−1 )(τ − tm ) dτ × 2!  2  ∂  α  ≤ sup  2 h(τ, u(τ )) AB(α)(α) ∂τ τ ∈[tm ,τ ]

k  

tm+1

τ =γτ m=0 tm

(tk+1 − τ )α−1

(τ − tm−1 )(τ − tm ) dτ 2!  2  ∂  α sup  2 h(τ, u(τ )) ≤ AB(α)(α) τ ∈[tm ,tm+1 ] ∂τ τ =γτ ! k  tm+1 2  α−1 τ − (tm−1 + tm )τ + tm−1 tm dτ × (tk+1 − τ ) 2 tm ×

m=0

where γτ is a constant obtained from the Taylor series approximation for the error. We evaluate since tk+1 > τ . Then  tk+1  tm+1 (tk+1 − τ )α−1 τ 2 dτ ≤ (tk+1 − τ )α−1 τ 2 dτ (7.17) tm

0

Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel

 2 ≤ tk+1

1

159

(1 − y)α−1 y 2 dτ

0

2 ≤ tk+1 B(3, α)

≤ (t)α+2 (k + 1)α+2 B(3, α). Then 

tm+1 tm



2 (tk+1 − τ )α−1 τ dτ ≤ tk+1 B(3, α)

(7.18)

≤ (t)α+1 (k + 1)α+1 B(2, α)   tm+1 (k − m + 1)α (k − m)α − . (tk+1 − τ )α−1 dτ ≤ (t)α α α tm

Therefore, we have the following  2  ∂  α  sup  2 h(τ, u(τ )) AB(α)(α) τ ∈[0,tk+1 ] ∂τ τ =γτ ⎧ α+2 α+2 (t) (k + 1) B(3, α) k ⎪ ⎨  −(2m − 1)(t)α+2(k + 1)α+1 B(2, α)  × α ⎪ (k−m+1)α − (k−m) m=0 ⎩ +m(m − 1)(t)α+2 α α  2  ∂  α(t)α+2 ≤ (α, k). sup  2 h(τ, u(τ )) AB(α)(α) τ ∈[0,tk+1 ] ∂τ τ =γτ

|Rτα (γτ )| ≤

(7.19) ⎫ ⎪ ⎬ ⎪ ⎭

where ⎫ (k + 1)α+2 B(3, α) ⎪ ⎬ α+1 B(2, α) −(2m − 1)(k + 1) . (α, k) =   ⎪ (k−m+1)α (k−m)α (α) ⎪ ⎭ − m=0 ⎩ +m(m − 1) α α ⎧

k ⎪ ⎨ 

7.2

(7.20)

Numerical illustrations

To demonstrate the efficiency of the proposed numerical scheme, we shall now give numerical simulations for the solution of some differential equations which contain the Atangana–Baleanu fractional derivative. Example 7.1. We deal with the following test problem with the Atangana–Baleanu fractional derivative: AB α 0 Dt u (t) = sin (t) ,

u (0) = 0.1.

(7.21)

160

New Numerical Scheme With Newton Polynomial

Figure 7.1 Numerical simulation for the AB fractional derivative for α = 0.33.

Thus, the following scheme is given uk+1 = u0 + +

 1−α  h tk , uk AB (α)

k     α (t)α h tm−2 , um−2 (k − m + 1)α − (k − m)α AB (α)  (α + 1) m=2

k       α (t)α h tm−1 , um−1 − h tm−2 , um−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

+

+

(7.22)

 

k   α (t)α h (tm , um ) − 2h tm−1 ,um−1 +h tm−2 , um−2 2AB (α)  (α + 3) 

m=2

 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥ ⎦

 

where h (tm , um ) = sin (tm ), h tm−1 , um−1 = sin (tm−1 ) and h tm−2 , um−2 = sin (tm−2 ). The numerical simulation is depicted in Fig. 7.1 for α = 0.33. Example 7.2. Let us consider the following problem with the Atangana–Baleanu fractional derivative: AB α 0 Dt u (t) = u (t) + ln (t

u (0) = 0.01.

+ 1) ,

(7.23)

Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel

161

Figure 7.2 Numerical simulation for the AB fractional derivative for α = 0.92.

Applying the suggested numerical scheme to this example yields uk+1 = u0 + +

 1−α  h tk , uk AB (α)

k     α (t)α h tm−2 , um−2 (k − m + 1)α − (k − m)α AB (α)  (α + 1) m=2

k       α (t)α h tm−1 , um−1 − h tm−2 , um−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

+

+

(7.24)

 

k   α (t)α h (tm , um ) − 2h tm−1 ,um−1 +h tm−2 , um−2 2AB (α)  (α + 3) 

m=2

 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎦



where h (tm , um ) = u (tm ) + ln (tm + 1), h tm−1 , um−1 = u (tm−1 ) + ln (tm−1 + 1)

 and h tm−2 , um−2 = u (tm−2 ) + ln (tm−2 + 1). The numerical simulation is given in Fig. 7.2 for α = 0.92. Example 7.3. We consider the following problem with the Atangana–Baleanu fractional derivative: AB α 0 Dt u (t) = u (t) cos (t) ,

(7.25)

162

New Numerical Scheme With Newton Polynomial

Figure 7.3 Numerical simulation for the AB fractional derivative for α = 0.42.

u (0) = 0.1. For a numerical solution of the above equation, we can obtain the following scheme:  1−α  h tk , uk uk+1 = u0 + AB (α) +

k     α (t)α h tm−2 , um−2 (k − m + 1)α − (k − m)α AB (α)  (α + 1) m=2

k       α (t)α h tm−1 , um−1 − h tm−2 , um−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

+

+

(7.26)

 

k   α (t)α h (tm , um ) − 2h tm−1 ,um−1 +h tm−2 , um−2 2AB (α)  (α + 3) 

m=2

 ⎤ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12  

where h (tm , um ) = um cos (tm ), h tm−1 , um−1 = um−1 cos (tm−1 ) and h tm−2 , um−2 = um−2 cos (tm−2 ). We depict the numerical simulation in Fig. 7.3 for α = 0.42. ⎡

Example 7.4. We present the fractional differential equation with the Mittag-Leffler kernel AB α 0 Dt u (t) = −6t,

(7.27)

Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel

163

u (0) = 0.5. The following scheme for the numerical solution of such a problem can be written: uk+1 = u0 + +

 1−α  h tk , uk AB (α)

k     α (t)α h tm−2 , um−2 (k − m + 1)α − (k − m)α AB (α)  (α + 1) m=2

k       α (t)α h tm−1 , um−1 − h tm−2 , um−2 AB (α)  (α + 2) m=2   α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

+

(7.28)

 

k   α (t)α h (tm , um ) − 2h tm−1 ,um−1 + +h tm−2 , um−2 2AB (α)  (α + 3) 

m=2

 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎦

  where h (tm , um ) = −6tm , h tm−1 , um−1 = −6tm−1 and h tm−2 , um−2 = −6tm−2 . We provide the numerical simulation for the above equation in Fig. 7.4 for α = 0.93.

Figure 7.4 Numerical simulation for the AB fractional derivative for α = 0.93.

Example 7.5. The Lotka–Volterra model, which is a pair of first order non-linear differential equations, expresses the dynamics of a system where two or more species interacting. Also this model is known as the predator–prey model [48]. We consider

164

New Numerical Scheme With Newton Polynomial

the following Lotka–Volterra 3D model with the Atangana–Baleanu fractional derivative: ⎧ AB D α u (t) = u (4 − v) + aw, ⎨ t 0 

AB D α v (t) = −v 1 − u2 + sin(u), (7.29) t 0 ⎩ AB α D w = −u − w) + cw, (t) (b t 0 with the initial conditions u (0) = 0.1, v (0) = 0.5, w (0) = 0.6. Applying the suggested numerical scheme to this example yields   1−α h1 tk , uk , v k , w k AB (α) α (t)α + AB (α)  (α + 1) k     × h1 tm−2 , um−2 , v m−2 , w m−2 (k − m + 1)α − (k − m)α

uk+1 = u0 +

m=2

  k   α (t)α h1 tm−1 , um−1 , v m−1 , w m−1  −h1 tm−2 , um−2 , v m−2 , w m−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × (7.30) − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ m , v m , wm ) k 

h1 (tm , um−1  α (t)α ⎣ −2h1 tm−1 , u , v m−1 , w m−1 ⎦ +

2AB (α)  (α + 3) m−2 +h1 tm−2 , u , v m−2 , w m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12 +

v k+1 = v 0 +

  1−α h2 tk , uk , v k , w k AB (α)

α (t)α AB (α)  (α + 1) k     × h2 tm−2 , um−2 , v m−2 , w m−2 (k − m + 1)α − (k − m)α +

m=2

+

  k   α (t)α h2 tm−1 , um−1 , v m−1 , w m−1  −h2 tm−2 , um−2 , v m−2 , w m−2 AB (α)  (α + 2) m=2

Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel

165

 (k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k h2 (tm , um , v m , w m ) α 

 α (t) ⎣ −2h2 tm−1 , um−1 , v m−1 , w m−1 ⎦ +

 2AB (α)  (α + 3) +h2 tm−2 , um−2 , v m−2 , w m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 

×

w k+1 = w 0 +

  1−α h3 tk , uk , v k , w k AB (α)

α (t)α AB (α)  (α + 1) k     × h3 tm−2 , um−2 , v m−2 , w m−2 (k − m + 1)α − (k − m)α +

m=2

  k   α (t)α h3 tm−1 , um−1 , v m−1 , w m−1  + −h3 tm−2 , um−2 , v m−2 , w m−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ m , v m , wm ) k 

h3 (tm , um−1  α (t)α ⎣ −2h3 tm−1 , u , v m−1 , w m−1 ⎦ +

2AB (α)  (α + 3) m−2 +h3 tm−2 , u , v m−2 , w m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 where



 h1 tm , um , v m , w m = um 4 − v m + aw m , 

2 

 h2 tm , um , v m , w m = −v m 1 − um + sin(um ),



 h3 tm , um , v m , w m = −um b − w m + cw m ,     h1 tm−1 , um−1 , v m−1 , w m−1 = um−1 4 − v m−1 + aw m−1 ,

(7.31)

166

New Numerical Scheme With Newton Polynomial

Figure 7.5 Numerical simulation for the Lotka–Volterra 3D model with the Atangana–Baleanu fractional derivative for α = 1.

   2 ! + sin(um−1 ), h2 tm−1 , um−1 , v m−1 , w m−1 = −v m−1 1 − um−1     h3 tm−1 , um−1 , v m−1 , w m−1 = −um−1 b − w m−1 + cw m−1 , and

(7.32)

    h1 tm−2 , um−2 , v m−2 , w m−2 = um−2 4 − v m−2 + aw m−2 ,   2 !  m−2 m−2 m−2 m−2 m−2 h2 tm−2 , u = −v ,v ,w 1− u + sin(um−2 ),     h3 tm−2 , um−2 , v m−2 , w m−2 = −um−2 b − w m−2 + cw m−2 .

(7.33)

With the parameters a = 0.3, b = 1.5 and c = −0.05, we present the numerical simulations in Fig. 7.5 and Fig. 7.6.

Numerical method for a fractional differential equation with the generalized Mittag-Leffler kernel

167

Figure 7.6 Numerical simulation for the Lotka–Volterra 3D model with the Atangana–Baleanu fractional derivative for α = 0.98.

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Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel

8

In this section, we give the numerical algorithm for a problem where the derivative is the Caputo–Fabrizio fractal–fractional derivative. Assume that u (t) is continuous and fractal differentiable on an open interval (a, b) with order β, then the fractal–fractional derivative of u (t) with order α in the Riemann-Liouville sense having exponentially decaying type kernel is given by F F E α,β Dt u (t) = 0

M (α) d (1 − α) dt β

 0

t

  α exp − (t − τ ) u (τ ) dτ 1−α

(8.1)

where α > 0, β ≤ n ∈ N and M (0) = M (1) = 1. The fractal–fractional integral of u (t) with order α having exponentially decaying type kernel is given by F F E α,β Jt u (t) = 0

αβ M (α)



t

τ β−1 u (τ ) dτ +

0

β (1 − α) t β−1 u (t) . M (α)

(8.2)

We now address the following Cauchy problem with new exponential decay kernel: F F E α,β Dt u (t) = h (t, u (t)) . 0

(8.3)

Using the definition of the Caputo–Fabrizio fractal–fractional integral, we can convert the above equation into u (t) − u (0) =

1 − α β−1 αβ βt h (t, u (t)) + M (α) M (α)



t

τ β−1 h (τ, u (τ )) dτ. (8.4)

0

Here, we shall take g (τ, u (τ )) = βτ β−1 h (τ, u (τ )) and write Eq. (8.4) at the point tk+1 = (k + 1) t u (tk+1 ) − u (0) =

1−α α g (tk , u (tk )) + M (α) M (α)



tk+1

g (τ, u (τ )) dτ

(8.5)

g (τ, u (τ )) dτ.

(8.6)

0

and at the point tk = kt, we write u (tk ) − u (0) =

1−α α g (tk−1 , u (tk−1 )) + M (α) M (α)



tk

0

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00014-7 Copyright © 2021 Elsevier Inc. All rights reserved.

170

New Numerical Scheme With Newton Polynomial

From Eqs. (8.5) and (8.6), we obtain the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)  tk+1 α g (τ, u (τ )) dτ. + M (α) tk

u (tk+1 ) − u (tk ) =

(8.7)

Now, we insert the Newton polynomial into Eq. (8.8) to obtain get the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α) ⎧

g tk−2 , uk−2 ⎪ ⎪

⎪ ⎪ g tk−1 ,uk−1 −g tk−2 ,uk−2 ⎪  tk+1 ⎪ + ⎨ t α × (τ − tk−2 ) +



k −2g t k−1 +g t k−2 M (α) tk ⎪ g t ,u ,u ⎪ k k−1 k−2 ,u ⎪ + ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

uk+1 − uk =

(8.8) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

dτ

and we organize this as follows; 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)

⎧ g tk−2 , uk−2 t

⎪ ⎪ ⎪ g tk−1 ,uk−1 −g tk−2 ,uk−2 ⎪ ⎪ +  ⎪ t t α ⎨ × tkk+1 (τ − tk−2 ) dτ +





M (α) ⎪ g tk ,uk −2g tk−1 ,uk−1 +g tk−2 ,uk−2 ⎪ ⎪ + ⎪ 2 ⎪ 2(t) ⎪ t ⎩ × tkk+1 (τ − tk−2 ) (τ − tk−1 ) dτ

uk+1 − uk =

(8.9) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

Adding calculations of the integrals on the right hand side of the above equation, we have the following: 1−α (8.10) [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)

⎫ ⎧ k−2 t g t , u ⎬ ⎨ k−2



  α k−1 − g t k−2 5 t + . , u , u + g t k−1 k−2 2





 23 ⎭ M (α) ⎩  k k−1 k−2 + g tk−2 , u + g tk , u − 2g tk−1 , u 12 t

uk+1 = uk +

We know that g (t, u (t)) = βt β−1 h (t, u (t)), then we can re-order the above equation as follows: uk+1 = uk +

 1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) M (α)

Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel 171

+

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

α M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

β−1 k−2 t  βtk−2 h tk−2 , u

 β−1 βtk−1 h tk−1 , uk−1

52 t + β−1 −βtk−2 h tk−2 , uk−2 

 β−1 β−1 βtk h tk , uk − 2βtk−1 h tk−1 , uk−1 23

+ β−1 12 t +βtk−2 h tk−2 , uk−2

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(8.11) Thus, we have the following iteration:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) M (α) ⎧ ⎫

23 β−1 ⎪ ⎪ βtk h tk , uk t ⎨ ⎬ 12

α β−1 4 k−1 . + t − 3 βtk−1 h tk−1 , u ⎪

M (α) ⎪ ⎩ ⎭ β−1 5 k−2 + 12 βtk−2 h tk−2 , u t

uk+1 = uk +

8.1

(8.12)

Predictor–corrector method for fractal–fractional derivative with the exponential decay kernel

Let us consider the following Cauchy problem with the Caputo–Fabrizio fractal– fractional derivative, F F E α,β Dt u (t) = h (t, u (t)) , 0

(8.13)

u (0) = u0 . If we integrate above equation, we have the following: αβ (1 − α) βt β−1 h (t, u (t)) + u (t) = u0 + M (α) M (α)



t

τ β−1 h (τ, u (τ )) dτ. (8.14)

0

At the point tk+1 = (k + 1) t, we have (1 − α) β−1 p

βt h tk+1 , uk+1 M (α) k+1 k  αβ  tm+1 β−1 + τ h (τ, u (τ )) dτ. M (α) tm

u (tk+1 ) = u0 +

m=0

In the interval [tm , tm+1 ], the function h (τ, u (τ )) is approximated as P2 (τ ) = h (tm+1 , um+1 ) + × (τ − tm+1 )

h (tm+1 , um+1 ) − h (tm , um ) t

(8.15)

172

New Numerical Scheme With Newton Polynomial

+

h (tm+1 , um+1 ) − 2h (tm , um ) + h (tm−1 , um−1 )

(8.16)

2 (t)2 × (τ − tm+1 ) (τ − tm ) .

If we put this polynomial into the above equation, we can write

uk+1 = uk +

αβ M (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ k−1 ⎨ ⎪ ⎪ m=0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

 tm+1 β−1 τ dτ

tm h tm+1 ,um+1 −h(tm ,um ) t +t × tmm+1 τ β−1 (τ − tk+1 ) dτ



−2h(tm ,um )+h tm−1 ,um−1 h t ,u + m+1 m+1 2(t)2 t × tmm+1 τ β−1 (τ − tk+1 ) (τ − tk ) dτ h (tm+1 , um+1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(8.17)

.

Here, we write (1 − α) αβ β−1 p

tk+1 h tk+1 , uk+1 M (α) ⎧

β ⎪ h (tm+1 , um+1 ) (t) (m + 1)β − mβ ⎪ β ⎪

⎪ ⎪ h tm+1 ,um+1 −h(tm ,um ) ⎪ ⎪ + ⎪ t  ⎪

 ⎪ (t)β+1 mβ (m+1+β)−(m+1)β+1 k−1 ⎪ × αβ  ⎨ β(β+1) +



⎪ M (α) −2h(t h t ,u m ,um )+h tm−1 ,um−1 m+1 m+1 m=0 ⎪ ⎪ + ⎪ 2 ⎪   2(t)    ⎪  ⎪ ⎪ 2(t)β+2 m− β2 (m+1)β+1 −mβ+1 m+1+ β2 ⎪ ⎪ × ⎪ ⎩ β(β+1)(β+2)

uk+1 = uk +

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(8.18) Also, we can write (1 − α) αβ β−1 p

tk+1 h tk+1 , uk+1 M (α) ⎧

β β β ⎪ h (tm+1 , um+1 ) (t) ⎪ β (m + 1) − m ⎪ ⎪ ⎪ h tm+1 ,um+1 −h(tm ,um ) ⎪ ⎪ ⎪ t  + ⎪

 ⎪ (t)β+1 mβ (m+1+β)−(m+1)β+1 k−1 ⎪ ⎨  × αβ β(β+1) +



⎪ M (α) h tm+1 ,um+1 −2h(tm ,um )+h tm−1 ,um−1 ⎪ m=0 ⎪ + ⎪ 2 ⎪    2(t)    ⎪ ⎪ ⎪ 2(t)β+2 m− β2 (m+1)β+1 −mβ+1 m+1+ β2 ⎪ ⎪ ⎪ ⎩ × β(β+1)(β+2)

uk+1 = uk +

+

β

αβ p (t) h tk+1 , uk+1 (k + 1)β − k β M (α) β

(8.19) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel 173

where p

uk+1 = u0 +

k αβ  (1 − α) β β−1 tk+1 h (tk , uk ) + h (tm , um ) M (α) M (α)

(8.20)

m=0

×

(t)β (m + 1)β − mβ . β

We write the first iteration u1 =

8.2

αβ (1 − α) β (t)β h (t0 , u0 ) . (t)β−1 h (t0 , u0 ) + M (α) M (α) β

(8.21)

Error analysis with the Caputo–Fabrizio fractal–fractional derivative

Let the following be a general Cauchy problem: F F E α,β Dt u (t) = h (t, u (t)) ; 0

(8.22)

this is a problem with the Caputo–Fabrizio differential operator. At the point t = tk+1 , after applying the associated fractional integral operator, we have  tk+1 αβ 1 − α β−1 βt h (t, u (t)) + τ β−1 h (τ, u (τ )) dτ M (α) M (α) 0 1 − α β−1 = βt h (t, u (t)) M (α) k  αβ  tm+1 β−1 + τ h (τ, u (τ )) dτ. (8.23) M (α) tm

u (tk+1 ) − u (0) =

m=0

The function h(τ, u(τ )) can be approximated within the interval [tm , tm+1 ] using the Newton polynomial (8.24) h(τ, u(τ )) = Pm (τ ) + R(τ ) = h(tm−2 , u(tm−2 )) h(tm−1 , u(tm−1 )) − h(tm−2 , u(tm−2 )) (τ − tm−2 ) + t h(tm , u(tm )) − 2h(tm−1 , u(tm−1 )) + h(tm−2 , u(tm−2 )) + 2(t)2 (τ − tm−1 )(τ − tm ) ∂ 2 × (τ − tm−2 )(τ − tm−1 ) + h(τ, u(τ ))|τ =γτ . 2! ∂τ 2

174

New Numerical Scheme With Newton Polynomial

Then, we write the error Rτα (γτ ) =

  k  (τ − tm−1 )(τ − tm ) αβ  tm+1 β−1 ∂ 2 τ h(τ, u(τ ))| dτ τ =γm M(α) 2! ∂τ 2 tm m=0

(8.25) and |Rτα (γτ )| ≤

  k   αβ  tm+1 β−1  ∂ 2  τ h(τ, u(τ ))| τ =γ m  2 2M(α) ∂τ tm

(8.26)

m=0

× (τ 2 − (tm−1 + tm )τ + tm−1 tm )dτ  2  ∂  αβ  sup  2 h(τ, u(τ )) ≤ 2M(α) ∂τ τ ∈[0,tk+1 ]

×

k  tm+1 

τ =γτ

  τ β+2 − t (2m − 1)τ β+1 + m(m − 1)τ β (t)2

m=0 tm

 2  ∂  αβ  sup  2 h(τ, u(τ )) ≤ 2M(α) τ ∈[0,tk+1 ] ∂τ τ =γτ  k   tm+1  τ β+2 − t (2m − 1)τ β+1 + m(m − 1)τ β (t)2 × m=0 tm

where γτ is a constant obtained from the Taylor series approximation for the error. Then, we calculate  tm+1  (t)β+2  (m + 1)β+2 − mβ+2 (8.27) τ β+2 dτ ≤ β +2 tm  tm+1   (t)β+2 t (2m − 1)τ β+1 dτ ≤ (2m − 1) (m + 1)β+1 − mβ+1 β +1 tm  tm+1

(t)β+2 (m + 1)β − mβ . m(m − 1)τ β (t)2 dτ ≤ m(m − 1) β tm Therefore, we have the following  2  ∂  αβ  sup  2 h(τ, u(τ )) 2M(α) τ ∈[0,tk+1 ] ∂τ τ =γτ ⎧

β+2 (t) β+2 ⎪ (m + 1) − mβ+2 ⎪ k β+2 ⎨ 

β+2 × − (t) (2m − 1) (m + 1)β+1 − mβ+1 β+1 ⎪ m=0 ⎪ ⎩ +m(m − 1) (t)β+2 (m + 1)β − mβ

β  2  β+2 ∂  α(t)  sup h(τ, u(τ )) (β, k) ≤ 2M(α) τ ∈[0,tk+1 ]  ∂τ 2 τ =γτ

|Rτα (γτ )| ≤

(8.28) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel 175

where ⎧ ⎪



(m + 1)β+2 − mβ+2

β+2 β+1 − mβ+1 (β, k) = − (t) β+1 (2m − 1) (m + 1) ⎪ m=0 ⎪ ⎩ +m(m − 1) (t)β+2 (m + 1)β − mβ

β (t)β+2 β+2

k ⎪ ⎨ 

8.3

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

.

(8.29)

Numerical illustrations

In this section, we give a numerical simulation for the solution of some differential equations involving the Caputo–Fabrizio fractal–fractional derivative. Example 8.1. Let us consider the following problem with the Caputo–Fabrizio fractal–fractional derivative: F F E α,β Dt u (t) = t 0

cos (t) ,

(8.30)

u (0) = 1. The following scheme is presented:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) M (α) ⎧ ⎫

23 β−1 ⎪ ⎪ βtk h tk , uk t ⎨ ⎬ 12

α β−1 4 k−1 + − 3 βtk−1 h tk−1 , u t ⎪

M (α) ⎪ ⎩ ⎭ β−1 5 + 12 βtk−2 h tk−2 , uk−2 t

uk+1 = uk +

(8.31)





where h tk , uk = tk cos (tk ), h tk−1 , uk−1 = tk−1 cos (tk−1 ) and h tk−2 , uk−2 = tk−2 cos (tk−2 ). The numerical simulation is presented in Fig. 8.1 for α = 0.9, β = 0.8. Example 8.2. We next consider the following fractal–fractional differential equation with exponential decay kernel: F F E α,β Dt u (t) = −3u (t) + 5t, 0

(8.32)

u (0) = 0.5. We have the following scheme for a numerical solution of the above problem:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) M (α) ⎧ ⎫

23 β−1 ⎪ ⎪ βtk h tk , uk t ⎨ ⎬ 12

α β−1 4 k−1 + t − 3 βtk−1 h tk−1 , u ⎪

M (α) ⎪ ⎩ ⎭ β−1 5 + 12 βtk−2 h tk−2 , uk−2 t

uk+1 = uk +

(8.33)

176

New Numerical Scheme With Newton Polynomial

Figure 8.1 Numerical simulation for the Caputo–Fabrizio fractal–fractional derivative for α = 0.9, β = 0.8.

Figure 8.2 Numerical simulation for the Caputo–Fabrizio fractal–fractional derivative for α = 0.8, β = 0.6.





where h tk , uk = −3uk + 5tk , h tk−1 , uk−1 = −3uk−1 + 5tk−1 and h tk−2 , uk−2 = −3uk−2 + 5tk−2 . We give the numerical simulation in Fig. 8.2 for α = 0.8, β = 0.6. Example 8.3. We consider the following test problem with the Caputo–Fabrizio fractal–fractional derivative: F F E α,β Dt u (t) = u4 (t) − u2 (t) , 0

(8.34)

u (0) = −0.2. The following scheme is given:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) uk+1 = uk + M (α) ⎧ ⎫

23 β−1 k t ⎪ ⎪ βt h t , u k ⎨ ⎬ 12 k

α β−1 4 k−1 + − 3 βtk−1 h tk−1 , u t ⎪

M (α) ⎪ ⎩ ⎭ β−1 5 + 12 βtk−2 h tk−2 , uk−2 t

(8.35)

Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel 177

Figure 8.3 Numerical simulation for the Caputo–Fabrizio fractal–fractional derivative for α = 0.96, β = 0.5.



where h tk , uk = u4 (tk ) − u2 (tk ), h tk−1 , uk−1 = u4 (tk−1 ) − u2 (tk−1 ) and

h tk−2 , uk−2 = u4 (tk−2 ) − u2 (tk−2 ). The numerical simulation is given in Fig. 8.3 for α = 0.96, β = 0.5. Example 8.4. We present the following differential equation with the Caputo– Fabrizio fractal–fractional derivative: F F E α,β Dt u (t) = u3 (t) t, 0

(8.36)

u (0) = 0.1. Thus the following scheme is given:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) M (α) ⎧ ⎫

23 β−1 ⎪ ⎪ βtk h tk , uk t ⎨ ⎬ 12

α β−1 + − 43 βtk−1 h tk−1 , uk−1 t ⎪

M (α) ⎪ ⎩ ⎭ β−1 5 + 12 βtk−2 h tk−2 , uk−2 t

uk+1 = uk +

(8.37)





where h tk , uk = u3 (tk ) tk , h tk−1 , uk−1 = u3 (tk−1 ) tk−1 and h tk−2 , uk−2 = u3 (tk−2 ) tk−2 . The numerical simulation is provided in Fig. 8.4 for α = 0.9, β = 0.6. Example 8.5. The Arneodo chaotic model, which displays chaotic behavior, was introduced by Arneodo, Coullet and Spiegel [49]. This system provides a description of the dynamics of triple convection. We consider the following Arneodo system with the Caputo–Fabrizio fractal–fractional derivative: ⎧ F F E D α,β u (t) = v, ⎪ ⎨ t 0 F F E D α,β v (t) = w, (8.38) t 0 ⎪ ⎩ F F E α,β 3 Dt w (t) = au + bv + cw − u , 0

178

New Numerical Scheme With Newton Polynomial

Figure 8.4 Numerical simulation for the Caputo–Fabrizio fractal–fractional derivative for α = 0.9, β = 0.6.

with the initial conditions u (0) = 0.1, v (0) = 0, w (0) = 0.1.

(8.39)

For this model, we take a = 0.8, b = −1.1 and c = −0.45. Applying the suggested numerical scheme to this example yields  1 − α  β−1  βtk h1 tk , uk , v k , w k M (α)   β−1 −βtk−1 h1 tk−1 , uk−1 , v k−1 , w k−1 ⎧

23 β−1 k k k ⎪ 12 βtk h1 tk , u , v , w t

α ⎨ β−1 + − 43 βtk−1 h1 tk−1 , uk−1 , v k−1 , w k−1 t ⎪

M (α) ⎩ β−1 5 + 12 βtk−2 h1 tk−2 , uk−2 , v k−2 , w k−2 t

uk+1 = uk +

 1 − α  β−1  βtk h2 tk , uk , v k , w k M (α)   β−1 −βtk−1 h2 tk−1 , uk−1 , v k−1 , w k−1 ⎧

23 β−1 ⎪ βtk h2 tk , uk , v k , w k t ⎨ 12

α β−1 + − 43 βtk−1 h2 tk−1 , uk−1 , v k−1 , w k−1 t

M (α) ⎪ ⎩ β−1 5 + 12 βtk−2 h2 tk−2 , uk−2 , v k−2 , w k−2 t

(8.40)

⎫ ⎪ ⎬ ⎪ ⎭

,

v k+1 = v k +

 1 − α  β−1  βtk h3 tk , uk , v k , w k M (α)   β−1 −βtk−1 h3 tk−1 , uk−1 , v k−1 , w k−1

w k+1 = w k +

⎫ ⎪ ⎬ ⎪ ⎭

,

Numerical method for a fractal–fractional ordinary differential equation with exponential decay kernel 179

Figure 8.5 Numerical simulation for the Caputo–Fabrizio fractal–fractional derivative for α = 1, β = 1.

+

⎧ ⎪ ⎨

α M (α) ⎪ ⎩



23 β−1 k k k 12 βtk h3 tk , u , v , w t

β−1 − 43 βtk−1 h3 tk−1 , uk−1 , v k−1 , w k−1 t

β−1 5 + 12 βtk−2 h3 tk−2 , uk−2 , v k−2 , w k−2 t

⎫ ⎪ ⎬ ⎪ ⎭

.

where   h1 tk , uk , v k , w k = v k ,   h2 tk , uk , v k , w k = w k ,    3 h3 tk , uk , v k , w k = auk + bv k + cw k − uk ,

(8.41)

  h1 tk−1 , uk−1 , v k−1 , w k−1 = v k−1 ,   h2 tk−1 , uk−1 , v k−1 , w k−1 = w k−1 ,   3  h3 tk−1 , uk−1 , v k−1 , w k−1 = auk−1 + bv k−1 + cw k−1 − uk−1 ,

(8.42)

180

New Numerical Scheme With Newton Polynomial

Figure 8.6 Numerical simulation for the Caputo–Fabrizio fractal–fractional derivative for α = 0.98, β = 0.96.

and

  h1 tk−2 , uk−2 , v k−2 , w k−2 = v k−2 ,   h2 tk−2 , uk−2 , v k−2 , w k−2 = w k−2 ,   3  h3 tk−2 , uk−2 , v k−2 , w k−2 = auk−2 + bv k−2 + cw k−2 − uk−2 .

The numerical simulations are presented in Fig. 8.5 and Fig. 8.6.

(8.43)

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

9

In this section, to derive the numerical algorithm, we consider a general Cauchy problem with the Caputo fractal–fractional derivative. If the function u (t) is continuous and fractal differentiable on an open interval (a, b) with order τ , then the fractal– fractional derivative of u (t) with order α in the sense of Riemann-Liouville having a power law type kernel is of the form  t d 1 F F P α,β D u = (9.1) (t) (t − τ )n−α−1 u (τ ) dτ t 0  (n − α) dt β 0 where n − 1 < α, β ≤ n ∈ N. Also the fractal–fractional integral of u (t) with order α having a power law type kernel is given by  t β F F P α,β J u = (9.2) (t) (t − τ )α−1 τ β−1 u (τ ) dτ. t 0  (α) 0 We consider the following problem: F F P α,β Dt u (t) = h (t, u (t)) . 0

Eq. (9.3) can be reorganized as follows:  t β u (t) − u (0) = τ β−1 h (τ, u (τ )) (t − τ )α−1 dτ.  (α) 0 For convenience, if we write g (t, u (t)) = βt β−1 h (t, u (t)), then we get  t 1 g (τ, u (τ )) (t − τ )α−1 dτ. u (t) − u (0) =  (α) 0 At the point tk+1 = (k + 1) t, we can write the following:  tk+1 1 g (τ, u (τ )) (tk+1 − τ )α−1 dτ. u (tk+1 ) − u (0) =  (α) 0

(9.3)

(9.4)

(9.5)

(9.6)

Thus, we have u (tk+1 ) = u (0) +

k  1  tm+1 g (τ, u (τ )) (tk+1 − τ )α−1 dτ.  (α) tm m=2

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00015-9 Copyright © 2021 Elsevier Inc. All rights reserved.

(9.7)

182

New Numerical Scheme With Newton Polynomial

Replacing the Newton polynomial into Eq. (9.7), we have ⎧  m−2 g t , u ⎪ m−2 ⎪ ⎪ ⎪ g t ,um−1 −g tm−2 ,um−2 k  (τ − tm−2 ) 1  tm+1 ⎨ + m−1 k+1 t   = u0 + u g(tm ,um )−2g tm−1 ,um−1 +g tm−2 ,um−2 ⎪  (α) ⎪ + m=2 tm ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ.

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ (9.8)

Thus, we can rearrange the above equation as follows:

uk+1 = u0 +

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k ⎨ ⎪ m=2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

tm+1  g tm−2 , um−2 (tk+1 − τ )α−1 dτ tm  

t g t ,um−1 −g tm−2 ,um−2 + tmm+1 m−1 t × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

tm+1 g(tm ,um )−2g tm−1 ,um−1 +g tm−2 ,um−2 + tm 2 2(t)

× (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(9.9)

and we have k   tm+1 1   g tm−2 , um−2 (tk+1 − τ )α−1 dτ  (α) t m m=2   k m−1 − g tm−2 , um−2 1  g tm−1 , u +  (α) t m=2  tm+1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

uk+1 = u0 +

(9.10)

tm

  k 1  g (tm , um ) − 2g tm−1 , um−1 + g tm−2 , um−2  (α) 2 (t)2 m=2  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ. +

tm

We can calculate the above integrals in Eq. (9.10) as follows:  

tm+1 tm tm+1

(tk+1 − τ )α−1 dτ =

(τ − tm−2 ) (tk+1 − τ )α−1 dτ

tm

=

 (t)α  (k − m + 1)α − (k − m)α , α

(t)α+1 α (α + 1)



(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α)

 ,

(9.11)

Numerical method for a fractal–fractional ordinary differential equation with power law kernel



(t)α+2 α (α + 1) (α + 2)   ⎤ 2 2 (k − m) + (3α + 10) (k − m) α (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥. 2 ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

tm+1 tm

⎡

⎢ +⎢ ⎣

183

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ =

If we plug these calculations into Eq. (9.10), we have the following scheme: k  (t)α   g tm−2 , um−2  (α + 1) m=2   α × (k − m + 1) − (k − m)α

uk+1 = u0 +

k    (t)α    g tm−1 , um−1 − g tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

+

  k  (t)α  g (tm , um ) − 2g tm−1 , um−1  + +g tm−2 , um−2 2 (α + 3) m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

(9.12)

⎤ ⎥ ⎥. ⎦

Thus, replacing the function g (t, u (t)) by its value, one has the following:

uk+1 = u0 +

k   β (t)α  β−1  tm−2 h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

k    β (t)α   β−1  β−1 tm−1 h tm−1 , um−1 − tm−2 h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   k β−1 β−1  β (t)α  tm h (tm , um ) − 2tm−1 h tm−1 , um−1 + β−1  2 (α + 3) +tm−2 h tm−2 , um−2 m=2

+

(9.13)

184

New Numerical Scheme With Newton Polynomial

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

9.1

⎤ ⎥ ⎥. ⎦

Predictor–corrector method for fractal–fractional derivative with power law kernel

In this subsection, we present the predictor–corrector method for the following Cauchy problem having the fractal–fractional derivative with power law kernel: F F P α,β Dt u (t) = h (t, u (t)) , 0

(9.14)

u (0) = u0 . If we apply the fractal–fractional integral, we obtain the following: u (t) − u (0) =

β  (α)



t

τ β−1 (t − τ )α−1 h (τ, u (τ )) dτ.

(9.15)

0

We have the following u (tk+1 ) = u0 +

β  (α)



tk+1

τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ

(9.16)

0

at the point tk+1 = (k + 1) t and we can write it as follows: k  β  tm+1 β−1 τ uk+1 = u0 + (tk+1 − τ )α−1 h (τ, u (τ )) dτ.  (α) tm

(9.17)

m=0

For the approximation of the function h (τ, u (τ )), we write the Newton polynomial which is given by h (tm+1 , um+1 ) − h (tm , um ) (τ − tm+1 ) t h (tm+1 , um+1 ) − 2h (tm , um ) + h (tm−1 , um−1 )

P2 (τ ) = h (tm+1 , um+1 ) + +

2 (t)2 × (τ − tm+1 ) (τ − tm ) .

If we put this polynomial into the above equation, we can write

(9.18)

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

β uk+1 = u0 +  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k−1 ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

m=0 ⎪ ⎪ ⎪

t h (tm+1 , um+1 ) × tmm+1 τ β−1 (tk+1 − τ )α−1 dτ

185

 h tm+1 ,um+1 −h(tm ,um ) t

t × tmm+1 τ β−1 (tk+1 − τ )α−1 (τ − tm+1 ) dτ   −2h(tm ,um )+h tm−1 ,um−1 h t ,u + m+1 m+1 2 2(t)

t × tmm+1 τ β−1 (tk+1 − τ )α−1 (τ − tm+1 ) (τ − tm ) dτ

+

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(9.19) Here α,β

A1,m = α,β

A2,m = α,β

A3,m =



tm+1

tm  tm+1 tm  tm+1

τ β−1 (tk+1 − τ )α−1 dτ, τ β−1 (tk+1 − τ )α−1 (τ − tm+1 ) dτ,

(9.20)

τ β−1 (tk+1 − τ )α−1 (τ − tm+1 ) (τ − tm ) dτ,

tm

and we write

⎧ ⎫  h tm+1 ,um+1 −h(tm ,um ) α,β ⎬ α,β k A β  ⎨ h (tm+1, um+1 ) A 1,m + 2,m  t uk+1 = u0 + . h tm+1 ,um+1 −2h(tm ,um )+h tm−1 ,um−1 α,β ⎩ ⎭  (α) + A 2 3,m m=0 2(t)

(9.21) Here α,β A1,m

((k + 1) t)α−1 = β



   ((m + 1) t)β hypergeom [β, 1 − α] , [1 + β] , m+1 k+1 ,  − (mt)β hypergeom [β, 1 − α] , [1 + β] , mk (9.22) ⎡

α,β

A2,m =

and

((k + 1) t)α−1 β (β + 1)

⎤ β((m + 1) t)β+1  ⎢ ×hypergeom [1 + β, 1 − α] , [2 + β] , m+1 ⎥ ⎢ ⎥ k+1 ⎢ ⎥ β+1 ⎢ ⎥ − + 1) t) + β) ((m (1 ⎢  ⎥  ⎢ ⎥ m+1 ⎢ ×hypergeom [β, 1 − α] , [1 + β] , k ⎥ ⎢ ⎥, β+1 ⎢ ⎥ −β (mt) ⎢ ⎥   ⎢ ⎥ m ⎢ ×hypergeom [1 + β, 1 − α] , [2 + β] , k+1 ⎥ ⎢ ⎥ β ⎣ ⎦ +t (mt) + β) + m) (1 (1  m ×hypergeom [β, 1 − α] , [1 + β] , k (9.23)

186

New Numerical Scheme With Newton Polynomial

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ α−1 ⎢ ⎢ ((k + 1) t) α,β ⎢ A3,m = β (β + 1) (β + 2) ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

β (β + 1) ((m + 1) t)β+2



⎤

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ β+1 ⎥ +mt (β + 1) + 2) + 1) t) (β ((m  ⎥  ⎥ m+1 ×hypergeom [β, 1 − α] , [1 + β] , k ⎥ ⎥.   ⎥ +2β (β + 2) m + 12 t (mt)β+1 ⎥  ⎥  ⎥ m ×hypergeom [1 + β, 1 − α] , [2 + β] , k+1 ⎥ ⎥ ⎥ −β(β + 1) (mt)β+2  ⎥ ⎥ m ⎥ ×hypergeom [2 + β, 1 − α] , [3 + β] , k+1 ⎥ β+1 ⎥ −t (β + 1) (β + 2) + 1) (m (mt)  ⎦  m ×hypergeom [β, 1 − α] , [1 + β] , k+1

×hypergeom [2 + β, 1 − α] , [3 + β] , m+1 k+1   1 β+1 −2β (β + 2) m + 2 t ((m + 1) t)   ×hypergeom [1 + β, 1 − α] , [2 + β] , m+1 k+1

(9.24) So, we write the following:

uk+1 = u0 +

= u0 +

+

β  (α)

+

β  (α)

β  (α)

⎧ ⎪ k ⎪ ⎨ 

⎫ ⎪ ⎪ ⎬

α,β

h (tm+1 , u m+1 ) A1,m 

h t ,um+1 −h(tm ,um ) α,β + m+1 A2,m (9.25)   ⎪ h tm+1 ,um+1 −2h(tt ⎪ α,β ⎪ m ,um )+h tm−1 ,um−1 m=0 ⎪ ⎩ + ⎭ A3,m 2(t)2 ⎧ ⎫ α,β ⎪ ⎪ h (tm+1 , u m+1 ) A1,m ⎪ ⎪ k−1  ⎬ β ⎨ h tm+1 ,um+1 −h(tm ,um ) α,β + A 2,m t ⎪ htm+1 ,um+1 −2h(tm ,um )+htm−1 ,um−1 α,β ⎪  (α) ⎪ m=0 ⎪ ⎩ + A3,m ⎭ 2(t)2  p  β h tk+1 , uk+1 − h (tk , uk ) α,β p α,β h tk+1 , uk+1 A1,k + A2,k

 (α) t  p h tk+1 , uk+1 − 2h (tk , uk ) + h (tk−1 , uk−1 ) 2

2 (t)

α,β

A3,k

where p

uk+1 = u0 +

k  β  tm+1 β−1 τ (tk+1 − τ )α−1 h (τ, u (τ )) dτ  (α) tm m=0

= u0 +

 tm+1 k β  h (tm , um ) τ β−1 (tk+1 − τ )α−1 dτ  (α) tm m=0

= u0 +

k β  α,β h (tm , um ) A1,m .  (α) m=0

(9.26)

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

187

Thus, we have the following: β  (α)

⎧ ⎪ k ⎪ ⎨ 

α,β

h (tm+1 , u m+1 ) A1,m 

h t

,u

−h(t ,u )

α,β

⎫ ⎪ ⎪ ⎬

m m m+1 + m+1 A2,m t   ⎪ ⎪ m=0 ⎪ ⎩ + h tm+1 ,um+1 −2h(tm ,um )+h tm−1 ,um−1 Aα,β ⎪ ⎭ 2 3,m 2(t) ⎧ ⎫ α,β ⎪ ⎪ h (tm+1 , u m+1 ) A1,m ⎪ ⎪ k−1  ⎨ ⎬ β  h tm+1 ,um+1 −h(tm ,um ) α,β + A = u0 + (9.27) 2,m  ⎪ htm+1 ,um+1 −2h(tt ⎪  (α) α,β ⎪ m ,um )+h tm−1 ,um−1 m=0 ⎪ ⎩ + ⎭ A3,m 2(t)2  p  β β h tk+1 , uk+1 − h (tk , uk ) α,β p α,β + h tk+1 , uk+1 A1,k + A2,k  (α)  (α) t  p β h tk+1 , uk+1 − 2h (tk , uk ) + h (tk−1 , uk−1 ) α,β + A3,k .  (α) 2 (t)2

uk+1 = u0 +

9.2

Error analysis with Caputo fractal–fractional derivative

We consider the following general Cauchy problem to calculate the error of the suggested scheme: F F P α,β Dt y (t) = h (t, u (t)) , 0

(9.28)

which has the fractal–fractional differential operator with power-law kernel. At the point t = tk+1 , when applying the associated fractional integral operator, we obtain the following:  tm+1 β u (tk+1 ) − u (0) = τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ (9.29)  (α) 0 k  β  tm+1 β−1 = τ (tk+1 − τ )α−1 h (τ, u (τ )) dτ.  (α) tm m=0

We can have the approximation for the function h(τ, u(τ )) within the interval [tm , tm+1 ] using the Newton polynomial h(τ, u(τ )) = Pm (τ ) + R(τ ) = h(tm−2 , u(tm−2 )) h(tm−1 , u(tm−1 )) − h(tm−2 , u(tm−2 )) + (τ − tm−2 ) t h(tm , u(tm )) − 2h(tm−1 , u(tm−1 )) + h(tm−2 , u(tm−2 )) + (9.30) 2(t)2 (τ − tm−1 )(τ − tm ) ∂ 2 × (τ − tm−2 )(τ − tm−1 ) + h(τ, u(τ ))|τ =γτ . 2! ∂τ 2

188

New Numerical Scheme With Newton Polynomial

Then, we write the error k  β  tm+1 β−1 τ (tk+1 − τ )α−1 (τ − tm−1 )(τ − tm ) (9.31) 2(α) m=0 tm  2  ∂ × h(τ, u(τ ))|τ =γm dτ ∂τ 2 k  β  tm+1 β−1 τ (tk+1 − τ )α−1 (τ 2 − (tm−1 + tm )τ + tm−1 tm ) = 2(α) t m m=0   2 ∂ h(τ, u(τ ))| × τ =γ m dτ ∂τ 2 k    β  tm+1 = (tk+1 − τ )α−1 τ β+1 − (tm−1 + tm )τ β + tm−1 tm τ β−1 2(α) m=0 tm   2 ∂ × h(τ, u(τ ))| τ =γm dτ ∂τ 2

Rτα (γτ ) =

and k  β  tm+1 β−1 τ (tk+1 − τ )α−1 (τ − tm−1 )(τ − tm ) (9.32) 2(α) t m m=0   2 ∂ h(τ, u(τ ))| × τ =γm dτ ∂τ 2 k  β  tm+1 β−1 = τ (tk+1 − τ )α−1 (τ 2 − (tm−1 + tm )τ + tm−1 tk ) 2(α) t m=0 m   2 ∂ × h(τ, u(τ ))|τ =γm dτ ∂τ 2 k    β  tm+1 = (tk+1 − τ )α−1 τ β+1 − (tm−1 + tm )τ β + tm−1 tm τ β−1 2(α) m=0 tm   2 ∂ × h(τ, u(τ ))|τ =γm dτ ∂τ 2

Rτα (γτ ) =

Thus we write ∂2 β sup (9.33) h(τ, u(τ )) 2(α) τ ∈[0,tk+1 ] ∂τ 2 τ =γτ k  tm+1  (tk+1 − τ )α−1 (τ β+1 − (tm−1 + tm )τ β + tm−1 tm τ β−1 )dτ ×

|Rτα (γτ )| ≤

m=0 tm

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

≤

∂2 β h(τ, u(τ )) sup 2(α) τ ∈[0,tk+1 ] ∂τ 2 τ =γτ ⎧

tm+1 α−1 − τ ) τ β+1 dτ k ⎪ ⎨ tm (tk+1 

tm+1 −(tm−1 + tm ) tm (tk+1 − τ )α−1 τ β dτ ×

⎪ m=0 ⎩ +tm−1 tm tm+1 (tk+1 − τ )α−1 τ β−1 dτ tm

189

⎫ ⎪ ⎬ ⎪ ⎭

where γτ is a constant obtained from the Taylor series approximation for the error. Now we have to evaluate each integral separately  tm+1  tk+1 α−1 β+1 (tk+1 − τ ) τ dτ ≤ (tk+1 − τ )α−1 τ β+1 dτ (9.34) 0

tm



β+1+α

≤ tk+1

1

(1 − y)α−1 y β+1+α dτ

0

β+1+α

≤ tk+1

B(β + 2 + α, α)

≤ (t) (k + 1)β+1+α B(β + 2 + α, α) ,  tm+1  tk+1 (tk+1 − τ )α−1 τ β dτ ≤ (tk+1 − τ )α−1 τ β dτ (9.35) β+1+α

0

tm

β+α

≤ tk+1



1

(1 − y)α−1 y β+α dτ

0

β+α

≤ tk+1 B(β + 1 + α, α) ≤ (t)β+α (k + 1)β+α B(β + 1 + α, α) , and



tm+1

 (tk+1 − τ )

α−1 β−1

τ

tk+1

dτ ≤

(tk+1 − τ )α−1 τ β+1 dτ

0

tm

β−1+α

≤ tk+1

1

(1 − y)α−1 y β−1+α dτ

0

β−1+α

≤ tk+1



(9.36)

B(β + α, α)

≤ (t)β−1+α (k + 1)β−1+α B(β + α, α). Replacing the above to the error, we get |Rτα (γτ )| ≤

≤

∂2 β h(τ, u(τ )) sup 2(α) τ ∈[0,tk+1 ] ∂τ 2 τ =γτ ⎧ β+1+α β+1+α k ⎨ (k + 1) B(β + 2 + α, α) (t)  × −(2m − 1)(t)β+α+1 (k + 1)β+α B(β + 1 + α, α) ⎩ m=0 +m(m − 1)(t)β−1+α (k + 1)β−1+α B(β + α, α) β(t)β+1+α (k + 1)β+α 2(α)

sup

τ ∈[0,tk+1 ]

∂2 h(τ, u(τ )) ∂τ 2

τ =γτ

(9.37) ⎫ ⎬ ⎭

190

New Numerical Scheme With Newton Polynomial

×

m=0

≤

≤

⎫ (k + 1)B(β + 2 + α, α) ⎬ −(2m − 1)B(β + 1 + α, α) ⎭ ⎩ +m(m − 1)(k + 1)−1 B(β + α, α)

⎧ k ⎨ 

∂2 β(t)β+1+α (k + 1)β+α h(τ, u(τ )) sup 2 2(α) τ ∈[0,tk+1 ] ∂τ τ =γτ ⎧ ⎫ 2 k ⎨ (k + 1) B(β + 2 + α, α) ⎬  −((k + 1)k − (k + 1))B(β + 1 + α, α) × ⎩ k(k+1)k ⎭ m=0 + 2 (k + 1)−1 B(β + α, α) β(t)β+1+α (k + 1)β+α 2(α)

sup

τ ∈[0,tk+1 ]

∂2 h(τ, u(τ )) ∂τ 2

(α, β, k) τ =γτ

where ⎧ k ⎨ 

⎫ (k + 1)2 B(β + 2 + α, α) ⎬ −(k + 1)(k − 1)B(β + 1 + α, α) . (α, β, k) = ⎩ ⎭ m=0 + k2 B(β + α, α)

9.3

(9.38)

Numerical illustrations

We shall now discuss the efficiency and accuracy of our numerical scheme by taking some differential equations which have the Caputo fractal–fractional derivative. Example 9.1. Let us consider the following problem with the Caputo fractal– fractional derivative: F F P α,β Dt u (t) = t 2 sin (t) , 0

(9.39)

u (0) = 1. For the solution of the above equation, we get the following scheme: uk+1 = u0 +

k   β (t)α  β−1  tm−2 h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

k    β (t)α   β−1  β−1 tm−1 h tm−1 , um−1 − tm−2 h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   k β−1 β−1  β (t)α  tm h (tm , um ) − 2tm−1 h tm−1 , um−1 + β−1  2 (α + 3) +tm−2 h tm−2 , um−2 m=2

+

(9.40)

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

191

Figure 9.1 Numerical simulation for the Caputo fractal–fractional derivative for α = 0.8, β = 0.9.



 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎦

  2 sin (t ), h t m−1 = t 2 m−2 = where h (tm , um ) = tm m m−1 , u m−1 sin (tm−1 ) and h tm−2 , u 2 tm−2 sin (tm−2 ). The numerical simulation is depicted in Fig. 9.1 for α = 0.8, β = 0.9.

Example 9.2. We next consider the following problem with the Caputo fractal– fractional derivative: F F P α,β Dt u (t) = exp (−t) , 0

(9.41)

u (0) = 0.2. Therefore, the following scheme is given:

uk+1 = u0 +

k   β (t)α  β−1  tm−2 h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

k    β (t)α   β−1  β−1 tm−1 h tm−1 , um−1 − tm−2 h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   k β−1 β−1  β (t)α  tm h (tm , um ) − 2tm−1 h tm−1 , um−1 + β−1  2 (α + 3) +tm−2 h tm−2 , um−2 m=2

+

(9.42)

192

New Numerical Scheme With Newton Polynomial

Figure 9.2 Numerical simulation for the Caputo fractal–fractional derivative for α = 0.8, β = 0.5.



 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎦

  where h (tm , um ) = exp (−tm ), h tm−1 , um−1 = exp (−tm−1 ) and h tm−2 , um−2 = exp (−tm−2 ). We provide the numerical simulation in Fig. 9.2 for α = 0.8, β = 0.5.

Example 9.3. We deal with the fractal–fractional differential equation with power law kernel F F P α,β Dt u (t) = u (t) − cos (t) , 0

(9.43)

u (0) = 1. The following scheme for the solution of the problem (9.43) can be presented:

uk+1 = u0 +

k   β (t)α  β−1  tm−2 h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

k    β (t)α   β−1  β−1 tm−1 h tm−1 , um−1 − tm−2 h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   k β−1 β−1  β (t)α  tm h (tm , um ) − 2tm−1 h tm−1 , um−1 + β−1  2 (α + 3) +tm−2 h tm−2 , um−2 m=2

+

(9.44)

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

193

Figure 9.3 Numerical simulation for the Caputo fractal–fractional derivative for α = 0.7, β = 0.6.



 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎦

 where h (tm , um ) = um − cos (tm ), h tm−1 , um−1 = um−1 − cos (tm−1 ) and  h tm−2 , um−2 = um−2 − cos (tm−2 ). The numerical simulation is given in Fig. 9.3 for α = 0.7, β = 0.6.

Example 9.4. We consider the following test problem: F F P α,β Dt u (t) = cos 0

  t3 ,

(9.45)

u (0) = 0.1, which has a singular and nonlocal kernel. Thus the following scheme is given: uk+1 = u0 +

k   β (t)α  β−1  tm−2 h tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

k    β (t)α   β−1  β−1 tm−1 h tm−1 , um−1 − tm−2 h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   k β−1 β−1  β (t)α  tm h (tm , um ) − 2tm−1 h tm−1 , um−1 + β−1  2 (α + 3) +tm−2 h tm−2 , um−2 m=2

+

(9.46)

194

New Numerical Scheme With Newton Polynomial

Figure 9.4 Numerical simulation for the Caputo fractal–fractional derivative for α = 0.87, β = 0.67.

  ⎤ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12  3   3 , h tm−1 , um−1 = cos tm−1 and h tm−2 , um−2 = where h (tm , um ) = cos tm 3 cos tm−2 . The numerical simulation is presented in Fig. 9.4 for α = 0.87, β = 0.67. ⎡

Example 9.5. For this example, we shall introduce the Rikitake system, which explains the reversal of the Earth’s magnetic field. This model describes the currents of two coupled dynamo disks [50]. Let us assume the Rikitake system can be given by the following system of differential equations: ⎧ F F P D α,β x (t) = yz − ax, ⎪ ⎨ t 0 F F P D α,β y (t) = (z − b) x − ay, (9.47) t 0 ⎪ ⎩ F F P D α,β z (t) = 1 − xy, t 0 where the derivative is the Caputo fractal–fractional derivative. The initial conditions are as follows: x (0) = 0.8, y (0) = 0.9, z (0) = 0.5.

(9.48)

For this model, we take a = 1, b = 1. Applying the suggested numerical scheme to this example yields β (t)α (9.49)  (α + 1) k     β−1 × tm−2 h1 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α

x k+1 = x0 +

m=2 k β (t)α  +  (α + 2)

m=2



  β−1 tm−1 h1 tm−1 , x m−1 , y m−1 , zm−1  β−1 −tm−2 h1 tm−2 , x m−2 , y m−2 , zm−2

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

 (k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 tm h1 (tm , x m , y m , zm ) k α   ⎥ β (t) ⎢ β−1 + ⎣ −2tm−1 h1 tm−1 , x m−1 , y m−1 , zm−1 ⎦  2 (α + 3) β−1 m=2 +tm−2 h1 tm−2 , x m−2 , y m−2 , zm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎥ ⎢ (k − m + 1) +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎦ ⎣ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 α β (t) y k+1 = y0 +  (α + 1) k   β−1   × tm−2 h2 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α 

×

m=2

   k β−1 β (t)α  tm−1 h2 tm−1 , x m−1 , y m−1 , zm−1  + β−1  (α + 2) −tm−2 h2 tm−2 , x m−2 , y m−2 , zm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 tm h2 (tm , x m , y m , zm ) k α   β (t) ⎢ ⎥ β−1 + ⎣ −2tm−1 h2 tm−1 , x m−1 , y m−1 , zm−1 ⎦  2 (α + 3) β−1 m=2 +tm−2 h2 tm−2 , x m−2 , y m−2 , zm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎥ ⎢ (k − m + 1) +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎦ ⎣ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12 β (t)α  (α + 1) k     β−1 × tm−2 h3 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α

zk+1 = z0 +

m=2

   k β−1 β (t)α  tm−1 h3 tm−1 , x m−1 , y m−1 , zm−1  + β−1  (α + 2) −tm−2 h3 tm−2 , x m−2 , y m−2 , zm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 tm h3 (tm , x m , y m , zm ) k α   ⎥ β (t) ⎢ β−1 + ⎣ −2tm−1 h3 tm−1 , x m−1 , y m−1 , zm−1 ⎦  2 (α + 3) β−1 m=2 +tm−2 h3 tm−2 , x m−2 , y m−2 , zm−2

195

196

New Numerical Scheme With Newton Polynomial

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

where

and

 h1 tm , x m , y m , zm = y m zm − ax m ,   h2 tm , x m , y m , zm = zm − b x m − ay m ,  h3 tm , x m , y m , zm = 1 − x m y m ,   h1 tm−1 , x m−1 , y m−1 , zm−1 = y m−1 zm−1 − ax m−1 ,     h2 tm−1 , x m−1 , y m−1 , zm−1 = zm−1 − b x m−1 − ay m−1 ,   h3 tm−1 , x m−1 , y m−1 , zm−1 = 1 − x m−1 y m−1 ,

⎤ ⎥ ⎥, ⎦

(9.50)

(9.51)

  h1 tm−2 , x m−2 , y m−2 , zm−2 = y m−2 zm−2 − ax m−2 ,

Figure 9.5 Chaotic attractor of the Rikitake system with the Caputo fractal–fractional derivative for α = 1,

β = 1.

Numerical method for a fractal–fractional ordinary differential equation with power law kernel

197

Figure 9.6 Chaotic attractor of the Rikitake system with the Caputo fractal–fractional derivative for α = 1,

β = 0.99.

    h2 tm−2 , x m−2 , y m−2 , zm−2 = zm−2 − b x m−2 − ay m−2 ,   h3 tm−2 , x m−2 , y m−2 , zm−2 = 1 − x m−2 y m−2 . The numerical simulations are given in Fig. 9.5 and Fig. 9.6.

(9.52)

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Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

10

In this chapter, we aim to obtain a numerical scheme for the general Cauchy problem with the Atangana–Baleanu fractal–fractional derivative in the sense of Caputo. We have    t α AB (α) d F F M α,β α − u (τ ) dτ. (10.1) D u = E − τ (t) (t ) α t 0 1−α (1 − α) dt β 0 α Here, u (t) ∈ W21 (0, l), α ∈ [0, 1] and AB (α) = 1 − α + (α) . If the function u (t) is continuous on the interval (a, b), then the fractal–fractional integral of u (t) is of the form  t αβ F F M α,β J u = τ β−1 u (τ ) (t − τ )α−1 dτ (10.2) (t) t 0 AB (α)  (α) 0 β (1 − α) t β−1 u (t) . + AB (α)

Here, the integral has the Mittag-Leffler kernel. Now we consider the Cauchy problem with the Mittag-Leffler kernel  F F M D α,β u (t) = h (t, u (t)) , t 0 (10.3) u (0) = u0 . We transform Eq. (10.3) into 1−α βt β−1 h (t, u (t)) AB (α)  t αβ τ β−1 h (τ, u (τ )) (t − τ )α−1 dτ. + AB (α)  (α) 0

u (t) − u (0) =

(10.4)

At the point tk+1 = (k + 1) t and taking g (t, u (t)) = βt β−1 h (t, u (t)), we have the following: 1−α g (t, u (t)) AB (α)  tk+1 α + g (τ, u (τ )) (tk+1 − τ )α−1 dτ AB (α)  (α) 0

u (tk+1 ) − u (0) =

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00016-0 Copyright © 2021 Elsevier Inc. All rights reserved.

(10.5)

200

New Numerical Scheme With Newton Polynomial

and we write  1−α  g tk , uk AB (α) k  tm+1  α + g (τ, u (τ )) (tk+1 − τ )α−1 dτ. AB (α)  (α) tm

u (tk+1 ) = u (0) +

(10.6)

m=2

After putting the Newton polynomial into Eq. (10.6), the above equation can be written as follows: uk+1 = u0 +

+

 1−α  g tk , uk AB (α)

α AB (α)  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  k  tm+1 ⎨

m=2 tm

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

 g tm−2, um−2



g tm−1 ,um−1 −g tm−2 ,um−2 t × (τ − tm−2 )  g(t ,um )−2g tm−1 ,um−1 +g tm−2 ,um−2 + m 2(t)2 × (τ − tm−2 ) (τ − tm−1 )

+

(10.7) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we have  1−α  g tk , uk AB (α)  ⎧  tm+1 g tm−2 , um−2 (tk+1 − τ )α−1 dτ ⎪ tm ⎪

⎪  tm+1 g tm−1 ,um−1 −g tm−2 ,um−2  ⎪ ⎪ ⎪ + k ⎨ tm t  α + × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

 ⎪ t AB (α)  (α) m m−1 +g t m−2 m−2 ,u m+1 g(tm ,u )−2g tm−1 ,u m=2 ⎪ ⎪ ⎪ + t ⎪ 2 m 2(t) ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

uk+1 = u0 +

(10.8) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

and we have uk+1 = u0 +  ×

k     α 1−α  g tk , uk + g tm−2 , um−2 AB (α) AB (α)  (α) m=2

tj +1

tj

(tk+1 − τ )

α−1

dτ





k  g tm−1 , um−1 − g tm−2 , um−2 α + AB (α)  (α) t m=2  tm+1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ tm

(10.9)

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

201





k  g (tm , um ) − 2g tm−1 , um−1 + g tm−2 , um−2 α + AB (α)  (α) 2 (t)2 m=2  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ. tm

As we perform the calculations for the above integrals in Eq. (10.9), we obtain the following approximation:  1−α  uk+1 = u0 + g tk , uk AB (α) k    α (t)α g tm−2 , um−2 + AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α k       α (t)α g tm−1 , um−1 − g tm−2 , um−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)  

k   α (t)α g (tm , um ) − 2g tm−1 ,um−1 + +g tm−2 , um−2 2AB (α)  (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+

(10.10)

Thus, we have uk+1 = u0 +

  1−α β−1 βtk h tk , uk AB (α)

k    α (t)α β−1 βtm−2 h tm−2 , um−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α    k β−1  α (t)α βtm−1 h tm−1 , um−1  + β−1 AB (α)  (α + 2) −βtm−2 h tm−2 , um−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 m) βt h , u (t k m m α ⎢  ⎥ α (t) β−1 + ⎣ −2βtm−1 h tm−1 , um−1 ⎦  2AB (α)  (α + 3) β−1 m=2 +βtm−2 h tm−2 , um−2

+

(10.11)

202

New Numerical Scheme With Newton Polynomial

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

10.1

⎤ ⎥ ⎥. ⎦

Predictor–corrector method for fractal–fractional derivative with the generalized Mittag-Leffler kernel

Let us consider the following Cauchy problem: F F M α,β Dt u (t) = h (t, u (t)) , 0

(10.12)

u (0) = u0 , where the derivative is the Atangana–Baleanu fractal–fractional derivative. Integrating the above equation, we obtain the following: 1−α βt β−1 h (t, u (t)) AB (α)  t αβ τ β−1 (t − τ )α−1 h (τ, u (τ )) dτ. + AB (α)  (α) 0

u (t) − u (0) =

(10.13)

We have the initial condition 1−α βt β−1 h (t, u (t)) AB (α)  t αβ τ β−1 h (τ, u (τ )) (t − τ )α−1 dτ. + AB (α)  (α) 0

u (t) = u0 +

(10.14)

At the point tk+1 = (k + 1) t, we have 1−α β−1 βt h (tk+1 , u (tk+1 )) (10.15) AB (α) k+1 k  tm+1  αβ τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ. + AB (α)  (α) tm

u (tk+1 ) = u0 +

m=0

For the approximation of the function h (τ, u (τ )), we write the Newton polynomial which is given by h (tm+1 , um+1 ) − h (tm , um ) (τ − tm+1 ) h h (tm+1 , um+1 ) − 2h (tm , um ) + h (tm−1 , um−1 ) + 2h2 × (τ − tm+1 ) (τ − tm ) .

P2 (τ ) = h (tm+1 , um+1 ) +

(10.16)

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

203

If we put this polynomial into the above equation, we have 1−α p  β−1 βt h tk+1 , uk+1 AB (α) k+1 ⎧ α,β ⎪ h (tm+1 , u m+1 ) A1,m k−1 ⎪

⎨  αβ h t ,um+1 −h(tm ,um ) α,β + m+1 A2,m  + h



⎪ AB (α)  (α) m=0 ⎪ ⎩ + h tm+1 ,um+1 −2h(tm ,um )+h tm−1 ,um−1 Aα,β

uk+1 = u0 +

2(t)2

3,m

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

(10.17)

 αβ p α,β + h tk+1 , uk+1 A1,k AB (α)  (α)

p  h tk+1 , uk+1 − h (tk , uk ) α,β αβ A2,k + AB (α)  (α) t

p  h tk+1 , uk+1 − 2h (tk , uk ) + h (tk−1 , uk−1 ) α,β αβ + A3,k . AB (α)  (α) 2 (t)2 Here p

1−α β−1 βtk+1 h (tk , uk ) AB (α) k  tm+1  αβ τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ. + AB (α)  (α) tm

uk+1 = u0 +

(10.18)

m=0

Also, we can write p

uk+1 = u0 +

1−α β−1 βt h (tk , uk ) AB (α) k+1

(10.19)

 tm+1 k  αβ h (tm , um ) τ β−1 (tk+1 − τ )α−1 dτ + AB (α)  (α) tm m=0

and p

uk+1 = u0 +

k  1−α αβ β−1 α,β βtk+1 h (tk , uk ) + h (tm , um ) J1,m . (10.20) AB (α) AB (α)  (α) m=0

We write the first iteration u1 = u0 +

1−α αβ α,β βhβ−1 h (t0 , u0 ) + h (t0 , u0 ) J0,0 AB (α) AB (α)  (α)

(10.21)

where α,β

J0,0 =

 (1 + β)  (α) (t)α+β−1 . β (α + β)

(10.22)

204

10.2

New Numerical Scheme With Newton Polynomial

Error analysis with the Atangana–Baleanu fractal–fractional derivative

We consider the general Cauchy problem F F M α,β Dt u (t) = h (t, u (t)) 0

(10.23)

where the operator is the Atangana–Baleanu fractal–fractional differential operator. At the point t = tk+1 , after applying the associated fractional integral operator, we have 1−α β−1 βt h (tk , u (tk )) (10.24) AB (α)  (α) k  tk+1 αβ τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ + AB (α)  (α) 0 1−α β−1 βt = h (tk , u (tk )) AB (α)  (α) k k  tm+1  αβ + τ β−1 (tk+1 − τ )α−1 h (τ, u (τ )) dτ. AB (α)  (α) tm

u (tk+1 ) − u (0) =

m=0

We can approximate for the function h(τ, u(τ )) within the interval [tm , tm+1 ] using the Newton polynomial h(τ, u(τ )) = Pm (τ ) + R(τ ) = h(tm−2 , u(tm−2 )) (10.25) h(tm−1 , u(tm−1 )) − h(tm−2 , u(tm−2 )) (τ − tm−2 ) + t h(tm , u(tm )) − 2h(tm−1 , u(tm−1 )) + h(tm−2 , u(tm−2 )) + 2(t)2 (τ − tm−1 )(τ − tm ) ∂ 2 × (τ − tm−2 )(τ − tm−1 ) + h(τ, u(τ ))|τ =γτ . 2! ∂τ 2 Then, we write the error k  tm+1  β τ β−1 (tk+1 − τ )α−1 (τ − tm−1 )(τ − tm ) 2AB(α)(α) t m m=0   2 ∂ (10.26) h(τ, u(τ ))|τ =γm dτ × ∂τ 2 ⎛ ⎞ k  tm+1 τ2  β β−1 α−1 ⎝ = τ (tk+1 − τ ) −(tm−1 + tm )τ ⎠ 2AB(α)(α) t m +tm−1 tm m=0  2  ∂ × h(τ, u(τ ))|τ =γm dτ ∂τ 2

Rτα (γτ ) =

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

=

205

β 2AB(α)(α) k  tm+1  (tk+1 − τ )α−1 (τ β+1 − (tm−1 + tm )τ β + tm−1 tm τ β−1 ) × ×

m=0 tm  2 ∂

∂τ

 h(τ, u(τ ))| τ =γ m dτ. 2

Thus we write $ 2 $ $∂ $ β sup $$ 2 h(τ, u(τ ))$$ (10.27) 2AB(α)(α) τ ∈[0,tk+1 ] ∂τ τ =γτ k  tm+1  (tk+1 − τ )α−1 (τ β+1 − (tm−1 + tm )τ β + tm−1 tm τ β−1 )dτ ×

|Rτα (γτ )| ≤

m=0 tm

$ 2 $ $∂ $ β sup $$ 2 h(τ, u(τ ))$$ 2AB(α)(α) τ ∈[0,tk+1 ] ∂τ τ =γτ ⎧  tm+1 α−1 β+1 ⎪ − τ) τ dτ k ⎨ tm (tk+1  t −(tm−1 + tm ) tmm+1 (tk+1 − τ )α−1 τ β dτ ×  ⎪ m=0 ⎩ +tm−1 tm tm+1 (tk+1 − τ )α−1 τ β−1 dτ tm ≤

⎫ ⎪ ⎬ ⎪ ⎭

where γτ is a constant obtained from the Taylor series approximation for the error. Now we have to evaluate each integral separately 

tm+1

(tk+1 − τ )α−1 τ β+1 dτ ≤ (t)β+1+α (k + 1)β+1+α B(β + 2 + α, α)

tm





tm+1

(tk+1 − τ )α−1 τ β dτ ≤ (t)β+α (k + 1)β+α B(β + 1 + α, α)

(10.28)

tm tm+1

(tk+1 − τ )α−1 τ β−1 dτ ≤ (t)β−1+α (k + 1)β−1+α B(β + α, α).

tm

Replacing the above to the error, we get $ 2 $ $∂ $ β(t)β+1+α (k + 1)β+α sup $$ 2 h(τ, u(τ ))$$ (10.29) 2AB(α)(α) τ ∈[0,tk+1 ] ∂τ τ =γτ ⎧ ⎫ k ⎨ (k + 1)2 B(β + 2 + α, α) ⎬  −((k + 1)k − (k + 1))B(β + 1 + α, α) × ⎩ ⎭ −1 m=0 + (k+1)k 2 (k + 1) B(β + α, α) $ 2 $ $∂ $ β(t)β+1+α (k + 1)β+α sup $$ 2 h(τ, u(τ ))$$ (α, β, k) ≤ 2AB(α)(α) ∂τ

|Rτα (γτ )| ≤

τ ∈[0,tk+1 ]

τ =γτ

206

where

10.3

New Numerical Scheme With Newton Polynomial

⎧ k ⎨ 

⎫ (k + 1)2 B(β + 2 + α, α) ⎬ −(k + 1)(k − 1)B(β + 1 + α, α) . (α, β, k) = ⎩ ⎭ m=0 + k2 B(β + α, α)

(10.30)

Numerical illustrations

In this section, we give numerical simulations for some problems involving the Atangana–Baleanu fractal–fractional derivative. Example 10.1. We consider the following problem with the Atangana–Baleanu fractal–fractional derivative: F F M α,β Dt u (t) = 5u (t) , 0

(10.31)

u (0) = 0.1. Thus, we can solve the above equation numerically with the following scheme:   1−α β−1 βtk h tk , uk uk+1 = u0 + AB (α) k    α (t)α β−1 βtm−2 h tm−2 , um−2 + AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α    k β−1  α (t)α βtm−1 h tm−1 , um−1  + β−1 AB (α)  (α + 2) −βtm−2 h tm−2 , um−2 m=2   (k − m + 1)α (k − m + 3 + 2α) (10.32) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 βtm h (tm , um ) k   ⎥ α (t)α ⎢ β−1 + ⎣ −2βtm−1 h tm−1 , um−1 ⎦  2AB (α)  (α + 3) β−1 m=2 +βtm−2 h tm−2 , um−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎥ ⎢ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎦ ⎣ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12  

where h (tm , um ) = 5um , h tm−1 , um−1 = 5um−1 and h tm−2 , um−2 = 5um−2 . The numerical simulation is presented in Fig. 10.1 for α = 0.85, β = 0.75. Example 10.2. We consider the following test problem with the Atangana–Baleanu fractal–fractional derivative:

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

207

Figure 10.1 Numerical simulation for the Atangana–Baleanu fractal–fractional derivative for α = 0.85,

β = 0.75.

F F M α,β Dt u (t) = −3t 0

+ 17, u (0) = −0.1.

(10.33)

Thus, the following scheme is obtained:   1−α β−1 βtk h tk , uk uk+1 = u0 + AB (α) k    α (t)α β−1 βtm−2 h tm−2 , um−2 AB (α)  (α + 1) m=2   α × (k − m + 1) − (k − m)α    k β−1  α (t)α βtm−1 h tm−1 , um−1  + β−1 m−2 AB (α)  (α + 2) −βt h t , u m−2 m−2 m=2   (k − m + 1)α (k − m + 3 + 2α) (10.34) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 βtm h (tm , um ) k α   ⎥ α (t) ⎢ β−1 + ⎣ −2βtm−1 h tm−1 , um−1 ⎦

 2AB (α)  (α + 3) β−1 m=2 +βtm−2 h tm−2 , um−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎥ ⎢ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎦ ⎣ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12  

where h (tm , um ) = −3tm + 17, h tm−1 , um−1 = −3tm−1 + 17 and h tm−2 , um−2 = −3tm−2 + 17. The numerical simulation is depicted in Fig. 10.2 for α = 0.75, β = 0.5.

+

Example 10.3. We deal with the following fractal–fractional differential equation with the Mittag-Leffler kernel:

208

New Numerical Scheme With Newton Polynomial

Figure 10.2 Numerical simulation for the Atangana–Baleanu fractal–fractional derivative for α = 0.75,

β = 0.5.

F F M α,β Dt u (t) = u (t) + tan (t) , 0

(10.35)

u (0) = 1.1. Then, we can get the following scheme: uk+1 = u0 +

  1−α β−1 βtk h tk , uk AB (α)

k    α (t)α β−1 βtm−2 h tm−2 , um−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α    k β−1  α (t)α βtm−1 h tm−1 , um−1  + β−1 AB (α)  (α + 2) −βtm−2 h tm−2 , um−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 βtm h (tm , um ) k 

 α (t)α ⎢ ⎥ β−1 + ⎣ −2βtm−1 h tm−1 , um−1 ⎦ 

2AB (α)  (α + 3) β−1 m=2 +βtm−2 h tm−2 , um−2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

+

(10.36)

⎤ ⎥ ⎥ ⎦



where h (tm , um ) = um + tan (tm ), h tm−1 , um−1 = um−1 + tan (tm−1 ) and 

h tm−2 , um−2 = um−2 + tan (tm−2 ). We give a numerical simulation in Fig. 10.3 for α = 0.75, β = 0.9.

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

209

Figure 10.3 Numerical simulation for the Atangana–Baleanu fractal–fractional derivative for α = 0.75,

β = 0.9.

Example 10.4. We consider the following problem with the Atangana–Baleanu fractal–fractional derivative: F F M α,β Dt u (t) = cot (2t) , 0

(10.37)

u (0) = −1. Eq. (10.37) can be solved by the following scheme:   1−α β−1 uk+1 = u0 + βtk h tk , uk AB (α) k    α (t)α β−1 βtm−2 h tm−2 , um−2 + AB (α)  (α + 1) m=2   α × (k − m + 1) − (k − m)α    k β−1  α (t)α βtm−1 h tm−1 , um−1  + β−1 m−2 AB (α)  (α + 2) −βt h t , u m−2 m−2 m=2   (k − m + 1)α (k − m + 3 + 2α) (10.38) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 βtm h (tm , um ) k   ⎥ α (t)α ⎢ β−1 + ⎣ −2βtm−1 h tm−1 , um−1 ⎦ 

2AB (α)  (α + 3) β−1 m=2 +βtm−2 h tm−2 , um−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12  

where h (tm , um ) = cot (2tm ), h tm−1 , um−1 = cot (2tm−1 ) and h tm−2 , um−2 = cot (2tm−2 ). The numerical simulation is depicted in Fig. 10.4 for α = 0.65, β = 0.9.

210

New Numerical Scheme With Newton Polynomial

Figure 10.4 Numerical simulation for the Atangana–Baleanu fractal–fractional derivative for α = 0.65,

β = 0.9.

Example 10.5. The three-dimensional Lorenz system which is non-linear and nonperiodic relates the properties of a two-dimensional fluid layer uniformly warmed from below and cooled from above [51]. We consider the following Lorenz attractor with the Atangana–Baleanu fractal–fractional derivative: ⎧ F F M D α,β x (t) = a (y − x) , ⎪ ⎨ 0 t F F M D α,β y (t) = (b − z) x − y, (10.39) t 0 ⎪ ⎩ F F M D α,β z (t) = xy − cz, t 0 with the initial conditions x (0) = −2, y (0) = 1, z (0) = −1.

(10.40)

This model displays chaotic behavior with the parameters a = 10, b = 28 and c = 8/3. Applying the suggested numerical scheme to this example yields   1−α β−1 βtk h1 tk , x k , y k , zk (10.41) AB (α) k    α (t)α β−1 + βtm−2 h1 tm−2 , x m−2 , y m−2 , zm−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α 

  k β−1  α (t)α βtm−1 h1 tm−1 , x m−1 , y m−1 , zm−1

 + β−1 AB (α)  (α + 2) −βtm−2 h1 tm−2 , x m−2 , y m−2 , zm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 m , y m , zm ) βt h , x (t k 1 m m α ⎢

 ⎥ α (t) β−1 + ⎣ −2βtm−1 h1 tm−1 , x m−1 , y m−1 , zm−1 ⎦

 2AB (α)  (α + 3) β−1 m=2 +βtm−2 h1 tm−2 , x m−2 , y m−2 , zm−2

x k+1 = x0 +

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

211

⎤ ⎥ ⎥, ⎦

  1−α β−1 βtk h2 tk , x k , y k , zk AB (α) k    α (t)α β−1 + βtm−2 h2 tm−2 , x m−2 , y m−2 , zm−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α 

  k β−1  α (t)α βtm−1 h2 tm−1 , x m−1 , y m−1 , zm−1

 + β−1 AB (α)  (α + 2) −βtm−2 h2 tm−2 , x m−2 , y m−2 , zm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 m , y m , zm ) βt h , x (t k 2 m m α ⎢

 ⎥ α (t) β−1 + ⎣ −2βtm−1 h2 tm−1 , x m−1 , y m−1 , zm−1 ⎦

 2AB (α)  (α + 3) β−1 m=2 +βtm−2 h2 tm−2 , x m−2 , y m−2 , zm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

y k+1 = y0 +

  1−α β−1 βtk h3 tk , x k , y k , zk AB (α) k    α (t)α β−1 + βtm−2 h3 tm−2 , x m−2 , y m−2 , zm−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α 

  k β−1  α (t)α βtm−1 h3 tm−1 , x m−1 , y m−1 , zm−1

 + β−1 AB (α)  (α + 2) −βtm−2 h3 tm−2 , x m−2 , y m−2 , zm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 m , y m , zm ) βt h , x (t k 3 m m α ⎢

 ⎥ α (t) β−1 + ⎣ −2βtm−1 h3 tm−1 , x m−1 , y m−1 , zm−1 ⎦

 2AB (α)  (α + 3) β−1 m=2 +βtm−2 h3 tm−2 , x m−2 , y m−2 , zm−2

zk+1 = z0 +

212

New Numerical Scheme With Newton Polynomial

Figure 10.5 Numerical simulation for the Atangana–Baleanu fractal–fractional derivative for α = 1, β = 1.

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥, ⎦

where



 h1 tm , x m , y m , zm = a y m − x m ,

  h2 tm , x m , y m , zm = b − zm x m − y m ,

 h3 tm , x m , y m , zm = x m y m − czm ,     h1 tm−1 , x m−1 , y m−1 , zm−1 = a y m−1 − x m−1 ,     h2 tm−1 , x m−1 , y m−1 , zm−1 = b − zm−1 x m−1 − y m−1 ,   h3 tm−1 , x m−1 , y m−1 , zm−1 = x m−1 y m−1 − czm−1 , and

(10.42)

(10.43)

Numerical method for a fractal–fractional ordinary differential equation with Mittag-Leffler kernel

213

Figure 10.6 Numerical simulation for the Lorenz system with the Atangana–Baleanu fractal–fractional derivative for α = 1, β = 0.96.

    h1 tm−2 , x m−2 , y m−2 , zm−2 = a y m−2 − x m−2 ,     h2 tm−2 , x m−2 , y m−2 , zm−2 = b − zm−2 x m−2 − y m−2 ,   h3 tm−2 , x m−2 , y m−2 , zm−2 = x m−2 y m−2 − czm−2 . For this model, the numerical simulations are given in Fig. 10.5 and Fig. 10.6.

(10.44)

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Numerical method for a fractal–fractional ordinary differential equation with variable order with exponential decay kernel

11

To derive the numerical scheme for the fractal–fractional differential operator with constant order and variable fractal dimension, we consider the general Cauchy problem with non-linear function. We start with the case where the kernel is the exponential kernel. This differential operator is the extension of the fractal–fractional derivative based on the well-known Caputo–Fabrizio differential operator F F E α,β(t) Dt u (t) = h (t, u (t)) 0

(11.1)

with the initial condition u (0) = u0 .

(11.2)

We can convert Eq. (11.1) as follows:   β (t) 1 − α β(t) β (t) ln (t) + t h (t, u (t)) u (t) = M (α) t    t β (s) β(s) α + s h (s, u (s)) β (s) ln (s) + ds. M (α) 0 s

(11.3)

We write at the point tk+1 = (k + 1) t   β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) u (tk+1 ) = h (tk , u (tk )) t ln tk + M (α) k t tk    tk+1 β (s) β(s) α s ds h (s, u (s)) β (s) ln (s) + + M (α) 0 s

(11.4)

and at the point tk = kt, we have    β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 u (tk ) = ln tk−1 + M (α) t tk−1 × h (tk−1 , u (tk−1 ))    tk β (s) β(s) α s h (s, u (s)) β (s) ln (s) + ds. + M (α) 0 s New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00017-2 Copyright © 2021 Elsevier Inc. All rights reserved.

(11.5)

216

New Numerical Scheme With Newton Polynomial

Taking the difference of these equations, we have the following: ⎡

β(tk )

tk



  β tk+1 −β(tk ) t

ln tk + ×h (t k , u (tk ))

β(tk ) tk

⎢ 1−α ⎢ ⎢ 

 u (tk+1 ) − u (tk ) = β tk−1 β(tk )−β tk−1 M (α) ⎢ ⎣ −tk−1 ln tk−1 + t ×h (tk−1 , u (tk−1 )) +

α M (α)



tk+1





⎤



β tk−1 tk−1



⎥ ⎥ ⎥ ⎥ ⎦ (11.6)



h (s, u (s)) β (s) ln (s) +

tk

β (s) β(s) s ds. s

For simplicity, we consider   β (t) β(t) t H (t, u (t)) = h (t, u (t)) β (t) ln (t) + t

(11.7)

and we have 1−α [H (tk , u (tk )) − H (tk−1 , u (tk−1 ))] M (α)  tk+1 α H (s, u (s)) ds. + M (α) tk

u (tk+1 ) − u (tk ) =

(11.8)

Here, we shall use the Newton polynomial as the approximation of the function H (t, u (t)), H (t, u (t))  H (tk−2 , u (tk−2 )) H (tk−1 , u (tk−1 )) − H (tk−2 , u (tk−2 )) + (s − tk−2 ) t H (tk , u (tk )) − 2H (tk−1 , u (tk−1 )) + H (tk−2 , u (tk−2 )) + 2 (t)2 × (s − tk−2 ) (s − tk−1 ) ,

(11.9)

and put this polynomial into Eq. (11.8), and we have uk+1 − uk =

+

1−α [H (tk , u (tk )) − H (tk−1 , u (tk−1 ))] M (α) ⎧ H (tk−2, u (tk−2 )) ⎪    ⎪  tk+1 ⎪ H tk−1 ,u tk−1 −H tk−2 ,u tk−2 ⎨ + (s − t α t

M (α)

tk







H (t ,u(t ))−2H tk−1 ,u tk−1 +H tk−2 ,u ⎪ + k k ⎪ ⎪ 2(t)2 ⎩ × (s − tk−2 ) (s − tk−1 )

(11.10)

)

 k−2 tk−2

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

ds.

Numerical method for a fractal–fractional ordinary differential equation

217

Let us organize the above equation such that 1−α [H (tk , u (tk )) − H (tk−1 , u (tk−1 ))] M (α) ⎧ H (tk−2 , u(tk−2 )) t  ⎪    ⎪ ⎪ H tk−1 ,u tk−1 −H tk−2 ,u tk−2 ⎪ ⎪ + ⎪ t t α ⎨ × tkk+1 (s − tk−2 ) ds +       M (α) ⎪ H (tk ,u(tk ))−2H tk−1 ,u tk−1 +H tk−2 ,u tk−2 ⎪ ⎪ + ⎪ 2 ⎪ 2(t) ⎪ t ⎩ × tkk+1 (s − tk−2 ) (s − tk−1 ) ds

uk+1 − uk =

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

. (11.11)

We can perform the following calculations for the integrals located on the right hand side of Eq. (11.11):  tk+1 5 (11.12) (s − tk−2 ) ds = (t)2 , 2 tk  tk+1 23 (s − tk−2 ) (s − tk−1 ) ds = (t)3 . 6 tk If we put these computations into the above scheme, we get the following scheme: 1−α [H (tk , u (tk )) − H (tk−1 , u (tk−1 ))] M (α) ⎧ ⎪ ⎪  H (tk−2 , u (tk−2 )) t ⎪ ⎪ H (tk−1 , u (tk−1 )) ⎪ 5 ⎪ α ⎨ +⎡ −H (tk−2 , u (tk−2 )) ⎤2 t + H (tk , u (tk )) ⎪ M (α) ⎪ ⎪ ⎪ ⎪ + ⎣ −2H (tk−1 , u (tk−1 )) ⎦ 23 ⎪ 12 t ⎩ +H (tk−2 , u (tk−2 ))

uk+1 − uk =

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(11.13)

.

We can reorganize the above scheme as follows: 1−α [H (tk , u (tk )) − H (tk−1 , u (tk−1 ))] M (α) ⎧ ⎫ 23 H (tk , u (tk )) t ⎨ ⎬ 12 α + . − 43 H (tk−1 , u (tk−1 )) t ⎭ M (α) ⎩ 5 + 12 H (tk−2 , u (tk−2 )) t

uk+1 − uk =

(11.14)

Thus, we have the following scheme:

   ⎡ β(t ) β tk+1 −β(tk ) tk k ln tk + β(ttkk ) t ⎢ 1−α ⎢ ×h (t k , u (tk )) k+1 k ⎢ 

   u =u + β tk−1 β(tk )−β tk−1 β t M (α) ⎢ ⎣ −t ln tk−1 + k−1 k−1

t

×h (tk−1 , u (tk−1 ))

tk−1

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

(11.15)

218

New Numerical Scheme With Newton Polynomial

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

23 β(tk ) 12 tk



  β tk+1 −β(tk ) t

ln tk + ×h (tk ,u (tk )) t

β(tk ) tk





   β tk−1 β(tk )−β tk−1 β tk−1 α − 43 tk−1 ln t + k−1 t tk−1 + M (α) ⎪ , u ×h t (t (t )) ⎪ k−1 k−1 ⎪         ⎪ ⎪ β tk−2 ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ t ln t + + k−2 ⎪ t tk−2 ⎪ ⎩ 12 k−2 ×h (tk−2 , u (tk−2 )) t

11.1

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

Numerical illustrations

In this section, in order to show that the suggested method is effective and accurate, we present some illustrations where we will consider some problems with constant fractional order and variable fractal dimension. Example 11.1. We consider the following Cauchy problem with the exponential decay kernel: F F E α,β(t) Dt u (t) = 4u (t) − 2, 0

(11.16)

u (0) = −1, where the fractional order is constant and the fractal dimension is variable. Thus the following scheme is given: ⎡

β(tk )

tk



  β tk+1 −β(tk ) t

ln tk + ×h (t k , u (tk ))

β(tk ) tk



⎤

⎢ ⎥ ⎥ 1−α ⎢ ⎢ 

  ⎥  ⎢ ⎥ β t β(t β t t )−β k k−1 k−1 M (α) ⎣ −t k−1 ⎦ ln tk−1 + tk−1 k−1 t ×h (tk−1 , u (tk−1 ))

   ⎧ ⎫ β tk+1 −β(tk ) β(t ) β(tk ) 23 k ⎪ ⎪ t ln t + ⎪ ⎪ k k 12 t t ⎪ ⎪ k ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ×h , u t (t (t )) ⎪ ⎪ k k      ⎪ ⎪

 ⎪ ⎪ β tk−1 β(tk )−β tk−1 ⎨ ⎬ β tk−1 4 α − 3 tk−1 ln t + k−1 t t k−1 + ⎪ M (α) ⎪ ×h (t  , u (tk−1 )) t ⎪ ⎪ ⎪   k−1    ⎪  ⎪ ⎪ ⎪ ⎪ β tk−2 β tk−1 −β tk−2 β tk−2 ⎪ ⎪ 5 ⎪ ⎪ + 12 tk−2 ln tk−2 + tk−2 ⎪ ⎪ t ⎪ ⎪ ⎩ ⎭ ×h (tk−2 , u (tk−2 )) t

uk+1 = uk +

(11.17)

      where h tk , uk = 4uk − 2, h tk−1 , uk−1 = 4uk−1 − 2 and h tk−2 , uk−2 = 4uk−2 − 2. A numerical simulation is depicted in Fig. 11.1. Example 11.2. We consider the following Cauchy problem with constant fractional order and variable fractal dimension:

Numerical method for a fractal–fractional ordinary differential equation

219

Figure 11.1 Numerical simulation for fractal–fractional derivative with variable order with the exponential decay kernel for α = 0.8, β = 1 − t. F F E α,β(t) Dt u (t) = sin (u (t)) , 0

(11.18)

u (0) = 0.1. The following scheme is given: ⎡

β(tk )

tk

k+1

u



  β tk+1 −β(tk ) t

ln tk + ×h (t k , u (tk ))

β(tk ) tk



⎤

⎢ ⎥ ⎥ 1−α ⎢ ⎢ ⎥    

 =u + ⎢ β t M (α) ⎣ −t k−1 β(tk )−β tk−1 ln tk−1 + β tk−1 ⎥ ⎦ k−1 t tk−1 ×h (tk−1 , u (tk−1 ))

   ⎧ ⎫ β tk+1 −β(tk ) β(t ) β(tk ) 23 k ⎪ ⎪ t ln t + ⎪ ⎪ k 12 k t tk ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ , u ×h t (t (t )) ⎪ k k     ⎪   ⎪ ⎪

⎪ ⎪ β t ⎨ ⎬ β(t β t t )−β k−1 k k−1 k−1 α − 43 tk−1 ln tk−1 + tk−1 t + ⎪ ⎪ M (α) ⎪ ×h (t  , u (tk−1 )) t ⎪ ⎪   k−1    ⎪  ⎪ ⎪ ⎪ ⎪ β tk−2 β tk−1 −β tk−2 β tk−2 ⎪ ⎪ 5 ⎪ ⎪ + t ln t + k−2 ⎪ ⎪ k−2 12 t t ⎪ ⎪ k−2 ⎩ ⎭ ×h (tk−2 , u (tk−2 )) t k

(11.19)

            where h tk , uk = sin uk , h tk−1 , uk−1 = sin uk−1 and h tk−2 , uk−2 = sin uk−2 . We give the numerical simulation in Fig. 11.2. Example 11.3. We deal with the following Cauchy problem with the exponential decay kernel: F F E α,β(t) Dt u (t) = t 2 u (t) , 0

(11.20)

u (0) = 0.1, where the fractional order is constant and the fractal dimension is variable. We can solve our problem numerically as follows:

220

New Numerical Scheme With Newton Polynomial

Figure 11.2 Numerical simulation for fractal–fractional derivative with variable order with the exponential decay kernel for α = 0.95, β = 0.1t.

Figure 11.3 Numerical simulation for fractal–fractional with variable order with the exponential decay kernel for α = 0.75, β = sin 0.02t.

⎡

β(tk )

tk

k+1

u



  β tk+1 −β(tk ) t

ln tk + ×h (t k , u (tk ))

β(tk ) tk



⎤

⎢ ⎥ ⎥ 1−α ⎢ ⎢    

 ⎥ =u + ⎢ ⎥ β t β(t β t t )−β k−1 k k−1 k−1 M (α) ⎣ −t ⎦ ln tk−1 + tk−1 k−1 t ×h (tk−1 , u (tk−1 ))

   ⎧ ⎫ β(tk ) 23 β(tk ) β tk+1 −β(tk ) ⎪ ⎪ t ln t + ⎪ ⎪ k 12 k t tk ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ×h , u t (t (t )) ⎪ ⎪ k k       ⎪ ⎪

 ⎪ ⎪ β t ⎨ ⎬ β(t β t t )−β k−1 k k−1 k−1 4 α − 3 tk−1 ln t + k−1 t tk−1 + ⎪ M (α) ⎪ ×h (t  , u (tk−1 )) t ⎪ ⎪ ⎪   k−1    ⎪  ⎪ ⎪ ⎪ ⎪ β t −β t β t β t k−2 k−1 k−2 k−2 ⎪ +5t ⎪ ⎪ ⎪ ln t + k−2 ⎪ ⎪ k−2 12 t tk−2 ⎪ ⎪ ⎩ ⎭ ×h (tk−2 , u (tk−2 )) t k

(11.21)

      2 uk−1 and h t k−2 = t 2 uk−2 . A where h tk , uk = tk2 uk , h tk−1 , uk−1 = tk−1 k−2 , u k−2 numerical simulation is presented in Fig. 11.3.

Numerical method for a fractal–fractional ordinary differential equation

221

Figure 11.4 Numerical simulation for fractal–fractional derivative with variable order with the exponential decay kernel for α = 0.8, β = 1 − t.

Example 11.4. We deal with the following Cauchy problem with a new differential operator having constant fractional order and variable fractal dimension: F F E α,β(t) Dt u (t) = cos (u (t) + 5) , 0

(11.22)

u (t) = 1. Applying the suggested numerical scheme to this example yields ⎡

β(tk )

tk



  β tk+1 −β(tk ) t

ln tk + ×h (t k , u (tk ))

β(tk ) tk



⎤

⎢ ⎥ ⎥ 1−α ⎢ ⎢ 

  ⎥  ⎢ ⎥ β t β(t β t t )−β k k−1 k−1 M (α) ⎣ −t k−1 ⎦ ln tk−1 + tk−1 k−1 t ×h (tk−1 , u (tk−1 ))

   ⎧ ⎫ β tk+1 −β(tk ) β(t ) β(tk ) 23 k ⎪ ⎪ t ln t + ⎪ ⎪ k k 12 t t ⎪ ⎪ k ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ , u ×h t (t (t )) ⎪ ⎪ k k      ⎪ ⎪

 ⎪ ⎪ β tk−1 β(tk )−β tk−1 ⎨ ⎬ β tk−1 4 α − 3 tk−1 ln t + k−1 t t k−1 + ⎪ M (α) ⎪ ×h (t  , u (tk−1 )) t ⎪ ⎪ ⎪   k−1    ⎪  ⎪ ⎪ ⎪ ⎪ β t β tk−1 −β tk−2 β tk−2 k−2 ⎪ ⎪ 5 ⎪ ⎪ + 12 tk−2 ln tk−2 + tk−2 ⎪ ⎪ t ⎪ ⎪ ⎩ ⎭ ×h (tk−2 , u (tk−2 )) t

uk+1 = uk +

(11.23)

          where h tk , uk = cos uk + 5 , h tk−1 , uk−1 = cos uk−1 + 5 and h tk−2 , uk−2 =   cos uk−2 + 5 . For this equation, we provide the numerical simulation in Fig. 11.4. Example 11.5. The Chua circuit, which is a simple electronic circuit, represents a non-periodic oscillator. Namely, unlike that the electronic oscillator, it generates an oscillating waveform [52]. We consider the following Chua circuit model with con-

222

New Numerical Scheme With Newton Polynomial

stant fractional order and fractal variable dimension: ⎧ ⎪ ⎨ ⎪ ⎩

F F E D α,β(t) x (t) = γ (y − a sin (2πbx)) , t 0 F F E D α,β(t) y (t) = x − y + z, t 0 F F E D α,β(t) z (t) = −λy. t 0

(11.24)

Here, the initial conditions are given as x (0) = −0.2, y (0) = 0.5, z (0) = 0.2.

(11.25)

For this model, we take γ = 10.814, λ = 14, a = 0.2 and b = 0.15. Applying the suggested numerical scheme to this example yields    ⎤ β tk+1 −β(tk ) β(tk ) ln t + k t t k ⎢ ⎥   ⎥ 1−α ⎢ ×h1 tk, x k , y k , zk k+1 k ⎢ ⎥    

 x =x + ⎢ M (α) ⎣ −t β tk−1 β(tk )−β tk−1 ln tk−1 + β tk−1 ⎥ ⎦ k−1 t  tk−1 k−1 k−1 k−1 ×h1 tk−1 , x , y , z

   ⎧ ⎫ β(tk ) 23 β(tk ) β tk+1 −β(tk ) ⎪ ⎪ t ln t + k ⎪ ⎪ 12 k tk ⎪ ⎪  tk k k  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ×h t t , x , y , z 1 k ⎪ ⎪       ⎪ ⎪

 ⎪ ⎪ β t β(t β t t )−β ⎨ ⎬ k−1 k k−1 k−1 4 α − 3 tk−1 ln t + k−1 t t k−1 , +   k−1 , y k−1 , zk−1 t ⎪ M (α) ⎪ ×h t , x ⎪ ⎪ 1 k−1 ⎪ ⎪        ⎪  ⎪ ⎪ ⎪ β tk−2 ⎪ ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ ⎪ + t ln t + k−2 ⎪ ⎪ k−2 12 t tk−2 ⎪ ⎪   ⎩ ⎭ ×h1 tk−2 , x k−2 , y k−2 , zk−2 t

⎡

β(tk )



tk

   ⎤ β tk+1 −β(tk ) β(tk ) ln t + k ⎥ ⎢ t  tk ⎥ 1−α ⎢ ×h2 tk, x k , y k , zk k+1 k ⎢    ⎥ 

y =y + ⎥ ⎢ β t β(t β t t )−β k−1 k k−1 k−1 M (α) ⎣ −t ⎦ ln tk−1 + tk−1 k−1 t   k−1 k−1 k−1 ×h2 tk−1 , x , y , z

   ⎧ ⎫ 23 β(tk ) β tk+1 −β(tk ) ⎪ ⎪ t ln tk + β(ttkk ) ⎪ ⎪ 12 k t ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ k k k ⎪ ⎪ ×h t t , x , y , z 2 k ⎪ ⎪       ⎪ ⎪

 ⎪ ⎪ β tk−1 β(tk )−β tk−1 β tk−1 ⎨ ⎬ 4 α − 3 tk−1 ln t + k−1 t t k−1 , +   M (α) ⎪ ×h t  , xk−1, y k−1 , zk−1 t   ⎪ ⎪ ⎪ ⎪ ⎪  2  k−1   ⎪ ⎪ ⎪ ⎪ β tk−2 ⎪ ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ ⎪ + 12 tk−2 ln tk−2 + tk−2 ⎪ ⎪ t ⎪ ⎪   ⎩ ⎭ k−2 k−2 k−2 ×h2 tk−2 , x , y , z t

⎡

β(tk )

tk



(11.26)

Numerical method for a fractal–fractional ordinary differential equation

223

Figure 11.5 Numerical simulation for fractal–fractional derivative with variable order with the exponential decay kernel for α = 0.98, β = 0.01 + 0.01t.    β tk+1 −β(tk ) ln tk + β(ttkk ) t ⎢   1−α ⎢ ×h3 tk, x k , y k , zk k+1 k ⎢    

z =z + β tk−1 β(tk )−β tk−1 β tk−1 M (α) ⎢ ⎣ −tk−1 ln t + k−1 t  tk−1 ×h3 tk−1 , x k−1 , y k−1 , zk−1

   ⎧ β(tk ) 23 β(tk ) β tk+1 −β(tk ) ⎪ t ln t + k ⎪ 12 k tk ⎪  tk k k  ⎪ ⎪ ⎪ ×h , x , y , z t t 3 k ⎪ 

     ⎪ ⎪ β t β tk−1 tk−1 α ⎨ − 43 tk−1k−1 β(tk )−β ln t + k−1 t +   tk−1 k−1 , y k−1 , zk−1 t M (α) ⎪ ×h t , x ⎪ 3 k−1 ⎪         ⎪ ⎪ β tk−2 ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ + t ln t + k−2 ⎪ t ⎪  tk−2 ⎩ 12 k−2  ×h3 tk−2 , x k−2 , y k−2 , zk−2 t

⎡

β(tk )



tk

where

     h1 tk , x k , y k , zk = γ y k − a sin 2πbx k ,   h2 tk , x k , y k , zk = x k − y k + zk ,

⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

,

(11.27)

224

New Numerical Scheme With Newton Polynomial

Figure 11.6 Numerical simulation for fractal–fractional derivative with variable order with the exponential decay kernel for α = 1, β = 1.

  h3 tk , x k , y k , zk = −λy k ,      h1 tk−1 , x k−1 , y k−1 , zk−1 = γ y k−1 − a sin 2πbx k−1 ,   h2 tk−1 , x k−1 , y k−1 , zk−1 = x k−1 − y k−1 + zk−1 ,   h3 tk−1 , x k−1 , y k−1 , zk−1 = −λy k−1 , and

     h1 tk−2 , x k−2 , y k−2 , zk−2 = γ y k−2 − a sin 2πbx k−2 ,   h2 tk−2 , x k−2 , y k−2 , zk−2 = x k−2 − y k−2 + zk−2 ,   h3 tk−2 , x k−2 , y k−2 , zk−2 = −λy k−2 .

Numerical simulations are depicted in Fig. 11.5 and Fig. 11.6.

(11.28)

(11.29)

Numerical method for a fractal–fractional ordinary differential equation with variable order with power-law kernel

12

In this chapter, we aim to extend the newly suggested numerical scheme for solving non-linear differential equations with the new fractal–fractional operator involving constant fractional order and variable fractal dimension. To achieve our goal, we deal with the general Cauchy problem with the extension of the fractal–fractional based on the well-known Caputo differential operator. Such problem is given by F F P α,β(t) Dt u (t) = h (t, u (t)) , 0

(12.1)

u (0) = u0 . Applying the new integral with the power-law kernel, we convert the above equation into    t β (s) β(s) 1 α−1 β (s) ln (s) + u (t) = u0 + s ds. (12.2) h (s, u (s)) (t − s)  (α) 0 s At the point tk+1 = (k + 1) t, we get 1 u (tk+1 ) = u0 +  (α)



tk+1

 h (s, u (s)) (tk+1 − s)

0

α−1

 β (s) β(s) β (s) ln (s) + s ds. s (12.3)

For simplicity, we put   β (s) β(s) H (s, u (s)) = h (s, u (s)) β (s) ln (s) + s . s Then, we write the following  tk+1 1 u (tk+1 ) = u0 + H (s, u (s)) (tk+1 − s)α−1 ds.  (α) 0

(12.4)

(12.5)

We obtain u (tk+1 ) = u0 +

k  1  tm+1 H (s, u (s)) (tk+1 − s)α−1 ds.  (α) tm m=2

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00018-4 Copyright © 2021 Elsevier Inc. All rights reserved.

(12.6)

226

New Numerical Scheme With Newton Polynomial

With the help of the Newton polynomial, the function H (t, u (t)) can be approximated as follows: P2 (τ ) = H (tm−2 , u (tm−2 )) H (tm−1 , u (tm−1 )) − H (tm−2 , u (tm−2 )) + (s − tm−2 ) t H (tm , u (tm )) − 2H (tm−1 , u (tm−1 )) + H (tm−2 , u (tm−2 )) + 2 (t)2 × (s − tm−2 ) (s − tm−1 ) .

(12.7)

Putting this approximation into Eq. (12.6), we get the following: ⎧ H (t m−2 , u (t m−2 ))  ⎪

⎪ ⎪ H tm−1 ,u tm−1 −H tm−2 ,u tm−2 ⎪ ⎪ + k  t 1  tm+1 ⎨ × (s − tm−2)

uk+1 = u0 + 



⎪  (α) H (tm ,u(tm ))−2H tm−1 ,u tm−1 +H tm−2 ,u tm−2 ⎪ m=2 tm ⎪ + ⎪ 2 ⎪ 2(t) ⎩ × (s − tm−2 ) (s − tm−1 ) × (tk+1 − s)α−1 ds

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(12.8)

and we can reorder the above equation:

uk+1 = u0 +

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k ⎨ ⎪ m=2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

 tm+1

H (tm−2 , u (tm−2 )) (tk+1 − s)α−1 ds







t H tm−1 ,u tm−1 −H tm−2 ,u tm−2 + tmm+1 t × (s − tm−2 ds ) (t k+1 − s)α−1 

 tm+1 H tm−1 ,u tm−1 −2H tm−1 ,u tm−1 +H tm−2 ,u tm−2  + tm 2(t)2 × (s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds tm



⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(12.9) Thus, we write the following: k+1

u

k   tm+1 1   m−2 t = u0 + H tm−2 , u (tk+1 − s)α−1 ds  (α) t m m=2 



k  H tm−1 , um−1 − H tm−2 , um−2 1 +  (α) t m=2  tm+1 × (s − tm−2 ) (tk+1 − s)α−1 ds tm





k 1  H (tm , um ) − 2H tm−1 , um−1 + H tm−2 , um−2 +  (α) 2 (t)2 m=2  tm+1 × (s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds. tm

(12.10)

Numerical method for a fractal–fractional ordinary differential equation

227

We know that the following calculations are valid:  tm+1  (t)α  (tk+1 − s)α−1 ds = (k − m + 1)α − (k − m)α , α tm  tm+1 (s − tm−2 ) (tk+1 − s)α−1 ds tm

  (t)α+1 (k − m + 1)α (k − m + 3 + 2α) , = − (k − m)α (k − m + 3 + 3α) α (α + 1)  tm+1 (t)α+2 (s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds = α (α + 1) (α + 2) tm   ⎤ ⎡ 2 2 (k − m) + (3α + 10) (k − m) α ⎥ ⎢ (k − m + 1) +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎦ ⎣ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12

(12.11)

Let us plug these equalities into the above scheme and obtain the following scheme: k+1

u

k   (t)α   = u0 + H tm−2 , um−2 (k − m + 1)α − (k − m)α  (α + 1) m=2

    (t) H tm−1 , um−1 − H tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)  

k  (t)α  H (tm , um ) − 2H tm−1 , um−1

 + +H tm−2 , um−2 2 (α + 3) m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 +

α

k  

⎤ ⎥ ⎥. ⎦

(12.12)

Replacing the function H (t, u (t)) by its value, we get the following  k 

β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) k+1 ln tm−2 + u = u0 + tm−2  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t t m−1 ⎥ k ⎢ ⎥ (t)α  ⎢ ×h (tm−1 , u (t m−1 )) ⎢ ⎥ 

 



  + ⎢ ⎥ β t β tm−1 −β tm−2 β t  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 ⎦ m=2

m−2

t

×h (tm−2 , u (tm−2 ))

tm−2

228

New Numerical Scheme With Newton Polynomial

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡ 



  β tm+1 −β(tm ) ln tm t + β(ttmm ) ×h (tm , u (tm )) ⎡ 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

⎤

β(t ) tm m

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  (t) ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

12.1

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥. ⎦

(12.13)

Numerical illustrations

In this section, we will deal with some problems with constant fractional order and variable fractal dimension in order to demonstrate that the proposed method is effective and accurate. Example 12.1. Let us consider the following Cauchy problem with the new differential operator where the fractional order is constant and the fractal dimension is variable: F F P α,β(t) Dt u (t) = (t 0

+ 2) sin (u (t)) , u (0) = 0.1.

(12.14)

Thus, the following scheme is given: k+1

u

 k 

β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) = u0 + tm−2 ln tm−2 +  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t t m−1 ⎥ k ⎢ ⎥ (t)α  ⎢ ×h (tm−1 , u (t m−1 )) ⎢ ⎥ 

 



  + ⎢ ⎥ β t β tm−1 −β tm−2 β t  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 ⎦ m=2

m−2

t

×h (tm−2 , u (tm−2 ))

tm−2

Numerical method for a fractal–fractional ordinary differential equation

229

Figure 12.1 Numerical simulation for the fractal–fractional derivative with variable order with the powerlaw kernel for α = 0.53, β = t + 2.

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡ 



  β tm+1 −β(tm ) ln t m t + β(ttmm ) ×h (tm , u (tm )) ⎡ 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

⎤

β(t ) tm m

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  (t) ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥ ⎦

(12.15)

 

where h (tm , um ) = (tm + 2) sin (um ), h tm−1 , um−1 = (tm−1 + 2) sin um−1 and  



h tm−2 , um−2 = (tm−2 + 2) sin um−2 . We present the numerical simulation in Fig. 12.1.

Example 12.2. We consider the following Cauchy problem with power-law kernel: F F P α,β(t) Dt u (t) = u (t) + 2t, 0

u (0) = −0.2,

(12.16)

230

New Numerical Scheme With Newton Polynomial

which has constant fractional order and variable fractal dimension. The following scheme for the solution of above equation is given as k+1

u

 k

 β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + = u0 + tm−2  (α + 1) t tm−2 m=2   α α × h (tm−2 , u (tm−2 )) (k − m + 1) − (k − m) 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t t m−1 ⎥ k ⎢ ⎥ (t)α  ⎢ ×h (tm−1 , u (t m−1 )) ⎢ ⎥

  



  + ⎢ ⎥ β t β tm−1 −β tm−2 β t  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 ⎦ m=2

m−2

t

tm−2

×h (tm−2 , u (tm−2 ))   α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡    ⎤ β tm+1 −β(tm ) ln t β(tm ) m t tm ⎢ ⎥ ⎢ ⎥ + β(ttmm ) ⎢ ⎥ ⎢ ⎥ ×h , u (t (t )) m m ⎢ ⎥ ⎡ ⎤ 

⎢ ⎥ β(tm )−β tm−1

 ⎢ ⎥ ln t β tm−1 k ⎢ t

 m−1 ⎦ ⎥ α ⎣  −2t (t) β tm−1 ⎢ ⎥ m−1 + + tm−1 ⎢ ⎥ ⎢ ⎥ 2 (α + 3) m=2 ⎢ ⎥ ×h , u (t (t )) ⎢ ⎡ m−1  m−1 ⎤ ⎥ ⎢ ⎥ β tm−1 −β tm−2 

⎢ ln tm−2 ⎦ ⎥ t

 ⎢ +t β tm−2 ⎣ ⎥ β tm−2 ⎢ ⎥ m−2 + tm−2 ⎣ ⎦ ×h (tm−2 , u (tm−2 ))   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ (12.17) 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12  



where h (tm , um ) = um + 2tm , h tm−1 , um−1 = um−1 + 2tm−1 and h tm−2 , um−2 = um−2 + 2tm−2 . A numerical simulation is given in Fig. 12.2. Example 12.3. We consider the following Cauchy problem with power-law kernel: F F P α,β(t) Dt u (t) = (u (t))3 0

+ sin (t) ,

(12.18)

u (0) = 0.2, where the fractional order is constant and the fractal dimension is variable. Thus, the following scheme can be obtained: uk+1 = u0 +

 k

 β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + tm−2  (α + 1) t tm−2 m=2

Numerical method for a fractal–fractional ordinary differential equation

231

Figure 12.2 Numerical simulation for the fractal–fractional derivative with variable order with the powerlaw kernel for α = 0.93, β = exp (−t).

  × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α 

 



⎡ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 k ⎢ (t)α  ⎢ ×h , u (t (t )) m−1 m−1 ⎢ 



 

+ ⎢ β t β tm−1 −β tm−2 β t  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 m=2

m−2

t

×h (tm−2 , u (tm−2 ))   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡    β tm+1 −β(tm ) ln tm β(tm ) t tm ⎢ ⎢ + β(ttmm ) ⎢ ⎢ ×h (tm , u (tm )) ⎢ ⎡ 

⎢ β(tm )−β tm−1 

⎢ ln t β tm−1 k t

 m−1 ⎣ (t)α  ⎢ β tm−1 ⎢ −2tm−1 + + tm−1 ⎢ ⎢ 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

tm−2

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

⎤

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥ ⎦

(12.19)



3

where h (tm , um ) = (um )3 + sin (tm ), h tm−1 , um−1 = um−1 + sin (tm−1 ) and 



3 h tm−2 , um−2 = um−2 + sin (tm−2 ). A numerical simulation is depicted in Fig. 12.3.

232

New Numerical Scheme With Newton Polynomial

Figure 12.3 Numerical simulation for the fractal–fractional derivative with variable order with the powerlaw kernel for α = 0.85, β = −t.

Example 12.4. We deal with the following Cauchy problem: F F P α,β(t) Dt u (t) = tan (t) , 0

(12.20)

u (0) = 1, where the differential operator is of constant fractional order and variable fractal dimension. Thus the following scheme is given  k 

β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) tm−2 ln tm−2 +  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 ⎥ k ⎢ ⎥ (t)α  ⎢ , u ×h (t (t )) m−1 m−1 ⎢ 

 ⎥

   + ⎢ ⎥ β t −β t β t β t m−2 m−1 m−2 m−2  (α + 2) ⎣ −t ⎦ ln tm−2 +

uk+1 = u0 +

m=2

m−2

t

×h (tm−2 , u (tm−2 ))   α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡    β tm+1 −β(tm ) ln t β(tm ) m t tm ⎢ β(tm ) ⎢ + t ⎢ m ⎢ ×h (tm , u (tm )) ⎢ ⎡ 

⎢ β(tm )−β tm−1 

⎢ ln t β t m−1 k t

 m−1 ⎣ (t)α  ⎢ β tm−1 ⎢ −2tm−1 + + tm−1 ⎢ ⎢ 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))

tm−2

⎤

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Numerical method for a fractal–fractional ordinary differential equation

233

Figure 12.4 Numerical simulation for the fractal–fractional derivative with variable order with the powerlaw kernel for α = 0.45, β = cos t.

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥ ⎦

(12.21)

 



where h (tm , um ) = tan (tm ), h tm−1 , um−1 = tan (tm−1 ) and h tm−2 , um−2 = tan (tm−2 ). A numerical simulation is depicted in Fig. 12.4.

Example 12.5. The Rabinovich–Fabrikant system shows the stochasticity arising from the modulation instability in a nonequilibrium dissipative medium. Some physical systems can be modeled with this system; for example waves in hydrodynamic flows, wind waves on water, and Langmuir waves in plasma [53]. We present the following Rabinovich–Fabrikant system with constant fractional order and fractal variable dimension: ⎧ ⎪ ⎨ ⎪ ⎩



F F P D α,β(t) x (t) = y z − 1 + x 2 + θ x, t 0

 F F P D α,β(t) y (t) = x 3z − 1 − x 2 + θy, t 0 F F P D α,β(t) z (t) = −2z (b + xy) . t 0

(12.22)

Here, the initial conditions are given by x (0) = −1, y (0) = 0, z (0) = 0.5.

(12.23)

Applying the suggested numerical scheme to this example yields x k+1 = x0 +

 k

 β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + tm−2  (α + 1) t tm−2 m=2

234

New Numerical Scheme With Newton Polynomial

   × h1 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α ⎡ +

k ⎢ (t)α  ⎢ ⎢ ⎢  (α + 2) m=2 ⎣

 ×



 



β tm−1 β(tm )−β tm−1 β tm−1 ln t + m−1 t tm−1

 m−1 , y m−1 , zm−1 ×h t , x 1 m−1 



 

β tm−2 β tm−1 −β tm−2 β tm−2 −tm−2 ln t + m−2 t

 tm−2 ×h1 tm−2 , x m−2 , y m−2 , zm−2

tm−1

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) 

⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  (t) ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎦



  β tm+1 −β(tm ) ln t m t + β(ttmm ) m m , zm ) ×h⎡ 1 (tm , x , y 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

 m−1 , y m−1 , zm−1 ×h1 t⎡ , x m−1



 β tm−1 −β tm−2 

ln tm−2 β tm−2 t

 +tm−2 ⎣ β tm−2 + tm−2

 m−2 ×h1 tm−2 , x , y m−2 , zm−2

⎤

β(t ) tm m

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

(12.24)

 k 

β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + tm−2  (α + 1) t tm−2 m=2    × h2 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α

y k+1 = y0 +

⎡ +

k ⎢ (t)α  ⎢ ⎢ ⎢  (α + 2) m=2 ⎣

 ×



 



β tm−1 β(tm )−β tm−1 β tm−1 ln t + m−1 t tm−1

 m−1 , y m−1 , zm−1 ×h t , x 2 m−1 



 

β tm−2 β tm−1 −β tm−2 β tm−2 −tm−2 ln t + m−2 t

 tm−2 ×h2 tm−2 , x m−2 , y m−2 , zm−2

tm−1

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α)



⎤ ⎥ ⎥ ⎥ ⎥ ⎦

Numerical method for a fractal–fractional ordinary differential equation

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  (t) ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 



 β tm+1 −β(tm ) ln tm t + β(ttmm ) m m , zm ) ×h⎡ 2 (tm , x , y 

β(tm )−β tm−1 

ln t β tm−1 t  m−1

−2tm−1 ⎣ β tm−1 + tm−1

 m−1 m−1 , zm−1 ×h2 t⎡ m−1 , x  , y

 β tm−1 −β tm−2 

ln tm−2 β tm−2 t 

+tm−2 ⎣ β tm−2 + tm−2

 ×h2 tm−2 , x m−2 , y m−2 , zm−2



⎤

β(t ) tm m

 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

235

α

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

 k 

β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + tm−2  (α + 1) t tm−2 m=2    × h3 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α

zk+1 = z0 +

⎡ k ⎢ (t)α  ⎢ ⎢ + ⎢  (α + 2) m=2 ⎣

 ×





 

β tm−1 β(tm )−β tm−1 β tm−1 ln t + m−1 t tm−1

 m−1 , y m−1 , zm−1 ×h t , x  3 m−1

 

β tm−2 β tm−1 −β tm−2 β tm−2 −tm−2 ln t + m−2 t

 tm−2 ×h3 tm−2 , x m−2 , y m−2 , zm−2

tm−1

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) 

⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k α ⎢ (t) ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣



  β tm+1 −β(tm ) ln t m t + β(ttmm ) m m , zm ) ×h⎡ 3 (tm , x , y 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

 m−1 m−1 , zm−1 ×h3 t⎡ m−1 , x  , y

 β tm−1 −β tm−2 

ln tm−2 β tm−2 t

 +tm−2 ⎣ β tm−2 + tm−2

 m−2 ×h3 tm−2 , x , y m−2 , zm−2

⎤

β(t ) tm m

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

236

New Numerical Scheme With Newton Polynomial

Figure 12.5 Numerical simulation for the Rabinovich–Fabrikant system with fractal–fractional derivative with variable order with the power-law kernel for α = 1, β = 1.002 − exp (−t).

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

where





2  + θ xm, h1 tm , x m , y m , zm = y m zm − 1 + x m 



2  + θy m , h2 tm , x m , y m , zm = x m 3zm − 1 − x m



 h3 tm , x m , y m , zm = −2zm b + x m y m ,

⎤ ⎥ ⎥, ⎦

(12.25)

   2   + θ x m−1 , h1 tm−1 , x m−1 , y m−1 , zm−1 = y m−1 zm−1 − 1 + x m−1    2   + θy m−1 , h2 tm−1 , x m−1 , y m−1 , zm−1 = x m−1 3zm−1 − 1 − x m−1 (12.26)

Numerical method for a fractal–fractional ordinary differential equation

237

Figure 12.6 Numerical simulation for the Rabinovich–Fabrikant system with the fractal–fractional derivative with variable order with the power-law kernel for α = 1, β = 1.

    h3 tm−1 , x m−1 , y m−1 , zm−1 = −2zm−1 b + x m−1 y m−1 , and

   2   + θ x m−2 , h1 tm−2 , x m−2 , y m−2 , zm−2 = y m−2 zm−2 − 1 + x m−2    2   + θy m−2 , h2 tm−2 , x m−2 , y m−2 , zm−2 = x m−2 3zm−2 − 1 − x m−2 







(12.27)

h3 tm−2 , x m−2 , y m−2 , zm−2 = −2zm−2 b + x m−2 y m−2 . For this model, we provide the numerical simulations with the constant parameters θ = 0.87, b = 1.1 in Fig. 12.5 and Fig. 12.6.

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Numerical method for a fractal–fractional ordinary differential equation with variable order with the generalized Mittag-Leffler kernel

13

The generalized Mittag-Leffler function has been documented in much literature as the queen of fractional calculus as it appears to be the fundamental solution of the evolution equation associating the fractional derivative with a power-law kernel. Thus, the fractal–fractional differential operator with the generalized Mittag-Leffler function was suggested as the fractal dimension was replaced by variable dimension, this giving birth to a new class of differential and integral operators with the crossover property. In this section, by the newly introduced numerical scheme, we present a derivation of the numerical solution of the new class of Cauchy problems, as follows: F F M α,β(t) Dt u (t) = h (t, u (t)) , 0

(13.1)

u (0) = u0 . Using the new fractional integral with the Mittag-Leffler kernel, we can reformulate the above equation thus:   1 − α β(t) β (t) u (t) = u (0) + β (t) ln (t) + t h (t, u (t)) AB (α) t  t α + h (s, u (s)) (t − s)α−1 (13.2) AB (α)  (α) 0   β (s) β(s) s ds. × β (s) ln (s) + s At the point tk+1 = (k + 1) t, we can get the following:   β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) tk u (tk+1 ) = u (0) + ln tk + AB (α) t tk × h (tk , u (tk ))  tk+1 α h (s, u (s)) (tk+1 − s)α−1 + AB (α)  (α) 0 New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00019-6 Copyright © 2021 Elsevier Inc. All rights reserved.

(13.3)

240

New Numerical Scheme With Newton Polynomial

  β (s) β(s) s ds. × β (s) ln (s) + s For convenience, we put   β (s) β(s) H (s, u (s)) = h (s, u (s)) β (s) ln (s) + s s

(13.4)

and then we have

  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) t ln tk + u (tk+1 ) = u0 + AB (α) k t tk (13.5)  tk+1 α + H (s, u (s)) (tk+1 − s)α−1 ds. AB (α)  (α) 0

Also, we write

  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) t ln tk + u (tk+1 ) = u0 + AB (α) k t tk

+

α AB (α)  (α)

n  tm+1 

(13.6) H (s, u (s)) (tk+1 − s)α−1 ds.

m=2 tm

When replacing the Newton polynomial into the above equation, the function H (t, u (t)) can be approximated as follows: P2 (τ ) = H (tm−2 , u (tm−2 )) H (tm−1 , u (tm−1 )) − H (tm−2 , u (tm−2 )) + (s − tm−2 ) t H (tm , u (tm )) − 2H (tm−1 , u (tm−1 )) + H (tm−2 , u (tm−2 )) + 2 (t)2 × (s − tm−2 ) (s − tm−1 ) .

(13.7)

Putting this approximation in Eq. (13.6), we get the following:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) k+1 u t ln tk + h (tk , u (tk )) = u0 + AB (α) k t tk α (13.8) + AB (α)  (α) ⎧ ⎫ H (t m−2 , u (t m−2 ))  ⎪ ⎪

⎪ ⎪ ⎪ ⎪ H tm−1 ,u tm−1 −H tm−2 ,u tm−2 ⎪ ⎪ ⎪ ⎪  + k ⎬ t  tm+1 ⎨ × − t (s ) ×



m−2



⎪ ⎪ +H tm−2 ,u tm−2 ⎪ ⎪ m=2 tm ⎪ + H (tm ,u(tm ))−2H tm−1 ,u tm−1 ⎪ ⎪ ⎪ 2 ⎪ ⎪ 2(t) ⎩ ⎭ × (s − tm−2 ) (s − tm−1 )

Numerical method for a fractal–fractional ordinary differential equation

241

× (tk+1 − s)α−1 ds and we can reorder the above equation thus:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) tk ln tk + uk+1 = u0 + AB (α) t tk α (13.9) + AB (α)  (α)  tm+1 ⎧ ⎫ H (tm−2 , u (tm−2 )) (tk+1 − s)α−1 ds ⎪ ⎪ tm ⎪ ⎪ 





⎪ ⎪  ⎪ ⎪ tm+1 H tm−1 ,u tm−1 −H tm−2 ,u tm−2 ⎪ ⎪ ⎪ ⎪ + k ⎨ ⎬ tm t  α−1 . × × − t − s) ds (s ) (t m−2 k+1 



 ⎪





⎪ t ⎪ m+1 H tm−1 ,u tm−1 −2H tm−1 ,u tm−1 +H tm−2 ,u tm−2 m=2 ⎪ ⎪ ⎪ ⎪ + tm ⎪ ⎪ ⎪ 2(t)2 ⎪ ⎪ ⎩ ⎭ α−1 × (s − tm−2 ) (s − tm−1 ) (tk+1 − s) ds Thus, we may write the following:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) uk+1 = u0 + h (tk , u (tk )) tk ln tk + AB (α) t tn k   tm+1   α + H tm−2 , um−2 t (tk+1 − s)α−1 ds AB (α)  (α) t m m=2 



k m−1  H tm−1 , u − H tm−2 , um−2 α (13.10) + AB (α)  (α) t m=2  tm+1 × (s − tm−2 ) (tk+1 − s)α−1 ds tm





k  H (tm , um ) − 2H tm−1 , um−1 + H tm−2 , um−2 α + AB (α)  (α) 2 (t)2 m=2  tm+1 × (s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds. tm

We know that the following calculations are valid;  

tm+1 tm tm+1

(tk+1 − s)α−1 ds =

 (t)α  (k − m + 1)α − (k − m)α α

(s − tm−2 ) (tk+1 − s)α−1 ds =

tm



× 

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α)

tm+1 tm

(t)α+1 α (α + 1) 

(s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds =

(t)α+2 α (α + 1) (α + 2)

(13.11)

242

New Numerical Scheme With Newton Polynomial

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥. ⎦

Let us replace these equalities in the above scheme and obtain the following scheme:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) tk ln tk + AB (α) t tk k   α (t)α   + H tm−2 , um−2 (k − m + 1)α − (k − m)α AB (α)  (α + 1)

uk+1 = u0 +

m=2

k       α (t) H tm−1 , um−1 − H tm−2 , um−2 AB (α)  (α + 2) m=2   α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)  

k  α (t)α  H (tm , um ) − 2H tm−1 , um−1

 + +H tm−2 , um−2 AB (α) 2 (α + 3) m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+

α

(13.12)

If we replace the function H (t, u (t)) by its value, we can have the following scheme:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) tk ln tk + AB (α) t tk  k

 α β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) + ln tm−2 + tm−2 AB (α)  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 ⎥ k ⎢ ⎥ α (t)α  ⎢ , u ×h (t (t )) m−1 m−1 ⎢ 

 ⎥

   + ⎢ ⎥ β t −β t β t β t m−2 m−1 m−2 m−2 AB (α)  (α + 2) ⎦ ln tm−2 + tm−2 m=2 ⎣ −tm−2 t ×h (tm−2 , u (tm−2 )) (13.13)   α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

uk+1 = u0 +

Numerical method for a fractal–fractional ordinary differential equation

⎡



 β tm+1 −β(tm ) ln tm t + β(ttmm ) ×h (tm , u (tm )) ⎡ 

β(tm )−β tm−1 

ln t β tm−1 t  m−1

−2tm−1 ⎣ β tm−1 + tm−1

243



⎤

β(t ) tm m

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))   ⎤ ⎡ 2 2 (k − m) + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α 2 +6α + 18α + 12

13.1

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Numerical illustrations

In this section, we present some illustrations of problems with constant fractional order and variable fractal dimension. Example 13.1. We consider the following Cauchy problem with the Mittag-Leffler kernel: F F M α,β(t) Dt u (t) = 8u (t) , 0

(13.14)

u (0) = 0.5, where the fractional order is constant and the fractal dimension is variable. Thus, the following scheme can be presented:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) tk ln tk + uk+1 = u0 + h (tk , u (tk )) AB (α) t tk  k 

α β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) + ln tm−2 + tm−2 AB (α)  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 ⎥ k ⎢ ⎥ α (t)α  ⎢ ×h , u (t (t )) m−1 m−1 ⎢ 



  ⎥

+ ⎢ ⎥ β t −β t β t β t m−1 m−2 AB (α)  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 ⎦ m=2

m−2

t

×h (tm−2 , u (tm−2 ))

tm−2

(13.15)

244

New Numerical Scheme With Newton Polynomial

Figure 13.1 Numerical simulation for the fractal–fractional derivative with variable order with the MittagLeffler kernel for α = 0.77, β = tan 0.1t.

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡

 



 β tm+1 −β(tm ) ln t m t + β(ttmm ) ×h (tm , u (tm )) ⎡ 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

⎤

β(t ) tm m

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦



  where h (tm , um ) = 8um , h tm−1 , um−1 = 8um−1 and h tm−2 , um−2 = 8um−2 . A numerical simulation is depicted in Fig. 13.1.

Example 13.2. We consider the following Cauchy problem with the new fractal– fractional differential operator: F F M α,β(t) Dt u (t) = 2u (t) − 1, 0

u (0) = 1,

(13.16)

Numerical method for a fractal–fractional ordinary differential equation

245

where the fractional order is constant and the fractal dimension is variable. Applying the suggested numerical scheme to this example yields   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) t ln tk + = u0 + u AB (α) k t tk  k 

β (tm−2 ) α (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) tm−2 + ln tm−2 + AB (α)  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α

  



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 ⎥ k ⎢ ⎥ α (t)α  ⎢ ×h , u (t (t )) m−1 m−1 ⎢ 



  ⎥

+ ⎢ ⎥ β t β tm−1 −β tm−2 β t AB (α)  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 ⎦ k+1

m=2

m−2

t

×h (tm−2 , u (tm−2 ))  ×

(k − m + 1) (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡ α

tm−2

(13.17)

 



 β tm+1 −β(tm ) ln t m t + β(ttmm ) ×h (tm , u (tm )) ⎡ 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

⎤

β(t ) tm m

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

 



where h (tm , um ) = 2um − 1, h tm−1 , um−1 = 2um−1 − 1 and h tm−2 , um−2 = 2um−2 − 1. A numerical simulation is given in Fig. 13.2. Example 13.3. We consider the following Cauchy problem with the Mittag-Leffler kernel: F F M α,β(t) Dt u (t) = 3t 2 0

− t,

u (0) = −0.4,

(13.18)

246

New Numerical Scheme With Newton Polynomial

Figure 13.2 Numerical simulation for the fractal–fractional derivative with variable order with the MittagLeffler kernel for α = 0.87, β = 1 + 0.03t.

which has constant fractional order and variable fractal dimension. Thus, the above scheme is solved by the following approximation:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) k+1 u h (tk , u (tk )) t ln tk + = u0 + AB (α) k t tk  k 

α β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) + ln tm−2 + tm−2 AB (α)  (α + 1) t tm−2 m=2   α α × h (tm−2 , u (tm−2 )) (k − m + 1) − (k − m) 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln tm−1 + tm−1 t ⎥ k ⎢ ⎥ α (t)α  ⎢ ×h (tm−1 , u (t m−1 )) ⎢ 

 



  ⎥ + ⎢ ⎥ β t −β t β t β t m−2 m−1 m−2 m−2 AB (α)  (α + 2) ⎦ ln tm−2 + t m=2 ⎣ −tm−2 t ×h (tm−2 , u (tm−2 ))  ×

(k − m + 1) (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡ α

m−2

(13.19)

 



 β tm+1 −β(tm ) ln t m t + β(ttmm ) ×h (tm , u (tm )) ⎡ 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

⎤

β(t ) tm m

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k α ⎢ α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Numerical method for a fractal–fractional ordinary differential equation

247

Figure 13.3 Numerical simulation for the fractal–fractional derivative with variable order with the MittagLeffler kernel for α = 0.71, β = t 2 .

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥ ⎦

 



2 −t , h t m−1 = 3t 2 m−2 = where h (tm , um ) = 3tm m m−1 , u m−1 −tm−1 and h tm−2 , u 2 3tm−2 − tm−2 . We present a numerical simulation in Fig. 13.3. Example 13.4. We consider the following Cauchy problem with the Mittag-Leffler kernel: F F M α,β(t) Dt u (t) = exp (u (t) + 1) , 0

(13.20)

u (0) = 0.3, where the fractional order is constant and the fractal dimension is variable. Then the following scheme is given:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) uk+1 = u0 + tk ln tk + h (tk , u (tk )) AB (α) t tk  k 

α β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) + ln tm−2 + tm−2 AB (α)  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 ⎥ k ⎢ ⎥ α (t)α  ⎢ ×h , u (t (t )) m−1 m−1 ⎢ 



  ⎥

+ ⎢ ⎥ β t −β t β t β t m−1 m−2 AB (α)  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 ⎦ m=2

m−2

t

×h (tm−2 , u (tm−2 ))

tm−2

(13.21)

248

New Numerical Scheme With Newton Polynomial

Figure 13.4 Numerical simulation for the fractal–fractional derivative with variable order with the MittagLeffler kernel for α = 0.48, β = 0.001t.

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡

 



 β tm+1 −β(tm ) ln t m t + β(ttmm ) ×h (tm , u (tm )) ⎡ 

β(tm )−β tm−1 

ln t β tm−1 t  m−1

−2tm−1 ⎣ β tm−1 + tm−1

⎤

β(t ) tm m

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ×h ⎢ ⎡ (t m−1 , u (t m−1)) ⎢ β tm−1 −β tm−2 

⎢ ln tm−2 t

 ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣ ×h (tm−2 , u (tm−2 ))   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥ ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦





 and where h (tm , um ) = exp (um + 1), h tm−1 , um−1 = exp um−1 + 1 

m−2 

m−2 = exp u + 1 . A numerical simulation is depicted in Fig. 13.4. h tm−2 , u Example 13.5. The Wang–Sun system, which shows a 3-D four-wing smooth autonomous chaotic attractor, leads us to analyze the dynamical behavior of multi-wing chaotic systems [54]. We deal with the following four scroll Wang–Sun chaotic attractor with constant fractional order and fractal variable dimension: ⎧ F F M D α,β(t) x (t) = ax + byz, ⎪ ⎨ 0 t F F M D α,β(t) y (t) = cx + dy − xz, (13.22) t 0 ⎪ ⎩ F F M α,β(t) Dt z (t) = exy + f z, 0

Numerical method for a fractal–fractional ordinary differential equation

249

with the initial conditions x (0) = −1, y (0) = 0, z (0) = 0.5.

(13.23)

Applying the suggested numerical scheme to this example yields x

k+1

    1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h1 tk , x k , y k , zk tk ln tk + = x0 + AB (α) t tk

 k 

α β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) + tm−2 ln tm−2 + AB (α)  (α + 1) t tm−2 m=2    × h1 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α ⎡ +

k ⎢ α (t)α  ⎢ ⎢ ⎢ AB (α)  (α + 2) m=2 ⎣



 



β tm−1 β(tm )−β tm−1 β tm−1 ln t + m−1 t tm−1

 m−1 , y m−1 , zm−1 ×h t , x 1 m−1 



 

β tm−2 β tm−1 −β tm−2 β tm−2 −tm−2 ln t + m−2 t

 tm−2 ×h1 tm−2 , x m−2 , y m−2 , zm−2

tm−1

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

(13.24)  ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

 



 β tm+1 −β(tm ) ln t m t + β(ttmm ) m m , zm ) ×h⎡ 1 (tm , x , y 

β(tm )−β tm−1 

ln t β tm−1 t  m−1

−2tm−1 ⎣ β tm−1 + tm−1

 m−1 , y m−1 , zm−1 ×h1 t⎡ , x m−1



 β tm−1 −β tm−2

 ln tm−2 β tm−2 t 

+tm−2 ⎣ β tm−2 + tm−2

 ×h1 tm−2 , x m−2 , y m−2 , zm−2

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

y k+1 = y0 +

⎤

β(t ) tm m

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

    1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h2 tk , x k , y k , zk tk ln tk + AB (α) t tk

250

New Numerical Scheme With Newton Polynomial

 k 

α β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) tm−2 ln tm−2 + AB (α)  (α + 1) t tm−2 m=2    × h2 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α

+

⎡ k ⎢ α (t)α  ⎢ ⎢ + ⎢ AB (α)  (α + 2) m=2 ⎣

 ×



 



β tm−1 β(tm )−β tm−1 β tm−1 ln t + m−1 t tm−1

 m−1 , y m−1 , zm−1 ×h t , x 2 m−1 



 

β tm−2 β tm−1 −β tm−2 β tm−2 −tm−2 ln t + m−2 t

 tm−2 ×h2 tm−2 , x m−2 , y m−2 , zm−2

tm−1

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

zk+1 = z0 +

⎥ ⎥ ⎥ ⎥ ⎦







 β tm+1 −β(tm ) ln t m t + β(ttmm ) m m , zm ) ×h⎡ 2 (tm , x , y 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

 m−1 m−1 , zm−1 ×h2 t⎡ m−1 , x  , y

 β tm−1 −β tm−2

 ln tm−2 β tm−2 t

 ⎣ +tm−2 β tm−2 + tm−2

 m−2 ×h2 tm−2 , x , y m−2 , zm−2

⎤

β(t ) tm m

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎤

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

    1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) tk h3 tk , x k , y k , zk ln tk + AB (α) t tk

 k 

α β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + tm−2 AB (α)  (α + 1) t tm−2 m=2    × h3 tm−2 , x m−2 , y m−2 , zm−2 (k − m + 1)α − (k − m)α 

 



⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 ⎥

 k ⎢ m−1 , y m−1 , zm−1 ⎥ α (t)α  ⎢ ×h , x t 3 m−1 ⎢ 



  ⎥

+ ⎢ ⎥ β t β tm−1 −β tm−2 β tm−2 m−2 AB (α)  (α + 2) ⎦ ln tm−2 + tm−2 m=2 ⎣ −tm−2 t

 m−2 m−2 m−2 ×h3 tm−2 , x ,y ,z +

Numerical method for a fractal–fractional ordinary differential equation

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  α (t) ⎢ + ⎢ ⎢ AB (α) 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 

 



 β tm+1 −β(tm ) ln tm t + β(ttmm ) m m , zm ) ×h⎡ 3 (tm , x , y 

β(tm )−β tm−1 

ln t β tm−1 t

 m−1 −2tm−1 ⎣ β tm−1 + tm−1

 m−1 m−1 , zm−1 ×h3 t⎡ m−1 , x  , y

 β tm−1 −β tm−2 

ln tm−2 β tm−2 t 

+tm−2 ⎣ β tm−2 + tm−2

 ×h3 tm−2 , x m−2 , y m−2 , zm−2

α

⎤

β(t ) tm m

 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

251

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

Figure 13.5 Numerical simulation for the fractal–fractional derivative with variable order with the MittagLeffler kernel for α = 1, β = 0.05 + 0.01t.

252

New Numerical Scheme With Newton Polynomial

Figure 13.6 Numerical simulation for the fractal–fractional derivative with variable order with the MittagLeffler kernel for α = 1, β = 1.

where

and

 h1 tm , x m , y m , zm = ax m + by m zm ,

 h2 tm , x m , y m , zm = cx m + dy m − x m zm ,

 h3 tm , x m , y m , zm = ex m y m + f zm ,   h1 tm−1 , x m−1 , y m−1 , zm−1 = ax m−1 + by m−1 zm−1 ,   h2 tm−1 , x m−1 , y m−1 , zm−1 = cx m−1 + dy m−1 − x m−1 zm−1 ,   h3 tm−1 , x m−1 , y m−1 , zm−1 = ex m−1 y m−1 + f zm−1 ,   h1 tm−2 , x m−2 , y m−2 , zm−2 = ax m−2 + by m−2 zm−2 ,   h2 tm−2 , x m−2 , y m−2 , zm−2 = cx m−2 + dy m−2 − x m−2 zm−2 ,   h3 tm−2 , x m−2 , y m−2 , zm−2 = ex m−2 y m−2 + f zm−2 .

(13.25)

(13.26)

(13.27)

For this model, we shall choose the parameters a = 0.2, b = 1, c = −0.01, d = −0.4, e = −1 and f = −1. Numerical simulations are provided in Fig. 13.5 and Fig. 13.6.

Numerical scheme for partial differential equations with integer and non-integer order

14

In this chapter, we present a new numerical scheme for partial differential equations with integer and non-integer order.

14.1

Numerical scheme with classical derivative

In this section, to construct the numerical scheme that will help to solve partial differential equations, we consider the following advection–diffusion equation: ut = uxx − ux + h (x, t)

(14.1)

where we can take H (u, x, t) = uxx − ux + h (x, t). We convert Eq. (14.1) into  t u (x, t) − u (x, 0) = H (u, x, τ ) dτ. (14.2) 0

At the point tk+1 = (k + 1) t  u (x, tk+1 ) − u (x, 0) =

tk+1

H (u, x, τ ) dτ

(14.3)

0

and at the point tk = kt, we have  tk u (x, tk ) − u (x, 0) = H (u, x, τ ) dτ.

(14.4)

0

If we take the difference of Eqs. (14.3) and (14.4), we can get the following:  tk+1 H (u, x, τ ) dτ. u (x, tk+1 ) − u (x, tk ) =

(14.5)

tk

Here, we suggest the usage of the Newton polynomial as the approximation of the function H (u, x, t),   P2 (τ ) = H uk−2 , x, tk−2     H uk−1 , x, tk−1 − H uk−2 , x, tk−2 + (τ − tk−2 ) t New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00020-2 Copyright © 2021 Elsevier Inc. All rights reserved.

254

New Numerical Scheme With Newton Polynomial

+

      H uk , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 2 (t)2

(14.6)

× (τ − tk−2 ) (τ − tk−1 ) . Let us insert this polynomial into Eq. (14.5); we have   (14.7) uk+1 − uk = H uk−2 , x, tk−2 t   k−2   tk+1  k−1 H u , x, tk−1 − H u , x, tk−2 + (τ − tk−2 ) dτ t tk        tk+1 H uk , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 + 2 (t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ and we obtain   (14.8) uk+1 − uk = H uk−2 , x, tk−2 t  k−1   k−2  t k+1 H u , x, tk−1 − H u , x, tk−2 + (τ − tk−2 ) dτ t tk       H uk , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 + 2 (t)2  tk+1 (τ − tk−2 ) (τ − tk−1 ) dτ. tk

We can compute the integrals on the right hand side of Eq. (14.8); thus,  

tk+1

(τ − tk−2 ) dτ =

tk tk+1 tk

(τ − tk−2 ) (τ − tk−1 ) dτ =

5 (t)2 , 2

(14.9)

23 (t)3 . 6

When we place the above calculations into Eq. (14.8), at the point xn we obtain the following:   uk+1 (14.10) = ukn + H uk−2 , xn , tk−2 t n    5   t + H uk , xn , tk − H uk−1 , xn , tk−1 2        23 t + H uk , xn , tk − 2H uk−1 , xn , tk−1 + H uk−2 , xn , tk−2 12

Numerical scheme for partial differential equations with integer and non-integer order

255

and we get   4  5  k−2 H u , xn , tk−2 t − H uk−1 , xn , tk−1 t 12 3  23  k + H u , xn , tk t. 12

uk+1 = ukn + n

(14.11)

Replacing the function H (u, x, t) by its value, the considered equation can be solved numerically as follows: uk+1 n

k−2 k−2 − uk−2 uk−2 5 un+1 − 2uk−2 n n−1 n+1 − un−1 k−2 + hn + − t 12 2x (x)2

k−1 k−1 uk−1 − uk−1 4 un+1 − 2uk−1 n n−1 n+1 − un−1 k−1 + hn − − t (14.12) 3 2x (x)2

23 ukn+1 − 2ukn − ukn−1 ukn+1 − ukn−1 k − + + hn t. 12 2x (x)2 = ukn

14.1.1 Numerical illustrations To test the suggested numerical method, we consider the following advection– diffusion equation with classical derivative Example 14.1. We consider the following advection–diffusion equation with classical derivative. ut = uxx − aux + h (x, t)

(14.13)

where h (x, t) = 0, a = 1.5. Thus, the following scheme is given: uk+1 n

k−2 k−2 uk−2 − uk−2 5 un+1 − 2uk−2 n n−1 n+1 − un−1 k−2 + hn + − t 12 2x (x)2

k−1 k−1 − uk−1 uk−1 4 un+1 − 2uk−1 n n−1 n+1 − un−1 k−1 + hn − − t (14.14) 3 2x (x)2

23 ukn+1 − 2ukn − ukn−1 ukn+1 − ukn−1 + hnk t. − + 12 2x (x)2 = ukn

The numerical simulation is presented in Fig. 14.1.

256

New Numerical Scheme With Newton Polynomial

Figure 14.1 Numerical simulation of advection–diffusion equation with classical derivative for α = 0.92.

14.2

Numerical scheme with fractal derivative

To derive the numerical algorithm, we deal with the advection–diffusion equation which is given by CH β 0 D u = uxx

− ux + h (x, t)

(14.15)

where we can take h (u, x, t) = uxx − ux + h (x, t). We convert Eq. (14.15) into  t u (x, t) − u (x, 0) = βτ β−1 h (u, x, τ ) dτ. (14.16) 0

At the point tk+1 = (k + 1) t and with H (u, x, τ ) = βτ β−1 h (u, x, τ )  tk+1 u (x, tk+1 ) − u (x, 0) = H (u, x, τ ) dτ

(14.17)

0

and at the point tk = kt, we have  tk u (x, tk ) − u (x, 0) = H (u, x, τ ) dτ. 0

(14.18)

Numerical scheme for partial differential equations with integer and non-integer order

257

If we take the difference of Eqs. (14.17) and (14.18), we can get the following:  u (x, tk+1 ) − u (x, tk ) =

tk+1

H (u, x, τ ) dτ.

(14.19)

tk

Here, we suggest the usage of the Newton polynomial as the approximation of the function H (u, x, t),  H uk−1 , x, tk−1  − H uk−2 , x, tk−2   k−2 P2 (τ ) = H u , x, tk−2 + (τ − tk−2 ) t   k−1   k−2   k H u , x, tk − 2H u , x, tk−1 + H u , x, tk−2 + (14.20) 2 (t)2 × (τ − tk−2 ) (τ − tk−1 ) . Let us insert this polynomial into Eq. (14.19); we have   uk+1 = uk + H uk−2 , x, tk−2 t (14.21)   k−2   tm+1  k−1 H u , x, tk−1 − H u , x, tk−2 + (τ − tk−2 ) dτ t tm       tm+1  k H u , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 + 2 (t)2 tm × (τ − tk−2 ) (τ − tk−1 ) dτ and at the point xn , we obtain   uk+1 = ukn + H uk−2 , xn , tk−2 t n     t k+1 H uk−1 , xn , tk−1 − H uk−2 , xn , tk−2 + (τ − tk−2 ) dτ t tk       H uk , xn , tk − 2H uk−1 , xn , tk−1 + H uk−2 , xn , tk−2 + 2 (t)2  tk+1 × (τ − tk−2 ) (τ − tk−1 ) dτ.

(14.22)

tk

We can compute the integrals on the right hand side of Eq. (14.22); thus,  

tk+1

(τ − tk−2 ) dτ =

tk tk+1 tk

(τ − tk−2 ) (τ − tk−1 ) dτ =

5 (t)2 , 2 23 (t)3 . 6

(14.23)

258

New Numerical Scheme With Newton Polynomial

When placing the above calculations into Eq. (14.23), we obtain the following:   uk+1 = ukn + H uk−2 , xn , tk−2 t n    H uk−1 , xn , tk−1  5 t + −H uk−2 , xn , tk−2 2      23 H uk , xn , tk − 2H uk−1 , xn , tk−1 + t k−2 +H u , xn , tk−2 12

(14.24)

and we get   4  5  k−2 H u , xn , tk−2 t − H uk−1 , xn , tk−1 t 12 3  23  k + H u , xn , tk t. 12

= ukn + uk+1 n

(14.25)

Thus we get the following approximation:  5 β−1  k−2 βtk−2 h u , xn , tk−2 t 12  4 β−1  k−1 − βtk−1 h u , xn , tk−1 t 3  23 β−1  + βtk h uk , xn , tk t. 12

= uk + uk+1 n

(14.26)

Substituting all terms into the above, we have the following uk+1 n

k−2

k−2 − uk−2 uk−2 5 β−1 un+1 − 2uk−2 n n−1 n+1 − un−1 k−2 + βtk−2 − t + hn 12 2x (x)2 k−1

k−1 uk−1 − uk−1 4 β−1 un+1 − 2uk−1 n n−1 n+1 − un−1 k−1 + hn − − βtk−1 t (14.27) 3 2x (x)2

23 β−1 ukn+1 − 2ukn − ukn−1 ukn+1 − ukn−1 + hkn t. − + βtk 12 2x (x)2 = ukn

14.2.1 Numerical illustrations We present a numerical simulation for the advection–diffusion equation with fractal derivative. Example 14.2. We consider the following advection–diffusion equation with fractal derivative: CH β 0 Dt u = uxx

− aux + h (x, t)

(14.28)

Numerical scheme for partial differential equations with integer and non-integer order

259

Figure 14.2 Numerical simulation of advection–diffusion equation with fractal derivative for α = 0.94.

where h (x, t) = 0, a = 2. Thus, the following scheme is given: k−2

k−2 − uk−2 uk−2 5 β−1 un+1 − 2uk−2 n n−1 n+1 − un−1 k+1 k k−2 + hn un = un + βtk−2 − t 12 2x (x)2 k−1

k−1 − uk−1 uk−1 4 β−1 un+1 − 2uk−1 n n−1 n+1 − un−1 k−1 + hn − − βtk−1 t (14.29) 3 2x (x)2

23 β−1 ukn+1 − 2ukn − ukn−1 ukn+1 − ukn−1 k + hn t. − + βtk 12 2x (x)2 The numerical simulation is depicted in Fig. 14.2.

14.3

Numerical scheme with the Atangana–Baleanu fractional operator

In this section, we present a newly established numerical approximation for the advection–diffusion equation having the Atangana–Baleanu fractional operator. This

260

New Numerical Scheme With Newton Polynomial

numerical approximation established with the Newton polynomial is used for solving linear and non-linear partial differential equations. We firstly consider the advection–diffusion equation including the Atangana– Baleanu fractional operator, ABC α Dt u (x, t) = 0

∂2 ∂ u (x, t) + h (x, t) u (x, t) − ∂x ∂x 2

(14.30)

with the initial condition u (x, 0) = g (x) , u (x, t) |∂ = f (t) .

(14.31)

We shall write the above equation ABC α Dt u (x, t) = H 0

(14.32)

(u, x, t)

where H (u, x, t) = uxx − ux + h (x, t). Integrating on the left and right side, we have the following:   ABC α ABC α J D u t) =ABC Jtα [H (x, t, u)] . (14.33) (x, t 0 t 0 0 We transform Eq. (14.30) into u (x, t)−u (x, 0) =

α 1−α H (x, t, u) + AB (α) AB (α)  (α)



t

H (x, τ, u) (t − τ )α−1 dτ.

0

(14.34) At the point tk+1 = (k + 1) t, we have the following:   1−α H x, tk , uk (14.35) AB (α)  tk+1 α + H (x, τ, u) (tk+1 − τ )α−1 dτ, AB (α)  (α) 0

u (x, tk+1 ) − u (x, 0) =

and we write   1−α H x, tk , uk AB (α) k  tm+1

α H (x, τ, u) (tk+1 − τ )α−1 dτ. + AB (α)  (α) tm

u (x, tk+1 ) = u (x, 0) +

(14.36)

m=2

After putting the Newton polynomial into Eq. (14.36), the above equation can be written as follows:   α 1−α H x, tk , uk + (14.37) uk+1 = u0 + AB (α) AB (α)  (α)

Numerical scheme for partial differential equations with integer and non-integer order

261

⎧   H x, tm−2 , um−2  ⎪ ⎪   ⎪ k  tm+1 ⎪ ⎨ + H x,tm−1 ,um−1 −H x,tm−2 ,um−2 (τ − t

m−2 ) t    × H (x,tm ,um )−2H x,tm−1 ,um−1 +H x,tm−2 ,um−2 ⎪ + ⎪ m=2 tm ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we have   α 1−α H x, tk , uk + uk+1 = u0 + AB (α) AB (α)  (α)  tm+1   ⎧ m−2 H x, tm−2 , u (tk+1 − τ )α−1 dτ ⎪ tm ⎪     ⎪ m−1 −H x,t m−2  ⎪ H x,t ,u tm+1 m−1 m−2 ,u ⎪ ⎪ + k ⎨ tm t

× × (τ − tm−2 )(tk+1 − τ )α−1 dτ    ⎪ t m m−1 +H x,t m−2 m−2 ,u m+1 H (x,tm ,u )−2H x,tm−1 ,u m=2 ⎪ ⎪ ⎪ + ⎪ tm 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

(14.38) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

Now, at the point xn , we set   1−α H xn , tk , uk AB (α) k   tm+1 

α m−2 H xn , tm−2 , u + (tk+1 − τ )α−1 dτ AB (α)  (α) t m m=2     k

H xn , tm−1 , um−1 − H xn , tm−2 , um−2 α + AB (α)  (α) t m=2  tm+1 α (14.39) × (τ − tm−2 ) (tk+1 − τ )α−1 dτ + AB (α)  (α) tm     k

H (xn , tm , um ) − 2H xn , tm−1 , um−1 + H xn , tm−2 , um−2 × 2 (t)2 m=2  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

uk+1 = u0n + n

tm

We insert the calculations for the above integrals into Eq. (14.39), and we obtain the following approximation:   1−α 0 k H x uk+1 = u + , t , u n k n n AB (α) k  

α (t)α + H xn , tm−2 , um−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α

262

New Numerical Scheme With Newton Polynomial

  k 

α (t)α H xn , tm−1 , um−1  −H xn , tm−2 , um−2 AB (α)  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H(xn , tm , um ) 

α (t)α ⎣ −2H xn , tm−1 , um−1 ⎦ +   2AB (α)  (α + 3) +H xn , tm−2 , um−2 m=2  ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 +

(14.40)

⎤ ⎥ ⎥. ⎦

In our case, we have H (u, x, t) = uxx − ux + h (x, t) .

(14.41)

So, we have the following equalities: m m m  um  um n+1 − 2un − un−1 n+1 − un−1 + hnm , H um − (14.42) n , x n , tm = 2x (x)2  um−1 − 2um−1 − um−1 um−1 − um−1  n n−1 n−1 + hnm−1 , H um−1 , xn , tm−1 = n+1 − n+1 n 2 2x (x)

and   um−2 − 2um−2 − um−2 um−2 − um−2 n n−1 n−1 H um−2 + hnm−2 . (14.43) , xn , tm−2 = n+1 − n+1 n 2x (x)2 Inserting the above into our equation, we have

ukn+1 − 2ukn − ukn−1 ukn+1 − ukn−1 k+1 0 k + hn un = un + − 2x (x)2 ⎧ ⎫ m−2 −um−2 um−2 k ⎨ ⎬ n+1 −2un n−1

α (t)α+1 2 (x) + m−2 m−2 ⎩ un+1 −un−1 ⎭ AB (α)  (α + 1) m=2 − 2x + hnm−2   × (k − m + 1)α − (k − m)α ⎡  m−1 m−1 m−1  ⎤ m−1 un+1 −2un −un−1 um−1 n+1 −un−1 m−1 k − + h 2

⎢ n ⎥ 2x α (t)α ⎢  m−2 (x)  ⎥ + m−2 m−2 m−2 m−2 ⎦ ⎣ u −2un −un−1 u −u AB (α)  (α + 2) − n+1 − n+12x n−1 + hnm−2 m=2 2 (x)  α (k − m + 1) (k − m + 3 + 2α) (14.44) × − (k − m)α (k − m + 3 + 3α)

Numerical scheme for partial differential equations with integer and non-integer order

⎡

m m um n+1 −2un −un−1

263

!

m um n+1 −un−1 + hnm 2x  m−1 m−1 un+1 −un−1 m−1 − + hn 2x

−

⎢  m−1 m−1 m−1 k ⎢

α (t)α ⎢ −2 un+1 −2un 2 −un−1 + ⎢ ⎢ 2AB (α)  (α + 3)   m−2 (x) m−2 m=2 ⎣ un+1 −2um−2 −um−2 um−2 n n−1 n+1 −un−1 m−2 + − + hn 2x (x)2  ⎡ ⎤ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ 2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 (x)2

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

14.3.1 Numerical illustrations In this subsection, we deal with the numerical solution of the advection–diffusion equation which has the Atangana–Baleanu fractional derivative. Example 14.3. We consider the following advection–diffusion equation with the Mittag-Leffler kernel: AB α 0 Dt u = uxx

− aux + h (x, t)

(14.45)

where h (x, t) = 0, a = 2.2. Thus, the following scheme is given:

ukn+1 − 2ukn − ukn−1 ukn+1 − ukn−1 k+1 0 k − un = un + + hn 2x (x)2 ⎧ ⎫ m−2 −um−2 um−2 k ⎨ ⎬ n+1 −2un n−1 α+1

α (t) (x)2 + m−2 m−2 ⎩ un+1 −un−1 ⎭ AB (α)  (α + 1) m=2 − 2x + hnm−2   × (k − m + 1)α − (k − m)α ⎡  +

α (t)α AB (α)  (α + 2) 

×

k

m=2

⎢ ⎢ ⎣

m−1 −um−1 um−1 n+1 −2un n−1

 −

(x)2

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡

−

m−2 −um−2 um−2 n+1 −2un n−1 (x)2

m−1 um−1 n+1 −un−1 2x

−

2x

⎥  ⎥ ⎦ m−2

+ hn

!

m um n+1 −un−1 + hnm 2x  um−1 −um−1 − n+12x n−1 + hnm−1

−

2 ⎢  m−1 (x) k ⎢ m−1 −um−1

u −2u n α (t)α n+1 n−1 ⎢ −2 + ⎢ (x)2 ⎢ 2AB (α)  (α + 3)  m−2 m−2 m−2 m=2 ⎣ u −2un −un−1 + n+1 − 2

(x)

 ⎤ + hnm−1

m−2 um−2 n+1 −un−1



m m um n+1 −2un −un−1

(14.46)

m−2 um−2 n+1 −un−1 2x

⎤

⎥ ⎥ ⎥ ⎥  ⎥ ⎦ m−2

+ hn

264

New Numerical Scheme With Newton Polynomial

Figure 14.3 Numerical simulation of advection–diffusion equation with the Atangana–Baleanu derivative for α = 0.96.

 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ 2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥. ⎦

The numerical simulation is presented in Fig. 14.3.

14.4

Numerical scheme with the Caputo fractional operator

We firstly consider the advection–diffusion equation including the Caputo fractional operator C α 0 Dt u (x, t) =

∂2 ∂ u (x, t) + h (x, t) , u (x, t) − ∂x ∂x 2

(14.47)

with the initial condition u (x, 0) = g (x) , u (x, t) |∂ = f (t) .

(14.48)

Numerical scheme for partial differential equations with integer and non-integer order

265

We shall write the above equation C α 0 Dt u (x, t) = H

(14.49)

(u, x, t)

where H (u, x, t) = uxx − ux + h (x, t). Applying the Caputo integral on both sides, we get   C α C α C α (14.50) 0 Jt 0 Dt u (x, t) =0 Jt [H (x, t, u)] . We transform Eq. (14.47) into u (x, t) − u (x, 0) =

1  (α)



t

H (x, τ, u) (t − τ )α−1 dτ.

(14.51)

0

At the point tk+1 = (k + 1) t, we have the following:  tk+1 1 u (x, tk+1 ) − u (x, 0) = H (x, τ, u) (tk+1 − τ )α−1 dτ,  (α) 0

(14.52)

and we write k  1 tm+1 u (x, tk+1 ) = u (x, 0) + H (x, τ, u) (tk+1 − τ )α−1 dτ. (14.53)  (α) tm m=2

After replacing by its Newton polynomial, we write the above equation as follows: ⎧ ⎫   H x, tm−2 , um−2 ⎪ ⎪ ⎪ ⎪    ⎪ ⎪ ⎪ H x,tm−1 ,um−1 −H x,tm−2 ,um−2 ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ t ⎪ ⎪ ⎪ ⎪  k ⎨ ⎬ × − t (τ ) t

m+1 m−2 1   k+1 0 m m−1 H (x,tm ,u )−2H x,tm−1 ,u (14.54) u =u + ⎪ ⎪  (α) ⎪ + ⎪ 2(t)2 m=2 tm ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ H x,tm−2 ,um−2 ⎪ ⎪ ⎪ ⎪ + 2 ⎪ ⎪ 2(t) ⎪ ⎪ ⎩ ⎭ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ. Thus, we have

uk+1 = u0 +

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k

⎨ ⎪ ⎪ ⎪ ⎩

m=2 ⎪ ⎪ ⎪

 tm+1

  H x, tm−2 , um−2 (tk+1 − τ )α−1 dτ

    H x,tm−1 ,um−1 −H x,tm−2 ,um−2 t × (τ − tm−2 ) (tk+1 − τ )α−1 dτ  tm+1 H (x,tm ,um )−2H x,tm−1 ,um−1 +H x,tm−2 ,um−2  + tm 2(t)2 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ tm

+

 tm+1 tm

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(14.55)

266

New Numerical Scheme With Newton Polynomial

Now, at the point xn , we set k   tm+1 1  H xn , tm−2 , um−2 (tk+1 − τ )α−1 dτ  (α) tm m=2     k 1 H xn , tm−1 , um−1 − H xn , tm−2 , um−2 +  (α) t

uk+1 = u0n + n

 × ×

k 1  (α) tm m=2     H (xn , tm , um ) − 2H xn , tm−1 , um−1 + H xn , tm−2 , um−2

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

(14.56)

2 (t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

We insert the calculations for the above integrals into Eq. (14.56), and we obtain the following approximation: uk+1 n

= u0n

k  (t)α  + H xn , tm−2 , um−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α k    (t)α   H xn , tm−1 , um−1 − H xn , tm−2 , um−2  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × (14.57) − (k − m)α (k − m + 3 + 3α)   k  (t)α H (xn , tm , um ) − 2H xn , tm−1 , um−1   + +H xn , tm−2 , um−2 2 (α + 3) m=2  ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+

Replacing the above in our equation, we have

ukn+1 − 2ukn − ukn−1 ukn+1 − ukn−1 k+1 0 k un = u + + hn − 2x (x)2 ⎧ ⎫ m−2 −um−2 um−2 k ⎬ n+1 −2un n−1 (t)α+1 ⎨ 2 (x) + m−2 m−2 ⎩ un+1 −un−1 ⎭  (α + 1) m=2 − 2x + hnm−2

Numerical scheme for partial differential equations with integer and non-integer order

  × (k − m + 1)α − (k − m)α ⎫ ⎤ ⎡ ⎧ m−1 m−1 m−1 ⎨ un+1 −2un −un−1 ⎬ (x)2 ⎢ ⎥ m−1 ⎥ k ⎢ ⎩ um−1 n+1 −un−1 m−1 ⎭ ⎥ α (t)α ⎢ − + h n 2x ⎢ ⎥ ⎧ ⎫ + m−2 −um−2 ⎢ ⎨ um−2  (α + 2) ⎬ ⎥ n+1 −2un n−1 ⎢ ⎥ m=2 ⎣ ⎦ (x)2 − m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn  α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎧ ⎫ ⎡ ⎤ m m m ⎨ un+1 −2un −un−1 ⎬ 2 (x) ⎢ ⎥ ⎢ ⎥ ⎩ − umn+1 −umn−1 + h m ⎭ ⎢ ⎥ n 2x ⎧ ⎫ ⎢ ⎥ m−1 m−1 −um−1 u −2u ⎢ k n ⎨ ⎬ ⎥ n+1 n−1 ⎥ α (t)α ⎢ 2 (x) ⎢ −2 ⎥ + m−1 m−1 ⎢ ⎥ u −u ⎩ ⎭ 2 (α + 3) n+1 n−1 m−1 ⎢ ⎥ hn m=2 ⎢ ⎧ − m−22x m−2 + m−2 ⎫ ⎥ ⎢ ⎨ un+1 −2un −un−1 ⎬ ⎥ ⎢ ⎥ 2 ⎣ + ⎦ (x) m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn  ⎡ 2 2 (k − m) + (3α + 10) (k − m) α ⎢ (k − m + 1) 2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

267

(14.58)

⎤ ⎥ ⎥. ⎦

14.4.1 Numerical illustrations As a numerical simulation, we present a partial differential equation having the Caputo fractional derivative. Example 14.4. We consider the following advection–diffusion equation with the Caputo fractional derivative: C α 0 Dt u = uxx

− aux + h (x, t)

(14.59)

where h (x, t) = 0, a = 5. Thus, the following scheme is given: uk+1 n

=u + 0

+

ukn+1 − 2ukn − ukn−1

(t)α+1  (α + 1)

ukn+1 − ukn−1

− 2x (x)2 ⎧ ⎫ m−2 m−2 m−2 un+1 −2un −un−1 k ⎨ ⎬

m=2

⎩

(x)2 m−2 um−2 −u − n+12x n−1 + hnm−2

⎭

+ hkn

268

New Numerical Scheme With Newton Polynomial

  × (k − m + 1)α − (k − m)α ⎫ ⎤ ⎡ ⎧ m−1 m−1 m−1 ⎨ un+1 −2un −un−1 ⎬ (x)2 ⎢ ⎥ m−1 ⎥ k ⎢ ⎩ um−1 n+1 −un−1 m−1 ⎭ ⎥ α (t)α ⎢ − + h n 2x ⎢ ⎥ ⎧ ⎫ + m−2 −um−2 ⎢ ⎨ um−2  (α + 2) ⎬ ⎥ n+1 −2un n−1 ⎢ ⎥ m=2 ⎣ ⎦ (x)2 − m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn  α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎧ ⎫ ⎡ ⎤ m m m ⎨ un+1 −2un −un−1 ⎬ 2 (x) ⎢ ⎥ ⎢ ⎥ ⎩ − umn+1 −umn−1 + h m ⎭ ⎢ ⎥ n 2x ⎧ ⎫ ⎢ ⎥ m−1 m−1 −um−1 u −2u ⎢ k n ⎨ ⎬ ⎥ n+1 n−1 ⎥ α (t)α ⎢ 2 (x) ⎢ −2 ⎥ + m−1 m−1 ⎢ ⎥ u −u ⎩ ⎭ 2 (α + 3) n+1 n−1 m−1 ⎢ ⎥ hn m=2 ⎢ ⎧ − m−22x m−2 + m−2 ⎫ ⎥ ⎢ ⎨ un+1 −2un −un−1 ⎬ ⎥ ⎢ ⎥ 2 ⎣ + ⎦ (x) m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn  ⎡ 2 2 (k − m) + (3α + 10) (k − m) α ⎢ (k − m + 1) 2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

(14.60)

⎤ ⎥ ⎥. ⎦

The numerical simulation is given in Fig. 14.4.

14.5

Numerical scheme with the Caputo–Fabrizio fractional operator

In this section, we offer a numerical scheme for the advection–diffusion equation with the Caputo–Fabrizio operator, namely we deal with the following equation: ∂2 ∂ u (x, t) + h (x, t) u (x, t) − ∂x ∂x 2 u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) .

CF α 0 Dt u (x, t) =

(14.61)

We write the above equation as follows: CF α 0 Dt u (x, t) = H

(u, x, t) .

(14.62)

Here, H (u, x, t) = uxx − ux + h (x, t). If we integrate Eq. (14.62), we obtain   CF α CF α α =CF J D u t) (14.63) (x, t 0 t 0 0 Jt [H (u, x, t)] .

Numerical scheme for partial differential equations with integer and non-integer order

269

Figure 14.4 Numerical simulation of advection–diffusion equation with the Caputo derivative for α = 0.94.

This yields the following: α 1−α H (u, x, t) + u (x, t) − u (x, 0) = M (α) M (α)



t

H (u, x, τ ) dτ.

(14.64)

0

Then at tk+1 and tk , we obtain    1−α   k H u , x, tk − H uk−1 , x, tk−1 M (α)  tk+1 α + H (u, x, τ ) dτ. M (α) tk

u (x, tk+1 ) =

(14.65)

Using the Newton polynomial, we have the following approximation: at xn ,    1−α   k H u , xn , tk − H uk−1 , xn , tk−1 M (α)    k−1  23  k  4 α − t H u , xn , tk t H u , x , t n k−1 12 3   + . 5 H uk−2 , xn , tk−2 t + 12 M (α)

= u0n + uk+1 n

(14.66)

270

New Numerical Scheme With Newton Polynomial

If we replace the function H (u, x, t) by its value, we can obtain the following scheme: ⎧ ⎫  k  un+1 −2ukn −ukn−1 ukn+1 −ukn−1 ⎪ ⎪ k ⎪ ⎪ − 2x + hn ⎬ 1−α ⎨  (x)2 k+1 0  (14.67) un = un + k−1 k−1 uk−1 −2uk−1 uk−1 ⎪ M (α) ⎪ n −un−1 n+1 −un−1 k−1 ⎪ ⎪ − + h ⎩ − n+1 (x) ⎭ 2 n 2x ⎧ ⎫   k un+1 −2ukn −ukn−1 ukn+1 −ukn−1 ⎪ ⎪ 23 k ⎪ ⎪ − 2x + hn ⎪ ⎪ ⎪ ⎪ 12 t (x)2 ⎪ ⎪ ⎪ ⎪   k−1 k−1 k−1 ⎨ ⎬ k−1 k−1 α u −2u −u u −u n 4 n+1 n−1 n+1 n−1 k−1 . + − 3 t − + h 2 n 2x ⎪ M (α) ⎪ ⎪ ⎪   k−2 (x) ⎪ ⎪ ⎪ ⎪ u −2uk−2 −uk−2 uk−2 −uk−2 ⎪ ⎪ 5 ⎪ t n+1 n 2 n−1 − n+12x n−1 + hnk−2 ⎪ ⎩ + 12 ⎭ (x)

14.5.1 Numerical illustration Example 14.5. We consider the following advection–diffusion equation with the Caputo–Fabrizio fractional derivative: CF α 0 Dt u = uxx

− aux + h (x, t)

(14.68)

where h (x, t) = 0, a = 5. For this equation, we have the following scheme: ⎧ ⎫  k  un+1 −2ukn −ukn−1 ukn+1 −ukn−1 ⎪ ⎪ k ⎪ ⎪ − 2x + hn ⎨ ⎬ 2 1 − α (x) k+1 0   (14.69) un = un + k−1 k−1 k−1 k−1 u −2uk−1 u −u ⎪ M (α) ⎪ n −un−1 ⎪ − n+12x n−1 + hnk−1 ⎪ ⎩ − n+1 (x) ⎭ 2 ⎧ ⎫   k un+1 −2ukn −ukn−1 ukn+1 −ukn−1 ⎪ ⎪ 23 k ⎪ ⎪ − 2x + hn ⎪ ⎪ ⎪ ⎪ 12 t (x)2 ⎪ ⎪ ⎪ ⎪   ⎨ ⎬ k−1 k−1 k−1 k−1 k−1 −u α u −2u u −u n 4 n+1 n−1 n+1 n−1 k−1 . + − 3 t − 2x + hn 2 M (α) ⎪ ⎪ ⎪  ⎪  k−2 (x) ⎪ ⎪ k−2 k−2 k−2 k−2 ⎪ ⎪ u −2u −u u −u ⎪ ⎪ 5 ⎪ t n+1 n 2 n−1 − n+12x n−1 + hnk−2 ⎪ ⎩ + 12 ⎭ (x) The numerical simulation is presented in Fig. 14.5.

14.6

Numerical scheme with the Atangana–Baleanu fractal–fractional operator

In this section, we establish new numerical approximation for solving our equation which has the Atangana–Baleanu fractal–fractional operator. Namely, we consider FFM α Dt u (x, t) = 0

∂2 ∂ u (x, t) − u (x, t) + h (x, t) 2 ∂x ∂x

(14.70)

Numerical scheme for partial differential equations with integer and non-integer order

271

Figure 14.5 Numerical simulation of advection–diffusion equation with the Caputo–Fabrizio derivative for α = 0.93.

under the conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) ,

(14.71)

where h (u, x, t) = uxx − ux + h (x, t). For convenience, we have ABC α Dt u (x, t) = βt β−1 h (u, x, t) . 0

(14.72)

Here, we shall take H (u, x, t) = βt β−1 h (u, x, t). We transform Eq. (14.72) into 1−α H (x, t, u) AB (α)  t α + H (x, τ, u) (t − τ )α−1 dτ. AB (α)  (α) 0

u (x, t) − u (x, 0) =

(14.73)

At the point tk+1 = (k + 1) t, we have the following:   1−α H x, tk , uk (14.74) u (x, tk+1 ) − u (x, 0) = AB (α)  tk+1 α + H (x, τ, u) (tk+1 − τ )α−1 dτ AB (α)  (α) 0

272

New Numerical Scheme With Newton Polynomial

and we write   1−α H x, tk , uk AB (α) k  tm+1

α H (x, τ, u) (tk+1 − τ )α−1 dτ. + AB (α)  (α) tm

u (x, tk+1 ) = u (x, 0) +

(14.75)

m=2

With the help of the Newton polynomial, Eq. (14.75) can be approximated as follows:   α 1−α H x, tk , uk + AB (α) AB (α)  (α) ⎧   H x, tm−2 , um−2  ⎪ ⎪    ⎪ k  tm+1 ⎪ ⎨ + H x,tm−1 ,um−1 −H x,tm−2 ,um−2 (τ − t

m−2 ) t    × m )−2H x,t m−1 +H x,t m−2 H ,u ,u ,u (x,t m m−1 m−2 ⎪ + ⎪ 2 m=2 tm ⎪ 2(t) ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 = u0 +

(14.76) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we have   α 1−α H x, tk , uk + uk+1 = u0 + AB (α) AB (α)  (α)  tm+1   ⎧ H x, tm−2 , um−2 (tk+1 − τ )α−1 dτ ⎪ tm ⎪  ⎪  tm+1 H x,tm−1 ,um−1 −H x,tm−2 ,um−2  ⎪ ⎪ ⎪ + k ⎨ tm t

× × (τ − tm−2 )(tk+1 − τ )α−1 dτ    ⎪ t m m−1 +H x,t m−2 m−2 ,u m+1 H (x,tm ,u )−2H x,tm−1 ,u m=2 ⎪ ⎪ ⎪ + t ⎪ 2 m 2(t) ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

(14.77) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

Now, at the point xn , we set   1−α H xn , tk , uk AB (α) k   tm+1 

α m−−2 H xn , tm−2 , u + (tk+1 − τ )α−1 dτ AB (α)  (α) t m m=2     k

H xn , tm−1 , um−1 − H xn , tm−2 , um−2 α + AB (α)  (α) t

= u0n + uk+1 n

 × ×

m=2

k

α AB (α)  (α) tm m=2     m m−1 + H xn , tm−2 , um−2 H (xn , tm , u ) − 2H xn , tm−1 , u tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 (t)2

(14.78)

Numerical scheme for partial differential equations with integer and non-integer order

 ×

tm+1

273

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

Putting the calculations for the above integrals into Eq. (14.78), we obtain the following approximation:   1−α H xn , tk , uk AB (α) k  

α (t)α H xn , tm−2 , um−2 + AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α   k 

α (t)α H xn , tm−1 , um−1  + −H xn , tm−2 , um−2 AB (α)  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H(xn , tm , um ) 

α (t)α ⎣ −2H xn , tm−1 , um−1 ⎦ +   2AB (α)  (α + 3) +H xn , tm−2 , um−2 m=2  ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

uk+1 = u0n + n

(14.79)

⎤ ⎥ ⎥. ⎦

Replacing the above into our equation, we have ⎧ ⎨ 1 − α β−1 βtk uk+1 = u0 + ⎩ AB (α)

ukn+1 −2ukn −ukn−1

−

(x)2 ukn+1 −ukn−1 2x

+ hkn

⎫ ⎬ ⎭

⎧ ⎫ m−2 m−2 m−2 k ⎨ un+1 −2un −un−1 ⎬

α (t)α+1 β−1 (x)2 + βtm−2 −um−2 ⎩ um−2 ⎭ AB (α)  (α + 1) m=2 − n+12x n−1 + hnm−2   × (k − m + 1)α − (k − m)α ⎧ ⎫ ⎡ m−1 m−1 m−1 ⎨ un+1 −2un −un−1 ⎬ (x)2 ⎢ βt β−1 m−1 m−1 ⎩ k ⎢ um−1 ⎭ n+1 −un−1

⎢ α (t)α − + hnm−1 ⎫ 2x ⎢ ⎧ + m−2 m−2 m−2 ⎢ AB (α)  (α + 2) ⎨ un+1 −2un −un−1 ⎬ m=2 ⎢ ⎣ −βt β−1 (x)2 m−2 m−2 m−2 ⎩ u −u ⎭ − n+12x n−1 + hnm−2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(14.80)

274

New Numerical Scheme With Newton Polynomial

⎡

⎧ ⎨

m m um n+1 −2un −un−1

⎫ ⎬

⎤

⎢ ⎢ ⎩ − + hnm ⎭ ⎢ 2x ⎧ ⎫ ⎢ m−1 m−1 m−1 k ⎢ ⎨ un+1 −2un −un−1 ⎬

⎢ α (t)α 2 β−1 (x) ⎢ −2βt + m−1 m−1 ⎩ ⎢ um−1 ⎭ 2AB (α)  (α + 3) n+1 −un−1 hnm−1 ⎫ m=2 ⎢ ⎢ ⎧ − m−22x m−2 + m−2 ⎢ ⎨ un+1 −2un −un−1 ⎬ ⎢ ⎣ +βt β−1 (x)2 m−2 ⎩ um−2 −um−2 ⎭ − i+12x n−1 + hnm−2  ⎡ ⎤ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ 2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 β−1 βtm

14.7

(x)2 m um n+1 −un−1

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Numerical scheme with the Caputo fractal–fractional operator

In this section, we present a newly suggested numerical scheme for our equation involving the Caputo fractal–fractional operator. Here, we deal with the following equation: FFP α Dt u (x, t) = 0

∂2 ∂ u (x, t) + h (x, t) , u (x, t) − ∂x ∂x 2

(14.81)

with the conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) .

(14.82)

Then, we can have C α 0 Dt u (x, t) = H

(14.83)

(u, x, t)

where H (u, x, t) = βt β−1 h (u, x, t). After integrating, we can write Eq. (14.81) as follows: u (x, t) − u (x, 0) =

1  (α)



t

H (x, τ, u) (t − τ )α−1 dτ.

(14.84)

0

At the point tk+1 = (k + 1) t, we have the following: u (x, tk+1 ) − u (0) =

1  (α)

 0

tk+1

H (x, τ, u) (tk+1 − τ )α−1 dτ

(14.85)

Numerical scheme for partial differential equations with integer and non-integer order

275

and we write u (x, tk+1 ) = u (x, 0) +

k  1 tm+1 H (x, τ, u) (tk+1 − τ )α−1 dτ. (14.86)  (α) tm m=2

After putting the Newton polynomial into Eq. (14.86), the above equation can be written as follows: ⎫ ⎧   m−2 H x, t , u ⎪ ⎪ m−2 ⎪ ⎪     ⎪ ⎪ ⎪ ⎪ H x,tm−1 ,um−1 −H x,tm−2 ,um−2 ⎪ ⎪ ⎪ ⎪  + k ⎬ ⎨ t t

m+1 1 k+1 0 × (τ − tm−2)  un = un +  ⎪  (α) H (x,tm ,um )−2H x,tm−1 ,um−1 +H x,tm−2 ,um−2 ⎪ ⎪ ⎪ m=2 tm ⎪ ⎪ ⎪ ⎪ + 2 ⎪ ⎪ 2(t) ⎪ ⎪ ⎭ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (14.87) × (tk+1 − τ )α−1 dτ. Thus, we have

uk+1 = u0n + n

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ k ⎪ ⎨

⎪ ⎪ ⎪ ⎩

m=2 ⎪ ⎪ ⎪

 tm+1

  H x, tm−2 , um−2 (tk+1 − τ )α−1 dτ

    t H x,tm−1 ,um−1 −H x,tm−2 ,um−2 + tmm+1 t × (τ − tm−2 )(tk+1 − τ )α−1 dτ  tm+1 H (x,tm ,um )−2H x,tm−1 ,um−1 +H x,tm−2 ,um−2  + tm 2(t)2 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ tm

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(14.88) Now, at the point xn , we set k   tm+1 1  H xn , tm−2 , um−2 (tk+1 − τ )α−1 dτ  (α) t m m=2     k m−1 − H xn , tm−2 , um−2 1 H xn , tm−1 , u +  (α) t

uk+1 = u0n + n

 × ×

k 1  (α) tm m=2     m m−1 + H xn , tm−2 , um−2 H (xn , tm , u ) − 2H xn , tm−1 , u

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 (t)2 tm+1

tm

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

(14.89)

276

New Numerical Scheme With Newton Polynomial

Rearranging the above equation, we obtain the following scheme: k  (t)α  H xn , tm−2 , um−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

= u0n + uk+1 n

k    (t)α   H xn , tm−1 , um−1 − H xn , tm−2 , um−2  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   k  (t)α H (xn , tm , um ) − 2H xn , tm−1 , um−1   + +h xn , tm−2 , um−2 2 (α + 3) m=2  ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+

(14.90)

Replacing H (x, t, u) by its value in our equation, we obtain

= u0n + uk+1 n

α+1

k

⎧ ⎨

m−2 −um−2 um−2 n+1 −2un n−1

⎫ ⎬

(t) β−1 (x)2 βtm−2 m−2 −u ⎩ um−2 ⎭  (α + 1) m=2 − n+12x n−1 + hnm−2   × (k − m + 1)α − (k − m)α ⎧ ⎫ ⎤ ⎡ m−1 m−1 m−1 ⎨ un+1 −2un −un−1 ⎬ 2 β−1 (x) ⎢ βt ⎥ m−1 m−1 m−1 ⎩ k ⎢ un+1 −un−1 ⎭ ⎥ α m−1 ⎢ ⎥ (t) + hn 2x ⎢ ⎧ − m−2 ⎫ ⎥ + m−2 m−2 ⎢  (α + 2) ⎨ un+1 −2un −un−1 ⎬ ⎥ ⎥ m=2 ⎢ ⎣ −βt β−1 ⎦ (x)2 m−2 m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn  α (k − m + 1) (k − m + 3 + 2α) × (14.91) − (k − m)α (k − m + 3 + 3α) ⎧ ⎫ ⎡ ⎤ m m m ⎨ un+1 −2un −un−1 ⎬ 2 β−1 (x) ⎢ ⎥ βtm ⎢ ⎥ ⎩ − umn+1 −umn−1 + h m ⎭ ⎢ ⎥ n 2x ⎧ ⎫ ⎢ ⎥ m−1 m−1 −um−1 u −2u ⎢ ⎥ k n ⎨ ⎬ n+1 n−1 α

⎢ ⎥ (t) (x)2 ⎢ −2βt β−1 ⎥ + m−1 m−1 m−1 ⎩ ⎢ ⎥ un+1 −un−1 2 (α + 3) m−1 ⎭ ⎥ ⎢ − + h m=2 ⎢ n 2x ⎧ ⎫ ⎥ m−2 m−2 m−2 ⎢ ⎨ un+1 −2un −un−1 ⎬ ⎥ ⎢ ⎥ ⎣ +βt β−1 ⎦ (x)2 m−2 m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn

Numerical scheme for partial differential equations with integer and non-integer order

 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ 2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

14.8

277

⎤ ⎥ ⎥. ⎦

Numerical scheme for Caputo–Fabrizio fractal–fractional operator

In this section, we present a numerical scheme for the advection–diffusion equation involving the Caputo–Fabrizio fractal–fractional operator. Namely, we take the following equation: FFE α Dt u (x, t) = 0

∂2 ∂ u (x, t) + h (x, t) , u (x, t) − ∂x ∂x 2

(14.92)

with the initial conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = h (t) .

(14.93)

We shall take h (u, x, t) = uxx − ux + h (x, t). We have CF α 0 Dt u (x, t) = H

(u, x, t)

where H (u, x, t) = βt β−1 h (u, x, t), and   CF α CF α α =CF J D u t) (x, t 0 t 0 0 Jt [H (x, t, u)] When applying the associated integral, we can have the following:  t α (1 − α) u (x, t) − u (x, 0) = H (u, x, t) + H (u, x, τ ) dτ. M (α) M (α) 0 As before, we have the following:    ! 1−α H uk , x, tk − H uk−1 , x, tk−1 u (x, tk+1 ) = M (α)   k ⎧ ⎫ 23 H u , x, tk t ⎨ ⎬ 12  α + . − 43 H uk−1 , x, tk−1 t ⎭ M (α) ⎩ 5 k−2 + 12 H u , x, tk−2 t

(14.94)

(14.95)

(14.96)

(14.97)

Using the Newton polynomial, we write the following approximation: at xn , ⎧ ⎫  k  un+1 −2ukn −ukn−1 ukn+1 −ukn−1 ⎪ ⎪ k ⎪ ⎪ − + h ⎨ ⎬ n 2x 1 − α β−1 (x)2 k+1   k−1 k−1 k−1 (14.98) βt = u k−1 u −2un −un−1 uk−1 ⎪ ⎪ M (α) n+1 −un−1 k−1 ⎪ ⎪ − + h ⎩ − n+1 (x) ⎭ 2 n 2x

278

New Numerical Scheme With Newton Polynomial

⎧ ⎫ ⎧ k k k ⎨ un+1 −2un −un−1 ⎬ ⎪ ⎪ 2 β−1 23 ⎪ (x) ⎪ t ⎪ 12 tk ⎪ ⎩ ukn+1 −ukn−1 ⎪ k ⎭ ⎪ − + h ⎪ n 2x ⎧ ⎫ ⎪ k−1 k−1 ⎪ k−1 ⎪ ⎨ un+1 −2un −un−1 ⎬ αβ ⎨ β−1 (x)2 t − 43 tk−1 + k−1 ⎩ uk−1 M (α) ⎪ n+1 −un−1 k−1 ⎭ ⎪ − + h ⎪ n ⎪ 2x ⎧ ⎫ ⎪ k−2 k−2 ⎪ uk−2 ⎪ ⎨ ⎬ n+1 −2un −un−1 ⎪ ⎪ 5 β−1 ⎪ (x)2 ⎪ + t t ⎪ ⎩ 12 k−2 ⎩ uk−2 −uk−2 ⎭ − n+12x n−1 + hk−2 n

14.9

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

New scheme with fractal–fractional with variable order with exponential decay kernel

To derive the numerical scheme where the fractal–fractional differential operator occurs with constant order and variable fractal dimension, we consider the general Cauchy problem with a non-linear function. We start with the case where the kernel is the exponential decay kernel. This differential operator is the extension of the fractal– fractional derivative based on the well-known Caputo–Fabrizio differential operator. Here, we deal with the following equation: FFE α Dt u (x, t) = 0

∂2 ∂ u (x, t) + h (x, t) u (x, t) − 2 ∂x ∂x

(14.99)

with the conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) . For brevity, h (x, t, u) = Eq. (14.99) as follows:

∂2 u (x, t) ∂x 2

−

∂ ∂x u (x, t)

(14.100) + h (x, t). We can reformulate

 β (t) 1 − α β(t) β (t) ln (t) + t h (x, t, u) M (α) t   t β (s) β(s) α + s ds. h (x, s, u) β (s) ln (s) + M (α) 0 s

u (x, t) =

(14.101)

We write at the point tk+1 = (k + 1) t   β (tk )  1 − α β(tk ) β (tk+1 ) − β (tk ) h x, tk , uk tk ln tk + u (x, tk+1 ) = M (α) t tk (14.102)   tk+1 α β (s) β(s) + h (x, s, u) β (s) ln (s) + ds s M (α) 0 s

Numerical scheme for partial differential equations with integer and non-integer order

279

and at the point tk = kt, we have   β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + M (α) t tk−1   × h x, tk−1 , uk−1   tk β (s) β(s) α s ds. + h (x, s, u) β (s) ln (s) + M (α) 0 s

u (x, tk ) =

(14.103)

Taking the difference of these equations, we have the following: ⎡

 ⎤ ln tk + β(ttkk ) ⎢ ⎥  ⎢ ⎥ ×h x, tk , uk ⎢ ⎥ ⎡ ⎤   1−α ⎢ ⎥ β(tk )−β tk−1   u (x, tk+1 ) − u (x, tk ) = ⎢ ⎥ ln t β t t   k−1 ⎦ ⎥ M (α) ⎢ −t k−1 ⎣ ⎢ ⎥ β tk−1 k−1 + tk−1 ⎣ ⎦   k−1 ×h x, tk−1 , u (14.104)   tk+1 α β (s) β(s) + s ds. h (x, s, u) β (s) ln (s) + M (α) tk s β(tk )

tk



  β tk+1 −β(tk ) t

For simplicity, we consider 

β (t) β(t) t H (x, t, u) = h (x, t, u) β (t) ln (t) + t

(14.105)

and we have    1−α   k H u , x, tk − H uk−1 , x, tk−1 (14.106) M (α)  tk+1 α + H (x, s, u) ds. M (α) tk

u (x, tk+1 ) − u (x, tk ) =

As before, we have the following:    ! 1−α H uk , x, tk − H uk−1 , x, tk−1 (14.107) u (x, tk+1 ) = u (x, tk ) + M (α)   ⎧ ⎫ 23 uk , x, tk t ⎬ 12 H  k−1 αβ ⎨ 4 + . − 3 H u , x, tk−1 t ⎭ M (α) ⎩ 5 k−2 + 12 H u , x, tk−2 t

280

New Numerical Scheme With Newton Polynomial

Then ⎡

β(tk )



tk

  β tk+1 −β(tk ) t

ln tk +  ×H  x, tk , uk

β(tk ) tk



⎤

⎢ ⎥ ⎥ 1−α ⎢ ⎢    ⎥  ⎢ ⎥ β t β(t β t t )−β k k−1 k−1 M (α) ⎣ −t k−1 ⎦ ln tk−1 + tk−1 k−1 t   k−1 ×H x, tk−1 , u     ⎧ ⎫ β tk+1 −β(tk ) β(t ) β(tk ) 23 k ⎪ ⎪ t ln t + k ⎪ ⎪ k 12 tk ⎪ ⎪  t  ⎪ ⎪ ⎪ ⎪ k t ⎪ ⎪ ×H x, t , u k ⎪     ⎪   ⎪ ⎪  ⎪ ⎪ β tk−1 β(tk )−β tk−1 β tk−1 ⎨ ⎬ 4 α − 3 tk−1 ln tk−1 + tk−1 t . +   ⎪ M (α) ⎪ ×H x,tk−1 , uk−1 t ⎪ ⎪ ⎪ ⎪        ⎪ ⎪ ⎪ ⎪ β tk−1 −β tk−2 β tk−2 ⎪ ⎪ 5 β tk−2 ⎪ ⎪ + t ln t + k−2 ⎪ ⎪ k−2 12 t t ⎪ ⎪ k−2   ⎩ ⎭ k−2 ×H x, tk−2 , u t

uk+1 = uk +

(14.108)

Then, we can obtain the following scheme:    ⎤ β tk+1 −β(tk ) β(tk ) ln t + k t tk ⎢  ⎥ ⎥ ⎢ ukn+1 −2ukn −ukn−1 ukn+1 −ukn−1 k ⎥ ⎢ × − + h 2 n 2x ⎥ ⎢ (x) 1−α ⎢ k     ⎥   uk+1 = u + n n ⎥ ⎢ β t β(t β t t k )−β k−1 k−1 M (α) ⎢ −t k−1 ⎥ ln tk−1 + tk−1 t ⎢ k−1  ⎥ k−1 k−1 k−1 k−1 ⎦ ⎣ k−1 u −2u −u u −u × n+1 n 2 n−1 − n+12x n−1 + hk−1 n (x)     ⎧ ⎫ β(tk ) 23 β(tk ) β tk+1 −β(tk ) ⎪ ⎪ t ln t + ⎪ ⎪ k k 12 t t ⎪ ⎪ k   k ⎪ ⎪ ⎪ ⎪ k −uk k −uk ⎪ ⎪ u −2u u ⎪ ⎪ n n+1 n−1 n+1 n−1 k ⎪ ⎪ − + h × t ⎪ ⎪ 2 n 2x ⎪ ⎪ (x) ⎪ ⎪       ⎪ ⎪   ⎪ ⎪ β t ⎪ ⎪ β(t β t t )−β k−1 k k−1 k−1 4 ⎪ − t ⎪ ln t + ⎨ ⎬ k−1 k−1 3 t t α k−1  k−1 k−1 k−1 k−1 . + k−1 u −2u −u u −u M (α) ⎪ t ⎪ × n+1 n 2 n−1 − n+12x n−1 + hk−1 ⎪ ⎪ n ⎪ ⎪ (x) ⎪ ⎪ ⎪         ⎪ ⎪ ⎪ ⎪ ⎪ β tk−2 β tk−1 −β tk−2 β tk−2 ⎪ ⎪ 5 ⎪ ⎪ + 12 tk−2 ln tk−2 + tk−2 ⎪ ⎪ t ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ k−2 k−2 k−2 k−2 k−2 ⎪ un+1 −2un −un−1 un+1 −un−1 ⎪ ⎪ k−2 t ⎪ ⎩ × ⎭ − + h 2 n 2x (x)

⎡

β(tk )



t k

14.10

(14.109)

New scheme with fractal–fractional with variable order with the Mittag-Leffler kernel

The generalized Mittag-Leffler function has been documented in many papers as the queen of fractional calculus as it appears to be the fundamental solution of the evolution equation associating the fractional derivative with the power-law kernel. Thus,

Numerical scheme for partial differential equations with integer and non-integer order

281

the fractal–fractional differential operator with the generalized Mittag-Leffler function was suggested as the fractal dimension was replaced by a variable dimension, this giving birth to a new class of differential and integral operators with the crossover property. In this section, by the newly introduced numerical scheme, we present the derivation of a numerical solution of the new class of Cauchy problems; we deal with the following equation: FFM α Dt u (x, t) = 0

∂2 ∂ u (x, t) − u (x, t) + h (x, t) , ∂x ∂x 2

(14.110)

with the conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) .

(14.111)

2

∂ ∂ Taking h (x, t, u) = ∂x 2 u (x, t) − ∂x u (x, t) + h (x, t) and using the new fractional integral with the Mittag-Leffler kernel, we can reformulate the above equation; thus,  β (t) 1 − α β(t) β (t) ln (t) + t h (x, t, u) u (x, t) = AB (α) t  t α + h (x, s, u) (t − s)α−1 (14.112) AB (α)  (α) 0  β (s) β(s) s ds. × β (s) ln (s) + s

At the point tk+1 = (k + 1) t, we can get the following:  β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) t ln tk + u (x, tk+1 ) − u (x, 0) = AB (α) k t tk    t k+1 α h (x, s, u) (14.113) × h x, tk , uk + AB (α)  (α) 0  β (s) β(s) s ds. × (tk+1 − s)α−1 β (s) ln (s) + s For convenience, we put  β (s) β(s) s H (x, s, u) = h (x, s, u) β (s) ln (s) + . s

(14.114)

Then, we have   1−α H x, tk , uk AB (α) k  tm+1

α H (x, s, u) (tk+1 − s)α−1 ds. + AB (α)  (α) tm

u (x, tk+1 ) = u (x, 0) +

m=2

(14.115)

282

New Numerical Scheme With Newton Polynomial

After putting the Newton polynomial into Eq. (14.115), the above equation can be written as follows: k  

α 1−α H x, tk , uk + AB (α) AB (α)  (α) m=2 ⎧   m−2 H x, tm−2 ,u ⎪ ⎪    ⎪  tm+1 ⎪ ⎨ + H x,tm−1 ,um−1 −H x,tm−2 ,um−2 (s − t m−2 ) t    × H (x,tm ,um )−2H x,tm−1 ,um−1 +H x,tm−2 ,um−2 ⎪ + tm ⎪ ⎪ 2(t)2 ⎪ ⎩ × (s − tm−2 ) (s − tm−1 )

uk+1 = u0 +

(14.116) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − s)α−1 ds. Putting the calculations into Eq. (14.116), we obtain the following approximation: at xn   1−α H xn , tk , uk AB (α) k  

α (t)α + H xn , tm−2 , um−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α   k 

α (t)α H xn , tm−1 , um−1  + −H xn , tm−2 , um−2 AB (α)  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H(xn , tm , um )  α

α (t) ⎣ −2H xn , tm−1 , um−1 ⎦ +   2AB (α)  (α + 3) +h xn , tm−2 , um−2 m=2  ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

uk+1 = u0n + n

(14.117)

⎤ ⎥ ⎥. ⎦

Replacing H (x, t, u) by its value, we can get the following: k+1

u

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) t ln tk + =u + AB (α) k t tk

k k k k k un+1 − 2un − un−1 un+1 − un−1 × + hkn − 2x (x)2 0

(14.118)

Numerical scheme for partial differential equations with integer and non-integer order

⎡    ⎤  β tm−1 −β tm−2 k  

α (t)α ln t β tm−2 m−2 t   ⎦ + tm−2 ⎣ β tm−2 AB (α)  (α + 1) + m=2 tm−2 m−2

m−2 m−2 m−2 m−2 un+1 − 2un − un−1 un+1 − un−1 m−2 × − + hn 2x (x)2  ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎥ ⎢ (k − m + 1) +2α 2 + 9α + 12  ⎥ ×⎢ 2 ⎦ ⎣ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12   × (k − m + 1)α − (k − m)α ⎡ ⎤ ⎤   ⎡ β(tm )−β tm−1   ln t β t t   m−1 ⎦ ⎢ tm−1m−1 ⎣ ⎥ β tm−1 ⎢ ⎥ + tm−1 ⎢ ⎥ ⎧ ⎫ ⎢ ⎥ m−1 m−1 m−1 u −2u −u ⎢ ⎥ n ⎨ ⎬ n+1 n−1 ⎢ ⎥ 2 (x) ⎢ ⎥ × m−1 m−1 ⎢ ⎥ k u −u ⎩ ⎭ α n+1 n−1 m−1

⎢ ⎥ α (t) − ⎢ ⎡ 2x  + hn ⎤ ⎥ + ⎢ ⎥ β tm−1 −β tm−2   AB (α)  (α + 2) ln tm−2 ⎦ ⎥ m=2 ⎢ t   ⎢ −t β tm−2 ⎣ ⎥ β tm−2 m−2 ⎢ ⎥ + tm−2 ⎢ ⎥ ⎢ ⎥ ⎧ ⎫ m−2 m−2 −um−2 ⎢ ⎥ u −2u n ⎨ ⎬ n+1 n−1 ⎢ ⎥ 2 ⎣ ⎦ (x) × m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn  α (k − m + 1) (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)  "  # ⎡ β tm+1 −β(tm ) ln t β(tm ) m t +tm ⎢ ⎢ + β(ttmm ) ⎧ ⎫ ⎢ m m m ⎢ ⎨ un+1 −2un −un−1 ⎬ ⎢ 2 (x) ⎢ × ⎢ ⎩ − umn+1 −umn−1 + h m ⎭ ⎢ ⎡ 2x   n ⎤ ⎢ ⎢ β(tm )−β tm−1   ⎢ ln t β tm−1 m−1 t   ⎦ ⎢ −2tm−1 ⎣ β tm−1 ⎢ + k ⎢ α tm−1

⎢ α (t) ⎧ ⎫ m−1 −um−1 + ⎢ um−1 ⎨ ⎬ n+1 −2un n−1 ⎢ 2AB (α)  (α + 3) 2 (x) m=2 ⎢ × m−1 ⎢ ⎩ um−1 n+1 −un−1 m−1 ⎭ ⎢ − ⎢ ⎡ 2x  + hn ⎤ ⎢ β tm−1 −β tm−2   ⎢ ln t β t m−2 ⎦ t   ⎢ +t m−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎢ ⎢ ⎧ ⎫ m−2 m−2 m−2 ⎢ ⎨ un+1 −2un −un−1 ⎬ ⎢ ⎣ (x)2 × −um−2 ⎩ um−2 ⎭ − n+12x n−1 + hnm−2

283

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

284

New Numerical Scheme With Newton Polynomial

14.11

New scheme with fractal–fractional with variable order with power-law kernel

In this section, by the newly introduced numerical scheme, we present the derivation of a numerical solution of the new class of Cauchy problems; we deal with the following equation: FFP α Dt u (x, t) = 0

∂2 ∂ u (x, t) + h (x, t) , u (x, t) − ∂x ∂x 2

(14.119)

with the conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) ,

(14.120)

2

∂ ∂ where h (x, t, u) = ∂x 2 u (x, t) − ∂x u (x, t) + h (x, t). Using the new fractional integral with power-law kernel, we can convert the above equation into  t 1 h (x, s, u) (t − s)α−1 (14.121) u (x, t) =  (α) 0  β (s) β(s) × β (s) ln (s) + s ds. s

At the point tk+1 = (k + 1) t, we can get the following:  tk+1 1 u (x, tk+1 ) − u (x, 0) = h (x, s, u) (tk+1 − s)α−1  (α) 0  β (s) β(s) s × β (s) ln (s) + ds. s

(14.122)

For convenience, we put

 β (s) β(s) . H (x, s, u) = h (x, s, u) β (s) ln (s) + s s

(14.123)

Then, we have k  1 tm+1 u (x, tk+1 ) = u (x, 0) + H (x, s, u) (tk+1 − s)α−1 ds. (14.124)  (α) tm m=2

Putting the Newton polynomial into Eq. (14.124), the following equation can be obtained: ⎫ ⎧   H x, tm−2 , um−2  ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ m−1 −H x,t m−2 ⎪ ⎪  H x,t ,u ,u k m−1 m−2 ⎬ ⎨ t

m+1 + − t (s ) 1 m−2 k+1 0 t     u =u + m m−1 m−2 +H x,tm−2 ,u H (x,tm ,u )−2H x,tm−1 ,u ⎪ ⎪  (α) ⎪ ⎪ + m=2 tm ⎪ ⎪ 2(t)2 ⎪ ⎪ ⎭ ⎩ × (s − tm−2 ) (s − tm−1 ) × (tk+1 − s)α−1 ds.

(14.125)

Numerical scheme for partial differential equations with integer and non-integer order

285

Performing the calculations for the above integrals in Eq. (14.125), we obtain the following approximation: at xn k  (t)α  H xn , tm−2 , um−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α   k  (t)α H xn , tm−1 , um−1  + −H xn , tm−2 , um−2  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H(xn , tm , um )  α

(t) ⎣ −2H xn , tm−1 , um−1 ⎦ +   2 (α + 3) +H xn , tm−2 , um−2 m=2  ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

uk+1 = u0n + n

(14.126)

⎤ ⎥ ⎥. ⎦

Thus, we have

k+1

u

⎡     ⎤ β tm−1 −β tm−2 k   ln t (t)α β tm−2 ⎣ m−2 t   ⎦ =u + tm−2 β tm−2  (α + 1) + m=2 tm−2 m−2

m−2 m−2 m−2 m−2 un+1 − 2un − un−1 un+1 − un−1 + hnm−2 × − 2x (x)2   × (k − m + 1)α − (k − m)α ⎡ ⎤   ⎡ β(tm )−β tm−1   ln t β tm−1 t   m−1 ⎦ ⎢ tm−1 ⎣ β tm−1 ⎢ + ⎢ tm−1 ⎧ ⎫ ⎢ m−1 −um−1 um−1 ⎢ ⎨ ⎬ n+1 −2un n−1 ⎢ (x)2 ⎢ × m−1 k ⎢ ⎩ um−1 ⎭ n+1 −un−1 (t)α ⎢ − + hnm−1 2x ⎢ ⎡ ⎤     + ⎢ β tm−1 −β tm−2    (α + 2) ⎢ ln tm−2 ⎦ β tm−2 m=2 ⎢ t   ⎣ β tm−2 ⎢ −tm−2 + ⎢ tm−2 ⎢ ⎧ ⎫ m−2 m−2 −um−2 ⎢ u −2u n ⎨ ⎬ n+1 n−1 ⎢ ⎣ (x)2 × m−2 m−2 ⎩ un+1 −un−1 ⎭ − 2x + hnm−2 0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (14.127)

286

New Numerical Scheme With Newton Polynomial

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) " ⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α

(t) ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎢ ×⎢ ⎣

β(t ) +tm m



  β tm+1 −β(tm ) ln tm t + β(ttmm ) ⎫ m m um ⎬ n+1 −2un −un−1

#

⎤

⎥ ⎥ ⎥ ⎥ ⎥ (x)2 ⎥ × m −um u ⎥ ⎩ − n+1 n−1 + h m ⎭ ⎥ ⎡ 2x  ⎤ ⎥ n ⎥ β(tm )−β tm−1   ⎥ ln t β tm−1 t   m−1 ⎦ ⎥ −2tm−1 ⎣ β tm−1 ⎥ + tm−1 ⎥ ⎧ ⎫ ⎥ m−1 m−1 m−1 ⎥ ⎨ un+1 −2un −un−1 ⎬ ⎥ (x)2 ⎥ × m−1 m−1 ⎥ ⎩ un+1 −un−1 m−1 ⎭ ⎥ − + h n 2x ⎡    ⎤ ⎥  ⎥ β tm−1 −β tm−2   ⎥ ln t , β tm−2 t   m−2 ⎦ ⎥ +tm−2 ⎣ β tm−2 ⎥ + tm−2 ⎥ ⎥ ⎧ ⎫ m−2 m−2 −um−2 ⎥ u −2u n ⎨ ⎬ n+1 n−1 ⎥ 2 ⎦ (x) × m−2 m−2 ⎩ un+1 −un−1 ⎭ m−2 − 2x + hn  ⎤ 2 2 (k − m) + (3α + 10) (k − m) (k − m + 1)α ⎥ +2α 2 + 9α + 12  ⎥. 2 ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎧ ⎨

Application to linear ordinary differential equations

15

In this chapter, we will deal with linear ordinary differential equations which can be used in modeling simple real life problems.

15.1

Linear ordinary differential equations with integer and non-integer orders

In this section, we present applications as regards the numerical solution for linear ordinary differential equations having classical and fractional operators.

15.1.1 A non-homogeneous linear differential equation In this subsection, we consider a linear differential equation with classical derivative and solve this equation by using the suggested scheme. To do this, we consider du (t) = t 2 u (t) + sin t dt

(15.1)

with the condition u (0) = 0.1.

(15.2)

For simplicity, we take h (t, u (t)) = t 2 u + sin t. Integrating the above equation, we can have  t h (τ, u (τ )) dτ. (15.3) u (t) − u (0) = 0

At the point tk+1 = (k + 1) t  tk+1 u (tk+1 ) − u (0) = h (τ, u (τ )) dτ

(15.4)

0

and at the point tk = kt, we obtain  tk u (tk ) − u (0) = h (τ, u (τ )) dτ.

(15.5)

0

If we take the difference of Eqs. (15.4) and (15.5), we obtain the following:  tk+1 u (tk+1 ) − u (tk ) = h (τ, u (τ )) dτ. tk New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00021-4 Copyright © 2021 Elsevier Inc. All rights reserved.

(15.6)

288

New Numerical Scheme With Newton Polynomial

As the approximation of the function h (t, u (t)), if we put its Newton polynomial into Eq. (15.6) uk+1 − uk = h (tk−2 , u (tk−2 )) t (15.7)  tk+1 h (tk−1 , u (tk−1 )) − h (tk−2 , u (tk−2 )) + (τ − tk−2 ) dτ t tk  tk+1 h (tk , u (tk )) − 2h (tk−1 , u (tk−1 )) + h (tk−2 , u (tk−2 )) + 2 (t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ. Arranging the above, we have k+1

u

      h tk−1 , uk−1 − h tk−2 , uk−2 k−2 t + − u = h tk−2 , u t  tk+1 × (τ − tk−2 ) dτ k

(15.8)

tk

+

      h tk , uk − 2h tk−1 , uk−1 + h tk−2 , uk−2 

×

2 (t)2 tk+1

(τ − tk−2 ) (τ − tk−1 ) dτ.

tk

Replacing the calculations for the above integrals into Eq. (15.8), we have the following:       5  t uk+1 = uk + h tk−2 , uk−2 t + h tk−1 , uk−1 − h tk−2 , uk−2 2 (15.9)      23   t + h tk , uk − 2h tk−1 , uk−1 + h tk−2 , uk−2 12 and we can reorder uk+1 = uk +

   4  5  23  h tk , uk t − h tk−1 , uk−1 t + h tk−2 , uk−2 t. (15.10) 12 3 12

Replacing h (t, u (t)) by its value, our problem can be solved numerically with the help of the following scheme:   4  2 k−1 23  2 k uk+1 = uk + tk u + sin tk t − tk−1 u + sin tk−1 t 12 3  5  2 k−2 + t u + sin tk−2 t. 12 k−2 The numerical simulation for the considered problem is depicted in Fig. 15.1.

(15.11)

Application to linear ordinary differential equations

289

Figure 15.1 Numerical simulation with the classical derivative.

15.1.2 Non-homogeneous linear differential equation with the Atangana–Baleanu derivative We modify our problem thus: ABC α Dt u (t) = h (t, u (t)) , 0

(15.12)

u (0) = 0.2, where h (t, u (t)) = − (2t + 5) − t 3 . We convert Eq. (15.12) to u (t) − u (0) =

α 1−α h (t, u (t)) + AB (α) AB (α)  (α)



t

h (τ, u (τ )) (t − τ )α−1 dτ.

0

(15.13) At the point tk+1 = (k + 1) t, we get the following: 1−α h (t, u (t)) AB (α)  tk+1 α + h (τ, u (τ )) (tk+1 − τ )α−1 dτ AB (α)  (α) 0

u (tk+1 ) − u (0) =

(15.14)

and we write  1−α  h tk , uk AB (α) k  tm+1 α + h (τ, u (τ )) (tk+1 − τ )α−1 dτ. AB (α)  (α) tm

u (tk+1 ) = u (0) +

(15.15)

m=2

After replacing the Newton polynomial into Eq. (15.15), the above equation can be revised as follows:

290

New Numerical Scheme With Newton Polynomial k  α 1−α  h tk , uk + AB (α) AB (α)  (α) m=2 ⎧   m−2 h tm−2 ,u ⎪ ⎪    ⎪  tm+1 ⎪ ⎨ + h tm−1 ,um−1 −h tm−2 ,um−2 (τ − t m−2 ) t    × h(tm ,um )−2h tm−1 ,um−1 +h tm−2 ,um−2 ⎪ + tm ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 = u0 +

(15.16) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we have k  α 1−α  h tk , uk + AB (α) AB (α)  (α) m=2  tm+1   α−1 m−2 (t h t , u − τ dτ ) m−2 k+1 tm  tm+1 htm−1 ,um−1 −htm−2 ,um−2  + tm t × (τ − tm−2 ) (tk+1 − τ )α−1 dτ  tm+1 h(tm ,um )−2h tm−1 ,um−1 +htm−2 ,um−2  + tm 2

uk+1 = u0 +

×

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

2(t)

× (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

(15.17) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

and we arrange this as uk+1 = u0 +  ×

k    α 1−α  h tk , uk + h tm−2 , um−2 AB (α) AB (α)  (α) m=2

tm+1

(tk+1 − τ )α−1 dτ

tm

+

    k h tm−1 , um−1 − h tm−2 , um−2 α AB (α)  (α) t 

× ×

k α AB (α)  (α) tm m=2     h (tm , um ) − 2h tm−1 , um−1 + h tm−2 , um−2

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

(15.18)

2 (t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

With the help of the calculations for the above integrals in Eq. (15.18), we can have the following scheme: uk+1 = u0 +

 1−α  h tk , uk AB (α)

Application to linear ordinary differential equations

291

Figure 15.2 Numerical simulation with Atangana–Baleanu fractional derivative for α = 0.91. k   α (t)α h tm−2 , um−2 AB (α)  (α + 1) m=2   α × (k − m + 1) − (k − m)α    k  α (t)α h tm−1 , um−1  + −h tm−2 , um−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k h(tm , um )  α (t)α ⎣ −2h tm−1 , um−1 ⎦ +   2AB (α)  (α + 3) +h tm−2 , um−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

+

(15.19)

⎤ ⎥ ⎥. ⎦

For the above problem, the numerical simulation is presented in Fig. 15.2.

15.1.3 Non-homogeneous linear differential equation with the Caputo derivative We consider the problem having the Caputo fractional derivative C α 3 0 Dt u (t) = −t

+ u (t) sin (11t) ,

(15.20)

u (0) = 0.2, and for convenience, we take h (t, u (t)) = −t 3 + u (t) sin (11t) .

(15.21)

292

New Numerical Scheme With Newton Polynomial

We write Eq. (15.20) thus: u (t) − u (0) =

1  (α)



t

h (τ, u (τ )) (t − τ )α−1 dτ.

(15.22)

0

At the point tk+1 = (k + 1) t, we have the following:  tk+1 1 u (tk+1 ) − u (0) = h (τ, u (τ )) (tk+1 − τ )α−1 dτ  (α) 0

(15.23)

and we write u (tk+1 ) = u (0) +

k  1 tm+1 h (τ, u (τ )) (tk+1 − τ )α−1 dτ.  (α) tm

(15.24)

m=2

Replacing the Newton polynomial into Eq. (15.24), we write the following: ⎧   m−2 h t , u ⎪ m−2 ⎪ ⎪ ⎪ ht ,um−1 −htm−2 ,um−2  k  (τ − tm−2 ) 1 tm+1 ⎨ + m−1 k+1 0 t    u =u + h(tm ,um )−2h tm−1 ,um−1 +h tm−2 ,um−2 ⎪  (α) ⎪ + m=2 tm ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ and

uk+1 = u0 +

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ k ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

m=2 ⎪ ⎪ ⎪

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ (15.25)

 tm+1   h tm−2 , um−2 (tk+1 − τ )α−1 dτ tm     t h t ,um−1 −h tm−2 ,um−2 + tmm+1 m−1 t × (τ − tm−2 ) (tk+1 − τ )α−1 dτ  tm+1 h(tm ,um )−2h tm−1 ,um−1 +htm−2 ,um−2  + tm 2 2(t)

× (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

. (15.26)

Thus, we have uk+1 = u0 +  ×

k  1  h tm−2 , um−2  (α) m=2

tm+1

tm

 ×

tm+1

tm

(tk+1 − τ )

α−1

    k 1 h tm−1 , um−1 − h tm−2 , um−2 dτ +  (α) t m=2

(τ − tm−2 ) (tk+1 − τ )α−1 dτ

    k 1 h (tm , um ) − 2h tm−1 , um−1 + h tm−2 , um−2 +  (α) 2 (t)2 m=2

(15.27)

Application to linear ordinary differential equations

293

Figure 15.3 Numerical simulation with the Caputo derivative for α = 0.81.

 ×

tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

Doing the same routine, Eq. (15.27) can be revised as follows: k  (t)α  h tm−2 , um−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

uk+1 = u0 +

k    (t)α   h tm−1 , um−1 − h tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)    k  (t)α h (tm , um ) − 2h tm−1 , um−1   + +h tm−2 , um−2 2 (α + 3) m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

+

(15.28)

⎤ ⎥ ⎥. ⎦

The numerical simulation is presented in Fig. 15.3.

15.1.4 Non-homogeneous linear differential equation with fractal–fractional with the exponential law We next consider the following problem with the exponential decay kernel: F F E α,β Dt u (t) = h (t, u (t)) , 0

u (0) = 1,

(15.29)

294

New Numerical Scheme With Newton Polynomial

where h (t, u (t)) = − exp (5t) u (t) + t + 2.

(15.30)

Using the definition of the Caputo–Fabrizio fractal–fractional integral, we can transform the above equation into  t α 1−α g (t, u (t)) + g (τ, u (τ )) dτ, (15.31) u (t) − u (0) = M (α) M (α) 0 where g (τ, u (τ )) = βτ β−1 h (τ, u (τ )). We write Eq. (15.31) at the point tk+1 = (k + 1) t  tk+1 1−α α g (tk , u (tk )) + u (tk+1 ) − u (0) = g (τ, u (τ )) dτ (15.32) M (α) M (α) 0 and at the point tk = kt, we get u (tk ) − u (0) =

1−α α g (tk−1 , u (tk−1 )) + M (α) M (α)



tk

(15.33)

g (τ, u (τ )) dτ. 0

From Eqs. (15.32) and (15.33), we obtain the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)  tk+1 α g (τ, u (τ )) dτ. + M (α) tk

u (tk+1 ) − u (tk ) =

(15.34)

Now, we place the Newton polynomial in Eq. (15.34), to get the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α) ⎧   g tk−2 , uk−2  ⎪ ⎪   ⎪  tk+1 ⎪ ⎨ + g tk−1 ,uk−1 −g tk−2 ,uk−2 (τ − t ) α k−2  t     + k k−1 k−2 ⎪ + g tk ,u −2g tk−1 ,u 2 +g tk−2 ,u M (α) tk ⎪ ⎪ 2(t) ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

uk+1 − uk =

(15.35) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

dτ

and we organize this as follows: uk+1 − uk =

+

1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)   ⎧ g  tk−2 , uk−2 t ⎪    ⎪ k−1 −g t k−2  ⎪ g t ,u ,u ⎪ t k−1 k−2 k+1 (τ − tk−2 ) dτ α ⎨ + tk t

⎫ ⎪ ⎪ ⎪ ⎪ ⎬

M (α) ⎪ ⎪ ⎪ ⎪ ⎩

⎪ ⎪ ⎪ ⎪ ⎭

      g tk ,uk −2g tk−1 ,uk−1 +g tk−2 ,uk−2 2 2(t) t × tkk+1 (τ − tk−2 ) (τ − tk−1 ) dτ

+

(15.36)

.

Application to linear ordinary differential equations

295

Plugging the calculations into the above equation, we get the following: 1−α (15.37) [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)   ⎫ ⎧ t g tk−2 , uk−2 ⎬      α ⎨ k−1 − g t k−2 5 t + . , u , u + g t k−1 k−2 2      ⎭ M (α) ⎩   t + g tk , uk − 2g tk−1 , uk−1 + g tk−2 , uk−2 23 6

uk+1 = uk +

We know that g (t, u (t)) = βt β−1 h (t, u (t)), then we can reorder the above equation as follows:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) (15.38) M (α) ⎧ ⎫  β−1  ⎪ ⎪ βtk−2 h tk−2 , uk−2 t ⎪ ⎪  ⎪ ⎪  ⎪ ⎪ β−1  k−1 ⎪ ⎪ ⎪ ⎪ βtk−1 h tk−1 , u 5 ⎨ ⎬   + t α β−1 2 k−2 −βtk−2 h tk−2 , u . +  ⎪ M (α) ⎪    ⎪ ⎪ β−1  β−1  k k−1 ⎪ ⎪ ⎪ ⎪ βtk h tk , u − 2βtk−1 h tk−1 , u 23 ⎪ ⎪ ⎪ ⎪  ⎩ + β−1  6 t ⎭ k−2 +βtk−2 h tk−2 , u

uk+1 = uk +

Thus, we have the following iteration:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) M (α) ⎧ ⎫   23 β−1 k t ⎪ ⎪ βt h t , u k ⎨ ⎬ 12 k  α β−1  . + − 43 βtk−1 h tk−1 , uk−1 t ⎪  M (α) ⎪ ⎩ ⎭ β−1  5 k−2 + 12 βtk−2 h tk−2 , u t

uk+1 = uk +

(15.39)

Then, our problem can be solved by the following:

k+1

u

   β−1  1−α − exp (tk ) uk + tk + 2 βtk  =u + β−1  M (α) −βtk−1 − exp (tk−1 ) uk−1 + tk−1 + 2 ⎧ ⎫   23 β−1 ⎪ ⎪ − exp (tk ) uk + tk + 2 t βtk ⎨ ⎬ 12  α β−1  4 k−1 . + − 3 βtk−1 − exp (tk−1 ) u + tk−1 + 2 t ⎪  M (α) ⎪ ⎩ ⎭ β−1  5 k−2 + 12 βtk−2 − exp (tk−2 ) u + tk−2 + 2 t k

The numerical simulation is given in Fig. 15.4.

(15.40)

296

New Numerical Scheme With Newton Polynomial

Figure 15.4 Numerical simulation with the classical derivative for α = 0.96, β = 0.8.

15.1.5 Non-homogeneous linear differential equation with fractal–fractional derivative with the Mittag-Leffler kernel Let us consider the following problem: F F M α,β(t) Dt u (t) = u (t) + cosh (t) , 0

(15.41)

u (0) = 0.1, where the fractional order is constant and the fractal dimension is variable. Using the new fractional integral with the Mittag-Leffler kernel, we can reformulate the above equation, thus:   1 − α β(t) β (t) t u (t) = β (t) ln (t) + h (t, u (t)) (15.42) AB (α) t    t β (s) β(s) α s h (s, u (s)) (t − s)α−1 β (s) ln (s) + ds, + AB (α)  (α) 0 s where h (t, u (t)) = u + cosh (t). At the point tk+1 = (k + 1) t, we can get the following:   β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) h (tk , u (tk )) t ln tk + u (tk+1 ) − u (0) = AB (α) k t tk  tk+1 α (15.43) h (s, u (s)) (tk+1 − s)α−1 + AB (α)  (α) 0   β (s) β(s) s × β (s) ln (s) + ds. s For simplicity, we take   β (s) β(s) H (s, u (s)) = h (s, u (s)) β (s) ln (s) + s s

(15.44)

Application to linear ordinary differential equations

297

and then we have   β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) tk ln tk + AB (α) t tk × h (tk , u (tk ))  tk+1 α H (s, u (s)) (tk+1 − s)α−1 ds. + AB (α)  (α) 0

u (tk+1 ) = u (0) +

(15.45)

Also, we write   β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) tk ln tk + AB (α) t tk × h (tk , u (tk )) k  tm+1 α + H (s, u (s)) (tk+1 − s)α−1 ds. AB (α)  (α) tm

u (tk+1 ) = u (0) +

(15.46)

m=2

As before, we obtain the following: k+1

u

  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) t ln tk + = u (0) + AB (α) k t tk α (15.47) + AB (α)  (α) ⎧ ⎫ H  (tm−2 , u (tm−2 )) ⎪ ⎪    ⎪ ⎪ ⎪ k  tm+1 ⎪ ⎨ + H tm−1 ,u tm−1 −H tm−2 ,u tm−2 (s − t ⎬ ) m−2 t      × H (t ,u(t ))−2H tm−1 ,u tm−1 +H tm−2 ,u tm−2 ⎪ ⎪ + m m ⎪ ⎪ m=2 tm ⎪ ⎪ 2(t)2 ⎩ ⎭ × (s − tm−2 ) (s − tm−1 ) × (tk+1 − s)α−1 ds

and we can reorder the above equation such that uk+1 = u (0) +

  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) tk ln tk + h (tk , u (tk )) AB (α) t tk

α AB (α)  (α)  tm+1 ⎧ H (tm−2 , u (tm−2 )) (tk+1 − s)α−1 ds ⎪ tm ⎪       ⎪  ⎪ tm+1 H tm−1 ,u tm−1 −H tm−2 ,u tm−2 ⎪ ⎪ + k ⎨ tm t × × − t − s)α−1 ds (s ) (t m−2 k+1        ⎪  ⎪ −2H t H t ,u t ,u tm−1 +H tm−2 ,u tm−2 t m−1 m−1 m−1 m+1 m=2 ⎪ ⎪ + tm 2 ⎪ 2(t) ⎪ ⎩ × (s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds +

(15.48) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

298

New Numerical Scheme With Newton Polynomial

Thus, we write the following:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) h (tk , u (tk )) tk ln tk + AB (α) t tk k   tm+1  α + H tm−2 , um−2 t (tk+1 − s)α−1 ds AB (α)  (α) t m m=2     k m−1 H tm−1 , u − H tm−2 , um−2 α (15.49) + AB (α)  (α) t

uk+1 = u (0) +

 × ×

k α AB (α)  (α) tm m=2     m m−1 m−2 + H tm−2 , u H (tm , u ) − 2H tm−1 , u

 ×

m=2

tm+1

(s − tm−2 ) (tk+1 − s)α−1 ds +

2 (t)2 tm+1

(s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds

tm

and obtain the following scheme:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) tk ln tk + AB (α) t tk × h (tk , u (tk ))

uk+1 = u0 +

k  α (t)α  H tm−2 , um−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α    k  α (t)α H tm−1 , um−1  + −H tm−2 , um−2 AB (α)  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H(tm , um )  α (t)α ⎣ −2H tm−1 , um−1 ⎦ + AB (α) 2 (α + 3) +H tm−2 , um−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

+

(15.50)

⎤ ⎥ ⎥. ⎦

Application to linear ordinary differential equations

299

If we replace the function H (t, u (t)) by its value, we can have the following scheme:   1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) k+1 0 u h (tk , u (tk )) t =u + ln tk + AB (α) k t tk ⎡    ⎤  β tm−1 −β tm−2 k   α α ln tm−2 ⎦ (t) β tm−2 t   tm−2 ⎣ + β tm−2 AB (α)  (α + 1) + m=2

tm−2

  × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α ⎡ ⎡  



 β(tm )−β tm−1 ln t t   m−1 β tm−1 + tm−1

⎤

⎤

⎢ t β tm−1 ⎣ ⎦ ⎥ m−1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ k ⎥ α (t)α ⎢ ×h , u (t (t )) m−1 m−1 ⎢ ⎥ ⎡ ⎤     + ⎢ ⎥ −β t β t   m−1 m−2 AB (α)  (α + 2) ⎢ ln tm−2 ⎦ ⎥ β tm−2 m=2 ⎢ t   ⎥ ⎣ −t β tm−2 m−2 ⎢ ⎥ + tm−2 ⎣ ⎦ ×h (tm−2 , u (tm−2 ))   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡     ⎤ β tm+1 −β(tm ) ln t β(tm ) m t tm ⎢ ⎥ β(tm ) ⎢ ⎥ + t ⎢ ⎥ m ⎢ ⎥ ×h (tm , u (tm )) ⎢ ⎡ ⎤ ⎥   ⎢ ⎥ β(tm )−β tm−1   ⎢ ln tm−1 ⎦ ⎥ β tm−1 k t   ⎢ ⎥ α ⎣ ⎢ −2tm−1 α (t) β tm−1 ⎥ + + tm−1 ⎢ ⎥ ⎢ ⎥ AB (α) 2 (α + 3) m=2 ⎢ ⎥ ×h ⎢ ⎡ (tm−1 , u (tm−1)) ⎤ ⎥ ⎢ ⎥ β tm−1 −β tm−2   ⎢ ln tm−2 ⎦ ⎥ t   ⎢ +t β tm−2 ⎣ ⎥ β tm−2 ⎢ ⎥ m−2 + tm−2 ⎣ ⎦ ×h (tm−2 , u (tm−2 ))   ⎤ ⎡ 2 2 (k − m) + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ (15.51) 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 Thus, we get the following scheme for our problem:    1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk )  k u + cosh (tk ) tk ln tk + uk+1 = AB (α) t tk ⎡    ⎤  β tm−1 −β tm−2 k   α ln t (t)α β tm−2 ⎣ m−2 t   ⎦ + tm−2 β t AB (α)  (α + 1) + m−2 m=2

tm−2

300

New Numerical Scheme With Newton Polynomial

Figure 15.5 Numerical simulation with the AB derivative for α = 0.67, β = 1 + 0.01t.

  × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α ⎡ ⎡ 

⎤

⎢ ⎢ ⎢ k ⎢ α (t)α ⎢ ⎢ + ⎢ AB (α)  (α + 2) m=2 ⎢ ⎢ ⎢ ⎣

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

 ×

⎤  β(tm )−β tm−1 ln t β tm−1 t   m−1 ⎦ tm−1 ⎣ β tm−1 + tm−1   × ⎡um−1 + cosh (tm−1 ) ⎤     β tm−1 −β tm−2   ln tm−2 ⎦ β tm−2 t   −tm−2 ⎣ β tm−2 + tm−2   × um−2 + cosh (tm−2 ) 

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡









β t



−β(t )



m m+1 ln tm β(t ) t tm m ⎢ β(t m) ⎢ + tm ⎢ m + cosh (t )) ⎢ × (u ⎢ ⎡  m ⎢ β(tm )−β tm−1   ⎢ ln t β tm−1 k t   m−1 ⎣ α (t)α ⎢ β tm−1 ⎢ −2tm−1 + tm−1 + ⎢ ⎢ AB (α) 2 (α + 3)  m−1  m=2 ⎢ × ⎡u + cosh (tm−1 ) ⎢     ⎢ β tm−1 −β tm−2   ⎢ ln tm−2 t   ⎢ +t β tm−2 ⎣ β tm−2 ⎢ m−2 + tm−2 ⎣  m−2  × u + cosh (tm−2 )   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12

For the above problem, the numerical simulation is depicted in Fig. 15.5.

(15.52) ⎤

⎤ ⎦ ⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Application to non-linear ordinary differential equations

16

In this chapter, we aim to present the numerical approximation for solving nonlinear ordinary differential equations which model more complex real life problems.

16.1

Non-linear ordinary differential equations with integer and non-integer orders

In this section, some non-linear ordinary differential equations will be considered to get their numerical scheme.

16.1.1 Non-homogeneous nonlinear differential equation with classical derivative In this section, we consider a non-linear differential equation with classical derivative and solve this equation by using the suggested scheme. To do this, we consider du (t) = sin (u (t)) + t 3 dt

(16.1)

with the conditions u (0) = 1.

(16.2)

For simplicity, we take h (t, u (t)) = sin (u) + t 3 . Integrating the above equation, we can have  t u (t) − u (0) = h (τ, u (τ )) dτ. (16.3) 0

At the point tk+1 = (k + 1) t 

tk+1

u (tk+1 ) − u (0) =

h (τ, u (τ )) dτ

(16.4)

0

and at the point tk = kt, we obtain  u (tk ) − u (0) =

tk

h (τ, u (τ )) dτ. 0

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00022-6 Copyright © 2021 Elsevier Inc. All rights reserved.

(16.5)

302

New Numerical Scheme With Newton Polynomial

Subtracting Eq. (16.4) from Eq. (16.5), we obtain the following:  tk+1 u (tk+1 ) − u (tk ) = h (τ, u (τ )) dτ.

(16.6)

tk

We replace its Newton polynomial as the approximation of the function h (t, u (t)), uk+1 − uk = h (tk−2 , u (tk−2 )) t (16.7)  tk+1 h (tk−1 , u (tk−1 )) − h (tk−2 , u (tk−2 )) + (τ − tk−2 ) dτ t tk  tk+1 h (tk , u (tk )) − 2h (tk−1 , u (tk−1 )) + h (tk−2 , u (tk−2 )) + 2 (t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ. Rearranging the above, we have k+1

u

      h tk−1 , uk−1 − h tk−2 , uk−2 k−2 t + − u = h tk−2 , u t  tk+1 × (τ − tk−2 ) dτ k

(16.8)

tk

+

      h tk , uk − 2h tk−1 , uk−1 + h tk−2 , uk−2 

×

2 (t)2 tk+1

(τ − tk−2 ) (τ − tk−1 ) dτ.

tk

Plugging the calculations for the above integrals into Eq. (16.8), we have the following:       5  uk+1 = uk + h tk−2 , uk−2 t + h tk−1 , uk−1 − h tk−2 , uk−2 t 2        23 t (16.9) + h tk , uk − 2h tk−1 , uk−1 + h tk−2 , uk−2 12 and we can reorder    4  5  23  uk+1 = uk + h tk , uk t − h tk−1 , uk−1 t + h tk−2 , uk−2 t. (16.10) 12 3 12 Replacing h (t, u (t)) by its value, our problem can be solved numerically,

  2 23 tk uk + uk + tk t uk+1 = uk + 12

 2 4 tk−1 uk−1 + uk−1 + tk−1 t − 3

 2 5 tk−2 uk−2 + uk−2 + tk−2 t. + 12

(16.11)

Application to non-linear ordinary differential equations

303

Figure 16.1 Numerical simulation with the classical derivative.

The numerical simulation for the considered problem is depicted in Fig. 16.1.

16.1.2 Non-homogeneous non-linear differential equation with Caputo–Fabrizio derivative We next consider the following problem with the exponential decay kernel: CF α 0 Dt u (t) = h (t, u (t)) ,

(16.12)

u (0) = −1. where h (t, u (t)) = u (t) sin (u (t)) + 2t + 3.

(16.13)

Using the definition of the Caputo–Fabrizio fractal–fractional integral, we can transform the above equation into u (t) − u (0) =

α 1−α h (t, u (t)) + M (α) M (α)



t

h (τ, u (τ )) dτ.

(16.14)

0

We write Eq. (16.14) at the point tk+1 = (k + 1) t u (tk+1 ) − u (0) =

1−α α h (tk , u (tk )) + M (α) M (α)



tk+1

h (τ, u (τ )) dτ

(16.15)

h (τ, u (τ )) dτ.

(16.16)

0

and at the point tk = kt, we have u (tk ) − u (0) =

1−α α h (tk−1 , u (tk−1 )) + M (α) M (α)

 0

tk

304

New Numerical Scheme With Newton Polynomial

From Eqs. (16.15) and (16.16), we obtain the following: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)  tk+1 α h (τ, u (τ )) dτ. + M (α) tk

u (tk+1 ) − u (tk ) =

(16.17)

With the same routine, we can get 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α) ⎧   h tk−2 , uk−2  ⎪ ⎪   ⎪  tk+1 ⎪ ⎨ + h tk−1 ,uk−1 −h tk−2 ,uk−2 (τ − t ) α k−2  t     + h tk ,uk −2h tk−1 ,uk−1 +h tk−2 ,uk−2 M (α) tk ⎪ + ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

uk+1 − uk =

(16.18) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

dτ

and uk+1 − uk =

+

1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)   ⎧ h  tk−2 , uk−2 t ⎪    ⎪ k−1 −h t k−2  ⎪ h t ,u ,u ⎪ t k−1 k−2 k+1 (τ − tk−2 ) dτ α ⎨ + tk t

⎫ ⎪ ⎪ ⎪ ⎪ ⎬

M (α) ⎪ ⎪ ⎪ ⎪ ⎩

⎪ ⎪ ⎪ ⎪ ⎭

      h tk ,uk −2h tk−1 ,uk−1 +h tk−2 ,uk−2 2 2(t) t × tkk+1 (τ − tk−2 ) (τ − tk−1 ) dτ

+

(16.19)

.

Substituting the calculations into the above equation, we get the following: 1−α (16.20) [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)   ⎫ ⎧ t h tk−2 , uk−2 ⎬      α ⎨ k−1 − h t k−2 5 t + . , u , u + h t k−1 k−2 2      ⎭ M (α) ⎩   t + h tk , uk − 2h tk−1 , uk−1 + h tk−2 , uk−2 23 6

uk+1 = uk +

Then, we can reorder the above equation as follows: 1−α [h (tk , u (tk )) − h (tk−1 , u (tk−1 ))] M (α)     23   k−1 t α h tk , uk t − 43 h tk−1 , u 12  + . 5 h tk−2 , uk−2 t + 12 M (α)

uk+1 = uk +

(16.21)

Application to non-linear ordinary differential equations

305

Figure 16.2 Numerical simulation with the Caputo–Fabrizio derivative for α = 0.96.

Replacing h (t, u (t)) by its value, we get the following:    k k   1−α u  u + 2tk + 3  sin  k−1  uk+1 = uk + uk−1 + 2tk−1 + 3 M (α) − sin u     ⎧ ⎫ 23 uk uk + 2tk + 3 t  ⎬ 12 sin   α ⎨ + . − 43 sin uk−1 uk−1 + 2tk−1 + 3 t ⎭ M (α) ⎩ 5 sin uk−2 uk−2 + 2tk−2 + 3 t + 12

(16.22)

The numerical simulation is given in Fig. 16.2.

16.1.3 Non-homogeneous non-linear differential equation with fractal derivative We consider the following problem: CH β 0 Dt u (t) = h (t, u (t))

(16.23)

where h (t, u (t)) = u3 + 2ut. After integrating Eq. (16.23), we obtain the following: t u (t) − u (0) = β

τ β−1 h (τ, u (τ )) dτ.

(16.24)

0

At the point tk+1 = (k + 1) t  tk+1 u (tk+1 ) − u (0) = β τ β−1 h (τ, u (τ )) dτ

(16.25)

0

and at the point tk = kt, we get the following:  tk τ β−1 h (τ, u (τ )) dτ. u (tk ) − u (0) = β 0

(16.26)

306

New Numerical Scheme With Newton Polynomial

From Eqs. (16.25) and (16.26), we obtain the following:  u (tk+1 ) − u (tk ) = β

tk+1

τ β−1 h (τ, u (τ )) dτ.

(16.27)

tk

If we put the Newton polynomial into Eq. (16.27) and we take g (τ, u (τ )) = βτ β−1 h (τ, u (τ )), we obtain the following: uk+1 − uk = g (tk−2 , u (tk−2 )) t  tk+1 g (tk−1 , u (tk−1 )) − g (tk−2 , u (tk−2 )) + (τ − tk−2 ) dτ (16.28) t tk  tk+1 g (tk , u (tk )) − 2g (tk−1 , u (tk−1 )) + g (tk−2 , u (tk−2 )) + 2 (t)2 tk (τ − tk−2 ) (τ − tk−1 ) dτ. If we order the above equation, we may write k+1

u





    g tk−1 , uk−1 − g tk−2 , uk−2 t + t

− u = g tk−2 , u  tk+1 × (τ − tk−2 ) dτ tk       g tk , uk − 2g tk−1 , uk−1 + g tk−2 , uk−2 + 2 (t)2  tk+1 × (τ − tk−2 ) (τ − tk−1 ) dτ. k

k−2

(16.29)

tk

If these calculations are substituted in Eq. (16.29), the following scheme is written:   uk+1 = uk + g tk−2 , uk−2 t    5   + g tk−1 , uk−1 − g tk−2 , uk−2 t 2      23   t + g tk , uk − 2g tk−1 , uk−1 + g tk−2 , uk−2 12

(16.30)

and it can be reorganized as   4 β−1  23 β−1  βtk h tk , uk t − βtk−1 h tk−1 , uk−1 t 12 3  5 β−1  + βtk−2 h tk−2 , uk−2 t. 12

uk+1 = uk +

(16.31)

Application to non-linear ordinary differential equations

307

Figure 16.3 Numerical simulation with the fractal derivative for β = 0.92.

If we replace h (t, u (t)) by its value, we obtain the following numerical scheme:

23 β−1  k 3 u βtk + 2uk tk t 12

4 β−1  k−1 3 u − βtk−1 + 2uk−1 tk−1 t 3

5 β−1  k−2 3 u + βtk−2 + 2uk−2 tk−2 t. 12

uk+1 = uk +

(16.32)

The numerical simulation is presented in Fig. 16.3.

16.1.4 Non-homogeneous non-linear differential equation with the fractal–fractional derivative with exponential law We next consider the following problem with the exponential decay kernel: F F E α,β Dt u (t) = h (t, u (t)) , 0

(16.33)

u (0) = −1, where h (t, u (t)) = exp (u (t)) u (t) − exp (t) .

(16.34)

Using the definition of the Caputo–Fabrizio fractal–fractional integral, we can transform the above equation into u (t) − u (0) =

α 1−α g (t, u (t)) + M (α) M (α)



t

g (τ, u (τ )) dτ 0

(16.35)

308

New Numerical Scheme With Newton Polynomial

where g (τ, u (τ )) = βτ β−1 h (τ, u (τ )). We write Eq. (16.35) at the point tk+1 = (k + 1) t  tk+1 1−α α g (tk , u (tk )) + u (tk+1 ) − u (0) = g (τ, u (τ )) dτ (16.36) M (α) M (α) 0 and at the point tk = kt, one can have the following 1−α α g (tk−1 , u (tk−1 )) + u (tk ) − u (0) = M (α) M (α)



tk

g (τ, u (τ )) dτ.

(16.37)

0

From Eqs. (16.36) and (16.37), we obtain the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)  tk+1 α g (τ, u (τ )) dτ. + M (α) tk

u (tk+1 ) − u (tk ) =

(16.38)

Now, we set the Newton polynomial in Eq. (16.38), and we can get the following: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α) ⎧   g tk−2, uk−2 ⎪ ⎪    ⎪ ⎪ g tk−1 ,uk−1 −g tk−2 ,uk−2 ⎪  tk+1 ⎪ + ⎨ t α ×(τ − tk−2 ) +      k −2g t k−1 +g t k−2 M (α) tk ⎪ g t ,u ,u ⎪ k k−1 k−2 ,u ⎪ + ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

uk+1 − uk =

(16.39) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

dτ

and we organize this as follows: 1−α [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)   ⎧ g tk−2 , uk−2 t  ⎪   ⎪ ⎪ g tk−1 ,uk−1 −g tk−2 ,uk−2 ⎪ ⎪ +  ⎪ t t α ⎨ × tkk+1 (τ − tk−2 ) dτ +       k k−1 k−2 M (α) ⎪ ⎪ ⎪ + g tk ,u −2g tk−1 ,u 2 +g tk−2 ,u ⎪ ⎪ 2(t) ⎪ t ⎩ × tkk+1 (τ − tk−2 ) (τ − tk−1 ) dτ

uk+1 − uk =

(16.40) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

Thus, we get the following: 1−α (16.41) [g (tk , u (tk )) − g (tk−1 , u (tk−1 ))] M (α)   ⎫ ⎧ k−2 t , u g t ⎬ ⎨ k−2      α k−1 − g t k−2 5 t + . , u , u + g t k−1 k−2 2      ⎭ M (α) ⎩   t + g tk , uk − 2g tk−1 , uk−1 + g tk−2 , uk−2 23 6

uk+1 = uk +

Application to non-linear ordinary differential equations

309

We know that g (t, u (t)) = βt β−1 h (t, u (t)), then we can reorder the above equation as follows:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) uk+1 = uk + (16.42) M (α) ⎧ ⎫  β−1  ⎪ ⎪ βtk−2 h tk−2 , uk−2 t ⎪ ⎪  ⎪ ⎪  ⎪ ⎪ β−1  k−1 ⎪ ⎪ ⎪ ⎪ βt h t , u k−1 5 ⎨ ⎬ k−1   + t α β−1 2 k−2 −βtk−2 h tk−2 , u . +   ⎪ M (α) ⎪     ⎪ ⎪ β−1 β−1 k k−1 ⎪ ⎪ ⎪ ⎪ βtk h tk , u − 2βtk−1 h tk−1 , u 23 ⎪ ⎪ ⎪ ⎪  ⎩ + β−1  6 t ⎭ k−2 +βtk−2 h tk−2 , u Thus, we have the following iteration:  1 − α  β−1 β−1 βtk h (tk , u (tk )) − βtk−1 h (tk−1 , u (tk−1 )) M (α) ⎧ ⎫   23 β−1 ⎪ ⎪ βtk h tk , uk t ⎨ ⎬ 12  α β−1  4 k−1 + . − 3 βtk−1 h tk−1 , u t ⎪  M (α) ⎪ ⎩ ⎭ β−1  5 + 12 βtk−2 h tk−2 , uk−2 t

uk+1 = uk +

Replacing h (t, u (t)) by its value, the following scheme can be obtained      β−1  1−α exp uk uk − exp (tk ) βtk k+1 k    u =u + β−1  M (α) −βtk−1 exp uk−1 uk−1 − exp (tk−1 ) ⎧ ⎫     23 β−1 ⎪ ⎪ exp uk uk − exp (tk ) t βtk ⎨ ⎬ 12   k−1  k−1  α 4 β−1 . + u − exp (tk−1 ) t − 3 βtk−1 exp u ⎪  k−2  k−2  M (α) ⎪ ⎩ ⎭ β−1  5 u − exp (tk−2 ) t + 12 βtk−2 exp u

(16.43)

(16.44)

The numerical simulation is given in Fig. 16.4.

Figure 16.4 Numerical simulation with the Caputo–Fabrizio fractal–fractional derivative for α = 0.95, β = 0.95.

310

New Numerical Scheme With Newton Polynomial

16.1.5 Non-homogeneous non-linear differential equation with fractal–fractional with variable order with power law Let us consider the following problem: F F P α,β(t) Dt u (t) = log (u (t) + 5) − t, 0

(16.45)

u (0) = 1, where the fractional order is constant and the fractal dimension is variable. Using the new fractional integral with the power-law kernel, we can obtain the following equation    t β (s) β(s) 1 α−1 β (s) ln (s) + s ds, (16.46) h (s, u (s)) (t − s) u (t) =  (α) 0 s where h (t, u (t)) = log (u (t) + 5) − t. At the point tk+1 = (k + 1) t, we can get the following:  tk+1 1 h (s, u (s)) (tk+1 − s)α−1 (16.47) u (tk+1 ) − u (0) =  (α) 0   β (s) β(s) × β (s) ln (s) + ds. s s For convenience, we put   β (s) β(s) H (s, u (s)) = h (s, u (s)) β (s) ln (s) + s s

(16.48)

and then we have u (tk+1 ) = u0 +

1  (α)



tk+1

H (s, u (s)) (tk+1 − s)α−1 ds.

(16.49)

0

Also, we write u (tk+1 ) = u0 +

k  1  tm+1 H (s, u (s)) (tk+1 − s)α−1 ds.  (α) tm

(16.50)

m=2

When replacing the Newton polynomial into the above equation, we get the following: ⎧ ⎫ H  (tm−2 , u (tm−2 )) ⎪ ⎪    ⎪ ⎪ ⎪ ⎪ H tm−1 ,u tm−1 −H tm−2 ,u tm−2 k  tm+1 ⎨ ⎬  + − t (s ) 1 m−2 k+1 t      u = u0 + H (t ,u(t ))−2H tm−1 ,u tm−1 +H tm−2 ,u tm−2 ⎪ ⎪  (α) + m m ⎪ ⎪ m=2 tm ⎪ ⎪ 2(t)2 ⎩ ⎭ × (s − tm−2 ) (s − tm−1 ) (16.51) × (tk+1 − s)α−1 ds

Application to non-linear ordinary differential equations

311

and we can reorder the above equation thus:

uk+1 = u0 +

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ k ⎪ ⎨  ⎪ ⎪ ⎪ ⎩

m=2 ⎪ ⎪ ⎪

 tm+1

H (tm−2 , u (tm−2 )) (tk+1 − s)α−1 ds

      t H tm−1 ,u tm−1 −H tm−2 ,u tm−2 + tmm+1 t × − t − s)α−1 ds (s ) (t m−2 k+1  tm+1 H tm−1 ,utm−1 −2H tm−1 ,utm−1 +H tm−2 ,utm−2  + tm 2(t)2 × (s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds tm

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(16.52) Thus, we write the following: k   tm+1 1   H tm−2 , um−2 t (tk+1 − s)α−1 ds  (α) tm m=2     k 1  H tm−1 , um−1 − H tm−2 , um−2 +  (α) t

uk+1 = u0 +

 × ×

k 1   (α) tm m=2     H (tm , um ) − 2H tm−1 , um−1 + H tm−2 , um−2

 ×

m=2

tm+1

(s − tm−2 ) (tk+1 − s)α−1 ds +

(16.53)

2 (t)2 tm+1

(s − tm−2 ) (s − tm−1 ) (tk+1 − s)α−1 ds

tm

and obtain the following scheme: k  (t)α   H tm−2 , um−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

uk+1 = u0 +

k    (t)α    H tm−1 , um−1 − H tm−2 , um−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)    k  (t)α  H (tm , um ) − 2H tm−1 , um−1   + +H tm−2 , um−2 2 (α + 3) m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

+

(16.54)

⎤ ⎥ ⎥. ⎦

312

New Numerical Scheme With Newton Polynomial

If we replace the function H (t, u (t)) by h (t, u (t)), we can have the following scheme:  k   β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + tm−2  (α + 1) t tm−2 m=2   × h (tm−2 , u (tm−2 )) (k − m + 1)α − (k − m)α       ⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 t tm−1 ⎥ k ⎢ ⎥ (t)α  ⎢ , u ×h (t (t )) m−1 m−1 ⎢        ⎥  + ⎢ ⎥ β t −β t β t β t m−1 m−2  (α + 2) ⎣ −t m−2 ln tm−2 + m−2 ⎦

uk+1 = u0 +

m=2

m−2

t

tm−2

×h (tm−2 , u (tm−2 ))   α (k − m + 1) (k − m + 3 + 2α) × (16.55) − (k − m)α (k − m + 3 + 3α)     ⎡ ⎤ β(t ) β tm+1 −β(tm ) tm m ln tm + β(ttmm ) t ⎢ ⎥ ⎢ ⎥ ×h (tm , u (tm ))    ⎥  ⎢ k ⎢ β tm−1 β(tm )−β tm−1 ⎥ β tm−1 α  (t) ln tm−1 + tm−1 ⎢ −2tm−1 ⎥ t + ⎢ ⎥ ⎢ ⎥ 2 (α + 3) ×h (tm−1 , u (t )) m=2 ⎢     m−1  ⎥  ⎢ ⎥ β t β tm−1 −β tm−2 β tm−2 ⎣ +tm−2m−2 ⎦ ln tm−2 + tm−2 t ×h (tm−2 , u (tm−2 ))   ⎤ ⎡ 2 2 (k − m) + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 and finally replacing h (t, u (t)) by its value, we obtain the following approximation:  k   β (tm−2 ) (t)α  β tm−2 β (tm−1 ) − β (tm−2 ) ln tm−2 + tm−2  (α + 1) t tm−2 m=2      × log um−2 + 5 − tm−2 (k − m + 1)α − (k − m)α       ⎡ ⎤ β tm−1 β(tm )−β tm−1 β tm−1 tm−1 ln t + m−1 ⎥  tm−1  tm−1  k ⎢ ⎥ (t)α  ⎢ × log u + 5 − t m−1 ⎢        ⎥  + ⎢ ⎥ β t β tm−1 −β tm−2 β tm−2 m−2  (α + 2) ⎦ ln tm−2 + tm−2 m=2 ⎣ −tm−2 t   m−2   × log u + 5 − tm−2   α (k − m + 1) (k − m + 3 + 2α) (16.56) × − (k − m)α (k − m + 3 + 3α)

uk+1 = u0 +

Application to non-linear ordinary differential equations

313

Figure 16.5 Numerical simulation with the Caputo fractal–fractional derivative for α = 0.95, β = exp (0.05t).

⎡ ⎢ ⎢ ⎢ k (t)α  ⎢ ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ⎢ ⎣

   β tm+1 −β(tm ) β(tm ) ln t + m t tm m + 5) − t ) × (log (u m       β tm−1 β(tm )−β tm−1 β tm−1 −2tm−1 ln t + m−1 t   m−1   tm−1 × log u + 5 − t m−1         β tm−2 β tm−1 −β tm−2 β tm−2 +tm−2 ln t + m−2   t  tm−2  × log um−2 + 5 − tm−2  ⎤ 2 β(tm )



tm

 2 (k − m) + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 ⎡

For the above problem, the numerical simulation is depicted in Fig. 16.5.

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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Application to linear partial differential equations

17

Partial differential equations show us how to model the rate of change of physical events with respect to space and time variables. In this chapter, we will address linear partial differential equations whose well-known examples are the wave equation, the heat equation and the transport equation.

17.1

Linear partial differential equations with integer and non-integer orders

In this section, the heat equation will be considered for which we will obtain the numerical scheme to solve such equations.

17.1.1 Linear partial differential equation with the classical derivative We deal with the following heat equation: ut = uxx + h (x, t)

(17.1)

where we can take H (u, x, t) = uxx + h (x, t). We convert Eq. (17.1) into 

t

u (x, t) − u (x, 0) =

H (u, x, τ ) dτ.

(17.2)

0

At the point tk+1 = (k + 1) t 

tk+1

u (x, tk+1 ) − u (x, 0) =

H (u, x, τ ) dτ

(17.3)

0

and at the point tk = kt, we have  u (x, tk ) − u (x, 0) =

tk

H (u, x, τ ) dτ.

(17.4)

0

Taking the difference of Eqs. (17.3) and (17.4), we can get the following:  u (x, tk+1 ) − u (x, tk ) =

tk+1

H (u, x, τ ) dτ. tk

New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00023-8 Copyright © 2021 Elsevier Inc. All rights reserved.

(17.5)

316

New Numerical Scheme With Newton Polynomial

If we insert the Newton polynomial into Eq. (17.5), we have   uk+1 − uk = H uk−2 , x, tk−2 t (17.6)   k−2   tk+1  k−1 H u , x, tk−1 − H u , x, tk−2 + (τ − tk−2 ) dτ t tk       tk+1  k H u , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 + 2 (t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ and we obtain       H uk−1 , x, tk−1 − H uk−2 , x, tk−2 uk+1 − uk = H uk−2 , x, tk−2 t + t  tk+1 × (17.7) (τ − tk−2 ) dτ tk

      H uk , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2

+ 

2 (t)2 tk+1

(τ − tk−2 ) (τ − tk−1 ) dτ.

tk

When we plug the well-known calculations into Eq. (17.7), we obtain the following at point xn :   k k−2 t (17.8) uk+1 = u + H u , x , t n k−2 n n    5   t + H uk , xn , tk − H uk−1 , xn , tk−1 2        23 t + H uk , xn , tk − 2H uk−1 , xn , tk−1 + H uk−2 , xn , tk−2 12 and we get   4  5  k−2 H u , xn , tk−2 t − H uk−1 , xn , tk−1 t 12 3  23  k−1 + H u , xn , tk−1 t. 12

uk+1 = ukn + n

Thus, the considered equation can be solved numerically as follows:

k−2 − uk−2 5 un+1 − 2uk−2 n n−1 k+1 k k−2 un = un + + hn t 12 (x)2

k−1 − uk−1 4 un+1 − 2uk−1 n n−1 + hnk−1 t − 3 (x)2

(17.9)

(17.10)

Application to linear partial differential equations

317

Figure 17.1 Numerical simulation for the heat equation with the classical derivative.

23 + 12



ukn+1 − 2ukn − ukn−1 (x)2

+ hnk

t.

The numerical simulation for the heat equation is presented in Fig. 17.1.

17.1.2 Linear partial differential equation with the fractal derivative We deal with the heat equation which is given by CH β 0 D u (x, t) = uxx

+ h (x, t)

(17.11)

where we can take H (u, x, t) = uxx + h (x, t). We convert Eq. (17.11) into  u (x, t) − u (x, 0) = 0

t

βτ β−1 h (u, x, τ ) dτ.

(17.12)

318

New Numerical Scheme With Newton Polynomial

At the point tk+1 = (k + 1) t and H (u, x, τ ) = βτ β−1 h (u, x, τ ) 

tk+1

u (x, tk+1 ) − u (x, 0) =

H (u, x, τ ) dτ

(17.13)

0

and at the point tk = kt, we have  u (x, tk ) − u (x, 0) =

tk

H (u, x, τ ) dτ.

(17.14)

0

Subtracting Eqs. (17.14) and (17.13), we can get the following:  u (x, tk+1 ) − u (x, tk ) =

tk+1

H (u, x, τ ) dτ.

(17.15)

tk

Let us insert this polynomial into Eq. (17.15); we have   (17.16) uk+1 = uk + H uk−2 , x, tk−2 t      tm+1 H uk−1 , x, tk−1 − H uk−2 , x, tk−2 + (τ − tk−2 ) dτ t tm       tm+1  k H u , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 + 2 (t)2 tm × (τ − tk−2 ) (τ − tk−1 ) dτ and we obtain   uk+1 = uk + H uk−2 , x, tk−2 t     t k+1 H uk−1 , x, tk−1 − H uk−2 , x, tk−2 + (τ − tk−2 ) dτ t t   k−1  k k−2   k H u , x, tk − 2H u , x, tk−1 + H u , x, tk−2 + 2 (t)2  tk+1 × (τ − tk−2 ) (τ − tk−1 ) dτ.

(17.17)

tk

When substituting the calculations into Eq. (17.17), we obtain the following at xn : uk+1 n

     H uk−1 , xn , tk−1  5 k−2 t (17.18) + H u , xn , tk−2 t + −H uk−2 , xn , tk−2 2  k   k−1   23 H u , xn , tk − 2H u , xn , tk−1 + t k−2 +H u , xn , tk−2 12 = ukn

Application to linear partial differential equations

319

and we get   4  5  k−2 H u , xn , tk−2 t − H uk−1 , xn , tk−1 t 12 3  23  k + H u , xn , tk t. 12

uk+1 = ukn + n

(17.19)

Thus, we get the following approximation:  5 β−1  k−2 βtk−2 h u , xn , tk−2 t 12  4 β−1  k−1 − βtk−1 h u , xn , tk−1 t 3  23 β−1  + βtk h uk , xn , tk t. 12

uk+1 = ukn + n

Plugging all this into the above equation, we obtain the following: k−2

− uk−2 5 β−1 un+1 − 2uk−2 n n−1 k+1 k k−2 un = un + βtk−2 + hn t 12 (x)2 k−1

− uk−1 4 β−1 un+1 − 2uk−1 n n−1 k−1 + hn − βtk−1 t 3 (x)2

23 β−1 ukn+1 − 2ukn − ukn−1 k + hn t. + βtk 12 (x)2

(17.20)

(17.21)

17.1.3 Linear partial differential equation with the Caputo fractional derivative We firstly consider the heat equation including the Caputo fractional operator, ∂2 u (x, t) + h (x, t) ∂x 2

(17.22)

u (x, 0) = g (x) , u (x, t) |∂ = f (t) .

(17.23)

C α 0 Dt u (x, t) =

with initial condition

We shall write the above equation C α 0 Dt u (x, t) = h (u, x, t)

where h (u, x, t) = uxx + h (x, t). Integrating on the left and right side,   C α C α α =C J D u t) (x, 0 t 0 t 0 Jt [h (x, t, u)] .

(17.24)

(17.25)

320

New Numerical Scheme With Newton Polynomial

We transform Eq. (17.24) into u (x, t) − u (x, 0) =

1  (α)



t

h (x, τ, u) (t − τ )α−1 dτ.

(17.26)

0

At the point tk+1 = (k + 1) t, we have the following  tk+1 1 u (x, tk+1 ) − u (x, 0) = h (x, τ, u) (tk+1 − τ )α−1 dτ  (α) 0

(17.27)

and we write u (x, tk+1 ) = u (x, 0) +

k  1 tm+1 h (x, τ, u) (tk+1 − τ )α−1 dτ.  (α) tm

(17.28)

m=2

After putting the Newton polynomial into Eq. (17.28), the above equation can be written as follows: ⎫ ⎧   h x, tm−2 , um−2  ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ h x,tm−1 ,um−1 −h x,tm−2 ,um−2 k  tm+1 ⎪ ⎬ ⎨

+ − t (τ ) 1 m−2 k+1 0 t     =u + u m )−2h x,t m−1 +h x,t m−2 h(x,t ,u ,u ,u m m−1 m−2 ⎪ ⎪  (α) + ⎪ ⎪ m=2 tm ⎪ ⎪ 2(t)2 ⎪ ⎪ ⎭ ⎩ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ.

(17.29)

Thus, we have

uk+1 = u0 +

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k

⎨ ⎪ + ⎪ ⎪ ⎩

m=2 ⎪ ⎪ ⎪

 tm+1   h x, tm−2 , um−2 (tk+1 − τ )α−1 dτ tm     t h x,tm−1 ,um−1 −h x,tm−2 ,um−2 + tmm+1 t × (τ − tm−2 ) (tk+1 − τ )α−1 dτ  tm+1 h(x,tm ,um )−2h x,tm−1 ,um−1 +h x,tm−2 ,um−2  tm

2(t)2

× (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(17.30) Now, at the point xn , we have k   tm+1 1  h xn , tm−2 , um−2 (tk+1 − τ )α−1 dτ  (α) t m m=2     k m−1 − h xn , tm−2 , um−2 1 h xn , tm−1 , u +  (α) t

uk+1 = u0n + n

 ×

m=2

tm+1

tm

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

k 1  (α)

m=2

(17.31)

Application to linear partial differential equations

×

321

    h (xn , tm , um ) − 2h xn , tm−1 , um−1 + h xn , tm−2 , um−2 

×

2 (t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

We put the calculations for the above integrals into Eq. (17.31), and we obtain the following approximation: k  (t)α  h xn , tm−2 , um−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

uk+1 = u0n + n

k    (t)α   h xn , tm−1 , um−1 − h xn , tm−2 , um−2  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) (17.32) × − (k − m)α (k − m + 3 + 3α)   k  (t)α h (xn , tm , um ) − 2h xn , tm−1 , um−1   + +h xn , tm−2 , um−2 2 (α + 3) m=2  ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+

Replacing h (x, t, u) by its value, we can present the following scheme: = u0n + uk+1 n

ukn+1 − 2ukn − ukn−1

+ hkn (x)2 m−2 k

− um−2 un+1 − 2um−2 n n−1

(t)α+1 +  (α + 1) (x)2 m=2   × (k − m + 1)α − (k − m)α ⎡  m−1 m−1 +

un+1 −2un

k α ⎢

α (t)  (α + 2) 

×



m=2

⎢ ⎣

 −

−um−1 n−1

(x)2

 ⎤ + hnm−1

m−2 −um−2 um−2 n+1 −2un n−1 (x)2

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α)

+ hnm−2

⎥  ⎥ ⎦ m−2

+ hn



(17.33)

322

New Numerical Scheme With Newton Polynomial

⎡

m m um n+1 −2un −un−1

+ hnm

!

⎤

⎢  ⎥  m−1 m−1 m−1 k ⎥ α (t)α ⎢ ⎢ −2 un+1 −2un 2 −un−1 + h m−1 ⎥ n + ⎥ ⎢ (x) ⎢ 2 (α + 3)  ⎥  m−2 m−2 m−2 m=2 ⎣ ⎦ un+1 −2un −un−1 + + hnm−2 (x)2  ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ 2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 (x)2

17.1.4 Linear partial differential equation with the fractal–fractional derivative with the exponential law We present a numerical scheme for the heat equation involving the Caputo–Fabrizio fractal–fractional operator. Namely, we take the following equation: FFE α Dt u (x, t) = 0

∂2 u (x, t) + h (x, t) ∂x 2

(17.34)

under initial conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) .

(17.35)

We shall take h (u, x, t) = uxx + h (x, t) and we shall write the above as follows CF α 0 Dt u (x, t) = H

(17.36)

(u, x, t)

where H (u, x, t) = βt β−1 h (u, x, t). Applying the associated integral, we have 



CF α CF α 0 Jt 0 Dt u (x, t)

α =CF 0 Jt [H (x, t, u)]

(17.37)

and u (x, t) − u (x, 0) =

α (1 − α) H (u, x, t) + M (α) M (α)



t

H (u, x, τ ) dτ.

(17.38)

0

As before, we have the following:    ! 1−α H uk , x, tk − H uk−1 , x, tk−1 M (α)   k ⎧ ⎫ 23 H u , x, tk t ⎨ ⎬ 12  α + . − 43 H uk−1 , x, tk−1 t ⎭ M (α) ⎩ 5 k−2 + 12 H u , x, tk−2 t

u (x, tk+1 ) =

(17.39)

Application to linear partial differential equations

323

      Replacing H uk , x, tk by h uk , x, tk and then h uk , x, tk by its value, we write the following approximation at xn : ⎧ ⎫  k  k k ukn+1 −ukn−1 ⎪ ⎪ β−1 un+1 −2un −un−1 k ⎪ ⎪ βt − + h ⎬ n k 2x 1−α ⎨ (x)2 k+1  k−1 k−1 k−1  (17.40) un = k−1 k−1 un+1 −un−1 ⎪ β−1 un+1 −2un −un−1 M (α) ⎪ k−1 ⎪ ⎪ −βt − + h ⎩ ⎭ n k−1 2x (x)2 ⎧ ⎫  k  k k ukn+1 −ukn−1 ⎪ ⎪ 23 β−1 un+1 −2un −un−1 k t ⎪ ⎪ t − + h ⎪ ⎪ n ⎪ ⎪ 12 k 2x (x)2 ⎪ ⎪ ⎪ ⎪   ⎨ ⎬ k−1 k−1 k−1 k−1 k−1 αβ un+1 −un−1 4 β−1 un+1 −2un −un−1 k−1 . + t − 3 tk−1 − + h 2 n 2x (x) ⎪ M (α) ⎪ ⎪ ⎪   ⎪ ⎪ k−2 k−2 k−2 ⎪ ⎪ uk−2 −uk−2 ⎪ ⎪ 5 β−1 un+1 −2un −un−1 ⎪ tk−2 − n+12x n−1 + hk−2 t ⎪ ⎩ + 12 ⎭ 2 n (x)

17.1.5 Linear partial differential equation with the fractal–fractional with variable order with the Mittag-Leffler kernel We deal with the following heat equation: F F M α,β(t) Dt u (x, t) = 0

∂2 u (x, t) + h (x, t) , ∂x 2

(17.41)

with the conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) .

(17.42)

2

∂ We have h (x, t, u) = ∂x 2 u (x, t) + h (x, t). Using the new fractional integral with the Mittag-Leffler kernel, we can write the above equation as  β (t) 1 − α β(t) β (t) ln (t) + t h (x, t, u) u (x, t) = AB (α) t  t α + h (x, τ, u) (t − τ )α−1 (17.43) AB (α)  (α) 0  β (τ ) β(τ ) τ × β (τ ) ln (τ ) + dτ. τ

At the point tk+1 = (k + 1) t, we can get the following: "   #   1 − α β(tk ) β tk+1 −β(tk ) ln tk t tk u (x, tk+1 ) − u (x, 0) = h x, tk , uk β(tk ) AB (α) + tk  tk+1 α + h (x, τ, u) (tk+1 − τ )α−1 (17.44) AB (α)  (α) 0  β (τ ) β(τ ) τ × β (τ ) ln (τ ) + dτ. τ

324

New Numerical Scheme With Newton Polynomial

For convenience, we put 

β (τ ) β(τ ) H (x, τ, u) = h (x, τ, u) β (τ ) ln (τ ) + τ . τ

(17.45)

Then, we have   1−α H x, tk , uk AB (α) k  tm+1

α + H (x, τ, u) (tk+1 − τ )α−1 dτ. AB (α)  (α) tm

u (x, tk+1 ) = u (x, 0) +

(17.46)

m=2

Eq. (17.46) can be approximated with the Newton polynomial as follows: k  

α 1−α H x, tk , uk + AB (α) AB (α)  (α) m=2 ⎧   m−2 H x, tm−2 ,u ⎪ ⎪    ⎪  tm+1 ⎪ ⎨ + H x,tm−1 ,um−1 −H x,tm−2 ,um−2 (τ − t m−2 ) t    × H (x,tm ,um )−2H x,tm−1 ,um−1 +H x,tm−2 ,um−2 ⎪ + tm ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 = u0 +

(17.47) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ and we obtain the following approximation:   1−α H xn , tk , uk AB (α) k  

α (t)α + H xn , tm−2 , um−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α   k 

α (t)α H xn , tm−1 , um−1  + −H xn , tm−2 , um−2 AB (α)  (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H(xn , tm , um )  α

α (t) ⎣ −2H xn , tm−1 , um−1 ⎦ +   2AB (α)  (α + 3) +H xn , tm−2 , um−2 m=2  ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

= u0n + uk+1 n

(17.48)

⎤ ⎥ ⎥. ⎦

Application to linear partial differential equations

325

Thus, we get after replacing H (x, t, u) by its value  1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) (17.49) tk ln tk + AB (α) t tk

ukn+1 − 2ukn − ukn−1 k × + hn (x)2 ⎡    ⎤  β tm−1 −β tm−2 k  

α (t)α ln t β tm−2 m−2 ⎦ t   + tm−2 ⎣ β tm−2 AB (α)  (α + 1) + m=2 tm−2 m−2

un+1 − 2um−2 − um−2 n n−1 × + hnm−2 (x)2   × (k − m + 1)α − (k − m)α ⎡ ⎡ ⎤ ⎤   β(tm )−β tm−1   ln t t   m−1 ⎦ ⎥ ⎢ t β tm−1 ⎣ β tm−1 m−1 ⎥ ⎢ + tm−1 ⎥ ⎢ ⎥ ⎢   m−1 m−1 m−1 ⎥ ⎢ un+1 −2un −un−1 m−1 ⎥ ⎢ k × + hn

⎥ ⎢ α (t)α (x)2 ⎥ ⎢ ⎡ ⎤     + ⎥ ⎢ −β t β t   m−1 m−2 AB (α)  (α + 2) ⎢ ln tm−2 ⎦ ⎥ β tm−2 m=2 ⎢ t   ⎥ ⎣ −t β tm−2 m−2 ⎥ ⎢ + tm−2 ⎥ ⎢ ⎥ ⎢   m−2 m−2 m−2 ⎦ ⎣ un+1 −2un −un−1 m−2 × + hn 2

= u0n + uk+1 n

(x)

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α) ⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k α

⎢ α (t) ⎢ + ⎢ 2AB (α)  (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

"

  β tm+1 −β(tm ) t + β(ttmm ) m m um n+1 −2un −un−1

β(t ) +tm m

×

# !

+ hnm (x)2   β(tm )−β tm−1   ln t β tm−1 t   m−1 −2tm−1 ⎣ β tm−1 +   m−1 m−1 m−1tm−1 un+1 −2un −un−1 m−1 × + h 2 n ⎡ (x)     β tm−1 −β tm−2   ln tm−2 β tm−2 t   +tm−2 ⎣ β tm−2 + tm−2 ⎧ ⎫ m−2 m−2 −um−2 u −2u n ⎨ ⎬ n+1 n−1 ×

⎩

⎤

ln tm

⎡

(x)2 m−2 um−2 −u − n+12x n−1 + hnm−2

⎭

⎤ ⎦

⎤ ⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

326

New Numerical Scheme With Newton Polynomial

 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥. ⎦

Application to non-linear partial differential equations

18

In this section, we deal with a numerical solution of non-linear partial differential equations with integer and non-integer orders. To achieve this, we consider Fisher’s equation, which is one of the semi-linear equations that belongs to the class of reaction–diffusion equations.

18.1

Non-linear partial differential equations with integer and non-integer orders

In this section, we will consider Fisher’s equations with integer and non-integer orders in order to present a numerical algorithm for non-linear partial differential equations.

18.1.1 Non-linear partial differential equation with the classical derivative We consider the following Fisher equation: ut = ρuxx − μu (1 − u) .

(18.1)

For simplicity, we shall take H (u, x, t) = ρuxx − μu (1 − u). We write Eq. (18.1) as follows:  t u (x, t) − u (x, 0) = H (u, x, τ ) dτ. (18.2) 0

At the point tk+1 = (k + 1) t  u (x, tk+1 ) − u (x, 0) =

tk+1

H (u, x, τ ) dτ

(18.3)

0

and at the point tk = kt, we have  tk u (x, tk ) − u (x, 0) = H (u, x, τ ) dτ.

(18.4)

0

If we take the difference of Eqs. (18.3) and (18.4), we can get the following:  tk+1 u (x, tk+1 ) − u (x, tk ) = H (u, x, τ ) dτ. tk New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00024-X Copyright © 2021 Elsevier Inc. All rights reserved.

(18.5)

328

New Numerical Scheme With Newton Polynomial

Inserting the Newton polynomial into Eq. (18.5), we have   uk+1 − uk = H uk−2 , x, tk−2 t (18.6)      tk+1 H uk−1 , x, tk−1 − H uk−2 , x, tk−2 + (τ − tk−2 ) dτ t tk       tk+1  k H u , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 + 2 (t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ and we obtain       H uk−1 , x, tk−1 − H uk−2 , x, tk−2 uk+1 − uk = H uk−2 , x, tk−2 t + t  tk+1 × (18.7) (τ − tk−2 ) dτ tk

      H uk , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2

+ 

2 (t)2 tk+1

(τ − tk−2 ) (τ − tk−1 ) dτ.

tk

As before, we obtain the following:   (18.8) = ukn + H uk−2 , xn , tk−2 t uk+1 n    5   t + H uk , xn , tk − H uk−1 , xn , tk−1 2        23 + H uk , xn , tk − 2H uk−1 , xn , tk−1 + H uk−2 , xn , tk−2 t 12 and we get   4  5  k−2 H u , xn , tk−2 t − H uk−1 , xn , tk−1 t 12 3  23  k + H u , xn , tk t. 12

uk+1 = ukn + n

Thus, the considered equation can be solved numerically as follows: k−2

  − uk−2 un+1 − 2uk−2 5 n n−1 k+1 k k−2 k−2 1 − un − μun un = un + ρ t 12 (x)2 k−1

  − uk−1 un+1 − 2uk−1 4 n n−1 1 − uk−1 − μuk−1 − ρ t n n 3 (x)2

(18.9)

(18.10)

Application to non-linear partial differential equations

329



  ukn+1 − 2ukn − ukn−1 23 k k + − μun 1 − un t. ρ 12 (x)2

18.1.2 Non-linear partial differential equation with the fractal derivative We deal with the Fisher equation which is given by CH β 0 D u = ρuxx

− μu (1 − u)

(18.11)

where we can take H (u, x, t) = ρuxx − μu (1 − u) + h (x, t). Eq. (18.11) can be converted to  t u (x, t) − u (x, 0) = βτ β−1 h (u, x, τ ) dτ. (18.12) 0

At the point tk+1 = (k + 1) t and H (u, x, τ ) = βτ β−1 h (u, x, τ )  tk+1 u (x, tk+1 ) − u (x, 0) = H (u, x, τ ) dτ

(18.13)

0

and at the point tk = kt, we have  tk H (u, x, τ ) dτ. u (x, tk ) − u (x, 0) =

(18.14)

0

The difference of Eqs. (18.13) and (18.14) yields  tk+1 u (x, tk+1 ) − u (x, tk ) = H (u, x, τ ) dτ.

(18.15)

tk

Let us insert the Newton polynomial into Eq. (18.15); then we have   (18.16) uk+1 = uk + H uk−2 , x, tk−2 t   k−2   tm+1  k−1 H u , x, tk−1 − H u , x, tk−2 + (τ − tk−2 ) dτ t tm        tm+1 H uk , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 + 2 (t)2 tm × (τ − tk−2 ) (τ − tk−1 ) dτ and we obtain

  uk+1 = uk + H uk−2 , x, tk−2 t     t k+1 H uk−1 , x, tk−1 − H uk−2 , x, tk−2 + (τ − tk−2 ) dτ t tk

(18.17)

330

New Numerical Scheme With Newton Polynomial

+

      H uk , x, tk − 2H uk−1 , x, tk−1 + H uk−2 , x, tk−2 

×

2 (t)2 tk+1

(τ − tk−2 ) (τ − tk−1 ) dτ.

tk

Doing the same routine, we obtain the following: at xn      H uk−1 , xn , tk−1  5 k k−2 t (18.18) t + uk+1 = u + H u , x , t n k−2 n n −H uk−2 , xn , tk−2 2      23 H uk , xn , tk − 2H uk−1 , xn , tk−1 + t +H uk−2 , xn , tk−2 12 and we get   4  5  k−2 H u , xn , tk−2 t − H uk−1 , xn , tk−1 t 12 3  23  k + H u , xn , tk t. 12

uk+1 = ukn + n

(18.19)

Thus, we get the following approximation:  5 β−1  k−2 βtk−2 h u , xn , tk−2 t 12  4 β−1  k−1 − βtk−1 h u , xn , tk−1 t 3  23 β−1  + βtk h uk , xn , tk t. 12

uk+1 = ukn + n

(18.20)

Putting all this into the above, we have the following: k−2

  − uk−2 5 β−1 un+1 − 2uk−2 n n−1 k+1 k k−2 k−2 un = un + βtk−2 1 − un − μun t 12 (x)2 k−1

  − uk−1 un+1 − 2uk−1 4 β−1 n n−1 k−1 k−1 1 − un − μun − βtk−1 ρ t (18.21) 3 (x)2

  ukn+1 − 2ukn − ukn−1 23 β−1 k k − μun 1 − un t. ρ + βtk 12 (x)2

18.1.3 Non-linear partial differential equation with the Caputo fractional derivative We firstly consider Fisher’s equation including the Caputo fractional operator, C α 0 Dt u (x, t) = ρuxx

− μu (1 − u) .

(18.22)

Application to non-linear partial differential equations

331

Here, the initial condition is given as u (x, 0) = g (x) , u (x, t) |∂ = f (t) .

(18.23)

We shall write the above equation C α 0 Dt u (x, t) = h (u, x, t)

(18.24)

where h (u, x, t) = ρuxx − μu (1 − u) + h (x, t). Integrating on the left and right side, we have   C α C α C α (18.25) 0 Jt 0 Dt u (x, t) =0 Jt [h (x, t, u)] . Eq. (18.24) can be written as 1 u (x, t) − u (x, 0) = (α)



t

h (x, τ, u) (t − τ )α−1 dτ.

(18.26)

0

At the point tk+1 = (k + 1) t, we have the following:  tk+1 1 u (x, tk+1 ) − u (x, 0) = h (x, τ, u) (tk+1 − τ )α−1 dτ (α) 0

(18.27)

and we write u (x, tk+1 ) = u (x, 0) +

k  1 tm+1 h (x, τ, u) (tk+1 − τ )α−1 dτ. (α) tm

(18.28)

m=2

With the same routine, we find ⎧   h x, tm−2 , um−2  ⎪ ⎪  ⎪ m−1 m−2 ⎪ h x,tm−1 ,u −h x,tm−2 ,u k  (τ − tm−2) 1 tm+1 ⎨ + t    uk+1 = u0 + h(x,tm ,um )−2h x,tm−1 ,um−1 +h x,tm−2 ,um−2 ⎪ (α) + ⎪ m=2 tm ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ.

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

(18.29)

Thus, we have

uk+1 = u0 +

1 (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k

⎨ ⎪ + ⎪ ⎪ ⎩

m=2 ⎪ ⎪ ⎪

 tm+1   h x, tm−2 , um−2 (tk+1 − τ )α−1 dτ tm     t h x,tm−1 ,um−1 −h x,tm−2 ,um−2 + tmm+1 t × (τ − tm−2 ) (tk+1 − τ )α−1 dτ  tm+1 h(x,tm ,um )−2hx,tm−1 ,um−1 +hx,tm−2 ,um−2  tm

2(t)2

× (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

(18.30)

332

New Numerical Scheme With Newton Polynomial

Now, at the point xn , we get k   tm+1 1  h xn , tm−2 , um−2 (tk+1 − τ )α−1 dτ (α) t m m=2     k 1 h xn , tm−1 , um−1 − h xn , tm−2 , um−2 + (α) t

uk+1 = u0n + n

 × ×

k 1 (α) tm m=2     m m−1 + h xn , tm−2 , um−2 h (xn , tm , u ) − 2h xn , tm−1 , u

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

(18.31)

2 (t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

After same manipulations, we obtain the following scheme:

uk+1 n

= u0n

k  (t)α  + h xn , tm−2 , um−2 (α + 1) m=2   × (k − m + 1)α − (k − m)α k    (t)α   h xn , tm−1 , um−1 − h xn , tm−2 , um−2 (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) (18.32) × − (k − m)α (k − m + 3 + 3α)   k  (t)α h (xn , tm , um ) − 2h xn , tm−1 , um−1   + +h xn , tm−2 , um−2 2 (α + 3) m=2  ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

+

Plugging the above into our equation, we have

uk+1 n

m−2

k   − um−2 un+1 − 2um−2 (t)α+1 n n−1 m−2 m−2 1 − un + − μun ρ (α + 1) (x)2 m=2   × (k − m + 1)α − (k − m)α = u0n

Application to non-linear partial differential equations

⎡

333



 ⎤ m−1 −um−1   um−1 n+1 −2un n−1 m−1 1 − um−1 ρ − μu 2 n n ⎢ ⎥ α (t) ⎢  m−2 (x)  ⎥ + m−2 m−2 ⎣ ⎦   u −2un −un−1 (α + 2) − −μunm−2 1 − um−2 − ρ n+1 m=2 n (x)2  (k − m + 1)α (k − m + 3 + 2α) × (18.33) − (k − m)α (k − m + 3 + 3α) ⎡ ⎤  ! um −2um −um ρ n+1 n 2 n−1 − μum 1 − um n n ⎢  ⎥  m−1 (x) k m−1 m−1   ⎥ α (t)α ⎢ ⎢ −2 ρ un+1 −2un 2 −un−1 − μum−1 1 − um−1 ⎥ n n + ⎢ ⎥ (x) ⎢  ⎥ 2 (α + 3)  m−2 m−2 m−2 m=2 ⎣   ⎦ un+1 −2un −un−1 m−2 m−2 1 − un − −μun + ρ (x)2  ⎡ ⎤ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ 2α 2 + 9α + 12  ⎥. ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α 2 +6α + 18α + 12 α

k

18.1.4 Non-linear partial differential equation with the fractal–fractional with exponential law We present numerical scheme for Fisher’s equation involving Caputo–Fabrizio fractal–fractional operator. Namely, we take the following equation: FFE α Dt u (x, t) = ρuxx 0

− μu (1 − u) + h (x, t)

(18.34)

under the initial conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) .

(18.35)

We shall take h (u, x, t) = ρuxx − μu (1 − u) + h (x, t) and we have CF α 0 Dt u (x, t) = H

(18.36)

(u, x, t)

where H (u, x, t) = βt β−1 h (u, x, t). We have u (x, t) − u (x, 0) =

α (1 − α) H (u, x, t) + M (α) M (α)



t

H (u, x, τ ) dτ

(18.37)

0

and    H uk , x, tk  −H uk−1 , x, tk−1   k ⎧ ⎫ 23 H u , x, tk t ⎨ ⎬ 12  αβ + . − 43 H uk−1 , x, tk−1 t ⎭ M (α) ⎩ 5 k−2 + 12 H u , x, tk−2 t

u (x, tk+1 ) =

1−α M (α)



(18.38)

334

New Numerical Scheme With Newton Polynomial

As before, we write the following approximation at xn : ⎧  k    un+1 −2ukn −ukn−1 ⎪ β−1 k k ⎪ βtk ρ − μun 1 − un 2 1−α ⎨  k−1 (x)  = uk+1 k−1 n k−1   un+1 −2un −un−1 β−1 M (α) ⎪ k−1 k−1 ⎪ 1 − un − μun ⎩ −βtk−1 ρ (x)2

+

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

αβ M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

(18.39) ⎫  k    un+1 −2ukn −ukn−1 ⎪ 23 β−1 k k ⎪ − μun 1 − un t ρ ⎪ ⎪ 12 tk (x)2 ⎪ ⎪  k−1 k−1 k−1  ⎬   u −2u −u n 4 β−1 n+1 n−1 k−1 k−1 . t 1 − un − μun − 3 tk−1 ρ 2 ⎪ ⎪  k−2 (x)  ⎪ k−2 ⎪   u −2uk−2 −u ⎪ 5 β−1 t ⎪ 1 − uk−2 tk−2 ρ n+1 n 2 n−1 − μuk−2 + 12 ⎭ n n (x)

18.1.5 Non-linear partial differential equation with the fractal–fractional with variable order with the Mittag-Leffler kernel We deal with the following equation: F F M α,β(t) Dt u (x, t) = ρuxx 0

− μu (1 − u)

(18.40)

with the conditions u (x, 0) = ϕ (x) , u (x, t) |∂ = f (t) .

(18.41)

We have h (x, t, u) = ρuxx − μu (1 − u). Integrating the above equation, we can reformulate it as  β (t) 1 − α β(t) β (t) ln (t) + t h (x, t, u) u (x, t) − u (x, 0) = AB (α) t  t α + h (x, τ, u) (t − τ )α−1 (18.42) AB (α) (α) 0  β (τ ) β(τ ) τ × β (τ ) ln (τ ) + dτ. τ At the point tk+1 = (k + 1) t, we can get the following:  β (tk ) 1 − α β(tk ) β (tk+1 ) − β (tk ) t ln tk + u (x, tk+1 ) − u (x, 0) = AB (α) k t tk   k × h x, tk , u

Application to non-linear partial differential equations

335

 tk+1 α h (x, τ, u) (tk+1 − τ )α−1 AB (α) (α) 0  β (τ ) β(τ ) τ × β (τ ) ln (τ ) + dτ. τ

+

(18.43)

For simplicity, we take  β (τ ) β(τ ) H (x, τ, u) = h (x, τ, u) β (τ ) ln (τ ) + τ . τ

(18.44)

Then, we have   1−α H x, tk , uk (18.45) AB (α) k  tm+1

α H (x, τ, u (τ )) (tk+1 − τ )α−1 dτ. + AB (α) (α) tm

u (x, tk+1 ) = u (x, 0) +

m=2

Replacing its Newton polynomial into Eq. (18.45), the above equation can be written as follows: k  

α 1−α H x, tk , uk + AB (α) AB (α) (α) m=2 ⎧   m−2 H x, tm−2 ,u ⎪ ⎪    ⎪  tm+1 ⎪ ⎨ + H x,tm−1 ,um−1 −H x,tm−2 ,um−2 (τ − t m−2 ) t    × H (x,tm ,um )−2H x,tm−1 ,um−1 +H x,tm−2 ,um−2 ⎪ + tm ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 = u0 +

(18.46) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we obtain the following approximation: at xn   1−α H xn , tk , uk AB (α) k  

α (t)α H xn , tm−2 , um−2 + AB (α) (α + 1) m=2   α × (k − m + 1) − (k − m)α   k 

α (t)α H xn , tm−1 , um−1  + −H xn , tm−2 , um−2 AB (α) (α + 2) m=2  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

uk+1 = u0n + n

(18.47)

336

New Numerical Scheme With Newton Polynomial

⎤ ⎡ k H(xn , tm , um ) 

α (t)α ⎣ −2H xn , tm−1 , um−1 ⎦ +   2AB (α) (α + 3) +H xn , tm−2 , um−2 m=2  ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12  ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

⎤ ⎥ ⎥. ⎦

We have

uk+1 n

 1 − α β(tk ) β (tk+1 ) − β (tk ) β (tk ) t + ln tk + AB (α) k t tk

  uk − 2ukn − ukn−1 × ρ n+1 − μukn 1 − ukn 2 (x) ⎡     ⎤ β tm−1 −β tm−2 k  

α (t)α ln t β tm−2 m−2 t   ⎦ tm−2 ⎣ + β t AB (α) (α + 1) + m−2 = u0n

m=2

× ρ

m−2 − um−2 um−2 n+1 − 2un n−1

tm−2

− μum−2 n



1 − unm−2





(x)2   (18.48) × (k − m + 1)α − (k − m)α  ⎡ ⎤ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12  ⎥ ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12 ⎡ ⎡ ⎤ ⎤   β(tm )−β tm−1   ln t t  ⎢ t β tm−1 ⎣ ⎥  m−1 ⎦ ⎢ ⎥ β tm−1 m−1 + tm−1 ⎢ ⎥ ⎢

⎥ m−1 −um−1 ⎢ ⎥ um−1 −2u n n−1 ⎢ ⎥ ρ n+1 2 ⎢ ⎥ × (x)   k ⎢ ⎥ α

⎢ α (t) 1 − um−1 + hm−1 −μu⎡m−1 n n n ⎤ ⎥     + ⎢ ⎥ β tm−1 −β tm−2   ⎢ ⎥ AB (α) (α + 2) ln t β tm−2 m=2 ⎢ m−2 t   ⎦ ⎥ ⎢ −tm−2 ⎣ ⎥ β t m−2 ⎢ ⎥ + tm−2 ⎢ ⎥

⎢ ⎥ m−2 −um−2 um−2 ⎢ ⎥ n+1 −2un n−1 ρ ⎣ × ⎦ 2 (x)   m−2 m−2 m−2 1 − un + hn −μun  (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

Application to non-linear partial differential equations

337

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α

α (t) ⎢ + ⎢ ⎢ 2AB (α) (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

" β(t ) +tm m

×

  β tm+1 −β(tm ) ln tm t + β(ttmm )

m m um n+1 −2un −un−1

ρ (x)  2  m 1 − um −μu n n ⎡

  β(tm )−β tm−1   ln t β tm−1 t   m−1 −2tm−1 ⎣ β tm−1 + tm−1

m−1 un+1 −2um−1 −um−1 n n−1

×

ρ

⎤

#

⎤ ⎦

(x)  2  m−1 1 − um−1 −μu ⎡ n   n ⎤ β tm−1 −β tm−2   ln t β tm−2 m−2 ⎦ t   +tm−2 ⎣ β tm−2 + tm−2

×

ρ

m−2 −um−2 um−2 n+1 −2un n−1

(x)  2  −μunm−2 1 − um−2 n

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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Application to a system of ordinary differential equations

19

In this chapter, we deal with a system of differential equations which include several linear differential equations with several unknown function. When the literature is examined, many systems of ordinary differential equations can be found; for instance the Chua attractor, a model of malaria transmission, a tumor growth model, and the Hartley oscillator.

19.1

System of ordinary differential equations with integer and non-integer orders

In this section, we present a numerical scheme for a system of ordinary differential equations such as a hybrid attractor with the classical derivative, the Shaw oscillator, and the Dengue, HIV and Ebola models with various fractional and fractal–fractional operators.

19.1.1 A hybrid attractor with the classical derivative We deal with the strange hybrid attractor including the classical derivative ⎧  ⎨ u (t) = v + 2uv + uw, v  (t) = 1 − 2u2 + vw, ⎩ w  (t) = u − u2 − v 2 . For the sake of simplicity, we take ⎧  ⎨ u (t) = U (u, v, w, t) , v  (t) = V (u, v, w, t) , ⎩  w (t) = W (u, v, w, t) . If we integrate Eq. (19.2), we can have the following equality:  t U (u, v, w, τ ) dτ, u (t) = u (0) + 0  t v (t) = v (0) + V (u, v, w, τ ) dτ, 0  t W (u, v, w, τ ) dτ. w (t) = w (0) + 0 New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00025-1 Copyright © 2021 Elsevier Inc. All rights reserved.

(19.1)

(19.2)

(19.3)

340

New Numerical Scheme With Newton Polynomial

We rewrite the equality (19.3) at point t = tk+1 

tk+1

u (tk+1 ) = u (tk ) +

U (u, v, w, τ ) dτ, 

0 tk+1

v (tk+1 ) = v (tk ) +

(19.4)

V (u, v, w, τ ) dτ, 0



tk+1

w (tk+1 ) = w (tk ) +

W (u, v, w, τ ) dτ, 0

and at point t = tk 

tk

u (tk ) = u (tk−1 ) +

U (u, v, w, τ ) dτ, 

0 tk

v (tk ) = v (tk−1 ) +

(19.5)

V (u, v, w, τ ) dτ, 0



tk

w (tk ) = w (tk−1 ) +

W (u, v, w, τ ) dτ. 0

If we take the difference of Eqs. (19.4) and (19.5), we have the following: 

tk+1

u (tk ) = u (tk−1 ) + v (tk ) = v (tk−1 ) +

U (u, v, w, τ ) dτ, tk  tk+1

(19.6)

V (u, v, w, τ ) dτ, tk

 w (tk ) = w (tk−1 ) +

tk+1

W (u, v, w, τ ) dτ. tk

Now we use a two step Newton polynomial, so we can obtain the following scheme: u (tk+1 ) = u (tk ) ⎧   U uk−2, v k−2 , w k−2 , tk−2  ⎪ ⎪  k−2 k−2 k−2  k−1 k−1 k−1 ⎪ ⎪ ⎪ + U u ,v ,w ,tk−1 −U u ,v ,w ,tk−2 (τ − t ) ⎪  tk+1 ⎪   ⎤ k−2 ⎡  k k k t ⎨ U u ,v ,w ,tk −2U uk−1 ,v k−1 ,w k−1 ,tk−1 + 2 ⎦ ⎪   k−2 2(t) +⎣ tk ⎪ k−2 ,w k−2 ,t ⎪ U u ,v k−2 ⎪ ⎪ + 2 ⎪ 2(t) ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

dτ,

(19.7)

Application to a system of ordinary differential equations

341

v (tk+1 ) = v (tk ) ⎧   V uk−2, v k−2 , w k−2 , tk−2  ⎪ ⎪   ⎪ V uk−1 ,v k−1 ,wk−1 ,tk−1 −V uk−2 ,v k−2 ,wk−2 ,tk−2 ⎪ ⎪ + −t ) (τ ⎪  tk+1 ⎪   ⎤ k−2 ⎡  k k k t ⎨ V u ,v ,w ,tk −2V uk−1 ,v k−1 ,w k−1 ,tk−1 + 2(t)2 ⎦ ⎪   +⎣ tk ⎪ ⎪ V uk−2 ,v k−2 ,w k−2 ,tk−2 ⎪ ⎪ + ⎪ 2(t)2 ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

w (tk+1 ) = w (tk ) ⎧   W uk−2, v k−2 , w k−2 , t ⎪ ⎪  k−2 k−2 k−2   k−1 k−1 k−1 ⎪ ⎪ W u ,v ,w ,tk−1 −W u ,v ,w k−2 ,tk−2 ⎪ ⎪ + −t ) (τ ⎪  tk+1 ⎨   ⎤ k−2 ⎡  k k k t W u ,v ,w ,tk −2W uk−1 ,v k−1 ,w k−1 ,tk−1 + 2 ⎦ ⎪   k−2 2(t) +⎣ tk ⎪ k−2 ,w k−2 ,t ⎪ W u ,v k−2 ⎪ ⎪ + 2 ⎪ 2(t) ⎪ ⎩ × (τ − tk−2 ) (τ − tk−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

dτ,

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

dτ.

Thus, a numerical solution of the hybrid attractor is given by the following scheme: 

 5 4 23 Un−2 t − Un−1 t + Un t , 12 3 12   5 4 23 Vn−2 t − Vn−1 t + Vn t , vn+1 = vn + 12 3 12   5 4 23 Wn−2 t − Wn−1 t + Wn t . wn+1 = wn + 12 3 12

un+1 = un +

(19.8)

Replacing the functions u, v, w by their values, we obtain ⎡ un+1 = un + ⎣

5 12 (vn−2 + 2un−2 vn−2 + un−2 wn−2 ) t − 43 (vn−1 + 2un−1 vn−1 + un−1 wn−1 ) t + 23 12 (vn + 2un vn + un wn ) t

  ⎤ 1 − 2u2n−2 + vn−2 wn−2 t   vn+1 = vn + ⎣ 1 − 2u2n−1 + vn−1 wn−1 t ⎦ ,   2 + 23 12 1 − 2un + vn wn t  ⎤ ⎡ 5  2 2 12 un−2 − un−2 − vn−2 t 2 wn+1 = wn + ⎣ − 43 un−1 − u2n−1 − vn−1 t ⎦ .   23 2 2 + 12 un − un − vn t ⎡

5 12 − 43

The numerical simulation is given in Fig. 19.1.

⎤ ⎦,

(19.9)

342

New Numerical Scheme With Newton Polynomial

Figure 19.1 Dynamical behavior of the hybrid attractor with the classical derivative for u (0) = 2, v (0) = 1, w (0) = 1.

19.1.2 Shaw oscillator with the Caputo fractional derivative We consider a Shaw oscillator having the Caputo fractional derivative ⎧ C α   ⎨ 0 Dt x (t) = 0.7y + 10x 0.1 − y 2 , C Dα y (t) = −x + 0.25 sin (1.57z) , ⎩ 0 t C Dα z 0 t (t) = 1. We convert Eq. (19.10) to ⎧ C α ⎨ 0 Dt x (t) = X (x, y, z, t) , C Dα y (t) = Y (x, y, z, t) , ⎩ 0C tα D z = Z (x, y, z, t) . (t) 0 t Integrating the above equation, we have the following equations:  t 1 X (x, y, z, τ ) (t − τ )α−1 dτ, x (t) = x (0) +  (α) 0  t 1 Y (x, y, z, τ ) (t − τ )α−1 dτ, y (t) = y (0) +  (α) 0

(19.10)

(19.11)

(19.12)

Application to a system of ordinary differential equations

z (t) = z (0) +

1  (α)



t

343

Z (x, y, z, τ ) (t − τ )α−1 dτ.

0

We can have at point t = tk+1

 tk+1 1 X (x, y, z, τ ) (tk+1 − τ )α−1 dτ,  (α) 0  tk+1 1 y (tk+1 ) = y (0) + Y (x, y, z, τ ) (tk+1 − τ )α−1 dτ,  (α) 0  tk+1 1 Z (x, y, z, τ ) (tk+1 − τ )α−1 dτ, z (tk+1 ) = z (0) +  (α) 0

x (tk+1 ) = x (0) +

(19.13)

and we have k  1  tm+1 x (tk+1 ) = x (0) + X (x, y, z, τ ) (tk+1 − τ )α−1 dτ,  (α) tm m=2

k  1  tm+1 y (tk+1 ) = y (0) + Y (x, y, z, τ ) (tk+1 − τ )α−1 dτ,  (α) tm

(19.14)

m=2

k  1  tm+1 z (tk+1 ) = z (0) + Z (x, y, z, τ ) (tk+1 − τ )α−1 dτ.  (α) tm m=2

Based on the Newton polynomial, we reorder Eq. (19.14) as follows: x (tk+1 ) =x (0)

⎧   X x m−2 , y m−2 , zm−2 , tm−2 ⎪ ⎪    m−1 m−1 m−1 ⎪ m−2 m−2 ⎪ X x ,y ,z ,tm−1 −X x ,y ,zm−2 ,tm−2 ⎪ ⎪ + ⎪ t ⎪ ⎪ k  × (τ − tm−2 ) 1  tm+1 ⎨ ⎡  ⎤ X(x m ,y m ,zm ,tm )−2X x m−1 ,y m−1 ,zm−1 ,tm−1 + ⎪  (α) 2 ⎪ +⎣ 2(t) ⎦ m=2 tm ⎪   ⎪ ⎪ X x m−2 ,y m−2 ,zm−2 ,tm−2 ⎪ ⎪ + ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

× (tk+1 − τ )α−1 dτ, y (tk+1 ) =y (0)

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(19.15)

⎧   Y x m−2 , y m−2 , zm−2 , tm−2 ⎪ ⎪    ⎪ ⎪ Y x m−1 ,y m−1 ,zm−1 ,tm−1 −Y x m−2 ,y m−2 ,zm−2 ,tm−2 ⎪ ⎪ + ⎪ t ⎪ ⎪ k  × (τ − tm−2 ) 1  tm+1 ⎨ ⎡  ⎤ Y (x m ,y m ,zm ,tm )−2Y x m−1 ,y m−1 ,zm−1 ,tm−1 + ⎪  (α) ⎪ 2(t)2 ⎦ m=2 tm ⎪   +⎣ ⎪ ⎪ Y x m−2 ,y m−2 ,zm−2 ,tm−2 ⎪ ⎪ + ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

344

New Numerical Scheme With Newton Polynomial

× (tk+1 − τ )α−1 dτ,

z (tk+1 )=z (0) +

1  (α)

⎧   Z x m−2 , y m−2 , zm−2 , tm−2 ⎪ ⎪    ⎪ ⎪ Z x m−1 ,y m−1 ,zm−1 ,tm−1 −Z x m−2 ,y m−2 ,zm−2 ,tm−2 ⎪ ⎪ + ⎪ t ⎪ k  tm+1 ⎪ ⎨ × (τ − tm−2 )   ⎤ ⎡ Z(x m ,y m ,zm ,tm )−2Z x m−1 ,y m−1 ,zm−1 ,tm−1 × ⎪ ⎪ 2(t)2 ⎦ m=2 tm ⎪   +⎣ ⎪ ⎪ Z x m−2 ,y m−2 ,zm−2 ,tm−2 ⎪ ⎪ + 2 ⎪ 2(t) ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ, and thus, we obtain the following: x k+1 = x0 +

k   tm+1 1   m−2 m−2 m−2 X x ,y ,z , tm−2 (tk+1 − τ )α−1 dτ  (α) tm m=2

k 

(19.16)   m−2 m−2 m−2   m−1 m−1 m−1 X x ,y ,z , tm−1 − X x ,y ,z , tm−2 t

1  (α) m=2  tm+1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

+

tm k 

⎡

  X(x m ,y m ,zm ,tm )−2X x m−1 ,y m−1 ,zm−1 ,tm−1 2(t)2   X x m−2 ,y m−2 ,zm−2 ,tm−2 + 2 2(t)

1 ⎣  (α) m=2  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ, +

⎤ ⎦

tm k   tm+1 1   m−2 m−2 m−2 Y x ,y ,z , tm−2 (tk+1 − τ )α−1 dτ  (α) tm m=2   m−2 m−2 m−2   m−1 m−1 m−1 k ,y ,z , tm−1 − Y x ,y ,z , tm−2 1 Y x +  (α) t m=2  tm+1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

y k+1 = y0 +

tm

1 +  (α)

k  m=2

⎡ ⎣

  Y (x m ,y m ,zm ,tm )−2Y x m−1 ,y m−1 ,zm−1 ,tm−1 2(t)2   Y x m−2 ,y m−2 ,zm−2 ,tm−2 + 2(t)2

⎤ ⎦

Application to a system of ordinary differential equations

 ×

tm+1

345

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

tm k   tm+1 1   m−2 m−2 m−2 Z x ,y ,z , tm−2 (tk+1 − τ )α−1 dτ  (α) tm m=2   m−2 m−2 m−2   m−1 m−1 m−1 k ,y ,z , tm−1 − Z x ,y ,z , tm−2 1 Z x +  (α) t m=2  tm+1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ

zk+1 = z0 +

tm k 

⎡

  Z(x m ,y m ,zm ,tm )−2Z x m−1 ,y m−1 ,zm−1 ,tm−1 2(t)2   Z x m−2 ,y m−2 ,zm−2 ,tm−2 + 2(t)2

1 ⎣ +  (α) m=2  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

⎤ ⎦

tm

Thus, the numerical algorithm for the hybrid attractor is of the form of (t)α  (α + 1) k     × X x m−2 , y m−2 , zm−2 , tm−2 (k − m + 1)α − (k − m)α (19.17)

x k+1 = x0 +

m=2

   k  (t)α  X x m−1 , y m−1 , zm−1 , tm−1  −X x m−2 , y m−2 , zm−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k (x m , y m , zm , tm )  X (t)α  ⎣ ⎦ −2X x m−1 , y m−1 , zw m−1 , tm−1 +  2 (α + 3) m−2 m−2 m−2 +X x ,y ,z , tm−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12 +

(t)α  (α + 1) k     × Y x m−2 , y m−2 , zm−2 , tm−2 (k − m + 1)α − (k − m)α

y k+1 = y0 +

m=2

346

New Numerical Scheme With Newton Polynomial

   k  (t)α  Y x m−1 , y m−1 , zm−1 , tm−1  −Y x m−2 , y m−2 , zm−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k (x m , y m , zm , tm )   Ym−1 (t)α  ⎣ , y m−1 , zm−1 , tm−1 ⎦ −2Y x + 2 (α + 3) m−2 +Y x , y m−2 , zm−2 , tm−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

+

⎤ ⎥ ⎥, ⎦

(t)α  (α + 1) k     × Z x m−2 , y m−2 , zm−2 , tm−2 (k − m + 1)α − (k − m)α

zk+1 = z0 +

m=2

   k  (t)α  Z x m−1 , y m−1 , zm−1 , tm−1  + −Z x m−2 , y m−2 , zm−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k Z (x m , y m , zm , tm ) α    (t) ⎣ −2Z x m−1 , y m−1 , zm−1 , tm−1 ⎦ +   2 (α + 3) +Z x m−2 , y m−2 , zm−2 , tm−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

⎤ ⎥ ⎥. ⎦

Replacing the functions X, Y , Z by their values, we present the following scheme: k  (t)α   m−2 m−2 m−2 X x ,y ,z , tm−2  (α + 1) m=2   α × (k − m + 1) − (k − m)α    k  (t)α  X x m−1 , y m−1 , zm−1 , tm−1  + −X x m−2 , y m−2 , zm−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)

x k+1 = x0 +

(19.18)

Application to a system of ordinary differential equations

⎤ ⎡ k (x m , y m , zm , tm )  Xm−1 (t)α  ⎣ , y m−1 , zm−1 , tm−1 ⎦ −2X x + 2 (α + 3) m−2 +X x , y m−2 , zm−2 , tm−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 k  (t)α   m−2 m−2 m−2 Y x ,y ,z , tm−2  (α + 1) m=2   α × (k − m + 1) − (k − m)α    k  (t)α  Y x m−1 , y m−1 , zm−1 , tm−1  + −Y x m−2 , y m−2 , zm−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k (x m , y m , zm , tm )   Ym−1 (t)α  ⎣ −2Y x , y m−1 , zm−1 , tm−1 ⎦ + 2 (α + 3) m−2 +Y x , y m−2 , zm−2 , tm−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

347

⎤ ⎥ ⎥, ⎦

y k+1 = y0 +

k  (t)α   m−2 m−2 m−2 Z x ,y ,z , tm−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α    k  (t)α  Z x m−1 , y m−1 , zm−1 , tm−1  + −Z x m−2 , y m−2 , zm−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k Z (x m , y m , zm , tm ) α    (t) ⎣ −2Z x m−1 , y m−1 , zm−1 , tm−1 ⎦ +   2 (α + 3) +Z x m−2 , y m−2 , zm−2 , tm−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

⎤ ⎥ ⎥, ⎦

zk+1 = z0 +

⎤ ⎥ ⎥. ⎦

348

New Numerical Scheme With Newton Polynomial

Figure 19.2 Dynamical behavior of the Shaw oscillator with the Caputo derivative for x (0) = 0.1, y (0) = 0.1, z (0) = 0.1 and α = 0.96.

The numerical scheme is depicted in Fig. 19.2.

19.1.3 Dengue model with the Atangana–Baleanu fractional derivative We deal with the following dengue problem: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

bK βA SA IK ABC D α S − μA SA , t A (t) = A − 0 NA bK βA SA IK ABC D α I − (δA + γA + μA ) IA , t A (t) = 0 NA ABC D α R = γ (t) A IZA − μA RA , t A 0 bK βK SK IA ABC D α S − μK SK , t K (t) = K − 0 NA bK βK SK IA ABC D α I − μ = (t) K IK , t K 0 NA

(19.19)

where the derivative is the Atangana–Baleanu fractional derivative. We convert Eq. (19.19) to

Application to a system of ordinary differential equations

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

ABC D α S t A (t) = S (SA , IA , RA , SK , IK , t) , 0 ABC D α I t A (t) = I (SA , IA , RA , SK , IK , t) , 0 ABC D α R t A (t) = R (SA , IA , RA , SK , IK , t) , 0 ABC D α S  t K (t) = S (SA , IA , RA , SK , IK , t) , 0 ABC D α I = I(SA , IA , RA , SK , IK , t) . (t) t K 0

349

(19.20)

Using the Atangana–Baleanu integral, we have the following equations: 1−α S (SA , IA , RA , SK , IK , t) AB (α)  t α + S (SA , IA , RA , SK , IK , τ ) (t − τ )α−1 dτ, AB (α)  (α) 0 1−α I (SA , IA , RA , SK , IK , t) IA (t) = IA (0) + AB (α)  t α I (SA , IA , RA , SK , IK , τ ) (t − τ )α−1 dτ, + AB (α)  (α) 0 1−α RA (t) = RA (0) + R (SA , IA , RA , SK , IK , t) AB (α)  t α R (SA , IA , RA , SK , IK , τ ) (t − τ )α−1 dτ, + AB (α)  (α) 0 1−α  S (SA , IA , RA , SK , IK , t) SK (t) = SK (0) + AB (α)  t α  + S (SA , IA , RA , SK , IK , τ ) (t − τ )α−1 dτ, AB (α)  (α) 0 1−α  I (SA , IA , RA , SK , IK , t) IK (t) = IK (0) + AB (α)  t α + I(SA , IA , RA , SK , IK , τ ) (t − τ )α−1 dτ. AB (α)  (α) 0 SA (t) = SA (0) +

(19.21)

We can have at point t = tk+1 1−α S (SA , IA , RA , SK , IK , t) (19.22) AB (α)  tk+1 α S (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, + AB (α)  (α) 0 1−α IA (tk+1 ) = IA (0) + I (SA , IA , RA , SK , IK , t) AB (α)  tk+1 α I (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, + AB (α)  (α) 0

SA (tk+1 ) = SA (0) +

350

New Numerical Scheme With Newton Polynomial

1−α R (SA , IA , RA , SK , IK , t) AB (α)  tk+1 α + R (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, AB (α)  (α) 0 1−α  S (SA , IA , RA , SK , IK , t) SK (tk+1 ) = SK (0) + AB (α)  tk+1 α  + S (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, AB (α)  (α) 0 1−α  I (SA , IA , RA , SK , IK , t) IK (tk+1 ) = IK (0) + AB (α)  tk+1 α + I(SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ. AB (α)  (α) 0

RA (tk+1 ) = RA (0) +

and we have SA (tk+1 ) = SA (0) +

1−α S (SA , IA , RA , SK , IK , t) AB (α)

α AB (α)  (α) k  tm+1  × S (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, +

m=2 tm

IA (tk+1 ) = IA (0) + +

1−α I (SA , IA , RA , SK , IK , t) AB (α)

α AB (α)  (α) k  tm+1  I (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, m=2 tm

RA (tk+1 ) = RA (0) +

1−α R (SA , IA , RA , SK , IK , t) AB (α)

α AB (α)  (α) k  tm+1  × R (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, +

m=2 tm

SK (tk+1 ) = SK (0) +

1−α  S (SA , IA , RA , SK , IK , t) AB (α)

α AB (α)  (α) k  tm+1   × S (SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ, +

m=2 tm

(19.23)

Application to a system of ordinary differential equations

IK (tk+1 ) = IK (0) +

351

1−α  I (SA , IA , RA , SK , IK , t) AB (α)

α AB (α)  (α) k  tm+1  × I(SA , IA , RA , SK , IK , τ ) (tk+1 − τ )α−1 dτ. +

m=2 tm

As before, we reorder Eq. (19.23) as follows: SA (tk+1 ) = SA (0) + +

×

α AB (α)  (α) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  k  tm+1 ⎨

1−α  k k k k k  S SA , IA , RA , SK , IK , tk AB (α)

(19.24)

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬

  m−2 m−2 m−2 S SAm−2 , IAm−2 , RA , SK , IK , tm−2 ⎡   ⎤ ⎢ +⎣

m−1 m−1 m−1 m−1 m−1 S SA ,IA ,RA ,SK ,IK ,tm−1

− ⎡

t   m−2 m−2 m−2 m−2 m−2 S SA ,IA ,RA ,SK ,IK ,tm−2

⎥ ⎦

t

× (τ− tm−2 )

⎤ ⎪  m m m m m  ⎪ m−1 m−1 m−1 m−1 m−1 ⎪ S SA ,IA ,RA ,SK ,IK ,tm −2S SA ,IA ,RA ,SK ,IK ,tm−1 ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ 2(t)2 ⎪ ⎥   +⎢ ⎪ m−2 m−2 m−2 m−2 m−2 ⎪ ⎣ ⎦ S S ,I ,R ,S ,I ,t ⎪ m−2 A A A K K ⎪ ⎪ + ⎪ 2 ⎪ 2(t) ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ, m=2 tm

IA (tk+1 ) = IA (0) + +

×

α AB (α)  (α) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  k tm+1 ⎨  m=2 tm

⎡ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎪ ⎪ +⎢ ⎪ ⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

× (tk+1 − τ )α−1 dτ,

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

1−α  k k k k k  I SA , IA , RA , SK , IK , tk AB (α)   m−2 m−2 m−2 I SAm−2 , IAm−2 , RA , SK , IK , tm−2 ⎡   ⎤ ⎢ +⎣

m−1 m−1 m−1 m−1 m−1 I SA ,IA ,RA ,SK ,IK ,tm−1

−

t   m−2 m−2 m−2 m−2 m−2 I SA ,IA ,RA ,SK ,IK ,tm−2

⎥ ⎦

t

× (τ − tm−2 )

  m m m m m  m−1 m−1 m−1 m−1 m−1 I SA ,IA ,RA ,SK ,IK ,tm −2I SA ,IA ,RA ,SK ,IK ,tm−1

+

2(t)2   m−2 m−2 m−2 m−2 m−2 I SA ,IA ,RA ,SK ,IK ,tm−2 2(t)2

× (τ − tm−2 ) (τ − tm−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎥⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

352

New Numerical Scheme With Newton Polynomial

α 1−α  k k k k k  R SA , IA , RA , SK , IK , tk + RA (tk+1 ) = RA (0) + AB (α) AB (α)  (α) ⎧   m−2 m−2 m−2 m−2 m−2 ⎪ R S , I , R , S , I , t ⎪ m−2 A K K ⎪ ⎪ ⎡A  A  ⎤ ⎪ m−1 m−1 m−1 m−1 m−1 ⎪ ⎪ R S ,I ,R ,S ,I ,tm−1 A A A K K ⎪ ⎪ ⎪ ⎢ t ⎪  ⎥  +⎣ ⎪ ⎦ m−2 m−2 m−2 m−2 m−2 ⎪ R S ,I ,R ,S ,I ,t ⎪ m−2 A A A K K ⎪ ⎪  − k ⎨  tm+1 t × (τ− tm−2 ) × ⎡  ⎪  m m m m m  ⎪ m−1 m−1 m−1 m−1 m−1 m=2 tm ⎪ R SA ,IA ,RA ,SK ,IK ,tm −2R SA ,IA ,RA ,SK ,IK ,tm−1 ⎪ ⎪ ⎪ ⎢ ⎪ 2(t)2 ⎪   +⎢ ⎪ m−2 m−2 m−2 m−2 m−2 ⎪ ⎣ R S ,I ,R ,tm−2 ⎪ A A A ,SK ,IK ⎪ ⎪ + ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎥⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ,

α 1 − α  k k k k k  S SA , IA , RA , SK , IK , tk + SK (tk+1 ) = SK (0) + AB (α) AB (α)  (α) ⎧   m−2 m−2 m−2 m−2 m−2 ⎪  S S , I , R , S , I , t ⎪ m−2 A K K ⎪ ⎪ ⎡A  A  ⎤ ⎪ m−1 m−1 m−1 m−1 m−1 ⎪  ⎪ S SA ,IA ,RA ,SK ,IK ,tm−1 ⎪ ⎪ ⎪ ⎢ t ⎪  ⎥ + ⎣  m−2 m−2 m−2 ⎪ ⎦ m−2 m−2 ⎪ S S ,I ,R ,S ,I ,t ⎪ m−2 A A A K K ⎪ ⎪ − k  tm+1 ⎨  t × (τ− tm−2 ) × ⎡  ⎪  m m m m m  ⎪ m−1 m−1 m−1 m−1 m−1  m=2 tm ⎪ S SA S SA ,IA ,RA ,SK ,IK ,tm −2 ,IA ,RA ,SK ,IK ,tm−1 ⎪ ⎪ ⎢ ⎪ ⎪ 2(t)2 ⎪   +⎢ ⎪ m−2 m−2 m−2 m−2 m−2 ⎪ ⎣  S S ,I ,R ,tm−2 ⎪ A A A ,SK ,IK ⎪ ⎪ + ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎥⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ,

α 1 − α  k k k k k  I SA , IA , RA , SK , IK , tk + IK (tk+1 ) = IK (0) + AB (α) AB (α)  (α) ⎧   m−2 m−2 m−2 ⎪ I SAm−2 , IAm−2 , RA , SK , IK , tm−2 ⎪ ⎪ ⎪ ⎡   ⎤ ⎪ m−1 m−1 m−1 m−1 m−1 ⎪ ⎪ I SA ,IA ,RA ,SK ,IK ,tm−1 ⎪ ⎪ ⎪ ⎢ t ⎪  ⎥ + ⎣  m−2 m−2 m−2 ⎪ ⎦ m−2 m−2 ⎪ I SA ,IA ,RA ,SK ,IK ,tm−2 ⎪ ⎪ ⎪  − k tm+1 ⎨  t × − t (τ × m−2 ) ⎡    ⎪  ⎪  S m ,I m ,R m ,S m ,I m ,tm −2I S m−1 ,I m−1 ,R m−1 ,S m−1 ,I m−1 ,tm−1 m=2 tm ⎪ I ⎪ A A A K K A A A K K ⎪ ⎪ ⎢ ⎪ 2(t)2 ⎪   +⎢ ⎪ m−2 m−2 m−2 m−2 m−2 ⎪ ⎣  I SA ,IA ,RA ,SK ,IK ,tm−2 ⎪ ⎪ ⎪ + ⎪ ⎪ 2(t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ,

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎥⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Application to a system of ordinary differential equations

353

and we can have the following: 1−α  k k k k k  S SA , IA , RA , SK , IK , tk (19.25) AB (α) k    α m−2 m−2 m−2 + S SAm−2 , IAm−2 , RA , SK , IK , tm−2 AB (α)  (α) m=2  tm+1 × (tk+1 − τ )α−1 dτ tm ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 S SA ,IA ,RA ,SK ,IK ,tm−1 k  α ⎢ t   ⎥ + ⎣ ⎦ m−2 m−2 m−2 m−2 m−2 S SA ,IA ,RA ,SK ,IK ,tm−2 AB (α)  (α) m=2 − t  tm+1 α × (τ − tm−2 ) (tk+1 − τ )α−1 dτ + AB (α)  (α) tm ⎡    ⎤ 

SAk+1 = SA (0) +

k  ⎢ ⎢ × ⎣

m ,I m ,R m ,S m ,I m ,t −2S S m−1 ,I m−1 ,R m−1 ,S m−1 ,I m−1 ,t S SA m−1 A A K K m A A A K K



+

m=2

 ×

tm+1

S

2(t)2 m−2 m−2 m−2 m−2 m−2 SA ,IA ,RA ,SK ,IK ,tm−2 2(t)2



⎥ ⎥ ⎦

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

tm

1−α  k k k k k  I SA , IA , RA , SK , IK , tk AB (α) k    α m−2 m−2 m−2 I SAm−2 , IAm−2 , RA , SK , IK , tm−2 + AB (α)  (α) m=2  tm+1 × (tk+1 − τ )α−1 dτ tm ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 I SA ,IA ,RA ,SK ,IK ,tm−1 k  α ⎢ t   ⎥ + ⎣ ⎦ m−2 m−2 m−2 m−2 m−2 I SA ,IA ,RA ,SK ,IK ,tm−2 AB (α)  (α) m=2 − t  tm+1 α × (τ − tm−2 ) (tk+1 − τ )α−1 dτ + AB  (α) (α) tm ⎡    ⎤ 

IAk+1 = IA (0) +

k  ⎢ ⎢ × ⎣ m=2

 ×

tm+1

tm

m ,I m ,R m ,S m ,I m ,t −2I S m−1 ,I m−1 ,R m−1 ,S m−1 ,I m−1 ,t I SA m−1 A A K K m A A A K K

+

2(t)2   m−2 m−2 m−2 m−2 m−2 I SA ,IA ,RA ,SK ,IK ,tm−2 2(t)2

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

⎥ ⎥ ⎦

354

New Numerical Scheme With Newton Polynomial

k+1 RA = RA (0) +

1−α  k k k k k  R SA , IA , RA , SK , IK , tk AB (α)

k    α m−2 m−2 m−2 R SAm−2 , IAm−2 , RA , SK , IK , tm−2 AB (α)  (α) m=2  tm+1 × (tk+1 − τ )α−1 dτ

+

tm

+

α AB (α)  (α) 

×

tm+1

tm

×

⎡

k  ⎢ ⎢ ⎣

×

tm+1

  m−1 m−1 m−1 m−1 m−1 R SA ,IA ,RA ,SK ,IK ,tm−1

−

m=2

t   m−2 m−2 m−2 m−2 m−2 R SA ,IA ,RA ,SK ,IK ,tm−2

⎤ ⎥ ⎦

t

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

α AB (α)  (α)

   m m m m m  m−1 m−1 m−1 m−1 m−1 R SA ,IA ,RA ,SK ,IK ,tm −2R SA ,IA ,RA ,SK ,IK ,tm−1

+

m=2



⎡ k  ⎢ ⎣

2(t)2   m−2 m−2 m−2 m−2 m−2 R SA ,IA ,RA ,SK ,IK ,tm−2

⎤ ⎥ ⎥ ⎦

2(t)2

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

tm k+1 SK = SK (0) +

1 − α  k k k k k  S SA , IA , RA , SK , IK , tk AB (α)

k    α m−2 m−2 m−2  , SK , IK , tm−2 S SAm−2 , IAm−2 , RA AB (α)  (α) m=2  tm+1 × (tk+1 − τ )α−1 dτ

+

tm

+

α AB (α)  (α) 

×

tm+1

tm

×

k  m=2

 ×

⎡ ⎢ ⎢ ⎣

tm+1

tm

⎡ k  ⎢ ⎣ m=2

  m−1 m−1 m−1 m−1 m−1  S SA ,IA ,RA ,SK ,IK ,tm−1

−

t   m−2 m−2 m−2 m−2 m−2  S SA ,IA ,RA ,SK ,IK ,tm−2

⎤ ⎥ ⎦

t

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

α AB (α)  (α)

   m m m m m  m−1 m−1 m−1 m−1 m−1  S SA ,IA ,RA ,SK ,IK ,tm −2 ,IA ,RA ,SK ,IK ,tm−1 S SA

+

2(t)2   m−2 m−2 m−2 m−2 m−2  S SA ,IA ,RA ,SK ,IK ,tm−2 2(t)2

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

⎤ ⎥ ⎥ ⎦

Application to a system of ordinary differential equations

355

1 − α  k k k k k  I SA , IA , RA , SK , IK , tk AB (α) k    α m−2 m−2 m−2 , SK , IK , tm−2 + I SAm−2 , IAm−2 , RA AB (α)  (α) m=2  tm+1 × (tk+1 − τ )α−1 dτ tm ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 I SA ,IA ,RA ,SK ,IK ,tm−1 k  α ⎢ t  ⎥ + ⎣  m−2 m−2 m−2 ⎦ m−2 m−2 I SA ,IA ,RA ,SK ,IK ,tm−2 AB (α)  (α) m=2 − t  tm+1 α × (τ − tm−2 ) (tk+1 − τ )α−1 dτ + AB  (α) (α) tm ⎡   ⎤  

IKk+1 = IK (0) +

k  ⎢ ⎢ × ⎣

m ,I m ,R m ,S m ,I m ,t −2I  S m−1 ,I m−1 ,R m−1 ,S m−1 ,I m−1 ,tm−1 I SA A A K K m A A A K K

+

m=2

 ×

tm+1

2(t)2   m−2 m−2 m−2 m−2 m−2 I SA ,IA ,RA ,SK ,IK ,tm−2

⎥ ⎥ ⎦

2(t)2

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

Thus, the numerical algorithm for the dengue model is of the form of 1−α  k k k k k  S SA , IA , RA , SK , IK , tk (19.26) AB (α) k  α (t)α   m−2 m−2 m−2 m−2 m−2 + S SA , IA , RA , SK , IK , tm−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 k α (t)α  ⎣ S SA , IA , RA , SK , IK , tm−1  ⎦ + m−2 m−2 m−2 AB (α)  (α + 2) , SK , IK , tm−2 −S SAm−2 , IAm−2 , RA m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   ⎡ ⎤ m, Sm, I m, t S SAm , IAm , RA K K m   k ⎥ m−1 m−1 m−1 α (t)α  ⎢ , SK , IK , tm−1 ⎥ ⎢ −2S SAm−1 , IAm−1 , RA + ⎣   ⎦ AB (α) 2 (α + 3) m−2 m−2 m−2 m=2 +S SAm−2 , IAm−2 , RA , SK , IK , tm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12

SAk+1 = SA (0) +

356

New Numerical Scheme With Newton Polynomial

1−α  k k k k k  I SA , IA , RA , SK , IK , tk AB (α) k  α (t)α   m−2 m−2 m−2 m−2 m−2 I SA , IA , RA , SK , IK , tm−2 + AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 k α  I S , I , R , S , I , t m−1 α (t) A A K K ⎣  ⎦ A + m−2 m−2 m−2 m−2 m−2 AB (α)  (α + 2) −I SA , IA , RA , SK , IK , tm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   ⎡ ⎤ m, Sm, I m, t I SAm , IAm , RA m K K   k ⎥ m−1 m−1 m−1 α (t)α  ⎢ , SK , IK , tm−1 ⎥ ⎢ −2I SAm−1 , IAm−1 , RA + ⎣ ⎦   AB (α) 2 (α + 3) m−2 m−2 m−2 m=2 +I SAm−2 , IAm−2 , RA , SK , IK , tm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

IAk+1 = IA (0) +

1−α  k k k k k  R SA , IA , RA , SK , IK , tk AB (α) k  α (t)α   m−2 m−2 m−2 m−2 m−2 + R SA , IA , RA , SK , IK , tm−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 k α (t)α  ⎣ R SA , IA , RA , SK , IK , tm−1  ⎦ + m−2 m−2 m−2 AB (α)  (α + 2) , SK , IK , tm−2 −R SAm−2 , IAm−2 , RA m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α)   ⎡ ⎤ m, Sm, I m, t R SAm , IAm , RA K K m   k ⎥ m−1 m−1 m−1 α (t)α  ⎢ , SK , IK , tm−1 ⎥ ⎢ −2R SAm−1 , IAm−1 , RA + ⎣   ⎦ AB (α) 2 (α + 3) m−2 m−2 m−2 m=2 +R SAm−2 , IAm−2 , RA , SK , IK , tm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12

k+1 RA = RA (0) +

Application to a system of ordinary differential equations

357

1 − α  k k k k k  S SA , IA , RA , SK , IK , tk AB (α) k  α (t)α   m−2 m−2 m−2 m−2 m−2 + S SA , IA , RA , SK , IK , tm−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 k  α  S S , I , R , S , I , t m−1 α (t) A A K K ⎣  ⎦ A + m−2 m−2 m−2 m−2 m−2  AB (α)  (α + 2) −S SA , IA , RA , SK , IK , tm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡   m, Sm, I m, t  S SAm , IAm , RA m K K   k ⎥ m−1 m−1 m−1 α (t)α  ⎢ S SAm−1 , IAm−1 , RA , SK , IK , tm−1 ⎥ ⎢ −2 + ⎦ ⎣   AB (α) 2 (α + 3) m−2 m−2 m−2 m=2 + S SAm−2 , IAm−2 , RA , SK , IK , tm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

k+1 SK = SK (0) +

1 − α  k k k k k  I SA , IA , RA , SK , IK , tk AB (α) k  α (t)α   m−2 m−2 m−2 m−2 m−2 + I SA , IA , RA , SK , IK , tm−2 AB (α)  (α + 1) m=2   × (k − m + 1)α − (k − m)α ⎡   ⎤ m−1 m−1 m−1 m−1 m−1 k  α  I S , I , R , S , I , t m−1 α (t) A A K K ⎣  ⎦ A + m−2 m−2 m−2 m−2 m−2  AB (α)  (α + 2) , I , R , S , I , t − I S m−2 A A A K K m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡   m, Sm, I m, t I SAm , IAm , RA m K K   k ⎥ m−1 m−1 m−1 α (t)α  ⎢ , SK , IK , tm−1 ⎥ ⎢ −2I SAm−1 , IAm−1 , RA + ⎦ ⎣   AB (α) 2 (α + 3) m−2 m−2 m−2 m=2 , SK , IK , tm−2 +I SAm−2 , IAm−2 , RA   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

IKk+1 = IK (0) +

358

New Numerical Scheme With Newton Polynomial

19.1.4 HIV model with fractal–fractional derivative with power law We deal with the following HIV problem: ⎧  ⎪ ⎨ F0 F P Dtα T (t) = p − rT + s 1 − ⎪ ⎩

T +I Tmax



− kV T ,

(t) = kV T − βI, (t) = NβI − γ V ,

F F P Dα I t 0 F F P Dα V t 0

(19.27)

where the derivative is the Caputo fractal–fractional derivative. We write Eq. (19.27) as ⎧ FFP α ⎨ 0 Dt T (t) = T (T , I, V , t) , F F P D α I (t) = I (T , I, V , t) , (19.28) ⎩ F0F P αt  D V = V , I, V , t) . (t) (T t 0 Here we take U (T , I, V , t) = βt β−1 T (T , I, V , t) , Y (T , I, V , t) = βt β−1 I(T , I, V , t) , W (T , I, V , t) = βt

(19.29)

β−1 

V (T , I, V , t) .

If we integrate above equation, we have the following equations:  t 1 U (T , I, V , τ ) (t − τ )α−1 dτ, T (t) = T (0) +  (α) 0  t 1 I (t) = I (0) + Y (T , I, V , τ ) (t − τ )α−1 dτ,  (α) 0  t 1 W (T , I, V , τ ) (t − τ )α−1 dτ. V (t) = V (0) +  (α) 0

(19.30)

We can have at point t = tk+1  tk+1 1 U (T , I, V , τ ) (tk+1 − τ )α−1 dτ,  (α) 0  tk+1 1 Y (T , I, V , τ ) (tk+1 − τ )α−1 dτ, I (tk+1 ) = I (0) +  (α) 0  tk+1 1 V (tk+1 ) = V (0) + W (T , I, V , τ ) (tk+1 − τ )α−1 dτ,  (α) 0 T (tk+1 ) = T (0) +

and we have T (tk+1 ) = T (0) +

k  1  tm+1 U (T , I, V , τ ) (tk+1 − τ )α−1 dτ,  (α) tm m=2

(19.31)

Application to a system of ordinary differential equations

I (tk+1 ) = I (0) +

k  1  tm+1 Y (T , I, V , τ ) (tk+1 − τ )α−1 dτ,  (α) tm

359

(19.32)

m=2

V (tk+1 ) = V (0) +

k  1  tm+1 W (T , I, V , τ ) (tk+1 − τ )α−1 dτ.  (α) tm m=2

Based on the Newton polynomial, we reorder Eq. (19.32) as follows: T (tk+1 ) = T0 +

×

1  (α)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  k tm+1 ⎨  m=2 tm

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(19.33) ⎫   U T m−2 , I m−2 , Vm−2 , tm−2 ⎪ ⎪    m−1 m−1 m−1 ⎪ U T ,I ,V ,tm−1 −U T m−2 ,I m−2 ,V m−2 ,tm−2 ⎪ ⎪ ⎪ + ⎪ t ⎪ ⎪ ⎬ × (τ − tm−2 )  ⎡ ⎤ m m m m−1 m−1 m−1 U (T ,I ,V ,tm )−2U T ,I ,V ,tm−1 ⎪ 2(t)2 ⎦ ⎪ ⎪   +⎣ ⎪ ⎪ U T m−2 ,I m−2 ,V m−2 ,tm−2 ⎪ ⎪ + 2 ⎪ 2(t) ⎪ ⎭ × (τ − tm−2 ) (τ − tm−1 )

× (tk+1 − τ )α−1 dτ, I (tk+1 ) = I0 +

1  (α)

⎧ ⎫   Y T m−2 , I m−2 , Vm−2 , tm−2 ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ Y T m−1 ,I m−1 ,V m−1 ,tm−1 −Y T m−2 ,I m−2 ,V m−2 ,tm−2 ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ t ⎪ ⎪ ⎪ ⎪  k ⎨ ⎬ × − t (τ ) t  m+1 m−2   ⎡ ⎤ m m m m−1 m−1 m−1 Y (T ,I ,V ,tm )−2Y T ,I ,V ,tm−1 × ⎪ ⎪ ⎪ 2(t)2 ⎦⎪ m=2 tm ⎪ ⎪   +⎣ ⎪ ⎪ ⎪ ⎪ Y T m−2 ,I m−2 ,V m−2 ,tm−2 ⎪ ⎪ ⎪ ⎪ + 2 ⎪ ⎪ 2(t) ⎪ ⎪ ⎩ ⎭ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ, V (tk+1 ) = V0 +

1  (α)

⎧ ⎫   W T m−2 , I m−2 , Vm−2 , tm−2 ⎪ ⎪ ⎪ ⎪    m−1 m−1 m−1 ⎪ ⎪ ⎪ W T ,I ,V ,tm−1 −W T m−2 ,I m−2 ,V m−2 ,tm−2 ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ t ⎪ ⎪ ⎪ ⎪  k ⎬ × (τ −tm−2 )  tm+1 ⎨ ⎡  ⎤ W (T m ,I m ,V m ,tm )−2W T m−1 ,I m−1 ,V m−1 ,tm−1 × ⎪ ⎪ 2 ⎪ ⎦ ⎪ m=2 tm ⎪ ⎪   m−2 2(t) +⎣ ⎪ ⎪ m−2 ,V m−2 ,t ⎪ ⎪ W T ,I m−2 ⎪ ⎪ ⎪ ⎪ + 2 ⎪ ⎪ 2(t) ⎪ ⎪ ⎩ ⎭ × (τ − tm−2 ) (τ − tm−1 ) × (tk+1 − τ )α−1 dτ,

360

New Numerical Scheme With Newton Polynomial

and we get the following approximation:

T k+1 = T0 +

k   tm+1 1   m−2 m−2 m−2 U T ,I ,V , tm−2 (tk+1 − τ )α−1 dτ  (α) tm m=2

k 



T m−1 , I m−1 , V m−1 , t

(19.34)  m−2 m−2 m−2  −U T ,I ,V , tm−2 t



U 1 m−1  (α) m=2  tm+1 1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ +  (α) tm  ⎤  ⎡ U (T m ,I m ,V m ,tm )−2U T m−1 ,I m−1 ,V m−1 ,tm−1 k  2(t)2 ⎣ ⎦   × U T m−2 ,I m−2 ,V m−2 ,tm−2 + 2 m=2 2(t)  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ, +

tm

k   tm+1 1   m−2 m−2 m−2 Y T ,I ,V , tm−2 (tk+1 − τ )α−1 dτ  (α) t m m=2   m−2 m−2 m−2   m−1 m−1 m−1 k ,I ,V , tm−1 − Y T ,I ,V , tm−2 1 Y T +  (α) t m=2  tm+1 1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ +  (α) tm  ⎤  ⎡ Y (T m ,I m ,V m ,tm )−2Y T m−1 ,I m−1 ,V m−1 ,tm−1 k  2(t)2 ⎣ ⎦   × Y T m−2 ,I m−2 ,V m−2 ,tm−2 + m=2 2 2(t)  tm+1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

I k+1 = I0 +

tm

k   tm+1 1   m−2 m−2 m−2 W T ,I ,V , tm−2 (tk+1 − τ )α−1 dτ  (α) t m m=2     m−1 m−1 m−1 k ,I ,V , tm−1 − W T m−2 , I m−2 , V m−2 , tm−2 1 W T +  (α) t m=2  tm+1 1 × (τ − tm−2 ) (tk+1 − τ )α−1 dτ +  (α) tm

V k+1 = V0 +

Application to a system of ordinary differential equations

×

k 

⎡ ⎣

m=2



×

tm+1

  W (T m ,I m ,V m ,tm )−2W T m−1 ,I m−1 ,V m−1 ,tm−1 2 2(t)   W T m−2 ,I m−2 ,V m−2 ,tm−2 + 2(t)2

361

⎤ ⎦

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

Thus, the numerical algorithm for the HIV model can be written as k  (t)α   m−2 m−2 m−2 U T ,I ,V , tm−2  (α + 1) m=2   α × (k − m + 1) − (k − m)α    k  (t)α  U T m−1 , I m−1 , V m−1 , tm−1  + −U T m−2 , I m−2 , V m−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k (T m , I m , V m , tm )  U m−1 (t)α  ⎣ , I m−1 , V m−1 , tm−1 ⎦ −2U T + 2 (α + 3) m−2 +U T , I m−2 , V m−2 , tm−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

T k+1 = T0 +

k  (t)α   m−2 m−2 m−2 Y T ,I ,V , tm−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α    k  (t)α  Y T m−1 , I m−1 , V m−1 , tm−1  + −Y T m−2 , I m−2 , V m−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k (T m , I m , V m , tm )   Y m−1 (t)α  ⎣ , I m−1 , V m−1 , tm−1 ⎦ −2Y T + 2 (α + 3) m−2 +Y T , I m−2 , V m−2 , tm−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α 2 +6α + 18α + 12

I k+1 = I0 +

(19.35)

362

New Numerical Scheme With Newton Polynomial k  (t)α   m−2 m−2 m−2 W T ,I ,V , tm−2  (α + 1) m=2   α × (k − m + 1) − (k − m)α    k  (t)α  W T m−1 , I m−1 , V m−1 , tm−1  + −W T m−2 , I m−2 , V m−2 , tm−2  (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k (T m , I m , V m , tm )  W m−1 (t)α  ⎣ , I m−1 , V m−1 , tm−1 ⎦ −2W T + 2 (α + 3) m−2 +W T , I m−2 , V m−2 , tm−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12

V k+1 = V0 +

Replacing the functions U (T , I, V , t), Y (T , I, V , t) and W (T , I, V , t) by their values (t)α (19.36)  (α + 1) k     β−1 × βtm−2 T T m−2 , I m−2 , V m−2 , tm−2 (k − m + 1)α − (k − m)α

T k+1 = T0 +

m=2

   k β−1  (t)α  βtm−1 T T m−1 , I m−1 , V m−1 , tm−1  + β−1   (α + 2) −βtm−2 T T m−2 , I m−2 , V m−2 , tm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 βtm T (T m , I m , V m , tm ) k α   ⎥ (t) ⎢ β−1  + ⎣ −2βtm−1 T T m−1 , I m−1 , V m−1 , tm−1 ⎦   2 (α + 3) β−1 m=2 +βtm−2 T T m−2 , I m−2 , V m−2 , tm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎥ ⎢ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎦ ⎣ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 (t)α  (α + 1) k   β−1   × βtm−2 I T m−2 , I m−2 , V m−2 , tm−2 (k − m + 1)α − (k − m)α

I k+1 = I0 +

m=2

Application to a system of ordinary differential equations

363

   k β−1  (t)α  βtm−1 I T m−1 , I m−1 , V m−1 , tm−1  + β−1   (α + 2) −βtm−2 I T m−2 , I m−2 , V m−2 , tm−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 βtm I(T m , I m , V m , tm ) k α   ⎥ (t) ⎢ β−1  + ⎣ −2βtm−1 I T m−1 , I m−1 , V m−1 , tm−1 ⎦   2 (α + 3) β−1 m=2 +βtm−2 I T m−2 , I m−2 , V m−2 , tm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎥ ⎢ (k − m + 1) +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎦ ⎣ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

(t)α  (α + 1) k   β−1    T m−2 , I m−2 , V m−2 , tm−2 (k − m + 1)α − (k − m)α × βtm−2 V

V k+1 = V0 +

m=2

   k β−1   m−1 m−1 (t)α  ,I , V m−1 , tm−1 T βtm−1 V  + β−1   m−2 m−2  (α + 2) −βtm−2 V ,I , V m−2 , tm−2 T m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1  βtm V (T m , I m , V m , tm ) k α    (t) ⎢ ⎥ β−1  + T m−1 , I m−1 , V m−1 , tm−1 ⎦ ⎣ −2βtm−1 V   2 (α + 3) β−1  m=2 +βtm−2 V T m−2 , I m−2 , V m−2 , tm−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎥ ⎢ (k − m + 1) +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎦ ⎣ 2 − m) + + 10) − m) (k (5α (k − (k − m)α +6α 2 + 18α + 12  by their values If we replace the functions T, I, V

T

k+1

    m−2 m−2 k (t)α  β−1 p − rT m−2 + s 1 − T T +I max = T0 + βtm−2  (α + 1) −kV m−2 T m−2 m=2   × (k − m + 1)α − (k − m)α (19.37)

364

New Numerical Scheme With Newton Polynomial



⎡ k ⎢ (t)α  ⎢ ⎢ + ⎢  (α + 2) m=2 ⎣



β−1 βtm−1 β−1

−βtm−2



 p − rT m−1 + s 1 −

T m−1 +I m−1 Tmax −kV m−1 T m−1

 p − rT m−2 + s 1 − 

  ⎤

⎥ ⎥ ⎥   ⎥ T m−2 +I m−2 ⎦ Tmax

−kV m−2 T m−2

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α)     ⎡ m +I m p − rT m + s 1 − T Tmax β−1 βtm ⎢ ⎢ −kV mT m ⎢    m−1 +I m−1 k (t)α  ⎢ p − rT m−1 + s 1 − T Tmax ⎢ β−1 + ⎢ −2βtm−1 m−1 ⎢ 2 (α + 3) −kV m−1 m=2 ⎢  T m−2 m−2    ⎢ +I p − rT m−2 + s 1 − T Tmax ⎣ β−1 +βtm−2 −kV m−2 T m−2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12 ×

I

k+1

k  (t)α  β−1  m−2 m−2 = I0 + βtm−2 kV T − βI m−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α    k β−1  (t)α  βtm−1 kV m−1 T m−1 − βI m−1  + β−1   (α + 2) −βtm−2 kV m−2 T m−2 − βI m−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ β−1 βtm (kV m T m − βI m ) k α ⎢  (t) β−1  + ⎣ −2βtm−1 kV m−1 T m−1 − βI m−1  2 (α + 3) β−1  m=2 +βtm−2 kV m−2 T m−2 − βI m−2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

k  (t)α  β−1  βtm−2 NβI m−2 − γ V m−2  (α + 1) m=2   × (k − m + 1)α − (k − m)α

V k+1 = V0 +

⎤ ⎥ ⎦ ⎤ ⎥ ⎥, ⎦

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Application to a system of ordinary differential equations

   k β−1  (t)α  βtm−1 NβI m−1 − γ V m−1  + β−1   (α + 2) −βtm−2 NβI m−2 − γ V m−2 m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎤ β−1 βtm (NβI m − γ V m ) k α   ⎥ (t) ⎢ β−1  + ⎣ −2βtm−1 NβI m−1 − γ V m−1 ⎦   2 (α + 3) β−1 m=2 +βtm−2 NβI m−2 − γ V m−2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

365

⎤ ⎥ ⎥. ⎦

19.1.5 Ebola model with fractal–fractional derivative with variable order with the exponential law We deal with the Ebola model having a fractal–fractional derivative with the exponential decay kernel with constant fractional order and variable fractal dimension, F F E α,β(t) Dt S (t) = −iSI + sR − γ N, 0 α,β(t) FFE Dt I (t) = −iSI − dI − rI, 0 F F E α,β(t) Dt R (t) = rI − sR, 0 F F E α,β(t) Dt D (t) = dI + γ N, 0

(19.38)

with initial conditions S (0) = S0 I (0) = I0 , R (0) = R0 , D (0) = D0 . We can convert Eq. (19.38) as follows:   β (t)  1 − α β(t) β (t) ln (t) + t S (t) = S (S, I, R, D, t) M (α) t    t β (s) β(s) α  s ds, S (S, I, R, D, s) β (s) ln (s) + + M (α) 0 s   1 − α β(t) β (t)  β (t) ln (t) + t I (S, I, R, D, t) M (α) t    t β (s) β(s) α s ds, I(S, I, R, D, s) β (s) ln (s) + + M (α) 0 s

I (t) =

R (t) =

  β (t)  1 − α β(t) β (t) ln (t) + t R (S, I, R, D, t) M (α) t

(19.39)

(19.40)

366

New Numerical Scheme With Newton Polynomial

+

α M (α)

 0

t

  (S, I, R, D, s) β (s) ln (s) + β (s) s β(s) ds, R s

  1 − α β(t) β (t)  t β (t) ln (t) + D (S, I, R, D, t) M (α) t    t α  (S, I, R, D, s) β (s) ln (s) + β (s) s β(s) ds. D + M (α) 0 s

D (t) =

We write at the point tk+1 = (k + 1) t   β (tk )  k k k k  1 − α β(tk ) β (tk+1 ) − β (tk ) t ln tk + S (tk+1 ) = S S , I , R , D , tk M (α) k t tk (19.41)    tk+1 β (s) β(s) α  s ds, + S (S, I, R, D, s) β (s) ln (s) + M (α) 0 s   β (tk )  k k k k  1 − α β(tk ) β (tk+1 ) − β (tk ) I (tk+1 ) = t ln tk + I S , I , R , D , tk M (α) k t tk    tk+1 β (s) β(s) α + s ds, I(S, I, R, D, s) β (s) ln (s) + M (α) 0 s   β (tk )  k k k k  1 − α β(tk ) β (tk+1 ) − β (tk ) tk ln tk + R S , I , R , D , tk M (α) t tk    tk+1 α (S, I, R, D, s) β (s) ln (s) + β (s) s β(s) ds, + R M (α) 0 s

R (tk+1 ) =

  β (tk )   k k k k  1 − α β(tk ) β (tk+1 ) − β (tk ) tk ln tk + D S , I , R , D , tk M (α) t tk    tk+1 α  (S, I, R, D, s) β (s) ln (s) + β (s) s β(s) ds, + D M (α) 0 s

D (tk+1 ) =

and at the point tk = kt, we have    β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) t ln tk−1 + S (tk ) = M (α) k−1 t tk−1   × S S k−1 , I k−1 , R k−1 , D k−1 , tk−1    tk α β (s) β(s)  + ds, S (S, I, R, D, s) β (s) ln (s) + s M (α) 0 s

(19.42)

Application to a system of ordinary differential equations

367

   β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + M (α) t tk−1   k−1 k−1 k−1 k−1 × I S , I , R , D , tk−1    tk β (s) β(s) α  s ds, + I (S, I, R, D, s) β (s) ln (s) + M (α) 0 s

I (tk ) =

   β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) t R (tk ) = ln tk−1 + M (α) k−1 t tk−1    S k−1 , I k−1 , R k−1 , D k−1 , tk−1 ×R    tk α (S, I, R, D, s) β (s) ln (s) + β (s) s β(s) ds, + R M (α) 0 s    β (tk−1 ) 1 − α β tk−1 β (tk ) − β (tk−1 ) tk−1 ln tk−1 + M (α) t tk−1   k−1 k−1 k−1 k−1  S ,I ×D , R , D , tk−1    tk β (s) β(s) α  s + ds. D (S, I, R, D, s) β (s) ln (s) + M (α) 0 s

D (tk ) =

Taking the difference of these equations, we have the following:    β tk+1 −β(tk ) ln tk + β(ttkk ) t   × S S k , Ik , Rk , D k , tk      β tk−1 β(tk )−β tk−1 β tk−1 −tk−1 ln tk−1 + tk−1 t   × S S k−1 , I k−1 , R k−1 , D k−1 , tk−1

⎡

β(tk )

⎢ 1−α ⎢ ⎢ S (tk+1 ) − S (tk ) = M (α) ⎢ ⎣

+

α M (α)



tk+1

tk

β(tk )

tk+1 tk

⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 

(19.43)

β (s) β(s)  s ds, S (S, I, R, D, s) β (s) ln (s) + s

tk

⎢ 1−α ⎢ ⎢ I (tk+1 ) − I (tk ) = M (α) ⎢ ⎣ 



   β tk+1 −β(tk ) ln tk + β(ttkk ) t   ×I S k , Ik , Rk , D k , tk      β tk−1 β(tk )−β tk−1 β tk−1 −tk−1 ln tk−1 + tk−1 t   ×I S k−1 , I k−1 , R k−1 , D k−1 , tk−1

⎡

α + M (α)



tk



⎤ ⎥ ⎥ ⎥ ⎥ ⎦

  β (s) β(s)  s ds, I (S, I, R, D, s) β (s) ln (s) + s

368

New Numerical Scheme With Newton Polynomial

⎡ ⎢ 1−α ⎢ ⎢ R (tk+1 ) − R (tk ) = M (α) ⎢ ⎣

   β tk+1 −β(tk ) ln tk + β(ttkk ) t    S k , I k , R k , D k , tk ×R       β tk−1 β(tk )−β tk−1 β tk−1 −tk−1 ln t + k−1 t tk−1   S k−1 , I k−1 , R k−1 , D k−1 , tk−1 ×R β(tk )



tk

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

  β (s) β(s)  s ds, R (S, I, R, D, s) β (s) ln (s) + s tk     ⎡ ⎤ β(t ) β tk+1 −β(tk ) β(tk ) tk k ln t + k ⎢ ⎥  t tk ⎥  S k , I k , R k , D k , tk 1−α ⎢ × D ⎢     ⎥   D (tk+1 ) − D (tk ) = ⎢ ⎥ β t β(t β t t )−β k k−1 k−1 M (α) ⎣ −t k−1 ⎦ ln tk−1 + tk−1 k−1 t  k−1 k−1 k−1 k−1   ×D S , I , R , D , tk−1    tk+1 α  (S, I, R, D, s) β (s) ln (s) + β (s) s β(s) ds. + D M (α) tk s α + M (α)



tk+1

For simplicity, we consider   β (t) β(t) t , U (S, I, R, D, t) =  S (S, I, R, D, t) β (t) ln (t) + t   β (t) β(t) t , V (S, I, R, D, t) = I(S, I, R, D, t) β (t) ln (t) + t   (S, I, R, D, t) β (t) ln (t) + β (t) t β(t) , W (S, I, R, D, t) = R t   β  (S, I, R, D, t) β (t) ln (t) + (t) t β(t) , Z (S, I, R, D, t) = D t

(19.44)

and we have

    1−α U S k , I k , R k , D k , tk  k−1  M (α) −U S , I k−1 , R k−1 , D k−1 , tk−1  tk+1 α + U (S, I, R, D, s) ds, M (α) tk     1−α V S k , I k , R k , D k , tk  k−1  I (tk+1 ) − I (tk ) = M (α) −V S , I k−1 , R k−1 , D k−1 , tk−1  tk+1 α + V (S, I, R, D, s) ds, M (α) tk

S (tk+1 ) − S (tk ) =

 k k k k    1−α , I , R , D , t W S k   R (tk+1 ) − R (tk ) = M (α) −W S k−1 , I k−1 , R k−1 , D k−1 , tk−1  tk+1 α + W (S, I, R, D, s) ds, M (α) tk

(19.45)

Application to a system of ordinary differential equations

369

    1−α Z S k , I k , R k , D k , tk  k−1  M (α) −Z S , I k−1 , R k−1 , D k−1 , tk−1  tk+1 α + Z (S, I, R, D, s) ds, M (α) tk

D (tk+1 ) − D (tk ) =

and put this polynomial into Eq. (19.45), so we have 1−α M (α) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  tk+1 ⎪ ⎨



S k+1 − S k =

+

α M (α)

I k+1 − I k =

α + M (α)



1−α M (α) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨



⎡ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ +⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

1−α M (α) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  tk+1 ⎪ ⎨



R k+1 − R k =

+

α M (α)

tk

   V S k , I k , R k , D k , tk  k−1  −V S , I k−1 , R k−1 , D k−1 , tk−1   V S k−2 , I k−2 , R k−2 , D k−2 , tk−2    k−1 k−1 k−1 k−1

V S ,I ,R ,D ,tk−1   k−2 k−2t k−2 k−2 V S ,I ,R ,D ,tk−2 − t × (s − tk−2 )  k−1   k k k k  V S ,I ,R ,D ,tk −2V S ,I k−1 ,R k−1 ,D k−1 ,tk−1 2(t)2   V S k−2 ,I k−2 ,R k−2 ,D k−2 ,tk−2 + 2 2(t) × (s − tk−2 ) (s − tk−1 )

+

tk+1

tk

U S ,I ,R ,D ,tk−1   k−2 k−2t k−2 k−2 U S ,I ,R ,D ,tk−2 − t × (s − tk−2 )  k−1   k k k k  U S ,I ,R ,D ,tk −2U S ,I k−1 ,R k−1 ,D k−1 ,tk−1 2(t)2   U S k−2 ,I k−2 ,R k−2 ,D k−2 ,tk−2 + 2 2(t) × (s − tk−2 ) (s − tk−1 )

+

⎡ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ +⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

tk

   U S k , I k , R k , D k , tk  k−1  −U S , I k−1 , R k−1 , D k−1 , tk−1   U S k−2 , I k−2 , R k−2 , D k−2 , tk−2    k−1 k−1 k−1 k−1

⎡ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ +⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

 k k k k   S , I , R , D , tk  W  −W S k−1 , I k−1 , R k−1 , D k−1 , tk−1   W S k−2 , I k−2 , R k−2 , D k−2 , tk−2    k−1 k−1 k−1 k−1

W S ,I ,R ,D ,tk−1   k−2 k−2t k−2 k−2 W S ,I ,R ,D ,tk−2 − t × (s − tk−2 )   k k k k  W S ,I ,R ,D ,tk −2W S k−1 ,I k−1 ,R k−1 ,D k−1 ,tk−1 2(t)2   W S k−2 ,I k−2 ,R k−2 ,D k−2 ,tk−2 + 2 2(t) × (s − tk−2 ) (s − tk−1 )

+

(19.46) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤ ⎪ ds, ⎪ ⎪ ⎪ ⎦ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤ ⎪ ds, ⎪ ⎪ ⎪ ⎦ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤ ⎪ ds, ⎪ ⎪ ⎪ ⎦ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

370

New Numerical Scheme With Newton Polynomial

D k+1 − D k =

α + M (α)



tk+1

tk



1−α M (α) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

   Z S k , I k , R k , D k , tk  k−1  −Z S , I k−1 , R k−1 , D k−1 , tk−1   Z S k−2 , I k−2 , R k−2 , D k−2 , tk−2    k−1 k−1 k−1 k−1

Z S ,I ,R ,D ,tk−1   k−2 k−2 tk−2 k−2 Z S ,I ,R ,D ,tk−2 − t × (s − tk−2 )  k−1   k k k k  Z S ,I ,R ,D ,tk −2Z S ,I k−1 ,R k−1 ,D k−1 ,tk−1 2(t)2   Z S k−2 ,I k−2 ,R k−2 ,D k−2 ,tk−2 + 2 2(t) × (s − tk−2 ) (s − tk−1 )

+

⎡ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ +⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤ ⎪ ds. ⎪ ⎪ ⎪ ⎦ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Let us organize the above equation such that     1−α U S k , I k , R k , D k , tk  k−1  (19.47) M (α) −U S , I k−1 , R k−1 , D k−1 , tk−1   ⎧ ⎫ U S k−2 , I k−2 , R k−2 , D k−2 , tk−2 ⎪ ⎪  ⎪ ⎪   ⎪ ⎪ U S k−1 ,I k−1 ,R k−1 ,D k−1 ,tk−1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ t ⎪ ⎪   + ⎪ ⎪ k−2 k−2 k−2 k−2 ⎪ ⎪ U S ,I ,R ,D ,tk−2 ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ t ⎨ ⎬ tk+1 α × − t ds (s ) k−2 t , + k  ⎤ ⎡  k k k k  ⎪ M (α) ⎪ U S ,I ,R ,D ,tk −2U S k−1 ,I k−1 ,R k−1 ,D k−1 ,tk−1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2(t)2 ⎦ ⎪   k−2 k−2 ⎪ ⎪ +⎣ k−2 ,D k−2 ,t ⎪ ⎪ U S ,I ,R ⎪ ⎪ k−2 ⎪ ⎪ + ⎪ ⎪ 2 ⎪ ⎪ 2(t) ⎪ ⎪ ⎩ ⎭ tk+1 × tk (s − tk−2 ) (s − tk−1 ) ds

S k+1 − S k =

   α V S k , I k , R k , D k , tk   I × + −V S k−1 , I k−1 , R k−1 , D k−1 , tk−1 M (α)   ⎧ ⎫ V S k−2 , I k−2 , R k−2, D k−2 , tk−2 ⎪ ⎪   k−1 k−1 k−1 k−1 ⎪ ⎪ k−2 k−2 k−2 k−2 ⎪ ⎪ V S ,I ,R ,D ,tk−1 −V S ,I ,R ,D ,tk−2 ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ t ⎨ ⎬ tk+1 × tk (s − tk−2 ) ds      ,  ⎪ V S k ,I k ,R k ,D k ,tk −2V S k−1 ,I k−1 ,R k−1 ,D k−1 ,tk−1 +V S k−2 ,I k−2 ,R k−2 ,D k−2 ,tk−2 ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ ⎪ ⎪ 2(t)2 ⎪ ⎪ ⎩ ⎭ tk+1 × tk (s − tk−2 ) (s − tk−1 ) ds k+1

1−α −I = M (α)



k

R k+1 − R k =

1−α M (α)



 k k k k   S , I , R , D , tk  W  −W S k−1 , I k−1 , R k−1 , D k−1 , tk−1

Application to a system of ordinary differential equations

+

D

k+1

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

α ⎡ M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ +⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

  W S k−2 , I k−2 , R k−2 , D k−2 , tk−2    k−1 k−1 k−1 k−1

W S ,I ,R ,D ,tk−1   k−2 k−2t k−2 k−2 W S ,I ,R ,D ,tk−2 − t t × k+1 (s − tk−2 ) ds  k−1 k−1 k−1 k−1   k k k k tk  W S ,I ,R ,D ,tk −2W S ,I ,R ,D ,tk−1 2(t)2   W S k−2 ,I k−2 ,R k−2 ,D k−2 ,tk−2 + 2(t)2 t × tkk+1 (s − tk−2 ) (s − tk−1 ) ds

+

371

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎤⎪ , ⎪ ⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

 k k k k    1−α , I , R , D , t Z S k   −D = M (α) −Z S k−1 , I k−1 , R k−1 , D k−1 , tk−1   ⎧ ⎫ Z S k−2 , I k−2 , R k−2 , D k−2 , tk−2 ⎪ ⎪  ⎪ ⎪   ⎪ ⎪ Z S k−1 ,I k−1 ,R k−1 ,D k−1 ,tk−1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   k−2 k−2 tk−2 k−2 + ⎪ ⎪ ⎪ ⎪ Z S ,I ,R ,D ,t k−2 ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ t ⎨ ⎬ tk+1 α × − t ds (s ) k−2 t . + k     ⎪ ⎡ Z S k ,I k ,R k ,D k ,tk −2Z S k−1 ,I k−1 ,R k−1 ,D k−1 ,tk−1 ⎤⎪ M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪+ ⎣ ⎪ 2(t)2 ⎦⎪   k−2 k−2 ⎪ ⎪ k−2 ,D k−2 ,t ⎪ ⎪ Z S ,I ,R ⎪ ⎪ k−2 ⎪ ⎪ + ⎪ ⎪ 2 ⎪ ⎪ 2(t) ⎪ ⎪ ⎩ ⎭ tk+1 × tk (s − tk−2 ) (s − tk−1 ) ds k

We can reorganize the above scheme as follows:     1−α U S k , I k , R k , D k , tk  k−1  S k+1 − S k = (19.48) M (α) −U S , I k−1 , R k−1 , D k−1 , tk−1  k k k k  ⎧ ⎫ 23 ⎬ 12 U S , I , R , D , tk t   α ⎨ + , − 43 U S k−1 , I k−1 , R k−1 , D k−1 , tk−1 t ⎭ M (α) ⎩ 5 U S k−2 , I k−2 , R k−2 , D k−2 , tk−2 t + 12

I

 k k k k   , I , R , D , tk V  k−1 S k−1  −V S , I , R k−1 , D k−1 , tk−1   ⎧ ⎫ 23 k k k k ⎬ 12 V S , I , R , D , tk t   α ⎨ + , − 43 V S k−1 , I k−1 , R k−1 , D k−1 , tk−1 t ⎭ M (α) ⎩ 5 + 12 V S k−2 , I k−2 , R k−2 , D k−2 , tk−2 t 

k+1

1−α −I = M (α)

k+1

1−α −R = M (α)

R

k

k

 k k k k   S , I , R , D , tk  W  −W S k−1 , I k−1 , R k−1 , D k−1 , tk−1  k k k k  ⎧ ⎫ 23 t W S , I , R , D , t ⎨ ⎬ k 12   α + , − 43 W S k−1 , I k−1 , R k−1 , D k−1 , tk−1 t ⎭ M (α) ⎩ 5 + 12 W S k−2 , I k−2 , R k−2 , D k−2 , tk−2 t 

372

New Numerical Scheme With Newton Polynomial

   Z S k , I k , R k , D k , tk  k−1  −Z S , I k−1 , R k−1 , D k−1 , tk−1  k k k k  ⎧ ⎫ 23 ⎬ 12 Z S , I , R , D , tk t   α ⎨ + . − 43 Z S k−1 , I k−1 , R k−1 , D k−1 , tk−1 t ⎭ M (α) ⎩ 5 Z S k−2 , I k−2 , R k−2 , D k−2 , tk−2 t + 12

D k+1 − D k =

1−α M (α)



Thus, we have the following scheme:    ⎤ β tk+1 −β(tk ) ln tk + β(ttkk ) t ⎢ ⎥   1−α ⎢ × S S k , Ik , Rk , D k , tk   ⎥ k+1 k ⎢     ⎥ =S + S ⎥ β tk−1 β(tk )−β tk−1 β tk−1 M (α) ⎢ ⎣ −tk−1 ⎦ ln t + k−1 t tk−1  k−1 k−1 k−1 k−1 × S S ,I , R , D , tk−1     ⎧ ⎫ β(tk ) 23 β(tk ) β tk+1 −β(tk ) ⎪ ⎪ t ln t + k ⎪ ⎪ 12 k t t k ⎪ ⎪  k k k k  ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ × S S , I , R , D , t t k ⎪ ⎪       ⎪ ⎪   ⎪ ⎪ β t β(t β t t )−β ⎨ ⎬ k−1 k k−1 k−1 4 α − 3 tk−1 ln t + k−1 t t k−1 , +  k−1 k−1 k−1 k−1  M (α) ⎪ × S  S  , I , R  ,D , tk−1 t  ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ β tk−2 ⎪ ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ ⎪ + t ln t + k−2 ⎪ ⎪ 12 k−2 t t ⎪ ⎪ k−2  k−2 k−2 k−2 k−2  ⎩ ⎭  ×S S , I , R , D , tk−2 t

⎡

β(tk )



tk

   ⎤ β tk+1 −β(tk ) ln tk + β(ttkk ) t ⎢ ⎥   1−α ⎢ ×I S k , Ik , Rk , D k , tk   ⎥ k+1 k ⎢ ⎥     I =I + ⎥ β tk−1 β(tk )−β tk−1 β tk−1 M (α) ⎢ ⎣ −tk−1 ⎦ ln t + k−1 t tk−1  k−1 k−1 k−1 k−1  ×I S , I , R , D , tk−1     ⎧ ⎫ −β(t β t β(tk ) k) k+1 23 β(tk ) ⎪ ⎪ t ln t + k ⎪ ⎪ 12 k ⎪ ⎪  k kt k k  tk ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ × I S t , I , R , D , t k ⎪ ⎪       ⎪ ⎪   ⎪ ⎪ β t β(t β t t )−β ⎨ ⎬ k−1 k k−1 k−1 4 α − 3 tk−1 ln t + k−1 t t k−1 , +   k−1 , R k−1 , D k−1 , t ⎪ M (α) ⎪ ×I S k−1 t , I ⎪ ⎪ k−1 ⎪ ⎪       ⎪  ⎪ ⎪ ⎪ β tk−2 ⎪ ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ ⎪ + t ln t + k−2 ⎪ ⎪ k−2 12  t tk−2 ⎪ ⎪  ⎩ ⎭ ×I S k−2 , I k−2 , R k−2 , D k−2 , tk−2 t

⎡

β(tk )



tk

⎡ R

k+1

⎢ 1−α ⎢ ⎢ =R + M (α) ⎢ ⎣ k

   β tk+1 −β(tk ) β(tk ) ln t + k  t tk  S k , I k , R k , D k , tk × R       β tk−1 β(tk )−β tk−1 β tk−1 −tk−1 ln t + k−1 t tk−1   S k−1 , I k−1 , R k−1 , D k−1 , tk−1 ×R β(tk )

tk



⎤ ⎥ ⎥ ⎥ ⎥ ⎦

(19.49)

Application to a system of ordinary differential equations

+

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

α M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

    23 β(tk ) β tk+1 −β(tk ) ln tk + β(ttkk ) 12 tk t    S k , I k , R k , D k , tk t ×R       β tk−1 β(tk )−β tk−1 β tk−1 − 43 tk−1 ln t + k−1 t t k−1   S k−1 , I k−1 , R k−1 , D k−1 , tk−1 t ×R         β tk−2 5 β tk−2 β tk−1 −β tk−2 + 12 tk−2 ln t + k−2 t  tk−2  S k−2 , I k−2 , R k−2 , D k−2 , tk−2 t ×R

373

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

,

   ⎤ β tk+1 −β(tk ) ln tk + β(ttkk ) t ⎢ ⎥   ⎥  S k , I k , R k , D k , tk 1−α ⎢ ×D k+1 k ⎢ ⎥         D =D + ⎢ ⎥ β t β(t β t t )−β k−1 k k−1 k−1 M (α) ⎣ −t ⎦ ln t + k−1 k−1 t tk−1  k−1 k−1 k−1 k−1  S ,I ×D , R , D , tk−1     ⎧ ⎫ β(tk ) 23 β(tk ) β tk+1 −β(tk ) ⎪ ⎪ t ln t + k ⎪ ⎪ 12 k ⎪ ⎪  k kt k k  tk ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ × D S t , I , R , D , t k ⎪ ⎪       ⎪ ⎪   ⎪ ⎪ β t β(t β t t )−β ⎨ ⎬ k−1 k k−1 k−1 4 α − 3 tk−1 ln t + k−1 t t k−1 . +   ⎪ M (α) ⎪  S k−1 , I k−1 , R k−1 , D k−1 , tk−1 t ×D ⎪ ⎪ ⎪ ⎪        ⎪  ⎪ ⎪ ⎪ β tk−2 ⎪ ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ ⎪ + t ln t + k−2 ⎪ ⎪ 12 k−2 t t ⎪ ⎪ k−2  k−2 k−2 k−2 k−2  ⎩ ⎭  ×D S , I , R , D , tk−2 t

⎡

β(tk )



tk

D  by its values, we get the numerical scheme Replacing the functions  S, I, R,     ⎡ ⎤ β(t ) β tk+1 −β(tk ) β(tk ) tk k ln t + k t ⎢ ⎥  tk k −γN ⎥ 1−α ⎢ × iS k I k + sR k+1 k ⎢     ⎥   S (19.50) =S + ⎢ ⎥ β t β(t β t t )−β k−1 k k−1 k−1 M (α) ⎣ −t ⎦ ln tk−1 + tk−1 k−1 t   × iS k−1 I k−1 + sR k−1 − γ N     ⎧ ⎫ β(tk ) 23 β(tk ) β tk+1 −β(tk ) ⎪ ⎪ t ln t + k ⎪ ⎪ k 12 t t ⎪ ⎪  k k  k ⎪ ⎪ ⎪ ⎪ k ⎪ ⎪ × iS I + sR − γ N t ⎪ ⎪       ⎪ ⎪   ⎪ ⎪ β tk−1 β(tk )−β tk−1 β tk−1 ⎨ ⎬ 4 α − 3 tk−1 ln t + k−1 t t k−1 , +  k−1 k−1  M (α) ⎪ × iS I  + sR k−1 − γ N t  ⎪ ⎪ ⎪ ⎪ ⎪    ⎪ ⎪ ⎪ ⎪ β tk−2 ⎪ ⎪ 5 β tk−2 β tk−1 −β tk−2 ⎪ ⎪ + 12 tk−2 ln tk−2 + tk−2 ⎪ ⎪ t ⎪ ⎪  k−2 k−2  ⎩ ⎭ k−2 × iS I + sR − γ N t ⎡ I

k+1

⎢ 1−α ⎢ ⎢ =I + M (α) ⎢ ⎣ k

   β tk+1 −β(tk ) β(tk ) ln t + k  k t tk k − dI k − rI k × iS I       β tk−1 β(tk )−β tk−1 β tk−1 −tk−1 ln t + k−1 t tk−1   × iS k−1 I k−1 − dI k−1 − rI k−1 β(tk )

tk



⎤ ⎥ ⎥ ⎥ ⎥ ⎦

374

New Numerical Scheme With Newton Polynomial

+

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

α M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

   β tk+1 −β(tk ) ln tk + β(ttkk ) t  ×  iS k I k −dI k− rI k t      β tk−1 β(tk )−β tk−1 β tk−1 − 43 tk−1 ln tk−1 + tk−1 t   k−1 − rI k−1 t ×  iS k−1 I k−1 − dI       β tk−2 5 β tk−2 β tk−1 −β tk−2 + 12 tk−2 ln t + k−2 t   tk−2 × iS k−2 I k−2 − dI k−2 − rI k−2 t 23 β(tk ) 12 tk 

⎡ R

k+1

⎢ 1−α ⎢ ⎢ =R + M (α) ⎢ ⎣ k

+

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

α M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

D

+

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

α M (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

   β tk+1 −β(tk ) ln tk + β(ttkk ) t   × rI k − sR k     β tk−1 β(tk )−β tk−1 β tk−1 −tk−1 ln t + k−1 tk−1  t  × rI k−1 − sR k−1 β(tk )



   β tk+1 −β(tk ) β(tk ) ln t + k tk  kt  k t × rI − sR       β tk−1 β(tk )−β tk−1 β tk−1 − 43 tk−1 ln t + k−1 t tk−1  k−1  k−1 t × rI − sR         β tk−2 5 β tk−2 β tk−1 −β tk−2 + 12 tk−2 ln t + k−2 t tk−2   × rI k−2 − sR k−2 t

⎢ 1−α ⎢ ⎢ =D + M (α) ⎢ ⎣ k

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬





⎥ ⎥ ⎥ ⎥ ⎦ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

   β tk+1 −β(tk ) β(tk ) ln t + k tk t k  × dI + γ N       β tk−1 β(tk )−β tk−1 β tk−1 −tk−1 ln tk−1 + tk−1 t   × dI k−1 + γ N β(tk )

tk

   β tk+1 −β(tk ) ln tk + β(ttkk ) t   × dI k + γN t     β tk−1 β(tk )−β tk−1 β tk−1 − 43 tk−1 ln t + k−1 t tk−1  k−1  × dI + γ N t         β tk−2 5 β tk−2 β tk−1 −β tk−2 + 12 tk−2 ln t + k−2 tk−2  t  × dI k−2 + γ N t 23 β(tk ) 12 tk



,

⎤

tk

23 β(tk ) 12 tk

⎡ k+1



,

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

Application to system of non-linear partial differential equations

20

In this chapter, we construct an algorithm to solve the system of non-linear partial differential equations numerically.

20.1

System of non-linear partial differential equations

In this section, we deal with the Burgers equation which appears in many fields of applied mathematics; for instance, fluid mechanics, gas dynamics, non-linear acoustics and turbulent convection.

20.1.1 System of non-linear partial differential equations with the classical derivative We consider the following non-linear Burgers equation: ut = −γ uxx − ηux u − κ (uvx + vux ) , vt = −ρvxx − μvx v − ξ (uvx + vux ) ,

(20.1)

where the derivative is the classical one. Also we can take H1 (u, v, x, t) = −γ uxx − ηux u − κ (uvx + vux ), H2 (u, v, x, t) = −ρvxx − μvx v − ξ (uvx + vux ). Here the initial condition can be taken as u (x, 0) = g1 (x), u (x, t) |∂ = f1 (t), v (x, 0) = g2 (x), v (x, t) |∂ = f2 (t). We convert Eq. (20.1) into  t H1 (u, v, x, τ ) dτ, (20.2) u (x, t) − u (x, 0) = 0  t v (x, t) − v (x, 0) = H2 (u, v, x, τ ) dτ. 0

At the point tk+1 = (k + 1) t  tk+1 H1 (u, v, x, τ ) dτ, u (x, tk+1 ) − u (x, 0) = 0  tk+1 H2 (u, v, x, τ ) dτ, v (x, tk+1 ) − v (x, 0) = 0 New Numerical Scheme With Newton Polynomial. https://doi.org/10.1016/B978-0-32-385448-1.00026-3 Copyright © 2021 Elsevier Inc. All rights reserved.

(20.3)

376

New Numerical Scheme With Newton Polynomial

and at the point tk = k t, we have  tk u (x, tk ) − u (x, 0) = H1 (u, v, x, τ ) dτ, 0  tk v (x, tk ) − v (x, 0) = H2 (u, v, x, τ ) dτ.

(20.4)

0

If we take the difference of Eqs. (20.3) and (20.4), we can get the following:  tk+1 u (x, tk+1 ) − u (x, tk ) = H1 (u, v, x, τ ) dτ, u (x, tk+1 ) − u (x, tk ) =

(20.5)

tk  tk+1

H2 (u, v, x, τ ) dτ. tk

Let us arrange the above equation, then we have   (20.6) uk+1 − uk = H1 uk−2 , v k−2 , x, tk−2 t  k−1 k−1   k−2 k−2   tk+1 H1 u , v , x, tk−1 − H1 u , v , x, tk−2 + (τ − tk−2 ) dτ,

t tk       tk+1  k k H1 u , v , x, tk −2H1 uk−1 , v k−1 , x, tk−1 +H1 uk−2 , v k−2 , x, tk−2 + 2 ( t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ,   v k+1 − v k = H2 uk−2 , v k−2 , x, tk−2 t      tk+1 H2 uk−1 , v k−1 , x, tk−1 − H2 uk−2 , v k−2 , x, tk−2 + (τ − tk−2 ) dτ

t tk       tk+1  k k H2 u , v , x, tk −2H2 uk−1 , v k−1 , x, tk−1 +H2 uk−2 , v k−2 , x, tk−2 + 2 ( t)2 tk × (τ − tk−2 ) (τ − tk−1 ) dτ, and we obtain

  (20.7) uk+1 − uk = H1 uk−2 , v k−2 , x, tk−2 t  k−1 k−1   k−2 k−2  t k+1 H1 u , v , x, tk−1 − H1 u , v , x, tk−2 + (τ − tk−2 ) dτ,

t tk       H1 uk , v k , x, tk − 2H1 uk−1 , v k−1 , x, tk−1 + H1 uk−2 , v k−2 , x, tk−2 + 2 ( t)2  tk+1 × (τ − tk−2 ) (τ − tk−1 ) dτ, tk

Application to system of non-linear partial differential equations

377

  v k+1 − v k = H2 uk−2 , v k−2 , x, tk−2 t     t k+1 H2 uk−1 , v k−1 , x, tk−1 − H2 uk−2 , v k−2 , x, tk−2 + (τ − tk−2 ) dτ

t tk       tk+1  k k H2 u , v , x, tk −2H2 uk−1 , v k−1 , x, tk−1 +H2 uk−2 , v k−2 , x, tk−2 + 2 ( t)2 tk  tk+1 × (τ − tk−2 ) (τ − tk−1 ) dτ. tk

If we do the same routine, we obtain the following:   uk+1 = uk + H1 uk−2 , v k−2 , x, tk−2 t     H1 uk−1 , v k−1 , x, tk−1  5

t + −H1 uk−2 , v k−2 , x, tk−2 2       23 H1 uk , v k , x, tk − 2H1 uk−1 , v k−1 , x, tk−1  +

t, +H1 uk−2 , v k−2 , x, tk−2 12

(20.8)

  v k+1 = v k + H2 uk−2 , v k−2 , x, tk−2 t     H2 uk−1 , v k−1 , x, tk−1  5

t + −H2 uk−2 , v k−2 , x, tk−2 2       23 H2 uk , v k , x, tk − 2H2 uk−1 , v k−1 , x, tk−1  +

t, +H2 uk−2 , v k−2 , x, tk−2 12 and we get    4  5 H1 uk−2 , v k−2 , x, tk−2 t − H1 uk−1 , v k−1 , x, tk−1 t 12 3  23  k k + H1 u , v , x, tk t, (20.9) 12     4 5 v k+1 = v k + H2 uk−2 , v k−2 , x, tk−2 t − H2 uk−1 , v k−1 , x, tk−1 t 12 3  23  k k + H2 u , v , x, tk t. 12

uk+1 = uk +

Thus, our equation can be solved numerically as follows: ⎧ k−2 uk−2 −2uk−2 −uk−2 uk−2 n+1 −un−1 ⎪ ⎨ −γ n+1 n 2 n−1 − ηuk−2 n 2 x 5 ( x) 

uk+1 = uk + k−2 k−2 k−2 k−2 vn+1 −vn−1 k−2 un+1 −un−1 12 ⎪ ⎩ −κ uk−2 + v n n 2 x 2 x

⎫ ⎪ ⎬ ⎪ ⎭

t

378

New Numerical Scheme With Newton Polynomial

⎧ k−1 k−1 k−1 ⎫ k−1 uk−1 n+1 −2un −un−1 k−1 un+1 −un−1 ⎪ ⎪ ⎬ ⎨ − ηu −γ 2 n 2 x  4 ( x)

t − k−1 k−1 k−1 k−1 −u vn+1 −vn−1 u 3⎪ ⎭ ⎩ −κ uk−1 + vnk−1 n+12 x n−1 ⎪ n 2 x ⎫ ⎧ k k ukn+1 −2ukn −ukn−1 k un+1 −un−1 ⎪ ⎪ ⎬ ⎨ − ηu −γ 2 n 2 x  23

( x) +

t, k −v k k −uk v u 12 ⎪ ⎭ ⎩ −κ ukn n+12 xn−1 + vnk n+12 x n−1 ⎪ ⎧ v k−2 −2v k−2 −v k−2 v k−2 −v k−2 ⎪ ⎨ −ρ n+1 n 2 n−1 − μvnk−2 n+12 xn−1 5 ( x) 

v k+1 = v k + k−2 k−2 k−2 k−2 ⎪ −ξ uk−2 vn+1 −vn−1 + v k−2 un+1 −un−1 12 ⎩ n

2 x

n

2 x

(20.10)

⎫ ⎪ ⎬ ⎪ ⎭

t

⎧ k−1 k−1 k−1 ⎫ k−1 uk−1 n+1 −2un −un−1 k−1 vn+1 −vn−1 ⎪ ⎪ ⎬ ⎨ − μv −ρ n 2 x  4 ( x)2

t − k−1 k−1 k−1 k−1 vn+1 −vn−1 u −u 3⎪ ⎭ ⎩ −ξ uk−1 + vnk−1 n+12 x n−1 ⎪ n 2 x ⎫ ⎧ k −2v k −v k k k vn+1 n n−1 k vn+1 −vn−1 ⎪ ⎪ ⎬ ⎨ − μv −ρ 2 n 2 x  23

( x) +

t. k k k k v u −v −u 12 ⎪ ⎭ ⎩ −ξ ukn n+12 xn−1 + vnk n+12 x n−1 ⎪

20.1.2 System of non-linear partial differential equations with the Atangana–Baleanu derivative We deal with the Burgers equation including the Atangana–Baleanu fractional operator, AB α 0 Dt u (x, t) = −γ uxx − ηux u − κ (uvx + vux ) , AB α 0 Dt v (x, t) = −ρvxx − μvx v − ξ (uvx + vux ) ,

(20.11)

where we can take H1 (u, v, x, t) = −γ uxx − ηux u − κ (uvx + vux ), H2 (u, v, x, t) = −ρvxx − μvx v − ξ (uvx + vux ). Here, the initial condition can be taken as u (x, 0) = g1 (x) , u (x, t) |∂ = f1 (t) ,

(20.12)

v (x, 0) = g2 (x) , v (x, t) |∂ = f2 (t) . We shall write the above equation as follows: 1−α H1 (x, t, u, v) AB (α)  t α H1 (x, τ, u, v) (t − τ )α−1 dτ, + AB (α) (α) 0

u (x, t) − u (x, 0) =

(20.13)

Application to system of non-linear partial differential equations

379

1−α H2 (x, t, u, v) AB (α)  t α H2 (x, τ, u, v) (t − τ )α−1 dτ. + AB (α) (α) 0

v (x, t) − v (x, 0) =

At the point tk+1 = (k + 1) t, we have the following   1−α H1 x, tk , uk , v k , (20.14) AB (α)  tk+1 α + H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, AB (α) (α) 0   1−α H2 x, tk , uk , v k v (x, tk+1 ) − v (x, 0) = AB (α)  tk+1 α + H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ, AB (α) (α) 0

u (x, tk+1 ) − u (x, 0) =

and we write   1−α H1 x, tk , uk , v k (20.15) AB (α) k  tm+1  α H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, + AB (α) (α) tm

u (x, tk+1 ) = u (x, 0) +

m=2

  1−α H2 x, tk , uk , v k AB (α) k  tm+1  α H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ. + AB (α) (α) tm

v (x, tk+1 ) = v (x, 0) +

m=2

After putting the Newton polynomial into Eqs. (20.15), the above equation can be written as follows: k    α 1−α H1 x, tk , uk , v k + (20.16) AB (α) AB (α) (α) m=2 ⎫ ⎧   m−2 , v m−2 H x, t , u ⎪ ⎪ 1 m−2 ⎪ ⎪     ⎪ ⎪ ⎪  tm+1 ⎪ ⎬ ⎨ + H1 x,tm−1 ,um−1 ,v m−1 −H1 x,tm−2 ,um−2 ,v m−2 (τ − t ) m−2

t     × m m m−1 m−1 m−2 m−2 x,t +H x,t H ,u ,v ,u ,v ,u ,v (x,t )−2H m 1 1 m−1 1 m−2 ⎪ ⎪ + tm ⎪ ⎪ ⎪ ⎪ 2( t)2 ⎪ ⎪ ⎭ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 = u0 +

× (tk+1 − τ )α−1 dτ,

380

New Numerical Scheme With Newton Polynomial k    α 1−α H2 x, tk , uk , v k + AB (α) AB (α) (α) m=2 ⎧   m−2 m−2 H2 x, tm−2, u ,v ⎪ ⎪   ⎪  tm+1 ⎪ ⎨ + H2 x,tm−1 ,um−1 ,v m−1 −H2 x,tm−2 ,um−2 ,v m−2 (τ − t m−2 )   t   × H2 (x,tm ,um ,v m )−2H2 x,tm−1 ,um−1 ,v m−1 +H2 x,tm−2 ,um−2 ,v m−2 ⎪ + tm ⎪ ⎪ 2( t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

v k+1 = v0 +

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we have uk+1 = u0 +

×

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

⎪ t ⎪ ⎪ ⎪ + tmm+1 ⎪ ⎪ ⎩

v k+1 = v0 +

×

k    α 1−α H1 x, tk , uk , v k + AB (α) AB (α) (α) m=2  tm+1   α−1 m−2 m−2 H1 x, tm−2 , u ,v dτ (tk+1 − τ ) tm  tm+1 H1 x,tm−1 ,um−1 ,v m−1 −H1 x,tm−2 ,um−2 ,v m−2  + tm

t × (τ − tm−2 − τ )α−1 dτ ) (t k+1  

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

H1 (x,tm ,um ,v m )−2H1 x,tm−1 ,um−1 ,v m−1 +H1 x,tm−2 ,um−2 ,v m−2 2( t)2 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

k    α 1−α H2 x, tk , uk , v k + AB (α) AB (α) (α) m=2  tm+1   α−1 m−2 , v m−2 (t x, t H , u − τ dτ ) 2 m−2 k+1 tm  tm+1 H2 x,tm−1 ,um−1 ,v m−1 −H2 x,tm−2 ,um−2 ,v m−2  + tm

t × (τ − tm−2 dτ ) (tk+1 − τ )α−1  

⎪ t ⎪ ⎪ ⎪ + tmm+1 ⎪ ⎪ ⎩

H2 (x,tm ,um ,v m )−2H2 x,tm−1 ,um−1 ,v m−1 +H2 x,tm−2 ,um−2 ,v m−2 2( t)2 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

(20.17) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ 

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

,

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ 

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

Thus, we write the following:   1−α H1 x, tk , uk , v k , (20.18) AB (α) k    tm+1  α H1 x, tm−2 , um−2 , v m−2 + (tk+1 − τ )α−1 dτ AB (α) (α) t m m=2     k m−1 m−1  H1 x, tm−1 , u − H1 x, tm−2 , um−2 , v m−2 ,v α + AB (α) (α)

t

uk+1 = u0 +

 ×

m=2

tm+1

tm

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

k  α AB (α) (α)

m=2

Application to system of non-linear partial differential equations

×

381

    H1 (x, tm , um , v m ) − 2H1 x, tm−1 , um−1 , v m−1 + H1 x, tm−2 , um−2 , v m−2 

×

2 ( t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

tm

  1−α H2 x, tk , uk , v k AB (α) k    tm+1  α m−2 m−2 H2 x, tm−2 , u ,v + (tk+1 − τ )α−1 dτ AB (α) (α) t m m=2     k  H2 x, tm−1 , um−1 , v m−1 − H2 x, tm−2 , um−2 , v m−2 α + AB (α) (α)

t

v k+1 = v0 +

 × ×

k  α AB (α) (α) tm m=2     H2 (x, tm , um , v m ) − 2H2 x, tm−1 , um−1 , v m−1 + H2 x, tm−2 , um−2 , v m−2

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 ( t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

As we did before, we can have   1−α H1 x, tk , uk , v k , AB (α) k    α ( t)α + H1 x, tm−2 , um−2 , v m−2 AB (α) (α + 1) m=2   × (k − m + 1)α − (k − m)α    k   α ( t)α H1 x, tm−1 , um−1 , v m−1  + −H1 x, tm−2 , um−2 , v m−2 AB (α) (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H1 (x, tm , um , v m )   α ( t)α ⎣ −2H1 x, tm−1 , um−1 , v m−1 ⎦ +   2AB (α) (α + 3) +H1 x, tm−2 , um−2 , v m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 − m) + + 10) − m) (k (5α (k − (k − m)α 2 +6α + 18α + 12

uk+1 = u0 +

(20.19)

382

New Numerical Scheme With Newton Polynomial

  1−α H2 x, tk , uk , v k AB (α) k    α ( t)α H2 x, tm−2 , um−2 , v m−2 + AB (α) (α + 1) m=2   × (k − m + 1)α − (k − m)α    k   α ( t)α H2 x, tm−1 , um−1 , v m−1  + −H2 x, tm−2 , um−2 , v m−2 AB (α) (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H2 (x, tm , um , v m ) α   α ( t) ⎣ −2H2 x, tm−1 , um−1 , v m−1 ⎦ +   2AB (α) (α + 3) +H2 x, tm−2 , um−2 , v m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

v k+1 = v0 +

Putting all this in the above equation, at the point xn the following scheme can be achieved: ⎧ uk −uk uk −2uk −uk ⎪ ⎨ −γ n+1 n 2 n−1 − ηukn n+12 x n−1 1 − α 

( x) uk+1 = u0 + k −v k vn+1 ukn+1 −ukn−1 n−1 k k AB (α) ⎪ ⎩ −κ un 2 x + vn 2 x ⎧

+

α+1

α ( t) AB (α) (α + 1)

k ⎪ ⎨ 

−γ

⎪ −κ

m=2 ⎩

m−2 −um−2 um−2 n+1 −2un n−1

( x)2 m−2 m−2 vn+1 −vn−1 um−2 n 2 x

⎫ ⎪ ⎬ ⎪ ⎭

m−2 um−2 n+1 −un−1 2 x  um−2 −um−2 + vnm−2 n+12 x n−1

− ηum−2 n

  × (k − m + 1)α − (k − m)α α ( t)α AB (α) (α + 2) ⎡ um−1 −um−1 um−1 −2um−1 −um−1 n n−1 −γ n+1 − ηunm−1 n+12 x n−1 2 ( x) ⎢

 m−1 m−1 m−1 m−1 ⎢ ⎢ −κ um−1 vn+1 −vn−1 + v m−1 un+1 −un−1 k ⎢ n n 2 x 2 x ⎢ × m−2 m−2 m−2 m−2 ⎢ u −2un −un−1 um−2 n+1 −un−1 −γ n+1 − ηum−2 m=2 ⎢ 2 n 2 x ⎢ ( x) 

⎣ − m−2 m−2 m−2 m−2 vn+1 −vn−1 m−2 un+1 −un−1 −κ um−2 + v n n 2 x 2 x +

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(20.20) ⎫ ⎪ ⎬ ⎪ ⎭

Application to system of non-linear partial differential equations

 ×

(k − m + 1)α (k − m + 3 + 2α) − (k − m)α (k − m + 3 + 3α)

383



α ( t)α 2AB (α) (α + 3) ⎡ ⎧ ⎫ m m m m m n−1 m un+1 −un−1 ⎬ ⎨ −γ un+1 −2un −−u − ηu 2 n 2 x ⎢   ( x) m −v m ⎢ −um vn+1 um n−1 n+1 n−1 m m ⎩ ⎭ ⎢ −κ un 2 x + vn 2 x ⎢ ⎧ m−1 m−1 ⎫ m−1 −um−1 ⎢ um−1 n+1 −2un n−1 m−1 un+1 −un−1 ⎪ ⎪ ⎢ ⎨ ⎬ −γ − ηu k 2 n 2 x  ⎢ ( x)

⎢ −2 m−1 m−1 m−1 m−1 × −u vn+1 −vn−1 u ⎢ ⎪ ⎩ −κ um−1 ⎭ + vnm−1 n+12 x n−1 ⎪ ⎢ n 2 x m=2 ⎢ ⎧ ⎫ ⎢ m−2 um−2 um−2 −2um−2 −um−2 ⎢ n n−1 n+1 −un−1 ⎪ ⎪ ⎨ −γ n+1 ⎬ − ηum−2 ⎢ 2 n 2 x ( x) ⎢ + 

m−2 m−2 m−2 m−2 ⎣ −u vn+1 −vn−1 u ⎪ ⎩ −κ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x +

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

⎧ v k −2v k −v k v k −v k ⎪ ⎨ −ρ n+1 n 2 n−1 − μvnk n+12 xn−1 1 − α 

( x) v k+1 = v 0 + k −v k vn+1 ukn+1 −ukn−1 n−1 k k AB (α) ⎪ ⎩ −ξ un 2 x + vn 2 x

⎫ ⎪ ⎬ ⎪ ⎭

⎧

m−2 m−2 m−2 m−2 ⎫ vn+1 −2vnm−2 −vn−1 m−2 vn+1 −vn−1 ⎪ ⎬ −ρ − μv n 2 x  α ( t) ( x)2

+ m−2 m−2 m−2 m−2 m−2 vn+1 −vn−1 + v m−2 un+1 −un−1 ⎪ ⎪ AB (α) (α + 1) ⎭ m=2 ⎩ −ξ un n 2 x 2 x

α+1

k ⎪ ⎨ 

  × (k − m + 1)α − (k − m)α ⎡

v m−1 −2v m−1 −v m−1

v m−1 −v m−1

n n−1 −ρ n+1 − μvnm−1 n+12 xn−1 ( x)2 

m−1 m−1 m−1 vn+1 um−1 −vn−1 n+1 −un−1 m−1 m−1 −ξ un + vn 2 x 2 x

⎢ ⎢ α ⎢ α ( t) ⎢ + m−2 ⎢ v m−2 −v m−2 v m−2 −2vnm−2 −vn−1 AB (α) (α + 2) −ρ n+1 − μvnm−2 n+12 xn−1 m=2 ⎢ 2 ⎢ ( x)

 ⎣ − m−2 m−2 m−2 vn+1 um−2 −vn−1 n+1 −un−1 m−2 m−2 −ξ un + vn 2 x 2 x k ⎢ 



(k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) α ( t)α + 2AB (α) (α + 3)



⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

384

New Numerical Scheme With Newton Polynomial

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢  ⎢ × ⎢ ⎢ m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎧ ⎫ m m m m m ⎨ −ρ vn+1 −2vn 2−vn−1 − μv m vn+1 −vn−1 ⎬ n 2 x  ( x) m m m m  ⎩ −ξ um vn+1 −vn−1 + v m un+1 −un−1 ⎭ n n 2 x 2 x ⎧ m−1 m−1 m−1 m−1 ⎫ vn+1 vn+1 −2vnm−1 −vn−1 −vn−1 m−1 ⎪ ⎪ ⎨ −ρ ⎬ − μv n 2 x  ( x)2

−2 m−1 m−1 m−1 m−1 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−1 ⎭ + vnm−1 n+12 x n−1 ⎪ n 2 x ⎧ ⎫ m−2 v m−2 −v m−2 v m−2 −2vnm−2 −vn−1 ⎪ ⎨ −ρ n+1 ⎬ − μvnm−2 n+12 xn−1 ⎪ 2 ( x) 

+ m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x 

 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥. ⎦

20.1.3 System of non-linear partial differential equations with the Caputo fractional derivative We consider the Burgers equation including the Caputo fractional operator, C α 0 Dt u (x, t) = −γ uxx − ηux u − κ (uvx + vux ) , C α 0 Dt v (x, t) = −ρvxx − μvx v − ξ (uvx + vux ) .

(20.21)

Here, for simplicity, we put H1 (u, v, x, t) = −γ uxx − ηux u − κ (uvx + vux ), H2 (u, v, x, t) = −ρvxx − μvx v − ξ (uvx + vux ). Also, we have the following initial condition: u (x, 0) = g1 (x) , u (x, t) |∂ = f1 (t) , v (x, 0) = g2 (x) , v (x, t) |∂ = f2 (t) .

(20.22)

We shall write the above equation as follows:  t 1 u (x, t) − u (x, 0) = H1 (x, τ, u, v) (t − τ )α−1 dτ, (α) 0  t 1 H2 (x, τ, u, v) (t − τ )α−1 dτ. v (x, t) − v (x, 0) = (α) 0

(20.23)

At the point tk+1 = (k + 1) t, we have the following:  tk+1 1 u (x, tk+1 ) − u (x, 0) = H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, (α) 0  tk+1 1 H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ, v (x, tk+1 ) − v (x, 0) = (α) 0

(20.24)

Application to system of non-linear partial differential equations

385

and we write u (x, tk+1 ) = u (x, 0) +

k  1  tm+1 H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, (α) tm m=2

(20.25) v (x, tk+1 ) = v (x, 0) +

k  1  tm+1 H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ. (α) tm m=2

After some manipulations, we can rearrange the above equation thus: 1 (20.26) (α) ⎫ ⎧   m−2 , v m−2 H x, t , u ⎪ ⎪ 1 m−2 ⎪ ⎪     ⎪ ⎪ ⎪ k  tm+1 ⎪ ⎬ ⎨ + H1 x,tm−1 ,um−1 ,v m−1 −H1 x,tm−2 ,um−2 ,v m−2 (τ − t  ) m−2

t     × m ,v m )−2H x,t m−1 ,v m−1 +H x,t m−2 ,v m−2 H ,u ,u ,u (x,t 1 m−1 1 m−2 ⎪ ⎪ + 1 m ⎪ ⎪ m=2 tm ⎪ ⎪ 2( t)2 ⎪ ⎪ ⎭ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 = u0 +

× (tk+1 − τ )α−1 dτ,

1 (α) ⎧   H2 x, tm−2, um−2 , v m−2  ⎪ ⎪  ⎪ k  tm+1 ⎪ ⎨ + H2 x,tm−1 ,um−1 ,v m−1 −H2 x,tm−2 ,um−2 ,v m−2 (τ − t  m−2 )   t   × m ,v m )−2H x,t m−1 ,v m−1 +H x,t m−2 ,v m−2 H ,u ,u ,u (x,t m 2 2 m−1 2 m−2 ⎪ ⎪ + m=2 tm ⎪ 2( t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

v k+1 = v0 +

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. Thus, we have 1 uk+1 = u0 + (α)  tm+1   ⎧ ⎫ H1 x, tm−2 , um−2 , v m−2 (tk+1 − τ )α−1 dτ ⎪ ⎪ tm ⎪ ⎪     ⎪ ⎪  tm+1 H1 x,tm−1 ,um−1 ,v m−1 −H1 x,tm−2 ,um−2 ,v m−2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + k ⎬ tm

t ⎨ α−1 , × × (τ − tm−2 − τ dτ ) ) (t k+1     ⎪ ⎪ m ,v m )−2H x,t m−1 ,v m−1 +H x,t m−2 ,v m−2  ⎪ ⎪ H ,u ,u ,u (x,t t m 1 1 m−1 1 m−2 m+1 m=2 ⎪ ⎪ + ⎪ ⎪ ⎪ ⎪ tm 2( t)2 ⎪ ⎪ ⎩ ⎭ α−1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ ) dτ (20.27)

386

New Numerical Scheme With Newton Polynomial

1 v k+1 = v 0 + (α)  tm+1   ⎧ α−1 m−2 , v m−2 (t x, t H , u dτ ⎪ 2 m−2 k+1 − τ ) t ⎪ m     ⎪ m−1 ,v m−1 −H x,t m−2 ,v m−2  ⎪ x,t H ,u ,u tm+1 2 m−1 2 m−2 ⎪ ⎪ + k ⎨ tm

t  × × (τ − tm−2 dτ ) (tk+1 − τ )α−1    ⎪ t m m m−1 ,v m−1 +H x,t m−2 ,v m−2 2 m−2 ,u m+1 H2 (x,tm ,u ,v )−2H2 x,tm−1 ,u m=2 ⎪ ⎪ ⎪ + ⎪ tm 2( t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

We order the above equation as follows:

uk+1 = u0 +

k    tm+1 1  H1 x, tm−2 , um−2 , v m−2 (tk+1 − τ )α−1 dτ, (α) tm m=2

+

1 (α) 

× ×

k  m=2

    H1 x, tm−1 , um−1 , v m−1 − H1 x, tm−2 , um−2 , v m−2

t

(20.28)

k 1  (α) tm m=2     m m m−1 m−1 + H1 x, tm−2 , um−2 , v m−2 ,v H1 (x, tm , u , v ) − 2H1 x, tm−1 , u

 ×

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 ( t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

tm

k    tm+1 1  H2 x, tm−2 , um−2 , v m−2 (tk+1 − τ )α−1 dτ (α) t m m=2     k m−1 m−1 − H2 x, tm−2 , um−2 , v m−2 ,v 1  H2 x, tm−1 , u + (α)

t

v k+1 = v0 +

 × ×

k 1  (α) tm m=2     m m m−1 m−1 + H2 x, tm−2 , um−2 , v m−2 ,v H2 (x, tm , u , v ) − 2H2 x, tm−1 , u

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 ( t)2 tm+1

tm

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

Application to system of non-linear partial differential equations

387

The following approximation can be obtained with the same routine: uk+1 = u0 +

k   ( t)α  H1 x, tm−2 , um−2 , v m−2 (α + 1) m=2   × (k − m + 1)α − (k − m)α    k  ( t)α  H1 x, tm−1 , um−1 , v m−1  + −H1 x, tm−2 , um−2 , v m−2 (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H1 (x, tm , um , v m ) α   ( t) ⎣ −2H1 x, tm−1 , um−1 , v m−1 ⎦ +   2 (α + 3) +H1 x, tm−2 , um−2 , v m−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

k   ( t)α  H2 x, tm−2 , um−2 , v m−2 (α + 1) m=2   α × (k − m + 1) − (k − m)α    k  ( t)α  H2 x, tm−1 , um−1 , v m−1  + −H2 x, tm−2 , um−2 , v m−2 (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H2 (x, tm , um , v m )  ( t)α  ⎣ −2H2 x, tm−1 , um−1 , v m−1 ⎦ + 2 (α + 3) +H2 x, tm−2 , um−2 , v m−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

(20.29)

⎤ ⎥ ⎥, ⎦

v k+1 = v0 +

⎤ ⎥ ⎥. ⎦

Replacing the above into our equation, we have ⎧ m−2 m−2 m−2 −um−2 um−2 n+1 −2un n−1 m−2 un+1 −un−1 k ⎪ ⎨ −γ − ηu α+1  2 n 2 x  ( t) ( x)

uk+1 = u0 + m−2 m−2 m−2 vn+1 um−2 −vn−1 n+1 −un−1 m−2 m−2 ⎪ (α + 1) + vn m=2 ⎩ −κ un 2 x 2 x

⎫ ⎪ ⎬ ⎪ ⎭

(20.30)

388

New Numerical Scheme With Newton Polynomial

  × (k − m + 1)α − (k − m)α ⎡ ⎤ m−1 um−1 um−1 −2um−1 −um−1 n n−1 n+1 −un−1 −γ n+1 − ηum−1 2 n 2 x ( x) ⎢

 ⎥ m−1 m−1 m−1 ⎢ ⎥ vn+1 um−1 −vn−1 n+1 −un−1 m−1 m−1 ⎢ ⎥ k −κ un + vn α ⎢ ⎥ 2 x 2 x ( t) ⎢ ⎥ + m−2 m−2 m−2 m−2 m−2 ⎢ ⎥ un+1 −2un −un−1 (α + 2) m−2 un+1 −un−1 ⎢ ⎥ −γ − ηu m=2 ⎢ 2 n 2 x ( x)  ⎥

⎣ − ⎦ m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u −κ um−2 + vnm−2 n+12 x n−1 n 2 x   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎡ ⎧ ⎫ m m m m m ⎨ −γ un+1 −2un 2−un−1 − ηum un+1 −un−1 ⎬ n 2 x ⎢  ( x) m m m m  ⎢ ⎩ −κ um vn+1 −vn−1 + v m un+1 −un−1 ⎭ ⎢ n n 2 x 2 x ⎢ ⎧ m−1 ⎫ m−1 −um−1 ⎢ um−1 um−1 n+1 −2un n−1 n+1 −un−1 m−1 ⎪ ⎪ ⎢ ⎨ ⎬ −γ − ηun k 2 x  ( x)2

( t)α  ⎢ ⎢ −2 m−1 m−1 m−1 m−1 + −u vn+1 −vn−1 u ⎢ ⎪ ⎩ −κ um−1 ⎭ + vnm−1 n+12 x n−1 ⎪ ⎢ 2 (α + 3) n 2 x m=2 ⎢ ⎧ ⎢ m−2 m−2 m−2 m−2 ⎫ m−2 ⎢ ⎪ −γ un+1 −2un −un−1 − ηum−2 un+1 −un−1 ⎬ ⎪ ⎨ ⎢ n 2 x  ( x)2 ⎢ +

m−2 m−2 m−2 m−2 ⎣ vn+1 −vn−1 u −u ⎪ ⎩ −κ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x   2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12 ⎡

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

⎧ m−2 v m−2 −2vnm−2 −vn−1 v m−2 −v m−2 k ⎪ ⎨ −ρ n+1 − μvnm−2 n+12 xn−1  2 ( t) ( x)

 v k+1 = v0 + m−2 m−2 m−2 m−2 m−2 vn+1 −vn−1 + v m−2 un+1 −un−1 ⎪ (α + 1) ⎩ −ξ u m=2 n n 2 x 2 x   × (k − m + 1)α − (k − m)α ⎡ ⎤ m−1 m−1 m−1 v m−1 −2vnm−1 −vn−1 m−1 vn+1 −vn−1 −ρ n+1 − μv n 2 x  ( x)2 ⎢ ⎥

m−1 m−1 ⎢ ⎥ vn+1 um−1 −vn−1 −um−1 n+1 n−1 m−1 m−1 ⎢ ⎥ k −ξ u + v α ⎢ n n ⎥ 2 x 2 x ( t) ⎢ ⎥ + m−2 m−2 m−2 m−2 m−2 ⎢ ⎥ vn+1 −2vn −vn−1 (α + 2) m−2 vn+1 −vn−1 ⎥ −ρ − μv m=2 ⎢ 2 n 2 x ⎢ ( x)  ⎥

⎣ − ⎦ m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u −ξ um−2 + vnm−2 n+12 x n−1 n 2 x   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) α+1

⎤

⎫ ⎪ ⎬ ⎪ ⎭

Application to system of non-linear partial differential equations

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢ α  ( t) ⎢ + ⎢ ⎢ 2 (α + 3) m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎧ ⎫ m m m m m ⎨ −ρ vn+1 −2vn 2−vn−1 − μv m vn+1 −vn−1 ⎬ n 2 x  ( x) m m m m  ⎩ −ξ um vn+1 −vn−1 + v m un+1 −un−1 ⎭ n n 2 x 2 x ⎧ m−1 m−1 m−1 m−1 ⎫ vn+1 vn+1 −2vnm−1 −vn−1 −vn−1 m−1 ⎪ ⎪ ⎨ −ρ ⎬ − μv n 2 x  ( x)2

−2 m−1 m−1 m−1 m−1 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−1 ⎭ + vnm−1 n+12 x n−1 ⎪ n 2 x ⎧ ⎫ m−2 v m−2 −v m−2 v m−2 −2vnm−2 −vn−1 ⎪ ⎨ −ρ n+1 ⎬ − μvnm−2 n+12 xn−1 ⎪ 2 ( x) 

+ m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x



 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

389

α

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥. ⎦

20.1.4 System of non-linear partial differential equations with the fractal–fractional with the Mittag-Leffler kernel We establish a new numerical approximation for solving our equation which has the Atangana–Baleanu fractal–fractional operator. Namely, we consider FFM α Dt u (x, t) = −γ uxx − ηux u − κ (uvx + vux ) , 0 FFM α Dt v (x, t) = −ρvxx − μvx v − ξ (uvx + vux ) , 0

(20.31)

u (x, 0) = g1 (x) , u (x, t) |∂ = f1 (t) ,

(20.32)

and

v (x, 0) = g2 (x) , v (x, t) |∂ = f2 (t) . We shall write the above equation as follows: 1−α H1 (x, t, u, v) , AB (α)  t α H1 (x, τ, u, v) (t − τ )α−1 dτ, + AB (α) (α) 0 1−α v (x, t) − v (x, 0) = H2 (x, t, u, v) AB (α)  t α H2 (x, τ, u, v) (t − τ )α−1 dτ, + AB (α) (α) 0

u (x, t) − u (x, 0) =

(20.33)

where we can take H1 (u, v, x, t) = βt β−1 [−γ uxx − ηux u − κ (uvx + vux )], H2 (u, v, x, t) = βt β−1 [−ρvxx − μvx v − ξ (uvx + vux )]. At the point tk+1 = (k + 1) t, we have the following:

390

New Numerical Scheme With Newton Polynomial

  1−α H1 x, tk , uk , v k , (20.34) AB (α)  tk+1 α + H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, AB (α) (α) 0   1−α H2 x, tk , uk , v k v (x, tk+1 ) − v (x, 0) = AB (α)  tk+1 α + H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ, AB (α) (α) 0

u (x, tk+1 ) − u (x, 0) =

and we write

  1−α H1 x, tk , uk , v k , (20.35) AB (α) k  tm+1  α + H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, AB (α) (α) tm

u (x, tk+1 ) = u (x, 0) +

m=2

  1−α H2 x, tk , uk , v k AB (α) k  tm+1  α H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ. + AB (α) (α) tm

v (x, tk+1 ) = v (x, 0) +

m=2

After putting their Newton polynomials into Eqs. (20.35), the above equation can be written as follows: k    α 1−α H1 x, tk , uk , v k + (20.36) AB (α) AB (α) (α) m=2 ⎫ ⎧   H1 x, tm−2, um−2 , v m−2  ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪  tm+1 ⎪ H1 x,tm−1 ,um−1 ,v m−1 −H1 x,tm−2 ,um−2 ,v m−2 ⎬ ⎨ + − t (τ ) m−2

t     × m m m−1 m−1 m−2 m−2 +H1 x,tm−2 ,u H1 (x,tm ,u ,v )−2H1 x,tm−1 ,u ,v ,v ⎪ ⎪ tm ⎪ ⎪ + ⎪ ⎪ 2( t)2 ⎪ ⎪ ⎭ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 = u0 +

× (tk+1 − τ )α−1 dτ, k    α 1−α H2 x, tk , uk , v k + AB (α) AB (α) (α) m=2 ⎧   m−2 , v m−2 H x, t , u ⎪ 2 m−2 ⎪    ⎪  tm+1 ⎪ H x,t ,um−1 ,v m−1 −H2 x,tm−2 ,um−2 ,v m−2 ⎨ + 2 m−1 (τ − tm−2 )   t   × m ,v m )−2H x,t m−1 ,v m−1 +H x,t H ,u ,u ,um−2 ,v m−2 (x,t m 2 2 m−1 2 m−2 ⎪ tm ⎪ + 2 ⎪ 2( t) ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

v k+1 = v0 +

× (tk+1 − τ )α−1 dτ.

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

Application to system of non-linear partial differential equations

391

Thus, we have

uk+1 = u0 +

×

v

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

⎪ t ⎪ ⎪ ⎪ + tmm+1 ⎪ ⎪ ⎩

k+1

×

k    α 1−α H1 x, tk , uk , v k + AB (α) AB (α) (α) m=2  tm+1   α−1 m−2 m−2 H1 x, tm−2 , u ,v dτ (tk+1 − τ ) tm  tm+1 H1 x,tm−1 ,um−1 ,v m−1 −H1 x,tm−2 ,um−2 ,v m−2  + tm

t × (τ − tm−2 − τ )α−1 dτ ) (t k+1  

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

H1 (x,tm ,um ,v m )−2H1 x,tm−1 ,um−1 ,v m−1 +H1 x,tm−2 ,um−2 ,v m−2 2( t)2 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

(20.37) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ 

k    α 1−α k k H2 x, tk , u , v + = v0 + AB (α) AB (α) (α) m=2  tm+1   α−1 m−2 , v m−2 (t x, t H , u − τ dτ ) 2 m−2 k+1 tm  tm+1 H2 x,tm−1 ,um−1 ,v m−1 −H2 x,tm−2 ,um−2 ,v m−2  + tm

t × (τ − tm−2 dτ ) (tk+1 − τ )α−1   tm+1 H2 (x,tm ,um ,v m )−2H2 x,tm−1 ,um−1 ,v m−1 +H2 x,tm−2 ,um−2 ,v m−2  + tm 2 2( t)

× (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

,

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

We can arrange the above system as follows:   1−α H1 x, tk , uk , v k , (20.38) AB (α) k    tm+1  α + H1 x, tm−2 , um−2 , v m−2 (tk+1 − τ )α−1 dτ, AB (α) (α) t m m=2     k m−1 m−1  H1 x, tm−1 , u − H1 x, tm−2 , um−2 , v m−2 ,v α + AB (α) (α)

t

uk+1 = u0 +

 × ×

k  α AB (α) (α) tm m=2     m m m−1 m−1 + H1 x, tm−2 , um−2 , v m−2 ,v H1 (x, tm , u , v ) − 2H1 x, tm−1 , u

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 ( t)2 tm+1

tm

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

392

New Numerical Scheme With Newton Polynomial

  1−α H2 x, tk , uk , v k AB (α) k    tm+1  α m−2 m−2 H2 x, tm−2 , u ,v + (tk+1 − τ )α−1 dτ AB (α) (α) t m m=2     k  H2 x, tm−1 , um−1 , v m−1 − H2 x, tm−2 , um−2 , v m−2 α + AB (α) (α)

t

v k+1 = v0 +

 × ×

k  α AB (α) (α) tm m=2     m m m−1 m−1 + H2 x, tm−2 , um−2 , v m−2 ,v H2 (x, tm , u , v ) − 2H2 x, tm−1 , u

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 ( t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ.

tm

Thus, we have the following algorithm:   1−α H1 x, tk , uk , v k , AB (α) k    α ( t)α H1 x, tm−2 , um−2 , v m−2 + AB (α) (α + 1) m=2   × (k − m + 1)α − (k − m)α    k   α ( t)α H1 x, tm−1 , um−1 , v m−1  + −H1 x, tm−2 , um−2 , v m−2 AB (α) (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H1 (x, tm , um , v m )   α ( t)α ⎣ −2H1 x, tm−1 , um−1 , v m−1 ⎦ +   2AB (α) (α + 3) +H1 x, tm−2 , um−2 , v m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ ⎥ +2α 2 + 9α + 12   ⎥, ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

uk+1 = u0 +

  1−α H2 x, tk , uk , v k AB (α) k    α ( t)α H2 x, tm−2 , um−2 , v m−2 + AB (α) (α + 1)

v k+1 = v0 +

m=2

(20.39)

Application to system of non-linear partial differential equations

393

  × (k − m + 1)α − (k − m)α    k   α ( t)α H2 x, tm−1 , um−1 , v m−1  + −H2 x, tm−2 , um−2 , v m−2 AB (α) (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H2 (x, tm , um , v m )   α ( t)α ⎣ −2H2 x, tm−1 , um−1 , v m−1 ⎦ +   2AB (α) (α + 3) +H2 x, tm−2 , um−2 , v m−2 m=2   ⎤ ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) ⎥ +2α 2 + 9α + 12   ⎥. ×⎢ 2 ⎣ ⎦ 2 (k − m) + (5α + 10) (k − m) − (k − m)α 2 +6α + 18α + 12

Plugging the above into our equation, at the point xn , we obtain the following:

⎧ ⎫ k k ukn+1 −2ukn −ukn−1 k un+1 −un−1 ⎪ ⎪ ⎨ ⎬ −γ − ηu 2 n 2 x  1−α β−1

( x) βtk uk+1 = u0 + (20.40) k −v k k −uk v u ⎪ AB (α) ⎩ −κ ukn n+12 xn−1 + vnk n+12 x n−1 ⎪ ⎭ ⎧ m−2 ⎫ m−2 −um−2 um−2 um−2 n+1 −2un n−1 n+1 −un−1 m−2 ⎪ ⎪ k ⎨ ⎬ −γ − ηun  β−1 2 x  α ( t)α+1 ( x)2

βtm−2 + m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ AB (α) (α + 1) ⎩ −κ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ m=2 n 2 x   × (k − m + 1)α − (k − m)α α ( t)α + AB (α) (α + 2) ⎧ ⎡ m−1 m−1 ⎫ ⎤ um−1 −2um−1 −um−1 n n−1 m−1 un+1 −un−1 ⎪ ⎪ ⎨ −γ n+1 ⎬ − ηu n 2 x  ( x)2 ⎥ ⎢ βt β−1

m−1 m−1 m−1 m−1 ⎥ ⎢ m−1 ⎪ vn+1 −vn−1 un+1 −un−1 m−1 m−1 ⎪ ⎥ ⎢ k ⎩ ⎭ −κ un + vn ⎢ ⎥ 2 x 2 x ⎥ ⎢ ⎧ ⎫ × m−2 m−2 m−2 m−2 m−2 ⎥ ⎢ un+1 −2un −un−1 m−2 un+1 −un−1 ⎪ ⎪ ⎥ ⎢ ⎨ ⎬ −γ − ηu m=2 ⎢ 2 n 2 x ⎥ ( x) β−1 

⎦ ⎣ −βtm−2 m−2 m−2 m−2 m−2 v u −v −u n+1 n−1 ⎪ ⎩ −κ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) α ( t)α + 2AB (α) (α + 3)

394

New Numerical Scheme With Newton Polynomial

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢  ⎢ × ⎢ ⎢ m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎫ ⎧ m m m m m ⎨ −γ un+1 −2un 2−un−1 − ηum un+1 −un−1 ⎬ n 2 x β−1  ( x) βtm m m m m  ⎩ −κ um vn+1 −vn−1 + v m un+1 −un−1 ⎭ n n 2 x 2 x ⎧ m−1 ⎫ m−1 −um−1 um−1 um−1 n+1 −2un n−1 n+1 −un−1 m−1 ⎪ ⎪ ⎨ −γ ⎬ − ηu n 2 x  ( x)2 β−1

−2βtm−1 m−1 m−1 m−1 m−1 vn+1 −vn−1 u −u ⎪ ⎩ −κ um−1 ⎭ + vnm−1 n+12 x n−1 ⎪ n 2 x ⎧ ⎫ m−2 um−2 um−2 −2um−2 −um−2 n n−1 n+1 −un−1 ⎪ ⎪ ⎨ −γ n+1 ⎬ − ηum−2 2 n 2 x ( x) β−1 

+βtm−2 m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ ⎩ −κ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x 

 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥, ⎦

⎧ v k −v k v k −2v k −v k ⎪ ⎨ −ρ n+1 n 2 n−1 − μvnk n+12 xn−1 1 − α β−1 

( x) v k+1 = v0 + βt k −v k vn+1 ukn+1 −ukn−1 n−1 k k AB (α) k ⎪ ⎩ −ξ un 2 x + vn 2 x

⎫ ⎪ ⎬ ⎪ ⎭

α ( t)α+1 AB (α) (α + 1) ⎧ m−2 m−2 m−2 m−2 ⎫ vn+1 −2vnm−2 −vn−1 −vn−1 vn+1 m−2 ⎪ ⎪ k ⎨ ⎬ −ρ − μvn  β−1 2 x  ( x)2

× βtm−2 m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ m=2 n 2 x   × (k − m + 1)α − (k − m)α α ( t)α + AB (α) (α + 2) ⎧ ⎡ m−1 m−1 m−1 ⎫ v m−1 −2vnm−1 −vn−1 m−1 vn+1 −vn−1 ⎪ ⎪ ⎨ −ρ n+1 ⎬ − μv n 2 x  ( x)2 ⎢ βt β−1

m−1 m−1 m−1 m−1 ⎢ m−1 ⎪ vn+1 −vn−1 u −u k ⎢ ⎩ −ξ um−1 ⎭ + vnm−1 n+12 x n−1 ⎪  n ⎢ 2 x ⎢ ⎧ × m−2 m−2 m−2 m−2 ⎫ ⎢ vn+1 −2vnm−2 −vn−1 m−2 vn+1 −vn−1 ⎪ ⎪ ⎢ ⎨ ⎬ −ρ − μv m=2 ⎢ n 2 x  ( x)2 β−1

⎣ −βtm−2 m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) α ( t)α + 2AB (α) (α + 3) +

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Application to system of non-linear partial differential equations

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ k ⎢  ⎢ × ⎢ ⎢ m=2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

395

⎫ ⎧ m m m m m ⎨ −ρ vn+1 −2vn 2−vn−1 − μv m vn+1 −vn−1 ⎬ n 2 x β−1  ( x) βtm m m m m  ⎩ −ξ um vn+1 −vn−1 + v m un+1 −un−1 ⎭ n n 2 x 2 x ⎧ m−1 m−1 m−1 m−1 ⎫ vn+1 vn+1 −2vnm−1 −vn−1 −vn−1 m−1 ⎪ ⎪ ⎨ −ρ ⎬ − μv n 2 x  ( x)2 β−1

−2βtm−1 m−1 m−1 m−1 m−1 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−1 ⎭ + vnm−1 n+12 x n−1 ⎪ n 2 x ⎧ ⎫ m−2 v m−2 −v m−2 v m−2 −2vnm−2 −vn−1 ⎪ ⎨ −ρ n+1 ⎬ − μvnm−2 n+12 xn−1 ⎪ 2 ( x) β−1 

+βtm−2 m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ ⎩ −ξ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ n 2 x 

 2 (k − m)2 + (3α + 10) (k − m) ⎢ (k − m + 1) 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

α

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥. ⎦

20.1.5 System of non-linear partial differential equations with fractal–fractional with the power law We construct a new numerical scheme to achieve the solution of our equation which has the Caputo fractal–fractional operator. Namely, we consider FFP α Dt u (x, t) = −γ uxx − ηux u − κ (uvx + vux ) , 0 FFP α Dt v (x, t) = −ρvxx − μvx v − ξ (uvx + vux ) , 0

(20.41)

u (x, 0) = g1 (x) , u (x, t) |∂ = f1 (t) , v (x, 0) = g2 (x) , v (x, t) |∂ = f2 (t) .

(20.42)

and

We convert the above equation to  t 1 H1 (x, τ, u, v) (t − τ )α−1 dτ, u (x, t) − u (x, 0) = (α) 0  t 1 v (x, t) − v (x, 0) = H2 (x, τ, u, v) (t − τ )α−1 dτ, (α) 0

(20.43)

where we can take H1 (u, v, x, t) = βt β−1 [−γ uxx − ηux u − κ (uvx + vux )], At the point H2 (u, v, x, t) = βt β−1 [−ρvxx − μvx v − ξ (uvx + vux )]. tk+1 = (k + 1) t, we write the following:  tk+1 1 H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, (20.44) u (x, tk+1 ) − u (x, 0) = (α) 0  tk+1 1 v (x, tk+1 ) − v (x, 0) = H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ, (α) 0

396

New Numerical Scheme With Newton Polynomial

and k  1  tm+1 H1 (x, τ, u, v) (tk+1 − τ )α−1 dτ, u (x, tk+1 ) = u (x, 0) + (α) tm m=2

(20.45) k  1  tm+1 v (x, tk+1 ) = v (x, 0) + H2 (x, τ, u, v) (tk+1 − τ )α−1 dτ. (α) tm m=2

Based on the Newton polynomial, the above equation can be reformulated as follows: 1 (20.46) (α) ⎫ ⎧   H1 x, tm−2, um−2 , v m−2  ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ m−1 ,v m−1 −H x,t m−2 ,v m−2 ⎪ ⎪  x,t H ,u ,u k 1 m−1 1 m−2 ⎬ ⎨ t  m+1 + − t (τ ) m−2

t     × m m m−1 m−1 m−2 m−2 +H1 x,tm−2 ,u H (x,t ,u ,v )−2H1 x,tm−1 ,u ,v ,v ⎪ ⎪ + 1 m ⎪ ⎪ m=2 tm ⎪ ⎪ 2( t)2 ⎪ ⎪ ⎭ ⎩ × (τ − tm−2 ) (τ − tm−1 )

uk+1 =u0 +

× (tk+1 − τ )α−1 dτ,

1 (α) ⎧   H2 x, tm−2, um−2 , v m−2  ⎪ ⎪  ⎪ m−1 m−1 m−2 m−2 −H2 x,tm−2 ,u ,v k  tm+1 ⎪ ⎨ + H2 x,tm−1 ,u ,v  (τ − tm−2 )   t   × m m m−1 m−1 +H2 x,tm−2 ,um−2 ,v m−2 H2 (x,tm ,u ,v )−2H2 x,tm−1 ,u ,v ⎪ + ⎪ m=2 tm ⎪ 2( t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 )

v k+1 = v 0 +

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

× (tk+1 − τ )α−1 dτ. So, we get the following: 1 (20.47) uk+1 = u0 + (α)  tm+1   ⎧ ⎫ H1 x, tm−2 , um−2 , v m−2 (tk+1 − τ )α−1 dτ ⎪ ⎪ tm ⎪ ⎪     ⎪ ⎪  tm+1 H1 x,tm−1 ,um−1 ,v m−1 −H1 x,tm−2 ,um−2 ,v m−2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + k ⎨ ⎬ tm

t  α−1 , × × (τ − tm−2 − τ dτ ) ) (t k+1     ⎪ ⎪ t m m m−1 ,v m−1 +H x,t m−2 ,v m−2 ⎪ 1 m−2 ,u m+1 H1 (x,tm ,u ,v )−2H1 x,tm−1 ,u m=2 ⎪ ⎪ ⎪ ⎪ + t ⎪ ⎪ ⎪ m 2( t)2 ⎪ ⎪ ⎩ ⎭ α−1 × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ ) dτ

Application to system of non-linear partial differential equations

397

1 v k+1 = v 0 + (α)  tm+1   ⎧ α−1 m−2 , v m−2 (t x, t H , u dτ ⎪ 2 m−2 k+1 − τ ) t ⎪ m     ⎪ m−1 ,v m−1 −H x,t m−2 ,v m−2  ⎪ x,t H ,u ,u tm+1 2 m−1 2 m−2 ⎪ ⎪ + k ⎨ tm

t  × × (τ − tm−2 dτ ) (tk+1 − τ )α−1    ⎪ t m m m−1 ,v m−1 +H x,t m−2 ,v m−2 2 m−2 ,u m+1 H2 (x,tm ,u ,v )−2H2 x,tm−1 ,u m=2 ⎪ ⎪ ⎪ + ⎪ tm 2( t)2 ⎪ ⎩ × (τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

.

With the same routine, one can obtain the following:

uk+1 = u0 +

k    tm+1 1  H1 x, tm−2 , um−2 , v m−2 (tk+1 − τ )α−1 dτ, (α) tm m=2

+

1 (α) 

× ×

k  m=2

    H1 x, tm−1 , um−1 , v m−1 − H1 x, tm−2 , um−2 , v m−2

t

(20.48)

k 1  (α) tm m=2     m m m−1 m−1 + H1 x, tm−2 , um−2 , v m−2 ,v H1 (x, tm , u , v ) − 2H1 x, tm−1 , u

 ×

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 ( t)2 tm+1

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

tm

k    tm+1 1  H2 x, tm−2 , um−2 , v m−2 (tk+1 − τ )α−1 dτ (α) t m m=2     k m−1 m−1 − H2 x, tm−2 , um−2 , v m−2 ,v 1  H2 x, tm−1 , u + (α)

t

v k+1 = v0 +

 × ×

k 1  (α) tm m=2     m m m−1 m−1 + H2 x, tm−2 , um−2 , v m−2 ,v H2 (x, tm , u , v ) − 2H2 x, tm−1 , u

 ×

m=2

tm+1

(τ − tm−2 ) (tk+1 − τ )α−1 dτ +

2 ( t)2 tm+1

tm

(τ − tm−2 ) (τ − tm−1 ) (tk+1 − τ )α−1 dτ,

398

New Numerical Scheme With Newton Polynomial

and the following scheme is in order: uk+1 = u0 +

k   ( t)α  H1 x, tm−2 , um−2 , v m−2 (α + 1) m=2   × (k − m + 1)α − (k − m)α

(20.49)

   k  ( t)α  H1 x, tm−1 , um−1 , v m−1  + −H1 x, tm−2 , um−2 , v m−2 (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H1 (x, tm , um , v m )  ( t)α  ⎣ −2H1 x, tm−1 , um−1 , v m−1 ⎦ + 2 (α + 3) +H1 x, tm−2 , um−2 , v m−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) − (k − m)α +6α 2 + 18α + 12

⎤ ⎥ ⎥, ⎦

k   ( t)α  H2 x, tm−2 , um−2 , v m−2 (α + 1) m=2   α × (k − m + 1) − (k − m)α

v k+1 = v0 +

   k  ( t)α  H2 x, tm−1 , um−1 , v m−1  + −H2 x, tm−2 , um−2 , v m−2 (α + 2) m=2   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ⎤ ⎡ k H2 (x, tm , um , v m )  ( t)α  ⎣ −2H2 x, tm−1 , um−1 , v m−1 ⎦ + 2 (α + 3) +H2 x, tm−2 , um−2 , v m−2 m=2   ⎡ 2 (k − m)2 + (3α + 10) (k − m) α ⎢ (k − m + 1) +2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12

⎤ ⎥ ⎥. ⎦

After putting all this in the above equation, we can give the following numerical approximation:

Application to system of non-linear partial differential equations

399

( t)α+1 (α + 1) ⎧ m−2 m−2 ⎫ um−2 −2um−2 −um−2 n n−1 m−2 un+1 −un−1 ⎪ ⎪ k ⎨ −γ n+1 ⎬ − ηu  n 2 x  ( x)2 β−1

× βtm−2 (20.50) m−2 m−2 m−2 m−2 vn+1 −vn−1 u −u ⎪ ⎩ −κ um−2 ⎭ + vnm−2 n+12 x n−1 ⎪ m=2 n 2 x   × (k − m + 1)α − (k − m)α ( t)α + (α + 2) ⎧ ⎡ m−1 ⎫ ⎤ um−1 um−1 −2um−1 −um−1 n n−1 n+1 −un−1 ⎪ ⎪ ⎨ −γ n+1 ⎬ − ηum−1 2 n 2 x  ( x) ⎥ ⎢ βt β−1

m−1 m−1 m−1 m−1 ⎥ ⎢ m−1 ⎪ vn+1 −vn−1 un+1 −un−1 m−1 m−1 ⎪ ⎢ k ⎩ −κ un ⎭ ⎥ + vn  ⎥ ⎢ 2 x 2 x ⎥ ⎢ ⎧ × m−2 m−2 m−2 ⎫ ⎥ m−2 −um−2 ⎢ u u −2u −u n n+1 n−1 n+1 n−1 m−2 ⎪ ⎪ ⎢ ⎬ ⎥ ⎨ −γ − ηun m=2 ⎢ 2 x  ⎥ ( x)2 β−1

⎦ ⎣ −βtm−2 m−2 m−2 m−2 −vn−1 vn+1 um−2 n+1 −un−1 m−2 m−2 ⎪ ⎪ ⎭ ⎩ −κ un + vn 2 x 2 x   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ( t)α + 2 (α + 3) ⎤ ⎡ ⎫ ⎧ m m m m m ⎨ −γ un+1 −2un 2−un−1 − ηum un+1 −un−1 ⎬ n 2 x β−1 ⎥ ⎢   ( x) βtm m −v m ⎥ ⎢ −um vn+1 um n−1 n+1 n−1 m m ⎭ ⎩ ⎥ ⎢ −κ un 2 x + vn 2 x ⎥ ⎢ ⎧ m−1 m−1 m−1 ⎫ ⎥ m−1 −um−1 ⎢ u u −2u −u n n+1 n−1 n+1 n−1 m−1 ⎪ ⎪ ⎢ ⎨ −γ ⎬ ⎥ − ηun k ⎢ ⎥ 2 x   ( x)2

⎥ ⎢ −2βt β−1 m−1 m−1 m−1 m−1 ⎪ × −vn−1 vn+1 um−1 ⎥ ⎢ n+1 −un−1 m−1 m−1 ⎪ ⎩ −κ un ⎭ + v ⎥ ⎢ n 2 x 2 x m=2 ⎢ ⎥ ⎧ ⎫ ⎥ ⎢ m−2 m−2 m−2 m−2 un+1 −2um−2 −un−1 un+1 −un−1 ⎪ ⎥ ⎢ n m−2 ⎪ ⎨ −γ ⎬ − ηun ⎥ ⎢ 2 2 x  ( x) ⎥ ⎢ +βt β−1

m−2 m−2 m−2 m−2 ⎦ ⎣ m−2 ⎪ vn+1 −vn−1 m−2 un+1 −un−1 ⎪ ⎩ −κ um−2 ⎭ + v n n 2 x 2 x

uk+1 = u0 +

  2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥, ⎦

( t)α+1 (α + 1) ⎧ m−2 v m−2 −2vnm−2 −vn−1 v m−2 −v m−2 ⎪ k ⎨ −ρ n+1 − μvnm−2 n+12 xn−1  2 ( x) β−1 

× βtm−2 m−2 m−2 m−2 m−2 m−2 vn+1 −vn−1 + v m−2 un+1 −un−1 ⎪ ⎩ −ξ u m=2 n n 2 x 2 x

v k+1 = v0 +

⎫ ⎪ ⎬ ⎪ ⎭

400

New Numerical Scheme With Newton Polynomial

  × (k − m + 1)α − (k − m)α ( t)α + (α + 2) ⎧ ⎡ m−1 m−1 m−1 ⎫ ⎤ v m−1 −2vnm−1 −vn−1 m−1 vn+1 −vn−1 ⎪ ⎪ ⎨ −ρ n+1 ⎬ − μv n 2 x  ( x)2 ⎢ βt β−1 ⎥

m−1 m−1 m−1 m−1 ⎢ ⎥ m−1 ⎪ m−1 vn+1 −vn−1 + v m−1 un+1 −un−1 ⎪ ⎢ ⎥ k ⎩ ⎭ −ξ u ⎢ n n ⎥ 2 x 2 x ⎢ ⎥ ⎧ ⎫ × m−2 m−2 m−2 m−2 −v m−2 ⎢ ⎥ v −2v −v v n n+1 n−1 n−1 m−2 n+1 ⎪ ⎪ ⎥ ⎨ ⎬ − μv −ρ m=2 ⎢ 2 n 2 x ⎢ ⎥ ( x) β−1 

⎣ −βtm−2 ⎦ m−2 m−2 m−2 m−2 vn+1 −vn−1 un+1 −un−1 m−2 m−2 ⎪ ⎪ ⎩ −ξ un ⎭ + vn 2 x 2 x   (k − m + 1)α (k − m + 3 + 2α) × − (k − m)α (k − m + 3 + 3α) ( t)α + 2 (α + 3) ⎤ ⎡ ⎫ ⎧ m m m m m ⎨ −ρ vn+1 −2vn 2−vn−1 − μv m vn+1 −vn−1 ⎬ n 2 x β−1 ⎥ ⎢   ( x) βtm m −v m ⎥ ⎢ −um vn+1 um n−1 n+1 n−1 m m ⎭ ⎩ ⎥ ⎢ −ξ un 2 x + vn 2 x ⎥ ⎢ ⎧ m−1 m−1 m−1 ⎫ ⎥ m−1 −v m−1 ⎢ v v −2v −v n n+1 n−1 n+1 n−1 m−1 ⎪ ⎪ ⎢ ⎬ ⎥ ⎨ −ρ − μvn k ⎢ ⎥ 2 x   ( x)2

⎥ ⎢ −2βt β−1 m−1 m−1 m−1 m−1 ⎪ × vn+1 um−1 −vn−1 ⎥ ⎢ n+1 −un−1 m−1 m−1 ⎪ ⎭ ⎩ −ξ un + v ⎥ ⎢ n 2 x 2 x m=2 ⎢ ⎥ ⎫ ⎧ ⎥ ⎢ m−2 m−2 m−2 m−2 vn+1 −2vnm−2 −vn−1 −vn−1 vn+1 ⎥ ⎢ m−2 ⎪ ⎪ ⎬ ⎨ −ρ − μv ⎥ ⎢ n 2 x  ( x)2 ⎥ ⎢ +βt β−1

m−2 m−2 m−2 m−2 ⎦ ⎣ m−2 ⎪ vn+1 −vn−1 m−2 un+1 −un−1 ⎪ ⎭ ⎩ −ξ um−2 + v n n 2 x 2 x   2 (k − m)2 + (3α + 10) (k − m) α − m + 1) (k ⎢ 2α 2 + 9α + 12   ×⎢ ⎣ 2 (k − m)2 + (5α + 10) (k − m) α − (k − m) +6α 2 + 18α + 12 ⎡

⎤ ⎥ ⎥. ⎦

Appendix

A

AS_Method_for_Chaotic_with_AB_Fractal-Fractional.m % h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts x(2)=x(1)+h.*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h.*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h.*z1(t(1),x(1),y(1),z(1)); x(3)=x(2)+(h/2).*(3.*x1(t(2),x(2),y(2),z(2))-x1(t(1),x(1),y(1),z(1))); y(3)=y(2)+(h/2).*(3.*y1(t(2),x(2),y(2),z(2))-y1(t(1),x(1),y(1),z(1))); z(3)=z(2)+(h/2).*(3.*z1(t(2),x(2),y(2),z(2))-z1(t(1),x(1),y(1),z(1))); for n=3:N j=3:n; x(n+1)=x(1)+((1-alpha)./AB).*beta.*t(n).^(beta-1).*x1(t(n),x(n),y(n), z(n))+((h.^alpha).*alpha./(AB.*gamma(alpha+1))). *sum(((n+1-j).^alpha-(n-j).^alpha).*beta.*t(j-2).^(beta-1). *x1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+2))).*sum((beta.*t(j-1). ^(beta-1).*x1(t(j-1),x(j-1),y(j-1),z(j-1))-beta.*t(j-2).^(beta-1). *x1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha. (n-j).^alpha.*(n-j+3+3*alpha))) *(n-j+3+2*alpha)-... +((h.^alpha).*alpha./(2.*gamma(alpha+3))). *sum((beta.*t(j).^(beta-1).*x1(t(j),x(j),y(j),z(j))-2.*beta. *t(j-1).^(beta-1).*x1(t(j-1),x(j-1),y(j-1),z(j-1))+... beta.*t(j-2).^(beta-1).*x1(t(j-2),x(j-2),y(j-2),z(j-2))). *((n-j+1).^alpha.*(2.*(n-j).^2+(3.*alpha+10).*(n-j)+2. *(alpha).^2+9.*alpha+12)-...(n-j).^alpha.*(2.*(n-j).^2 +(5.*alpha+10).*(n-j)+6.*alpha.^2+18.*alpha+12))); y(n+1)=y(1)+((1-alpha)./AB).*beta.*t(n).^(beta-1). *y1(t(n),x(n),y(n),z(n))+((h.^alpha).*alpha./(AB.*gamma(alpha+1))). *sum(((n+1-j).^alpha-(n-j).^alpha).*beta.*t(j-2).^(beta-1). *y1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+2))).*sum((beta.*t(j-1). ^(beta-1).*y1(t(j-1),x(j-1),y(j-1),z(j-1))-beta.*t(j-2).^(beta-1). *y1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha. *(n-j+3+2*alpha)-... (n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha).*alpha./

402

Appendix

(2.*AB.*gamma(alpha+3))).*sum((beta.*t(j).^(beta-1). *y1(t(j),x(j),y(j),z(j))-2.*beta.*t(j-1).^(beta-1). *y1(t(j-1),x(j-1),y(j-1),z(j-1))+... beta.*t(j-2).^(beta-1).*y1(t(j-2),x(j-2),y(j-2),z(j-2))). *((n-j+1).^alpha.*(2.*(n-j).^2+(3.*alpha+10).*(n-j)+2.*(alpha). ^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10). *(n-j)+6.*alpha.^2+18.*alpha+12))); z(n+1)=z(1)+((1-alpha)./AB).*beta.*t(n).^(beta-1). *z1(t(n),x(n),y(n),z(n))+((h.^alpha).*alpha./ (AB.*gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j).^alpha). *beta.*t(j-2).^(beta-1).*z1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+2))).*sum((beta.*t(j-1). ^(beta-1).*z1(t(j-1),x(j-1),y(j-1),z(j-1))-beta.*t(j-2). ^(beta-1).*z1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha. *(n-j+3+2*alpha)-... (n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha).*alpha./ (2.*AB.*gamma(alpha+3))).*sum((beta.*t(j).^(beta-1). *z1(t(j),x(j),y(j),z(j))-2.*beta.*t(j-1).^(beta-1). *z1(t(j-1),x(j-1),y(j-1),z(j-1))+... beta.*t(j-2).^(beta-1).*z1(t(j-2),x(j-2),y(j-2),z(j-2))). *((n-j+1).^alpha.*(2.*(n-j).^2+(3.*alpha+10).*(n-j)+2. *(alpha).^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha. ^2+18.*alpha+12))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AS_Method_for_Chaotic_with_AB_FractalFractional_with_Variable_Order.m %h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts x(2)=x(1)+h.*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h.*y1(t(1),x(1),y(1),z(1));

Appendix

403

z(2)=z(1)+h.*z1(t(1),x(1),y(1),z(1)); x(3)=x(2)+(h/2).*(3.*x1(t(2),x(2),y(2),z(2))-x1(t(1),x(1),y(1),z(1))); y(3)=y(2)+(h/2).*(3.*y1(t(2),x(2),y(2),z(2))-y1(t(1),x(1),y(1),z(1))); z(3)=z(2)+(h/2).*(3.*z1(t(2),x(2),y(2),z(2))-z1(t(1),x(1),y(1),z(1))); for n=3:N j=3:n; x(n+1)=x(1)+t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h).*log(t(n)) +beta(t(n))./t(n)).*(1-alpha)./AB.*x1(t(n),x(n),y(n),z(n))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+1))).*sum(((n+1-j). ^alpha-(n-j).^alpha).*x1(t(j-2),x(j-2),y(j-2),z(j-2)).*(t(j-2). ^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))))+((h.^alpha).*alpha./(AB.*gamma(alpha+2))). *sum((x1(t(j-1),x(j-1),y(j-1),z(j-1)).*... (t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-x1(t(j-2),x(j-2), y(j-2),z(j-2)).*(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha. *(n-j+3+3*alpha)))+... ((h.^alpha).*alpha./(2.*AB.*gamma(alpha+3))). *sum((x1(t(j),x(j),y(j),z(j)).*(t(j).^beta(t(j)).*(((beta(t(j+1)) -beta(t(j)))./h).*log(t(j))+beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*x1(t(j-1),x(j-1),y(j-1), z(j-1))+(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))).*x1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha. *(2.*(n-j).^2+(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2 +(5.*alpha+10).*(n-j)+6.*alpha^2+18.*alpha+12))); y(n+1)=y(1)+t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h).*log(t(n)) +beta(t(n))./t(n)).*(1-alpha)./AB.*y1(t(n),x(n),y(n),z(n))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+1))).*sum(((n+1-j).^alpha -(n-j).^alpha).*y1(t(j-2),x(j-2),y(j-2),z(j-2)).*(t(j-2). ^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))))+((h.^alpha).*alpha./(AB.*gamma(alpha+2))). *sum((y1(t(j-1),x(j-1),y(j-1),z(j-1)).*... (t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-y1(t(j-2),x(j-2),y(j-2), z(j-2)).*(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha.

404

Appendix

*(n-j+3+3*alpha)))+... ((h.^alpha).*alpha./(2.*AB.*gamma(alpha+3))). *sum((y1(t(j),x(j),y(j),z(j)).*(t(j).^beta(t(j)).*(((beta(t(j+1)) -beta(t(j)))./h).*log(t(j))+beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*y1(t(j-1),x(j-1),y(j-1), z(j-1))+(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))).*y1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha. *(2.*(n-j).^2+(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10). *(n-j)+6.*alpha^2+18.*alpha+12))); z(n+1)=z(1)+t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+beta(t(n))./t(n)).*(1-alpha)./AB. *z1(t(n),x(n),y(n),z(n))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+1))).*sum(((n+1-j). ^alpha-(n-j).^alpha).*z1(t(j-2),x(j-2),y(j-2),z(j-2)).*(t(j-2). ^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))))+((h.^alpha).*alpha./(AB.*gamma(alpha+2))). *sum((z1(t(j-1),x(j-1),y(j-1),z(j-1)).*... (t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-z1(t(j-2),x(j-2),y(j-2), z(j-2)).*(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha. *(n-j+3+3*alpha)))+... ((h.^alpha).*alpha./(2.*AB.*gamma(alpha+3))).*sum((z1(t(j),x(j), y(j),z(j)).*(t(j).^beta(t(j)).*(((beta(t(j+1))-beta(t(j)))./h). *log(t(j))+beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*z1(t(j-1),x(j-1),y(j-1), z(j-1))+(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))).*z1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha. *(2.*(n-j).^2+(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2 +(5.*alpha+10).*(n-j)+6.*alpha^2+18.*alpha+12))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

Appendix

405

%After the required functions are added, code can be used.

AS_Method_for_Chaotic_with_AB_Fractional.m %h,t(1),x(1),y(1),z(1),alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts x(2)=x(1)+h.*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h.*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h.*z1(t(1),x(1),y(1),z(1)); x(3)=x(2)+(h/2).*(3.*x1(t(2),x(2),y(2),z(2))-x1(t(1),x(1),y(1),z(1))); y(3)=y(2)+(h/2).*(3.*y1(t(2),x(2),y(2),z(2))-y1(t(1),x(1),y(1),z(1))); z(3)=z(2)+(h/2).*(3.*z1(t(2),x(2),y(2),z(2))-z1(t(1),x(1),y(1),z(1))); for n=3:N j=3:n; x(n+1)=x(1)+((1-alpha)./AB).*x1(t(n),x(n),y(n),z(n))+((h.^alpha). *alpha./(AB.*gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j).^alpha). *x1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+2))).*sum((x1(t(j-1),x(j-1), y(j-1),z(j-1))-x1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j). ^alpha.*(n-j+3+2*alpha)-... (n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha).*alpha./ (2.*AB.*gamma(alpha+3))). *sum((x1(t(j),x(j),y(j),z(j)) -2.*x1(t(j-1),x(j-1),y(j-1),z(j-1))+... x1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha.*(2.*(n-j).^2 +(3.*alpha+10).*(n-j)+2.*(alpha).^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2+18. *alpha+12))); y(n+1)=y(1)+((1-alpha)./AB).*y1(t(n),x(n),y(n),z(n))+((h.^alpha).*alpha. /(AB.*gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j).^alpha). *y1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+2))).*sum((y1(t(j-1),x(j-1), y(j-1),z(j-1))-y1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha. *(n-j+3+2*alpha)-... (n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha).*alpha./(2.*AB. *gamma(alpha+3))).*sum((y1(t(j),x(j),y(j),z(j))-2.*y1(t(j-1), x(j-1),y(j-1),z(j-1))+... y1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha.*(2.*(n-j).^2 +(3.*alpha+10).*(n-j)+2.*(alpha).^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2+18.

406

Appendix

*alpha+12))); z(n+1)=z(1)+((1-alpha)./AB).*z1(t(n),x(n),y(n),z(n))+((h.^alpha).*alpha. /(AB.*gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j).^alpha). *z1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+2))).*sum((z1(t(j-1),x(j-1), y(j-1),z(j-1))-z1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha. *(n-j+3+2*alpha)-... (n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha).*alpha./(2.*AB. *gamma(alpha+3))).*sum((z1(t(j),x(j),y(j),z(j))-2.*z1(t(j-1), x(j-1),y(j-1),z(j-1))+... z1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha.*(2.*(n-j).^2 +(3.*alpha+10).*(n-j)+2.*(alpha).^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2+18. *alpha+12))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AS_Method_for_Chaotic_with_Caputo_FractalFractional_with_Variable_Order.m %h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts x(2)=x(1)+h.*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h.*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h.*z1(t(1),x(1),y(1),z(1)); x(3)=x(2)+(h/2).*(3.*x1(t(2),x(2),y(2),z(2))-x1(t(1),x(1),y(1),z(1))); y(3)=y(2)+(h/2).*(3.*y1(t(2),x(2),y(2),z(2))-y1(t(1),x(1),y(1),z(1))); z(3)=z(2)+(h/2).*(3.*z1(t(2),x(2),y(2),z(2))-z1(t(1),x(1),y(1),z(1))); for n=3:N j=3:n; x(n+1)=x(1)+((h.^alpha)./(gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j). ^alpha).*x1(t(j-2),x(j-2),y(j-2),z(j-2)).*(t(j-2).^beta(t(j-2)).*

Appendix

407

...(((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))))+((h.^alpha)./(gamma(alpha+2))).*sum((x1(t(j-1),x(j-1), y(j-1),z(j-1)).*... (t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-x1(t(j-2),x(j-2),y(j-2), z(j-2)).*(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha. *(n-j+3+3*alpha)))+... ((h.^alpha)./(2.*gamma(alpha+3))).*sum((x1(t(j),x(j),y(j),z(j)). *(t(j).^beta(t(j)).*(((beta(t(j+1))-beta(t(j)))./h).*log(t(j)) +beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*x1(t(j-1),x(j-1),y(j-1), z(j-1))+(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))).*x1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha. *(2.*(n-j).^2+(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10). *(n-j)+6.*alpha^2+18.*alpha+12))); y(n+1)=y(1)+((h.^alpha)./(gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j). ^alpha).*y1(t(j-2),x(j-2),y(j-2),z(j-2)).*(t(j-2).^beta(t(j-2)).* ...(((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))))+((h.^alpha)./(gamma(alpha+2))).*sum((y1(t(j-1),x(j-1), y(j-1),z(j-1)).*... (t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-y1(t(j-2),x(j-2),y(j-2), z(j-2)).*(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha. *(n-j+3+3*alpha)))+... ((h.^alpha)./(2.*gamma(alpha+3))).*sum((y1(t(j),x(j),y(j),z(j)). *(t(j).^beta(t(j)).*(((beta(t(j+1))-beta(t(j)))./h).*log(t(j)) +beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*y1(t(j-1),x(j-1),y(j-1), z(j-1))+(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))).*y1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha. *(2.*(n-j).^2+(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10). *(n-j)+6.*alpha^2+18.*alpha+12))); z(n+1)=z(1)+((h.^alpha)./(gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j). ^alpha).*z1(t(j-2),x(j-2),y(j-2),z(j-2)).*(t(j-2).^beta(t(j-2)).* ...(((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./

408

Appendix

t(j-2))))+((h.^alpha)./(gamma(alpha+2))).*sum((z1(t(j-1),x(j-1), y(j-1),z(j-1)).*... (t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-z1(t(j-2),x(j-2),y(j-2), z(j-2)).*(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha. *(n-j+3+3*alpha)))+... ((h.^alpha)./(2.*gamma(alpha+3))).*sum((z1(t(j),x(j),y(j),z(j)). *(t(j).^beta(t(j)).*(((beta(t(j+1))-beta(t(j)))./h).*log(t(j)) +beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*z1(t(j-1),x(j-1),y(j-1), z(j-1))+(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))).*z1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha. *(2.*(n-j).^2+(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10). *(n-j)+6.*alpha^2+18.*alpha+12))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AS_Method_for_Chaotic_with_Caputo_Fractional.m %h,t(1),x(1),y(1),z(1),alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.\vspace{-2pt}

%Then algorithm starts x(2)=x(1)+h.*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h.*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h.*z1(t(1),x(1),y(1),z(1)); x(3)=x(2)+(h/2).*(3.*x1(t(2),x(2),y(2),z(2))-x1(t(1),x(1),y(1),z(1))); y(3)=y(2)+(h/2).*(3.*y1(t(2),x(2),y(2),z(2))-y1(t(1),x(1),y(1),z(1))); z(3)=z(2)+(h/2).*(3.*z1(t(2),x(2),y(2),z(2))-z1(t(1),x(1),y(1),z(1))); for n=3:N j=3:n;

Appendix

409

x(n+1)=x(1)+((h.^alpha)./(gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j). ^alpha).*x1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha)./(gamma(alpha+2))).*sum((x1(t(j-1),x(j-1),y(j-1), z(j-1))-x1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha. *(n-j+3+2*alpha)-... (n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha)./(2.*gamma(alpha+3))). *sum((x1(t(j),x(j),y(j),z(j))-2.*x1(t(j-1),x(j-1),y(j-1),z(j-1))+ ...x1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha.*(2.*(n-j). ^2+(3.*alpha+10).*(n-j)+2.*(alpha).^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2+18. *alpha+12))); y(n+1)=y(1)+((h.^alpha)./(gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j). ^alpha).*y1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha)./(gamma(alpha+2))).*sum((y1(t(j-1),x(j-1),y(j-1),z(j-1)) -y1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha.*(n-j+3+2*alpha)...(n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha)./(2.*gamma(alpha+3))). *sum((y1(t(j),x(j),y(j),z(j))-2.*y1(t(j-1),x(j-1),y(j-1),z(j-1))+... y1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha.*(2.*(n-j).^2 +(3.*alpha+10).*(n-j)+2.*(alpha).^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2+18. *alpha+12))); z(n+1)=z(1)+((h.^alpha)./(gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j). ^alpha).*z1(t(j-2),x(j-2),y(j-2),z(j-2)))+... ((h.^alpha)./(gamma(alpha+2))).*sum((z1(t(j-1),x(j-1),y(j-1), z(j-1))-z1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n+1-j).^alpha. *(n-j+3+2*alpha)-... (n-j).^alpha.*(n-j+3+3*alpha)))+((h.^alpha)./(2.*gamma(alpha+3))). *sum((z1(t(j),x(j),y(j),z(j))-2.*z1(t(j-1),x(j-1),y(j-1),z(j-1))+ ...z1(t(j-2),x(j-2),y(j-2),z(j-2))).*((n-j+1).^alpha.*(2.*(n-j).^2 +(3.*alpha+10).*(n-j)+2.*(alpha).^2+9.*alpha+12)-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2+18. *alpha+12))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

410

Appendix

AS_Method_for_Chaotic_with_CF_FractalFractional_with_Variable_Order.m %h,t(1),x(1),y(1),z(1),beta,alpha, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),M=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts x(2)=x(1)+h*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h*z1(t(1),x(1),y(1),z(1)); x(3)=x(2)+(h/2).*(3.*x1(t(2),x(2),y(2),z(2))-x1(t(1),x(1),y(1),z(1))); y(3)=y(2)+(h/2).*(3.*y1(t(2),x(2),y(2),z(2))-y1(t(1),x(1),y(1),z(1))); z(3)=z(2)+(h/2).*(3.*z1(t(2),x(2),y(2),z(2))-z1(t(1),x(1),y(1),z(1))); for n=3:N x(n+1)=x(n)+((1-alpha)/M)*(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n))) ./h).*log(t(n))+(beta(t(n))./t(n))).*x1(t(n),x(n),y(n),z(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h).*log(t(n-1)) +(beta(t(n-1))./t(n-1))).*x1(t(n-1),x(n-1),y(n-1),z(n-1)))+... alpha/M.*h.*(5/12.*t(n-2).^beta(t(n-2)).*(((beta(t(n-1)) -beta(t(n-2)))./h).*(log(t(n-2))+(beta(t(n-2)))./t(n-2))). *x1(t(n-2),x(n-2),y(n-2),z(n-2))-... 4/3.* t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h). *log(t(n-1))+(beta(t(n-1))./t(n-1))).*x1(t(n-1),x(n-1),y(n-1), z(n-1))+... 23/12.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+(beta(t(n))./t(n))).*x1(t(n),x(n),y(n),z(n))); y(n+1)=y(n)+((1-alpha)/M)*(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n))) ./h).*log(t(n))+(beta(t(n))./t(n))).*y1(t(n),x(n),y(n),z(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h).*log(t(n-1)) +(beta(t(n-1))./t(n-1))).*y1(t(n-1),x(n-1),y(n-1),z(n-1)))+... alpha/M.*h.*(5/12.*t(n-2).^beta(t(n-2)).*(((beta(t(n-1)) -beta(t(n-2)))./h).*(log(t(n-2))+(beta(t(n-2)))./t(n-2))). *y1(t(n-2),x(n-2),y(n-2),z(n-2))-... 4/3.* t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h). *log(t(n-1))+(beta(t(n-1))./t(n-1))).*y1(t(n-1),x(n-1), y(n-1),z(n-1))+... 23/12.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+(beta(t(n))./t(n))).*y1(t(n),x(n),y(n),z(n))); z(n+1)=z(n)+((1-alpha)/M)*(t(n).^beta(t(n)).*(((beta(t(n+1)) -beta(t(n)))./h).*log(t(n))+(beta(t(n))./t(n))). *z1(t(n),x(n),y(n),z(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h).

Appendix

411

*log(t(n-1))+(beta(t(n-1))./t(n-1))).*z1(t(n-1),x(n-1), y(n-1),z(n-1)))+... alpha/M.*h.*(5/12.*t(n-2).^beta(t(n-2)).*(((beta(t(n-1)) -beta(t(n-2)))./h).*(log(t(n-2))+(beta(t(n-2)))./t(n-2))). *z1(t(n-2),x(n-2),y(n-2),z(n-2))-... 4/3.* t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h). *log(t(n-1))+(beta(t(n-1))./t(n-1))).*z1(t(n-1),x(n-1), y(n-1),z(n-1))+... 23/12.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+(beta(t(n))./t(n))).*z1(t(n),x(n),y(n),z(n))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AS_Method_for_Chaotic_with_CF_Fractional.m %h,t(1),x(1),y(1),z(1),alpha tfinal, t=t(1):h:tfinal, M=1-alpha+alpha/gamma(alpha), N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts x(2)=x(1)+h*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h*z1(t(1),x(1),y(1),z(1)); x(3)=x(2)+(h/2).*(3.*x1(t(2),x(2),y(2),z(2))-x1(t(1),x(1),y(1),z(1))); y(3)=y(2)+(h/2).*(3.*y1(t(2),x(2),y(2),z(2))-y1(t(1),x(1),y(1),z(1))); z(3)=z(2)+(h/2).*(3.*z1(t(2),x(2),y(2),z(2))-z1(t(1),x(1),y(1),z(1))); for n=3:N x(n+1)=x(n)+(1-alpha)/M.*(x1(t(n),x(n),y(n),z(n))-x1(t(n-1),x(n-1), y(n-1),z(n-1)))+(alpha.*h./M).*(23/12.*x1(t(n),x(n),y(n),z(n))-... 4/3.*x1(t(n-1),x(n-1),y(n-1),z(n-1))+5/12.*x1(t(n-2),x(n-2), y(n-2),z(n-2))); y(n+1)=y(n)+(1-alpha)/M.*(y1(t(n),x(n),y(n),z(n))-y1(t(n-1),x(n-1), y(n-1),z(n-1)))+(alpha.*h./M).*(23/12.*y1(t(n),x(n),y(n),z(n))-... 4/3.*y1(t(n-1),x(n-1),y(n-1),z(n-1))+5/12.*y1(t(n-2),x(n-2), y(n-2),z(n-2)));

412

Appendix

z(n+1)=z(n)+(1-alpha)/M.*(z1(t(n),x(n),y(n),z(n))-z1(t(n-1),x(n-1), y(n-1),z(n-1)))+(alpha.*h./M).*(23/12.*z1(t(n),x(n),y(n),z(n))-... 4/3.*z1(t(n-1),x(n-1),y(n-1),z(n-1))+5/12.*z1(t(n-2),x(n-2), y(n-2),z(n-2))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AS_Method_for_Differential_Equation_with_AB_FractalFractional.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), AB=1-alpha+alpha/gamma(alpha) are parameters. %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(1)+((1-alpha)/AB)*beta.*t(n).^(beta-1).*f(t(n),u(n))+((h^(alpha). *alpha)./AB*gamma(alpha+1)).*sum(beta.*t(j-2).^(beta-1). *f(t(j-2),u(j-2)).*((n-j+1).^alpha-(n-j).^alpha))+... ((h.^(alpha).*alpha)./AB*gamma(alpha+2)).*sum((beta.*t(j-1). ^(beta-1).*f(t(j-1),u(j-1))-beta.*t(j-2).^(beta-1). *f(t(j-2),u(j-2))).*((n-j+1).^alpha.*(n-j+3+2.*alpha)-(n-j). ^alpha.*(n-j+3+3.*alpha)))+... ((h.^(alpha).*alpha)./AB*(2.*gamma(alpha+3))).*sum(((beta.*t(j). ^(beta-1).*f(t(j),u(j))-2.*beta.*t(j-1).^(beta-1). *f(t(j-1),u(j-1))+beta.*t(j-2).^(beta-1).*f(t(j-2),u(j-2)))). *(((n-j+1).^alpha.*(2.*(n-j).^2+(3.*alpha+10).*(n-j)+2. *alpha.^2+9.*alpha+12))-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6. *alpha.^2+18.*alpha+12)));

Appendix

413

t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_AB_FractalFractional_with_Variable_Order.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), AB=1-alpha+alpha/gamma(alpha) are parameters. %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(1)+t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+beta(t(n))./t(n)).*(1-alpha)./AB.*f(t(n),u(n))+... ((h.^alpha).*alpha./(AB.*gamma(alpha+1))).*sum(((n+1-j). ^alpha-(n-j).^alpha).*f(t(j-2),u(j-2)).*(t(j-2).^beta(t(j-2)).* ... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2))))+((h.^alpha).*alpha./(AB.*gamma(alpha+2))). *sum((f(t(j-1),u(j-1)).*... (t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-f(t(j-2),u(j-2)).*(t(j-2). ^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./ t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha. *(n-j+3+3*alpha)))+... ((h.^alpha).*alpha./(2.*AB.*gamma(alpha+3))).*sum((f(t(j),u(j)). *(t(j).^beta(t(j)).*(((beta(t(j+1))-beta(t(j)))./h).*log(t(j)) +beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*f(t(j-1),u(j-1))+(t(j-2). ^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2))./

414

Appendix

t(j-2))).*f(t(j-2),u(j-2))).*((n-j+1).^alpha.*(2.*(n-j).^2 +(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10). *(n-j)+6.*alpha^2+18.*alpha+12))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_AB_Fractional.m clc;clear;close all; %h,t(1),u(1),alpha, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), AB=1-alpha+alpha/gamma(alpha) are parameters. %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(1)+((1-alpha)/AB).*f(t(n),u(n))+((h^(alpha).*alpha)./AB.* gamma(alpha+1)).*sum(f(t(j-2),u(j-2)).*((n-j+1).^alpha-(n-j).^alpha)) +((h.^(alpha).*alpha)./AB*gamma(alpha+2)).*sum((f(t(j-1),u(j-1)) -f(t(j-2),u(j-2))).*((n-j+1).^alpha.*(n-j+3+2.*alpha)-(n-j). ^alpha.*(n-j+3+3.*alpha)))+... ((h.^(alpha).*alpha)./AB*(2.*gamma(alpha+3))).*sum(((f(t(j),u(j)) -2.*f(t(j-1),u(j-1))+f(t(j-2),u(j-2)))).*(((n-j+1).^alpha. *(2.*(n-j).^2+(3.*alpha+10).*(n-j)+2.*alpha.^2+9.*alpha+12))-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2 +18.*alpha+12))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

Appendix

415

AS_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters. %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(1)+(h^(alpha)/gamma(alpha+1)).*sum(beta.*t(j-2).^(beta-1). *f(t(j-2),u(j-2)).*((n-j+1).^alpha-(n-j).^alpha))+... (h.^(alpha)./gamma(alpha+2)).*sum((beta.*t(j-1).^(beta-1). *f(t(j-1),u(j-1))-beta.*t(j-2).^(beta-1).*f(t(j-2),u(j-2))). *((n-j+1).^alpha.*(n-j+3+2.*alpha)-(n-j).^alpha. *(n-j+3+3.*alpha)))+... (h.^(alpha)./(2.*gamma(alpha+3))).*sum(((beta.*t(j).^(beta-1). *f(t(j),u(j))-2.*beta.*t(j-1).^(beta-1).*f(t(j-1),u(j-1)) +beta.*t(j-2).^(beta-1).*f(t(j-2),u(j-2)))).*(((n-j+1).^alpha. *(2.*(n-j).^2+(3.*alpha+10).*(n-j)+2.*alpha.^2+9.*alpha+12))-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6. *alpha.^2+18.*alpha+12))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_Caputo_Fractional.m clc;clear;close all; %h,t(1),u(1),alpha,tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters. %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value

416

Appendix

u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(1)+(h^(alpha)/gamma(alpha+1)).*sum(f(t(j-2),u(j-2)). *((n-j+1).^alpha-(n-j).^alpha))+... (h.^(alpha)./gamma(alpha+2)).*sum((f(t(j-1),u(j-1)) -f(t(j-2),u(j-2))).*((n-j+1).^alpha.*(n-j+3+2.*alpha)-(n-j). ^alpha.*(n-j+3+3.*alpha)))+... (h.^(alpha)./(2.*gamma(alpha+3))).*sum(((f(t(j),u(j)) -2.*f(t(j-1),u(j-1))+f(t(j-2),u(j-2)))).*(((n-j+1).^alpha. *(2.*(n-j).^2+(3.*alpha+10).*(n-j)+2.*alpha.^2+9.*alpha+12))-... (n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10).*(n-j)+6.*alpha.^2 +18.*alpha+12))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional_with_Variable_Order.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters. %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(1)+((h.^alpha)./(gamma(alpha+1))).*sum(((n+1-j).^alpha-(n-j). ^alpha).*f(t(j-2),u(j-2)).*(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2)). /t(j-2))))+((h.^alpha)./(gamma(alpha+2))).*sum((f(t(j-1), u(j-1)).*...

Appendix

417

(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1)))-f(t(j-2),u(j-2)).*(t(j-2). ^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2)). /t(j-2)))).*((n+1-j).^alpha.*(n-j+3+2*alpha)-(n-j).^alpha. *(n-j+3+3*alpha)))+... ((h.^alpha)./(2.*gamma(alpha+3))).*sum((f(t(j),u(j)).*(t(j). ^beta(t(j)).*(((beta(t(j+1))-beta(t(j)))./h).*log(t(j)) +beta(t(j))./t(j)))-... 2.*(t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h). *log(t(j-1))+beta(t(j-1))./t(j-1))).*f(t(j-1),u(j-1)) +(t(j-2).^beta(t(j-2)).*... (((beta(t(j-1))-beta(t(j-2)))./h).*log(t(j-2))+beta(t(j-2)). /t(j-2))).*f(t(j-2),u(j-2))).*((n-j+1).^alpha.*(2.*(n-j). ^2+(3*alpha+10).*(n-j)+... 2.*(alpha).^2+9*alpha+12)-(n-j).^alpha.*(2.*(n-j).^2+(5.*alpha+10). *(n-j)+6.*alpha^2+18.*alpha+12))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_CF_FractalFractional.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), M=1-alpha+alpha/gamma(alpha) are parameters %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(1)+((1-alpha)/M)*(beta.*t(n)^(beta-1)*f(t(n),u(n))-beta. *t(n-1)^(beta-1).*f(t(n-1),u(n-1)))+alpha/M.*h.*(5/12.*beta. *t(n-2)^(beta-1).*f(t(n-2),u(n-2))-... 4/3.*beta.*t(n-1)^(beta-1).* f(t(n-1),u(n-1))+ 23/12.

418

Appendix

*beta.*t(n)^(beta-1).*f(t(n),u(n))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_CF_Fractional.m clc;clear;close all; %h,t(1),u(1),alpha, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), M=1-alpha+alpha/gamma(alpha) are parameters. %Then algorithm starts %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method. for n=3:N j=3:n; u(n+1)=((1-alpha)/M)*(f(t(n),u(n))-f(t(n-1),u(n-1))) +alpha/M.*h.*(5/12.*f(t(n-2),u(n-2)) - 4/3.* f(t(n-1),u(n-1))+ 23/12.* f(t(n),u(n))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_CF_FractalFractional_with_Variable_Order.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), M=1-alpha+alpha/gamma(alpha) are parameters. %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1)));

Appendix

419

%Two-step Adams-Bashforth method.

%Then algorithm starts for n=3:N j=3:n; u(n+1)=u(n)+((1-alpha)/M).*(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n))). /h).*log(t(n))+(beta(t(n))./t(n))).*f(t(n),u(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h).*log(t(n-1)) +(beta(t(n-1))./t(n-1))).*f(t(n-1),u(n-1)))+... alpha/M.*h.*(5/12.*t(n-2).^beta(t(n-2)).*(((beta(t(n-1)) -beta(t(n-2)))./h).*(log(t(n-2))+(beta(t(n-2)))./t(n-2))). *f(t(n-2),u(n-2))-... 4/3.* t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h). *log(t(n-1))+(beta(t(n-1))./t(n-1))).*f(t(n-1),u(n-1))+... 23/12.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+(beta(t(n))./t(n))).*f(t(n),u(n))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AS_Method_for_Differential_Equation_with_Classical.m clc;clear;close all; %h,t(1),u(1),alpha, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method. for n=3:N j=3:n; u(n+1)=u(n)+h.*(5/12.*f(t(n-2),u(n-2))- 4/3.* f(t(n-1),u(n-1)) + 23/12.* f(t(n),u(n))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

420

Appendix

AS_Method_for_Differential_Equation_with_Fractal.m clc;clear;close all; %h,t(1),u(1),beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts %Second value u(2)=u(1)+h*f(t(1),u(1)); %One step Euler method %Third value u(3)=u(2)+(h/2)*(3*f(t(2),u(2))-f(t(1),u(1))); %Two-step Adams-Bashforth method. for n=3:N j=3:n; u(n+1)=(beta.*t(n)^(beta-1)*u(n))+h.*(5/12.*beta.*t(n-2).^(beta-1). *f(t(n-2),u(n-2))-4/3.*beta.*t(n-1)^(beta-1). *f(t(n-1),u(n-1))+ 23/12.*beta.*t(n)^(beta-1). *f(t(n),u(n))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Chaotic_with_AB_Fractal-Fractional.m %h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts for n=1:N j=2:n; x(n+1)=x(1)+(beta.*t(n).^(beta-1)*(1-alpha)/AB)*x1(t(n),x(n),y(n),z(n)) +(beta/AB)*(h^alpha.*alpha./gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*x1(t(j),x(j),y(j),z(j)).*t(j).^(beta-1)-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)).*x1(t(j-1), x(j-1),y(j-1),z(j-1)).*t(j-1).^(beta-1)); y(n+1)=y(1)+(beta.*t(n).^(beta-1)*(1-alpha)/AB)*y1(t(n),x(n),y(n),z(n)) +(beta/AB).*(h^alpha.*alpha./gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha .

Appendix

421

*(n-j+2+2*alpha)).*y1(t(j),x(j),y(j),z(j)).*t(j).^(beta-1)-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *y1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^(beta-1)); z(n+1)=z(1)+(beta*t(n)^(beta-1)*(1-alpha)/AB)*z1(t(n),x(n),y(n),z(n)) +(beta/AB)*(h^alpha.*alpha./gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*z1(t(j),x(j),y(j),z(j)).*t(j).^(beta-1)-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *z1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^(beta-1)); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AT_Method_for_Chaotic_with_AB_FractalFractional_with_Variable_Order.m %h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts for n=1:N j=2:n; x(n+1)=x(1)+(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+beta(t(n))./t(n)).*(1-alpha)/AB). *x1(t(n),x(n),y(n),z(n))+... (beta*alpha/AB).*(h.^alpha./gamma(alpha+2)).*sum(((n+1-j). ^alpha .* (n-j+2+alpha)-(n-j).^alpha .* (n-j+2+2.*alpha)).*... x1(t(j),x(j),y(j),z(j)).*t(j).^beta(t(j)). *(((beta(t(j+1))-beta(t(j)))./h).*log(t(j))+beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *x1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^beta(t(j-1)). *(((beta(t(j))-beta(t(j-1)))./h).*... log(t(j-1))+beta(t(j-1))./t(j-1))); y(n+1)=y(1)+(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+beta(t(n))./t(n)).*(1-alpha)/AB). *y1(t(n),x(n),y(n),z(n))+...

422

Appendix

(beta*alpha/AB).*(h.^alpha./gamma(alpha+2)). *sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2.*alpha)).*... y1(t(j),x(j),y(j),z(j)).*t(j).^beta(t(j)). *(((beta(t(j+1))-beta(t(j)))./h).*log(t(j)) +beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *y1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^beta(t(j-1)). *(((beta(t(j))-beta(t(j-1)))./h).*... log(t(j-1))+beta(t(j-1))./t(j-1))); z(n+1)=z(1)+(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+beta(t(n))./t(n)).*(1-alpha)/AB). *z1(t(n),x(n),y(n),z(n))+... (beta*alpha/AB).*(h.^alpha./gamma(alpha+2)).*sum(((n+1-j). ^alpha .* (n-j+2+alpha)-(n-j).^alpha .* (n-j+2+2.*alpha)).*... z1(t(j),x(j),y(j),z(j)).*t(j).^beta(t(j)).*(((beta(t(j+1)) -beta(t(j)))./h).*log(t(j))+beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *z1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^beta(t(j-1)). *(((beta(t(j))-beta(t(j-1)))./h).*... log(t(j-1))+beta(t(j-1))./t(j-1))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AT_Method_for_Chaotic_with_AB_Fractional.m %h,t(1),u(1),alpha tfinal,AB=1-alpha+alpha/gamma(alpha), t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts for n=1:N j=2:n; x(n+1)=x(1)+((1-alpha)/AB)*x1(t(n),x(n),y(n),z(n))+(h^alpha.*alpha. /gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha .

Appendix

*(n-j+2+2*alpha)).*x1(t(j),x(j),y(j),z(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .*(n-j+1+alpha)). *x1(t(j-1),x(j-1),y(j-1),z(j-1))); y(n+1)=y(1)+((1-alpha)/AB)*y1(t(n),x(n),y(n),z(n))+(beta/AB). *(h^alpha.*alpha./gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*y1(t(j),x(j),y(j),z(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *y1(t(j-1),x(j-1),y(j-1),z(j-1))); z(n+1)=z(1)+((1-alpha)/AB)*z1(t(n),x(n),y(n),z(n))+(beta/AB) *(h^alpha.*alpha./gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*z1(t(j),x(j),y(j),z(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *z1(t(j-1),x(j-1),y(j-1),z(j-1))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AT_Method_for_Chaotic_with_Caputo_FractalFractional_with_Variable_Order.m %h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts for n=1:N j=2:n; x(n+1)=x(1)+(h.^alpha./gamma(alpha+2)).*sum(((n+1-j).^alpha . *(n-j+2+alpha)-(n-j).^alpha .*(n-j+2+2.*alpha)).*... x1(t(j),x(j),y(j),z(j)).*t(j).^beta(t(j)).*(((beta(t(j+1)) -beta(t(j)))./h).*log(t(j))+beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *x1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^beta(t(j-1)). *(((beta(t(j))-beta(t(j-1)))./h).*... log(t(j-1))+beta(t(j-1))./t(j-1)));

423

424

Appendix

y(n+1)=y(1)+(h.^alpha./gamma(alpha+2)).*sum(((n+1-j).^alpha . *(n-j+2+alpha)-(n-j).^alpha .*(n-j+2+2.*alpha)).*... y1(t(j),x(j),y(j),z(j)).*t(j).^beta(t(j)). *(((beta(t(j+1))-beta(t(j)))./h).*log(t(j)) +beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *y1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^beta(t(j-1)). *(((beta(t(j))-beta(t(j-1)))./h).*... log(t(j-1))+beta(t(j-1))./t(j-1))); z(n+1)=z(1)+(h.^alpha./gamma(alpha+2)).*sum(((n+1-j).^alpha . *(n-j+2+alpha)-(n-j).^alpha .* (n-j+2+2.*alpha)).*... z1(t(j),x(j),y(j),z(j)).*t(j).^beta(t(j)).*(((beta(t(j+1)) -beta(t(j)))./h).*log(t(j))+beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *z1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^beta(t(j-1)). *(((beta(t(j))-beta(t(j-1)))./h).*... log(t(j-1))+beta(t(j-1))./t(j-1))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AT_Method_for_Chaotic_with_Caputo_FractalFractional.m %h,t(1),u(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts for n=1:N j=2:n; x(n+1)=x(1)+(beta)*(h^alpha/gamma(alpha+2)).*... sum(((n+1-j).^alpha .*(n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*x1(t(j),x(j),y(j),z(j)).*t(j).^(beta-1)-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *x1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^(beta-1)); y(n+1)=y(1)+(beta).*(h^alpha/gamma(alpha+2)).*...

Appendix

sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*y1(t(j),x(j),y(j),z(j)).*t(j).^(beta-1)-... ((n+1-j).^(alpha+1)-(n-j).^alpha .*(n-j+1+alpha)). *y1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^(beta-1)); z(n+1)=z(1)+(beta)*(h^alpha/gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*z1(t(j),x(j),y(j),z(j)).*t(j).^(beta-1)-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *z1(t(j-1),x(j-1),y(j-1),z(j-1)).*t(j-1).^(beta-1)); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AT_Method_for_Chaotic_with_Caputo_Fractional.m %h,t(1),u(1),alpha, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts for n=1:N j=2:n; x(n+1)=x(1)+(h^alpha/gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*x1(t(j),x(j),y(j),z(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *x1(t(j-1),x(j-1),y(j-1),z(j-1))); y(n+1)=y(1)+(h^alpha/gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*y1(t(j),x(j),y(j),z(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *y1(t(j-1),x(j-1),y(j-1),z(j-1))); z(n+1)=z(1)+(h^alpha/gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2*alpha)).*z1(t(j),x(j),y(j),z(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *z1(t(j-1),x(j-1),y(j-1),z(j-1)));

425

426

Appendix

t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AT_Method_for_Chaotic_with_CF_Fractal-Fractional.m %h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, M=1-alpha+alpha/gamma(alpha), N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts x(2)=x(1)+h*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h*z1(t(1),x(1),y(1),z(1)); for n=2:N x(n+1)=x(n)+(1-alpha)/M*(beta.*t(n).^(beta-1)*x1(t(n),x(n),y(n),z(n)) -beta.*t(n-1).^(beta-1)*x1(t(n-1),x(n-1),y(n-1),z(n-1)))+... alpha.*M.*h.*(3/2.*x1(t(n),x(n),y(n),z(n))-1/2. *x1(t(n-1),x(n-1),y(n-1),z(n-1))); y(n+1)=y(n)+(1-alpha)/M*(beta.*t(n).^(beta-1)*y1(t(n),x(n),y(n),z(n)) -y1(t(n-1),x(n-1),y(n-1),z(n-1)))+... alpha.*M.*h.*(3/2.*beta.*t(n).^(beta-1)*y1(t(n),x(n),y(n),z(n)) -1/2.*beta.*t(n-1).^(beta-1)*y1(t(n-1),x(n-1),y(n-1),z(n-1))); z(n+1)=z(n)+(1-alpha)/M*(beta.*t(n).^(beta-1)*z1(t(n),x(n),y(n),z(n)) -z1(t(n-1),x(n-1),y(n-1),z(n-1)))+... alpha.*M.*h.*(3/2.*beta.*t(n).^(beta-1)*z1(t(n),x(n),y(n),z(n)) -1/2.*beta.*t(n-1).^(beta-1)*z1(t(n-1),x(n-1),y(n-1),z(n-1))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

Appendix

AT_Method_for_Chaotic_with_CF_FractalFractional_with_Variable_Order.m %h,t(1),x(1),y(1),z(1),beta,alpha tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),M=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts x(2)=x(1)+h*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h*z1(t(1),x(1),y(1),z(1)); for n=2:N x(n+1)=x(n)+((1-alpha)/M)*(t(n).^beta(t(n)).*(((beta(t(n+1)) -beta(t(n)))./h).*log(t(n))+(beta(t(n))./t(n))). *x1(t(n),x(n),y(n),z(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h). *log(t(n-1))+(beta(t(n-1))./t(n-1))).*x1(t(n-1),x(n-1), y(n-1),z(n-1)))+... alpha/M.*h.*(-1/2.* t(n-1).^beta(t(n-1)).*(((beta(t(n)) -beta(t(n-1)))./h).*log(t(n-1))+(beta(t(n-1))./t(n-1))). *x1(t(n-1),x(n-1),y(n-1),z(n-1))+... 3/2.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+(beta(t(n))./t(n))).*x1(t(n),x(n),y(n),z(n))); y(n+1)=y(n)+((1-alpha)/M)*(t(n).^beta(t(n)).*(((beta(t(n+1)) -beta(t(n)))./h).*log(t(n))+(beta(t(n))./t(n))). *y1(t(n),x(n),y(n),z(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h). *log(t(n-1))+(beta(t(n-1))./t(n-1))).*y1(t(n-1),x(n-1), y(n-1),z(n-1)))+... alpha/M.*h.*(-1/2.* t(n-1).^beta(t(n-1)).*(((beta(t(n)) -beta(t(n-1)))./h).*log(t(n-1))+(beta(t(n-1))./t(n-1))). *y1(t(n-1),x(n-1),y(n-1),z(n-1))+... 3/2.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+(beta(t(n))./t(n))).*y1(t(n),x(n),y(n),z(n))); z(n+1)=z(n)+((1-alpha)/M)*(t(n).^beta(t(n)).*(((beta(t(n+1)) -beta(t(n)))./h).*log(t(n))+(beta(t(n))./t(n))). *z1(t(n),x(n),y(n),z(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h). *log(t(n-1))+(beta(t(n-1))./t(n-1))).*z1(t(n-1),x(n-1), y(n-1),z(n-1)))+... alpha/M.*h.*(-1/2.* t(n-1).^beta(t(n-1)).*(((beta(t(n)) -beta(t(n-1)))./h).*log(t(n-1))+(beta(t(n-1))./t(n-1))). *z1(t(n-1),x(n-1),y(n-1),z(n-1))+... 3/2.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h).

427

428

Appendix

*log(t(n))+(beta(t(n))./t(n))).*z1(t(n),x(n),y(n),z(n))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

AT_Method_for_Chaotic_with_CF_Fractional.m %h,t(1),u(1),alpha tfinal, t=t(1):h:tfinal,M=1-alpha+alpha/gamma(alpha), N=ceil((tfinal-t(1))/h) are parameters. %First value x(2)=x(1)+h*x1(t(1),x(1),y(1),z(1)); y(2)=y(1)+h*y1(t(1),x(1),y(1),z(1)); z(2)=z(1)+h*z1(t(1),x(1),y(1),z(1));

%Then algorithm starts for n=2:N x(n+1)=x(n)+(1-alpha)/M.*(x1(t(n),x(n),y(n),z(n))-x1(t(n-1),x(n-1), y(n-1),z(n-1)))+... alpha.*M.*h.*(3/2.*x1(t(n),x(n),y(n),z(n))-1/2.*x1(t(n-1),x(n-1), y(n-1),z(n-1))); y(n+1)=y(n)+(1-alpha)/M.*(y1(t(n),x(n),y(n),z(n))-y1(t(n-1),x(n-1), y(n-1),z(n-1)))+... alpha.*M.*h.*(3/2.*y1(t(n),x(n),y(n),z(n))-1/2.*y1(t(n-1),x(n-1), y(n-1),z(n-1))); z(n+1)=z(n)+(1-alpha)/M.*(z1(t(n),x(n),y(n),z(n))-z1(t(n-1),x(n-1), y(n-1),z(n-1)))+... alpha.*M.*h.*(3/2.*z1(t(n),x(n),y(n),z(n))-1/2.*z1(t(n-1),x(n-1), y(n-1),z(n-1))); t(n+1)=t(n)+h; end curve=animatedline; for k=1:length(t) addpoints(curve,x(k),y(k),z(k)) drawnow end

%After the required functions are added, code can be used.

Appendix

429

AT_Method_for_Differential_Equation_with_AB_FractalFractional.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts for n=1:N j=2:n; u(n+1)=u(1)+(beta.*t(n).^(beta-1).*(1-alpha)/AB).*f(t(n),u(n)) +(beta/AB).*((h.^alpha).*alpha./gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2.*alpha)).*f(t(j),u(j)).*t(j).^(beta-1)-... ((n+1-j).^(alpha+1)-(n-j).^alpha .*(n-j+1+alpha)). *f(t(j-1),u(j-1)).*t(j-1).^(beta-1)); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_AB_FractalFractional_with_Variable_Order.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts for n=1:N j=2:n; u(n+1)=u(1)+(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h). *log(t(n))+beta(t(n))./t(n)).*(1-alpha)/AB).*f(t(n),u(n))+... (beta*alpha/AB).*(h.^alpha./gamma(alpha+2)).*sum(((n+1-j).^alpha. *(n-j+2+alpha)-(n-j).^alpha .* (n-j+2+2.*alpha)).*... f(t(j),u(j)).*t(j).^beta(t(j)).*(((beta(t(j+1))-beta(t(j)))./h). *log(t(j))+beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)).*f(t(j-1), u(j-1)).\*t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h).*...

430

Appendix

log(t(j-1))+beta(t(j-1))./t(j-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_AB_Fractional.m clc;clear;close all; %h,t(1),u(1),alpha, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h),AB=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts for n=1:N j=2:n; u(n+1)=u(1)+((1-alpha)/AB).*f(t(n),u(n))+(alpha/AB). *(h.^alpha./gamma(alpha+2)).*... sum(((n+1-j).^alpha .* (n-j+2+alpha)-(n-j).^alpha . *(n-j+2+2.*alpha)).*f(t(j),u(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .*(n-j+1+alpha)). *f(t(j-1),u(j-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional_with_Variable_Order.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts for n=1:N j=2:n;

Appendix

431

u(n+1)=u(1)+(h.^alpha./gamma(alpha+2)).*sum(((n+1-j).^alpha . *(n-j+2+alpha)-(n-j).^alpha .* (n-j+2+2.*alpha)).*... f(t(j),u(j)).*t(j).^beta(t(j)).*(((beta(t(j+1)) -beta(t(j)))./h).*log(t(j))+beta(t(j))./t(j))-... ((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)).*f(t(j-1), u(j-1)).*t(j-1).^beta(t(j-1)).*(((beta(t(j))-beta(t(j-1)))./h).* ...log(t(j-1))+beta(t(j-1))./t(j-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_Caputo_Fractal-Fractional.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts for n=1:N j=2:n; u(n+1)=u(1)+(h.^alpha./gamma(alpha+2)).*sum(((n+1-j).^alpha . *(n-j+2+alpha)-(n-j).^alpha .* (n-j+2+2.*alpha)).*f(t(j),u(j))...((n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *f(t(j-1),u(j-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_Caputo_Fractional.m clc;clear;close all; %h,t(1),u(1),alpha, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

432

Appendix

%Then algorithm starts for n=1:N j=2:n; u(n+1)=u(1)+(h.^alpha./gamma(alpha+2)).*sum(((n+1-j).^alpha . *(n-j+2+alpha)-(n-j).^alpha .* (n-j+2+2.*alpha)).*f(t(j),u(j))...(n+1-j).^(alpha+1)-(n-j).^alpha .* (n-j+1+alpha)). *f(t(j-1),u(j-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_CF_FractalFractional.m %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), M=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts for n=2:N u(n+1)=u(n)+((1-alpha)/M)*(beta.*t(n).^(beta-1).*f(t(n),u(n))-beta.*t(n1). ^(beta-1).*f(t(n-1),u(n-1)))+alpha.*beta/M.*h.*(3/2.*t(n).^(beta1). *f(t(n),u(n))-... 1/2.*t(n-1).^(beta-1).* f(t(n-1),u(n-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_CF_FractalFractional_with_Variable_Order.m clc;clear;close all; %h,t(1),u(1),alpha,beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h), M=1-alpha+alpha/gamma(alpha) are parameters.

Appendix

433

%Then algorithm starts for n=2:N u(n+1)=u(n)+((1-alpha)/M)*(t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n))). /h).*log(t(n))+(beta(t(n))./t(n))).*f(t(n),u(n))-... t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n-1)))./h).*log(t(n-1)) +(beta(t(n-1))./t(n-1))).*f(t(n-1),u(n-1)))+... alpha/M.*h.*(-1/2.* t(n-1).^beta(t(n-1)).*(((beta(t(n))-beta(t(n1))) ./h).*log(t(n-1))+(beta(t(n-1))./t(n-1))).*f(t(n-1),u(n-1))+... 3/2.* t(n).^beta(t(n)).*(((beta(t(n+1))-beta(t(n)))./h).*log(t(n)) +(beta(t(n))./t(n))).*f(t(n),u(n))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_CF_Fractional.m %h,t(1),u(1),alpha,tfinal, t=t(1):h:tfinal; N=ceil((tfinal-t(1))/h), M=1-alpha+alpha/gamma(alpha) are parameters.

%Then algorithm starts for n=2:N u(n+1)=u(n)+((1-alpha)/M)*(f(t(n),u(n))-f(t(n-1),u(n-1))) +alpha./M.*h.*(3/2.*f(t(n),u(n))-1/2.*f(t(n-1),u(n-1))); t(n+1)=t(n)+h; end plot(t,u,’r-’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_Classical.m %h,t(1),u(1), tfinal, t=t(1):h:tfinal; N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts

434

Appendix

for n=2:N u(n+1)=u(n)+h.*(3/2.*f(t(n),u(n))-1/2.* f(t(n-1),u(n-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

AT_Method_for_Differential_Equation_with_Fractal.m %h,t(1),u(1),beta, tfinal, t=t(1):h:tfinal, N=ceil((tfinal-t(1))/h) are parameters.

%Then algorithm starts for n=2:N u(n+1)=u(n)+beta.*h.*(3/2.*t(n).^(beta-1).*f(t(n),u(n))-1/2.*t(n-1). ^(beta-1).* f(t(n-1),u(n-1))); t(n+1)=t(n)+h; end plot(t,u,’-r’) hold on

%After the required function is added, code can be used.

A.1

Supplementary material

Supplementary material related to this chapter can be found online at https://doi.org/ 10.1016/B978-0-32-385448-1.00029-9.

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Index

A Accuracy, 24, 52, 93, 143, 190 Advection–diffusion equation, 259, 268, 277 Arneodo chaotic model, 177 system, 177 Atangana–Baleanu derivative, 43–46, 48, 49, 264, 289 fractal–fractional, 68, 73–76, 78, 79, 202, 206, 209 fractional, 38, 39, 43, 46, 155, 157, 159–161, 263, 291, 348 integral, 349 fractal–fractional, 68, 70 operator, 38 fractal–fractional, 204, 270, 389 fractional, 259, 378 Attractor, 65 ACT, 20 asymptotic, 149 chaotic 3-D four-wing smooth autonomous, 248 four scroll Wang–Sun, 248 Chua, 149, 339 Dadras–Momeni, 36, 38, 39 Dequan Li, 65, 67, 68 Halvorsen, 58, 59 hybrid, 339, 341, 345 King–Cobra, 129 Labyrinth, 138–140 Lorenz, 65, 210 Nose–Hoover, 76, 78, 79 problem Hadley, 46 Halvorsen, 56 B B-spline, 7 polynomial, 7 Basis spline function, 7

Bernstein polynomial, 3, 4 basis, 4 Bi-cubic interpolation, 3 Boussinesq fluid, 109 Burgers equation, 375, 378, 384 error, 6 C Caputo, 38, 155, 199 derivative, 34–36, 38, 39, 143, 269, 291, 293, 348 fractal–fractional, 49, 51, 53–59, 187, 194, 313, 358 fractional, 29, 33–35, 141, 143, 146–150, 152, 153, 267, 291, 319, 330, 342, 384 integral, 265 operator differential, 101, 225 fractal–fractional, 52, 53, 274 fractional, 29, 33, 264, 319, 330, 384 Caputo–Fabrizio case, 26 derivative, 25–27, 29, 30, 271, 303, 305 fractal–fractional, 58, 60, 63–65, 67, 68, 169, 171, 173, 175–180, 309 fractional, 22, 28, 131, 133, 135–140, 270 integral, 132 fractal–fractional, 59, 169, 294, 303, 307 operator, 22 differential, 173 fractal–fractional, 60, 62, 277, 322, 333 fractional, 26, 268 Cauchy problem, 11–13, 17, 18, 39, 49, 51, 58, 60, 67, 80, 113, 115, 121, 123, 132, 133, 143, 155, 157, 169, 171, 173, 184, 187, 199, 202, 218, 228–230, 232, 243–245, 247, 281, 284

440

Chaotic attractor 3-D four-wing smooth, 248 four scroll Wang–Sun, 248 model, 36 Arneodo, 177 problem, 98, 118 system, 36, 125 King Cobra, 128 Chebyshev polynomial, 3, 8, 9 Chen–Lee system, 28–30 Chua attractor, 149, 339 circuit, 149, 221 model, 152, 153 Chua–Hartley system, 87 Constant lateral, 109 order, 101, 215, 278 fractional, 79, 85, 89, 96, 97, 101, 107, 218, 222, 228, 230, 232, 233, 243, 246, 248, 365 parameters, 66, 101, 237 Continuous function, 48, 79, 89, 101 Convection problem, 109 Crossover property, 89, 239, 281 Cubic equation, 7 formula, 7 nonlinearity, 87 polynomial, 7 D Dadras–Momeni attractor, 36, 38, 39 system, 36 Dengue model, 339, 348, 355 problem, 348 Dequan Li attractor, 65, 67, 68 Derivation, 101, 102, 239, 281, 284 Derivative, 22 Atangana–Baleanu, 43–46, 48, 49, 264, 289 fractal–fractional, 68, 73–76, 78, 79, 202, 209 fractional, 291 Caputo, 34–36, 38, 39, 143, 269, 291, 293, 348

Index

fractal–fractional, 49, 51, 53–58, 187, 194, 313, 358 Caputo fractal–fractional, 59 Caputo–Fabrizio, 25–27, 29, 30, 271, 303, 305 fractal–fractional, 58, 60, 63–65, 67, 68, 169, 171, 173, 175–180, 309 classical, 12–16, 115, 117–119, 256, 289, 296, 301, 303, 327, 339 extension, 11 fractal, 16–20, 121, 123, 125, 126, 128, 256, 258, 305, 307, 317, 329 fractal–fractional, 202, 229, 231–233, 236, 237, 244, 246–248, 251, 252, 296, 307, 358, 365 fractional, 23, 79, 89, 101, 110, 132, 239, 280 Atangana–Baleanu, 38, 39, 43, 46, 155, 157, 159–161, 263, 348 Caputo, 29, 33–35, 141, 143, 146–150, 152, 153, 267, 291, 319, 330, 342, 384 Caputo–Fabrizio, 22, 28, 131, 133, 135–140, 270 Differential function, 79, 89, 101 operator, 17, 25, 34, 38, 43, 79, 89, 90, 93, 95, 101, 106, 121, 146, 215, 221, 228, 232, 239, 278, 281 Atangana–Baleanu fractal–fractional, 204, 206 Caputo, 101, 225 Caputo fractal–fractional, 52 Caputo–Fabrizio, 173, 215, 278 Caputo–Fabrizio fractal–fractional, 60 fractal–fractional, 51, 70, 87, 187, 244 Differential equation linear partial, 315, 317, 319, 322, 323 non-homogeneous linear, 287, 289, 291, 293, 296 non-linear, 301, 303, 305, 307, 310 non-linear ordinary, 301 partial, 260, 327, 329, 330, 333, 334, 375, 378, 384, 389, 395 ordinary, 339 Dimension fractal, 79, 89, 218, 219, 228, 230, 239, 243, 245, 247, 281, 296, 310

Index

variable, 79, 85, 89, 96, 97, 101, 107, 215, 218, 221, 222, 225, 228, 230, 232, 233, 243, 246, 248, 278, 365 E Ebola model, 339, 365 Equation advection–diffusion, 255, 256, 258–260, 263, 264, 268–271, 277 Burgers, 375, 378, 384 cubic, 7 differential, 22, 29, 38 classical, 11 fractal, 16, 127 partial, 1, 253, 260, 267, 315, 327, 375, 378, 384, 389, 395 Fisher, 327, 329 heat, 315, 317, 319, 322, 323 Error analysis, 24, 32, 41, 51, 60, 70, 82, 93, 104, 115, 123, 133, 143, 157, 173, 187, 204 Burgers, 6 Exponential kernel, 24, 215 decay, 22, 25, 57, 58, 62, 79, 80, 82, 84, 87, 131, 132, 169, 171, 175, 278, 293, 303, 307, 365 law, 293, 307, 322, 333, 365 F Fisher equation, 327, 329 Formula Lox–de Boor recursion, 7 Sylvester, 10 Fractal calculus, 121 case, 18, 20 derivative, 16–20, 121, 123, 125, 126, 128, 256, 258, 305, 307, 317, 329 differential equation, 16, 127 differentiation, 48 dimension, 79, 89, 218, 219, 228, 230, 239, 243, 245, 247, 281, 296, 310 variable, 79, 89, 101, 222, 233, 248 integral, 17, 121 order, 19 properties, 16, 48, 57, 66

441

Fractal–fractional derivative, 202, 229, 231–233, 236, 237, 244, 246–248, 251, 252, 296, 307, 358, 365 Atangana–Baleanu, 68, 73–76, 78, 79, 202, 206, 209 Caputo, 49, 51, 53–59, 187, 194, 313, 358 Caputo–Fabrizio, 58, 60, 63–65, 67, 68, 169, 171, 173, 175–180, 309 integral, 184 Atangana–Baleanu, 68, 70 Caputo–Fabrizio, 59, 169, 294, 303, 307 operator Atangana–Baleanu, 204, 270, 389 Caputo, 52, 53, 274 Caputo–Fabrizio, 60, 62, 277, 322, 333 differential, 51, 70, 187, 244 Fractional calculus, 11, 131, 141, 155 derivative, 23, 79, 89, 101, 110, 132, 239, 280 Atangana–Baleanu, 43, 46, 157, 159–161, 263, 348 Caputo, 29, 33–35, 141, 143, 146–150, 152, 153, 267, 291, 319, 330, 342, 384 Caputo–Fabrizio, 22, 28, 131, 133, 135–140, 270 differential equation, 162 non-linear, 24, 32 differentiation, 48, 66 integral, 22, 29, 39, 80, 90, 102, 103, 132, 141, 155, 239, 281, 284, 296, 310, 323 operator Atangana–Baleanu, 259, 378 Caputo, 29, 33, 264, 319, 330, 384 Caputo–Fabrizio, 26, 268 order, 218, 219, 221, 225, 228, 230, 243, 245, 247, 296, 310 Frobenius covariants, 10 Function basis spline, 7 continuous, 48, 79, 89, 101 differential, 79, 89, 101 Mittag-Leffler, 239 nonlinear, 1, 6 normalization, 22, 131

442

Fundamental solution, 89, 239, 280 H Hadley attractor problem, 46 Halvorsen attractor, 58, 59 problem, 56 circulant system, 55 Hartley oscillator, 339 Heat equation, 315, 317, 319, 322, 323 Hermite interpolating polynomial, 10 interpolation, 3, 6 HIV model, 339, 358, 361 Hybrid attractor, 339, 341, 345 I Integral Atangana–Baleanu, 349 fractal–fractional, 68, 70 Caputo, 29, 141, 265 Caputo–Fabrizio, 132 fractal–fractional, 59, 169, 294, 303, 307 fractal, 17, 121 fractal–fractional, 184 fractional, 22, 29, 39, 80, 90, 102, 103, 132, 141, 155, 239, 281, 284, 296, 310, 323 operator, 38, 43, 57, 79, 89, 239, 281 Interpolation bi-cubic, 3 Hermite, 3, 6 Lagrange, 11 Lagrange–Sylvester, 3, 9 monotone cubic, 3 multivariate, 3 Newton, 4, 113 polynomial, 1–3 spline, 3 tricubic, 3 K Karatsuba multiplication, 1 Kernel exponential, 24, 215 decay, 22, 25, 57, 58, 62, 79, 80, 82, 84, 87, 131, 132, 169, 171, 175, 278, 293, 303, 307, 365

Index

Mittag-Leffler, 39, 41, 45, 66, 75, 89, 90, 93, 95–99, 101, 102, 155, 162, 199, 202, 243–248, 251, 252, 263, 281, 296, 323, 334 power-law, 29, 32, 48, 49, 55, 101, 106–109, 141, 181, 184, 187, 192, 225, 229, 231–233, 236, 237, 284, 310 King Cobra chaotic system, 128 King–Cobra attractor, 129 Knots, 7 L Labyrinth attractor, 138–140 system, 138 Lagrange form, 3 interpolation, 11 polynomial, 3, 4, 11, 12, 17, 23, 31, 40, 50, 51, 60, 61, 69, 71, 81, 83, 91, 103, 113 basis, 3 Lagrange–Sylvester interpolation, 3, 9 Legendre polynomial, 8 Linear bijection, 1 differential equations, 287, 339 ordinary, 287 partial, 315, 317, 319, 322, 323 problem, 44 Lorenz attractor, 65, 210 system, 210 Lotka–Volterra 3D model, 164, 166, 167 Lox–de Boor recursion formula, 7 M Mittag-Leffler function, 239 kernel, 39, 41, 45, 66, 75, 89, 90, 93, 95–99, 101, 102, 155, 162, 199, 202, 243–248, 251, 252, 263, 281, 296, 323, 334 Model Arneodo chaotic, 177 Chua, 152, 153 dengue, 339, 348, 355 Ebola, 339, 365 HIV, 339, 358, 361

Index

Lotka–Volterra 3D, 164, 166, 167 Moore–Spiegel, 16 Rucklidge, 109, 111, 112 Monotone cubic interpolation, 3 Moore–Spiegel model, 16 system, 15 Multiplication Karatsuba, 1 sub-quadratic, 1 Toom–Cook, 1 Multivariate interpolation, 3 N Newton polynomial, 3, 4, 6, 113–115, 122, 124, 132, 134, 142, 144, 156, 170, 182, 184, 200, 202, 204, 216, 226, 240, 253, 257, 260, 265, 269, 272, 275, 277, 282, 284, 288, 289, 292, 294, 302, 306, 308, 310, 316, 320, 324, 328, 329, 335, 343, 359, 379, 390, 396 interpolation, 4, 113 Nonlinear differential equation, 82 function, 1, 6 problem, 26 Normalization function, 22, 131 Nose–Hoover attractor, 76, 78, 79 system, 76 Numerical algorithm, 58, 60, 169, 181, 256, 327, 345, 355, 361 analysis, 1, 3, 7 approximation, 41, 66, 259, 260, 270, 301, 389, 398 simulation, 16 solution, 12, 33, 38, 90, 101, 102, 125, 144, 146, 155, 162, 163, 175, 239, 263, 281, 284, 287, 327, 341 O Operator Atangana–Baleanu, 38 Caputo fractal–fractional, 274 Caputo–Fabrizio, 22

443

differential, 17, 25, 34, 90, 93, 95, 101, 106, 121, 146, 215, 221, 228, 232, 278 Atangana–Baleanu fractal–fractional, 204, 206 Caputo, 101, 225 Caputo fractal–fractional, 52 Caputo–Fabrizio, 173, 215, 278 Caputo–Fabrizio fractal–fractional, 60 fractal–fractional, 51, 70, 87, 187, 244 fractal–fractional Atangana–Baleanu, 270, 389 Caputo, 53 Caputo–Fabrizio, 62, 277, 322, 333 fractional Atangana–Baleanu, 259, 378 Caputo, 29, 33, 264, 319, 330, 384 Caputo–Fabrizio, 26, 268 integral, 38, 43, 57, 79, 89, 239, 281 Order constant, 101, 215, 278 fractional, 79, 85, 89, 96, 97, 101, 107, 218, 219, 221, 222, 225, 228, 230, 232, 233, 243, 245–248, 296, 310, 365 variable, 229, 231–233, 246–248, 251, 252, 310 Oscillator Hartley, 339 non-periodic, 221 Shaw, 339, 342, 348 P Partial differential equations, 1, 253, 267, 315 non-linear, 260, 327, 375, 378, 384, 389, 395 Polynomial B-spline, 7 Bernstein, 3, 4 basis, 4 Chebyshev, 3, 8, 9 cubic, 7 degree, 1, 2, 8 Hermite interpolating, 10 interpolation, 1–3 Lagrange, 3, 4, 11, 12, 17, 23, 31, 40, 50, 51, 60, 61, 69, 71, 81, 83, 91, 103, 113 basis, 3 Legendre, 8

444

Newton, 3, 4, 6, 113–115, 122, 124, 132, 134, 142, 144, 156, 170, 182, 184, 200, 202, 204, 216, 226, 240, 253, 257, 260, 265, 269, 272, 275, 277, 282, 284, 288, 289, 292, 294, 302, 306, 308, 310, 316, 320, 324, 328, 329, 335, 343, 359, 379, 390, 396 Power-law, 310, 358, 395 kernel, 29, 32, 48, 49, 55, 101, 106–109, 141, 181, 184, 187, 192, 225, 229, 231–233, 236, 237, 284, 310 Predictor–corrector method, 171, 184, 202 Problem Cauchy, 11–13, 17, 18, 39, 49, 51, 58, 60, 67, 80, 113, 115, 121, 123, 132, 133, 143, 155, 157, 169, 171, 173, 184, 187, 199, 202, 218, 228–230, 232, 243–245, 247, 281, 284 chaotic, 98, 118 convection, 109 dengue, 348 Hadley attractor, 46 linear, 44 nonlinear, 26 Property crossover, 89, 239, 281 fractal, 16, 48, 57, 66

Index

Shaw oscillator, 339, 342, 348 Simulation numerical, 16 Spline interpolation, 3 Sylvester formula, 10 System Arneodo, 177 chaotic, 36, 125 King Cobra, 128 three-dimensional quadratic autonomous, 65 Chen–Lee, 28–30 Chua–Hartley, 87 Dadras–Momeni, 36 Halvorsen circulant, 55 Labyrinth, 138 Lorenz, 210 Moore–Spiegel, 15 Nose–Hoover, 76 Rabinovich–Fabrikant, 233, 236, 237 Rikitake, 194 Rucklidge, 109 Wang–Sun, 248 T Toom–Cook multiplication, 1 Tricubic interpolation, 3

R Rabinovich–Fabrikant system, 233, 236, 237 Rikitake system, 194 Rucklidge model, 109, 111, 112 system, 109

V Vandermonde matrix, 1, 2 Variable fractal dimension, 85, 96, 97, 101, 107, 215, 218, 221, 225, 228, 230, 232, 243, 246, 278, 365 order, 229, 231–233, 246–248, 251, 252, 310

S Scaling factor, 7

W Wang–Sun system, 248