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Integral transforms and their applications by L. Debnath and D. Bhatta (2007)

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A published graduate and senior undergraduate textbook from Chapman & Hall/CRC, by Debnath and Bhatta, not Phil's own work. The preface says it covers Fourier, Laplace and other classical transforms, plus new chapters on Radon transforms, wavelets and fractional calculus. It applies them to differential and integral equations, quantum mechanics, fluid mechanics, and probability, with over 600 worked examples and exercises.

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© 2007 by Taylor & Francis Group, LLCIntegral Transforms and Their Applications Second Edition © 2007 by Taylor & Francis Group, LLCLokenath Debnath Dambaru BhattaIntegral Transforms and Their Applications Second Edition © 2007 by Taylor & Francis Group, LLCChapman & Hall/CRC Taylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742 © 2007 by Taylor & Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper10 9 8 7 6 5 4 3 2 1 International Standard Book Number-10: 1-58488-575-0 (Hardcover) International Standard Book Number-13: 978-1-58488-575-7 (Hardcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the conse-quences of their use. N o part o f this boo k ma y be r e p rin ted, r e p r od u ced, trans mi tted, o r u tilized in an y f o rm b y an y electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www. copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data Debnath, Lokenath. Integral transforms and their applications. -- 2nd ed. / Lokenath Debnath and Dambaru Bhatta. p. cm. Includes bibliographical references and index. ISBN 1-58488-575-0 (acid-free paper)1. Integral transforms. I. Bhatta, Dambaru. II. Title. QA432.D36 2006515’.723--dc22 2006045638 Visit the Taylor & Francis Web site athttp://www.taylorandfrancis.com and the CRC Press Web site at http:/ /www.crcpress.com © 2007 by Taylor & Francis Group, LLC To my wife Sadhana and granddaughter Princess Maya Lokenath Debnath To my wife Bisruti and sons Rohit andAmit Dambaru Bhatta © 2007 by Taylor & Francis Group, LLC Preface to the Second Edition “A teacher can never truly teach unless he is still learning himself. A lamp can never light another lamp unless it continues to burn its own ﬂame. The teacher who has come to the end of his subject,who has no living traﬃc with his knowledge but merely repeats his lessons to his students, can only load their minds; he cannot quicken them.” Rabindranath Tagore When the ﬁrst edition of this book was published in 1995 under the sole authorship of Lokenath Debnath, it w as well received, and has been used as a senior undergraduate or graduate level text and research reference in the United States and abroad for the last ten years. We received many comments and suggestions from many students and faculty around the world. These comments and criticisms have been very helpful, beneﬁcial, and encouraging. This second edition is the result of that input. Another reason for adding this second edition to the literature is the fact that there have been major discoveries of several integral transforms including the Radon transform, the Gabor transform, the inverse scattering transform, and wavelet transforms in the twentieth century. It is becoming even moredesirable for mathematicians, scientists and engineers to pursue study and research on these and related topics. So what has changed, and will continue to change, is the nature of the topics that are of interest in mathematics, science and engineering, the evolution of books such as this one is a history of these shifting concerns. This new and revised edition preserves the basic content and style of the ﬁrst edition. As with the previous edition, this book has been revised primar-ily as a comprehensive text for senior undergraduates or beginning graduate students and a research reference for p rofessionals in mathematics, science, and engineering, and other applied sciences. The main goal of this book is on the development of the required analytical skills on the part of the reader, rather than the importance of more abstract formulation with full mathe-matical rigor. Indeed, our major emphasis is to provide an accessible working knowledge of the analytical methods with proofs required in pure and applied mathematics, physics, and engineering. © 2007 by Taylor & Francis Group, LLC We have made many additions and changes in order to modernize the con- tents and to improve the clarity of the previous edition. We have also taken advantage of this new edition to update the bibliography and correct typo- graphical errors, to include additional topics, examples of applications, exer- cises, comments, and observations, and i n some cases, to entirely rewrite whole section. This edition contains a collection of over 600 challenging worked ex-amples and exercises with answers and hints to selected exercises. There is plenty of material in the book for a year-long course. Some of the material need not be covered in a course work a nd can be left for the readers to study on their own in order to prepare them for further study and research. Some of the major changes, additions, and highlights in this edition and the most signiﬁcant diﬀerence from the ﬁrst edition include the following: 1. Chapter 1 on Integral Transforms has been completely revised and some new material on brief historical introduction was added to provide new information about the historical d evelopments of the subject. These changes have been made to provide the reader to see the direction inwhich the subject has developed and ﬁnd those contributed to its devel- opments. 2. Chapter 2 on Fourier Transforms has been completely revised and new material added, including new sections on Fourier transforms of general-ized functions, the Poisson summation formula, the Gibbs phenomenon, and the Heisenberg uncertainty prin ciple. Many sections have been com- pletely rewritten with new ex amples of applications. 3. Four entirely new chapters on Radon Transforms, and Wavelets and Wavelet Transforms, Fractional Calculus and its applications to ordinary and partial diﬀerential equations have been added to modernize thecontents of the book. A new section on the transfer function and the impulse response function with examples of applications was included in Chapters 2 and 4. 4. The book oﬀers a detailed and clea r explanation of every concept and method that is introduced, accompan ied by carefully selected worked examples, with special emphasis bei ng given to those topics in which students experience diﬃculty. 5. A wide variety of modern examples of applications has been selected from areas of ordinary and partial diﬀerential equations, quantum me-chanics, integral equations, ﬂuid mechanics and elasticity, mathematical statistics, fractional ordinary and partial diﬀerential equations, and spe- cial functions. 6. The book is organized with suﬃcient ﬂexibility to enable instructors to select chapters appropriate to courses of diﬀering lengths, emphases,and levels of diﬃculty. © 2007 by Taylor & Francis Group, LLC 7. A wide spectrum of exercises has been carefully chosen and included at the end of each chapter so the reader may further develop both analyticalskills in the theory and applications of transform methods and a deeper insight into the subject. 8. Answers and hints to selected exer cises are provided at the end of the book to provide additional help to students. All ﬁgures have been re-drawn and many new ﬁgures have been added for a clear understanding of physical explanations. 9. All appendices, tables of integral transforms, and the bibliography have been completely revised and updated. Many new research papers and standard books have been added to the bibliography to stimulate new interest in future study and research. Index of the book has also beencompletely revised in order to include a wide variety of topics. 10. The book provides information that puts the reader at the forefront of current research. With the improvements and many challenging worked problems and exer- cises, we hope this edition will continue to be a useful textbook for studentsas well as a research reference for profe ssionals in mathematics, science and engineering. It is our pleasure to express our grateful thanks to many friends, colleagues, and students around the world who oﬀered their suggestions and help at various stages of the preparation of the book. We express our sincere thanks to Veronica Martinez and Maria Lisa Cisneros for typing the ﬁnal manuscript with constant changes. In spite of the be st eﬀorts of everyone involved, some typographical errors doubtless remain. Finally, we wish to express our specialthanks to Bob Stern, Executive Editor, and the staﬀ of CRC/Chapman Hall for their help and cooperation. Lokenath Debnath Dambaru Bhatta The University of Texas-Pan American © 2007 by Taylor & Francis Group, LLC Preface to the First Edition Historically, the concept of an integra l transform originated from the cele- brated Fourier integral formula. The importance of integral transforms is that they provide powerful operational methods for solving initial value problemsand initial-boundary value problems for linear diﬀerential and integral equa- tions. In fact, one of the main impulses for the development of the operational calculus of integral transforms was the study of diﬀerential and integral equa- tions arising in applied mathematics, mathematical physics, and engineering science; it was in this setting that integral transforms arose and achievedtheir early successes. With ever greate r demand for mathematical methods to provide both theory and applications for science and engineering, the u- tility and interest of integral transforms seems more clearly established thanever. In spite of the fact that integral transforms have many mathematical and physical applications, their use is still predominant in advanced study and research. Keeping these features in mind, our main goal in this book is to provide a systematic exposition of the basic properties of various integral transforms and their applications to the solution of boundary and initial valueproblems in applied mathematics, mathematical physics, and engineering. In addition, the operational calculus of integral transforms is applied to integral equations, diﬀerence equations, fractional integrals and fractional derivatives,summation of inﬁnite series, evaluation of deﬁnite integrals, and problems of probability and statistics. There appear to be many books available for students studying integral transforms with applications. Some are excellent but too advanced for the beginner. Some are too elementary or have limited scope. Some are out ofprint. While teaching transform methods, operational mathematics, and/or mathematical physics with applications, the author has had diﬃculty choosing textbooks to accompany the lectures. This book, which was developed as aresult of many years of experience teaching advanced undergraduates and ﬁrst-year graduate students in mathematics, physics, and engineering, is an attempt to meet that need. It is based essentially on a set of mimeographed lecture notes developed for courses given by the author at the University of Central Florida, East Carolina University, and the University of Calcutta. This book is designed as an introduction to theory and applications of inte- gral transforms to problems in linear diﬀerential equations, and to boundary and initial value problems in partial diﬀerential equations. It is appropriate © 2007 by Taylor & Francis Group, LLC for a one-semester course. There are two basic prerequisites for the course: a standard calculus sequence and ordinar y diﬀerential equations. The book as- sumes only a limited knowledge of complex variables and contour integration, partial diﬀerential equations, and continuum mechanics. Many new examples of applications dealing with problems in applied mathematics, physics, chem- istry, biology, and engineering are included. It is notessential for the reader to know everything about these topics, but limited knowledge of at least some of them would be useful. Besides, the book is intended to serve as a reference work for those seriously interested in advanced study and research in the sub- ject, whether for its own sake or for its applications to other ﬁelds of appliedmathematics, mathematical physics, and engineering. The ﬁrst chapter gives a brief historical introduction and the basic ideas of integral transforms. The second chapter deals with the theory and applicationsof Fourier transforms, and of Fourier cosine and sine transforms. Important examples of applications of interest in applied mathematics, physics statis- tics, and engineering are included. The theory and applications of Laplace transforms are discussed in Chapters 3 and 4 in considerable detail. The ﬁfth chapter is concerned with the operational calculus of Hankel transforms withapplications. Chapter 6 gives a detailed treatment of Mellin transforms and its various applications. Included are Mellin transforms of the Weyl fractional in- tegral, Weyl fractional derivatives, and generalized Mellin transforms. Hilbertand Stieltjes transforms and their applications are discussed in Chapter 7. Chapter 8 provides a short introduction to ﬁnite Fourier cosine and sine transforms and their basic operational properties. Applications of these trans- forms are also presented. The ﬁnite Laplace transform and its applications to boundary value problems are included in Chapter 9. Chapter 10 deals with adetailed theory and applications of Z transforms. Chapter 12 is devoted to the operational calculus of Legendre transforms and their applications to boundary value problems in potential theory. Jacobiand Gegenbauer transforms and their applications are included in Chapter 13. Chapter 14 deals with the theory and applications of Laguerre transforms. The ﬁnal chapter is concerned with the Hermite transform and its basic operational properties including the Convolution Theorem. Most of the material of these chapters has been developed since the early sixties and appears here in bookform for the ﬁrst time. The book includes two important appendices. The ﬁrst one deals with sev- eral special functions and their basic properties. The second appendix includesthirteen short tables of integral transforms. Many standard texts and reference books and a set of selected classic and recent research papers are included in the Bibliography that will be very useful for the reader interested in learning more about the subject. The book contains 750 worked examples , applications, and exercises which include some that have been chosen from many standard books as well asrecent papers. It is hoped that they will serve as helpful self-tests for under- standing of the theory and mastery of the transform methods. These exam- © 2007 by Taylor & Francis Group, LLC ples of applications and exercises were chosen from the areas of diﬀerential and diﬀerence equations, electric circu its and networks, vibration and wave propagation, heat conduction in solids, qu antum mechanics, fractional calcu- lus and fractional diﬀerential equations, dynamical systems, signal processing, integral equations, physical chemistry, mathematical biology, probability and statistics, and solid and ﬂuid mechanics. This varied number of examples and exercises should provide something of int erest for everyone. The exercises tru- ly complement the text and range fro m the elementary to the challenging. Answers and hints to many selected exer cises are provided at the end of the book. This is a textand a reference book designed for use by the student and the reader of mathematics, science, and e ngineering. A serio us attempt has been made to present almost all the standard material, and some new materialas well. Those interested in more advanced rigorous treatment of the topic- s covered may consult standard books and treatises by Churchill, Doetsch, Sneddon, Titchmarsh, and Widder listed in the Bibliography. Many ideas, results, theorems, methods, problems, and exercises presented in this book are either motivated by or borrowed from the works cited in the Bibliography.The author wishes to acknowledge his gratitude to the authors of these works. This book is designed as a new source for both classical and modern topics dealing with integral transforms and their applications for the future devel-opment of this useful subject. Its main features are: 1. A systematic mathematical treatment of the theory and method of integral transforms that gives the reader a clear understanding of the subject and its varied applications. 2. A detailed and clear explanation of every concept and method that is in- troduced, accompanied by carefully se lected worked examples, with special emphasis being given to those topics in which students experience diﬃculty. 3. A wide variety of diverse examples of applications carefully selected from areas of applied mathematics, mathema tical physics, and engineering sci- ence to provide motivation, and to illustrate how operational methods can be applied eﬀectively to solve them. 4. A broad coverage of the essential standard material on integral transforms and their applications together with some new material that is notusually covered in familiar texts or reference books. 5. Most of the recent developments in the subject since the early sixties appear here in book form for the ﬁrst time. 6. A wide spectrum of exercises has been carefully selected and included at the end of each chapter so that the reader may further develop both manip- ulative skills in the applications of integral transforms and a deeper insight into the subject. © 2007 by Taylor & Francis Group, LLC 7. Two appendices have been included in order to make the book self-contained. 8. Answers and hints to selected exerci ses are provided at the end of the book for additional help to students. 9. An updated Bibliography is included to stimulate new interest in future study and research. In preparing the book, the author has been encouraged by and has beneﬁt- ed from the helpful comments and criticism of a number of graduate students and faculty of several universities in the United States, Canada, and India.The author expresses his grateful thanks to all these individuals for their in- terest in the book. My special thanks to Jackie Callahan and Ronee Trantham who typed the manuscript and cheerfully put up with constant changes andrevisions. In spite of the best eﬀorts of everyone involved, some typographical errors doubtlessly remain. I do hope that these are both few and obvious, and will cause minimal confusion. The author also wishes to thank his friends and colleagues including Drs. Sudipto Roy Choudhury and Carroll A. Webber for their interest and help during the preparation of the book. Finally, the authorwishes to express his special thanks to Dr. Wayne Yuhasz, Executive Editor, and the staﬀ of CRC Press for their help and cooperation. I am also deeply indebted to my wife, Sadhana, for all her understanding and tolerance whilethe book was being written. Lokenath Debnath University of Central Florida © 2007 by Taylor & Francis Group, LLC About the Authors Lokenath Debnath is Professor and Chair of the Department of Mathe- matics at the University of Texas-Pan American, Edinburg, Texas. Professor Debnath received his M.Sc . and Ph.D. degrees in pure mathemat- ics from the University of Calcutta, and obtained his D.I.C. and Ph.D. degrees in applied mathematics from the Imperial College of Science and Technolo- gy, London. He was a Senior Research Fellow at the Department of Applied Mathematics and Theoretical Physics at the University of Cambridge and has had several visiting appointments at the University of Oxford, Florida StateUniversity, University of Maryland, and the University of Calcutta. He served the University of Central Florida as Professor and Chair of Mathematics and as Professor of Mechanical and Aerospace Engineering from 1983 to 2001. Hewas Acting Chair of the Department of Statistics at the University of Central Florida and has served as Professor of M athematics and Professor of Physics at East Carolina University for a period of ﬁfteen years. Among many other honors and awards, he has received a Senior Fulbright Fellowship and an NSF Scientist Award to v isit India for lectures and research. He was a University Grants Commission Re search Professor at the University of Calcutta and was elected President o f the Calcutta Mathematical Society for a period of three years. He has serv ed as a Lecturer of the SIAM Visiting Lecturer Program and as a Visiting Speaker of the Mathematical Association of America (MAA) from 1990. He also has served as organizer of several professional meetings and conferences at regional, national, and international levels; and as Director of six NSF-CBMS r esearch conferences at the University of Central Florida, and East Carolina University and the University of Texas-Pan American. He has received many g rants from NSF and state agencies of North Carolina, Florid a, and Texas. He has also received many university awards for teaching, research, services and leadership. Dr. Debnath is author or co-author of ten graduate level books and research monographs, including the third edition of Introduction to Hilbert Spaces with Applications, Nonlinear Water Waves, Continuum Mechanics published by A- cademic Press, the fourth edition of Linear Partial Diﬀerential Equations for Scientists and Engineers published by Birkhauser Verlag, the second edition ofNonlinear Partial Diﬀerential Equations for Scientists and Engineers, and Wavelet Transforms and Their Applications published by Birkhauser Verlag. He has also edited eleven research monographs including Nonlinear Waves © 2007 by Taylor & Francis Group, LLC published by Cambridge University Press. He is an author or co-author of over 300 research papers in pure and applied mathematics, including appliedpartial diﬀerential equations, integral transforms and special functions, histo- ry of mathematics, mathematical inequalities, wavelet transforms, solid and ﬂuid mechanics, linear and nonlinear waves, solitons, mathematical physics, magnetohydrodynamics, unsteady boundary layers, dynamics of oceans, and stability theory. Professor Debnath is a member of many scientiﬁc organizations at both na- tional and international levels. He has been an Associate Editor and a member of the editorial boards of many refereed journals and he currently serves on theEditorial Board of many ref ereed journals including Journal of Mathematical Analysis and Applications, Indian Journal of Pure and Applied Mathematics, Fractional Calculus and Applied Analysis, Bulletin of the Calcutta Mathemat-ical Society, Integral Transforms and Special Functions, International Journal of Engineering Science, andInternational Journal of Mathematical Education in Science and Technology . He is the current and founding Managing Editor of the International Journal of Mathematics and Mathematical Sciences . Dr. Debnath delivered twenty ﬁve invit ed lectures at national and interna- tional conferences, presented over 200 research papers at national and inter- national professional meetings, and given over 250 seminar and colloquium lectures at universities and institutes in the United States and abroad. Dambaru Bhatta is an Assistant Professor of Mathematics at the University of Texas-Pan American, Edinburg, Texas. Dr. Bhatta received his Ph.D. degre e in applied mathematics from Dal- housie University, Halifax, Canada and M.Sc. degree in mathematics from the University of Delhi. He had worked with various companies in Montre-al, Ottawa, and Atlanta. His research interests include wave-structure inter- action, computational mathematics, ﬁnite element method, nonlinear partial diﬀerential equations, fractional calculus, and fractional diﬀerential equations. © 2007 by Taylor & Francis Group, LLC Contents 1 Integral Transforms 1 1 . 1 B r i e f H i s t o r i c a l I n t r o d u c t i o n ................... 1 1 . 2 ................. 2 Fourier Transforms and Their Applications 9 2 . 1 I n t r o d u c t i o n ............................ 2 . 2 2.3 Deﬁnition of the Fourier Transform and Examples . . . . . .2.4 Fourier Transforms of Generalized Functions . . . . . . . . . .2.5 Basic Properties of Fourier Transforms . . . . . . . . . . . . . 282 . 62.7 The Shannon Sampling Theorem . . . . . . . . . . . . . . . . 442 . 8 G i b b s ’ P h e n o m e n o n2.9 Heisenberg’s Uncertainty Principle . . . . . . . . . . . . . . . 572.10 Applications of Fourier Transforms to Ordinary Diﬀerential E q u a t i o n s ............................. 6 0 2.11 Solutions of Integral Equations . . . . . . . . . . . . . . . . . 652.12 Solutions of Partial Diﬀerential Equations . . . . . . . . . . . 682.13 Fourier Cosine and Sine Transforms with Examples . . . . . . 91 2.14 Properties of Fourier Cosine and Sine Transforms . . . . . . . 93 2.15 Applications of Fourier Cosine and Sine Transforms to Partial D i ﬀ e r e n t i a l E q u a t i o n s 2.16 Evaluation of Deﬁnite Integrals . . . . . . . . . . . . . . . . . 1002.17 Applications of Fourier Transforms in Mathematical Statistics 1032.18 Multiple Fourier Transforms and Their Applications . . . . . 109 3 Laplace Transforms and Their Basic Properties 133 3 . 13.2 Deﬁnition of the Laplace Transform and Examples . . . . . . 1343.3 Existence Conditions for the Laplace Transform . . . . . . . . 1393.4 Basic Properties of Laplace Transforms . . . . . . . . . . . . . 1403.5 The Convolution Theorem and Properties of Convolution . . 1453.6 Diﬀerentiation and Integration of Laplace Transforms . . . . 1513.7 The Inverse Laplace Transform and Examples . . . . . . . . . 1543.8 Tauberian Theorems and Watson’s Lemma . . . . . . . . . . 168BasicConceptsandDeﬁnitions TheFourierIntegralFormulas..................106 17129 Poisson’sSummationFormula..................37 .......................54 ......................96 Introduction............................1332.19Exercises..............................119 © 2007 by Taylor & Francis Group, LLC 3 . 9 E x e r c i s e s..............................1 7 3 4 Applications of Laplace Transforms 181 4 . 1 I n t r o d u c t i o n ............................1 8 1 4.2 Solutions of Ordinary Diﬀerential Equations . . . . . . . . . . 182 4.3 Partial Diﬀerential Equations, Initial and Boundary Value P r o b l e m s .............................2 0 7 4 . 4 S o l u t i o n s o f I n t e g r a l E q u a t i o n s .................2 2 24.5 Solutions of Boundary Value Problems . . . . . . . . . . . . . 2254.6 Evaluation of Deﬁnite Integrals . . . . . . . . . . . . . . . . . 228 4.7 Solutions of Diﬀerence and Diﬀerential-Diﬀerence Equations . 230 4.8 Applications of the Joint Laplace and Fourier Transform . . . 2374.9 Summation of Inﬁnite Series . . . . . . . . . . . . . . . . . . . 2484.10 Transfer Function and Impulse Response Function of a Linear S y s t e m ...............................2 5 1 4 . 1 1 E x e r c i s e s..............................2 5 6 5 Fractional Calculus and Its Applications 269 5 . 1 I n t r o d u c t i o n ............................2 6 95 . 2 H i s t o r i c a l C o m m e n t s .......................2 7 05.3 Fractional Derivatives and Integrals . . . . . . . . . . . . . . . 2725.4 Applications of Fractional Calculus . . . . . . . . . . . . . . . 2795 . 5 E x e r c i s e s..............................2 8 2 6 Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 2836 . 1 I n t r o d u c t i o n ............................2 8 36.2 Laplace Transforms of Fractional Integrals and Fractional D e r i v a t i v e s ............................2 8 4 6.3 Fractional Ordinary Diﬀerential Equations . . . . . . . . . . . 287 6 . 4 F r a c t i o n a l I n t e g r a l E q u a t i o n s ..................2 9 0 6.5 Initial Value Problems for Fractional Diﬀerential Equations . 2956.6 Green’s Functions of Fractional Diﬀerential Equations . . . . 2986.7 Fractional Partial Diﬀerential Equations . . . . . . . . . . . . 2996 . 8 E x e r c i s e s..............................3 1 2 7 Hankel Transforms and Their Applications 315 7 . 1 I n t r o d u c t i o n ............................3 1 57.2 The Hankel Transform and Examples . . . . . . . . . . . . . . 3167.3 Operational Properties of the Hankel Transform . . . . . . . . 3197.4 Applications of Hankel Transforms to Partial Diﬀerential E q u a t i o n s .............................3 2 2 7 . 5 E x e r c i s e s..............................3 3 1 © 2007 by Taylor & Francis Group, LLC 8 Mellin Transforms and Their Applications 339 8 . 1 I n t r o d u c t i o n ............................3 3 9 8.2 Deﬁnition of the Mellin Transform and Examples . . . . . . . 3408.3 Basic Operational Properties of Mellin Transforms . . . . . . 3438.4 Applications of Mellin Transforms . . . . . . . . . . . . . . . 349 8.5 Mellin Transforms of the Weyl Fractional Integral and t h e W e y l F r a c t i o n a l D e r i v a t i v e .................3 5 3 8.6 Application of Mellin Transforms to Summation of Series . . 358 8.7 Generalized Mellin Transforms . . . . . . . . . . . . . . . . . 361 8 . 8 E x e r c i s e s..............................3 6 5 9 Hilbert and Stieltjes Transforms 371 9 . 1 I n t r o d u c t i o n ............................3 7 1 9.2 Deﬁnition of the Hilbert Transform and Examples . . . . . . 3729.3 Basic Properties of Hilbert Transforms . . . . . . . . . . . . . 3759 . 4 H i l b e r t T r a n s f o r m s i n t h e C o m p l e x P l a n e ...........3 7 8 9 . 5 A p p l i c a t i o n s o f H i l b e r t T r a n s f o r m s ...............3 8 0 9.6 Asymptotic Expansions of One-Sided Hilbert Transforms . . . 3889.7 Deﬁnition of the Stieltjes Transform and Examples . . . . . . 3919.8 Basic Operational Properties of Stieltjes Transforms . . . . . 3949.9 Inversion Theorems for Stieltjes Transforms . . . . . . . . . . 396 9.10 Applications of Stieltjes Transforms . . . . . . . . . . . . . . . 399 9 . 1 1 T h e G e n e r a l i z e d S t i e l t j e sT r a n s f o r m ..............4 0 19.12 Basic Properties of the Generalized Stieltjes Transform . . . . 4039 . 1 3 E x e r c i s e s..............................4 0 4 10 Finite Fourier Sine and Cosine Transforms 407 1 0 . 1 I n t r o d u c t i o n ............................4 0 7 10.2 Deﬁnitions of the Finite Fourier Sine and Cosine Transforms a n d E x a m p l e s ...........................4 0 8 10.3 Basic Properties of Finite Fourier Sine and Cosine Transforms 410 10.4 Applications of Finite Fourier Sine and Cosine Transforms . . 416 10.5 Multiple Finite Fourier Transforms and Their Applications . 422 1 0 . 6 E x e r c i s e s..............................4 2 5 11 Finite Laplace Transforms 429 1 1 . 1 I n t r o d u c t i o n ............................4 2 911.2 Deﬁnition of the Finite Laplace Transform and Examples . . 43011.3 Basic Operational Properties of the Finite Laplace Transform 436 11.4 Applications of Finite Laplace Transforms . . . . . . . . . . . 439 11.5 Tauberian Theorems . . . . . . . . . . . . . . . . . . . . . . . 4431 1 . 6 E x e r c i s e s..............................4 4 3 © 2007 by Taylor & Francis Group, LLC 12ZTransforms 445 1 2 . 1 I n t r o d u c t i o n ............................4 4 5 12.2 Dynamic Linear Systems and Impulse Response . . . . . . . . 44512.3 Deﬁnition of the ZTransform and Examples . . . . . . . . . . 449 12.4 Basic Operational Properties of ZT r a n s f o r m s .........4 5 3 12.5 The Inverse ZTransform and Examples . . . . . . . . . . . . 459 12.6 Applications of ZTransforms to Finite Diﬀerence Equations . 463 12.7 Summation of Inﬁnite Series . . . . . . . . . . . . . . . . . . . 466 1 2 . 8 E x e r c i s e s..............................4 6 9 13 Finite Hankel Transforms 473 1 3 . 1 I n t r o d u c t i o n ............................4 7 3 13.2 Deﬁnition of the Finite Hankel Transform and Examples . . . 47313.3 Basic Operational Properties . . . . . . . . . . . . . . . . . . 47613.4 Applications of Finite Hankel Transforms . . . . . . . . . . . 476 1 3 . 5 E x e r c i s e s..............................4 8 1 14 Legendre Transforms 485 1 4 . 1 I n t r o d u c t i o n ............................4 8 5 14.2 Deﬁnition of the Legendre Transform and Examples . . . . . 48614.3 Basic Operational Properties of Legendre Transforms . . . . . 48914.4 Applications of Legendre Transforms to Boundary Value P r o b l e m s .............................4 9 7 1 4 . 5 E x e r c i s e s..............................4 9 8 15 Jacobi and Gegenbauer Transforms 501 1 5 . 1 I n t r o d u c t i o n ............................5 0 115.2 Deﬁnition of the Jacobi Transform and Examples . . . . . . . 501 15.3 Basic Operational Properties . . . . . . . . . . . . . . . . . . 504 15.4 Applications of Jacobi Transforms to the Generalized Heat Conduction Problem . . . . . . . . . . . . . . . . . . . . . . . 505 15.5 The Gegenbauer Transform and Its Basic Operational P r o p e r t i e s .............................5 0 7 15.6 Application of the Gegenbauer Transform . . . . . . . . . . . 510 16 Laguerre Transforms 511 1 6 . 1 I n t r o d u c t i o n ............................5 1 1 16.2 Deﬁnition of the Laguerre Transform a n d E x a m p l e s ...........................5 1 1 16.3 Basic Operational Properties . . . . . . . . . . . . . . . . . . 516 16.4 Applications of Laguerre Transforms . . . . . . . . . . . . . . 5201 6 . 5 E x e r c i s e s..............................5 2 3 © 2007 by Taylor & Francis Group, LLC 17 Hermite Transforms 525 1 7 . 1 I n t r o d u c t i o n ............................5 2 5 17.2 Deﬁnition of the Hermite Transform and Examples . . . . . . 52617.3 Basic Operational Properties . . . . . . . . . . . . . . . . . . 5291 7 . 4 E x e r c i s e s..............................5 3 8 18 The Radon Transform and Its Applications 539 1 8 . 1 I n t r o d u c t i o n ............................5 3 91 8 . 2 T h e R a d o n T r a n s f o r m ......................5 4 118.3 Properties of the Radon Transform . . . . . . . . . . . . . . . 5451 8 . 4 T h e R a d o n T r a n s f o r m o f D e r i v a t i v e s ..............5 5 018.5 Derivatives of the Radon Transform . . . . . . . . . . . . . . 55118.6 Convolution Theorem for the Radon Transform . . . . . . . . 553 18.7 Inverse of the Radon Transform and the Parseval Relation . . 554 18.8 Applications of the Radon Transform . . . . . . . . . . . . . . 5601 8 . 9 E x e r c i s e s..............................5 6 1 19 Wavelets and Wavelet Transforms 563 1 9 . 1 B r i e f H i s t o r i c a l R e m a r k s .....................5 6 3 19.2 Continuous Wavelet Transforms . . . . . . . . . . . . . . . . . 565 1 9 . 3 T h e D i s c r e t e W a v e l e t T r a n s f o r m ................5 7 319.4 Examples of Orthonormal Wavelets . . . . . . . . . . . . . . . 5751 9 . 5 E x e r c i s e s..............................5 8 4 Appendix A Some Special Functions and Their Properties 587 A-1 Gamma, Beta, and Error Functions . . . . . . . . . . . . . . . 587A - 2 B e s s e l a n d A i r yF u n c t i o n s ....................5 9 2A-3 Legendre and Associated Legendre Functions . . . . . . . . . 598A-4 Jacobi and Gegenbauer Polynomials . . . . . . . . . . . . . . 601A-5 Laguerre and Associated Laguerre Functions . . . . . . . . . . 605A-6 Hermite Polynomials and Weber-Hermite Functions . . . . . . 607 A - 7 M i t t a g L e ﬄ e r F u n c t i o n......................6 0 9 Appendix B Tables of Integral Transforms 611 B - 1 F o u r i e rT r a n s f o r m s ........................6 1 1 B - 2 F o u r i e r C o s i n e T r a n s f o r m s ....................6 1 5B - 3 F o u r i e r S i n e T r a n s f o r m s .....................6 1 7 B - 4 L a p l a c e T r a n s f o r m s........................6 1 9 B - 5 H a n k e l T r a n s f o r m s ........................6 2 4B-6 Mellin Transforms . . . . . . . . . . . . . . . . . . . . . . . . 627B - 7 H i l b e r t T r a n s f o r m s ........................6 3 0B - 8 S t i e l t j e sT r a n s f o r m s .......................6 3 3 B - 9 F i n i t e F o u r i e r C o s i n e T r a n s f o r m s ................6 3 6 B - 1 0F i n i t e F o u r i e r S i n e T r a n s f o r m s .................6 3 8B - 1 1F i n i t e L a p l a c e T r a n s f o r m s ....................6 4 0 © 2007 by Taylor & Francis Group, LLC B - 1 2ZT r a n s f o r m s ...........................6 4 2 B - 1 3F i n i t e H a n k e l T r a n s f o r m s ....................6 4 4 Answers and Hints to Selected Exercises 645 2 . 1 9 E x e r c i s e s..............................6 4 53 . 9 E x e r c i s e s..............................6 5 14 . 1 1 E x e r c i s e s..............................6 5 5 6 . 8 E x e r c i s e s..............................6 6 2 7 . 5 E x e r c i s e s..............................6 6 2 8 . 8 E x e r c i s e s..............................6 6 3 9 . 1 3 E x e r c i s e s..............................6 6 41 0 . 6 E x e r c i s e s..............................6 6 5 1 1 . 6 E x e r c i s e s..............................6 6 7 1 2 . 8 E x e r c i s e s..............................6 6 71 3 . 5 E x e r c i s e s..............................6 7 0 1 6 . 5 E x e r c i s e s..............................6 7 0 1 7 . 4 E x e r c i s e s..............................6 7 0 1 8 . 9 E x e r c i s e s..............................6 7 1 1 9 . 5 E x e r c i s e s..............................6 7 1 Bibliography 673 © 2007 by Taylor & Francis Group, LLC 1 Integral Transforms “The thorough study of nature is the most fertile ground for math- ematical discoveries.” Joseph Fourier “If you wish to foresee the future of mathematics our proper course is to study the history and present condition of the science.” Henri Poincar´ e “The tool which serves as interme diary between theory and prac- tice, between thought and observation, is mathematics, it is math- ematics which builds the linking bridges and gives the ever more reliable forms. From this it has come about that our entire contem- porary culture, in as much as it is based the intellectual penetrationand the exploitation of nature, has its foundations in mathematic- s.” David Hilbert 1.1 Brief Historical Introduction Integral transformations have been su ccessfully used for almost two centuries in solving many problems in applied mathematics, mathematical physics, and engineering science. Historically, the or igin of the integral transforms includ- ing the Laplace and Fourier transforms can be traced back to celebrated workof P. S. Laplace (1749–1827) on probability theory in the 1780s and to mon- umental treatise of Joseph Fourier (1768–1830) on La Th´eorie Analytique de la Chaleur published in 1822. In fact, Laplace’s classic book on La Th´ eorie Analytique des Probabilities includes some basic results of the Laplace trans- form which is one of the oldest and most commonly used integral transformsavailable in the mathematical literature. This has eﬀectively been used in ﬁnd- ing the solution of linear diﬀerential equations and integral equations. On the other hand, Fourier’s treatise provided the modern mathematical theory of 1 © 2007 by Taylor & Francis Group, LLC 2 INTEGRAL TRANSFORMS and THEIR APPLICATIONS heat conduction, Fourier series, and Fourier integrals with applications. In his treatise, Fourier stated a remarkable result that is universally known as theFourier Integral Theorem . He gave a series of exampl es before stating that an arbitrary function deﬁned on a ﬁnite interval can be expanded in terms of trigonometric series which is now universally known as the Fourier series .I n an attempt to extend his new ideas to functions deﬁned on an inﬁnite interval, Fourier discovered an integral transform and its inversion formula which arenow well known as the Fourier transform and the inverse Fourier transfor- m. However, this celebrated idea of Fourier was known to Laplace and A. L. Cauchy (1789–1857) as some of their earlier work involved this transforma-tion. On the other hand, S. D. Poisson (1781–1840) also independently used the method of transform in his research on the propagation of water waves. However, it was G. W. Leibniz (1646–1716) who ﬁrst introduced the idea of a symbolic method in calculus. Subs equently, both J. L. Lagrange (1736–1813) and Laplace made considerable contributions to symbolic methods which be- came known as operational calculus. Although both the Laplace and the Fouri- er transforms have been discovered in th e nineteenth century, it was the British electrical engineer Oliver Heaviside (1850–1925) who made the Laplace trans-form very popular by using it to solve ordinary diﬀerential equations of elec- trical circuits and systems, and then to develop modern operational calculus. It may be relevant to point out that the Laplace transform is essentially aspecial case of the Fourier transform for a class of functions deﬁned on the positive real axis, but it is more simple than the Fourier transform for the following reasons. First, the question of convergence of the Laplace transform is much less delicate because of its exponentially decaying kernel exp ( −st), where Re s>0a n d t>0. Second, the Laplace transform is an analytic func- tion of the complex variable and its properties can easily be studied with the knowledge of the theory of complex variable. Third, the Fourier integral for- mula provided the deﬁnitions of the Laplace transform and the inverse Laplacetransform in terms of a complex contour integral that can be evaluated with the help the Cauchy residue theory and deformation of contour in the complex plane. It was the work of Cauchy that contained the exponential form of the Fourier Integral Theorem as f(x)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞eik(x−y)f(y)dydk. (1.1.1) Cauchy’s work also contained the following formula for functions of the oper- atorD: φ(D)f(x)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞φ(ik)eik(x−y)f(y)dydk. (1.1.2) This essentially led to the modern form of the operational calculus. His famous treatise entitled Memoire sur l’Emploi des Equations Symboliques provided a © 2007 by Taylor & Francis Group, LLC Integral Transforms 3 fairly rigorous description of symbolic methods. The deep signiﬁcance of the Fourier Integral Theorem was recogni zed by mathematicians and mathemati- cal physicists of the nineteenth and twentieth centuries. Indeed, this theorem is regarded as one of the most fundamental results of modern mathematical analysis and has widespread physical and engineering applications. The gen- erality and importance of the theorem is well expressed by Kelvin and Tait who said: ”...Fourier’s Theorem, which is not only one of the most beautifulresults of modern analysis, but may be said to furnish an indispensable instru- ment in the treatment of nearly every recondite question in modern physics. To mention only sonorous vibrations, the propagation of electric signals alonga telegraph wire, and the conduction of heat by the earth’s crust, as subjects in their generality intractable without it, is to give but a feeble idea of its importance.” During the late nineteenth century, it was Oliver Heaviside (1850–1925) who recognized the power and success of ope rational calculus and ﬁrst used the operational method as a powerful and eﬀective tool for the solutions of tele- graph equation and the second order hyperbolic partial diﬀerential equations with constant coeﬃcients. In his two p apers entitled “On Operational Meth- ods in Physical Mathematics,” Parts I and II, published in The Proceedings of the Royal Society , London, in 1892 and 1893, Heaviside developed opera- tional methods. His 1899 book on Electromagnetic Theory also contained the use and application of the operational methods to the analysis of electrical circuits or networks. Heaviside re placed the diﬀerential operator D≡ d dtby pand treated the latter as an element of the ordinary laws of algebra. The development of his operational methods paid little attention to questions of mathematical rigor. The widespread use of the Heaviside method prior to itsvindication by the theory of the Fourier or Laplace transform created a lot of controversy. This was similar to the controversy put forward against the widespread use of the delta function as one of the most useful mathematicaldevices in Dirac’s logical formulation of quantum mechanics during the 1920s. In fact, P. A. M. Dirac (1902–1984) said: “Al l electrical engineers are familiar with the idea of a pulse, and the δ-function is just a way of expressing a pulse mathematically.” Dirac’s study of Heaviside’s operator calculus in electromag- netic theory, his training as an electrical engineer, and his deep knowledge ofthe modern theory of electrical pulses seemed to have a tremendous impact on his ingenious development of modern quantum mechanics. Apparently, the ideas of operational methods originated from the classic work of Laplace, Fourier, and Cauchy. Inspired by this remarkable work, Heav- iside developed his new but less rigorous operational mathematics. In spite of the striking success of Heaviside’s ca lculus as one of the most useful math- ematical methods, contemporary math ematicians hardly recognized Heavi- side’s work in his lifetime, primarily due to lack of mathematical rigor. In hislecture on Heaviside and Operational Calculus at the Birth Centenary of Oliv- er Heaviside, J. L. B. Cooper (1952) revealed some of the controversial issues surrounding Heaviside’s celebrated work, and declared: “As a mathematician © 2007 by Taylor & Francis Group, LLC 4 INTEGRAL TRANSFORMS and THEIR APPLICATIONS he was gifted with manipulative skill and with a genius for ﬁnding convenient methods of calculation. He simpliﬁed Maxwell’s theory enormously; accordingto Hertz, the four equations known as Maxwell’s were ﬁrst given by Heaviside. He is one of the founders of vector analysis....” Reviewing the history of Heav- iside’s calculus, Cooper gave a fairly complete account of early history of the subject along with mathematicians’ varying opinions about Heaviside’s con- tributions to operational calculus. According to Cooper, a widely publicizedstory that operational calculus was dis covered by Heaviside remained contro- versial. In spite of the controversies, it is generally believed that Heaviside’s real achievement was to develop operational calculus, which is one of the mostuseful mathematical devices in applied m athematics, mathematical physics, and engineering science. In this context Lord Rayleigh’s following quotation seems to be most appropriate from a physical point of view: “In the mathemat-ical investigation I have usually employed such methods as present themselves naturally to a physicist. The pure mathematician will complain, and (it must be confessed) sometimes with justice, of deﬁcient rigor. But to this question there are two sides. For, however important it may be to maintain a uniformly high standard in pure mathematics, the physicist may occasionally do well torest content with arguments which are fairly satisfactory and conclusive from his point of view. To his mind, exercised in a diﬀerent order of ideas, the more severe procedure of the pure mathematician may appear not more but lessdemonstrative. And further, in many cases of diﬃculty to insist upon highest standard would mean the exclusion of t he subject altogether in view of the space that would be required.” With the exception of a group of pure mathematicians, everyone has found Heaviside’s work a remarkable achievement even though he did not providea rigorous demonstration of his operational calculus. In defense of Heaviside, Richard P. Feynman’s thought seems to be worth quoting. “However, the em- phasis should be somewhat more on how to do the mathematics quickly andeasily, and what formulas are true, rather than the mathematicians’ interest in methods of rigorous proof.” The development of operational calculus was somewhat similar to that of calculus o f the seventeenth century. Mathemati- cians who invented the calculus did not provide a rigorous formulation of it. The rigorous formulation came only in the nineteenth century, even thoughin the transition the non-rigorous demonstration of the calculus that is still admired. It is well known that twentieth-century mathematicians have pro- vided a rigorous foundation of the Heaviside operational calculus. So, by anystandard, Heaviside deserves a lot of credit for his remarkable work. The next phase of the development of op erational calculu s is characterized by the eﬀort to provide justiﬁcations of the heuristic methods by rigorous proofs. In this phase, T. J. Bromwich (1875-1930) ﬁrst successfully introduced the theory of complex functions to give formal justiﬁcation of Heaviside’scalculus. In addition to his many contributions to this subject, he gave the formal derivation of the Heaviside expansion theorem and the correct inter- pretation of Heaviside’s operational results. After Bromwich’s work, notable © 2007 by Taylor & Francis Group, LLC Integral Transforms 5 contributions to rigorous formulation of operational calculus were made by J. R. Carson, B. van der Pol, G. Doetsch, and many others. In concluding our discussion on the historical development of operational calculus, we should add a note of caution against the controversial evaluation of Heaviside’s work. From an applied mathematical point of view, Heavi- side’s operational calculus was an important achievement. In support of his statement, an assessment of Heaviside’s work made by E. T. Whittaker inHeaviside’s obituary is recorded below: “Looking back..., we should place the operational calculus with Poincar´ e’s discovery of automorphic functions and Ricci’s discovery of the tensor calculus as the three most important math-ematical advances of the last quarter of the nineteenth century.” Although Heaviside paid little attention to questions of mathematical rigor, he recog- nized that operational calculus is one of the most eﬀective and useful mathe- matical methods in applied mathemati cal sciences. This has led naturally to rigorous mathematical analysis of integral transforms. Indeed, the Fourier or Laplace transform methods based on the rigorous mathematical foundation are essentially equivalent to the modern operational calculus. There are many other integral transformations including the Mellin trans- form, the Hankel transform, the Hilbert transform and the Stieltjes transform which are widely used to solve initial and boundary value problems involving ordinary and partial diﬀerential equations and other problems in mathematics,science and engineering. Although , Mellin ( 1854–1933) presented an elaborate discussion of his transform and its inversion formula, it was G. Bernhard Rie- mann (1826–1866) who ﬁrs t recognized the Mellin transform and its inversion formula in his famous memoir on prime numbers. Hermann Hankel (1839– 1873), a student of G. B. Riemann, introduced the Hankel transform with theBessel function as its kernel, and this transform can easily be derived from the two-dimensional Fourier transform when circular symmetry is assumed. The Hankel transform arises naturally in solving boundary value problems incylindrical polar coordinates. Although the Hilbert transform was named after one of the greatest mathe- maticians of the twentieth century, David Hilbert (1862–1943), this transform and its properties are basically studied by G. H. Hardy (1877-1947) and E. C. Titchmarsh (1899-1963). The Dutch mathematician, T. J. Stieltjes (1856–1894) introduced the Stieltjes transform in his study of continued fractions. Both the Hilbert and Stieltjes transforms arise in many problems in mathe- matics, science and engineering. The f ormer is used to solve problems in ﬂuid mechanics, signal processing, and electronics, while the latter arises in solving the integral equations and moment problems. We would like to conclude this section by making some comments on the history of the Radon transform, the Gabor transform and the wavelet trans- form. The Radon transform is introduced by Johann Radon (1887–1956) in1917 and has enormous useful applications to medical imaging, and comput- er assisted tomography (CAT). The wav elet transform is discovered by Jean Morlet, a French geophysical engineer , as a new mathematical tool to study © 2007 by Taylor & Francis Group, LLC 6 INTEGRAL TRANSFORMS and THEIR APPLICATIONS seismic signal analysis in 1982. It is one of the most versatile linear integral transformations and can be applied to solve a wide variety of problems inmathematics, science and engineering. The reader is referred to Chapter 19 of this book for more detailed information on wavelets and wavelet transforms. 1.2 Basic Concepts and Deﬁnitions Theintegral transform of a function f(x) deﬁned in a≤x≤bis denoted by I{f(x)}=F(k), and deﬁned by I{f(x)}=F(k)=b⎪integraldisplay aK(x, k)f(x)dx, (1.2.1) where K(x, k), given function of two variables xandk, is called the kernel of the transform. The operator Iis usually called an integral transform operator or simply an integral transformation . The transform function F(k)i so f t e n referred to as the image of the given object function f(x), and kis called the transform variable . Similarly, the integral transform of a function of several variables is deﬁned by I{f(x)}=F(κ)=⎪integraldisplay SK(x,κ)f(x)dx, (1.2.2) where x=(x1,x2,...,x n),κ=(k1,k2,...,k n), and S⊂Rn. A mathematical theory of transformations of this type can be developed by using the properties of Banach spaces . From a mathematical point of view, such a program would be of great interest, but it may notbe useful for prac- tical applications. Our goal here is to study integral transforms as operationalmethods with special emphasis to applications. The idea of the integral transform operator is somewhat similar to that of the well-known linear diﬀerential operator, D≡ d dx, which acts on a function f(x) to produce another function f/prime(x), that is, Df(x)=f/prime(x). (1.2.3) Usually, f/prime(x) is called the derivative or the image of f(x) under the linear transformation D. Evidently, there are a number of important integral transforms including Fourier ,Laplace ,Hankel ,a n dMellin transforms. They are deﬁned by choosing diﬀerent kernels K(x, k) and diﬀerent values for aandbinvolved in (1.2.1). © 2007 by Taylor & Francis Group, LLC Integral Transforms 7 Obviously, Iis alinear operator since it satisﬁes the property of linearity : I{αf(x)+βg(x)}=b⎪integraldisplay a{αf(x)+βg(x)}K(x, k)dx =αI{f(x)}+βI{g(x)}, (1.2.4) where αandβare arbitrary constants. In order to obtain f(x)f r o mag i v e n F(k)=I{f(x)}, we introduce the inverse operator I−1such that I−1{F(k)}=f(x). (1.2.5) Accordingly I−1I=II−1=1which is the identity operator. It can be proved that I−1is also a linear operator as follows I−1{αF(k)+βG(k)}=I−1{αIf(x)+βIg(x)} =I−1{I[αf(x)+βg(x)]} =αf(x)+βg(x) =αI−1{F(k)}+βI−1{G(k)}. It can also be proved that the integral transform is unique. In other words, ifI{f(x)}=I{g(x)},t h e n f(x)=g(x) under suitable conditions. This is known as the uniqueness theorem . We close this section by adding the basic scope and applications of integral transformation from a general point of view. It follows from the above dis-cussion that an integral transformation simply means a unique mathematical operation through which a real or complex-valued function fis transformed into another new function F=If, or into a set of data that can be measured (or observed) experimentally. Thus, the importance of the integral transform is that it transforms a diﬃcult mathematical problem to an relatively easy problem, which can easily be solved. In the study of initial-boundary value problem involving diﬀerential equations, the diﬀerential operators are replaced by much simpler algebraic operations involving F, which can readily be solved. The solution of the original problem is then obtained in the original variables by the inverse transformation. So, the next basic problem leads to the com- putation of the inverse integral transform exactly or approximately. Indeed,in order to make the integral transform method eﬀective, it is essential to reconstruct ffromIf=Fwhich is, in general, a diﬃcult step in practice. However, this diﬃculty can be resolved in many diﬀerent ways. In application- s, often the transform function Fitself has some physical meaning and needs to be studied in its own right. For example , in electrical engineering problems, the original function f(t) may represent a signal that is a function of time t. The Fourier transform F(ω)o ff(t) represents the frequency spectrum of the signal f(t) and it is physically useful as the time representation of the signal © 2007 by Taylor & Francis Group, LLC 8 INTEGRAL TRANSFORMS and THEIR APPLICATIONS itself. Indeed, it is often more important to work with Frather than with f. Conversely, given the frequency spectrum, F(ω), the original signal f(t)c a n be reconstructed by the inverse Fourier transform. Other important and major examples include the Gabor transform and the wavelet transform both of which transform a signal f(t)i nt h et i m e - frequency domain ( t−ωplane). In other words, these new transforms convey essential information about the nature and structure of a signal in the time-frequency domain simultaneously. In 1946, Dennis Gabor, a Hungarian-British physicist and engineer and a 1971 Nobel Prize winner in physics, introduced thewindowed Fourier transform (or the Gabor transform ) of a signal f(t) with respect to a window function g, denoted by ⎪tildewidef g(t, ω) and deﬁned by G[f](t, ω)=⎪tildewidefg(t, ω)=∞⎪integraldisplay −∞f(τ)g(τ−t)e−iωtdτ =⎪angbracketleftbig f, gt,ω⎪angbracketrightbig , (1.2.6) where fandg∈L2(R) with the inner product /angbracketleftf,g/angbracketright. Gabor (1900–1979) ﬁrst recognized the major weaknesses of the Fourier transform analysis of signals, and also realized the great importance of lo- calized time and frequency concentratio ns in signal processing. All these mo- tivated him to formulate a fundamental method of the Gabor transform for decomposition of signal in terms of elementary signals (or wave transforms).Gabor’s pioneering approach has now become one of the standard model- s for time-frequency signal analysis. It is also important to point out that the Gabor transform ⎪tildewidef g(t, ω) is referred to as the ca nonical coherent state representation of fin quantum mechanics. In the 1960s, the term “coherent states” was ﬁrst used in quantum optics. For more information on the Gabor and the wavelet transforms and their basic properties, the reader is referred to Debnath (2002). © 2007 by Taylor & Francis Group, LLC 2 Fourier Transforms and Their Applications “The profound study of nature is the most fertile source of math- ematical discoveries.” Joseph Fourier “The theory of Fourier series and integrals has always had ma- jor diﬃculties and necessitated a large mathematical apparatus indealing with questions of convergence. It engendered the develop- ment of methods of summation, although these did not lead to a completely satisfactory solution of the problem. .... For the Fourier transform, the introduction of distributions (hence, the space S) is inevitable either in an explicit or hidden form. .... As a resultone may obtain all that is desired from the point of view of the continuity and inversion of the Fourier transform.” Laurent Schwartz 2.1 Introduction Many linear boundary value and initial value problems in applied mathemat- ics, mathematical physics, and engineer ing science can be eﬀectively solved by the use of the Fourier transform, the Fourier cosine transform, or the Fouriersine transform. These transforms are very useful for solving diﬀerential or in- tegral equations for the following reasons. First, these equations are replaced by simple algebraic equations, which enable us to ﬁnd the solution of the transform function. The solution of the given equation is then obtained in the original variables by inverting the transform solution. Second, the Fouri-er transform of the elementary source term is used for determination of the fundamental solution that illustrates the basic ideas behind the construction and implementation of Green’s functions. Third, the transform solution com-bined with the convolution theorem provides an elegant representation of the solution for the boundary value and initial value problems. We begin this chapter with a formal derivation of the Fourier integral for- 9 © 2007 by Taylor & Francis Group, LLC 10 INTEGRAL TRANSFORMS and THEIR APPLICATIONS mulas. These results are then used to d eﬁne the Fourier, Fourier cosine, and Fourier sine transforms. This is followed by a detailed discussion of the basicoperational properties of these transforms with examples. Special attention is given to convolution and its main properties. Sections 2.10 and 2.11 deal with applications of the Fourier transform to the solution of ordinary diﬀerential equations and integral equations. In Section 2.12, a wide variety of partial diﬀerential equations are solved by the use of the Fourier transform method.The technique that is developed in this and other sections can be applied with little or no modiﬁcation to diﬀerent kinds of initial and boundary value problems that are encountered in applications. The Fourier cosine and sinetransforms are introduced in Section 2.13. The properties and applications of these transforms are discussed in Sections 2.14 and 2.15. This is followed by evaluation of deﬁnite integrals with the aid of Fourier transforms. Section2.17 is devoted to applications of Fourier transforms in mathematical statis- tics. The multiple Fourier transforms and their applications are discussed in Section 2.18. 2.2 The Fourier Integral Formulas A function f(x) is said to satisfy Dirichlet’s conditions in the interval −a< x<a,i f (i)f(x) has only a ﬁnite number of ﬁnite discontinuities in −a<x<a and has no inﬁnite discontinuities. (ii)f(x) has only a ﬁnite number of maxima and minima in −a<x<a . From the theory of Fourier series we know that if f(x) satisﬁes the Dirichlet conditions in −a<x<a , it can be represented as the complex Fourier series f(x)=∞⎪summationdisplay n=−∞anexp(inπx/a ), (2.2.1) where the coeﬃcients are an=1 2aa⎪integraldisplay −af(ξ)exp (−inπξ/a )dξ. (2.2.2) This representation is evid ently periodic of period 2 ain the interval. However, the right hand side of (2.2.1) cannot represent f(x)outside the interval −a< x<a unless f(x) is periodic of period 2 a. Thus, problems on ﬁnite intervals lead to Fourier series, and problems on the whole line −∞<x< ∞lead to the © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 11 Fourier integrals. We now attempt to ﬁnd an integral representation of a non- periodic function f(x)i n(−∞,∞) by letting a→∞. As the interval grows (a→∞)t h ev a l u e s kn=nπ abecome closer together and form a dense set. If we write δk=(kn+1−kn)=π aand substitute coeﬃcients aninto (2.2.1), we obtain f(x)=1 2π∞⎪summationdisplay n=−∞(δk)⎡ ⎣a⎪integraldisplay −af(ξ)exp (−iξkn)dξ⎤ ⎦exp(ixkn). (2.2.3) In the limit as a→∞,knbecomes a continuous variable kandδkbecomes dk. Consequently, the sum can be repla ced by the integral in the limit and (2.2.3) reduces to the result f(x)=1 2π∞⎪integraldisplay −∞⎡ ⎣∞⎪integraldisplay −∞f(ξ)e−ikξdξ⎤ ⎦eikxdk. (2.2.4) This is known as the celebrated Fourier integral formula . Although the above arguments do not constitute a rigorous proof of (2.2.4), the formula is correct and valid for functions that are piecewise c ontinuously diﬀerentiable in every ﬁnite interval and is absolutely integrable on the whole real line. A function f(x)i ss a i dt ob e absolutely integrable on (−∞,∞)i f ∞⎪integraldisplay −∞|f(x)|dx <∞ (2.2.5) exists. It can be shown that the formula (2.2.4) is valid under more general condi- tions. The result is contained in the following theorem: THEOREM 2.2.1 Iff(x) satisﬁes Dirichlet’s conditions in ( −∞,∞), and is absolutely inte- grable on ( −∞,∞), then the Fourier integral (2.2.4) converges to the function 1 2[f(x+0 )+ f(x−0)] at a ﬁnite discontinuity at x.I no t h e rw o r d s , 1 2[f(x+0 )+ f(x−0)]=1 2π∞⎪integraldisplay −∞eikx⎡ ⎣∞⎪integraldisplay −∞f(ξ)e−ikξdξ⎤ ⎦dk. (2.2.6) This is usually called the Fourier integral theorem . If the function f(x) is continuous at point x,t h e n f(x+0 )= f(x−0)= f(x), then (2.2.6) reduces to (2.2.4). The Fourier integral theorem was originally stated in Fourier’s famous trea- tise entitled La Th ´eorie Analytique da la Chaleur (1822), and its deep signiﬁ- cance was recognized by mathematicians and mathematical phy sicists. Indeed, © 2007 by Taylor & Francis Group, LLC 12 INTEGRAL TRANSFORMS and THEIR APPLICATIONS this theorem is one of the most monumental results of modern mathematical analysis and has widespread physical and engineering applications. We express the exponential factor exp[ ik(x−ξ)] in (2.2.4) in terms of trigonometric functions and use the even and odd nature of the cosine and the sine functions respectively as functions of kso that (2.2.4) can be written as f(x)=1 π∞⎪integraldisplay 0dk∞⎪integraldisplay −∞f(ξ)cosk(x−ξ)dξ. (2.2.7) This is another version of the Fourier integral formula .I nm a n yp h y s i c a l problems, the function f(x) vanishes very rapidly as |x|→∞ ,w h i c he n s u r e s the existence of the repeated integrals as expressed. We now assume that f(x) is an even function and expand the cosine function in (2.2.7) to obtain f(x)=f(−x)=2 π∞⎪integraldisplay 0coskxdk∞⎪integraldisplay 0f(ξ)coskξ dξ. (2.2.8) This is called the Fourier cosine integral formula . Similarly, for an odd function f(x), we obtain the Fourier sine integral formula f(x)=−f(−x)=2 π∞⎪integraldisplay 0sinkxdk∞⎪integraldisplay 0f(ξ)sinkξ dξ. (2.2.9) These integral formulas were discovered independently by Cauchy in his work on the propagation of waves on the surface of water. 2.3 Deﬁnition of the Fourier Transform and Examples We use the Fourier integral formula (2.2.4) to give a formal deﬁnition of the Fourier transform. DEFINITION 2.3.1 The Fourier transform of f(x)is denoted by F{f(x)}= F(k),k∈R, and deﬁned by the integral F{f(x)}=F(k)=1 √ 2π∞⎪integraldisplay −∞e−ikxf(x)dx, (2.3.1) whereFis called the Fourier transform operator or the Fourier transfor- mation and the factor1 √ 2πis obtained by splitting the factor1 2πinvolved in © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 13 (2.2.4). This is often called the complex Fourier transform. A suﬃcient condi- tion for f(x)to have a Fourier transform is that f(x)is absolutely integrable on(−∞,∞). The convergence of the integral (2.3.1) follows at once from the fact that f(x)is absolutely integrable. In fact, the integral converges uniformly with respect to k. Thus, the deﬁnition of the Fourier transform is restricted to absolutely inte- grable functions. This restriction is too strong for many physical applications.Many simple and common functions, such as constant function, trigonometric functions sin ax,c o sax, exponential functions, and x nH(x) do not have Fouri- er transforms, even though they occur frequently in applications. The integralin (2.3.1) fails to converge when f(x) is one of the above elementary function- s. This is a very unsatisfactory feature of the theory of Fourier transforms. However, this unsatisfactory feature can be resolved by means of a naturalextension of the deﬁnition of the Fourier transform of a generalized function, f(x) in (2.3.1). We follow Lighthill (1958) and Jones (1982) to discuss brieﬂy the theory of the Fourier transforms of good functions. The inverse Fourier transform, denoted by F −1{F(k)}=f(x), is deﬁned by F−1{F(k)}=f(x)=1 √ 2π∞⎪integraldisplay −∞eikxF(k)dk, (2.3.2) whereF−1is called the inverse Fourier transform operator. Clearly, both FandF−1are linear integral operators. In applied math- ematics, xusually represents a space variable and k(=2π λ)i saw a v e n u m - ber variable where λis the wavelength. However, in electrical engineering, x is replaced by the time variable tandkis replaced by the frequency vari- ableω(= 2πν)w h e r e νis the frequency in cycles per second. The function F(ω)=F{f(t)}is called the spectrum of the time signal function f(t). In electrical engineering lit erature, the Fourier transform pairs are deﬁned s- lightly diﬀerently by F{f(t)}=F(ν)=∞⎪integraldisplay −∞f(t)e−2πνitdt, (2.3.3) and F−1{F(ν)}=f(t)=∞⎪integraldisplay −∞F(ν)e2πiνtdν=1 2π∞⎪integraldisplay −∞F(ω)eiωtdω, (2.3.4) where ω=2πνis called the angular frequency . The Fourier integral formula implies that any function of time f(t) that has a Fourier transform can be equally speciﬁed by its spectrum. Physically, the signal f(t) is represented as an integral superposition of an inﬁnite number of sinusoidal oscillations with © 2007 by Taylor & Francis Group, LLC 14 INTEGRAL TRANSFORMS and THEIR APPLICATIONS diﬀerent frequencies ωand complex amplitudes1 2πF(ω). Equation (2.3.4) is called the spectral resolution of the signal f(t), andF(ω) 2πis called the spectral density . In summary, the Fourier transform maps a function ( or signal) of time tto a function of frequency ω. In the same way as the Fourier series expansion of a periodic function decomposes the function into harmonic components, the Fourier transform generates a function (or signal) of a continuous variable whose value represents the frequency co ntent of the original signal. This led to the successful use of the Fourier transform to analyze the form of time-varying signals in electrical eng ineering and seismology. Next we give examples of Fourier transforms. Example 2.3.1 Find the Fourier transform of exp( −ax2). In fact, we prove F(k)=F{exp(−ax2)}=1 √ 2aexp⎪parenleftbigg −k2 4a⎪parenrightbigg ,a > 0. (2.3.5) Here we have, by deﬁnition, F(k)=1 √ 2π∞⎪integraldisplay −∞e−ikx−ax2dx =1 √ 2π∞⎪integraldisplay −∞exp⎪bracketleftBigg −a⎪parenleftbigg x+ik 2a⎪parenrightbigg2 −k2 4a⎪bracketrightBigg dx =1 √ 2πexp(−k2/4a)∞⎪integraldisplay −∞e−ay2dy=1 √ 2aexp⎪parenleftbigg −k2 4a⎪parenrightbigg , in which the change of variable y=x+ik 2ais used. The above result is correct, but the change of variable can be justiﬁed by the method of complex analysis because ( ik/2a)i sc o m p l e x .I f a=1 2 F{e−x2/2}=e−k2/2. (2.3.6) This shows F{f(x)}=f(k). Such a function is said to be self-reciprocal un- der the Fourier transformation. Graphs of f(x)=e x p ( −ax2)a n di t sF o u r i e r transform is shown in Figure 2.1 for a=1. Example 2.3.2 Find the Fourier transform of exp( −a|x|), i.e., F{exp(−a|x|)}=⎪radicalbigg 2 π·a (a2+k2),a > 0. (2.3.7) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 15 Figure 2.1 Graphs of f(x)=e x p ( −ax2)a n d F(k)w i t h a=1. Here we can write F⎪braceleftBig e−a|x|⎪bracerightBig =1 √ 2π∞⎪integraldisplay −∞e−a|x|−ikxdx =1 √ 2π⎡ ⎣∞⎪integraldisplay 0e−(a+ik)xdx+0⎪integraldisplay −∞e(a−ik)xdx⎤⎦ =1 √ 2π⎪bracketleftbigg1 a+ik+1 a−ik⎪bracketrightbigg =⎪radicalbigg 2 πa (a2+k2). We note that f(x)=e x p ( −a|x|) decreases rapidly at inﬁnity, it is not diﬀer- entiable at x= 0. Graphs of f(x)=e x p ( −a|x|) and its Fourier transform is displayed in Figure 2.2 for a=1. Example 2.3.3 Find the Fourier transform of f(x)=⎪parenleftbigg 1−|x| a⎪parenrightbigg H⎪parenleftbigg 1−|x| a⎪parenrightbigg , where H(x)i st h e Heaviside unit step function deﬁned by H(x)=⎪braceleftbigg 1,x > 0 0,x < 0⎪bracerightbigg . (2.3.8) Or, more generally, H(x−a)=⎪braceleftbigg 1,x > a 0,x < a⎪bracerightbigg , (2.3.9) © 2007 by Taylor & Francis Group, LLC 16 INTEGRAL TRANSFORMS and THEIR APPLICATIONS - 6 - 4 - 2 024600.20.40.60.81 kF(k) - 6 - 4 - 2 024600.51 xf(x) Figure 2.2 Graphs of f(x)=e x p ( −a|x|)a n d F(k)w i t h a=1. where ais a ﬁxed real number. So the Heaviside function H(x−a) has a ﬁnite discontinuity at x=a. F{f(x)}=1 √ 2πa⎪integraldisplay −ae−ikx⎪parenleftbigg 1−|x| a⎪parenrightbigg dx=2 √ 2πa⎪integraldisplay 0⎪parenleftBig 1−x a⎪parenrightBig coskxdx =2a √ 2π1⎪integraldisplay 0(1−x)cos(akx)dx=2a √ 2π1⎪integraldisplay 0(1−x)d dx⎪parenleftbiggsinakx ak⎪parenrightbigg dx =2a √ 2π1⎪integraldisplay 0sin(akx) akdx=a √ 2π1⎪integraldisplay 0d dx⎡ ⎢⎢⎢⎣sin2⎪parenleftbiggakx 2⎪parenrightbigg ⎪parenleftbiggak 2⎪parenrightbigg2⎤ ⎥⎥⎥⎦dx =a √ 2πsin2⎪parenleftbiggak 2⎪parenrightbigg ⎪parenleftbiggak 2⎪parenrightbigg2. (2.3.10) Example 2.3.4 Find the Fourier transform of the characteristic function χ[−a,a](x), where χ[−a,a](x)=H(a−|x|)=⎪braceleftbigg 1,|x|<a 0,|x|>a⎪bracerightbigg . (2.3.11) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 17 We have Fa(k)=F{χ[−a,a](x)}=1 √ 2π∞⎪integraldisplay −∞e−ikxχ[−a,a](x)dx =1 √ 2πa⎪integraldisplay −ae−ikxdx=⎪radicalbigg 2 π⎪parenleftbiggsinak k⎪parenrightbigg .(2.3.12) Graphs of f(x)=χ[−a,a](x) and its Fourier transform are shown in Figure 2.3 fora=1. -15 -10 -5 0 5 10 1500.20.40.60.8 kFa(k) a -a [-a,a](x) x1 Figure 2.3 Graphs of χ[−a,a](x)a n d Fa(k)w i t h a=1 . 2.4 Fourier Transforms of Generalized Functions The natural way to deﬁne the Fourier transform of a generalized function, is to treat f(x) in (2.3.1) as a generalized function. The advantage of this is that every generalized function has a Fo urier transform and an inverse Fourier transform, and that the ordinary functions whose Fourier transforms are ofinterest form a subset of the generalized functions. We would not go into great detail, but refer to the famous books of Lighthill (1958) and Jones (1982) for © 2007 by Taylor & Francis Group, LLC 18 INTEGRAL TRANSFORMS and THEIR APPLICATIONS the introduction to the subject of generalized functions. Agood function ,g(x) is a function in C∞(R) that decays suﬃciently rapidly thatg(x) and all of its derivatives decay to zero faster than |x|−Nas|x|→∞ for all N>0. DEFINITION 2.4.1 Suppose a real or complex valued function g(x)is deﬁned for all x∈Rand is inﬁnitely diﬀerentiable everywhere, and suppose that each derivative tends to zero as |x|→∞ faster that any positive power of⎪parenleftbig x−1⎪parenrightbig ,or in other words, suppose that for each positive integer Nandn, lim |x|→∞xNg(n)(x)=0, theng(x)is called a good function. Usually, the class of good functions is represented by S. The good functions play an important role in Fourier analysis because the inversion, convolution, and diﬀerentiation theorems as well as many others take simple forms with noproblem of convergence. The rapid decay and inﬁnite diﬀerentiability proper- ties of good functions lead to the fact that the Fourier transform of a good function is also a good function. Good functions also play an important role in the theory of generalized func- tions. A good function of bounded support is a special type of good function that also plays an important part in the theory of generalized functions. Goodfunctions also have the following important properties. The sum (or diﬀerence) of two good functions is also a good function. The product and convolution of two good functions are good functions. The derivative of a good functionis a good function; x ng(x) is a good function for all non-negative integers nwhenever g(x) is a good function. A good function belongs to Lp(a class ofpthpower Lebesgue integrable functions) for every pin 1≤p≤∞.T h e integral of a good function is not necessarily good. However, if φ(x) is a good function, then the function gdeﬁned for all xby g(x)=⎪integraldisplayx −∞φ(t)dt is a good function if and only if⎪integraltext∞ −∞φ(t)dtexists. Good functions are not only continuous, but are also uniformly continuous inRand absolutely continuous in R. However, a good function cannot be necessarily represented by a Taylor seri es expansion in every interval. As an example, consider a good function of bounded support g(x)=⎪braceleftbigg exp[−(1−x2)−1], if|x|<1 0, if|x|≥1⎪bracerightbigg . © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 19 The function gis inﬁnitely diﬀerentiable at x=±1, as it must be in order to be good. It does not have a Taylor series e xpansion in every interval, because a Taylor expansion based on the various derivatives of gfor any point having |x|>1 would lead to zero value for all x. For example, exp( −x2),xexp(−x2),⎪parenleftbig 1+x2⎪parenrightbig−1exp(−x2), and sech2xare good functions, while exp( −|x|) is not diﬀerentiable at x= 0, and the function⎪parenleftbig 1+x2⎪parenrightbig−1is not a good function as it decays too slowly as |x|→∞ . A sequence of good functions, {fn(x)}is called regular if, for any good function g(x), lim n→∞⎪integraldisplay∞ −∞fn(x)g(x)dx (2.4.1) exists. For example, fn(x)=1 nφ(x) is a regular sequence for any good function φ(x), if lim n→∞⎪integraldisplay∞ −∞fn(x)g(x)dx= lim n→∞1 n⎪integraldisplay∞ −∞φ(x)g(x)dx=0. Two regular sequences of good functions are equivalent if, for any good func- tiong(x), the limit (2.4.1) exists and is the same for each sequence. Ageneralized function ,f(x), is a regular sequence of good functions, and two generalized functions are equal if t heir deﬁning sequences are equivalent. Generalized functions are, therefore, only deﬁned in terms of their action onintegrals of good functions if /angbracketleftf, g/angbracketright=⎪integraldisplay ∞ −∞f(x)g(x)dx= lim n→∞⎪integraldisplay∞ −∞fn(x)g(x)dx= lim n→∞/angbracketleftfn,g/angbracketright(2.4.2) for any good function, g(x), where the symbol /angbracketleftf, g/angbracketrightis used to denote the action of the generalized function f(x) on the good function g(x), or/angbracketleftf, g/angbracketright represents the number that fassociates with g.I ff(x) is an ordinary function such that⎪parenleftbig 1+x2⎪parenrightbig−Nf(x)i si n t e g r a b l ei n( −∞,∞)f o rs o m e N, then the generalized function f(x) equivalent to the ordinary function is deﬁned as any sequence of good functions {fn(x)}such that, for any good function g(x), lim n→∞⎪integraldisplay∞ −∞fn(x)g(x)dx=⎪integraldisplay∞ −∞f(x)g(x)dx (2.4.3) For example, the generalized function equivalent to zero can be represented by either of the sequences⎪braceleftBig φ(x) n⎪bracerightBig and⎪braceleftBig φ(x) n2⎪bracerightBig . The unit function, I(x), is deﬁned by ⎪integraldisplay∞ −∞I(x)g(x)dx=⎪integraldisplay∞ −∞g(x)dx (2.4.4) © 2007 by Taylor & Francis Group, LLC 20 INTEGRAL TRANSFORMS and THEIR APPLICATIONS for any good function g(x). A very important and useful good function that deﬁnes the unit function is⎪braceleftBig exp⎪parenleftBig −x2 4n⎪parenrightBig⎪bracerightBig . Thus, the unit function is the gen- eralized function that is equivalent to the ordinary function f(x)=1 . TheHeaviside function ,H(x), is deﬁned by ⎪integraldisplay∞ −∞H(x)g(x)dx=⎪integraldisplay∞ 0g(x)dx. (2.4.5) The generalized function H(x) is equivalent to the ordinary unit function H(x)=⎪braceleftbigg0,x < 0 1,x > 0(2.4.6) since generalized functions are deﬁned through the action on integrals of good functions, the value of H(x)a tx= 0 does not have signiﬁcance here. Thesign function , sgn(x), is deﬁned by ⎪integraldisplay∞ −∞sgn(x)g(x)dx =⎪integraldisplay∞ 0g(x)dx−⎪integraldisplay0 −∞g(x)dx (2.4.7) for any good function g(x). Thus, sgn(x) can be identiﬁed with the ordinary function sgn(x) =⎪braceleftbigg−1,x < 0, +1,x > 0.(2.4.8) In fact, sgn(x) = 2 H(x) −I(x) can be seen as follows: ⎪integraldisplay∞ −∞sgn(x)g(x)dx =⎪integraldisplay∞ −∞[2H(x)−I(x)]g(x)dx =2⎪integraldisplay∞ −∞H(x)g(x)dx−⎪integraldisplay∞ −∞I(x)g(x)dx =2⎪integraldisplay∞ 0g(x)dx−⎪integraldisplay∞ −∞g(x)dx =⎪integraldisplay∞ 0g(x)dx−⎪integraldisplay0 −∞g(x)dx In 1926, Dirac introduced the delta function, δ(x), having the following properties δ(x)=0,x /negationslash=0, (2.4.9)∞⎪integraldisplay −∞δ(x)dx=1. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 21 The Dirac delta function, δ(x) is deﬁned so that for any good function φ(x), ∞⎪integraldisplay −∞δ(x)φ(x)dx=φ(0). There is no ordinary function equivalent to the delta function. The properties (2.4.9) cannot be satisﬁed by any ordinary functions in clas- sical mathematics. Hence, the delta func tion is not a function in the classical sense. However, it can be treated as a function in the generalized sense, and in fact, δ(x) is called a generalized function ordistribution . The concept of the delta function is clear and simple in modern mathematics. It is very useful in physics and engineering. Physically, the delta function represents a point mass, that is a particle of unit mass located at the origin. In this context, it may be called a mass-density function. This leads to the result for a point particle that can be considered as the limit of a sequence of continuous dis- tributions which become more and more concentrated. Even though δ(x)i s not a function in the classical sense, it can be approximated by a sequence ofordinary functions. As an example, we consider the sequence δ n(x)=⎪radicalbigg n πexp(−nx2),n=1,2,3,.... (2.4.10) Clearly, δn(x)→0a sn→∞ for any x/negationslash=0 a n d δn(0)→∞ asn→∞ as shown in Figure 2.4. Also, for all n=1,2,3,..., ∞⎪integraldisplay −∞δn(x)dx=1 and lim n→∞∞⎪integraldisplay −∞δn(x)dx=∞⎪integraldisplay −∞δ(x)dx=1 as expected. So the delta function can be considered as the limit of a sequence of ordinary functions, and we write δ(x) = lim n→∞⎪radicalbigg n πexp(−nx2). (2.4.11) Sometimes, the delta function δ(x) is deﬁned by its fundamental property ∞⎪integraldisplay −∞f(x)δ(x−a)dx=f(a), (2.4.12) © 2007 by Taylor & Francis Group, LLC 22 INTEGRAL TRANSFORMS and THEIR APPLICATIONS -4 -2 0 2 400.20.40.60.81 xn(x) n=4 n=3 n=2 n=1 Figure 2.4 The sequence of delta functions, δn(x). where f(x) is continuous in any interval containing the point x=a. Clearly, ∞⎪integraldisplay −∞f(a)δ(x−a)dx=f(a)∞⎪integraldisplay −∞δ(x−a)dx=f(a). (2.4.13) Thus, (2.4.12) and (2.4.13) lead to the result f(x)δ(x−a)=f(a)δ(x−a). (2.4.14) The following results are also true xδ(x) = 0 (2.4.15) δ(x−a)=δ(a−x). (2.4.16) Result (2.4.16) shows that δ(x) is an even function. Clearly, the result x⎪integraldisplay −∞δ(y)dy=⎪braceleftBigg 1,x > 0 0,x < 0⎪bracerightBigg =H(x) shows thatd dxH(x)=δ(x). (2.4.17) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 23 The Fourier transform of the Dirac delta function is F{δ(x)}=1 √ 2π∞⎪integraldisplay −∞e−ikxδ(x)dx=1 √ 2π. (2.4.18) Hence, δ(x)=F−1⎪braceleftbigg1 √ 2π⎪bracerightbigg =1 2π∞⎪integraldisplay −∞eikxdk. (2.4.19) This is an integral representation of the delta function extensively used in quantum mechanics. Also, (2.4.19) can be rewritten as δ(k)=1 2π∞⎪integraldisplay −∞eikxdx. (2.4.20) The Dirac delta function, δ(x), is deﬁned so that for any good function g(x), /angbracketleftδ, g/angbracketright=⎪integraldisplay∞ −∞δ(x)g(x)dx=g(0). (2.4.21) Derivatives of generalized functions are deﬁned by the derivatives of any equivalent sequences of good function s. We can integrate by parts using any member of the sequences and assuming g(x) vanishes at inﬁnity. We can obtain this deﬁnition as follows: /angbracketleftf/prime,g/angbracketright=⎪integraldisplay∞ −∞f/prime(x)g(x)dx =[f(x)g(x)]∞ −∞−⎪integraldisplay∞ −∞f(x)g/prime(x)dx=−/angbracketleftf, g/prime/angbracketright. The derivative of a generalized function fis the generalized function f/primedeﬁned by /angbracketleftf/prime,g/angbracketright=−/angbracketleftf, g/prime/angbracketright (2.4.22) for any good function g. The diﬀerential calculus of generali zed functions can easily be developed with locally integrable functions. To every locally integrable function f,t h e r e corresponds a generalized function (ordistribution ) deﬁned by /angbracketleftf, φ/angbracketright=⎪integraldisplay∞ −∞f(x)φ(x)dx (2.4.23) where φis a test function in R→Cwith bounded support ( φis inﬁnitely diﬀerentiable with its derivatives of all orders exist and are continuous). © 2007 by Taylor & Francis Group, LLC 24 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The derivative of a generalized function fis the generalized function f/prime deﬁned by /angbracketleftf/prime,φ/angbracketright=−/angbracketleftf, φ/prime/angbracketright (2.4.24) for all test functions φ.This deﬁnition follows from the fact that /angbracketleftf/prime,φ/angbracketright=⎪integraldisplay∞ −∞f/prime(x)φ(x)dx =[f(x)φ(x)]∞ −∞−⎪integraldisplay∞ −∞f(x)φ/prime(x)dx=−/angbracketleftf, φ/prime/angbracketright which was obtained from integration by parts and using the fact that φvan- ishes at inﬁnity. It is easy to check that H/prime(x)=δ(x), for /angbracketleftH/prime,φ/angbracketright=⎪integraldisplay∞ −∞H/prime(x)φ(x)dx=−⎪integraldisplay∞ −∞H(x)φ/prime(x)dx =−⎪integraldisplay∞ 0φ/prime(x)dx=−[φ(x)]∞ 0=φ(0) =/angbracketleftδ, φ/angbracketright. Another result is /angbracketleftδ/prime,φ/angbracketright=−⎪integraldisplay∞ −∞δ(x)φ/prime(x)dx=−φ/prime(0). It is easy to verify f(x)δ(x)=f(0)δ(x). We next deﬁne |x|=xsgn(x) and calculate its derivative as follows. We have d dx|x|=d dx{xsgn(x) }=xd dx{sgn(x) }+s g n ( x )dx dx =xd dx{2H(x)−I(x)}+s g n ( x ) =2xδ(x) + sgn(x) = sgn(x) (2.4.25) which is, by sgn(x) = 2 H(x) −I(x) and xδ(x)=0 . Similarly, we can show that d dx{sgn(x) }=2H/prime(x)=2δ(x). (2.4.26) If we can show that (2.3.1) holds for good functions, it follows that it holds for generalized functions. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 25 THEOREM 2.4.1 The Fourier transform of a good function is a good function. PROOF The Fourier transform of a good function f(x)e x i s t sa n di sg i v e n by F{f(x)}=F(k)=1 √ 2π⎪integraldisplay∞ −∞e−ikxf(x)dx. (2.4.27) Diﬀerentiating F(k)ntimes and integrating Ntimes by parts, we get ⎪vextendsingle⎪vextendsingle⎪vextendsingleF(n)(k)⎪vextendsingle⎪vextendsingle⎪vextendsingle≤⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle(−1)N (−ik)N1 √ 2π⎪integraldisplay∞ −∞e−ikxdN dxN{(−ix)nf(x)}dx⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle ≤1 |k|N1 √ 2π⎪integraldisplay∞ −∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingled N dxN{xnf(x)}⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingledx. Evidently, all derivatives tend to zero as fast as |k| −Nas|k|→∞ for any N>0 and hence, F(k) is a good function. THEOREM 2.4.2 Iff(x) is a good function with the Fourier transform (2.4.27), then the inverse Fourier transform is given by f(x)=1 √ 2π⎪integraldisplay∞ −∞eikxF(k)dk. (2.4.28) PROOF For any /epsilon1>0, we have F⎪braceleftBig e−/epsilon1x2F(−x)⎪bracerightBig =1 2π⎪integraldisplay∞ −∞e−ikx−/epsilon1x2⎪braceleftbigg⎪integraldisplay∞ −∞eixtf(t)dt⎪bracerightbigg dx. Since fis a good function, the order of integration can be interchanged to obtain F⎪braceleftBig e−/epsilon1x2F(−x)⎪bracerightBig =1 2π⎪integraldisplay∞ −∞f(t)dt⎪integraldisplay∞ −∞e−i(k−t)x−/epsilon1x2dx which is, by similar calculation used in Example 2.3.1, =1 √ 4π/epsilon1⎪integraldisplay∞ −∞exp⎪bracketleftbigg −(k−t)2 4/epsilon1⎪bracketrightbigg f(t)dt . Using the fact that 1 √ 4π/epsilon1⎪integraldisplay∞ −∞exp⎪bracketleftbigg −(k−t)2 4/epsilon1⎪bracketrightbigg dt=1, © 2007 by Taylor & Francis Group, LLC 26 INTEGRAL TRANSFORMS and THEIR APPLICATIONS we can write F⎪braceleftBig e−/epsilon1x2F(−x)⎪bracerightBig −f(k).1 =1 √ 4π/epsilon1⎪integraldisplay∞ −∞[f(t)−f(k)] exp⎪bracketleftbigg −(k−t)2 4/epsilon1⎪bracketrightbigg dt.(2.4.29) Since fis a good function, we have ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglef(t)−f(k) t−k⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle≤max x∈R|f/prime(x)|. It follows from (2.4.29) that ⎪vextendsingle⎪vextendsingle⎪vextendsingleF⎪braceleftBig e−/epsilon1x2F(−x)⎪bracerightBig −f(k)⎪vextendsingle⎪vextendsingle⎪vextendsingle ≤1 √ 4π/epsilon1max x∈R|f/prime(x)|⎪integraldisplay∞ −∞|t−k|exp⎪bracketleftbigg −(t−k)2 4/epsilon1⎪bracketrightbigg dt =1 √ 4π/epsilon1max x∈R|f/prime(x)|4/epsilon1⎪integraldisplay∞ −∞|α|e−α2dα→0 as/epsilon1→0, where α=t−k 2√ /epsilon1. Consequently, f(k)=F{F(−x)}=1 √ 2π⎪integraldisplay∞ −∞e−ikxF(−x)dx =1 √ 2π⎪integraldisplay∞ −∞eikxF(x)dx =1 2π⎪integraldisplay∞ −∞eikxdx⎪integraldisplay∞ −∞e−iξxf(ξ)dξ. Interchanging kwithx, this reduces to the Fourier integral formula (2.2.4) and hence, the theorem is proved. Example 2.4.1 The Fourier transform of a constant function cis F{c}=√ 2π.c.δ(k). (2.4.30) In the ordinary sense F{c}=c √ 2π⎪integraldisplay∞ −∞e−ikxdx is not a well deﬁned (divergent) integr al. However, treated as a generalized function, c=cI(x) and we consider⎪braceleftBig exp⎪parenleftBig −x2 4n⎪parenrightBig⎪bracerightBig as an equivalent sequence © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 27 to the unit function, I(x). Thus, F⎪braceleftbigg cexp⎪parenleftbigg −x2 4n⎪parenrightbigg⎪bracerightbigg =c √ 2π⎪integraldisplay∞ −∞exp⎪parenleftbigg −ikx−x2 4n⎪parenrightbigg dx which is, by Example 2.3.1, =c√ 2nexp(−nk2)=√ 2π.c.⎪radicalbigg n πexp(−nk2) =√ 2π.c.δ n(k)=√ 2π.c.δ(k)a s n→∞, since{δn(k)}=⎪braceleftbig⎪radicalbig n πexp⎪parenleftbig −nk2⎪parenrightbig⎪bracerightbig is a sequence equivalent to the delta func- tion deﬁned by (2.4.10). Example 2.4.2 Show that F{e−axH(x)}=1 √ 2π(ik+a),a > 0. (2.4.31) We have, by deﬁnition, F{e−axH(x)}=1 √ 2π∞⎪integraldisplay 0exp{−x(ik+a)}dx=1 √ 2π(ik+a). Example 2.4.3 By considering the function (see Figure 2.5) fa(x)=e−axH(x)−eaxH(−x),a > 0, (2.4.32) ﬁnd the Fourier transform of sgn(x). In Figure 2.5, the vertical axis (y-axis)represents f a(x) and the horizontal axis represents the x-axis. We have, by deﬁnition, F{fa(x)}=−1 √ 2π0⎪integraldisplay −∞exp{(a−ik)x}dx +1 √ 2π∞⎪integraldisplay 0exp{−(a+ik)x}dx =1 √ 2π⎪bracketleftbigg1 a+ik−1 a−ik⎪bracketrightbigg =⎪radicalbigg 2 π·(−ik) a2+k2. © 2007 by Taylor & Francis Group, LLC 28 INTEGRAL TRANSFORMS and THEIR APPLICATIONS -101 xfa(x) Figure 2.5 Graph of the function fa(x). In the limit as a→0,fa(x)→sgn(x) and then F{sgn(x) }=⎪radicalbigg 2 π·1 ik. Or, F⎪braceleftbigg⎪radicalbigg π 2isgn(x)⎪bracerightbigg =1 k. 2.5 Basic Properties of Fourier Transforms THEOREM 2.5.1 IfF{f(x)}=F(k),then © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 29 (a) (Shifting) F{f(x−a)}=e−ikaF{f(x)}, (2.5.1) (b) (Scaling) F{f(ax)}=1 |a|F(k a), (2.5.2) (c) (Conjugate) F{ f(−x)}= F{f(x)}, (2.5.3) (d) (Translation) F{eiaxf(x)}=F(k−a), (2.5.4) (e) (Duality) F{F(x)}=f(−k), (2.5.5) (f) (Composition)∞⎪integraldisplay −∞F(k)g(k)eikxdk=∞⎪integraldisplay −∞f(ξ)G(ξ−x)dξ,(2.5.6) where G(k)=F{g(x)}. PROOF (a) We obtain, from the deﬁnition, F{f(x−a)}=1 √ 2π∞⎪integraldisplay −∞e−ikxf(x−a)dx =1 √ 2π∞⎪integraldisplay −∞e−ik(ξ+a)f(ξ)dξ, (x−a=ξ) =e−ikaF{f(x)}. The proofs of results (b)–(d) follow easily from the deﬁnition of the Fourier transform. We give a proof of the duality (e) and composition (f). We have, by deﬁnition, f(x)=1 √ 2π∞⎪integraldisplay −∞eikxF(k)dk=F−1{F(k)}. Interchanging xandk, and then replacing kby−k,w eo b t a i n f(−k)=1 √ 2π∞⎪integraldisplay −∞e−ikxF(x)dx=F{F(x)}. © 2007 by Taylor & Francis Group, LLC 30 INTEGRAL TRANSFORMS and THEIR APPLICATIONS To prove (f), we have ∞⎪integraldisplay −∞F(k)g(k)eikxdk=∞⎪integraldisplay −∞g(k)eikxdk1 √ 2π∞⎪integraldisplay −∞e−ikξf(ξ)dξ =∞⎪integraldisplay −∞f(ξ)dξ1 √ 2π∞⎪integraldisplay −∞e−ik(ξ−x)g(k)dk =∞⎪integraldisplay −∞f(ξ)G(ξ−x)dξ. In particular, when x=0 ∞⎪integraldisplay −∞F(k)g(k)dk=∞⎪integraldisplay −∞f(ξ)G(ξ)dξ. THEOREM 2.5.2 Iff(x) is piecewise continuously diﬀerenti able and absolutely integrable, then (i)F(k) is bounded for −∞<k< ∞, (ii)F(k) is continuous for −∞<k< ∞. PROOF It follows from the deﬁnition that |F(k)|≤1 √ 2π∞⎪integraldisplay −∞|e−ikx||f(x)|dx =1 √ 2π∞⎪integraldisplay −∞|f(x)|dx=c √ 2π, where c=∞⎪integraltext −∞|f(x)|dx= constant .This proves result (i). To prove (ii), we have |F(k+h)−F(k)|≤1 √ 2π∞⎪integraldisplay −∞|e−ihx−1||f(x)|dx ≤⎪radicalbigg 2 π∞⎪integraldisplay −∞|f(x)|dx. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 31 Since lim h→0|e−ihx−1|=0f o ra l l x∈R,w eo b t a i n lim h→0|F(k+h)−F(k)|≤lim h→01 √ 2π∞⎪integraldisplay −∞|e−ihx−1||f(x)|dx=0. This shows that F(k) is continuous. THEOREM 2.5.3 (Riemann-Lebesgue Lemma ). IfF(k)=F{f(x)},then lim |k|→∞|F(k)|=0. (2.5.7) PROOF Since e−ikx=−e−ikx−iπ,w eh a v e F(k)=−1 √ 2π∞⎪integraldisplay −∞e−ik(x+π k)f(x)dx =−1 √ 2π∞⎪integraldisplay −∞e−ikxf⎪parenleftBig x−π k⎪parenrightBig dx. Hence, F(k)=1 2⎧ ⎨ ⎩1 √ 2π⎡ ⎣∞⎪integraldisplay −∞e−ikxf(x)dx−∞⎪integraldisplay −∞e−ikxf⎪parenleftBig x−π k⎪parenrightBig dx⎤ ⎦⎫ ⎬ ⎭ =1 21 √ 2π∞⎪integraldisplay −∞e−ikx⎪bracketleftBig f(x)−f⎪parenleftBig x−π k⎪parenrightBig⎪bracketrightBig dx. Therefore, |F(k)|≤1 2√ 2π∞⎪integraldisplay −∞⎪vextendsingle⎪vextendsingle⎪vextendsinglef(x)−f⎪parenleftBig x−π k⎪parenrightBig⎪vextendsingle⎪vextendsingle⎪vextendsingledx. Thus, we obtain lim |k|→∞|F(k)|≤1 2√ 2πlim |k|→∞∞⎪integraldisplay −∞⎪vextendsingle⎪vextendsingle⎪vextendsinglef(x)−f⎪parenleftBig x−π k⎪parenrightBig⎪vextendsingle⎪vextendsingle⎪vextendsingledx=0. © 2007 by Taylor & Francis Group, LLC 32 INTEGRAL TRANSFORMS and THEIR APPLICATIONS THEOREM 2.5.4 Iff(x) is continuously diﬀerentiable and f(x)→0a s|x|→∞ ,t h e n F{f/prime(x)}=(ik)F{f(x)}=ik F(k). (2.5.8) PROOF We have, by deﬁnition, F{f/prime(x)}=1 √ 2π∞⎪integraldisplay −∞e−ikxf/prime(x)dx which is, integrating by parts, =1 √ 2π⎪bracketleftbig f(x)e−ikx⎪bracketrightbig∞ −∞+ik √ 2π∞⎪integraldisplay −∞e−ikxf(x)dx =(ik)F(k). Iff(x) is continuously n-times diﬀerentiable and f(k)(x)→0as|x|→∞ for k=1,2,...,(n−1), then the Fourier transform of the nth derivative is F{f(n)(x)}=(ik)nF{f(x)}=(ik)nF(k). (2.5.9) A repeated application of Theorem 2.5.4 to higher derivatives gives the result. The operational results similar to those of (2.5.8) and (2.5.9) hold for partial derivatives of a function of two or more independent variables. For example, ifu(x, t) is a function of space variable xand time variable t,t h e n F⎪braceleftbigg∂u ∂x⎪bracerightbigg =ik U(k,t),F⎪braceleftbigg∂2u ∂x2⎪bracerightbigg =−k2U(k,t), F⎪braceleftbigg∂u ∂t⎪bracerightbigg =dU dt, F⎪braceleftbigg∂2u ∂t2⎪bracerightbigg =d2U dt2, where U(k,t)=F{u(x, t)}. DEFINITION 2.5.1 The convolution of two integrable functions f(x)and g(x), denoted by (f∗g)(x), is deﬁned by (f∗g)(x)=1 √ 2π∞⎪integraldisplay −∞f(x−ξ)g(ξ)dξ, (2.5.10) provided the integral in (2.5.10) exists, where the factor1 √ 2πis a matter of choice. In the study of convolution, this factor is often omitted as this factor © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 33 does not aﬀect the properties of convolution. We will include or exclude the factor1 √ 2πfreely in this book. We give some examples of convolution. Example 2.5.1 Find the convolution of (a)f(x)=c o s xand g(x)=e x p ( −a|x|),a > 0, (b)f(x)=χ[a,b](x)a n d g(x)=x2, where χ[a,b](x) is the characteristic function of the interval [ a, b]⊆Rdeﬁned by χ[a,b](x)=⎪braceleftBigg 1,a≤x≤b 0,otherwise⎪bracerightBigg . (a) We have, by deﬁnition, (f∗g)(x)=∞⎪integraldisplay −∞f(x−ξ)g(ξ)dξ=∞⎪integraldisplay −∞cos(x−ξ)e−a|ξ|dξ =0⎪integraldisplay −∞cos(x−ξ)eaξdξ+∞⎪integraldisplay 0cos(x−ξ)e−aξdξ =∞⎪integraldisplay 0cos(x+ξ)e−aξdξ+∞⎪integraldisplay 0cos(x−ξ)e−aξdξ =2c o s x∞⎪integraldisplay 0cosξe−aξdξ=2acosx (1 +a2). (b) We have (f∗g)(x)=∞⎪integraldisplay −∞f(x−ξ)g(ξ)dξ=∞⎪integraldisplay −∞χ[a,b](x−ξ)g(ξ)dξ =⎪integraldisplayb aξ2dξ=1 3⎪parenleftbig b3−a3⎪parenrightbig . THEOREM 2.5.5 (Convolution Theorem ). IfF{f(x)}=F(k)a n dF{g(x)}=G(k), then F{f(x)∗g(x)}=F(k)G(k), (2.5.11) © 2007 by Taylor & Francis Group, LLC 34 INTEGRAL TRANSFORMS and THEIR APPLICATIONS or, f(x)∗g(x)=F−1{F(k)G(k)}, (2.5.12) or, equivalently, ∞⎪integraldisplay −∞f(x−ξ)g(ξ)dξ=∞⎪integraldisplay −∞eikxF(k)G(k)dk. (2.5.13) PROOF We have, by the deﬁnition of the Fourier transform, F{f(x)∗g(x)}=1 2π∞⎪integraldisplay −∞e−ikxdx∞⎪integraldisplay −∞f(x−ξ)g(ξ)dξ =1 2π∞⎪integraldisplay −∞e−ikξg(ξ)dξ∞⎪integraldisplay −∞e−ik(x−ξ)f(x−ξ)dx =1 2π∞⎪integraldisplay −∞e−ikξg(ξ)dξ∞⎪integraldisplay −∞e−ikηf(η)dη=G(k)F(k), where, in this proof, the factor1 √ 2πis included in the deﬁnition of the convo- lution. This completes the proof. The convolution has the following algebraic properties: f∗g=g∗f(Commutative) , (2.5.14) f∗(g∗h)=(f∗g)∗h(Associative) , (2.5.15) (αf+βg)∗h=α(f∗h)+β(g∗h) (Distributive) , (2.5.16) f∗√ 2πδ=f=√ 2πδ∗f(Identity) , (2.5.17) where αandβare constants. We give proofs of (2.5.15) and (2.5.16). If f∗(g∗h) exists, then [f∗(g∗h)] (x)=∞⎪integraldisplay −∞f(x−ξ)(g∗h)(ξ)dξ =∞⎪integraldisplay −∞f(x−ξ)∞⎪integraldisplay −∞g(ξ−t)h(t)dt dξ © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 35 =∞⎪integraldisplay −∞⎡ ⎣∞⎪integraldisplay −∞f(x−ξ)g(ξ−t)dξ⎤ ⎦h(t)dt =∞⎪integraldisplay −∞⎡ ⎣∞⎪integraldisplay −∞f(x−t−η)g(η)dη⎤ ⎦h(t)dt(putξ−t=η) =∞⎪integraldisplay −∞[(f∗g)(x−t)]h(t)dt =[ (f∗g)∗h](x), where, in the above proof, under suitable assumptions, the interchange of the order of integration can be justiﬁed. Similarly, we prove (2.5.16) using the right-hand side of (2.5.16), that is, α(f∗h)+β(g∗h)=α∞⎪integraldisplay −∞f(x−ξ)h(ξ)dξ+β∞⎪integraldisplay −∞g(x−ξ)h(ξ)dξ =∞⎪integraldisplay −∞[αf(x−ξ)+βg(x−ξ)]h(ξ)dξ =[ (αf+βg)∗h](x). In view of the commutative property of the convolution, (2.5.13) can be writ- ten as∞⎪integraldisplay −∞f(ξ)g(x−ξ)dξ=∞⎪integraldisplay −∞eikxF(k)G(k)dk. (2.5.18) This is valid for all real x, and hence, putting x=0g i v e s ∞⎪integraldisplay −∞f(ξ)g(−ξ)dξ=∞⎪integraldisplay −∞f(x)g(−x)dx=∞⎪integraldisplay −∞F(k)G(k)dk. (2.5.19) We substitute g(x)= f(−x)t oo b t a i n G(k)=F{g(x)}=F⎪braceleftBig f(−x)⎪bracerightBig = F{f(x)}= F(k). Evidently, (2.5.19) becomes ∞⎪integraldisplay −∞f(x) f(x)dx=∞⎪integraldisplay −∞F(k) F(k)dk (2.5.20) © 2007 by Taylor & Francis Group, LLC 36 INTEGRAL TRANSFORMS and THEIR APPLICATIONS or, ∞⎪integraldisplay −∞|f(x)|2dx=∞⎪integraldisplay −∞|F(k)|2dk. (2.5.21) This is well known as Parseval’s relation. For square integrable functions f(x)a n d g(x), the inner product /angbracketleftf, g/angbracketrightis deﬁned by /angbracketleftf, g/angbracketright=∞⎪integraldisplay −∞f(x) g(x)dx (2.5.22) so the norm /bardblf/bardbl2is deﬁned by /bardblf/bardbl2 2=/angbracketleftf, f/angbracketright=∞⎪integraldisplay −∞f(x) f(x)dx=∞⎪integraldisplay −∞|f(x)|2dx. (2.5.23) The function space L2(R) of all complex-valued Lebesgue square integrable functions with the inner product deﬁned by (2.5.22) is a complete normed space with the norm (2.5.23). In terms of the norm, the Parseval relationtakes the form /bardblf/bardbl 2=/bardblF/bardbl2=/bardblFf/bardbl2. (2.5.24) This means that the Fourier transform action is unitary . Physically, the quan- tity/bardblf/bardbl2is a measure of energy and /bardblF/bardbl2represents the power spectrum of f. THEOREM 2.5.6 (General Parseval’s Relation ). IfF{f(x)}=F(k)a n dF{g(x)}=G(k)t h e n ∞⎪integraldisplay −∞f(x) g(x)dx=∞⎪integraldisplay −∞F(k) G(k)dk. (2.5.25) PROOF We proceed formally to obtain ∞⎪integraldisplay −∞F(k) G(k)dk=∞⎪integraldisplay −∞dk·1 2π∞⎪integraldisplay −∞e−ikyf(y)dy ∞⎪integraldisplay −∞e−ikxg(x)dx =1 2π∞⎪integraldisplay −∞f(y)dy∞⎪integraldisplay −∞ g(x)dx∞⎪integraldisplay −∞eik(x−y)dk =∞⎪integraldisplay −∞ g(x)dx∞⎪integraldisplay −∞δ(x−y)f(y)dy=∞⎪integraldisplay −∞f(x) g(x)dx. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 37 In particular, when g(x)=f(x), the above result agrees with (2.5.20). We now use an indirect method to obtain the Fourier transform of sgn(x), that is, F{sgn(x) }=⎪radicalbigg 2 π1 ik. (2.5.26) From (2.4.26), we ﬁnd F⎪braceleftbiggd dxsgn(x)⎪bracerightbigg =F{2H/prime(x)}=2F{δ(x)}=⎪radicalbigg 2 π, which is, by (2.5.8), ikF{sgn(x) }=⎪radicalbigg 2 π, or F{sgn(x) }=⎪radicalbigg 2 π·1 ik. The Fourier transform of H(x) follows from (2.4.30) and (2.5.26): F{H(x)}=1 2F{1+s g n ( x ) }=1 2[F{1}+F{sgn(x) }] =⎪radicalbigg π 2⎪bracketleftbigg δ(k)+1 iπk⎪bracketrightbigg . (2.5.27) 2.6 Poisson’s Summation Formula A class of functions designated as Lp(R) is of great importance in the theory of Fourier transformations, where p(≥1) is any real number. We denote the vector space of all complex-valued functions f(x) of the real variable x.I ff is a locally integrable function such that |f|p∈L(R), then we say fisp-th power Lebesgue integrable. The set of all such functions is written Lp(R). The number ||f||pis called the Lp-norm of fand is deﬁned by ||f||p=⎪bracketleftbigg⎪integraldisplay∞ −∞|f(x)|pdx⎪bracketrightbigg1 p <∞. (2.6.1) Suppose fis a Lebesgue integrable function on R. Since exp( −ikx)i sc o n t i n - uous and bounded, the product exp( −ikx)f(x) is locally integrable for any k∈R.A l s o , |exp(−ikx)|≤1 for all kandxonR. Consider the inner product ⎪angbracketleftbig f, eikx⎪angbracketrightbig =⎪integraldisplay∞ −∞f(x)e−ikxdx, k ∈R. (2.6.2) © 2007 by Taylor & Francis Group, LLC 38 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Clearly, ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪integraldisplay ∞ −∞f(x)e−ikxdx⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle≤⎪integraldisplay ∞ −∞|f(x)|dx=||f||1<∞. (2.6.3) This means that integral (2.6.2) exists for all k∈R, and was used to deﬁne the Fourier transform, F(k)=F{f(x)}without the factor1 √ 2π. Although the theory of Fourier series is a very important subject, a detailed study is beyond the scope of this book. Without rigorous analysis, we can establish a simple relation between the Fourier transform of functions in L1(R) and the Fourier series of related periodic functions in L1(−a, a)o fp e r i o d2 a. Iff(x)∈L1(−a, a) and is deﬁned by f(x)=∞⎪summationdisplay n=−∞cneinx,(−a≤x≤a), (2.6.4) where the Fourier coeﬃcients cnis given by cn=1 2a⎪integraldisplaya −af(x)e−ikxdx. (2.6.5) THEOREM 2.6.1 Iff(x)∈L1(R), then the series ∞⎪summationdisplay n=−∞f(x+2na) (2.6.6) converges absolutely for almost all xin (−a, a) and its sum g(x)∈L1(−a, a) withg(x+2a)=g(x)f o rx∈R. Ifandenotes the Fourier coeﬃcient of a function g,t h e n an=1 2a⎪integraldisplaya −ag(x)e−inxdx=1 2a⎪integraldisplay∞ −∞f(x)e−inxdx=1 2aF(n). PROOF We have ∞⎪summationdisplay n=−∞⎪integraldisplaya −a|f(x+2na)|dx= lim N→∞N⎪summationdisplay n=−N⎪integraldisplaya −a|f(x+2na)|dx = lim N→∞N⎪summationdisplay n=−N⎪integraldisplay(2n+1)a (2n−1)a|f(t)|dt = lim N→∞⎪integraldisplay(2N+1)a −(2N+1)a|f(t)|dt =⎪integraldisplay∞ −∞|f(t)|dt <∞. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 39 It follows from Lebesgue’s theorem on monotone convergence that ⎪integraldisplaya −a⎪bracketleftBigg∞⎪summationdisplay n=−∞|f(x+2na)|⎪bracketrightBigg dx=∞⎪summationdisplay n=−∞⎪integraldisplaya −a|f(x+2na)|dx < ∞. Hence, the series⎪summationtext∞ n=−∞f(x+2na) converges absolutely for almost all x in (−a, a). IfgN(x)=⎪summationtextN n=−Nf(x+2na), lim N→∞gN(x)=g(x), where g∈ L1(−a, a), and g(x+2a)=g(x). Moreover, ||g||1=⎪integraldisplaya −a|g(x)|dx=⎪integraldisplaya −a⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle∞⎪summationdisplay n=−∞f(x+2na)⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingledx ≤⎪integraldisplay a −a∞⎪summationdisplay n=−∞|f(x+2na)|dx =∞⎪summationdisplay n=−∞⎪integraldisplaya −a|f(x+2na)|dx =⎪integraldisplay∞ −∞|f(x)|dx=||f||1. We consider the Fourier series of g(x)g i v e nb y g(x)=∞⎪summationdisplay m=−∞cmexp(imπx/a ), (2.6.7) where the coeﬃcients cmform=0,±1,±2, ...are given by cm=1 2aa⎪integraldisplay −ag(x)exp(−imπx/a )dx. (2.6.8) We replace g(x) by the limit of the sum g(x) = lim N→∞N⎪summationdisplay n=−Nf(x+2na), (2.6.9) © 2007 by Taylor & Francis Group, LLC 40 INTEGRAL TRANSFORMS and THEIR APPLICATIONS so that (2.6.8) reduces to cm=1 2alim N→∞N⎪summationdisplay n=−Na⎪integraldisplay −af(x+2na)exp(−imπx/a )dx =1 2alim N→∞N⎪summationdisplay n=−N(2n+1)a⎪integraldisplay (2n−1)af(y)exp(−imπy/a )dy =1 2alim N→∞(2N+1)a⎪integraldisplay −(2N+1)af(x)exp(−imπx/a )dx =√ 2π 2aF⎪parenleftBigmπ a⎪parenrightBig , (2.6.10) where F⎪parenleftbigmπ a⎪parenrightbig is the discrete Fourier transform of f(x). Evidently, ∞⎪summationdisplay n=−∞f(x+2na)=g(x)=∞⎪summationdisplay n=−∞√ 2π 2aF⎪parenleftBignπ a⎪parenrightBig exp(inπx/a ).(2.6.11) We let x= 0 in (2.6.11) to obtain the Poisson summation formula ∞⎪summationdisplay n=−∞f(2na)=∞⎪summationdisplay n=−∞√ 2π 2aF⎪parenleftBignπ a⎪parenrightBig . (2.6.12) When a=π, this formula becomes ∞⎪summationdisplay n=−∞f(2πn)=1 √ 2π∞⎪summationdisplay n=−∞F(n). (2.6.13) When 2 a= 1, formula (2.6.12) becomes ∞⎪summationdisplay n=−∞f(n)=√ 2π∞⎪summationdisplay n=−∞F(2nπ). (2.6.14) To obtain a more general formula, we assume that ais a given positive constant, and write g(x)=f(ax) for all x.T h e n f⎪parenleftbigg a.2πn a⎪parenrightbigg =g⎪parenleftbigg2πn a⎪parenrightbigg , © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 41 and we deﬁne the Fourier transform of f(x) without the factor1 √ 2πso that F(n)=⎪integraldisplay∞ −∞e−inxf(x)dx=⎪integraldisplay∞ −∞e−inxf⎪parenleftBig a.x a⎪parenrightBig dx =⎪integraldisplay∞ −∞e−inxg⎪parenleftBigx a⎪parenrightBig dx =a⎪integraldisplay∞ −∞e−i(an)yg(y)dy =aG(an). Consequently, equality (2.6.13) reduces to ∞⎪summationdisplay n=−∞g(2πn a)=a √ 2π∞⎪summationdisplay n=−∞G(an). (2.6.15) Putting b=2π ain (2.6.15) gives ∞⎪summationdisplay n=−∞g(bn)=√ 2πb−1∞⎪summationdisplay n=−∞G(2πb−1n). (2.6.16) When b=2π, result (2.6.16) becomes (2.6.13). We apply these formulas to prove the following series (a)∞⎪summationdisplay n=−∞1 (n2+b2)=π bcoth(πb), (2.6.17) (b)∞⎪summationdisplay n=−∞exp(−πn2t)=1 √ t∞⎪summationdisplay n=−∞exp⎪parenleftbigg −πn2 t⎪parenrightbigg , (2.6.18) (c)∞⎪summationdisplay n=−∞1 (x+nπ)2=c o s e c2(x). (2.6.19) To prove (a), we write f(x)=(x2+b2)−1so that F(k)=⎪radicalbig π 21 bexp(−b|k|). We now use (2.6.14) to derive ∞⎪summationdisplay n=−∞1 (n2+b2)=π b∞⎪summationdisplay n=−∞exp(−2|n|πb) =π b⎪bracketleftBigg∞⎪summationdisplay n=0exp(−2nπb)+∞⎪summationdisplay n=1exp(2nπb)⎪bracketrightBigg which is, by writing r=e x p ( −2πb), =π b⎪bracketleftBigg∞⎪summationdisplay n=0rn+∞⎪summationdisplay n=1⎪parenleftbigg1 r⎪parenrightbiggn⎪bracketrightBigg =π b⎪parenleftbiggr 1−r+1 1−r⎪parenrightbigg =π b⎪parenleftbigg1+r 1−r⎪parenrightbigg =π bcoth(πb). © 2007 by Taylor & Francis Group, LLC 42 INTEGRAL TRANSFORMS and THEIR APPLICATIONS It follows from (2.6.14) that ∞⎪summationdisplay n=−∞1 (n2+b2)=π b⎪parenleftbig 1+e−2πb⎪parenrightbig (1−e−2πb). Or, 2∞⎪summationdisplay n=11 (n2+b2)+1 b2=π b⎪parenleftbig 1+e−2πb⎪parenrightbig (1−e−2πb). It turns out that ∞⎪summationdisplay n=11 (n2+b2)=π 2b⎪bracketleftBigg⎪parenleftbig 1+e−2πb⎪parenrightbig (1−e−2πb)−1 πb⎪bracketrightBigg =π2 x⎪bracketleftbigg(1 +e−x) (1−e−x)−2 x⎪bracketrightbigg , (2πb=x) =π2 x2⎪bracketleftbiggx(1 +e−x)−2( 1−e−x) (1−e−x)⎪bracketrightbigg =⎪parenleftBigπ x⎪parenrightBig2⎪bracketleftBigg x3⎪parenleftbig1 2−1 3⎪parenrightbig −x4 12+.... x−x2 2!+x3 3!−....⎪bracketrightBigg . In the limit as b→0(x→0), we obtain the well-known result ∞⎪summationdisplay n=11 n2=π2 6. (2.6.20) To prove (b), we assume f(x)=e x p ( −πtx2)s ot h a t F(k)=1 √ 2πtexp⎪parenleftBig −k2 4πt⎪parenrightBig . Thus, the Poisson formula (2.6.14) gives ∞⎪summationdisplay n=−∞exp(−πtn2)=1 √ t∞⎪summationdisplay n=−∞exp(−πn2/t). This identity plays an important role in number theory and in the theory of elliptic functions. The Jacobi theta function Θ(s) is deﬁned by Θ(s)=∞⎪summationdisplay n=−∞exp(−πsn2),s > 0, (2.6.21) so that (2.6.16) gives the functional equation for the theta function √ sΘ(s)=Θ⎪parenleftbigg1 s⎪parenrightbigg . (2.6.22) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 43 The theta function Θ( s) also extends to complex values of swhen Re(s)>0 and the functional equation is still valid for complex s. The theta function is closely related to the Riemann zeta function ζ(s) deﬁned for Re(s)>1b y ζ(s)=∞⎪summationdisplay n=11 ns. (2.6.23) An integral representation of ζ(s) can be found from the result ⎪integraldisplay∞ 0xs−1e−nxdx=Γ(s) ns,R e (s)>0, where the gamma function Γ(s) is deﬁned by Γ(s)=⎪integraldisplay∞ 0e−tts−1dt, Re (s)>0. Summing both sides of this result and interchanging the order of summation and integration, which is permissible for Re(s)>1, gives Γ(s)ζ(s)=⎪integraldisplay∞ 0xs−1dx ex−1,R e (s)>1. (2.6.24) It turns out that ζ(s), Θ(s), and Γ( s) are related by the following identity: ζ(s)Γ(s/2) =1 2πs/2⎪integraldisplay∞ 0xs/2−1[Θ(x)−1]dx, Re (s)>1.(2.6.25) Considering the complex integra l in a suitable closed contour C I=1 2πi⎪integraldisplay Czs−1 e−z−1dz, and using the Cauchy residue theorem with all zeros of ( e−z−1) atz=2πin, n=±1,±2, ...,±Ngives I=−2s i n⎪parenleftBigπs 2⎪parenrightBig∞⎪summationdisplay n=1(2πn)s−1. To prove (c), we use the Fourier transform of the function f(x)=( 1 −|x|) H(1−|x|) to obtain the result. In the limit as N→∞, the sum of the residues is convergent so that the int egral gives the relation 2sπs−1sin⎪parenleftBigπs 2⎪parenrightBig ζ(1−s)=ζ(s) Γ(1−s). (2.6.26) In view of another relation for the gamma function, Γ(1 + z)Γ(−z)=−π sinπz, the relation (2.6.26) leads to a famous functional relation for ζ(s)i nt h ef o r m πsζ(1−s)=21−sΓ(s)cos⎪parenleftBigπs 2⎪parenrightBig ζ(s). (2.6.27) © 2007 by Taylor & Francis Group, LLC 44 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 2.7 The Shannon Sampling Theorem An analog signal f(t) is a continuous function of time tdeﬁned in −∞<t<∞, with the exception of perhaps a countable number of jump discontinuities. Almost all analog signals f(t) of interest in engineerin g have ﬁnite energy. By this we mean that f∈L2(−∞,∞). The norm of fdeﬁned by ||f||=⎪bracketleftbigg⎪integraldisplay∞ −∞|f(x)|2dx⎪bracketrightbigg1 2 (2.7.1) represents the square root of the total energy content of the signal f(t). The spectrum of a signal f(t) is represented by its Fourier transform F(ω), where ωis called the frequency . The frequency is measured by ν=ω 2πin terms of Hertz. A continuous signal f(t) is called band limited if its Fourier transform F(ω) is zero except in a ﬁnite interval, that is, if Fa(ω)=0 f o r |ω|>a . (2.7.2) Then a(>0) is called the cutoﬀ frequency . In particular, if F(ω)=⎪braceleftbigg1,|ω|≤a 0,|ω|>a⎪bracerightbigg (2.7.3) thenF(ω) is called a gate function and is denoted by Fa(ω), and the band limited signal is denoted by fa(t). Ifais the smallest value for which (2.7.2) holds, it is called the bandwidth of the signal. Even if an analog signal f(t)i s not band-limited, we can reduce it to a band-limited signal by what is called anideal low-pass ﬁltering . To reduce f(t) to a band-limited signal fa(t)w i t h bandwidth less than or equal to a,w ec o n s i d e r Fa(ω)=⎪braceleftbigg F(ω),|ω|≤a 0,|ω|>a⎪bracerightbigg (2.7.4) and ﬁnd the low-pass ﬁlter function fa(t) by the inverse Fourier transform fa(t)=1 2π⎪integraldisplay∞ −∞eiωtFa(ω)dω=1 2π⎪integraldisplaya −aeiωtFa(ω)dω. (2.7.5) This function fa(t) is called the Shannon sampling function .W h e n a=π, fπ(t) is called the Shannon scaling function . The band-limited signal fa(t)i s given by fa(t)=1 2π∞⎪integraldisplay −∞F(ω)eiωtdω=1 2πa⎪integraldisplay −aeiωtdω=sinat πt. (2.7.6) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 45 -4 -2 0 2 400.20.40.6 tfa(t) aF() -a1 Figure 2.6 The gate function and its Fourier transform. BothF(ω)a n d fa(t) are shown in Figure 2.6 for a=2 . Consider the limit as a→∞ of the Fourier integral for −∞<ω< ∞ 1 = lim a→∞∞⎪integraldisplay −∞e−iωtfa(t)dt= lim a→∞∞⎪integraldisplay −∞e−iωtsinat πtdt =∞⎪integraldisplay −∞e−iωt⎪bracketleftbigg lim a→∞sinat πt⎪bracketrightbigg dt=∞⎪integraldisplay −∞e−iωtδ(t)dt. Clearly, the delta function δ(t) can be thought of as the limit of the sequence of functions fa(t). More precisely, δ(t) = lim a→∞⎪parenleftbiggsinat πt⎪parenrightbigg . (2.7.7) We next consider the band-limited signal fa(t)=1 2πa⎪integraldisplay −aF(ω)eiωtdω=1 2π∞⎪integraldisplay −∞F(ω)Fa(ω)eiωtdω, which is, by the Convolution Theorem, fa(t)=∞⎪integraldisplay −∞f(τ)fa(t−τ)dτ=∞⎪integraldisplay −∞sina(t−τ) π(t−τ)f(τ)dτ. (2.7.8) This integral represents the sampling integral representation of the band- limited signal fa(t). © 2007 by Taylor & Francis Group, LLC 46 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 2.7.1 (Synthesis and Resolution of a Signal; Physical Interpretation of Convolu- tion). In electrical engineering problems, a time-dependent electric, optical or electromagnetic pulse is usually called a signal. Such a signal can be consid- ered as a superposition of plane waves of all real frequencies so that it can be represented by the inverse Fourier transform f(t)=F−1{F(ω)}=1 2π∞⎪integraldisplay −∞F(ω)eiωtdω, (2.7.9) where F(ω)=F{f(t)},t h ef a c t o r( 1 /2π) is introduced because the angular frequency ωis related to linear frequency νbyω=2πν, and negative fre- quencies are introduced for mathemati cal convenience so that we can avoid dealing with the cosine and sine functions separately. Clearly, F(ω)c a nb e represented by the Fourier transform of the signal f(t)a s F(ω)=∞⎪integraldisplay −∞f(t)e−iωtdt. (2.7.10) This represents the resolution of the signal into its angular frequency compo- nents, and (2.7.9) gives a synthesis of the signal from its individual compo- nents. Consider a simple electrical device such as an ampliﬁer with an input signal f(t), and an output signal g(t). For an input of a single frequency ω,f(t)= eiωt. The amplifer will change the amplitude and may also change the phase so that the output can be expressed in terms of the input, the amplitude and the phase modifying function Φ( ω)a s g(t)=Φ ( ω)f(t), (2.7.11) where Φ( ω) is usually known as the transfer function and is, in general, a complex function of the real variable ω. This function is generally independent of the presence or absence of any other frequency components. Thus, the totaloutput may be found by integrating over the entire input as modiﬁed by the ampliﬁer g(t)=1 2π∞⎪integraldisplay −∞Φ(ω)F(ω)eiωtdω. (2.7.12) Thus, the total output signal can readily be calculated from any given input signal f(t). On the other hand, the transfer function Φ( ω) is obviously charac- teristic of the ampliﬁer device and can, in general, be obtained as the Fouriertransform of some function φ(t)s ot h a t Φ(ω)=∞⎪integraldisplay −∞φ(t)e−iωtdt. (2.7.13) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 47 The Convolution Theorem 2.5.5 allows us to rewrite (2.7.12) as g(t)=F−1{Φ(ω)F(ω)}=f(t)∗φ(t)=∞⎪integraldisplay −∞f(τ)φ(t−τ)dτ. (2.7.14) Physically, this result represents an output signal g(t) as the integral superpo- sition of an input signal f(t)m o d i ﬁ e db y φ(t−τ). Linear translation invariant systems, such as sensors andﬁlters, are modeled by the convolution equations g(t)=f(t)∗φ(t), where φ(t) is the system impulse response function. In fact (2.7.14) is the most general mathematical representation of an output (eﬀect)function in terms of an input (cause) function modiﬁed by the ampliﬁer where tis the time variable. Assuming the principle of causality, that is, every eﬀect has a cause, we must require τ<t. The principle of causality is imposed by requiring φ(t−τ)=0 w h e n τ>t . (2.7.15) Consequently, (2.7.14) gives g(t)=t⎪integraldisplay −∞f(τ)φ(t−τ)dτ. (2.7.16) In order to determine the signiﬁcance of φ(t), we use an impulse function f(τ)=δ(τ) so that (2.7.16) becomes g(t)=t⎪integraldisplay −∞δ(τ)φ(t−τ)dτ=φ(t)H(t). (2.7.17) This recognizes φ(t) as the output corresponding to a unit impulse at t=0 , and the Fourier transform of φ(t)i s Φ(ω)=F{φ(t)}=∞⎪integraldisplay 0φ(t)e−iωtdt, (2.7.18) withφ(t)=0f o r t<0. Example 2.7.2 (The Series Sampling Expansion of a Bandlimited Signal ). Consider a band- limited signal fa(t) with Fourier transform F(ω)=0f o r |ω|>a .We write the Fourier series expansion of F(ω)o nt h ei n t e r v a l −a<ω<a in terms of the orthogonal set of functions⎪braceleftbig exp⎪parenleftbig −inπω a⎪parenrightbig⎪bracerightbig in the form F(ω)=∞⎪summationdisplay n=−∞anexp⎪parenleftbigg −inπ aω⎪parenrightbigg , (2.7.19) © 2007 by Taylor & Francis Group, LLC 48 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where the Fourier coeﬃcients anare given by an=1 2aa⎪integraldisplay −aF(ω)exp⎪parenleftbigginπ aω⎪parenrightbigg dω=1 2afa⎪parenleftBignπ a⎪parenrightBig . (2.7.20) Thus, the Fourier series expansion (2.7.19) becomes F(ω)=1 2a∞⎪summationdisplay n=−∞fa⎪parenleftBignπ a⎪parenrightBig exp⎪parenleftbigg −inπ aω⎪parenrightbigg . (2.7.21) The signal function fa(t) is obtained by multiplying (2.7.21) by eiωtand in- tegrating over ( −a, a)s ot h a t fa(t)=a⎪integraldisplay −aF(ω)eiωtdω =1 2aa⎪integraldisplay −aeiωtdω⎪bracketleftBigg∞⎪summationdisplay n=−∞fa⎪parenleftBignπ a⎪parenrightBig exp⎪parenleftbigg −inπ aω⎪parenrightbigg⎪bracketrightBigg =1 2a∞⎪summationdisplay n=−∞fa⎪parenleftBignπ a⎪parenrightBiga⎪integraldisplay −aexp⎪bracketleftBig iω⎪parenleftBig t−nπ a⎪parenrightBig⎪bracketrightBig dω =∞⎪summationdisplay n=−∞fa⎪parenleftBignπ a⎪parenrightBigsina⎪parenleftbig t−nπ a⎪parenrightbig a⎪parenleftbig t−nπ a⎪parenrightbig =∞⎪summationdisplay n=−∞fa⎪parenleftBignπ a⎪parenrightBigsin(at−nπ) (at−nπ). (2.7.22) This result is the main content of the sampling theorem. It simply states that a band-limited signal fa(t) can be reconstructed from the inﬁnite set of dis- crete samples of fa(t)a tt=0,±π a,.... . In practice, a discrete set of samples is useful in the sense that most systems receive discrete samples {f(tn)}as an input. The sampling theorem can be realized physically. Modern telephoneequipment employs sampling to send messages over wires. In fact, it seems that sampling is audible on some transoceanic cable calls. Result (2.7.22) can be obtained from the convolution theorem by using discrete input samples ∞⎪summationdisplay n=−∞π afa⎪parenleftBignπ a⎪parenrightBig δ⎪parenleftBig t−nπ a⎪parenrightBig =f(t). (2.7.23) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 49 Hence, the sampling expansion (2.7.8) gives the band-limited signal fa(t)=∞⎪integraldisplay −∞sina(t−τ) π(t−τ)⎪bracketleftBigg∞⎪summationdisplay n=−∞π afa⎪parenleftBignπ a⎪parenrightBig δ⎪parenleftBig τ−nπ a⎪parenrightBig⎪bracketrightBigg dτ =∞⎪summationdisplay n=−∞fa⎪parenleftBignπ a⎪parenrightBig∞⎪integraldisplay −∞sina(t−τ) a(t−τ)δ⎪parenleftBig τ−nπ a⎪parenrightBig dτ =∞⎪summationdisplay n=−∞fa⎪parenleftBignπ a⎪parenrightBigsina⎪parenleftbig t−nπ a⎪parenrightbig a⎪parenleftbig t−nπ a⎪parenrightbig. (2.7.24) In general, the output can be best descr ibed by taking the Fourier transform of (2.7.14) so that G(ω)=F(ω)Φ(ω), (2.7.25) where Φ( ω) is called the transfer function of the system. Thus, the output can be calculated from (2.7.25) by the Fourier inversion formula g(t)=1 2π⎪integraldisplay∞ −∞F(ω)Φ(ω)eiωtdω, (2.7.26) Obviously, the transfer function Φ( ω) is a characteristic of a linear system. A linear system is a ﬁlterif it possesses signals of certain frequencies and attenuates others. If the transfer function Φ(ω)=0 |ω|≥ω0, (2.7.27) thenφ(t), the Fourier inverse of Φ( ω), is called a low-pass ﬁlter . On the other hand, if the transfer function Φ(ω)=0 |ω|≤ω1, (2.7.28) thenφ(t)i sahigh-pass ﬁlter .Abandpass ﬁlter possesses a band ω0≤|ω|≤ω1. It is often convenient to express the system transfer function Φ( ω)i nt h e complex form Φ(ω)=A(ω)exp [−iθ(ω)], (2.7.29) where A(ω) is called the amplitude andθ(ω) is called the phase of the transfer function. Obviously, the system impulse response φ(t) is given by the inverse Fourier transform φ(t)=1 2π⎪integraldisplay∞ −∞A(ω)exp [i{ωt−θ(ω)}]dω. (2.7.30) © 2007 by Taylor & Francis Group, LLC 50 INTEGRAL TRANSFORMS and THEIR APPLICATIONS For a unit step function as the input f(t)=H(t), we have F(ω)=ˆH(ω)=⎪parenleftbigg πδ(ω)+1 iω⎪parenrightbigg , where ˆH(ω)=F{H(t)}and the associated output g(t)i st h e ng i v e nb y g(t)=1 2π⎪integraldisplay∞ −∞Φ(ω)ˆH(ω)eiωtdω =1 2π⎪integraldisplay∞ −∞⎪parenleftbigg πδ(ω)+1 iω⎪parenrightbigg A(ω)exp [i{ωt−θ(ω)}]dω =1 2A(0) +1 2π⎪integraldisplay∞ −∞A(ω) ωexp⎪bracketleftBig i⎪braceleftBig ωt−θ(ω)−π 2⎪bracerightBig⎪bracketrightBig dω .(2.7.31) We next give another characterization of a ﬁlter in terms of the amplitude of the transfer function. A ﬁlter is called distortionless if its output g(t) to an arbitrary input f(t) has the same form as the input, that is, g(t)=A0f(t−t0). (2.7.32) Evidently, G(ω)=A0e−iωt0F(ω)=Φ ( ω)F(ω) where Φ(ω)=A0e−iωt0 represents the transfer function of the distortionless ﬁlter. It has a constant amplitude A0and a linear phase shift θ(ω)=ωt0. However, in general, the amplitude A(ω) of a transfer function is not con- stant, and the phase θ(ω) is not a linear function. A ﬁlter with constant amplitude, |θ(ω)|=A0is called an all-pass ﬁlter .I t follows from Parseval’s formula that the energy of the output of such a ﬁlter is proportional to the energy of its input. A ﬁlter whose amplitude is constant for |ω|<ω0and zero for |ω|>ω0is called an ideal low-pass ﬁlter . More explicitly, the amplitude is given by A(ω)=A0ˆH(ω0−|ω|)=A0ˆχω0(ω), (2.7.33) where ˆ χω0(ω) is a rectangular pulse. So, the transfer function of the low-pass ﬁlter is Φ(ω)=A0ˆχω0(ω)exp (−iωt0). (2.7.34) Finally, the ideal high-pass ﬁlter is characterized by its amplitude given by A(ω)=A0ˆH(|ω|−ω0)=A0ˆχω0(ω), (2.7.35) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 51 where A0is a constant. Its transfer function is given by Φ(ω)=A0[1−ˆχω0(ω)] exp( −iωt0). (2.7.36) Example 2.7.3 (Bandwidth and Bandwidth Equation ). The Fourier spectrum of a signal (or waveform) gives an indication of the fre quencies that exist during the total duration of the signal (or waveform ). From the knowledge of the frequencies that are present, we can calculate the average frequency and the spread aboutthat average. In particular, if the signal is represented by f(t), we can deﬁne its Fourier spectrum by F(ν)=⎪integraldisplay ∞ −∞e−2πiνtf(t)dt. (2.7.37) Using |F(ν)|2for the density in frequency, the average frequency is denoted by<ν> a n dd e ﬁ n e db y <ν>=⎪integraldisplay∞ −∞ν|F(ν)|2dν. (2.7.38) The bandwidth is then the root mean square (RMS) deviation at about the average, that is, B2=⎪integraldisplay∞ −∞(ν−<ν>)2dν. (2.7.39) Expressing the signal in terms of its amplitude and phase f(t)=a(t)e x p{iθt}, (2.7.40) the instantaneous frequency, ν(t) is the frequency at a particular time deﬁned by ν(t)=1 2πθ/prime(t). (2.7.41) Substituting (2.7.37) and (2.7.40) into (2.7.38) gives <ν>=1 2π⎪integraldisplay∞ −∞θ/prime(t)a2(t)dt=⎪integraldisplay∞ −∞ν(t)a2(t)dt. (2.7.42) This formula states that the average frequency is the average value of the in- stantaneous frequency weighted by the square of the amplitude of the signal. We next derive the bandwidth equation in terms of the amplitude and phase of the signal in the form B2=1 (2π)2⎪integraldisplay∞ −∞⎪bracketleftbigga/prime(t) a(t)⎪bracketrightbigg2 a2(t)dt+⎪integraldisplay∞ −∞⎪bracketleftbigg1 2πθ/prime(t)−<ν>⎪bracketrightbigg2 a2(t)dt. (2.7.43) © 2007 by Taylor & Francis Group, LLC 52 INTEGRAL TRANSFORMS and THEIR APPLICATIONS A straightforward but lengthy way to derive it is to substitute (2.7.40) into (2.7.39) and simplify. However, we give an elegant derivation of (2.7.43) byrepresenting the frequency by the operator ν=1 2πid dt. (2.7.44) We calculate the average by sandwiching the operator between the complex conjugate of the signal and the signal. Thus, <ν> =⎪integraldisplay∞ −∞ν|F(ν)|2dν=⎪integraldisplay∞ −∞¯f(t)⎪bracketleftbigg1 2πid dt⎪bracketrightbigg f(t)dt =1 2π⎪integraldisplay∞ −∞a(t){−ia/prime(t)+a(t)θ/prime(t)}dt =1 2π⎪integraldisplay∞ −∞−1 2i⎪bracketleftbiggd dta2(t)⎪bracketrightbigg dt+1 2π⎪integraldisplay∞ −∞a2(t)θ/prime(t)dt(2.7.45) =1 2π⎪integraldisplay∞ −∞θ/prime(t)a2(t)dt (2.7.46) provided the ﬁrst integral in (2.7.44) vanishes if a(t)→0a s|t|→∞ . It follows from the deﬁnition (2.7.39) of the bandwidth that B2=⎪integraldisplay∞ −∞(ν−<ν>)2|F(ν)|2dν =⎪integraldisplay∞ −∞¯f(t)⎪bracketleftbigg1 2πid dt−<ν>⎪bracketrightbigg2 f(t)dt =⎪integraldisplay∞ −∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪bracketleftbigg1 2πid dt−<ν>⎪bracketrightbigg f(t)⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle2 dt =⎪integraldisplay∞ −∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle1 2πia/prime(t) a(t)+1 2πθ/prime(t)−<ν>⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle2 a2(t)dt =1 4π2⎪integraldisplay∞ −∞⎪bracketleftbigga/prime(t) a(t)⎪bracketrightbigg2 a2(t)dt+⎪integraldisplay∞ −∞⎪bracketleftbigg1 2πθ/prime(t)−<ν>⎪bracketrightbigg2 a2(t)dt. This completes the derivation. Physically, the second term in equation (2.7.43) gives averages of all of the deviations of the instantaneous frequency from the average frequency. In elec- trical engineering literature, the spread of frequency about the instantaneous frequency , which is deﬁned as an average of the frequencies that exist at a particular time, is called instantaneous bandwidth ,g i v e nb y σ2 ν/t=1 (2π)2⎪bracketleftbigga/prime(t) a(t)⎪bracketrightbigg2 . (2.7.47) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 53 In the case of a chirp with a Gaussian envelope f(t)=⎪parenleftBigα π⎪parenrightBig1 4exp⎪bracketleftbigg −1 2αt2+1 2iβαt2+2πiν0t⎪bracketrightbigg , (2.7.48) where its Fourier spectrum is given by F(ν)=(απ)1 4⎪parenleftbigg1 α−iβ⎪parenrightbigg1 2 exp⎪bracketleftbig −2π2(ν−ν0)2/(α−iβ)⎪bracketrightbig . (2.7.49) Theenergy density spectrum of the signal is |F(ν)|2=2⎪parenleftbiggαπ α2+β2⎪parenrightbigg1 2 exp⎪bracketleftbigg −4απ2(ν−ν0)2 α2+β2⎪bracketrightbigg . (2.7.50) Finally, the average frequency <ν> and the bandwidth square are respec- tively given by <ν>=ν0and B2=1 8π2⎪parenleftbigg α+β2 α⎪parenrightbigg . (2.7.51) A large bandwidth can be achieved in two very qualitatively diﬀerent ways. The amplitude modulation can be made large by taking αlarge, and the frequency modulation can be small by letting β→0. It is possible to make the frequency modulation large by making βlarge and αvery small. These two extreme situations are physically very diﬀerent even though they produce the same bandwidth. Example 2.7.4 Find the transfer function and the corresponding i m p u l s er e s p o n s ef u n c t i o n of the RLC circuit governed by the diﬀerential equation Ld2q dt2+Rdq dt+1 Cq=e(t) (2.7.52) where q(t) is the charge, R,L,Care constants, and e(t) is the given voltage (input). Equation (2.7.25) provides the deﬁnition of the transfer function in the frequency domain Φ(ω)=G(ω) F(ω)=F{g(t)} F{f(t)}, (2.7.53) where φ(t)=F−1{Φ(ω)}is called the impulse response function . Taking the Fourier transfrom of (2.7.52) gives ⎪parenleftbigg −Lω2+Riω+1 C⎪parenrightbigg Q(ω)=E(ω). (2.7.54) © 2007 by Taylor & Francis Group, LLC 54 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thus, the transfer function is Φ(ω)=Q(ω) E(ω)=−C LCω2−iRCω −1 =i 2Lβ⎪bracketleftbigg1 ω−i(α+β)−1 ω−i(α−β)⎪bracketrightbigg , (2.7.55) where α=R 2Land β=⎪bracketleftBigg⎪parenleftbiggR 2L⎪parenrightbigg2 −1 LC⎪bracketrightBigg1 2 . (2.7.56) The inverse Fourier transform of (2.7.55) yields the impulse response func- tion φ(t)=1 2βL⎪parenleftbig eβt−e−βt⎪parenrightbig e−αtH(t). (2.7.57) 2.8 Gibbs’ Phenomenon We now examine the so-called the Gibbs jump phenomenon which deals with the limiting behavior of a band-limited signal fω0(t) represented by the sam- pling integral representation (2.7.8) at a point of discontinuity of f(t).This phenomenon reveals the intrinsic overshoot near a jump discontinuity of a function associated with the Fourier se r i e s .M o r ep r e c i s e l y ,t h ep a r t i a ls u m s of the Fourier series overshoot the function near the discontinuity, and the overshoot continues no matter how many terms are taken in the partial sum. However, the Gibbs phenomenon does not occur if the partial sums are re-placed by the Cesaro means, the average of the partial sums. In order to demonstrate the Gibbs phenomenon, we rewrite (2.7.8) in the convolution form f ω0(t)=⎪integraldisplay∞ −∞f(τ)sinω0(t−τ) π(t−τ)dτ=(f∗δω0)(t), (2.8.1) where δω0(t)=sinω0t πt. (2.8.2) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 55 Clearly, at every point of continuity of f(t),we have lim ω0→∞fω0(t) = lim ω0→∞(f∗δω0)(t) = lim ω0→∞⎪integraldisplay∞ −∞f(τ)sinω0(t−τ) π(t−τ)dτ =⎪integraldisplay∞ −∞f(τ)⎪bracketleftbigg lim ω0→∞sinω0(t−τ) π(t−τ)⎪bracketrightbigg dτ =⎪integraldisplay∞ −∞f(τ)δ(t−τ)dτ=f(t). (2.8.3) We now consider the limiting behavior of fω0(t)a tt h ep o i n to fd i s c o n t i n u i t y t=t0. To simplify the calculation, we set t0= 0 so that we can write f(t)a s a sum of a continuous function, fc(t) and a suitable step function f(t)=fc(t)+[f(0+)−f(0−)]H(t). (2.8.4) Replacing f(t) by the right hand side of (2.8.4) in Equation (2.8.1) yields fω0(t)=⎪integraldisplay∞ −∞fc(τ)sinω0(t−τ) π(t−τ)dτ +[f(0+)−f(0−)]⎪integraldisplay∞ −∞H(τ)sinω0(t−τ) π(t−τ)dτ =fc(t)+[f(0+)−f(0−)]Hω0(t), (2.8.5) where Hω0(t)=⎪integraldisplay∞ −∞H(τ)sinω0(t−τ) π(t−τ)dτ=⎪integraldisplay∞ 0sinω0(t−τ) π(t−τ)dτ =⎪integraldisplayω0t −∞sinx πxdx (putting ω0(t−τ)=x) =⎪parenleftbigg⎪integraldisplay0 −∞+⎪integraldisplayω0t 0⎪parenrightbigg⎪parenleftbiggsinx πx⎪parenrightbigg dx=⎪parenleftbigg⎪integraldisplay∞ 0+⎪integraldisplayω0t 0⎪parenrightbigg⎪parenleftbiggsinx πx⎪parenrightbigg dx =1 2+1 πsi(ω0t), (2.8.6) and the function si(t) is deﬁned by si(t)=⎪integraldisplayt 0sinx xdx. (2.8.7) Note that Hω0⎪parenleftbiggπ ω0⎪parenrightbigg =1 2+⎪integraldisplayπ 0sinx πxdx >1,H ω0⎪parenleftbigg −π ω0⎪parenrightbigg =1 2−⎪integraldisplayπ 0sinx πxdx <0. C l e a r l y ,f o raﬁ x e d ω0,1 πsi(ω0t) attains its maximum at t=π ω0in (0,∞)a n d minimum at t=−π ω0, since for a larger tthe integrand oscillates with decreas- ing amplitudes. The function Hω0(t) is shown in Figure 2.7 since Hω0(0) =1 2 © 2007 by Taylor & Francis Group, LLC 56 INTEGRAL TRANSFORMS and THEIR APPLICATIONS andfc(0) =f(0−)a n d fω0(0) = fc(0) +1 2[f(0+)−f(0−)] =1 2[f(0+) + f(0−)]. 0.51 tH 0(t) -4 0-3 0-2 0- 0002 03 04 0 Figure 2.7 Graph of Hω0(t). Thus, the graph of Hω0(t) shows that as ω0increases, the time scale changes, and the ripples remain the same. In the limit ω0→∞,the convergence of Hω0(t)=(H∗δω0)(t)t oH(t) exhibits the intrinsic overshoot leading to the classical Gibbs phenomenon. Example 2.8.1 (The Square Wave Function and the Gibbs Phenomenon ). Consider the single- pulse square function deﬁned by f(x)=⎧ ⎨ ⎩1,−a<x<a 1 2,x =±a 0, |x|>a⎫ ⎬ ⎭. The graph of f(x) is given in Figure 2.8. Thus, F(k)=F{f(x)}=⎪radicalbigg 2 π⎪parenleftbiggsinak k⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 57 a -a f(x) x1 0 Figure 2.8 T h es q u a r ew a v ef u n c t i o n . We next deﬁne a function fλ(x) by the integral fλ(x)=⎪integraldisplayλ −λF(k)eikxdk . As|λ|→∞ ,fλ(x) will tend pointwise to f(x) for all x.Convergence occurs even at x=±abecause the function f(x) is deﬁned to have a value “half way up the step” at these points. Let us examine the behavior of fλ(x)a s|λ|→∞ in a region just one side of one of the discontinuities, that is, for x∈(0,a).For aﬁ x e d λ,the diﬀerence, fλ(x)−f(x),oscillates above and below the value 0a sx→a,attaining a maximum positive value at some point, say x=xλ. Then the quantity fλ(xλ)−f(xλ) is called the overshoot . As|λ|→∞ ,so the period of the oscillations tends to zero and so also xλ→a; however, the value of the overshoot fλ(xλ)−f(xλ) does not tend to zero but instead tends to a ﬁnite limit. The existence of this non-zero, ﬁnite, limitingv a l u ef o rt h eo v e r s h o o ti sk n o w na st h e Gibbs phenomenon . This phenomenon also occurs in an almost identical manner in the Fourier synthesis of periodic functions using Fourier series. 2.9 Heisenberg’s Uncertainty Principle Iff∈L2(R), then fandF(k)=F{f(x)}cannot both be essentially localized. In other words, it is not possible that the widths of the graphs of |f(x)|2and |F(k)|2can both be made arbitrarily small. This fact underlines the Heisen- berg uncertainty principle in quantum mechanics and the bandwidth theorem in signal analysis. If |f(x)|2and|F(k)|2are interpreted as weighting functions, © 2007 by Taylor & Francis Group, LLC 58 INTEGRAL TRANSFORMS and THEIR APPLICATIONS then the weighted means (averages) <x> and<k> ofxandkare given by <x> =1 ||f||2 2⎪integraldisplay∞ −∞x|f(x)|2dx, (2.9.1) <k> =1 ||F||2 2⎪integraldisplay∞ −∞k|F(k)|2dk. (2.9.2) Corresponding measures of the widths of these weight functions are given by the second moments about the respectiv e means. Usually, it is convenient to deﬁne widths /trianglexand/trianglekby (/trianglex)2=1 ||f||2 2⎪integraldisplay∞ −∞(x−<x>)2|f(x)|2dx, (2.9.3) (/trianglek)2=1 ||F||2 2⎪integraldisplay∞ −∞(k−<k>)2|F(k)|2dk. (2.9.4) T h ee s s e n c eo ft h e Heisenberg principle and the bandwidth theorems lies in the fact that the product ( /trianglex)(/trianglek) will never less than1 2. Indeed, (/trianglex)(/trianglek)≥1 2, (2.9.5) where equality in (2.9.5) holds only if f(x) is a Gaussian function given by f(x)=Cexp(−ax2),a>0. We next state the Heisenberg inequality theorem as follows: THEOREM 2.9.1 (Heisenberg Inequality ). Iff(x),xf(x)a n dkF(k)b e l o n gt o L2(R)a n d√ x|f(x) |→0a s|x|→∞ ,then (/trianglex)2(/trianglek)2≥1 4, (2.9.6) where ( /trianglex)2and (/trianglek)2are deﬁned by (2.9.3) and (2.9.4) respectively. Equal- ity in (2.9.6) holds only if f(x)i saGaussian function given by f(x)=Ce−ax2, a>0. PROOF If the averages are <x> and<k>, then the average location of exp(−i<k>x )f(x+<x>) is zero. Hence, it is suﬃcient to prove the theorem around the zero mean values, that is, <x> =<k> =0. Since ||f||2=||F||2, we have ||f||4 2(/trianglex)2(/trianglek)2=⎪integraldisplay∞ −∞|xf(x)|2dx⎪integraldisplay∞ −∞|kF(k)|2dk. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 59 Using ikF(k)=F{f/prime(x)}and the Parseval formula ||f/prime(x)||2=||ikF(k)||2,w e obtain ||f||4 2(/trianglex)2(/trianglek)2=⎪integraldisplay∞ −∞|xf(x)|2dx⎪integraldisplay∞ −∞|f/prime(x)|2dx ≥⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪integraldisplay ∞ −∞⎪braceleftBig xf(x) f/prime(x)⎪bracerightBig dx⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle2 ,(see Debnath(2002)) ≥⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪integraldisplay ∞ −∞x.1 2⎪braceleftBig f/prime(x) f(x)+ f/prime(x)f(x)⎪bracerightBig⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle2 =1 4⎪bracketleftbigg⎪integraldisplay∞ −∞x⎪parenleftbiggd dx|f|2⎪parenrightbigg dx⎪bracketrightbigg2 =1 4⎪braceleftbigg⎪bracketleftbig x|f(x)|2⎪bracketrightbig∞ −∞−⎪integraldisplay∞ −∞|f|2dx⎪bracerightbigg2 =1 4||f||4 2. in which√ xf(x)→0a s|x|→∞ was used to eliminate the integrated term. This completes the proof. If we assume f/prime(x) is proportional to xf(x), that is, f/prime(x)=bxf(x), where b is a constant of proportionality, this leads to the Gaussian signals f(x)=Cexp(−ax2), where Cis a constant of integration and a=−b 2>0. In 1924, Heisenberg ﬁrst formulated the uncertainty principle between the position and momentum in quantum mechanics. This principle has an impor-tant interpretation as an uncertainty of both the position and momentum of a particle described by a wave function ψ∈L 2(R). In other words, it is not possible to determine the position and momentum of a particle exactly and simultaneously. In signal processing, time and frequency concentrations of energy of a signal fare also governed by the Heisenberg un certainty principle. The average or expectation values of time tand frequency ω, are respectively deﬁned by <t>=1 ||f||2 2⎪integraldisplay∞ −∞t|f(t)|2dt, < ω > =1 ||F||22⎪integraldisplay∞ −∞ω|F(ω)|2dω,(2.9.7) where the energy of a signal f(t) is well localized in time, and its Fourier transform F(ω) has an energy concentrated in a small frequency domain. The variances around these average values are given respectively by σ2 t=1 ||f||2 2⎪integraldisplay∞ −∞(t−<t>)2|f(t)|2dt, (2.9.8) σ2 ω=1 2π||F||2 2⎪integraldisplay∞ −∞(ω−<ω> )2|F(ω)|2dω. © 2007 by Taylor & Francis Group, LLC 60 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Remarks: 1. In a time-frequency analysis of signals, the measure of the resolution of a signal fin the time or frequency domain is given by σtandσω. Then, the joint resolution is given by the product ( σt)(σω)w h i c hi s governed by the Heisenberg uncerta inty principle. In other words, the product ( σt)(σω) cannot be arbitrarily small and is always greater than the minimum value1 2which is attained for the Gaussian signal. 2. In many applications in science and engineering, signals with a high con- centration of energy in the time an d frequency domains are of special interest. The uncertainty principle can also be interpreted as a mea-sure of this concentratio n of the second moment of f 2(t) and its energy spectrum F2(ω). 2.10 Applications of Fourier Transforms to Ordinary Diﬀerential Equations We consider the nth order linear ordinary diﬀer ential equation with constant coeﬃcients Ly(x)=f(x), (2.10.1) where Lis the nth order diﬀerential operator given by L≡anDn+an−1Dn−1+···+a1D+a0, (2.10.2) where an,an−1,...,a 1,a0are constants, D≡d dxandf(x) is a given function. Application of the Fourier transform to both sides of (2.10.1) gives [an(ik)n+an−1(ik)n−1+···+a1(ik)+a0]Y(k)=F(k), whereF{y(x)}=Y(k)a n dF{f(x)}=F(k). Or, equivalently P(ik)Y(k)=F(k), where P(z)=n⎪summationdisplay r=0arzr. Thus, Y(k)=F(k) P(ik)=F(k)Q(k), (2.10.3) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 61 where Q(k)=1 P(ik). Applying the Convolution Theorem 2.5.5 to (2.10.3) gives the formal solu- tion y(x)=F−1{F(k)Q(k)}=1 √ 2π∞⎪integraldisplay −∞f(ξ)q(x−ξ)dξ, (2.10.4) provided q(x)=F−1{Q(k)}is known explicitly. In order to give a physical interpretation of the solution (2.10.4), we consider the diﬀerential equation with a suddenly applied impulse function f(x)=δ(x) so that L{G(x)}=δ(x). (2.10.5) The solution of this equation can be written from the inversion of (2.10.3) in the form G(x)=F−1⎪braceleftbigg1 √ 2πQ(k)⎪bracerightbigg =1 √ 2πq(x). (2.10.6) Thus, the solution (2.10.4) takes the form y(x)=∞⎪integraldisplay −∞f(ξ)G(x−ξ)dξ. (2.10.7) Clearly, G(x) behaves like a Green’s function , that is, it is the response to a u- nit impulse . In any physical system, f(x) usually represents the input function , while y(x)i sr e f e r r e dt oa st h e output obtained by the superposition principle. The Fourier transform of {√ 2πG(x)}=q(x) is called the admittance .I no r d e r to ﬁnd the reponse to a given input, we determine the Fourier transform of the input function, multiply the result by the admittance, and then apply the inverse Fourier transform to the product so obtained. We illustrate these ideas by solving a simple problem in the electrical circuit theory. Example 2.10.1 (Electric Current in a Simple Circuit ). The current I(t)i nas i m p l ec i r c u i t containing the resistance Rand inductance Lsatisﬁes the equation LdI dt+RI=E(t), (2.10.8) where E(t) is the applied electromagnetic force and RandLare constants. With E(t)=E0exp(−a|t|),we use the Fourier transform with respect to timetto obtain (ikL+R)ˆI(k)=E0⎪radicalbigg 2 πa (a2+k2). © 2007 by Taylor & Francis Group, LLC 62 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Or, ˆI(k)=aE0 iL⎪radicalbigg 2 π1 ⎪parenleftbig k−Ri L⎪parenrightbig (k2+a2), whereF{I(t)}=ˆI(k). The inverse Fourier transform gives I(t)=aE0 iπL∞⎪integraldisplay −∞exp(ikt)dk ⎪parenleftbig k−Ri L⎪parenrightbig (k2+a2). (2.10.9) This integral can be evaluated by the Cauchy Residue Theorem. For t>0 I(t)=aE0 iπL·2πi⎪bracketleftbigg Residue at k=Ri L+ Residue at k=ia⎪bracketrightbigg =2aE0 L⎪bracketleftBigg e−R Lt ⎪parenleftbig a2−R2 L2⎪parenrightbig−e−at 2a⎪parenleftbig a−R L⎪parenrightbig⎪bracketrightBigg =E0⎪bracketleftBigg e−at R−aL−2aLe−R Lt R2−a2L2⎪bracketrightBigg . (2.10.10) Similarly, for t<0, the Residue Theorem gives I(t)=−aE0 iπL·2πi[Residue at k=−ia] =−2aE0 L⎪bracketleftbigg−Leat (aL+R)2a⎪bracketrightbigg =E0eat (aL+R). (2.10.11) Att= 0, the current is continuous and therefore, I(0) = lim t→0I(t)=E0 R+aL. IfE(t)=δ(t), then ˆE(k)=1 √ 2πand the solution is obtained by using the inverse Fourier transform I(t)=1 2πiL∞⎪integraldisplay −∞eikt k−iR Ldk, which is, by the Theorem of Residues, =1 L[Residue at k=iR/L] =1 Lexp⎪parenleftbigg −Rt L⎪parenrightbigg . (2.10.12) Thus, the current tends to zero as t→∞ as expected. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 63 Example 2.10.2 Find the solution of the ordinary diﬀerential equation −d2u dx2+a2u=f(x),−∞<x< ∞ (2.10.13) by the Fourier transform method. Application of the Fourier transform to (2.10.13) gives U(k)=F(k) k2+a2. This can readily be inverted by the Convolution Theorem 2.5.5 to obtain u(x)=1 √ 2π∞⎪integraldisplay −∞f(ξ)g(x−ξ)dξ, (2.10.14) where g(x)=F−1⎪braceleftBig 1 k2+a2⎪bracerightBig =1 a⎪radicalbig π 2exp(−a|x|) by Example 2.3.2. Thus, the ﬁnal solution is u(x)=1 2a∞⎪integraldisplay −∞f(ξ)e−a|x−ξ|dξ. (2.10.15) Example 2.10.3 (The Bernoulli-Euler Beam Equation ). We consider the v ertical deﬂection u(x) of an inﬁnite beam on an elastic foundation under the action of a pre- scribed vertical load W(x).The deﬂection u(x) satisﬁes the ordinary diﬀer- ential equation EId4u dx4+κu=W(x),−∞<x< ∞. (2.10.16) where EIis the ﬂexural rigidity and κis the foundation modulus of the beam. We ﬁnd the solution assuming that W(x) has a compact support and u,u/prime,u/prime/prime,u/prime/prime/primeall tend to zero as |x|→∞ . We ﬁrst rewrite (2.10.16) as d4u dx4+a4u=w(x) (2.10.17) where a4=κ/EI andw(x)=W(x)/EI.Use of the Fourier transform to (2.10.17) gives U(k)=W(k) k4+a4. © 2007 by Taylor & Francis Group, LLC 64 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The inverse Fourier transform gives the solution u(x)=1 √ 2π∞⎪integraldisplay −∞W(k) k4+a4eikxdk =1 2π∞⎪integraldisplay −∞eikx k4+a4dk∞⎪integraldisplay −∞w(ξ)e−ikξdξ =∞⎪integraldisplay −∞w(ξ)G(ξ,x)dξ, (2.10.18) where G(ξ,x)=1 2π∞⎪integraldisplay −∞eik(x−ξ) k4+a4dk=1 π∞⎪integraldisplay 0cosk(x−ξ)dk k4+a4. (2.10.19) The integral can be evaluated by the Theorem of Residues or by using the table of Fourier integrals. We simply state the result G(ξ,x)=1 2a3exp⎪parenleftbigg −a √ 2|x−ξ|⎪parenrightbigg sin⎪bracketleftbigga(x−ξ) √ 2+π 4⎪bracketrightbigg . (2.10.20) In particular, we ﬁnd the explicit solution due to a concentrated load of unit strength acting at some point x0,that is, w(x)=δ(x−x0).Then the solution for this case becomes u(x)=∞⎪integraldisplay −∞δ(ξ−x0)G(x, ξ)dξ=G(x, x0). (2.10.21) Thus, the kernel G(x, ξ) involved in the solution (2.10.18) has the physical signiﬁcance of being the deﬂection, as a function of x,due to a unit point load acting at ξ.Thus, the deﬂection due to a point load of strength w(ξ)dξatξis w(ξ)dξ·G(x, ξ),and hence, (2.10.18) represents the superposition of all such incremental deﬂections. The reader is referred to a more general dynamic problem of an inﬁnite Bernoulli-Euler beam with damping and elastic foundation that has beensolved by Stadler and Shreeves (1970), and also by Sheehan and Debnath (1972). These authors used the Fourier-Laplace transform method to deter- mine the steady state and the transient solutions of the beam problem. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 65 2.11 Solutions of Integral Equations The method of Fourier transforms can be used to solve simple integral equa- tions of the convolution type. We illustrate the method by examples. We ﬁrst solve the Fredholm integral equation with convolution kernel in the form∞⎪integraldisplay −∞f(t)g(x−t)dt+λf(x)=u(x), (2.11.1) where g(x)a n d u(x) are given functions and λis a known parameter. Application of the Fourier transform to (2.11.1) gives √ 2πF(k)G(k)+λF(k)=U(k). Or, F(k)=U(k) √ 2πG(k)+λ. (2.11.2) The inverse Fourier transform leads to a formal solution f(x)=1 √ 2π∞⎪integraldisplay −∞U(k)eikxdk √ 2πG(k)+λ. (2.11.3) In particular, if g(x)=1 xso that G(k)=−i⎪radicalbigg π 2sgnk, then the solution becomes f(x)=1 √ 2π∞⎪integraldisplay −∞U(k)eikxdk λ−iπsgnk. (2.11.4) Ifλ=1 a n d g(x)=1 2⎪parenleftBig x |x|⎪parenrightBig so that G(k)=1 √ 2π1 (ik),solution (2.11.3) reduces to the form f(x)=1 √ 2π∞⎪integraldisplay −∞(ik)U(k)eikxdk (1 +ik) =1 √ 2π∞⎪integraldisplay −∞F{u/prime(x)}F{√ 2πe−x}eikxdk =u/prime(x)∗√ 2πe−x=∞⎪integraldisplay −∞u/prime(ξ)exp (ξ−x)dξ. (2.11.5) © 2007 by Taylor & Francis Group, LLC 66 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 2.11.1 Find the solution of the integral equation ∞⎪integraldisplay −∞f(x−ξ)f(ξ)dξ=1 x2+a2. (2.11.6) Application of the Fourier transform gives √ 2πF(k)F(k)=⎪radicalbigg π 2e−a|k| a. Or, F(k)=1 √ 2aexp⎪braceleftbigg −1 2a|k|⎪bracerightbigg . (2.11.7) The inverse Fourier transform gives the solution f(x)=1 √ 2π1 √ 2a∞⎪integraldisplay −∞exp⎪parenleftbigg ikx−1 2a|k|⎪parenrightbigg dk =1 2√ πa⎡ ⎣∞⎪integraldisplay 0exp⎪braceleftBig −k⎪parenleftBiga 2+ix⎪parenrightBig⎪bracerightBig dk+∞⎪integraldisplay 0exp⎪braceleftBig −k⎪parenleftBiga 2−ix⎪parenrightBig⎪bracerightBig dk⎤ ⎦ =1 2√ πa⎪bracketleftbigg4a (4x2+a2)⎪bracketrightbigg =⎪radicalbigg a π·2 (4x2+a2). Example 2.11.2 Solve the integral equation ∞⎪integraldisplay −∞f(t)dt (x−t)2+a2=1 (x2+b2),b >a > 0. (2.11.8) Taking the Fourier transform, we obtain √ 2πF(k)F⎪braceleftbigg1 x2+a2⎪bracerightbigg =⎪radicalbigg π 2e−b|k| b, or, √ 2πF(k)⎪radicalbigg π 2·e−a|k| a=⎪radicalbigg π 2e−b|k| b. Thus, F(k)=1 √ 2π⎪parenleftBiga b⎪parenrightBig exp{−|k|(b−a)}. (2.11.9) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 67 The inverse Fourier transform leads to the solution f(x)=a 2πb∞⎪integraldisplay −∞exp[ikx−|k|(b−a)]dk =a 2πb⎡ ⎣∞⎪integraldisplay 0exp[−k{(b−a)+ix}]dk+∞⎪integraldisplay 0exp[−k{(b−a)−ix}]⎤ ⎦dk =a 2πb⎪bracketleftbigg1 (b−a)+ix+1 (b−a)−ix⎪bracketrightbigg =⎪parenleftBiga πb⎪parenrightBig(b−a) (b−a)2+x2. (2.11.10) Example 2.11.3 Solve the integral equation f(t)+4∞⎪integraldisplay −∞e−a|x−t|f(t)dt=g(x). (2.11.11) Application of the Fourier transform gives F(k)+4√ 2πF(k)·2a √ 2π(a2+k2)=G(k) F(k)=(a2+k2) a2+k2+8aG(k). (2.11.12) The inverse Fourier transform gives f(x)=1 √ 2π∞⎪integraldisplay −∞(a2+k2)G(k) a2+k2+8aeikxdk. (2.11.13) In particular, if a=1a n d g(x)=e−|x|so that G(k)=⎪radicalBig 2 π1 1+k2, then solution (2.11.13) becomes f(x)=1 π∞⎪integraldisplay −∞eikx k2+32dk. (2.11.14) Forx>0, we use a semicircular closed contour in the lower half of the complex plane to evaluate (2.11.14). It turns out that f(x)=1 3e−3x. (2.11.15) © 2007 by Taylor & Francis Group, LLC 68 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Similarly, for x<0, a semicircular closed contour in the upper half of the complex plane is used to evaluate (2.11.14) so that f(x)=1 3e3x,x < 0. (2.11.16) Thus, the ﬁnal solution is f(x)=1 3exp(−3|x|). (2.11.17) 2.12 Solutions of Partial Diﬀerential Equations In this section we illustrate how the Fourier transform method can be used to obtain the solution of boundary value and initial value problems for linear partial diﬀerential equa tions of diﬀerent kinds. Example 2.12.1 (Dirichlet’s Problem in the Half-Plane ). We consider the solution of the Laplace equation in the half-plane uxx+uyy=0,−∞<x< ∞,y≥0, (2.12.1) with the boundary conditions u(x,0) =f(x),−∞<x< ∞, (2.12.2) u(x, y)→0a s|x|→∞ ,y→∞. (2.12.3) We introduce the Fourier transform with respect to x U(k,y)=1 √ 2π∞⎪integraldisplay −∞e−ikxu(x, y)dx (2.12.4) so that (2.12.1)–(2.12.3) becomes d2U dy2−k2U=0, (2.12.5) U(k,0) =F(k),U(k,y)→0a s y→∞. (2.12.6ab) Thus, the solution of this transformed system is U(k,y)=F(k)e−|k|y. (2.12.7) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 69 Application of the Convolution Theorem 2.5.5 gives the solution u(x, y)=1 √ 2π∞⎪integraldisplay −∞f(ξ)g(x−ξ)dξ, (2.12.8) where g(x)=F−1{e−|k|y}=⎪radicalbigg 2 πy (x2+y2). (2.12.9) Consequently, the solution (2.12.8) becomes u(x, y)=y π∞⎪integraldisplay −∞f(ξ)dξ (x−ξ)2+y2,y > 0. (2.12.10) This is the well-known Poisson integral formula in the half-plane. It is noted that lim y→0+u(x, y)=∞⎪integraldisplay −∞f(ξ)⎪bracketleftbigg lim y→0+y π·1 (x−ξ)2+y2⎪bracketrightbigg dξ=∞⎪integraldisplay −∞f(ξ)δ(x−ξ)dξ, (2.12.11) where Cauchy’s deﬁnition of the delta function is used, that is, δ(x−ξ) = lim y→0+y π·1 (x−ξ)2+y2. (2.12.12) This may be recognized as a solution of the Laplace equation for a dipole source at ( x, y)=(ξ,0). In particular, when f(x)=T0H(a−|x|) (2.12.13) the solution (2.12.10) reduces to u(x, y)=yT0 πa⎪integraldisplay −adξ (ξ−x)2+y2 =T0 π⎪bracketleftbigg tan−1⎪parenleftbiggx+a y⎪parenrightbigg −tan−1⎪parenleftbiggx−a y⎪parenrightbigg⎪bracketrightbigg =T0 πtan−1⎪parenleftbigg2ay x2+y2−a2⎪parenrightbigg . (2.12.14) The curves in the upper half-plane for which the steady state temperature is constant are known as isothermal curves. In this case, these curves represent a family of circular arcs x2+y2−αy=a2(2.12.15) © 2007 by Taylor & Francis Group, LLC 70 INTEGRAL TRANSFORMS and THEIR APPLICATIONS axy -a Figure 2.9 A family of circular arcs. with centers on the y-axis and the ﬁxed end points on the x-axis at x=±a. The graphs of the arcs are is displayed in Figure 2.9. Another special case deals with f(x)=δ(x). (2.12.16) The solution for this case follows from (2.12.10) and is u(x, y)=y π∞⎪integraldisplay −∞δ(ξ)dξ (x−ξ)2+y2=y π1 (x2+y2). (2.12.17) Further, we can readily deduce the solution of the Neumann problem in the half-plane from the solution of the Dirichlet problem. Example 2.12.2(Neumann’s Problem in the Half-Plane ). Find a solution of the Laplace equa- tion u xx+uyy=0,−∞<x< ∞,y > 0, (2.12.18) with the boundary condition uy(x,0) =f(x),−∞<x< ∞. (2.12.19) This condition speciﬁes the normal derivative on the boundary, and physically, it describes the ﬂuid ﬂow or, heat ﬂux at the boundary. We deﬁne a new function υ(x, y)=uy(x, y)s ot h a t u(x, y)=y⎪integraldisplay υ(x, η)dη, (2.12.20) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 71 where an arbitrary constant can be added to the right-hand side. Clearly, the function υsatisﬁes the Laplace equation ∂2υ ∂x2+∂2υ ∂y2=∂2uy ∂x2+∂2uy ∂y2=∂ ∂y(uxx+uyy)=0, with the boundary condition υ(x,0) =uy(x,0) =f(x)f o r−∞<x< ∞. Thus, υ(x, y) satisﬁes the Laplace equation with the Dirichlet condition on the boundary. Obviously, the solution is given by (2.12.10); that is, υ(x, y)=y π∞⎪integraldisplay −∞f(ξ)dξ (x−ξ)2+y2. (2.12.21) Then the solution u(x, y) can be obtained from (2.12.20) in the form u(x, y)=y⎪integraldisplay υ(x, η)dη=1 πy⎪integraldisplay ηd η∞⎪integraldisplay −∞f(ξ)dξ (x−ξ)2+η2 =1 π∞⎪integraldisplay −∞f(ξ)dξy⎪integraldisplayηd η (x−ξ)2+η2,y > 0 =1 2π∞⎪integraldisplay −∞f(ξ)log[( x−ξ)2+y2]dξ, (2.12.22) where an arbitrary constant can be added to this solution. In other words, the solution of any Neumann problem is uniquely determined up to an arbitrary constant. Example 2.12.3 (The Cauchy Problem for the Diﬀusion Equation ). We consider the initial value problem for a one-dimensional diﬀusion equation with no sources or sinks ut=κuxx,−∞<x< ∞,t > 0, (2.12.23) where κis a diﬀusivity constant with the initial condition u(x,0)=f(x),−∞<x< ∞. (2.12.24) We solve this problem using the Fourier transform in the space variable x deﬁned by (2.12.4). Application of this transform to (2.12.23)–(2.12.24) gives Ut=−κk2U, t > 0, (2.12.25) U(k,0) =F(k). (2.12.26) © 2007 by Taylor & Francis Group, LLC 72 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The solution of the transformed system is U(k,t)=F(k)e−κk2t. (2.12.27) The inverse Fourier transform gives the solution u(x, t)=1 √ 2π∞⎪integraldisplay −∞F(k)exp [( ikx−κk2t)]dk which is, by the Convolution Theorem 2.5.5, =1 √ 2π∞⎪integraldisplay −∞f(ξ)g(x−ξ)dξ, (2.12.28) where g(x)=F−1{e−κk2t}=1 √ 2κtexp⎪parenleftbigg −x2 4κt⎪parenrightbigg ,by (2.3.5). Thus, solution (2.12.28) becomes u(x, t)=1 √ 4πκt∞⎪integraldisplay −∞f(ξ)exp⎪bracketleftbigg −(x−ξ)2 4κt⎪bracketrightbigg dξ. (2.12.29) The integrand involved in the solution consists of the initial value f(x)a n d Green’s function (or,elementary solution )G(x−ξ,t) of the diﬀusion equation for the inﬁnite interval: G(x−ξ,t)=1 √ 4πκtexp⎪bracketleftbigg −(x−ξ)2 4κt⎪bracketrightbigg . (2.12.30) So, in terms of G(x−ξ,t), solution (2.12.29) can be written as u(x, t)=∞⎪integraldisplay −∞f(ξ)G(x−ξ,t)dξ (2.12.31) so that, in the limit as t→0+, this formally becomes u(x,0) =f(x)=∞⎪integraldisplay −∞f(ξ) lim t→0+G(x−ξ,t)dξ. The limit of G(x−ξ,t) represents the Dirac delta function δ(x−ξ) = lim t→0+1 2√ πκtexp⎪bracketleftbigg −(x−ξ)2 4κt⎪bracketrightbigg . (2.12.32) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 73 -3 -2 -1 0 1 2 300.40.81.21.622.4 xt = 1/60 t = 1/16 t = 1/4 Figure 2.10 Graphs of G(x, t) against x. Graphs of G(x, t) are shown in Figure 2.10 for diﬀerent values of κt. It is important to point out that the integrand in (2.12.31) consists of the initial temperature distribution f(x) and Green’s function G(x−ξ,t)w h i c h represents the temperature response along the rod at time tdue to an initial unit impulse of heat at x=ξ. The physical meaning of the solution (2.12.31) is that the initial temperature distribution f(x) is decomposed into a spec- trum of impulses of magnitude f(ξ)a te a c hp o i n t x=ξto form the resulting temperature f(ξ)G(x−ξ,t). Thus, the resulting temperature is integrated to ﬁnd solution (2.12.31). This is called the principle of integral superposition. We make the change of variable ξ−x 2√ κt=ζ, dζ =dξ 2√ κt to express solution (2.12.29) in the form u(x, t)=1 √ π∞⎪integraldisplay −∞f(x+2√ κtζ)exp(−ζ2)dζ. (2.12.33) The integral solution (2.12.33) or (2.12.29) is called the Poisson integral rep- resentation of the temperature distribution. This integral is convergent for all timet>0, and the integrals obtained from (2.12.33) by diﬀerentiation under the integral sign with respect to xandtare uniformly convergent in the neigh- borhood of the point ( x, t). Hence, the solution u(x, t) and its derivatives of all orders exist for t>0. Finally, we consider a special case involving discontinuous initial condition in the form f(x)=T0H(x), (2.12.34) © 2007 by Taylor & Francis Group, LLC 74 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where T0is a constant. In this case, solution (2.12.29) becomes u(x, t)=T0 2√ πκt∞⎪integraldisplay 0exp⎪bracketleftbigg −(x−ξ)2 4κt⎪bracketrightbigg dξ. (2.12.35) Introducing the change of variable η=ξ−x 2√ κt, we can express solution (2.12.35) in the form u(x, t)=T0 √ π∞⎪integraldisplay −x/2√ κte−η2dη=T0 2erfc⎪parenleftbigg −x 2√ κt⎪parenrightbigg =T0 2⎪bracketleftbigg 1+erf⎪parenleftbiggx 2√ κt⎪parenrightbigg⎪bracketrightbigg . (2.12.36) The solution given by equation (2.12.36) with T0= 1 is shown in Figure 2.11. - 4 - 3 - 2 - 1 0123400.511.5 xu(x,t)t = 0.1 t = 0.5 t = 2.0 Figure 2.11 The time development of solution (2.12.36). Iff(x)=δ(x), then the fundamental solution (2.7.29) is given by u(x, t)=1 √ 4πκtexp⎪parenleftbigg −x2 4κt⎪parenrightbigg . Example 2.12.4 (The Cauchy Problem for the Wave Equation ). Obtain the d’Alembert solu- tion of the initial value problem for the wave equation utt=c2uxx,−∞<x< ∞,t > 0, (2.12.37) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 75 with the arbitrary but ﬁxed initial data u(x,0)=f(x),u t(x,0) =g(x),−∞<x< ∞. (2.12.38ab) Application of the Fourier transform F{u(x, t)}=U(k,t)t ot h i ss y s t e mg i v e s d2U dt2+c2k2U=0, U(k,0)=F(k),⎪parenleftbiggdU dt⎪parenrightbigg t=0=G(k). The solution of the transformed system is U(k,t)=Aeickt+Be−ickt, where AandBare constants to be determined from the transformed data so thatA+B=F(k)a n d A−B=1 ikcG(k). Solving for AandB,w eo b t a i n U(k,t)=1 2F(k)(eickt+e−ickt)+G(k) 2ick(eickt−e−ickt). (2.12.39) Thus, the inverse Fourier transform of (2.12.39) yields the solution u(x, t)=1 2⎡ ⎣1 √ 2π∞⎪integraldisplay −∞F(k){eik(x+ct)+eik(x−ct)}dk⎤ ⎦ +1 2c⎡ ⎣1 √ 2π∞⎪integraldisplay −∞G(k) ik{eik(x+ct)−eik(x−ct)}dk⎤ ⎦.(2.12.40) We use the following results f(x)=F−1{F(k)}=1 √ 2π∞⎪integraldisplay −∞eikxF(k)dk, g(x)=F−1{G(k)}=1 √ 2π∞⎪integraldisplay −∞eikxG(k)dk, to obtain the solution in the ﬁnal form u(x, t)=1 2[f(x−ct)+f(x+ct)] +1 2c1 √ 2π∞⎪integraldisplay −∞G(k)dkx+ct⎪integraldisplay x−cteikξdξ =1 2[f(x−ct)+f(x+ct)] +1 2cx+ct⎪integraldisplay x−ctdξ⎡ ⎣1 √ 2π∞⎪integraldisplay −∞eikξG(k)dk⎤ ⎦ =1 2[f(x−ct)+f(x+ct)] +1 2cx+ct⎪integraldisplay x−ctg(ξ)dξ. (2.12.41) © 2007 by Taylor & Francis Group, LLC 76 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This is the well known d’Alembert’s solution of the wave equation. The method and the form of the solution reveal several important features of the wave equation. First, the method of solution essentially proves the existence of the d’Alembert solution and the solution is unique provided f(x) is twice continuously diﬀerentiable and g(x) is continuously diﬀerentiable. Second, the terms involving f(x±ct) in (2.12.41) show that disturbances are propagated along the characteristics with constant velocity c.B o t ht e r m s combined together suggest that the value of the solution at position xand at timetdepends only on the initial values of f(x)a tx−ctandx+ctand the values of g(x) between these two points. The interval ( x−ct, x+ct) is called thedomain of dependence of the variable ( x, t). Finally, the solution depends continuously on the initial data, that is, the problem is well posed. In other words, a small change in either f(x)o rg(x) results in a correspondingly small change in the solution u(x, t). In particular, if f(x)=e x p ( −x2)a n d g(x)≡0, the time development of solution (2.12.41) with c= 1 is shown in Figure 2.12. In this case, the solution becomes u(x, t)=1 2[e−(x−t)2+e−(x+t)2]. (2.12.42) As shown in Figure 2.12, the initial form f(x)=e x p ( −x2) is found to split into two similar waves propagating in opposite direction with unit velocity. Example 2.12.5 (The Schr ¨odinger Equation in Quantum Mechanics ). The time-dependent Schr¨odinger equation of a particle of mass mis i/planckover2pi1ψt=⎪bracketleftbigg V(x)−/planckover2pi12 2m∇2⎪bracketrightbigg ψ=Hψ, (2.12.43) where h=2π/planckover2pi1is thePlanck constant ,ψ(x,t) is the wave function, V(x)i st h e potential, ∇2=∂2 ∂x2+∂2 ∂y2+∂2 ∂z2is the three-dimensional Laplacian ,a n d His theHamiltonian . IfV(x)=c o n s t a n t= V, we can seek a plane wave solution of the form ψ(x,t)=Aexp[i(κ·x−ωt)], (2.12.44) where Ais a constant amplitude, κ=(k,l,m ) is the wavenumber vector, and ωis the frequency. Substituting this solution into (2.12.43), we conclude that this solution is possible provided the following relation is satisﬁed: i/planckover2pi1(−iω)=V−/planckover2pi12 2m(iκ)2,κ2=k2+l2+m2. Or, /planckover2pi1ω=V+/planckover2pi12κ2 2m. (2.12.45) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 77 -4 -2 0 2 400.51 xu(x,0) -4 -2 0 2 400.20.40.60.8 xu(x,1) -4 -2 0 2 400.20.40.60.8 xu(x,2) -4 -2 0 2 400.20.40.60.8 xu(x,3) Figure 2.12 The time development of solution (2.12.42). This is called the dispersion relation and shows that the sum of the potential energy Vand the kinetic energy(/planckover2pi1κ)2 2mis equal to the total energy /planckover2pi1ω.F u r t h e r , the kinetic energy K.E.=1 2m(/planckover2pi1κ)2=p2 2m, (2.12.46) where p=/planckover2pi1κis the momentum of the particle. The phase velocity, Cpand the group velocity, Cgof the wave are deﬁned by Cp=ω κˆκ,C g=∇κω(κ), (2.12.47ab) where κis the wavenumber vector and κ=|κ|andˆκis the unit wavenumber vector. In the one-dimensional case, the phase velocity is Cp=ω k(2.12.48) and the group velocity is Cg=∂ω ∂k=/planckover2pi1k m=p m=mυ υ=υ. (2.12.49) This shows that the group velocity is equal to the classical particle velocity υ. © 2007 by Taylor & Francis Group, LLC 78 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We now use the Fourier transform method to solve the one-dimensional Schr¨odinger equation for a free particle ( V≡0), that is, i/planckover2pi1ψt=−/planckover2pi12 2mψxx,−∞<x< ∞,t > 0, (2.12.50) ψ(x,0) = ψ0(x),−∞<x< ∞, (2.12.51) ψ(x, t)→0a s |x|→∞ . (2.12.52) Application of the Fourier transform to (2.12.50)–(2.12.52) gives Ψt=−i/planckover2pi1k2 2mΨ,Ψ(k,0)= Ψ 0(k). (2.12.53) The solution of this transformed system is Ψ(k,t)=Ψ 0(k)exp (−iαk2t),α=/planckover2pi1 2m. (2.12.54) The inverse Fourier transform gives the formal solution ψ(x, t)=1 √ 2π∞⎪integraldisplay −∞Ψ0(k)exp{ik(x−αkt)}dk =1 2π∞⎪integraldisplay −∞e−ikyψ(y,0)dy∞⎪integraldisplay −∞exp{ik(x−αkt)}dk =1 2π∞⎪integraldisplay −∞ψ(y,0)dy∞⎪integraldisplay −∞exp{ik(x−y−αkt)}dk. (2.12.55) We rewrite the integrand of the second integral in (2.12.55) as follows exp[ik(x−y−αkt)] =e x p⎪bracketleftBigg −iαt⎪braceleftBigg k2−2k·x−y 2αt+⎪parenleftbiggx−y 2αt⎪parenrightbigg2 −⎪parenleftbiggx−y 2αt⎪parenrightbigg2⎪bracerightBigg⎪bracketrightBigg =e x p⎪bracketleftBigg −iαt⎪braceleftbigg k−x−y 2αt⎪bracerightbigg2⎪bracketrightBigg exp⎪bracketleftbiggi(x−y)2 4αt⎪bracketrightbigg =e x p⎪bracketleftbiggi(x−y)2 4αt⎪bracketrightbigg exp(−iαtξ2),ξ=k−x−y 2αt. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 79 Using this result in (2.12.55), we obtain ψ(x, t)=1 2π∞⎪integraldisplay −∞exp⎪bracketleftbiggi(x−y)2 4αt⎪bracketrightbigg ψ(y,0)dy∞⎪integraldisplay −∞exp(−iαtξ2)dξ =1 2π⎪radicalbigg π 2αt(1−i)∞⎪integraldisplay −∞exp⎪bracketleftbiggi(x−y)2 4αt⎪bracketrightbigg ψ(y,0)dy =(1−i) 2√ 2απt∞⎪integraldisplay −∞exp⎪bracketleftbiggi(x−y)2 4αt⎪bracketrightbigg ψ(y,0)dy. (2.12.56) This is the integral solution of the problem. Example 2.12.6 (Slowing Down of Neutrons ). We consider the problem of slowing down neu- trons in an inﬁnite medium with a source of neutrons governed by ut=uxx+δ(x)δ(t),−∞<x< ∞,t > 0, (2.12.57) u(x,0) =δ(x),−∞<x< ∞, (2.12.58) u(x, t)→0a s |x|→∞ fort>0, (2.12.59) where u(x, t) represents the number of neutrons per unit volume per unit time, which reach the age t,a n d δ(x)δ(t) is the source function. Application of the Fourier transform method gives dU dt+k2U=1 √ 2πδ(t), U(k,0) =1 √ 2π. The solution of this transformed system is U(k,t)=1 √ 2πe−k2t, and the inverse Fourier transform gives the solution u(x, t)=1 2π∞⎪integraldisplay −∞eikx−k2tdk=1 √ 2πF−1⎪braceleftBig e−k2t⎪bracerightBig =1 √ 4πtexp⎪parenleftbigg −x2 4t⎪parenrightbigg . (2.12.60) © 2007 by Taylor & Francis Group, LLC 80 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 2.12.7 (One-Dimensional Wave Equation ). Obtain the solution of the one-dimensional wave equation utt=c2uxx,−∞<x< ∞,t > 0, (2.12.61) u(x,0) = 0 ,u t(x,0) =δ(x),−∞<x< ∞. (2.12.62ab) Making reference to Example 2.12.4, we ﬁnd f(x)≡0a n d g(x)=δ(x)s o thatF(k)=0a n d G(k)=1 √ 2π.T h es o l u t i o nf o r U(k,t)i sg i v e nb y U(k,t)=1 2c√ 2π⎪bracketleftbiggeickt ik−e−ickt ik⎪bracketrightbigg . Thus, the inverse Fourier transform gives u(x, t)=1 2c√ 2πF−1⎪braceleftbiggeickt ik−e−ickt ik⎪bracerightbigg =1 2c√ 2π⎪bracketleftbigg⎪radicalbigg π 2{sgn(x + ct) −sgn(x−ct)}⎪bracketrightbigg =1 4c[sgn(x + ct) −sgn(x −ct)] =⎧ ⎪⎪⎨ ⎪⎪⎩1−1 4c=0,|x|>c t> 0 1+1 4c=1 2c,|x|<c t . In other words, the solution can be written in the form u(x, t)=1 2cH(c2t2−x2). Example 2.12.8 (Linearized Shallow Water Equations in a Rotating Ocean ). The horizontal equations of motion of a uniformly rotating inviscid homogeneous ocean of constant depth hare ut−fυ=−gηx, (2.12.63) υt+fu=0, (2.12.64) ηt+hux=0, (2.12.65) where f=2 Ωs i n θis the Coriolis parameter, which is constant in the present problem, gis the acceleration due to gravity, η(x, t)i st h ef r e es u r f a c ee l e v a - tion,u(x, t)a n d υ(x, t) are the velocity ﬁelds. The wave motion is generated © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 81 by the prescribed free surface elevation at t= 0 so that the initial conditions are u(x,0) = 0 = υ(x,0),η(x,0) =η0H(a−|x|), (2.12.66abc) and the velocity ﬁelds and free surface elevation function vanish at inﬁnity. We apply the Fourier transform with respect to xdeﬁned by F{f(x, t)}=F(k,t)=1 √ 2π∞⎪integraldisplay −∞e−ikxf(x, t)dx (2.12.67) to the system (2.12.63)–(2.12.65) so that the system becomes dU dt−fV=−gikE dV dt+fU=0 dE dt=−hikU U(k,0)= 0 = V(k,0),E(k,0) =⎪radicalbigg 2 πη0⎪parenleftbiggsinak k⎪parenrightbigg , (2.12.68abc) where E(k,t)=F{η(x, t)}. Elimination of UandVfrom the transformed system gives a single equation forE(k,t)a s d3E dt3+ω2dE dt=0, (2.12.69) where ω2=(f2+c2k2)a n d c2=gh. The general solution of (2.12.69) is E(k,t)=A+Bcosωt+Csinωt, (2.12.70) where A,B,a n d Care arbitrary constants to be determined from (2.12.68c) and⎪parenleftbiggd2E dt2⎪parenrightbigg t=0=−c2k2E(k,0) =−c2k2·⎪radicalbigg 2 πη0sinak k, which gives B=⎪radicalbigg 2 πη0⎪parenleftbiggsinak k⎪parenrightbigg ·⎪parenleftbiggc2k2 ω2⎪parenrightbigg . Also⎪parenleftbigdE dt⎪parenrightbig t=0=0 g i v e s C≡0 and (2.12.68c) implies A+B=⎪radicalBig 2 πη0sinak k. Consequently, the solution (2.12.70) becomes E(k,t)=⎪radicalbigg 2 πη0⎪parenleftbiggsinak k⎪parenrightbiggf2+c2k2cosωt (f2+c2k2). (2.12.71) © 2007 by Taylor & Francis Group, LLC 82 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Similarly U(k,t)=⎪radicalbigg 2 πη0sinak ih·c2sinωt ⎪radicalbig c2k2+f2, (2.12.72) V(k,t)=1 f⎪parenleftbiggdU dt+gikE⎪parenrightbigg . (2.12.73) The inverse Fourier transform gives the formal solution for η(x, t) η(x, t)=⎪parenleftBigη0 π⎪parenrightBig∞⎪integraldisplay −∞sinak k·f2+c2k2cosωt (f2+c2k2)eikxdk. (2.12.74) Similar integral expressions for u(x, t)a n d υ(x, t) can be obtained. Example 2.12.9 (Sound Waves Induced by a Spherical Body ). We consider propagation of sound waves in an unbounded ﬂuid medium generated by an impulsive radialacceleration of a sphere of radius a. Such waves are assumed to be spherically symmetric and the associated velocity potential on the pressure ﬁeld p(r, t) satisﬁes the wave equation ∂ 2p ∂t2=c2⎪bracketleftbigg1 r2∂ ∂r⎪parenleftbigg r2∂p ∂r⎪parenrightbigg⎪bracketrightbigg , (2.12.75) where cis the speed of sound. The boundary condition required for the prob- lem is 1 ρ0⎪parenleftbigg∂p ∂r⎪parenrightbigg =−a0δ(t)o n r=a, (2.12.76) where ρ0is the mean density of the ﬂuid and a0is a constant. Application of the Fourier transform of p(r, t) with respect to time tgives 1 r2d dr⎪parenleftbigg r2dP dr⎪parenrightbigg =−k2P(r, ω), (2.12.77) dP dr=−a0ρ0 √ 2π,onr=a, (2.12.78) whereF{p(r, t)}=P(r, ω)a n d k2=ω2 c2. The general solution of (2.12.77)–(2.12.78) is P(r, ω)=A reikr+B re−ikr, (2.12.79) where AandBare arbitrary constants. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 83 The inverse Fourier transform gives the solution p(r, t)=1 √ 2π∞⎪integraldisplay −∞⎪bracketleftbiggA rei(ωt+kr)+B rei(ωt−kr)⎪bracketrightbigg dω. (2.12.80) The ﬁrst term of the integrand represents incoming spherical waves generated at inﬁnity and the second term corresponds to outgoing spherical waves due to the impulsive radial acceleration of the sphere. Since there is no disturbance at inﬁnity, we impose the Sommerfeld radiation condition at inﬁnity to eliminate the incoming waves so that A=0 ,a n d Bis calculated using (2.12.78). Thus, the inverse Fourier transform gives the formal solution p(r, t)=⎪parenleftbigga0ρ0a2 2πr⎪parenrightbigg∞⎪integraldisplay −∞exp⎪bracketleftbig iω⎪braceleftbig t−r−a c⎪bracerightbig⎪bracketrightbig dω ⎪parenleftbig 1+iωa c⎪parenrightbig . (2.12.81) We next choose a closed contour with a semicircle in the upper half plane and the real ω-axis. Using the Cauchy theory of residues, we calculate the residue contribution from the pole at ω=ic/a. Finally, it turns out that the ﬁnal solution is u(r, t)=⎪parenleftBigρ0a0ca r⎪parenrightBig exp⎪bracketleftbigg −c a⎪parenleftbigg t−r−a c⎪parenrightbigg⎪bracketrightbigg H⎪parenleftbigg t−r−a c⎪parenrightbigg .(2.12.82) Example 2.12.10 (The Linearized Korteweg-de Vries Equation ). The linearized KdV equation for the free surface elevation η(x, t) in an inviscid water of constant depth h is ηt+cηx+ch2 6ηxxx=0,−∞<x< ∞,t > 0, (2.12.83) where c=√ ghis the shallow water speed. Solve equation (2.12.83) with the initial condition η(x,0) =f(x),−∞<x< ∞. (2.12.84) Application of the Fourier transform F{η(x, t)}=E(k,t)t ot h eK d Vs y s - tem gives the solution for E(k,t)i nt h ef o r m E(k,t)=F(k)exp⎪bracketleftbigg ikct⎪parenleftbiggk2h2 6−1⎪parenrightbigg⎪bracketrightbigg . The inverse transform gives η(x, t)=1 √ 2π∞⎪integraldisplay −∞F(k)exp⎪bracketleftbigg ik⎪braceleftbigg (x−ct)+⎪parenleftbiggcth2 6⎪parenrightbigg k2⎪bracerightbigg⎪bracketrightbigg dk. (2.12.85) © 2007 by Taylor & Francis Group, LLC 84 INTEGRAL TRANSFORMS and THEIR APPLICATIONS In particular, if f(x)=δ(x), then (2.12.85) reduces to the Airy integral η(x, t)=1 π∞⎪integraldisplay 0cos⎪bracketleftbigg k(x−ct)+⎪parenleftbiggcth2 6⎪parenrightbigg k3⎪bracketrightbigg dk (2.12.86) which is, in terms of the Airy function, =⎪parenleftbiggcth2 2⎪parenrightbigg−1 3 Ai⎪bracketleftBigg⎪parenleftbiggcth2 2⎪parenrightbigg−1 3 (x−ct)⎪bracketrightBigg , (2.12.87) where the Airy function Ai(az) is deﬁned by Ai(az)=1 2πa∞⎪integraldisplay −∞exp⎪bracketleftbigg i⎪parenleftbigg kz+k3 3a3⎪parenrightbigg⎪bracketrightbigg dk=1 πa∞⎪integraldisplay 0cos⎪parenleftbigg kz+k3 3a3⎪parenrightbigg dk. (2.12.88) Example 2.12.11 (Biharmonic Equation in Fluid Mechanics ). Usually, the biharmonic equation arises in ﬂuid mechanics and in elasticity. The equation can readily be solved by using the Fourier transform method. We ﬁrst derive a biharmonic equation from the Navier-Stokes equations of motion in a viscous ﬂuid which is given by ∂u ∂t+(u·∇)u=F−1 ρ∇p+ν∇2u, (2.12.89) where u=(u,υ,w ) is the velocity ﬁeld, Fis the external force per unit mass of the ﬂuid, pis the pressure, ρis the density and νis the kinematic viscosity of the ﬂuid. The conservation of mass of an incompressible ﬂuid is described by the continuity equation divu=0. (2.12.90) In terms of some representative length scale Land velocity scale U,i ti s convenient to introduce the nondimensional ﬂow variables x/prime=x L,t/prime=Ut L,u/prime=u U,p/prime=p ρU2. (2.12.91) In terms of these nondimensional variables, equation (2.12.89) without the external force can be written, dropping the primes, as ∂u ∂t+(u·∇)u=−∇p+1 R∇2u, (2.12.92) where R=UL/ν is called the Reynolds number . Physically, it measures the ratio of inertial forces of the order U2/Lto viscous forces of the order νU/L2, © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 85 and it has special dynamical signiﬁcance. This is one of the most fundamental nondimensional parameters for the speciﬁcation of the dynamical state ofviscous ﬂow ﬁelds. In the absence of the external force, F=0, it is preferable to write the Navier-Stokes equations (2.12.89) in the form (since u×ω= 1 2∇u2−u·∇u) ∂u ∂t−u×ω=−∇⎪parenleftbiggp ρ+1 2u2⎪parenrightbigg −ν∇2u, (2.12.93) where ω=c u r l uis the vorticity vector andu2=u·u. We can eliminate the pressure pfrom (2.12.93) by taking the curl of (2.12.93), giving ∂ω ∂t−curl(u×ω)=ν∇2ω (2.12.94) which becomes, by div u=0a n dd i v ω=0 , ∂ω ∂t=(ω·∇)u−(u·∇)ω+ν∇2ω. (2.12.95) This is universally known as the vorticity transport equation . The left hand- side represents the rate of change of vorticity. The ﬁrst two terms on the right-hand side represent the rate of change of vorticity due to stretching andtwisting of vortex lines. The last term describes the diﬀusion of vorticity by molecular viscosity. In case of two-dimensional ﬂow, ( ω·∇)u= 0, equation (2.12.95) becomes Dω dt=∂ω ∂t+(u·∇)ω=ν∇2ω, (2.12.96) whereu=(u,υ,0) and ω=( 0,0,ζ), and ζ=υx−uy. Equation (2.12.96) shows that only convection and conduction occur. In terms of the stream functionψ(x, y)w h e r e u=ψ y,υ=−ψx,ω=−∇2ψ, (2.12.97) which satisfy (2.12.90) identically, equation (2.12.96) assumes the form ∂ ∂t⎪parenleftbig ∇2ψ⎪parenrightbig +⎪parenleftbigg∂ψ ∂y∂ ∂x−∂ψ ∂x∂ ∂y⎪parenrightbigg ∇2ψ=ν∇4ψ. (2.12.98) In case of slow motion (velocity is s mall) or in case of a very viscous ﬂuid (νvery large), the Reynolds number Ris very small. For a steady ﬂow in such cases of an incompressible viscous ﬂuid,∂ ∂t≡0, while ( u·∇)ωis negligible in comparison with the viscous term. Consequently, (2.12.98) reduces to the standard biharmonic equation ∇4ψ=0. (2.12.99) Or, more explicitly, ∇2(∇2)ψ≡ψxxxx+2ψxxyy+ψyyyy=0. (2.12.100) © 2007 by Taylor & Francis Group, LLC 86 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We solve this equation in a semi-inﬁnite viscous ﬂuid bounded by an in- ﬁnite horizontal plate at y= 0, and the ﬂuid is introduced normally with a prescribed velocity through a strip −a<x<a of the plate. Thus, the required boundary conditions are u≡∂ψ ∂y=0,υ≡∂ψ ∂x=H(a−|x|)f(x)o n y=0, (2.12.101ab) where f(x) is a given function of x. Furthermore, the ﬂuid is assumed to be at rest at large distances from the plate, that is, (ψx,ψy)→(0,0) as y→∞ for−∞<x< ∞. (2.12.102) To solve the biharmonic equation (2 .12.100) with the boundary conditions (2.12.101ab) and (2.12.102), we introduce the Fourier transform with respect tox Ψ(k,y)=1 √ 2π∞⎪integraldisplay −∞e−ikxψ(x, y)dx. (2.12.103) Thus, the Fourier transformed problem is ⎪parenleftbiggd2 dy2−k2⎪parenrightbigg2 Ψ(k,y)=0, (2.12.104) dΨ dy=0,(ik)Ψ =F(k),y=0, (2.12.105ab) where F(k)=1 √ 2πa⎪integraldisplay −ae−ikxf(x)dx. (2.12.106) In view of the Fourier transform o f (2.12.102), the bounded solution of (2.12.104) is Ψ(k,y)=(A+B|k|y)exp (−|k|y), (2.12.107) where AandBcan be determined from (2.12.105ab) so that A=B=(ik)−1F(k). Consequently, the solution (2.12.107) becomes Ψ(k,y)=(ik)−1(1 +|k|y)F(k)exp (−|k|y). (2.12.108) The inverse Fourier transform gives the formal solution ψ(x, y)=1 √ 2π∞⎪integraldisplay −∞F(k)G(k)exp(ikx)dk, (2.12.109) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 87 where G(k)=(ik)−1(1 +|k|y)exp (−|k|y) so that g(x)=F−1{G(k)}=F−1{(ik)−1exp(−|k|y)} +yF−1{(ik)−1|k|exp(−|k|y)} =F−1 s{k−1exp(−ky)}+yF−1 s{e−ky}, which is, by (2.13.7) and (2.13.8), =⎪radicalbigg 2 πtan−1⎪parenleftbiggx y⎪parenrightbigg +⎪radicalbigg 2 πxy (x2+y2). (2.12.110) Using the Convolution Theorem 2.5.5 in (2.12.109) gives the ﬁnal solution ψ(x, y)=1 π∞⎪integraldisplay −∞f(x−ξ)⎪bracketleftbigg tan−1⎪parenleftbiggξ y⎪parenrightbigg +yξ ξ2+y2⎪bracketrightbigg dξ. (2.12.111) In particular, if f(x)=δ(x), then solution (2.12.111) becomes ψ(x, y)=1 π⎪bracketleftbigg tan−1⎪parenleftbiggx y⎪parenrightbigg +xy x2+y2⎪bracketrightbigg . (2.12.112) The velocity ﬁelds uandυcan be determined from (2.12.112). Example 2.12.12 (Biharmonic Equation in Elasticity ). We derive the biharmonic equation in elasticity from the two-dimensional equilibrium equations and the compati- bility condition. In two-dimensional elastic medium, the strain componentse xx,exy,eyyin terms of the displacement functions ( u,υ,0) are exx=∂u ∂x,e yy=∂υ ∂y,e xy=1 2⎪parenleftbigg∂u ∂y+∂υ ∂x⎪parenrightbigg . (2.12.113) Diﬀerentiating these results gives the compatibility condition ∂2exx ∂y2+∂2eyy ∂x2=2∂2exy ∂x∂y. (2.12.114) In terms of the Poisson ratio νandYoung’s modulus Eof the elastic ma- terial, the strain component in the zdirection is expressed in terms of stress components Eezz=σzz−ν(σxx+σyy). (2.12.115) In the case of plane strain, ezz=0 ,s ot h a t σzz=ν(σxx+σyy). (2.12.116) © 2007 by Taylor & Francis Group, LLC 88 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Substituting this result in other stress-strain relations, we obtain the strain components exx,exy,eyythat are related to stress components σxx,σxy,σyy by Eexx=σxx−ν(σyy+σzz)=( 1 −ν2)σxx−ν(1 +ν)σyy,(2.12.117) Eeyy=σyy−ν(σxx+σzz)=( 1 −ν2)σyy−ν(1 +ν)σxx,(2.12.118) Eexy=( 1+ ν)σxy. (2.12.119) Putting (2.12.117)-(2.12.119) into (2.12.114) gives ∂2 ∂y2[σxx−ν(σyy+σzz)] +∂2 ∂x2[σyy−ν(σxx+σzz)] =2 ( 1+ ν)∂2σxy ∂x∂y. (2.12.120) The basic diﬀerential equations for the stress components σxx,σyy,σxyin the medium under the action of body forces XandYare ∂σxx ∂x+∂σxy ∂y+ρX=ρ∂2u ∂t2, (2.12.121) ∂σxy ∂x+∂σyy ∂y+ρY=ρ∂2υ ∂t2, (2.12.122) where ρis the mass density of the elastic material. The equilibrium equations follow fro m (2.12.121)–(2.12. 122) in the absence of the body forces ( X=Y=0 )a s ∂ ∂xσxx+∂ ∂yσxy=0, (2.12.123) ∂ ∂xσxy+∂ ∂yσyy=0. (2.12.124) It is obvious that the expressions σxx=∂2χ ∂y2,σ xy=−∂2χ ∂x∂y,σ yy=∂2χ ∂x2(2.12.125) satisfy the equilibrium equations for any arbitrary function χ(x, y). Substi- tuting from equations (2.12.125) into th e compatibility condition (2.12.120), we see that χmust satisfy the biharmonic equation ∂4χ ∂x4+2∂4χ ∂x2∂y2+∂4χ ∂y4=0, (2.12.126) which may be written symbolically as ∇4χ=0. (2.12.127) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 89 The function χwas ﬁrst introduced by Airy in 1862 and is known as the Airy stress function . We determine the stress distribution in a semi-inﬁnite elastic medium boun- ded by an inﬁnite plane at x= 0 due to an external pressure to its surface. Thex-axis is normal to this plane and assumed positive in the direction into the medium. We assume that th e external surface pressure pvaries along the surface so that the boundary conditions are σxx=−p(y),σ xy=0 o n x=0 f o ra l l yin (−∞,∞). (2.12.128) We derive solutions so that stress components σxx,σyy,a n d σxyall vanish asx→∞. In order to solve the biharmonic equation (2.12.127), we introduce the Fouri- er transform ˜ χ(x, k)o ft h e Airy stress function with respect to yso that (2.12.127)–(2.12.128) reduce to ⎪parenleftbiggd2 dx2−k2⎪parenrightbigg2 ˜χ=0, (2.12.129) k2˜χ(0,k)=˜p(k),(ik)⎪parenleftbiggd˜χ dx⎪parenrightbigg x=0=0, (2.12.130) where ˜ p(k)=F{p(y)}. The bounded solution of the transformed problem is ˜χ(x, k)=(A+Bx)exp(−|k|x), (2.12.131) where AandBare constants of integration to be determined from (2.12.130). It turns out that A=˜p(k)/k2andB=˜p(k)/|k|and hence, the solution be- comes ˜χ(x, k)=˜p(k) k2{1+|k|x}exp(−|k|x). (2.12.132) The inverse Fourier transform yields the formal solution χ(x, y)=1 √ 2π∞⎪integraldisplay −∞˜p(k) k2(1 +|k|x)exp(iky−|k|x)dk. (2.12.133) The stress components are obtained from (2.12.125) in the form σxx(x, y)=−1 √ 2π∞⎪integraldisplay −∞k2˜χ(x, k)exp (iky)dk, (2.12.134) σxy(x, y)=−1 √ 2π∞⎪integraldisplay −∞(ik)⎪parenleftbiggd˜χ dx⎪parenrightbigg exp(iky)dk, (2.12.135) σyy(x, y)=1 √ 2π∞⎪integraldisplay −∞d2˜χ dx2exp(iky)dk, (2.12.136) © 2007 by Taylor & Francis Group, LLC 90 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where ˜ χ(x, k) are given by (2.12.132). In particular, if p(y)=Pδ(y)s ot h a t ˜p(k)=P(2π)−1 2. Consequently, from (2.12.133)–(2.12.136) we obtain χ(x, y)=P 2π∞⎪integraldisplay −∞k−2(1 +|k|x)exp (iky−|k|x)dk =P π∞⎪integraldisplay 0k−2(1 +kx)coskyexp(−kx)dk. (2.12.137) σxx=−P π∞⎪integraldisplay 0(1 +kx)e−kxcosky dk=−2Px3 π(x2+y2)2. (2.12.138) σxy=−Px π∞⎪integraldisplay 0ksinkyexp(−kx)dk=−2Px2y π(x2+y2)2. (2.12.139) σyy=−P π∞⎪integraldisplay 0(1−kx)exp (−kx)cosky dk=−2Pxy2 π(x2+y2)2.(2.12.140) Another physically realistic pressure distribution is p(y)=PH(|a|−y), (2.12.141) where Pis a constant, so that ˜p(k)=⎪radicalbigg 2 πP ksinak. (2.12.142) Substituting this value for ˜ p(k) into (2.12.133)–(2.12.136), we obtain the in- tegral expression for χ, σxx,σxy,a n d σyy. It is noted here that if a point force of magnitude P0acts at the origin located on the boundary, then we put P=(P0/2a) in (2.12.142) and ﬁnd ˜p(k) = lim a→0⎪radicalbigg 2 πP0 2⎪parenleftbiggsinak ak⎪parenrightbigg =P0 √ 2π. (2.12.143) Thus, the stress components can also be written in this case. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 91 2.13 Fourier Cosine and Sine Transforms with Examples The Fourier cosine integral formula (2.2.8) leads to the Fourier cosine trans- formand its inverse deﬁned by Fc{f(x)}=Fc(k)=⎪radicalbigg 2 π∞⎪integraldisplay 0coskxf(x)dx, (2.13.1) F−1 c{Fc(k)}=f(x)=⎪radicalbigg 2 π∞⎪integraldisplay 0coskxF c(k)dk, (2.13.2) whereFcis the Fourier cosine transform operator and F−1 cis its inverse operator. Similarly, the Fourier sine integral formula (2.2.9) leads to the Fourier sine transform a n di t si n v e r s ed e ﬁ n e db y Fs{f(x)}=Fs(k)=⎪radicalbigg 2 π∞⎪integraldisplay 0sinkxf(x)dx, (2.13.3) F−1 s{Fs(k)}=f(x)=⎪radicalbigg 2 π∞⎪integraldisplay 0sinkx F s(k)dk, (2.13.4) whereFsis the Fourier sine transform operator and F−1 sis its inverse. Example 2.13.1 Show that (a)Fc{e−ax}=⎪radicalbigg 2 πa (a2+k2),(a>0). (2.13.5) (b)Fs{e−ax}=⎪radicalbigg 2 πk (a2+k2),(a>0). (2.13.6) We have Fc{e−ax}=⎪radicalbigg 2 π∞⎪integraldisplay 0e−axcoskxdx =1 2⎪radicalbigg 2 π∞⎪integraldisplay 0[e−(a−ik)x+e−(a+ik)x]dx Fc{e−ax}=1 2⎪radicalbigg 2 π⎪bracketleftbigg1 a−ik+1 a+ik⎪bracketrightbigg =⎪radicalbigg 2 πa (a2+k2). © 2007 by Taylor & Francis Group, LLC 92 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The proof of the other result is similar and hence left to the reader. Example 2.13.2 Show that F−1 s⎪braceleftbigg1 kexp(−sk)⎪bracerightbigg =⎪radicalbigg 2 πtan−1⎪parenleftBigx s⎪parenrightBig . (2.13.7) We have the standard deﬁnite integral ⎪radicalbigg π 2F−1 s{exp(−sk)}=∞⎪integraldisplay 0exp(−sk)sinkxdk =x s2+x2. (2.13.8) Integrating both sides with respect to sfromsto∞gives ∞⎪integraldisplay 0e−sk ksinkxdk =∞⎪integraldisplay sxds x2+s2=⎪bracketleftBig tan−1s x⎪bracketrightBig∞ s =π 2−tan−1⎪parenleftBigs x⎪parenrightBig =t a n−1⎪parenleftBigx s⎪parenrightBig . (2.13.9) Thus, F−1 s⎪braceleftbigg1 kexp(−sk)⎪bracerightbigg =⎪radicalbigg 2 π∞⎪integraldisplay 01 kexp(−sk)sinxk dk =⎪radicalbigg 2 πtan−1⎪parenleftBigx s⎪parenrightBig . Example 2.13.3 Show that Fs{erfc(ax)}=⎪radicalbigg 2 π1 k⎪bracketleftbigg 1−exp⎪parenleftbigg −k2 4a2⎪parenrightbigg⎪bracketrightbigg . (2.13.10) We have Fs{erfc(ax)}=⎪radicalbigg 2 π∞⎪integraldisplay 0erfc(ax)sinkxdx =2√ 2 π∞⎪integraldisplay 0sinkxdx∞⎪integraldisplay axe−t2dt. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 93 Interchanging the order of integration, we obtain Fs{erf(ax)}=2√ 2 π∞⎪integraldisplay 0exp(−t2)dtt/a⎪integraldisplay 0sinkxdx =2√ 2 πk∞⎪integraldisplay 0exp(−t2)⎪braceleftbigg 1−cos⎪parenleftbiggkt a⎪parenrightbigg⎪bracerightbigg dt =2√ 2 πk⎪bracketleftbigg√ π 2−√ π 2exp⎪parenleftbigg −k2 4a2⎪parenrightbigg⎪bracketrightbigg . Thus, Fs{erfc(ax)}=⎪radicalbigg 2 π1 k⎪bracketleftbigg 1−exp⎪parenleftbigg −k2 4a2⎪parenrightbigg⎪bracketrightbigg . 2.14 Properties of Fourier Cosine and Sine Transforms THEOREM 2.14.1 IfFc{f(x)}=Fc(k)a n dFs{f(x)}=Fs(k), then Fc{f(ax)}=1 aFc⎪parenleftbiggk a⎪parenrightbigg ,a > 0. (2.14.1) Fs{f(ax)}=1 aFs⎪parenleftbiggk a⎪parenrightbigg ,a > 0. (2.14.2) Under appropriate conditions, the following properties also hold: Fc{f/prime(x)}=kFs(k)−⎪radicalbigg 2 πf(0), (2.14.3) Fc{f/prime/prime(x)}=−k2Fc(k)−⎪radicalbigg 2 πf/prime(0), (2.14.4) Fs{f/prime(x)}=−kFc(k), (2.14.5) Fs{f/prime/prime(x)}=−k2Fs(k)+⎪radicalbigg 2 πkf(0). (2.14.6) These results can be gener alized for the cosine and sine transforms of higher order derivatives of a function. They are left as exercises. © 2007 by Taylor & Francis Group, LLC 94 INTEGRAL TRANSFORMS and THEIR APPLICATIONS THEOREM 2.14.2 (Convolution Theorem for the Fourier Cosine Transform ). IfFc{f(x)}= Fc(k)andFc{g(x)}=Gc(k), then F−1 c{Fc(k)Gc(k)}=1 √ 2π∞⎪integraldisplay 0f(ξ)[g(x+ξ)+g(|x−ξ|)]dξ. (2.14.7) Or, equivalently, ∞⎪integraldisplay 0Fc(k)Gc(k)coskxdk =1 2∞⎪integraldisplay 0f(ξ)[g(x+ξ)+g(|x−ξ|)]dξ. (2.14.8) PROOF Using the deﬁnition of the inverse Fourier cosine transform, we have F−1 c{Fc(k)Gc(k)}=⎪radicalbigg 2 π∞⎪integraldisplay 0Fc(k)Gc(k)coskxdk =⎪parenleftbigg2 π⎪parenrightbigg∞⎪integraldisplay 0Gc(k)coskxdk∞⎪integraldisplay 0f(ξ)coskξ dξ. Hence, F−1 c{Fc(k)Gc(k)}=⎪parenleftbigg2 π⎪parenrightbigg∞⎪integraldisplay 0f(ξ)dξ∞⎪integraldisplay 0coskxcoskξ G c(k)dk =1 2⎪radicalbigg 2 π∞⎪integraldisplay 0f(ξ)dξ⎪radicalbigg 2 π∞⎪integraldisplay 0[cosk(x+ξ)+c o s k(|x−ξ|)]Gc(k)dk =1 √ 2π∞⎪integraldisplay 0f(ξ)[g(x+ξ)+g(|x−ξ|)]dξ, in which the deﬁnition of the inverse Fourier cosine transform is used. This proves (2.14.7). It also follows from the proof of Theorem 2.14.2 that ∞⎪integraldisplay 0Fc(k)Gc(k)coskxdk =1 2∞⎪integraldisplay 0f(ξ)[g(x+ξ)+g(|x−ξ|)]dξ. This proves result (2.14.8). Putting x= 0 in (2.14.8), we obtain ∞⎪integraldisplay 0Fc(k)Gc(k)dk=∞⎪integraldisplay 0f(ξ)g(ξ)dξ=∞⎪integraldisplay 0f(x)g(x)dx. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 95 Substituting g(x)= f(x) gives, since Gc(k)= Fc(k), ∞⎪integraldisplay 0|Fc(k)|2dk=∞⎪integraldisplay 0|f(x)|2dx. (2.14.9) This is the Parseval relation for the Fourier cosine transform. Similarly, we obtain ∞⎪integraldisplay 0Fs(k)Gs(k)coskxdk =⎪radicalbigg 2 π∞⎪integraldisplay 0Gs(k)coskxdk∞⎪integraldisplay 0f(ξ)sinkξ dξ which is, by interchanging the order of integration, =⎪radicalbigg 2 π∞⎪integraldisplay 0f(ξ)dξ∞⎪integraldisplay 0Gs(k)sinkξcoskxdk =1 2∞⎪integraldisplay 0f(ξ)dξ⎪radicalbigg 2 π∞⎪integraldisplay 0Gs(k)[sink(ξ+x)+s i n k(ξ−x)]dk =1 2∞⎪integraldisplay 0f(ξ)[g(ξ+x)+g(ξ−x)]dξ, in which the inverse Fourier sine transform is used. Thus, we ﬁnd ∞⎪integraldisplay 0Fs(k)Gs(k)coskxdk =1 2∞⎪integraldisplay 0f(ξ)[g(ξ+x)+g(ξ−x)]dξ. (2.14.10) Or, equivalently, F−1 c{Fs(k)Gs(k)}=1 √ 2π∞⎪integraldisplay 0f(ξ)[g(ξ+x)+g(ξ−x)]dξ. (2.14.11) Result (2.14.10) or (2.14.11) is also called the Convolution Theorem of the Fourier cosine transform. Putting x= 0 in (2.14.10) gives ∞⎪integraldisplay 0Fs(k)Gs(k)dk=∞⎪integraldisplay 0f(ξ)g(ξ)dξ=∞⎪integraldisplay 0f(x)g(x)dx. © 2007 by Taylor & Francis Group, LLC 96 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Replacing g(x)b y f(x)g i v e st h e Parseval relation for the Fourier sine trans- form ∞⎪integraldisplay 0|Fs(k)|2dk=∞⎪integraldisplay 0|f(x)|2dx. (2.14.12) 2.15 Applications of Fourier Cosine and Sine Transforms to Partial Diﬀerential Equations Example 2.15.1 (One-Dimensional Diﬀusion Equation on a Half Line ). Consider the initial- boundary value problem for the one-dimensional diﬀusion equation in 0 <x< ∞with no sources or sinks: ∂u ∂t=κ∂2u ∂x2,0<x< ∞,t > 0, (2.15.1) where κis a constant, with the initial condition u(x,0)= 0 ,0<x< ∞, (2.15.2) and the boundary conditions (a)u(0,t)=f(t),t≥0,u(x, t)→0a s x→∞, (2.15.3) or, (b)ux(0,t)=f(t),t≥0,u(x, t)→0a s x→∞. (2.15.4) This problem with the boundary conditions (2.15.3) is solved by using the Fourier sine transform Us(k,t)=⎪radicalbigg 2 π∞⎪integraldisplay 0sinkxu(x, t)dx. Application of the Fourier sine transform gives dUs dt=−κk2Us(k,t)+⎪radicalbigg 2 πκkf(t), (2.15.5) Us(k,0) = 0 . (2.15.6) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 97 The bounded solution of this diﬀerential system with Us(k,0)= 0 is Us(k,t)=⎪radicalbigg 2 πκkt⎪integraldisplay 0f(τ)exp[−κ(t−τ)k2]dτ. (2.15.7) The inverse transform gives the solution u(x, t)=⎪radicalbigg 2 πκt⎪integraldisplay 0f(τ)F−1 s{kexp[−κ(t−τ)k2]}dτ =x √ 4πκt⎪integraldisplay 0f(τ)exp⎪bracketleftbigg −x2 4κ(t−τ)⎪bracketrightbiggdτ (t−τ)3/2(2.15.8) in which F−1 s{kexp(−tκk2)}=x 2√ 2·exp(−x2/4κt) (κt)3/2is used. In particular, f(t)=T0= constant, (2.15.7) reduces to Us(k,t)=⎪radicalbigg 2 πT0 k[1−exp(−κtk2)]. (2.15.9) Inversion gives the solution u(x, t)=⎪parenleftbigg2T0 π⎪parenrightbigg∞⎪integraldisplay 0sinkx k[1−exp(−κtk2)]dk. (2.15.10) Making use of the integral ∞⎪integraldisplay 0e−k2a2sinkx kdk=π 2erf⎪parenleftBigx 2a⎪parenrightBig , (2.15.11) the solution becomes u(x, t)=2T0 π⎪bracketleftbiggπ 2−π 2erf⎪parenleftbiggx 2√ κt⎪parenrightbigg⎪bracketrightbigg =T0erfc⎪parenleftbiggx 2√ κt⎪parenrightbigg , (2.15.12) where the error function ,erf(x) is deﬁned by erf(x)=2 √ πx⎪integraldisplay 0e−α2dα, (2.15.13) © 2007 by Taylor & Francis Group, LLC 98 INTEGRAL TRANSFORMS and THEIR APPLICATIONS so that erf(0) = 0 ,erf(∞)=2 √ π∞⎪integraldisplay 0e−α2dα=1,anderf(−x)=−erf(x), and the complementary error function ,erfc(x) is deﬁned by erfc(x)=1−erf(x)=2 √ π∞⎪integraldisplay xe−α2dα, (2.15.14) so that erfc(x)=1−erf(x),erfc(0) = 1 ,erfc(∞)=0, and erfc(−x)=1−erf(−x)=1+ erf(x)=2−erfc(x). Equation (2.15.1) with boundary condition (2.15.4) is solved by the Fourier cosine transform Uc(k,t)=⎪radicalbigg 2 π∞⎪integraldisplay 0coskxu(x, t)dx. Application of this transform to (2.15.1) gives dUc dt+κk2Uc=−⎪radicalbigg 2 πκf(t). (2.15.15) The solution of (2.15.15) with Uc(k,0) = 0 is Uc(k,t)=−⎪radicalbigg 2 πκt⎪integraldisplay 0f(τ)exp [−k2κ(t−τ)]dτ. (2.15.16) Since F−1 c{exp(−tκk2)}=1 √ 2κtexp⎪parenleftbigg −x2 4κt⎪parenrightbigg , (2.15.17) the inverse Fourier cosine transform gives the ﬁnal form of the solution u(x, t)=−⎪radicalbigg κ πt⎪integraldisplay 0f(τ) √ t−τexp⎪bracketleftbigg −x2 4κ(t−τ)⎪bracketrightbigg dτ. (2.15.18) Example 2.15.2 (The Laplace Equation in the Quarter Plane ). Solve the Laplace equation uxx+uyy=0,0<x , y< ∞, (2.15.19) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 99 with the boundary conditions u(0,y)=a, u (x,0) = 0 , (2.15.20a) ∇u→0a s r=⎪radicalbig x2+y2→∞, (2.15.20b) where ais a constant. We apply the Fourier sine transform with respect to xto ﬁnd d2Us dy2−k2Us+⎪radicalbigg 2 πka=0. The solution of this inhomogeneous equation is Us(k,y)=Ae−ky+⎪radicalbigg 2 π·a k, where Ais a constant to be determined from Us(k,0) = 0. Consequently, Us(k,y)=a k⎪radicalbigg 2 π(1−e−ky). (2.15.21) The inverse transformation gives the formal solution u(x, y)=2a π∞⎪integraldisplay 01 k(1−e−ky)sinkxdk Or, u(x, y)=2a π⎡ ⎣∞⎪integraldisplay 0sinkx kdk−∞⎪integraldisplay 01 ke−kysinkxdk⎤ ⎦ =a−2a π⎪parenleftBigπ 2−tan−1y x⎪parenrightBig =2a πtan−1⎪parenleftBigy x⎪parenrightBig ,(2.15.22) in which (2.13.9) is used. Example 2.15.3 (The Laplace Equation in a Semi-Inﬁnite Strip with the Dirichlet Data ). Solve the Laplace equation uxx+uyy=0,0<x< ∞,0<y<b , (2.15.23) with the boundary conditions u(0,y)=0,u(x, y)→0a s x→∞ for 0<y<b (2.15.24) u(x, b)=0,u(x,0)=f(x)f o r0 <x< ∞. (2.15.25) © 2007 by Taylor & Francis Group, LLC 100 INTEGRAL TRANSFORMS and THEIR APPLICATIONS In view of the Dirichlet data, the Fourier sine transform with respect to xcan be used to solve this problem. Applying the Fourier sine transform to (2.15.23)–(2.15.25) gives d2Us dy2−k2Us=0, (2.15.26) Us(k,b)=0,U s(k,0)=Fs(k). (2.15.27) The solution of (2.15.26) with (2.15.27) is Us(k,y)=Fs(k)sinh[k(b−y)] sinhkb. (2.15.28) The inverse Fourier sine transform gives the formal solution u(x, y)=⎪radicalbigg 2 π∞⎪integraldisplay 0Fs(k)sinh[k(b−y)] sinhkbsinkxdk =2 π∞⎪integraldisplay 0⎡ ⎣∞⎪integraldisplay 0f(l)sinkldl⎤ ⎦sinh[k(b−y)] sinhkbsinkxdk. (2.15.29) In the limit as kb→∞,sinh[k(b−y)] sinhkb∼exp(−ky), hence the above problem re- duces to the corresponding problem in the quarter plane, 0 <x< ∞,0<y< ∞. Thus, solution (2.15.29) becomes u(x, y)=2 π∞⎪integraldisplay 0f(l)dl∞⎪integraldisplay 0sinklsinkxexp(−ky)dk =1 π∞⎪integraldisplay 0f(l)dl∞⎪integraldisplay 0{cosk(x−l)−cosk(x+l)}exp(−ky)dk =1 π∞⎪integraldisplay 0f(l)⎪bracketleftbiggy (x−l)2+y2−y (x+l)2+y2⎪bracketrightbigg dl. (2.15.30) This is the exact integral solution of the problem. If f(x) is an odd function ofx, then solution (2.15.30) reduces to the solution (2.12.10) of the same problem in the half plane. 2.16 Evaluation of Deﬁnite Integrals The Fourier transform can be employed to evaluate certain deﬁnite integrals.Although the method of evaluation may not be very rigorous, it is quite simple and straightforward. The method can be illustrated by means of examples. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 101 Example 2.16.1 Evaluate the integral I(a, b)=∞⎪integraldisplay −∞dx (x2+a2)(x2+b2),a > 0,b>0. (2.16.1) If we write f(x)=e−a|x|andg(x)=e−b|x|thenF(k)=⎪radicalBig 2 πa (k2+a2),G(k)=⎪radicalBig 2 πb (k2+b2). The Convolution Theorem 2.5.5 gives (2.5.19), that is, ∞⎪integraldisplay −∞F(k)G(k)dk=∞⎪integraldisplay −∞f(x)g(−x)dx. Or, equivalently, ∞⎪integraldisplay −∞dk (k2+a2)(k2+b2)=π 2ab∞⎪integraldisplay −∞e−|x|(a+b)dx =π ab∞⎪integraldisplay 0e−(a+b)xdx=π ab(a+b).(2.16.2) This is the desired result. Further∞⎪integraldisplay 0dx (x2+a2)(x2+b2)=π 2ab(a+b). (2.16.3) Example 2.16.2 Show that∞⎪integraldisplay 0x−pdx (a2+x2)=π 2a−(p+1)sec⎪parenleftBigπp 2⎪parenrightBig . (2.16.4) We write f(x)=e−axso that Fc(k)=⎪radicalbigg 2 πa (a2+k2). g(x)=xp−1so that Gc(k)=⎪radicalbigg 2 πk−pΓ(p)cos⎪parenleftBigπp 2⎪parenrightBig . Using Parseval’s result for the Fourier cosine transform gives ∞⎪integraldisplay 0Fc(k)Gc(k)dk=∞⎪integraldisplay 0f(x)g(x)dx. © 2007 by Taylor & Francis Group, LLC 102 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Or, 2a πcos⎪parenleftBigπp 2⎪parenrightBig Γ(p)∞⎪integraldisplay 0k−pdk k2+a2=∞⎪integraldisplay 0xp−1e−axdx =1 ap∞⎪integraldisplay 0e−ttp−1dt=Γ(p) ap,(ax=t). Thus, ∞⎪integraldisplay 0k−pdk a2+k2=π 2ap+1sec⎪parenleftBigπp 2⎪parenrightBig . Example 2.16.3 Ifa>0,b>0,show that ∞⎪integraldisplay 0x2dx (a2+x2)(b2+x2)=π 2(a+b). (2.16.5) We consider Fs{e−ax}=⎪radicalbigg 2 πk k2+a2=Fs(k) Fs{e−bx}=⎪radicalbigg 2 πk k2+b2=Gs(k). Then the Convolution Theorem for the Fourier cosine transform gives ∞⎪integraldisplay 0Fs(k)Gs(k)coskxdk =1 2∞⎪integraldisplay 0g(ξ)[f(ξ+x)+f(ξ−x)]dξ. Putting x=0 g i v e s ∞⎪integraldisplay 0Fs(k)Gs(k)dk=∞⎪integraldisplay 0g(ξ)f(ξ)dξ, or, ∞⎪integraldisplay 0k2dk (k2+a2)(k2+b2)=π 2∞⎪integraldisplay 0e−(a+b)ξdξ=π 2(a+b). © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 103 Example 2.16.4 Show that∞⎪integraldisplay 0x2dx (x2+a2)4=π (2a)5,a > 0. (2.16.6) We write f(x)=1 2(x2+a2)so that f/prime(x)=−x (x2+a2)2,a n dF{f(x)}=F(k)=⎪radicalbig π 2⎪parenleftbig1 2a⎪parenrightbig exp(−a|k|). Making reference to the Parseval relation (2.4.19), we obtain ∞⎪integraldisplay −∞|f/prime(x)|2dx=∞⎪integraldisplay −∞|F{f/prime(x)}|2dk=∞⎪integraldisplay −∞|(ik)F{f(x)}|2dk. Thus, ∞⎪integraldisplay −∞x2 (x2+a2)4dx=π 2∞⎪integraldisplay −∞k2·1 (2a)2exp(−2a|k|)dk =π (2a)2∞⎪integraldisplay 0k2exp(−2ak)dk=2π (2a)5. This gives the desired result. 2.17 Applications of Fourier Transforms in Mathematical Statistics In probability theory and mathematical statistics, the characteristic function of a random variable is deﬁned by the Fourier transform or by the Fourier- Stieltjes transform of the distribution function of a random variable. Many important results in probability theory and mathematical statistics can be obtained, and their proofs can be simpliﬁed with rigor by using the methods of characteristic functions. Thus, the Fourier transforms play an importantrole in probability theory and mathematical statistics. DEFINITION 2.17.1 (Distribution Function). The distribution function F(x)of a random variable Xis deﬁned as the probability, that is, F(x)= P(X<x )for every real number x. It is immediately evident from this deﬁnition that the distribution function satisﬁes the following properties: © 2007 by Taylor & Francis Group, LLC 104 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (i)F(x) is a non-decreasing function, that is, F(x1)≤F(x2)i fx1<x2. (ii)F(x) is continuous only from the left at a point x,t h a ti s , F(x−0)= F(x), but F(x+0 )/negationslash=F(x). (iii)F(−∞)=0a n d F(+∞)=1. IfXis a continuous variable and if there exists a non-negative function f(x) such that for every real xthe following relation holds: F(x)=x⎪integraldisplay −∞f(x)dx, (2.17.1) where F(x) is the distribution function of the random variable X, then the function f(x) is called the probability density or simply the density function of the random variable X. It is immediately obvious that every density function f(x) satisﬁes the following properties: (i) F(+∞)=∞⎪integraldisplay −∞f(x)dx=1. (2.17.2a) (ii) For every real aandbwhere a<b, P(a≤X≤b)=F(b)−F(a)=b⎪integraldisplay af(x)dx. (2.17.2b) (iii) If f(x) is continuous at some point x,t h e n F/prime(x)=f(x). It is noted that every real function f(x) which is non-negative, and in- tegrable over the whole real line and satisﬁes (2.17.2ab), is the probability density function of a continuous random variable X. On the other hand, the function F(x) deﬁned by (2.17.1) satisﬁes all properties of a distribution func- tion. DEFINITION 2.17.2 (Characteristic Function). If Xis a continuous random variable with the density function f(x), then the characteristic func- tion,φ(t)of the random variable Xor of the distribution function F(x)is deﬁned by the formula φ(t)=E(exp(itX)) =∞⎪integraldisplay −∞f(x)exp (itx)dx, (2.17.3) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 105 where E[g(X)] is called the expected value of the random variable g(X). In problems of mathematical statistics, it is convenient to deﬁne the Fourier transform of f(x) and its inverse in a slightly diﬀerent way by F{f(x)}=φ(t)=∞⎪integraldisplay −∞exp(itx)f(x)dx, (2.17.4) F−1{φ(t)}=f(x)=1 2π∞⎪integraldisplay −∞exp(−itx)φ(t)dt. (2.17.5) Evidently, the characteristic function of F(x) is the Fourier transform of the density function f(x). The Fourier transform of the distribution function fol- lows from the fact that F{F/prime(x)}=F{f(x)}=φ(t), or, F{F(x)}=it−1φ(t). (2.17.6) Thecomposition of two distribution functions F1(x)a n d F2(x) is deﬁned by F(x)=F1(x)∗F2(x)=∞⎪integraldisplay −∞F1(x−y)F/prime 2(y)dy. (2.17.7) Thus, the Fourier transform of (2.17.7) gives it−1φ(t)=F⎧ ⎨ ⎩∞⎪integraldisplay −∞F1(x−y)F/prime 2(y)dy⎫ ⎬ ⎭ =F{F1(x)}F{f2(x)}=it−1φ1(t)φ2(t), whence an important result follows: φ(t)=φ1(t)φ2(t), (2.17.8) where φ1(t)a n d φ2(t) are the characteristic functions of the distribution func- tionsF1(x)a n d F2(x) respectively. Thenth moment of a random variable Xis deﬁned by mn=E[Xn]=∞⎪integraldisplay −∞xnf(x)dx, n =1,2,3,.... (2.17.9) provided this integral exists. The ﬁrst moment m1(or simply m) is called the expectation ofXand has the form m=E(X)=∞⎪integraldisplay −∞xf(x)dx. (2.17.10) © 2007 by Taylor & Francis Group, LLC 106 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thus, the moment of any order nis calculated by evaluating the integral (2.17.9). However, the evaluation of the integral is, in general, a diﬃcult task.This diﬃculty can be resolved with the help of the characteristic function deﬁned by (2.17.4). Diﬀerentiating (2.17.4) ntimes and putting t=0 g i v e sa fairly simple formula m n=∞⎪integraldisplay −∞xnf(x)dx=(−i)nφ(n)(0), (2.17.11) where n=1,2,3,.... When n= 1, the expectation of a random variable Xbecomes m1=E(X)=∞⎪integraldisplay −∞xf(x)dx=(−i)φ/prime(0). (2.17.12) Thus, the simple formula (2.17.11) involving the derivatives of the character- istic function provides for the existe nce and the computation of the moment of any arbitrary order. Similarly, the variance σ2of a random variable is given in terms of the characteristic function as σ2=∞⎪integraldisplay −∞(x−m)2f(x)dx=m2−m2 1 ={φ/prime(0)}2−φ/prime/prime(0). (2.17.13) Example 2.17.1 Find the moments of the normal distribution deﬁned by the density function f(x)=1 σ√ 2πexp⎪braceleftbigg −(x−m)2 2σ2⎪bracerightbigg . (2.17.14) The characteristic function of the normal distribution is the Fourier transform off(x), which is φ(t)=1 σ√ 2π∞⎪integraldisplay −∞eitxexp⎪bracketleftbigg −(x−m)2 2σ2⎪bracketrightbigg dx. We substitute x−m=yand use Example 2.3.1 to obtain φ(t)=exp(itm) σ√ 2π∞⎪integraldisplay −∞eityexp⎪parenleftbigg −y2 2σ2⎪parenrightbigg dy=e x p⎪parenleftbigg itm−1 2t2σ2⎪parenrightbigg .(2.17.15) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 107 Thus, m1=(−i)φ/prime(0) =m, m2=−φ/prime/prime(0) = ( m2+σ2), m3=m(m2+3σ2). Finally, the variance of the normal distribution is m2−m2 1=σ2. (2.17.16) The above discussion reveals that characteristic functions are very useful for investigation of certain problems in mathematical statistics. We close this section by discussing more properties of characteristic functions. THEOREM 2.17.1 (Addition Theorem ). The characteristic function of the sum of a ﬁnite number of independent random variables is equal to the product of their characteristic functions. PROOF Suppose X1,X2,...,X narenindependent random variables and Z=X1+X2+···+Xn. Further, suppose φ1(t),φ2(t),...,φ n(t), and φ(t)a r e the characteristic functions of X1,X2,...,X nandZrespectively. Then we have φ(t)=E[exp(itZ)] =E[exp{it(X1+X2+···+Xn)}], which is, by the independence of the random variables, =E(eitX1)E(eitX2)···E(eitXn) =φ1(t)φ2(t)···φn(t). (2.17.17) This proves the Addition Theorem . Example 2.17.2 Find the expected value and the standard deviation of the sum of nindepen- dent normal random variables. Suppose X1,X2,...,X narenindependent random variables with the nor- mal distributions N(mr,σr), where r=1,2,...,n . The respective character- istic functions of these distributions are φr(t)=e x p⎪bracketleftbigg itmr−1 2t2σ2 r⎪bracketrightbigg ,r=1,2,3,...,n. (2.17.18) © 2007 by Taylor & Francis Group, LLC 108 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Because of the independence of X1,X2,...,X n, the random variable Z=X1+ X2+···+Xnhas the characteristic function φ(t)=φ1(t)φ2(t)···φn(t) =e x p⎪bracketleftbigg it(m1+m2+···+mn)−1 2(σ2 1+σ2 2+···+σ2 n)t2⎪bracketrightbigg .(2.17.19) This represents the characteristic function of the normal distribution N(m1+ ···+mn,⎪radicalbig σ2 1+···+σ2n). Thus, the expected value of Zis (m1+m2+···+ mn) and its standard deviation is ( σ2 1+σ2 2+···+σ2 n)1 2. Finally, we state the fundamental Central Limit Theorems without proof. THEOREM 2.17.2 (The L ´evy-Cram ´er Theorem ). Suppose {Xn}is a sequence of random vari- ables, Fn(x)a n d φn(t) are respectively the distribution and characteristic functions of Xn. Then the sequence {Fn(x)}is convergent to a distribution function F(x) if and only if the sequence {φn(t)}is convergent at every point ton the real line to a function φ(t) continuous in some neighborhood of the origin. The limit function φ(t) is then the characteristic function of the limit distribution function F(x), and the convergence φn(t)→φ(t) is uniform in every ﬁnite interval on the t-axis. THEOREM 2.17.3 (The Central Limit Theorem in Probability ). Supose f(x) is a nonnegative absolutely integrable function in Rand has the following properties: ∞⎪integraldisplay −∞f(x)dx=1,∞⎪integraldisplay −∞xf(x)dx=1,∞⎪integraldisplay −∞x2f(x)dx=1. Iffn=f∗f∗...∗fis the convolution product of fwith itself ntimes, then lim n→∞b√ n⎪integraldisplay a√ nfn(x)dx=1 √ 2πb⎪integraldisplay ae−x2dx−∞<a<b< ∞.(2.17.20) For a proof of the theorem, we refer the reader to Chandrasekharan (1989). All these ideas developed in this section can be generalized for the multi- dimensional distribution functions by t he use of multiple Fourier transforms. We refer interested readers to Lukacs (1960). © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 109 2.18 Multiple Fourier Transforms and Their Applications DEFINITION 2.18.1 Under the assumptions on f(x)similar to those made for the one dimensional case, the multiple Fourier transform of f(x), wherex=(x1,x2,...,x n)is the n-dimensional vector, is deﬁned by F{f(x)}=F(κ)=1 (2π)n/2∞⎪integraldisplay −∞···∞⎪integraldisplay −∞exp{−i(κ·x)}f(x)dx,(2.18.1) where κ=(k1,k2,...,k n)is the n-dimensional transform vector and κ·x= (k1x1+k2x2+···+knxn). The inverse Fourier transform is similarly deﬁned by F−1{F(κ)}=f(x)=1 (2π)n/2∞⎪integraldisplay −∞···∞⎪integraldisplay −∞exp{i(κ·x)}F(κ)dκ.(2.18.2) In particular, the double Fourier transform is deﬁned by F{f(x, y)}=F(k,/lscript)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp{−i(κ·r)}f(x, y)dxdy, (2.18.3) wherer=(x, y)andκ=(k,/lscript). The inverse Fourier transform is given by F−1{F(k,/lscript)}=f(x, y)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp{i(κ·r)}F(k,/lscript)dk d/lscript. (2.18.4) Similarly, the three-dimensional Fourier transform and its inverse are de- ﬁned by the integrals F{f(x, y, z )}=F(k,/lscript,m ) =1 (2π)3/2∞⎪integraldisplay −∞∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp{−i(κ·r)}f(x, y, z )dxdy dz, (2.18.5) F−1{F(k,/lscript,m )}=f(x, y, z ) 1 (2π)3/2∞⎪integraldisplay −∞∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp{i(κ·r)}F(k,/lscript,m )dk d/lscript dm. (2.18.6) © 2007 by Taylor & Francis Group, LLC 110 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The operational properties of these multiple Fourier transforms are similar to those of the one-dimensional case. In particular, results (2.4.7) and (2.4.8)relating the Fourier transforms of derivatives to the Fourier transforms of given functions are valid for the higher dimensional case as well. In higher dimensions, they are applied to the transforms of partial derivatives of f(x) under the assumptions that fand its partial derivatives vanish at inﬁnity. We illustrate the multiple Fourier transform method by the following ex- amples of applications: Example 2.18.1 (The Dirichlet Problem for the Three-Dimensional Laplace Equation in the Half-Space ). The boundary value problem for u(x, y, z ) satisﬁes the following equation and boundary conditions: ∇ 2u≡uxx+uyy+uzz=0,−∞<x ,y< ∞,z > 0,(2.18.7) u(x, y,0)=f(x, y) −∞<x ,y< ∞ (2.18.8) u(x, y, z )→0a s r=⎪radicalbig x2+y2+z2→∞. (2.18.9) We use the double Fourier transform deﬁned by (2.18.3) to the system (2.18.7)–(2.18.9) which reduces to d2U dz2−κ2U=0 f o r z>0,(κ2=k2+l2) U(k,/lscript,0) =F(k,/lscript). Thus, the solution of this transformed problem is U(k,/lscript,z)=F(k,/lscript)exp (−|κ|z)=F(k,/lscript)G(k,/lscript), (2.18.10) where κ=(k,/lscript)a n d G(k,/lscript)=e x p ( −|κ|z)s ot h a t g(x, y)=F−1{exp(−|κ|z)}=z (x2+y2+z2)3/2. (2.18.11) Applying the Convolution Theorem to (2.18.10), we obtain the formal solution u(x, y, z )=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞f(ξ,η)g(x−ξ,y−η,z)dξ dη =z 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞f(ξ,η)dξ dη [(x−ξ)2+(y−η)2+z2]3/2.(2.18.12) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 111 Example 2.18.2 (The Two-Dimensional Diﬀusion Equation ). We solve the two-dimensional diﬀusion equation ut=K∇2u,−∞<x ,y< ∞,t > 0, (2.18.13) with the initial and boundary conditions u(x, y,0)=f(x, y)−∞<x ,y< ∞, (2.18.14) u(x, y, t)→0a s r=⎪radicalbig x2+y2→∞, (2.18.15) where Kis the diﬀusivity constant. The double Fourier transform of u(x, y, t) deﬁned by (2.18.3) is used to reduce the system (2.18.13)–(2.18.14) into the form dU dt=−κ2KU, t> 0, U(k,/lscript,0) =F(k,/lscript). The solution of this system is U(k,/lscript,t)=F(k,/lscript)exp (−tKκ2)=F(k,/lscript)G(k,/lscript), (2.18.16) where G(k,/lscript)=e x p ( −Kκ2t), so that g(x, y)=F−1{exp(−tKκ2)}=1 2Ktexp⎪parenleftbigg −x2+y2 4Kt⎪parenrightbigg . (2.18.17) Finally, the Convolution Theorem gives the formal solution u(x, y, t)=1 4πKt∞⎪integraldisplay −∞∞⎪integraldisplay −∞f(ξ,η)exp⎪bracketleftbigg −(x−ξ)2+(y−η)2 4Kt⎪bracketrightbigg dξ dη. (2.18.18) Or, equivalently, u(x, y, t)=1 4πKt∞⎪integraldisplay −∞∞⎪integraldisplay −∞f(r/prime)exp⎪braceleftbigg −|r−r/prime|2 4Kt⎪bracerightbigg dr/prime, (2.18.19) where r/prime=(ξ,η). We make the change of variable ( r/prime−r)=√ 4KtRto reduce (2.18.19) in the form u(x, y, t)=1 π√ 4Kt∞⎪integraldisplay −∞∞⎪integraldisplay −∞f(r+√ 4KtR)exp (−R2)dR. (2.18.20) © 2007 by Taylor & Francis Group, LLC 112 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Similarly, the formal solution of the initial value problem for the three-dimensional diﬀusion equation ut=K(uxx+uyy+uzz),−∞<x ,y,z< ∞,t > 0 (2.18.21) u(x, y, z, 0)=f(x, y, z ),−∞<x ,y,z< ∞ (2.18.22) is given by u(x, y, z, t )=1 (4πKt)3/2⎪integraldisplay∞⎪integraldisplay −∞⎪integraldisplay f(ξ,η,ζ)exp⎪parenleftbigg −r2 4Kt⎪parenrightbigg dξ dη dζ, (2.18.23) where r2=(x−ξ)2+(y−η)2+(z−ζ)2. Or, equivalently, u(x, y, z, t )=1 (4πKt)3/2⎪integraldisplay∞⎪integraldisplay −∞⎪integraldisplay f(r/prime)exp⎪braceleftbigg −|r−r/prime|2 4Kt⎪bracerightbigg dξ dη dζ, (2.18.24) where r=(x, y, z )a n dr/prime=(ξ,η,ζ). Making the change of variable r/prime−r=√ 4tKR,solution (2.18.24) reduces to u(x, y, z, t )=1 π3/24Kt⎪integraldisplay∞⎪integraldisplay −∞⎪integraldisplay f(r+√ 4KtR)exp (−R2)dR.(2.18.25) This is known as the Fourier solution. Example 2.18.3 (The Cauchy Problem for the Two-Dimensional Wave Equation ). The initial value problem for the wave equation in two dimensions is governed by utt=c2(uxx+uyy),−∞<x ,y< ∞,t > 0, (2.18.26) with the initial data u(x, y,0)= 0 ,u t(x, y,0) =f(x, y),−∞<x ,y< ∞, (2.18.27ab) where cis a constant. We assume that uand its ﬁrst partial derivatives vanish at inﬁnity. We apply the two-dimensional Fourier transform deﬁned by (2.18.3) to the system (2.18.26)–(2.18.27ab), which becomes d2U dt2+c2κ2U=0,κ2=k2+/lscript2, U(k,/lscript,0) = 0 ,⎪parenleftbiggdU dt⎪parenrightbigg t=0=F(k,/lscript). © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 113 The solution of this transformed system is U(k,/lscript,t)=F(k,/lscript)sin(cκt) cκ. (2.18.28) The inverse Fourier transform gives the formal solution u(x, y, t)=1 2πc∞⎪integraldisplay −∞⎪integraldisplay exp(iκ·r)sin(cκt) κF(κ)dκ (2.18.29) =1 4iπc∞⎪integraldisplay −∞∞⎪integraldisplay −∞F(κ) κ⎪bracketleftBig exp⎪braceleftBig iκ⎪parenleftBigκ·r κ+ct⎪parenrightBig⎪bracerightBig −exp⎪braceleftBig iκ⎪parenleftBigκ·r κ−ct⎪parenrightBig⎪bracerightBig⎪bracketrightBig dκ.(2.18.30) The form of this solution reveals an interesting feature of the wave equation. The exponential terms exp⎪braceleftbig iκ⎪parenleftbig ct±κ·r κ⎪parenrightbig⎪bracerightbig involved in the integral solution (2.18.30) represent plane wave solutions of the wave equation (2.18.26). Thus, the solutions remain constant on the planes κ·r= constant that move par- allel to themselves with velocity c. Evidently, solution (2.18.30) represents a superposition of the plane wave solutions traveling in all possible directions. Similarly, the solution of the Cauchy problem for the three-dimensional wave equation utt=c2(uxx+uyy+uzz),−∞<x ,y,z< ∞,t >0, (2.18.31) u(x, y, z, 0)= 0 ,ut(x, y, z, 0)=f(x, y, z ),−∞<x ,y,z< ∞ (2.18.32ab) is given by u(r,t)=1 2ic(2π)3/2⎪integraldisplay∞⎪integraldisplay −∞⎪integraldisplayF(κ) κ⎪bracketleftBigg exp⎪braceleftBig iκ⎪parenleftBigκ·r κ+ct⎪parenrightBig⎪bracerightBig −exp⎪braceleftBig iκ⎪parenleftBigκ·r κ−ct⎪parenrightBig⎪bracerightBig⎪bracketrightBigg dκ,(2.18.33) where r=(x, y, z )a n d κ=(k,/lscript,m ). In particular, when f(x, y, z )=δ(x)δ(y)δ(z)s ot h a t F(κ)=( 2π)−3/2,s o l u - tion (2.18.33) becomes u(r,t)=1 (2π)3⎪integraldisplay∞⎪integraldisplay −∞⎪integraldisplay⎪parenleftbiggsincκt cκ⎪parenrightbigg exp(i(κ·r))dκ. (2.18.34) In terms of the spherical polar coordinates ( κ,θ,φ ) where the polar axis (the z-axis) is taken along the rdirection with κ·r=κrcosθ, we write (2.18.34) © 2007 by Taylor & Francis Group, LLC 114 INTEGRAL TRANSFORMS and THEIR APPLICATIONS in the form u(r, t)=1 (2π)32π⎪integraldisplay 0dφπ⎪integraldisplay 0dθ∞⎪integraldisplay 0exp(iκrcosθ)sincκt cκ·κ2sinθd κ =1 2π2cr∞⎪integraldisplay 0sin(cκt)sin (κr)dκ =1 8π2cr∞⎪integraldisplay −∞[eiκ(ct−r)−eiκ(ct+r)]dκ. Or, u(r, t)=1 4πcr[δ(ct−r)−δ(ct+r)]. (2.18.35) Fort>0,c t+r>0s ot h a t δ(ct+r) = 0 and hence, u(r,t)=1 4πcrδ(ct−r)=1 4πc2rδ(t−r c). (2.18.36) Example 2.18.4 (The Three-Dimensional Poisson Equation ). The solution of the Poisson e- quation −∇2u=f(r), (2.18.37) where r=(x, y, z )i sg i v e nb y u(r)=∞⎪integraldisplay −∞∞⎪integraldisplay −∞∞⎪integraldisplay −∞G(r,ξ)f(ξ)dξ, (2.18.38) where the Green’s function G(r,ξ) of the operator, −∇2,i s G(r,ξ)=1 4π1 |r−ξ|. (2.18.39) To obtain the fundamental solution, we need to solve the equation −∇2G(r,ξ)=δ(x−ξ)δ(y−η)δ(z−ζ),r/negationslash=ξ. (2.18.40) Application of the three-dimensional Fourier transform deﬁned by (2.18.5) to (2.18.40) gives κ2ˆG(κ,ξ)=1 (2π)3/2exp(−iκ·ξ), (2.18.41) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 115 where ˆG(κ,ξ)=F{G(r,ξ)}andκ=(k,/lscript,m ). The inverse Fourier transform gives the formal solution G(r,ξ)=1 (2π)3∞⎪integraldisplay −∞∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp{iκ·(r−ξ)}dκ κ2 =1 (2π)3∞⎪integraldisplay −∞∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp(iκ·x)dκ κ2, (2.18.42) where x=|r−ξ|. We evaluate this integral using polar coordinates in the κ-space with the axis along the x-axis. In terms of spherical polar coordinates ( κ,θ,φ )s ot h a t κ·x=κRcosθwhere R=|x|. Thus, (2.18.42) becomes G(r,ξ)=1 (2π)32π⎪integraldisplay 0dφπ⎪integraldisplay 0dθ∞⎪integraldisplay 0exp(iκRcosθ)κ2sinθ·dκ κ2 =1 (2π)2∞⎪integraldisplay 02sin (κR) κRdκ=1 4πR=1 4π|r−ξ|, (2.18.43) provided R>0. In electrodynamics, the fundamental solution (2.18.43) has a well-known interpretation. Physically, it represents the potential at point rgenerated by the unit point charge distribution at point ξ. This is what can be expected because δ(r−ξ) is the charge density corresponding to a unit point charge at ξ. The solution of (2.18.37) is then given by u(r)=∞⎪integraldisplay −∞∞⎪integraldisplay −∞∞⎪integraldisplay −∞G(r,ξ)f(ξ)dξ=1 4π∞⎪integraldisplay −∞∞⎪integraldisplay −∞∞⎪integraldisplay −∞f(ξ)dξ |r−ξ|. (2.18.44) The integrand in (2.18.44) consists of the given charge distribution f(r)a t r=ξand Green’s function G(r,ξ). Physically, G(r,ξ)f(ξ) represents the re- sulting potentials due to elementary po int charges, and the total potential due to a given charge distribution f(r) is then obtained by the integral superpo- sition of the resulting potentials. This is called the principle of superposition . Example 2.18.5 (The Two-Dimensional Helmholtz Equation ). To ﬁnd the fundamental solu- tion of the two-dimensional Helmholtz equation −∇2G+α2G=δ(x−ξ)δ(y−η),−∞<x ,y< ∞. (2.18.45) © 2007 by Taylor & Francis Group, LLC 116 INTEGRAL TRANSFORMS and THEIR APPLICATIONS It is convenient to make the change of variables x−ξ=x∗,y−η=y∗. Consequently, (2.18.45) reduces to the form, dropping the asterisks, Gxx+Gyy−α2G=−δ(x)δ(y). (2.18.46) Application of the double Fourier transform ˆG(κ)=F{G(x, y)}to (2.18.46) gives ˆG(κ)=1 2π1 (κ2+α2), (2.18.47) where κ=(k,/lscript)a n d κ2=k2+/lscript2. The inverse Fourier transform yields the solution G(x, y)=1 4π2∞⎪integraldisplay −∞∞⎪integraldisplay −∞(κ2+α2)−1exp(iκ·x)dk d/lscript. (2.18.48) In terms of polar coordinates ( x, y)=r(cosθ,sinθ),(k,/lscript)=ρ(cosφ,sinφ), the integral solution (2.18.48) becomes G(x, y)=1 4π2∞⎪integraldisplay 0ρdρ (ρ2+α2)2π⎪integraldisplay 0exp{irρcos(φ−θ)}dφ, which is, replacing the second integral by 2 πJ0(rρ), =1 2π∞⎪integraldisplay 0ρJ0(rρ)dρ (ρ2+α2). (2.18.49) In terms of the original coordinates, the fundamental solution of (2.18.45) is given by G(r,ξ)=1 2π∞⎪integraldisplay 0ρJ0⎪bracketleftBig ρ⎪braceleftbig (x−ξ)2+(y−η)2⎪bracerightbig1 2⎪bracketrightBig dρ (ρ2+α2). (2.18.50) Accordingly, the solution of the inhomogeneous equation (∇2−α2)u=−f(x, y) (2.18.51) is u(x, y)=⎪integraldisplay∞ −∞⎪integraldisplay G(r,ξ)f(ξ)dξ, (2.18.52) where G(r,ξ) is given by (2.18.50). © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 117 Since the integral solution (2.18.49) does not exist for α= 0, Green’s func- tion for the two-dimensional Poisson equation (2.18.51) cannot be derivedfrom (2.18.49). Instead, we diﬀerentiate (2.18.49) with respect to rto obtain ∂G ∂r=1 2π∞⎪integraldisplay 0ρ2J/prime 0(rρ)dρ (ρ2+α2) which is, for α=0 , ∂G ∂r=1 2π∞⎪integraldisplay 0J/prime 0(rρ)dρ=−1 2πr. Integrating this result gives G(r, θ)=−1 2πlogr. In terms of the original coordinates, the Green’s function becomes G(r,ξ)=−1 4πlog[(x−ξ)2+(y−η)2]. (2.18.53) This is Green’s function for the two-dimensional Poisson equation ∇2=−f(x, y). Thus, the solution of the Poisson equation is u(x, y)=∞⎪integraldisplay −∞∞⎪integraldisplay −∞G(r,ξ)f(ξ)dξ. (2.18.54) Example 2.18.6 (Diﬀusion of Vorticity from a Vortex Sheet ). We solve the two-dimensional vorticity equation in the x, yplane given by ζt=ν∇2ζ (2.18.55) with the initial condition ζ(x, y,0) =ζ0(x, y), (2.18.56) where ζ=υx−uy. Application of the double Fourier transform deﬁned by ˆζ(k,/lscript,t)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp[−i(kx+/lscripty)]ζ(x, y, t)dxdy © 2007 by Taylor & Francis Group, LLC 118 INTEGRAL TRANSFORMS and THEIR APPLICATIONS to (2.18.55)–(2.18.56) gives dˆζ dt=−ν(k2+/lscript2)ˆζ, ˆζ(k,/lscript,0) =ˆζ0(k,/lscript). Thus, the solution of the transformed system is ˆζ(k,/lscript,t)=ˆζ0(k,/lscript)exp[−ν(k2+/lscript2)t]. (2.18.57) The inversion theorem for Fourier transform gives the formal solution ζ(x, y, t)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞ˆζ0(k,/lscript)exp [i(κ·r)−νκ2t]dk d/lscript, (2.18.58) where κ=(k,/lscript)a n d κ2=k2+/lscript2. In particular, if ζ0(x, y)=Vδ(x) represents a vortex sheet of constant strength Vper unit width in the plane x= 0, we ﬁnd ˆζ0(k,/lscript)=Vδ(/lscript) and hence, ζ(x, y, t)=V 2π∞⎪integraldisplay −∞exp{ikx−νk2t}dk =V 2√ πνtexp⎪parenleftbigg −x2 4νt⎪parenrightbigg . (2.18.59) Apart from a constant, the velocity ﬁeld is given by u(x, t)=0,υ(x, t)=V √ πerf⎪parenleftbiggx 2√ νt⎪parenrightbigg . (2.18.60) © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 119 2.19 Exercises 1. Find the Fourier transforms of each of the following functions: (a)f(x)=1 1+x2, (b)f(x)=x 1+x2, (c)f(x)=δ(n)(x), (d)f(x)=xexp(−a|x|),a >0, (e)f(x)=exexp(−ex), (f)f(x)=xexp⎪parenleftbigg −ax2 2⎪parenrightbigg ,a >0, (g)f(x)=x2exp⎪parenleftbigg −1 2x2⎪parenrightbigg , (h)f(x)=⎪braceleftBigg1−|x|,|x|≤1 0, |x|>1⎪bracerightBigg , (i)f(x)=⎪braceleftBigg 1−x2,|x|≤1 0, |x|>1⎪bracerightBigg ,(j)hn(x)=(−1)nexp⎪parenleftbigg1 2x2⎪parenrightbigg ×⎪parenleftbiggd dx⎪parenrightbiggn exp(−x2), (k)f(x)=χ[a,b](x)eiαx, (l)f(x)=cos sin(ax2). 2. Show that (a)F{δ(x−ct)+δ(x+ct)}=⎪radicalbigg 2 πcos(kct), (b)F{H(ct−|x|)}=⎪radicalbigg 2 πsinkct k, (c)F⎪braceleftBig f⎪parenleftBigx a+b⎪parenrightBig⎪bracerightBig =aexp(iabk)F(ak), (d)F{eibxf(ax)}=1 aF⎪parenleftbiggk+b a⎪parenrightbigg . 3. Show that (a)id dkF(k)=F{xf(x)}, (b)indn dknF(k)=F{xnf(x)}. 4. Use exercise 3(b) to ﬁnd the Fourier transform of f(x)=x2exp(−ax2). 5. Prove the following: (a)F⎪braceleftBig (a2−x2)−1 2H(a−|x|)⎪bracerightBig =⎪radicalbigg π 2J0(ak),a >0. © 2007 by Taylor & Francis Group, LLC 120 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b)F{Pn(x)H(1−|x|)}=(−i)n1 √ kJn+1 2(k), where Pn(x) is the Legendre polynomial of degree n. (c) If f(x) has a ﬁnite discontinuity at a point x=a,t h e n F{f/prime(x)}=(ik)F(k)−1 √ 2πexp(−ika)[f]a, where [ f]a=f(a+0 )−f(a−0). Generalize this result for F{f(n)(x)}. 6. Find the convolution ( f∗g)(x)i f (a)f(x)=eax,g(x)=χ[0,∞](x),a/negationslash=0, (b)f(x)=s i n bx, g (x)=e x p ( −a|x|),a >0, (c)f(x)=χ[a,b](x),g(x)=x2, (d)f(x)=e x p ( −x2),g(x)=e x p ( −x2). 7. Prove the following results for the convolution: (a)δ(x)∗f(x)=f(x), (b)δ/prime(x)∗f(x)=f/prime(x), (c)d dx{f(x)∗g(x)}=f/prime(x)∗g(x)=f(x)∗g/prime(x), (d)⎪integraldisplay∞ −∞(f∗g)(x)dx=⎪integraldisplay∞ −∞f(u)du⎪integraldisplay∞ −∞g(v)dv, (e)d2 dx2(f∗g)(x)=(f/prime∗g/prime)(x)=(f/prime/prime∗g)(x), (f) (f∗g)(n+l)(x)=f(n)(x)∗g(l)(x), (g) If fandgare both even or both odd ,then ( f∗g)(x)i se v e n , (h) If fis even or gis odd ,or vice versa ,then ( f∗g)(x)i so d d , (i) If g(x)=1 2aH(a−|x|),then ( f∗g)(x) is the average of the function f(x)i n[x−a, x+a], (j) If Gt(x)=1 √ 4πkt∞⎪integraldisplay −∞f(ξ)exp⎪bracketleftbigg −(x−ξ)2 4kt⎪bracketrightbigg dξ, thenGt(x)∗Gs(x)=Gt+s(x). 8. Use the Fourier transform to solve the following ordinary diﬀerential equations in −∞<x< ∞: (a)y/prime/prime(x)−y(x)+2f(x)=0,where f(x)=0 w h e n x<−aand when x>a,a n d y(x) and its derivatives vanish at x=±∞, © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 121 (b) 2y/prime/prime(x)+xy/prime(x)+y(x)=0, (c)y/prime/prime(x)+xy/prime(x)+y(x)=0, (d)y/prime/prime(x)+xy/prime(x)+xy(x)=0,(e) ¨y(t)+2α˙y(t)+ω2y(t)=f(t). 9. Solve the following integral equations for an unknown function f(x): (a)∞⎪integraldisplay −∞φ(x−t)f(t)dt=g(x). (b)∞⎪integraldisplay −∞exp(−at2)f(x−t)dt=e x p ( −bx2),a > b > 0. (c)∞⎪integraldisplay −∞f(x−t)f(t)dt=b (x2+b2). (d)∞⎪integraldisplay −∞f(t)dt (x−t2)+a2=√ 2π (x2+b2)forb>a> 0. (e)1 π∞⎪contintegraldisplay −∞f(t)dt x−t=φ(x), where the integral in (e) is treated as the Cauchy Principal value. 10. Solve the Cauchy problem for the Klein-Gordon equation utt−c2uxx+a2u=0,−∞<x< ∞,t >0. u(x,0) =f(x),⎪parenleftbigg∂u ∂t⎪parenrightbigg t=0=g(x)f o r −∞<x< ∞. 11. Solve the telegraph equation utt−c2uxx+ut−aux=0,−∞<x< ∞,t >0. u(x,0) =f(x),⎪parenleftbigg∂u ∂t⎪parenrightbigg t=0=g(x)f o r −∞<x< ∞. Show that the solution is unstable when c2<a2.I fc2>a2, show that the bounded integral solution is u(x, t)=1 √ 2π∞⎪integraldisplay −∞A(k)exp [−k2(c2−a2)t+ik(x+at)]dk where A(k) is given in terms of the transformed functions of the initial data. Hence, deduce the asymptotic solution as t→∞ in the form u(x, t)=A(0)⎪radicalbigg π 2(c2−a2)texp⎪bracketleftbigg −(x+at)2 4(c2−a2)t⎪bracketrightbigg . © 2007 by Taylor & Francis Group, LLC 122 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 12. Solve the equation utt+uxxxx=0,−∞<x< ∞,t >0 u(x,0) =f(x),ut(x,0) = 0 for −∞<x< ∞. 13. Find the solution of the dissipative wave equation utt−c2uxx+αut=0,−∞<x< ∞,t >0, u(x,0) =f(x),⎪parenleftbigg∂u ∂t⎪parenrightbigg t=0=g(x)f o r −∞<x< ∞, where α>0 is the dissipation parameter. 14. Obtain the Fourier cosine transforms of the following functions: (a)f(x)=xexp(−ax),a >0, (b)f(x)=e−axcosx, a > 0, (c)f(x)=1 x, (d)K0(ax), where K0(ax)i st h e modiﬁed Bessel function . 15. Find the Fourier sine transform of the following functions: (a)f(x)=xexp(−ax),a >0, (b)f(x)=1 xexp(−ax),a >0, (c)f(x)=1 x, (d)f(x)=x a2+x2. 16. (a) If F(k)=F{exp(−ax2)},a >0,show that F(k) satisﬁes the dif- ferential equation 2adF dk+kF(k)=0 w i t h F(0) =1 √ 2a. (b) If Fc(k)=Fc{exp(−ax2)},show that Fc(k) satisﬁes the equation dFc dk+⎪parenleftbiggk 2a⎪parenrightbigg Fc=0 w i t h Fc(0) = 1 . 17. Prove the following for the Fourier sine transform (a)∞⎪integraldisplay 0Fs(k)Gc(k)sinkx dk =1 2∞⎪integraldisplay 0g(ξ)[f(ξ+x)−f(ξ−x)]dξ, (b)∞⎪integraldisplay 0Fc(k)Gs(k)sinkx dk =1 2∞⎪integraldisplay 0f(ξ)[g(ξ+x)−g(ξ−x)]dξ. 18. Solve the integral equation ∞⎪integraldisplay 0f(x)sinkx dk =⎪braceleftBigg1−k,0≤k<1 0,k > 1⎪bracerightBigg . © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 123 19. Solve Example 2.15.1 with the boundary data u(0,t)=0,u(x, t)→0a s x→∞,fort>0. 20. Apply the Fourier cosine transform to ﬁnd the solution u(x, y)o ft h e problem uxx+uyy=0,0<x< ∞,0<y< ∞, u(x,0) =H(a−x),a>x ;ux(0,y)=0,0<x ,y< ∞. 21. Use the Fourier cosine (or sine) transform to solve the following integral equation: (a)∞⎪integraldisplay 0f(x)coskx dx =⎪radicalbigg π 2k,(b)∞⎪integraldisplay 0f(x)sinkx dx =a a2+k2, (c)∞⎪integraldisplay 0f(x)sinkxdx=π 2J0(ak),(d)∞⎪integraldisplay 0f(x)coskx dx =sinak k. 22. Solve the diﬀusion equation in the semi-inﬁnite line ut=κuxx,0≤x<∞,t >0, with the boundary and initial data u(0,t)=0 f or t>0, u(x, t)→0a s x→∞ fort>0, u(x,0) =f(x)f o r0 <x< ∞. 23. Use the Parseval formula to evaluate the following integrals with a>0 andb>0: (a)∞⎪integraldisplay −∞dx (x2+a2)2, (b)∞⎪integraldisplay −∞sinax x(x2+b2)dx(c)∞⎪integraldisplay −∞sin2ax x2dx, (d)∞⎪integraldisplay −∞exp(−bx2)dx (x2+a2). 24. Show that ∞⎪integraldisplay 0sinaxsinbx x2dx=π 2min(a, b). © 2007 by Taylor & Francis Group, LLC 124 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 25. If f(x)=e x p ( −ax)a n d g(x)=H(t−x), show that ∞⎪integraldisplay 0sintx x(x2+a2)dx=π 2a2[1−exp(−at)]. 26. Use the Poisson summation formula to ﬁnd the sum of each of the fol- lowing series with non-zero a: (a)∞⎪summationdisplay n=−∞1 (1 +n2a2), (b)∞⎪summationdisplay n=1sinan n, (c)∞⎪summationdisplay n=1sin2an n2, (d)∞⎪summationdisplay n=−∞a n2+a2. 27. The Fokker-Planck equation (Reif, 1965) is used to describe the evolu- tion of probability distribution functions u(x, t) in nonequilibrium sta- tistical mechanics and has the form ∂u ∂t=∂ ∂x⎪parenleftbigg∂ ∂x+x⎪parenrightbigg u. The fundamental solution of this equation is deﬁned by the equation ⎪bracketleftbigg∂ ∂t−∂ ∂x⎪parenleftbigg∂ ∂x+x⎪parenrightbigg⎪bracketrightbigg G(x, ξ;t, τ)=δ(x−ξ)δ(t−τ). Show that the fundamental solution is G(x, ξ;t, τ)=[ 2π{1−exp[−2(t−τ)]}]−1 2exp⎪bracketleftbigg −{x−ξexp[−(t−τ)]}2 2[1−exp{−2(t−τ)}]⎪bracketrightbigg . Hence, derive lim t→∞G(x, ξ;t, τ)=1 √ 2πexp⎪parenleftbigg −1 2x2⎪parenrightbigg . With the initial condition u(x,0)=f(x),show that the function u(x, t) tends to the normal distribution as t→∞,that is, lim t→∞u(x, t)=1 √ 2πexp⎪parenleftbigg −1 2x2⎪parenrightbigg∞⎪integraldisplay −∞f(ξ)dξ. 28. The transverse vibration of an inﬁnite elastic beam of mass mper unit length and the bending stiﬀness EIis governed by utt+a2uxxxx=0,⎪parenleftbigg a2=EI m⎪parenrightbigg ,−∞<x< ∞,t >0. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 125 Solve this equation subject to the boundary and initial data u(0,t)= 0 for all t>0, u(x,0) =φ(x),ut(x,0) =ψ/prime/prime(x)f o r0 <x< ∞. Show that the Fourier transform solution is U(k,t)=Φ(k)cos (atk2)−Ψ(k)sin(atk2). Find the integral solution for u(x, t). 29. Solve the Lamb (1904) problem in geophysics that satisﬁes the Helmholtz equation in an inﬁnite elastic half-space uxx+uzz+ω2 c22u=0,−∞<x< ∞,z >0, where ωis the frequency and c2is the shear wave speed. At the surface of the half-space ( z=0 ),the boundary condition relating the surface stress to the impulsive point load distribution is μ∂u ∂z=−Pδ(x)a t z=0, where μis one of the Lam´ e’s constants, Pis a constant and u(x, z)→0a s z→∞ for−∞<x< ∞. Show that the solution in terms of polar coordinates is u(x, z)=P 2iμH0(2)⎪parenleftbiggωr c2⎪parenrightbigg ∼P 2iμ⎪parenleftbigg2c2 πωr⎪parenrightbigg1 2 exp⎪parenleftbiggπi 4−iωr c2⎪parenrightbigg forωr>>c 2. 30. Find the solution of the Cauchy-Poisson problem (Debnath, 1994, p. 83) in an inviscid water of inﬁnite depth which is governed by φxx+φzz=0,−∞<x< ∞,−∞<z≤0,t >0, φz−ηt=0 φt+gη=0⎪bracerightBigg onz=0,t >0, φz→0a s z→− ∞ , φ(x,0,0) = 0 and η(x,0) =Pδ(x), where φ=φ(x, z, t) is the velocity potential, η(x, t)i st h ef r e es u r f a c e elevation, and Pis a constant. Derive the asymptotic solution for the free surface elevation in the limit ast→∞. © 2007 by Taylor & Francis Group, LLC 126 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 31. Obtain the solutions for the velocity potential φ(x, z, t)a n dt h ef r e e surface elevation η(x, t) involved in the two-dimensional surface waves in water of ﬁnite (or inﬁnite) depth h. The governing equation, boundary, and free surface conditions and initial conditions (see Debnath 1994, p. 92) are φxx+φzz=0,−h≤z≤0,−∞<x< ∞,t >0 φt+gη=−P ρp(x)exp( iωt) φz−ηt=0⎫ ⎬ ⎭z=0,t >0 φ(x, z,0) = 0 = η(x,0) for all xandz. 32. Solve the steady-state surface wave problem (Debnath, 1994, p. 47) on a running stream of inﬁnite depth due to an external steady pressure applied to the free surface. The governing equation and the free surfaceconditions are φ xx+φzz=0,−∞<x< ∞,−∞<z< 0,t >0, φx+Uφx+gη=−P ρδ(x)exp( /epsilon1t) ηt+Uηx=φz⎫ ⎬ ⎭z=0,(/epsilon1>0), φz→0a s z→− ∞ . where Uis the stream velocity, φ(x, z, t) is the velocity potential, and η(x, t) is the free surface elevation. 33. Use the Fourier sine transform to solve the following initial and bound- ary value problem for the wave equation: utt=c2uxx,0<x< ∞,t >0, u(x,0) = 0 ,ut(x,0) = 0 for 0 <x< ∞, u(0,t)=f(t)f o r t>0, where f(t) is a given function. 34. Solve the following initial and boundary value problem for the wave equation using the Fourier cosine transform: utt=c2uxx,0<x< ∞,t >0, u(0,t)=f(t)f o r t>0, u(x,0)= 0 ,ut(x,0) = 0 for 0 <x< ∞, where f(t) is a known function. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 127 35. Apply the Fourier transform to solve the initial value problem for the dissipative wave equation utt=c2uxx+αuxxt,−∞<x< ∞,t >0, u(x,0) =f(x),ut(x,0) =αf/prime/prime(x)f o r −∞<x< ∞, where αis a positive constant. 36. Use the Fourier sine transform to solve the initial and boundary value problem for free vibrations of a semi-inﬁnite string: utt=c2uxx,0<x< ∞,t >0, u(0,t)=0,t≥0, u(x,0)=f(x)a n d ut(x,0) =g(x)f o r0 <x< ∞. 37. The static deﬂection u(x, y) in a thin elastic disk in the form of a quad- rant satisﬁes the boundary value problem uxxxx+2uxxyy+uyyyy=0,0<x< ∞,0<y< ∞, u(0,y)=uxx(0,y)=0 f or 0 <y< ∞, u(x,0)=ax 1+x2,uyy(x,0) = 0 for 0 <x< ∞, where ais a constant, and u(x, y) and its derivatives vanish as x→∞ andy→∞. Use the Fourier sine transform to show that u(x, y)=a 2∞⎪integraldisplay 0(2 +ky)exp[−(1 +y)k]si nkx dx =ax x2+( 1+ y)2+axy(1 +y) [x2+( 1+ y)2]2 38. In exercise 37, replace the conditions on y= 0 with the conditions u(x,0) = 0 ,uyy(x,0) =ax (1 +x2)2for 0 <x< ∞. Show that the solution is u(x, y)=−ax 4∞⎪integraldisplay 0exp[−(1 +y)k]si nkx dk =−1 4axy [x2+( 1+ y)2]. © 2007 by Taylor & Francis Group, LLC 128 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 39. In exercise 37, solve the biharmonic equation in 0 <x< ∞,0<y<b with the boundary conditions u(0,y)=asiny, u xx(0,y)=0 f or 0 <y<b , u(x,0) =uyy(x,0) =u(x, b)=uyy(x, b)=0 f o r 0 <x< ∞, andu(x, y),ux(x, y)v a n i s ha s x→∞. 40. Use the Fourier transform to solve the boundary value problem uxx+uyy=−xexp(−x2),−∞<x< ∞,0<y< ∞, u(x,0) = 0 ,for−∞<x< ∞,uand its derivative vanish as y→∞. Show that u(x, y)=1 √ 4π∞⎪integraldisplay 0[1−exp(−ky)]sinkx kexp⎪parenleftbigg −k2 4⎪parenrightbigg dk. 41. Using the deﬁnition of the characteristic function for the discrete random variable X φ(t)=E[exp(itX)] =⎪summationdisplay rprexp(itxr) where pr=P(X=xr),show that the characteristic function of the bi- nomial distribution pr=⎪parenleftbiggn r⎪parenrightbigg pr(1−p)n−r is φ(t)=[ 1+ p(eit−1)]n. Find the moments. 42. Show that the characteristic function of the Poisson distribution pr=P(X=r)=λr r!e−λ,r=0,1,2,... is φ(t)=e x p [ λ(eit−1)]. Find the moments. 43. Find the characteristic function of (a) The gamma distribution whose density function is f(x)=ap Γ(p)xp−1e−axH(x), © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 129 (b) The beta distribution whose density function is f(x)=⎧ ⎪⎨ ⎪⎩xp−1(1−x)q−1 B(p,q)for 0 <x< 1, 0f o r x<0a n d x>1⎫ ⎪⎬ ⎪⎭, (c) The Cauchy distribution whose density function is f(x)=1 πλ [λ2+(x−μ)2], (d) The Laplace distribution whose density function is f(x)=1 2λexp⎪parenleftbigg −|x−u| λ⎪parenrightbigg ,λ > 0. 44. Find the density function of the random variable Xwhose characteristic function is φ(t)=( 1 −|t|)H(1−|t|). 45. Find the characteristic function of uniform distribution whose density function is f(x)=⎧ ⎨ ⎩0,x < 0 1,0≤x≤a 0,x > a⎫ ⎬ ⎭. 46. Solve the initial value problem (Debnath, 1994, p. 115) for the two- dimensional surface waves at the free surface of a running stream of velocity U.The problem satisﬁes the equation, boundary, and initial conditions φxx+φzz=0,−∞<x< ∞,−h≤z≤0,t > 0, φx+Uφx+gη=−P ρδ(x)exp(iωt) ηt+Uηx−φz=0⎫ ⎬ ⎭onz=0,t >0, φ(x, z,0) =η(x,0) = 0 ,for all xandz. 47. Apply the Fourier tranform to solve the equation uxxxx+uyy=0,−∞<x< ∞,y≥0, satisfying the conditions u(x,0)=f(x),uy(x,0) = 0 for −∞<x< ∞, u(x, y) and its partial derivatives vanish as |x|→∞ . © 2007 by Taylor & Francis Group, LLC 130 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 48. The transverse vibration of a thin membrane of great extent satisﬁes the wave equation c2(uxx+uyy)=utt,−∞<x ,y< ∞,t >0, with the initial and boundary conditions u(x, y, t)→0a s |x|→∞ ,|y|→∞ for all t≥0, u(x, y,0)=f(x, y),ut(x, y,0) = 0 for all x, y. Apply the double Fourier transform method to solve this problem. 49. Solve the diﬀusion problem with a source q(x, t) ut=κuxx+q(x, t),−∞<x< ∞,t >0, u(x,0) = 0 for −∞<x< ∞. Show that the solution is u(x, t)=1 √ 4πtκt⎪integraldisplay 0(t−τ)−1 2dτ∞⎪integraldisplay −∞q(k,τ)exp⎪bracketleftbigg −(x−k)2 4κ(t−τ)⎪bracketrightbigg dk. 50. The function u(x, t) satisﬁes the diﬀusion problem in a half-line ut=κuxx+q(x, t),0≤x<∞,t >0, u(x,0) = 0 ,u(0,t)=0 f o r x≥0a n d t>0. Show that u(x, t)=⎪radicalbigg 2 πt⎪integraldisplay 0dτ∞⎪integraldisplay 0Qs(k,τ)exp[−κk2(t−τ)] sinkx dk, where Qs(k,t) is the Fourier sine transform of q(x, t). 51. Apply the triple Fourier transform to solve the initial value problem ut=κ(uxx+uyy+uzz),−∞<x , y, z< ∞,t >0, u(x,0) =f(x) for all x, y, z, where x=(x, y, z ). 52. Use the Fourier transform with respect to tand Laplace transform with respect to xto solve the telegraph equation utt+aut+bu=c2uxx,0<x< ∞,−∞<t<∞, u(0,t)=f(t),ux(0,t)=g(t),for−∞<t<∞, where a, b, c are constants and f(t)a n d g(t) are arbitrary functions of timet. © 2007 by Taylor & Francis Group, LLC Fourier Transforms and Their Applications 131 53. Determine the steady-state temperature distribution in a disk occupying the semi-inﬁnite strip 0 <x< ∞,0<y< 1i ft h ee d g e s x=0 a n d y=0 are insulated, and the edge y= 1 is kept at a constant temperature T0H(a−x).Assuming that the disk loses heat due to its surroundings according to Newton’s law with proportionality constant h,s o l v et h e boundary value problem uxx+uyy−hu=0,0<x< ∞,0<y< 1, u(x,1) =T0H(a−x),for 0 <x< ∞, ux(0,y)=0= uy(x,0) for 0 <x< ∞,0<y< 1. 54. Use the double Fourier transform to solve the following equations: (a)uxxxx−uyy+2u=f(x, y), (b)uxx+2uyy+3ux−4u=f(x, y), where f(x, y) is a given function. 55. Use the Fourier transform to solve the Rossby wave problem in an in- viscid β-plane ocean bounded by walls at y=0a n d y=1w h e r e yandx represent vertical and horizontal directions. The ﬂuid is initially at rest and then, at t=0 +,an arbitrary disturbance localized to the vicinity ofx= 0 is applied to generate Rossby w aves. This problem satisﬁes the Rossby wave equation ∂ ∂t[(∇2−κ2)ψ]+βψx=0,−∞<x< ∞,0≤y≤1,t >0, with the boundary and initial conditions ψx(x, y)=0 f o r 0 <x< ∞,y=0 a n d y=1, ψ(x, y, t)=ψ0(x, y)a tt=0 f o ra l l xandy. 56. Find the transfer function and the corresponding impulse response func- tion of the input and output of the RCcircuit governed by the equation Rdq dt+1 Cq(t)=e(t), where R,Care constants, q(t) is the electric charge and e(t)i st h e given voltage. 57. Prove the Poisson summation formula for the Fourier cosine transform Fc{f(x)}=Fc(k)i nt h ef o r m √ a⎪bracketleftBigg 1 2f(0) +∞⎪summationdisplay n=1f(na)⎪bracketrightBigg =√ b⎪bracketleftBigg 1 2Fc(0) +∞⎪summationdisplay n=1Fc(nb)⎪bracketrightBigg , where ab=2πanda>0. Apply this formula to the following examples: © 2007 by Taylor & Francis Group, LLC 132 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (a)f(x)=e−x,F c(k)=⎪radicalbigg 2 π(1 +k2)−1, (b)f(x)=e x p ( −1 2x2),F c(k)=e x p ( −1 2k2), (c)f(x)=e x p ( −1 2x2)cosαx , F c(k)=e x p⎪bracketleftbigg −1 2(α2+k2)⎪bracketrightbigg cosh(kα), (d)f(x)=⎪braceleftBigg 21 2−ν Γ(ν+1 2)(1−x2)ν−1 2,0≤x<1, 0,x ≥1. Fc(k)=k−νJν(k),k > 0;Fc(0) =1 2νΓ(ν+1 ). © 2007 by Taylor & Francis Group, LLC 3 Laplace Transforms and Their Basic Properties “What we know is not much. What we do not know is immense.” Pierre-Simon Laplace “The algebraic analysis soon makes us forget the main object [of our research] by focusing our attention on abstract combinations and it is only at the end that we return to the original objective. But in abandoning oneself to the operations of analysis, one is led to the generality of this method and the inestimable advantage oftransforming the reasoning by mechanical procedures to results of- ten inaccessible by geometry....No other language has the capacity for the elegance that arises from a long sequence of expressionslinked one to the other and all stemming from one fundamental idea.” Pierre-Simon Laplace “... For Laplace, on the contrary, mathematical analysis was an instrument that he bent to his purposes for the most varied appli- cations, but always subordinating the method itself to the content of each question. Perhaps posterity will....” Simeon-Denis Poisson 3.1 Introduction In this chapter, we present the formal deﬁnition of the Laplace transform andcalculate the Laplace transforms of some elementary functions directly from the deﬁnition. The existence conditions for the Laplace transform are stated in Section 3.3. The basic operational properties of the Laplace transforms in-cluding convolution and its properties, and the diﬀerentiation and integration of Laplace transforms are discussed in some detail. The inverse Laplace trans- form is introduced in Section 3.7, and four methods of evaluation of the inverse 133 © 2007 by Taylor & Francis Group, LLC 134 INTEGRAL TRANSFORMS and THEIR APPLICATIONS transform are developed with examples. The Heaviside Expansion Theorem and the Tauberian theorems for the Laplace transform are discussed. 3.2 Deﬁnition of the Laplace Transform and Examples We start with the Fourier Integral Formula (2.2.4), which expresses the rep- resentation of a function f1(x) deﬁned on −∞<x< ∞in the form f1(x)=1 2π∞⎪integraldisplay −∞eikxdk∞⎪integraldisplay −∞e−iktf1(t)dt. (3.2.1) We next set f1(x)≡0i n−∞<x< 0a n dw r i t e f1(x)=e−cxf(x)H(x)=e−cxf(x),x > 0, (3.2.2) where cis a positive ﬁxed number, so that (3.2.1) becomes f(x)=ecx 2π∞⎪integraldisplay −∞eikxdk∞⎪integraldisplay 0exp{−t(c+ik)}f(t)dt. (3.2.3) With a change of variable, c+ik=s, i dk =dswe rewrite (3.2.3) as f(x)=ecx 2πic+i∞⎪integraldisplay c−i∞exp{(s−c)x}ds∞⎪integraldisplay 0e−stf(t)dt. (3.2.4) Thus, the Laplace transform off(t) is formally deﬁned by L{f(t)}=¯f(s)=∞⎪integraldisplay 0e−stf(t)dt, Res>0, (3.2.5) where e−stis thekernel of the transform and sis thetransform variable which is a complex number. Under broad conditions on f(t), its transform ¯f(s)i s analytic in sin the half-plane, where Re s>a. Result (3.2.4) then gives the formal deﬁnition of the inverse Laplace trans- form L−1{¯f(s)}=f(t)=1 2πic+i∞⎪integraldisplay c−i∞est¯f(s)ds, c > 0. (3.2.6) Obviously, LandL−1are linear integral operators. © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 135 Using the deﬁnition (3.2.5), we can calculate the Laplace transforms of some simple and elementary functions. Example 3.2.1 Iff(t)=1 f o r t>0, then ¯f(s)=L{1}=∞⎪integraldisplay 0e−stdt=1 s. (3.2.7) Example 3.2.2 Iff(t)=eat,w h e r e ais a constant, then L{eat}=¯f(s)=∞⎪integraldisplay 0e−(s−a)tdt=1 s−a,s > a . (3.2.8) Example 3.2.3 Iff(t)=s i n at,w h e r e ais a real constant, then L{sinat}=∞⎪integraldisplay 0e−stsinat dt=1 2i∞⎪integraldisplay 0[e−t(s−ia)−e−t(s+ia)]dt(3.2.9) =1 2i⎪bracketleftbigg1 s−ia−1 s+ia⎪bracketrightbigg =a s2+a2. Similarly, L{cosat}=s s2+a2. (3.2.10) Example 3.2.4 Iff(t)=s i n h ator cosh at,w h e r e ais a real constant, then L{sinhat}=∞⎪integraldisplay 0e−stsinhat dt=a s2−a2, (3.2.11) L{coshat}=∞⎪integraldisplay 0e−stcoshat dt=s s2−a2. (3.2.12) © 2007 by Taylor & Francis Group, LLC 136 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 3.2.5 Iff(t)=tn,w h e r e nis a positive integer, then ¯f(s)=L{tn}=n! sn+1. (3.2.13) We recall (3.2.7) and formally diﬀerentiate it with respect to s.T h i sg i v e s ∞⎪integraldisplay 0te−stdt=1 s2, (3.2.14) which means that L{t}=1 s2. (3.2.15) Diﬀerentiating (3.2.14) with respect to sgives L{t2}=∞⎪integraldisplay 0t2e−stdt=2 s3. (3.2.16) Similarly, diﬀerentiation of (3.2.7) ntimes yields L{tn}=∞⎪integraldisplay 0tne−stdt=n! sn+1. (3.2.17) Example 3.2.6 Ifa(>−1) is a real number, then L{ta}=Γ(a+1 ) sa+1,(s>0). (3.2.18) We have L{ta}=∞⎪integraldisplay 0tae−stdt, which is, by putting st=x, =1 sa+1∞⎪integraldisplay 0xae−xdx=Γ(a+1 ) sa+1, © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 137 where Γ( a)r e p r e s e n t st h e gamma function deﬁned by the integral Γ(a)=∞⎪integraldisplay 0xa−1e−xdx, a > 0. (3.2.19) It can be shown that the gamma function satisﬁes the relation Γ(a+1 )= aΓ(a). (3.2.20) Obviously, result (3.2.18) is an extension of (3.2.17). The latter is a special case of the former when ais a positive integer. In particular, when a=−1 2, result (3.2.18) gives L⎪braceleftbigg1 √ t⎪bracerightbigg =Γ⎪parenleftbig1 2⎪parenrightbig √ s=⎪radicalbigg π s,where Γ⎪parenleftbigg1 2⎪parenrightbigg =√ π. (3.2.21) Similarly, L⎪braceleftBig√ t⎪bracerightBig =Γ⎪parenleftbig3 2⎪parenrightbig s3/2=√ π 21 s3/2, (3.2.22) where Γ⎪parenleftbigg3 2⎪parenrightbigg =Γ⎪parenleftbigg1 2+1⎪parenrightbigg =1 2Γ⎪parenleftbigg1 2⎪parenrightbigg =√ π 2. Example 3.2.7 Iff(t)=erf⎪parenleftbigga 2√ t⎪parenrightbigg ,t h e n L⎪braceleftbigg erf⎪parenleftbigga 2√ t⎪parenrightbigg⎪bracerightbigg =1 s(1−e−a√ s), (3.2.23) where erf(t)i st h e error function deﬁned by (2.5.13). To prove (3.2.23), we begin with the deﬁnition (3.2.5) so that L⎪braceleftbigg erf⎪parenleftbigga 2√ t⎪parenrightbigg⎪bracerightbigg =∞⎪integraldisplay 0e−st⎡ ⎢⎣2 √ πa/2√ t⎪integraldisplay 0e−x2dx⎤ ⎥⎦dt, which is, by putting x=a 2√ tort=a2 4x2and interchanging the order of inte- © 2007 by Taylor & Francis Group, LLC 138 INTEGRAL TRANSFORMS and THEIR APPLICATIONS gration, =2 √ π∞⎪integraldisplay 0e−x2dxa2/4x2⎪integraldisplay 0e−stdt =2 √ π∞⎪integraldisplay 0e−x21 s⎪braceleftbigg 1−exp⎪parenleftbigg −a2s 4x2⎪parenrightbigg⎪bracerightbigg dx =1 s·2 √ π⎡ ⎣∞⎪integraldisplay 0e−x2dx−∞⎪integraldisplay 0exp⎪braceleftbigg −⎪parenleftbigg x2+sa2 4x2⎪parenrightbigg⎪bracerightbigg dx⎤ ⎦, where the integral ∞⎪integraldisplay 0exp⎪braceleftbigg −⎪parenleftbigg x2+α2 x2⎪parenrightbigg⎪bracerightbigg dx=1 2⎡ ⎣∞⎪integraldisplay 0⎪parenleftBig 1−α x2⎪parenrightBig exp⎪bracketleftbigg −⎪parenleftBig x+α x⎪parenrightBig2 +2α⎪bracketrightbigg +∞⎪integraldisplay 0⎪parenleftBig 1+α x2⎪parenrightBig exp⎪bracketleftbigg −⎪parenleftBig x−α x⎪parenrightBig2 −2α⎪bracketrightbigg⎤ ⎦dx, which is, by putting y=⎪parenleftBig x±α x⎪parenrightBig ,dy=⎪parenleftBig 1∓α x2⎪parenrightBig dx, and observing that the ﬁrst integral vanishes, =1 2e−2α∞⎪integraldisplay −∞e−y2dy=√ π 2e−2α,α =a√ s 2. Consequently, L⎪braceleftbigg erf⎪parenleftbigga 2√ t⎪parenrightbigg⎪bracerightbigg =1 s2 √ π⎪bracketleftbigg√ π 2−√ π 2e−a√ s⎪bracketrightbigg =1 s[1−e−a√ s]. We use (3.2.23) to ﬁnd the Laplace transform of the complementary error function deﬁned by (2.10.14) and obtain L⎪braceleftbigg erfc⎪parenleftbigga 2√ t⎪parenrightbigg⎪bracerightbigg =1 se−a√ s. (3.2.24) The proof of this result follows from erfc(x)=1−erf(x)a n dL{1}=1 s. Example 3.2.8 Iff(t)=J0(at)i saBessel function of order zero, then L{J0(at)}=1 √ s2+a2. (3.2.25) © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 139 Using the series representation of J0(at), we obtain L{J0(at)}=L⎪bracketleftbigg 1−a2t2 22+a4t4 22·42−a6t6 22·42··62+···⎪bracketrightbigg =1 s−a2 222! s3+a4 22·42·4! s5−a6 22·42·62·6! s7+··· =1 s⎪bracketleftbigg 1−1 2⎪parenleftbigga2 s2⎪parenrightbigg +1·3 2·4⎪parenleftbigga4 s4⎪parenrightbigg −1·3·5 2·4·6⎪parenleftbigga6 s6⎪parenrightbigg +···⎪bracketrightbigg =1 s⎪bracketleftBigg⎪parenleftbigg 1+a2 s2⎪parenrightbigg−1 2⎪bracketrightBigg =1 √ a2+s2. 3.3 Existence Conditions for the Laplace Transform A function f(t)i ss a i dt ob eo f exponential order a(>0) on 0 ≤t<∞if there exists a positive constant Ksuch that for all t>T |f(t)|≤Keat, (3.3.1) and we write this symbolically as f(t)=O(eat)a s t→∞. (3.3.2) Or, equivalently, lim t→∞e−bt|f(t)|≤Klim t→∞e−(b−a)t=0,b > a . (3.3.3) Such a function f(t) is simply called an exponential order ast→∞,a n d clearly, it does not grow faster than Keatast→∞. THEOREM 3.3.1 If a function f(t) is continuous or piecewise continuous in every ﬁnite interval (0,T), and of exponential order eat, then the Laplace transform of f(t)e x i s t s for all sprovided Re s>a. PROOF We have |¯f(s)|=⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle∞⎪integraldisplay 0e−stf(t)dt⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle≤∞⎪integraldisplay 0e−st|f(t)|dt (3.3.4) ≤K∞⎪integraldisplay 0e−t(s−a)dt=K s−a,for Re s>a . © 2007 by Taylor & Francis Group, LLC 140 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thus, the proof is complete. It is noted that the conditions as stated in Theorem 3.3.1 are suﬃcient rather than necessary conditions. It also follows from (3.3.4) that lim s→∞|¯f(s)|= 0, that is, lim s→∞¯f(s)=0 . This result can be regarded as the limiting property of the Laplace transform. However, ¯f(s)=sors2is not the Laplace transform of any continuous (or piecewise continuous) function because ¯f(s) does not tend to zero as s→∞. Further, a function f(t)=e x p ( at2),a>0 cannot have a Laplace transform even though it is continuous but is notof the exponential order because lim t→∞exp(at2−st)=∞. 3.4 Basic Properties of Laplace Transforms THEOREM 3.4.1 (Heaviside’s First Shifting Theorem). IfL{f(t)}=¯f(s), then L{e−atf(t)}=¯f(s+a), (3.4.1) where ais a real constant. PROOF We have, by deﬁnition, L{e−atf(t)}=∞⎪integraldisplay 0e−(s+a)tf(t)dt=¯f(s+a). Example 3.4.1 The following results readily follow from (3.4.1) L{tne−at}=n! (s+a)n+1, (3.4.2) L{e−atsinbt}=b (s+a)2+b2, (3.4.3) L{e−atcosbt}=s+a (s+a)2+b2. (3.4.4) © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 141 THEOREM 3.4.2 IfL{f(t)}=¯f(s), then the Second Shifting property holds: L{f(t−a)H(t−a)}=e−as¯f(s)=e−asL{f(t)},a > 0.(3.4.5) Or, equivalently, L{f(t)H(t−a)}=e−asL{f(t+a)}. (3.4.6) where H(t−a) is the Heaviside unit step function deﬁned by (2.3.9). It follows from the deﬁnition that L{f(t−a)H(t−a)}=∞⎪integraldisplay 0e−stf(t−a)H(t−a)dt =∞⎪integraldisplay ae−stf(t−a)dt, which is, by putting t−a=τ, =e−sa∞⎪integraldisplay 0e−sτf(τ)dτ=e−sa¯f(s). We leave it to the reader to prove (3.4.6). In particular, if f(t)=1 ,t h e n L{H(t−a)}=1 sexp(−sa). (3.4.7) Example 3.4.2 Use the shifting property (3.4.5) or (3.4.6) to ﬁnd the Laplace transform of (a)f(t)=⎧ ⎨ ⎩1,0<t<1 −1,1<t<2 0,t > 2⎫ ⎬ ⎭, (b)g(t)=s i n tH(t−π). To ﬁnd L{f(t)},w ew r i t e f(t)a s f(t)=1−2H(t−1) +H(t−2). Hence, ¯f(s)=L{f(t)}=L{1}−2L{H(t−1)}+L{H(t−2)} =1 s−2e−s s+e−2s s. © 2007 by Taylor & Francis Group, LLC 142 INTEGRAL TRANSFORMS and THEIR APPLICATIONS To obtain L{g(t)}, we use (3.4.6) so that ¯g(s)=L{sintH(t−π)}=−e−πsL{cost}=−se−πs s2+1. Scaling Property: L{f(at)}=1 |a|¯f⎪parenleftBigs a⎪parenrightBig ,a/negationslash=0. (3.4.8) Example 3.4.3 Show that the Laplace transform of the square wave function f(t) deﬁned by f(t)=H(t)−2H(t−a)+2H(t−2a)−2H(t−3a)+··· (3.4.9) is ¯f(s)=1 stanh⎪parenleftBigas 2⎪parenrightBig . (3.4.10) The graph of f(t) is shown in Figure 3.1. f(t)=H(t)−2H(t−a)=1−2·0=1,0<t<a f(t)=H(t)−2H(t−a)+2H(t−2a) =1−2·1+2·0=−1,0<a<t< 2a. Thus, ¯f(s)=1 s−2·e−as s+2·e−2as s−2·e−3as s+··· =1 s[1−2r(1−r+r2−···)],where r=e−as =1 s⎪bracketleftbigg 1−2r 1+r⎪bracketrightbigg =1 s⎪bracketleftbigg 1−2e−as 1+e−as⎪bracketrightbigg =1 s⎪parenleftbigg1−e−as 1+e−as⎪parenrightbigg =1 s⎪parenleftbiggesa 2−e−as 2 esa 2+e−as 2⎪parenrightbigg =1 stanh⎪parenleftBigas 2⎪parenrightBig . Example 3.4.4(The Laplace Transform of a Periodic Function ). Iff(t) is a periodic function of period a,a n di f L{f(t)}exists, show that L{f(t)}=[ 1−exp(−as)] −1a⎪integraldisplay 0e−stf(t)dt. (3.4.11) © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 143 -101 tf(t) 2a 3a 4a 5a a Figure 3.1 Square wave function. We have, by deﬁnition, L{f(t)}=∞⎪integraldisplay 0e−stf(t)dt=a⎪integraldisplay 0e−stf(t)dt+∞⎪integraldisplay ae−stf(t)dt. Letting t=τ+ain the second integral gives ¯f(s)=a⎪integraldisplay 0e−stf(t)dt+e x p ( −sa)∞⎪integraldisplay 0e−sτf(τ+a)dτ, which is, due to f(τ+a)=f(τ) and replacing the dummy variable τbytin the second integral, =a⎪integraldisplay 0e−stf(t)dt+e x p ( −sa)∞⎪integraldisplay 0e−stf(t)dt. Finally, combining the second term with the left hand side, we obtain (3.4.11). In particular, we calculate the Laplace transform of a rectiﬁed sine wave, that is, f(t)=|sinat|. This is a periodic function with periodπ a.W eh a v e π a⎪integraldisplay 0e−stsinat dt=⎪bracketleftbigge−st(−acosat−ssinat) (s2+a2)⎪bracketrightbiggπ a 0=a⎪braceleftbig 1+e x p⎪parenleftbig −sπ a⎪parenrightbig⎪bracerightbig (s2+a2). © 2007 by Taylor & Francis Group, LLC 144 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Clearly, the property (3.4.11) gives L{f(t)}=a (s2+a2)·1+e x p⎪parenleftBig −sπ a⎪parenrightBig 1−exp⎪parenleftBig −sπ a⎪parenrightBig =a (s2+a2)⎡ ⎢⎢⎣exp⎪parenleftBigsπ 2a⎪parenrightBig +e x p⎪parenleftBig −sπ 2a⎪parenrightBig exp⎪parenleftbigg2π 2a⎪parenrightbigg −exp⎪parenleftBig −sπ 2a⎪parenrightBig⎤ ⎥⎥⎦ =a s2+a2coth⎪parenleftBigπs 2a⎪parenrightBig . THEOREM 3.4.3 (Laplace Transforms of Derivatives). IfL{f(t)}=¯f(s), then L{f/prime(t)}=sL{f(t)}−f(0) =s¯f(s)−f(0), (3.4.12) L{f/prime/prime(t)}=s2L{f(t)}−sf(0)−f/prime(0) =s2¯f(s)−sf(0)−f/prime(0).(3.4.13) More generally, L{f(n)(t)}=sn¯f(s)−sn−1f(0)−sn−2f/prime(0)−···− sf(n−2)(0)−f(n−1)(0), (3.4.14) where f(r)(0) is the value of f(r)(t)a tt=0 ,r=0,1, ...,(n−1). PROOF We have, by deﬁnition, L{f/prime(t)}=∞⎪integraldisplay 0e−stf/prime(t)dt, which is, integrating by parts, =⎪bracketleftbig e−stf(t)⎪bracketrightbig∞ 0+s∞⎪integraldisplay 0e−stf(t)dt =s¯f(s)−f(0), in which we assumed f(t)e−st→0a st→∞. Similarly, L{f/prime/prime(t)}=sL{f/prime(t)}−f/prime(0),by (3.4.12) =s[s¯f(s)−f(0)]−f/prime(0) =s2¯f(s)−sf(0)−f/prime(0), © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 145 where we have assumed e−stf/prime(t)→0a st→∞. A similar procedure can be used to prove the general result (3.4.14).It may be noted that similar results hold when the Laplace transform is ap- plied to partial derivatives of a function of two or more independent variables. For example, if u(x, t) is a function of two variables xandt,t h e n L⎪braceleftbigg∂u ∂t⎪bracerightbigg =s¯u(x, s)−u(x,0), (3.4.15) L⎪braceleftbigg∂2u ∂t2⎪bracerightbigg =s2¯u(x, s)−su(x,0)−⎪bracketleftbigg∂u ∂t⎪bracketrightbigg t=0, (3.4.16) L⎪braceleftbigg∂u ∂x⎪bracerightbigg =d¯u dx,L⎪braceleftbigg∂2u ∂x2⎪bracerightbigg =d2¯u dx2. (3.4.17) Results (3.4.12) to (3.4.14) imply that the Laplace transform reduces the op- eration of diﬀerentiation into algebraic operation. In view of this, the Laplace transform can be used eﬀectively to sol ve ordinary or partial diﬀerential e- quations. Example 3.4.5 Use (3.4.14) to ﬁnd L{tn}. Heref(t)=tn,f/prime(t)=ntn−1,···,f(n)(t)=n!a n d f(0) =f/prime(0) =···= f(n−1)(0) = 0. Thus, L{n!}=snL{tn}. Or, L{tn}=n! snL{1}=n! sn+1. 3.5 The Convolution Theorem and Properties of Convolution THEOREM 3.5.1 (Convolution Theorem) . IfL{f(t)}=¯f(s)a n dL{g(t)}=¯g(s), then L{f(t)∗g(t)}=L{f(t)}L{g(t)}=¯f(s)¯g(s). (3.5.1) Or, equivalently, L−1{¯f(s)¯g(s)}=f(t)∗g(t), (3.5.2) © 2007 by Taylor & Francis Group, LLC 146 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where f(t)∗g(t) is called the convolution off(t)a n d g(t) and is deﬁned by the integral f(t)∗g(t)=t⎪integraldisplay 0f(t−τ)g(τ)dτ. (3.5.3) The integral in (3.5.3) is often referred to as the convolution integral (or Faltung ) and is denoted simply by ( f∗g)(t). PROOF We have, by deﬁnition, L{f(t)∗g(t)}=∞⎪integraldisplay 0e−stdtt⎪integraldisplay 0f(t−τ)g(τ)dτ, (3.5.4) where the region of integration in the τ−tplane is as shown in Figure 3.2. The integration in (3.5.4) is ﬁ rst performed with respect to τfromτ=0 t o τ=tof the vertical strip and then from t=0 t o ∞by moving the vertical strip from t= 0 outwards to cover the whole region under the line τ=t. =00 t=t _<t< Figure 3.2 Region of integration. We now change the order of integration so that we integrate ﬁrst along the horizontal strip from t=τto∞and then from τ=0 t o ∞by moving the © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 147 horizontal strip vertically from τ= 0 upwards. Evidently, (3.5.4) becomes L{f(t)∗g(t)}=∞⎪integraldisplay 0g(τ)dτ∞⎪integraldisplay t=τe−stf(t−τ)dτ, which is, by the change of variable t−τ=x, L{f(t)∗g(t)}=∞⎪integraldisplay 0g(τ)dτ∞⎪integraldisplay 0e−s(x+τ)f(x)dx =∞⎪integraldisplay 0e−sτg(τ)dτ∞⎪integraldisplay 0e−sxf(x)dx=¯g(s)¯f(s). This completes the proof. PROOF (Second Proof.) We have, by deﬁnition, ¯f(s)¯g(s)=∞⎪integraldisplay 0e−sσf(σ)dσ∞⎪integraldisplay 0e−sμg(μ)dμ =∞⎪integraldisplay 0∞⎪integraldisplay 0e−s(σ+μ)f(σ)g(μ)dσ dμ, (3.5.5) where the double integral is taken over the entire ﬁrst quadrant Rof the σ−μ plane bounded by σ=0 a n d μ= 0 as shown in Figure 3.3(a). =00 t=t S(b) =00R(a)=0 Figure 3.3 Regions of integration. We make the change of variables μ=τ,σ=t−μ=t−τso that the axes σ=0 andμ= 0 transform into the lines τ=0 a n d τ=t, respectively, as shown in © 2007 by Taylor & Francis Group, LLC 148 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Figure 3.3(b) in the τ−tplane. Consequently, (3.5.5) becomes ¯f(s)¯g(s)=∞⎪integraldisplay 0e−stdtτ=t⎪integraldisplay τ=0f(t−τ)g(τ)dτ =L⎧ ⎨ ⎩t⎪integraldisplay 0f(t−τ)g(τ)dτ⎫ ⎬ ⎭ =L{f(t)∗g(t)}. This proves the theorem. Note: A more rigorous proof of the convolution theorem can be found in any standard treatise (see Doetsch, 1950) on Laplace transforms. The convolution operation has the following properties: f(t)∗{g(t)∗h(t)}={f(t)∗g(t)}∗h(t), (Associative), (3.5.6) f(t)∗g(t)=g(t)∗f(t), (Commutative), (3.5.7) f(t)∗{ag(t)+bh(t)}=af(t)∗g(t)+bf(t)∗h(t),(Distributive), (3.5.8) f(t)∗{ag(t)}={af(t)}∗g(t)=a{f(t)∗g(t)}, (3.5.9) L{f1∗f2∗f3∗···∗fn}=¯f1(s)¯f2(s)···¯fn(s), (3.5.10) L{f∗n}={¯f(s)}n, (3.5.11) where aandbare constants. f∗n=f∗f∗···∗fis sometimes called the nth convolution . Remark: By virtue of (3.5.6) and (3.5.7), it is clear that the set of all Laplace transformable functions forms a commutative semigroup with respect to theoperation ∗. The set of all Laplace transformable functions does not form a group because f∗g −1does not, in general, have a Laplace transform. We now prove the associative property. We have f(t)∗{g(t)∗h(t)}=t⎪integraldisplay 0f(τ)t−τ⎪integraldisplay 0g(t−σ−τ)h(σ)dσ dτ (3.5.12) =t⎪integraldisplay 0h(σ)t−σ⎪integraldisplay 0g(t−τ−σ)f(τ)dτ dσ =h(t)∗{f(t)∗g(t)}={f(t)∗g(t)}∗h(t),(3.5.13) where (3.5.13) is obtained from (3.5.12) by interchanging the order of integra- tion combined with the fact that 0 ≤σ≤t−τand 0≤τ≤timply 0 ≤τ≤t−σ and 0≤σ≤t. Properties (3.5.10) and (3.5.11) follow immediately from the as- sociative law of the convolution. © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 149 To prove (3.5.7), we recall the deﬁnition of the convolution and make a change of variable t−τ=t/prime.T h i sg i v e s f(t)∗g(t)=t⎪integraldisplay 0f(t−τ)g(τ)dτ=t⎪integraldisplay 0g(t−t/prime)f(t/prime)dt/prime=g(t)∗f(t). The proofs of (3.5.8)–(3.5.9) are very simple and hence, may be omitted. Example 3.5.1 Obtain the convolutions (a)t∗eat, (d) 1∗a 2e−a2/4t √ πt3,(b) (sin at∗sinat), (e) cos t∗e2t,(c)1 √ πt∗eat, (f)t∗t∗t. We have (a)t∗eat=t⎪integraldisplay 0τea(t−τ)dτ=eatt⎪integraldisplay 0τe−aτdτ=1 a2(eat−at−1). (b) sin at∗sinat=t⎪integraldisplay 0sinaτsina(t−τ)dτ=1 2a(sinat−atcosat). (c)1 √ πt∗eat=1 √ πt⎪integraldisplay 01 √ τea(t−τ)dτ, which is, by putting√ aτ=x, 1 √ πt∗eat=2eat √ πa√ at⎪integraldisplay 0e−x2dx=eat √ aerf⎪parenleftBig√ at⎪parenrightBig . (d) We have 1∗a 2e−a2/4t √ πt3=a 2√ πt⎪integraldisplay 0e−a2/4τ τ3/2dτ, which is, by lettinga 2√ τ=x, =2 √ π∞⎪integraldisplay a 2√ te−x2dx=erfc⎪parenleftbigga 2√ t⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC 150 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (e) cos t∗e2t=t⎪integraldisplay 0cos(t−τ)e2τdτ=1 2t⎪integraldisplay 0e2τ⎪braceleftBig ei(t−τ)+e−i(t−τ)⎪bracerightBig dτ =⎪bracketleftbiggei(t−τ)+2τ 2(2−i)+e−i(t−τ)+2τ 2(2 + i)⎪bracketrightbigg =2 5e2t+1 5(sint−2c o st). (f) (t∗t)∗t=⎡ ⎣t⎪integraldisplay 0(t−τ)τd τ⎤ ⎦∗t=1 6t3∗t =1 6t⎪integraldisplay 0(t−τ)τ3dτ=t5 5!. Example 3.5.2 Using the Convolution Theorem 3.5.1, prove that B(m, n)=Γ(m)Γ(n) Γ(m+n), (3.5.14) where Γ( m) is the gamma function, and B(m, n) is the beta function deﬁned by B(m, n)=1⎪integraldisplay 0xm−1(1−x)n−1dx, (m>0,n>0). (3.5.15) To prove (3.5.14), we consider f(t)=tm−1(m>0) and g(t)=tn−1,(n>0). Evidently, ¯f(s)=Γ(m) smand ¯g(s)=Γ(n) sn. We have f∗g=t⎪integraldisplay 0τm−1(t−τ)n−1dτ=L−1{¯f(s)¯g(s)} =Γ (m)Γ(n)L−1{s−(m+n)} =Γ(m)Γ(n) Γ(m+n)tm+n−1. Letting t= 1, we derive the result 1⎪integraldisplay 0τm−1(1−τ)n−1dτ=Γ(m)Γ(n) Γ(m+n), which proves the result (3.5.14). © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 151 3.6 Diﬀerentiation and Integration of Laplace Transforms THEOREM 3.6.1 Iff(t)=O(eat)a st→∞, then the Laplace integral ∞⎪integraldisplay 0e−stf(t)dt, (3.6.1) is uniformly convergent with respect to sprovided s≥a1where a1>a. PROOF Since |e−stf(t)|≤Ke−t(s−a)≤Ke−t(a1−a)for all s≥a1 and∞⎪integraldisplay 0e−t(a1−a)dtexists for a1>a, by Weierstrass’ test, the Laplace integral is uniformly convergent for all s>a 1where a1>a. This completes the proof. In view of the uniform convergence of (3.6.1), diﬀerentiation of (3.2.5) with respect to swithin the integral sig n is permissible. Hence, d ds¯f(s)=d ds∞⎪integraldisplay 0e−stf(t)dt=∞⎪integraldisplay 0∂ ∂se−stf(t)dt =−∞⎪integraldisplay 0tf(t)e−stdt=−L{tf(t)}. (3.6.2) Similarly, we obtain d2 ds2¯f(s)=(−1)2L{t2f(t)}, (3.6.3) d3 ds3¯f(s)=(−1)3L{t3f(t)}. (3.6.4) More generally, dn dsn¯f(s)=(−1)nL{tnf(t)}. (3.6.5) © 2007 by Taylor & Francis Group, LLC 152 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Results (3.6.5) can be stated in the following theorem: THEOREM 3.6.2 (Derivatives of the Laplace Transform). IfL{f(t)}=¯f(s), then L{tnf(t)}=(−1)ndn dsn¯f(s), (3.6.6) where n= 0, 1, 2, 3,.... Example 3.6.1 Show that (a)L{tne−at}=n! (s+a)n+1, (c)L{tsinat}=2as (s2+a2)2,(b)L{tcosat}=s2−a2 (s2+a2)2, (d)L{tf/prime(t)}=−⎪braceleftbigg sd ds¯f(s)+¯f(s)⎪bracerightbigg . (a) Application of Theorem 3.6.2 gives L{tne−at}=(−1)ndn dsn.1 (s+a)=(−1)2nn! (s+a)n+1. (b) L{tcosat}=(−1)d ds⎪parenleftbiggs s2+a2⎪parenrightbigg =s2−a2 (s2+a2)2. Results (c) and (d) can be proved similarly. THEOREM 3.6.3 (Integral of the Laplace Transform). IfL{f(t)}=¯f(s), then L⎪braceleftbiggf(t) t⎪bracerightbigg =∞⎪integraldisplay s¯f(s)ds. (3.6.7) PROOF In view of the uniform convergence of (3.6.1), ¯f(s)c a nb ei n t e - © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 153 grated with respect to sin (s,∞)s ot h a t ∞⎪integraldisplay s¯f(s)ds=∞⎪integraldisplay sds∞⎪integraldisplay 0e−stf(t)dt =∞⎪integraldisplay 0f(t)dt∞⎪integraldisplay se−stds =∞⎪integraldisplay 0f(t) te−stdt=L⎪braceleftbiggf(t) t⎪bracerightbigg . This proves the theorem. Example 3.6.2 Show that (a)L⎪braceleftbiggsinat t⎪bracerightbigg =t a n−1⎪parenleftBiga s⎪parenrightBig ,(b)L⎪braceleftBigg e−a2/4t √ πt3⎪bracerightBigg =2 aexp(−a√ s). (a) Using (3.6.7), we obtain L⎪braceleftbiggsinat t⎪bracerightbigg =a∞⎪integraldisplay sds s2+a2=π 2−tan−1⎪parenleftBigs a⎪parenrightBig =t a n−1⎪parenleftBiga s⎪parenrightBig . (b)L⎪braceleftBigg 1 t·e−a2/4t √ πt⎪bracerightBigg =∞⎪integraldisplay s¯f(s)ds=∞⎪integraldisplay se−a√ s √ sds, by Table B-4 of Laplace transforms, which is, by putting a√ s=x, =2 a∞⎪integraldisplay a√ se−xdx=2 aexp(−a√ s). THEOREM 3.6.4 ( The Laplace Transform of an Integral). IfL{f(t)}=¯f(s), then L⎧ ⎨ ⎩t⎪integraldisplay 0f(τ)dτ⎫ ⎬ ⎭=¯f(s) s. (3.6.8) © 2007 by Taylor & Francis Group, LLC 154 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF We write g(t)=t⎪integraldisplay 0f(τ)dτ so that g(0) = 0 and g/prime(t)=f(t). Then it follows from (3.4.10) that ¯f(s)=L{f(t)}=L{g/prime(t)}=s¯g(s)=sL⎪braceleftbigg⎪integraldisplayt 0f(τ)dτ⎪bracerightbigg . Dividing both sides by s, we obtain (3.6.8). It is noted that the Laplace transform of an integral corresponds to the division of the transform of its integrand by s. Result (3.6.8) can be used for evaluation of the inverse Laplace transform. Example 3.6.3 Use result (3.6.8) to ﬁnd (a)L⎧ ⎨ ⎩t⎪integraldisplay 0τne−aτdτ⎫ ⎬ ⎭,( b ) L{Si(at)}=L⎧ ⎨ ⎩t⎪integraldisplay 0sinaτ τdτ⎫ ⎬ ⎭. (a) We know L{tne−at}=n! (s+a)n+1. It follows from (3.6.8) that L⎧ ⎨ ⎩t⎪integraldisplay 0τne−aτdτ⎫ ⎬ ⎭=n! s(s+a)n+1. (b) Using (3.6.8) and Example 3.6.2(a), we obtain L⎧ ⎨ ⎩t⎪integraldisplay 0sinaτ τdτ⎫ ⎬ ⎭=1 stan−1⎪parenleftBiga s⎪parenrightBig . 3.7 The Inverse Laplace Transform and Examples It has already been demonstrated that the Laplace transform ¯f(s)o fag i v e n function f(t) can be calculated by direct integration. We now look at the © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 155 inverse problem. Given a Laplace transform ¯f(s) of an unknown function f(t), how can we ﬁnd f(t)? This is essentially concerned with the solution of the integral equation ∞⎪integraldisplay 0e−stf(t)dt=¯f(s). (3.7.1) At this stage, it is rather diﬃcult to handle the problem as it is. However, in simple cases, we can ﬁnd the inverse transform from Table B-4 of Laplace transforms. For example L−1⎪braceleftbigg1 s⎪bracerightbigg =1,L−1⎪braceleftbiggs s2+a2⎪bracerightbigg =c o s at. In general, the inverse Laplace transform can be determined by using four methods: (i) Partial Fraction Decomposition, (ii) the Convolution Theorem, (iii) Contour Integration of the Laplace Inversion Integral, and (iv) Heavi- side’s Expansion Theorem. (i)Partial Fraction Decomposition Method If ¯f(s)=¯p(s) ¯q(s), (3.7.2) where ¯ p(s)a n d¯ q(s)a r ep o l y n o m i a l si n s,a n dt h ed e g r e eo f¯ p(s)i sl e s st h a n that of ¯ q(s), the method of partial fractions may be used to express ¯f(s)a st h e sum of terms which can be inverted by using a table of Laplace transforms. We illustrate the method by means of simple examples. Example 3.7.1 To ﬁnd L−1⎪braceleftbigg1 s(s−a)⎪bracerightbigg , where aisa constant, we write L−1⎪braceleftbigg1 s(s−a)⎪bracerightbigg =L−1⎪bracketleftbigg1 a⎪braceleftbigg1 s−a−1 s⎪bracerightbigg⎪bracketrightbigg =1 a⎪bracketleftbigg L−1⎪braceleftbigg1 s−a⎪bracerightbigg −L−1⎪braceleftbigg1 s⎪bracerightbigg⎪bracketrightbigg =1 a(eat−1). Example 3.7.2Show that L −1⎪braceleftbigg1 (s2+a2)(s2+b2)⎪bracerightbigg =1 b2−a2⎪parenleftbiggsinat a−sinbt b⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC 156 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We write L−1⎪braceleftbigg1 (s2+a2)(s2+b2)⎪bracerightbigg =1 b2−a2⎪bracketleftbigg L−1⎪braceleftbigg1 s2+a2−1 s2+b2⎪bracerightbigg⎪bracketrightbigg =1 (b2−a2)⎪parenleftbiggsinat a−sinbt b⎪parenrightbigg . Example 3.7.3 Find L−1⎪braceleftbiggs+7 s2+2s+5⎪bracerightbigg . We have L−1⎪braceleftbiggs+7 (s+1 )2+4⎪bracerightbigg =L−1⎪braceleftbiggs+1+6 (s+1 )2+22⎪bracerightbigg =L−1⎪braceleftbiggs+1 (s+1 )2+22⎪bracerightbigg +3L−1⎪braceleftbigg2 (s+1 )2+22⎪bracerightbigg =e−tcos2t+3e−tsin2t. Example 3.7.4 Evaluate the following inverse Laplace transform L−1⎪braceleftbigg2s2+5s+7 (s−2)(s2+4s+ 13)⎪bracerightbigg . We have L−1⎪braceleftbigg2s2+5s+7 (s−2)(s2+4s+ 13)⎪bracerightbigg =L−1⎪braceleftbigg1 s−2+s+2 (s+2 )2+32+1 (s+2 )2+32⎪bracerightbigg =L−1⎪braceleftbigg1 s−2⎪bracerightbigg +L−1⎪braceleftbiggs+2 (s+2 )2+32⎪bracerightbigg +1 3L−1⎪braceleftbigg3 (s+2 )2+32⎪bracerightbigg =e2t+e−2tcos3t+1 3e−2tsin3t. (ii)Convolution Theorem We shall apply the convolution theorem for calculation of inverse Laplace transforms. © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 157 Example 3.7.5 L−1⎪braceleftbigg1 s(s−a)⎪bracerightbigg =1∗eat=t⎪integraldisplay 0eaτdτ=(eat−1) a. Example 3.7.6 L−1⎪braceleftbigg1 s2(s2+a2)⎪bracerightbigg =t∗sinat a =1 at⎪integraldisplay 0(t−τ)sinaτ dτ =t at⎪integraldisplay 0sinaτ dτ−1 at⎪integraldisplay 0τsinaτ dτ =1 a2⎪parenleftbigg t−1 asinat⎪parenrightbigg . Example 3.7.7 L−1⎪braceleftbigg1 (s2+a2)2⎪bracerightbigg =sinat a∗sinat a =1 a2t⎪integraldisplay 0sinaτsina(t−τ)dτ =1 2a3(sinat−atcosat). © 2007 by Taylor & Francis Group, LLC 158 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 3.7.8 L−1⎪braceleftbigg1 √ s(s−a)⎪bracerightbigg =1 √ πt∗eat,(a>0) =1 √ πt⎪integraldisplay 01 √ τea(t−τ)dτ =2eat √ πa√ at⎪integraldisplay 0e−x2dx,⎪parenleftbig putting√ aτ=x⎪parenrightbig =eat √ aerf(√ at). (3.7.3) Example 3.7.9 Show that L−1⎪braceleftbigg1 se−a√ s⎪bracerightbigg =erfc⎪parenleftbigga 2√ t⎪parenrightbigg . (3.7.4) In view of Example 3.6.2(b), and the Convolution Theorem 3.5.1, we obtain L−1⎪braceleftbigg1 se−a√ s⎪bracerightbigg =1∗a 2e−a2/4t √ πt3 =a 2√ πt⎪integraldisplay 0e−a2/4τ τ3/2dτ, which is, by puttinga 2√ τ=x, =2 √ π∞⎪integraldisplay a 2√ te−x2dx=erfc⎪parenleftbigga 2√ t⎪parenrightbigg . Example 3.7.10 Show that L−1⎪braceleftbigg1 √ s+a⎪bracerightbigg =1 √ πt−aexp(ta2)erfc(a√ t). (3.7.5) © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 159 We have L−1⎪braceleftbigg1 √ s+a⎪bracerightbigg =L−1⎪braceleftbigg1 √ s−a √ s(√ s+a)⎪bracerightbigg =L−1⎪braceleftbigg1 √ s⎪bracerightbigg −aL−1⎪braceleftbigg√ s−a √ s(s−a2)⎪bracerightbigg =L−1⎪braceleftbigg1 √ s⎪bracerightbigg −aL−1⎪braceleftbigg1 s−a2⎪bracerightbigg +a2L−1⎪braceleftbigg1 √ s(s−a2)⎪bracerightbigg =1 √ πt−aexp(a2t)+aexp(a2t)erf(a√ t),by (3.7.3) =1 √ πt−aexp(a2t)erfc(a√ t). Example 3.7.11 Iff(t)=L−1{¯f(s)},t h e n L−1⎪braceleftbigg1 s¯f(s)⎪bracerightbigg =t⎪integraldisplay 0f(x)dx. (3.7.6) We have, by the Convolution Theorem with g(t)=1s ot h a t¯ g(s)=1 s, L−1⎪braceleftbigg1 s¯f(s)⎪bracerightbigg =t⎪integraldisplay 0f(t−τ)dτ, which is, by putting t−τ=x, =t⎪integraldisplay 0f(x)dx. (iii)Contour Integration of the Laplace Inversion Integral In Section 3.2, in inverse Laplace transform is deﬁned by the complex integral formula L−1{¯f(s)}=f(t)=1 2πic+i∞⎪integraldisplay c−i∞est¯f(s)ds, (3.7.7) where cis a suitable real constant and ¯f(s) is an analytic function of the complex variable sin the right half-plane Re s>a. The details of evaluation of (3.7.7) depend on the nature of the singularities of¯f(s). Usually, ¯f(s) is a single valued function with a ﬁnite or enumerably © 2007 by Taylor & Francis Group, LLC 160 INTEGRAL TRANSFORMS and THEIR APPLICATIONS inﬁnite number of polar singularities. Often it has branch points. The path of integration is the straight line L(see Figure 3.4(a)) in the complex s-plane with equation s=c+iR,−∞<R< ∞,R es=cbeing chosen so that all the singularities of the integrand of (3.7.7) lie to the left of the line L. This line is called by Bromwich Contour . In practice, the Bromwich Contour is closed by an arc of a circle of radius Ras shown in Figure 3.4(a), and then the limit asR→∞ is taken to expand the contour of integration to inﬁnity so that all the singularities of ¯f(s) lie inside the contour of integration. When ¯f(s) has a branch point at the origin, we draw the modiﬁed contour of integration by making a cut along the negative real axis and a small semicircle γsurrounding the origin as shown in Figure 3.4(b). cRe sIm s AB c-iRc+iR LR (a)0 cRe sIm s AB c-iRc+iR LR (b)0L1 L2 Figure 3.4 The Bromwich contour and the contour of integration. In either case, the Cauchy Residue Theorem is used to evaluate the integral ⎪integraldisplay Lest¯f(s)ds+⎪integraldisplay Γest¯f(s)ds=⎪integraldisplay Cest¯f(s)ds =2πi×[sum of the residues of est¯f(s)a tt h ep o l e si n s i d e C]. (3.7.8) Letting R→∞, the integral over Γtends to zero, and this is true in most problems of interest. Consequently, result (3.7.7) reduces to the form lim R→∞1 2πic+iR⎪integraldisplay c−iRest¯f(s)ds= sum of the residues of est¯f(s)a tt h ep o l e so f ¯f(s). (3.7.9) We illustrate the above method of evaluation by simple examples. © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 161 Example 3.7.12 If¯f(s)=s s2+a2, show that f(t)=1 2πic+i∞⎪integraldisplay c−i∞est¯f(s)ds=c o sat. Clearly, the integrand has two simple poles at s=±iaand the residues at these poles are R1= Residue of est¯f(s)a ts=ia = lim s→ia(s−ia)sest (s2+a2)=1 2eiat. R2= Residue of est¯f(s)a ts=−ia = lim s→−ia(s+ia)sest (s2+a2)=1 2e−iat. Hence, f(t)=1 2πic+i∞⎪integraldisplay c−i∞est¯f(s)ds=R1+R2=1 2(eiat+e−iat)=c o s at, as obtained earlier. If ¯g(s)=est¯f(s) has a pole of order nats=z, then the residue R1of ¯g(s) at this pole is given by the formula R1= lim s→z1 (n−1)!dn−1 dsn−1[(s−z)n¯g(s)]. (3.7.10) This is obviously true for a simple pole ( n= 1) and for a double pole ( n=2 ) . Example 3.7.13 Evaluate L−1⎪braceleftbiggs (s2+a2)2⎪bracerightbigg . Clearly ¯g(s)=est¯f(s)=sest (s2+a2)2 has double poles at s=±ia. The residue formula (3.7.10) for double poles gives R1= lim s→iad ds⎪bracketleftbigg (s−ia)2sest (s2+a2)2⎪bracketrightbigg = lim s→iad ds⎪bracketleftbiggsest (s+ia)2⎪bracketrightbigg =teiat 4ia. © 2007 by Taylor & Francis Group, LLC 162 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Similarly, the residue at the double pole at s=−iais (−te−iat)/4ia. Thus, f(t) = Sum of the residues=t 4ia(eiat−e−iat)=t 2asinat, (3.7.11) as given in Table B-4 of Laplace transforms. Example 3.7.14 Evaluate L−1⎪braceleftbiggcosh(αx) scosh(α/lscript)⎪bracerightbigg ,α =⎪radicalbigg s a. We have f(t)=1 2πic+i∞⎪integraldisplay c−i∞estcosh(αx) cosh(α/lscript)ds s. Clearly, the integrand has simple poles at s=0a n d s=sn=−(2n+1 )2aπ2 4/lscript2, where n= 0,1,2,.... R1= Residue at the pole s=0i s1 ,and Rn= Residue at the pole s=snis exp(−snt)cosh⎪braceleftBig i(2n+1 )πx 2/lscript⎪bracerightBig ⎪bracketleftbigg sd ds⎪braceleftbigg coshl⎪radicalbigg s a⎪bracerightbigg⎪bracketrightbigg s=sn =4(−1)n+1 (2n+1 )πexp⎪bracketleftBigg −⎪braceleftbigg(2n+1 )π 2/lscript⎪bracerightbigg2 at⎪bracketrightBigg cos⎪braceleftBig (2n+1 )πx 2/lscript⎪bracerightBig . Thus, f(t) = Sum of the residues at the poles =1+4 π∞⎪summationdisplay n=0(−1)n+1 (2n+1 )exp⎪bracketleftbigg −(2n+1 )2π2at 4/lscript2⎪bracketrightbigg ×cos⎪braceleftBig (2n+1 )πx 2/lscript⎪bracerightBig , (3.7.12) as given later by the Heaviside Expansion Theorem. © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 163 Example 3.7.15 Show that f(t)=L−1⎪braceleftBigg e−a√ s s⎪bracerightBigg =1 2πic+i∞⎪integraldisplay c−i∞1 sexp(st−a√ s)ds =erfc⎪parenleftbigga 2√ t⎪parenrightbigg . (3.7.13) The integrand has a branch point at s= 0. We use the contour of integration as shown in Figure 3.4(b) which excludes the branch point at s=0 .T h u s ,t h e Cauchy Fundamental Theorem gives 1 2πi⎡ ⎣⎪integraldisplay L+⎪integraldisplay Γ+⎪integraldisplay L1+⎪integraldisplay L2+⎪integraldisplay γ⎤⎦exp(st−a√ s)ds s=0. (3.7.14) It is shown that the integral on Γtends to zero as R→∞, and that on L gives the Bromwich integral. We now evaluate the remaining three integrals in (3.7.14). On L1,w eh a v e s=reiπ=−rand ⎪integraldisplay L1exp(st−a√ s)ds s=0⎪integraldisplay −∞exp(st−a√ s)ds s=−∞⎪integraldisplay 0exp{−(rt+ia√ r)}dr r. OnL2,s=re−iπ=−rand ⎪integraldisplay L2exp(st−a√ s)ds s=−∞⎪integraldisplay 0exp(st−a√ s)ds s=∞⎪integraldisplay 0exp{−rt+ia√ r}dr r. Thus, the integrals along L1andL2combined yield −2i∞⎪integraldisplay 0e−rtsin(a√ r)dr r=−4i−∞⎪integraldisplay 0e−x2tsinax xdx, (√ r=x).(3.7.15) Integrating the following standard integral with respect to β ∞⎪integraldisplay 0e−x2α2cos(2βx)dx=√ π 2αexp⎪parenleftbigg −β2 α2⎪parenrightbigg , (3.7.16) © 2007 by Taylor & Francis Group, LLC 164 INTEGRAL TRANSFORMS and THEIR APPLICATIONS we obtain 1 2∞⎪integraldisplay 0e−x2α2sin2βx xdx=√ π 2αβ⎪integraldisplay 0exp⎪parenleftbigg −β2 α2⎪parenrightbigg dβ =√ π 2β/α⎪integraldisplay 0e−u2du, (β=αu) =π 4erf⎪parenleftbiggβ α⎪parenrightbigg . (3.7.17) In view of (3.7.17), result (3.7.15) becomes −4i∞⎪integraldisplay 0exp(−tx2)sinax xdx=−2πierf⎪parenleftbigga 2√ t⎪parenrightbigg . (3.7.18) Finally, on γ,w eh a v e s=reiθ,ds=ireiθdθ,a n d ⎪integraldisplay γ|exp(st−a√ s)|ds s=i−π⎪integraldisplay πexp⎪parenleftbigg rtcosθ−a√ rcosθ 2⎪parenrightbigg dθ =iπ⎪integraldisplay −πdθ=2πi, (3.7.19) in which the limit as r→0 is used and integration from πto−πis interchanged to make γin the counterclockwise direction. Thus, the ﬁnal result follows from (3.7.14), (3.7.18), and (3.7.19) in the form L−1⎪braceleftBigg e−a√ s s⎪bracerightBigg =1 2πic+i∞⎪integraldisplay c−i∞exp(st−a√ s)ds s =⎪bracketleftbigg 1−erf⎪parenleftbigga 2√ t⎪parenrightbigg⎪bracketrightbigg =erfc⎪parenleftbigga 2√ t⎪parenrightbigg . (iv)Heaviside’s Expansion Theorem Suppose ¯f(s) is the Laplace transform of f(t), which has a Maclaurin power series expansion in the form f(t)=∞⎪summationdisplay r=0artr r!. (3.7.20) Taking the Laplace transform, it is possible to write formally ¯f(s)=∞⎪summationdisplay r=0ar sr+1. (3.7.21) © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 165 Conversely, we can derive (3.7.20) from a given expansion (3.7.21). This kind of expansion is useful for determining the behavior of the solution for smalltime. Further, it provides an alternating way to prove the Tauberian theorems. THEOREM 3.7.1 (Heaviside’s Expansion Theorem ). If¯f(s)=¯p(s) ¯q(s),w h e r e¯ p(s)a n d¯q(s)a r ep o l y - nomials in sand the degree of ¯ q(s) is higher than that of ¯ p(s), then L−1⎪braceleftbigg¯p(s) ¯q(s)⎪bracerightbigg =n⎪summationdisplay k=1¯p(αk) ¯q/prime(αk)exp(tαk), (3.7.22) where αkare the distinct roots of the equation ¯ q(s)=0 . PROOF Without loss of generality, we ca n assume that the leading coef- ﬁcient of ¯ q(s) is unity and write distinct factors of ¯ q(s)s ot h a t ¯q(s)=(s−α1)(s−α2)···(s−αk)···(s−αn). (3.7.23) Using the rules of partial fraction decomposition, we can write ¯f(s)=¯p(s) ¯q(s)=n⎪summationdisplay k=1Ak (s−αk), (3.7.24) where Akare arbitrary constants to be det ermined. In view of (3.7.23), we ﬁnd ¯p(s)=n⎪summationdisplay k=1Ak(s−α1)(s−α2)···(s−αk−1)(s−αk+1)···(s−αn). Substitution of s=αkgives ¯p(αk)=Ak(αk−α1)(αk−α2)···(αk−αk+1)···(αk−αn),(3.7.25) where k=1,2,3,...,n . Diﬀerentiation of (3.7.23) yields ¯q/prime(s)=n⎪summationdisplay k=1(s−α1)(s−α2)···(s−αk−1)(s−αk+1)···(s−αn), whence it follows that ¯q/prime(αk)=(αk−α1)(αk−α2)···(αk−αk−1)(αk−αk+1)···(αk−αn). (3.7.26) © 2007 by Taylor & Francis Group, LLC 166 INTEGRAL TRANSFORMS and THEIR APPLICATIONS From (3.7.25) and (3.7.26), we ﬁnd Ak=¯p(αk) ¯q/prime(αk), and hence, ¯p(s) ¯q(s)=n⎪summationdisplay k=1¯p(αk) ¯q/prime(αk)1 (s−αk). (3.7.27) Inversion gives immediately L−1⎪braceleftbigg¯p(s) ¯q(s)⎪bracerightbigg =n⎪summationdisplay k=1¯p(αk) ¯q/prime(αk)exp(tαk). This proves the theorem. We give some examples of this theorem. Example 3.7.16 We consider L−1⎪braceleftbiggs s2−3s+2⎪bracerightbigg . Here ¯p(s)=s,a n d¯ q(s)=s2−3s+2=( s−1)(s−2). Hence, L−1⎪braceleftbiggs s2−3s+2⎪bracerightbigg =¯p(2) ¯q/prime(2)e2t+¯p(1) ¯q/prime(1)et=2e2t−et. Example 3.7.17 Use Heaviside’s power series expansion to evaluate L−1⎪braceleftbigg1 ssinhx√ s sinh√ s⎪bracerightbigg ,0<x< 1,s > 0. We have 1 ssinhx√ s sinh√ s=1 s⎪parenleftBigg ex√ s−e−x√ s e√ s−e−√ s⎪parenrightBigg =1 se−(1−x)√ s−e−(1+x)√ s 1−e−2√ s =1 s⎪bracketleftBig e−(1−x)√ s−e−(1+x)√ s⎪bracketrightBig⎪parenleftBig 1−e−2√ s⎪parenrightBig−1 =1 s⎪bracketleftBig e−(1−x)√ s−e−(1+x)√ s⎪bracketrightBig∞⎪summationdisplay n=0exp(−2n√ s) =1 s∞⎪summationdisplay n=0⎪bracketleftbig exp{−(1−x+2n)√ s}−exp{−(1 +x+2n)√ s}⎪bracketrightbig . © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 167 Hence, L−1⎪braceleftbigg1 ssinhx√ s sinh√ s⎪bracerightbigg =L−1⎪braceleftBigg 1 s∞⎪summationdisplay n=0⎪bracketleftbig exp{−(1−x+2n)√ s}−exp{−(1 +x+2n)√ s}⎪bracketrightbig⎪bracerightBigg =∞⎪summationdisplay n=0⎪bracketleftbigg erfc⎪parenleftbigg1−x+2n 2√ t⎪parenrightbigg −erfc⎪parenleftbigg1+x+2n 2√ t⎪parenrightbigg⎪bracketrightbigg . Example 3.7.18 Ifα=⎪radicalbigg s a, show that L−1⎪bracketleftbiggcoshαx scoshα/lscript⎪bracketrightbigg =1−4 π∞⎪summationdisplay k=0(−1)kcos⎪braceleftbigg⎪parenleftbigg k+1 2⎪parenrightbiggπx /lscript⎪bracerightbigg exp⎪bracketleftbigg −(2k+1 )2aπ2t 4/lscript2⎪bracketrightbigg (2k+1 ). (3.7.28) In this case, we write L−1{¯f(s)}=L−1⎪braceleftbigg¯p(s) ¯q(s)⎪bracerightbigg =L−1⎪braceleftbiggcoshαx scoshα/lscript⎪bracerightbigg . Clearly, the zeros of ¯f(s)a r ea t s= 0 and at the roots of cosh α/lscript=0 ,t h a ti s , ats=sk=a⎪parenleftbigg k+1 2⎪parenrightbigg2⎪parenleftbiggπi /lscript⎪parenrightbigg2 ,k=0,1,2,....Thus, αk=⎪radicalbigg sk a=⎪parenleftbigg k+1 2⎪parenrightbiggπi /lscript,k =0,1,2,.... Here ¯p(s)=c o s h ( αx), ¯q(s)=scosh(α/lscript). In order to apply the Heaviside Ex- pansion Theorem, we need ¯q/prime(s)=d ds(scoshα/lscript)=c o s h ( α/lscript)+1 2α/lscriptsinh(α/lscript). For the zero s=0 , ¯q/prime(0) = 1, and for the zeros at s=sk, ¯q/prime(sk)=1 2⎪parenleftbigg k+1 2⎪parenrightbigg πi·sinh⎪bracketleftbigg⎪parenleftbigg k+1 2⎪parenrightbigg πi⎪bracketrightbigg =( 2k+1 )πi 4·isin⎪bracketleftbigg⎪parenleftbigg k+1 2⎪parenrightbigg π⎪bracketrightbigg =−(2k+1 )π 4·coskπ=(−1)k+1(2k+1 )π 4. © 2007 by Taylor & Francis Group, LLC 168 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Consequently, L−1⎪braceleftbiggcoshαx scoshα/lscript⎪bracerightbigg =1+4 π∞⎪summationdisplay k=0(−1)k+1 (2k+1 )cosh⎪bracketleftbigg (2k+1 )πix 2/lscript⎪bracketrightbigg exp(tsk) =1−4 π∞⎪summationdisplay k=0(−1)k (2k+1 )cos⎪bracketleftBig (2k+1 )πx 2/lscript⎪bracketrightBig ×exp⎪bracketleftBigg −⎪parenleftbigg k+1 2⎪parenrightbigg2π2at /lscript2⎪bracketrightBigg . 3.8 Tauberian Theorems and Watson’s Lemma These theorems give the behavior of object functions in terms of the behavior of transform functions. Particularly, they determine the value of the object functions f(t) for large and small values of time t. Tauberian theorems are extremely useful and have frequent applications. THEOREM 3.8.1 ( The Initial Value Theorem ). IfL{f(t)}=¯f(s) exists, then lim s→∞¯f(s)=0. (3.8.1) In addition, if f(t) and its derivatives exist as t→0, we obtain the Initial Value Theorem : (i) lim s→∞[s¯f(s)] = lim t→0f(t)=f(0) (3.8.2) (ii) lim s→∞[s2¯f(s)−sf(0)] = lim t→0f/prime(t)=f/prime(0),and (3.8.3) (iii) lim s→∞[sn+1¯f(s)−sn¯f(s)−···− sf(n−1)(0)] = f(n)(0). (3.8.4) Results (3.8.2)–(3.8.4), which are true under fairly general conditions, de- termine the initial values f(0),f/prime(0),...,f(n)(0) of the function f(t)a n di t s derivatives from the Laplace transform ¯f(s). PROOF To prove (3.8.1), we use the fact that the Laplace integral (3.2.5) is uniformly convergent with respect to the parameter s. Hence, it is permissible © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 169 to take the limit s→∞ under the sign of integration so that lim s→∞¯f(s)=∞⎪integraldisplay 0( lim s→∞e−st)f(t)dt=0. N e x t ,w eu s et h es a m ea r g u m e n tt oo b t a i n lim s→∞L{f/prime(t)}=∞⎪integraldisplay 0( lim s→∞e−st)f/prime(t)dt=0. Then it follows from result (3.4.10) that lim s→∞[s¯f(s)−f(0)] = 0 , and hence, we obtain (3.8.2), that is, lim s→∞[s¯f(s)] =f(0) = lim t→0f(t). A similar argument combined with Theorem 3.4.2 leads to (3.8.3) and (3.8.4). Example 3.8.1 Verify the truth of Theorem 3.8.1 for ¯f(s)=(n+1 ) !s−(n+1)where nis a positive integer. Clearly, f(t)=tn.T h u s ,w eh a v e lim s→∞¯f(s) = lim s→∞(n+1 ) ! sn+1=0, lim s→∞s¯f(s)=0= f(0). Example 3.8.2 Findf(0) and f/prime(0) when (a)¯f(s)=1 s(s2+a2),( b ) ¯f(s)=2s s2−2s+5. (a) It follows from (3.8.2) and (3.8.3) that f(0) = lim s→∞[s¯f(s)] = lim s→∞1 s2+a2=0. f/prime(0) = lim s→∞[s2¯f(s)−sf(0)] = lim s→∞s s2+a2=0. © 2007 by Taylor & Francis Group, LLC 170 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b)f(0) = lim s→∞2s2 s2−2s+5=2. f/prime(0) = lim s→∞[s2¯f(s)−sf(0)] = lim s→∞⎪bracketleftbigg2s3 s2−2s+5−2s⎪bracketrightbigg =4. THEOREM 3.8.2 ( The Final Value Theorem ). If¯f(s)=¯p(s) ¯q(s),w h e r e¯ p(s)a n d¯ q(s)a r ep o l y n o m i a l si n s,a n dt h ed e g r e eo f ¯p(s)i sl e s st h a nt h a to f¯ q(s), and if all roots of ¯ q(s) = 0h a v en e g a t i v er e a l parts with the possible exception of one root which may be at s=0 ,t h e n (i) lim s→0¯f(s)=∞⎪integraldisplay 0f(t)dt, and (3.8.5) (ii) lim s→0[s¯f(s)] = lim t→∞f(t), (3.8.6) provided the limits exist. Result (3.8.6) is true under more general conditions, and known as the Final Value Theorem . This theorem determines the ﬁnal value of f(t) at inﬁnity from its Laplace transform at s= 0. However, if ¯f(s) is more general than the rational function as stated above, a statement of a more general theorem is needed with appropriate conditions under which it is valid. PROOF To prove (i), we use the same argument as employed in Theorem 3.8.1 and ﬁnd lim s→0¯f(s)=∞⎪integraldisplay 0⎪parenleftBig lim s→0exp(−st)⎪parenrightBig f(t)dt=∞⎪integraldisplay 0f(t)dt. As before, we can use result (3.4.12) to obtain lim s→0L{f/prime(t)}= lim s→0[s¯f(s)−f(0)] =∞⎪integraldisplay 0⎪parenleftBig lim s→0exp(−st)⎪parenrightBig f/prime(t)dt =∞⎪integraldisplay 0f/prime(t)dt=f(∞)−f(0) = lim t→∞[f(t)−f(0)]. Thus, it follows immediately that lim s→0[s¯f(s)] = lim t→∞f(t)=f(∞). © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 171 Example 3.8.3 Findf(∞), if it exists, from the following functions: (a)¯f(s)=1 s(s2+2s+2 ), (c)¯f(s)=s+a s2+b2,(b/negationslash=0 ) ,(b)¯f(s)=1 s−a, (d)¯f(s)=s s−2. (a) Clearly, ¯ q(s) = 0 has roots at s=0a n d s=−1±i, and the conditions of Theorem 3.8.2 are satisﬁed. Thus, lim s→0[s¯f(s)] = lim s→01 s2+2s+2=1 2=f(∞). (b) Here ¯ q(s)=0h a sar e a lp o s i t i v er o o ta t s=aifa>0, and a real negative root if a<0. Thus, when a<0 lim s→0[s¯f(s)] = lim s→0s s−a=0= f(∞). Ifa>0, the Final Value Theorem does not apply. In fact, f(t)=L−1⎪braceleftbigg1 s−a⎪bracerightbigg =eat→∞ ast→∞. (c) Here ¯ q(s) = 0 has purely imaginary roots at s=±ibwhich do not have negative real parts. The Final Value Theorem does not apply. In fac-t,f(t)=c o s bt+ a bsinbtand lim t→∞f(t)d o e sn o te x i s t .H o w e v e r , f(t)i s bounded and oscillatory for all t>0. (d) The Final Value Theorem does not apply as ¯ q(s)=0h a sap o s i t i v er o o t ats=2 . Watson’s Lemma. If (i)f(t)=O(eat)a st→∞,t h a ti s , |f(t)|≤Kexp(at) fort>T where KandTare constants, and (ii) f(t) has the expansion f(t)=tα⎪bracketleftBiggn⎪summationdisplay r=0artr+Rn+1(t)⎪bracketrightBigg for 0<t<T andα>−1, (3.8.7) where |Rn+1(t)|<Atn+1for 0<t<T andAis a constant, then the Laplace transform ¯f(s)h a st h e asymptotic expansion ¯f(s)∼n⎪summationdisplay r=0arΓ(α+r+1 ) sα+r+1+O⎪parenleftbigg1 sα+n+2⎪parenrightbigg ass→∞. (3.8.8) © 2007 by Taylor & Francis Group, LLC 172 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF We have, for s>a, ¯f(s)=T⎪integraldisplay 0e−stf(t)dt+∞⎪integraldisplay Te−stf(t)dt =T⎪integraldisplay 0e−sttα⎪parenleftBiggn⎪summationdisplay r=0artr⎪parenrightBigg dt+T⎪integraldisplay 0e−sttαRn+1(t)dt +∞⎪integraldisplay Te−stf(t)dt. (3.8.9) The general term of the ﬁrst integral in (3.8.9) can be written as T⎪integraldisplay 0are−sttα+rdt=∞⎪integraldisplay 0are−sttα+rdt−∞⎪integraldisplay Tare−sttα+rdt =arΓ(α+r+1 ) sα+r+1+O(e−Ts). (3.8.10) Ass→∞, the second integral in(3.8.9) is less in magnitude than AT⎪integraldisplay 0e−sttα+n+1dt=O⎪parenleftbigg1 sα+n+2⎪parenrightbigg , (3.8.11) and the magnitude of the third integral in (3.8.9) is ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle∞⎪integraldisplay Te−stf(t)dt⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle≤K∞⎪integraldisplay Te−(s−a)tdt=Kexp[−(s−a)T], (3.8.12) which is exponentially small as s→∞. Finally, combining (3.8.10), (3.8.11), and (3.8.12), we obtain ¯f(s)∼n⎪summationdisplay r=0arΓ(α+r+1 ) sα+r+1+O⎪parenleftbigg1 sα+n+2⎪parenrightbigg ass→∞. This completes the proof of Watson’s lemma. This lemma is one of the most widely used methods for ﬁnding asymptotic expansions. In order to further expand its applicability, this lemma has sub-sequently been generalized and its conv erse has also been proved. The reader is referred to Erd´ elyi (1956), Copson (1965), Wyman (1964), Watson (1981), Ursell (1990), and Wong (1989). © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 173 Example 3.8.4 Find the asymptotic expansion of the parabolic cylinder function Dν(s), which is valid for Re( ν)<0, given by Dν(s)=exp⎪parenleftbigg −s2 4⎪parenrightbigg Γ(−ν)∞⎪integraldisplay 0exp⎪bracketleftbigg −⎪parenleftbigg st+t2 2⎪parenrightbigg⎪bracketrightbiggdt tν+1. (3.8.13) To ﬁnd the asymptotic behavior of Dν(s)a ss→∞, we expand exp⎪parenleftbigg −1 2t2⎪parenrightbigg as a power series in tin the form exp⎪parenleftbigg −1 2t2⎪parenrightbigg =∞⎪summationdisplay n=0(−1)nt2n 2nn!. (3.8.14) A c c o r d i n gt oW a t s o n ’ sl e m m a ,a s s→∞, Dν(s)∼exp⎪parenleftbigg −s2 4⎪parenrightbigg Γ(−ν)∞⎪summationdisplay n=0(−1)n 2nn!∞⎪integraldisplay 0t2n−ν−1e−stdt =exp⎪parenleftbigg −s2 4⎪parenrightbigg Γ(−ν)∞⎪summationdisplay n=0(−1)n 2nn!Γ(2n−ν) s2n−ν. (3.8.15) This result is also valid for Re( ν)≥0. 3.9 Exercises 1. Find the Laplace transforms of the following functions: (a) 2t+asinat, (b) (1 −2t)exp (−2t), (c)tcosat, (d)t3/2, (e)H(t−3)exp( t−3), (f)H(t−a) sinh( t−a), (g) (t−3)2H(t−3), (h)tH(t−a), (i) (1 + 2 at)t−1 2exp(at), (j)acos2ωt. 2. Ifnis a positive integer, show that L{t−n}does not exist. 3. Use result (3.4.12) to ﬁnd (a) L{cosat}and (b) L{sinat}. 4. Use the Maclaurin series for sin atand cos atto ﬁnd the Laplace trans- forms of these functions. © 2007 by Taylor & Francis Group, LLC 174 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 5. Show that L⎪bracketleftbigg1 t{exp(−at)−exp(−bt)}⎪bracketrightbigg =l o g⎪parenleftbiggs+b s+a⎪parenrightbigg . 6. Show that L⎧ ⎨ ⎩t⎪integraldisplay 0s(u) udu⎫ ⎬ ⎭=1 s∞⎪integraldisplay s¯f(x)dx. 7. Obtain the inverse Laplace transforms of the following functions: (a)s (s2+a2)(s2+b2), (d)1 (s−1)2(s−2), (g)1 s(s−a)2,(b)1 s2(s2+c2), (e)1 s2+2s+5, (h)1 s2(s−a)2,(c)1 s2exp(−as), (f)1 s2(s+1 ) (s+2 ), (i)1 s2(s−a). 8. Use the Convolution Theorem to ﬁnd the inverse Laplace transforms of the following functions: (a)s2 (s2+a2)2, (b)1 s√ s+4, (c)¯f(s) s, (d)s (s2+a2)2, (e)⎪parenleftbiggω s2+ω2⎪parenrightbigg ¯f(s),(f)1 (s2+a2)2, (g)s (s−a)(s2+b2),(h)1 (s+1 )2, (i)1 sexp(−a√ s), (j)1 s2(s2+a2), (k)(s2−a2) (s2+a2)2, (l)1 2ln (1 +a2 s2). 9. Show that (a)L{exp(−t2)}=√ π 2exp⎪parenleftbiggs2 4⎪parenrightbigg⎪parenleftBig 1−erfs 2⎪parenrightBig , (b)L−1⎪braceleftbigg1 √ s−√ a⎪bracerightbigg =√ aexp(at)+1 √ πt+√ aexp(at)erf(√ at), (c)L−1⎧ ⎪⎪⎨ ⎪⎪⎩sinh⎪parenleftBigsx a⎪parenrightBig s2cosh⎪parenleftbiggsb 2a⎪parenrightbigg⎫ ⎪⎪⎬ ⎪⎪⎭=x a+∞⎪summationdisplay n=0(−1)n+1⎪parenleftbigg4b aπ2⎪parenrightbigg (2n+1 )−2 ×⎪bracketleftbigg sin⎪braceleftBig (2n+1 )πx b⎪bracerightBig cos⎪braceleftbigg (2n+1 )πat b⎪bracerightbigg⎪bracketrightbigg , (d)L−1⎪braceleftbigg1 √ s2+a2⎪bracerightbigg =1 π⎪integraldisplay1 −1eiatx √ 1−x2dx. © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 175 10. Show that (a)L⎪braceleftbigg1 t(sinat−atcosat)⎪bracerightbigg =t a n−1⎪parenleftBiga s⎪parenrightBig −as s2+a2, (b)L⎧ ⎨ ⎩t⎪integraldisplay 01 τ(sinaτ−aτcosaτ)dτ⎫ ⎬ ⎭=1 s⎪bracketleftbigg tan−1⎪parenleftBiga s⎪parenrightBig −as s2+a2⎪bracketrightbigg . 11. Using the Heaviside power series expansion, evaluate the inverse Laplace transforms of the following functions: (a)1 √ s2+a2, (d)1 scosech( x√ s),(b) tan−1⎪parenleftBiga s⎪parenrightBig , (e)1 sexp⎪parenleftbigg −1 s⎪parenrightbigg ,(c) sinh−1⎪parenleftbigg1 s⎪parenrightbigg , (f) sin−1⎪parenleftBiga s⎪parenrightBig . 12. IfL{f(t)}=¯f(s), show that (i)L−1⎪braceleftbigg¯f(s) s⎪bracerightbigg =t⎪integraldisplay 0f(τ)dτ, (ii)L−1⎪braceleftbigg¯f(s) s2⎪bracerightbigg =t⎪integraldisplay 0⎧ ⎨ ⎩t1⎪integraldisplay 0f(τ)dτ⎫ ⎬ ⎭dt1=t⎪integraldisplay 0(t−τ)f(τ)dτ, (iii)L−1⎪braceleftbigg¯f(s) s3⎪bracerightbigg =t⎪integraldisplay 0t1⎪integraldisplay 0t2⎪integraldisplay 0f(τ)dτdt1dt2=t⎪integraldisplay 01 2(t−τ)2f(τ)dτ, and in general (iv)L−1⎪braceleftbigg¯f(s) sn⎪bracerightbigg =t⎪integraldisplay 0t1⎪integraldisplay 0t2⎪integraldisplay 0···tn−1⎪integraldisplay 0f(τ)dτ dt 1···dtn−1 =t⎪integraldisplay 0(t−τ)n−1 (n−1)!f(τ)dτ. 13. The staircase function f(t)=[t] represents the greatest integer less than or equal to t. Find its Laplace transform. 14. Use the convolution theorem to prove the identity t⎪integraldisplay 0J0(τ)J0(t−τ)dτ=s i nt. © 2007 by Taylor & Francis Group, LLC 176 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 15. Show that (a)L{tH(t−a)}=⎪parenleftbigg1 s2+a s⎪parenrightbigg exp(−sa), (b)L{tnexp(at)}=n!(s−a)−(n+1). 16. IfL{f(t)}=¯f(s)a n d f(t) has a ﬁnite discontinuity at t=a, show that L{f/prime(t)}=s¯f(s)−f(0)−exp(−sa)[f]a, where [ f]a=f(a+0 )−f(a−0). 17. If f(t)=H⎪parenleftBig t−π 2⎪parenrightBig sint, ﬁnd its Laplace transforms. 18. Establish the following results: (a)L{sin2at}=2a2 s(s2+4a2), (b)L{I0(x)}=1 √ s2+a2, (c)L{|sinat|}=a s2+a2coth⎪parenleftBigπs 2a⎪parenrightBig ,s >0, (d)L⎧ ⎨ ⎩t⎪integraldisplay 0sinax xdx⎫ ⎬ ⎭=1 stan−1⎪parenleftBiga s⎪parenrightBig , (e)L⎪braceleftbiggd dt(f∗g)⎪bracerightbigg =g(0)¯f(s)+L{f∗g/prime}=s¯f(s)¯g(s) =L{f/prime∗g}+f(0)¯g(s). 19. Establish the following results: (a)L{t2f/prime/prime(t)}=s2d2 ds2¯f(s)+4sd ds¯f(s)+2¯f(s), (b)L{tmf(n)(t)}=(−1)mdm dsm⎪bracketleftBig sn¯f(s)−sn−1f(0)−···− f(n−1)(0)⎪bracketrightBig . 20. (a) Show that f(t)=s i n ( a√ t) satisﬁes the diﬀerential equation 4tf/prime/prime(t)+2f/prime(t)+a2f(t)=0. Use this diﬀerential equation to show that (b)L{sin√ t}=1 2Γ⎪parenleftbigg1 2⎪parenrightbigg s−3/2exp⎪parenleftbigg −1 4s⎪parenrightbigg ,s >0, (c)L⎪braceleftbiggcos√ t √ t⎪bracerightbigg =Γ⎪parenleftbigg1 2⎪parenrightbigg1 √ sexp⎪parenleftbigg −1 4s⎪parenrightbigg ,s >0. © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 177 21. Establish the following results: (a)L⎧ ⎨ ⎩∞⎪integraldisplay tf(x) xdx⎫ ⎬ ⎭=1 ss⎪integraldisplay 0¯f(x)dx, (b)L⎧ ⎨ ⎩∞⎪integraldisplay 0f(x) xdx⎫ ⎬ ⎭=1 s∞⎪integraldisplay 0¯f(x)dx. 22. Use exercise 21(a) to ﬁnd the Laplace transform of (a) the cosine integral deﬁned by Ci(t)=t⎪integraldisplay ∞cosx xdx, t > 0, (b) the exponential integral deﬁned by Ei(t)=∞⎪integraldisplay te−x xdx, t > 0. 23. Show that (a)L{te−btcosat}=(s+b)2−a2 [(s+b)2+a2]2, (b)L⎪braceleftbiggcosat−cosbt t⎪bracerightbigg =1 2log⎪parenleftbiggs2+a2 s2+b2⎪parenrightbigg , (c)L{Ln(t)}=1 s⎪parenleftbiggs−1 s⎪parenrightbiggn ,where Ln(t) are the Laguerre polyno- mials of degree n. 24. IfL{f(t)}=¯f(s)a n dL{g(x, t)}=¯h(s)exp{−x¯h(s)},p r o v et h a t (a)L⎧ ⎨ ⎩∞⎪integraldisplay 0g(x, t)f(x)dx⎫ ⎬ ⎭=¯h(s)¯f{¯h(s)}. (b)L⎧ ⎨ ⎩∞⎪integraldisplay 0J0(2√ xt)f(x)dx⎫ ⎬ ⎭=1 s¯f⎪parenleftbigg1 s⎪parenrightbigg ,w h e n g(x, t)=J0(2√ xt). 25. Use Exercise 24(b) to show that (a)∞⎪integraldisplay 0J0(2√ xt)sin⎪parenleftBigx a⎪parenrightBig dx=acosat, (a/negationslash=0 ), © 2007 by Taylor & Francis Group, LLC 178 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b)∞⎪integraldisplay 0J0(2√ xt)e−xxndx=n!e−tLn(t). 26. Find the Laplace transform of the triangular wave function deﬁned over (0,2a)b y f(t)=⎪braceleftbigg t, 0<t<a 2a−t, a < t < 2a⎪bracerightbigg . 27. Use the Initial Value Theorem to ﬁnd f(0), and f/prime(0) from the following functions: (a)¯f(s)=s s2−5s+1 2, (c)¯f(s)=exp(−sa) s2+3s+5,a >0,(b)¯f(s)=1 s(s2+a2), (d)¯f(s)=s2−1 (s2+1 ). 28. Use the Final Value Theorem to ﬁnd f(∞), if it exists, from the following functions: (a)¯f(s)=1 s(s2+as+b), (c)¯f(s)=1 1+as,(b)¯f(s)=s+2 s2+4, (d)¯f(s)=3 (s2+4 )2. 29. IfL{f(t)}=¯f(s)a n dL{g(t)}=¯g(s), establish Duhamel’s integrals: L−1{s¯f(s)¯g(s)}=⎧ ⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎩f(0)g(t)+ t⎪integraldisplay 0f/prime(τ)g(t−τ)dτ g(0)f(t)+t⎪integraldisplay 0g/prime(τ)f(t−τ)dτ⎫ ⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎭. 30. Using Watson’s lemma, ﬁnd the asymptotic expansion of (a)¯f(s)= ∞⎪integraldisplay 0(1 +t2)−1exp(−st)dt,ass→∞, (b)K0(s)=∞⎪integraldisplay 0(t2−1)−1 2exp(−st)dt,ass→∞, where K0(s)i st h e modiﬁed Bessel function. 31. Find the asymptotic expansion of ¯f(s)a ss→∞ when f(t)i sg i v e nb y © 2007 by Taylor & Francis Group, LLC Laplace Transforms and Their Basic Properties 179 (a) (1 + t)−1, (c) log(1 + t),(b) sin2√ t, (d)J0(at). 32. Use the shifting property (3.4.5) or (3.4.6) to obtain the Laplace trans- form of the following functions: (a)f(t)=(t−a)nH(t−a),(b)f(t)=t2H(t−a), (c)f(t)=⎪braceleftBigg t,0≤t≤a 0,t ≥a⎪bracerightBigg ,(d)f(x)=⎪braceleftBigg w0⎪parenleftbig 1−2x l⎪parenrightbig ,0<x<l 2 0,l 2<x<l⎪bracerightBigg , (e)f(t)=c o s2 tH(t−π), (f)f(t)=⎪braceleftBigg2,0≤t≤a −2,t≥a⎪bracerightBigg . 33. For the square wave function f(t)g i v e nb y f(t)=aH(t)−aH(t−a), show that ¯f(s)=a s(1 +e−as). 34. If f(t)=aH(t)−2aH(t−1) +aH(t−2),show that ¯f(s)=a s⎪parenleftbig 1−2e−s+e−2s⎪parenrightbig . 35. If f(t)=⎪braceleftBigg sint t,t/negationslash=0 1,t =0⎪bracerightBigg ,show that ¯f(s)=t a n−1⎪parenleftbig1 s⎪parenrightbig 36. If fp(t)=tp−1e−tH(t),show that ( fp∗fq)(t) exists if and only if pand qare both positive. Hence, derive the following results (a) (fp∗fq)(t)=B(p, q)fp+q(t). (b)f/prime p(t)=(p−1)fp−1(t)−fp(t). (c) (fp∗fq)/prime(t)=(p−1)B(p−1,q)fp+q−1(t)−B(p, q)fp+q(t). (d) (fp∗fq)/prime(t)=B(p, q)[(p+q−1)fp+q−1(t)−fp+q(t)]. 37. A family {hp(t):p>0}of functions on Ris called a convolution semi- group ifhp∗hq=hp+qfor all p, q > 0.Show that hp(t)=fp(t) Γ(p)deﬁnes a convolution semi-group where fp(t) i sd e ﬁ n e di nE x e r c i s e3 6 . 38. Using the change of variables, s=c+iω,show that the inverse Laplace transformation is a Fourier transformation, that is, (i)f(t)=L−1⎪braceleftbig¯f(s)⎪bracerightbig =ect 2π∞⎪integraltext −∞¯f(c+iω)eiωtdω. © 2007 by Taylor & Francis Group, LLC 180 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (ii)f(t)=1 πectRe∞⎪integraltext 0¯f(c+i ω)eiωtdω. Hence, for real f(t),show that (iii)Fc{ectf(t)}=2 R e⎪bracketleftbig¯f(c+i ω)⎪bracketrightbig , (iv)Fs{ectf(t)}=2 I m⎪bracketleftbig¯f(c+i ω)⎪bracketrightbig . © 2007 by Taylor & Francis Group, LLC 4 Applications of Laplace Transforms “Mathematical sciences have attrac ted special attention since great antiquity, they are attracting still more attention today because of their inﬂuence on industry and the arts. The agreement of theory and practice brings mos t beneﬁcial results, an d it is not exclusively the practical side which gains; science is advancing under its inﬂu- ence as it discovers new object s of study and new aspects of the mathematical sciences....” P. L. Chebyshev “... partial diﬀerential equations are the basis of all physical the- orems. In the theory of sound in gases, liquids and solids, in theinvestigations of elasticity, in opt ics, everywhere partial diﬀeren- tial equations formulate basic laws of nature which can be checked against experiments.” Bernhard Riemann 4.1 Introduction Many problems of physical interest are described by ordinary or partial d-iﬀerential equations with appropriate initial or boundary conditions. Theseproblems are usually formulated as initial value problems, boundary value problems, orinitial-boundary value problems that seem to be mathematically more rigorous andphysically realistic in applied and engineering sciences. The Laplace transform method is particularly useful for ﬁnding solutions of these problems. The method is very eﬀective for the solution of the response of alinear system governed by an ordin ary diﬀerential equation to the initial data and/or to an external disturbance (orexternal input function ). More precisely, we seek the solution of a linear system for its state at subsequent time t>0 due to the initial state at t= 0 and/or to the disturbance applied for t>0. This chapter deals with the solutions of ordinary and partial diﬀerential equations that arise in mathematical, p hysical, and engineering sciences. The 181 © 2007 by Taylor & Francis Group, LLC 182 INTEGRAL TRANSFORMS and THEIR APPLICATIONS applications of Laplace transforms to the solutions of certain integral equa- tions and boundary value problems are also discussed in this chapter. It isshown by examples that the Laplace transform can also be used eﬀectively for evaluating certain deﬁnite integrals. We also give a few examples of solutions of diﬀerence and diﬀerential equations using the Laplace transform technique. The eﬀective use of the joint Laplace and Fourier transform is illustrated by solving several initial-boundary value problems. Application of Laplace trans-forms to the problem of summation of inﬁnite series in closed form is presented with examples. Finally, it is noted that the examples given in this chapter are only representative of a wide variety of problems which can be solved by theuse of the Laplace transform method. 4.2 Solutions of Ordinary Diﬀerential Equations As stated in the introduction of this chapter, the Laplace transform can beused as an eﬀective tool for analyzing the basic characteristi cs of a linear sys- tem governed by the diﬀerential equation in response to initial data and/or to an external disturbance. The following examples illustrate the use of theLaplace transform in solving certain initial value problems described by ordi- nary diﬀerential equations. Example 4.2.1 (Initial Value Problem ). We consider the ﬁrst-order o rdinary diﬀerential equa- tiondx dt+px=f(t),t > 0, (4.2.1) with the initial condition x(t=0 )= a, (4.2.2) where pandaare constants and f(t)isan external input function so that its Laplace transform exists. Application of the Laplace transform ¯ x(s) of the function x(t)g i v e s s¯x(s)−x(0) +p¯x(s)=¯f(s), or ¯x(s)=a s+p+¯f(s) s+p. (4.2.3) The inverse Laplace transform together with the Convolution Theorem leads to the solution x(t)=ae−pt+t⎪integraldisplay 0f(t−τ)e−pτdτ. (4.2.4) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 183 Thus, the solution naturally splits into two terms—the ﬁrst term corresponds to the response of the initial condition and the second term is entirely due tothe external input function f(t). In particular, if f(t)=q= constant, then the solution (4.2.4) becomes x(t)=q p+⎪parenleftbigg a−q p⎪parenrightbigg e−pt. (4.2.5) The ﬁrst term of this solution is independent of time tand is usually called thesteady-state solution . The second term depends on time ta n di sc a l l e dt h e transient solution . In the limit as t→∞, the transient solution decays to zero ifp>0 and the steady-state solution is attained. On the other hand, when p<0,the transient solution grows exponentially as t→∞,and the solution becomes unstable. Equation (4.2.1) describes the law of natural growth or decay process with an external forcing function f(t) according as p>0o r<0. In particular, if f(t)=0 a n d p>0,the resulting equation (4.2 .1) occurs very frequently in chemical kinetics. Such an equation des cribes the rate of che mical reactions. Example 4.2.2 (Second Order Ordinary Diﬀerential Equation ). The second order linear ordi- nary diﬀerential equation has the general form d2x dt2+2pdx dt+qx=f(t),t > 0. (4.2.6) The initial conditions are x(t)=a,dx dt=˙x(t)=batt=0, (4.2.7ab) where p,q,a andbare constants. Application of the Laplace transform to this general initial value problem gives s2¯x(s)−sx(0)−˙x(0) + 2 p{s¯x(s)−x(0)}+q¯x(s)=¯f(s). The use of (4.2.7ab) leads to the solution for ¯ x(s)a s ¯x(s)=(s+p)a+(b+pa)+¯f(s) (s+p)2+n2,n2=q−p2. (4.2.8) The inverse transform gives the solutio n in three distinct forms depending on © 2007 by Taylor & Francis Group, LLC 184 INTEGRAL TRANSFORMS and THEIR APPLICATIONS q>=<p2,and they are x(t)=ae−ptcosnt+1 n(b+pa)e−ptsinnt +1 nt⎪integraldisplay 0f(t−τ)e−pτsinnτdτ, when n2=q−p2>0,(4.2.9) x(t)=ae−pt+(b+pa)te−pt +t⎪integraldisplay 0f(t−τ)τe−pτdτ, when n2=q−p2=0, (4.2.10) x(t)=ae−ptcoshmt+1 m(b+pa)e−ptsinhmt +1 mt⎪integraldisplay 0f(t−τ)e−pτsinhmτdτ, when m2=p2−q>0.(4.2.11) Example 4.2.3 (Higher Order Ordinary Diﬀerential Equations ). We solve the linear equation of order nwith constant coeﬃcients as f(D){x(t)}≡Dnx+a1Dn−1x+a2Dn−2x+···+anx=φ(t),t > 0, (4.2.12) with the initial conditions x(t)=x0,D x (t)=x1,D2x(t)=x2,...,Dn−1x(t)=xn−1,att=0, (4.2.13) where D=d dtis the diﬀerential operator and x0,x1,...,x n−1are constants. We take the Laplace transform of (4.2.12) to get (sn¯x−sn−1x0−sn−2x1−···− sxn−2−xn−1⎪parenrightbig +a1⎪parenleftbig sn−1¯x−sn−2x0−sn−3x1−···− xn−2⎪parenrightbig +a2⎪parenleftbig xn−2¯x−sn−3x0−···− xn−3⎪parenrightbig +···+an−1(s¯x−x0)+an¯x=¯φ(s). (4.2.14) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 185 Or, (sn+a1sn−1+a2sn−2+···+an)¯x(s) =¯φ(s)+(sn−1+a1sn−2+···+an−1)x0 +(sn−2+a1sn−3+···+an−2)x1+···+(s+a1)xn−2+xn−1 =¯φ(s)+¯ψ(s), (4.2.15) where ¯ψ(s) is made up of all terms on the right hand side of (4.2.15) except ¯φ(s),and is a polynomial in sof degree ( n−1). Hence, ¯f(s)¯x(s)=¯φ(s)+¯ψ(s), where ¯f(s)=sn+a1sn−1+···+an. Thus, the Laplace transform solution, ¯ x(s)i s ¯x(s)=¯φ(s)+¯ψ(s) ¯f(s). (4.2.16) Inversion of (4.2.16) yields x(t)=L−1⎪braceleftbigg¯φ(s) ¯f(s)⎪bracerightbigg +L−1⎪braceleftbigg¯ψ(s) ¯f(s)⎪bracerightbigg . (4.2.17) The inverse operation on the right can be carried out by partial fraction de- composition, by the Heaviside Expansion Theorem, or by contour integration. Example 4.2.4(Third Order Ordinary Diﬀerential Equations ). We solve (D 3+D2−6D)x(t)=0,D ≡d dt,t > 0, (4.2.18) with the initial data x(0) = 1 ,˙x(0) = 0 ,and ¨x(0) = 5 . (4.2.19) The Laplace transform of equation (4.2.18) gives [s3¯x−s2x(0)−s˙x(0)−¨x(0)] + [ s2¯x−sx(0)−˙x(0)]−6[s¯x−x(0)] = 0 . In view of the initial conditions, we ﬁnd ¯x(s)=s2+s−1 s(s2+s−6)=s2+s−1 s(s+3 ) (s−2). © 2007 by Taylor & Francis Group, LLC 186 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Or, ¯x(s)=1 6·1 s+1 3·1 s+3+1 2·1 s−2. Inverting gives the solution x(t)=1 6+1 3e−3t+1 2e2t. (4.2.20) Example 4.2.5 (System of First Order Ordinary Diﬀerential Equations ). Consider the system dx1 dt=a11x1+a12x2+b1(t) dx2 dt=a21x1+a22x2+b2(t)⎫ ⎪⎪⎬ ⎪⎪⎭(4.2.21ab) with the initial data x 1(0) =x10and x2(0) =x20; (4.2.22ab) where a11,a12,a21,a22are constants. Introducing the matrices x≡⎪parenleftBigg x1 x2⎪parenrightBigg ,dx dt≡⎛ ⎜⎜⎝dx1 dt dx2 dt⎞ ⎟⎟⎠,A ≡⎪parenleftBigg a11a12 a21a22⎪parenrightBigg , b(t)≡⎪parenleftBigg b1(t) b2(t)⎪parenrightBigg and x0=⎪parenleftBigg x10 x20⎪parenrightBigg , we can write the above system in a matrix diﬀerential system as dx dt=Ax+b(t),x (0) =x0. (4.2.23ab) We take the Laplace transform of the system with the initial conditions to get (s−a11)¯x1−a12¯x2=x10+¯b1(s), −a21¯x1+(s−a22)¯x2=x20+b2(s). The solutions of this algebraic system are ¯x1(s)=⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglex 10+¯b1(s)−a12 x20+¯b2(s)s−a22⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingles−a 11−a12 −a21s−a22⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle,¯x 2(s)=⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingles−a 11x10+¯b1(s) −a21x20+¯b2(s)⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingles−a 11−a12 −a21s−a22⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle. (4.2.24ab) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 187 Expanding these determinants, results for ¯ x1(s)a n d¯ x2(s) can readily be inverted, and the solutions for x1(t)a n d x2(t) can be found in closed forms. Example 4.2.6 Solve the matrix diﬀerential system dx dt=Ax, x (0) =⎪parenleftbigg0 1⎪parenrightbigg , (4.2.25) where x=⎪parenleftbiggx1 x2⎪parenrightbigg and A=⎪parenleftBigg 01 −23⎪parenrightBigg . This system is equivalent to dx1 dt−x2=0, dx2 dt+2x1−3x2=0, with x1(0) = 0 and x2(0) = 1 . Taking the Laplace transform of the co upled system with the given initial data, we ﬁnd s¯x1−¯x2=0, 2¯x1+(s−3)¯x2=1. This system has the solutions ¯x1(s)=1 s2−3s+2=1 s−2−1 s−1, ¯x2(s)=s s2−3s+2=2 s−2−1 s−1. Inverting these results, we obtain x1(t)=e2t−et,x 2(t)=2e2t−et. In matrix notation, the solution is x(t)=⎪parenleftbigge2t−et 2e2t−et⎪parenrightbigg . (4.2.26) © 2007 by Taylor & Francis Group, LLC 188 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 4.2.7 (Second Order Coupled Diﬀerential System ). Solve the system d2x1 dt2−3x1−4x2=0 d2x2 dt2+x1+x2=0⎫ ⎪⎪⎬ ⎪⎪⎭t>0, (4.2.27) with the initial conditions x 1(t)=x2(t)=0 ;dx1 dt=2 a n ddx2 dt=0a t t=0. (4.2.28) The use of the Laplace transform to (4.2.27) with (4.2.28) gives (s2−3)¯x1−4¯x2=2 ¯x1+(s2+1 )¯x2=0. Then ¯x1(s)=2(s2+1 ) (s2−1)2=(s+1 )2+(s−1)2 (s2−1)2=1 (s−1)2+1 (s+1 )2. Hence, the inversion yields x1(t)=t(et+e−t). (4.2.29) ¯x2(s)=−2 (s2−1)2=1 2⎪bracketleftbigg1 s−1−1 s+1−1 (s−1)2−1 (s+1 )2⎪bracketrightbigg , which can be readily inverted to ﬁnd x2(t)=1 2(et−e−t−tet−te−t). (4.2.30) Example 4.2.8 (The Harmonic Oscillator in a Non-Resisting Medium ). The diﬀerential e- quation of the oscillator in the presence of an external driving force Ff(t) is d2x dt2+ω2x=Ff(t), (4.2.31) where ωis the frequency and Fis a constant. The initial conditions are x(t)=a,˙x(t)=U att=0, (4.2.32) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 189 where aandUare constants. Taking the Laplace transform of (4.2.31) with the initial conditions, we obtain (s2+ω2)¯x(s)=sa+U+F¯f(s). Or, ¯x(s)=as s2+ω2+U s2+ω2+F¯f(s) s2+ω2. (4.2.33) Inversion together with the convolution theorem yields x(t)=acosωt+U ωsinωt+F ωt⎪integraldisplay 0f(t−τ)sinωτdτ (4.2.34) =Acos(ωt−φ)+F ωt⎪integraldisplay 0f(t−τ)s i nωτdτ, (4.2.35) where A=⎪parenleftbigg a2+U2 ω2⎪parenrightbigg1/2 and φ=t a n−1⎪parenleftbiggU ωa⎪parenrightbigg . The solution (4.2.35) consists of two terms. The ﬁrst term represents the response to the initial data, and it describes free oscillations with amplitude A,phase φ,and frequency ω,which is called the natural frequency of the oscillator. The second term arises in response to the external force, and hence, it represents the forced oscillations. In order to investigate some interesting features of solution (4.2.35), we sel ect the following cases of interest: (i) Zero Forcing Function . In this case, solution (4.2.35) reduces to x(t)=Acos(ωt−φ). (4.2.36) This represents simple harmonic motion with amplitude A,frequency ωand phase φ.Evidently, the motion is oscillatory. (ii) Steady Forcing Function, that is ,f(t)=1. In this case, solution (4.2.35) becomes x−F ω2=Acos(ωt−φ)−F ω2cosωt. (4.2.37) In particular, when the particle is released from rest, U=0,(4.2.37) takes the form x−F ω2=⎪parenleftbigg a−F ω2⎪parenrightbigg cosωt. (4.2.38) This corresponds to free oscillations with the natural frequency ωand displays a shift in the equilibrium position from the origin to the pointF ω2. © 2007 by Taylor & Francis Group, LLC 190 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (iii) Periodic Forcing Function, that is ,f(t)=c o s ω0t. The transform solution can readily be found form (4.2.33) in the form ¯x(s)=as s2+ω2+U s2+ω2+Fs (s2+ω2 0)(s2+ω2) =as s2+ω2+U s2+ω2+Fs (ω2 0−ω2)⎪parenleftbigg1 s2+ω2−1 s2+ω2 0⎪parenrightbigg .(4.2.39) Inversion yields the solution x(t)=acosωt+U ωsinωt+F (ω2 0−ω2)(cosωt−cosω0t) (4.2.40) =Acos(ωt−φ)+F (ω2 0−ω2)cosω0t, (4.2.41) where A=⎪braceleftBigg⎪parenleftbigg a+F ω2 0−ω2⎪parenrightbigg2 +U2 ω2⎪bracerightBigg1/2 and tan φ=U ω÷⎪parenleftbigg a+F ω2 0−ω2⎪parenrightbigg . It is noted that solution (4.2.41) consists of free oscillations of period⎪parenleftbigg2π ω⎪parenrightbigg and forced oscillations of period⎪parenleftbigg2π ω0⎪parenrightbigg ,which is the same as that of the external periodic force. If ω0<ω,the phase of the forced oscillations is the same as that of the external periodic force. If ω0>ω,the forced term suﬀers from a phase change by an amount π. In other words, the forced motion is in phase or 180◦out of phase with the exter nal force according as ω>or<ω0. When ω=ω0,result (4.2.40) can be written as x(t)=acosωt+U ωsinωt+Ft (ω0+ω)⎡ ⎢⎢⎣sin⎪braceleftbigg1 2(ω−ω0)t⎪bracerightbigg sin⎪braceleftbigg1 2(ω+ω0)t⎪bracerightbigg 1 2(ω0−ω)t⎤ ⎥⎥⎦ =acosωt+U ωsinωt+Ft 2ωsinωt=Acos(ωt−φ)+Ft 2ωsinωt, (4.2.42) where A2=⎪parenleftbigg a2+U2 ω2⎪parenrightbigg and tan φ=U aω. This solution clearly shows that the a mplitude of the forced motion increases witht. Thus, if the natural frequency is equal to the forcing frequency, the oscillations become unbounded, which is physically undesirable. This phe- nomenon is usually called resonance, and the corresponding frequency ω=ω0 is referred to as the resonant frequency of the system. It may be emphasized that at the resonant frequency, the solution of the problem becomes mathe- matically invalid for large times, and hence, it is physically unrealistic. In most © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 191 dynamical systems, this kind of situation is resolved by including dissipating and/or nonlinear eﬀects. Example 4.2.9 (Harmonic Oscillator in a Resisting Medium ). The diﬀerential equation of the oscillator in a resisting medium where the resistance is proportional to velocity is given by d2x dt2+2kdx dt+ω2x=Ff(t), (4.2.43) where k(>0) is a constant of proportionality and the right hand side repre- sents the external driving force. The initial state of the system is x(t)=a,dx dt=U att=0. (4.2.44) In view of the initial conditions, the Laplace transform solution of equation (4.2.43) is obtained as ¯x(s)=a(s+2k)+U+F¯f(s) (s2+2ks+ω2) =a(s+k)+(U+ak)+¯F(s) (s+k)2+n2, (4.2.45) where n2=ω2−k2. T h r e ep o s s i b l ec a s e sd e s e r v ea t t e n t i o n : (i)k<ω (small damping ). In this case, n2=ω2−k2>0 and the inversion of (4.2.45) along with the Convolution Theorem yields x(t)=ae−ktcosnt+(U+ak) ne−ktsinnt+F nt⎪integraldisplay 0f(t−τ)e−kτsinnτdτ. (4.2.46) This is the most general solution of the problem for an arbitrary form of the external driving force. (ii)k=ω(critical damping )so that n2=0 . The solution for this case can readily be obtained from (4.2.45) by inversion and has the form x(t)=ae−kt+(U+ak)te−kt+Ft⎪integraldisplay 0f(t−τ)τe−ktdτ. (4.2.47) (iii)k>ω (large damping ). Setn2=−(k2−ω2)=−m2so that m2=k2−ω2>0. © 2007 by Taylor & Francis Group, LLC 192 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The transformed solution (4.2.45) assumes the form ¯x(s)=a(s+k)+(U+ak)+F¯f(s) (s+k)2−m2. (4.2.48) After inversion, it turns out that x(t)=ae−ktcoshmt+⎪parenleftbiggU+ak m⎪parenrightbigg e−ktsinhmt +F mt⎪integraldisplay 0f(t−τ)e−kτsinhmτdτ. (4.2.49) In order to examine the characteristic features of the problem, it is necessary to specify the nature and functional form of f(t) involved in the external force term. Suppose the external driving force is zero. The solution can readily bewritten down in all three cases. For 0<k<ω, the solution is x(t)=e −kt⎪parenleftbigg acosnt+U+ak nsinnt⎪parenrightbigg =Ae−ktcos(nt−φ),(4.2.50) where A=⎪braceleftBig a2+(U+ak)2 n2⎪bracerightBig1/2 and φ=t a n−1⎪parenleftbiggU+ak an⎪parenrightbigg . Like the harmonic oscillator in a vacuum, the motion is oscillatory with the time-dependent amplitude Ae−ktand the modiﬁed frequency n=(ω2−k2)1/2=ω⎪parenleftbigg 1−1 2k2 ω2+···⎪parenrightbigg ,0<k<ω. This means that, when the resistance is small, the modiﬁed frequency (or theundamped natural frequency) is obviously smaller than the natural frequen- cy,ω.Although the small resistance pro duces an insigniﬁcant eﬀect on the frequency, the amplitude is radically modiﬁed. It should also be noted thatthe amplitude decays exponentially to zero as time t→∞.The phase of the motion is also changed by the small resistance. Thus, the motion is called the damped oscillatory motion, and depicted by Figure 4.1. At the critical case, ω=k,and hence, n=0.The solution can readily be found from (4.2.47) with F=0,and has the form x(t)=ae −kt+(ak+U)te−kt. (4.2.51) The motion ceases to be oscillatory and decays very rapidly as t→∞. If damping is large with no external force, solution (4.2.49) reduces to x(t)=ae−ktcoshmt+⎪parenleftBigg ak+U m⎪parenrightBigg e−ktsinhmt. (4.2.52) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 193 tx(t) Figure 4.1 Damped oscillatory motion. Usingcosh sinhmt=1 2(emt±e−mt),we can write the solution as x(t)=Ae−(k−m)t+Be−(k+m)t, (4.2.53) where A=1 2⎪parenleftbig a+ak+U m⎪parenrightbig and B=1 2⎪parenleftbigg a−ak+U m⎪parenrightbigg . The above solution suggests that the motion is no longer oscillatory and in fact, it decays very rapidly as t→∞. Example 4.2.10 (Harmonic Oscillator in a Resisting Medium with an External Periodic Force ), The motion is governed by the equation d2x dt2+2kdx dt+ω2x=Fcosω0t, k > 0 (4.2.54) with the initial data x(0) =aand ˙ x(0) =U. The transformed solution for the case of small damping ( k<ω)i s ¯x(s)=a(s+k)+(U+ak) (s+k)2+n2+Fs {(s+k)2+n2}(s2+ω2 0) =a(s+k)+(U+ak) (s+k)2+n2+F⎪bracketleftbiggAs−B (s+k)2+n2−As−C s2+ω2 0⎪bracketrightbigg ,(4.2.55) where A=ω2 0−ω2 (ω2−ω2 0)2+4k2ω2 0,B=2kω2 (ω2−ω2 0)2+4k2ω2 0, © 2007 by Taylor & Francis Group, LLC 194 INTEGRAL TRANSFORMS and THEIR APPLICATIONS and C=2kω2 0 (ω2−ω2 0)2+4k2ω2 0withω2=n2+k2. The expression for ¯ x(s) can be inverted to obtain the solution x(t)=(a+FA)e−ktcosnt+1 n(U+ak−FAk−FB)e−ktsinnt −AFcosω0t+CF ω0sinω0t. (4.2.56) It is convenient to write it in the form x(t)=A1cos(ω0t−φ1)+A2e−ktcos(nt−φ2), (4.2.57) where A2 1=F2⎪parenleftbigg A2+C2 ω2 0⎪parenrightbigg =F2 (ω2−ω2 0)2+4k2ω2 0, (4.2.58) tanφ1=−C Aω0=2kω0 ω2−ω2 0, (4.2.59) A2 2=(a+FA)2+1 n2(U+ak−kFa−FB)2,(4.2.60) and tanφ2=U+ak−kFA−FB n(a+FB). (4.2.61) This form of solution (4.2.57) lends itself to some interesting physical inter- pretations. First, t he displacement ﬁeld x(t) essentially consists of the steady state and the transient terms, which are independently modiﬁed by the damp- ing and driving forces involved in the equation of motion. In the limit as t→∞, the latter decays exponentially to zero. Consequently, the ultimate steady s- tate is attained in the limit, and represented by the ﬁrst term of (4.2.57). In fact, the steady-state solution is denoted by xst(t)a n dg i v e nb y xst(t)=A1cos(ω0t−φ1), (4.2.62) where A1is the amplitude, ω0is the frequency, and φ1represents the phase lag given by φ1=t a n−1⎪braceleftBigg 2kω0 ⎪parenleftbig ω2−ω2 0⎪parenrightbig⎪bracerightBigg when ω0<ω, =π−tan−1⎪braceleftBigg 2kω0 ⎪parenleftbig ω2 0−ω2⎪parenrightbig⎪bracerightBigg when ω0>ω, =π 2asω0→ω. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 195 It should be noted that the frequency of the steady-state solution is the same as that of the external driving force, but the amplitude and the phase aremodiﬁed by the parameters ω, kandω 0.It is of interest to examine the nature of the amplitude and the phase with respect to the forcing frequency ω0.For a low frequency ( ω0→0),A1=F ω2andφ1=0.Asω0→ω,the amplitude of the motion is still bounded and equal to⎪parenleftbiggF 2kω⎪parenrightbigg ifk/negationslash=0.The displacement suﬀers from a phase lag of π/2.Further, we note that dA1 dω0=2ω0F⎪parenleftbig ω2−ω2 0−2k2⎪parenrightbig ⎪braceleftbig (ω2−ω2 0)2+4k2ω2 0⎪bracerightbig3/2. (4.2.63) It follows that A1has a minimum at ω0= 0 with minimum valueF ω2,and a maximum at ω0=(ω2−2k2)1/2with maximum valueF 2k(ω2−2k2)1/2pro- vided 2 k2<ω2.If 2k2>ω2,A1has no maximum and gradually decreases. The non-dimensional amplitude A∗=⎪parenleftbigg2A1ω2 F⎪parenrightbigg is plotted against the non- dimensional frequency ω∗=ω0 ωfor a given value ofk ω(<1) in Figure 4.2. 1205k=1 10 *A* Figure 4.2 Amplitude versus frequency with damping. In the absence of the damping term, the amplitude A1becomes A1=F ⎪vextendsingle⎪vextendsingleω2−ω2 0⎪vextendsingle⎪vextendsingle, which is unbounded at ω 0=ωand shown in Figure 4.3. This situation has already been encountered earlier, and the frequency ω0= ωwas deﬁned as the resonant frequency . The diﬃculty for the resonant case has been resolved by the inclu sion of small damping eﬀect. © 2007 by Taylor & Francis Group, LLC 196 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 1205k=1 10 *A* Figure 4.3 Amplitude versus frequency without damping. At the critical case ( k2=ω2),the solution is found from (4.2.55) by inversion and has the form x(t)=A1cos(ω0t−φ)+(a+FA)e−kt +t(U+ak−FAk−FB)e−kt.(4.2.64) The transient term of this solution decays as t→∞ and the steady state is attained. The solution for the case of high damping ( k2>ω2) is obtained from (4.2.55) as x(t)=(a+FA)e−ktcoshmt+1 m(U+ak−FAk−FB)e−ktsinhmt −AFcosω0t+CF ω0sinω0t (4.2.65) where m2=−n2=k2−ω2>0.This result is somewhat similar to that of (4.2.56) or (4.2.57) with the exception that the transient term decays very rapidly as t→∞. Like previous cases, the steady state is reached in the limit. Example 4.2.11Obtain the solution of the Bessel equation td 2x dt2+dx dt+a2tx(t)=0,x (0) = 1 . (4.2.66) Application of the Laplace transform gives L⎪braceleftbigg td2x dt2⎪bracerightbigg +L⎪braceleftbiggdx dt⎪bracerightbigg +a2L{tx(t)}=0. Or, −d ds⎪bracketleftbigg L⎪braceleftbiggd2x dt2⎪bracerightbigg⎪bracketrightbigg +s¯x(s)−x(0)−a2d¯x ds=0. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 197 Or, −d ds[s2¯x−sx(0)−˙x(0)] + s¯x(s)−1−a2d¯x ds=0. Thus, (s2+a2)d¯x ds+s¯x=0. Or, d¯x ¯x=−sd s s2+a2. Integration gives the solution for ¯ x(s) ¯x(s)=A √ s2+a2, where Ais an integrating constant. By the inverse transformation, we obtain the solution x(t)=AJ0(at). Example 4.2.12 Find the solution of the initial value problem d2x dt2+tdx dt−2x=2,x (0) = ˙x(0) = 0 . Taking the Laplace transform yields L⎪braceleftbiggd2x dt2⎪bracerightbigg +L⎪braceleftbigg tdx dt⎪bracerightbigg −2¯x(s)=2 s. Or, s2¯x−d ds{s¯x(s)}−2¯x=2 s d¯x ds+⎪parenleftbigg3 s−s⎪parenrightbigg ¯x=−2 s2. This is a ﬁrst order linear equation, which can be solved by the method of the integrating factor. The integrating factor is s3exp⎪parenleftbigg −1 2s2⎪parenrightbigg .Multiplying the equation by the integrating factor and integrating, it turns out that ¯x(s)=2 s3+A s3exp⎪parenleftbiggs2 2⎪parenrightbigg , © 2007 by Taylor & Francis Group, LLC 198 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where Ais an integrating constant. As ¯ x(s)→∞ ass→∞,we must have A≡0.Thus, ¯ x(s)=2 s3.Inverting, we get the solution x(t)=t2. Example 4.2.13 (Current and Charge in a Simple Electric Circuit ). The current in a circuit (see Figure 4.4) containing inductance L,resistance R,and capacitance C with an applied voltage E(t) is governed by the equation LdI dt+RI+1 Ct⎪integraldisplay 0Idt=E(t), (4.2.67) where L,R,a n dCare constants and I(t) is the current that is related to the accumulated charge Qon the condenser at time tby Q(t)=t⎪integraldisplay 0I(t)dt so thatdQ dt=I(t). (4.2.68) L RC E(t)I(t) Q(t) Figure 4.4 Simple electric circuit. If the circuit is without a condenser ( C→∞),equation (4.2.67) reduces to LdI dt+RI=E(t),t > 0. (4.2.69) This can easily be solved with the initial condition I(t=0 )= I0. However, we solve the system (4.2.67)–(4.2.68) with the initial data I(t=0 )=0 ,Q (t=0 )=0 . (4.2.70) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 199 Then, in the limit C→∞,the solution of the system reduces to that of (4.2.69). Application of the Laplace transform to (4.2.67) with (6.2.70) gives ¯I(s)=1 Ls¯E(s) ⎪parenleftbig s2+R Ls+1 CL⎪parenrightbig=1 L·(s+k−k)¯E(s) (s+k)2+n2, (4.2.71) where k=R 2L,ω2=1 LCandn2=ω2−k2. Inversion of (4.2.71) gives the current ﬁeld for three cases: I(t)=1 Lt⎪integraldisplay 0E(t−τ)⎪parenleftbigg cosnτ−k nsinnτ⎪parenrightbigg e−kτdτ, ifω2>k2(4.2.72) =1 Lt⎪integraldisplay 0E(t−τ)(1−kτ)e−kτdτ, ifω2=k2(4.2.73) =1 Lt⎪integraldisplay 0E(t−τ)⎪parenleftbigg coshmτ−k msinhmτ⎪parenrightbigg e−kτdτ,ifk2>ω2(4.2.74) where m2=−n2. In particular, if E(t)= c o n s t a n t = E0,then the solution can be obtained directly from (4.2.71) by inversion as I(t)=E0 nLexp⎪parenleftbigg −Rt 2L⎪parenrightbigg sinnt, ifn2=1 CL−⎪parenleftbiggR 2L⎪parenrightbigg2 >0,(4.2.75) =E0 Ltexp⎪parenleftbigg −Rt 2L⎪parenrightbigg , if⎪parenleftbiggR 2L⎪parenrightbigg2 =1 CL, (4.2.76) =E0 mLexp⎪parenleftbigg −Rt 2L⎪parenrightbigg sinhmt, ifm2=⎪parenleftbiggR 2L⎪parenrightbigg2 −1 CL>0.(4.2.77) It may be observed that the solution for the case of low resistance ( R2C<4L), or small damping, describes a damped sinusoidal current with slowly decaying amplitude. In fact, the rate of damping is proportional toR L,and when this quantity is large, the attenuation of the current is very rapid. The frequency of the oscillating current ﬁeld is n=⎪parenleftbigg1 CL−R2 4L2⎪parenrightbigg1/2 , which is called the natural frequency of the current ﬁeld. IfR2 4L2<<1 CL,the frequency nis approximately equal to n∼1 √ CL. © 2007 by Taylor & Francis Group, LLC 200 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The case,R2 4L2=1 CL,corresponds to critical damping , and the solution for this case decays exponentially with time. The last case, R2C>4L,corresponds to high resistance or high damping. The current related to this case has the form I(t)=E0 2mL⎪bracketleftBig e−(R 2L−m)t−e−(R 2L+m)t⎪bracketrightBig . (4.2.78) It may be recognized that the solution is no longer oscillatory and decays exponentially to zero as t→∞. This is really expected in an electrical circuit with a very high resistance. If C→∞,the circuit is free from a condenser and m→R 2L.Consequently, solution (4.2.77) reduces to I(t)=E0 R⎪bracketleftbigg 1−exp⎪parenleftbigg −Rt L⎪parenrightbigg⎪bracketrightbigg . (4.2.79) This is identical with the solution of equation (4.2.69). We consider another special case where the alternating voltage is applied to the circuit so that E(t)=E0sinω0t. (4.2.80) The transformed solution for ¯I(s) follows from (4.2.71) as ¯I(s)=⎪parenleftbiggE0ω0 L⎪parenrightbiggs {(s+k)2+n2}⎪parenleftbig s2+ω2 0⎪parenrightbig. (4.2.81) Using the rules of partial fractions, it turns out that ¯I(s)=⎪parenleftbiggE0ω0 L⎪parenrightbigg⎪bracketleftbiggAs−B (s+k)2+n2−As−C s2+ω2 0⎪bracketrightbigg , (4.2.82) where ( A, B, C )≡(ω2 0−ω2,2kω2,2kω2 0) (ω2−ω2 0)2+4k2ω2 0. The inversion of (4.2.82) can be completed by Table B-4 of Laplace trans- forms, and the solution for I(t) assumes three distinct forms according to ω2>=<k2. The solution for the case of low resistance ( ω2>k2)i s I(t)=⎪parenleftbiggE0ω0 L⎪parenrightbigg⎪bracketleftBigg Ae−ktcosnt−1 n(Ak+B)e−ktsinnt −Acosω0t+C ω0sinω0t⎪bracketrightBigg ,(4.2.83) which has the equivalent form I(t)=A1sin(ω0t−φ1)+A2e−ktcos(nt−φ2), (4.2.84) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 201 where A2 1=E2 0 L2⎪parenleftbig A2ω2 0+C2⎪parenrightbig =E2 0ω2 0 L2⎪braceleftBig⎪parenleftbig ω2−ω2 0⎪parenrightbig2+4k2ω2 0⎪bracerightBig,tanφ1=Aω0 C,(4.2.85) A2 2=⎪parenleftbiggE2 0ω2 0 L2⎪parenrightbigg⎪bracketleftbigg A2+1 n2(Ak+B)2⎪bracketrightbigg and tan φ2=−(Ak+B) An.(4.2.86) The current ﬁeld consists of the steady-state and transient components. The latter decays exponentially in a time scale of the orderL R.Consequently, the steady current ﬁeld is established in t he electric circuit and describes the sinusoidal current with constant amplitude and phase lagging by an angle φ1. The frequency of the steady oscillating current is the same as that of the applied voltage. In the critical situation ( ω2=k2),the current ﬁeld is derived from (4.2.82) by inversion and has the form I(t)=A1sin(ω0t−φ1)+⎪parenleftbiggE0ω0 L⎪parenrightbigg⎪bracketleftbig Ae−kt−(Ak+B)te−kt⎪bracketrightbig .(4.2.87) This result suggests that the transient component of the current dies out exponentially in the limit as t→∞.Eventually, the steady oscillating current is set up in the circuit and described by the ﬁrst term of (4.2.87). Finally, thesolution related to the case of high resistance ( ω 2<k2) can be found by direct inversion of (4.2.82) and is given by I(t)=A1sin(ω0t−φ1) +⎪parenleftbiggE0ω0 L⎪parenrightbigg⎪bracketleftbigg Acoshmt−1 m(Ak+B)sin h mt⎪bracketrightbigg e−kt.(4.2.88) This solution is somewhat similar to (4.2.84) with the exception of the form of the transient term which, of course, decays very rapidly as t→∞.C o n s e - quently, the steady current ﬁeld is estalished in the circuit and has the samevalue as in (4.2.84). Finally, we close this example by suggesting a similarity between this elec- tric circuit system and the mechanical s ystem as described in Example 4.2.9. Diﬀerentiation of (4.2.67) with respect to tgives a second order equation for the current ﬁeld as Ld 2I dt2+RdI dt+I C=dE dt. (4.2.89) Also, an equation for the charge ﬁeld Q(t) can be found from (4.2.67) and (4.2.68) as Ld2Q dt2+RdQ dt+Q C=E(t). (4.2.90) © 2007 by Taylor & Francis Group, LLC 202 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Writing 2 k=R Landω2=1 LC,the above equation can be put into the form ⎪parenleftbiggd2 dt2+2kd dt+ω2⎪parenrightbigg⎪parenleftbiggI Q⎪parenrightbigg =1 L⎛ ⎝dE dt E⎞ ⎠. (4.2.91ab) These equations are very similar to equation (4.2.43) for a harmonic oscillator. Example 4.2.14 (Current and Charge in an Electrical Network ). An electrical network is a combination of several interrelated simple electric circuits. Consider a more general network consisting of two electric circuits coupled by the mutual in- ductance Mwith resistances R1andR2,capacitances C1andC2,and self- inductances L1andL2as shown in Figure 4.5. A time-dependent voltage E(t) is applied to the ﬁrst circuit at time t=0,when charges and currents are zero. R1 R2L1 L2MI1(t) I2(t) _+C1 C2Q1 Q2 Figure 4.5 Two coupled electric circuits. The charge and current ﬁelds in the network are governed by the system of ordinary diﬀerential equations L1dI1 dt+R1I1+MdI2 dt+Q1 C1=E(t),t > 0 (4.2.92) MdI1 dt+L2dI2 dt+R2I2+Q2 C2=0,t > 0 (4.2.93) withdQ1 dt=I1anddQ2 dt=I2. The initial conditions are I1=0,Q1=0,I2=0,Q2=0 a t t=0. (4.2.94) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 203 Eliminating the currents from (4.2.92) and (4.2.93), we obtain ⎪parenleftbigg L1D2+R1D+1 C1⎪parenrightbigg Q1+MD2Q2=E(t), (4.2.95) MD2Q1+⎪parenleftbigg L2D2+R2D+1 C2⎪parenrightbigg Q2=0, (4.2.96) where D≡d dt. The Laplace transform can be used to solve this system for Q1andQ2. Similarly, we can ﬁnd solutions for the current ﬁelds I1andI2independently or from the charge ﬁelds. We leave it as an exercise for the reader. In the absence of the external voltage ( E=0 )w i t h R1=R2=0,L1=L2= LandC1=C2=C,addition and subtraction of (4.2.95) and (4.2.96) give ¨Q++α2Q+=0,¨Q−+β2Q−=0, (4.2.97ab) where Q+=Q1+Q2,Q −=Q1−Q2, α2=[C(L+M)]−1,and β2=[ C ( L −M)]−1. Clearly, the system executes uncoupled simple harmonic oscillations with frequencies αandβ. Hence, the normal modes can be generated in this freely oscillatory electrical system. Finally, in the absence of capacitances ( C1→∞,C2→∞),the above net- work reduces to a simple one that consis ts of two electric circuits coupled by the mutual inductance Mwith inductances L1andL2,and resistances R1 andR2. As shown in Figure 4.6, an external voltage is applied to the ﬁrst circuit at time t=0. R1 R2L1 L2MI1I2 _+E(t) Figure 4.6 Two coupled electric circuit s without capacitances. © 2007 by Taylor & Francis Group, LLC 204 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The current ﬁelds in the network are governed by a pair of coupled ordinary diﬀerential equations L1dI1 dt+R1I1+MdI2 dt=E(t),t > 0, (4.2.98) MdI1 dt+L2dI2 dt+R2I2=0,t > 0, (4.2.99) where I1(t)a n d I2(t) are the currents in the ﬁrst and the second circuits, respectively. The initial conditions are I1(0) =I2(0) = 0 . (4.2.100) We shall not pursue the problem further because the transform method of solution is a simple exercise. Example 4.2.15 (Linear Dynamical Systems and Signals . In physical and engineering sciences, a large number of linear dynamical systems with a time dependent input signal f(t) that generates an output signal x(t) can be described by the ordinary diﬀerential equation with constant coeﬃcients (Dn+an−1Dn−1+···+a0)x(t)=(Dm+bm−1Dm−1+···+b0)f(t), (4.2.101) where D≡d dtis the diﬀerential operator, arandbrare constants. We apply the Laplace transform to ﬁnd the output x(t) so that (4.2.101) becomes ¯pn(s)¯x(s)−¯Rn−1=¯qm(s)¯f(s)−¯Sm−1, (4.2.102) where ¯pn(s)=sn+an−1sn−1+···+a0,¯qm(s)=sm+am−1sm−1+···+b0, ¯Rn−1(s)=n−1⎪summationdisplay r=0sn−r−1x(r)(0), ¯Sm−1(s)=m−1⎪summationdisplay r=0sm−r−1f(r)(0). It is convenient to express (4.2.102) in the form ¯x(s)=¯h(s)¯f(s)+¯g(s), (4.2.103) where ¯h(s)=¯qm(s) ¯pn(s)and ¯ g(s)=¯Rn−1(s)−¯Sm−1(s) ¯pn(s), (4.2.104ab) and¯h(s) is usually called the transfer function. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 205 The inverse Laplace transform comb ined with the Convolution Theorem leads to the formal solution x(t)=⎪integraldisplayt 0f(t−τ)h(τ)dτ+g(t). (4.2.105) With zero initial data, ¯ g(s)=0,the transfer function takes the simple form ¯h(s)=¯x(s) ¯f(s). (4.2.106) Iff(t)=δ(t)s ot h a t ¯f(s)=1,then the output function is x(t)=⎪integraldisplayt 0δ(t−τ)h(τ)dτ=h(t), (4.2.107) andh(t)i sk n o w na st h e impulse response . Example 4.2.16 (Delay Diﬀerential Equations ). In many problems, the derivatives of the un- known function x(t) are related to its value at diﬀerent times t−τ.This leads us to consider diﬀerential equations of the form dx dt+ax(t−τ)=f(t), (4.2.108) where ais a constant and f(t) is a given function. Equations of this type are called delay diﬀerential equations . In general, initial value problems for these equations involve the speciﬁcation of x(t)i nt h ei n t e r v a l t0−τ≤t<t0,and this information combined with the equation itself is suﬃcient to determine x(t)f o rt>t0. We show how equation (4.2.108) can be solved by the Laplace transform when t0=0 a n d x(t)=x0fort≤0. In view of the initial condition, we can write x(t−τ)=x(t−τ)H(t−τ) so equation (4.2.108) is equivalent to dx dt+ax(t−τ)H(t−τ)=f(t). (4.2.109) Application of the Laplace transform to (4.2.109) gives s¯x(s)−x0+aexp(−τs)¯x(s)=¯f(s). © 2007 by Taylor & Francis Group, LLC 206 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Or, ¯x(s)=x0+¯f(s) {s+aexp(−τs)}(4.2.110) =1 s{x0+¯f(s)}⎪bracketleftBig 1+a sexp(−τs)⎪bracketrightBig−1 =1 s{x0+¯f(s)}∞⎪summationdisplay n=0(−1)n⎪parenleftBiga s⎪parenrightBign exp(−nτs). (4.2.111) The inverse Laplace transform gives the formal solution x(t)=L−1⎪bracketleftBigg 1 s{x0+¯f(s)}∞⎪summationdisplay n=0(−1)n⎪parenleftBiga s⎪parenrightBign exp(−nτs)⎪bracketrightBigg . (4.2.112) In order to write an explicit solution, we choose x0=0 a n d f(t)=t,and hence, (4.2.112) becomes x(t)=L−1⎪bracketleftBigg 1 s3∞⎪summationdisplay n=0(−1)n⎪parenleftBiga s⎪parenrightBign exp(−nτs)⎪bracketrightBigg =∞⎪summationdisplay n=0(−1)nan(t−nτ)n+2 (n+2 ) !H(t−nτ),t > 0.(4.2.113) Example 4.2.17 (The Renewal Equation in Statistics ). The random function X(t)o ft i m e t represents the number of times some event has occurred between time 0 and timet,and is usually referred to as a counting process . A random variable Xnthat records the time it assumes for Xto get the value nfrom the n−1 is referred to as an inter-arrival time . If the random variables X1,X2,X3, ... are independent and identically distributed, then the counting process X(t)i s called a renewal process . We represent their common probability distribution function by F(t) and the density function by f(t)s ot h a t F/prime(t)=f(t). The renewal function is deﬁned by the expected number of times the event being counted occurs by time tand is denoted by r(t)s ot h a t r(t)=E{X(t)}=∞⎪integraldisplay 0E{X(t)|X1=x}f(x)dx, (4.2.114) where E{X(t)|X1=x}is the conditional expected value of X(t) under the condition that X1=xand has the value E{X(t)|X1=x}=[ 1+ r(t−x)]H(t−x). (4.2.115) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 207 Thus, r(t)=t⎪integraldisplay 0{1+r(t−x)}f(x)dx. Or, r(t)=F(t)+t⎪integraldisplay 0r(t−x)f(x)dx. (4.2.116) This is called the renewal equation in mathematical statistics. We solve the equation by taking the Laplace transform with respect to t,and the Laplace transformed equation is ¯r(s)= F(s)+¯r(s)¯f(s). Or, ¯r(s)= F(s) 1−¯f(s). (4.2.117) The inverse transform gives the formal solution of the renewal function r(t)=L−t⎪braceleftbigg F(s) 1−¯f(s)⎪bracerightbigg . (4.2.118) 4.3 Partial Diﬀerential Equations, Initial and Boundary Value Problems The Laplace transform method is very useful in solving a variety of partial diﬀerential equations with assigned initial and boundary conditions. The fol- lowing examples illustrate the use of the Laplace transform method. Example 4.3.1 (First-Order Initial-Boundary Value Problem ). Solve the equation ut+xux=x, x > 0,t > 0, (4.3.1) with the initial and boundary conditions u(x,0) = 0 for x>0, (4.3.2) u(0,t)=0 f o r t>0. (4.3.3) © 2007 by Taylor & Francis Group, LLC 208 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We apply the Laplace transform of u(x, t) with respect to tto obtain s¯u(x, s)+xd¯u dx=x s, ¯u(0,s)=0. Using the integrating factor xs,the solution of this transformed equation is ¯u(x, s)=Ax−s+x s(s+1 ), where Ais a constant of integration. Since ¯ u(0,s)=0,A= 0 for a bounded solution. Consequently, ¯u(x, s)=x s(s+1 )=x⎪parenleftbigg1 s−1 s+1⎪parenrightbigg . The inverse Laplace transform gives the solution u(x, t)=x(1−e−t). (4.3.4) Example 4.3.2 Find the solution of the equation xut+ux=x, x > 0,t > 0 (4.3.5) with the same initial and boundary conditions (4.3.2) and (4.3.3). Application of the Laplace transform with respect to tto (4.3.5) with the initial conditon gives d¯u dx+xs¯u=x s. Using the integrating factor exp⎪parenleftbigg1 2x2s⎪parenrightbigg gives the solution ¯u(x, s)=1 s2+Aexp⎪parenleftbigg −1 2sx2⎪parenrightbigg , where Ais an integrating constant. Since ¯ u(0,s)=0,A=−1 s2and hence, the solution is ¯u(x, s)=1 s2⎪bracketleftbigg 1−exp⎪parenleftbigg −1 2x2s⎪parenrightbigg⎪bracketrightbigg . (4.3.6) Finally, we obtain the solution by inversion u(x, t)=t−⎪parenleftbigg t−1 2x2⎪parenrightbigg H⎪parenleftbigg t−x2 2⎪parenrightbigg . (4.3.7) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 209 Or, equivalently, u(x, t)=⎧ ⎨ ⎩t,2t<x2 1 2x2,2t>x2⎫ ⎬ ⎭. (4.3.8) Example 4.3.3 (The Heat Conduction Equation in a Semi-Inﬁnite Medium ). Solve the equa- tion ut=κuxx,x > 0,t > 0 (4.3.9) with the initial and boundary conditions u(x,0) = 0 ,x > 0 (4.3.10) u(0,t)=f(t),t > 0 (4.3.11) u(x, t)→0a s x→∞,t > 0. (4.3.12) Application of the Laplace transform with respect to tto (4.3.9) gives d2¯u dx2−s κ¯u=0. (4.3.13) The general solution of this equation is ¯u(x, s)=Aexp⎪parenleftbigg −x⎪radicalbigg s κ⎪parenrightbigg +Bexp⎪parenleftbigg x⎪radicalbigg s κ⎪parenrightbigg . (4.3.14) where AandBare integrating constants. For a bounded solution, B≡0, and using ¯ u(0,s)=¯f(s), we obtain the solution ¯u(x, s)=¯f(s)exp⎪parenleftbigg −x⎪radicalbigg s κ⎪parenrightbigg . (4.3.15) The inversion theorem gives the solution u(x, t)=x 2√ πκt⎪integraldisplay 0f(t−τ)τ−3/2exp⎪parenleftbigg −x2 4κτ⎪parenrightbigg dτ, (4.3.16) which is, by putting λ=x 2√ κτ,o r ,dλ=−x 4√ κτ−3/2dτ, =2 √ π∞⎪integraldisplay x 2√ κtf⎪parenleftbigg t−x2 4κλ2⎪parenrightbigg e−λ2dλ. (4.3.17) © 2007 by Taylor & Francis Group, LLC 210 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This is the formal solution of the problem. In particular, if f(t)=T0= constant, solution (4.3.17) becomes u(x, t)=2T0 √ π∞⎪integraldisplay x κte−λ2dλ=T0erfc⎪parenleftbiggx 2√ κt⎪parenrightbigg . (4.3.18) Clearly, the temperature distribution tends asymptotically to the constant value T0ast→∞. We consider another physical problem that is concerned with the determi- nation of the temperature distribution in a semi-inﬁnite solid when the rate of ﬂow of heat is prescribed at the end x= 0. Thus, the problem is to solve diﬀusion equation (4.3.9) subject to conditions (4.3.10) and (4.3.12) −k⎪parenleftbigg∂u ∂x⎪parenrightbigg =g(t)a tx=0,t > 0, (4.3.19) where kis a constant that is called thermal conductivity . Application of the Laplace transform gives the solution of the transformed problem ¯u(x, s)=1 k⎪radicalbigg κ s¯g(s)exp⎪parenleftbigg −x⎪radicalbigg s κ⎪parenrightbigg . (4.3.20) The inverse Laplace transform yields the solution u(x, t)=1 k⎪radicalbigg κ πt⎪integraldisplay 0g(t−τ)τ−1 2exp⎪parenleftbigg −x2 4κt⎪parenrightbigg dτ, (4.3.21) which is, by the change of variable λ=x 2√ κτ, =x k√ π∞⎪integraldisplay x √ 4κtg⎪parenleftbigg t−x2 4κλ2⎪parenrightbigg λ−2e−λ2dλ. (4.3.22) In particular, if g(t)=T0= constant, the solution becomes u(x, t)=⎪parenleftbiggT0x k√ π⎪parenrightbigg∞⎪integraldisplay x √ 4κtλ−2e−λ2dλ. Integrating this result by parts gives the solution u(x, t)=T0 κ⎪bracketleftBigg 2⎪radicalbigg kt πexp⎪parenleftbigg −x2 4κt⎪parenrightbigg −xerfc⎪parenleftbiggx 2√ κt⎪parenrightbigg⎪bracketrightBigg . (4.3.23) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 211 Alternatively, the heat conduction problem (4.3.9)–(4.3.12) can be solved byusingfractionalderivatives(seeChapter5orDebnath,1978).Werecall(4.3.15) and rewrite it ∂¯u ∂x=−⎪radicalbigg s κ¯u. (4.3.24) In view of (3.9.21), this can be expressed in terms of fractional derivative of order1 2as ∂u ∂x=−1 √ κL−1⎪braceleftbig√ s¯u(x, s)⎪bracerightbig =−1 √ κ0D1 2 tu(x, t). (4.3.25) Thus, the heat ﬂux is expressed in terms of the fractional derivative. In par- ticular, when u(0,t)= con st an t= T0, then the heat ﬂux at the surface is −k⎪parenleftbigg∂u ∂x⎪parenrightbigg x=0=k √ κD1 2 tT0=kT0 √ πκt. (4.3.26) Example 4.3.4 (Diﬀusion Equation in a Finite Medium ). Solve the diﬀusion equation ut=κuxx,0<x<a , t> 0, (4.3.27) with the initial and boundary conditions u(x,0) = 0 ,0<x<a , (4.3.28) u(0,t)=U, t > 0, (4.3.29) ux(a,t)=0,t > 0, (4.3.30) where Uis a constant. We introduce the Laplace transform of u(x, t) with respect to tto obtain d2¯u dx2−s κ¯u=0,0<x<a , (4.3.31) ¯u(0,s)=U s,⎪parenleftbiggd¯u dx⎪parenrightbigg x=a=0. (4.3.32ab) The general solution of (4.3.31) is ¯u(x, s)=Acosh⎪parenleftbigg x⎪radicalbigg s κ⎪parenrightbigg +Bsinh⎪parenleftbigg x⎪radicalbigg s κ⎪parenrightbigg , (4.3.33) © 2007 by Taylor & Francis Group, LLC 212 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where AandBare constants of integration. Using (4.3.32ab), we obtain the values of AandBso that the solution (4.3.33) becomes ¯u(x, s)=U s·cosh⎪bracketleftbigg (a−x)⎪radicalbigg s κ⎪bracketrightbigg cosh⎪parenleftbigg a⎪radicalbigg s κ⎪parenrightbigg. (4.3.34) The inverse Laplace transform gives the solution u(x, t)=UL−1⎧ ⎪⎪⎨ ⎪⎪⎩cosh(a−x)⎪radicalbigg s κ scosh⎪parenleftbigg a⎪radicalbigg s κ⎪parenrightbigg⎫ ⎪⎪⎬ ⎪⎪⎭. (4.3.35) The inversion can be carried out by the Cauchy Residue Theorem to obtain u(x, t)=U⎪bracketleftBigg 1+4 π∞⎪summationdisplay n=1(−1)n 2n−1cos⎪braceleftbigg(2n−1)(a−x)π 2a⎪bracerightbigg ×exp⎪braceleftbigg −(2n−1)2⎪parenleftBigπ 2a⎪parenrightBig2 κt⎪bracerightbigg⎪bracketrightbigg ,(4.3.36) which is, by expanding the cosine term, =U⎪bracketleftBigg 1−4 π∞⎪summationdisplay n=11 (2n−1)sin⎪braceleftbigg⎪parenleftbigg2n−1 2a⎪parenrightbigg πx⎪bracerightbigg ×exp⎪braceleftbigg −(2n−1)2⎪parenleftBigπ 2a⎪parenrightBig2 κt⎪bracerightbigg⎪bracketrightbigg .(4.3.37) This result can be obtained by the method of separation of variables. Example 4.3.5 (Diﬀusion in a Finite Medium ). Solve the one-dimensional diﬀusion equation in a ﬁnite medium 0 <z<a , where the concentration function C(z,t)s a t i s ﬁ e s the equation Ct=κCzz,0<z<a , t> 0, (4.3.38) and the initial and boundary data C(z,0) = 0 for 0 <z<a , (4.3.39) C(z,t)=C0forz=a, t > 0, (4.3.40) ∂C ∂z=0 f o r z=0,t > 0, (4.3.41) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 213 where C0is a constant. Application of the Laplace transform of C(z,t) with respect to tgives d2¯C dz2−⎪parenleftBigs κ⎪parenrightBig ¯C=0,0<z<a , ¯C(a, s)=C0 s,⎪parenleftbiggd¯C dz⎪parenrightbigg z=0=0. The solution of this system is ¯C(z,s)=C0cosh⎪parenleftbigg z⎪radicalbigg s κ⎪parenrightbigg scosh⎪parenleftbigg a⎪radicalbigg s κ⎪parenrightbigg, (4.3.42) which is, by writing α=⎪radicalbigg s κ, =C0 s(eαz+e−αz) (eαa+e−αa) =C0 s[exp{−α(a−z)}+e x p{−α(a+z)}]∞⎪summationdisplay n=0(−1)nexp(−2nαa) =C0 s⎪braceleftBigg∞⎪summationdisplay n=0(−1)nexp[−α{(2n+1 )a−z}] +∞⎪summationdisplay n=0(−1)nexp[−α{(2n+1 )a+z}]⎪bracerightBigg . (4.3.43) Using the result (3.7.4), we obtain the ﬁnal solution C(z,t)=C0⎪braceleftBigg∞⎪summationdisplay n=0(−1)n⎪bracketleftbigg erfc⎪braceleftbigg(2n+1 )a−z 2√ κt⎪bracerightbigg +erfc⎪braceleftbigg(2n+1 )a+z 2√ κt⎪bracerightbigg⎪bracketrightbigg⎪bracerightbigg .(4.3.44) This solution represents as inﬁnite series of complementary error functions. The successive terms of this series are in fact the concentrations at depths a−z,a+z,3a−z,3a+z, ...in the medium. The series converges rapidly for all except large values of⎪parenleftbiggκt a2⎪parenrightbigg . Example 4.3.6 (The Wave Equation for the Transverse Vibration of a Semi-Inﬁnite String ). Find the displacement of a semi-inﬁnite string which is initially at rest in its © 2007 by Taylor & Francis Group, LLC 214 INTEGRAL TRANSFORMS and THEIR APPLICATIONS equilibrium position. At time t=0 , t h e e n d x= 0 is constrained to move so that the displacement is u(0,t)=Af(t)f o rt≥0, where Ais a constant. The problem is to solve the one-dimensional wave equation utt=c2uxx,0≤x<∞,t > 0, (4.3.45) with the boundary and initial conditions u(x, t)=Af(t)a t x=0,t ≥0, (4.3.46) u(x, t)→0a s x→∞,t≥0, (4.3.47) u(x, t)=0=∂u ∂tatt=0 f o r0 <x< ∞. (4.3.48ab) Application of the Laplace transform of u(x, t) with respect to tgives d2¯u dx2−s2 c2¯u=0,for 0≤x<∞, ¯u(x, s)=A¯f(s)a t x=0, ¯u(x, s)→0a s x→∞. The solution of this diﬀerential system is ¯u(x, s)=A¯f(s)exp⎪parenleftBig −xs c⎪parenrightBig . (4.3.49) Inversion gives the solution u(x, t)=Af⎪parenleftBig t−x c⎪parenrightBig H⎪parenleftBig t−x c⎪parenrightBig . (4.3.50) In other words, the solution is u(x, t)=⎡ ⎣Af⎪parenleftBig t−x c⎪parenrightBig ,t >x c 0,t <x c⎤ ⎦. (4.3.51) This solution represents a wave propagating at a velocity cwith the charac- teristic x=ct. Example 4.3.7 (Potential and Current in an Electric Transmission Line ). We consider a transmission line which is a model of co-axial cable containing resistance R, inductance L, capacitance C, and leakage conductance G. The current I(x, t) and potential V(x, t)a tap o i n t xand time tin the line satisfy the coupled equations © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 215 L∂I ∂t+RI=−∂V ∂x, (4.3.52) C∂V ∂t+GV=−∂I ∂x. (4.3.53) IfIorVis eliminated from these equations, both IandVsatisfy the same equation in the form 1 c2utt−uxx+aut+bu= 0 (4.3.54) where c2=(LC)−1,a=LG+RC,a n db=RG. Equation (4.3.54) is called the telegraph equation. Or, equivalently, the telegraph equation can be written in the form utt=c2uxx−(p+q)ut−pqu (4.3.55) where ac2=R C+G C=p+qandbc2=pq. For a lossless transmission line, R=0a n d G=0 ,IorVsatisﬁes the classical wave equation utt=c2uxx. (4.3.56) The solution of this equation with the initial and boundary data is obtained from Example 4.3.6 using the boundary conditions in the potential V(x, t): (i)V(x, t)=V0f(t)a t x=0,t >0. (4.3.57) This corresponds to a signal at the end x=0 f o r t>0, and V(x, t)→0a s x→∞ fort>0. A special case when f(t)=H(t) is also of interest. The solution for this special case is given by V(x, t)=V0f⎪parenleftBig t−x c⎪parenrightBig H⎪parenleftBig t−x c⎪parenrightBig . (4.3.58) This represents a wave propagating at a speed cwith the characteristic x=ct. Similarly, the solution associated with the boundary data (ii) V(x ,t) =V0cosωt at x = 0 for t >0 (4.3.59) V(x, t)→0a s x →∞ for t>0 (4.3.60) can readily be obtained from Example 4.3.6. For ideal submarine cable (or the Kelvin ideal cable ),L=0 a n d G=0 e - quation (4.3.54) reduces to the classical diﬀusion equation ut=κuxx, (4.3.61) © 2007 by Taylor & Francis Group, LLC 216 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where κ=a−1=(RC)−1. The method of solution is similar to that discussed in Example 4.3.3. Using the boundary data (i), the solution for the potential V(x, t)i sg i v e nb y V(x, t)=V0erfc⎪parenleftbiggx 2√ κt⎪parenrightbigg . (4.3.62) The current ﬁeld is given by I(x, t)=−1 R⎪parenleftbigg∂V ∂x⎪parenrightbigg =V0 R(πκt)−1/2exp⎪parenleftbigg −x2 4κt⎪parenrightbigg . (4.3.63) For very large x, the asymptotic representation of the complementary error function is erfc(x)∼1 x√ πexp(−x2),x→∞. (4.3.64) In view of this asymptotic representation, solution (4.3.62) becomes V(x, t)∼2V0 x⎪parenleftbiggκt π⎪parenrightbigg1/2 exp⎪parenleftbigg −x2 4κt⎪parenrightbigg . (4.3.65) For any t>0, no matter how small, solution (4.3.62) reveals that V(x, t)>0 for all x>0, even though V(x, t)→0a sx→∞ Thus, the signal applied at t= 0 propagates with the inﬁnite speed alth ough its amplitude is very small for largex. Physically, the inﬁnite speed is unrealistic and is essentially caused by the neglect of the ﬁrst term in equation (4.3.54). In a real cable, the presence of some inductance would set a limit to the speed of propagation. Instead of the Kelvin cable, a non-inductive leady cable ( L=0a n d G/negationslash=0 ) is of interest. The equation for this case is obtained from (4.3.54) in the form Vxx−aVt−bV=0, (4.3.66) with zero initial conditions, and with the boundary data V(0,t)=H(t)a n d V(x, t)→0a sx→∞. (4.3.67ab) The Laplace transformed problem is d2 V dx2=(sa+b) V, (4.3.68) V(0,s)=1 s, V(x, s)→0a s x→∞. (4.3.69ab) Thus, the solution is given by V(x, s)=1 sexp[−x(sa+b)1/2]. (4.3.70) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 217 With the aid of a standard table of the inverse Laplace transform, the solution is given by V(x, t)=1 2ex√ berfc⎪parenleftBigg x 2⎪radicalbigg a t+⎪radicalbigg bt a⎪parenrightBigg +1 2e−x√ berfc⎪parenleftbiggx 2⎪radicalbigg a t−bt a⎪parenrightbigg . (4.3.71) When G=0(b= 0), the solution becomes identical with (4.3.62). For the Heaviside distortionless cable,R L=G C=k= constant, the potential V(x, t) and the current I(x, t) satisﬁes the same equation utt+2kut+k2u=c2uxx,0≤x<∞,t > 0. (4.3.72) We solve this equation with the init ial data (4.3.48ab) and the boundary condition (4.3.57). Application of the Laplace transform with respect to tto (4.3.72) gives d2 V dx2=⎪parenleftbiggs+k c⎪parenrightbigg2 V. (4.3.73) The solution for V(x, s) with the transformed boundary condition (4.3.56) is V(x, s)=V0¯f(s)exp⎪bracketleftbigg −⎪parenleftbiggs+k c⎪parenrightbigg x⎪bracketrightbigg . (4.3.74) This can easily be inverted to obtain the ﬁnal solution V(x, t)=V0exp⎪parenleftbigg −kx c⎪parenrightbigg f⎪parenleftBig t−x c⎪parenrightBig H⎪parenleftBig t−x c⎪parenrightBig . (4.3.75) This solution represents the signal that propagates with velocity c=(LC)−1/2 with exponentially decaying amplitude, but with no distortion. Thus, the sig- nals can propagate along the Heaviside distortionless line over long distances if appropriate boosters are placed at regular intervals in order to increase thestrength of the signal so as to counteract the eﬀects of attenuation. Example 4.3.8Find the bounded solution of the axisymmetric heat conduction equation u t=κ⎪parenleftbigg urr+1 rur⎪parenrightbigg ,0≤r<a , t> 0, (4.3.76) with the initial and boundary data u(r,0) = 0 for 0 <r<a , (4.3.77) u(r, t)=f(t)a t r=afort>0, (4.3.78) where κandT0are constants. © 2007 by Taylor & Francis Group, LLC 218 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Application of the Laplace transform to (4.3.76) gives d2¯u dr2+1 rd¯u dr−s κ¯u=0. Or, r2d2¯u dr2+rd¯u dr−r2⎪parenleftBigs κ⎪parenrightBig ¯u=0. (4.3.79) This is the standard Bessel equation with the solution ¯u(r, s)=AI0⎪parenleftbigg r⎪radicalbigg s κ⎪parenrightbigg +BK0⎪parenleftbigg r⎪radicalbigg s κ⎪parenrightbigg , (4.3.80) where AandBare constants of integration and I0(x)a n d K0(x)a r et h e modiﬁed Bessel functions of zero order. Since K0(αr) is unbounded at r= 0, for the bounded solution B≡0, and hence, the solution is ¯u(r, s)=AI0(kr),k=⎪radicalbigg s κ. In view of the transformed boundary condition ¯ u(a, s)=¯f(s), we obtain ¯u(r, s)=¯f(s)I0(kr) I0(ka)=¯f(s)¯g(s), (4.3.81) where ¯ g(s)=I0(kr) I0(ka). By Convolution Theorem 3.5.1, the solution takes the form u(r, t)=t⎪integraldisplay 0f(t−τ)g(τ)dτ, (4.3.82) where g(t)=1 2πic+i∞⎪integraldisplay c−i∞estI0(kr) I0(ka)ds. (4.3.83) This complex integral can be evaluated by the theory of residues where the poles of the integrand are at the points s=sn=−κα2 n,n=1,2,3, ...andαn are the roots of J0(aα)=0.The residue at pole s=snis ⎪parenleftbigg2iκαn a⎪parenrightbiggI0(irαn) I/prime 0(iaαn)exp(−κtα2 n)=⎪parenleftbigg2καn a⎪parenrightbiggJ0(rαn) J1(aαn)exp(−κtα2 n), © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 219 so that g(t)=⎪parenleftbigg2κ a⎪parenrightbigg∞⎪summationdisplay n=1αnJ0(rαn) J1(aαn)exp(−κtα2 n). Thus, solution (4.3.82) becomes u(r, t)=⎪parenleftbigg2κ a⎪parenrightbigg∞⎪summationdisplay n=1αnJ0(rαn) J1(aαn)t⎪integraldisplay 0f(t−τ)exp(−κτα2 n)dτ, (4.3.84) where the summation is taken over the positive roots of J0(aα)=0. In particular, if f(t)=T0,then the solution (4.3.84) reduces to u(r, t)=⎪parenleftbigg2T0 a⎪parenrightbigg∞⎪summationdisplay n=1J0(rαn) αnJ1(aαn)(1−e−κtα2 n) =T0⎪bracketleftBigg 1−2 a∞⎪summationdisplay n=1J0(rαn) αnJ1(aαn)e−κtα2 n⎪bracketrightBigg . (4.3.85) Example 4.3.9 (Inhomogeneous Partial Diﬀerential Equation ). We solve the inhomogeneous problem uxt=−ωsinωt, t> 0 (4.3.86) u(x,0)=x, u (0,t)=0. (4.3.87ab) Application of the Laplace transform with respect to tgives d¯u dx=s s2+ω2, which admits the general solution ¯u(x, s)=sx s2+ω2+A, where Ais a constant. Since ¯ u(0,s)=0,A= 0 and hence, the solution is ob- tained by inversion as u(x, t)=xcosωt. (4.3.88) Example 4.3.10 (Inhomogeneous Wave Equation ). Find the solution of 1 c2utt−uxx=ksin⎪parenleftBigπx a⎪parenrightBig ,0<x<a ,t> 0, (4.3.89) u(x,0) = 0 = ut(x,0),0<x<a , (4.3.90) u(0,t)=0= u(a, t),t > 0, (4.3.91) © 2007 by Taylor & Francis Group, LLC 220 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where c, k,andaare constants. Application of the Laplace transforms gives d2¯u dx2−s2 c2¯u=−k ssin⎪parenleftBigπx a⎪parenrightBig , (4.3.92) ¯u(0,s)=0=¯ u(a, s). (4.3.93) The general solution of equation (4.3.92) is ¯u(x, s)=Aexp⎪parenleftBigsx c⎪parenrightBig +Bexp⎪parenleftBig −sx c⎪parenrightBig +ksin⎪parenleftBigπx a⎪parenrightBig a2s⎪parenleftbigg s2+π2c2 a2⎪parenrightbigg. (4.3.94) In view of (4.3.93), A=B=0,and hence, the solution (4.3.94) becomes ¯u(x, s)=k π2c2sin⎪parenleftBigπx a⎪parenrightBig⎡ ⎢⎣1 s−s s2+π2c2 a2⎤ ⎥⎦, (4.3.95) which, by inversion, gives the solution, u(x, t)=k (πc)2⎪bracketleftbigg 1−cos⎪parenleftbiggπct a⎪parenrightbigg⎪bracketrightbigg sin⎪parenleftBigπx a⎪parenrightBig . (4.3.96) Example 4.3.11 (The Stokes Problem and the Rayleigh Problem in Fluid Dynamics ). Solve the Stokes problem, which is concerned with the unsteady boundary layer ﬂows induced in a semi-inﬁnite viscous ﬂuid bounded by an inﬁnite horizontal disk atz= 0 due to non-torsional oscillations of the disk in its own plane with a given frequency ω. We solve the boundary layer equation in ﬂuid dynamics ut=νuzz,z > 0,t > 0, (4.3.97) with the boundary and initial conditions u(z,t)=U0eiωtonz=0,t>0, (4.3.98) u(z,t)→0a s z→∞,t >0, (4.3.99) u(z,t)→0a t t≤0 for all z>0, (4.3.100) where u(z,t) is the velocity of ﬂuid of kinematic viscosity νandU0is a constant. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 221 The Laplace transform solution of the problem with the transformed bound- ary conditions is u(z,s)=U0 (s−iω)exp⎪parenleftbigg −z⎪radicalbigg s ν⎪parenrightbigg . (4.3.101) Using a standard table of inverse Laplace transforms, we obtain the solution u(z,t)=U0 2eiωt[exp(−λz)erfc(ζ−√ iωt) +e x p ( λz)erfc(ζ+√ iωt)], (4.3.102) where ζ=z/(2√ νt) is called the similarity variable of the viscous boundary layer theory and λ=(iω/ν)1 2. The result (4.3.101) describes the unsteady boundary layer ﬂow. In view of the asymptotic formula for the complementary error function erfc(ζ∓√ iωt)∼(2,0) as t→∞, (4.3.103) the above solution for u(z,t) has the asymptotic representation u(z,t)∼U0exp(iωt−λz)=U0exp⎪bracketleftbigg iωt−⎪parenleftBigω 2ν⎪parenrightBig1 2(1 +i)z⎪bracketrightbigg .(4.3.104) This is called the Stokes steady-state solution . This represents the propaga- tion of shear waves which spread out from the oscillating disk with velocity (ω/k)=√ 2νωand exponentially decaying amplitude. The boundary layer as- sociated with the solution has thickness of the order⎪radicalbig ν/ωin which the shear oscillations imposed by the disk decay exponentially with distance zfrom the disk. This boundary layer is called the Stokes layer . In other words, the thick- ness of the Stokes layer is equal to the depth of penetration of vorticity which is essentially conﬁned to the immediate vicinity of the disk for high frequencyω. The Stokes problem with ω= 0 becomes the Rayleigh problem .I no t h e r words, the motion is generated in th e ﬂuid from rest by moving the disk impulsively in its own plane with constant velocity U 0. In this case, the Laplace transformed solution is u(z,s)=U0 sexp⎪parenleftbigg −z⎪radicalbigg s ν⎪parenrightbigg . (4.3.105) Hence, the inversion gives the Rayleigh solution u(z,t)=U0erfc⎪parenleftbiggz 2√ νt⎪parenrightbigg . (4.3.106) This describes the growth of a boundary layer adjacent to the disk. The as- sociated boundary layer is called the Rayleigh layer of thickness of the order © 2007 by Taylor & Francis Group, LLC 222 INTEGRAL TRANSFORMS and THEIR APPLICATIONS δ∼√ νt, which grows with increasing time. The rate of growth is of the order dδ/dt∼⎪radicalbig ν/t, which diminishes with increasing time. The vorticity of the unsteady ﬂow is given by ∂u ∂z=U0 √ πνtexp(−ζ2), (4.3.107) which decays exponentially to zero as z> >δ . Note that the vorticity is everywhere zero at t= 0. This implies that it is generated at the disk and diﬀuses outward within the Rayleigh layer. The total viscous diﬀusion time is Td∼⎪parenleftbig δ2/ν⎪parenrightbig . Another physical quantity related to the Stokes and Rayleigh problems is theskin friction on the disk deﬁned by τ0=μ⎪parenleftbigg∂u ∂z⎪parenrightbigg z=0, (4.3.108) where μ=νρis the dynamic viscosity and ρis the density of the ﬂuid. The skin friction can readily be calculated from the ﬂow ﬁeld given by (4.3.104) or (4.3.106). 4.4 Solutions of Integral Equations DEFINITION 4.4.1 An equation in which the unknown function occurs under an integral is called an integral equation. An equation of the form f(t)=h(t)+λb⎪integraldisplay ak(t, τ)f(τ)dτ, (4.4.1) in which fis the unknown function, h(t),k(t, τ); and the limits of integration aandbare known; and λis a constant, is called the linear integral equation of the second kind or the linear Volterra integral equation. The function k(t, τ)is called the kernel of the equation. Such an equation is said to be homogeneous or inhomogeneous according to h(t)=0orh(t)/negationslash=0. If the kernel of the equation has the form k(t, τ)=g(t−τ), the equation is referred to as the convolution integral equation. In this section, we show how the Laplace transform method can be applied successfully to solve the convolution integral equations. This method is simple and straightforward, and can be illustrated by examples. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 223 To solve the convolution integral equation of the form f(t)=h(t)+λt⎪integraldisplay 0g(t−τ)f(τ)dτ, (4.4.2) we take the Laplace transform of this equation to obtain ¯f(s)=¯h(s)+λL⎧ ⎨ ⎩t⎪integraldisplay 0g(t−τ)f(τ)dτ⎫ ⎬ ⎭, which is, by the Convolution Theorem, ¯f(s)=¯h(s)+λ¯f(s)¯g(s). Or, ¯f(s)=¯h(s) 1−λ¯g(s). (4.4.3) Inversion gives the formal solution f(t)=L−1⎪braceleftbigg¯h(s) 1−λ¯g(s)⎪bracerightbigg . (4.4.4) In many simple cases, the right-hand side can be inverted by using partial fractions or the theory of residues. Hence, the solution can readily be found. Example 4.4.1 Solve the integral equation f(t)=a+λt⎪integraldisplay 0f(τ)dτ. (4.4.5) We take the Laplace transform of (4.4.5) to ﬁnd ¯f(s)=a s−λ, whence, by inversion, it follows that f(t)=aexp(λt). (4.4.6) Example 4.4.2 Solve the integro-diﬀerential equation f(t)=asint+2t⎪integraldisplay 0f/prime(τ)sin(t−τ)dτ, f (0) = 0 . (4.4.7) © 2007 by Taylor & Francis Group, LLC 224 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Taking the Laplace transform, we obtain ¯f(s)=a s2+1+2L{f/prime(t)}L{sint} Or, ¯f(s)=a s2+1+2{s¯f(s)−f(0)} s2+1. Hence, by the initial condition, ¯f(s)=a (s−1)2. Inversion yields the solution f(t)=atexp(t). (4.4.8) Example 4.4.3 Solve the integral equation f(t)=atn−e−bt−c⎪integraldisplayt 0f(τ)ec(t−τ)dτ. (4.4.9) Taking the Laplace transform, we obtain ¯f(s)=an! sn+1−1 s+b−¯f(s)c s−c so that we have ¯f(s)=⎪parenleftbiggs−c s⎪parenrightbigg⎪bracketleftbiggan! sn+1−1 s+b⎪bracketrightbigg =an! sn+1−(ac)n! sn+2−1 s⎪bracketleftbiggs+b−c−b s+b⎪bracketrightbigg =an! sn+1−(ac)n! sn+2−1 s+c+b b⎪bracketleftbigg1 s−1 s+b⎪bracketrightbigg =an! sn+1−(ac)n! sn+2−1 s+⎪parenleftBig 1+c b⎪parenrightBig1 s−⎪parenleftBig 1+c b⎪parenrightBig1 s+b =an! sn+1−(ac)n! sn+2+c bs−⎪parenleftBig 1+c b⎪parenrightBig1 s+b Inversion yields the solution f(t)=atn−n!ac (n+1 ) !tn+1+c b−⎪parenleftBig 1+c b⎪parenrightBig e−bt. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 225 4.5 Solutions of Boundary Value Problems The Laplace transform technique is also very useful in ﬁnding solutions of certain simple boundary value problems that arise in many areas of applied mathematics and engineering sciences. We illustrate the method by solving boundary value problems in the theory of deﬂection of elastic beams. A horizontal beam experiences a ver tical deﬂection due to the combined eﬀect of its own weight and the applied load on the beam. We consider a beam of length /lscriptand its equilibrium position is taken along the horizontal x-axis. Example 4.5.1 (Deﬂection of Beams ). The diﬀerential equation for the vertical deﬂection y(x) of a uniform beam under the action of a transverse load W(x)p e ru n i tl e n g t h at a distance xfrom the origin on the x-axis of the beam is Eld4y dx4=W(x),for 0 <x</lscript , (4.5.1) where EisYoung’s modulus ,Iis the moment of inertia of the cross section about an axis normal to the plane of bending and EIis called the ﬂexural rigidity of the beam. Some physical quantities associated with the problem are y/prime(x),M(x)= EIy/prime/prime(x)a n dS(x)=M/prime(x)=EIy/prime/prime/prime(x), which respectively represent the slope, bending moment, and shear at a point. It is of interest to ﬁnd the solution of (4.5.1) subject to a given loading func- tion and simple boundary conditions involving the deﬂection, slope, bending moment and shear. We consider the following cases: (i) Concentrated load on a clamped beam of length /lscript,t h a ti s , W(x)≡Wδ(x−a), y(0) =y/prime(0) = 0 and y(/lscript)=y/prime(/lscript)=0, where Wis a constant and 0 <a</lscript . (ii) Distributed load on a uniform beam of length /lscriptclamped at x=0 a n d unsupported at x=/lscript,t h a ti s , W(x)=WH(x−a), y(0) =y/prime(0) = 0, and M(/lscript)=S(/lscript)=0 . (iii) A uniform semi-inﬁnite beam freely hinged at x= 0 resting horizontally on an elastic foundation and carrying a load Wper unit length. © 2007 by Taylor & Francis Group, LLC 226 INTEGRAL TRANSFORMS and THEIR APPLICATIONS In order to solve the problem, we use the Laplace transform ¯ y(s)o fy(x) deﬁned by ¯y(s)=∞⎪integraldisplay 0e−sxy(x)dx. (4.5.2) In view of this transformation, equation (4.5.1) becomes EI[s4¯y(s)−s3y(0)−s2y/prime(0)−sy/prime/prime(0)−y/prime/prime/prime(0)] = W(s). (4.5.3) The solution of the transformed deﬂection function ¯ y(s)f o rc a s e( i )i s ¯y(s)=y/prime/prime(0) s3+y/prime/prime/prime(0) s4+W EIe−as s4. (4.5.4) Inversion gives y(x)=y/prime/prime(0)x2 2+y/prime/prime/prime(0)x3 6+W 6EI(x−a)3H(x−a). (4.5.5) y/prime(x)=y/prime/prime(0)x+1 2x2y/prime/prime/prime(0) +W 2EI(x−a)2H(x−a). (4.5.6) The conditions y(/lscript)=y/prime(/lscript)=0r e q u i r et h a t y/prime/prime(0)/lscript2 2+y/prime/prime/prime(0)/lscript3 6+W 6EI(/lscript−a)3=0, y/prime/prime(0)/lscript+y/prime/prime/prime(0)/lscript2 2+W 2EI(/lscript−a)2=0. These algebraic equation s determine the value of y/prime/prime(0) and y/prime/prime/prime(0). Solving these equations, it turns out that y/prime/prime(0) =Wa(/lscript−a)2 EI /lscript2andy/prime/prime/prime(0) =−W(/lscript−a)2(/lscript+2a) EI /lscript3. Thus, the ﬁnal solution for case (i) is y(x)=W 2EI⎪bracketleftbigga(/lscript−a)2x2 /lscript2−(/lscript−a)2(/lscript+2a)x3 3/lscript3+(x−a)3H(x−a) 3⎪bracketrightbigg .(4.5.7) It is now possible to calculate the bending moment and shear at any point of the beam, and, in particular, at the ends. The solution for case (ii) follows directly from (4.5.3) in the form y(s)=y/prime/prime(0) s3+y/prime/prime/prime(0) s4+W EIe−as s5. (4.5.8) The inverse transformation yields y(x)=1 2y/prime/prime(0)x2+1 6y/prime/prime/prime(0)x3+W 24EI(x−a)4H(x−a), (4.5.9) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 227 where y/prime/prime(0) and y/prime/prime/prime(0) are to be determined from the remaining boundary conditions M(/lscript)=S(/lscript) = 0, that is, y/prime/prime(/lscript)=y/prime/prime/prime(/lscript)=0 . From (4.5.9) with y/prime/prime(/lscript)=y/prime/prime/prime(/lscript) = 0, it follows that y/prime/prime(0) +y/prime/prime/prime(0)/lscript+W 2EI(/lscript−a)2=0 y/prime/prime/prime(0) +W EI(/lscript−a)=0 which give y/prime/prime(0) =W(/lscript−a)(/lscript+a) 2EIandy/prime/prime/prime(0) =−W EI(/lscript−a). Hence, the solution for y(x) for case (ii) is y(x)=W 2EI⎪bracketleftbigg(/lscript2−a2)x2 2−(/lscript−a)x3 3+W 12(x−a)4H(x−a)⎪bracketrightbigg .(4.5.10) The shear, S, and the bending moment, M, at the origin, can readily be calculated from the solution. The diﬀerential equation for case (iii) takes the form EId4y dx4+ky=W, x > 0, (4.5.11) where the second term on the left-hand side represents the eﬀect of elastic foundation and kis a positive constant. Writing⎪parenleftbiggk EI⎪parenrightbigg =4ω4, equation (4.5.11) becomes ⎪parenleftbiggd4 dx4+4ω4⎪parenrightbigg y(x)=W EI,x > 0. (4.5.12) This has to be solved subject to the boundary conditions y(0) =y/prime/prime(0) = 0 , (4.5.13) y(x) is ﬁnite as x→∞. (4.5.14) Using the Laplace transform with respect to xto (4.5.12), we obtain (s4+4ω4)¯y(s)=⎪parenleftbiggW EI⎪parenrightbigg1 s+sy/prime(0) +y/prime/prime/prime(0). (4.5.15) In view of the Tauberian Theorem 3.8.2 (ii), that is, lim s→0s¯y(s) = lim x→∞y(x), © 2007 by Taylor & Francis Group, LLC 228 INTEGRAL TRANSFORMS and THEIR APPLICATIONS it follows that ¯ y(s) must be of the form ¯y(s)=W EI1 s(s4+4ω4), (4.5.16) which gives lim x→∞y(x)=W k. (4.5.17) We now write (4.5.16) as ¯y(s)=W EI1 4ω4⎪bracketleftbigg1 s−s3 s4+4ω4⎪bracketrightbigg . (4.5.18) Using the standard table of inverse Laplace transforms, we obtain y(x)=W k(1−cosωxcoshωx) =W k⎪bracketleftbigg 1−1 2e−ωxcosωx−1 2eωxcosωx⎪bracketrightbigg . (4.5.19) In view of (4.5.17), the ﬁnal solution is y(x)=W k⎪parenleftbigg 1−1 2e−ωxcosωx⎪parenrightbigg . (4.5.20) 4.6 Evaluation of Deﬁnite Integrals The Laplace transform can be employ ed to evaluate eas ily certain deﬁnite integrals containing a parameter. Although the method of evaluation may not be very rigorous, it is quite simple and straightforward. The method is essen-tially based upon the permissibility of interchange of the order of integration, that is, Lb⎪integraldisplay af(t, x)dx=b⎪integraldisplay aLf(t, x)dx, (4.6.1) and may be well described by considering some important integrals. Example 4.6.1 Evaluate the integral f(t)=∞⎪integraldisplay 0sintx x(a2+x2)dx. (4.6.2) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 229 We take the Laplace transform of (4.6.2) with respect to tand interchange the order of integration, which is permissible due to uniform convergence, toobtain ¯f(s)=∞⎪integraldisplay 0dx x(a2+x2)∞⎪integraldisplay 0e−stsintxdt =∞⎪integraldisplay 0dx (a2+x2)(x2+s2) =1 s2−a2∞⎪integraldisplay 0⎪parenleftbigg1 a2+x2−1 x2+s2⎪parenrightbigg dx =1 s2−a2⎪parenleftbigg1 a−1 s⎪parenrightbiggπ 2 =π 21 s(s+a)=π 2⎪parenleftbigg1 s−1 s+a⎪parenrightbigg . Inversion gives the value of the given integral f(t)=π 2a(1−e−at). (4.6.3) Example 4.6.2 Evaluate the integral f(t)=∞⎪integraldisplay 0sin2tx x2dx. (4.6.4) A procedure similar to the above integral with 2 sin2tx=1−cos(2tx)g i v e s ¯f(s)=1 2∞⎪integraldisplay 01 x2⎪parenleftbigg1 s−s 4x2+s2⎪parenrightbigg dx=2 s∞⎪integraldisplay 0dx 4x2+s2 =1 s∞⎪integraldisplay 0dy y2+s2=1 s2⎪bracketleftBig tan−1y s⎪bracketrightBig∞ 0=±π 2s2 according as s>or<0. The inverse transform yields f(t)=πt 2sgnt. (4.6.5) © 2007 by Taylor & Francis Group, LLC 230 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 4.6.3 Show that ∞⎪integraldisplay 0xsinxt x2+a2dx=π 2e−at,(a, t >0). (4.6.6) Suppose f(t)=∞⎪integraldisplay 0xsinxt x2+a2dx. Taking the Laplace transform with respect to tgives ¯f(s)=∞⎪integraldisplay 0x2dx (x2+a2)(x2+s2) =∞⎪integraldisplay 0dx x2+s2−a2 s2−a2∞⎪integraldisplay 0⎪parenleftbigg1 x2+a2−1 x2+s2⎪parenrightbigg dx =π 2s⎪parenleftbigg 1−a s+a⎪parenrightbigg =π 21 (s+a). Taking the inverse transform, we obtain f(t)=π 2e−at. 4.7 Solutions of Diﬀerence and Diﬀerential-Diﬀerence E- quations Like diﬀerential equations, the diﬀerence and diﬀerential-diﬀerence equations describe mechanical, electrical, and e lectronic systems of interest. These e- quations also arise frequently in pro blems of economics and business, and particularly in problems concerning interest, annuities, amortization, loan- s, and mortgages. Thus, for the study of the above systems or problems, it is often necessary to solve diﬀerence or diﬀerential-diﬀerence equations withprescribed initial data. This section is essentially devoted to the solution of simple diﬀerence and diﬀerential-diﬀerence equations by the Laplace transfor- m technique. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 231 Suppose {ur}∞ r=1is a given sequence. We introduce the diﬀerence operators Δ, Δ2,Δ3,...,Δndeﬁned by Δur=ur+1−ur, (4.7.1) Δ2ur=Δ ( Δ ur)=Δ ( ur+1−ur)=ur+2−2ur+1+ur,(4.7.2) Δ3ur=Δ2(ur+1−ur)=ur+3−3ur+2+3ur+1−ur. (4.7.3) More generally, Δnur=Δn−1(ur+1−ur)=n⎪summationdisplay k=0(−1)k⎪parenleftbiggn k⎪parenrightbigg ur+n−k. (4.7.4) These expressions are usually called the ﬁrst, second, third, andnth ﬁnite diﬀerences respectively. Any equatio n expressing a relation between ﬁnite d- iﬀerences is called a diﬀerence equation . The highest order ﬁnite diﬀerence involved in the equation is referred to as its order. A diﬀerence equation containing the derivatives of the unknown function is called the diﬀerential- diﬀerence equation . Thus, the diﬀerential-diﬀer ence equation has two distinct orders—one is related to the highest order ﬁnite diﬀerence and the other isassociated with the highest order derivatives. Equations Δu r−ur=0, (4.7.5) Δ2ur−2Δur=0, (4.7.6) are the examples of diﬀerence equations of the ﬁrst and second order, respec- tively. The most general linear nth order diﬀerence equation has the form a0Δnur+a1Δn−1ur+···+an−1Δur+anur=f(n), (4.7.7) where a0,a1,...,a nandf(n) are either constants or functions of non-negative integer n. Like ordinary diﬀerential equations, (4.7.7) is called a homogeneous orinhomogeneous according to f(n)=0o r /negationslash=0 . The following equations u/prime(t)−u(t−1)= 0 , (4.7.8) u/prime(t)−au(t−1)=f(t), (4.7.9) are the examples of the diﬀerenti al-diﬀerence equations, where f(t)i sag i v e n function of t. The study of the above equation is facilitated by introducing the function Sn(t)=H(t−n)−H(t−n−1),n≤t<n+1, (4.7.10) where nis a non-negative integer and H(t) is the Heaviside unit step function. © 2007 by Taylor & Francis Group, LLC 232 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The Laplace transform of Sn(t)i sg i v e nb y Sn(s)=L{Sn(t)}=∞⎪integraldisplay 0e−st{H(t−n)−H(t−n−1)}dt =n+1⎪integraldisplay ne−stdt=1 s(1−e−s)e−ns= S0(s)exp (−ns),(4.7.11) where S0(s)i se q u a lt o1 s(1−e−s). We next deﬁne the function u(t)b yas e r i e s u(t)=∞⎪summationdisplay n=0unSn(t), (4.7.12) where {un}∞ n=0is a given sequence. It follows that u(t)=uninn≤t<n+1 and represents a staircase function. Further u(t+1 )=∞⎪summationdisplay n=0unSn(t+1 )=∞⎪summationdisplay n=0un[H(t+1−n)−H(t−n)] =∞⎪summationdisplay n=1unSn−1(t)=∞⎪summationdisplay n=0un+1Sn(t). (4.7.13) Similarly, u(t+2 )=∞⎪summationdisplay n=0un+2Sn(t). (4.7.14) More generally, u(t+k)=∞⎪summationdisplay n=0un+kSn(t). (4.7.15) The Laplace transform of u(t)i sg i v e nb y ¯u(s)=L{u(t)}=∞⎪integraldisplay 0e−stu(t)dt=∞⎪summationdisplay n=0un∞⎪integraldisplay 0e−stSn(t)dt =1 s(1−e−s)∞⎪summationdisplay n=0unexp(−ns). Thus, ¯u(s)=1 s(1−e−s)ζ(s)=¯S0(s)ζ(s), (4.7.16) where ζ(s)r e p r e s e n t st h e Dirichlet function deﬁned by ζ(s)=∞⎪summationdisplay n=0unexp(−ns). (4.7.17) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 233 We thus deduce u(t)=L−1{¯S0(s)ζ(s)}. (4.7.18) In particular, if un=anis a geometric sequence, then ζ(s)=∞⎪summationdisplay n=0(ae−s)n=1 1−ae−s=es es−a. (4.7.19) Thus, we obtain from (4.7.16) that L{an}=¯S0(s)ζ(s)=¯S0(s)es es−a, (4.7.20) so that L−1⎪braceleftbigg ¯S0(s)es es−a⎪bracerightbigg =an. (4.7.21) From the identity, ∞⎪summationdisplay n=0(n+1 ) (ae−s)n=( 1−ae−s)−2, (4.7.22) it further follows that L{(n+1 )an}=¯S0(s)(1−ae−s)−2=e2s¯S0(s) (es−a)2. (4.7.23) Thus, L−1⎪braceleftbigge2s¯S0(s) (es−a)2⎪bracerightbigg =(n+1 )an. (4.7.24) We deduce from (4.7.22) that ∞⎪summationdisplay n=0nane−ns=aes (1−ae−s)2. (4.7.25) Hence, L{nan}=¯S0(s)aes (es−a)2. (4.7.26) Therefore, L−1⎪braceleftbigga¯S0(s)es (es−a)2⎪bracerightbigg =nan. (4.7.27) THEOREM 4.7.1 If ¯u(s)=L{u(t)},t h e n L{u(t+1 )}=es[¯u(s)−u0¯S0(s)],u 0=u(0). (4.7.28) © 2007 by Taylor & Francis Group, LLC 234 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF We have L{u(t+1 )}=∞⎪integraldisplay 0e−stu(t+1 )dt=es∞⎪integraldisplay 1e−sτu(τ)dτ =es⎡ ⎣¯u(s)−1⎪integraldisplay 0e−sτu(τ)dτ⎤ ⎦ =es⎡ ⎣¯u(s)−u(0)1⎪integraldisplay 0e−sτdτ⎤ ⎦=es[¯u(s)−u0¯S0(s)]. This proves the theorem. In view of this theorem, we derive L{u(t+2 )}=es[L{u(t+1 )}−u(1)¯S0(s)] =e2s[¯u(s)−u(0)¯S0(s)]−esu1¯S0(s) =e2s[¯u(s)−(u0+u1e−s)¯S0(s)],u (1) =u1.(4.7.29) Similarly, L{u(t+3 )}=e3s[¯u(s)−(u0+u1e−s+u2e−2s)¯S0(s)]. (4.7.30) More generally, if kis an integer, L{u(t+k)}=eks⎪parenleftBigg ¯u(s)−¯S0(s)k−1⎪summationdisplay r=0ure−rs⎪parenrightBigg . (4.7.31) Example 4.7.1 Solve the diﬀerence equation Δun−un=0, (4.7.32) with the initial condition u0=1 . We take the Laplace transform of the equation to obtain L{un+1}−2L{un}=0, which is, by (4.7.28), es[¯u(s)−u0¯S0(s)]−2¯u(s)=0. Thus, ¯u(s)=es¯S0(s) es−2. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 235 Inversion with (4.7.21) gives the solution un=2n. (4.7.33) Example 4.7.2 Show that the solution of the diﬀerence equation Δ2un−2Δun= 0 (4.7.34) is un=A+B3n, (4.7.35) where A=1 2(3u0−u1)a n d B=1 2(u1−u0). The given equation is un+2−4un+1+3un=0. Taking the Laplace transform, we obtain e2s[¯u(s)−(u0+u1e−s)¯S0(s)]−4es[¯u(s)−u0¯S0(s)] + 3¯u(s)=0 or, (e2s−4es+3 )¯u(s)=[u0(e2s−4es)+u1es]¯S0(s). Hence, ¯u(s)=¯S0(s)⎪bracketleftbiggu0(e2s−4es)+u1es (es−1)(es−3)⎪bracketrightbigg =¯S0(s)⎪bracketleftbigg(3u0−u1)es 2(es−1)+(u1−u0)es 2(es−3)⎪bracketrightbigg . The inverse Laplace transform combined with (4.7.21) gives un=A+B3n. Example 4.7.3 Solve the diﬀerence equation un+2−2λun+1+λ2un=0, (4.7.36) withu0=0a n d u1=1 . The Laplace transformed equation is e2s[¯u(s)−e−s¯S0(s)]−2λ¯u(s)es+λ2¯u(s)=0 © 2007 by Taylor & Francis Group, LLC 236 INTEGRAL TRANSFORMS and THEIR APPLICATIONS or, ¯u(s)=es¯S0(s) (es−λ)2. The inverse transform gives the solution un=1 λnλn=nλn−1. (4.7.37) Example 4.7.4 Solve the diﬀerential-diﬀerence equation u/prime(t)=u(t−1),u(0) = 1 . (4.7.38) Application of the Laplace transform gives s¯u(s)−u(0) =e−s[¯u(s)−u(0)¯S0(s)], or, ¯u(s)(s−e−s)=1+e−s s(e−s−1). Or, ¯u(s)=⎪braceleftbigg1 s−e−s−e−s s(s−e−s)⎪bracerightbigg +e−2s s(s−e−s) =1 s+e−2s s2⎪parenleftbigg 1−e−s s⎪parenrightbigg−1 =1 s+e−2s s2+e−3s s3+e−4s s4+···+e−ns sn+···. In view of the result L−1⎪braceleftbigge−as sn⎪bracerightbigg =(t−a)n−1 Γ(n)H(t−a), (4.7.39) we obtain the solution u(t)=1+(t−2) 1!+(t−3)2 2!+···+(t−n)n−1 (n−1)!,t > n . (4.7.40) Example 4.7.5 Solve the diﬀerential-diﬀerence equation u/prime(t)−αu(t−1) =β, u (0) = 0 . (4.7.41) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 237 Application of the Laplace transform yields s¯u(s)−u(0)−αe−s[¯u(s)−u(0)¯S0(s)] =β s. Or, ¯u(s)=β s(s−αe−s)=β s2⎪parenleftBig 1−α se−s⎪parenrightBig−1 =β⎪bracketleftbigg1 s2+αe−s s3+α2e−2s s4+···+αne−ns sn+2+···⎪bracketrightbigg . Inverting with the help of (4.7.39), we obtain the solution u(t)=β⎪bracketleftbigg t+α(t−1)2 Γ(3)+α2(t−2)3 Γ(4)+···+αn(t−n)n+1 Γ(n+2 )⎪bracketrightbigg ,t>n . (4.7.42) 4.8 Applications of the Joint Laplace and Fourier Trans- form Example 4.8.1 (The Inhomogeneous Cauchy Problem for the Wave Equation ). Use the joint Fourier and Laplace transform method to solve the Cauchy problem for the wave equation as stated in Example 2.12.4. with an inhomogeneous term, q(x, t). We deﬁne the joint Fourier and Laplace transform of u(x, t)b y ¯U(k,s)=1 √ 2π∞⎪integraldisplay −∞e−ikxdx∞⎪integraldisplay 0e−stu(x, t)dt. (4.8.1) The transformed inhomogeneous Cauchy problem has the solution in the form ¯U(k,s)=sF(k)+G(k)+¯Q(k,s) (s2+c2k2), (4.8.2) where ¯Q(k,s) is the joint transform of the inhomogeneous term, q(x, t)p r e s e n t on the right side of the wave equation. © 2007 by Taylor & Francis Group, LLC 238 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The joint inverse transform gives the solution as u(x, t)=1 √ 2π∞⎪integraldisplay −∞eikxL−1⎪bracketleftbiggsF(k)+G(k)+¯Q(k,s) s2+c2k2⎪bracketrightbigg dk =1 √ 2π∞⎪integraldisplay −∞⎪bracketleftbigg F(k)cosckt+G(k) cksinckt⎪bracketrightbigg eikxdk +1 ck⎪integraldisplayt 0sinck(t−τ)Q(k,τ)dτ =1 2√ 2π∞⎪integraldisplay −∞F(k)(eickt+e−ickt)eikxdk +1 2√ 2π∞⎪integraldisplay −∞G(k) ick(eickt−e−ickt)eikxdk +1 √ 2π1 2c⎪integraldisplayt 0dτ⎪integraldisplay∞ −∞Q(k,τ) ik⎪bracketleftBig eick(t−τ)+e−ick(t−τ)⎪bracketrightBig eikxdk =1 2[f(x−ct)+f(x+ct)] +1 √ 2π1 2c∞⎪integraldisplay −∞G(k)dkx+ct⎪integraldisplay x−cteikξdξ +1 2c⎪integraldisplayt 0dτ⎪integraldisplay∞ −∞1 √ 2πQ(k,τ)dk⎪integraldisplayx+c(t−τ) x−c(t−τ)eikξdξ =1 2[f(x−ct)+f(x+ct)] +1 2cx+ct⎪integraldisplay x−ctg(ξ)dξ +1 2c⎪integraldisplayt 0dτ⎪integraldisplayx+c(t−τ) x−c(t−τ)q(ξ,τ)dξ. (4.8.3) This is identical with the d’Alembert solution (2.12.41) when q(x, t)≡0. Example 4.8.2 (Dispersive Long Water Waves in a Rotating Ocean ). We use the joint Laplace and Fourier transform to solve the linearized horizontal equations of motion and the continuity equation in a rotating inviscid ocean. These equations in a rotating coordinate system (see Proudman, 1953; Debnath and Kulchar, 1972) are given by ∂u ∂t+fˆk×u=−1 ρ∇p+1 ρhτττ, (4.8.4) ∇·u=−1 h∂ζ ∂t, (4.8.5) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 239 where u=(u,υ) is the horizontal velocity ﬁeld, ˆkis the unit vector normal to the horizontal plane, f=2Ωs i n φis the constant Coriolis parameter, ρ is the constant density of water, ζ(x, t) is the vertical free surface elevation, τ=(τx,τy) represents the components of wind stress in the xandydirections, and the pressure is given by the hydrostatic equation p=p0+gρ(ζ−z), (4.8.6) where zis the depth of water below the mean free surface and gis the accel- eration due to gravity. Equation (4.8.4)–(4.8.5) combined with (4.8.6) reduce to the form ∂u ∂t−fυ=−g∂ζ ∂x+τx ρh, (4.8.7) ∂υ ∂t+fu=−g∂ζ ∂y+τy ρh, (4.8.8) ∂u ∂x+∂υ ∂y=−1 h∂ζ ∂t. (4.8.9) It follows from (4.8.7)–(4.8.8) that Du=−g⎪parenleftbigg∂2 ∂x∂t+f∂ ∂y⎪parenrightbigg ζ+1 ρh⎪parenleftbigg∂τx ∂t+fτy⎪parenrightbigg , (4.8.10) Dυ=−g⎪parenleftbigg∂2 ∂y∂t−f∂ ∂x⎪parenrightbigg ζ+1 ρh⎪parenleftbigg∂τy ∂t−fτx⎪parenrightbigg , (4.8.11) where the diﬀerential operator Dis D≡⎪parenleftbigg∂2 ∂t2+f2⎪parenrightbigg . (4.8.12) Elimination of uandυfrom (4.8.9)–(4.8.11) gives ⎪parenleftbigg ∇2−1 c2D⎪parenrightbigg ζt=E(x, y, t), (4.8.13) where c2=ghand∇2is the horizontal Laplacian, and E(x, y, t)i sak n o w n forcing function given by E(x, y, t)=1 ρc2⎪bracketleftbigg∂2τx ∂x∂t+∂2τy ∂y∂t+f⎪parenleftbigg∂τy ∂x−∂τx ∂y⎪parenrightbigg⎪bracketrightbigg . (4.8.14) Further, we assume that the conditions are uniform in the ydirection and the wind stress acts only in the xdirection so that τxandEare given functions ofxandtonly. Consequently, equation (4.8.13) becomes ⎪bracketleftbigg∂2 ∂x2−1 c2⎪parenleftbigg∂2 ∂t2+f2⎪parenrightbigg⎪bracketrightbigg ζt=1 ρc2⎪parenleftbigg∂2τx ∂x∂t⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC 240 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Integrating this equation with respect to tgives ⎪bracketleftbigg∂2 ∂x2−1 c2⎪parenleftbigg∂2 ∂t2+f2⎪parenrightbigg⎪bracketrightbigg ζ=1 ρc2⎪parenleftbigg∂τx ∂x⎪parenrightbigg . (4.8.15) Similarly, the velocity u(x, t) satisﬁes the equation ⎪bracketleftbigg∂2 ∂x2−1 c2⎪parenleftbigg∂2 ∂t2+f2⎪parenrightbigg⎪bracketrightbigg u=−1 ρhc2⎪parenleftbigg∂τx ∂t⎪parenrightbigg . (4.8.16) If the right-hand side of equations (4.8.15) and (4.8.16) is zero, these equa- tions are known as the Klein-Gordon equations , which have received extensive attention in quantum mechanics and in applied mathematics. Equation (4.8.15) is to be solved subject to the following boundary and initial conditions |ζ|is bounded as |x|→∞ , (4.8.17) ζ(x, t)=0 a t t=0 f o r a l l r e a l x. (4.8.18) Before we solve the initial value problem, we seek a plane wave solution of the homogeneous equation (4.8.15) in the form ζ(x, t)=Aexp{i(ωt−kx)}, (4.8.19) where Ais a constant amplitude, ωis the frequency, and kis the wavenumber. Such a solution exists provided the dispersion relation ω2=c2k2+f2(4.8.20) is satisﬁed. Thus, the phase and the group velocities of waves are given by Cp=ω k=⎪parenleftbigg c2+f2 k2⎪parenrightbigg1 2 ,C g=∂ω ∂k=c2k (c2k2+f2)1 2. (4.8.21ab) Thus, the waves are dispersive in a rotating ocean ( f/negationslash=0 ) .H o w e v e r ,i nan o n - rotating ocean ( f= 0) all waves would propagate with constant velocity c, and they are non-dispersive shallow water waves. Further, CpCg=c2whence it follows that the phase velocity has a minimum of cand the group velocity a maximum. The short waves will be observed ﬁrst at a given point, even though they have the smallest phase velocity. Application of the joint Laplace and Fourier transform to (4.8.15) together with (4.8.17)–(4.8.18) give the transformed solution ˜ ζ(k,s)=−Ac2 (s2+a2)˜ f(k,s),a2=(c2k2+f2), (4.8.22) where f(x, t)=1 ρc2⎪parenleftbigg∂τx ∂x⎪parenrightbigg H(t). (4.8.23) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 241 The inverse transforms combined with the Convolution Theorem of the Laplace transform lead to the formal solution ζ(x, t)=−Ac √ 2π∞⎪integraldisplay −∞⎪parenleftbigg k2+f2 c2⎪parenrightbigg−1 2 eikxdkt⎪integraldisplay 0˜f(k,t−τ)sinaτ dτ. (4.8.24) In general, this integral cannot be evaluated unless f(x, t) is prescribed. Even if some particular form of fis given, an exact evaluation of (4.8.24) is almost a formidable task. Hence, it is necessary to resort to asymptotic methods (seeDebnath and Kulchar, 1972). To investigate the solution, we choose a particular form of the wind stress distribution τ x ρc2=AeiωtH(t)H(−x), (4.8.25) where Ais a constant and ωis the frequency of the applied disturbance. Thus, 1 ρc2⎪parenleftbigg∂τx ∂x⎪parenrightbigg =−AeiωtH(t)δ(−x). (4.8.26) In this case, solution (4.8.24) reduces to the form ζ(x, t)=Ac √ 2πt⎪integraldisplay 0eiω(t−τ)H(t−τ)F−1⎡ ⎣sinaτ ⎪radicalBig k2+f2 c2⎤⎦dτ =Ac 2t⎪integraldisplay 0eiω(t−τ)H(t−τ)J0⎪braceleftbiggf c(c2τ2−x2)1 2⎪bracerightbigg ×H(cτ−|x|)dτ, (4.8.27) where J0(z) is the zero-order Bessel function of the ﬁrst kind. When ω≡0, this solution is identical with that of Crease (1956) who ob- tained the solution using the Green’s function method. In this case, the solu- tion becomes ζ=Ac 2t⎪integraldisplay 0H(t−τ)J0⎪bracketleftBigg f⎪braceleftbigg τ2−x2 c2⎪bracerightbigg1 2⎪bracketrightBigg H⎪parenleftbigg τ−|x| c⎪parenrightbigg dτ. (4.8.28) In terms of non-dimensional parameters fτ=α, ft=a,a n dfx c=b,s o l u t i o n (4.8.28) assumes the form ⎪parenleftbigg2f Ac⎪parenrightbigg ζ=a⎪integraldisplay 0H(a−α)J0⎪bracketleftBig (α2−b2)1 2⎪bracketrightBig H(α−|b|)dα. (4.8.29) © 2007 by Taylor & Francis Group, LLC 242 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Or, equivalently, ⎪parenleftbigg2f Ac⎪parenrightbigg ζ=d⎪integraldisplay |b|J0⎪bracketleftBig (α2−b2)1 2⎪bracketrightBig dα, (4.8.30) where d=m a x ( |b|,a). This is the basic solution of the problem. In order to ﬁnd the solution of (4.8.16), we ﬁrst choose 1 ρc2⎪parenleftbigg∂τx ∂t⎪parenrightbigg =Aδ(t)H(−x), (4.8.31) so that the joint Laplace and Fourier transform of this result is AF{H(−x)}. Thus, the transformed solution of (4.8.16) is ¯˜u(k,s)=Ac2 hF{H(−x)}1 (s2+ω2),ω2=(ck)2+f2. (4.8.32) The inverse transforms combined with the Convolution Theorem lead to the solution u(x, t)=Ac 2h∞⎪integraldisplay −∞H(−ξ)J0⎡ ⎣f⎪braceleftBigg t2−⎪parenleftbiggx−ξ c⎪parenrightbigg2⎪bracerightBigg1 2⎤ ⎦ ×H⎪parenleftbigg t−(x−ξ) c⎪parenrightbigg dξ, (4.8.33) which is, by the change of variable ( x−ξ)f=cα,w i t h a=ftandb=(fx/c), =Ac2 2hf∞⎪integraldisplay bJ0⎪bracketleftBig (a2−α2)1 2⎪bracketrightBig H(a−|α|)dα. (4.8.34) For the case b>0, solution (4.8.34) becomes u(x, t)=Ac2 2hfH(a−b)a⎪integraldisplay bJ0⎪braceleftBig (a2−α2)1 2⎪bracerightBig dα. (4.8.35) When b<0, the velocity ﬁeld is u(x, t)=Ac2 2hf⎡ ⎣a⎪integraldisplay −aJ0⎪braceleftBig (a2−α2)1 2⎪bracerightBig dα−H(a−|b|)b⎪integraldisplay −aJ0⎪braceleftBig (a2−α2)1 2⎪bracerightBig dα⎤ ⎦ =gA 2f⎡ ⎢⎣2s i na−H(a−|b|)a⎪integraldisplay |b|J0⎪braceleftBig (a2−α2)1 2⎪bracerightBig dα⎤ ⎥⎦, (4.8.36) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 243 which is, for a<|b|, u(x, t)=⎪parenleftbigggA 2f⎪parenrightbigg sina. (4.8.37) Finally, it can be shown that the velocity transverse to the direction of propagation is υ=⎪parenleftbigg −gA 2f⎪parenrightbigga⎪integraldisplay 0dβ∞⎪integraldisplay bJ0⎪braceleftBig (β2−α2)1 2⎪bracerightBig H(β−|α|)dα. (4.8.38) Ifb>0, that is, xis outside the generating region, then ⎪parenleftbigg2f gA⎪parenrightbigg υ=−H(a−b)a⎪integraldisplay bdββ⎪integraldisplay bJ0⎪braceleftBig (β2−α2)1 2⎪bracerightBig dα, which becomes, after some simpliﬁcation, =−⎡ ⎣(1−cosa)−b⎪integraldisplay 0dαa⎪integraldisplay αJ0⎪braceleftBig (β2−α2)1 2⎪bracerightBig⎤ ⎦H(a−b). (4.8.39) Forb<0, it is necessary to consider two cases: (i) a<|b|and (ii) a>|b|.I n the former case, (4.8.38) takes the form ⎪parenleftbigg2f gA⎪parenrightbigg υ=−a⎪integraldisplay 0dββ⎪integraldisplay −βJ0⎪braceleftBig (β2−α2)1 2⎪bracerightBig dα=−2(1−cosb). (4.8.40) In the latter case, the ﬁnal form of the solution is ⎪parenleftbigg2f gA⎪parenrightbigg υ=−(1−cosb)+|b|⎪integraldisplay 0dαa⎪integraldisplay αJ0⎪braceleftBig (β2−α2)1 2⎪bracerightBig dβ. (4.8.41) Finally, the steady-state solutions are obtained in the limit as t→∞(b→∞) ζ=Ac 2fexp(−|b|), u=Ag 2fsinft, υ=Ag 2f⎪bracketleftBigg cosft−exp(−b),b > 0 cosft+e x p ( −|b|)−2,b < 0⎪bracketrightBigg . (4.8.42) Thus, the steady-state solutions are attained in a rotating ocean. This shows a striking contrast with the corresponding solutions in the non-rotating ocean © 2007 by Taylor & Francis Group, LLC 244 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where an ever-increasing free surface elevation is found. The terms sin ftand cosftinvolved in the steady-state velocity ﬁeld represent inertial oscillations with frequency f. Example 4.8.3 (One-Dimensional Diﬀusion Equation on a Half Line ). Solve the equation ut=κuxx,0<x< ∞,t > 0, (4.8.43) with the boundary data u(x, t)=f(t)f o r x=0 u(x, t)→0a s x→∞⎪bracerightBigg t>0 (4.8.44ab) and the initial condition u(x, t)=0 a t t=0 f o r 0 <x< ∞. (4.8.45) We use the joint Fourier sine and Laplace transform deﬁned by Us(k,s)=⎪radicalbigg 2 π∞⎪integraldisplay 0e−stdt∞⎪integraldisplay 0u(x, t)sinkxdx, (4.8.46) so that the solution of the transformed problem is Us(k,s)=⎪radicalbigg 2 π(κk)¯f(s) (s+k2κ). (4.8.47) The inverse transform yields the solution u(x, t)=⎪parenleftbigg2κ π⎪parenrightbigg∞⎪integraldisplay 0ksinkxdkt⎪integraldisplay 0f(t−τ)exp(−κτk2)dτ. In particular, if f(t)=T0= constant, then the solution becomes u(x, t)=2T0 π∞⎪integraldisplay 0sinkx k(1−e−κk2t)dk. (4.8.48) Making use of the integral (2.15.11) gives the solution u(x, t)=2T0 π⎪bracketleftbiggπ 2−π 2erf⎪parenleftbiggx 2√ κt⎪parenrightbigg⎪bracketrightbigg =T0erfc⎪parenleftbiggx 2√ κt⎪parenrightbigg . (4.8.49) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 245 This is identical with (2.15.12). Example 4.8.4 (The Bernoulli-Euler Equation on an Elastic Foundation ). Solve the equation EI∂4u ∂x4+κu+m∂2u ∂t2=Wδ(t)δ(x),−∞<x< ∞,t > 0,(4.8.50) with the initial data u(x,0) = 0 and ut(x,0) = 0 . (4.8.51) We use the joint Laplace and Fourier transform (4.8.1) to ﬁnd the solution of the transformed problem in the form U(k,s)=W m√ 2π1 (s2+a2k4+ω2), (4.8.52) where a2=EI mand ω2=κ m. The inverse Laplace transform gives U(k,t)=W m√ 2π⎪parenleftbiggsinαt α⎪parenrightbigg ,α=(a2k4+ω2)1 2. (4.8.53ab) Then the inverse Fourier transform yields the formal solution u(x, t)=W 2πm∞⎪integraldisplay −∞eikx⎪parenleftbiggsinαt α⎪parenrightbigg dk. (4.8.54) Example 4.8.5 (The Cauchy-Poisson Wave Problem in Fluid Dynamics ). We consider the two-dimensional Cauchy-Poisson problem for an inviscid liquid of inﬁnite depth with a horizontal free surface. We assume that the liquid has con- stant density ρand negligible surface tension . Waves are generated on the surface of water initially at rest for time t<0 by the prescribed free surface displacement at t=0 . In terms of the velocity potential φ(x, z, t) and the free surface elevation η(x, t), the linearized surface wave motion in Cartesian coordinates ( x, y, z )i s governed by the following equation and free surface and boundary conditions: ∇2φ=φxx+φzz=0,−∞<z≤0,−∞<x< ∞,t>0, (4.8.55) © 2007 by Taylor & Francis Group, LLC 246 INTEGRAL TRANSFORMS and THEIR APPLICATIONS φz−ηt=0 φt+gη=0⎪bracerightbigg onz=0,t>0, (4.8.56ab) φz→0a s z→− ∞ . (4.8.57) The initial conditions are φ(x,0,0) = 0 and η(x,0) =η0(x), (4.8.58) where η0(x) is a given function with compact support. We introduce the Laplace transform with respect to tand the Fourier trans- form with respect to xdeﬁned by [˜ φ(k,z,s),˜ η(k,s)] =1 √ 2π∞⎪integraldisplay −∞e−ikxdx∞⎪integraldisplay 0e−st[φ, η]dt. (4.8.59) The use of joint transform to the above system gives ˜ φzz−k2˜ φ=0,−∞<z≤0, (4.8.60) ˜ φz=s˜ η−˜η0(k) s˜ φ+g˜ η=0⎫ ⎬ ⎭onz=0, (4.8.61ab) ˜ φz→0a s z→− ∞ . (4.8.62) The bounded solution of (4.8.60) is ˜ φ(k,s)=¯Aexp(|k|z) (4.8.63) where A= A(s) is an arbitrary function of s,a n d˜ η0(k)=F{η0(x)}. Substituting (4.8.63) into (4.8.61ab) and eliminating ˜ ηfrom the resulting equations gives ¯A. Hence, the solutions for˜ φand˜ ηare [˜ φ,˜ η]=⎪bracketleftbigg −g˜η0exp(|k|z) s2+ω2,s˜η0 s2+ω2⎪bracketrightbigg , (4.8.64ab) where the dispersion relation for deep water waves is ω2=g|k|. (4.8.65) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 247 The inverse Laplace and Fourier transforms give the solutions φ(x, z, t)=−g √ 2π∞⎪integraldisplay −∞sinωt ωexp(ikx+|k|z)˜η0(k)dk (4.8.66) η(x, t)=1 √ 2π∞⎪integraldisplay −∞˜η0(k)cosωteikxdk =1 √ 2π∞⎪integraldisplay 0˜η0(k)[ei(kx−ωt)+ei(kx+ωt)]dk, (4.8.67) in which ˜ η0(−k)=˜η0(k) is assumed. Physically, the ﬁrst and second integrals of (4.8.67) represent waves trav- eling in the positive and negative directions of xrespectively with phase ve- locity⎪parenleftBigω k⎪parenrightBig . These integrals describe superposition of all such waves over the wavenumber spectrum 0 <k< ∞. For the classical Cauchy-Poisson wave problem, η(x)=aδ(x)w h e r e δ(x) is the Dirac delta function so that ˜ η0(k)=⎪parenleftbig a/√ 2π⎪parenrightbig . Thus, solution (4.8.67) becomes η(x, t)=a 2π∞⎪integraldisplay 0⎪bracketleftBig ei(kx−ωt)+ei(kx+ωt)⎪bracketrightBig dk. (4.8.68) The wave integrals (4.8.66) and (4.8.67) represent the exact solution for the velocity potential φand the free surface elevation ηfor all xandt>0. However, they do not lend any physical interpretations. In general, the exact evaluation of these integrals is almost a formidable task. So it is necessary to resort to asymptotic methods. It would be suﬃcient for the determination of the principal features of the wave motions to investigate (4.8.67) or (4.8.68) asymptotically for large time tand large distance xwith ( x/t)h e l dﬁ x e d .T h e asymptotic solution for this kind of problem is available in many standardbooks (for example, see Debnath, 1994, p 85). We state the stationary phase approximation of a typical wave integral, for t→∞, η(x, t)=b⎪integraldisplay af(k)exp [itW(k)]dk (4.8.69) ∼f(k1)⎪bracketleftbigg2π t|W/prime/prime(k1)|⎪bracketrightbigg1 2 exp⎪bracketleftBig i⎪braceleftBig tW(k1)+π 4sgn W/prime/prime(k1)⎪bracerightBig⎪bracketrightBig ,(4.8.70) where W(k)=kx t−ω(k),x>0a n d k=k1is a stationary point that sat- isﬁes the equation W/prime(k1)=x t−ω/prime(k1)=0,a < k 1<b . (4.8.71) © 2007 by Taylor & Francis Group, LLC 248 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Application of (4.8.70) to (4.8.67) shows that only the ﬁrst integral in (4.8.67) has a stationary point for x>0. Hence, the stationary phase ap- proximation gives the asymptotic solution, as t→∞,x>0, η(x, t)∼⎪bracketleftbigg1 t|ω/prime/prime(k1)|⎪bracketrightbigg1 2 ˜η0(k1)exp [i{(k1x−tω(k1)} +iπ 4sgn{−ω/prime/prime(k1)}],(4.8.72) where k1=(gt2/4x2) is the root of the equation ω/prime(k)=x t. On the other hand, when x<0, only the second integral of (4.8.67) has a stationary point k1=(gt2/4x2), and hence, the same result (4.8.70) can be used to obtain the asymptotic solution for t→∞ andx<0a s η(x, t)∼⎪bracketleftbigg1 t|ω/prime/prime(k1)|⎪bracketrightbigg1 2 ˜η0(k1)exp [i{tω(k1)−k1|x|} +iπ 4sgn ω/prime/prime(k1)]. (4.8.73) In particular, for the classical Cauchy-Poisson solution (4.8.68), the asymp- totic representation for η(x, t) follows from (4.8.73) in the form η(x, t)∼at 2√ 2π√ g x3/2cos⎪parenleftbigggt2 4x⎪parenrightbigg ,g t2>>4x (4.8.74) and a similar result for x<0a n d t→∞. 4.9 Summation of Inﬁnite Series With the aid of Laplace transforms, Wheelon (1954) ﬁrst developed a direct method to the problem of summing inﬁnite series in closed form. His method is essentially based on the operation that is contained in the summation ofboth sides of a Laplace transform with respect to the transform variable s, which is treated as the dummy index of summation n.T h i si sf o l l o w e db ya n interchange of summation and integration that leads to the desired sum as theintegral of a geometric or exponential series, which can be summed in closed form. We next discuss this procedure in some detail. If¯f(s)=L{f(x)},t h e n ∞⎪summationdisplay n=1an¯f(n)=∞⎪summationdisplay n=1an∞⎪integraldisplay 0f(x)e−nxdx. (4.9.1) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 249 In many cases, it is possible to interchange the order of summation and integration so that (4.9.1) gives ∞⎪summationdisplay n=1an¯f(n)=∞⎪integraldisplay 0f(t)b(t)dt, (4.9.2) where b(t)=∞⎪summationdisplay n=1anexp(−nt). (4.9.3) We now assume f(t)=1 Γ(p)tp−1exp(−xt)s ot h a t ¯f(n)=(n+x)−p.C o n s e - quently, (4.9.2) becomes ∞⎪summationdisplay n=1an¯f(n)=∞⎪summationdisplay n=1an (n+x)p=1 Γ(p)∞⎪integraldisplay 0b(t)tp−1exp(−xt)dt. (4.9.4) This shows that a general series has been expressed in terms of an integral. We next illustrate the method by simple examples. Example 4.9.1 Show that the sum of the series ∞⎪summationdisplay n=11 n2=π2 6. (4.9.5) Putting x=0,p=2 ,a n d an=1 f o ra l l n, we ﬁnd, from (4.9.3) and (4.9.4), b(t)=∞⎪summationdisplay n=1exp(−nt)=1 et−1, (4.9.6) and ∞⎪summationdisplay n=11 n2=∞⎪integraldisplay 0td t et−1=ζ(2) =π2 6, (4.9.7) in which the following standard result is used ∞⎪integraldisplay 0tp−1 eat−1dt=Γ(p) apζ(p), (4.9.8) where ζ(p)i st h e Riemann zeta function deﬁned below by (4.9.10). Similarly, we can show ∞⎪summationdisplay n=11 n3=1 Γ(3)∞⎪integraldisplay 0t2dt et−1=ζ(3). (4.9.9) © 2007 by Taylor & Francis Group, LLC 250 INTEGRAL TRANSFORMS and THEIR APPLICATIONS More generally, we obtain, from (4.9.8), ∞⎪summationdisplay n=11 np=1 Γ(p)∞⎪integraldisplay 0tp−1dt et−1=ζ(p). (4.9.10) Example 4.9.2 Show that∞⎪summationdisplay n=11 nexp(−an)=−log(1−e−a). (4.9.11) We put x=0,p=1 ,a n d an=e x p ( −an)s ot h a t b(t)=∞⎪summationdisplay n=1exp[−n(t+a)] =1 ea+t−1. (4.9.12) Then result (4.9.4) gives ∞⎪summationdisplay n=11 nexp(−an)=∞⎪integraldisplay 0dt ea+t−1,exp(−t)=x, =1⎪integraldisplay 0dx ea−x=−log(1−e−a). Example 4.9.3 Show that∞⎪summationdisplay n=11 (n2+x2)=1 2x2(πxcothπx−1). (4.9.13) We set f(t)=1 xsinxt,¯f(n)=1 n2+x2,andan=1 f o ra l l n. Clearly b(t)=∞⎪summationdisplay n=1exp(−nt)=1 et−1. Thus, ∞⎪summationdisplay n=11 (n2+x2)=1 x∞⎪integraldisplay 0sinxt et−1dt=1 2x2(πxcothπx−1). © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 251 4.10 Transfer Function and Impulse Response Function of a Linear System Many science and engineering systems are described by initial value problems that are governed by linear ordinary diﬀerential equations. In general, a linear system is governed by an nth order linear ordinary d iﬀerential equation with constant coeﬃcients in the form L(D)[x(t)]≡anx(n)(t)+an−1x(n−1)(t)+...+a0x(t)=f(t),(4.10.1) where an,an−1,...,a0are real constants with an/negationslash= 0 and the initial conditions are x(0) = x0,x/prime(0) =x1, ..., x(n−1)(0)=xn−1. (4.10.2) The solution, x(t) of the system (4.10.1)–(4.10.2) is called the output or the response function ,a n dt h eg i v e n f(t) is called the input function (ordriving function )o ft i m e t. Thetransfer function h(s) of a linear system is deﬁned as the ratio of the Laplace transform of the output function x(t) to the Laplace transform of the input function f(t), under the assumption that all initial conditions are zero. More generally, however, the Laplace transform of the system (4.10.1)– (4.10.2) gives an⎪bracketleftBig sn x(s)−sn−1x(0)−...−x(n−1)(0)⎪bracketrightBig +an−1⎪bracketleftBig sn−1 x(s)−sn−2x(0)−...−x(n−2)⎪bracketrightBig +...+a1[s x(s)−x(0)] + a0 x(s)= f(s). (4.10.3) Or, equivalently, ⎪parenleftbig ansn+an−1sn−1+...+a0⎪parenrightbig x(s)= f(s)+ g(s), or, pn(s) x(s)= f(s)+ g(s), (4.10.4) where pn(s)=⎪parenleftbig ansn+an−1sn−1+...+a1s+a0⎪parenrightbig (4.10.5) is a polynomial of degree n, g(s) is a polynomial of degree less than or equal to (n−1) consisting of the various products of the coeﬃcients ar(r=1,2,...,n ) and the given initial conditions x0,x1,...,xn−1. © 2007 by Taylor & Francis Group, LLC 252 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thetransfer function (orsystem function ) is denoted by h(s) and deﬁned by h(s)=1 pn(s)=1 ansn+an−1sn−1+...+a0. (4.10.6) Consequently, equation (4.10.4) becomes x(s)= f(s) pn(s)+ g(s) pn(s)= h(s)⎪bracketleftbig f(s)+ g(s)⎪bracketrightbig . (4.10.7) The inverse Laplace transform of (4.10.7) provides the response function x(t) of the system which is the superposition of two responses as follows: x(t)=L−1⎪braceleftbig h(s) g(s)⎪bracerightbig +L−1⎪braceleftbig h(s) f(s)⎪bracerightbig (4.10.8) =⎪integraldisplayt 0h(t−τ)g(τ)dτ+⎪integraldisplayt 0h(t−τ)f(τ)dτ (4.10.9) =x0(t)+x1(t), (4.10.10) where x0(t)=L−1⎪braceleftbig h(s) g(s)⎪bracerightbig ,x1(t)=L−1⎪braceleftbig h(s) f(s)⎪bracerightbig , and h(t)=L−1⎪braceleftbig h(s)⎪bracerightbig =L−1⎪braceleftbigg1 pn(s)⎪bracerightbigg , (4.10.11) are often called the i m p u l s er e s p o n s ef u n c t i o n of the linear system. If the input is f(t)≡0, the solution of the problem is x0(t), which is called thezero-input response of the system. On the other hand, x1(t) is the output due to the input f(t) and is called the zero-state response of the system. If all initial conditions are zero, that is, x0=x1=...=xn−1=0 , t h e n g(s)=0 and so, the unique solution of the nonhomogeneous equation (4.10.1) is x1(t). For example, h(t)=L−1⎪braceleftbig h(s)⎪bracerightbig describes the solution for a mass-spring system when it is struck by a hammer. Fo r an electric circuit, the function z(s)=⎪bracketleftbig s h(s)⎪bracketrightbig−1is called the impendence of the circuit. The polynomial pn(s)=⎪parenleftbig ansn+an−1sn−1+...+a0⎪parenrightbig insof degree nis called the characteristic polynomial of the system, and pn(s) = 0 is called the characteristic equation of the system. Since the coeﬃcients of pn(s) are real, it follows that roots of the characteristic equation are all real or, if complex, they must occur in complex conjugate pairs. If h(s) is expressed in partial fractions, the system is said to be stable provided all roots of the characteristic equation have negative real parts. Fro m a physical point of view, when every root of pn(s) = 0 has a negative real part, any bounded input to a system that is stable will lead to an output that is also bounded for all time t. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 253 We close this section by adding the following examples: Example 4.10.1 Find the transfer function for each of the following linear systems. Determine the order of each system and ﬁnd which is stable. (a) LdI dt+RI+1 C⎪integraldisplayt 0I(τ)dτ=E(t), (4.10.12) (b) x/prime/prime(t)+2x/prime(t)+5x(t)=3f/prime(t)+2f(t), (4.10.13) (c) x/prime/prime/prime(t)+x/prime/prime(t)+3x/prime(t)−5x(t)=6f/prime/prime(t)−13f/prime(t)+6f(t), where L, R, andCare constants . (4.10.14) (a) This current equation is solved in Example 4.2.13. The Laplace trans- formed equation with the zero initial condition is given by ⎪parenleftbigg Ls+R+1 Cs⎪parenrightbigg I(s)= E(s) so that the transfer equation is h(s)=1 ⎪parenleftbig Ls+R+1 Cs⎪parenrightbig=1 Ls ⎪parenleftbig s2+R Ls+1 CL⎪parenrightbig. T h es y s t e mi so fo r d e r2a n di t sc haracteristic equation is s2+R Ls+1 CL=0. Or, (s+k)2+n2=0 where k=R 2L,n2=1 CL−R2 4L2. The roots of the characteristic equation are complex and they are s=−k+ in with the negative real part. So, the system is stable. (b) We take the Laplace transform of the equation (4.10.13) with zero initial conditions so that ⎪parenleftbig s2+2s+5⎪parenrightbig x(s)=( 3 s+2 ) f(s). Thus, the transfer function is h(s)= x(s) f(s)=(3s+2 ) s2+2s+5. © 2007 by Taylor & Francis Group, LLC 254 INTEGRAL TRANSFORMS and THEIR APPLICATIONS T h es y s t e mi so fo r d e r2a n di t sc haracteristic equation is s2+2s+5=0 with complex roots s=−1+ 2i. Since the real part of these roots is negative, the system is stable. (c) Similarly, h(s)= x(s) f(s)=⎪parenleftbig 6s2−13s+6⎪parenrightbig (s3+s2+3s−5). T h es y s t e mi so fo r d e r3a n di t sc haracteristic equation is s3+s2+3s−5=0 with roots s1=1 ,s2,s3=−1+ 2i. Since the real parts of all roots are not negative, the system is unstable . Example 4.10.2 Find the transfer function, the impulse response function, and the solution of a linear system described by x/prime/prime(t)+2ax/prime(t)+⎪parenleftbig a2+4⎪parenrightbig x(t)=f(t) (4.10.15) x(0) = 1 ,x/prime(0) = −a. (4.10.16ab) According to formula (4.10.4), the transfer function of this system is h(s)=1 (s2+2as+a2+4 )=1 (s+a)2+22. The inverse Laplace transform of the transform function h(s) is the impulse response function h(t)=L−1⎪braceleftbig h(s)⎪bracerightbig =1 2L−1⎪braceleftBigg 2 (s2+a)2+22⎪bracerightBigg =1 2e−atsin2t.(4.10.17) Solving the homogeneous initial value problem gives x0(t)=e−atcos(2t). (4.10.18) The solution of the problem (4.10.15)–(4.10.16ab) is x(t)=x0(t)+h(t)∗f(t) =e−atcos(2t)+⎪integraldisplayt 0e−atf(t−τ)s i n2τd τ . (4.10.19) © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 255 Example 4.10.3 Consider a linear system governed by the diﬀerential equation a2x/prime/prime(t)+a1x/prime(t)+a0x(t)=H(t), (4.10.20) where H(t) is the Heaviside unit step function. Derive Duhamel’s formulas (a)x(t)=⎪integraldisplayt 0A/prime(t−τ)f(τ)dτ, (4.10.21) (b) x(t)=⎪integraldisplayt 0A(τ)f/prime(t−τ)dτ+A(t)f(0). (4.10.22) The transfer function for this system (4.10.20) is h(s)= x(s) f(s)=s x(s). (4.10.23) Or, x(s)= h(s) s. (4.10.24) The output function in this special case is called the indicial admittance and is denoted by A(t)s ot h a t A(s)= h(s) s. (4.10.25) We next derive Duhamel’s formulas. We have, from (4.10.7) with g(s)=0 , x(s)=s⎪bracketleftbigg h(s) s⎪bracketrightbigg f(s)=s A(s) f(s). (4.10.26) Using the convolution theorem gives x(t)=L−1⎪braceleftbig s A(s)· f(s)⎪bracerightbig =⎪integraldisplayt 0A/prime(t−τ)f(τ)dτ=d dt⎪integraldisplayt 0A(τ)f(t−τ)dτ which is, by Leibniz’s rule, =⎪integraldisplayt 0A(τ)f/prime(t−τ)dτ+A(t)f(0), where the initial conditions A(0)=A/prime(0) = 0 are used. © 2007 by Taylor & Francis Group, LLC 256 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 4.11 Exercises 1. Using the Laplace transform, solve the following initial value problems (a)dx dt+ax=e−bt,t>0,a/negationslash=bwithx(0) = 0 . (b)dx dt−x=t2,t>0,x(0) = 0 . (c)dx dt+2x=c o st, t >0,x(0) = 1 . (d)dx dt−2x=4,t>0,x(0) = 0 . 2. Solve the initial value problem for the radioactive decay of an element dx dt=−kx,(k>0),t >0,x(0) =x0. Prove that the half-life time Tof the element, which is deﬁned as the time taken for half a given amount of the element to decay, is T=1 klog 2. 3. Find the solutions of the following systems of equations with the initial data: (a)dx dt=x−2y,dy dt=y−2x, x (0) = 1 ,y(0) = 0 . (b)dx1 dt=x1+2x2+t,dx2 dt=x2+2x1+t;x1(0) = 2 ,x2(0) = 4 . (c)dx dt=6x−7y+4z,dy dt=3x−4y+2z,dz dt=−5x+5y−3z, withx(0) = 5 ,y(0) =z(0) = 0 . (d)dx dt=2x−3y,dy dt=y−2x;x(0) = 2 ,y(0) = 1 . (e)dx dt+x=y,dy dt−y=x, x (0) =y(0) = 1 . (f)dx dt+dy dt+x=0,dx dt+2dy dt−x=e−at,x(0) =y(0) = 1 . 4. Solve the matrix diﬀerential system dx dt=Axwithx(0) =⎪parenleftbiggx0 0⎪parenrightbigg , where x(t)=⎪parenleftbigg x1(t) x2(t)⎪parenrightbigg andA=⎪parenleftbigg −3−2 32⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 257 5. Find the solution of the autonomous system described by dx dt=x,dy dt=x+2ywith x(0) =x0,y(0) =y0. 6. Solve the diﬀerential systems (a) d2x dt2−2kdy dt+lx=0 d2y dt2+2kdx dt+ly=0⎫ ⎪⎪⎬ ⎪⎪⎭t>0 with the initial conditions x(0) =a,˙x(0) = 0; y(0) = 0 ,˙y(0) =υ, where k,l,a, andυare constants. (b) d2x dt2=y−2x d2y dt2=x−2y⎫ ⎪⎪⎬ ⎪⎪⎭t>0 with the initial conditions x(0) =y(0) = 1 ,and ˙ x(0) = ˙y(0) = 0 . 7. The glucose concentration in the blood during continuous intravenous injection of glucose is C(t), which is in excess of the initial value at the start of the infusion. The function C(t) satisﬁes the initial value problem dC dt+kC=α V,t > 0,C(0) = 0 , where kis the constant velocity of elimination, αi st h er a t eo fi n f u s i o n (in mg/min), and Vis the volume in which glucose is distributed. Solve this problem. 8. The blood is pumped into the aorta by the contraction of the heart. The pressure p(t) in the aorta satisﬁes the initial value problem dp dt+c kp=cAsinωt, t> 0;p(0) =p0 where c, k, A, andp0are constants. Solve this initial value problem. © 2007 by Taylor & Francis Group, LLC 258 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 9. The zero-order chemical reaction satisﬁes the initial value problem dc dt=−k0,t > 0,with c=c0att=0 where k0is a positive constant and c(t) is the concentration of a reacting substance at time t. Show that c(t)=c0−k0t. 10. Solve the equation governing the ﬁrst order chemical reaction dc dt=−k1cwith c(t)=c0att=0 ( k1>0). 11. Obtain the solutions of the systems of diﬀerential equations governing the consecutive chemical reactions of the ﬁrst order dc1 dt=−k1c1,dc2 dt=k1c1−k2c2,dc3 dt=k2c2,t > 0, with the initial conditions c1(0) =c1,c2(0) =c3(0) = 0 , where c1(t) is the concentration of a substance Aat time t, which breaks down to form a new substance A2with concentration c2(t), and c3(t)i s the concentration of a new element originated from A2. 12. Solve the following initial value problems (a) ¨x+ω2x=c o snt,(ω/negationslash=n)x(0) = 1 ,˙x(0) = 0 . (b) ¨x+x=s i n2 t, x(0) = ˙x(0) = 0 . (c)d3x dt3+d2x dt2=3e−4t,x(0) = 0 ,˙x(0) =−1,¨x(0) = 1 . (d)d4x dt4=1 6x, x(t)=¨x(t)=0,˙x(t)=˙ ¨x(t)=1a t t=0 . (e) (D4+2D3−D2−2D+ 10)x(t)=0,t>0, x(0) =−1,˙x(0) = 3 ,¨x(0) =−1,˙¨x(0) = 4 . (f)d2x dt2+bdx dt=δ(t−a),x(0) =α,˙x(0) =β. (g)Cd2v dt2+1 Rdv dt+v L=di dt,v(0) = ˙v(0) = 0; i(t)=H(t−1)−H(t), where R, L, andCare constants . (h)d2x dt2+2tdx dt−4x=2,x(0) = 0 = ˙ x(0). © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 259 (i)d2x dt2−2adx dt+a2x=t−(t−a)H(t−a)−aH(t−a), x(0) = 0 = ˙ x(0). 13. Solve the following systems of equations: (a) ¨x−2˙y−x=0,¨y+2˙x−y=0, x(t)=y(t)=0,˙x(t)= ˙y(t)=1a t t=0. (b) ¨x1+3˙x1−2x1+˙x2−3x2=2e−t,2˙x1−x1+˙x2−2x2=0 , withx1(0) = ˙x1(0) = 0 and x2(0) = 4 . 14. With the aid of the Laplace transform, investigate the motion of a par- ticle governed by the equations of motion ¨ x−ω˙y=0,¨y+ω˙x=ω2aand the initial conditions x(0) =y(0) = ˙x(0) = ˙y(0) = 0 . 15. Show that the solution of the equation d2y dx2+(a+b)dy dx+aby=e−ax,x>0 with the initial data y(x)=1 a2anddy dx=0 a t x=0i s y(x)=1 a2(a−b)(ae−bx−be−ax−xa2e−ax)+e−bx−e−ax (a−b)2. 1 6 .T h em o t i o no fa ne l e c t r o no fc h a r g e −ein a static electric ﬁeld E= (E,0,0) and a static magnetic ﬁeld H=( 0,0,H) is governed by the vector equation m¨r=−eE+e c(˙r×H),t > 0, with zero initial velocity and displacement ( r=˙r=0att=0 )w h e r e r= (x, y, z )a n d cis the velocity of light. Show that the displacement ﬁelds are x(t)=eE mω2(cosωt−1),y(t)=eE mω2(sinωt−ωt),z(t)=0, where ω=eH mc. Hence, calculate the velocity ﬁeld. 17. An electron of mass mand charge −eis acted on by a periodic electric ﬁeldEsinω0talong the x-axis and a constant magnetic ﬁeld Halong the z-axis. Initially, the electron is emitted at the origin with zero velocity. With the same ωas given in exercise 16, show that x(t)=eE mω(ω2−ω2 0)(ω0sinωt−ωsinω0t), y(t)=eE mω(ω2−ω2 0)ω0⎪braceleftbig (ω2−ω2 0)+⎪parenleftbig ω2 0cosωt−ω2cosω0t⎪parenrightbig⎪bracerightbig . © 2007 by Taylor & Francis Group, LLC 260 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 18. The stress-strain relation and equation of motion for a viscoelastic rod in the absence of external force are ∂e ∂t=1 E∂σ ∂t+σ η,∂σ ∂x=ρ∂2u ∂t2, where eis the strain, ηis the coeﬃcient of viscosity, and the displacement u(x, t) is related to the strain by e=∂u ∂x. Prove that the stress σ(x, t) satisﬁes the equation ∂2σ ∂x2−ρ η∂σ ∂t=1 c2∂2σ ∂t2. Show that the stress distribution in a semi-inﬁnite viscoelastic rod sub- ject to the boundary and initial conditions ˙u(0,t)=UH(t),σ(x, t)→0a s x→∞, σ(x,0) = 0 ,˙u(x,0) = 0 ,for 0 <x< ∞, is given by σ(x, t)=−Uρcexp⎪parenleftbigg −Et 2η⎪parenrightbigg I0⎪bracketleftBigg E 2η⎪parenleftbigg t2−x2 c2⎪parenrightbigg1/2⎪bracketrightBigg H⎪parenleftBig t−x c⎪parenrightBig . 19. An elastic string is stretched between x=0a n d x=/lscriptand is initially at rest in the equilibrium position. Find the Laplace transform solution forthe displacement subject to the boundary conditions y(0,t)=f(t)a n d y(l,t)=0,t >0. 20. The end x= 0 of a semi-inﬁnite submarine cable is maintained at a potential V 0H(t). If the cable has no initial current and potential, de- termine the potential V(x, t)a tap o i n t xand at time t. 21. A semi-inﬁnite lossless transmission line has no initial current or poten- tial. A time-dependent electromagnetic force, V0(t)H(t) is applied at the endx= 0. Find the potential V(x, t). Hence, determine the potential for cases (i) V0(t)=V0= constant, and (ii) V0(t)=V0cosωt. 22. Solve the Blasius problem of an unsteady boundary layer ﬂow in a semi- inﬁnite body of viscous ﬂuid enclosed by an inﬁnite horizontal disk atz= 0. The governing equation and the boundary and initial conditions are ∂u ∂t=ν∂2u ∂z2,z > 0,t > 0, u(z,t)=Ut onz=0,t > 0, u(z,t)→0a s z→∞,t > 0, u(z,t)=0 a t t≤0,z > 0. Explain the signiﬁcance of the solution. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 261 23. Obtain the solution of the Stokes-Ekman problem of an unsteady bound- ary layer ﬂow in a semi-inﬁnite body of viscous ﬂuid bounded by aninﬁnite horizontal disk at z= 0, when both the ﬂuid and the disk ro- tate with a uniform angular velocity Ω about the z-axis. The governing boundary layer equation, the boundary and the initial conditions are ∂q ∂t+2 Ωiq=ν∂2q ∂z2,z > 0, q(z,t)=aeiωt+be−iωtonz=0,t>0, q(z,t)→0a s z→∞,t >0, q(z,t)=0 a t t≤0 for all z>0, where q=u+iυ, ω is the frequency of oscillations of the disk and a, b are complex constants. Hence, deduce the steady-state solution and de- termine the structure of the associated boundary layers. 24. Show that, when ω= 0 in exercise 23, the steady ﬂow ﬁeld is given by q(z,t)∼(a+b)exp⎪braceleftBigg⎪parenleftbigg −2iΩ ν⎪parenrightbigg1/2 z⎪bracerightBigg . Hence, determine the thickness of the Ekman layer. 25. Solve the following integral and integro-diﬀerential equations: (a)f(t)=s i n2 t+t⎪integraldisplay 0f(t−τ)sinτd τ. (b)f(t)=t 2sint+t⎪integraldisplay 0f(τ)sin(t−τ)dτ. (c)t⎪integraldisplay 0f(τ)J0[a(t−τ)]dτ=s i nat. (d)f(t)=s i n t+t⎪integraldisplay 0f(τ)sin{2(t−τ)}dτ. (e)f(t)=t2+t⎪integraldisplay 0f/prime(t−τ)exp(−aτ)dτ, f (0) = 0 . (f)x(t)=1+ a2t⎪integraldisplay 0(t−τ)x(τ)dτ. © 2007 by Taylor & Francis Group, LLC 262 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (g)x(t)=t+1 at⎪integraldisplay 0(t−τ)3x(τ)dτ. 26. Prove that the solution of the integro-diﬀerential equation f(t)=2 √ π⎡ ⎣√ t+√ at⎪integraldisplay 0(t−τ)1/2f/prime(τ)dτ⎤ ⎦,f(0) = 0 is f(t)=eat √ a[1 +erf√ at]−1 √ a. 27. Solve the integro-diﬀerential equations (a)d2x dt2=e x p ( −2t)−t⎪integraltext 0exp{−2(t−τ)}⎪parenleftbigdx dτ⎪parenrightbig dτ, x (0) = 0 and ˙ x(0)= 0 . (b)dx dt=t⎪integraltext 0x(τ)cos(t−τ)dτ, x (0) = 1 . 28. Using the Laplace transform, evaluate the following integrals: (a)∞⎪integraldisplay 0sintx x(x2+a2)dx,(a, t >0), (c)∞⎪integraldisplay −∞costx x2+a2dx,(a, t >0), (e)∞⎪integraldisplay 0exp(−tx2)dx, t > 0,(b)∞⎪integraldisplay 0sintx xdx, (d)∞⎪integraldisplay −∞xsinxt x2+a2dx,(a, t >0), (f)∞⎪integraldisplay 0cos(tx2)dx. 29. Show that (a)∞⎪integraldisplay 0e−ax⎪parenleftbiggcospx−cosqx x⎪parenrightbigg dx=1 2log⎪parenleftbigga2+q2 a2+p2⎪parenrightbigg ,(a>0). (b)∞⎪integraldisplay 0e−ax⎪parenleftbiggsinqx−sinpx x⎪parenrightbigg dx=t a n−1⎪parenleftBigq a⎪parenrightBig −tan−1⎪parenleftBigp a⎪parenrightBig ,a > 0. 30. Establish the following results: (a)∞⎪integraldisplay −∞costxdx (x2+a2)(x2+b2)=π a2−b2⎪parenleftbigge−bt b−e−at a⎪parenrightbigg , a ,b,t> 0. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 263 (b)∞⎪integraldisplay 0sin(πt x) x(1 +x2)dx=π 2(1−e−πt),t > 0. (c)∞⎪integraldisplay 0cos(tu2)du=∞⎪integraldisplay 0sin(tu2)du=1 2⎪parenleftBigπ 2t⎪parenrightBig1/2 ,t > 0. 31. In Example 4.5.1(i), write the solution when the point load is applied at the mid point of the beam. 32. A uniform horizontal beam of length 2 /lscriptis clamped at the end x=0a n d freely supported at x=2/lscript. It carries a distributed load of constant value Win/lscript 2<x<3/lscript 2and zero elsewhere. Obtain the deﬂection of the beam which satisﬁes the boundary value problem EId4y dx4=W⎪bracketleftbigg H⎪parenleftbigg x−/lscript 2⎪parenrightbigg −H⎪parenleftbigg x−3/lscript 2⎪parenrightbigg⎪bracketrightbigg ,0<x< 2/lscript, y(0) = 0 = y/prime(0),y/prime/prime(2/lscript)=0= y/prime/prime/prime(2/lscript). 33. Solve exercise 32 if the beam carries a constant distributed load Wper u n i tl e n g t hi n0 <x</lscript and zero in /lscript<x< 2/lscript. Find the bending moment and shear at x=/lscript 2. 34. A horizontal cantilever beam of length 2 /lscriptis deﬂected under the combined eﬀect of its own constant weight Wand a point load of magnitude Plocated at the midpoint. Obtain the deﬂection of the beam which satisﬁes the boundary value problem EId4y dx4=W[H(x)−H(x−2/lscript)] +Pδ(x−/lscript),0<x< 2/lscript, y(0) = 0 = y/prime(0),y/prime/prime(2/lscript)=0= y/prime/prime/prime(2/lscript). Find the bending moment and shear at x=/lscript 2. 35. Using the Laplace transform, solve the following diﬀerence equations: (a) Δ un−2un=0,u 0=1, (b) Δ2un−2un+1+3un=0,u0=0 a n d u1=1, (c)un+2−4un+1+4un=0,u 0=1 a n d u1=4, (d)un+2−5un+1+6un=0,u 0=1 a n d u1=4, (e) Δ2un+3un=0,u0=0,u1=1, (f)un+2−4un+1+3un=0, (g)un+2−9un=0,u 0=1a n d u1=3, © 2007 by Taylor & Francis Group, LLC 264 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (h) Δ un−(a−1)un=0,u0= constant . 36. Show that the solution of the diﬀerence equation un+2+4un+1+un=0,with u0=0 a n d u1=1, is un=1 2√ 3⎪bracketleftBig⎪parenleftBig√ 3−2⎪parenrightBign +(−1)n+1⎪parenleftBig 2+√ 3⎪parenrightBign⎪bracketrightBig . 37. Show that the solution of the diﬀ erential-diﬀerence equation ˙u(t)−u(t−1) = 2 ,u(0) = 0 is u(t)=2⎪bracketleftbigg t−(t−1)2 2!+(t−2)3 3!+···+(t−n)n+1 (n+1 ) !⎪bracketrightbigg ,t > n . 38. Obtain the solution of the diﬀerential-diﬀerence equation ˙u=u(t−1),u(0) = 1 ,0<t<∞withu(t)=1w h e n −1≤t<0. 39. Use the Laplace transform to solve the initial-boundary value problem utt−uxx=k2uxxtt,0<x< ∞,t > 0, u(x,0) = 0 ,⎪parenleftbigg∂u ∂x⎪parenrightbigg t=0=0,forx>0, u(x, t)→0a s x→∞,t > 0, u(0,t)=1 f o r t>0. Hence, show that⎪parenleftbigg∂u ∂x⎪parenrightbigg x=0=−1 kJ0⎪parenleftbiggt k⎪parenrightbigg . 40. Solve the telegraph equation utt−c2uxx+2aut=0,−∞<x< ∞,t > 0, u(x,0) = 0 ,u t(x,0) =g(x). 41. Use the joint Laplace and Fourier transform to solve Example 2.12.3 in Chapter 2. 42. Use the Laplace transform to solve the initial-boundary value problem ut=c2uxx,0<x<a , t> 0, u(x,0)=x+s i n⎪parenleftbigg3πx a⎪parenrightbigg for 0 <x<a , u(0,t)=0= u(a,t)f o r t>0. © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 265 43. Solve the diﬀusion equation ut=kuxx,−a<x<a , t> 0, u(x,0) = 1 for −a<x<a , u(−a,t)=0= u(a,t)f o r t>0. 44. Use the joint Laplace and Fourier transform to solve the initial value problem for water waves which satisﬁes (see Debnath, 1994, p. 92) ∇2φ=φxx+φzz=0,−∞<z< 0,−∞<x< ∞,t > 0 φz=ηt φt+gη=−P ρp(x)eiωt⎫ ⎪⎬ ⎪⎭onz=0,t > 0, φ(x, z,0) = 0 = η(x,0) for all xandz, where Pandρare constants. 45. Show that (a)∞⎪summationdisplay n=0an √ n2+x2=∞⎪integraldisplay 0b(t)J0(xt)dt,where b(t)i sg i v e nb y( 4 .9.3). (b)∞⎪summationdisplay n=01 n2−a2=1 2a2(1−πacotπa). 46. Show that (a)∞⎪summationdisplay n=1(−1)ncosnx (n2−a2)=1 2a2⎪bracketleftBig 1−πacosax sinaπ⎪bracketrightBig . (b)∞⎪summationdisplay n=1log⎪parenleftbigg 1+a2 n2⎪parenrightbigg =l o g⎪parenleftbiggsinhπa πa⎪parenrightbigg . 47. (a) If f(t) = 1 in Example 4.3.3, show that u(x, t)=x √ 4πκ⎪integraldisplayt oτ−3 2exp⎪parenleftbigg −x2 4κτ⎪parenrightbigg dτ=u0(x, t)( s a y ) (b) Hence or otherwise derive the Duhamel/primesf o r m u l a from (4 .3.16): u(x, t)=⎪integraldisplayt of(t−τ)⎪parenleftbigg∂u0 ∂τ⎪parenrightbigg dτ, where∂u0 ∂t=x √ 4πκτ−3 2exp⎪parenleftbigg −x2 4κt⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC 266 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 48. Consider a progressive plane wave solution that propagates to the right with the phase velocity⎪parenleftbigω k⎪parenrightbig of the telegraph equation (4.3.55) (a) Derive the dispersion relation ω2+i(p+q)ω−⎪parenleftbig c2k2+pq⎪parenrightbig =0. (b) If 4 pq/negationslash=(p+q)2, show that the plane wave solution is given by u(x, t)=Aexp⎪bracketleftbigg −1 2(p+q)t⎪bracketrightbigg exp[i(kx±σt)], where σ=1 2⎪radicalbig 4c2k2+4pq−(p+q)2. (c) If 4 pq=(p+q)2, show that the plane wave solution is given by u(x, t)=Aexp⎪bracketleftbigg −1 2(p+q)t⎪bracketrightbigg exp[ik(x±ct)]. Explain the physical signiﬁcance of the solutions given in cases (b) and (c). 49. (a) Use the substitution v(x, t)=e x p⎪bracketleftbig1 2(p+q)t⎪bracketrightbig u(x, t) into (4.3.55) to show that v(x, t) satisﬁes the wave equation vtt−c2vxx=1 4(p−q)2v. (b) Show that the undistorted wave solution exists if p=qand that a progressive wave of the form exp( −at)f(x±ct) propagates in either direction where fis an arbitrary twice diﬀerentiable function of its argument. 50. (a) Use the joint Laplace and Fourier transform to solve the inhomo- geneous diﬀusion problem ut−κuxx=q(x, t)x∈R,t > 0, u(x,0) =f(x),for all x∈R. (b) Solve the initial-boundary value problem for the diﬀusion equation ut−κuxx=0,0<x<l, t> 0 u(x,0) = 0 ,and u(0,t)=1= u(l,t). 51. Use the Laplace transform to solve for the small displacement y(x, t)o f a semi-inﬁnite string ﬁxed at x= 0 under the action of gravity gthat © 2007 by Taylor & Francis Group, LLC Applications of Laplace Transforms 267 satisﬁes the wave equation and the initial-boundary conditions ∂2y ∂t2−c2∂2y ∂x2=−g,0<x< ∞,t > 0, y(x,0) = 0 = yt(x,0),x≥0, ∂y ∂x→0a sx →∞. 52. Use the Laplace transform to solve the boundary layer equation (4.3.97) subject to the boundary and initial conditions u(z, t)=U0f(t),on z = 0 ,t>0, u(z, t)→0a s z→∞,t > 0, u(z, t)→0a t t≤0 for all z>0. Consider the special case where f(t)=s i n ωt. 53. Find the transfer function, the impulse response function and a formula for the solution of the following systems: (a)x/prime/prime(t)+2x/prime(t)+5x(t)=f(t),x(0) = 2 ,x/prime(0) =−2. (b)x/prime/prime(t)−2x/prime(t)+5x(t)=f(t),x(0) = 0 ,x/prime(0) = 2 . (c)x/prime/prime(t)+9x/prime(t)=f(t),x (0) = 2 ,x/prime(0) =−3. (d)x/prime/prime(t)−2x/prime(t)+5x(t)=f(t),x(0) =x0,x/prime(0) =x1. 54. Determine the transfer function for each of the following systems. Obtain the order of each system and ﬁnd which is stable. (a)x/prime/prime(t)+2x/prime(t)+2x(t)=3f/prime(t)+2f(t). (b) 4 x/prime/prime(t)+1 6 x/prime(t)+2 5 x(t)=2f/prime(t)+3f(t). (c) 36 x/prime/prime(t)+1 2 x/prime(t)+3 7 x(t)=2f/prime/prime(t)+f/prime(t)−6f(t). (d)x/prime/prime(t)−6x/prime(t)+1 0 x(t)=2f/prime(t)+5f(t). 55. Examine the stability of a system for real constants aandbwith zero initial data x/prime/prime/prime(t)−ax/prime/prime(t)+b2x/prime(t)−ab2x(t)=f(t), where x(t) is the output corresponding to input f(t). Discuss three cases: (a) a>0, (b) a≤0,b/negationslash=0 , ( c ) a/negationslash=0 , b=0 . © 2007 by Taylor & Francis Group, LLC 5 Fractional Calculus and Its Applications “In his discovery of calculus, Leibniz ﬁrst introduced the idea of as y m b o l i cm e t h o da n du s e dt h es y m b o ldny dxn=Dnyfor the nth derivative, where nis a non-negative integer. L’Hospital asked Leibniz about the possibility that nbe a fraction. ‘What if n=1 2.’ Leibniz (1695) replied, ‘It will lead to a paradox.’ But he added prophetically, ‘From this apparent paradox, one day useful conse-quences will be drawn’.” Gottfried Wilhelm Leibniz “The mathematician’s best work is art, a high perfect art, as dar- ing as the most secret dreams of imagination, clear and limpid. Mathematical genius and artistic genius touch one another.” G˙osta Mittag-Leﬄer 5.1 Introduction This chapter deals with fractional derivatives and fractional integrals and their basic properties. Several method s including the Laplace transform are discussed to introduce the Riemann-Liouville fractional integrals. Attention is given to the Weyl fractional integral and its properties. Finally, the fractional derivative is applied to solve the celebrated Abel integral equation. This is followed by brief comments on the Heaviside operational calculus and modernapplications of fractional calculus to science and engineering. This chapter is based on two articles of Debnath (2003, 2004) and hence, the reader is referred to these articles for all references cited in this chapter. 269 © 2007 by Taylor & Francis Group, LLC 270 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 5.2 Historical Comments Historically, Isaac Newton (1642-1727) and Gottfried Wihelm Leibniz (1646- 1716) independently discovered calculus in the seventieth century. In recogni- tion of this remarkable discovery, John Von Neumann’s (1903-1957) thought seems to worth quoting. “...the calculus was the ﬁrst achievement of modernmathematics and it is diﬃcult to overestimate its importance. I think it de- ﬁnes more equivocally than anything else the inception of modern mathemat- ics, and the system of mathematical analysis, which is its logical development, still constitute the greatest technical advance in exact thinking.” In his dis- covery of calculus, Leibniz ﬁrst introduced the idea of a symbolic method andused the symbol dny dxn=Dnyfor the nth derivative, where nis a non-negative integer. L’Hospital asked Leibniz about the possibility that nbe a fraction. “What if n=1 2.” Leibniz (1695) replied, “It will lead to a paradox.” But he added prophetically, “From this apparent paradox, one day useful conse- quences will be drawn.” Can the meani ng of derivatives of integral order Dny be extended to have meaning where nis any number — rational, irrational, or complex? In his 700-page long book on calculus published in 1819, Lacroix developed the formula for the nth derivative of y=xm,mis a positive integer, Dny=m! (m−n)!xm−n, (5.2.1) where n(≤m) is an integer. Replacing the f actorial symbol by the gamma function, he further obtained the formula for the fractional derivative Dαxβ=Γ(β+1 ) Γ(β−α+1 )xβ−α, (5.2.2) where αandβare fractional numbers. In particular, he calculated D1 2x=Γ( 2) Γ⎪parenleftbig3 2⎪parenrightbigx1 2=2⎪radicalbigg x π. (5.2.3) On the other hand, in 1832, Joseph Liouville (1809-1882) formally extended the formula for the derivative of integral order n Dneax=aneax(5.2.4) to the derivative of arbitrary order α Dαeax=aαeax. (5.2.5) Using the series expansion of a function f(x), Liouville derived the formula Dαf(x)=∞⎪summationdisplay n=0cnaα neanx, (5.2.6) © 2007 by Taylor & Francis Group, LLC Fractional Calculus and Its Applications 271 where f(x)=∞⎪summationdisplay n=0cnexp(anx),R e a n>0. (5.2.7) Formula (5.2.6) is referred to as Liouville’s ﬁrst formula for fractional deriva- tive. It can be used as a formula for derivative of arbitrary order α,w h i c hm a y be rational, irrational or complex. However, it can only be used for functions of the form (5.2.7). In order to extend his ﬁrst deﬁnition (5.2.6), Liouville formulated another deﬁnition of a fractional derivative based on the gammafunction (see Debnath and Speight (1971)) Γ(β)x −β=⎪integraldisplay∞ 0tβ−1e−xtdt, β > 0, (5.2.8) Dαx−β=(−1)αΓ(α+β) Γ(β)x−α−β,β > 0. (5.2.9) This is called the Liouville’s second deﬁnition of fractional derivative. He suc- cessfully applied both his deﬁnitions to p roblems in potential theory. However, Liouville’s ﬁrst deﬁnition is restricted to a certain class of function in the form (5.2.7), and his second deﬁnition is useful only for rational functions. Neither of his deﬁnitions was found to be suitable for a wide class of functions. Ac- cording to (5.2.9), the derivative of a constant function ( β= 0) is zero because Γ( 0)= ∞. On the other hand, the Lacroix deﬁnition (5.2.2) gives a nonzero value for the fractional derivative of a constant function ( β=0 )i nt h ef o r m Dα1=x−α Γ( 1−α)/negationslash=0. (5.2.10) Peacock (1833) favored Lacroix formula (5.2.2) for fractional derivatives, but other mathematicians preferred Liouville’s deﬁnitions. This led to a discrep-ancy between the two deﬁnitions of a fractional derivative. In spite of a lot of subsequent progress on the subject of fractional calculus, this controversy has hardly been resolved. In 1822, Fourier obtained the following integral representations for f(x) and its derivatives. f(x)=1 2π⎪integraldisplay∞ −∞f(ξ)dξ⎪integraldisplay∞ −∞cost(x−ξ)dt, (5.2.11) and Dnf(x)=1 2π⎪integraldisplay∞ −∞f(ξ)dξ⎪integraldisplay∞ −∞tncos⎪braceleftBig t(x−ξ)+nπ 2⎪bracerightBig dt.(5.2.12) Replacing integer nby arbitrary real αyields formally Dαf(x)=1 2π⎪integraldisplay∞ −∞f(ξ)dξ⎪integraldisplay∞ −∞tαcos⎪braceleftBig t(x−ξ)+πα 2⎪bracerightBig dt.(5.2.13) © 2007 by Taylor & Francis Group, LLC 272 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Greer (1858-1859) derived formulas for the fractional derivatives of trigono- metric functions based on (5.2.4) in the form Dαeiax=iαaαeiax=iαaα(cosax+isinax) =aα⎪parenleftBig cosπα 2+isinπα 2⎪parenrightBig (cosax+isinax) (5.2.14) so that the fractional derivatives of trigonometric functions are given by Dα(cosax)=aα⎪parenleftBig cosπα 2cosax−sinπα 2sinax⎪parenrightBig =aαcos⎪parenleftBig ax+πα 2⎪parenrightBig , (5.2.15) Dα(sinax)=aα⎪parenleftBig cosaxsinπα 2+s i naxcosπα 2⎪parenrightBig =aαsin⎪parenleftBig ax+πα 2⎪parenrightBig . (5.2.16) When α=1 2anda= 1, Greer’s formulas are as follows: D1 2cosx=c o s⎪parenleftBig x+π 4⎪parenrightBig ,D1 2sinx=s i n⎪parenleftBig x+π 4⎪parenrightBig . (5.2.17) Similarly, fractional derivatives for hyperbolic functions can be obtained. 5.3 Fractional Derivatives and Integrals The idea of fractional derivative or fractional integral can be described in diﬀerent ways. First, we consider a linear nonhomogenerous nth order ordinary diﬀerential equation Dny=f(x),b≤x≤c. (5.3.1) Then⎪braceleftbig 1,x ,x2, ...xn−1⎪bracerightbig is a fundamental set of the corresponding homoge- neous equation, Dny=0 .I f f(x) is any continuous on b≤x≤c, then for any a∈(b,c), y(x)=⎪integraldisplayx a(x−t)n−1 (n−1)!f(t)dt (5.3.2) is the unique solution of equation (5.3.1) with the initial data y(k)(a)=0,0≤k≤n−1. Or, equivalently, y=aD−n xf(x)=1 Γ(n)⎪integraldisplayx a(x−t)n−1f(t)dt. (5.3.3) © 2007 by Taylor & Francis Group, LLC Fractional Calculus and Its Applications 273 Replacing nbyα,w h e r e Reα> 0 in the above formula, we obtain the Riemann- Liouville deﬁnition of fractional integral that was reported by Liouville in 1832 and by Riemann in 1876 as aD−α xf(x)=aJα xf(x)=1 Γ(α)⎪integraldisplayx a(x−t)α−1f(t)dt, (5.3.4) where aD−α x=aJα xis the Riemann-Liouville integral operator. When a=0 , (5.3.4) is the Riemann deﬁnition of fractional integral, and if a=−∞, (5.3.4) represents the Liouville deﬁnition. Integrals of this type were found to arise in the theory of linear ordinary diﬀer ential equations where they are known as Euler transforms of the ﬁrst kind. Ifa=0 a n d x>0, then the Laplace transform solution of the initial value problem Dny(x)=f(x),x > 0,y(k)(0) = 0 ,0≤k≤n−1, is y(s)=s−n f(s), where y(s) is the Laplace transform of y(x) deﬁned by (3.2.5). The inverse Laplace transform gives the solution of the initial value problem y(x)=0D−n xf(x)=L−1⎪braceleftbig s−n f(s)⎪bracerightbig =1 Γ(n)⎪integraldisplayx 0(x−t)n−1f(t)dt. This is the Riemann-Liouville integral formula for an integer n. Replacing n by real αgives the Riemann-Liouville fractional integral (5.3.4) with a=0 . We consider a deﬁnite integral in the form fn(x)=1 (n−1)!⎪integraldisplayx a(x−t)n−1f(t)dt (5.3.5) withf0(x)=f(x)s ot h a t aDxfn(x)=1 (n−2)!⎪integraldisplayx a(x−t)n−2f(t)dt=fn−1(x),(5.3.6) and hence, fn(x)=⎪integraldisplayx afn−1(t)dt=aJxfn−1(x)=aJ2 xfn−2(x) =...=aJn xf(x). (5.3.7) Thus, for a positive integer n, it follows that aJn xf(x)=1 (n−1)!⎪integraldisplayx a(x−t)n−1f(t)dt. (5.3.8) © 2007 by Taylor & Francis Group, LLC 274 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Replacing nbyαwhere Reα> 0 in (5.3.8) leads to the deﬁnition of the Riemann-Liouville fractional integral aJα xf(x)=1 Γ(α)⎪integraldisplayx a(x−t)α−1f(t)dt. (5.3.9) Thus, this fractional-order integral formula is a natural extension of an iterat- ed integral. The fractional integral formula (5.3.9) can also be obtained from the Euler integral formula ⎪integraldisplayx 0(x−t)rtsdt=Γ(r+1 )Γ( s+1 ) Γ(r+s+2 )xr+s+1,r , s > −1.(5.3.10) Replacing rbyn−1a n d sbyngives ⎪integraldisplayx 0(x−t)n−1tndt=Γ(n) (n+1 )...(2n)x2n=Γ(n)0D−n xxn.(5.3.11) Consequently, (5.3.8) follows from (5.3.11) when f(t)=tnanda=0 .I n g e n - eral, (5.3.11) gives (5.3.8) replacing tnbyf(t). Hence, when nis replaced by α, we derive (5.3.9). It may be interesting to point out that Euler’s integral expression for the 2F1(a,b,c;x) hypergeometric series can now be expressed as a fractional in- tegral of order ( c−b)a s 2F1(a,b,c;x)=Γ(c)x1−c Γ(b)Γ(c−b)⎪integraldisplayx 0tb−1(x−t)c−b−1(1−t)−adt =Γ(c) Γ(b)x1−c 0Jc−b xf(x), (5.3.12) where f(t)=tb−1(1−t)−a. In complex analysis, the Cauchy integral formula for the nth derivative of an analytic function f(z)i sg i v e nb y Dnf(z)=n! 2πi⎪integraldisplay Cf(t)dt (t−z)n+1, (5.3.13) where Cis a closed contour on which f(z)i sa n a l y t i c ,a n d t=zis any point inside C,a n d t=zis a pole. Ifnis replaced by an arbitrary number αandn!b yΓ ( α+ 1), then a derivative of arbitrary order αc a nb ed e ﬁ n e db y Dαf(z)=Γ(α+1 ) 2πi⎪integraldisplay Cf(t)dt (t−z)α+1, (5.3.14) where t=zis no longer a pole but a branch point. In (5.3.14) Cis no longer an appropriate contour, and it is necessary to make a branch cut along the © 2007 by Taylor & Francis Group, LLC Fractional Calculus and Its Applications 275 real axis from the point z=x>0 to negative inﬁnity. Thus, we can deﬁne a derivative of arbitrary order αby a loop integral aDα xf(z)=Γ(α+1 ) 2πi⎪integraldisplayx a(t−z)−α−1f(t)dt, (5.3.15) where ( t−z)−α−1=e x p[ −(α+1 )ln(t−z)] and ln(t−z)i sr e a lw h e n t− z>0. Using the classical method of contour integration along the branch cut contour D,i tc a nb es h o w nt h a t 0Dα zf(z)=Γ(α+1 ) 2πi⎪integraldisplay D(t−z)−α−1f(t)dt =Γ(α+1 ) 2πi[1−exp{−2πi(α+1 )}]⎪integraldisplayz 0(t−z)−α−1f(t)dt =1 Γ(−α)⎪integraldisplayz 0(t−z)−α−1f(t)dt (5.3.16) which agrees with the Riemann-Liouville deﬁnition (5.3.4) with z=x,a n d a=0w h e n αis replaced −α. On the other hand, the Weyl fractional integral of order αwas introduced by Weyl (1917) by xW−α ∞f(x)=1 Γ(α)⎪integraldisplay∞ x(t−x)α−1f(t)dt, Re α > 0.(5.3.17) The major diﬀerence between this deﬁnition and the Riemann-Liouville deﬁ- nition are the limits of integration with the kernel here being ( t−x)α−1. IfDny=f(x)i st h e nth order nonhomogeneous diﬀerential equations and its adjoint equation is ( −1)nDny=f(x) whose solution with the initial con- ditions Dky(c)=0 ,0 ≤k≤n−1i sg i v e nb y y(x)= xW−n cf(x)=1 Γ(n)⎪integraldisplayc x(t−x)n−1f(t)dt. (5.3.18) Replacing nbyαso that Reα> 0, and c=∞, we can deﬁne the Weyl adjoint fractional integral by (5.3.18). For a class of good functions Gconsisting of functions fwhich are everywhere diﬀerentiable any number of times and all of its derivatives are 0⎪parenleftbig x−N⎪parenrightbig asx→∞for all N(see Lighthill, 1958), xW−α ∞f(x) deﬁned by (5.3.17) exists. Putting t−x=ξin (5.3.17) gives xW−α ∞f(x)=1 Γ(α)⎪integraldisplay∞ 0ξα−1f(ξ+x)dξ. (5.3.19) Application of the operator Dnto both sides of (5.3.19), dropping the sub- scripts xand∞in the Weyl operator, gives DnW−αf(x)=W−αDnf(x). (5.3.20) © 2007 by Taylor & Francis Group, LLC 276 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Similarly, we can prove EnW−αf(x)=W−αEnf(x), (5.3.21) where En=(−1)nDn. Iff∈G,a nn-fold integration of (5.3.17) by parts gives W−αf(x)=W−(α+n)[Enf(x)] (5.3.22) =En⎪bracketleftBig W−(α+n)f(x)⎪bracketrightBig by (5.3.21) . (5.3.23) In order to deﬁne the Weyl fractio nal derivatives, we assume that ν>0 andnis the smallest integer greater than νso that β=n−ν>0. If, for any function f,W−βfexists and has continuous derivatives. We then deﬁne the Weyl fractional derivative offof order νby Wνf(x)=W−(β−n)f(x)=En⎪bracketleftBig W−(β−n+n)f(x)⎪bracketrightBig ,by (5.3.23) =En⎪bracketleftbig W−βf(x)⎪bracketrightbig =En⎪bracketleftBig W−(n−ν)f(x)⎪bracketrightBig . (5.3.24) =En⎪bracketleftbigg1 Γ(n−ν)⎪integraldisplay∞ x(t−x)n−ν−1f(t)dt⎪bracketrightbigg . (5.3.25) It may be relevant to mention that Liouville’s classical problem of potential theory can be described by an integral equation involving the Weyl fractional integral as (πx)1 2W−1 2f(x)=W−1f(x), (5.3.26) where the force ﬁeld φ(r)=rf⎪parenleftbig r2⎪parenrightbig ,r=√ x. Using a series expansion f(x), Liouville (1832) obtained the desired law of force as φ(r)=a r2, (5.3.27) where ais any constant. Gr¨unwald (1867) introduced the idea of fractional derivative as the limit of as u mg i v e nb y Dαf(x) = lim h→01 hαn⎪summationdisplay r=0(−1)rΓ(α+1 )f(x−rh) Γ(r+1 )Γ( α−r+1 )(5.3.28) provided the limit exists. Using the identity (−1)rΓ(α+1 ) Γ(α−r+1 )=Γ(r−α) Γ(−α), (5.3.29) the result (5.3.28) becomes Dαf(x) = lim h→0h−α Γ(−α)n⎪summationdisplay r=0Γ(r−α) Γ(r+1 )f(x−rh). (5.3.30) © 2007 by Taylor & Francis Group, LLC Fractional Calculus and Its Applications 277 When αis equal to an integer m, deﬁnition (5.3.28) reduces to the derivative of integral order mas Dmf(x) = lim h→01 hmn⎪summationdisplay r=0(−1)r⎪parenleftbiggm r⎪parenrightbigg f(x−rh), (5.3.31) where⎪parenleftbiggm r⎪parenrightbigg is the usual binomial coeﬃcient. Result (5.3.31) follows from the classical deﬁnitions of f/prime(x),f/prime/prime(x), ... as Df(x) = lim h→0f(x)−f(x−h) h= lim h→0Δf(x) h, (5.3.32) D2f(x) = lim h→0f/prime(x)−f/prime(x−h) h= lim h→0Δ2f(x) h2, (5.3.33) where Δf(x−rh)=f(x−rh)−f(x−(r+1 )h). (5.3.34) On the other hand, Marchaud (1927) formulated the fractional derivative of arbitrary order αin the form Dαf(x)=f(x) Γ( 1−α)xα+α Γ( 1−α)⎪integraldisplayx 0f(x)−f(t) (x−t)α+1dt, (5.3.35) where 0 <α< 1. It has been shown by Samko et al. (1987) that (5.3.35) and (5.3.28) are equivalent. Replacing mby−min (5.3.31), it can be shown inductively that 0D−m xf(x) = lim h→0hmn⎪summationdisplay r=0⎪bracketleftbiggm r⎪bracketrightbigg f(t−rh) =1 Γ(m)⎪integraldisplayx 0(x−t)m−1f(t)dt, (5.3.36) where ⎪bracketleftbigg m r⎪bracketrightbigg =m(m+1 )...(m+r+1 ) r!. (5.3.37) It is important to point out that Hargreave (1848) extended the Leibniz product rule for the nth derivative to the fractional order n=αin the form Dα[f(x)g(x)] =∞⎪summationdisplay r=0Γ(α+1 ) r!Γ(α−r+1 )Dα−rf(x)Drg(x),(5.3.38) provided the series converges, where Dris the diﬀerential operator of integral order randDα−ris a fractional operator. © 2007 by Taylor & Francis Group, LLC 278 INTEGRAL TRANSFORMS and THEIR APPLICATIONS In a series of papers, Osler (1970, 1971, 1972) and others thoroughly studied the Leibniz product rule of derivatives of arbitrary order αand proved a general result Dα[f(x)g(x)] =∞⎪summationdisplay r=−∞Γ(α+1 ) Γ(α−γ−r+1 )Γ( γ+r+1 ) ×Dα−γ−rf(x)Dr+γg(x),(5.3.39) where γis arbitrary. When γ= 0, Osler’s result (5.3.39) reduces to (5.3.38). Osler also proved a generalization of the Leibniz product rule (5.3.38) in theintegral form D α[f(x)g(x)] =⎪integraldisplay∞ −∞Γ(α+1 ) Γ(α−γ−r+1 )Γ( γ+r+1 ) ×Dα−γ−rf(r)Dγ+rg(r)dr. (5.3.40) This is a very useful formula for evaluating many deﬁnite integrals including a generalized version of Parseva l’s formula in Fourier analysis. Using the Cauchy integral formula for fractional derivatives, Nishimoto (1991) gave a new proof of the Leibniz product formula Dα z[f(z)g(z)] for analytic functions f(z)a n d g(z). Watanabe (1931) also derived the Leibniz product rule by using formula (5.3.38). It may be important to point out Nishi- moto’s (1991) formula for the fractional derivatives and integrals of logarithm function Dα(logaz)=−e−iπαΓ(α)z−α,D−α⎪parenleftbig z−α⎪parenrightbig =−e−iπα Γ(α)logz,(5.3.41) where a/negationslash=0 ,|arga|<π 2,zandαare complex numbers. In addition to deﬁnition (5.3.15), it is revelant to mention Nishimoto’s def- inition (1991) and properties of fractional calculus of functions of a single complex variable and several complex variables. On the other hand, in their paper, Hardy and Littlewood (1928, 1932) proved the formula Dαf(z)=Γ(α+1 ) 2πi⎪integraldisplay Cf(t) (t−z)α+1dt, (5.3.42) where |z|<1a n d Cis the Hardy-Littlewood loop from t= 0 round t=zin a positive sense. The Riemann-Liouville integral operator aDα xdeﬁned by (5.3.9) satisﬁes the following properties: 0D0 xf(x)=If(x)=f(x) (Identity), (5.3.43) aDα x[cf(x)+dg(x)] =caDα xf(x)+daDα xg(x) (Linearity), (5.3.44) where canddare arbitrary constants. © 2007 by Taylor & Francis Group, LLC Fractional Calculus and Its Applications 279 12345 00.511.5200551010151520202525 xvaluesvaluesDerivative Figure 5.1 Fractional derivative of Dαx2for 0≤α≤2. Computational results of fractional derivative of x2for 0≤α≤2a r es h o w n in the Figure 5.1. Computational results of fractional derivative of sin xand cos xare shown in Figure 5.2 and Figure 5.3, respectively (Bhatta 2006). 5.4 Applications of Fractional Calculus It may be important to point out that the ﬁrst application of fractional calcu- lus was made by Abel (1802-1829) in the solution of an integral equation that arises in the formulation of the tautochronous problem . This problem deals with the determination of the shape of a f rictionless plane curve through the origin in a vertical plane along which a particle of mass mcan fall in a time that is independent of the starting position. If the sliding time is constant T, then the Abel integral equation (1823) is ⎪radicalbig 2gT=⎪integraldisplayη 0(η−y)−1 2f/prime(y)dy, (5.4.1) where gis the acceleration due to gravity, ( ξ,η) is the initial position and s= f(y) is the equation of the sliding curve. It turns out that (5.4.1) is equivalent © 2007 by Taylor & Francis Group, LLC 280 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 0 1 2 3 4 5 600.20.40.60.81 -1-1 -0.5-0.5 00 0.50.5 11 xvaluesvaluesFractional Derivative of sin x Figure 5.2 Fractional derivative of Dαsinxfor 0≤α≤1. to the fractional integral equation T⎪radicalbig 2g=Γ⎪parenleftbigg1 2⎪parenrightbigg 0D−1 2ηf/prime(η). (5.4.2) Or, equivalently, f/prime(η)=T⎪radicalbigg 2g π0D1 2η1=⎪radicalbigg 2a η, (5.4.3) where a=⎪parenleftBig gT2 π2⎪parenrightBig . Finally, the solution is f(η)=⎪radicalbig 8aη=4asinψ, (5.4.4) wheredη ds=s i nψ. This curve is the cycloid with the vertex at the origin and the tangent at the vertex as the x-axis. The solution of the Abel problem is based on the fact that the derivative o f a constant is not always equal to zero. During the last decades of the nineteenth century, Heaviside successfully developed his operational calculus without rigorous mathematical arguments. In 1892 he introduced the idea of fractional derivatives in his study of elec- tric transmission lines. Based on the symbolic operator form solution of heat equation due to Gregory (1846), Heaviside introduced the letter pfor the diﬀerential operatord dtand gave the solution of the diﬀusion equation ∂2u ∂x2=a2p (5.4.5) © 2007 by Taylor & Francis Group, LLC Fractional Calculus and Its Applications 281 0 1 2 3 4 5 600.20.40.60.81 -1-1 -0.5-0.5 00 0.50.5 11 xvaluesvaluesFractional Derivative of cos x Figure 5.3 Fractional derivative of Dαcosxfor 0≤α≤1. for the temperature distribution u(x, t) in the symbolic form u(x, t)=Aexp(ax√ p)+Bexp(−ax√ p), (5.4.6) in which p≡d dxwas treated as constant, where a,A,a n dBare also constants. Indeed, Heaviside gave an interpretation of√ p=D1 2so that 0D1 2 t1=1 √ πt, which is in complete agreement with (5.2.10) The development of the Heavi- side operational calculus was somewhat similar to that of calculus. Both New- ton and Leibniz who discovered calculus did not provide a rigorous formulation of it. The rigorous theory had been developed in the nineteenth century, eventhough it is the transition of the non-rigorous development of the calculus that is still admired. It is well known that twentieth-century mathematicians have provided a rigorous foundation of the Heaviside operational calculus. In his book, Davis (1936) described the theory of linear operators with fractional calculus and its applications. He also states “The period of the formal devel-opment of operational methods may be regarded as having ended by 1900. The theory of integral equations was just beginning to stir the imagination of mathematicians and to reveal the possibilities of operational methods.” During the second half of the twentieth century, considerable amount of research in fractional calculus was published in engineering literature. Indeed, recent advances of fractional calculus are dominated by modern examples of applications in diﬀerential and integral equations, physics, signal processing, ﬂuid mechanics, viscoelasticity, mathe matical biology and electrochemistry. There is no doubt that fractional calculus has become an exciting new math-ematical method of solution of diverse problems in mathematics, science, and engineering. In a recent article by Debnath (2003), he presented numerous new and recent applications of fractional calc ulus in mathematics, science, and en- © 2007 by Taylor & Francis Group, LLC 282 INTEGRAL TRANSFORMS and THEIR APPLICATIONS gineering. For more details, the reader is referred to the paper by Debnath (2003). 5.5 Exercises 1. Show that (a)⎪integraldisplay∞ 0√ xe−x3dx=√ π 3(b)⎪integraldisplay∞ 0x4e−x3dx=1 3Γ⎪parenleftbigg5 3⎪parenrightbigg 2. Iff(x)=√ x,show that (a)d1 2f dx1 2=√ π 2(b)d−1 2f dx−1 2=√ πx 2 3. Iff(x)=x,show that (a)d1 2f dx1 2=2⎪radicalbigg x π(b)d−1 2f dx−1 2=4x3/2 3√ π 4. Iff(x)=x3/2,show that (a)d1 2f dx1 2=3 4√ πx (b)d−1 2f dx−1 2=3 8√ πx2 5. Iff(x)=x2,show that (a)d1 2f dx1 2=8x3/2 3√ π(b)d−1 2f dx−1 2=16x5/2 15√ π 6. Iff(x)=s i n (√ x),show that (a)d1 2f dx1 2=√ π 2J0(√ x)( b)d−1 2f dx−1 2=√ πxJ1(√ x) 7. Iff(x) = sinh(√ x),show that (a)d1 2f dx1 2=√ π 2I0(√ x)( b)d−1 2f dx−1 2=√ πxI1(√ x) 8. Iff(x)=J0(√ x),show that (a)d1 2f dx1 2=cos(√ x) √ πx(b)d−1 2f dx−1 2=2s i n (√ x) √ π 6 Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations “In every mathematical investigation, the question will arise whether we can apply our mathematical results to the real world.” V. I. Arnold “All of Abel’s works carry the imprint of an ingenuity and force ofthought which is unusual and sometimes amazing, even if the youth of the author is not taken into consideration. One may say that he was able to penetrate all obstacles down to the very foundations of the problems, with a force which a ppeared irresistible; he attacked the problems with extraordinary energy; he regarded them from above and was able to soar so high over their present state that all diﬃculties seemed to vanish under the victorious onslaught ofhis genius.... But it was not only his great talent which created the respect for Abel and made his loss inﬁnitely regrettable. He distin- guished himself equally by the purity and nobility of his character and by a rare modesty which made his person cherished to the same unusual degree as was his genius.” August Leopold Crelle 6.1 Introduction In the proceeding chapter, the basic ideas of fractional calculus and its applica- tions have been presented. This chapter is essentially devoted to applicationsof Laplace, Fourier and Hankel transforms to fractional integral equations, fractional ordinary and partial diﬀerential equations. Many examples of ap- plications are presented in some detail. Included are also Green’s functions offractional diﬀerential equations. Most of the applications involving the partial diﬀerential equations are based on auth ors’ recent papers, which are listed in the Bibliography. 283 © 2007 by Taylor & Francis Group, LLC 284 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 6.2 Laplace Transforms of Fractional Integrals and Fractional Derivatives TheRiemann-Liouville fractional integral is usually deﬁned by D−αf(t)=0D−α tf(t)=1 Γ(α)t⎪integraldisplay 0(t−x)α−1f(x)dx, Reα>0.(6.2.1) Clearly, D−αis a linear integral operator. A simple change of variable ( t−x)α=uin (6.2.1) allows us to prove the following result D[D−αf(t)] =D−α[Df(t)] +f(0)tα−1 Γ(α). (6.2.2) Clearly, the integral in (6.2.1) is a convolution, and hence, the Laplace trans- form of (6.2.1) gives L{D−αf(t)}=L{f(t)∗g(t)}=L{f(t)}L{g(t)}, (6.2.3) =s−α¯f(s),α > 0. (6.2.4) where g(t)=tα−1 Γ(α)and ¯g(s)=s−α. The result (6.2.4) is also valid for α=0 ,a n d lim α→0L⎪braceleftbiggtα−1 Γ(α)⎪bracerightbigg = lim α→0s−α=1. (6.2.5) Using (6.2.4), it can readily be veriﬁed that the fractional integral operatorsatisﬁes the laws of exponents D −α[D−βf(t)] =D−(β+α)f(t)=D−β[D−αf(t)]. (6.2.6) Formula (6.2.4) can be used for evaluating the fractional integral of a given function using the inverse Laplace transform. The following examples illus-trate this point. L{D −αtβ}=Γ(β+1 ) sα+β+1,β > −1. (6.2.7) Or, equivalently, D−αtβ=L−1⎪braceleftbiggΓ(β+1 ) sα+β+1⎪bracerightbigg =Γ(β+1 ) Γ(α+β+1 )tα+β. (6.2.8) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 285 In particular, if α=1 2andβ(=n) is an integer, then (6.2.8) gives D−1/2tn=Γ(n+1 ) Γ⎪parenleftbigg n+1 2+1⎪parenrightbigg·tn+1 2,n > −1. (6.2.9) It also follows from (6.2.4) that L{D−αeat}=1 sα(s−a),a>0. (6.2.10) Or, D−αeat=L−1⎪braceleftbigg1 sα(s−a)⎪bracerightbigg (6.2.11) =L−1⎪braceleftbigg1 sα+1⎪parenleftbigg 1+a s−a⎪parenrightbigg⎪bracerightbigg =tα Γ(α+1 )+aE(t, α+1,a), (6.2.12) where E(t, α, a) is deﬁned by E(t, α, a)=1 Γ(α)t⎪integraldisplay 0ξα−1exp{a(t−ξ)}dξ. (6.2.13) In particular, if α=1 2then D−1/2eat=L−1⎪braceleftbigg1 √ s(s−a)⎪bracerightbigg =1 √ πt∗eat, which is, by Example 3.7.8, =eat √ aerf(√ at). (6.2.14) The following results follow readily from (6.2.4): L{D−αsinat}=a sα(s2+a2),α > 0. (6.2.15) L{D−αcosat}=s sα(s2+a2),α > 0. (6.2.16) L{D−αeattβ−1}=Γ(β) sα(s−a)β,α > 0,β>0. (6.2.17) The inverse Laplace transforms of these results combined with the Convo- lution Theorem lead to fractional i ntegrals of functions involved. © 2007 by Taylor & Francis Group, LLC 286 INTEGRAL TRANSFORMS and THEIR APPLICATIONS One of the consequences of (6.2.1) is that lim α→0D−αf(t)=f(t). (6.2.18) This follows from the inverse Laplace transform of (6.2.4) combined with the limit as α→0. We now evaluate the Laplace transform of the fractional integral of the derivative and then the Laplace transform of the derivative of the integral. Inview of (6.2.4), it follows that L{D −α[Df(t)]}=s−αL{Df(t)} =s−α[s¯f(s)−f(0)],α > 0. (6.2.19) Although this result is proved for α>0, it is valid even if α=0 . On the other hand, the Laplace transform of (6.2.2) gives L{D[D−αf(t)]}=L{D−αD[f(t)]}+f(0)L⎪braceleftbiggtα−1 Γ(α)⎪bracerightbigg =s−α[s¯f(s)−f(0)] + s−αf(0) =s1−α¯f(s),α≥0. (6.2.20) Obviously, if α= 0, this result does not agree with that obtained from (6.2.19) asα→0. This disagreement is due to the fact that “ L”a n d“ l i m ”d on o t commute, as is seen from (6.2.5). Another consequence of (6.2.1) is that the fractional derivative Dαf(t)c a n be deﬁned as the solution φ(t) of the integral equation D−αφ(t)=f(t). (6.2.21) The Laplace transform of this result gives the solution for ¯φ(s)a s ¯φ(s)=sα¯f(s). (6.2.22) Inversion gives the fractional derivative of f(t)a s φ(t)=Dαf(t)=L−1{sα¯f(s)}, (6.2.23) leading to the result Dαf(t)=1 Γ(−α)t⎪integraldisplay 0(t−x)−α−1f(x)dx, α > 0. (6.2.24) This is the Cauchy integral formula , which is often used to deﬁne the fractional derivative. However, formula (6.2.23) can be used for ﬁnding the fractional derivatives. If f(t)=tβ, it is seen from (6.2.23) that Dαtβ=L−1⎪braceleftbiggΓ(β+1 ) sβ−α+1⎪bracerightbigg =Γ(β+1 ) Γ(β−α+1 )tβ−α. (6.2.25) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 287 In particular, if α=1 2andβ(=n) is an integer, D1/2tn=Γ(n+1 ) Γ⎪parenleftbig n+1 2⎪parenrightbigtn−1 2,n > −1. D1/2eat=L−1⎪braceleftbigg√ s s−a⎪bracerightbigg =L−1⎪braceleftbigg1 √ s+a √ s(s−a)⎪bracerightbigg ,(6.2.26) which is, by (3.2.21) and (3.7.3), =1 √ πt+√ aexp(at)erf(√ at). (6.2.27) Example 6.2.1 Show that D−αJ0(a√ t)=⎪parenleftbigg2 a⎪parenrightbiggα tα/2Jα(a√ t). (6.2.28) We apply the Laplace transform to the left-hand side of (6.2.28) and use (6.2.22) to obtain L⎪braceleftBig D−αJ0(a√ t)⎪bracerightBig =s−αL⎪braceleftBig J0(a√ t)⎪bracerightBig =s−(1+α)exp⎪parenleftbigg −a2 4s⎪parenrightbigg . The inverse Laplace transform gives D−αJ0(a√ t)=L−1⎪braceleftbigg s−(1+α)exp⎪parenleftbigg −a2 4s⎪parenrightbigg⎪bracerightbigg =⎪parenleftbigg2 a⎪parenrightbiggα tα/2Jα(a√ t). 6.3 Fractional Ordinary Diﬀerential Equations We ﬁrst deﬁne a fractional diﬀerential equation with constant coeﬃcients of order ( n,q)a s [Dnα+an−1D(n−1)α+···+a0D0]x(t)=0,t≥0, (6.3.1) where α=1 q.Ifq=1,thenα= 1a n dt h i se q u a t i o ni ss i m p l ya no r d i n a r y diﬀerential equation of order n.Symbolically, we write (6.3.1) as f(Dα)x(t)=0, (6.3.2) © 2007 by Taylor & Francis Group, LLC 288 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where f(Dα) is a fractional diﬀerential operator. We next use the Laplace transform method to solve a simple fractional diﬀerential equation with constant coeﬃcients of order (2, 2) in the form f⎪parenleftBig D1 2⎪parenrightBig x(t)=⎪parenleftBig D1+a1D1 2+a0D0⎪parenrightBig x(t)=0. (6.3.3) Application of the Laplace transform to this equation gives [s¯x(s)−x(0)] + a1⎪bracketleftBig√ s¯x(s)−D−1 2x(0)⎪bracketrightBig +a0¯x(s)=0. Or, ¯x(s)=x(0) +a1D−1 2x(0) (s+a1√ s+a0)=A f(√ s), (6.3.4) where f(x)=x2+a1x+a0is an associated indicial equation andAis as- sumed to be a non-zero ﬁnite constant deﬁned by A=x(0) +a1D−1 2{x(0)}. (6.3.5) We next write the following partial fractions for the right hand side of (6.3.4) so that ¯x(s)=A a−b⎪parenleftbigg1 √ s−a−1 √ s−b⎪parenrightbigg =A a−b⎪parenleftbigg√ s s−a2+a s−a2−√ s s−b2−b s−b2⎪parenrightbigg , (6.3.6) where aandbare two distinct roots of f(x)=0 . Using the inverse Laplace transform formula (6.2.11) with α=−1 2andα= 0,we invert (6.3.6) to obtain the formal solution x(t)=A a−b⎪bracketleftbigg E⎪parenleftbigg t,−1 2,a2⎪parenrightbigg +aE(t,0,a2) −E⎪parenleftbigg t,−1 2,b2⎪parenrightbigg −bE(t,0,b2)⎪bracketrightbigg .(6.3.7) For equal roots ( a=b)o ff(x)=0,we ﬁnd ¯x(s)=A (√ s−a)2=A⎪bracketleftbigg√ s (s−a2)+a (s−a2)⎪bracketrightbigg2 . (6.3.8) In view of the result L−1⎪braceleftbigg1 sα(s−a)2⎪bracerightbigg =tE(t, α, a)−αE(t, α+1,a), (6.3.9) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 289 the inverse Laplace transform of (6.3.8) gives the solution as x(t)=A⎪bracketleftbig (1 + 2 a2t)E(t,0,a2) +aE⎪parenleftbigg t,1 2,a2⎪parenrightbigg +2at E⎪parenleftbigg t,−1 2,a2⎪parenrightbigg⎪bracketrightbigg . (6.3.10) Kempﬂe and Gaul (1996) developed the criteria for the existence, conti- nuity and causality of global solutions of the fractional order linear ordinary diﬀerential equation of the form L(D)x(t)=f(t),t∈R (6.3.11) with given initial or boundary data, where L(D) is a fractional diﬀerential operator L(D)≡Dαn+an−1Dαn−1+...+a1Dα1+a0, (6.3.12) where D≡d dt,0≤α1≤...≤αn,αkare non-integers, an−1,an−2, ...,a1,a0 are real constants and f(t) is a given forcing function. Application of the Fourier transform of x(t) with respect to tgives the physical solution in the form x(t)={L(D)}−1·f(t)=G(t)∗f(t) =⎪integraldisplay∞ −∞f(t−τ)G(τ)dτ =1 2π⎪integraldisplay∞ −∞f(t−τ)dτ⎪integraldisplay∞ −∞eiωτ ˜p(ω)dω, (6.3.13) where ∗denotes the Fourier convolution, G(t) is the impulse response function given by G(t)=1 2π⎪integraldisplay∞ −∞eiωt ˜p(ω)dω, ω ∈R, (6.3.14) and ˜p(ω)=(iω)αn+an−1(iω)αn−1+...+a1(iω)α1+a0.(6.3.15) The solution (6.3.14) exists provided1 ˜p(ω)∈L2(R)o r˜p(ω) has no real zeros and deg L>1 2. According to the well-known stability criteria for a linear system ( iω→s), equation (6.3.13) has the solution provided q(s)=˜p(−is)=sαn+an−1sαn−1+...+a1sα1+a0 (6.3.16) has no zeros in the right half s-plane. For simplicity, if q(s) is restricted to the principal branch with its zeros at sk=−σk+ iΩk, integral (6.3.14) can be © 2007 by Taylor & Francis Group, LLC 290 INTEGRAL TRANSFORMS and THEIR APPLICATIONS evaluated by using the Cauchy residue theory so that (6.3.14) yields G(t)=∞⎪summationdisplay k=1Ake−σktcos(Ω kt+εk)−1 π⎪integraldisplay∞ 0e−rtdr q(r), (6.3.17) where the ﬁrst series solution represents phase-shifted oscillations and the second integral term is the relaxation function which becomes dominant ast→∞. 6.4 Fractional Integral Equations (a)Abel’s integral equation of the ﬁrst kind is given by ⎪integraldisplayt 0(t−τ)α−1f(τ)dτ=g(t),0<α< 1, (6.4.1) where g(t) is given function. This equation can be expressed in terms of fractional integral Γ(α)0Jα tf(t)=g(t). (6.4.2) Application of the Laplace transform to (6.4.1) or (6.4.2) gives the solution f(s)=1 Γ(α)sα g(s)=1 Γ(α)s⎪bracketleftbigg1 s1−α· g(s)⎪bracketrightbigg , (6.4.3) which leads to the solution of (6.4.1) in the form f(t)=1 Γ(α)1 Γ( 1−α)d dt⎪integraldisplayt 0(t−τ)−αg(τ)d(τ). (6.4.4) (b)Abel’s integral equation of the second kind is given by f(t)+a Γ(α)⎪integraldisplayt 0(t−τ)α−1f(τ)dτ=g(t) α>0, (6.4.5) where ais real or complex parameter and g(t) is a given function. Application of the Laplace transform to (6.4.5) leads to the transform so- lution f(s)=⎪parenleftbiggsα sα+a⎪parenrightbigg g(s)=⎪bracketleftbigg s·sα−1 sα+a· g(s)⎪bracketrightbigg (6.4.6) whence the inverse Laplace gives the solution f(t)=d dt⎪integraldisplayt 0Eα,1(−aτα)g(t−τ)dτ, (6.4.7) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 291 where the Mittag-Leﬄer function, Eα,β(z)i sg i v e nb y Eα,β(z)=∞⎪summationdisplay n=0zn Γ(αn+β),α , β > 0. (6.4.8) (c)Poisson’s integral equation is given by ⎪integraldisplayπ 2 0φ(rcosθ)s i n2α+1θd θ=h(r). (6.4.9) Substituting x=rcosθinto (6.4.9) gives ⎪integraldisplayr 0⎪parenleftbigg 1−x2 r2⎪parenrightbiggα φ(x)dx=rh(r), which is, by replacing1 rby√ z,a n d1 √ zh⎪parenleftBig 1 √ z⎪parenrightBig by Ψ ( z), ⎪integraldisplay 1 √ z 0⎪parenleftbigg1 z−x2⎪parenrightbiggα φ(x)dx=z−αΨ(z). Invoking substitution of x2=τand1 z=tyields the Abel integral equation ⎪integraldisplay√ t 0(t−τ)αf(τ)dτ=g(t), (6.4.10) where f(τ)=φ(√ τ) √ τandg(t)=2tαΨ⎪parenleftbigg1 t⎪parenrightbigg . Thus, the solution of (6.4.10) is f(t)=1 Γ(α+1 )0Dα tg(t) (6.4.11) so that the solution of the Poisson equation (6.4.9) is φ⎪parenleftBig√ t⎪parenrightBig =2√ t Γ( 1+ α)0Dα ttα+1 2h⎪parenleftBig√ t⎪parenrightBig . (6.4.12) Example 6.4.1 Consider the Abel integral equation g(t)=t⎪integraldisplay 0f/prime(t)(t−τ)−αdτ,0<α< 1. (6.4.13) © 2007 by Taylor & Francis Group, LLC 292 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Application of the Laplace transform gives ¯g(s)=L{f/prime(t)}L{t−a}. Or, ¯f(s)=f(0) s+¯g(s) Γ(1−α)s−a. Inverting, we ﬁnd f(t)=f(0) +1 Γ(α)Γ(1−α)t⎪integraldisplay 0g(τ)(t−τ)α−1dτ. (6.4.14) This is the required solution of Abel’s equation. Example 6.4.2 Solve the Abel integral equation g(t)=t⎪integraldisplay 0(t−x)−αf(x)dx, 0<α< 1. (6.4.15) Clearly, it follows from (6.2.1) that g(t)=Γ ( 1 −α)Dα−1f(t). Or, D1−αg(t)=Γ ( 1 −α)f(t). Hence, f(t)=1 Γ(1−α)D·D−αg(t) =1 Γ(1−α)·1 Γ(α)·Dt⎪integraldisplay 0(t−x)α−1g(x)dx =1 Γ(α)Γ(1−α)·d dtt⎪integraldisplay 0(t−x)α−1g(x)dx. (6.4.16) Example 6.4.3 (Abel’s Problem of Tautochronous Motion ). The problem is to determine the form of a frictionless plane curve through the origin in a vertical plane along © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 293 which a particle of mass mcan fall in a time that does not depend on the initial position. Suppose the particle is placed on a curve at the point ( ξ,η), where ηis measured positive upward. Let the particle be allowed to fall to the origin under the action of gravity. Suppose ( x, y) is any position of the particle during its descent as shown in Figure 6.1. Accord ing to the principle of conservation of energy, the sum of the kinetic and potential energies is constant, that is, 1 2mυ2+mgy=mgη= constant . (6.4.17) This gives the velocity of the particle at any position ( x, y) υ2=⎪parenleftbiggds dt⎪parenrightbigg2 =2g(η−y), (6.4.18) where sis the length of the arc of the curve measured from the origin and t is the time. 0 xy g(x, y)(,)P Figure 6.1 Abel’s problem. Integrating (6.4.18) from y=ηto 0 gives τ⎪integraldisplay 0dt=−1 √ 2g0⎪integraldisplay ηds √ η−y=1 √ 2gη⎪integraldisplay 0f/prime(y)dy √ η−y, where s=f(y) represents the equation of the curve with f(0) = 0 .Thus, we obtain ⎪radicalbig 2gT=η⎪integraldisplay 0f/prime(y)dy √ η−y. (6.4.19) © 2007 by Taylor & Francis Group, LLC 294 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This is the Abel integral equation (6.4.13) with α=1 2andg(η)=T√ 2g= constant. Thus, the solution (6.4.14) becomes f(η)=T√ 2g πη⎪integraldisplay 0(η−y)1 2−1dy, which is, putting y=ηsin2θanda=gT2 π2, f(η)=2⎪radicalbig 2aη . (6.4.20) Ifψis the angle made by the tangent to the curve at a point ( x, y), then dy ds=s i nψanddx ds=c o sψso that cosecψ=ds dy=f/prime(y)=⎪radicalbigg 2a y. Or, s=f(y)=2⎪radicalbig 2ay=4asinψ. (6.4.21) This is the equation of the curve and represents the cycloid with the vertex at the origin and the tangent at the vertex as the x-axis. Alternatively, equation (6.4.19) can be expressed in terms of a half-order fractional derivative of one as given by (5.4.3) that was solved by using frac- tional derivatives with solution (5.4.4). The solution is the same as (6.4.21). Example 6.4.4 (Abel’s Equation and Fractional Derivatives in a Problem of Fluid Flow ). We consider the ﬂow of water along the xdirection through a symmetric dam in a vertical yzplane with the y-axis along the face of the dam. The problem is to determine the form y=f(z) of the opening of the dam so that the quantity of water per unit time is proportional to a given power of the depth of the stream. It follows from Bernoulli’s equation of ﬂuid mechanics that the ﬂuidvelocity υat a given height zabove the base of the dam is given by υ 2=2g(h−z). (6.4.22) The volume ﬂux dQthrough an elementary cross section dAof the opening of the dam is dQ=υdA=2√ 2g(h−z)1/2f(z)dzso that the total volume ﬂux is Q(h)=ah⎪integraldisplay 0(h−z)1 2f(z)dz, a =2⎪radicalbig 2g. (6.4.23) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 295 This is the Abel integral equation and hence can be written as the fractional integral Q(h)=aΓ⎪parenleftbigg3 2⎪parenrightbigg D−3/2 hf(h). (6.4.24) Multiplying this result by D3/2 hwith a given Q(h)=hβ, we obtain, by (6.2.24), f(h)=1 aΓ⎪parenleftbig3 2⎪parenrightbigD3/2 hhβ=2a √ πΓ(β+1 ) Γ⎪parenleftbig β−1 2⎪parenrightbighβ−3/2, (6.4.25) where β>−1. In particular, the shape y=f(z) is either a parabola or a rectangle depending on whether, β=7 2or3 2. There are also other shapes of the opening of the dam depending on the value of β. 6.5 Initial Value Problems for Fractional Diﬀerential Equations (a) We consider the following fractional diﬀerential equation 0Dα ty(t)+ω2y(t)=f(t),t > 0 (6.5.1) with the initial conditions ⎪bracketleftbig 0Dα−k ty(t)⎪bracketrightbig t=0=ck,k=1,2, ..., n, (6.5.2) where n−1<α<n . Application of the Laplace transform to (6.5.1)–(6.5.2) gives ⎪parenleftbig sα+ω2⎪parenrightbig y(s)= f(s)+n⎪summationdisplay k=1cksk−1. (6.5.3) Thus, the Laplace transform solution is y(s)=n⎪summationdisplay k=1cksk−1 (sα+ω2)+ f(s) sα+ω2. (6.5.4) The inverse Laplace transform gives the solution of the initial value problem y(t)=n⎪summationdisplay k=1cktα−kEα,α−k+1⎪parenleftbig −ω2tα⎪parenrightbig +⎪integraldisplayt 0f(t−τ)τα−1Eα,α⎪parenleftbig −ω2τα⎪parenrightbig dτ, (6.5.5) © 2007 by Taylor & Francis Group, LLC 296 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where Eα,β(z) is the Mittag-Leﬄer type function deﬁned by the series Eα,β(z)=∞⎪summationdisplay m=0zm Γ(αm+β),α > 0,β > 0, (6.5.6) and the inverse Laplace transform is L−1⎪braceleftBigg m!sα−β (sα¯+a)m+1⎪bracerightBigg =tαm+β−1E(m) α,β(+ atα), (6.5.7) with E(m) α,β(z)=dm dzmEα,β(z). (6.5.8) (i) When α=1 ,n= 1 so that the initial condition is y(0)=c1and the solution (6.5.5) reduces to the form y(t)=c1E1,1⎪parenleftbig −ω2t⎪parenrightbig +⎪integraldisplayt 0f(t−τ)E1,1⎪parenleftbig −ω2τ⎪parenrightbig dτ, (6.5.9) where E1,1(z)=ez. Consequently, the solution assumes the standard form y(t)=c1e−ω2t+⎪integraldisplayt 0f(t−τ)e−ω2τdτ. (6.5.10) (ii) When α=2 ,n= 2 so that the initial data are y(0) =c2andy/prime(0)=c1. In this case, the solution (6.5.5) reduces to the form y(t)=c1tE2,2⎪parenleftbig i2ω2t2⎪parenrightbig +c2E2,1⎪parenleftbig i2ω2t2⎪parenrightbig +⎪integraldisplayt 0f(t−τ)τE2,2⎪parenleftbig i2ω2τ2⎪parenrightbig dτ, (6.5.11) where E2,2⎪parenleftbig i2z2⎪parenrightbig =sinh (iz) iz=sinz iz, E2,1⎪parenleftbig i2z2⎪parenrightbig =c o s h( iz)=cos z. Consequently, the solution (6.5.11) reduces to the standard form y(t)=c1 ωsinωt+c2cosωt+1 ω⎪integraldisplayt 0f(t−τ)s i nωτdτ. (6.5.12) It is noted that equation (6.5.1) describes fractional relaxation when 0 < α≤1 and fractional oscillation when 1 <α≤2. It is easy to recognize the remarkable diﬀerence between the classical solutions for cases α=1 a n d α= 2. On the other hand, the solution of the fractional equation (6.5.1) show © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 297 remarkably diﬀerent features. The classical solution corresponding to α=1 decays exponentially as t→∞, and the fractional solution (0 <α< 1) exhibits af a s t e rd e c a ya s t→0+ and much slower decay (algebraic decay compared to exponential decay) as t→∞. (b)Fractional Simple Harmonic Oscillator The initial value problem for the fractional simple harmonic oscillator is given by d2y dt2+bdαy dtα+ω2y(t)= f(t),t > 0,0<α< 2,(6.5.13) y(0) =c0andy/prime 0(0) =c1, (6.5.14) where b,ω,c0andc1are constants anddαy dtαrepresents Caputo’s fractional derivative (see Caputo, 1967). Two special cases are of interest: (i) 0 <α< 1 and (ii) 1 <α< 2a n d α=1 . Application of the Laplace transform gives the following transform solu- tions: (i) y(s)=c0 y0(s)+c1 yδ(s)+ f(s). yδ(s),0<α< 1 (6.5.15) (ii) y(s)=c0 y0(s)+c1 y0(s) s+ f(s). yδ(s),1<α< 2 (6.5.16) where y0(s)=⎪parenleftbig s+bsα−1⎪parenrightbig g(s), yδ(s)=1 g(s),0<α< 2, (6.5.17) g(s)=⎪parenleftbig s2+bsα+ω2⎪parenrightbig , y0(s) s=⎪parenleftbig 1+bsα−2⎪parenrightbig yδ(s).(6.5.18) Using the following properties of the Laplace transform y0(0) = lim s→∞s y0(s)=1, (6.5.19) yδ(s)=−1 ω2[s y0(s)−1] =−1 ω2L{y/prime 0(t)}, (6.5.20) L⎪braceleftbigg⎪integraldisplayt 0y0(τ)dτ⎪bracerightbigg =1 s y0(s), (6.5.21) the inverse Laplace transform of (6.5.15)–(6.5.16) yields the closed form solu-tions: (i)y(t)=c 0y0(t)−c1 ω2y/prime 0(t)+⎪integraldisplayt 0f(t−τ)yδ(τ)dτ,0<α< 1 (6.5.22) (ii)y(t)=c0y0(t)+c1⎪integraldisplayt 0y0(τ)dτ+⎪integraldisplayt 0f(t−τ)yδ(τ)dτ,1<α< 2 (6.5.23) where yδ(t)=−1 ω2y/prime 0(t)r e p r e s e n t st h e impulse response solution of the e- quation (6.5.13) and this solution can be obtained from (6.5.23) by puttingc 0=c1=0a n d f(t)=δ(t). © 2007 by Taylor & Francis Group, LLC 298 INTEGRAL TRANSFORMS and THEIR APPLICATIONS In particular, when α= 1, equation (6.5.13) represents the classical solution for the damped simple harmonic oscillator. In order to simplify the solution, we write b=2kand ﬁnd y0(s)=(s+2k) (s+k)2+(ω2−k2), yδ(s)=1 (s+k)2+(ω2−k2).(6.5.24) The inverse Laplace transform yields the solution in three distinct cases: (i)ω>k, (ii)ω=k, and (iii) ω<k. The ﬁnal closed form solutions are given by y(t)=c 0e−kt⎪parenleftbigg cosσt+k σsinσt⎪parenrightbigg +c1 σe−ktsinσt +1 σ⎪integraldisplayt 0f(t−τ)e−kτsinστdτ, ω >k (6.5.25) y(t)=c0(1 +kt)e−kt+c1te−kt+⎪integraldisplayt 0f(t−τ)τc−kτdτ, ω=k(6.5.26) y(t)=c0e−kt⎪parenleftbigg coshμt+k μsinhμt⎪parenrightbigg +c1 μe−ktsinhμt +1 μ⎪integraldisplayt 0f(t−τ)e−kτsinhμτ dτ, ω < k, (6.5.27) where σ2=ω2−k2andμ2=k2−ω2. As expected, all solutions exhibit an exponential decay as t→∞. 6.6 Green’s Functions of Fractional Diﬀerential Equations (a) We consider the linear system governed by the fractional order diﬀerentialequation with constant coeﬃcien ts and zero initial conditions 0Dα ty(t)=f(t). (6.6.1) Application of the Laplace transform gives y(s)=s−α f(s). (6.6.2) So, the solution given by y(t)=L−1⎪braceleftbig s−α f(s)⎪bracerightbig =⎪integraldisplayt 0G(t−τ)f(τ)dτ, (6.6.3) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 299 where G(t)=L−1⎪braceleftbigg1 sα⎪bracerightbigg =tα−1 Γ(α)(6.6.4) is called the Green’s function of (6.6.1). (b) We next consider a more general fractional order diﬀerential equation with constant coeﬃcients and zero i nitial conditions in the form 0Dα ty(t)+ω2y(t)=f(t). (6.6.5) Application of the Laplace transform to (6.6.5) gives the solution y(t)=L−1⎪braceleftbigg1 sα+ω2· f(s)⎪bracerightbigg =⎪integraldisplayt 0G(t−τ)f(τ)dτ, (6.6.6) where the Green’s function G(t)i sg i v e nb y G(t)=L−1⎪braceleftbigg1 sα+ω2⎪bracerightbigg =tα−1Eα,α⎪parenleftbig −ω2tα⎪parenrightbig . (6.6.7) Similarly, the Green’s function for the n-term fractional- order diﬀerential equation with constant coeﬃcients an d zero initial conditions in the form [anDαn+an−1Dαn−1+...+a1Dα1+a0Dα0]y(t)=f(t),(6.6.8) can be obtained from the inverse Laplace transform G(t)=L−1⎡ ⎣⎪parenleftBiggn⎪summationdisplay k=oaksαn⎪parenrightBigg−1⎤⎦. (6.6.9) 6.7 Fractional Partial Diﬀerential Equations (a) The Fractional Diﬀusion Equation is given by ∂αu ∂tα=κ∂2u ∂x2,x∈R, t > 0, (6.7.1) with the boundary and initial conditions u(x, t)→0a s|x|→∞ , (6.7.2)⎪bracketleftbig 0Dα−1 tu(x, t)⎪bracketrightbig t=0=f(x)f o r x∈R, (6.7.3) where κis a diﬀusivity constant and 0 <α≤1. © 2007 by Taylor & Francis Group, LLC 300 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Application of the Fourier transform to (6.7.1) with respect xand using the boundary condition (6.7.2) yields Dα t˜u(k,t)=−κk2˜u, (6.7.4)⎪bracketleftbig 0Dα−1 t˜u(x, t)⎪bracketrightbig t=0=˜f(k), (6.7.5) where ˜ u(k,t) is the Fourier transform of u(x, t) is deﬁned by ˜u(k,t)=1 √ 2π⎪integraldisplay∞ −∞e−ikxu(x, t)dx. (6.7.6) The Laplace transform solution of (6.7.4) and (6.7.5) is ˜¯u(k,s)=˜f(k) (sα+κk2). (6.7.7) The inverse Laplace transform of (6.7.7) gives ˜u(k,t)=˜f(k)tα−1Eα,α⎪parenleftbig −κk2tα⎪parenrightbig , (6.7.8) where Eα,βis the Mittag-Leﬄer type function deﬁned by (6.5.6). Finally, the inverse Fourier transform leads to the solution of the diﬀusion problem as u(x, t)=⎪integraldisplay∞ −∞G(x−ξ,t)f(ξ)dξ, (6.7.9) where G(x, t)=1 π⎪integraldisplay∞ −∞tα−1Eα,α⎪parenleftbig −κk2tα⎪parenrightbig coskxdk. (6.7.10) This integral for G(x, t) can be evaluated by using the Laplace transform of G(x, t)a s G(x, s)=1 π⎪integraldisplay∞ −∞coskxdk sα+κk2=1 √ 4κs−α/2exp⎪parenleftbigg −|x| √ κsα/2⎪parenrightbigg ,(6.7.11) whence the inverse Laplace trans form gives the explicit solution G(x, t)=1 √ 4κtα 2−1W⎪parenleftBig −ξ,−α 2,α 2⎪parenrightBig , (6.7.12) where ξ=|x| √ κtα/2,a n d W(z,α,β ) is the Wright function deﬁned by W(z,α,β )=∞⎪summationdisplay n=0zn n!Γ(αn+β). (6.7.13) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 301 -3 -2 -1 0 1 2 300.40.81.21.622.4 xt = 1/60 t = 1/16 t = 1/4 Figure 6.2 Graphs of (6.7.14) for diﬀerent κt. It is important to note that when α= 1, the initial value problem (6.7.1)– (6.7.3) reduces to the classical diﬀusion problem and solution (6.7.9) reduces to the classical fundamental solution because G(x, t)=1 √ 4κtW⎪parenleftbigg −x √ κt,−1 2,1 2⎪parenrightbigg =1 √ 4πκtexp⎪parenleftbigg −x2 4κt⎪parenrightbigg .(6.7.14) It is noted that the order αof the derivative with respect to time tin equa- tion (6.7.1) can be of arbitrary real order including α=2 s o t h a t i t m a y be called the fractional diﬀusion-wave equation .F o rα= 2, it becomes the classi- cal wave equation. The equation (6.7.1) with 1 <α≤2 will be solved next in some detail. (b) The Nonhomogeneous Fractional Wave Equation is given by ∂αu ∂tα−c2∂2u ∂x2=q(x, t),x∈R, t > 0 (6.7.15) with the initial condition u(x,0) =f(x),ut(x,0) =g(x),x∈R, (6.7.16) where cis a constant and 1 <α≤2. Application of the joint Laplace transform with respect to tand Fourier transform with respect to xgives the transform solution ˜¯u(k,s)= f(k)sα−1 sα+c2k2+˜g(k)sα−2 sα+c2k2+˜¯q(k,s) sα+c2k2, (6.7.17) © 2007 by Taylor & Francis Group, LLC 302 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where kis the Fourier transform variable and sis the Laplace transform variable. The inverse Laplace transform produces the following result ˜u(k,t)=˜f(k)L−1⎪braceleftbiggsα−1 sα+c2k2⎪bracerightbigg +˜g(k)L−1⎪braceleftbiggsα−2 sα+c2k2⎪bracerightbigg +L−1⎪braceleftbigg˜¯q(k,s) sα+c2k2⎪bracerightbigg (6.7.18) which is, by (6.5.7), =˜f(k)Eα,1⎪parenleftbig −c2k2tα⎪parenrightbig +˜g(k)tEα,2⎪parenleftbig −c2k2tα⎪parenrightbig +⎪integraldisplayt 0˜q(k,t−τ)τα−1Eα,α⎪parenleftbig −c2k2τα⎪parenrightbig dτ. (6.7.19) Finally, the inverse Fourier transform gives the formal solution u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −c2k2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplay∞ −∞t˜g(k)Eα,2⎪parenleftbig −c2k2τα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α⎪parenleftbig −c2k2tα⎪parenrightbig eikxdk.(6.7.20) In particular, when α= 2, the fractional wave equation (6.7.15) reduces to the classical wave equation. In this particular case, we obtain E2,1⎪parenleftbig −c2k2tα⎪parenrightbig =c o s h( ickt)=cos( ckt), (6.7.21) tE2,2⎪parenleftbig −c2k2tα⎪parenrightbig =t·sinh (ickt) ickt=1 cksin(ckt). (6.7.22) Consequently, solution (6.7.20) reduces to the classical solution (see Debnath, 2005) of the wave equation (6.7.15) with α=2i nt h ef o r m u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)cos(ckt)eikxdk+1 √ 2π⎪integraldisplay∞ −∞˜g(k)sin(ckt) ckeikxdk +1 √ 2πc⎪integraldisplayt 0dτ⎪integraldisplay∞ −∞˜q(k,τ)sinck(t−τ) keikxdk (6.7.23) =1 2[f(x−ct)+f(x+ct)] +1 2c⎪integraldisplayx+ct x−ctg(ξ)dξ+1 2c⎪integraldisplayt 0dτ⎪integraldisplayx+c(t−τ) x−c(t−τ)q(ξ,τ)dξ. (6.7.24) We now derive the solution of the inhomogeneous fractional diﬀusion e- quation (6.7.15) with c2=κandg(x)≡0. In this case, the joint transform © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 303 solutions (6.7.17) becomes ˜¯u(k,s)=˜f(k)sα−1 (sα+κk2)+˜¯q(k,s) (sα+κk2), (6.7.25) which is inverted by (6.5.7) to obtain ˜u(k,t)=˜f(k)Eα,1⎪parenleftbig −κk2tα⎪parenrightbig +⎪integraldisplayt 0(t−τ)α−1Eα,α⎪parenleftbig −κk2(t−τ)α⎪parenrightbig ˜q(k,τ)dτ. (6.7.26) Finally, the inverse Fourier transform gives the exact solution for the temper- ature distribution u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −κk2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0dτ⎪integraldisplay∞ −∞(t−τ)α−1Eα,α⎪parenleftbig −κk2(t−τ)α⎪parenrightbig ×˜q(k,τ)eikxdk. (6.7.27) Application of the convolution theorem of the Fourier transform gives the ﬁnal solution in the form u(x, t)=⎪integraldisplay∞ −∞G1(x−ξ,t)f(ξ)dξ +⎪integraldisplayt 0(t−τ)α−1dτ⎪integraldisplay∞ −∞G2(x−ξ,t−τ)q(ξ,τ)dξ,(6.7.28) where G1(x, t)=1 2π⎪integraldisplay∞ −∞eikxEα,1⎪parenleftbig −κk2tα⎪parenrightbig dk, (6.7.29) and G2(x, t)=1 2π⎪integraldisplay∞ −∞eikxEα,α⎪parenleftbig −κk2tα⎪parenrightbig dk. (6.7.30) In particular, when α= 1, the classical solution of the nonhomogeneous diﬀusion equation is obtained in the form u(x, t)=⎪integraldisplay∞ −∞G1(x−ξ,t)f(ξ)dξ +⎪integraldisplayt 0dτ⎪integraldisplay∞ −∞G2(x−ξ,t−τ)·q(ξ,τ)dξ, (6.7.31) where G1(x, t)=G2(x, t)=1 √ 4πκtexp⎪parenleftbigg −x2 4κt⎪parenrightbigg . (6.7.32) © 2007 by Taylor & Francis Group, LLC 304 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (c) We consider the fractional-order diﬀusion equation in a semi-inﬁnite medi- umx>0, when the boundary is kept at a temperature u0f(t) and the initial temperature is zero in the whole medium. Thus, the initial-boundary valueproblem is described by the equation ∂ αu ∂tα=κ∂2u ∂x2,0<x< ∞,t > 0, (6.7.33) with u(x, t=0 )=0 ,x > 0, (6.7.34) u(x=0,t)=u0f(t),t > 0a n d u(x, t)→0a sx→∞.(6.7.35) Application of the Laplace transform with respect to tgives d2 u dx2+⎪parenleftbiggsα κ⎪parenrightbigg u(x, s),x > 0, (6.7.36) u(x=0,s)=u0 f(s), u(x, s)→0a s x→∞. (6.7.37) Evidently, the solution of this transformed boundary value problem is u(x, s)=u0 f(s)exp( −ax),a =(sα/κ)1 2. (6.7.38) Thus, the solution is given by u(x, t)=u0⎪integraldisplayt 0f(t−τ)g(x, τ)dτ=u0f(t)∗g(x, t),(6.7.39) where g(x, t)=L−1{exp(−ax)}. In this case, α=1a n d f(t) = 1, solution (6.7.38) becomes u(x, s)=⎪parenleftBigu0 s⎪parenrightBig exp⎪parenleftbigg −x⎪radicalbigg s κ⎪parenrightbigg , (6.7.40) which yields the classical solution in t erms of the complementary error func- tion u(x, t)=u0erfc⎪parenleftbiggx 2√ κt⎪parenrightbigg . (6.7.41) In the classical case ( α= 1) and the more general solution is given by u(x, t)=u0⎪integraldisplayt 0f(t−τ)g(x, τ)dτ=u0f(t)∗g(x, t),(6.7.42) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 305 where g(x, t)=L−1⎪braceleftbigg exp⎪parenleftbigg −x⎪radicalbigg s κ⎪parenrightbigg⎪bracerightbigg =x 2√ πκt3exp⎪parenleftbigg −x2 4κt⎪parenrightbigg .(6.7.43) (d)The Fractional Stokes and Rayleigh Problems in Fluid Dynamics The classical Stokes problem deals with the unsteady boundary layer ﬂows induced in a semi-inﬁnite viscous ﬂuid bounded by an inﬁnite horizontal disk atz= 0 due to non-torsional oscillations of the disk in its own plane with a given frequency ω.W h e n ω= 0, the Stokes problem reduces to the classical Rayleigh problem where the unsteady boundary layer ﬂow is generated in theﬂuid from rest by moving the disk impulsively in its own plane with constant velocity U. We consider the unsteady fractional boundary layer equation (see Debnath (2003)) for the ﬂuid velocity u(z,t) ∂ αu ∂tα=ν∂2u ∂t2,0<z< ∞,t > 0, (6.7.44) with the given boundary and initial conditions u(0,t)=Uf(t),u(z,t)→0a s z→∞,t > 0,(6.7.45) u(z,0)= 0 for all z>0, (6.7.46) where νis the kinematic viscosity, Uis a constant velocity and f(t)i sa n arbitrary function of time t. Application of the Laplace transform with respect to tgives sα u(z,s)=νd2 u dz2,0<z< ∞, (6.7.47) u(0,s)=U f(s), u(z,s)→0a s z→∞. (6.7.48) Using the Fourier sine transform with respect to zyields Us(k,s)=⎪parenleftBigg⎪radicalbigg 2 πνU⎪parenrightBigg k f(s) (sα+νk2). (6.7.49) The inverse Fourier sine transform of (6.7.49) leads to the solution u(z,s)=⎪parenleftbigg2 πνU⎪parenrightbigg f(s)⎪integraldisplay∞ 0ksinkz (sα+νk2)dk, (6.7.50) and the inverse Laplace transform gives the solution for the velocity ﬁeld u(z,t)=⎪parenleftbigg2 πνU⎪parenrightbigg⎪integraldisplay∞ 0ksinkz dk ×⎪integraldisplayt 0f(t−τ)τα−1Eα,α⎪parenleftbig −νk2τα⎪parenrightbig dτ. (6.7.51) © 2007 by Taylor & Francis Group, LLC 306 INTEGRAL TRANSFORMS and THEIR APPLICATIONS When f(t)=e x p( iωt), the solution of the fractional Stokes problem is u(z,t)=⎪parenleftbigg2νU π⎪parenrightbigg eiωt⎪integraldisplay∞ 0ksinkzdk ×⎪integraldisplayt 0e−iωττα−1Eα,α⎪parenleftbig −νk2τα⎪parenrightbig dτ. (6.7.52) When α= 1, solution (6.7.52) reduces to the classical Stokes solution in the form u(z,t)=⎪parenleftbigg2νU π⎪parenrightbigg⎪integraldisplay∞ 0⎪parenleftBig 1−e−νtk2⎪parenrightBigksinkz (iω+νk2)dk. (6.7.53) For the fractional Rayleigh problem, f(t) = 1 and the solution follows from (6.7.51) in the form u(z,t)=⎪parenleftbigg2νU π⎪parenrightbigg⎪integraldisplay∞ 0ksinkzdk⎪integraldisplayt 0τα−1Eα,α⎪parenleftbig −νk2τα⎪parenrightbig dτ.(6.7.54) This solution reduces to the classical Rayleigh solution when α=1a s u(z,t)=⎪parenleftbigg2νU π⎪parenrightbigg⎪integraldisplay∞ 0ksinkzdk⎪integraldisplayt 0E1,1⎪parenleftbig −ντk2⎪parenrightbig dτ =⎪parenleftbigg2νU π⎪parenrightbigg⎪integraldisplay∞ 0ksinkzdk⎪integraldisplayt 0exp⎪parenleftbig −ντk2⎪parenrightbig dτ =⎪parenleftbigg2U π⎪parenrightbigg⎪integraldisplay∞ 0⎪parenleftBig 1−e−νtk2⎪parenrightBigsinkz kdk, which is, by (2.15.10), =⎪parenleftbigg2U π⎪parenrightbigg⎪bracketleftbiggπ 2−π 2erf⎪parenleftbiggz 2√ νt⎪parenrightbigg⎪bracketrightbigg =Uerfc⎪parenleftbiggz 2√ νt⎪parenrightbigg , (6.7.55) where erf c(x) is the complimentary error function. (e)The Fractional Unsteady Couette Flow We consider the unsteady viscous ﬂuid ﬂow between the plate at z=0a tr e s t and the plate z=hin motion parallel to itself with a variable velocity U(t) in the x-direction. The ﬂuid velocity u(z,t) satisﬁes the fractional equation of motion (see Debnath(2003)) ∂αu ∂tα=P(t)+ν∂2u ∂t2,0≤z≤h, t > 0, (6.7.56) with the boundary and initial conditions u(0,t)=0an d u(h,t)=U(t),t > 0, (6.7.57) u(z,t)=0 at t≤0f o r0 ≤z≤h, (6.7.58) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 307 where −1 ρpx=P(t)a n d νis the kinematic viscosity of the ﬂuid. We apply the joint Laplace transform with respect to tand the ﬁnite Fourier sine transform with respect to zdeﬁned by ¯˜us(n,s)=⎪integraldisplay∞ 0e−stdt⎪integraldisplayh 0u(z,t)sin⎪parenleftBignπz h⎪parenrightBig dz, (6.7.59) to the system (6.7.56) - (6.7.58) so that the transform solution is ¯˜us(n,s)= P(s)1 a[1−(−1)n] (sα+νa2)+νa(−1)n+1 U(s) (sα+νa2), (6.7.60) where a=⎪parenleftbignπ h⎪parenrightbig ,nis the ﬁnite Fourier sine transform variable. Thus, the inverse Laplace transform yields ˜us(n,t)=1 a[1−(−1)n]⎪integraldisplayt 0P(t−τ)τα−1Eα,α⎪parenleftbig −νa2τα⎪parenrightbig dτ +νa(−1)n+1⎪integraldisplayt 0U(t−τ)τα−1Eα,α⎪parenleftbig −νa2τα⎪parenrightbig dτ.(6.7.61) Finally, the inverse ﬁnite Fourier sine transform leads to the solution u(z,t)=2 h∞⎪summationdisplay n=1˜us(n,t)sin⎪parenleftBignπz h⎪parenrightBig . (6.7.62) If, in particular, P(t)= c o n s t a n t a n d U(t) =constant, then solution (6.7.62) reduces to the solution of the generalized Couette ﬂow. (f)Fractional Axisymmetric Wave-Diﬀusion Equation The fractional axisymmetric equation in an inﬁnite domain ∂αu ∂tα=a⎪parenleftbigg∂2u ∂r2+1 r∂u ∂r⎪parenrightbigg ,0<r< ∞,t > 0, (6.7.63) is called the diﬀusion orwave equation according as a=κora=c2. For the fractional diﬀusion equation, we prescribe the initial condition u(r,0) =f(r),0<r<∞. (6.7.64) Application of the joint Laplace transform with respect to tand Hankel trans- form (7.4.3) of zero order (s ee Chapter 7) with respect to rto (6.7.63)–(6.7.64) gives the transform solution ¯˜u(k,s)=sα−1˜f(k) (sα+κk2), (6.7.65) where k,sare the Hankel and Laplace transf orm variables, respectively. © 2007 by Taylor & Francis Group, LLC 308 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The joint inverse transform leads to the solution u(r, t)=⎪integraldisplay∞ 0rJ0(kr)˜f(k)Eα,1⎪parenleftbig −κk2tα⎪parenrightbig dk, (6.7.66) where J0(kr) is the Bessel function of the ﬁrst kind of order zero and ˜f(k)i s the Hankel transform of f(r). On the other hand, we can solve the wave equation (6.7.63) with a=c2and the initial conditions u(r,0) =f(r),u t(r,0) =g(r)f o r 0 <r< ∞, (6.7.67) provided the Hankel transforms of f(r)a n d g(r)e x i s t . Application of the joint Laplace and Hankel transform leads to the trans- form solution ¯˜u(k,s)=sα−1˜f(k) (sα+c2k2)+sα−2˜g(k) (sα+c2k2). (6.7.68) The joint inverse transformation gives the solution u(r, t)=⎪integraldisplay∞ 0kJ0(k,r)˜f(k)Eα,1⎪parenleftbig −c2k2tα⎪parenrightbig dk +⎪integraldisplay∞ 0kJ0(k,r)˜g(k)tEα,2⎪parenleftbig −c2k2tα⎪parenrightbig dk. (6.7.69) When α= 2, solution (6.7.69) is in total agreement with that of the classical axisymmetric wave equation (see Example 7.4.1 in Chapter 7). In a ﬁnite domain 0 ≤r≤a, the fractional diﬀusion equation (6.7.63) can be solved by using the joint Laplace and ﬁnite Hankel transform with the boundary and initial data u(r, t)=f(t)o nr=a, t > 0, (6.7.70) u(r,0) = 0 for all rin (0,a). (6.7.71) Application of the joint Laplace and ﬁni te Hankel transform of zero order (see Chapter 13) yields the solution u(r, t)=2 a2∞⎪summationdisplay i=1˜u(ki,t)J0(rki) J2 1(aki), (6.7.72) where ˜u(ki,t)=(aκ k i)J1(aki)⎪integraldisplayt 0f(t−τ)τα−1Eα,α⎪parenleftbig −κk2 iτα⎪parenrightbig dτ.(6.7.73) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 309 Similarly, the fractional wave equation (6.7.63) with a=c2in a ﬁnite domain 0≤r≤awith the boundary and initial data u(r, t)=0 o n r=a, t > 0, (6.7.74) u(r,0) =f(r)a n d ut(r,0)=g(r)f o r 0 <r<a , (6.7.75) can be solved by means of the joint Laplace and the ﬁnite Hankel transform (13.2.8). The solution of this problem is u(r, t)=2 a2∞⎪summationdisplay i=1˜u(ki,t)J0(rki) J2 1(aki), (6.7.76) where ˜u(ki,t)=˜f(ki)Eα,1⎪parenleftbig −c2ki2tα⎪parenrightbig +˜g(ki)Eα,2⎪parenleftbig −c2k2 itα⎪parenrightbig .(6.7.77) (g)The Fractional Schr¨ odinger Equation in Quantum Mechanics The one-dimensional fractional Schr¨ odinger equation (see Debnath(2003)) for a free particle of mass mis i/planckover2pi1∂αψ ∂tα=−/planckover2pi12 2m∂2ψ ∂x2,−∞<x< ∞,t > 0, (6.7.78) ψ(x,0) =ψ0(x), −∞<x< ∞, (6.7.79) ψ(x, t)→0a s |x|→∞ , (6.7.80) where ψ(x, t) is the wave function, h=2π/planckover2pi1=6.625×10−27ergsec = 4 .14× 10−21MeV sec is the Planck constant and ψ0(x) is an arbitrary function. Application of the joint Laplace and Fourier transform to (6.7.78)–(6.7.80) gives the solution in the transform space in the form ˜ ψ(k,s)=sα−1˜ψ0(k) sα+ak2,⎪parenleftbigg a=i/planckover2pi1 2m⎪parenrightbigg , (6.7.81) where k,srepresent the Fourier and the Laplace transforms variables. The use of the joint inverse transform yields the solution ψ(x, t)=1 √ 2π⎪integraldisplay∞ −∞eikx˜ψ0(k)Eα,1⎪parenleftbig −ak2tα⎪parenrightbig dk (6.7.82) =F−1⎪braceleftBig ˜ψ0(k)Eα,1⎪parenleftbig −ak2tα⎪parenrightbig⎪bracerightBig , (6.7.83) which is, by the convolution theorem of the Fourier transform, =⎪integraldisplay∞ −∞G(x−ξ, t)ψ0(ξ)dξ. (6.7.84) © 2007 by Taylor & Francis Group, LLC 310 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where G(x, t)=1 √ 2πF−1⎪braceleftbig Eα,1⎪parenleftbig −ak2tα⎪parenrightbig⎪bracerightbig =1 2π⎪integraldisplay∞ −∞eikxEα,1⎪parenleftbig −ak2tα⎪parenrightbig dk. (6.7.85) When α= 1, the solution of the Schr¨ odinger equation (6.7.78) becomes ψ(x, t)=⎪integraldisplay∞ −∞G(x−ξ, t)ψ0(ξ)dξ, (6.7.86) where the Green’s function G(x, t)i sg i v e nb y G(x, t)=1 2π⎪integraldisplay∞ −∞eikxE1,1⎪parenleftbig −ak2t⎪parenrightbig dk =1 2π⎪integraldisplay∞ −∞exp⎪parenleftbig ikx−atk2⎪parenrightbig =1 √ 4πatexp⎪parenleftbigg −x2 4at⎪parenrightbigg .(6.7.87) (h)Linear Inhomogeneous Fractional Evolution Equation The fairly general linear inhomogeneous fractional evolution equation is given by (see Debnath and Bhatta (2004)) ∂αu ∂tα+c∂u ∂x−ν∂2u ∂x2+b3∂3u ∂x3+....+bn∂nu ∂xn=q(x, t),x∈R, t > 0,(6.7.88) where c,ν,a n d b3, ..., b nare constants and 0 <α≤1. We solve this fractional evolution equation with the following initial and boundary conditions u(x,0) =f(x) x∈R (6.7.89) u(x, t)→0as|x|→∞ ,t >0. (6.7.90) Application of the joint Laplace transform with respect to tand Fourier transform with respect to xgives the transform solution ˜¯u(k,s)= f(k)sα−1 sα+a2+˜¯q(k,s) sα+a2, (6.7.91) where kis the Fourier transform variable sis the Laplace transform variable, anda2is given by a2=ikc+k2ν+(ik)3b3+(ik)4b4+....+(ik)nbn. (6.7.92) We use the following inverse Laplace transform formula L−1⎪braceleftBigg m!sα−β (sα¯+a)m+1⎪bracerightBigg =tαm+β−1E(m) α,β(+ atα), (6.7.93) © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 311 where Eα,β(z)a n d E(m) α,β(z) are deﬁned by (6.5.6) and (6.5.8) respectively. The inverse Laplace transform yields the following result ˜u(k,t)=˜f(k)L−1⎪braceleftbiggsα−1 sα+a2⎪bracerightbigg +L−1⎪braceleftbigg˜¯q(k,s) sα+a2⎪bracerightbigg ,(6.7.94) which can be written as ˜u(k,t)=˜f(k)Eα,1⎪parenleftbig −a2tα⎪parenrightbig +⎪integraldisplayt 0˜q(k,t−τ)τα−1Eα,α⎪parenleftbig −a2τα⎪parenrightbig dτ.(6.7.95) Finally, the inverse Fourier transform gives the formal solution u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −a2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α⎪parenleftbig −a2tα⎪parenrightbig eikxdk,(6.7.96) where a2is given by (6.7.92). The solution in equation (6.7.96) is fairly general and contains solutions of many special evolution equations including the frac- tional wave equation, Korteweg de Vries (KdV) equation, and KdV-Burgers equation (see Debnath and Bhatta, 2004). (i)Linear Inhomogeneous Fractional Telegraph Equation Here we solve the one-dimensional linear inhomogeneous fractional telegraph equation given by ∂αu ∂tα−c2∂2u ∂x2+a∂u ∂t+bu=q(x, t),x∈R, t > 0 (6.7.97) where a,b,a n d care constants and 1 <α≤2. We solve this fractional evolution equation with the following initial and boundary conditions u(x, t)=f(x),∂u(x, t) ∂t=g(x)at t=0,x∈R(6.7.98) u(x, t)→0as|x|→∞ ,t >0 (6.7.99) Applying the joint transform of Laplace and Fourier and taking the inverseLaplace tansform, we have ˜u(k,t)=˜f(k)E α,1⎪parenleftbig −λ2tα⎪parenrightbig +t˜g(k)Eα,1⎪parenleftbig −λ2tα⎪parenrightbig +⎪integraldisplayt 0˜q(k,t−τ)τα−1Eα,α⎪parenleftbig −λ2τα⎪parenrightbig dτ, (6.7.100) where λ2=c2k2+aik+b. (6.7.101) © 2007 by Taylor & Francis Group, LLC 312 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Application of the inverse Fourier transform to equation (6.7.100) yields the solution u(x, t)a s u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −λ2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplay∞ −∞t˜g(k)Eα,2⎪parenleftbig −λ2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α⎪parenleftbig −λ2tα⎪parenrightbig eikxdk.(6.7.102) In the limit as a→0, the telegraph equation re duces to the Klein-Gordon equation and its solution is in perfect agreement with each other (see Debnath and Bhatta, 2004). 6.8 Exercises 1. Consider linear inhomogeneous f ractional diﬀer ential equation ∂αu ∂tα+c∂u ∂x=q(x, t),x∈R, t > 0 where cis a constant, 0 <α≤1, and q, source term, is a function of x andt. Assuming the initial and boundary conditions u(x,0) =f(x),x ∈R u(x, t)→0 as|x|→∞ ,t>0 show that the solution u(x, t)i sg i v e nb y u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1(−icktα)eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α(−icktα)eikxdk. In particular, when α= 1, the solution becomes u(x, t)=f(x−ct)+1 √ 2π⎪integraldisplayt 0dτ⎪integraldisplay∞ −∞˜q(k,τ)eik{x−c(t−τ)}dk =f(x−ct)+⎪integraldisplayt 0q{x−c(t−τ),τ}dτ. © 2007 by Taylor & Francis Group, LLC Applications of Integral Transforms to Fractional Diﬀerential and Integral Equations 313 2. Consider linear inhomogeneous fractional Burgers equation ∂αu ∂tα+c∂u ∂x−ν∂2u ∂x2=q(x, t),x∈R, t > 0, where cis a constant, 0 <α≤1,νis the kinematic viscosity and q(x, t) is a source term. Assuming the initial and boundary conditions u(x,0) =f(x),x ∈R, u(x, t)→0as|x|→∞ ,t >0, show that the solution u(x, t)i sg i v e nb y u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −a2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α⎪parenleftbig −a2tα⎪parenrightbig eikxdk, where a2=⎪parenleftbig ick+νk2⎪parenrightbig . 3. Consider linear inhomogeneous fractional KdV equation ∂αu ∂tα+c∂u ∂x+b∂3u ∂x3=q(x, t),x∈R, t > 0 where bandcare constants, 0 <α≤1. Assuming the initial and boundary conditions u(x,0) =f(x),x ∈R, u(x, t)→0as|x|→∞ ,t >0, show that the solution u(x, t)i sg i v e nb y u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −a2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α⎪parenleftbig −a2tα⎪parenrightbig eikxdk, where a2=⎪parenleftbig ick−ik3b⎪parenrightbig . 4. Consider linear inhomogeneous fractional KdV-Burgers equation ∂αu ∂tα+c∂u ∂x−ν∂2u ∂x2+b∂3u ∂x3=q(x, t),x∈R, t > 0, © 2007 by Taylor & Francis Group, LLC 314 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where b,candνare constants, 0 <α≤1. Assuming the initial and boundary conditions u(x,0) =f(x),x ∈R, u(x, t)→0as|x|→∞ ,t >0, show that the solution u(x, t)i sg i v e nb y u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −a2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α⎪parenleftbig −a2tα⎪parenrightbig eikxdk, where a2=⎪parenleftbig ick+k2ν−ik3b⎪parenrightbig . 5. Consider linear inhomogeneous fractional Klein-Gordon equation ∂αu ∂tα−c2∂2u ∂x2+d2u=q(x, t),x∈R, t > 0, where canddare constants, 1 <α≤2. Assuming the initial and boundary conditions u(x, t)=f(x),∂u(x, t) ∂t=g(x)at t=0,x∈R, u(x, t)→0,a s |x|→∞ ,t >0, show that the solution u(x, t)i sg i v e nb y u(x, t)=1 √ 2π⎪integraldisplay∞ −∞˜f(k)Eα,1⎪parenleftbig −a2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplay∞ −∞t˜g(k)Eα,2⎪parenleftbig −a2tα⎪parenrightbig eikxdk +1 √ 2π⎪integraldisplayt 0τα−1dτ⎪integraldisplay∞ −∞˜q(k,t−τ)Eα,α⎪parenleftbig −a2tα⎪parenrightbig eikxdk, where a2=⎪parenleftbig c2k2+d2⎪parenrightbig . © 2007 by Taylor & Francis Group, LLC © 2007 by Taylor & Francis Group, LLC 7 Hankel Transforms and Their Applications “In most sciences one generation tears down what another has built, and what one has established, another undoes. In mathemat- ics alone each generation adds a n ew storey to the old structure.” Hermann Hankel “I have always regarded mathematics as an object of amusement rather than of ambition, and I can assure you that I enjoy the works of others much more than my own.” Joseph-Louis Lagrange 7.1 Introduction Hermann Hankel (1839-1873), a German mathematician, is remembered forhis numerous contributions to mathematical analysis including the Hankel transformation, which occurs in the study of functions which depend only on the distance from the origin. He also studied functions, now named Hankelfunctions or Bessel functions of the third kind. The Hankel transform involv- ing Bessel functions as the kernel arises naturally in axisymmetric problems formulated in cylindrical polar coordinates. This chapter deals with the deﬁni- tion and basic operational properties of the Hankel transform. A large number of axisymmetric problems in cylindrical polar coordinates are solved with theaid of the Hankel transform. The use of the joint Laplace and Hankel trans- forms is illustrated by several examples of applications to partial diﬀerential equations. 315 © 2007 by Taylor & Francis Group, LLC 316 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 7.2 The Hankel Transform and Examples We introduce the deﬁnition of the Hankel transform from the two-dimensional Fourier transform and its inverse given by F{f(x, y)}=F(k,l)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp{−i(κ·r)}f(x, y)dxdy, (7.2.1) F−1{F(k,l)}=f(x, y)=1 2π∞⎪integraldisplay −∞∞⎪integraldisplay −∞exp{i(κ·r)}F(k,l)dk dl, (7.2.2) where r=(x, y)a n d κ=(k,l). Introducing polar coordinates ( x, y)=r(cosθ, sinθ)a n d( k,l)=κ(cosφ,sinφ), we ﬁnd κ·r=κrcos(θ−φ)a n dt h e n F(κ,φ)=1 2π∞⎪integraldisplay 0rd r2π⎪integraldisplay 0exp[−iκrcos(θ−φ)]f(r, θ)dθ. (7.2.3) We next assume f(r, θ)=e x p ( inθ)f(r), which is not a very severe restric- tion, and make a change of variable θ−φ=α−π 2to reduce (7.2.3) to the form F(κ,φ)=1 2π∞⎪integraldisplay 0rf(r)dr ×2π+φ0⎪integraldisplay φ0exp⎪bracketleftBig in⎪parenleftBig φ−π 2⎪parenrightBig +i(nα−κrsinα)⎪bracketrightBig dα, (7.2.4) where φ0=⎪parenleftBigπ 2−φ⎪parenrightBig . Using the integral representation of the Bessel function of order n Jn(κr)=1 2π2π+φ0⎪integraldisplay φ0exp[i(nα−κrsinα)]dα (7.2.5) integral (7.2.4) becomes F(κ,φ)=e x p⎪bracketleftBig in⎪parenleftBig φ−π 2⎪parenrightBig⎪bracketrightBig∞⎪integraldisplay 0rJn(κr)f(r)dr (7.2.6) =e x p⎪bracketleftBig in⎪parenleftBig φ−π 2⎪parenrightBig⎪bracketrightBig ˜fn(κ), (7.2.7) © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 317 where ˜fn(κ) is called the Hankel transform off(r) and is deﬁned formally by Hn{f(r)}=˜fn(κ)=∞⎪integraldisplay 0rJn(κr)f(r)dr. (7.2.8) Similarly, in terms of the polar variables with the assumption f(x, y)= f(r, θ)=einθf(r) with (7.2.7), the inverse Fourier transform (7.2.2) becomes einθf(r)=1 2π∞⎪integraldisplay 0κd κ2π⎪integraldisplay 0exp[iκrcos(θ−φ)]F(κ,φ)dφ =1 2π∞⎪integraldisplay 0κ˜fn(κ)dκ2π⎪integraldisplay 0exp⎪bracketleftBig in⎪parenleftBig φ−π 2⎪parenrightBig +iκrcos(θ−φ)⎪bracketrightBig dφ, which is, by the change of variables θ−φ=−⎪parenleftBig α+π 2⎪parenrightBig andθ0=−⎪parenleftBig θ+π 2⎪parenrightBig , =1 2π∞⎪integraldisplay 0κ˜fn(κ)dκ2π+θ0⎪integraldisplay θ0exp[in(θ+α)−iκrsinα]dα =einθ∞⎪integraldisplay 0κJn(κr)˜fn(κ)dκ, by (7.2.5). (7.2.9) Thus, the inverse Hankel transform is deﬁned by H−1 n⎪bracketleftBig ˜fn(κ)⎪bracketrightBig =f(r)=∞⎪integraldisplay 0κJn(κr)˜fn(κ)dκ. (7.2.10) Instead of ˜fn(κ), we often simply write ˜f(κ) for the Hankel transform speci- fying the order. Integrals (7.2.8) and (7.2.10) exist for certain large classes of functions, which usually occur in physical applications. Alternatively, the famous Hankel integral formula (Watson, 1944, p. 453) f(r)=∞⎪integraldisplay 0κJn(κr)dκ∞⎪integraldisplay 0pJn(κp)f(p)dp, (7.2.11) can be used to deﬁne the Hankel transform (7.2.8) and its inverse (7.2.10). In particular, the Hankel transforms of the zero order ( n= 0) and of order one (n= 1) are often useful for the solution of problems involving Laplace’s equation in an axisymmetric cylindrical geometry. Example 7.2.1 Obtain the zero-order Hankel transforms of © 2007 by Taylor & Francis Group, LLC 318 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (a)r−1exp(−ar), (b)δ(r) r, (c)H(a−r), where H(r) is the Heaviside unit step function. We have (a)˜f(κ)=H0⎪braceleftbigg1 rexp(−ar)⎪bracerightbigg =∞⎪integraldisplay 0exp(−ar)J0(κr)dr=1 √ κ2+a2. (b)˜f(κ)=H0⎪braceleftbiggδ(r) r⎪bracerightbigg =∞⎪integraldisplay 0δ(r)J0(κr)dr=1. (c)˜f(κ)=H0{H(a−r)}=a⎪integraldisplay 0rJ0(κr)dr=1 κ2aκ⎪integraldisplay 0pJ0(p)dp =1 κ2[pJ1(p)]aκ 0=a κJ1(aκ). Example 7.2.2 Find the ﬁrst order Hankel transforms of (a)f(r)=e−ar, (b)f(r)=1 re−ar,( c ) f(r)=sinar r. We can write (a)˜f(κ)=H1{e−ar}=∞⎪integraldisplay 0re−arJ1(κr)dr=κ (a2+κ2)3 2. (b)˜f(κ)=H1⎪braceleftbigga−ar r⎪bracerightbigg =∞⎪integraldisplay 0e−arJ1(κr)dr=1 κ⎪bracketleftBig 1−a(κ2+a2)−1 2⎪bracketrightBig . (c)˜f(κ)=H1⎪braceleftbiggsinar r⎪bracerightbigg =∞⎪integraldisplay 0sinar J1(κr)dr=aH(κ−a) κ(κ2−a2)1 2. Example 7.2.3 Find the nth (n>−1) order Hankel transforms of (a)f(r)=rnH(a−r), (b)f(r)=rnexp(−ar2). © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 319 Here we have, for n>−1, (a)˜f(κ)=Hn[rnH(a−r)] =a⎪integraldisplay 0rn+1Jn(κr)dr=an+1 κJn+1(aκ). (b)˜f(κ)=Hn⎪bracketleftbig rnexp(−ar2)⎪bracketrightbig =∞⎪integraldisplay 0rn+1Jn(κr)exp(−ar2)dr =κn (2a)n+1exp⎪parenleftbigg −κ2 4a⎪parenrightbigg . 7.3 Operational Properties of the Hankel Transform THEOREM 7.3.1 (Scaling ). IfHn{f(r)}=˜fn(κ), then Hn{f(ar)}=1 a2˜fn⎪parenleftBigκ a⎪parenrightBig ,a > 0. (7.3.1) PROOF We have, by deﬁnition, Hn{f(ar)}=∞⎪integraldisplay 0rJn(κr)f(ar)dr =1 a2∞⎪integraldisplay 0sJn⎪parenleftBigκ as⎪parenrightBig f(s)ds=1 a2˜fn⎪parenleftBigκ a⎪parenrightBig . THEOREM 7.3.2 (Parseval’s Relation ). If˜f(κ)=Hn{f(r)}and ˜g(κ)=Hn{g(r)},t h e n ∞⎪integraldisplay 0rf(r)g(r)dr=∞⎪integraldisplay 0κ˜f(κ)˜g(κ)dκ. (7.3.2) © 2007 by Taylor & Francis Group, LLC 320 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF We proceed formally to obtain ∞⎪integraldisplay 0κ˜f(κ)˜g(κ)dκ=∞⎪integraldisplay 0κ˜f(κ)dκ∞⎪integraldisplay 0rJn(κr)g(r)dr, which is, interchanging the order of integration, =∞⎪integraldisplay 0rg(r)dr∞⎪integraldisplay 0κJn(κr)˜f(κ)dκ =∞⎪integraldisplay 0rg(r)f(r)dr. THEOREM 7.3.3 (Hankel Transforms of Derivatives )I f˜fn(κ)=Hn{f(r)},t h e n Hn{f/prime(r)}=κ 2n⎪bracketleftBig (n−1)˜fn+1(κ)−(n+1 )˜fn−1(κ)⎪bracketrightBig ,n≥1,(7.3.3) H1{f/prime(r)}=−κ˜f0(κ), (7.3.4) provided [ rf(r)] vanishes as r→0a n d r→∞. PROOF We have, by deﬁnition, Hn{f/prime(r)}=∞⎪integraldisplay 0rJn(κr)f/prime(r)dr which is, integrating by parts, =[rf(r)Jn(κr)]∞ 0−∞⎪integraldisplay 0f(r)d dr[rJn(κr)]dr. (7.3.5) We now use the properties of the Bessel function d dr[rJn(κr)] =Jn(κr)+rκJ/prime n(κr)=Jn(κr)+rκJn−1(κr)−nJn(κr) =( 1−n)Jn(κr)+rκJn−1(κr). (7.3.6) In view of the given condition, the ﬁrst term of (7.3.5) vanishes as r→0 andr→∞, and the derivative within the integral in (7.3.5) can be replaced © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 321 by (7.3.6) so that (7.3.5) becomes Hn{f/prime(r)}=(n−1)∞⎪integraldisplay 0f(r)Jn(κr)dr−κ˜fn−1(κ). (7.3.7) We next use the standard recurrence relation for the Bessel function Jn(κr)=κr 2n[Jn−1(κr)+Jn+1(κr)]. (7.3.8) Thus, (7.3.7) can be rewritten as Hn[f/prime(r)] =−κ˜fn−1(κ)+κ⎪parenleftbiggn−1 2n⎪parenrightbigg⎡ ⎣∞⎪integraldisplay 0rf(r){Jn−1(κr)+Jn+1(κr)}dr⎤ ⎦ =−κ˜fn−1(κ)+κ⎪parenleftbiggn−1 2n⎪parenrightbigg⎪bracketleftBig ˜fn−1(κ)+˜fn+1(κ)⎪bracketrightBig =⎪parenleftBigκ 2n⎪parenrightBig⎪bracketleftBig (n−1)˜fn+1(κ)−(n+1 )˜fn−1(κ)⎪bracketrightBig . In particular, when n= 1, (7.3.4) follows immediately. Similarly, repeated applications of (7.3.3) lead to the following result Hn{f/prime/prime(r)}=κ 2n⎪bracketleftbig (n−1)Hn+1{f/prime(r)}−(n+1 )Hn−1{f/prime(r)}⎪bracketrightbig =κ2 4⎪bracketleftbigg⎪parenleftbiggn+1 n−1⎪parenrightbigg ˜fn−2(κ)−2⎪parenleftbiggn2−3 n2−1⎪parenrightbigg ˜fn(κ) +⎪parenleftbiggn−1 n+1⎪parenrightbigg ˜fn+2(κ)⎪bracketrightbigg . (7.3.9) THEOREM 7.3.4 IfHn{f(r)}=˜fn(κ), then Hn⎪braceleftbigg⎪parenleftbigg ∇2−n2 r2⎪parenrightbigg f(r)⎪bracerightbigg =Hn⎪braceleftbigg1 rd dr⎪parenleftbigg rdf dr⎪parenrightbigg −n2 r2f(r)⎪bracerightbigg =−κ2˜fn(κ), (7.3.10) provided both rf/prime(r)a n d rf(r)v a n i s ha s r→0a n d r→∞. PROOF We have, by deﬁnition (7.2.8), Hn⎪braceleftbigg1 rd dr⎪parenleftbigg rdf dr⎪parenrightbigg −n2 r2f(r)⎪bracerightbigg =∞⎪integraldisplay 0Jn(κr)⎪bracketleftbiggd dr⎪parenleftbigg rdf dr⎪parenrightbigg⎪bracketrightbigg dr −∞⎪integraldisplay 0n2 r2[rJn(κr)]f(r)dr, © 2007 by Taylor & Francis Group, LLC 322 INTEGRAL TRANSFORMS and THEIR APPLICATIONS which is, invoking integration by parts, =⎪bracketleftbigg⎪parenleftbigg rdf dr⎪parenrightbigg Jn(κr)⎪bracketrightbigg∞ 0−κ∞⎪integraldisplay 0rdf drJ/prime n(κr)dr−∞⎪integraldisplay 0n2 r2[rJn(κr)]f(r)dr, which is, by replacing the ﬁrst term with zero because of the given assumption, and by invoking integration by parts again, =−⎪bracketleftbig κrf(r)J/prime n(κr)⎪bracketrightbig∞ 0+∞⎪integraldisplay 0d dr⎪bracketleftbig κrJ/prime n(κr)⎪bracketrightbig f(r)dr−∞⎪integraldisplay 0n2 r2[rJn(κr)]f(r)dr. We use the given assumptions and Bessel’s diﬀerential equation, d dr⎪bracketleftbig κr J/prime n(κr)⎪bracketrightbig +r⎪parenleftbigg κ2−n2 r2⎪parenrightbigg Jn(κr)=0, (7.3.11) to obtain Hn⎪braceleftbigg⎪parenleftbigg ∇2−n2 r2⎪parenrightbigg f(r)⎪bracerightbigg =−∞⎪integraldisplay 0⎪parenleftbigg κ2−n2 r2⎪parenrightbigg rf(r)Jn(κr)dr −∞⎪integraldisplay 0n2 r2[rf(r)]Jn(κr)dr =−κ2∞⎪integraldisplay 0rJn(κr)f(r)dr=−κ2Hn[f(r)] =−κ2˜fn(κ). This proves the theorem. In particular, when n=0 a n d n=1 ,w eo b t a i n H0⎪braceleftbigg1 rd dr⎪parenleftbigg rdf dr⎪parenrightbigg⎪bracerightbigg =−κ2˜f0(κ), (7.3.12) H1⎪braceleftbigg1 rd dr⎪parenleftbigg rdf dr⎪parenrightbigg −1 r2f(r)⎪bracerightbigg =−κ2˜f1(κ). (7.3.13) Results (7.3.10), (7.3.12), and (7.3.13) are widely used for ﬁnding solutions of partial diﬀerential equations in axisymmetric cylindrical conﬁgurations. Weillustrate this point by considering several examples of applications. 7.4 Applications of Hankel Transforms to Partial Diﬀerential Equations The Hankel transforms are extremely useful in solving a variety of partial diﬀerential equations in cylindrical polar coordinates. The following examples © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 323 illustrate applications of the Hankel transforms. The examples given here are only representative of a whole variety o f physical problems that can be solved in a similar way. Example 7.4.1 (Free Vibration of a Large Circular Membrane ). Obtain the solution of the free vibration of a large circular elastic membrane governed by the initial value problem c2⎪parenleftbigg∂2u ∂r2+1 r∂u ∂r⎪parenrightbigg =∂2u ∂t2,0<r<∞,t > 0, (7.4.1) u(r,0)=f(r),u t(r,0) =g(r),for 0≤r<∞, (7.4.2ab) where c2=(T/ρ)= con st an t , Tis the tension in the membrane, and ρis the surface density of the membrane. Application of the zero-order Hankel transform with respect to r ˜u(κ,t)=∞⎪integraldisplay 0rJ0(κr)u(r, t)dr, (7.4.3) to (7.4.1)–(7.4.2ab) gives d2˜u dt2+c2κ2˜u=0, (7.4.4) ˜u(κ,0) =˜f(κ), ˜ut(κ,0)= ˜g(κ). (7.4.5ab) The general solution of this transformed system is ˜u(κ,t)=˜f(κ)cos(cκt)+(cκ)−1˜g(κ)sin (cκt). (7.4.6) The inverse Hankel transf orm leads to the solution u(r, t)=∞⎪integraldisplay 0κ˜f(κ)cos (cκt)J0(κr)dκ +1 c∞⎪integraldisplay 0˜g(κ)sin (cκt)J0(κr)dκ. (7.4.7) In particular, we consider u(r,0)=f(r)=Aa(r2+a2)−1 2,u t(r,0) =g(r)=0, (7.4.8ab) so that ˜ g(κ)≡0a n d ˜f(κ)=Aa∞⎪integraldisplay 0r(a2+r2)−1 2J0(κr)dr=Aa κe−aκ, by Example 7.2.1(a). © 2007 by Taylor & Francis Group, LLC 324 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thus, the formal solution (7.4.7) becomes u(r, t)=Aa∞⎪integraldisplay 0e−aκJ0(κr)cos(cκt)dκ=AaRe∞⎪integraldisplay 0exp[−κ(a+ict)]J0(κr)dκ =AaRe⎪braceleftbig r2+(a+ict)2⎪bracerightbig−1 2,by Example 7.2.1(a). (7.4.9) Example 7.4.2 (Steady Temperature Distribution in a Semi-Inﬁnite Solid with a Steady Heat Source ). Find the solution of the Laplace equation for the steady temperature distribution u(r, z) with a steady and symmetric heat source Q0q(r): urr+1 rur+uzz=−Q0q(r),0<r<∞,0<z< ∞,(7.4.10) u(r,0) = 0 ,0<r<∞, (7.4.11) where Q0is a constant. This boundary condition represents zero temperature at the boundary z=0 . Application of the zero-order Hankel transform to (7.4.10) and (7.4.11) gives d2˜u dz2−κ2˜u=−Q0˜q(κ),˜u(κ,0)= 0 . The bounded general solution of this system is ˜u(κ,z)=Aexp(−κz)+Q0 κ2˜q(κ), where Ais a constant to be determined from the transformed boundary con- dition. In this case A=−Q0 κ2˜q(κ). Thus, the formal solution is ˜u(κ,z)=Q0˜q(κ) κ2(1−e−κz). (7.4.12) The inverse Hankel transform yie lds the exact integral solution u(r, z)=Q0∞⎪integraldisplay 0˜q(κ) κ(1−e−κz)J0(κr)dκ. (7.4.13) © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 325 Example 7.4.3 (Axisymmetric Diﬀusion Equation ). Find the solution of the axisymmetric diﬀusion equation ut=κ⎪parenleftbigg urr+1 rur⎪parenrightbigg ,0<r<∞,t > 0, (7.4.14) where κ(>0) is a diﬀusivity constant and u(r,0)=f(r),for 0 <r<∞. (7.4.15) We apply the zero-order Hankel transform deﬁned by (7.4.3) to obtain d˜u dt+k2κ˜u=0, ˜u(k,0) =˜f(k), where kis the Hankel transform variable. The solution of this transformed system is ˜u(k,t)=˜f(k)exp (−κk2t). (7.4.16) Application of the inverse Hankel transform gives u(r, t)=∞⎪integraldisplay 0k˜f(k)J0(kr)e−κk2tdk=∞⎪integraldisplay 0k⎡ ⎣∞⎪integraldisplay 0lJ0(kl)f(l)dl⎤ ⎦e−κk2tJ0(kr)dk which is, interchanging the order of integration, =∞⎪integraldisplay 0lf(l)dl∞⎪integraldisplay 0kJ0(kl)J0(kr)exp(−κk2t)dk. (7.4.17) Using a standard table of integrals involving Bessel functions, we state ∞⎪integraldisplay 0kJ0(kl)J0(kr)exp(−k2κt)dk=1 2κtexp⎪bracketleftbigg −(r2+l2) 4κt⎪bracketrightbigg I0⎪parenleftbiggrl 2κt⎪parenrightbigg ,(7.4.18) where I0(x) is the modiﬁed Bessel function and I0(0) = 1. In particular, when l=0,J0(0) = 1 and integral (7.4.18) becomes ∞⎪integraldisplay 0kJ0(kr)exp (−k2κt)dk=1 2κtexp⎪parenleftbigg −r2 4κt⎪parenrightbigg . (7.4.19) We next use (7.4.18) to rewrite (7.4.17) as u(r, t)=1 2κt∞⎪integraldisplay 0lf(l)I0⎪parenleftbiggrl 2κt⎪parenrightbigg exp⎪bracketleftbigg −(r2+l2) 4κt⎪bracketrightbigg dl. (7.4.20) © 2007 by Taylor & Francis Group, LLC 326 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We now assume f(r) to represent a heat source concentrated in a circle of radius aand allow a→0 so that the heat source is concentrated at r=0a n d lim a→02πa⎪integraldisplay 0rf(r)dr=1. Or, equivalently, f(r)=1 2πδ(r) r, where δ(r) is the Dirac delta function. Thus, the ﬁnal solution due to the concentrated heat source at r=0 i s u(r, t)=1 4πκt∞⎪integraldisplay 0δ(l)I0⎪parenleftbiggrl 2κt⎪parenrightbigg exp⎪bracketleftbigg −r2+l2 4κt⎪bracketrightbigg dl =1 4πκtexp⎪parenleftbigg −r2 4κt⎪parenrightbigg . (7.4.21) Example 7.4.4 (Axisymmetric Acoustic Radiation Problem ). Obtain the solution of the wave equation c2⎪parenleftbigg urr+1 rur+uzz⎪parenrightbigg =utt,0<r<∞,z > 0,t > 0,(7.4.22) uz=F(r, t)o n z=0, (7.4.23) where F(r, t) is a given function and cis a constant. We also assume that the solution is bounded and behaves as outgoing spherical waves. We seek a steady-state solution for the acoustic radiation potential u= eiωtφ(r, z)w i t h F(r, t)=eiωtf(r), so that φsatisﬁes the Helmholtz equation φrr+1 rφr+φzz+⎪parenleftbiggω2 c2⎪parenrightbigg φ=0,0<r<∞,z > 0, (7.4.24) with the boundary condition φz=f(r)o n z=0, (7.4.25) where f(r) is a given function of r. Application of the Hankel transform H0{φ(r, z)}=˜φ(k,z) to (7.4.24)-(7.4.25) gives ˜φzz=κ2˜φ, z > 0, ˜φz=˜f(k),onz=0, © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 327 where κ=⎪parenleftbigg k2−ω2 c2⎪parenrightbigg1 2 . The solution of this diﬀerential system is ˜φ(k,z)=−1 κ˜f(k)exp (−κz), (7.4.26) where κis real and positive for k>ω/ c , and purely imaginary for k<ω / c . The inverse Hankel transform yields the formal solution φ(r, z)=−∞⎪integraldisplay 0k κ˜f(k)J0(kr)exp(−κz)dk. (7.4.27) Since the exact evaluation of this integral is diﬃcult for an arbitrary ˜f(k), we choose a simple form of f(r)a s f(r)=AH(a−r), (7.4.28) where Ais a constant, and hence, ˜f(k)=Aa kJ1(ak). Thus, the solution (7.4.27) takes the form φ(r, z)=−Aa∞⎪integraldisplay 01 κJ1(ak)J0(kr)exp(−κz)dk. (7.4.29) For an asymptotic evaluation of this integral, it is convenient to express (7.4.29) in terms of Rwhich is the distance from the z-axis so that R2=r2+z2 andz=Rcosθ. Using the asymptotic result for the Bessel function J0(kr)∼⎪parenleftbigg2 πkr⎪parenrightbigg1 2 cos⎪parenleftBig kr−π 4⎪parenrightBig asr→∞, (7.4.30) where r=Rsinθ. Consequently, (7.4.29) combined with u=e x p ( iωt)φbe- comes u∼−Aa√ 2eiωt √ πRsinθ∞⎪integraldisplay 01 κ√ kJ1(ak)cos⎪parenleftBig kRsinθ−π 4⎪parenrightBig exp(−κz)dk. This integral can be evaluated asymptotically for R→∞using the stationary phase approximation formula to obtain the ﬁnal result u∼−Aac ωRsinθJ1(ak1)exp⎪bracketleftbigg i⎪parenleftbigg ωt−ωR c⎪parenrightbigg⎪bracketrightbigg , (7.4.31) © 2007 by Taylor & Francis Group, LLC 328 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where k1=ω/(csinθ) is the stationary point. Physically, this solution repre- sents outgoing spherical waves with constant velocity cand decaying ampli- tude as R→∞. Example 7.4.5 (Axisymmetric Biharmonic Equation ). We solve the axisymmetric boundary value problem ∇4u(r, z)=0, 0≤r<∞,z > 0, (7.4.32) with the boundary data u(r,0) =f(r),0≤r<∞, (7.4.33) ∂u ∂z=0 o n z=0,0≤r<∞, (7.4.34) u(r, z)→0a s r→∞, (7.4.35) where the axisymmetric biharmonic operator is ∇4=∇2(∇2)=⎪parenleftbigg∂2 ∂r2+1 r∂ ∂r+∂2 ∂z2⎪parenrightbigg⎪parenleftbigg∂2 ∂r2+1 r∂ ∂r+∂2 ∂z2⎪parenrightbigg .(7.4.36) The use of the Hankel transform H0{u(r, z)}=˜u(k,z) to this problem gives ⎪parenleftbiggd2 dz2−k2⎪parenrightbigg2 ˜u(k,z)=0,z > 0, (7.4.37) ˜u(k,0) =˜f(k),d˜u dz=0 o n z=0. (7.4.38) The bounded solution of (7.4.37) is ˜u(k,z)=(A+zB)exp (−kz), (7.4.39) where AandBare integrating constants to be determined by (7.4.38) as A=˜f(k)a n d B=k˜f(k). Thus, solution (7.4.39) becomes ˜u(k,z)=( 1+ kz)˜f(k)exp (−kz). (7.4.40) The inverse Hankel transform gives the formal solution u(r, z)=∞⎪integraldisplay 0k(1 +kz)˜f(k)J0(kr)exp (−kz)dk. (7.4.41) © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 329 Example 7.4.6 (The Axisymmetric Cauchy-Poisson Water Wave Problem ). We consider the initial value problem for an inviscid water of ﬁnite depth hwith a free hori- zontal surface at z=0 , a n d t h e z-axis positive upward. We assume that the liquid has constant density ρwith no surface tension. The surface waves are generated in water, which is initially at rest for t<0 by the prescribed free surface elevation. In cylindrical polar coordinates ( r, θ, z), the axisymmetric water wave equations for the velocity potential φ(r, z, t) and the free surface elevation η(r, t)a r e ∇2φ=φrr+1 rφr+φzz=0,0≤r<∞,−h≤z≤0,t > 0,(7.4.42) φz−ηt=0 φt+gη=0⎫ ⎬ ⎭onz=0,t > 0, (7.4.43ab) φz=0 o n z=−h, t > 0. (7.4.44) The initial conditions are φ(r,0,0)= 0 and η(r,0) =η0(r),for 0≤r<∞, (7.4.45) where gis the acceleration due to gravity and η0(r) is the given free surface elevation. We apply the joint Laplace and the zero-order Hankel transform deﬁned by ˜ φ(k,z,s)=∞⎪integraldisplay 0e−stdt∞⎪integraldisplay 0rJ0(kr)φ(r, z, t)dr, (7.4.46) to (7.4.42)–(7.4.44) so that these equations reduce to ⎪parenleftbiggd2 dz2−k2⎪parenrightbigg ˜ φ=0, d˜ φ dz−s˜ η=−˜η0(k) s˜ φ+g˜ η=0⎫ ⎪⎪⎬ ⎪⎪⎭onz=0, ˜ φz=0 o n z=−h, where ˜ η0(k) is the Hankel transform of η0(r) of order zero. The solutions of this system are ˜ φ(k,z,s)=−g˜η0(k) (s2+ω2)coshk(z+h) coshkh, (7.4.47) ˜ η(k,s)=s˜η0(k) (s2+ω2), (7.4.48) © 2007 by Taylor & Francis Group, LLC 330 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where ω2=gktanh(kh), (7.4.49) is the famous dispersion relation between frequency ωand wavenumber kfor water waves in a liquid of depth h. Physically, this dispersion relation describes the interaction between the inertial and gravitational forces. Application of the inverse transforms gives the integral solutions φ(r, z, t)=−g∞⎪integraldisplay 0kJ0(kr)˜η0(k)⎪parenleftbiggsinωt ω⎪parenrightbiggcoshk(z+h) coshkhdk,(7.4.50) η(r, t)=∞⎪integraldisplay 0kJ0(kr)˜η0(k)cosωtdk. (7.4.51) These wave integrals repr esent exact solutions for φandηat any randt, but the physical features of the wave moti ons cannot be described by them. In general, the exact evaluation of the integrals is almost a formidable task. In order to resolve this diﬃculty, it is necessary and useful to resort to asymptoticmethods. It will be suﬃcient for the determination of the basic features of the wave motions to evaluate (7.4.50) or (7.4.51) asymptotically for a large time and distance with ( r/t) held ﬁxed. We now replace J 0(kr) by its asymptotic formula (7.4.30) for kr→∞, so that (7.4.51) gives η(r, t)∼⎪parenleftbigg2 πr⎪parenrightbigg1 2∞⎪integraldisplay 0√ k˜η0(k)cos⎪parenleftBig kr−π 4⎪parenrightBig cosωtdk =( 2πr)−1 2Re∞⎪integraldisplay 0√ k˜η0(k)exp⎪bracketleftBig i⎪parenleftBig ωt−kr+π 4⎪parenrightBig⎪bracketrightBig dk.(7.4.52) Application of the stationary phase method to (7.4.52) yields the solution η(r, t)∼⎪bracketleftbiggk1 rt|ω/prime/prime(k1)|⎪bracketrightbigg1 2 ˜η0(k1)cos [tω(k1)−k1r], (7.4.53) where the stationary point k1=⎪parenleftbig gt2/4r2⎪parenrightbig is the root of the equation ω/prime(k)=r t. (7.4.54) For suﬃciently deep water, kh→∞, the dispersion relation becomes ω2=gk. (7.4.55) The solution of the axisymmetric Cauchy-Poisson problem is based on a pre- scribed initial displacement of unit volume that is concentrated at the origin, © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 331 which means that η0(r)=(a/2πr)δ(r)s ot h a t˜ η0(k)=a 2π.T h u s ,t h ea s y m p - totic solution is obtained from (7.4.53) in the form η(r, t)∼agt2 4π√ 2r3cos⎪parenleftbigggt2 4r⎪parenrightbigg ,g t2>>4r. (7.4.56) It is noted that solution (7.4.53) is no longer valid when ω/prime/prime(k1)=0 . T h i s case can be handled by a modiﬁcation of the asymptotic evaluation (see Deb- nath, 1994, p. 91). A wide variety of other physical problems solved by the Hankel transform, and/or by the joint Hankel and Laplace transform are given in books by Sneddon (1951, 1972) and by Debnath (1994), and in research papers by Debnath (1969, 1983, 1989), Mohanti (1979), and Debnath and Rollins (1992) listed in the Bibliography. 7.5 Exercises 1. Show that (a)H0{(a2−r2)H(a−r)}=4a κ3J1(κa)−2a2 κ2J0(aκ), (b)Hn{rne−ar}=a √ π·2n+1Γ⎪parenleftbigg n+3 2⎪parenrightbigg κn(a2+κ2)−(n+3 2), (c)Hn⎪braceleftbigg2n rf(r)⎪bracerightbigg =kHn−1{f(r)}+kHn+1{f(r)}. 2. (a) Show that the solution of the boundary value problem urr+1 rur+uzz=0,0<r<∞,0<z< ∞, u(r, z)=1 √ a2+r2on z = 0 ,0<r<∞, is u(r, z)=∞⎪integraldisplay 0e−κ(z+a)J0(κr)dκ=1 ⎪radicalbig (z+a)2+r2. (b) Obtain the solution of the equation in 2( a)w i t h u(r,0)=f(r)= H(a−r),0<r< ∞. © 2007 by Taylor & Francis Group, LLC 332 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 3. (a) The axisymmetric initial value problem is governed by ut=κ⎪parenleftbigg urr+1 rur⎪parenrightbigg +δ(t)f(r),0<r<∞,t >0, u(r,0) = 0 for 0 <r<∞. Show that the formal solution of this problem is u(r, t)=∞⎪integraldisplay 0kJ0(kr)˜f(k)exp (−k2κt)dk. (b) For the special case when f(r)=⎪parenleftbiggQ πa2⎪parenrightbigg H(a−r), show that the solution is u(r, t)=⎪parenleftbiggQ πa⎪parenrightbigg∞⎪integraldisplay 0J0(kr)J1(ak)exp (−k2κt)dk. 4. Iff(r)=A(a2+r2)−1 2where Ais a constant, show that the solution of the biharmonic equation described in Example 7.4.5 is u(r, z)=A{r2+(z+a)(2z+a)} [r2+(z+a)2]3/2. 5. Show that the solution of the boundary value problem urr+1 rur+uzz=0,0≤r<∞,z > 0, u(r,0) =u0for 0≤r≤a, u 0is a constant, u(r, z)→0a s z→∞, is u(r, z)=au0∞⎪integraldisplay 0J1(ak)J0(kr)exp(−kz)dk. Find the solution of the problem when u0is replaced by an arbitrary function f(r), and aby inﬁnity. 6. Solve the axisymmetric biharmonic equation for the small-amplitude free vibration of a thin elastic disk b2⎪parenleftbigg∂2 ∂r2+1 r∂ ∂r⎪parenrightbigg2 u+utt=0,0<r< ∞,t > 0, u(r,0) =f(r),u t(r,0) = 0 for 0 <r<∞, © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 333 where b2=⎪parenleftbiggD 2σh⎪parenrightbigg is the ratio of the ﬂexural rigidity of the disk and its mass 2 hσper unit area. 7. Show that the zero-order Hankel transform solution of the axisymmetric Laplace equation urr+1 rur+uzz=0,0<r< ∞,−∞<z< ∞, with the boundary data lim r→0(r2u)=0,lim t→0(2πr)ur=−f(z),−∞<z< ∞, is ˜u(k,z)=1 4πk∞⎪integraldisplay −∞exp{−k|z−ζ|}f(ζ)dζ. Hence, show that u(r, z)=1 4π∞⎪integraldisplay −∞⎪braceleftbig r2+(z−ζ)2⎪bracerightbig−1 2f(ζ)dζ. 8. Solve the nonhomogeneous diﬀusion problem ut=κ⎪parenleftbigg urr+1 rur⎪parenrightbigg +Q(r, t),0<r<∞,t > 0, u(r,0)=f(r)f o r 0 <r<∞, where κis a constant. 9. Solve the problem of the electriﬁed unit disk in the x−yplane with center at the origin. The electric potential u(r, z) is axisymmetric and satisﬁes the boundary value problem urr+1 rur+uzz=0,0<r<∞,0<z< ∞, u(r,0)=u0,0≤r<a , ∂u ∂z=0,onz=0 f o r a<r< ∞, u(r, z)→0a s z→∞ for all r, where u0is constant. Show that the solution is u(r, z)=⎪parenleftbigg2au0 π⎪parenrightbigg∞⎪integraldisplay 0J0(kr)⎪parenleftbiggsinak k⎪parenrightbigg e−kzdk. © 2007 by Taylor & Francis Group, LLC 334 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 10. Solve the axisymmetric surface wave problem in deep water due to an oscillatory surface pressure. The governing equations are ∇2φ=φrr+1 rφr+φzz=0,0≤r<∞,−∞<z≤0, φt+gη=−P ρp(r)exp(iωt) φz−ηt=0⎫ ⎬ ⎭onz=0,t > 0, φ(r, z,0)= 0 = η(r,0),for 0≤r<∞,−∞<z≤0. 11. Solve the Neumann problem for the Laplace equation urr+1 rur+uzz=0,0<r<∞,0<z< ∞ uz(r,0) =−1 πa2H(a−r),0<r<∞ u(r, z)→0a s z→∞ for 0<r<∞. Show that lim a→0u(r, z)=1 2π(r2+z2)−1 2. 12. Solve the Cauchy problem for the wave equation in a dissipating medium utt+2κut=c2⎪parenleftbigg urr+1 rur⎪parenrightbigg ,0<r<∞,t > 0, u(r,0)=f(r),u t(r,0) =g(r)f o r 0 <r<∞, where κis a constant. 13. Use the joint Laplace and Hankel transform to solve the initial-boundary value problem c2⎪parenleftbigg urr+1 rur+uzz⎪parenrightbigg =utt,0<r<∞,0<z< ∞,t > 0, uz(r,0,t)=H(a−r)H(t),0<r< ∞,t > 0, u(r, z, t)→0a s r→∞ andu(r, z, t)→0a s z→∞, u(r, z,0)= 0 = ut(r, z,0), and show that ut(r, z, t)=−ac H⎪parenleftBig t−z c⎪parenrightBig∞⎪integraldisplay 0J1(ak)J0⎪braceleftBigg ck⎪radicalbigg t2−z2 c2⎪bracerightBigg J0(kr)dk. © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 335 14. Find the steady temperature u(r, z)i nab e a m0 ≤r<∞,0≤z≤awhen the face z= 0 is kept at temperature u(r,0) = 0, and the face z=ais insulated except that heat is supplied through a circular hole such that uz(r, a)=H(b−r). The temperature u(r, z) satisﬁes the axisymmetric equation urr+1 rur+uzz=0,0≤r<∞,0≤z≤a. 15. Find the integral solution of the initial-boundary value problem urr+1 rur+uzz=ut,0≤r<∞,0≤z<∞,t > 0, u(r, z,0)= 0 for all randz, ⎪parenleftbigg∂u ∂r⎪parenrightbigg r=0=0,for 0 ≤z<∞,t > 0, ⎪parenleftbigg∂u ∂z⎪parenrightbigg z=0=−H(a−r) √ a2+r2,for 0 <r<∞,0<t<∞, u(r, z, t)→0a s r→∞ orz→∞. 16. Heat is supplied at a constant rate Qper unit area per unit time over a circular area of radius ain the plane z= 0 to an inﬁnite solid of thermal conductivity K, the rest of the plane is kept at zero temperature. Solve for the steady temperature ﬁeld u(r, z) that satisﬁes the Laplace equation urr+1 rur+uzz=0,0<r< ∞,−∞<z< ∞, with the boundary conditions u→0a sr→∞,u→0a s |z|→∞ , −Kuz=⎪parenleftbigg2Q πa2⎪parenrightbigg H(a−r)w h e n z=0. 17. The velocity potential φ(r, z) for the ﬂow of an inviscid ﬂuid through a circular aperture of unit radius in a plane rigid screen satisﬁes the Laplace equation φrr+1 rφr+φzz=0,0<r<∞ with the boundary conditions φ=1 f o r 0 <r< 1 φz=0 f o r r>1⎪bracerightBigg onz=0. Obtain the solution of this boundary value problem. © 2007 by Taylor & Francis Group, LLC 336 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 18. Solve the Cauchy-Poisson wave problem (Debnath, 1989) for a viscous liquid of ﬁnite or inﬁnite depth governed by the equations, free surface,boundary, and initial conditions φ rr+1 rφr+φzz=0, ψt=ν⎪parenleftbigg ψrr+1 rψr−1 r2ψ+ψzz⎪parenrightbigg , where φ(r, z, t)a n d ψ(r, z, t) represent the potential and stream func- tions, respectively, 0 ≤r<∞,−h≤z≤0( o r−∞<z≤0) and t>0. The free surface conditions are ηt−w=0 μ(uz+wr)=0 φt+gη+2νwz=0⎫ ⎪⎪⎪⎬ ⎪⎪⎪⎭onz=0,t >0 where η=η(r, t) is the free surface elevation, u=φ r+ψzandw=φz− ψ r−ψrare the radial and vertical velocity components of liquid par- ticles, μ=ρνis the dynamic viscosity, ρis the density, and νis the kinematic viscosity of the liquid. The boundary conditions at the rigid bottom are u=φr+ψz=0 w=φz−1 r(rψ)r=0⎫ ⎪⎬ ⎪⎭onz=−h. The initial conditions are η=aδ(r) r,φ=ψ=0 a t t=0, where ais a constant and δ(r) is the Dirac delta function. If the liquid is of inﬁnite depth, the bottom boundary conditions are (φ, ψ)→(0,0) as z→−∞ . 19. Use the joint Hankel and Laplace transform method to solve the initial- boundary value problem urr+1 rur−utt−2εut=aδ(r) rδ(t),0<r<∞,t > 0, u(r, t)→0a s r→∞, u(0,t) is ﬁnite for t>0, u(r,0) = 0 = ut(r,0) for 0 <r<∞. © 2007 by Taylor & Francis Group, LLC Hankel Transforms and Their Applications 337 20. Surface waves are generated in an inviscid liquid of inﬁnite depth due to an explosion (Sen, 1963) above it, which generates the pressure ﬁeldp(r, t). The velocity potential u=φ(r, z, t) satisﬁes the Laplace equation u rr+1 rur+uzz=0,0<r<∞,t > 0, and the free surface condition utt+guz=1 ρ⎪parenleftbigg∂p ∂t⎪parenrightbigg [H(r)−H{r, r0(t)}]o n z=0, where ρis the constant density of the liquid, r0(t) is the extent of the blast, and the liquid is initially at rest. Solve this problem. 21. The electrostatic potential u(r, z) generated in the space between two horizontal disks at z=±aby a point charge qatr=z=0i s d e s c r i be d by a singular function at r=z=0 i s u(r, z)=φ(r, z)+q(r2+z2)−1 2, where φ(r, z) satisﬁes the Laplace equation φrr+1 rφr+φzz=0,0<r<∞ with the boundary conditions φ(r, z)=−q(r2+z2)−1 2atz=±a. Obtain the solution for φ(r, z)a n dt h e n u(r, z). 22. Show that (a)Hn⎪bracketleftbig e−arf(r)⎪bracketrightbig =L{rf(r)Jn(kr)}, (b)H0⎪bracketleftBig e−ar2J0(br)⎪bracketrightBig =a 2exp⎪parenleftbiggk2−b2 4a⎪parenrightbigg I0⎪parenleftbiggbk 2a⎪parenrightbigg , (c)Hn⎪bracketleftbig rn−1e−ar⎪bracketrightbig =(2k)n(n−1 2)! √ π(k2+a2)n+1 2, (d)Hn⎪bracketleftbiggf(r) r⎪bracketrightbigg =⎪parenleftbiggk 2n⎪parenrightbigg⎪bracketleftBig ˜fn−1(k)+˜fn+1(k)⎪bracketrightBig , (e)Hn⎪bracketleftbigg rn−1d dr⎪braceleftbig r1−nf(r)⎪bracerightbig⎪bracketrightbigg =−k˜fn−1(k), (f)Hn⎪bracketleftbigg r−(n+1)d dr⎪braceleftbig rn+1f(r)⎪bracerightbig⎪bracketrightbigg =k˜fn+1(k). © 2007 by Taylor & Francis Group, LLC 8 Mellin Transforms and Their Applications “One cannot understand ... the universality of laws of nature, the relationship of things, without an understanding of mathematics. There is no other way to do it.” Richard P. Feynman “The research worker, in his eﬀorts to express the fundamental laws of Nature in mathematical form, should strive mainly for mathe-matical beauty. He should take simplicity into consideration in a subordinate way to beauty. ... It often happens that the require- ments of simplicity and beauty are the same, but where they clashthe latter must take precedence.” Paul Dirac 8.1 Introduction This chapter deals with the theory and applications of the Mellin transform. We derive the Mellin transform and its inverse from the complex Fourier trans-form. This is followed by several examples and the basic operational properties of Mellin transforms. We discuss several applications of Mellin transforms to boundary value problems and to summation of inﬁnite series. The Weyl trans- form and the Weyl fractional derivatives with examples are also included. Historically, Riemann (1876) ﬁrst recognized the Mellin transform in his famous memoir on prime numbers. Its explicit formulation was given by C- ahen (1894). Almost simultaneousl y, Mellin (1896, 1902) gave an elaborate discussion of the Mellin transform and its inversion formula. 339 © 2007 by Taylor & Francis Group, LLC 340 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 8.2 Deﬁnition of the Mellin Transform and Examples We derive the Mellin transform and its inverse from the complex Fourier transform and its inverse, whi ch are deﬁned respectively by F{g(ξ)}=G(k)=1 √ 2π∞⎪integraldisplay −∞e−ikξg(ξ)dξ, (8.2.1) F−1{G(k)}=g(ξ)=1 √ 2π∞⎪integraldisplay −∞eikξG(k)dk. (8.2.2) Making the changes of variables exp( ξ)=xandik=c−p,w h e r e cis a constant, in results (8.2.1) and (8.2.2) we obtain G(ip−ic)=1 √ 2π∞⎪integraldisplay 0xp−c−1g(logx)dx, (8.2.3) g(logx)=1 √ 2πc+i∞⎪integraldisplay c−i∞xc−pG(ip−ic)dp. (8.2.4) We now write1 √ 2πx−cg(logx)≡f(x)a n d G(ip−ic)≡˜f(p) to deﬁne the Mellin transform off(x)a n dt h e inverse Mellin transform as M{f(x)}=˜f(p)=∞⎪integraldisplay 0xp−1f(x)dx, (8.2.5) M−1{˜f(p)}=f(x)=1 2πic+i∞⎪integraldisplay c−i∞x−p˜f(p)dp, (8.2.6) where f(x) is a real valued function deﬁned on (0 ,∞) and the Mellin transform variable pis a complex number. Sometimes, the Mellin transform of f(x)i s denoted explicitly by ˜f(p)=M[f(x),p]. Obviously, MandM−1are linear integral operators. Example 8.2.1 (a) If f(x)=e−nx,w h e r e n>0, then M{e−nx}=˜f(p)=∞⎪integraldisplay 0xp−1e−nxdx, © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 341 which is, by putting nx=t, =1 np∞⎪integraldisplay 0tp−1e−tdt=Γ(p) np. (8.2.7) (b) If f(x)=1 1+x,t h e n M⎪braceleftbigg1 1+x⎪bracerightbigg =˜f(p)=∞⎪integraldisplay 0xp−1·dx 1+x, which is, by substituting x=t 1−tort=x 1+x, =1⎪integraldisplay 0tp−1(1−t)(1−p)−1dt=B(p,1−p)=Γ( p)Γ(1−p), which is, by a well-known result for the gamma function, =πcosec( pπ),0<Re(p)<1. (8.2.8) (c) If f(x)=(ex−1)−1,t h e n M⎪braceleftbigg1 ex−1⎪bracerightbigg =˜f(p)=∞⎪integraldisplay 0xp−11 ex−1dx, which is, by using∞⎪summationdisplay n=0e−nx=1 1−e−xand hence,∞⎪summationdisplay n=1e−nx=1 ex−1, =∞⎪summationdisplay n=1∞⎪integraldisplay 0xp−1e−nxdx=∞⎪summationdisplay n=1Γ(p) np=Γ (p)ζ(p), (8.2.9) where ζ(p)=∞⎪summationdisplay n=11 np,( R ep>1) is the famous Riemann zeta function . (d) If f(x)=2 e2x−1,t h e n M⎪braceleftbigg2 e2x−1⎪bracerightbigg =˜f(p)=2∞⎪integraldisplay 0xp−1dx e2x−1=2∞⎪summationdisplay n=1∞⎪integraldisplay 0xp−1e−2nxdx =2∞⎪summationdisplay n=1Γ(p) (2n)p=21−pΓ(p)∞⎪summationdisplay n=11 np=21−pΓ(p)ζ(p).(8.2.10) © 2007 by Taylor & Francis Group, LLC 342 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (e) If f(x)=1 ex+1,t h e n M⎪braceleftbigg1 ex+1⎪bracerightbigg =( 1−21−p)Γ(p)ζ(p). (8.2.11) This follows from the result ⎪bracketleftbigg1 ex−1−1 ex+1⎪bracketrightbigg =2 e2x−1 combined with (8.2.9) and (8.2.10). (f) Iff(x)=1 (1 +x)n,t h e n M⎪braceleftbigg1 (1 +x)n⎪bracerightbigg =∞⎪integraldisplay 0xp−1(1 +x)−ndx, which is, by putting x=t 1−tort=x 1+x, =1⎪integraldisplay 0tp−1(1−t)n−p−1dt =B(p,n−p)=Γ(p)Γ(n−p) Γ(n), (8.2.12) where B(p,q) is the standard beta function. Hence, M−1{Γ(p)Γ(n−p)}=Γ(n) (1 +x)n. (g) Find the Mellin transform of cos kxand sin kx. It follows from Example 8.2.1(a) that M[e−ikx]=Γ(p) (ik)p=Γ(p) kp⎪parenleftBig cospπ 2−isinpπ 2⎪parenrightBig . Separating real and imaginary parts, we ﬁnd M[coskx]=k−pΓ(p)cos⎪parenleftBigπp 2⎪parenrightBig , (8.2.13) M[sinkx]=k−pΓ(p)sin⎪parenleftBigπp 2⎪parenrightBig . (8.2.14) These results can be used to calculate the Fourier cosine and Fourier sine transforms of xp−1. Result (8.2.13) can be written as ∞⎪integraldisplay 0xp−1coskxdx =Γ(p) kpcos⎪parenleftBigπp 2⎪parenrightBig . © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 343 Or, equivalently, Fc⎪braceleftbigg⎪radicalbigg π 2xp−1⎪bracerightbigg =Γ(p) kpcos⎪parenleftBigπp 2⎪parenrightBig . Or, Fc{xp−1}=⎪radicalbigg 2 πΓ(p) kpcos⎪parenleftBigπp 2⎪parenrightBig . (8.2.15) Similarly, Fs{xp−1}=⎪radicalbigg 2 πΓ(p) kpsin⎪parenleftBigπp 2⎪parenrightBig . (8.2.16) 8.3 Basic Operational Properties of Mellin Transforms IfM{f(x)}=˜f(p), then the following operational properties hold: (a) (Scaling Property ). M{f(ax)}=a−p˜f(p),a >0. (8.3.1) PROOF By deﬁnition, we have, M{f(ax)}=∞⎪integraldisplay 0xp−1f(ax)dx, which is, by substituting ax=t, =1 ap∞⎪integraldisplay 0tp−1f(t)dt=˜f(p) ap. (b) (Shifting Property ). M[xaf(x)] =˜f(p+a). (8.3.2) Its proof follows from the deﬁnition. (c) M{f(xa)}=1 a˜f⎪parenleftBigp a⎪parenrightBig , (8.3.3) © 2007 by Taylor & Francis Group, LLC 344 INTEGRAL TRANSFORMS and THEIR APPLICATIONS M⎪braceleftbigg1 xf⎪parenleftbigg1 x⎪parenrightbigg⎪bracerightbigg =˜f(1−p), (8.3.4) M{(logx)nf(x)}=dn dpn˜f(p),n=1,2,3,.... (8.3.5) The proofs of (8.3.3) and (8.3.4) are easy and hence, left to the reader. Result (8.3.5) can easily be proved by using the result d dpxp−1=( l o g x)xp−1. (8.3.6) (d) (Mellin Transforms of Derivatives ). M[f/prime(x)] =−(p−1)˜f(p−1), (8.3.7) provided [ xp−1f(x)] vanishes as x→0a n da s x→∞. M[f/prime/prime(x)] = (p−1)(p−2)˜f(p−2). (8.3.8) More generally, M[f(n)(x)] = (−1)nΓ(p) Γ(p−n)˜f(p−n) =(−1)nΓ(p) Γ(p−n)M[f(x),p−n], (8.3.9) provided xp−r−1f(r)(x)=0a s x→0f o rr=0,1,2,...,(n−1). PROOF We have, by deﬁnition, M[f/prime(x)] =∞⎪integraldisplay 0xp−1f/prime(x)dx, which is, integrating by parts, =[xp−1f(x)]∞ 0−(p−1)∞⎪integraldisplay 0xp−2f(x)dx =−(p−1)˜f(p−1). The proofs of (8.3.8) and (8.3.9) are similar and left to the reader. (e) IfM{f(x)}=˜f(p), then M{xf/prime(x)}=−p˜f(p), (8.3.10) © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 345 provided xpf(x)v a n i s h e sa t x=0a n da s x→∞. M{x2f/prime/prime(x)}=(−1)2p(p+1 )˜f(p). (8.3.11) More generally, M{xnf(n)(x)}=(−1)nΓ(p+n) Γ(p)˜f(p). (8.3.12) PROOF We have, by deﬁnition, M{xf/prime(x)}=∞⎪integraldisplay 0xpf/prime(x)dx, which is, integrating by parts, =[xpf(x)]∞ 0−p∞⎪integraldisplay 0xp−1f(x)dx=−p˜f(p). Similar arguments can be used to prove results (8.3.11) and (8.3.12). (f) (Mellin Transforms of Diﬀerential Operators ). IfM{f(x)}=˜f(p), then M⎪bracketleftBigg⎪parenleftbigg xd dx⎪parenrightbigg2 f(x)⎪bracketrightBigg =M[x2f/prime/prime(x)+xf/prime(x)] = (−1)2p2˜f(p),(8.3.13) and more generally, M⎪bracketleftbigg⎪parenleftbigg xd dx⎪parenrightbiggn f(x)⎪bracketrightbigg =(−1)npn˜f(p). (8.3.14) PROOF We have, by deﬁnition, M⎪bracketleftBigg⎪parenleftbigg xd dx⎪parenrightbigg2 f(x)⎪bracketrightBigg =M[x2f/prime/prime(x)+xf/prime(x)] =M[x2f/prime/prime(x)] +M[xf/prime(x)] =−p˜f(p)+p(p+1 )˜f(p) by (8.3.10) and (8.3.11) =(−1)2p2˜f(p). © 2007 by Taylor & Francis Group, LLC 346 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Similar arguments can be used to pro ve the general result (8.3.14). (g) (Mellin Transforms of Integrals ). M⎧ ⎨ ⎩x⎪integraldisplay 0f(t)dt⎫ ⎬ ⎭=−1 p˜f(p+1 ). (8.3.15) In general, M{Inf(x)}=M⎧ ⎨ ⎩x⎪integraldisplay 0In−1f(t)dt⎫ ⎬ ⎭=(−1)nΓ(p) Γ(p+n)˜f(p+n),(8.3.16) where Inf(x)i st h e nth repeated integral of f(x) deﬁned by Inf(x)=x⎪integraldisplay 0In−1f(t)dt. (8.3.17) PROOF We write F(x)=x⎪integraldisplay 0f(t)dt so that F/prime(x)=f(x)w i t h F(0) = 0. Application of (8.3.7) with F(x) as deﬁned gives M{f(x)=F/prime(x),p}=−(p−1)M⎧ ⎨ ⎩x⎪integraldisplay 0f(t)dt, p−1⎫ ⎬ ⎭, which is, replacing pbyp+1 , M⎧ ⎨ ⎩x⎪integraldisplay 0f(t)dt, p⎫ ⎬ ⎭=−1 pM{f(x),p+1}=−1 p˜f(p+1 ). An argument similar to this can be used to prove (8.3.16). (h) (Convolution Type Theorems ). IfM{f(x)}=˜f(p)a n dM{g(x)}=˜g(p), then M[f(x)∗g(x)] =M⎡ ⎣∞⎪integraldisplay 0f(ξ)g⎪parenleftbiggx ξ⎪parenrightbiggdξ ξ⎤ ⎦=˜f(p)˜g(p),(8.3.18) M[f(x)◦g(x)] =M⎡ ⎣∞⎪integraldisplay 0f(xξ)g(ξ)dξ⎤ ⎦=˜f(p)˜g(1−p).(8.3.19) © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 347 PROOF We have, by deﬁnition, M[f(x)∗g(x)] =M⎡ ⎣∞⎪integraldisplay 0f(ξ)g⎪parenleftbiggx ξ⎪parenrightbiggdξ ξ⎤ ⎦ =∞⎪integraldisplay 0xp−1dx∞⎪integraldisplay 0f(ξ)g⎪parenleftbiggx ξ⎪parenrightbiggdξ ξ =∞⎪integraldisplay 0f(ξ)dξ ξ∞⎪integraldisplay 0xp−1g⎪parenleftbiggx ξ⎪parenrightbigg dx,⎪parenleftbiggx ξ=η⎪parenrightbigg , =∞⎪integraldisplay 0f(ξ)dξ ξ∞⎪integraldisplay 0(ξη)p−1g(η)ξd η =∞⎪integraldisplay 0ξp−1f(ξ)dξ∞⎪integraldisplay 0ηp−1g(η)dη=˜f(p)˜g(p). Similarly, we have M[f(x)◦g(x)] =M⎡ ⎣∞⎪integraldisplay 0f(xξ)g(ξ)dξ⎤ ⎦ =∞⎪integraldisplay 0xp−1dx∞⎪integraldisplay 0f(xξ)g(ξ)dξ,(xξ=η), =∞⎪integraldisplay 0g(ξ)dξ∞⎪integraldisplay 0ηp−1ξ1−pf(η)dη ξ =∞⎪integraldisplay 0ξ1−p−1g(ξ)dξ∞⎪integraldisplay 0ηp−1f(η)dη=˜g(1−p)˜f(p). Note that, in this case, the operation ◦is not commutative. Clearly, putting x=s, M−1{˜f(1−p)˜g(p)}=∞⎪integraldisplay 0g(st)f(t)dt. Putting g(t)=e−tand ˜g(p)=Γ ( p), we obtain the Laplace transform of f(t) M−1{˜f(1−p)Γ(p)}=∞⎪integraldisplay 0e−stf(t)dt=L{f(t)}=¯f(s). (8.3.20) © 2007 by Taylor & Francis Group, LLC 348 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (i) (Parseval’s Type Property ). IfM{f(x)}=˜f(p)a n dM{g(x)}=˜g(p), then M[f(x)g(x)] =1 2πic+i∞⎪integraldisplay c−i∞˜f(s)˜g(p−s)ds. (8.3.21) Or, equivalently, ∞⎪integraldisplay 0xp−1f(x)g(x)dx=1 2πic+i∞⎪integraldisplay c−i∞˜f(s)˜g(p−s)ds. (8.3.22) In particular, when p= 1, we obtain the Parseval formula for the Mellin trans- form, ∞⎪integraldisplay 0f(x)g(x)dx=1 2πic+i∞⎪integraldisplay c−i∞˜f(s)˜g(1−s)ds. (8.3.23) PROOF By deﬁnition, we have M[f(x)g(x)] =∞⎪integraldisplay 0xp−1f(x)g(x)dx =1 2πi∞⎪integraldisplay 0xp−1g(x)dxc+i∞⎪integraldisplay c−i∞x−s˜f(s)ds =1 2πic+i∞⎪integraldisplay c−i∞˜f(s)ds∞⎪integraldisplay 0xp−s−1g(x)dx =1 2πic+i∞⎪integraldisplay c−i∞˜f(s)˜g(p−s)ds. When p= 1, the above result becomes (8.3.23). © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 349 8.4 Applications of Mellin Transforms Example 8.4.1 Obtain the solution of the boundary value problem x2uxx+xux+uyy=0,0≤x<∞,0<y< 1 (8.4.1) u(x,0) = 0 ,u(x,1) =⎧ ⎨ ⎩A, 0≤x≤1 0,x > 1⎫ ⎬ ⎭, (8.4.2) where Ais a constant. We apply the Mellin transform of u(x, y) with respect to xdeﬁned by ˜u(p,y)=∞⎪integraldisplay 0xp−1u(x, y)dx to reduce the given system into the form ˜uyy+p2˜u=0,0<y< 1 ˜u(p,0)= 0 ,˜u(p,1) =A1⎪integraldisplay 0xp−1dx=A p. The solution of the transformed problem is ˜u(p,y)=A psinpy sinp,0<Rep<1. The inverse Mellin transform gives u(x, y)=A 2πic+i∞⎪integraldisplay c−i∞x−p psinpy sinpdp, (8.4.3) where ˜ u(p,y) is analytic in the vertical strip 0 <Re (p)=c<π. The integrand of (8.4.3) has simple poles at p=nπ,n=1,2,3,...which lie inside a semi- circular contour in the right half plane. Evaluating (8.4.3) by theory of residues gives the solution for x>1a s u(x, y)=A π∞⎪summationdisplay n=11 n(−1)nx−nπsinnπy. (8.4.4) © 2007 by Taylor & Francis Group, LLC 350 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 8.4.2 (Potential in an Inﬁnite Wedge ). Find the potential φ(r, θ) that satisﬁes the Laplace equation r2φrr+rφr+φθθ= 0 (8.4.5) in an inﬁnite wedge 0 <r<∞,−α<θ<α a ss h o w ni nF i g u r e8 . 1w i t ht h e boundary conditions φ(r, α)=f(r),φ(r,−α)=g(r)0≤r<∞, (8.4.6ab) φ(r, θ)→0a s r→∞ for all θin−α<θ<α . (8.4.7) 0 xy -= =- Figure 8.1 An inﬁnite wedge. We apply the Mellin transform of the potential φ(r, θ) deﬁned by M[φ(r, θ)] =˜φ(p,θ)=∞⎪integraldisplay 0rp−1φ(r, θ)dr to the diﬀerential system (8.4.5)–(8.4.7) to obtain d2˜φ dθ2+p2˜φ=0, (8.4.8) ˜φ(p,α)=˜f(p), ˜φ(p,−α)=˜g(p). (8.4.9ab) The general solution of th e transformed equation is ˜φ(p,θ)=Acospθ+Bsinpθ, (8.4.10) © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 351 where AandBare functions of pandα. The boundary conditions (8.4.9ab) determine AandB, which satisfy Acospα+Bsinpα=˜f(p), Acospα−Bsinpα=˜g(p). These give A=˜f(p)+˜g(p) 2c o spα,B =˜f(p)−˜g(p) 2s i npα. Thus, solution (8.4.10) becomes ˜φ(p,θ)=˜f(p).sinp(α+θ) sin(2pα)+˜g(p)sinp(α−θ) sin(2pα) =˜f(p)˜h(p,α+θ)+˜g(p)˜h(p,α−θ), (8.4.11) where ˜h(p,θ)=sinpθ sin(2pα). Or, equivalently, h(r, θ)=M−1⎪braceleftbiggsinpθ sin2pα⎪bracerightbigg =⎪parenleftbigg1 2α⎪parenrightbiggrnsinnθ (1 + 2 rncosnθ+r2n),(8.4.12) where n=π 2αor,2α=π n. Application of the inverse Mellin transform to (8.4.11) gives φ(r, θ)=M−1⎪braceleftBig ˜f(p)˜h(p,α+θ)⎪bracerightBig +M−1⎪braceleftBig ˜g(p)˜h(p,α−θ)⎪bracerightBig , which is, by the convolution property (8.3.18), φ(r, θ)=rncosnθ 2α⎡ ⎣∞⎪integraldisplay 0ξn−1f(ξ)dξ ξ2n−2(rξ)nsinnθ+r2n +∞⎪integraldisplay 0ξn−1g(ξ)dξ ξ2n+2 (rξ)nsinnθ+r2n⎤ ⎦,|α|<π 2n.(8.4.13) This is the formal solution of the problem. In particular, when f(r)=g(r), solution (8.4.11) becomes ˜φ(p,θ)=˜f(p)cospθ cospα=˜f(p)˜h(p,θ), (8.4.14) where ˜h(p,θ)=cospθ cospα=M{h(r, θ)}. © 2007 by Taylor & Francis Group, LLC 352 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Application of the inverse Mellin transform to (8.4.14) combined with the convolution property (8.3.18) yields the solution φ(r, θ)=∞⎪integraldisplay 0f(ξ)h⎪parenleftbiggr ξ,θ⎪parenrightbiggdξ ξ, (8.4.15) where h(r, θ)=M−1⎪braceleftbiggcospθ cospα⎪bracerightbigg =⎪parenleftbiggrn α⎪parenrightbigg(1 +r2n)cos(nθ) (1 + 2 r2ncos2nθ+r2n), (8.4.16) andn=π 2α. Some applications of the Mellin transform to boundary value problems are given by Sneddon (1951) and Tranter (1966). Example 8.4.3 Solve the integral equation ∞⎪integraldisplay 0f(ξ)k(xξ)dξ=g(x),x > 0. (8.4.17) Application of the Mellin transform with respect to xto equation (8.4.17) combined with (8.3.19) gives ˜f(1−p)˜k(p)=˜g(p), which gives, replacing pby 1−p, ˜f(p)=˜g(1−p)˜h(p), where ˜h(p)=1 ˜k(1−p). The inverse Mellin transform combined with (8.3.19) leads to the solution f(x)=M−1⎪braceleftBig ˜g(1−p)˜h(p)⎪bracerightBig =∞⎪integraldisplay 0g(ξ)h(xξ)dξ, (8.4.18) provided h(x)=M−1⎪braceleftBig ˜h(p)⎪bracerightBig exists. Thus, the problem is formally solved. If, in particular, ˜h(p)=˜k(p), then the solution of (8.4.18) becomes f(x)=∞⎪integraldisplay 0g(ξ)k(xξ)dξ, (8.4.19) © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 353 provided ˜k(p)˜k(1−p)=1 . Example 8.4.4 Solve the integral equation ∞⎪integraldisplay 0f(ξ)g⎪parenleftbiggx ξ⎪parenrightbiggdξ ξ=h(x), (8.4.20) where f(x) is unknown and g(x)a n d h(x) are given functions. Applications of the Mellin transform with respect to xgives ˜f(p)=˜h(p)˜k(p),˜k(p)=1 ˜g(p). Inversion, by the convolution property (8.3.18), gives the solution f(x)=M−1⎪braceleftBig ˜h(p)˜k(p)⎪bracerightBig =∞⎪integraldisplay 0h(ξ)k⎪parenleftbiggx ξ⎪parenrightbiggdξ ξ. (8.4.21) 8.5 Mellin Transforms of the Weyl Fractional Integral and the Weyl Fractional Derivative DEFINITION 8.5.1 The Mellin transform of the Weyl fractional integral off(x)is deﬁned by W−α[f(x)] =1 Γ(α)∞⎪integraldisplay x(t−x)α−1f(t)dt,0<Reα<1,x > 0.(8.5.1) Often xW−α ∞is used instead of W−αto indicate the limits to integration. Result (8.5.1) can be interpreted as the Weyl transform of f(t), deﬁned by W−α[f(t)] =F(x, α)=1 Γ(α)∞⎪integraldisplay x(t−x)α−1f(t)dt. (8.5.2) We ﬁrst give some simple examples of the Weyl transform. © 2007 by Taylor & Francis Group, LLC 354 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Iff(t)=exp ( −at), Re a>0, then the Weyl transform of f(t)i sg i v e nb y W−α[exp(−at)] =1 Γ(α)∞⎪integraldisplay x(t−x)α−1exp(−at)dt, which is, by the change of variable t−x=y, =e−ax Γ(α)∞⎪integraldisplay 0yα−1exp(−ay)dy which is, by letting ay=t, W−α[f(t)] =e−ax aα1 Γ(α)∞⎪integraldisplay 0tα−1e−tdt=e−ax aα. (8.5.3) Similarly, it can be shown that W−α[t−μ]=Γ(μ−α) Γ(μ)xα−μ,0<Reα<Reμ. (8.5.4) Making reference to Gradshteyn and Ryzhik (2000, p. 424), we obtain W−α[sinat]=a−αsin⎪parenleftBig ax+πα 2⎪parenrightBig , (8.5.5) W−α[cosat]=a−αcos⎪parenleftBig ax+πα 2⎪parenrightBig , (8.5.6) where 0 <Reα<1a n d a>0. It can be shown that, for any two positive numbers αandβ,t h eW e y l fractional integral satisﬁes the laws of exponents W−α[W−βf(x)] =W−(β+α)[f(x)] =W−β[W−αf(x)]. (8.5.7) Invoking a change of variable t−x=yin (8.5.1), we obtain W−α[f(x)] =1 Γ(α)∞⎪integraldisplay 0yα−1f(x+y)dy. (8.5.8) We next diﬀerentiate (8.5.8) to obtain, D=d dx, D[W−αf(x)] =1 Γ(α)∞⎪integraldisplay 0tα−1∂ ∂xf(x+t)dt =1 Γ(α)∞⎪integraldisplay 0tα−1Df(x+t)dt =W−α[Df(x)]. (8.5.9) © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 355 A similar argument leads to a more general result Dn[W−αf(x)] =W−α[Dnf(x)], (8.5.10) where nis a positive integer. Or, symbolically, DnW−α=W−αDn. (8.5.11) We now calculate the Mellin transform of the Weyl fractional integral by putting h(t)=tαf(t)a n dg⎪parenleftBigx t⎪parenrightBig =1 Γ(α)⎪parenleftbig 1−x t⎪parenrightbigα−1H⎪parenleftbig 1−x t⎪parenrightbig ,w h e r e H⎪parenleftBig 1−x t⎪parenrightBig is the Heaviside unit step function so that (8.5.1) becomes F(x, α)=∞⎪integraldisplay 0h(t)g⎪parenleftBigx t⎪parenrightBigdt t, (8.5.12) which is, by the convolution property (8.3.18), ˜F(p,α)=˜h(p)˜g(p), where ˜h(p)=M{xαf(x)}=˜f(p+α), and ˜g(p)=M⎪braceleftbigg1 Γ(α)(1−x)α−1H(1−x)⎪bracerightbigg =1 Γ(α)1⎪integraldisplay 0xp−1(1−x)α−1dx=B(p,α) Γ(α)=Γ(p) Γ(p+α). Consequently, ˜F(p,α)=M[W−αf(x),p]=Γ(p) Γ(p+α)˜f(p+α). (8.5.13) It is important to note that this result is an obvious extension of result 7(b) in Exercise 8.8 DEFINITION 8.5.2 Ifβis a positive number and nis the smallest integer greater than βsuch that n−β=α>0, the Weyl fractional derivative of a function f(x)is deﬁned by Wβ[f(x)] =EnW−(n−β)[f(x)] =(−1)n Γ(n−β)dn dxn∞⎪integraldisplay x(t−x)n−β−1f(t)dt, (8.5.14) © 2007 by Taylor & Francis Group, LLC 356 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where E=−D. Or, symbolically, Wβ=EnW−α=EnW−(n−β). (8.5.15) It can be shown that, for any β, W−βWβ=I=WβW−β. (8.5.16) And, for any βandγ, the Weyl fractional derivative satisﬁes the laws of exponents Wβ[Wγf(x)] =Wβ+γ[f(x)] =Wγ[Wβf(x)]. (8.5.17) We now calculate the Weyl fractional derivative of some elementary functions. Iff(x)=e x p ( −ax),a>0, then the deﬁnition (8.5.14) gives Wβe−ax=En[W−(n−β)e−ax]. (8.5.18) Writing n−β=α>0 and using (8.5.3) yields Wβe−ax=En[W−αe−ax]=En[a−αe−ax] =a−α(ane−ax)=aβe−ax. (8.5.19) Replacing βby−αin (8.5.19) leads to result (8.5.3) as expected. Similarly, we obtain Wβx−μ=Γ(β+μ) Γ(μ)x−(β+μ). (8.5.20) It is easy to see that Wβ(cosax)=E[W−(1−β)cosax], which is, by (8.5.6), =aβcos⎪parenleftbigg ax−1 2πβ⎪parenrightbigg . (8.5.21) Similarly, Wβ(sinax)=aβsin⎪parenleftbigg ax−1 2πβ⎪parenrightbigg , (8.5.22) provided αandβlie between 0 and 1. Ifβis replaced by −α, result (8.5.20)–(8.5.22) reduce to (8.5.4)–(8.5.6) respectively. Finally, we calculate the Mellin transform of the Weyl fractional derivative with the help of (8.3.9) and ﬁnd M[Wβf(x)] =M[EnW−(n−β)f(x)] = (−1)nM[DnW−(n−β)f(x)] =Γ(p) Γ(p−n)M[W−(n−β)f(x),p−n], © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 357 which is, by result (8.5.13), =Γ(p) Γ(p−n)·Γ(p−n) Γ(p−β)˜f(p−β) =Γ(p) Γ(p−β)M[f(x),p−β] =Γ(p) Γ(p−β)˜f(p−β). (8.5.23) Example 8.5.1 (The Fourier Transform of the Weyl Fractional Integral ). F{W−αf(x)}=e x p⎪parenleftbigg −πiα 2⎪parenrightbigg k−αF{f(x)}. (8.5.24) We have, by deﬁnition, F{W−αf(x)}=1 √ 2π1 Γ(α)∞⎪integraldisplay −∞e−ikxdx∞⎪integraldisplay x(t−x)α−1f(t)dt =1 √ 2π∞⎪integraldisplay −∞f(t)dt·1 Γ(α)t⎪integraldisplay −∞exp(−ikx)(t−x)α−1dx. Thus, F{W−αf(x)}=1 √ 2π∞⎪integraldisplay −∞e−iktf(t)dt·1 Γ(α)∞⎪integraldisplay 0eikττα−1dτ,(t−x=τ) =F{f(x)}1 Γ(α)M{eikτ} =e x p⎪parenleftbigg −πiα 2⎪parenrightbigg k−αF{f(x)}. In the limit as α→0 lim α→0F{W−αf(x)}=F{f(x)}. This implies that W0{f(x)}=f(x). We conclude this section by proving a general property of the Riemann- Liouville fractional integral operator D−α,a n dt h eW e y lf r a c t ional integral © 2007 by Taylor & Francis Group, LLC 358 INTEGRAL TRANSFORMS and THEIR APPLICATIONS operator W−α. It follows from the deﬁnition (6.2.1) that D−αf(t)c a nb e expressed as the convolution D−αf(x)=gα(t)∗f(t), (8.5.25) where gα(t)=tα−1 Γ(α),t >0. Similarly, W−αf(x) can also be written in terms of the convolution W−αf(x)=gα(−x)∗f(x). (8.5.26) Then, under suitable conditions, M[D−αf(x)] =Γ(1−α−p) Γ(1−p)˜f(p+α), (8.5.27) M[W−αf(x)] =Γ(p) Γ(α+p)˜f(p+α). (8.5.28) Finally, a formal computation gives ∞⎪integraldisplay 0{D−αf(x)}g(x)dx=1 Γ(α)∞⎪integraldisplay 0g(x)dxx⎪integraldisplay 0(x−t)α−1f(t)dt =∞⎪integraldisplay 0f(t)dt·1 Γ(α)∞⎪integraldisplay t(x−t)α−1g(x)dx =∞⎪integraldisplay 0f(t)[W−αg(t)]dt, which is, using the inner product notation, /angbracketleftD−αf, g/angbracketright=/angbracketleftf, W−αg/angbracketright. (8.5.29) This show that D−αandW−αbehave like adjoint operators. Obviously, this result can be used to deﬁne fractional integrals of distributions. This result istaken from Debnath and Grum (1988). 8.6 Application of Mellin Transforms to Summation of Series In this section we discuss a method of summation of series that is particularly associated with the work of Macfarlane (1949). © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 359 THEOREM 8.6.1 IfM{f(x)}=˜f(p), then ∞⎪summationdisplay n=0f(n+a)=1 2πic+i∞⎪integraldisplay c−i∞˜f(p)ξ(p,a)dp, (8.6.1) where ξ(p,a)i st h e Hurwitz zeta function deﬁned by ξ(p,a)=∞⎪summationdisplay n=01 (n+a)p,0≤a≤1,Re(p)>1. (8.6.2) PROOF If follows from the inverse Mellin transform that f(n+a)=1 2πic+i∞⎪integraldisplay c−i∞˜f(p)(n+a)−pdp. (8.6.3) Summing this over all ngives ∞⎪summationdisplay n=0f(n+a)=1 2πic+i∞⎪integraldisplay c−i∞˜f(p)ξ(p,a)dp. This completes the proof. Similarly, the scaling property (8.3.1) gives f(nx)=M−1{n−p˜f(p)}=1 2πic+i∞⎪integraldisplay c−i∞x−pn−p˜f(p)dp. Thus, ∞⎪summationdisplay n=1f(nx)=1 2πic+i∞⎪integraldisplay c−i∞x−p˜f(p)ζ(p)dp=M−1{˜f(p)ζ(p)}, (8.6.4) where ζ(p)=∞⎪summationdisplay n=1n−pis the Riemann zeta function . When x= 1, result (8.6.4) reduces to ∞⎪summationdisplay n=1f(n)=1 2πic+i∞⎪integraldisplay c−i∞˜f(p)ζ(p)dp. (8.6.5) This can be obtained from (8.6.1) when a=0 . © 2007 by Taylor & Francis Group, LLC 360 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 8.6.1 Show that∞⎪summationdisplay n=1(−1)n−1n−p=( 1−21−p)ζ(p). (8.6.6) Using Example 8.2.1(a), we can write the left-hand side of (8.6.6) multiplied bytnas ∞⎪summationdisplay n=1(−1)n−1n−ptn=∞⎪summationdisplay n=1(−1)n−1tn·1 Γ(p)∞⎪integraldisplay 0xp−1e−nxdx =1 Γ(p)∞⎪integraldisplay 0xp−1dx∞⎪summationdisplay n=1(−1)n−1tnxe−nx =1 Γ(p)∞⎪integraldisplay 0xp−1·te−x 1+te−x·dx =1 Γ(p)∞⎪integraldisplay 0xp−1·t ex+tdx. In the limit as t→1, the above result gives ∞⎪summationdisplay n=1(−1)n−1n−p=1 Γ(p)∞⎪integraldisplay 0xp−11 ex+1dx =1 Γ(p)M⎪braceleftbigg1 ex+1⎪bracerightbigg =( 1−21−p)ζ(p), in which result (8.2.11) is used. Example 8.6.2 Show that∞⎪summationdisplay n=1⎪parenleftbiggsinan n⎪parenrightbigg =1 2(π−a),0<a< 2π. (8.6.7) The Mellin transform of f(x)=⎪parenleftbiggsinax x⎪parenrightbigg gives M⎪bracketleftbiggsinax x⎪bracketrightbigg =∞⎪integraldisplay 0xp−2sinaxdx =Fs⎪braceleftbigg⎪radicalbigg π 2xp−2⎪bracerightbigg =−Γ(p−1) ap−1cos⎪parenleftBigπp 2⎪parenrightBig . © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 361 Substituting this result into (8.6.5) gives ∞⎪summationdisplay n=1⎪parenleftbiggsinan n⎪parenrightbigg =−1 2πic+i∞⎪integraldisplay c−i∞Γ(p−1) ap−1ζ(p)c o s⎪parenleftBigπp 2⎪parenrightBig dp. (8.6.8) We next use the well-known functional equation for the zeta function (2π)pζ(1−p)=2Γ( p)ζ(p)cos⎪parenleftBigπp 2⎪parenrightBig (8.6.9) in the integrand of (8.6.8) to obtain ∞⎪summationdisplay n=1⎪parenleftbiggsinan n⎪parenrightbigg =−a 2·1 2πic+i∞⎪integraldisplay c−i∞⎪parenleftbigg2π a⎪parenrightbiggpζ(1−p) p−1dp. The integral has two simple poles at p=0a n d p= 1 with residues 1 and −π/a, respectively, and the complex integral i s evaluated by calculating the residues at these poles. Thus, the sum of the series is ∞⎪summationdisplay n=1⎪parenleftbiggsinan n⎪parenrightbigg =1 2(π−a). 8.7 Generalized Mellin Transforms In order to extend the applicability of the classical Mellin transform, Naylor (1963) generalized the method of Mellin integral transforms. This generalized Mellin transform is useful for ﬁnding solutions of boundary value problems in regions bounded by the natural coordinate surfaces of a spherical or cylindrical coordinate system. They can be used to solve boundary value problems inﬁnite regions or in inﬁnite regions bounded internally. Thegeneralized Mellin transform of a function f(r) deﬁned in a<r< ∞is introduced by the integral M −{f(r)}=F−(p)=∞⎪integraldisplay a⎪parenleftbigg rp−1−a2p rp+1⎪parenrightbigg f(r)dr. (8.7.1) The inverse transform is given by M−1 −{F−(p)}=f(r)=1 2πi⎪integraldisplay Lr−pF(p)dp, r > a, (8.7.2) © 2007 by Taylor & Francis Group, LLC 362 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where Lis the line Re p=c,a n dF(p) is analytic in the strip |Re(p)|=|c|<γ. By integrating by parts, we can show that M−⎪bracketleftbigg r2∂2f ∂r2+r∂f ∂r⎪bracketrightbigg =p2F−(p)+2papf(a), (8.7.3) provided f(r) is appropriately behaved at inﬁnity. More precisely, lim r→∞⎪bracketleftbig (rp−a2pr−p)rfr−p(rp+a2pr−p)f⎪bracketrightbig =0. (8.7.4) Obviously, this generalized transform seems to be very useful for ﬁnding the solution of boundary value problems in which f(r) is prescribed on the internal boundary at r=a. On the other hand, if the derivative of f(r) is prescribed at r=a,i ti s convenient to deﬁne the associated integral transform by M+[f(r)] =F+(p)=∞⎪integraldisplay a⎪parenleftbigg rp−1+a2p rp+1⎪parenrightbigg f(r)dr,|Re(p)|<r , (8.7.5) and its inverse given by M−1 +[f(p)] =f(r)=1 2πi⎪integraldisplay Lr−pF+(p)dp, r > a. (8.7.6) In this case, we can show by integration by parts that M+⎪bracketleftbigg r2∂2f ∂r2+r∂f ∂r⎪bracketrightbigg =p2F+(p)−2ap+1f/prime(a), (8.7.7) where f/prime(r)e x i s t sa t r=a. THEOREM 8.7.1 (Convolution ). IfM+{f(r)}=F+(p), andM+{g(r)}=G+(p), then M+{f(r)g(r)}=1 2πi⎪integraldisplay LF+(ξ)G+(p−ξ)dξ. (8.7.8) Or, equivalently, f(r)g(r)=M−1 +⎡ ⎣1 2πi⎪integraldisplay LF+(ξ)G+(p−ξ)dξ⎤ ⎦. (8.7.9) © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 363 PROOF We assume that F+(p)a n d G+(p) are analytic in some strip |Re(p)|<γ.T h e n M+{f(r)g(r)}=∞⎪integraldisplay a⎪parenleftbigg rp−1+a2p rp+1⎪parenrightbigg f(r)g(r)dr =∞⎪integraldisplay arp−1f(r)g(r)dr+∞⎪integraldisplay aa2p rp+1f(r)g(r)dr.(8.7.10) =1 2πi⎪integraldisplay LF+(ξ)dξ∞⎪integraldisplay arp−ξ−1g(r)dr +1 2π∞⎪integraldisplay aa2p rp+1g(r)dr⎪integraldisplay Lr−ξF+(ξ)dξ.(8.7.11) Replacing ξby−ξin the ﬁrst integral term and using F+(ξ)=a2ξF+(−ξ), which follows from the deﬁnition (8.7.5), we obtain ⎪integraldisplay Lr−ξF+(ξ)dξ=⎪integraldisplay Lrξa−2ξF+(ξ)dξ. (8.7.12) The path of integration L,R e (ξ)=c, becomes Re( ξ)=−c, but these paths can be reconciled if F(ξ) tends to zero for large Im( ξ). In view of (8.7.11), we have rewritten ∞⎪integraldisplay aa2p rp+1f(r)g(r)dr=1 2πi⎪integraldisplay LF+(ξ)dξ∞⎪integraldisplay aa2p−2ξ rp−ξ+1g(r)dr. (8.7.13) This result is used to rewrite (8.7.10) as M+{f(r)g(r)}=∞⎪integraldisplay a⎪parenleftbigg rp−1+a2p rp+1⎪parenrightbigg f(r)g(r)dr =∞⎪integraldisplay arp−1f(r)g(r)dr+∞⎪integraldisplay aa2p rp+1f(r)g(r)dr =1 2πi⎪integraldisplay LF+(ξ)dξ∞⎪integraldisplay arp−ξ−1g(r)dr +1 2πi⎪integraldisplay LF+(ξ)dξ∞⎪integraldisplay aa2p−2ξ rp−ξ+1g(r)dr =1 2πi⎪integraldisplay LF+(ξ)G+(p−ξ)dξ. © 2007 by Taylor & Francis Group, LLC 364 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This completes the proof. If the range of integration is ﬁnite, then we deﬁne the generalized ﬁnite Mellin transform by Ma −{f(r)}=Fa −(p)=a⎪integraldisplay 0⎪parenleftbigg rp−1−a2p rp+1⎪parenrightbigg f(r)dr, (8.7.14) where Re p<γ. The corresponding inverse transform is given by f(r)=−1 2πi⎪integraldisplay L⎪parenleftBigr a2⎪parenrightBigp Fa −(p)dp,0<r<a , which is, by replacing pby−pand using Fa −(−p)=−a−2pFa −(p), =1 2πi⎪integraldisplay Lr−pFa −(p)dp,0<r<a , (8.7.15) where the path Lis Re p=−cwith|c|<γ. It is easy to verify the result Ma −{r2frr+rf−r}=a⎪integraldisplay 0⎪parenleftbigg rp−1−a2p rp+1⎪parenrightbigg {r2frr+rfr}dr =p2Fa −(p)−2papf(a). (8.7.16) This is a useful result for applications. Similarly, we deﬁne the generalized ﬁnite Mellin transform-pair by Ma +{f(r)}=Fa +(p)=a⎪integraldisplay 0⎪parenleftbigg rp−1+a2p rp+1⎪parenrightbigg f(r)dr, (8.7.17) f(r)=⎪parenleftbig Ma +⎪parenrightbig−1⎪bracketleftbig Fa +(p)⎪bracketrightbig =1 2πi⎪integraldisplay Lr−pFa +(p)dp, (8.7.18) where |Rep|<γ. For this ﬁnite transform, we can also prove Ma +⎪bracketleftbig r2frr+rfr⎪bracketrightbig =a⎪integraldisplay 0⎪parenleftbigg rp−1+a2p rp+1⎪parenrightbigg⎪parenleftbig r2frr+rfr⎪parenrightbig dr =p2Fa +(p)+2ap−1f/prime(a). (8.7.19) This result also seems to be useful for applications. The reader is referred to Naylor (1963) for applications of the above results to boundary value prob- lems. © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 365 8.8 Exercises 1. Find the Mellin transform of each of the following functions: (a)f(x)=H(a−x),a>0, (c)f(x)=1 1+x2, (e)f(x)=xzH(x−x0), (g)f(x)=Ei(x),(b)f(x)=xme−nx,m, n > 0, (d)f(x)=J2 0(x), (f)f(x)=[H(x−x0)−H(x)]xz, (h)f(x)=exEi(x), where the exponential integral is deﬁned by Ei(x)=∞⎪integraldisplay xt−1e−tdt=∞⎪integraldisplay 1ξ−1e−ξxdξ. 2. Derive the Mellin transform-pairs from the bilateral Laplace transform and its inverse given by ¯g(p)=∞⎪integraldisplay −∞e−ptg(t)dt, g (t)=1 2πic+i∞⎪integraldisplay c−i∞ept¯g(p)dp. 3. Show that M⎪bracketleftbigg1 ex+e−x⎪bracketrightbigg =Γ (p)L(p), where L(p)=1 1p−1 3p+1 5p−··· is the Dirichlet L-function . 4. Show that M⎪braceleftbigg1 (1 +ax)n⎪bracerightbigg =Γ(p)Γ(n−p) apΓ(n). 5. Show that M{x−nJn(ax)}=1 2⎪parenleftBiga 2⎪parenrightBign−pΓ⎪parenleftBigp 2⎪parenrightBig Γ⎪parenleftBig n−p 2+1⎪parenrightBig,a > 0,n >−1 2. 6. Show that (a)M−1⎪bracketleftBig cos⎪parenleftBigπp 2⎪parenrightBig Γ(p)˜f(1−p)⎪bracketrightBig =Fc⎪braceleftbigg⎪radicalbigg π 2f(x)⎪bracerightbigg , © 2007 by Taylor & Francis Group, LLC 366 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b)M−1⎪bracketleftBig sin⎪parenleftBigπp 2⎪parenrightBig Γ(p)˜f(1−p)⎪bracketrightBig =Fs⎪braceleftbigg⎪radicalbigg π 2f(x)⎪bracerightbigg . 7. IfI∞ nf(x) denotes the nth repeated integral of f(x) deﬁned by I∞ nf(x)=∞⎪integraldisplay xI∞ n−1f(t)dt, show that (a)M⎡ ⎣∞⎪integraldisplay xf(t)dt, p⎤ ⎦=1 p˜f(p+1 ), (b)M[I∞ nf(x)] =Γ(p) Γ(p+n)˜f(p+n). 8. Show that the integral equation f(x)=h(x)+∞⎪integraldisplay 0g(xξ)f(ξ)dξ has the formal solution f(x)=1 2πic+i∞⎪integraldisplay c−i∞⎪bracketleftBigg˜h(p)+˜g(p)˜h(1−p) 1−˜g(p)˜g(1−p)⎪bracketrightBigg x−pdp. 9. Find the solution of the Laplace integral equation ∞⎪integraldisplay 0e−xξf(ξ)dξ=1 (1 +x)n. 10. Show that the integral equation f(x)=h(x)+∞⎪integraldisplay 0f(ξ)g⎪parenleftbiggx ξ⎪parenrightbiggdξ ξ has the formal solution f(x)=1 2πic+i∞⎪integraldisplay c−i∞x−p˜h(p) 1−˜g(p)dp. © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 367 11. Show that the solution of the integral equation f(x)=e−ax+∞⎪integraldisplay 0exp⎪parenleftbigg −x ξ⎪parenrightbigg f(ξ)dξ ξ is f(x)=1 2πic+i∞⎪integraldisplay c−i∞(ax)−p⎪braceleftbiggΓ(p) 1−Γ(p)⎪bracerightbigg dp. 12. Assuming (see Harrington, 1967) M⎪bracketleftbig f(reiθ)⎪bracketrightbig =∞⎪integraldisplay 0rp−1f(reiθ)dr, p is real , and putting reiθ=ξ,M{f(ξ)}=F(p) show that (a)M[f(reiθ);r→p]=e xp ( −ipθ)F(p). Hence, deduce (b)M−1{F(p)c o s pθ}=R e [f(reiθ)], (c)M−1{F(p)s i npθ}=−Im[f(reiθ)]. 13. (a) If M[exp(−r)] = Γ( p), show that M⎪bracketleftbig exp(−reiθ)⎪bracketrightbig =Γ (p)e−ip θ, (b) IfM[log(1 + r)] =π psinπp, then show that M⎪bracketleftbig Re log (1 + reiθ)⎪bracketrightbig =πcospθ psinπp. 14. Use M−1⎪braceleftbiggπ sinpπ⎪bracerightbigg =1 1+x=f(x), and Exercises 12(b) and 12(c), re- spectively, to show that (a)M−1⎪braceleftbiggπcospθ sinpπ;p→r⎪bracerightbigg =1+rcosθ 1+2rcosθ+r2, (b)M−1⎪braceleftbiggπsinpθ sinpπ;p→r⎪bracerightbigg =rsinθ 1+2rcosθ+r2. 15. Find the inverse Mellin transforms of (a) Γ( p)cospθ, where −π 2<θ<π 2, (b) Γ( p)sinpθ. © 2007 by Taylor & Francis Group, LLC 368 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 16. Obtain the solution of Example 8.4.2 with the boundary data (a)φ(r, α)=φ(r,−α)=H(a−r). (b) Solve equation (8.4.5) in 0 <r< ∞,0<θ<α with the boundary conditions φ(r,0) = 0 and φ(r, α)=f(r). 17. Show that (a)∞⎪summationdisplay n=1coskn n2=⎪bracketleftbiggk2 4−πk 2+π2 6⎪bracketrightbigg ,and (b)∞⎪summationdisplay n=11 n2=π2 6. 18. If f(x)=∞⎪summationtext n=1ane−nx,show that M{f(x)}=˜f(p)=Γ ( p)g(p), where g(p)=∞⎪summationdisplay n=1ann−pis the Dirichlet series. Ifan=1f o ra l l n,d e r i v e ˜f(p)=Γ ( p)ζ(p). Show that M⎪braceleftbiggexp(−ax) 1−e−x⎪bracerightbigg =Γ (p)ξ(p,a). 19. Show that (a)∞⎪summationdisplay n=1(−1)n−1 np=( 1−21−p)ζ(p). Hence, deduce (b)∞⎪summationdisplay n=1(−1)n−1 n2=π2 12, (c)∞⎪summationdisplay n=1(−1)n−1 n4=⎪parenleftbigg7 8⎪parenrightbiggπ4 90. 20. Find the sum of the following series (a)∞⎪summationdisplay n=1(−1)n−1 n2coskn, (b)∞⎪summationdisplay n=1(−1)n−1 nsinkn. 21. Show that the solution of the boundary value problem r2φrr+rφr+φθθ=0,0<r<∞,0<θ<π φ(r,0) =φ(r, π)=f(r), © 2007 by Taylor & Francis Group, LLC Mellin Transforms and Their Applications 369 is φ(r, θ)=1 2πic+i∞⎪integraldisplay c−i∞r−p˜f(p)c o s⎪braceleftBig p⎪parenleftBig θ−π 2⎪parenrightBig⎪bracerightBig dp cos⎪parenleftBigπp 2⎪parenrightBig . 22. Evaluate∞⎪summationdisplay n=1cosan n3=1 12(a3−3πa2+2π2a). 23. Prove the following results: (a)M⎡ ⎣∞⎪integraldisplay 0ξnf(xξ)g(ξ)dξ⎤ ⎦=˜f(p)˜g(1 +n−p), (b)M⎡ ⎣∞⎪integraldisplay 0ξnf⎪parenleftbiggx ξ⎪parenrightbigg g(ξ)dξ⎤ ⎦=˜f(p)˜g(p+n+1 ). 24. Show that (a)W−α[e−x]=e−x,α > 0, (b)W1 2⎪bracketleftbigg1 √ xexp⎪parenleftbig −√ x⎪parenrightbig⎪bracketrightbigg =K1(√ x) √ πx,x > 0, where K1(x) is the modiﬁed Bessel function of the second kind and order one. 25. (a) Show that the integral (Wong, 1989, pp. 186–187) I(x)=π/2⎪integraldisplay 0J2 ν(xcosθ)dθ, ν > −1 2, can be written as a Mellin convolution I(x)=∞⎪integraldisplay 0f(xξ)g(ξ)dξ, where f(ξ)=J2 ν(ξ)a n dg ( ξ)=⎪braceleftBigg (1−ξ2)−1 2,0<ξ< 1 0,ξ ≥1⎪bracerightBigg . (b) Prove that the integration contour in the Parseval identity I(x)=1 2πic+i∞⎪integraldisplay c−i∞x−p˜f(p)˜g(1−p)dp, −2ν<c< 1, cannot be shifted to the right beyond the vertical line Re p=2 . © 2007 by Taylor & Francis Group, LLC 370 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 26. If f(x)=∞⎪integraldisplay 0exp(−x2t2)·sint t2J1(t)dt, show that M{f(x)}=Γ⎪parenleftbigg p+3 2⎪parenrightbigg Γ⎪parenleftbigg1−p 2⎪parenrightbigg pΓ(p+3 ). 27. Prove the following relations to the Laplace and the Fourier transforms: (a)M[f(x),p]=L[f(e−t),p], (b)M[f(x);a+iω]=F[f(e−t)e−at;ω], whereLis the two-sided Laplace transform and Fis the Fourier trans- form without the factor (2 π)−1 2. 28. Prove the following properties of convolution: (a)f∗g=g∗f, (c)f(x)∗δ(x−1) =f(x),(b) (f∗g)∗h=f∗(g∗h), (d)δ(x−a)∗f(x)=a−1f⎪parenleftBigx a⎪parenrightBig , (e)δn(n−1)∗f(x)=⎪parenleftbiggd dx⎪parenrightbiggn (xnf(x)), (f)⎪parenleftbigg xd dx⎪parenrightbiggn (f∗g)=⎪bracketleftbigg⎪parenleftbigg xd dx⎪parenrightbiggn f⎪bracketrightbigg ∗g=f∗⎪bracketleftbigg⎪parenleftbigg xd dx⎪parenrightbiggn g⎪bracketrightbigg . 29. IfM{f(r, θ)}=˜f(p,θ)a n d ∇2f(r, θ)=frr+1 rfr+1 r2fθθ, show that M⎪braceleftbig ∇2f(r, θ)⎪bracerightbig =⎪bracketleftbiggd2 dθ2+(p−2)2⎪bracketrightbigg ˜f(p−2,θ). © 2007 by Taylor & Francis Group, LLC 9 Hilbert and Stieltjes Transforms “The organic unity of mathematics is inherent in the nature of this science, for mathematics is the foundation of all exact knowledge of natural phenomena.” David Hilbert “Mathematics knows no races or geographic boundaries; for math- ematics the cultural world is one country.” David Hilbert 9.1 Introduction In his 1912 famous paper on integral equations, David Hilbert (1862 −1943) introduced an integral transformation, which is now known as the Hilbert transform . Although it was named after Hilbert, the Hilbert transform and its basic properties were developed mainly by G.H. Hardy (1924) and simultane- ously by E.C. Titchmarsh during 1925-1930. On the other hand, T.J. Stieltjes(1856−1894) introduced the Stieltjes transform in his studies on continued fractions. This transform was also involved in Stieltjes’ moment problems. Both the Hilbert and Stieltjes transforms arise in many problems in ap- plied mathematics, mathematical physi cs, and engineering science. The former plays an important role in ﬂuid mechanics, aerodynamics, signal processing, and electronics, while the latter arises in the moment problem. This chap- ter deals with deﬁnitions of Hilbert and Stieltjes transforms with examples.This is followed by a discussion of basic operational properties of these trans- forms. Finally, examples of applications of Hilbert and Stieltjes transforms to physical problems are discussed. 371 © 2007 by Taylor & Francis Group, LLC 372 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 9.2 Deﬁnition of the Hilbert Transform and Examples Iff(t) is deﬁned on the real line −∞<t<∞,i t sHilbert transform , denoted byˆfHHH(x), is deﬁned by HHH{f(t)}=ˆfHHH(x)=1 π∞⎪contintegraldisplay −∞f(t) t−xdt, (9.2.1) where xis real and the integral is treated as a Cauchy principal value, that is, ∞⎪contintegraldisplay −∞f(t)dt t−x= lim ε→0⎡ ⎣x−ε⎪integraldisplay −∞+∞⎪integraldisplay x+ε⎤ ⎦f(t)dt t−x. (9.2.2) To derive the inverse Hilbert transform, we rewrite (9.2.1) as ˆfHHH(x)=1 √ 2π∞⎪integraldisplay −∞f(t)g(x−t)dt=(f∗g)(x), (9.2.3) where g(x)=⎪radicalbigg 2 π⎪parenleftbigg −1 x⎪parenrightbigg .Application of the Fourier transform with respect toxgives F(k)=ˆFHHH(k) G(k),G(k)=is gnk. (9.2.4) Taking the inverse Fourier transform, we obtain the solution for f(x)a s f(x)=−1 √ 2π∞⎪integraldisplay −∞(is gnk )ˆFHHH(k)exp(ikx)dk which is, by the Convolution Theorem 2.5.5, =1 π∞⎪contintegraldisplay −∞ˆfHHH(ξ) x−ξdξ=−HHH⎪braceleftBig ˆfHHH(ξ)⎪bracerightBig . (9.2.5) Obviously, −HHH2{f(t)}=−HHH[HHH{f(t)}]=f(x) and hence, HHH−1=−HHH.T h u s , theinverse Hilbert transform is given by f(t)=HHH−1⎪braceleftBig ˆfHHH(x)⎪bracerightBig =−HHH⎪braceleftBig ˆfHHH(x)⎪bracerightBig =−1 π∞⎪contintegraldisplay −∞ˆfHHH(x)dx x−t. (9.2.6) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 373 Example 9.2.1 Find the Hilbert transform of a rectangular pulse given by f(t)=⎪braceleftBigg 1,for|t|<a 0,for|t|>a⎪bracerightBigg . (9.2.7) We have, by deﬁnition, ˆfHHH(x)=1 πa⎪integraldisplay −adt t−x. If|x|<a, the integrand has a singularity at t=x, and hence, ˆfHHH(x)=1 πlim ε→0⎡ ⎣x−ε⎪integraldisplay −adt t−x+a⎪integraldisplay x+εdt t−x⎤ ⎦ =1 πlim ε→0⎪braceleftBig [log|t−x|]x−ε −a+ [log |t−x|]a x+ε⎪bracerightBig =1 πlim ε→0{log|ε|−log|a+x|+l o g|a−x|−log|ε|} =1 πlog⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglea−x a+x⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglefor|x|<a . On the other hand, if |x|>a, the integrand has no singularity in −a<t<a , and hence, ˆfHHH(x)=1 πa⎪integraldisplay −adt t−x=1 π[log|t−x|]a −a=1 πlog⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglea−x a+x⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglefor|x|>a . Finally, we obtain the Hilbert transform of f(t) deﬁned by (9.2.7) as ˆf HHH(x)=1 πlog⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglea−x a+x⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle. (9.2.8) Example 9.2.2 Find the Hilbert transform of f(t)=t (t2+a2),a > 0. (9.2.9) © 2007 by Taylor & Francis Group, LLC 374 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We have, by deﬁnition, ˆfHHH(x)=1 π∞⎪contintegraldisplay −∞td t (t2+a2)(t−x) =1 π(a2+x2)∞⎪contintegraldisplay −∞⎪bracketleftbigga2 t2+a2+x t−x−xt t2+a2⎪bracketrightbigg dt =1 π1 (a2+x2)⎡ ⎣a2∞⎪integraldisplay −∞dt t2+a2+x∞⎪contintegraldisplay −∞dt (t−x)−x∞⎪integraldisplay −∞td t (t2+a2)⎤ ⎦. The second and third integrals as the Cauchy principal value vanish and hence, only the ﬁrst integral makes a non-zero contribution. Thus, we obtain ˆfHHH(x)=1 π1 (a2+x2)·(aπ)=a (a2+x2). (9.2.10) Example 9.2.3 Find the Hilbert transform of (a)f(t)=c o s ωtand ( b)f(t)=s i n ωt. It follows from the deﬁnition of the Hilbert transform that ˆfHHH(x)=1 π∞⎪integraldisplay −∞cosωt (t−x)dt=1 π∞⎪integraldisplay −∞cos{ω(t−x)+ωx}dt (t−x) =1 π∞⎪integraldisplay −∞(t−x)−1[cosω(t−x)cosωx−sinω(t−x)sinωx]dt =cosωx π∞⎪integraldisplay −∞cosω(t−x) t−xdt−sinωx π∞⎪integraldisplay −∞sinω(t−x)dt t−x, which is, in terms of the new variable T=t−x, =cosωx π∞⎪integraldisplay −∞cosωT TdT−sinωx π∞⎪integraldisplay −∞sinωT TdT. (9.2.11) Obviously, the ﬁrst integral vanishes because its integrand is an odd function ofT. On the other hand, the second integral makes a non-zero contribution so that (9.2.11) gives HHH{cosωt}=ˆfHHH(x)=−sinωx π·π=−sinωx. (9.2.12) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 375 Similarly, it can be shown that HHH{sinωt}=c o sωx. (9.2.13) 9.3 Basic Properties of Hilbert Transforms THEOREM 9.3.1 IfHHH{f(t)}=ˆfHHH(x), then the following properties hold: (a)HHH{f(t+a)}=ˆfHHH(x+a), (9.3.1) (b)HHH{f(at)}=ˆfHHH(ax),a > 0, (9.3.2) (c)HHH{f(−at)}=−ˆfHHH(−ax), (9.3.3) (d)HHH{f/prime(t)}=d dxˆfHHH(x), (9.3.4) (e)HHH{tf(t)}=xˆfHHH(x)+1 π∞⎪integraldisplay −∞f(t)dt, (9.3.5) (f)F[HHH{f(t)}]=(−is gnk )F{f(x)}, (9.3.6) (g)/bardblHHH{f(t)}/bardbl=/bardblf(t)/bardbl, (9.3.7) where /bardblf/bardbl=√ <f,f> denotes the norm in L2(R), (h)HHH[f](x)=ˆfHHH(x),HHH[ˆfHHH](x)=−f(Reciprocity relations) ,(9.3.8) (i)<f,HHHg> =<−HHHf, g> and<HHHf, g> =<f,−HHHg>, (9.3.9) (Parseval’s formulas). PROOF (a) We have, by deﬁnition, HHH{f(t+a)}=1 π∞⎪contintegraldisplay −∞f(t+a)dt t−x(t+a=u) =1 π∞⎪contintegraldisplay −∞f(u)du u−(x+a)=ˆfHHH(x+a). © 2007 by Taylor & Francis Group, LLC 376 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b) HHH{f(at)}=1 π∞⎪contintegraldisplay −∞f(at)dt t−x(at=u, a > 0) =1 π∞⎪contintegraldisplay −∞f(u)du u−ax=ˆfHHH(ax). Similarly, result (c) can be proved. (d) HHH{f/prime(t)}=1 π∞⎪contintegraldisplay −∞f/prime(t)dt t−x which is, integrating by parts, =1 π⎪bracketleftbiggf(t) t−x⎪bracketrightbigg∞ −∞+1 π∞⎪contintegraldisplay −∞f(t)dt (t−x)2=d dxˆfHHH(x). Proofs of (e)–(i) are similar and hence, are left to the reader. THEOREM 9.3.2 Iff(t)i sa ne v e nf u n c t i o no f t, then, an alternative form of the Hilbert trans- form is ˆfHHH(x)=x π∞⎪contintegraldisplay −∞f(t)−f(x) (t2−x2)dt. (9.3.10) PROOF As the Cauchy principal value, we have ∞⎪contintegraldisplay −∞dt t−x=0. Consequently, ˆfHHH(x)=1 π∞⎪contintegraldisplay −∞f(t)−f(x) t−xdt=1 π∞⎪contintegraldisplay −∞(t+x){f(t)−f(x)} (t2−x2)dt =1 π∞⎪contintegraldisplay −∞t{f(t)−f(x)} (t2−x2)dx+x π∞⎪contintegraldisplay −∞{f(t)−f(x)}dt (t2−x2).(9.3.11) Since f(t) is an even function, the integrand of the ﬁrst integral of (9.3.11) is an odd function; hence, the ﬁrst integral vanishes, and (9.3.11) gives (9.3.10). Since result (9.2.3) reveals that the Hilbert transform can be written as a convolution transform, we state the following. © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 377 THEOREM 9.3.3 Iffandg∈L1(R) are such that their Hilbert transforms are also in L1(R), then HHH(f∗g)(x)=(HHHf∗g)(x)=(f∗HHHg)(x) (9.3.12) and (f∗g)(x)=−(HHHf∗HHHg)(x). (9.3.13) PROOF We have, by deﬁnition HHH(f∗g)(x)=1 π1 √ 2π∞⎪integraldisplay −∞dt t−x⎡ ⎣∞⎪integraldisplay −∞f(y)g(t−y)dy⎤ ⎦,(t−y=ξ), =1 π√ 2π∞⎪integraldisplay −∞f(y)dy⎡ ⎣∞⎪integraldisplay −∞g(ξ)dξ ξ−(x−y)⎤ ⎦, =1 √ 2π∞⎪integraldisplay −∞f(y)(HHHg)(x−y)dy=(f∗HHHg)(x). Similarly, we can prove the second result in (9.3.12). To prove (9.3.13), we replace gbyHHHgin (9.3.12) and then use HHH2g=−g. Another version of the Hilbert transform and its inversion formula is stated in the following theorem: THEOREM 9.3.4 Iff∈L2(R), then the Hilbert transform ( HHHf)(x)∈L2(R)i sg i v e nb y (HHHf)(x)=−1 πd dx∞⎪integraldisplay −∞f(t)ln⎪parenleftBig 1−x t⎪parenrightBig dt (9.3.14) almost everywhere. Further, the following inversion formula f(t)=1 πd dt∞⎪integraldisplay −∞(HHHf)(x)ln⎪parenleftbigg 1−t x⎪parenrightbigg dx, (9.3.15) holds everywhere with /bardblf/bardbl=/bardblHHHf/bardbl,t h a ti s , ∞⎪integraldisplay −∞|f(x)|2dx=∞⎪integraldisplay −∞|(HHHf)(x)|2dx. (9.3.16) © 2007 by Taylor & Francis Group, LLC 378 INTEGRAL TRANSFORMS and THEIR APPLICATIONS If the diﬀerentiation is performed under the integral signs in (9.3.14) and (9.3.15), we obtain the Hilbert transform pair (9.2.1) and (9.2.6). We close this section by adding a comment. A more rigorous mathematical treatment of classical Hilbert transforms can be found in a treatise by Titch-marsh (1959). Further results and references of related work on Hilbert trans- forms and their applications are given by Kober (1943a,b), Gakhov (1966), Newcomb (1962), and Muskhelishvili (1953). Several authors including Okikiolu (1965) and Kober (1967) introduced the modiﬁed Hilbert transform of a function f(t) ,w h i c hi sd e ﬁ n e db y HHH α[f(t)] =ˆfHHHα(x)=cosec⎪parenleftBigπα 2⎪parenrightBig 2Γ (α)∞⎪contintegraldisplay −∞(t−x)α−1f(t)dt, (9.3.17) where xis real and 0 <α< 1, and the integral is treated as the Cauchy princi- pal value. Obviously, HHHα[f(t)] is closely related to the W eyl fractional integral W−αso that 2s i n⎪parenleftBigπα 2⎪parenrightBig HHHα[f(t)] =W−α[f(t),x]−Wα[f(−t),−x]. (9.3.18) Several properties of HHHα[f(t)] and W−α[f(t)] are investigated by Kober (1967). He also proved the following results, which is stated below without proof. THEOREM 9.3.5 (Parseval’s Relation ). IfHHHα[f(t)] =ˆfHHHα(x), then <HHHαf, g> =−<f,HHHαg>. Or, equivalently, ∞⎪integraldisplay −∞HHHα[f(t),x]g(x)dx=−∞⎪integraldisplay −∞HHHα[g(t),x]f(x)dx. (9.3.19) 9.4 Hilbert Transforms in the Complex Plane In communication and coherence proble ms in electrical engineering (see, for example, Tuttle, 1958), the Hilbert transform in the complex plane plays an important role. In order to deﬁne such a transform, we ﬁrst consider the function f0(z) of a complex variable z=x+iygiven by f0(z)=1 π∞⎪contintegraldisplay −∞f(t)dt t−z,y > 0. (9.4.1) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 379 Application of the Fourier transform deﬁned by (2.3.1) and its inverse by (2.3.2) to (9.2.1) and (9.4.1) gives ˆFHHH(ω)=is gn(ω)F(ω), (9.4.2) F0(ω)=2iexp(−ωy)HHH(ω)F(ω). (9.4.3) In view of (9.4.2), (9.4.3) can be written as F0(ω)=2e x p ( −ωy)HHH(ω)ˆFHHH(ω). (9.4.4) Taking the inverse Fourier transform, we obtain f0(z)=i π∞⎪contintegraldisplay −∞ˆfHHH(t) t−zdt. Or, ∞⎪contintegraldisplay −∞f(t)dt t−z=i∞⎪contintegraldisplay −∞ˆfHHH(t)dt t−z,Im (z) = y >0. (9.4.5) Since, lim y→0+1 t−z=1 t−x+πiδ(t−x), we have, from (9.4.5), f0(z) = lim y→0+1 π∞⎪contintegraldisplay −∞f(t)dt t−z=1 π∞⎪contintegraldisplay −∞f(t)dt t−x+if(x)=ˆfHHH(x)+if(x). This gives a relation between f0(z) and Hilbert transforms. We now deﬁne a complex analytic signal fc(x) from a real signal f(x)b y fc(x)=1 π∞⎪integraldisplay −∞F(ω)HHH(ω)e x p ( iωx)dω. (9.4.6) Since F(ω)HHH(ω)=1 2[F(ω)+sgn(ω)F(ω)] =1 2⎪bracketleftBig F(ω)−iˆFHHH(ω)⎪bracketrightBig , fc(x)=1 2π∞⎪integraldisplay −∞⎪bracketleftBig F(ω)−iˆFHHH(ω)⎪bracketrightBig exp(iωx)dω=f(x)−iˆfHHH(x).(9.4.7) Since f(x)i sr e a l ,R e {fc(x)}=f(x)a n d Im{fc(x)}=−ˆfHHH(x)=1 π∞⎪integraldisplay −∞f(t)dt x−t. © 2007 by Taylor & Francis Group, LLC 380 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thus, it follows from the inverse Hilbert transform that f(t)=−1 π∞⎪contintegraldisplay −∞Im{fc(x)}dx x−t=1 π∞⎪contintegraldisplay −∞fHHH(x)dx x−t. (9.4.8) 9.5 Applications of Hilbert Transforms Example 9.5.1 (Boundary Value Problems ). Solve the Laplace equation uxx+uyy=0,−∞<x< ∞,y > 0, (9.5.1) with the boundary conditions ux(x, y)=f(x)o n y=0,for−∞<x< ∞, (9.5.2) u(x, y)→0a s r=(x2+y2)1 2→∞. (9.5.3) Application of the Fourier transform deﬁned by (2.3.1) with respect to x gives the solution for U(k,y)a s U(k,y)=F(k) ikexp(−|k|y)=F(k)G(k), (9.5.4) where G(k)=(ik)−1exp(−|k|y)s ot h a t g(x)=⎪radicalbigg 2 πtan−1⎪parenleftbiggx y⎪parenrightbigg . Using the Convolution Theorem 2.5.5 gives the formal solution u(x, y)=1 √ 2π∞⎪integraldisplay −∞f(t)g(x−t)dt =1 π∞⎪integraldisplay −∞f(t)t an−1⎪parenleftbiggx−t y⎪parenrightbigg dt. (9.5.5) Obviously, it follows from (9.5.5) that uy(x,0) =1 π∞⎪integraldisplay −∞f(t)dt t−x=HHH{f(t)}. (9.5.6) Thus, the Hilbert transform of the tangential derivative ux(x,0) =f(x)i st h e normal derivative uy(x,0) on the boundary at y=0 . © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 381 Example 9.5.2 (Nonlinear Internal Waves ). We consider a linear homogeneous partial diﬀer- ential equation with consta nt coeﬃcients in the form P⎪parenleftbigg∂ ∂t,∂ ∂x,∂ ∂y,∂ ∂z⎪parenrightbigg u(x,t)=0, (9.5.7) where Pis a polynomial in partial derivatives, and x=(x, y, z )a n dt i m e t>0. We seek a three-dimensional plane wave solution of (9.5.7) in the form u(x,t)=aexp[i(κ·x−ωt)], (9.5.8) where ais the amplitude, κ=(k,/lscript,m ) is the wavenumber vector, and ωis the frequency. If this solution (9.5.8) is substituted into (9.5.7), partial derivatives ∂ ∂t,∂ ∂x,∂ ∂y,a n d∂ ∂zwill be replaced by −iω,ik,i/lscript,a n d imrespectively. Hence, the solution of (9.5.7) exis ts provided the algebraic equation P(−iω,ik,i/lscript,im )= 0 (9.5.9) is satisﬁed. This relation is universally known as the dispersion relation .P h y s - ically, this gives the frequency ωin terms of wavenumbers k,/lscript,andm.F u r t h e r , the above analysis shows that there is a direct correspondence between thegoverning equation (9.5.7) and the dispersion relation (9.5.9) given by ∂ ∂t↔−iω,⎪parenleftbigg∂ ∂x,∂ ∂y,∂ ∂z⎪parenrightbigg ↔(ik,i/lscript,im ). (9.5.10) Clearly, the dispersion relation can be derived from the governing equation and vice versa by using (9.5.10). In many physical problems, the dispersion relation can be written explicitly in terms of the wavenumbers as ω=W(k,/lscript,m ). (9.5.11) The phase and the group velocities of the waves are deﬁned by Cp(κ)=ω κˆκ, C g(κ)=∇κω, (9.5.12ab) where ˆ κis the unit vector in the direction of the wavevector κ.I nt h eo n e - dimensional problem, (9.5.11)–(9.5.12ab) reduce to ω=W(k),C p=ω k,C g=dω dk. (9.5.13abc) Thus, the one-dimensional waves given by (9.5.8) are called dispersive if the group velocity Cg=ω/prime(k) is not constant, that is, ω/prime/prime(k)/negationslash= 0. Physically, © 2007 by Taylor & Francis Group, LLC 382 INTEGRAL TRANSFORMS and THEIR APPLICATIONS as time progresses, the diﬀerent waves disperse in the medium with the result that a single hump breaks into a series of wavetrains. We consider a simple model of internal solitary waves in an inviscid, stably stratiﬁed two-ﬂuid system between rigid horizontal planes at z=h1andz= h2. The upper ﬂuid of depth h1and density ρ1lies over the heavier lower ﬂuid of depth h2and density ρ2(>ρ1). Both ﬂuids are subj ected to a vertical gravitational force g, and the eﬀects of surface tension are neglected. With z= η(x, t) as the internal wave displacement ﬁeld, the linear dispersion relation for the two-ﬂuid system is ω2=gk(ρ2−ρ1) (ρ1cothkh1+ρ2cothkh2), (9.5.14) where ω(k)a n d kare frequency and wavenumber for a small amplitude sinu- soidal disturbance at the interface of the two ﬂuids. Several important limiting cases of (9.5.14) are of interest. Case (i) :Deep-Water Theory (Benjamin, 1967; Ono, 1975). In this case, the depth of the lower ﬂuid is assumed to be inﬁnite ( h2→∞), and waves are long compared with the depth h1of the upper ﬂuid. This leads to the double limit in the form lim k→0lim h2→∞ω2=c2 0k2−2αc0k3(sgn k +···), (9.5.15) where k→0i su s e dw i t hﬁ x e d h1, and the limit h2→∞ is taken with kand h1ﬁxed, and c2 0=⎪parenleftbiggρ2−ρ1 ρ1⎪parenrightbigg gh1and α=⎪parenleftbiggρ2 ρ1⎪parenrightbigg⎪parenleftbiggh1c0 2⎪parenrightbigg . (9.5.16ab) We consider internal waves propagating only in one direction and retain the ﬁrst dispersive term so that the associated dispersion relation becomes ω=c0k−αk|k|. (9.5.17) This enables us to deﬁne the appropriate space and time scales associated with this limiting case as ξ=β(x−c0t),τ =β2t, (9.5.18ab) where β(<<1) is the long wave parameter deﬁned as the ratio of the wave- guide scale to the wavelength. The linear evolution equation associated with (9.5.17) is ηt+coηx+αHHH{ηxx}=β2[ητ+αHHH{ηξξ}]=0, (9.5.19) where HHH{η(x/prime,t)}is the Hilbert transform of η(x/prime,t) deﬁned by HHH{η(x/prime,t)}=1 π∞⎪contintegraldisplay −∞η(x/prime,t)dx/prime (x/prime−x). (9.5.20) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 383 Equation (9.5.19) is often called the linear Benjamin −Ono equation .B e n - jamin (1967) and Ono (1975) investigated nonlinear internal wave motion anddiscovered the following nonlinear equation (η t+coηx)+c1ηηx+αHHH{ηxx}=0, (9.5.21) where c1andαare constants, which are the characteristics of speciﬁc ﬂows. This equation is usually known as the Benjamin −Ono equation . The solitary wave solution of (9.5.21) has the form (Benjamin, 1967) η(x−ct)=aλ2 (x−ct)2+λ2, (9.5.22) where c=c0+1 2ac1andaλ=−4α c1. It is noted here that the Benjamin −Ono equation is one of the model non- linear evolution equations and it arises in a large variety of physical wave systems. Case (ii) :Shallow-Water Theory (Benjamin, 1966). In this case, long wave ( k→0) disturbances with the length scale h=(h1+ h2)ﬁ x e dl e a dt ot h er e s u l t lim k→0(ω2)=c2 0k2−2c0γk4, (9.5.23) where c2 0=g(ρ2−ρ1)h1h2 (ρ1h2+ρ2h1)and γ=c0h1h2⎪parenleftbiggρ1h1+ρ2h2 ρ1h2+ρ2h1⎪parenrightbigg . (9.5.24ab) If we retain only the ﬁrst dispersive term in (9.5.23) and assume that the wave propagates only to the right, it turns out that ω(k)=c0k−γk3. (9.5.25) The evolution equation associated with (9.6.25) is the well-known linear KdV equation ηt+c0ηx+γηxxx=0. (9.5.26) In terms of a slow time scale τand a slow spatial modulation ξin a coordi- nate system moving at the linear wave velocity deﬁned by ξ=β(x−c0t)a n d τ=β2t, the above equation (9.5.26) reduces to the linear KdV equation ητ+γηξξξ=0. (9.5.27) The standard nonlinear KdV equation is given by ηt+c0ηx+αηη x+γηxxx=0. (9.5.28) © 2007 by Taylor & Francis Group, LLC 384 INTEGRAL TRANSFORMS and THEIR APPLICATIONS It is well known that this equation admits the soliton solution in the form η(x−ct)=asech2⎪parenleftbiggx−ct λ⎪parenrightbigg , (9.5.29) where c=c0+aα 3andaλ2=12γ α. A similar argument can be employed to determine the integrodiﬀerential nonlinear evolution equation associated with an arbitary dispersion relationω(k)=kc(k)i nt h ef o r m ∂η ∂t+c1ηηx+∞⎪integraldisplay −∞K(x−ζ)⎪parenleftbigg∂η ∂ζ⎪parenrightbigg dζ=0, (9.5.30) where the kernel K(x) is a given function. The linearized version of (9.5.30) admits the plane wavelike solution η(x, t)=Aexp[i(kx−ωt)], (9.5.31) provided the following dispersion relation holds, (−iω)exp (ikx)+i∞⎪integraldisplay −∞K(x−ζ)kexp(ikζ)dζ=0. Substituting x−ζ=ξ,t h i sc a nb er e w r i t t e ni nt h ef o r m ω=k∞⎪integraldisplay −∞K(ξ)e x p ( −ikξ)dξ=kc(k), (9.5.32) where c(k) is the Fourier transform of the given kernel K(x)s ot h a t K(x)= F−1{c(k)}. This means that any phase velocity c(k)=F{K(x)}can be ob- tained by choosing the kernel K(x). In particular, if K(x)=c0δ(x)+γδ/prime/prime(x),c(k)=c0+γk2, (9.5.33) equation (9.5.30) reduces to the linear KdV equation ηt+c0ηx+γηxxx=0. (9.5.34) Combining the general dispersion relation of the integral form with typical nonlinearity, we obtain ηt+c0ηx+αηη x+∞⎪integraldisplay −∞K(x−ζ)⎪parenleftbigg∂η ∂ζ⎪parenrightbigg dζ=0. (9.5.35) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 385 Using (9.5.33) in (9.5.35), we can derive the KdV equation (9.5.28). On the other hand, if c(k)=c0(1−α|k|), we can deduce the Benjamin −Ono equation (9.5.21) from (9.5.30). Case (iii) :Finite-Depth Water Wave Theory (Kubota et al., 1978). In this case, h2>> h 1,t h a ti s ,( h1/h2)=O(β), but kh1=O(β)a n d kh2= O(1). This dispersion relation appropriate for this case is ω=c0k−1 2⎪parenleftbiggρ2 ρ1⎪parenrightbigg c0h1k2coth(kh2), (9.5.36) where c2 0=⎪parenleftbiggρ2−ρ1 ρ1⎪parenrightbigg gh1. (9.5.37) We can use (9.5.18ab) for the appropriate space and time scales to investi- gate this case. Thus, the ﬁnite-depth evolution equation can be derived from (9.5.36) and has the form (see Kubota et al., 1978) ηt+c0ηx+c1ηηx+c2∂2 ∂x2⎡ ⎣∞⎪integraldisplay −∞η(x/prime,t) ×⎪braceleftbigg cothπ(x−x/prime) 2h−sgn⎪parenleftbiggx−x/prime h⎪parenrightbigg⎪bracerightbigg⎪bracketrightbigg dx/prime.(9.5.38) The solitary wave solution of this equation was obtained by Joseph and Adams (1981). It is noted that the ﬁnite-depth equation reduces to the Benjamin −Ono equation and the KdV equation in the deep- and shallow-water limits, respec-tively. Finally, all of the above theories ca n be formulated in the framework of a generalized evolution equ ation usually known as the Whitham equation (Whitham, 1967) in the form ∂η ∂t+c1ηηx+∂ ∂x⎡ ⎣∞⎪integraldisplay −∞η(x/prime,t)dx/prime ×1 2π∞⎪integraldisplay −∞exp{ik(x−x/prime)}c(k)dk⎤⎦=0. (9.5.39) Subsequently, Maslowe and Redekop p (1980) have generalized the theory of long nonlinear waves in stratiﬁed shear ﬂows. They have obtained the gov- erning nonlinear evolution equation, which involves the Hilbert transform.In their analysis, the evolution equation contains a damping term describing energy loss by radiation, which can be used to determine the persistence of solitary waves or nonlinear wave packets in physically realistic situations. © 2007 by Taylor & Francis Group, LLC 386 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 9.5.3 (Airfoil Design ). An example of application arises in the design of airfoil which is a symmetric body designed to produce a desired lifting force when it moves in an air medium. A typical example of an airfoil is an airplane wing. We denote the x-coordinates of the leading and trailing edges by xL=a andxT=b, respectively. The normal component of the induced velocity at a point ( ξ, η) on the surface of the airfoil is given by (vi)n=1 2πb⎪integraldisplay af(x)dx x−ξ, (9.5.40) for some function f, which depends on the curl of the velocity vector. Evident- ly, (vi)n(ξ) is the ﬁnite Hilbert transform of fin [a, b]. The normal component of the uniform stream velocity is found to be (v∞)n=v∞sin⎪bracketleftbigg α−tan−1⎪parenleftbiggdz dx⎪parenrightbigg⎪bracketrightbigg , where αis the angle made by the uniform stream with the x-axis and⎪parenleftbigdz dx⎪parenrightbig is the slope of the tangent line to the mean camber line at ( ξ, η). Since the sum of the normal components is zero, we have 1 2πb⎪integraldisplay af(x)dx ξ−x=v∞sin⎪bracketleftbigg α−tan−1⎪parenleftbiggdz dx⎪parenrightbigg⎪bracketrightbigg atx=ξ, (9.5.41) together with the boundary condition f(xT=b)= 0 which is known as the Kutta boundary condition . The major problem of a thin airfoil is to solve the integral equation for f. For small αand⎪parenleftbigdz dx⎪parenrightbig , equation (9.5.41) becomes 1 2πb⎪integraldisplay af(x)dx ξ−x=v∞⎪parenleftbigg α−dz dx⎪parenrightbigg x=ξ, (9.5.42) which, for a symmetrical airfoil with z=constant, reduces to 1 2πb⎪integraldisplay af(x)dx ξ−x=αv∞. (9.5.43) We can solve (9.5.43) explicitly. Without loss of generality, we set b=0 w i t h b−a=cas the length of the main chord of the airfoil. We also assume x=1 2c(1−cosθ)a n d ξ=1 2c(1−cosθ0) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 387 so that (9.5.43) can be transformed into the form 1 2ππ⎪integraldisplay 0f(θ)sinθd θ (cosθ−cosθ0)=αv∞,f(π)=0. (9.5.44) In view of the fact π⎪integraldisplay 0cosnθ dθ (cosθ−cosθ0)=π⎪parenleftbiggsinnθ0 sinθ0⎪parenrightbigg , (9.5.45) the solution of (9.5.44) is f(θ)=2αv∞⎪parenleftbigg1+c o s θ sinθ⎪parenrightbigg . (9.5.46) The lift per unit span is given by L=C⎪integraldisplay 0ρv∞f(θ)dx=ρv∞π⎪integraldisplay 02αv∞⎪parenleftbigg1+c o s θ sinθ⎪parenrightbigg1 2csinθd θ=παcρv2 ∞, (9.5.47) where ρis the constant air density. The general solution of (9.5.42) can b e represented by a sum of two terms. T h eﬁ r s tt e r mh a st h ef o r m f(θ) for the symmetric airfoil given by (9.5.46) and the second term can be represented by a Fourier sine series. Thus, we have f(θ)=2v∞⎪bracketleftBigg a0⎪parenleftbigg1+c o s θ sinθ⎪parenrightbigg +∞⎪summationdisplay n=1ansinnθ⎪bracketrightBigg . (9.5.48) Substituting (9.5.48) into (9.5.42) gives 1 π⎡ ⎣a0π⎪integraldisplay 0(1 + cos θ) (cosθ−cosθ0)dθ+∞⎪summationdisplay n=1anπ⎪integraldisplay 0sinnθsinθdθ (cosθ−cosθ0)⎤ ⎦=⎪parenleftbigg α−dz dx⎪parenrightbigg x=ξ, (9.5.49) where constants an(n=0,1,2,...) are to be determined. Using the relation (9.5.45) and the identity 2s i nnθsinθ=[ c o s ( n−1)θ−cos(n+1 )θ], we obtain ⎪parenleftbiggdz dx⎪parenrightbigg x=ξ=(α−a0)+∞⎪summationdisplay n=1ancosnθ0. © 2007 by Taylor & Francis Group, LLC 388 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Sincedz dxis known, we calculate the coeﬃcients as a0=α−1 ππ⎪integraldisplay 0⎪parenleftbiggdz dx⎪parenrightbigg dθ, a n=2 ππ⎪integraldisplay 0⎪parenleftbiggdz dx⎪parenrightbigg cosnθ dθ. Thus, the problem is completely solved. This example of application is taken from Zayed (1996). Theﬁnite Hilbert transform was deﬁned by Tricomi (1951) as HHH{f(t),a ,b}=ˆfHHH(x, a, b)=1 πb⎪integraldisplay af(t) t−xdt. (9.5.50) Such transforms arise naturally in aerodynamics. Tricomi (1951) studied the ﬁnite Hankel transform and its applications to airfoil theory. Subsequent-ly, considerable attention has been given to the methods of solution of the singular integral equation for the unknown function f(t)a n dk n o w n ˆf HHH(x)a s 1 π1⎪integraldisplay −1f(t) t−xdt=ˆfHHH(x),−1<x< 1, (9.5.51) where f(x)a n d ˆfHHH(x)s a t i s f yt h eH ¨ older conditions on ( −1,1). This equa- tion arises in boundary value problems in elasticity and in other areas. Sev- eral authors including Muskhelishvili (1963), Gakhov (1966), Peters (1972), Chakraborty (1980, 1988), Chakrabor ty and Williams (1980), Williams (1978), Comninou (1977), Gaute sen and Dunders (1987ab), and Pennline (1976) have studied the methods of the solution of (9.5.41) and its various generalizations.The readers are referred to these papers for details. 9.6 Asymptotic Expansions of One-Sided Hilbert Transforms A two-sided Hilbert transform can be written as the sum of two one-sided transforms∞⎪contintegraldisplay −∞f(t) t−xdt=∞⎪contintegraldisplay 0f(t) t−xdt−∞⎪integraldisplay 0f(−t) t+xdt, (9.6.1) when x>0 (with a similar expression for x<0) where the second integral is actually a Stieltjes transform of [ −f(−t)], which has been deﬁned by (9.7.4) in Section 9.7. © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 389 We examine the one-sided Hilbert transform , which is deﬁned by HHH+{f(t)}=ˆf+ HHH(x)=⎪integraldisplay∞ 0f(t) t−xdt. (9.6.2) The Mellin transform of HHH+{f(t)}is M[HHH+{f(t)}]=∞⎪integraldisplay 0xp−1⎡ ⎣∞⎪integraldisplay 0f(t) t−xdt⎤ ⎦dx=∞⎪integraldisplay 0f(t)⎡ ⎣∞⎪contintegraldisplay 0xp−1 t−xdx⎤ ⎦dt M{HHH+{f(t)}}=πcot(πp)∞⎪integraldisplay 0tp−1f(t)dt =πcot(πp)M{f(t)}=πcot(πp)˜f(p). Taking the inverse Mellin transform, we obtain HHH+{f(t)}=ˆf+ HHH(x)=1 2πic+i∞⎪integraldisplay c−i∞x−pπcot(πp)˜f(p)dp. (9.6.3) Example 9.6.1 (Asymptotic Expansion of One-Sided Hilbert Transforms ). ∞⎪contintegraldisplay 0cosωt t−xdt∼−πsinωx−∞⎪summationdisplay n=0n! (ωx)n+1cos⎪braceleftBig (n+1 )π 2⎪bracerightBig ,asx→∞,(9.6.4) ∞⎪contintegraldisplay 0sinωt t−xdt∼πcosωx−∞⎪summationdisplay n=0n! (ωx)n+1sin⎪braceleftBig (n+1 )π 2⎪bracerightBig ,asx→∞.(9.6.5) We have∞⎪contintegraldisplay 0exp(iωt) t−xdt=πiexp(iωx)+∞⎪integraldisplay 0exp(iωt) t−xdt, where in the integral on the right the contour of integration passes above the polet=x. The contour can be deformed into the positive imaginary axis on which t=iuwithu>0. Thus, ∞⎪integraldisplay 0exp(iωt) t−xdt=−i∞⎪integraldisplay 0exp(−ωu) x−iudu ∼−∞⎪summationdisplay n=0in+1 xn+1∞⎪integraldisplay 0unexp(−ωu)du, by Watson’s lemma =−∞⎪summationdisplay n=0n! (ωx)n+1exp⎪braceleftBig i(n+1 )π 2⎪bracerightBig . (9.6.6) © 2007 by Taylor & Francis Group, LLC 390 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Separating the real and imaginary parts, we obtain the desired results. These results are due to Ursell (1983). THEOREM 9.6.1 (Ursell, 1983 ). Iff(t) is analytic for real t,0≤t<∞, and if it has the asymp- totic expansion in the form f(t)∼∞⎪summationdisplay n=1an tn+c o sωt∞⎪summationdisplay n=1An tn+s i nωt∞⎪summationdisplay n=1Bn tnast→∞, (9.6.7) where the coeﬃcients an,AnandBnare known and ω>0, then the one-sided Hilbert transform ˆf+(x) has the following asymptotic expansion ˆf+ HHH(x)=HHH+{f(t)}=∞⎪contintegraldisplay 0f(t) t−xdt∼∞⎪summationdisplay 1cn xn−logx∞⎪summationdisplay 1an xn +⎪parenleftBigg∞⎪summationdisplay 1An xn⎪parenrightBigg∞⎪contintegraldisplay 0cosωt t−xdt+⎪parenleftBigg∞⎪summationdisplay 1Bn xn⎪parenrightBigg∞⎪contintegraldisplay 0sinωt t−xdtasx→∞,(9.6.8) where cnis given by cn=dn−n−1⎪summationdisplay r=1Γ(n−r) ωn−r⎪bracketleftBig Arcos⎪braceleftBigπ 2(n−r)⎪bracerightBig +Brsin⎪braceleftBigπ 2(n−r)⎪bracerightBig⎪bracketrightBig (9.6.9) and dn= lim p→n⎪bracketleftbigg M{f(t),p}+an p−n⎪bracketrightbigg . (9.6.10) Note that when an= 0, (9.6.10) becomes dn=M{f(t),n}=∞⎪integraldisplay 0tn−1f(t)dt. (9.6.11) Substituting (9.6.9) into (9.6.8) and using (9.6.4) and (9.6.5), we obtain the following theorem: THEOREM 9.6.2 (Ursell, 1983 ). Under the same conditions of Theorem 9.6.1, the one-sided Hilbert transform ˆf+(x) has the asymptotic expansion ˆf+ HHH(x)=∞⎪integraldisplay 0f(t) t−xdt∼−∞⎪summationdisplay 1dn xn−logx∞⎪summationdisplay 1an xn−(πsinωx)∞⎪summationdisplay 1An xn +(πcosωx)∞⎪summationdisplay 1Bn xnasx→∞,(9.6.12) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 391 where dnis given by (9.6.10). The reader is referred to Ursell (1983) for a detailed discussion of proof of Theorems 9.6.1 and 9.6.2. 9.7 Deﬁnition of the Stieltjes Transform and Examples We use the Laplace transform of ¯f(s)=L{f(t)}with respect to sto deﬁne the Stieltjes transform of f(t). Clearly, L{¯f(s)}=˜f(z)=∞⎪integraldisplay 0e−sz¯f(s)ds =∞⎪integraldisplay 0e−szds∞⎪integraldisplay 0e−stf(t)dt. (9.7.1) Interchanging the order of integration and evaluating the inner integral, we obtain ˜f(z)=∞⎪integraldisplay 0f(t) t+zdt. (9.7.2) TheStieltjes transform of a locally integrable function f(t)o n0 ≤t<∞ is denoted by ˜f(z) and deﬁned by S{f(t)}=˜f(z)=∞⎪integraldisplay 0f(t) t+zdt, (9.7.3) where zis a complex variable in the cut plane |argz|<π. Ifz=xis real and positive, then S{f(t)}=˜f(x)=∞⎪integraldisplay 0f(t) t+xdt. (9.7.4) Diﬀerentiating (9.7.4) with respect to x,w eo b t a i n dn dxn˜f(x)=(−1)nn!∞⎪integraldisplay 0f(t) (t+x)n+1dt, n =1,2,3, .... (9.7.5) We now state the inversion theorem for the Stieltjes transform without proof. © 2007 by Taylor & Francis Group, LLC 392 INTEGRAL TRANSFORMS and THEIR APPLICATIONS THEOREM 9.7.1 Iff(t) is absolutely integrable in 0 ≤t≤Tfor every positive Tand is such that the integral (9.7.4) converges for x>0, then ˜f(z)e x i s t sf o rc o m p l e x z(z/negationslash=0 ) not lying on the negative real axis and lim ε→0+1 2πi[˜f(−x−iε)−˜f(−x+iε)] =1 2[f(x+0 )+ f(x−0)] (9.7.6) for any positive xat which f(x+0 )a n d f(x−0) exist. For a rigorous proof of this theorem the reader is referred to Widder (1941, pp. 340–341). Example 9.7.1 Find the Stieltjes transform of each of the following functions: (a)f(t)=(t+a)−1,(b)f(t)=tα−1. (a) We have, by deﬁnition, ˜f(z)=∞⎪integraldisplay 0dt (t+a)(t+z)=1 (a−z)⎡ ⎣∞⎪integraldisplay 0⎪parenleftbigg1 t+z−1 t+a⎪parenrightbigg dt⎤ ⎦ =1 (a−z)log⎪vextendsingle⎪vextendsingle⎪vextendsinglea z⎪vextendsingle⎪vextendsingle⎪vextendsingle. (9.7.7) (b)˜f(z)=S{tα−1}=∞⎪integraldisplay 0tα−1 t+zdt=z−1∞⎪integraldisplay 0⎪parenleftbigg 1+t z⎪parenrightbigg−1 tα−1dt,⎪parenleftbiggt z=x⎪parenrightbigg , =zα−1∞⎪integraldisplay 0xα−1dx 1+x=zα−1M⎪braceleftbigg1 1+x⎪bracerightbigg which is, by Example 8.2.1(b), =zα−1πcosec( πα). (9.7.8) Example 9.7.2 Obtain the Stieltjes transform of J2 ν(t). We have ˜f(x)=S{J2 ν(t)}=∞⎪integraldisplay 0J2 ν(t)dt t+x(9.7.9) © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 393 satisﬁes the Parseval relation ˜f(x)=1 2πic+i∞⎪integraldisplay c−i∞x−p˜f(p)˜g(1−p)dp. (9.7.10) We write t=xuso that (9.7.9) becomes ˜f(x)=∞⎪integraldisplay 0f(xu)g(u)du, (9.7.11) where f(u)=J2 ν(u)a n d g(u)=( 1+ u)−1. Taking the Mellin transform of (9.7.11) with respect to x,w eo b t a i n M{˜f(x),p}=˜f(p)˜g(1−p) where, from Oberhettinger (1974, p. 98), ˜f(p)=2p−1Γ⎪parenleftbig ν+p 2⎪parenrightbig πcosec( πp) ⎪braceleftbig Γ⎪parenleftbig 1−p 2⎪parenrightbig⎪bracerightbig2Γ⎪parenleftbig 1+ν−p 2⎪parenrightbig Γ(p), ˜g(1−p)=πcosec( πp). Thus, the inverse Mellin transform gives the desired result. Example 9.7.3 Show that S{sin(k√ t)}=πexp(−k√ z),k > 0. (9.7.12) We have, by deﬁnition, S{sin(k√ t)}=∞⎪integraldisplay 0sin(k√ t) t+zdt, (√ t=u), =2∞⎪integraldisplay 0usinku (u2+z)du=πexp(−k√ z) by (2.13.6) . © 2007 by Taylor & Francis Group, LLC 394 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 9.8 Basic Operational Properties of Stieltjes Transforms The following properties hold for the Stieltjes transform: (a)S{f(t+a)}=˜f(z−a), (9.8.1) (b)S{f(at)}=˜f(az),a > 0, (9.8.2) (c)S{tf(t)}=−z˜f(z)+∞⎪integraldisplay 0f(t)dt, (9.8.3) provided the integral on t he right hand side exists. (d)S⎪braceleftbiggf(t) t+a⎪bracerightbigg =1 a−z⎪bracketleftBig ˜f(z)−˜f(a)⎪bracketrightBig , (9.8.4) (e)S⎪braceleftbigg1 tf⎪parenleftBiga t⎪parenrightBig⎪bracerightbigg =1 z˜f⎪parenleftBiga z⎪parenrightBig ,a > 0. (9.8.5) PROOF (a) We have, by deﬁnition, S{f(t+a)}=∞⎪integraldisplay 0f(t+a) t+zdt which is, by the change of variable t+a=τ, S{f(t+a)}=∞⎪integraldisplay 0f(τ) τ+(z−a)dτ=˜f(z−a). (b) We have, by deﬁnition, S{f(at)}=∞⎪integraldisplay 0f(at) t+zdt, at =τ, =∞⎪integraldisplay 0f(τ) τ+azdτ=˜f(az). (c) We have from the deﬁnition S{tf(t)}=∞⎪integraldisplay 0tf(t) t+zdt=∞⎪integraldisplay 0(t+z−z)f(t) t+zdt =∞⎪integraldisplay 0f(t)dt−z∞⎪integraldisplay 0f(t) t+zdt. © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 395 This gives the desired result. (d) We have, by deﬁnition, S⎪braceleftbiggf(t) t+a⎪bracerightbigg =∞⎪integraldisplay 0f(t) (t+a)(t+z)dt =1 a−z⎡ ⎣∞⎪integraldisplay 0⎪braceleftbigg1 t+z−1 t+a⎪bracerightbigg f(t)dt⎤ ⎦ =1 a−z⎪bracketleftBig ˜f(z)−˜f(a)⎪bracketrightBig . (e) We have, by deﬁnition, S⎪braceleftbigg1 tf⎪parenleftBiga t⎪parenrightBig⎪bracerightbigg =∞⎪integraldisplay 01 t(t+z)f⎪parenleftBiga t⎪parenrightBig dt,⎪parenleftBiga t=τ⎪parenrightBig , =1 z∞⎪integraldisplay 0f(τ) ⎪parenleftbig τ+a z⎪parenrightbigdτ=1 z˜f⎪parenleftBiga z⎪parenrightBig . THEOREM 9.8.1 (Stieltjes Transforms of Derivatives ). IfS{f(t)}=˜f(z), then S{f/prime(t)}=−1 zf(0)−d dz˜f(z), (9.8.6) S{f/prime/prime(t)}=−⎪bracketleftbigg1 zf/prime(0) +1 z2f(0)⎪bracketrightbigg −d2 dz2˜f(z). (9.8.7) More generally, S{f(n)(t)}=−⎪bracketleftbigg1 zf(n−1)(0) +1 z2f(n−2)(0) +···+1 znf(0)⎪bracketrightbigg −dn dzn˜f(z).(9.8.8) PROOF Using the deﬁnition and integrating by parts, we obtain S{f/prime(t)}=∞⎪integraldisplay 0f/prime(t) t+zdt =⎪bracketleftbiggf(t) t+z⎪bracketrightbigg∞ 0+∞⎪integraldisplay 0f(t)dt (t+z)2=−1 zf(0)−d dz˜f(z). © 2007 by Taylor & Francis Group, LLC 396 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This proves result (9.8.6). Similarly, other results can readily be proved. 9.9 Inversion Theorems for Stieltjes Transforms We ﬁrst introduce the following diﬀerential operator that can be used to es- tablish inversion theorems for the Stieltjes transform. A diﬀerential operator is deﬁned for any real positive number tby the following equations: Lk,t[f(x)] = (−1)k−1cktk−1D(2k−1) t [tkf(t)], (9.9.1) L0,t[f(x)] =f(t), (9.9.2) L1,t[f(x)] =Dt[tf(t)], (9.9.3) where k=2,3,...,c k=[k!(k−2)!]−1,Dt≡d dt,a n df(x)h a sd e r i v a t i v e so fa l l orders. We state a basic theorem due to Widder (1941) without proof. THEOREM 9.9.1 IfS{f(t)}=˜f(x) exists and is deﬁned by ˜f(x)=∞⎪integraldisplay 0f(t) t+xdt (9.9.4) then, for all positive t, (i) Lk,t[˜f(x)] = (2 k−1)!cktk−1∞⎪integraldisplay 0ukf(u) (t+u)2kdu, (9.9.5) (ii) lim k→∞Lk,t[˜f(x)] =f(t). (9.9.6) Obviously, tk t+u=tk−(−u)k t+u+(−u)k t+u=tk−1−utk−2+···±uk−1+(−u)k t+u. In view of this result, we can ﬁnd Lk,t[˜f(x)] = (−1)k−1cktk−1D(2k−1) t [tk˜f(t)] =cktk−1(2k−1)!∞⎪integraldisplay 0uk (t+u)2kf(u)du. © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 397 THEOREM 9.9.2 If˜f(x) is the Laplace transform of ¯f(s)=L{f(t)}so that ˜f(x)=L{˜f(s)}=∞⎪integraldisplay 0f(t) t+xdtfors>0,then (i) Lk,x[˜f(x)] = (−1)kckxk−1∞⎪integraldisplay 0e−xtt2k−1f(k)(t)dt, (9.9.7) (ii) lim k→∞Lk,x[˜f(x)] =f(x),for all positive x. (9.9.8) PROOF We have, by deﬁnition of the operator deﬁned by (9.9.1), Lk,x[˜f(x)] = (−1)k−1xk−1ckD(2k−1) x [xk˜f(x)]. (9.9.9) We use the result (Widder, 1941, p. 350) xk−1D(2k−1) x [xkf(x)] =Dk x[x2k−1f(k−1)(x)], (9.9.10) where f(x) is any function that has derivatives of all orders. This can easily be veriﬁed by computing both sides of (9.9.10). Each of both sides is equal to k⎪summationdisplay n=0(2k−1)! (2k−n−1)!⎪parenleftbiggk n⎪parenrightbigg x2k−n−1f(2k−n−1)(x). In view of (9.9.10), result (9.9.9) becomes Lk,x[˜f(x)] = (−1)k−1ckDk x[x2k−1˜f(k−1)(x)]. (9.9.11) We next show that (−1)k−1˜f(k−1)(x)=(−1)k−1dk−1 dxk−1∞⎪integraldisplay 0e−xtf(t)dt =(−1)2(k−1)∞⎪integraldisplay 0e−xttk−1f(t)dt =1 x∞⎪integraldisplay 0e−u⎪parenleftBigu x⎪parenrightBigk−1 f⎪parenleftBigu x⎪parenrightBig du. (9.9.12) Using (9.9.12) in (9.9.11), we obtain Lk,x[˜f(x)] =ck∞⎪integraldisplay 0e−uuk−1Dk x⎪braceleftBig xk−1f⎪parenleftBigu x⎪parenrightBig⎪bracerightBig du, © 2007 by Taylor & Francis Group, LLC 398 INTEGRAL TRANSFORMS and THEIR APPLICATIONS which is, due to Lemma 25 (Widder 1941, p. 385), =ckx−(k+1)∞⎪integraldisplay 0e−uuk−1(−u)kf(k)⎪parenleftBigu x⎪parenrightBig du. (9.9.13) We again set u=xtin (9.9.13) to obtain the desired result Lk,x[˜f(x)] =ck(−1)kxk−1∞⎪integraldisplay 0e−xtt2k−1f(k)(t)dt. We next take the limit as k→∞ and use Widder’s result (9.9.6) to derive (9.9.8). Thus, the proof is complete. It is important to note that result (9.9.7) depends on the values of all deriva- tives of f(x) in the domain (0 ,∞). This seems to be a very severe restriction on the formula (9.9.7). This restriction can be eliminated by applying theoperator L k,xto the Laplace integral directly. Then we prove the following theorem. THEOREM 9.9.3 (Widder, 1941 ). Under the same conditions of Theorem 9.9.2, the following results hold (i) Lk,x[˜f(x)] =∞⎪integraldisplay 0e−xsP2k−1(xs)¯f(s)ds, (9.9.14) (ii) lim k→∞Lk,x[˜f(x)] =f(x), (9.9.15) where P2k−1(t)=(−1)k−1ck(2k−1)!k⎪summationdisplay n=0⎪parenleftbiggk n⎪parenrightbigg(−t)2k−n−1 (2k−n−1)!. (9.9.16) PROOF We apply the operator Lk,xto the Laplace integral directly to obtain Lk,x[˜f(x)] =Lk,x[L{¯f(s)}]=Lk,x⎡ ⎣∞⎪integraldisplay 0e−sx¯f(s)ds⎤ ⎦ Lk,x[˜f(x)] =∞⎪integraldisplay 0Lk,x[e−sx]¯f(s)ds © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 399 which is, after direct computation of Lk,x[exp(−sx)], =∞⎪integraldisplay 0e−xsP2k−1(xs)¯f(s)ds. Taking the limit as k→∞ and using result (9.9.8), we obtain lim k→∞Lk,x[˜f(x)] = lim k→∞∞⎪integraldisplay 0e−xsP2k−1(xs)¯f(s)ds=f(x) for all x>0. The signiﬁcance of this result lies in the fact that the integral representation forf(x) depends onlyon the values of ¯f(s)i n( 0 ,∞)a n d noton any of its derivatives. 9.10 Applications of Stieltjes Transforms Example 9.10.1 (Moment Problem ). Iff(t) has an exponential rate of decay as t→∞,t h e n all of the moments exist and are given by mr=∞⎪integraldisplay 0trf(t)dt, r =0,1,2,.... (9.10.1) Then it can easily be shown from (9.7.4) that ˜f(x)=n−1⎪summationdisplay r=0(−1)rmrx−(r+1)+εn(x), (9.10.2) where |εn(x)|≤x−(n+1)sup 0<t<∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglet⎪integraldisplay 0τnf(τ)dτ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle. (9.10.3) The Stieltjes transform is found to arise in the problems of moments for the semi-inﬁnite interval. The reader is referred to Tamarkin and Shohat (1943). © 2007 by Taylor & Francis Group, LLC 400 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 9.10.2 (Solution of Integral Equations ). Find the solution of the integral equation λ∞⎪integraldisplay 0f(t) t+xdt=f(x), (9.10.4) where λis a real parameter. Case (i): Suppose λ/negationslash=1 π. In this case, we show that the solution of (9.10.4) is f(t)=At−α+Btα−1, (9.10.5) where AandBare arbitrary constants and αis a root of the equation sin απ= λπbetween zero and unity if λ<1 π, and with real part1 2ifλ>1 π. If 0<Reα<1, then λ∞⎪integraldisplay 0t−α t+xdt=λ∞⎪integraldisplay 0tp−1 t+xdt, (p=1−α) =λπ xp−1cosec ( πp) =x−α⎪parenleftbiggλπ sinπα⎪parenrightbigg =x−α, (9.10.6) so that x−αis a solution of (9.10.4). Obviously, equation (9.10.6) holds if α is replaced by 1 −α, and hence, tα−1is also a solution. Thus, (9.10.5) is a solution of equation (9.10.4). Case (ii): λ=1 π. In this case, we show that1 √ t,a n d1 √ tlogtare solutions of (9.10.4). f(x)=1 π∞⎪integraldisplay 0dt √ t(t+x)=1 π∞⎪integraldisplay 01 (t+x)t1 2−1dt =1 πx1 2−1πcosec⎪parenleftBigπ 2⎪parenrightBig =1 √ x, by Example 9.7.1(b) . Thus,1 √ tis a solution of the integral equation (9.10.4). © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 401 To show that f(t)=1 √ tlogtis a solution, we write f(x)=1 π∞⎪integraldisplay 0logt √ t(t+x)dt (logt=u) =1 π∞⎪integraldisplay −∞u (x+eu)exp⎪parenleftBigu 2⎪parenrightBig du. Replacing xbyexand multiplying both sides by exp⎪parenleftBigx 2⎪parenrightBig , we ﬁnd exp⎪parenleftBigx 2⎪parenrightBig f(ex)=1 π∞⎪integraldisplay −∞⎪parenleftbiggu ex+eu⎪parenrightbigg exp⎪parenleftbiggx+u 2⎪parenrightbigg du =1 2π∞⎪integraldisplay −∞usech⎪parenleftbiggx−u 2⎪parenrightbigg du, (x−u=t). exp⎪parenleftBigx 2⎪parenrightBig f(ex)=1 2π∞⎪integraldisplay −∞(x−t)sec h⎪parenleftbiggt 2⎪parenrightbigg dt=x 2π∞⎪integraldisplay −∞sech⎪parenleftbiggt 2⎪parenrightbigg dt=x, or, f(ex)=xexp⎪parenleftBig −x 2⎪parenrightBig . Thus, f(t)=1 √ tlogt is a solution, and hence,1 √ t(A+Blogt) is also a solution of (9.10.4). 9.11 The Generalized Stieltjes Transform Thegeneralized Stieltjes transform of a function f(t) is deﬁned by Sg{f(t)}=˜f(z,ρ)=∞⎪integraldisplay 0f(t) (t+z)ρdt, (9.11.1) provided the integral exists and |argz|<π. © 2007 by Taylor & Francis Group, LLC 402 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 9.11.1 If Re a>0, ﬁnd the generalized Stieltjes transform of (a)ta−1, (b) exp( −at). (a) We have, by deﬁnition, Sg{ta−1}=∞⎪integraldisplay 0ta−1 (t+z)ρdt =z−ρ∞⎪integraldisplay 0ta−1⎪parenleftbigg 1+t z⎪parenrightbigg−ρ dt, (t=zu), =za−ρ∞⎪integraldisplay 0ua−1(1 +u)−ρdu. (9.11.2) Substituting x=u 1+uoru=x 1−xinto integral (9.11.2), we obtain Sg{ta−1}=za−ρ1⎪integraldisplay 0xa−1(1−x)ρ−a−1dx =za−ρB(a, ρ−a)=Γ(a)Γ(ρ−a) Γ(ρ)za−ρ. (9.11.3) (b) We have, by deﬁnition, Sg{exp(−at)}=∞⎪integraldisplay 0exp(−at) (t+z)ρdt, (t+z = u ), =e x p ( az)∞⎪integraldisplay 0e−auu−ρdu. Substituting au=xinto this integral, we obtain Sg{exp(−at)}=aρ−1exp(az)∞⎪integraldisplay 0e−xx1−ρ−1dx =aρ−1exp(az)Γ( 1−ρ). (9.11.4) The reader is referred to Erd´ elyi et al. (1954, pp. 234–235) where there is a table for generalized Stieltjes transforms. © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 403 9.12 Basic Properties of the Generalized Stieltjes Transform The generalized Stieltjes transform satisﬁes the following properties: (a)Sg{f(at)}=aρ−1˜f(az),a > 0 (9.12.1) (b)Sg{tf(t)}=˜f(z,ρ−1)−z˜f(z,ρ), (9.12.2) (c)Sg{f/prime(t)}=ρ˜f(z,ρ+1 )−z−ρf(0),f(t)→0a st→∞.(9.12.3) (d)Sg⎧ ⎨ ⎩t⎪integraldisplay 0f(x)dx⎫ ⎬ ⎭=(ρ−1)−1˜f(z,ρ−1),Reρ>1.(9.12.4) PROOF (a) We have, by deﬁnition, Sg{f(at)}=∞⎪integraldisplay 0f(at)dt (t+z)ρ,(at=x), =aρ−1∞⎪integraldisplay 0f(x)dx (x+az)ρ=aρ−1˜f(az). (b) It follows from the deﬁnition that Sg{tf(t)}=∞⎪integraldisplay 0tf(t) (t+z)ρdt=∞⎪integraldisplay 0(t+z−z)f(t) (t+z)ρdt =∞⎪integraldisplay 0f(t) (t+z)ρ−1dt−z∞⎪integraldisplay 0f(t)dt (t+z)ρ =˜f(z,ρ−1)−z˜f(z,ρ). (c) We have, by deﬁnition, Sg{f/prime(t)}=∞⎪integraldisplay 0f/prime(t) (t+z)ρdt which is, by integrating by parts, Sg{f/prime(t)}=⎪bracketleftbiggf(t) (t+z)ρ⎪bracketrightbigg∞ 0+ρ∞⎪integraldisplay 0f(t) (t+z)ρ+1dt =ρ˜f(z,ρ+1 )−z−ρf(0). © 2007 by Taylor & Francis Group, LLC 404 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (d) We write g(t)=t⎪integraldisplay 0f(x)dx so that g/prime(t)=f(t)a n d g(0) = 0. Thus, Sg{f(t),ρ}=Sg{g/prime(t),ρ} which is, by (9.12.3), =ρ˜g(z,ρ+1 )−z−ρg(0) = ρSg{g(t),ρ+1}. Replacing ρbyρ−1, we obtain (9.12.4). 9.13 Exercises 1. Find the Hilbert transform of each of the following functions: (a)f(t)=1 (a2+t2),Rea>0, (c)f(t)=e x p ( −at),(b)f(t)=tα (t+a),|Reα|<1, (d)f(t)=sint t, (e)f(t)=1 tsin(a√ t),a > 0, (f)f(t)=t−αexp(−at),Rea>0,Reα<1. 2. Show that (a)S⎪braceleftbigg1 √ tcos(a√ t)⎪bracerightbigg =π √ zexp(−a√ z),z > 0. (b)S{sin(a√ t)J0(b√ t)}=πexp(−a√ z)I0(b√ z),0<b<a . 3. If˜f(z)=S{f(t)}andf(t)=S{g(u)}, then show that ˜f(z)=∞⎪integraldisplay 0K(z,u)g(u)du, where K(z,u)=(z−u)−1log⎪parenleftBigz u⎪parenrightBig . 4. Show that © 2007 by Taylor & Francis Group, LLC Hilbert and Stieltjes Transforms 405 (a)Sg⎪braceleftBig tρ−2f⎪parenleftBiga t⎪parenrightBig⎪bracerightBig =aρ−1z−ρ˜f⎪parenleftBiga z⎪parenrightBig ,a > 0. (b)Sg⎧ ⎨ ⎩1 Γ(α)t⎪integraldisplay 0f(x)(t−x)α−1dx⎫ ⎬ ⎭=Γ(ρ−α) Γ(ρ)˜f(z,ρ−α), where 0 <Reα<Reρ. 5. Show that the dispersion relation associated with the linearized Ben- jamin−Ono equation ut+HHH{uxx}=0 i s ω=−k|k|. 6. Find the Stieltjes transforms of each of the following functions: (a)f(t)=tα−1 t+a, (b)f(t)=1 t2+a2, (c)f(t)=t t2+a2. 7. Show that (a)S[f(teiπ)−f(te−iπ)] = 2πi˜f(z), (b)S[f(√ t)] =˜f(i√ z)+˜f(−i√ z). 8. Suppose f(t) is ia locally integrable function on (0 ,∞) and has the asymptotic representation (Wong, 1989) f(t)∼∞⎪summationdisplay r=0artαrast→0+ where Re αr↑+∞asr→∞,R eα0>−1, and f(t)=O(t−a),a>1. Show that the generalized Stieltjes transform ˜f(x)=∞⎪integraldisplay 0f(t)dt (t+x)ρ,ρ > 0 has the asymptotic representation, as x→0+, ˜f(x)∼∞⎪summationdisplay r=0arΓ(1 + αr)Γ(ρ−1−αr) Γ(ρ)x1+αr−ρ +∞⎪summationdisplay r=0(−1)rΓ(r+ρ)M[f;1−ρ−r] r!Γ (ρ)xr provided 1 + αr/negationslash=ρ+nfor all non-negative integers randn. 9. Show that the one-sided Hilbert transform involved in research on water waves by Hulme (1981) ˆfHHH(x)=∞⎪integraldisplay 0J2 0(t)dt t−x © 2007 by Taylor & Francis Group, LLC 406 INTEGRAL TRANSFORMS and THEIR APPLICATIONS satisﬁes the Parseval relation ˆfHHH(x)=1 2πic+i∞⎪integraldisplay c−i∞x−pM[J2 0(x);p]πcotπp dp. 10. Prove the following asymptotic expansions (Ursell, 1983): ∞⎪contintegraldisplay 0J2 0(t) t−xdt∼−1 πx(logx+γ+3l o g2 )+1 xcos2x+1 4x2sin2x +1 8πx3⎪parenleftbigg logx+γ+3l o g2 −5 2⎪parenrightbigg −5 32x3cos2x,asx→∞, and ∞⎪contintegraldisplay 0J2 0(t) t−xdt∼−π 2J0(x)Y0(x)−√ π∞⎪summationdisplay r=0cos(πr)Γ(r+1 )x2r+1 ⎪braceleftbig Γ⎪parenleftbig r+3 2⎪parenrightbig⎪bracerightbig3,asx→0. 11. If λ=1 √ π, show that f(t)=A √ t,w h e r e Ais a constant, is the only solution of the integral equation f(s)=λ∞⎪integraldisplay 0e−stf(t)dt. 12. If λ=−1 √ π, show that f(t)=A⎪bracketleftbiggΓ/prime(1 2) √ πt−2l o gt √ t⎪bracketrightbigg , where Ais a constant, is the only solution of the integral equation as stated in Exercise 11. 13. Show that Lk,t[(x+a)−1]=ck(2k−1)!tk−1ak(t+a)−2k,(a>0,t>0,k=2,3,...). 14. Prove that lim ε→0+1 π∞⎪integraldisplay ε⎪bracketleftbiggf(x+u)−f(x−u) u⎪bracketrightbigg du=1 π∞⎪contintegraldisplay −∞f(t) (t−x)dt. 15. Prove Parseval’s formulas (9.3.9). © 2007 by Taylor & Francis Group, LLC 10 Finite Fourier Sine and Cosine Transforms “Mathematics compares the most diverse phenomena and discovers the secret analogies that unite them.” Joseph Fourier “In the mathematical investigation I have usually employed such methods as present themselves naturally to a physicist. The pure mathematician will complain, and (it must be confessed) some-times with justice, of deﬁcient rigor. But to this question there are two sides. For, however important it may be to maintain a uniformly high standard in pure mathematics, the physicist may occasionally do well to rest content with arguments which are fair- ly satisfactory and conclusive from his point of view. To his mind,exercised in a diﬀerent order of ideas, the more severe procedure of the pure mathematician may appear not more but less demonstra- tive. And further, in many cases of diﬃculty to insist upon higheststandard would mean the exclusion of the subject altogether in view of the space that would be required.” Lord Rayleigh 10.1 Introduction This chapter deals with the theory and applications of ﬁnite Fourier sineand cosine transforms. The basic operational properties including convolution theorem of these transforms are discussed in some detail. Special attention is given to the use of these transforms to the solutions of boundary value and initial-boundary value problems. The ﬁnite Fourier sine transform was ﬁrst introduced by Doetsch (1935). Subsequently, the method has been developed and generalized by several au- thors including Kneitz (1938), Koschmieder (1941), Roettinger (1947), and Brown (1944). 407 © 2007 by Taylor & Francis Group, LLC 408 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 10.2 Deﬁnitions of the Finite Fourier Sine and Cosine Transforms and Examples Both ﬁnite Fourier sine and cosine transforms are deﬁned from the corre- sponding Fourier sine and Fourier cosine series. DEFINITION 10.2.1 (The Finite Fourier Sine Transform). If f(x)is a continuous or piecewise continuous function on a ﬁnite interval 0<x<a ,t h e ﬁnite Fourier sine transform of f(x)is deﬁned by Fs{f(x)}=˜fs(n)=a⎪integraldisplay 0f(x)sin⎪parenleftBignπx a⎪parenrightBig dx, (10.2.1) where n=1,2,3,.... It is a well-known result of the theory of Fourier series that the Fourier sine series for f(x)i n0 <x<a 2 a∞⎪summationdisplay n=1˜fs(n)sin⎪parenleftBignπx a⎪parenrightBig (10.2.2) converges to the value f(x) at each point of continuity in the interval 0 < x<a and to the value1 2[f(x+0 )+ f(x−0)] at each point xof the ﬁnite discontinuity in 0 <x<a . In view of the deﬁnition (10.2.1), the inverse Fourier sine transform is given by F−1 s⎪braceleftBig ˜fs(n)⎪bracerightBig =f(x)=2 a∞⎪summationdisplay n=1˜fs(n)sin⎪parenleftBignπx a⎪parenrightBig . (10.2.3) Clearly, both FsandF−1 sare linear transformations. DEFINITION 10.2.2 (The Finite Fourier Cosine Transform). If f(x)is a continuous or piecewise continuous function on a ﬁnite interval 0<x<a , the ﬁnite Fourier cosine transform of f(x)is deﬁned by Fc{f(x)}=˜fc(n)=a⎪integraldisplay 0f(x)cos⎪parenleftBignπx a⎪parenrightBig dx, (10.2.4) where n=0,1,2,.... © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 409 It is also a well-known result of the theory of Fourier series that the Fourier cosine series for f(x)i n0 <x<a 1 a˜fc(0) +2 a∞⎪summationdisplay n=1˜fc(n)cos⎪parenleftBignπx a⎪parenrightBig (10.2.5) converges to f(x) at each point of continuity in 0 <x<a ,a n dt o1 2[f(x+0 )+ f(x−0)] at each point xof ﬁnite discontinuity in 0 <x<a .B yv i r t u eo ft h e deﬁnition (10.2.4), the inverse Fourier cosine transform is given by F−1 c⎪braceleftBig ˜fc(n)⎪bracerightBig =f(x)=1 a˜fc(0) +2 a∞⎪summationdisplay n=1˜fc(n)cos⎪parenleftBignπx a⎪parenrightBig . (10.2.6) Clearly, both FcandF−1 care linear transformations. When a=π, the ﬁnite Fourier sine and cosine transforms are deﬁned, re- spectively, by (10.2.1) and (10.2.4) on the interval 0 <x<π . The correspond- ing inverse transforms are given by the same results (10.2.3) and (10.2.6) with a=π. The transform of a function deﬁned over an interval 0 <x<a can be written easily in terms of a transform on the standard interval 0 <x<π .W e substitute ξ=πx ato write (10.2.1) and (10.2.4) as follows: ˜fs(n)=a⎪integraldisplay 0sin⎪parenleftBignπx a⎪parenrightBig f(x)dx=a ππ⎪integraldisplay 0sin(nξ)f⎪parenleftbiggaξ π⎪parenrightbigg dξ=a πFs⎪braceleftBig f⎪parenleftBigax π⎪parenrightBig⎪bracerightBig ˜fc(n)=a⎪integraldisplay 0cos⎪parenleftBignπx a⎪parenrightBig f(x)dx=a ππ⎪integraldisplay 0cos(nξ)f⎪parenleftbiggaξ π⎪parenrightbigg dξ=a πFc⎪braceleftBig f⎪parenleftBigax π⎪parenrightBig⎪bracerightBig . Example 10.2.1 Find the ﬁnite Fourier sine and cosine transforms of (a)f(x)=1 a n d ( b)f(x)=x. (a) We have Fs(1) = ˜fs(n)=a⎪integraldisplay 0sin⎪parenleftBignπx a⎪parenrightBig dx=a nπ[1−(−1)n], (10.2.7) Fc{1}=˜fc(n)=a⎪integraldisplay 0cos⎪parenleftBignπx a⎪parenrightBig dx=⎪braceleftBigga, n =0 0,n/negationslash=0⎪bracerightBigg . (10.2.8) © 2007 by Taylor & Francis Group, LLC 410 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b) Fs{x}=a⎪integraldisplay axsin⎪parenleftBignπx a⎪parenrightBig dx=(−1)n+1a2 nπ. (10.2.9) Fc{x}=a⎪integraldisplay 0xcos⎪parenleftBignπx a⎪parenrightBig dx=⎧ ⎪⎨ ⎪⎩a2 2,n =0 ⎪parenleftBiga nπ⎪parenrightBig2 [(−1)n−1],n/negationslash=0⎫ ⎪⎬ ⎪⎭.(10.2.10) 10.3 Basic Properties of Finite Fourier Sine and Cosine Transforms As a preliminary to the solution of diﬀerential equations by the ﬁnite Fourier sine and cosine transforms, we now establish the transforms of derivatives of f(x). Fs{f/prime(x)}=−⎪parenleftBignπ a⎪parenrightBig ˜fc(n), (10.3.1) Fs{f/prime/prime(x)}=−⎪parenleftBignπ a⎪parenrightBig2˜fs(n)+⎪parenleftBignπ a⎪parenrightBig [f(0) + ( −1)n+1f(a)],(10.3.2) Fc{f/prime(x)}=⎪parenleftBignπ a⎪parenrightBig ˜fs(n)+(−1)nf(a)−f(0), (10.3.3) Fc{f/prime/prime(x)}=−⎪parenleftBignπ a⎪parenrightBig2˜fc(n)+(−1)nf/prime(a)−f/prime(0). (10.3.4) Similar results can be obtained for the ﬁnite Fourier sine and cosine transforms of higher derivatives of f(x). Results (10.3.1)–(10.3.4) can be proved by integrating by parts. For exam- ple, we have Fs{f/prime(x)}=a⎪integraldisplay 0f/prime(x)sin⎪parenleftBignπx a⎪parenrightBig dx, which is, integrating by parts, =⎪bracketleftBig f(x)sin⎪parenleftBignπx a⎪parenrightBig⎪bracketrightBiga 0−nπ aa⎪integraldisplay 0f(x)cos⎪parenleftBignπx a⎪parenrightBig dx =−⎪parenleftBignπ a⎪parenrightBig ˜fc(n). © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 411 Similarly, we ﬁnd that Fc{f/prime(x)}=a⎪integraldisplay 0f/prime(x)cos⎪parenleftBignπx a⎪parenrightBig dx =⎪bracketleftBig f(x)cos⎪parenleftBignπx a⎪parenrightBig⎪bracketrightBiga 0+nπ aa⎪integraldisplay 0f(x)sin⎪parenleftBignπx a⎪parenrightBig dx =(−1)nf(a)−f(0) +nπ a˜fs(n). This proves the result (10.3.3). Results (10.3.2) and (10.3.4) and the results for higher derivatives can be obtained by the repeated application of the fundamental results (10.3.1) and (10.3.3). DEFINITION 10.3.1 (Odd Periodic Extension). A function f1(x)is said to be the odd periodic extension of the function f(x), with period 2πif f1(x)=⎪braceleftBigg f(x) for0<x<π −f(−x)for−π<x< 0⎪bracerightBigg . (10.3.5) Or, equivalently, f1(x)=f(x)when 0<x<π, f1(−x)=−f1(x),f 1(x+2π)=f1(x)for−∞<x< ∞.(10.3.6) Similarly, the even periodic extension f2(x)off(x), with period 2πis de- ﬁned in −π<x<π by the equations f2(x)=⎪braceleftBigg f(x) for0<x<π f(−x)for−π<x< 0⎪bracerightBigg . (10.3.7) Or, equivalently, f2(x)=f(x)when 0<x<π, f2(−x)=f2(x),f 2(x+2π)=f2(x)for−∞<x< ∞.(10.3.8) THEOREM 10.3.1 Iff1(x) is the odd periodic extension of f(x) with period 2 π, then, for any constant α, Fs{f1(x−α)+f1(x+α)}=2c o s nαFs{f(x)}. (10.3.9) In particular, when α=π,a n d n=1,2,3,..., Fs{f(x−π)}=(−1)nFs{f(x)}. (10.3.10) © 2007 by Taylor & Francis Group, LLC 412 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Similarly, we obtain Fc{f1(x+α)−f1(x−α)}=2s i n ( nα)Fs{f(x)}. (10.3.11) PROOF To prove (10.3.9), we follow Churchill (1972) and write the right hand side of (10.3.9) as 2c o snα˜fs(n)=2c o s ( nα)π⎪integraldisplay 0sin(nx)f(x)dx =2π⎪integraldisplay 0cosnαsinnxf1(x)dx =π⎪integraldisplay 0[sinn(x+α)+s i n n(x−α)]f1(x)dx, which is, since the integrand is even function of x, =1 2π⎪integraldisplay −π[sinn(x+α)+s i n n(x−α)]f1(x)dx, which is, by putting x+α=tandx−α=t, =1 2π+α⎪integraldisplay −π+αsinnt f1(t−α)dt+1 2π−α⎪integraldisplay −(π+α)sinnt f1(t+α)dt, which is, since the integrands are periodic function of twith period 2 π,a n d hence, the limits of integrati on can be replaced with limits −πtoπ, =1 2π⎪integraldisplay −πsinnt f1(t−α)dt+1 2π⎪integraldisplay −πsinnt f1(t+α)dt =1 2⎡ ⎣0⎪integraldisplay −π+π⎪integraldisplay 0⎤ ⎦{sinnt f1(t−α)}dt +1 2⎡ ⎣0⎪integraldisplay −π+π⎪integraldisplay 0⎤ ⎦{sinnt f1(t+α)}dt. (10.3.12) Furthermore, 0⎪integraldisplay −πsinnt f1(t−α)dt=π⎪integraldisplay 0sinnxf1(x+α)dx © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 413 in which f1(−x−α)=−f1(x+α)i su s e d . Making a similar change of variables in the third integral of (10.3.12), we obtain the formula 2c o snα˜fs(n)=π⎪integraldisplay 0sinnt f1(t−α)dt+π⎪integraldisplay 0sinnxf1(x+α)dx, which gives the desired result (10.3.9). Finally, f1(x+π)=f1(2π+x−π)=f1(x−π)=−f1(π−x) ,a n dw h e n0 < x<π,f1(π−x)=f(π−x). Thus, when α=π, result (10.3.9) becomes ˜fs(n)c o snπ=π⎪integraldisplay 0sinnxf(x−π)dx=Fs{f(x−π)}, which reduces to (10.3.10). The proof of (10.3.11) is similar to that of (10.3.9), and hence, is left to the reader. THEOREM 10.3.2 Iff2(x) is the even periodic extension of f(x) with period 2 π, then, for any constant α, Fc{f2(x−α)+f2(x+α)}=2c o s nαFc{f(x)}, (10.3.13) Fc{f2(x−α)−f2(x+α)}=2s i n nαFc{f(x)}. (10.3.14) This theorem is very much similar to that of Theorem 10.3.1, and hence, the proof is left to the reader. In the notation of Churchill (1972), we introduce the convolution of two sectionally continuous periodic functions f(x)a n d g(x) deﬁned in −π<x<π by f(x)∗g(x)=π⎪integraldisplay −πf(x−u)g(u)du. (10.3.15) Clearly, f(x)∗g(x) is continuous and periodic with period 2 π. The convolution is symmetric, that is, f∗g=g∗f. Furthermore, the convolution is an even function if f(x)a n d g(x) are both even or both odd. It is odd if either f(x) org(x) is even or the other odd. We next prove the convolution theorem. THEOREM 10.3.3 (Convolution ). Iff1(x)a n d g1(x) are the odd periodic extensions of f(x)a n d g(x) respectively on 0 <x<π ,a n di f f2(x)a n d g2(x)a r et h ee v e np e r i o d i c © 2007 by Taylor & Francis Group, LLC 414 INTEGRAL TRANSFORMS and THEIR APPLICATIONS extensions of f(x)a n d g(x) respectively on 0 <x<π ,t h e n Fc{f1(x)∗g1(x)}=−2˜fs(n)˜gs(n), (10.3.16) Fc{f2(x)∗g2(x)}=2˜fc(n)˜gc(n), (10.3.17) Fs{f1(x)∗g2(x)}=2˜fs(n)˜gc(n), (10.3.18) Fs{f2(x)∗g1(x)}=2˜fc(n)˜gs(n). (10.3.19) Or, equivalently, F−1 c⎪braceleftBig ˜fs(n)˜gs(n)⎪bracerightBig =−1 2{f1(x)∗g1(x)}, (10.3.20) F−1 c⎪braceleftBig ˜fc(n)˜gc(n)⎪bracerightBig =1 2{f2(x)∗g2(x)}, (10.3.21) F−1 s⎪braceleftBig ˜fs(n)˜gc(n)⎪bracerightBig =1 2{f1(x)∗g2(x)}, (10.3.22) F−1 s⎪braceleftBig ˜fc(n)˜gs(n)⎪bracerightBig =1 2{f2(x)∗g1(x)}. (10.3.23) PROOF To prove (10.3.16), we consider the product 2˜fs(n)˜gs(n)=2π⎪integraldisplay 0˜fs(n)s i nnu g(u)du, which is, by using (10.3.11), =π⎪integraldisplay 0g(u)[Fc{f1(x+u)−f1(x−u)}]du =π⎪integraldisplay 0g(u)⎡ ⎣π⎪integraldisplay 0{f1(x+u)−f1(x−u)}cosnx⎤ ⎦du, which is, by interchanging the order of integration, =π⎪integraldisplay 0cos(nx)⎡ ⎣π⎪integraldisplay 0{f1(x+u)−f1(x−u)}g(u)du⎤ ⎦dx. (10.3.24) Using the deﬁnition of convolution (10.3.15), introducing new variables of integration, and invoking the odd extension properties of f1(x)a n d g1(x), we obtain f1(x)∗g1(x)=π⎪integraldisplay 0[f1(x−u)−f1(x+u)]g(u)du (10.3.25) =I1−I2−I3+I4, (10.3.26) © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 415 where I1=x⎪integraldisplay 0f(u)g(x+u)du, I 2=π⎪integraldisplay xf(u)g(u−x)du, (10.3.27ab) I3=π−x⎪integraldisplay 0f(u)g(x+u)du, I 4=π⎪integraldisplay xf(u)g(2π−x−u)du.(10.3.28ab) In view of (10.3.25), we thus obtain the desired result (10.3.16) from (10.3.24). This completes the proof. The other results included in Theorem 10.3.3 can be proved by the above method of proof. As an example of convolution theorem, we evaluate the inverse cosine Fouri- er transform of ( n2−a2)−1.W ew r i t e ,f o r n/negationslash=0 , 1 (n2−a2)=n(−1)n+1 (n2−a2)·(−1)n+1 n=˜fs(n)˜gs(n), where ˜fs(n)=n(−1)n+1(n2−a2)−1and ˜ gs(n)=(−1)n+1 nso that f(x)=⎪parenleftbiggsinax sinaπ⎪parenrightbigg and g(x)=x π. Evidently, 1 (n2−a2)=˜fs(n)˜gs(n)=Fs⎪braceleftbiggsinax sinaπ⎪bracerightbigg Fs⎪braceleftBigx π⎪bracerightBig . According to result (10.3.20), F−1 c⎪braceleftbigg1 (n2−a2)⎪bracerightbigg =F−1 c⎪braceleftBig ˜fs(n)˜gs(n)⎪bracerightBig =−1 2f1(x)∗g1(x),(10.3.29) where f1(x) is the periodic extension of the odd function f(x) with period 2 π andg1(x)=x π. Thus, it turns out that F−1 c⎪braceleftbigg1 (n2−a2)⎪bracerightbigg =−1 2π⎪integraldisplay −πf1(x−u)g1(u)du=−1 2ππ⎪integraldisplay −πf1(x−u)ud u . This integral can easily be evaluated by splitting up the interval of integration or, by using (10.3.26), and hence, F−1 c⎪braceleftbigg1 (n2−a2)⎪bracerightbigg =−cos{a(π−x)} asinaπ. (10.3.30) © 2007 by Taylor & Francis Group, LLC 416 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 10.4 Applications of Finite Fourier Sine and Cosine Transforms In this section we illustrate the use of ﬁnite Fourier sine and cosine transforms to the solutions of boundary value and initial-boundary value problems. Example 10.4.1 (Heat Conduction Problem in a Finite Domain with the Dirichlet Data at the Boundary ). We began by considering the solution of the temperature distri- bution u(x, t) of the diﬀusion equation ut=κuxx,0≤x≤a, t > 0, (10.4.1) with the boundary and initial conditions u(0,t)=0= u(a, t), (10.4.2ab) u(x,0) =f(x)f o r 0 ≤x≤a. (10.4.3) Application of the ﬁnite Fourier sine transform (10.2.1) to this diﬀusion problem gives the initial value problem d˜us dt+κ⎪parenleftBignπ a⎪parenrightBig2 ˜us=0, (10.4.4) ˜us(n,0) =˜fs(n). (10.4.5) The solution of (10.4.4)–(10.4.5) is ˜us(n, t)=˜fs(n)e x p⎪braceleftbigg −κ⎪parenleftBignπ a⎪parenrightBig2 t⎪bracerightbigg . (10.4.6) The inverse ﬁnite Fourier sine transform (10.2.3) leads to the solution u(x, t)=2 a∞⎪summationdisplay n=1˜fs(n)e x p⎪braceleftbigg −κ⎪parenleftBignπ a⎪parenrightBig2 t⎪bracerightbigg sin⎪parenleftBignπx a⎪parenrightBig u(x, t)=2 a∞⎪summationdisplay n=1exp⎪braceleftbigg −κ⎪parenleftBignπ a⎪parenrightBig2 t⎪bracerightbigg sin⎪parenleftBignπx a⎪parenrightBig ×a⎪integraldisplay 0f(ξ)sin⎪parenleftbiggnπξ a⎪parenrightbigg dξ. (10.4.7) If, in particular, f(x)=T0= constant, then (10.4.7) becomes u(x, t)=⎪parenleftbigg2T0 π⎪parenrightbigg∞⎪summationdisplay n=11 n[1−(−1)n]e xp⎪braceleftbigg −κ⎪parenleftBignπ a⎪parenrightBig2 t⎪bracerightbigg sin⎪parenleftBignπx a⎪parenrightBig .(10.4.8) © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 417 This series solution can be evaluated numerically using the Fast Fourier transform which is an algorithm for the eﬃcient calculation of the ﬁnite Fourier transform. Example 10.4.2 (Heat Conduction Problem in a Finite Domain with the Neumann Data at the Boundary ). We consider the solution of the diﬀusion equation (10.4.1) with the prescribed heat ﬂux at x=0a n d x=a, and the associated boundary and initial data are ux(0,t)=0= ux(a, t)f o r t>0, (10.4.9) u(x,0)=f(x)f o r 0 ≤x≤a. (10.4.10) In this case, it is appropriate to use the ﬁnite Fourier cosine transform (10.2.4). So, the application of this transform gives the initial value problem d˜uc dt+κ⎪parenleftBignπ a⎪parenrightBig2 ˜uc=0, (10.4.11) ˜uc(n,0)=˜fc(n). (10.4.12) The solution of this problem is ˜uc(n, t)=˜fc(n)exp⎪braceleftbigg −κ⎪parenleftBignπ a⎪parenrightBig2 t⎪bracerightbigg . (10.4.13) The inverse ﬁnite cosine transform (10.2.5) gives the formal solution u(x, t)=1 a˜fc(0) +2 a∞⎪summationdisplay n=1˜fc(n)exp⎪braceleftbigg −κ⎪parenleftBignπ a⎪parenrightBig2 t⎪bracerightbigg cos⎪parenleftBignπx a⎪parenrightBig =1 aa⎪integraldisplay 0f(ξ)dξ+2 a∞⎪summationdisplay n=1⎡ ⎣a⎪integraldisplay 0f(ξ)cos⎪parenleftbiggnπξ a⎪parenrightbigg dξ⎤ ⎦ ×exp⎪braceleftbigg −κ⎪parenleftBignπ a⎪parenrightBig2 t⎪bracerightbigg cos⎪parenleftBignπx a⎪parenrightBig .(10.4.14) Example 10.4.3 (The Static Deﬂection of a Uniform Elastic Beam ). We consider the static deﬂection y(x) of a uniform elastic beam of ﬁnite length /lscriptwhich satisﬁes the equilibrium equation d4y dx4=W(x) EI=w(x),0≤x≤/lscript, (10.4.15) © 2007 by Taylor & Francis Group, LLC 418 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where W(x) is the applied load per unit length of the beam, Eis the Young’s modulus of the beam, and Iis the moment of inertia of the cross section of the beam. If the beam is freel y hinged at its ends, then y(x)=y/prime/prime(x)=0 a t x=0 a n d x=/lscript. (10.4.16) Application of the ﬁnite Fourier sine transform of y(x) to (10.4.15) and (10.4.16) gives ˜ys(n)=⎪parenleftbigg/lscript nπ⎪parenrightbigg4 ˜ws(n). (10.4.17) Inverting this result, we ﬁnd y(x)=2/lscript3 π4∞⎪summationdisplay n=11 n4sin⎪parenleftBignπx /lscript⎪parenrightBig ˜ws(n) =2/lscript3 π4∞⎪summationdisplay n=11 n4sin⎪parenleftBignπx /lscript⎪parenrightBig/lscript⎪integraldisplay 0w(ξ)sin⎪parenleftbiggnπξ /lscript⎪parenrightbigg dξ. (10.4.18) In particular, if the applied load of magnitude W0is conﬁned to the point x=α,w h e r e0 <α</lscript ,t h e n w(x)=W0δ(x−α)w h e r e W0is a constant. Consequently, the static deﬂection is y(x)=2/lscript3W0 π4∞⎪summationdisplay n=11 n4sin⎪parenleftBignπx /lscript⎪parenrightBig sin⎪parenleftBignπα /lscript⎪parenrightBig . (10.4.19) Example 10.4.4 (Transverse Displacement of an Elastic Beam of Finite Length ). We consider the transverse displacement of an elastic beam at a point xin the down- ward direction where the equilibrium position of the beam is along the x-axis. With the applied load W(x, t) per unit length of the beam, the displacement function y(x, t) satisﬁes the equation of motion ∂4y ∂x4+1 a2∂2y ∂t2=W(x, t) EI,0≤x≤/lscript, t> 0, (10.4.20) where a2=EI/(ρα),αis the cross-sectional area and ρis the line density of the beam. If the beam is freely hinged at its ends, then y(x, t)=∂2y ∂x2=0 a t x=0 a n d x=/lscript. (10.4.21) The initial conditions are y(x, t)=f(x),∂y ∂t=g(x)a t t=0 f o r 0 <x</lscript . (10.4.22) © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 419 We use the joint Laplace transform with respect to tand the ﬁnite Fourier sine transform with respect to xdeﬁned by ¯˜us(n, s)=∞⎪integraldisplay 0e−stdt/lscript⎪integraldisplay 0u(x, t)s i n⎪parenleftBignπx /lscript⎪parenrightBig dx. (10.4.23) Application of the double transform to (10.4.20)–(10.4.22) gives the solution for¯˜ys(n, s)a s ¯˜ys(n, s)=s˜fs(n)+˜gs(n) (s2+c2)+⎪parenleftbigga2 EI⎪parenrightbigg ˜Ws(n, s) (s2+c2), (10.4.24) where c=a⎪parenleftBignπ /lscript⎪parenrightBig2 . T h ei n v e r s eL a p l a c et r a n s f o r mg i v e s ˜ys(n, t)=˜fs(n)c o s (ct)+˜gs(n) csin(ct) +⎪parenleftbigga2 EI⎪parenrightbigg1 ct⎪integraldisplay 0sinc(t−τ)˜Ws(n, τ)dτ.(10.4.25) Thus, the inverse ﬁnite Fourier sine transform yields the formal solution as y(x, t)=2 /lscript∞⎪summationdisplay n=1ys(n,t)sin⎪parenleftBigπnx /lscript⎪parenrightBig , =2 /lscript∞⎪summationdisplay n=1sin⎪parenleftBignπx /lscript⎪parenrightBig⎪bracketleftbigg⎪braceleftbigg ˜fs(n)cos (ct)+˜gs(n) csin(ct)⎪bracerightbigg +⎪parenleftbigga2 EI⎪parenrightbigg1 ct⎪integraldisplay 0sinc(t−τ)˜Ws(n, τ)dτ⎤ ⎦,(10.4.26) where ˜fs(n)=/lscript⎪integraldisplay 0f(ξ)sin⎪parenleftbiggnπξ /lscript⎪parenrightbigg dξ,˜gs(n)=/lscript⎪integraldisplay 0g(ξ)sin⎪parenleftbiggnπξ /lscript⎪parenrightbigg dξ.(10.4.27ab) The case of free vibrations is of interest. In this case, W(x, t)≡0 and hence, ˜Ws(n, t)≡0. Consequently, solution (10.4.26) reduces to a simple form y(x, t)=2 /lscript∞⎪summationdisplay n=1⎪bracketleftbigg ˜fs(n)cosct+˜gs(n) csinct⎪bracketrightbigg sin⎪parenleftBignπx /lscript⎪parenrightBig , (10.4.28) where ˜fs(n)a n d˜ gs(n) are given by (10.4.27ab). © 2007 by Taylor & Francis Group, LLC 420 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 10.4.5 (Free Transverse Vibrations of an Elastic String of Finite Length ). We con- sider the free vibration of a string of length /lscriptstretched to a constant tension Tbetween two points (0 ,0) and (0 ,/lscript)l y i n go nt h e x-axis. The free transverse displacement function u(x, t) satisﬁes the wave equation ∂2u ∂t2=c2∂2u ∂x2,0≤x≤/lscript, t > 0, (10.4.29) where c2=T ρandρis the line density of the string. The initial and boundary conditions are u(x, t)=f(x),∂u ∂t=g(x)a t t=0 f o r 0 ≤x≤/lscript,(10.4.30ab) u(x, t)=0 a t x=0 a n d x=/lscriptfort>0. (10.4.31ab) Application of the joint Laplace transform with respect to tand the ﬁnite Fourier sine transform with respect to xdeﬁned by a similar result (10.4.23) to (10.4.29)–(10.4.31ab) gives ¯˜us(n, s)=s˜fs(n) (s2+a2)+˜gs(n) (s2+a2), (10.4.32) where a2=⎪parenleftBignπc /lscript⎪parenrightBig2 . T h ei n v e r s eL a p l a c et r a n s f o r mg i v e s ˜us(n,t)=˜fs(n)cosat+˜gs(n) asinat. (10.4.33) The inverse ﬁnite Fourier sine transform leads to the solution for u(x, t)a s u(x, t)=2 /lscript∞⎪summationdisplay n=1⎪bracketleftbigg ˜fs(n)cosat+˜gs(n) asinat⎪bracketrightbigg sin⎪parenleftBignπx /lscript⎪parenrightBig , (10.4.34) where ˜fs(n)a n d˜ gs(n) are given by (10.4.27ab). Example 10.4.6 (Two-Dimensional Unsteady Couette Flow ). We consider two-dimensional un- steady viscous ﬂow between the plate at z= 0 at rest and the plate z=hin motion parallel to itself with a variable velocity U(t)i nt h e xdirection. The ﬂuid velocity u(z,t) satisﬁes the equation of motion ∂u ∂t=−P(t) ρ+ν∂2u ∂z2,0≤z≤h, t > 0, (10.4.35) © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 421 and the boundary and initial conditions u(z, t)=0 o n z=0,t >0; (10.4.36) u(z, t)=U(t)o n z=h, t > 0; (10.4.37) u(z, t)=0 a t t≤0,for 0≤z≤h; (10.4.38) where the pressure gradient px=P(t)a n d νis the kinematic viscosity of the ﬂuid. Application of the double transform deﬁned by (10.4.23) to this initial boundary value problem gives the solution for ¯˜us(n,s)a s ⎪parenleftbigg s+νn2π2 h2⎪parenrightbigg ¯˜us(n, t)=−h¯P(s) nπρ[1 + (−1)n+1] +νn π h(−1)n+1¯U(s).(10.4.39) The inverse Laplace transform yields ˜us(n, t)=−h nπρ[1 + (−1)n+1]t⎪integraldisplay 0P(t−τ)exp⎪parenleftbigg −νn2π2τ h2⎪parenrightbigg dτ +νn π h(−1)n+1t⎪integraldisplay 0U(t−τ)exp⎪parenleftbigg −νn2π2τ h2⎪parenrightbigg dτ.(10.4.40) Finally, the inverse ﬁnite Fourier sine transform gives the formal solution u(z, t)=2 h∞⎪summationdisplay n=1˜us(n, t)sin⎪parenleftBignπz h⎪parenrightBig . (10.4.41) If, in particular, P(t)=c o n s t a n t PandU(t)=c o n s t a n t= U, then (10.4.41) reduces to u(z, t)=−2P μh∞⎪summationdisplay n=1⎪parenleftbiggh nπ⎪parenrightbigg3 [1 + (−1)n+1]si n⎪parenleftBignπz h⎪parenrightBig⎪bracketleftbigg 1−exp⎪parenleftbigg −νn2π2t h2⎪parenrightbigg⎪bracketrightbigg +2U h∞⎪summationdisplay n=1(−1)n+1⎪parenleftbiggh nπ⎪parenrightbigg sin⎪parenleftBignπz h⎪parenrightBig⎪bracketleftbigg 1−exp⎪parenleftbigg −νn2π2t h2⎪parenrightbigg⎪bracketrightbigg .(10.4.42) This solution for the velocity ﬁeld cons ists of both steady-state and transient components. In the limit as t→∞, the transient component decays to zero, and the steady state is attained in the form u(z, t)=−2P μh∞⎪summationdisplay n=1⎪parenleftbiggh nπ⎪parenrightbigg3 [1 + (−1)n+1]si n⎪parenleftBignπz /planckover2pi1⎪parenrightBig +2U h2∞⎪summationdisplay n=1(−1)n+1⎪parenleftbiggh2 nπ⎪parenrightbigg sin⎪parenleftBignπz h⎪parenrightBig .(10.4.43) © 2007 by Taylor & Francis Group, LLC 422 INTEGRAL TRANSFORMS and THEIR APPLICATIONS In view of the inverse ﬁnite Fourier sine transforms F−1 s⎪braceleftBigg 2⎪parenleftbiggh nπ⎪parenrightbigg3 [1 + (−1)n+1]⎪bracerightBigg =z(h−z), (10.4.44) F−1 s⎪braceleftbigg (−1)n+1⎪parenleftbiggh2 nπ⎪parenrightbigg⎪bracerightbigg =z, (10.4.45) solution (10.4.43) can be rewritten in the closed form u(z, t)=Uz h−h 2μ⎪parenleftbigg∂p ∂x⎪parenrightbigg⎪parenleftBig 1−z h⎪parenrightBig z. (10.4.46) This is known as the generalized Couette ﬂow . In the absence of the pressure gradient term, solution (10.4.46) reduces to the linear proﬁle of simple Couette ﬂow. On the other hand, if U(t)≡0a n d P(t)/negationslash= 0, the solution (10.4.46) rep- resents the parabolic proﬁle of Poiseuille ﬂow between two parallel stationary plates due to an imposed pressure gradient. 10.5 Multiple Finite Fourier Transforms and Their Applications The above analysis for the ﬁnite Fourier sine and cosine transforms of a func- tion of one independent variable can readily be extended to a function of several independent variables. In particular, if f(x, y) is a function of two independent variables xandy,d e ﬁ n e di nar e g i o n0 ≤x≤a,0≤y≤b,i t s double ﬁnite Fourier sine transform is deﬁned by Fs{f(x, y)}=˜fs(m, n)=a⎪integraldisplay 0b⎪integraldisplay 0sin⎪parenleftBigmπx a⎪parenrightBig sin⎪parenleftBignπy b⎪parenrightBig dxdy. (10.5.1) Theinverse transform is given by the double series F−1 s⎪braceleftBig ˜fs(m, n)⎪bracerightBig =f(x, y)=⎪parenleftbigg4 ab⎪parenrightbigg∞⎪summationdisplay m=1∞⎪summationdisplay n=1˜fs(m, n)s i n⎪parenleftBigmπx a⎪parenrightBig ×sin⎪parenleftBignπy b⎪parenrightBig .(10.5.2) Similarly, we can deﬁne the double ﬁnite Fourier cosine transform and its inverse. The double Fourier sine transforms of the partial derivatives of f(x, y)c a n easily be obtained. If f(x, y) vanishes on the boundary of the rectangular © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 423 region D{0≤x≤a,0≤y≤b},t h e n Fs⎪bracketleftbigg∂2f ∂x2+∂2f ∂y2⎪bracketrightbigg =−π2⎪parenleftbiggm2 a2+n2 b2⎪parenrightbigg ˜fs(m, n). (10.5.3) Example 10.5.1 (Free Vibrations of a Rectangular Elastic Membrane ). The initial value prob- lem for the transverse displacement ﬁeld u(x, y, t) satisﬁes the following equa- tion and the boundary and initial data c2⎪parenleftbigg∂2u ∂x2+∂2u ∂y2⎪parenrightbigg =∂2u ∂t2,for all ( x, y)i nD, t> 0, (10.5.4) u(x, y, t) = 0 on the boundary ∂Dfor all t>0, (10.5.5) u(x, y, t)=f(x, y),ut(x, y, t)=g(x, y)a t t=0,for (x, y)∈D.(10.5.6ab) Application of the double ﬁnite Fourier sine transform deﬁned by ˜us(m, n)=a⎪integraldisplay 0b⎪integraldisplay 0u(x, y)sin⎪parenleftBigmπx a⎪parenrightBig sin⎪parenleftBignπy b⎪parenrightBig dxdy, (10.5.7) to the system (10.5.4)–(10.5.6ab) gives d2˜us dt2+c2π2⎪parenleftbiggm2 a2+n2 b2⎪parenrightbigg ˜us=0,t > 0 (10.5.8) ˜us(m,n,0)=˜fs(m, n),⎪parenleftbiggd˜us dt⎪parenrightbigg t=0=˜gs(m, n). (10.5.9) The solution of this transformed problem is ˜us(m,n,t )=˜fs(m, n)cos (cπωmnt) +(cπωmn)−1˜gs(m, n)sin (cπωmnt), (10.5.10) where ωmn=⎪parenleftbiggm2 a2+n2 b2⎪parenrightbigg1 2 . (10.5.11) The inverse transform gives the formal solution for u(x, y, t)i nt h ef o r m u(x, y, t)=⎪parenleftbigg4 ab⎪parenrightbigg∞⎪summationdisplay m=1∞⎪summationdisplay n=1sin⎪parenleftBigmπx a⎪parenrightBig sin⎪parenleftBignπy b⎪parenrightBig⎪bracketleftbig˜fs(m, n)cos (cπωmnt) +(cπωmn)−1˜gs(m, n)sin (cπωmnt)⎪bracketrightbig , (10.5.12) © 2007 by Taylor & Francis Group, LLC 424 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where ˜fs(m, n)=a⎪integraldisplay 0b⎪integraldisplay 0f(ξ,η)sin⎪parenleftbiggmπξ a⎪parenrightbigg sin⎪parenleftBignπη b⎪parenrightBig dξdη, (10.5.13) ˜gs(m, n)=a⎪integraldisplay 0b⎪integraldisplay 0g(ξ,η)sin⎪parenleftbiggmπξ a⎪parenrightbigg sin⎪parenleftBignπη b⎪parenrightBig dξdη. (10.5.14) Example 10.5.2 (Deﬂection of a Simply Supported Rectangular Elastic Plate ). The deﬂection u(x, y) of the plate satisﬁes the biharmonic equation ∇4u≡∂4u ∂x4+2∂4u ∂x2∂y2+∂4u ∂y4=w(x, y) D, inD{0≤x≤a,0≤y≤b},(10.5.15) where w(x, y) represents the applied load at a point ( x, y)a n dD=2Eh3 3(1−σ2) is the constant ﬂexural rigidity of the plate. On the edge of the simply supported plate the deﬂection and bending mo- ments are zero; hence, equation (10.5.15) has to be solved subject to the boundary conditions u(x, y)=0 on x=0 a n d x=a u(x, y)=0 on y=0 a n d y=b ∂2u ∂x2=0 o n x=0 a n d x=a ∂2u ∂y2=0 o n y=0 a n d y=b⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎪⎭. (10.5.16) We ﬁrst solve the problem due to a concentrated load W 0at the point ( ξ,η) inside Dso that w(x, y)=Pδ(x−ξ)δ(y−η), where Pis a constant. Application of the double ﬁnite Fourier sine transform (10.5.7) to (10.5.15)– (10.5.16) gives π4⎪parenleftbiggm2 a2+n2 b2⎪parenrightbigg2 ˜us(m, n)=⎪parenleftbiggP D⎪parenrightbigg sin⎪parenleftbiggmπξ a⎪parenrightbigg sin⎪parenleftBignπη b⎪parenrightBig , or, ˜us(m, n)=⎪parenleftbiggP Dπ4ω4mn⎪parenrightbigg sin⎪parenleftbiggmπξ a⎪parenrightbigg sin⎪parenleftBignπη b⎪parenrightBig , (10.5.17) © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 425 where ωmnis deﬁned by (10.5.11). The inverse transform gives the formal solution u(x, y)=⎪parenleftbigg4P π4abD⎪parenrightbigg∞⎪summationdisplay m=1∞⎪summationdisplay n=1⎪bracketleftbigg ω−4 mnsin⎪parenleftbiggmπξ a⎪parenrightbigg sin⎪parenleftBignπη b⎪parenrightBig⎪bracketrightbigg ×sin⎪parenleftBigmπx a⎪parenrightBig sin⎪parenleftBignπy b⎪parenrightBig .(10.5.18) For an arbitrary load w(x, y) over the region α≤x≤β, γ≤y≤δinside the region D, we can replace Pbyw(ξ,η)dξdηand integrate over the rectangle α≤ξ≤β,γ≤η≤δ. Consequently, the formal solution is obtained from (10.5.18) and has the form u(x, y)=⎪parenleftbigg4 π4abD⎪parenrightbigg∞⎪summationdisplay m=1∞⎪summationdisplay n=1⎧ ⎨ ⎩β⎪integraldisplay αδ⎪integraldisplay γw(ξ,η)sin⎪parenleftbiggmπξ a⎪parenrightbigg sin⎪parenleftBignπη b⎪parenrightBig dξdη⎫ ⎬ ⎭ ×ω−4 mnsin⎪parenleftBigmπx a⎪parenrightBig sin⎪parenleftBignπy b⎪parenrightBig .(10.5.19) 10.6 Exercises 1. Find the ﬁnite Fourier cosine transform of f(x)=x2. 2. Use the result (10.3.2) to prove (a)Fs{x2}=a3 nπ(−1)n+1−2⎪parenleftBiga nπ⎪parenrightBig3 [1 + (−1)n+1], (b)Fs{x3}=(−1)na4 π2⎪parenleftbigg6 n3π3−1 nπ⎪parenrightbigg . 3. Solve the initial-boundary value problem in a ﬁnite domain ut=κuxx,0≤x≤a, t > 0, u(x,0) = 0 for 0 ≤x≤a, u(0,t)=f(t)f o r t>0, u(a, t)=0 f o r t>0. 4. Solve Exercise 3 above by replacing the only condition at x=awith the radiation condition ux+hu=0 a t x=a, © 2007 by Taylor & Francis Group, LLC 426 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where his a constant. 5. Solve the heat conduction problem ut=κuxx, 0≤x≤a, t > 0, ux(0,t)=f(t) ux(a, t)+hu=0⎪bracerightBigg fort>0, u(x,0)= 0 for 0 ≤x≤a. 6. Solve the diﬀusion equation (10.4.1) with the following boundary and initial data ux(0,t)=f(t),u x(a, t)=0 f o r t>0 u(x,0)= 0 for 0 ≤x≤a. 7. Solve the problem of free vibrations described in Example 10.4.4, when the beam is at rest in its equilibrium position at time t=0 , a n d a n impulse Iis applied at x=η,t h a ti s , f(x)≡0a n d g(x)=⎪parenleftbiggI ρα⎪parenrightbigg δ(x−η). 8. Find the solution of the problem in Example 10.4.4 when (i)W(x, t)=W0φ(t)δ(x−η),0<η</lscript ; (ii) a concentrated applied load is moving along the beam with a con- stant speed U,t h a ti s , W(x, t)=W0φ(t)δ(x−Ut)H(Ut−/lscript), where W0is a constant. 9. Find the solution of the forced vibration of an elastic string of ﬁnite length /lscriptwhich satisﬁes the forced wave equation 1 c2∂2u ∂t2=∂2u ∂x2+F(x, t),0≤x≤/lscript, t> 0, with the initial and boundary data u(x, t)=f(x),ut=g(x)a t t=0 f o r0 ≤x≤/lscript, u(0,t)=0= u(/lscript, t)f o r t>0. Derive the solution for special cases when f(x)=0= g(x)w i t h (i) an arbitrary non-zero F(x, t), and (ii)F(x, t)=P(t) Tδ(x−a),0≤a≤/lscript,w h e r e Tis a constant. © 2007 by Taylor & Francis Group, LLC Finite Fourier Sine and Cosine Transforms 427 10. For the ﬁnite Fourier sine transform deﬁned over (0 ,π), show that (a)Fs⎪braceleftBigx 2(π−x)⎪bracerightBig =1 n3[1 + (−1)n+1] (b)Fs⎪braceleftbiggsinha(π−x) sinhaπ⎪bracerightbigg =n (n2+a2),a/negationslash=0 . 11. For the ﬁnite Fourier cosine transform deﬁned over (0 ,π), show that (a)Fc{(π−x)2}=2π n2forn=1,2,...;Fs{(π−x)2}=π3 3forn=0. (b)Fc{cosha(π−x)}=asinh(aπ) (n2+a2)fora/negationslash=0. 12. Use the ﬁnite Fourier sine transform to solve the problem of diﬀusion of electricity along a cable of length a.T h ep o t e n t i a l V(x, t)a ta n yp o i n t xof the cable of resistance Rand capacitance Cper unit length satisﬁes the diﬀusion equation Vt=κVxx,0≤x≤a, t > 0, where κ=(RC)−1and the boundary conditions (the ends of the cable are earthed) V(0,t)=0= V(a, t)f o r t>0, and the initial conditions V(x,0) =⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩⎪parenleftbigg2V 0 a⎪parenrightbigg x, 0≤x≤a 2 ⎪parenleftbigg2V0 a⎪parenrightbigg (a−x),a 2≤x≤a⎫ ⎪⎪⎪⎬ ⎪⎪⎪⎭, where V 0is a constant. 13. Establish the following results (a)Fs⎪bracketleftbiggd dx{f1(x)∗g1(x)}⎪bracketrightbigg =2n˜fs(n)˜gs(n), (b)Fs⎡ ⎣x⎪integraldisplay 0{f1(u)∗g1(u)}du⎤ ⎦=2 n˜fs(n)˜gs(n). 14. If pis not necessarily an integer, we write ˜fc(p)=π⎪integraldisplay 0f(x)cospxdx and ˜fs(p)=π⎪integraldisplay 0f(x)sinpxdx. Show that, for any constant α, © 2007 by Taylor & Francis Group, LLC 428 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (a)Fc{2f(x)cosαx}=˜fc(n−α)+˜fc(n+α), (b)Fc{2f(x)sinαx}=˜fs(n+α)−˜fs(n−α), (c)Fs{2f(x)cosαx}=˜fs(n+α)+˜fs(n−α), (d)Fs{2f(x)sinαx}=˜fc(n−α)−˜fc(n+α). 15. Solve the problem in Example 10.5.2 for uniform load W0over the region α≤x≤βandγ≤y≤δ. 16. Solve the problem of free oscillations of a rectangular elastic plate of density ρbounded by the two parallel planes z=±h. The deﬂection u(x, y, t) satisﬁes the equation D∇4u+ρh u tt=0,0≤x≤a,0≤y≤b, t> 0, where the deﬂection and the bending moments are all zero at the edges. © 2007 by Taylor & Francis Group, LLC 11 Finite Laplace Transforms “The genius of Laplace was a perfect sledgehammer in bursting purely mathematical obstacles, but like that useful instrument, it gave neither ﬁnish nor beauty to the results ... nevertheless, Laplace never attempted the investigation of a subject without leaving up-on it the marks of diﬃculties conquered: sometimes clumsily, some- times indirectly, but still his end is obtained and the diﬃculty is conquered.” Anonymous “It seems to be one of the fundamental features of nature that fundamental physics laws are described in terms of great beautyand power. As time goes on, it becomes increasingly evident that the rules that the mathematicians ﬁnd interesting are the same as those that nature has chosen.” Paul Dirac 11.1 Introduction The Laplace transform method is normally used to ﬁnd the response of alinear system at any time tto the initial data at t= 0 and the disturbance f(t) acting for t≥0. If the disturbance or input function is f(t)=e x p ( at 2),a >0, the usual Laplace transform cannot be used to ﬁnd the solution of an initial value problem because the Laplace transform of f(t) does not exist. From a physical point of view, there seems to be no reason at all why the function f(t) cannot be used as an acceptable distu rbance for a system. It is often true that the solution at times later than twould not aﬀect the state at time t. This leads to the idea of introducing the ﬁnite Laplace transform in 0≤t≤T in order to extend the power and usefulness of the usual Laplace transform in0≤t<∞. This chapter deals with the deﬁnition and basic operational properties of the ﬁnite Laplace transform. In Section 11.4, the method of the ﬁnite Laplace 429 © 2007 by Taylor & Francis Group, LLC 430 INTEGRAL TRANSFORMS and THEIR APPLICATIONS transform is used to solve the initial value problems and the boundary value problems. This chapter is essentially based on papers by Debnath and Thomas(1976) and Dunn (1967). 11.2 Deﬁnition of the Finite Laplace Transform and Examples Theﬁnite Laplace transform of a continuous (or an almost piecewise contin- uous) function f(t)i n( 0 ,T) is denoted by ST{f(t)}=¯f(s, T), and deﬁned by ST{f(t)}=¯f(s, T)=T⎪integraldisplay 0f(t)e−stdt, (11.2.1) where sis a real or complex number and Tis a ﬁnite number that may be positive or negative so that (11.2.1) can be deﬁned in any interval ( −T1,T2). Clearly, STis a linear integral transformation. Theinverse ﬁnite Laplace transform is deﬁned by the complex integral f(t)=S−1 T{¯f(s, T)}=1 2πi⎪integraldisplayc+i∞ c−i∞¯f(s, T)estds, (11.2.2) where the integral is taken over any open contour Γ joining any two points c−iRandc+iRin the ﬁnite complex splane as R→∞. Iff(t) is almost piecewise continuous, that is, it has at most a ﬁnite number of simple discontinuities in 0 ≤t≤T. Moreover, in the intervals where f(t) is continuous, it satisﬁes a Lipschitz condition of order α>0. Under these conditions, it can be shown that the inversion integral (11.2.2) is equal to 1 2πi⎪integraldisplay Γ¯f(s, T)estds=1 2[f(t−0) +f(t+0 ) ], (11.2.3) where Γ is an arbitrary open contour that terminates with ﬁnite constant c asR→∞. This is due to the fact that ¯f(s, T) is an entire function of s. It follows from (11.2.1) that if ⎪integraldisplay f(t)e−stdt=−F(s, t)e−st, (11.2.4) then ¯f(s, T)=F(s,0)−F(s, T)e−sT(11.2.5) =¯f(s)−F(s, T)e−sT, (11.2.6) © 2007 by Taylor & Francis Group, LLC Finite Laplace Transforms 431 where ¯f(s) is the usual Laplace transform deﬁned by (3.2.1) and hence, ¯f(s)=F(s,0) =∞⎪integraldisplay 0e−stf(t)dt. (11.2.7) Further, using (11.2.2) and (11.2.6), the inversion formula can be written as f(t)=1 2πi⎪integraldisplay ΓF(s,0)estds−1 2πi⎪integraldisplay ΓF(s, T)es(t−T)ds. (11.2.8) It is noted that the ﬁrst integral may be closed in the left half of the complex plane. On the other hand, for t<T, the contour of the second integral must be closed in the right half-plane. We select Γ so that all poles of F(s,0) lie to the left of Γ. Thus, the ﬁrst integral represents the solution of the initial value problem, and for t<T, the second integral vanishes. When t>T, the second integral may be closed in the left half of the complex plane so that f(t)=0f o r t>T. Thus, for the solution of the initial value problem, there is no need to consider the second integral, and this ca se is identical with the usual Laplace transform. So, unlike the usual Laplace transform of a function f(t), there is no re- striction needed on the transform variable sfor the existence of the ﬁnite Laplace transform ST{f(t)}=¯f(s, T). Further, the existence of (11.2.1) does not require the exponential order property of f(t). If a function f(t)h a st h e usual Laplace transform, then it also has the ﬁnite Laplace transform. In oth- er words, if ¯f(s)=S{f(t)}exists, then ST{f(t)}=¯f(s, T)e x i s t sa ss h o w n below. We have ¯f(s)=T⎪integraldisplay 0e−stf(t)dt+∞⎪integraldisplay Te−stf(t)dt. (11.2.9) Since ¯f(s) exists, both the integrals on the right of (11.2.9) exist. Hence, the ﬁrst integral in (11.2.9) exists and deﬁnes ¯f(s, T). However, the converse of this result is not necessarily true. This can be shown by an example. It is well known that the ususal Laplace transform off(t)=e x p ( at 2),a >0, does not exist. But the ﬁnite Laplace transform of this function exists as shown below. ¯f(s, T)=ST{exp(at2)}=T⎪integraldisplay 0exp(−st+at2)dt =e x p⎪parenleftbigg −s2 4a⎪parenrightbiggT⎪integraldisplay 0exp⎪bracketleftbigg −⎪parenleftbigg√ at−s 2√ a⎪parenrightbigg i⎪bracketrightbigg2 dt =1 2i⎪parenleftBigπ a⎪parenrightBig1 2exp⎪parenleftbigg −s2 4a⎪parenrightbigg⎪bracketleftbigg erf⎪braceleftbigg⎪parenleftbigg√ aT−s 2√ a⎪parenrightbigg i⎪bracerightbigg +erf⎪parenleftbiggsi 2√ a⎪parenrightbigg⎪bracketrightbigg . (11.2.10) © 2007 by Taylor & Francis Group, LLC 432 INTEGRAL TRANSFORMS and THEIR APPLICATIONS In the limit as T→∞, (11.2.10) does not exist as seen below. We use the result (see Carslaw and Jeager, 1953, p. 48), to obtain erf(z)=e x p ( −z2)⎡ ⎣1+2i πz⎪integraldisplay 0ex2dx⎤⎦→∞ asz→∞, (11.2.11) where z=⎪parenleftbigg T√ a−s 2√ a⎪parenrightbigg i. This ensures that the right-hand side of (11.2.10) tends to inﬁnity as T→∞. Thus, the usual Laplace transform of exp( at2) does not exist as expected. The solution of the ﬁnal value problem is denoted by ffia n dd e ﬁ n e db y ffi(t)=1 2πi⎪integraldisplay ΓF(s, T)es(t−T)ds, (11.2.12) where the contour Γ lies to the left of the singularities of F(s, t)o rF(s,0). THEOREM 11.2.1 The solution of an initial value problem is identical with that of the ﬁnal value problem. PROOF Suppose finis the solution of the initial value problem, and it is given by fin(t)=1 2πi⎪integraldisplay BrF(s,0)estds, (11.2.13) where Bris theBromwich contour extending from c−iRtoc+iRasR→∞. We next reverse the direction of Γ in (11.2.12) and then subtract (11.2.13)from (11.2.12) to obtain f in(t)−ffi(t)=1 2πi⎪integraldisplay C{F(s,0)−F(s, T)e−sT}estds =1 2πi⎪integraldisplay C¯f(s, T)estds, (11.2.14) where Cis a closed contour which contains all the singularities of F(s,0) orF(s, T). Thus, the integrand of (11.2.14) is an entire function of sand hence, the integral around a contour Cmust vanish by Cauchy’s Fundamental Theorem . Hence, fin(t)=ffi(t)=f(t). (11.2.15) © 2007 by Taylor & Francis Group, LLC Finite Laplace Transforms 433 This completes the proof. We next calculate the ﬁnite Laplace transform of several elementary func- tions: Example 11.2.1 Iff(t)=1 ,t h e n ST{1}=¯f(s, T)=T⎪integraldisplay 0e−stdt=1 s(1−e−sT). (11.2.16) Example 11.2.2 Iff(t)=eat,t h e n ST{eat}=¯f(s, T)=T⎪integraldisplay 0e−(s−a)tdt=1−e−(s−a)T (s−a). (11.2.17) Example 11.2.3 Iff(t)=s i n ator cos at,t h e n ST{sinat}=T⎪integraldisplay 0sinat e−stdt =a s2+a2−e−sT s2+a2(ssinaT+acosaT).(11.2.18) ST{cosat}=s s2+a2+e−st s2+a2(asinaT−scosaT).(11.2.19) Example 11.2.4 Iff(t)=t,t h e n ST{t}=T⎪integraldisplay 0te−stdt=1 s2−e−sT s⎪parenleftbigg1 s+T⎪parenrightbigg . (11.2.20) © 2007 by Taylor & Francis Group, LLC 434 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 11.2.5 Iff(t)=t2,t h e n ST{t2}=T⎪integraldisplay 0t2e−stdt=2 s3−e−sT s⎪parenleftbigg T2+2T s+2 s2⎪parenrightbigg . (11.2.21) More generally, if f(t)=tn,t h e n ST{tn}=T⎪integraldisplay 0tne−stdt=n! sn+1−e−sT s ×⎪braceleftbigg Tn+n sTn−1+n(n−1) s2Tn−2+···+n!T sn−1+n! sn⎪bracerightbigg .(11.2.22) Example 11.2.6 Iff(t)=ta,a(>−1) is a real number, then ST{ta}=s−(a+1)γ(a+1,sT), (11.2.23) where γ(α, x) is called the incomplete gamma function a n di sd e ﬁ n e db y γ(α, x)=x⎪integraldisplay 0e−uuα−1du. (11.2.24) We have ST{ta}=T⎪integraldisplay 0tae−stdt,(u=st) =s−(a+1)sT⎪integraldisplay 0uae−udu=s−(a+1)γ(a+1,s T). Example 11.2.7 If 0<a<T andf(t)=0i n −a<t< 0, then ST{f(t−a)}=e−asT−a⎪integraldisplay 0e−τsf(τ)dτ=e−as¯f(s, T−a). (11.2.25) © 2007 by Taylor & Francis Group, LLC Finite Laplace Transforms 435 In particular, ST{H(t−a)}=T⎪integraldisplay ae−stdt=1 s(e−as−e−sT). (11.2.26) Example 11.2.8 ST{erf at}=s−1exp⎪parenleftbiggs2 4a2⎪parenrightbigg⎪bracketleftBig erf⎪parenleftBig aT+s 2a⎪parenrightBig −erf⎪parenleftBigs 2a⎪parenrightBig⎪bracketrightBig −e−sT serf(aT).(11.2.27) We have, by deﬁnition, ST{erf(at)}=T⎪integraldisplay 0erf(at)e−stdt which is, integrating by parts and using the deﬁnition of erf(at), =−[s−1e−sterf(at)]T 0+1 s2 √ πT⎪integraldisplay 0exp[−(st+a2t2)]dt =−e−sT serf(aT)+1 sexp⎪parenleftbiggs2 4a2⎪parenrightbigg2 √ πaT+s 2a⎪integraldisplay s 2ae−u2du =−e−sT serf(aT)+1 sexp⎪parenleftbiggs2 4a2⎪parenrightbigg⎪bracketleftBig erf⎪parenleftBig aT+s 2a⎪parenrightBig −erf⎪parenleftBigs 2a⎪parenrightBig⎪bracketrightBig . Example 11.2.9 Iff(t) is a periodic function with period ω,t h e n ¯f(s, T)=ST{f(t)}=(1−e−sT) (1−e−sω)¯f(s, ω), (11.2.28) where T=nω,a n d nis a ﬁnite positive integer. © 2007 by Taylor & Francis Group, LLC 436 INTEGRAL TRANSFORMS and THEIR APPLICATIONS By deﬁnition, we have ST{f(t)}=T⎪integraldisplay 0f(t)e−stdt =ω⎪integraldisplay 0f(t)e−stdt+⎪integraldisplay2ω ωf(t)e−stdt+···+nω⎪integraldisplay (n−1)ωf(t)e−stdt, which is, substituting t=u+ω, t=u+2ω,···,t=u+(n−1)ωin the sec- ond, third, and the last integral, respectively, =⎪integraldisplayω 0f(u)e−sudu+e−sω⎪integraldisplayω 0f(u)e−sudu+···+e−s(n−1)ω⎪integraldisplayω 0f(u)e−sudu =⎪bracketleftBig 1+e−sω+···+e−sω(n−1)⎪bracketrightBig⎪integraldisplayω 0e−suf(u)du =(1−e−nsω) (1−e−sω)¯f(s, ω). In the limit as n→∞ (T→∞), (11.2.28) reduces to the known result ¯f(s)=( 1 −e−sω)−1ω⎪integraldisplay 0e−suf(u)du. (11.2.29) 11.3 Basic Operational Properties of the Finite Laplace Transform THEOREM 11.3.1 IfST{f(t)}=¯f(s, T), then (a) (Shifting) ST{e−atf(t)=¯f(s+a, T), (11.3.1) (b) (Scaling) ST{f(at)}=1 a¯f⎪parenleftBigs a,a T⎪parenrightBig . (11.3.2) The proofs are easy exercises for the reader. © 2007 by Taylor & Francis Group, LLC Finite Laplace Transforms 437 THEOREM 11.3.2 (Finite Laplace Transforms of Derivatives ). IfST{f(t)}=¯f(s, T), then ST{f/prime(t)}=s¯f(s, T)−f(0) +e−sTf(T), (11.3.3) ST{f/prime/prime(t)}=s2¯f(s, T)−sf(0)−f/prime(0) +sf(T)e−sT+f/prime(T)esT.(11.3.4) More generally, ST{f(n)(t)}=sn¯f(s, T)−n⎪summationdisplay k=1sn−kf(k−1)(0) +e−sTn⎪summationdisplay k=1sn−kf(k−1)(T). (11.3.5) PROOF We have, integrating by parts, ST{f/prime(t)}=T⎪integraldisplay 0f/prime(t)e−stdt=[f(t)e−st]T 0+sT⎪integraldisplay 0f(t)e−stdt =s¯f(s, T)−f(0) +f(T)e−sT. Repeating this process gives (11.3.4). By induction, we can prove (11.3.5). THEOREM 11.3.3 (Finite Laplace Transform of Integrals ). If F(t)=t⎪integraldisplay 0f(u)du (11.3.6) so that F/prime(t)=f(t) for all t,t h e n ST⎧ ⎨ ⎩t⎪integraldisplay 0f(u)du⎫ ⎬ ⎭=1 s{¯f(s, T)−e−sTF(T)}. (11.3.7) PROOF We have from (11.3.3) ST{F/prime(t)}=sST{F(t)}−F(0) +e−sTF(T). Or, ¯f(s, T)=sST⎧ ⎨ ⎩t⎪integraldisplay 0f(u)du⎫ ⎬ ⎭+e−sTF(T). © 2007 by Taylor & Francis Group, LLC 438 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Hence, ST⎧ ⎨ ⎩t⎪integraldisplay 0f(u)du⎫ ⎬ ⎭=1 s[¯f(s, T)−e−sTF(T)]. THEOREM 11.3.4 IfST{f(t)}=¯f(s, T), then d ds¯f(s, T)=ST{(−t)f(t)}, (11.3.8) d2 ds2¯f(s, T)=ST{(−t)2f(t)}. (11.3.9) More generally, dn dsn¯f(s, T)=ST{(−t)nf(t)}. (11.3.10) The proofs of these results are easy exercises. These results can be used to ﬁnd the ﬁnite Laplace transform of the product tnand any derivatives of f(t)i nt e r m so f ¯f(s, T). In other words, ST{tnf/prime(t)}=(−1)ndn dsn[ST{f/prime(t)}] =(−1)ndn dsn[s¯f(s, T)−f(0) +f(T)e−sT]. Similarly, we obtain a more general result ST{tnf(m)(t)}=(−1)ndn dsn[ST{f(m)(t)}], which is, by (11.3.5), =(−1)ndn dsn⎡ ⎣sm¯f(s, T)−m⎪summationdisplay k=1sm−kf(k−1)(0) +e−sTm⎪summationdisplay k=1sm−kf(k−1)(T)⎤ ⎦. Finally, we can ﬁnd T⎪integraldisplay s¯f(s, T)ds=T⎪integraldisplay sdsT⎪integraldisplay 0f(t)e−stdt=T⎪integraldisplay 0f(t)dtT⎪integraldisplay se−stds =T⎪integraldisplay 0f(t) te−stdt−T⎪integraldisplay 0f(t) te−Ttdt=¯g(s, T)−¯g(T,T), © 2007 by Taylor & Francis Group, LLC Finite Laplace Transforms 439 where g(t)=f(t) tand the existence of ¯ g(s, T) is assumed. 11.4 Applications of Finite Laplace Transforms Example 11.4.1 Use the ﬁnite Laplace transform to solve the initial value problem dx dt+αx=At, 0≤t≤T, (11.4.1) x(t=0 )= a. (11.4.2) Application of the ﬁnite Laplace transform gives s¯x(s, T)−x(0) +e−sTx(T)+α¯x(s, T)=A⎪bracketleftbigg1 s2−1 se−sT⎪parenleftbigg1 s+T⎪parenrightbigg⎪bracketrightbigg . Or, ¯x(s, T)=a s+α−e−sTx(T) s+α+A s+α⎪bracketleftbigg1 s2−1 se−sT⎪parenleftbigg1 s+T⎪parenrightbigg⎪bracketrightbigg .(11.4.3) This is not an entire function, but it becomes an entire function by setting ¯x(T)=AT α−A α2+A α2e−αT+ae−αT(11.4.4) so that ¯x(s, T)=⎪parenleftbigg a+A α2⎪parenrightbigg⎪bracketleftbigg1−e−(s+α)T s+α⎪bracketrightbigg +A α⎪bracketleftbigg1 s2(1−e−sT)−T se−sT⎪bracketrightbigg −A α2⎪bracketleftbigg1 s(1−e−sT)⎪bracketrightbigg . (11.4.5) Using Table B-11 of ﬁnite Laplace transforms gives the ﬁnal solution x(t)=⎪parenleftbigg a+A α2⎪parenrightbigg e−αt+At α−A α2. (11.4.6) Example 11.4.2Solve the simple harmonic oscillator governed by d 2x dt2+ω2x=F, (11.4.7) © 2007 by Taylor & Francis Group, LLC 440 INTEGRAL TRANSFORMS and THEIR APPLICATIONS x(t=0 )= a,˙x(t=0 )= u, (11.4.8ab) where F,a,a n d uare constants. Application of the ﬁnite Laplace transform gives the solution ¯x(s, T)=as s2+ω2+u s2+ω2−se−sTx(T) s2+ω2 −e−sT˙x(T) s2+ω2+F s(s2+ω2)(1−e−sT),(11.4.9) Since ¯ x(s, T) is not an entire function, we choose x(T) such that x(T)=⎪parenleftbigg a−F ω2⎪parenrightbigg cosωT+u ωsinωT+F ω2(11.4.10) ¯x(s, T) becomes an entire function. Consequently, (11.4.9) becomes ¯x(s, T)=⎪parenleftbigg a−F ω2⎪parenrightbigg⎪bracketleftbiggs s2+ω2+e−sT s2+ω2{ωsinωT−scosωT}⎪bracketrightbigg +u ω⎪bracketleftbiggω s2+ω2−e−sT s2+ω2{ssinωT+ωcosωT}⎪bracketrightbigg +F ω2⎪braceleftbigg1−e−sT s⎪bracerightbigg . (11.4.11) Using Table B-11 of ﬁnite Laplace transforms, we invert (11.4.11) so that the solution becomes x(t)=⎪parenleftbigg a−F ω2⎪parenrightbigg cosωt+u ωsinωt+F ω2. (11.4.12) Example 11.4.3(Boundary Value Problem ). The equation for the upward displacement of a taut string caused by a concentrated or distributed load W(x) normalized with respect to the tension of the string of length Lis d 2y dx2=W(x),0≤x≤L (11.4.13) and the associated boundary conditions are y(0) =y(L)=0. (11.4.14) We solve this boundary value problem due to a concentrated load of unitmagnitude at a point awhere W(x)=δ(x−a),0<a<L . © 2007 by Taylor & Francis Group, LLC Finite Laplace Transforms 441 The use of the ﬁnite Laplace transform deﬁned by ¯y(s, L)=L⎪integraldisplay 0y(x)e−sxdx (11.4.15) leads to the solution of (11.4.13)–(11.4.14) in the form ¯y(s, L)=1 s2[e−sa+y/prime(0)−e−sLy/prime(L)], (11.4.16) where y/prime(x) denotes the derivative of y(x) with respect to x. The function ¯y(s, L) is not an entire function of sunless the condition y/prime(0) =y/prime(L)−1i s satisﬁed. Using this condition, solution (11.4.16) can be put in the form ¯y(s, L)=y/prime(0) s2⎪bracketleftbig 1−esL−sLe−sL⎪bracketrightbig +e−sa s2⎪bracketleftBig 1−e−s(L−a)+y/prime(0)sLe−s(L−a)⎪bracketrightBig .(11.4.17) In order to complete the inversion of (11.4.17), we set Ly/prime(0) =a−Lso that the inversion gives the solution y(x)=xy/prime(0) + ( x−a)H(x−a). (11.4.18) Example 11.4.4 (Transient Current in a Simple Circuit ). The current I(t)i nas i m p l ec i r - cuit (see Figure 4.4) containing a resistance R, and an inductance Lwith an oscillating voltage E(t)=E0cosωtis given by LdI dt+RI=E0cosωt, 0≤t≤T, (11.4.19) I(t)=0 a t t=0. (11.4.20) Application of the ﬁnite Laplace transform to (11.4.19)–(11.4.20) gives s¯I(s, T)+e−sTI(T)+R L¯I(s, T) =E0 L⎪bracketleftbiggs s2+ω2+e−sT s2+ω2(ωsinωT−scosωT)⎪bracketrightbigg . Or, ¯I(s, T)=−e−sTI(T) ⎪parenleftbig s+R L⎪parenrightbig+E0 L⎪parenleftbig s+R L⎪parenrightbig ×⎪bracketleftbiggs s2+ω2+e−sT s2+ω2(ωsinωT−scosωT)⎪bracketrightbigg .(11.4.21) © 2007 by Taylor & Francis Group, LLC 442 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Since ¯I(s, T) is not an entire function, we make it entire by setting I(T)=E0 Lω ⎪parenleftbig ω2+R2 L2⎪parenrightbig⎪parenleftbiggR ωLcosωT+s i nωT−R ωLe−RT L⎪parenrightbigg . (11.4.22) Putting this into (11.4.21) gives ¯I(s, T)=ωE0 L⎪parenleftbig ω2+R2 L2⎪parenrightbig⎪bracketleftbiggR ωL⎪braceleftbiggs s2+ω2+e−sT s2+ω2(ωsinωT−scosωT)⎪bracerightbigg⎪bracketrightbigg +ωE0 L⎪parenleftbig ω2+R2 L2⎪parenrightbig⎪bracketleftbigg⎪braceleftbiggω s2+ω2−e−sT s2+ω2(ssinωT+ωcosωT)⎪bracerightbigg⎪bracketrightbigg −ωE0 L⎪parenleftbig ω2+R2 L2⎪parenrightbig⎡ ⎣R ωL⎪parenleftBig s+R L⎪parenrightBig⎪braceleftBig 1−e−(s+R L)T⎪bracerightBig⎤ ⎦. (11.4.23) Using the table of the ﬁnite Laplace transform, we can invert (11.4.23) to obtain the solution I(t)=ωE0 L⎪parenleftbig ω2+R2 L2⎪parenrightbig⎪braceleftbiggR ωLcosωt+s i n ωt−R ωLexp⎪parenleftbigg −Rt L⎪parenrightbigg⎪bracerightbigg .(11.4.24) Obviously, the ﬁrst two terms in the curly brackets represent the steady-state current ﬁeld, and the last term represents the transient current. In the limitt→∞, the transient term decays and t he steady state is attained. Example 11.4.5 (Moments of a Random Variable ). Find the nth order moments of a random variable Xwith the density function f(x)i n0 ≤x≤T. It follows from deﬁnition (11.2.1) that the ﬁnite Laplace transform ¯f(s, T)o f the density function f(x) can be interpreted as the mathematical expectation of exp( sX). In other words. ¯f(s, T)=ST{f(x)}=E{exp(sX)}=T⎪integraldisplay 0esxf(x)dx, (11.4.25) where sis a real parameter. Consequently, d ds¯f(s, T)=T⎪integraldisplay 0xesxf(x)dx. This result gives the deﬁnition of the expectation of Xas m1=T⎪integraldisplay 0xf(x)dx=⎪bracketleftbiggd ds¯f(s, T)⎪bracketrightbigg s=0. (11.4.26) © 2007 by Taylor & Francis Group, LLC Finite Laplace Transforms 443 This implies that the mean of Xis expressed in terms of the derivative of the ﬁnite Laplace transform of the density function f(x). Similarly, diﬀerentiating (11.4.25) ntimes with respect to s,w eo b t a i n mn=T⎪integraldisplay 0xnf(x)dx=⎪bracketleftbiggdn dsn¯f(s, T)⎪bracketrightbigg s=0. (11.4.27) In view of the result, the standard deviation and the variance of Xcan be obtained in terms of the derivatives of the ﬁnite Laplace transform of thedensity function. 11.5 Tauberian Theorems THEOREM 11.5.1 IfST{f(t)}=¯f(s, T) exists, then lim s→∞¯f(s, T)=0. (11.5.1) If, in addition, ST{f/prime(t)}exists, then lim s→∞[s¯f(s, T)] = lim t→0f(t). (11.5.2) THEOREM 11.5.2 IfST{f(t)}=¯f(s, T) exists, then lim s→0¯f(s, T)=T⎪integraldisplay 0f(t)dt. (11.5.3) If, in addition, ST{f/prime(t)}exists, then lim s→0s¯f(s, T)=0. (11.5.4) The proofs of these theorems are similar to those for the usual Laplace transforms discussed in Section 3.8. 11.6 Exercises 1. Find the ﬁnite Laplace transform of each of the following functions: © 2007 by Taylor & Francis Group, LLC 444 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (a) cosh at, (c) exp( −at2),a > 0,(b) sinh at, (d)H(t), (e)tne−at,a > 0, (f)e−atsinbt. 2. Iff(t) has a ﬁnite discontinuity at t=a,w h e r e0 <a<T , show that ST{f/prime(t)}=s¯f(s, T)+f(T)e−sT−f(0)−e−sa[f]a, where [ f]a=f(a+0 )−f(a−0). Generalize this result if f(t) has a ﬁnite number of ﬁnite discontinuities att=a1,a2,...,a nin [0,T]. 3. Verify the result (11.3.7) when f(t)=s i n at. 4. Verify the Tauberian theorems for the function f(t)=e x p ( −at),a >0. 5. Solve the initial value problem d2x dt2+ω2x=Aexp(αt2),(α>0),0≤t≤T x(0) =a,˙x(0) =u, where ω,A,α,a anduare constants. © 2007 by Taylor & Francis Group, LLC 12 ZT r a n s f o r m s “Don’t just read it; ﬁght it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis neces-sary? Is the converse true? What h appens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?” Paul R. Halmos “The shortest path between two truths in the real domain passes through the complex domain.” Jacques Hadamard 12.1 Introduction We begin this chapter with a brief introduction to the input-output charac-teristics of a linear dynamic system. Some special features of linear dynamicsystems are brieﬂy discussed. Analogous to the Fourier and Laplace trans- forms applied to the continuous linear systems, the Ztransform applicable to linear time-invariant discrete-time systems is studied in this chapter. The basic operational properties including the convolution theorem, initial and ﬁ- nal value theorems, the Ztransform of partial derivatives, and the inverse Z transform are presented in some detail. Applications of the Ztransform to diﬀerence equations and to the summation of inﬁnite series are discussed with examples. 12.2 Dynamic Linear Systems and Impulse Response In physical and engineering problems, a system is referred to as a physical device that can transform a forcing orinput function (input signal or simply 445 © 2007 by Taylor & Francis Group, LLC 446 INTEGRAL TRANSFORMS and THEIR APPLICATIONS signal)f(t)i n t oa n output function (output signal orresponse )g(t)w h e r e tis an independent time variable. In other words, the output is simply the response of the input due to the action of the physical device. Both input and output signals are functions of the continuous time variable. These may include steps or impulses. However, the input or the output or both may be sequences in the sense that they can assume values deﬁned only for discrete values of time t. One of the essential features of a system is that the output g(t) is completely determined by the given input function f(t), and the char- acteristics of the system, and in some instances by the initial data. Usually,the action of the system is math ematically represented by g(t)=Lf(t), (12.2.1) where Lis a transformation (or operator) that transforms the input signal f(t) to the output signal g(t). The system is called linear if its operator Lis linear, that is, Lsatisﬁes the principle of superposition . Obvious examples of linear operators are integral transformations. Another fundamental characteristic of linear systems is that the response to an arbitrary input can be found by analyzing the input components of standard type and adding the responses to the individual components. The very nature of the delta function, δ(t), suggests that it can be used to represent theunit impulse function . In Section 2.4 it was shown that δ(t) satisﬁes the following fundamental property f(t)δ(t−t n)=f(tn)δ(t−tn), (12.2.2) where tn(nis an integer) is any particular value of tandf(t) is a continuous function in any interval containing the point t=tn. Result (12.2.2) is very important in the theory of sampling systems. Sampling of signals is very com-mon in communication and digital systems. It is also used in pulse modulation systems and in all kinds of feedback sy stems where a digit al computer is one of the common elements. Summing (12.2.2) over all integral ngives the sampled function f ∗(t)a s f∗(t)=f(t)∞⎪summationdisplay n=−∞δ(t−tn)=∞⎪summationdisplay n=−∞f(tn)δ(t−tn). (12.2.3) Thus, the sampled function is approximately represented by a train of impulsefunctions, each having an area equal to the function at the sampling instant. Witht n=nT,t h es e r i e s∞⎪summationtext n=−∞δ(t−nT) is called the impulse train as shown in Figure 12.1. As the sampling period Tassumes a small value dτ, the function f(t)i n - © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 447 t 0 T 2T 3T 4T 5T-T -2T-3T-4T-5T Figure 12.1 The impulse train∞⎪summationdisplay n=−∞δ(t−nT). volved in (12.2.3) can be written in the form f(t)=∞⎪summationtext n=−∞f(nd τ)δ(t−nd τ) ∞⎪summationtext n=−∞δ(t−nd τ). (12.2.4) Multiplying the numerator and the denominator of (12.2.4) by ndτand re- placing nd τbydτ,w eo b t a i n f(t)=∞⎪integraltext −∞f(τ)δ(t−τ)dτ ∞⎪integraltext −∞δ(t−τ)dτ=∞⎪integraldisplay −∞f(t)δ(t−τ)dτ. (12.2.5) Denoting the impulse response of the system to the input δ(t)b yh(t), the output is mathematically represented by the Fourier convolution as g(t)=∞⎪integraldisplay −∞f(τ)h(t−τ)dτ=f(t)∗h(t). (12.2.6) The convolution of f(t) and the impulse train is g(t)s ot h a t g(t)=f(t)∗∞⎪summationdisplay n=−∞δ(t−nT) =∞⎪summationdisplay n=−∞∞⎪integraldisplay −∞f(τ)δ(t−nT−τ)dτ =∞⎪summationdisplay n=−∞f(t−nT). (12.2.7) This represents the superposition of all translations of f(t)b ynT. © 2007 by Taylor & Francis Group, LLC 448 INTEGRAL TRANSFORMS and THEIR APPLICATIONS If the input function f(τ) is the impulse δ(τ−τn) located at τn,t h e n h(t− τn) is the response (output) of the system to the above impulse. This follows from (12.2.6) as g(t)=∞⎪integraldisplay −∞h(t−τ)δ(τ−τn)dτ=h(t−τn). If the impulse is located at τ0=0 ,t h e n g(t)=h(t). This explains why the term impulse response of h(t) was coined in the system’s analysis, for the system’s response to this particular case of an impulse located at τ0=0 . T h e m o s t important result in this section is (12.2.6), which gives the output g(t)a st h e Fourier convolution product of the input signal f(t) and the impulse response of the system h(t). This shows an application of Fourier integral analysis to the analysis of linear dynamic systems. Usually, the input is applied only for t≥0, and h(t)=0 f o r t<0. Hence, the output represented by (12.2.6) re duces to the Laplace convolution as g(t)=t⎪integraldisplay 0f(τ)h(t−τ)dτ=f(t)∗h(t). (12.2.8) Physically, this represents the response for any input when the impulse re- sponse of any linear time invariant system is known. We consider a certain waveform f(t) shown in Figure 12.2 which is sampled periodically by a switch. tf(t) 0T 2 T3 T4 T5 Ttf*(t) 0T 2 T3 T4 T5 T Figure 12.2 Input and sampled functions. We have seen earlier that the delta function takes on the value of the func- tion at the instant at which it is applied, the sampled function f∗(t)c a nb e expressed as f∗(t)=f(t)∞⎪summationdisplay n=0δ(t−nT)=∞⎪summationdisplay n=0f(nT)δ(t−nT). (12.2.9) © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 449 Result (12.2.9) can be considered as the amplitude modulation of unit im- pulses by the waveform f(t). Evidently, this result is very useful for analyzing the systems where signals are sampled at a time interval T.T h u s ,t h ea b o v e discussion enables us to introduce the Ztransform in the next section. 12.3 Deﬁnition of the ZTransform and Examples We take the Laplace transform of the sampled function given by (12.2.9) so that L{f∗(t)}=¯f∗(s)=∞⎪summationdisplay n=0f(nT)exp (−nsT). (12.3.1) It is convenient to make a change of variable z=e x p ( sT) so that (12.3.1) becomes L{f∗(t)}=F(z)=∞⎪summationdisplay n=0f(nT)z−n. (12.3.2) Thus, F(z) is called the Ztransform off(nT). Since the interval Tbetween the samples has no eﬀect on the properties and the use of the Ztransform, it is convenient to set T= 1. We now deﬁne the Ztransform of a sequence {f(n)}as the function F(z) of a complex variable zdeﬁned by Z{f(n)}=F(z)=∞⎪summationdisplay n=0f(n)z−n. (12.3.3) Thus, Zis a linear transformation and can be considered as an operator map- ping sequences of scalars into functions of the complex variable z.I ti sa s - sumed in this chapter that there exists an Rsuch that (12.3.3) converges for|z|>R.S i n c e |z|=|exp(sT)|=|exp(σ+iμ)T|=|exp(σT)|, it follows that, when σ<0 (that is, in the left half of the complex splane), |z|<1, and thus, t h el e f th a l fo ft h e splane corresponds to the interior of the unit circle in the complex zplane. Similarly, the right half of the splane corresponds to the exterior ( |z|>1) of the unit circle in the zplane. And σ=0 i n t h e splane corresponds to the unit circle in the zplane. The inverse Ztransform is given by the complex integral Z−1{F(z)}=f(n)=1 2πi⎪contintegraldisplay CF(z)zn−1dz, (12.3.4) where Cis a simple closed contour enclosing the origin and lying outside the circle |z|=R. The existence of the inverse imposes restrictions on f(n)f o r uniqueness. We require that f(n)=0f o r n<0. © 2007 by Taylor & Francis Group, LLC 450 INTEGRAL TRANSFORMS and THEIR APPLICATIONS To obtain the inversion integral, we consider F(z)=∞⎪summationdisplay n=0f(n)z−n =f(0) +f(1)z−1+f(2)z−2+···+f(n)z−n+f(n+1 )z−(n+1)+···. Multiplying both sides by (2 πi)−1zn−1and integrating along the closed con- tourC, which usually encloses all singularities of F(z), we obtain 1 2πi⎪contintegraldisplay CF(z)zn−1dz=1 2πi⎪bracketleftbigg⎪contintegraldisplay Cf(0)zn−1dz+⎪contintegraldisplay Cf(1)zn−2dz +···+⎪contintegraldisplay Cf(n)z−1dz+⎪contintegraldisplay Cf(n+1 )z−2dz+···⎪bracketrightbigg . By Cauchy’s Fundamental Theorem all integrals on the right vanish except 1 2πi⎪contintegraldisplay Cf(n)dz z=f(n). This leads to the inversion integral for the Ztransform in the form Z−1{F(z)}=f(n)=1 2πi⎪contintegraldisplay CF(z)zn−1dz. Similarly, we can deﬁne the so called bilateral Ztransform by Z{f(n)}=F(z)=∞⎪summationdisplay n=−∞f(n)z−n, (12.3.5) for all complex numbers zfor which the series converges. This reduces to the unilateral Ztransform (12.3.3) if f(n)=0f or n<0. The inverse Ztransform is given by a complex integral similar to (12.3.4). Substituting z=reiθin (12.3.5), we obtain the Ztransform evaluated at r=1 F{f(n)}=F(θ)=∞⎪summationdisplay n=−∞f(n)e−inθ. This is known as the Fourier transform of the sequence {f(n)}∞ −∞. Example 12.3.1 Iff(n)=an,n≥0, then Z{an}=∞⎪summationdisplay n=0⎪parenleftBiga z⎪parenrightBign =1 1−a z=z z−a,|z|>a . (12.3.6) © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 451 When a=1 ,w eo b t a i n Z{1}=∞⎪summationdisplay n=0z−n=z z−1,|z|>1. (12.3.7) Iff(n)=nanforn≥0, then Z{nan}=∞⎪summationdisplay n=0nanz−n=az (z−a)2,|z|>|a|. (12.3.8) Example 12.3.2 Iff(n)=e x p ( inx), then Z{exp(inx)}=z z−exp(ix). (12.3.9) This follows immediately from (12.3.6). Furthermore, Z{cosnx}=z(z−cosx) z2−2zcosx+1,Z{sinnx}=zsinx z2−2zcosx+1.(12.3.10) These follow readily from (12.3.9) by writing exp( inx)=c o s nx+isinnx. Example 12.3.3 Iff(n)=n,t h e n Z{n}=∞⎪summationdisplay n=0nz−n=z∞⎪summationdisplay n=0nz−(n+1) =−zd dz⎪parenleftBigg∞⎪summationdisplay n=0z−n⎪parenrightBigg =z (z−1)2,|z|>1. (12.3.11) Example 12.3.4 Iff(n)=1 n!,t h e n Z⎪braceleftbigg1 n!⎪bracerightbigg =∞⎪summationdisplay n=01 n!z−n=e x p⎪parenleftbigg1 z⎪parenrightbigg for all z. (12.3.12) © 2007 by Taylor & Francis Group, LLC 452 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 12.3.5 Iff(n)=c o s h nx,t h e n Z{coshnx}=z(z−coshx) z2−2zcoshx+1. (12.3.13) We have Z{coshnx}=1 2Z{enx+e−nx} =1 2⎪bracketleftbiggz z−ex+z z−e−x⎪bracketrightbigg =z(z−coshx) z2−2zcoshx+1. Example 12.3.6 Show that Z{n2}=z(z+1 ) (z−1)3. (12.3.14) We have, from (12.4.13) in section 12.4, Z{n·n}=−zd dzZ{n}=−zd dzz (z−1)2=z(z+1 ) (z−1)3. Example 12.3.7 Iff(n) is a periodic sequence of integral period N,t h e n F(z)=Z{f(n)}=zN zN−1F1(z), where F1(z)=N−1⎪summationdisplay k=0f(k)z−k. (12.3.15) We have, by deﬁnition, F(z)=Z{f(n)}=∞⎪summationdisplay n=0f(n)z−n=zN∞⎪summationdisplay n=0f(n+N)z−(n+N) =zN∞⎪summationdisplay k=Nf(k)z−k,(n+N=k) =zN⎪bracketleftBigg∞⎪summationdisplay k=0f(k)z−k−N−1⎪summationdisplay k=0f(k)z−k⎪bracketrightBigg =⎪braceleftbig zNF(z)−zNF1(z)⎪bracerightbig . © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 453 Thus, F(z)=zN (zN−1)F1(z). 12.4 Basic Operational Properties of ZTransforms THEOREM 12.4.1 (Translation ). IfZ{f(n)}=F(z)a n d m≥0, then Z{f(n−m)}=z−m⎪bracketleftBigg F(z)+−1⎪summationdisplay r=−mf(r)z−r⎪bracketrightBigg , (12.4.1) Z{f(n+m)}=zm⎪bracketleftBigg F(z)−m−1⎪summationdisplay r=0f(r)z−r⎪bracketrightBigg . (12.4.2) In particular, if m=1,2,3,...,then Z{f(n−1)}=z−1F(z)−f(−1)z. (12.4.3) Z{f(n−2)}=z−2⎪bracketleftBigg F(z)+−1⎪summationdisplay r=−2f(r)z−r⎪bracketrightBigg . (12.4.4) and so on. Similarly, it follows from (12.4.2) that Z{f(n+1 )}=z{F(z)−f(0)}, (12.4.5) Z{f(n+2 )}=z2{F(z)−f(0)}−zf(1), (12.4.6) Z{f(n+3 )}=z3{F(z)−f(0)}−z2f(1)−zf(2). (12.4.7) More generally, for m>0, Z{f(n+m)}=zm{F(z)−f(0)}−zm−1f(1)−···− zf(m−1).(12.4.8) All these results are widely used for the solution of initial value problems involving diﬀerence equations. Result (12.4.8) is somewhat similar to (3.4.12) for this Laplace transform, and has been used to solve initial value problems involving diﬀerential equations. © 2007 by Taylor & Francis Group, LLC 454 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF We have, by deﬁnition, Z{f(n−m)}=∞⎪summationdisplay n=0f(n−m)z−n,(n−m=r), =z−m∞⎪summationdisplay r=−mf(r)z−r=z−m∞⎪summationdisplay r=0f(r)z−r+z−m−1⎪summationdisplay r=−mf(r)z−r. When m= 1, we get (12.4.3). Iff(r) = 0 for all r<0, then Z{f(n−m)}=z−m∞⎪summationdisplay r=0f(r)z−r. (12.4.9) When m= 1, this result gives Z{f(n−1)}=z−1F(z). (12.4.10) Similarly, we prove (12.4.2) by writing Z{f(n+m)}=∞⎪summationdisplay n=0f(n+m)z−n,(n+m=r), =zm∞⎪summationdisplay r=mf(r)z−r=zm∞⎪summationdisplay r=0f(r)z−r−zmm−1⎪summationdisplay r=0f(r)z−r =zm⎪bracketleftBigg F(z)−m−1⎪summationdisplay r=0f(r)z−r⎪bracketrightBigg . When m=1,2,3,...,results (12.4.5)–(12.4.7) follow immediately. THEOREM 12.4.2 (Multiplication ). IfZ{f(n)}=F(z), then Z{anf(n)}=F⎪parenleftBigz a⎪parenrightBig ,|z|>|a|. (12.4.11) Z{e−nbf(n)}=F(zeb),|z|>|e−b|. (12.4.12) Z{nf(n)}=−zd dzF(z). (12.4.13) More generally, Z⎪bracketleftbig nkf(n)⎪bracketrightbig =(−1)k⎪parenleftbigg zd dz⎪parenrightbiggk F(z),k=0,1,2,..., (12.4.14) © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 455 where ⎪parenleftbigg zd dz⎪parenrightbiggk F(z)=⎪parenleftbigg zd dz⎪parenrightbigg(k−1)⎪parenleftbigg zd dz⎪parenrightbigg F. PROOF Result (12.4.11) follows immediately from the deﬁnition (12.3.3), and (12.4.12) follows from (12.4.11) by writing a=e−b. Iff(n)=1s ot h a t Z{f(n)}=z z−1,a n di f a=eb, then (12.4.1) gives Z{(eb)n}=ze−b ze−b−1=z z−eb,|z|>|eb|. (12.4.15) Putting b=ixalso gives (12.3.9) To prove (12.4.13), we use the deﬁnition (12.3.3) to obtain Z{nf(n)}=∞⎪summationdisplay n=0nf(n)z−n=z∞⎪summationdisplay n=0nf(n)z−(n+1) =z∞⎪summationdisplay n=0f(n)⎪braceleftbigg −d dzz−n⎪bracerightbigg =−zd dz⎪braceleftBigg∞⎪summationdisplay n=0f(n)z−n⎪bracerightBigg =−zd dzF(z). THEOREM 12.4.3 (Division ). Z⎪braceleftbiggf(n) n+m⎪bracerightbigg =−zm⎪integraldisplayz 0F(ξ)dξ ξm+1. (12.4.16) PROOF We have Z⎪braceleftbiggf(n) n+m⎪bracerightbigg =∞⎪summationdisplay n=0f(n) n+mz−n,(m≥0), =−zm∞⎪summationdisplay n=0f(n)⎪bracketleftbigg −⎪integraldisplayz 0ξ−(n+m+1)dξ⎪bracketrightbigg =−zm⎪integraldisplayz 0ξ−(m+1)⎪bracketleftBigg∞⎪summationdisplay n=0f(n)ξ−n⎪bracketrightBigg dξ =−zm⎪integraldisplayz 0ξ−(m+1)F(ξ)dξ. When m=0,1,2,...,several particular results follow from (12.4.16). © 2007 by Taylor & Francis Group, LLC 456 INTEGRAL TRANSFORMS and THEIR APPLICATIONS THEOREM 12.4.4 (Convolution ). IfZ{f(n)}=F(z)a n d Z{g(n)}=G(z), then the Ztransform of the convolution f(n)∗g(n)i sg i v e nb y Z{f(n)∗g(n)}=Z{f(n)}Z{g(n)}, (12.4.17) where the convolution is deﬁned by f(n)∗g(n)=∞⎪summationdisplay m=0f(n−m)g(m). (12.4.18) Or, equivalently, Z−1{F(z)G(z)}=∞⎪summationdisplay m=0f(n−m)g(m). (12.4.19) PROOF We proceed formally to obtain Z{f(n)∗g(n)}=∞⎪summationdisplay n=0z−n∞⎪summationdisplay m=0f(n−m)g(m), which is, interchanging the order of summation, =∞⎪summationdisplay m=0g(m)∞⎪summationdisplay n=0f(n−m)z−n. Substituting n−m=r,w eo b t a i n Z{f(n)∗g(n)}=∞⎪summationdisplay m=0g(m)z−m∞⎪summationdisplay r=−mf(r)z−r, which is, in view of f(r)=0f o r r<0, =∞⎪summationdisplay m=0g(m)z−m∞⎪summationdisplay r=0f(r)z−r =Z{f(n)}Z{g(n)}. This proves the theorem. More generally, the convolution f(n)∗g(n) is deﬁned by f(n)∗g(n)=∞⎪summationdisplay m=−∞f(n−m)g(m). (12.4.20) If we assume f(n)=0= g(n)f o rn<0, then (12.4.20) becomes (12.4.18). © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 457 However, the Ztransform of (12.4.20) gives Z{f(n)∗g(n)}=∞⎪summationdisplay n=−∞z−n∞⎪summationdisplay m=−∞f(n−m)g(m), which is, interchanging the order of summation, =∞⎪summationdisplay m=−∞g(m)∞⎪summationdisplay n=−∞f(n−m)z−n =∞⎪summationdisplay m=−∞z−mg(m)∞⎪summationdisplay n=−∞f(n−m)z−(n−m) =∞⎪summationdisplay m=−∞z−mg(m)∞⎪summationdisplay r=−∞f(r)z−r,(r=n−m) =Z{f(n)}Z{g(n)}. (12.4.21) This is the convolution theorem for the bilateral Ztransform. TheZtransform of the product f(n)g(n)i sg i v e nb y Z{f(n)g(n)}=1 2πi⎪contintegraldisplay CF(w)G⎪parenleftBigz w⎪parenrightBigdw w, (12.4.22) where Cis a closed contour enclosing the origin in the domain of convergence ofF(w)a n d G⎪parenleftbigz w⎪parenrightbig . THEOREM 12.4.5 (Parseval’s Formula ). IfF(z)=Z{f(n)}andG(z)=Z{g(n)},t h e n ∞⎪summationdisplay n=−∞f(n) g(n)=1 2π⎪integraldisplayπ −πF(eiθ) G(eiθ)dθ. (12.4.23) In particular, ∞⎪summationdisplay n=−∞|f(n)|2=1 2π⎪integraldisplayπ −π|F(eiθ)|2dθ. (12.4.24) THEOREM 12.4.6 (Initial Value Theorem ). IfZ{f(n)}=F(z), then f(0) = lim z→∞F(z). (12.4.25) Also, if f(0) = 0, then f(1) = lim z→∞zF(z). (12.4.26) © 2007 by Taylor & Francis Group, LLC 458 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF We have, by deﬁnition, F(z)=∞⎪summationdisplay n=0f(n)z−n=f(0) +f(1) z+f(2) z2+···. (12.4.27) The initial value of f(n)a tn= 0 is obtained from (12.4.27) by letting z→∞, and hence f(0) = lim z→∞F(z). Iff(0) = 0, then (12.4.27) gives f(1) = lim z→∞zF(z). This proves the theorem. THEOREM 12.4.7 (Final Value Theorem ). IfZ{f(n)}=F(z), then lim n→∞f(n) = lim z→1{(z−1)F(z)}. (12.4.28) provided the limits exist. PROOF We have, from (12.3.3) and (12.4.5), Z{f(n+1 )−f(n)}=z{F(z)−f(0)}−F(z). Or, equivalently, ∞⎪summationdisplay n=0[f(n+1 )−f(n)]z−n=(z−1)F(z)−zf(0). In the limit as z→1, we obtain lim z→1∞⎪summationdisplay n=0[f(n+1 )−f(n)]z−n= lim z→1(z−1)F(z)−f(0). Or, lim n→∞[f(n+1 )−f(0)] = f(∞)−f(0) = lim z→1(z−1)F(z)−f(0) Thus, lim n→∞f(n) = lim z→1(z−1)F(z), provided the limits exist. This proves the theorem. The reader is referred to Zadeh and Desoer (1963) for a rigorous proof. © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 459 Example 12.4.1 Verify the initial value theorem for the function F(z)=z (z−a)(z−b). We have f(0) = lim z→∞z (z−a)(z−b)=0,f (1) = lim z→∞zF(z)=1. THEOREM 12.4.8 (TheZTransform of Partial Derivatives ). Z⎪braceleftbigg∂ ∂af(n,a)⎪bracerightbigg =∂ ∂a[Z{f(n,a)}]. (12.4.29) PROOF Z⎪braceleftbigg∂ ∂af(n,a)⎪bracerightbigg =∞⎪summationdisplay n=0⎪bracketleftbigg∂ ∂af(n,a)⎪bracketrightbigg z−n =∂ ∂a⎪bracketleftBigg∞⎪summationdisplay n=0f(n,a)z−n⎪bracketrightBigg =∂ ∂a[Z{f(n,a)}]. As an example of this result, we show Z{nean}=Z⎪braceleftbigg∂ ∂aena⎪bracerightbigg =∂ ∂aZ{ena}=∂ ∂a⎪parenleftbiggz z−ea⎪parenrightbigg =zea (z−ea)2. 12.5 The Inverse ZTransform and Examples The inverse Ztransform is given by the complex integral (12.3.4), which can be evaluated by using the Cauchy residue theorem. However, we discuss other simple ways of ﬁnding the inverse transform of a given F(z). These include a method from the deﬁnition (12.3.3), which leads to the expansion of F(z)a s a series of inverse powers of zin the form F(z)=f(0) +f(1)z−1+f(2)z−2+···+f(n)z−n+···. (12.5.1) © 2007 by Taylor & Francis Group, LLC 460 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The coeﬃcient of z−nin this expansion is f(n)=Z−1{F(z)}. (12.5.2) IfF(z) is given by a series F(z)=∞⎪summationdisplay n=−∞anz−n,r 1<z<r 2, then its inverse Ztransform is unique and is equal to {f(n)=an}for all n. If the domain of analyticity of F(z) contains the unit circle |z|=1 , a n d i f F is single valued therein, then F(eiθ) is a periodic function with period 2 πand hence, it can be expanded in a Fourier series. The coeﬃcients of this series represent the inverse Ztransform of F(z)a n da r eg i v e nb y Z−1{F(z)}=f(n)=1 2π⎪integraldisplayπ −πF(eiθ)einθdθ. Example 12.5.1 Find the inverse Ztransform of F(z)=z(z−a)−1. We have F(z)=z z−a=⎪parenleftBig 1−a z⎪parenrightBig−1 =1+ az−1+a2z−2+···+anz−n+···. so that f(0) = 1 ,f(1) =a, f(2) =a2, ...,f (n)=an, .... Obviously, f(n)=Z−1⎪braceleftbiggz (z−a)⎪bracerightbigg =an. Example 12.5.2 FindZ−1⎪braceleftbigg exp⎪parenleftbigg1 z⎪parenrightbigg⎪bracerightbigg . Obviously, exp⎪parenleftbigg1 z⎪parenrightbigg =1+1 ·z−1+1 2!z−2+···+1 n!z−n+···. This gives f(n)=1 n!=Z−1⎪braceleftbigg exp⎪parenleftbigg1 z⎪parenrightbigg⎪bracerightbigg . © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 461 Other methods for inversion use partial fractions and the Convolution The- orem 12.4.4. We illustrate these methods by the following examples. Example 12.5.3 Find the inverse Ztransform of F(z)=z z2−6z+8. We write F(z)=z (z−2)(z−4)=1 2⎪parenleftbiggz z−4−z z−2⎪parenrightbigg . It follows from the table of Ztransforms that f(n)=Z−1{F(z)}=1 2⎪bracketleftbigg Z−1⎪braceleftbiggz z−4⎪bracerightbigg −Z−1⎪braceleftbiggz z−2⎪bracerightbigg⎪bracketrightbigg =1 2(4n−2n). Example 12.5.4 Use the Convolution Theorem 12.4.4 to ﬁnd the inverse ofz2 (z−a)(z−b). We set F(z)=z z−a,G(z)=z z−b so that f(n)=Z−1{F(z)}=an,g(n)=Z−1{G(z)}=bn. Thus, the convolution theorem gives Z−1{F(z)G(z)}=n⎪summationdisplay m=0an−mbm=ann⎪summationdisplay m=0⎪parenleftbiggb a⎪parenrightbiggm =an·⎪braceleftBigg 1−⎪parenleftbigb a⎪parenrightbign+1 1−b a⎪bracerightBigg =an+1 (a−b)⎪braceleftBigg 1−⎪parenleftbiggb a⎪parenrightbiggn+1⎪bracerightBigg . Example 12.5.5 Find the inverse Ztransform of F(z)=3z2−z (z−1)(z−2)2. We write F(z) as partial fractions F(z)=3z2−z (z−1)(z−2)2=2·z (z−1)−2·z (z−2)+5 2·2z (z−2)2 © 2007 by Taylor & Francis Group, LLC 462 INTEGRAL TRANSFORMS and THEIR APPLICATIONS so that its inverse is f(n)=Z−1⎪braceleftbigg2z (z−1)⎪bracerightbigg −Z−1⎪braceleftbigg 2·z (z−2)⎪bracerightbigg +5 2Z−1⎪braceleftbigg2z (z−2)2⎪bracerightbigg which is, by (12.3.6), and (12.4.13) with f(n)=2n, =2−2n+1+5 2·n2n=2−2n+1+5·n2n−1. Example 12.5.6 Use the Convolution Theorem to show that Z−1⎪braceleftbiggz(z+1 ) (z−1)3⎪bracerightbigg =n2. We write z(z+1 ) (z−1)3=z (z−1)2⎪parenleftbiggz+1 z−1⎪parenrightbigg =z (z−1)2⎪bracketleftbiggz z−1+1 z−1⎪bracketrightbigg . Letting F(z)=z (z−1)2and G(z)=z z−1+1 z−1, we obtain f(n)=nand g(n)=H(n)+H(n−1). Thus, Z−1⎪braceleftbiggz(z+1 ) (z−1)3⎪bracerightbigg =f(n)∗g(n)=n⎪summationdisplay m=0m[H(n−m)+H(n−m−1)] =n2. Example 12.5.7 (Reconstruction of a Sequence from its ZTransform ). Suppose F(z)=z z−1,|z|>1a n d G(z)=z z−1,|z|<1,(12.5.3) Z−1{F(z)}=f(n)=⎧ ⎨ ⎩1,n≥0, 0,n < 0,⎫ ⎬ ⎭andZ−1{G(z)}=g(n)=⎧ ⎨ ⎩1,n≤0, 0,n≥0,⎫ ⎬ ⎭. (12.5.4) This shows that the inverse Ztransform of z(z−1)−1is not unique. In general, the inverse Ztransform is not unique, unless its region of convergence is speciﬁed. © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 463 12.6 Applications of ZTransforms to Finite Diﬀerence Equations Example 12.6.1 (First Order Diﬀerence Equation ). Solve the initial value problem for the diﬀerence equation f(n+1 )−f(n)=1,f (0) = 0 . (12.6.1) Application of the Ztransform to (12.6.1) combined with (12.4.5) gives z[F(z)−f(0)]−F(z)=z z−1. Or,F (z)=z (z−1)2. The inverse Ztransform (see result (12.3.11)) gives the solution f(n)=Z−1⎪braceleftbiggz (z−1)2⎪bracerightbigg =n. (12.6.2) Example 12.6.2 (First Order Diﬀerence Equation ). Solve the equation f(n+1 )+2 f(n)=n, f (0) = 1 . (12.6.3) The use of the Ztransform to this problem gives z{F(z)−f(0)}+2F(z)=z (z−1)2. Or,F(z)=z z+2+z (z+2 ) (z−1)2 =z z+2+1 9·z z+2+3 9·z (z−1)2−1 9·z z−1 =⎪parenleftbigg10 9⎪parenrightbiggz (z+2 )+3 9·z (z−1)2−1 9z (z−1). The inverse Ztransform yields the solution f(n)=1 9[10(−2)n+3n−1]. (12.6.4) © 2007 by Taylor & Francis Group, LLC 464 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 12.6.3 (The Fibonacci Sequence ). The Fibonacci sequence is deﬁned as a sequence in which every term is the sum of the two proceeding terms. So it satisﬁes the diﬀerence equation un+1=un+un−1,u 1=u(0) = 1 . (12.6.5) Application of the Ztransform gives U(z)=z2 z2−z−1,where U(z)=Z{un}. Thus, the inverse transform leads to the solution un=Z−1⎪braceleftbiggz2 z2−z−1⎪bracerightbigg =Z−1⎪braceleftbiggz2 (z−a)(z−b)⎪bracerightbigg , where a=1 2(1 +√ 5) and b=1 2(1−√ 5). Using Example 12.5.4, the Fibonacci sequence is un=an+1−bn+1 (a−b),n=0,1,2,... (12.6.6) More explicitly, the Fibonacci sequence is given by 1 ,1,2,3,5,.... Example 12.6.4 (Second Order Diﬀerence Equation ). Solve the initial value problem f(n+2 )−3f(n+1 )+2 f(n)=0,f(0) = 1 ,f(1) = 2 . (12.6.7) Application of the Ztransform gives z2{F(z)−f(0)}−zf(1)−3[z{F(z)−f(0)}]+2F(z)=0. Or, (z2−3z+2 )F(z)=(z2−z). Hence, F(z)=z (z−2). Thus, the inversion gives the solution f(n)=Z−1⎪braceleftbiggz (z−2)⎪bracerightbigg =2n. (12.6.8) © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 465 Example 12.6.5 (Periodic Solution ). Find the solution of the initial value problem u(n+2 )−u(n+1 )+ u(n)=0, (12.6.9) u(0) = 1 and u(1) = 2 . (12.6.10) TheZtransform of (12.6.9)–(12.6.10) gives {z2U(z)−z2−2z}−{zU(z)−z}+U(z)=0. Or, U(z)=z2+z (z2−z+1 )=⎪parenleftbig z2−1 2z⎪parenrightbig z2−z+1+√ 3⎪parenleftBig√ 3 2z⎪parenrightBig z2−z+1. (12.6.11) Writing x=π 3in (12.3.10), the inverse Ztransform of (12.6.11) gives the periodic solution u(n)=c o s⎪parenleftBignπ 3⎪parenrightBig +√ 3s i n⎪parenleftBignπ 3⎪parenrightBig . (12.6.12) Example 12.6.6 (Second Order Nonhomogeneous Diﬀerence Equation ). Solve the initial value problem u(n+2 )−5u(n+1 )+6 u(n)=2n,u(0) = 1 ,u(1) = 0 . (12.6.13) TheZtransform of (12.6.13) yields (z2−5z+6 )U(z)=z2−5z+z z−2. (12.6.14) Or, U(z)=z⎪bracketleftbiggz−5 (z−2)(z−3)+1 (z−2)2(z−3)⎪bracketrightbigg =z⎪bracketleftbigg⎪parenleftbigg3 z−2−2 z−3⎪parenrightbigg +⎪parenleftbigg1 z−3−1 z−2−1 (z−2)2⎪parenrightbigg⎪bracketrightbigg =z⎪bracketleftbigg2 z−2−1 z−3−1 (z−2)2⎪bracketrightbigg . (12.6.15) The inverse Ztransform of (12.6.15) gives the solution u(n)=2n+1−3n−n2n−1. (12.6.16) © 2007 by Taylor & Francis Group, LLC 466 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 12.6.7 (Chebyshev Polynomials ). Solve the second orde r diﬀerence equation un+2−2xun+1+un=0,|x|≤1, (12.6.17) u(0) =u0and u(1) =u1, (12.6.18) where u0andu1are constants. TheZtransform of equation (12.6.17) with (12.6.18) gives z2U(z)−z2u0−zu1−2x[zU(z)−zu0]+U(z)=0. Or, U(z)=u0⎪bracketleftbiggz2−zx z2−2xz+1⎪bracketrightbigg +(u1−xu0)⎪bracketleftbiggz z2−2xz+1⎪bracketrightbigg (12.6.19) =u0⎪bracketleftbiggz2−zx z2−2xz+1⎪bracketrightbigg +(u1−xu0) √ 1−x2⎪bracketleftBigg z√ 1−x2 z2−2xz+1⎪bracketrightBigg =u0⎪bracketleftbiggz2−zx z2−2xz+1⎪bracketrightbigg +v0⎪bracketleftBigg z√ 1−x2 z2−2xz+1⎪bracketrightBigg , (12.6.20) where v0=(u1−xu0)(1−x2)−1 2is independent of z. Since |x|≤1, we may write x=c o sta n dt h e nt a k et h ei n v e r s e Ztransform with the aid of (12.3.10) to obtain the solution un=u0cosnt+v0sinnt (12.6.21) =u0cos(ncos−1x)+v0sin(ncos−1x). (12.6.22) Usually, the function Tn(x)=c o s ( ncos−1x) (12.6.23) is called the Chebyshev polynomial of the ﬁrst kind of degree n. The properties of this polynomial are presented in Appendix A-4. This polynomial plays an important role in the theory of special functions, and isfound to be extremely useful in approximation theory and modern numericalanalysis. 12.7 Summation of Inﬁnite Series THEOREM 12.7.1 IfZ{f(n)}=F(z), then (i)n⎪summationdisplay k=1f(k)=Z−1⎪braceleftbiggz z−1F(z)⎪bracerightbigg , (12.7.1) © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 467 and (ii)∞⎪summationdisplay k=1f(k) = lim z→1F(z)=F(1). (12.7.2) PROOF We write g(n)=n⎪summationdisplay k=0f(k)s o t h a t g(n)=f(n)+g(n−1). Application of the Ztransform gives G(z)=F(z)+z−1G(z) so that G(z)=z (z−1)F(z). Or, Z{g(n)}=Z⎪braceleftBiggn⎪summationdisplay k=0f(k)⎪bracerightBigg =z (z−1)F(z). In the limit as z→1 together with the Final Value Theorem 12.4.7 gives lim n→∞n⎪summationdisplay k=0f(k) = lim z→1(z−1)·z z−1F(z)=F(1). This proves the theorem. Example 12.7.1 Use the Ztransform to show that ∞⎪summationdisplay n=0xn n!=ex. (12.7.3) We have, from (12.4.11), Z{xnf(n)}=F⎪parenleftBigz x⎪parenrightBig . Setting f(n)=1 n!so that F(z)=e x p⎪parenleftbigg1 z⎪parenrightbigg , we ﬁnd Z⎪braceleftbiggxn n!⎪bracerightbigg =e x p⎪parenleftBigx z⎪parenrightBig . © 2007 by Taylor & Francis Group, LLC 468 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The use of Theorem 12.7.1(ii) gives ∞⎪summationdisplay n=0xn n!= lim z→1exp⎪parenleftBigx z⎪parenrightBig =ex. Example 12.7.2 Show that ∞⎪summationdisplay n=0(−1)nxn+1 n+1= log(1 + x). (12.7.4) Using (12.3.6), we ﬁnd Z{xn+1}=zx z−x whence, in view of 4(b) in 12.8 Exercises, Z⎪braceleftbiggxn+1 n+1⎪bracerightbigg =z⎪integraldisplay∞ zzx (z−x)·dz z2 =xz⎪integraldisplay∞ zdz z(z−x) =xz⎪bracketleftbigg1 xlog⎪parenleftbiggz−x z⎪parenrightbigg⎪bracketrightbigg∞ z =−zlog⎪parenleftbiggz−x z⎪parenrightbigg . Replacing xby (−x) in this result, we obtain Z⎪braceleftbigg (−1)nxn+1 n+1⎪bracerightbigg =zlog⎪parenleftbiggz+x z⎪parenrightbigg . Application of Theorem 12.7.1(ii) gives ∞⎪summationdisplay n=0(−1)n·xn+1 n+1= lim z→1zlog⎪parenleftbiggz+x z⎪parenrightbigg =l o g( 1+ x). Example 12.7.3 Find the sum of the series ∞⎪summationdisplay n=0ansinnx. © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 469 We know from (12.3.10) and (12.4.11) that Z{f(n)}=Z{sinnx}=zsinx z2−2zcosx+1, Z{ansinnx}=F⎪parenleftBigz a⎪parenrightBig =azsinx a2−2azcosx+z2. Hence, Theorem 12.7.1(ii) gives ∞⎪summationdisplay n=0ansinnx= lim z→1F⎪parenleftBigz a⎪parenrightBig =asinx a2−2acosx+1. (12.7.5) 12.8 Exercises 1. Find the Ztransform of the following functions: (a)n3, (b)an n!, (c)nexp{(n−1)α}, (d)H(n)−H(n−2),(e)n2an,(f)δ(n)=⎪braceleftbigg 1,n =0, 0,otherwise⎪bracerightbigg . 2. Show that (a)Z{sinhna}=z(sinh a) z2−2zcosha+1, (b)Z{exp(−an)cosbn}=z(z−e−acosb) z2−2ze−acosb+e−2a. (c)Z⎪braceleftbig e−ansinbn⎪bracerightbig =eazsinb e2az2−2eazcosb+1,|z|>e−a. 3. Show that Z{nanf(n)}=−zd dz⎪braceleftBig F⎪parenleftBigz a⎪parenrightBig⎪bracerightBig . 4. Prove that (a)Z⎪braceleftbiggf(n) n⎪bracerightbigg =⎪integraldisplay∞ zF(z) zdz, (b)Z⎪braceleftbiggf(n) n+m⎪bracerightbigg =zm⎪integraldisplay∞ zF(z)dz zm+1. © 2007 by Taylor & Francis Group, LLC 470 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Hence, deduce that Z⎪braceleftbigg1 n+1⎪bracerightbigg =zlog⎪parenleftbiggz z−1⎪parenrightbigg . 5. Show that (a)Z{nan−1}=z (z−a)2, (b)Z⎪braceleftbiggn(n−1)···(n−m+1 ) m!an−m⎪bracerightbigg =z (z−a)m+1. 6. Find the inverse Ztransform of the following functions: (a)z2 (z−2)(z−3),(b)z2−1 z2+1, (c)z (z−1)2, (d)z (z−a)2, (e)1 (z−a)2, (f)1 (z−1)2(z−2), (g)z+3 (z+1 ) (z+2 ),(h)z3 (z2−1)(z−2),(i)z2 (z−1)⎪parenleftbig z−1 2⎪parenrightbig. (j)F(z)=z2 (z−e−a)(z−e−b),a,bare constants. (k)F(z)=(z−a)−k,k =1,2,..., |z|>|a|>0. (l)F(z)=z4+5 (z−1)2(z−2),|z|>2,(m)F(z)=(z−1) (z+2)(z−1 2),|z|>2. 7. Solve the following diﬀerence equations: (a)f(n+1 )+3 f(n)=n, f (0) = 1 . (b)f(n+1 )−5f(n)=s i n n, f (0) = 0 . (c)f(n+1 )−af(n)=an,f(0) =x0. (d)f(n+1 )−f(n)=a[1−f(n)],f(0) =x0. (e)f(n+2 )−f(n+1 )−6f(n)=0,f(0) = 0 ,f(1) = 3 . (f)f(n+2 )+4 f(n+1 )+3 f(n)=0,f(0) = 1 ,f(1) = 1 . (g)f(n+2 )−f(n+1 )−6f(n)=s i n⎪parenleftBignπ 2⎪parenrightBig (n≥2),f(0) = 0 ,f(1) = 3 . (h)f(n+2 )−2f(n+1 )+ f(n)=0,f(0) = 2 ,f(1) = 0 . (i)f(n+2 )−2af(n+1 )+ a2f(n)=0,f(0) = 0 ,f(1) =a. (j)f(n+3 )−f(n+2 )−f(n+1 )+ f(n)=0,f(0) = 1 ,f(1) =f(2) = 0 . (k)f(n)=f(n−1) + 2 f(n−2),f(0) = 1 ,f(1) = 2 . (l)f(n)−af(n−1) = 1 ,f(−1)= 2 . © 2007 by Taylor & Francis Group, LLC ZT r a n s f o r m s 471 (m)f(n+2 )+3 f(n+1 )+2 f(n)=0,f(0) = 1 ,f(1) = 2 . (n)f(n+1 )−2f(n)=0,f(0) = 3 . 8. Show that the solution of the resistive ladder network governed by the diﬀerence equation for the current ﬁeld i(n) i(n+2 )−3i(n+1 )+ i(n)=0,i(0) = 1 ,i(1) = 2 i(0)−V R is i(n)=c o s h ( xn)+2 √ 5⎪parenleftbigg1 2−V R⎪parenrightbigg sinh(nx), where cosh x=3 2and sinh x=√ 5 2. 9. Use the Initial Value Theorem to ﬁnd f(0) for F(z)g i v e nb y (a)z z−α, (b)z (z−α)(z−β), (c)z(z−cosx) z2−2zcosx+1, (d)1 (z−a)m. 10. Use the Final Value Theorem to ﬁnd lim n→∞f(n)f o rF(z): (a)F(z)=z z−a, (b)F(z)=z2−zcosa (z2−2zcosa+1 ). 11. Find the sum of the following series using the Ztransform: (a)∞⎪summationdisplay n=0aneinx,(b)∞⎪summationdisplay n=0(−1)ne−n n+1,(c)∞⎪summationdisplay n=0exp[−x(2n+1 ) ]. 12. Solve the second orde r diﬀerence equation 3f(n+2 )−2f(n+1 )−f(n)=0,f(0) = 1 ,f(1) = 2 and then show that f(n)→7 4asn→∞. 13. Solve the simultaneous diﬀerence equations u(n+1 )=2 υ(2) + 2 , υ(n+1 )=2 u(n)−1, with the initial data u(0) =υ(0) = 0. © 2007 by Taylor & Francis Group, LLC 472 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 14. Show that the solution of the third order diﬀerence equation u(n+3 )−3u(n+2 )+3 u(n+1 )−u(n)=0, u(0) = 1 ,u(1) = 0 ,u(2) = 1 , is u(n)=(n−1)2. 15. Show that the solution of the initial value problem u(n+2 )−4u(n+1 )+3 u(n)=0,u(0) =u0andu(1) =u1 is un=1 2(3u0−u1)+1 2(u1−u0)3n. 16. Find the solution of the following initial value problems: (a)un+2+2un+1−3un=0,u0=1,u 1=0, (b) 3 un+2−5un+1+2un=0,u 0=1,u1=0, (c)un+2−4un+1+5un=0,u0=1 2,u 1=3. © 2007 by Taylor & Francis Group, LLC 13 Finite Hankel Transforms “No human investigation can be called real science if it cannot be demonstrated mathematically.” Leonardo da Vinci “The mathematician’s patterns, like the painter’s or the poet’s must be beautiful; the ideas, like the colors or the words, must ﬁttogether in a harmonious way. Beauty is the ﬁrst test: there is no permanent place in this world for ugly mathematics.” Godfrey H. Hardy 13.1 Introduction This chapter is devoted to the study of the ﬁnite Hankel transform and its basic operational properties. The usefulness of this transform is shown bysolving several initial-boundary problems of physical interest. The method of ﬁnite Hankel transforms was ﬁrst introduced by Sneddon (1946). 13.2 Deﬁnition of the Finite Hankel Transform and Examples Just as problems on ﬁnite invervals −a<x<a lead to Fourier series, problems on ﬁnite intervals 0 <r<a , where ris the cylindrical polar coordinate, lead to the Fourier-Bessel series representation of a function f(r)w h i c hc a nb e stated in the following theorem: 473 © 2007 by Taylor & Francis Group, LLC 474 INTEGRAL TRANSFORMS and THEIR APPLICATIONS THEOREM 13.2.1 Iff(r)i sd e ﬁ n e di n0 ≤r≤aand ˜fn(ki)=⎪integraldisplaya orf(r)Jn(rki)dr, (13.2.1) thenf(r) can be represented by the Fourier-Bessel series as f(r)=2 a2∞⎪summationdisplay i=1˜fn(ki)Jn(rki) J2 n+1(aki), (13.2.2) where ki(0<k1<k2<···) are the roots of the equation Jn(aki)=0,that means J/prime n(aki)=Jn−1(aki)=−Jn+1(aki), (13.2.3) due to the standard recurrence relations among J/prime n(x),Jn−1(x),andJn+1(x). PROOF We write formally the Bessel series expansion of f(r)a s f(r)=∞⎪summationdisplay i=1ciJn(rki), (13.2.4) where the summation is taken over all the positive zeros k1,k2,...of the Bessel function Jn(aki).Multiplying (13.2.4) by rJn(rki),integrating the both sides of the result from 0 to a, and then using the orthogonal property of the Bessel functions, we obtain a⎪integraldisplay orf(r)Jn(rki)dr=cia⎪integraldisplay orJ2 n(rki)dr. Or, ˜fn(ki)=a2ci 2J2 n+1(aki), hence, we obtain ci=2 a2˜fn(ki) J2 n+1(aki). (13.2.5) Substituting the value of ciinto (13.2.4) gives (13.2.2). DEFINITION 13.2.1 The ﬁnite Hankel transform of order nofafunction f(r)is denoted by Hn{f(r)}=˜fn(ki)a n di sd e ﬁ n e db y Hn{f(r)}=˜fn(ki)=⎪integraldisplaya 0rf(r)Jn(rki)dr. (13.2.6) © 2007 by Taylor & Francis Group, LLC Finite Hankel Transforms 475 The inverse ﬁnite Hankel transform is then deﬁned by H−1 n⎪braceleftBig ˜fn(ki)⎪bracerightBig =f(r)=2 a2∞⎪summationdisplay i=1˜fn(ki)Jn(rki) J2 n+1(aki), (13.2.7) where the summation is taken over all positive roots of Jn(ak)=0. The zero-order ﬁnite Hankel transform and its inverse are deﬁned by H0{f(r)}=˜f0(ki)=⎪integraldisplaya 0rf(r)J0(rki)dr, (13.2.8) H−1 0⎪braceleftBig ˜f0(ki)⎪bracerightBig =f(r)=2 a2∞⎪summationdisplay i=1˜f0(ki)J0(rki) J2 1(aki), (13.2.9) where the summation is taken over the positive roots of J0(ak)=0. Similarly, the ﬁrst-order ﬁnite Hankel transform and its inverse are H1{f(r)}=˜f1(ki)=⎪integraldisplaya 0rf(r)J1(rki)dr, (13.2.10) H−1 1⎪braceleftBig ˜f1(ki)⎪bracerightBig =f(r)=2 a2∞⎪summationdisplay i=1˜f1(ki)J1(rki) J2 2(aki), (13.2.11) where kiis chosen as a positive root of J1(ak)=0. We now give examples of ﬁnite Hankel transforms of some functions. Example 13.2.1 Iff(r)=rn,t h e n Hn{rn}=⎪integraldisplaya 0rn+1Jn(rki)dr=an+1 kiJn+1(aki). (13.2.12) When n=0, H0{1}=a kiJ1(aki). (13.2.13) Example 13.2.2 Iff(r)=(a2−r2), then H0{(a2−r2)}=⎪integraldisplaya 0r(a2−r2)J0(aki)dr=4a k3 iJ1(aki)−2a2 k2 iJ0(aki). © 2007 by Taylor & Francis Group, LLC 476 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Since kiare the roots of J0(ak)=0,we ﬁnd H0{(a2−r2)}=4a k3 iJ1(aki). (13.2.14) 13.3 Basic Operational Properties We state the following operational properties of the ﬁnite Hankel transforms: Hn{f/prime(r)}=ki 2n[(n−1)Hn+1{f(r)} −(n+1 )Hn−1{f(r)}],n≥1, (13.3.1) provided f(r) is ﬁnite at r=0 . When n=1,we obtain the ﬁnite Hankel transform of derivatives H1{f/prime(r)}=−kiH0{f(r)}=−ki˜f0(ki). (13.3.2) Hn⎪bracketleftbigg1 rd dr{rf/prime(r)}−n2 r2f(r)⎪bracketrightbigg =−k2 i˜fn(ki)−akif(a)J/prime n(aki).(13.3.3) When n=0 H0⎪bracketleftbigg f/prime/prime(r)+1 rf/prime(r)⎪bracketrightbigg =−k2 i˜f0(ki)+akif(a)J1(aki). (13.3.4) Ifn=1,(13.3.3) becomes H1⎪bracketleftbigg f/prime/prime(r)+1 rf/prime(r)−1 r2f(r)⎪bracketrightbigg =−k2 i˜f1(ki)−akif(a)J/prime 1(aki).(13.3.5) Results (13.3.4) and (13.3.5) are very useful for ﬁnding solutions of diﬀer- ential equations in cylindrical polar coordinates. The proofs of the above results are elementary exercises for the reader. 13.4 Applications of Finite Hankel Transforms Example 13.4.1(Temperature Distribution in a Long Circular Cylinder ). Find the solution of © 2007 by Taylor & Francis Group, LLC Finite Hankel Transforms 477 the axisymmetric heat conduction equation ∂u ∂t=κ⎪parenleftbigg∂2u ∂r2+1 r∂u ∂r⎪parenrightbigg ,0≤r≤a, t > 0 (13.4.1) with the boundary and initial conditions u(r, t)=f(t)o n r=a, t > 0 (13.4.2) u(r,0) = 0 ,0≤r≤a. (13.4.3) Application of the ﬁnite Hankel transform deﬁned by ˜u(ki,t)=H0{u(r, t)}=⎪integraldisplaya 0rJo(rki)u(r, t)dr, (13.4.4) yields the given system with the boundary condition ˜ut+κk2 i˜u=κakiJ1(aki)f(t), ˜u(ki,0) = 0 . (13.4.5ab) The solution of the ﬁrst order system is ˜u(ki,t)=κakiJ1(aki)⎪integraldisplayt 0f(τ)exp⎪braceleftbig −κk2 i(t−τ)⎪bracerightbig dτ. (13.4.6) The inverse transform gives the formal solution u(r, t)=⎪parenleftbigg2κ a⎪parenrightbigg∞⎪summationdisplay i=1kiJ0(rki) J1(aki)⎪integraldisplayt 0f(τ)exp⎪braceleftbig −κk2 i(t−τ)⎪bracerightbig dτ. (13.4.7) In particular, if f(t)=T0= constant, u(r, t)=⎪parenleftbigg2T0 a⎪parenrightbigg∞⎪summationdisplay i=1J0(rki) kiJ1(aki)⎪bracketleftbig 1−exp⎪parenleftbig −κk2 it⎪parenrightbig⎪bracketrightbig . (13.4.8) Using the inverse version of (13.2.7) gives the ﬁnal solution u(r, t)=T0−⎪parenleftbigg2T0 a⎪parenrightbigg∞⎪summationdisplay i=1J0(rki) kiJ1(aki)exp⎪parenleftbig −κk2 it⎪parenrightbig . (13.4.9) This solution representing the temperature distribution consists of the steady- state term, and the transient term which decays to zero as t→∞.C o n s e q u e n t - ly, the steady temperature is attained in the limit as t→∞. Example 13.4.2 (Unsteady Viscous Flow in a Rotating Long Circular Cylinder ). The axisym- metric unsteady motion of a viscous ﬂuid in an inﬁnitely long circular cylinder of radius ais governed by ut=ν⎪parenleftbigg urr+1 rur−u r2⎪parenrightbigg ,0≤r≤a, t > 0, (13.4.10) © 2007 by Taylor & Francis Group, LLC 478 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where u=u(r, t) is the tangential ﬂuid velocity and νis the constant kinematic viscosity of the ﬂuid. The cylinder is initially at rest at t= 0, and it is then allowed to rotate with constant angular velocity Ω. Thus, the boundary and initial conditions are u(r, t)=aΩo n r=a, t > 0, (13.4.11) u(r, t)=0 a t t=0 f o r 0 <r<a . (13.4.12) We solve the problem by using the joint Laplace and the ﬁnite Hankel transform of order one deﬁned by ˜¯u(ki,s)=⎪integraldisplay∞ 0e−stdt⎪integraldisplaya 0rJ1(kir)u(r, t)dr, (13.4.13) where kiare the positive roots of J1(aki)=0. Application of the joint transform gives s˜¯u(ki,s)=−νk2 i˜¯u(ki,s)−νa2Ωki sJ/prime 1(aki). Or, ˜¯u(ki,s)=−νa2ΩkiJ/prime 1(aki) s(s+νk2 i). (13.4.14) T h ei n v e r s eL a p l a c et r a n s f o r mg i v e s ˜u(ki,t)=−a2Ω kiJ/prime 1(aki)⎪bracketleftbig 1−exp⎪parenleftbig −νtk2 i⎪parenrightbig⎪bracketrightbig . (13.4.15) Thus, the ﬁnal solution is found from (13.4.15) by using the inverse Hankel transform withJ/prime 1(aki)=−J2(aki)i nt h ef o r m u(r, t)=2 Ω∞⎪summationdisplay i=1J1(rki) kiJ2(aki)⎪bracketleftbig 1−exp⎪parenleftbig −νtk2 i⎪parenrightbig⎪bracketrightbig . (13.4.16) This solution is the sum of the steady-state and the transient ﬂuid velocities. In view of (13.2.12) for n=1 ,w ec a nw r i t e r=H−1 1⎪braceleftbigga2 kiJ2(aki)⎪bracerightbigg =2∞⎪summationdisplay i=1J1(rki) kiJ2(aki). (13.4.17) This result is used to simplify (13.4.16) so that the ﬁnal solution for u(r, t) takes the form u(r, t)=rΩ−2Ω∞⎪summationdisplay i=1J1(rki) kiJ2(aki)exp⎪parenleftbig −νtk2 i⎪parenrightbig . (13.4.18) © 2007 by Taylor & Francis Group, LLC Finite Hankel Transforms 479 In the limit as t→∞, the transient velocity component decays to zero, and the ultimate steady state ﬂow is attained in the form u(r, t)=rΩ. (13.4.19) Physically, this represents the rigid body rotation of the ﬂuid inside the cylin- der. Example 13.4.3 (Vibrations of a Circular Membrane ). The free symmetric vibration of a thin circular membrane of radius ais governed by the wave equation utt=c2⎪parenleftbigg urr+1 rur⎪parenrightbigg ,0<r<a , t> 0 (13.4.20) with the initial and boundary data u(r, t)=f(r),∂u ∂t=g(r)a t t=0 f o r 0 <r<a , (13.4.21ab) u(a, t)= 0 for all t>0. (13.4.22) Application of the zero-order ﬁnite Hankel transform ofu(r, t) deﬁned by (13.4.4) to (13.4.20)–(13.4.22) gives d2˜u dt2+c2k2 i˜u=0, (13.4.23) ˜u=˜f(ki)a n d⎪parenleftbiggd˜u dt⎪parenrightbigg t=0=˜g(ki). (13.4.24ab) The solution of this system is ˜u(ki,t)=˜f(ki)cos (ctki)+˜g(ki) ckisin(ctki). (13.4.25) The inverse transform yields the formal solution u(r, t)=2 a2∞⎪summationdisplay i=1f(ki)cos (ctki)J0(rki) J2 1(aki) +2 ca2∞⎪summationdisplay i=1g(ki)sin (ctki)J0(rki) kiJ2 1(aki), (13.4.26) where the summation is taken over all positive roots of J0(aki)=0. We consider a more general form of the ﬁnite Hankel transform associated with a more general boundary condition f/prime(r)+hf(r)=0 a t r=a, (13.4.27) © 2007 by Taylor & Francis Group, LLC 480 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where his a constant. We deﬁne the ﬁnite Hankel transform off(r)b y Hn{f(r)}=˜fn(ki)=⎪integraldisplaya 0rJn(rki)f(r)dr, (13.4.28) where kiare the roots of the equation kiJ/prime n(aki)+hJn(aki)=0. (13.4.29) The corresponding inverse transform is given by f(r)=H−1 n{˜fn(ki)}=2∞⎪summationdisplay i=1k2 i˜fn(ki)Jn(rki) {(k2 i+h2)a2−n2}J2n(aki). (13.4.30) Thisﬁnite Hankel transform has the following operational property Hn⎪bracketleftbigg1 rd dr{rf/prime(r)}−n2 r2f(r)⎪bracketrightbigg =−k2 i˜fn(ki) +a[f/prime(a)+hf(a)]Jn(aki),(13.4.31) which is, by (13.4.29) =−k2 i˜fn(ki)−aki h[f/prime(a)+hf(a)]J/prime n(aki). (13.4.32) Thus, result (13.4.32) involves f/prime(a)+hf(a) as the boundary condition. We apply this more general ﬁnite Hankel transform pairs (13.4.28) and (13.4.30) to solve the following axisymmetric initial-boundary value problem. Example 13.4.4 (Temperature Distribution of Cooling of a Circular Cylinder ). Solve the ax- isymmetric heat conduction problem for an inﬁnitely long circular cylinder of radius r=awith the initial constant temperature T0, and the cylinder is cooling by radiation of heat from its boundary surface at r=ato the out- side medium at zero temperature accord ing to Newton’s law of cooling, which satisﬁes the boundary condition ∂u ∂r+hu=0 a t r=a, t > 0, (13.4.33) where his a constant. The problem is governed by the axisymmetric heat conduction equation ut=κ(urr+1 rur),0≤r≤a, t > 0, (13.4.34) © 2007 by Taylor & Francis Group, LLC Finite Hankel Transforms 481 with the boundary condition (13.4.33) and the initial condition u(r,0)=T0att=0,for 0<r<a . (13.4.35) Application of the zero-order Hankel transform (13.4.28) with (13.4.29) to the system (13.4.33)–(13.4.35) gives d˜u dt+κk2 i˜u=0,t > 0 (13.4.36) ˜u(ki,0) =T0⎪integraldisplaya 0rJ0(rki)dr=aT0 kiJ1(aki). (13.4.37) The solution of (13.4.36)–(13.4.37) is ˜u(ki,t)=⎪parenleftbiggaT0 ki⎪parenrightbigg J1(aki)exp (−κtk2 i). (13.4.38) The inverse transform (13.4.30) with n=0a n d kiJ/prime 0(aki)+hJ0(aki) = 0, that is,kiJ1(aki)=hJ0(aki), leads to the formal solution u(r, t)=⎪parenleftbigg2hT0 a⎪parenrightbigg∞⎪summationdisplay i=1J0(rki)exp (−κtk2 i) (k2 i+h2)J0(aki), (13.4.39) where the summation is taken over all the positive roots of kiJ1(aki)=hJ0(aki). 13.5 Exercises 1. Find the zero-order ﬁnite Hankel transform of (a)f(r)=r2, (b)f(r)=J0(αr), (c)f(r)=(a2−r2). 2. Show that Hn⎪braceleftbiggJn(αr) Jn(αa)⎪bracerightbigg =aki (α2−k2 i)J/prime n(aki) 3. IfHn{f(r)}is the ﬁnite Hankel transform off(r) deﬁned by (13.2.6), and if n>0, show that (a)Hn{r−1f/prime(r)}=1 2ki⎪bracketleftbig Hn+1{r−1f(r)}−Hn−1{r−1f(r)}⎪bracketrightbig , (b)H0{r−1f/prime(r)}=kiH1{r−1f(r)}−f(a). © 2007 by Taylor & Francis Group, LLC 482 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 4. Solve the initial-boundary value problem c2(urr+1 rur)=utt,0<r<a ,t> 0, u(r,0) = 0 ,u t(r,0) =u0,for all r, u(a, t)=0,t > 0, where u0is a constant. 5. Obtain a solution of the initial-boundary problem κ⎪parenleftbigg urr+1 rur⎪parenrightbigg =ut,0<r<a , t> 0, u(r,0) =f(r),for 0 <r<a u(a, t)=0. 6. If we deﬁne the ﬁnite Hankel transform off(r)b y Hn{f(r)}=˜fn(ki)=⎪integraldisplayb arf(r)An(rki)dr, b > a, where An(rki)=Jn(rki)Yn(aki)−Yn(rki)Jn(aki), andYn(x) is the Bessel function of the second kind of order n,s h o w that the inverse transform is H−1 n{˜fn(ki)}=f(r)=π2 2∞⎪summationdisplay i=1k2 i˜fn(ki)An(rki)J2 n(bki) J2n(aki)−J2n(bki), where kiare the positive roots of An(bki)=0. 7. For the transform deﬁned in problem 6, show that Hn⎪bracketleftbigg f/prime/prime(r)+1 rf/prime(r)−n2f(r) r2⎪bracketrightbigg =−k2 i˜fn(ki)+2 π⎪bracketleftbigg f(b)Jn(aki) Jn(bki)−f(a)⎪bracketrightbigg . 8. Viscous ﬂuid of kinematic viscosity νis bounded between two inﬁnitely long concentric circular cylinders of radii aandb. The inner cylinder is stationary and the outer cylinder begins to rotate with uniform angular velocity Ω at t= 0. The axisymmetric ﬂow is governed by (13.4.10) with υ(a,0)= 0 and υ(b,0)= Ω b. show that υ(r, t)=(πbΩ)∞⎪summationdisplay i=1J1(aki)J1(bki)A1(rki)⎪bracketleftbig 1−exp(−νtk2 i)⎪bracketrightbig J2 1(aki)−J2 1(bki), where A1(rki)=Jn(rki)Yn(aki)−Yn(rki)Jn(aki), andkiare the positive roots of the equation A1(bki)=0 . © 2007 by Taylor & Francis Group, LLC Finite Hankel Transforms 483 9. Find the solution of the forced symmetric vibrations of a thin elastic membrane that satisfy the initial-boundary value problem urr+1 rur−1 c2utt=−p(r, t) T0, where p(r, t) is the applied pressure which produces vibrations, and the membrane is stretched by a constant tension T0. The membrane is set into motion from rest in its equilibrium position so that u(r, t)=0=⎪parenleftbigg∂u dt⎪parenrightbigg att=0. 10. Use the joint Hankel and Laplace transform method to the axisymmetric diﬀusion problem in an inﬁnitely long circular cylinder of radius a: ut=κ⎪parenleftbigg urr+1 rur⎪parenrightbigg +Q(r, t),0<r<a , t> 0, u(a, t)=0 f o r t>0, u(r,0) = 0 for 0 <r≤a, where Q(r, t) represents a heat source inside the cylinder. Find the ex- plicit solution for two special cases: (a)Q(r, t)=κQ0 k, (b)Q(r, t)=Q0δ(r) rf(t), where Q0,κ,a n d kare constants. © 2007 by Taylor & Francis Group, LLC 14 Legendre Transforms “Legendre, who for so many reasons is considered the founder of elliptic functions, greatly smoothed the way for his successors; it is the fact of the double periodicity of the inverse function, immedi-ately discovered by Abel and Jacobi, that is missing and that gave such a restrained analytical c haracter to his treatise.” Charles Hermite “First causes are not known to us, but they are subjected to simple and constant laws that can be studied by observation and whose study is the goal of Natural Philosophy. ... Heat penetrates, asdoes gravity, all the substances of the universe; its rays occupy all regions of space. The aim of our work is to expose the mathematical laws that this element follows. ... The diﬀerential equations for the propagation of heat express the most general conditions and reduce physical questions to problems in pure Analysis that is properly theobject of the theory.” Clerk Maxwell 14.1 Introduction We consider in this chapter the Legendre transform with a Legendre polynomi- al as kernel and discuss basic operational properties including the ConvolutionTheorem. Legendre transforms are then used to solve boundary value prob- lems in potential theory. This chapter is based on papers by Churchill (1954) and Churchill and Dolph (1954) listed in the Bibliography. 485 © 2007 by Taylor & Francis Group, LLC 486 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 14.2 Deﬁnition of the Legendre Transform and Examples Churchill (1954) deﬁned the Legendre transform of a function f(x) deﬁned in −1<x< 1 by the integral Tn{f(x)}=˜f(n)=1⎪integraldisplay −1Pn(x)f(x)dx, (14.2.1) provided the integral exists and where Pn(x)i st h e Legendre polynomial of degree n(≥0). Obviously Tnis a linear integral transformation. When x=c o sθ, (14.2.1) becomes Tn{f(cosθ)}=˜f(n)=π⎪integraldisplay 0Pn(cosθ)f(cosθ)sinθdθ. (14.2.2) Theinverse Legendre transform is given by f(x)=T−1 n{˜f(n)}=∞⎪summationdisplay n=0⎪parenleftbigg2n+1 2⎪parenrightbigg ˜f(n)Pn(x). (14.2.3) This follows from the expansion of any function f(x)i nt h ef o r m f(x)=∞⎪summationdisplay n=0anPn(x), (14.2.4) where the coeﬃcient ancan be determined from the orthogonal property of Pn(x). It turns out that an=⎪parenleftbigg2n+1 2⎪parenrightbigg1⎪integraldisplay −1Pn(x)f(x)dx=⎪parenleftbigg2n+1 2⎪parenrightbigg ˜f(n), (14.2.5) and hence, result (14.2.3) follows. Example 14.2.1 Tn{exp(iαx)}=⎪parenleftbigg2π α⎪parenrightbigg1/2 inJn+1/2(α), (14.2.6) where Jν(x) is the Bessel function. © 2007 by Taylor & Francis Group, LLC Legendre Transforms 487 We have, by deﬁnition, Tn{exp(iαx)}=1⎪integraldisplay −1exp(iαx)Pn(x)dx, which is, by a result in Copson (1935, p. 341), =⎪radicalbigg 2π αinJn+1/2(α). Similarly, Tn{exp(αx)}=⎪radicalbigg 2π αIn+1/2(α), (14.2.7) where Iν(x) is the modiﬁed Bessel function of the ﬁrst kind. Example 14.2.2 (a) Tn{(1−x2)−1/2}=πP2 n(0) (14.2.8) (b) Tn⎪braceleftbigg1 2(t−x)⎪bracerightbigg =Qn(t), |t|>1, (14.2.9) where Qn(t) is the Legendre function of the second kind given by Qn(t)=1 21⎪integraldisplay −1(t−x)−1Pn(x)dx. These results are easy to verify with the aid of results given in Copson (1935, p. 292 and p. 310). Example 14.2.3 If|r|≤1, then (a)Tn{(1−2rx+r2)−1/2}=2rn (2n+1 ), (14.2.10) (b)Tn{1−2rx+r2)−3/2}=2rn (1−r2). (14.2.11) We have, from the generating function of Pn(x), (1−2rx+r2)−1/2=∞⎪summationdisplay n=0rnPn(x),|r|<1. © 2007 by Taylor & Francis Group, LLC 488 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Multiplying this result by Pn(x) and using the orthogonality condition of the Legendre polynomial gives 1⎪integraldisplay −1(1−2rx+r2)−1/2Pn(x)dx=2rn (2n+1 ). (14.2.12) In particular, when r=1 ,w eo b t a i n Tn{(1−x)−1/2}=2√ 2 (2n+1 ). (14.2.13) Diﬀerentiating (14.2.12) with respect to rgives 1 21⎪integraldisplay −1(1−2rx+r2)−3/2(2rx−2r2)Pn(x)dx=2nrn (2n+1 ), so that −Tn{(1−2rx+r2)−1/2}+( 1−r2)Tn{(1−2rx+r2)−3/2}=2nrn (2n+1 ). Using (14.2.10), we obtain (14.2.11). Example 14.2.4 If|r|<1a n d α>0, then Tn⎧ ⎨ ⎩r⎪integraldisplay 0tα−1dt (1−2xt+t2)1/2⎫ ⎬ ⎭=2rn+α (2n+1 ) (n+α). (14.2.14) We replace rbytin (14.2.10) and multiply the result by tα−1to obtain Tn{tα−1(1−2xt+t2)−1/2}=2tn+α−1 (2n+1 ). Integrating this result on (0 ,r) we ﬁnd (14.2.14). Example 14.2.5 IfH(x) is a Heaviside unit step function, then Tn{H(x)}=⎧ ⎪⎨ ⎪⎩1,n =0 Pn−1(0)−Pn+1(0) (2n+1 ),n ≥1⎫ ⎪⎬ ⎪⎭. (14.2.15) © 2007 by Taylor & Francis Group, LLC Legendre Transforms 489 Obviously, Tn{H(x)}=1⎪integraldisplay 0Pn(x)dx=1 w h e n n=0. However, for n>1, we use the recurrence relation forPn(x)a s (2n+1 )Pn(x)=P/prime n+1(x)−P/prime n−1(x) (14.2.16) to derive Tn{H(x)}=1 (2n+1 )1⎪integraldisplay 0[P/prime n+1(x)−P/prime n−1(x)]dx =1 2n+1[Pn−1(0)−Pn+1(0)]. Debnath and Harrel (1976) introduced the associated Legendre transform deﬁned by Tn,m{f(x)}=˜f(n,m)=1⎪integraldisplay −1(1−x2)−m/2Pm n(x)f(x)dx, (14.2.17) where Pm n(x)i st h e associated Legendre function of the ﬁrst kind. The inverse transform is given by f(x)=T−1 n,m{˜f(n,m)}=∞⎪summationdisplay n=0(2n+1 ) 2(n−m)! (n+m)! ×˜f(n,m)(1−x2)m/2Pm n(x).(14.2.18) The reader is referred to Debnath and Harrel (1976) for a detailed discussion of this transform. 14.3 Basic Operational Properties of Legendre Transforms THEOREM 14.3.1 Iff/prime(x) is continuous and f/prime/prime(x) is bounded and integrable in each subinterval of−1≤x≤1, and if Tn{f(x)}exists and lim |x|→1(1−x2)f(x) = lim |x|→1(1−x2)f/prime(x)=0, (14.3.1) © 2007 by Taylor & Francis Group, LLC 490 INTEGRAL TRANSFORMS and THEIR APPLICATIONS then Tn{R[f(x)]}=−n(n+1 )˜f(n), (14.3.2) where R[f(x)] is a diﬀerential form given by R[f(x)] =d dx⎪bracketleftbigg (1−x2)d dxf(x)⎪bracketrightbigg ,n > 0. (14.3.3) PROOF We have, by deﬁnition, Tn{R[f(x)]}=1⎪integraldisplay −1d dx⎪bracketleftbigg (1−x2)d dxf(x)⎪bracketrightbigg Pn(x)dx which is, by integrating by parts together with (14.3.1), =−1⎪integraldisplay −1(1−x2)P/prime n(x)d dxf(x)dx. Integrating this result by parts again, we obtain Tn{R[f(x)]}=−[(1−x2)]P/prime n(x)f(x)]1 −1+1⎪integraldisplay −1d dx[(1−x2)]P/prime n(x)]f(x)dx. Using (14.3.1) and the diﬀerential equation for the Legendre polynomial d dx⎪bracketleftbigg (1−x2)dy dx⎪bracketrightbigg +n(n+1 )y=0, (14.3.4) we obtain the desired result Tn{R[f(x)]}=−n(n+1 )˜f(n). We may extend this result to evaluate the Legendre transforms of the dif- ferential forms R2[f(x)],R3[f(x)],...,Rk[f(x)]. Clearly Tn{R2[f(x)]}=Tn{R[R[f(x)]]} =−n(n+1 )Tn{R[f(x)]}=n2(n+1 )2˜f(n),(14.3.5) provided f/prime(x)a n d f/prime/prime(x) satisfy the conditions of Theorem 14.3.1. Similarly, Tn{R3[f(x)]}=(−1)3n3(n+1 )3˜f(n). (14.3.6) © 2007 by Taylor & Francis Group, LLC Legendre Transforms 491 More generally, for a positive integer k, Tn{Rk[f(x)]}=(−1)knk(n+1 )k˜f(n). (14.3.7) COROLLARY 14.3.1 IfTn{R[f(x)]}=−n(n+1 )˜f(n), then Tn⎪braceleftbigg1 4f(x)−R[f(x)]⎪bracerightbigg =⎪parenleftbigg n+1 2⎪parenrightbigg2 ˜f(n). (14.3.8) PROOF We replace n(n+1 )b y⎪parenleftbigg n+1 2⎪parenrightbigg2 −1 4in (14.3.2) to obtain Tn{R[f(x)]}=−⎪bracketleftBigg⎪parenleftbigg n+1 2⎪parenrightbigg2 −1 4⎪bracketrightBigg ˜f(n). (14.3.9) Rearranging the terms in (14.3.9) gives Tn⎪braceleftbigg1 4f(x)−R[f(x)]⎪bracerightbigg =⎪parenleftbigg n+1 2⎪parenrightbigg2 ˜f(n). In general, this result can be written as (−1)kTn{Rk[f(x)]−4−kf(x)}=k−1⎪summationdisplay r=0(−1)r⎪parenleftbiggk r⎪parenrightbigg⎪bracketleftBigg 4−r⎪parenleftbigg n+1 2⎪parenrightbigg2k−2r⎪bracketrightBigg ˜f(n). (14.3.10) The proof of (14.3.10) follows from (14.3.7) by replacing n(n+1 )w i t h⎪parenleftbig n+1 2⎪parenrightbig2−1 4and using the binomial expansion. Example 14.3.1 Tn{log(1−x)}=⎧ ⎨ ⎩2(log 2 −1),n =0 −2 n(n+1 ),n > 0⎫ ⎬ ⎭. (14.3.11) Clearly, R[log(1 −x)] =d dx⎪bracketleftbigg (1−x2)d dxlog(1−x)⎪bracketrightbigg =−1. Althoughd dxlog(1−x) does not satisfy the conditions of Theorem 14.3.1, we integrate by parts to obtain © 2007 by Taylor & Francis Group, LLC 492 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Tn{R[log(1 −x)]}=1⎪integraldisplay −1R[log(x)]Pn(x)dx =[−(1 +x)Pn(x)]1 −1+1⎪integraldisplay −1(1 +x)P/prime n(x)dx, which is, since (1 + x)=−(1−x2)d dxlog(1−x), and by integrating by parts, =−2+1⎪integraldisplay −1log(1−x)d dx[(1−x2)]P/prime n(x)]dx. (14.3.12) By integrating by parts twice, result (14.3.12) gives Tn{R[log(1 −x)]}=−2+1⎪integraldisplay −1d dx⎪bracketleftbigg (1−x2)d dxlog(1−x)⎪bracketrightbigg Pn(x)dx, which is, by (14.3.2), =−2−n(n+1 )˜f(n), (14.3.13) where ˜f(n)=Tn{log(1−x)}. However, R[log(1 −x)] =−1s ot h a t Tn{R[log(1 −x)]}=0f o ra l l n>0a n d hence, result (14.3.13) gives Tn[log(1 −x)] =˜f(n)=−2 n(n+1 ). On the other hand, since P0(x)=1 ,w eh a v e T0{[log(1 −x)]}=1⎪integraldisplay −1log(1−x)dx, which is, by direct integration, =−[(1−x){log(1−x)−x}]1 −1=2 ( l o g2 −1). © 2007 by Taylor & Francis Group, LLC Legendre Transforms 493 THEOREM 14.3.2 Iff(x)a n d f/prime(x) are piecewise continuous in −1<x< 1,R−1[f(x)] =h(x), andf(0) =1⎪integraltext −1f(x)dx=0 ,t h e n T−1 n⎪braceleftBigg˜f(n) n(n+1 )⎪bracerightBigg =A−x⎪integraldisplay 0ds (1−s2)s⎪integraldisplay −1f(t)dt, (14.3.14) where Ais an arbitrary constant of integration. PROOF We have R[h(x)] =f(x) or, d dx⎪bracketleftbigg (1−x2)d dxh(x)⎪bracketrightbigg =f(x). Integrating over ( −1,x)g i v e s x⎪integraldisplay −1f(t)dt=( 1−x2)d dxh(x), (14.3.15) which is a continuous function of xin|x|<1 with limit zero as |x|→1. Integration of (14.3.15) gives h(x)=x⎪integraldisplay 0ds (1−s2)s⎪integraldisplay −1f(t)dt−A, where Ais an arbitrary constant. Clearly, h(x) satisﬁes the conditions of Theorem 14.3.1, and there exists a positive real constant m<1 such that |h(x)|=O{(1−x2)−m}as|x|→1. Hence, Tn{R[h(x)]}exists, and by Theorem 14.3.1, it follows that Tn{R[h(x)]}=−n(n+1 )Tn{h(x)}=−n(n+1 )Tn{R−1[f(x)]},(14.3.16) whence it turns out that Tn{R−1{f(x)}}=−˜f(n) n(n+1 ). (14.3.17) Inversion leads to the result T−1 n⎪braceleftbiggf(n) n(n+1 )⎪bracerightbigg =−R−1{f(x)}=−h(x) =A−x⎪integraldisplay 0ds 1−s2s⎪integraldisplay −1f(t)dt. (14.3.18) © 2007 by Taylor & Francis Group, LLC 494 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This proves the theorem. THEOREM 14.3.3 Iff(x) is continuous in each subinterval of ( −1,1) and a continuous function g(x) is deﬁned by g(x)=x⎪integraldisplay −1f(t)dt, (14.3.19) then Tn{g/prime(x)}=˜f(n)=g(1)−1⎪integraldisplay −1g(x)P/prime n(x)dx. (14.3.20) PROOF We have, by deﬁnition, Tn{g/prime(x)}=1⎪integraldisplay −1g/prime(x)Pn(x)dx, which is, by integrating by parts, =[Pn(x)g(x)]1 −1−1⎪integraldisplay −1g(x)P/prime n(x)dx. Since Pn(1) = 1 and g(−1)= 0, the preceding result becomes (14.3.20). COROLLARY 14.3.2 If result (14.3.20) is true and g(x) is given by (14.3.19), then Tn{g(x)}=f(0)−f(1) when n=0 =˜f(n−1)−˜f(n+1 ) (2n+1 )when n>1⎫ ⎪⎬ ⎪⎭.(14.3.21) PROOF We write ˜f(n−1) and ˜f(n+ 1) using (14.3.20) and then subtract so that the resulting expression gives (14.3.21) with the help of (14.2.16). COROLLARY 14.3.3 Ifg/prime(x) is a sectionally continuous function and g(x) is the continuous function © 2007 by Taylor & Francis Group, LLC Legendre Transforms 495 given by (14.3.19), then Tn{g/prime(x)}=g(1),when n=0 =g(1)−(2n−1) ˜g(n−1)−(2n−5) ˜g(n−3)−···− g(0) when n=1,3,5,... =g(1)−2(2n−1)˜g(n−1)−(2n−5)˜g(n−3)−···− 3g(1) when n=2,4,6,...⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎪⎭. (14.3.22) These results can readily be veriﬁed using (14.3.20) and (14.2.16). THEOREM 14.3.4 (Convolution ). IfT n{f(x)}=˜f(n)a n d Tn{g(x)}=˜g(n), then Tn{f(x)}∗g(x)=˜f(n)˜g(n), (14.3.23) where the convolution f(x)∗g(x)i sg i v e nb y f(x)∗g(x)=h(x)=1 ππ⎪integraldisplay 0f(cosμ)sinμd μπ⎪integraldisplay 0g(cosλ)dβ, (14.3.24) with x=c o s vand cos λ=c o sμcosv+s i nμsinvcosβ. (14.3.25) PROOF We have, by deﬁnition (14.2.2), ˜f(n)˜g(n)=π⎪integraldisplay 0f(cosμ)Pn(cosμ)sinμd μπ⎪integraldisplay 0g(cosλ)Pn(cosλ)sinλdλ =π⎪integraldisplay 0f(cosμ)sinμ⎡ ⎣π⎪integraldisplay 0g(cosλ)Pn(cosλ)Pn(cosμ)sinλdλ⎤⎦dμ, (14.3.26) where f(x)=f(cosμ)a n d g(x)=g(cosλ). With the aid of an addition formula (see Sansone, 1959, p. 169) given as P n(cosλ)Pn(cosμ)=1 ππ⎪integraldisplay 0Pn(cosv)dα, (14.3.27) © 2007 by Taylor & Francis Group, LLC 496 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where cos v=c o sλcosμ+s i nλsinμcosα, the product can be rewritten in the form ˜f(n)˜g(n)=1 ππ⎪integraldisplay 0f(cosμ)sinμ ×⎡ ⎣π⎪integraldisplay 0π⎪integraldisplay 0g(cosμ)Pn(cosμ)sinλd αd λ⎤ ⎦dμ.(14.3.28) We next use Churchill and Dolph’s (1954, pp. 94–96) geometrical arguments to replace the double integral inside the square bracket by π⎪integraldisplay 0π⎪integraldisplay 0g(cosμcosv+s i nμsinvcosβ)Pn(cosv)sinvd v . (14.3.29) Substituting this result in (14.3.26) and changing the order of integration, we obtain ˜f(n)˜g(n)=1 ππ⎪integraldisplay 0Pn(cosv)sinv⎡ ⎣π⎪integraldisplay 0π⎪integraldisplay 0f(cosμ)sinμg(cosλ)dμ dβ⎤ ⎦dv =π⎪integraldisplay 0h(cosv)Pn(cosv)sinvd v , (14.3.30) where cosλ=c o sμcosv+s i nμsinvcosβ, (14.3.31) and h(cosv)=1 ππ⎪integraldisplay 0f(cosμ)sinμd μπ⎪integraldisplay 0g(cosλ)dβ. This proves the theorem. In particular, when v= 0, result (14.3.24) becomes h(1) =1⎪integraldisplay −1f(t)g(−t)dt, (14.3.32) and when v=π, (14.3.24) gives h(−1) =1⎪integraldisplay −1f(t)g(−t)dt. (14.3.33) © 2007 by Taylor & Francis Group, LLC Legendre Transforms 497 14.4 Applications of Legendre Transforms to Boundary Value Problems We solve the Dirichlet problem for the potential u(r, θ) inside a unit sphere r= 1, which satisﬁes the Laplace equation ∂ ∂r⎪bracketleftbigg r2∂u ∂r⎪bracketrightbigg +∂ ∂x⎪bracketleftbigg (1−x2)∂u ∂x⎪bracketrightbigg =0,0<r< 1, (14.4.1) with the boundary condition ( x=c o sθ) u(1,x)=f(x),−1<x< 1. (14.4.2) We introduce the Legendre transform ˜u(r, n)=Tn{u(r, θ)}deﬁned by (14.2.1). Application of this transform to (14.4.1)–(14.4.2) gives r2d2˜u(r, n) dr2+2rd˜u dr−n(n+1 )˜u(r, n)=0, (14.4.3) ˜u(1,n)=˜f(n), (14.4.4) where ˜ u(r, n) is to be continuous function for rfor 0≤r<1. The bounded solution of (14.4.3)–(14.4.4) is ˜u(r, n)=˜f(n)rn,0≤r<1,forn=0,1,2,3,... (14.4.5) Thus, the solution for u(r, x) can be found by the inverse transform so that u(r, x)=∞⎪summationdisplay n=0⎪parenleftbigg n+1 2⎪parenrightbigg ˜f(n)rnPn(x)f o r 0 <r≤1,|x|<1.(14.4.6) The Convolution Theorem allows us to give another representation of the solutiuon. In view of (14.2.11), we ﬁnd T−1 n{rn}=1 2(1−r2)(1−2rx+r2)−3/2. Thus, it follows from (14.4.5) that u(r,cosθ)=T−1 n{˜f(n)rn} =1 2ππ⎪integraldisplay 0f(cosμ)sinμd μπ⎪integraldisplay 0(1−r2)dλ (1−2rcosv+r2)3/2,(14.4.7) where cosv=c o sμcosθ+s i nμsinθcosλ. (14.4.8) © 2007 by Taylor & Francis Group, LLC 498 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Integral (14.4.7) is called the Poisson integral formula for the potential inside the unit sphere for the Dirichlet problem. On the other hand, for the Dirichlet exterior problem, the potential w(r,cosθ) outside the unit sphere ( r>1) can be obtained with the boundary condition w(1,cosθ)=f(cosθ). The solution of the Legendre transformed problem is ˜w(r, n)=1 r˜f(n)r−n,n =0,1,2,..., (14.4.9) which is, in terms of w, w(r,cosθ)=1 rw⎪parenleftbigg1 r,cosθ⎪parenrightbigg ,r > 1 (14.4.10) =1 2ππ⎪integraldisplay 0f(cosμ)sinμd μπ⎪integraldisplay 0(r2−1)dλ (1−2rcosv+r2)3/2,(14.4.11) where cos vis given by (14.4.8). 14.5 Exercises 1. Show that, if |r|<1, (a)Tn{xn}=2n+1(n!)2 (2n+1 ) !. (b)Tn⎪bracketleftbigg log⎪braceleftbiggr−x+( 1−2rx+r2)1/2 1−x⎪bracerightbigg⎪bracketrightbigg =2rn+1 (n+ 1)(2 n+1 ). (c)Tn⎪bracketleftbigg⎪braceleftBig 2r(1−rx+r2)−1/2⎪bracerightBig −log⎪braceleftbiggr−x+( 1−2rx+r2)1/2 1−x⎪bracerightbigg⎪bracketrightbigg =2rn+1 (n+1 ). (d)Tn⎪bracketleftbigg −log1 2{1−rx+( 1−2rx+r2)1/2}⎪bracketrightbigg =⎧ ⎨ ⎩0,n =0 2rn n(2n+1 ),n > 0⎫ ⎬ ⎭. (e)Tn⎪bracketleftbigg (1−2rx+r2)−1 2−1 2log⎪braceleftbigg1−rx+( 1−2rx+r2)1/2 2⎪bracerightbigg⎪bracketrightbigg =rn n. 2. Using the recurrence relation for the Legendre polynomials, show that Tn[xf(x)] = (2 n+1 )−1[(n+1 )˜f(n+1 )+ n˜f(n−1)]. Hence, ﬁnd Tn{x2f(x)}. © 2007 by Taylor & Francis Group, LLC Legendre Transforms 499 3. Use the deﬁnition of the even Legendre-transform pairs (Tranter, 1966) T2n{f(x)}=˜f(2n)=1⎪integraldisplay 0f(x)P2n(x)dx, n =0,1,2,... f(x)=T−1 2n{˜f(2n)}=∞⎪summationdisplay n=0(4n+1 )˜f(2n)P2n(x),0<x< 1, to show that T2n⎪bracketleftbiggd dx{(1−x2)f/prime(x)}⎪bracketrightbigg =−2n(2n+1 )˜f(2n)−f/prime(0)P2n(0). Hence, deduce T2n{x}=−P2n(0) (2n−1)(2n+2 ). 4. Use the deﬁnition of the odd Legendre-transform pairs (Tranter, 1966) T2n+1=˜f(2n+1 )=1⎪integraldisplay 0P2n+1(x)f(x)dx, n =0,1,2,.... f(x)=T−1 2n+1{˜f(2n+1 )}=∞⎪summationdisplay n=0(4n+3 )P2n+1(x)˜f(2n+1 ), to prove the result T2n+1⎪bracketleftbiggd dx{(1−x2)f/prime(x)}⎪bracketrightbigg =−(2n+ 1)(2 n+2 )˜f(2n+1 ) +f(0)P/prime 2n+1(0). Hence, derive T2n+1{1}=P/prime 2n+1(0) (2n+ 1)(2 n+2 ). 5. From the deﬁnition of the even Legendre transform , show that T2n{x2r}=22n(2r)!(r+n)! (2r+2n+1 ) ! ( r−n)!. 6. Show that the Legendre transform solution of the Dirichlet boundary value problem for u(r, θ) urr+1 rur+( 1−x2)uxx−2xux=0,0≤r≤a,0≤θ≤π u(a, θ)=f(x),0≤θ≤π, © 2007 by Taylor & Francis Group, LLC 500 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where x=c o s θ,i s ˜u(r, n)=⎪parenleftBigr a⎪parenrightBign˜f(n). Obtain the solution for u(r, θ) with the help of (14.2.11) and the Con- volution Theorem 14.3.4. 7. Solve the problem of the electriﬁed disk for the potential u(ξ,η)w h i c h satisﬁes the equation (see Tranter, 1966, p. 99) ∂ ∂ξ⎪bracketleftbigg (1−ξ2)∂u ∂ξ⎪bracketrightbigg +∂ ∂η⎪bracketleftbigg (1−η2)∂u ∂η⎪bracketrightbigg =0, and the boundary data u(ξ,η)=0 o n η=0,and∂u ∂ξ=0 o n ξ=0, where ( ξ,η) are the oblate spheroidal corrdinates related to the cylin- drical polar coordinates ( r, z)b yr=( 1−ξ2)1/2(1−η2)1/2andz=ξη. © 2007 by Taylor & Francis Group, LLC 15 Jacobi and Gegenbauer Transforms “The real end of science is the honor of the human mind.” Carl Jacobi “... Jacobi possessed not only the impulse to acquire pure scientiﬁc knowledge, but also the desire to impart it. ...” Felix Klein 15.1 Introduction This chapter deals with Jacobi and Gegenbauer transforms and their basic op-erational properties. The former is a fairly general ﬁnite integral transform inthe sense that both Gegenbauer and Lege ndre transforms follow as special cas- es of the Jacobi transform. Some applications of both Jacobi and Gegenbauertransforms are discussed. This chapter is based on papers by Debnath (1963, 1967), Scott (1953), Conte (1955), and Lakshmanarao (1954). In Chapters 12–15, we discussed several special transforms with orthogonal polynomialsas kernels. All these special transforms have been uniﬁed by Eringen (1954) in his paper on the ﬁnite Sturm-Liouville transform. 15.2 Deﬁnition of the Jacobi Transform and Examples Debnath (1963) introduced the Jacobi transform of a function F(x) deﬁned in−1<x< 1 by the integral J{F(x)}=f(α,β)(n)=1⎪integraldisplay −1(1−x)α(1 +x)βP(α,β) n(x)F(x)dx, (15.2.1) where P(α,β) n(x) is the Jacobi polynomial of degree nand orders α(>−1) and β(>−1). 501 © 2007 by Taylor & Francis Group, LLC 502 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We assume that F(x) admits the following series expansion F(x)=∞⎪summationdisplay n=1anP(α,β) n(x). (15.2.2) In view of the orthogonal relation 1⎪integraldisplay −1(1−x)α(1 +x)βP(α,β) n(x)P(α,β) m(x)dx=δnδmn, (15.2.3) where δnmis the Kronecker delta symbol, δn=2α+β+1Γ(n+α+1 ) Γ ( n+β+1 ) n!(α+β+2n+1 ) Γ ( n+α+β+1 ), (15.2.4) and the coeﬃcients anin (15.2.2) are given by an=1 δn1⎪integraldisplay −1(1−x)α(1 +x)βF(x)P(α,β) n(x)dx=f(α,β)(n) δn. (15.2.5) Thus, the inverse Jacobi transform is given by J−1{f(α,β)(n)}=F(x)=∞⎪summationdisplay n=0(δn)−1f(α,β)(n)P(α,β) n(x). (15.2.6) Note that both JandJ−1are linear transformations. Example 15.2.1 IfF(x) is a polynomial of degree m<n ,t h e n J{F(x)}=0. (15.2.7) Example 15.2.2 J{P(α,β) m(x)}=δmn. (15.2.8) Example 15.2.3 From the uniformly convergent expansion of the generating function for |z|<1 2α+βQ−1(1−z+Q)−α(1 +z+Q)−β=∞⎪summationdisplay n=0znP(α,β) n(x), (15.2.9) © 2007 by Taylor & Francis Group, LLC Jacobi and Gegenbauer Transforms 503 where Q=⎪parenleftbig 1−2xz+z2⎪parenrightbig1 2,it turns out that J{2α+βQ−1(1−z+Q)−α(1 +z+Q)−β} =∞⎪summationdisplay n=0zn1⎪integraldisplay −1(1−x)α(1 +x)βP(α,β) n(x)P(α,β) n(x)dx =∞⎪summationdisplay n=0(δn)zn. (15.2.10) Example 15.2.4 J{xn}=1⎪integraldisplay −1(1−x)α(1 +x)βP(α,β) n(x)xndx =2n+α+β+1Γ(n+α+1 ) Γ ( n+β+1 ) Γ(n+α+β+1 ). (15.2.11) Example 15.2.5 Ifp>β−1, then J{(1 +x)p−β}=1⎪integraldisplay −1(1−x)α(1 +x)pP(α,β) n(x)dx =⎪parenleftbiggn+α n⎪parenrightbigg 2α+p+1Γ(p+1 ) Γ ( α+1 ) Γ ( p−β+1 ) Γ(α+p+n+2 ) Γ ( p−β+n+1 ).(15.2.12) In particular, when α=β= 0, the above results reduce to the corresponding results for the Legendre transform deﬁned by (14.2.1) so that Tn{(1 +x)p}=1⎪integraldisplay −1(1 +x)pPn(x)dx =2p+1{Γ(1 + p)}2 Γ(p+n+2 ) Γ ( p+n+1 ),(p>−1).(15.2.13) © 2007 by Taylor & Francis Group, LLC 504 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 15.2.6 If Re σ>−1, then J{(1−x)σ−α}=1⎪integraldisplay −1(1−x)σ(1 +x)βP(α,β) n(x)dx, Reσ>−1, =2σ+β+1 n!Γ (α−σ)·Γ(σ+1 ) Γ ( n+β+1 ) Γ ( α−σ+n) Γ(β+σ+n+2 ),(15.2.14) Example 15.2.7 If Re σ>−1, then J{(1 +x)σ−βP(α,σ) m(x)}=1⎪integraldisplay −1(1−x)α(1 +x)σP(α,β) n(x)P(α,σ) m(x)dx =2α+σ+1Γ(n+α+1 ) Γ ( α+β+m+n+1 ) Γ ( σ+m+1 ) m!(n−m)! Γ(α+β+n+1 ) Γ ( α+σ+m+n+2 ) ×Γ(σ−β+1 ) Γ(α−β+m+1 ). (15.2.15) 15.3 Basic Operational Properties THEOREM 15.3.1 IfJ{F(x)}=f(α,β)(n), lim |x|→1(1−x)α+1(1 +x)β+1F(x)=0, (15.3.1a) lim |x|→1(1−x)α+1(1 +x)β+1F/prime(x)=0, (15.3.1b) R[F(x)] = (1 −x)−α(1 +x)−βd dx⎪bracketleftbigg (1−x)α+1(1 +x)β+1d dxF(x)⎪bracketrightbigg ,(15.3.2) thenJ{R[F(x)]}exists and is given by J{R[F(x)]}=−n(n+α+β+1 )f(α,β)(n), (15.3.3) where n=0,1,2,3,.... © 2007 by Taylor & Francis Group, LLC Jacobi and Gegenbauer Transforms 505 PROOF We have, by deﬁnition, J{R[F(x)]}=1⎪integraldisplay −1d dx⎪bracketleftbigg (1−x)α+1(1 +x)β+1dF dx⎪bracketrightbigg P(α,β) n(x)dx, which is, by integrating by parts and using the orthogonal relation (15.2.3), =−n(n+α+β+1 )1⎪integraldisplay −1(1−x)α(1 +x)βP(α,β) n(x)F(x)dx =−n(n+α+β+1 )f(α,β)(n). This completes the proof. IfF(x)a n d R[F(x)] satisfy the conditions of Theorem 15.3.1, then J{[R[F(x)]]}exists and is given by J{R2[F(x)]}=J{R[R[F(x)]]}=(−1)2n2(n+α+β+1 )2f(α,β)(n).(15.3.4) More generally, if F(x)a n d Rk[F(x)] satisfy the conditions of Theorem 15.3.1, where k=1,2,···,m−1, and mis a positive integer then J{Rm[F(x)]}=(−1)mnm(n+α+β+1 )mf(α,β)(n). (15.3.5) When α=β=0,P(0,0) n(x) becomes the Legendre polynomial Pn(x)a n dt h e Jacobi transform pairs (15.2.1) and (15.2.5) reduce to the Legendre transform pairs (14.2.1) and (14.2.3). All results for the Jacobi transform also reduce to those given in Chapter 14. 15.4 Applications of Jacobi Transforms to the Generalized Heat Conduction Problem The one-dimensional generalized heat equation for temperature u(x, t)i s ∂ ∂x⎪bracketleftbigg κ∂u ∂x⎪bracketrightbigg +Q(x, t)=ρc∂u ∂t, (15.4.1) where κis the thermal conductivity, Q(x, t) is a continuous heat source within the medium, ρandcare density and speciﬁc heat respectively. If the thermal conductivity is κ=a(1−x2), where ais a real constant, and the source is Q(x, t)=(μx+ν)∂u ∂x, then the heat equation (15.4.1) reduces to ∂ ∂x⎪bracketleftbigg (1−x2)∂u ∂x⎪bracketrightbigg +⎪parenleftbiggμx+ν a⎪parenrightbigg∂u ∂x=⎪parenleftBigρc a⎪parenrightBig∂u ∂t. (15.4.2) © 2007 by Taylor & Francis Group, LLC 506 INTEGRAL TRANSFORMS and THEIR APPLICATIONS We consider a non-homogeneous beam with ends at x=±1w h o s el a t e r a l surface is insulated. Since κ= 0 at the ends, the ends of the beam are also insulated. We assume the initial conditions as u(x,0) =G(x) for all −1<x< 1, (15.4.3) where G(x) is a suitable function so that J{G(x)}exists. If we writeμ a=−(α+β)a n dν a=β−αso that ( α, β)=−⎪parenleftbiggμ+ν 2a,μ−ν 2a⎪parenrightbigg , the left-hand side of (15.4.2) becomes ∂ ∂x⎪bracketleftbigg (1−x2)∂u ∂x⎪bracketrightbigg +[ (β−α)−(β+α)x]∂u ∂x =∂ ∂x⎪bracketleftbigg (1−x2)∂u ∂x⎪bracketrightbigg +[ ( 1−x)β−(1 +x)α]∂u ∂x =( 1−x)−α(1 +x)−β⎪braceleftbigg (1−x)α(1 +x)β∂ ∂x⎪bracketleftbigg (1−x2)∂u ∂x⎪bracketrightbigg +⎪bracketleftbig β(1 +x)β(1−x)α+1−α(1−x)α(1 +x)β+1⎪bracketrightbig∂u ∂x⎪bracerightbigg =( 1−x)−α(1 +x)−β⎪braceleftbigg∂ ∂x⎪bracketleftbigg (1−x)α+1(1 +x)β+1∂u ∂x⎪bracketrightbigg⎪bracerightbigg =R[u(x, t)]. Thus, equation (15.4.2) reduces to R[u(x, t)] =⎪parenleftbigg1 d⎪parenrightbigg∂u ∂t,d=⎪parenleftbigga ρc⎪parenrightbigg . (15.4.4) Application of the Jacobi transform to (15.4.4) and (15.4.3) gives d dtu(α,β)(n,t)=−dn(n+α+β+1 )u(α,β)(n,t), (15.4.5) u(α,β)(n,0) =g(α,β)(n). (15.4.6) The solution of this system is u(α,β)(n,t)=g(α,β)(n)e x p [−n(n+α+β+1 )td]. (15.4.7) Theinverse Jacobi transform gives the formal solution u(x, t)=∞⎪summationdisplay n=0δ−1 ng(α,β)(n)P(α,β) n(x)e x p [−n(n+α+β+1 )td],(15.4.8) where α=−1 2a(μ+ν)a n d β=1 2a(μ−ν). © 2007 by Taylor & Francis Group, LLC Jacobi and Gegenbauer Transforms 507 15.5 The Gegenbauer Transform and Its Basic Operational Properties When α=β=ν−1 2,t h eJacobi polynomial P(α,β) n(x) becomes the Gegenbauer polynomial Cν n(x) which satisﬁes the self-adjoint diﬀerential form d dx⎪bracketleftbigg (1−x2)ν+1 2dy dx⎪bracketrightbigg +n(n+2ν)(1−x2)ν−1y=0, (15.5.1) and the orthogonal relation 1⎪integraldisplay −1(1−x2)ν−1 2Cν m(x)Cν n(x)dx=δnδmn, (15.5.2) where δn=21−2νπΓ(n+2ν) n!(n+ν)[Γ(ν)]2. (15.5.3) Thus, when α=β=ν−1 2, the Jacobi transform pairs (15.2.1) and (15.2.6) reduce to the Gegenbauer transform pairs, in the form G{F(x)}=f(ν)(n)=1⎪integraldisplay −1(1−x2)ν−1 2Cν n(x)F(x)dx, (15.5.4) G−1{f(ν)(n)}=F(x)=∞⎪summationdisplay n=0δ−1 nCν n(x)f(ν)(n),−1<x< 1.(15.5.5) Obviously, GandG−1stand for the Gegenbauer transformation and its inverse respectively. They are linear i ntegral transformations. When α=β=ν−1 2, the diﬀerential form (15.3.2) becomes R[F(x)] = (1 −x2)d2F dx2−(2ν+1 )xdF dx, (15.5.6) which can be expressed as R[F(x)] = (1 −x2)1 2−νd dx⎪bracketleftbigg (1−x2)ν+1 2dF dx⎪bracketrightbigg . (15.5.7) Under the Gegenbauer transformation G, the diﬀerential form (15.5.6) is reduced to the algebraic form G{R[F(x)]}=−n(n+2ν)f(ν)(n). (15.5.8) © 2007 by Taylor & Francis Group, LLC 508 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This follows directly from the relation (15.3.3). Similarly, we obtain G{R2[F(x)]}=(−1)2n2(n+2ν)2f(ν)(n). (15.5.9) More generally, G{Rk[F(x)]}=(−1)knk(n+2ν)kf(ν)(n), (15.5.10) where k=1,2,.... Convolution Theorem 15.5.1 IfG{F(x)}=f(ν)(n)a n d G{G(x)}= g(ν)(n), then f(ν)(n)g(ν)(n)=G{H(x)}=h(ν)(n), (15.5.11) where H(x)=G−1{h(ν)(n)}=G−1{f(ν)(n)g(ν)(n)}=F(x)∗G(x),(15.5.12) andH(x)i sg i v e nb y H(cosψ)=A(sinψ)1−2νπ⎪integraldisplay 0π⎪integraldisplay 0F(cosθ)G(cosφ)(sinθ)2ν ×(sinφ)2ν−1(sinλ)2ν−1dθ dα, (15.5.13) where αis deﬁned by (15.5.19). PROOF We have, by deﬁnition, f(ν)(n)g(ν)(n)=1⎪integraldisplay −1F(x)(1−x2)v−1 2Cν n(x)dx ×1⎪integraldisplay −1G(x)(1−x2)ν−1 2Cν n(x)dx =π⎪integraldisplay 0F(cosθ)(sinθ)2νCν n(cosθ)dθ ×π⎪integraldisplay 0G(cosφ)(sinφ)2νCν n(cosφ)dφ =π⎪integraldisplay 0F(cosθ)(sinθ)2ν[π⎪integraldisplay 0G(cosφ)Cν n(cosθ) ×Cν n(cosφ)(sinφ)2νdφ]dθ.(15.5.14) © 2007 by Taylor & Francis Group, LLC Jacobi and Gegenbauer Transforms 509 The addition formula for the Gegenbauer polynomial (see Erd´ elyi, 1953, p. 177) is Cν n(cosθ)Cν n(cosφ)=Aπ⎪integraldisplay 0Cν n(cosψ)(sinλ)2ν−1dλ, (15.5.15) where A={Γ(n+2ν)/n!22ν−1Γ2(ν)}, (15.5.16) and cosψ=c o sθcosφ+s i nθsinφcosλ. (15.5.17) In view of this formula, result (15.5.14) assumes the form f(ν)(n)g(ν)(n) =Aπ⎪integraldisplay 0F(cosθ)(sinθ)2ν[π⎪integraldisplay 0π⎪integraldisplay 0G(cosφ)Cν n(cosψ) ×(sinφ)2ν(sinλ)2ν−1dλ dφ ]dθ.(15.5.18) We next introduce a new variable αdeﬁned by the relation cosφ=c o sθcosψ+s i nθsinψcosα. (15.5.19) Thus, under transformation of coordinates deﬁned by (15.5.17) and (15.5.19), the elementary area dλ dφ =( s i n ψ/sinφ)dψ dα ,w h e r e( s i n ψ/sinφ)i st h e Jacobian of the transformation. In view of this transformation, the squareregion of the φ-λplane given by (0 ≤φ≤π,0≤λ≤π) transforms into a square region of the same dimension in the ψ-αplane. Consequently, the double integral inside the square bracket in (15.5.18) reduces to π⎪integraldisplay 0π⎪integraldisplay 0G(cosφ)Cν n(cosψ)(sinφ)2ν−1(sinλ)2ν−1sinψd ψd α , (15.5.20) where cos ψis deﬁned by (15.5.17) and cos φis deﬁned by (15.5.19). If the double integral (15.5.20) is substituted into (15.5.18), and if the order of in- tegration is interchanged, (15.5.18) becomes f(ν)(n)g(ν)(n)=π⎪integraldisplay 0(sinψ)2νCν n(cosψ)H(cosψ)dψ=G{H(cosψ)},(15.5.21) where H(cosψ)=A(sinψ)1−2νπ⎪integraldisplay 0π⎪integraldisplay 0F(cosθ)G(cosφ)(sinθ)2ν ×(sinφ)2ν−1(sinλ)2ν−1dθ dα. (15.5.22) © 2007 by Taylor & Francis Group, LLC 510 INTEGRAL TRANSFORMS and THEIR APPLICATIONS When ν=1 2,C1 2n(x) becomes the Legendre polynomial, the Gegenbauer trans- form pairs (15.5.4) and (15.5.5) reduce to the Legendre transform pairs (14.2.1)and (14.2.3), and the Convolution Theorem 15.5.1 reduces to the correspond- ing Convolution Theorem 14.3.4 for the Legendre transform. 15.6 Application of the Gegenbauer Transform The generalized one-dimensional heat equation in a non-homogeneous solid beam for the temperature u(x, t)i s ∂ ∂x⎪bracketleftbigg (1−x2)∂u ∂x⎪bracketrightbigg −(2ν+1 )x∂u ∂x=1 d∂u ∂t, (15.6.1) where κ=( 1−x2) is the thermal conductivity, d=⎪parenleftbigga ρc⎪parenrightbigg , and the second term on the left hand side represents the continuous source of heat within the solid beam. We assume that the beam is bounded by the planes at x=±1 and its lateral surfaces are insulated. The initial condition is u(x,0)=G(x)f o r −1<x< 1, (15.6.2) where G(x) is a given function so that its Gegenbauer transform exists. Application of the Gegenbauer transform to (15.6.1) and (15.6.2) and the use of (15.5.8) gives d dtu(ν)(n,t)=−dn(n+2ν)u(ν)(n,t), (15.6.3) u(ν)(n,0) =g(ν)(n). (15.6.4) This solution of this system is u(ν)(n,t)=g(ν)(n)e x p [−n(n+2ν)td]. (15.6.5) The inverse transform gives the formal solution u(x, t)=∞⎪summationdisplay n=0δ−1 nCν n(x)g(ν)(n)e x p [−n(n+2ν)td], (15.6.6) where δnis given by (15.5.3). © 2007 by Taylor & Francis Group, LLC 16 Laguerre Transforms “The search for truth is more precious than its possession.” Albert Einstein “Nature is an inﬁnite sphere of which the center is everywhere and the circumference nowhere.” Blaise Pascal “Mathematics is the tool specially suited for dealing with abstract concepts of any kind and there is no limit to its power in this ﬁeld.” Paul A. M. Dirac 16.1 Introduction This chapter is devoted to the study of the Laguerre transform and its basicoperational properties. It is shown that the Laguerre transform can be usedeﬀectively to solve the heat conduction problem in a semi-inﬁnite medium with variable thermal conductivity in the presence of a heat source within the medium. This chapter is based on a series of papers by Debnath (1960–1962) and McCully (1960) listed in the Bibliography. 16.2 Deﬁnition of the Laguerre Transform and Examples Debnath (1960) introduced the Laguerre transform of a function f(x) deﬁned in 0≤x<∞by means of the integral L{f(x)}=˜fα(n)=∞⎪integraldisplay 0e−xxαLα n(x)f(x)dx, (16.2.1) 511 © 2007 by Taylor & Francis Group, LLC 512 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where Lα n(x)i st h e Laguerre polynomial of degree n(≥0) and order α(>−1), which satisﬁes the ordinary diﬀerential equation expressed in the self-adjointform d dx⎪bracketleftbigg e−xxα+1d dxLα n(x)⎪bracketrightbigg +ne−xxαLα n(x)=0. (16.2.2) In view of the orthogonal property of the Laguerre polynomials ∞⎪integraldisplay 0e−xxαLα n(x)Lα m(x)dx=⎪parenleftbiggn+α n⎪parenrightbigg Γ(α+1 )δmn=δnδnm,(16.2.3) where δmnis the Kronecker delta symbol, and δnis given by δn=⎪parenleftbiggn+α n⎪parenrightbigg Γ(α+1 ). (16.2.4) Theinverse Laguerre transform is given by f(x)=L−1{˜fα(n)}=∞⎪summationdisplay n=0(δn)−1˜fα(n)Lα n(x). (16.2.5) When α= 0, the Laguerre transform pairs due to McCully (1960) follow from (16.2.1) and (16.2.5) in the form L{f(x)}=˜f0(n)=∞⎪integraldisplay 0e−xLn(x)f(x)dx, (16.2.6) L−1{˜f0(n)}=f(x)=∞⎪summationdisplay n=0˜f0(n)Ln(x), (16.2.7) where Ln(x) is the Laguerre polynomial of degree nand order zero. Obviously, LandL−1are linear integral transformations. The following examples (Debnath, 1960) illustrate th e Laguerre transfor m of some simple functions. Example 16.2.1 Iff(x)=Lα m(x)t h e n L{Lα m(x)}=δnδnm. (16.2.8) This follows directly from the deﬁnitions, (16.2.1) and (16.2.3). Example 16.2.2 Iff(x)=xs−1where sis a positive real number, then L{xs−1}=∞⎪integraldisplay 0e−xxα+s−1Lα n(x)dx=Γ(s+α)Γ(n−s+1 ) n!Γ ( 1−s), (16.2.9) © 2007 by Taylor & Francis Group, LLC Laguerre Transforms 513 in which a result due to Howell (1938) is used. Example 16.2.3 Ifa>−1, and f(x)=e−ax,t h e n L{e−ax}=∞⎪integraldisplay 0e−x(1+a)xαLα n(x)dx=Γ(n+α+1 )an n!(a+1 )n+α+1,(16.2.10) where result in Erd´ elyiet al. (1954, vol. 2, p 191) is used. Example 16.2.4 Iff(x)=e−axLα m(x), then L{e−axLα m(x)}=∞⎪integraldisplay 0e−x(a+1)xαLα n(x)Lα m(x)dx, which is, due to Howell (1938), =1 n!m!Γ(n+α+1 ) Γ ( m+α+1 ) Γ(1 + α)·(a−1)n−m+α+1 an+m+2α+2 ×2F1⎪parenleftbigg n+α+1,m+a+1 a+1,1 a2⎪parenrightbigg ,(16.2.11) where 2F1(x, α, β ) is the hypergeometric function. Example 16.2.5 L{f(x)xβ−α}=∞⎪integraldisplay 0e−xxβLα n(x)f(x)dx. We use a result from Erd´ elyi (1953, vol. 2, p. 192) as Lα n(x)=n⎪summationdisplay m=0(m!)−1(α−β)mLβ n−m(x) (16.2.12) to obtain the following result: L{f(x)xβ−α}=n⎪summationdisplay m=0(m!)−1(α−β)m˜fβ(n−m). (16.2.13) In particular, when β=α−1, we obtain L⎪braceleftbiggf(x) x⎪bracerightbigg =n⎪summationdisplay m=0(m!)−1˜fα−1(n−m). © 2007 by Taylor & Francis Group, LLC 514 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 16.2.6 L{exx−αΓ(α, x)}=∞⎪summationdisplay n=0δn (n+1 ),−1<α< 0. (16.2.14) We use a result from Erd´ elyi (1953, vol. 2, p. 215) as exx−αΓ(α, x)=∞⎪summationdisplay n=0(n+1 )−1Lα n(x),(α>−1,x>0), in the deﬁnition (16.2.1) to derive (16.2.14). Example 16.2.7 Ifβ>0, then L{xβ}=Γ (α+β+1 )∞⎪summationdisplay n=0(−β)nδn Γ(n+α+1 ). (16.2.15) Using the result from Erd´ elyi (1953, vol. 2, p. 214) xβ=Γ (α+β+1 )∞⎪summationdisplay n=0(−β)n Γ(n+α+1 )Lα n(x), where −β<1+m i n⎪parenleftbigg α,α 2−1 4⎪parenrightbigg ,x > 0,α>−1, we can easily obtain (16.2.15). Example 16.2.8 If|z|<1a n d α≥0, then (a)L⎪braceleftbigg (1−z)−(α+1)exp⎪parenleftbiggxz z−1⎪parenrightbigg⎪bracerightbigg =∞⎪summationdisplay n=0δnzn, (16.2.16) (b)L⎪braceleftBig (xz)−α 2ezJα⎪bracketleftBig 2(xz)1 2⎪bracketrightBig⎪bracerightBig =∞⎪summationdisplay n=0δnzn Γ(n+α+1 ). (16.2.17) We have the following generating functions (Erd´ elyi, 1953, vol. 2, p. 189) (1−z)−(α+1)exp⎪parenleftbiggxz z−1⎪parenrightbigg =∞⎪summationdisplay n=0Lα n(x)zn,|z|<1, (xz)−α/2ezJα[2√ xz]=∞⎪summationdisplay n=0znLα n(x) Γ(n+α+1 ),|z|<1. In view of these results combined with the orthogonality relation (16.2.3), we obtain (16.2.16) and (16.2.17). © 2007 by Taylor & Francis Group, LLC Laguerre Transforms 515 Example 16.2.9 (Recurrence Relations ). (a)˜fα+1(n)=(n+α+1 )˜fα(n)−(n+1 )˜fα(n+1 ), (16.2.18) (b)n!˜fm−n(n)=(−1)n−mm!m⎪summationdisplay k=0(k!)−1(2n−2m)k˜fm−n(m−k).(16.2.19) We have ˜fα+1(n)=∞⎪integraldisplay 0e−xxα+1Lα+1 n(x)f(x)dx, which is, by using the recurrence relation for the Laguerre polynomial, =∞⎪integraldisplay 0e−xxα[(n+α+1 )Lα n(x)−(n+1 )Lα n+1(x)]f(x)dx =(n+α+1 )˜fα(n)−(n+1 )˜fα(n+1 ). Similarly, we ﬁnd n!˜fm−n(n)=∞⎪integraldisplay 0e−xxm−nn!Lm−n n(x)f(x)dx. We next use the following result due to Howell (1938) n!Lm−n n(x)=(−1)n−mm!Ln−m m(x) to obtain n!˜fm−n(n)=(−1)n−mm!∞⎪integraldisplay 0e−xxm−nLn−m m(x)f(x)dx =(−1)n−mm!m⎪summationdisplay k=0(k!)−1(2n−2m)k˜fm−n(m−k). © 2007 by Taylor & Francis Group, LLC 516 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 16.3 Basic Operational Properties We obtain the Laguerre transform of derivatives of f(x)a s L{f/prime(x)}=˜fα(n)−αn⎪summationdisplay k=0fα−1(k)+n−1⎪summationdisplay k=0fα(k), (16.3.1) L{f/prime/prime(x)}=˜fα(n)−2αn⎪summationdisplay m=0˜fα−1(n−m)+2n−1⎪summationdisplay m=0˜fα(n−m−1) −2αn−1⎪summationdisplay m=0(m+1 )˜fα+1(n−m−1) +α(α−1)n⎪summationdisplay m=0(m+1 )fα−2(n−m) +n−2⎪summationdisplay m=0(m+1 )˜fα(n−m−2), (16.3.2) and so on for the Laguerre transforms of higher derivatives. We have, by deﬁnition, L{f/prime(x)}=∞⎪integraldisplay 0e−xxαLα n(x)f/prime(x)dx =⎪bracketleftbig e−xnαLα n(x)f(x)⎪bracketrightbig∞ 0+∞⎪integraldisplay 0e−xxαLα n(x)f(x)dx −α∞⎪integraldisplay 0e−xxα−1Lα n(x)f(x)dx−∞⎪integraldisplay 0e−xxα⎪bracketleftbiggd dxLα n(x)⎪bracketrightbigg f(x)dx, which is, due to Erd´ elyi (1954, vol. 2, p. 192), =˜fα(n)−αn⎪summationdisplay k=0˜fα−1(k)+n−1⎪summationdisplay k=0fα(k). Similarly, we can derive (16.3.2). THEOREM 16.3.1 Ifg(x)=x⎪integraldisplay 0f(t)dtso that g(x) is absolutely continuous and g/prime(x) exists, and ifg/prime(x) is bounded and integrable, then ˜fα(n)−˜fα(n−1)= ˜gα(n)−α˜gα−1(n), (16.3.3) © 2007 by Taylor & Francis Group, LLC Laguerre Transforms 517 and L⎧ ⎨ ⎩x⎪integraldisplay 0f(t)dt⎫ ⎬ ⎭=˜f0(n)−˜f0(n−1), (16.3.4) where Lstands for the zero-order Laguerre transform deﬁned by (16.2.6). PROOF We have ˜fα(n)=∞⎪integraldisplay 0e−xxαLα n(x)g/prime(x)dx, which is, by integrating by parts, =∞⎪integraldisplay 0e−xxαLα n(x)g(x)dx−αα⎪integraldisplay 0e−xxα−1Lα n(x)g(x)dx −∞⎪integraldisplay 0e−xxα⎪bracketleftbiggd dxLα n(x)⎪bracketrightbigg g(x)dx. Thus, ˜fα(n)−˜fα(n+1 )=∞⎪integraldisplay 0e−xxα⎪bracketleftbig Lα n(x)−Lα n+1(x)⎪bracketrightbig g(x)dx +α∞⎪integraldisplay 0e−xxα⎪bracketleftbig Lα n+1(x)−Lα n(x)⎪bracketrightbig g(x)dx −∞⎪integraldisplay 0e−xxαd dx⎪bracketleftbig Lα n(x)−Lα n+1(x)⎪bracketrightbig g(x)dx. Thus, ˜fα(n)−˜fα(n+1 )=∞⎪integraldisplay 0e−xxα⎪bracketleftbig Lα n(x)−Lα n+1(x)⎪bracketrightbig g(x)dx +α∞⎪integraldisplay 0e−xxαLα−1 n+1(x)g(x)dx−∞⎪integraldisplay 0e−xxαLα n(x)g(x)dx =−˜gα(n+1 )+ α˜gα−1(n+1 ). This proves (16.3.3). Putting α= 0, and replacing nbyn−1g i v e s ˜g0(n)=˜f0(n)−˜f0(n−1). © 2007 by Taylor & Francis Group, LLC 518 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Or, L⎧ ⎨ ⎩x⎪integraldisplay 0f(t)dt⎫ ⎬ ⎭=˜f0(n)−˜f0(n−1). THEOREM 16.3.2 IfL{f(x)}=˜fα(n) exists, then L{R[f(x)]}=−n˜fα(n), (16.3.5) where R[f(x)] is the diﬀerential operator given by R[f(x)] =exx−αd dx⎪bracketleftbigg e−xxα+1d dxf(x)⎪bracketrightbigg . (16.3.6) PROOF We have, by deﬁnition, L{R[f(x)]}=∞⎪integraldisplay 0Lα n(x)d dx⎪bracketleftbigg e−xxα+1df dx⎪bracketrightbigg dx, which is, by integrating by parts and using (16.2.2), =−n∞⎪integraldisplay 0e−xxαLα n(x)f(x)dx=−n˜fα(n). This completes the proof of the basic operational property. This result can easily be extended as follows: L{R2[f(x)]}=L{R[R[f(x)]]}=(−1)2n2˜fα(n). (16.3.7) More generally, L{Rm[f(x)]}=(−1)mnm˜fα(n), (16.3.8) where mis a non-negative integer. The Convolution Theorem for the Laguerre transform can be stated as follows: THEOREM 16.3.3 (Convolution Theorem ). IfL{f(x)}=˜fα(n)a n d L{g(x)}=˜gα(n), then L−1{˜fα(n)˜gα(n)}=h(x), (16.3.9) © 2007 by Taylor & Francis Group, LLC Laguerre Transforms 519 where h(x) is given by the following repeated integral h(x)=Γ(n+α+1 ) √ πΓ(n+1 )∞⎪integraldisplay 0e−ttαf(t)dtπ⎪integraldisplay 0exp(−√ xtcosφ) ×sin2αφg(x+t+2√ xtcosφ)Jα−1 2(√ xtsinφ)dφ ⎪bracketleftbigg1 2(√ xtsinφ)⎪bracketrightbiggα−1 2.(16.3.10) In order to avoid long proof of this Convolution Theorem 16.3.3, we will not present the proof here, but refer the reader to the article of Debnath (1969). However, when α=0a n d φis replaced by ( π−θ), and the standard result J−1 2(x)=⎪radicalbigg 2 πxcosx (16.3.11) is used, the Convolution Theorem 16.3.3 reduces to that of McCully’s (1960).We now state and prove McCully’s Convolution Theorem as follows: THEOREM 16.3.4 (McCully’s Theorem). If L{f(x)}=˜f 0(n)a n d L{g(x)}=˜g0(n), then L−1{˜f0(n)˜g0(n)}=h(x), (16.3.12) where h(x) is given by the formula h(x)=1 π∞⎪integraldisplay 0e−tf(t)dtπ⎪integraldisplay 0exp(√ xtcosθ)cos(√ xtsinθ) ×g(x+t−2√ xtcosθ)dθ. (16.3.13) PROOF We have, by deﬁnition, ˜f0(n)˜g0(n)=∞⎪integraldisplay 0e−xLn(x)f(x)dx∞⎪integraldisplay 0e−yLn(y)g(y)dy =∞⎪integraldisplay 0e−xf(x)dx∞⎪integraldisplay 0e−yLn(x)Ln(y)g(y)dy. (16.3.14) T h i sc a nb ew r i t t e ni nt h ef o r m ˜f0(n)˜g0(n)=L{h(t)}=∞⎪integraldisplay 0e−tLn(t)h(t)dt. © 2007 by Taylor & Francis Group, LLC 520 INTEGRAL TRANSFORMS and THEIR APPLICATIONS This shows that his the convolution of fandgand has the representation h(x)=f(x)∗g(x). (16.3.15) It follows from a formula of Bateman (1944, p. 457) that Ln(x)Ln(y)=1 ππ⎪integraldisplay 0e√ xycosθcos(√ xysinθ)Ln(x+y−2√ xycosθ)dθ. (16.3.16) In view of this result, (16.3.14) becomes π˜f0(n)˜g0(n)=∞⎪integraldisplay 0e−xf(x)dx⎡ ⎣∞⎪integraldisplay 0e−yg(y)π⎪integraldisplay 0exp(√ xycosθ) ×cos(√ xysinθ)Ln(x+y−2√ xycosθ)dθdy].(16.3.17) Using√ yas the variable of integration combined with polar coordinates, the integral inside the square bracket in (16.3.17) can be reduced to the form ∞⎪integraldisplay 0e−tLn(t)dtπ⎪integraldisplay 0exp(√ xtcosφ)cos(√ xtsinφ) ×g(x+t−2√ xtcosφ)dφ, (16.3.18) so that (16.3.17) becomes ˜f0(n)˜g0(n)=L{h(t)}=∞⎪integraldisplay 0e−tLn(t)h(t)dt, where h(x)i sg i v e nb y h(x)=1 π∞⎪integraldisplay 0e−tf(t)dtπ⎪integraldisplay 0exp(√ xtcosθ)cos(√ xtsinθ) ×g(x+t−2√ xtcosθ)dθ. (16.3.19) This proves the McCully’s Theorem for the Laguerre transform (16.2.6). 16.4 Applications of Laguerre Transforms Example 16.4.1 (Heat Conduction Problem ). The diﬀusion equation for one-dimensional linear ﬂow of heat in a semi-inﬁnite medium 0 ≤x<∞with a source Q(x, t)i nt h e © 2007 by Taylor & Francis Group, LLC Laguerre Transforms 521 medium is ∂ ∂x⎪bracketleftbigg κ∂u ∂x⎪bracketrightbigg +Q(x, t)=ρc∂u ∂t,t > 0, (16.4.1) where κ=κ(x)=λe−xxβis the variable thermal conductivity; Q(x, t)=μ e−xxβ/prime∂u ∂x;ρ=νe−xxβ/prime;λ, μ, ν, andcare constants; and β≥1a n d β−β/prime= 1. Thus, the above equation reduces to ∂ ∂x⎪bracketleftbigg e−xxβ∂u ∂x⎪bracketrightbigg +μ λe−xxβ/prime∂u ∂x=νc λe−xxβ/prime∂u ∂t. (16.4.2) The initial condition is u(x,0) =g(x),0≤x<∞. (16.4.3) Clearly, equation (16.4.2) assumes the form exx−α∂ ∂x⎪parenleftbigg e−xxα+1∂u ∂x⎪parenrightbigg =γ∂u ∂t, (16.4.4) where α=μ λ+β−1a n d γ=νc λ. Application of the Laguerre transform to (16.4.4) gives d dtuα(n,t)=−n γuα(n,t),u α(n,0)=gα(n). Thus, the solution of this system is uα(n,t)=gα(n)exp⎪parenleftbigg −nt γ⎪parenrightbigg . (16.4.5) The inverse transform (16.2.5) gives the formal solution u(x, t)=∞⎪summationdisplay n=0(δn)−1gα(n)Lα n(x)exp⎪parenleftbigg −nt γ⎪parenrightbigg , (16.4.6) where δnis given by (16.2.4). Example 16.4.2 (Diﬀusion Equation ). Solve equation (16.4.1) with κ=xe−x,Q(x, t)=e−xf(t),andρc=e−x. In this case, the diﬀusion equation (16.4.1) becomes ∂u ∂t=ex∂ ∂x⎪parenleftbigg xe−x∂u ∂x⎪parenrightbigg +f(t),0≤x<∞,t >0, (16.4.7) © 2007 by Taylor & Francis Group, LLC 522 INTEGRAL TRANSFORMS and THEIR APPLICATIONS has to be solved with the initial-boundary data u(x,0) =g(x), 0≤x<∞ ∂ ∂tu(x, t)=f(x),att=0,forx>0⎫ ⎬ ⎭. (16.4.8) Application of the Laguerre transform L{u(x, t)}=˜u0(n,t) to (16.4.7)– (16.4.8) gives ˜u0(n,t)=g0(n)e−nt,n=1,2,3,... (16.4.9) ˜u0(0,t)=g0(0) +t⎪integraldisplay 0f(τ)dτ. (16.4.10) The inverse Laguerre transform (16.2.5) leads to the formal solution u(x, t)=g0(0) +t⎪integraldisplay 0f(τ)dτ+∞⎪summationdisplay n=1g0(n)e−ntLn(x) =t⎪integraldisplay 0f(τ)dτ+∞⎪summationdisplay n=0g0(n)e−ntLn(x). (16.4.11) In view of the Convolution Theorem 16.3.4, this result takes the form u(x, t)=t⎪integraldisplay 0f(τ)dτ+1 π∞⎪integraldisplay 0e−τ(eτ−1)−1exp⎪parenleftbigg−τ et−1⎪parenrightbigg ×π⎪integraldisplay 0exp(√ xτcosθ)cos (√ xτsinθ)g(x+τ−2√ xτcosθ)dθdτ. (16.4.12) This result is obtained by McCully (1960). Another application of the Laguerre transform to the problem of oscillations of a very long and heavy chain with variable tension was discussed by Debnath (1961). We conclude this chapter by adding references of recent work on the Laguerre- Pinney transformation and the Wiener-Laguerre transformation by Glaeske(1981, 1986). For more details, the reader is referred to these papers. © 2007 by Taylor & Francis Group, LLC Laguerre Transforms 523 16.5 Exercises 1. Find the zero-order Laguerre transform of each of the following func- tions: (a)H(x−a)f o rc o n s t a n t a≥0, (c)ALm(x), (d) xm,(b)e−ax(a>−1), (e)Ln(x). 2. IfL{f(x)}=f0(n)=∞⎪integraldisplay 0e−xLn(x)f(x)dx,a n d a>0, show that (a)L{sinax}=an (1 +a2)n+1 2sin⎪bracketleftbigg ntan−1⎪parenleftbigg1 a⎪parenrightbigg +t a n−1(−a)⎪bracketrightbigg , (b)L{cosax}=an (1 +a2)n+1 2cos⎪bracketleftbigg ntan−1⎪parenleftbigg1 a⎪parenrightbigg +t a n−1(−a)⎪bracketrightbigg . 3. IfL{f(x)}=˜f0(n)=∞⎪integraldisplay 0e−xLn(x)f(x)dx, prove the following properties: (a)L{xf/prime(x)}=−(n+1 )˜f0(n+1 )+ n˜f0(n), (b)L⎪bracketleftbigg exd dx{xe−xf/prime(x)}⎪bracketrightbigg =−n˜f0(n), (c)L⎪bracketleftbigg e−xd dx{xexf/prime(x)}⎪bracketrightbigg =n˜f0(n)−2(n+1 )˜f0(n+1 ) , (d)L⎪bracketleftbiggd dx{xf/prime(x)}⎪bracketrightbigg =−(n+1 )˜f0(n+1 ) . 4. Show that (a)˜fα(n)=L{Lα n(x)}=Γ(n+α+1 ) n!forα>−1. (b)˜fα(n)=L{xLα n(x)}=Γ(n+α+1 ) n!(2n+α+1 ) f o r α>−1. © 2007 by Taylor & Francis Group, LLC 17 Hermite Transforms “We are servants rather than masters in mathematics.” Charles Hermite “Success [in teaching] depends ... to a great extent upon the teach- er’s leading the student continually to some research. This howeverdoes not occur by chance ... but chieﬂy as follows ... through his arrangement of the material and em phasis, the teach er’s presenta- tion of lectures on a discipline let s the student discern leading ideas appropriately. In these ways, the f ully conversant thinker logical- ly advances from mature and previous research and attains newresults or better foundations than exist. Next the teacher should not fail to designate boundaries not yet crossed by science and to point out some positions from which further advances would thenbe possible. A university teacher should also not deny the student a deeper insight into the progress of his own investigations, nor should he remain silent about his own past errors and disappoint- ments.” Karl Weierstrass 17.1 Introduction In this chapter we introduce the Hermite transform with a kernel involvinga Hermite polynomial and discuss its basic operational properties, includingthe convolution theorem. Debnath (1964) ﬁrst introduced this transform and proved some of its basic operational properties. This chapter is based on pa- pers by Debnath (1964, 1968) and Dimovski and Kalla (1988). 525 © 2007 by Taylor & Francis Group, LLC 526 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 17.2 Deﬁnition of the Hermite Transform and Examples Debnath (1964) deﬁned the Hermite transform of a function F(x) deﬁned in −∞<x< ∞by the integral H{F(x)}=fH(n)=∞⎪integraldisplay −∞exp(−x2)Hn(x)F(x)dx, (17.2.1) where Hn(x) is the well-known Hermite polynomial of degree n. Theinverse Hermite transform is given by H−1{fH(n)}=F(x)=∞⎪summationdisplay n=0(δn)−1fH(n)Hn(x), (17.2.2) where δnis given by δn=√ πn!2n. (17.2.3) This follows from the expansion of any function F(x)i nt h ef o r m F(x)=∞⎪summationdisplay n=0anHn(x), (17.2.4) where the coeﬃcients ancan be determined from the orthogonal relation of the Hermite polynomial Hn(x)a s ∞⎪integraldisplay −∞exp(−x2)Hn(x)Hn(x)dx=δnmδn. (17.2.5) Multiplying (17.2.4) by exp( −x2)Hm(x) and integrating over ( −∞,∞)a n d using (17.2.4), we obtain an=δ−1 nfH(n) (17.2.6) so that (17.2.2) follows immediately. Example 17.2.1 IfF(x) is a polynomial of degree m,t h e n fH(n)=0 f o r n>m . (17.2.7) © 2007 by Taylor & Francis Group, LLC Hermite Transforms 527 Example 17.2.2 IfF(x)=Hm(x), then H{Hm(x)}=∞⎪integraldisplay −∞exp(−x2)Hn(x)Hm(x)dx=δnδnm. (17.2.8) Example 17.2.3 If exp(2 xt−t2)=∞⎪summationdisplay n=0tn n!Hn(x) (17.2.9) is the generating function of Hn(x), then H{exp(2xt−t2)}=√ π∞⎪summationdisplay n=0(2t)n,|t|<1 2. (17.2.10) We have, by deﬁnition, H{exp(2xt−t2)}=∞⎪summationdisplay n=0tn n!∞⎪integraldisplay −∞exp(−x2)H2 n(x)ds =∞⎪summationdisplay n−0δntn n!=√ π∞⎪summationdisplay n=0(2t)n, |t|<1 2. Example 17.2.4 IfF(x)=Hm(x)Hp(x), then H{Hm(x)Hp(x)}=⎧ ⎪⎪⎨ ⎪⎪⎩√ π2km!n!p! (k−m)!(k−n)!(k−p)!,m+n+p=2k, k≥m, n, p 0, otherwise⎫ ⎪⎪⎬ ⎪⎪⎭. (17.2.11) This follows from a result proved by Bailey (1939). Example 17.2.5IfF(x)=H 2 m(x)Hn(x), then H⎪braceleftbig Hm2(x)Hn(x)⎪bracerightbig =2mδnn⎪summationdisplay k=0⎪parenleftbiggm k⎪parenrightbigg⎪parenleftbiggn k⎪parenrightbigg⎪parenleftbigg2k k⎪parenrightbigg ,ifm>n . (17.2.12) Using a result proved by Feldheim (1938), (17.2.12) follows immediately. © 2007 by Taylor & Francis Group, LLC 528 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 17.2.6 IfF(x)=Hn+p+q(x)Hp(x)Hq(x), then H{F(x)}=δn+p+q. (17.2.13) We have, by deﬁnition, H{F(x)}=∞⎪integraldisplay −∞exp(−x2)Hn+p+q(x)Hp(x)Hq(x)dx=δn+p+q, where a result due to Bailey (1939) is used and δnis given by (17.2.3). Example 17.2.7IfF(x)=e x p ( ax), then H{exp(ax)}=√ π⎪summationdisplay anexp⎪parenleftbigg1 4a2⎪parenrightbigg . (17.2.14) This result follows from the standard result ∞⎪integraldisplay −∞exp(−x2+2bx)Hn(x)dx=√ π(2b)nexp(b2). Example 17.2.8If|2z|<1, show that H{exp(z 2)sin (√ 2xz)}=⎧ ⎪⎪⎨ ⎪⎪⎩0,n /negationslash=2m+1 √ π∞⎪summationdisplay m=0(−1)m(2z)2m+1,n=2m+1⎫ ⎪⎪⎬ ⎪⎪⎭. (17.2.15) We have, by deﬁnition, H{exp(z2)sin (√ 2xz)}=∞⎪integraldisplay −∞exp(z2−x2)Hn(x)sin(√ 2xz)dx. We use a result (see Erd´ elyi et al., 1954, vol. 2, p. 194) exp(z2)sin (√ 2xz)=∞⎪summationdisplay m=0(−1)mH2m+1(x)z2m+1 (2m+1 ) !, (17.2.16) © 2007 by Taylor & Francis Group, LLC Hermite Transforms 529 to derive H{exp(z2)sin (√ 2xz)} =∞⎪summationdisplay m=0(−1)mz2m+1 (2m+1 ) !∞⎪integraldisplay −∞exp(−x2)Hn(x)H2m+1(x)dx =⎧ ⎪⎪⎨ ⎪⎪⎩√ π∞⎪summationdisplay m=0(−1)m(2z)2m+1,n=2m+1 0,n /negationslash=2m+1⎫ ⎪⎪⎬ ⎪⎪⎭. Example 17.2.9 H⎪bracketleftbigg (1−z2)−1 2exp⎪braceleftbigg2xyz−(x2+y2)z2 (1−z2)⎪bracerightbigg⎪bracketrightbigg =√ π∞⎪summationdisplay m=0zmHm(y)δmn. (17.2.17) We use a result (see Erd´ elyi et al., 1954, vol. 2, p. 194) (1−z2)−1 2exp⎪braceleftbigg2xyz−(x2+y2)z2 (1−z2)⎪bracerightbigg =∞⎪summationdisplay m=0⎪parenleftbigg1 2z⎪parenrightbiggm1 m!Hm(x)Hm(y) to derive H⎪bracketleftbigg (1−z2)−1 2exp⎪braceleftbigg2xyz−(x2+y2)z2 (1−z2)⎪bracerightbigg⎪bracketrightbigg =∞⎪summationdisplay m=0⎪parenleftbigg1 2z⎪parenrightbiggm1 m!Hm(y)∞⎪integraldisplay −∞exp(−x2)Hn(x)Hm(x)dx =∞⎪summationdisplay m=0⎪parenleftbigg1 2z⎪parenrightbiggm1 m!Hm(y)δmδmn=√ π∞⎪summationdisplay m=0zmHm(y)δmn. 17.3 Basic Operational Properties THEOREM 17.3.1 IfF/prime(x) is continuous and F/prime/prime(x) is bounded and locally integrable in the © 2007 by Taylor & Francis Group, LLC 530 INTEGRAL TRANSFORMS and THEIR APPLICATIONS interval −∞<x< ∞,a n di f H{F(x)}=fH(n), then H{R[F(x)]}=−2nfH(n), (17.3.1) where R[F(x)] is the diﬀerential form given by R[F(x)] = exp( x2)d dx⎪bracketleftbigg exp(−x2)dF dx⎪bracketrightbigg . (17.3.2) PROOF We have, by deﬁnition, H{R[F(x)]}=∞⎪integraldisplay −∞d dx⎪bracketleftbigg exp(−x2)dF dx⎪bracketrightbigg Hn(x)dx which is, by integrating by parts and using the orthogonal relation (17.2.8), =−2n∞⎪integraldisplay −∞exp(−x2)Hn(x)F(x)dx=−2nfH(n). Thus, the theorem is proved. IfF(x)a n d R[F(x)] satisfy the conditions of Theorem 17.3.1, then H{R2[F(x)]}=H{R[R[F(x)]]}=(−1)2(2n)2fH(n).(17.3.3) H{R3[F(x)]}=(−1)3(2n)3fH(n). (17.3.4) More generally, H{Rm[F(x)]}=(−1)m(2n)mfH(n), (17.3.5) where m=1,2,...,m −1. THEOREM 17.3.2 IfF(x) is bounded and locally integrable in −∞<x< ∞,a n dfH(0) = 0, then H{F(x)}=fH(n) exists and for each constant C, H−1⎪braceleftbigg −fH(n) 2n⎪bracerightbigg =R−1[F(x)] =x⎪integraldisplay 0exp(s2)s⎪integraldisplay −∞exp(−t2)F(t)dt ds+C,(17.3.6) where R−1is the inverse of the diﬀerential operator Randnis a positive integer. PROOF We write R−1[F(x)] =Y(x) © 2007 by Taylor & Francis Group, LLC Hermite Transforms 531 so that Y(x) is a solution of the diﬀerential equation R[Y(x)] =F(x). (17.3.7) Since fH(0)= 0, and H0(x)=1 ,t h e n ∞⎪integraldisplay −∞exp(−x2)F(x)dx=0. The ﬁrst integral of (17.3.7) is exp(−x2)Y/prime(x)=x⎪integraldisplay −∞exp(−t2)F(t)dt, which is a continuous function of xand tends to zero as |x|→∞ . The second integral Y(x)=x⎪integraldisplay 0exp(s2)s⎪integraldisplay −∞exp(−t2)F(t)dt ds+C, where Cis an arbitrary constant, is also continuous. Evidently, lim |x|→∞exp(−x2)Y(x)=0 provided Y(x) is bounded. Then H{Y(x)}exists and H{R[Y(x)]}=−2nH{Y(x)}. Or, H[F(x)] =−2nH{Y(x)}. Hence, fH(n)=−2nH{R−1[F(x)]}. Thus, for any positive integer n, H{R−1[F(x)]}=−fH(n) 2n. THEOREM 17.3.3 IfF(x) has bounded derivatives of order mand if H{F(x)}=fH(n) exists, then H{F(m)(x)}=fH(n+m). (17.3.8) © 2007 by Taylor & Francis Group, LLC 532 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF We have, by deﬁnition, H{F/prime(x)}=∞⎪integraldisplay −∞exp(−x2)Hn(x)F/prime(x)dx, which is, by integrating by parts, =[ e x p ( −x2)F(x)Hn(x)]∞ −∞−∞⎪integraldisplay −∞F(x)d dx⎪bracketleftBig e−x2Hn(x)⎪bracketrightBig dx =2∞⎪integraldisplay −∞xexp(−x2)Hn(x)F(x)−∞⎪integraldisplay −∞F(x)exp(−x2)H/prime n(x)dx.(17.3.9) We use recurrence relations (A-6.5)–(A-6.6) for the Hermite polynomial to rewrite (17.3.9) in the form H{F/prime(x)}=∞⎪integraldisplay −∞exp(−x2)[Hn+1(x)+2nHn−1(x)]F(x)dx −2n∞⎪integraldisplay −∞exp(−x2)Hn−1(x)F(x)dx =∞⎪integraldisplay −∞exp(−x2)Hn+1(x)F(x)dx=fH(n+1 ). Proceeding in a similar manner, we can prove H{F(m)(x)}=fH(n+m). Thus, the theorem is proved. THEOREM 17.3.4 If the Hermite transforms of F(x)a n d xF(m−1)(x)e x i s t ,t h e n H{xF(m)(x)}=nfH(m+n−1) +1 2fH(m+n+1 ). (17.3.10) PROOF We have, by deﬁnition, H{xF(m)(x)}=∞⎪integraldisplay −∞exp(−x2)Hn(x)⎪braceleftbigg xdmF(x) dxm⎪bracerightbigg dx =⎪bracketleftBig xexp(−x2)Hn(x)F(m−1)(x)⎪bracketrightBig∞ −∞ −∞⎪integraldisplay −∞d dx⎪bracketleftbig xexp(−x2)Hn(x)⎪bracketrightbig F(m−1)(x)dx. © 2007 by Taylor & Francis Group, LLC Hermite Transforms 533 Thus, H{xF(m)(x)}=∞⎪integraldisplay −∞2x2exp(−x2)Hn(x)F(m−1)(x)dx −∞⎪integraldisplay −∞exp(−x2)Hn(x)F(m−1)(x)dx −n∞⎪integraldisplay −∞2xexp(−x2)Hn−1(x)F(m−1)(x)dx, which is, by the recurrence relations (17.3.10)–(17.3.11), and (17.3.8), =∞⎪integraldisplay −∞xexp(−x2)[Hn+1(x)+2nHn−1(x)]F(m−1)(x)dx −n∞⎪integraldisplay −∞exp(−x2)[Hn(x)+2 (n−1)Hn−2(x)]F(m−1)(x)dx −fH(n+m+1 ) =1 2∞⎪integraldisplay −∞exp(−x2)[Hn+2(x)+2 (n+1 )Hn(x)]F(m−1)(x)dx +n∞⎪integraldisplay −∞exp(−x2)[Hn(x)+2 (n−1)Hn−2(x)]F(m−1)(x)dx −nfH(n+m−1)−2n(n−1)fH(n+m−3)−fH(n+m+1 ) =1 2fH(n+m+1 )+( n+1 )fH(n+m−1) +n[fH(n+m−1) + 2( n−1)fH(n+m−3)] −nfH(n+m−1)−2n(n−1)fH(n+m−3)−fH(n+m+1 ) =nfH(n+m−1) +1 2fH(n+m+1 ). In particular, when m=1a n d m=2 ,w eo b t a i n H{xF/prime(x)}=nfH(n)+1 2fH(n+2 ), (17.3.11) H{xF/prime/prime(x)}=nfH(n+1 )+1 2fH(n+3 ). (17.3.12) The reader is referred to a paper by Debnath (1968) for other results similar to those of (17.3.11)–(17.3.12). © 2007 by Taylor & Francis Group, LLC 534 INTEGRAL TRANSFORMS and THEIR APPLICATIONS DEFINITION 17.3.1 (Generalized Convolution). The generalized convo- lution of F(x)andG(x)for the Hermite transform deﬁned by H{F(x)∗G(x)}=μnH{F(x)}H{G(x)}=μnfH(n)gH(n),(17.3.13) where μnis a non-zero quantity given by μn=√ π(−1)n⎪braceleftbigg 22n+1Γ⎪parenleftbigg n+3 2⎪parenrightbigg⎪bracerightbigg−1 . (17.3.14) Debnath (1968) ﬁrst proved the convolution theorem of the Hermite trans- form for odd functions. However, Dimovski and Kalla (1988) extended thetheorem for both odd and even functions. We follow Dimovski and Kalla to state and prove the convolution theorem of the Hermite transform. Before we discuss the theorem, it is observed that, if F(x) is an odd function, then H{F(x); 2n}=f H(2n)=∞⎪integraldisplay −∞exp(−x2)H2n(x)F(x)dx=0, (17.3.15) but H{F(x); 2n+1}=fH(2n+1 )/negationslash=0. (17.3.16) On the other hand, if F(x) is an even function, then H{F(x); 2n+1}=fH(2n+1 )=0 , (17.3.17) but H{F(x); 2n}=fH(2n)/negationslash=0. (17.3.18) THEOREM 17.3.5 (Convolution of the Hermite Transform for Odd Functions ). IfF(x)a n d G(x) are odd functions and nis an odd positive integer, then H{F(x)◦ ∗G(x); 2n+1}=μnfH(2n+1 )gH(2n+1 ), (17.3.19) where◦ ∗denotes the convolution operation for odd functions and is given by F(x)◦ ∗G(x)=x π∞⎪integraldisplay −∞exp(−t2)tF(t)dtπ⎪integraldisplay 0exp(−xtcosφ)sinφ ×π⎪integraldisplay 0G[(x2+t2+2xtcosφ)]1 2 (x2+t2+2xtcosφ)1 2J0(xtsinφ)dφ,(17.3.20) andJ0(z) is the Bessel function of the ﬁrst kind of order zero. © 2007 by Taylor & Francis Group, LLC Hermite Transforms 535 PROOF We have, by deﬁnition, fH(2n+1 )=∞⎪integraldisplay −∞exp(−x2)H2n+1(x)F(x)dx. (17.3.21) We replace H2n+1(x)b yu s i n gar e s u l tf o rE r d ´ elyi (1953, vol. 2, p. 1993) H2n+1(x)=(−1)n22n+1n!xL1 2n(x2), (17.3.22) where Lα n(x) is the Laguerre polynomial of degree nand order αso that (17.3.21) reduces to the form fH(2n+1 )=( −1)n22n+2n!∞⎪integraldisplay 0xexp(−x2)L1 2n(x2)F(x)dx. (17.3.23) Invoking the change of variable x2=t,w eo b t a i n H{F(x); 2n+1}=(−1)n22n+1n!∞⎪integraldisplay 0√ texp(−t)L1 2n(t)F(√ t) √ tdt.(17.3.24) It is convenient to introduce the transformation Tby (TF)(t)=F(√ t) √ t,0≤t<∞ (17.3.25) so that the inverse of Tis given by T−1(Φ)(x)=xΦ(x2). (17.3.26) Consequently, (17.3.24) takes the form H{F(x); 2n+1}=(−1)n22n+1n!L{TF(x)}, (17.3.27) where Lis the Laguerre transformation of degree nand order α=1 2deﬁned by (16.2.1) in Chapter 16. The use of (17.3.27) allows us to write the product of two Hermite trans- forms as the product of two Laguerre transforms as fH(2n+1 )gH(2n+1 )=24n+2(n!)2L{TF(x)}L{TG(x)}.(17.3.28) We now apply the Convolution Theorem for the Laguerre transform (when α= 0) proved by Debnath (1969) in the form L{F˜∗G(x)}=n!√ π Γ⎪parenleftbigg n+3 2⎪parenrightbiggL{F(x)}L{G(x)}, (17.3.29) © 2007 by Taylor & Francis Group, LLC 536 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where F˜∗Gis given by F˜∗G(x)=∞⎪integraldisplay 0exp(−τ)√ τF(τ)dτπ⎪integraldisplay 0exp(−√ tτcosφ)sinφ ×G(t+τ+2√ tτcosφ)J0(√ tτsinφ)dφ.(17.3.30) Substituting (17.3.29) into (17.3.28), we obtain fH(2n+1 )gH(2n+1 )= π−1 224n+2n!Γ⎪parenleftbigg n+3 2⎪parenrightbigg L{TF˜∗TG}, which is, by (17.3.27), =22n+1Γ⎪parenleftbigg n+3 2⎪parenrightbigg (−1)n√ πH{T−1(TF˜∗TG)}. (17.3.31) Or, equivalently, H{F◦ ∗G(x); 2n+1}=μnH{F(x)}H{G(x)}, (17.3.32) where F◦ ∗G(x)=T−1{TF◦ ∗TG(x)}. (17.3.33) This coincides with (17.3.20). Thus, the proof is complete. THEOREM 17.3.6 (Convolution of the Hermite Transform for Even Functions ). IfF(x)a n dG(x) are even functions and nis an even positive integer, then H{F(x)e ∗G(x); 2n}=μnH{F(x); 2n}H{G(x); 2n}. (17.3.34) PROOF We use result (17.3.8), that is, H{F/prime(x);n}=H{F(x);n+1} so that H{IF(x); 2n+1}=H{F(x),2n}, (17.3.35) where IF(x)=x⎪integraldisplay 0F(t)dtand [ IF(x)]/prime=F(x). © 2007 by Taylor & Francis Group, LLC Hermite Transforms 537 Obviously, H⎪braceleftbig F(x)e ∗G(x); 2n⎪bracerightbig =H⎪braceleftBig⎪bracketleftbig IF(x)e ∗IG(x)⎪bracketrightbig/prime;2n⎪bracerightBig =H⎪braceleftbig IF(x)◦ ∗IG(x); 2n+1⎪bracerightbig =μnH{IF(x); 2n+1}H{IG(x); 2n+1} =μnH{F(x); 2n}H{G(x); 2n}. This proves the theorem. THEOREM 17.3.7 IfF(x)a n d G(x) are two arbitrary functions such that their Hermite trans- forms exist, then H{F(x)∗G(x);n}=μ[n/2]H{F(x);n}H{G(x);n}, (17.3.36) where F(x)∗G(x)=F0(x)◦ ∗G0(x)+Fe(x)e ∗Ge(x), (17.3.37) and F0(x)=1 2[F(x)−F(−x)] and Fe(x)=1 2[F(x)+F(−x)].(17.3.38) PROOF We ﬁrst note that arbitrary functions F(x)a n d G(x)c a nb e expressed as sums of even and odd functions, that is, F(x)=F0(x)+Fe(x) andG(x)=G0(x)+Ge(x) so that result (17.3.38) follows. Suppose nis odd. Then H{F(x);n}=H{F0(x);n},H{G(x);n}=H{G0(x);n}, and H{F(x)+G(x);n}=H{F0(x)+G0(x);n}. Clearly, H{F(x)∗G(x); 2n+1} =H{F0(x)◦ ∗G0(x); 2n+1}+H{Fe(x)e ∗Ge(x); 2n+1} =μnH{F0(x)}H{G0(x)}=μnH{F(x)}H{G(x)}. Similarly, the case for even ncan be handled without any diﬃculty. We conclude this chapter by citing some recent work on the generalized Hermite transformation by Glaeske (1983, 1986, 1987). These papers includesome interesting discussion on operational properties and convolution struc- ture of the generalized Hermite transformations. For more details, the reader is referred to these papers. © 2007 by Taylor & Francis Group, LLC 538 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 17.4 Exercises 1. Find the Hermite transform of the following functions: (a) exp( −x2)Hn(x), (b)xm, (c)x2Hn(x). 2. Show that H{xn}=√ πn!Pn(1), where Pn(x) is the Legendre polynomial. 3. Show that H⎪braceleftbig H2 n(x)⎪bracerightbig =√ πn⎪summationdisplay r=0⎪parenleftbiggn r⎪parenrightbigg 2r+n(2r)!n!. © 2007 by Taylor & Francis Group, LLC 18 The Radon Transform and Its Applications “This struck me as a typical nineteenth century piece of mathe- matics which a Cauchy or a Riemann might have dashed oﬀ in alight moment, but a diligent search of standard texts on analysis failed to reveal it, so I had to solve the problem myself. I still felt that the problem must have been solved, so I contacted mathe-maticians on three continents to see if they knew about it, but to no avail.” Allan MacLeod Cormack “We live in an age in which mathematics plays a more and more important role, to the extent that it is hard to think of an aspect of human life to which it either has not provided, or does not have the potential to provide, crucial insights. Mathematics is the languagein which quantitative models of the world around us are described. As subjects become more understood, they become more mathe- matical. A good example is medic ine, where the Radon transform is what makes X-ray tomography work, where statistics form the basis of evaluating the success or f ailure of treatments, and where mathematical models of organs such as the heart, of tumor growth, and of nerve impulses are of key importance.” John Ball 18.1 Introduction The origin of the Radon transform can be traced to Johann Radon ’s 1917 celebrated work “On the det ermination of functions from integrals along cer- tain manifolds.” In his seminal work, Radon demonstrated how to construct a function of two variables from its integrals over all straight lines in theplane. He also made other generalizations of this transform involving the re- construction of a function from its integrals over other smooth curves as well as the reconstruction of a function of nvariables from its integrals over all 539 © 2007 by Taylor & Francis Group, LLC 540 INTEGRAL TRANSFORMS and THEIR APPLICATIONS hyperplanes. Although the Radon transform had some direct ramiﬁcations on solutions of hyperbolic partial diﬀerential equations with constant coeﬃcients,it did not receive much attention fro m mathematicians and scientists. In the 1960s, the Radon transform played a major role in tomography which is widely used method to reconstruct cross-sections of the interior structure of an object without having to cut or damage the object. Through the inter- action of a physical organ or “probe” — varying from X-rays, gamma rays,visible light, electrons, or neutron s to ultrasound waves — with the object, we usually obtain line or (hyper) plane integrals of the internal distribution to be reconstructed. Ther e is a close relation between the Radon transform and the development of X-ray scans (or CAT scans ) in medical imaging. In practice, X-ray scans provide a picture of an internal organ of a human or animal body, and hence help detect and locate many types of abnormalities.Thus, one of the most prominent examples of applications of computer assist- ed tomography occurs in diagnostic med icine, where the method is employed to generate images of the interior of human organs. The central problem of reconstruction and the introduction of new algorithms and faster electronic computers led to a rapid developm ent of computerized tomography. More than ﬁfty years later, Allan Cormack , a young South African physi- cist, became interested in ﬁnding a set of maps of absorption coeﬃcients for diﬀerent sections of the human body. In order to make X-ray radiotherapymore eﬀective, he quickly recognized th e importance of the Radon transform which is similar to measurements of the absorption of X-rays along lines in their sections of the human body. Since the logarithm of the ratio of incident to reﬂected X-ray intensities along a give n straight line is just the line integral of the absorption coeﬃcient along that line, the problem is mathematically e-quivalent to ﬁnding a function from the values of its integrals along all or some lines in the plane. As early as 1963, Cormack already obtained three alterna- tive solutions of this major problem. At the same time, Godfrey Hounsﬁeld , a young British biomedical engineer, realized the unique importance of the major ideas of Radon and Cormack and then used them to develop a new X-ray machine that totally revolutionized the ﬁeld of medical imaging. Soon after that Cormack and Hounsﬁeld joined together to work on the reﬁnement of the solution of medical imaging. Their joint work led to the major discoveryof the CT-scanning technique and then culminated in winning the 1979 Nobel Prize in Physiology and Medicine. In their Nobel Prize addresses Cormack and Hounsﬁeld acknowledged the pioneering work of Radon in 1917. The Radon transform is found to be very useful in many diverse ﬁelds of sci- ence and engineering including medical imaging, astronomy, crystallography, electron microscopy, geophy sics, material science, an d optics. It is importan- t to mention that the Radon transform has been used in computer assisted tomography (CAT) heavily. The problem of determining internal structure of an object by observations or projections is closely associated with the Radon transform. In this chapter, we introduce the Radon transform, its basic prop- erties, its inverse and the relationshi p between the Radon transform and Fouri- © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 541 er transform. Included are Parseval theorems and applications of the Radon transform. 18.2 The Radon Transform DEFINITION 18.2.1 (The Radon Transform). If Lis any straight line in thex-yplane (or, in R2)a n d dsis the arc length along L(see Figure 18.1),the Radon transform of a function f(x, y)of two real variables is deﬁned by its integral along Las ˆf(p,φ)=R{f(x, y)}=⎪integraldisplay Lf(x, y)ds. (18.2.1) In other words, the totality of all these line integrals constitutes the Radon transform of f(x, y) and each line integral is called a sample of the Radon transform of f(x, y). Thus, the Radon transform ˆfoffc a nb ev i e w e da sa function deﬁned on all straight lines in the plane and the value of ˆf(p,φ)a t ag i v e n Lis the integral of f(x, y)o v e rt h a tl i n e . xy or uPQ(x,y) u p L Figure 18.1 Graph of the line L. Making references to Figure 18.1, we write the equation of the line Lin the © 2007 by Taylor & Francis Group, LLC 542 INTEGRAL TRANSFORMS and THEIR APPLICATIONS formp=xcosφ+ysinφ,w h e r e pis the length of the perpendicular from the origin to Landφis the angle that the perpendicular makes with the positive x-axis. If we rotate the coordinate system by an angle φ, and label the new axes by pands,t h e n x=pcosφ−ssinφ,y=psinφ+scosφ.C o n s e q u e n t l y , the Radon transform (18.2.1) can be deﬁned by R{f(x, y)}=ˆf(p,φ)=⎪integraldisplay∞ −∞f(pcosφ−ssinφ, psinφ+scosφ)ds. (18.2.2) This deﬁnition is very practical in two dimensions. However, it does not lend itself readily to higher dimensions. In order to generalize the above deﬁnition in higher dimensions, we intro- duce the unit vectors u=( c o s φ,sinφ)a n du⊥=(−sinφ,cosφ), so that x= (x, y)=(r, θ)=pu+tu⊥for some scalar parameter t,w h e r e randθare the usual polar coordinates. The equation of the line Lcan now be written in terms of the unit vector uasp=x.u=xcosφ+ysinφ. Using the deﬁnition of Dirac delta function, we express (18.2.1) in the form ˆf(p,φ)=⎪integraldisplay∞ −∞f⎪parenleftbig pu+tu⊥⎪parenrightbig dt=⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x)δ(p−x.u)dx.(18.2.3) It is noted that the integral is taken over a line orthogonal to the line θ=φ and that ˆf(−p,φ)=ˆf(p,φ+π) so that negative values for pcan be assigned andφmay be restricted to [0 ,π]. DEFINITION 18.2.2 (The Radon Transform in Higher Dimensions). Inndimensional Euclidean space Rn,x=(x1,x2, ....,xn)andf(x)= f(x1,x2, ...., x n). Let us introduce a unit vector u=(u1,u2, ..., u n)inRn that deﬁnes the orientation of a hyperplane with the equation p=x.u=x1u1+......+xnun. (18.2.4) Then the Radon transform of a function f(x)is deﬁned by ˆf(p,u)=R{f(x)}(p,u)=⎪integraldisplay∞ −∞f(x)δ(p−x.u)dx, (18.2.5) where the integration is taken over dx=dx1dx2...dx n. Therefore, the Radon transform of a function of nvariables is the totality of all integrals of fover all hyperplanes in Rn. In other words, the Radon transform ˆf(p,u)o ff(x) is a function deﬁned on all hyperplanes in Rn,a n d the value of that function at any hyperplane is the integral of f(x)o v e rt h a t hyperplane. The inverse Radon transform is equivalent to ﬁnding f(x)f r o m the values of its integrals over all hyperplanes. © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 543 The Radon transform has been further generalized to use this idea in the ﬁeld of medical imaging, where the integral along a line represents a measure-ment of the intensity of the X-ray beam at the detector after passing through the object to be radiographed. This is essentially the idea of the X-ray trans- form where a two-dimensional object lying in the plane R 2which may be considered as a planer cross section o f a human or animal organ. After send- ing the X-ray beam through the object along a line L, we calculate the integral in (18.2.1), which is also called the X-ray transform (or the Radon transform). Thus, this transform assigns to each suitable function fonR2another unique function ˆf=R{f}which domain is the set of lines in R2.F r o map r a c t i c a l point of view, the major interest lies essentially in internal structure of the object, and hence, the centr al problem is to reconstruct f(or, to ﬁnd the inverse) from the given ˆf. This is called the reconstruction problem which has a deﬁnite solution based on the general mathematical theory. However, in practice, this problem can be solved using sampling procedures, numericalapproximations, or computer algorithms. To explain this idea, we observe that in R 2, if a function f(x, y)i si n t e g r a t e d over a line Lθwith direction θ, and then if the line is moved parallel to itself, we obtain a function Lθf=P1fdeﬁned on a line Lθ⊥orthogonal to Lθ.T h e value of this function at any point xonLθ⊥is equal to the integral of f(x) over the line with direction θwhich intersects Lθ⊥at the point x. Similarly, in R3, if a function f(x)i si n t e g r a t e do v e rap l a n e P2,a n dt h e n if the plane is moved parallel to itself, we obtain a function P2fdeﬁned on a straight line orthogonal to the plane P2. The value of this function at any point xon this line is equal to the integral over the plane passing through xand parallel to P2.I ff(x) represents the density at x,t h e na nX - r a yt a k e ni nt h e θdirection generates a function Lθf=P1fdeﬁned on the plane orthogonal toθ.T h ev a l u eo f( Lθf)a tt h ep o i n t xon this plane is equal to the integral offalong a line through xin the direction θ,t h a ti s , [Lθf](x)=⎪integraldisplay∞ −∞f(x+tθ)dt, (18.2.6) where x∈θ⊥that represents the plane orthogonal to θ. DEFINITION 18.2.3 (The k-Plane Transform). If Πis ak-dimensional subspace of Rndetermined by the direction θ, then the k-plane transform of f(x)wherex∈Rnin the direction θat the point ξξξ∈Π⊥is deﬁned by ⎪parenleftbig Pkf⎪parenrightbig (Π,ξξξ)=⎪integraldisplay Πf(ξξξ, ηηη)dηηη, (18.2.7) wherex=(ξξξ, ηηη),ηηη∈Π,a n d ξξξ∈Π⊥. Or, equivalently, ⎪parenleftbig Pkf⎪parenrightbig (θ, ξξξ)=⎪integraldisplay f(ξξξ+u.θ)du, (18.2.8) © 2007 by Taylor & Francis Group, LLC 544 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where u.θ=(u1θ1+u2θk+...+ukθk)andξξξ∈θ⊥. The 1-plane transform⎪parenleftbig P1f⎪parenrightbig is called the X-ray transform and (n−1)- plane transform is the Radon transform. The k-plane transformation is a linear transformation. Example 18.2.1 Show that R⎪bracketleftbig exp⎪braceleftbig −a2(x2+y2)⎪bracerightbig⎪bracketrightbig =√ π aexp⎪parenleftbig −a2p2⎪parenrightbig ,a > 0. (18.2.9) Using m=u1x+u2y,n=−u2x+u1y,w eh a v e x2+y2=m2+n2 f(m, n)=e x p⎪braceleftbig −a2(m2+n2)⎪bracerightbig . So it turns out that ˆf(p,u)=⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞exp⎪braceleftbig −a2⎪parenleftbig m2+n2⎪parenrightbig⎪bracerightbig δ(p−m)dm dn =⎪integraldisplay∞ −∞exp(−a2m2)δ(m−p)dm⎪integraldisplay∞ −∞exp(−a2n2)dn =√ π aexp(−a2p2). In other words, R⎪bracketleftbig exp⎪braceleftbig −a2(x2+y2)⎪bracerightbig⎪bracketrightbig =√ π aexp⎪parenleftbig −a2p2⎪parenrightbig . Alternatively, we can use (18.2.2) to obtain ˆf(p, φ)=⎪integraldisplay∞ −∞exp⎪bracketleftbig −a(p2+s2)⎪bracketrightbig ds=e−ap2⎪integraldisplay∞ −∞e−as2ds=⎪radicalbigg π ae−ap2. When a= 1, the above result yields R⎪bracketleftbig exp(−x2−y2)⎪bracketrightbig =√ πe−p2. (18.2.10) Example 18.2.2 Find the Radon transform of the following functions: (a)f(x, y)=xexp⎪bracketleftbig −a(x2+y2)⎪bracketrightbig ,a > 0, (b)g(x, y)=yexp⎪bracketleftbig −a(x2+y2)⎪bracketrightbig ,a > 0. © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 545 (a) We have, by deﬁnition (18.2.2), ˆf(p, φ)=⎪integraldisplay∞ −∞(pcosφ−ssinφ)exp⎪bracketleftbig −a(p2+s2)⎪bracketrightbig ds =⎪radicalbigg π apcosφexp(−ap2). (18.2.11) (b) Similarly, ˆg(p, φ)=⎪integraldisplay∞ −∞(pcosφ+ssinφ)exp⎪bracketleftbig −a(p2+s2)⎪bracketrightbig ds =⎪radicalbigg π apsinφexp(−ap2). (18.2.12) Combining these two results gives R⎪braceleftbig (x+y)e x p⎪bracketleftbig −a(x2+y2)⎪bracketrightbig⎪bracerightbig =ˆf(p, φ)+iˆg(p, φ)=⎪radicalbigg π apeiφexp(−ap2). (18.2.13) 18.3 Properties of the Radon Transform (Relation between the Fourier Transform and the Radon Transform ). Consider two-dimensional Fourier transform deﬁned by ˜f(k)=F{f(x, y)}=⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞e−i(k.x)f(x, y)dxdy (18.3.1) where k=(k,l)a n dx=(x, y). The kernel exp[ −i(k.x)] of the Fourier transform can be written as e−i(k.x)=⎪integraldisplay∞ −∞e−itδ(t−k.x)dt (18.3.2) so that (18.3.1) becomes ˜f(k)=⎪integraldisplay∞ −∞e−itdt⎪integraldisplay∞ −∞f(x)δ(t−k.x)dx. (18.3.3) Substituting k=suandt=spin (18.3.3) where sis real and uis a unit vector © 2007 by Taylor & Francis Group, LLC 546 INTEGRAL TRANSFORMS and THEIR APPLICATIONS yields ˜f(su)=⎪integraldisplay∞ −∞e−ispdp⎪integraldisplay∞ −∞f(x)δ(p−u.x)dx =⎪integraldisplay∞ −∞e−ispˆf(p,u)dp =F⎪braceleftBig ˆf(p,u)⎪bracerightBig . (18.3.4) In other words, ˜f=F(Rf). This means that ˜fis the two-dimensional Fourier transform of f,w h e r e a s F(Rf) is the one-dimensional Fourier transform of Rf.T h u s , ˆf(p,u)=R{f(x)}. Or, equivalently, F−1⎪braceleftBig ˜f(su)⎪bracerightBig =1 2π⎪integraldisplay∞ −∞eips˜f(su)ds. (18.3.5) THEOREM 18.3.1 Linearity : R[af(x)+bg(x)] =aR[f(x)] +bR[g(x)].(18.3.6) Shifting : (a) If R{f(x, y)}=ˆf(p,u1,u2),then R{f(x−a, y−b)}=ˆf(p−au1−bu2,u).(18.3.6a) (b) In general , ˆf(x−a)=ˆf(p−a.u,u). (18.3.6b) Scaling : (a) If R{f(x, y)}=ˆf(p,u1,u2),then R{f(ax, by )}=1 |ab|ˆf⎪parenleftBig p,u1 a,u2 b⎪parenrightBig . (18.3.7a) (b)ˆf(ax)=1 anˆf⎪parenleftBig p,x a⎪parenrightBig =1 an−1ˆf(ap,x).(18.3.7b) Symmetry : (a) If ˆf(p,u)=R[f(x, y)],then,ifa/negationslash=0, ˆf(ap,au)=|a|−1ˆf(p,u). (18.3.8a) (b) ˆf(p, au)=|a|−1ˆf⎪parenleftBigp a,u⎪parenrightBig . (18.3.8b) Hereaandbare two constants. © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 547 PROOF Linearity: R[af(x)+bg(x)] =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞[af(x)+bg(x)]δ(p−x.u)dx =a⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x)δ(p−x.u)dx +b⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞g(x)δ(p−x.u)dx =aR[f(x)] +bR[g(x)]. This property is also true for any x=(x1,x2,,... ,xn)∈Rn,n≥2. Shifting: (a) Putting x−a=ξandy−b=η,w ec a nw r i t e R{f(x−a, y−b)}=⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x−a, y−b)δ(p−xu1−yu2)dxdy =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(ξ,η)δ(p−(a+ξ)u1−(b+η)u2)dξdη =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(ξ,η)δ(p−au1−bu2−ξu1−ηu2)dξdη =ˆf(p−au1−bu2,u). Similarly, results (18.3.6b) can be proved. Scaling: (a) Let a>0,b>0. Putting ax=ξ,by=η,w eh a v e R[f(ax, by )] =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(ax, by )δ(p−xu1−yu2)dxdy =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(ξ,η)δ⎪parenleftBig p−u1 aξ−u2 bη⎪parenrightBigdξdη ab =1 ab⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(ξ,η)δ⎪parenleftBig p−u1 aξ−u2 bη⎪parenrightBig =1 abˆf⎪parenleftBig p,u1 a,u2 b⎪parenrightBig . Ifaorb<0, then R[f(ax, by )] =−1 abˆf⎪parenleftBig p,u1 a,u2 b⎪parenrightBig . Or, R[f(ax, by )] =1 |ab|ˆf⎪parenleftBig p,u1 a,u2 b⎪parenrightBig . Similarly, results (18.3.7b) can be proved. © 2007 by Taylor & Francis Group, LLC 548 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Symmetry: (a) By deﬁnition, we have ˆf(ap , au)=⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x, y)δ(ap−axu1−ayu2)dxdy, which is ,by using δ(ap−axu1−ayu2)=1 |a|δ(p−xu1−yu2), =1 |a|⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x, y)δ(p−xu1−yu2)dxdy =1 |a|ˆf(p,u). This implies that the Radon transform is an even homogeneous function of degree −1. In particular if a=−1, we have ˆf(−p,−u)=ˆf(p,u), that is, ˆfis an even function. (b) Another form of symmetry property is ˆf(p,au)=|a|−1ˆf⎪parenleftBigp a,u⎪parenrightBig . We have, by deﬁnition, ˆf(p,au)=⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x, y)δ(p−axu1−ayu2)dxdy =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x, y)δ(p a.a−axu1−ayu2)dxdy =1 |a|⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞f(x, y)δ(p a−xu1−yu2)dxdy =1 |a|ˆf⎪parenleftBigp a,u⎪parenrightBig . In general, there is an important relation between the n-dimensional Fourier transform and the Radon transform given by ˜f(su)=1 (2π)n 2⎪integraldisplay∞ −∞e−ips˜f(p,u)dp. (18.3.9) Then-dimensional Fourier transform of f(x)i s ˜f(k)=1 (2π)n 2⎪integraldisplay∞ −∞e−i(k.x)˜f(x)dx. (18.3.10) © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 549 Invoking the hyperspherical polar coordinates allows us to write k=suwhere u∈Sn−1which is the generalized unit sphere⎪summationtextn k=1x2 k=1.Thus, (18.3.10) becomes ˜f(k)=˜f(su)= =1 (2π)n 2⎪integraldisplay∞ −∞e−is(u.x)˜f(x)dx. (18.3.11) We put g(x)=e x p[ −is(u.x)]f(x)s ot h a t ˆg(p,ηηη)=⎪integraldisplay Lexp[−is(u.x)]f(x)ds, (18.3.12) where Lis the hyperplane x.ηηη=panddsis the ( n−1)-dimensional surface area in Rn. Hence, ˆg(p,u)=e x p ( −ips)⎪integraldisplay u.x=pf(x)ds=e−ipsˆf(p,u). (18.3.13) Using the result (18.4.12), we have ⎪integraldisplay∞ −∞ˆg(p,u)dp=⎪integraldisplay∞ −∞e−ipsˆf(p,u)dp=⎪integraldisplay∞ −∞g(x)dx =⎪integraldisplay∞ −∞e−is(u.x)f(x)dx=( 2π)n 2˜f(su) =1 (2π)n−1 21 √ 2π⎪integraldisplay∞ −∞e−ipsˆf(p,u)dp. Consequently, this yields the following result ˜f(su)=1 (2π)n 2⎪integraldisplay∞ −∞e−ipsˆf(p,u)dp. (18.3.14) Denoting the one-dimensional Fourier transform along the radial direction by Fr, equation (18.3.14) can be written as F{f}=1 (2π)n−1 2Fr{ˆf} (18.3.15) In view of (18.3.14) and the inverse Fourier transform, we ﬁnd the relation ˆf(p,u)=( 2 π)n−1 21 √ 2π⎪integraldisplay∞ −∞eips˜f(su)ds =( 2π)n−2 2⎪integraldisplay∞ −∞˜f(su)eispds. (18.3.16) © 2007 by Taylor & Francis Group, LLC 550 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 18.4 The Radon Transform of Derivatives Since ∂f ∂x= lim h→0⎡ ⎣f⎪parenleftBig x+h u1,y⎪parenrightBig −f(x, y) h u1⎤ ⎦, we can write R⎪bracketleftbigg∂f ∂x⎪bracketrightbigg =u1lim h→0⎡ ⎣R⎪braceleftBig f⎪parenleftBig x+h u1,y⎪parenrightBig⎪bracerightBig −R{f(x, y)} h⎤ ⎦ =u1lim h→0ˆf(p+h,u)−ˆf(p,u) h =u1∂ ∂pˆf(p,u). Similarly partial derivative with respect to y R⎪bracketleftbigg∂f ∂y⎪bracketrightbigg =u2∂ ∂pˆf(p,u). The Radon transform of the ﬁrst derivatives, that is, R⎪bracketleftBiggn⎪summationdisplay k=1ak∂f ∂xk⎪bracketrightBigg (p,u)=(a.u)∂ ∂pˆf(p,u), (18.4.1) where a=(a1,a2, ..., a n). Or, equivalently, R[(a.∇)f](p,u)=(a.u)∂ ∂pˆf(p,u), (18.4.2) where ∇=⎪parenleftBig ∂ ∂x1,∂ ∂x2, ...,∂ ∂xn⎪parenrightBig is the gradient operator. In particular, R⎪bracketleftbigg∂f ∂xk⎪bracketrightbigg (p,u)=uk∂ ∂pˆf(p,u). (18.4.3) We leave proofs of the above results to the reader. The Radon transform of the second order derivatives are given by R⎪bracketleftbigg∂2f ∂x2⎪bracketrightbigg =R⎪bracketleftbigg∂ ∂x⎪parenleftbigg∂f ∂x⎪parenrightbigg⎪bracketrightbigg =u1∂ ∂pR⎪bracketleftbigg∂f ∂x⎪bracketrightbigg =u2 1∂2 ∂p2ˆf(p,u), R⎪bracketleftbigg∂2f ∂x∂y⎪bracketrightbigg =u1u2∂2 ∂p2ˆf(p,u), R⎪bracketleftbigg∂2f ∂y2⎪bracketrightbigg =u2 2∂2 ∂p2ˆf(p,u). © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 551 We state more general results: IfL=L⎪parenleftBig ∂ ∂x1,∂ ∂x2, ...,∂ ∂xn⎪parenrightBig is a linear diﬀerential operator with constant coeﬃcients, then R⎪bracketleftbigg L⎪parenleftbigg∂ ∂x1,∂ ∂x2, ...,∂ ∂xn⎪parenrightbigg f(x)⎪bracketrightbigg (p,u) =L⎪parenleftbigg u1∂ ∂p,u2∂ ∂p, ..., u n∂ ∂p⎪parenrightbigg ˆf(p,u). (18.4.4) In particular, R⎪bracketleftBiggn⎪summationdisplay k=1n⎪summationdisplay l=1(akbl)∂2f ∂xk∂xl⎪bracketrightBigg (p,u)=(a.u)(b.u)∂2ˆf(p,u) ∂p2, (18.4.5) R⎪bracketleftbigg∂2f(x) ∂xk∂xl⎪bracketrightbigg (p,u)=(ukul)∂2ˆf(p,u) ∂p2. (18.4.6) Ifakbl=δkl(the Kronecker delta), then the operator involved is the Laplace operator ∇2=∂2 ∂x2 1+∂2 ∂x2 2+....+∂2 ∂x2n, and R⎪braceleftbig ∇2f(x)⎪bracerightbig =|u|2∂2ˆf(p,u) ∂p2=∂2ˆf(p,u) ∂p2, (18.4.7) where |u|2= 1 is used to obtain the last result. This is a very important result that is employed to solve partial diﬀerential equations. 18.5 Derivatives of the Radon Transform In order to calculate the derivative of the radon transform, the following for-mulas of the derivative of the Dirac delta function are needed and statedas∂ ∂xδ(x−y)=−∂ ∂yδ(x−y). (18.5.1) Ifyis replaced by by,t h e n ∂ ∂(by)δ(x−by)=1 b∂ ∂yδ(x−by)=−∂ ∂xδ(x−by). (18.5.2) Similarly, an n-dimensional result is ∂ ∂yjδ(x−y)=−∂ ∂xjδ(x−y). (18.5.3) © 2007 by Taylor & Francis Group, LLC 552 INTEGRAL TRANSFORMS and THEIR APPLICATIONS It follows from the above results that ∂ ∂ξjδ(p−ξξξ.x)=−xj∂ ∂pδ((p−ξξξ.x), (18.5.4) where ξξξis not necessarily a unit vector in the term ξξξ.x. Formula (18.5.4) reveals that the derivatives are calculated with respect to components of ξξξand then evaluated at ξξξ=u. Using the deﬁnition of (18.2.5), we obtain ∂ˆf ∂uk=⎪bracketleftBigg ∂ˆf(p,ξξξ) ∂ξk⎪bracketrightBigg ξξξ=u=⎪bracketleftbigg⎪integraldisplay f(x)∂ ∂ξkδ(p−ξξξ.x)dx⎪bracketrightbigg ξξξ=u =−∂ ∂p⎪integraldisplay xkf(x)δ(p−u.x)dx. (18.5.5) Consequently, we obtain the formula for the derivative of the Radon transform ∂ˆf(p,u) ∂uk=⎪bracketleftbigg∂ ∂ξkR{f(x)}⎪bracketrightbigg ξξξ=u=−∂ ∂pR{xkf(x)}. (18.5.6) More generally, we state the derivatives of the Radon transform as ⎪parenleftbigg a.∂ ∂u⎪parenrightbigg R{f(x)}(p,u)=−∂ ∂pR[(a.x)f(x)] (p,u), (18.5.7) where a.∂ ∂u=⎪summationtextn k=1ak∂ ∂uk. Formula (18.5.6) can be generalized for h igher order derivatives as follows: ∂2ˆf(p,u) ∂ul∂uk=(−1)2∂2 ∂p2R{xlxkf(x)}, (18.5.8) ∂3ˆf(p,u) ∂ul∂2uk=(−1)3∂3 ∂p3R⎪braceleftbig xlx2 kf(x)⎪bracerightbig , (18.5.9) where + or −sign is used for even or odd order derivatives, respectively. More generally, n⎪summationdisplay k,l=1(akbl)∂2ˆf(p,u) ∂uk∂ul=∂2 ∂p2R{(a.x)(b.x)f(x)}. (18.5.10) For a two dimensional function, f(x)=f(x, y)w eo b t a i n ∂k+lˆf(p,u) ∂uk 1∂ul2=⎪parenleftbigg −∂ ∂p⎪parenrightbiggk+l R⎪braceleftbig xkxlf(x)⎪bracerightbig (p,u). (18.5.11) Finally, the property involving the integration of the radon transform with respect to pcan be stated as follows: ⎪integraldisplay∞ −∞ˆf(p,u)dp=⎪integraldisplay∞ −∞f(x)dx. (18.5.12) © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 553 Example 18.5.1 The Radon transform of Hermite polynomials is given by R⎪braceleftbig Hl(x)Hk(y)exp (−(x2+y2)⎪bracerightbig =√ π(cosφ)l(sinφ)ke−p2Hl+k(p), (18.5.13) where Hn(x) is the Hermite polynomials of degree deﬁned by Rodrigues for- mulas e−x2Hn(x)=(−1)n⎪parenleftbigg∂ ∂x⎪parenrightbiggn e−x2. (18.5.14) Obviously, exp⎪bracketleftbig −(x2+y2)⎪bracketrightbig Hl(x)Hk(x)=(−1)l+k⎪parenleftbigg∂ ∂x⎪parenrightbiggl⎪parenleftbigg∂ ∂y⎪parenrightbiggk exp⎪bracketleftbig −(x2+y2)⎪bracketrightbig . (18.5.15) It follows from the above formulas R⎪bracketleftBigg⎪parenleftbigg∂ ∂x⎪parenrightbiggl⎪parenleftbigg∂ ∂y⎪parenrightbiggk f(x, y)⎪bracketrightBigg =( c o s φ)l(sinφ)k⎪parenleftbigg∂ ∂p⎪parenrightbiggl+k ˆf(p,u), (18.5.16) that the Radon transform of (18.5.15) is given by R⎪bracketleftbig exp{−(x2+y2)}Hl(x)Hk(x)⎪bracketrightbig =( c o s φ)l(sinφ)k⎪parenleftbigg∂ ∂p⎪parenrightbiggl+k√ πe−p2, (18.5.17) where ˆf(p,u)=R⎪bracketleftbig exp{−(x2+y2)}⎪bracketrightbig =√ πe−p2is used. 18.6 Convolution Theorem for the Radon Transform THEOREM 18.6.1 (Convolution). Ifˆf(p,u)=R{f(x)}and ˆg(p,u)=R{g(x)},then R{(f⋆g)(x)}=⎪parenleftBig ˆf⋆ˆg⎪parenrightBig (p,u). (18.6.1) © 2007 by Taylor & Francis Group, LLC 554 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF Leth(x) be the convolution of f(x)a n d g(x). So, we have h(x)=(f⋆g)(x)=⎪integraldisplay∞ −∞f(x)g(x−y)dy, (18.6.2) where x=(x1,x2, ..., x n)a n dy=(y1,y2, ..., y n). Taking the Radon transform of (18.5.3) yields the following ˆh(p,u)=R{h(x)}=R{(f⋆g)(x)} =⎪integraldisplay∞ −∞h(x)δ(p−u.x)dx =⎪integraldisplay∞ −∞f(y)dy⎪integraldisplay∞ −∞g(x−y)δ(p−u.x)dx =⎪integraldisplay∞ −∞f(y)dy⎪integraldisplay∞ −∞g(z)δ(p−u.y−u.z)dz, (z=x−y) =⎪integraldisplay∞ −∞f(y)ˆg(p−u.y,u)dy =⎪integraldisplay∞ −∞f(y)dy⎪integraldisplay∞ −∞ˆg(p−s,u)δ(s−u.y)ds =⎪integraldisplay∞ −∞ˆg(p−s,u)ds⎪integraldisplay∞ −∞f(y)δ(s−u.y)dy =⎪integraldisplay∞ −∞ˆf(s,u)ˆg(p−s,u)ds =⎪bracketleftBig⎪parenleftBig ˆf∗ˆg⎪parenrightBig (p,u)⎪bracketrightBig , where ˆg(p−u.y,u)=⎪integraldisplay∞ −∞ˆg(p−s,u)δ(s−u.y)ds. Or, ˆh(p,u)=R{f⋆g}(x)=ˆf(p,u)⋆ˆg(p,u). 18.7 Inverse of the Radon Transform and the Parseval Relation We consider the n-dimensional Fourier transform ˜f(k)o ff(x) deﬁned by ˜f(k)=F{f(x)}=1 (2π)n/2⎪integraldisplay∞ −∞e−ik.xf(x)dx. (18.7.1) © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 555 Using the hyperspherical polar coordinates, we can put k=ρuwhereu∈Sn−1 which is the generalized unit sphere⎪summationtextn r=1x2 r=1 .C o n s e q u e n t l y , ˜f(k)=˜f(ρu)=1 (2π)n/2⎪integraldisplay∞ −∞e−iρ(u.x)f(x)dx. (18.7.2) For ﬁxed ρandu,w es e t F(x)= e x p [ −iρ(u.x)]f(x), so that ˆF(p,u)=⎪integraldisplay Le−iρ(u.x)f(x)ds, (18.7.3) where Lis the hyperplane u/prime.x=panddsis (n−1)-dimensional surface area measure in Rn.T h u s , ˆF(p,u)=e−iρp⎪integraldisplay u/prime.x=pf(x)ds=e−iρpˆf(p,u), (18.7.4) which is integrated by using (18.5.12) so that ⎪integraldisplay∞ −∞ˆF(p,u)dp=⎪integraldisplay∞ −∞e−iρpˆf(p,u)dp. (18.7.5) Therefore, result (18.7.2) becomes ˜f(ρu)=1 (2π)n/2⎪integraldisplay∞ −∞e−iρpˆf(p,u)dp. (18.7.6) We next denote the one-dimensional Fourier transform along the radial direction by Fr, so that (18.7.6) can be written as F{f(ρu)}=1 (2π)n−1 2Fr⎪bracketleftBig ˆf(p,u)⎪bracketrightBig . (18.7.7) Invoking the inverse Fourier transform in (18.7.6) ˆf(p,u)=1 (2π)n−1 2⎪integraldisplay∞ −∞˜f(ρu)eiρpdρ. (18.7.8) We next consider the inverse Fourier transform with (18.7.6) to obtain f(x)=1 (2π)n/2⎪integraldisplay∞ −∞eik.x˜f(k)dk (18.7.9) =1 (2π)n/2⎪integraldisplay∞ 0ρn−1dρ⎪integraldisplay |u|=1eiρ(x.u)˜f(ρu)du =1 (2π)n/2⎪integraldisplay∞ 0ρn−1dρ⎪integraldisplay |u|=1du⎪integraldisplay∞ −∞eiρ(x.u)ˆf(p,u)e−iρpdp =⎪integraldisplay |u|=1h(x.u,u)du, © 2007 by Taylor & Francis Group, LLC 556 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where h(t,u)=1 (2π)n/2⎪integraldisplay∞ 0dρ⎪integraldisplay∞ −∞ρn−1e−iρ(p−t)ˆf(p,u)dp. (18.7.10) We thus obtain a theorem for the inverse Radon transform: THEOREM 18.7.1 (Inversion Theorem). Ifˆfis the Radon transform of f(x),then f(x)=⎪integraldisplay |u|=1h(x.u,u)du, (18.7.11) where h(t,u)=⎧ ⎪⎪⎨ ⎪⎪⎩a n∂n−1 ∂tn−1ˆf(t,u), for odd n anHHH⎪bracketleftbigg∂n−1 ∂pn−1ˆf(p,u)⎪bracketrightbigg (t),for even n⎫ ⎪⎪⎬ ⎪⎪⎭,(18.7.12ab) where HHHstands for the Hilbert transform with respect to p, a n=⎧ ⎪⎪⎨ ⎪⎪⎩in−1 2( 2π)n−1,for odd n in 2( 2π)n−1,for even n⎫ ⎪⎪⎬ ⎪⎪⎭. (18.7.13ab) THEOREM 18.7.2 (Two Dimensional Inversion Theorem). Ifˆf(p,u) is the Radon transform of f(x)=f(x, y),then f(x, y)=−1 4π2⎪integraldisplay |u|=1du⎪integraldisplay∞ −∞ˆfp(p,u) p−x.udp. (18.7.14) In this case n= 2, it follows from (18.7.12ab)–(18.7.13ab) that h(t,u)=−1 2( 2π)1 π⎪integraldisplay∞ −∞ˆfp(p,u) p−tdp, (18.7.15) where ˆfp(p,u)=∂ ∂pˆf(p,u). Consequently, the formula (18.7.11) reduces to (18.7.14). On the other hand, using u=( c o s φ,sinφ), (18.7.14) gives f(x, y)=−1 π⎪integraldisplayπ 0dφ⎪integraldisplay∞ −∞ˆfp(p,u) p−x.udp. (18.7.16) © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 557 A simple change of variables x=rcosθ,y=rsinθleads to the inversion formula in the polar form f(r, θ)=−1 4π2⎪integraldisplay2π 0dφ⎪integraldisplay∞ −∞ˆfp(p, φ) p−rcos(φ−θ)dp. (18.7.17) We discuss the three-dimensional inverse Radon transform independently be- cause it does not involve the Hilbert transform. THEOREM 18.7.3 (Three Dimensional Inversion Theorem). Ifˆf(p,u)=R{f(x, y, z )},then f(x)=−∇2⎪integraldisplay |u|=1ˆf(p,x.u)du, (18.7.18) where x=(x, y, z )∈R3. PROOF We begin with the inverse of the three-dimensional Fourier trans- form in the form f(x)=F−1 3˜f(su)=⎪integraldisplay∞ 0s2ds⎪integraldisplay |u|=1˜f(qu)eis(x.u)du where the integral over the unit sphere is stated as follows: ⎪integraldisplay |u|=1du=⎪integraldisplay2π 0dφ⎪integraldisplayπ 0sinθd θ , where u=( s i n θcosφ,sinθsinφ,cosθ),θis the polar angle and φis the az- imuthal angle. Invoking the symmetry of the Radon transform ˆf,w h e r e F{ˆf}=˜f,t h e integral over qfrom 0 to ∞can be replaced by one-ha lf the integral from −∞ to∞and hence, f(x)=1 2⎪integraldisplay |u|=1du⎪bracketleftbigg⎪integraldisplay∞ −∞s2˜f(su)eisp⎪bracketrightbigg p=x.uds =1 2⎪integraldisplay |u|=1F−1⎪bracketleftBig {s2˜f(su)}⎪bracketrightBig p=x.udu which is, by the Fourier transform of the second derivatives, or, =−1 2⎪integraldisplay |u|=1⎪bracketleftBig ˆfpp(p,u)⎪bracketrightBig p=x.udu. (18.7.19) This is an inversion formula for the three-dimensional Radon transform. In view of the fact that, for any f(x,u), ∇2f(x,u)=|u|2[fpp(p)]p=x.u=[fpp(p)]p=x.u, (18.7.20) © 2007 by Taylor & Francis Group, LLC 558 INTEGRAL TRANSFORMS and THEIR APPLICATIONS we obtain another form of the inversion formula for the Radon transform f(x)=−1 2∇2⎪integraldisplay |u|=1ˆf(x.u,u)du. (18.7.21) We next introduce the adjoint Radon transform from the deﬁnition of the inner product as /angbracketleftφ,R[f]/angbracketright=⎪integraldisplay∞ −∞dp⎪integraldisplay |u|=1φ(p,u) (Rf)(p,u)du =⎪integraldisplay∞ −∞dp⎪integraldisplay |u|=1φ(p,u)du⎪integraldisplay∞ −∞¯f(x)δ(p−x.u)dx =⎪integraldisplay∞ −∞¯f(x)dx⎪bracketleftBigg⎪integraldisplay |u|=1du⎪integraldisplay∞ −∞φ(p,u)δ(p−x.u)dp⎪bracketrightBigg =⎪integraldisplay∞ −∞⎪bracketleftBigg⎪integraldisplay |u|=1φ(x.u,u)du⎪bracketrightBigg ¯f(x)dx =⎪integraldisplay∞ −∞(R∗[φ])¯f(x)dx=/angbracketleftR∗[φ],f/angbracketright (18.7.22) where the adjoint R∗is deﬁned by R∗[φ](x)=⎪integraldisplay |u|=1φ(x.u,u)du. (18.7.23) This means that the action of the adjoint R∗onφcorresponds to the integra- tion of φover all hyperplanes passing through a given point. We use (18.7.12ab) to introduce the operator Kas follows: Kφ(p,u)=⎧ ⎪⎪⎨ ⎪⎪⎩an∂n−1 ∂pn−1φ(p,u), for odd n anHHH⎪bracketleftbigg∂n−1 ∂pn−1φ(p,u)⎪bracketrightbigg ,for even n⎫ ⎪⎪⎬ ⎪⎪⎭,(18.7.24) where anis deﬁned by (18.7.13ab) and HHHstands for the Hilbert transform. Clearly, it follows from (18.7.12ab) that Kˆf(x.u,u)=h(x.u,u) (18.7.25) and hence, by (18.7.23), R∗[Kˆf(x.u,u)] =R∗[h(x.u,u)] =⎪integraldisplay |u|=1h(x.u,u)du=f(x).(18.7.26) © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 559 This means that the inversion formula (18.7.11) can be written as f=R∗K[ˆf]. (18.7.27) THEOREM 18.7.4 (Parseval’s Theorem). IfR{f(x)}=ˆf(p,u)a n dR{g(x)}=ˆg(p,u),then (a) for even n /angbracketleftf,g/angbracketright=⎪integraldisplay∞ −∞f(x)¯g(x)dx =an⎪integraldisplay |u|=1du⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞ˆf(p,u)¯ˆg(q,u)(p−q)−ndp dq, (18.7.28) where an=(−1)n 2(2π)−n(n−1)!, (b) for odd n /angbracketleftf,g/angbracketright=⎪integraldisplay∞ −∞f(x)¯g(x)dx =(−1)n−1 2 2(2π)n−1⎪integraldisplay |u|=1du⎪integraldisplay∞ −∞ˆf(p,u)¯ˆg(n−1) p(p,u)dp (18.7.29) =1 2(2π)n−1⎪integraldisplay |u|=1du⎪integraldisplay∞ −∞ˆfp(m)(p,u)¯ˆg(m) p(p,u)dp, (18.7.30) where m=n−1 2. We next introduce an operator Hdue to Ludwig (1966) by H[f](p)=1 √ 21 (2π)m∂mf(p) ∂pm, (18.7.31) where m=n−1 2. Consequently, the Parseval’s formula (18.7.30) reduces to the form /angbracketleftf, g/angbracketright=/angbracketleftHˆf, Hˆg/angbracketright. (18.7.32) Ludwig (1966) proved that HRis a unitary transformation from L2(Rn) ontoL2(R×Sn−1). For a proof of the above Parseval’s formulas, the reader is referred to Ludwig (1966) and Zayed (1996). © 2007 by Taylor & Francis Group, LLC 560 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 18.8 Applications of the Radon Transform We prove a remarkable relation between the Radon transform and the solution of the Cauchy problem involving the solution of the wave equation utt=c2∇2u, x∈R3,t >0, (18.8.1) u(x,0)=f(x)ut(x,0) =g(x), (18.8.2ab) where cis a constant and ∇2is the three-dimensional Laplacian. We apply the radon transform of u(x,t)b y ˆu(p,ξξξ,t)=R{u(x,t)}=⎪integraldisplay∞ −∞u(x,t)δ(p−x.ξξξ)dξξξ, (18.8.3) where ξξξ=(ξ1,ξ2,ξ3) is the three-dimensional unit vector in R3so that |ξξξ|= ξ2 1+ξ2 2+ξ2 3=1 . Application of (18.8.3) to (18.8.1)-(18.8.2ab) gives ˆutt=c2⎪parenleftbig ξ2 1+ξ2 2+ξ2 3⎪parenrightbig ˆupp=c2ˆupp, (18.8.4) ˆu(p, ξξξ,0) =ˆf(p, ξξξ),⎪bracketleftbiggdˆu(p, ξξξ, t) dt⎪bracketrightbigg t=0=ˆg(p,ξξξ). (18.8.5) Thus, the radon transform ˆ u(p,ξξξ,t) satisﬁes the Cauchy problem (18.8.4)– (18.8.5). We solve this problem by the application of the Fourier transform of ˆu(p,ξξξ,t)s ot h a t d2ˆU dt2=−c2k2ˆU, (18.8.6) ˆU(k, ξξξ,0) =ˆF(k, ξξξ),⎪parenleftBigg dˆU dt⎪parenrightBigg t=0=ˆG(k, ξξξ), (18.8.7) where ˆU(k,ξξξ,t)=F{ˆu(p, ξξξ, t)}=1 √ 2π⎪integraldisplay∞ −∞e−ikpˆu(p, ξξξ, t)dp (18.8.8) The solution of this transformed problem is obtained in Chapter 2, and the solution of (18.8.6)-(18.8.7) gives ˆU(k, ξξξ, t)=ˆF(k, ξξξ)cos (ckt)+ˆG(k, ξξξ) 2icksin(ckt). © 2007 by Taylor & Francis Group, LLC The Radon Transform and Its Applications 561 Following the method presented in Section 2.12, the D’Alembert solution is obtained in the form ˆu(p, ξξξ, t)=1 2⎪bracketleftBig ˆf(p−ct)+ˆf(p+ct)⎪bracketrightBig +1 2c⎪integraldisplayp+ct p−ctˆg(α, ξξξ)dα.(18.8.9) The inverse Radon transform yields the solution of the Cauchy problem in the form u(x,t)=R−1{ˆu(p, ξξξ, t)}=−∇2⎪integraldisplay |ξξξ|=1ˆu(x.ξξξ, ξξξ, t)dξξξ.(18.8.10) It is noted that the Radon transform transformed the (1+3)-dimensional wave equation (18.8.1) to the (1+1)-dimensional wave equation (18.8.4) which can be solved by using standard methods. In general, the Radon transformreduces problems with ( n+ 1) independent variables to problems with two independent variables. In other words, as stated in equation (18.4.4), if Lis a diﬀerential operator of ( n+ 1)-dimensions with cons tant coeﬃcients, then its Radon transform is R⎪bracketleftbigg L⎪parenleftbigg∂ ∂x1,∂ ∂x2, ...,∂ ∂xn;∂ ∂t⎪parenrightbigg f(x,t)⎪bracketrightbigg =L⎪parenleftbigg u1∂ ∂p,u2∂ ∂p, ..., u n∂ ∂p;∂ ∂t⎪parenrightbigg ˆf(p, ξξξ, t).(18.8.11) This fundamental property help solve hyperbolic partial diﬀerential equations with constant coeﬃcients. 18.9 Exercises 1. Show that (a)R⎪braceleftbig x2exp(−x2−y2)⎪bracerightbig =√ π 2⎪parenleftbig 2p2cos2φ+s i n2φ⎪parenrightbig e−p2. (b)R⎪braceleftbig y2exp(−x2−y2)⎪bracerightbig =√ π 2⎪parenleftbig 2p2sin2φ+c o s2φ⎪parenrightbig e−p2. (c)R⎪braceleftbig (x2+y2)e x p ( −x2−y2)⎪bracerightbig =√ π 2⎪parenleftbig 2p2+1⎪parenrightbig e−p2. 2. Verify that ∂f ∂u1=−∂ ∂p⎪bracketleftBig R⎪braceleftBig xe−x2−y2⎪bracerightBig⎪bracketrightBig . © 2007 by Taylor & Francis Group, LLC 562 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 3. Iff(x, y)=e x p⎪parenleftbig −x2−y2⎪parenrightbig ,show that ∂ˆf ∂uk=√ πuk(2p2−1)e−p2. 4. Prove result (18.5.16). 5. IfAis nonsingular m×nmatrix, show that (a)R{f(Ax)}=ˆf(p,Au), where x=(x1,x2, ..., x n),and Auis a unit vector. (b)R{f(cx)}=c−nˆf⎪parenleftBig p,u c⎪parenrightBig =c1−nˆf(cp,u),A=cI. 6. Iff(x, y)=e x p ( −x2−y2),n=2,c=(σ√ 2)−1, use 5(b) and f(Ax)=e x p⎪parenleftbigg −x2+y2 2σ2⎪parenrightbigg and1 cˆf(cp,u)=σ√ 2πexp⎪parenleftbigg −p2 2σ2⎪parenrightbigg , show that the Radon transform of the symmetric Gaussian probability density function is given by R⎪braceleftbigg1 2πσ2exp⎪parenleftbigg −x2+y2 2σ2⎪parenrightbigg⎪bracerightbigg =1 σ√ 2πexp⎪parenleftbigg −p2 2σ2⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC 19 Wavelets and Wavelet Transforms “Wavelets are without doubt an ex citing and intuitive concept. The concept brings with it a new way of thinking, which is ab- solutely essential and was entirely missing in previously existing algorithms.” Yves Meyer “Today the boundaries between mathematics and signal and im- age processing have faded, and mathematics has beneﬁted from the rediscovery of wavelets by experts from other disciplines. The detour through signal and image processing was the most direct path leading from Haar basis to Daubechies’s wavelets.” Yves Meyer 19.1 Brief Historical Remarks The concept of “wavelets” or “ondelettes” started to appear in the literatureonly in the early 1980s. This new concept can be viewed as a synthesis of var-ious ideas which originated from diﬀerent disciplines including mathematics, physics and engineering. In 1982 Jean Morlet, a French geophysical engineer, ﬁrst introduced the idea of wavelet transform as a new mathematical tool forseismic signal analysis. It was Alex Grossmann, a French theoretical physicist, who quickly recognized the importance of the Morlet wavelet transform which is something similar to coherent states formalism in quantum mechanics, and developed an exact inversion formula for the wavelet transform. In 1984 the joint venture of Morlet and Grossmann l ed to a detailed mat hematical study of the continuous wavelet transforms and their various applications. It has become clear from their work that, analogous to the Fourier expansions, the wavelet theory has provided a new method for decomposing a function or asignal. In 1985 Yves Meyer, a French pure mathematician, recognized immediately the deep connection between the Calder´ on formula in harmonic analysis and 563 © 2007 by Taylor & Francis Group, LLC 564 INTEGRAL TRANSFORMS and THEIR APPLICATIONS the new algorithm discovered by Morlet and Grossmann. Using the knowledge of the Calder´ on-Zygmund operators and the L ittlewood-Paley theory, Meyer was able to give a mathematical foundation for wavelet theory. The ﬁrst ma- jor achievement of wavelet analysis was Daubechies, Grossmann and Meyer’s (1986) construction of a “painless” non-orthogonal wavelet expansion. During 1985-1986, further work of Meyer and Lemari´ e on the ﬁrst construction of a smooth orthonormal wavelet basis on RandRNmarked the beginning of their famous contributions to the wavelet theory. At the same time, St´ ephans Mallat recognized that some quadratic mirror ﬁlters play an important role for the construction of orthogonal wavelet bases generalizing the Haar sys-tem. Meyer (1986) and Mallat (1988) realized that the orthogonal wavelet bases could be constructed systemati cally from a general formalism. Their collaboration culminated with the remarkable discovery by Mallat (1989 a,b)of a new formalism, which is the so called multiresolution analysis .I tw a s also Mallat who constructed the wavelet decomposition and reconstruction algorithms using the multiresolution analysis. Mallat’s brilliant work was the major source of many new developments in wavelets. A few months later, G. Battle (1987) and Lamari´ e (1988) independently proposed the construction of spline orthogonal wavelets with exponential decay. Inspired by the work of Meyer, Ingrid Daubechies (1988) made a new re- markable contribution to wavelet theory by constructing families of compactly supported orthonormal wavelets with some degree of smoothness. Her 1988 paper had a tremendous positive impact on the study of wavelets and their diverse applications. This work signiﬁcantly explained the connection between the continuous wavelets on R, and the discrete wavelets on ZorZ N,w h e r e the latter has become useful for digital signal analysis. The idea of frames wasintroduced by Duﬃn and Schaeﬀer (1952) and subsequently studied in some detail by Daubechies (1990, 1992). In spite of tremendous success, expert- s in wavelet theory recognized that it is diﬃcult to construct wavelets that are symmetric, orthogonal and compactly supported. In order to overcome this diﬃculty, Cohen et al. (1992a,b) studied bi-orthogonal wavelets in some detail. Chui and Wang (1991, 1992) introduced compactly supported spline wavelets, and semi-orthogonal wavelet analysis. On the other hand, Beylkin, Coifman and Rokhlin (1991), and Beylkin (1992) have successfully applied themultiresolution analysis generated by a completely orthogonal scaling func- tion to study a wide variety of integral operators on L 2(R) by a matrix in a wavelet basis. This work culminated with the remarkable discovery of new algorithms in numerical analysis. Consequently, some signiﬁcant progress has been made in boundary element methods, ﬁnite element methods, and numer- ical solutions of partial diﬀerential equations using wavelet analysis. For more detailed historical introduction, the reader is referred to Debnath (2002). We close this historical introduction by citing some of the applications which include addressing problems in signal processing, computer vision, seismology, turbulence, computer graphics, image processing, structures of the galaxies in the Universe, digital communication, pattern recognition, approximation © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 565 theory, quantum optics, biomedical engineering, sampling theory, matrix the- ory, operator theory, diﬀerential equations, numerical analysis, statistics andmultiscale segmentation of well logs, natural scenes and mammalian visual systems. Wavelets allow complex information such as music, speech, images, patterns, etc., to be decomposed into elem entary form, called building blocks (wavelets). 19.2 Continuous Wavelet Transforms An integral transform is an operator Ton a space of functions on some Ω ⊂RN w h i c hi sd e ﬁ n e db y (Tf)(y)=⎪integraldisplay ΩK(x, y)f(x)dx. The properties of the transform depend on the function K, which is called thekernel of the transform. For example, in the case of the Fourier transform K(x, y)=e−ixy.N o t et h a t ycan be interpreted as a s caling factor. We take the exponential function ϕ(x)=eixand then generate a one parameter family of functions by taking scaled copies of ϕ,t h a ti s ϕα(x)=e−iαx, for all α∈R. The continuous wavelet transform is similar to the Fourier transform in the sense that it is based on a single function ψand that this function is scaled. But unlike the Fourier transform, we also shift the function, thus, generatinga two parameter family of functions ψ a,b. It is convenient to deﬁne ψa,bas follows: ψa,b(x)=|a|−1 2ψ⎪parenleftbiggx−b a⎪parenrightbigg . Then the continuous wavelet transform is deﬁned by (Wψf)(a, b)=⎪integraldisplay∞ −∞f(t) ψa,b(t)dt=|a|−1 2⎪integraldisplay∞ −∞f(t) ψ⎪parenleftbiggt−b a⎪parenrightbigg dt. The continuous wavelet transform is not a single transform as the Fourier transform, but any transform obtained in this way. Properties of a particular transform will depend on the choice of ψ. One of the ﬁrst properties we expect of any integral transform is that the original function can be reconstructed from the transform. We will prove a theorem which gives conditions on ψthat guarantee invertibility of the transform. First we need to deﬁne the object of our study more precisely. DEFINITION 19.2.1 (Wavelet) By a wavelet we mean a function ψ∈ © 2007 by Taylor & Francis Group, LLC 566 INTEGRAL TRANSFORMS and THEIR APPLICATIONS L2(R)satisfying the admissibility condition ⎪integraldisplay∞ −∞|⎪hatwideψ(ω)|2 |ω|dω <∞, (19.2.1) where⎪hatwideψis the Fourier transform ψ, i.e., ⎪hatwideψ(ω)=1 √ 2π⎪integraldisplay∞ −∞e−iωxψ(x)dx. Ifψ∈L2(R), then ψa,b(x)∈L2(R) for all a, b. Indeed, /bardblψa,b(t)/bardbl2=|a|−1⎪integraldisplay∞ −∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingleψ⎪parenleftbiggx−b a⎪parenrightbigg⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle2 dt=⎪integraldisplay∞ −∞|ψ(t)|2dt=/bardblψ/bardbl2.(19.2.2) The Fourier transform of ψa,b(x)i sg i v e nb y ˆψa,b(ω)=|a|−1 21 √ 2π⎪integraldisplay∞ −∞e−iωxψ⎪parenleftbiggx−b a⎪parenrightbigg dx=⎪radicalbig |a|e−ibωˆψ(aω).(19.2.3) 1 -10 1 Figure 19.1 The Haar wavelet. Example 19.2.1 (The Haar Wavelet )L e t ψ(x)=⎧ ⎪⎨ ⎪⎩10 ≤x<1 2 −11 2≤x<1 0o t h e r w i s e . The Haar wavelet is shown in Figure 19.1. © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 567 Then the Fourier transform ⎪hatwideψ(ω)=F{ψ(x)}is ⎪hatwideψ(ω)=1 ⎪radicalbig (2π)(sinω 4)2 ω 4e−i(ω−π)/2=1 ⎪radicalbig (2π)⎪parenleftbigg4i ω⎪parenrightbigg e−iω 2sin2⎪parenleftBigω 4⎪parenrightBig . and ⎪integraldisplay∞ −∞|⎪hatwideψ(ω)|2 |ω|dω=8 π⎪integraldisplay∞ −∞|sinω 4|4 |ω|3dω <∞. Figure 19.2 The absolute value of the Fourier transform of the Haar wavelet. The Haar wavelet is one of the classic examples. It is well-localized in the time domain, but it is not continuous. The absolute value of the Fourier trans- form of the Haar wavelet, |ˆψ(ω)|, is plotted in Figure 19.2. This ﬁgure clearly indicates that the Haar wavelet has poor frequency localization, since it does not have compact support in the frequency domain. The function |ˆψ(ω)|is even and attains its maximum at the frequency ω0∼4.662. The rate of decay asω→∞ is asω−1. The reason for the slow decay is discontinuity of ψ.I t s discontinuous nature is a serious weakness in many applications. However, theHaar wavelet is one of the most fundamental examples that illustrate major features of the general wavelet theory. THEOREM 19.2.1 Letψbe a wavelet and let ϕbe a bounded integrable function. Then the function ψ∗ϕis a wavelet. © 2007 by Taylor & Francis Group, LLC 568 INTEGRAL TRANSFORMS and THEIR APPLICATIONS PROOF Since ⎪integraldisplay∞ −∞|ψ∗ϕ(x)|2dx=⎪integraldisplay∞ −∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪integraldisplay ∞ −∞ψ(x−u)ϕ(u)du⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle2 dx ≤⎪integraldisplay∞ −∞⎪parenleftbigg⎪integraldisplay∞ −∞|ψ(x−u)||ϕ(u)|du⎪parenrightbigg2 dx =⎪integraldisplay∞ −∞⎪parenleftbigg⎪integraldisplay∞ −∞|ψ(x−u)||ϕ(u)|1/2|ϕ(u)|1/2du⎪parenrightbigg2 dx ≤⎪integraldisplay∞ −∞⎪parenleftbigg⎪integraldisplay∞ −∞|ψ(x−u)|2|ϕ(u)|du⎪integraldisplay∞ −∞|ϕ(u)|du⎪parenrightbigg dx ≤⎪integraldisplay∞ −∞|ϕ(u)|du⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞|ψ(x−u)|2|ϕ(u)|dxdu =⎪parenleftbigg⎪integraldisplay∞ −∞|ϕ(u)|du⎪parenrightbigg2⎪integraldisplay∞ −∞|ψ(x)|2dx <∞, we have ψ∗ϕ∈L2(R). Moreover, ⎪integraldisplay∞ −∞|/hatwideψ∗ϕ(ω)|2 |ω|dω=⎪integraldisplay∞ −∞|⎪hatwideψ(ω)⎪hatwideϕ(ω)|2 |ω|dω=⎪integraldisplay∞ −∞|⎪hatwideψ(ω)|2 |ω||⎪hatwideϕ(ω)|2dω ≤sup|⎪hatwideϕ(ω)|2⎪integraldisplay∞ −∞|⎪hatwideψ(ω)|2 |ω|dω <∞ Thus, the function ψ∗ϕis a wavelet. Figure 19.3 A continuous wavelet. Example 19.2.2 Theorem 19.2.1 can be used to generate examples of wavelets. For example, © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 569 if we take the Haar wavelet and convolve it with the following function ϕ(x)=⎧ ⎪⎨ ⎪⎩0x<0 10≤x≤1, 0x≥1, then we obtain a simple continuous function (see Figure 19.3). If we convolve the Haar wavelet with ϕ(x)=e−x2, then the obtained wavelet is smooth (see Figure 19.4). Figure 19.4 A smooth wavelet. DEFINITION 19.2.2 (Continuous Wavelet Transform). Let ψ∈L2(R) and let, for a,b∈R,a/negationslash=0, ψa,b(x)=|a|−1 2ψ⎪parenleftbiggx−b a⎪parenrightbigg . The integral transform Wψdeﬁned on L2(R)by (Wψf)(a,b)=⎪integraldisplay∞ −∞f(t) ψa,b(t)dt=/angbracketleftf,ψa,b/angbracketright (19.2.4) is called a continuous wavelet transform. The function ψis often called the mother wavelet or, the analyzing wavelet . The parameter bcan be interpreted as th et i m et r a n s l a t i o na n d ais a scaling parameter which measures the degree of compression. © 2007 by Taylor & Francis Group, LLC 570 INTEGRAL TRANSFORMS and THEIR APPLICATIONS LEMMA 19.2.1 For any f∈L2(R), we have F{(Wψf)(a, b)}=⎪radicalbig 2π|a|ˆf(ω) ˆψ(aω). (19.2.5) PROOF Using the Parseval formula for the Fourier transform, it follows from (19.2.4) that (Wψf)(a, b)=/angbracketleftf,ψa,b/angbracketright=⎪angbracketleftBig ˆf,ˆψa,b⎪angbracketrightBig =1 √ 2π⎪integraldisplay∞ −∞⎪braceleftBig⎪radicalbig 2π|a|ˆf(ω) ˆψ(aω)⎪bracerightBig eibωdω. (19.2.6) This means that F{(Wψf)(a, b)}=1 √ 2π⎪integraldisplay∞ −∞e−ibω(Wψf)(a, b)db =⎪radicalbig 2π|a|ˆf(ω) ˆψ(aω). (19.2.7) THEOREM 19.2.2 (Parseval’s Relation for Wavelet Transforms ). Suppose that ψ∈L2(R)w h i c h satisfy the admissibility condition Cψ=2π⎪integraldisplay∞ −∞|⎪hatwideψ(ω)|2 |ω|dω <∞. (19.2.8) Then, for any f,g∈L2(R), we have ⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞(Wψf)(a, b) (Wψg)(a, b)dbda a2=Cψ/angbracketleftf,g/angbracketright. (19.2.9) PROOF From (19.2.6) we get (Wψf)(a, b)=⎪radicalbig |a|⎪integraldisplay∞ −∞⎪hatwidef(ω)eibω ⎪hatwideψ(aω)dω (19.2.10) and (Wψg)(a, b)=⎪radicalbig |a|⎪integraldisplay∞ −∞ ⎪hatwideg(σ)e−ibσ⎪hatwideψ(aσ)dσ. (19.2.11) Substituting (19.2.10) and (19.2.11) in the left hand-side of (19.2.9) gives ⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞(Wψf)(a, b) (Wψg)(a, b)dbda a2 =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞dbda a2⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞|a|⎪hatwidef(ω) ⎪hatwideg(σ) ⎪hatwideψ(aω)⎪hatwideψ(aσ)eib(ω−σ)dωdσ © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 571 which is, by interchanging the order of integration, =2π⎪integraldisplay∞ −∞da |a|⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞⎪hatwidef(ω) ⎪hatwideg(σ) ⎪hatwideψ(aω)⎪hatwideψ(aσ)dωdσ1 2π⎪integraldisplay∞ −∞eib(ω−σ)db =2π⎪integraldisplay∞ −∞da |a|⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞⎪hatwidef(ω) ⎪hatwideg(σ) ⎪hatwideψ(aω)⎪hatwideψ(aσ)δ(σ−ω)dωdσ =2π⎪integraldisplay∞ −∞da |a|⎪integraldisplay∞ −∞⎪hatwidef(ω) ⎪hatwideg(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideψ(aω)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 dω and ﬁnally, again interchanging the order of integration and putting aω=x, =2π⎪integraldisplay∞ −∞⎪hatwidef(ω) ⎪hatwideg(ω)dω⎪integraldisplay∞ −∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideψ(x)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 |x|dx=Cψ⎪angbracketleftBig ⎪hatwidef,⎪hatwideg⎪angbracketrightBig =Cψ/angbracketleftf,g/angbracketright.(19.2.12) Iff=g, then (19.2.9) assumes the form ⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞|(Wψf)(a, b)|2dbda a2=Cψ/bardblf/bardbl2. (19.2.13) THEOREM 19.2.3 (Inversion formula ). Iff∈L2(R), then f(x)=1 Cψ⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞(Wψf)(a, b)ψa,b(x)dbda a2, (19.2.14) where the equality holds almost everywhere. PROOF For any g∈L2(R), we have Cψ/angbracketleftf,g/angbracketright=⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞(Wψf)(a, b) (Wψg)(a, b)dbda a2 =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞(Wψf)(a, b) ⎪integraldisplay∞ −∞g(t) ψa,b(t)dtdbda a2 =⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞(Wψf)(a, b)ψa,b(t)dbda a2 g(t)dt =⎪angbracketleftbigg⎪integraldisplay∞ −∞⎪integraldisplay∞ −∞(Wψf)(a, b)ψa,bdbda a2,g⎪angbracketrightbigg . Since gis an arbitrary element of L2(R), the inversion formula 19.2.14 follows. © 2007 by Taylor & Francis Group, LLC 572 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The following theorem su mmarizes some elementary properties of the con- tinuous wavelet transform. Proofs are straightforward, hence, are left as exer- cises. THEOREM 19.2.4 Suppose ψandϕare wavelets and let f,g∈L2(R). (i) (Wψ(αf+βg))(a,b)=α(Wψf)(a,b)+β(Wψg)(a,b) for any α, β∈C, (ii) (Wψ(Tcf))(a,b)=(Wψf)(a,b−c), where Tcis the translation operator deﬁned by Tcf(t)=f(t−c), (iii) (Wψ(Dcf))(a,b)=1 √ c(Wψf)⎪parenleftbiga c,b c⎪parenrightbig ,w h e r e cis a positive number and Dcis the dilation operator deﬁned by Dcf(t)=1 cf(t c), (iv) (Wψϕ)(a,b)= (Wϕψ)⎪parenleftbig1 a,−b a⎪parenrightbig ,a/negationslash=0 , (v) (Wαψ+βϕf)(a,b)= α(Wψf)(a,b)+ β(Wϕf)(a,b) for any α, β∈C, (vi) (WPψPf)(a,b)=(Wψf)(a,−b), where Pis the parity operator deﬁned byPf(t)=f(−t), (vii) ( WTcψf)(a,b)=(Wψf)(a,b+ca), (viii) ( WDcψf)(a,b)=1 √ c(Wψf)(ac, b),c>0. For the wavelets to be useful analyzing functions, the mother wavelet must have certain properties. One such property is deﬁned by the admissibilitycondition (19.2.1) which guarantees existence of the inversion formula for the continuous wavelet transform. If ψ∈L 1(R), then its Fourier transform ⎪hatwideψis continuous. If ⎪hatwideψis continuous, Cψcan be ﬁnite only if ⎪hatwideψ(0) = 0, or, equiv- alently,⎪integraltext∞ −∞ψ(t)dt= 0. This means that ψmust be an oscillatory function with zero mean. Condition (19.2.1) also imposes a restriction on the rate ofdecay of |⎪hatwideψ(ω)| 2. In addition to the admissibility condition (19.2.1), there are other prop- erties that may be useful in particular applications. For example, it may be necessary to require that ψbentimes continuously diﬀerentiable or inﬁnitely diﬀerentiable. If the Haar wavelet is convolved ( n+ 1) times with the function ϕgiven in Example 19.2.2, then the resulting function ψ∗ϕ∗···∗ ϕis ann times diﬀerentiable wavelet. The function in Figure 19.4 is an inﬁnitely dif-ferentiable wavelet. The so-called “Mexican hat wavelet” is another example of an inﬁnitely diﬀerentiable wavelet. Example 19.2.3 (Mexican Hat Wavelet ). This wavelet is deﬁned by ψ(t)=( 1 −t 2)e−at2/2 © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 573 Figure 19.5 Mexican hat wavelet. and shown in Figure 19.5 with a=1 . Another desirable property of wavelets is the so called “localization prop- erty.” We want ψto be well localized in both time and frequency domains. In other words, ψand its derivatives must decay very rapidly. For frequency localization, ⎪hatwideψ(ω) must decay suﬃciently rapidly as ω→∞ and⎪hatwideψ(ω) should be ﬂat in the neighborhood of ω= 0. The ﬂatness at ω= 0 is associated with the number of vanishing moments of ψ.T h e k-th moment of ψis deﬁned by mk=⎪integraldisplay∞ −∞tkψ(t)dt. A wavelet is said to have nvanishing moments if ⎪integraldisplay∞ −∞tkψ(t)dt=0 f o r k=0,1,...,n. Or, equivalently, ⎪bracketleftBigg dk⎪hatwideψ(ω) dωk⎪bracketrightBigg ω=0=0 f o r k=0,1,...,n. Wavelets with a larger number of vanishing moments result in more ﬂatness when frequency ωis small. 19.3 The Discrete Wavelet Transform While the continuous wavelet transform is compared to the Fourier transform, which requires calculating the integral⎪integraltext∞ −∞e−iωxf(x)dxfor all (or, almost all) © 2007 by Taylor & Francis Group, LLC 574 INTEGRAL TRANSFORMS and THEIR APPLICATIONS ω∈R, the discrete wavelet transform can be compared to the Fourier series, which requires calculating the integral⎪integraltext2π 0e−inxf(x)dxfor integer values of n. Since the continuous wavelet transf orm is a two parameter representation of a function (Wψf)(a, b)=|a|−1 2⎪integraldisplay∞ −∞f(t) ψ⎪parenleftbiggt−b a⎪parenrightbigg dt, we can discretize it by assuming that aandbtake only integer values. It turns out that it is better to discretize it in a diﬀerent way. First we ﬁx two positive constants a0andb0and then deﬁne ψm,n(x)=a−m/2 0ψ(a−m 0x−nb0), (19.3.1) where mandnrange over Z.B yt h e discrete wavelet coeﬃcients off∈L2(R) we mean the numbers /angbracketleftf,ψm,n/angbracketright,w h e r e m, n∈Z. The fundamental question here is whether it is possible to reconstruct ffrom those coeﬃcients. The weakest interpretation o f this problem is whether /angbracketleftf,ψm,n/angbracketright=/angbracketleftg,ψm,n/angbracketrightfor all m, n∈Zimplies f=g. In practice we expect much more than that: we want /angbracketleftf,ψm,n/angbracketrightand/angbracketleftg,ψm,n/angbracketrightto be “close” if fandgare “close.” This will be guaranteed if there exists a B>0, such that ∞⎪summationdisplay m,n=−∞|/angbracketleftf,ψm,n/angbracketright|2≤B/bardblf/bardbl2 for all f∈L2(R). Similarly, we want fandgto be “close” if /angbracketleftf,ψm,n/angbracketrightand /angbracketleftg,ψm,n/angbracketrightare “close.” This is important because we want to be sure that when we neglect some small terms in the representation of fin terms of /angbracketleftf,ψm,n/angbracketright, then the reconstructed function will not diﬀer much from f.T h e representation will have this property if there exists an A>0, such that A/bardblf/bardbl2≤∞⎪summationdisplay n=1|/angbracketleftf,ψm,n/angbracketright|2 for all f∈L2(R). These two requirements are best investigated in terms of the so-called frames. DEFINITION 19.3.1 (Frame). A sequence (ϕ1,ϕ2,···)in a Hilbert s- paceHis called a frame if there exist A, B > 0such that A||f||2≤∞⎪summationdisplay n=1|/angbracketleftf, ϕ n/angbracketright|2≤B||f||2(19.3.2) for all f∈H. The constants AandBare called frame bounds. If A=B,t h e n the frame is called tight. © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 575 If (ϕn) is an orthonormal basis, then it is a tight frame since⎪summationtext∞ n=1|/angbracketleftf, ϕ n/angbracketright|2 =||f||2for all f∈H. The vectors (1 ,0), (−1 2,√ 3 2), (−1 2,−√ 3 2) form a tight frame in C2which is not a basis. 19.4 Examples of Orthonormal Wavelets Since the discovery of wavelets, orthon ormal wavelets pla ya ni m p o r t a n tr o l e in the wavelet theory and have a variet y of applications. In this section we discuss several examples of orthonormal wavelets. DEFINITION 19.4.1 (Orthonormal Wavelet). A wavelet ψ∈L2(R)is called orthonormal if the family of functions ψm,ngenerated from ψby ψm,n(x)=2m/2ψ⎪parenleftBig 2m⎪parenleftBig x−n 2m⎪parenrightBig⎪parenrightBig =2m/2ψ(2mx−n),m , n ∈Z,(19.4.1) is orthonormal, that is, /angbracketleftψm,n,ψk,/lscript/angbracketright=⎪integraldisplay∞ −∞ψm,n(x)ψk,/lscript(x)dx=δm,kδn,/lscript, (19.4.2) for all m,n,k,/lscript ∈Z. The following lemma is often useful when dealing with orthogonality of wavelets. LEMMA 19.4.1 Ifψ,ϕ∈L2(R),t h e n /angbracketleftψm,k,ϕm,/lscript/angbracketright=/angbracketleftψn,k,ϕn,/lscript/angbracketright, (19.4.3) for all m,n,k,/lscript ∈Z. PROOF We have /angbracketleftψm,k,ϕm,/lscript/angbracketright=⎪integraldisplay∞ −∞2mψ(2mx−k)ϕ(2mx−/lscript)dx, which is, by assuming 2mx=2nt, /angbracketleftψm,k,ϕm,/lscript/angbracketright=⎪integraldisplay∞ −∞2nψ(2nt−k)ϕ(2nt−/lscript)dx=/angbracketleftψn,k,ϕn,/lscript/angbracketright. © 2007 by Taylor & Francis Group, LLC 576 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Example 19.4.1 (The Haar Wavelet ). The simplest example of an orthonormal wavelet is the classic Haar wavelet. We consider the scaling function ϕ=χ[0,1). The function ϕsatisﬁes the dilation equation ϕ(x)=√ 2∞⎪summationdisplay n=−∞cnϕ(2x−n), (19.4.4) where the coeﬃcients cnare given by cn=√ 2⎪integraldisplay∞ −∞ϕ(x)ϕ(2x−n)dx. (19.4.5) Evaluating this integral with ϕ=χ[0,1)givescnas follows: c0=c1=1 √ 2and cn=0 f o r n>1. Consequently, the dilation equation becomes ϕ(x)=ϕ(2x)+ϕ(2x−1). (19.4.6) This means that ϕ(x) is a linear combination of the even and odd translates ofϕ(2x) and satisﬁes a very simple two-scale relation (19.4.6), as shown in Figure 19.6. Figure 19.6 Two-scale relation of ϕ(x)=ϕ(2x)+ϕ(2x−1). The Haar mother wavelet is obtained as a simple two-scale relation ψ(x)=ϕ(2x)−ϕ(2x−1) (19.4.7) =χ[0,1 2](x)−χ[1 2,1](x) =⎧ ⎨ ⎩1i f 0 ≤x<1 2, −1i f1 2≤x<1, 0o t h e r w i s e .⎫ ⎬ ⎭. (19.4.8) © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 577 For any m, n∈Z,w eh a v e ψm,n(t)=2−m/2ψ⎪parenleftbig 2−mt−n⎪parenrightbig =⎧ ⎪⎨ ⎪⎩2−m/22mn≤t<2mn+2m−1, −2−m/22mn+2m−1≤t<2mn+2m, 0o t h e r w i s e . Clearly, /bardblψm,n/bardbl2=/bardblψ/bardbl2= 1, for all m, n∈Z.T ov e r i f yt h a t {ψm,n}is an or- thonormal system, we observe that /angbracketleftψm,n,ψk,/lscript/angbracketright=⎪integraldisplay∞ −∞2m/2ψ(2mx−n)2k/2ψ⎪parenleftbig 2kx−/lscript⎪parenrightbig dx, which gives, by the change of variables 2mx−n=t, /angbracketleftψm,n,ψk,/lscript/angbracketright=2k/22−m/2⎪integraldisplay∞ −∞ψ(t)ψ⎪parenleftbig 2k−m(t+n)−/lscript⎪parenrightbig dt. (19.4.9) Form=k,w eo b t a i n /angbracketleftψm,n,ψm,/lscript/angbracketright=⎪integraldisplay∞ −∞ψ(t)ψ(t+n−/lscript)dt=δ0,n−/lscript=δn,/lscript, (19.4.10) where ψ(t)/negationslash=0i n0 ≤t<1a n d ψ⎪parenleftbig t− /lscript−n⎪parenrightbig /negationslash=0i n /lscript−n≤t<1+/lscript−n,a n d these intervals are disjoint from each other unless n=/lscript. We now consider the case m/negationslash=k. In view of symmetry, it suﬃces to consider the case m>k . Putting r=m−k>0 in (19.4.9), we obtain, for m>k , /angbracketleftψm,n,ψm,/lscript/angbracketright=2r/2⎪integraldisplay∞ −∞ψ(t)ψ(2rt+s)dt, (19.4.11) where s=2rn−/lscript. Thus, it suﬃces to show that ⎪integraldisplay1 2 0ψ(2rt+s)dt−⎪integraldisplay1 1 2ψ(2rt+s)dt=0. Using a simple change of variables 2rt+s=x, we ﬁnd ⎪integraldisplay1 2 0ψ(2rt+s)dt−⎪integraldisplay1 1 2ψ(2rt+s)dt=⎪integraldisplaya sψ(x)dx−⎪integraldisplayb aψ(x)dx,(19.4.12) where a=s+2r−1andb=s+2r. Since the interval [ s, a] contains the support [0,1] ofψ, the ﬁrst integral in (19.4.12) is zero. Similarly, the second integral is also zero. Example 19.4.2 (The Shannon Wavelet ). The function ψwhose Fourier transform satisﬁes ⎪hatwideψ(ω)=χI(ω), (19.4.13) © 2007 by Taylor & Francis Group, LLC 578 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where I=[−2π,−π]∪[π,2π], is called the Shannon wavelet . The function ψ can directly be obtained from th e inverse Fourier transform of ⎪hatwideψso that ψ(t)=1 2π⎪integraldisplay∞ −∞eiωt⎪hatwideψ(ω)dω =1 2π⎪bracketleftbigg⎪integraldisplay−π −2πeiωtdω+⎪integraldisplay2π πeiωtdω⎪bracketrightbigg =1 πt(sin2πt−sinπt)=sin⎪parenleftbigπt 2⎪parenrightbig ⎪parenleftbigπt 2⎪parenrightbigcos⎪parenleftbigg3πt 2⎪parenrightbigg . (19.4.14) This function is orthonormal to its translates by integers. Indeed, by Parseval’s relation, /angbracketleftψ(t),ψ(t−n)/angbracketright=1 2π⎪angbracketleftBig ⎪hatwideψ,einω⎪hatwideψ⎪angbracketrightBig =1 2π⎪integraldisplay∞ −∞⎪hatwideψ(ω)einω ⎪hatwideψ(ω)dω =1 2π⎪integraldisplay2π −2πeinωdω=δ0,n. The wavelet basis is now given by ψm,n(t)=2−m/2ψ⎪parenleftbigg 2−mt−n−1 2⎪parenrightbigg ,m , n ∈Z or, ψm,n(t)=2−m 2sin⎪braceleftbigπ 2(2−mt−n)⎪bracerightbig π 2(2−mt−n)cos⎪braceleftbigg3π 2⎪parenleftbig 2−mt−n⎪parenrightbig⎪bracerightbigg .(19.4.15) For any ﬁxed n∈Z, the functions ψm,n(t) form a basis for the space of functions supported on the interval ⎪bracketleftbig −2−m+1π,−2−mπ⎪bracketrightbig ∪⎪bracketleftbig 2−mπ,2−m+1π⎪bracketrightbig . The system {ψm,n(t)},m, n∈Z, is an orthonormal basis for L2(R). Both ψ(t)a n d⎪hatwideψ(ω) are shown in Figure 19.7. The Fourier transform of ψm,nis ⎪hatwideψm,n(ω)=⎪braceleftBigg 2m/2exp(−iωn2m)i f 2−mπ<|ω|<2−m+1π, 0o t h e r w i s e .(19.4.16) Evidently, ⎪hatwideψm,nand⎪hatwideψk,/lscriptdo not overlap for m/negationslash=k. Hence, by the Parseval relation [(equation (3.4.37), Debnath, 2002)], it turns out that, for m/negationslash=k, /angbracketleftψm,n,ψk,/lscript/angbracketright=1 2π⎪angbracketleftBig ⎪hatwideψm,n,⎪hatwideψk,/lscript⎪angbracketrightBig =0. (19.4.17) © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 579 Figure 19.7 The Shannon wavelet and its Fourier transform. Form=k,w eh a v e /angbracketleftψm,n,ψk,/lscript/angbracketright=1 2π⎪angbracketleftBig ⎪hatwideψm,n,⎪hatwideψm,/lscript⎪angbracketrightBig =1 2π2−m⎪integraldisplay∞ −∞exp⎪braceleftbig −iω2−m(n−/lscript)⎪bracerightbig⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideψ⎪parenleftbig 2−mω⎪parenrightbig⎪vextendsingle⎪vextendsingle⎪vextendsingle2 dω =1 2π⎪integraldisplay∞ −∞exp{−iσ(n−/lscript)}dσ=δn,/lscript. (19.4.18) This shows that {ψm,n(t)}is an orthonormal system. Example 19.4.3 (The Daubechies Wavelets and Algorithms ). Daubechies (1988, 1992) ﬁrst developed the theory and construction of continuous orthonormal wavelets with compact support. Wavelets with compact support have many interestingproperties. They can be constructed to have a given number of derivatives and to have a given number of vanishing moments. We assume that the scaling function ϕsatisﬁes the dilation equation ϕ(x)=√ 2∞⎪summationdisplay n=−∞cnϕ(2x−n), (19.4.19) where cn=/angbracketleftϕ, ϕ1,n/angbracketrightand⎪summationtext∞ n=−∞|cn|2≤∞. If the scaling function ϕhas compact support, then only a ﬁnite number of cnhave nonzero values. The associated generating function ⎪hatwidem, ⎪hatwidem(ω)=1 √ 2∞⎪summationdisplay n=−∞cne−iωn(19.4.20) © 2007 by Taylor & Francis Group, LLC 580 INTEGRAL TRANSFORMS and THEIR APPLICATIONS is a trigonometric polynomial and it satisﬁes the identity [(see equation (7.3.4), Debnath, 2002)] with special values ⎪hatwidem(0)=1 and ⎪hatwidem(π) = 0. If coeﬃcients cn are real, then the corresponding scaling function, as well as the mother wavelet ψ, will also be real-valued. The mother wavelet ψcorresponding to ϕis given by the formula [(see equation (7.3.24), Debnath, 2002)] with |⎪hatwideϕ(0)|=1 .T h e Fourier transform ⎪hatwideψ(ω)i sm-times continuously diﬀerentiable and it satisﬁes the moment condition ⎪hatwideψ(k)(0) = 0 for k=0,1,...,m . (19.4.21) It follows that ψ∈Cm(R) implies that ⎪hatwidem0has a zero at ω=πof order ( m+1 ) . In other words, ⎪hatwidem0(ω)=⎪parenleftbigg1+e−iω 2⎪parenrightbiggm+1 ⎪hatwideL(ω), (19.4.22) where⎪hatwideLis a trigonometric polynomial. In addition to the orthogonality condition [(see equation (7.3.4), Debnath, 2002)], we assume ⎪hatwidem0(ω)=⎪parenleftbigg1+e−iω 2⎪parenrightbiggN ⎪hatwideL(ω), (19.4.23) where⎪hatwideL(ω)i s2π-periodic and ⎪hatwideL∈CN−1(R). Evidently, |⎪hatwidem0(ω)|2=⎪hatwidem0(ω)⎪hatwidem0(−ω) (19.4.24) =⎪parenleftbigg1+e−iω 2⎪parenrightbiggN⎪parenleftbigg1+eiω 2⎪parenrightbiggN ⎪hatwideL(ω)⎪hatwideL(−ω) =⎪parenleftBig cos2ω 2⎪parenrightBigN⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 , (19.4.25) where⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle 2 is a polynomial in cos ω,t h a ti s , ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle 2 =Q(cosω). Since cos ω=1−2s i n2⎪parenleftbigω 2⎪parenrightbig , it is convenient to introduce x=s i n2⎪parenleftbigω 2⎪parenrightbig so that (19.4.25) reduces to the form |⎪hatwidem0(ω)|2=⎪parenleftBig cos2ω 2⎪parenrightBigN Q(1−2x)=( 1 −x)NP(x), (19.4.26) where P(x) is a polynomial in x. We next use the fact that cos2⎪parenleftbiggω+π 2⎪parenrightbigg =s i n2⎪parenleftBigω 2⎪parenrightBig =x © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 581 and ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω+π)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 =Q(−cosω)=Q(2x−1), =Q(1−2( 1−x)) =P(1−x) (19.4.27) to express the identity [(see equation (7.3.4), Debnath, 2002)] in terms of x so that it becomes (1−x)NP(x)+xNP(1−x)=1. (19.4.28) Since (1 −x)NandxNare two polynomials of degree Nwhich are relatively prime, then, by Bezout’s theorem (Daubechies, 1992), there exists a unique polynomial PNof degree ≤N−1 such that (19.4.28) holds. An explicit solu- tion for PN(x)i sg i v e nb y PN(x)=N−1⎪summationdisplay k=0⎪parenleftbiggN+k−1 k⎪parenrightbigg xk, (19.4.29) which is positive for 0 <x< 1s ot h a t PN(x) is at least a possible candidate for⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 . There also exist higher degree polynomial solutions PN(x)o f (19.4.28) which can be written as PN(x)=N−1⎪summationdisplay k=0⎪parenleftbiggN+k−1 k⎪parenrightbigg xk+xNR⎪parenleftbigg x−1 2⎪parenrightbigg , (19.4.30) where Ris an odd polynomial. Since PN(x) is a possible candidate for⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle 2 and ⎪hatwideL(ω)⎪hatwideL(−ω)=⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle 2 =Q(cosω)=Q(1−2x)=PN(x), (19.4.31) the next problem is how to ﬁnd out ⎪hatwideL(ω). This can be done by the following lemma. LEMMA 19.4.2 (Riesz’s Lemma for Spectral Factorization). If ⎪hatwideA(ω)=n⎪summationdisplay k=0akcoskω, (19.4.32) where ak∈Randan/negationslash=0,a n di f⎪hatwideA(ω)≥0for all ω∈Rwith⎪hatwideA(0) = 1 ,t h e n there exists a trigonometric polynomial ⎪hatwideL(ω)=n⎪summationdisplay k=0bke−ikω(19.4.33) © 2007 by Taylor & Francis Group, LLC 582 INTEGRAL TRANSFORMS and THEIR APPLICATIONS with real coeﬃcients bksuch that ⎪hatwideL(0) = 1 and ⎪hatwideA(ω)=⎪hatwideL(ω)⎪hatwideL(−ω)=⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 (19.4.34) for all ω∈R. We refer to Daubechies (1992) for a proof of the Riesz Lemma. We also point out that the factorization of ⎪hatwideA(ω) given in (19.4.34) is not unique. F o rag i v e n N, if we select P=PN,t h e n⎪hatwideA(ω) becomes a polynomial of de- greeN−1i nc o s ωand⎪hatwideL(ω) is a polynomial of degree ( N−1) in exp( −iω). Therefore, the generating function ⎪hatwidem0given by (19.4.23) is of degree (2 N−1) in exp ( −iω). The interval [0 ,2N−1] becomes the support of the correspond- ing scaling function Nϕ.T h em o t h e rw a v e l e t Nψobtained from Nψis called theDaubechies wavelet . ForN= 2, it follows from (19.4.29) that P2(x)=1⎪summationdisplay k=0⎪parenleftbiggk+1 k⎪parenrightbigg xk=1+2 x and hence, (19.4.31) gives ⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideL2(ω)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 =P2(x)=P2⎪parenleftBig sin2ω 2⎪parenrightBig =1+2s i n2ω 2=2−cosω. Using (19.4.33), we obtain that ⎪hatwideL(ω) is a polynomial of degree N−1=1a n d ⎪hatwideL(ω)⎪hatwideL(−ω)=2−1 2⎪parenleftbig eiω+e−iω⎪parenrightbig . It follows from (19.4.33) that ⎪parenleftbig b0+b1e−iω⎪parenrightbig⎪parenleftbig b0+b1eiω⎪parenrightbig =2−1 2⎪parenleftbig eiω+e−iω⎪parenrightbig . (19.4.35) Equating the coeﬃcients in this identity gives b2 0+b21=1 a n d 2 b0b1=−1. (19.4.36) These equations admit solutions b0=1 2⎪parenleftBig 1+√ 3⎪parenrightBig and b1=1 2⎪parenleftBig 1−√ 3⎪parenrightBig . (19.4.37) Thus, the generating function (19.4.21) takes the form ⎪hatwidem0(ω)=⎪parenleftbigg1+e−iω 2⎪parenrightbigg2⎪parenleftbig b0+b1e−iω⎪parenrightbig =1 8⎪bracketleftBig⎪parenleftBig 1+√ 3⎪parenrightBig +⎪parenleftBig 3+√ 3⎪parenrightBig e−iω+⎪parenleftBig 3−√ 3⎪parenrightBig e−2iω+⎪parenleftBig 1−√ 3⎪parenrightBig e−3iω⎪bracketrightBig (19.4.38) © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 583 Figure 19.8 The Daubechies scaling function 2ϕ(x). with⎪hatwidem0(0) = 1. Comparing coeﬃcients of (19.4.38) with the Equation (7.3.3) (Debnath, 2002) gives hn=cnas c0=1 4√ 2⎪parenleftbig 1+√ 3⎪parenrightbig ,c1=1 4√ 2⎪parenleftbig 3+√ 3⎪parenrightbig c2=1 4√ 2⎪parenleftbig 3−√ 3⎪parenrightbig ,c3=1 4√ 2⎪parenleftbig 1−√ 3⎪parenrightbig .(19.4.39) Consequently, the Daubechies scaling function 2ϕ(x) takes the form, dropping the subscript, ϕ(x)=√ 2[c0ϕ(2x)+c1ϕ(2x−1) +c2ϕ(2x−2) +c3ϕ(2x−3)]. (19.4.40) Using the equation (7.3.31) (Debnath, 2002) with N= 2, we obtain the Daubechies wavelet 2ψ(x), dropping the subscript, ψ(x)=√ 2[d0ϕ(2x)+d1ϕ(2x−1) +d2ϕ(2x−2) +d3ϕ(2x−3)] =√ 2[−c3ϕ(2x)+c2ϕ(2x−1)−c1ϕ(2x−2) +c0ϕ(2x−3)], (19.4.41) where the coeﬃcients in (19.4.41) are the same as for the scaling function ϕ(x), but in reverse order and with alternate terms having their signs changed from plus to minus. On the other hand, the use of the equation (7.3.29) (see Debnath, 2002) with the equation (7.3.34) also gives the Daubechies wavelet 2ψ(x)i nt h e form 2ψ(x)=√ 2[−c0ϕ(2x−1) +c1ϕ(2x)−c2ϕ(2x+1 )+ c3ϕ(2x+2 ) ]. The wavelet has the same coeﬃcients as ψgiven in (19.4.41) except that the wavelet is reversed in sign and runs from x=−1 to 2 instead of starting from © 2007 by Taylor & Francis Group, LLC 584 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Figure 19.9 The Daubechies wavelet 2ψ(x). x= 0. It is often referred to as the Daubechies D4 wavelet since it is generated by four coeﬃcients. Both Daubechies’ scaling function 2ϕand Daubechies’ wavelet 2ψare shown in Figures 19.8 and 19.9, respectively. ForN= 1, it follows from (19.4.29) that P1(x)≡1 ,a n dt h i si nt u r nl e a d s to the fact that Q(cosω)=1 ,⎪hatwideL(ω) = 1 so that the generating function is ⎪hatwidem0(ω)=1 2⎪parenleftbig 1+e−iω⎪parenrightbig . (19.4.42) This corresponds to the generating function for the Haar wavelet. 19.5 Exercises 1. For the Haar wavelet deﬁned in Example 19.2.1, show that (a)F{ψ(2mx−n)}=⎪parenleftbigg4i ω⎪parenrightbigg e−iωnexp⎪parenleftbigg −iω 2.2m⎪parenrightbigg sin2⎪parenleftBigω 4.2m⎪parenrightBig (b)|F{ψ(2mx−n)}|=4 ωsin2⎪parenleftBigω 4.2m⎪parenrightBig . Explain the signiﬁcance of this result. 2. Find the Fourier transforms of the following wavelets: (a)ψ(t)=⎪braceleftbigg 1−|t|,0≤t≤1 0, otherwise⎪bracerightbigg (piecewise linear spline wavelet) (b)ψ(t)=( 1 −t2)exp⎪parenleftbigg −t2 2⎪parenrightbigg =−d2 dt2exp⎪parenleftbigg −t2 2⎪parenrightbigg (Mexcican hat wavelet) © 2007 by Taylor & Francis Group, LLC Wavelets and Wavelet Transforms 585 (c)ψ(t)=e x p⎪parenleftbigg iω0t−t2 2⎪parenrightbigg (Morlet wavelet). 3. Iffis a homogeneous function of degree n, show that (Wψf)(λa, λb )=λn+1 2(Wψf)(a, b). 4. In the proof of Theorem 19.2.2 we do not address the diﬃculty that arises from the fact that atakes both positive and negative values. Provide a more detailed proof that removes the diﬃculty. 5. Prove Theorem 19.2.4. 6. Show that ⎪integraldisplay∞ −∞sinπx πxsinπ(2x−n) π(2x−n)dx=1 2πnsin⎪parenleftBigπn 2⎪parenrightBig . 7. Let ψm,n(t)=2−m/2ψ(2−mt−n), where ψ(t) is the Haar wavelet, that is, ψm,n(t)=⎧ ⎪⎨ ⎪⎩2−m/2if 2mn<t< 2mn+2m−1, −2−m/2if 2mn+2m−1<t<2mn+2m, 0o t h e r w i s e . and f(t)=⎧ ⎪⎨ ⎪⎩aif 0<t<1 2, bif1 2<t<1, 0o t h e r w i s e . (a) Find /angbracketleftf,ψm,0/angbracketright. (b) Show that⎪summationtext∞ m=0/angbracketleftf, ψ m,0/angbracketrightψm,0(t)=⎪braceleftBigg aif 0<t<1 2, bif1 2<t<1. 8. Consider the cardinal B-splines Bn(x)o fo r d e r nare deﬁned by B1(x)=χ[0,1](x), Bn(x)=B1(x)∗B1(x)∗...∗B1(x)=B1(x)∗Bn−1(x),n≥2, where nfactors are involved in the convolution product. (a) Show that Bn(x)=⎪integraldisplay∞ −∞Bn−1(x−t)B1(t)dt =⎪integraldisplay1 0Bn−1(x−t)dt=⎪integraldisplayx x−1Bn−1(t)dt. © 2007 by Taylor & Francis Group, LLC 586 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b) Find B2(x),B3(x),B4(x) explicitly. (c) Show that ⎪hatwideB1(ω)=⎪parenleftbigg2 ω⎪parenrightbigg exp⎪parenleftbigg −iω 2⎪parenrightbigg sin⎪parenleftBigω 2⎪parenrightBig =⎪integraldisplay1 0e−iωtdt. 9. Use the Fourier transform ⎪hatwideB1(ω)o fB1(x)t op r o v e ∞⎪summationdisplay k=−∞⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪hatwideB n(2ω+2πk)⎪vextendsingle⎪vextendsingle⎪vextendsingle2 =−sin2n(ω) (2n−1)!d2n−1 dω2n−1(cotω). 10. The Franklin wavelet is generated by the second-order ( n= 2) splines. Show that the Fourier transform ⎪hatwideϕ(ω) of this wavelet is ⎪hatwideϕ(ω)=sin2ω 2 ⎪parenleftbigω 2⎪parenrightbig2⎪parenleftbigg 1−2 3sin2ω 2⎪parenrightbigg−1 2 . 11. The Gabor window (or the Gaussian window) function f∈L2(R)i s deﬁned by f(t)=⎪parenleftbigg2a π⎪parenrightbigg1 4 exp⎪parenleftbig −at2⎪parenrightbig . Show that ||f||=1 a n d ˆf(ω)=1 √ 2af⎪parenleftbigω 2a⎪parenrightbig . Discuss the time and frequency characteristics of the Gabor windows. Draw graphs of f(t)a n d ˆf(ω). 12. For a triangular window f∈L2(R) is deﬁned by f(t)=⎪radicalbigg 3 2a3⎪bracketleftBig χ[−a 2,a 2](t)∗χ[−a 2,a 2](t)⎪bracketrightBig =⎪radicalbigg 3 2a3(a−|t|)χ[−a,a](t). Show that (a)||f||=1, (b)ˆf(ω)=⎪parenleftbigg2 π⎪parenrightbigg⎪radicalbigg 3a 2⎪parenleftbiggsinaω 2 aω⎪parenrightbigg2 . Examine its time and frequency characteristics. © 2007 by Taylor & Francis Group, LLC Appendix A Some Special Functions and Their Properties The main purpose of this appendix is to introduce several special functions and to state their basic properties that are most frequently used in the theory and applications of integral transforms. The subject is, of course, too vast to be treated adequately in so short a space, so that only the more importantresults will be stated. For a fuller discussion of these topics and of further properties of these functions the reader is referred to the standard treatises on the subject. A-1 Gamma, Beta, and Error Functions Thegamma function (also called the factorial function ) is deﬁned by a deﬁnite integral in which a variable appears as a parameter Γ(x)=∞⎪integraldisplay 0e−ttx−1dt, x > 0. (A-1.1) The integral (A-1.1) is uniformly convergent for all xin [a, b] where 0<a≤b<∞, and hence, Γ( x) is a continuous function for all x>0. Integrating (A-1.1) by parts, we obtain the fundamental property of Γ( x) Γ(x)=[−e−ttx−1]∞ 0+(x−1)∞⎪integraldisplay 0e−ttx−2dt =(x−1)Γ(x−1),forx−1>0. Then we replace xbyx+ 1 to obtain the fundamental result Γ(x+1 )= xΓ(x). (A-1.2) In particular, when x=nis a positive integer, we make repeated use of 587 © 2007 by Taylor & Francis Group, LLC 588 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (A-1.2) to obtain Γ(n+1 )= nΓ(n)=n(n−1)Γ(n−1)=··· =n(n−1)(n−2)···3·2·1Γ ( 1 ) = n!, (A-1.3) where Γ(1) = 1. We put t=u2in (A-1.1) to obtain Γ(x)=2∞⎪integraldisplay 0exp(−u2)u2x−1du, x > 0. (A-1.4) Letting x=1 2, we ﬁnd Γ⎪parenleftbigg1 2⎪parenrightbigg =2∞⎪integraldisplay 0exp(−u2)du=2√ π 2=√ π. (A-1.5) Using (A-1.2), we deduce Γ⎪parenleftbigg3 2⎪parenrightbigg =1 2Γ⎪parenleftbigg1 2⎪parenrightbigg =√ π 2. (A-1.6) Similarly, we can obtain the values of Γ⎪parenleftbigg5 2⎪parenrightbigg ,Γ⎪parenleftbigg7 2⎪parenrightbigg ,...,Γ⎪parenleftbigg2n+1 2⎪parenrightbigg . The gamma function can also be deﬁned for negative values of xby the rewritten form of (A-1.2) as Γ(x)=Γ(x+1 ) x,x/negationslash=0,−1,−2,.... (A-1.7) For example Γ⎪parenleftbigg −1 2⎪parenrightbigg =Γ⎪parenleftbigg1 2⎪parenrightbigg −1 2=−2Γ⎪parenleftbigg1 2⎪parenrightbigg =−2√ π, (A-1.8) Γ⎪parenleftbigg −3 2⎪parenrightbigg =Γ⎪parenleftbigg −1 2⎪parenrightbigg −3 2=4 3√ π. (A-1.9) We diﬀerentiate (A-1.1) with respect to xto obtain d dxΓ(x)=Γ/prime(x)=∞⎪integraldisplay 0d dx(tx)e−t tdt =∞⎪integraldisplay 0d dx[exp(xlogt)]e−t tdt=∞⎪integraldisplay 0tx−1(logt)e−tdt.(A-1.10) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 589 Atx=1 ,t h i sg i v e s Γ/prime(1) =∞⎪integraldisplay 0e−tlogtd t=−γ, (A-1.11) where γis called the Euler constant and has the value 0.5772. The graph of the gamma function is shown in Figure A.1. -4 -2 0 2 4 -15-10-5051015 x(x) Figure A.1 The gamma function. The volume, Vnand the surface area, Snof a sphere of radius rinRnare given by Vn=⎪braceleftbig Γ(1 2)⎪bracerightbignrn Γ(n 2+1 ),S n=⎪braceleftbig Γ(1 2)⎪bracerightbignrn−1 Γ(n 2) Thus, dVn=Sn. In particular, when n=2,3, ....,,V2=πr2,S2=2πr;V3=4 3πr3,S3=4πr2; .... Legendre Duplication Formula Several useful properties of the gamma function are recorded below for refer-ence without proof. 2 2x−1Γ(x)Γ⎪parenleftbigg x+1 2⎪parenrightbigg =√ πΓ(2x). (A-1.12) In particular, when x=n(n=0,1,2,...) Γ⎪parenleftbigg n+1 2⎪parenrightbigg =√ π(2n)! 22nn!. (A-1.13) © 2007 by Taylor & Francis Group, LLC 590 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The following properties also hold for Γ( x): Γ(x)Γ(1−x)=πcosec πx, x is a noninteger, (A-1.14) Γ(x)=px∞⎪integraldisplay 0exp(−pt)tx−1dt, (A-1.15) Γ(x)=∞⎪integraldisplay −∞exp(xt−et)dt. (A-1.16) Γ(x+1 )−√ 2πexp(−x)xx+1 2 for large x, (A-1.17) n!∼√ 2πexp(−n)xn+1 2 for large n. (A-1.18) Theincomplete gamma function ,γ(x, a), is deﬁned by the integral γ(a, x)=x⎪integraldisplay 0e−tta−1dt, a > 0. (A-1.19) Thecomplementary incomplete gamma function, Γ( a, x), is deﬁned by the integral Γ(a, x)=∞⎪integraldisplay xe−tta−1dt, a > 0. (A-1.20) Thus, it follows that γ(a, x)+Γ ( a, x)=Γ ( a). (A-1.21) Thebeta function , denoted by B(x, y) is deﬁned by the integral B(x, y)=t⎪integraldisplay 0tx−1(1−t)y−1dt, x > 0,y > 0. (A-1.22) The beta function B(x, y)i ssymmetric with respect to its arguments xand y,t h a ti s , B(x, y)=B(y,x). (A-1.23) This follows from (A-1.22) by the change of variable 1 −t=u,t h a ti s , B(x, y)=1⎪integraldisplay 0uy−1(1−u)x−1du=B(y,x). If we make the change of variable t=u/(1 +u) in (A-1.22), we obtain another integral representation of the beta function B(x, y)=∞⎪integraldisplay 0ux−1(1 +u)−(x+y)du=∞⎪integraldisplay 0uy−1(1 +u)−(x+y)du, (A-1.24) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 591 Putting t=c o s2θin (A-1.22), we derive B(x, y)=2π/2⎪integraldisplay 0cos2x−1θsin2y−1θd θ . (A-1.25) Several important results are record ed below for ready reference without proof. B(1,1) = 1 ,B⎪parenleftbigg1 2,1 2⎪parenrightbigg =π, (A-1.26) B(x, y)=⎪parenleftbiggx−1 x+y−1⎪parenrightbigg B(x−1,y), (A-1.27) B(x, y)=Γ(x)Γ(y) Γ(x+y), (A-1.28) B⎪parenleftbigg1+x 2,1−x 2⎪parenrightbigg =πsec⎪parenleftBigπx 2⎪parenrightBig ,0<x< 1. (A-1.29) Theerror function ,e r f (x) is deﬁned by the integral erf(x)=2 √ πx⎪integraldisplay 0exp(−t2)dt,−∞<x< ∞. (A-1.30) Clearly it follows from (A-1.30) that erf(−x)=−erf(x), (A-1.31) d dx[erf(x)] =2 √ πexp(−x2), (A-1.32) erf(0) = 0 ,erf(∞)=1. (A-1.33) Thecomplementary error function ,e r f c ( x) is deﬁned by the integral erfc(x)=2 √ π∞⎪integraldisplay xexp(−t2)dt. (A-1.34) Clearly it follows that erfc(x)=1−erf(x), (A-1.35) erfc(0)=1, erfc( ∞)=0. (A-1.36) erfc(x)∼1 x√ πexp(−x2)f o r l a r g e x. (A-1.37) The graphs of erf( x)a n de r f c ( x) are shown in Figure A.2. © 2007 by Taylor & Francis Group, LLC 592 INTEGRAL TRANSFORMS and THEIR APPLICATIONS - 3 - 2 - 1 0123 -1012 xy erf(x) erfc(x) Figure A.2 The error function and the complementary error function. Closely associated with the error functi on are the Fresnel integrals, which are deﬁned by C(x)=x⎪integraldisplay 0cos⎪parenleftbiggπt2 2⎪parenrightbigg dtand S(x)=x⎪integraldisplay 0sin⎪parenleftbiggπt2 2⎪parenrightbigg dt. (A-1.38) These integrals arises in diﬀraction problems in optics, in water waves and in elasticity and elsewhere. Clearly it follows from (A-1.38) that C(0) = 0 = S(0) (A-1.39) C(∞)=S(∞)=π 2, (A-1.40) d dxC(x)=c o s⎪parenleftbiggπx2 2⎪parenrightbigg ,d dxS(x)=s i n⎪parenleftbiggπx2 2⎪parenrightbigg . (A-1.41) It also follows from (A-1.38) that C(x) has extrema at the points where x2=( 2n+1 ),n=0,1,2,3,...,andS(x) has extrema at the points where x2=2n,n=1,2,3,....The largest maxima occur ﬁrst and are found to be C(1) = 0 .7799 and S(√ 2) = 0 .7139. We also infer that both C(x)a n d S(x) are oscillatory about the line y=0.5. The graphs of C(x)a n d S(x) for non- negative real xare shown in Figure A.3. A-2 Bessel and Airy Functions The Bessel function of the ﬁrst kind of order orv(non-negative real number) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 593 0123400.51 xy S(x) C(x) Figure A.3 The Fresnel integrals C(x)a n d S(x). is denoted by Jv(x), and deﬁned by Jv(x)=xv∞⎪summationdisplay r=0(−1)rx2r 22r+vr!Γ (r+v+1 ). (A-2.1) This series is convergent for all x. The Bessel function y=Jv(x) satisﬁes the Bessel equation x2y/prime/prime+xy/prime+(x2−v2)y=0. (A-2.2) When visnota positive integer or zero, Jv(x)a n d J−v(x) are two linearly independent solutions so that y=AJv(x)+BJ−v(x) (A-2.3) is the general solution of (A-2.2), where AandBare arbitrary constants. However, when v=n,w h e r e nis apositive integer orzero,Jn(x)a n d J−n(x) are no longer independent, but are related by the equation J−n(x)=(−1)nJn(x). (A-2.4) Thus, when nis a positive integer or zero, equation (A-2.2) has only one solution given by Jn(x)=∞⎪summationdisplay r=0(−1)r r!(n+r)!⎪parenleftBigx 2⎪parenrightBign+2r . (A-2.5) A second solution, known as Neumann’s or Webber’s solution ,Yn(x)i sg i v e n by Yn(x) = lim v→nYv(x), (A-2.6) where Yv(x)=(cosvπ)Jv(x)−J−v(x) sinvπ. (A-2.7) © 2007 by Taylor & Francis Group, LLC 594 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thus, the general solution of (A-2.2) is y(x)=AJn(x)+BYn(x), (A-2.8) where A and B are arbitrary constants. In particular, from (A-2.5), J0(x)=∞⎪summationdisplay r=0(−1)r (r!)2⎪parenleftBigx 2⎪parenrightBig2r , (A-2.9) J1(x)=∞⎪summationdisplay r=0(−1)r r!(r+1 ) !⎪parenleftBigx 2⎪parenrightBig2r+1 . (A-2.10) Clearly, it follows from (A-2.9) and (A-2.10) that J/prime 0(x)=−J1(x). (A-2.11) Bessel’s equation may not always arise in the standard form given in (A- 2.2), but more frequently as x2y/prime/prime+xy/prime+(k2x2−v2)y= 0 (A-2.12) with the general solution y(x)=AJv(kx)+BYv(kx). (A-2.13) Therecurrence relations are recorded below for easy reference without proof. Jv+1(x)=⎪parenleftBigv x⎪parenrightBig Jv(x)−J/prime v(x), (A-2.14) Jv−1(x)=⎪parenleftBigv x⎪parenrightBig Jv(x)+J/prime v(x), (A-2.15) Jv−1(x)+Jv+1(x)=⎪parenleftbigg2v x⎪parenrightbigg Jv(x), (A-2.16) Jv−1(x)−Jv+1(x)=2J/prime v(x). (A-2.17) We have, from (A-2.5), xnJn(x)=∞⎪summationdisplay r=0(−1)r2−(n+2r) r!(n+r)!x2n+2r. Diﬀerentiating both sides of this result with respect to xand using the fact that 2( n+r)/(n+r)! = 2/(n+r−1)!, it turns out that d dx⎪bracketleftbig xnJn(x)⎪bracketrightbig =∞⎪summationdisplay r=0(−1)r2−(n+2r+1) r!(n+r−1)!x2n+2r−1=xnJn−1(x).(A-2.18) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 595 Similarly, we can show d dx⎪bracketleftbig x−nJn(x)⎪bracketrightbig =−x−nJn+1(x). (A-2.19) The generating function for the Bessel function is exp⎪bracketleftbigg1 2x⎪parenleftbigg t−1 t⎪parenrightbigg⎪bracketrightbigg =∞⎪summationdisplay n=−∞tnJn(x). (A-2.20) The integral representation of Jn(x)i s Jn(x)=1 ππ⎪integraldisplay 0cos(nθ−xsinθ)dθ. (A-2.21) The following are known as the Lommel integrals : a⎪integraldisplay 0xJn(px)Jn(qx)dx =a (q2−p2)[pJn(qa)J/prime n(pa)−qJn(pa)J/prime n(qa)],p/negationslash=q,(A-2.22) anda⎪integraldisplay 0xJ2 n(px)dx=a2 2⎪bracketleftbigg J/prime2 n(pa)+⎪parenleftbigg 1−n2 p2a2⎪parenrightbigg J2 n(pa)⎪bracketrightbigg . (A-2.23) When n=±1 2, J1 2(x)=⎪radicalbigg 2 πxsinx, J−1 2(x)=⎪radicalbigg 2 πxcosx. (A-2.24) A rough idea of the shape of the Bessel functions when xis large may be obtained from equation (A-2.2). Substitution of y=x−1 2u(x) eliminates the ﬁrst derivative, and hence, gives the equation u/prime/prime+⎪parenleftbigg 1−4n2−1 4x2⎪parenrightbigg u=0. (A-2.25) For large x, this equation approximately becomes u/prime/prime+u=0. (A-2.26) This equation admits the solution u(x)=Acos(x+ε)t h a ti s , y=A √ xcos(n+ε). (A-2.27) © 2007 by Taylor & Francis Group, LLC 596 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 02468 1 0 -0.500.51 xy J0(x) J1(x) J2(x) Figure A.4 Graphs of y=J0(x),J1(x)a n d J2(x). 02468 -10123 xy J1/2(x) J-1/2(x) Figure A.5 Graphs of J1 2(x)a n d J−1 2(x). This suggests that Jn(x) is oscillatory and has an inﬁnite number of zeros. It also tends to zero as x→∞. The graphs of Jn(x)f o rn=0,1,2a n df o r n=±1 2are shown in Figure A.4 and Figure A.5, respectively. An important special case arises in particular physical problems when k2= −1 in equation (A-2.12). we then have the modiﬁed Bessel equation x2y/prime/prime+xy/prime−(x2+v2)y=0, (A-2.28) with the general solution y=AJv(ix)+BYv(ix). (A-2.29) We now deﬁne a new function Iv(x)=i−vJv(ix), (A-2.30) and then use the series (A-2.1) for Jv(x)s ot h a t Iv(x)=i−v∞⎪summationdisplay r=0(−1)r r!Γ(r+v+1 )⎪parenleftbiggix 2⎪parenrightbiggv+2r =∞⎪summationdisplay r=01 r!Γ(r+v+1 )⎪parenleftBigx 2⎪parenrightBigv+2r . (A-2.31) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 597 Similarly, we can ﬁnd the second solution, Kv(x) of the modiﬁed Bessel equa- tion (A-2.28). Usually, Iv(x)a n d Kv(x) are called modiﬁed Bessel functions and their properties can be obtained in a similar way to those of Jv(x)a n d Yv(x). The graphs of Y0(x),Y1(x)a n d Y2(x) are shown in Figure A.6. 02468 1 0 1 2 -1-0.500.5 xy Y0(x) Y1(x) Y2(x) Figure A.6 Graphs of y=Y0(x),Y1(x)a n d Y2(x). We state a few important inﬁnite integrals involving Bessel functions which arise frequently in the application of Hankel transforms. ∞⎪integraldisplay 0exp(−at)Jv(bt)tvdt=(2b)vΓ⎪parenleftbigg v+1 2⎪parenrightbigg √ π(a2+b2)v+1 2,v > −1 2,(A-2.32) ∞⎪integraldisplay 0exp(−at)Jv(bt)tv+1dt=2a(2b)vΓ⎪parenleftbigg v+3 2⎪parenrightbigg √ π(a2+b2)v+3 2,v > −1,(A-2.33) ∞⎪integraldisplay 0exp(−a2t2)Jv(bt)tv+1dt=bv (2a2)v+1exp⎪parenleftbigg −b2 4a2⎪parenrightbigg ,v > −1,(A-2.34) ∞⎪integraldisplay 0exp(−a2t2)Jv(bt)Jv(ct)td t=1 2a2exp⎪parenleftbigg −b2+c2 4a2⎪parenrightbigg Iv⎪parenleftbiggbc 2a2⎪parenrightbigg ,v >−1, (A-2.35) ∞⎪integraldisplay 0t2μ−v−1Jv(t)dt=22μ−v−1Γ(μ) Γ(v−μ+1 ),0<μ<1 2,v >−1 2.(A-2.36) © 2007 by Taylor & Francis Group, LLC 598 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TheAiry function ,y=Ai(x) is the ﬁrst solution of the diﬀerential equation y/prime/prime−xy=0. (A-2.37) The second solution is denoted by Bi(x). Then these functions are given by Ai(x)=⎪radicalbigg x 3⎪bracketleftbigg I−1 3⎪parenleftbigg2 3x3/2⎪parenrightbigg −I1 3⎪parenleftbigg2 3x3/2⎪parenrightbigg⎪bracketrightbigg , (A-2.38) Bi(x)=⎪radicalbigg x 3⎪bracketleftbigg I−1 3⎪parenleftbigg2 3x3/2⎪parenrightbigg +I1 3⎪parenleftbigg2 3x3/2⎪parenrightbigg⎪bracketrightbigg . (A-2.39) The integral representation for Ai(x)i s Ai(x)=1 π∞⎪integraldisplay 0cos⎪parenleftbigg1 3t3+xt⎪parenrightbigg dt. (A-2.40) The graph of y=Ai(x) is shown in Figure A.7. -8 -6 -4 -2 0 201 xy Ai(x) Figure A.7 The Airy function. A-3 Legendre and Associated Legendre Functions The Legendre polynomials Pn(x) are deﬁned by the Rodrigues formula Pn(x)=1 2nn!dn dxn(x2−1)n. (A-3.1) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 599 The seven Legendre polynomials are P0(x)=1 P1(x)=x P2(x)=1 2(3x2−1) P3(x)=1 2(5x3−3x) P4(x)=1 8(35x4−30x2+3 ) P5(x)=1 8(63x5−70x3+1 5x) P6(x)=1 16(231x6−315x4+ 105 x2−5). The generating function for the Legendre polynomial is (1−2xt+t2)−1 2=∞⎪summationdisplay n=0tnPn(x). (A-3.2) This function provides more information about the Legendre polynomials. For example Pn(1) = 1 ,P n(−1)= (−1)n, (A-3.3) P2n(0) = ( −1)n1·3·5···(2n−1) 2nn!=(−1)n(2n−1)!! (2n)!!,(A-3.4) P2n+1(0) = 0 ,n=0,1,2,..., (A-3.5) Pn(−x)=(−1)nPn(x),dn dxnPn(x)=(2n)! 2nn!, (A-3.6) where the double factorial is deﬁned by (2n−1)!!= 1 ·3·5···(2n−1) and (2 n)!! = 2 ·4·6···(2n). The graphs of the ﬁrst four Legendre polynomials are shown in Figure A.8. The recurrence relations for the Legendre polynomials are (n+1 )Pn+1(x)=( 2n+1 )xPn(x)−nPn−1(x), (A-3.7) P/prime n+1(x)−P/prime n−1(x)=( 2n+1 )Pn(x), (A-3.8) (1−x2)P/prime n(x)=nPn−1(x)−nxP n(x), (A-3.9) (1−x2)P/prime n(x)=(n+1 )xPn(x)−(n+1 )Pn+1(x).(A-3.10) The Legendre polynomials y=Ln(x) satisfy the Legendre d iﬀerential equation (1−x2)y/prime/prime−2xy/prime+n(n+1 )y=0. (A-3.11) © 2007 by Taylor & Francis Group, LLC 600 INTEGRAL TRANSFORMS and THEIR APPLICATIONS -1 0 1 -101 xy P0(x) P1(x)P2(x) P3(x) Figure A.8 Graphs of y=P0(x),P1(x),P2(x)a n d P3(x). Ifnisnotan integer, both solutions of (A-3.11) diverge at x=±1. The orthogonal relation is 1⎪integraldisplay −1Pn(x)Pm(x)dx=2 (2n+1 )δnm. (A-3.12) Theassociated Legendre functions are deﬁned by Pm n(x)=( 1 −x2)m 2dm dxmPn(x)=1 2nn!(1−x2)m 2dm+n dxm+n(x2−1)n,(A-3.13) where 0 ≤m≤n. Clearly, it follows that P0 n(x)=Pn(x), (A-3.14) Pm n(−x)=(−1)n+mPm n(x),P−m n(x)=(−1)m(n−m)! (n+m)!Pm n(x).(A-3.15) The generating function for Pm n(x)i s (2m)!(1−x2)m 2 2mm!(1−2tx+t2)m+1 2=∞⎪summationdisplay r=0Pm r+m(x)tr. (A-3.16) The recurrence relations are (2n+1 )xPm n(x)=(n+m)Pm n−1(x)+(n−m+1 )Pm n+1(x),(A-3.17) 2(1−x2)1 2d dxPm n(x)=Pm+1 n(x)−(n+m)(n−m+1 )Pm−1 n(x).(A-3.18) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 601 The associated Legendre functions Pm n(x) are solutions of the diﬀerential e- quation (1−x2)y/prime/prime−2xy/prime+⎪bracketleftbigg n(n+1 )−m2 (1−x2)⎪bracketrightbigg y=0. (A-3.19) This reduces to the Legendre equation when m=0 . Listed below are few associat ed Legendre functions with x=c o sθ: P1 1(x)=( 1 −x2)1 2=s i nθ, P1 2(x)=3x(1−x2)1 2=3c o s θsinθ P2 2(x)=3 ( 1 −x2)=3s i n2θ P1 3(x)=3 2(5x2−1)(1−x2)1 2=3 2(5 cos2θ−1)sinθ P2 3(x)=1 5 x(1−x2)=1 5c o s θsin2θ P3 3(x) = 15(1 −x2)3/2=1 5s i n3θ. The orthogonal relations are 1⎪integraldisplay −1Pm n(x)Pm t(x)dx=2 (2/lscript+1 )·(/lscript+m)! (/lscript−m)!δn/lscript, (A-3.20) 1⎪integraldisplay −1(1−x2)−1Pm n(x)P/lscript n(x)dx=(n+m)! m(n−m)!δm/lscript. (A-3.21) A-4 Jacobi and Gegenbauer Polynomials TheJacobi polynomials P(α,β) n(x)o fd e g r e e nare deﬁned by the Rodrigues formula P(α,β) n(x)=(−1)n 2nn!(1−x)−α(1 +x)−βdn dxn[(1−x)α+n(1 +x)β+n],(A-4.1) where α>−1a n d β>−1. When α=β= 0, the Jacobi polynomials become Legendre polynomials, that is, Pn(x)=P(0,0) n(x),n =0,1,2,.... (A-4.2) On the other hand, the associated Laguerre functions arise as the limit Lα n(x) = lim β→∞P(α,β) n⎪parenleftbigg 1−2x β⎪parenrightbigg . (A-4.3) © 2007 by Taylor & Francis Group, LLC 602 INTEGRAL TRANSFORMS and THEIR APPLICATIONS The recurrence relations for P(α,β) n(x)a r e 2(n+1 ) (α+β+n+1 ) (α+β+2n)P(α,β) n+1(x) =(α+β+2n+1 ) [ ( α2−β2)+x(α+β+2n+2 ) (α+β+2n)]P(α,β) n(x) −2(α+n)(β+n)(α+β+2n+2 )P(α,β) n−1(x),(A-4.4) where n=1,2,3,...,a n d P(α,β−1) n (x)−P(α−1,β) n (x)=P(α,β) n−1(x). (A-4.5) The generating function for Jacobi polynomials is 2(α+β)R−1(1−t+R)−α(1 +t+R)−β=∞⎪summationdisplay n=0P(α,β) n(x)tn, (A-4.6) where R=( 1−2xt+t2)1 2. The Jacobi polynomials, y=P(α,β) n(x), satisfy the diﬀerential equation (1−x2)y/prime/prime+[ (β−α)−(α+β+2 )x]y/prime+n(n+α+β+1 )y=0.(A-4.7) The orthogonal relation is 1⎪integraldisplay −1(1−x)α(1 +x)βP(α,β) n(x)P(α,β) m(x)dx=⎪braceleftbigg0,n /negationslash=m δn,n =m⎪bracerightbigg ,(A-4.8) where δn=2α+β+1Γ(n+α+1 ) Γ ( n+β+1 ) n!(α+β+2n+1 ) Γ ( α+β+n+1 ). (A-4.9) When α=β=v−1 2, the Jacobi polynomials reduce to the Gegenbauer poly- nomials Cv n(x), which are deﬁned by the Rodrigues formula Cv n(x)=(−1)n 2nn!(1−x2)v−1 2dn dxn⎪bracketleftBig (1−x2)v+n−1 2⎪bracketrightBig . (A-4.10) The generating function for Cv n(x)o fd e g r e e nis (1−2xt+t2)−v=∞⎪summationdisplay n=0Cv n(x)tn,|t|<1,|x|≤1,v > −1 2.(A-4.11) The recurrence relations are (n+1 )Cv n+1(x)−2(v+n)xCv n(x)+( 2v+n−1)Cv n−1(x)=0,(A-4.12) (n+1 )Cv n+1(x)−2vCv+1 n(x)+2vCv+1 n−1(x)=0, (A-4.13) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 603 d dx[Cv n(x)] = 2vCv+1 n+1(x). (A-4.14) The diﬀerential equation satisﬁed by y=Cv n(x)i s (1−x2)y/prime/prime−(2v+1 )xy/prime+n(n+2v)y=0. (A-4.15) The orthogonal property is 1⎪integraldisplay −1(1−x2)v−1 2Cv n(x)Cv m(x)dx=δnδnm, (A-4.16) where δn=21−2vnΓ(n+2v) n!(n+v)[Γ(v)]2. (A-4.17) When v=1 2, the Gegenbauer polynomials reduce to Legendre polynomials, that is, C1 2n(x)=Pn(x). (A-4.18) The Hermite polynomials can also be obtained from the Gegenbauer polyno- mials as the limit Hn(x)=n! lim v→∞v−n/2Cv n⎪parenleftbiggx √ v⎪parenrightbigg . (A-4.19) Finally, when α=β=1 2, the Gegenbauer polynomials reduce to the well- known Chebyshev polynomials ,Tn(x), which are deﬁned by a solution of the second order diﬀerence equation (see Example 12.6.7) un+2−2xun+1+un=0,|x|≤1 (A-4.20) u(0) =u0andu(1) =u1. (A-4.21) The generating function for Tn(x)i s (1−t2) (1−2xt+t2)=T0(x)+2∞⎪summationdisplay n=1Tn(x)tn,|x|≤1t<1. (A-4.22) The ﬁrst seven Chebyshev polynomials of degree nof the ﬁrst kind are T0(x)=1 T1(x)=x T2(x)=2x2−1 T3(x)=4x3−3x T4(x)=8x4−8x2+1 T5(x)=1 6 x5−20x3+5x T6(x)=3 2 x6−48x4+1 8x2−1. © 2007 by Taylor & Francis Group, LLC 604 INTEGRAL TRANSFORMS and THEIR APPLICATIONS -1 0 1 -101 xy T0(x) T1(x) T2(x)T3(x) Figure A.9 Chebyshev polynomials y=Tn(x). The graphs of the ﬁrst four Chebyshev polynomials are shown in Figure A.9. The Chebyshev polynomials y=Tn(x) satisfy the diﬀerential equation (1−x2)y/prime/prime−xy/prime+n2y=0. (A-4.23) It follows from (A-4.22) that Tn(x) satisﬁes the recurrence relations Tn+1(x)−2xTn(x)+Tn−1(x)=0, (A-4.24) Tn+m(x)−2Tn(x)Tm(x)+Tn−m(x)=0, (A-4.25) (1−x2)T/prime n(x)+nxT n(x)−nTn−1(x)=0. (A-4.26) The parity relation for Tn(x)i s Tn(−x)=(−1)nTn(x). (A-4.27) The Rodrigues formula is Tn(x)=√ π(−1)n(1−x2)1 2 2n⎪parenleftbig n−1 2⎪parenrightbig !·dn dxn⎪bracketleftBig (1−x2)n−1 2⎪bracketrightBig . (A-4.28) The orthogonal relation for Tn(x)i s 1⎪integraldisplay −1(1−x2)−1 2Tm(x)Tn(x)dx=⎧ ⎨ ⎩0,m /negationslash=n π 2,m =n π, m =n=0⎫ ⎬ ⎭.(A-4.29) TheChebyshev polynomials of the second kind ,Un(x), are deﬁned by Un(x)=( 1 −x2)−1 2sin[(n+1 )c o s−1x],−1≤x≤1. (A-4.30) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 605 The generating function for Un(x)i s (1−2xt+t2)−1=∞⎪summationdisplay n=0Un(x)tn,|x|<1,|t|<1. (A-4.31) The ﬁrst seven Chebyshev polynomials Un(x)a r eg i v e nb y U0(x)=1 U1(x)=2x U2(x)=4x2−1 U3(x)=8x3−4x U4(x)=1 6 x4−12x2+1 U5(x)=3 2 x5−32x3+6x U6(x)=6 4 x6−80x4+2 4x2−1. The diﬀerential equation for y=Un(x)i s (1−x2)y/prime/prime−3xy/prime+n(n+2 )y=0. (A-4.32) The recurrence relations are Un+1(x)−2xUn(x)+Un−1(x)=0. (A-4.33) (1−x2)U/prime n(x)+nxU n(x)−(n+1 )Un−1(x)=0. (A-4.34) The parity relation is Un(−x)=(−1)nUn(x). (A-4.35) The Rodrigues formula is Un(x)=√ π(−1)n(n+1 ) 2n+1⎪parenleftbig n+1 2⎪parenrightbig !(1−x2)1 2dn dxn⎪bracketleftBig (1−x2)n+1 2⎪bracketrightBig . (A-4.36) The orthogonal relation for Un(x)i s 1⎪integraldisplay −1(1−x2)1 2Um(x)Un(x)dx=π 2δmn. (A-4.37) A-5 Laguerre and Associated Laguerre Functions The Laguerre polynomials Ln(x) are deﬁned by the Rodrigues formula Ln(x)=exdn dxn(xne−x), (A-5.1) © 2007 by Taylor & Francis Group, LLC 606 INTEGRAL TRANSFORMS and THEIR APPLICATIONS where n=0,1,2,3,.... The ﬁrst seven Laguerre polynomials are L0(x)=1 L1(x)=1−x L2(x)=2−4x+x2 L3(x)=6−18x+9x2−x3 L4(x)=2 4 −96x+7 2x2−16x3+x4 L5(x) = 120 −600x+ 600 x2−200x3+2 5x4−x5 L6(x) = 720 −4320x+ 5400 x2−2400x3+ 450 x4−36x5+x6. The generating function is (1−t)−1exp⎪parenleftbiggxt 1−t⎪parenrightbigg =∞⎪summationdisplay n=0tnLn(x). (A-5.2) In particular Ln(0) = 1 . (A-5.3) The orthogonal relation for the Laguerre polynomial is ∞⎪integraldisplay 0e−xLm(x)Ln(x)dx=(n!)2δnm. (A-5.4) The recurrence relations are (n+1 )Ln+1(x)=( 2n+1−x)Ln(x)−nLn−1(x), (A-5.5) xL/prime n(x)=nLn(x)−nLn−1(x), (A-5.6) L/prime n(x)=L/prime n−1(x)−Ln−1(x). (A-5.7) The Laguerre polynomials y=Ln(x)s a t i s f yt h e Laguerre diﬀerential equation xy/prime/prime+( 1−x)y/prime+ny=0. (A-5.8) Theassociated Laguerre polynomials are deﬁned by Lm n(x)=dm dxmLn(x)f o r n≥m. (A-5.9) The generating function for Lm n(x)i s (1−z)−(m+1)exp⎪parenleftbigg −xz 1−z⎪parenrightbigg =∞⎪summationdisplay n=0Lm n(x)zn,|z|<1. (A-5.10) © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 607 It follows from this that Lm n(0) =(n+m)! n!m!. (A-5.11) The associated Laguerre function satisﬁes the recurrence relation (n+1 )Lm n+1(x)=( 2n+m+1−x)Lm n(x)−(n+m)Lm n−1(x),(A-5.12) xd dxLm n(x)=nLm n(x)−(n+m)Lm n−1(x). (A-5.13) The associated Laguerre function y=Lm n(x) satisﬁes the associated Laguerre diﬀerential equation xy/prime/prime+(m+1−x)y/prime+ny=0. (A-5.14) The Rodrigues formula for Lm n(x)i s Lm n(x)=exx−m n!dn dxn(e−xxn+m). (A-5.15) Theorthogonal relation for Lm n(x)i s ∞⎪integraldisplay 0e−xxmLm n(x)Lm l(x)dx=(n+m)! n!δnl. (A-5.16) A-6 Hermite Polynomials and Weber-Hermite Function- s The Hermite polynomials Hn(x) are deﬁned by the Rodrigues formula Hn(x)=(−1)nexp(x2)dn dxn[exp(−x2)], (A-6.1) where n=0,1,2,3,.... The ﬁrst seven Hermite polynomials are H0(x)=1 H1(x)=2x H2(x)=4x2−2 H3(x)=8x3−12x H4(x)=1 6 x4−48x2+1 2 H5(x)=3 2 x5−16x3+ 120 x H6(x)=6 4 x6−480x4+ 720 x2−120. © 2007 by Taylor & Francis Group, LLC 608 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Thegenerating function is exp(2xt−t2)=∞⎪summationdisplay n=0tn n!Hn(x). (A-6.2) It follows from (A-6.2) that Hn(x) satisﬁes the parity relation Hn(−x)=(−1)nHn(x). (A-6.3) Also, it follows from (A-6.2) that H2n+1(0) = 0 ,H 2n(0) = ( −1)n(2n)! n!. (A-6.4) Therecurrence relations for Hermite polynomials are Hn+1(x)−2xHn(x)+2nHn−1(x)=0, (A-6.5) H/prime n(x)=2xHn−1(x). (A-6.6) The Hermite polynomials, y=Hn(x), are solutions of the Hermite diﬀerential equation y/prime/prime−2xy/prime+2ny=0. (A-6.7) The orthogonal property of Hermite polynomials is ∞⎪integraldisplay −∞exp(−x2)Hn(x)Hm(x)dx=2nn!√ πδmn. (A-6.8) With repeated use of integration by parts, it follows from (A-6.1) that ∞⎪integraldisplay −∞exp(−x2)Hn(x)xmdx=0,m=0,1,...,(n−1), (A-6.9) ∞⎪integraldisplay −∞exp(−x2)Hn(x)xndx=√ πn!. (A-6.10) TheWeber-Hermite function or, simply, Hermite functions y=hn(x)=e x p⎪parenleftbigg −x2 2⎪parenrightbigg Hn(x) (A-6.11) satisﬁes the Hermite diﬀerential equation y/prime/prime+(λ−x2)y=0,x∈R (A-6.12) where λ=2n+1 .I f λ/negationslash=2n+1 ,t h e n yis not ﬁnite as |x|→∞ . © 2007 by Taylor & Francis Group, LLC Some Special Functions and Their Properties 609 -6 -4 -2 0 2 4 6x0(x) -6 -4 -2 0 2 4 6x1(x) -6 -4 -2 0 2 4 6x2(x) -6 -4 -2 0 2 4 6x3(x) Figure A.10 The normalized Weber-Hermite functions. The Hermite functions {hn(x)}∞ 0form an orthogonal basis for the Hilbert space L2(R) with weight function 1. They satisfy the following fundamental properties: h/prime n(x)+xhn(x)−2nhn−1(x)=0, h/prime n(x)−xhn(x)+hn+1(x)=0, h/prime/prime n(x)−x2hn(x)+( 2n+1 )hx=0, F{hn(x)}=˜hn(k)=(−i)nhn(k). The normalized Weber-Hermite functions are given by ψn(x)=2−n/2π−1 4(n!)−1 2exp⎪parenleftbigg −x2 2⎪parenrightbigg Hn(x). (A-6.13) Physically, they represent quantum mechanical oscillator wave functions. The graphs of these functions are shown in Figure A.10. A-7 Mittag Leﬄer Function Another important function that has widespred use in fractional calculus andfractional diﬀerential equation is the Mittag-Leﬄer function . The Mittag- © 2007 by Taylor & Francis Group, LLC 610 INTEGRAL TRANSFORMS and THEIR APPLICATIONS Leﬄer function is an entire function deﬁned by the series Eα(z)=∞⎪summationdisplay n=0zn Γ(αn+1 ),α > 0. (A-7.1) -50 -40 -30 -20 -10 0 10 -20246 xEE1 E2 E3 E5 E4E2 E3 Figure A.11 Graph of the Mittag-Leﬄer function Eα(x). The graph of the Mittag-Leﬄer function is shown in Figure A.11. The generalized Mittag-Leﬄer function ,Eα,β(z), is deﬁned by Eα,β(z)=∞⎪summationdisplay n=0zn Γ(αn+β),α , β > 0. (A-7.2) Also the inverse Laplace transform yields L−1⎪braceleftBigg m!sα−β (sα¯+a)m+1⎪bracerightBigg =tαm+β−1E(m) α,β(+ atα), (A-7.3) where E(m) α,β(z)=dm dzmEα,β(z). (A-7.4) Obviously, Eα,1(z)=Eα(z),E 1,1(z)=E1(z)=ez. (A-7.5) © 2007 by Taylor & Francis Group, LLC Appendix B Tables of Integral Transforms In this appendix we provide a set of short tables of integral transforms of the functions that are either cited in the text or in most common use in math-ematical, physical, and engineering applications. In these tables no attempt is made to give complete lists of transforms. For exhaustive lists of integral transforms, the reader is referred to Erd´ elyi et al. (1954), Campbell and Foster (1948), Ditkin and Prudnikov (1965) , Doetsch (1950–1956, 1970), Marichev (1983), and Oberhettinger (1972, 1974). TABLE B-1 Fourier Transforms f(x) F(k)=1 √ 2π∞⎪integraldisplay −∞exp(−ikx)f(x)dx 1 exp(−a|x|),a > 0 ⎪parenleftBigg⎪radicalbigg 2 π⎪parenrightBigg a(a2+k2)−1 2 xexp(−a|x|),a > 0 ⎪parenleftBigg⎪radicalbigg 2 π⎪parenrightBigg (−2aik)(a2+k2)−2 3 exp(−ax2),a > 0 1 √ 2aexp⎪parenleftbigg −k2 4a⎪parenrightbigg 4 (x2+a2)−1,a > 0 ⎪radicalbigg π 2exp(−a|k|) a 5 x(x2+a2)−1,a > 0 ⎪radicalbigg π 2⎪parenleftbiggik 2a⎪parenrightbigg exp(−a|k|) 6 ⎪braceleftBigg c, a≤x≤b 0,outside⎪bracerightBigg ic √ 2π1 k(e−ibk−e−iak) 7 |x|exp(−a|x|),a > 0 ⎪radicalbigg 2 π(a2−k2)(a2+k2)−2 611 © 2007 by Taylor & Francis Group, LLC 612 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(x) F(k)=1 √ 2π∞⎪integraldisplay −∞exp(−ikx)f(x)dx 8 sinax x ⎪radicalbigg π 2H(a−|k|) 9 exp{−x(a−iω)}H(x) 1 √ 2πi (ω−k+ia) 10 (a2−x2)−1 2H(a−|x|) ⎪radicalbigg π 2J0(ak) 11 sin⎪bracketleftBig b(x2+a2)1 2⎪bracketrightBig (x2+a2)1 2 ⎪radicalbigg π 2J0⎪parenleftBig a⎪radicalbig b2−k2⎪parenrightBig H(b−|k|) 12 cos⎪parenleftbig b√ a2−x2⎪parenrightbig (a2−x2)1 2H(a−|x|) ⎪radicalbigg π 2J0⎪parenleftBig a⎪radicalbig b2+k2⎪parenrightBig 13 e−axH(x),a > 0 1 √ 2π(a−ik)(a2+k2)−1 14 1 ⎪radicalbig |x|exp(−a|x|) (a2+k2)−1 2⎪bracketleftBig a+(a2+k2)1 2⎪bracketrightBig1 2 15 δ(x) 1 √ 2π 16 δ(n)(x) 1 √ 2π(ik)n 17 δ(x−a) 1 √ 2πexp(−iak) 18 δ(n)(x−a) 1 √ 2π(ik)nexp(−iak) 19 exp(iax) √ 2πδ(k−a) 20 1 √ 2πδ(k) 21 x √ 2πiδ/prime(k) 22 xn √ 2πinδ(n)(k) © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 613 f(x) F(k)=1 √ 2π∞⎪integraldisplay −∞exp(−ikx)f(x)dx 23 H(x) ⎪radicalbigg π 2⎪bracketleftbigg1 iπk+δ(k)⎪bracketrightbigg 24 H(x−a) ⎪radicalbigg π 2⎪bracketleftbiggexp(−ika) πik+δ(k)⎪bracketrightbigg 25 H(x)−H(−x) ⎪radicalbigg 2 π⎪parenleftbigg −i k⎪parenrightbigg 26 xnexp(iax) √ 2πinδ(n)(k−a) 27 |x|−1 1 √ 2π(A−2l o g|k|),Ais a constant 28 log(|x|) −⎪radicalbigg π 21 |k| 29 H(a−|x|) ⎪radicalbigg 2 π⎪parenleftbiggsinak k⎪parenrightbigg 30 |x|α(α<1,not a negative integer) ⎪radicalbigg 2 πΓ(α+1 )|k|−(1+α) ×cos⎪bracketleftBigπ 2(α+1 )⎪bracketrightBig 31 sgn x ⎪radicalbigg 2 π1 (ik) 32 x−n−1sgn x 1 √ 2π(−ik)n n!(A−2l o g|k|) 33 1 x −i⎪radicalbigg π 2sgn k 34 1 xn −i⎪radicalbigg π 2⎪bracketleftbigg(−ik)n−1 (n−1)!sgn k⎪bracketrightbigg 35 xnexp(iax) √ 2πinδ(n)(k−a) © 2007 by Taylor & Francis Group, LLC 614 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(x) F(k)=1 √ 2π∞⎪integraldisplay −∞exp(−ikx)f(x)dx 36 xαH(x),(αnot an integer) Γ(α+1 ) √ 2π|k|−(α+1) ×exp⎪bracketleftbigg −⎪parenleftbiggπi 2⎪parenrightbigg (α+1 )sgn k⎪bracketrightbigg 37 xnexp(iax)H(x) ⎪radicalbig π 2⎪bracketleftBig n! iπ(k−a)n+1+inδ(n)(k−a)⎪bracketrightBig 38 exp(iax)H(x−b) ⎪radicalbigg π 2⎪bracketleftbiggexp[−ib(k−a)] iπ(k−a)+δ(k−a)⎪bracketrightbigg 39 1 x−a −i⎪radicalbigg π 2exp(−iak)sgn k 40 1 (x−a)n −i⎪radicalbigg π 2exp(−iak)(−ik)n−1 (n−1)!sgn k 41 eiax (x−b) i⎪radicalbigg π 2exp[ib(a−k)][1−2H(k−a)] 42 eiax (x−b)n i⎪radicalbigg π 2[1−2H(k−a)] ×exp{ib(a−k)} (n−1)![−i(k−a)]n−1 43 |x|αsgn x (αnot integer) ⎪radicalbigg 2 π(−i)Γ(α+1 ) |k|α+1cos⎪parenleftBigπα 2⎪parenrightBig sgn k 44 xnf(x) (−i)ndn dknF(k) 45 dn dxnf(x) (ik)nF(k) 46 eiaxf(bx) 1 bF⎪parenleftbiggk−a b⎪parenrightbigg 47 sin cos⎪parenleftbig ax2⎪parenrightbig 1 √ 2asin cos⎪parenleftBig k2 4a−π 4⎪parenrightBig © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 615 TABLE B-2 Fourier Cosine Transforms f(x) Fc(k)=⎪radicalbigg 2 π∞⎪integraldisplay 0cos(kx)f(x)dx 1 exp(−ax),a > 0 ⎪parenleftBigg⎪radicalbigg 2 π⎪parenrightBigg a(a2+k2)−1 2 xexp(−ax),a > 0 ⎪parenleftBigg⎪radicalbigg 2 π⎪parenrightBigg (a2−k2)(a2+k2)−2 3 exp(−a2x2) 1 |a|√ 2exp⎪parenleftbigg −k2 4a2⎪parenrightbigg 4 H(a−x) ⎪radicalbigg 2 π⎪parenleftbiggsinak k⎪parenrightbigg 5 xa−1,0<a< 1 ⎪radicalbigg 2 πΓ(a)k−acos⎪parenleftBigaπ 2⎪parenrightBig 6 cos(ax2) 1 2√ a⎪bracketleftbigg cos⎪parenleftbiggk2 4a⎪parenrightbigg +s i n⎪parenleftbiggk2 4a⎪parenrightbigg⎪bracketrightbigg 7 sin(ax2),a > 0 1 2√ a⎪bracketleftbigg cos⎪parenleftbiggk2 4a⎪parenrightbigg −sin⎪parenleftbiggk2 4a⎪parenrightbigg⎪bracketrightbigg 8 (a2−x2)v−1 2H(a−x),v>−1 2 2v−1 2Γ⎪parenleftbigg v+1 2⎪parenrightbigg⎪parenleftBiga k⎪parenrightBigv Jv(ak) 9 (a2+x2)−1J0(bx),a , b > 0 ⎪radicalbig π 2a−1e−akI0(ab),b < k < ∞ 10 x−vJv(ax),v > −1 2 (a2−k2)v−1 2H(a−k) 2v−1 2avΓ⎪parenleftbigg v+1 2⎪parenrightbigg 11 (x2+a2)−1 2e−b(x2+a2)1 2 K0⎪bracketleftBig a(k2+b2)1 2⎪bracketrightBig ,a >0,b>0 12 (2ax−x2)v−1 2H(2a−x),v>−1 2 √ 2Γ⎪parenleftbigg v+1 2⎪parenrightbigg⎪parenleftbigg2a k⎪parenrightbiggv ×cos(ak)Jv(ak) © 2007 by Taylor & Francis Group, LLC 616 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(x) Fc(k)=⎪radicalbigg 2 π∞⎪integraldisplay 0cos(kx)f(x)dx 13 xν−1e−ax,ν > 0,a >0 ⎪radicalBig 2 πΓ(ν)r−νcosνθ,w h e r e r=(a2+k2)1 2,θ=t a n−1⎪parenleftbigk a⎪parenrightbig 14 2 xe−xsinx ⎪radicalbigg 2 πtan−1⎪parenleftbigg2 k2⎪parenrightbigg 15 sin⎪bracketleftBig a(b2−x2)1 2⎪bracketrightBig H(b−x) ⎪radicalbigg π 2(ab)(a2+k2)−1 2 ×J1⎪bracketleftBig b(a2+k2)1 2⎪bracketrightBig 16 (1−x2) (1 +x2)2 ⎪radicalbigg π 2kexp(−k) 17 x−α,0<α< 1 ⎪radicalbigg π 2kα−1 Γ(α)sec⎪parenleftBigπα 2⎪parenrightBig 18 ⎪parenleftbigg1 a+x⎪parenrightbigg e−ax ⎪radicalbigg 2 π2a2 (a2+k2)2 19 log⎪parenleftbigg 1+a2 x2⎪parenrightbigg ,a > 0 √ 2π(1−e−ak) k 20 log⎪parenleftbigga2+x2 b2+x2⎪parenrightbigg ,a , b > 0 √ 2π(e−bk−e−ak) k 21 a(x2+a2)−1,a > 0 ⎪radicalbigg π 2exp(−ak),k > 0 22 (a2−x2)−1 ⎪radicalbigg π 2sin(ak) k 23 e−bxsin(ax) 1 √ 2π⎪bracketleftbigga+k b2+(a+k)2+a−k b2+(a−k)2⎪bracketrightbigg 24 e−bxcos(ax) b √ 2π⎪bracketleftbigg1 b2+(a−k)2+1 b2+(a+k)2⎪bracketrightbigg © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 617 TABLE B-3 Fourier Sine Transforms f(x) Fs(k)=⎪radicalbigg 2 π∞⎪integraldisplay 0sin(kx)f(x)dx 1 exp(−ax),a > 0 ⎪radicalbigg 2 πk(a2+k2)−1 2 xexp(−ax),a > 0 ⎪radicalbigg 2 π(2ak)(a2+k2)−2 3 xα−1,0<α< 1 ⎪radicalbigg 2 πk−αΓ(α)sin⎪parenleftBigπα 2⎪parenrightBig 4 1 √ x 1 √ k,k > 0 5 xα−1e−ax,α > −1,a >0 ⎪radicalbigg 2 πΓ(α)r−αsin(αθ), where r=(a2+k2)1 2,θ=t a n−1⎪parenleftbigk a⎪parenrightbig 6 x−1e−ax,a > 0 ⎪radicalbigg 2 πtan−1⎪parenleftbiggk a⎪parenrightbigg ,k > 0 7 xexp(−a2x2) 2−3/2⎪parenleftbiggk a3⎪parenrightbigg exp⎪parenleftbigg −k2 4a2⎪parenrightbigg 8 erfc(ax) ⎪radicalbigg 2 π1 k⎪bracketleftbigg 1−exp⎪parenleftbigg −k2 4a2⎪parenrightbigg⎪bracketrightbigg 9 x(a2+x2)−1 ⎪radicalbigg π 2exp(−ak),a > 0 10 x(a2+x2)−2 1 √ 2π⎪parenleftbiggk a⎪parenrightbigg exp(−ak),(a>0) 11 x(a2−x2)v−1 2H(a−x), 2v−1 2av+1k−vΓ⎪parenleftbig v+1 2⎪parenrightbig v>−1 2 ×Jv+1(ak) 12 tan−1⎪parenleftBigx a⎪parenrightBig ⎪radicalbigg π 2k−1exp(−ak) © 2007 by Taylor & Francis Group, LLC 618 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(x) Fs(k)=⎪radicalbigg 2 π∞⎪integraldisplay 0sin(kx)f(x)dx 13 x−vJv+1(ax),v > −1 2 k(a2−k2)v−1 2 2v−1 2av+1Γ⎪parenleftbigg v+1 2⎪parenrightbiggH(a−k) 14 x−1J0(ax) ⎧ ⎪⎨ ⎪⎩⎪radicalbigg 2 πsin−1⎪parenleftbiggk a⎪parenrightbigg ,0<k<a ⎪radicalbig π 2,a < k < ∞⎫ ⎪⎬ ⎪⎭ 15 x(a2+x2)−1J0(bx),a>0,b>0 ⎪radicalbig π 2e−akI0(ab),a < k < ∞ 16 J0(a√ x),a > 0 ⎪radicalbigg 2 π1 kcos⎪parenleftbigga2 4k⎪parenrightbigg 17 (x2−a2)v−1 2H(x−a),|v|<1 2 2v−1 2⎪parenleftbiga k⎪parenrightbigvΓ⎪parenleftbig v+1 2⎪parenrightbig J−v(ak) 18 x1−v(x2+a2)−1Jv(ax), ⎪radicalbigg π 2a−vexp(−ak)Iv(ab), v>−3 2,a , b > 0 a<k< ∞ 19 H(a−x),a > 0 ⎪radicalbigg 2 π1 k(1−cosak) 20 erfc(ax) ⎪radicalbigg 2 π1 k⎪bracketleftbigg 1−exp⎪parenleftbigg −k2 4a2⎪parenrightbigg⎪bracketrightbigg 21 x−α,0<α< 2 Γ(1−α)kα−1cos⎪parenleftBigαπ 2⎪parenrightBig 22 (ax−x2)α−1 2H(a−x),α >−1 2 √ 2Γ⎪parenleftbigg α+1 2⎪parenrightbigg⎪parenleftBiga k⎪parenrightBigα ×sin⎪parenleftbiggak 2⎪parenrightbigg Jα⎪parenleftbiggak 2⎪parenrightbigg 23 e−bxsin(ax) b √ 2π⎪bracketleftbigg1 b2+(a−k)2−1 b2+(a+k)2⎪bracketrightbigg 24 ln⎪vextendsingle⎪vextendsingle⎪vextendsinglea+x b−x⎪vextendsingle⎪vextendsingle⎪vextendsingle √ 2πsin(ak) k © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 619 TABLE B-4 Laplace Transforms f(t) ¯f(s)=∞⎪integraldisplay 0exp(−st)f(t)dt 1 tn(n=0,1,2,3,...) n! sn+1 2 eat 1 s−a 3 cosat s s2+a2 4 sinat a s2+a2 5 coshat s s2−a2 6 sinhat a s2−a2 7 tne−at Γ(n+1 ) (s+a)n+1 8 ta(a>−1) Γ(a+1 ) sa+1 9 eatcosbt s−a (s−a)2+b2 10 eatsinbt b (s−a)2+b2 11 (eat−ebt) a−b (s−a)(s−b) 12 1 (a−b)(aeat−bebt) s (s−a)(s−b) 13 tsinat 2as (s2+a2)2 14 tcosat s2−a2 (s2+a2)2 15 sinatsinhat 2sa2 (s4+4a4) © 2007 by Taylor & Francis Group, LLC 620 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(t) ¯f(s)=∞⎪integraldisplay 0exp(−st)f(t)dt 16 (sinhat−sinat) 2a3 (s4−a4) 17 (coshat−cosat) 2a2s (s4−a4) 18 cosat−cosbt (b2−a2)(a2/negationslash=b2) s (s2+a2)(s2+b2) 19 1 √ t ⎪radicalbigg π s 20 2√ t 1 s⎪radicalbigg π s 21 tcoshat (s2+a2)(s2−a2)−2 22 tsinhat 2as(s2−a2)−2 23 sin(at) t tan−1⎪parenleftBiga s⎪parenrightBig 24 t−1/2exp⎪parenleftBig −a t⎪parenrightBig ⎪radicalbigg π sexp(−2√ as) 25 t−3/2exp⎪parenleftBig −a t⎪parenrightBig ⎪radicalbigg π aexp(−2√ as) 26 1 √ πt(1 + 2 at)eat s (s−a)√ s−a 27 (1 +at)eat s (s−a)2 28 1 2√ πt3(ebt−eat) √ s−a−√ s−b 29 exp(a2t)erf(a√ t) a √ s(s−a2) 30 exp(a2t)erfc(a√ t) 1 √ s(√ s+a) 31 1 √ πt+aexp(a2t)erf(a√ t) √ s (s−a2) © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 621 f(t) ¯f(s)=∞⎪integraldisplay 0exp(−st)f(t)dt 32 1 √ πt−aexp(a2t)erfc(a√ t) 1 √ s+a 33 exp(−at) √ b−aerf⎪parenleftBig⎪radicalbig (b−a)t⎪parenrightBig 1 (s+a)√ s+b 34 1 2eiωt⎪bracketleftbig e−λzerfc(ζ−√ iωt) (s−iω)−1e−z√ s v +e x p ( λz)erfc(ζ+√ iωt)⎪bracketrightbig , where ζ=z/2√ vt, λ=⎪radicalBig iω v. 35 1 2⎪bracketleftbigg e−aberfc⎪parenleftbiggb−2at 2√ t⎪parenrightbigg e−b(s+a2)1 2 +e x p ( ab)erfc⎪parenleftbiggb+2at 2√ t⎪parenrightbigg⎪bracketrightbigg 36 Si(t)=t⎪integraldisplay 0sinx xdx 1 scot−1(s) 37 Ci(t)=−∞⎪integraldisplay tcosx xdx −1 2slog(1 + s2) 38 −Ei(−t)=∞⎪integraldisplay te−x xdx 1 slog(1 + s) 39 J0(at) (s2+a2)−1 2 40 I0(at) (s2−a2)−1 2 41 tα−1exp(−at),a > 0 Γ(α)(s+a)−α 42 √ π Γ⎪parenleftbigg v+1 2⎪parenrightbigg⎪parenleftbiggt 2a⎪parenrightbiggv Jv(at) (s2+a2)−(v+1 2),Rev>−1 2 43 t−1Jv(at) av v(√ s2+a2+s)v,Rev>−1 2 44 J0(a√ t) 1 sexp⎪parenleftbigg −a2 4s⎪parenrightbigg © 2007 by Taylor & Francis Group, LLC 622 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(t) ¯f(s)=∞⎪integraldisplay 0exp(−st)f(t)dt 45 ⎪parenleftbigg2 a⎪parenrightbiggv tv/2Jv(a√ t) s−(v+1)exp⎪parenleftBig −a2 4s⎪parenrightBig ,Rev>−1 2 46 a 2t√ πtexp⎪parenleftbigg −a2 4t⎪parenrightbigg exp(−a√ s),a > 0 47 1 √ πtexp⎪parenleftbigg −a2 4t⎪parenrightbigg 1 √ sexp(−a√ s),a≥0 48 exp⎪parenleftbigg −a2t2 4⎪parenrightbigg √ π aexp⎪parenleftbiggs2 a2⎪parenrightbigg erfc⎪parenleftBigs a⎪parenrightBig ,a >0 49 (t2−a2)−1 2H(t−a) K0(as),a > 0 50 δ(t−a) exp(−as),a≥0 51 H(t−a) 1 sexp(−as),a≥0 52 δ/prime(t−a) se−as,a≥0 53 δ(n)(t−a) snexp(−as) 54 |sinat|,(a>0) a (s2+a2)coth⎪parenleftBigπs 2a⎪parenrightBig 55 1 √ πtcos(2√ at) 1 √ sexp⎪parenleftBig −a s⎪parenrightBig 56 1 √ πtsin(2√ at) 1 s√ sexp⎪parenleftBig −a s⎪parenrightBig 57 1 √ πacosh(2√ at) 1 √ sexp⎪parenleftBiga s⎪parenrightBig 58 1 √ πasinh(2√ at) 1 s√ sexp⎪parenleftBiga s⎪parenrightBig 59 erf⎪parenleftbiggt 2a⎪parenrightbigg 1 sexp(a2s2)erfc(as),a > 0 © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 623 f(t) ¯f(s)=∞⎪integraldisplay 0exp(−st)f(t)dt 60 erfc⎪parenleftbigga 2√ t⎪parenrightbigg 1 sexp(−a√ s),a≥0 61 ⎪radicalbigg 4t πe−a2 4t−aerfc⎪parenleftbigga 2√ t⎪parenrightbigg 1 s√ sexp(−a√ s),a≥0 62 ea(b+at)erfc⎪parenleftbigg a√ t+b 2√ t⎪parenrightbigg exp(−b√ s) √ s(√ s+a),a≥0 63 J0⎪parenleftbig a√ t2−ω2⎪parenrightbig H(t−ω) (s2+a2)−1 2exp⎪braceleftbig −ω√ s2+a2⎪bracerightbig 64 1 t(ebt−eat) log⎪parenleftbiggs−a s−b⎪parenrightbigg 65 {π(t+a)}−1 2 1 √ sexp(as)erfc(√ as),a >0 66 1 πtsin(2a√ t) erf⎪parenleftbigga √ s⎪parenrightbigg 67 1 √ πtexp(−2a√ t),a≥0 1 √ sexp⎪parenleftbigga2 s⎪parenrightbigg erfc⎪parenleftbigga √ s⎪parenrightbigg 68 C(t)=1 √ 2πt⎪integraldisplay 0cosu √ udu 1 2s⎪bracketleftbigg1 √ 1+s2+s 1+s2⎪bracketrightbigg1 2 69 S(t)=1 √ 2πt⎪integraldisplay 0sinu √ udu 1 2s⎪bracketleftbigg1 √ 1+s2−s 1+s2⎪bracketrightbigg1 2 70 I(t)=1+2∞⎪summationdisplay n=1exp(−n2πt) (√ stanh√ s)−1 71 tmα+β−1E(m) α,β(±at) m!sα−β (sα∓a)m+1 72 1+2at √ πt s+a s√ s © 2007 by Taylor & Francis Group, LLC 624 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TABLE B-5 Hankel Transforms f(r) order ˜fn(k)=∞⎪integraldisplay 0rJn(kr)f(r)dr 1 H(a−r) 0 a kJ1(ak) 2 exp(−ar) 0 a(a2+k2)−3 2 3 1 rexp(−ar) 0 (a2+k2)−1 2 4 (a2−r2)H(a−r) 0 4a k3J1(ak)−2a2 k2J0(ak) 5 a(a2+r2)−3 2 0 exp(−ak) 6 1 rcos(ar) 0 (k2−a2)−1 2H(k−a) 7 1 rsin(ar) 0 (a2−k2)−1 2H(a−k) 8 1 r2(1−cosar) 0 cosh−1⎪parenleftBiga k⎪parenrightBig H(a−k) 9 1 rJ1(ar) 0 1 aH(a−k),a > 0 10 Y0(ar) 0 ⎪parenleftbigg2 π⎪parenrightbigg (a2−k2)−1 11 K0(ar) 0 (a2+k2)−1 12 δ(r) r 0 1 13 (r2+b2)−1 2 0 (k2+a2)−1 2 ×exp⎪braceleftBig −a(r2+b2)1 2⎪bracerightBig ×exp⎪braceleftBig −b(k2+a2)1 2⎪bracerightBig 14 sinr r2 0 ⎧ ⎪⎨ ⎪⎩π 2,k < 1 sin−1⎪parenleftbigg1 k⎪parenrightbigg ,k > 1⎫ ⎪⎬ ⎪⎭ © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 625 f(r) order ˜fn(k)=∞⎪integraldisplay 0rJn(kr)f(r)dr 15 (r2+a2)−1 2 0 1 kexp(−ak) 16 exp(−ar) 1 k(a2+k2)−3/2 17 sinar r 1 aH(k−a) k(k2−a2)1 2 18 1 rexp(−ar) 1 1 k⎪bracketleftbigg 1−a (k2+a2)1 2⎪bracketrightbigg 19 1 r2exp(−ar) 1 1 k⎪bracketleftBig (k2+a2)1 2−a⎪bracketrightBig 20 rnH(a−r) >−1 1 kan+1Jn+1(ak) 21 rnexp(−ar),Rea>0 >−1 1 √ π2n+1Γ⎪parenleftbigg n+3 2⎪parenrightbigg akn (a2+k2)n+3 2 22 rnexp(−ar2) >−1 kn (2a)n+1exp⎪parenleftbigg −k2 4a⎪parenrightbigg 23 ra−1 >−1 2aΓ⎪bracketleftbigg1 2(a+n+1 )⎪bracketrightbigg ka+1Γ⎪bracketleftbigg1 2(1−a+n)⎪bracketrightbigg 24 rn(a2−r2)m−n−1 >−1 2m−n−1Γ(m−n)am ×H(a−r) ×kn−mJm(ak) 25 rmexp(−r2/a2) >−1 1F1⎪parenleftbig 1+m 2+n 2;n+1 ;−1 4a2k2⎪parenrightbig ×knam+n+2 2n+1Γ(n+1)Γ⎪parenleftbig 1+m 2+n 2⎪parenrightbig 26 1 rJn+1(ar) >−1 kna−(n+1)H(a−k),a >0 27 rn(a2−r2)mH(a−r), >−1 2manΓ(m+1 )⎪parenleftbiga k⎪parenrightbigm+1 m>−1 ×Jn+m+1(ak) © 2007 by Taylor & Francis Group, LLC 626 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(r) order ˜fn(k)=∞⎪integraldisplay 0rJn(kr)f(r)dr 28 1 r2Jn(ar) >1 2 ⎧ ⎪⎪⎨ ⎪⎪⎩1 2n⎪parenleftbiggk a⎪parenrightbiggn ,0<k≤a 1 2n⎪parenleftBiga k⎪parenrightBign ,a < k < ∞⎫ ⎪⎪⎬ ⎪⎪⎭ 29 rn (a2+r2)m+1,a > 0 >−1 ⎪parenleftbiggk 2⎪parenrightbiggman−m Γ(m+1 )Kn−m(ak) 30 exp(−p2r2)Jn(ar), >−1 (2p2)−1exp⎪parenleftbigg −a2+k2 4p2⎪parenrightbigg ×In⎪parenleftBig ak 2p2⎪parenrightBig 31 1 rexp(−ar) >−1 ⎪braceleftBig (k2+a2)1 2−a⎪bracerightBign kn(k2+a2)1 2 32 rn (r2+a2)n+1 >−1 ⎪parenleftbiggk 2⎪parenrightbiggnK0(ak) Γ(n+1 ) 33 rn (a2−r2)n+1 2H(a−r) <1 1 √ π⎪parenleftbiggk 2⎪parenrightbiggn Γ⎪parenleftbigg1 2−n⎪parenrightbigg⎪parenleftbiggsinak k⎪parenrightbigg 34 1 √ rJn−1(ar) >−1 ⎧ ⎪⎨ ⎪⎩0, 0<k≤a 1 √ a⎪parenleftBiga k⎪parenrightBign−1 2,a < k < ∞⎫ ⎪⎬ ⎪⎭ 35 1 r√ rJn(ar) >0 ⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩√ a 2n⎪parenleftbiggk a⎪parenrightbiggn+1 2 ,0<k≤a √ a 2n⎪parenleftBiga k⎪parenrightBign+1 2,a < k < ∞⎫ ⎪⎪⎪⎬ ⎪⎪⎪⎭ 36 1 √ rJn+1(ar) >−3 2 ⎧ ⎪⎨ ⎪⎩1 √ a⎪parenleftbiggk a⎪parenrightbiggn+1 2 ,0<k≤a 0,a < k < ∞⎫ ⎪⎬ ⎪⎭ 37 rn−1e−ar >−1 (2k)n(n−1 2)! √ π(k2+a2)n+1 2 38 e−ar2J0(br) 0 a 2exp⎪parenleftbiggk2−b2 4a⎪parenrightbigg I0⎪parenleftbiggbk 2a⎪parenrightbigg © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 627 TABLE B-6 Mellin Transforms f(x) ˜f(p)=∞⎪integraldisplay 0xp−1f(x)dx 1 exp(−nx) n−pΓ(p),Rep>0 2 exp(−ax2),a > 0 1 2a−(p/2)Γ⎪parenleftBigp 2⎪parenrightBig ,Rep>0 3 cos(ax) a−pΓ(p)cos⎪parenleftBigπp 2⎪parenrightBig ,0<Rep<1 4 sin(ax) a−pΓ(p)sin⎪parenleftBigπp 2⎪parenrightBig ,0<Rep<1 5 (a+x)−1,|arga|<π πap−1cosec( πp),0<Rep<1 6 (a−x)−1 πap−1cot(πp),0<Rep<1 7 (1 +x)−a,Rea>0 Γ(p)Γ(a−p) Γ(a) 8 (1 +xa)−s Γ(p/a)Γ(s−p/a) aΓ(s) 9 (a2+x2)−1 π 2a(p−2)cosec⎪parenleftBigπp 2⎪parenrightBig 10 ⎪braceleftBigg 1,0≤x≤a 0,x > a⎪bracerightBigg p−1ap 11 Ci(x)=−∞⎪integraldisplay xcost tdt −p−1Γ(p)cos⎪parenleftBigpπ 2⎪parenrightBig ,0<Rep<1 12 Si(x)=x⎪integraldisplay 0sint tdt −p−1Γ(p)sin⎪parenleftBigπp 2⎪parenrightBig ,−1<Rep<0 13 ⎪braceleftBigg (1−x)a−1,0<x< 1 0,x ≥1⎪bracerightBigg Γ(a)Γ(p) Γ(a+p) 14 ⎪braceleftBigg 0, 0<x≤1 (x−1)−a,x > 1⎪bracerightBigg Γ(a−p)Γ( 1−a) Γ(1−p) © 2007 by Taylor & Francis Group, LLC 628 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(x) ˜f(p)=∞⎪integraldisplay 0xp−1f(x)dx 15 exp(−ax)H(x−b) a−pΓ(p,ab) 16 exp(−ax)H(b−x) a−pγ(p,ab) 17 2F1(a,b,c;−x) Γ(p)Γ(a−p)Γ(b−p)Γ(c) Γ(c−p)Γ(a)Γ(b) 18 x1 2Jv(x) 2p−1 2Γ⎪bracketleftbigg1 2⎪parenleftbigg p+v+1 2⎪parenrightbigg⎪bracketrightbigg Γ⎪bracketleftbigg1 2⎪parenleftbigg v−p+3 2⎪parenrightbigg⎪bracketrightbigg 19 x−vJv(ax) 2p−v−1av−pΓ⎪parenleftBigp 2⎪parenrightBig Γ⎪parenleftbigg v−1 2p+1⎪parenrightbigg 20 Pn(x)H(1−x) Γ⎪parenleftBigp 2⎪parenrightBig Γ⎪parenleftbiggp 2+1 2⎪parenrightbigg 2Γ⎪parenleftbiggp 2−n 2+1 2⎪parenrightbigg Γ⎪parenleftBigp 2+n 2+1⎪parenrightBig 21 ⎪braceleftBigg log⎪parenleftBiga x⎪parenrightBig ,x < a 0,x ≥a⎪bracerightBigg ap p2 22 x−1log(1 + x) π(1−p)−1cosec( πp) 23 (ex−1)−1 Γ(p)ζ(p) 24 (ex+e−x)−1 L(p)Γ(p) 25 log⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle1+x 1−x⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle ⎪parenleftbiggπ p⎪parenrightbigg tan⎪parenleftBigpπ 2⎪parenrightBig 26 (1 +x)−mPm−1⎪parenleftBig 1−x 1+x⎪parenrightBig Γ(p){Γ(m−p)}2 Γ(1−p){Γ(m)}2 © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 629 f(x) ˜f(p)=∞⎪integraldisplay 0xp−1f(x)dx 27 xa(1 +x)−b Γ(a+p)Γ(b−a−p) Γ(b) 28 x−2vJv(x)Kv(x) 2p−2v−2Γ⎪parenleftBigp 4⎪parenrightBig Γ⎪parenleftBigp 2−v⎪parenrightBig Γ⎪parenleftBig 1+v−p 4⎪parenrightBig 29 log(1 + ax),|arga|<π π pa−pcosec( πp),−1<Rep<0 30 xv+1Jv(ax) 2p+va−(p+v+1)Γ⎪parenleftbiggp 2+v+1 2⎪parenrightbigg Γ⎪parenleftbigg1−p 2⎪parenrightbigg 31 (1 +x2)−(1+α)H(x−1) Γ⎪parenleftBigp 2⎪parenrightBig Γ⎪parenleftBig α+1−p 2⎪parenrightBig 2Γ (α+1 ) 32 cos(xα) 1 αΓ⎪parenleftBigp α⎪parenrightBig cos⎪parenleftBigpπ 2α⎪parenrightBig 33 sin(xα) 1 αΓ⎪parenleftBigp α⎪parenrightBig sin⎪parenleftBigpπ 2α⎪parenrightBig 34 (1 +ax)−n Γ(p)Γ(n−p) apΓ(n),0<Rep<n 35 e−x(logx)n dn dpnΓ(p)Rep > 0 36 e−axIv(ax),R e;a>0 Γ⎪parenleftbigg1 2−p⎪parenrightbigg Γ(v+p) 2pap√ πΓ(1 + v−p) 37 e−axKv(ax),R e a > 0 √ πΓ(p+v)Γ (p−v) 2papΓ⎪parenleftbigg p+1 2⎪parenrightbigg 38 erfc(x) Γ(p+1 2) p√ π © 2007 by Taylor & Francis Group, LLC 630 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TABLE B-7 Hilbert Transforms f(t) ˆfH(x)=1 π∞⎪contintegraldisplay −∞f(t) (t−x)dt 1 1 0 2 ⎧ ⎪⎨ ⎪⎩0,−∞<t<a 1, a<t<b 0,b < t < ∞⎫ ⎪⎬ ⎪⎭ 1 πlog⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingleb−x a−x⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle 3 (t+a)−1,Ima>0 i(x+a)−1 4 (t+a)−1,Ima<0 −i(x+a)−1 5 ⎧ ⎪⎨ ⎪⎩0, −∞<t<0 (at+b)−1,0<t<∞ a, b > 0⎫ ⎪⎬ ⎪⎭ 1 π(ax+b)−1log⎪vextendsingle⎪vextendsingleb ax⎪vextendsingle⎪vextendsingle,a x/negationslash=−b 6 t (t2+a2),Rea>0 a (x2+a2) 7 1 (t2+a2),Rea>0 −x a(x2+a2) 8 αt+βa (t2+a2),Rea>0 αa−βx (x2+a2) 9 exp(iat),a > 0 iexp(iax) 10 cos(at),a > 0 −sin(ax) 11 sin(at),a > 0 cos(ax) 12 a a2+(t+b)2,a > 0 −(b+x) a2+(b+x)2 13 ⎧ ⎪⎨ ⎪⎩0, −∞<t<−a (a2−t2)−1 2,−a<t<a 0,a < t < ∞⎫ ⎪⎬ ⎪⎭ ⎧ ⎪⎪⎨ ⎪⎪⎩(x2−a2)−1 2,−∞<x< −a 0, −a<x<a −(x2−a2)−1 2,a < x < ∞⎫ ⎪⎪⎬ ⎪⎪⎭ 14 H(t−a)−H(t−b),b > a > 0 1 πlog⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglex−b x−a⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 631 f(t) ˆfH(x)=1 π∞⎪contintegraldisplay −∞f(t) (t−x)dt 15 1 tH(t−a),a > 0 1 πxlog⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglea x−a⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle,x/negationslash=0,x/negationslash=a 16 ⎧ ⎪⎪⎨ ⎪⎪⎩−(t2−a2)−1 2,−∞<t<−a 0, −a<t<a (t2−a2)−1 2,a < t < ∞⎫ ⎪⎪⎬ ⎪⎪⎭ ⎧ ⎪⎨ ⎪⎩0, −∞<x< −a (a2−x2)−1 2,−a<x<a 0,a < x < ∞⎫ ⎪⎬ ⎪⎭ 17 sinat t,a > 0 1 x(cosax−1) 18 ⎪braceleftBigg0, −∞<t<0 sin(a√ t),0<t<∞,a >0⎪bracerightBigg ⎪braceleftBigg exp⎪parenleftBig −a⎪radicalbig |x|⎪parenrightBig ,−∞<x< 0 cos(a√ x),0<x< ∞⎪bracerightBigg 19 sgntsin⎪parenleftBig a⎪radicalbig |t|⎪parenrightBig ,a > 0 cos⎪parenleftBig a⎪radicalbig |x|⎪parenrightBig +e x p⎪parenleftBig −a⎪radicalbig |x|⎪parenrightBig 20 1 t(1−cosat),a > 0 1 x(sinax) 21 Jn(t)s i n (t−x),n=0,1,... Jn(x) 22 sgnt|t|vJv(a|t|), where −|x|vYv(a|x|) a>0,−1 2<Rev<3 2 23 sin(at)J1(at),a > 0 cos(ax)J1(ax) 24 sin(at)Jn(bt), where cos(ax)Jn(bx) 0<b<a , n =0,1,2,... 25 cos(at)J1(at),a > 0 −sin(ax)J1(ax) 26 cos(at)Jn(bt),0<b<a −sin(ax)Jn(bx) where n=0,1,2,... 27 exp(−at)I0(at)H(t),a>0 1 πexp(−ax)K0(a|x|) © 2007 by Taylor & Francis Group, LLC 632 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(t) ˆfH(x)=1 π∞⎪contintegraldisplay −∞f(t) (t−x)dt 28 exp(−a|t|)I0(at),a >0 −2 πsinh(ax)K0(a|x|) 29 sgntexp(−a|t|)I0(at),a >0 2 πcosh(ax)K0(a|x|) 30 exp(at)K0(a|t|),a > 0 πexp(ax)I0(ax)H(−x) 31 |t|vYv(a|t|),a >0, |x|vJv(a|x|)sgnx −1 2<Rev<3 2 32 sinh(at)K0(a|t|),a > 0 π 2exp(−a|x|)I0(ax) 33 cosh(at)K0(a|t|),a > 0 −π 2exp(−a|x|)I0(ax)sgnx 34 log⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingleb−t t−a⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle,a < b ⎧ ⎪⎨ ⎪⎩0,−∞<x<a −π, a<x<b 0,b < x < ∞⎫ ⎪⎬ ⎪⎭ 35 log⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsinglet 2−a2 t2−b2⎪vextendsingle⎪vextendsingle⎪vextendsingle⎪vextendsingle,0<a<b ⎧ ⎪⎨ ⎪⎩−π,−b<x<a π, a<x<b 0,elsewhere⎫ ⎪⎬ ⎪⎭ 36 1−cosat t,a > 0 sinax x © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 633 TABLE B-8 Stieltjes Transforms f(t) ˜f(x)=∞⎪integraldisplay 0f(t) (t+x)dt 1 (a+t)−1,|arga|<π (a−x)−1log⎪parenleftBiga x⎪parenrightBig 2 1 (a2+t2),Rea>0 (a2+x2)−1⎪bracketleftBigπx 2a−log⎪parenleftBigx a⎪parenrightBig⎪bracketrightBig 3 t (a2+t2),Rea>0 (a2+x2)−1⎪bracketleftBigπa 2+xlog⎪parenleftBigx a⎪parenrightBig⎪bracketrightBig 4 tv,−1<Rev<0 −πxvcosec( πv) 5 ⎧ ⎪⎨ ⎪⎩−1,2n<x< 2n+1 +1,2n+1<x< 2n+2 n=0,1,2,3...⎫ ⎪⎬ ⎪⎭ log⎡ ⎣x 2⎪braceleftBigg Γ⎪parenleftBigx 2⎪parenrightBig⎪slashBigg Γ⎪parenleftbiggx+1 2⎪parenrightbigg⎪bracerightBigg2⎤ ⎦ 6 tv (a+t),|arga|<π, (a−x)−1π(av−xv)cosec ( πv) where −1<Rev<1 7 ⎪parenleftbiggtv−av t−a⎪parenrightbigg ,−1<Rev<1 ⎪parenleftbiggπ a+x⎪parenrightbigg⎪bracketleftBigg xvcosec( vπ) −avctn(vπ)+av πlog⎪parenleftBiga x⎪parenrightBig⎪bracketrightBigg 8 tv−1(a+t)1−μ,|arga|<π, Γ(v)Γ(μ−v) Γ(μ)⎪parenleftbiggxv−1 aμ−1⎪parenrightbigg 0<Rev<Reμ ×2F1⎪parenleftBig μ−1,v,μ;1−x a⎪parenrightBig 9 t−ρ(a+t)−σ,|arga|<π, πcosec( ρπ)x−ρ(a−x)−σ −Reσ<Reρ<1 ×I(1−x a)(σ, ρ) 10 exp(−at),Rea>0 −exp(ax)Ei(−ax) 11 ⎪braceleftBigg exp(−at),0<t<b 0,b < t < ∞⎪bracerightBigg eax[Ei(−ab−ax)−Ei(−ax)] © 2007 by Taylor & Francis Group, LLC 634 INTEGRAL TRANSFORMS and THEIR APPLICATIONS f(t) ˜f(x)=∞⎪integraldisplay 0f(t) (t+x)dt 12 ⎧ ⎪⎨ ⎪⎩0, 0<t<b exp(−at),b < t < ∞ Rea>0⎫ ⎪⎬ ⎪⎭ −exp(−ax)Ei(−ab−ax) 13 1 √ texp(−at),Rea>0 π √ xexp(ax)erfc(√ ax) 14 √ texp(−at),Rea>0 ⎪radicalbigg π a−π√ xexp(ax)erfc(√ ax) 15 t−vexp(−at),Rea>0, Rev<1 Γ(1−v)x−vexp(ax)Γ(v,ax) 16 tv−1exp⎪parenleftBig −a t⎪parenrightBig ,Rea>0, Rev<1 Γ(1−v)xv−1exp⎪parenleftBiga x⎪parenrightBig Γ⎪parenleftBig v,a x⎪parenrightBig 17 exp(−a√ t),Rea>0 2⎪bracketleftbig cos(a√ x)Ci(a√ x) −sin(a√ x)Si(a√ x)] 18 1 √ texp(−a√ t),Rea>0 −2 √ x⎪bracketleftbig sin(a√ x)Ci(a√ x) +c o s ( a√ x)Si(a√ x)] 19 (a+t)−1log⎪parenleftbigt a⎪parenrightbig ,|arga|<π 1 2(x−a)−1⎪bracketleftBig log⎪parenleftBigx a⎪parenrightBig⎪bracketrightBig2 20 (t−a)−1log⎪parenleftbiggt a⎪parenrightbigg ,a>0 1 2(x+a)−1⎪bracketleftbigg π2+⎪braceleftBig log⎪parenleftBigx a⎪parenrightBig⎪bracerightBig2⎪bracketrightbigg 21 1 √ tlog(at+b),Rea>0, Reb>0 2π √ xlog⎪parenleftBig√ ax+√ b⎪parenrightBig 22 tvlogt,−1<Rev<0 −πxvcosec( vπ)[logx−πctn(vπ)] 23 sinat, a > 0 −[sin(ax)Ci(ax)+c o s ( ax)Si(ax)] 24 sin(a√ t),a > 0 πexp(−a√ x) © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 635 f(t) ˜f(x)=∞⎪integraldisplay 0f(t) (t+x)dt 25 t−1sin(a√ t),a > 0 ⎪parenleftBigπ x⎪parenrightBig⎪bracketleftbig 1−exp(−a√ x)⎪bracketrightbig 26 t−αsin(a√ t+απ), ⎪parenleftBigπ xα⎪parenrightBig exp(−a√ x) where a>0,−1 2<Reα<1 27 sin⎪parenleftbigg a√ t−b √ t⎪parenrightbigg ,a , b > 0 πexp⎪bracketleftbigg −⎪parenleftbigg a√ x+b √ x⎪parenrightbigg⎪bracketrightbigg 28 1 √ tsin2(a√ t),a > 0 ⎪parenleftbiggπ 2√ x⎪parenrightbigg⎪bracketleftbig 1−exp(−2a√ x)⎪bracketrightbig 29 cos(at),a > 0 cos(ax)Ci(ax)−sin(ax)Si(ax) 30 1 √ tcos(a√ t),a > 0 ⎪parenleftbiggπ √ x⎪parenrightbigg exp(−a√ x) 31 1 √ tcos⎪parenleftBig a√ t−b √ t⎪parenrightBig ,a , b> 0 ⎪parenleftbiggπ √ x⎪parenrightbigg exp⎪bracketleftbigg −⎪parenleftbigg a√ x+b √ x⎪parenrightbigg⎪bracketrightbigg 32 1 √ tcos(a√ t)cos (b√ t), π √ xexp(−a√ x)cosh ( b√ x) a≥b>0 33 t(v 2+k)Jv(a√ t) 2(−1)kx(1 2v+k)Kv(a√ x) 34 sin(a√ t)J0(b√ t),0<b<a πexp(−a√ x)I0(b√ x) 35 1 √ tsin(a√ t)J0(b√ t), 2 √ xsinh(a√ x)K0(b√ x) 0<a<b 36 cos(a√ t)J0(b√ t),0<a<b 2c o s h ( a√ x)K0(b√ x) 37 1 √ tcos(a√ t)J0(b√ t), π √ xexp(−a√ x)I0(b√ x) 0<b<a 38 J2 v(at),a > 0 2Iv(a√ x)Kv(a√ x) © 2007 by Taylor & Francis Group, LLC 636 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TABLE B-9 Finite Fourier Cosine Transforms f(x) ˜fc(n)=a⎪integraldisplay 0f(x)cos⎪parenleftBignπx a⎪parenrightBig dx 1 1 ⎪braceleftBigg a, n =0 0,n/negationslash=0⎪bracerightBigg 2 x ⎧ ⎪⎨ ⎪⎩a2 2,n =0 ⎪parenleftBiga nπ⎪parenrightBig2 [(−1)n−1],n/negationslash=0⎫ ⎪⎬ ⎪⎭ 3 x2 ⎧ ⎪⎨ ⎪⎩1 3a3,n =0 2a⎪parenleftBiga nπ⎪parenrightBig2 (−1)n,n=1,2,...⎫ ⎪⎬ ⎪⎭ 4 x3 ⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩1 4a4n=0 3a4(−1)n (nπ)2+6⎪parenleftbiga nπ⎪parenrightbig4[(−1)n−1], n=1,2,3,...⎫ ⎪⎪⎪⎬ ⎪⎪⎪⎭ 5 ⎧ ⎪⎨ ⎪⎩1,0<x<a 2 −1,1 2a<x<a⎫ ⎪⎬ ⎪⎭ ⎧ ⎪⎨ ⎪⎩0,n =0 ⎪parenleftbigg2a nπ⎪parenrightbigg sin⎪parenleftBignπ 2⎪parenrightBig ,n=1,2,3,...⎫ ⎪⎬ ⎪⎭ 6 ⎪parenleftBig 1−x a⎪parenrightBig2 ⎧ ⎪⎪⎨ ⎪⎪⎩1 3a, n =0 2a (nπ)2,n=1,2,...⎫ ⎪⎪⎬ ⎪⎪⎭ 7 sin(bx) ba2 (nπ)2−(ab)2[(−1)ncos(ab)−1], nπ/negationslash=ab 8 cos(bx) (−1)nba2sin(ab) (ab)2−(nπ)2,n π /negationslash=ab 9 sin⎪parenleftbigmπx a⎪parenrightbig ,man integer ⎧ ⎪⎨ ⎪⎩0,n =m mπ[(−1)n+m−1] π(n2−m2),n/negationslash=m⎫ ⎪⎬ ⎪⎭ © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 637 f(x) ˜fc(n)=a⎪integraldisplay 0f(x)cos⎪parenleftBignπx a⎪parenrightBig dx 10 exp(bx) (a2b)⎪bracketleftbigg(−1)nexp(ab)−1 (nπ)2+(ba)2⎪bracketrightbigg 11 x−1 2(a2−x2)−1 2 ⎪parenleftBigπ 2⎪parenrightBig3/2⎪parenleftBignπ a⎪parenrightBig1 2⎪braceleftBig J−1/4⎪parenleftBignπ 2⎪parenrightBig⎪bracerightBig2 12 (a2−x2)v−1 2 √ π2v−1Γ⎪parenleftbig v+1 2⎪parenrightbig⎪parenleftBig a2 nπ⎪parenrightBigv Jv(nπ) 13 sin⎪braceleftBig b(a2−x2)1 2⎪bracerightBig ⎪parenleftbiggπab 2⎪parenrightbigg⎪parenleftbigg b2+n2π2 a2⎪parenrightbigg−1 2 ×J1⎪bracketleftBig {(ab)2+(nπ)2}1 2⎪bracketrightBig 14 (a2−x2)−1 2 ⎪parenleftBigπ 2⎪parenrightBig J0⎪bracketleftBig {(ab)2+(nπ)2}1 2⎪bracketrightBig ×cos⎪braceleftBig b(a2−x2)1 2⎪bracerightBig 15 J0⎪braceleftBig b(a2−x2)1 2⎪bracerightBig ⎪parenleftbigg b2+n2π2 a2⎪parenrightbigg−1 2 ×sin⎪bracketleftBig {(ab)2+(nπ)2}1 2⎪bracketrightBig © 2007 by Taylor & Francis Group, LLC 638 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TABLE B-10 Finite Fourier Sine Transforms f(x) ˜fs(n)=a⎪integraldisplay 0sin⎪parenleftBignπx a⎪parenrightBig f(x)ds 1 1 ⎪parenleftBiga nπ⎪parenrightBig [1−(−1)n] 2 x (−1)n+1⎪parenleftbigga2 nπ⎪parenrightbigg 3 x2 a3(−1)n−1 nπ−2a3⎪bracketleftbig 1+(−1)n+1⎪bracketrightbig (nπ)3 4 x3 (−1)na4 π5⎪parenleftbigg6 n3−π2 n⎪parenrightbigg 5 ⎪parenleftbigga−x a⎪parenrightbigg ⎪parenleftBiga nπ⎪parenrightBig 6 x(a−x) 2⎪parenleftBiga nπ⎪parenrightBig3⎪bracketleftbig 1+(−1)n+1⎪bracketrightbig 7 x(a2−x2) (−1)n+16a⎪parenleftBiga nπ⎪parenrightBig3 8 exp(bx) nπa (nπ)2+(ab2)⎪bracketleftbig 1+(−1)n+1exp(ab)⎪bracketrightbig 9 cos(bx) nπa (nπ)2−(ab)2⎪bracketleftbig 1+(−1)n+1cos(ab)⎪bracketrightbig , nπ/negationslash=ab 10 sin(bx) (−1)nanπsin(ab) (nπ)2−(ab)2,n π /negationslash=ab 11 cosh(bx) nπa [(nπ)2+(ab)2][1 + (−1)n+1cosh(ab)] 12 sin⎪parenleftBigmπx a⎪parenrightBig ,minteger ⎪braceleftBigg0,n /negationslash=m 1 2a, n =m⎪bracerightBigg 13 cos⎪parenleftBigmπx a⎪parenrightBig ,minteger ⎪braceleftBiggna π(n2−m2)⎪bracketleftbig 1+(−1)n+m+1⎪bracketrightbig ,n/negationslash=m 0,n =m⎪bracerightBigg 14 x−1 Si(nπ) © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 639 f(x) ˜fs(n)=a⎪integraldisplay 0sin⎪parenleftBignπx a⎪parenrightBig f(x)ds 15 x−1 2(x2−a2)−1 2 ⎪parenleftBigπ 2⎪parenrightBig3/2⎪parenleftBignπ a⎪parenrightBig1 2⎪braceleftBig J1/4⎪parenleftBignπ 2⎪parenrightBig⎪bracerightBig2 16 x(a2−x2)α−1 2 √ π2α−1aα+1Γ⎪parenleftbigg α+1 2⎪parenrightbigg ×⎪parenleftBignπ a⎪parenrightBig−α Jα+1(nπ) 17 (a2−x2)−1 2T2n+1⎪parenleftbigx a⎪parenrightbig ⎪parenleftBigπ 2⎪parenrightBig (−1)nJ2n+1(nπ) 18 (ax−x2)α−1 2 √ πΓ⎪parenleftbigg α+1 2⎪parenrightbigg⎪parenleftbigga2 nπ⎪parenrightbiggα Jα⎪parenleftBignπ 2⎪parenrightBig © 2007 by Taylor & Francis Group, LLC 640 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TABLE B-11 Finite Laplace Transforms f(t) LT{f(t)}=¯f(s, T)=T⎪integraldisplay 0e−stf(t)dt 1 1 1 s(1−e−sT) 2 t 1 s2−1 se−sT⎪parenleftbigg1 s+T⎪parenrightbigg 3 tn n! sn+1−e−sT sn+1⎪bracketleftbig (sT)n+n(sT)n−1 +n(n−1)(sT)n−2+···+n!⎪bracketrightbig 4 ta,(a>−1) 1 sa+1γ(a+1,s T) 5 exp(−at),a > 0 (s+a)−1[1−exp{−T(s+a)}] 6 tnexp(−at),a > 0 n! (s+a)n+1−e−(a+s)T (s+a)n+1[{(s+a)T}n +n{T(s+a)}n−1 +n(n−1){T(s+a)}n−2+···+n!⎪bracketrightbig 7 H(t−a),a > 0 1 s⎪bracketleftbig e−sa−e−sT⎪bracketrightbig H(T−a) 8 cos(at) s (s2+a2)+e−sT (s2+a2) ×(asinaT−scosaT) 9 sin(at) a (s2+a2)−e−sT (s2+a2) ×(ssinaT+acosaT) 10 e−atsin(bt) b (s+a)2+b2−exp(−sT) (s+a)2+b2 ×(ssinbT+asinbT+bcosbT) 11 e−atcos(bt) s+a (s+a)2+b2+exp(−sT) (s+a)2+b2 ×(bsinbT−scosbT−acosbT) © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 641 f(t) LT{f(t)}=¯f(s, T)=T⎪integraldisplay 0e−stf(t)dt 12 sinh(at) a (s2−a2)−exp(−sT) (s2−a2)(acoshaT+ssinhaT) 13 cosh(at) s (s2−a2)−exp(−sT) (s2−a2)(scoshaT+asinhaT) 14 t1 2 −√ Texp(−sT) s+√ π 2erf(√ sT) s3/2 15 t−1 2 π serf(√ sT) 16 erfc⎪parenleftbiggt 2a⎪parenrightbigg 1 s{1−exp(a2s2)erfc(as)}−e−sT serfc⎪parenleftbiggT 2a⎪parenrightbigg +exp(a2s2) serfc⎪parenleftbiggT 2a+as⎪parenrightbigg 17 erfc(bt) −1 sexp⎪parenleftBig s2 4b2⎪parenrightBig erf⎪parenleftbigs 2b⎪parenrightbig −exp⎪parenleftBig s2 4b2⎪parenrightBig serfc⎪parenleftBigs 2b⎪parenrightBig +1 sexp⎪parenleftbiggs2 4b2⎪parenrightbigg erfc⎪parenleftBig bT+s 2b⎪parenrightBig −e−sT serf(bT) 18 erf(t) −es2 4 serf⎪parenleftBigs 2⎪parenrightBig +es2 4 serf⎪parenleftBig T+s 2⎪parenrightBig −e−sT serf(T) 19 erf(√ t) erf(√ T) s(s+1 )−exp(−sT)erf(√ T) s 20 ebterf(√ bT) √ berf(√ sT) √ s(s−b)+e−(s−b)Terf(√ bT) (s−b) 21 ebterfc(√ bT) 1 (s−b)⎪braceleftBigg 1−√ b √ serf(√ sT)⎪bracerightBigg −e−(s−b)Terf(√ bT) (s−b) © 2007 by Taylor & Francis Group, LLC 642 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TABLE B-12 Z Transforms f(n) F(z)=∞⎪summationdisplay n=0f(n)z−n 1 ⎪braceleftBigg 1,n=0 0,n/negationslash=0⎪bracerightBigg 1 2 1 z z−1 3 an z z−a 4 n z (z−1)2 5 n2 z(z+1 ) (z−1)3 6 1 n! exp⎪parenleftbigg1 z⎪parenrightbigg 7 cosnx z(z−cosx) z2−2zcosx+1 8 sinnx zsinx z2−2zcosx+1 9 exp(±nx) z z−exp(±x) 10 nkenx ∂ ∂xk⎪parenleftbiggz z−ex⎪parenrightbigg 11 ne−nx zexp(−x) (z−e−x)2 12 n2e−nx z(z+e−x)e−x (z−e−x)3 13 exp(−nx)sin (an) zexp(−x)s i na z2−2ze−xcosa+e−2x 14 exp(−nx)cos (an) z(z−e−xcosa) z2−2ze−xcosa+e−2x © 2007 by Taylor & Francis Group, LLC Tables of Integral Transforms 643 f(n) F(z)=∞⎪summationdisplay n=0f(n)z−n 15 sinh (nx) zsinhx z2−2zcoshx+1 16 cosh(nx) z(z−coshx) z2−2zcoshx+1 17 H(n−1) 1 z−1 18 H(n)−H(n−1) 1 19 H(n−m),m=1,2,3 1 zm−1(z−1) 20 H(n−1)−H(n−2) 1 z 21 H(n−m)−H[n−(m+1 ) ] 1 zm 22 m(m−1)···(m−n+1 ) n! ⎪parenleftbigg 1+1 z⎪parenrightbiggm 23 1 (2n+1 ) ! √ zsinh⎪parenleftbigg1 √ z⎪parenrightbigg 24 1 (2n)! cosh⎪parenleftbigg1 √ z⎪parenrightbigg 25 an (2n+1 ) ! ⎪radicalbigg z asinh⎪parenleftbigg⎪radicalbigg a z⎪parenrightbigg 26 an (2n)! cosh⎪parenleftbigg⎪radicalbigg a z⎪parenrightbigg 27 ansinh (nx) zasinhx z2−2zacoshx+a2 28 ancosh(nx) z(z−acoshx) z2−2zacoshx+a2 © 2007 by Taylor & Francis Group, LLC 644 INTEGRAL TRANSFORMS and THEIR APPLICATIONS TABLE B-13 Finite Hankel Transforms order f(r) n ˜fn(ki)=a⎪integraltext 0rJn(rki)f(r)dr 1 c,w h e r e cis a constant 0 ⎪parenleftbiggac ki⎪parenrightbigg J1(aki) 2 (a2−r2) 0 4a k3 iJ1(aki) 3 (a2−r2)−1 2 0 k−1 isin(aki) 4 J0(αr) J0(αa) 0 −aki ⎪parenleftbig α2−k2 i⎪parenrightbigJ1(aki) 5 1 r 1 k−1 i{1−J0(aki)} 6 r−1(a2−r2)−1 2 1 (1−cosaki) (aki) 7 rn >−1 an+1 kiJn+1(aki) 8 Jv(αr) Jv(αa) >−1 aki ⎪parenleftbig α2−k2 i⎪parenrightbigJ/prime v(aki) 9 r−n(a2−r2)−1 2 >−1 π 2⎪braceleftbigg Jn 2⎪parenleftbiggaki 2⎪parenrightbigg⎪bracerightbigg2 10 rn(a2−r2)−(n+1 2) <1 2 Γ⎪parenleftbigg1 2−n⎪parenrightbigg √ π2nkn−1 isin(aki) 11 rn−1(a2−r2)n−1 2 >−1 2 √ π 2Γ⎪parenleftbigg n+1 2⎪parenrightbigg⎪parenleftbigg2 ki⎪parenrightbiggn ×a2nJ2 n⎪parenleftbiggaki 2⎪parenrightbigg © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 2.19 Exercises 1.(a)⎪radicalbigg π 2exp(−|k|). (c)1 √ 2π(ik)n. (e) Hint : Put ex=y, F (k)=Γ(1−ik) √ 2π. (f) Hint : f(x)=−1 ad dxexp⎪parenleftbigg −1 2ax2⎪parenrightbigg . (h)F(k)=1 √ 2π⎪parenleftbiggk 2⎪parenrightbigg−2 sin2⎪parenleftbiggk 2⎪parenrightbigg . (j)F{hn(x)}=(−i)nhn(k), hn(k) is an eigenfunction for the Fourier transform . (k)F(k)=i √ 2π1 (α−k)⎪bracketleftBig eia(α−k)−eib(α−k)⎪bracketrightBig . (l)1 √ 2acos sin⎪parenleftbiggk2 4a∓π 4⎪parenrightbigg . 5. (b) Hint: Use Pn(x)=1 2nn!dn dxn(x2−1)n. 6. (a) ( f∗g)(x)=eax∞⎪integraldisplay 0e−aydy=1 aeax. (b) (f∗g)(x)=∞⎪integraldisplay −∞sinb(x−y)e−a|y|dy =∞⎪integraldisplay 0[sinb(x+y)+s i n b(x−y)]e−aydy =2s i n bx∞⎪integraldisplay 0e−aycosby dy=2asinbx a2+b2. 645 © 2007 by Taylor & Francis Group, LLC 646 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 7. (j) Verify this result using the Fourier transform of convolution. 8. (a) y(x)=e−xx⎪integraldisplay −aeαf(α)dα+exa⎪integraldisplay xe−αf(α)dα. For (b)–(c) use Exercises 3(a) and 3(b). (b)y(x)=Aexp⎪parenleftbigg −1 4x2⎪parenrightbigg ,w h e r e Ais a constant. (e)y(t)=1 √ 2π∞⎪integraldisplay −∞F(k)exp (ikt)dk (ω2−k2+2iαk). 9. (b) f(x)=a ⎪radicalbig π(a−b)exp⎪parenleftbigg −abx2 a−b⎪parenrightbigg . (c)f(x)=⎪radicalbigg 2 π⎪braceleftbiggb a(a−b)⎪bracerightbigg1 (a−b)2+x2. (d)f(x)=⎪radicalbigg 2 π·ac b(x2+c2),c=b−a. (e)F(k)=1 π·iπΦ(k) sgnk,f(x)=−1 π∞⎪integraldisplay −∞(x−t)−1φ(t)dt. 10.u(x, t)=1 √ 2π∞⎪integraldisplay −∞{A(k)exp [i(kx+ωt)] +B(k)exp [i(kx−ωt)]}dk, where A(k)=1 2⎪bracketleftbigg F(k)+1 iωG(k)⎪bracketrightbigg ,B(k)=1 2⎪bracketleftbigg F(k)−1 iωG(k)⎪bracketrightbigg andω2=⎪parenleftbig c2k2+a2⎪parenrightbig . 11. Hint: u(x, t)=2∞⎪integraldisplay 0A(k)exp (−k2bt)cos{(x+at)k}dk ≈⎪radicalbigg π btA(0) exp⎪bracketleftbigg −(x+at)2 4bt⎪bracketrightbigg ast→∞, where A(k) is expanded in Taylor series and only the ﬁrst term is re- tained at k=0 . 12. Hint: F−1{cos(k2t)}=1 √ 2tcos⎪parenleftbiggx2 4t−π 4⎪parenrightbigg . © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 647 14. (a) Hint: Diﬀerentiate both sides of the integral ∞⎪integraldisplay 0e−axsinkxdx =a k2+a2with respect to ato obtain Fs(k)=⎪radicalbigg a π2ak (k2+a2)2. (c)Fc⎪braceleftbigg1 x⎪bracerightbigg does not exist. (d)⎪radicalbigg π 2(a2+k2)−1 2.H i n t :U s e K0(ax)=∞⎪integraldisplay 0exp(−axcoshu)duand in- terchange the order of integration. 15. (a) Diﬀerentiate both sides of the integral∞⎪integraldisplay 0e−axsinkxdx =k k2+a2 with respect to ato obtain the result. (b) Integrate the above integral with respect to afromato∞to obtain the answer. (c)⎪radicalbigg π 2(−isgnk). (d)⎪radicalbigg π 2e−ak. 16. (a) Hint: If f(x)=e x p ( −ax2),thenf(x) satisﬁes the equation f/prime(x)+2axf(x)=0. We take the Fourier transform and use 3(a) to obtain 2aF/prime(k)+kF(k)=0,F(0) =1 √ 2a. Solving this equation yields F(k)=1 √ 2aexp⎪parenleftbigg −k2 4a⎪parenrightbigg . (b) Use the deﬁnition of the Fourier cosine transform and integrate by parts. 17. Hint: Use the Parseval formula for the gate function.19. Hint: U s(k,t)=Fs(k)Gc(k,t), where Gc(k,t)=e x p ( −κk2t). 20.u(x, y)=−2 π∞⎪integraldisplay 01 ksinakcoskxe−kydk. © 2007 by Taylor & Francis Group, LLC 648 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 21. (a) f(x)=1 √ x, (c)f(x)=H(x−a) √ x2−a2,(b)f(x)=e x p ( −ax), (d)H(a−x). 22. Hint: Use the Fourier sine transform. u(x, t)=1 √ 4πκ t∞⎪integraldisplay 0f(ξ)⎪bracketleftbigg exp⎪braceleftbigg −(x−ξ)2 4κt⎪bracerightbigg −exp⎪braceleftbigg −(x+ξ)2 4κt⎪bracerightbigg⎪bracketrightbigg dξ. 23. (a)πa3 2, (b)π b2(1−e−ab), (c)πa. 24. Hint: Use the Convolution Theorem for the Fourier cosine transform. 25. Hint: Use the Convolution Theorem for the Fourier cosine transform. 26. (a)1 2(π−a), (d)π⎪parenleftbiggexp(2πa)+1 exp(2πa)−1⎪parenrightbigg . 28. Hint: F−1⎪braceleftbigg cos sin(atk2)⎪bracerightbigg =1 2√ at⎪bracketleftbigg cos⎪parenleftbiggx2 4at⎪parenrightbigg ±sin⎪parenleftbiggx2 4at⎪parenrightbigg⎪bracketrightbigg . 29.u(x, z)=P 2πμ∞⎪integraldisplay −∞1 αexp(ikx−αz)dx, α =⎪radicalBigg k2−ω2 c2 2. Hint: Write ( x, y)=r(cosθ,sinθ)a l o n gw i t h k=ω c2cosφ andα=iω c2sinφto obtain u(x, z)=P 2πiμπ+i∞⎪integraldisplay 0−i∞exp⎪bracketleftbigg −iωr c2sin(θ+φ)⎪bracketrightbigg dφ. 30.φ(x, z, t)=−Pg 2π∞⎪integraldisplay −∞sinωt ωexp(ikx+|k|z)dk, η(x, t)=P 2π∞⎪integraldisplay −∞cosωtexp(ikx)dk,where ω2=g|k|, η(x, t)≈Pt 2√ 2π√ g x3/2cos⎪parenleftbigggt2 4x⎪parenrightbigg forgt2>>4x. © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 649 32.φ(x, z, t)=iPexp(/epsilon1t) 2πρ∞⎪integraldisplay −∞(Uk−i/epsilon1)exp (|k|z+ikx) (Uk−i/epsilon1)2−g|k|dk, η(x, t)=Pexp(/epsilon1t) 2πρ∞⎪integraldisplay −∞|k|exp(ikx)dk (Uk−i/epsilon1)2−g|k|. 36.u(x, t)=1 2[f(x−ct)+f(x+ct)] +1 2cx+ct⎪integraldisplay x−ctg(ξ)dξforx>c t . Similar result for x<c t . 37. Hint: Fs{uxxxx}=⎪radicalbigg 2 π[k4Us(k,y)−k3u(0,y)], Fs{uxxyy}=⎪radicalbigg 2 π∂2 ∂y2[−k2Us(k,y)+ku(0,y)]. 43. (a) φ(t)=⎪parenleftbigg 1−it a⎪parenrightbigg−p . (b)φ(t)=e x p ( iμt−λ|t|). (c)φ(t)=( 1+ λ2t2)−1exp(iμt). 44.f(x)=1−cosx πx2. 45.φ(t)=1 it[exp(ita)−1]. 47.U(k,y)=F(k)cos (k2y),u(x, y)=1 √ 2π∞⎪integraldisplay −∞F(k)cos (k2y)exp (ikx)dk. 48.u(x, y, t)=1 2π⎪integraldisplay∞⎪integraldisplay −∞F(k,l)cos⎪bracketleftBig c(k2+l2)1 2t⎪bracketrightBig exp[i(kx+ly)]dk dl. 52.u(x, t)=1 2π∞⎪integraldisplay −∞⎪bracketleftbigg F(k)cos (xα)+G(k) αsinxα⎪bracketrightbigg exp(ikt)dk, where −α2=b+ika−k2 c2. 53.u(x, y)=2T0 π∞⎪integraldisplay 0sinakcosxkcoshyα kcoshαdk, α =⎪radicalbig h+k2. 54. (a) u(x, y)=1 4π2∞⎪integraldisplay −∞⎪integraldisplayF(k,l)exp{i(kx+ly)} (k4+l2+2 )dk dl. © 2007 by Taylor & Francis Group, LLC 650 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b)u(x, y)=−1 4π2∞⎪integraldisplay −∞⎪integraldisplayF(k,l)exp{i(kx+ly)}dk dl (k2+2l2−3ik+4 ). 55. Hint: Seek a solution of the form ψ=φn(x, t)sinnπywithψ0(x, y)= ψ0n(x)sinnπyso that φnsatisﬁes the equation ∂ ∂t⎪bracketleftbigg∂2 ∂x2φn−α2φn⎪bracketrightbigg +β∂φn ∂x=0,α2=(nπ)2+κ2. Apply the Fourier transform of φn(x, t) with respect to xand use Ψn(k,0)=F{ψ0n(x)}. φn(x, t)=1 √ 2π∞⎪integraldisplay −∞Ψn(k,0)exp[ i{kx−ω(k)t}]dk, where ω(k)=−βk(k2+α2)−1. Examine the case for ψ0n(x)=1 a√ 2exp⎪braceleftbigg ik0x−⎪parenleftBigx a⎪parenrightBig2⎪bracerightbigg . 56. Hint: The Fourier transform of the equation gives ⎪parenleftbigg iωR+1 C⎪parenrightbigg Q(ω)=E(ω), so that the transfer function in the frequency domain is Φ(ω)=Q(ω) E(ω)=C (1 +iωRC ). The inverse Fourier transform gives the impulse response function φ(t)=F−1{Φ(ω)}=1 Rexp⎪parenleftbigg −t RC⎪parenrightbigg H(t). 57. (a) See Titchmarsh (1959) pages 60-61. (b)√ a⎪bracketleftBigg 1 2+∞⎪summationdisplay n=1e−na⎪bracketrightBigg =⎪radicalbigg 2b π⎪bracketleftBigg 1 2+∞⎪summationdisplay n=1⎪parenleftbig 1+n2b2⎪parenrightbig−1⎪bracketrightBigg . (c)√ a⎪bracketleftBigg 1 2+∞⎪summationdisplay n=1exp⎪parenleftbigg −1 2a2n2⎪parenrightbigg⎪bracketrightBigg =√ b⎪bracketleftBigg 1 2+∞⎪summationdisplay n=1exp⎪parenleftbigg −1 2n2b2⎪parenrightbigg⎪bracketrightBigg . (d)√ a⎪bracketleftBigg 1 2+∞⎪summationdisplay n=1exp⎪parenleftbigg −1 2a2n2⎪parenrightbigg cos(αan)⎪bracketrightBigg =√ bexp⎪parenleftbigg −1 2α2⎪parenrightbigg⎪bracketleftBigg 1 2+∞⎪summationdisplay n=1exp⎪parenleftbigg −1 2b2n2⎪parenrightbigg cosh(αbn)⎪bracketrightBigg . © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 651 (e)⎪radicalbigg a bf(0)⎡ ⎣1 2+∞⎪summationdisplay na≤1⎪parenleftbig 1−a2n2⎪parenrightbigν−1 2⎤ ⎦=1 2Fc(0) +∞⎪summationdisplay n=1(nb)−νJν(nb), where, in the case ν=1 2,t h et e r m na=1 ,i fi ti sp r e s e n t ,i st ob e halved. 3.9 Exercises 1. (a)2 s2+a2 s2+a2, (c)s2−a2 (s2+a2)2, (g)2 s3exp(−3s), (i)s√ π(s−a)−3/2,(b)s(s+2 )−2, (e)exp(−3s) (s−1), (h) (1 + sa)s−2exp(−as), (j)a(s2+2ω2) s(s2+4ω2). 2. Hint:∞⎪integraldisplay 0t−nexp(−st)dt≥e−s1⎪integraldisplay 0t−ndt+∞⎪integraldisplay 1t−nexp(−st)dt, since exp( −st)≥exp(−s)f o r0 ≤t≤1. But1⎪integraldisplay 0t−ndtdoes not exist. 5. Hint: Use (3.6.7). 6. Hint: Use deﬁnition 3.2.5 and result (3.6.7). 7. (a)1 (a2−b2)(cosbt−cosat), (c) (t−a)H(t−a), (e)1 2exp(−t)sin2t,(b)⎪parenleftbiggt c2−sinct c3⎪parenrightbigg , (d) exp(2 t)−(t+1 )e x p ( t), (f) Hint:1 s2(s+1 ) (s+2 )=1 2s2−3 4s+1 s+1−1 4(s+2 ), (g)1 a2⎪bracketleftbig 1+(at−1)eat⎪bracketrightbig , (i)1 a2⎪parenleftbig eat−at−1⎪parenrightbig .(h)1 a3⎪bracketleftbig 2+at(at−2)eat⎪bracketrightbig , © 2007 by Taylor & Francis Group, LLC 652 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 8. (a)1 2a(sinat+atcosat), (b)1 2erf(2√ t), (c)t⎪integraldisplay 0f(τ)dτ, (d)t 2asinat, (e)t⎪integraldisplay 0f(t−τ)sinωτ dτ, (f)1 2a3(sinat−atcosat), (g)1 (a2+b2)(bsinbt−acosbt+aeat),(i)erfc⎪parenleftbigga 2√ t⎪parenrightbigg , (j)1 a3(at−sinat),(k)tcosat, (l)Use(3 .6.7).⎪parenleftbigg1−cosat t⎪parenrightbigg . 9. (b) Hint:1 (√ s−√ a)=1 √ s⎪parenleftbigga s−a+1⎪parenrightbigg +√ a s−a. (c) Hint: ¯f(s) has simple poles at s=0a n da t s=±(2n+1 )aπi b=±sn. The residue at s=0i sx a, and the residue at s=snis sinh⎪braceleftbigg (2n+1 )πix a⎪bracerightbigg exp⎪braceleftbigg (2n+1 )πiax b⎪bracerightbigg ⎪parenleftbiggb 2a⎪parenrightbigg⎪braceleftbigg (2n+1 )πai b⎪bracerightbigg2 sinh⎪braceleftbigg (2n+1 )πi 2⎪bracerightbigg. Grouping the residues at s=±sntogether and using sinh⎪braceleftbigg (2n+1 )πi 2⎪bracerightbigg =isin(2n+1 )π 2=i(−1)n, we obtain the result. (d) Hint: L−1⎪braceleftBig 1 √ s+ia⎪bracerightBig =e−iat √ πt L−1⎪braceleftbigg1 √ s2+a2⎪bracerightbigg =L−1⎪braceleftbigg1 √ s+ia√ s−ia⎪bracerightbigg , =1 π⎪integraldisplayt 0τ−1 2e−iaτ(t−τ)−1 2eia(t−τ)dτ =1 π⎪integraldisplayt 01 ⎪radicalbig τ(t−τ)eia(t−2τdτ, =1 π⎪integraldisplay1 0eiat(1−2v) ⎪radicalbig v(1−v)dv (τ=tv) =1 π⎪integraldisplay1 −1eiatx √ 1−x2dx (x=1−2v). 10. (a) Use result (3.6.7). © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 653 (b) Use 10(a) and result (3.7.6). 11. (a) J0(at), (b)1 tsinhat, (c) 1 +1 2·t2 3!+1.3 2.4·t4 5!+···, (d) 2∞⎪summationdisplay n=0erfc⎪bracketleftbigg(2n+1 )x 2√ t⎪bracerightbigg , (e)J0(2√ t), (f)a⎪bracketleftbigg 1+(at)2 22·3+(at)4 22·42·52+···⎪bracketrightbigg . 12. Hint: (i) L−1⎪braceleftbigg¯f(s) s⎪bracerightbigg =t⎪integraldisplay 0f(τ)dτ=g(t), (ii)L−1⎪braceleftbiggL{g(t)} s⎪bracerightbigg =L−1⎪braceleftbigg¯f(s) s2⎪bracerightbigg =t⎪integraldisplay 0g(t1)dt1 =t⎪integraldisplay 0⎧ ⎨ ⎩t1⎪integraldisplay 0f(τ)dτ⎫ ⎬ ⎭dt1=t⎪integraldisplay 0t1⎪integraldisplay 0f(τ)dτ dt 1. 13.1 s{exp(s)−1}−1. 15. (b) Hint: Use Example 3.6.1(a). 17.s(s2+1 )−1exp⎪parenleftBig −πs 2⎪parenrightBig . 18. (f) Use result (3.6.7). 22. (a) −1 2slog(1 + s2), (b)1 slog(1 + s). 23. (a) Hint: Use (3.6.2) and then the shifting property (3.4.1). (c)Ln(t)=n⎪summationdisplay r=0⎪parenleftbiggn r⎪parenrightbigg(−t)r r!. 24. (a) Hint: Use the deﬁnition and then interchange the order of integra- tion. 26.¯f(s)=s−2tanh⎪parenleftBigas 2⎪parenrightBig ,s > 0; Hint: f(t+2a)=f(t). 27. (a) f(0) = 1 ,f/prime(0) = 5 . © 2007 by Taylor & Francis Group, LLC 654 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 29. Hint: Use the identities s¯f(s)¯g(s)=f(0)¯g(s)+{s¯f(s)−f(0)}¯g(s) =f(0)¯g(s)+L{f/prime(t)}L{g(t)}. s¯f(s)¯g(s)=g(0)¯f(s)+{s¯g(s)−g(0)}¯f(s). 30. (a) ¯f(s)∼1 s⎪parenleftbigg 1−2! s2+4! s4−···⎪parenrightbigg . (b) Hint: Put t=x+ 1 and then write the binomial expansion of ( x2+ 2x)1 2for|x|<2. K0(s)∼e−s √ π∞⎪summationdisplay n=0(−1)n n!⎪braceleftbigg Γ⎪parenleftbigg n+1 2⎪parenrightbigg⎪bracerightbigg2 (2s)n+1 2ass→∞. 31. (a)∞⎪summationdisplay n=0(−1)nΓ(n+1 ) sn+1, Hint : (1 + t)−1=∞⎪summationdisplay n=0(−1)ntn, (b)∞⎪summationdisplay n=0(−1)n22n+1Γ⎪parenleftbigg n+3 2⎪parenrightbigg (2n+1 ) !sn+3 2,sin(2√ t)=∞⎪summationdisplay n=0(−1)n22n+1t(n+1 2) (2n+1 ) !. (c)∞⎪summationdisplay n=0(−1)n nΓ(n+1 ) sn+1, Hint : log(1 + t)=∞⎪summationdisplay n=1(−1)n−1tn n, (d)∞⎪summationdisplay n=0(−1)na2nΓ(2n+1 ) {22·42·····(2n)2}s2n+1,Hint : J0(at)=∞⎪summationdisplay n=0(−1)n(at)2n 22·42·····(2n)2. 32. (a) L{(t−a)nH(t−a)}=e−asL{tn}=e−asn! sn+1. (b)L⎪braceleftbig t2H(t−a)⎪bracerightbig =e−asL⎪braceleftbig (t+a)2⎪bracerightbig =e−asL⎪braceleftbig t2+2at+a2⎪bracerightbig =e−as⎪parenleftBig 2 s3+2a s2+a2 s⎪parenrightBig . (c)f(t)=t−tH(t−a).Hence ,¯f(s)=1 s2−L{tH(t−a)} =1 s2−e−asL{t+a}=1 s2−e−as⎪parenleftbigg1 s2+a s⎪parenrightbigg . (d)f(x)=w0⎪parenleftbigg 1−2x l⎪parenrightbigg −w0⎪parenleftbigg 1−2x l⎪parenrightbigg H⎪parenleftbigg x−l 2⎪parenrightbigg =2w0 l⎪bracketleftbigg⎪parenleftbiggl 2−x⎪parenrightbigg +⎪parenleftbigg x−l 2⎪parenrightbigg H⎪parenleftbigg x−l 2⎪parenrightbigg⎪bracketrightbigg ¯f(s)=2w0 l⎪bracketleftbigg⎪parenleftbiggl 2s−1 s2⎪parenrightbigg +1 s2exp⎪parenleftbigg −sl 2⎪parenrightbigg⎪bracketrightbigg . © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 655 (e)e−πsL{cos2(t+π)}=e−πsL{cos2t}=e−πs⎪parenleftbiggs s2+4⎪parenrightbigg . (f)f(t)=2−4H(t−a),¯f(s)=2 s−4 se−as. 33.⎪braceleftbigg a,0≤t≤a 0,a < t < 2a⎪bracerightbigg 34.⎧ ⎨ ⎩a,0≤t≤a −a,1<t<2 0,t < 2⎫ ⎬ ⎭ 36. (a) ( fp∗fq)(t)=t⎪integraldisplay 0fp(t−τ)fq(τ)dτ=tp+q−1e−tB(p, q),x=1−τ t =fp+q(t)B(p,q). (c) (fp∗fq)/prime(t)=f/prime p(t)∗fq(t)=[ (p−1)fp−1(t)−fp(t)]∗fq(t) =(p−1)B(p−1,q)fp+q−1(t)−B(p,q)fp+q(t). 4.11 Exercises 1. (a)1 (a−b)(e−bt−e−at), (c)1 5(2 cos t+s i nt+3e−2t),(b) 2e−t−t2−2t−2, (d) 2( e2t−1). 2.x(t)=x0exp(−kt). 3. (a) x(t)=1 2(e3t+e−t),y(t)=1 2(e3t−e−t). (b)x1=28 9e3t−e−1−t 3−1 9,x2=28 9e3t+e−t−t 3−1 9. (c)x=1 5c o s t+2 0s i n t−10e−t, y=1 0c o s t+5s i n t−10e−t, z=−25 sin t. (d)x=1 5(7e−t+3e4t),y=1 5(7e−t−2e4t). 4.x(t)=⎪parenleftbigg x1 x2⎪parenrightbigg =x0⎪parenleftbigg 3e−t−2 3−3e−t⎪parenrightbigg . 5.x(t)=x0et,y(t)=(x0+y0)e2t−x0et. 6. (a) Write s4+2s2(/lscript+2k2)+/lscript2=(s2+α2)(s2+β2)s ot h a t α2+β2= 2(/lscript+2k2),(αβ)2=/lscript2andα=⎪radicalbig k2+/lscript+k, β=⎪radicalbig k2+/lscript−k. © 2007 by Taylor & Francis Group, LLC 656 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b) x(s)= y(s)=s(s2+3 ) (s2+2 )2−1. 7.C(t)=⎪parenleftBigα kV⎪parenrightBig (1−e−kt). 8.p(t)=p0exp⎪parenleftbigg −ct k⎪parenrightbigg +Ac ω⎪parenleftbigg ω2+c2 k2⎪parenrightbigg−1⎪bracketleftbiggc ωksinωt−cosωt+e x p⎪parenleftbigg −ct k⎪parenrightbigg⎪bracketrightbigg . 10.c(t)=c0exp(−k1t). 11. Hint:d dt(c1+c2+c3)=0a n ds o c1+c2+c3=c1(0). c1(t)=c1e−k1t,c2(t)=k1c1 k2−k1(e−k1t−e−k2t)a n d c3(t)=c1(0)−c1(t)−c2(t). 12. (a) x(t)=⎪parenleftbigg 1+1 n2−ω2⎪parenrightbigg cosωt−cosnt n2−ω2. (b)x(t)=1 3(sint−sin2t). (c) 2e−1+1 16e−4t+3t 4−31 16. (d)1 16(3 sin 2 t+5s i n h2 t). (e) Hint: ¯ x(s)=1 (s−1)2+1−(s+2 ) (s+2 )2+1. (f) ¯x(s)=e−as+α(s+b)+β s(s+b) =α s+β b⎪parenleftBig 1 s−1 s+b⎪parenrightBig +e−as b⎪parenleftBig 1 s−1 s+b⎪parenrightBig x(t)=α+β b⎪parenleftbig 1−e−bt⎪parenrightbig +1 bH(t−a)⎪parenleftbig 1−e−bt⎪parenrightbig . (g)v(t)=1 CL−1⎪braceleftbigg (e−as−1)1 (s+1 2RC)2+ω2⎪bracerightbigg , where ω2=1 LC−1 4R2C2. (i) ¯x(s)=1 s2(s−a)2−e−as s2(s−a)2−ae−as s(s−a)2. Inversion yields the solution as x(t)=f(t)−f(t−a)H(t−a)−ag(t−a)H(t−a),where f(t)=L−1⎪braceleftbigg1 s2(s−a)2⎪bracerightbigg =1 a3⎪bracketleftbig 2+at+(at−2)eat⎪bracketrightbig , © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 657 g(t)=L−1⎪braceleftbigg1 s(s−a)2⎪bracerightbigg =1 a2⎪bracketleftbig 1+(at−1)eat⎪bracketrightbig . 14.x(t)=a(ωt−sinωt),y(t)=a(1−cosωt). 16. ˙x(t)=eE mωsinωt,˙y(t)=eE mω(cosωt−1),˙z=0. 19. ¯y(x, s)=¯f(s)sinh⎪braceleftBigs c(l−x)⎪bracerightBig sinh⎪parenleftbiggsl c⎪parenrightbigg. 20.V(x, t)=V0erfc⎪parenleftbiggx 2√ κt⎪parenrightbigg . 21.V(x, t)=V0⎪parenleftBig t−x c⎪parenrightBig H⎪parenleftBig t−x c⎪parenrightBig , (i)V=V0H⎪parenleftBig t−x c⎪parenrightBig , (ii)V=V0cos⎪braceleftBig ω⎪parenleftBig t−x c⎪parenrightBig⎪bracerightBig H⎪parenleftBig t−x c⎪parenrightBig . 22.u(z,t)=Ut⎪bracketleftbigg (1 + 2 ζ2)erfc(ζ)−2ζ √ πe−ζ2⎪bracketrightbigg where ζ=z 2√ νt. 23.q(z,t)=a 2eiωt⎪bracketleftBig e−λ1zerfc{ζ−[it(2Ω + ω)]1/2} +eλ1zerfc{ζ+[it(2Ω + ω)]1/2}⎪bracketrightBig +b 2e−iωt[e−λ2zerfc{ζ−[it(2Ω−ω)]1/2}+eλ2zerfc{ζ+[it(2Ω−ω)]1/2}], where λ1,2=⎪braceleftbiggi(2Ω±ω) ν⎪bracerightbigg1/2 . q(z,t)∼aexp(iωt−λ1z)+bexp(−iωt−λ2z),δ1,2=⎪braceleftbiggν |2Ω±ω|⎪bracerightbigg1/2 . 24.⎪parenleftBigν 2Ω⎪parenrightBig1/2 . 25. (a)1 2⎪parenleftbigg t+3 2sin2t⎪parenrightbigg , (b) (1 −cost), (c)aJ0(at), (d) 3 sin t−√ 2s i n (√ 2t), (e)⎪parenleftbigg t2+2t a⎪parenrightbigg . (f) ¯x(s)=s s2−a2,x(t)=c o s h at. 26. Hint: f(s)=1 s(√ s−a). 27. 1−(1 +t)e−t. © 2007 by Taylor & Francis Group, LLC 658 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 28. (a)π 2a2(1−e−at), (b)π 2sgn t, (c)π ae−at, (d)πe−at, (e)⎪radicalbigg π 4t, (f)⎪radicalbigg π 8t. 29. Hint: Use the Laplace transform of sine and cosine functions. 31.EIs4¯y(s)=Wexp(−as)+As+B, where A=EI y/prime/prime(0) and B=EI y/prime/prime/prime(0). EI y(x)=W 6(x−a)3H(x−a)+A 2x2+B 6x3, y(/lscript)=0= y/prime/prime(/lscript)g i v e s A=Wa /lscript−2(/lscript−a)2and B=−W/lscript−3(/lscript−a)2(/lscript+2a). 32.EI s4¯y(s)=W s⎪bracketleftbigg exp⎪parenleftbigg −/lscripts 2⎪parenrightbigg −exp⎪parenleftbigg −3/lscripts 2⎪parenrightbigg⎪bracketrightbigg +As+B, where A=EI y/prime/prime(0) and B=EI y/prime/prime/prime(0). EI y(x)=W 24⎪bracketleftBigg⎪parenleftbigg x−/lscript 2⎪parenrightbigg H⎪parenleftbigg x−/lscript 2⎪parenrightbigg −⎪parenleftbigg x−3/lscript 2⎪parenrightbigg4 H⎪parenleftbigg x−3/lscript 2⎪parenrightbigg⎪bracketrightBigg +Ax2 2+Bx3 6. y/prime/prime(2/lscript)=0= y/prime/prime/prime(2/lscript)g i v e s A=W/lscript2,B=−W/lscript. 33.EI y(IV)(x)=W[1−H(x−/lscript)],0<x< 2/lscript. EI y(x)=W 8⎪bracketleftbigg9 8(/lscriptx)2−19 16/lscriptx3+1 3{x4−(x−/lscript)4H(x−/lscript)}⎪bracketrightbigg . 34.EI s4¯y(s)=W s[1−exp(−2/lscripts)] +Pexp(−/lscripts)+As+B, where A=EI y/prime/prime(0) and B=EI y/prime/prime/prime(0). EI y(x)=W 24[x4−(x−2/lscript)4H(x−2/lscript)] +P 6(x−/lscript)3H(x−/lscript)+A 2x2+B 6x3. The second term inside the square bracket in y(x) does not contribute © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 659 because the beam extends over 0 ≤x<2/lscript.T h u s EI y(x)=W 24x4+A 2x2+B 6x3, 0≤x</lscript , =W 24x4+P 6(x−/lscript)3+A 2x2+B 6x3,/lscript < x ≤2/lscript. y/prime/prime(2/lscript)=0= y/prime/prime/prime(2/lscript)g i v e s A=/lscript(2W/lscript+P)a n d B=−(A//lscript). M⎪parenleftbigg/lscript 2⎪parenrightbigg =EI y/prime/prime⎪parenleftbigg/lscript 2⎪parenrightbigg =/lscript 8(9W/lscript+4P)a n d S⎪parenleftbigg/lscript 2⎪parenrightbigg =EI y/prime/prime/prime⎪parenleftbigg/lscript 2⎪parenrightbigg =−⎪parenleftbigg3 2W/lscript+P⎪parenrightbigg . 35. (a) un=3n, (c)un=(n+1 ) 2n, (e)un=n2n,(b)un=n2n−1, (d)un=2 ( 3n−2n−1), (f)un=A3n+B2n, where A=(u1−2u0)a n d B=( 3u0−u1), (g)un=3n,( h ) un=can. 37.u(t)=1+ t+(t−1)3 3+···. 38.u(x, t)=1 2πi⎪integraldisplayc+i∞ c−i∞s−1exp⎪bracketleftbigg st−sx √ 1+k2s2⎪bracketrightbigg ds. 39. Hint: L−1⎪bracketleftbigg1 √ s2−α2exp{−β(s2−α2)1/2}⎪bracketrightbigg =I0[α(t2−β2)1/2]H(t−α). 42.u(x, t)=x+e x p⎪bracketleftBigg −⎪parenleftbigg3πc a⎪parenrightbigg2 t⎪bracketrightBigg sin⎪parenleftbigg3πx a⎪parenrightbigg −a⎪bracketleftBigg∞⎪summationdisplay n=0erfc⎪braceleftbigg(2n+1 )a+x 3c√ t⎪bracerightbigg −∞⎪summationdisplay n=0erfc⎪braceleftbigg(2n+1 )a−x 2c√ t⎪bracerightbigg⎪bracketrightBigg . 50. (a) u(x, t)=1 √ 4πκt⎪integraldisplay∞ −∞exp⎪bracketleftbigg −(x−ξ)2 4κt⎪bracketrightbigg f(ξ)dξ +⎪integraldisplayt 0dτ⎪integraldisplay∞ −∞q(ξ,τ)G(x, t;ξ,τ)dξ, where G(x, t;ξ,τ)=1 ⎪radicalbig 4πκ(t−τ)exp⎪bracketleftbigg −(x−ξ)2 4κ(t−τ)⎪bracketrightbigg . © 2007 by Taylor & Francis Group, LLC 660 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b) The Laplace transform solution is ¯x(x, s)=1 s(1 +e−al)⎪bracketleftBig e−ax+e−a(l−x)⎪bracketrightBig , where a=⎪radicalbig s κ. Expanding the denominator, the solution is u(x, t)=∞⎪summationdisplay n=0(−1)n⎪bracketleftbigg erfc⎪parenleftbiggx+nl √ 4κt⎪parenrightbigg −erfc⎪parenleftbigg(n+1 )l−x √ 4κt⎪parenrightbigg⎪bracketrightbigg . 52. ¯u(z,s)=U0¯f(s)exp⎪parenleftbigg −z⎪radicalbigg s ν⎪parenrightbigg =U0sf(s)1 sexp⎪parenleftbigg −z⎪radicalbigg s ν⎪parenrightbigg =U0[L{f/prime(t)+f(0)}]L⎪bracketleftbigg erfc⎪parenleftbiggz √ 4νt⎪parenrightbigg⎪bracketrightbigg Using the convolution theorem, we obtain u(z,t)=U0⎪integraldisplayt 0[f/prime(t−τ)+f(0)]erfc⎪parenleftbiggz √ 4νt⎪parenrightbigg dτ. For the special case, we obtain u(z,t)=U0 2πi⎪integraldisplayc+i∞ c−i∞⎪parenleftbiggω s2+ω2⎪parenrightbigg exp⎪parenleftbigg st−z⎪radicalbigg s ν⎪parenrightbigg ds, c > 0 =⎪parenleftbiggU0ω π⎪parenrightbigg⎪integraldisplay∞ 0e−σtsin⎪parenleftbig⎪radicalbig σ νz⎪parenrightbig σ2+ω2dσ +U0exp⎪parenleftbigg −z⎪radicalbigg ω 2ν⎪parenrightbigg sin⎪parenleftbigg ωt−z⎪radicalbigg ω 2ν⎪parenrightbigg , where the ﬁrst integral is due to the branch cut of the Bromwich integral and it tends to zero as t→∞, and represents the initial transient term that occurs because the disk starts from rest. The second term comes from the residues at the poles at s=±iω. It represents the oscillatory motion of the viscous ﬂuid whose amplitude decays exponentially withzand whose phase changes with z. 53. (a) ¯h(s)=⎪parenleftbig s 2+2s+5⎪parenrightbig−1,h(t)=1 2e−tsin2t, x(t)=2e−tcos2t+(h∗f)(t). (b)¯h(s)=⎪parenleftbig s2−2s+5⎪parenrightbig−1,h(t)=1 2etsin2t, x(t)=etsin2t+(h∗f)(t). © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 661 (c)¯h(s)=⎪parenleftbig s2+32⎪parenrightbig−1,h(t)=1 3sin3t, x(t)=( 2c o s3 t−sin3t)+(h∗f)(t). (d)¯h(s)=⎪parenleftbig s2−2s+5⎪parenrightbig−1,h(t)=1 2etsin2t, x(t)=L−1⎪braceleftbiggx0s+(x1−2x0) s2−2s+5⎪bracerightbigg =L−1⎪braceleftbiggx0(s−1) + (x1−x0) (s−1)2+22⎪bracerightbigg +1 2t⎪integraldisplay 0et−τsin2(t−τ)f(τ)dτ x(t)=et⎪braceleftbigg x0cos2t+1 2(x1−x0)sin2t⎪bracerightbigg +(h∗f)(t). 54. (a) ¯h(s)=¯x(s) ¯f(s)=3s+2 s2+2s+2. The system is of order 2, and its characteristic equation is s2+ 2s+2=0w i t hr o o t s s=−1±i.Since the real parts are negative, the system is stable. (b)¯h(s)=¯x(s) ¯f(s)=2s+3 4s2+1 6s+2 5. Order 2, characteristic equation is 4 s2+1 6s+2 5=0 , roots s=−2±3 2i.Stable. (c)¯h(s)=2s2+s−6 36s2+1 2s+3 7. Order 2, characteristic equation is 36 s2+1 2s+3 7=0 , roots s=1 6±i.Unstable. (d)¯h(s)=2s−1 s2−6s+1 0. Order 2, characteristic equation is s2−6s+1 0=0 , roots s=3±i.Unstable. 55. h(s)=1 s3−as2+b2s−ab2=1 (s−a)(s2+b2) =(a2+b2)−1⎪bracketleftbigg1 (s−a)−(s+a) (s2+b2)⎪bracketrightbigg . It has simple poles at s=aands=±ib(b/negationslash= 0). The system is always unstable. (a) The system has a pole in the right-half plane and hence, is unstable. (b) The poles are at ±ibwhich lie on the imaginary axis. The system is unstable (or marginally unstable). (c) The pole zero is of second-order and the system is unstable. h(t)=1 (a2+b2)⎪parenleftBig eat−cosbt−a bsinbt⎪parenrightBig . © 2007 by Taylor & Francis Group, LLC 662 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 6.8 Exercises 1. Hint: E1,1(x)=ex. 7.5 Exercises 2. (b) u(r, z)=∞⎪integraldisplay 0ke−kzJ0(kr)⎪tildewidef(k)dk=a∞⎪integraldisplay 0e−kzJ0(kr)J1(ak)dk. 3. (b) Hint: ˜f(k)=⎪parenleftbiggQ πak⎪parenrightbigg J1(ak). 6.u(r, t)=∞⎪integraldisplay 0k˜f(k)cos(btk2)J0(kr)dk. 9. Hint: The solution of the dual integral equations ∞⎪integraldisplay 0kJ0(kr)A(k)dk=u0,0≤r≤a, ∞⎪integraldisplay 0k2J0(kr)A(k)dk=0,a < r < ∞, isA(k)=⎪parenleftbigg2u0 π⎪parenrightbiggsin(ak) k2. 10. Hint: See Debnath, 1994, pp. 103–105. 11.u(r, z)=1 πa∞⎪integraldisplay 0k−1J1(ak)J0(kr)e x p ( −kz)dk. 13. Hint: L−1⎡ ⎣exp⎪braceleftBig −k(s2+a2)1 2⎪bracerightBig (s2+a2)1 2⎤⎦=H(t−k)J 0(a⎪radicalbig t2−k2). 14.u(r, z)=b∞⎪integraldisplay 0k−1⎪parenleftbiggsinhkz coshka⎪parenrightbigg J1(bk)J0(kr)dk. © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 663 15. Hint: H0⎪bracketleftbiggH(a−r) √ a2−r2⎪bracketrightbigg =sinak kand L−1⎪braceleftbiggexp(−√ sk) √ s(√ s−a)⎪bracerightbigg =e x p ( −ak−a2t)erfc⎪braceleftbiggk 2√ t−a√ t⎪bracerightbigg . 16.u(r, z)=⎪parenleftbiggQ πaK⎪parenrightbigg∞⎪integraldisplay 0k−1e−|k|zJ1(ak)J0(kr)dr. 17. Use the hint in exercise 9 with a=1a n d u0=1. 18. Hint: Use the joint Hankel and Laplace transform method. 20. Hint: Use the Hankel transform. u(r, z, t)=1 ρ∞⎪integraldisplay 0kexp(kz)J0(kz)t⎪integraldisplay 0⎛ ⎜⎝r0(τ)⎪integraldisplay 0αp(α, τ)J0(kα)dα⎞ ⎟⎠ ×cos[ω(t−τ)dτ]dk, where ω2=gk. 21. ⎪tildewideφ(k,z)=−q kcosh(kz) cosh(ka)e−ak,φ(r, z)=−q∞⎪integraldisplay 0e−akJ0(kr)cosh(kz) cosh(ka)dk. 8.8 Exercises 1. (e)⎪tildewidef(p)=−xp+z 0 p+z,Re(p) <−Re(z). (f)⎪tildewidef(p)=−xp+z 0 p+z,Re(p) >−Re(z). (g)⎪tildewidef(p)=p−1Γ(p), (h)⎪tildewidef(p)=(−p)!{Γ(p)}2. 2. Hint: Substitute e−t=xandg(−logx)=f(x). 3. Hint: Similar to Example 6.2.1(d). 4. Hint: Use (6.2.12) and the scaling property of the Mellin transform. 5. Hint: Use Fc{x−nJn(ax)}andFc{xp−1}and then the Parseval relation for the Fourier cosine transform. © 2007 by Taylor & Francis Group, LLC 664 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 16. (a) Hint: Use (8.4.8) and ˜φ(p,±α)=ap p,Re(p) >0. The solution of (8.4.8) is ˜φ(p,θ)=A(p)eipθ+B(p)e−ipθ. Hence, A=B=ap 2pcos(pα),and˜φ(p,θ)=apcospθ pcos(pα). Inversion gives the solution. (b)φ(r, θ)=M−1⎪braceleftBig ˜f(p)sinpθ sinpα⎪bracerightBig . 17. Hint:∞⎪summationdisplay n=1coskn n2=−k2 2·1 2πic+i∞⎪integraldisplay c−i∞⎪parenleftbigg2π k⎪parenrightbiggpζ(1−p) (p−1)(p−2)dp, and the integrand has three simple poles at p=0,1,2 with residues −1 2,π k,−π2 3k2. 18. Hint:∞⎪summationdisplay n=1e−nx=1 (1−e−x). 27. Hint: (a) Put x=−et,dx=−e−tdtin (8.2.5) to obtain ⎪tildewidef(p)=M{f(x);p}=∞⎪integraldisplay −∞e−ptf(e−t)dt=L⎪braceleftbig f(e−t);p⎪bracerightbig . (b) Put p=a+iωto obtain ⎪tildewidef(p)=M{f(x);p}=∞⎪integraldisplay −∞f(e−t)e−ate−iωtdt=F⎪braceleftbig f(e−t)e−at;ω⎪bracerightbig . 9.13 Exercises 1. (a) ( a2+z2)−1⎪bracketleftBigπz 2a−log⎪parenleftBigz a⎪parenrightBig⎪bracketrightBig . (b) (a−z)−1(aα−zα)πcosecπα. (c)−exp(az)Ei(−az). (d) Γ(1 −α)z−αexp(az)Γ(α, az). (e)⎪parenleftBigπ z⎪parenrightBig [1−exp(−a√ z)]. (f)z−1(cosz−1). 6. (a) ( a−z)−1(zα−1−aα−1)πcosec ( απ). (b) (a2+z2)−1⎪bracketleftBig⎪parenleftBigπz 2a⎪parenrightBig −log⎪parenleftBigz a⎪parenrightBig⎪bracketrightBig . (c) (a2+z2)−1⎪bracketleftBig⎪parenleftBigπa 2⎪parenrightBig +zlog⎪parenleftBigz a⎪parenrightBig⎪bracketrightBig . © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 665 9. Hint: Use t=xuin the transform solution and then apply the convolu- tion theorem for the Mellin transform. 14. Hint:1 π∞⎪contintegraldisplay −∞f(t) t−xdt= lim ε→0⎡ ⎣1 π⎧ ⎨ ⎩x−ε⎪integraldisplay −∞+∞⎪integraldisplay x+ε⎫ ⎬ ⎭f(t) t−xdt⎤ ⎦and then put t−x=u. 15. Hint: Use general Parseval’s relation⎪integraldisplay∞ −∞f1(x)f2(x)dx=⎪integraldisplay∞ −∞(HHHf1)(x)(HHHf2)(x)dx, where f1∈Lp(R)a n d f2∈Lq(R)w i t h( p−1+q−1)=1 . Putf1(x)=f(x)a n d f2(x)=(HHHg)(x)t oo b t a i n ⎪integraldisplay∞ −∞f(x)(HHHg)(x)dx=⎪integraldisplay∞ −∞(HHHf)(x)HHH[HHHg(x)] (x)dx =−⎪integraldisplay∞ −∞(HHHf)(x)g(x)dx. Thus, (9.3.9) follows. 10.6 Exercises 1.1 3a3when n=0 ,a n d2⎪parenleftBiga nπ⎪parenrightBig2 a(−1)n,n=1,2,3,.... 3.u(x, t)=⎪parenleftbigg2πκ a⎪parenrightbigg∞⎪summationdisplay n=1nsin⎪parenleftBignπx a⎪parenrightBigt⎪integraldisplay 0f(τ)exp⎪bracketleftbigg −κ(t−τ)⎪parenleftBignπ a⎪parenrightBig2⎪bracketrightbigg dτ. 4. Hint: ˜fs(n)=a⎪integraldisplay 0f(x)sin(ξnx)dx f(x)=F−1 s{˜fs(n)}=2 a∞⎪summationdisplay n=0(h2+ξ2 n)˜fs(n)sin (xξn) h+(h2+ξ2n) where ξnis the root of the equation ξcot(aξ)+h=0. u(x, t)=⎪parenleftbigg2 a⎪parenrightbigg∞⎪summationdisplay n=1ξn(h2+ξ2 n) h+(h2+ξ2n)t⎪integraldisplay 0f(ξ)exp [−κξn(t−ξ)] sin(xξn)dξ. © 2007 by Taylor & Francis Group, LLC 666 INTEGRAL TRANSFORMS and THEIR APPLICATIONS 5. Hint: Use ˜fc(n)=a⎪integraldisplay 0f(x)cos(xξn)dx, f(x)=2∞⎪summationdisplay n=1(h2+ξ2 n)˜fc(n)cos (xξn) h+a(h2+ξ2n), where ξnis the root of the equation ξtan(aξ)=h. 6. Hint: Apply the ﬁnite Fourier cosine transform. 8. Hint: ˜Ws(n,t)=W0φ(t)H(Ut−/lscript)/lscript⎪integraldisplay 0sin⎪parenleftBigπnx /lscript⎪parenrightBig δ(x−Ut)dx =W0φ(t)H⎪parenleftbigg t−/lscript U⎪parenrightbigg sin⎪parenleftbiggnπUt /lscript⎪parenrightbigg . 12. Hint:d˜Vs dt+κ⎪parenleftBignπ a⎪parenrightBig2˜Vs=0,˜Vs(n,t)=Aexp⎪parenleftbigg −κn2π2t a2⎪parenrightbigg , A=˜Vs(n,0)=4aV0 n2π2sin⎪parenleftBignπ 2⎪parenrightBig =(−1)r4aV0 (2r+1 )2π2 where n=( 2r+1 ),r=0,1,2,... V(x, t)=⎪parenleftbigg8V0 π2⎪parenrightbigg∞⎪summationdisplay r=0(−1)r (2r+1 )2sin⎪braceleftBig (2r+1 )nx a⎪bracerightBig exp⎪braceleftbigg −κ(2r+1 )2π2t a2⎪bracerightbigg . 15. Hint: Replace PbyW0dξ dη and integrate with respect to ξandηover the region α≤ξ≤β, γ≤η≤δ. u(x, y)=⎪parenleftbigg4W0 Dπ6⎪parenrightbigg∞⎪summationdisplay m=1∞⎪summationdisplay n=1⎪bracketleftbigg⎪braceleftbigg cos⎪parenleftBigmπα a⎪parenrightBig −cos⎪parenleftbiggmπβ a⎪parenrightbigg⎪bracerightbigg ×⎪braceleftbigg cos⎪parenleftBignπγ b⎪parenrightBig −cos⎪parenleftbiggnπδ b⎪parenrightbigg⎪bracerightbiggsin⎪parenleftBigmπx a⎪parenrightBig sin⎪parenleftBignπy b⎪parenrightBig mn ω4mn⎤ ⎦. 16. Hint:d2˜us(m,n,t ) dt2+Ω2 mn˜us(m,n,t )=0,where Ω2 mn=Dπ4ω4 mn ρh. u(x, y, t)=⎪parenleftbigg4 ab⎪parenrightbigg∞⎪summationdisplay m=1∞⎪summationdisplay n=1{Amncos(Ω mnt)+Bmnsin(Ω mnt)} ×sin⎪parenleftBigmπx a⎪parenrightBig sin⎪parenleftBignπy b⎪parenrightBig . © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 667 11.6 Exercises 1. (a)s (s2−a2)−exp(−sT) (s2−a2)(scoshaT+asinhaT). (d)1 s(1−e−sT)H(T). 12.8 Exercises 1. (a)z3+4z2+z (z−1)4, use (12.4.13) and (12.3.14). (b) exp( a/z),(c)z (z−ea)2,(d)⎪parenleftbigg 1+1 z⎪parenrightbigg ,(e)z(z+a) (z−a)2. 2. Hint: Put b=ixin (12.4.14). 6. (a) Z−1⎪braceleftbiggz z−2·z z−3⎪bracerightbigg =n⎪summationdisplay m=02m3n−m=3nn⎪summationdisplay m=0⎪parenleftbigg2 3⎪parenrightbiggm , =( 3n+1−2n+1), (d)nan−1, (f) (2n−1−n), (h)1 6⎪bracketleftbig (−1)n+2n+3−3(1)n⎪bracketrightbig ,(e) (n−1)an−2H(n−1), (g) 2( −1)n−(−2)n, (i) Hint: U(z) z=2 (z−1)−1 ⎪parenleftbigg z−1 2⎪parenrightbigg,U(z)=2z (z−1)−z ⎪parenleftbigg z−1 2⎪parenrightbigg u(n)=( 2 −2−n). (j)f(n)=Z−1⎪braceleftbiggz z−e−a⎪bracerightbigg =e−an,g(n)=Z−1⎪braceleftbiggz z−e−b⎪bracerightbigg =e−bn, h(n)=f(n)∗g(n)=∞⎪summationdisplay m=0e−ame−b(n−m) =e−bn∞⎪summationdisplay m=0e−(a−b)m=e−bn⎪bracketleftbigg1−e(b−a)(n+1) 1−e(b−a)⎪bracketrightbigg . © 2007 by Taylor & Francis Group, LLC 668 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (k) Divide (12.3.6) by zand diﬀerentiate both sides with respect to z Z−1⎪braceleftbig (z−a)−k⎪bracerightbig =(n−k+1 ) k−1an−kH(n−k) (k−1)!,|z|>a> 0, where ( a)n=a(a+1 )...(a+n−1),(a)0=1,n=1,2,..., andZ−1⎪braceleftbig z−k⎪bracerightbig =δ(n−k). (l)F(z)=(z+4 )−1 (z−1)2−5 (z−1)+16 (z−2). Use 6(k) to obtain Z−1{F(z)}=δ(n+1 )+4 δ(n)−(n−1)H(n−2)−5H(n−1) +16.2n−1H(n−1). Since f(−1)= 1 ,f(0) = 4 ,f(1) = 11 ,f(n)=−(n+4 )+1 62n−1,n≥2. (m)f(n)=1 5Z−1⎪parenleftbigg6 z+2−1 z−1 2⎪parenrightbigg =6 5(−2)n−1−1 5⎪parenleftbigg1 2⎪parenrightbiggn−1 ,n≥1. 7. (a)1 16[17(−3)n+4n−1], (c)x0an+nan−1, (d)x0(1−a)n+1−(1−a)n, (e)3 5[3n−(−2)n], (i)nan. (k) From the given equation and initial data, f(−1)=1 2andf(−2)=1 4. TheZtransform is F(z)=−1⎪summationdisplay k=−1f(k)z−(k+1)+z−1F(z) +2⎪bracketleftBigg−1⎪summationdisplay k=−2f(k)z−(k+2)+z−2F(z)⎪bracketrightBigg =f(−1) +z−1F(z)+2⎪bracketleftbig f(−2) +f(−1)z−1+z−2F(z)⎪bracketrightbig F(z)=1+ z−1+z−1F(z)+2z−2F(z) F(z)=z2+z z2−z−2=z z−2.Hence ,f(n)=2n. (l)F(z)−af(−1)−az−1F(z)=z(z−1)−1 F(z)=2a (1−a z)+1 (1−a)1 (1−1 z)+⎪parenleftbigga a−1⎪parenrightbigg1 (1−a z) f(n)=2an+1+( 1−a)−1+(a−1)−1an+1 =1 1−a+⎪parenleftbigg2a−1 a−1⎪parenrightbigg an+1. © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 669 (m)F(z)=z2+5z (z+1 ) (z+2 )=4z z+1−3z z+2.f(n)=4 (−1)n−3(−2)n. (n)F(z)=3z z−2,f (n)=3.2n,n =0,1,2,.... 9. (a) 1 , (b) 0, (c) 1, (d)f(0) = 0 ,m > 0;f(0) = 1 ,m=0. 11. (a) (1 −aeix)−1, (b)elog⎪parenleftbigg 1+1 e⎪parenrightbigg , (c) (2 sinh x)−1. 12.7 4−3 4⎪parenleftbigg −1 3⎪parenrightbiggn . 13.U(z)=2z z2−4,u(n)=2n−1[1 + (−1)n+1],ν(n)=2n−1[1 + (−1)n]−1, where n=0,1,2,.... 14. Apply the Ztransform to obtain z3⎪bracketleftbig U(z)−u(0)−u(1)z−1−u(2)z−2⎪bracketrightbig −3z2⎪bracketleftbig U(z)−u(0)−u(1)z−1⎪bracketrightbig +3z[U(z)−u(0)]−U(z)=0. U(z)=1 (z−1)3(z3−3z2+4z)=(1−3z−1+4z−2) (1−z−1)3. Use Z⎪braceleftbig nan−1⎪bracerightbig =z−1 (1−az−1)2,Z⎪braceleftbig n(n−1)an−2⎪bracerightbig =2z−2 (1−az−2), zN−1 pN(z)=z2 (z−1)3=z−2 (1−z−1)3. Inversion gives u(n)=Z−1⎪braceleftbigg1 (1−z−1)3⎪bracerightbigg −Z−1⎪braceleftbigg3z−1 (1−z−1)3⎪bracerightbigg +Z−1⎪braceleftbigg4z−2 (1−z−1)3⎪bracerightbigg =1 2(n+1 )(n+2 )−3 2n(n+1 )+4 2n(n−1) = (n−1)2. 16. (a) z2⎪bracketleftbig U(z)−u(0)−u(1)z−1⎪bracketrightbig +2[U(z)−u(1)]−3U(z)=0 U(z)=z(z+2 ) (z2+2z−3)=1 4⎪parenleftbigg1 1+3z−1+3 1−z−1⎪parenrightbigg un=1 4[(−3)n−3.1n]=3 4⎪bracketleftbig 1−(−3)n−1⎪bracketrightbig , © 2007 by Taylor & Francis Group, LLC 670 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (b)un=2⎪bracketleftBigg⎪parenleftbigg2 3⎪parenrightbiggn−1 −1⎪bracketrightBigg , (c)un=5n/2⎪parenleftbigg 2s i nnx+1 2cosnx⎪parenrightbigg ,where x=t a n−1⎪parenleftbigg1 2⎪parenrightbigg . 13.5 Exercises 1. (a)a2 k2 i⎪parenleftbigg aki−4 aki⎪parenrightbigg J1(aki), (b)aki (α2−k2 i)J0(aα)J1(aki). 10.u(r, t)=2 a2∞⎪summationdisplay i=1J0(rki) J2 1(aki)t⎪integraldisplay 0˜Q(ki,τ)e x p [−(t−τ)k2 i]dτ. (a)u(r, t)=Q0 4k(a2−r2)−⎪parenleftbigg2Q0 ak⎪parenrightbigg∞⎪summationdisplay i=1J0(rki) k3 iJ1(aki)exp(−tκk2 i). (b)u(r, t)=2κQ0 ka2∞⎪summationdisplay i=1J0(rki) J2 1(aki)t⎪integraldisplay 0f(τ)e x p [−κ(t−τ)k2 i]dτ. 16.5 Exercises 1. (a) f0(n)=e x p ( −a),n=0, f0(n)exp(−a)[Ln(a)−Ln−1(a)],n≥1. (b)an(1 +a)n+1,( c ) Aδmn, (d) 0,n > m ;(−1)n⎪parenleftBigm n⎪parenrightBig m!,m≥n, (e) 1 for n=0,1,2,3,.... 17.4 Exercises 1. (a) 2n−1 2Γ⎪parenleftbigg n+1 2⎪parenrightbigg , (b) 0; m=0,1,2,...,n −1, (c)⎪parenleftbigg n+1 2⎪parenrightbigg δn. © 2007 by Taylor & Francis Group, LLC Answers and Hints to Selected Exercises 671 3. Hint: Use Feldheim’s result (1938). H2 n(x)=n!2nn⎪summationdisplay r=0⎪parenleftBign r⎪parenrightBigH2r(x) 22r!. 18.9 Exercises 1. Hint: Use the same method employed in Example 18.2.2 to solve (a) and (b). To solve (c) apply the linearity property. 2. Hint:⎪hatwidef(p,u)=√ πexp(−p2).Usev=su,s=⎪radicalbig v2 1+v2 2a n dt h e nu s e (18.3.8b) to obtain ⎪hatwidef(p,v)=⎪hatwidef(p,su)=√ π sexp(−p2/s2). Apply∂ ∂vk=∂s ∂vk∂ ∂s,(k=1,2) to ﬁnd ∂⎪hatwidef ∂vk=√ π⎪parenleftBigvk s⎪parenrightBig∂ ∂s⎪bracketleftbigg1 sexp(−p2/s2)⎪bracketrightbigg =√ π⎪parenleftBigvk s⎪parenrightBig (2p2−s2)exp (−p2/s2). Finally, replace vanduwiths=1 t og e t ∂⎪hatwidef ∂uk=√ πuk(2p2−1)exp( −p2). 19.5 Exercises 1. (a) F{ψ(x)}=⎪integraldisplay∞ −∞e−iω xψ(x)dx=⎪integraldisplay1 2 0e−iω xdx−⎪integraldisplay1 1 2e−iω xdx =4i ωexp(−iω 2)s i n2(iω 4). Using scaling and shifting properties of the Fourier transform gives F{ψ(2mx−n)}. (b) The amplitude spectrum decays like ω−1and hence, tends to zero s- lowly as |ω|→∞ . This shows that the Haar wavelet has poor frequency localization. But it has a very good time localization. 2. (a) ˆψ(ω)=1 √ 2π⎪bracketleftbiggsin (ω 2) (ω 2)⎪bracketrightbigg2 , (b)ˆψ(ω)=√ 2πω2exp⎪parenleftbigg −ω2 2⎪parenrightbigg , © 2007 by Taylor & Francis Group, LLC 672 INTEGRAL TRANSFORMS and THEIR APPLICATIONS (c)ˆψ(ω)=√ 2πexp⎪bracketleftbigg −1 2(ω−ω0)2⎪bracketrightbigg . 8. (a)–(c) See pages 434-437 of a book by Debnath (2002). 9. See Debnath (2002) pages 434-438. 10. See Debnath (2002) pages 440-441. 11. A window function is af∈L2(R) with norm unity. /bardblf/bardbl2=⎪integraldisplay∞ −∞|f(t)|2dt=⎪parenleftbigg2a π⎪parenrightbigg1 2⎪integraldisplay∞ −∞e−2at2dt=⎪parenleftbigg2a π⎪parenrightbigg1 2 .⎪radicalbigg π 2a=1. ˆf(ω)=⎪parenleftbigg2a π⎪parenrightbigg1 41 √ 2π⎪integraldisplay∞ −∞e−iωtexp(−at2)dt=⎪parenleftbigg2a π⎪parenrightbigg1 4 F⎪braceleftbig exp(−at2)⎪bracerightbig . =⎪parenleftbigg2a π⎪parenrightbigg1 41 √ 2πexp⎪parenleftbigg −ω2 4a⎪parenrightbigg =1 √ 2af⎪parenleftBigω 2a⎪parenrightBig . The Gabor windows have optimal time and frequency localization prop- erties. 12. (a)/bardblf/bardbl2=⎪parenleftbigg3 2a3⎪parenrightbigg⎪integraldisplay∞ −∞|(a−|t|)2|χ[−a,a](t)dt =⎪parenleftbigg3 2a3⎪parenrightbigg⎪integraldisplaya −a|(a−|t|)|2dt=⎪parenleftbigg3 2a3⎪parenrightbigg 2⎪integraldisplaya 0(a−t)2dt=1. (b) ˆf(ω)=⎪radicalbigg 3 2a3F⎪braceleftBig χ[−a 2,a 2](t)∗χ[−a 2,a 2](t)⎪bracerightBig , =⎪radicalbigg 3 2a3⎪bracketleftBig F⎪braceleftBig χ[−a 2,a 2](t)⎪bracerightBig⎪bracketrightBig2 ,by convolution theorem =⎪radicalbigg 3 2a4⎪parenleftbigg2 π⎪parenrightbigg⎪parenleftbiggsinaω 2 ω⎪parenrightbigg2 =⎪radicalbigg 3a 2⎪parenleftbigg2 π⎪parenrightbigg⎪parenleftbiggsinaω 2 aω⎪parenrightbigg2 . It is better localized in the frequency domain, as ˆf(ω) decays like |ω|−2 as|ω|→∞ . (c)∇2 f=⎪integraldisplay∞ −∞t2|f(t)|2dt=⎪integraldisplaya −at2⎪parenleftBigg⎪radicalbigg 3 2a3(a−|t|)⎪parenrightBigg2 dt =3 a3⎪integraldisplaya 0t2(a−t)2dt=1 10a2 ∇2 ˆf=1 4π2/bardblf/prime/bardbl2=3 8π2a3⎪bracketleftbigg⎪integraldisplay0 −a12dt+⎪integraldisplaya 0(−1)2dt⎪bracketrightbigg =3 4π2a3 (4∇f∇ˆf)=1 π⎪radicalbigg 6 5>1 π. © 2007 by Taylor & Francis Group, LLC Bibliography The following bibliography is not by any means a complete one for the subject. 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