Special functions and generalized Sturm-Liouville problems 9783030328191, 9783030328207

Table of contents :
Preface......Page 6
Contents......Page 10
1.1 Introduction......Page 13
1.2 Pearson Distribution Families and Gauss Hypergeometric Functions......Page 14
1.3 Six Sequences of Hypergeometric Orthogonal Polynomials......Page 24
1.3.2 Laguerre Polynomials......Page 25
1.3.3 Hermite Polynomials......Page 26
1.3.4 First Finite Sequence of Hypergeometric Orthogonal Polynomials......Page 27
1.3.5 Second Finite Sequence of Hypergeometric Orthogonal Polynomials......Page 29
1.3.6 Third Finite Sequence of Hypergeometric Orthogonal Polynomials......Page 32
1.4 A Generic Polynomial Solution for the Classical Hypergeometric Differential Equation......Page 37
1.4.1 Six Special Cases of the Generic Polynomials n(x;a,b,c,d,e)......Page 48
1.4.2 How to Find the Initial Vector If a Special Case of the Main Weight Function Is Given?......Page 52
1.5 Fourier Transforms of Finite Sequences of Hypergeometric Orthogonal Polynomials......Page 54
1.6 A Symmetric Generalization of Sturm–Liouville Problems......Page 59
1.7 A Basic Class of Symmetric Orthogonal Polynomials......Page 61
1.7.1 A Direct Relationship Between n(x;a,b,c,d,e) and n(x;p,q,r,s)......Page 66
1.7.2 Four Special Cases of the Symmetric Polynomials Sn(x;p,q,r,s)......Page 68
1.7.3 A Unified Approach for the Classification of n(x;p,q,r,s)......Page 78
1.8 Fourier Transforms of Symmetric Orthogonal Polynomials......Page 81
1.8.1 Fourier Transform of Generalized Ultraspherical Polynomials......Page 82
1.8.2 Fourier Transform of Generalized Hermite Polynomials......Page 84
1.8.3 Fourier Transform of the First Finite Sequence of Symmetric Orthogonal Polynomials......Page 88
1.8.4 Fourier Transform of the Second Finite Sequence of Symmetric Orthogonal Polynomials......Page 92
1.9 A Class of Symmetric Orthogonal Functions......Page 95
1.9.1 Four Special Cases of Sn(θ)(x;p,q,r,s)......Page 100
1.10 An Extension of Sn(θ)(x;p,q,r,s)......Page 105
1.10.1 Four Special Cases of Sn(α,β)(x;p,q,r,s)......Page 111
1.11 A Generalization of Fourier Trigonometric Series......Page 115
1.11.1 A Generalization of Trigonometric Orthogonal Sequences......Page 116
1.11.2 Application to Function Expansion Theory......Page 120
1.12 Another Extension for Trigonometric Orthogonal Sequences......Page 124
1.13.1 Incomplete Symmetric Orthogonal Polynomials of Jacobi Type......Page 130
1.13.2 Incomplete Symmetric Orthogonal Polynomials of Laguerre Type......Page 136
1.14 A Class of Hypergeometric-Type Orthogonal Functions......Page 140
1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville Problems......Page 146
1.15.1 A Relationship Between the Chebyshev Polynomials of the Third and Fourth Kinds and the Associated Legendre Differential Equation......Page 156
Further Reading......Page 164
2.1 Introduction......Page 168
2.2 A Finite Sequence of Hahn-Type Orthogonal Polynomials......Page 174
2.3 Classical Symmetric Orthogonal Polynomials of a Discrete Variable......Page 179
2.3.1 Classification......Page 182
2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete Spaces......Page 189
2.4.1 Some Illustrative Examples......Page 194
2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete Variable......Page 208
2.5.1 Two Infinite Types of Sn*(x;a,b,c,d)......Page 217
2.5.2 Moments of the Two Introduced Infinite Sequences......Page 227
2.5.3 A Special Case of Sn*(x;a,b,c,d) Generating All Classical Symmetric Orthogonal Polynomials of a Discrete Variable......Page 228
2.5.4 Two Finite Types of Sn*(x;a,b,c,d)......Page 230
Further Reading......Page 240
3.1 Introduction......Page 243
3.2.1 The q-Shifted Factorial......Page 246
3.2.2 q-Hypergeometric Series......Page 247
3.2.3 q-Binomial Coefficients and the q-Binomial Theorem......Page 248
3.2.5 q-Analogues of Some Special Functions......Page 249
3.2.6 q-Difference Operators......Page 250
3.2.7 q-Integral Operators......Page 252
3.2.8 An Analytical Solution for the q -Pearson DifferenceEquation......Page 255
3.2.9 Difference Equations of q -Hypergeometric Series......Page 256
3.2.10 q-Analogues of Jacobi Polynomials......Page 258
3.2.11 q-Analogues of Laguerre Polynomials......Page 260
3.2.12 q-Analogue of Hermite Polynomials......Page 263
3.2.13 A Biorthogonal Exponential Sequence......Page 264
Some Remarks on Theorem 3.2......Page 269
3.3 Three Finite Classes of q-Orthogonal Polynomials......Page 272
3.3.1 First Finite Sequence of q-Orthogonal Polynomials Corresponding to the Inverse Gamma Distribution......Page 273
3.3.2 Second Finite Sequence of q-Orthogonal Polynomials Corresponding to the Fisher Distribution......Page 280
3.3.3 Third Finite Sequence of q-Orthogonal Polynomials Corresponding to Student's t-Distribution......Page 285
3.3.4 A Characterization of Three Introduced Finite Sequences......Page 291
Comparison with the Third Finite Sequence......Page 292
3.4 A Symmetric Generalization of Sturm–Liouville Problems in q-Difference Spaces......Page 293
3.4.1 Some Illustrative Examples......Page 296
3.5 A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free Parameters......Page 300
3.5.1 Two Finite Sequences Based on Ramanujan's Identity......Page 308
References......Page 316
Further Reading......Page 317
Index......Page 320

Citation preview

Frontiers in Mathematics

Mohammad Masjed-Jamei

Special Functions

and Generalized

Sturm-Liouville Problems

Frontiers in Mathematics

Advisory Editorial Board Leonid Bunimovich (Georgia Institute of Technology, Atlanta) William Y. C. Chen (Nankai University, Tianjin) Benoît Perthame (Sorbonne Université, Paris) Laurent Saloff-Coste (Cornell University, Ithaca) Igor Shparlinski (The University of New South Wales, Sydney) Wolfgang Sprößig (TU Bergakademie Freiberg) Cédric Villani (Institut Henri Poincaré, Paris)

More information about this series at http://www.springer.com/series/5388

Mohammad Masjed-Jamei

Special Functions and Generalized Sturm-Liouville Problems

Mohammad Masjed-Jamei Department of Mathematics K.N. Toosi University of Technology Tehran, Iran

ISSN 1660-8046 ISSN 1660-8054 (electronic) Frontiers in Mathematics ISBN 978-3-030-32819-1 ISBN 978-3-030-32820-7 (eBook) https://doi.org/10.1007/978-3-030-32820-7 Mathematics Subject Classification (2010): 34B24, 33C45, 33C47 © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com, by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

In recent years, Sturm–Liouville theory as an attractive field of research has received considerable attention by authors, since it appears in solving many problems of engineering, physics, biology, and the social sciences in a natural manner. Such problems generally lead to some eigenvalue problems for ordinary and partial differential equations. For instance, the associated Legendre functions, Bessel functions, ultraspherical functions, Hermite functions, and trigonometric sequences related to Fourier analysis are particular solutions of some Sturm–Liouville problems. Most of these functions are symmetric and have found various applications in physics and engineering. An extensive class of special functions includes orthogonal polynomials, whose history goes back to the eighteenth century and are closely linked with solutions of some eigenvalue problems. One of those problems was related to Newton’s theory of gravity. While solving a problem in this area, A. M. Legendre introduced a sequence of orthogonal polynomials, which are known today as Legendre polynomials. Since then, many other families of continuous orthogonal polynomials have been introduced from their direct applications in various problems. Clearly, all these families can be introduced by diverse approaches. For instance, in 1929, S. Bochner found all families of polynomials satisfying a second-order differential equation, which led to the classical continuous orthogonal polynomials. In this sense, some famous classical orthogonal polynomials associated with the names of Jacobi, Laguerre, Hermite, and Bessel have been studied in the literature in detail, and their weight functions are known nowadays as Pearson distributions. More precisely, the boundary value problems corresponding to the above-mentioned cases have been considered as examples of a singular Sturm–Liouville problem to introduce classical orthogonal polynomials as solutions of a second-order linear hypergeometric differential equation. On the other hand, in modern physics we encounter mathematical operators, eigenvalue equations, and properties of orthogonal functions. The quantum harmonic oscillator necessitates constructing an eigenvalue problem and the appearance of the discrete nature of physical quantities in quantum physics, which is considered a special case of orthogonal functions arising in the context of Sturm–Liouville theory. In other words, many important special functions are solutions of a regular or singular Sturm–Liouville problem that satisfy

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an orthogonality relation, which can be considered in different continuous and discrete spaces and also in one particular case of them, i.e., q-spaces. The present book deals mainly with introducing and classifying some generalized Sturm–Liouville problems in three different continuous, discrete, and q-discrete spaces. In this direction, some conditions under which the usual Sturm–Liouville problems with symmetric solutions can be extended to a larger class are presented. Of course, the main motivation to write this monograph comes not only from various applications of Sturm–Liouville problems in mathematical physics, but also from the necessity of gathering and collecting the theoretical aspects of the subject and its generalization in a unified structure in one place, with special attention to the solutions of the generalized problems as new orthogonal sequences of continuous or discrete functions. This book includes three chapters. The first chapter is devoted to introducing a principal class of symmetric orthogonal functions and their generalizations by presenting a generalized Sturm–Liouville theorem for symmetric functions. Some sequences of orthogonal functions of a discrete variable are introduced in the second chapter by presenting a similar theorem, which is now expressed in discrete spaces. New classes of q-orthogonal polynomials that are generated by some q-Sturm–Liouville problems are considered in Chap. 3. In Chap. 1, to study six sequences of hypergeometric orthogonal polynomials, we first show that the classical Pearson distributions and Gauss’s hypergeometric function satisfy a unique differential equation of hypergeometric type. Three polynomials, namely Jacobi, Laguerre, and Hermite, are known as infinite classical orthogonal polynomials, whereas three other sequences are called finite hypergeometric orthogonal polynomials, which are orthogonal with respect to the generalized T, inverse gamma, and F distributions respectively. Besides reviewing these classical cases in detail, a class of symmetric orthogonal polynomials with four free parameters is introduced in Chap. 1 and all its standard properties, such as a generic second-order differential equation together with its explicit polynomial solution, a generic orthogonality relation, an analogue of Pearson distributions as their weight function, and a generic three-term recurrence relation, are obtained. Essentially, four sequences of symmetric orthogonal polynomials can be extracted from the introduced class. They are respectively the generalized ultraspherical polynomials, generalized Hermite polynomials, and two other sequences of symmetric polynomials, which are finitely orthogonal on the real line. For instance Chebyshev polynomials of the fifth and sixth kinds can be defined as special cases of these generalizations. We know that four kinds of trigonometric orthogonal polynomials, i.e., the first, second, third, and fourth kinds of Chebyshev polynomials have been investigated in the literature up to now. In Chap. 1, we introduce two further kinds of half-trigonometric orthogonal polynomials and call them the fifth and sixth kinds of Chebyshev polynomials, since they are generated by employing the Chebyshev polynomials of the first and second kinds. In connection to the aforementioned symmetric orthogonal class with four free parameters and according to the fact that some orthogonal polynomial systems are mapped onto each other by the Fourier transform, we compute the Fourier transforms of four

Preface

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introduced sequences of symmetric orthogonal polynomials and obtain their orthogonality relations via Parseval’s identity. In the sequel, using the generalized Sturm–Liouville theorem for symmetric functions, a basic class of symmetric orthogonal functions and also its generalization with six parameters are introduced. Then four orthogonal special cases of these classes are represented, and their properties are studied in detail. It must be mentioned that further important cases of these classes can still be found, which becomes a motivation for further investigation into the orthogonal sequences. Just as an example, two cases that generalize Fourier trigonometric sequences and are orthogonal with respect to the constant weight function are introduced. They are useful in finding some new trigonometric series. As another application of the key theorem related to the extension of Sturm–Liouville problems with symmetric solutions, two specific differential equations are considered, one of whose basis solutions yields two classes of incomplete symmetric orthogonal polynomials. Such systems do not contain polynomials of every degree and therefore do not have all properties as in the classical cases. However, they can be directly applied to approximation theory of functions, since their norm square values are explicitly computed. In Chap. 2, orthogonal functions of a discrete variable are considered as solutions of a discrete Sturm–Liouville problem that satisfies a set of discrete boundary conditions. Interest in discrete Sturm–Liouville operators comes from a number of problems in mathematical physics, scattering theory, Toda and Langmuir chains, spectral properties of operators and so on. Similar to the continuous case, special functions of a discrete variable have found various applications in mathematics and physics. For instance, Gram polynomials and symmetric Hahn polynomials play an important role in some branches of numerical analysis, while symmetric Kravchuk polynomials appear in the so-called Fourier–Kravchuk transform used in optics and quantum mechanics as approximations of harmonic oscillator wave functions. Also, some applications of classical orthogonal polynomials of a discrete variable and their zeros can be observed in the least-squares method of approximation, queueing theory, and cross-directional control on paper machines and codes. It is proved in Chap. 2 that the previous extensions for the continuous case also hold for discrete variables and particularly for a homogeneous second-order difference equation. Hence, by using a generalization of Sturm–Liouville problems in discrete spaces, a symmetric class of orthogonal polynomials of a discrete variable, which generalizes all classical discrete symmetric orthogonal polynomials, is introduced, and its standard properties, such as a second-order difference equation, an explicit form for polynomials, a three-term recurrence relation, and an orthogonality relation, are obtained. As mentioned earlier, one of the important classes of special functions is q-orthogonal polynomials, which can be considered solutions of a q-Sturm–Liouville problem of regular or singular type. This type of polynomial has found various applications in quantum mechanics, the q-Schrödinger equation, q-harmonic oscillators, and algebraic combinatorics, including coding theory and theories of group representation. Therefore, it seems worthwhile to investigate new developments of the main ideas applied in Chaps. 1

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and 2 in this field, too. To reach this goal, in Chap. 3, some new classes of q-orthogonal polynomials are defined through new q-Sturm–Liouville problems, and then their basic properties are studied. It should be mentioned that each chapter provides valuable references in the section “Further Reading” for those who need some additional sources and more detailed materials. It is hoped that this work will be useful as a coherent scientific reference for mathematicians, physicists, and engineers to carry out their research projects and to find new applications that effectively develop previous results. In this sense, any comments, suggestions, and corrections would be much appreciated. Finally, I would like to present my special thanks to Zahra Moalemi for carefully reading this book and for her valuable assistance in typesetting. As a Humboldtian fellow for 18 months, I would also like to thank my scientific host Wolfram Koepf for his warm hospitality. The finalization of this project was made possible by the financial support of the Alexander von Humboldt foundation under the grant No. Ref 3.4-IRN-1128637-GF-E. Tehran, Iran December 2018

Mohammad Masjed-Jamei

Contents

1

Special Functions Generated by Generalized Sturm–Liouville Problems in Continuous Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.1 Introduction . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.2 Pearson Distribution Families and Gauss Hypergeometric Functions . . . . . 1.3 Six Sequences of Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . . . 1.3.1 Jacobi Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.3.2 Laguerre Polynomials.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.3.3 Hermite Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.3.4 First Finite Sequence of Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.3.5 Second Finite Sequence of Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.3.6 Third Finite Sequence of Hypergeometric Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.4 A Generic Polynomial Solution for the Classical Hypergeometric Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.4.1 Six Special Cases of the Generic Polynomials P¯n (x; a, b, c, d, e) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.4.2 How to Find the Initial Vector If a Special Case of the Main Weight Function Is Given? . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.5 Fourier Transforms of Finite Sequences of Hypergeometric Orthogonal Polynomials.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.6 A Symmetric Generalization of Sturm–Liouville Problems .. . . . . . . . . . . . . . . 1.7 A Basic Class of Symmetric Orthogonal Polynomials . . . . . .. . . . . . . . . . . . . . . 1.7.1 A Direct Relationship Between P¯n (x; a, b, c, d, e) and S¯n (x; p, q, r, s) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.7.2 Four Special Cases of the Symmetric Polynomials Sn (x ; p, q, r, s). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.7.3 A Unified Approach for the Classification of S¯n (x; p, q, r, s) . . . 1.8 Fourier Transforms of Symmetric Orthogonal Polynomials.. . . . . . . . . . . . . . . 1.8.1 Fourier Transform of Generalized Ultraspherical Polynomials .. .

1 1 2 12 13 13 14 15 17 20 25 36 40 42 47 49 54 56 66 69 70 ix

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1.8.2 1.8.3

Fourier Transform of Generalized Hermite Polynomials . . . . . . . . . Fourier Transform of the First Finite Sequence of Symmetric Orthogonal Polynomials . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.8.4 Fourier Transform of the Second Finite Sequence of Symmetric Orthogonal Polynomials . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.9 A Class of Symmetric Orthogonal Functions . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . (θ) 1.9.1 Four Special Cases of Sn (x; p, q, r, s) . . . . . . . . . . . .. . . . . . . . . . . . . . . (θ) 1.10 An Extension of Sn (x; p, q, r, s).. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . (α,β) 1.10.1 Four Special Cases of Sn (x; p, q, r, s) . . . . . . . . . .. . . . . . . . . . . . . . . 1.11 A Generalization of Fourier Trigonometric Series . . . . . . . . . . .. . . . . . . . . . . . . . . 1.11.1 A Generalization of Trigonometric Orthogonal Sequences . . . . . . 1.11.2 Application to Function Expansion Theory.. . . . . . . .. . . . . . . . . . . . . . . 1.12 Another Extension for Trigonometric Orthogonal Sequences .. . . . . . . . . . . . . 1.13 Incomplete Symmetric Orthogonal Polynomials.. . . . . . . . . . . .. . . . . . . . . . . . . . . 1.13.1 Incomplete Symmetric Orthogonal Polynomials of Jacobi Type .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.13.2 Incomplete Symmetric Orthogonal Polynomials of Laguerre Type .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.14 A Class of Hypergeometric-Type Orthogonal Functions . . . .. . . . . . . . . . . . . . . 1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville Problems . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 1.15.1 A Relationship Between the Chebyshev Polynomials of the Third and Fourth Kinds and the Associated Legendre Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . References . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2

Special Functions Generated by Generalized Sturm–Liouville Problems in Discrete Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.1 Introduction . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.2 A Finite Sequence of Hahn-Type Orthogonal Polynomials .. . . . . . . . . . . . . . . 2.3 Classical Symmetric Orthogonal Polynomials of a Discrete Variable . . . . . 2.3.1 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.4.1 Some Illustrative Examples.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete Variable.. . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.5.1 Two Infinite Types of Sn∗ (x ; a, b, c, d) .. . . . . . . . . . . .. . . . . . . . . . . . . . . 2.5.2 Moments of the Two Introduced Infinite Sequences . . . . . . . . . . . . . .

72 76 80 83 88 93 99 103 104 108 112 118 118 124 129 134

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A Special Case of Sn∗ (x ; a, b, c, d) Generating All Classical Symmetric Orthogonal Polynomials of a Discrete Variable .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 217 2.5.4 Two Finite Types of Sn∗ (x ; a, b, c, d) . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 219 References . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 229 2.5.3

3

Special Functions Generated by Generalized Sturm–Liouville Problems in q-Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.1 Introduction . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2 Some Preliminaries and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.1 The q-Shifted Factorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.2 q-Hypergeometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.3 q-Binomial Coefficients and the q-Binomial Theorem . . . . . . . . . . . 3.2.4 q-Gamma and q-Beta Functions . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.5 q-Analogues of Some Special Functions . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.6 q-Difference Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.7 q-Integral Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.8 An Analytical Solution for the q-Pearson Difference Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.9 Difference Equations of q-Hypergeometric Series .. . . . . . . . . . . . . . . 3.2.10 q-Analogues of Jacobi Polynomials . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.11 q-Analogues of Laguerre Polynomials . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.12 q-Analogue of Hermite Polynomials . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2.13 A Biorthogonal Exponential Sequence . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.3 Three Finite Classes of q-Orthogonal Polynomials .. . . . . . . . .. . . . . . . . . . . . . . . 3.3.1 First Finite Sequence of q-Orthogonal Polynomials Corresponding to the Inverse Gamma Distribution . . . . . . . . . . . . . . . 3.3.2 Second Finite Sequence of q-Orthogonal Polynomials Corresponding to the Fisher Distribution . . . . . . . . . . .. . . . . . . . . . . . . . . 3.3.3 Third Finite Sequence of q-Orthogonal Polynomials Corresponding to Student’s t-Distribution .. . . . . . . . .. . . . . . . . . . . . . . . 3.3.4 A Characterization of Three Introduced Finite Sequences .. . . . . . . 3.4 A Symmetric Generalization of Sturm–Liouville Problems in q-Difference Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.4.1 Some Illustrative Examples.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.5 A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free Parameters .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.5.1 Two Finite Sequences Based on Ramanujan’s Identity . . . . . . . . . . . References . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

233 233 236 236 237 238 239 239 240 242 245 246 248 250 253 254 262 263 270 275 281 283 286 290 298 306

Index . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 311

1

Special Functions Generated by Generalized Sturm–Liouville Problems in Continuous Spaces

1.1

Introduction

Consider the second-order differential equation d dx

    dy k(x) + λρ(x) − q(x) y = 0, dx

where

k(x) > 0 and ρ(x) > 0,

(1.1)

on an open interval, say (a, b), with the following boundary conditions, α1 y(a) + β1 y  (a) = 0, (1.2)

α2 y(b) + β2 y  (b) = 0,

in which α1 , α2 and β1 , β2 are given constants and k(x) , k  (x) , q(x), and ρ(x) in (1.1) are to be assumed continuous for x ∈ [a, b]. The boundary value problem (1.1)–(1.2) is called a regular Sturm–Liouville problem, and if one of the points a and b is singular (i.e., k(a) = 0 or k(b) = 0), it is called a singular Sturm–Liouville problem. In the latter case, boundary conditions (1.2) can be ignored. Let yn (x) and ym (x) be two solutions (eigenfunctions) of Eq. (1.1). Following Sturm– Liouville theory, these functions are orthogonal with respect to the positive weight function ρ(x) on (a, b) under the given conditions (1.2), i.e., 

b a



b

ρ(x)yn (x)ym (x) dx = a

 ρ(x)yn2 (x) dx

δn,m ,

© Springer Nature Switzerland AG 2020 M. Masjed-Jamei, Special Functions and Generalized Sturm–Liouville Problems, Frontiers in Mathematics, https://doi.org/10.1007/978-3-030-32820-7_1

(1.3)

1

2

1 Generalized Sturm–Liouville Problems in Continuous Spaces

where  δn,m =

0 1

(n = m), (n = m).

Many special functions in theoretical and mathematical physics are solutions of a regular or singular Sturm–Liouville problem satisfying the orthogonality condition (1.3). For instance, the associated Legendre functions, Bessel functions, trigonometric sequences related to Fourier analysis, ultraspherical functions, and Hermite functions are particular solutions of a Sturm–Liouville problem. Most of these functions are symmetric and have found interesting applications in physics and engineering. Hence if one can extend them symmetrically and preserve their orthogonality property, new applications can be derived that logically extend the previous known applications. In this chapter, by achieving this goal, we first review some primary definitions by considering a hypergeometric-type differential equation and study a relationship between classical Pearson distributions and Gauss hypergeometric functions. Then, by introducing a generalized Sturm–Liouville problem of symmetric type in continuous spaces, we extend some classical symmetric orthogonal functions and obtain their orthogonality properties.

1.2

Pearson Distribution Families and Gauss Hypergeometric Functions

Consider the homogeneous differential equation σ (x) y  (x) + τ (x) y  (x) + λ y(x) = 0,

(1.4)

in which σ (x) = ax 2 + bx + c, τ (x) = dx + e; a, b, c, d, e are free parameters, and λ is a constant depending on a, b, c, d, e. This equation is called a hypergeometric-type differential equation, because its particular solutions can be indicated in terms of Gauss hypergeometric functions. In other words, if in (1.4), σ (x) = x(1 − x)

and τ (x) = γ − (α + β + 1) x,

the Gauss hypergeometric differential equation x(1 − x) y  (x) + (γ − (α + β + 1)x) y  (x) − αβ y(x) = 0

(1.5)

1.2 Pearson Distribution Families and Gauss Hypergeometric Functions

3

appears, where α, β, and γ are real parameters. The indicial equation corresponding to Eq. (1.5) is r 2 − (1 − γ ) r = 0, with two roots r1 = 0 and r2 = 1 − γ . Using the Frobenius method, we obtain a series solution of Eq. (1.5) for r1 = 0 as follows: y1 (x) = 1 +

α(α+1)β(β+1) x 2 α(α+1)(α+2)β(β+1)(β + 2) x 3 αβ x + + + ··· , γ 1! γ (γ + 1) 2! γ (γ + 1)(γ + 2) 3! (1.6)

where γ = 0, −1, −2, −3, . . . and the series is convergent for all x ∈ [−1, 1]. The series (1.6) isknown in the literature as a Gauss hypergeometric series, and its αβ sum, denoted by 2 F1 x or 2 F1 (α, β, γ ; x), is called the Gauss hypergeometric γ function. Hence, we have  2 F1



∞  (α)k (β)k x k αβ , x = (γ )k k! γ k=0

(1.7)

in which (α)n =

n−1

(α + j ).

j =0

Notice that (1.7) is a special case of the generalized hypergeometric function  p Fq



∞  (a1 )k (a2 )k · · · (ap )k zk a1 a2 . . . ap , z = (b1 )k (b2 )k · · · (bq )k k! b1 b2 . . . bq k=0

(1.8)

in which z may be a complex variable. The infinite sum (1.8) is indeed a Taylor series ∞

expansion of a function, say f , namely ck∗ zk with ck∗ = f (k) (0)/k!, for which the ratio k=0

of successive terms can be written as ∗ ck+1

ck∗

=

(k + a1 )(k + a2 ) · · · (k + ap ) . (k + b1 )(k + b2 ) · · · (k + bq )(k + 1)

According to the ratio test, the hypergeometric series (1.8) is convergent for all p ≤ q + 1. In a more precise expression, it converges in |z| < 1 for p = q + 1, converges everywhere

4

1 Generalized Sturm–Liouville Problems in Continuous Spaces

for p < q + 1, and converges nowhere (z = 0) for p > q + 1. Moreover, for p = q + 1 it converges absolutely for |z| = 1 if the condition ⎞ ⎛ q q+1   A∗ = Re ⎝ bj − aj ⎠ > 0 j =1

j =1

holds and is conditionally convergent for |z| = 1 and z = 1 if −1 < A∗ ≤ 0, and is divergent for |z| = 1 and z = 1 if A∗ ≤ −1. Since the Gauss hypergeometric function has an integral representation of the form  2 F1



 1 Γ (c) a,b t b−1 (1 − t)c−b−1 (1 − t x)−a dt , x = Γ (c − b) Γ (b) 0 c

(1.9)

according to Beta integral 

1

B(λ1 , λ2 ) =

x

λ1 −1

λ2 −1

(1 − x)

0



π/2

=2

 dx=

1

−1

x

2λ1 −1

sin(2λ1−1) x cos(2λ2−1) x dx =

0

2 λ2 −1

(1−x )



∞

dx= 0

x λ1 −1 (1+x)λ1 +λ2

dx

Γ (λ1 )Γ (λ2 ) = B(λ2 , λ1 ), Γ (λ1 + λ2 )

where Γ (λ) denotes the well-known gamma function 

∞

Γ (λ) =

x λ−1 e−x dx

for Re (λ) > 0,

0

we can conclude that  2F 1



Γ (c) Γ (c − b − a) a, b . 1 = Γ (c − b) Γ (c − a) c

(1.10)

One of the properties of the series (1.7) is that many elementary functions or special functions of mathematical physics can be directly expressed in terms of it. For instance, we have ln(1 + x) = x 2 F1 (1, 1, 2; −x),   1 2 , 1, 1; sin x , sec x = 2 F1 2   1 1 3 2 , , ;x , arcsin x = x 2 F1 2 2 2   3 1 arctan x = x 2 F1 , 1, ; −x 2 , 2 2

1.2 Pearson Distribution Families and Gauss Hypergeometric Functions

5

and/or the incomplete beta function defined by  x 1 t a−1 (1 − t)b−1 dt = x a 2 F1 (a, 1 − b, a + 1; x). Bx (a, b) = a 0 Another important special function having a direct relationship with the Gauss hypergeometric function (1.7) is the family of Pearson distributions, which plays a key role in the theory of classical orthogonal polynomials. If Eq. (1.4) is written in a self-adjoint form, then the Pearson differential equation d (σ (x)W (x)) = τ (x)W (x) dx

(1.11)

appears, whose solutions are indeed the weight functions of classical orthogonal polynomial solutions of Eq. (1.4). By solving the first-order Eq. (1.11), the Pearson distributions family appears as  W (x) = W



  (d − 2a)x + e − b  de dx , x = K exp ax 2 + bx + c a b c

(1.12)

where K is a normalizing constant. For convenience, if in (1.12) we put d − 2a = d ∗

and e − b = e∗ ,

then (ax 2 + bx + c)y  − (d ∗ x + e∗ )y = 0

   d ∗ x + e∗ d ∗ e∗ ∗ dx . ⇒y=W x = K exp ax 2 + bx + c a b c

(1.13)

Now taking the derivative on both sides of Eq. (1.13) yields   (ax 2 + bx + c)y  + (2a − d ∗ )x + (b − e∗ ) y  − d ∗ y = 0,

(1.14)

which can be transformed to a special case of the Gauss equation (1.5) if we suppose x = pt + q and replace it in Eq. (1.14) to get   aq 2 + bq + c d 2 y 2aq + b 2 t+ t + ap ap2 dt 2   (2a − d ∗ )q + b − e∗ dy d∗ 2a − d ∗ t+ − y = 0, + a ap dt a

(1.15)

6

1 Generalized Sturm–Liouville Problems in Continuous Spaces

provided that ap = 0. Moreover, if in (1.15) we take aq 2 + bq + c = 0 and

2aq + b = −1, ap

(1.16)

we obtain p=−

√ Δ a

and

q=

√ −b + Δ , 2a

(1.17)

where Δ = b2 − 4ac. Hence, the differential equation (1.15) is simplified as √   2ae∗ − bd ∗ + (d ∗ − 2a) Δ  dy d ∗ d 2y  d∗ t+ − y = 0. t (t − 1) 2 + 2− √ a dt a dt 2a Δ

(1.18)

In conclusion, according to relations (1.13)–(1.17), one of the general solutions of Eq. (1.18) must be  y= W

∗

√ √

−b + Δ Δ d ∗ e∗ t+ . − a 2a a b c

(1.19)

On the other hand, we mentioned that Eq. (1.18) could be a special case of the equation t (t − 1)

 dy d 2y  + αβ y = 0, + (α + β + 1) t − γ 2 dt dt

(1.20)

with the general solution  y = A 2 F1



 αβ α + 1 − γ, β + 1 − γ 1−γ t +B t 2 F1 γ 2−γ



t ,

in which γ , α−β, and γ − α−β are all nonintegers and A, B are two constants. Therefore, to find the parameters α, β, and γ in Eq. (1.20) in terms of the parameters a, b, c and d ∗ , e∗ in Eq. (1.18), we should just equate the two equations, which leads to the system α+β =1−

d∗ a

and α β = −

d∗ . a

By solving the above system and comparing (1.18) with (1.20), we get α=1 , β=−

d∗ d∗ 2ae∗ − bd ∗ √ . , and γ = 1 − − a 2a 2a Δ

(1.21)

1.2 Pearson Distribution Families and Gauss Hypergeometric Functions

7

Note that since (1.21) is a symmetric system, the second solution reads as α=−

d∗ 2ae∗ − bd ∗ d∗ . , β = 1, and γ = 1 − − √ a 2a 2a Δ

Consequently, the general solution of Eq. (1.18) is ⎛ y = A∗ 2 F1 ⎝

+ B ∗t

1, −d ∗ /a 1−

d∗ 2a

2ae∗√ −bd ∗ 2a Δ

−

∗√ ∗ ∗ √ −bd ) ( d Δ + 2ae 2a Δ

⎛

2 F1 ⎝

⎞ t⎠

1+

d∗ 2a

+

√ 2ae∗√ −bd ∗ −d ∗ Δ +√2ae∗ −bd ∗ , 2a Δ 2a Δ

1+

d∗ 2a

+

2ae∗√ −bd ∗ 2a Δ

⎞ t⎠ .

(1.22)

But (1.22) can be further simplified, because in the second term of the right-hand side, we generally have  2 F1



∞  (v)k k uv x = (1 − x)−v . x = k! u k=0

So the final form would be ⎛ 1, −d ∗ /a y = A∗ 2 F1 ⎝ ∗ ∗ −bd ∗ 1 − d2a − 2ae √ 2a Δ

⎞ √ ∗√ ∗ ∗ ∗ ∗ d ∗ Δ+ 2ae √ −bd ) √ −bd ) ) ( d Δ−( 2ae t⎠ + B∗ t( 2a Δ 2a Δ (1 − t) . (1.23)

On the other hand, we observed that the function (1.19) was also a solution of Eq. (1.18). This means that there must be a direct relationship between (1.23) and (1.19). To find it, without loss of generality, we assume in (1.13) that ax 2 + bx + c = a(x + θ1 )(x + θ2 ), in which √ √ b− Δ b+ Δ and θ2 = with Δ = b2 − 4ac . θ1 = 2a 2a The latter relation implies that  W

∗



d ∗ e∗ x = R (x + θ1 )r (x + θ2 )s , a b c

(1.24)

where R is a constant and √ d ∗ Δ + 2ae∗ − bd ∗ r= √ 2a Δ

√ d ∗ Δ − (2ae∗ − bd ∗ ) and s = √ . 2a Δ

(1.25)

8

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Equality (1.24) is valid because the logarithmic derivative of the function ρ(x) = (x + θ1 )r (x + θ2 )s equals the logarithmic derivative of W ∗ (a, b, c, d ∗ , e∗ ; x) defined in (1.13), and since v  (x) u (x) = ⇔ u(x) = R v(x), u(x) v(x)

(1.26)

equality (1.24) holds. Now, by noting (1.24)–(1.26), the function (1.19) is simplified as  W∗

√

√ √ Δ Δ r Δ d ∗ e∗ t − θ1 = R(− t) (− t + θ 2 − θ 1 )s − a a a a b c √ Δ r+s r r ) = R(−1) ( t (1 − t)s , a

which gives exactly the second term of (1.23) for B ∗ = R(−1)r ( This means that y1 (t) = t

(d

∗ √Δ+ 2ae∗ −bd ∗ √ ) 2a Δ

⎛ y2 (t) = 2 F1 ⎝

(1 − t)

(d

∗ √Δ−( 2ae∗ −bd ∗ ) √ ) 2a Δ

d∗ 2a

−

,

⎞ t⎠ ,

1, −d ∗ /a 1−

√ Δ r+s . a )

2ae∗√ −bd ∗ 2a Δ

(1.27)

constitute a solution basis for Eq. (1.18). The final point is that according to the theory of ordinary differential equations, if Y (x) satisfies a homogeneous second-order differential equation of the form a(x) Y (x) + b(x) Y  (x) + c(x) Y (x) = 0, such that it has the basis solutions y1 (x) and y2 (x), then Y (x) = c1 y1 (x) + c2 y2 (x), where y2 (x) = N ∗ y1 (x)



x λ

(y1 (t))−2 exp

 −

 b(t) dt dt, a(t)

N ∗ is an arbitrary constant, and λ is a parameter independent of x.

(1.28)

1.2 Pearson Distribution Families and Gauss Hypergeometric Functions

9

In this sense, note that the values N ∗ and λ can be explicitly obtained. For this purpose, it is enough to put x = λ in (1.28) to get y2 (λ) = N ∗ y1 (λ)



λ

(y1 (t))−2 exp

 −

λ

 b(t) dt dt = 0. a(t)

Therefore, λ must be one of the known roots of the second basis solution. To obtain N ∗ , since N∗ = y1 (x)

x

y2−1 (0)

y2 (x) (y1 (t))−2 exp



b(t ) − a(t ) dt



, dt

the above equality should be valid for all values of x = θ where θ ∈ S = {x| x ∈ D y1 (x) ∩ D y2 (x) & y1 (x) = 0 & y2 (x) = 0 }.

(1.29)

All these results eventually simplify (1.28) as y2 (x) =

y2 (θ )

  y1 (x) b(t ) (y1 (t))−2 exp − a(t dt dt )   b(t) −2 dt dt, (y1 (t)) exp − a(t) y2−1 (0)

y1 (θ )  x ×

θ

y2−1 (0)

(1.30)

for all θ ∈ S defined in (1.29). Using the latter result and (1.27), we can now determine direct relationships between the Pearson distributions family and Gauss hypergeometric functions. According to Table 1.1, there are six special cases of the Pearson distributions family that appear in the theory of classical orthogonal polynomials as the corresponding weight functions. For example, consider the generalized T weight function W1 (x; u, v) = (1 + x 2 )−u exp(2v arctan x), Table 1.1 Special cases of the Pearson distributions family

Distribution name

Definition

Interval

1. Shifted beta 2. Gamma 3. Normal 4. Fisher F 5. Student’s-t 5*. Generalized t 6. Inverse gamma

(1 − x)u (1 + x)v x u exp(−x) exp(−x 2 ) x v / (1 + x)u+v (1 + x 2 )−u (1 + x 2 )−u exp(v arctan x) x −u exp(−1/x)

[−1, 1] [0, ∞) (−∞, ∞) [0, ∞) (−∞, ∞) (−∞, ∞) [0, ∞)

10

1 Generalized Sturm–Liouville Problems in Continuous Spaces

for u > 0 and v ∈ R, where x ∈ (−∞, ∞). Taking the logarithmic derivative on both sides and then comparing with (1.13) eventually yields a = 1 , b = 0 , c = 1 , d ∗ = −2u, and e∗ = 2v. Hence  W1 (x; u, v) = K1 W

∗



−2u 2v x , 1 0 1

which, according to (1.14), satisfies the equation (1 + x 2 ) W1 (x; u, v) + ((2u + 2)x − 2v) W1 (x; u, v) + 2u W1 (x; u, v) = 0.

(1.31)

On the other hand, since in this example √ √ Δ = b2 − 4ac = −4 , p = − −4 = −2i, and q = −4/2 = i , by applying the change of variable x = −2 i t + i in (1.31) (or equivalently t = in the Gauss hypergeometric equation (1.20)), it turns out from (1.23) that W1 (x; u, v) = (1 + x 2 )−u exp(2v arctan x) 

1 1 1 i i 1, 2u i = A 2 F1 + B ( x + )−u−iv (− x + )−u+iv . x+ 2 2 2 2 2 1 + u + iv 2

i 2

x+

1 2

(1.32)

Since the second term on the right-hand side of (1.32) can be simplified as 

i 1 x+ 2 2

−u−iv   1 −u+iv i − x+ = 22u (1 + x 2 )−u exp(2v arctan x), 2 2

(1.33)

it eventually follows that  2 −u

y1 (x) = (1 + x )

exp(2v arctan x) and y2 (x) = 2 F1



1 1, 2u i x+ 2 1 + u + iv 2 (1.34)

are two basis solutions of Eq. (1.31). It is interesting to observe that the Taylor expansion of the generalized T distribution function is explicitly computable, because if we reconsider (1.33) in the form (1 + ix)−u−iv (1 − ix)−u+iv = (1 + x 2 )−u exp(2varctan x),

(1.35)

1.2 Pearson Distribution Families and Gauss Hypergeometric Functions

11

then according to the Leibniz rule 

n f (k) (x) g (n−k) (x), k k=0

d n (f (x)g(x))  = dx n n

the following identity is valid: d n ((1 + ix)−u−iv (1 − ix)−u+iv ) dx n



n =i (−1) (u + iv)k (u − iv)n−k k k=0

 −n u + iv n = i (u − iv)n 2 F1 −1 . 1 − n − u + iv n

x=0

n 

k

Therefore 2 −u

(1 + x )

 n ∞  



(ix)n n exp(2v arctan x) = (−1) (u + iv)k (u − iv)n−k n! k n=0 k=0

 ∞  (ix)n −n u + iv . = − 1 (u − iv)n 2 F1 n! 1 − n − u + iv k

n=0

(1.36) Moreover, if in (1.35), ix = z, iv → v, −u + v = α, and −u − v = β, then (1.36) changes to  n 

∞   zn α β k n (1 − z) (1 + z) = (−1) (−β)k (−α)n−k n! k n=0 k=0

 ∞  zn −n −β (1.37) = − 1 (−α)n . 2 F1 n! 1−n+α n=0 Hence, integrating both sides of (1.37) on [−1, 1] gives   1 ∞  −n −β (1 − z)α (1 + z)β dz = 2 F1 1−n+α −1 n=0



1 + (−1)n , − 1 (−α)n (n + 1)!

which is simplified as ∞  n=0

 2 F1



 1 1 (−α)2n −2n −β = (1 − z)α (1 + z)β dz −1 (2n + 1)! 2 −1 1 − 2n + α = 2α+β

Γ (α + 1)Γ (β + 1) . Γ (α + β + 2)

12

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Although (1.37) shows the explicit form of the Taylor expansion of the shifted beta distribution at z = 0, the real beta distribution function is given as W2 (x; u, v) = x u (1 − x)v

(0 ≤ x ≤ 1),

which is a particular case of the general distribution (1.13) for a = 1, b = −1, c = 0, d ∗ = u + v, and e∗ = −u.

(1.38)

The relation (1.38) implies that  W2 (x; u, v) = K2 W

∗



u + v −u x . 1 −1 0

Consequently, if the aforementioned parameters are substituted into (1.14), then the equation   x(x − 1) y  + (2 − u − v)x + v − 1 y  − (u + v) y = 0 has the following two basis solutions:  y1 (x) = W2 (1 − x; u, v) = (1 − x) x

u v

1.3

and y2 (x) = 2 F1



1, −u − v x . 1−v

Six Sequences of Hypergeometric Orthogonal Polynomials

Let us begin this section with a special case of Eq. (1.4) in the form σ (x)yn (x) + τ (x)yn (x) − λn yn (x) = 0,

(1.39)

where σ (x) = ax 2 + bx + c and τ (x) = dx + e are independent of n and   λn = n d + (n − 1)a is the eigenvalue depending on n = 0, 1, 2, . . .. It is well known in the literature that the classical hypergeometric polynomials of Jacobi, Laguerre, and Hermite are infinite types of polynomial solutions of Eq. (1.39) that

1.3 Six Sequences of Hypergeometric Orthogonal Polynomials

13

are infinitely orthogonal. However, there are three further sequences of hypergeometric polynomials that are solutions of the above equation but finitely orthogonal with respect to three specific weight functions. This means that there are altogether six sequences of hypergeometric orthogonal polynomials as follows.

1.3.1

Jacobi Polynomials

If σ (x) = 1 − x 2 and τ (x) = −(α + β + 2)x + (β − α) are substituted into (1.39), we obtains • Differential equation   (1 − x 2 )yn (x) − (α + β + 2) x + α − β yn (x) + n (α + β + n + 1)yn (x) = 0, with the polynomial solution Pn(α,β) (x)

  n  1  n+α n+β = n (x − 1)n−k (x + 1)k . k n−k 2 k=0

• Weight function: w(x) = (1 − x)α (1 + x)β

for x ∈ [−1, 1] and α, β > −1.

• Rodrigues representation: Pn(α,β) (x) =

 dn  (−1)n (1 − x)n+α (1 + x)n+β . α β dx n n 2 (1 − x) (1 + x)

• Orthogonality relation: 

1

−1

1.3.2

(1 − x)α (1 + x)β Pn(α,β) (x) Pm(α,β) (x) dx =

2α+β+1 Γ (n + α + 1) Γ (n + β + 1) δn,m . (2n+α+β + 1) n!Γ (2n + α + β + 1)

Laguerre Polynomials

If σ (x) = x and τ (x) = α + 1 − x are substituted into (1.39), we obtain: • Differential equation: x yn (x) + (−x + α + 1)yn (x) + n yn (x) = 0,

(1.40)

14

1 Generalized Sturm–Liouville Problems in Continuous Spaces

with the polynomial solution L(α) n (x) =

n 

(−1)k

k=0

  n+α 1 k x . n − k k!

• Weight function: w(x) = x α e−x for x ∈ [0, ∞) and α > −1. • Rodrigues representation: L(α) n (x) =

1 n!x α e−x

d n  n+α −x  x . e dx n

• Orthogonality relation: 

∞ 0

1.3.3

Γ (n + α + 1) δn,m . n!

(α) x α e−x L(α) m (x)Ln (x) dx =

Hermite Polynomials

If σ (x) = 1 and τ (x) = −2x are substituted into (1.39), we obtain: • Differential equation: yn (x) − 2xyn (x) + 2nyn (x) = 0, with the polynomial solution Hn (x) = n!

[n/2] 

(−1)k

k=0

(2x)n−2k . k! (n − 2k)!

• Weight function: w(x) = e−x

2

for x ∈ (−∞, ∞).

• Rodrigues representation: Hn (x) = (−1)n ex

2

d n  −x 2  e . dx n

(1.41)

1.3 Six Sequences of Hypergeometric Orthogonal Polynomials

15

• Orthogonality relation: 

∞ −∞

1.3.4

e−x Hn (x) Hm (x) dx = 2

√ n π 2 n! δn,m .

First Finite Sequence of Hypergeometric Orthogonal Polynomials

The first finite sequence is defined for σ (x) = x 2 + x

and τ (x) = (2 − p)x + q + 1

in (1.39). Thus, the differential equation   (x 2 + x)yn (x) + (2 − p)x + q + 1 yn (x) − n(n + 1 − p)yn (x) = 0

(1.42)

has a polynomial solution (p, q) Mn (x)

 n   p−n−1 q +n = (−1) n! (−x)k . k n−k n

(1.43)

k=0

Now we prove that the polynomials (1.43) are finitely orthogonal with respect to the weight function W1 (x; p, q) = x q (1 + x)−(p+q) on [0, ∞) if and only if p > 2{max n} + 1 and q > −1. To prove this, we first consider the self-adjoint form of Eq. (1.42) as   x 1+q (1 + x)1−p−q yn (x) = n(n + 1 − p)x q (1 + x)−(p+q)yn (x), and for the index m as    (x) = m(m + 1 − p)x q (1 + x)−(p+q) ym (x), x 1+q (1 + x)1−p−q ym (p,q)

where yn (x) = Mn get 

(1.44)

(x). Then we multiply by ym (x) and yn (x) in (1.44) and subtract to

 ∞   y (x)y (x) − y (x)y (x) m n n m p+q−1

x q+1 (1 + x)

 (1) = (λ(1) n − λm )

0

0

∞

xq (p,q) (p,q) Mn (x)Mm (x) dx, (1 + x)p+q

(1.45)

16

1 Generalized Sturm–Liouville Problems in Continuous Spaces (1)

where λn = n(n + 1 − p). Since  (x)yn (x)} = m + n − 1, max deg {yn (x)ym (x) − ym

if q > −1 and p > 2N + 1 for N = max{m, n}, the left-hand side of (1.45) tends to zero, and we will have  ∞ xq (p,q) (p,q) Mn (x)Mm (x) dx = 0 ⇔ m = n, p > 2N + 1 and q > −1 . (1 + x)p+q 0 (p>2N+1,q>−1)

(p,q)

N 2n + 1 is a necessary condition for orthogonality (p,q) of the polynomials Mn (x). Consequently, we have  ∞ n ! (p − n − 1)! (q + n)! xq (p,q) (p,q) (x)Mm (x) dx = δn,m p+q Mn (p − 2n − 1) (p + q − n − 1)! (1 + x) 0 (1.47) if and only if m, n = 0, 1, 2, . . . , N < For example, the polynomial set

p−1 2

and q > −1.

  n  100  100   201 − n n (202,0) n Mn (x) = (−1) n! (−x)k n=0 n=0 k k k=0

is finitely orthogonal with respect to the weight function W1 (x, 202, 0) = (1 + x)−202 on [0, ∞), and we have 

∞

0

(202,0) (1 + x)−202 Mn(202,0)(x)Mm (x) dx =

(n!)2 δn,m ⇔ m, n ≤ 100. 201 − 2n

Finally, we point out that due to the form of the weight function and relation (1.46), (p,q) there is a limit relation between Mn (x) and the Laguerre polynomials as follows: (p,q)

lim Mn

p→∞

d n (t n+q (1 + t/p)n−p−q ) t t ( ) = lim (−1)n t −q (1 + )p+q p→∞ p p dt n = (−1)n t −q et

1.3.5

d n (t n+q e−t ) (q) = (−1)n n! Ln (t). dt n

Second Finite Sequence of Hypergeometric Orthogonal Polynomials

The second finite sequence is directly connected to the Bessel polynomials for σ (x) = x 2 and τ (x) = 2x + 2, which were first studied by Krall and Frink in 1949. Of course, they established a complex orthogonality of these polynomials on the unit circle. Then in 1973, the generalized Bessel polynomials were studied by Grosswald, which are special solutions of Eq. (1.39) for σ (x) = x 2

and τ (x) = (2 + α)x + 2; α = −2, −3, . . . ,

leading to B¯ n(α) (x)

=2

n

n  k=0



n Γ (n + k + α + 1)  x k , Γ (2n + α + 1) 2 k

as the monic type of generalized Bessel polynomial.

18

1 Generalized Sturm–Liouville Problems in Continuous Spaces

The second finite sequence can now be defined for σ (x) = x 2

and τ (x) = (2 − p)x + 1

to obtain the differential equation   x 2 yn (x) + (2 − p)x + 1 yn (x) − n(n + 1 − p)yn (x) = 0,

(1.48)

with the polynomial solution (p)

Nn (x) = (−1)n

   n  p−n−1 n k! (−x)k . k n−k

(1.49)

k=0

If Eq. (1.48) is written in the self-adjoint form       1 1 x −p+2 exp − yn (x) = n(n + 1 − p)x −p exp − yn (x), x x and for the index m as       1 1  x −p+2 exp − (x) = m(m + 1 − p)x −p exp − ym ym (x), x x

(1.50)

(p)

where yn (x) = Nn (x), then on multiplying by ym (x) and yn (x) in (1.50) and subtracting, we get  ∞   1  (2) (x)yn (x) = (λ(2) − λ ) x −p+2 e− x yn (x)ym(x) − ym n m 0

∞

1

x −p e− x Nn (x)Nm (x) dx, (p)

(p)

0

(1.51) (2)

where λn = n(n + 1 − p). Once again, since  max deg{yn (x)ym (x) − ym (x)yn (x)} = m + n − 1,

the condition p > 2N + 1 for N = max{m, n} causes the left-hand side of (1.51) to tend to zero, and therefore 

∞

x −p e− x Nn (x)Nm (x) dx = 0 ⇔ m = n and p > 2N + 1. 1

(p)

(p)

0 (p>2N+1)

(p)

N 1, then 

∞



∞

W2 (x; p)dx =

0

1

x −p e− x dx = Γ (p − 1).

0

Therefore, 

∞ 0

2 1  (p) n!Γ (p − n) , x −p e− x Nn (x) dx = p − (2n + 1)

which gives the final orthogonality relation as 

∞

1

x −p e− x Nn (x)Nm (x) dx = (p)

(p)

0

Nn(202)(x)

100 n=0

(1.53)

p−1 2 .

if and only if m, n = 0, 1, 2, . . . , N < For example, the polynomial set 

n!Γ (p − n) δn,m p − (2n + 1)

100    n  201 − n n k = (−1) k! (−x) k n−k 

n

k=0

n=0

is finitely orthogonal with respect to the weight function W2 (x; 202) = x −202 exp(−1/x) on [0, ∞). There is a direct relation between recent polynomials and the generalized Bessel polynomials as (p) Nn (x)

n! = n 2



p−1−n n

(−p) B¯ n (2x),

20

1 Generalized Sturm–Liouville Problems in Continuous Spaces

as well as the Laguerre polynomials as 

(p) Nn (x)

= n! x

n

p−(2n+1) Ln

  1 . x

(1.54)

The latter relation is useful for generating a definite integral for the Laguerre polynomials, because substituting (1.54) into (1.53) yields 

∞ 0

1.3.6

 2 1 (n + p)! (p) . x p−1 e−x Ln (x) dx = p n!

Third Finite Sequence of Hypergeometric Orthogonal Polynomials

In this part, we study the last finite sequence of hypergeometric polynomials, which is finitely orthogonal with respect to the generalized Student’s-t weight function  −p exp(q arctan x), W (p,q) (x) = 1 + x 2 on (−∞, ∞). This function can also be considered an important statistical distribution, because by means of it we can generalize the usual Student’s-t distribution and prove that it tends to the normal distribution just like the Student’s-t sampling distribution. Before studying the third finite case, we consider a particular case of Eq. (1.39) for σ (x) = 1 + x 2

and τ (x) = (3 − 2p)x,

which corresponds to the same as the usual Student’s-t weight function, i.e., (1 + x 2 )yn (x) + (3 − 2p)x yn (x) − n(n + 2 − 2p)yn (x) = 0, with the symmetric polynomial solution (p) In (x)

= n!

[n/2]  k=0

 k

(−1)

p−1 n−k



n−k k

(2x)n−2k ,

which is finitely orthogonal with respect to the weight function 1

ρ(x, p) = (1 + x 2 )−(p− 2 ) on (−∞, ∞).

(1.55)

1.3 Six Sequences of Hypergeometric Orthogonal Polynomials

21

To prove this claim, it is enough to transform Eq. (1.55) to a Sturm–Liouville equation and apply the same technique that we applied for the first and second kinds of finite classical orthogonal polynomials. This proccess eventually leads to 

∞ −∞

(1 + x 2 )

−(p− 12 ) (p) (p) In (x)Im (x) dx

= 0 ⇔ m = n and p > N + 1.

On the other hand, since (p) In (x)

n−(p− 12 )

n 2 1 d ((1 + x ) (−2)n (p − n)n = (1 + x 2 )p− 2 (2p − 2n − 1)n dx n

)

,

n = 0, 1, 2, . . . ,

the orthogonality relation finally takes the form 

∞ −∞

1

(1 + x 2 )−(p− 2 ) In (x)Im (x) dx (p)

(p)

√ n! 22n−1 π Γ 2 (p) Γ (2p − 2n) δn,m = (p − n − 1) Γ (p − n) Γ (p − n + 1/2) Γ (2p − n − 1)

if and only if m, n = 0, 1, 2, . . . , N < p − 1. As we mentioned, the weight function of the above orthogonality relation corresponds to the well-known Student’s-t distribution   Γ ((n + 1)/2) x n T (x; n) = √ ρ √ ; + 1 for x ∈ (−∞, ∞) and n ∈ N . nπ Γ (n/2) n 2 However, that is not the end of story, and there is still a more extensive polynomial sequence that is finitely orthogonal on (−∞, ∞) and generalizes the Student’s-t distribution as its weight function. For this purpose, we first replace σ (x) = 1 + x 2

and τ (x) = 2(1 − p)x + q

in Eq. (1.39) to obtain   (1 + x 2 ) yn (x) + 2(1 − p)x + q yn (x) − n(n + 1 − 2p) yn (x) = 0 .

(1.56)

The polynomial solution of above equation can be represented in terms of a Gauss hypergeometric function as (p,q) Jn (x)

= (−i) (n + 1 − 2p)n n

n  k=0





 n k − n, p − n − iq/2 F 2 (−ix)k . 2 1 k 2p − 2n (1.57)

22

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Now write Eq. (1.56) in the self-adjoint form 

1 + x2

1−p

exp(q arctan x)yn (x)



−p  = n(n + 1 − 2p) 1 + x 2 exp(q arctan x)yn (x)

and apply the Sturm–Liouville theorem on (−∞, ∞) to get  ∞ 1−p  1 + x2 exp(q arctan x)(yn (x)ym (x) − ym (x) yn (x)) 

(3) = λ(3) n − λm



−∞

∞

−∞

 −p (p,q) (p,q) 1 + x2 exp(q arctan x) Jn (x)Jm (x) dx,

(1.58)

(3)

where λn = n(n + 1 − 2p). Since  max deg{yn (x)ym (x) − ym (x)yn (x)} = n + m − 1,

if the conditions p > N + 1/2 for N = max{m, n} and q ∈ R hold, the left-hand side of (1.58) tends to zero, and 

∞

(1 + x 2 )−p exp(q arctan x) Jn

(p,q)

−∞

(p,q)

(x)Jm

(x) dx = 0

⇔ m = n , p > N + 1/2 and q ∈ R.

(1.59)

To compute the norm square value, it is enough to first substitute the Rodrigues representation (p,q) Jn (x)

 n−p  p d n( 1 + x 2 exp(q arctan x)) 2 = (−1) 1 + x exp(−q arctan x) dx n n

into 

∞ −∞

(1 + x 2 )

n−p

(p,q)

exp(q arctan x) (Jn  = (−1)

∞

n

−∞

2

(x)) dx

d (p,q) Jn (x)

n(



1 + x2

n−p

Then integration by parts yields  (−1)

n

∞

−∞

d (p,q) Jn (x)

n(



1 + x2

n−p

exp(q arctan x) ) dx dx n

exp(q arctan x) ) dx. dx n

1.3 Six Sequences of Hypergeometric Orthogonal Polynomials



n!Γ (2p − n) = Γ (2p − 2n)

∞ −∞

23

 n−p 1 + x2 exp(q arctan x) dx =

n! Γ (2p − n) Γ (2p − 2n)



π 2

− π2

(cos θ )2p−2n−2 eqθ dθ.

Therefore 

∞ −∞

(p,q)

W (p,q) (x) Jn

(p,q)

(x) Jm =

(x) dx

 n!Γ (2p − n) 

π/2

Γ (2p − 2n)

−π/2

 (cos θ )2p−2n−2 eqθ dθ δn,m

(1.60)

if and only if m, n = 0, 1, 2, . . . , N < p − 1/2 and q ∈ R. Note that (1.60) can be simplified if 2p is a natural number, because 

π/2 −π/2

m

cos θ e

qθ

dθ = 2

−m

m    m k=0

  π π 1 e((m−2k) i+q) 2 − e−((m−2k) i+q) 2 . k (m − 2k) i + q

Remark 1.1 If Re(a) > 0, Re(b) > 0 and Re(c + d) > 1, then the Cauchy beta integral 1 2π



∞ −∞

dt Γ (c + d − 1) (a + b)1−(c+d) = (a + it)c (b − it)d Γ (c)Γ (d)

appears. One of the direct consequences of Cauchy’s formula is the definite integral 

π 2

− π2

est cosr t dt =

π 2−r Γ (r + 1) Γ (1 +

r+is r−is 2 ) Γ (1 + 2 )

,

which can be directly applied for evaluating the integral in (1.60). For instance, the polynomial set S = {Jn(4,1)(x)}3n=0 = { 1 , 6x − 1 , 20x 2 − 10x − 3 , 2 4x 3 − 36x 2 − 12x + 7 } is finitely orthogonal with respect to the weight function (1 + x 2 )−4 exp(arctan x) on (−∞, ∞), so that 

∞

exp(arctan x)

−∞

(1 + x 2 )

4

Jn(4,1) (x)Jm(4,1) (x) dx =

n!(7 − n)!(2 sinh π2 ) (7 − 2n)

3−n  k=0

δn,m

(1 + (6 − 2n − 2k) ) 2

⇔

m, n ≤ 3.

24

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Table 1.2 Six sequences of hypergeometric orthogonal polynomials

Kind

Notation

a

(α,β) Infinite Pn (x) (α) Infinite Ln (wx)

b c d

−1 0 1 −α − β − 2 −α + β 1 0 −w 0 1 −2w2

0 Infinite Hn (wx + v/2w) 0 Finite Finite Finite

(p,q)

Mn

e

v 0 (−p + 2)w (q + 1)v

((w/v)x) w

(p) Nn (wx) (p,q) Jn (x)

α+1 −v

w

0 0 (−p + 2)w 1

1

0 1 2 − 2p

q

Table 1.3 General properties of six sequences of hypergeometric orthogonal polynomials Shifted polynomials

Weight function

Interval

Parametric conditions

v ) Hn (wx + 2w

exp(−x(v + w2 x))

(−∞, ∞)

∀ n, w = 0, v ∈ R

(p)

In (wx + v)

(1 + (wx +

1 v)2 )−(p− 2 )

(x)

(1 + x 2 )−p exp(q arctan x)

(−∞, ∞)

(wx)

x α e−w x

[0, ∞)

(p,q) w ( v x)

x q (wx + v)−(p+q)

[0, ∞)

(p)

 1  x −p exp − wx

[0, ∞)

(1 − x)α (1 + x)β

[−1, 1]

(p,q)

Jn

(α)

Ln

Mn

Nn (wx) (α,β)

Pn

(x)

max n < p − 1 w = 0 , v ∈ R max n < (p − 1)/2 q∈R

(−∞, ∞)

∀n, w > 0 , α > −1 max n < (p − 1)/2 q > −1 , w > 0 , v > 0 max n < (p − 1)/2 w>0 ∀n, α > −1 , β > −1

By referring to the set S, some rather complicated integrals can be explicitly evaluated. For instance, we have  ∞ exp(arctan x) π 2 (20x 2 − 10x − 3) dx = 32 sinh . 4 2 2 −∞ (1 + x ) (p,q)

Also there is a limit relation between Jn

(x) and the Hermite polynomials as follows:   2 n n d n (1 + xp ) W (p,q) ( √xp ) (−1) x (p,q) lim Jn ( √ ) = lim p→∞ p→∞ W (p,q) ( √x ) p dx n p n +x 2 d

= (−1) e

n (e −x 2 )

dx n

= Hn (x).

At the end of this part, we conclude that there are altogether six sequences of hypergeometric orthogonal polynomials that are generated by the main Eq. (1.39). The parameters a, b, c, d, and e corresponding to the equations of each these six sequences are shown in Table 1.2. Also, Table 1.3 shows the basic properties of the sequences.

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

1.4

25

A Generic Polynomial Solution for the Classical Hypergeometric Differential Equation

In this section, we reconsider the expanded form of Eq. (1.39) as   (ax 2 + bx + c)yn (x) + (dx + e)yn (x) − n d + (n − 1)a yn (x) = 0  and suppose that its polynomial solution is denoted by yn (x) = Pn

(1.61)



de x or in the a b c

text as yn (x) = pn (x; a, b, c, d, e). In 1929, Bochner classified all polynomial-type solutions of Eq. (1.61) and showed that the only polynomial systems up to a linear change of variable arising as eigenfunctions of Eq. (1.61) are: • • • •

(x)}∞ for α, β, α + β + 1 ∈ / {−1, −2, . . .}, Jacobi polynomials {Pn n=0 (α) ∞ Laguerre polynomials {Ln (x)}n=0 for α ∈ / {−1, −2, . . .}, Hermite polynomials {Hn (x)}∞ , and n=0 Bessel polynomials {Bn(α) (x)}∞ for α ∈ / {−2, −3, . . .}. n=0 (α,β)



de Then in 1988, Nikiforov and Uvarov obtained some general properties of Pn x a b c based only on its Rodrigues representation. We obtain here a generic representation for the polynomial solution of Eq. (1.61) and use the following algebraic identity to reach this goal. If a, b, and Ck (k = 0, 1, . . . , n) are real numbers, then 

n  k=0

Ck (ax + b) = k

n−k 

n   n−i k=0

k

i=0

=

b

n−i

Cn−i

(

ax k ) b

 k 

n   n−i k=0

i=0

k−i

b

k−i

Cn−i

(ax)n−k .

Theorem 1.1 The monic polynomial solution of Eq. (1.61) is given by

 

n  n de k ¯ yn (x) = Pn G(n) x = k (a, b, c, d, e) x , k a b c k=0

(1.62)

(1.63)

26

1 Generalized Sturm–Liouville Problems in Continuous Spaces

where  G(n) k =

b+

2a √ b2 − 4ac

k−n

⎛

, ⎜k − n 2 F1 ⎝

2ae−bd √ 2a b2 −4ac

+1−

d 2a

2 − d/a − 2n

⎞ − n 2√b2 − 4ac ⎟ √ ⎠. b + b2 − 4ac

For the special case a = 0, the equality can be adapted by limit considerations and gives (1.63) in the form  2

b b k−n + 1 − n k − n, cd−be (n) (n) 2 b Gk (0, b, c, d, e) = lim Gk (a, b, c, d, e) = ( ) 2 F0 cd , a→0 c − which is valid for c, d = 0, leading to  P¯n

⎛     n −n eb − cd b de ⎝ F x = 1 1 d b2 eb−cd 0 b c n

⎞ d cd − x − ⎠. b b2

b2

Also for a = b = 0 and d = 0, (1.63) is transformed to  P¯n





   e n − n2 , − n−1 de d e 2 ¯ x = lim Pn x = x+ 2 F0 a→0 d − 0 0 c a b c b→0

2cd (dx + e)2 .

Proof First change Eq. (1.61) by means of x = pt + q to   aq 2 + bq + c  d 2 y  d e + dq  dy n 2aq + b t+ t+ − d + (n − 1)a y = 0. t2 + + 2 2 ap ap dt a pa dt a (1.64) 2 If aq + bq + c = 0 and ( 2aq + b)/ap = −1 in (1.64), then √ b2 − 4ac p=∓ a

and

q=

−b ±

√ b2 − 4ac . 2a

Therefore, we have √  2ae − bd ± d b2 − 4ac  dy n  d 2y  d − d + (n − 1)a y = 0. t (t − 1) 2 + t + √ 2 dt a dt a ∓2a b − 4ac

(1.65)

Equation (1.65) is a special case of the Gauss equation t (t − 1)

 dy d 2y  + (α + β + 1) t − γ + αβ y = 0, 2 dt dt

for α = −n, β = n − 1 + d/a, and γ =

√

2ae−bd±d ∓2a

√

b2 −4ac

b2 −4ac

.

(1.66)

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

27



de Hence, by considering Pn x as a preassigned solution of Eq. (1.61) and a b c comparing the two relations (1.65) and (1.66), we obtain 

 Pn

⎛ √

√ −n, n − 1 + d/a 2 2 −b ± b − 4ac b − 4ac de ⎜ √ t+ = K 2 F1 ⎝ ∓ 2ae−bd±d b2 −4ac a 2a a b c √ ±2a

b2 −4ac



⎞ ⎟ t⎠ .

(1.67) Relation (1.67) can also be written in terms of the variable x as  Pn

⎛

−n, n − 1 + d/a de ⎜ √ x = K 2 F1 ⎝ 2ae−bd±d b2 −4ac a b c √ ±2a

b2 −4ac

⎞ √ b ∓ b2 − 4ac ⎟ ax + √ )⎠ . ∓( √ b2 − 4ac 2 b2 − 4ac (1.68)

From relation (1.68) the following two subcases are derived:  (i) P¯n

⎛

−n, n − 1 + d/a de ⎜ ∗ √ x = K 2 F1 ⎝ 2ae−bd+d b2 −4ac a b c √ 2a

b2 −4ac

⎞ √ −ax b − b2 − 4ac ⎟ − √ √ ⎠, 2 b − 4ac 2 b2 − 4ac (1.69)

 (ii) P¯n

⎛

−n, n − 1 + d/a d e ⎜ ∗∗ √ x = K 2 F1 ⎝ −(2ae−bd)+d b2 −4ac a b c √ 2a

b2 −4ac

⎞ √ 2 b + b − 4ac ⎟ ax + √ √ ⎠. b2 − 4ac 2 b2 − 4ac

Note that the latter relations differ only by a minus sign in the argument of the second formula not affect Eq. (1.61). In other words, if in (i) we consider the case  (ii), which does −d −e P¯n x , we will reach the second formula of (1.69). This means that only −a − b − c formula (ii) must be considered as the main solution. To compute K ∗∗ , it is sufficient to obtain the leading coefficient of 2 F1 (. . . | . . . ) in (ii), which is given by K

∗∗

√ √ √ n n! ( b2 − 4ac) ((bd − 2ae + d b2 − 4ac)/(2a b2 − 4ac))n = . a n (−n)n (n − 1 + d/a)n

On the other hand, according to the algebraic identity (1.62), we have  2 F1





 n (−1)n (p)n  n −n, p k − n, 1 − q − n n−k s 2 F1 rx + s = (q)n k q 1−p−n k=0



1 (rx)k . s (1.70)

28

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Hence, by considering the main solution (1.69) and assuming p =n−1+

d , a

√ bd − 2ae + d b2 − 4ac q= √ , 2a b2 − 4ac a r= √ , b2 − 4ac √ b + b2 − 4ac √ , s= 2 b2 − 4ac relation (1.70) changes to ⎛

−n, n − 1 + d/a ⎜ √ 2 F1 ⎝ bd−2ae+d b2 −4ac √ 2a

b2 −4ac

⎞ √ 2 ax b + b − 4ac ⎟ + √ √ ⎠ b2 − 4ac 2 b2 − 4ac

n (−1)n ( da + n − 1)n  = √ 2 √ b −4ac ) k=0 ( bd−2ae+d 2 2a

b −4ac



√ n−k k a n  b + b2 − 4ac   √ √ k 2 b2 − 4ac b2 − 4ac

n

⎛

⎜ k − n, × 2 F1 ⎝

√

2ae−bd−(d+(2n−2)a) 2a

√

b2 −4ac

b2 −4ac

−d/a − 2n + 2

⎞ √ 2 b2 − 4ac ⎟ k √ ⎠x . b + b2 − 4ac

Simplifying the above relation and substituting K ∗∗ finally gives the monic polynomial solution of Eq. (1.61) in the form (1.63). To deduce the limiting case when a → 0, we can use the limit relation   d r lim a − + 2 − 2n = (−d)r a→0 a r and the identity  1 F1





 n Γ (1 − p − n)  n −n k − n, 1 − p − n rx + s = 2 F0 Γ (1 − p) p k − k=0

which is a special case of identity (1.62) for Ck = (n)

(−n)k (p)k k! .



1  r k x , − s s

 

The general formula Gk (a, b, c, d, e) in (1.63) is a suitable tool to compute the coefficients of x k for any fixed degree k and arbitrary a. For example, if the coefficient

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

 x n−1 is needed in P¯n



de x , it is enough to calculate the term a b c 

(n)

Gn−1 (a, b, c, d, e) =  =

29

2a b+Δ

⎛

−1

⎜

2 F1 ⎝

−1,

2ae−bd−(d+(2n−2)a)Δ 2aΔ

−



d+(2n−2)a a



⎞ 2Δ ⎟ ⎠ b+Δ

  2ae − bd − (d + (2n − 2)a)Δ 2Δ a b+Δ  1+ 2a 2aΔ b + Δ d + (2n − 2)a =

e + (n − 1)b , d + (2n − 2)a

(1.71)

√ in which Δ = b2 − 4ac. Note in (1.71) that all parameters are free and can assume any value including zero, because it is easy to see that neither both values a and d nor both values b and e can vanish together in (1.63). After simplifying G(n) k (a, b, c, d, e) for k = n − 1, n − 2, . . ., we eventually obtain  P¯n





n e + (n − 1)b n−1 de x x = xn + 1 d + (2n − 2)a a b c 

n (e + (n − 1)b)(e + (n − 2)b) + c(d + (2n − 2)a) n−2 x + + ··· (d + (2n − 2)a)(d + (2n − 3)a) 2

⎛ 

√   2 b + b − 4ac n n ⎜ −n, + 2 F1 ⎝ 2a n

√

2ae−bd−(d+(2n−2)a)

√

b2 −4ac

2a b2 −4ac d+(2n−2)a −( ) a

⎞ √ 2 b2 − 4ac ⎟ √ ⎠. b+ b2 −4ac

In particular, the above relation gives  P¯n



 b + √b2 − 4ac n de 0 = 2a a b c ⎛ √ 2ae−bd−(d+(2n−2)a) b2 −4ac √ −n, ⎜ 2a b2 −4ac × 2 F1 ⎝ −( d+(2n−2)a ) a

⎞ 2√b2 − 4ac ⎟ √ ⎠. b + b2 − 4ac

30

1 Generalized Sturm–Liouville Problems in Continuous Spaces

For instance, we have

d e P¯0 x =1, a b c

 e d e P¯1 x =x+ , d a b c 

c(d + 2a) + e(e + b) e+b d e | x = x2 + 2 P¯2 x+ , d + 2a (d + 2a)(d + a) a b c

 e + 2b 2 c(d + 4a) + (e + b)(e + 2b) d e P¯3 x +3 x x = x3 + 3 d + 4a (d + 4a)(d + 3a) a b c 

+

2c(d + 3a)(e + 2b) + ce(d + 4a) + e(e + b)(e + 2b) . (d + 4a)(d + 3a)(d + 2a)

Remark 1.2 We can apply the Gauss summation formula and take

2

√ 2 √b −4ac = 1 in 2

b+

b −4ac

(1.63) to conclude that ac = 0. In this case, if c = 0, the following special case is derived: ⎞ ⎛



2ae−bd d n + 1 − − n k − n,  k−n a 2ab 2a n d e 1⎠ x k ¯ ⎝ ( ) 2 F1 Pn x = b k a b 0 2 − d/a − 2n k=0 

n  a k−n Γ (2 − 2n − d/a)Γ (1 − k − e/b) k n x = ( ) b Γ (2 − n − k − d/a)Γ (1 − n − e/b) k k=0 

b n Γ (2 − 2n − d/a)Γ (1 − e/b) −n, n − 1 + d/a a (1.72) = n − x . 2 F1 a Γ (2 − n − d/a)Γ (1 − n − e/b) b e/b 

Moreover, if we refer to the Nikiforov and Uvarov approach and consider Eq. (1.61) as a self-adjoint form, the Rodrigues representation of the monic polynomials is derived as  P¯n



de x = n a b c 

1 

d + (n + k − 2)a W

k=1

dn ×





de x a b c 

 n (ax 2 + bx + c) W dx n



 de x a b c

,

(1.73)

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

31

where  W



  (d − 2a)x + e − b  de dx . x = exp ax 2 + bx + c a b c

(1.74)

Now if once again c = 0 in (1.73), by referring to (1.72), we obtain   exp − (

n 

(d−2a)x+e−b dx ax 2 +bx



  n d n (ax 2 + bx) exp



 (d−2a)x+e−b dx 2 ax +bx

dx n

d + (n + k − 2)a)

k=1

Γ (2 − 2n − d/a)  = Γ (1 − n − e/b) n

k=0



n  a k−n Γ (1 − k − e/b) xk. b Γ (2 − n − k − d/a) k

Remark 1.3 There is another representation for P¯n (x; a, b, c, d, e) in (1.63). If the weight function (1.74) is written in the same form as (1.24), i.e.,

de x = K(x + θ1 )A (x + θ2 )B , a b c

 W where K is a constant, A=

2ae − bd d −1+ √ , 2a 2a b2 − 4ac

B=

2ae − bd d −1− √ , 2a 2a b2 − 4ac

and θ1 =

b−

√

b2 − 4ac , 2a

θ2 =

b+

√ b2 − 4ac , 2a

(1.75)

then replacing it in the Rodrigues representation (1.73) yields  P¯n



de x = a b c

dn (Ka n (x dx n  n 

+ θ1 )n (x + θ2 )n (x + θ1 )A (x + θ2 )B )  d + (n + k − 2)a (x + θ1 )A (x + θ2 )B

K

k=1

=

 dn  n+A n+B (x + θ . ) (x + θ ) 1 2 (n − 1 + d/a)n (x + θ1 )A (x + θ2 )B dx n 1

(1.76)

32

1 Generalized Sturm–Liouville Problems in Continuous Spaces

But the Leibniz rule   d n (x + θ1 )n+A (x + θ2 )n+B dx n



n (−n − A)k (−n − B)n−k (x + θ1 )n+A−k (x + θ2 )B+k , = (−1) k k=0 n

n 

implies that relation (1.76) is simplified as

1 d e P¯n x = (2 − 2n − d/a)n a b c 

n  2ae − bd 2ae − bd d d n +1− √ +1+ √ × )k (−n − )n−k (−n − 2 2a 2a k 2a b − 4ac 2a b2 − 4ac k=0 √ √  b + b2 − 4ac k b − b2 − 4ac n−k  x+ × x+ , (1.77) 2a 2a 

which is another representation for the polynomial solution of Eq. (1.61). By combining (1.63) and (1.77), we straightforwardly arrive at the identity 

  2ae − bd 2ae − bd d d n +1− )k −n − +1+ (−n − 2a 2aΔ 2a 2aΔ k n−k k=0 n−k  k

 b−Δ b+Δ x+ x+ 2a 2a 

k−n  n  2a n = (2 − 2n − d/a)n b+Δ k k=0 ⎞ ⎛ d k −n, 2ae−bd + 1 − − n 2aΔ 2a 2Δ ⎠ x k , × 2 F1 ⎝ b + Δ 2 − d/a − 2n

n 

√ where Δ = b2 − 4ac. Relation (1.77) can also be represented in terms of the hypergeometric form  P¯n



(−n − de x = a b c

d 2a

+1+

2ae−bd 2aΔ )n (x

(2 − 2n − d/a)n

+ θ 1 )n

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

⎛ × 2 F1 ⎝

−n, −n − d 2a

d 2a

−

+1−

33 2ae−bd 2aΔ

2ae−bd 2aΔ

⎞ x + θ2 ⎠ x + θ1 ,

where θ1 and θ2 are defined by (1.75). Remark 1.4 Using the explicit representation (1.69), one can compute the generic values of the polynomials at their boundary points of definition, −θ1 and −θ2 , so that if  √ b + b2 − 4ac 0 + √ = √ 2 2 1 b − 4ac 2 b − 4ac ax

then x=

−b −

√ b2 − 4ac = −θ2 2a

and

x=

−b +

√ b2 − 4ac = −θ1 . 2a

Hence by substituting the above values into (1.69), we respectively obtain 



√ √ n ( b2 − 4ac) ((d/2a) − (2ae − bd)/(2a b2 − 4ac))n de −θ2 = (−a)n (n − 1 + d/a)n a b c





√ √ n ( b2 − 4ac) ((d/2a) + (2ae − bd)/(2a b2 − 4ac))n de . −θ1 = a n (n − 1 + d/a)n a b c

P¯n and P¯n

For instance, for the monic Jacobi orthogonal polynomials P¯n (a, b, c, d, e) = (−1, 0, 1, −α − β − 2, −α + β), we have

(α,β)

P¯n(α,β) (+1) = 2n

(x) with the initial vector

(α + 1)n Γ (n + 1 + α)Γ (n + 1 + α + β) , = 2n (n + 1 + α + β)n Γ (α + 1)Γ (2n + 1 + α + β)

and P¯n(α,β) (−1) = (−2)n

(β + 1)n Γ (n + 1 + β)Γ (n + 1 + α + β) . = (−2)n (n + 1 + α + β)n Γ (β + 1)Γ (2n + 1 + α + β)

Some Other Properties of P¯n (x; a, b, c, d, e)

34

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Property 1 Using the representation (1.73), a linear change of variable can be applied for the monic polynomials so that if x = wt + v, then we have  P¯n



de wt + v a b c wn d n

 n (aw2 t 2 + (2awv + bw)t + (av 2 + bv + c)) W

=

n 



d + (n + k − 2)a W

k=1





de wt + v a b c



 de wt + v a b c

. dt n (1.78)

But since  W



   (d − 2a)(wt + v) + e − b de w dt wt + v = exp 2 2 2 aw t + (2av + b)wt + av + bv + c a b c  =W



dw2 , (dv + e)w t , aw2 , (2av + b)w, av 2 + bv + c

relation (1.78) is eventually transformed to  P¯n



de wt + v = wn P¯n a b c aw2 ,



dw2 , (dv + e)w t . 2 (2av + b)w, av + bv + c

For instance, we can straightforwardly conclude that  P¯n





 de d −e n ¯ −t = (−1) Pn t . a b c a −b c

Property 2 The second formula of (1.69) is a suitable tool to compute a generic three-term recurrence equation for the polynomials (1.63), since it can be applied along with various identities of the Gauss hypergeometric function for generating a recurrence relation. For example, by noting the well-known identity (p − q) ((p − q − 1)(p − q + 1)t + 2pq + r(1 − p − q)) 2 F1 (p, q, r ; t) + q(r − p)(p − q + 1)2 F1 (p − 1, q + 1, r ; t) + p(r − q)(p − q − 1)2 F1 (p + 1, q − 1, r ; t) = 0,

(1.79)

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

35

if we assume in (1.79) that p = −n, q=

d + n − 1, a

√ bd − 2ae + d b2 − 4ac r= √ , 2a b2 − 4ac and √ b + b2 − 4ac x+ √ , t= √ b2 − 4ac 2 b2 − 4ac a

then after some computations, we obtain   2n(n + 1)ab + (d − 2a)(e + 2nb) ¯ P¯n+1 (x) = x + Pn (x) + n(d + (n − 2)a) (d + 2na)(d + (2n − 2)a)   c(d + (2n − 2)a)2 − nb 2 (d + (n − 2)a) + (e − b)(a(e + b) − bd) × P¯n−1 (x), (d + (2n − 3)a)(d + (2n − 2)a)2 (d + (2n − 1)a) in which P¯n (x) has the same meaning as the monic polynomials of (1.63) with values e P¯0 (x) = 1 and P¯1 (x) = x + . d Property 3 As the last property, we can compute the norm square value of the monic polynomials. Let [L, U ] be a predetermined orthogonality interval that (other than for finite families) consists of the zeros of σ (x) = ax 2 + bx + c or ±∞. By noting the Rodrigues representation (1.73), we have  2 P¯n  =



U L

 ×

U L

 P¯n2 

P¯n



de de x W x dx = n  a b c a b c

1 d + (n + k − 2)a

k=1





  dn  n de de 2 (ax + bx + c) W dx. x x n dx a b c a b c

(1.80)

36

1 Generalized Sturm–Liouville Problems in Continuous Spaces

So integrating by parts on the right-hand side of (1.80) yields  2 P¯n  =

n! (−1)n n 

d + (n + k − 2)a



U

n

(ax + bx + c) 2

L



 exp

(d − 2a)x + e − b  dx dx. ax 2 + bx + c

k=1

(1.81)

Six Special Cases of the Generic Polynomials P¯n (x; a, b, c, d, e)

1.4.1

As we have shown in Tables 1.2 and 1.3, there are six sequences of hypergeometric orthogonal polynomials. In this section, we apply previously obtained formulas to determine the basic properties of each of the six sequences. 1. For Jacobi polynomials we obtain • Initial vector: (a, b, c, d, e) = (−1, 0, 1, −α − β − 2, −α + β). • Generic representation of polynomials: (n + α + β + 1)n ¯ Pn(α,β) (x) = Pn 2n n!





−α − β − 2, −α + β x −1, 0, 1

n (n + α + β + 1)n  = (−1)k−n 2n n! k=0



 n k − n, −n − α 2 F1 k −2n − α − β



2 xk .

• Hypergeometric representation:  P¯n



 −n, −α − n −α − β − 2, −α + β n n x = (−1) (1 − x) 2 F1 −α − β − 2n −1, 0, 1



2 . 1−x

• Weight function:  W



  −(α + β)x + β − α  −α − β − 2, −α + β dx x = exp −x 2 + 1 −1, 0, 1 = (1 − x)α (1 + x)β .

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

37

Jacobi polynomials include some special cases, such as: • Gegenbauer polynomials:

Cn(λ) (x)

2n (λ)n ¯ = Pn n!



−2λ − 1, 0 x −1 0 1



2n (λ)n = (−1)n (1 − x)n 2 F1 n!





−n, 1/2 − λ − n 2 , 1 − 2λ − 2n 1 − x

• Legendre polynomials: Pn (x) =



(2n)! (n!)2 2n

P¯n





 (2n)! −n, −n 2 n n (−1) (1 − x) 2 F1 , x = (n!)2 2n −2n 1−x 0, 1

−2, 0 −1,

• Chebyshev polynomials of the first kind:  Tn (x) = 2

n−1

P¯n





 −n, 1/2 − n 2 n−1 n n , x = 2 (−1) (1 − x) 2 F1 1 − 2n 1−x 0, 1

−1, 0 −1,

• Chebyshev polynomials of the second kind:  Un (x) = 2 P¯n n





−n, −1/2 − n 2 −3, 0 n n n . x = 2 (−1) (1 − x) 2 F1 −1 − 2n 1−x −1, 0, 1

2. For the Laguerre polynomials we obtain • Initial vector: (a, b, c, d, e) = (0, 1, 0, −1, α + 1). • Generic and hypergeometric representations: (−1)n ¯ L(α) Pn n (x) = n!





 (α + 1)n −n −1, α + 1 x = 1 F1 n! α+1 0, 1, 0

• Weight function:  W

−1, α + 1 0, 1, 0



  −x + α  dx = x α e−x . x = exp x



x .

38

1 Generalized Sturm–Liouville Problems in Continuous Spaces

3. For Hermite polynomials we obtain • Initial vector: (a, b, c, d, e) = (0, 0, 1, −2, 0). • Generic and hypergeometric representations:  Hn (x) = 2n P¯n





n n−1 1 , − − −2, 0 n 2 2 − 2 . x = (2x) 2 F0 x − 0, 0, 1

• Weight function:  W



  −2, 0 x = exp −2x dx = exp(−x 2 ). 0, 0, 1

4. For the first finite sequence of hypergeometric orthogonal polynomials we respectively find • Initial vector: (a, b, c, d, e) = (1, 1, 0, −p + 2, q + 1). In [4], this type of polynomials are called Romanovski-Jacobi polynomials, see also (p,q) [5, 8], while in [6] they are indicated by Mn (x). • Generic and hypergeometric representations:

−p + 2, q + 1 = (−1) (n + 1 − p)n P¯n x 1, 1, 0  −n, n + 1 − p = (−1)n (q + 1)n 2 F1 q +1 

(p,q) Mn (x)

n



−x .

By referring to the general formula (1.81), we can recompute their norm square value as follows 

∞ 0

xq P¯ 2 (1 + x)p+q n





−p + 2, q + 1 x dx 1, 1, 0  ∞ xq n ! (−1)n n (x 2 + x) dx, = n  (1 + x)p+q 0 (−p + n + k) k=1

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

39

which yields 

∞

 (p,q) 2 xq n ! (p − n − 1)! (q + n)! . (x) dx = p+q Mn (p − 2n − 1) (p + q − n − 1)! (1 + x)

0

5. For the second finite sequence of hypergeometric orthogonal polynomials we obtain • Initial vector: (a, b, c, d, e) = (1, 0, 0, −p + 2, 1). In [4], such polynomials are called Romanovski–Bessel polynomials, while in [6] (p) they are indicated by Nn (x). • Generic and hypergeometric representations:  (p) Nn (x)

= (−1) (n+1 − p)n P¯n n





 −n, 1 − p+n −p + 2, 1 n −x . x = (−1) 2 F0 − 1, 0, 0

Once again, we can refer to (1.81) and recompute their norm square value as 

∞

 x −p e

− x1

0

P¯n2



 ∞ 1 n ! (−1)n −p + 2, 1 x 2n x −p e− x dx, x dx = n  1, 0, 0 (−p + n + k) 0 k=1

which yields 

∞

x −p e

−1 x



(p)

Nn (x)

2

0

dx =

n! (p − (n + 1))! . p − (2n + 1)

6. For the third finite sequence of hypergeometric orthogonal polynomials we obtain • Initial vector:   (a, b, c, d, e) = 1, 0, 1, 2(1 − p), q . In [4], such polynomials are called Romanovski–Pseudo–Jacobi polynomials, while (p,q) in [7] they are indicated by Jn (x). • Generic and hypergeometric representations:  (p,q) (x) Jn

= (−1)n (n + 1 − 2p)n P¯n



2(1 − p), q x 1, 0, 1

40

1 Generalized Sturm–Liouville Problems in Continuous Spaces

 = (−1)n (n + 1 − 2p)n (x + i)n 2 F1



−n, −n + p − q2 i 2i . x+i 2p − 2n (1.82)

Taking into account the identity



 n k − n, 1 − q ∗ − n 1  r ∗ k F x ∗ 2 1 s s∗ k 1 − p∗ − n k=0

n 

(q ∗ )n = 2 F1 (−s ∗ )n (p∗ )n





−n, p∗ ∗ r x + s∗ , q∗

we can simplify relation (1.82) as  (p,q) (x) Jn

= (2i) (1 − p + iq/2)n 2 F1 n

(−p+iq/2,−p−iq/2)

= n!(2i)n Pn



−n, n + 1 − 2p 1 − ix 1 − p + iq/2 2

(ix),

(α,β)

where Pn (x) are the same as the Jacobi orthogonal polynomials. • Weight function:  W

1.4.2



2(1 − p), q x = (1 + x 2 )−p exp(q arctan x). 1, 0, 1

How to Find the Initial Vector If a Special Case of the Main Weight Function Is Given?

The easiest way to find initial parameters a, b, c, d, e is to first compute the logarithmic derivative of the given weight function and then match the pattern with (d − 2a)x + e − b W  (x) = . W (x) ax 2 + bx + c We clarify this technique with some particular examples. Example 1.1 Consider the weight function W (x) = (−x 2 + 3x − 2)10 on x ∈ [1, 2]. By computing the logarithmic derivative of it, we get −20x + 30 (d − 2a)x + e − b W  (x) = = ⇒ (a, b, c, d, e) = (−1, 3, −2, −22, 33). 2 W (x) −x + 3x − 2 ax 2 + bx + c

1.4 A Generic Polynomial Solution for the Classical Hypergeometric. . .

41



−22, 33 Therefore, the related monic polynomials are P¯n x , which are orthog−1, 3, −2 onal with respect to the given weight function on [1, 2] for every value n. This means that it is no longer necessary to know that they are shifted Jacobi polynomials on the interval [1, 2], because they can be explicitly expressed in terms of the generic polynomials (1.63). 

Example 1.2 The weight function W (x) = (2x 2 + 2x + 1)−10 is given on x ∈ (−∞, ∞). Thus −40x − 20 W  (x) = 2 ⇒ (a, b, c, d, e) = (2, 2, 1, −38, −18), W (x) 2x + 2x + 1

−38, −18 and the related monic polynomials are P¯n x . Note that they are finitely 2, 2, 1 orthogonal for n ≤ 9, because according to (1.59), N = max{n} < 10 − (1/2). In other words, the finite set 

{ P¯n (x ; 2, 2, 1, −38, −18)}9n=0 is orthogonal with respect to the weight function (2x 2 + 2x + 1)−10 on (−∞, ∞).   Example 1.3 Consider the weight function W (x) = exp x(θ − x) on x ∈ (−∞, ∞), where θ ∈ R. Then W  (x) = −2x + θ ⇒ (a, b, c, d, e) = (0, 0, 1, −2, θ ), W (x)  and the related monic orthogonal polynomials are P¯n



−2, θ x for every value 0, 0, 1

n. √

x Example 1.4 For the weight function W (x) = on x ∈ [0, ∞), which is a special (x+2)8 case of the second kind of the beta distribution, we have

W  (x) −15x + 2 = ⇒ (a, b, c, d, e) = (2, 4, 0, −11, 6). W (x) 2x 2 + 4x

−11, 6 Therefore, the related monic orthogonal polynomials are P¯n x , valid for 2, 4, 0 n ≤ 3 . This means that the finite set { P¯n (x ; 2, 4, 0, −11, 6) }3n=0 is orthogonal with √ respect to the weight function x/(x + 2)8 on [0, ∞). 

42

1.5

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Fourier Transforms of Finite Sequences of Hypergeometric Orthogonal Polynomials

It is known that some orthogonal polynomial systems are mapped onto each other by the Fourier transform. The best-known examples of this type are Hermite functions, i.e., Hermite polynomials Hn (x) multiplied by exp(−x 2/2), which are eigenfunctions of a Fourier transform. In this section, we introduce three examples of this type using finite sequences of hypergeometric orthogonal polynomials and obtain their orthogonality relations in the sequel. (p,q) (p) To derive the Fourier transform of polynomials Mn (x) and Nn (x) defined in (1.43) and (1.49), we will use the two beta integrals 

1

B(a, b) =



∞

x a−1(1 − x)b−1 dx =

0

t a−1 (1 + t)

0

a+b

dt =

Γ (a)Γ (b) . Γ (a + b)

The Fourier transform of a function, say g(x), is defined as  F(s) = F(g(x)) =

∞

−∞

e−isx g(x) dx,

and for the inverse transform we have g(x) =

1 2π



∞

−∞

eisx F(s) ds.

Also for g, h ∈ L2 (R), Parseval’s identity is given by 

∞ −∞

g(x)h(x) dx =

1 2π



∞ −∞

F(g(x))F(h(x)) ds.

(1.83)

We now define the specific functions g(x) =

eqx M (r,u)(ex ) (1 + ex )p+q n

and h(x) =

eax (1 + ex )a+b

(c,d) x Mm (e )

(p,q)

in terms of the polynomials Mn (x). For both functions, the Fourier transform exists. For instance, for g(x) we have   F g(x) =



∞

−∞

e−isx

eqx M (r,u)(ex ) dx = (1 + ex )p+q n



∞ 0

t −is−1

tq M (r,u)(t) dt (1 + t)p+q n

1.5 Fourier Transforms of Finite Sequences of Hypergeometric. . .

 = (−1)n n!

43

 ∞ q−is−1+k   n t u + n  (−1)k (−n)k (n + 1 − r)k p+q dt (u + 1)k k! n (1 + t) 0 k=0

(u + n)!  (−1)k (−n)k (n + 1 − r)k Γ (q − is + k)Γ (p + is − k) u! (u + 1)k k! Γ (p + q) n

= (−1)n

k=0

(−1)n Γ (u + n + 1)Γ (q − is)Γ (p + is) = 3 F2 Γ (u + 1)Γ (p + q)





−n, n + 1 − r, q − is 1 , u + 1, −p + 1 − is (1.84)

in which the following identities have been used: Γ (a + k) = Γ (a)(a)k

and Γ (a − k) =

(−1)k Γ (a) . (1 − a)k

By substituting (1.84) into Parseval’s identity (1.83), we obtain  2π

∞

−∞

e(q+a)x

(c,d) x Mn(r,u)(ex )Mm (e )dx (1 + ex )(p+q+a+b)  ∞ t q+a−1 (c,d) = 2π Mn(r,u) (t)Mm (t)dt (1 + t)p+q+a+b 0

(−1)n+m Γ (u + n + 1)Γ (d + m + 1) Γ (u + 1)Γ (p + q)Γ (d + 1)Γ (a + b)   ∞ −n, n + 1 − r, q − is × Γ (q − is)Γ (p + is)Γ (a − is)Γ (b + is) 3 F2 u + 1, −p + 1 − is −∞

 −m, m + 1 − c, a − is × 3 F2 1 ds. d + 1, −b + 1 − is =



1 (1.85)

On the other hand, if on the left-hand side of (1.85) we take u=d =q +a−1

and

r = c = p + b + 1,

then according to the orthogonality relation (1.47), Eq. (1.85) reads Γ 2 (q + a)Γ (p + q)Γ (a + b) (2π)n!(p + b − n)!(q + a − 1 + n)! δn,m (p + b − 2n)(p + q + a + b − n − 1)! (−1)n+m Γ (q + a + n)Γ (q + a + m)

44

1 Generalized Sturm–Liouville Problems in Continuous Spaces



−n, n − p − b, q − is = Γ (q − is)Γ (p + is)Γ (a − is)Γ (b + is) 3 F2 1 q + a, −p + 1 − is −∞

 −m, m − p − b, a − is × 3 F2 1 ds. q + a, −b + 1 − is 



∞

Hence, the special function Γ (a + d + n) An (x; a, b, c, d) = 3 F2 Γ (a + d)



−n, n − b − c, d − x a + d, −c + 1 − x



1

satisfies a finite orthogonality relation of the form 1 2π



∞

−∞

Γ (a + ix)Γ (b − ix)Γ (c + ix)Γ (d − ix)An (ix; a, b, c, d)Am (−ix; d, c, b, a)dx =

Γ (a + d + n)Γ (b + c + 1 − n)Γ (c + d)Γ (a + b) n! δn,m , (b + c − 2n) Γ (a + b + c + d − n)

where a + d > −n, b + c > 2n, a + b > 0, and c + d > 0. (p) This approach can be similarly applied to the finite orthogonal polynomials Nn (x). For this purpose, we first define the specific functions 1 (q) u(x) = exp(−px − e−x )Nn (ex ) 2

1 and v(x) = exp(−rx − e−x )Nm(u) (ex ), 2

and take the Fourier transform of, e.g., u(x) to get   F u(x) =



∞ −∞

e

−isx −(px+ 21 e−x )

e

(q) Nn (ex )dx



∞

= 0

= (−1)n n!(q − n − 1)!

n  k=0

k

−1

t −is−1−p e 2t Nn (t)dt

(−1) (q − n − 1 − k)!k!(n − k)!



∞

(q)

−1

t −is−1−p+k e 2t dt



0

k n  (−n)k (n + 1 − q)k (2−1 ) (1 − p − is)k k! k=0 

−n, n + 1 − q 1 n p+is = (−1) 2 Γ (p + is)2 F1 . 1 − p − is 2

= (−1)n 2p+is Γ (p + is)

In (1.86), we have used the definite integral 

∞ 0

−1

t −is−1−p+k e 2t dt = 2p+is−k Γ (p + is − k).

(1.86)

1.5 Fourier Transforms of Finite Sequences of Hypergeometric. . .

45

Now apply Parseval’s identity once again to obtain  2π

∞ −∞

−x

e−(p+r)x e−e Nn (ex )Nm(u) (ex )dx = 2π (q)



∞

t −(p+r+1) e

0

−1 t

(q)

Nn (t)Nm(u) (t)dt

= (−1)n+m 2p+r  ×



∞ −∞

Γ (p + is)Γ (r + is) 2 F1

−n, n + 1 − q 1 − p − is



1 −m, m + 1 − u 2 2 F1 1 − r − is



1 2 ds.

(1.87) If q = u = p + r + 1 in (1.87), on noting the orthogonality relation (1.53), it can be concluded that the function

 −n, n − a − b 1 Bn (x; a, b) = 2 F1 −a + 1 − x 2 satisfies a finite orthogonality relation of the form 1 2π



∞ −∞

Γ (a + ix)Γ (b − ix) Bn (ix; a, b)Bm (−ix; b, a) dx =

n!Γ (a + b + 1 − n) δn,m , (a + b − 2n) 2a+b

provided that a + b > 2n. Finally, the aforementioned approach can be applied to the third finite sequence (p,q) (x) defined in (1.57). We now consider the specific functions Jn g(x) = (1 + x 2 )−β exp(α arctan x)Jn(c,d)(x) and h(x) = (1 + x 2 )−u exp(l arctan x)Jm(v,w)(x), in which α, β, c, d, and l, u, v, w are all real parameters. For instance, for the function g, we get   F g(x) =



∞

−∞

e−isx (1 + x 2 )−β exp(α arctan x)Jn(c,d)(x)dx

= (2i)n (1 − c + id/2)n  n

 ∞  (−n)k (n + 1 − 2c)k −isx −β+i α2 −β−i α2 k × e (1 − ix) (1 + ix) (1 − ix) dx (1 − c + id/2)k k!2k −∞ k=0

= (2i)n (1 − c + id/2)n

46

1 Generalized Sturm–Liouville Problems in Continuous Spaces

×

 n  α α (−n)k (n + 1 − 2c)k ∞ −isx e (1 − ix)−β+k+i 2 (1 + ix)−β−i 2 dx. (1 − c + id/2)k k!2k −∞

(1.88)

k=0

So it remains to evaluate in (1.88) the integral A∗k (s; α, β)

 =

∞ −∞

α

α

e−isx (1 − ix)−β+k+i 2 (1 + ix)−β−i 2 dx.

(1.89)

In general, if Re(p + q) > 1, then according to [3] we have 

∞ −∞

e

−isx

−p

(1 − ix)

−q

(1 + ix)

⎧ p+q π 2 −1 W p−q 1−p−q (2s) ⎪ ⎪ ⎨ Γ (p) (s/2) 2 , 2 dx=

⎪ ⎪ ⎩

π Γ (q)

(−s/2)

p+q 2 −1

(s > 0),

W q−p , 1−p−q (−2s) 2

(s < 0),

2

where Wa,b (s) denotes Whittaker functions of the second kind, defined by  Wa,b (s) =

s 1/2 e−s/2

 Γ (−2b) b Γ (1/2−b−a) s 1 F1



1/2 + b − a s 2b + 1

 Γ (2b) s −b 1 F1 + Γ (1/2+b−a)



1/2 − b − a , s −2b + 1

2b ∈ / Z.

Hence, for Re(2β − k) < 1 and k + 1 − 2β ∈ / Z, (1.89) would be equal to

A∗k (s; α, β)

=

⎧ π β−1−k/2 W− k+iα , k+1 −β (2s) ⎪ ⎨ Γ (β−k−iα/2) (s/2) 2 2 ⎪ ⎩

π Γ (β+iα/2)

(−s/2)β−1−k/2 W k+iα , k+1 −β (−2s) 2

2

(s > 0), (s < 0),

and relation (1.88) can be simplified as F(g(x)) = 2n i n

Γ (1 − c + n + id/2) Cn (s; α, β, c, d) , Γ (1 − c + id/2)

(1.90)

where we have defined Cn (s; p1 , p2 , p3 , p4 ) =

n  (−n)k (n + 1 − 2p3 )k ∗ A (s; p1 , p2 ). (1 − p3 + ip4 /2)k k!2k k

(1.91)

k=0

By referring to (1.83), another consequence is that F(h(x)) =

(−i)m 2m Γ (1 − v + m − iw/2) Cm (s; −l, u, v, −w) . Γ (1 − v − iw/2)

(1.92)

1.6 A Symmetric Generalization of Sturm–Liouville Problems

47

Therefore, on substituting (1.90) and (1.92) into Parseval’s identity (1.83), we obtain 

∞ −∞

(1 + x 2 )−(β+u) exp((α + l) arctan x) Jn(c,d)(x) Jm(v,w)(x) dx =

(−1)m i n+m 2m+n Γ (1 − c + n + id/2) Γ (1 − v + m − iw/2) Γ (1 − c + id/2) Γ (1 − v − iw/2)  ∞ Cn (s; α, β, c, d) Cm (s; −l, u, v, −w) ds . (1.93) × −∞

If c = v = β + u and d = w = α + l in (1.93), then according to the orthogonality relation (1.60), we conclude that the sequence of functions {Cn (s; p1 , p2 , p3 , p4 )}n≥0 , defined in (1.91) satisfies the following finite orthogonality relation: 

∞ −∞

=

Cn (s; α, β, ν, ω) Cm (s; α − ω, ν − β, ν, −ω) ds

Γ (2ν − n) Γ 2 (1 − ν − iω/2) π 22−2ν n! δm,n , (2ν − 2n − 1) Γ (ν − n + iw/2)Γ (ν − n − iw/2)Γ 2 (1 − ν + n + iω/2)

for m, n = 0, 1, . . . , N = max{m, n} < 2β −1 ≤ ν −1/2, Re(2β −n) < 1, n+1−2β ∈ / Z, and α, ω ∈ R.

1.6

A Symmetric Generalization of Sturm–Liouville Problems

In this section, we present some conditions under which ordinary Sturm–Liouville problems with symmetric solutions are extended to a larger class. The main advantage of this extension is that the corresponding solutions preserve the orthogonality property. Let Φn (x) be a sequence of symmetric functions that satisfies the second-order differential equation   1 − (−1)n E(x) Φn (x) = 0, (1.94) A(x)Φn (x)+B(x)Φn (x)+ λn C(x)+D(x) + 2

where A(x) , B(x) , C(x) , D(x), and E(x) are independent functions and {λn } is a sequence of constants. Clearly, choosing E(x) = 0 in (1.94) leads to an ordinary Sturm– Liouville equation. Since Φn (x) is symmetric, by substituting the symmetry property Φn (−x) = (−1)n Φn (x),

48

1 Generalized Sturm–Liouville Problems in Continuous Spaces

into (1.94), we can immediately conclude that A(x) , C(x) , D(x), and E(x) are even functions, while B(x) must be odd. Now we write Eq. (1.94) in a self-adjoint form to get d  d Φn (x)  A(x) R(x) dx dx   1 − (−1)n E(x)R(x) Φn (x) , = − λn C(x) + D(x) R(x) Φn (x) − 2

(1.95)

where R(x) = exp

  B(x)    B(x) − A (x)  1 dx = exp dx . A(x) A(x) A(x)

Since A(x)R(x) is an even function, the orthogonality interval should be also symmetric, say [−θ, θ ]. Hence if x = θ is a root of the function A(x)R(x), applying the Sturm– Liouville theorem to (1.95) yields    θ A(x) R(x) Φn (x)Φm (x) − Φ  m (x)Φn (x)

−θ

 = (λm − λn ) −(

θ

−θ

C(x) R(x) Φn (x) Φm (x) dx

(−1)m − (−1)n ) 2



θ −θ

E(x) R(x) Φn (x) Φm (x) dx .

(1.96)

Obviously, the left-hand side of (1.96) is zero. So to prove the orthogonality property, it is enough to show that the value (−1)m − (−1)n F (n, m) = 2



θ −θ

E(x) R(x) Φn (x) Φm (x) dx

is always equal to zero for every m, n ∈ Z+ . For this purpose, four cases should be generally considered for m and n: If both m and n are even (or odd), then clearly F (n, m) = 0. If one of the two values is odd and the other one is even (or conversely), then  F (2i , 2j + 1) = −

θ −θ

E(x) R(x) Φ2i (x) Φ2j +1 (x) dx.

(1.97)

Since in the above integral E(x), R(x), and Φ2i (x) are even functions and Φ2j +1 (x) is odd, the integrand of (1.97) is an odd function, and consequently F (2i , 2j + 1) = 0. This issue also holds for the case n = 2i + 1 and m = 2j , i.e., F (2i + 1 , 2j ) = 0.

1.7 A Basic Class of Symmetric Orthogonal Polynomials

49

Theorem 1.2 The symmetric sequence Φn (x) = (−1)n Φn (−x), as a specific solution of differential equation (1.94), satisfies the orthogonality relation 

θ −θ

W ∗ (x) Φn (x) Φm (x) dx =



θ −θ

W ∗ (x) Φn2 (x) dx

 δn,m ,

where 

∗

W (x) = C(x)R(x) = C(x) exp(

B(x) − A (x) dx) A(x)

(1.98)

denotes the corresponding weight function and is a positive and even function on [−θ, θ ]. Remark 1.5 Although some special functions such as Bessel functions are symmetric and satisfy a differential equation whose coefficients are alternately even and odd, they do not satisfy the conditions of the above theorem. For instance, if in Eq. (1.94) we choose A(x) = x 2 ,

B(x) = x,

C(x) = 1,

D(x) = x 2 ,

E(x) = 0,

and λn = −n2 ,

we obtain the Bessel differential equation x 2 Φn (x) + x Φn (x) + (x 2 − n2 ) Φn (x) = 0, with the symmetric solution (−1)n Jn (−x) = Jn (x) =

∞  k=0

x n+2k (−1)k ( ) k!(n + k)! 2

(n ∈ Z+ ).

However, since the orthogonality interval of Bessel functions is not symmetric (i.e., [0, 1]), the theorem cannot be applied to Jn (x) unless there exists a specific even function E(x) for Eq. (1.94) such that the corresponding solution has infinitely many zeros, identical to the case of the usual Bessel functions of order n.

1.7

A Basic Class of Symmetric Orthogonal Polynomials

In the previous section, we determined how to generalize ordinary Sturm–Liouville problems with symmetric solutions. In this section, using Theorem 1.2, we introduce a basic class of symmetric orthogonal polynomials with four free parameters and obtain its standard properties. Then we show that four main sequences of symmetric orthogonal polynomials can be extracted from the introduced class. They are respectively the generalized ultraspherical polynomials, generalized Hermite polynomials, and two other sequences of symmetric polynomials, which are finitely orthogonal on the real line.

50

1 Generalized Sturm–Liouville Problems in Continuous Spaces

So referring to differential equation (1.94), let p, q, r, s ∈ R and choose A(x) = x 2 (px 2 + q), B(x) = x(rx 2 + s), C(x) = x 2 , D(x) = 0, E(x) = −s,

(1.99)

leading to the second-order equation x 2 (px 2 + q) Φn (x) + x(rx 2 + s) Φn (x)   − n(r + (n − 1)p)x 2 + (1 − (−1)n ) s/2 Φn (x) = 0,

(1.100)

whose polynomial solution is

r s Φn (x) = Sn x pq ⎞ 

⎛[n/2]−(k+1)   [n/2]  [n/2] 2i + (−1)n+1 + 2 [n/2] p + r ⎝ ⎠ x n−2k , =   n+1 k 2i + (−1) + 2 q + s k=0 i=0 

(1.101)

where neither both q and s nor p and r can vanish together. These polynomials cover almost all known symmetric orthogonal polynomials, such as Legendre polynomials, the first and second kinds of Chebyshev polynomials, ultraspherical polynomials, generalized ultraspherical polynomials, Hermite polynomials, and generalized Hermite polynomials. Furthermore, there are two symmetric sequences of finite orthogonal polynomials that are special cases of the general representation (1.101). We define the monic type of polynomials (1.101) as  S¯n

[n/2]−1 

  2i + (−1)n+1 + 2 q + s r s r s S   x = x . n pq pq 2i + (−1)n+1 + 2 [n/2] p + r i=0

(1.102)

A straightforward result from (1.101) and (1.102) is that  S¯2n+1



 r s r + 2p, s + 2q x = x S¯2n pq p, q



x ,

(1.103)

1.7 A Basic Class of Symmetric Orthogonal Polynomials

51

and accordingly  Φ2n (x) = S¯2n



 r s r + 2p, s + 2q and Φ2n+1 (x) = x S¯2n x pq p, q



x . (1.104)

For instance, we have r s S¯0 pq  r s S¯1 pq  r s S¯2 pq  r s S¯3 pq  r s S¯4 pq  r s ¯ S5 pq 

= 1,

x

=x,

x

x

= x2 +

q+s , p+r

= x3 +

3q + s x, 3p + r

x

x

= x4 + 2

3q + s 2 (3q + s)(q + s) x + , 5p + r (5p + r)(3p + r)

= x5 + 2

5q + s 3 (5q + s)(3q + s) x + x. 7p + r (7p + r)(5p + r)

x

(1.105)

The hypergeometric representation of monic polynomials (1.102) is  S¯n

⎛

−[n/2] , (q − s)/2q − [(n + 1)/2] r s x = x n 2 F1 ⎝ pq −(r + (2n − 3)p)/2p

⎞ − q ⎠. px 2

Therefore, on referring to the relation (1.10), we obtain  S¯n

$

q r s = 3 − p pq (2 −

n

(− pq ) 2 Γ ( 32 − r 2p

− n) Γ ( 32 − n

r s r 2p )Γ (1 + 2q − 2p ) r n s r 2p + [ 2 ] − n) Γ (1 + 2q − 2p

− [ n2 ])

.

Also, a three-term recurrence relation can be derived for them as  S¯n+1 (x) = x S¯n (x) + Cn

r s pq

S¯n−1 (x)

(n ∈ N),

(1.106)

52

1 Generalized Sturm–Liouville Problems in Continuous Spaces

in which 

r s pq

Cn

  pq n2 + (r − 2p)q − (−1)n ps n + (r − 2p) s(1 − (−1)n )/2 . = (2pn + r − p)(2pn + r − 3p) (1.107)

Since the recurrence relation (1.106) is explicitly known, to determine the norm square value of the polynomials, we can now use Favard’s theorem by noting that there is orthogonality with respect to a weight function. According to this theorem, if {Pn (x)}∞ n=0 is defined by x Pn (x) = An Pn+1 (x) + Bn Pn (x) + Cn Pn−1 (x)

(n = 0, 1, 2, . . .),

where P−1 (x) = 0 , P0 (x) = 1 , An , Bn , Cn real, and An Cn+1 > 0 for n = 0, 1, . . ., then there exists a weight function W (x) such that 

∞ −∞

W (x)Pn (x) Pm (x) dx =

n−1 Ci+1  i=0

Ai



∞ −∞

W (x) dx

δn,m .

(1.108)

Moreover, if the positive condition An Cn+1 > 0 holds only for n = 0, 1, . . . , N then the orthogonality relation (1.108) holds for only a finite number of m, n. The latter note will help us in the next sections obtain two new subclasses of S¯n (x ; p, q, r, s) that are finitely orthogonal on (−∞, ∞). It is clear that Favard’s theorem is also valid for (1.106), in which An = 1 , Bn = 0, and Cn = − Cn (p, q, r, s). Moreover, the condition − Cn (p, q, r, s) > 0 must always be satisfied if we want to apply Favard’s theorem to (1.106). By noting (1.98), the weight function corresponding to symmetric polynomials (1.101) is derived as

   (r − 4p)x 2 + (s − 2q)    (r − 2p)x 2 + s  r s W dx = exp dx . x = x 2 exp x(px 2 + q) x(px 2 + q) pq (1.109) Hence, a generic orthogonality relation can be designed for (1.101) as 



θ −θ

W





 r s r s r s x S¯n x S¯m x dx pq pq pq



   n θ r s r s n Ci W = (−1) x dx δn,m , pq pq −θ i=1

(1.110)

where θ might be the standard values 1, ∞ and the function (px 2 + q) W (x; p, q, r, s) must vanish at x = θ in order that the main orthogonality relation (1.110) be valid.

1.7 A Basic Class of Symmetric Orthogonal Polynomials

53

The positive function (1.109) can be investigated from a statistical point of view, too. In fact, this function is an analogue of the Pearson family of distributions  ρ



  (d − 2a)x + e − b  de dx , x = exp ax 2 + bx + c a b c

(1.111)

satisfying the first-order differential equation  d  (ax 2 + bx + c) ρ(x) = (dx + e) ρ(x). dx

(1.112)

Therefore, similar to Eq. (1.112), the weight function (1.109) satisfies a first-order differential equation of the form x

d ((px 2 + q) W (x)) = (rx 2 + s) W (x), dx

(1.113)

which is equivalent to d 2 (x (px 2 + q) W (x)) = x (r ∗ x 2 + s ∗ ) W (x), dx

(1.114)

for r ∗ = r + 2p and s ∗ = s + 2q. It is deduced from (1.113) or (1.114) that W (x; p, q, r, s) is an analytic integrable function, and since it is positive, its probability density function is available. In general, there are four main subclasses of the family of distributions (1.109) (and consequently subsolutions of Eq. (1.113)) whose explicit probability density functions are as follows:  K1 W



Γ (a + b + 3/2) −2a − 2b − 2, 2a x 2a (1 − x 2 )b , x = Γ (a + 1/2)Γ (b + 1) −1, 1 for x ∈ [−1, 1] and a + 1/2 > 0, b + 1 > 0,  K2 W



1 2 −2, 2a x 2a e−x , x = Γ (a + 1/2) 0, 1 for x ∈ (−∞, ∞) and a + 1/2 > 0,

 K3 W



x −2a Γ (b) −2a − 2b + 2, −2a , x = Γ (b + a − 1/2)Γ (−a + 1/2) (1 + x 2 )b 1, 1 for x ∈ (−∞, ∞) and b + a > 1/2, a < 1/2, b > 0,

54

1 Generalized Sturm–Liouville Problems in Continuous Spaces

 K4 W



1 −2a + 2, 2 − 1 x −2a e x 2 , x = Γ (a − 1/2) 1, 0 for x ∈ (−∞, ∞) and a > 1/2.

The values {Ki }4i=1 play a normalizing constant role in the above distributions. Moreover, it is observed that the value of the distribution vanishes at x = 0 in each of the four cases, i.e., W (0; p, q, r, s) = 0 for s = 0. Note that if s = 0 in (1.113), the aforementioned equation will be reduced to a special case of the Pearson differential equation (1.112). Hence, we hereinafter suppose that s = 0. Since the explicit forms of Sn (x ; p, q, r, s) in (1.101), Cn (p, q, r, s) in (1.107), and W (x; p, q, r, s) in (1.109) are all known, another probability density function can be defined by referring to the orthogonality relation (1.110) as follows:  Dm



r s x = m pq  i=1

 Ci

r s pq

(−1)m θ

−θ

 

W



W r s x dx pq



r s x pq 

 ×

1.7.1

S¯m



2 r s . x pq

A Direct Relationship Between P¯n (x; a, b, c, d, e) and S¯ n (x; p, q, r, s)

There is a direct relation between the first kind of hypergeometric orthogonal polynomials and Sn (x ; p, q, r, s). To find it, we recall the differential equation (ax 2 + bx + c)yn (x) + (dx + e)yn (x) − n((n − 1)a + d)yn (x) = 0,

(1.115)

together with the Rodrigues formula  P¯n



1 de

 x = n a b c  de ( d + (n + k − 2)a) ρ x a b c k=1 dn  × n (ax 2 + bx + c)n ρ dx

where ρ(x; a, b, c, d, e) is defined in (1.111).





 de x , a b c

1.7 A Basic Class of Symmetric Orthogonal Polynomials

55

Since S¯2n (x ; p, q, r, s) is an even function, taking x = wt 2 + v in (1.115) gives t 2 (aw2 t 4 + w(2av + b)t 2 + av 2 + bv + c)yn (t) + t ((2d − a)w2 t 4 + (2wv(d − a) + w(2e − b))t 2 − (av 2 + bv + c))yn (t) − 4w2 n(d + (n − 1)a)t 4 yn (t) = 0 .

(1.116)

If (1.116) is equated with (1.100), we should have av + bv + c = 0 2

or

v=

−b ±

√ b2 − 4ac . 2a

(1.117)

The condition (1.117) simplifies Eq. (1.116) as t (awt 2 ±

% b2 − 4ac)yn (t) + ((2d − a)wt 2 + (

% d − 1)(−b ± b2 − 4ac) + 2e − b)yn (t) a

− 4wn(d + (n − 1)a)t yn (t) = 0 

√ 2 − 4ac b −b ± d e 2 ⇔ yn (t) = P¯n . wt + 2a a b c

(1.118)

Equation (1.118) is clearly a special case of (1.100), so that we have  Pn



√ d e 2 −b ± b2 − 4ac wt + 2a a b c ⎛ √ (2d − a)w, ( da − 1)(−b ± b2 − 4ac) + 2e − b = K S2n ⎝ √ aw, ± b2 − 4ac

⎞ t⎠ ,

(1.119)

where K is the leading coefficient of the left-hand-side polynomial of relation (1.119) divided by the leading coefficient of the right-hand-side polynomial. As we observe, there exist five free parameters a, b, c, d, e on the left-hand side of (1.119). So one of them must be preassigned in order to obtain the explicit form of S¯2n (x ; p, q, r, s) in terms of P¯n (wt 2 + v ; a, b, c, d, e). For this purpose, if, for instance, c = 0 in (1.119), the following two cases appear:  S¯2n

⎛

r+p s+q 2w , 2 r s −n ¯ ⎝ t = w Pn pq p/w, q, 0

⎞ 2 wt ⎠

(w = 0),

56

1 Generalized Sturm–Liouville Problems in Continuous Spaces

and ⎛ r+p sp−rq

2w , 2p r s t = w−n P¯n ⎝ pq p/w, −q, 0

 S¯2n

⎞ 2 q wt + w ⎠ p

(p, w = 0).

Moreover, if (1.103) is applied to the latter two relations, we respectively obtain  S¯2n+1

⎛

r+3p s+3q 2w , 2 r s −n ¯ ⎝ t = w t Pn pq p/w, q, 0

⎞ 2 wt ⎠

(w = 0)

and  S¯2n+1

1.7.2

⎛ r+3p sp−rq

2w , 2p r s t = w−n t P¯n ⎝ pq p/w, −q, 0

⎞ 2 q wt + w ⎠ p

(p, w = 0).

Four Special Cases of the Symmetric Polynomials Sn (x ; p, q, r, s)

In this section, we study four special cases of Sn (x ; p, q, r, s) that are known as the generalized ultraspherical polynomials, generalized Hermite polynomials, and two other sequences of finite symmetric polynomials orthogonal on the real line. 1. For the generalized ultraspherical polynomials (GUP) we have • Initial vector: (p, q, r, s) = (−1 , 1 , −2a − 2b − 2 , 2a). • Differential equation:   x 2 (−x 2 + 1) Φn (x) − 2x (a + b + 1)x 2 − a Φn (x)   + n(2a + 2b + n + 1)x 2 + ((−1)n − 1)a Φn (x) = 0. • Explicit form of monic GUP:  S¯n

[n/2]−1 2i + 2a + 2 − (−1)n −2a − 2b − 2, 2a x = −2i − (2b + 2a + 2 − (−1)n + 2 [n/2]) −1, 1 i=0

1.7 A Basic Class of Symmetric Orthogonal Polynomials

×

[n/2] 



k=0

57

⎞

⎛[ n ]−(k+1) 2 n −2i − (2b + 2a + 2 − (−1) + 2 [n/2]) ⎠ n−2k [n/2] ⎝ . x 2i + 2a + 2 − (−1)n k i=0

• Recurrence relation of monic polynomials:  S¯n+1 (x) = x S¯n (x) + Cn

−2a − 2b − 2, 2a −1, 1

S¯n−1 (x),

where  Cn

−2a − 2b − 2, 2a −1, 1

=

−n2 − (2b + 2(1 − (−1)n )a)n−2a(a+b)(1 − (−1)n ) (2n + 2a + 2b − 1)(2n+2a+2b + 1) =

−(n + (1 − (−1)n )a)(n + (1 − (−1)n )a + 2b) . (2n + 2a + 2b − 1)(2n + 2a + 2b + 1)

• Weight function:  W



−2a − 2b − 2, 2a b x = x 2a (1 − x 2 ) . −1, 1

• Orthogonality relation: 

1 −1

 x (1 − x ) S¯n 2 b

2a

 = (−1)

n

n i=1

 Ci





 −2a − 2b − 2, 2a −2a − 2b − 2, 2a x S¯m x dx −1, 1 −1, 1

−2a − 2b − 2, 2a −1, 1



1

−1

2 b

x (1 − x ) dx 2a

δn,m ,

(1.120)

where 

Γ (a + 1/2)Γ (b + 1) 1 b . x 2a (1 − x 2 ) dx = B(a + , b + 1) = 2 Γ (a + b + 3/2) −1 1

(1.121)

Relation (1.121) shows that the constraints on the parameters a and b are respectively a + 1/2 > 0, (−1)2a = 1, and b + 1 > 0. Since the ultraspherical, Legendre, and Chebyshev polynomials of the first and second kinds are all special cases of GUP, they can be expressed in terms of

58

1 Generalized Sturm–Liouville Problems in Continuous Spaces

S¯n (x ; p, q, r, s) directly, so that we have • Ultraspherical polynomials: Cna (x)

2n (a)n ¯ = Sn n!





−2a − 1, 0 x . −1, 1

• Legendre polynomials: Pn (x) =

(2n)! (n!)2 2n

 S¯n



−2 0 x . −1 1

• Chebyshev polynomials of the first kind:  Tn (x) = 2n−1 S¯n



−1 0 x . −1 1

(1.122)

• Chebyshev polynomials of the second kind:  Un (x) = 2 S¯n n



−3 0 x . −1 1

Remark 1.6 Up to now, four kinds of trigonometric orthogonal polynomials, i.e., first, second, third, and fourth kinds of Chebyshev polynomials, have been well known in the literature. For x = cos θ , the first and second kinds are defined as Tn (x) = 2n−1

 n  (2k − 1)π = cos(nθ ) x − cos 2n

k=1

and  n  kπ sin((n + 1)θ ) , Un (x) = 2 x − cos = n+1 sin θ n

k=1

while for x = cos 2θ , the third and fourth kinds are defined as Vn (x) = 2n

 n  (2k − 1)π cos((2n + 1)θ ) x − cos = 2n + 1 cos(θ )

k=1

1.7 A Basic Class of Symmetric Orthogonal Polynomials

59

and Wn (x) = 2n

 n  2kπ sin((2n + 1)θ ) x − cos = . 2n + 1 sin(θ )

k=1

Here we would like to add that there exist two further kinds of half-trigonometric orthogonal polynomials, which are particular cases of S¯n (x ; p, q, r, s) and are generated by the first and second kinds of Chebyshev polynomials. To generate them, we first refer to relation (1.103). According to (1.122), the initial vector of the first kind of Chebyshev polynomials is (p, q, r, s) = (−1, 1, −1, 0). If this vector is replaced in (1.103), then we obtain



  −1 0 −3 2 S¯2n+1 x = T¯2n+1 (x) = x S¯2n x . −1 1 −1 1 Hence, the secondary vector (p, q, r, s) = (−1, 1, −3, 2) appears, and we can define the half-trigonometric polynomials ⎧ (−1)n/2 cos((n + 1)θ ) ⎪  ⎪ ⎨ (n = 2m), −3 2 n+1 cos θ (1.123) Xn (x) = Sn x = ⎪ −1 1 ⎪ ⎩ Sn (x ; −1, 1, −3, 2) (n = 2m + 1), in which x = cos θ . According to (1.123) and (1.106), X¯ n (x) satisfies the recurrence relation 

−3 2 X¯ n−1 (x), X¯ n+1 (x) = x X¯ n (x) + Cn −1 1 where  Cn

−3 2 −1 1

=

−(n − (−1)n )(n + 1 − (−1)n ) . 4n (n + 1)

By substituting the initial vector (−1, 1, −3, 2) into the generic relation (1.110), we obtain 



1

−1

W



−3 2 x X¯ n (x)X¯ m (x)dx −1 1 

   n 1 −3 2 −3 2 n = (−1) Ci W −1 1 −1 1 −1 i=1



x dx δn,m .

(1.124)

60

1 Generalized Sturm–Liouville Problems in Continuous Spaces

On the other hand, since 



1 −1

W



   1 π x2 3 1 −3 2 , = , dx = B √ x dx = 2 2 2 −1 1 1 − x2 −1

(1.124) is simplified as 

1

−1

 

n π −3 2 n ¯ ¯ δn,m . Xn (x)Xm (x)dx = (−1) √ Ci 2 2 −1 1 1−x i=1 x2

Clearly X¯ 2n (x) is decomposable as X¯ 2n (x) =

2n  k=1

 (2k − 1)π x − cos . 2(2n + 1)

We can similarly follow the above-mentioned approach for the second kind Chebyshev polynomials with (p, q, r, s) = (−1, 1, −3, 0). If this vector is substituted into (1.103), then  S¯2n+1



 −3 0 −5 2 ¯ ¯ x = U2n+1 (x) = x S2n −1 1 −1 1



x ,

which gives the secondary vector as (p, q, r, s) = (−1, 1, −5, 2). By assuming x = cos θ , we can now define the polynomials  Yn (x) = Sn

⎧ (−1)n/2 sin((n + 2)θ ) ⎪ ⎪ ⎨ (n = 2m), −5 2 n + 2 cos θ sin θ x = ⎪ −1 1 ⎪ ⎩ Sn (x ; −1, 1, −5, 2) (n = 2m + 1),

satisfying the recurrence relation  Y¯n+1 (x) = x Y¯n (x) + Cn

−5 2 −1 1

Y¯n−1 (x),

where  Cn

−5 2 −1 1

=

−(n + 1 − (−1)n )(n + 2 − (−1)n ) , 4(n + 1) (n + 2)

1.7 A Basic Class of Symmetric Orthogonal Polynomials

61

and having the orthogonality relation 



1

−1

W



−5 2 x Y¯n (x)Y¯m (x) dx −1 1   

 n 1 −5 2 −5 2 n = (−1) Ci W −1 1 −1 1 −1 i=1



x dx δn,m ,

(1.125)

where 



1 −1

W

−5 2 −1 1



   1 % π 3 3 2 2 x dx = = . , x 1 − x dx = B 2 2 8 −1

(1.126)

The relation (1.126) simplifies (1.125) as 

1 −1

x

2

%

 1 − x 2 Y¯n (x)Y¯m (x)dx

= (−1)

n

n

 Ci

i=1

−5 2 −1 1



π δn,m . 8

Similar to the previous case, Y¯2n (x) is decomposable as Y¯2n (x) =

2n  x − cos k=1

 kπ . 2n + 2

Let us add that there are still two other sequences of half-trigonometric polynomials that are not orthogonal but can be expressed in terms of S¯n (x; p, q, r, s). These sequences are defined as   S¯n (x; −1, 1, 1, −2) (n = 2m), 1 −2 S¯n x = x T¯n−1 (x) (n = 2m + 1), −1 1 and  S¯n

 (n = 2m), −1 −2 S¯n (x; −1, 1, −1, −2) x = ¯ x Un−1 (x) (n = 2m + 1). −1 1

However, since 



1 −1

W

1 −2 −1 1



 1 dx x dx = √ = +∞ 2 1 − x2 −1 x

62

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Table 1.4 Chebyshev polynomials in terms of S¯n (x ; p, q, r, s) Type

Notation

First kind

T¯n (x)

Second kind

U¯ n (x)

Third kind

V¯n (x)

Fourth kind

W¯ n (x)

Fifth kind

X¯ n (x)

Sixth kind

Y¯n (x)

Definition

 −1 0 ¯ Sn x −1 1

 −3 0 S¯n x −1 1 $

 1+x −3 2 2n S¯2n 2 −1 1 $

 1−x −3 2 n 2 S¯2n 2 −1 1

 −3 2 S¯n x −1 1

 −5 2 ¯ Sn x −1 1

and 



1 −1

W

Weight % %

1 1 − x2

1 − x2

$ $

%

1+x 1−x 1−x 1+x x2 1 − x2

% x2 1 − x2



 1 √ 1 − x2 −1 −2 dx = +∞, x dx = x2 −1 1 −1

they cannot fall into the half-trigonometric orthogonal polynomials category. In this sense, Table 1.4 shows some properties of the first kind to the sixth kind of monic Chebyshev polynomials orthogonal on [−1, 1]. 2. For generalized Hermite polynomials (GHP) we have • Initial vector: (p, q, r, s) = (0 , 1 , −2 , 2a). • Differential equation:   x 2 Φn (x) − 2x(x 2 − a) Φn (x) + 2n x 2 + ((−1)n − 1)a Φn (x) = 0. • Explicit form of monic GHP:  S¯n



n (−1)n −2 2a )n x = (−1)[ 2 ] (a + 1 − 2 [2] 0 1 ⎞ 

⎛[n/2]−(k+1) [n/2]  [n/2] −2 ⎠ x n−2k . ⎝ × n+1 k 2i + (−1) + 2 + 2a k=0 i=0

1.7 A Basic Class of Symmetric Orthogonal Polynomials

63

• Recurrence relation of monic polynomials:  S¯n+1 (x) = x S¯n (x) + Cn

−2 2a 0 1

S¯n−1 (x),

where  Cn

−2 2a 0 1



1 − (−1)n 1 a. =− n− 2 2

• Weight function:  W



−2, 2a x = x 2a exp(−x 2 ). 0, 1

• Orthogonality relation: 

∞ −∞

 x 2a e

−x 2

S¯n





 −2 2a −2 2a x S¯m x dx 0 1 0 1 

n 1 1 i = (1 − (−1) )a + i Γ (a + ) δn,m , n 2 2 i=1

which is valid for a + 1/2 > 0 and (−1)2a = 1. 3. For the first finite sequence of symmetric orthogonal polynomials we have • Initial vector: (p, q, r, s) = (1 , 1 , −2a − 2b + 2 , −2a). • Differential equation:   x 2 (x 2 + 1) Φn (x) − 2x (a + b − 1)x 2 + a Φn (x)   + n(2a + 2b − (n + 1)) x 2 + (1 − (−1)n )a Φn (x) = 0. • Explicit form of polynomials:  Sn



−2a − 2b + 2, −2a x 1, 1

(1.127)

64

1 Generalized Sturm–Liouville Problems in Continuous Spaces

=

[n/2] 



⎞

⎛[ n ]−(k+1) 2 n+1 2i + 2[n/2] + (−1) + 2 − 2a − 2b ⎠ n−2k ⎝ . x n+1 k 2i + (−1) + 2 − 2a i=0

[ n2 ]

k=0

• Recurrence relation of monic polynomials:  S¯n+1 (x) = x S¯n (x) + Cn where  Cn

−2a − 2b + 2, −2a 1, 1

=

−2a − 2b + 2, −2a 1, 1

S¯n−1 (x),

(n − (1 − (−1)n ) a) (n − (1 − (−1)n ) a − 2b) . (2n − 2a − 2b + 1)(2n − 2a − 2b − 1)

• Weight function:  W



x −2a −2a − 2b + 2, −2a . x = b 1, 1 (1 + x 2 )

• Orthogonality relation:



  −2a − 2b + 2, −2a −2a−2b + 2, −2a ¯ ¯ S x Sm x dx b n 1, 1 1, 1 −∞ (1 + x 2 ) 

 n Γ (b + a − 1/2)Γ (−a + 1/2) −2a − 2b + 2, −2a Ci δn,m , = (−1)n Γ (b + a) 1, 1 i=1



∞

x −2a

(1.128) which is valid if  − Cn+1

−2a − 2b + 2, −2a 1,

1

> 0 for all n ∈ Z+ , b + a > 1/2, a < 1/2, and b > 0.

Here we explain how to determine these conditions in order to establish the orthogonality property (1.128). For this purpose, we reconsider Eq. (1.127) and write it in a self-adjoint form to get 

 x −2a (1 + x 2 )−b+1 (Φn (x)Φm (x) − Φm (x)Φn (x))

∞ −∞

= 0.

(1.129)

Since  max deg{Φn (x)Φm (x) − Φm (x)Φn (x)} = n + m − 1,

(1.130)

1.7 A Basic Class of Symmetric Orthogonal Polynomials

65

relations (1.129) and (1.130) yield −2a + 2(−b + 1) + n + m − 1 ≤ 0, which gives the final result as −2a − 2b + n + m + 1 ≤ 0

for a < 1/2 and b > 0 .

In other words, (1.128) holds if and only if m, n = 0, 1, . . . , N ≤ a + b − 1/2, a < 1/2,

and b > 0.

Corollary 1.1 The symmetric polynomial set {Sn (x ; 1, 1, −2a − 2b + 2, −2a)}N n=0 is finitely orthogonal with respect to the weight function x −2a (1 + x 2 )−b on (−∞, ∞) if and only if N ≤ a + b − 1/2, a < 1/2, b > 0, and (−1)2a = 1. 4. For the second finite sequence of symmetric orthogonal polynomials we have • Initial vector: (p, q, r, s) = (1 , 0 , −2a + 2 , 2). • Differential equation:     x 4 Φn (x)+2x (1−a)x 2 +1 Φn (x)− n(n + 1 − 2a) x 2 + 1 − (−1)n Φn (x) = 0. • Explicit form of polynomials:  Sn



−2a + 2, 2 x 1, 0 =

[n/2]  k=0



[n/2] k

⎞

⎛[ n ]−(k+1) 2 n+1 2i + 2[n/2] + (−1) + 2 − 2a ⎠ x n−2k . ⎝ 2 i=0

• Recurrence relation of monic polynomials:  S¯n+1 (x) = x S¯n (x) + Cn

−2a + 2, 2 1, 0

S¯n−1 (x),

66

1 Generalized Sturm–Liouville Problems in Continuous Spaces

where  Cn

−2a + 2, 2 1, 0

=

−2 (−1)n (n − a) − 2a . (2n − 2a + 1)(2n − 2a − 1)

• Weight function:  W

−2a + 2, 2 1, 0



  1 −2a x = x exp − . x2

• Orthogonality relation: 





 −2a + 2, 2 −2a + 2, 2 x e S¯n x S¯m x dx 1, 0 1, 0 −∞  

  n 1 −2a + 2, 2 n = (−1) δn,m . Ci Γ a− 2 1, 0 i=1 

∞

1 −2a − x 2

(1.131)

Relation (1.131) is valid only if 

x 2−2a exp(−

∞ 1    ) Φn (x)Φm (x) − Φm (x)Φn (x) = 0, 2 −∞ x

which is equivalent to 2 − 2a + n + m − 1 ≤ 0

⇔

N ≤a−

1 for N = max{m, n}. 2

Corollary 1.2 The symmetric polynomial set {Sn (x ; 1, 0, −2a + 2, 2)}N n=0 is finitely 2 −2a −1/x on (−∞, ∞) if and only if orthogonal with respect to the weight function x e N ≤ a − 1/2 and (−1)2a = 1.

1.7.3

A Unified Approach for the Classification of S¯ n (x; p, q, r, s)

We observed that each of the four introduced classes of symmetric orthogonal polynomials was directly determined by S¯n (x ; p, q, r, s), and it was sufficient only to obtain the initial vector corresponding to them. On the other hand, since the orthogonality interval of all classes, except the first one (GUP), is (−∞, ∞), applying a linear transformation preserves the orthogonality interval. This means that by having the initial vector, we can have access to all other standard properties and design a unified approach for some cases that may occur. Here we consider two special cases of this approach.

1.7 A Basic Class of Symmetric Orthogonal Polynomials

67

First Case: How to Find Initial Parameters If a Special Case of the Weight Function is Given An easy way to derive p, q, r, s is to first compute the logarithmic derivative of the weight function as before and then equate the pattern with (1.109), as the following general example shows. Example 1.5 The weight functions (i) (ii) (iii)

W1 (x) = −x 6 + 4x 4 W2 (x) = (16x 2 − 8x + 1) exp(2x(1 − 2x)) W3 (x) = (2x + 1)−2 (2x 2 + 2x + 1)−5

x ∈ [−2, 2], x ∈ (−∞, ∞), x ∈ (−∞, ∞),

with their orthogonality intervals are given. To find initial vectors corresponding to each given weight function, we first compute the logarithmic derivative of the first weight function as 6x 2 − 16 (r − 2p)x 2 + s W  1 (x) = = ⇒ (p, q, r, s) = (1, −4, 8, −16). W1 (x) x(x 2 − 4) x(px 2 + q) & '∞ Hence the related monic polynomials are S¯n (x; 1, −4, 8, −16) n=0 , which are orthogonal with respect to x 4 (4 − x 2 ) on [−2, 2] for every n. Note that it is not necessary to know that they are the same as shifted GUP on [−2, 2], because they can be explicitly and independently expressed by S¯n (x ; p, q, r, s). For the weight function W2 (x), it differs somewhat, so that we have W2 (x) = 4e 4 (2x − 12 )2 exp(−(2x − 12 )2 ) ⇒ 14 e− 4 W2 ( 2t4+1 ) = t 2 e−t = W2∗ (t) 1

→

1

2

W ∗2  (t) −2t 2 + 2 = ⇒ (p, q, r, s) = (0, 1, −2, 2). ∗ W2 (t) t

 Hence the related orthogonal polynomials are Sn (2x − Finally, for the third one we have W3 (x) = 25 →

(2x+1)−2 (1+(2x+1)2 )

5

1 2

∞ ; 0, 1, −2, 2) .

⇒ 2−5 W3 ( t −1 2 )=

n=0

t −2 5 (1+t 2 )

= W3∗ (t)

W ∗3  (t) −12t 2 − 2 = ⇒ (p, q, r, s) = (1, 1, −10, −2). W3∗ (t) t (t 2 + 1)

Consequently, the finite set {Sn (2x + 1 ; 1, 1, −10, −2)}5n=0 is orthogonal with respect to W3 (x) on (−∞, ∞), and the upper bound of this set is determined based on the condition N ≤ b + 1/2 for b = 5.

68

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Second Case: How to Find Initial Parameters If a Special Case of the Main ThreeTerm Recurrence Equation Is Given? There are two ways to determine a special case of S¯n (x ; p, q, r, s) when a three-term recurrence equation is given. The first way is to directly compare the given recurrence equation with (1.106). This leads to a system of polynomial equations in terms of the parameters p, q, r, s. The second way is to equate the first four terms of the two recurrence equations, which leads to a polynomial system with four equations the four unknowns p, q, r, and s. The following example illustrates these methods. Example 1.6 The recurrence equation 6 + (−1)n (n − 6) ¯ Sn−1 (x) S¯n+1 (x) = x S¯n (x) − 2 (2n − 11)(2n − 13) with S¯0 (x) = 1 and S¯1 (x) = x is given. Find its explicit polynomial solution, differential equation of polynomials, the related weight function, and orthogonality relation of polynomials. First Method If the above recurrence equation is directly compared with the main Eq. (1.106), we can directly obtain the values (p, q, r, s) = (1, 0, −10, 2). Hence, the explicit solution of the above recurrence equation is S¯n (x ; 1, 0, −10, 2), satisfying the differential equation   x 4 Φn (x) + x (−10 x 2 + 2) Φn (x) − n (n − 11) x 2 + 1 − (−1)n Φn (x) = 0. Moreover, by substituting the initial vector into the main weight function (1.109) as  W

−10, 2 1, 0



 1  −12x 2 + 2  −12 − x 2 dx = x e , x = exp x3

we find out that the related polynomials are a particular case of the fourth introduced class, so that 



∞ −∞

x

1 −12 − x 2

e

S¯n





 −10, 2 −10, 2 x S¯m x dx 1, 0 1, 0  

n 11 −10, 2 = (−1)n Ci Γ ( ) δ n,m ⇔ m, n ≤ 5. 2 1, 0 i=1

1.8 Fourier Transforms of Symmetric Orthogonal Polynomials

69

Table 1.5 Four special cases of Sn (x; p, q, r, s) Definition

 −2a − 2b − 2, 2a Sn x −1, 1

 −2, 2a Sn x 0, 1

 −2a − 2b + 2, −2a Sn x 1, 1

 −2a + 2, 2 Sn x 1, 0

Weight function

Interval and kind

x 2a (1 − x 2 )b

[−1, 1], Infinite

x 2a exp(−x 2 )

(−∞, ∞), Infinite a > − 12

x −2a (1 + x 2 )

b

(−∞, ∞), Finite

x −2a exp(−1/x 2 ) (−∞, ∞), Finite

Parameter constraint a > −1/2 b > −1

N ≤ a + b − 1/2 a < 1/2, b > 0 N ≤ a − 12

Second Method If the given recurrence relation is expanded only for n = 2, 3, 4, 5 and then equated with (1.105), the following system will be derived: q +s 2 2 q+s ⇒ =− , S¯2 (x) = x 2 − = x 2 + 9 p+r p+r 9 4 3q + s x, S¯3 (x) = x S¯2 (x) − S¯1 (x) = x 3 + 63 3p + r 18 3q + s 2 (3q + s)(q + s) S¯4 (x) = x S¯3 (x) − S¯2 (x) = x 4 + 2 x + , 35 5p + r (5p + r)(3p + r) 2

4 5q + s =− . 7p + r 3

Solving this system again results in (p, q, r, s) = (1, 0, −10, 2). Finally, Table 1.5 shows the four polynomials in terms of Sn (x ; p, q, r, s) as well as their weight functions, kind of polynomials (finite or infinite), orthogonality interval, and constraint on the parameters. Note that since all weights are even functions in this table, the condition (−1)2a = 1 must be always satisfied. Therefore they can be considered in the forms |x|2a (1 − x 2 )b , 2 2 |x|2a e−x , |x|−2a (1 + x 2 )−b and |x|−2a e−1/x respectively.

1.8

Fourier Transforms of Symmetric Orthogonal Polynomials

Similar to Sect. 1.5, in this section we introduce four classes of orthogonal functions as the Fourier transforms of the introduced cases of Sn (x; p, q, r, s) and then obtain their orthogonality relations using Parseval’s identity.

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1 Generalized Sturm–Liouville Problems in Continuous Spaces

1.8.1

Fourier Transform of Generalized Ultraspherical Polynomials

For the generalized ultraspherical polynomials  Un(a,b) (x)

= x 2 F1 n



−[n/2] , −a + 1/2 − [(n + 1)/2] 1 2 , x −a − b − n + 1/2

let us define the specific functions g(x) = (tanh x)2α (1 − tanh2 x)β Un(c,d) (tanh x) with (−1)2α = 1 and

(1.132) h(x) = (tanh x)2l (1 − tanh2 x)u Um(v,w) (tanh x)

with (−1)2l = 1.

The Fourier transform of the function, e.g., g(x) is computed as   F g(x) =  =

1

−1



∞ −∞

(1 + t)− 2 (1 − t) 2 t 2α (1 − t 2 ) is



1

= 22β−1 0

is

β−1

Un(c,d) (t) dt

(1 − z)− 2 z 2 (1 − 2z)2α zβ−1 (1 − z)β−1 Un(c,d) (1 − 2z) dz is

is

⎛

 =2

β

e −isx (tanh x)2α (1 − tanh2 x) Un(c,d) (tanh x) dx

1

2β−1

β−1− is2 β−1+ is2

(1−z)

z

[n/2]  2α+n ⎝

(1−2z)

0

= 22β−1

k=0

[n/2] 

(−[ n2 ] )k ( 12 − c − [ n+1 2 ] )k

k=0

(−c − d − n + 1/2)k k!



1

⎞

(−[ n2 ] )k ( 12 − c − [ n+1 2 ] )k

1

(−c−d−n+1/2)k k!

(1−2z)2k

⎠ dz

 is is (1 − z)β−1− 2 zβ−1+ 2 (1 − 2z)2α+n−2k dz .

0

(1.133) According to the integral representation of Gauss’s hypergeometric function (1.9), for Re β > 0 and Re (2α + n − 2k) > −1, relation (1.133) is simplified as     is is F g(x) = 22β−1 B β + , β − 2 2

1.8 Fourier Transforms of Symmetric Orthogonal Polynomials

×

[n/2] 

(−[ n2 ] )k ( 12 − c − [ n+1 2 ] )k

k=0

(−c − d − n + 1/2)k k!

 2 F1

71

2k − n − 2α β + 2β

is 2



2 .

(1.134)

If for simplicity we define Kn (x; p1 , p2 , p3 , p4 ) =

[n/2] 

(−[ n2 ] )k ( 12 −p3 −[ n+1 2 ] )k

k=0

(−p3 − p4 − n + 1/2)k k!

 2 F1



2k−n−2p1 p2 + x2 2 , 2p2 (1.135)

then (1.134) can be written as       22β−1 is is F g(x) = Γ β+ Γ β− Kn (is; α, β, c, d). Γ (2β) 2 2

(1.136)

Now, by substituting (1.136) in Parseval’s identity (1.83) and noting (1.132), we get  2π

∞

β+u

(tanh x)2(α+l) (1 − tanh2 x)

−∞

 = 2π

 ×

∞

−∞

Γ (β+

1

−1

Un(c,d) (tanh x) Um(v,w) (tanh x) dx

β+u−1

t 2(α+l) (1 − t 2 )

Un(c,d) (t) Um(v,w) (t) dt =

22β+2u−2 Γ (2β)Γ (2u)

is is is is )Γ (β− ) Γ (u+ )Γ (u − ) Kn (is; a, b, c, d) Km (is; l, u, v, w) ds . 2 2 2 2 (1.137)

If on the left-hand side of (1.137) we take c =v =α+l

and

d = w = β + u − 1,

then according to the orthogonality relation (1.120), Eq. (1.137) finally reads as 1 2π



∞ −∞

Γ (β +

is is is is )Γ (β − )Γ (u + )Γ (u − ) 2 2 2 2

Kn (is; α, β, α + l, β + u − 1)Km (−is; l, u, α + l, β + u − 1)ds Γ (2β)Γ (2u)Γ (α + l + 1/2)Γ (β + u) 22β+2u+2n−2 Γ (α + β + l + u + 1/2)        n j + 1 − (−1)j (α + l) j + 1 − (−1)j (α + l) + 2β + 2u − 2 =

×

j =1

(j + α + β + l + u − 3/2)(j + α + β + l + u − 1/2)

δn,m , (1.138)

provided that β, u > 0 and α + l > −1/2.

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1 Generalized Sturm–Liouville Problems in Continuous Spaces

Remark 1.7 The weight function corresponding to (1.138) can be simplified using the Cauchy beta integral 1 2π



∞

dt

−∞

(a + it) (b − it) c

d

=

Γ (c + d − 1) (a + b)1−(c+d) Γ (c)Γ (d)

and one of its consequences, i.e., 22−2p π Γ (2p − 1) Γ (p + iq) Γ (p − iq) =  π/2 , 2qx cos2p−2 x dx e −π/2 which results in         is is is 24−2β−2u π 2 Γ (2β−1)Γ (2u−1) is . Γ β− Γ u+ Γ u− =  π/2 Γ β+  sx 2β−2 x dx π/2 e sx cos2u−2 x dx 2 2 2 2 −π/2 e cos −π/2

Therefore, if we define W ∗ (x; λ) =



π/2 −π/2

exθ cos2λ−2 θ dθ,

then the function Kn (x; p1, p2 , p3 , p4 ) defined in (1.135) satisfies the orthogonality relation 

∞ −∞

Kn (ix; α, β, α + l, β + u − 1) Km (−ix; l, u, α + l, β + u − 1) dx W ∗ (x; β) W ∗ (x; u)

(2u − 1)(2β − 1) Γ (α + l + 1/2) Γ (β + u) Γ (α + β + l + u + 1/2) π 22n+1     n j + (1 − (−1)j )(α + l) j + (1 − (−1)j )(α + l) + 2β + 2u − 2 × δn,m , (j + α + β + l + u − 3/2) (j + α + β + l + u − 1/2) =

j =1

where β, u > 1/2, α + l > −1/2.

1.8.2

Fourier Transform of Generalized Hermite Polynomials

The above-mentioned technique can be similarly applied to the generalized Hermite polynomials  Hn(a)(x) = x n 2 F0



1 −[n/2] , −[n/2] − a + (−1)n /2 − 2 . x −

1.8 Fourier Transforms of Symmetric Orthogonal Polynomials

73

First define the specific functions u(x) = x 2a e− 2 x Hn(b)(x) 1 2

with (−1)2a = 1

and

(1.139) v(x) = x 2c e− 2 x Hm(d)(x) 1 2

with (−1)2c = 1.

The Fourier transform of the function, e.g., u(x) is computed as   F u(x) =  =

=

∞ −∞

[n/2]  k=0



∞ −∞

e−isx e

1 2

e−isx x 2a e− 2 x Hn(b) (x) dx ⎛ − 12 x 2

x 2a+n ⎝

[n/2]  k=0

⎞ (−[n/2] )k (−[n/2] − b + (−1)n /2 )k k (−x −2 ) ⎠ dx k!

(−[n/2] )k (−[n/2] − b + (−1)n /2 )k (−1)k k!



∞ −∞

e

−isx − 12 x 2 2a+n−2k

e

x

 dx .

(1.140) So it remains in (1.140) to evaluate the integral  In,k (s; a) =

∞ −∞

1 2

e−isx e− 2 x x 2a+n−2k dx.

There are two ways to compute this integral. In the first method, by noting that (−1)2a = 1, we can directly compute In,k (s; a) for n = 2m as follows:  I2m,k (s; a) =

=

∞ −∞

∞  j =0

=

⎛ ⎞ ∞ j  1 (−isx) ⎝ ⎠ x 2a+2m−2k e− 2 x 2 dx j! j =0

(−1)j i j s j j!



∞ −∞

x

j +2a+2m−2k − 12 x 2

e

 dx

   ∞ ∞  1 2 (−1)r s 2r x 2r+2a+2m−2k e− 2 x dx 2 (2r) ! 0 r=0

=

∞  (−1)r s 2r r=0

1 1 2r+a+m−k+ 2 Γ (r + a + m − k + ) (2r) ! 2

74

1 Generalized Sturm–Liouville Problems in Continuous Spaces ∞  (−1)r s 2r 2r+a+m−k+ 2 Γ (a + m − k + 12 ) (a + m − k + 12 )r 1

=

(1/2)r 22r r !  a + m − k + 1/2 a+m−k+1/2 Γ (a + m − k + 1/2) 1 F1 =2 1/2 r=0



s2 . − 2 (1.141)

This technique can be also applied to I2m+1,k (s; a). We should note that 

∞ −∞

x j +2a+2m+1−2k e−x

2 /2

dx = 0 for all j = 0, 2, 4, . . . .

After some computations we obtain  I2m+1,k (s; a) = (−is) 2a+m−k+3/2Γ (a + m − k + 3/2) 1 F1



a+m−k + 3/2 s 2 . − 2 3/2 (1.142)

By combining relations (1.141) and (1.142) and using the well-known identity [

n+1 n 1 − (−1)n ]−[ ]= , 2 2 2

we finally obtain In,k (s; a) = 2a−k+ 2 +[ 2 ]    1−(−1)n n+1 1 ] (−is) 2 1 F1 ×Γ a−k+ +[ 2 2 1

n+1

1 2



2 s + a − k + [ n+1 ] 2 . − 2 1 − (−1)n /2 (1.143)

The second way to compute In,k (s; a) is that we consider n = 2m and n = 2m + 1 and 2 directly apply the cosine and sine Fourier transforms of the function e−x /2 x 2a+n−2k . So, by noting that (−1)2a = 1, we have  I2m,k (s; a) = =2

∞

cos(sx) x 2a+2m−2k e− 2 x dx − i

−∞  ∞

1 2

cos(sx)x 2a+2m−2k e

− 21 x 2

=2

∞ −∞

sin(sx) x 2a+2m−2k e− 2 x dx 1 2

dx

0 a+m−k+ 21



1 Γ (a + m − k + ) 1 F1 2



a+m−k+ 1/2

1 2



s2 − 2

1.8 Fourier Transforms of Symmetric Orthogonal Polynomials

75

and  I2m+1,k (s; a) =

∞

1 2

−∞

cos(sx) x 2a+2m+1−2k e− 2 x dx−i



∞

= −2i



∞

1 2

−∞

sin(sx) x 2a+2m+1−2k e− 2 x dx

sin(sx)x 2a+2m+1−2k e− 2 x dx 1 2

0 a+m−k+ 23

= (−is)2

3 Γ (a + m − k + ) 1 F1 2



a+m−k+ 3/2

3 2



s2 . − 2

Consequently, the result (1.143) simplifies (1.140) as   1 n+1 n+1 1 ] ) 2a+ 2 +[ 2 ] F u(x) = Γ (a + + [ 2 2 × (−is)

1−(−1)n 2

[n/2] 

(−[ n2 ] )k ( 12 − b − [ n+1 2 ] )k

k=0

k ( 12 − a − [ n+1 2 ] )k 2 k!

 1 F1

1 2



s2 + a − k + [ n+1 2 ] − , 2 1 − (−1)n /2

and according to Kummer’s formula  1 F1



 a c−a x x = e 1 F1 c c



−x ,

it will be transformed to   1 n+1 1 2 1 n+1 F u(x) = Γ (a + + [ ] ) 2a+ 2 +[ 2 ] e− 2 s 2 2 × (−is)

1−(−1)n 2

[n/2] 

(−[ n2 ] )k ( 12 − b − [ n+1 2 ] )k

k=0

k ( 12 − a − [ n+1 2 ] )k 2 k!



−a + k − [ n2 ] 1 2 . s 1 − (−1)n /2 2

 1 F1

If for simplicity we define Jn (x; q1, q2 ) = x

1−(−1)n 2

[n/2] 

(−[ n2 ] )k ( 12 − q2 − [ n+1 2 ] )k

k=0

k ( 12 − q1 − [ n+1 2 ] )k 2 k!

 1 F1



−q1 + k − [ n2 ] 1 2 x , 1 − (−1)n /2 2

then by referring to definitions (1.139) and applying Parseval’s identity, we get  2π = i

∞

−∞

x 2(a+c)e−x Hn(b) (x) Hm(d)(x)dx

(−1)n −(−1)m 2

2

Γ (a+ 12 +[ n+1 2 ] ) Γ (c + 2

1 m+1 2 +[ 2 ]) 1 n+1 1 m+1 −(a+ 2 +[ 2 ]) −(c+ 2 +[ 2 ])

2



∞ −∞

e−s Jn (s; a, b) Jm (s; c, d)ds. 2

76

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Finally, taking b = d = a + c and noting the orthogonality relation, we obtain 

∞ −∞

e−x Jn (x; a, b) Jm (x; b − a, b) dx = 2

π2−b−2 [ Γ (a +

1 2

n+1 ] 2

Γ ([ n2 ] + 1)Γ (b+ 12 +[ n+1 2 ]) δn,m , n+1 + [ 2 ])Γ (b − a + 12 + [ n+1 2 ])

where a, b > −1/2 and (−1)2b = 1.

1.8.3

Fourier Transform of the First Finite Sequence of Symmetric Orthogonal Polynomials

To derive the Fourier transform of the first finite sequence of symmetric polynomials  A(a,b) (x) n

= S¯n



−2a − 2b + 2, −2a x 1, 1  −[n/2] , a + 1/2 − [(n + 1)/2] n = x 2 F1 a + b − n + 1/2



1 − 2 , x

we first define g(x) = x −2α (1 + x 2 )−β A(c,d) (x) n

with (−1)2α = 1

and

(1.144) h(x) = x −2l (1 + x 2 )−u A(v,w) (x) with m

(−1)2l = 1.

Then the Fourier transform of the function, e.g., g(x) is computed as  F(g(x)) =  =

=

∞ −∞

[n/2]  k=0

∞ −∞

−β −2α

e−isx (1 + x 2 )

x

A(c,d) (x) dx n

⎛

[n/2]  2 −β −2α+n ⎝

e−isx (1 + x )

x

k=0

⎞ (−[n/2] )k (c + 1/2 − [(n + 1)/2] )k (−1)k ⎠ dx (c + d − n + 1/2)k k! x 2k

(−[n/2] )k (c+1/2 − [(n+1)/2] )k (−1)k (c + d − n + 1/2)k k!



∞ −∞

−β −2α+n−2k

e−isx (1+x 2 )

x

 dx .

(1.145)

1.8 Fourier Transforms of Symmetric Orthogonal Polynomials

77

Now it remains in (1.145) to evaluate the integral  In,k (s; α, β) =

∞ −∞

e−isx (1 + x 2 )

−β −2α+n−2k

x

(1.146)

dx.

Once again, there are two ways to compute the integral (1.146). In the first method, by noting that (−1)2α = 1, we can directly compute In,k (s; α, β) for n = 2m as follows:  I2m,k (s; α, β) =

=

∞ −∞

⎛ ⎞ ∞ j  (−isx) ⎝ ⎠ x −2α+2m−2k (1 + x 2 )−β dx j! j =0

 ∞  (−1)j i j s j

∞

j!

−∞

j =0

x

j −2α+2m−2k

2 −β

(1 + x )

 dx

   ∞ ∞  (−1)r s 2r 2r−2α+2m−2k 2 −β = x (1 + x ) dx 2 (2r) ! 0 r=0

=

  1 1 B r −α +m−k + ;β −r + α −m+k − , (2r) ! 2 2

∞  (−1)r s 2r r=0

(1.147) where we have used the definition of the beta integral of the third kind. Since the last sum in (1.147) can be represented in terms of a hypergeometric function, we have I2m,k (s; α, β) =

Γ (−α + m − k + 1/2) Γ (β + α − m + k − 1/2) Γ (β)

 s2 −α + m − k + 1/2 . × 1 F2 1/2, −β − α + m − k + 3/2 4

By knowing that 

∞ −∞

x j −2α+2m+1−2k (1 + x 2 )

−β

dx = 0

for all

j = 0, 2, 4, . . . ,

(1.148)

78

1 Generalized Sturm–Liouville Problems in Continuous Spaces

this method can be similarly applied to I2m+1,k (s; α, β), so that after some computations we obtain I2m+1,k (s; α, β) = (−is)

Γ (−α + m − k + 3/2) Γ (β + α − m + k − 3/2) Γ (β)

 s2 −α + m − k + 3/2 . × 1 F2 3/2, −β − α + m − k + 5/2 4

(1.149)

Hence, combining relations (1.148) and (1.149) gives the final form of (1.146) as       1 n+1 1 n+1 −α − k + + Γ β +α+k− − 2 2 2 2

 n 1−(−1) s2 (−is) 2 1/2 − α − k + [(n + 1)/2] × F . n 1 2 4 Γ (β) + [(n + 1)/2] , −β − α − k + 3/2 1 − (−1) 2

In,k (s; α, β) = Γ

(1.150) The second way of computing In,k (s; α, β) is by taking n = 2m and n = 2m + 1 as before and applying the cosine and sine Fourier transforms to the function (1 + x 2 )−β x −2α+n−2k . In this sense, by noting that (−1)2α = 1, we have I2m,k (s; α, β)  ∞  −β = cos(sx) (1 + x 2 ) x −2α+2m−2k dx − i =2

−∞  ∞

∞

−∞

cos(sx) (1 + x 2 )

−β −2α+2m−2k

x

sin(sx) (1 + x 2 )

−β −2α+2m−2k

x

dx

0

=

Γ (−α + m − k + 1/2) Γ (β + α − m + k − 1/2) Γ (β)

 s2 −α + m − k + 1/2 , × 1 F2 1/2, −β − α + m − k + 3/2 4

as well as I2m+1,k (s; α, β)  ∞  ∞ −β −β = cos(sx) (1 + x 2 ) x −2α+2m+1−2k dx−i sin(sx)(1+x 2 ) x −2α+2m+1−2k dx −∞

= (−2i)

 ∞ 0

−∞

−β sin(sx) (1 + x 2 ) x −2α+2m+1−2k dx

dx

1.8 Fourier Transforms of Symmetric Orthogonal Polynomials  = (−is)Γ

    −α + m − k + 3/2 3 3 −α+m−k+ F Γ β+α−m+k − 1 2 2 2 3/2, −β − α + m − k + 5/2

79

s2 . 4

Therefore, the result (1.150) simplifies (1.145) as F(g(x)) =

1 Γ Γ (β)

×

      1−(−1)n 1 n+1 n+1 1 −α + + Γ β+α− − (−is) 2 2 2 2 2

[n/2] 

(−[n/2] )k (c + 1/2 − [(n + 1)/2] )k (β + α − 1/2 − [(n + 1)/2] )k (c + d − n + 1/2)k (1/2 + α − [(n + 1)/2] )k k! k=0

 s2 1/2 − α − k + [(n + 1)/2] × 1 F2 . (1.151) (−1)n 4 + [(n + 1)/2] 1 − 2 , −β − α − k + 3/2

If in (1.151) we define the symmetric function An (x; p1 , p2 , p3 , p4 ) = x ×

[n/2]  k=0

1−(−1)n 2

(−[n/2])k (p3 + 1/2 − [(n + 1)/2])k (p1 + p2 − 1/2 − [(n + 1)/2])k (p3 + p4 − n + 1/2)k (1/2 + p1 − [(n + 1)/2])k k!

 x2 1/2 − p1 − k + [(n + 1)/2] , × 1 F2 (−1)n 4 + [(n + 1)/2] 1 − 2 , −p1 − p2 − k + 3/2

then clearly F(g(x)) = (−i)

1−(−1)n 2

      1 n+1 n+1 1 1 Γ −α+ + Γ β+α − − An (s; α, β, c, d). Γ (β) 2 2 2 2 (1.152)

By substituting (1.152) in Parseval’s identity and noting (1.144), we obtain  ∞ 2π

−∞

x −2(α+l) (1 + x 2 )

−(β+u) (c,d) (v,w) An (x) Am (x) dx



  n+1 1 + 2 2          1 1 1 1 n+1 m+1 m+1 Γ −l + + Γ u+l− − × Γ β +α− − Γ (β)Γ (u) 2 2 2 2 2 2 =i

(−1)n −(−1)m 2

Γ

−α +

80

1 Generalized Sturm–Liouville Problems in Continuous Spaces

×

 ∞ −∞

An (s; α, β, c, d) Am (s; l, u, v, w) ds .

(1.153)

Now, if on the left-hand side of (1.153) we take c =v =α+l

and

d = w = β + u,

then according to the orthogonality relation (1.128) we finally obtain 1 2π



∞

−∞

An (x; α, β, p, q ) Am (x; p − α, q − β, p, q) dx =

  n  −j + (1 − (−1)j )p j − (1 − (−1)j )p − 2q (2j − 2p − 2q + 1) (2j − 2p − 2q − 1)

j =1

×

Γ (β) Γ (q−β) Γ (p+q − 1/2) Γ (−p + 1/2)δn,m        , 1 n+1 1 n+1 1 n+1 Γ (q)Γ −α+ 2 +[ 2 ] Γ α+β− 12 −[ n+1 2 ] Γ α−p+ 2 +[ 2 ] Γ p+q−α−β− 2 −[ 2 ]

where m, n = 0, 1, . . . , N = max{m, n} ≤ p+q−1/2, p < 1/2, (−1)2p = 1, q > β > 0, 0 < α < 1/2, and α + β > 1/2.

1.8.4

Fourier Transform of the Second Finite Sequence of Symmetric Orthogonal Polynomials

To derive the Fourier transform of the second finite sequence of symmetric orthogonal polynomials  Bn(a)(x) = S¯n



 −2a + 2 2 −[n/2] n x = x 1 F1 1 0 a + (−1)n /2



1 2 , x

let us define the specific functions −1

u(x) = x −2a e 2x 2 Bn(b) (x) with

(−1)2a = 1

and

(1.154) −1

(d) v(x) = x −2c e 2x 2 Bm (x) with

(−1)2c = 1.

1.8 Fourier Transforms of Symmetric Orthogonal Polynomials

81

The Fourier transform of, e.g., u(x) is computed as  F(u(x)) =

∞

−1

e−isx x −2a e 2x 2 Bn(b) (x) dx

−∞

 =

∞ −∞

⎛ e−isx e

−1 2x 2

[n/2] 

x −2a+n ⎝

k=0

=

[n/2]  k=0

⎞ (−[n/2] )k x −2k ⎠ dx (b + (−1)n /2 )k k!

(−[n/2] )k (b + (−1)n /2 )k k!



∞ −∞

 −1 e−isx e 2x 2 x −2a+n−2k dx .

So the following integral should be evaluated:  ∞ −1 Rn,k (s; a) = e−isx e 2x 2 x −2a+n−2k dx. −∞

We have  R2m,k (s; a) =

=

∞

⎛ ⎝

−∞

j =0

j!

⎞ −1

⎠ e 2x 2 x −2a+2m−2k dx

 ∞  (−1)j i j s j

∞

j!

−∞

j =0

=

∞  (−isx)j

−1

e 2x 2 x j −2a+2m−2k dx



  ∞  ∞  −1 (−1)r s 2r 2 x 2r−2a+2m−2k e 2x 2 dx (2r) ! 0 r=0

  1 2 Γ −r + a − m + k − = (2r) ! 2 r=0    1 − a−m+k− 12 =2 Γ a−m+k− 0 F2 1 3 2 2, 2 − a + m − k ∞  (−1)r s 2r

−r+a−m+k− 21



s2 , 8

as well as R2m+1,k (s; a) =

 ∞  (−1)j i j s j

∞

j!

−∞

j =0

−1

e 2x 2 x j −2a+2m+1−2k dx



   ∞ ∞  −1 (−1)r s 2r x 2r−2a+2m−2k+2e 2x 2 dx 2 (2r + 1) ! 0 r=0    3 − a−m+k− 23 = (−is) 2 Γ a−m+k− 0 F2 3 5 2 , − a +m−k 2 2

= (−is)



s2 . 8

82

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Consequently, Rn,k (s; a) = 2

a+k− 21 −[ n+1 2 ]

   1−(−1)n n+1 1 Γ a+k− − (−is) 2 2 2

 s2 − ×0 F2 n 3 n+1 8 , −a − k + + [ ] 1 − (−1) 2 2 2

and    1−(−1)n 1 n+1 n+1 1 F(u(x)) = Γ a − − 2a− 2 −[ 2 ] (−is) 2 2 2  [n/2] k  (−[ n2 ] ) (a − 12 − [ n+1 − 2 ] )k 2 k × n 0 F2 n (−1) (b + (−1) /2 )k k! 1 − 2 , −a − k + k=0

3 2



s2 . 8 + [ n+1 ] 2 (1.155)

If in (1.155) we define the symmetric function Bn (x; q1 , q2 ) = x ×

[n/2] 

1−(−1)n 2

(−[ n2 ])k (q1 −

k − [ n+1 2 ])k 2 0 F2 (q2 + (−1) /2)k k! 1 2



n

k=0

1−

(−1)n 2 ,

− −q1 − k +

3 2

(1.156)

x2 , n+1 8 +[ 2 ]

then by referring to definitions (1.154) and applying Parseval’s identity, we get  2π =i

∞

−∞

−1

(d) x −2(a+c)e x 2 Bn(b) (x)Bm (x) dx

(−1)n −(−1)m 2

Γ (a − 2

1 n+1 1 m+1 2 − [ 2 ])Γ (c − 2 − [ 2 ]) 1 n+1 1 m+1 −a+ 2 +[ 2 ] −c+ 2 +[ 2 ]

2



∞ −∞

Bn (s; a, b)Bm (s; c, d) ds . (1.157)

Now if in (1.157) we take b = d = a + c and then refer to the finite orthogonality relation (1.131), we conclude that the special function Bn (x; q1, q2 ) defined in (1.156) satisfies the orthogonality relation 1 2π



∞ −∞

Bn (x; a, b)Bm(x; b − a, b) dx = 2−b+1+2[

n+1 2 ]

n j =1

×

2(−1)j (j − b) + 2b (2j −2b+1) (2j −2b−1)

Γ (b − 1/2) Γ (a −

1 2

− [ n+1 2 ])Γ (b

−a−

for m, n = 0, 1, . . . , N = max{m, n} ≤ b − 12 , (−1)2b = 1, and

1 2

1 2

− [ n+1 2 ])

< a < b − 12 .

δn,m ,

1.9 A Class of Symmetric Orthogonal Functions

1.9

83

A Class of Symmetric Orthogonal Functions

In Sect. 1.7, we introduced a class of symmetric orthogonal polynomials satisfying the equation x 2 (px 2 + q) Φn (x) + x(rx 2 + s) Φn (x)   − n(r + (n − 1)p)x 2 + (1 − (−1)n ) s/2 Φn (x) = 0. In this section, we introduce a class of symmetric orthogonal functions as an extension of Sn (x; p, q, r, s) and obtain its standard properties. We show that this new class satisfies the equation   x 2 (px 2 + q) Φn (x) + x(rx 2 + s) Φn (x) − αn x 2 + (1 − (−1)n ) β/2 Φn (x) = 0, (1.158) in which β is a free parameter and −αn denotes the corresponding eigenvalues. We then consider four cases of the introduced class and study their properties in detail. Let us replace the following options in the generic differential equation (1.94): A(x) = x 2 (px 2 + q), B(x) = x(rx 2 + s), C(x) = x 2 > 0, D(x) = 0, E(x) = −β ∈ R,

(1.159)

to reach Eq. (1.158), where λn = −αn is the corresponding eigenvalue to be derived. Although there is a small difference between the above two differential equations, we must solve Eq. (1.158) separately. Since it is independent of β for n = 2m (m ∈ Z+ ), without loss of generality we can assume that its solution is almost similar to relation (1.104) as  Φ2n (x) = S¯2n





 r s r ∗ s∗ θ ¯ for (θ ∈ R − {0}), x x and Φ2n+1 (x) = x S2n p∗ q ∗ pq (1.160)

and try to obtain the parameters p∗ , q ∗ , r ∗ , s ∗ , and θ in terms of the five known parameters p, q, r, s, and β to somehow get to Eq. (1.158) and determine the eigenvalues −αn too. In this sense, the condition (−1)θ = −1 is necessary in the second definition of (1.160) because Φn (x) must be symmetric and Φ2n+1 (−x) = − Φ2n+1 (x). To solve this problem, as Eq. (1.100) and its generic solution show, if n = 2m, then   (x) + (rx 2 + s) Φ2m (x) − 2m (r + (2m − 1)p) x Φ2m (x) = 0 x(px 2 + q) Φ2m

84

1 Generalized Sturm–Liouville Problems in Continuous Spaces

has the general solution  Φ2m (x) = S2m

⎞

 ⎛m−(k+1) m  r s m ⎝ (2j − 1 + 2m)p + r ⎠ 2m−2k . x x = (2j + 1) q + s pq k k=0 j =0

Hence the second equality of (1.160) must satisfy the equation x(p∗ x 2 + q ∗ )

d 2 −θ d (x −θ Φ2m+1 (x)) (x Φ2m+1 (x)) + (r ∗ x 2 + s ∗ ) dx 2 dx − 2m (r ∗ + (2m − 1)p∗ ) x 1−θ Φ2m+1 (x) = 0 .

(1.161)

After some calculations, (1.161) is simplified as x 2 (p∗ x 2 + q ∗ ) Φ  2m+1 (x) + x((r ∗ − 2θ p∗ )x 2 + s ∗ − 2θ q ∗ ) Φ  2m+1 (x)   + (−2m(r ∗ + (2m − 1)p∗ ) − θ (r ∗ − (θ + 1)p∗ )) x 2 + θ ((θ + 1)q ∗ − s ∗ ) Φ2m+1 (x) = 0.

(1.162)

If (1.162) is compared with the special case of (1.158) for n = 2m + 1, i.e.,     x 2 (px 2 + q) Φ2m+1 (x) + x(rx 2 + s) Φ2m+1 (x) − α2m+1 x 2 + β Φ2m+1 (x) = 0, (1.163) then equating Eqs. (1.162) and (1.163) yields p∗ = p , q ∗ = q , r ∗ = r + 2θ p, and s ∗ = s + 2θ q. Moreover, the values −α2m+1 and −β in (1.163) will be −α2m+1 = −2m(r + 2 θ p + (2m − 1)p) − θ (r + 2 θ p − (θ + 1)p) = −( θ + 2m)(r + ( θ + 2m − 1)p) and −β = θ ((θ + 1)q − (s + 2θ q)) = −θ (s + (θ − 1)q) . All these results eventually lead to the following corollary.

1.9 A Class of Symmetric Orthogonal Functions

Corollary 1.3 Let σn =  Φn (x) =

Sn(θ)

1−(−1)n . 2

85

Then the symmetric sequence



 r s r + 2σn θ p, s + 2σn θ q θσn ¯ n S2 [ 2 ] x =x p, q pq



x ,

having the explicit definition [n/2]−1 (2j + 1 + 2σn θ ) q + s r s x (θ−1)σn x = (2j − 1 + n + (2θ − 1)σn ) p + r pq j =0 ⎞ ⎛ 

[n/2] [n/2]−(k+1)  [n/2] − 1 + n + (2θ − 1)σ p + r (2j ) n ⎠ x n−2k , ⎝ × (2j + 1 + 2σn θ ) q + s k k=0 j =0 

Sn(θ)

(1.164)

satisfies the equation x 2 (px 2 + q) Φn (x) + x(rx 2 + s) Φn (x) − (αn x 2 + β σn ) Φn (x) = 0 ,

(1.165)

where αn = (n + (θ − 1)σn ) (r + (n − 1 + (θ − 1)σn )p) and β = θ (s + (θ − 1)q). (θ) There is a direct relationship between Sn (x; p, q, r, s) and S¯n (x; p, q, r, s), namely

 Sn(θ)



 r s r + 2σn (θ − 1)p, s + 2σn (θ − 1) q (θ−1)σn ¯ Sn x =x p, q pq



x ,

which leads to the hypergeometric representation  Sn(θ)



r s x pq

= x (n+σn (θ−1)) 2 F1



−[n/2] ,

2−θ+(−1)n (θ−1) s − 2q − [(n + 2 4−θ−2n+(−1)n (θ−1) r − 2p 2



1)/2] q . − px 2

86

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Also, they satisfy a three-term recurrence relation as  (θ) Sn+1







  

  r s r s r s r s (θ) 1+(−1)n (θ−1) (θ) (θ) Sn Sn−1 x = x x + Cn x , pq pq pq pq

where  Cn(θ)

r s pq

=

A(θ) n (p, q, r, s)   , n ((−1) (θ − 1) + 2n + θ − 2)p + r ((−1)n (1 − θ ) + 2n + θ − 4)p + r (1.166) with   n 2 A(θ) n (p, q, r, s) = 1 + (−1) (θ − 1) pq n      + θ − 3 + 3(−1)n (1 − θ ) pq + 1 + (−1)n (θ − 1) qr − (−1)n θ ps n     + (1 − θ )(p − r) q + (θ − 3)p + r s (1 − (−1)n )/2. For instance, we have

r s x = 1, pq

 r s (θ) S1 x = xθ , pq

 q+s r s (θ) , S2 x = x2 + p+r pq

   (2θ + 1)q + s r s (θ) θ 2 S3 , x =x x + (2θ + 1)p + r pq

 3q + s 2 (3q + s)(q + s) r s S4(θ) x + , x = x4 + 2 5p + r (5p + r)(3p + r) pq

   (2θ + 3)q + s 2 ((2θ + 3)q + s)((2θ + 1)q + s) r s (θ) S5 x + . x = xθ x4 + 2 (2θ + 5)p + r ((2θ + 5)p + r)((2θ + 3)p + r) pq 

(θ) S0

(1.167)

1.9 A Class of Symmetric Orthogonal Functions

87

(θ)

Relations (1.167) show that the S2n (x ; p, q, r, s) are independent of θ . Hence the condition (−1)θ = −1 must be satisfied only for odd n. Moreover, for θ = 1 in (1.164) and (1.166) we respectively obtain  Sn(1)





 r s r s x = S¯n x pq pq

 and Cn(1)



r s pq

= Cn

r s pq

.

(θ)

It can be verified that the weight function corresponding to Sn (x; p, q, r, s) is the same W (x; p, q, r, s) as defined in (1.109) (i.e., independent of θ ). Therefore, we can design a generic orthogonality relation as 



α −α

W







  r s r s r s (θ) x Sn(θ) x Sm x dx = Nn δn,m , pq pq pq

(1.168)

where  Nn =



α −α

W



2   r s r s (θ) Sn dx x x pq pq

and (px 2 + q) W (x; p, q, r, s) vanishes at x = α. To compute the norm square value, we can directly use the orthogonality relation (1.110), so that for n = 2m we have  N2m =



α −α

W



2   r s r s (θ) S2m dx x x pq pq 

2    α r s r s S¯2m W dx = x x pq pq −α

  2m   α r s r s = Ci W x dx , pq pq −α i=1

(1.169)

while for n = 2m + 1 we get

2   r s r s (θ) S2m+1 = W dx x x pq pq −α

    α r + 2θ p, s + 2θ q r s 2θ ¯ = W S2m x x p, q pq −α 

N2m+1

α





2 dx . x

88

1 Generalized Sturm–Liouville Problems in Continuous Spaces

On the other hand, since  x 2θ W



  (r − 2p)x 2 + s 2θ r s dx) exp( dx) x = exp( x x(px 2 + q) pq

 = exp

(r + 2θ p − 2p)x 2 + s + 2θ q dx x(px 2 + q)



 =W



r + 2θ p, s + 2θ q x , p, q

by noting (1.169), N2m+1 is simplified as 

2  r + 2θ p, s + 2θ q r + 2θ p, s + 2θ q S¯2m N2m+1 = W dx x x p, q p, q −α

  2m 

 α r + 2θ p, s + 2θ q r + 2θ p, s + 2θ q = Ci W (1.170) x dx . p, q p, q −α i=1 



α

Combining relations (1.169) and (1.170) finally gives

Nn =

2[n/2] i=1

1.9.1

 Ci

r + 2σn θ p, s + 2σn θ q p, q   α r + 2σn θ p, s + 2σn θ q W × p, q −α



x dx .

(1.171)

(θ )

Four Special Cases of Sn (x; p, q, r, s)

Case 1 Consider the equation   x 2 (1 − x 2 ) Φn (x) − 2x (a + b + 1)x 2 − a Φn (x)  + (n + (θ − 1)σn )(n + 2a + 2b + 1 + (θ − 1)σn ) x 2  − θ (θ + 2a − 1)σn Φn (x) = 0,

(1.172)

together with its solution  Φn (x) =

Sn(θ)



−2a − 2b − 2, 2a x . −1, 1

(1.173)

1.9 A Class of Symmetric Orthogonal Functions

89

According to the generalized Sturm–Liouville theorem for symmetric functions, the sequence (1.173) satisfies the orthogonality relation 

1 −1

 2 b

x (1 − x ) 2a

=

2[n/2] i=1

Sn(θ)



 −2a − 2b − 2, 2a −2a − 2b − 2, 2a (θ) x Sm −1, 1 −1, 1



x dx

(i + 2σi (a + θ σn )) (i + 2σi (a + θ σn ) + 2b) (2i + 2(a + θ σn ) + 2b − 1) (2i + 2(a + θ σn ) + 2b + 1)   1 × B a + θ σn + ; b + 1 δn,m , 2

provided that a + 1/2 > 0, a + θ + 1/2 > 0, b + 1 > 0, (−1)2a = 1, and (−1)θ = −1. Case 2 Consider the equation   x 2 Φn (x) − 2x(x 2 − a) Φn (x) + 2 (n + (θ − 1)σn ) x 2 − θ (2a + θ − 1)σn Φn (x) = 0, together with its solution  Φn (x) =

Sn(θ)



−2, 2a x . 0, 1

(1.174)

The solution (1.174) satisfies the orthogonality relation 



∞ −∞

x

2a

exp(−x

2

) Sn(θ)



 −2 2a −2 2a (θ) x Sm 0 1 0 1



x dx

2[n/2]    1 δn,m , i + 2σi (a + θ σn ) Γ a + θ σn + = 2−2 [n/2] 2 i=1

provided that a + 1/2 > 0, a + θ + 1/2 > 0, (−1)2a = 1, and (−1)θ = −1. Case 3 Consider the equation   x 2 (x 2 + 1) Φn (x) − 2x (a + b − 1)x 2 + a Φn (x)  − (n + (θ − 1)σn )(n − 2a − 2b + 1 + (θ − 1)σn ) x 2  + θ (−2a + θ − 1)σn Φn (x) = 0,

(1.175)

90

1 Generalized Sturm–Liouville Problems in Continuous Spaces

together with its solution  Φn (x) =

Sn(θ)



−2a − 2b + 2, −2a x , 1, 1

(1.176)

which satisfies the orthogonality relation 

∞

x −2a





 −2a − 2b + 2, −2a −2a − 2b + 2, −2a (θ) x Sm x dx 1, 1 1, 1 

2[n/2] −2a − 2b + 2 + 2θ σn , −2a + 2θ σn = Ci 1, 1 i=1 

S (θ) b n 2 −∞ (1 + x )

×

Γ (b + a − θ σn − 1/2)Γ (−a + θ σn + 1/2) δn,m . Γ (b)

(1.177)

Now the question is how to determine the parameter constraint in (1.177). For this purpose, we write Eq. (1.175) in a self-adjoint form to get 

 x −2a (1 + x 2 )−b+1 (Φn (x)Φm (x) − Φm (x)Φn (x))

∞ −∞

= 0.

(1.178)

Since Φn (x) defined in (1.176) is a function of degree at most n + (θ − 1)σn , we have  max deg{(Φn (x)Φm (x) − Φm (x)Φn (x))} = n + m − 1 + (θ − 1)(σn + σm ).

(1.179)

Hence it is deduced from (1.178) and (1.179) that − 2a + 2(−b + 1) + n + m − 1 + (θ − 1)(σn + σm ) ≤ 0.

(1.180)

Furthermore, the right-hand side of (1.177) shows that b + a − 1/2 > 0, b + a − 1/2 − θ > 0, −a + 1/2 > 0, −a + 1/2 + θ > 0, and b > 0, which are equivalent to b+a−

1 1 1 1 > 0, a < , b > 0, and a − < θ < b + a − . 2 2 2 2

By considering the simplified form of (1.180) as − 2a − 2b + n + m + 1 + (θ − 1)(σn + σm ) ≤ 0,

(1.181)

1.9 A Class of Symmetric Orthogonal Functions

91

four cases may occur for n and m as follows:  (i)



n = 2i m = 2j + 1

(ii)

n = 2i + 1 m = 2j

 (iii)

n = 2i m = 2j

 (iv)

n = 2i + 1 m = 2j + 1 (1.182)

If each of the above cases is replaced in (1.181), then by taking N = max{m, n} we get N ≤ b + a − θ/2,

N ≤ b + a − 1/2,

and

N ≤ b + a − θ + 1/2 .

(1.183)

Finally, by taking (b+a) , min {b + a − θ/2 , b + a − 1/2 , b + a − θ + 1/2} = Mθ

the following corollary is derived. Corollary 1.4 The finite set of symmetric functions 

N Sn(θ) (x; 1, 1, −2a − 2b + 2, −2a)

n=0

is orthogonal with respect to the weight function x −2a (1 + x 2 )−b on (−∞, ∞) if and only if N ≤ Mθ(b+a), a < 1/2, b > 0, b + a − 1/2 > 0, a − 1/2 < θ < b + a − 1/2 , (−1)2a = 1, and (−1)θ = −1. For example, suppose that the even weight function x −2/3 (1 + x 2 )−10 is given on (−∞, ∞). Therefore a = 1/3 < 1/2, b = 10 > 0, −1/6 < θ < 59/6, and (31/3)

Mθ

= min {62/6 − θ/2, 61/6, 63/6 − θ } ,

for all −1/6 < θ < 59/6 and (−1)θ = −1. This means that the finite set of symmetric functions N≤ M (31/3)  θ Sn(θ) (x; 1, 1, −56/3, −2/3) n=0

is orthogonal with respect to the weight function x −2/3 (1 + x 2 )−10 on (−∞, ∞). For instance, if θ = 1/3 ∈ (−1/6 , 59/6), then (−1)1/3 = −1, (31/3)

M1/3

= min {61/6 , 61/6 , 61/6} = 61/6,

92

1 Generalized Sturm–Liouville Problems in Continuous Spaces

and N = 10, which yields 

∞ −∞



 −56/3, −2/3 −56/3, −2/3 (1/3) (1/3) S √ x Sm 3 2 10 n 2 1, 1 1, 1 x (1 + x ) 

2[n/2] −(55 + (−1)n )/3, −(1 + (−1)n )/3 Ci = 1, 1 i=1 

1

Γ ((58 + (−1)n )/6)Γ ((2 − (−1)n )/6) δn,m 9!

×



x dx

⇔ m, n ≤ 10.

Case 4 Consider the equation   x 4 Φn (x) + 2x (1 − a)x 2 + 1 Φn (x)−   (n + (θ − 1)σn )(n + 1 − 2a + (θ − 1)σn )x 2 + θ σn Φn (x) = 0, together with its solution  Φn (x) =

Sn(θ)



−2a + 2, 2 x , 1 0

which satisfies the orthogonality relation 

∞ −∞

x

−2a





  1 (θ) −2a + 2, 2 −2a + 2, 2 (θ) exp(− 2 ) Sn x Sn x dx x 1, 0 1, 0

   2[n/2] 1 −2a + 2 + 2θ σn , 2 = Γ a − − θ σn δn,m . Ci 2 1, 0 i=1

(1.184)

Once again, to determine the parameter constraint in (1.184), we should apply the described technique for the first finite case, i.e., we must have −2a + n + m + 1 + (θ − 1)(σn + σm ) ≤ 0, which is the same condition as (1.181) for b = 0. Therefore, by referring to (1.182) and (1.183), if we define (a)

Mθ

= min {a − θ/2 , a − 1/2 , a − θ + 1/2} ,

the following corollary is deduced.

(θ)

1.10 An Extension of Sn (x; p, q, r, s)

93

 N Corollary 1.5 The finite set of symmetric functions Sn(θ) (x; 1, 0, −2a + 2, 2) x −2a

n=0

is

on (−∞, ∞) if and only orthogonal with respect to the weight function if N ≤ Mθ(a) , a − 1/2 > 0, a − θ − 1/2 > 0, (−1)2a = 1, and (−1)θ = −1.

1.10

exp(−1/x 2)

An Extension of Sn(θ) (x; p, q, r, s)

As we observe in relations (1.99) and (1.159), the function D(x) is equal to zero in both cases. Here one may ask what happens if D(x) = 0 in (1.99) or (1.159). In this section, we answer this question by considering the following options in the generic differential equation (1.94): A(x) = x 2 (px 2 + q), B(x) = x(rx 2 + s), C(x) = x 2 > 0, D(x) = −(c + d) = −α (s + (α − 1) q),   E(x) = 2c = (α − β)s + α(α − 1) − β(β − 1) q, where p, q, r, s and α, β are free parameters. We show that the solution of above equation is another class of symmetric orthogonal (θ) functions that generalizes Sn (x; p, q, r, s). For this purpose, we first showed in relation (1.164) that the symmetric sequence  Φ2n (x) = S¯2n



r s x pq

 and Φ2n+1 (x) = x S¯2n θ



r + 2θ p, s + 2θ q x p, q (1.185)

is a basis solution for the generalized Sturm–Liouville equation (1.165). Now, without loss of generality, consider the symmetric sequence  Φ2n (x) = x α S¯2n



r1 s1 x p1 q1

 and Φ2n+1 (x) = x β S¯2n



r2 s2 x , p2 q2

(1.186)

for (−1)α = 1 and (−1)β = −1. It is clear that (1.186) is an extension of (1.185) for α = 0. According to Eq. (1.100) and its generic solution, if n = 2m, then the equation     x(p1 x 2 + q1 ) y2m (x) + (r1 x 2 + s1 ) y2m (x) − 2m r1 + (2m − 1)p1 x y2m (x) = 0

94

1 Generalized Sturm–Liouville Problems in Continuous Spaces

has the general solution  y2m (x) = S2m

⎞

 ⎛m−(k+1) m  r1 s1 m ⎝ (2j − 1 + 2m)p1 + r1 ⎠ 2m−2k . x x = (2j + 1) q1 + s1 p1 q1 k k=0 j =0

Therefore, the first equality of (1.186) must satisfy the equation x(p1 x 2 + q1 )

d 2 −α d (x −α Φ2m (x)) (x Φ2m (x)) + (r1 x 2 + s1 ) 2 dx dx − 2m (r1 + (2m − 1)p1 ) x 1−α Φ2m (x) = 0 .

(1.187)

After some calculations, (1.187) is simplified as     x 2 (p1 x 2 + q1 ) Φ2m (x) + x (r1 − 2α p1 )x 2 + s1 − 2α q1 Φ2m (x)  + (−2m(r1 + (2m − 1)p1 ) − α (r1 − (α + 1)p1 )) x 2  + α ((α + 1)q1 − s1 ) Φ2m (x) = 0 .

(1.188)

Similarly, for the second equality of (1.186) we get x(p2 x 2 + q2 )

d 2 −β d (x −β Φ2m+1 (x)) (x Φ2m+1 (x)) + (r2 x 2 + s2 ) dx 2 dx − 2m (r2 + (2m − 1)p2 ) x 1−β Φ2m+1 (x) = 0 ,

which is simplified as     x 2 (p2 x 2 + q2 ) Φ2m+1 (x) + x (r2 − 2β p2 )x 2 + s2 − 2β q2 Φ2m+1 (x)  + (−2m(r2 + (2m − 1)p2 ) − β (r2 − (β + 1)p2 )) x 2  + β ((β + 1) q2 − s2 ) Φ2m+1 (x) = 0 . (1.189) By comparing and then equating the differential equations (1.188) and (1.189), it can be concluded that p2 = p1 , q2 = q1 , r2 = r1 + 2p1 (β − α), and s2 = s1 + 2q1 (β − α). If these results are substituted into (1.186), the following corollary is derived.

(θ)

1.10 An Extension of Sn (x; p, q, r, s)

Corollary 1.6 Let σn =  Sn(α,β)

1−(−1)n . 2

95

Then the symmetric sequence



r s x pq 

= x α+(β−α) σn S¯2 [ n2 ]



r + 2p(α + (β − α) σn ), s + 2q(α + (β − α) σn ) x , p, q

which is equivalent to the explicit definition  Sn(α,β)

=



r s x pq [n/2]−1 j =0

×

[n/2] 



k=0

[n/2] k

(2j + 1 + 2α + 2 (β − α) σn ) q + s x α+(β−α−1) σn (2j − 1 + n + 2α + (2β − 2α − 1) σn ) p + r

⎞

⎛[n/2]−(k+1) − 1 + n + 2α + (2β − 2α − 1) σ p + r (2j ) n ⎠ x n−2k , ⎝ (2j + 1 + 2α + 2 (β − α) σn ) q + s j =0

(1.190) satisfies the equation   x 2 (px 2 +q) Φn (x)+x(rx 2 +s) Φn (x)− an x 2 + (−1)n c + d Φn (x) = 0,

(1.191)

where an = (n + α + (β − α − 1) σn ) (r + (n − 1 + α + (β − α − 1) σn )p) , 2c = (α − β)s + (α(α − 1) − β(β − 1))q , and 2d = (α + β)s + (α(α − 1) + β(β − 1))q . (α,β)

There is a direct relationship between Sn  Sn(α,β)  × S¯n

(x; p, q, r, s) and S¯n (x; p, q, r, s) as follows:



r s x = x α+(β−α−1) σn pq



r + 2p(α + (β − α − 1) σn ), s + 2 q (α + (β − α − 1) σn ) x , p, q

(1.192)

96

1 Generalized Sturm–Liouville Problems in Continuous Spaces

which leads to the hypergeometric representation  Sn(α,β)



r s x = x n+α+(β−α−1) σn pq  n+1 −[ n2 ] , q−s 2q − (α + (β − α − 1) σn ) − [ 2 ] × 2 F1 r − 2p − (α + (β − α − 1) σn ) − n + 3/2



q . − px 2

Also, they satisfy a three-term recurrence relation as



   n r s r s x = x 1+(−1) (β−α−1) Sn(α,β) x pq pq



 r s r s (α,β) (α,β) + Cn Sn−1 x , pq pq

 (α,β)

Sn+1

(1.193)

where  Cn(α,β)

r s pq

= (α,β)



An (p, q, r, s)  , ((−1)n (β−α−1) + 2n+β + α − 2)p + r ((−1)n (1 − β + α) + 2n + β+α−4)p + r

with   n pq n2 (p, q, r, s) = 1 + (β − α − 1)(−1) A(α,β) n       + (1 − β + α) (1 − 2α)p − r q + (α + β − 3)p + r (s + 2α q) σn   + α + β − 3 + (−1)n (α + 3 − 3β) pq    + 1 + (−1)n (β − α − 1) qr + (−1)n (α − β) ps n. For instance, we have r s pq  r s (α,β) S1 pq 

(α,β) S0

= xα ,

x

x

= xβ ,

(θ)

1.10 An Extension of Sn (x; p, q, r, s)

r s pq  r s (α,β) S3 pq  r s (α,β) S4 pq  r s (α,β) S5 pq 

(α,β) S2



  (2α + 1)q + s , = xα x2 + (2α + 1)p + r



  (2β + 1)q + s = xβ x2 + , (2β + 1)p + r



  (2α + 3)q + s 2 ((2α + 3)q + s)((2α + 1)q + s) x + , = xα x4 + 2 (2α + 5)p + r ((2α + 5)p + r)((2α + 3)p + r)



  (2β + 3)q + s 2 ((2β + 3)q + s)((2β + 1)q + s) = xβ x4 + 2 x + . (2β + 5)p + r ((2β + 5)p + r)((2β + 3)p + r)

x

x

x

x

97

Note that both conditions (−1)α = 1 and (−1)β = −1 are necessary, because Sn(α,β) (−x; p, q, r, s) = (−1)n Sn(α,β) (x; p, q, r, s). An interesting case for the symmetric functions (1.190) occurs when β − α = 1 and (−1)α = 1, because in this case, (1.193) is reduced to a recurrence relation of polynomial type as  (α,α+1) Sn+1







 

 r s r s r s (α,α+1) (α,α+1) (α,α+1) r s Sn−1 x = x Sn x + Cn x , pq pq pq pq (α,α+1)

(α,α+1)

though the corresponding initial values are S0 (x) = x α and S1 (x) = x α+1 if α and only if (−1) = 1. On the other hand, by referring to the key identity (1.192), we have  Sn(α,α+1)



 r s r + 2pα, s + 2 q α x = x α S¯n pq p, q



x .

Therefore, the polynomials S¯n (x ; p, q, r + 2pα, s + 2 q α) are orthogonal with respect to the weight function x 2α W (x; p, q, r, s). As Eq. (1.191) and relation (1.98) show, the weight function corresponding to symmetric functions (1.190) is the same W (x; p, q, r, s) as defined in (1.109) (i.e., independent of α, β). Hence we can design a generic orthogonality relation as follows: 



v −v

W







  r s r s r s (α,β) x Sn(α,β) x Sm x dx = Nn δn,m , pq pq pq

(1.194)

98

1 Generalized Sturm–Liouville Problems in Continuous Spaces

where  Nn =



v −v

W



2   r s r s (α,β) dx, Sn x x pq pq

(1.195)

and (px 2 + q) W (x; p, q, r, s) vanishes at x = v. To compute the norm square value, we can directly use the orthogonality relation (1.110), so that for n = 2m in (1.195) we have  N2m =



v −v

W



2  r s r s (α,β) S2m dx x x pq pq

    v r + 2α p, s + 2α q r s 2α ¯ = W S2m x x p, q pq −v



2 dx . x

On the other hand, since

    (r − 2p)x 2 + s 2α r s dx exp dx x W x = exp x x(px 2 + q) pq    (r + 2α p − 2p)x 2 + s + 2α q r + 2α p, s + 2α q = exp dx = W x(px 2 + q) p, q 

2α



x ,

by noting (1.110), N2m is simplified as 

2  r + 2α p, s + 2α q r + 2α p, s + 2α q = W dx S¯2m x x p, q p, q −v

 2m  

 v r + 2α p, s + 2α q r + 2α p, s + 2α q = Ci W x dx , p, q p, q −v i=1 

N2m



v

(1.196) in which Ci (p, q, r, s) = Ci(0,1)(p, q, r, s). Similarly, for n = 2m + 1, we get

2   r + 2β p, s + 2β q r + 2β p, s + 2β q = W dx S¯2m x x p, q p, q −v



  2m  v r + 2β p, s + 2β q r + 2β p, s + 2β q Ci W = x dx . p, q p, q −v i=1 

N2m+1

v



(1.197)

(θ)

1.10 An Extension of Sn (x; p, q, r, s)

99

Combining relations (1.196) and (1.197) eventually gives 

r + 2p(α + (β − α − 1) σn ), s + 2 q (α + (β − α − 1) σn ) Nn = Ci p, q i=1

  v r + 2p(α + (β − α − 1) σn ), s + 2 q (α + (β − α − 1) σn ) W × x dx . p, q −v 2[n/2]

(α,β)

1.10.1 Four Special Cases of Sn

(x; p, q, r, s)

Case 1 Consider the differential equation    x 2 (1 − x 2 ) Φn (x) − 2x (a + b + 1)x 2 − a Φn (x) + (n + α + (β − α − 1) σn )× (n + 2a + 2b + 1 + α + (β − α − 1) σn ) x 2 + −

β(β − 1 + 2a) − α(α − 1 + 2a) (−1)n 2

β(β − 1 + 2a) + α(α − 1 + 2a)  Φn (x) = 0, 2

with the solution  Φn (x) =

Sn(α,β)



−2a − 2b − 2, 2a x . −1, 1

(1.198)

According to the generalized Sturm–Liouville theorem for symmetric functions, the sequence (1.198) satisfies the orthogonality relation 

1

−1

=

 2 b

x (1 − x ) 2a

2[n/2] i=1

Sn(α,β)





 −2a − 2b − 2, 2a −2a − 2b − 2, 2a (α,β) x Sm x dx −1, 1 −1, 1

(i + 2σi (a + α + (β − α) σn )) (i + 2σi (a + α + (β − α) σn ) + 2b) (2i + 2(a + α + (β − α) σn ) + 2b − 1) (2i + 2(a + α + (β − α) σn ) + 2b + 1) ×B

  1 a + α + (β − α) σn + ; b + 1 δn,m , 2

provided that a + α + 1/2 > 0, a + β + 1/2 > 0, b + 1 > 0, (−1)2a = 1, (−1)α = 1, and (−1)β = −1.

100

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Case 2 Consider the differential equation  x 2 Φn (x) − 2x(x 2 − a)Φn (x) + 2(n + α + (β − α − 1) σn ) x 2 +

β(β − 1 + 2a) − α(α − 1 + 2a) (−1)n 2 −

β(β − 1 + 2a) + α(α − 1 + 2a)  Φn (x) = 0, 2

with the solution  Φn (x) = Sn(α,β)



−2, 2a x . 0, 1

This solution satisfies the orthogonality relation



 −2 2a −2 2a (α,β) x exp(−x x Sm x dx 0 1 0 1 −∞ ⎞ ⎛   2[n/2] 1 1 i + 2 σi (a + α + (β − α) σn )⎠ Γ a + α + (β − α) σn + δn,m , = ⎝ n−σ 2 n 2



∞



2a

2

) Sn(α,β)

i=1

provided that a +α+1/2 > 0, a +β +1/2 > 0, (−1)2a = 1, (−1)α = 1, and (−1)β = −1.

Case 3 Consider the differential equation x 2 (x 2 + 1) Φn (x) − 2x((a + b − 1)x 2 + a) Φn (x)  − (n + α + (β − α − 1) σn ) (n − 2a − 2b + 1 + α + (β − α − 1) σn ) x 2 +

α(α − 1 − 2a) − β(β − 1 − 2a) (−1)n 2 +

α(α − 1 − 2a) + β(β − 1 − 2a)  Φn (x) = 0, 2

(1.199)

with the solution  Φn (x) =

Sn(α,β)



−2a − 2b + 2, −2a x , 1, 1

(1.200)

(θ)

1.10 An Extension of Sn (x; p, q, r, s)

101

which satisfies the orthogonality relation



 −2a − 2b + 2, −2a −2a − 2b + 2, −2a (α,β) (α,β) x S S x dx n m b 2 1, 1 1, 1 −∞ (1 + x ) 

2[n/2] −2a − 2b + 2 + 2(α + (β − α − 1) σn ), −2a + 2(α + (β − α − 1) σn ) = Ci 1, 1 i=1 

×

∞

x −2a



Γ (b + a − (α + (β − α − 1) σn ) − 1/2)Γ (−a + α + (β − α − 1) σn + 1/2) δn,m . Γ (b) (1.201)

To determine the parameter constraint in (1.201), we write Eq. (1.199) in a self-adjoint form to obtain 

∞   x −2a (1 + x 2 )−b+1 Φn (x)Φm (x) − Φm (x)Φn (x) = 0.

(1.202)

−∞

Since Φn (x) defined in (1.200) is a symmetric sequence of degree at most n + α + (β − α − 1) σn , we have  (x)Φn (x)} = n + m − 1 + 2α + (β − α − 1) (σn + σm ). max deg{Φn (x)Φm (x) − Φm

(1.203) Therefore, it is deduced from (1.202) and (1.203) that − 2a + 2(−b + 1) + n + m − 1 + 2α + (β − α − 1) (σn + σm ) ≤ 0.

(1.204)

Moreover, the right-hand side of (1.201) shows that b + a − α − 1/2 > 0, b + a − β − 1/2 > 0, −a + α + 1/2 > 0, −a + β + 1/2 > 0, and b > 0. Once again, four cases may occur for n and m in inequality (1.204):  (i)

n = 2i m = 2j + 1

 (ii)

n = 2i + 1 m = 2j

 (iii)

n = 2i m = 2j

 (iv)

n = 2i + 1 m = 2j + 1 (1.205)

If each of the above cases is replaced in (1.204), then by taking N = max{m, n}, we obtain N ≤ b + a − (α + β)/2 , N ≤ b + a − α − 1/2, and N ≤ b + a − β + 1/2 .

(1.206)

102

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Finally, by taking (b+a) min {b + a − (α + β)/2 , b + a − α − 1/2 , b + a − β + 1/2} = Mα,β ,

we obtain the following corollary. Corollary 1.7 The finite set of symmetric functions N  Sn(α,β) (x; 1, 1, −2a − 2b + 2, −2a)

n=0

is orthogonal with respect to the weight function x −2a (1 + x 2 )−b on (−∞, ∞) if and only (b+a) if N ≤ Mα,β , b + a − α − 1/2 > 0, b + a − β − 1/2 > 0, −a + α + 1/2 > 0, −a + β + 1/2 > 0, b > 0, (−1)2a = 1, (−1)α = 1, and (−1)β = −1. Case 4 Consider the differential equation   x 4 Φn (x) + 2x (1 − a)x 2 + 1 Φn (x)−  (n + α + (β − α − 1) σn )(n + 1 − 2a + α + (β − α − 1) σn ) x 2  + (α − β)(−1)n + α + β Φn (x) = 0, with the solution  Φn (x) = Sn(α,β)



−2a + 2, 2 x , 1 0

which satisfies the orthogonality relation



 −2a + 2, 2 −2a + 2, 2 (α,β) x x Sm x dx 1, 0 1, 0 −∞ 

2[n/2] 1 −2a + 2 + 2(α + (β − α) σn ), 2 = Ci Γ (a − − (α + (β − α) σn )) δn,m . 2 1, 0 i=1 

∞

−2a

1 exp(− 2 ) Sn(α,β) x



(1.207) To determine the parameter constraint in (1.207), noting 

x 2−2a exp(−

∞ 1   )(Φ (x)Φ (x) − Φ (x)Φ (x)) = 0, m n n m −∞ x2

1.11 A Generalization of Fourier Trigonometric Series (α,β)

Table 1.6 Four special cases of Sn

Definition

 (α,β) −2a − 2b − 2, 2a Sn x −1, 1

 (α,β) −2, 2a Sn x 0, 1

 (α,β) −2a − 2b + 2, −2a Sn x 1, 1

 (α,β) −2a + 2, 2 Sn x 1, 0

103

(x; p, q, r, s) Weight function

Interval and kind

x 2a (1 − x 2 )b

[−1, 1], Infinite

x 2a exp(−x 2 )

(−∞, ∞), Infinite

x −2a (1 + x 2 )

(−∞, ∞), Finite

b

x −2a exp(− 12 ) x

(−∞, ∞), Finite

then (1.203) yields −2a + n + m + 1 + 2α + (β − α − 1) (σn + σm ) ≤ 0, which is exactly the same condition as (1.204) for b = 0. Therefore, by referring to (1.205) and (1.206), if we define (a) Mα,β = min {a − (α + β)/2 , a − α − 1/2 , a − β + 1/2} ,

the following corollary is deduced.  N (α,β) Corollary 1.8 The finite set of symmetric functions Sn (x; 1, 0, −2a + 2, 2)

n=0

is

orthogonal with respect to the weight function x −2a exp(−1/x 2) on (−∞, ∞) if and only (a) if N ≤ Mα,β , a − α − 1/2 > 0, a − β − 1/2 > 0, (−1)2a = 1, (−1)α = 1, and (−1)β = −1. Table 1.6 shows the principal properties of the four classes.

1.11

A Generalization of Fourier Trigonometric Series (θ)

Although we introduced four main cases of Sn (x; p, q, r, s) in Sect. 1.9, further important cases can be still found for the specific values of p, q, r, s, and θ . In this section, using the generalized Sturm–Liouville theorem for symmetric functions, we introduce one of these samples, which generalizes Fourier trigonometric sequences and is orthogonal with respect to the same constant weight function on [0, π]. One of the advantages of this generalization is its use in finding some new trigonometric series.

104

1 Generalized Sturm–Liouville Problems in Continuous Spaces

1.11.1 A Generalization of Trigonometric Orthogonal Sequences ∞ It is known that the trigonometric sequences {sin nx}∞ n=1 and {cos nx}n=0 are orthogonal solutions of a usual Sturm–Liouville equation of the form

Φn (x) + n2 Φn (x) = 0 , x ∈ [0, π],

(1.208)

where [0, π] can be transformed to any other arbitrary interval, say [−l, l], with period 2l, by a simple linear transformation. To extend such trigonometric sequences, we start with a special case of Eq. (1.165) as   ) 2 x 2 (px 2 +q)Φn (x)+x(rx 2 + s)Φn (x) + λ(θ x − θ (s+(θ − 1)q)σ Φn (x) = 0, (1.209) n n

for      λ(θ) = − n + (θ − 1)σ r + n − 1 + (θ − 1)σ p , n n n and then assume that p = −1 and q = 1. In this case, by the change of variable x = cos t, we obtain the modified differential equation (1.209) as 2

cos t

Φn (t)

 +

 −(r + 1)cos3 t − s cos t Φn (t) sin t   2 + λ(θ) n cos t − θ (s + θ − 1)σn Φn (t) = 0.

(1.210)

If r + 1 = 0 and s = 0 in (1.210), a generalization of differential equation (1.208) is derived for θ = 1. Consequently, by substituting the initial vector (p, q, r, s, θ ) = (−1, 1, −1, 0, θ ) into Eq. (1.209) and applying the change of variable x = cos t, we get Φn (t)

+

 

n + (θ − 1)σn

2

θ (θ − 1) − σn cos2 t

 Φn (t) = 0.

(1.211)

By referring to the comments described in Sect. 1.9, Eq. (1.211) has a basic solution as  Φn (t) = Sn(θ)



[n/2]−1 2j + 1 + 2θ σn −1 0 (cos t)(θ−1)σn cos t = 2j + n + (2θ − 1)σn −1 1 j =0

1.11 A Generalization of Fourier Trigonometric Series

×

[n/2] 

 (−1)k

k=0

105

⎞

⎛[n/2]−(k+1) 2j + n + (2θ − 1)σn ⎠ [n/2] ⎝ cosn−2k t, 2j + 1 + 2θ σn k j =0

which can be simplified to an easier form, because for n = 2m we have  (θ) S2m m−1 j =0



−1 0 cos t = −1 1

m j + 1/2  (−1)k j +m



k=0

⎞

⎛m−(k+1) j + m m ⎝ ⎠ cos2m−2k t = 1 cos 2mt, j + 1/2 22m−1 k j =0

while for n = 2m + 1, it reads as  (θ) S2m+1 m−1 j =0



−1 0 cos t = −1 1

m  j + θ + 1/2 cos (θ−1) t (−1)k j +θ +m



k=0

m k

⎞

⎛m−(k+1) j + θ + m ⎠ cos2m+1−2k t , ⎝ j + θ + 1/2 j =0

(1.212) which still has a complicated form. To simplify (1.212), let us assume that  j A(m) j (θ ) = (−1)

Therefore we have  −1 0 (θ) S2m+1 −1 1

m j

⎞

⎛m−(j +1) i + θ + m ⎠. ⎝ i + θ + 1/2 i=0



m  ( θ + 1/2)m 2m+1−2j cos (θ−1) t A(m) t, cos t = j (θ ) cos ( θ + m)m j =0

(1.213) where (a)m =

m−1  j =0

(a + j ). On the other hand, since

2n+1

cos

n 1  x = 2n 2 k=0



2n + 1 k

cos(2n + 1 − 2k) x,

106

1 Generalized Sturm–Liouville Problems in Continuous Spaces

the sum in (1.213) is simplified as m 

2m+1−2j A(m) t= j (θ ) cos

j =0

=

m 

m 

2k+1 A(m) t m−k (θ ) cos

k=0

 (m) Am−k (θ )

k=0

k 1  22k



i=0

2k + 1 i



cos(2k + 1 − 2i) t

=

m 

(m)

Bk (θ ) cos(2m+1−2k) t ,

k=0

in which

(m) Bk (θ )

=

k A(m) (θ )  j j =0



22(m−j )

2m + 1 − 2j k−j

k  (−1)j = 22(m−j ) j =0

 m j



⎞

⎛m−(j +1)

i + θ + m 2m + 1 − 2j ⎠ ⎝ . i + θ + 1/2 k−j i=0

(m)

So it remains to compute Bk (θ ). After some calculations, it is simplified as

Bk(m) (θ ) =

k  j =0

4j Γ (θ + 1/2)Γ (θ + 2m)Γ (2m + 2 − 2j ) + m)Γ (θ + m + 1/2)Γ (2m + 2 − k)k!

22m Γ (θ

×

(−k)j (k − 2m − 1)j (−m)j (1/2 − θ − m)j (1 − θ − 2m)j j !

.

Since Bk(m) (θ ) is not still in a hypergeometric form, applying the duplication Legendre formula 22z−1 Γ (z) Γ Γ (2z) = √ π

  1 z+ 2

yields 2 Γ (θ + 1/2)Γ (θ + 2m) Γ (m + 3/2) m! Bk(m) = √ Γ (θ + m)Γ (θ + m + 1/2)Γ (2m + 2 − k)k! π ×

k  (−k)j (k − 2m − 1)j (1/2 − θ − m)j 1j (−m − 1/2)j (1 − θ − 2m)j j! j =0

(1.214)

1.11 A Generalization of Fourier Trigonometric Series

107

2 Γ (θ + 1/2)Γ (θ + 2m) Γ (m + 3/2) m! = √ π Γ (θ + m)Γ (θ + m + 1/2)Γ (2m + 2 − k)k!  × 3 F2



−k, k − 2m − 1, 1/2 − θ − m 1 . −m − 1/2, 1 − θ − 2m

(1.215)

The hypergeometric term of (1.215) is a special case of Saalschutz’s theorem [1], which says that if c is a negative integer and a + b + c + 1 = d + e, then  3 F2



(d − a)|c| (d − b)|c| abc . 1 = (d)|c| (d − a − b)|c| de

Therefore, after some calculations, (1.213) is simplified as

−1 0 cos t −1 1 

m  1 (θ − 1)k (−m + 1/2)k 2m + 1 = 2m cos (θ−1) t cos(2m + 1 − 2k) t . 2 (1 − θ − 2m)k (−m − 1/2)k k k=0 

(θ) S2m+1

(θ)

Corollary 1.9 The trigonometric sequence Φn (t) defined by (θ)

Φ2n (t) = (θ) Φ2n+1 (t)

1 cos 2nt , 22n−1

n  1 = 2n cos (θ−1) t 2 k=0



2n + 1 k



(θ − 1)k (−n + 1/2)k cos(2n+1−2k) t , (1 − θ − 2n)k (−n − 1/2)k

satisfies the differential equation (1.211). Clearly Φn(1) (t) leads to the usual cosine sequence. To compute the norm square value of this sequence, we should refer to relations (1.109), (1.168), and (1.171). Hence on substituting (p, q, r, s, θ ) = (−1, 1, −1, 0, θ ) and α = 1 into (1.168), we get 

1 −1

√

1 1 − x2

 Sn(θ)





  π −1 0 −1 0 (θ) (θ) Φn(θ) (t) Φm (t) dt x Sm x dx = −1 1 −1 1 0 

 2[n/2] −1 − 2θ σn , 2θ σn Ci = −1 1 i=1

108

1 Generalized Sturm–Liouville Problems in Continuous Spaces

 ×



1 −1

W



 −1 − 2θ σn , 2θ σn x dx δn,m . −1 1

(1.216)

Now let n → 2n + 1 in (1.216). In this case, the orthogonality (1.216) changes to 

π 0

(θ) (θ) Φ2n+1 (t) Φ2m+1 (t) dt

=

 2n



 Ci

i=1

−1 − 2 θ, 2 θ −1 1



1 −1

2 −1/2

x (1 − x ) 2θ

2n √ Γ (θ + 1/2) (i + 2θ σi )(−i + 1 − 2θ σi ) = π Γ (θ + 1) 4 (i + θ )(i + θ − 1)

dx

δn,m

δn,m ,

i=1

if and only if θ + 1 > 0 and (−1)θ = −1. This result leads to the following final corollary. (a)

Corollary 1.10 The modified trigonometric sequence Cn (x) defined as (a) C2n (x) = cos 2nx , (a) C2n+1 (x)



= cosa x

n   k=0



2n + 1 k



(a)k (−n + 1/2)k cos(2n + 1 − 2k) x , (−a − 2n)k (−n − 1/2)k

satisfies the orthogonality relation



π 0

(a)

(a)

Ci (x) Cj (x) dx =

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ π ⎪ ⎪ ⎪ ⎪ ⎨ π/2

if i = j , if i = j = 0 , if i = j = 2n ,

⎪  2n √  Γ (a + 3/2 + n)Γ (a + 1 + n) ⎪ ⎪ ⎪ 2 (2n)! π ⎪ ⎪ Γ (a + 2 + 2n)Γ (a + 1 + 2n) ⎪ ⎪ ⎪ ⎩ if i = j = 2n + 1 , (1.217)

provided that a > −2 and (−1)a = 1. Note that using (1.214), the last equality of (1.217) is equal to π/2 if a = 0.

1.11.2 Application to Function Expansion Theory Let f (x) be a periodic function satisfying Dirichlet conditions and suppose for convenience that (a) (x) = (cosa x) Dn(a) (x). C2n+1

1.11 A Generalization of Fourier Trigonometric Series

109

According to [2], Courant and Hilbert were probably the first to prove that every sequence of eigenfunctions of a Sturm–Liouville problem is a complete set of orthogonal functions. Hence {Cn(a)(x)}∞ n=0 is a complete set over [0, π], and the periodic function f (x) can be expanded in terms of them as f (x) =

∞

∞

∞

k=1

k=1

k=0

   1 1 bk Ck(a) (x)= b0 + b2k cos 2kx + (cosa x) b2k+1 Dk(a) (x), b0 C0(a) (x)+ 2 2

(1.218) in which b2k =

2 π



π

cos 2kx f (x) dx, 0

Γ (a + 2 + 2k)Γ (a + 1 + 2k) b2k+1 = √ 2k π 2 (2k)! Γ (a + 3/2 + k)Γ (a + 1 + k)

 0

π

(a)

f (x) cosa x Dk (x) dx , (1.219)

and Dk(a) (x) =

k  j =0

=

k  j =0

djk (a) cos(2k + 1 − 2j ) x 

2k + 1 j



(a)j (−k + 1/2)j (−a − 2k)j (−k − 1/2)j

cos(2k + 1 − 2j ) x .

As (1.219) shows, the integral in b2k+1 can be simplified as (a)

Rk



π

= 0

(a)

f (x) cosa x Dk (x) dx =

k  j =0

 djk (a)

π

f (x) cosa x cos(2k + 1 − 2j )x dx.

0

(1.220) One of the interesting choices for a in (1.220), by noting that a > −2 and (−1)a = 1, is an even integer, because in this case, 

m 1  2m 2m cos(2m − 2k) x (m ∈ N), (1.221) cos x = 2m−1 2 k k=0 and therefore we have (2m)

Rk =

1 22m−1

k  j =0

djk (2m)

 m 



 2m  π f (x) cos(2m − 2r)x cos(2k + 1 − 2j )x dx r 0 r=0

110

1 Generalized Sturm–Liouville Problems in Continuous Spaces

=

k 1 

22m

djk (2m)

m 

j =0

r=0

 2m r

 

π

f (x) cos(2m − 2r + 2k + 1 − 2j )x dx

0



π

+

f (x) cos(2m − 2r − 2k − 1 + 2j ) x dx



.

0

Let us consider a practical example. It can be straightforwardly verified that the usual cosine series of the function f (x) = x 2 over [0, π] is as follows: x2 =

∞ ∞ ∞   π2 π2  1 (−1)k 1 +4 + cos kx = cos 2kx − 4 cos(2k + 1)x. 3 k2 3 k2 (2k + 1)2 k=1

k=1

k=0

(1.222) Now, to derive the generalized cosine series of type (1.218), if we suppose a = 2m, then we get Rk(2m) = k 

djk (2m)

−π × 22m−1

 m   2m 

j =0

(2m − 2r+2k + 1 − 2j )2

r

r=0

1

+



1 (2m − 2r − (2k+1 − 2j ))2

.

Hence the corresponding odd coefficients take the form (m)

b2k+1 = −22m+2

(2m + 2k + 1)! (2m + 2k)! (4m + 2k + 1)! (2k)! 

k  (2m)j (−k + 1/2)j 2k + 1 × S (m) (j, k), (−2m − 2k)j (−k − 1/2)j j j =0

where S

(m)

(j, k) =

m  r=0

 2m r



1 (2m − 2r + (2k + 1 − 2j ))2

+



1 (2m − 2r − (2k + 1 − 2j ))2

.

This gives the desired series as x2 =

∞ ∞  π2  1 (m) (2m) 2m + cos 2kx + (cos x) b2k+1 Dk (x), 3 k2 k=1

k=0

(1.223)

1.11 A Generalization of Fourier Trigonometric Series

111

which is a generalization of (1.222) for m = 0, because Dk(0) (x) = cos(2k + 1)x

(0) and b2k+1 /2 = −2 S (0) (0, k) = −4/(2k + 1)2 . (0)

In this sense, note that the identity (1.221) is not valid for m = 0, and b2k+1 /2 should be (0) considered in (1.223) not b2k+1 , though for the rest of the values of m, (1.223) is valid. For instance, if m = 1, then

(1) b2k+1

k (k + 1) (2k + 1)  = −16 (k + 2) (2k + 5)



j =0

2k + 1 j



(2)j (−k + 1/2)j (−2 − 2k)j (−k − 1/2)j

S (1) (j, k),

where S (1) (j, k) =

1 (2k + 3 − 2j )

2

+

4 (2k + 1 − 2j )

2

+

1 (2k − 1 − 2j )2

.

There exists a similar case for a generalization of the Fourier sine series such that if we replace the initial data (p, q, r, s, θ ) = (−1, 1, −3, 0, θ ) in the main orthogonality relation (1.168) for α = 1 as 

1

−1

%





 −3 0 −3 0 (θ) x Sm x dx −1 1 −1 1  



 π −3 0 −3 0 2 (θ) (θ) (sin t) Sn = cos t dt cos t Sm −1 1 −1 1 0 

 2[n/2] −3 − 2θ σn , 2θ σn Ci = −1 1 i=1

  1  −3 − 2θ σn , 2θ σn W × x dx δn,m , −1 1 −1 

1 − x 2 Sn(θ)

then by defining the trigonometric sequence  (a) Sn+1 (x)

=2

n+1

sin x

Sn(a+1)



−3 0 cos x , −1 1

112

1 Generalized Sturm–Liouville Problems in Continuous Spaces

we can verify (for example) that  (a) (x) S2n+1

1.12

=2

2n+1

sin x

(a+1) S2n



−3 0 cos x = sin(2n + 1)x. −1 1

Another Extension for Trigonometric Orthogonal Sequences

As we pointed out in Sect. 1.10, if A(x) = x 2 (px 2 + q) , B(x) = x(rx 2 + s) , C(x) = x 2 > 0 , D(x) = −α (s + (α − 1) q) , E(x) = (α − β)s + (α(α − 1) − β(β − 1))q , and = (n + α + (β − α − 1) σn ) (r + (n − 1 + α + (β − α − 1) σn )p) λ(α,β) n are substituted into Eq. (1.94), we obtain the second-order equation   x 2 + (−1)n c∗ + d ∗ Φn (x) = 0, x 2 (px 2 + q) Φn (x) + x(rx 2 + s) Φn (x) − λ(α,β) n (1.224) where 2c∗ = (α − β)s + (α(α − 1) − β(β − 1))q and 2d ∗ = (α + β)s + (α(α − 1) + β(β − 1))q. (a)

To introduce a sequence that generalizes Cn (x), we first suppose in Eq. (1.224) that p = −1 and q = 1. Then by the change of variable x = cos t, we obtain

2

cos t

Φn (t)

 +

 −(r + 1)cos3 t − s cos t Φn (t) sin t   + λ(α,β) cos2 t + (−1)n c∗ + d ∗ Φn (t) = 0. n

(1.225)

1.12 Another Extension for Trigonometric Orthogonal Sequences

113

If r + 1 = 0 and s = 0 in (1.225), an extension of Eq. (1.211) is derived for α = 0. Consequently, by substituting the initial vector (p, q, r, s, α, β) = (−1, 1, −1, 0, α, β) into Eq. (1.224) and applying the change of variable x = cos t, we get Φn (t)   (α(α − 1) − β(β − 1)) σn − α(α − 1) + (n + α + (β − α − 1) σn )2 + Φn (t) = 0. cos2 t (1.226) According to (1.190), Eq. (1.226) has a basis solution as  Φn (t) =

Sn(α,β)

−1 0 −1 1



cos t

(α + (β − α) σn + 1/2)[n/2] (cos t)α+(β−α−1) σn (α + (β − α − 1/2) σn + n/2)[n/2] ⎞ 

⎛[n/2]−(k+1) [n/2]  2j + n + 2α + (2β − 2α − 1) σ [n/2] n ⎠ cosn−2k t , ⎝ × (−1)k 2j + 1 + 2α + 2 (β − α) σn k =

j =0

k=0

which can be represented as  Sn(α,β)



−1 0 cos t = −1 1

(cos t)n+α+(β−α−1) σn 2 F1





−[n/2], −[n/2] + 1/2 − α + (α − β)σn 1 . cos2 t −n + 1 − α + (α − β + 1)σn (1.227)

The above representation can be further simplified. For this purpose, if we take Am k (β) =

(−m)k (−m − β + 1/2)k , (−2m − β + 1)k k!

then (1.227) changes to  (α,β) S2m



m  −1 0 2m−2k Am t cos t = cosα t k (α) cos −1 1 k=0

(1.228)

114

1 Generalized Sturm–Liouville Problems in Continuous Spaces

and  (α,β)

S2m+1



m  −1 0 2m+1−2k A(m) t. cos t = cos(β−1) t k (β) cos −1 1 k=0

(1.229)

Since in general, 2n+1

cos

 n  1  2n + 1 cos(2n + 1 − 2k)x, x = 2n k 2 k=0

the sum in (1.229) is simplified as m 

(m)

Ak (β)cos2m+1−2k t =

k=0

m 

(m)

Am−k (β)cos2k+1 t

k=0

=

m  k=0

⎛

k 1  (m) Am−k (β) ⎝ 2k 2 j =0



2k + 1 j

⎞



cos(2k + 1 − 2j )t ⎠

=

m 

Bk(m) (β) cos(2m + 1 − 2k)t,

k=0

where (m) Bk (β)

=

k A(m) (β)  j j =0

=

22(m−j )



2m + 1 − 2j k−j



k  (−m)j (−m − β + 1/2)j j =0

22(m−j ) (−2m − β + 1)j j !



2m + 1 − 2j , k−j

which is equivalent to k 2 m! Γ (m + 3/2)  (−m − β + 1/2)j (−k)j (k − 2m − 1)j Bk(m) (β) = √ (−m − 1/2)j (−2m − β + 1)j j ! π Γ (2m + 2 − k) k! j =0

 2 m! Γ (m + 3/2) −k, k − 2m − 1, 1/2 − β − m F = √ (1.230) 1 . 3 2 π Γ (2m + 2 − k) k! −m − 1/2, 1 − β − 2m

1.12 Another Extension for Trigonometric Orthogonal Sequences

115

Therefore, (1.229) reads as  (α,β) S2m+1



−1 0 cos t −1 1

m cos(β−1) t  = 22m k=0



2m + 1 k



(β − 1)k (−m + 1/2)k cos(2m + 1 − 2k)t. (1 − β − 2m)k (−m − 1/2)k

Similarly, this method can be applied to the function (1.228), so that by noting the general identity 1 cos x = 2n 2



2n

2

n  k=0



2n k

 cos(2n − 2k) x −



2n n

,

the sum in (1.228) is simplified as m 

(m) Ak (α)cos2m−2k t

k=0

=

m  k=0

=

m  k=0

=

1 (m) Am−k (α) 2k 2 m 

(m)

Am−k (α)cos2k t ⎛ ⎝2

k  j =0



2k j

Ck(m) (α) cos(2m − 2k)t

k=0

 cos(2k − 2j )t −

−

m  k=0

1 A(m) m−k (α) 2k 2

⎞ 2k ⎠ k 

2k k

,

in which Ck(m) (α)

=

(m) k  Aj (α) j =0

22(m−j )−1



2m − 2j k−j



 2 m! Γ (m + 1/2) −k k − 2m 1/2 − α − m = √ 3 F2 π Γ (2m + 1 − k) k! −m + 1/2 1 − α − 2m 

1 (α)k 2m = 2m−1 , 2 (1 − α − 2m)k k



1

(1.231)

116

1 Generalized Sturm–Liouville Problems in Continuous Spaces

where we have once again used the Saalschutz identity. Therefore, (1.228) is simplified as

−1 0 (α,β) S2m cos t −1 1  m 





1 2m (α)k (α)m cosα t  2m . cos(2m − 2k)t − = 2m−1 2 (1 − α − 2m)k 2 m (1 − α − 2m)m k k=0 

Corollary 1.11 The trigonometric sequence Φn (t; α, β) = (cos t)α+(β−1−α)σn × ⎛ 

[n/2]  n (α + (β − 1 − α)σn )k (−[n/2] + 1/2)k cos(n − 2k)t 1 ⎝   n 2n−1 k −[n/2] + (−1) (1 − α − n + (α − β + 1)σ ) /2 n k k k=0 −

(1/2)[n/2] (α)[n/2] [n/2]! (1 − α − n)[n/2]



σn+1

satisfies the differential equation (1.226). To compute the norm square value of Φn (t; α, β), it is enough to refer to relations (1.194) and (1.109) and substitute (p, q, r, s, α, β) = (−1, 1, −1, 0, α, β) and v = 1 into (1.194) to get 

1



1

√ Sn(α,β) 1 − x2 −1





 −1 0 −1 0 (α,β) x Sm x dx −1 1 −1 1 

π

=

Φn (t; α, β) Φm (t; α, β) dt 0

=

2[n/2]

⎛ Cj ⎝

j =1

−1 − 2 (α + (β − 1 − α)σn ) , 2 (α + (β − 1 − α)σn ) −1

⎞ ⎠

1

⎞ −1 − 2 (α + (β − 1 − α)σn ) , 2 (α + (β − 1 − α)σn ) x ⎠ dx δn,m . × W⎝ −1 −1 1 

⎛

1

(1.232)

1.12 Another Extension for Trigonometric Orthogonal Sequences

117

Let n → 2n. Then (1.232) changes to 

π

Φ2n (t; α, β) Φ2m (t; α, β) dt 0

⎛ =⎝

2n

 Cj

j =1

−2 α − 1, 2 α −1 1

⎞



1 −1

2 −1/2

x 2α (1 − x )

dx ⎠ δn,m

⎞ 2n √ Γ (α + 1/2) (j + 2α σj )(j + 2α σj − 1) ⎠ δn,m =⎝ π Γ (α + 1) 4 (j + α)(j + α − 1) ⎛

j =1

if and only if α + 1/2 > 0 and (−1)2α = 1. Similarly, for n → 2n + 1 in (1.232), we obtain 

π

Φ2n+1 (t; α, β) Φ2m+1 (t; α, β) dt 0

⎛ =⎝

2n j =1

 Cj

−2β + 1 , 2β − 2 −1 1



⎞ 1 −1

2 −1/2

x 2β−2 (1 − x )

dx ⎠ δn,m

⎞ ⎛ 2n √ Γ (β − 1/2) (j + 2(β − 1) σj )(j + 2(β − 1)σj − 1) ⎠ δn,m =⎝ π Γ (β) 4 (j + β − 1)(j + β − 2) j =1

if and only if β − 1/2 > 0 and (−1)2β = 1. Corollary 1.12 The modified trigonometric sequence Φ¯ 2m (t; α, β) = cosα t 





m  1 2m (α)k (α)m 2m × − + cos(2m − 2k) t , 2 m (1 − α − 2m)m (1 − α − 2m)k k k=0 Φ¯ 2m+1 (t; α, β) = cosβ−1 t 

m  (β − 1)k (−m + 1/2)k 2m + 1 × cos(2m + 1 − 2k) t , (1 − β − 2m)k (−m − 1/2)k k k=0

118

1 Generalized Sturm–Liouville Problems in Continuous Spaces

in which Φ¯ n (t; 0, 1) = cos nt, satisfies the orthogonality relation



π

Φ¯ i (t; α, β)Φ¯ j (t; α, β)dt =

0

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

π Γ (2n + 2α)Γ (2n + 1) 2α+1 Γ (2n + α)Γ (2n + α + 1) 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ π Γ (2n + 2β − 2)Γ (2n + 1) ⎪ ⎪ ⎩ 22β−1 Γ (2n + β − 1)Γ (2n + β)

if i = j, if i = j = 2n , if i=j =2n + 1,

if and only if α + 1/2 > 0, β − 1/2 > 0, (−1)2α = 1, and (−1)2β = 1.

1.13

Incomplete Symmetric Orthogonal Polynomials

In the classical case, systems of orthogonal polynomials are such that the nth polynomial has the exact degree n. Such systems are most often complete and form a basis of the space of arbitrary polynomials. In this section, we introduce some incomplete sets of orthogonal polynomials that do not contain all degrees but are solutions of some symmetric generalized Sturm–Liouville problems. Although such polynomials do not possess all the properties as in the classical cases, they can nevertheless be applied to approximation of functions, since we will compute their explicit norm square values.

1.13.1 Incomplete Symmetric Orthogonal Polynomials of Jacobi Type Here we introduce incomplete symmetric orthogonal polynomials of Jacobi type as specific solutions of a generalized Sturm–Liouville equation. For this purpose, we first consider the shifted Jacobi polynomials on [0, 1] as 



n  n + α + β + k n + α (α,β) xk, (−1)k Pn,+ (x) = k n − k k=0 which satisfy the equation x(1 − x)y  − ((α + β + 2)x − (α+1)) y  + n(n + α + β + 1)y = 0 with y = Pn,+ (x), (1.233) (α,β)

and the orthogonality relation 

1 0

(α,β)

(α,β)

x α (1 − x)β Pn,+ (x)Pm,+ (x) dx =

Γ (n + α + 1)Γ (n + β + 1) δn,m . (2n + α + β + 1) Γ (n + 1) Γ (n + α + β + 1)

(1.234)

1.13 Incomplete Symmetric Orthogonal Polynomials

119

By referring to the main Theorem 1.2 of Sect. 1.6 and noting the Jacobi differential equation (1.233), a differential equation of type (1.94) can be constructed whose solutions are orthogonal with respect to an even weight function, say |x|2a (1 − x 2m )b on the symmetric interval [−1, 1]. Hence, suppose in the generic equation (1.94) that A(x) = x 2 (1 − x 2m ),   B(x) = −2x (a + mb + 1)x 2m − a + m − 1 , C(x) = x 2m > 0, D(x) = −2s(2s + 2a − 2m + 1), E(x) = 2s(2s + 2a − 2m + 1) − 2(2r + 1)(r + a − m + 1), and λn = (mn + 2s + (2r + 1 − m − 2s)σn )(mn + 2s + 2a + 1 + 2mb + (2r + 1 − m − 2s)σn ), n

where σn = 1−(−1) and a, b, m, r, s are free parameters. 2 In this case, the following differential equation appears:   x 2 (1 − x 2m ) Φn (x) − 2x (a + mb + 1)x 2m − a + m − 1 Φn (x) + (αn x 2m + β + σn γ ) Φn (x) = 0,

(1.235)

in which β = −2s(2s + 2a − 2m + 1), γ = 2s(2s + 2a − 2m + 1) − 2(2r + 1)(r + a − m + 1) and   αn = mn + 2s + (r − s + (m − 1)/2)(1 − (−1)n )   × mn + 2s + 2a + 1 + 2mb + (r − s + (m − 1)/2)(1 − (−1)n ) . We prove that Eq. (1.235) has a polynomial solution of the form ( a+1−m+s+r +(−1)n 2s−2r−1 , b) 2m

Φn(r,s) (x; a, b, m) = (x 2s + (x 2r+1 − x 2s ) σn ) P[n/2],+m

(x 2m ) , (1.236)

which satisfies the orthogonality relation 

1 −1

b

(r,s)

x 2a (1 − x 2m ) Φn(r,s)(x; a, b, m)Φk  =

1

−1

x 2a (1 − x 2m )

(x; a, b, m) dx

b



2  Φn(r,s)(x; a, b, m) dx δn,k .

(1.237)

120

1 Generalized Sturm–Liouville Problems in Continuous Spaces

To prove this claim, we first substitute (α,β)

g(x) = x λ Pn,+ (x θ ) for λ , θ ∈ R into Eq. (1.233) to obtain the differential equation    x 2 (1 − x θ ) g  (x) + x 2λ − (α + β + 1) θ − 1 x θ − 2λ + αθ + 1 g  (x)   + (θ 2 n(n + α + β + 1) + (α + β + 1) θ λ − λ2 ) x θ + λ2 − αθ λ g(x) = 0 . (1.238) For convenience, let α=

q + 2λ − 1 θ

and

β =−

p+q −1 θ

in Eq. (1.238), which changes to   x 2 (1 − x θ ) g  (x) + x p x θ + q g  (x)   + (θ n + λ)(θ n + λ − p − 1) x θ − λ(λ + q − 1) g(x) = 0.

(1.239)

Now, by noting Theorem 1.2, we define the following odd and even polynomial sequences: ( q+4s−1 ,− p+q 2m −1)

Φ2n (x) = x 2s Pn,+2m

(x 2m ) for λ = 2s , s ∈ Z+ and θ = 2m , m ∈ N,

and ( q+4r+1 ,− p+q 2m −1)

Φ2n+1 (x) = x 2r+1 Pn,+2m

(x 2m ) for λ = 2r + 1 , r ∈ Z+ and θ = 2m , m ∈ N. (1.240)

Since in general, u−w u+w + (−1)n = u + (w − u) σn = 2 2



u if n = 2k, w if n = 2k + 1,

the polynomial sequence Φn (x) defined in (1.240) can be written in a unique form as  ( q+2s+2r+(−1)n (2s−2r−1) ,− p+q  2m −1) (x 2m ) . Φn (x) = x 2s + (x 2r+1 − x 2s ) σn P[n/2],+ 2m

1.13 Incomplete Symmetric Orthogonal Polynomials

121

According to definitions (1.240) and differential equation (1.239), Φ2n (x) should satisfy the equation     (x) + x px 2m + q Φ2n (x) x 2 (1 − x 2m ) Φ2n   + (2m n + 2s)(2m n + 2s − p − 1) x 2m − 2s(2s + q − 1) Φ2n (x) = 0 , and Φ2n+1 (x) should satisfy     x 2 (1 − x 2m ) Φ2n+1 (x) + x px 2m + q Φ2n+1 (x)   + (2mn + 2r + 1)(2mn + 2r − p) x 2m − (2r + 1)(2r + q) Φ2n+1 (x) = 0 . By combining these two equations, we finally obtain   x 2 (1 − x 2m ) Φn (x) + x px 2m + q Φn (x)+  (mn + 2s + (2r + 1 − m − 2s) σn )(mn + 2s − p − 1 + (2r + 1 − m − 2s) σn ) x 2m  − 2s(2s + q − 1) − ((2r + 1)(2r + q) − 2s(2s + q − 1)) σn Φn (x) = 0 , (1.241) which is a special case of the generalized Sturm–Liouville equation (1.94). Also, the weight function corresponding to (1.241) takes the form  W (x) = x 2m exp(

x(px 2m + q) − (2x − (2m + 2)x 2m+1) dx) x 2 (1 − x 2m ) = K x 2m+q−2 (1 − x 2m )−

p+q 2m −1

.

(1.242)

Without loss of generality, we can assume that K = 1, and since W (x) must be positive, the weight function (1.242) can be considered as W (x) = |x| 2a (1 − x 2m )b , for 2a = 2m + q − 2 and b = −1 − (p + q)/2m. To compute the norm square value of (1.237) we directly use the orthogonality relation (1.234), so that for n = 2j we have  N2j =

1 −1

 2 b (r,s) x 2a (1 − x 2m ) Φ2j (x; a, b, m) dx

122

1 Generalized Sturm–Liouville Problems in Continuous Spaces

 =

1 −1



1 = m =

x 2a+4s (1 − x 2m ) 1

t

2a+4s+1−2m 2m

0

b

 2a+4s+1−2m 2 ( ,b) 2m Pj,+ 2m (x ) dx

2  2a+4s+1−2m ( ,b) 2m (1 − t) Pj,+ (t) dt b

Γ (j + (2mj + a + 2s +

2a+4s+1 ) Γ (j 2m

+ b + 1)

1 2

+ mb)Γ (j + 1) Γ (j +

b



2a+4s+1 2m

+ b)

(1.243)

,

and for n = 2j + 1 we get  N2j +1 =

−1

 =

1

−1

1 = m =

1

x 2a (1 − x 2m )

2 (r,s) Φ2j (x; a, b, m) dx +1

x 2a+4r+2(1 − x 2m )



1

t

2a+4r+3−2m 2m

b

 

(1 − t)

0

Γ (j + (2mj + a + 2r +

3 2

( 2a+4r+3−2m , b) 2m

Pj,+ b

2 (x 2m )

( 2a+4r+3−2m ,b) (t) Pj ,+ 2m

2a+4r+3 ) Γ (j 2m

dx

2 dt

+ 1 + b)

+ mb)Γ (j + 1) Γ (j +

2a+4r+3 2m

+ b)

(1.244)

.

By combining both relations (1.243) and (1.244), the following value is derived: Nn = 2a+4s+1 + 2r+1−2s σ ) Γ ( n−σn + b + 1) n Γ ( n−σ n m 2 + 2m 2 . n−σn 1 2a+4s+1 +b+ 2r+1−2s σ ) n (m(n−σn )+a+2s+ 2 +mb+(2r+1−2s)σn ) Γ ( n−σ +1) Γ ( n m 2 2 + 2m

This value shows that the orthogonality (1.237) is valid if and only if b > −1, 2a+4s+1 > 0, 2a+4s+1+2mb > 0, m ∈ N, 2a+4r +3 > 0, 2a+4r +3+2mb > 0,

and (−1)2a = 1. Example 1.7 Find the standard properties√of incomplete symmetric polynomials orthogonal with respect to the weight function x 4 1 − x 4 on [−1, 1]. To solve the problem, it is sufficient in (1.236) to choose m = a = 2 and b = 12 to get the polynomials   n 2s−2r−1 , 1 )   1 ( 1+s+r 2 +(−1) 4 2 (x 4 ) Φn(r,s) x; 2, , 2 = x 2s + (x 2r+1 − x 2s σn ) P[n/2],+ 2

r, s ∈ Z+ , (1.245)

1.13 Incomplete Symmetric Orthogonal Polynomials

123

that satisfy the differential equation   x 2 (1 − x 4 ) Φn (x) + 2x 1 − 4x 4 Φn (x)+  (2n + 2s + (2r − 1 − 2s) σn )(2n + 2s + 7 + (2r − 1 − 2s) σn ) x 4  − 2s(2s + 1) + (2s(2s + 1) − 2(2r + 1)(r + 1)) σn Φn (x) = 0 and the orthogonality relation 

1 −1

    % 1 1 x 4 1 − x 4 Φn(r,s) x; 2, , 2 Φk(r,s) x; 2, , 2 dx = 2 2 n Γ ( n−σ 2 +s+

(2(n − σn ) +

7 2

+ 2s + (2r − 2s +

5 4

+

2r−2s+1 2 n 1) σn ) Γ ( n−σ 2

σn ) Γ ( n−σ2n +3 ) n + 1) Γ ( n−σ 2 +s+

7 4

+

2r−2s+1 2

σn )

δn,k .

As (1.245) shows, Φn(r,s) (x; 2, 1/2, 2) are incomplete symmetric polynomials with degrees respectively {2s, 2r + 1, 2s + 4, 2r + 5, 2s + 8, 2r + 9, . . .}. For instance, we have degrees

of Φn(1,0)

  1 x; 2, , 2 = {0, 3, 4, 7, 8, 11, . . .}, 2

  1 degrees of Φn(2,0) x; 2, , 2 = {0, 5, 4, 9, 8, 13, . . .}, 2 and degrees of

Φn(1,1)

  1 x; 2, , 2 = {2, 3, 6, 7, 10, 11, . . .}. 2

Example 1.8 If m = 1 is considered in (1.236), a generalization of generalized ultraspherical polynomials as (a+s+r+(−1)n (s−r− 21 ) , b)

Φn(r,s)(x; a, b, 1) = (x 2s + (x 2r+1 − x 2s ) σn ) P[n/2],+

(x 2 )

(1.246)

is derived for r = s = 0 that satisfies the differential equation   x 2 (1 − x 2 ) Φn (x) − 2x (a + b + 1) x 2 − a Φn (x)+  (n + 2s + (2r − 2s)σn )(n + 2s + 2a + 1 + 2b + (2r − 2s) σn ) x 2  − 2s(2s + 2a − 1) + (2s(2s + 2a − 1) − 2(2r + 1)(r + a)) σn Φn (x) = 0

124

1 Generalized Sturm–Liouville Problems in Continuous Spaces

and has an orthogonality relation of the form 

1 −1

b

x 2a (1 − x 2 ) Φn(r,s) (x; a, b, 1) Φk(r,s) (x; a, b, 1) dx = δn,k × n + (2r + 1 − 2s) σn ) Γ ( n−σ Γ ( n−σn +2a+4s+1 2 2 + b + 1)

(n−σn +a+2s+ 12

n−σn +2a+4s+1 n + b + (2r + 1 − 2s) σn )Γ ( n−σ + b + (2r+1−2s) σn ) 2 +1)Γ ( 2

.

(r,s)

As (1.246) shows, Φn (x; a, b, 1) are incomplete symmetric polynomials with degrees respectively D (r,s) = {2s, 2r + 1, 2s + 2, 2r + 3, 2s + 4, 2r + 5, . . .} though it is complete for r = s = 0, because in this case we have D (0,0) = {0, 1, 2, 3, . . .}.

1.13.2 Incomplete Symmetric Orthogonal Polynomials of Laguerre Type In this part, we introduce incomplete symmetric orthogonal polynomials of Laguerre type as specific solutions of a generalized Sturm–Liouville equation. First, we suppose in the generic equation (1.94) that A(x) = x 2 ,   B(x) = −2x m x 2m − a + m − 1 , C(x) = x 2m > 0, D(x) = −2s(2s + 2a − 2m + 1), E(x) = 2s(2s + 2a − 2m + 1) − 2(2r + 1)(r + a − m + 1), and λn = 2m (mn + 2s + (2r + 1 − m − 2s) σn ), where a, m, r, s are free parameters. Replacing these options in (1.94) leads to the differential equation x

2

Φn (x)−2x

 mx

2m

   1 − (−1)n  2m γ Φn (x) = 0, − a + m − 1 Φn (x)+ αn x + β + 2 (1.247)

in which β = −2s(2s + 2a − 2m + 1), γ = 2s(2s + 2a − 2m + 1) − 2(2r + 1)(r + a − m + 1) and   αn = 2m mn + 2s + (r − s + (m − 1)/2)(1 − (−1)n ) .

1.13 Incomplete Symmetric Orthogonal Polynomials

125

Once again, we prove that Eq. (1.247) has a polynomial solution as ( a+1−m+s+r +(−1)n 2s−2r−1 ) 2m

Φn(r,s)(x; a, m) = (x 2s + (x 2r+1 − x 2s ) σn ) L [n/2] m

(x 2m ) ,

(1.248)

which satisfies the orthogonality relation 

∞ −∞

x 2a exp(−x 2m )Φn(r,s)(x; a, m)Φk(r,s)(x; a, m) dx  =

∞ −∞

x

2a

exp(−x

2m

 2  (r,s) ) Φn (x; a, m) dx δn,k .

(1.249)

For this purpose, we first substitute θ g(x) = x λ L(α) n (x ) for λ , θ ∈ R

into Eq. (1.40) to obtain the differential equation     x 2 g  (x) + x −2λ + αθ + 1 − θ x θ g  (x) + θ (n θ + λ) x θ + λ(λ − αθ ) g(x) = 0 . (1.250) If −2λ + αθ + 1 = q,

or equivalently

α=

q + 2λ − 1 , θ

Equation (1.250) is transformed to     x 2 g  (x) + x q − θ x θ g  (x) + θ (θ n + λ) x θ − λ(λ + q − 1) g(x) = 0.

(1.251)

By referring to Theorem 1.2, we now define the following odd and even polynomial sequences: ( q+4s−1 2m )

Φ2n (x) = x 2s Ln

(x 2m )

for λ = 2s , s ∈ Z+ , and θ = 2m , m ∈ N,

and ( q+4r+1 2m )

Φ2n+1 (x) = x 2r+1 Ln

(x 2m ) for λ = 2r + 1 , r ∈ Z+ , and θ = 2m , m ∈ N, (1.252)

which can be written in the unique form n (2s−2r−1)

( q+2s+2r+(−1) 2m

Φn (x) = (x 2s + (x 2r+1 − x 2s ) σn ) L [n/2]

)

(x 2m ) .

126

1 Generalized Sturm–Liouville Problems in Continuous Spaces

According to the two definitions of (1.252) and Eq. (1.251), Φ2n (x) should satisfy the equation       (x)−x 2m x 2m −q Φ2n (x)+ 4m (m n + s) x 2m − 2s(2s + q − 1) Φ2n (x) = 0 , x 2 Φ2n while Φ2n+1 (x) should satisfy     x 2 Φ2n+1 (x) − x 2m x 2m − q Φ2n+1 (x)   + 2m (2mn + 2r + 1) x 2m − (2r + 1)(2r + q) Φ2n+1 (x) = 0 . Combining these two equations finally gives    x 2 Φn (x) − x 2m x 2m − q Φn (x) + 2m (mn + 2s + (2r + 1 − m − 2s) σn ) x 2m  − 2s(2s + q − 1) − ((2r + 1)(2r + q) − 2s(2s + q − 1)) σn Φn (x) = 0 , (1.253) which is a special case of the generalized Sturm–Liouville equation (1.94). The weight function corresponding to Eq. (1.253) takes the form  W (x) = x 2m exp(

−x(2m x 2m − q) − 2x dx) = K x 2m+q−2 exp(−x 2m ), x2

with K = 1 without loss of generality. Moreover, since W (x) is positive, it can be considered as |x| 2a exp(−x 2m ) for 2a = 2m + q − 2. To compute the norm square value of (1.249) we use the Laguerre orthogonality relation in (1.41), so that for n = 2j we have  N2j =  =

∞ −∞ ∞ −∞

1 = m



 2 (r,s) x 2a exp(−x 2m ) Φ2j (x; a, m) dx  2a+4s+1−2m 2 ( ) 2m x 2a+4s exp(−x 2m ) L j 2m (x ) dx ∞

t 0

2a+4s+1−2m 2m

 2a+4s+1−2m 2 ( ) 2m Lj e (t) dt = −t

and for n = 2j + 1 we get  N2j +1 =

∞ −∞

 2 (r,s) x 2a exp(−x 2m ) Φ2j (x; a, m) dx +1

1 Γ m j!

  2a + 4s + 1 j+ , 2m (1.254)

1.13 Incomplete Symmetric Orthogonal Polynomials

 =

∞ −∞

1 = m



127

 2a+4r+3−2m 2 ( ) 2m x 2a+4r+2 exp(−x 2m ) L j 2m (x ) dx ∞

t

2a+4r+3−2m 2m

0

2  2a+4r+3−2m ( ) 2m e (t) dt = Lj −t

1 Γ m j!

  2a + 4r + 3 j+ . 2m (1.255)

Combining relations (1.254) and (1.255) finally gives Nn =

1 Γ m ((n − σn )/2) !



 n − σn 2a + 4s + 1 2r + 1 − 2s + + σn , 2 2m m

which shows that the orthogonality (1.249) is valid if and only if 2a + 4s + 1 > 0, 2a + 4r + 3 > 0

and (−1)2a = 1.

Example 1.9 Find the standard properties of incomplete symmetric polynomials orthogonal with respect to the weight function x 4 exp(−x 4 ) on (−∞, ∞). To find the solution, it is sufficient in (1.248) to choose m = a = 2 to get the polynomials ( 1+s+r +(−1)n 2s−2r−1 ) 4

2 Φn(r,s) (x; 2, 2) = (x 2s + (x 2r+1 − x 2s ) σn ) L [n/2]

(x 4 )

r, s ∈ Z+ , (1.256)

which satisfy the equation    x 2 Φn (x) − 2x 2x 4 − 1 Φn (x) + 4 (2n + 2s + (2r − 1 − 2s) σn ) x 4  − 2s(2s + 1) + (2s(2s + 1) − 2(2r + 1)(r + 1)) σn Φn (x) = 0 and the orthogonality relation 

∞ −∞

x 4 exp(−x 4 ) Φn(r,s)(x; 2, 2) Φk(r,s)(x; 2, 2) dx =

1 Γ 2 ((n − σn )/2)!



n − σn 5 2r − 2s + 1 +s+ + σn 2 4 2

(r,s)

 δn,k .

As (1.256) shows, Φn (x; 2, 2) are incomplete symmetric polynomials with degrees respectively {2s, 2r + 1, 2s + 4, 2r + 5, 2s + 8, 2r + 9, . . .}. For instance, we have degrees of Φn(1,0) (x; 2, 2) = {0, 3, 4, 7, 8, 11, . . .}, degrees of Φn(2,0) (x; 2, 2) = {0, 5, 4, 9, 8, 13, . . .},

128

1 Generalized Sturm–Liouville Problems in Continuous Spaces

and degrees of Φn(1,1)(x; 2, 2) = {2, 3, 6, 7, 10, 11, . . .}. Example 1.10 If we set m = 1 in (1.248), a generalization of generalized Hermite polynomials as (a+s+r+(−1)n (s−r− 21 ))

Φn(r,s) (x; a, 1) = (x 2s + (x 2r+1 − x 2s ) σn ) L [n/2]

(x 2 )

(1.257)

is derived for r = s = 0 that satisfies the equation    x 2 Φn (x) − 2x x 2 − a Φn (x) + 2 (n + 2s + (2r − 2s)σn ) x 2 − 2s(2s + 2a − 1)  + (2s(2s + 2a − 1) − 2(2r + 1)(r + a)) σn Φn (x) = 0 and the orthogonality relation 

∞ −∞

(r,s)

x 2a exp(−x 2 ) Φn(r,s)(x; a, 1) Φk

=

(x; a, 1) dx + (2r + 1 − 2s) σn ) Γ ( n−σn +2a+4s+1 2 δn,k . ((n − σn )/2)!

As (1.257) shows, Φn(r,s)(x; a, 1) are incomplete symmetric polynomials with degrees respectively D (r,s) = {2s, 2r + 1, 2s + 2, 2r + 3, 2s + 4, 2r + 5, . . .} though they are complete for r = s = 0, because in this case we have D (0,0) = {0, 1, 2, 3, . . .}. Remark 1.8 For the finite sequences of hypergeometric orthogonal polynomials, we can also construct their incomplete type just as in the previous two sections. The weight functions corresponding to the finite type of incomplete symmetric orthogonal polynomials are clearly symmetric on the real line and can be defined as W1 (x; m, p) = x

−2p

  1 exp − 2m , x

x ∈ (−∞, ∞),

and W2 (x; m, p, q) =

x 2q , (1 + x 2m )p

x ∈ (−∞, ∞).

1.14 A Class of Hypergeometric-Type Orthogonal Functions

1.14

129

A Class of Hypergeometric-Type Orthogonal Functions

Here we introduce a sequence of hypergeometric series that satisfies a specific type of Sturm–Liouville equation. For this purpose, by recalling that the Gauss hypergeometric function

 ∞  (a)k (b)k zk a, b (1.258) y(z) = 2 F1 z = (c)k k! c k=0

satisfies the differential equation   z(z − 1)y  + (a + b + 1)z − c y  + aby = 0,

(1.259)

we first define the following hypergeometric function:   y(x) = y x; a(x), b(x), p, q, r = a(x) 2F1





p + iq, p − iq b(x) , r

in which a(x) and b(x) are two predetermined functions, p, q, r ∈ R, and i = Since (p + iq)k (p − iq)k =

k−1



(1.260) √

−1.

 (p + j )2 + q 2 ≥ 0

j =0

is always a positive real value, the function (1.260) is real, so that we have k ∞  k−1     2 2 b(x) . y x; a(x), b(x), p, q, r = a(x) (p + j ) + q (r)k k! k=0

(1.261)

j =0

To obtain the differential equation of the function (1.260), we can use relations (1.258) and (1.259) such that if z = b(x), a = p + iq, b = p − iq, and c = r are replaced in (1.259), then     b(x) b(x) − 1 d 2 2 F1 p + iq, p − iq, r ; b(x)  2 2 dx b (x) +

  b(x)(b(x) − 1)  d p + iq, p − iq, r ; b(x) F 2 1   3 b (x) dx b  (x)   + (p2 + q 2 ) 2 F1 p + iq, p − iq, r ; b(x) = 0. (1.262)  (2p + 1)b(x) − r

−

130

1 Generalized Sturm–Liouville Problems in Continuous Spaces

On the other hand, relation (1.260) shows that  2 F1



  y x; a(x), b(x), p, q, r p + iq, p − iq . b(x) = a(x) r

(1.263)

Therefore, by substituting (1.263) into (1.262), we finally obtain   b(b − 1)  (2p + 1)b − r   a b  − b y y − + 2 a b b2 b3 

  b(b − 1) a  b − a  b (2p + 1)b − r a  a 2 b 2 2 y = 0, + + q − + 2 + p a b a a2 b3

b(b − 1)



which is a particular case of the Sturm–Liouville equation   A(x)yq (x) + B(x)yq (x) + λq + C(x) yq (x) = 0,

(1.264)

for   b(x) b(x) − 1 A(x) = , 2  b (x)

  b(x) b(x) − 1   (2p + 1)b(x) − r a  (x)b (x)  − B(x) = b (x) + 2 ,   3 b (x) a(x) b  (x)     2  b(x) b(x) − 1  a  (x)b (x) − a  (x)b (x) a (x)  + 2b (x) C(x) =  3  a(x) a(x) b (x) −

(2p + 1)b(x) − r a  (x) , b (x) a(x)

and λq = q 2 + p2 . This means that if A(x) is a real function that vanishes at the boundary points of an interval, say [α, β], then the solution of Eq. (1.264), which is indeed the series (1.261), is orthogonal with respect to the weight function  w(x) = exp

= exp

 

B(x) − A (x) dx A(x)

(2p + 1)

b (x) b  (x) a  (x) A (x)  b (x)  −  −r −2 − dx b(x) − 1 b (x) a(x) A(x) b(x) b(x) − 1

1.14 A Class of Hypergeometric-Type Orthogonal Functions

131

  2  b(x) b(x) − 1 a(x)  b(x) − 1 − ln = exp (2p + 1) ln(b(x) − 1) − r ln b(x) b = 

2p−r  r−1 b (x)  b(x) , 2 b(x) − 1 a(x)

which must be positive on the interval [α, β]. Example As the simplest case, assume in (1.262) that a(x) = 1 and b(x) = x. Hence the series

 p + iq, p − iq yq (x) = 2 F1 (1.265) x r satisfies the equation   x(x − 1)yq + (2p + 1)x − r yq + (p2 + q 2 )yq = 0.

(1.266)

Equation (1.266) shows that the orthogonality interval is [0, 1], and the corresponding weight function is denoted by w(x) = x r−1 (1 − x)2p−r , which is a particular case of the shifted beta distribution on [0, 1]. Therefore we have 

1 0

x r−1 (1 − x)2p−r yq1 (x)yq2 (x) dx =

 0

1

 x r−1 (1 − x)2p−r yq21 (x) dx δq1 ,q2 . (1.267)

In order to obtain the norm square value of the series (1.265), we need to evaluate the general integral I∗ =



1

 x u−1 (1 − x)v−1 2 F1 2

0



a, b x dx. c

For this purpose, we first have ∗

I =

 ∞  (a)k (b)k k=0

(c)k k!



1

x 0

u−1+k

(1 − x)

v−1

2 F1



a, b x dx. c

132

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Since 

1





 ∞  (a)j (b)j 1 u−1+k+j a, b x (1 − x)v−1 dx x dx = (c)j j ! 0 c j =0

 (u)k a, b, u + k = B(u, v) 1 , 3 F2 (u + v)k c, u + v + k





a, b x dx c

x u−1+k (1 − x)v−1 2 F1

0

we eventually obtain 

1

x u−1 (1 − x)v−12 F1 2

0

∞  (a)k (b)k (u)k = B(u, v) 3 F2 (c)k (u + v)k k!



k=0



a, b, u + k 1 . c, u + v + k

(1.268)

Now, it is just enough to replace u = r, v = 2p − r + 1, a = p + iq, b = p − iq, and c = r in (1.268) to get 

1 0

x r−1 (1 − x)2p−r yq2 (x) dx = B(r, 2p + 1 − r)

∞  (p + iq)k (p − iq)k k=0

(2p + 1)k k!

 3 F2



p + iq, p − iq, r + k 1 , r, 2p + 1 + k

which is positive if 2p + 1 > r > 0 and q ∈ R. Hence the orthogonality relation (1.267) reads as 

1 0

x r−1 (1 − x)2p−r yq1 (x)yq2 (x) dx ∞

Γ (r)Γ (2p + 1 − r)  (p + iq)k (p − iq)k = 3 F2 Γ (2p + 1) (2p + 1)k k! k=0





p + iq, p − iq, r + k 1 δq1 ,q2 . r, 2p + 1 + k

For q = k ∈ Z+ , one of the advantages of the above relation is the following expansion: f (x) =

∞  k=0

 ck 2 F1



p + ik, p − ik x , r

1.14 A Class of Hypergeometric-Type Orthogonal Functions

133

where

p + ik, p − ik x r−1 (1 − x)2p−r f (x) 2 F1 x dx r 0 .  ck = ∞ (p + ik) (p − ik)

p + ik, p − ik, r + j j j B(r, 2p + 1 − r) 1 3 F2 (2p + 1)j j ! r, 2p + 1 + j j =0 

1

This expansion is especially useful when f (x) is considered as a generalized hypergeometric function. The symmetric case of the orthogonal series (1.265) is defined as ⎞ p + 12 σn + i[ n2 ], p + 12 σn − i[ n2 ] x2⎠ , gn (x) = x σn 2 F1 ⎝ r + σn ⎛

which satisfies the equation      n  x 2 (x 2 − 1)gn + x (4p + 1)x 2 + 1 − 2r gn + 4 p2 + [ ]2 x 2 + (2r − 1)σn gn = 0. 2 (1.269) Equation (1.269) shows that the orthogonality interval is [−1, 1], and the corresponding weight function is given by   x (4p + 1)x 2 + 1 − 2r − 2x(2x 2 − 1) w(x) = x exp dx x 2 (x 2 − 1)   (4p + 1)x 3 − 2r 4x  = x 2 exp + − 2 dx 2 2 x −1 x(x − 1) x − 1  4p + 1    1 ln |x 2 − 1| + (3 − 2r) − ln |x| + ln |x 2 − 1| − 2 ln |x 2 − 1| = x 2 exp 2 2     = x 2 exp ln (1 − x 2 )2p−r |x|2r−3 , 

2

which can be denoted by w(x) = x 2r−1 (1 − x 2 )2p−r = w(−x)

⇔

(−1)2r = −1.

Consequently, according to the generalized Sturm–Liouville theorem for symmetric functions, we have 

1 −1

x

2r−1

2 2p−r

(1 − x )

gn (x)gm (x) dx =



1 −1

 x 2r−1(1 − x 2 )2p−r gn2 (x) dx δn,m .

134

1.15

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Application of Zero Eigenvalue in Solving Some Sturm–Liouville Problems

Usually, the zero eigenvalue is ignored in the solution of some Sturm–Liouville problems. This case also occurs in calculating the eigenvalues of a matrix, so that we would often like to find nonzero solutions of the linear system AX = λX

when λ = 0.

But if in the above system we take λ = 0, then det A = 0 for X = 0, and the dimension of the matrix A is reduced by at least one. Now, if in a boundary value problem we similarly take one of the eigenvalues of the equations related to the main problem to be equal to zero, then one of the solutions will be specified in advance. In view of this fact, in this section we define a class of special functions and apply them to the classical potential, heat, and wave equations in spherical coordinates. We start by defining the two sequences Cn (z; a(z)) =

(a(z))n + (a(z))−n = cosh (n ln(a(z))) , 2

Sn (z; a(z)) =

(a(z))n − (a(z))−n = sinh (n ln(a(z))) , 2

(1.270)

in which a(z) can be a complex (or real) function and n is a positive integer. It is not difficult to verify that both above sequences satisfy a unique second-order differential equation of the form     2 3 a 2 (z)a  (z)y  + a(z) a  (z) − a 2 (z)a  (z) y  − n2 a  (z) y = 0,

(1.271)

provided that a 2 (z)a  (z) = 0. Clearly, Eq. (1.271) is a particular case of the Sturm–Liouville equation (1.264) for A(z) = B(z) =

a 2 (z) (a  (z))2

,

a(z)(a (z))2 − a 2 (z)a  (z) (a  (z))3

C(z) = 0, and λn = −n2 .

,

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

135

There are various cases for sequences (1.270). The first case may lead to the Chebyshev polynomials if one takes a(z) = exp(i arccos z) and uses the well-known Euler identity. In other words, we have Cn (z; exp(iarccos z)) = cos(narccos z) = Tn (z), % Sn (z; exp(iarccos z)) = i sin(narccos z) = i 1 − z2 Un−1 (z). Also replacing a(z) = exp(i arccos z) in (1.271) gives the well-known differential equation of Chebyshev polynomials of the first kind (1 − z2 )y  − zy  + n2 y = 0. Rational Chebyshev functions are another case that can be derived by choosing a(z) = exp(i arccot z) in (1.270), leading to Cn (z; exp(iarccot z)) = cos(narccot z), Sn (z; exp(iarccot z)) = i sin(narccot z).

(1.272)

Also, since in this case −i exp(3iarccot z) , 1 + z2 − exp(3iarccot z) a(z)(a  (z))2 = , 2 (1 + z2 ) a 2 (z)a  (z) =

a 2 (z)a  (z) =

(2iz − 1) exp(3iarccot z) (1 + z2 )

2

,

and (a  (z))3 =

i exp(3iarccot z) (1 + z2 )

3

,

the functions (1.272) satisfy the equation (1 + z2 )2 y  + 2z(1 + z2 )y  + n2 y = 0. Notice that the explicit forms of the real functions (1.272) can be directly obtained by de Moivre’s formula, because it is enough to substitute θ = arccot x into the above formula to get (x + i)n √ n = cos(n arccot x) + i sin(n arccot x). ( 1 + x 2)

136

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Consequently, we have ⎛

[n/2] 

Cn (x; exp(iarccot x)) = ⎝

 (−1)k

k=0

n 2k

⎛

[n/2] 

− i Sn+1 (x; exp(iarccot x))= ⎝

% x n−2k ⎠ /( 1 + x 2 )n = Tn∗ (x),

 (−1)k

k=0

⎞

n+1 2k + 1



⎞

% x n−2k ⎠ /( 1 + x 2 )n+1 = Un∗ (x).

We can prove that the Chebyshev rational functions Tn∗ (x) and Un∗ (x) are orthogonal with respect to the weight function W (x) =

1 1 + x2

on (−∞, ∞),

so that we have 

∞ −∞

Tn∗ (x)Tn∗ (x) dx 1 + x2

⎧ ⎪ ⎨0 = π/2 ⎪ ⎩ π

( m = n), (m = n), (m = n = 0),

and 

∞ −∞

Un∗ (x)Un∗ (x) dx = 1 + x2



(m = n), (m = n).

0 π/2

As the last particular example for sequences (1.270), we know that the associated Legendre differential equation  (1 − x )y (x) − 2xy (x) + p − 2





q 1 − x2

 y(x) = 0

has been solved for the following three cases up to now: (a) p = 0, q = 0 (generating the associated Legendre functions), (b) p =  0, q = 0 (generating the Legendre polynomials), (c) p = 0, q = 0, which is reduced to the simple equation (1 − x 2 )y  (x) − 2xy (x) = 0, having the general solution y(x) = c1 ln

1+x + c2 . 1−x

(1.273)

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

137

Hence, the fourth case p = 0, q = 0 remains to be studied. To derive the solution of the corresponding equation, we first substitute  a(z) =

1−z 1+z

1 2

into Eq. (1.271) to get (1 − z2 )y  − 2zy  −

n2 y = 0, 1 − z2

(1.274)

which is a particular case of Eq. (1.273) for p = 0, q = n2 . By noting (1.270), the solutions of Eq. (1.274) are of the form    1

n   −n

1−z 2 1 1−z 2 1−z 2 = , + Cn z; 1+z 2 1+z 1+z    1

n   −n

1−z 2 1 1−z 2 1−z 2 = , Sn z; − 1+z 2 1+z 1+z and will appear in Helmholtz’s equation in spherical coordinates. In other words, if the equation ∇ 2 U ( r, θ , Φ ) = k 2 U ( r, θ , Φ ) is separated to three ordinary differential equations, one of the equations takes the form 1 d sin θ dθ

   dy n2  sin θ + m(m + 1) − y(θ ) = 0, dθ sin2 θ

(1.275)

which is the same as Eq. (1.273) for x = cos θ , p = m(m + 1), and q = n2 . Therefore, if a(z) = tg 2z in (1.271), a special case of Eq. (1.275) for m = 0, i.e., y  + (cot z) y  −

n2 sin2 z

y = 0,

(1.276)

would have the following solutions:    z n  z −n z 1  = tan , Cn z; tan + tan 2 2 2 2    z 1  z n  z −n Sn z; tan = tan − tan . 2 2 2 2

(1.277)

138

1 Generalized Sturm–Liouville Problems in Continuous Spaces

• Application of the functions (1.277) for the potential, heat, and wave equations in spherical coordinates Most of boundary value problems related to the wave, heat, and potential equations are somehow connected to the Helmholtz partial differential equation ∇ 2 U ( r, θ , Φ ) = k 2 U ( r, θ , Φ ) in spherical coordinates. For instance, k = 0 in this equation leads to the potential equation ∂ 2U 1 ∂2 U 1 2 ∂U cot θ ∂ U ∂2 U + + + + = 0, r ∂ r ∂ r2 r2 ∂ θ 2 r2 ∂ θ r 2 sin2 θ ∂ Φ 2

(1.278)

and by separating the variables as U ( r, θ , Φ ) = R(r) A(θ ) B(Φ), Equation (1.278) is transformed into three ordinary differential equations (i) r 2 R  + 2rR  − λ 1 R = 0 , (ii) B  − λ2 B = 0 ,

(1.279)

λ2  A = 0. (iii) A + (cot θ ) A + λ1 + sin2 θ 

Since the solution of a potential equation is generally determined when the boundary conditions are known, if λ1 = 0 and λ2 = −k 2 = 0 are assumed in (1.279), then for the variable r we get R(r) = −c12

1 + c2 r

(c1 and c2 are constant),

and for the third equation, we get the same form as Eq. (1.276). Hence, the general solution corresponding to third equation is as         θ θ θ θ A(θ ) = A1 Ck θ ; tan + A2 Sk θ ; tan = a1 tank + a2 tan−k , 2 2 2 2 where A1 and A2 and a1 and a2 are all constant values. This means that the general solution of the potential equation ∇ 2 U ( r, θ , Φ ) = 0 with 2 the predetermined condition R(r) = −cr1 + c2 is  U (r, θ, Φ) =

      −c1 2 θ θ + c2 (b1 cos k Φ + b2 sin k Φ ) a1 tank . + a2 tan−k r 2 2 (1.280)

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

139

The solution (1.280) reveals the sensitivity of the potential equation with respect to the variable r, because lim U (r, θ, Φ) = ∞. r→0

For example, let us consider the equation ∇ 2 U (r, θ, Φ) = 0

for 0 < r < a

in spherical coordinates when the variable r takes the preassigned form R(r) =

−c1 2 + c2 r

and the following initial and boundary conditions hold: lim U (r, θ , Φ ) = ∞ ,

r→0

U ( a2 , θ , Φ ) = 0 , U (r, π2 , Φ ) = 0 , U (a, π3 , Φ ) = Φ . The general solution of this problem, according to the given conditions and assuming A k = a1 b1 c2 and B k = a2 b1 c2 for simplicity, is denoted by U (r, θ, Φ) =



Uk (r, θ, Φ) ,

k

     θ a  k θ − tan−k (Ak cos k Φ + Bk sin k Φ ) . tan Uk (r, θ, Φ) = 1 − 2r 2 2 Now by employing the last condition in the above solution, we get 2Φ =

 √ −k √ +k  (Ak cos k Φ + Bk sin k Φ ), 3 − 3 k

where Ak and Bk are calculated as  π √ −k √ +k  4 4((−1)k − 1) Ak = 3 − 3 Φ cos kΦ dΦ = , π 0 π k2  π √ −k √ +k  4 4(−1)k . 3 − 3 Φ sin kΦ dΦ = Bk = π 0 k

140

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Therefore, the solution under the given conditions is  a U (r, θ, Φ) = 1 − 2r 

  ∞ k  4((−1) − 1)k −2 4(−1)k k −1 kθ −k θ cos k Φ + −k/2 sin k Φ tan − tan . × π (3−k/2 − 3k/2 ) (3 − 3k/2) 2 2 k=1

This approach can be similarly stated for the classical heat and wave equations. For instance, if the heat equation ∇ 2U =

∂ U ∂ t

is considered, separating the variables as U (r, θ , Φ , t) = S(r, θ , Φ)T (t), where S(r, θ , Φ) = R(r)A(θ )B(Φ), yields ⎧ 2 2 2 ⎪ ⎨∇ U = ∇ (S T ) = T ∇ S ⎪ ⎩ ∂ U = ∂ (S T ) = S ∂ T ∂ t ∂ t ∂ t

⇒

T  (t) ∇ 2S = = α. S T (t)

Hence the corresponding differential equations appear as (i) T  − α T = 0 , (ii) r 2 R  + 2rR  − (α r 2 + λ 1 )R = 0 , (1.281)

(iii) B  − λ2 B = 0 ,  λ2  A = 0. (iv) A + (cot θ ) A + λ1 + sin2 θ Once again, if λ1 = 0, λ2 = −n2 = 0,

and α = −k 2 = 0

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

141

are considered in (1.281), then the general solution, when R (r) =

c1 J1/2 (k r) + c2 J−1/2 (k r) , √ kr

takes the form  e−k t U (r, θ, Φ, t) = √ (c1 J1/2(k r) + c2 J−1/2 (k r)) b1 tann kr 2

    θ θ + b2 tan−n 2 2

× (a1 cos n Φ + a2 sin n Φ ) ,

(1.282)

where J1/2 (x) and J−1/2 (x) are two particular cases of the well-known Bessel functions Jp (x). For example, let us consider the heat equation ∇ 2 U (r, θ, Φ, t) =

∂ U ∂t

for 0 < r < a

in spherical coordinates when the variable r takes the preassigned form R (r) =

c1 J1/2 (k r) + c2 J−1/2 (k r) , √ kr

and the following initial and boundary conditions hold: lim U (r, θ , Φ, t ) < M ,

r→0

 π  U r, , Φ , t = 0 , 2 U (r, θ , 0, t ) = 0 ,   π U r, , Φ , 0 = f (r, Φ). 3 By referring to the general solution of the problem and using the given conditions, we first conclude in (1.282) that c2 = a1 = 0 and b1 + b2 = 0. Hence if A k , n = a2 b1 c1 , then U (r, θ, Φ, t) =

 k

Uk , n (r, θ, Φ, t) ,

n

2  e−k t θ θ sin n Φ , Uk , n (r, θ, Φ, t) = Ak,n √ J1/2 (kr) tann − tan−n 2 2 kr

is the general solution.

142

1 Generalized Sturm–Liouville Problems in Continuous Spaces

On the other hand, by noting the orthogonality relation of the Bessel functions Jp (x) as 

a 0

 a2 2 x  x Jp Z(p,n) x dx = Jp+1 Jp Z(p,m) (Z(p,m)) δn,m , a a 2

where Z(p,m) is the mth zero of Jp (x), i.e., Jp (Z(p,m) ) = 0, it is better for the eigenvalues Z k to be considered as k = (1/2,m) . Thus, the general solution becomes a U (r, θ, Φ, t) =

Z(1/2,m) 2  ) t) θ ∗ exp(−( a J 1 (Z(1/2,m)r/a) tann An,m % 2 2 Z(1/2,m)r/a m

 n

in which A∗n,m = A

n,

Z(1/2,m) a

− tan−n

θ sin n Φ, 2

.

In the sequel, it is sufficient to compute the coefficients A∗n,m . To this end, substituting the last given condition into the above solution yields $

   A∗n,m (3 2 − 3 2 ) r r f (r, Φ) = sin (n Φ). % J1/2 Z(1/2,m) a a Z(1/2,m) n m −n

n

Therefore, using the orthogonality relation of the Bessel functions and also the orthogo∗ nality property of the sequence {sin(nΦ)}∞ n=1 on [0, π], An,m is computed as A∗n,m

=

a π % 3 4 Z(1/2,m) 0 0 f (r, Φ) J1/2 (Z(1/2,m) ar ) sin (n Φ)r 2 dr dΦ π (3

−n 2

n

3

2 (Z − 3 2 ) a 2 J3/2 (1/2,m) )

A similar problem can be stated for the classical wave equation ∇ 2U =

∂ 2U . ∂t 2

First we have ⎧ 2 2 2 ⎪ ⎨∇ U = ∇ (S T ) = T ∇ S 2 2 2 ⎪ ⎩ ∂ U = ∂ (S T ) = S ∂ T 2 2 ∂ t ∂ t ∂ t2

⇒

T  (t) ∇ 2S = = α, S T (t)

.

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

143

which results in four ordinary differential equations (i) T  − α T = 0 , (ii) r 2 R  + 2rR  − (α r 2 + λ1 )R = 0 , (iii) B  − λ2 B = 0 ,

(1.283)

 λ2  (iv) A + (cot θ )A + λ1 + A = 0. sin2 θ Now, if λ1 = 0, λ2 = −k 2 = 0 and α = −n2 = 0 in (1.283), then the general solution of the wave equation when R (r) =

c1 J1/2 (n r) + c2 J−1/2 (n r) √ , nr

is U (r, θ, Φ, t) = (d1 cos n t + d2 sin n t) (b1 cos k Φ + b2 sin k Φ )    c J (n r) + c J  1 1/2 2 −1/2 (n r) k θ −k θ + a2 tan × a1 tan √ . 2 2 nr

(1.284)

For example, let us consider the classical wave equation in spherical coordinates when the variable r has the preassigned form R (r) =

c1 J1/2 (n r) + c2 J−1/2 (n r) , √ nr

together with the following initial and boundary conditions: lim U (r, θ , Φ, t ) < M,

r→0

 π  U r, , Φ , t = 0, 2 U (r, θ , 0, t ) = 0, U (r, θ , Φ, 0 ) = 0,   π U r, , Φ , q = g(r, Φ). 3

144

1 Generalized Sturm–Liouville Problems in Continuous Spaces

To solve this problem, substituting the given conditions into the general solution (1.284) gives c2 = b1 = d1 = 0 and a1 + a2 = 0. Now if B k , n = a1 b2 c1 d2 , then (1.284) is simplified as  Uk , n (r, θ, Φ, t) , U (r, θ, Φ, t) = k

n

  J1/2 (nr) θ θ tank − tan−k sin k Φ . Uk , n (r, θ, Φ, t) = Bk,n sin(nt) √ 2 2 nr Similar to the previous example, if n = U (r, θ, Φ, t) =

 m

k

Z(1/2,m) , a

then

    t J1/2 (Z(1/2,m) r/a) θ θ ∗ tank − tan−k Bk,m sin Z(1/2,m) % sin k Φ a 2 2 Z(1/2,m) r/a

∗ =B is the general solution of the given problem, where Bk,m

k,

Z(1/2,m) a

.

Finally, by replacing the last condition in the above solution, i.e., $

∗ (3   Bk,m r g(r, Φ) = a m k

−k 2

 − 3 2 ) sin(Z(1/2,m) qa ) r % J1/2 Z(1/2,m) sin (k Φ), a Z(1/2,m) k

and using the orthogonality relation of the Bessel functions J1/2 (Z(1/2,m) ar ) on [0, a], the ∗ will be derived as unknown coefficients Bk,m ∗ Bk,m

=

a π % 3 4 Z(1/2,m) 0 0 g (r, Φ) J1/2 (Z(1/2,m) ar ) sin (k Φ)r 2 dr dΦ π sin(Z(1/2,m) qa ) (3

−k 2

k

3

2 (Z − 3 2 ) a 2 J3/2 (1/2,m))

.

1.15.1 A Relationship Between the Chebyshev Polynomials of the Third and Fourth Kinds and the Associated Legendre Differential Equation In this section, we provide a short survey on the Chebyshev polynomials of the third and fourth kinds and show that their differential equations are particular cases of Eq. (1.273). (α,β) (x) that are orthogonal with respect to Let us start with the Jacobi polynomials y = Pn α β the weight function (1 − x) (1 + x) for α, β > −1 on [−1, 1] and satisfy the differential equation (1 − x 2 )y  − ((α + β + 2)x + (α − β)) y  + n(n + α + β + 1)y = 0 .

(1.285)

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

145

If β

α

An (x) = (1 − x) 2 (1 + x) 2 Pn(α,β) (x), Then Eq. (1.285) changes to the self-adjoint form 

(α + β)(α + β + 2) + 4n(n + α + β + 1) 2 x 4  2(α + β) − (α − β)2 + 4n(n + α + β + 1) (α + β)(α − β) x− + A = 0. 2 4 (1.286)

(1 − x 2 )2 A − 2x(1 − x 2 )A −

There are two cases in Eq. (1.286) that are transformed to the associated Legendre differential equation (1 − x 2 )y  (x) − 2x y  (x) + (p −

q )y(x) = 0. 1 − x2

(1.287)

The first case is α = β, which leads to the particular equation  (1 − x 2 )A − 2xA + (n + α)(n + α + 1) −

α2  A = 0, 1 − x2

where α

An (x) = (1 − x 2 ) 2 Pn(α,α) (x)

for α > −1.

The second case, which generates the third and fourth kinds of Chebyshev polynomials, is α = −β. In this case, the equation  (1 − x 2 )A − 2xA + n(n + 1) −

α2  A=0 1 − x2

(1.288)

is a special case of Eq. (1.287) for p = n(n + 1) and q = α 2 , where An (x) =

1 − x α 2

1+x

Pn(α,−α) (x) for − 1 < α < 1.

Note that −1 < α < 1 is a necessary condition for orthogonality of the polynomials (α,−α) (x), because we must simultaneously have α + β = 0 and α, β > −1. This means Pn that the parameter q = α 2 should lie in [0, 1), and we can, for instance, consider the eigenvalues q = 1/m , m ∈ N, corresponding to this case. Hence Eq. (1.288) can be

146

1 Generalized Sturm–Liouville Problems in Continuous Spaces

directly applied in the potential theory when α + β = 0. It is clear that the polynomials generated by this condition are orthogonal on [−1, 1], since 

1 −1

 1 − x α 1+x

Pn(α,−α) (x)Pm(α,−α) (x) dx =

2 (n + α)!(n − α)! δ n,m , 2n + 1 (n!)2

and they satisfy the differential equation (1 − x 2 )y  − 2(x + α)y  + n(n + 1)y = 0 for y = Pn(α,−α) (x). There are two subsequences of polynomials Pn(α,−α) (x) that have trigonometric forms just like the ultraspherical polynomials Pn(α,α) (x) that generate the first and second kinds of Chebyshev polynomials for α = − 12 and α = 12 respectively. To introduce these two subsequences, it is enough to substitute 2θ = arccos x and m = 2n + 1 into de Moivre’s formula to get  cos

 arccos x  2

+ i sin

 arccos x 2n+1 2

= cos

      1 1 n+ arccos x + i sin n+ arccos x . 2 2

(1.289) Since cos

 arccos x  2

$ =

1+x 2

and

sin

 arccos x  2

$ =

1−x , 2

relation (1.289) changes to $

1+x +i 2

$

1−x 2

2n+1

      1 1 = cos n+ arccos x + i sin n+ arccos x . 2 2

Consequently, we have $

  n 1 − x k  1 + x n−k 1+x  k 2n + 1 (−1) 2k 2 2 2 k=0 

$ n  1 − x k  1 + x n−k 1−x  k 2n + 1 +i (−1) 2 2 2 2k + 1 k=0       1 1 = cos n+ arccos x + i sin n+ arccos x , 2 2

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

147

which yields $ Vn (x) =

2 cos 1+x

      n 1 2n + 1  1 − x k  1 + x n−k (−1)k , n+ arccos x = 2 2 2 2k k=0

and $ Wn (x) =

2 sin 1−x



    n  1 − x k  1 + x n−k 1 k 2n + 1 n+ arccos x = (−1) , 2 2 2 2k + 1 k=0

as the explicit representations of the third and fourth kinds of Chebyshev polynomials. In this direction, note that since Pn(α,−α) (−x) = (−1)n Pn(−α,α) (x), it follows that Vn (−x) = (−1)n Wn (x). Corollary 1.13 The two sequences  An (x) =

1−x 1+x

− 1



4

Vn (x)

and An (−x) = (−1)

n

1−x 1+x

1 4

Wn (x)

satisfy the following self-adjoint equation according to the general Eq. (1.288):  (1 − x 2 )A − 2xA + n(n + 1) −

 1 A = 0. 4(1 − x 2 )

Moreover, the norm square value of the two sequences is equal to π, because 

1 −1

$

1+x Vn (x)Vm (x)dx = 2 1−x



1

−1 π



=2

0 π

 = 0

cos((n + 1/2) arccos x) cos((m + 1/2) arccos x) √ dx 1 − x2 cos(n + 1/2)θ cos(m + 1/2)θ dθ  cos(n + m + 1)θ + cos(n − m)θ dθ = π δm,n ,

148

1 Generalized Sturm–Liouville Problems in Continuous Spaces

and for Wn (x) we similarly have 

$

1 −1

1−x Wn (x)Wm (x)dx = 2 1+x  =



π

0 π

sin(n + 1/2)θ sin(m + 1/2)θ dθ  cos(n − m)θ − cos(n + m + 1)θ dθ = π δm,n .

0

Therefore 

$

1+x dx = Vn (x)Vm (x) 1−x −1 1



$

1 −1

Wn (x)Wm (x)

1−x dx = π δm,n . 1+x

(1.290)

From the result (1.290), one can conclude that the sequences   1 π x fn (x) = cos n + 2 l

  1 π and gn (x) = sin n + x 2 l

are orthogonal on [0, l], so that we have 

l 0

1 fn (x)fm (x)dx = 2

 l

π π  l cos(n + m + 1) x + cos(n − m) x dx = δm,n l l 2

0

and  0

l

1 gn (x)gm (x)dx = 2

 l 0

π  l π cos(n − m) x − cos(n + m + 1) x dx = δm,n . l l 2

The two latter relations show that A = {fn (x)}∞ n=0

and B = {gn (x)}∞ n=0

are complete orthogonal sets in [0, l]. On the other hand, by recalling that the sequences An (x) = cos

nπ x l

and Bn (x) = sin

nπ x l

are also orthogonal on [0, l], the following expansions can be derived, which are various combinations of the above four trigonometric sequences: f (x) =

∞  k=0

(1) ak 2 cos

  1 πx , k+ 2 l

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

149

where (1)

ak 2 =

2 l



l

0

f (x) =

  1 πx dx, f (x) cos k + 2 l

∞  k=0

(1) bk 2 sin

  1 πx , k+ 2 l

where (1) bk 2

f (x) =

∞ 

2 = l

(1) ak 2 cos

k=0



l 0

  1 πx dx, f (x) sin k + 2 l

    1 πx 1 πx ( 21 ) + bk sin k + , k+ 2 l 2 l

(1.291)

where ⎧ 1 (2) ⎪ ⎪ ⎨ ak =

 1 l l −l

f (x) cos(k + 12 ) πx l dx ,

⎪ ⎪ ⎩

 1 l l −l

f (x) sin(k + 12 ) πx l dx ,

(1)

bk 2 =

f (x) =

∞  k=0

  1 πx kπx (1) + bk sin , ak 2 cos k + 2 l l

(1.292)

where ⎧ 1 ( ) ⎪ ⎨ ak 2 = ⎪ ⎩

bk =

f (x) =

 1 2l l 0

 1 l l −l

∞  k=0

f (x) cos(k + 12 ) πx l dx ,

f (x) sin kπx l dx,

ak cos

kπx 1 πx (1) + bk 2 sin(k + ) , l 2 l

where ⎧ l ⎪ a0 = 2l1 −l f (x) dx , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ l ak = 1l −l f (x) cos kπx l dx , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ( 21 ) 1  l bk = l −l f (x) sin(k + 12 ) πx l dx .

(1.293)

150

1 Generalized Sturm–Liouville Problems in Continuous Spaces

Note that relations (1.291)–(1.293) also have complex forms. For instance, the complex form of relation (1.291) is (1)

2 e−i f (x) = C−0

πx 2l

+

∞ 

(1)

1 πx l

Ck 2 ei(k+ 2 )

,

k=−∞

where ⎧ 1 (2) ⎪ ⎪ ⎨ Ck = ⎪ ⎪ ⎩

(1)

2 C−0 =

1 2l 1 2l

l

−l

l

−l

f (x) e−i(k+ 2 )

1 πx l

f (x) e

iπ x 2l

dx ,

dx .

Furthermore, Bessel’s inequality, Parseval’s identity, and the Dirikhlet kernel are some other items that can be applied to relations (1.291)–(1.293). For example, the Dirikhlet kernel corresponding to relation (1.291) for l = π is derived as           n  1  π 1 1 1 1 f (t) cos k + t cos k + x + sin k+ t sin k+ x dt π 2 2 2 2 k=0 −π  n

  π  1 sin(n + 1)(t − x) 1 π 1 f (t) cos(k + )(t − x) dt = f (t) dt. = π −π 2 2π −π sin t−x 2

Sn (x) =

k=0

Hence Dn+1/2 (x) =

sin(n + 1)x sin x2

is the Dirichlet kernel corresponding to relation (1.291). Moreover, since Dn (x) =

n  k=−n

eikx =

sin(n + 12 )x , sin x2

it follows that   sin (n + 12 ) arccos x = Wn (x) Dn (arccos x) = x sin( arccos ) 2

⇔

Wn (cos x) = Dn (x).

It is not difficult to verify that  1 1−x Vn (x) =2 F1 − n, n + 1, ; 2 2

1.15 Application of Zero Eigenvalue in Solving Some Sturm–Liouville. . .

151

and  3 1−x , Wn (x) = (2n + 1) 2 F1 − n, n + 1, ; 2 2 in which 2 F1 (a, b, c; x) denotes the hypergeometric function of order (2, 1). To compute the generating functions of the polynomials Vn (x) and Wn (x), we can start with the relation ∞  n=0

=

    θ ∞ ∞    iθ n 1 1 ei 2 n i θ2 te t cos n + t sin n + = θ +i θ =e 2 2 1 − teiθ n

n=0

n=0

(1 − t cos θ ) cos(θ/2) − t sin θ sin(θ/2) (1 − t cos θ ) sin(θ/2) + t sin θ cos(θ/2) +i , 1 + t 2 − 2t cos θ 1 + t 2 − 2t cos θ

which yields ∞  cos(n + 1/2)θ n=0

cos(θ/2)

tn =

1−t 1 + t 2 − 2t cos θ

(1.294)

and ∞  sin(n + 1/2)θ n=0

sin(θ/2)

tn =

1+t . − 2t cos θ

1 + t2

(1.295)

Thus, by setting x = cos θ in (1.294) and (1.295), the generating functions of the third and fourth kinds of Chebyshev polynomials will be respectively derived as follows: ∞ 

Vn (x) t n =

1−t 1 + t 2 − 2tx

Wn (x) t n =

1+t , 1 + t 2 − 2tx

n=0

and ∞  n=0

which are special cases of the Jacobi polynomial generating function for α = −β = − 12 and α = −β = 12 . Finally, since the Chebyshev polynomials of the first and second kinds have trigonometric forms, by applying different trigonometric identities, we can derive various relationships. For example, the identity 1 arccos x = arccos 2

$

1+x 2

152

1 Generalized Sturm–Liouville Problems in Continuous Spaces

implies that $ Vn (x) =

$ 1 + x  2 T2n+1 1+x 2

and Wn (x) = U2n

$ 1 + x  2

,

where Tn (x) and Un (x) denote the first and second kinds of Chebyshev polynomials respectively.

References 1. G.E. Andrews, R. Askey, R. Roy, Special functions, in Encyclopedia of Mathematics and Its Applications, vol. 71 (Cambridge University Press, Cambridge, 1999) 2. G. Arfken, Mathematical Methods for Physicists (Academic, Cambridge, 1985) 3. A. Erdelyi, W. Magnus, F. Oberhettinger, F.G. Tricomi, Tables of Integral Transforms, vol. 1 (McGraw-Hill, New York, 1954) 4. W. Lesky, Endliche und unendliche systeme von kontinuierlichen klassischen orthogonalpolynomen. Z. Angew. Math. Mech. 76, 181–184 (1996) 5. P. Lesky, Eine Charakterisierung der Klassischen Kontinuierlichen, Diskreten und qOrthogonalpolynome (Shaker, Aachen, 2005) 6. M. Masjed-Jamei, Three finite classes of hypergeometric orthogonal polynomials and their application in functions approximation. Integral Transform. Spec. Funct. 13(2), 169–190 (2002) 7. M. Masjed-Jamei, Classical orthogonal polynomials with weight function ) ; x ∈ (−∞, ∞) and a generalization of ((ax + b)2 + (cx + d)2 )−p exp(q arctan ax+b cx+d T and F distributions. Integral Transform. Spec. Funct. 15(2), 137–153 (2004) 8. V. Romanovski, Sur quelques classes nouvelles de polynômes orthogonaux. Comptes Rendues Acad. Sci. Paris. 188, 1023–1025 (1929)

Further Reading • M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th Printing (Dover, New York, 1972) • W. Al-Salam, W.R. Allaway, R. Askey, Sieved ultraspherical polynomials. Trans. Am. Math. Soc. 284, 39–55 (1984). • R. Askey, Orthogonal polynomials old and new, and some combinatorial connections, in Enumeration and Design, ed. by D.M. Jacson, S.A. Vanstone (Academic, New York, 1984) • R. Askey, Continuous Hahn polynomials. J. Phys. A 18(16), L1017–L1019 (1985)

Further Reading

153

• R. Askey, An integral of Ramanujan and orthogonal polynomials. J. Indian Math. Soc. 51, 27–36 (1987) • S. Bochner, Über Sturm–Liouvillesche polynomsysteme. Math. Zeit. 29, 730–736 (1929) • A. Branquinho, F. Marcellán, J.C. Petronilho, Classical orthogonal polynomials: a functional approach. Acta Appl. Math. 34 , 283–303 (1994) • A.L. Cauchy, Sur les integrales définies prises entre des limites imaginaires, Bulletin de Ferussoc, T. III (1825), 214–221, in Oeuvres de A. L. Cauchy, 2 serie, T. II, GauthierVillars. Paris, 59–65 (1958) • J.A. Charris, M.E.H. Ismail, On sieved orthogonal polynomials II: random walk polynomials. Canad. J. Math. 38, 397–415 (1986) • J.A. Charris, M.E.H. Ismail, Sieved orthogonal polynomials VII: generalized polynomial mappings. Trans. Am. Math. Soc. 340, 71–93 (1993) • S.A. Chihara, An Introduction to Orthogonal Polynomials (Gordon and Breach, New York, 1978) • H. Dette, Characterizations of generalized Hermite and sieved ultraspherical polynomials. Proc. Am. Math. Soc. 384, 691–711 (1996) • Digital Library of Mathematical Functions, DLMF. Available at: http://dlmf.nist.gov/ • L. Fox, I.B. Parker, Chebyshev Polynomials in Numerical Analysis (Oxford University Press, London, 1968) • E. Grosswald, Bessel Polynomials. Lecture Notes Mathematics, vol. 698 (Springer, New York, 1973) • M.E.H. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable. Encyclopedia of Mathematics, vol. 98 (Cambridge University Press, Cambridge, 2005) • R. Koekoek, P.A. Lesky, R.F. Swarttouw, Hypergeometric Orthogonal Polynomials and Their q-Analogues. Springer Monographs in Mathematics (Springer, Berlin, 2010) • W. Koepf, Hypergeometric Summation (Vieweg, Braunschweig/Wiesbaden, 1988) • W. Koepf, M. Masjed-Jamei, A generic formula for the values at the boundary points of monic classical orthogonal polynomials. J. Comput. Appl. Math. 191, 98–105 (2006) • W. Koepf, M. Masjed-Jamei, A generic polynomial solution for the differential equation of hypergeometric type and six sequences of classical orthogonal polynomials related to it. Integral Transform. Spec. Funct. 17, 559–576 (2006) • W. Koepf, M. Masjed-Jamei, Two classes of special functions using Fourier transforms of some finite classes of classical orthogonal polynomials. Proc. Am. Math. Soc. 135, 3599–3606 (2007) • W. Koepf, D. Schmersau, Representation of orthogonal polynomials. J. Comput. Appl. Math. 90, 57–94 (1998) • W. Koepf, D. Schmersau, Recurrence equations and their classical orthogonal polynomial solutions. Appl. Math. Comput. 128(2–3), 303–327 (2002) • H.T. Koelink, On Jacobi and continuous Hahn polynomials. Proc. Am. Math. Soc. 124, 997–898 (1996)

154

1 Generalized Sturm–Liouville Problems in Continuous Spaces

• T. H. Koornwinder, Special orthogonal polynomial systems mapped onto each other by the Fourier–Jacobi transform, in Polynomes Orthogonaux et Applications, ed. by C. Brezinski, A. Draux, A.P. Magnus, P. Maroni, A. Ronveaux. Lecture Notes Mathematics, vol. 1171 (Springer, Berlin, 1985), pp. 174–183 • H.L. Krall, O. Frink, A new class of orthogonal polynomials: the Bessel polynomials. Trans. Am. Math. Soc. 65, 100–115 (1949) • W. Lesky, Uber Polynomlösungen von Differentialgleichungen und Differenzengleichungen zweiter Ordnung. Anz. Österr. Akad. Wiss., math. nat. Kl. 121, 29–33 (1985) • F. Marcellán, R. Sfaxi, Orthogonal polynomials and second-order pseudo-spectral linear differential equations. Integral Transform. Spec. Funct. 21, 487–501 (2010) • M. Masjed-Jamei, A generalization of classical symmetric orthogonal functions using a symmetric generalization of Sturm–Liouville problems. Integral Transform. Spec. Funct. 18, 871–883 (2007) • M. Masjed-Jamei, A basic class of symmetric orthogonal polynomials using the extended Sturm–Liouville theorem for symmetric functions. J. Math. Anal. Appl. 325, 753–775 (2007) • M. Masjed-Jamei, A basic class of symmetric orthogonal functions using the extended Sturm–Liouville theorem for symmetric functions. J. Comput. Appl. Math. 216, 128– 143 (2008) • M. Masjed-Jamei, Biorthogonal exponential sequences with weight function exp(ax 2 + ibx) on the real line and an orthogonal sequence of trigonometric functions. Proc. Am. Math. Soc. 136, 409–417 (2008) • M. Masjed-Jamei, W. Koepf, Two classes of special functions using Fourier transforms of generalized ultraspherical and generalized Hermite polynomials. Proc. Am. Math. Soc. 140, 2053–2063 (2012) • M. Masjed-Jamei, F. Marcellan, E.J. Huertas, A finite class of orthogonal functions generated by Routh–Romanovski polynomials. Complex Var. Elliptic Equ. 59(2), 162– 171 (2014) • A.F. Nikiforov, V.B. Uvarov, Special Functions of Mathematical Physics (Birkhäuser, Berlin, 1988) • W.H. Press, B.P. Flannery, S.A. Teukolsky, W.T. Vetterling, Beta Function, T Student distribution and F-Distribution, in Numerical Recipes in Fortran: The Art of Scientific Computing, Section 6.2, 2nd edn. (Cambridge University Press, Cambridge, 1992), pp. 219–223 • S. Ramanujan, A class of definite integrals. Quarterly J. Math, 48, 294–310 (1920) • T. Rivlin, Chebyshev Polynomials: From Approximation Theory to Algebra and Number Theory, 2nd edn. (Wiley, New York, 1990) • E. Routh, On some properties of certain solutions of a differential equation of the second order. Proc. Lond. Math. Soc. 16, 245–261 (1884) • J. Shohat, A differential equation for orthogonal polynomials. Duke Math. J. 5, 401– 417 (1939)

References

155

• Y. Sosov, C.E. Theodosiou, On the complete solution of the Sturm–Liouville problem d 2 X/dx 2 + λ 2 X = 0 over a closed interval. J. Math. Phys. 43, 2831–2843 (2002) • C. Truesdell, An Assay Toward a Unified Theory of Special Functions (Princeton University Press, Princeton, 1948) • E.W. Weisstein, Saalschütz’s Theorem From MathWorld: A Wolfram Web Resource. http://mathworld.wolfram.com/SaalschuetzsTheorem.html • E.T. Whittaker, G.N. Watson, A Course of Modern Analysis, 4th edn. (Cambridge University Press, Cambridge, 1927)

2

Special Functions Generated by Generalized Sturm–Liouville Problems in Discrete Spaces

2.1

Introduction

Orthogonal functions of a discrete variable may be solutions of a discrete-type Sturm– Liouville problem of the form     Δ k ∗ (x)∇yn (x) + λn w(x) − q ∗ (x) yn (x) = 0 for k ∗ (x) > 0 and w(x) > 0, (2.1) where Δf (x) = ∇f (x + 1) = f (x + 1) − f (x), and Eq. (2.1) satisfies a set of discrete boundary conditions as α1 y(a) + β1 ∇y(a) = 0, α2 y(b) + β2 ∇y(b) = 0. This means that if yn (x) and ym (x) are two eigenfunctions of Eq. (2.1), then they are orthogonal with respect to the weight function w(x) on a discrete set of points. Similar to the continuous case, many special functions of a discrete variable are solutions of a regular or singular Sturm–Liouville problem of type (2.1). In this chapter, we extend discrete-type Sturm–Liouville problems to symmetric functions whose corresponding solutions preserve the orthogonality property. Let us start with the second-order difference equation of hypergeometric type σ (x)Δ∇y(x) + τ (x)Δy(x) + λy(x) = 0, © Springer Nature Switzerland AG 2020 M. Masjed-Jamei, Special Functions and Generalized Sturm–Liouville Problems, Frontiers in Mathematics, https://doi.org/10.1007/978-3-030-32820-7_2

(2.2) 157

158

2 Generalized Sturm–Liouville Problems in Discrete Spaces

where σ (x) = ax 2 + bx + c

and τ (x) = px + q

with

p = 0

are polynomials of degree at most 2 and 1, respectively, and λ is a constant. Then Eq. (2.2) can be written as   Δ σ (x)w(x)∇y(x) + λw(x)y(x) = 0, in which w(x) satisfies the Pearson-type difference equation   Δ σ (x)w(x) = τ (x)w(x).

(2.3)

The solutions of Eq. (2.2) with   λ ≡ λn = −n (n − 1)a + d are polynomials of degree n, say y(x) = yn (x), and usually called hypergeometric-type discrete polynomials. They are orthogonal with respect to the weight function w(x) on the counter set x = A, A + 1, . . . , B as B 

 yn (x)ym (x)w(x) =

x=A

B 

yn2 (x)w(x) δn,m ,

x=A

provided that w(x) > 0 for A ≤ x ≤ B and σ (x)w(x)x k

x=A,B+1

= 0, ∀k ≥ 0.

For y(x) = yn (x), Eqs. (2.2) and (2.3) can be represented as σ1 (x)(Δ2 yn )(x) + τ1 (x)(Δyn )(x) + λn yn (x + 1) = 0 and   Δ σ1 (x − 1)w(x) = τ1 (x)w(x + 1), where σ1 (x) = σ (x + 1) + τ (x + 1) and τ1 (x) = τ (x + 1),

2.1 Introduction

159

or as σ2 (x)(∇ 2 yn )(x) + τ2 (x)(∇yn )(x) + λn yn (x − 1) = 0 and   Δ σ2 (x + 1)w(x) = τ2 (x + 1)w(x), where σ2 (x) = σ (x − 1) and τ2 (x) = τ (x − 1). According to Hahn’s classification, Eq. (2.2) has four orthogonal polynomial solutions, which are known in the literature, respectively, as follows: • The CHARLIER polynomials:  Cn (x; a) = 2 F0



−n, −x 1 − , − a

orthogonal with respect to the Poisson weight function as ∞  ak k=0

k!

Cn (k; a)Cm (k; a) = a −n ea n! δn,m (a > 0).

• The MEIXNER polynomials:  Mn (x; β, c) = 2 F1



−n, −x 1 , 1 − c β

orthogonal with respect to the Pascal weight function as ∞  (β)k ck k=0

k!

Mn (k; β, c)Mm (k; β, c) =

c−n n! δn,m (β > 0, 0 < c < 1). (β)n (1 − c)β

• The KRAVCHUK polynomials:  Kn (x; p, N) = 2 F1



−n, −x 1 , n = 0, 1, 2, . . . , N, −N p

160

2 Generalized Sturm–Liouville Problems in Discrete Spaces

orthogonal with respect to the binomial weight function as N    N

k

k=0

pk (1 − p)N−k Kn (k; p, N)Km (k; p, N) =

(−1)n n! (−N)n



 1−p n δn,m (0 < p < 1), p

• the HAHN polynomials:  Qn (x; α, β, N) = 3 F2



−n, −x, n + α + β + 1 1 , n = 0, 1, 2, . . . , N, −N, α + 1

orthogonal with respect to the hypergeometric weight function as   N   α+k β +N −k Qn (k; α, β, N)Qm (k; α, β, N) k N −k k=0

=

(−1)n (n + α + β + 1)N+1 (β + 1)n n! δn,m (α > −1, β > −1). (2n + α + β + 1)(α + 1)n (−N)n N!

Table 2.1 shows the data σ (x) and τ (x) for each of the above four polynomial families. By referring to this table, we observe that all polynomial coefficients σ and τ have real zeros. However, there is still the main case of the real difference equation (2.2), whose coefficients have complex zeros with four free parameters. To obtain this case, we first expand the Pearson equation (2.3) as follows: σ (x) + τ (x) σ1 (x − 1) w(x + 1) = = . w(x) σ (x + 1) σ1 (x) − τ1 (x)

(2.4)

It is clear that if in (2.4) σ (x) has degree < 2, it leads to the Charlier, Meixner, and Kravchuk polynomials, while if σ (x) has the exact degree 2, the Hahn polynomials are derived if both the numerator σ (x) + τ (x) and the denominator σ (x + 1) have real factorizations. Another case in (2.4) is that in which the polynomials σ (x) + τ (x) and σ (x + 1) are real and have complex zeros. In other words, consider Eq. (2.4) in the expanded form ax 2 + (b + d)x + c + e w(x + 1) = . w(x) ax 2 + (2a + b)x + a + b + c

(2.5)

Table 2.1 Four cases of classical discrete orthogonal polynomials Symbol

Qn (x; α, β, N)

Mn (x; β, c)

Kn (x; p, N)

Cn (x; a)

σ (x)

x(N + α − x)

x

x

τ (x)

(β + 1)(N − 1) − (α + β + 2)x

(c − 1)x + βc

x Np − x 1−p

a−x

2.1 Introduction

161

Then only two cases can occur for the parameter a in (2.5), i.e., a = 0 or a = 0. If a = 0, the result is well known. So by assuming a = 1 = 0, we will obtain the simplified equation x 2 + (b + d)x + c + e w(x + 1) = 2 . w(x) x + (b + 2)x + b + c + 1

(2.6)

Now, four cases can occur for Eq. (2.6): 1. Both numerator and denominator have real zeros, namely w(x + 1) (x + p)(x + q) = w(x) (x + r)(x + s)

(p, q, r, s ∈ R).

(2.7)

2. The numerator has real zeros but the denominator has complex roots, namely (x + p)(x + q) w(x + 1) = (p, q, r, s ∈ R). w(x) (x + r + is)(x + r − is)

(2.8)

3. The numerator has complex zeros but the denominator has real roots, namely (x + p + iq)(x + p − iq) w(x + 1) = (p, q, r, s ∈ R). w(x) (x + r)(x + s)

(2.9)

4. Finally, both numerator and denominator have complex zeros, namely (x + p + iq)(x + p − iq) w(x + 1) = (p, q, r, s ∈ R). w(x) (x + r + is)(x + r − is)

(2.10)

Here we should remark that the solutions of Eqs. (2.7)–(2.10) are expressible in terms of the gamma function Γ (z) = lim

n! nz

n n→∞ 

,

(2.11)

(z + k)

k=0

which implies that Γ (p + iq) Γ (p − iq) = Γ 2 (p)

∞ k=0

(p + k)2 (p + k)2 + q 2

(2.12)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

is always a real positive value for all p > 0 and q ∈ R. One of the consequences of (2.12) is that the multiplicative term n−1 

(p + iq)n (p − iq)n =

q 2 + (p + k)2



(p, q ∈ R)

k=0

is also a real positive value. Moreover, when p = m ∈ N, we have Γ (m + iq) Γ (m − iq) =

m−1   πq q 2 + (m − k)2 . sinh(πq) k=1

By noting the above remark, the solutions of Eqs. (2.7)–(2.10) can be, respectively, represented as w(x) = w1 (x; p, q, r, s) =

Γ (x + p)Γ (x + q) , Γ (x + r)Γ (x + s)

(2.13)

w(x) = w2 (x; p, q, r, s) =

Γ (x + p)Γ (x + q) , Γ (x + r + is)Γ (x + r − is)

(2.14)

w(x) = w3 (x; p, q, r, s) =

Γ (x + p + iq)Γ (x + p − iq) , Γ (x + r)Γ (x + s)

(2.15)

w(x) = w4 (x; p, q, r, s) =

Γ (x + p + iq)Γ (x + p − iq) . Γ (x + r + is)Γ (x + r − is)

(2.16)

and

On the other hand, since a weight function must be positive, relation (2.13) implies that p = q and r = s, leading to w1 (x; p, p, r, r) =

Γ 2 (x + p) > 0. Γ 2 (x + r)

(2.17)

Also, by referring to the important relation (2.12), the function (2.14) is positive when p = q, i.e., w2 (x; p, p, r, s) =

Γ 2 (x + p) > 0. Γ (x + r + is)Γ (x + r − is)

(2.18)

Similarly, for the case (2.15), we must have r = s, i.e., w3 (x; p, q, r, r) =

Γ (x + p + iq)Γ (x + p − iq) > 0. Γ 2 (x + r)

(2.19)

2.2 A Finite Sequence of Hahn-Type Orthogonal Polynomials

163

Finally, for the case (2.16), we observe that no restriction is required and w4 (x; p, q, r, s) is always positive. A short look at relations (2.17)–(2.19) shows that they are just particular cases of the positive weight function (2.16). This means that we are dealing with a main sequence of hypergeometric orthogonal polynomials of a discrete variable with four free parameters that is finitely orthogonal on the real line. In the next section, we introduce such a sequence and consider three particular cases of it corresponding to the weight functions (2.17)–(2.19).

2.2

A Finite Sequence of Hahn-Type Orthogonal Polynomials

For p, q, r, s ∈ R, suppose that σ (x) + τ (x) = (x + p)2 + q 2 and σ (x + 1) = (x + r)2 + s 2 , which leads to the real equation     (x + p + 1)2 + q 2 (Δ2 yn )(x) + 2(p + 1 − r)x + (p + 1)2 + q 2 − r 2 − s 2 (Δyn )(x) − n(n + 1 + 2p − 2r)yn (x + 1) = 0,

(2.20)

or   (x + r − 2)2 + s 2 (∇ 2 yn )(x)   + 2(p + 1 − r)x + p2 + q 2 − r 2 − s 2 − 2p + 4r − 3 (∇yn )(x) − n(n + 1 + 2p − 2r)yn (x − 1) = 0. We look for a polynomial solution of Eq. (2.20) of the form yn (x) =

n  k=0

an,k

  x − 1 + r + is , an,n = 0, k

which is a particular case of the general solution yn (x) =

∞  k=0

an,k

  x+c . k

(2.21)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

Since   1 x+c (−1)k (−x − c)k = (x + c)[k] = k! k! k for x [k] = x(x − 1)(x − 2) · · · (x − k + 1) and we have the relations Δ(x +c)[n] = n(x +c)[n−1] and x(x +c)[n] = (x +c)[n+1] +(n−c)(x +c)[n],

(2.22)

we can directly prove the following theorem. Theorem 2.1 The monic polynomial solution of the difference equation (2.20) is yn (x) = R¯ n (x; p, q, r, s) =

(−r + p + 1 − i(q + s))n (−r + p + 1 + i(q − s))n (n + 2p − 2r + 1)n 

× 3 F2

(2.23)

−n, n + 1 + 2p − 2r, −x + 1 − r − is 1 −r + p + 1 − i(q + s), −r + p + 1 + i(q − s)

(−r + p + 1 + i(q + s))n (−r + p + 1 + i(q − s))n (n + 2p − 2r + 1)n

 −n, n + 1 + 2p − 2r, x + p + iq × 3 F2 1 . −r + p + 1 + i(q + s), −r + p + 1 + i(q − s) = (−1)n

Proof For c = r + is − 1, if in (2.20) we take yn (x) = R¯ n (x; p, q, r, s) =

∞ 

an,k (x + c)[k]

k=0

and use properties (2.22), we eventually obtain the recurrence equation  (k − n) (k + n + 1 + 2 p − 2 r) an,k + (k + 1) −s 2 + 2i (p + k − r + 1) s + p2  + (4 k − 2 n − 2 r + 4) p + 2 k 2 − 4 kr − n2 + 2 nr + q 2 + r 2 + 5 k − n − 4 r + 3 an,k+1 − (k + 1) (k + 2) (iq + is + k + p − r + 2) (iq − is − k − p + r − 2) an,k+2 = 0,

which has a solution of the hypergeometric form (2.23).

2.2 A Finite Sequence of Hahn-Type Orthogonal Polynomials

165

The second representation of R¯ n comes from (2.21) following the same approach and using the basis (−x + c)[k] for c = −p − iq with the property ∇(−x + c)[k] = −k(−x + c)[k−1] . We can show that both representations of R¯ n are solutions of the following three-term recurrence relation: R¯ n+1 (x; p, q, r, s) = (x − cn )R¯ n (x; p, q, r, s) − dn R¯ n−1 (x; p, q, r, s), where cn = −

  (p + r − 1) n2 + (p + r − 1) (1 + 2 p − 2 r) n + (−r + p) p 2 + q 2 − r 2 − s 2 + 2 r − 1 2 (n + p − r) (n + 1 + p − r)

and dn =

   n (2 r − 2 p − n) (n + p − r)2 + (q − s)2 (n + p − r)2 + (q + s)2 4 (n + p − r)2 (2 n + 1 + 2 p − 2 r) (2 n − 1 + 2 p − 2 r)

, (2.24)

with the unique initial conditions p2 + q 2 − r 2 − s 2 + 2 r − 1 . R¯ 0 (x; p, q, r, s) = 1 and R¯ 1 (x; p, q, r, s) = x + 2(p + 1 − r)

 

There is a direct relationship between R¯ n (x; p, q, r, s) and continuous Hahn polynomials defined by

 −n, n + a + b + c + d − 1, a + ix i n (a + c)n (a + d)n Pn (x; a, b, c, d) = 1 , 3 F2 n! a + c, a + d so that we have R¯ n (x; p, q, r, s) = P¯n (ix; 1 − r − is, 1 − r + is, p − iq, p + iq). We now prove that the monic polynomials (2.23) are finitely orthogonal on the real line. For this purpose, we first reconsider Eq. (2.20) and write it in a self-adjoint form to obtain 

w4 (x; p, q, r, s)(x + p + iq)(x + p − iq)

 ∞  × R¯ m (x; p, q, r, s)R¯ n (x + 1; p, q, r, s) − R¯ n (x; p, q, r, s)R¯ m (x + 1; p, q, r, s)

−∞

= (n(n + 1 + 2p − 2r) − m(m + 1 + 2p − 2r)) ×

∞  x=−∞

w4 (x; p, q, r, s)R¯ m (x; p, q, r, s)R¯ n (x; p, q, r, s).

(2.25)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

In order to show that the left-hand side of (2.25) is equal to zero when m = n, we use the following limit relations: lim

x→∞

Γ (x + a) = 1 and Γ (x)x a

lim

x→−∞

Γ (x + a) = (−1)a (∀a ∈ C), Γ (x)x a

(2.26)

which can be proved directly via the limit definition (2.11). Since   deg R¯ m (x; p, q, r, s)R¯ n (x + 1; p, q, r, s) − R¯ n (x; p, q, r, s)R¯ m (x + 1; p, q, r, s) = n + m − 1,

the left-hand side of (2.25) is equal to zero for n = m if and only if Γ (x + p + iq)Γ (x + p − iq) k+2 x = 0 for all k = 0, 1, . . . , n + m − 1. x→±∞ Γ (x + r + is)Γ (x + r − is) (2.27) By noting (2.26), it is now straightforward to verify that (2.27) is equivalent to lim

lim x k+2−2r+2p = 0 and

x→∞

lim (−x)k+2−2r+2p = 0 for any k = 0, 1, . . . , n + m − 1.

x→−∞

(2.28) Finally, if in (2.28) we take max{m, n} = N, the left-hand side of (2.25) will be equal to zero if and only if 2N + 1 − 2r + 2p < 0

⇔

1 N 0. Γ (r − p + i(s − q))Γ (r − p − i(s − q))Γ (r − p − i(s + q))Γ (r − p + i(s + q))

Theorem 2.2 The polynomial set {R¯ n (x; p, q, r, s)}n=0 is finitely orthogonal with respect to the weight function w4 (x; p, q, r, s) on the real line, so that we have N 0 for xk ∈ {A, A + 1, . . . , B − 2, B − 1}, it is necessary and sufficient that (A) > 0

and (x + 1)/(x) > 0

for x = A, A + 1, . . . , B − 2.

First Case: a = 0 When a = 0 in (2.32), we can assume, without loss of generality, that b = 1 and therefore σ (x) = x + c ,

τ (x) = −2x,

and λn = 2n

(c = 0) .

Hence relation (2.37) can be rewritten as c−x (x + 1) = , (x) x+c+1 with the explicit solution (x) =

1 , Γ (c − x + 1) Γ (c + x + 1)

for x ∈ {−c, −c + 1, . . . , c − 1, c} as support of orthogonality. Moreover, the condition 2c ∈ N implies c ∈ N or c + 1/2 ∈ N. The corresponding monic symmetric polynomial solutions of Eq. (2.30) is obtained from (2.35), with Cn =

(1 + 2 c − n) n 4

(1 ≤ n ≤ 2c),

or from its representation in terms of hypergeometric series as  Pn (x) = 2

−n

(−2c)n 2 F1

−n , −c − x 2 −2c

(1/2)

= kn

(x + c; 2c)

(0 ≤ n ≤ 2c),

(p)

where kn (x; N) are monic Kravchuk polynomials for 0 < p < 1 and N ∈ N. In this sense, the factorial moments are computed as c  k=−c

22c−n (k + c)[n] = (2c)[n] Γ (c − k + 1) Γ (c + k + 1) Γ (1 + 2c)

(0 ≤ n ≤ 2c) .

2.3 Classical Symmetric Orthogonal Polynomials of a Discrete Variable

173

Second Case: a = 0 In this case, we can assume, without loss of generality, that a = 1 in (2.32). As shown in (2.38), the roots of σ (x) fix the lower bound of the support A. In order to have real roots for σ (x), the condition b2 − 4c ≥ 0 is needed. Representation of the polynomials in terms of hypergeometric series must therefore be given in each particular case. • If b2 − 4c = 0 (b = 0), then we have   b 2 , σ (x) = x + 2

τ (x) = −2bx,

and λn = n (1 + 2b − n) .

Hence (2.37) takes the form (x + 1) (x − b/2)2 = , (x) (x + b/2 + 1)2 with the symmetric solution (x) =   Γ x+

1 b 2



 + 1 Γ −x +

b 2

+1

 2 ,

and the support of orthogonality is {−b/2, −b/2 + 1, . . . , b/2 − 1, b/2} where b ∈ N. Now, the symmetric polynomial solutions corresponding to Eq. (2.30) are obtained from (2.35), with Cn =

(1 + b − n)2 (2 + 2 b − n) n 4 (1 + 2 b − 2 n) (3 + 2 b − 2 n)

(1 ≤ n ≤ b),

and they can be represented as ⎞ ⎛ b ((−b)n )2 −n , n − 2b − 1 , − − x ⎠ Pn (x) = 1 2 3 F2 ⎝ (n − 2b − 1)n −b , −b = h˜ n(0,0)(x + b/2; b + 1)

(0 ≤ n ≤ b) ,

where h˜ n (x; N) are monic Hahn–Eberlein polynomials [1] for μ > −1, ν > −1, and N ∈ N, generally defined by (μ,ν)

(−1)n (N − n)n (1 − N − ν)n (x; N) = h˜ (μ,ν) n (n + 1 − 2N − μ − ν)n

 −n, −x, n − 2N − μ − ν + 1 × 3 F2 1 1 − N − ν, 1 − N

(0 ≤ n ≤ N − 1) .

(2.39)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

In this sense, the factorial moments are computed as b/2 

(k + b/2)[n]  2 k=−b/2 Γ (k + b/2 + 1) Γ (−k + b/2 + 1) 22b−n Γ (b + 1/2) b[n] (b − E((n + 1)/2))[E(n/2)] = √ π (Γ (b + 1))3 (b − 1/2)[E(n/2)]

(0 ≤ n ≤ b) ,

where E(x) denotes the largest integer less than or equal to x. • If b2 − 4c > 0 (b = 0), it is possible to write σ (x) = x 2 + bx + c = (x − δ1 )(x − δ2 ), where δ1 = − and d =

b+d , 2

δ2 =

d −b , 2

√ b2 − 4c > 0. Thus δ1 < δ2 .

Moreover, we have τ (x) = 2 (δ1 + δ2 ) x,   λn = −n n + 2(δ1 + δ2 ) − 1 , and (x + 1) (x + δ1 )(x + δ2 ) = , (x) (x − δ1 + 1)(x − δ2 + 1) as well as Cn =

− (n (−1 + 2 δ1 + n) (−1 + 2 δ2 + n) (2 (−1 + δ1 + δ2 ) + n)) . 4 (−3 + 2 δ1 + 2 δ2 + 2 n) (−1 + 2 δ1 + 2 δ2 + 2 n)

Conditions (2.38) and (xk ) > 0 with xk ∈ {A, A + 1, . . . , B − 2, B − 1} will be satisfied in some cases. The combination of the four roots of σ (x) and σ (−x) (δ1 , δ2 and −δ1 , −δ2 , respectively) as endpoints of the support generates five situations, in which the case [δ1 , −δ1 ] gives two different possibilities.

2.3 Classical Symmetric Orthogonal Polynomials of a Discrete Variable

175

For the case {[δ1, −δ1 ], 1 }, the support of orthogonality is {δ1 , δ1 + 1, . . . , −δ1 − 1, −δ1 }

with δ1 < δ2 < 0 and δ2 < δ1 + 1.

In this situation, −2δ1 ∈ N, and the orthogonality weight appears as 1 (x) =

1 . Γ (−x − δ1 + 1) Γ (−x − δ2 + 1) Γ (x − δ1 + 1) Γ (x − δ2 + 1)

For the case {[δ1, −δ1 ], 2 }, the support of orthogonality is {δ1 , δ1 + 1, . . . , −δ1 − 1, −δ1 }

with

δ2 > −δ1 .

In this case, −2δ1 ∈ N, and the orthogonality weight is denoted by 2 (x) =

Γ (−x + δ2 ) Γ (x + δ2 ) . Γ (−x − δ1 + 1) Γ (x − δ1 + 1)

For the case {[δ2, −δ2 ], 1 }, the support is {δ2 , δ2 + 1, . . . , −δ2 − 1, −δ2 } with the weight function 1 (x). Also −2δ2 ∈ N. For the case {[δ2, −δ1 ], 1 }, the support of orthogonality is {δ2 , δ2 + 1, . . . , −δ1 − 1, −δ1 }

with δ1 − δ2 + 1 > 0,

and the weight 1 (x), where −δ1 − δ2 ∈ N. For the case {[δ1, −δ2 ], 1 }, the support of orthogonality is {δ1 , δ1 + 1, . . . , −δ2 − 1, −δ2 }

with

δ1 − δ2 + 1 > 0.

In this situation, −δ1 − δ2 ∈ N, and the orthogonality weight is 1 (x). Note that the inequality conditions in each case come from a careful analysis of the positivity of (x + 1)/(x)

for x ∈ {A, A + 1, . . . , B − 2}

and an appropriate choice of the argument in the gamma function inside the eight possible representations of (x). In order to finish the classification, we present in each of these five cases the factorial moments for the weights as well as the representation in terms of hypergeometric series of the polynomial solutions of Eq. (2.30) assuming that a = 1 and b2 − 4c > 0.

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

For the case {[δ1, −δ1 ], 1 }, the symmetric polynomial solutions of Eq. (2.30) can be written in terms of a hypergeometric series as (δ1 + δ2 )n (2δ1 )n Pn (x) = 3 F2 (n + 2(δ1 + δ2 ) − 1)n



−n , n + 2(δ1 + δ2 ) − 1 , δ1 − x 1 δ1 + δ2 , 2δ1

1 −δ2 ,δ1 −δ2 ) (x − δ ; 1 − 2δ ) = h˜ (δ 1 1 n



(0 ≤ n ≤ −2δ1 ) ,

where h˜ n (x; N) are monic Hahn–Eberlein polynomials defined in (2.39). Also, the factorial moments are computed as (μ,ν)

−δ1 

1 (k)(k − δ1 )[n] =

k=δ1

22(δ1 +δ2 ) ×

Γ (1/2 − δ1 − δ2 ) √ π Γ (1 − 2δ2 ) Γ (1 − δ1 − δ2 ) Γ (1 − 2δ1 )

(−1)n δ1 (1 + 2δ1 )n−1 (δ1 + δ2 + E((n + 1)/2))E(n/2) , (δ1 + δ2 + 1/2)E(n/2) 2n−1

valid for n = 0, 1, 2, . . . , −2δ1 . For the case {[δ1, −δ1 ], 2 }, the symmetric polynomial solutions of Eq. (2.30) can be written as (δ1 + δ2 )n (2δ1 )n Pn (x) = 3 F2 (n + 2(δ1 + δ2 ) − 1)n



−n , n + 2(δ1 + δ2 ) − 1 , δ1 − x 1 δ1 + δ2 , 2δ1

1 +δ2 −1,δ1 +δ2 −1) (x − δ ; 1 − 2δ ) = h(δ 1 1 n

(α,β)

where hn by



(0 ≤ n ≤ −2δ1 ) ,

(2.40)

(x; N) are monic Hahn polynomials for α > −1, β > −1, and N ∈ N defined

(x; N) = h(α,β) n

(−1)n (N − n)n (β + 1)n (α + β + n + 1)n 

−n , −x , n + α + β + 1 × 3 F2 1 1 − N ,β + 1

(0 ≤ n ≤ N − 1) .

In this sense, note that if δ1 + δ2 = 1, the polynomials (2.40) are reduced to the monic Gram polynomials.

2.3 Classical Symmetric Orthogonal Polynomials of a Discrete Variable

177

The moments corresponding to this case are computed as −δ1 

2 (k)(k − δ1 )

k=δ1

[n]

√ π Γ (2δ2 ) Γ (δ1 + δ2 ) = 2(δ +δ )−1 1 2 2 Γ (1 − 2δ1 ) Γ (δ1 + δ2 + 1/2) ×

(−1)n δ1 (1 + 2δ1 )n−1 (δ1 + δ2 + E((n + 1)/2))E(n/2) , (δ1 + δ2 + 1/2)E(n/2) 2n−1

valid for n = 0, 1, 2, . . . , −2δ1 . For the case {[δ2, −δ2 ], 1 }, the symmetric polynomial solutions of Eq. (2.30) can be represented as

 (δ1 + δ2 )n (2δ2 )n −n , n + 2(δ1 + δ2 ) − 1 , δ2 − x 1 Pn (x) = 3 F2 (n + 2(δ1 + δ2 ) − 1)n δ1 + δ2 , 2δ2 = h˜ n(δ2 −δ1 ,δ2 −δ1 ) (x − δ2 ; 1 − 2δ2 )

(0 ≤ n ≤ −2δ2 ) ,

where h˜ n (x; N) are monic Hahn–Eberlein polynomials. The factorial moments are computed as (μ,ν)

−δ2 

1 (k)(k − δ2 )[n] =

k=δ2

2δ1 +δ2 ×

Γ (1/2 − δ1 − δ2 ) √ π Γ (1 − 2δ2 ) Γ (1 − δ1 − δ2 ) Γ (1 − 2δ1)

(−1)n δ2 (1 + 2δ2 )n−1 (δ1 + δ2 + E((n + 1)/2))E(n/2) , (δ1 + δ2 + 1/2)E(n/2) 2n−1

valid for n = 0, 1, 2, . . . , −2δ2 . For the case {[δ2, −δ1 ], 1 }, the symmetric polynomial solutions of Eq. (2.30) can be represented as

 (δ1 + δ2 )n (2δ1 )n −n , n + 2(δ1 + δ2 ) − 1 , δ2 − x 1 Pn (x) = 3 F2 (n + 2(δ1 + δ2 ) − 1)n δ1 + δ2 , 2δ1 1 −δ2 ,δ2 −δ1 ) (x − δ ; 1 − δ − δ ) = h˜ (δ 2 1 2 n

(0 ≤ n ≤ −δ1 − δ2 ) .

Also, the factorial moments are computed as −δ1  k=δ2

1 (k)(k − δ2 )[n] =

22(δ1 +δ2 ) ×

Γ (1/2 − δ1 − δ2 ) √ π Γ (1 − 2δ2 ) Γ (1 − δ1 − δ2 ) Γ (1 − 2δ1 )

(−1)n δ2 (1 + 2δ2 )n−1 (δ1 + δ2 + E((n + 1)/2))E(n/2) , (δ1 + δ2 + 1/2)E(n/2) 2n−1

valid for n = 0, 1, 2, . . . , −δ1 − δ2 .

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

Finally, for the case {[δ1 , −δ2 ], 1 }, the symmetric polynomial solutions of Eq. (2.30) can be written in terms of a hypergeometric series as (δ1 + δ2 )n (2δ1 )n Pn (x) = 3 F2 (n + 2(δ1 + δ2 ) − 1)n



−n , n + 2(δ1 + δ2 ) − 1 , δ1 − x 1 δ1 + δ2 , 2δ1

1 −δ2 ,δ2 −δ1 ) (x − δ ; 1 − δ − δ ) = h˜ (δ 1 1 2 n



(0 ≤ n ≤ −δ1 − δ2 ) ,

and we have −δ2 

1 (k)(k − δ1 )[n] =

k=δ1

22(δ1 +δ2 ) ×

Γ (1/2 − δ1 − δ2 ) √ π Γ (1 − 2δ2 ) Γ (1 − δ1 − δ2 ) Γ (1 − 2δ1 )

(−1)n δ1 (1 + 2δ1 )n−1 (δ1 + δ2 + E((n + 1)/2))E(n/2) , (δ1 + δ2 + 1/2)E(n/2) 2n−1

which is valid for n = 0, 1, 2, . . . , −δ1 − δ2 .

2.4

A Symmetric Generalization of Sturm–Liouville Problems in Discrete Spaces

In this section, we generalize the difference equation (2.1) symmetrically and show that the corresponding orthogonality property is preserved. For this purpose, we first consider the following limit relations: y(x + h) − 2y(x) + y(x − h) h→0 h2

y  (x) = lim and

y  (x) = lim

h→0

y(x + h) − y(x) . h

Suppose in Eq. (1.94) that Φn (x) = y(x). We can approximate it up to the second order in terms of h in order to obtain a second-order difference equation as

A(x)

y(x + h) − y(x) y(x + h) − 2y(x) + y(x − h) + B(x) 2 h h   + λn C(x) + D(x) + σn E(x) y(x) = 0.

(2.41)

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

179

In general, two cases can be considered for the step size h = xi+1 − xi in (2.41). They are respectively uniform step size for h, which can be fixed equal to one without loss of generality, and nonuniform lattices. Although the extension of Sturm–Liouville problems is possible for both cases, we here consider the first one. If we assume in what follows that h = 1, then for y(x) = φn (x), (2.41) takes the form A(x)Δ∇φn (x) + B(x)Δφn (x) + (λn C(x) + D(x) + σn E(x)) φn (x) = 0,

(2.42)

where Δ∇ = Δ − ∇

  and σn = 1 − (−1)n /2.

In order to have a symmetric solution for the difference equation (2.42), some initial conditions should be considered for its coefficients. If Eq. (2.42) is expanded as 

   A(x) + B(x) φn (x + 1) − 2A(x) + B(x) φn (x) + A(x)φn (x − 1)   = λn C(x) + D(x) + σn E(x) φn (x),

(2.43)

then by taking into account that the solution is symmetric, i.e., φn (−x) = (−1)n φn (x), we conclude from (2.43) that A(x) + B(x) = A(−x), C(−x) = C(x), D(−x) = D(x),

(2.44)

and E(−x) = E(x). The first condition of (2.44) implies that B(x) = A(−x) − A(x) is an odd function. Therefore, Eqs. (2.42) and (2.43) are respectively simplified as     A(x)Δ∇φn (x) + A(−x) − A(x) Δφn (x) + λn C(x) + D(x) + σn E(x) φn (x) = 0 (2.45) and A(−x)φn (x + 1) + A(x)φn (x − 1)   = − λn C(x) + D(x) − A(x) − A(−x) + σn E(x) φn (x), leading to the following theorem.

(2.46)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

Theorem 2.3 Let φn (x) = (−1)n φn (−x) be a sequence of symmetric functions that satisfies the difference equation (2.45) (or equivalently (2.46)). If A(x) is a free real function and C(x), D(x), and E(x) are real even functions, then β−1  x=α

W ∗ (x)φn (x)φm (x) =

 β−1 

 W ∗ (x)φn2 (x) δn,m ,

x=α

where W ∗ (x) = C(x)W (x)

(2.47)

and W (x) is a symmetric solution of the Pearson difference equation     Δ A(x)W (x) = A(−x) − A(x) W (x),

(2.48)

which is equivalent to A(−x) W (x + 1) = . W (x) A(x + 1)

(2.49)

Moreover, the weight function defined in (2.47) must be even over one of the following symmetric counter sets: (i) (ii) (iii) (iv)

S1 S2 S3 S4

= {−a − n, −a − n + 1, . . . , −a − 1, −a, a, a + 1, a + n − 1, a + n}, a ∈ R, = S1 ∪ {0} (since every odd function is equal to zero at x = 0), = {. . . , −a − n, −a − n + 1, . . . , −a − 1, −a, a, a + 1, a + n − 1, a + n, . . . } = S3 ∪ {0},

and the function A(x)W (x) must also vanish at x = α and x = 1 − β, where [α, β − 1] ∈ {S1 , S2 , S3 , S4 }. Proof If Eq. (2.45) is written in a self-adjoint form, then     Δ A(x)W (x)∇φn (x) + λn C(x) + D(x) + σn E(x) W (x)φn (x) = 0

(2.50)

    Δ A(x)W (x)∇φm (x) + λm C(x) + D(x) + σm E(x) W (x)φm (x) = 0.

(2.51)

and

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

181

By multiplying (2.50) by φm (x) and (2.51) by φn (x) and subtracting one from the other and also noting that     φm (x)Δ A(x)W (x)∇φn (x) − φn (x)Δ A(x)W (x)∇φm (x)    = Δ A(x)W (x) φm (x)∇φn (x) − φn (x)∇φm (x) , we obtain   Δ A(x)W (x) (φm (x)∇φn (x) − φn (x)∇φm (x)) + (λn − λm ) C(x)W (x)φn (x)φm (x) +

(−1)m − (−1)n E(x)W (x)φn (x)φm (x) = 0. 2 (2.52)

Now, a simple but important idea can appear here: “The sum of any odd summand on a symmetric counter set is equal to zero.” This set can be one of the four introduced sets S1 , S2 , S3 , S4 . Consequently, summation on both sides of (2.52) gives β−1 

   Δ A(x)W (x) φm (x)∇φn (x) − φn (x)∇φm (x)

x=α

+(λn − λm )

β−1  x=α

C(x)W (x)φn (x)φm (x)+

β−1 (−1)m − (−1)n  E(x)W (x)φn (x)φm (x) = 0, 2 x=α

(2.53) in which [α, β − 1] ∈ {S1 , S2 , S3 , S4 } is assumed to be the main support. On the other hand, W (x) is a symmetric solution for the Pearson equation (2.49), and therefore the condition W (x + 1)A(x + 1) − W (x)A(−x) = 0 is valid for all x ∈ [α, β − 1]. Hence without loss of generality, we can assume [α, β − 1] = [−θ, θ ], i.e., α = −θ and β = θ + 1. By noting the identity φm (x)∇φn (x) − φn (x)∇φm (x) = φn (x)φm (x − 1) − φm (x)φn (x − 1),

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

the first sum of (2.53) is simplified as θ 

Δ (A(x)W (x) (φm (x)∇φn (x) − φn (x)∇φm (x)))

x=−θ

  x=θ+1 = A(x)W (x) φn (x)φm (x − 1) − φm (x)φn (x − 1) x=−θ

= A(θ + 1)W (θ + 1) (φn (θ + 1)φm (θ ) − φm (θ + 1)φn (θ )) − A(−θ )W (−θ ) (φn (−θ )φm (−θ − 1) − φm (−θ )φn (−θ − 1)) .

(2.54)

Since W (−x) = W (x), the solutions are symmetric, and the Pearson equation (2.49) is valid for x = θ , i.e., A(θ + 1)W (θ + 1) = A(−θ )W (θ ), relation (2.54) changes to θ  x=−θ

Δ (A(x)W (x) (φm (x)∇φn (x) − φn (x)∇φm (x)))   = A(−θ )W (θ ) 1 + (−1)n+m (φm (θ )φn (θ + 1) − φn (θ )φm (θ + 1)) .

(2.55)

By noting that φm (θ )φn (θ + 1) − φn (θ )φm (θ + 1) = 0, two cases can occur for relation (2.55): 1. If n + m is odd, then 1 + (−1)n+m = 0, and the right-hand side of equality (2.55) is automatically zero. This case is clear, since the sum of any odd summand on a symmetric counter set is equal to zero. 2. If A(−θ )W (θ ) = 0, then W (θ ) = 0 or A(−θ ) = 0,

i.e., A(α) = A(1 − β) = 0.

In this case, the right-hand side of equality (2.55) is again equal to zero. Now, in order to prove the orthogonality property, it remains to show that F (m, n) =

β−1 (−1)m − (−1)n  E(x)W (x)φn (x)φm (x) = 0. 2 x=α

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

183

For this purpose, four cases should be considered for values m, n respectively as follows: 1. If both m and n are even (or odd), then it is obvious that F (n, m) = 0. 2. If m is even and n is odd (or conversely), then we have F (2i, 2j + 1) =

β−1 

E(x)W (x)φ2j +1 (x)φ2i (x).

x=α

Since E(x), W (x), and φ2i (x) are assumed to be even functions and φ2j +1 (x) is odd, we are computing a sum of an odd function on a symmetric set of points, which means that F (2i, 2j + 1) = 0.  

2.4.1

Some Illustrative Examples

Although there are many examples of difference equation (2.45) whose solutions are symmetric functions, symmetric solutions of polynomial type are of special importance in physics and engineering. In this section, we first consider an example for the infinite case of Theorem 2.3 whose solution is a sequence of hypergeometric functions and then consider five examples of the extended Eq. (2.45) whose solutions are symmetric orthogonal polynomials. In each example, we obtain the corresponding orthogonality relation whose support for the first example is the infinite set x ∈ {. . . , −2, −1, 0, 1, 2, . . . } , and for the five further examples is x ∈ {−N, . . . , 0, 1, . . . , N} . Example 2.1 Consider the difference equation (x + b)Δ∇φn (x) − 2xΔφn (x) + (2n + eσn )φn (x) = 0 as a special case of (2.45) for A(x) = x + b, C(x) = 1, D(x) = 0, E(x) = e, λn = 2n.

(2.56)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

We introduce the symmetric basis ϑn (x) = (−1)[n/2] x σn (σn − x)[n/2] (σn + x)[n/2] = (−1)n ϑn (−x),

(2.57)

satisfying the following properties: n(n − 1)(n − 2) n(n − 1) ϑn−2 (x) + σn+1 ϑn−3 (x), 2 4 (2.58)  n 2 xϑn (x) = ϑn+1 (x) + σn+1 ϑn−1 (x), (2.59) 2

Δϑn (x) = nϑn−1 (x) +

and Δ∇ϑn (x) = n(n − 1)ϑn−2 (x).

(2.60)

If in (2.56), φn (x) =

∞ 

aj (n)ϑj (x),

j =0

then the recurrence relation of the coefficients will be derived as aj +2 (n) =

−eσn + 2j − 2n − 4   aj −2 (n),  +b (j − 1)j 1−j 2

j ≥ 0,

with two initial values a0 (n) and a1 (n). Therefore, we have a2j (n) =

  (−4)j a0 (n) − 4e − n2 j (2j )!(b − j + 1)j

j = 1, 2, . . . ,

,

and

a2j +1 (n) =

(−4)j a1 (n)



1 4 (−e

− 2n + 2)

(2j + 1)!(b − j )j

where (a)j =

Γ (a + j ) . Γ (a)

 j

,

j = 1, 2, . . . ,

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

185

The above two relations imply that for odd n we get  φn (x) = a0 (n) 3 F2

−(e + 2n)/4, −x, x 1 1/2, −b 

−(e + 2n − 2)/4, 1 − x, 1 + x 1 . + x a1 (n) 3 F2 3/2, 1 − b

(2.61)

But since φn (−x) = (−1)n φn (x), it follows that φ2m+1 (0) = 0, and consequently a0 (n) = 0 in (2.61). Similarly, for even n we obtain  φn (x) = a0 (n) 3 F2

−n/2, −x, x 1 1/2, −b



+ xa1 (n) 3 F2



(1 − n)/2, 1 − x, 1 + x 1 3/2, 1 − b

.

(2.62)

= (−1)n φn (−x; b, e)

(2.63)

By combining the two formulas (2.61) and (2.62), one can conclude that ⎛

⎞ 1 ((2 − e)σ − 2n), σ − x, σ + x n n n φn (x; b, e) = x σn 3 F2 ⎝ 4 1⎠ σn + 1/2, σn − b

is a symmetric solution for Eq. (2.56). In this sense, if e = 4k ∈ N in (2.63), the sequence will be reduced to a finite polynomial sequence. From Theorem 2.3, it is known that the sequence {φn (x; b, e)} satisfies an orthogonality relation of the form ⎛ ⎞ β−1 β−1   W1∗ (x)φn (x; b, e)φm(x; b, e) = ⎝ W1∗ (x)φn2 (x; b, e)⎠ δn,m , x=α

x=α

in which W1∗ (x) = C(x)W1 (x) = W1 (x)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

is the main weight function and W1 (x) satisfies the equation A(−x) b−x W1 (x + 1) = = . W1 (x) A(x + 1) b+x+1

(2.64)

Up to a periodic function of period 1, a symmetric solution of Eq. (2.64) is W1 (x) =

1 . Γ (b + 1 − x)Γ (b + 1 + x)

(2.65)

It is clear that the convergence of the hypergeometric series (2.63) depends on the value of x and the parameters b and e. In order to have a full convergence, we can impose some restrictions on b and e to reduce the hypergeometric series (2.63) to a polynomial function that is convergent everywhere (except infinity). For this purpose, as we pointed out, if e = 4k ∈ N, then φ¯n (x; b, 4k)



= (σn + 1/2)[n/2] (σn − b)[n/2] x σn 3 F2

−[n/2] − kσn , σn − x, σn + x 1 σn + 1/2, σn − b

(2.66)

is a monic finite sequence of symmetric polynomials orthogonal with respect to the weight function (2.65) with support Z. For instance, for n = 0, 1, 2, 3, 4, and 5, (2.66) takes the forms φ¯ 0 (x; b, 4k) = 1,



φ¯ 1 (x; b, 4k) = x 3 F2

−k, 1 − x, 1 + x 1 3/2, 1 − b

b φ¯ 2 (x; b, 4k) = x 2 − , 2 3(1 − b) φ¯ 3 (x; b, 4k) = x 3 F2 2



,

−k − 1, 1 − x, 1 + x 1 3/2, 1 − b

,

3 φ¯ 4 (x; b, 4k) = x 4 + (2 − 3b)x 2 + (b − 1)b, 4 

15 −k − 2, 1 − x, 1 + x φ¯ 5 (x; b, 4k) = (1 − b)(2 − b) x 3 F2 1 . 4 3/2, 1 − b To compute the norm square value of the polynomials (2.66), for n = 2m we obtain ∞ 

22b−4m (2m)! (φ¯ 2m (x; b, 4k))2 (−4b − 1)2m ∗ = = N2m , 2 2 Γ (b + 1 − x) Γ (b + 1 + x) Γ (2b + 1) ((1/2) ) ((−b) ) m m x=−∞ (2.67)

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

187

where b ∈ / Z+ , b > −1/2, and for n = 2m + 1 we have ∞ 

(m + k)!(b − (m + k + 1))Γ (m + k + 5/2) (φ¯ 2m+1 (x; b, 4k))2 = √ Γ (b + 1 − x) Γ (b + 1 + x) π Γ (2b + 1) x=−∞ ×

22b+1 (5/2 − b)m+k ((3/2)m+k )2 (1 − b)m+k

∗ = N2m+1

(b ∈ / Z+ , b > −1/2). (2.68)

By combining the two relations (2.67) and (2.68), the final form of the orthogonality relation is derived as ∞ 

φ¯ n (x; b, 4k) φ¯ m(x; b, 4k) = Nn∗ δn,m Γ (b + 1 − x) Γ (b + 1 + x) x=−∞

(b ∈ / Z+ , b > −1/2).

Example 2.2 Consider the difference equation   (2x−7)(2x+1)(5x − 9)(x + N)Δ∇φn (x) − 2x 4(5N − 24)x 2 + 73N + 63 Δφn (x)     (2.69) + n 1 − 4x 2 (5n − 10N + 43) + 78(2N + 1)σn φn (x) = 0, which is a special case of Eq. (2.45) for A(x) = (2x − 7)(2x + 1)(5x − 9)(x + N), C(x) = 1 − 4x 2 , D(x) = 0, E(x) = 78(2N + 1), λn = n(5n − 10N + 43). We are interested in computing the symmetric polynomial solution of Eq. (2.69). Hence, by taking φn (x) = x n + δn x n−2 + · · · and replacing it in (2.69), the solution satisfies a three-term recurrence relation as φn+1 (x) = xφn (x) − γn φn−1 (x) with φ0 (x) = 1 and φ1 (x) = x,

(2.70)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

where δn =

5n4 + (66 − 20N)n3 + (1 − 228N)n2 − (220N + 306)n + 234(2N + 1)σn . 24(10n − 10N + 33)

Since in general, γn = δn − δn+1 ,

(2.71)

the explicit form of γn (for even and odd n) is derived as γ2n = −

n(10n + 43)(5n − 5N + 4)(2n − 2N + 5) (20n − 10N + 33)(20n − 10N + 43)

(2.72)

and γ2n+1 = −

(2n + 7)(5n + 9)(10n − 10N + 43)(n − N) . (20n − 10N + 43)(20n − 10N + 53)

(2.73)

Some particular solutions of Eq. (2.69) are φ2 (x) = x 2 +

63N , 10N − 53

φ3 (x) = x 3 +

63 − 116N x, 63 − 10N

φ4 (x) = x 4 +

179 − 242N 2 7938(N − 1)N x + . 73 − 10N (10N − 73)(10N − 63)

By using the well-known Navima algorithm, a solvable recurrence relation can be obtained for the coefficients of the polynomial solution of Eq. (2.69), which eventually leads to a hypergeometric series as

φn (x) =

(σn + 9/5)[n/2] (σn + 7/2)[n/2] (σn − N)[n/2]

× x σn

(−N + [n/2] + σn + 43/10)[n/2] 

−[n/2], σn − x, σn + x, −N + [n/2] + σn + 43/10 1 . 4 F3 σn + 9/5, σn + 7/2, σn − N

(2.74)

Since the support of orthogonality is [α, β] = [−N, N + 1], the recurrence relation (2.70) helps us directly compute the norm square value of the polynomials (2.74) as N  x=−N

W2∗ (x)φn (x)φm (x) =

n  k=1

γk

N   x=−N

 W2∗ (x) δn,m ,

(2.75)

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

189

in which W2∗ (x) = (1/4 − x 2 )W2 (x) is the main weight function and W2 (x) satisfies the equation W2 (x + 1) A(−x) (2x − 1)(2x + 7)(5x + 9)(x − N) = = . W2 (x) A(x + 1) (2x − 5)(2x + 3)(5x − 4)(N + x + 1)

(2.76)

Up to a periodic function of period 1, a symmetric solution of Eq. (2.76) is          5 5 4 4 W2 (x) = Γ − − x Γ +x Γ − −x Γ − +x 2 2 5 5     −1 3 3 ×Γ −x Γ + x Γ (N + 1 − x)Γ (N + 1 + x) . 2 2 Therefore,    5   4   4  5 +x Γ − −x Γ − +x W2∗ (x) = Γ − − x Γ 2 2 5 5 ×Γ

−1  1  −x Γ + x Γ (N + 1 − x)Γ (N + 1 + x) 2 2

1

(2.77)

and N  x=−N

W2∗ (x)

=2

N 

W2∗ (x) − W2∗ (0)

x=0

    225 2 4 F3 1, 95 , 72 , −N; − 52 , − 45 , N + 1; 1 − 1   = S2∗ . = 4 2 2 2 64π Γ − 5 Γ (N + 1)

(2.78)

By substituting the results (2.72)–(2.74), (2.77), and (2.78) into (2.75), the final form of the orthogonality relation is derived as N 

W2∗ (x)φn (x)φm (x) = S2∗ (−1)σn n! δn,m

x=−N

×

(−N)n+σn (9/5)n+σn (7/2)n+σn (9/5 − N)n (7/2 − N)n (53/10)n . (53/10 − N)2n+σn (n − N + 53/10)n

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

Example 2.3 Consider the difference equation (11 − 2x)2(2x + 1)(x + N)Δ∇φn (x) − 2x(4(2N − 21)x 2 + 198N + 121)Δφn (x)     + 2n(n − 2N + 20) 1 − 4x 2 + 200(2N + 1)σn φn (x) = 0, (2.79) which is a special case of (2.45) for A(x) = (11 − 2x)2 (2x + 1)(x + N), C(x) = 1 − 4x 2 , D(x) = 0, E(x) = 200(2N + 1), λn = 2n(n − 2N + 20). Similar to the previous example, the polynomial solution of Eq. (2.79) can be written as φn (x) = x n + δn x n−2 + · · · , satisfying a recurrence relation of the form (2.70), in which δn =

n4 − 4(N − 9)n3 + (185 − 114N)n2 + (482N + 522)n + 300(2N + 1)σn , 48(n − N + 9)

and from (2.71) the explicit form of γn (for even and odd n) is obtained as n(n + 10)(2n − 2N + 9)2 4(2n − N + 9)(2n − N + 10)

(2.80)

(2n + 11)2 (n − N)(n − N + 10) . 4(2n − N + 10)(2n − N + 11)

(2.81)

γ2n = − and γ2n+1 = −

Some particular solutions of Eq. (2.79) are φ2 (x) = x 2 +

121N , 4(N − 11)

φ3 (x) = x 3 +

11(15N − 11) x, 4(N − 12)

φ4 (x) = x 4 +

167N − 143 2 20449(N − 1)N x + . 2(N − 13) 16(N − 13)(N − 12)

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

191

Using the symmetric basis (2.57), considering the properties (2.58)–(2.60), and using the Navima algorithm, we can again obtain a solvable recurrence relation for the coefficients of the polynomial solution of Eq. (2.79), which eventually leads to the hypergeometric representation   (σn + 11/2)[n/2] 2 (σn − N)[n/2] φn (x) = (−N + [n/2] + σn + 10)[n/2] 

− [n/2] , σn − x, x + σn , −N + [n/2] + σn + 10 σn × x 4 F3 1 . σn + 11/2, σn + 11/2, σn − N

(2.82)

For these polynomials, W3∗ (x) = (1/4 − x 2 )W3 (x) is the main weight function, and W3 (x) satisfies the equation A(−x) (2x − 1)(2x + 11)2 (x − N) W3 (x + 1) = = . W3 (x) A(x + 1) (9 − 2x)2(2x + 3)(N + x + 1)

(2.83)

Up to a periodic function of period 1, a symmetric solution of Eq. (2.83) is given by    − x Γ 11 + x 2        W3 (x) =  . 9 9 3 3 Γ − 2 − x Γ − 2 + x Γ 2 − x Γ 2 + x Γ (N + 1 − x)Γ (N + 1 + x) (2.84) Therefore     11 9 9 N , , −N; − , − , N + 1; 1 − 1 797493650625 2 4 F3 1, 11  2 2 2 2 = S3∗ . W3∗ (x) = 1048576πΓ 2 (N + 1) 

Γ

11 2

x=−N

(2.85) By substituting the results (2.80)–(2.82), (2.84), and (2.85) into (2.75), the final form of the orthogonality relation is derived as N  x=−N

×

W3∗ (x)φn (x)φm (x) =

4S3∗ (−1)σn+1 n! N(n − N)σn (n − N + 10)σn δn,m 12658629375 π

Γ (n + 11)Γ 2 (11 − N)Γ 2 (n + σn + 11/2) (1−N)n−1 (11−N)n−1 ((11/2 − N)n )2 . Γ (2n − N + 11)Γ (2n − N + 10 + 2σn )

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

Example 2.4 Consider the difference equation   (2x − 13)(2x + 1)(x + N)2 Δ∇φn (x) − 2x −2N(13 + 12N) + 8(N − 3)x 2 Δφn (x)     + n(n − 4N + 11) 1 − 4x 2 − 12(2N + 1)2 σn φn (x) = 0,

(2.86)

which is a special case of (2.45) for A(x) = (2x − 13)(2x + 1)(x + N)2 , C(x) = 1 − 4x 2 , D(x) = 0, E(x) = −12(2N + 1)2 , λn = n(n − 4N + 11). By noting the two previous examples, we respectively obtain n(n − 2N − 1)(2n − 2N + 11)2 (4n − 4N + 9)(4n − 4N + 11)

(2.87)

(2n + 13)(2n − 4N + 11)(n − N)2 . (4n − 4N + 11)(4n − 4N + 13)

(2.88)

γ2n = − and γ2n+1 = −

Some particular solutions of Eq. (2.86) are 13N 2 , 13 − 4N (26 − 15N)N φ3 (x) = x 3 + x, 4N − 15

φ2 (x) = x 2 +

φ4 (x) = x 4 +

30N 2 − 56N + 13 2 195(N − 1)2 N 2 x + , 17 − 4N (4N − 17)(4N − 15)

which are indeed particular cases of the hypergeometric polynomial

φn (x) =

  (σn + 13/2)[n/2] (σn − N)[n/2] 2

(−2N + [n/2] + σn + 11/2)[n/2]

 − x, x + σ , −2N + + σ + 11/2 − , σ [n/2] [n/2] n n n × x σn 4 F3 1 . σn + 13/2, σn − N, σn − N

(2.89)

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

193

Once again, W4∗ (x) = (1/4 − x 2 )W4 (x) is the main weight function, and W4 (x) satisfies the equation A(−x) (2x − 1)(2x + 13)(N − x)2 W4 (x + 1) = = . W4 (x) A(x + 1) (2x − 11)(2x + 3)(N + x + 1)2

(2.90)

Up to a periodic function of period 1, a symmetric solution of Eq. (2.90) is given by

W4 (x) =

        13 Γ − 12 − x Γ − 12 + x Γ 13 − x Γ + x 2 2 Γ 2 (N + 1 − x)Γ 2 (N + 1 + x)

(2.91)

.

Therefore N 

W4∗ (x) =

x=−N

    11 108056025π 2 2 4 F3 1, 13 2 , −N, −N; − 2 , N + 1, N + 1; 1 − 1 4096Γ 4 (N + 1)

= S4∗ .

(2.92)

By substituting the results (2.87)–(2.89), (2.91), and (2.92) into (2.75), the final form of the orthogonality relation is derived as N 

W4∗ (x)φn (x)φm (x)

x=−N

= ×

64S4∗ (−1)σn n! N 2 (n − N)2σn (−2N)n ((1 − N)n−1 )2 δn,m √ 10395 π

Γ (13/2 − 2N)Γ 2 (13/2 − N + n)Γ (n + σn + 13/2)Γ (n + σn + 11/2 − 2N) . Γ (2n − 2N + 13/2)Γ 2 (13/2 − N)Γ (2n − 2N + 11/2 + 2σn )

Example 2.5 Consider the difference equation   (2x + 1)(2x + 7)(N + x)Δ∇φn (x) − 2x 16N + 4x 2 + 7 Δφn (x)     + −2n 1 − 4x 2 + 16(2N + 1)σn φn (x) = 0,

(2.93)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

which is a special case of (2.45) for A(x) = (2x + 1)(2x + 7)(N + x), C(x) = 1 − 4x 2 , D(x) = 0, E(x) = 16(1 + 2N), λn = −2n. By referring to the previous examples, we can eventually obtain that γn =

1 (n(−n + 2N + 9) − 8(2N + 1)σn ). 4

(2.94)

Some particular solutions of Eq. (2.93) are φ2 (x) = x 2 +

7N , 2

1 φ3 (x) = x 3 + (5N − 7)x, 2 φ4 (x) = x 4 + (5N − 6)x 2 +

35 (N − 1)N, 4

which are particular cases of the hypergeometric polynomial  φn (x) =

Γ

1 2

 (2 [n/2] + 2σn − 7) Γ (−N + [n/2] + σn )   Γ 12 (2σ n) − 7) Γ (σn − N)

 − [n/2] , σn − x, x + σn σn × x 3 F2 1 . σn − 7/2, σn − N

The function W5∗ (x) = (1/4 − x 2 )W5 (x) is the main weight function for this example, where W5 (x) satisfies the equation A(−x) (2x − 7)(2x − 1)(N − x) W5 (x + 1) = = . W5 (x) A(x + 1) (2x + 3)(2x + 9)(N + x + 1)

(2.95)

2.4 A Symmetric Generalization of Sturm–Liouville Problems in Discrete. . .

195

By solving the above equation, we finally obtain W5∗ (x) =

 Γ

1 2



−x Γ



1 2



+x Γ



9 2



−x Γ

1 

9 2

 . + x Γ (N + 1 − x)Γ (N + 1 + x) (2.96)

Therefore we have N 

W5∗ (x) =

x=−N

22N+9 (N + 1)   . 105π 2 4N 2 + 24N + 35 Γ (2N + 4)

(2.97)

By substituting the results (2.94)–(2.97) into (2.75), the final form of the orthogonality relation is derived as N 

W5∗ (x)φn (x)φm (x)

x=−N

=

(−1)σn+1 n! 22N+5 N(N + 1)(1 − N)n−σn+1 (−N − 7/2)n Γ (n + σn − 7/2) δn,m . √ π 2 π (4N 2 + 24N + 35) Γ (2N + 4)

Example 2.6 Consider the difference equation   (2x + 1)(N + x)2 Δ∇φn (x) − 4x N 2 + N + x 2 Δφn (x)     + −n 1 − 4x 2 + (2N + 1)2 σn φn (x) = 0, which is a special case of (2.45) for A(x) = (2x + 1)(N + x)2 , C(x) = 1 − 4x 2, D(x) = 0, E(x) = (2N + 1)2 , λn = −n. After some computations we get γn =

 1 n(−n + 4N + 2) − (2N + 1)2 σn 4

(2.98)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

and φ2 (x) = x 2 + N 2 , φ3 (x) = x 3 + N(N − 2)x, φ4 (x) = x 4 + (2(N − 2)N + 1)x 2 + (N − 1)2 N 2 , which are particular cases of the hypergeometric polynomial Γ 2 (−N + [n/2] + σn ) σn x 3 F2 φn (x) = Γ 2 (σn − N)



− [n/2] , σn − x, x + σn 1 σn − N, σn − N

(2.99)

.

Since W6∗ (x) = (1/4 − x 2 )W6 (x) is the main weight function and W6 (x) satisfies the equation A(−x) (2x − 1)(N − x)2 W6 (x + 1) = =− , W6 (x) A(x + 1) (2x + 3)(N + x + 1)2 solving the above equation eventually yields  W6∗ (x) =

Γ

1 2

   − x Γ 12 + x

Γ 2 (N + 1 − x)Γ 2 (N + 1 + x)

.

(2.100)

Hence N 

W6∗ (x) =

x=−N

π (2 3 F2 (1, −N, −N; N + 1, N + 1; −1) − 1) = S6∗ . Γ 4 (N + 1)

(2.101)

By substituting the results (2.98)–(2.101) into (2.75), the final form of the orthogonality relation is derived as N 

 2 W6∗ (x)φn (x)φm (x) = (−1)σn n! N 2 S6∗ (1 − N)n−σn+1 (−2N)n δn,m .

x=−N

In the next section, we will introduce a main sequence of symmetric orthogonal polynomials of a discrete variable such that all examples introduced in this section are just particular cases of it.

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

2.5

197

A Basic Class of Symmetric Orthogonal Polynomials of a Discrete Variable

Using the extended Sturm–Liouville theorem in discrete spaces, we will introduce a basic class of symmetric orthogonal polynomials of a discrete variable with four free parameters that generalizes all classical symmetric orthogonal polynomials presented in Sect. 2.3. We then obtain standard properties of these polynomials such as a secondorder difference equation, an explicit form for the polynomials, a three-term recurrence relation, and a generic orthogonality relation. We show that two infinite types and two finite types of hypergeometric orthogonal sequences with different weight functions can be extracted from this class, and moments corresponding to the derived weight functions can be explicitly computed. We also present a particular example containing all classical discrete symmetric orthogonal polynomials described in Sect. 2.3. Let us recall the differential equation (1.100) as x 2 (p∗ x 2 + q ∗ )Φn (x) + x(r ∗ x 2 + s ∗ )Φn (x) − (n(r ∗ + (n − 1)p∗ )x 2 + σn s ∗ )Φn (x) = 0.

(2.102)

We observe in (2.102) that A(x) = x 2 (p∗ x 2 + q ∗ ) is a polynomial of degree at most 4, C(x) = x 2 is a symmetric quadratic polynomial, D(x) = 0, and E(x) = s ∗ . Since discrete orthogonal polynomials have a direct relationship with continuous polynomials, motivated by (2.102), we suppose in the main Eq. (2.45) that A(x) =

4 

ai x i ,

i=0

C(x) = c2 x 2 + c0 , D(x) = 0,

(2.103)

and E(x) = e0 . We are interested in obtaining a symmetric orthogonal polynomial solution as φn (x) = x n + δn x n−2 + · · · ,

(2.104)

which satisfies the three-term recurrence relation φn+1 (x) = xφn (x) − γn φn−1 (x)

with φ0 (x) = 1 and φ1 (x) = x.

(2.105)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

From (2.105), (2.104), and (2.103), equating the coefficient in x n+2 gives the eigenvalues as λn =

n (2a3 − a4 (n − 1)) , c2

(2.106)

provided that c2 = 0 and |a3 | + |a4 | = 0. By considering λn in (2.106) and equating the coefficient in x n , we obtain   &  δn = 6c2 4a1 n − 2a2(n − 1)n + e0 (−1)n − 1 +a4 (n − 1)n (12c0 − c2 (n − 3)(n − 2)) +4a3n (c2 (n − 2)(n − 1) − 6c0 )} / {24c2 (a4 (3 − 2n) + 2a3 )} . Also from (2.104) and (2.105) we have     x n+1 + δn+1 x n−1 + · · · = x x n + δn x n−2 + · · · − γn x n−1 + δn−1 x n−3 + · · · , which implies γn = δn − δn+1 .

(2.107)

Therefore, in order that φn (x) be a solution of Eq. (2.103), the following extra conditions must be considered for the initial values of n: e0 = 2a1 − a0 =

2a3 c0 , c2

c0 ((2a3 − a4 ) c0 + (a2 − 2a1 ) c2 ) , c22

and c2 = −4c0 . Summarizing the data, we get     1 a3 a4 1 A(x) = a4 x 4 + a3 x 3 + a2 x 2 − + a1 x + + − , 2 8 16  4  B(x) = A(−x) − A(x) = −2x a3 x 2 + a1 ,   C(x) = c0 1 − 4x 2 , D(x) = 0, 1 E(x) = (4a1 + a3 ), 2 n (a4 (n − 1) − 2a3 ) , λn = 4c0

(2.108)

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

199

and   &   δn = 12a1 2n + (−1)n − 1 + 3a3 (−1)n − 1 +n (−12a2(n − 1) + 2a3(2(n − 3)n + 7) −a4 (n − 1)((n − 5)n + 9))} / {24a4(3 − 2n) + 48a3} .

(2.109)

For simplicity, if we set a4 = 2a,

a3 = a + 2b,

a2 = b + 2c,

a1 = c + 2d,

then (2.108) changes to A(x) = (2x + 1)(ax 3 + bx 2 + cx + d),   B(x) = −2x (a + 2b)x 2 + c + 2d , C(x) =

1 − x 2, 4

D(x) = 0, E(x) =

1 (a + 2b + 4c + 8d), 2

λn = 2n(an − 2(a + b)) for |a| + |b| = 0. Consequently, the following basic difference equation appears:   (2x + 1)(ax 3 + bx 2 + cx + d)Δ∇φn (x) − 2x x 2 (a + 2b) + c + 2d Δφn (x)     a 1 + 2n(an − 2(a + b)) − x 2 + σn + b + 2c + 4d φn (x) = 0. (2.110) 4 2 Since the polynomial solution of Eq. (2.110) is symmetric, we use the notation φn (x) =

Sn∗



ab x cd

for mathematical display formulas and Sn∗ (x ; a, b, c, d) in the text. This means that we are dealing with just one characteristic vector V = (a, b, c, d) for any given subcase.

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

If Sn∗ (x ; a, b, c, d) satisfies a three-term recurrence relation of type (2.105), then by referring to (2.107) and (2.109), we obtain 

4 i ab i=0 Ki (a, b, c, d)n , (2.111) γn = γn = 32(b − a(n − 2))(b − a(n − 1)) cd where K4 (a, b, c, d) = −2a 2, K3 (a, b, c, d) = 4a(3a + 2b),   K2 (a, b, c, d) = −8 3a 2 + a(4b + c) + b2 , K1 (a, b, c, d) = 2(3a + 2b)(3a + 4(b + c)) − 2a(−1)n(a + 2b + 4c + 8d),   K0 (a, b, c, d) = (−1)n − 1 (3a + 2b)(a + 2b + 4c + 8d). For n = 2m and n = 2m + 1, γn in (2.111) is decomposed as     m −a 2 (m − 1)3 + a 2b(m − 1)2 + c(1 − m) − d + b(b(1 − m) + c) γ2m = (b − 2a(m − 1))(b − a(2m − 1)) and γ2m+1

  (a(m − 1) − b) −am3 + b m2 − c m + d . = (2a m − (a + b)) (2a m − b)

Theorem 2.4 The explicit form of the polynomial Sn∗ (x ; a, b, c, d) is denoted by Sn∗



  [n/2]  [n/2] ab x = x σn (−1)j j cd j =0 ⎛

[n/2]−1

×⎝

i=j

⎞ a(i + σn )3 − b(i + σn )2 + c(i + σn ) − d ⎠ (σn − x)j (σn + x)j , a(i + [n/2] − σn+1 ) − b (2.112)

where [x] denotes the integer part of x and

−1 

(.) = 1.

i=0

Moreover, since (−x)j (x)j = (−1)j

j −1 k=0

(x 2 − k 2 ),

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

201

for n = 2m and n = 2m + 1, (2.112) respectively changes to ⎛ ⎞ 

j −1 m   m−1 3 2  m ⎝ ai − bi + ci − d ⎠ 2 ab ∗ S2m x = (x − k 2 ), a(i + m − 1) − b j cd j =0

i=j

k=0

and ∗ S2m+1



ab x = cd x

m    m

j

j =0

⎛ ⎝

m−1

⎞

a(i

i=j

+ 1)3

− b(i + c(i + 1) − d ⎠ 2 (x − k 2 ). a(i + m) − b + 1)2

j

k=1

Proof Once again, we consider the symmetric basis ϑn (x) = (−1)[n/2] x σn (σn − x)[n/2] (σn + x)[n/2] = (−1)n ϑn (−x),

(2.113)

together with the following straightforward properties: n(n − 1) n(n − 1)(n − 2) ϑn−2 (x) + σn+1 ϑn−3 (x), 2 4  n 2 xϑn (x) = ϑn+1 (x) + σn+1 ϑn−1 (x), 2

Δϑn (x) = nϑn−1 (x) +

Δ∇ϑn (x) = n(n − 1)ϑn−2 (x).

(2.114) (2.115) (2.116)

If Sn∗ (x ; a, b, c, d) is expanded as Sn∗ (x ;

a, b, c, d) =

[n/2] 

cj (n)ϑ2j +σn (x)

j =0

=

⎧ m  ⎪ ⎪ ⎪ (−1)j cj (2m)(−x)j (x)j , ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

j =0 m 

n = 2m, (2.117)

(−1)j cj (2m + 1)x(1 − x)j (1 + x)j , n = 2m + 1,

j =0

then using the Navima algorithm, we reach a solvable recurrence relation for the connection coefficients cj (n). So, using the properties (2.114)–(2.116) and substituting (2.117) into (2.110), we obtain 0=

[n/2]  j =0

  cj (n) Ej (n) + ϑ2j +σn +2 + Fj (n)ϑ2j +σn + Gj (n)ϑ2j +σn −2 ,

202

2 Generalized Sturm–Liouville Problems in Discrete Spaces

where Ej (n) = 2 (2j + σn − n) (a(2j + n − 2) + aσn − 2b) , Fj (n) = 4a(j − 1)2 j (2j − 1)σn−1 1 + σn+2 (2j + σn + 1) 2 (2j + σn − n) (a(2j + n − 2) + aσn − 2b) 2     + 2j σn+1 a 4j 3 − 6j 2 − j (n − 3)(n + 1) − 1 + 2b(j (−4j + n + 3) − 1) 1 (σn (σn (σn (σn (8aj + aσn − 4a − 2b) + a(24(j − 1)j + 5) + 4b(1 − 3j )) 2   +2a(2j − 1)(8(j − 1)j + 1) + 8b(2 − 3j )j + 4c) + a 4(j − 1)j (1−2j )2 + 1

+

+4j (b(1 − 4(j − 1)j ) + 4c) − 4c) − 2n(a + b) + an2  −4j (b + 4(c + d)) + 8j 2 (b + 2c) ,

and    Gj (n) = (j − 1)j (2j − 1) 2(j − 1)σn−1 2a(j − 1)2 σn−1 + b + 2c   −2σn+1 (j − 1)2 (a + 2b)σn−1 + c + 2d 1 (2j + σn − 1) (2j + σn ) (σn (2j + σn − 1) (σn (2 ((j − 1) (−4j (a + b) 8   +4aj 2 + a + 2c + σn (σn (6aj + aσn − 4a − 2b)

+

+ (2j − 1)(a(6j − 5) − 4b))) + b(4j − 2) − 8(c + d) + 8cj ) + 8d) . But since ϑn (x) is linearly independent, the coefficients cj (n) satisfy the relation Ej −1 (n)cj −1 (n) + Fj (n)cj (n) + Gj +1 (n)cj +1 (n) = 0, which is explicitly solvable with the initial conditions cm (n) = 0 for m > [n/2] and   c[n/2] (n) = 1, providing (2.112).

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

203

Since the recurrence relation (2.105) is now explicitly known, the complete form of the orthogonality relation can be designed as θ  x=−θ

  



ab ∗ a b ∗ a b W x Sn x Sm x cd cd cd ∗

  θ 

 ab ∗ a b γk W x δn,m , = cd cd x=−θ k=1 n

(2.118)

where

   1 ab 2 W − x W ∗ (x) x = 4 cd ∗

is the original weight function and W ∗ (x) satisfies the difference equation (1/2) − x W ∗ (x + 1) −ax 3 + bx 2 − cx + d = . W ∗ (x) (3/2) + x a(x + 1)3 + b(x + 1)2 + c(c + 1) + d

(2.119)

By noting that A(x) = (2x + 1)(ax 3 + bx 2 + cx + d) for |a| + |b| = 0, we see that two cases can generally occur for the parameter a in (2.119), i.e., when a = 0 and b arbitrary or a = 0 and b = 0. In the first case, since every polynomial of degree 3 has at least one real root, say x = p ∈ R, the aforementioned A(x) can be decomposed in three different forms, i.e.,   A(x) = (2x + 1) (x − p) a x 2 + u x + v ⎧   2 2 ⎪ ⎪ ⎨(2x + 1) (x − p) a x + u x + v , (u < 4av), = (2x + 1) a(x − p)(x − q)(x − r), (u2 > 4av), ⎪ ⎪ ⎩ (2x + 1) a(x − p)(x − q)2 , (u2 = 4av).

(2.120)

Similarly, in the second case, when a = 0 and b = 0, A(x) can be decomposed as ⎧  2  ⎪ (2x + 1) b x + c x + d , ⎪ ⎨   2 A(x) = (2x + 1) b x + c x + d = (2x + 1) b(x − p)(x − q), ⎪ ⎪ ⎩ (2x + 1) b(x − p)2 ,

(c2 < 4bd), (c2 > 4bd), (c2 = 4bd). (2.121)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

For the two subcases   A(x) = (2x + 1) (x − p) a x 2 + u x + v in (2.120) and   A(x) = (2x + 1) b x 2 + c x + d in (2.121), the difference equations corresponding to (2.119) respectively take the forms (1/2) − x p + x a x2 − u x + v W ∗ (x + 1) = ∗ W (x) (3/2) + x p − x − 1 a (x + 1)2 + u (x + 1) + v

(u2 < 4av) (2.122)

and W ∗ (x + 1) (1/2) − x b x2 − c x + d = W ∗ (x) (3/2) + x b (x + 1)2 + c (x + 1) + d

(c2 < 4bd) .

(2.123)

Since the denominators of the two fractions (2.122) and (2.123) are not decomposable in the real line, we should separately consider them in finite types of symmetric orthogonal polynomials in the next section. Note that the cases analyzed in this part allow us to recover all classical symmetric orthogonal polynomials of a discrete variable in Sect. 2.5.3. For the second subcase of (2.120), without loss of generality take a = 1 to get A(x) = (2x + 1)(x − p)(x − q)(x − r), where p, q, r ∈ R, which indeed covers the third subcase of (2.120) too. For the second subcase of (2.121) we can similarly consider A(x) = (2x + 1)(x − p)(x − q), where p, q ∈ R. This means that there exist only two infinite types of orthogonal sequences of Sn∗ (x ; a, b, c, d) when A(x) is decomposable. In other words, when A(x) is a polynomial of degree four, a three-parameter family appears, and when A(x) is of degree three, a twoparameter family will appear. Finally, when A(x) is of degree 2, the well-known classical symmetric discrete families appear. The question is now how to determine restrictions on the parameter θ in the orthogonality support [−θ, θ ] in (2.118). To answer, we should reconsider the main difference

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

205

equation (2.110) on [α, β − 1] = [−θ, θ ] and write it in a self-adjoint form to obtain   Δ A(x)W ∗ (x) (φm (x)∇φn (x) − φn (x)∇φm (x))

θ  x=−θ

 θ   1 2 + (λn − λm ) − x W ∗ (x)φn (x)φm (x) 4 x=−θ

+

θ   (−1)m − (−1)n  a + b + 2c + 4d W ∗ (x)φn (x)φm (x) = 0. 2 2

(2.124)

x=−θ

On the other hand, the identity φm (x)∇φn (x) − φn (x)∇φm (x) = φn (x)φm (x − 1) − φm (x)φn (x − 1) simplifies the first sum of (2.124) as θ 

  Δ A(x)W ∗ (x) (φm (x)∇φn (x) − φn (x)∇φm (x))

x=−θ

  x=θ+1 = A(x)W ∗ (x) φn (x)φm (x − 1) − φm (x)φn (x − 1) x=−θ

∗

= A(θ + 1)W (θ + 1) (φn (θ + 1)φm (θ ) − φm (θ + 1)φn (θ )) − A(−θ )W ∗ (−θ ) (φn (−θ )φm (−θ − 1) − φm (−θ )φn (−θ − 1)) .

(2.125)

By taking into account that all weight functions are even, i.e., W ∗ (−x) = W ∗ (x), the polynomials are symmetric, i.e., φn (x) = (−1)n φn (−x), and the corresponding Pearson difference equation is also valid for x = θ , i.e., A(θ + 1)W ∗ (θ + 1) = A(−θ )W ∗ (θ ), relation (2.125) may finally be simplified as θ 

  Δ A(x)W ∗ (x) (φm (x)∇φn (x) − φn (x)∇φm (x))

x=−θ

  = A(−θ )W ∗ (θ ) 1 + (−1)n+m (φm (θ )φn (θ + 1) − φn (θ )φm (θ + 1)) .

(2.126)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

Since φm (θ )φn (θ + 1) − φn (θ )φm (θ + 1) = 0, two cases can in general occur for the right-hand side of (2.126): 1. If n + m is odd, then 1 + (−1)n+m = 0, and (2.126) is automatically zero. 2. If simultaneously A(−θ ) = 0 and W ∗ (θ ) = 0, then (2.126) is again equal to zero.

2.5.1

Two Infinite Types of Sn∗ (x ; a, b, c, d)

Here we introduce two infinite hypergeometric sequences of symmetric orthogonal polynomials that are particular cases of Sn∗ (x ; a, b, c, d) and obtain all possible weight functions together with orthogonality supports for them. First Sequence For p, q, r ∈ R, if the characteristic vector (a, b, c, d) = (1, −(p + q + r), pq + pr + qr, −pqr) is replaced in Eq. (2.110), then (2x + 1)(x − p)(x − q)(x − r)Δ∇φn (x)   − 2x x 2 (1 − 2p − 2q − 2r) + pq + pr + qr − 2pqr Δφn (x)    1 2 −x + 2n(n + 2(p + q + r − 1)) 4  σn − (2p − 1)(2q − 1)(2r − 1) φn (x) = 0 2 has a basis hypergeometric-type solution as  φn (x) =

Sn∗

1 −(p + q + r) x pq + pr + qr −pqr =

 × x σn 4 F3



(p + σn )[n/2] (q + σn )[n/2] (r + σn )[n/2] ([n/2] + p + q + r − 1 + σn )[n/2]

−[n/2], [n/2] + p + q + r − 1 + σn , σn − x, σn + x 1 p + σn , q + σn , r + σn

which satisfies the recurrence relation

,

(2.127)

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

 φn+1 (x) = xφn (x) − γn

1 −(p + q + r) pq + pr + qr −pqr

207

φn−1 (x),

(φ0 (x) = 1,

φ1 (x) = x),

(2.128)

where  γn

 1 −(p + q + r) = −2n4 − 4(−3 + 2p + 2q + 2r)n3 pq + pr + qr −pqr   −8 p2 + p(3q + 3r − 4) + 3qr + (q − 4)q + r 2 − 4r + 3 n2  + 2(−1)n (−1 + 2p)(−1 + 2q)(−1 + 2r) −2(−3 + 2p + 2q + 2r)(3 + 4q(−1 + r) − 4r + 4p(−1 + q + r))) n ' +(−1 + (−1)n )(−1 + 2p)(−1 + 2q)(−1 + 2r)(−3 + 2p + 2q + 2r) / {32(n + p + q + r − 2)(n + p + q + r − 1)} ,

(2.129)

dividing to γ2n = −

n(n + p + q − 1)(n + p + r − 1)(n + q + r − 1) (2n + p + q + r − 2)(2n + p + q + r − 1)

and γ2n+1 = −

(n + p)(n + q)(n + r)(n + p + q + r − 1) . (2n + p + q + r − 1)(2n + p + q + r)

The latter representations will allow us to analyze the sign of γn in terms of the values of p, q, and r. By noting the relations (2.128) and (2.129), the orthogonality relation of the first sequence can be expressed as θ  x=−θ





1 −(p + q + r) (1) (1) x Sn (x)Sm (x) pq + pr + qr −pqr 

n 1 −(p + q + r) = γk pq + pr + qr −pqr k=1 

 θ  1 −(p + q + r) ∗ W x δn,m , × pq + pr + qr −pqr x=−θ

 W

∗

208

2 Generalized Sturm–Liouville Problems in Discrete Spaces

in which  Sn(1) (x)

=

Sn∗

1 −(p + q + r) x pq + pr + qr −pqr



and  W

∗

1 −(p + q + r) x pq + pr + qr −pqr



 =

 1 2 − x W ∗ (x) 4

denotes the original weight function corresponding to the hypergeometric polynomial (2.127), and finally W ∗ (x) satisfies a particular case of the difference equation (2.119) in the form −x + (1/2) −x − p −x − q −x − r W ∗ (x + 1) = . ∗ W (x) x + (3/2) x + 1 − p x + 1 − q x + 1 − r

(2.130)

It is interesting that there exist 16 symmetric solutions for Eq. (2.130), as follows: W1 (x; p, q, r) = (Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + x)Γ (1 − q − x) ×Γ (1 − r + x)Γ (1 − r − x)Γ (3/2 + x)Γ (3/2 − x))−1 , W2,1 (x; p, q, r) =

Γ (p + x)Γ (p − x) , Γ (1 − q + x)Γ (1 − q − x)Γ (1 − r + x)Γ (1 − r − x)Γ (3/2 + x)Γ (3/2 − x)

W2,2 (x; p, q, r) = W2,1 (x; q, p, r) =

Γ (q + x)Γ (q − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − r + x)Γ (1 − r − x)Γ (3/2 + x)Γ (3/2 − x)

W2,3 (x; p, q, r) = W2,1 (x; r, q, p) =

Γ (r + x)Γ (r − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + x)Γ (1 − q − x)Γ (3/2 + x)Γ (3/2 − x)

W3,1 (x; p, q, r) =

Γ (p + x)Γ (p − x)Γ (q + x)Γ (q − x) , Γ (1 − r + x)Γ (1 − r − x)Γ (3/2 + x)Γ (3/2 − x)

W3,2 (x; p, q, r) = W3,1 (x; p, r, q) =

Γ (p + x)Γ (p − x)Γ (r + x)Γ (r − x) , Γ (1 − q + x)Γ (1 − q − x)Γ (3/2 + x)Γ (3/2 − x)

W3,3 (x; p, q, r) = W3,2 (x; r, q, p) =

Γ (q + x)Γ (q − x)Γ (r + x)Γ (r − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (3/2 + x)Γ (3/2 − x)

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

W4,1 (x; p, q, r) =

209

Γ (p + x)Γ (p − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − q + x)Γ (1 − q − x)Γ (1 − r + x)Γ (1 − r − x)

W4,2 (x; p, q, r) = W4,1 (x; q, p, r) =

Γ (q + x)Γ (q − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − r + x)Γ (1 − r − x)

W4,3 (x; p, q, r) = W4,1 (x; r, q, p) =

Γ (r + x)Γ (r − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + x)Γ (1 − q − x)

W5,1 (x; p, q, r) =

Γ (p + x)Γ (p − x)Γ (q + x)Γ (q − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − r + x)Γ (1 − r − x)

W5,2 (x; p, q, r) = W5,1 (x; p, r, q) =

Γ (p + x)Γ (p − x)Γ (r + x)Γ (r − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − q + x)Γ (1 − q − x)

W5,3 (x; p, q, r) = W5,1 (x; r, q, p) =

Γ (q + x)Γ (q − x)Γ (r + x)Γ (r − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)

W6 (x; p, q, r) =

Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + x)Γ (1 − q − x)Γ (1 − r + x)Γ (1 − r − x)

W7 (x; p, q, r) =

Γ (p + x)Γ (p − x)Γ (q + x)Γ (q − x)Γ (r + x)Γ (r − x) , Γ (3/2 + x)Γ (3/2 − x)

W8 (x; p, q, r) = Γ (p + x)Γ (p − x)Γ (q + x)Γ (q − x) × Γ (r + x)Γ (r − x)Γ (−1/2 + x)Γ (−1/2 − x).

210

2 Generalized Sturm–Liouville Problems in Discrete Spaces

Since the original weight functions corresponding to all 16 cases are 

 1 2 − x Wk (x), 4

the two following identities are remarkable in this direction: (1/4 − x 2 )Γ (−1/2 + x)Γ (−1/2 − x) = Γ (1/2 + x)Γ (1/2 − x), and 1 1/4 − x 2 = . Γ (3/2 + x)Γ (3/2 − x) Γ (1/2 + x)Γ (1/2 − x) For example, for the last given case, the original weight function becomes 



 1 2 − x W8 (x; p, q, r) = 4     1 1 = Γ (p + x)Γ (p − x)Γ (q + x)Γ (q − x)Γ (r + x)Γ (r − x)Γ +x Γ −x . 2 2 W8∗

1 −(p + q + r) x pq + pr + qr −pqr



Table 2.2 shows the orthogonality supports of each weight function together with their parameter restrictions in which Z− = {0, −1, −2, . . . }. Table 2.2 Orthogonality supports for the first hypergeometric sequence Wk (x)

Support

Parameter restrictions

W1 (x; p, q, r)

[−p, p] [−q, q] [−r, r] [−q, q] [−r, r] [−r, r] [−q, q] [−r, r] [−r, r] [−p, p] [−q, q] [−r, r] – –

p ∈ Z− , 1 − q ± p ∈ Z− , 1 − r ± p ∈ Z− . q ∈ Z− , 1 − p ± q ∈ Z− , 1 − r ± q ∈ Z− . r ∈ Z− , 1 − q ± r ∈ Z− , 1 − p ± r ∈ Z− . q ∈ Z− , p ± q ∈ Z− , 1 − r ± q ∈ Z− . r ∈ Z− , p ± r ∈ Z− , 1 − q ± r ∈ Z− . r ∈ Z− , p ± r ∈ Z− , q ± r ∈ Z− . q ∈ Z− , p ± q ∈ Z− , 1 − r ± q ∈ Z− . r ∈ Z− , p ± r ∈ Z− , 1 − q ± r ∈ Z− . r ∈ Z− , p ± r ∈ Z− , q ± r ∈ Z− . p ∈ Z− , 1 − q ± p ∈ Z− , 1 − r ± q ∈ Z− . q ∈ Z− , 1 − r ± q ∈ Z− , 1 − p ± q ∈ Z− . r ∈ Z− , 1 − q ± r ∈ Z− , 1 − p ± r ∈ Z− . – –

W2,1 (x; p, q, r) W3,1 (x; p, q, r) W4,1 (x; p, q, r) W5,1 (x; p, q, r) W6 (x; p, q, r)

W7 (x; p, q, r) W8 (x; p, q, r)

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

211

Note that since some weight functions are symmetric with respect to the parameters p, q, and r, e.g., W4,2 (x; p, q, r) = W4,1 (x; q, p, r) and W4,3 (x; p, q, r) = W4,1 (x; r, q, p), their orthogonality supports and parameter restrictions can be directly derived via the rows of Table 2.2 by just interchanging the parameters. Also W7 (x; p, q, r) and W8 (x; p, q, r) have no valid orthogonality support. Therefore, there are 24 eligible orthogonality supports altogether for the first sequence. • A numerical example for the first infinite sequence Let us replace p = −9, q = −10, and r = −11 in (2.127) to get  Sn∗



(−9 + σn )[n/2] (−10 + σn )[n/2] (−11 + σn )[n/2] (−31 + [n/2] + σn )[n/2]

 −[n/2], −31 + [n/2] + σn , σn − x, σn + x σn × x 4 F3 1 . −9 + σn , −10 + σn , −11 + σn

1 30 x 299 990

=

(2.131)

Since p ∈ Z− , 1 − q ± p ∈ Z− , and 1 − r ± p ∈ Z− , referring to Table 2.2 shows that there are two possible weight functions for the selected parameters that are orthogonal with respect to the sequence Sn∗ (x ; 1, 30, 299, 990) on the support {−9, −8, . . . , 8, 9}, i.e.,

 W1∗

1 30 x 299 990

 =

 1 − x 2 W1 (x; −9, −10, −11) 4

= (Γ (10 + x)Γ (10 − x)Γ (11 + x)Γ (11 − x) × Γ (12 + x)Γ (12 − x)Γ (1/2 + x)Γ (1/2 − x))−1 and  W6∗



1 30 x 299 990 =

 =

 1 − x 2 W6 (x; −9, −10, −11) 4

Γ (1/2 + x)Γ (1/2 − x) . Γ (10 + x)Γ (10 − x)Γ (11 + x)Γ (11 − x)Γ (12 + x)Γ (12 − x)

212

2 Generalized Sturm–Liouville Problems in Discrete Spaces

Hence there are two orthogonality relations corresponding to the polynomials (2.131) respectively as follows: 9 





1 30 x 299 990

Wi∗

x=−9

∗ Sn∗ (x ; 1, 30, 299, 990)Sm (x ; 1, 30, 299, 990)

36183421612800000(−1)n(−29)) n−1 * 2

= αi

−1

(−19) n −2 (−18) n −2 (−17) n −2 2

(−31)n (−30)n × (−9)) n−1 * 2

−1

(−8)) n−1 * 2

−1

(−7)) n−1 * 2

−1

Γ

2

) n * 2

2

 + 1 δn,m ,

where x denotes the floor function, and αi =

9 

Wi∗

x=−9

in which β =



 1 30 x 299 990

⎧ ⎨β/π, i = 1, = ⎩βπ, i = 6,

667 . 1998530094928466929986605067918114816000000000

Second Sequence For p, q ∈ R, if the characteristic vector (a, b, c, d) = (0, 1, −p − q, pq) is replaced in Eq. (2.110), then   (2x + 1)(x − p)(x − q)Δ∇φn (x) − 2x 2x 2 + p(2q − 1) − q Δφn (x)    1 − (−1)n 1 (2p − 1)(2q − 1) φn (x) = 0 + − 4n − x 2 + 4 2 has a basis solution as  φn (x) =

Sn∗

0 1 x −p − q pq

= (p + σn )[n/2] (q + σn )[n/2]  ×x

σn

3 F2

−[n/2], σn − x, σn + x 1 p + σn , q + σn

,

(2.132)

satisfying a recurrence relation of type (2.105) with 

   1 0 1 γn = −2n2 − 4n(p + q − 1) + (−1)n − 1 (2p − 1)(2q − 1) , 8 −p − q pq

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

213

which implies γ2n = −n(−1 + n + p + q) and γ2n+1 = −(n + p)(n + q). Hence the orthogonality relation corresponding to this sequence takes the form θ 





 0 1 x −p − q pq

W∗

x=−θ

=

n



 Sn∗



0 1 x −p − q pq



0 1 −p − q pq

γk

k=1

θ  x=−θ

∗ Sm

0 1 x −p − q pq

 W

∗







0 1 x −p − q pq

δn,m ,

where  W

∗

0 1 x −p − q pq



 =

 1 − x 2 W ∗ (x) 4

is the original weight function and W ∗ (x) satisfies a particular case of the difference equation (2.119) as 1/2 − x −x − p −x − q W ∗ (x + 1) = . ∗ W (x) 3/2 + x x + 1 − p x + 1 − q

(2.133)

There exist eight symmetric solutions for Eq. (2.133) respectively as follows: W9 (x; p, q) =

1 , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + x)Γ (1 − q − x)Γ (3/2 + x)Γ (3/2 − x)

W10,1 (x; p, q) =

Γ (p + x)Γ (p − x) , Γ (1 − q + x)Γ (1 − q − x)Γ (3/2 + x)Γ (3/2 − x)

W10,2 (x; p, q) =

Γ (q + x)Γ (q − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (3/2 + x)Γ (3/2 − x)

W11 (x; p, q) =

Γ (p + x)Γ (p − x)Γ (q + x)Γ (q − x) , Γ (3/2 + x)Γ (3/2 − x)

214

2 Generalized Sturm–Liouville Problems in Discrete Spaces

W12 (x; p, q) =

Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + x)Γ (1 − q − x)

W13,1 (x; p, q) =

Γ (p + x)Γ (p − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − q + x)Γ (1 − q − x)

W13,2 (x; p, q) =

Γ (q + x)Γ (q − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)

W14 (x; p, q) = Γ (p + x)Γ (p − x)Γ (q + x)Γ (q − x)Γ (−1/2 + x)Γ (−1/2 − x). Table 2.3 shows the orthogonality supports of each of the above-mentioned weights and their parameter restrictions. As the main orthogonality relation (2.118) shows, an important part that we have to compute in norm square value is θ 

W ∗ (x; a, b, c, d),

x=−θ

where [−θ, θ ] is the same orthogonality support as determined in Tables 2.2 and 2.3. Since {Wk (x; a, b, c, d)}k=1 are all even functions, the aforementioned sums can be simplified on their orthogonality supports using the two identities Γ (p + x) = Γ (p)(p)x

and Γ (p − x) =

Γ (p)(−1)x (−p)x

and the fact that θ 

∗

W (x; a, b, c, d) = 2

x=−θ

Table 2.3 Orthogonality supports for the second hypergeometric sequence

θ 

W ∗ (x; a, b, c, d) − W ∗ (0; a, b, c, d).

x=0

Wk (x)

Support

Parameter restrictions

W9 (x; p, q)

[−p, p] [−q, q] [−q, q] – [−p, p] [−q, q] [−q, q] –

p ∈ Z− , 1 − q ± p ∈ Z− . q ∈ Z− , 1 − p ± q ∈ Z− . q ∈ Z− , p ± q ∈ Z− . – p ∈ Z− , 1 − q ± p ∈ Z− . q ∈ Z− , 1 − p ± q ∈ Z− . q ∈ Z− , p ± q ∈ Z− . –

W10,1 (x; p, q) W11 (x; p, q) W12 (x; p, q) W13,1 (x; p, q) W14 (x; p, q)

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

215

For example, for the first given weight function W1 (x; p, q, r) on, e.g., [−p, p] we have  p   1 − x 2 W1 (x; p, q, r) 4 x=−p

 p  (p)x (q)x (r)x 1 −1 . 2 = (1 − p)x (1 − q)x (1 − r)x πΓ 2 (1 − p)Γ 2 (1 − q)Γ 2 (1 − r) x=0

• A numerical example for the second infinite sequence Let us replace p = 14 and q = −10 in (2.132) to get

 Sn∗

0 1 x −4 −140

= (14 + σn )[n/2] (−10 + σn )[n/2]  ×x

σn

3 F2

−[n/2], σn − x, σn + x 1 14 + σn , −10 + σn

.

(2.134)

Since q ∈ Z− and p ± q ∈ Z− , referring to Table 2.3 shows that there are two possible weight functions for the selected parameters that are orthogonal with respect to the sequence Sn∗ (x ; 0, 1, −4, −140) on the support {−10, −9, . . . , 9, 10}, i.e.,  ∗ W10,1



0 1 x −4 −140

 =

=

 1 2 − x W10,1 (x; 14, −10) 4

(13 − x)(12 − x)(11 − x)(x + 11)(x + 12)(x + 13) cos(πx) π

and

 ∗ W13,1

0 1 x −4 −140

 =

 1 2 − x W13,1 (x; 14, −10) 4

= π(13 − x)(12 − x)(11 − x)(x + 11)(x + 12)(x + 13) sec(πx). Hence there are two orthogonality relations corresponding to the polynomials (2.134) respectively as follows: 10  x=−10



 ∗ Wj,1

0 1 x −4 −140

∗ Sn∗ (x ; 0, 1, −4, −140)Sm (x ; 0, 1, −4, −140)

(−1)n+1 (−9)) n−1 * Γ = α˜ j

2

)

n−1 2

*

 + ,  + ,  + 15 Γ n2 + 1 Γ n2 + 4

3736212480

,

216

2 Generalized Sturm–Liouville Problems in Discrete Spaces

where α˜ j =

10 

∗ Wj,1

x=−10

2.5.2



 0 1 x −4 −140

=

⎧ ⎨10296/π,

j = 10,

⎩10296π,

j = 13.

Moments of the Two Introduced Infinite Sequences

To compute the moments of a continuous distribution, different bases are usually considered. For example, for the normal distribution, the classical basis {x j }j is used to get 

⎧ ⎪ ⎨0, ∞ e   xn √ dx = 2m+1/2 1 ⎪ 2π −∞ Γ m+ , ⎩ √ 2 2π −x 2 /2

n = 2m + 1, n = 2m,

while for the Jacobi weight function (1 − x)α (1 + x)β as the shifted beta distribution on [−1, 1], using one of the two bases {(1 − x)j }j or {(1 + x)j }j is appropriate for this purpose. This matter similarly holds for the moments of discrete orthogonal polynomials. For instance, in the negative hypergeometric distribution corresponding to Hahn polynomials, it is more convenient to use the Pochhammer basis {(−x)n }n , instead of the classical basis, to get N−1  x=0

Γ (N)Γ (α + β + 2)Γ (α + N − x)Γ (β + x + 1) (−x)n Γ (α + 1)Γ (β + 1)Γ (α + β + N + 1)Γ (N − x)Γ (x + 1) = (−1)n

(1 − N)n (β + 1)n . (α + β + 2)n

Following this approach, for the weight functions i (x) appearing in the two introduced hypergeometric sequences, we can compute the moments of the form (i )n =

θ  x=−θ

ϑn (x)i (x) =

θ  x=−θ

 ϑn (x)

 1 − x 2 Wi (x), 4

where the basis ϑn (x) is defined in (2.113). Since ϑ2n+1 (x) are odd polynomials, clearly all odd moments with respect to this basis are zero. Moreover, from definitions (2.127) and (2.132) and using their orthogonality

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

217

property, it can be proved by induction that the even moments corresponding to the first and second sequences respectively satisfy the following recurrence relations: (i )2n = −

(p + n − 1)(q + n − 1)(r + n − 1) (i )2n−2 p+q +r +n−1

and (j )2n = −(n + p − 1)(n + q − 1)(j )2n−2 . Therefore, if these weight functions are normalized with the first moment equal to one, we eventually obtain θ  x=−θ



1 − x2 4



⎧ ⎪ ⎨0, n = 2m + 1, Wi (x)ϑn (x) = (−1)m (p)m (q)m (r)m ⎪ , n = 2m, ⎩ (p + q + r)m

for the weight functions of the first sequence, and ⎧  θ  ⎨0,  n = 2m + 1, 1 − x 2 Wj (x)ϑn (x) = ⎩(−1)m (p)m (q)m , n = 2m, 4 x=−θ for the weight functions of the second sequence.

2.5.3

A Special Case of Sn∗ (x ; a, b, c, d) Generating All Classical Symmetric Orthogonal Polynomials of a Discrete Variable

In this part, we introduce a particular example of Sn∗ (x ; a, b, c, d) that generates all the classical symmetric orthogonal polynomials of a discrete variable studied in Sect. 2.3 and is different from the two introduced infinite hypergeometric sequences. If a = −2(b + 2c + 4d) is replaced in the main difference equation (2.110), after simplification of a common factor we get   x 2 (b + 2c + 4d) + x(c + 2d) + d Δ∇φn (x) − 2x(c + 2d)Δφn (x) + n(−(b(n − 1) + 2(n − 2)(c + 2d)))φn (x) = 0, which is the same equation as analyzed in Sect. 2.3, i.e., ˆ + c)Δ∇y ˆ ˆ (ax ˆ 2 + bx ˆ + n(a(1 ˆ − n) + 2b)y(x) = 0, n (x) − 2bxΔy(x)

218

2 Generalized Sturm–Liouville Problems in Discrete Spaces

for d = c, ˆ

c = bˆ − 2c, ˆ

ˆ b = aˆ − 2b,

and a = −2a. ˆ

This means that the particular polynomial  yn (x) =

Sn∗

−2aˆ aˆ − 2bˆ x ˆb − 2cˆ cˆ

=x

σn

  [n/2] (−1) j

[n/2]  j =0

j

 ⎞ [n/2]−1 ˆ + σn ) − cˆ ˆ + σn )2 + b(i (2i + 2σn + 1) −a(i ⎠ (σn − x)j (σn + x)j   ×⎝ ˆ 2 b + a(σ ˆ − i − [n/2]) − a ˆ n+1 i=j ⎛

(2.135) generates all cases of classical symmetric orthogonal polynomials of a discrete variable as follows: ˆ + cˆ = x + cˆ has Case 1 If aˆ = 0, bˆ = 1, and cˆ is free in (2.135), then σ (x) = ax ˆ 2 + bx one real root, and the symmetric Kravchuk polynomials are derived as 

0 −2 x 1 − 2cˆ cˆ

Sn∗



 = 2−n (−2c) ˆ n 2 F1

−n, −cˆ − x 2 −2cˆ

(1/2)

= kn

(x + c; ˆ 2c), ˆ

which are orthogonal with respect to the weight function 1 Γ (cˆ − x + 1)Γ (cˆ + x + 1)

(x) =

for x ∈ {−c, ˆ −cˆ + 1, . . . , cˆ − 1, c} ˆ when 2cˆ ∈ N.

ˆ cˆ = (x+b/2) ˆ 2 Case 2 If aˆ = 1, bˆ is free, and cˆ = bˆ 2 /4 in (2.135), then σ (x) = ax ˆ 2 +bx+ has a double real root and the symmetric Hahn–Eberlein polynomials  Sn∗

−2 1 − 2bˆ x ˆb(2 − b)/2 ˆ bˆ 2 /4  × 3 F2

=

ˆ n )2 ((−b) (n − 2bˆ − 1)n

ˆ −n, n − 2bˆ − 1, −x − b/2 1 ˆ −bˆ −b,

ˆ = h˜ n(0,0)(x + b/2, bˆ + 1)

are orthogonal with respect to the weight function (x) =

1 Γ 2 (x

2 (−x + 1 + b/2) ˆ ˆ + 1 + b/2)Γ

ˆ ˆ + 1, . . . , b/2 ˆ − 1, b/2} ˆ for x ∈ {−b/2, −b/2 when b ∈ N.

,

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

219

Case 3 If aˆ = 1, bˆ = −δ1 − δ2 , and cˆ = δ1 δ2 in (2.135), then σ (x) has two different real roots δ1 and δ2 , and depending on the values δ1 and δ2 , the symmetric Hahn–Eberlein polynomials are derived as 

−2 1 + 2(δ1 + δ2 ) x δ1 δ2 −δ2 − δ1 (1 + 2δ2 ) 

(δ1 + δ2 )n (2δ2 )n −n, n + 2(δ1 + δ2 ) − 1, δ2 − x 1 = 3 F2 (n + 2(δ1 + δ2 ) − 1)n δ1 + δ2 , 2δ2

Sn∗

2 −δ1 ,δ2 −δ1 ) (x − δ ; 1 − 2δ ), = h˜ (δ 2 2 n

when −2δ2 ∈ N, while the symmetric Hahn polynomials

−2 1 + 2(δ1 + δ2 ) x δ1 δ2 −δ2 − δ1 (1 + 2δ2 )

 (δ1 + δ2 )n (2δ1 )n −n, n + 2(δ1 + δ2 ) − 1, δ1 − x = 1 3 F2 (n + 2(δ1 + δ2 ) − 1)n δ1 + δ2 , 2δ1

 Sn∗

1 +δ2 −1,δ1 +δ1 −1) (x − δ ; 1 − 2δ ), = h(δ 1 1 n

are derived when −2δ1 ∈ N.

2.5.4

Two Finite Types of Sn∗ (x ; a, b, c, d)

In this section, we introduce two finite sequences of hypergeometric symmetric orthogonal polynomials of a discrete variable and obtain their basic properties. To define such polynomials, we need to recall that the limit definition Γ (z) = lim

n→∞

n! nz n  z+k k=0

implies that Γ (p + iq) Γ (p − iq) = Γ 2 (p)

∞ k=0

(p + k)2 (p + k)2 + q 2

is always a positive real value for every p > 0 and q ∈ R.

(2.136)

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2 Generalized Sturm–Liouville Problems in Discrete Spaces

First Finite Sequence For p, q, r ∈ R, consider the equation (2x + 1)(x − p)(x 2 − 2qx + q 2 + r 2 )Δ∇φn (x)   − 2x (1 − 2p − 4q)x 2 + 2pq + (q 2 + r 2 )(1 − 2p) Δφn (x)    1   σn  + 2n n + 2(p + 2q − 1) − x 2 − (2p − 1) 4r 2 + (2q − 1)2 φn (x) = 0, 4 2 having a monic polynomial solution  Sn∗

1, −(p + 2q) x 2 2 2pq + q + r , −p(q 2 + r 2 ) = 

× x

σn

4 F3

= φ¯n (x; p, q, r)

(p + σn )[n/2] (q + ir + σn )[n/2] (q − ir + σn )[n/2] ([n/2] + p + 2q − 1 + σn )[n/2]

−[n/2], [n/2] + p + 2q − 1 + σn , σn − x, σn + x 1 p + σn , q + ir + σn , q − ir + σn

(2.137)

that satisfies the recurrence relation φ¯ n+1 (x; p, q, r) = x φ¯ n (x; p, q, r) − γn (p, q, r) φ¯ n−1 (x; p, q, r),

(2.138)

in which 32(n + p + 2q − 2)(n + p + 2q − 1)γn (p, q, r) =

(2.139)

− 2n4 + 4(3 − 2p − 4q)n3 − 8(p2 + 5q 2 + r 2 + 6pq − 4p − 8q + 3)n2      + 2 (3 − 2p − 4q) 3 − 4p − 8q(1 − p) + 4q 2 + 4r 2 + (−1)n (2p − 1) 4r 2 + (2q − 1)2 n     − 1 − (−1)n (2p − 1)(2p + 4q − 3) 4r 2 + (2q − 1)2 .

Relation (2.139) can be decomposed as 1 16(n + p + 2q − 2)(n + p + 2q − 1)    × n + (2p − 1)σn n + 2q + 2ri − 1 + (2p − 1)(1 − σn )    × n + 2q − 2ri − 1 + (2p − 1)(1 − σn ) n + 4q − 2 + (2p − 1)σn .

γn (p, q, r) = −

(2.140)

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

221

By noting the relations (2.137)–(2.140), the orthogonality relation of the first finite sequence now takes the general form θ  x=−θ

 W

∗

1 −(p + 2q) x 2pq + q 2 + r 2 −p(q 2 + r 2 ) =



n

γk (p, q, r)

θ  x=−θ

k=1

φ¯ n (x; p, q, r) φ¯ m (x; p, q, r)

 W

∗

1 −(p + 2q) x 2pq + q 2 + r 2 −p(q 2 + r 2 )



δn,m ,

in which  W

∗

1 −(p + 2q) x 2 2 2pq + q + r −p(q 2 + r 2 )

=

1 4

 − x 2 W ∗ (x)

denotes the original weight function and W ∗ (x) satisfies the difference equation (1/2) − x −x − p −x − q − ir −x − q + ir W ∗ (x + 1) = . ∗ W (x) (3/2) + x x + 1 − p x + 1 − q − ir x + 1 − q + ir

(2.141)

There are 16 symmetric solutions for Eq. (2.141) as follows:  W1 (x) = Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q − ir + x)Γ (1 − q − ir − x) × Γ (1 − q + ir + x)Γ (1 − q + ir − x)Γ (3/2 + x)Γ (3/2 − x)

−1

,

W2 (x) = Γ (p + x)Γ (p − x) , Γ (1 − q − ir + x)Γ (1 − q − ir − x)Γ (1 − q + ir + x)Γ (1 − q + ir − x)Γ (3/2 + x)Γ (3/2 − x) W3 (x) = Γ (q + ir + x)Γ (q + ir − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + ir + x)Γ (1 − q + ir − x)Γ (3/2 + x)Γ (3/2 − x) W4 (x) = Γ (q − ir + x)Γ (q − ir − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q − ir + x)Γ (1 − q − ir − x)Γ (3/2 + x)Γ (3/2 − x)

222

2 Generalized Sturm–Liouville Problems in Discrete Spaces

W5 (x) =

Γ (p + x)Γ (p − x)Γ (q + ir + x)Γ (q + ir − x) , Γ (1 − q + ir + x)Γ (1 − q + ir − x)Γ (3/2 + x)Γ (3/2 − x)

W6 (x) =

Γ (p + x)Γ (p − x)Γ (q − ir + x)Γ (q − ir − x) , Γ (1 − q − ir + x)Γ (1 − q − ir − x)Γ (3/2 + x)Γ (3/2 − x)

W7 (x) =

Γ (q + ir + x)Γ (q + ir − x)Γ (q − ir + x)Γ (q − ir − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (3/2 + x)Γ (3/2 − x)

W8 (x) =

Γ (p + x)Γ (p − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − q − ir + x)Γ (1 − q − ir − x)Γ (1 − q + ir + x)Γ (1 − q + ir − x)

W9 (x) =

Γ (q + ir + x)Γ (q + ir − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q + ir + x)Γ (1 − q + ir − x)

W10 (x) =

Γ (q − ir + x)Γ (q − ir − x)Γ (−1/2+x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q − ir + x)Γ (1 − q−ir − x)

W11 (x) =

Γ (p + x)Γ (p − x)Γ (q + ir + x)Γ (q+ir − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − q+ir + x)Γ (1−q + ir − x)

W12 (x) =

Γ (p + x)Γ (p − x)Γ (q − ir + x)Γ (q − ir−x)Γ (−1/2+x)Γ (−1/2−x) , Γ (1 − q − ir + x)Γ (1 − q − ir−x)

W13 (x) = Γ (q + ir + x)Γ (q + ir − x)Γ (q − ir + x)Γ (q − ir − x)Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)

W14 (x) = Γ (−1/2 + x)Γ (−1/2 − x) , Γ (1 − p + x)Γ (1 − p − x)Γ (1 − q − ir + x)Γ (1 − q − ir − x)Γ (1 − q + ir + x)Γ (1 − q + ir − x)

W15 (x) =

Γ (p + x)Γ (p − x)Γ (q + ir + x)Γ (q + ir − x)Γ (q − ir + x)Γ (q − ir − x) , Γ (3/2 + x)Γ (3/2 − x)

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

223

and W16 (x) = Γ (p + x)Γ (p − x)Γ (q + ir + x)Γ (q + ir − x) × Γ (q − ir + x)Γ (q − ir − x)Γ (−1/2 + x)Γ (−1/2 − x). Although {Wk (x)}16 k=1 are all symmetric, only the three weight functions W1 , W2 , and W14 are applicable, since they are real-valued functions according to (2.136). On the other hand, using the two identities Γ (p + x) = Γ (p)(p)x

and Γ (p − x) =

Γ (p)(−1)x , (1 − p)x

and the fact that  θ    ∗ 1 1 2 − x W (x) = 2 − x W ∗ (x) − W ∗ (0), 4 4 4

θ  1 x=−θ

2

x=0

if we assume  B(p, q, r) = 2 4 F3

q + ir, q − ir, p, 1 1 1 − q + ir, 1 − q − ir, 1 − p

− 1,

then after some computations we obtain ∞  1 x=−∞

4

 − x 2 W1 (x) =

π

B(p, q, r) , − ir)Γ 2 (1 − q + ir)

Γ 2 (1 − p)Γ 2 (1 − q

∞  1  Γ 2 (p) − x 2 W2 (x) = B(p, q, r), 2 4 π Γ (1 − q − ir) Γ 2 (1 − q + ir) x=−∞

and ∞  1 x=−∞

4

 − x 2 W14 (x) =

π B(p, q, r) . Γ 2 (1 − p)Γ 2 (1 − q − ir)Γ 2 (1 − q + ir)

The above computations show that there exists a unique representation for the original weight function as W (x; p, q, r) =

(p)x (q + ir)x (q − ir)x . (1 − p)x (1 − q + ir)x (1 − q − ir)x

(2.142)

224

2 Generalized Sturm–Liouville Problems in Discrete Spaces

Also, by noting the identity (p)−x =

(−1)x , (1 − p)x

(2.143)

we see that (2.142) is symmetric on the support (−∞, ∞), i.e., using (2.143) we can prove that W (−x; p, q, r) = W (x; p, q, r). Now, to prove the finite orthogonality, since in the first sequence A(x) = (2x + 1)(x − p)(x 2 − 2qx + q 2 + r 2 ) and W ∗ (x) =

W (x; p, q, r) K Γ (p + x) Γ (q + ir + x) Γ (q − ir + x) , = 2 2 (1/4) − x 1 − 4x Γ (1 − p + x) Γ (1 − q + ir + x) Γ (1 − q − ir + x)

where K=4

Γ (1 − p) Γ (1 − q + ir) Γ (1 − q − ir) , Γ (p) Γ (q + ir) Γ (q − ir)

the key condition lim A(−θ )W ∗ (θ ) θ n+m = 0

θ→∞

(2.144)

implies that we have (2θ − 1)(θ + p)(θ 2 + 2qθ + q 2 + r 2 )Γ (p + θ )Γ (q + ir + θ )Γ (q − ir + θ ) n+m θ = 0. θ →∞ (1 − 4θ 2 )Γ (1 − p + θ )Γ (1 − q + ir + θ )Γ (1 − q − ir + θ ) lim

(2.145)

On the other hand, since lim

θ→∞

1 (2θ − 1)(θ + p) =− 1 − 4 θ2 2

and lim

θ→∞

Γ (p + θ )Γ (q + ir + θ )Γ (q − ir + θ ) 1 ln = 2p + 4q − 3, ln θ Γ (1 − p + θ )Γ (1 − q + ir + θ )Γ (1 − q − ir + θ )

2.5 A Basic Class of Symmetric Orthogonal Polynomials of a Discrete. . .

225

relation (2.145) is equivalent to lim θ 2+n+m+2p+4q−3 = 0.

θ→∞

(2.146)

If in (2.146) we assume that max{m, n} = N, then it yields 2N + 2p + 4q − 1 < 0 or

N
0 and w(x; q) > 0, (3.2) with the boundary conditions α1 y(a; q) + β1 Dq y(a; q) = 0, (3.3) α2 y(b; q) + β2 Dq y(b; q) = 0, in which Dq is the q-difference operator defined by Dq f (x) =

f (qx) − f (x) (q − 1)x

(x = 0, q = 1),

with Dq f (0) := f  (0) (provided f  (0) exists).

© Springer Nature Switzerland AG 2020 M. Masjed-Jamei, Special Functions and Generalized Sturm–Liouville Problems, Frontiers in Mathematics, https://doi.org/10.1007/978-3-030-32820-7_3

233

234

3 Generalized Sturm–Liouville Problems in q-Spaces

Two eigenfunctions of the q-Sturm–Liouville problem (3.1)–(3.3) are orthogonal with respect to the positive weight function w(x; q) on (a, b). This result is a consequence of the following fact, which is a q-analogue of integration by parts. Let us consider the identity  x2 x1

  x2 (f Lg − gLf )(x; q)dq x = r(x; q) (Dq g)(q −1 x; q)f (x; q) − (Dq f )(q −1 x; q)g(x; q) , x1

(3.4) where the q-integral operator is defined by 

x

f (t)dq t = (1 − q)x

0

∞ 

q j f (q j x),

j =0

and for two nonnegative numbers x1 and x2 with x1 < x2 we have 

x2

 f (x)dq x = 0

x1

x2



x1

f (x)dq x −

f (x)dq x. 0

Let ym (x; q) and yn (x; q) be solutions of Eq. (3.1) corresponding to two eigenvalues λm,q and λn,q . If x1 = a, x2 = b, f (x; q) = ym (x; q), and g(x; q) = yn (x; q) in (3.4), then 

b

(ym L[yn ] − yn L[ym ])(x; q)dq x = 0.

a

This means that yn (x; q) and ym (x; q) as eigenfunctions of Eq. (3.1) are orthogonal with respect to the weight function w(x; q), and we have 

b

ym (x; q) yn(x; q) w(x; q)dq x = 0,

(λm = λn ).

a

Hence, one would be able to obtain some q-orthogonal polynomials through the solutions of a q-Sturm–Liouville problem. For instance, classical q-orthogonal polynomials are discrete versions of continuous classical orthogonal polynomials. In 1884, Markov found a q-analogue of Chebyshev polynomials that was indeed the first example of qorthogonal polynomial families. In 1949, Hahn studied a problem similar to the case considered by Bochner, i.e., finding the sequences of q-orthogonal polynomials that are solutions of a linear second-order q-difference equation. Today they are known as the Hahn class of q-orthogonal polynomials. He also obtained the most general q-orthogonal polynomial on the exponential lattice, i.e., big q-Jacobi polynomials.

3.1 Introduction

235

Afterward, many authors considered q-polynomials with different aspects and found various applications in theoretical and mathematical physics such as continued fractions, Eulerian series, algebras and quantum groups, discrete mathematics, algebraic combinatorics (coding theory, design theory, various theories of group representation), the q-Schrödinger equation and q-harmonic oscillators. The theory of q-special functions has received a new impulse with the introduction of quantum algebras and groups. Indeed, real interpretations of these functions have been derived from representation theory of quantum algebras in similar ways to how Lie algebras are used to describe classical special functions. The q-classical polynomials satisfy a q-difference equation of hypergeometric type as σ (x)(Dq2 yn )(x; q) + τ (x)(Dq yn )y(x; q) + λn,q yn (qx; q) = 0,

(3.5)

where σ (x) and τ (x) are polynomials of degree at most 2 and 1, respectively, and λn,q is computed as λn,q = −

[n]q  (σ [n − 1]q + τ  ), qn

n = 0, 1, . . . .

To prove the orthogonality relation of q-polynomials satisfied by Eq. (3.5), we can first assume in Eq. (3.1) that r(x; q) = σ (x)w(x; q) and then write Eq. (3.5) in the self-adjoint form    Dq σ wDq yn (x; q) + w(x; q)λn,q yn (qx; q) = 0 to conclude that they are orthogonal with respect to the weight function w as 

b a

Pm (x; q)Pn (x; q)w(x; q)dq x = dn2 δm ,

where dn shows the norm square value and w is a solution of the q-Pearson equation   Dq σ (q −1 x)w(x; q) = w(qx; q)τ (x),

(3.6)

such that the following boundary conditions always holds b σ (x)w(x; q) a = 0.

(3.7)

Thus, similar to the continuous and discrete cases, if some extera constraints are required for satisfying the boundary conditions (3.7), then finite classes of q-orthogonal polynomials will appear.

236

3 Generalized Sturm–Liouville Problems in q-Spaces

In this chapter, we introduce some finite classes of q-orthogonal polynomials via introducing some q-Sturm–Liouville problems. We also concentrate on their general properties and compare the results with the corresponding continuous cases. Next, using a symmetric generalization of Sturm–Liouville problems in q-difference spaces, we introduce some classes of symmetric q-orthogonal polynomials and obtain their standard properties. For this purpose, we should first state some basic definitions and necessary preliminaries related to q-calculus that are less well known.

3.2

Some Preliminaries and Definitions

3.2.1

The q-Shifted Factorial

One of the basic concepts of the q-calculus is the limit relation 1 − qα = α. q→1 1 − q lim

Based on this limit,  1 − qn qk = 1−q n−1

[n]q =

with

[0]q = 1

(3.8)

k=0

is called a q-number. The q-shifted factorial is defined by (a; q)k = (1 − a)(1 − aq) · · · (1 − aq k−1)

with

(a; q)0 = 1.

So it is clear that (q α ; q)k = (α)k . q→1 (1 − q)k lim

For negative indices, the q-shifted factorial is defined by (a; q)−k = k

1

−i i=1 (1 − aq )

,

where a = q, q 2, q 3 , . . . , q k

and k = 1, 2, 3, . . . .

Also, as an extension we have (a; q)∞ =

∞ k=0

(1 − aq k )

for 0 < |q| < 1.

3.2 Some Preliminaries and Definitions

237

This implies that for every complex number λ, (a; q)λ =

(a; q)∞ , (aq λ; q)∞

0 < |q| < 1,

where the principal value of q λ is taken. The following relations hold for q-shifted factorials: (a; q)∞ = (a; q 2)∞ (aq; q 2)∞ ,

0 < |q| < 1,

(a 2 ; q 2)∞ = (a; q)∞(−a; q)∞ ,

0 < |q| < 1.

and

3.2.2

q-Hypergeometric Series

Since a usual hypergeometric series is defined as

 r Fs

a1 , . . . , ar z b1 , . . . , bs

:=

∞  (a1 , . . . , ar )k zk , (b1 , . . . , bs )k k! k=0

where (a1 , . . . , ar )k := (a1 )k · · · (ar )k , a basic or q-hypergeometric series is defined as 

∞   k(k−1) 1+s−r (a1 , . . . , ar ; q)k zk  a1 , . . . , ar , (−1)k q 2 q; z := r φs (b1 , . . . , bs ; q)k (q; q)k b1 , . . . , bs k=0 where (a1 , . . . , ar ; q)k := (a1 ; q)k · · · (ar ; q)k , and a1 , a2 , . . . , ar , b1 , b2 , . . . , bs , z ∈ C. In this case, we must assume that b1 , b2 , . . . , bs = q −k (k = 0, 1, . . . )

238

3 Generalized Sturm–Liouville Problems in q-Spaces

in order to have a well-defined series. By noting these assumptions, we see that the following limit relation holds:  



, . . . , a a q a1 , . . . , q ar 1 r z . lim r φs b1 q; (q − 1)1+s−r z = r Fs q→1 q , . . . , q bs b1 , . . . , bs

3.2.3

q-Binomial Coefficients and the q-Binomial Theorem

A q-binomial coefficient is defined as     n (q; q)n n = = (q; q)k (q; q)n−k n−k q k q

for k = 0, 1, 2, . . . , n,

where n is a nonnegative integer. Moreover, for all complex α and β and 0 < |q| < 1, we have   Γq (α + 1) α (q β+1 ; q)∞ (q α−β+1 ; q)∞ = = . β q Γq (β + 1)Γq (α − β + 1) (q; q)∞ (q α+1 ; q)∞ Note that     α α α! lim . = = q→1 β q β!(α − β)! β The q-binomial theorem, 

∞  (a; q)n n (az; q)∞ a z = q; z = 1 φ0 (q; q)n (z; q)∞ − n=0

for 0 < |q| < 1

and |z| < 1, (3.9)

is indeed a q-analogue of the usual binomial theorem

 1 F0

a z −

=

∞  (a)n n=0

n!

zn = (1 − z)−a

for |z| < 1.

If in (3.9) we set a = q −n , then we obtain 

q −n q; z = (zq −n ; q)n 1 φ0 −

for all n = 0, 1, 2, . . . .

3.2 Some Preliminaries and Definitions

3.2.4

239

q-Gamma and q-Beta Functions

The q-gamma function defined by Γq (x) =

(q; q)∞ (1 − q)1−x (q x ; q)∞

(3.10)

holds for 0 < |q| < 1 if the principal values of q x and (1 − q)1−x are considered. For q > 1, it is defined as Γq (x) =

(q −1 ; q −1 )∞ (x ) q 2 (q − 1)1−x . (q −x ; q −1 )∞

Naturally we have lim Γq (x) = Γ (x).

q→1

Also, there exists a q-extension of the functional equation of the ordinary gamma function as Γq (z + 1) =

1 − qz Γq (z) with 1−q

Γq (1) = 1.

The q-beta function defined by Bq (t, s) =

Γq (s)Γq (t) Γq (s + t)

satisfies the following q-integral representation, which is a q-analogue of Euler’s formula:  Bq (t, s) =

1

x t −1 (qx; q)s−1 dq x

for all t, s > 0.

0

3.2.5

q-Analogues of Some Special Functions

The exponential function has two different q-extensions as follows: eq (z) =

∞  (1 − q)n n=0

(q; q)n

zn =

1 ((1 − q)z; q)∞

for |z| < 1

(3.11)

240

3 Generalized Sturm–Liouville Problems in q-Spaces

and Eq (z) =

n ∞  q (2) (1 − q)n

n=0

(q; q)n

zn = (−(1 − q)z; q)∞ ,

which are related to each other by eq (z)Eq (z) = 1. In the limit case we have lim eq (z) = lim Eq (z) = ez .

q→1

q→1

By assuming |z| < 1, we can now define ∞

sinq (z) =

eq (iz) − eq (−iz)  (−1)n (1 − q)2n+1 z2n+1 = 2i (q; q)2n+1 n=0

and ∞

eq (iz) + eq (−iz)  (−1)n (1 − q)2n z2n = . cosq (z) = 2 (q; q)2n n=0

Another q-analogue of the trigonometric functions may be defined as Sinq (z) =

Eq (iz) − Eq (−iz) 2i

Cosq (z) =

Eq (iz) + Eq (−iz) , 2

and

where eq (iz) = cosq (z) + i sinq (z) and Eq (iz) = Cosq (z) + i Sinq (z).

3.2.6

q-Difference Operators

A linear q-difference operator Dq defined by (Dq f )(x) =

f (qx) − f (x) , (q − 1)x

q ∈ R \ {−1, 0, 1},

x ∈ C \ {0},

3.2 Some Preliminaries and Definitions

241

is valid for arbitrary complex-valued functions f whose domain contains qx ∈ C, x ∈ C, and satisfies the following relations: Dq Dq −1 = q −1 Dq −1 Dq , Dq = Dq −1 + (q − 1)xDq Dq −1 , Dq f (q −1 x) = Dq −1 f (x), and Dq Dq −1 f (x) = q −1 Dq2 f (q −1 x). Moreover, for two functions f1 and f2 we have (Dq (f1 f2 ))(x) = (Dq f1 )(x)f2 (x) + f1 (qx)(Dq f2 )(x). It is important to note that if we apply the operator Dq to the function f (· + c), where c ∈ R is a constant, then we have   f (qx + c) − f (x + c) Dq f (· + c) (x) = , (q − 1)x while if we first apply the operator Dq to the function f and then replace the argument x by x + c, we have   f (qx + c) − f (x + c) . Dq f (·) (x + c) = (q − 1)(x + c) For simplicity, if we take (Dq f )(x) = Dq f (x) =

f (qx) − f (x) = p(x), (q − 1)x

then we get (Dq2 f )(x) = (Dq p)(x) =

f (q 2 x) − (1 + q)f (qx) + qf (x) p(qx) − p(x) = , (q − 1)x q(q − 1)2 x 2 (3.12)

as well as (Dq −1 Dq f )(x) =

f (qx) − (1 + q)f (x) + qf (q −1 x) . (q − 1)2 x 2

242

3 Generalized Sturm–Liouville Problems in q-Spaces

For example, we have  (qx)n − x n (qx)n−1−k x k = (q − 1)x n−1

Dq (x n ) =

k=0

= x n−1

n−1 

q k + r(x) = [n]q x n−1 + r(x)

for all n = 2, 3, 4, . . . ,

k=0

where r is a polynomial of degree at most n − 2. If the operator Lq is defined as (Lq f )(x) = f (qx),

(3.13)

 m  Lm q f (x) = f q x

(3.14)

we can conclude that

and (L−1 q f )(x) = f

3.2.7

  x . q

q-Integral Operators

A q-integral, as the inverse of a q-difference operator, is defined as 

x

f (t)dq t = (1 − q)x

0

∞ 

q j f (q j x),

j =0

provided that the series converges. For two nonnegative numbers a and b with a < b, the definition (3.15) yields 

b



b

f (t)dq t =



0

a

a

f (t)dq t −

f (t)dq t. 0

Furthermore, we have 

∞ 0

f (t)dq t = (1 − q)

∞  n=−∞

q n f (q n )

(3.15)

3.2 Some Preliminaries and Definitions

243

and 

∞ −∞

∞ 

f (t)dq t = (1 − q)

  q n f (q n ) + f (−q n ) .

n=−∞

A function f that is defined on a q-geometric set A with 0 ∈ A is said to be q-regular at zero if lim f (xq n ) = f (0)

n→∞

for every x ∈ A.

The rule of q-integration by parts is given by 

a



a

g(x)Dq f (x)dq x = (fg)(a) − lim (fg)(aq ) − n

n→∞

0

Dq g(x)f (qx)dq x.

(3.16)

0

Note that if f, g are q-regular at zero, then limn→∞ (fg)(aq n ) on the right-hand side of (3.16) can be replaced by (fg)(0). For 0 < R ≤ ∞, let ΩR denote the disk {z ∈ C : |z| < R}. The q-analogue of the fundamental theorem of calculus says that if f : ΩR → C is q-regular at zero and θ ∈ ΩR is fixed, then the function 

x

F (x) =

f (t)dq t

(x ∈ ΩR )

(3.17)

θ

is q-regular at zero, Dq F (x) exists for all x ∈ ΩR , and Dq F (x) = f (x). Conversely, if a, b ∈ ΩR , we have 

b

Dq f (t)dq t = f (b) − f (a).

a

Relation (3.15) can be obtained via (3.17) and applying the properties (3.13) and (3.14), because the equality (Dq F )(x) = f (x) implies that (Lq − I )F (x) F (qx) − F (x) = = f (x). (q − 1)x (q − 1)x Hence we have F (x) =

∞

∞

j =0

j =0

   j  1  (1 − q)xf (x) = Lq ((1 − q)xf (x) = (1 − q)x q j f (q j x). I − Lq

244

3 Generalized Sturm–Liouville Problems in q-Spaces

Finally, if u(x) = αx β , the following q-integral relation holds: 

u(b)



b

f (u)dq u =

u(a)

f (u(x))D

1

qβ

a

u(x)d

1

qβ

x.

(3.18)

• Some Definite q-Integrals b For |q| < 1, |a| > |q|, |b| < 1, and | | < |x| < 1, Ramanujan’s celebrated formula is a given by   q n+1 (1 − (1 − axq n ) 1− ∞ ∞  (a; q)n n ax Ψ (a, b; q; x) = x = .    (b; q)n bq n q n+1 n=−∞ n=0 (1 − bq n ) 1 − 1− (1 − xq n ) a ax 

bq n 1− a



q n+1 )

(3.19) Now consider the integrals 

∞/A 0

Bq (t, s) x t −1 dq x = (−x; q)t +s K(A, t)

(3.20)

and 

∞/(1−q)

0

x t −1 eq−x dq x =

Γq (t) , K(1, t)

(3.21)

where 

∞/A 0

 n ∞  q qn f f (t)dq t = (1 − q) A A n=−∞

and K(x, t) =

xt 1 (− ; q)t (−x; q)1−t . 1+x x

For instance, the integral (3.21) can be directly derived by Ramanujan’s formula as follows: 

∞ 1−q

0

x t −1eq−x dq x

∞ 

qn = (1 − q) 1−q n=−∞ = (1 − q)1−t

∞  n=−∞

qn



qn 1−q

t −1

n

q − 1−q

eq

∞ qn  q nt − 1−q q nt 1−t e = (1 − q) q n q (−q n ; q)∞ n=−∞

3.2 Some Preliminaries and Definitions

3.2.8

245

=

∞ (1 − q)1−t  q nt (−1; q)n (1 − q)1−t = Ψ (−1, 0; q; q t ) (−1; q)∞ n=−∞ (0; q)n (−1; q)∞

=

(1 − (−1; q)∞

=

Γq (t) (1 − (q; q)∞ (−q 1−t ; q)∞ (−q t ; q)∞ . = t (−1; q)∞ (−q; q)∞(q ; q)∞ K(1, t)

q)1−t

n+1 n+1 )(1 + q )(1 + q t q n ) ∞ (1 − q qt (1 + q n+1 )(1 − q t q n )

n=0

q)1−t

An Analytical Solution for the q-Pearson Difference Equation

If Eq. (3.6) is expanded, a general q-difference equation will appear as W (qx) = A(x; q). W (x)

(3.22)

Theorem 3.1 The solution of the generic equation (3.22) can be represented as W (x) =

∞ j =0

1 A(q j x; q)

.

Proof Taking the logarithm on both sides of (3.22) first yields ln A(x; q) ln W (qx) − ln W (x) = . (q − 1)x (q − 1)x Now by referring to the definition 

x

f (t) dq t = (1 − q)x

0

∞ 

q j f (xq j ),

j =0

replace f (t) =

ln W (qt) − ln W (t) (q − 1)t

(3.23)

246

3 Generalized Sturm–Liouville Problems in q-Spaces

in (3.23) to get  0

x

∞

 ln W (q j +1 x) − ln W (q j x) ln W (qt) − ln W (t) dq t = (1 − q)x qj (q − 1)t (q − 1)xq j j =0

= lim

n→∞

n  j =0

= lim ln n→∞

W (q j x) W (q j x) = lim ln W (q j +1 x) n→∞ W (q j +1 x) n

ln

j =0

W (x) W (x) . = ln n+1 W (0) W (q x)

Therefore, we have 

x 0

ln A(t; q) W (x) dq t = ln . (q − 1)t W (0)

Since W (0) = 0 has no effect on the initial q-difference equation, we finally obtain W (x) = exp



x 0

   ln A(q j x; q)  ln A(t; q) dq t = exp (1 − q)x qj (q − 1)t (q − 1)xq j

 = exp − ln

∞

j =0

∞

 A(q j x; q) =

j =0

∞ j =0

1 . A(q j x; q)  

One of the direct consequences of this theorem is that the q-difference equation W (x) = B(x; q) W (qx) has a solution of the form W (x) =

∞

B(q j x; q).

j =0

3.2.9

Difference Equations of q-Hypergeometric Series

Let

 a1 , . . . , ar y(z) = r φs q; z . b1 , . . . , bs

3.2 Some Preliminaries and Definitions

247

Then the above-mentioned series satisfies a general difference equation as ΔΔb1 /q Δb2 /q · · · Δbs /q y(z) = z(Δa1 · · · Δar )y(zq 1+s−r ),

(3.24)

where Δb f (x) = bf (qx) − f (x). For instance, suppose in (3.24) that r = 2 and s = 1. Then the equation Δ(Δc/q y(x)) − xΔa (Δb y(x)) = 0 is expanded as q −1 cy(q 2 x) − y(qx) − q −1 cy(qx) + y(x)   − x aby(q 2x) − ay(qx) − by(qx) + y(x) = 0. This means that the explicit form of the q-difference equation of 2 φ1 is (cq −1 − abx)y(q 2x) − (cq −1 + 1 − (a + b)x)y(qx) + (1 − x)y(x) = 0, which is equivalent to  (1 − q)2 x(c − abqx)Dq2 y(x) + (1 − q) 1 − c − (a + b − ab − abq)x)Dq y(x) − (1 − a)(1 − b)y(x) = 0. Similarly, by taking r = s = 1 in (3.24), the equation Δ(Δc/q y(x)) − x(Δa y(qx)) = 0 is expanded as   q −1 cy(q 2x) − y(qx) − q −1 cy(qx) + y(x) − x ay(q 2x) − y(qx) = 0. Therefore, the explicit form of the q-difference equation of 1 φ1 is (cq −1 − ax)y(q 2x) − (cq −1 + 1 − x)y(qx) + y(x) = 0, which is equivalent to  q(1 − q)2 x(c − aqx)Dq2 y(x) + (1 − q) q(1 − c) − q(1 − a − aq)x)Dq y(x) + q(1 − a)y(x) = 0.

248

3 Generalized Sturm–Liouville Problems in q-Spaces

It is not difficult to verify that

 

a, b x a = 1 φ1 q; x . lim 2 φ1 q; b→∞ b c c

3.2.10 q-Analogues of Jacobi Polynomials There are two q-analogues of Jacobi orthogonal polynomials whose standard properties are represented as follows. 1. Big q-Jacobi Polynomials • Basic hypergeometric representation:

 q −n , abq n+1 , x Pn (x; a, b, c; q) = 3 φ2 q; q . aq, cq

(3.25)

• Orthogonality relation: For 0 < aq < 1, 0 ≤ bq < 1, and c < 0 we have 

aq cq

(a −1 x, c−1 x; q)∞ Pm (x; a, b, c; q)Pn (x; a, b, c; q)dq x (x, bc−1 x; q)∞ = aq(1 − q) ×

(q, abq 2 , a −1 c, ac−1 q; q)∞ (aq, bq, cq, abc−1 q; q)∞

n (1 − abq) (q, bq, abc−1 q; q)n (−acq 2 )n q (2) δm,n . (1 − abq 2n+1 ) (aq, abq, cq; q)n

• Recurrence relation (x − 1)Pn (x; a, b, c; q) = An Pn+1 (x; a, b, c; q) − (An + Cn )Pn (x; a, b, c; q) + Cn Pn−1 (x; a, b, c; q), where An =

(1 − aq n+1 )(1 − abq n+1)(1 − cq n+1 ) (1 − abq 2n+1)(1 − abq 2n+2)

3.2 Some Preliminaries and Definitions

249

and Cn = −acq n+1

(1 − q n )(1 − abc−1q n )(1 − bq n ) . (1 − abq 2n)(1 − abq 2n+1)

• q-Difference equation:   q −n (1 − q n )(1 − abq n+1 )x 2 y(x) = B(x)y(qx) − B(x) + D(x) y(x) + D(x)y(q −1 x),

where y(x) = Pn (x; a, b, c; q), B(x) = aq(x − 1)(bx − c),

and D(x) = (x − aq)(x − cq).

• Limit relation: If we set c = 0, a = q α , and b = q β in (3.25) and then let q → 1, we obtain the usual shifted Jacobi polynomials as (α,β)

lim Pn (x; q α , q β , 0; q) =

q→1

Pn

(2x − 1)

(α,β) Pn (1)

,

and if c = −q γ for arbitrary real γ instead of c = 0, then we get (α,β)

lim Pn (x; q α , q β , −q γ ; q) =

q→1

Pn

(x)

(α,β) Pn (1)

.

Note that the big q-Legendre polynomials are a particular case of the big q-Jacobi polynomials for a = b = 1. 2. Little q-Jacobi Polynomials • Basic hypergeometric representation: 

q −n , abq n+1 pn (x; a, b; q) = 2 φ1 q; qx . aq

(3.26)

• Orthogonality relation: For 0 < aq < 1 and 0 ≤ bq < 1 we have ∞  (bq; q)k (aq)k pm (q k ; a, b; q)Pn(q k ; a, b; q) (q; q)k k=0

=

(abq 2; q)∞ (1 − abq)(aq)n (q, bq; q)n δm,n . (aq; q)∞ (1 − abq 2n+1) (aq, abq; q)n

250

3 Generalized Sturm–Liouville Problems in q-Spaces

• Recurrence relation: − xpn (x; a, b; q) = An pn+1 (x; a, b; q) − (An + Cn )pn (x; a, b; q) + Cn pn−1 (x; a, b; q),

where An = q n

(1 − aq n+1 )(1 − abq n+1) (1 − abq 2n+1)(1 − abq 2n+2)

and Cn = aq n

(1 − q n )(1 − bq n ) . (1 − abq 2n)(1 − abq 2n+1)

• q-Difference equation:   q −n (1−q n )(1−abq n+1)xy(x) = B(x)y(qx)− B(x)+D(x) y(x)+D(x)y(q −1x), where y(x) = pn (x; a, b; q), B(x) = a(bqx − 1),

and D(x) = x − 1.

• Limit relation: The little q-Jacobi polynomials tend to the usual continuous Laguerre polynomials, so that  lim pn

q→1

1 (1 − q)x; q α , −q β ; q 2

 =

L(α) n (x) (α)

.

Ln (0)

Moreover, the little q-Legendre polynomials are a particular case of the little qJacobi polynomials for a = b = 1.

3.2.11 q-Analogues of Laguerre Polynomials There are three q-analogues of the Laguerre orthogonal polynomials, whose standard properties are represented as follows. 1. Big q-Laguerre Polynomials • Basic hypergeometric representation:  



1 q −n , 0, x q −n , aqx −1 x Ln (x; a, b; q) = 3 φ2 . q; q; q = −1 −n 2 φ1 b aq aq, bq (b q ; q)n (3.27)

3.2 Some Preliminaries and Definitions

251

• Orthogonality relation: For 0 < aq < 1 and b < 0 we have 

aq bq

(a −1 x, b −1 x; q)∞ Lm (x; a, b; q)Ln(x; a, b; q)dq x (x; q)∞ = aq(1 − q)

n (q, a −1b, ab−1q; q)∞ (q; q)n (−abq 2)n q (2) δm,n . (aq, bq; q)∞ (aq, bq; q)n

• Recurrence relation: (x−1)Ln (x; a, b; q) = An Ln+1 (x; a, b; q)−(An +Cn )Ln (x; a, b; q)+Cn Ln−1 (x; a, b; q),

where An = (1 − aq n+1 )(1 − bq n+1 ) and Cn = −abq n+1 (1 − q n ). • q-Difference equation   q −n (1 − q n )x 2 y(x) = B(x)y(qx) − B(x) + D(x) y(x) + D(x)y(q −1 x), where y(x) = Ln (x; a, b; q), B(x) = abq(1 − x),

and D(x) = (x − aq)(x − bq).

• Limit relations: If b = 0, a = q α , and c = q β are replaced in (3.25), the big q-Laguerre polynomials given by (3.27) are derived, i.e., Pn (x; a, 0, c; q) = Ln (x; a, c; q). Also, the shifted Laguerre polynomials can be obtained from the limit relation lim Ln (x; q α , (1 − q)−1 q β ; q) =

q→1

L(α) n (x − 1) (α)

Ln (0)

.

252

3 Generalized Sturm–Liouville Problems in q-Spaces

2. Little q-Laguerre Polynomials • Basic hypergeometric representation:

 

1 q −n , 0 q −n , x −1 x ln (x; a; q) = 2 φ1 . q; q; qx = −1 −n 2 φ0 a − aq (a q ; q)n (3.28) • Orthogonality relation: ∞  (aq)k (aq)n (q; q)n lm (q k ; a; q)ln(q k ; a; q) = δm,n , 0 < aq < 1. (q; q)k (aq; q)∞ (aq; q)n k=0

• Recurrence relation: − xln (x; a; q) = An ln+1 (x; a; q) − (An + Cn )ln (x; a; q) + Cn ln−1 (x; a; q), where An = q n (1 − aq n+1)

and

Cn = aq n (1 − q n ).

• q-Difference equation: q −n (1−q n )xy(x) = ay(qx)+(x−a−1)y(x)+(1−x)y(q −1 x)

with

y(x) = ln (x; a; q).

• Limit relations: The little q-Laguerre polynomials (3.28) are a limit case of the big q-Laguerre polynomials, so that lim Ln (bqx; a, b; q) = ln (x; a; q).

b→−∞

In this sense, note that they are also a particular case of the little q-Jacobi polynomials for b = 0. Also, if a = q α and x → (1 − q)x in (3.28), then lim ln ((1 − q)x; q α ; q) =

q→1

L(α) n (x) (α)

Ln (0)

.

3.2 Some Preliminaries and Definitions

253

3. q-Laguerre Polynomials • Basic hypergeometric representation: L(α) n (x; q)



(q α + 1; q)n q −n n+α+1 = x q; −q 1 φ1 (q; q)n q α+1 

1 q −n , −x n+α+1 = . q; q 2 φ1 (q; q)n 0

• Orthogonality relation for α > −1:  ∞ 0

xα (q −α ; q)∞ (q α+1 ; q)n (α) (α) Lm (x; q)Ln (x; q)dx = Γ (−α)Γ (α + 1)δm,n . (−x; q)∞ (q; q)∞ (q; q)n q n

• Recurrence relation: − q 2n+α+1 xL(α) n (x; q) = n+1 n+α (α) (1−q n+1 )L(α) )+q(1−q n+α )]L(α) )Ln−1 (x; q). n (x; q)+q(1−q n+1 (x; q)−[(1−q

(α)

• q-Difference equation for y(x) = Ln (x; q): −q α (1 − q n )xy(x) = q α (1 + x)y(qx) − [1 + q α (1 + x)]y(x) + y(q −1 x). • Limit relations: If we substitute a = q α and x → −b−1q −1 x in (3.26) and then take b → −∞, we obtain the q-Laguerre polynomials as lim pn (−b −1 q −1 x; q α , b; q) =

b→−∞

(q; q)n L(α) n (x; q). (q α+1 ; q)n

3.2.12 q-Analogue of Hermite Polynomials • Basic hypergeometric representation:  



−n , x −1 −n , q −n+1 n q 2n−1 q q n 2 ( ) . hn (x; q) = q 2 2 φ1 q; −qx = x 2 φ0 q ; x2 0 −

254

3 Generalized Sturm–Liouville Problems in q-Spaces

• Orthogonality relation: 

1 −1

n

(qx, −qx; q)∞ hm (x; q)hn (x; q)dq x = (1 − q)(q; q)∞ (q, −1, −q; q; q)∞ q (2) δm,n .

• Recurrence relation: xhn (x; q) = hn+1 (x; q) + q n−1 (1 − q n )hn−1 (x; q). • q-Difference equation: −q −n+1 x 2 y(x) = y(qx) − (1 + q)y(x) + q(1 − x 2)y(q −1 x) with

y(x) = hn (x; q).

• Limit relation: The usual continuous Hermite polynomials can be found via the following limit relation: % Hn (x) hn (x 1 − q 2 ; q) = . lim n q→1 2n (1 − q 2 ) 2

3.2.13 A Biorthogonal Exponential Sequence Using the Fourier transform of the Stieltjes–Wigert polynomials, in this section we derive a sequence of exponential functions that are biorthogonal with respect to a complex weight function like   exp q1 (ix + p1 ) 2 + q2 (ix + p2 ) 2

on (−∞, ∞).

Then we restrict these biorthogonal functions to a special case to obtain a real trigonometric sequence of orthogonal functions with respect to the weight function exp(−qx 2)

on (−∞, ∞).

In 1895, Stieltjes [6] proved that 

∞ 0

x n x − ln x sin(2π ln x) dx = 0 ,

n ∈ Z.

3.2 Some Preliminaries and Definitions

255

Hence, independent of λ, we have 1 √ π



∞ 0

1 x n x − ln x (1 + λ sin(2π ln x)) dx = √ π



∞

x n x − ln x dx = exp



0

 1 (n + 1)2 , 4

which shows that for all λ ∈ [−1, 1], the densities 1 Wλ (x) = √ x − ln x (1 + λ sin(2π ln x)) , x > 0, π

(3.29)

have the same moments. Using this result, one can conclude that there exist many different weight functions for orthogonal polynomials corresponding to positive measures (3.29), known later as Stieltjes–Wigert polynomials. In this sense, it was Wigert [7] who first succeeded in 1923 in finding the explicit form of the orthogonal polynomials corresponding to the log-normal weight function γ 1 γ 2 , ρ(x; q) = √ exp(−γ 2 ln2 x) = √ x (−γ ln x) for − γ 2 = ln q 2 π π

(3.30)

which is a more general case of (3.29) for λ = 0. The Wigert polynomials are defined as - . 1 n (−1)k q k(k+ 2 ) x k Wn (x; q) = k k=0 n 

for 0 < q < 1,

(3.31)

q

in which, as before, - . i−1 1 − q n−j (q ; q)n n = = (q ; q)i (q ; q)n−i 1 − q j +1 i j =0

and (a ; q)n =

q

n−1

1 − aq i .

i=0

According to [2], they satisfy the orthogonality relation  0

∞

  ρ(x; q) Wn (x; q) Wm (x; q) dx = q −(n+1/2)(q; q)n δn,m .

(3.32)

Although Stieltjes considered the weight function (3.30) only for γ = 1 (or equivalently λ = 0 in (3.29)) as an example leading to an indeterminate Stieltjes moment problem [6], the author in [3] has considered such a problem associated with the Stieltjes–Wigert polynomials in a general case. Consequently, there should be various weight functions for

256

3 Generalized Sturm–Liouville Problems in q-Spaces

polynomials (3.31). For instance, the shifted polynomials Sn (x; q) =

1 1 Wn (q − 2 x; q) (q; q)n

(3.33)

satisfy the orthogonality relation 

∞ 0

  ln q (q; q)∞ ρ (x; q) Sn (x; q) Sm (x; q) dx = − n δn,m , q (q; q)n ∗

(3.34)

where [4] ρ ∗ (x; q) =

∞

1 1 . = −1 (−x; q)∞ (−qx ; q)∞ (1 + xq i )(1 + x −1 q i+1 )

(3.35)

i=0

Now if (3.33) is replaced in (3.34), we get 

∞ 0

  ln q ρ (q x; q) Wn (x; q) Wm (x; q) dx = − n+1/2 (q; q)n (q; q)∞ δn,m . q ∗

1 2

Notice that the shape of ρ ∗ (q 1/2x; q), by noting the definition (3.35), is completely different from the log-normal function, though both of them play the role of a weight function for the Stieltjes–Wigert polynomials. Hence, let us define the shifted polynomials Vn (x; p, q) = Wn (q

p− 21

- . n x ; q) = (−1)k q k (k+p) x k , k k=0 n 

(3.36)

q

in which p is a free real parameter and q ∈ (0, 1). If (3.36) is replaced in (3.32), then 

∞ 0

  1 ρ(q p− 2 x ; q) Vn (x; p, q) Vm (x; p, q) dx = q −(p+n) (q; q)n δn,m .

(3.37)

Since the weight function of (3.37) takes the form   1 1 1 ρ(q p− 2 x ; q) = √ exp ((p − 1/2) ln q + ln x)2 2 ln q (−2π) ln q   (2p−1)2 1 1 1 ln2 x , = √ q 8 x p− 2 exp 2 ln q (−2π) ln q

(3.38)

3.2 Some Preliminaries and Definitions

257

relation (3.38) simplifies (3.37) in the form 

∞

x

p− 12

 exp

0

1 ln q 2



 Vn (x ; p, q) Vm (x ; p, q) dx

2

ln x =

 % 1 2 −2π ln q q −n− 8 (2p+1) (q; q)n δn,m .

(3.39)

We can now compute the Fourier transform of Vn (x; p, q) and substitute it in Parseval’s identity to generate a new sequence of biorthogonal exponential functions. For this purpose, we first define the specific functions g(x) = exp(p1 x) exp( ln1q1 x 2 ) Vn (ex ; a1 , b1 ) , 0 < b1 , q1 < 1 , a1 , p1 ∈ R , and h(x) = exp(p2 x) exp( ln1q2 x 2 ) Vm (ex ; a2 , b2 ) , 0 < b2 , q2 < 1 , a2 , p2 ∈ R ,

(3.40) in terms of the polynomials (3.36). Clearly, for both functions defined in (3.40), the Fourier transform exists. For example, for the function g we get 

 1 2 e exp(p1 x) exp x Vn (ex ; a1 , b1 ) dx F g(s) = ln q1 −∞ - .    ∞ n 1 k(k+a1 ) k −is+p1 −1 2 k n = t exp ln t (−1) b1 t dt ln q1 k 0 k=0 b1 - .    ∞ n  1 k(k+a1 ) k n −is+p1 −1+k 2 = (−1) b1 t exp ln t dt ln q1 k 0 k=0 b1 - . n −(−is+p1 +k)2  % k n 4 = −π ln q1 (−1) b1k(k+a1) q1 , (3.41) k k=0 

∞

−isx

b1

where we have used the general definite integral 

∞ 0

$ x α exp(β ln2 x) dx =

−

(α + 1)2 π exp(− ), β 4β

β < 0.

258

3 Generalized Sturm–Liouville Problems in q-Spaces

Now, by substituting (3.41) into Parseval’s identity, we have   1 1 2π x 2 Vn (ex ; a1 , b1 ) Vm (ex ; a2 , b2 ) dx exp((p1 + p2 ) x) exp + ln q1 ln q2 −∞     ∞ 1 1 t −1+p1 +p2 exp + = 2π ln2 t Vn (t; a1 , b1 ) Vm (t; a2 , b2 ) dt ln q1 ln q2 0 / = π 2 ln q1 ln q2  ∞ −(p −is)2 /4 −(p2 −is)2 /4 × q1 1 q2 Jn (is; a1, b1 , p1 , q1 ) Jm (is; a2 , b2 , p2 , q2 ) ds , 



∞

−∞

(3.42) in which 0 < q1 , q2 < 1; p1 , p2 ∈ R, and - . 1 x k n Jn (x; a, b, c, d) = (−1) bk(k+a) d − 4 k(k+2c) (d 2 ) . k k=0 n 

k

b

On the other hand, if on the left-hand side of (3.42) we take 

1 a1 = a2 = p1 + p2 − 2

and

ln q1 ln q2 b1 = b2 = exp 2 ln q1 q2

 ,

then according to orthogonality (3.39), relation (3.42) changes to 

∞

2π

t −1+p1 +p2 exp



0

1 1 + ln q1 ln q2



 ln2 t

   1 ln q1 ln q2 Vn t; p1 + p2 − , exp 2 2 ln q1 q2

   1 ln q1 ln q2 × Vm t; p1 + p2 − , exp dt 2 2 ln q1 q2 = 2πδn,m × 0 −π

=

       

ln q1 ln q2 1 ln q1 ln q2 ln q1 ln q2 ln q1 ln q2 exp − n + (p1 + p2 )2 exp ; exp ln q1 q2 2 2 ln q1 q2 2 ln q1 q2 2 ln q1 q2 n

 / π 2 ln q1 ln q2

∞ −∞

−(p1 −is)2 /4

q1

−(p2 −is)2 /4

q2

    ln q1 ln q2 1 Jn is; p1 + p2 − , exp , p1 , q1 2 2 ln q1 q2

    1 ln q1 ln q2 × Jm is; p1 + p2 − , exp , p2 , q2 ds . 2 2 ln q1 q2

(3.43)

3.2 Some Preliminaries and Definitions

259

For convenience, we suppose in (3.43) that q1 → exp(−4q1 ), q2 → exp(−4q2),

and p1 → −p1 ,

to eventually obtain the following theorem. Theorem 3.2 The exponential functions En (x; q1 , p1 ; q2 , p2 ) = - .   n  q1 k k n (−1) exp ((q1 − q2 )k + q2 − 2(q1 p1 + q2 p2 )) exp(−2q1 k x) q1 + q2 k exp( −2q1 q2 ) k=0 q1 +q2

satisfy a biorthogonality relation 

∞

−∞

  exp q1 (ix + p1 ) 2 + q2 (ix + p2 ) 2 En (ix; q1 , p1 ; q2 , p2 )Em (−ix; q2 , −p2 ; q1 , −p1 ) dx

$

=

       π q1 q2 −2q1 q2 −2q1 q2 exp (2n + (p2 − p1 )2 ) exp ; exp δn,m , q1 + q2 q1 + q2 q1 + q2 q1 + q2 n

(3.44) provided that q1 , q2 > 0 and p1 , p2 ∈ R.

Some Remarks on Theorem 3.2 1. The equality (3.44) is in fact a biorthogonality relation in complex weighted spaces, and En (ix; q1 , p1 ; q2 , p2 ) in this relation are complex-valued functions that are   biorthogonal with respect to exp q1 (ix + p1 ) 2 + q2 (ix + p2 ) 2 on (−∞, ∞). In general, complex orthogonal functions are defined on a rectifiable curve like C such that    2 W (z)Φn (z) Φm (z) dz = W (z) |Φn (z)| dz δn,m , C

C

where |Φn (z)|2 = Φn (z) Φn (z), and W (z) is a positive function of the complex variable z = x + iy defined on C.

260

3 Generalized Sturm–Liouville Problems in q-Spaces

2. According to Euler’s identity exp(it) = cos(t) + i sin(t), the real and imaginary parts of the sequence En (ix) are respectively defined as Re En (ix; q1, p1 ; q2 , p2 ) = Cn (x; q1 , p1 ; q2, p2 ) - .   n  q1 k k n = (−1) exp ((q1 − q2 )k + q2 − 2(q1 p1 + q2 p2 )) q1 + q2 k −2q1 q2 k=0

exp(

q1 +q2

)

× cos(2q1 k x)

(3.45)

and Im En (ix; q1 , p1 ; q2 , p2 ) = Sn (x; q1 , p1 ; q2 , p2 ) - .   n  q1 k k+1 n (−1) exp ((q1 − q2 )k + q2 − 2(q1 p1 + q2 p2 )) = q1 + q2 k −2q1 q2 k=0

exp(

q1 +q2

)

× sin(2q1 k x) ,

(3.46)

where q1 , q2 > 0 and p1 , p2 ∈ R. 3. The density of biorthogonality relation (3.44) is expandable as W (z; q1 , p1 ; q2 , p2 ) = exp(q1 (iz + p1 )2 + q2 (iz + p2 )2 ) = exp(q1 p12 + q2 p22 ) exp(−(q1 + q2 )z2 ) exp(2 i (q1 p1 + q2 p2 )z) = K ∗ exp(az2 + ibz),

(3.47)

where a = −(q1 + q2 )

and b = 2(q1 p1 + q2 p2 ).

Hence, by noting that |exp(it)| = 1, we have |W (z)| = K ∗ exp(−(q1 + q2 )z2 ) ,

z ∈ R,

in which K ∗ = exp(q1 p12 + q2 p22 ) is a real constant that has no effect on the weight function. This means that the complex weight function (3.47) is real in the weighted L2 -space if and only if b2 = q1 p1 + q2 p2 = 0.

3.2 Some Preliminaries and Definitions

261

By noting this important comment, let us reconsider the orthogonality relation (3.44) and suppose that q1 = q2 = q/2 , q > 0,

and p1 = −p2 , p1 ∈ R.

So the special sequence (q) En (x)

= En



  q q −x; , p1 ; , −p1 = (−1)k 2 2 n

k=0

which is independent of p1 , satisfies  ∞  2 (q) (q) e−q x En (ix)Em (−ix) dx = −∞

∞ −∞

$ =

- . n k

e

q

−

q 2

ek 4 ekqx ,

e−q x En (−ix)Em (ix) dx 2

(q)

(q)

π n q −q −q e 2 (e 2 ; e 2 )n q

 δn,m ,

(3.48)

for q > 0. By noting the definitions (3.45) and (3.46), we can now define the following trigonometric sequences, - . n    q q q (q) k n Cn (x) = Cn −x; , p1 ; , −p1 = (−1) ek 4 cos(kqx), q 2 2 k − 2 k=0 e

and (q) Sn (x)

n    q q = Sn −x; , p1 ; , −p1 = (−1)k 2 2 k=0

- . n k

q

q − e 2

ek 4 sin(kqx),

and verify that (q)

(q)

(q)

En (ix) = Cn (x) + i Sn (x)

and

(q)

(q)

(q)

En (−ix) = Cn (x) − i Sn (x).

(3.49)

If the relations in (3.49) are replaced in (3.48), then we get 

∞ −∞

e−q x

2

  (q) (q) (q) (q) Cn (x)Cm (x) + Sn (x)Sm (x) dx  +i

∞ −∞

e−q x

2



 (q) (q) (q) (q) Sn (x)Cm (x) − Sm (x)Cn (x) dx =

%

 π/q enq/2 (e−q/2 ; e−q/2)n δn,m .

(3.50)

262

3 Generalized Sturm–Liouville Problems in q-Spaces (q)

(q)

But Sn (x) is always an odd sequence, while Cn (x) is even in (3.50), i.e., 

∞ −∞

e−q x Sn (x)Cm (x) dx = 2

(q)

(q)



∞ −∞

e−q x Sm (x)Cn (x) dx = 0. 2

(q)

(q)

Corollary 3.1 Two trigonometric sequences defined as Sn (x) + Cn (x)  k = = (−eq/4) √ 2 k=0

- . n k

Sn (x) − Cn (x)  k = √ (−eq/4) = 2 k=0

- . n k

(q)

(q) Mn,+ (x)

n

(q)

e

−

q 2

 π sin kqx + 4

and (q)

(q) Mn,− (x)

n

(q)

e

q −2

 π sin kqx − 4

respectively satisfy the orthogonality relations 

∞ −∞

e

−q x 2

$ (q) (q) Mn,+ (x) Mm,+ (x) dx

=

π n q −q −q e 2 (e 2 ; e 2 )n 4q

 δn,m

(3.51)

δn,m ,

(3.52)

and 

∞ −∞

e

−q x 2

$ (q) (q) Mn,− (x) Mm,− (x) dx

(q)

=

π n q −q −q e 2 (e 2 ; e 2 )n 4q



(q)

in which q > 0. However, since Mn,− (−x) = −Mn,+ (x) if x = −t in (3.52), only one orthogonality relation between (3.51) and (3.52) should be considered the main relation.

3.3

Three Finite Classes of q-Orthogonal Polynomials

In this section, we introduce three ordinary q-Sturm–Liouville problems and prove that their polynomial solutions are finitely orthogonal with respect to three weight functions that correspond to the inverse gamma, Fisher, and Student-t distributions as q → 1. We obtain the general properties of these polynomials and show that all results in three finite continuous cases are recovered again as q → 1. Let ϕ(x) and ψ(x) be two polynomials of degree at most 2 and 1 respectively as follows: ϕ(x) = ax 2 + bx + c

and ψ(x) = dx + e

(a, b, c, d, e ∈ C, d = 0).

3.3 Three Finite Classes of q-Orthogonal Polynomials

263

If {yn (x; q)}n is a sequence of polynomials that satisfies the q-difference equation ϕ(x)Dq2 yn (x; q) + ψ(x)Dq yn (x; q) + λn,q yn (qx; q) = 0,

(3.53)

where Dq2 is defined in (3.12), then the following orthogonality relation holds: 

b



b

w(x; q)yn(x; q)ym(x; q)dq x =

a

a

 w(x; q)yn2(x; q)dq x δn,m ,

in which w(x; q) > 0 is a solution of the Pearson q-difference equation   Dq w(x; q)ϕ(q −1x) = w(qx; q)ψ(x), and w(q −1 x; q)ϕ(q −2x)x k for k ∈ N0 must vanish at x = a, b. If Pn (x) = x n + · · · is a monic polynomial solution of Eq. (3.53), the eigenvalue λn,q can be directly obtained by equating the coefficients of x n in (3.53) as λn,q = −

3.3.1

[n]q (a[n − 1]q + d). qn

First Finite Sequence of q-Orthogonal Polynomials Corresponding to the Inverse Gamma Distribution

Consider the q-difference equation  x(q x + 2

1)Dq2 yn (x; q) −

1 q2 − qp x+ 1−q 1−q

 Dq yn (x; q) + λn,q yn (qx; q) = 0, (3.54)

with λn,q = −

[n]q qn

  q2 − qp q 2 [n − 1]q − , 1−q

for all n = 0, 1, 2, . . . and q ∈ R \ {−1, 0, 1}, where lim λn,q = −n(n + 1 − p).

q→1

264

3 Generalized Sturm–Liouville Problems in q-Spaces

If Eq. (3.54) is expanded as 

yn (q 2 x) − (1 + q)yn (qx) + qyn (x) q(q − 1)2 x 2



    2 p [n]q yn (qx) − yn (x) q2 − qp 1 2 [n − 1] − q − q − x+ yn (qx) = 0, q − q 1−q 1−q (q − 1)x qn 1−q x(q 2 x + 1)

then its symmetric form is derived as   (qx + 1)yn (qx) − (q n+1 + q p−n )x + 1 yn (x) + q p x yn (q −1 x) = 0. The following theorem holds for the polynomial solution of Eq. (3.54). (p)

Theorem 3.3 Let {Nn (x; q)}n be a sequence of polynomials that satisfies Eq. (3.54). For n = 0, 1, . . . , N we have 

∞ 0

 (p)

(p)

∞

(p)

w1 (x; q)Nn (x; q)Nm (x; q)dq x =

0

  2 (p) (p) w1 (x; q) Nn (x; q) dq x δn,m ,

1 (p) where q > 1, N = max{m, n} < (p − 1), p ∈ R, and w1 (x; q) is the solution of the 2 q-Pearson equation  p    1 q − q2 (p) (p) x− Dq w1 (x; q)(x 2 + q −1 x) = w1 (qx; q) , 1−q 1−q which is equivalent to (p)

w1 (x; q) (p)

w1 (qx; q)

=

qp . 1 + q −1 x −1

(3.55)

Proof First, we can verify that a solution of the q-Pearson equation (3.55) is (p)

w1 (x; q) =

  x −p −1 | < 1 . p ∈ R and 0 < |q (−q −1 x −1 ; q −1 )∞

In this sense, note that lim (q − 1)−p w1 ((q − 1)−1 x; q) = x −p e−1/x . (p)

q→1

(3.56)

3.3 Three Finite Classes of q-Orthogonal Polynomials

265

Now we write Eq. (3.54) in the self-adjoint form   (p) (p) (p) (p) Dq w1 (x; q)(x 2 +q −1 x)Dq Nn (x; q) +λn,q w1 (qx; q)Nn (qx; q) = 0,

(3.57)

and for m as   (p) (p) (p) (p) Dq w1 (x; q)(x 2 + q −1 x)Dq Nm (x; q) + λm,q w1 (qx; q)Nm (qx; q) = 0. (3.58) By multiplying (3.57) by Nm (qx; q) and (3.58) by Nn (qx; q) and subtracting one from the other, we get (p)

(p)

(p)

(λm,q − λn,q )w1 (x; q)Nm (x; q)Nn (x; q)   (p) (p) (p) = q 2 Dq w1 (q −1 x; q)(q −2x 2 + q −2 x) Dq Nn (q −1 x; q) Nm (x; q)   (p) (p) (p) (3.59) − q 2 Dq (w1 (q −1 x; q)(q −2x 2 + q −2 x) Dq Nm (q −1 x; q) Nn (x; q). Hence q-integration by parts on both sides of (3.59) over [0, ∞) yields 

∞

(λm,q − λn,q ) 

∞

= 0



0

(p)

(p)

(p)

w1 (x; q)Nm (x; q)Nn (x; q)dq x



 (p) (p) (p) q 2 Dq w1 (q −1 x; q)(q −2x 2 + q −2 x) Dq Nn (q −1 x; q) Nm (x; q)

  (p)  (p) (p) − Dq w1 (q −1 x; q)(q −2x 2 + q −2 x) Dq Nm (q −1 x; q) Nn (x; q) dq x  (p) = q 2 w1 (q −1 x; q)(q −2x 2 + q −2 x) ∞  (p) (p) (p) (p) . × Dq Nn (q −1 x; q)Nm (x; q) − Dq Nm (q −1 x; q)Nn (x; q) 0

(3.60)

But since   (p) (p) (p) (p) max deg Dq Nn (q −1 x; q)Nm (x; q) − Dq Nm (q −1 x; q)Nn (x; q) = m + n − 1, if N
0 for all n = 1, 2, . . . , N < 12 (p − 1), Favard’s theorem can be applied, so that we have  n

 ∞  ∞ (p) (p) (p) (p) w (x; q)N¯ m (x; q)N¯ n (x; q)dq x = dk w (x; q)dq x δn,m , 1

0

k=1

0

1

(p)

and w1 (x; q) is given by (3.56). In order to compute 

∞ 0

(p)

w1 (x; q)dq x =



∞ 0

x −p dq x, (−q −1 x −1 ; q −1 )∞

we use relation (3.18) for u(x) = q −1 x −1 to obtain 

∞ 0

x −p dq x = q p (−q −1 x −1 ; q −1 )∞



∞ 0

x (p−2) d −1 x. (−x; q −1)∞ q

(3.64)

3.3 Three Finite Classes of q-Orthogonal Polynomials

269

But the integral on the right-hand side of (3.64) can be computed by the Ramanujan identity (3.19) as follows:  q

∞

p 0

∞  x (p−2) (−1; q −1)n p −1 −n(p−1) d ) q −1 x = q (1 − q q (−x; q −1)∞ (−1; q −1)∞ (0; q −1)n n=−∞

=

∞ q p (1 − q −1 )  −n(p−1) (−1; q −1)n q p (1 − q −1 ) q = Ψ (−1, 0; q −1; q 1−p ) (−1; q −1)∞ n=−∞ (0; q −1)n (−1; q −1)∞

=

∞ q p (1 − q −1 ) (1 − q −n−1 )(1 + q p−n−2 )(1 + q 1−n−p ) (−1; q −1)∞ (1 + q −n−1 )(1 − q 1−n−p ) n=0

=

q p−1 (q − 1)(q −1; q −1 )2−p . (−1; q −1)1−p (−q −1 ; q −1)p−1

(3.65)

Hence the norm square value is explicitly computed as 

∞ 0

x −p (p) (N¯ n (x; q))2dq x (−q −1 x −1 ; q −1 )∞ (−1)n (q − 1)(q, q 1−p ; q)n (q −1 ; q −1 )2−p q 2 (3n −4np+n+2p−2) (q 1−p , q 2−p , q 2−p , q 3−p ; q 2 )n (−1; q −1)1−p (−q −1 ; q −1)p−1 1

=

2

⇐⇒ n = 0, 1, . . . , N
12 (p − 1) in (3.66), the above moments will be divergent and consequently the norm square value will be divergent too.

270

3 Generalized Sturm–Liouville Problems in q-Spaces

3.3.2

Second Finite Sequence of q-Orthogonal Polynomials Corresponding to the Fisher Distribution

Consider the q-difference equation x(qx + 1)Dq2 yn (x; q) − (q [s − 2]q x + [−t − 1]q )Dq yn (x; q) + λn,q yn (qx; q) = 0, (3.67) with λn,q = [n]q [s − n − 1]q , for all n = 0, 1, 2, . . . and q ∈ R \ {−1, 0, 1}, where lim λn,q = n(s − n − 1).

q→1

If Eq. (3.67) is expanded as x(qx + 1)(

yn (q 2 x) − (1 + q)yn (qx) + qyn (x) ) q(q − 1)2 x 2

− (q [s − 2]q x + [−t − 1]q )(

yn (qx) − yn (x) ) + [n]q [s − n − 1]q yn (qx) = 0, (q − 1)x

then its symmetric form is derived as   (x +1)yn (qx; q)− (q s−1−n + q n )x + (q −t + 1) yn (x; q)+(q s−1 x +q −t ) yn (q −1 x; q) = 0.

Theorem 3.4 Let {Mn(s,t )(x; q)}n be a sequence of polynomials that satisfies Eq. (3.67). For n = 0, 1, . . . , N we have  ∞ 0

(s,t )

w2

(s,t )

(x; q)Mn

(s,t )

(x; q)Mm

(x; q)dq x =

 ∞ 0

(s,t )

w2

  2 (s,t ) (x; q) Mn (x; q) dq x δn,m , (s,t )

where 0 < q < 1, t > −1, N = max{m, n} < 12 (s − 1), and w2 the q-Pearson equation

(x; q) is the solution of

    (s,t ) (s,t ) Dq w2 (x; q)(q −1x 2 + q −1 x) = −w2 (qx; q) q [s − 2]q x + [−t − 1]q ,

3.3 Three Finite Classes of q-Orthogonal Polynomials

271

which is equivalent to (s,t )

w2

(x; q)

(s,t ) w2 (qx; q)

=

q s+t x + 1 −t q . x+1

(3.68)

Proof It can be verified that (s,t )

w2

(x; q) =

xt (−x; q)s+t

(s, t ∈ R and 0 < |q| < 1)

(3.69)

is a solution of the Pearson equation (3.68), where lim w2(s,t )(x; q) =

q→1

xt . (1 + x)s+t

Now we write Eq. (3.67) in the self-adjoint form   (s,t ) (s,t ) Dq w2 (x; q)(q −1x 2 + q −1 x)Dq Mn(s,t )(x; q) + λn,q w2 (qx; q)Mn(s,t )(qx; q) = 0, (3.70) and for m as   (s,t ) (s,t ) (s,t ) (s,t ) (x; q) + λm,q w2 (qx; q)Mm (qx; q) = 0. Dq w2 (x; q)(q −1x 2 + q −1 x)Dq Mm (3.71) (s,t ) Multiply (3.70) by Mm (qx; q) and (3.71) by Mn(s,t )(qx; q) and subtract one from the other to get (s,t )

(λm,q − λn,q )w2

(s,t ) (x; q)Mm (x; q)Mn(s,t )(x; q)

  (s,t ) (x; q) = q 2 Dq w2(s,t )(q −1 x; q)(q −3x 2 + q −2 x)Dq Mn(s,t )(q −1 x; q) Mm   (s,t ) −1 (q x; q) Mn(s,t )(x; q). − q 2 Dq w2(s,t )(q −1 x; q)(q −3x 2 + q −2 x)Dq Mm

(3.72)

Using q-integration by parts on both sides of (3.72) over [0, ∞), we obtain 

∞

(λm,q − λn,q ) 

∞

= 0



0

(s,t )

w2

(s,t ) (x; q)Mm (x; q)Mn(s,t )(x; q)dq x

  (s,t ) q 2 Dq w2(s,t )(q −1 x; q)(q −3x 2 + q −2 x)Dq Mn(s,t )(q −1 x; q) Mm (x; q)

272

3 Generalized Sturm–Liouville Problems in q-Spaces

   (s,t ) −1 − Dq w2(s,t )(q −1 x; q)(q −3x 2 + q −2 x)Dq Mm (q x; q) Mn(s,t )(x; q) dq x   (s,t ) (s,t ) (x; q) = q 2 w2 (q −1 x; q)(q −3x 2 + q −2 x) Dq Mn(s,t )(q −1 x; q)Mm ∞ (s,t ) −1 − Dq Mm (q x; q)Mn(s,t )(x; q) . (3.73) 0

Since   (s,t ) (s,t ) −1 (x; q)−Dq Mm (q x; q)Mn(s,t )(x; q) = m+n−1, max deg Dq Mn(s,t )(q −1 x; q)Mm if t > −1 and N < 12 (s − 1) for N = max{m, n}, the following boundary conditions hold in (3.73): (s,t )

lim w2

x→0

(q −1 x; q)(q −3x 2 + q −2 x)x 2N−1 = 0

and (s,t )

lim w2

x→∞

(q −1 x; q)(q −3x 2 + q −2 x)x 2N−1 = 0.

Under these conditions, the right-hand side of (3.73) tends to zero and 

∞ 0

(s,t ) w2(s,t )(x; q)Mm (x; q)Mn(s,t )(x; q)dq x = 0

if and only if m = n,

t > −1 and N = max{m, n}
−1 is orthogonal with respect xt (s,t ) to the weight function w2 (x; q) = on [0, ∞). (−x; q)s+t The monic polynomial solution of Eq. (3.67) can be represented as 1

M¯ n(s,t )(x; q)

=

q2





(q t +1; q)n q −n , q n−s+1 s+t q; −q x , 2 φ1 (q n−s+1 ; q)n q t +1

 n2 +(1−2s−2t )n

(3.74)

3.3 Three Finite Classes of q-Orthogonal Polynomials

273

because it is enough to first expand the polynomial solution as Mn(s,t )(x; q) =

n 

an,k

k=0

  an,n = 0, n = 0, 1, 2, . . . .

xk [k]q !

Then the coefficients {an,k }nk=0 satisfy the two-term recurrence relation [n − k]q (q s−1 − q n+k )an,k = (q −(t +1) − q k )q n−k an,k+1 , which is directly determined up to the normalizing constant an,n = 0 as

an,k

n−k

q i (q −(t +1) − q n−i ) = an,n , k = 0, 1, . . . , n − 1. [i]q (q s−1 − q 2n−i ) i=1

Once again, by choosing an,n = (1 − q)−n (q; q)n , the monic polynomial solution is finally derived as M¯ n(s,t )(x; q) = q

1 2 n(n+1)

n  (q −n ; q)k k=0

(q; q)k

n−k

(q −(t +1) − q n−i ) (−x)k . (q s−1 − q 2n−i ) i=1

Remark 3.2 From (3.74), it can be concluded that q− 2 1

lim

 2  n +(1−2s−2t )n

q→1

(q n−s+1 ; q)n ¯ (s,t ) Mn (x; q) = Mn(s,t )(x), (q t +1; q)n

in which Mn(s,t )(x)



  t +n −n, n + 1 − s = (−1) n! −x 2 F1 n t +1 n

is the same as the first finite sequence of hypergeometric orthogonal polynomials. The Rodrigues representation of the monic polynomials is 1

 −1 −1  (1 − q)n (−x; q)s+t n t +n (−q x; q )n x , × D q x t (q 2n−s+2 ; q −1)n (−q −n x; q)s+t

2 +(1−2s−2t )n)

q 2 (n M¯ n(s,t )(x; q) =

which can be directly derived by the formula (3.63) when w(x; q) =

xt (−x; q)s+t

and ϕ(x) = qx 2 + x.

274

3 Generalized Sturm–Liouville Problems in q-Spaces

Computation of the Norm Square Value By noting the explicit representation (3.74), we see that the monic polynomials satisfy the recurrence relation (s,t ) (s,t ) (x; q) = (x − cn ) M¯ n(s,t )(x; q) − dn M¯ n−1 (x; q), M¯ n+1

where cn = −

q n+s−1(1 − q n − q n+1 + q 2n−s+1 ) + q 2n−t (q s−n−1 − q −1 − 1 + q n ) q 2s−2(1 − q 2n−s )(1 − q 2n−s+2)

and dn = q 2n−2s−t +1

[n]q [n − s]q [n − t − s]q [n + t]q . [2n − s − 1]q ([2n − s]q )2 [2n − s + 1]q

with the initial terms (s,t ) (s,t ) M¯ 0 (x; q) = 1 and M¯ 1 (x; q) = x − c0 .

Since dn > 0 for all n = 1, 2, . . . , N < 12 (s − 1), applying Favard’s theorem yields 

∞ 0

 (s,t ) (s,t ) ¯m ¯ n(s,t )(x; q)dq x w2 (x; q)M (x; q)M

=

n k=1



∞

dk 0

(s,t ) w2 (x; q)dq x

δn,m .

To compute 

∞ 0

(s,t ) w2 (x; q)dq x



∞

= 0

xt dq x, (−x; q)s+t

we can directly use (3.20) to get 

∞ 0

2 Γq (t + 1)Γq (s − 1) xt . dq x = (−x; q)s+t (−1; q)t +1(−1; q)−t Γq (s + t)

(3.75)

Hence the norm square value is computed as 

∞ 0

xt (M¯ (s,t )(x; q))2dq x (−x; q)s+t n 



2 q n +(2−t −2s)n (q, q 1−s , q 1−s−t , q 1+t ; q)n Γq (t + 1)Γq (s − 1) = , (q 1−s , q 2−s , q 2−s , q 3−s ; q 2 )n (−1; q)t +1(−1; q)−t Γq (s + t) 2

which is valid for n = 0, 1, . . . , N
12 (s − 1) in (3.76), the above moments will be divergent, and the norm square value will be divergent too.

3.3.3

Third Finite Sequence of q-Orthogonal Polynomials Corresponding to Student’s t-Distribution

As another special case of Eq. (3.53), let us consider the q-difference equation (q 2 x 2 + 1)Dq2 yn (x; q) − q 2 [2p − 3]q xDq yn (x; q) + λn,q yn (qx; q) = 0,

(3.77)

with λn,q = q [n]q [2p − n − 2]q , for all n = 0, 1, 2, . . . and q ∈ R \ {−1, 0, 1}, where lim λn,q = −n(n + 2 − 2p).

q→1

The symmetric form of Eq. (3.77) is denoted by   (x 2 +1)yn (qx; q)− (q 2p−2−n + q n )x 2 + (q + 1) yn (x; q)+(q 2p−2 x 2 +q) yn (q −1 x; q) = 0.

276

3 Generalized Sturm–Liouville Problems in q-Spaces (p)

Theorem 3.5 Let {In (x; q)}n be a sequence of symmetric polynomials that satisfies Eq. (3.77). For n = 0, 1, . . . , N we have  ∞   ∞  2 (p) (p) (p) (p) (p) w3 (x; q)In (x; q)Im (x; q)dq x = w3 (x; q) In (x; q) dq x δn,m , −∞

−∞

where q > 1, N = max{m, n} < p − 1, (−1)2p = −1, and the symmetric function (p) w3 (x; q) is the solution of the q-Pearson equation   (p) (p) Dq w3 (x; q)(x 2 + 1) = −q 2 [2p − 3]q w3 (qx; q), which is equivalent to (p)

w3 (x; q) (p)

w3 (qx; q)

=

q 2p−1 x 2 + 1 . x2 + 1

(3.78)

Proof First, it can be verified that (p)

w3 (x; q) =

x 1−2p (−x −2 ; q −2)p− 1



0 < |q −2 | < 1

 (3.79)

2

is a solution of Eq. (3.78) with (−1)2p = −1, where 1

lim w3 (x; q) = (1 + x 2 )−(p− 2 ) . (p)

q→1

Now write Eq. (3.77) in the self-adjoint form   (p) (p) (p) (p) Dq w3 (x; q)(x 2 + 1)Dq In (x; q) + λn,q w3 (qx; q)In (qx; q) = 0,

(3.80)

and for m as   (p) (p) (p) (p) Dq w3 (x; q)(x 2 + 1)Dq Im (x; q) + λm,q w3 (qx; q)Im (qx; q) = 0. (p)

(3.81)

(p)

By multiplying (3.80) by Im (qx; q) and (3.81) by In (qx; q) and subtracting one from the other, we get (p)

(p)

(p)

(λm,q − λn,q )w3 (x; q)Im (x; q)In (x; q)   (p) (p) (p) = q 2 Dq w3 (q −1 x; q)(q −2x 2 + 1)Dq In (q −1 x; q) Im (x; q)   (p) (p) (p) − q 2 Dq w3 (q −1 x; q)(q −2x 2 + 1)Dq Im (q −1 x; q) In (x; q).

(3.82)

3.3 Three Finite Classes of q-Orthogonal Polynomials

277

Hence q-integration by parts on both sides of (3.82) over (−∞, ∞) yields  (λm,q − λn,q )

∞

(p)

−∞

 =

(p)

(p)

w3 (x; q)Im (x; q)In (x; q)dq x

∞

−∞

   (p) (p) (p) q 2 Dq w3 (q −1 x; q)(q −2 x 2 + 1)Dq In (q −1 x; q) Im (x; q)

   (p) (p) (p) − Dq w3 (q −1 x; q)(q −2 x 2 + 1)Dq Im (q −1 x; q) In (x; q) dq x   ∞ (p) (p) (p) (p) (p) = q 2 w3 (q −1 x; q)(q −2 x 2 + 1) Dq In (q −1 x; q)Im (x; q) − Dq Im (q −1 x; q)In (x; q) . −∞

(3.83) Since   (p) (p) (p) (p) max deg Dq In (q −1 x; q)Im (x; q) − Dq Im (q −1 x; q)In (x; q) = m + n − 1, if N < p − 1 for N = max{m, n}, the following boundary condition holds in (3.83): lim w3 (q −1 x; q)(q −2x 2 + 1)x 2N−1 = 0. (p)

x→∞

Therefore, the right-hand side of (3.83) tends to zero, and 

∞ −∞

(p)

(p)

(p)

w3 (x; q)Im (x; q)In (x; q)dq x = 0

if and only if m = n, (−1)2p = −1 and N = max{m, n} < p − 1. (p)

N 0 for all n = 1, 2, . . . , N < p − 1, applying Favard’s theorem yields 

∞ −∞

 (p) (p) (p) w3 (x; q)I¯m (x; q)I¯n (x; q)dq x

=

n

k=1

dk∗



∞ −∞

(p) w3 (x; q)dq x

δn,m .

In order to compute 

∞ −∞

(p) w3 (x; q)dq x

 =

∞ −∞

x 1−2p dq x, (−x −2 ; q −2)p− 1 2

1

we can use (3.18) for u(x) = x − 2 to obtain 

∞

−∞

2q 2 x 1−2p d x = q (−x −2 ; q −2 )p− 1 (q + 1) 2



∞ 0

x p−2 dq −2 x. (−x; q −2 )p− 1 2

(3.86)

280

3 Generalized Sturm–Liouville Problems in q-Spaces

Since the integral on the right-hand side of (3.86) is directly computable as 

∞ −∞

x 1−2p dq x (−x −2 ; q −2 )p− 1 2

=

Γq −2 (p − 1)Γq −2 ( 12 ) 4q 2 , (q + 1)(−1; q −2)p−1 (−1; q −2)2−p Γq −2 (p − 12 )

(3.87)

the norm square value is eventually computed as 

∞ −∞

=

 2 x 1−2p ¯n(p) (x; q) dq x I (−x −2 ; q −2 )p− 1 2

1 2 4(−1)n q 2 (n +5n−4np+4) (q; q)

(q

2−2p ; q) n (q n + 1)(−1; q −2)p−1 (−1; q −2)2−p (q 2−2p ; q 2)n (q 4−2p ; q 2 )n

Γq −2 (p − 1)Γq −2 ( 12 ) Γq −2 (p − 12 )

⇐⇒ n = 0, 1, . . . , N < p − 1.

(3.88)

23

( ) For instance, the set {I¯n 2 (x; q)}10 n=0 is a finite sequence of q-orthogonal polynomials with x −22 the weight function w(x; q) = that satisfies the relation (−x −2 ; q −2 )11



∞

23 23 x −22 ¯m( 2 ) (x; q)I¯n( 2 ) (x; q)dq x I −2 −2 −∞ (−x ; q )11 ⎛ ⎞ 1 2 1 4(−1)n q 2 (n −41n+4) (q; q)n (q −21 ; q)n Γq −2 ( 21 )Γ ( ) −2 q 2 2 ⎠ δm,n =⎝ (q + 1)(−1; q −2) 21 (−1; q −2)− 19 (q −21 ; q 2)n (q −19 ; q 2)n Γq −2 (11) 2

2

⇐⇒ m, n < 10. Also, relation (3.87) helps us compute the moments corresponding to the weight function (3.79) as  μk =

=

∞ −∞

x k+1−2p dq x (−x −2 ; q −2 )p− 1 2

1 q −2 (k + p − 1)Γq −2 ( 2 − k) + 1)(−1; q −2)k+p−1 (−1; q −2)2−p−k Γq −2 (p

4q 2 Γ

(q

− 12 )

.

Notice that if n > p − 1 in (3.88), the above moments will be divergent, and the norm square value will be divergent too.

3.3 Three Finite Classes of q-Orthogonal Polynomials

3.3.4

281

A Characterization of Three Introduced Finite Sequences

In 2001, the authors in [1] classified all orthogonal polynomial families of the q-Hahn tableau and compared their scheme with the q-Askey scheme and Nikiforov–Uvarov tableau. In fact, they considered a second-order q-difference equation of the form φ(x)Dq Dq −1 y(x; q) + ψ(x)Dq −1 y(x; q) + λn,q y(x; q) = 0,

(3.89)

in which φ(x) = a2 x 2 + a1 x + a0

and ψ(x) = b1 x + b0

with

b1 = 0.

Now consider the function φ ∗ (x) = q −1 φ(x) + (q −1 − 1)xψ(x),

(3.90)

which can be derived via the Pearson equation Dq −1 (φw) = qψw or its expanded form Dq −1 φ(x)w(q −1 x) + Dq −1 w(x)φ(x) = qψw.

(3.91)

After some computations, Eq. (3.91) changes to   φ(q −1 x)w(q −1 x) = φ(x) + q(q −1 − 1)xψ(x) w(x), which is finally equivalent to w(q −1 x) q −1 φ(x) + (q −1 − 1)xψ(x) φ ∗ (x) = = . w(x) q −1 φ(q −1 x) q −1 φ(q −1 x) The purpose of the authors in [1] was to classify all classical q-orthogonal polynomials using the roots of φ and φ ∗ . In this regard, if x = 0 is a root of φ (i.e., a0 = 0), then it is also a root of φ ∗ and conversely. Hence, all families of q-orthogonal polynomials can be devided to two main groups of 0-families and -families. Afterward, the authors classified the two above-mentioned families based on the degree of the polynomials φ and φ ∗ and multiplicity of the 0-families. Our purpose in this section is to compare the three introduced finite q-orthogonal polynomials with the authors’ characterization in [1].

282

3 Generalized Sturm–Liouville Problems in q-Spaces

Comparison with the First Finite Sequence Reconsider the equation  x(q 2x + 1)Dq2 yn (x; q) −

q2 − qp 1 x+ 1−q 1−q

 Dq yn (x; q) + λn,q yn (qx; q) = 0,

which can be written as q 1−p x(x + q −1 )Dq Dq −1 y(x; q) −



q −p q 1−p − q −1 x+ 1−q 1−q

 Dq −1 y(x; q)

+ q 1−p λn,q y(x; q) = 0.

(3.92)

By comparing Eq. (3.92) with (3.89) and (3.90), we obtain φ(x) = q 1−p x(x + q −1 )

and φ ∗ (x) = q −2 x 2 .

This means that the polynomial solution of Eq. (3.92) is a particular case of the 0Jacobi/Bessel polynomials jn (x; a, b) in [1] with a = q 1−p and b = −q −1 .

Comparison with the Second Finite Sequence The q-difference equation of the second sequence, i.e., x(qx + 1)Dq2 yn (x; q) − (q [s − 2]q x + [−t − 1]q )Dq yn (x; q) + λn,q yn (qx; q) = 0, can be written as  (x 2 + x)(x)Dq Dq −1 y(x; q) −

q s−2 − 1 q −t−1 − 1 x+ q−1 q−1



Dq −1 y(x; q) + qλn,q y(x; q) = 0.

(3.93) By comparing Eq. (3.93) with (3.89) and (3.90), we obtain φ(x) = x(x + 1) and φ ∗ (x) = q −2 x(q s−1x + q −t ). Although the polynomial solution of Eq. (3.93) can be considered a 0-Jacobi/Jacobi polynomial, it is not similar to any other polynomial sequence given in [1], i.e., a new type of 0-Jacobi/Jacobi polynomial is presented in Sect. 3.3.2.

Comparison with the Third Finite Sequence The q-difference equation of the third sequence, i.e., (q 2 x 2 + 1)Dq2 yn (x; q) − q 2 [2p − 3]q xDq yn (x; q) + λn,q yn (qx; q) = 0,

3.4 A Symmetric Generalization of Sturm–Liouville Problems in q-. . .

283

is equivalent to 

1 − q 2p−3 (x + 1)(x)Dq Dq −1 y(x; q) + q −1



2

x Dq −1 y(x; q) + λn,q y(x; q) = 0.

(3.94)

By comparing Eq. (3.94) with (3.89) and (3.90), we find that φ(x) = x 2 + 1 and φ ∗ (x) = q 2p−4x 2 + q −1 . Once again, although the polynomial solution of Eq. (3.94) can be considered a Jacobi/Jacobi polynomial, it is not similar to any other polynomial sequence given in [1], i.e., a new type of -Jacobi/Jacobi polynomial is presented.

3.4

A Symmetric Generalization of Sturm–Liouville Problems in q-Difference Spaces

Theorem 3.6 Let φn (x; q) = (−1)n φn (−x; q) be a sequence of symmetric functions that satisfies the q-difference equation   A(x)Dq Dq −1 φn (x; q) + B(x)Dq φn (x; q) + λn,q C(x) + D(x) + σn E(x) φn (x; q) = 0,

(3.95) n

where A(x), B(x), C(x), D(x), and E(x) are real functions, σn = 1−(−1) , and λn,q is a 2 sequence of constants. If A(x), (C(x) > 0), D(x), and E(x) are even functions and B(x) is odd, then 

α −α

∗



W (x; q)φn (x; q)φm(x; q)dq x =

α −α

W

∗

 (x; q)φn2(x; q)dq x

δn,m ,

where W ∗ (x; q) = C(x)W (x; q),

(3.96)

and W (x; q) is a solution of the Pearson q-difference equation Dq (A(x)W (x; q)) = B(x)W (x; q), which is equivalent to (q − 1)xB(x) + A(x) W (qx; q) = . W (x; q) A(qx)

(3.97)

284

3 Generalized Sturm–Liouville Problems in q-Spaces

Of course, the weight function defined in (3.96) must be positive and even, and A(x)W (x; q) must vanish at x = α. Proof If Eq. (3.95) is written in a self-adjoint form, then     Dq A(x)W (x; q)Dq −1 φn (x; q) + λn,q C(x) + D(x) + σn E(x) W (x; q)φn (x; q) = 0,

(3.98) and for m we have     Dq A(x)W (x; q)Dq −1 φm (x; q) + λm,q C(x) + D(x) + σm E(x) W (x; q)φm (x; q) = 0.

(3.99) By multiplying (3.98) by φm (x; q) and (3.99) by φn (x; q) and subtracting one from the other, we get     φm (x; q)Dq A(x)W (x)Dq −1 φn (x; q) − φn (x; q)Dq A(x)W (x)Dq −1 φm (x; q)   + λn,q − λm,q C(x)W (x; q)φn (x; q)φm(x; q) +

(−1)m − (−1)n E(x)W (x; q)φn (x; q)φm (x; q) = 0. 2

(3.100)

Since the q-integral of an odd integrand over a symmetric interval is equal to zero, qintegrating both sides of (3.100) over the symmetric interval [−α, α] yields 

α −α

  φm (x; q)Dq A(x)W (x; q)Dq −1 φn (x; q) dq x  −

α −α

  φn (x; q)Dq A(x)W (x; q)Dq −1 φm (x; q) dq x

  + λn,q − λm,q +

(−1)m − (−1)n 2





α

−α α −α

C(x)W (x; q)φn (x; q)φm (x; q)dq x

E(x)W (x; q)φn (x; q)φm (x; q)dq x = 0.

Now, using the rule of q-integration by parts, relation (3.101) is transformed to 

α A(x)W (x; q)φm(x; q)Dq −1 φn (x; q) −α  α − A(qx)W (qx; q)Dq −1 φn (qx; q)Dq φm (x; q)dq x −α

(3.101)

3.4 A Symmetric Generalization of Sturm–Liouville Problems in q-. . .

285

 α − A(x)W (x; q)φn(x; q)Dq −1 φm (x; q)  +

−α

α

−α

A(qx)W (qx; q)Dq −1 φm (qx; q)Dq φn (x; q)dq x

  + λn,q − λm,q (−1)m − (−1)n + 2





α

−α α −α

C(x)W (x; q)φn (x; q)φm (x; q)dq x

E(x)W (x; q)φn (x; q)φm (x; q)dq x = 0.

(3.102)

Since Dq −1 f (qx) = Dq f (x), relation (3.102) is simplified as 

  α A(x)W (x; q) φm (x; q)Dq −1 φn (x; q) − φn (x; q)Dq −1 φm (x; q) −α    α + λn,q − λm,q C(x)W (x; q)φn (x; q)φm (x; q)dq x +

(−1)m

− (−1)n



2

−α α

−α

E(x)W (x; q)φn (x; q)φm (x; q)dq x = 0.

(3.103)

On the other hand, W (x; q) is a symmetric solution of the Pearson q-difference equation (3.97). Hence if in (3.103) we take A(−α)W (−α; q) = A(α)W (α; q) = 0, then to prove the orthogonality property it remains to show that F ∗ (m, n) =

(−1)m − (−1)n 2



α

−α

E(x)W (x; q)φn (x; q)φm (x; q)dq x = 0.

For this purpose, four cases should be considered for values m, n as follows: 1. If both m and n are even (or odd), then F ∗ (n, m) = 0, because we have F ∗ (2i, 2j ) = F ∗ (2i + 1, 2j + 1) = 0. 2. If one of the two mentioned values is odd and the other one is even, then ∗

F (2i, 2j + 1) =



α −α

E(x)W (x; q)φ2j +1(x; q)φ2i (x; q)dq x.

(3.104)

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3 Generalized Sturm–Liouville Problems in q-Spaces

Since E(x), W (x; q) and φ2i (x; q) are assumed to be even functions and φ2j +1 (x; q) is odd in (3.104), its integrand will be an odd function, and therefore F ∗ (2i, 2j + 1) = 0. These results similarly hold for the case m = 2i + 1

i.e., F ∗ (2i + 1, 2j ) = 0.

and n = 2j,

 

3.4.1

Some Illustrative Examples

Example 3.1 Consider the q-difference equation   x 2 (1 − x 2 )Dq Dq −1 φn (x; q) + qx −(q 2 + q + 1)x 2 + q + 1 Dq φn (x; q)     + [n]q q(q 2 + q + 1) − [1 − n]q x 2 − q(1 + q)σn φn (x; q) = 0 as a special case of (3.95). If we set φn (x; q) =

∞ 

aj (n)x j ,

j =0

then a solvable recurrence relation for the coefficients {aj (n)}∞ j =0 is derived, giving rise eventually to the following representation φn (x; q) = x

σn



  q σn −n , q n+σn −1 q 4 − q + 1 2 2 2   q ;q x . 2 φ1 q 2σn +1 q 3 − q + 1

(3.105)

According to Theorem 3.6, the above sequence satisfies the orthogonality relation 

1 −1

W1∗ (x; q)φn (x; q)φm(x; q)dq x =



1 −1

 W1∗ (x; q)φn2(x; q)dq x δn,m ,

in which W1∗ (x; q) = C(x)W1 (x; q) is the main weight function and W1 (x; q) satisfies the equation W1 (qx; q) (q 4 − q + 1)x 2 − q 3 + q − 1   = . W1 (x; q) q 2 q 2x 2 − 1

(3.106)

3.4 A Symmetric Generalization of Sturm–Liouville Problems in q-. . .

287

Up to a periodic function, a solution of Eq. (3.106) is of the form

W1 (x; q) =

  log(x 2 ) (q 2 x 2 ; q 2 )∞ q 3 − q + 1 2 log(q) x 2 ( qq 3 −q+1 x 2 ; q 2 )∞ −q+1 4

= W1 (−x; q).

In this sense, note that x2 , lim W1∗ (x; q) = lim x 2 W1 (x; q) = √ q↑1 q↑1 1 − x2 which indeed gives the weight function of the Chebyshev polynomials of the fifth kind. To compute the norm square value of the symmetric polynomials (3.105), we can use Favard’s theorem [2] in q-spaces, which says that if {Pn (x; q)} satisfies the recurrence relation xPn (x; q) = An Pn+1 (x; q) + Bn Pn (x; q) + Cn Pn−1 (x; q),

n = 0, 1, 2, . . . ,

where P−1 (x; q) = 0, P0 (x; q) = 1, An , Bn , Cn real, and An Cn+1 > 0 for n = 0, 1, 2, . . . , then there exists a weight function W ∗ (x; q) such that 

α −α

∗

W (x; q)Pn (x; q)Pm (x; q)dq x =

n−1 Ci+1  i=0

Ai

α −α

∗

W (x; q)dq x δn,m .

It is clear that Favard’s theorem also holds for the monic type of symmetric q-polynomials in which An = 1 and Bn = 0. So, if φ¯ n (x; q) is considered the monic form of the symmetric q-polynomials (3.105), then after some calculations, they satisfy the following three-term recurrence relation: φ¯ n+1 (x; q) = x φ¯ n (x; q) − γn φ¯n−1 (x; q) with φ¯ 0 (x; q) = 1 and φ¯ 1 (x; q) = x, (3.107) in which γ2m

       q 2m+1 q 2m − 1 q 4 − q + 1 q 2m + −q 3 + q − 1 q 2 =  2    q 4 − q + 1 q 8m − q 2 + 1 q 4 − q + 1 q 4m+1 + q 4

and γ2m+1

      q 2m q 3 − q + 1 q 2m+1 − 1 q 4 − q + 1 q 2m − q . =  2    q 4 − q + 1 q 8m+1 − q 2 + 1 q 4 − q + 1 q 4m + q

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3 Generalized Sturm–Liouville Problems in q-Spaces

Therefore, the norm square value takes the form 

1 −1

φ¯ n2 (x; q)W1∗ (x; q)dq x = dn2



1

  log(x 2 ) (q 2 x 2 ; q 2 )∞ q 3 − q + 1 2 log(q)

−1

( qq 3 −q+1 x 2 ; q 2 )∞ −q+1 4

dq x,

where   m+1   (q − 1)2 q 2 + q + 1 q 3 + q 2 + q − 1 q m(2m−1)−2 q 3 − q + 1     = q q4 − q + 1 ; q4 m      2 2  3   4 −q+1  2 q ; q m q − 1 + q1 ; q 2 m q q 3 − q + 1 ; q 2 m qq5 −q m+1 3 +q 2 ; q      ×  q 4 − q 2 + 1 q − q12 + q13 ; q 4 m+1 q 3 − 1 + q1 ; q 4 m q 3 − 1 + q1 ; q 4 m+1 2 d2m

and     m+1 2 (q − 1)2 q 2 + q + 1 q 3 + q 2 + q − 1 q 2m +m−2 q 3 − q + 1     = q q 4 − q + 1 ; q 4 m+1   4       2 2  3 −q+1 2 ; q q ; q m q − 1 + q1 ; q 2 m+1 q q 3 − q + 1 ; q 2 m+1 qq5 −q m+1 3 +q 2     × .   1 1 1 q 4 − q 2 + 1 q − q 2 + q 3 ; q 4 m+1 q 3 − 1 + q ; q 4 m+1 2

2 d2m+1

Example 3.2 Consider the q-difference equation   x 2 (1 − x 2 )Dq Dq −1 φn (x; q) + qx −[5]q x 2 + q + 1 Dq φn (x; q)   + [n]q (q[5]q − [1 − n]q )x 2 − q(q + 1)σn φn (x; q) = 0

(3.108)

as a special case of (3.95). Following the approach of Example 3.1, we can obtain the polynomial solution of Eq. (3.108) as φn (x; q) = x

σn



  q σn −n , q n+σn −1 q 6 − q + 1 2 2 2   q ;q x . 2 φ1 q 2σn +1 q 3 − q + 1

Again, this sequence satisfies an orthogonality relation of the form 

1 −1

W2∗ (x; q)φn (x; q)φm(x; q)dq x =



1 −1

 W2∗ (x; q)φn2(x; q)dq x δn,m ,

3.4 A Symmetric Generalization of Sturm–Liouville Problems in q-. . .

289

where W2∗ (x; q) = C(x)W2 (x; q) is the main weight function and W2 (x; q) satisfies the equation (q 6 − q + 1)x 2 − q 3 + q − 1 W2 (qx; q)   . = W2 (x; q) q 2 q 2x 2 − 1

(3.109)

Up to a periodic function, a solution of Eq. (3.109) is denoted by 

2  log(x )   q 2 x 2 ; q 2 ∞ q 3 − q + 1 2 log(q)  6  W2 (x; q) = = W2 (−x; q), 2; q 2 x 2 qq 3 −q+1 x ∞ −q+1

where % lim W2∗ (x; q) = lim x 2 W2 (x; q) = x 2 1 − x 2 , q↑1

q↑1

which indeed shows the weight function of the Chebyshev polynomials of the sixth kind. The monic polynomial solution of Eq. (3.108) satisfies a three-term recurrence relation of type (3.107) with γ2m =

     q 2m+1 (q m − 1) (q m + 1) q 6 − q + 1 q 2m + −q 3 + q − 1 q 2  2    q 6 − q + 1 q 8m − q 2 + 1 q 6 − q + 1 q 4m+1 + q 4

and       q 2m q 3 − q + 1 q 2m+1 − 1 q 6 − q + 1 q 2m − q . γ2m+1 =  2    q 6 − q + 1 q 8m+1 − q 2 + 1 q 6 − q + 1 q 4m + q Thus, the norm square value is derived as 

1 −1

φ¯ n2 (x; q)W2∗ (x; q)dq x = dn2



1 −1

 2 2 2  3  log(x 2 ) q x ; q ∞ q − q + 1 2 log(q)  6  dq x, q −q+1 2 2 x ; q ∞ 3 q −q+1

where    m+1  (1 − q) q 5 + q 4 + q 3 − 1 [6]q − 2 q m(2m−1)−2 q 3 − q + 1     = q q6 − q + 1 ; q4 m   6       −q+1 2 (−1; q)m+1(q; q)m q q 3 − q + 1 ; q 2 m q 5 − 1 + q1 ; q 2 m qq5 −q m+1 3 +q 2 ; q     6  ×   2 q 5 + q 2 − 1 q 5 − 1 + q1 ; q 4 m q 5 − 1 + q1 ; q 4 m+1 q −q+1 ; q 4 m+1 q3 2 d2m

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3 Generalized Sturm–Liouville Problems in q-Spaces

and   m+1   2 2 d2m+1 = (1 − q) q 5 + q 4 + q 3 − 1 [6]q − 2 q 2m +m−2 q 3 − q + 1    6      −q+1 2 (−1; q)m+1 (q; q)m q q 3 − q + 1 ; q 2 m+1 q 5 − 1 + q1 ; q 2 m+1 qq5 −q m+1 3 +q 2 ; q    6  ×  .       2 q5 + q2 − 1 q 5 − 1 + q1 ; q 4 m+1 2 q −q+1 ; q 4 m+1 q q 6 − q + 1 ; q 4 m+1 q3

3.5

A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free Parameters

In this section, using the generalized Sturm–Liouville Theorem 3.6 in q-difference spaces, we introduce a class of symmetric q-orthogonal polynomials with four free parameters and obtain its standard properties, such as a generic second-order q-difference equation, the explicit form of the polynomials in terms of q-hypergeometric series, a generic three-term recurrence relation, and a general orthogonality relation. We also present some particular examples in the sequel. Motivated by Eq. (3.95), let us consider the q-difference equation       x 2 ax 2 + b Dq Dq −1 φn (x; q) + x cx 2 + d Dq φn (x; q) + λn,q x 2 − σn d φn (x; q) = 0.

(3.110) To find a symmetric monic q-polynomial solution for Eq. (3.110), let φ¯ n (x; q) = x n + δn,q x n−2 + · · ·

(3.111)

satisfy the well-known three-term recurrence relation φ¯ n+1 (x; q) = x φ¯ n (x; q) − Cn,q φ¯ n−1 (x; q) with

φ¯ 0 (x; q) = 1 and φ¯ 1 (x; q) = x. (3.112)

From (3.110) and (3.111), equating the coefficient in x n+2 gives   λn,q = −[n]q c − [1 − n]q a , provided that |a| + |c| = 0.

(3.113)

3.5 A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free. . .

291

Using the eigenvalue λn,q given in (3.113) and equating the coefficient in x n , we obtain in (3.111) that δn,q

  q 2 −b(q − 1)q[n − 1]q [n]q − dq 2n + dq n (qσn + σn−1 )   = . (q + 1) aq 3 − q 2n (a + c(q − 1))

Also from (3.111) and (3.112) we have     x n+1 +δn+1,q x n−1 +· · · = x x n + δn,q x n−2 + · · · −Cn,q x n−1 + δn−1,q x n−3 + · · · , which implies Cn,q = δn,q − δn+1,q =

∗ Bn,q

  , a 2 q 4 + q 4n (a + c(q − 1))2 − a q 3 + q q 2n (a + c(q − 1)) (3.114)

where   ∗ = q n+1 q 2n (a + c(q − 1))((d − dq)σn − b) Bn,q        +q n a b q 2 + 1 + d(q − 1)q 2 + bc(q − 1) − aq 2 (b + d(q − 1)σn−1 ) . The limit case of (3.114) is computed as lim Cn,q = q↑1

n(a(b(2 − n) + d) − bc) − dσn (2a(n − 1) + c) , (a(2n − 3) + c)(a(2n − 1) + c)

and for the eigenvalue (3.113) we have lim λn,q = −n(c − (1 − n)a), q↑1

which gives exactly the same result as in the continuous case [5] by taking into account that the three-term recurrence relation (3.112) has a minus sign in the coefficients Cn,q . Since the polynomial solution of Eq. (3.110) is symmetric, we use the notation  φn (x; q) = Sn



cd q; x ab

for mathematical display formulas and Sn (x; q; a, b, c, d) in the text. This means that we have to deal with just one characteristic vector (a, b, c, d) for any given subclass.

292

3 Generalized Sturm–Liouville Problems in q-Spaces

For n = 2m and n = 2m + 1, the Cn,q in (3.114) are simplified as C2m,q

  q 2m+1 [2m]q (q − 1) bq 2m (a + c(q − 1)) − aq 2(b + d(q − 1))   =− 2 4 a q + q 8m (a + c(q − 1))2 − a q 3 + q q 4m (a + c(q − 1))

and C2m+1,q

   q 2m q 2m (a + c(q − 1)) − aq q 2m+1 (b + d(q − 1)) − b   =− 2 . a q + q 8m+1 (a + c(q − 1))2 − a q 2 + 1 q 4m (a + c(q − 1))

Theorem 3.7 The explicit form of the polynomial Sm (x; q; a, b, c, d) is  Sm

=



cd q; x ab [ m2 ] 

q

(k−1)k m−2k

x

1 m 2 [ m2 ]−k−1 2

k

k=0

q2

j =0

a[2j + σm + m − 1]q + cq 2j +σm +m−1 ,   m+1 b[ 2j + (−1)m+1 + 2 ]q + dq 2j +(−1) +2

where the q-number [n]q has been defined in (3.8) and the q-binomial coefficient is denoted by   (q; q)n n = . (q; q)m (q; q)n−m m q Moreover, if a, b = 0, then  Sm

cd q; x ab

=x

σm



m+σm−1 2x 2 aq q −m+σm , (a+c(q−1))q 2 a . q ;− 2 φ1 (b+d(q−1))q 2σm+1 b

(3.115)

b

Proof Despite the degrees of A(x), B(x), C(x), and E(x), the proof can be done in a similar way as in [4, Section 10.2] for classical q-orthogonal polynomials.   It is easy to check that  lim Sm q↑1

cd q; x ab



 = Sm



cd x , ab

where the right-hand-side polynomial was introduced in (1.101).

3.5 A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free. . .

293

Also, the monic form of the polynomials (3.115) is represented as  S¯m



cd q; x ab

=q

σm −m

2σm+1  m  (q 2 ; q −mσm ; (b+d(q−1))q ; q 2 )[ m ] b [2] b 2 − (a+c(q−1))q m+σm−1 −m −(m−1)σ a m (q ; q ; ; q 2 )[ m ] a 2 

m+σm−1 (a+c(q−1))q 2 2 q −m+σm , 2 aq x a . × x σ m 2 φ1 q ;− (b+d(q−1))q 2σm+1 b

b

By noting Theorem 3.6, since W (qx; q) A(x) + (q − 1)xB(x) x 2 (a + c(q − 1)) + b + d(q − 1)   , = = W (x; q) A(qx) q 2 aq 2x 2 + b

(3.116)

if a, b = 0, then the solution of Eq. (3.116) will be  W (x; q) =

d(q − 1) +1 b

 log x 2

2 log q

2 2

(− aqbx ; q 2 )∞ 2

2 x 2 (− (a+c(q−1))x b+d(q−1) ; q )∞

= W (−x; q),

in which some restrictions on the parameters must be considered in order to have convergence for the infinite products. We are now in a position to analyze some particular cases of the q-difference equation (3.110) that provide q-analogues of different families of continuous orthogonal polynomials. Case 1: A Generalization of q-Ultraspherical Polynomials Consider the q-difference equation     x 2 1 − x 2 Dq Dq −1 φn (x; q) + (q + 1)qx α − x 2 (α + β + 1) Dq φn (x; q)   + −[n]q (−(1 + α + β)q(1 + q) + [1 − n]q )x 2 − αq(1 + q)σn φn (x; q) = 0 (3.117) as a special case of (3.110) with the polynomial solution 

−q(q + 1)(α + β + 1) αq(q + 1) φn (x; α, β|q) = Sn q; x −1 1 

    q σn −n , q n+σn −1 (α + β + 1)q q 2 − 1 + 1 2 2 2 σn     = x 2 φ1 q ;q x . q 2σn +1 αq q 2 − 1 + 1

(3.118)

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3 Generalized Sturm–Liouville Problems in q-Spaces

The sequence (3.118) satisfies an orthogonality relation of the form 

1 −1

W1∗ (x; α, β|q)φn (x; α, β|q)φm (x; α, β|q)dq x  =

1

−1

W1∗ (x; α, β|q)φn2(x; α, β|q)dq x

 δn,m ,

in which W1∗ (x; α, β|q) = x 2 W1 (x; α, β|q) is the main weight function, and the function W1 (x; α, β|q) satisfies the equation    q q 2 − 1 x 2 (α + β + 1) − α + x 2 − 1 W1 (qx; α, β|q)   . = W1 (x; α, β|q) q 2 q 2x 2 − 1

(3.119)

Up to a periodic function, a solution of Eq. (3.119) is of the form W1 (x; α, β|q) = B

log x 2 log(q)

( )



 q 2x 2; q 2 ∞  2 2  = W1 (−x; α, β|q), x 2 − ABx2 ; q 2 ∞

where A=

/  q − q 3 (α + β + 1) − 1

and

B=

/   α q 3 − q + 1.

Notice that lim W1∗ (x; α, β|q) = lim x 2 W1 (x; α, β|q) = x 2α (1 − x 2 )β q↑1

q↑1

looks the same as the weight function of generalized ultraspherical polynomials. The monic type of polynomials (3.118) satisfies a three-term recurrence relation of type (3.112) with C2m,q

          q 2m+2 q 2m − 1 q 2m q q 2 − 1 ϑ + 1 + q 2 α q − q 3 − 1 =          2 − q 2 + 1 q 4m+2 q q 2 − 1 ϑ + 1 + q 8m+1 q q 2 − 1 ϑ + 1 + q 5

3.5 A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free. . .

295

and C2m+1,q              q 2m q 2m − αq 5 + (β + 1)q 3 + q 2 − qϑ + 1 + αq q 2 − 1 + 1 q 4m+1 q q 2 − 1 ϑ + 1 + q , =          2 − q 2 + 1 q 4m q q 2 − 1 ϑ + 1 + q 8m+1 q q 2 − 1 ϑ + 1 + q

where ϑ = α + β + 1. Hence the norm square value takes the form 

1 −1

2 φ¯ n2 (x; α, β|q)W1∗ (x; α, β|q)dq x = dn,α,β



1 −1

W1∗ (x; α, β|q)dq x,

where        q αq q 2 − 1 + 1 ; q 2 m m 2    d2m,α,β =      (α+β+1)q q 2 −1 +1 4 (α+β+1)q q 2 −1 +1 4 ;q ;q m 3 q 

q2; q2

q

m+1

   m+1 (q − 1)q m(2m−1)−2 (q(q + 1)(α + β) − 1)(q(q + 1)(α + β + 1) − 1) αq q 2 − 1 + 1   × (q + 1) −q(α + β + 1) + αq 3 + 1         (α+β+1)q q 2 −1 +1 2 (α+β+1)q q 2 −1 +1 2   ;q   ;q 2 2 q q αq q −1 +1 m m+1  ×   2       (α+β+1)q q −1 +1 4 2 q (α + β + 1)q q − 1 + 1 ; q 4 m ;q q m+1

and        q 2 ; q 2 m q αq q 2 − 1 + 1 ; q 2 m+1     =  q (α + β + 1)q q 2 − 1 + 1 ; q 4 m+1 

2 d2m+1,α,β

×

(q − 1)q 2m

2 +m−2

   m+1 (q(q + 1)(α + β) − 1)(q(q + 1)(α + β + 1) − 1) αq q 2 − 1 + 1   (q + 1) −q(α + β + 1) + αq 3 + 1         (α+β+1)q q 2 −1 +1 (α+β+1)q q 2 −1 +1 2 2 ;q ;q q q 2 (αq (q 2 −1)+1) m+1 m+1 ×  2 .   2 2 (α+β+1)q (q −1)+1 (α+β+1)q (q −1)+1 4 4 ;q ;q q3

m+1

q

m+1

A special case of Eq. (3.117) generating q-Chebyshev polynomials of the fifth kind is α = 1 and β = [3]q /[2]q − 2, which gives the monic polynomial solution      [3]q −q q 2 + q + 1 q(q + 1) − 2|q = S¯n φ¯n x; 1, [2]q −1 1



q; x ,

296

3 Generalized Sturm–Liouville Problems in q-Spaces

satisfying the orthogonality relation 

1 −1

W1∗

      [3]q [3]q [3]q ¯ ¯ x; 1, − 2|q φn x; 1, − 2|q φm x; 1, − 2|q dq x [2]q [2]q [2]q   1 [3]q = d 2 [3]q W1∗ (x; 1, − 2|q)dq x δn,m , n,1, [2]q −2 [2]q −1

in which 2)  2log(x  3     2x2; q 2) log(q) − q + 1 (q q [3] [3]q ∞ q ∗ W1∗ x; 1, − 2|q = = W − 2|q . −x; 1, 1 4 [2]q [2]q ( q −q+1 x 2 ; q 2 )

q 3 −q+1

∞

The second special case of Eq. (3.117) generating q-Chebyshev polynomials of the sixth kind is α = 1 and β = [5]q /[2]q − 2, which gives the monic polynomial solution    [5]q −q[5]q q(q + 1) φ¯ n x; 1, − 2|q = S¯n [2]q −1 1



q; x ,

satisfying the orthogonality relation 

      [5]q [5]q [5]q W1∗ x; 1, − 2|q φ¯ n x; 1, − 2|q φ¯ m x; 1, − 2|q dq x [2]q [2]q [2]q −1   1 [5]q W1∗ (x; 1, − 2|q)dq x, δn,m , = d 2 [5]q n,1, [2]q −2 [2]q −1 1

in which W1∗

 log(x 2 )   2 2 2  3    q x ; q ∞ q − q + 1 2 log(q) [5]q [5]q  6  − 2|q = = W1∗ −x; 1, − 2|q . x; 1, q −q+1 2 [2]q [2]q x ; q2 q 3 −q+1

∞

Case 2: A Generalization of q-Hermite Polynomials Consider the q-difference equation x2



    1 − q 2 x 2 − 1 Dq Dq −1 φn (x; q) + (q + 1)x x 2 + p Dq φn (x; q)   + q[−n]q x 2 − σn p φn (x; q) = 0

3.5 A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free. . .

297

as a special case of (3.110) with the polynomial solution 

1 + q p(1 + q) q; x φn (x; p|q) = Sn 1 − q2 −1 

  q σn −n , 0 σn 2 2 2 2  q ;q 1 − q x . = x 2 φ1 2σn +1  −pq 2 + p + 1 q

(3.120)

The sequence (3.120) satisfies an orthogonality relation 

α −α

W2∗ (x; p|q)φn (x; p|q)φm (x; p; q)dq x =



α −α

 W2∗ (x; p|q)φn2 (x; p|q)dq x δn,m ,

% in which α = 1/ 1 − q 2 and W2∗ (x; p|q) = x 2 W2 (x; p|q) is the main weight function, where W2 (x; p|q) satisfies the equation −pq 2 + p + 1 W2 (qx; p|q)  = 2  2 . W2 (x; p|q) q + q − 1 q 4x 2

(3.121)

Up to a periodic function, a solution of Eq. (3.121) is of the form 2   log(x )     p 1 − q 2 + 1 2 log(q) (q 2 1 − q 2 x 2 ; q 2 )∞ W2 (x; p|q) = = W2 (−x; p|q), x2

where lim W2∗ (x; p|q) = lim x 2 W2 (x; p|q) = x −2p e−x , 2

q↑1

q↑1

which is the weight function of generalized Hermite polynomials. The monic type of polynomials (3.120) satisfies a three-term recurrence relation of type (3.112) with C2m,q

  2     p q − 1 − 1 q 2m−1 q 2m − 1 =− q2 − 1

and  C2m+1,q =

 −pq 2 + p + 1 q 4m+1 − q 2m . q2 − 1

298

3 Generalized Sturm–Liouville Problems in q-Spaces

Consequently, the norm square value takes the form 

α −α

2 φ¯ n2 (x; p|q)W2∗ (x; p|q)dq x = dn,p



α −α

W2∗ (x; p|q) dq x,

where 2 d2m,p

 m   1 m(2m−1) −pq 2 + p + 1 = q (−1; q)m+1(q; q)m −pq 3 + pq + q; q 2 2m  m 2 q2 − 1

and 2 d2m+1,p

 m   1 m(2m+1) −pq 2 + p + 1 3 2 = q . 2m+1 (−1; q)m+1(q; q)m −pq + pq + q; q  m+1 2 q2 − 1

We point out that if p = 0, the weight function of discrete q-Hermite polynomials appears as 1   , W2∗ (x; 0|q) =  2 1 − q x 2; q 2 ∞ and therefore we have / φn (x; 0|q) = kn hn (x 1 − q 2 |q), in which hn (x|q) denotes the discrete q-Hermite polynomials and kn is a normalizing constant.

3.5.1

Two Finite Sequences Based on Ramanujan’s Identity

Case 3: First Finite Sequence For u, v ∈ R, consider a special case of Eq. (3.110),     x 2 x 2 + 1 Dq Dq −1 φn (x; q) − 2x (u + v − 1)x 2 + u Dq φn (x; q)     + [n]q 2u + 2v − 2 + [1 − n]q x 2 + 2uσn φn (x; q) = 0,

(3.122)

whose monic polynomial solution can be represented as φ¯ n (x; q, u, v) 

q −n+σn , (1 − 2(q − 1)(u + v − 1))q n+σn −1 2 σn 2 2 = K1 x 2 φ1 q ;−q x , (1 − 2u(q − 1))q 2σn+1

(3.123)

3.5 A Basic Class of Symmetric q-Orthogonal Polynomials with Four Free. . .

299

where K1 =

q [n/2]([n/2]−1)(q n+σn −1 (1 − 2(u + v − 1)(q − 1)); q 2 )[n/2] . n (q (−1) +2 (1 − 2u(q − 1)); q 2 )[n/2]

In order to prove the orthogonality of the finite set {φ¯ n (x; q, u, v)}N n=0 on (−∞, ∞), it is necessary to impose a specific condition, N