Handbook of Integral Equations

Author: Andrei D. Polyanin | Alexander V. Manzhirov

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HANDBOOK OF

INTEGRAL EQUATIONS

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Andrei D. Polyanin and Alexander V. Manzhirov

HANDBOOK OF

INTEGRAL EQUATIONS

Page iv

Library of Congress Cataloging-in-Publication Data Polyanin, A. D. (Andrei Dmitrievich) Handbook of integral equations/Andrei D. Polyanin, Alexander V. Manzhirov. p. cm. Includes bibliographical references (p. - ) and index. ISBN 0-8493-2876-4 (alk. paper) 1. Integral equations—Handbooks, manuals, etc. I. Manzhirov. A. V. (Aleksandr Vladimirovich) II. Title. QA431.P65 1998 515’.45—dc21

98-10762 CIP

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 consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage or retrieval system, without prior permission in writing from the publisher. The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from CRC Press LLC for such copying. Direct all inquiries to CRC Press LLC, 2000 N.W. Corporate Blvd., Boca Raton, Florida 33431. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation, without intent to infringe.

Visit the CRC Press Web site at www.crcpress.com © 1998 by CRC Press LLC No claim to original U.S. Government works International Standard Book Number 0-8493-2876-4 Library of Congress Card Number 98-10762 Printed in the United States of America 3 4 5 6 7 8 Printed on acid-free paper

9

0

ANNOTATION More than 2100 integral equations with solutions are given in the first part of the book. A lot of new exact solutions to linear and nonlinear equations are included. Special attention is paid to equations of general form, which depend on arbitrary functions. The other equations contain one or more free parameters (it is the reader’s option to fix these parameters). Totally, the number of equations described is an order of magnitude greater than in any other book available. A number of integral equations are considered which are encountered in various fields of mechanics and theoretical physics (elasticity, plasticity, hydrodynamics, heat and mass transfer, electrodynamics, etc.). The second part of the book presents exact, approximate analytical and numerical methods for solving linear and nonlinear integral equations. Apart from the classical methods, some new methods are also described. Each section provides examples of applications to specific equations. The handbook has no analogs in the world literature and is intended for a wide audience of researchers, college and university teachers, engineers, and students in the various fields of mathematics, mechanics, physics, chemistry, and queuing theory.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page iii

FOREWORD Integral equations are encountered in various fields of science and numerous applications (in elasticity, plasticity, heat and mass transfer, oscillation theory, fluid dynamics, filtration theory, electrostatics, electrodynamics, biomechanics, game theory, control, queuing theory, electrical engineering, economics, medicine, etc.). Exact (closed-form) solutions of integral equations play an important role in the proper understanding of qualitative features of many phenomena and processes in various areas of natural science. Lots of equations of physics, chemistry and biology contain functions or parameters which are obtained from experiments and hence are not strictly fixed. Therefore, it is expedient to choose the structure of these functions so that it would be easier to analyze and solve the equation. As a possible selection criterion, one may adopt the requirement that the model integral equation admit a solution in a closed form. Exact solutions can be used to verify the consistency and estimate errors of various numerical, asymptotic, and approximate methods. More than 2100 integral equations and their solutions are given in the first part of the book (Chapters 1–6). A lot of new exact solutions to linear and nonlinear equations are included. Special attention is paid to equations of general form, which depend on arbitrary functions. The other equations contain one or more free parameters (the book actually deals with families of integral equations); it is the reader’s option to fix these parameters. Totally, the number of equations described in this handbook is an order of magnitude greater than in any other book currently available. The second part of the book (Chapters 7–14) presents exact, approximate analytical, and numerical methods for solving linear and nonlinear integral equations. Apart from the classical methods, some new methods are also described. When selecting the material, the authors have given a pronounced preference to practical aspects of the matter; that is, to methods that allow effectively “constructing” the solution. For the reader’s better understanding of the methods, each section is supplied with examples of specific equations. Some sections may be used by lecturers of colleges and universities as a basis for courses on integral equations and mathematical physics equations for graduate and postgraduate students. For the convenience of a wide audience with different mathematical backgrounds, the authors tried to do their best, wherever possible, to avoid special terminology. Therefore, some of the methods are outlined in a schematic and somewhat simplified manner, with necessary references made to books where these methods are considered in more detail. For some nonlinear equations, only solutions of the simplest form are given. The book does not cover two-, three- and multidimensional integral equations. The handbook consists of chapters, sections and subsections. Equations and formulas are numbered separately in each section. The equations within a section are arranged in increasing order of complexity. The extensive table of contents provides rapid access to the desired equations. For the reader’s convenience, the main material is followed by a number of supplements, where some properties of elementary and special functions are described, tables of indefinite and definite integrals are given, as well as tables of Laplace, Mellin, and other transforms, which are used in the book. The first and second parts of the book, just as many sections, were written so that they could be read independently from each other. This allows the reader to quickly get to the heart of the matter.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page v

We would like to express our deep gratitude to Rolf Sulanke and Alexei Zhurov for fruitful discussions and valuable remarks. We also appreciate the help of Vladimir Nazaikinskii and Alexander Shtern in translating the second part of this book, and are thankful to Inna Shingareva for her assistance in preparing the camera-ready copy of the book. The authors hope that the handbook will prove helpful for a wide audience of researchers, college and university teachers, engineers, and students in various fields of mathematics, mechanics, physics, chemistry, biology, economics, and engineering sciences. A. D. Polyanin A. V. Manzhirov

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page vi

SOME REMARKS AND NOTATION 1. In Chapters 1–11 and 14, in the original integral equations, the independent variable is denoted by x, the integration variable by t, and the unknown function by y = y(x). 2. For a function of one variable f = f (x), we use the following notation for the derivatives: fx =

df , dx

= fxx

d2 f , dx2

fxxx =

d3 f , dx3

fxxxx =

d4 f , dx4

and

fx(n) =

dn f for n ≥ 5. dxn

Occasionally, we use the similar notation for partial derivatives of a function of two variables, ∂ for example, Kx (x, t) = K(x, t). ∂x d n 3. In some cases, we use the operator notation f (x) g(x), which is defined recursively by dx n n–1 d d d f (x) f (x) g(x) = f (x) g(x) . dx dx dx 4. It is indicated in the beginning of Chapters 1–6 that f = f (x), g = g(x), K = K(x), etc. are arbitrary functions, and A, B, etc. are free parameters. This means that: (a) f = f (x), g = g(x), K = K(x), etc. are assumed to be continuous real-valued functions of real arguments;* (b) if the solution contains derivatives of these functions, then the functions are assumed to be sufficiently differentiable;** (c) if the solution contains integrals with these functions (in combination with other functions), then the integrals are supposed to converge; (d) the free parameters A, B, etc. may assume any real values for which the expressions occurring A in the equation and the solution make sense (for example, if a solution contains a factor , 1–A then it is implied that A ≠ 1; as a rule, this is not specified in the text). 5. The notations Re z and Im z stand, respectively, for the real and the imaginary part of a complex quantity z. 6. In the first part of the book (Chapters 1–6) when referencing a particular equation, we use a notation like 2.3.15, which implies equation 15 from Section 2.3. 7. To highlight portions of the text, the following symbols are used in the book: indicates important information pertaining to a group of equations (Chapters 1–6); indicates the literature used in the preparation of the text in specific equations (Chapters 1–6) or sections (Chapters 7–14). * Less severe restrictions on these functions are presented in the second part of the book. ** Restrictions (b) and (c) imposed on f = f (x), g = g(x), K = K(x), etc. are not mentioned in the text.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page vii

AUTHORS Andrei D. Polyanin, D.Sc., Ph.D., is a noted scientist of broad interests, who works in various fields of mathematics, mechanics, and chemical engineering science. A. D. Polyanin graduated from the Department of Mechanics and Mathematics of the Moscow State University in 1974. He received his Ph.D. degree in 1981 and D.Sc. degree in 1986 at the Institute for Problems in Mechanics of the Russian (former USSR) Academy of Sciences. Since 1975, A. D. Polyanin has been a member of the staff of the Institute for Problems in Mechanics of the Russian Academy of Sciences. Professor Polyanin has made important contributions to developing new exact and approximate analytical methods of the theory of differential equations, mathematical physics, integral equations, engineering mathematics, nonlinear mechanics, theory of heat and mass transfer, and chemical hydrodynamics. He obtained exact solutions for several thousands of ordinary differential, partial differential, and integral equations. Professor Polyanin is an author of 17 books in English, Russian, German, and Bulgarian. His publications also include more than 110 research papers and three patents. One of his most significant books is A. D. Polyanin and V. F. Zaitsev, Handbook of Exact Solutions for Ordinary Differential Equations, CRC Press, 1995. In 1991, A. D. Polyanin was awarded a Chaplygin Prize of the USSR Academy of Sciences for his research in mechanics. Alexander V. Manzhirov, D.Sc., Ph.D., is a prominent scientist in the fields of mechanics and applied mathematics, integral equations, and their applications. After graduating from the Department of Mechanics and Mathematics of the Rostov State University in 1979, A. V. Manzhirov attended a postgraduate course at the Moscow Institute of Civil Engineering. He received his Ph.D. degree in 1983 at the Moscow Institute of Electronic Engineering Industry and his D.Sc. degree in 1993 at the Institute for Problems in Mechanics of the Russian (former USSR) Academy of Sciences. Since 1983, A. V. Manzhirov has been a member of the staff of the Institute for Problems in Mechanics of the Russian Academy of Sciences. He is also a Professor of Mathematics at the Bauman Moscow State Technical University and a Professor of Mathematics at the Moscow State Academy of Engineering and Computer Science. Professor Manzhirov is a member of the editorial board of the journal “Mechanics of Solids” and a member of the European Mechanics Society (EUROMECH). Professor Manzhirov has made important contributions to new mathematical methods for solving problems in the fields of integral equations, mechanics of solids with accretion, contact mechanics, and the theory of viscoelasticity and creep. He is an author of 3 books, 60 scientific publications, and two patents.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page ix

CONTENTS Annotation Foreword Some Remarks and Notation

Part I. Exact Solutions of Integral Equations 1. Linear Equations of the First Kind With Variable Limit of Integration 1.1. Equations Whose Kernels Contain Power-Law Functions 1.1-1. Kernels Linear in the Arguments x and t 1.1-2. Kernels Quadratic in the Arguments x and t 1.1-3. Kernels Cubic in the Arguments x and t 1.1-4. Kernels Containing Higher-Order Polynomials in x and t 1.1-5. Kernels Containing Rational Functions 1.1-6. Kernels Containing Square Roots 1.1-7. Kernels Containing Arbitrary Powers 1.2. Equations Whose Kernels Contain Exponential Functions 1.2-1. Kernels Containing Exponential Functions 1.2-2. Kernels Containing Power-Law and Exponential Functions 1.3. Equations Whose Kernels Contain Hyperbolic Functions 1.3-1. Kernels Containing Hyperbolic Cosine 1.3-2. Kernels Containing Hyperbolic Sine 1.3-3. Kernels Containing Hyperbolic Tangent 1.3-4. Kernels Containing Hyperbolic Cotangent 1.3-5. Kernels Containing Combinations of Hyperbolic Functions 1.4. Equations Whose Kernels Contain Logarithmic Functions 1.4-1. Kernels Containing Logarithmic Functions 1.4-2. Kernels Containing Power-Law and Logarithmic Functions 1.5. Equations Whose Kernels Contain Trigonometric Functions 1.5-1. Kernels Containing Cosine 1.5-2. Kernels Containing Sine 1.5-3. Kernels Containing Tangent 1.5-4. Kernels Containing Cotangent 1.5-5. Kernels Containing Combinations of Trigonometric Functions 1.6. Equations Whose Kernels Contain Inverse Trigonometric Functions 1.6-1. Kernels Containing Arccosine 1.6-2. Kernels Containing Arcsine 1.6-3. Kernels Containing Arctangent 1.6-4. Kernels Containing Arccotangent

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

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1.7. Equations Whose Kernels Contain Combinations of Elementary Functions 1.7-1. Kernels Containing Exponential and Hyperbolic Functions 1.7-2. Kernels Containing Exponential and Logarithmic Functions 1.7-3. Kernels Containing Exponential and Trigonometric Functions 1.7-4. Kernels Containing Hyperbolic and Logarithmic Functions 1.7-5. Kernels Containing Hyperbolic and Trigonometric Functions 1.7-6. Kernels Containing Logarithmic and Trigonometric Functions 1.8. Equations Whose Kernels Contain Special Functions 1.8-1. Kernels Containing Bessel Functions 1.8-2. Kernels Containing Modified Bessel Functions 1.8-3. Kernels Containing Associated Legendre Functions 1.8-4. Kernels Containing Hypergeometric Functions 1.9. Equations Whose Kernels Contain Arbitrary Functions 1.9-1. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + g2 (x)h2 (t) 1.9-2. Equations With Difference Kernel: K(x, t) = K(x – t) 1.9-3. Other Equations 1.10. Some Formulas and Transformations 2. Linear Equations of the Second Kind With Variable Limit of Integration 2.1. Equations Whose Kernels Contain Power-Law Functions 2.1-1. Kernels Linear in the Arguments x and t 2.1-2. Kernels Quadratic in the Arguments x and t 2.1-3. Kernels Cubic in the Arguments x and t 2.1-4. Kernels Containing Higher-Order Polynomials in x and t 2.1-5. Kernels Containing Rational Functions 2.1-6. Kernels Containing Square Roots and Fractional Powers 2.1-7. Kernels Containing Arbitrary Powers 2.2. Equations Whose Kernels Contain Exponential Functions 2.2-1. Kernels Containing Exponential Functions 2.2-2. Kernels Containing Power-Law and Exponential Functions 2.3. Equations Whose Kernels Contain Hyperbolic Functions 2.3-1. Kernels Containing Hyperbolic Cosine 2.3-2. Kernels Containing Hyperbolic Sine 2.3-3. Kernels Containing Hyperbolic Tangent 2.3-4. Kernels Containing Hyperbolic Cotangent 2.3-5. Kernels Containing Combinations of Hyperbolic Functions 2.4. Equations Whose Kernels Contain Logarithmic Functions 2.4-1. Kernels Containing Logarithmic Functions 2.4-2. Kernels Containing Power-Law and Logarithmic Functions 2.5. Equations Whose Kernels Contain Trigonometric Functions 2.5-1. Kernels Containing Cosine 2.5-2. Kernels Containing Sine 2.5-3. Kernels Containing Tangent 2.5-4. Kernels Containing Cotangent 2.5-5. Kernels Containing Combinations of Trigonometric Functions 2.6. Equations Whose Kernels Contain Inverse Trigonometric Functions 2.6-1. Kernels Containing Arccosine 2.6-2. Kernels Containing Arcsine 2.6-3. Kernels Containing Arctangent 2.6-4. Kernels Containing Arccotangent

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page x

2.7. Equations Whose Kernels Contain Combinations of Elementary Functions 2.7-1. Kernels Containing Exponential and Hyperbolic Functions 2.7-2. Kernels Containing Exponential and Logarithmic Functions 3.7-3. Kernels Containing Exponential and Trigonometric Functions 2.7-4. Kernels Containing Hyperbolic and Logarithmic Functions 2.7-5. Kernels Containing Hyperbolic and Trigonometric Functions 2.7-6. Kernels Containing Logarithmic and Trigonometric Functions 2.8. Equations Whose Kernels Contain Special Functions 2.8-1. Kernels Containing Bessel Functions 2.8-2. Kernels Containing Modified Bessel Functions 2.9. Equations Whose Kernels Contain Arbitrary Functions 2.9-1. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t) 2.9-2. Equations With Difference Kernel: K(x, t) = K(x – t) 2.9-3. Other Equations 2.10. Some Formulas and Transformations 3. Linear Equation of the First Kind With Constant Limits of Integration 3.1. Equations Whose Kernels Contain Power-Law Functions 3.1-1. Kernels Linear in the Arguments x and t 3.1-2. Kernels Quadratic in the Arguments x and t 3.1-3. Kernels Containing Integer Powers of x and t or Rational Functions 3.1-4. Kernels Containing Square Roots 3.1-5. Kernels Containing Arbitrary Powers 3.1-6. Equation Containing the Unknown Function of a Complicated Argument 3.1-7. Singular Equations 3.2. Equations Whose Kernels Contain Exponential Functions 3.2-1. Kernels Containing Exponential Functions 3.2-2. Kernels Containing Power-Law and Exponential Functions 3.3. Equations Whose Kernels Contain Hyperbolic Functions 3.3-1. Kernels Containing Hyperbolic Cosine 3.3-2. Kernels Containing Hyperbolic Sine 3.3-3. Kernels Containing Hyperbolic Tangent 3.3-4. Kernels Containing Hyperbolic Cotangent 3.4. Equations Whose Kernels Contain Logarithmic Functions 3.4-1. Kernels Containing Logarithmic Functions 3.4-2. Kernels Containing Power-Law and Logarithmic Functions 3.4-3. An Equation Containing the Unknown Function of a Complicated Argument 3.5. Equations Whose Kernels Contain Trigonometric Functions 3.5-1. Kernels Containing Cosine 3.5-2. Kernels Containing Sine 3.5-3. Kernels Containing Tangent 3.5-4. Kernels Containing Cotangent 3.5-5. Kernels Containing a Combination of Trigonometric Functions 3.5-6. Equations Containing the Unknown Function of a Complicated Argument 3.5-7. A Singular Equation 3.6. Equations Whose Kernels Contain Combinations of Elementary Functions 3.6-1. Kernels Containing Hyperbolic and Logarithmic Functions 3.6-2. Kernels Containing Logarithmic and Trigonometric Functions

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page xi

3.7. Equations Whose Kernels Contain Special Functions 3.7-1. Kernels Containing Bessel Functions 3.7-2. Kernels Containing Modified Bessel Functions 3.7-3. Other Kernels 3.8. Equations Whose Kernels Contain Arbitrary Functions 3.8-1. Equations With Degenerate Kernel 3.8-2. Equations Containing Modulus 3.8-3. Equations With Difference Kernel: K(x, t) = K(x – t) b 3.8-4. Other Equations of the Form a K(x, t)y(t) dt = F (x) b 3.8-5. Equations of the Form a K(x, t)y(· · ·) dt = F (x) 4. Linear Equations of the Second Kind With Constant Limits of Integration 4.1. Equations Whose Kernels Contain Power-Law Functions 4.1-1. Kernels Linear in the Arguments x and t 4.1-2. Kernels Quadratic in the Arguments x and t 4.1-3. Kernels Cubic in the Arguments x and t 4.1-4. Kernels Containing Higher-Order Polynomials in x and t 4.1-5. Kernels Containing Rational Functions 4.1-6. Kernels Containing Arbitrary Powers 4.1-7. Singular Equations 4.2. Equations Whose Kernels Contain Exponential Functions 4.2-1. Kernels Containing Exponential Functions 4.2-2. Kernels Containing Power-Law and Exponential Functions 4.3. Equations Whose Kernels Contain Hyperbolic Functions 4.3-1. Kernels Containing Hyperbolic Cosine 4.3-2. Kernels Containing Hyperbolic Sine 4.3-3. Kernels Containing Hyperbolic Tangent 4.3-4. Kernels Containing Hyperbolic Cotangent 4.3-5. Kernels Containing Combination of Hyperbolic Functions 4.4. Equations Whose Kernels Contain Logarithmic Functions 4.4-1. Kernels Containing Logarithmic Functions 4.4-2. Kernels Containing Power-Law and Logarithmic Functions 4.5. Equations Whose Kernels Contain Trigonometric Functions 4.5-1. Kernels Containing Cosine 4.5-2. Kernels Containing Sine 4.5-3. Kernels Containing Tangent 4.5-4. Kernels Containing Cotangent 4.5-5. Kernels Containing Combinations of Trigonometric Functions 4.5-6. A Singular Equation 4.6. Equations Whose Kernels Contain Inverse Trigonometric Functions 4.6-1. Kernels Containing Arccosine 4.6-2. Kernels Containing Arcsine 4.6-3. Kernels Containing Arctangent 4.6-4. Kernels Containing Arccotangent 4.7. Equations Whose Kernels Contain Combinations of Elementary Functions 4.7-1. Kernels Containing Exponential and Hyperbolic Functions 4.7-2. Kernels Containing Exponential and Logarithmic Functions 4.7-3. Kernels Containing Exponential and Trigonometric Functions 4.7-4. Kernels Containing Hyperbolic and Logarithmic Functions

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

Page xii

4.7-5. Kernels Containing Hyperbolic and Trigonometric Functions 4.7-6. Kernels Containing Logarithmic and Trigonometric Functions 4.8. Equations Whose Kernels Contain Special Functions 4.8-1. Kernels Containing Bessel Functions 4.8-2. Kernels Containing Modified Bessel Functions 4.9. Equations Whose Kernels Contain Arbitrary Functions 4.9-1. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t) 4.9-2. Equations With Difference Kernel: K(x, t) = K(x – t) b 4.9-3. Other Equations of the Form y(x) + a K(x, t)y(t) dt = F (x) b 4.9-4. Equations of the Form y(x) + a K(x, t)y(· · ·) dt = F (x) 4.10. Some Formulas and Transformations 5. Nonlinear Equations With Variable Limit of Integration 5.1. Equations With Quadratic Nonlinearity That Contain Arbitrary Parameters x 5.1-1. Equations of the Form 0 y(t)y(x – t) dt = F (x) x 5.1-2. Equations of the Form 0 K(x, t)y(t)y(x – t) dt = F (x) x 5.1-3. Equations of the Form 0 G(· · ·) dt = F (x) x 5.1-4. Equations of the Form y(x) + a K(x, t)y 2 (t) dt = F (x) x 5.1-5. Equations of the Form y(x) + a K(x, t)y(t)y(x – t) dt = F (x) 5.2. Equations With Quadratic Nonlinearity That Contain Arbitrary Functions x 5.2-1. Equations of the Form a G(· · ·) dt = F (x) x 5.2-2. Equations of the Form y(x) + a K(x, t)y 2 (t) dt = F (x) x 5.2-3. Equations of the Form y(x) + a G(· · ·) dt = F (x) 5.3. Equations With Power-Law Nonlinearity 5.3-1. Equations Containing Arbitrary Parameters 5.3-2. Equations Containing Arbitrary Functions 5.4. Equations With Exponential Nonlinearity 5.4-1. Equations Containing Arbitrary Parameters 5.4-2. Equations Containing Arbitrary Functions 5.5. Equations With Hyperbolic Nonlinearity 5.5-1. Integrands With Nonlinearity of the Form cosh[βy(t)] 5.5-2. Integrands With Nonlinearity of the Form sinh[βy(t)] 5.5-3. Integrands With Nonlinearity of the Form tanh[βy(t)] 5.5-4. Integrands With Nonlinearity of the Form coth[βy(t)] 5.6. Equations With Logarithmic Nonlinearity 5.6-1. Integrands Containing Power-Law Functions of x and t 5.6-2. Integrands Containing Exponential Functions of x and t 5.6-3. Other Integrands 5.7. Equations With Trigonometric Nonlinearity 5.7-1. Integrands With Nonlinearity of the Form cos[βy(t)] 5.7-2. Integrands With Nonlinearity of the Form sin[βy(t)] 5.7-3. Integrands With Nonlinearity of the Form tan[βy(t)] 5.7-4. Integrands With Nonlinearity of the Form cot[βy(t)] 5.8. Equations With Nonlinearity ofGeneral Form x 5.8-1. Equations of the Form a G(· · ·) dt = F (x)

x 5.8-2. Equations of the Form y(x) + a K(x, t)G y(t) dt = F (x)

x 5.8-3. Equations of the Form y(x) + a K(x, t)G t, y(t) dt = F (x) 5.8-4. Other Equations

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC

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6. Nonlinear Equations With Constant Limits of Integration 6.1. Equations With Quadratic Nonlinearity That Contain Arbitrary Parameters b 6.1-1. Equations of the Form a K(t)y(x)y(t) dt = F (x) b 6.1-2. Equations of the Form a G(· · ·) dt = F (x) b 6.1-3. Equations of the Form y(x) + a K(x, t)y 2 (t) dt = F (x) b 6.1-4. Equations of the Form y(x) + a K(x, t)y(x)y(t) dt = F (x) b 6.1-5. Equations of the Form y(x) + a G(· · ·) dt = F (x) 6.2. Equations With Quadratic Nonlinearity That Contain Arbitrary Functions b 6.2-1. Equations of the Form a G(· · ·) dt = F (x) b 6.2-2. Equations of the Form y(x) + a K(x, t)y 2 (t) dt = F (x) b Knm (x, t)y n (x)y m (t) dt = F (x), n + m ≤ 2 6.2-3. Equations of the Form y(x) + a b 6.2-4. Equations of the Form y(x) + a G(· · ·) dt = F (x) 6.3. Equations With Power-Law Nonlinearity b 6.3-1. Equations of the Form a G(· · ·) dt = F (x) b 6.3-2. Equations of the Form y(x) + a K(x, t)y β (t) dt = F (x) b 6.3-3. Equations of the Form y(x) + a G(· · ·) dt = F (x) 6.4. Equations With Exponential Nonlinearity 6.4-1. Integrands With Nonlinearity of the Form exp[βy(t)] 6.4-2. Other Integrands 6.5. Equations With Hyperbolic Nonlinearity 6.5-1. Integrands With Nonlinearity of the Form cosh[βy(t)] 6.5-2. Integrands With Nonlinearity of the Form sinh[βy(t)] 6.5-3. Integrands With Nonlinearity of the Form tanh[βy(t)] 6.5-4. Integrands With Nonlinearity of the Form coth[βy(t)] 6.5-5. Other Integrands 6.6. Equations With Logarithmic Nonlinearity 6.6-1. Integrands With Nonlinearity of the Form ln[βy(t)] 6.6-2. Other Integrands 6.7. Equations With Trigonometric Nonlinearity 6.7-1. Integrands With Nonlinearity of the Form cos[βy(t)] 6.7-2. Integrands With Nonlinearity of the Form sin[βy(t)] 6.7-3. Integrands With Nonlinearity of the Form tan[βy(t)] 6.7-4. Integrands With Nonlinearity of the Form cot[βy(t)] 6.7-5. Other Integrands 6.8. Equations With Nonlinearity of General Form b 6.8-1. Equations of the Form a G(· · ·) dt = F (x)

b 6.8-2. Equations of the Form y(x) + a K(x, t)G y(t) dt = F (x)

b 6.8-3. Equations of the Form y(x) + a K(x, t)G t, y(t) dt = F (x)

b 6.8-4. Equations of the Form y(x) + a G x, t, y(t) dt = F (x)

b

6.8-5. Equations of the Form F x, y(x) + a G x, t, y(x), y(t) dt = 0 6.8-6. Other Equations

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Part II. Methods for Solving Integral Equations 7 Main Definitions and Formulas. Integral Transforms 7.1. Some Definitions, Remarks, and Formulas 7.1-1. Some Definitions 7.1-2. The Structure of Solutions to Linear Integral Equations 7.1-3. Integral Transforms 7.1-4. Residues. Calculation Formulas 7.1-5. The Jordan Lemma 7.2. The Laplace Transform 7.2-1. Definition. The Inversion Formula 7.2-2. The Inverse Transforms of Rational Functions 7.2-3. The Convolution Theorem for the Laplace Transform 7.2-4. Limit Theorems 7.2-5. Main Properties of the Laplace Transform 7.2-6. The Post–Widder Formula 7.3. The Mellin Transform 7.3-1. Definition. The Inversion Formula 7.3-2. Main Properties of the Mellin Transform 7.3-3. The Relation Among the Mellin, Laplace, and Fourier Transforms 7.4. The Fourier Transform 7.4-1. Definition. The Inversion Formula 7.4-2. An Asymmetric Form of the Transform 7.4-3. The Alternative Fourier Transform 7.4-4. The Convolution Theorem for the Fourier Transform 7.5. The Fourier Sine and Cosine Transforms 7.5-1. The Fourier Cosine Transform 7.5-2. The Fourier Sine Transform 7.6. Other Integral Transforms 7.6-1. The Hankel Transform 7.6-2. The Meijer Transform 7.6-3. The Kontorovich–Lebedev Transform and Other Transforms x 8. Methods for Solving Linear Equations of the Form a K(x, t)y(t) dt = f (x) 8.1. Volterra Equations of the First Kind 8.1-1. Equations of the First Kind. Function and Kernel Classes 8.1-2. Existence and Uniqueness of a Solution 8.2. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t) 8.2-1. Equations With Kernel of the Form K(x, t) = g1 (x)h1 (t) + g2 (x)h2 (t) 8.2-2. Equations With General Degenerate Kernel 8.3. Reduction of Volterra Equations of the 1st Kind to Volterra Equations of the 2nd Kind 8.3-1. The First Method 8.3-2. The Second Method 8.4. Equations With Difference Kernel: K(x, t) = K(x – t) 8.4-1. A Solution Method Based on the Laplace Transform 8.4-2. The Case in Which the Transform of the Solution is a Rational Function 8.4-3. Convolution Representation of a Solution 8.4-4. Application of an Auxiliary Equation 8.4-5. Reduction to Ordinary Differential Equations 8.4-6. Reduction of a Volterra Equation to a Wiener–Hopf Equation

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8.5. Method of Fractional Differentiation 8.5-1. The Definition of Fractional Integrals 8.5-2. The Definition of Fractional Derivatives 8.5-3. Main Properties 8.5-4. The Solution of the Generalized Abel Equation 8.6. Equations With Weakly Singular Kernel 8.6-1. A Method of Transformation of the Kernel 8.6-2. Kernel With Logarithmic Singularity 8.7. Method of Quadratures 8.7-1. Quadrature Formulas 8.7-2. The General Scheme of the Method 8.7-3. An Algorithm Based on the Trapezoidal Rule 8.7-4. An Algorithm for an Equation With Degenerate Kernel 8.8. Equations With Infinite Integration Limit 8.8-1. An Equation of the First Kind With Variable Lower Limit of Integration 8.8-2. Reduction to a Wiener–Hopf Equation of the First Kind x 9. Methods for Solving Linear Equations of the Form y(x) – a K(x, t)y(t) dt = f (x) 9.1. Volterra Integral Equations of the Second Kind 9.1-1. Preliminary Remarks. Equations for the Resolvent 9.1-2. A Relationship Between Solutions of Some Integral Equations 9.2. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t) 9.2-1. Equations With Kernel of the Form K(x, t) = ϕ(x) + ψ(x)(x – t) 9.2-2. Equations With Kernel of the Form K(x, t) = ϕ(t) n + ψ(t)(t – x) 9.2-3. Equations With Kernel of the Form K(x, t) = m=1 ϕm (x)(x – t)m–1 n 9.2-4. Equations With Kernel of the Form K(x, t) = m=1 ϕm (t)(t – x)m–1 9.2-5. Equations With Degenerate Kernel of the General Form 9.3. Equations With Difference Kernel: K(x, t) = K(x – t) 9.3-1. A Solution Method Based on the Laplace Transform 9.3-2. A Method Based on the Solution of an Auxiliary Equation 9.3-3. Reduction to Ordinary Differential Equations 9.3-4. Reduction to a Wiener–Hopf Equation of the Second Kind 9.3-5. Method of Fractional Integration for the Generalized Abel Equation 9.3-6. Systems of Volterra Integral Equations 9.4. Operator Methods for Solving Linear Integral Equations 9.4-1. Application of a Solution of a “Truncated” Equation of the First Kind 9.4-2. Application of the Auxiliary Equation of the Second Kind 9.4-3. A Method for Solving “Quadratic” Operator Equations 9.4-4. Solution of Operator Equations of Polynomial Form 9.4-5. A Generalization 9.5. Construction of Solutions of Integral Equations With Special Right-Hand Side 9.5-1. The General Scheme 9.5-2. A Generating Function of Exponential Form 9.5-3. Power-Law Generating Function 9.5-4. Generating Function Containing Sines and Cosines 9.6. The Method of Model Solutions 9.6-1. Preliminary Remarks 9.6-2. Description of the Method 9.6-3. The Model Solution in the Case of an Exponential Right-Hand Side

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9.6-4. 9.6-5. 9.6-6. 9.6-7.

The Model Solution in the Case of a Power-Law Right-Hand Side The Model Solution in the Case of a Sine-Shaped Right-Hand Side The Model Solution in the Case of a Cosine-Shaped Right-Hand Side Some Generalizations

9.7. Method of Differentiation for Integral Equations 9.7-1. Equations With Kernel Containing a Sum of Exponential Functions 9.7-2. Equations With Kernel Containing a Sum of Hyperbolic Functions 9.7-3. Equations With Kernel Containing a Sum of Trigonometric Functions 9.7-4. Equations Whose Kernels Contain Combinations of Various Functions 9.8. Reduction of Volterra Equations of the 2nd Kind to Volterra Equations of the 1st Kind 9.8-1. The First Method 9.8-2. The Second Method 9.9. The Successive Approximation Method 9.9-1. The General Scheme 9.9-2. A Formula for the Resolvent 9.10. Method of Quadratures 9.10-1. The General Scheme of the Method 9.10-2. Application of the Trapezoidal Rule 9.10-3. The Case of a Degenerate Kernel 9.11. Equations With Infinite Integration Limit 9.11-1. An Equation of the Second Kind With Variable Lower Integration Limit 9.11-2. Reduction to a Wiener–Hopf Equation of the Second Kind b 10. Methods for Solving Linear Equations of the Form a K(x, t)y(t) dt = f (x) 10.1. Some Definition and Remarks 10.1-1. Fredholm Integral Equations of the First Kind 10.1-2. Integral Equations of the First Kind With Weak Singularity 10.1-3. Integral Equations of Convolution Type 10.1-4. Dual Integral Equations of the First Kind 10.2. Krein’s Method 10.2-1. The Main Equation and the Auxiliary Equation 10.2-2. Solution of the Main Equation 10.3. The Method of Integral Transforms 10.3-1. Equation With Difference Kernel on the Entire Axis 10.3-2. Equations With Kernel K(x, t) = K(x/t) on the Semiaxis 10.3-3. Equation With Kernel K(x, t) = K(xt) and Some Generalizations 10.4. The Riemann Problem for the Real Axis 10.4-1. Relationships Between the Fourier Integral and the Cauchy Type Integral 10.4-2. One-Sided Fourier Integrals 10.4-3. The Analytic Continuation Theorem and the Generalized Liouville Theorem 10.4-4. The Riemann Boundary Value Problem 10.4-5. Problems With Rational Coefficients 10.4-6. Exceptional Cases. The Homogeneous Problem 10.4-7. Exceptional Cases. The Nonhomogeneous Problem 10.5. The Carleman Method for Equations of the Convolution Type of the First Kind 10.5-1. The Wiener–Hopf Equation of the First Kind 10.5-2. Integral Equations of the First Kind With Two Kernels

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10.6. Dual Integral Equations of the First Kind 10.6-1. The Carleman Method for Equations With Difference Kernels 10.6-2. Exact Solutions of Some Dual Equations of the First Kind 10.6-3. Reduction of Dual Equations to a Fredholm Equation 10.7. Asymptotic Methods for Solving Equations With Logarithmic Singularity 10.7-1. Preliminary Remarks 10.7-2. The Solution for Large λ 10.7-3. The Solution for Small λ 10.7-4. Integral Equation of Elasticity 10.8. Regularization Methods 10.8-1. The Lavrentiev Regularization Method 10.8-2. The Tikhonov Regularization Method 11. Methods for Solving Linear Equations of the Form y(x) –

b a

K(x, t)y(t) dt = f (x)

11.1. Some Definition and Remarks 11.1-1. Fredholm Equations and Equations With Weak Singularity of the 2nd Kind 11.1-2. The Structure of the Solution 11.1-3. Integral Equations of Convolution Type of the Second Kind 11.1-4. Dual Integral Equations of the Second Kind 11.2. Fredholm Equations of the Second Kind With Degenerate Kernel 11.2-1. The Simplest Degenerate Kernel 11.2-2. Degenerate Kernel in the General Case 11.3. Solution as a Power Series in the Parameter. Method of Successive Approximations 11.3-1. Iterated Kernels 11.3-2. Method of Successive Approximations 11.3-3. Construction of the Resolvent 11.3-4. Orthogonal Kernels 11.4. Method of Fredholm Determinants 11.4-1. A Formula for the Resolvent 11.4-2. Recurrent Relations 11.5. Fredholm Theorems and the Fredholm Alternative 11.5-1. Fredholm Theorems 11.5-2. The Fredholm Alternative 11.6. Fredholm Integral Equations of the Second Kind With Symmetric Kernel 11.6-1. Characteristic Values and Eigenfunctions 11.6-2. Bilinear Series 11.6-3. The Hilbert–Schmidt Theorem 11.6-4. Bilinear Series of Iterated Kernels 11.6-5. Solution of the Nonhomogeneous Equation 11.6-6. The Fredholm Alternative for Symmetric Equations 11.6-7. The Resolvent of a Symmetric Kernel 11.6-8. Extremal Properties of Characteristic Values and Eigenfunctions 11.6-9. Integral Equations Reducible to Symmetric Equations 11.6-10. Skew-Symmetric Integral Equations 11.7. An Operator Method for Solving Integral Equations of the Second Kind 11.7-1. The Simplest Scheme 11.7-2. Solution of Equations of the Second Kind on the Semiaxis

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11.8. Methods of Integral Transforms and Model Solutions 11.8-1. Equation With Difference Kernel on the Entire Axis 11.8-2. An Equation With the Kernel K(x, t) = t–1 Q(x/t) on the Semiaxis 11.8-3. Equation With the Kernel K(x, t) = tβ Q(xt) on the Semiaxis 11.8-4. The Method of Model Solutions for Equations on the Entire Axis 11.9. The Carleman Method for Integral Equations of Convolution Type of the Second Kind 11.9-1. The Wiener–Hopf Equation of the Second Kind 11.9-2. An Integral Equation of the Second Kind With Two Kernels 11.9-3. Equations of Convolution Type With Variable Integration Limit 11.9-4. Dual Equation of Convolution Type of the Second Kind 11.10. The Wiener–Hopf Method 11.10-1. Some Remarks 11.10-2. The Homogeneous Wiener–Hopf Equation of the Second Kind 11.10-3. The General Scheme of the Method. The Factorization Problem 11.10-4. The Nonhomogeneous Wiener–Hopf Equation of the Second Kind 11.10-5. The Exceptional Case of a Wiener–Hopf Equation of the Second Kind 11.11. Krein’s Method for Wiener–Hopf Equations 11.11-1. Some Remarks. The Factorization Problem 11.11-2. The Solution of the Wiener–Hopf Equations of the Second Kind 11.11-3. The Hopf–Fock Formula 11.12. Methods for Solving Equations With Difference Kernels on a Finite Interval 11.12-1. Krein’s Method 11.12-2. Kernels With Rational Fourier Transforms 11.12-3. Reduction to Ordinary Differential Equations 11.13. The Method of Approximating a Kernel by a Degenerate One 11.13-1. Approximation of the Kernel 11.13-2. The Approximate Solution 11.14. The Bateman Method 11.14-1. The General Scheme of the Method 11.14-2. Some Special Cases 11.15. The Collocation Method 11.15-1. General Remarks 11.15-2. The Approximate Solution 11.15-3. The Eigenfunctions of the Equation 11.16. The Method of Least Squares 11.16-1. Description of the Method 11.16-2. The Construction of Eigenfunctions 11.17. The Bubnov–Galerkin Method 11.17-1. Description of the Method 11.17-2. Characteristic Values 11.18. The Quadrature Method 11.18-1. The General Scheme for Fredholm Equations of the Second Kind 11.18-2. Construction of the Eigenfunctions 11.18-3. Specific Features of the Application of Quadrature Formulas 11.19. Systems of Fredholm Integral Equations of the Second Kind 11.19-1. Some Remarks 11.19-2. The Method of Reducing a System of Equations to a Single Equation

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11.20. Regularization Method for Equations With Infinite Limits of Integration 11.20-1. Basic Equation and Fredholm Theorems 11.20-2. Regularizing Operators 11.20-3. The Regularization Method 12. Methods for Solving Singular Integral Equations of the First Kind 12.1. Some Definitions and Remarks 12.1-1. Integral Equations of the First Kind With Cauchy Kernel 12.1-2. Integral Equations of the First Kind With Hilbert Kernel 12.2. The Cauchy Type Integral 12.2-1. Definition of the Cauchy Type Integral 12.2-2. The H¨older Condition 12.2-3. The Principal Value of a Singular Integral 12.2-4. Multivalued Functions 12.2-5. The Principal Value of a Singular Curvilinear Integral 12.2-6. The Poincar´e–Bertrand Formula 12.3. The Riemann Boundary Value Problem 12.3-1. The Principle of Argument. The Generalized Liouville Theorem 12.3-2. The Hermite Interpolation Polynomial 12.3-3. Notion of the Index 12.3-4. Statement of the Riemann Problem 12.3-5. The Solution of the Homogeneous Problem 12.3-6. The Solution of the Nonhomogeneous Problem 12.3-7. The Riemann Problem With Rational Coefficients 12.3-8. The Riemann Problem for a Half-Plane 12.3-9. Exceptional Cases of the Riemann Problem 12.3-10. The Riemann Problem for a Multiply Connected Domain 12.3-11. The Cases of Discontinuous Coefficients and Nonclosed Contours 12.3-12. The Hilbert Boundary Value Problem 12.4. Singular Integral Equations of the First Kind 12.4-1. The Simplest Equation With Cauchy Kernel 12.4-2. An Equation With Cauchy Kernel on the Real Axis 12.4-3. An Equation of the First Kind on a Finite Interval 12.4-4. The General Equation of the First Kind With Cauchy Kernel 12.4-5. Equations of the First Kind With Hilbert Kernel 12.5. Multhopp–Kalandiya Method 12.5-1. A Solution That is Unbounded at the Endpoints of the Interval 12.5-2. A Solution Bounded at One Endpoint of the Interval 12.5-3. Solution Bounded at Both Endpoints of the Interval 13. Methods for Solving Complete Singular Integral Equations 13.1. Some Definitions and Remarks 13.1-1. Integral Equations With Cauchy Kernel 13.1-2. Integral Equations With Hilbert Kernel 13.1-3. Fredholm Equations of the Second Kind on a Contour 13.2. The Carleman Method for Characteristic Equations 13.2-1. A Characteristic Equation With Cauchy Kernel 13.2-2. The Transposed Equation of a Characteristic Equation 13.2-3. The Characteristic Equation on the Real Axis 13.2-4. The Exceptional Case of a Characteristic Equation

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13.2-5. The Characteristic Equation With Hilbert Kernel 13.2-6. The Tricomi Equation 13.3. Complete Singular Integral Equations Solvable in a Closed Form 13.3-1. Closed-Form Solutions in the Case of Constant Coefficients 13.3-2. Closed-Form Solutions in the General Case 13.4. The Regularization Method for Complete Singular Integral Equations 13.4-1. Certain Properties of Singular Operators 13.4-2. The Regularizer 13.4-3. The Methods of Left and Right Regularization 13.4-4. The Problem of Equivalent Regularization 13.4-5. Fredholm Theorems 13.4-6. The Carleman–Vekua Approach to the Regularization 13.4-7. Regularization in Exceptional Cases 13.4-8. The Complete Equation With Hilbert Kernel 14. Methods for Solving Nonlinear Integral Equations 14.1. Some Definitions and Remarks 14.1-1. Nonlinear Volterra Integral Equations 14.1-2. Nonlinear Equations With Constant Integration Limits 14.2. Nonlinear Volterra Integral Equations 14.2-1. The Method of Integral Transforms 14.2-2. The Method of Differentiation for Integral Equations 14.2-3. The Successive Approximation Method 14.2-4. The Newton–Kantorovich Method 14.2-5. The Collocation Method 14.2-6. The Quadrature Method 14.3. Equations With Constant Integration Limits 14.3-1. Nonlinear Equations With Degenerate Kernels 14.3-2. The Method of Integral Transforms 14.3-3. The Method of Differentiating for Integral Equations 14.3-4. The Successive Approximation Method 14.3-5. The Newton–Kantorovich Method 14.3-6. The Quadrature Method 14.3-7. The Tikhonov Regularization Method

Supplements Supplement 1. Elementary Functions and Their Properties 1.1. Trigonometric Functions 1.2. Hyperbolic Functions 1.3. Inverse Trigonometric Functions 1.4. Inverse Hyperbolic Functions Supplement 2. Tables of Indefinite Integrals 2.1. Integrals Containing Rational Functions 2.2. Integrals Containing Irrational Functions 2.3. Integrals Containing Exponential Functions 2.4. Integrals Containing Hyperbolic Functions 2.5. Integrals Containing Logarithmic Functions

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2.6. Integrals Containing Trigonometric Functions 2.7. Integrals Containing Inverse Trigonometric Functions Supplement 3. Tables of Definite Integrals 3.1. Integrals Containing Power-Law Functions 3.2. Integrals Containing Exponential Functions 3.3. Integrals Containing Hyperbolic Functions 3.4. Integrals Containing Logarithmic Functions 3.5. Integrals Containing Trigonometric Functions Supplement 4. Tables of Laplace Transforms 4.1. General Formulas 4.2. Expressions With Power-Law Functions 4.3. Expressions With Exponential Functions 4.4. Expressions With Hyperbolic Functions 4.5. Expressions With Logarithmic Functions 4.6. Expressions With Trigonometric Functions 4.7. Expressions With Special Functions Supplement 5. Tables of Inverse Laplace Transforms 5.1. General Formulas 5.2. Expressions With Rational Functions 5.3. Expressions With Square Roots 5.4. Expressions With Arbitrary Powers 5.5. Expressions With Exponential Functions 5.6. Expressions With Hyperbolic Functions 5.7. Expressions With Logarithmic Functions 5.8. Expressions With Trigonometric Functions 5.9. Expressions With Special Functions Supplement 6. Tables of Fourier Cosine Transforms 6.1. General Formulas 6.2. Expressions With Power-Law Functions 6.3. Expressions With Exponential Functions 6.4. Expressions With Hyperbolic Functions 6.5. Expressions With Logarithmic Functions 6.6. Expressions With Trigonometric Functions 6.7. Expressions With Special Functions Supplement 7. Tables of Fourier Sine Transforms 7.1. General Formulas 7.2. Expressions With Power-Law Functions 7.3. Expressions With Exponential Functions 7.4. Expressions With Hyperbolic Functions 7.5. Expressions With Logarithmic Functions 7.6. Expressions With Trigonometric Functions 7.7. Expressions With Special Functions

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Supplement 8. Tables of Mellin Transforms 8.1. General Formulas 8.2. Expressions With Power-Law Functions 8.3. Expressions With Exponential Functions 8.4. Expressions With Logarithmic Functions 8.5. Expressions With Trigonometric Functions 8.6. Expressions With Special Functions Supplement 9. Tables of Inverse Mellin Transforms 9.1. Expressions With Power-Law Functions 9.2. Expressions With Exponential and Logarithmic Functions 9.3. Expressions With Trigonometric Functions 9.4. Expressions With Special Functions Supplement 10. Special Functions and Their Properties 10.1. Some Symbols and Coefficients 10.2. Error Functions and Integral Exponent 10.3. Integral Sine and Integral Cosine. Fresnel Integrals 10.4. Gamma Function. Beta Function 10.5. Incomplete Gamma Function 10.6. Bessel Functions 10.7. Modified Bessel Functions 10.8. Degenerate Hypergeometric Functions 10.9. Hypergeometric Functions 10.10. Legendre Functions 10.11. Orthogonal Polynomials References

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Part I

Exact Solutions of Integral Equations

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Chapter 1

Linear Equations of the First Kind With Variable Limit of Integration Notation: f = f (x), g = g(x), h = h(x), K = K(x), and M = M (x) are arbitrary functions (these may be composite functions of the argument depending on two variables x and t); A, B, C, D, E, a, b, c, α, β, γ, λ, and µ are free parameters; and m and n are nonnegative integers. Preliminary remarks. For equations of the form x K(x, t)y(t) dt = f (x),

a ≤ x ≤ b,

a

where the functions K(x, t) and f (x) are continuous, the right-hand side must satisfy the following conditions: 1◦ . If K(a, a) ≠ 0, then we must have f (a) = 0 (for example, the right-hand sides of equations 1.1.1 and 1.2.1 must satisfy this condition). 2◦ . If K(a, a) = Kx (a, a) = · · · = Kx(n–1) (a, a) = 0, 0 < Kx(n) (a, a) < ∞, then the right-hand side of the equation must satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. For example, with n = 1, these are constraints for the right-hand side of equation 1.1.2. 3◦ . If K(a, a) = Kx (a, a) = · · · = Kx(n–1) (a, a) = 0, Kx(n) (a, a) = ∞, then the right-hand side of the equation must satisfy the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0. For example, with n = 1, this is a constraint for the right-hand side of equation 1.1.30. For unbounded K(x, t) with integrable power-law or logarithmic singularity at x = t and continuous f (x), no additional conditions are imposed on the right-hand side of the integral equation (e.g., see Abel’s equation 1.1.36). In Chapter 1, conditions 1◦ –3◦ are as a rule not specified.

1.1. Equations Whose Kernels Contain Power-Law Functions 1.1-1. Kernels Linear in the Arguments x and t x 1. y(t) dt = f (x). a

Solution: y(x) = fx (x).

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x

(x – t)y(t) dt = f (x).

2. a

3.

Solution: y(x) = fxx (x). x (Ax + Bt + C)y(t) dt = f (x). a

This is a special case of equation 1.9.5 with g(x) = x. 1◦ . Solution with B ≠ –A: x – B – A d y(x) = (A + B)t + C A+B ft (t) dt . (A + B)x + C A+B dx a 2◦ . Solution with B = –A:

A

1 d A x y(x) = exp exp – x t ft (t) dt . C dx C C a

1.1-2. Kernels Quadratic in the Arguments x and t

x

f (a) = fx (a) = fxx (a) = 0.

(x – t)2 y(t) dt = f (x),

4. a

5.

Solution: y(x) = 12 fxxx (x). x (x2 – t2 )y(t) dt = f (x),

f (a) = fx (a) = 0.

a

6.

This is a special case of equation 1.9.2 with g(x) = x2 . 1 Solution: y(x) = xfxx (x) – fx (x) . 2x2 x

Ax2 + Bt2 y(t) dt = f (x).

7.

For B = –A, see equation 1.1.5. This 1.9.4 with g(x) = x2 . is a special x case of equation 2A 2B 1 d Solution: y(x) = x– A+B t– A+B ft (t) dt . A + B dx a x

Ax2 + Bt2 + C y(t) dt = f (x).

a

a

8.

This is a special case of equation 1.9.5 with g(x) = x2 . Solution: x A B d – – y(x) = sign ϕ(x) |ϕ(x)| A+B |ϕ(t)| A+B ft (t) dt , ϕ(x) = (A + B)x2 + C. dx a x Ax2 + (B – A)xt – Bt2 y(t) dt = f (x), f (a) = fx (a) = 0. a

Differentiating with respect to x yields an equation of the form 1.1.3: x [2Ax + (B – A)t]y(t) dt = fx (x). a

Solution:

x 2A A–B 1 d – A+B A+B y(x) = t ftt (t) dt . x A + B dx a

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9.

x

Ax2 + Bt2 + Cx + Dt + E y(t) dt = f (x).

a

10.

This is a special case of equation 1.9.6 with g(x) = Ax2 + Cx and h(t) = Bt2 + Dt + E. x

Axt + Bt2 + Cx + Dt + E y(t) dt = f (x).

11.

This is a special case of equation 1.9.15 with g1 (x) = x, h1 (t) = At + C, g2 (x) = 1, and h2 (t) = Bt2 + Dt + E. x

Ax2 + Bxt + Cx + Dt + E y(t) dt = f (x).

a

a

This is a special case of equation 1.9.15 with g1 (x) = Bx + D, h1 (t) = t, g2 (x) = Ax2 + Cx + E, and h2 (t) = 1. 1.1-3. Kernels Cubic in the Arguments x and t

x

(x – t)3 y(t) dt = f (x),

12. a

13.

Solution: y(x) = 16 fxxxx (x). x (x3 – t3 )y(t) dt = f (x), a

f (a) = fx (a) = fxx (a) = fxxx (a) = 0.

f (a) = fx (a) = 0.

14.

This is a special case of equation 1.9.2 with g(x) = x3 . 1 Solution: y(x) = xfxxx (x) – 2fx (x) . 3 3x x

Ax3 + Bt3 y(t) dt = f (x).

15.

For B = –A, see equation 1.1.13. This is a special 1.9.4 with g(x) = x3 . case of equation x 3A 3B 1 d Solution with 0 ≤ a ≤ x: y(x) = t– A+B ft (t) dt . x– A+B A + B dx a x

Ax3 + Bt3 + C y(t) dt = f (x).

a

a

16.

This is a special case of equation 1.9.5 with g(x) = x3 . x (x2 t – xt2 )y(t) dt = f (x), f (a) = fx (a) = 0.

17.

2 This is a special case of equation 1.9.11 with g(x) = x and h(x) = x. 2 1 1 d Solution: y(x) = f (x) . x dx2 x x (Ax2 t + Bxt2 )y(t) dt = f (x).

a

a

For B = –A, see equation 1.1.16. This is a special case of equation 1.9.12 with g(x) = x2 and h(x) = x. Solution: x A B 1 d 1 – A+B d – A+B y(x) = t f (t) dt . x (A + B)x dx dt t a

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x

(Ax3 + Bxt2 )y(t) dt = f (x).

18. a

This is a special case of equation 1.9.15 with g1 (x) = Ax3 , h1 (t) = 1, g2 (x) = Bx, and h2 (t) = t2 .

x

(Ax3 + Bx2 t)y(t) dt = f (x).

19. a

This is a special case of equation 1.9.15 with g1 (x) = Ax3 , h1 (t) = 1, g2 (x) = Bx2 , and h2 (t) = t.

x

(Ax2 t + Bt3 )y(t) dt = f (x).

20. a

This is a special case of equation 1.9.15 with g1 (x) = Ax2 , h1 (t) = t, g2 (x) = B, and h2 (t) = t3 .

x

(Axt2 + Bt3 )y(t) dt = f (x).

21. a

This is a special case of equation 1.9.15 with g1 (x) = Ax, h1 (t) = t2 , g2 (x) = B, and h2 (t) = t3 . 22.

x

A3 x3 + B3 t3 + A2 x2 + B2 t2 + A1 x + B1 t + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A3 x3 + A2 x2 + A1 x + C and h(t) = B3 t3 + B2 t2 + B1 t. 1.1-4. Kernels Containing Higher-Order Polynomials in x and t

x

(x – t)n y(t) dt = f (x),

23.

n = 1, 2, . . .

a

It is assumed that the right-hand of the equation satisfies the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. 1 (n+1) Solution: y(x) = (x). f n! x Example. For f (x) = Axm , where m is a positive integer, m > n, the solution has the form y(x) =

x

(xn – tn )y(t) dt = f (x),

24. a

Solution: y(x) = 25.

x

Am! xm–n–1 . n! (m – n – 1)!

f (a) = fx (a) = 0,

n = 1, 2, . . .

1 d fx (x) . n dx xn–1

tn xn+1 – xn tn+1 y(t) dt = f (x),

n = 2, 3, . . .

a n+1 This is a special case of equation and h(x) = xn . 1.9.11 with g(x) = x 2 f (x) 1 d . Solution: y(x) = n 2 x dx xn

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1.1-5. Kernels Containing Rational Functions

x

26. 0

y(t) dt x+t

= f (x).

1◦ . For a polynomial right-hand side, f (x) =

N

An xn , the solution has the form

n=0

N An n y(x) = x , Bn

Bn = (–1)

n

n=0

2◦ . For f (x) = xλ

N

n (–1)k ln 2 + . k k=1

An xn , where λ is an arbitrary number (λ > –1), the solution has the

n=0

form y(x) = xλ

N An n x , Bn

1

tλ+n dt . 1+t

Bn = 0

n=0

N 3◦ . For f (x) = ln x An xn , the solution has the form n=0 N N An n An In n x + x , Bn Bn2 n=0 n=0 2 n n (–1)k (–1)k n n π Bn = (–1) ln 2 + . , In = (–1) + k 12 k2

y(x) = ln x

k=1

N

4◦ . For f (x) =

k=1

An ln x)n , the solution of the equation has the form

n=0

y(x) =

N

An Yn (x),

n=0

where the functions Yn = Yn (x) are given by n λ x d Yn (x) = , dλn I(λ) λ=0 N

5◦ . For f (x) =

An cos(λn ln x) +

n=1

N

I(λ) = 0

1

z λ dz . 1+z

Bn sin(λn ln x), the solution of the equation has the

n=1

form y(x) =

N

Cn cos(λn ln x) +

n=1

N

Dn sin(λn ln x),

n=1

where the constants Cn and Dn are found by the method of undetermined coefficients. 6◦ . For arbitrary f (x), the transformation x = 12 e2z ,

t = 12 e2τ ,

y(t) = e–τ w(τ ),

f (x) = e–z g(z)

leads to an integral equation with difference kernel of the form 1.9.26: z w(τ ) dτ = g(z). cosh(z – τ) –∞

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x

y(t) dt

27.

ax + bt

0

= f (x),

a > 0,

a + b > 0.

1◦ . For a polynomial right-hand side, f (x) =

N

An xn , the solution has the form

n=0

y(x) =

N An n x , Bn

N

tn dt . a + bt

0

n=0

2◦ . For f (x) = xλ

1

Bn =

An xn , where λ is an arbitrary number (λ > –1), the solution has the

n=0

form y(x) = xλ

N An n x , Bn

1

Bn = 0

n=0

tλ+n dt . a + bt

N 3◦ . For f (x) = ln x An xn , the solution has the form n=0

y(x) = ln x

N N An n An Cn n x – x , Bn Bn2 n=0

1

tn dt , a + bt

Bn = 0

n=0

1

Cn = 0

tn ln t dt. a + bt

4◦ . For some other special forms of the right-hand side (see items 4 and 5, equation 1.1.26), the solution may be found by the method of undetermined coefficients.

x

y(t) dt

28.

ax2 + bt2

0

= f (x),

a > 0,

a + b > 0.

1◦ . For a polynomial right-hand side, f (x) =

N

An xn , the solution has the form

n=0

y(x) =

N An n+1 x , Bn

1

Bn = 0

n=0

tn+1 dt . a + bt2

Example. For a = b = 1 and f (x) = Ax2 + Bx + C, the solution of the integral equation is: y(x) =

2◦ . For f (x) = xλ

N

2A 4B 2 2C x3 + x + x. 1 – ln 2 4–π ln 2

An xn , where λ is an arbitrary number (λ > –1), the solution has the

n=0

form y(x) = xλ

N An n+1 x , Bn

Bn = 0

n=0

1

tλ+n+1 dt . a + bt2

N 3◦ . For f (x) = ln x An xn , the solution has the form n=0 N N An n+1 An Cn n+1 y(x) = ln x x – x , Bn Bn2 n=0

n=0

Bn = 0

1

tn+1 dt , a + bt2

Cn = 0

1

tn+1 ln t dt. a + bt2

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x

y(t) dt

29.

axm + btm

0

= f (x),

a > 0,

a + b > 0,

1◦ . For a polynomial right-hand side, f (x) =

N

m = 1, 2, . . .

An xn , the solution has the form

n=0

N An m+n–1 x , Bn

y(x) =

0

n=0

2◦ . For f (x) = xλ

N

1

Bn =

tm+n–1 dt . a + btm

An xn , where λ is an arbitrary number (λ > –1), the solution has the

n=0

form y(x) = xλ

N An m+n–1 x , Bn

1

Bn = 0

n=0

tλ+m+n–1 dt . a + btm

N 3◦ . For f (x) = ln x An xn , the solution has the form n=0 N N An m+n–1 An Cn m+n–1 x – x , Bn Bn2 n=0 n=0 1 m+n–1 tm+n–1 dt t ln t , Cn = dt. m a + btm a + bt 0

y(x) = ln x Bn = 0

1

1.1-6. Kernels Containing Square Roots

x

30.

√

x – t y(t) dt = f (x).

a

Differentiating with respect to x, we arrive at Abel’s equation 1.1.36: a

x

y(t) dt √ = 2fx (x). x–t

Solution: 2 d2 y(x) = π dx2 31.

x √

x–

a

x

f (t) dt √ . x–t

√ t y(t) dt = f (x).

a

This is a special case of equation 1.1.44 with µ = 12 . d √ Solution: y(x) = 2 x fx (x) . dx 32.

x

√ √ A x + B t y(t) dt = f (x).

a

This is a special case of equation 1.1.45 with µ = 12 .

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33.

x

√

1 + b x – t y(t) dt = f (x).

a

Differentiating with respect to x, we arrive at Abel’s equation of the second kind 2.1.46: b y(x) + 2 34.

x

a

x

y(t) dt √ = fx (x). x–t

√ √ t x – x t y(t) dt = f (x).

a

√

This is a special case of equation 1.9.11 with g(x) = 35.

x

√ √ At x + Bx t y(t) dt = f (x).

a

√

This is a special case of equation 1.9.12 with g(x) =

x

36. a

x and h(x) = x.

x and h(t) = t.

y(t) dt = f (x). √ x–t

Abel’s equation. Solution: y(x) =

1 d π dx

x

a

f (t) dt 1 f (a) √ + = √ x–t π x–a π

x

a

ft (t) dt √ . x–t

• Reference: E. T. Whittacker and G. N. Watson (1958).

x

b+ √

37. a

1

y(t) dt = f (x).

x–t

Let us rewrite the equation in the form a

x

y(t) dt √ = f (x) – b x–t

x

y(t) dt. a

Assuming the right-hand side to be known, we solve this equation as Abel’s equation 1.1.36. After some manipulations, we arrive at Abel’s equation of the second kind 2.1.46: y(x) +

x

38. a

1 1 √ – √ x t

b π

a

x

y(t) dt √ = F (x), x–t

where

F (x) =

1 d π dx

a

x

f (t) dt √ . x–t

y(t) dt = f (x).

This is a special case of equation 1.1.44 with µ = – 12 . Solution: y(x) = –2 x3/2 fx (x) x , a > 0.

x

39. a

A B √ + √ y(t) dt = f (x). x t

This is a special case of equation 1.1.45 with µ = – 12 .

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40.

x

y(t) dt = f (x). √ x2 – t 2 a x 2 d tf (t) dt √ Solution: y = . π dx a x2 – t2 • Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

x

√

41. 0

y(t) dt

= f (x),

ax2 + bt2

a > 0,

a + b > 0.

1◦ . For a polynomial right-hand side, f (x) =

N

An xn , the solution has the form

n=0

y(x) =

N An n x , Bn

0

n=0

2◦ . For f (x) = xλ

N

1

Bn =

tn dt √ . a + bt2

An xn , where λ is an arbitrary number (λ > –1), the solution has the

n=0

form y(x) = xλ

N An n x , Bn

1

Bn = 0

n=0

tλ+n dt √ . a + bt2

N 3◦ . For f (x) = ln x An xn , the solution has the form n=0

N N An n An Cn n y(x) = ln x x – x , Bn Bn2 n=0

Bn = 0

n=0

N

4◦ . For f (x) =

1

tn dt √ , a + bt2

Cn = 0

1

tn ln t √ dt. a + bt2

An ln x)n , the solution of the equation has the form

n=0

y(x) =

N

An Yn (x),

n=0

where the functions Yn = Yn (x) are given by Yn (x) = N

5◦ . For f (x) =

λ x dn , n dλ I(λ) λ=0

An cos(λn ln x) +

n=1

N

I(λ) = 0

1

z λ dz √ . a + bz 2

Bn sin(λn ln x), the solution of the equation has the

n=1

form y(x) =

N n=1

Cn cos(λn ln x) +

N

Dn sin(λn ln x),

n=1

where the constants Cn and Dn are found by the method of undetermined coefficients.

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1.1-7. Kernels Containing Arbitrary Powers

x

(x – t)λ y(t) dt = f (x),

42.

f (a) = 0,

0 < λ < 1.

a

Differentiating with respect to x, we arrive at the generalized Abel equation 1.1.46: x y(t) dt 1 = fx (x). 1–λ (x – t) λ a Solution:

43.

d2 y(x) = k 2 dx

a

x

f (t) dt , (x – t)λ

k=

sin(πλ) . πλ

• Reference: F. D. Gakhov (1977). x (x – t)µ y(t) dt = f (x). a

For µ = 0, 1, 2, . . . , see equations 1.1.1, 1.1.2, 1.1.4, 1.1.12, and 1.1.23. For –1 < µ < 0, see equation 1.1.42. Set µ = n – λ, where n = 1, 2, . . . and 0 ≤ λ < 1, and f (a) = fx (a) = · · · = fx(n–1) (a) = 0. On differentiating the equation n times, we arrive at an equation of the form 1.1.46: x y(t) dτ Γ(µ – n + 1) (n) f (x), = λ (x – t) Γ(µ + 1) x a where Γ(µ) is the gamma function. Example. Set f (x) = Axβ , where β ≥ 0, and let µ > –1 and µ – β ≠ 0, 1, 2, . . . In this case, the solution has A Γ(β + 1) the form y(x) = xβ–µ–1 . Γ(µ + 1) Γ(β – µ)

44.

• Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971). x (xµ – tµ )y(t) dt = f (x). a

45.

This is a special case of equation 1.9.2 with g(x) = xµ . 1 1–µ Solution: y(x) = x fx (x) x . µ x

Axµ + Btµ y(t) dt = f (x). a

46.

For B = –A, see equation 1.1.44.This is aspecial case of equation 1.9.4 with g(x) = xµ . x Aµ Bµ 1 d Solution: y(x) = t– A+B ft (t) dt . x– A+B A + B dx a x y(t) dt = f (x), 0 < λ < 1. λ a (x – t) The generalized Abel equation. Solution: x x sin(πλ) d f (t) dt ft (t) dt sin(πλ) f (a) y(x) = . = + 1–λ π dx a (x – t)1–λ π (x – a)1–λ a (x – t) • Reference: E. T. Whittacker and G. N. Watson (1958).

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x

b+

47. a

1 (x – t)

y(t) dt = f (x), λ

0 < λ < 1.

Rewrite the equation in the form a

x

y(t) dt = f (x) – b (x – t)λ

x

y(t) dt. a

Assuming the right-hand side to be known, we solve this equation as the generalized Abel equation 1.1.46. After some manipulations, we arrive at Abel’s equation of the second kind 2.1.60: x b sin(πλ) x y(t) dt f (t) dt sin(πλ) d y(x) + = F (x), where F (x) = . 1–λ π (x – t) π dx (x – t)1–λ a a 48.

x √

x–

√ λ t y(t) dt = f (x),

0 < λ < 1.

a

Solution: 2 x k √ d f (t) dt √ y(x) = x √ λ , √ √ dx x a t x– t

x

49. a

y(t) dt √ λ = f (x), √ x– t

Solution: y(x) = 50.

x

k=

sin(πλ) . πλ

0 < λ < 1.

sin(πλ) d 2π dx

a

x

f (t) dt √ √ √ 1–λ . t x– t

Axλ + Btµ y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = Axλ and h(t) = Btµ . 51.

1 + A(xλ tµ – xλ+µ ) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.13 with g(x) = Axµ and h(x) = xλ . Solution: λ x

Aµ –λ d x y(x) = Φ(x) = exp – t f (t) t Φ(t) dt , xµ+λ . dx Φ(x) a µ+λ 52.

x

Axβ tγ + Bxδ tλ y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axβ , h1 (t) = tγ , g2 (x) = Bxδ , and h2 (t) = tλ . 53.

Axλ (tµ – xµ ) + Bxβ (tγ – xγ ) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.45 with g1 (x) = Axλ , h1 (x) = xµ , g2 (x) = Bxβ , and h2 (x) = xγ .

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54.

x

Axλ tµ + Bxλ+β tµ–β – (A + B)xλ+γ tµ–γ y(t) dt = f (x).

a

55.

This is a special case of equation 1.9.47 with g(x) = x. x tσ (xµ – tµ )λ y(t) dt = f (x), σ > –1, µ > 0,

λ > –1.

a

The transformation τ = tµ , z = xµ , w(τ ) = tσ–µ+1 y(t) leads to an equation of the form 1.1.42: z (z – τ )λ w(τ ) dτ = F (z), A µ

56.

1/µ

where A = a and F (z) = µf (z ). Solution with –1 < λ < 0: x µ sin(πλ) d µ–1 µ µ –1–λ y(x) = – t (x – t ) f (t) dt . πxσ dx a x y(t) dt = f (x). µ 0 (x + t) This is a special case of equation 1.1.57 with λ = 1 and a = b = 1. The transformation x = 12 e2z ,

57.

t = 12 e2τ ,

f (x) = e–µz g(z)

y(t) = e(µ–2)τ w(τ ),

leads to an equation with difference kernel of the form 1.9.26: z w(τ ) dτ = g(z). µ cosh (z – τ ) –∞ x y(t) dt = f (x), a > 0, a + b > 0. λ λ µ 0 (ax + bt ) 1◦ . The substitution t = xz leads to a special case of equation 3.8.45: 1 y(xz) dz = xλµ–1 f (x). λ µ 0 (a + bz ) n 2◦ . For a polynomial right-hand side, f (x) = Am xm , the solution has the form m=0

y(x) = xλµ–1

n m=0

Am m x , Im

Im = 0

1

(1)

z m+λµ–1 dz . (a + bz λ )µ

The integrals Im are supposed to be convergent. 3◦ . The solution structure for some other right-hand sides of the integral equation may be obtained using (1) and the results presented for the more general equation 3.8.45 (see also equations 3.8.26–3.8.32).

58.

4◦ . For a = b, the equation can be reduced, just as equation 1.1.56, to an integral equation with difference kernel of the form 1.9.26. √ √

2λ √

2λ x √ x+ x–t + x– x–t y(t) dt = f (x). √ 2tλ x – t a The equation can be rewritten in terms of the Gaussian hypergeometric functions in the form x x

y(t) dt = f (x), where γ = 12 . (x – t)γ–1 F λ, –λ, γ; 1 – t a See 1.8.86 for the solution of this equation.

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1.2. Equations Whose Kernels Contain Exponential Functions 1.2-1. Kernels Containing Exponential Functions

x

eλ(x–t) y(t) dt = f (x).

1. a

Solution: y(x) = fx (x) – λf (x). Example. In the special case a = 0 and f (x) = Ax, the solution has the form y(x) = A(1 – λx).

x

eλx+βt y(t) dt = f (x).

2. a

Solution: y(x) = e–(λ+β)x fx (x) – λf (x) . Example. In the special case a = 0 and f (x) = A sin(γx), the solution has the form y(x) = Ae–(λ+β)x × [γ cos(γx) – λ sin(γx)].

3.

x

eλ(x–t) – 1 y(t) dt = f (x),

a

Solution: y(x) = 4.

1 λ fxx (x)

f (a) = fx (a) = 0.

– fx (x).

x

eλ(x–t) + b y(t) dt = f (x).

a

For b = –1, see equation 1.2.3. Differentiating with respect to x yields an equation of the form 2.2.1: x λ f (x) y(x) + eλ(x–t) y(t) dt = x . b+1 a b+1 Solution:

5.

x

f (x) λ y(x) = x – b + 1 (b + 1)2

a

x

λb (x – t) ft (t) dt. exp b+1

eλx+βt + b y(t) dt = f (x).

a

For β = –λ, see equation 1.2.4. This is a special case of equation 1.9.15 with g1 (x) = eλx , h1 (t) = eβt , g2 (x) = 1, and h2 (t) = b. 6.

x

eλx – eλt y(t) dt = f (x),

a

f (a) = fx (a) = 0.

λx This is a special case of equation 1.9.2 with g(x) =e . 1 Solution: y(x) = e–λx f (x) – fx (x) . λ xx

7.

x

eλx – eλt + b y(t) dt = f (x).

a

For b = 0, see equation 1.2.6. This is a special case of equation 1.9.3 with g(x) = eλx . Solution: λt λx 1 λ λx x e –e y(x) = fx (x) – 2 e exp ft (t) dt. b b b a

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8.

x

Aeλx + Beλt y(t) dt = f (x).

a λx For B = –A, see equation 1.2.6. This is a special case 1.9.4 with g(x) =e . xof equation

Bλ 1 d Aλ exp – Solution: y(x) = exp – x t ft (t) dt . A + B dx A+B A +B a

9.

x

Aeλx + Beλt + C y(t) dt = f (x).

a

This is a special case of equation 1.9.5 with g(x) = eλx . 10.

x

Aeλx + Beµt y(t) dt = f (x).

a

For λ = µ, see equation 1.2.8. This is a special case of equation 1.9.6 with g(x) = Aeλx and h(t) = Beµt . 11.

x

eλ(x–t) – eµ(x–t) y(t) dt = f (x),

a

Solution: y(x) = 12.

f (a) = fx (a) = 0.

1 f – (λ + µ)fx + λµf , λ – µ xx

f = f (x).

x

Aeλ(x–t) + Beµ(x–t) y(t) dt = f (x).

a

For B = –A, see equation 1.2.11. This is a special case of equation 1.9.15 with g1 (x) = Aeλx , h1 (t) = e–λt , g2 (x) = Beµx , and h2 (t) = e–µt . Solution: y(x) = 13.

x eλx d f (t) dt e(µ–λ)x Φ(x) , A + B dx eµt t Φ(t) a

B(λ – µ) Φ(x) = exp x . A+B

x

Aeλ(x–t) + Beµ(x–t) + C y(t) dt = f (x).

a

This is a special case of equation 1.2.14 with β = 0. 14.

x

Aeλ(x–t) + Beµ(x–t) + Ceβ(x–t) y(t) dt = f (x).

a

Differentiating the equation with respect to x yields (A + B + C)y(x) +

x

Aλeλ(x–t) + Bµeµ(x–t) + Cβeβ(x–t) y(t) dt = fx (x).

a

Eliminating the term with eβ(x–t) with the aid of the original equation, we arrive at an equation of the form 2.2.10: x (A + B + C)y(x) + A(λ – β)eλ(x–t) + B(µ – β)eµ(x–t) y(t) dt = fx (x) – βf (x). a

In the special case A + B + C = 0, this is an equation of the form 1.2.12.

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15.

Aeλ(x–t) + Beµ(x–t) + Ceβ(x–t) – A – B – C y(t) dt = f (x),

x

a

f (a) = fx (a) = 0.

Differentiating with respect to x, we arrive at an equation of the form 1.2.14: x Aλeλ(x–t) + Bµeµ(x–t) + Cβeβ(x–t) y(t) dt = fx (x). a

16.

x

eλx+µt – eµx+λt y(t) dt = f (x),

a

17.

f (a) = fx (a) = 0.

This is a special case of equation 1.9.11 with g(x) = eλx and h(t) = eµt . Solution: f – (λ + µ)fx (x) + λµf (x) y(x) = xx . (λ – µ) exp[(λ + µ)x] x λx+µt

+ Beµx+λt y(t) dt = f (x). Ae a

For B = –A, see equation 1.2.16. This is a special case of equation 1.9.12 with g(x) = eλx and h(t) = eµt . Solution: x µ–λ

1 d d f (t) A B y(x) = dt , Φ(x) = exp (x) Φ (t) Φ x . (A + B)eµx dx dt eµt A+B a 18.

x

Aeλx+µt + Beβx+γt y(t) dt = f (x).

a

19.

This is a special case of equation 1.9.15 with g1 (x) = Aeλx , h1 (t) = eµt , g2 (x) = Beβx , and h2 (t) = eγt . x 2λx

Ae + Be2βt + Ceλx + Deβt + E y(t) dt = f (x). a

20.

This is a special case of equation 1.9.6 with g(x) = Ae2λx +Ceλx and h(t) = Be2βt +Deβt +E. x

λx+βt + Be2βt + Ceλx + Deβt + E y(t) dt = f (x). Ae a

21.

This is a special case of equation 1.9.15 with g1 (x) = eλx , h1 (t) = Aeβt + D, and g2 (x) = 1, h2 (t) = Be2βt + Deβt + E. x

2λx + Beλx+βt + Ceλx + Deβt + E y(t) dt = f (x). Ae a

22.

This is a special case of equation 1.9.15 with g1 (x) = Beλx + D, h1 (t) = eβt , and g2 (x) = Ae2λx + Ceλx + E, h2 (t) = 1. x 1 + Aeλx (eµt – eµx )y(t) dt = f (x). a

This is a special case of equation 1.9.13 with g(x) = eµx and h(x) = Aeλx . Solution: x d f (t) dt Aµ (λ+µ)x y(x) = . eλx Φ(x) , Φ(x) = exp e dx eλt t Φ(t) λ+µ a

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23.

Aeλx (eµx – eµt ) + Beβx (eγx – eγt ) y(t) dt = f (x).

x

a

24.

This is a special case of equation 1.9.45 with g1 (x) = Aeλx , h1 (t) = –eµt , g2 (x) = Beβx , and h1 (t) = –eγt . x A exp(λx + µt) + B exp[(λ + β)x + (µ – β)t] a – (A + B) exp[(λ + γ)x + (µ – γ)t] y(t) dt = f (x). This is a special case of equation 1.9.47 with g1 (x) = eλx .

25.

x

eλx – eλt

n

y(t) dt = f (x),

n = 1, 2, . . .

a

Solution: y(x) =

x

26.

√

1 λx 1 d n+1 e f (x). λn n! eλx dx

eλx – eλt y(t) dt = f (x),

λ > 0.

a

Solution: y(x) = 27.

x

√

y(t) dt

eλx – eλt Solution:

= f (x),

2 λx –λx d 2 e e π dx

x

a

eλt f (t) dt √ . eλx – eλt

λ > 0.

a

y(x) =

λ d π dx

x

eλt f (t) dt √ . eλx – eλt

a

x

(eλx – eλt )µ y(t) dt = f (x),

28.

λ > 0,

0 < µ < 1.

a

Solution: λx

y(x) = ke

x

29. a

y(t) dt (eλx – eλt )µ

–λx

e

= f (x),

d 2 dx

λ > 0,

x

a

eλt f (t) dt , (eλx – eλt )µ

sin(πµ) . πµ

0 < µ < 1.

Solution: y(x) =

k=

λ sin(πµ) d π dx

a

x

eλt f (t) dt . – eλt )1–µ

(eλx

1.2-2. Kernels Containing Power-Law and Exponential Functions 30.

x

A(x – t) + Beλ(x–t) y(t) dt = f (x).

a

Differentiating with respect to x, we arrive at an equation of the form 2.2.4: x By(x) + A + Bλeλ(x–t) y(t) dt = fx (x). a

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x

(x – t)eλ(x–t) y(t) dt = f (x),

31. a

32.

f (a) = fx (a) = 0.

Solution: y(x) = fxx (x) – 2λfx (x) + λ2 f (x). x (Ax + Bt + C)eλ(x–t) y(t) dt = f (x). a

The substitution u(x) = e–λx y(x) leads to an equation of the form 1.1.3: x (Ax + Bt + C)u(t) dt = e–λx f (x). a

x

(Axeλt + Bteµx )y(t) dt = f (x).

33. a

34.

This is a special case of equation 1.9.15 with g1 (x) = Ax, h1 (t) = eλt , and g2 (x) = Beµx , h2 (t) = t. x Axeλ(x–t) + Bteµ(x–t) y(t) dt = f (x). a

35.

This is a special case of equation 1.9.15 with g1 (x) = Axeλx , h1 (t) = e–λt , g2 (x) = Beµx , and h2 (t) = te–µt . x (x – t)2 eλ(x–t) y(t) dt = f (x), f (a) = fx (a) = fxx (a) = 0. a

Solution: y(x) =

1 2

(x) + 3λ2 fx (x) – λ3 f (x) . fxxx (x) – 3λfxx

x

(x – t)n eλ(x–t) y(t) dt = f (x),

36.

n = 1, 2, . . .

a

37.

It is assumed that f (a) = fx (a) = · · · = fx(n) (a) = 0. 1 λx dn+1 –λx e f (x) . Solution: y(x) = e n+1 n! dx x (Axβ + Beλt )y(t) dt = f (x). a

38.

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = Beλt . x (Aeλx + Btβ )y(t) dt = f (x). a

39.

This is a special case of equation 1.9.6 with g(x) = Aeλx and h(t) = Btβ . x (Axβ eλt + Btγ eµx )y(t) dt = f (x).

40.

This is a special case of equation 1.9.15 with g1 (x) = Axβ , h1 (t) = eλt , g2 (x) = Beµx , and h2 (t) = tγ . x √ eλ(x–t) x – t y(t) dt = f (x).

a

a

Solution:

2 d2 y(x) = eλx 2 π dx

a

x

e–λt f (t) dt √ . x–t

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x

41. a

eλ(x–t) y(t) dt = f (x). √ x–t

Solution: 1 d y(x) = eλx π dx

x

e–λt f (t) dt √ . x–t

a

x

(x – t)λ eµ(x–t) y(t) dt = f (x),

42.

0 < λ < 1.

a

Solution: µx

y(x) = ke

x

43. a

eλ(x–t) (x – t)µ

d2 dx2

y(t) dt = f (x),

x

a

e–µt f (t) dt , (x – t)λ

sin(πµ) λx d y(x) = e π dx 44.

x √

x–

sin(πλ) . πλ

0 < µ < 1.

Solution:

k=

√ λ µ(x–t) t e y(t) dt = f (x),

a

x

e–λt f (t) dt. (x – t)1–µ

0 < λ < 1.

a

The substitution u(x) = e–µx y(x) leads to an equation of the form 1.1.48:

x √

x–

√ λ t u(t) dt = e–µx f (x).

a

x

45. a

eµ(x–t) y(t) dt √ λ = f (x), √ x– t

0 < λ < 1.

The substitution u(x) = e–µx y(x) leads to an equation of the form 1.1.49:

x

a

46.

x

x

u(t) dt √ = e–µx f (x). √ ( x – t)λ

eλ(x–t) y(t) dt = f (x). √ x2 – t 2 a x 2 d te–λt √ Solution: y = eλx f (t) dt. π dx a x2 – t2 exp[λ(x2 – t2 )]y(t) dt = f (x).

47. a

Solution: y(x) = fx (x) – 2λxf (x).

x

[exp(λx2 ) – exp(λt2 )]y(t) dt = f (x).

48. a

This is a special case of equation with g(x) = exp(λx2 ). 1.9.2 1 d fx (x) Solution: y(x) = . 2λ dx x exp(λx2 )

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49.

A exp(λx2 ) + B exp(λt2 ) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.5 with g(x) = exp(λx2 ). 50.

A exp(λx2 ) + B exp(µt2 ) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A exp(λx2 ) and h(t) = B exp(µt2 ).

x

51.

√

x – t exp[λ(x2 – t2 )]y(t) dt = f (x).

a

Solution:

2 d2 exp(λx2 ) 2 π dx

y(x) =

x

52. a

x

exp(–λt2 ) √ f (t) dt. x–t

x

exp(–λt2 ) √ f (t) dt. x–t

a

exp[λ(x2 – t2 )] y(t) dt = f (x). √ x–t

Solution: 1 d exp(λx2 ) π dx

y(x) =

a

x

(x – t)λ exp[µ(x2 – t2 )]y(t) dt = f (x),

53.

0 < λ < 1.

a

Solution: d2 y(x) = k exp(µx ) 2 dx

x

2

a

exp(–µt2 ) f (t) dt, (x – t)λ

k=

sin(πλ) . πλ

x

exp[λ(xβ – tβ )]y(t) dt = f (x).

54. a

Solution: y(x) = fx (x) – λβxβ–1 f (x).

1.3. Equations Whose Kernels Contain Hyperbolic Functions 1.3-1. Kernels Containing Hyperbolic Cosine

x

cosh[λ(x – t)]y(t) dt = f (x).

1. a

Solution: y(x) = fx (x) – λ2

x

f (x) dx. a

2.

x

cosh[λ(x – t)] – 1 y(t) dt = f (x),

a

Solution: y(x) =

(x) = 0. f (a) = fx (a) = fxx

1 f (x) – fx (x). λ2 xxx

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3.

x

cosh[λ(x – t)] + b y(t) dt = f (x).

a

For b = 0, see equation 1.3.1. For b = –1, see equation 1.3.2. For λ = 0, see equation 1.1.1. Differentiating the equation with respect to x, we arrive at an equation of the form 2.3.16: x λ f (x) y(x) + sinh[λ(x – t)]y(t) dt = x . b+1 a b+1 1◦ . Solution with b(b + 1) < 0: f (x) λ2 y(x) = x – b + 1 k(b + 1)2

x

sin[k(x –

t)]ft (t) dt,

where

k=λ

a

–b . b+1

◦

2 . Solution with b(b + 1) > 0: f (x) λ2 y(x) = x – b + 1 k(b + 1)2

x

sinh[k(x –

t)]ft (t) dt,

where

k=λ

a

b . b+1

x

cosh(λx + βt)y(t) dt = f (x).

4. a

For β = –λ, see equation 1.3.1. Differentiating the equation with respect to x twice, we obtain x cosh[(λ + β)x]y(x) + λ sinh(λx + βt)y(t) dt = fx (x), (1) a x cosh[(λ + β)x]y(x) x + λ sinh[(λ + β)x]y(x) + λ2 cosh(λx + βt)y(t) dt = fxx (x). (2) a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the first-order linear ordinary differential equation wx + λ tanh[(λ + β)x]w = fxx (x) – λ2 f (x),

w = cosh[(λ + β)x]y(x).

(3)

Setting x = a in (1) yields the initial condition w(a) = fx (a). On solving equation (3) with this condition, after some manipulations we obtain the solution of the original integral equation in the form y(x) =

1 λ sinh[(λ + β)x] fx (x) – f (x) cosh[(λ + β)x] cosh2 [(λ + β)x] x λβ + f (t) coshk–2 [(λ + β)t] dt, coshk+1 [(λ + β)x] a

k=

λ . λ+β

x

[cosh(λx) – cosh(λt)]y(t) dt = f (x).

5. a

6.

This is a special case of equation 1.9.2 with g(x) = cosh(λx). 1 d fx (x) Solution: y(x) = . λ dx sinh(λx) x [A cosh(λx) + B cosh(λt)]y(t) dt = f (x). a

For B = –A, see equation 1.3.5. This = cosh(λx). is a special case ofx equation 1.9.4Bwith g(x) – – A 1 d Solution: y(x) = cosh(λt) A+B ft (t) dt . cosh(λx) A+B A + B dx a

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7.

x

A cosh(λx) + B cosh(µt) + C y(t) dt = f (x).

a

8.

This is a special case of equation 1.9.6 with g(x) = A cosh(λx) and h(t) = B cosh(µt) + C. x A1 cosh[λ1 (x – t)] + A2 cosh[λ2 (x – t)] y(t) dt = f (x).

9.

The equation is equivalent to the equation x B1 sinh[λ1 (x – t)] + B2 sinh[λ2 (x – t)] y(t) dt = F (x), a x A1 A2 B1 = , B2 = , F (x) = f (t) dt, λ1 λ2 a of the form 1.3.41. (Differentiating this equation yields the original equation.) x cosh2 [λ(x – t)]y(t) dt = f (x).

a

a

Differentiation yields an equation of the form 2.3.16: x y(x) + λ sinh[2λ(x – t)]y(t) dt = fx (x). a

Solution: y(x) = fx (x) –

10.

11.

2λ2 k

x

sinh[k(x – t)]ft (t) dt,

where

√ k = λ 2.

a

f (a) = fx (a) = 0. cosh2 (λx) – cosh2 (λt) y(t) dt = f (x), a 1 d fx (x) Solution: y(x) = . λ dx sinh(2λx) x A cosh2 (λx) + B cosh2 (λt) y(t) dt = f (x). x

a

12.

For B = –A, see equation 1.3.10. This is a special case of equation 1.9.4 with g(x) = cosh2 (λx). Solution: x – 2B – 2A 1 d A+B A+B y(x) = cosh(λt) cosh(λx) ft (t) dt . A + B dx a x A cosh2 (λx) + B cosh2 (µt) + C y(t) dt = f (x). a

13.

This is a special case of equation 1.9.6 with g(x) = A cosh2 (λx), and h(t) = B cosh2 (µt) + C. x cosh[λ(x – t)] cosh[λ(x + t)]y(t) dt = f (x). a

Using the formula cosh(α – β) cosh(α + β) = 12 [cos(2α) + cos(2β)],

α = λx,

β = λt,

we transform the original equation to an equation of the form 1.4.6 with A = B = 1: x [cosh(2λx) + cosh(2λt)]y(t) dt = 2f (x). a

Solution:

x d ft (t) dt 1 √ √ y(x) = . dx cosh(2λx) a cosh(2λt)

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x

[cosh(λx) cosh(µt) + cosh(βx) cosh(γt)]y(t) dt = f (x).

14. a

This is a special case of equation 1.9.15 with g1 (x) = cosh(λx), h1 (t) = cosh(µt), g2 (x) = cosh(βx), and h2 (t) = cosh(γt).

x

cosh3 [λ(x – t)]y(t) dt = f (x).

15. a

Using the formula cosh3 β = a

16.

3 4

cosh[3λ(x – t)] +

cosh β, we arrive at an equation of the form 1.3.8: 3 4

cosh[λ(x – t)] y(t) dt = f (x).

cosh3 (λx) – cosh3 (λt) y(t) dt = f (x),

Solution: y(x) =

17.

1 4

cosh 3β +

x

a

x

1 4

f (a) = fx (a) = 0.

1 d fx (x) . 3λ dx sinh(λx) cosh2 (λx)

A cosh3 (λx) + B cosh3 (λt) y(t) dt = f (x).

x

a

For B = –A, see equation 1.3.16. This is a special case of equation 1.9.4 with g(x) = cosh3 (λx). Solution: x – 3B – 3A 1 d A+B A+B y(x) = ft (t) dt . cosh(λt) cosh(λx) A + B dx a 18.

x

A cosh2 (λx) cosh(µt) + B cosh(βx) cosh2 (γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = A cosh2 (λx), h1 (t) = cosh(µt), g2 (x) = B cosh(βx), and h2 (t) = cosh2 (γt).

x

cosh4 [λ(x – t)]y(t) dt = f (x).

19. a

Let us transform the kernel of the integral equation using the formula cosh4 β =

1 8

cosh 4β +

1 2

cosh 2β + 38 ,

where

β = λ(x – t),

and differentiate the resulting equation with respect to x. Then we obtain an equation of the form 2.3.18: x 1 y(x) + λ 2 sinh[4λ(x – t)] + sinh[2λ(x – t)] y(t) dt = fx (x). a

x

[cosh(λx) – cosh(λt)]n y(t) dt = f (x),

20.

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 sinh(λx) 1 d Solution: y(x) = f (x). λn n! sinh(λx) dx

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x

21.

√

cosh x – cosh t y(t) dt = f (x).

a

Solution: y(x) = 22.

x

√

y(t) dt

cosh x – cosh t Solution:

1 2 d 2 x sinh t f (t) dt √ sinh x . π sinh x dx cosh x – cosh t a

= f (x).

a

1 d y(x) = π dx

x

√

a

sinh t f (t) dt . cosh x – cosh t

x

(cosh x – cosh t)λ y(t) dt = f (x),

23.

0 < λ < 1.

a

Solution: y(x) = k sinh x

1 d 2 sinh x dx

a

x

sinh t f (t) dt , (cosh x – cosh t)λ

k=

sin(πλ) . πλ

x

(coshµ x – coshµ t)y(t) dt = f (x).

24. a

25.

µ This is a special case of equation 1.9.2 with g(x) = cosh x. 1 d fx (x) Solution: y(x) = . µ dx sinh x coshµ–1 x x

A coshµ x + B coshµ t y(t) dt = f (x). a

For B = –A, see equation 1.3.24. This is a special case of equation 1.9.4 with g(x) = coshµ x. Solution: x – Bµ – Aµ 1 d y(x) = cosh(λt) A+B ft (t) dt . cosh(λx) A+B A + B dx a 26.

x

y(t) dt

(cosh x – cosh t)λ Solution:

= f (x),

0 < λ < 1.

a

y(x) =

sin(πλ) d π dx

x

(x – t) cosh[λ(x – t)]y(t) dt = f (x),

27. a

x

sinh t f (t) dt . (cosh x – cosh t)1–λ

a

f (a) = fx (a) = 0.

Differentiating the equation twice yields x x y(x) + 2λ sinh[λ(x – t)]y(t) dt + λ2 (x – t) cosh[λ(x – t)]y(t) dt = fxx (x). a

a

Eliminating the third term on the right-hand side with the aid of the original equation, we arrive at an equation of the form 2.3.16: x y(x) + 2λ sinh[λ(x – t)]y(t) dt = fxx (x) – λ2 f (x). a

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31.

√

cosh λ x – t y(t) dt = f (x). √ x–t a Solution: √

x cos λ x – t 1 d √ y(x) = f (t) dt. π dx a x–t √

x cosh λ x2 – t2 y(t) dt = f (x). √ x2 – t 2 0 Solution:

√ x cos λ x2 – t2 2 d √ y(x) = t f (t) dt. π dx 0 x2 – t2 √

∞ cosh λ t2 – x2 y(t) dt = f (x). √ t 2 – x2 x Solution:

√ ∞ cos λ t2 – x2 2 d √ y(x) = – t f (t) dt. π dx x t2 – x2 x Axβ + B coshγ (λt) + C]y(t) dt = f (x).

32.

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B coshγ (λt) + C. x A coshγ (λx) + Btβ + C]y(t) dt = f (x).

33.

This is a special case of equation 1.9.6 with g(x) = A coshγ (λx) and h(t) = Btβ + C. x

Axλ coshµ t + Btβ coshγ x y(t) dt = f (x).

28.

29.

30.

x

a

a

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = coshµ t, g2 (x) = B coshγ x, and h2 (t) = tβ . 1.3-2. Kernels Containing Hyperbolic Sine

x

sinh[λ(x – t)]y(t) dt = f (x),

34. a

Solution: y(x) = 35.

x

f (a) = fx (a) = 0.

1 f (x) – λf (x). λ xx

sinh[λ(x – t)] + b y(t) dt = f (x).

a

Differentiating the equation with respect to x, we arrive at an equation of the form 2.3.3: λ x 1 y(x) + cosh[λ(x – t)]y(t) dt = fx (x). b a b Solution: x 1 y(x) = fx (x) + R(x – t)ft (t) dt, b a √ λ λx λ λ 1 + 4b2 R(x) = 2 exp – sinh(kx) – cosh(kx) , k = . b 2b 2bk 2b

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x

sinh(λx + βt)y(t) dt = f (x).

36. a

For β = –λ, see equation 1.3.34. Assume that β ≠ –λ. Differentiating the equation with respect to x twice yields

x

cosh(λx + βt)y(t) dt = fx (x), x sinh[(λ + β)x]y(x) x + λ cosh[(λ + β)x]y(x) + λ2 sinh(λx + βt)y(t) dt = fxx (x). sinh[(λ + β)x]y(x) + λ

(1)

a

(2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the first-order linear ordinary differential equation wx + λ coth[(λ + β)x]w = fxx (x) – λ2 f (x),

w = sinh[(λ + β)x]y(x).

(3)

Setting x = a in (1) yields the initial condition w(a) = fx (a). On solving equation (3) with this condition, after some manipulations we obtain the solution of the original integral equation in the form y(x) =

1 λ cosh[(λ + β)x] f (x) f (x) – sinh[(λ + β)x] x sinh2 [(λ + β)x] x λβ – f (t) sinhk–2 [(λ + β)t] dt, k+1 sinh [(λ + β)x] a

x

[sinh(λx) – sinh(λt)]y(t) dt = f (x),

37. a

k=

λ . λ+β

f (a) = fx (a) = 0.

This is a special case of equation 1.9.2 with g(x) = sinh(λx). 1 d fx (x) Solution: y(x) = . λ dx cosh(λx)

x

[A sinh(λx) + B sinh(λt)]y(t) dt = f (x).

38. a

For B = –A, see equation 1.3.37. This is a special case equation 1.9.4 with g(x) = sinh(λx). of x – B – A 1 d Solution: y(x) = sinh(λt) A+B ft (t) dt . sinh(λx) A+B A + B dx a

x

[A sinh(λx) + B sinh(µt)]y(t) dt = f (x).

39. a

This is a special case of equation 1.9.6 with g(x) = A sinh(λx), and h(t) = B sinh(µt). 40.

x

µ sinh[λ(x – t)] – λ sinh[µ(x – t)] y(t) dt = f (x).

a It is assumed that f (a) = fx (a) = fxx (a) = fxxx (a) = 0. Solution: f – (λ2 + µ2 )fxx + λ2 µ2 f y(x) = xxxx , µλ3 – λµ3

f = f (x).

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41.

x

A1 sinh[λ1 (x – t)] + A2 sinh[λ2 (x – t)] y(t) dt = f (x),

a ◦

1 . Introduce the notation x I1 = sinh[λ1 (x – t)]y(t) dt, ax J1 = cosh[λ1 (x – t)]y(t) dt,

f (a) = fx (a) = 0.

x

I2 =

sinh[λ2 (x – t)]y(t) dt, ax

J2 =

a

cosh[λ2 (x – t)]y(t) dt. a

Let us successively differentiate the integral equation four times. As a result, we have (the first line is the original equation): A1 I1 + A2 I2 = f , f = f (x), A1 λ1 J1 + A2 λ2 J2 = fx , (A1 λ1 + A2 λ2 )y + A1 λ21 I1 + A2 λ22 I2 = fxx , 3 3 (A1 λ1 + A2 λ2 )yx + A1 λ1 J1 + A2 λ2 J2 = fxxx , (A1 λ1 +

A2 λ2 )yxx

+

(A1 λ31

+

A2 λ32 )y

+

(1) (2) (3) (4)

A1 λ41 I1

+

A2 λ42 I2

=

fxxxx .

(5)

Eliminating I1 and I2 from (1), (3), and (5), we arrive at the following second-order linear ordinary differential equation with constant coefficients: (A1 λ1 + A2 λ2 )yxx – λ1 λ2 (A1 λ2 + A2 λ1 )y = fxxxx – (λ21 + λ22 )fxx + λ21 λ22 f .

(6)

The initial conditions can be obtained by substituting x = a into (3) and (4): (A1 λ1 + A2 λ2 )y(a) = fxx (a),

(A1 λ1 + A2 λ2 )yx (a) = fxxx (a).

(7)

Solving the differential equation (6) under conditions (7) allows us to find the solution of the integral equation. 2◦ . Denote ∆ = λ1 λ2

A1 λ2 + A2 λ1 . A1 λ1 + A2 λ2

2.1. Solution for ∆ > 0: (A1 λ1 + A2 λ2 )y(x) = fxx (x) + Bf (x) + C

k=

√

∆,

B = ∆ – λ21 – λ22 ,

(A1 λ1 + A2 λ2 )y(x) = √

sinh[k(x – t)]f (t) dt, a

1 C = √ ∆2 – (λ21 + λ22 )∆ + λ21 λ22 . ∆

2.2. Solution for ∆ < 0:

k=

x

fxx (x)

+ Bf (x) + C

x

sin[k(x – t)]f (t) dt, a

–∆,

B = ∆ – λ21 – λ22 ,

C= √

1 2 ∆ – (λ21 + λ22 )∆ + λ21 λ22 . –∆

2.3. Solution for ∆ = 0: (A1 λ1 + A2 λ2 )y(x) =

fxx (x)

–

(λ21

+

λ22 )f (x)

+

λ21 λ22

x

(x – t)f (t) dt. a

2.4. Solution for ∆ = ∞: y(x) =

fxxxx – (λ21 + λ22 )fxx + λ21 λ22 f , A1 λ31 + A2 λ32

f = f (x).

In the last case, the relation A1 λ1 + A2 λ2 = 0 is valid, and the right-hand side of the integral equation is assumed to satisfy the conditions f (a) = fx (a) = fxx (a) = fxxx (a) = 0.

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42.

x

A sinh[λ(x – t)] + B sinh[µ(x – t)] + C sinh[β(x – t)] y(t) dt = f (x).

a

It assumed that f (a) = fx (a) = 0. Differentiating the integral equation twice yields x 2 (Aλ + Bµ + Cβ)y(x) + Aλ sinh[λ(x – t)] + Bµ2 sinh[µ(x – t)] y(t) dt a x + Cβ 2 sinh[β(x – t)]y(t) dt = fxx (x). a

Eliminating the last integral with the aid of the original equation, we arrive at an equation of the form 2.3.18: (Aλ + Bµ + Cβ)y(x) x + (x) – β 2 f (x). A(λ2 – β 2 ) sinh[λ(x – t)] + B(µ2 – β 2 ) sinh[µ(x – t)] y(t) dt = fxx a

43.

In the special case Aλ + Bµ + Cβ = 0, this is an equation of the form 1.3.41. x sinh2 [λ(x – t)]y(t) dt = f (x), f (a) = fx (a) = fxx (a) = 0. a

Differentiating yields an equation of the form 1.3.34: x 1 sinh[2λ(x – t)]y(t) dt = fx (x). λ a

Solution: y(x) = 12 λ–2 fxxx (x) – 2fx (x). x

45.

sinh2 (λx) – sinh2 (λt) y(t) dt = f (x), f (a) = fx (a) = 0. a 1 d fx (x) Solution: y(x) = . λ dx sinh(2λx) x A sinh2 (λx) + B sinh2 (λt) y(t) dt = f (x).

46.

For B = –A, see equation 1.3.44. This is a special case of equation 1.9.4 with g(x) = sinh2 (λx). Solution: x – 2B – 2A 1 d y(x) = sinh(λt) A+B ft (t) dt . sinh(λx) A+B A + B dx a x A sinh2 (λx) + B sinh2 (µt) y(t) dt = f (x).

44.

a

a

47.

This is a special case of equation 1.9.6 with g(x) = A sinh2 (λx) and h(t) = B sinh2 (µt). x sinh[λ(x – t)] sinh[λ(x + t)]y(t) dt = f (x). a

Using the formula sinh(α – β) sinh(α + β) = 12 [cosh(2α) – cosh(2β)],

α = λx,

β = λt,

we reduce the original equation to an equation of the form 1.3.5: x [cosh(2λx) – cosh(2λt)]y(t) dt = 2f (x). a

1 d fx (x) Solution: y(x) = . λ dx sinh(2λx)

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48.

x

A sinh(λx) sinh(µt) + B sinh(βx) sinh(γt) y(t) dt = f (x).

a

49.

This is a special case of equation 1.9.15 with g1 (x) = A sinh(λx), h1 (t) = sinh(µt), g2 (x) = B sinh(βx), and h2 (t) = sinh(γt). x sinh3 [λ(x – t)]y(t) dt = f (x), f (a) = fx (a) = fxx (a) = fxxx (a) = 0. a

Using the formula sinh3 β = 14 sinh 3β – 34 sinh β, we arrive at an equation of the form 1.3.41: x 1 3 4 sinh[3λ(x – t)] – 4 sinh[λ(x – t)] y(t) dt = f (x). a

50.

sinh3 (λx) – sinh3 (λt) y(t) dt = f (x),

x

a

51.

f (a) = fx (a) = 0.

This is a special case of equation 1.9.2 with g(x) = sinh3 (λx). x A sinh3 (λx) + B sinh3 (λt) y(t) dt = f (x). a

52.

This is a special case of equation 1.9.4 with g(x) = sinh3 (λx). Solution: x – 3B – 3A 1 d y(x) = sinh(λt) A+B ft (t) dt . sinh(λx) A+B A + B dx a x A sinh2 (λx) sinh(µt) + B sinh(βx) sinh2 (γt) y(t) dt = f (x). a

53.

This is a special case of equation 1.9.15 with g1 (x) = A sinh2 (λx), h1 (t) = sinh(µt), g2 (x) = B sinh(βx), and h2 (t) = sinh2 (γt). x sinh4 [λ(x – t)]y(t) dt = f (x). a It is assumed that f (a) = fx (a) = · · · = fxxxx (a) = 0. Let us transform the kernel of the integral equation using the formula

sinh4 β =

1 8

cosh 4β –

1 2

cosh 2β + 38 ,

where

β = λ(x – t),

and differentiate the resulting equation with respect to x. Then we arrive at an equation of the form 1.3.41: x 1 λ 2 sinh[4λ(x – t)] – sinh[2λ(x – t)] y(t) dt = fx (x). a

x

sinhn [λ(x – t)]y(t) dt = f (x),

54.

n = 2, 3, . . .

a

It is assumed that f (a) = fx (a) = · · · = fx(n) (a) = 0. Let us differentiate the equation with respect to x twice and transform the kernel of the resulting integral equation using the formula cosh2 β = 1 + sinh2 β, where β = λ(x – t). Then we have x x n 2 2 2 λ n sinh [λ(x – t)]y(t) dt + λ n(n – 1) sinhn–2 [λ(x – t)]y(t) dt = fxx (x). a

a

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Eliminating the first term on the left-hand side with the aid of the original equation, we obtain

x

sinhn–2 [λ(x – t)]y(t) dt = a

1 λ2 n(n

– 1)

fxx (x) – λ2 n2 f (x) .

This equation has the same form as the original equation, but the exponent of the kernel has been reduced by two. By applying this technique sufficiently many times, we finally arrive at simple integral equations of the form 1.1.1 (for even n) or 1.3.34 (for odd n).

x

55.

√

sinh λ x – t y(t) dt = f (x).

a

Solution: 2 d2 y(x) = πλ dx2

x

56.

√

x

a

√

cos λ x – t √ f (t) dt. x–t

sinh x – sinh t y(t) dt = f (x).

a

Solution: y(x) =

x

√

57. a

y(t) dt sinh x – sinh t

1 2 d 2 x cosh t f (t) dt √ cosh x . π cosh x dx sinh x – sinh t a

= f (x).

Solution: y(x) =

1 d π dx

x

a

cosh t f (t) dt √ . sinh x – sinh t

x

(sinh x – sinh t)λ y(t) dt = f (x),

58.

0 < λ < 1.

a

Solution: y(x) = k cosh x

1 d 2 cosh x dx

a

x

cosh t f (t) dt , (sinh x – sinh t)λ

k=

sin(πλ) . πλ

x

(sinhµ x – sinhµ t)y(t) dt = f (x).

59. a

This is a special case of equation 1.9.2 with g(x) = sinhµ x. 1 d fx (x) Solution: y(x) = . µ dx cosh x sinhµ–1 x 60.

A sinhµ (λx) + B sinhµ (λt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.4 with g(x) = sinhµ (λx). Solution with B ≠ –A: x – Bµ – Aµ 1 d A+B A+B y(x) = ft (t) dt . sinh(λt) sinh(λx) A + B dx a

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x

61. a

y(t) dt (sinh x – sinh t)λ

Solution:

= f (x),

0 < λ < 1.

sin(πλ) d y(x) = π dx

a

x

cosh t f (t) dt . (sinh x – sinh t)1–λ

f (a) = fx (a) = fxx (a) = 0.

(x – t) sinh[λ(x – t)]y(t) dt = f (x),

62.

x

a

Double differentiation yields x x 2λ cosh[λ(x – t)]y(t) dt + λ2 (x – t) sinh[λ(x – t)]y(t) dt = fxx (x). a

a

Eliminating the second term on the left-hand side with the aid of the original equation, we arrive at an equation of the form 1.3.1: x 1 fxx (x) – λ2 f (x) . cosh[λ(x – t)]y(t) dt = 2λ a Solution:

1 y(x) = f (x) – λfx (x) + 12 λ3 2λ xxx

x

f (t) dt. a

x

Axβ + B sinhγ (λt) + C]y(t) dt = f (x).

63.

a

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B sinhγ (λt) + C.

x

A sinhγ (λx) + Btβ + C]y(t) dt = f (x).

64.

a

This is a special case of equation 1.9.6 with g(x) = A sinhγ (λx) and h(t) = Btβ + C. 65.

x

Axλ sinhµ t + Btβ sinhγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = sinhµ t, g2 (x) = B sinhγ x, and h2 (t) = tβ . 1.3-3. Kernels Containing Hyperbolic Tangent 66.

x

tanh(λx) – tanh(λt) y(t) dt = f (x).

a

67.

This is a special case of equation 1.9.2 with g(x) = tanh(λx). 1 Solution: y(x) = cosh2 (λx)fx (x) x . λ x A tanh(λx) + B tanh(λt) y(t) dt = f (x). a

For B = –A, see equation 1.3.66. This is a special case equation 1.9.4 with g(x)= tanh(λx). of x – B – A 1 d Solution: y(x) = tanh(λt) A+B ft (t) dt . tanh(λx) A+B A + B dx a

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68.

x

A tanh(λx) + B tanh(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A tanh(λx) and h(t) = B tanh(µt) + C. 69.

tanh2 (λx) – tanh2 (λt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.2 with g(x) = tanh2 (λx). d cosh3 (λx)fx (x) Solution: y(x) = . dx 2λ sinh(λx) 70.

A tanh2 (λx) + B tanh2 (λt) y(t) dt = f (x).

x

a

For B = –A, see equation 1.3.69. This caseof equation 1.9.4 with g(x) = tanh2 (λx). is a special2A x – 2B – 1 d Solution: y(x) = tanh(λt) A+B ft (t) dt . tanh(λx) A+B A + B dx a 71.

x

A tanh2 (λx) + B tanh2 (µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A tanh2 (λx) and h(t) = B tanh2 (µt) + C. 72.

x

n

tanh(λx) – tanh(λt)

y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 1 d 2 Solution: y(x) = n (λx) f (x). cosh dx λ n! cosh2 (λx)

x

73.

√

tanh x – tanh t y(t) dt = f (x).

a

Solution: y(x) =

x

√

74. a

2 d 2 2 x cosh dx π cosh2 x

y(t) dt tanh x – tanh t

x

f (t) dt √ . cosh t tanh x – tanh t 2

a

= f (x).

Solution: 1 d y(x) = π dx

a

x

f (t) dt √ . cosh t tanh x – tanh t 2

x

(tanh x – tanh t)λ y(t) dt = f (x),

75.

0 < λ < 1.

a

Solution: sin(πλ) d 2 y(x) = cosh2 x 2 dx πλ cosh x

a

x

f (t) dt . cosh2 t (tanh x – tanh t)λ

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x

(tanhµ x – tanhµ t)y(t) dt = f (x).

76. a

77.

µ This is a special case of equation 1.9.2 with g(x) = tanh x. µ+1 1 d cosh xfx (x) . Solution: y(x) = µ dx sinhµ–1 x x

A tanhµ x + B tanhµ t y(t) dt = f (x). a

For B = –A, see equation 1.3.76. This is a special case of equation 1.9.4 with g(x) = tanhµ x. Solution: x – Bµ – Aµ 1 d A+B A+B y(x) = ft (t) dt . tanh(λt) tanh(λx) A + B dx a

x

78. a

79.

y(t) dt [tanh(λx) – tanh(λt)]µ

= f (x),

0 < µ < 1.

This is a special case of equation 1.9.42 with g(x) = tanh(λx) and h(x) ≡ 1. Solution: x λ sin(πµ) d f (t) dt y(x) = . 2 π dx a cosh (λt)[tanh(λx) – tanh(λt)]1–µ x Axβ + B tanhγ (λt) + C]y(t) dt = f (x). a

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B tanhγ (λt) + C.

x

A tanhγ (λx) + Btβ + C]y(t) dt = f (x).

80.

a

This is a special case of equation 1.9.6 with g(x) = A tanhγ (λx) and h(t) = Btβ + C. 81.

x

Axλ tanhµ t + Btβ tanhγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = tanhµ t, g2 (x) = B tanhγ x, and h2 (t) = tβ . 1.3-4. Kernels Containing Hyperbolic Cotangent 82.

x

coth(λx) – coth(λt) y(t) dt = f (x).

a

83.

This is a special case of equation 1.9.2 with g(x) = coth(λx). 1 d Solution: y(x) = – sinh2 (λx)fx (x) . λ dx x A coth(λx) + B coth(λt) y(t) dt = f (x). a

For B = –A, see equation 1.3.82. This is a special case with g(x) = coth(λx). xof equation 1.9.4 B A 1 d Solution: y(x) = tanh(λt) A+B ft (t) dt . tanh(λx) A+B A + B dx a

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84.

x

A coth(λx) + B coth(µt) + C y(t) dt = f (x).

a

85.

This is a special case of equation 1.9.6 with g(x) = A coth(λx) and h(t) = B coth(µt) + C. x coth2 (λx) – coth2 (λt) y(t) dt = f (x). a

86.

2 This is a special case of equation 1.9.2 with g(x) = coth (λx). 3 d sinh (λx)fx (x) Solution: y(x) = – . dx 2λ cosh(λx) x A coth2 (λx) + B coth2 (λt) y(t) dt = f (x). a

87.

2 For B = –A, see equation 1.3.85. This equation 1.9.4 with g(x) is a special2Acase = coth (λx). of x 2B 1 d Solution: y(x) = tanh(λt) A+B ft (t) dt . tanh(λx) A+B A + B dx a x A coth2 (λx) + B coth2 (µt) + C y(t) dt = f (x). a

88.

This is a special case of equation 1.9.6 with g(x) = A coth2 (λx) and h(t) = B coth2 (µt) + C. x n coth(λx) – coth(λt) y(t) dt = f (x), n = 1, 2, . . . a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 (–1)n d 2 Solution: y(x) = n (λx) f (x). sinh dx λ n! sinh2 (λx)

x

(cothµ x – cothµ t)y(t) dt = f (x).

89. a

90.

µ This is a special case of equation with g(x) = coth x. 1.9.2 µ+1 1 d sinh xfx (x) Solution: y(x) = – . µ dx coshµ–1 x x

A cothµ x + B cothµ t y(t) dt = f (x).

91.

For B = –A, see equation 1.3.89. This is a special case of equation 1.9.4 with g(x) = cothµ x. Solution: x Aµ Bµ 1 d tanh t A+B ft (t) dt . y(x) = tanh x A+B A + B dx a x Axβ + B cothγ (λt) + C]y(t) dt = f (x).

a

a

92.

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B cothγ (λt) + C. x A cothγ (λx) + Btβ + C]y(t) dt = f (x). a

This is a special case of equation 1.9.6 with g(x) = A cothγ (λx) and h(t) = Btβ + C.

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93.

x

Axλ cothµ t + Btβ cothγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = cothµ t, g2 (x) = B cothγ x, and h2 (t) = tβ . 1.3-5. Kernels Containing Combinations of Hyperbolic Functions 94.

x

cosh[λ(x – t)] + A sinh[µ(x – t)] y(t) dt = f (x).

a

Let us differentiate the equation with respect to x and then eliminate the integral with the hyperbolic cosine. As a result, we arrive at an equation of the form 2.3.16:

x

2

y(x) + (λ – A µ)

sinh[µ(x – t)]y(t) dt = fx (x) – Aµf (x).

a

95.

x

A cosh(λx) + B sinh(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A cosh(λx) and h(t) = B sinh(µt) + C. 96.

A cosh2 (λx) + B sinh2 (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = cosh2 (λx) and h(t) = B sinh2 (µt) + C.

x

sinh[λ(x – t)] cosh[λ(x + t)]y(t) dt = f (x).

97. a

Using the formula sinh(α – β) cosh(α + β) =

1 2

sinh(2α) – sinh(2β) ,

α = λx,

β = λt,

we reduce the original equation to an equation of the form 1.3.37:

x

sinh(2λx) – sinh(2λt) y(t) dt = 2f (x).

a

Solution: y(x) =

1 d fx (x) . λ dx cosh(2λx)

x

cosh[λ(x – t)] sinh[λ(x + t)]y(t) dt = f (x).

98. a

Using the formula cosh(α – β) sinh(α + β) =

1 2

sinh(2α) + sinh(2β) ,

α = λx,

β = λt,

we reduce the original equation to an equation of the form 1.3.38 with A = B = 1:

x

sinh(2λx) + sinh(2λt) y(t) dt = 2f (x).

a

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99.

x

A cosh(λx) sinh(µt) + B cosh(βx) sinh(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = A cosh(λx), h1 (t) = sinh(µt), g2 (x) = B cosh(βx), and h2 (t) = sinh(γt). x 100. sinh(λx) cosh(µt) + sinh(βx) cosh(γt) y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = sinh(λx), h1 (t) = cosh(µt), g2 (x) = sinh(βx), and h2 (t) = cosh(γt). x 101. cosh(λx) cosh(µt) + sinh(βx) sinh(γt) y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = cosh(λx), h1 (t) = cosh(µt), g2 (x) = sinh(βx), and h2 (t) = sinh(γt). x 102. A coshβ (λx) + B sinhγ (µt) y(t) dt = f (x). a

This is a special case of equation 1.9.6 with g(x) = A coshβ (λx) and h(t) = B sinhγ (µt). x 103. A sinhβ (λx) + B coshγ (µt) y(t) dt = f (x). a

This is a special case of equation 1.9.6 with g(x) = A sinhβ (λx) and h(t) = B coshγ (µt). x

104. Axλ coshµ t + Btβ sinhγ x y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = coshµ t, g2 (x) = B sinhγ x, and h2 (t) = tβ . x 105. (x – t) sinh[λ(x – t)] – λ(x – t)2 cosh[λ(x – t)] y(t) dt = f (x). a

Solution:

x

y(x) =

g(t) dt, a

where

6 t 5 d2 2 g(t) = –λ (t – τ ) 2 I 5 [λ(t – τ )] f (τ ) dτ . dt2 2 a x sinh[λ(x – t)] 106. – λ cosh[λ(x – t)] y(t) dt = f (x). x–t a π 1 2λ 64λ5

Solution: 1 y(x) = 2λ4 107.

3

d2 – λ2 dx2

x

sinh[λ(x – t)]f (t) dt. a

√

√

√ sinh λ x – t – λ x – t cosh λ x – t y(t) dt = f (x),

x

a

Solution: 4 d3 y(x) = – 3 3 πλ dx

a

x

f (a) = fx (a) = 0.

√

cos λ x – t √ f (t) dt. x–t

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108.

x

Axλ sinhµ t + Btβ coshγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = sinhµ t, g2 (x) = B coshγ x, and h2 (t) = tβ . 109.

x

A tanh(λx) + B coth(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A tanh(λx) and h(t) = B coth(µt) + C. 110.

A tanh2 (λx) + B coth2 (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = tanh2 (λx) and h(t) = B coth2 (µt). 111.

x

tanh(λx) coth(µt) + tanh(βx) coth(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = tanh(λx), h1 (t) = coth(µt), g2 (x) = tanh(βx), and h2 (t) = coth(γt). 112.

x

coth(λx) tanh(µt) + coth(βx) tanh(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = coth(λx), h1 (t) = tanh(µt), g2 (x) = coth(βx), and h2 (t) = tanh(γt). 113.

x

tanh(λx) tanh(µt) + coth(βx) coth(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = tanh(λx), h1 (t) = tanh(µt), g2 (x) = coth(βx), and h2 (t) = coth(γt). 114.

A tanhβ (λx) + B cothγ (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A tanhβ (λx) and h(t) = B cothγ (µt). 115.

A cothβ (λx) + B tanhγ (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cothβ (λx) and h(t) = B tanhγ (µt). 116.

x

Axλ tanhµ t + Btβ cothγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = tanhµ t, g2 (x) = B cothγ x, and h2 (t) = tβ . 117.

x

Axλ cothµ t + Btβ tanhγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = cothµ t, g2 (x) = B tanhγ x, and h2 (t) = tβ .

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1.4. Equations Whose Kernels Contain Logarithmic Functions 1.4-1. Kernels Containing Logarithmic Functions

x

(ln x – ln t)y(t) dt = f (x).

1. a

This is a special case of equation 1.9.2 with g(x) = ln x. Solution: y(x) = xfxx (x) + fx (x).

x

ln(x – t)y(t) dt = f (x).

2. 0

Solution:

x

ftt (t) dt

y(x) = –

0

∞

0

(x – t)z e–Cz dz – fx (0) Γ(z + 1)

∞

0

xz e–Cz dz, Γ(z + 1)

1 1 where C = lim 1 + + · · · + – ln k = 0.5772 . . . is the Euler constant and Γ(z) is k→∞ 2 k+1 the gamma function. • References: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971), A. G. Butkovskii (1979).

x

[ln(x – t) + A]y(t) dt = f (x).

3. a

Solution: y(x) = –

d dx

x

νA (x – t)f (t) dt,

νA (x) =

a

d dx

∞

0

xz e(A–C)z dz, Γ(z + 1)

where C = 0.5772 . . . is the Euler constant and Γ(z) is the gamma function. For a = 0, the solution can be written in the form y(x) = – 0

x

ftt (t) dt

0

∞

(x – t)z e(A–C)z dz – fx (0) Γ(z + 1)

0

∞

xz e(A–C)z dz, Γ(z + 1)

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

(A ln x + B ln t)y(t) dt = f (x).

4. a

This is a special case of equation 1.9.4 with g(x) = ln x. For B = –A, see equation 1.4.1. Solution: x – B – A sign(ln x) d A+B A+B y(x) = ft (t) dt . ln t ln x A + B dx a

x

(A ln x + B ln t + C)y(t) dt = f (x).

5. a

This is a special case of equation 1.9.5 with g(x) = x.

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6.

ln2 (λx) – ln2 (λt) y(t) dt = f (x),

x

f (a) = fx (a) = 0.

a

d xfx (x) Solution: y(x) = . dx 2 ln(λx) 7.

A ln2 (λx) + B ln2 (λt) y(t) dt = f (x).

x

a

For B = –A, see equation 1.4.7. This is a special case of equation 1.9.4 with g(x) = ln2 (λx). Solution: x 2B – 2A 1 d ln(λt)– A+B ft (t) dt . y(x) = ln(λx) A+B A + B dx a 8.

A ln2 (λx) + B ln2 (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = ln2 (λx) and h(t) = ln2 (µt) + C. 9.

x

n

ln(x/t)

y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 1 d Solution: y(x) = f (x). x n! x dx 10.

x

ln2 x – ln2 t

n

y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 ln x x d Solution: y(x) = n f (x). 2 n! x ln x dx

x

11.

ln a

x+b

t+b

y(t) dt = f (x).

This is a special case of equation 1.9.2 with g(x) = ln(x + b). Solution: y(x) = (x + b)fxx (x) + fx (x).

x

12.

ln(x/t) y(t) dt = f (x).

a

Solution:

x

13. a

2 x 2 f (t) dt d y(x) = x . πx dx a t ln(x/t)

y(t) dt = f (x). ln(x/t)

Solution: y(x) =

1 d π dx

a

x

f (t) dt . t ln(x/t)

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14.

lnµ (λx) – lnµ (λt) y(t) dt = f (x).

x

a

15.

This is a special case of equation 1.9.2 with g(x) = lnµ (λx). 1 d Solution: y(x) = x ln1–µ (λx)fx (x) . µ dx x A lnβ (λx) + B lnγ (µt) + C y(t) dt = f (x). a

This is a special case of equation 1.9.6 with g(x) = A lnβ (λx) and h(t) = B lnγ (µt) + C.

x

[ln(x/t)]λ y(t) dt = f (x),

16.

0 < λ < 1.

a

Solution: y(x) =

x

17. a

y(t) dt [ln(x/t)]λ

2 x k f (t) dt d , x λ x dx a t[ln(x/t)]

= f (x),

sin(πλ) . πλ

k=

0 < λ < 1.

This is a special case of equation 1.9.42 with g(x) = ln x and h(x) ≡ 1. Solution: x sin(πλ) d f (t) dt y(x) = . π dx a t[ln(x/t)]1–λ 1.4-2. Kernels Containing Power-Law and Logarithmic Functions

x

18.

(x – t) ln(x – t) + A y(t) dt = f (x).

a

Solution: y(x) = –

d2 dx2

x

νA (x – t)f (t) dt,

νA (x) =

a

d dx

0

∞

xz e(A–C)z dz, Γ(z + 1)

where C = 0.5772 . . . is the Euler constant and Γ(z) is the gamma function. • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

19. a

ln(x – t) + A (x – t)λ

y(t) dt = f (x),

0 < λ < 1.

Solution: x x sin(πλ) d F (t) dt , F (x) = νh (x – t)f (t) dt, π dx a (x – t)1–λ a ∞ z hz x e d νh (x) = dz, h = A + ψ(1 – λ), dx 0 Γ(z + 1)

y(x) = –

where Γ(z) is the gamma function and ψ(z) = Γ(z) z is the logarithmic derivative of the gamma function. • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

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20.

x

tβ lnλ x – xβ lnλ t)y(t) dt = f (x).

a

This is a special case of equation 1.9.11 with g(x) = lnλ x and h(t) = tβ . 21.

x

Atβ lnλ x + Bxµ lnγ t)y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = A lnλ x, h1 (t) = tβ , g2 (x) = Bxµ , and h2 (t) = lnγ t.

x

22.

ln a

xµ + b

ctλ + s

y(t) dt = f (x).

This is a special case of equation 1.9.6 with g(x) = ln(xµ + b) and h(t) = – ln(ctλ + s).

1.5. Equations Whose Kernels Contain Trigonometric Functions 1.5-1. Kernels Containing Cosine

x

cos[λ(x – t)]y(t) dt = f (x).

1. a

Solution: y(x) =

fx (x)

+λ

x

2

f (x) dx. a

2.

x

cos[λ(x – t)] – 1 y(t) dt = f (x),

f (a) = fx (a) = fxx (x) = 0.

a

Solution: y(x) = – 3.

x

1 f (x) – fx (x). λ2 xxx

cos[λ(x – t)] + b y(t) dt = f (x).

a

For b = 0, see equation 1.5.1. For b = –1, see equation 1.5.2. For λ = 0, see equation 1.1.1. Differentiating the equation with respect to x, we arrive at an equation of the form 2.5.16: y(x) –

λ b+1

x

sin[λ(x – t)]y(t) dt = a

fx (x) . b+1

1◦ . Solution with b(b + 1) > 0: f (x) λ2 y(x) = x + b+1 k(b + 1)2

x

sin[k(x –

t)]ft (t) dt,

where

k=λ

a

b . b+1

2◦ . Solution with b(b + 1) < 0: f (x) λ2 y(x) = x + b+1 k(b + 1)2

x

sinh[k(x – a

t)]ft (t) dt,

where

k=λ

–b . b+1

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x

cos(λx + βt)y(t) dt = f (x).

4. a

Differentiating the equation with respect to x twice yields x cos[(λ + β)x]y(x) – λ sin(λx + βt)y(t) dt = fx (x), (1) a x cos[(λ + β)x]y(x) x – λ sin[(λ + β)x]y(x) – λ2 cos(λx + βt)y(t) dt = fxx (x). (2) a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the first-order linear ordinary differential equation wx – λ tan[(λ + β)x]w = fxx (x) + λ2 f (x),

w = cos[(λ + β)x]y(x).

(3)

Setting x = a in (1) yields the initial condition w(a) = fx (a). On solving equation (3) under this condition, after some transformations we obtain the solution of the original integral equation in the form y(x) =

5.

1 λ sin[(λ + β)x] f (x) + f (x) cos[(λ + β)x] x cos2 [(λ + β)x] x λβ – f (t) cosk–2 [(λ + β)t] dt, cosk+1 [(λ + β)x] a

k=

λ . λ+β

x

cos(λx) – cos(λt) y(t) dt = f (x).

a

6.

This is a special case of equation 1.9.2 with g(x) = cos(λx). 1 d fx (x) Solution: y(x) = – . λ dx sin(λx) x A cos(λx) + B cos(λt) y(t) dt = f (x).

7.

This is a special case of equation 1.9.4 with g(x) = cos(λx). For B = –A, see equation 1.5.5. Solution with B ≠ –A: x B – A sign cos(λx) d cos(λt)– A+B ft (t) dt . y(x) = cos(λx) A+B A+B dx a x A cos(λx) + B cos(µt) + C y(t) dt = f (x).

a

a

8.

This is a special case of equation 1.9.6 with g(x) = A cos(λx) and h(t) = B cos(µt) + C. x A1 cos[λ1 (x – t)] + A2 cos[λ2 (x – t)] y(t) dt = f (x). a

The equation is equivalent to the equation x B1 sin[λ1 (x – t)] + B2 sin[λ2 (x – t)] y(t) dt = F (x), a x A1 A2 B1 = , B2 = , F (x) = f (t) dt. λ1 λ2 a which has the form 1.5.41. (Differentiation of this equation yields the original integral equation.)

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x

cos2 [λ(x – t)]y(t) dt = f (x).

9. a

Differentiating yields an equation of the form 2.5.16:

x

y(x) – λ

sin[2λ(x – t)]y(t) dt = fx (x).

a

Solution: y(x) = fx (x) + 10.

2λ2 k

x

sin[k(x – t)]ft (t) dt,

where

√ k = λ 2.

a

x

cos2 (λx) – cos2 (λt) y(t) dt = f (x),

a

f (a) = fx (a) = 0.

1 d fx (x) Solution: y(x) = – . λ dx sin(2λx) 11.

x

A cos2 (λx) + B cos2 (λt) y(t) dt = f (x).

a

For B = –A, see equation 1.5.10. This is a special case of equation 1.9.4 with g(x) = cos2 (λx). Solution: x – 2A – 2B 1 d y(x) = cos(λx) A+B cos(λt) A+B ft (t) dt . A + B dx a 12.

A cos2 (λx) + B cos2 (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cos2 (λx) and h(t) = B cos2 (µt) + C.

x

cos[λ(x – t)] cos[λ(x + t)]y(t) dt = f (x).

13. a

Using the trigonometric formula cos(α – β) cos(α + β) =

1 2

cos(2α) + cos(2β) ,

α = λx,

β = λt,

we reduce the original equation to an equation of the form 1.5.6 with A = B = 1:

x

cos(2λx) + cos(2λt) y(t) dt = 2f (x).

a

Solution with cos(2λx) > 0: y(x) = 14.

x d f (t) dt 1 √ √t . dx cos(2λx) a cos(2λt)

x

A cos(λx) cos(µt) + B cos(βx) cos(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = A cos(λx), h1 (t) = cos(µt), g2 (x) = B cos(βx), and h2 (t) = cos(γt).

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x

cos3 [λ(x – t)]y(t) dt = f (x).

15. a

Using the formula cos3 β =

x

a

16.

1 4

1 4

cos 3β +

3 4

cos β, we arrive at an equation of the form 1.5.8:

cos[3λ(x – t)] +

3 4

cos[λ(x – t)] y(t) dt = f (x).

cos3 (λx) – cos3 (λt) y(t) dt = f (x),

x

a

f (a) = fx (a) = 0.

1 d fx (x) Solution: y(x) = – . 3λ dx sin(λx) cos2 (λx) 17.

A cos3 (λx) + B cos3 (λt) y(t) dt = f (x).

x

a

For B = –A, see equation 1.3.16. This is a special case of equation 1.9.4 with g(x) = cos3 (λx). Solution: x – 3B – 3A 1 d y(x) = cos(λt) A+B ft (t) dt . cos(λx) A+B A + B dx a 18.

cos2 (λx) cos(µt) + cos(βx) cos2 (γt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.15 with g1 (x) = cos2 (λx), h1 (t) = cos(µt), g2 (x) = cos(βx), and h2 (t) = cos2 (γt).

x

cos4 [λ(x – t)]y(t) dt = f (x).

19. a

Let us transform the kernel of the integral equation using the trigonometric formula cos4 β = 1 1 3 8 cos 4β + 2 cos 2β + 8 , where β = λ(x – t), and differentiate the resulting equation with respect to x. Then we arrive at an equation of the form 2.5.18: y(x) – λ

x

a

20.

x

n

cos(λx) – cos(λt)

1 2

sin[4λ(x – t)] + sin[2λ(x – t)] y(t) dt = fx (x).

y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 (–1)n 1 d Solution: y(x) = n sin(λx) f (x). λ n! sin(λx) dx

x

21.

√

cos t – cos x y(t) dt = f (x).

a

This is a special case of equation 1.9.38 with g(x) = 1 – cos x. Solution: 1 d 2 x sin t f (t) dt 2 √ y(x) = sin x . π sin x dx cos t – cos x a

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22.

x

√

y(t) dt

cos t – cos x Solution:

= f (x).

a

1 d y(x) = π dx

x

sin t f (t) dt √ . cos t – cos x

a

x

(cos t – cos x)λ y(t) dt = f (x),

23.

0 < λ < 1.

a

Solution:

1 d 2 x sin t f (t) dt y(x) = k sin x , λ sin x dx a (cos t – cos x)

k=

sin(πλ) . πλ

x

(cosµ x – cosµ t)y(t) dt = f (x).

24. a

25.

µ This is a special case of equation with g(x) 1.9.2 = cos x. 1 d fx (x) Solution: y(x) = – . µ dx sin x cosµ–1 x x

A cosµ x + B cosµ t y(t) dt = f (x). a

For B = –A, see equation 1.5.24. This is a special case of equation 1.9.4 with g(x) = cosµ x. Solution: x Bµ – Aµ 1 d cos t– A+B ft (t) dt . y(x) = cos x A+B A + B dx a 26.

x

y(t) dt

(cos t – cos x)λ Solution:

= f (x),

0 < λ < 1.

a

sin(πλ) d y(x) = π dx

x

(x – t) cos[λ(x – t)]y(t) dt = f (x),

27. a

x

a

sin t f (t) dt . (cos t – cos x)1–λ

f (a) = fx (a) = 0.

Differentiating the equation twice yields x x y(x) – 2λ sin[λ(x – t)]y(t) dt – λ2 (x – t) cos[λ(x – t)]y(t) dt = fxx (x). a

a

Eliminating the third term on the left-hand side with the aid of the original equation, we arrive at an equation of the form 2.5.16: x y(x) – 2λ sin[λ(x – t)]y(t) dt = fxx (x) + λ2 f (x). a

√

x cos λ x – t y(t) dt = f (x). √ x–t a Solution: √

x cosh λ x – t 1 d √ y(x) = f (t) dt. π dx a x–t

28.

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31.

√ cos λ x2 – t2 y(t) dt = f (x). √ x2 – t 2 0 Solution:

√ x cosh λ x2 – t2 2 d √ y(x) = t f (t) dt. π dx 0 x2 – t2 √

∞ cos λ t2 – x2 y(t) dt = f (x). √ t 2 – x2 x Solution: √

∞ cosh λ t2 – x2 2 d √ y(x) = – t f (t) dt. π dx x t2 – x2 x Axβ + B cosγ (λt) + C]y(t) dt = f (x).

32.

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B cosγ (λt) + C. x A cosγ (λx) + Btβ + C]y(t) dt = f (x).

33.

This is a special case of equation 1.9.6 with g(x) = A cosγ (λx) and h(t) = Btβ + C. x

Axλ cosµ t + Btβ cosγ x y(t) dt = f (x).

29.

30.

x

a

a

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = cosµ t, g2 (x) = B cosγ x, and h2 (t) = tβ . 1.5-2. Kernels Containing Sine

x

sin[λ(x – t)]y(t) dt = f (x),

34. a

Solution: y(x) = 35.

x

f (a) = fx (a) = 0.

1 f (x) + λf (x). λ xx

sin[λ(x – t)] + b y(t) dt = f (x).

a

36.

Differentiating the equation with respect to x yields an equation of the form 2.5.3: λ x 1 y(x) + cos[λ(x – t)]y(t) dt = fx (x). b a b x sin(λx + βt)y(t) dt = f (x). a

For β = –λ, see equation 1.5.34. Assume that β ≠ –λ. Differentiating the equation with respect to x twice yields x sin[(λ + β)x]y(x) + λ cos(λx + βt)y(t) dt = fx (x), (1) a x sin(λx + βt)y(t) dt = fxx (x). (2) sin[(λ + β)x]y(x) x + λ cos[(λ + β)x]y(x) – λ2 a

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Eliminating the integral term from (2) with the aid of the original equation, we arrive at the first-order linear ordinary differential equation wx + λ cot[(λ + β)x]w = fxx (x) + λ2 f (x),

w = sin[(λ + β)x]y(x).

(3)

Setting x = a in (1) yields the initial condition w(a) = fx (a). On solving equation (3) under this condition, after some transformation we obtain the solution of the original integral equation in the form y(x) =

37.

1 λ cos[(λ + β)x] f (x) – f (x) sin[(λ + β)x] x sin2 [(λ + β)x] x λβ – f (t) sink–2 [(λ + β)t] dt, sink+1 [(λ + β)x] a

k=

λ . λ+β

x

sin(λx) – sin(λt) y(t) dt = f (x).

a

38.

This is a special case of equation 1.9.2with g(x) = sin(λx). 1 d fx (x) Solution: y(x) = . λ dx cos(λx) x A sin(λx) + B sin(λt) y(t) dt = f (x).

39.

This is a special case of equation 1.9.4 with g(x) = sin(λx). For B = –A, see equation 1.5.37. Solution with B ≠ –A: x B – A sign sin(λx) d sin(λt)– A+B ft (t) dt . y(x) = sin(λx) A+B A+B dx a x A sin(λx) + B sin(µt) + C y(t) dt = f (x).

40.

This is a special case of equation 1.9.6 with g(x) = A sin(λx) and h(t) = B sin(µt) + C. x µ sin[λ(x – t)] – λ sin[µ(x – t)] y(t) dt = f (x).

41.

It is assumed that f (a) = fx (a) = fxx (a) = fxxx (a) = 0. Solution: f + (λ2 + µ2 )fxx + λ2 µ2 f y(x) = xxxx , f = f (x). 3 3 λµ – λ µ x A1 sin[λ1 (x – t)] + A2 sin[λ2 (x – t)] y(t) dt = f (x), f (a) = fx (a) = 0.

a

a

a

a

This equation can be solved in the same manner as equation 1.3.41, i.e., by reducing it to a second-order linear ordinary differential equation with constant coefficients. Let A1 λ 2 + A2 λ 1 ∆ = –λ1 λ2 . A1 λ 1 + A2 λ 2 1◦ . Solution for ∆ > 0: x (A1 λ1 + A2 λ2 )y(x) = fxx (x) + Bf (x) + C sinh[k(x – t)]f (t) dt, k=

√

a

∆,

B = ∆ + λ21 + λ22 ,

1 C = √ ∆2 + (λ21 + λ22 )∆ + λ21 λ22 . ∆

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2◦ . Solution for ∆ < 0: (A1 λ1 + A2 λ2 )y(x) = fxx (x) + Bf (x) + C

k=

√

x

sin[k(x – t)]f (t) dt, a

–∆,

B = ∆ + λ21 + λ22 ,

C= √

1 2 ∆ + (λ21 + λ22 )∆ + λ21 λ22 . –∆

3◦ . Solution for ∆ = 0: (A1 λ1 + A2 λ2 )y(x) = fxx (x) + (λ21 + λ22 )f (x) + λ21 λ22

x

(x – t)f (t) dt. a

4◦ . Solution for ∆ = ∞: y(x) = –

fxxxx + (λ21 + λ22 )fxx + λ21 λ22 f , 3 3 A1 λ1 + A2 λ2

f = f (x).

In the last case, the relation A1 λ1 + A2 λ2 = 0 holds and the right-hand side of the integral equation is assumed to satisfy the conditions f (a) = fx (a) = fxx (a) = fxxx (a) = 0.

42.

Remark. The solution can be obtained from the solution of equation 1.3.41 in which the change of variables λk → iλk , Ak → –iAk , i2 = –1 (k = 1, 2), should be made. x A sin[λ(x – t)] + B sin[µ(x – t)] + C sin[β(x – t)] y(t) dt = f (x). a

It is assumed that f (a) = fx (a) = 0. Differentiating the integral equation twice yields x 2 (Aλ + Bµ + Cβ)y(x) – Aλ sin[λ(x – t)] + Bµ2 sin[µ(x – t)] y(t) dt a x – Cβ 2 sin[β(x – t)]y(t) dt = fxx (x). a

Eliminating the last integral with the aid of the original equation, we arrive at an equation of the form 2.5.18: x (Aλ + Bµ + Cβ)y(x) + A(β 2 – λ2 ) sin[λ(x – t)] a (x) + β 2 f (x). + B(β 2 – µ2 ) sin[µ(x – t)] y(t) dt = fxx

43.

In the special case Aλ + Bµ + Cβ = 0, this is an equation of the form 1.5.41. x sin2 [λ(x – t)]y(t) dt = f (x), f (a) = fx (a) = fxx (a) = 0. a

Differentiation yields an equation of the form 1.5.34: x 1 sin[2λ(x – t)]y(t) dt = fx (x). λ a Solution: y(x) = 12 λ–2 fxxx (x) + 2fx (x).

44.

sin2 (λx) – sin2 (λt) y(t) dt = f (x),

x

a

f (a) = fx (a) = 0.

1 d fx (x) Solution: y(x) = . λ dx sin(2λx)

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45.

A sin2 (λx) + B sin2 (λt) y(t) dt = f (x).

x

a

For B = –A, see equation 1.5.44. This is a special case of equation 1.9.4 with g(x) = sin2 (λx). Solution: x 2B – 2A 1 d sin(λt)– A+B ft (t) dt . y(x) = sin(λx) A+B A + B dx a 46.

x

A sin2 (λx) + B sin2 (µt) + C y(t) dt = f (x).

a

47.

This is a special case of equation 1.9.6 with g(x) = A sin2 (λx) and h(t) = B sin2 (µt) + C. x sin[λ(x – t)] sin[λ(x + t)]y(t) dt = f (x), f (a) = fx (a) = 0. a

Using the trigonometric formula sin(α – β) sin(α + β) =

1 2

cos(2β) – cos(2α) ,

α = λx,

β = λt,

we reduce the original equation to an equation of the form 1.5.5 with A = B = 1: x cos(2λx) – cos(2λt) y(t) dt = –2f (x). a

Solution: y(x) = 48.

1 d fx (x) . λ dx sin(2λx)

x

sin(λx) sin(µt) + sin(βx) sin(γt) y(t) dt = f (x).

a

49.

This is a special case of equation 1.9.15 with g1 (x) = sin(λx), h1 (t) = sin(µt), g2 (x) = sin(βx), and h2 (t) = sin(γt). x sin3 [λ(x – t)]y(t) dt = f (x). a It is assumed that f (a) = fx (a) = fxx (a) = fxxx (a) = 0. 1 3 3 Using the formula sin β = – 4 sin 3β + 4 sin β, we arrive at an equation of the form 1.5.41: x 1 – 4 sin[3λ(x – t)] + 34 sin[λ(x – t)] y(t) dt = f (x). a

50.

x

sin3 (λx) – sin3 (λt) y(t) dt = f (x),

a

51.

f (a) = fx (a) = 0.

This is a special case of equation 1.9.2 with g(x) = sin3 (λx). x A sin3 (λx) + B sin3 (λt) y(t) dt = f (x). a

This is a special case of equation 1.9.4 with g(x) = sin3 (λx). Solution: x – 3B – 3A sign sin(λx) d A+B A+B y(x) = ft (t) dt . sin(λt) sin(λx) A+B dx a

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52.

sin2 (λx) sin(µt) + sin(βx) sin2 (γt) y(t) dt = f (x).

x

a

53.

This is a special case of equation 1.9.15 with g1 (x) = sin2 (λx), h1 (t) = sin(µt), g2 (x) = sin(βx), and h2 (t) = sin2 (γt). x sin4 [λ(x – t)]y(t) dt = f (x). a It is assumed that f (a) = fx (a) = · · · = fxxxx (a) = 0. Let us transform the kernel of the integral equation using the trigonometric formula sin4 β = 18 cos 4β – 12 cos 2β + 38 , where β = λ(x – t), and differentiate the resulting equation with respect to x. Then we obtain an equation of the form 1.5.41: x 1 λ – 2 sin[4λ(x – t)] + sin[2λ(x – t)] y(t) dt = fx (x). a

x

sinn [λ(x – t)]y(t) dt = f (x),

54.

n = 2, 3, . . .

a

It is assumed that f (a) = fx (a) = · · · = fx(n) (a) = 0. Let us differentiate the equation with respect to x twice and transform the kernel of the resulting integral equation using the formula cos2 β = 1 – sin2 β, where β = λ(x – t). We have x x 2 2 n 2 –λ n sin [λ(x – t)]y(t) dt + λ n(n – 1) sinn–2 [λ(x – t)]y(t) dt = fxx (x). a

a

Eliminating the first term on the left-hand side with the aid of the original equation, we obtain x 1 sinn–2 [λ(x – t)]y(t) dt = 2 fxx (x) + λ2 n2 f (x) . λ n(n – 1) a

55.

This equation has the same form as the original equation, but the degree characterizing the kernel has been reduced by two. By applying this technique sufficiently many times, we finally arrive at simple integral equations of the form 1.1.1 (for even n) or 1.5.34 (for odd n). x √

sin λ x – t y(t) dt = f (x). a

Solution: 2 d2 y(x) = πλ dx2

x

56.

√

√

cosh λ x – t √ f (t) dt. x–t

x

a

sin x – sin t y(t) dt = f (x).

a

Solution: y(x) = 57.

x

√

y(t) dt

sin x – sin t Solution:

1 d 2 x cos t f (t) dt 2 √ . cos x π cos x dx sin x – sin t a

= f (x).

a

1 d y(x) = π dx

a

x

cos t f (t) dt √ . sin x – sin t

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x

(sin x – sin t)λ y(t) dt = f (x),

58.

0 < λ < 1.

a

Solution: 1 d 2 x cos t f (t) dt y(x) = k cos x , λ cos x dx a (sin x – sin t)

k=

sin(πλ) . πλ

x

(sinµ x – sinµ t)y(t) dt = f (x).

59. a

This is a special case of equation 1.9.2 with g(x) = sinµ x. 1 d fx (x) Solution: y(x) = . µ dx cos x sinµ–1 x 60.

x

A| sin(λx)|µ + B| sin(λt)|µ y(t) dt = f (x).

a

This is a special case of equation 1.9.4 with g(x) = | sin(λx)|µ . Solution: y(x) =

x

61. a

x – Aµ Bµ 1 d sin(λt)– A+B ft (t) dt . sin(λx) A+B A + B dx a

y(t) dt [sin(λx) – sin(λt)]µ

= f (x),

0 < µ < 1.

This is a special case of equation 1.9.42 with g(x) = sin(λx) and h(x) ≡ 1. Solution: x λ sin(πµ) d cos(λt)f (t) dt y(x) = . π dx a [sin(λx) – sin(λt)]1–µ

x

f (a) = fx (a) = fxx (a) = 0.

(x – t) sin[λ(x – t)]y(t) dt = f (x),

62. a

Double differentiation yields

x

cos[λ(x – t)]y(t) dt – λ

2λ a

2

x

(x – t) sin[λ(x – t)]y(t) dt = fxx (x).

a

Eliminating the second integral on the left-hand side of this equation with the aid of the original equation, we arrive at an equation of the form 1.5.1:

x

cos[λ(x – t)]y(t) dt = a

1 f (x) + λ2 f (x) . 2λ xx

Solution: y(x) = 63.

1 1 f (x) + λfx (x) + λ3 2λ xxx 2

x

f (t) dt. a

x

Axβ + B sinγ (λt) + C]y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B sinγ (λt) + C.

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x

A sinγ (λx) + Btβ + C]y(t) dt = f (x).

64.

a

65.

This is a special case of equation 1.9.6 with g(x) = A sinγ (λx) and h(t) = Btβ + C. x

Axλ sinµ t + Btβ sinγ x y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = sinµ t, g2 (x) = B sinγ x, and h2 (t) = tβ . 1.5-3. Kernels Containing Tangent 66.

x

tan(λx) – tan(λt) y(t) dt = f (x).

a

67.

This is a special case of equation 1.9.2 with g(x) = tan(λx). 1 d 2 Solution: y(x) = cos (λx)fx (x) . λ dx x A tan(λx) + B tan(λt) y(t) dt = f (x).

68.

For B = –A, see equation 1.5.66. This is a special case g(x) = tan(λx). x of equation B1.9.4 with – A – 1 d Solution: y(x) = tan(λx) A+B tan(λt) A+B ft (t) dt . A + B dx a x A tan(λx) + B tan(µt) + C y(t) dt = f (x).

69.

This is a special case of equation 1.9.6 with g(x) = A tan(λx) and h(t) = B tan(µt) + C. x 2 tan (λx) – tan2 (λt) y(t) dt = f (x).

a

a

a

70.

2 This is a special case of equation 1.9.2 with g(x) = tan (λx). 3 d cos (λx)fx (x) . Solution: y(x) = dx 2λ sin(λx) x A tan2 (λx) + B tan2 (λt) y(t) dt = f (x). a

71.

For B = –A, see equation 1.5.69. This is a special case 1.9.4 with g(x) = tan2 (λx). x of equation2B – – 2A 1 d tan(λt) A+B ft (t) dt . Solution: y(x) = tan(λx) A+B A + B dx a x A tan2 (λx) + B tan2 (µt) + C y(t) dt = f (x).

72.

This is a special case of equation 1.9.6 with g(x) = A tan2 (λx) and h(t) = B tan2 (µt) + C. x n n = 1, 2, . . . tan(λx) – tan(λt) y(t) dt = f (x),

a

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 1 d 2 Solution: y(x) = n f (x). cos (λx) λ n! cos2 (λx) dx

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x

73.

√

tan x – tan t y(t) dt = f (x).

a

Solution: y(x) = 74.

x

√

y(t) dt

tan x – tan t Solution:

2 d 2 2 x cos π cos2 x dx

x

a

f (t) dt √ . cos2 t tan x – tan t

= f (x).

a

1 d y(x) = π dx

a

x

cos2

f (t) dt √ . t tan x – tan t

x

(tan x – tan t)λ y(t) dt = f (x),

75.

0 < λ < 1.

a

Solution: y(x) =

sin(πλ) 2 d 2 cos x πλ cos2 x dx

a

x

f (t) dt . cos2 t(tan x – tan t)λ

x

(tanµ x – tanµ t)y(t) dt = f (x).

76. a

77.

This is a special case of equation 1.9.2 with g(x) = tanµ x. µ+1 1 d cos xfx (x) Solution: y(x) = . µ dx sinµ–1 x x

A tanµ x + B tanµ t y(t) dt = f (x). a

79.

For B = –A, see equation 1.5.76. This is a special case of equation 1.9.4 with g(x) = tanµ x. Solution: x – Bµ – Aµ 1 d A+B A+B y(x) = tan(λt) ft (t) dt . tan(λx) A + B dx a x y(t) dt = f (x), 0 < µ < 1. [tan(λx) – tan(λt)]µ a This is a special case of equation 1.9.42 with g(x) = tan(λx) and h(x) ≡ 1. Solution: x λ sin(πµ) d f (t) dt y(x) = . 2 π dx a cos (λt)[tan(λx) – tan(λt)]1–µ x Axβ + B tanγ (λt) + C]y(t) dt = f (x).

80.

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B tanγ (λt) + C. x A tanγ (λx) + Btβ + C]y(t) dt = f (x).

78.

a

a

81.

This is a special case of equation 1.9.6 with g(x) = A tanγ (λx) and h(t) = Btβ + C. x

Axλ tanµ t + Btβ tanγ x y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = tanµ t, g2 (x) = B tanγ x, and h2 (t) = tβ .

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1.5-4. Kernels Containing Cotangent 82.

x

cot(λx) – cot(λt) y(t) dt = f (x).

a

83.

This is a special case of equation 1.9.2 with g(x) = cot(λx). 1 d 2 Solution: y(x) = – sin (λx)fx (x) . λ dx x A cot(λx) + B cot(λt) y(t) dt = f (x). a

For B = –A, see equation 1.5.82. This is a special case of equation 1.9.4 with g(x) = cot(λx). x B A 1 d Solution: y(x) = tan(λt) A+B ft (t) dt . tan(λx) A+B A + B dx a 84.

x

A cot(λx) + B cot(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A cot(λx) and h(t) = B cot(µt) + C. 85.

cot2 (λx) – cot2 (λt) y(t) dt = f (x).

x

a 2 This is a special case of equation 3 1.9.2 withg(x) = cot (λx). d sin (λx)fx (x) Solution: y(x) = – . dx 2λ cos(λx)

86.

x

A cot2 (λx) + B cot2 (λt) y(t) dt = f (x).

a 2 For B = –A, see equation 1.5.85. This is a special case 1.9.4 with g(x) = cot (λx). x of equation 2A 2B 1 d tan(λt) A+B ft (t) dt . Solution: y(x) = tan(λx) A+B A + B dx a

87.

A cot2 (λx) + B cot2 (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cot2 (λx) and h(t) = B cot2 (µt) + C. 88.

x

n

cot(λx) – cot(λt)

y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 d (–1)n 2 f (x). Solution: y(x) = n sin (λx) λ n! sin2 (λx) dx

x

(cotµ x – cotµ t)y(t) dt = f (x).

89. a

µ This is a special case of equation with g(x) 1.9.2 = cot x. µ+1 1 d sin xfx (x) Solution: y(x) = – . µ dx cosµ–1 x

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90.

x

A cotµ x + B cotµ t y(t) dt = f (x).

a

For B = –A, see equation 1.5.89. This is a special case of equation 1.9.4 with g(x) = cotµ x. Solution: y(x) =

x Bµ Aµ 1 d tan t A+B ft (t) dt . tan x A+B A + B dx a

x

Axβ + B cotγ (λt) + C]y(t) dt = f (x).

91.

a

This is a special case of equation 1.9.6 with g(x) = Axβ and h(t) = B cotγ (λt) + C.

x

A cotγ (λx) + Btβ + C]y(t) dt = f (x).

92.

a

This is a special case of equation 1.9.6 with g(x) = A cotγ (λx) and h(t) = Btβ + C. 93.

x

Axλ cotµ t + Btβ cotγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = cotµ t, g2 (x) = B cotγ x, and h2 (t) = tβ .

1.5-5. Kernels Containing Combinations of Trigonometric Functions 94.

x

cos[λ(x – t)] + A sin[µ(x – t)] y(t) dt = f (x).

a

Differentiating the equation with respect to x followed by eliminating the integral with the cosine yields an equation of the form 2.3.16: 2

y(x) – (λ + A µ)

x

sin[µ(x – t)] y(t) dt = fx (x) – Aµf (x).

a

95.

x

A cos(λx) + B sin(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A cos(λx) and h(t) = B sin(µt) + C. 96.

x

A sin(λx) + B cos(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A sin(λx) and h(t) = B cos(µt) + C. 97.

A cos2 (λx) + B sin2 (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cos2 (λx) and h(t) = B sin2 (µt).

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x

sin[λ(x – t)] cos[λ(x + t)]y(t) dt = f (x),

98. a

f (a) = fx (a) = 0.

Using the trigonometric formula sin(α – β) cos(α + β) =

1 2

sin(2α) – sin(2β) ,

α = λx,

β = λt,

we reduce the original equation to an equation of the form 1.5.37:

x

sin(2λx) – sin(2λt) y(t) dt = 2f (x).

a

1 d fx (x) Solution: y(x) = . λ dx cos(2λx)

x

cos[λ(x – t)] sin[λ(x + t)]y(t) dt = f (x).

99. a

Using the trigonometric formula cos(α – β) sin(α + β) =

1 2

sin(2α) + sin(2β) ,

α = λx,

β = λt,

we reduce the original equation to an equation of the form 1.5.38 with A = B = 1:

x

sin(2λx) + sin(2λt) y(t) dt = 2f (x).

a

Solution with sin(2λx) > 0: y(x) = 100.

x d 1 f (t) dt √ √t . dx sin(2λx) a sin(2λt)

x

A cos(λx) sin(µt) + B cos(βx) sin(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = A cos(λx), h1 (t) = sin(µt), g2 (x) = B cos(βx), and h2 (t) = sin(γt). 101.

x

A sin(λx) cos(µt) + B sin(βx) cos(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = A sin(λx), h1 (t) = cos(µt), g2 (x) = B sin(βx), and h2 (t) = cos(γt). 102.

x

A cos(λx) cos(µt) + B sin(βx) sin(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = A cos(λx), h1 (t) = cos(µt), g2 (x) = B sin(βx), and h2 (t) = sin(γt). 103.

A cosβ (λx) + B sinγ (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cosβ (λx) and h(t) = B sinγ (µt).

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104.

A sinβ (λx) + B cosγ (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A sinβ (λx) and h(t) = B cosγ (µt). 105.

x

Axλ cosµ t + Btβ sinγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = cosµ t, g2 (x) = B sinγ x, and h2 (t) = tβ . 106.

x

Axλ sinµ t + Btβ cosγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = sinµ t, g2 (x) = B cosγ x, and h2 (t) = tβ . 107.

x

(x – t) sin[λ(x – t)] – λ(x – t)2 cos[λ(x – t)] y(t) dt = f (x).

a

Solution:

x

y(x) =

g(t) dt, a

where

g(t) =

x

108.

sin[λ(x – t)] x–t

a

π 1 2λ 64λ5

1 y(x) = 2λ4 109.

6

t

(t – τ )5/2 J5/2 [λ(t – τ )] f (τ ) dτ . a

– λ cos[λ(x – t)] y(t) dt = f (x).

Solution:

d2 + λ2 dt2

d2 + λ2 dx2

3

x

sin[λ(x – t)]f (t) dt. a

√

√

√ sin λ x – t – λ x – t cos λ x – t y(t) dt = f (x),

x

a

Solution: 4 d3 y(x) = πλ3 dx3 110.

a

x

f (a) = fx (a) = 0.

√

cosh λ x – t √ f (t) dt. x–t

x

A tan(λx) + B cot(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A tan(λx) and h(t) = B cot(µt) + C. 111.

A tan2 (λx) + B cot2 (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A tan2 (λx) and h(t) = B cot2 (µt). 112.

x

tan(λx) cot(µt) + tan(βx) cot(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = tan(λx), h1 (t) = cot(µt), g2 (x) = tan(βx), and h2 (t) = cot(γt).

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113.

x

cot(λx) tan(µt) + cot(βx) tan(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = cot(λx), h1 (t) = tan(µt), g2 (x) = cot(βx), and h2 (t) = tan(γt). 114.

x

tan(λx) tan(µt) + cot(βx) cot(γt) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = tan(λx), h1 (t) = tan(µt), g2 (x) = cot(βx), and h2 (t) = cot(γt). 115.

A tanβ (λx) + B cotγ (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A tanβ (λx) and h(t) = B cotγ (µt). 116.

A cotβ (λx) + B tanγ (µt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cotβ (λx) and h(t) = B tanγ (µt). 117.

x

Axλ tanµ t + Btβ cotγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = tanµ t, g2 (x) = B cotγ x, and h2 (t) = tβ . 118.

x

Axλ cotµ t + Btβ tanγ x y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = cotµ t, g2 (x) = B tanγ x, and h2 (t) = tβ .

1.6. Equations Whose Kernels Contain Inverse Trigonometric Functions 1.6-1. Kernels Containing Arccosine 1.

x

arccos(λx) – arccos(λt) y(t) dt = f (x).

a

2.

This is a special case of equation 1.9.2 with g(x) = arccos(λx). 1 d √ Solution: y(x) = – 1 – λ2 x2 fx (x) . λ dx x A arccos(λx) + B arccos(λt) y(t) dt = f (x). a

For B = –A, see equation 1.6.1. This is a special case of equation 1.9.4 with g(x) = arccos(λx). Solution: x – B – A 1 d A+B A+B y(x) = ft (t) dt . arccos(λt) arccos(λx) A + B dx a

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3.

x

A arccos(λx) + B arccos(µt) + C y(t) dt = f (x).

a

4.

This is a special case of equation 1.9.6 with g(x) = arccos(λx) and h(t) = B arccos(µt) + C. x n n = 1, 2, . . . arccos(λx) – arccos(λt) y(t) dt = f (x), a

5.

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. Solution: √ n+1 (–1)n d √ y(x) = 1 – λ2 x2 f (x). dx λn n! 1 – λ2 x2 x arccos(λt) – arccos(λx) y(t) dt = f (x). a

6.

7.

This is a special case of equation 1.9.38 with g(x) = 1 – arccos(λx). Solution: 2 x 2 ϕ(t)f (t) dt 1 1 d √ y(x) = ϕ(x) . , ϕ(x) = √ π ϕ(x) dx arccos(λt) – arccos(λx) 1 – λ2 x2 a x y(t) dt = f (x). √ arccos(λt) – arccos(λx) a Solution: x λ d ϕ(t)f (t) dt 1 √ y(x) = . , ϕ(x) = √ π dx a arccos(λt) – arccos(λx) 1 – λ2 x2 x µ 0 < µ < 1. arccos(λt) – arccos(λx) y(t) dt = f (x), a

Solution:

8.

2 x 1 d ϕ(t)f (t) dt y(x) = kϕ(x) , ϕ(x) dx [arccos(λt) – arccos(λx)]µ a 1 sin(πµ) ϕ(x) = √ . , k= πµ 1 – λ2 x2

x

arccosµ (λx) – arccosµ (λt) y(t) dt = f (x).

a

9.

This is a special case of equation 1.9.2√with g(x) = arccosµ (λx). 1 d fx (x) 1 – λ2 x2 Solution: y(x) = – . λµ dx arccosµ–1 (λx) x y(t) dt µ = f (x), 0 < µ < 1. arccos(λt) – arccos(λx) a Solution: λ sin(πµ) d y(x) = π dx

10.

a

x

ϕ(t)f (t) dt , [arccos(λt) – arccos(λx)]1–µ

ϕ(x) = √

1 1 – λ2 x2

.

A arccosβ (λx) + B arccosγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A arccosβ (λx) and h(t) = B arccosγ (µt)+C.

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1.6-2. Kernels Containing Arcsine 11.

x

arcsin(λx) – arcsin(λt) y(t) dt = f (x).

a

12.

This is a special case of equation 1.9.2 with g(x) = arcsin(λx). 1 d √ Solution: y(x) = 1 – λ2 x2 fx (x) . λ dx x A arcsin(λx) + B arcsin(λt) y(t) dt = f (x). a

For B = –A, see equation 1.6.11. This is a special case of equation 1.9.4 with g(x) = arcsin(λx). Solution: x B – A sign x d arcsin(λt)– A+B ft (t) dt . y(x) = arcsin(λx) A+B A + B dx a 13.

x

A arcsin(λx) + B arcsin(µt) + C y(t) dt = f (x).

a

14.

This is a special case of equation 1.9.6 with g(x) = A arcsin(λx) and h(t) = B arcsin(µt) + C. x n arcsin(λx) – arcsin(λt) y(t) dt = f (x), n = 1, 2, . . . a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. Solution: √ n+1 1 d √ y(x) = 1 – λ2 x2 f (x). dx λn n! 1 – λ2 x2

x

15.

arcsin(λx) – arcsin(λt) y(t) dt = f (x).

a

Solution: y(x) = 16.

x

√

2 x 2 ϕ(t)f (t) dt 1 d √ ϕ(x) , π ϕ(x) dx arcsin(λx) – arcsin(λt) a

ϕ(x) = √

1 1 – λ2 x2

.

y(t) dt

= f (x). arcsin(λx) – arcsin(λt) Solution: x λ d ϕ(t)f (t) dt 1 √ y(x) = . , ϕ(x) = √ π dx a arcsin(λx) – arcsin(λt) 1 – λ2 x2 x µ arcsin(λx) – arcsin(λt) y(t) dt = f (x), 0 < µ < 1. a

17.

a

Solution: 2 x 1 d ϕ(t)f (t) dt y(x) = kϕ(x) , µ ϕ(x) dx a [arcsin(λx) – arcsin(λt)] 1 sin(πµ) ϕ(x) = √ . , k= 2 2 πµ 1–λ x

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18.

arcsinµ (λx) – arcsinµ (λt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.2 with g(x) = arcsinµ (λx). √ 1 d fx (x) 1 – λ2 x2 Solution: y(x) = . λµ dx arcsinµ–1 (λx)

x

19. a

y(t) dt µ = f (x), arcsin(λx) – arcsin(λt)

0 < µ < 1.

Solution: y(x) = 20.

λ sin(πµ) d π dx

x

a

ϕ(t)f (t) dt , [arcsin(λx) – arcsin(λt)]1–µ

ϕ(x) = √

1 1 – λ2 x2

.

A arcsinβ (λx) + B arcsinγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A arcsinβ (λx) and h(t) = B arcsinγ (µt)+C.

1.6-3. Kernels Containing Arctangent 21.

x

arctan(λx) – arctan(λt) y(t) dt = f (x).

a

This is a special case of equation 1.9.2 with g(x) = arctan(λx). 1 d Solution: y(x) = (1 + λ2 x2 ) fx (x) . λ dx 22.

x

A arctan(λx) + B arctan(λt) y(t) dt = f (x).

a

For B = –A, see equation 1.6.21. This is a special case of equation 1.9.4 with g(x) = arctan(λx). Solution: y(x) = 23.

x B – A sign x d arctan(λt)– A+B ft (t) dt . arctan(λx) A+B A + B dx a

x

A arctan(λx) + B arctan(µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A arctan(λx) and h(t) = B arctan(µt) + C. 24.

x

n

arctan(λx) – arctan(λt)

y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. Solution: n+1 1 2 2 d y(x) = n f (x). (1 + λ x ) λ n! (1 + λ2 x2 ) dx

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x

25.

arctan(λx) – arctan(λt) y(t) dt = f (x).

a

Solution: y(x) =

x

√

26. a

2 x 2 ϕ(t)f (t) dt 1 d √ ϕ(x) , π ϕ(x) dx arctan(λx) – arctan(λt) a y(t) dt

ϕ(x) =

1 . 1 + λ2 x2

= f (x).

arctan(λx) – arctan(λt)

Solution: y(x) =

x

27.

√

a

λ d π dx

x

√

a

ϕ(t)f (t) dt , arctan(λx) – arctan(λt)

ϕ(x) =

1 . 1 + λ2 x2

x–t y(t) dt = f (x). t arctan t

The equation can be rewritten in terms of the Gaussian hypergeometric function in the form

x

a

x

(x – t)γ–1 F α, β, γ; 1 – y(t) dt = f (x), t

where

α = 12 ,

β = 1,

γ = 32 .

See 1.8.86 for the solution of this equation. 28.

x

µ

arctan(λx) – arctan(λt)

y(t) dt = f (x),

0 < µ < 1.

a

Solution:

2

x

ϕ(t)f (t) dt , [arctan(λx) – arctan(λt)]µ a 1 sin(πµ) ϕ(x) = , k= . 2 2 1+λ x πµ

y(x) = kϕ(x)

29.

1 d ϕ(x) dx

x

arctanµ (λx) – arctanµ (λt) y(t) dt = f (x).

a

This is a special case of equation g(x) = arctanµ (λx). 1.9.22 with 2 1 d (1 + λ x )fx (x) Solution: y(x) = . λµ dx arctanµ–1 (λx)

x

30. a

y(t) dt µ = f (x), arctan(λx) – arctan(λt)

0 < µ < 1.

Solution: λ sin(πµ) d y(x) = π dx 31.

a

x

ϕ(t)f (t) dt , [arctan(λx) – arctan(λt)]1–µ

ϕ(x) =

1 . 1 + λ2 x2

A arctanβ (λx) + B arctanγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A arctanβ (λx) and h(t) = B arctanγ (µt)+C.

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1.6-4. Kernels Containing Arccotangent 32.

x

arccot(λx) – arccot(λt) y(t) dt = f (x).

a

33.

This is a special case of equation 1.9.2 with g(x) = arccot(λx). 1 d Solution: y(x) = – (1 + λ2 x2 ) fx (x) . λ dx x A arccot(λx) + B arccot(λt) y(t) dt = f (x). a

For B = –A, see equation 1.6.32. This is a special case of equation 1.9.4 with g(x) = arccot(λx). Solution: x – B – A 1 d A+B A+B y(x) = ft (t) dt . arccot(λt) arccot(λx) A + B dx a 34.

x

A arccot(λx) + B arccot(µt) + C y(t) dt = f (x).

a

35.

This is a special case of equation 1.9.6 with g(x) = A arccot(λx) and h(t) = B arccot(µt) + C. x n n = 1, 2, . . . arccot(λx) – arccot(λt) y(t) dt = f (x), a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. Solution: n+1 (–1)n 2 2 d y(x) = n x ) f (x). (1 + λ λ n! (1 + λ2 x2 ) dx

x

36.

arccot(λt) – arccot(λx) y(t) dt = f (x).

a

Solution: y(x) = 37.

x

√

2 x 2 ϕ(t)f (t) dt 1 d √ ϕ(x) , π ϕ(x) dx arccot(λt) – arccot(λx) a

ϕ(x) =

1 . 1 + λ2 x2

y(t) dt

= f (x). arccot(λt) – arccot(λx) Solution: x λ d ϕ(t)f (t) dt 1 √ y(x) = . , ϕ(x) = π dx a 1 + λ2 x2 arccot(λt) – arccot(λx) x µ 0 < µ < 1. arccot(λt) – arccot(λx) y(t) dt = f (x), a

38.

a

Solution:

2

x

ϕ(t)f (t) dt , µ a [arccot(λt) – arccot(λx)] 1 sin(πµ) ϕ(x) = , k= . 1 + λ2 x2 πµ

y(x) = kϕ(x)

1 d ϕ(x) dx

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39.

arccotµ (λx) – arccotµ (λt) y(t) dt = f (x).

x

a

This is a special case of equation g(x) = arccotµ (λx). 1.9.22with 2 1 d (1 + λ x )fx (x) Solution: y(x) = – . λµ dx arccotµ–1 (λx)

x

40. a

y(t) dt µ = f (x), arccot(λt) – arccot(λx)

Solution: y(x) = 41.

λ sin(πµ) d π dx

a

x

0 < µ < 1.

ϕ(t)f (t) dt , [arccot(λt) – arccot(λx)]1–µ

ϕ(x) =

1 . 1 + λ2 x2

x

A arccotβ (λx) + B arccotγ (µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A arccotβ (λx) and h(t) = B arccotγ (µt)+C.

1.7. Equations Whose Kernels Contain Combinations of Elementary Functions 1.7-1. Kernels Containing Exponential and Hyperbolic Functions

x

1.

eµ(x–t) A1 cosh[λ1 (x – t)] + A2 cosh[λ2 (x – t)] y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.8: x A1 cosh[λ1 (x – t)] + A2 cosh[λ2 (x – t)] w(t) dt = e–µx f (x). a

x

eµ(x–t) cosh2 [λ(x – t)]y(t) dt = f (x).

2. a

Solution:

3.

2λ2 x µ(x–t) y(x) = ϕ(x) – e sinh[k(x – t)]ϕ(x) dt, k a x eµ(x–t) cosh3 [λ(x – t)]y(t) dt = f (x).

√ k = λ 2,

ϕ(x) = fx (x) – µf (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.15: x cosh3 [λ(x – t)]w(t) dt = e–µx f (x). a

x

eµ(x–t) cosh4 [λ(x – t)]y(t) dt = f (x).

4. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.19: x cosh4 [λ(x – t)]w(t) dt = e–µx f (x). a

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x

5.

n eµ(x–t) cosh(λx) – cosh(λt) y(t) dt = f (x),

n = 1, 2, . . .

a

Solution: n+1 1 µx 1 d y(x) = n e sinh(λx) Fµ (x), λ n! sinh(λx) dx

x

eµ(x–t)

6.

√

cosh x – cosh t y(t) dt = f (x),

Fµ (x) = e–µx f (x).

f (a) = 0.

a

Solution: y(x) =

7.

1 2 µx d 2 x e–µt sinh t f (t) dt √ . e sinh x π sinh x dx cosh x – cosh t a

x

eµ(x–t) y(t) dt = f (x). √ cosh x – cosh t a Solution: y(x) =

1 µx d e π dx

x

a

e–µt sinh t f (t) dt √ . cosh x – cosh t

x

eµ(x–t) (cosh x – cosh t)λ y(t) dt = f (x),

8.

0 < λ < 1.

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.23: x (cosh x – cosh t)λ w(t) dt = e–µx f (x). a

9.

Aeµ(x–t) + B coshλ x y(t) dt = f (x).

x

a

10.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B coshλ x, and h2 (t) = 1. x µ(x–t) + B coshλ t y(t) dt = f (x). Ae a

11.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = coshλ t. x eµ(x–t) (coshλ x – coshλ t)y(t) dt = f (x). a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.24: x (coshλ x – coshλ t)w(t) dt = e–µx f (x). a

x

12.

eµ(x–t) A coshλ x + B coshλ t y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.25: x

A coshλ x + B coshλ t w(t) dt = e–µx f (x). a

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13.

x

eµ(x–t) y(t) dt

(cosh x – cosh t)λ Solution:

= f (x),

0 < λ < 1.

a

y(x) =

x

14.

sin(πλ) µx d e π dx

a

x

e–µt sinh t f (t) dt . (cosh x – cosh t)1–λ

eµ(x–t) A1 sinh[λ1 (x – t)] + A2 sinh[λ2 (x – t)] y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.41: x A1 sinh[λ1 (x – t)] + A2 sinh[λ2 (x – t)] w(t) dt = e–µx f (x). a

x

eµ(x–t) sinh2 [λ(x – t)]y(t) dt = f (x).

15. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.43: x sinh2 [λ(x – t)]w(t) dt = e–µx f (x). a

x

eµ(x–t) sinh3 [λ(x – t)]y(t) dt = f (x).

16. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.49: x sinh3 [λ(x – t)]w(t) dt = e–µx f (x). a

x

eµ(x–t) sinhn [λ(x – t)]y(t) dt = f (x),

17.

n = 2, 3, . . .

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.54: x sinhn [λ(x – t)]w(t) dt = e–µx f (x). a

x

18.

√

eµ(x–t) sinh k x – t y(t) dt = f (x).

a

Solution: 2 µx d2 y(x) = e πk dx2

x

eµ(x–t)

19.

√

√

e–µt cos k x – t √ f (t) dt. x–t

x

a

sinh x – sinh t y(t) dt = f (x).

a

Solution: y(x) =

20.

1 2 µx d 2 x e–µt cosh t f (t) dt √ . e cosh x π cosh x dx sinh x – sinh t a

x

eµ(x–t) y(t) dt = f (x). √ sinh x – sinh t a Solution:

1 d y(x) = eµx π dx

a

x

e–µt cosh t f (t) dt √ . sinh x – sinh t

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x

eµ(x–t) (sinh x – sinh t)λ y(t) dt = f (x),

21.

0 < λ < 1.

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.58: x (sinh x – sinh t)λ w(t) dt = e–µx f (x). a

x

eµ(x–t) (sinhλ x – sinhλ t)y(t) dt = f (x).

22. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.59: x (sinhλ x – sinhλ t)w(t) dt = e–µx f (x). a

x

23.

eµ(x–t) A sinhλ x + B sinhλ t y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.60: x

A sinhλ x + B sinhλ t w(t) dt = e–µx f (x). a

24.

Aeµ(x–t) + B sinhλ x y(t) dt = f (x).

x

a

25.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B sinhλ x, and h2 (t) = 1. x µ(x–t) Ae + B sinhλ t y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = sinhλ t. 26.

x

eµ(x–t) y(t) dt

(sinh x – sinh t)λ Solution:

= f (x),

0 < λ < 1.

a

y(x) =

x

27.

sin(πλ) µx d e π dx

a

x

e–µt cosh t f (t) dt . (sinh x – sinh t)1–λ

eµ(x–t) A tanhλ x + B tanhλ t y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.77: x

A tanhλ x + B tanhλ t w(t) dt = e–µx f (x). a

x

28.

eµ(x–t) A tanhλ x + B tanhβ t + C y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.9.6 with g(x) = A tanhλ x, g(t) = B tanhβ t + C: x

A tanhλ x + B tanhβ t + C w(t) dt = e–µx f (x). a

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29.

Aeµ(x–t) + B tanhλ x y(t) dt = f (x).

x

a

30.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B tanhλ x, and h2 (t) = 1. x µ(x–t) Ae + B tanhλ t y(t) dt = f (x). a

31.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = tanhλ t. x

eµ(x–t) A cothλ x + B cothλ t y(t) dt = f (x). a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.3.90: x

A cothλ x + B cothλ t w(t) dt = e–µx f (x). a

x

32.

eµ(x–t) A cothλ x + B cothβ t + C y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.9.6 with g(x) = A cothλ x, h(t) = B cothβ t + C: x

A cothλ x + B cothβ t + C w(t) dt = e–µx f (x). a

33.

Aeµ(x–t) + B cothλ x y(t) dt = f (x).

x

a

34.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B cothλ x, and h2 (t) = 1. x µ(x–t) + B cothλ t y(t) dt = f (x). Ae a

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = cothλ t. 1.7-2. Kernels Containing Exponential and Logarithmic Functions

x

eλ(x–t) (ln x – ln t)y(t) dt = f (x).

35. a

Solution:

y(x) = eλx xϕxx (x) + ϕx (x) ,

ϕ(x) = e–λx f (x).

x

eλ(x–t) ln(x – t)y(t) dt = f (x).

36. 0

The substitution w(x) = e–λx y(x) leads to an equation of the form 1.4.2: x ln(x – t)w(t) dt = e–λx f (x). 0

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x

eλ(x–t) (A ln x + B ln t)y(t) dt = f (x).

37. a

The substitution w(x) = e–λx y(x) leads to an equation of the form 1.4.4: x (A ln x + B ln t)w(t) dt = e–λx f (x). a

x

38.

eµ(x–t) A ln2 (λx) + B ln2 (λt) y(t) dt = f (x).

a

The substitution w(x) = e–λx y(x) leads to an equation of the form 1.4.7: x A ln2 (λx) + B ln2 (λt) w(t) dt = e–λx f (x). a

x

39.

n eλ(x–t) ln(x/t) y(t) dt = f (x),

n = 1, 2, . . .

a

Solution:

x

eλ(x–t)

40.

n+1 1 λx d y(x) = x Fλ (x), e n! x dx

Fλ (x) = e–λx f (x).

ln(x/t) y(t) dt = f (x).

a

Solution:

x

41. a

2 x –λt 2eλx e f (t) dt d y(x) = . x πx dx a t ln(x/t)

eλ(x–t) y(t) dt = f (x). ln(x/t)

Solution: 1 d y(x) = eλx π dx 42.

a

x

e–λt f (t) dt . t ln(x/t)

x

Aeµ(x–t) + B lnν (λx) y(t) dt = f (x).

a

43.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B lnν (λx), and h2 (t) = 1. x µ(x–t) Ae + B lnν (λt) y(t) dt = f (x). a

44.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = lnν (λt). x eµ(x–t) [ln(x/t)]λ y(t) dt = f (x), 0 < λ < 1. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.4.16: x [ln(x/t)]λ w(t) dt = e–µx f (x). a

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x

45. a

eµ(x–t) [ln(x/t)]λ

y(t) dt = f (x),

Solution: y(x) =

0 < λ < 1.

sin(πλ) µx d e π dx

x

a

f (t) dt teµt [ln(x/t)]1–λ

.

1.7-3. Kernels Containing Exponential and Trigonometric Functions

x

eµ(x–t) cos[λ(x – t)]y(t) dt = f (x).

46. a

Solution: y(x) = fx (x) – µf (x) + λ2

x

eµ(x–t) f (t) dt. a

x

47.

eµ(x–t) A1 cos[λ1 (x – t)] + A2 cos[λ2 (x – t)] y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.8: x A1 cos[λ1 (x – t)] + A2 cos[λ2 (x – t)] w(t) dt = e–µx f (x). a

x

eµ(x–t) cos2 [λ(x – t)]y(t) dt = f (x).

48. a

49.

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.9. Solution: √ 2λ2 x µ(x–t) y(x) = ϕ(x) + e sin[k(x – t)]ϕ(t) dt, k = λ 2, ϕ(x) = fx (x) – µf (x). k a x eµ(x–t) cos3 [λ(x – t)]y(t) dt = f (x). a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.15: x cos3 [λ(x – t)]w(t) dt = e–µx f (x). a

x

eµ(x–t) cos4 [λ(x – t)]y(t) dt = f (x).

50. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.19: x cos4 [λ(x – t)]w(t) dt = e–µx f (x). a

x

51.

n eµ(x–t) cos(λx) – cos(λt) y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. Solution: n+1 (–1)n µx 1 d y(x) = n e sin(λx) Fµ (x), Fµ (x) = e–µx f (x). λ n! sin(λx) dx

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x

eµ(x–t)

52.

√

cos t – cos x y(t) dt = f (x).

a

1 d 2 x e–µt sin t f (t) dt 2 µx √ y(x) = e sin x . π sin x dx cos t – cos x a

Solution:

53.

x

eµ(x–t) y(t) dt = f (x). √ cos t – cos x a Solution:

1 d y(x) = eµx π dx

a

x

e–µt sin t f (t) dt √ . cos t – cos x

x

eµ(x–t) (cos t – cos x)λ y(t) dt = f (x),

54.

0 < λ < 1.

a

Solution: µx

y(x) = ke

1 d 2 x e–µt sin t f (t) dt sin x , λ sin x dx a (cos t – cos x)

k=

sin(πλ) . πλ

x

eµ(x–t) (cosλ x – cosλ t)y(t) dt = f (x).

55. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.24: x (cosλ x – cosλ t)w(t) dt = e–µx f (x). a

x

56.

eµ(x–t) A cosλ x + B cosλ t y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.25: x

A cosλ x + B cosλ t w(t) dt = e–µx f (x). a

x

57. a

eµ(x–t) y(t) dt (cos t – cos x)λ

= f (x),

0 < λ < 1.

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.26: x w(t) dt = e–µx f (x). λ a (cos t – cos x) 58.

Aeµ(x–t) + B cosν (λx) y(t) dt = f (x).

x

a

59.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B cosν (λx), and h2 (t) = 1. x µ(x–t) Ae + B cosν (λt) y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = cosν (λt).

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x

f (a) = fx (a) = 0.

eµ(x–t) sin[λ(x – t)]y(t) dt = f (x),

60. a

Solution: y(x) =

x

61.

1 λ

fxx (x) – 2µfx (x) + (λ2 + µ2 )f (x) .

eµ(x–t) A1 sin[λ1 (x – t)] + A2 sin[λ2 (x – t)] y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.41: x A1 sin[λ1 (x – t)] + A2 sin[λ2 (x – t)] w(t) dt = e–µx f (x). a

x

eµ(x–t) sin2 [λ(x – t)]y(t) dt = f (x).

62. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.43: x sin2 [λ(x – t)]w(t) dt = e–µx f (x). a

x

eµ(x–t) sin3 [λ(x – t)]y(t) dt = f (x).

63. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.49: x sin3 [λ(x – t)]w(t) dt = e–µx f (x). a

x

eµ(x–t) sinn [λ(x – t)]y(t) dt = f (x),

64.

n = 2, 3, . . .

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.54: x sinn [λ(x – t)]w(t) dt = e–µx f (x). a

x

65.

√

eµ(x–t) sin k x – t y(t) dt = f (x).

a

Solution: 2 µx d2 y(x) = e πk dx2

x

eµ(x–t)

66.

√

x

a

√

e–µt cosh k x – t √ f (t) dt. x–t

sin x – sin t y(t) dt = f (x).

a

Solution:

67.

1 d 2 x e–µt cos t f (t) dt 2 µx √ y(x) = e cos x . π cos x dx sin x – sin t a

x

eµ(x–t) y(t) dt = f (x). √ sin x – sin t a Solution:

1 d y(x) = eµx π dx

a

x

e–µt cos t f (t) dt √ . sin x – sin t

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x

eµ(x–t) (sin x – sin t)λ y(t) dt = f (x),

68.

0 < λ < 1.

a

Solution: µx

y(x) = ke

1 d 2 x e–µt cos t f (t) dt cos x , cos x dx (sin x – sin t)λ a

k=

sin(πλ) . πλ

x

eµ(x–t) (sinλ x – sinλ t)y(t) dt = f (x).

69. a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.59: x (sinλ x – sinλ t)w(t) dt = e–µx f (x). a

x

70.

eµ(x–t) A sinλ x + B sinλ t y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.60: x

A sinλ x + B sinλ t w(t) dt = e–µx f (x). a

71.

x

eµ(x–t) y(t) dt

= f (x), 0 < λ < 1. (sin x – sin t)λ The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.61: x w(t) dt = e–µx f (x). λ a (sin x – sin t) x µ(x–t) Ae + B sinν (λx) y(t) dt = f (x). a

72.

a

73.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B sinν (λx), and h2 (t) = 1. x µ(x–t) + B sinν (λt) y(t) dt = f (x). Ae

74.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = sinν (λt). x

eµ(x–t) A tanλ x + B tanλ t y(t) dt = f (x).

a

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.77: x

A tanλ x + B tanλ t w(t) dt = e–µx f (x). a

x

75.

eµ(x–t) A tanλ x + B tanβ t + C y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.9.6: x

A tanλ x + B tanβ t + C w(t) dt = e–µx f (x). a

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76.

Aeµ(x–t) + B tanν (λx) y(t) dt = f (x).

x

a

77.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B tanν (λx), and h2 (t) = 1. x µ(x–t) Ae + B tanν (λt) y(t) dt = f (x). a

78.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = tanν (λt). x

eµ(x–t) A cotλ x + B cotλ t y(t) dt = f (x). a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.5.90: x

A cotλ x + B cotλ t w(t) dt = e–µx f (x). a

x

79.

eµ(x–t) A cotλ x + B cotβ t + C y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 1.9.6: x

A cotλ x + B cotβ t + C w(t) dt = e–µx f (x). a

80.

x

Aeµ(x–t) + B cotν (λx) y(t) dt = f (x).

a

81.

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B cotν (λx), and h2 (t) = 1. x µ(x–t) Ae + B cotν (λt) y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = Aeµx , h1 (t) = e–µt , g2 (x) = B, and h2 (t) = cotν (λt). 1.7-4. Kernels Containing Hyperbolic and Logarithmic Functions 82.

x

A coshβ (λx) + B lnγ (µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A coshβ (λx) and h(t) = B lnγ (µt) + C. 83.

x

A coshβ (λt) + B lnγ (µx) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = B lnγ (µx) + C and h(t) = A coshβ (λt). 84.

A sinhβ (λx) + B lnγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A sinhβ (λx) and h(t) = B lnγ (µt) + C.

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85.

A sinhβ (λt) + B lnγ (µx) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = B lnγ (µx) and h(t) = A sinhβ (λt) + C. 86.

x

A tanhβ (λx) + B lnγ (µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A tanhβ (λx) and h(t) = B lnγ (µt) + C. 87.

A tanhβ (λt) + B lnγ (µx) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = B lnγ (µx) and h(t) = A tanhβ (λt) + C. 88.

A cothβ (λx) + B lnγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cothβ (λx) and h(t) = B lnγ (µt) + C. 89.

x

A cothβ (λt) + B lnγ (µx) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = B lnγ (µx) and h(t) = A cothβ (λt) + C. 1.7-5. Kernels Containing Hyperbolic and Trigonometric Functions 90.

x

A coshβ (λx) + B cosγ (µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A coshβ (λx) and h(t) = B cosγ (µt) + C. 91.

x

A coshβ (λt) + B sinγ (µx) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = B sinγ (µx) + C and h(t) = A coshβ (λt). 92.

x

A coshβ (λx) + B tanγ (µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A coshβ (λx) and h(t) = B tanγ (µt) + C. 93.

A sinhβ (λx) + B cosγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A sinhβ (λx) and h(t) = B cosγ (µt) + C. 94.

A sinhβ (λt) + B sinγ (µx) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = B sinγ (µx) and h(t) = A sinhβ (λt) + C. 95.

A sinhβ (λx) + B tanγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A sinhβ (λx) and h(t) = B tanγ (µt) + C.

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96.

A tanhβ (λx) + B cosγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A tanhβ (λx) and h(t) = B cosγ (µt) + C. 97.

x

A tanhβ (λx) + B sinγ (µt) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = A tanhβ (λx) and h(t) = B sinγ (µt) + C.

1.7-6. Kernels Containing Logarithmic and Trigonometric Functions 98.

A cosβ (λx) + B lnγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A cosβ (λx) and h(t) = B lnγ (µt) + C. 99.

x

A cosβ (λt) + B lnγ (µx) + C y(t) dt = f (x).

a

This is a special case of equation 1.9.6 with g(x) = B lnγ (µx) + C and h(t) = A cosβ (λt). 100.

A sinβ (λx) + B lnγ (µt) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = A sinβ (λx) and h(t) = B lnγ (µt) + C. 101.

A sinβ (λt) + B lnγ (µx) + C y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = B lnγ (µx) and h(t) = A sinβ (λt) + C.

1.8. Equations Whose Kernels Contain Special Functions 1.8-1. Kernels Containing Bessel Functions

x

J0 [λ(x – t)]y(t) dt = f (x).

1. a

Solution: 1 y(x) = λ

d2 + λ2 dx2

2

x

(x – t) J1 [λ(x – t)] f (t) dt. a

Example. In the special case λ = 1 and f (x) = A sin x, the solution has the form y(x) = AJ0 (x).

x

[J0 (λx) – J0 (λt)]y(t) dt = f (x).

2. a

d fx (x) Solution: y(x) = – . dx λJ1 (λx)

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x

[AJ0 (λx) + BJ0 (λt)]y(t) dt = f (x).

3. a

For B = –A, see equation 1.8.2. We consider the interval [a, x] in which J0 (λx) does not change its sign. Solution with B ≠ –A: x – A – B 1 d A+B A+B y(x) = ± ft (t) dt . J0 (λt) J0 (λx) A + B dx a

4.

Here the sign of J0 (λx) should be taken. x (x – t) J0 [λ(x – t)]y(t) dt = f (x). a

Solution:

x

y(x) =

g(t) dt, a

where g(t) =

1 λ

d2 + λ2 dt2

3

t

(t – τ ) J1 [λ(t – τ )] f (τ ) dτ . a

x

(x – t)J1 [λ(x – t)]y(t) dt = f (x).

5. a

Solution:

6.

x

x–t

2

1 3λ3

y(x) =

d2 + λ2 dx2

4

x

(x – t)2 J2 [λ(x – t)] f (t) dt. a

J1 [λ(x – t)]y(t) dt = f (x).

a

Solution:

x

y(x) =

g(t) dt, a

where 1 g(t) = 9λ3 7.

x

x–t

n

d2 + λ2 dt2

5

t

t–τ

2

J2 [λ(t – τ )] f (τ ) dτ .

a

Jn [λ(x – t)]y(t) dt = f (x),

n = 0, 1, 2, . . .

a

Solution: y(x) = A

d2 + λ2 dx2 A=

2n+2

x

(x – t)n+1 Jn+1 [λ(x – t)] f (t) dt, 2 λ

a

2n+1

n! (n + 1)! . (2n)! (2n + 2)!

If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(2n+1) (a) = 0 are satisfied, then the solution of the integral equation can be written in the form y(x) = A

x

(x – t) a

2n+1

J2n+1 [λ(x – t)]F (t) dt,

F (t) =

d2 + λ2 dt2

2n+2 f (t) dt.

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8.

x

x–t

n+1

Jn [λ(x – t)]y(t) dt = f (x),

n = 0, 1, 2, . . .

a

Solution:

x

y(x) =

g(t) dt, a

where

2n+3 t d2 2 g(t) = A +λ (t – τ )n+1 Jn+1 [λ(t – τ )] f (τ ) dτ , dt2 a 2 2n+1 n! (n + 1)! A= . λ (2n + 1)! (2n + 2)! If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(2n+2) (a) = 0 are satisfied, then the function g(t) defining the solution can be written in the form

t

g(t) = A

(t – τ )

n–ν–2

Jn–ν–2 [λ(t – τ )]F (τ ) dτ ,

F (τ ) =

a

d2 + λ2 dτ 2

2n+2 f (τ ).

x

(x – t)1/2 J1/2 [λ(x – t)]y(t) dt = f (x).

9. a

Solution: y(x) =

π 4λ2

d2 + λ2 dx2

3

x

(x – t)3/2 J3/2 [λ(x – t)] f (t) dt. a

x

(x – t)3/2 J1/2 [λ(x – t)]y(t) dt = f (x).

10. a

Solution:

x

y(x) =

g(t) dt, a

where g(t) =

π d2 2 + λ 8λ2 dt2

4

t

(t – τ )3/2 J3/2 [λ(t – τ )] f (τ ) dτ . a

x

(x – t)3/2 J3/2 [λ(x – t)]y(t) dt = f (x).

11. a

Solution:

√

π y(x) = 3/2 5/2 2 λ

d2 + λ2 dx2

3

x

sin[λ(x – t)] f (t) dt. a

x

(x – t)5/2 J3/2 [λ(x – t)]y(t) dt = f (x).

12. a

Solution:

x

y(x) =

g(t) dt, a

where π g(t) = 128λ4

d2 + λ2 dt2

6

t

(t – τ )5/2 J5/2 [λ(t – τ )] f (τ ) dτ . a

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x

(x – t)

13.

2n–1 2

J 2n–1 [λ(x – t)]y(t) dt = f (x),

a

n = 2, 3, . . .

2

Solution: √ y(x) = √

2λ

2n+1 2

π

(2n – 2)!!

d2 + λ2 dx2

n

x

sin[λ(x – t)] f (t) dt. a

x

[Jν (λx) – Jν (λt)]y(t) dt = f (x).

14. a

This is a special case of equation 1.9.2 with g(x) =Jν (λx). d xfx (x) Solution: y(x) = . dx νJν (λx) – λxJν+1 (λx)

x

[AJν (λx) + BJν (λt)]y(t) dt = f (x).

15. a

For B = –A, see equation 1.8.14. We consider the interval [a, x] in which Jν (λx) does not change its sign. Solution with B ≠ –A: x – A – B 1 d A+B A+B y(x) = ± ft (t) dt . Jν (λt) Jν (λx) A + B dx a Here the sign of Jν (λx) should be taken.

x

[AJν (λx) + BJµ (βt)]y(t) dt = f (x).

16. a

This is a special case of equation 1.9.6 with g(x) = AJν (λx) and h(t) = BJµ (βt).

x

(x – t)ν Jν [λ(x – t)]y(t) dt = f (x).

17. a

Solution:

n x d2 2 y(x) = A +λ (x – t)n–ν–1 Jn–ν–1 [λ(x – t)] f (t) dt, dx2 a 2 n–1 Γ(ν + 1) Γ(n – ν) A= , λ Γ(2ν + 1) Γ(2n – 2ν – 1) where – 12 < ν < n–1 2 and n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0 are satisfied, then the solution of the integral equation can be written in the form

x

(x – t)n–ν–1 Jn–ν–1 [λ(x – t)]F (t) dt,

y(x) = A a

F (t) =

d2 + λ2 dt2

n f (t).

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC Page 80

x

(x – t)ν+1 Jν [λ(x – t)]y(t) dt = f (x).

18. a

Solution:

y(x) =

x

g(t) dt, a

where

n t d2 2 g(t) = A +λ (t – τ )n–ν–2 Jn–ν–2 [λ(t – τ )] f (τ ) dτ , dt2 a 2 n–2 Γ(ν + 1) Γ(n – ν – 1) A= , λ Γ(2ν + 2) Γ(2n – 2ν – 3) where –1 < ν < n2 – 1 and n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0 are satisfied, then the function g(t) defining the solution can be written in the form

t

(t – τ )n–ν–2 Jn–ν–2 [λ(t – τ )]F (τ ) dτ ,

g(t) = A

F (τ ) =

a

d2 + λ2 dτ 2

n f (τ ).

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

19.

√ J0 λ x – t y(t) dt = f (x).

a

Solution:

20.

d2 y(x) = dx2

x

√ I0 λ x – t f (t) dt.

a

√ √ AJν λ x + BJν λ t y(t) dt = f (x).

x

a

√ We consider the interval [a, x] in which Jν λ x does not change its sign. Solution with B ≠ –A: x A √ – B 1 d √ – A+B A+B y(x) = ± Jν λ t ft (t) dt . Jν λ x A + B dx a √ Here the sign Jν λ x should be taken. 21.

√ √ AJν λ x + BJµ β t y(t) dt = f (x).

x

a

√ √ This is a special case of equation 1.9.6 with g(x) = AJν λ x and h(t) = BJµ β t .

x

22.

√

√ x – t J1 λ x – t y(t) dt = f (x).

a

Solution:

2 d3 y(x) = λ dx3

x

√ I0 λ x – t f (t) dt.

a

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x

23.

√ (x – t)1/4 J1/2 λ x – t y(t) dt = f (x).

a

Solution: y(x) =

x

24.

2 d2 πλ dx2

√

cosh λ x – t √ f (t) dt. x–t

x

a

√ (x – t)3/4 J3/2 λ x – t y(t) dt = f (x).

a

Solution:

23/2 d3 y(x) = √ 3/2 3 dx πλ

x

25.

x

a

√

cosh λ x – t √ f (t) dt. x–t

√

(x – t)n/2 Jn λ x – t y(t) dt = f (x),

n = 0, 1, 2, . . .

a

Solution: y(x) =

x

26.

x–t

2n–3 4

√

J 2n–3 λ x – t y(t) dt = f (x),

a

1 y(x) = √ π x

27.

n = 1, 2, . . .

2

Solution:

2 n dn+2 x √

I0 λ x – t f (t) dt. n+2 λ dx a

√

2n–3 n x 2 cosh λ x – t d 2 √ f (t) dt. λ dxn a x–t

√ (x – t)–1/4 J–1/2 λ x – t y(t) dt = f (x).

a

Solution: y(x) =

x

28.

λ d 2π dx

x

a

√

cosh λ x – t √ f (t) dt. x–t

√

(x – t)ν/2 Jν λ x – t y(t) dt = f (x).

a

2 n–2 dn x

n–ν–2 √ 2 In–ν–2 λ x – t f (t) dt, x–t n λ dx a where –1 < ν < n – 1, n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0 are satisfied, then the solution of the integral equation can be written in the form 2 n–2 x

n–ν–2 √ 2 y(x) = In–ν–2 λ x – t ft(n) (t) dt. x–t λ a x 2 2 –1/4 √

J–1/2 λ x2 – t2 y(t) dt = f (x). x –t

Solution:

y(x) =

29.

0

Solution: y(x) =

2λ d π dx

0

x

√

cosh λ x2 – t2 √ t f (t) dt. x2 – t2

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

© 1998 by CRC Press LLC

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30.

∞

t 2 – x2

–1/4

√

J–1/2 λ t2 – x2 y(t) dt = f (x).

x

Solution:

2λ d π dx

y(x) = – 31.

x

x2 – t 2

ν/2

∞

x

√ cosh λ t2 – x2 √ t f (t) dt. t2 – x2

√

Jν λ x2 – t2 y(t) dt = f (x),

–1 < ν < 0.

0

Solution:

d y(x) = λ dx

x

–(ν+1)/2 √

t x2 – t2 I–ν–1 λ x2 – t2 f (t) dt.

0

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993). 32.

∞

t 2 – x2

ν/2

√

Jν λ t2 – x2 y(t) dt = f (x),

–1 < ν < 0.

x

Solution: y(x) = –λ

d dx

∞

–(ν+1)/2 √

t t2 – x2 I–ν–1 λ t2 – x2 f (t) dt.

x

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

[Atk Jν (λx) + Bxm Jµ (λt)]y(t) dt = f (x).

33. a

This is a special case of equation 1.9.15 with g1 (x) = AJν (λx), h1 (t) = tk , g2 (x) = Bxm , and h2 (t) = Jµ (λt).

x

34. a

[AJν2 (λx) + BJν2 (λt)]y(t) dt = f (x).

Solution with B ≠ –A: y(x) = 35.

x – 2A 2B 1 d Jν (λt)– A+B ft (t) dt . Jν (λx) A+B A + B dx a

AJνk (λx) + BJµm (βt) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.6 with g(x) = AJνk (λx) and h(t) = BJµm (βt).

x

[Y0 (λx) – Y0 (λt)]y(t) dt = f (x).

36. a

Solution: y(x) = –

d fx (x) . dx λY1 (λx)

x

[Yν (λx) – Yν (λt)]y(t) dt = f (x).

37. a

d xfx (x) Solution: y(x) = . dx νYν (λx) – λxYν+1 (λx)

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x

[AYν (λx) + BYν (λt)]y(t) dt = f (x).

38. a

For B = –A, see equation 1.8.37. We consider the interval [a, x] in which Yν (λx) does not change its sign. Solution with B ≠ –A: x – A – B 1 d A+B A+B y(x) = ± ft (t) dt . Yν (λt) Yν (λx) A + B dx a

39.

Here the sign of Yν (λx) should be taken. x [Atk Yν (λx) + Bxm Yµ (λt)]y(t) dt = f (x).

40.

This is a special case of equation 1.9.15 with g1 (x) = AYν (λx), h1 (t) = tk , g2 (x) = Bxm , and h2 (t) = Yµ (λt). x [AJν (λx)Yµ (βt) + BJν (λt)Yµ (βx)]y(t) dt = f (x).

a

a

This is a special case of equation 1.9.15 with g1 (x) = AYν (λx), h1 (t) = Yµ (βt), g2 (x) = BYµ (βx), and h2 (t) = Jν (λt). 1.8-2. Kernels Containing Modified Bessel Functions

x

I0 [λ(x – t)]y(t) dt = f (x).

41. a

Solution: y(x) =

1 λ

d2 – λ2 dx2

2

x

a

43.

(x – t) I1 [λ(x – t)] f (t) dt. a

f (a) = fx (a) = 0.

[I0 (λx) – I0 (λt)]y(t) dt = f (x),

42.

x

d fx (x) Solution: y(x) = . dx λI1 (λx) x [AI0 (λx) + BI0 (λt)]y(t) dt = f (x). a

For B = –A, see equation 1.8.42. Solution with B ≠ –A: x – A – B 1 d A+B A+B y(x) = ± ft (t) dt . I0 (λt) I0 (λx) A + B dx a

44.

Here the sign of Iν (λx) should be taken. x (x – t)I0 [λ(x – t)]y(t) dt = f (x). a

Solution:

x

y(x) =

g(t) dt, a

where 1 g(t) = λ

d2 – λ2 dt2

3

t

(t – τ ) I1 [λ(t – τ )] f (τ ) dτ . a

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x

(x – t)I1 [λ(x – t)]y(t) dt = f (x).

45. a

Solution: 1 3λ3

y(x) =

d2 – λ2 dx2

4

x

(x – t)2 I2 [λ(x – t)] f (t) dt. a

x

(x – t)2 I1 [λ(x – t)]y(t) dt = f (x).

46. a

Solution:

x

y(x) =

g(t) dt, a

where 1 9λ3

g(t) =

d2 – λ2 dt2

5

t

t–τ

2

I2 [λ(t – τ )] f (τ ) dτ .

a

x

(x – t)n In [λ(x – t)]y(t) dt = f (x),

47.

n = 0, 1, 2, . . .

a

Solution: y(x) = A

d2 – λ2 dx2 A=

2n+2

x

(x – t)n+1 In+1 [λ(x – t)] f (t) dt, 2

a

2n+1

n! (n + 1)! . (2n)! (2n + 2)!

λ

If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(2n+1) (a) = 0 are satisfied, then the solution of the integral equation can be written in the form

x

(x – t)2n+1 I2n+1 [λ(x – t)]F (t) dt,

y(x) = A

F (t) =

a

d2 – λ2 dt2

2n+2 f (t).

x

(x – t)n+1 In [λ(x – t)]y(t) dt = f (x),

48.

n = 0, 1, 2, . . .

a

Solution: y(x) =

x

g(t) dt, a

where

2n+3 t d2 2 g(t) = A –λ (t – τ )n+1 In+1 [λ(t – τ )] f (τ ) dτ , dt2 a 2 2n+1 n! (n + 1)! A= . λ (2n + 1)! (2n + 2)! If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(2n+2) (a) = 0 are satisfied, then the function g(t) defining the solution can be written in the form 2n+2 2 t d n–ν–2 2 g(t) = A (t – τ ) In–ν–2 [λ(t – τ )]F (τ ) dτ , F (τ ) = –λ f (τ ). dτ 2 a

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x

(x – t)1/2 I1/2 [λ(x – t)]y(t) dt = f (x).

49. a

Solution: y(x) =

π 4λ2

d2 – λ2 dx2

3

x

(x – t)3/2 I3/2 [λ(x – t)] f (t) dt. a

x

(x – t)3/2 I1/2 [λ(x – t)]y(t) dt = f (x).

50. a

Solution:

x

y(x) =

g(t) dt, a

where π g(t) = 8λ2

d2 – λ2 dt2

4

t

(t – τ )3/2 I3/2 [λ(t – τ )] f (τ ) dτ . a

x

(x – t)3/2 I3/2 [λ(x – t)]y(t) dt = f (x).

51. a

Solution:

√

y(x) =

π 3/2 2 λ5/2

d2 – λ2 dx2

3

x

sinh[λ(x – t)] f (t) dt. a

x

(x – t)5/2 I3/2 [λ(x – t)]y(t) dt = f (x).

52. a

Solution:

x

y(x) =

g(t) dt, a

where g(t) = 53.

x

x–t

2n–1 2

π 128λ4

d2 – λ2 dt2

6

t

(t – τ )5/2 I5/2 [λ(t – τ )] f (τ ) dτ . a

I 2n–1 [λ(x – t)]y(t) dt = f (x),

a

n = 2, 3, . . .

2

Solution: √ y(x) = √

2n+1 2λ 2

π

(2n – 2)!!

d2 – λ2 dx2

n

x

sinh[λ(x – t)] f (t) dt. a

x

[Iν (λx) – Iν (λt)]y(t) dt = f (x).

54. a

This is a special case of equation 1.9.2 with g(x) = Iν (λx).

x

[AIν (λx) + BIν (λt)]y(t) dt = f (x).

55. a

Solution with B ≠ –A: x – A – B 1 d A+B A+B y(x) = ft (t) dt . Iν (λt) Iν (λx) A + B dx a

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x

[AIν (λx) + BIµ (βt)]y(t) dt = f (x).

56. a

This is a special case of equation 1.9.6 with g(x) = AIν (λx) and h(t) = BIµ (βt).

x

(x – t)ν Iν [λ(x – t)]y(t) dt = f (x).

57. a

Solution:

n x d2 2 y(x) = A –λ (x – t)n–ν–1 In–ν–1 [λ(x – t)] f (t) dt, dx2 a 2 n–1 Γ(ν + 1) Γ(n – ν) A= , λ Γ(2ν + 1) Γ(2n – 2ν – 1) where – 12 < ν < n–1 2 and n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0 are satisfied, then the solution of the integral equation can be written in the form

x

(x – t)n–ν–1 In–ν–1 [λ(x – t)]F (t) dt,

y(x) = A

F (t) =

a

d2 – λ2 dt2

n f (t).

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

(x – t)ν+1 Iν [λ(x – t)]y(t) dt = f (x).

58. a

Solution: y(x) =

x

g(t) dt, a

where

n t d2 2 g(t) = A –λ (t – τ )n–ν–2 In–ν–2 [λ(t – τ )] f (τ ) dτ , dt2 a 2 n–2 Γ(ν + 1) Γ(n – ν – 1) A= , λ Γ(2ν + 2) Γ(2n – 2ν – 3) where –1 < ν < n2 – 1 and n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0 are satisfied, then the function g(t) defining the solution can be written in the form n 2 t d n–ν–2 2 g(t) = A (t – τ ) In–ν–2 [λ(t – τ )]F (τ ) dτ , F (τ ) = –λ f (τ ). dτ 2 a • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

59.

√ I0 λ x – t y(t) dt = f (x).

a

Solution:

d2 y(x) = dx2

x

√ J0 λ x – t f (t) dt.

a

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60.

√ √ AIν λ x + BIν λ t y(t) dt = f (x).

x

a

Solution with B ≠ –A: y(x) = 61.

x A √ – B 1 d √ – A+B A+B f (t) dt . Iν λ x Iν λ t t A + B dx a

√ √ AIν λ x + BIµ β t y(t) dt = f (x).

x

a

√ √ This is a special case of equation 1.9.6 with g(x) = AIν λ x and h(t) = BIµ β t .

x

62.

√

√

x – t I1 λ x – t y(t) dt = f (x).

a

Solution:

x

63.

2 d3 y(x) = λ dx3

√

J0 λ x – t f (t) dt.

x

a

√

(x – t)1/4 I1/2 λ x – t y(t) dt = f (x).

a

Solution: y(x) =

x

64.

2 d2 πλ dx2

√

cos λ x – t √ f (t) dt. x–t

x

a

√ (x – t)3/4 I3/2 λ x – t y(t) dt = f (x).

a

Solution:

23/2 d3 y(x) = √ 3/2 3 dx πλ

x

65.

x

a

√ (x – t)n/2 In λ x – t y(t) dt = f (x),

√

cos λ x – t √ f (t) dt. x–t n = 0, 1, 2, . . .

a

Solution:

66.

x

x–t

2 n dn+2 x √

y(x) = J0 λ x – t f (t) dt. λ dxn+2 a

2n–3 4

√ I 2n–3 λ x – t y(t) dt = f (x),

a

Solution: 1 y(x) = √ π

x

67.

n = 1, 2, . . .

2

√

2n–3 n x 2 cos λ x – t d 2 √ f (t) dt. λ dxn a x–t

√

(x – t)–1/4 I–1/2 λ x – t y(t) dt = f (x).

a

Solution: y(x) =

λ d 2π dx

a

x

√

cos λ x – t √ f (t) dt. x–t

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x

68.

√ (x – t)ν/2 Iν λ x – t y(t) dt = f (x).

a

Solution: y(x) =

2 n–2 dn x

x–t n λ dx a

n–ν–2 2

√ Jn–ν–2 λ x – t f (t) dt,

where –1 < ν < n – 1, n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0 are satisfied, then the solution of the integral equation can be written in the form y(x) =

2 n–2 λ

x

x–t

n–ν–2 2

√

Jn–ν–2 λ x – t ft(n) (t) dt.

a

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993). 69.

x

x2 – t 2

–1/4

√

I–1/2 λ x2 – t2 y(t) dt = f (x).

0

Solution:

y(x) =

70.

∞

t 2 – x2

–1/4

2λ d π dx

x

0

√ cos λ x2 – t2 √ t f (t) dt. x2 – t2

√

I–1/2 λ t2 – x2 y(t) dt = f (x).

x

Solution:

y(x) = –

71.

x

x2 – t 2

ν/2

2λ d π dx

∞

x

√ cos λ t2 – x2 √ t f (t) dt. t2 – x2

√

Iν λ x2 – t2 y(t) dt = f (x),

–1 < ν < 0.

0

Solution: y(x) = λ

d dx

x

–(ν+1)/2 √

t x2 – t2 J–ν–1 λ x2 – t2 f (t) dt.

0

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

∞

72.

√

(t2 – x2 )ν/2 Iν λ t2 – x2 y(t) dt = f (x),

–1 < ν < 0.

x

Solution: y(x) = –λ

d dx

∞

√

t (t2 – x2 )–(ν+1)/2 J–ν–1 λ t2 – x2 f (t) dt.

x

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

[Atk Iν (λx) + Bxs Iµ (λt)]y(t) dt = f (x).

73. a

This is a special case of equation 1.9.15 with g1 (x) = AIν (λx), h1 (t) = tk , g2 (x) = Bxs , and h2 (t) = Iµ (λt).

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x

74. a

[AIν2 (λx) + BIν2 (λt)]y(t) dt = f (x).

Solution with B ≠ –A: y(x) =

x – 2A 2B 1 d Iν (λt)– A+B ft (t) dt . Iν (λx) A+B A + B dx a

x

75. a

[AIνk (λx) + BIµs (βt)]y(t) dt = f (x).

This is a special case of equation 1.9.6 with g(x) = AIνk (λx) and h(t) = BIµs (βt).

x

[K0 (λx) – K0 (λt)]y(t) dt = f (x).

76. a

77.

d fx (x) Solution: y(x) = – . dx λK1 (λx) x [Kν (λx) – Kν (λt)]y(t) dt = f (x).

78.

This is a special case of equation 1.9.2 with g(x) = Kν (λx). x [AKν (λx) + BKν (λt)]y(t) dt = f (x).

a

a

Solution with B ≠ –A: y(x) =

x – A – B 1 d Kν (λx) A+B Kν (λt) A+B ft (t) dt . A + B dx a

x

[Atk Kν (λx) + Bxs Kµ (λt)]y(t) dt = f (x).

79. a

80.

This is a special case of equation 1.9.15 with g1 (x) = AKν (λx), h1 (t) = tk , g2 (x) = Bxs , and h2 (t) = Kµ (λt). x [AIν (λx)Kµ (βt) + BIν (λt)Kµ (βx)]y(t) dt = f (x). a

This is a special case of equation 1.9.15 with g1 (x) = AIν (λx), h1 (t) = Kµ (βt), g2 (x) = BKµ (βx), and h2 (t) = Iν (λt). 1.8-3. Kernels Containing Associated Legendre Functions 81.

x

x

y(t) dt = f (x), 0 < a < ∞. t Here Pνµ (x) is the associated Legendre function (see Supplement 10). Solution: x n+µ–2 t dn y(x) = xn+µ–1 n x1–µ (x2 – t2 ) 2 t–n Pν2–n–µ f (t) dt , dx x a a

(x2 – t2 )–µ/2 Pνµ

where µ < 1, ν ≥ – 12 , and n = 1, 2, . . . • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

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82.

x

t

y(t) dt = f (x), 0 < a < ∞. x Here Pνµ (x) is the associated Legendre function (see Supplement 10). Solution: x x

n+µ–2 dn y(x) = (x2 – t2 ) 2 Pν2–n–µ f (t) dt, n dx a t (x2 – t2 )–µ/2 Pνµ

a

where µ < 1, ν ≥ – 12 , and n = 1, 2, . . . • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993). 83.

b

x

y(t) dt = f (x), 0 < b < ∞. t Here Pνµ (x) is the associated Legendre function (see Supplement 10). Solution: b n+µ–2 t dn y(x) = (–1)n xn+µ–1 n x1–µ (t2 – x2 ) 2 t–n Pν2–n–µ f (t) dt , dx x x x

(t2 – x2 )–µ/2 Pνµ

where µ < 1, ν ≥ – 12 , and n = 1, 2, . . . • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993). 84.

b

t

y(t) dt = f (x), 0 < b < ∞. x Here Pνµ (x) is the associated Legendre function (see Supplement 10). Solution:

n b n+µ–2 n d 2 2 2–n–µ x 2 y(x) = (–1) (t – x ) P f (t) dt, ν dxn x t x

(t2 – x2 )–µ/2 Pνµ

where µ < 1, ν ≥ – 12 , and n = 1, 2, . . . • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993). 1.8-4. Kernels Containing Hypergeometric Functions 85.

x

(x – t)b–1 Φ a, b; λ(x – t) y(t) dt = f (x).

s

Here Φ(a, b; z) is the degenerate hypergeometric function (see Supplement 10). Solution: x

dn (x – t)n–b–1 y(x) = Φ –a, n – b; λ(x – t) f (t) dt, n dx s Γ(b)Γ(n – b) where 0 < b < n and n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (s) = fx (s) = · · · = fx(n–1) (s) = 0 are satisfied, then the solution of the integral equation can be written in the form x

(x – t)n–b–1 y(x) = Φ –a, n – b; λ(x – t) ft(n) (t) dt. Γ(b)Γ(n – b) s • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

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x (x – t)c–1 F a, b, c; 1 – y(t) dt = f (x). t s Here Φ(a, b, c; z) is the Gaussian hypergeometric function (see Supplement 10). Solution: x n t

(x – t)n–c–1 –a d a y(x) = x F –a, n – b, n – c; 1 – f (t) dt , x dxn x s Γ(c)Γ(n – c)

86.

x

where 0 < c < n and n = 1, 2, . . . If the right-hand side of the equation is differentiable sufficiently many times and the conditions f (s) = fx (s) = · · · = fx(n–1) (s) = 0 are satisfied, then the solution of the integral equation can be written in the form x (x – t)n–c–1 t (n) y(x) = F –a, –b, n – c; 1 – f (t) dt. x t s Γ(c)Γ(n – c) • Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

1.9. Equations Whose Kernels Contain Arbitrary Functions 1.9-1. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + g2 (x)h2 (t)

x

g(x)h(t)y(t) dt = f (x).

1. a

2.

1 d f (x) g (x) 1 Solution: y = = fx (x) – 2 x f (x). h(x) dx g(x) g(x)h(x) g (x)h(x) x [g(x) – g(t)]y(t) dt = f (x).

3.

It is assumed that f (a) = fx (a) = 0 and fx /gx ≠ const. d fx (x) Solution: y(x) = . dx gx (x) x [g(x) – g(t) + b]y(t) dt = f (x).

a

a

Differentiation with respect to x yields an equation of the form 2.9.2: x 1 1 y(x) + gx (x) y(t) dt = fx (x). b b a Solution:

1 1 y(x) = fx (x) – 2 gx (x) b b

x

exp a

g(t) – g(x) b

ft (t) dt.

x

[Ag(x) + Bg(t)]y(t) dt = f (x).

4. a

For B = –A, see equation 1.9.2. Solution with B ≠ –A:

x – B – A sign g(x) d A+B A+B y(x) = ft (t) dt . g(t) g(x) A + B dx a

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x

[Ag(x) + Bg(t) + C]y(t) dt = f (x).

5. a

For B = –A, see equation 1.9.3. Assume that B ≠ –A and (A + B)g(x) + C > 0. Solution: x – B – A d A+B A+B y(x) = ft (t) dt . (A + B)g(t) + C (A + B)g(x) + C dx a

x

[g(x) + h(t)]y(t) dt = f (x).

6. a

Solution: y(x) = 7.

x d Φ(x) ft (t) dt , dx g(x) + h(x) a Φ(t)

Φ(x) = exp

x

a

ht (t) dt . g(t) + h(t)

x

g(x) + (x – t)h(x) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = g(x) + xh(x), h1 (t) = 1, g2 (x) = h(x), and h2 (t) = –t. Solution: d dt h(x) x f (t) y(x) = Φ(x) , dx g(x) a h(t) t Φ(t) 8.

Φ(t) = exp –

x

a

h(t) dt . g(t)

x

g(t) + (x – t)h(t) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = x, h1 (t) = h(t), g2 (x) = 1, and h2 (t) = g(t) – th(t). 9.

g(x) + (Axλ + Btµ )h(x) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.15 with g1 (x) = g(x) + Axλ h(x), h1 (t) = 1, g2 (x) = h(x), and h2 (t) = Btµ . 10.

g(t) + (Axλ + Btµ )h(t) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.15 with g1 (x) = Axλ , h1 (t) = h(t), g2 (x) = 1, and h2 (t) = g(t) + Btµ h(t).

x

[g(x)h(t) – h(x)g(t)]y(t) dt = f (x),

11. a

f (a) = fx (a) = 0.

For g = const or h = const, see equation 1.9.2. Solution: 1 d (f /h)x y(x) = , where f = f (x), h dx (g/h)x

g = g(x),

h = h(x).

Here Af + Bg + Ch ≡/ 0, with A, B, and C being some constants.

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x

[Ag(x)h(t) + Bg(t)h(x)]y(t) dt = f (x).

12. a

For B = –A, see equation 1.9.11. Solution with B ≠ –A: 1 d y(x) = (A + B)h(x) dx 13.

x

h(x) g(x)

A A+B

x a

h(t) g(t)

B A+B

d f (t) dt . dt h(t)

1 + [g(t) – g(x)]h(x) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = 1 – g(x)h(x), h1 (t) = 1, g2 (x) = h(x), and h2 (t) = g(t). Solution: y(x) = 14.

x

x d f (t) dt h(x)Φ(x) , dx h(t) t Φ(t) a

Φ(x) = exp

x

gt (t)h(t) dt .

a

e–λ(x–t) + eλx g(t) – eλt g(x) h(x) y(t) dt = f (x).

a

This is a special case of equation 1.9.15 with g1 (x) = eλx h(x), h1 (t) = g(t), g2 (x) = e–λx – g(x)h(x), and h2 (t) = eλt .

x

[g1 (x)h1 (t) + g2 (x)h2 (t)]y(t) dt = f (x).

15. a

For g2 /g1 = const or h2 /h1 = const, see equation 1.9.1. 1◦ . Solution with g1 (x)h1 (x) + g2 (x)h2 (x) ≡/ 0 and f (x) ≡/ const g2 (x): 1 d y(x) = h1 (x) dx where

g2 (x)h1 (x)Φ(x) g1 (x)h1 (x) + g2 (x)h2 (x)

x a

f (t) g2 (t)

dt , t Φ(t)

x h2 (t) g2 (t)h1 (t) dt Φ(x) = exp . h1 (t) t g1 (t)h1 (t) + g2 (t)h2 (t) a

(1)

(2)

If f (x) ≡ const g2 (x), the solution is given by formulas (1) and (2) in which the subscript 1 must be changed by 2 and vice versa. 2◦ . Solution with g1 (x)h1 (x) + g2 (x)h2 (x) ≡ 0: 1 d (f /g2 )x 1 d (f /g2 )x y(x) = =– , h1 dx (g1 /g2 )x h1 dx (h2 /h1 )x where f = f (x), g2 = g2 (x), h1 = h1 (x), and h2 = h2 (x).

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1.9-2. Equations With Difference Kernel: K(x, t) = K(x – t)

x

K(x – t)y(t) dt = f (x).

16. a ◦

1 . Let K(0) = 1 and f (a) = 0. Differentiating the equation with respect to x yields a Volterra equation of the second kind: x y(x) + Kx (x – t)y(t) dt = fx (x). a

The solution of this equation can be represented in the form x y(x) = fx (x) + R(x – t)ft (t) dt.

(1)

a

Here the resolvent R(x) is related to the kernel K(x) of the original equation by 1 ˜ R(x) = L–1 –1 , K(p) = L K(x) , ˜ pK(p) where L and L–1 are the operators of the direct and inverse Laplace transforms, respectively. ∞ c+i∞ 1 ˜ ˜ dp. ˜ K(p) = L K(x) = e–px K(x) dx, R(x) = L–1 R(p) epx R(p) = 2πi 0 c–i∞ 2◦ . Let K(x) have an integrable power-law singularity at x = 0. Denote by w = w(x) the solution of the simpler auxiliary equation (compared with the original equation) with a = 0 and constant right-hand side f ≡ 1, x K(x – t)w(t) dt = 1. (2) 0

17.

Then the solution of the original integral equation with arbitrary right-hand side is expressed in terms of w as follows: x x d y(x) = w(x – t)f (t) dt = f (a)w(x – a) + w(x – t)ft (t) dt. dx a a x K(x – t)y(t) dt = Axn , n = 0, 1, 2, . . . –∞

This is a special case of equation 1.9.19 with λ = 0. 1◦ . Solution with n = 0:

A y(x) = , B 2◦ . Solution with n = 1: A AC y(x) = x + 2 , B B

∞

K(z) dz.

B= 0

∞

B=

K(z) dz,

C=

∞

zK(z) dz.

0

0

3◦ . Solution with n = 2: A 2 AC AC 2 AD x +2 2 x+2 3 – 2 , B B B B ∞ ∞ ∞ B= K(z) dz, C = zK(z) dz, D = z 2 K(z) dz. y2 (x) =

0

0

0

◦

4 . Solution with n = 3, 4, . . . is given by: n λx e ∂ yn (x) = A , ∂λn B(λ) λ=0

B(λ) =

∞

K(z)e–λz dz.

0

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x

K(x – t)y(t) dt = Aeλx .

18. –∞

Solution:

A λx e , B

y(x) =

B=

∞

K(z)e–λz dz = L{K(z), λ}.

0

x

K(x – t)y(t) dt = Axn eλx ,

19.

n = 1, 2, . . .

–∞ ◦

1 . Solution with n = 1: A λx AC λx xe + 2 e , B B ∞ ∞ –λz B= K(z)e dz, C = zK(z)e–λz dz. y1 (x) =

0

0

It is convenient to calculate the coefficients B and C using tables of Laplace transforms according to the formulas B = L{K(z), λ} and C = L{zK(z), λ}. 2◦ . Solution with n = 2:

A 2 λx AC AC 2 AD x e + 2 2 xeλx + 2 3 – 2 eλx , B B B B ∞ ∞ ∞ B= K(z)e–λz dz, C = zK(z)e–λz dz, D = z 2 K(z)e–λz dz. y2 (x) =

0

0

0

◦

20.

3 . Solution with n = 3, 4, . . . is given by: λx ∂ ∂n e yn (x) = yn–1 (x) = A n , ∂λ ∂λ B(λ) x K(x – t)y(t) dt = A cosh(λx).

B(λ) =

∞

K(z)e–λz dz.

0

–∞

Solution: y(x) =

A λx 1 A A –λx 1 A A

A

cosh(λx) + sinh(λx), e + e = + – 2B– 2B+ 2 B– B+ 2 B– B+ ∞ ∞ B– = K(z)e–λz dz, B+ = K(z)eλz dz. 0

0

x

K(x – t)y(t) dt = A sinh(λx).

21. –∞

Solution: y(x) =

A λx 1 A A –λx 1 A A

A

cosh(λx) + sinh(λx), e – e = – + 2B– 2B+ 2 B– B+ 2 B– B+ ∞ ∞ B– = K(z)e–λz dz, B+ = K(z)eλz dz. 0

0

x

K(x – t)y(t) dt = A cos(λx).

22. –∞

Solution: A Bc cos(λx) – Bs sin(λx) , Bc2 + Bs2 ∞ ∞ Bc = K(z) cos(λz) dz, Bs = K(z) sin(λz) dz. y(x) =

0

0

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x

K(x – t)y(t) dt = A sin(λx).

23. –∞

Solution: A Bc sin(λx) + Bs cos(λx) , Bc2 + Bs2 ∞ ∞ Bc = K(z) cos(λz) dz, Bs = K(z) sin(λz) dz. y(x) =

0

0

x

K(x – t)y(t) dt = Aeµx cos(λx).

24. –∞

Solution: A eµx Bc cos(λx) – Bs sin(λx) , 2 + Bs ∞ ∞ Bc = K(z)e–µz cos(λz) dz, Bs = K(z)e–µz sin(λz) dz. y(x) =

Bc2

0

0

x

K(x – t)y(t) dt = Aeµx sin(λx).

25. –∞

Solution: A eµx Bc sin(λx) + Bs cos(λx) , 2 + Bs ∞ ∞ –µz Bc = K(z)e cos(λz) dz, Bs = K(z)e–µz sin(λz) dz. y(x) =

Bc2

0

0

x

K(x – t)y(t) dt = f (x).

26. –∞

1◦ . For a polynomial right-hand side of the equation, f (x) =

n

Ak xk , the solution has the

k=0

form y(x) =

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. The solution can also be obtained by the formula given in 1.9.17 (item 4◦ ). 2◦ . For f (x) = eλx

n

Ak xk , the solution has the form

k=0

y(x) = eλx

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. The solution can also be obtained by the formula given in 1.9.19 (item 3◦ ).

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n

3◦ . For f (x) =

Ak exp(λk x), the solution has the form

k=0

y(x) =

n Ak exp(λk x), Bk n

K(z) exp(–λk z) dz. 0

k=0

4◦ . For f (x) = cos(λx)

∞

Bk =

Ak xk , the solution has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n 5◦ . For f (x) = sin(λx) Ak xk , the solution has the form k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n 6◦ . For f (x) = Ak cos(λk x), the solution has the form k=0

Bck

n

Ak Bck cos(λk x) – Bsk sin(λk x) , 2 + Bsk ∞ ∞ = K(z) cos(λk z) dz, Bsk = K(z) sin(λk z) dz.

y(x) =

2 Bck k=0

0 n

7◦ . For f (x) =

0

Ak sin(λk x), the solution has the form

k=0

Bck

n

Ak Bck sin(λk x) + Bsk cos(λk x) , 2 + Bsk ∞ ∞ = K(z) cos(λk z) dz, Bsk = K(z) sin(λk z) dz.

y(x) =

2 Bck k=0

0

∞

27.

0

K(x – t)y(t) dt = Axn ,

n = 0, 1, 2, . . .

x

This is a special case of equation 1.9.29 with λ = 0. 1◦ . Solution with n = 0: y(x) =

A , B

∞

K(–z) dz.

B= 0

2◦ . Solution with n = 1: A AC y(x) = x – 2 , B B

B=

∞

K(–z) dz, 0

C=

∞

zK(–z) dz. 0

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3◦ . Solution with n = 2:

B=

y2 (x) = ∞

K(–z) dz,

A 2 AC AC 2 AD x –2 2 x+2 3 – 2 , B B B B ∞ ∞ C= zK(–z) dz, D = z 2 K(–z) dz.

0

0

0

4◦ . Solution with n = 3, 4, . . . is given by n λx e ∂ yn (x) = A , ∂λn B(λ) λ=0

∞

28.

∞

K(–z)eλz dz.

B(λ) = 0

K(x – t)y(t) dt = Aeλx .

x

Solution: y(x) =

29.

A λx e , B

∞

B=

K(–z)eλz dz.

0

The expression for B is the Laplace transform of the function K(–z) with parameter p = –λ and can be calculated with the aid of tables of Laplace transforms given (e.g., see Supplement 4). ∞ K(x – t)y(t) dt = Axn eλx , n = 1, 2, . . . x ◦

1 . Solution with n = 1: A λx AC λx xe – 2 e , B B ∞ ∞ B= K(–z)eλz dz, C = zK(–z)eλz dz. y1 (x) =

0

0

It is convenient to calculate the coefficients B and C using tables of Laplace transforms with parameter p = –λ. 2◦ . Solution with n = 2: A 2 λx AC λx AC 2 AD λx y2 (x) = x e – 2 2 xe + 2 3 – 2 e , B B B B ∞ ∞ ∞ λz λz B= K(–z)e dz, C = zK(–z)e dz, D = z 2 K(–z)eλz dz. 0

0

0

◦

3 . Solution with n = 3, 4, . . . is given by: λx ∂ ∂n e yn (x) = yn–1 (x) = A n , ∂λ ∂λ B(λ)

B(λ) =

∞

K(–z)eλz dz.

0

∞

K(x – t)y(t) dt = A cosh(λx).

30. x

Solution: y(x) =

A λx 1 A A –λx 1 A A

A

cosh(λx) + sinh(λx), e + e = + – 2B+ 2B– 2 B+ B– 2 B+ B– ∞ ∞ B+ = K(–z)eλz dz, B– = K(–z)e–λz dz. 0

0

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∞

K(x – t)y(t) dt = A sinh(λx).

31. x

Solution: y(x) =

A λx 1 A A –λx 1 A A

A

cosh(λx) + sinh(λx), e – e = – + 2B+ 2B– 2 B+ B– 2 B+ B– ∞ ∞ B+ = K(–z)eλz dz, B– = K(–z)e–λz dz. 0

0

∞

K(x – t)y(t) dt = A cos(λx).

32. x

Solution:

A Bc cos(λx) + Bs sin(λx) , 2 + Bs ∞ ∞ Bc = K(–z) cos(λz) dz, Bs = K(–z) sin(λz) dz. y(x) =

Bc2

0

0

∞

K(x – t)y(t) dt = A sin(λx).

33. x

Solution:

A Bc sin(λx) – Bs cos(λx) , 2 + Bs ∞ ∞ Bc = K(–z) cos(λz) dz, Bs = K(–z) sin(λz) dz. y(x) =

Bc2

0 ∞

34.

0

K(x – t)y(t) dt = Aeµx cos(λx).

x

Solution:

A eµx Bc cos(λx) + Bs sin(λx) , 2 + Bs ∞ ∞ Bc = K(–z)eµz cos(λz) dz, Bs = K(–z)eµz sin(λz) dz. y(x) =

Bc2

0 ∞

35.

0

K(x – t)y(t) dt = Aeµx sin(λx).

x

Solution:

A eµx Bc sin(λx) – Bs cos(λx) , 2 + Bs ∞ ∞ Bc = K(–z)eµz cos(λz) dz, Bs = K(–z)eµz sin(λz) dz. y(x) =

Bc2

0

0

∞

K(x – t)y(t) dt = f (x).

36. x

1◦ . For a polynomial right-hand side of the equation, f (x) =

n

Ak xk , the solution has the

k=0

form y(x) =

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. The solution can also be obtained by the formula given in 1.9.27 (item 4◦ ).

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2◦ . For f (x) = eλx

n

Ak xk , the solution has the form

k=0

y(x) = eλx

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. The solution can also be obtained by the formula given in 1.9.29 (item 3◦ ). n 3◦ . For f (x) = Ak exp(λk x), the solution has the form k=0

y(x) =

n Ak exp(λk x), Bk k=0

4◦ . For f (x) = cos(λx)

n

∞

Bk =

K(–z) exp(λk z) dz. 0

Ak xk , the solution has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n 5◦ . For f (x) = sin(λx) Ak xk , the solution has the form k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n 6◦ . For f (x) = Ak cos(λk x), the solution has the form k=0

Bck

n

Ak Bck cos(λk x) + Bsk sin(λk x) , 2 + Bsk ∞ ∞ = K(–z) cos(λk z) dz, Bsk = K(–z) sin(λk z) dz. y(x) =

2 Bck k=0

0 n

7◦ . For f (x) =

0

Ak sin(λk x), the solution has the form

k=0

Bck

n

Ak Bck sin(λk x) – Bsk cos(λk x) , 2 + B2 B ck sk k=0 ∞ ∞ = K(–z) cos(λk z) dz, Bsk = K(–z) sin(λk z) dz. y(x) =

0

0

8◦ . For arbitrary right-hand side f = f (x), the solution of the integral equation can be calculated by the formula c+i∞ ˜ 1 f (p) px y(x) = e dp, ˜ 2πi c–i∞ k(–p) ∞ ∞ ˜ f˜(p) = f (x)e–px dx, k(–p) = K(–z)epz dz. 0

0

˜ To calculate f˜(p) and k(–p), it is convenient to use tables of Laplace transforms, and to determine y(x), tables of inverse Laplace transforms.

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1.9-3. Other Equations

x

n

g(x) – g(t)

37.

y(t) dt = f (x),

n = 1, 2, . . .

a

The right-hand side of the equation is assumed to satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. n+1 1 1 d Solution: y(x) = gx (x) f (x). n! gx (x) dx

x

38.

g(x) – g(t) y(t) dt = f (x),

f (a) = 0.

a

Solution: y(x) =

x

√

39. a

y(t) dt g(x) – g(t)

2 x 2 1 d f (t)gt (t) dt √ gx (x) . π gx (x) dx g(x) – g(t) a gx > 0.

= f (x),

Solution: y(x) =

x

40. a

eλ(x–t) y(t) dt = f (x), √ g(x) – g(t)

1 d π dx

f (t)gt (t) dt √ . g(x) – g(t)

x

e–λt f (t)gt (t) √ dt. g(x) – g(t)

a

gx > 0.

Solution: 1 λx d e π dx

y(x) =

x

a

x

[g(x) – g(t)]λ y(t) dt = f (x),

41.

f (a) = 0,

0 < λ < 1.

a

Solution:

y(x) = kgx (x)

x

42. a

h(t)y(t) dt [g(x) – g(t)]λ

1 d gx (x) dx

= f (x),

2 a

gx > 0,

sin(πλ) d y(x) = πh(x) dx

x

K 0

t x

sin(πλ) . πλ

k=

0 < λ < 1.

Solution:

43.

gt (t)f (t) dt , [g(x) – g(t)]λ

x

x

a

f (t)gt (t) dt . [g(x) – g(t)]1–λ

y(t) dt = Axλ + Bxµ .

Solution: A λ–1 B µ–1 y(x) = x + x , Iλ Iµ

Iλ =

1

K(z)z 0

λ–1

dz,

1

K(z)z µ–1 dz.

Iµ = 0

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x

K

44. 0

t x

Pn (x) = xλ

y(t) dt = Pn (x),

n

A m xm .

m=0

Solution: y(x) = xλ

n Am m–1 x , Im

1

K(z)z λ+m–1 dz.

Im = 0

m=0

The integral I0 is supposed to converge. 45.

x

g1 (x) h1 (t) – h1 (x) + g2 (x) h2 (t) – h2 (x) y(t) dt = f (x).

a

This is a special case of equation 1.9.50 with g3 (x) = –g1 (x)h1 (x) – g2 (x)h2 (x) and h3 (t) = 1. x

The substitution Y (x) = equation of the form 1.9.15:

a

y(t) dt followed by integration by parts leads to an integral

x g1 (x) h1 (t) t + g2 (x) h2 (t) t Y (t) dt = –f (x). a

46.

x

g1 (x) h1 (t) – eλ(x–t) h1 (x) + g2 (x) h2 (t) – eλ(x–t) h2 (x) y(t) dt = f (x).

a

This is a special case of equation 1.9.50 with g3 (x) = –eλx g1 (x)h1 (x) + g2 (x)h2 (x) , and h3 (t) = e–λt . x

The substitution Y (x) = equation of the form 1.9.15:

a

e–λt y(t) dt followed by integration by parts leads to an integral

x g1 (x) eλt h1 (t) t + g2 (x) eλt h2 (t) t Y (t) dt = –f (x). a

47.

Ag λ (x)g µ (t) + Bg λ+β (x)g µ–β (t) – (A + B)g λ+γ (x)g µ–γ (t) y(t) dt = f (x).

x

a

This is a special case of equation 1.9.50 with g1 (x) = Ag λ (x), h1 (t) = g µ (t), g2 (x) = Bg λ+β (x), h2 (t) = g µ–β (t), g3 (x) = –(A + B)g λ+γ (x), and h3 (t) = g µ–γ (t).

x

Ag λ (x)h(x)g µ (t) + Bg λ+β (x)h(x)g µ–β (t)

48.

– (A + B)g λ+γ (x)g µ–γ (t)h(t) y(t) dt = f (x).

a

This is a special case of equation 1.9.50 with g1 (x) = Ag λ (x)h(x), h1 (t) = g µ (t), g2 (x) = Bg λ+β (x)h(x), h2 (t) = g µ–β (t), g3 (x) = –(A + B)g λ+γ (x), and h3 (t) = g µ–γ (t)h(t). 49.

x

Ag λ (x)h(x)g µ (t) + Bg λ+β (x)h(t)g µ–β (t)

a

– (A + B)g λ+γ (x)g µ–γ (t)h(t) y(t) dt = f (x).

This is a special case of equation 1.9.50 with g1 (x) = Ag λ (x)h(x), h1 (t) = g µ (t), g2 (x) = Bg λ+β (x), h2 (t) = g µ–β (t)h(t), g3 (x) = –(A + B)g λ+γ (x), and h3 (t) = g µ–γ (t)h(t).

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50.

g1 (x)h1 (t) + g2 (x)h2 (t) + g3 (x)h3 (t) y(t) dt = f (x),

x

a

g1 (x)h1 (x) + g2 (x)h2 (x) + g3 (x)h3 (x) ≡ 0.

where

x

The substitution Y (x) = h3 (t)y(t) dt followed by integration by parts leads to an integral a equation of the form 1.9.15: x

h1 (t) h3 (t)

g1 (x) a

+ g2 (x) t

h2 (t) h3 (t)

Y (t) dt = –f (x). t

x

Q(x – t)eαt y(ξ) dt = Aepx ,

51.

ξ = eβt g(x – t).

–∞

Solution: y(ξ) =

A p–α ξ β , q

q=

∞

Q(z)[g(z)]

p–α β e–pz

dz.

0

1.10. Some Formulas and Transformations 1. Let the solution of the integral equation

x

K(x, t)y(t) dt = f (x)

(1)

y(x) = F f (x) ,

(2)

a

have the form

where F is some linear integro-differential operator. Then the solution of the more complicated integral equation x K(x, t)g(x)h(t)y(t) dt = f (x) (3) a

has the form y(x) =

f (x) 1 F . h(x) g(x)

(4)

Below are formulas for the solutions of integral equations of the form (3) for some specific functions g(x) and h(t). In all cases, it is assumed that the solution of equation (1) is known and is determined by formula (2). (a) The solution of the equation

x

K(x, t)(x/t)λ y(t) dt = f (x) a

has the form

y(x) = xλ F x–λ f (x) .

(b) The solution of the equation

x

K(x, t)eλ(x–t) y(t) dt = f (x) a

has the form

y(x) = eλx F e–λx f (x) .

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2. Let the solution of the integral equation (1) have the form d

d x y(x) = L1 x, f (x) + L2 x, R(x, t)f (t) dt, dx dx a where L1 and L2 are some linear differential operators. The solution of the more complicated integral equation x

K ϕ(x), ϕ(t) y(t) dt = f (x),

(5)

(6)

a

where ϕ(x) is an arbitrary monotone function (differentiable sufficiently many times, ϕx > 0), is determined by the formula 1 d y(x) = ϕx (x)L1 ϕ(x), f (x) ϕx (x) dx x (7)

d 1 + ϕx (x)L2 ϕ(x), R ϕ(x), ϕ(t) ϕt (t)f (t) dt. ϕx (x) dx a Below are formulas for the solutions of integral equations of the form (6) for some specific functions ϕ(x). In all cases, it is assumed that the solution of equation (1) is known and is determined by formula (5). (a) For ϕ(x) = xλ , x

d d 1 1 2 λ–1 λ x y(x) = λxλ–1 L1 xλ , f (x) + λ x L , R xλ , tλ tλ–1 f (t) dt. 2 λ–1 λ–1 λx dx λx dx a (b) For ϕ(x) = eλx , x

1 d 1 d λx λx 2 λx λx y(x) = λe L1 e , f (x) + λ e L2 e , R eλx , eλt eλt f (t) dt. λx λx λe dx λe dx a (c) For ϕ(x) = ln(λx), x

1 d d 1 1 y(x) = L1 ln(λx), x f (x) + L2 ln(λx), x R ln(λx), ln(λt) f (t) dt. x dx x dx a t (d) For ϕ(x) = cos(λx), y(x) = –λ sin(λx)L1 cos(λx),

–1 d f (x) λ sin(λx) dx x

–1 d R cos(λx), cos(λt) sin(λt)f (t) dt. + λ2 sin(λx)L2 cos(λx), λ sin(λx) dx a

(e) For ϕ(x) = sin(λx), y(x) = λ cos(λx)L1 sin(λx),

1 d f (x) λ cos(λx) dx x

1 d 2 + λ cos(λx)L2 sin(λx), R sin(λx), sin(λt) cos(λt)f (t) dt. λ cos(λx) dx a

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Chapter 2

Linear Equations of the Second Kind With Variable Limit of Integration Notation: f = f (x), g = g(x), h = h(x), K = K(x), and M = M (x) are arbitrary functions (these may be composite functions of the argument depending on two variables x and t); A, B, C, D, a, b, c, α, β, γ, λ, and µ are free parameters; and m and n are nonnegative integers.

2.1. Equations Whose Kernels Contain Power-Law Functions 2.1-1. Kernels Linear in the Arguments x and t 1.

y(x) – λ

x

y(t) dt = f (x).

a

Solution:

x

eλ(x–t) f (t) dt.

y(x) = f (x) + λ a

2.

y(x) + λx

x

y(t) dt = f (x).

a

Solution:

3.

2

– x2 ) f (t) dt.

1

2

– x2 ) f (t) dt.

x exp a

y(x) + λ

1

x

y(x) = f (x) – λ

2 λ(t

x

ty(t) dt = f (x).

a

Solution: y(x) = f (x) – λ

t exp a

4.

y(x) + λ

x

2 λ(t

x

(x – t)y(t) dt = f (x).

a

This is a special case of equation 2.1.34 with n = 1. 1◦ . Solution with λ > 0: y(x) = f (x) – k

x

sin[k(x – t)]f (t) dt,

k=

√

λ.

a

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2◦ . Solution with λ < 0:

y(x) = f (x) + k

x

sinh[k(x – t)]f (t) dt,

k=

√

–λ.

a

5.

x

A + B(x – t) y(t) dt = f (x).

y(x) +

a ◦

1 . Solution with A2 > 4B:

x

y(x) = f (x) – R(x – t)f (t) dt, a 1 2B – A2 R(x) = exp – 2 Ax A cosh(βx) + sinh(βx) , 2β 2◦ . Solution with A2 < 4B:

a

6.

y(x) –

x

1 2 4A

– B.

x

y(x) = f (x) – R(x – t)f (t) dt, a 1 2B – A2 R(x) = exp – 2 Ax A cos(βx) + sin(βx) , 2β 3◦ . Solution with A2 = 4B: x y(x) = f (x) – R(x – t)f (t) dt,

β=

β=

B – 14 A2 .

R(x) = exp – 12 Ax A – 14 A2 x .

Ax + Bt + C y(t) dt = f (x).

a

For B = –A see equation 2.1.5. This is a special case of equation 2.9.6 with g(x) = –Ax and h(t) = –Bt – C. x

By differentiation followed by the substitution Y (x) =

y(t) dt, the original equation a

can be reduced to the second-order linear ordinary differential equation Yxx – (A + B)x + C Yx – AY = fx (x) under the initial conditions Y (a) = 0,

Yx (a) = f (a).

(1) (2)

A fundamental system of solutions of the homogeneous equation (1) with f ≡ 0 has the form Y1 (x) = Φ α, 12 ; kz 2 , Y2 (x) = Ψ α, 12 ; kz 2 , A A+B C α= , k= , z =x+ , 2(A + B) 2 A+B where Φ α, β; x and Ψ α, β; x are degenerate hypergeometric functions. Solving the homogeneous equation (1) under conditions (2) for an arbitrary function f = f (x) and taking into account the relation y(x) = Yx (x), we thus obtain the solution of the integral equation in the form x y(x) = f (x) – R(x, t)f (t) dt, a √

∂ 2 Y1 (x)Y2 (t) – Y2 (x)Y1 (t) 2 πk C 2 . R(x, t) = exp k t + , W (t) = ∂x∂t W (t) Γ(α) A+B

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2.1-2. Kernels Quadratic in the Arguments x and t 7.

x

x2 y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.1.50 with λ = 2 and µ = 0. Solution: x y(x) = f (x) – A x2 exp 13 A(t3 – x3 ) f (t) dt. a

8.

x

y(x) + A

xty(t) dt = f (x). a

This is a special case of equation 2.1.50 with λ = 1 and µ = 1. Solution: x y(x) = f (x) – A xt exp 13 A(t3 – x3 ) f (t) dt. a

9.

x

t2 y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.1.50 with λ = 0 and µ = 2. Solution: x y(x) = f (x) – A t2 exp 13 A(t3 – x3 ) f (t) dt. a

10.

y(x) + λ

x

(x – t)2 y(t) dt = f (x).

a

This is a special case of equation 2.1.34 with n = 2. Solution: x y(x) = f (x) – R(x – t)f (t) dt, a √ √ √ R(x) = 23 ke–2kx – 23 kekx cos 3 kx – 3 sin 3 kx , 11.

k=

1 1/3 . 4λ

x

(x2 – t2 )y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.5 with g(x) = Ax2 . Solution: x 1 y(x) = f (x) + u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a where the primes denote differentiation with respect to the argument specified in the parentheses; u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equation uxx + 2Axu = 0; and the functions u1 (x) and u2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of the parameter A: For A > 0,

√ √ 3/2 3/2 8 8 W = 3/π, u1 (x) = x J1/3 , u . A x (x) = x Y A x 2 1/3 9 9 For A < 0, W = – 32 , u1 (x) =

√

x I1/3

3/2 8 9 |A| x

, u2 (x) =

√

x K1/3

3/2 8 9 |A| x

.

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12.

x

(xt – t2 )y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.4 with g(t) = At. Solution: x A y(x) = f (x) + t y1 (x)y2 (t) – y2 (x)y1 (t) f (t) dt, W a where y1 (x), y2 (x) is a fundamental system of solutions of the second-order linear homo geneous ordinary differential equation yxx + Axy = 0; the functions y1 (x) and y2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of the parameter A: For A > 0, √ √ √ √ W = 3/π, y1 (x) = x J1/3 23 A x3/2 , y2 (x) = x Y1/3 23 A x3/2 . For A < 0,

13.

√ √ W = – 32 , y1 (x) = x I1/3 23 |A| x3/2 , y2 (x) = x K1/3 23 |A| x3/2 . x y(x) + A (x2 – xt)y(t) dt = f (x). a

This is a special case of equation 2.9.3 with g(x) = Ax. Solution: x A y(x) = f (x) + x y1 (x)y2 (t) – y2 (x)y1 (t) f (t) dt, W a where y1 (x), y2 (x) is a fundamental system of solutions of the second-order linear homo geneous ordinary differential equation yxx + Axy = 0; the functions y1 (x) and y2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of the parameter A: For A > 0, √ √ √ √ W = 3/π, y1 (x) = x J1/3 23 A x3/2 , y2 (x) = x Y1/3 23 A x3/2 . For A < 0,

14.

√ √ W = – 32 , y1 (x) = x I1/3 23 |A| x3/2 , y2 (x) = x K1/3 23 |A| x3/2 . x y(x) + A (t2 – 3x2 )y(t) dt = f (x).

15.

This is a special case of equation 2.1.55 with λ = 1 and µ = 2. x y(x) + A (2xt – 3x2 )y(t) dt = f (x).

16.

This is a special case of equation 2.1.55 with λ = 2 and µ = 1. x y(x) – (ABxt – ABx2 + Ax + B)y(t) dt = f (x).

a

a

a

This is a special case of equation 2.9.16 with g(x) = Ax and h(x) = B. Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x 1 2 2 2 R(x, t) = (Ax + B) exp 2 A(x – t ) + B exp 12 A(s 2 – t2 ) + B(x – s) ds. t

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17.

y(x) +

x

Ax2 – At2 + Bx – Ct + D y(t) dt = f (x).

a

18.

This is a special case of equation 2.9.6 with g(x) = Ax2 + Bx + D and h(t) = –At2 – Ct. Solution: x ∂ 2 Y1 (x)Y2 (t) – Y2 (x)Y1 (t) y(x) = f (x) + f (t) dt. W (t) a ∂x∂t Here Y1 (x), Y2 (x) is a fundamentalsystem of solutions of the second-order homogeneous ordinary differential equation Yxx + (B – C)x + D Yx + (2Ax + B)Y = 0 (see A. D. Polyanin and V. F. Zaitsev (1996) for details about this equation): Y1 (x) = exp(–kx)Φ α, 12 ; 12 (C – B)z 2 , Y2 (x) = exp(–kx)Ψ α, 12 ; 12 (C – B)z 2 , √ 2π(C – B) 2A W (x) = – exp 12 (C – B)z 2 – 2kx , k = , Γ(α) B–C 4A2 + 2AD(C – B) + B(C – B)2 4A + (C – B)D α=– , z =x– , 2(C – B)3 (C – B)2 where Φ α, β; x and Ψ α, β; x are degenerate hypergeometric functions and Γ(α) is the gamma function. x y(x) – Ax + B + (Cx + D)(x – t) y(t) dt = f (x). a

This is a special case of equation 2.9.11 with g(x) = Ax + B and h(x) = Cx + D. Solution with A ≠ 0: x f (t) y(x) = f (x) + Y2 (x)Y1 (t) – Y1 (x)Y2 (t) dt. W (t) a Here Y1 (x), Y2 (x) is a fundamental system of solutions of the second-order homogeneous – (Ax + B)Yx – (Cx + D)Y = 0 (see A. D. Polyanin and ordinary differential equation Yxx V. F. Zaitsev (1996) for details about this equation): Y1 (x) = exp(–kx)Φ α, 12 ; 12 Az 2 , Y2 (x) = exp(–kx)Ψ α, 12 ; 12 Az 2 , √ –1 W (x) = – 2πA Γ(α) exp 12 Az 2 – 2kx , k = C/A, α = 12 (A2 D – ABC – C 2 )A–3 , z = x + (AB + 2C)A–2 , where Φ α, β; x and Ψ α, β; x are degenerate hypergeometric functions, Γ(α) is the gamma function. x y(x) + At + B + (Ct + D)(t – x) y(t) dt = f (x).

19.

a

This is a special case of equation 2.9.12 with g(t) = –At – B and h(t) = –Ct – D. Solution with A ≠ 0: x f (t) y(x) = f (x) – Y1 (x)Y2 (t) – Y1 (t)Y2 (x) dt. W (x) a Here Y1 (x), Y2 (x) is a fundamental system of solutions of the second-order homogeneous ordinary differential equation Yxx – (Ax + B)Yx – (Cx + D)Y = 0 (see A. D. Polyanin and V. F. Zaitsev (1996) for details about this equation): Y1 (x) = exp(–kx)Φ α, 12 ; 12 Az 2 , Y2 (x) = exp(–kx)Ψ α, 12 ; 12 Az 2 , √ –1 W (x) = – 2πA Γ(α) exp 12 Az 2 – 2kx , k = C/A, α = 12 (A2 D – ABC – C 2 )A–3 , z = x + (AB + 2C)A–2 , where Φ α, β; x and Ψ α, β; x are degenerate hypergeometric functions and Γ(α) is the gamma function.

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2.1-3. Kernels Cubic in the Arguments x and t 20.

x

x3 y(t) dt = f (x).

y(x) + A a

Solution:

x

x3 exp

y(x) = f (x) – A

1

4

– x4 ) f (t) dt.

1

4

– x4 ) f (t) dt.

1

4

– x4 ) f (t) dt.

4 A(t

a

21.

x

x2 ty(t) dt = f (x).

y(x) + A a

Solution:

x

x2 t exp

y(x) = f (x) – A a

22.

4 A(t

x

xt2 y(t) dt = f (x).

y(x) + A a

Solution:

x

xt2 exp

y(x) = f (x) – A a

23.

4 A(t

x

t3 y(t) dt = f (x).

y(x) + A a

Solution:

x

t3 exp

y(x) = f (x) – A a

24.

y(x) + λ

1

4 A(t

4

– x4 ) f (t) dt.

x

(x – t)3 y(t) dt = f (x).

a

This is a special case of equation 2.1.34 with n = 3. Solution: x y(x) = f (x) –

R(x – t)f (t) dt, a

where k cosh(kx) sin(kx) – sinh(kx) cos(kx) , R(x) = 1 s = (–6λ)1/4 2 s sin(sx) – sinh(sx) , 25.

k=

3 1/4 2λ

for λ > 0, for λ < 0.

x

(x3 – t3 )y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.1.52 with λ = 3. 26.

x

y(x) – A

4x3 – t3 y(t) dt = f (x).

a

This is a special case of equation 2.1.55 with λ = 1 and µ = 3. 27.

x

(xt2 – t3 )y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.1.49 with λ = 2.

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28.

x

y(x) + A

x2 t – t3 y(t) dt = f (x).

a

The transformation z = x2 , τ = t2 , y(x) = w(z) leads to an equation of the form 2.1.4: z w(z) + 12 A (z – τ )w(τ ) dτ = F (z), F (z) = f (x). 29.

y(x) +

a2

x

Ax2 t + Bt3 y(t) dt = f (x).

a

The transformation z = x2 , τ = t2 , y(x) = w(z) leads to an equation of the form 2.1.6: z 1 1 w(z) + F (z) = f (x). 2 Az + 2 Bτ w(τ ) dτ = F (z),

a2

x

2x3 – xt2 y(t) dt = f (x).

30.

y(x) + B

31.

This is a special case of equation 2.1.55 with λ = 2, µ = 2, and B = –2A. x 3 y(x) – A 4x – 3x2 t y(t) dt = f (x).

32.

This is a special case of equation 2.1.55 with λ = 3 and µ = 1. x y(x) + ABx3 – ABx2 t – Ax2 – B y(t) dt = f (x).

a

a

a

This is a special case of equation 2.9.7 with g(x) = Ax2 and λ = B. Solution: x y(x) = f (x) + R(x – t)f (t) dt, a x R(x, t) = (Ax2 + B) exp 13 A(x3 – t3 ) + B 2 exp 13 A(s 3 – t3 ) + B(x – s) ds. 33.

y(x) +

t

x

ABxt2 – ABt3 + At2 + B y(t) dt = f (x).

a

This is a special case of equation 2.9.8 with g(t) = At2 and λ = B. Solution: x y(x) = f (x) + R(x – t)f (t) dt, a x R(x, t) = –(At2 + B) exp 13 A(t3 – x3 ) + B 2 exp 13 A(s 3 – x3 ) + B(t – s) ds. t

2.1-4. Kernels Containing Higher-Order Polynomials in x and t 34.

x

(x – t)n y(t) dt = f (x),

y(x) + A

n = 1, 2, . . .

a

1◦ . Differentiating the equation n + 1 times with respect to x yields an (n + 1)st-order linear ordinary differential equation with constant coefficients for y = y(x): yx(n+1) + An! y = fx(n+1) (x). This equation under the initial conditions y(a) = f (a), yx (a) = fx (a), . . . , yx(n) (a) = fx(n) (a) determines the solution of the original integral equation.

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2◦ . Solution:

x

y(x) = f (x) +

R(x – t)f (t) dt, a

R(x) =

n 1 exp(σk x) σk cos(βk x) – βk sin(βk x) , n+1 k=0

where the coefficients σk and βk are given by 1

2πk , n+1

2πk + π cos , n+1

σk = |An!| n+1 cos 1

σk = |An!| n+1 35.

∞

y(x) + A

1

2πk n+1

2πk + π sin n+1

βk = |An!| n+1 sin 1

βk = |An!| n+1

(t – x)n y(t) dt = f (x),

for

A < 0,

for

A > 0.

n = 1, 2, . . .

x

The Picard–Goursat equation. This is a special case of equation 2.9.62 with K(z) = A(–z)n . 1◦ . A solution of the homogeneous equation (f ≡ 0) is 1 λ = –An! n+1 ,

y(x) = Ce–λx ,

where C is an arbitrary constant and A < 0. This is a unique solution for n = 0, 1, 2, 3. The general solution of the homogeneous equation for any sign of A has the form y(x) =

s

Ck exp(–λk x).

(1)

k=1

Here Ck are arbitrary constants and λk are the roots of the algebraic equation λn+1 + An! = 0 that satisfy the condition Re λk > 0. The number of terms in (1) is determined by the inequality s ≤ 2 n4 + 1, where [a] stands for the integral part of a number a. For more details about the solution of the homogeneous Picard–Goursat equation, see Subsection 9.11-1 (Example 1). 2◦ . For f (x) =

m

ak exp(–βk x), where βk > 0, a solution of the equation has the form

k=1

y(x) =

m k=1

ak βkn+1 exp(–βk x), βkn+1 + An!

(2)

where βkn+1 + An! ≠ 0. For A > 0, this formula can also be used for arbitrary f (x) expandable into a convergent exponential series (which corresponds to m = ∞). 3◦ . For f (x) = e–βx

m

ak xk , where β > 0, a solution of the equation has the form

k=1

y(x) = e–βx

m

Bk xk ,

(3)

k=0

where the constants Bk are found by the method of undetermined coefficients. The solution can also be constructed using the formulas given in item 3◦ , equation 2.9.55.

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4◦ . For f (x) = cos(βx)

m

ak exp(–µk x), a solution of the equation has the form

k=1

y(x) = cos(βx)

m

Bk exp(–µk x) + sin(βx)

k=1

m

Ck exp(–µk x),

(4)

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. The solution can also be constructed using the formulas given in 2.9.60. 5◦ . For f (x) = sin(βx)

m

ak exp(–µk x), a solution of the equation has the form

k=1

y(x) = cos(βx)

m

Bk exp(–µk x) + sin(βx)

k=1

m

Ck exp(–µk x),

(5)

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. The solution can also be constructed using the formulas given in 2.9.61. 6◦ . To obtain the general solution in item 2◦ –5◦ , the solution (1) of the homogeneous equation must be added to each right-hand side of (2)–(5). 36.

x

(x – t)tn y(t) dt = f (x),

y(x) + A

n = 1, 2, . . .

a

This is a special case of equation 2.1.49 with λ = n. 37.

x

(xn – tn )y(t) dt = f (x),

y(x) + A

n = 1, 2, . . .

a

This is a special case of equation 2.1.52 with λ = n. 38.

y(x) +

x

ABxn+1 – ABxn t – Axn – B y(t) dt = f (x),

n = 1, 2, . . .

a

This is a special case of equation 2.9.7 with g(x) = Axn and λ = B. Solution: x y(x) = f (x) + R(x – t)f (t) dt, a x A n+1 n+1 A n+1 n+1 R(x, t) = (Axn + B) exp + B2 + B(x – s) ds. exp x –t s –t n+1 n+1 t 39.

y(x) +

x

ABxtn – ABtn+1 + Atn + B y(t) dt = f (x),

n = 1, 2, . . .

a

This is a special case of equation 2.9.8 with g(t) = Atn and λ = B. Solution: x y(x) = f (x) + R(x – t)f (t) dt, a x A n+1 n+1 A n+1 n+1 n 2 R(x, t) = –(At +B) exp exp +B +B(t–s) ds. t –x s –x n+1 n+1 t

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2.1-5. Kernels Containing Rational Functions 40.

y(x) + x

t 2Ax + (1 – A)t y(t) dt = f (x).

x

–3 a

This equation can be obtained by differentiating the equation x 2 Ax t + (1 – A)xt2 y(t) dt = F (x), F (x) = a

y(x) – λ

x

0

t3 f (t) dt,

a

which has the form 1.1.17: Solution: x 1 d y(x) = x–A tA–1 ϕt (t) dt , x dx a 41.

x

y(t) dt x+t

ϕ(x) =

1 x

x

t3 f (t) dt. a

= f (x).

Dixon’s equation. This is a special case of equation 2.1.62 with a = b = 1 and µ = 0. 1◦ . The solution of the homogeneous equation (f ≡ 0) is y(x) = Cxβ

(β > –1, λ > 0).

(1)

Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation λI(β) = 1,

where

1

I(β) = 0

z β dz . 1+z

(2)

2◦ . For a polynomial right-hand side, f (x) =

N

An xn

n=0

the solution bounded at zero is given by  N  An    for λ < λ0 , xn   1 – (λ/λn ) n=0 y(x) = N   An   xn + Cxβ for λ > λ0 and λ ≠ λn ,   1 – (λ/λn ) n=0 n (–1)m 1 n λn = , I(n) = (–1) ln 2 + , I(n) m m=1

where C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation (2). For special λ = λn (n = 1, 2, . . . ), the solution differs in one term and has the form y(x) =

n–1 m=0

N Am Am λ¯ n n xm + xm – An x ln x + Cxn , 1 – (λn /λm ) 1 – (λn /λm ) λn m=n+1

2 n (–1)k –1 π where λ¯ n = (–1)n+1 . + 12 k=1 k 2

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Remark. For arbitrary f (x), expandable into power series, the formulas of item 2◦ can

be used, in which one should set N = ∞. In this case, the radius of convergence of the solution y(x) is equal to the radius of convergence of f (x). 3◦ . For logarithmic-polynomial right-hand side, N f (x) = ln x An xn , n=0

the solution with logarithmic singularity at zero is given by  N N  An An Dn λ  n   + xn for λ < λ0 , x ln x   1 – (λ/λn ) [1 – (λ/λn )]2 n=0 n=0 y(x) = N N   An An Dn λ  n  ln x + xn + Cxβ for λ > λ0 and λ ≠ λn , x   1 – (λ/λn ) [1 – (λ/λn )]2 n=0 n=0 2 n n π (–1)k (–1)k 1 λn = . , I(n) = (–1)n ln 2 + , Dn = (–1)n+1 + I(n) k 12 k2 k=1

k=1

4◦ . For arbitrary f (x), the transformation x = 12 e2z ,

t = 12 e2τ ,

y(x) = e–z w(z),

f (x) = e–z g(z)

leads to an integral equation with difference kernel of the form 2.9.51: z w(τ ) dτ w(z) – λ = g(z). –∞ cosh(z – τ ) 42.

y(x) – λ

x

a

x+b t+b

y(t) dt = f (x).

This is a special case of equation 2.9.1 with g(x) = x + b. Solution: x x + b λ(x–t) y(x) = f (x) + λ f (t) dt. e a t+b 43.

y(x) =

2 (1 –

λ2 )x2

x

t

λx

1+t

y(t) dt.

This equation is encountered in nuclear physics and describes deceleration of neutrons in matter. 1◦ . Solution with λ = 0: y(x) =

C , (1 + x)2

where C is an arbitrary constant. 2◦ . For λ ≠ 0, the solution can be found in the series form y(x) =

∞

An xn .

n=0

• Reference: I. Sneddon (1951).

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2.1-6. Kernels Containing Square Roots and Fractional Powers 44.

x

y(x) + A

√ (x – t) t y(t) dt = f (x).

a

This is a special case of equation 2.1.49 with λ = 12 . 45.

y(x) + A

x √

x–

√ t y(t) dt = f (x).

a

This is a special case of equation 2.1.52 with λ = 12 . 46.

y(x) + λ

x

y(t) dt = f (x). √ x–t a Abel’s equation of the second kind. This equation is encountered in problems of heat and mass transfer. Solution: x 2 y(x) = F (x) + πλ exp[πλ2 (x – t)]F (t) dt, a

where F (x) = f (x) – λ

a

47.

x

f (t) dt √ . x–t

• References: H. Brakhage, K. Nickel, and P. Rieder (1965), Yu. I. Babenko (1986). x y(t) dt y(x) – λ = f (x), a > 0, b > 0. √ ax2 + bt2 0 1◦ . The solution of the homogeneous equation (f ≡ 0) is y(x) = Cxβ

(β > –1, λ > 0).

(1)

Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation 1 z β dz √ λI(β) = 1, where I(β) = . (2) a + bz 2 0 2◦ . For a polynomial right-hand side, f (x) =

N

An xn

n=0

the solution bounded at zero is given by  N  An    for λ < λ0 , xn   1 – (λ/λn ) n=0 y(x) = N   An   xn + Cxβ for λ > λ0 and λ ≠ λn ,   1 – (λ/λn ) n=0 √ 1 b z n dz 1 √ λ0 = . , I(n) = , λn = I(n) Arsinh b/a a + bz 2 0 Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation (2).

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3◦ . For special λ = λn , (n = 1, 2, . . . ), the solution differs in one term and has the form y(x) =

n–1 m=0

N λ¯ n n Am Am m x + xm – An x ln x + Cxn , 1 – (λn /λm ) 1 – (λn /λm ) λn m=n+1

–1 z n ln z dz √ . a + bz 2 0 4◦ . For arbitrary f (x), expandable into power series, the formulas of item 2◦ can be used, in which one should set N = ∞. In this case, the radius of convergence of the solution y(x) is equal to the radius of convergence of f (x). x y(t) dt y(x) + λ = f (x). 3/4 a (x – t) This equation admits solution by quadratures (see equation 2.1.60 and Section 9.4-2).

1

where λ¯ n =

48.

2.1-7. Kernels Containing Arbitrary Powers 49.

x

(x – t)tλ y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.4 with g(t) = Atλ . Solution: x A y(x) = f (x) + y1 (x)y2 (t) – y2 (x)y1 (t) tλ f (t) dt, W a where y1 (x), y2 (x) is a fundamental system of solutions of the second-order linear homo geneous ordinary differential equation yxx + Axλ y = 0; the functions y1 (x) and y2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of A: For A > 0, √ √ √ √ 2q λ+2 A q A q W = , y1 (x) = x J 1 x , y2 (x) = x Y 1 x , q= , π q q 2 2q 2q For A < 0, W = –q, y1 (x) = 50.

√

√ xI

1 2q

√ √ λ+2 |A| q |A| q x , y2 (x) = x K 1 x , q= . q q 2 2q

x

xλ tµ y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –Axλ and h(t) = tµ (λ and µ are arbitrary numbers). Solution: x y(x) = f (x) – R(x, t)f (t) dt, a  λ+µ+1 A  Axλ tµ exp – xλ+µ+1 for λ + µ + 1 ≠ 0, t λ+µ+1 R(x, t) =  λ–A µ+A Ax t for λ + µ + 1 = 0.

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51.

x

(x – t)xλ tµ y(t) dt = f (x).

y(x) + A a

The substitution u(x) = x–λ y(x) leads to an equation of the form 2.1.49: x u(x) + A (x – t)tλ+µ u(t) dt = f (x)x–λ . 52.

a x

(xλ – tλ )y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.5 with g(x) = Axλ . Solution: x 1 y(x) = f (x) + u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a where the primes denote differentiation with respect to the argument specified in the parentheses, and u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equation uxx + Aλxλ–1 u = 0; the functions u1 (x) and u2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of A: For Aλ > 0, √ √ √ √ 2q λ+1 Aλ q Aλ q W = , u1 (x) = x J 1 x , u2 (x) = x Y 1 x , q= , π q q 2 2q 2q For Aλ < 0, W = –q, u1 (x) = 53.

y(x) –

x

√

√ xI

1 2q

√ √ λ+1 |Aλ| q |Aλ| q x , u2 (x) = x λK 1 x , q= . q q 2 2q

Axλ tλ–1 + Bt2λ–1 y(t) dt = f (x).

a

The transformation z = xλ ,

54.

τ = tλ ,

y(x) = Y (z)

leads to an equation of the form 2.1.6: z A B Y (z) – z + τ Y (τ ) dτ = F (z), F (z) = f (x), b = aλ . λ λ b x y(x) – Axλ+µ tλ–µ–1 + Bxµ t2λ–µ–1 y(t) dt = f (x). a

The substitution y(x) = xµ w(x) leads to an equation of the form 2.1.53: x λ λ–1 w(x) – Ax t + Bt2λ–1 w(t) dt = x–µ f (x). 55.

y(x) + A

a x

λxλ–1 tµ – (λ + µ)xλ+µ–1 y(t) dt = f (x).

a

This equation can be obtained by differentiating equation 1.1.51: x λ µ λ+µ F (x) = 1 + A(x t – x ) y(t) dt = F (x), a

Solution: d y(x) = dx

x

f (x) dx.

a

xλ Φ(x)

a

x

–λ

t F (t) t Φ(t) dt ,

Aµ µ+λ . Φ(x) = exp – x µ+λ

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56.

y(x) +

x

ABxλ+1 – ABxλ t – Axλ – B y(t) dt = f (x).

a

This is a special case of equation 2.9.7. Solution:

x

y(x) = f (x) +

R(x – t)f (t) dt, x A λ+1 λ+1 A λ+1 λ+1 R(x, t) = (Axλ + B) exp + B2 + B(x – s) ds. x –t s –t exp λ+1 λ+1 t a

57.

y(x) +

x

ABxtλ – ABtλ+1 + Atλ + B y(t) dt = f (x).

a

This is a special case of equation 2.9.8. Solution:

x

y(x) = f (x) +

R(x – t)f (t) dt, x A λ+1 λ+1 A λ+1 λ+1 R(x, t) = –(Atλ +B) exp exp +B 2 +B(t–s) ds. t –x s –x λ+1 λ+1 t a

58.

y(x) – λ

x

x + b µ t+b

a

y(t) dt = f (x).

This is a special case of equation 2.9.1 with g(x) = (x + b)µ . Solution: x

x + b µ λ(x–t) y(x) = f (x) + λ e f (t) dt. t+b a 59.

y(x) – λ

x

a

xµ + b

y(t) dt = f (x).

tµ + b

This is a special case of equation 2.9.1 with g(x) = xµ + b. Solution: x µ x + b λ(x–t) y(x) = f (x) + λ f (t) dt. e µ a t +b 60.

y(x) – λ 0

x

y(t) dt (x – t)α

= f (x),

0 < α < 1.

Generalized Abel equation of the second kind. 1◦ . Assume that the number α can be represented in the form α=1–

m , n

where m = 1, 2, . . . ,

n = 2, 3, . . .

(m < n).

In this case, the solution of the generalized Abel equation of the second kind can be written in closed form (in quadratures): y(x) = f (x) +

x

R(x – t)f (t) dt, 0

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where R(x) =

n–1 m–1 λν Γν (m/n) (νm/n)–1 b x + εµ exp εµ bx Γ(νm/n) m ν=1

µ=0

m–1 x (νm/n)–1 λν Γν (m/n) εµ exp εµ bx t exp –εµ bt dt , Γ(νm/n) 0 ν=1 µ=0

2πµi b = λn/m Γn/m (m/n), εµ = exp , i2 = –1, µ = 0, 1, . . . , m – 1. m b m

+

n–1

2◦ . Solution with any α from 0 < α < 1: y(x) = f (x) +

x

R(x – t)f (t) dt,

where

0

61.

n ∞ λΓ(1 – α)x1–α . R(x) = xΓ n(1 – α) n=1

• References: H. Brakhage, K. Nickel, and P. Rieder (1965), V. I. Smirnov (1974). x y(t) dt λ y(x) – α = f (x), 0 < α ≤ 1. x 0 (x – t)1–α 1◦ . The solution of the homogeneous equation (f ≡ 0) is y(x) = Cxβ

(β > –1, λ > 0).

(1)

Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation λB(α, β + 1) = 1, where B(p, q) =

1 p–1 z (1 – z)q–1 dz 0

(2)

is the beta function.

◦

2 . For a polynomial right-hand side, f (x) =

N

An xn

n=0

the solution bounded at zero is given by  N  An    xn   1 – (λ/λn ) n=0 y(x) = N   An   xn + Cxβ   1 – (λ/λn )

for λ < α, for λ > α and λ ≠ λn ,

n=0

λn =

(α)n+1 , n!

(α)n+1 = α(α + 1) . . . (α + n).

Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation (2). For special λ = λn (n = 1, 2, . . . ), the solution differs in one term and has the form y(x) =

n–1 m=0

N λ¯ n n Am Am xm + xm – An x ln x + Cxn , 1 – (λn /λm ) 1 – (λn /λm ) λn

1

where λ¯ n =

m=n+1

–1 (1 – z)α–1 z n ln z dz

.

0

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3◦ . For arbitrary f (x), expandable into power series, the formulas of item 2◦ can be used, in which one should set N = ∞. In this case, the radius of convergence of the solution y(x) is equal to the radius of convergence of f (x). 4◦ . For f (x) = ln(kx)

N

An xn ,

n=0

a solution has the form y(x) = ln(kx)

N

Bn xn +

n=0

62.

N

Dn xn ,

n=0

where the constants Bn and Dn are found by the method of undetermined coefficients. To obtain the general solution we must add the solution (1) of the homogeneous equation. In Mikhailov (1966), solvability conditions for the integral equation in question were investigated for various classes of f (x). x λ y(t) dt y(x) – µ = f (x). x 0 (ax + bt)1–µ Here a > 0, b > 0, and µ is an arbitrary number. 1◦ . The solution of the homogeneous equation (f ≡ 0) is y(x) = Cxβ

(β > –1, λ > 0).

(1)

Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation 1 λI(β) = 1, where I(β) = z β (a + bz)µ–1 dz. (2) 0

2◦ . For a polynomial right-hand side, f (x) =

N

An xn

n=0

the solution bounded at zero is given by  N  An    for λ < λ0 , xn   1 – (λ/λn ) n=0 y(x) = N   An   xn + Cxβ for λ > λ0 and λ ≠ λn ,   1 – (λ/λn ) n=0 1 1 λn = , I(n) = z n (a + bz)µ–1 dz. I(n) 0 Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation (2). 3◦ . For special λ = λn (n = 1, 2, . . . ), the solution differs in one term and has the form N λ¯ n n Am Am xm + xm – An x ln x + Cxn , 1 – (λn /λm ) 1 – (λn /λm ) λn m=0 m=n+1 1 –1 n µ–1 ¯ z (a + bz) ln z dz . where λn =

y(x) =

n–1

0

4◦ . For arbitrary f (x) expandable into power series, the formulas of item 2◦ can be used, in which one should set N = ∞. In this case, the radius of convergence of the solution y(x) is equal to the radius of convergence of f (x).

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2.2. Equations Whose Kernels Contain Exponential Functions 2.2-1. Kernels Containing Exponential Functions 1.

x

eλ(x–t) y(t) dt = f (x).

y(x) + A a

Solution:

x

e(λ–A)(x–t) f (t) dt.

y(x) = f (x) – A a

2.

x

eλx+βt y(t) dt = f (x).

y(x) + A a

For β = –λ, see equation 2.2.1. This is a special case of equation 2.9.2 with g(x) = –Aeλx and h(t) = eβt . Solution: x A (λ+β)t (λ+β)x y(x) = f (x) – R(x, t)f (t) dt, R(x, t) = Aeλx+βt exp –e . e λ+β a 3.

y(x) + A

x

eλ(x–t) – 1 y(t) dt = f (x).

a ◦

1 . Solution with D ≡ λ(λ – 4A) > 0: 2Aλ y(x) = f (x) – √ D

x

R(x – t)f (t) dt,

R(x) = exp

1

a

2 λx

sinh

1√ 2

Dx .

2◦ . Solution with D ≡ λ(λ – 4A) < 0: 2Aλ y(x) = f (x) – √ |D|

x

R(x – t)f (t) dt,

R(x) = exp

a

1

2 λx

sin

1 2

|D| x .

3◦ . Solution with λ = 4A: y(x) = f (x) – 4A2

x

(x – t) exp 2A(x – t) f (t) dt.

a

4.

y(x) +

Aeλ(x–t) + B y(t) dt = f (x).

x

a

This is a special case of equation 2.2.10 with A1 = A, A2 = B, λ1 = λ, and λ2 = 0. 1◦ . The structure of the solution depends on the sign of the discriminant D ≡ (A – B – λ)2 + 4AB

(1)

µ2 + (A + B – λ)µ – Bλ = 0.

(2)

of the square equation 2◦ . If D > 0, then equation (2) has the real different roots √ √ µ1 = 12 (λ – A – B) + 12 D, µ2 = 12 (λ – A – B) – 12 D.

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In this case, the original integral equation has the solution y(x) = f (x) +

x

E1 eµ1 (x–t) + E2 eµ2 (x–t) f (t) dt,

a

where E1 = A

µ1 µ1 – λ +B , µ2 – µ1 µ2 – µ1

E2 = A

µ2 µ2 – λ +B . µ1 – µ2 µ1 – µ2

3◦ . If D < 0, then equation (2) has the complex conjugate roots µ1 = σ + iβ,

σ = 12 (λ – A – B),

µ2 = σ – iβ,

β=

1 2

√

–D.

In this case, the original integral equation has the solution y(x) = f (x) +

x E1 eσ(x–t) cos[β(x – t)] + E2 eσ(x–t) sin[β(x – t)] f (t) dt, a

where E1 = –A – B, 5.

E2 =

1 (–Aσ – Bσ + Bλ). β

x

(eλx – eλt )y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.5 with g(x) = Aeλx . Solution: x 1 u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, y(x) = f (x) + W a where the primes denote differentiation with respect to the argument specified in the parentheses, and u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equation uxx + Aλeλx u = 0; the functions u1 (x) and u2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of A: For Aλ > 0, √ √ λ 2 Aλ λx/2 2 Aλ λx/2 W = , u1 (x) = J0 , u2 (x) = Y0 , e e π λ λ For Aλ < 0, √ √ λ 2 |Aλ| λx/2 2 |Aλ| λx/2 W = – , u1 (x) = I0 , u2 (x) = K0 . e e 2 λ λ 6.

y(x) +

x

Aeλx + Beλt y(t) dt = f (x).

a

For B = –A, see equation 2.2.5. This is a special case of equation 2.9.6 with g(x) = Aeλx and h(t) = Beλt . x

Differentiating the original integral equation followed by substituting Y (x) = yields the second-order linear ordinary differential equation Yxx + (A + B)eλx Yx + Aλeλx Y = fx (x)

y(t) dt a

(1)

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under the initial conditions Y (a) = 0,

Yx (a) = f (a).

(2)

A fundamental system of solutions of the homogeneous equation (1) with f ≡ 0 has the form

A

A m m Y1 (x) = Φ , 1; – eλx , Y2 (x) = Ψ , 1; – eλx , m = A + B, m λ m λ where Φ α, β; x and Ψ α, β; x are degenerate hypergeometric functions. Solving the homogeneous equation (1) under conditions (2) for an arbitrary function f = f (x) and taking into account the relation y(x) = Yx (x), we thus obtain the solution of the integral equation in the form x y(x) = f (x) – R(x, t)f (t) dt, a

Γ(A/m) ∂ 2 m λt R(x, t) = Y1 (x)Y2 (t) – Y2 (x)Y1 (t) . exp e λ ∂x∂t λ 7.

y(x) + A

x

eλ(x+t) – e2λt y(t) dt = f (x).

a

The transformation z = eλx , τ = eλt leads to an equation of the form 2.1.4. 1◦ . Solution with Aλ > 0:

eλt sin k(eλx – eλt ) f (t) dt,

k=

eλt sinh k(eλx – eλt ) f (t) dt,

k=

x

y(x) = f (x) – λk

A/λ.

a

2◦ . Solution with Aλ < 0: y(x) = f (x) + λk

x

|A/λ|.

a

8.

y(x) + A

x

eλx+µt – e(λ+µ)t y(t) dt = f (x).

a

The transformation z = eµx , τ = eµt , Y (z) = y(x) leads to an equation of the form 2.1.52: A z k Y (z) + (z – τ k )Y (τ ) dτ = F (z), F (z) = f (x), µ b where k = λ/µ, b = eµa . 9.

y(x) + A

x

λeλx+µt – (λ + µ)e(λ+µ)x y(t) dt = f (x).

a

This equation can be obtained by differentiating an equation of the form 1.2.22: x x λx µt µx F (x) = f (t) dt. 1 + Ae (e – e ) y(t) dt = F (x), a

a

Solution: y(x) =

x d F (t) dt eλx Φ(x) , dx eλt t Φ(t) a

Aµ (λ+µ)x . Φ(x) = exp e λ+µ

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10.

y(x) +

x

A1 eλ1 (x–t) + A2 eλ2 (x–t) y(t) dt = f (x).

a ◦

1 . Introduce the notation I1 =

x

eλ1 (x–t) y(t) dt,

a

x

eλ2 (x–t) y(t) dt.

I2 = a

Differentiating the integral equation twice yields (the first line is the original equation) y + A1 I1 + A2 I2 = f , f = f (x), yx + (A1 + A2 )y + A1 λ1 I1 + A2 λ2 I2 = fx , yxx + (A1 + A2 )yx + (A1 λ1 + A2 λ2 )y + A1 λ21 I1 + A2 λ22 I2 = fxx .

(1) (2) (3)

Eliminating I1 and I2 , we arrive at the second-order linear ordinary differential equation with constant coefficients yxx + (A1 + A2 – λ1 – λ2 )yx + (λ1 λ2 – A1 λ2 – A2 λ1 )y = fxx – (λ1 + λ2 )fx + λ1 λ2 f .

(4)

Substituting x = a into (1) and (2) yields the initial conditions yx (a) = fx (a) – (A1 + A2 )f (a).

y(a) = f (a),

(5)

Solving the differential equation (4) under conditions (5), we can find the solution of the integral equation. 2◦ . Consider the characteristic equation µ2 + (A1 + A2 – λ1 – λ2 )µ + λ1 λ2 – A1 λ2 – A2 λ1 = 0

(6)

which corresponds to the homogeneous differential equation (4) (with f (x) ≡ 0). The structure of the solution of the integral equation depends on the sign of the discriminant D ≡ (A1 – A2 – λ1 + λ2 )2 + 4A1 A2 of the quadratic equation (6). If D > 0, the quadratic equation (6) has the real different roots √ √ µ1 = 12 (λ1 + λ2 – A1 – A2 ) + 12 D, µ2 = 12 (λ1 + λ2 – A1 – A2 ) – 12 D. In this case, the solution of the original integral equation has the form x y(x) = f (x) + B1 eµ1 (x–t) + B2 eµ2 (x–t) f (t) dt, a

where

µ1 – λ2 µ1 – λ1 µ2 – λ2 µ2 – λ1 + A2 , B2 = A1 + A2 . µ2 – µ1 µ2 – µ1 µ1 – µ2 µ1 – µ2 If D < 0, the quadratic equation (6) has the complex conjugate roots √ µ1 = σ + iβ, µ2 = σ – iβ, σ = 12 (λ1 + λ2 – A1 – A2 ), β = 12 –D. B1 = A1

In this case, the solution of the original integral equation has the form x y(x) = f (x) + B1 eσ(x–t) cos[β(x – t)] + B2 eσ(x–t) sin[β(x – t)] f (t) dt. a

where B1 = –A1 – A2 ,

B2 =

1 A1 (λ2 – σ) + A2 (λ1 – σ) . β

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11.

x

Aeλ(x+t) – Ae2λt + Beλt y(t) dt = f (x).

y(x) +

a

The transformation z = eλx , τ = eλt , Y (z) = y(x) leads to an equation of the form 2.1.5:

z

B1 (z – τ ) + A1 Y (τ ) dτ = F (z),

Y (z) +

F (z) = f (x),

b

where A1 = B/λ, B1 = A/λ, b = eλa . 12.

x

Aeλ(x+t) + Be2λt + Ceλt y(t) dt = f (x).

y(x) +

a

The transformation z = eλx , τ = eλt , Y (z) = y(x) leads to an equation of the form 2.1.6: Y (z) –

z

(A1 z + B1 τ + C1 )Y (τ ) dτ = F (z),

F (z) = f (x),

b

where A1 = –A/λ, B1 = –B/λ, C1 = –C/λ, b = eλa . 13.

x

λeλ(x–t) + A µeµx+λt – λeλx+µt y(t) dt = f (x).

y(x) +

a

This is a special case of equation 2.9.23 with h(t) = A. Solution: x F (t) e2λt d y(x) = λx Φ(x) dt , e dx eλt t Φ(t) a x λ – µ (λ+µ)x Φ(x) = exp A , F (x) = f (t) dt. e λ+µ a 1

14.

λe–λ(x–t) + A µeλx+µt – λeµx+λt y(t) dt = f (x).

x

y(x) –

a

This is a special case of equation 2.9.24 with h(x) = A. Assume that f (a) = 0. Solution: a

y(x) +

d e2λx x f (t) w(t) dt, w(x) = e Φ(t) dt , dx Φ(x) a eλt t λ – µ (λ+µ)x Φ(x) = exp A . e λ+µ –λx

y(x) =

15.

x

x

λeλ(x–t) + Aeβt µeµx+λt – λeλx+µt y(t) dt = f (x).

a

This is a special case of equation 2.9.23 with h(t) = Aeβt . Solution: x d F (t) e(2λ+β)t y(x) = e–(λ+β)x Φ(x) dt , dx eλt t Φ(t) a x λ – µ (λ+µ+β)x Φ(x) = exp A , F (x) = f (t) dt. e λ+µ+β a

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16.

x

λe–λ(x–t) + Aeβx µeλx+µt – λeµx+λt y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.24 with h(x) = Aeβx . Assume that f (a) = 0. Solution:

x

y(x) = a

17.

y(x) +

d e(2λ+β)x x f (t) w(t) dt, w(x) = e Φ(t) dt , dx Φ(x) a e(λ+β)t t λ – µ (λ+µ+β)x Φ(x) = exp A . e λ+µ+β –λx

x

ABe(λ+1)x+t – ABeλx+2t – Aeλx+t – Bet y(t) dt = f (x).

a

The transformation z = ex , τ = et , Y (z) = y(x) leads to an equation of the form 2.1.56: z Y (z) + ABz λ+1 – ABz λ τ – Az λ – B Y (τ ) dτ = F (z), b

where F (z) = f (x) and b = ea . 18.

y(x) +

x

ABex+λt – ABe(λ+1)t + Aeλt + Bet y(t) dt = f (x).

a

The transformation z = ex , τ = et , Y (z) = y(x) leads to an equation of the form 2.1.57 (in which λ is substituted by λ – 1): z Y (z) + ABzτ λ–1 – ABτ λ + Aτ λ–1 + B Y (τ ) dτ = F (z), b

where F (z) = f (x) and b = ea . 19.

y(x) +

n x

a

λk (x–t)

Ak e

y(t) dt = f (x).

k=1

1◦ . This integral equation can be reduced to an nth-order linear nonhomogeneous ordinary differential equation with constant coefficients. Set x Ik (x) = eλk (x–t) y(t) dt. (1) a

Differentiating (1) with respect to x yields Ik = y(x) + λk

x

eλk (x–t) y(t) dt,

(2)

a

where the prime stands for differentiation with respect to x. From the comparison of (1) with (2) we see that Ik = y(x) + λk Ik , Ik = Ik (x). (3) The integral equation can be written in terms of Ik (x) as follows: y(x) +

n

Ak Ik = f (x).

(4)

k=1

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Differentiating (4) with respect to x and taking account of (3), we obtain yx (x) + σn y(x) +

n

Ak λk Ik = fx (x),

σn =

k=1

n

Ak .

(5)

k=1

Eliminating the integral In from (4) and (5), we find that yx (x)

+ σn – λn )y(x) +

n–1

Ak (λk – λn )Ik = fx (x) – λn f (x).

(6)

k=1

Differentiating (6) with respect to x and eliminating In–1 from the resulting equation with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear n–2 1 differential operator (acting on y) with constant coefficients plus the sum Ak Ik . If we k=1

proceed with successively eliminating In–2 , In–3 , . . . , I1 with the aid of differentiation and formula (3), then we will finally arrive at an nth-order linear nonhomogeneous ordinary differential equation with constant coefficients. The initial conditions for y(x) can be obtained by setting x = a in the integral equation and all its derivative equations. 2◦ . The solution of the equation can be represented in the form y(x) = f (x) +

x n a

Bk eµk (x–t) f (t) dt.

(7)

k=1

The unknown constants µk are the roots of the algebraic equation n Ak + 1 = 0, z – λk

(8)

k=1

which is reduced (by separating the numerator) to the problem of finding the roots of an nth-order characteristic polynomial. After the µk have been calculated, the coefficients Bk can be found from the following linear system of algebraic equations: n k=1

Bk + 1 = 0, λm – µk

m = 1, . . . , n.

(9)

Another way of determining the Bk is presented in item 3◦ below. If all the roots µk of equation (8) are real and different, then the solution of the original integral equation can be calculated by formula (7). To a pair of complex conjugate roots µk,k+1 = α ± iβ of the characteristic polynomial (8) there corresponds a pair of complex conjugate coefficients Bk,k+1 in equation (9). In this case, µk (x–t) the corresponding terms + Bk+1 eµk+1 (x–t) Bk e α(x–t) in solution (7) can be written in the form α(x–t) Bk e cos β(x – t) + B k+1 e sin β(x – t) , where B k and B k+1 are real coefficients. 3◦ . For a = 0, the solution of the original integral equation is given by y(x) = f (x) –

x

R(x – t)f (t) dt,

R(x) = L–1 R(p) ,

(10)

0

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where L–1 R(p) is the inverse Laplace transform of the function R(p) =

K(p) , 1 + K(p)

K(p) =

n Ak . p – λk

(11)

k=1

The transform R(p) of the resolvent R(x) can be represented as a regular fractional function: Q(p) R(p) = , P (p) = (p – µ1 )(p – µ2 ) . . . (p – µn ), P (p) where Q(p) is a polynomial in p of degree < n. The roots µk of the polynomial P (p) coincide with the roots of equation (8). If all µk are real and different, then the resolvent can be determined by the formula R(x) =

n

Bk eµk x ,

Bk =

k=1

Q(µk ) , P (µk )

where the prime stands for differentiation. 2.2-2. Kernels Containing Power-Law and Exponential Functions 20.

x

xeλ(x–t) y(t) dt = f (x).

y(x) + A a

Solution:

21.

2

– x2 ) + λ(x – t) f (t) dt.

1

2

– x2 ) + λ(x – t) f (t) dt.

x exp a

1

x

y(x) = f (x) – A

2 A(t

x

teλ(x–t) y(t) dt = f (x).

y(x) + A a

Solution: y(x) = f (x) – A

x

t exp a

22.

2 A(t

x

(x – t)eλt y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.4 with g(t) = Aeλt . Solution: x A y(x) = f (x) + u1 (x)u2 (t) – u2 (x)u1 (t) eλt f (t) dt, W a where u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equation uxx + Aeλx u = 0; the functions u1 (x) and u2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on sign A: W =

λ , π

λ W =– , 2

√ √ 2 A λx/2 2 A λx/2 , u2 (x) = Y0 u1 (x) = J0 e e λ λ √ √ 2 |A| λx/2 2 |A| λx/2 , u2 (x) = K0 u1 (x) = I0 e e λ λ

for A > 0, for A < 0.

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23.

x

(x – t)eλ(x–t) y(t) dt = f (x).

y(x) + A a

1◦ . Solution with A > 0:

x

eλ(x–t) sin[k(x – t)]f (t) dt,

k=

eλ(x–t) sinh[k(x – t)]f (t) dt,

k=

y(x) = f (x) – k

√

A.

a

2◦ . Solution with A < 0:

x

y(x) = f (x) + k

√

–A.

a

24.

x

(x – t)eλx+µt y(t) dt = f (x).

y(x) + A a

The substitution u(x) = e–λx y(x) leads to an equation of the form 2.2.22: x u(x) + A (x – t)e(λ+µ)t u(t) dt = f (x)e–λx . a

25.

x

(Ax + Bt + C)eλ(x–t) y(t) dt = f (x).

y(x) – a

The substitution u(x) = e–λx y(x) leads to an equation of the form 2.1.6: x u(x) – A (Ax + Bt + C)u(t) dt = f (x)e–λx . a

26.

x

x2 eλ(x–t) y(t) dt = f (x).

y(x) + A a

Solution:

1

3

– x3 ) + λ(x – t) f (t) dt.

1

3

– x3 ) + λ(x – t) f (t) dt.

1

3

– x3 ) + λ(x – t) f (t) dt.

x

x2 exp

y(x) = f (x) – A

3 A(t

a

27.

x

xteλ(x–t) y(t) dt = f (x).

y(x) + A a

Solution:

x

y(x) = f (x) – A

xt exp

3 A(t

a

28.

x

t2 eλ(x–t) y(t) dt = f (x).

y(x) + A a

Solution:

x

t2 exp

y(x) = f (x) – A

3 A(t

a

29.

x

(x – t)2 eλ(x–t) y(t) dt = f (x).

y(x) + A a

Solution:

y(x) = f (x) –

x

R(x – t)f (t) dt, a

√ √ √ R(x) = 23 ke(λ–2k)x – 23 ke(λ+k)x cos 3 kx – 3 sin 3 kx ,

k=

1 1/3 . 4A

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30.

x

(x2 – t2 )eλ(x–t) y(t) dt = f (x).

y(x) + A 0

The substitution u(x) = e–λx y(x) leads to an equation of the form 2.1.11: x u(x) + A (x2 – t2 )u(t) dt = f (x)e–λx . 0

31.

x

(x – t)n eλ(x–t) y(t) dt = f (x),

y(x) + A

n = 1, 2, . . .

a

Solution:

x

y(x) = f (x) +

R(x – t)f (t) dt, a

R(x) =

n 1 λx exp(σk x) σk cos(βk x) – βk sin(βk x) , e n+1 k=0

where

2πk , n+1

2πk + π cos , n+1

1

σk = |An!| n+1 cos 1

σk = |An!| n+1 32.

y(x) + b

x

a

1

2πk n+1

2πk + π sin n+1

βk = |An!| n+1 sin 1

βk = |An!| n+1

for

A < 0,

for

A > 0.

exp[λ(x – t)] y(t) dt = f (x). √ x–t

Solution: λx

y(x) = e

F (x) + πb

x

2

2

exp[πb (x – t)]F (t) dt , a

where

F (x) = e–λx f (x) – b a

33.

x

e–λt f (t) √ dt. x–t

x

(x – t)tk eλ(x–t) y(t) dt = f (x).

y(x) + A a

The substitution u(x) = e–λx y(x) leads to an equation of the form 2.1.49: x u(x) + A (x – t)tk u(t) dt = f (x)e–λx . a

34.

x

(xk – tk )eλ(x–t) y(t) dt = f (x).

y(x) + A a

The substitution u(x) = e–λx y(x) leads to an equation of the form 2.1.52: x u(x) + A (xk – tk )u(t) dt = f (x)e–λx . a

35.

y(x) – λ 0

x

eµ(x–t) (x – t)α

Solution:

y(x) = f (x) +

y(t) dt = f (x),

x

R(x – t)f (t) dt, 0

0 < α < 1.

where

µx

R(x) = e

n ∞ λΓ(1 – α)x1–α . xΓ n(1 – α) n=1

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36.

x

y(x) + A

exp λ(x2 – t2 ) y(t) dt = f (x).

a

Solution:

x

y(x) = f (x) – A

exp λ(x2 – t2 ) – A(x – t) f (t) dt.

a

37.

x

y(x) + A

exp λx2 + βt2 y(t) dt = f (x).

a

In the case β = –λ, see equation 2.2.36. This is a special case of equation 2.9.2 with g(x) = –A exp λx2 ) and h(t) = exp βt2 . 38.

∞

y(x) + A

√ exp –λ t – x y(t) dt = f (x).

x

√ This is a special case of equation 2.9.62 with K(x) = A exp –λ –x . 39.

x

y(x) + A

exp λ(xµ – tµ ) y(t) dt = f (x),

µ > 0.

a

This is a special case of equation 2.9.2 with g(x) = –A exp λxµ and h(t) = exp –λtµ . Solution: x y(x) = f (x) – A exp λ(xµ – tµ ) – A(x – t) f (t) dt. a

40.

x

y(x) + k 0

t exp –λ y(t) dt = g(x). x x 1

This is a special case of equation 2.9.71 with f (z) = ke–λz . N For a polynomial right-hand side, g(x) = An xn , a solution is given by n=0

y(x) =

N n=0

An xn , 1 + kBn

n! 1 n! – e–λ . n+1 λ k! λn–k+1 n

Bn =

k=0

2.3. Equations Whose Kernels Contain Hyperbolic Functions 2.3-1. Kernels Containing Hyperbolic Cosine 1.

x

y(x) – A

cosh(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A cosh(λx) and h(t) = 1. Solution: y(x) = f (x) + A

A sinh(λx) – sinh(λt) f (t) dt. λ

x

cosh(λx) exp a

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2.

x

y(x) – A

cosh(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = cosh(λt). Solution: x A y(x) = f (x) + A sinh(λx) – sinh(λt) f (t) dt. cosh(λt) exp λ a 3.

x

y(x) + A

cosh[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.28 with g(t) = A. Therefore, solving the original integral equation is reduced to solving the second-order linear nonhomogeneous ordinary differential equation with constant coefficients yxx + Ayx – λ2 y = fxx – λ2 f ,

f = f (x),

under the initial conditions yx (a) = fx (a) – Af (a).

y(a) = f (a), Solution:

x

y(x) = f (x) + R(x – t)f (t) dt, a 2 1 A R(x) = exp – 2 Ax sinh(kx) – A cosh(kx) , k = λ2 + 14 A2 . 2k 4.

y(x) +

n x

a

Ak cosh[λk (x – t)] y(t) dt = f (x).

k=1

This equation can be reduced to an equation of the form 2.2.19 by using the identity cosh z ≡ 12 ez + e–z . Therefore, the integral equation in question can be reduced to a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. 5.

x

y(x) – A a

cosh(λx) cosh(λt)

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

cosh(λx) f (t) dt. cosh(λt)

eA(x–t)

cosh(λt) f (t) dt. cosh(λx)

a

6.

x

y(x) – A a

cosh(λt) cosh(λx)

y(t) dt = f (x).

Solution: y(x) = f (x) + A

a

7.

x

x

coshk (λx) coshm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A coshk (λx) and h(t) = coshm (µt).

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8.

x

y(x) + A

t cosh[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.28 with g(t) = At. 9.

x

tk coshm (λx)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –A coshm (λx) and h(t) = tk . 10.

x

xk coshm (λt)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = coshm (λt). 11.

x

A cosh(kx) + B – AB(x – t) cosh(kx) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = A cosh(kx). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x G(x) B2 A B(x–s) R(x, t) = [A cosh(kx) + B] + sinh(kx) . e G(s) ds, G(x) = exp G(t) G(t) t k 12.

x

A cosh(kt) + B + AB(x – t) cosh(kt) y(t) dt = f (x).

y(x) +

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A cosh(kt). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x G(t) B2 A R(x, t) = –[A cosh(kt) + B] + sinh(kx) . eB(t–s) G(s) ds, G(x) = exp G(x) G(x) t k 13.

∞

y(x) + A

√ cosh λ t – x y(t) dt = f (x).

x

√ This is a special case of equation 2.9.62 with K(x) = A cosh λ –x .

2.3-2. Kernels Containing Hyperbolic Sine 14.

x

y(x) – A

sinh(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A sinh(λx) and h(t) = 1. Solution: x A y(x) = f (x) + A sinh(λx) exp cosh(λx) – cosh(λt) f (t) dt. λ a

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15.

x

y(x) – A

sinh(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = sinh(λt). Solution: x A y(x) = f (x) + A sinh(λt) exp cosh(λx) – cosh(λt) f (t) dt. λ a 16.

x

y(x) + A

sinh[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.30 with g(x) = A. 1◦ . Solution with λ(A – λ) > 0: Aλ x y(x) = f (x) – sin[k(x – t)]f (t) dt, k a 2◦ . Solution with λ(A – λ) < 0: Aλ x y(x) = f (x) – sinh[k(x – t)]f (t) dt, k a

where

k=

where

k=

λ(A – λ).

λ(λ – A).

3◦ . Solution with A = λ:

x

y(x) = f (x) – λ2

(x – t)f (t) dt. a

17.

x

sinh3 [λ(x – t)]y(t) dt = f (x).

y(x) + A a

Using the formula sinh3 β = y(x) + a

18.

y(x) +

x

x

1 4

sinh 3β –

3 4

sinh β, we arrive at an equation of the form 2.3.18:

3λ(x – t) – 34 A sinh[λ(x – t)] y(t) dt = f (x).

1 4 A sinh

A1 sinh[λ1 (x – t)] + A2 sinh[λ2 (x – t)] y(t) dt = f (x).

a

1◦ . Introduce the notation x I1 = sinh[λ1 (x – t)]y(t) dt, ax J1 = cosh[λ1 (x – t)]y(t) dt, a

x

I2 =

sinh[λ2 (x – t)]y(t) dt, ax

J2 =

cosh[λ2 (x – t)]y(t) dt. a

Successively differentiating the integral equation four times yields (the first line is the original equation) y + A1 I1 + A2 I2 = f ,

f = f (x),

yx + A1 λ1 J1 + A2 λ2 J2 = fx , yxx + (A1 λ1 + A2 λ2 )y + A1 λ21 I1 + A2 λ22 I2 = fxx , 3 3 yxxx + (A1 λ1 + A2 λ2 )yx + A1 λ1 J1 + A2 λ2 J2 = fxxx , 3 3 4 yxxxx + (A1 λ1 + A2 λ2 )yxx + (A1 λ1 + A2 λ2 )y + A1 λ1 I1

(1) (2) (3) (4) +

A2 λ42 I2

=

fxxxx .

(5)

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Eliminating I1 and I2 from (1), (3), and (5), we arrive at a fourth-order linear ordinary differential equation with constant coefficients: yxxxx – (λ21 + λ22 – A1 λ1 – A2 λ2 )yxx + (λ21 λ22 – A1 λ1 λ22 – A2 λ21 λ2 )y = fxxxx – (λ21 + λ22 )fxx + λ21 λ22 f .

(6)

The initial conditions can be obtained by setting x = a in (1)–(4): y(a) = f (a),

yx (a) = fx (a),

yxx (a) = fxx (a) – (A1 λ1 + A2 λ2 )f (a), yxxx (a)

=

fxxx (a)

– (A1 λ1 +

(7)

A2 λ2 )fx (a).

On solving the differential equation (6) under conditions (7), we thus find the solution of the integral equation. 2◦ . Consider the characteristic equation z 2 – (λ21 + λ22 – A1 λ1 – A2 λ2 )z + λ21 λ22 – A1 λ1 λ22 – A2 λ21 λ2 = 0,

(8)

whose roots, z1 and z2 , determine the solution structure of the integral equation. Assume that the discriminant of equation (8) is positive: D ≡ (A1 λ1 – A2 λ2 – λ21 + λ22 )2 + 4A1 A2 λ1 λ2 > 0. In this case, the quadratic equation (8) has the real (different) roots √ √ z1 = 12 (λ21 + λ22 – A1 λ1 – A2 λ2 ) + 12 D, z2 = 12 (λ21 + λ22 – A1 λ1 – A2 λ2 ) – 12 D. Depending on the signs of z1 and z2 the following three cases are possible. Case 1. If z1 > 0 and z2 > 0, then the solution of the integral equation has the form (i = 1, 2): x √ y(x) = f (x) + {B1 sinh[µ1 (x – t)] + B2 sinh µ2 (x – t) f (t) dt, µi = zi , a

where B1 = A1

λ1 (µ21 – λ22 ) λ2 (µ21 – λ21 ) + A2 , 2 2 µ1 (µ2 – µ1 ) µ1 (µ22 – µ21 )

B2 = A1

λ1 (µ22 – λ22 ) λ2 (µ22 – λ21 ) + A2 . 2 2 µ2 (µ1 – µ2 ) µ2 (µ21 – µ22 )

Case 2. If z1 < 0 and z2 < 0, then the solution of the integral equation has the form x y(x) = f (x) + {B1 sin[µ1 (x – t)] + B2 sin µ2 (x – t) f (t) dt, µi = |zi |, a

where the coefficients B1 and B2 are found by solving the following system of linear algebraic equations: B1 µ1 B2 µ2 B1 µ1 B2 µ2 + + 1 = 0, + + 1 = 0. λ21 + µ21 λ21 + µ22 λ22 + µ21 λ22 + µ22 Case 3. If z1 > 0 and z2 < 0, then the solution of the integral equation has the form x y(x) = f (x) + {B1 sinh[µ1 (x – t)] + B2 sin µ2 (x – t) f (t) dt, µi = |zi |, a

where B1 and B2 are determined from the following system of linear algebraic equations: B1 µ1 B2 µ2 + + 1 = 0, λ21 – µ21 λ21 + µ22

B1 µ1 B2 µ2 + + 1 = 0. λ22 – µ21 λ22 + µ22

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19.

y(x) +

n x

a

Ak sinh[λk (x – t)] y(t) dt = f (x).

k=1

1◦ . This equation can be reduced to an equation of the form 2.2.19 with the aid of the formula sinh z = 12 ez – e–z . Therefore, the original integral equation can be reduced to a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. 2◦ . Let us find the roots zk of the algebraic equation n λk Ak + 1 = 0. z – λ2k k=1

(1)

By reducing it to a common denominator, we arrive at the problem of determining the roots of an nth-degree characteristic polynomial. Assume that all zk are real, different, and nonzero. Let us divide the roots into two groups z1 > 0, zs+1 < 0,

z2 > 0,

...,

zs > 0

(positive roots);

zs+2 < 0,

. . . , zn < 0

(negative roots).

Then the solution of the integral equation can be written in the form y(x) = f (x)+

x s a

n Bk sinh µk (x–t) + Ck sin µk (x–t) f (t) dt,

k=1

µk =

|zk |. (2)

k=s+1

The coefficients Bk and Ck are determined from the following system of linear algebraic equations: s n Bk µk Ck µk + + 1 = 0, 2 2 λm – µk k=s+1 λ2m + µ2k k=0

µk =

|zk |,

m = 1, . . . , n.

(3)

In the case of a nonzero root zs = 0, we can introduce the new constant D = Bs µs and proceed to the limit µs → 0. As a result, the term D(x – t) appears in solution (2) instead of Bs sinh µs (x – t) and the corresponding terms Dλ–2 m appear in system (3). 20.

x

y(x) – A a

sinh(λx) sinh(λt)

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

sinh(λx) f (t) dt. sinh(λt)

eA(x–t)

sinh(λt) f (t) dt. sinh(λx)

a

21.

x

y(x) – A a

sinh(λt) sinh(λx)

y(t) dt = f (x).

Solution: y(x) = f (x) + A

a

22.

x

x

sinhk (λx) sinhm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A sinhk (λx) and h(t) = sinhm (µt).

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23.

x

y(x) + A

t sinh[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.30 with g(t) = At. Solution: Aλ x y(x) = f (x) + t u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a where u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear ordinary differential equation uxx + λ(Ax – λ)u = 0, and W is the Wronskian. The functions u1 (x) and u2 (x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of Aλ, as follows: if Aλ > 0, then √ √ u1 (x) = ξ 1/2 J1/3 23 Aλ ξ 3/2 , u2 (x) = ξ 1/2 Y1/3 23 Aλ ξ 3/2 , W = 3/π, ξ = x – (λ/A); if Aλ < 0, then u1 (x) = ξ 1/2 I1/3

2√ 3

–Aλ ξ 3/2 ,

W = – 32 , 24.

u2 (x) = ξ 1/2 K1/3

2√ 3

–Aλ ξ 3/2 ,

ξ = x – (λ/A).

x

y(x) + A

x sinh[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.31 with g(x) = Ax and h(t) = 1. Solution: Aλ x y(x) = f (x) + x u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a

25.

where u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear ordinary differential equation uxx + λ(Ax – λ)u = 0, and W is the Wronskian. The functions u1 (x), u2 (x), and W are specified in 2.3.23. x y(x) + A tk sinhm (λx)y(t) dt = f (x). a

26.

This is a special case of equation 2.9.2 with g(x) = –A sinhm (λx) and h(t) = tk . x y(x) + A xk sinhm (λt)y(t) dt = f (x). a

27.

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = sinhm (λt). x y(x) – A sinh(kx) + B – AB(x – t) sinh(kx) y(t) dt = f (x). a

This is a special case of equation 2.9.7 with λ = B and g(x) = A sinh(kx). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x G(x) A B2 B(x–s) R(x, t) = [A sinh(kx) + B] e G(s) ds, G(x) = exp + cosh(kx) . G(t) G(t) t k

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28.

x

A sinh(kt) + B + AB(x – t) sinh(kt) y(t) dt = f (x).

y(x) +

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A sinh(kt). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x G(t) B2 A B(t–s) R(x, t) = –[sinh(kt) + B] + cosh(kx) . e G(s) ds, G(x) = exp G(x) G(x) t k 29.

√ sinh λ t – x y(t) dt = f (x).

∞

y(x) + A x

√ This is a special case of equation 2.9.62 with K(x) = A sinh λ –x .

2.3-3. Kernels Containing Hyperbolic Tangent 30.

x

y(x) – A

tanh(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A tanh(λx) and h(t) = 1. Solution: A/λ x cosh(λx) y(x) = f (x) + A tanh(λx) f (t) dt. cosh(λt) a 31.

x

y(x) – A

tanh(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = tanh(λt). Solution: A/λ x cosh(λx) y(x) = f (x) + A tanh(λt) f (t) dt. cosh(λt) a 32.

y(x) + A

x

tanh(λx) – tanh(λt) y(t) dt = f (x).

a

This is a special case of equation 2.9.5 with g(x) = A tanh(λx). Solution: x 1 y(x) = f (x) + Y1 (x)Y2 (t) – Y2 (x)Y1 (t) f (t) dt, W a where Y1 (x), Y2 (x) is a fundamental system of solutions of the second-order linear ordinary + AλY = 0, W is the Wronskian, and the primes stand for differential equation cosh2 (λx)Yxx the differentiation with respect to the argument specified in the parentheses. As shown in A. D. Polyanin and V. F. Zaitsev (1996), the functions Y1 (x) and Y2 (x) can be represented in the form

Y1 (x) = F α, β, 1;

eλx , 1 + eλx

Y2 (x) = Y1 (x) a

x

dξ , Y12 (ξ)

W = 1,

where F (α, β, γ; z) is the hypergeometric function, in which α and β are determined from the algebraic system α + β = 1, αβ = –A/λ.

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33.

x

y(x) – A

tanh(λx) tanh(λt)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

tanh(λx) f (t) dt. tanh(λt)

eA(x–t)

tanh(λt) f (t) dt. tanh(λx)

a

34.

x

y(x) – A

tanh(λt) tanh(λx)

a

y(t) dt = f (x).

Solution: y(x) = f (x) + A

x

a

35.

x

tanhk (λx) tanhm (µt)y(t) dt = f (x).

y(x) – A a

36.

This is a special case of equation 2.9.2 with g(x) = A tanhk (λx) and h(t) = tanhm (µt). x y(x) + A tk tanhm (λx)y(t) dt = f (x). a

37.

This is a special case of equation 2.9.2 with g(x) = –A tanhm (λx) and h(t) = tk . x y(x) + A xk tanhm (λt)y(t) dt = f (x). a

38.

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = tanhm (λt). ∞ y(x) + A tanh[λ(t – x)]y(t) dt = f (x).

39.

This is a special case of equation 2.9.62 with K(z) = A tanh(–λz). ∞ √ y(x) + A tanh λ t – x y(t) dt = f (x).

x

x

√ This is a special case of equation 2.9.62 with K(z) = A tanh λ –z .

x

A tanh(kx) + B – AB(x – t) tanh(kx) y(t) dt = f (x).

40.

y(x) –

41.

This is a special case of equation 2.9.7 with λ = B and g(x) = A tanh(kx). x y(x) + A tanh(kt) + B + AB(x – t) tanh(kt) y(t) dt = f (x).

a

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A tanh(kt). 2.3-4. Kernels Containing Hyperbolic Cotangent 42.

x

y(x) – A

coth(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A coth(λx) and h(t) = 1. Solution: x sinh(λx) A/λ y(x) = f (x) + A coth(λx) f (t) dt. sinh(λt) a

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43.

x

y(x) – A

coth(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = coth(λt). Solution: x sinh(λx) A/λ y(x) = f (x) + A coth(λt) f (t) dt. sinh(λt) a 44.

x

y(x) – A

coth(λt) coth(λx)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

coth(λt) f (t) dt. coth(λx)

eA(x–t)

coth(λx) f (t) dt. coth(λt)

a

45.

x

y(x) – A

coth(λx) coth(λt)

a

y(t) dt = f (x).

Solution: y(x) = f (x) + A

x

a

46.

x

cothk (λx) cothm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A cothk (λx) and h(t) = cothm (µt). 47.

x

tk cothm (λx)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –A cothm (λx) and h(t) = tk . 48.

x

xk cothm (λt)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = cothm (λt). 49.

∞

y(x) + A

coth[λ(t – x)]y(t) dt = f (x). x

This is a special case of equation 2.9.62 with K(z) = A coth(–λz). 50.

∞

y(x) + A

√ coth λ t – x y(t) dt = f (x).

x

√ This is a special case of equation 2.9.62 with K(z) = A coth λ –z . 51.

x

A coth(kx) + B – AB(x – t) coth(kx) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = A coth(kx). 52.

y(x) +

x

A coth(kt) + B + AB(x – t) coth(kt) y(t) dt = f (x).

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A coth(kt).

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2.3-5. Kernels Containing Combinations of Hyperbolic Functions 53.

x

coshk (λx) sinhm (µt)y(t) dt = f (x).

y(x) – A a

54.

This is a special case of equation 2.9.2 with g(x) = A coshk (λx) and h(t) = sinhm (µt). x y(x) – A + B cosh(λx) + B(x – t)[λ sinh(λx) – A cosh(λx)] y(t) dt = f (x).

55.

This is a special case of equation 2.9.32 with b = B and g(x) = A. x y(x) – A + B sinh(λx) + B(x – t)[λ cosh(λx) – A sinh(λx)] y(t) dt = f (x).

56.

This is a special case of equation 2.9.33 with b = B and g(x) = A. x y(x) – A tanhk (λx) cothm (µt)y(t) dt = f (x).

a

a

a

This is a special case of equation 2.9.2 with g(x) = A tanhk (λx) and h(t) = cothm (µt).

2.4. Equations Whose Kernels Contain Logarithmic Functions 2.4-1. Kernels Containing Logarithmic Functions 1.

x

y(x) – A

ln(λx)y(t) dt = f (x). a

2.

This is a special case of equation 2.9.2 with g(x) = A ln(λx) and h(t) = 1. Solution: x (λx)Ax y(x) = f (x) + A ln(λx)e–A(x–t) f (t) dt. (λt)At a x y(x) – A ln(λt)y(t) dt = f (x).

3.

This is a special case of equation 2.9.2 with g(x) = A and h(t) = ln(λt). Solution: x (λx)Ax y(x) = f (x) + A ln(λt)e–A(x–t) f (t) dt. (λt)At a x y(x) + A (ln x – ln t)y(t) dt = f (x).

a

a

This is a special case of equation 2.9.5 with g(x) = A ln x. Solution: x 1 y(x) = f (x) + u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a where the primes denote differentiation with respect to the argument specified in the parentheses; and u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equation uxx + Ax–1 u = 0, with u1 (x) and u2 (x) expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of A: √ √ √ √ W = π1 , u1 (x) = x J1 2 Ax , u2 (x) = x Y1 2 Ax for A > 0, √ √ √ √ 1 W = – 2 , u1 (x) = x I1 2 –Ax , u2 (x) = x K1 2 –Ax for A < 0.

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4.

x

y(x) – A

ln(λx) ln(λt)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

ln(λx) f (t) dt. ln(λt)

eA(x–t)

ln(λt) f (t) dt. ln(λx)

a

5.

x

y(x) – A

ln(λt) ln(λx)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A a

6.

x

lnk (λx) lnm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A lnk (λx) and h(t) = lnm (µt). 7.

y(x) + a

∞

ln(t – x)y(t) dt = f (x).

x

This is a special case of equation 2.9.62 with K(x) = a ln(–x). m For f (x) = Ak exp(–λk x), where λk > 0, a solution of the equation has the form k=1

y(x) =

m Ak exp(–λk x), Bk

Bk = 1 –

k=1

8.

a (ln λk + C), λk

where C = 0.5772 . . . is the Euler constant. ∞ y(x) + a ln2 (t – x)y(t) dt = f (x). x

This is a special case of equation 2.9.62 with K(x) = a ln2 (–x). m For f (x) = Ak exp(–λk x), where λk > 0, a solution of the equation has the form k=1

y(x) =

m Ak exp(–λk x), Bk k=1

Bk = 1 +

a 1 2 π + (ln λk + C)2 , 6 λk

where C = 0.5772 . . . is the Euler constant. 2.4-2. Kernels Containing Power-Law and Logarithmic Functions 9.

x

xk lnm (λt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = Axk and h(t) = lnm (λt). 10.

x

tk lnm (λx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A lnm (λx) and h(t) = tk .

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11.

x

A ln(kx) + B – AB(x – t) ln(kx) y(t) dt = f (x).

y(x) –

a

12.

This is a special case of equation 2.9.7 with λ = B and g(x) = A ln(kx). x y(x) + A ln(kt) + B + AB(x – t) ln(kt) y(t) dt = f (x).

13.

This is a special case of equation 2.9.8 with λ = B and g(t) = A ln(kt). ∞ y(x) + a (t – x)n ln(t – x)y(t) dt = f (x), n = 1, 2, . . .

a

x m

For f (x) =

Ak exp(–λk x), where λk > 0, a solution of the equation has the form

k=1 m Ak exp(–λk x), Bk

y(x) =

Bk = 1 +

k=1

14.

an! 1+ λn+1 k

1 2

+

1 3

+ ··· +

1 n

– ln λk – C ,

where C = 0.5772 . . . is the Euler constant. ∞ ln(t – x) y(x) + a y(t) dt = f (x). √ t–x x This is a special case of equation 2.9.62 with K(–x) = ax–1/2 ln x. m Ak exp(–λk x), where λk > 0, a solution of the equation has the form For f (x) = k=1 m Ak y(x) = exp(–λk x), Bk k=1

π Bk = 1 – a ln(4λk ) + C , λk

where C = 0.5772 . . . is the Euler constant.

2.5. Equations Whose Kernels Contain Trigonometric Functions 2.5-1. Kernels Containing Cosine 1.

x

y(x) – A

cos(λx)y(t) dt = f (x). a

2.

This is a special case of equation 2.9.2 with g(x) = A cos(λx) and h(t) = 1. Solution: x A y(x) = f (x) + A cos(λx) exp sin(λx) – sin(λt) f (t) dt. λ a x y(x) – A cos(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = cos(λt). Solution: x A y(x) = f (x) + A sin(λx) – sin(λt) f (t) dt. cos(λt) exp λ a

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3.

x

y(x) + A

cos[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.34 with g(t) = A. Therefore, solving this integral equation is reduced to solving the following second-order linear nonhomogeneous ordinary differential equation with constant coefficients: yxx + Ayx + λ2 y = fxx + λ2 f ,

f = f (x),

with the initial conditions yx (a) = fx (a) – Af (a).

y(a) = f (a), 1◦ . Solution with |A| > 2|λ|:

x

y(x) = f (x) + R(x – t)f (t) dt, a 2 1 A R(x) = exp – 2 Ax sinh(kx) – A cosh(kx) , k = 14 A2 – λ2 . 2k 2◦ . Solution with |A| < 2|λ|:

x

y(x) = f (x) +

R(x – t)f (t) dt, a

A2 R(x) = exp – 12 Ax sin(kx) – A cos(kx) , 2k

k=

λ2 – 14 A2 .

3◦ . Solution with λ = ± 12 A:

R(x) = exp – 12 Ax 12 A2 x – A .

x

y(x) = f (x) +

R(x – t)f (t) dt, a

4.

y(x) +

n x

a

Ak cos[λk (x – t)] y(t) dt = f (x).

k=1

This integral equation is reduced to a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. Set x Ik (x) = cos[λk (x – t)]y(t) dt. (1) a

Differentiating (1) with respect to x twice yields Ik Ik

x

= y(x) – λk =

yx (x)

sin[λk (x – t)]y(t) dt, a

–

λ2k

(2)

x

cos[λk (x – t)]y(t) dt, a

where the primes stand for differentiation with respect to x. Comparing (1) and (2), we see that Ik = yx (x) – λ2k Ik , Ik = Ik (x). (3)

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With the aid of (1), the integral equation can be rewritten in the form y(x) +

n

Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice taking into account (3) yields yxx (x) + σn yx (x) –

n

Ak λ2k Ik = fxx (x),

σn =

k=1

n

Ak .

(5)

k=1

Eliminating the integral In from (4) and (5), we obtain yxx (x) + σn yx (x) + λ2n y(x) +

n–1

Ak (λ2n – λ2k )Ik = fxx (x) + λ2n f (x).

(6)

k=1

Differentiating (6) with respect to x twice followed by eliminating In–1 from the resulting expression with the aid of (6) yields a similar equation whose left-hand side is a fourthn–2 order differential operator (acting on y) with constant coefficients plus the sum Bk Ik . k=1

Successively eliminating the terms In–2 , In–3 , . . . using double differentiation and formula (3), we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. The initial conditions for y(x) can be obtained by setting x = a in the integral equation and all its derivative equations. 5.

x

y(x) – A

cos(λx) cos(λt)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

cos(λx) f (t) dt. cos(λt)

eA(x–t)

cos(λt) f (t) dt. cos(λx)

a

6.

x

y(x) – A

cos(λt) cos(λx)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A a

7.

x

cosk (λx) cosm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A cosk (λx) and h(t) = cosm (µt). 8.

x

y(x) + A

t cos[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.34 with g(t) = At. 9.

x

tk cosm (λx)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –A cosm (λx) and h(t) = tk .

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10.

x

xk cosm (λt)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = cosm (λt). 11.

x

A cos(kx) + B – AB(x – t) cos(kx) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = A cos(kx). Solution:

x

y(x) = f (x) + R(x, t)f (t) dt, a x G(x) B2 A B(x–s) R(x, t) = [A cos(kx) + B] + sin(kx) . e G(s) ds, G(x) = exp G(t) G(t) t k 12.

x

A cos(kt) + B + AB(x – t) cos(kt) y(t) dt = f (x).

y(x) +

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A cos(kt). Solution:

x

y(x) = f (x) + R(x, t)f (t) dt, a x G(t) B2 R(x, t) = –[A cos(kt) + B] eB(t–s) G(s) ds, + G(x) G(x) t 13.

∞

y(x) + A

A G(x) = exp sin(kx) . k

√ cos λ t – x y(t) dt = f (x).

x

√ This is a special case of equation 2.9.62 with K(x) = A cos λ –x .

2.5-2. Kernels Containing Sine 14.

x

y(x) – A

sin(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A sin(λx) and h(t) = 1. Solution:

x

y(x) = f (x) + A

sin(λx) exp a

15.

A cos(λt) – cos(λx) f (t) dt. λ

x

y(x) – A

sin(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = sin(λt). Solution: y(x) = f (x) + A

A cos(λt) – cos(λx) f (t) dt. λ

x

sin(λt) exp a

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16.

x

y(x) + A

sin[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.36 with g(t) = A. 1◦ . Solution with λ(A + λ) > 0: Aλ x y(x) = f (x) – sin[k(x – t)]f (t) dt, k a 2◦ . Solution with λ(A + λ) < 0: Aλ x y(x) = f (x) – sinh[k(x – t)]f (t) dt, k a 3◦ . Solution with A = –λ:

y(x) = f (x) + λ2

where

k=

where

k=

λ(A + λ).

–λ(λ + A).

x

(x – t)f (t) dt. a

17.

x

sin3 [λ(x – t)]y(t) dt = f (x).

y(x) + A a

Using the formula sin3 β = – 14 sin 3β + 34 sin β, we arrive at an equation of the form 2.5.18: x 1 y(x) + – 4 A sin[3λ(x – t)] + 34 A sin[λ(x – t)] y(t) dt = f (x). a

18.

y(x) +

x

A1 sin[λ1 (x – t)] + A2 sin[λ2 (x – t)] y(t) dt = f (x).

a

This equation can be solved by the same method as equation 2.3.18, by reducing it to a fourth-order linear ordinary differential equation with constant coefficients. Consider the characteristic equation z 2 + (λ21 + λ22 + A1 λ1 + A2 λ2 )z + λ21 λ22 + A1 λ1 λ22 + A2 λ21 λ2 = 0,

(1)

whose roots, z1 and z2 , determine the solution structure of the integral equation. Assume that the discriminant of equation (1) is positive: D ≡ (A1 λ1 – A2 λ2 + λ21 – λ22 )2 + 4A1 A2 λ1 λ2 > 0. In this case, the quadratic equation (1) has the real (different) roots √ √ z1 = – 12 (λ21 + λ22 + A1 λ1 + A2 λ2 ) + 12 D, z2 = – 12 (λ21 + λ22 + A1 λ1 + A2 λ2 ) – 12 D. Depending on the signs of z1 and z2 the following three cases are possible. Case 1. If z1 > 0 and z2 > 0, then the solution of the integral equation has the form (i = 1, 2): x √ y(x) = f (x) + {B1 sinh[µ1 (x – t)] + B2 sinh µ2 (x – t) f (t) dt, µi = zi , a

where the coefficients B1 and B2 are determined from the following system of linear algebraic equations: B1 µ1 B2 µ2 B1 µ1 B2 µ2 + – 1 = 0, + – 1 = 0. λ21 + µ21 λ21 + µ22 λ22 + µ21 λ22 + µ22

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Case 2. If z1 < 0 and z2 < 0, then the solution of the integral equation has the form

x

y(x) = f (x) +

{B1 sin[µ1 (x – t)] + B2 sin µ2 (x – t) f (t) dt,

µi =

|zi |,

a

where B1 and B2 are determined from the system B1 µ1 B2 µ2 + – 1 = 0, λ21 – µ21 λ21 – µ22

B1 µ1 B2 µ2 + – 1 = 0. λ22 – µ21 λ22 – µ22

Case 3. If z1 > 0 and z2 < 0, then the solution of the integral equation has the form y(x) = f (x) +

x

{B1 sinh[µ1 (x – t)] + B2 sin µ2 (x – t) f (t) dt,

µi =

|zi |,

a

where B1 and B2 are determined from the system B1 µ1 B2 µ2 + – 1 = 0, λ21 + µ21 λ21 – µ22

B1 µ1 B2 µ2 + – 1 = 0. λ22 + µ21 λ22 – µ22

Remark. The solution of the original integral equation can be obtained from the solution of equation 2.3.18 by performing the following change of parameters:

λk → iλk , 19.

y(x) +

n x

a

µk → iµk ,

Ak → –iAk ,

Bk → –iBk ,

i2 = –1

(k = 1, 2).

Ak sin[λk (x – t)] y(t) dt = f (x).

k=1

1◦ . This integral equation can be reduced to a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. Set

x

Ik (x) =

sin[λk (x – t)]y(t) dt.

(1)

a

Differentiating (1) with respect to x twice yields Ik = λk

x

Ik = λk y(x) – λ2k

cos[λk (x – t)]y(t) dt, a

x

sin[λk (x – t)]y(t) dt,

(2)

a

where the primes stand for differentiation with respect to x. Comparing (1) and (2), we see that Ik = λk y(x) – λ2k Ik , Ik = Ik (x). (3) With aid of (1), the integral equation can be rewritten in the form y(x) +

n

Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice taking into account (3) yields yxx (x) + σn y(x) –

n k=1

Ak λ2k Ik = fxx (x),

σn =

n

Ak λk .

(5)

k=1

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Eliminating the integral In from (4) and (5), we obtain yxx (x) + (σn + λ2n )y(x) +

n–1

Ak (λ2n – λ2k )Ik = fxx (x) + λ2n f (x).

(6)

k=1

Differentiating (6) with respect to x twice followed by eliminating In–1 from the resulting expression with the aid of (6) yields a similar equation whose left-hand side is a fourthn–2 order differential operator (acting on y) with constant coefficients plus the sum Bk Ik . k=1

Successively eliminating the terms In–2 , In–3 , . . . using double differentiation and formula (3), we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. The initial conditions for y(x) can be obtained by setting x = a in the integral equation and all its derivative equations. 2◦ . Let us find the roots zk of the algebraic equation n λk Ak + 1 = 0. z + λ2k k=1

(7)

By reducing it to a common denominator, we arrive at the problem of determining the roots of an nth-degree characteristic polynomial. Assume that all zk are real, different, and nonzero. Let us divide the roots into two groups z1 > 0, zs+1 < 0,

z2 > 0,

...,

zs > 0

(positive roots);

zs+2 < 0,

. . . , zn < 0

(negative roots).

Then the solution of the integral equation can be written in the form x s n y(x) = f (x)+ Bk sinh µk (x–t) + Ck sin µk (x–t) f (t) dt, a

k=1

µk =

|zk |. (8)

k=s+1

The coefficients Bk and Ck are determined from the following system of linear algebraic equations: s n Bk µk Ck µk + – 1 = 0, 2 2 λm + µk k=s+1 λ2m – µ2k k=0

µk =

|zk |

m = 1, 2, . . . , n.

(9)

In the case of a nonzero root zs = 0, we can introduce the new constant D = Bs µs and proceed to the limit µs → 0. As a result, the term D(x – t) appears in solution (8) instead of Bs sinh µs (x – t) and the corresponding terms Dλ–2 m appear in system (9). 20.

x

y(x) – A a

sin(λx) sin(λt)

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

sin(λx) f (t) dt. sin(λt)

eA(x–t)

sin(λt) f (t) dt. sin(λx)

a

21.

x

y(x) – A a

sin(λt) sin(λx)

y(t) dt = f (x).

Solution: y(x) = f (x) + A

a

x

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22.

x

sink (λx) sinm (µt)y(t) dt = f (x).

y(x) – A a

23.

This is a special case of equation 2.9.2 with g(x) = A sink (λx) and h(t) = sinm (µt). x y(x) + A t sin[λ(x – t)]y(t) dt = f (x). a

This is a special case of equation 2.9.36 with g(t) = At. Solution: Aλ x y(x) = f (x) + t u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a where u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear ordinary differential equation uxx + λ(Ax + λ)u = 0, and W is the Wronskian. Depending on the sign of Aλ, the functions u1 (x) and u2 (x) are expressed in terms of Bessel functions or modified Bessel functions as follows: if Aλ > 0, then √ √ u1 (x) = ξ 1/2 J1/3 23 Aλ ξ 3/2 , u2 (x) = ξ 1/2 Y1/3 23 Aλ ξ 3/2 , W = 3/π, ξ = x + (λ/A); if Aλ < 0, then u1 (x) = ξ 1/2 I1/3 24.

2√ 3

–Aλ ξ 3/2 ,

W = – 32 ,

u2 (x) = ξ 1/2 K1/3

2√ 3

–Aλ ξ 3/2 ,

ξ = x + (λ/A).

x

y(x) + A

x sin[λ(x – t)]y(t) dt = f (x). a

25.

This is a special case of equation 2.9.37 with g(x) = Ax and h(t) = 1. Solution: Aλ x y(x) = f (x) + x u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a where u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear ordinary differential equation uxx + λ(Ax + λ)u = 0, and W is the Wronskian. The functions u1 (x), u2 (x), and W are specified in 2.5.23. x y(x) + A tk sinm (λx)y(t) dt = f (x).

26.

This is a special case of equation 2.9.2 with g(x) = –A sinm (λx) and h(t) = tk . x y(x) + A xk sinm (λt)y(t) dt = f (x).

a

a

27.

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = sinm (λt). x y(x) – A sin(kx) + B – AB(x – t) sin(kx) y(t) dt = f (x). a

This is a special case of equation 2.9.7 with λ = B and g(x) = A sin(kx). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x G(x) A B2 B(x–s) R(x, t) = [A sin(kx) + B] e G(s) ds, G(x) = exp – cos(kx) . + G(t) G(t) t k

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28.

x

A sin(kt) + B + AB(x – t) sin(kt) y(t) dt = f (x).

y(x) +

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A sin(kt). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x G(t) A B2 B(t–s) R(x, t) = –[A sin(kt) + B] e G(s) ds, G(x) = exp – cos(kx) . + G(x) G(x) t k 29.

√ sin λ t – x y(t) dt = f (x).

∞

y(x) + A x

√ This is a special case of equation 2.9.62 with K(x) = A sin λ –x .

2.5-3. Kernels Containing Tangent 30.

x

y(x) – A

tan(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A tan(λx) and h(t) = 1. Solution: x cos(λt) A/λ y(x) = f (x) + A tan(λx) f (t) dt. cos(λx) a 31.

x

y(x) – A

tan(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = tan(λt). Solution: x cos(λt) A/λ y(x) = f (x) + A tanh(λt) f (t) dt. cos(λx) a 32.

x

tan(λx) – tan(λt) y(t) dt = f (x).

y(x) + A

a

This is a special case of equation 2.9.5 with g(x) = A tan(λx). Solution: x 1 y(x) = f (x) + Y1 (x)Y2 (t) – Y2 (x)Y1 (t) f (t) dt, W a where Y1 (x), Y2 (x) is a fundamental system of solutions of the second-order linear ordinary + AλY = 0, W is the Wronskian, and the primes stand for differential equation cos2 (λx)Yxx the differentiation with respect to the argument specified in the parentheses. As shown in A. D. Polyanin and V. F. Zaitsev (1995, 1996), the functions Y1 (x) and Y2 (x) can be expressed via the hypergeometric function. 33.

x

y(x) – A a

tan(λx) tan(λt)

y(t) dt = f (x).

Solution:

x

eA(x–t)

y(x) = f (x) + A a

tan(λx) f (t) dt. tan(λt)

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34.

x

y(x) – A

tan(λt) tan(λx)

a

y(t) dt = f (x).

Solution:

x

eA(x–t)

y(x) = f (x) + A a

35.

tan(λt) f (t) dt. tan(λx)

x

tank (λx) tanm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A tank (λx) and h(t) = tanm (µt). 36.

x

tk tanm (λx)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –A tanm (λx) and h(t) = tk . 37.

x

xk tanm (λt)y(t) dt = f (x).

y(x) + A a

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = tanm (λt). 38.

x

A tan(kx) + B – AB(x – t) tan(kx) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = A tan(kx). 39.

y(x) +

x

A tan(kt) + B + AB(x – t) tan(kt) y(t) dt = f (x).

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A tan(kt). 2.5-4. Kernels Containing Cotangent 40.

x

y(x) – A

cot(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A cot(λx) and h(t) = 1. Solution: x sin(λx) A/λ y(x) = f (x) + A cot(λx) f (t) dt. sin(λt) a 41.

x

y(x) – A

cot(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = cot(λt). Solution: x sin(λx) A/λ y(x) = f (x) + A coth(λt) f (t) dt. sin(λt) a 42.

x

y(x) – A a

cot(λx) cot(λt)

y(t) dt = f (x).

Solution:

x

eA(x–t)

y(x) = f (x) + A a

cot(λx) f (t) dt. cot(λt)

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43.

x

y(x) – A

cot(λt) cot(λx)

a

y(t) dt = f (x).

Solution:

a

44.

x

eA(x–t)

y(x) = f (x) + A

cot(λt) f (t) dt. cot(λx)

x

tk cotm (λx)y(t) dt = f (x).

y(x) + A a

45.

This is a special case of equation 2.9.2 with g(x) = –A cotm (λx) and h(t) = tk . x y(x) + A xk cotm (λt)y(t) dt = f (x).

46.

This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = cotm (λt). x y(x) – A cot(kx) + B – AB(x – t) cot(kx) y(t) dt = f (x).

47.

This is a special case of equation 2.9.7 with λ = B and g(x) = A cot(kx). x y(x) + A cot(kt) + B + AB(x – t) cot(kt) y(t) dt = f (x).

a

a

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A cot(kt). 2.5-5. Kernels Containing Combinations of Trigonometric Functions 48.

x

cosk (λx) sinm (µt)y(t) dt = f (x).

y(x) – A a

49.

This is a special case of equation 2.9.2 with g(x) = A cosk (λx) and h(t) = sinm (µt). x y(x) – A + B cos(λx) – B(x – t)[λ sin(λx) + A cos(λx)] y(t) dt = f (x).

50.

This is a special case of equation 2.9.38 with b = B and g(x) = A. x y(x) – A + B sin(λx) + B(x – t)[λ cos(λx) – A sin(λx)] y(t) dt = f (x).

51.

This is a special case of equation 2.9.39 with b = B and g(x) = A. x y(x) – A tank (λx) cotm (µt)y(t) dt = f (x).

a

a

a

This is a special case of equation 2.9.2 with g(x) = A tank (λx) and h(t) = cotm (µt).

2.6. Equations Whose Kernels Contain Inverse Trigonometric Functions 2.6-1. Kernels Containing Arccosine 1.

x

y(x) – A

arccos(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A arccos(λx) and h(t) = 1.

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2.

x

y(x) – A

arccos(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = arccos(λt). 3.

x

y(x) – A

arccos(λx) arccos(λt)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

arccos(λx) f (t) dt. arccos(λt)

eA(x–t)

arccos(λt) f (t) dt. arccos(λx)

a

4.

x

y(x) – A

arccos(λt) arccos(λx)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A a

5.

x

A arccos(kx) + B – AB(x – t) arccos(kx) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = A arccos(kx). 6.

y(x) +

x

A arccos(kt) + B + AB(x – t) arccos(kt) y(t) dt = f (x).

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A arccos(kt).

2.6-2. Kernels Containing Arcsine 7.

x

y(x) – A

arcsin(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A arcsin(λx) and h(t) = 1. 8.

x

y(x) – A

arcsin(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = arcsin(λt). 9.

x

y(x) – A a

arcsin(λx) arcsin(λt)

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

arcsin(λx) f (t) dt. arcsin(λt)

eA(x–t)

arcsin(λt) f (t) dt. arcsin(λx)

a

10.

x

y(x) – A a

arcsin(λt) arcsin(λx)

y(t) dt = f (x).

Solution:

y(x) = f (x) + A a

x

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11.

x

A arcsin(kx) + B – AB(x – t) arcsin(kx) y(t) dt = f (x).

y(x) –

a

12.

This is a special case of equation 2.9.7 with λ = B and g(x) = A arcsin(kx). x y(x) + A arcsin(kt) + B + AB(x – t) arcsin(kt) y(t) dt = f (x). a

This is a special case of equation 2.9.8 with λ = B and g(t) = A arcsin(kt). 2.6-3. Kernels Containing Arctangent 13.

x

y(x) – A

arctan(λx)y(t) dt = f (x). a

14.

This is a special case of equation 2.9.2 with g(x) = A arctan(λx) and h(t) = 1. x y(x) – A arctan(λt)y(t) dt = f (x). a

15.

This is a special case of equation 2.9.2 with g(x) = A and h(t) = arctan(λt). x arctan(λx) y(x) – A y(t) dt = f (x). a arctan(λt) Solution: x arctan(λx) y(x) = f (x) + A f (t) dt. eA(x–t) arctan(λt) a

16.

x

y(x) – A

arctan(λt) arctan(λx)

a

y(t) dt = f (x).

Solution:

x

eA(x–t)

y(x) = f (x) + A a

17.

arctan(λt) f (t) dt. arctan(λx)

∞

y(x) + A

arctan[λ(t – x)]y(t) dt = f (x). x

18.

This is a special case of equation 2.9.62 with K(x) = A arctan(–λx). x y(x) – A arctan(kx) + B – AB(x – t) arctan(kx) y(t) dt = f (x). a

19.

This is a special case of equation 2.9.7 with λ = B and g(x) = A arctan(kx). x y(x) + A arctan(kt) + B + AB(x – t) arctan(kt) y(t) dt = f (x). a

This is a special case of equation 2.9.8 with λ = B and g(t) = A arctan(kt). 2.6-4. Kernels Containing Arccotangent 20.

x

y(x) – A

arccot(λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A arccot(λx) and h(t) = 1.

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21.

x

y(x) – A

arccot(λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = arccot(λt). 22.

x

y(x) – A

arccot(λx) arccot(λt)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

arccot(λx) f (t) dt. arccot(λt)

eA(x–t)

arccot(λt) f (t) dt. arccot(λx)

a

23.

x

y(x) – A

arccot(λt) arccot(λx)

a

y(t) dt = f (x).

Solution:

y(x) = f (x) + A

x

a

24.

∞

y(x) + A

arccot[λ(t – x)]y(t) dt = f (x). x

This is a special case of equation 2.9.62 with K(x) = A arccot(–λx). 25.

x

A arccot(kx) + B – AB(x – t) arccot(kx) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = A arccot(kx). 26.

x

A arccot(kt) + B + AB(x – t) arccot(kt) y(t) dt = f (x).

y(x) +

a

This is a special case of equation 2.9.8 with λ = B and g(t) = A arccot(kt).

2.7. Equations Whose Kernels Contain Combinations of Elementary Functions 2.7-1. Kernels Containing Exponential and Hyperbolic Functions 1.

x

eµ(x–t) cosh[λ(x – t)]y(t) dt = f (x).

y(x) + A a

Solution:

x

y(x) = f (x) + R(x – t)f (t) dt, a 2 A 1 R(x) = exp (µ – 2 A)x sinh(kx) – A cosh(kx) , 2k

k=

λ2 + 14 A2 .

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2.

x

eµ(x–t) sinh[λ(x – t)]y(t) dt = f (x).

y(x) + A a

1◦ . Solution with λ(A – λ) > 0: Aλ y(x) = f (x) – k

x

eµ(x–t) sin[k(x – t)]f (t) dt,

where k =

λ(A – λ).

a

2◦ . Solution with λ(A – λ) < 0: Aλ y(x) = f (x) – k

x

eµ(x–t) sinh[k(x – t)]f (t) dt,

where

k=

λ(λ – A).

a

3◦ . Solution with A = λ:

x

(x – t)eµ(x–t) f (t) dt.

y(x) = f (x) – λ2 a

3.

x

y(x) +

eµ(x–t) A1 sinh[λ1 (x – t)] + A2 sinh[λ2 (x – t)] y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.3.18:

x

w(x) +

A1 sinh[λ1 (x – t)] + A2 sinh[λ2 (x – t)] w(t) dt = e–µx f (x).

a

4.

x

teµ(x–t) sinh[λ(x – t)]y(t) dt = f (x).

y(x) + A a

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.3.23:

x

t sinh[λ(x – t)]w(t) dt = e–µx f (x).

w(x) + A a

2.7-2. Kernels Containing Exponential and Logarithmic Functions 5.

x

eµt ln(λx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A ln(λx) and h(t) = eµt . 6.

x

eµx ln(λt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = Aeµx and h(t) = ln(λt). 7.

x

eµ(x–t) ln(λx)y(t) dt = f (x).

y(x) – A a

Solution:

x

e(µ–A)(x–t) ln(λx)

y(x) = f (x) + A a

(λx)Ax f (t) dt. (λt)At

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8.

x

eµ(x–t) ln(λt)y(t) dt = f (x).

y(x) – A a

Solution:

a

9.

x

e(µ–A)(x–t) ln(λt)

y(x) = f (x) + A

(λx)Ax f (t) dt. (λt)At

x

eµ(x–t) (ln x – ln t)y(t) dt = f (x).

y(x) + A a

x 1 y(x) = f (x) + eµ(x–t) u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt, W a where the primes stand for the differentiation with respect to the argument specified in the parentheses, and u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equation uxx + Ax–1 u = 0, with u1 (x) and u2 (x) expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of A: √ √ √ √ W = π1 , u1 (x) = x J1 2 Ax , u2 (x) = x Y1 2 Ax for A > 0, √ √ √ √ 1 W = – 2 , u1 (x) = x I1 2 –Ax , u2 (x) = x K1 2 –Ax for A < 0. ∞ y(x) + a eλ(x–t) ln(t – x)y(t) dt = f (x).

Solution:

10.

x

This is a special case of equation 2.9.62 with K(x) = aeλx ln(–x). 2.7-3. Kernels Containing Exponential and Trigonometric Functions

x

eµt cos(λx)y(t) dt = f (x).

11.

y(x) – A

12.

This is a special case of equation 2.9.2 with g(x) = A cos(λx) and h(t) = eµt . x y(x) – A eµx cos(λt)y(t) dt = f (x).

13.

This is a special case of equation 2.9.2 with g(x) = Aeµx and h(t) = cos(λt). x y(x) + A eµ(x–t) cos[λ(x – t)]y(t) dt = f (x).

a

a

a ◦

1 . Solution with |A| > 2|λ|:

x

y(x) = f (x) + R(x – t)f (t) dt, a 2 A 1 R(x) = exp (µ – 2 A)x sinh(kx) – A cosh(kx) , k = 14 A2 – λ2 . 2k ◦ 2 . Solution with |A| < 2|λ|: x y(x) = f (x) + R(x – t)f (t) dt, a A2 R(x) = exp (µ – 12 A)x sin(kx) – A cos(kx) , k = λ2 – 14 A2 . 2k 3◦ . Solution with λ = ± 12 A: x y(x) = f (x) + R(x – t)f (t) dt, a

R(x) =

1

2A

2

x – A exp µ – 12 A x .

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14.

x

y(x) –

eµ(x–t) A cos(kx) + B – AB(x – t) cos(kx) y(t) dt = f (x).

a

Solution:

x

eµ(x–t) M (x, t)f (t) dt,

y(x) = f (x) + a

G(x) B2 M (x, t) = [A cos(kx) + B] + G(t) G(t) 15.

x

y(x) +

x

eB(x–s) G(s) ds,

G(x) = exp

t

A sin(kx) . k

eµ(x–t) A cos(kt) + B + AB(x – t) cos(kt) y(t) dt = f (x).

a

Solution:

x

eµ(x–t) M (x, t)f (t) dt, x G(t) A B2 M (x, t) = –[A cos(kt) + B] eB(t–s) G(s) ds, G(x) = exp + sin(kx) . G(x) G(x) t k y(x) = f (x) +

a

16.

x

eµt sin(λx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A sin(λx) and h(t) = eµt . 17.

x

eµx sin(λt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = Aeµx and h(t) = sin(λt). 18.

x

eµ(x–t) sin[λ(x – t)]y(t) dt = f (x).

y(x) + A a

1◦ . Solution with λ(A + λ) > 0: y(x) = f (x) –

Aλ k

x

eµ(x–t) sin[k(x – t)]f (t) dt,

where k =

λ(A + λ).

a

2◦ . Solution with λ(A + λ) < 0: Aλ y(x) = f (x) – k

x

eµ(x–t) sinh[k(x – t)]f (t) dt,

where

k=

–λ(λ + A).

a

3◦ . Solution with A = –λ: y(x) = f (x) + λ

x

(x – t)eµ(x–t) f (t) dt.

2 a

19.

x

eµ(x–t) sin3 [λ(x – t)]y(t) dt = f (x).

y(x) + A a

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.5.17:

x

sin3 [λ(x – t)]w(t) dt = e–µx f (x).

w(x) + A a

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20.

x

y(x) +

eµ(x–t) A1 sin[λ1 (x – t)] + A2 sin[λ2 (x – t)] y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.5.18: w(x) +

x

A1 sin[λ1 (x – t)] + A2 sin[λ2 (x – t)] w(t) dt = e–µx f (x).

a

21.

x µ(x–t)

y(x) +

e a

n

Ak sin[λk (x – t)] y(t) dt = f (x).

k=1

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.5.19: w(x) +

x n a

22.

Ak sin[λk (x – t)] w(t) dt = e–µx f (x).

k=1

x

teµ(x–t) sin[λ(x – t)]y(t) dt = f (x).

y(x) + A a

Solution: y(x) = f (x) +

Aλ W

x

teµ(x–t) u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt,

a

where u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear ordinary differential equation uxx + λ(Ax + λ)u = 0, and W is the Wronskian. Depending on the sign of Aλ, the functions u1 (x) and u2 (x) are expressed in terms of Bessel functions or modified Bessel functions as follows: if Aλ > 0, then u1 (x) = ξ 1/2 J1/3

2√

√ Aλ ξ 3/2 , u2 (x) = ξ 1/2 Y1/3 23 Aλ ξ 3/2 , W = 3/π, ξ = x + (λ/A);

3

if Aλ < 0, then u1 (x) = ξ 1/2 I1/3

2√ 3

–Aλ ξ 3/2 ,

W = – 32 , 23.

u2 (x) = ξ 1/2 K1/3

2√ 3

–Aλ ξ 3/2 ,

ξ = x + (λ/A).

x

xeµ(x–t) sin[λ(x – t)]y(t) dt = f (x).

y(x) + A a

Solution: Aλ y(x) = f (x) + W

x

xeµ(x–t) u1 (x)u2 (t) – u2 (x)u1 (t) f (t) dt,

a

where u1 (x), u2 (x) is a fundamental system of solutions of the second-order linear ordinary differential equation uxx + λ(Ax + λ)u = 0, and W is the Wronskian. The functions u1 (x), u2 (x), and W are specified in 2.7.22. 24.

∞

y(x) + A

√ eµ(t–x) sin λ t – x y(t) dt = f (x).

x

√ This is a special case of equation 2.9.62 with K(x) = Ae–µx sin λ –x .

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25.

x

y(x) –

eµ(x–t) A sin(kx) + B – AB(x – t) sin(kx) y(t) dt = f (x).

a

Solution:

x

eµ(x–t) M (x, t)f (t) dt,

y(x) = f (x) + a

G(x) B2 M (x, t) = [A sin(kx) + B] + G(t) G(t) 26.

x

y(x) +

x B(x–s)

e

G(s) ds,

t

A G(x) = exp – cos(kx) . k

eµ(x–t) A sin(kt) + B + AB(x – t) sin(kt) y(t) dt = f (x).

a

Solution:

x

eµ(x–t) M (x, t)f (t) dt,

y(x) = f (x) + a

M (x, t) = –[A sin(kt) + B] 27.

G(t) B2 + G(x) G(x)

x

eB(t–s) G(s) ds, t

A G(x) = exp – cos(kx) . k

x

eµt tan(λx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A tan(λx) and h(t) = eµt . 28.

x

eµx tan(λt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = Aeµx and h(t) = tan(λt). 29.

x

y(x) + A

eµ(x–t) tan(λx) – tan(λt) y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.5.32: w(x) + A

x

tan(λx) – tan(λt) w(t) dt = e–µx f (x).

a

30.

x

y(x) –

eµ(x–t) A tan(kx) + B – AB(x – t) tan(kx) y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.9.7 with λ = B and g(x) = A tan(kx):

x

A tan(kx) + B – AB(x – t) tan(kx) w(t) dt = e–µx f (x).

w(x) – a

31.

x

y(x) +

eµ(x–t) A tan(kt) + B + AB(x – t) tan(kt) y(t) dt = f (x).

a

The substitution w(x) = e–µx y(x) leads to an equation of the form 2.9.8 with λ = B and g(t) = A tan(kt): w(x) +

x

A tan(kt) + B + AB(x – t) tan(kt) w(t) dt = e–µx f (x).

a

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32.

x

eµt cot(λx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A cot(λx) and h(t) = eµt . 33.

x

eµx cot(λt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = Aeµx and h(t) = cot(λt).

2.7-4. Kernels Containing Hyperbolic and Logarithmic Functions 34.

x

coshk (λx) lnm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A coshk (λx) and h(t) = lnm (µt). 35.

x

coshk (λt) lnm (µx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = coshk (λt). 36.

x

sinhk (λx) lnm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A sinhk (λx) and h(t) = lnm (µt). 37.

x

sinhk (λt) lnm (µx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = sinhk (λt). 38.

x

tanhk (λx) lnm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A tanhk (λx) and h(t) = lnm (µt). 39.

x

tanhk (λt) lnm (µx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = tanhk (λt). 40.

x

cothk (λx) lnm (µt)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A cothk (λx) and h(t) = lnm (µt). 41.

x

cothk (λt) lnm (µx)y(t) dt = f (x).

y(x) – A a

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = cothk (λt).

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2.7-5. Kernels Containing Hyperbolic and Trigonometric Functions

x

coshk (λx) cosm (µt)y(t) dt = f (x).

42.

y(x) – A

43.

This is a special case of equation 2.9.2 with g(x) = A coshk (λx) and h(t) = cosm (µt). x y(x) – A coshk (λt) cosm (µx)y(t) dt = f (x).

a

a

44.

This is a special case of equation 2.9.2 with g(x) = A cosm (µx) and h(t) = coshk (λt). x y(x) – A coshk (λx) sinm (µt)y(t) dt = f (x). a

45.

This is a special case of equation 2.9.2 with g(x) = A coshk (λx) and h(t) = sinm (µt). x y(x) – A coshk (λt) sinm (µx)y(t) dt = f (x).

46.

This is a special case of equation 2.9.2 with g(x) = A sinm (µx) and h(t) = coshk (λt). x y(x) – A sinhk (λx) cosm (µt)y(t) dt = f (x).

a

a

47.

This is a special case of equation 2.9.2 with g(x) = A sinhk (λx) and h(t) = cosm (µt). x y(x) – A sinhk (λt) cosm (µx)y(t) dt = f (x).

48.

This is a special case of equation 2.9.2 with g(x) = A cosm (µx) and h(t) = sinhk (λt). x y(x) – A sinhk (λx) sinm (µt)y(t) dt = f (x).

a

a

49.

This is a special case of equation 2.9.2 with g(x) = A sinhk (λx) and h(t) = sinm (µt). x y(x) – A sinhk (λt) sinm (µx)y(t) dt = f (x).

50.

This is a special case of equation 2.9.2 with g(x) = A sinm (µx) and h(t) = sinhk (λt). x y(x) – A tanhk (λx) cosm (µt)y(t) dt = f (x).

a

a

51.

This is a special case of equation 2.9.2 with g(x) = A tanhk (λx) and h(t) = cosm (µt). x y(x) – A tanhk (λt) cosm (µx)y(t) dt = f (x). a

52.

This is a special case of equation 2.9.2 with g(x) = A cosm (µx) and h(t) = tanhk (λt). x y(x) – A tanhk (λx) sinm (µt)y(t) dt = f (x).

53.

This is a special case of equation 2.9.2 with g(x) = A tanhk (λx) and h(t) = sinm (µt). x tanhk (λt) sinm (µx)y(t) dt = f (x). y(x) – A

a

a

This is a special case of equation 2.9.2 with g(x) = A sinm (µx) and h(t) = tanhk (λt).

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2.7-6. Kernels Containing Logarithmic and Trigonometric Functions

x

cosk (λx) lnm (µt)y(t) dt = f (x).

54.

y(x) – A

55.

This is a special case of equation 2.9.2 with g(x) = A cosk (λx) and h(t) = lnm (µt). x y(x) – A cosk (λt) lnm (µx)y(t) dt = f (x).

56.

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = cosk (λt). x y(x) – A sink (λx) lnm (µt)y(t) dt = f (x).

a

a

a

57.

This is a special case of equation 2.9.2 with g(x) = A sink (λx) and h(t) = lnm (µt). x y(x) – A sink (λt) lnm (µx)y(t) dt = f (x). a

58.

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = sink (λt). x y(x) – A tank (λx) lnm (µt)y(t) dt = f (x).

59.

This is a special case of equation 2.9.2 with g(x) = A tank (λx) and h(t) = lnm (µt). x y(x) – A tank (λt) lnm (µx)y(t) dt = f (x).

60.

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = tank (λt). x y(x) – A cotk (λx) lnm (µt)y(t) dt = f (x).

a

a

a

61.

This is a special case of equation 2.9.2 with g(x) = A cotk (λx) and h(t) = lnm (µt). x y(x) – A cotk (λt) lnm (µx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A lnm (µx) and h(t) = cotk (λt).

2.8. Equations Whose Kernels Contain Special Functions 2.8-1. Kernels Containing Bessel Functions 1.

y(x) – λ

x

J0 (x – t)y(t) dt = f (x).

0

Solution: y(x) = f (x) +

x

R(x – t)f (t) dt, 0

where √ √ λ2 λ R(x) = λ cos 1 – λ2 x + √ sin 1 – λ2 x + √ 2 1–λ 1 – λ2

x

sin 0

√

J1(t) dt. t

1 – λ2 (x – t)

• Reference: V. I. Smirnov (1974).

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2.

x

y(x) – A

Jν (λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = AJν (λx) and h(t) = 1. 3.

x

y(x) – A

Jν (λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = Jν (λt). 4.

x

y(x) – A

Jν (λx) Jν (λt)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

Jν (λx) f (t) dt. Jν (λt)

eA(x–t)

Jν (λt) f (t) dt. Jν (λx)

a

5.

x

y(x) – A

Jν (λt) Jν (λx)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A a

6.

∞

y(x) + A

Jν [λ(t – x)]y(t) dt = f (x). x

This is a special case of equation 2.9.62 with K(x) = AJν (–λx). 7.

AJν (kx) + B – AB(x – t)Jν (kx) y(t) dt = f (x).

x

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = AJν (kx). 8.

y(x) +

x

AJν (kt) + B + AB(x – t)Jν (kt) y(t) dt = f (x).

a

This is a special case of equation 2.9.8 with λ = B and g(t) = AJν (kt). 9.

y(x) – λ

x

eµ(x–t) J0 (x – t)y(t) dt = f (x).

0

Solution: y(x) = f (x) +

x

R(x – t)f (t) dt, 0

where

µx

R(x) = e

10.

√ √ λ2 λ cos 1 – λ2 x + √ sin 1 – λ2 x + 2 1–λ x √ J1 (t) λ 2 √ dt . sin 1 – λ (x – t) t 1 – λ2 0

x

y(x) – A

Yν (λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = AYν (λx) and h(t) = 1.

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11.

x

y(x) – A

Yν (λt)y(t) dt = f (x). a

14.

This is a special case of equation 2.9.2 with g(x) = A and h(t) = Yν (λt). x Yν (λx) y(x) – A y(t) dt = f (x). a Yν (λt) Solution: x Yν (λx) y(x) = f (x) + A eA(x–t) f (t) dt. Yν (λt) a x Yν (λt) y(x) – A y(t) dt = f (x). Y ν (λx) a Solution: x Yν (λt) y(x) = f (x) + A eA(x–t) f (t) dt. Yν (λx) a ∞ y(x) + A Yν [λ(t – x)]y(t) dt = f (x).

15.

This is a special case of equation 2.9.62 with K(x) = AYν (–λx). x y(x) – AYν (kx) + B – AB(x – t)Yν (kx) y(t) dt = f (x).

16.

This is a special case of equation 2.9.7 with λ = B and g(x) = AYν (kx). x y(x) + AYν (kt) + B + AB(x – t)Yν (kt) y(t) dt = f (x).

12.

13.

x

a

a

This is a special case of equation 2.9.8 with λ = B and g(t) = AYν (kt). 2.8-2. Kernels Containing Modified Bessel Functions 17.

x

y(x) – A

Iν (λx)y(t) dt = f (x). a

18.

This is a special case of equation 2.9.2 with g(x) = AIν (λx) and h(t) = 1. x y(x) – A Iν (λt)y(t) dt = f (x). a

19.

20.

This is a special case of equation 2.9.2 with g(x) = A and h(t) = Iν (λt). x Iν (λx) y(x) – A y(t) dt = f (x). a Iν (λt) Solution: x Iν (λx) y(x) = f (x) + A eA(x–t) f (t) dt. Iν (λt) a x Iν (λt) y(x) – A y(t) dt = f (x). a Iν (λx) Solution: x Iν (λt) y(x) = f (x) + A f (t) dt. eA(x–t) I ν (λx) a

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21.

∞

y(x) + A

Iν [λ(t – x)]y(t) dt = f (x). x

This is a special case of equation 2.9.62 with K(x) = AIν (–λx). 22.

AIν (kx) + B – AB(x – t)Iν (kx) y(t) dt = f (x).

x

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = AIν (kx). 23.

AIν (kt) + B + AB(x – t)Iν (kt) y(t) dt = f (x).

x

y(x) +

a

This is a special case of equation 2.9.8 with λ = B and g(t) = AIν (kt). 24.

x

y(x) – A

Kν (λx)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = AKν (λx) and h(t) = 1. 25.

x

y(x) – A

Kν (λt)y(t) dt = f (x). a

This is a special case of equation 2.9.2 with g(x) = A and h(t) = Kν (λt). 26.

x

y(x) – A

Kν (λx) Kν (λt)

a

y(t) dt = f (x).

Solution:

x

y(x) = f (x) + A

eA(x–t)

Kν (λx) f (t) dt. Kν (λt)

eA(x–t)

Kν (λt) f (t) dt. Kν (λx)

a

27.

x

y(x) – A

Kν (λt) Kν (λx)

a

y(t) dt = f (x).

Solution:

y(x) = f (x) + A

x

a

28.

∞

y(x) + A

Kν [λ(t – x)]y(t) dt = f (x). x

This is a special case of equation 2.9.62 with K(x) = AKν (–λx). 29.

x

AKν (kx) + B – AB(x – t)Kν (kx) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.7 with λ = B and g(x) = AKν (kx). 30.

y(x) +

x

AKν (kt) + B + AB(x – t)Kν (kt) y(t) dt = f (x).

a

This is a special case of equation 2.9.8 with λ = B and g(t) = AKν (kt).

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2.9. Equations Whose Kernels Contain Arbitrary Functions 2.9-1. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t) 1.

y(x) – λ

x

g(x) g(t)

a

y(t) dt = f (x).

Solution:

x

eλ(x–t)

y(x) = f (x) + λ a

2.

g(x) f (t) dt. g(t)

x

y(x) –

g(x)h(t)y(t) dt = f (x). a

Solution:

y(x) = f (x) + 3.

x

R(x, t)f (t) dt,

where

a

R(x, t) = g(x)h(t) exp

x

g(s)h(s) ds .

t

x

y(x) +

(x – t)g(x)y(t) dt = f (x). a

This is a special case of equation 2.9.11. 1◦ . Solution: y(x) = f (x) +

1 W

x

Y1 (x)Y2 (t) – Y2 (x)Y1 (t) g(x)f (t) dt,

(1)

a

where Y1 = Y1 (x) and Y2 = Y2 (x) are two linearly independent solutions (Y1 /Y2 ≡/ const) of the second-order linear homogeneous differential equation Yxx + g(x)Y = 0. In this case, the Wronskian is a constant: W = Y1 (Y2 )x – Y2 (Y1 )x ≡ const.

4.

2◦ . Given only one nontrivial solution Y1 = Y1 (x) of the linear homogeneous differential equation Yxx + g(x)Y = 0, one can obtain the solution of the integral equation by formula (1) with x dξ W = 1, Y2 (x) = Y1 (x) , 2 Y b 1 (ξ) where b is an arbitrary number. x y(x) + (x – t)g(t)y(t) dt = f (x). a

This is a special case of equation 2.9.12. 1◦ . Solution: y(x) = f (x) +

1 W

x

Y1 (x)Y2 (t) – Y2 (x)Y1 (t) g(t)f (t) dt,

(1)

a

where Y1 = Y1 (x) and Y2 = Y2 (x) are two linearly independent solutions (Y1 /Y2 ≡/ const) of the second-order linear homogeneous differential equation Yxx + g(x)Y = 0. In this case, the Wronskian is a constant: W = Y1 (Y2 )x – Y2 (Y1 )x ≡ const. 2◦ . Given only one nontrivial solution Y1 = Y1 (x) of the linear homogeneous differential equation Yxx + g(x)Y = 0, one can obtain the solution of the integral equation by formula (1) with x dξ W = 1, Y2 (x) = Y1 (x) , 2 Y b 1 (ξ) where b is an arbitrary number.

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5.

y(x) +

x

g(x) – g(t) y(t) dt = f (x).

a ◦

1 . Differentiating the equation with respect to x yields yx (x) + gx (x)

x

y(t) dt = fx (x).

(1)

a

x

Introducing the new variable Y (x) = y(t) dt, we obtain the second-order linear ordinary a differential equation Yxx + gx (x)Y = fx (x), (2) which must be supplemented by the initial conditions Y (a) = 0,

Yx (a) = f (a).

(3)

Conditions (3) follow from the original equation and the definition of Y (x). For exact solutions of second-order linear ordinary differential equations (2) with various f (x), see E. Kamke (1977), G. M. Murphy (1960), and A. D. Polyanin and V. F. Zaitsev (1995, 1996). 2◦ . Let Y1 = Y1 (x) and Y2 = Y2 (x) be two linearly independent solutions (Y1 /Y2 ≡/ const) of the second-order linear homogeneous differential equation Yxx + gx (x)Y = 0, which follows from (2) for f (x) ≡ 0. In this case, the Wronskian is a constant: W = Y1 (Y2 )x – Y2 (Y1 )x ≡ const . Solving the nonhomogeneous equation (2) under the initial conditions (3) with arbitrary f = f (x) and taking into account y(x) = Yx (x), we obtain the solution of the original integral equation in the form y(x) = f (x) +

1 W

Y1 (x)Y2 (t) – Y2 (x)Y1 (t) f (t) dt,

x

(4)

a

where the primes stand for the differentiation with respect to the argument specified in the parentheses. 3◦ . Given only one nontrivial solution Y1 = Y1 (x) of the linear homogeneous differential equation Yxx + gx (x)Y = 0, one can obtain the solution of the nonhomogeneous equation (2) under the initial conditions (3) by formula (4) with W = 1,

x

Y2 (x) = Y1 (x) b

dξ , Y12 (ξ)

where b is an arbitrary number. 6.

y(x) +

x

g(x) + h(t) y(t) dt = f (x).

a

1◦ . Differentiating the equation with respect to x yields yx (x)

+ g(x) + h(x) y(x) + gx (x)

x

y(t) dt = fx (x).

a

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x

Introducing the new variable Y (x) = y(t) dt, we obtain the second-order linear ordinary a differential equation Yxx + g(x) + h(x) Yx + gx (x)Y = fx (x), (1) which must be supplemented by the initial conditions Y (a) = 0,

Yx (a) = f (a).

(2)

Conditions (3) follow from the original equation and the definition of Y (x). For exact solutions of second-order linear ordinary differential equations (1) with various f (x), see E. Kamke (1977), G. M. Murphy (1960), and A. D. Polyanin and V. F. Zaitsev (1995, 1996). 2◦ . Let Y1 = Y1 (x) and Y2 = Y2 (x) be two linearly independent solutions (Y1/Y2 ≡/ const) of the second-order linear homogeneous differential equation Yxx + g(x) + h(x) Yx + gx (x)Y = 0, which follows from (1) for f (x) ≡ 0. Solving the nonhomogeneous equation (1) under the initial conditions (2) with arbitrary f = f (x) and taking into account y(x) = Yx (x), we obtain the solution of the original integral equation in the form x y(x) = f (x) + R(x, t)f (t) dt, a ∂ 2 Y1 (x)Y2 (t) – Y2 (x)Y1 (t) R(x, t) = , W (x) = Y1 (x)Y2 (x) – Y2 (x)Y1 (x), ∂x∂t W (t) where W (x) is the Wronskian and the primes stand for the differentiation with respect to the argument specified in the parentheses. 7.

x

g(x) + λ – λ(x – t)g(x) y(t) dt = f (x).

y(x) –

a

This is a special case of equation 2.9.16 with h(x) = λ. Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x x G(x) λ2 R(x, t) = [g(x) + λ] eλ(x–s) G(s) ds, G(x) = exp g(s) ds . + G(t) G(t) t a 8.

x

g(t) + λ + λ(x – t)g(t) y(t) dt = f (x).

y(x) +

a

Solution: x y(x) = f (x) + R(x, t)f (t) dt, a x x G(t) λ2 R(x, t) = –[g(t) + λ] eλ(t–s) G(s) ds, G(x) = exp g(s) ds . + G(x) G(x) t a 9.

y(x) –

g1 (x) + g2 (x)t y(t) dt = f (x).

x

a

This equation can be rewritten in the form of equation 2.9.11 with g1 (x) = g(x) + xh(x) and g2 (x) = –h(x).

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g1 (t) + g2 (t)x y(t) dt = f (x).

x

10.

y(x) –

11.

This equation can be rewritten in the form of equation 2.9.12 with g1 (t) = g(t) + th(t) and g2 (t) = –h(t). x g(x) + h(x)(x – t) y(t) dt = f (x). y(x) –

a

a 1◦ . The solution of the integral equation can be represented in the form y(x) = Yxx , where Y = Y (x) is the solution of the second-order linear nonhomogeneous ordinary differential equation Yxx – g(x)Yx – h(x)Y = f (x), (1)

under the initial conditions

Y (a) = Yx (a) = 0.

(2)

◦

2 . Let Y1 = Y1 (x) and Y2 = Y2 (x) be two nontrivial linearly independent solutions of the second-order linear homogeneous differential equation Yxx –g(x)Yx –h(x)Y = 0, which follows from (1) for f (x) ≡ 0. Then the solution of the nonhomogeneous differential equation (1) under conditions (2) is given by x f (t) Y (x) = Y2 (x)Y1 (t) – Y1 (x)Y2 (t) dt, W (t) = Y1 (t)Y2 (t) – Y2 (t)Y1 (t), (3) W (t) a where W (t) is the Wronskian and the primes denote the derivatives. Substituting (3) into (1), we obtain the solution of the original integral equation in the form x 1 y(x) = f (x) + R(x, t)f (t) dt, R(x, t) = (4) [Y (x)Y1 (t) – Y1 (x)Y2 (t)]. W (t) 2 a 3◦ . Let Y1 = Y1 (x) be a nontrivial particular solution of the homogeneous differential equation (1) (with f ≡ 0) satisfying the initial condition Y1 (a) ≠ 0. Then the function x x W (t) Y2 (x) = Y1 (x) dt, W (x) = exp g(s) ds (5) 2 a [Y1 (t)] a is another nontrivial solution of the homogeneous equation. Substituting (5) into (4) yields the solution of the original integral equation in the form x y(x) = f (x) + R(x, t)f (t) dt, a W (x) Y1 (t) Y1 (t) x W (s) R(x, t) = g(x) ds, + [g(x)Y1 (x) + h(x)Y1 (x)] Y1 (x) W (t) W (t) t [Y1 (s)]2 x where W (x) = exp g(s) ds . 12.

y(x) –

a x

g(t) + h(t)(t – x) y(t) dt = f (x).

a

Solution:

y(x) = f (x) +

x

R(x, t)f (t) dt, a

R(x, t) = g(t)

Y (x)W (x) + Y (x)W (x)[g(t)Yt (t) + h(t)Y (t)] Y (t)W (t) t W (t) = exp g(t) dt ,

t

x

ds , W (s)[Y (s)]2

b

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where Y = Y (x) is an arbitrary nontrivial solution of the second-order homogeneous differential equation Yxx + g(x)Yx + h(x)Y = 0 satisfying the condition Y (a) ≠ 0. 13.

x

y(x) +

(x – t)g(x)h(t)y(t) dt = f (x). a

The substitution y(x) = g(x)u(x) leads to an equation of the form 2.9.4: u(x) +

x

(x – t)g(t)h(t)u(t) dt = f (x)/g(x). a

14.

y(x) –

x

g(x) + λxn + λ(x – t)xn–1 [n – xg(x)] y(t) dt = f (x).

a

This is a special case of equation 2.9.16 with h(x) = λxn . Solution: x y(x) = f (x) + R(x, t)f (t) dt, a

G(x) H(x) R(x, t) = [g(x) + λxn ] + λ(λx2n + nxn–1 ) G(t) G(t) where G(x) = exp a

15.

y(x) –

x

a

x

x

t

G(s) ds, H(s)

λ g(s) ds and H(x) = exp xn+1 . n+1

g(x) + λ + (x – t)[gx (x) – λg(x)] y(t) dt = f (x).

This is a special case of equation 2.9.16. Solution:

x

y(x) = f (x) +

R(x, t)f (t) dt, a

R(x, t) = [g(x) + λ]eλ(x–t) + [g(x)]2 + gx (x) G(x)

x

eλ(s–t) ds, G(s)

t

where G(x) = exp

x g(s) ds .

a

16.

y(x) –

x

a

g(x) + h(x) + (x – t)[hx (x) – g(x)h(x)] y(t) dt = f (x).

Solution: y(x) = f (x) +

x

R(x, t)f (t) dt, a

H(x) G(x) R(x, t) = [g(x) + h(x)] + {[h(x)]2 + hx (x)} G(t) G(t) where G(x) = exp a

x g(s) ds and H(x) = exp

t

x

G(s) ds, H(s)

x h(s) ds .

a

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x

17.

y(x) +

ϕx (x) ϕ(t)

a

+ ϕ(t)gx (x) – ϕx (x)g(t) h(t) y(t) dt = f (x).

1◦ . This equation is equivalent to the equation x a

x ϕ(x) f (x) dx, + ϕ(t)g(x) – ϕ(x)g(t) h(t) y(t) dt = F (x), F (x) = ϕ(t) a

(1)

obtained by differentiating the original equation with respect to x. Equation (1) is a special case of equation 1.9.15 with g1 (x) = g(x),

h1 (t) = ϕ(t)h(t),

g2 (x) = ϕ(x),

h2 (t) =

1 – g(t)h(t). ϕ(t)

2◦ . Solution: x 1 F (t) ϕ2 (t)h(t) d Ξ(x) dt , ϕ(x)h(x) dx ϕ(t) t Ξ(t) a x x g(t) 2 F (x) = f (x) dx, Ξ(x) = exp – ϕ (t)h(t) dt . ϕ(t) t a a y(x) =

18.

x

y(x) – a

ϕt (t) ϕ(x)

+ ϕ(x)gt (t) – ϕt (t)g(x) h(x) y(t) dt = f (x).

1◦ . Let f (a) = 0. The change

y(x) =

x

w(t) dt

(1)

a

followed by the integration by parts leads to the equation x a

ϕ(t) + ϕ(x)g(t) – ϕ(t)g(x) h(x) w(t) dt = f (x), ϕ(x)

(2)

which is a special case of equation 1.9.15 with g1 (x) =

1 – g(x)h(x), ϕ(x)

h1 (t) = ϕ(t),

g2 (x) = ϕ(x)h(x),

h2 (t) = g(t).

The solution of equation (2) is given by x 1 d dt f (t) 2 y(x) = ϕ (x)h(x)Φ(x) , ϕ(x) dx ϕ(t)h(t) t Φ(t) a x g(t) Φ(x) = exp ϕ2 (t)h(t) dt . ϕ(t) t a 2◦ . Let f (a) ≠ 0. The substitution y(x) = y(x) ¯ + f (a) leads to the integral equation y(x) ¯ with the right-hand side f¯(x) satisfying the condition f¯(a) = 0. Thus we obtain case 1◦ .

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19.

y(x) –

n x

a

gk (x)(x – t)

k–1

y(t) dt = f (x).

k=1

The solution can be represented in the form

x

y(x) = f (x) +

R(x, t)f (t) dt.

(1)

dn w , dxn

(2)

a

Here the resolvent R(x, t) is given by R(x, t) = wx(n) ,

wx(n) =

where w is the solution of the nth-order linear homogeneous ordinary differential equation wx(n) – g1 (x)wx(n–1) – g2 (x)wx(n–2) – 2g3 (x)wx(n–3) – · · · – (n – 1)! gn (x)w = 0

(3)

satisfying the following initial conditions at x = t: w

x=t

= wx

x=t

= · · · = wx(n–2)

x=t

= 0,

wx(n–1)

x=t

= 1.

(4)

Note that the differential equation (3) implicitly depends on t via the initial conditions (4). • References: E. Goursat (1923), A. F. Verlan’ and V. S. Sizikov (1987). 20.

y(x) –

n x

a

gk (t)(t – x)k–1 y(t) dt = f (x).

k=1

The solution can be represented in the form

x

y(x) = f (x) +

R(x, t)f (t) dt.

(1)

dn u , dtn

(2)

a

Here the resolvent R(x, t) is given by R(x, t) = –u(n) t ,

u(n) t =

where u is the solution of the nth-order linear homogeneous ordinary differential equation (n–1) u(n) + g2 (t)u(n–2) + 2g3 (t)u(n–3) + . . . + (n – 1)! gn (t)u = 0, t + g1 (t)ut t t

(3)

satisfying the following initial conditions at t = x: u

t=x

= ut

t=x

= · · · = u(n–2) t

t=x

= 0,

u(n–1) t

t=x

= 1.

(4)

Note that the differential equation (3) implicitly depends on x via the initial conditions (4).

21.

• References: E. Goursat (1923), A. F. Verlan’ and V. S. Sizikov (1987). x λx+µt y(x) + – eµx+λt g(t)y(t) dt = f (x). e a

Let us differentiate the equation twice and then eliminate the integral terms from the resulting relations and the original equation. As a result, we arrive at the second-order linear ordinary differential equation yxx – (λ + µ)yx + (λ – µ)e(λ+µ)x g(x) + λµ y = fxx (x) – (λ + µ)fx (x) + λµf (x), which must be supplemented by the initial conditions y(a) = f (a), yx (a) = fx (a).

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22.

eλx g(t) + eµx h(t) y(t) dt = f (x).

x

y(x) +

a

Let us differentiate the equation twice and then eliminate the integral terms from the resulting relations and the original equation. As a result, we arrive at the second-order linear ordinary differential equation yxx + eλx g(x) + eµx h(x) – λ – µ yx + eλx gx (x) + eµx hx (x) + (λ – µ)eλx g(x) + (µ – λ)eµx h(x) + λµ y = fxx (x) – (λ + µ)fx (x) + λµf (x), which must be supplemented by the initial conditions yx (a) = fx (a) – eλa g(a) + eµa h(a) f (a).

y(a) = f (a), Example. The Arutyunyan equation

x

y(x) –

ϕ(t) a

∂ ∂t

1 + ψ(t) 1 – e–λ(x–t) y(t) dt = f (x), ϕ(t)

can be reduced to the above equation. The former is encountered in the theory of viscoelasticity for aging solids. The solution of the Arutyunyan equation is given by

x

y(x) = f (x) – a

x 1 ∂ ϕ(t) – λψ(t)ϕ2 (t)eη(t) e–η(s) ds f (t) dt, ϕ(t) ∂t t

where

x η(x) = a

ϕ (t) λ 1 + ψ(t)ϕ(t) – ϕ(t)

dt.

• Reference: N. Kh. Arutyunyan (1966). 23.

λeλ(x–t) + µeµx+λt – λeλx+µt h(t) y(t) dt = f (x).

x

y(x) +

a

This is a special case of equation 2.9.17 with ϕ(x) = eλx and g(x) = eµx . Solution: x 1 F (t) e2λt h(t) d Φ(x) dt , eλx h(x) dx eλt t Φ(t) a x x F (x) = f (t) dt, Φ(x) = exp (λ – µ) e(λ+µ)t h(t) dt . y(x) =

a

24.

y(x) –

a

x

λe–λ(x–t) + µeλx+µt – λeµx+λt h(x) y(t) dt = f (x).

a

This is a special case of equation 2.9.18 with ϕ(x) = eλx and g(x) = eµx . Assume that f (a) = 0. Solution: y(x) =

x

d e2λx h(x) x f (t) w(x) = e–λx Φ(t) dt , dx Φ(x) eλt h(t) t a x (λ+µ)t Φ(x) = exp (λ – µ) e h(t) dt .

w(t) dt, a

a

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25.

x

y(x) –

g(x) + beλx + b(x – t)eλx [λ – g(x)] y(t) dt = f (x).

a

This is a special case of equation 2.9.16 with h(x) = beλx . Solution: x y(x) = f (x) + R(x, t)f (t) dt, a

G(x) H(x) R(x, t) = [g(x) + be ] + (b2 e2λx + bλeλx ) G(t) G(t)

x

λx

where G(x) = exp

t

G(s) ds, H(s)

x b λx . g(s) ds and H(x) = exp e λ a

26.

x

y(x) +

a

λeλ(x–t) + eλt gx (x) – λeλx g(t) h(t) y(t) dt = f (x).

This is a special case of equation 2.9.17 with ϕ(x) = eλx . 27.

x

y(x) –

a

λe–λ(x–t) + eλx gt (t) – λeλt g(x) h(x) y(t) dt = f (x).

This is a special case of equation 2.9.18 with ϕ(x) = eλx . 28.

x

y(x) +

cosh[λ(x – t)]g(t)y(t) dt = f (x). a

Differentiating the equation with respect to x twice yields yx (x) + g(x)y(x) + λ

x

sinh[λ(x – t)]g(t)y(t) dt = fx (x), a x yxx (x) + g(x)y(x) x + λ2 cosh[λ(x – t)]g(t)y(t) dt = fxx (x).

(1) (2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order linear ordinary differential equation yxx + g(x)y x – λ2 y = fxx (x) – λ2 f (x).

(3)

By setting x = a in the original equation and (1), we obtain the initial conditions for y = y(x): y(a) = f (a),

yx (a) = fx (a) – f (a)g(a).

(4)

Equation (3) under conditions (4) determines the solution of the original integral equation. 29.

x

y(x) +

cosh[λ(x – t)]g(x)h(t)y(t) dt = f (x). a

The substitution y(x) = g(x)u(x) leads to an equation of the form 2.9.28: u(x) +

x

cosh[λ(x – t)]g(t)h(t)u(t) dt = f (x)/g(x). a

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30.

x

y(x) +

sinh[λ(x – t)]g(t)y(t) dt = f (x). a

1◦ . Differentiating the equation with respect to x twice yields yx (x)

x

cosh[λ(x – t)]g(t)y(t) dt = fx (x), x 2 yxx (x) + λg(x)y(x) + λ sinh[λ(x – t)]g(t)y(t) dt = fxx (x). +λ

(1)

a

(2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order linear ordinary differential equation yxx + λ g(x) – λ y = fxx (x) – λ2 f (x).

(3)

By setting x = a in the original equation and (1), we obtain the initial conditions for y = y(x): y(a) = f (a),

yx (a) = fx (a).

(4)

For exact solutions of second-order linear ordinary differential equations (3) with various g(x), see E. Kamke (1977), G. M. Murphy (1960), and A. D. Polyanin and V. F. Zaitsev (1995, 1996). 2◦ . Let y1 = y1 (x) and y2 = y2 (x) be two linearly solutions (y1 /y2 ≡/ const) of independent the homogeneous differential equation yxx + λ g(x) – λ y = 0, which follows from (3) for f (x) ≡ 0. In this case, the Wronskian is a constant: W = y1 (y2 )x – y2 (y1 )x ≡ const . The solution of the nonhomogeneous equation (3) under conditions (4) with arbitrary f = f (x) has the form x λ y(x) = f (x) + (5) y1 (x)y2 (t) – y2 (x)y1 (t) g(t)f (t) dt W a and determines the solution of the original integral equation. 3◦ . Given only one nontrivial solution y1 = y1 (x) of the linear homogeneous differential equation yxx +λ g(x)–λ y = 0, one can obtain the solution of the nonhomogeneous equation (3) under the initial conditions (4) by formula (5) with W = 1,

y2 (x) = y1 (x) b

x

dξ , y12 (ξ)

where b is an arbitrary number. 31.

x

y(x) +

sinh[λ(x – t)]g(x)h(t)y(t) dt = f (x). a

The substitution y(x) = g(x)u(x) leads to an equation of the form 2.9.30: u(x) +

x

sinh[λ(x – t)]g(t)h(t)u(t) dt = f (x)/g(x). a

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32.

x

y(x) –

g(x) + b cosh(λx) + b(x – t)[λ sinh(λx) – cosh(λx)g(x)] y(t) dt = f (x).

a

This is a special case of equation 2.9.16 with h(x) = b cosh(λx). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a

H(x) G(x) 2 R(x, t) = [g(x) + b cosh(λx)] + b cosh2 (λx) + bλ sinh(λx) G(t) G(t) x b where G(x) = exp g(s) ds and H(x) = exp sinh(λx) . λ a 33.

x

y(x) –

x

t

G(s) ds, H(s)

g(x) + b sinh(λx) + b(x – t)[λ cosh(λx) – sinh(λx)g(x)] y(t) dt = f (x).

a

This is a special case of equation 2.9.16 with h(x) = b sinh(λx). Solution: x y(x) = f (x) + R(x, t)f (t) dt, a

H(x) G(x) 2 R(x, t) = [g(x) + b sinh(λx)] + b sinh2 (λx) + bλ cosh(λx) G(t) G(t) x b where G(x) = exp g(s) ds and H(x) = exp cosh(λx) . λ a 34.

x

t

G(s) ds, H(s)

x

y(x) +

cos[λ(x – t)]g(t)y(t) dt = f (x). a

Differentiating the equation with respect to x twice yields x yx (x) + g(x)y(x) – λ sin[λ(x – t)]g(t)y(t) dt = fx (x), a x yxx (x) + g(x)y(x) x – λ2 cos[λ(x – t)]g(t)y(t) dt = fxx (x).

(1) (2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order linear ordinary differential equation yxx + g(x)y x + λ2 y = fxx (x) + λ2 f (x).

(3)

By setting x = a in the original equation and (1), we obtain the initial conditions for y = y(x): y(a) = f (a), 35.

yx (a) = fx (a) – f (a)g(a).

(4)

x

y(x) +

cos[λ(x – t)]g(x)h(t)y(t) dt = f (x). a

The substitution y(x) = g(x)u(x) leads to an equation of the form 2.9.34: x u(x) + cos[λ(x – t)]g(t)h(t)u(t) dt = f (x)/g(x). a

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36.

x

y(x) +

sin[λ(x – t)]g(t)y(t) dt = f (x). a

1◦ . Differentiating the equation with respect to x twice yields yx (x) + λ

x

cos[λ(x – t)]g(t)y(t) dt = fx (x), x 2 yxx (x) + λg(x)y(x) – λ sin[λ(x – t)]g(t)y(t) dt = fxx (x).

(1)

a

(2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order linear ordinary differential equation yxx + λ g(x) + λ y = fxx (x) + λ2 f (x).

(3)

By setting x = a in the original equation and (1), we obtain the initial conditions for y = y(x): y(a) = f (a),

yx (a) = fx (a).

(4)

For exact solutions of second-order linear ordinary differential equations (3) with various f (x), see E. Kamke (1977) and A. D. Polyanin and V. F. Zaitsev (1995, 1996). 2◦ . Let y1 = y1 (x) and y2 = y2 (x) be two linearly solutions (y1 /y2 ≡/ const) of independent the homogeneous differential equation yxx + λ g(x) – λ y = 0, which follows from (3) for f (x) ≡ 0. In this case, the Wronskian is a constant: W = y1 (y2 )x – y2 (y1 )x ≡ const . The solution of the nonhomogeneous equation (3) under conditions (4) with arbitrary f = f (x) has the form x λ y(x) = f (x) + (5) y1 (x)y2 (t) – y2 (x)y1 (t) g(t)f (t) dt W a and determines the solution of the original integral equation. 3◦ . Given only solution y1 = y1 (x) of the linear homogeneous differential equa one nontrivial tion yxx + λ g(x) + λ y = 0, one can obtain the solution of the nonhomogeneous equation (3) under the initial conditions (4) by formula (5) with W = 1,

y2 (x) = y1 (x) b

x

dξ y12 (ξ)

,

where b is an arbitrary number. 37.

x

y(x) +

sin[λ(x – t)]g(x)h(t)y(t) dt = f (x). a

The substitution y(x) = g(x)u(x) leads to an equation of the form 2.9.36: u(x) +

x

sin[λ(x – t)]g(t)h(t)u(t) dt = f (x)/g(x). a

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38.

x

y(x) –

g(x) + b cos(λx) – b(x – t)[λ sin(λx) + cos(λx)g(x)] y(t) dt = f (x).

a

This is a special case of equation 2.9.16 with h(x) = b cos(λx). Solution:

x

y(x) = f (x) +

R(x, t)f (t) dt, a

R(x, t) = [g(x) + b cos(λx)] where G(x) = exp a

39.

x

y(x) –

x

H(x) G(x) 2 + b cos2 (λx) – bλ sin(λx) G(t) G(t)

x

t

G(s) ds, H(s)

b sin(λx) . g(s) ds and H(x) = exp λ

g(x) + b sin(λx) + b(x – t)[λ cos(λx) – sin(λx)g(x)] y(t) dt = f (x).

a

This is a special case of equation 2.9.16 with h(x) = b sin(λx). Solution:

x

y(x) = f (x) +

R(x, t)f (t) dt, a

R(x, t) = [g(x) + b sin(λx)] where G(x) = exp

H(x) G(x) 2 2 + b sin (λx) + bλ cos(λx) G(t) G(t)

t

x

G(s) ds, H(s)

x b g(s) ds and H(x) = exp – cos(λx) . λ a

2.9-2. Equations With Difference Kernel: K(x, t) = K(x – t) 40.

x

y(x) +

K(x – t)y(t) dt = f (x). a

Renewal equation. 1◦ . To solve this integral equation, direct and inverse Laplace transforms are used. The solution can be represented in the form

x

y(x) = f (x) –

R(x – t)f (t) dt.

(1)

a

Here the resolvent R(x) is expressed via the kernel K(x) of the original equation as follows: R(x) = ˜ R(p) =

1 2πi

˜ K(p) , ˜ 1 + K(p)

c+i∞

px ˜ R(p)e dp, c–i∞ ∞ ˜ K(p) = K(x)e–px dx. 0

• References: R. Bellman and K. L. Cooke (1963), M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971), V. I. Smirnov (1974).

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2◦ . Let w = w(x) be the solution of the simpler auxiliary equation with a = 0 and f ≡ 1:

x

w(x) +

K(x – t)w(t) dt = 1.

(2)

0

Then the solution of the original integral equation with arbitrary f = f (x) is expressed via the solution of the auxiliary equation (2) as y(x) =

d dx

x

w(x – t)f (t) dt = f (a)w(x – a) + a

x

w(x – t)ft (t) dt.

a

• Reference: R. Bellman and K. L. Cooke (1963). 41.

x

y(x) +

K(x – t)y(t) dt = 0. –∞

Eigenfunctions of this integral equation are determined by the roots of the following transcendental (algebraic) equation for the parameter λ:

∞

K(z)e–λz dz = –1.

(1)

0

The left-hand side of this equation is the Laplace transform of the kernel of the integral equation. 1◦ . For a real simple root λk of equation (1) there is a corresponding eigenfunction yk (x) = exp(λk x). 2◦ . For a real root λk of multiplicity r there are corresponding r eigenfunctions yk1 (x) = exp(λk x),

yk2 (x) = x exp(λk x),

...,

ykr (x) = xr–1 exp(λk x).

3◦ . For a complex simple root λk = αk + iβk of equation (1) there is a corresponding eigenfunction pair yk(1) (x) = exp(αk x) cos(βk x),

yk(2) (x) = exp(αk x) sin(βk x).

4◦ . For a complex root λk = αk +iβk of multiplicity r there are corresponding r eigenfunction pairs (1) yk1 (x) = exp(αk x) cos(βk x),

(2) (x) = exp(αk x) sin(βk x), yk1

(1) (x) = x exp(αk x) cos(βk x), yk2 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

(2) (x) = x exp(αk x) sin(βk x), yk2 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

(1) ykr (x) = xr–1 exp(αk x) cos(βk x),

(2) ykr (x) = xr–1 exp(αk x) sin(βk x).

The general solution is the combination (with arbitrary constants) of the eigenfunctions of the homogeneous integral equation.

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For equations 2.9.42–2.9.51, only particular solutions are given. To obtain the general solution, one must add the general solution of the corresponding homogeneous equation 2.9.41 to the particular solution. x 42. y(x) + K(x – t)y(t) dt = Axn , n = 0, 1, 2, . . . –∞

This is a special case of equation 2.9.44 with λ = 0. 1◦ . A solution with n = 0: y(x) = 2◦ . A solution with n = 1: A AC y(x) = x + 2 , B B

A , B

∞

B =1+

K(z) dz. 0

∞

B =1+

K(z) dz,

∞

C=

zK(z) dz.

0

0

3◦ . A solution with n = 2: A 2 AC AC 2 AD x +2 2 x+2 3 – 2 , B B B ∞ B ∞ ∞ B =1+ K(z) dz, C = zK(z) dz, D = z 2 K(z) dz. y2 (x) =

0

43.

0

4◦ . A solution with n = 3, 4, . . . is given by: n λx e ∂ yn (x) = A , ∂λn B(λ) λ=0 x y(x) + K(x – t)y(t) dt = Aeλx .

0

∞

B(λ) = 1 +

K(z)e–λz dz.

0

–∞

∞ A λx B =1+ K(z)e–λz dz. e , B 0 The integral term in the expression for B is the Laplace transform of K(z), which may be calculated using tables of Laplace transforms (e.g., see Supplement 4). x y(x) + K(x – t)y(t) dt = Axn eλx , n = 1, 2, . . .

A solution:

y(x) =

44.

–∞

1◦ . A solution with n = 1: A λx AC λx xe + 2 e , B B ∞ ∞ B =1+ K(z)e–λz dz, C = zK(z)e–λz dz. y1 (x) =

0

0

It is convenient to calculate B and C using tables of Laplace transforms. 2◦ . A solution with n = 2:

A 2 λx AC AC 2 AD x e + 2 2 xeλx + 2 3 – 2 eλx , B B B B ∞ ∞ ∞ B =1+ K(z)e–λz dz, C = zK(z)e–λz dz, D = z 2 K(z)e–λz dz. y2 (x) =

0

0

3◦ . A solution with n = 3, 4, . . . is given by: λx e ∂ ∂n yn (x) = yn–1 (x) = A n , ∂λ ∂λ B(λ)

0

B(λ) = 1 +

∞

K(z)e–λz dz.

0

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45.

x

y(x) +

K(x – t)y(t) dt = A cosh(λx). –∞

A solution: y(x) =

A λx A –λx 1 A A A 1 A e + e = + – cosh(λx) + sinh(λx), 2B– 2B+ 2 B– B+ 2 B– B+ ∞ ∞ B– = 1 + K(z)e–λz dz, B+ = 1 + K(z)eλz dz. 0

46.

0

x

y(x) +

K(x – t)y(t) dt = A sinh(λx). –∞

A solution: y(x) =

A λx 1 A A –λx 1 A A A cosh(λx) + sinh(λx), e – e = – + 2B– 2B+ 2 B– B+ 2 B– B+ ∞ ∞ B– = 1 + K(z)e–λz dz, B+ = 1 + K(z)eλz dz. 0

47.

0

x

y(x) +

K(x – t)y(t) dt = A cos(λx). –∞

A solution: A y(x) = 2 Bc cos(λx) – Bs sin(λx) , 2 Bc + Bs ∞ ∞ Bc = 1 + K(z) cos(λz) dz, Bs = K(z) sin(λz) dz. 0

48.

0

x

y(x) +

K(x – t)y(t) dt = A sin(λx). –∞

A solution: A y(x) = 2 Bc sin(λx) + Bs cos(λx) , 2 Bc + Bs ∞ ∞ Bc = 1 + K(z) cos(λz) dz, Bs = K(z) sin(λz) dz. 0

49.

0

x

K(x – t)y(t) dt = Aeµx cos(λx).

y(x) + –∞

A solution: A eµx Bc cos(λx) – Bs sin(λx) , 2 + Bs ∞ ∞ –µz Bc = 1 + K(z)e cos(λz) dz, Bs = K(z)e–µz sin(λz) dz. y(x) =

Bc2

0

50.

0

x

K(x – t)y(t) dt = Aeµx sin(λx).

y(x) + –∞

A solution: A eµx Bc sin(λx) + Bs cos(λx) , Bc2 + Bs2 ∞ ∞ –µz Bc = 1 + K(z)e cos(λz) dz, Bs = K(z)e–µz sin(λz) dz. y(x) =

0

0

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51.

x

y(x) +

K(x – t)y(t) dt = f (x). –∞

1◦ . For a polynomial right-hand side, f (x) =

n

Ak xk , a solution has the form

k=0

y(x) =

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. One can also make use of the formula given in item 4◦ of equation 2.9.42 to construct the solution. 2◦ . For f (x) = eλx

n

Ak xk , a solution of the equation has the form

k=0

λx

y(x) = e

n

Bk xk ,

k=0

where the Bk are found by the method of undetermined coefficients. One can also make use of the formula given in item 3◦ of equation 2.9.44 to construct the solution. 3◦ . For f (x) =

n

Ak exp(λk x), a solution of the equation has the form

k=0 n Ak y(x) = exp(λk x), Bk k=0

4◦ . For f (x) = cos(λx)

n

∞

Bk = 1 +

K(z) exp(–λk z) dz. 0

Ak xk , a solution of the equation has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. 5◦ . For f (x) = sin(λx)

n

Ak xk , a solution of the equation has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. 6◦ . For f (x) =

n

Ak cos(λk x), the solution of a equation has the form

k=0

y(x) = Bck = 1 + 0

n

Ak Bck cos(λk x) – Bsk sin(λk x) , 2 + Bsk ∞ K(z) cos(λk z) dz, Bsk = K(z) sin(λk z) dz.

2 Bck k=0 ∞

0

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7◦ . For f (x) =

n

Ak sin(λk x), a solution of the equation has the form

k=0

y(x) =

n

Ak Bck sin(λk x) + Bsk cos(λk x) , 2 + Bsk ∞ K(z) cos(λk z) dz, Bsk = K(z) sin(λk z) dz.

2 Bck k=0 ∞

Bck = 1 + 0

8◦ . For f (x) = cos(λx)

n

0

Ak exp(µk x), a solution of the equation has the form

k=0 n n Ak Bck Ak Bsk exp(µ x) – sin(λx) exp(µk x), k 2 + B2 2 2 B B ck sk ck + Bsk k=0 k=0 ∞ ∞ =1+ K(z) exp(–µk z) cos(λz) dz, Bsk = K(z) exp(–µk z) sin(λz) dz.

y(x) = cos(λx) Bck

0

9◦ . For f (x) = sin(λx)

0 n

Ak exp(µk x), a solution of the equation has the form

k=0

Bck

n n Ak Bck Ak Bsk y(x) = sin(λx) exp(µk x) + cos(λx) exp(µk x), 2 + B2 2 2 B B ck sk ck + Bsk k=0 k=0 ∞ ∞ =1+ K(z) exp(–µk z) cos(λz) dz, Bsk = K(z) exp(–µk z) sin(λz) dz.

52.

0

0

∞

y(x) +

K(x – t)y(t) dt = 0. x

Eigenfunctions of this integral equation are determined by the roots of the following transcendental (algebraic) equation for the parameter λ: ∞ K(–z)eλz dz = –1. (1) 0

The left-hand side of this equation is the Laplace transform of the function K(–z) with parameter –λ. 1◦ . For a real simple root λk of equation (1) there is a corresponding eigenfunction yk (x) = exp(λk x). ◦

2 . For a real root λk of multiplicity r there are corresponding r eigenfunctions yk1 (x) = exp(λk x),

yk2 (x) = x exp(λk x),

...,

ykr (x) = xr–1 exp(λk x).

3◦ . For a complex simple root λk = αk + iβk of equation (1) there is a corresponding eigenfunction pair yk(1) (x) = exp(αk x) cos(βk x),

yk(2) (x) = exp(αk x) sin(βk x).

4◦ . For a complex root λk = αk +iβk of multiplicity r there are corresponding r eigenfunction pairs (1) (2) yk1 (x) = exp(αk x) cos(βk x), (x) = exp(αk x) sin(βk x), yk1 (1) yk2 (x) = x exp(αk x) cos(βk x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

(2) yk2 (x) = x exp(αk x) sin(βk x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

(1) (2) ykr (x) = xr–1 exp(αk x) cos(βk x), ykr (x) = xr–1 exp(αk x) sin(βk x). The general solution is the combination (with arbitrary constants) of the eigenfunctions of the homogeneous integral equation.

© 1998 by CRC Press LLC

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For equations 2.9.53–2.9.62, only particular solutions are given. To obtain the general solution, one must add the general solution of the corresponding homogeneous equation 2.9.52 to the particular solution. ∞ 53. y(x) + K(x – t)y(t) dt = Axn , n = 0, 1, 2, . . . x

This is a special case of equation 2.9.55 with λ = 0. 1◦ . A solution with n = 0: y(x) = 2◦ . A solution with n = 1: A AC y(x) = x – 2 , B B ◦ 3 . A solution with n = 2:

A , B

∞

B =1+

K(–z) dz. 0

∞

B =1+

K(–z) dz,

∞

C=

zK(–z) dz.

0

0

A 2 AC AC 2 AD x –2 2 x+2 3 – 2 , B B B B ∞ ∞ B =1+ K(–z) dz, C = zK(–z) dz, D = y2 (x) =

0

54.

0

4◦ . A solution with n = 3, 4, . . . is given by: n λx e ∂ yn (x) = A , ∂λn B(λ) λ=0 ∞ y(x) + K(x – t)y(t) dt = Aeλx .

∞

z 2 K(–z) dz.

0

∞

B(λ) = 1 +

K(–z)eλz dz.

0

x

A solution:

∞ A λx e , B =1+ K(–z)eλz dz = 1 + L{K(–z), –λ}. B 0 The integral term in the expression for B is the Laplace transform of K(–z) with parameter –λ, which may be calculated using tables of Laplace transforms (e.g., see H. Bateman and A. Erd´elyi (vol. 1, 1954) and V. A. Ditkin and A. P. Prudnikov (1965)). ∞ y(x) + K(x – t)y(t) dt = Axn eλx , n = 1, 2, . . . y(x) =

55.

x

1◦ . A solution with n = 1: A λx AC λx xe – 2 e , B B ∞ ∞ B =1+ K(–z)eλz dz, C = zK(–z)eλz dz. y1 (x) =

0

0

It is convenient to calculate B and C using tables of Laplace transforms (with parameter –λ). 2◦ . A solution with n = 2:

A 2 λx AC λx AC 2 AD λx y2 (x) = x e – 2 2 xe + 2 3 – 2 e , B B B B ∞ ∞ ∞ λz λz K(–z)e dz, C = zK(–z)e dz, D = z 2 K(–z)eλz dz. B =1+ 0

0

3◦ . A solution with n = 3, 4, . . . is given by λx e ∂ ∂n yn (x) = yn–1 (x) = A n , ∂λ ∂λ B(λ)

0

B(λ) = 1 +

∞

K(–z)eλz dz.

0

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56.

∞

y(x) +

K(x – t)y(t) dt = A cosh(λx). x

A solution: y(x) =

A λx A –λx 1 A A A 1 A e + e = + – cosh(λx) + sinh(λx), 2B+ 2B– 2 B+ B– 2 B+ B– ∞ ∞ B+ = 1 + K(–z)eλz dz, B– = 1 + K(–z)e–λz dz. 0

57.

0

∞

y(x) +

K(x – t)y(t) dt = A sinh(λx). x

A solution: y(x) =

A λx 1 A A –λx 1 A A A cosh(λx) + sinh(λx), e – e = – + 2B+ 2B– 2 B+ B– 2 B+ B– ∞ ∞ B+ = 1 + K(–z)eλz dz, B– = 1 + K(–z)e–λz dz. 0

58.

0

∞

y(x) +

K(x – t)y(t) dt = A cos(λx). x

A solution: A y(x) = 2 Bc cos(λx) + Bs sin(λx) , 2 Bc + Bs ∞ ∞ Bc = 1 + K(–z) cos(λz) dz, Bs = K(–z) sin(λz) dz. 0

59.

0

∞

y(x) +

K(x – t)y(t) dt = A sin(λx). x

A solution: A y(x) = 2 Bc sin(λx) – Bs cos(λx) , 2 Bc + Bs ∞ ∞ Bc = 1 + K(–z) cos(λz) dz, Bs = K(–z) sin(λz) dz. 0

60.

∞

y(x) +

0

K(x – t)y(t) dt = Aeµx cos(λx).

x

A solution: A eµx Bc cos(λx) + Bs sin(λx) , 2 + Bs ∞ ∞ µz Bc = 1 + K(–z)e cos(λz) dz, Bs = K(–z)eµz sin(λz) dz. y(x) =

Bc2

0

61.

∞

y(x) +

0

K(x – t)y(t) dt = Aeµx sin(λx).

x

A solution: A eµx Bc sin(λx) – Bs cos(λx) , Bc2 + Bs2 ∞ ∞ µz Bc = 1 + K(–z)e cos(λz) dz, Bs = K(–z)eµz sin(λz) dz. y(x) =

0

0

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62.

∞

y(x) +

K(x – t)y(t) dt = f (x). x

1◦ . For a polynomial right-hand side, f (x) =

n

Ak xk , a solution has the form

k=0

y(x) =

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. One can also make use of the formula given in item 4◦ of equation 2.9.53 to construct the solution. 2◦ . For f (x) = eλx

n

Ak xk , a solution of the equation has the form

k=0

λx

y(x) = e

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. One can also make use of the formula given in item 3◦ of equation 2.9.55 to construct the solution. 3◦ . For f (x) =

n

Ak exp(λk x), a solution of the equation has the form

k=0 n Ak y(x) = exp(λk x), Bk k=0

n

4◦ . For f (x) = cos(λx)

∞

Bk = 1 +

K(–z) exp(λk z) dz. 0

Ak xk a solution of the equation has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. 5◦ . For f (x) = sin(λx)

n

Ak xk , a solution of the equation has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the Bk and Ck are found by the method of undetermined coefficients. 6◦ . For f (x) =

n

Ak cos(λk x), a solution of the equation has the form

k=0

y(x) = Bck = 1 + 0

n

Ak Bck cos(λk x) + Bsk sin(λk x) , 2 + Bsk ∞ K(–z) cos(λk z) dz, Bsk = K(–z) sin(λk z) dz.

2 Bck k=0 ∞

0

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7◦ . For f (x) =

n

Ak sin(λk x), a solution of the equation has the form

k=0 n

Ak Bck sin(λk x) – Bsk cos(λk x) , 2 + Bsk ∞ K(–z) cos(λk z) dz, Bsk = K(–z) sin(λk z) dz.

y(x) =

2 Bck k=0 ∞

Bck = 1 + 0

0

n

8◦ . For f (x) = cos(λx)

Ak exp(µk x), a solution of the equation has the form

k=0 n n Ak Bck Ak Bsk exp(µ x) + sin(λx) exp(µk x), k 2 + B2 2 2 B B ck sk ck + Bsk k=0 k=0 ∞ ∞ =1+ K(–z) exp(µk z) cos(λz) dz, Bsk = K(–z) exp(µk z) sin(λz) dz.

y(x) = cos(λx) Bck

0

0

9◦ . For f (x) = sin(λx)

n

Ak exp(µk x), a solution of the equation has the form

k=0 n n Ak Bck Ak Bsk exp(µ x) – cos(λx) exp(µk x), k 2 + B2 2 2 B B ck sk ck + Bsk k=0 k=0 ∞ ∞ =1+ K(–z) exp(µk z) cos(λz) dz, Bsk = K(–z) exp(µk z) sin(λz) dz.

y(x) = sin(λx) Bck

0

0

10◦ . In the general case of arbitrary right-hand side f = f (x), the solution of the integral equation can be represented in the form y(x) = f˜(p) =

∞

1 2πi –px

f (x)e

c+i∞

c–i∞

dx,

f˜(p) epx dp, ˜ 1 + k(–p) ∞ ˜ k(–p) = K(–z)epz dz.

0

0

˜ To calculate f˜(p) and k(–p), it is convenient to use tables of Laplace transforms, and to determine y(x), tables of inverse Laplace transforms. 2.9-3. Other Equations 63.

y(x) + 0

x

1 x

f

t x

y(t) dt = 0.

Eigenfunctions of this integral equation are determined by the roots of the following transcendental (algebraic) equation for the parameter λ:

1

f (z)z λ dz = –1.

(1)

0

1◦ . For a real simple root λk of equation (1) there is a corresponding eigenfunction yk (x) = xλk .

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Page 192

2◦ . For a real root λk of multiplicity r there are corresponding r eigenfunctions yk1 (x) = xλk ,

yk2 (x) = xλk ln x,

ykr (x) = xλk lnr–1 x.

...,

3◦ . For a complex simple root λk = αk + iβk of equation (1) there is a corresponding eigenfunction pair yk(1) (x) = xαk cos(βk ln x),

yk(2) (x) = xαk sin(βk ln x).

4◦ . For a complex root λk = αk +iβk of multiplicity r there are corresponding r eigenfunction pairs (1) (2) yk1 (x) = xαk cos(βk ln x), (x) = xαk sin(βk ln x), yk1 (1) yk2 (x) = xαk ln x cos(βk ln x),

(2) yk2 (x) = xαk ln x sin(βk ln x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (1) ykr (x)

= xαk lnr–1 x cos(βk ln x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (2) ykr (x) = xαk lnr–1 x sin(βk ln x).

The general solution is the combination (with arbitrary constants) of the eigenfunctions of the homogeneous integral equation. For equations 2.9.64–2.9.71, only particular solutions are given. To obtain the general solution, one must add the general solution of the corresponding homogeneous equation 2.9.63 to the particular solution. x 1 t 64. y(x) + f y(t) dt = Ax + B. x 0 x A solution: 1 1 A B y(x) = x+ , I0 = f (t) dt, I1 = tf (t) dt. 1 + I1 1 + I0 0 0 x 1 t 65. y(x) + f y(t) dt = Axβ . x 0 x A solution: 1 A β y(x) = x , B =1+ f (t)tβ dt. B 0 x 1 t 66. y(x) + f y(t) dt = A ln x + B. x 0 x A solution: y(x) = p ln x + q, where

1 A B AIl , q= – , I = f (t) dt, 0 1 + I0 1 + I0 (1 + I0 )2 0 x 1 t y(x) + f y(t) dt = Axβ ln x. x 0 x A solution: y(x) = pxβ ln x + qxβ , p=

67.

where A p= , 1 + I1

AI2 q=– , (1 + I1 )2

f (t)t dt, 0

1

f (t) ln t dt. 0

1 β

I1 =

Il =

1

f (t)tβ ln t dt.

I2 = 0

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Page 193

68.

x

y(x) +

1 x

0

f

t

y(t) dt = A cos(ln x).

x

A solution: AIc AIs cos(ln x) + 2 sin(ln x), Ic2 + Is2 Ic + Is2 1 1 Ic = 1 + f (t) cos(ln t) dt, Is = f (t) sin(ln t) dt. y(x) =

0

69.

x

y(x) +

1 x

0

f

t

0

y(t) dt = A sin(ln x).

x

A solution: AIs AIc cos(ln x) + 2 sin(ln x), Ic2 + Is2 Ic + Is2 1 1 Ic = 1 + f (t) cos(ln t) dt, Is = f (t) sin(ln t) dt. y(x) = –

0

70.

x

y(x) +

1 x

0

f

t

0

y(t) dt = Axβ cos(ln x) + Bxβ sin(ln x).

x

A solution: y(x) = pxβ cos(ln x) + qxβ sin(ln x), where p=

AIc – BIs , Ic2 + Is2

1

f (t)tβ cos(ln t) dt,

Ic = 1 +

AIs + BIc , Ic2 + Is2 1 Is = f (t)tβ sin(ln t) dt.

q=

0

71.

y(x) +

x

1

f

t

0

y(t) dt = g(x). x x 1◦ . For a polynomial right-hand side, 0

g(x) =

N

An xn

n=0

a solution bounded at zero is given by y(x) =

N An n x , 1 + fn n=0

1

f (z)z n dz.

fn = 0

Here its is assumed that f0 < ∞ and fn ≠ –1 (n = 0, 1, 2, . . . ). If for some n the relation fn = –1 holds, then a solution differs from the above case in one term and has the form 1 n–1 N Am m A m m An n ¯ y(x) = x + x + ¯ x ln x, fn = f (z)z n ln z dz. 1 + fm 1 + fm f n 0 m=0 m=n+1 For arbitrary g(x) expandable into power series, the formulas of item 1◦ can be used, in which one should set N = ∞. In this case, the convergence radius of the obtained solution y(x) is equal to that of the function g(x).

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2◦ . For g(x) = ln x

n

Ak xk , a solution has the form

k=0

y(x) = ln x

n

Bk xk +

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n 3◦ . For g(x) = Ak ln x)k , a solution of the equation has the form k=0 n

y(x) =

Bk ln x)k ,

k=0

where the Bk are found by the method of undetermined coefficients. n 4◦ . For g(x) = Ak cos(λk ln x), a solution of the equation has the form k=1

y(x) =

n

Bk cos(λk ln x) +

k=1

n

Ck sin(λk ln x),

k=1

where the Bk and Ck are found by the method of undetermined coefficients. n 5◦ . For g(x) = Ak sin(λk ln x) a solution of the equation has the form k=1

y(x) =

n

Bk cos(λk ln x) +

k=1

n

Ck sin(λk ln x),

k=1

where the Bk and Ck are found by the method of undetermined coefficients. 6◦ . For arbitrary right-hand side g(x), the transformation x = e–z ,

t = e–τ ,

y(x) = ez w(z),

f (ξ) = F (ln ξ),

g(x) = ez G(z)

leads to an equation with difference kernel of the form 2.9.62: ∞ w(z) + F (z – τ )w(τ ) dτ = G(z). z

7◦ . For arbitrary right-hand side g(x), the solution of the integral equation can be expressed via the inverse Mellin transform (see Section 7.3-1).

2.10. Some Formulas and Transformations Let the solution of the integral equation x y(x) + K(x, t)y(t) dt = f (x)

(1)

a

have the form y(x) = f (x) +

x

R(x, t)f (t) dt.

(2)

a

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Then the solution of the more complicated integral equation

x

y(x) +

K(x, t) a

has the form

g(x) y(t) dt = f (x) g(t)

(3)

g(x) f (t) dt. g(t)

(4)

x

y(x) = f (x) +

R(x, t) a

Below are formulas for the solutions of integral equations of the form (3) for some specific functions g(x). In all cases, it is assumed that the solution of equation (1) is known and is given by (2). 1◦ . The solution of the equation

x

K(x, t)(x/t)λ y(t) dt = f (x)

y(x) + a

has the form

x

R(x, t)(x/t)λ f (t) dt.

y(x) = f (x) + a

2◦ . The solution of the equation

x

K(x, t)eλ(x–t) y(t) dt = f (x)

y(x) + a

has the form

x

R(x, t)eλ(x–t) f (t) dt.

y(x) = f (x) + a

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Chapter 3

Linear Equation of the First Kind With Constant Limits of Integration Notation: f = f (x), g = g(x), h = h(x), K = K(x), and M = M (x) are arbitrary functions (these may be composite functions of the argument depending on two variables x and t); A, B, C, a, b, c, k, α, β, γ, λ, and µ are free parameters; and n is a nonnegative integer.

3.1. Equations Whose Kernels Contain Power-Law Functions 3.1-1. Kernels Linear in the Arguments x and t

1

|x – t| y(t) dt = f (x).

1. 0 ◦

1 . Let us remove the modulus in the integrand:

x

1

(x – t)y(t) dt +

(t – x)y(t) dt = f (x).

(1)

y(t) dt = fx (x).

(2)

x

0

Differentiating (1) with respect to x yields

x

1

y(t) dt – x

0

Differentiating (2) yields the solution y(x) = 12 fxx (x).

(3)

2◦ . Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy 1

certain relations. By setting x = 0 and x = 1 in (1), we obtain two corollaries 1

and

0

0

ty(t) dt = f (0)

(1 – t)y(t) dt = f (1), which can be rewritten in the form

1

ty(t) dt = f (0), 0

1

y(t) dt = f (0) + f (1).

(4)

0

In Section 3.1, we mean that kernels of the integral equations discussed may contain power-law functions or modulus of power-law functions.

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Substitute y(x) of (3) into (4). Integration by parts yields fx (1) = f (1)+f (0) and fx (1)–fx (0) = 2f (1) + 2f (0). Hence, we obtain the desired constraints for f (x): fx (1) = f (0) + f (1),

fx (0) + fx (1) = 0.

(5)

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation: f (x) = F (x) + Ax + B, A = – 12 Fx (1) + Fx (0) , B = 12 Fx (1) – F (1) – F (0) ,

2.

where F (x) is an arbitrary bounded twice differentiable function with bounded first derivative. b |x – t| y(t) dt = f (x), 0 ≤ a < b < ∞. a

This is a special case of equation 3.8.3 with g(x) = x. Solution: y(x) = 12 fxx (x). The right-hand side f (x) of the integral equation must satisfy certain relations. The general form of f (x) is as follows: A=

3.

– 12

f (x) = F (x) + Ax + B, Fx (a) + Fx (b) , B = 12 aFx (a) + bFx (b) – F (a) – F (b) ,

where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative). a |λx – t| y(t) dt = f (x), λ > 0. 0

Here 0 ≤ x ≤ a and 0 ≤ t ≤ a. 1◦ . Let us remove the modulus in the integrand: λx a (λx – t)y(t) dt + (t – λx)y(t) dt = f (x).

(1)

λx

0

Differentiating (1) with respect to x, we find that λx a λ y(t) dt – λ y(t) dt = fx (x). 0

(2)

λx

Differentiating (2) yields 2λ2 y(λx) = fxx (x). Hence, we obtain the solution 1 x y(x) = f . (3) 2λ2 xx λ 2◦ . Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = 0 in (1) and (2), we obtain two corollaries a a ty(t) dt = f (0), λ y(t) dt = –fx (0), (4) 0

0

Substitute y(x) from (3) into (4). Integrating by parts yields the desired constraints for f (x): (a/λ)fx (a/λ) = f (0) + f (a/λ),

fx (0) + fx (a/λ) = 0.

(5)

Conditions (5) make it possible to establish the admissible general form of the right-hand side of the integral equation: A=

– 12

f (x) = F (z) + Az + B, z = λx; 1 Fz (a) + Fz (0) , B = 2 aFz (a) – F (a) – F (0) ,

where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

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a

|x – λt| y(t) dt = f (x),

4.

λ > 0.

0

Here 0 ≤ x ≤ a and 0 ≤ t ≤ a. Solution:

y(x) = 12 λfxx (λx).

The right-hand side f (x) of the integral equation must satisfy the relations aλfx (aλ) = f (0) + f (aλ),

fx (0) + fx (aλ) = 0.

Hence, it follows the general form of the right-hand side: f (x) = F (x) + Ax + B, A = – 12 Fx (λa) + Fx (0) , B = 12 aλFx (aλ) – F (λa) – F (0) , where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative). 3.1-2. Kernels Quadratic in the Arguments x and t 5.

Ax + Bx2 – t y(t) dt = f (x),

a

A > 0,

B > 0.

0

This is a special case of equation 3.8.5 with g(x) = Ax + Bx2 . 6.

a

x – At – Bt2 y(t) dt = f (x),

A > 0,

B > 0.

0

This is a special case of equation 3.8.6 with g(x) = At + Bt2 . 7.

b

xt – t2 y(t) dt = f (x)

0 ≤ a < b < ∞.

a

The substitution w(t) = ty(t) leads to an equation of the form 1.3.2:

b

|x – t|w(t) dt = f (x). a

8.

b

x2 – t2 y(t) dt = f (x).

a

This is a special case of equation with g(x) = x2 . 3.8.3 d fx (x) Solution: y(x) = . The right-hand side f (x) of the equation must satisfy dx 4x certain constraints, given in 3.8.3. 9.

a

x2 – βt2 y(t) dt = f (x),

β > 0.

0

This is a special case of equation 3.8.4 with g(x) = x2 and β = λ2 . 10.

a

Ax + Bx2 – Aλt – Bλ2 t2 y(t) dt = f (x),

λ > 0.

0

This is a special case of equation 3.8.4 with g(x) = Ax + Bx2 .

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3.1-3. Kernels Containing Integer Powers of x and t or Rational Functions 11.

b

x – t3 y(t) dt = f (x).

a

Let us remove the modulus in the integrand:

x

b

3

(t – x)3 y(t) dt = f (x).

(x – t) y(t) dt + a

(1)

x

Differentiating (1) twice yields

x

b

(x – t)y(t) dt + 6

6 a

(t – x)y(t) dt = fxx (x).

x

This equation can be rewritten in the form 3.1.2:

b

a

|x – t| y(t) dt = 16 fxx (x).

(2)

Therefore the solution of the integral equation is given by y(x) =

1 12 yxxxx (x).

(3)

The right-hand side f (x) of the equation must satisfy certain conditions. To obtain these conditions, one must substitute solution (3) into (1) with x = a and x = b and into (2) with x = a and x = b, and then integrate the four resulting relations by parts. 12.

b

x3 – t3 y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = x3 . 13.

b

xt2 – t3 y(t) dt = f (x)

0 ≤ a < b < ∞.

a

The substitution w(t) = t2 y(t) leads to an equation of the form 3.1.2:

b

|x – t|w(t) dt = f (x). a

14.

b

x2 t – t3 y(t) dt = f (x).

a

The substitution w(t) = |t| y(t) leads to an equation of the form 3.1.8:

b

x2 – t2 w(t) dt = f (x).

a

15.

a

x3 – βt3 y(t) dt = f (x),

β > 0.

0

This is a special case of equation 3.8.4 with g(x) = x3 and β = λ3 .

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16.

b

x – t2n+1 y(t) dt = f (x),

n = 0, 1, 2, . . .

a

Solution: y(x) =

1 f (2n+2) (x). 2(2n + 1)! x

(1)

The right-hand side f (x) of the equation must satisfy certain conditions. To obtain these conditions, one must substitute solution (1) into the relations

b

b

(t – a)2n+1 y(t) dt = f (a), a

(–1)k+1 (k+1) fx (a), Ak k = 0, 1, . . . , 2n,

(t – a)2n–k y(t) dt = a

Ak = (2n + 1)(2n) . . . (2n + 1 – k); and then integrate the resulting equations by parts. 17. 0

∞

y(t) dt x+t

= f (x).

The left-hand side of this equation is the Stieltjes transform. 1◦ . By setting x = ez ,

t = eτ ,

y(t) = e–τ /2 w(τ ),

f (x) = e–z/2 g(z),

we obtain an integral equation with difference kernel of the form 3.8.15:

∞

–∞

w(τ ) dτ = g(z), 2 cosh 12 (z – τ )

whose solution is given by w(z) = √

1 2π 3

∞

iux cosh(πu) g(u)e ˜ du,

–∞

1 g(u) ˜ = √ 2π

∞

g(z)e–iuz dz.

–∞

• Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975). 2◦ . Under some assumptions, the solution of the original equation can be represented in the form 2n+1 (n) (n+1) (–1)n y(x) = lim fx (x) x , (1) x n→∞ (n – 1)! (n + 1)! which is the real inversion of the Stieltjes transform. An alternative form of the solution is y(x) = lim

n→∞

(–1)n e 2n 2n (n) (n) x fx (x) x . 2π n

(2)

To obtain an approximate solution of the integral equation, one restricts oneself to a specific value of n in (1) or (2) instead of taking the limit. • Reference: I. I. Hirschman and D. V. Widder (1955).

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3.1-4. Kernels Containing Square Roots 18.

a √

x–

√ t y(t) dt = f (x),

0 < a < ∞.

0

√ This is a special case of equation 3.8.3 with g(x) = x. Solution: d √ y(x) = x fx (x) . dx The right-hand side f (x) of the equation must satisfy certain conditions. The general form of the right-hand side is f (x) = F (x) + Ax + B,

A = –Fx (a),

B=

1 2

aFx (a) – F (a) – F (0) ,

where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative). 19.

√ x – β t y(t) dt = f (x),

a √

β > 0.

0

This is a special case of equation 3.8.4 with g(x) = 20.

√

x and β =

√

λ.

x – t y(t) dt = f (x).

a √

0

This is a special case of equation 3.8.5 with g(x) = 21.

a

x –

√

x (see item 3◦ of 3.8.5).

√ t y(t) dt = f (x).

0

This is a special case of equation 3.8.6 with g(t) =

a

22. 0

y(t) dt = f (x), √ |x – t|

√

t (see item 3◦ of 3.8.6).

0 < a ≤ ∞.

This is a special case of equation 3.1.29 with k = 12 . Solution: y(x) = –

∞

23. –∞

A d x1/4 dx

a

x

dt (t – x)1/4

0

t

f (s) ds , s 1/4 (t – s)1/4

A= √

1 . 8π Γ2 (3/4)

y(t) dt = f (x). √ |x – t|

This is a special case of equation 3.1.34 with λ = 12 . Solution: ∞ 1 f (x) – f (t) y(x) = dt. 4π –∞ |x – t|3/2

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3.1-5. Kernels Containing Arbitrary Powers

a

|xk – tk | y(t) dt = f (x),

24.

0 < a < ∞.

0 < k < 1,

0 ◦

1 . Let us remove the modulus in the integrand:

x

a

(xk – tk )y(t) dt +

(tk – xk )y(t) dt = f (x).

(1)

x

0

Differentiating (1) with respect to x yields

x

kxk–1

a

y(t) dt – kxk–1

y(t) dt = fx (x).

(2)

x

0

Let us divide both sides of (2) by kxk–1 and differentiate the resulting equation. As a result, we obtain the solution 1 d 1–k y(x) = (3) x fx (x) . 2k dx 2◦ . Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy a

certain relations. By setting x = 0 and x = a, in (1), we obtain two corollaries a

and

0

0

tk y(t) dt = f (0)

(ak – tk )y(t) dt = f (a), which can be rewritten in the form

a k

t y(t) dt = f (0),

k

a

0

a

y(t) dt = f (0) + f (a).

(4)

0

Substitute y(x) of (3) into (4). Integrating by parts yields the relations afx (a) = kf (a) + kf (0) and afx (a) = 2kf (a) + 2kf (0). Hence, the desired constraints for f (x) have the form f (0) + f (a) = 0,

fx (a) = 0.

(5)

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation: A = –Fx (a),

f (x) = F (x) + Ax + B,

25.

B=

1 2

aFx (a) – F (a) – F (0) ,

where F (x) is an arbitrary bounded twice differentiable function with bounded first derivative. The first derivative may be unbounded at x = 0, in which case the conditions x1–k Fx x=0 = 0 must hold. a |xk – βtk | y(t) dt = f (x), 0 < k < 1, β > 0. 0

This is a special case of equation 3.8.4 with g(x) = xk and β = λk .

a

|xk tm – tk+m | y(t) dt = f (x),

26.

0 < k < 1,

0 < a < ∞.

0

The substitution w(t) = tm y(t) leads to an equation of the form 3.1.24:

a

|xk – tk |w(t) dt = f (x). 0

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1

|xk – tm | y(t) dt = f (x),

27.

k > 0,

m > 0.

0

The transformation z = xk ,

τ = tm ,

w(τ ) = τ

leads to an equation of the form 3.1.1: 1 |z – τ |w(τ ) dτ = F (z),

1–m m

y(t)

F (z) = mf (z 1/k ).

0

b

|x – t|1+λ y(t) dt = f (x),

28.

0 ≤ λ < 1.

a

For λ = 0, see equation 3.1.2. Assume that 0 < λ < 1. 1◦ . Let us remove the modulus in the integrand: x b (x – t)1+λ y(t) dt + (t – x)1+λ y(t) dt = f (x). a

(1)

x

Let us differentiate (1) with respect to x twice and then divide both the sides by λ(λ + 1). As a result, we obtain x b 1 λ–1 f (x). (x – t) y(t) dt + (t – x)λ–1 y(t) dt = (2) λ(λ + 1) xx a x Rewrite equation (2) in the form b y(t) dt 1 = f (x), k λ(λ + 1) xx a |x – t|

k = 1 – λ.

(3)

See 3.1.29 and 3.1.30 for the solutions of equation (3) for various a and b. 2◦ . The right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries b b 1+λ (t – a) y(t) dt = f (a), (b – t)1+λ y(t) dt = f (b). (4) a

29.

a

On substituting the solution y(x) of (3) into (4) and then integrating by parts, we obtain the desired constraints for f (x). a y(t) dt = f (x), 0 < k < 1, 0 < a ≤ ∞. |x – t|k 0 1◦ . Solution:

1–2k t a k–1 d t 2 dt f (s) ds y(x) = –Ax 2 , 1–k 1–k 1–k dx x 0 s 2 (t – s) 2 (t – x) 2

–2 πk 1 1+k A= , cos Γ(k) Γ 2π 2 2 where Γ(k) is the gamma function. 2◦ . The transformation x = z 2 , t = ξ 2 , w(ξ) = 2ξy(t) leads to an equation of the form 3.1.31: √a w(ξ) dξ = f z 2 . 2 – ξ 2 |k |z 0

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30.

b

y(t) dt = f (x), 0 < k < 1. |x – t|k a It is assumed that |a| + |b| < ∞. Solution: x x 1 f (t) dt Z(t)F (t) d 1 2 1 y(x) = – cos ( πk) dt, cot( 12 πk) 2 1–k 2 1–k 2π dx a (x – t) π a (x – t) where Z(t) = (t – a)

31.

1+k 2

(b – t)

1–k 2

F (t) =

,

d dt

t

a

dτ (t – τ )k

• Reference: F. D. Gakhov (1977). a y(t) dt = f (x), 0 < k < 1, 0 < a ≤ ∞. 2 2k 0 |x – t | Solution: a 2–2k 2Γ(k) cos 12 πk k–1 d t F (t) dt y(x) = – x , 1+k 2 1–k dx x (t2 – x2 ) 2 π Γ 2

τ

b

f (s) ds . Z(s)(s – τ )1–k

t

F (t) = 0

s k f (s) ds (t2 – s 2 )

1–k 2

.

• Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975). 32.

b

y(t) dt = f (x), λ – tλ |k |x a 1◦ . The transformation

0 < k < 1,

z = xλ ,

τ = tλ ,

λ > 0.

w(τ ) = τ

1–λ λ

y(t)

leads to an equation of the form 3.8.30: B

w(τ ) dτ = F (z), |z – τ |k

A λ

λ

where A = a , B = b , F (z) = λf (z 1/λ ). 2◦ . Solution with a = 0: y(x) =

λ(k–1) –Ax 2

d dx

b

t

λ(3–2k)–2 2

x

dt

t

s

λ(k+1)–2 2

f (s) ds

1–k

0 (tλ – xλ ) 2 (tλ – s λ )

–2 λ2 πk 1+k A= , cos Γ(k) Γ 2π 2 2

1–k 2

,

where Γ(k) is the gamma function. 33.

1

y(t)

dt = f (x), 0 < k < 1, λ > 0, m > 0. – tm |k The transformation 1–m z = xλ , τ = tm , w(τ ) = τ m y(t) 0

|xλ

leads to an equation of the form 3.8.30: 1 w(τ ) dτ = F (z), |z – τ |k 0

F (z) = mf (z 1/λ ).

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34.

∞

y(t)

|x – t|1–λ Solution:

dt = f (x),

0 < λ < 1.

–∞

y(x) = It assumed that the condition

35.

36.

37.

38.

39.

πλ λ tan 2π 2 ∞ –∞

∞

–∞

f (x) – f (t) dt. |x – t|1+λ

|f (x)| dx < ∞ is satisfied for some p, 1 < p < 1/λ. p

• Reference: S. G. Samko, A. A. Kilbas, and A. A. Marichev (1993). ∞ y(t) dt = f (x), 0 < λ < 1. 3 1–λ –∞ |x – t| The substitution z = x3 leads to an equation of the form 3.1.34: ∞ y(t) dt = f z 1/3 . 1–λ |z – t| –∞ ∞ y(t) dt = f (x), 0 < λ < 1. 3 3 1–λ –∞ |x – t | The transformation z = x3 , τ = t3 , w(τ ) = τ –2/3 y(t) leads to an equation of the form 3.1.34: ∞ w(τ ) dτ = F (z), F (z) = 3f z 1/3 . 1–λ –∞ |z – τ | ∞ sign(x – t) y(t) dt = f (x), 0 < λ < 1. 1–λ –∞ |x – t| Solution: πλ ∞ f (x) – f (t) λ y(x) = cot sign(x – t) dt. 1+λ 2π 2 –∞ |x – t| • Reference: S. G. Samko, A. A. Kilbas, and A. A. Marichev (1993). ∞ a + b sign(x – t) y(t) dt = f (x), 0 < λ < 1. |x – t|1–λ –∞ Solution: ∞ λ sin(πλ) a + b sign(x – t) 1 1 y(x) = f (x) – f (t) dt. 1+λ 2 2 2 2 |x – t| 4π a cos 2 πλ + b sin 2 πλ –∞ • Reference: S. G. Samko, A. A. Kilbas, and A. A. Marichev (1993). ∞ y(t) dt = f (x), a > 0, b > 0, k > 0. (ax + bt)k 0 By setting 1 2z 1 2τ e , t= e , y(t) = be(k–2)τ w(τ ), f (x) = e–kz g(z). 2a 2b we obtain an integral equation with the difference kernel of the form 3.8.15: ∞ w(τ ) dτ = g(z). k cosh (z – τ ) –∞ x=

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∞

40.

tz–1 y(t) dt = f (z).

0

The left-hand side of this equation is the Mellin transform of y(t) (z is treated as a complex variable). Solution: c+i∞ 1 y(t) = t–z f (z) dz, i2 = –1. 2πi c–i∞ For specific f (z), one can use tables of Mellin and Laplace integral transforms to calculate the integral. • References: H. Bateman and A. Erd´elyi (vol. 2, 1954), V. A. Ditkin and A. P. Prudnikov (1965). 3.1-6. Equation Containing the Unknown Function of a Complicated Argument

1

y(xt) dt = f (x).

41. 0

Solution:

y(x) = xfx (x) + f (x). The function f (x) is assumed to satisfy the condition xf (x) x=0 = 0.

1

tλ y(xt) dt = f (x).

42. 0

x

The substitution ξ = xt leads to equation respect to x yields the solution

0

ξ λ y(ξ) dξ = xλ+1 f (x). Differentiating with

y(x) = xfx (x) + (λ + 1)f (x). The function f (x) is assumed to satisfy the condition xλ+1 f (x) x=0 = 0. 43.

1

Axk + Btm )y(xt) dt = f (x).

0

The substitution ξ = xt leads to an equation of the form 1.1.50: x k+m + Bξ m y(ξ) dξ = xm+1 f (x). Ax 0

44.

45.

1

1

y(xt) dt = f (x). √ 1–t 0 The substitution ξ = xt leads to Abel’s equation 1.1.36: x y(ξ) dξ √ √ = x f (x). x–ξ 0 y(xt) dt

= f (x), 0 < λ < 1. (1 – t)λ The substitution ξ = xt leads to the generalized Abel equation 1.1.46: x y(ξ) dξ = x1–λ f (x). (x – ξ)λ 0 0

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46.

1

tµ y(xt)

dt = f (x), 0 < λ < 1. (1 – t)λ The transformation ξ = xt, w(ξ) = ξ µ y(ξ) leads to the generalized Abel equation 1.1.46: x w(ξ) dξ = x1+µ–λ f (x). λ 0 (x – ξ) ∞ y(x + t) – y(x – t) dt = f (x). t 0 Solution: ∞ 1 f (x + t) – f (x – t) y(x) = – 2 dt. π 0 t 0

47.

• Reference: V. A. Ditkin and A. P. Prudnikov (1965). 3.1-7. Singular Equations In this subsection, all singular integrals are understood in the sense of the Cauchy principal value. ∞ y(t) dt 48. = f (x). –∞ t – x Solution: ∞ 1 f (t) dt y(x) = – 2 . π –∞ t – x The integral equation and its solution form a Hilbert transform pair (in the asymmetric form).

49.

• Reference: V. A. Ditkin and A. P. Prudnikov (1965). b y(t) dt = f (x). t–x a This equation is encountered in hydrodynamics in solving the problem on the flow of an ideal inviscid fluid around a thin profile (a ≤ x ≤ b). It is assumed that |a| + |b| < ∞. 1◦ . The solution bounded at the endpoints is b f (t) 1 dt √ y(x) = – 2 (x – a)(b – x) , π t (t – a)(b – t) – x a provided that b f (t) dt √ = 0. (t – a)(b – t) a 2◦ . The solution bounded at the endpoint x = a and unbounded at the endpoint x = b is b 1 x–a b – t f (t) y(x) = – 2 dt. π b–x a t–a t–x 3◦ . The solution unbounded at the endpoints is b √ 1 (t – a)(b – t) y(x) = – 2 √ f (t) dt + C , t–x π (x – a)(b – x) a b

where C is an arbitrary constant. The formula y(t) dt = C/π holds. a Solutions that have a singularity point x = s inside the interval [a, b] can be found in Subsection 12.4-3. • Reference: F. D. Gakhov (1977).

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3.2. Equations Whose Kernels Contain Exponential Functions 3.2-1. Kernels Containing Exponential Functions

b

–∞ < a < b < ∞.

eλ|x–t| y(t) dt = f (x),

1. a ◦

1 . Let us remove the modulus in the integrand: x b λ(x–t) e y(t) dt + eλ(t–x) y(t) dt = f (x). a

(1)

x

Differentiating (1) with respect to x twice yields x 2 λ(x–t) 2 2λy(x) + λ e y(t) dt + λ a

b

eλ(t–x) y(t) dt = fxx (x).

(2)

x

By eliminating the integral terms from (1) and (2), we obtain the solution 1 y(x) = (3) fxx (x) – λ2 f (x) . 2λ 2◦ . The right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries b b eλt y(t) dt = eλa f (a), e–λt y(t) dt = e–λb f (b). (4) a

a

On substituting the solution y(x) of (3) into (4) and then integrating by parts, we see that eλb fx (b) – eλa fx (a) = λeλa f (a) + λeλb f (b), e–λb fx (b) – e–λa fx (a) = λe–λa f (a) + λe–λb f (b). Hence, we obtain the desired constraints for f (x): fx (a) + λf (a) = 0,

fx (b) – λf (b) = 0.

(5)

The general form of the right-hand side satisfying conditions (5) is given by

2.

f (x) = F (x) + Ax + B, 1 1 A= Fx (a) + Fx (b) + λF (a) – λF (b) , B = – Fx (a) + λF (a) + Aaλ + A , bλ – aλ – 2 λ where F (x) is an arbitrary bounded, twice differentiable function. b λ|x–t| + Beµ|x–t| y(t) dt = f (x), Ae –∞ < a < b < ∞. a

Let us remove the modulus in the integrand and differentiate the resulting equation with respect to x twice to obtain b 2 λ|x–t| 2(Aλ + Bµ)y(x) + + Bµ2 eµ|x–t| y(t) dt = fxx (x). (1) Aλ e a

Eliminating the integral term with eµ|x–t| from (1) with the aid of the original integral equation, we find that b 2 2 2(Aλ + Bµ)y(x) + A(λ – µ ) eλ|x–t| y(t) dt = fxx (x) – µ2 f (x). (2) a

For Aλ + Bµ = 0, this is an equation of the form 3.2.1, and for Aλ + Bµ ≠ 0, this is an equation of the form 4.2.15. The right-hand side f (x) must satisfy certain relations, which can be obtained by setting x = a and x = b in the original equation (a similar procedure is used in 3.2.1).

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b

|eλx – eλt | y(t) dt = f (x),

3.

λ > 0.

a

4.

This is a special case of equation 3.8.3 with g(x) = eλx . Solution: 1 d –λx y(x) = e fx (x) . 2λ dx The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). a |eβx – eµt | y(t) dt = f (x), β > 0, µ > 0. 0

This is a special case of equation 3.8.4 with g(x) = eβx and λ = µ/β. 5.

b n a

Ak exp λk |x – t| y(t) dt = f (x),

–∞ < a < b < ∞.

k=1

1◦ . Let us remove the modulus in the kth summand of the integrand: b x b Ik (x) = exp λk |x – t| y(t) dt = exp[λk (x – t)]y(t) dt + exp[λk (t – x)]y(t) dt. (1) a

a

x

Differentiating (1) with respect to x twice yields x Ik = λk exp[λk (x – t)]y(t) dt – λk a

Ik

= 2λk y(x) +

b

exp[λk (t – x)]y(t) dt,

x

x

λ2k

exp[λk (x – t)]y(t) dt + a

(2)

b

λ2k

exp[λk (t – x)]y(t) dt, x

where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2), we find the relation between Ik and Ik : Ik = 2λk y(x) + λ2k Ik ,

Ik = Ik (x).

(3)

2◦ . With the aid of (1), the integral equation can be rewritten in the form n

Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice and taking into account (3), we obtain σ1 y(x) +

n

Ak λ2k Ik = fxx (x),

k=1

σ1 = 2

n

Ak λ k .

(5)

k=1

Eliminating the integral In from (4) and (5) yields σ1 y(x) +

n–1

Ak (λ2k – λ2n )Ik = fxx (x) – λ2n f (x).

(6)

k=1

Differentiating (6) with respect to x twice and eliminating In–1 from the resulting equation with the aid of (6), we obtain a similar equation whose right-hand side is a second-order n–2 linear differential operator (acting on y) with constant coefficients plus the sum Bk Ik . If k=1

we successively eliminate In–2 , In–3 , . . . , I1 with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n – 1) with constant coefficients.

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3◦ . The right-hand side f (x) must satisfy certain conditions. To find these conditions, one must set x = a in the integral equation and its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.)

b

6. a

y(t) dt |eλx – eλt |k

= f (x),

0 < k < 1.

The transformation z = eλx , τ = eλt , w(τ ) = e–λt y(t) leads to an equation of the form 3.1.30:

B

w(τ ) dτ = F (z), |z – τ |k A where A = eλa , B = eλb , F (z) = λf λ1 ln z . 7.

∞

y(t) dt

= f (x), λ > 0, k > 0. + eλt )k This equation can be rewritten as an equation with difference kernel in the form 3.8.16: ∞ w(t) dt = g(x), k 1 cosh 2 λ(x – t) 0 where w(t) = 2–k exp – 12 λkt y(t) and g(x) = exp 12 λkx f (x). (eλx

0

∞

8.

e–zt y(t) dt = f (z).

0

The left-hand side of the equation is the Laplace transform of y(t) (z is treated as a complex variable). 1◦ . Solution: 1 y(t) = 2πi

c+i∞

ezt f (z) dz,

i2 = –1.

c–i∞

For specific functions f (z), one may use tables of inverse Laplace transforms to calculate the integral (e.g., see Supplement 5). 2◦ . For real z = x, under some assumptions the solution of the original equation can be represented in the form y(x) = lim

n→∞

(–1)n n n+1 (n) n fx , n! x x

which is the real inversion of the Laplace transform. To calculate the solution approximately, one should restrict oneself to a specific value of n in this formula instead of taking the limit. • References: H. Bateman and A. Erd´elyi (vol. 1, 1954), I. I. Hirschman and D. V. Widder (1955), V. A. Ditkin and A. P. Prudnikov (1965). 3.2-2. Kernels Containing Power-Law and Exponential Functions 9.

keλx – k – t y(t) dt = f (x).

a

0

This is a special case of equation 3.8.5 with g(x) = keλx – k.

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10.

a

x – keλt – k y(t) dt = f (x).

0

11.

This is a special case of equation 3.8.6 with g(t) = keλt – k. b exp(λx2 ) – exp(λt2 )y(t) dt = f (x), λ > 0. a

12.

13.

This is a special case of equation 3.8.3 with g(x) = exp(λx2 ). Solution: 1 d 1 y(x) = exp(–λx2 )fx (x) . 4λ dx x The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). 2 ∞ 1 t y(t) dt = f (x). exp – √ πx 0 4x Applying the Laplace transformation to the equation, we obtain ∞ √ y( ˜ p) ˜ ˜ e–pt f (t) dt. = f (p), f (p) = √ p 0 Substituting p by p2 and solving for the transform y, ˜ we find that y(p) ˜ = pf˜(p2 ). The inverse Laplace transform provides the solution of the original integral equation: c+i∞ 1 –1 –1 2 ˜ y(t) = L {pf (p )}, L {g(p)} ≡ ept g(p) dp. 2πi c–i∞ ∞ exp[–g(x)t2 ]y(t) dt = f (x). 0

Assume that g(0) = ∞, g(∞) = 0, and gx < 0. 1 The substitution z = leads to equation 3.2.12: 4g(x) 2 ∞ 1 t √ y(t) dt = F (z), exp – 4z πz 0

2 1 where the function F (z) is determined by the relations F = √ f (x) g(x) and z = 4g(x) π by means of eliminating x.

3.3. Equations Whose Kernels Contain Hyperbolic Functions 3.3-1. Kernels Containing Hyperbolic Cosine 1.

b

cosh(λx) – cosh(λt) y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = cosh(λx). Solution: 1 d fx (x) y(x) = . 2λ dx sinh(λx) The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3).

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2.

a

cosh(βx) – cosh(µt) y(t) dt = f (x),

β > 0,

µ > 0.

0

This is a special case of equation 3.8.4 with g(x) = cosh(βx) and λ = µ/β. 3.

b

coshk x – coshk t| y(t) dt = f (x),

0 < k < 1.

a

This is a special case of equation 3.8.3 with g(x) = coshk x. Solution: 1 d fx (x) y(x) = . 2k dx sinh x coshk–1 x The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). 4.

b

y(t)

dt = f (x), 0 < k < 1. |cosh(λx) – cosh(λt)|k This is a special case of equation 3.8.7 with g(x) = cosh(λx) + β, where β is an arbitrary number. a

3.3-2. Kernels Containing Hyperbolic Sine

b

5.

sinh λ|x – t| y(t) dt = f (x),

–∞ < a < b < ∞.

a ◦

1 . Let us remove the modulus in the integrand: x b sinh[λ(x – t)]y(t) dt + sinh[λ(t – x)]y(t) dt = f (x). a

(1)

x

Differentiating (1) with respect to x twice yields x 2λy(x) + λ2 sinh[λ(x – t)]y(t) dt + λ2 a

b

sinh[λ(t – x)]y(t) dt = fxx (x).

(2)

x

Eliminating the integral terms from (1) and (2), we obtain the solution 1 y(x) = (3) fxx (x) – λ2 f (x) . 2λ 2◦ . The right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries b b sinh[λ(t – a)]y(t) dt = f (a), sinh[λ(b – t)]y(t) dt = f (b). (4) a

a

Substituting solution (3) into (4) and integrating by parts yields the desired conditions for f (x): sinh[λ(b – a)]fx (b) – λ cosh[λ(b – a)]f (b) = λf (a), sinh[λ(b – a)]fx (a) + λ cosh[λ(b – a)]f (a) = –λf (b).

(5)

The general form of the right-hand side is given by f (x) = F (x) + Ax + B,

(6)

where F (x) is an arbitrary bounded twice differentiable function, and the coefficients A and B are expressed in terms of F (a), F (b), Fx (a), and Fx (b) and can be determined by substituting formula (6) into conditions (5).

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6.

b A sinh λ|x – t| + B sinh µ|x – t| y(t) dt = f (x),

–∞ < a < b < ∞.

a

Let us remove the modulus in the integrand and differentiate the equation with respect to x twice to obtain b 2 2(Aλ + Bµ)y(x) + (x), (1) Aλ sinh λ|x – t| + Bµ2 sinh µ|x – t| y(t) dt = fxx a

Eliminating the integral term with sinh µ|x – t| from (1) yields b 2 2 2(Aλ + Bµ)y(x) + A(λ – µ ) sinh λ|x – t| y(t) dt = fxx (x) – µ2 f (x).

(2)

a

For Aλ + Bµ = 0, this is an equation of the form 3.3.5, and for Aλ + Bµ ≠ 0, this is an equation of the form 4.3.26. The right-hand side f (x) must satisfy certain relations, which can be obtained by setting x = a and x = b in the original equation (a similar procedure is used in 3.3.5). 7.

b

sinh(λx) – sinh(λt) y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = sinh(λx). Solution: 1 d fx (x) y(x) = . 2λ dx cosh(λx)

8.

The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). a sinh(βx) – sinh(µt) y(t) dt = f (x), β > 0, µ > 0. 0

This is a special case of equation 3.8.4 with g(x) = sinh(βx) and λ = µ/β.

b

9.

sinh3 λ|x – t| y(t) dt = f (x).

a

Using the formula sinh3 β = 14 sinh 3β – 34 sinh β, we arrive at an equation of the form 3.3.6: b 3 1 4 A sinh 3λ|x – t| – 4 A sinh λ|x – t| y(t) dt = f (x). a

10.

b n a

Ak sinh λk |x – t| y(t) dt = f (x),

–∞ < a < b < ∞.

k=1

1◦ . Let us remove the modulus in the kth summand of the integrand: b x b Ik (x) = sinh λk |x – t| y(t) dt = sinh[λk (x – t)]y(t) dt + sinh[λk (t – x)]y(t) dt. (1) a

a

x

Differentiating (1) with respect to x twice yields x Ik = λk cosh[λk (x – t)]y(t) dt – λk a

Ik

= 2λk y(x) +

λ2k

b

cosh[λk (t – x)]y(t) dt,

x

sinh[λk (x – t)]y(t) dt + a

x

(2)

b

λ2k

sinh[λk (t – x)]y(t) dt, x

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where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2), we find the relation between Ik and Ik : Ik = 2λk y(x) + λ2k Ik ,

Ik = Ik (x).

(3)

2◦ . With the aid of (1), the integral equation can be rewritten in the form n

Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice and taking into account (3), we find that σ1 y(x) +

n

Ak λ2k Ik = fxx (x),

σ1 = 2

k=1

n

Ak λ k .

(5)

k=1

Eliminating the integral In from (4) and (5) yields σ1 y(x) +

n–1

Ak (λ2k – λ2n )Ik = fxx (x) – λ2n f (x).

(6)

k=1

Differentiating (6) with respect to x twice and eliminating In–1 from the resulting equation with the aid of (6), we obtain a similar equation whose right-hand side is a second-order n–2 linear differential operator (acting on y) with constant coefficients plus the sum Bk Ik . k=1

If we successively eliminate In–2 , In–3 , . . . , with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n – 1) with constant coefficients. 3◦ . The right-hand side f (x) must satisfy certain conditions. To find these conditions, one should set x = a in the integral equation and its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.) 11.

sinhk x – sinhk t y(t) dt = f (x),

b

0 < k < 1.

0

This is a special case of equation 3.8.3 with g(x) = sinhk x. Solution: 1 d fx (x) y(x) = . 2k dx cosh x sinhk–1 x The right-hand side f (x) must satisfy certain conditions. As follows from item 3◦ of equation 3.8.3, the admissible general form of the right-hand side is given by f (x) = F (x) + Ax + B, A = –Fx (b), B = 12 bFx (b) – F (0) – F (b) , where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

b

12. a

y(t) |sinh(λx) – sinh(λt)|k

dt = f (x),

0 < k < 1.

This is a special case of equation 3.8.7 with g(x) = sinh(λx) + β, where β is an arbitrary number.

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13.

a

k sinh(λx) – t y(t) dt = f (x).

0

14.

This is a special case of equation 3.8.5 with g(x) = k sinh(λx). a x – k sinh(λt) y(t) dt = f (x). 0

This is a special case of equation 3.8.6 with g(x) = k sinh(λt). 3.3-3. Kernels Containing Hyperbolic Tangent 15.

b

tanh(λx) – tanh(λt) y(t) dt = f (x).

a

16.

This is a special case of equation 3.8.3 with g(x) = tanh(λx). Solution: 1 d y(x) = cosh2 (λx)fx (x) . 2λ dx The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). a tanh(βx) – tanh(µt) y(t) dt = f (x), β > 0, µ > 0. 0

This is a special case of equation 3.8.4 with g(x) = tanh(βx) and λ = µ/β.

b

| tanhk x – tanhk t| y(t) dt = f (x),

17.

0 < k < 1.

0

This is a special case of equation 3.8.3 with g(x) = tanhk x. Solution: 1 d y(x) = cosh2 x cothk–1 x fx (x) . 2k dx The right-hand side f (x) must satisfy certain conditions. As follows from item 3◦ of equation 3.8.3, the admissible general form of the right-hand side is given by f (x) = F (x) + Ax + B, A = –Fx (b), B = 12 bFx (b) – F (0) – F (b) , where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

b

y(t)

19.

dt = f (x), 0 < k < 1. |tanh(λx) – tanh(λt)|k This is a special case of equation 3.8.7 with g(x) = tanh(λx) + β, where β is an arbitrary number. a k tanh(λx) – t y(t) dt = f (x).

20.

This is a special case of equation 3.8.5 with g(x) = k tanh(λx). a x – k tanh(λt) y(t) dt = f (x).

18.

a

0

0

This is a special case of equation 3.8.6 with g(x) = k tanh(λt).

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3.3-4. Kernels Containing Hyperbolic Cotangent 21.

b

coth(λx) – coth(λt) y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = coth(λx). 22.

cothk x – cothk t y(t) dt = f (x),

b

0 < k < 1.

0

This is a special case of equation 3.8.3 with g(x) = cothk x.

3.4. Equations Whose Kernels Contain Logarithmic Functions 3.4-1. Kernels Containing Logarithmic Functions 1.

b

ln(x/t) y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = ln x. Solution: 1 d y(x) = xfx (x) . 2 dx The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3).

b

ln |x – t| y(t) dt = f (x).

2. a

Carleman’s equation. 1◦ . Solution with b – a ≠ 4: 1 y(x) = 2 √ π (x – a)(b – x)

√

b

a

(t – a)(b – t) ft (t) dt 1 1 + t–x π ln 4 (b – a)

a

b

f (t) dt √ . (t – a)(b – t)

2◦ . If b – a = 4, then for the equation to be solvable, the condition

b

f (t)(t – a)–1/2 (b – t)–1/2 dt = 0 a

must be satisfied. In this case, the solution has the form 1 y(x) = 2 √ π (x – a)(b – x)

a

b

√

(t – a)(b – t) ft (t) dt +C , t–x

where C is an arbitrary constant. • Reference: F. D. Gakhov (1977).

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3.

b

ln |x – t| + β y(t) dt = f (x).

a

By setting x = e–β z,

t = e–β τ ,

y(t) = Y (τ ),

we arrive at an equation of the form 3.4.2: B ln |z – τ | Y (τ ) dτ = g(z),

A = aeβ , B = beβ .

A

a

4.

f (x) = e–β g(z),

A

y(t) dt = f (x), –a ≤ x ≤ a. |x – t| This is a special case of equation 3.4.3 with b = –a. Solution with 0 < a < 2A: a 1 d y(x) = w(t, a)f (t) dt w(x, a) 2M (a) da –a ξ 1 a d 1 d – w(x, ξ) w(t, ξ)f (t) dt dξ 2 |x| dξ M (ξ) dξ –ξ ξ a 1 d w(x, ξ) – w(t, ξ) df (t) dξ, 2 dx |x| M (ξ) –ξ ln

–a

where

M (ξ) =

2A ln ξ

–1 ,

w(x, ξ) =

M (ξ) , π ξ 2 – x2

and the prime stands for the derivative.

5.

6.

7.

• Reference: I. C. Gohberg and M. G. Krein (1967). a x + t y(t) dt = f (x). ln x–t 0 Solution: a 2 d F (t) dt √ y(x) = – 2 , π dx x t2 – x2

F (t) =

d dt

0

t

sf (s) ds √ . t2 – s 2

• Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975). b 1 + λx ln 1 + λt y(t) dt = f (x). a This is a special case of equation 3.8.3 with g(x) = ln(1 + λx). Solution: 1 d y(x) = (1 + λx)fx (x) . 2λ dx The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). b β ln x – lnβ t y(t) dt = f (x), 0 < β < 1. a

8.

This is a special case of equation 3.8.3 with g(x) = lnβ x. b y(t) dt = f (x), 0 < β < 1. β a |ln(x/t)| This is a special case of equation 3.8.7 with g(x) = ln x + A, where A is an arbitrary number.

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3.4-2. Kernels Containing Power-Law and Logarithmic Functions 9.

a

k ln(1 + λx) – t y(t) dt = f (x).

0

10.

This is a special case of equation 3.8.5 with g(x) = k ln(1 + λx). a x – k ln(1 + λt) y(t) dt = f (x). 0

11.

This is a special case of equation 3.8.6 with g(x) = k ln(1 + λt). ∞ 1 x + t ln y(t) dt = f (x). t x–t 0 Solution: x d y(x) = 2 π dx

12.

∞

0

df (t) x2 ln1 – 2 dt. dt t

• Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975). ∞ ln x – ln t y(t) dt = f (x). x–t 0 The left-hand side of this equation is the iterated Stieltjes transform. Under some assumptions, the solution of the integral equation can be represented in the form e 4n 1 d y(x) = lim Dn x2n D2n x2n Dn f (x), D = . 2 4π n→∞ n dx To calculate the solution approximately, one should restrict oneself to a specific value of n in this formula instead of taking the limit. • Reference: I. I. Hirschman and D. V. Widder (1955).

b

ln |xβ – tβ | y(t) dt = f (x),

13.

β > 0.

a

The transformation z = xβ ,

τ = tβ ,

w(τ ) = t1–β y(t)

leads to Carleman’s equation 3.4.2: B ln |z – τ |w(τ ) dτ = F (z),

A = aβ ,

B = bβ ,

A

where F (z) = βf z 1/β .

1

ln |xβ – tµ | y(t) dt = f (x),

14.

β > 0, µ > 0.

0

The transformation z = xβ ,

τ = tµ ,

w(τ ) = t1–µ y(t)

leads to an equation of the form 3.4.2: 1 ln |z – τ |w(τ ) dτ = F (z),

F (z) = µf z 1/β .

0

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3.4-3. An Equation Containing the Unknown Function of a Complicated Argument 15.

1

A ln t + B)y(xt) dt = f (x).

0

The substitution ξ = xt leads to an equation of the form 1.9.3 with g(x) = –A ln x: x A ln ξ – A ln x + B y(ξ) dξ = xf (x). 0

3.5. Equations Whose Kernels Contain Trigonometric Functions 3.5-1. Kernels Containing Cosine

∞

cos(xt)y(t) dt = f (x).

1. 0

2 ∞ cos(xt)f (t) dt. π 0 Up to constant factors, the function f (x) and the solution y(t) are the Fourier cosine transform pair.

Solution: y(x) =

• References: H. Bateman and A. Erd´elyi (vol. 1, 1954), V. A. Ditkin and A. P. Prudnikov (1965). 2.

b

cos(λx) – cos(λt) y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = cos(λx). Solution: 1 d fx (x) y(x) = – . 2λ dx sin(λx)

3.

The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). a cos(βx) – cos(µt) y(t) dt = f (x), β > 0, µ > 0. 0

This is a special case of equation 3.8.4 with g(x) = cos(βx) and λ = µ/β. 4.

b

cosk x – cosk t y(t) dt = f (x),

0 < k < 1.

a

This is a special case of equation 3.8.3 with g(x) = cosk x. Solution: 1 d fx (x) y(x) = – . 2k dx sin x cosk–1 x The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3).

5.

b

y(t) dt = f (x), 0 < k < 1. |cos(λx) – cos(λt)|k a This is a special case of equation 3.8.7 with g(x) = cos(λx) + β, where β is an arbitrary number.

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3.5-2. Kernels Containing Sine

∞

sin(xt)y(t) dt = f (x).

6. 0

2 ∞ sin(xt)f (t) dt. π 0 Up to constant factors, the function f (x) and the solution y(t) are the Fourier sine transform pair.

Solution: y(x) =

• References: H. Bateman and A. Erd´elyi (vol. 1, 1954), V. A. Ditkin and A. P. Prudnikov (1965).

b

7.

sin λ|x – t| y(t) dt = f (x),

–∞ < a < b < ∞.

a

1◦ . Let us remove the modulus in the integrand:

x

b

sin[λ(x – t)]y(t) dt + a

sin[λ(t – x)]y(t) dt = f (x).

(1)

x

Differentiating (1) with respect to x twice yields 2λy(x) – λ

2

x

sin[λ(x – t)]y(t) dt – λ

b

2

a

sin[λ(t – x)]y(t) dt = fxx (x).

(2)

x

Eliminating the integral terms from (1) and (2), we obtain the solution y(x) =

1 f (x) + λ2 f (x) . 2λ xx

(3)

2◦ . The right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries

b

sin[λ(t – a)]y(t) dt = f (a), a

b

sin[λ(b – t)]y(t) dt = f (b).

(4)

a

Substituting solution (3) into (4) followed by integrating by parts yields the desired conditions for f (x): sin[λ(b – a)]fx (b) – λ cos[λ(b – a)]f (b) = λf (a), (5) sin[λ(b – a)]fx (a) + λ cos[λ(b – a)]f (a) = –λf (b). The general form of the right-hand side of the integral equation is given by f (x) = F (x) + Ax + B,

(6)

where F (x) is an arbitrary bounded twice differentiable function, and the coefficients A and B are expressed in terms of F (a), F (b), Fx (a), and Fx (b) and can be determined by substituting formula (6) into conditions (5).

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8.

b

A sin λ|x – t| + B sin µ|x – t| y(t) dt = f (x),

–∞ < a < b < ∞.

a

Let us remove the modulus in the integrand and differentiate the equation with respect to x twice to obtain b 2 2(Aλ + Bµ)y(x) – (x). (1) Aλ sin λ|x – t| + Bµ2 sin µ|x – t| y(t) dt = fxx a

Eliminating the integral term with sin µ|x – t| from (1) with the aid of the original equation, we find that b 2(Aλ + Bµ)y(x) + A(µ2 – λ2 ) sin λ|x – t| y(t) dt = fxx (x) + µ2 f (x). (2) a

For Aλ + Bµ = 0, this is an equation of the form 3.5.7 and for Aλ + Bµ ≠ 0, this is an equation of the form 4.5.29. The right-hand side f (x) must satisfy certain relations, which can be obtained by setting x = a and x = b in the original equation (a similar procedure is used in 3.5.7). 9.

b

sin(λx) – sin(λt) y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = sin(λx). Solution: 1 d fx (x) y(x) = . 2λ dx cos(λx)

10.

The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). a sin(βx) – sin(µt) y(t) dt = f (x), β > 0, µ > 0. 0

This is a special case of equation 3.8.4 with g(x) = sin(βx) and λ = µ/β.

b

11.

sin3 λ|x – t| y(t) dt = f (x).

a

Using the formula sin3 β = – 14 sin 3β + 34 sin β, we arrive at an equation of the form 3.5.8: b 1 – 4 A sin 3λ|x – t| + 34 A sin λ|x – t| y(t) dt = f (x). a

12.

b n a

Ak sin λk |x – t| y(t) dt = f (x),

–∞ < a < b < ∞.

k=1

1◦ . Let us remove the modulus in the kth summand of the integrand: b x b Ik (x) = sin λk |x – t| y(t) dt = sin[λk (x – t)]y(t) dt + sin[λk (t – x)]y(t) dt. (1) a

a

x

Differentiating (1) with respect to x yields x Ik = λk cos[λk (x – t)]y(t) dt – λk a

Ik

= 2λk y(x) –

λ2k

b

cos[λk (t – x)]y(t) dt,

x

sin[λk (x – t)]y(t) dt – a

x

(2)

b

λ2k

sin[λk (t – x)]y(t) dt, x

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where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2), we find the relation between Ik and Ik : Ik = 2λk y(x) – λ2k Ik ,

Ik = Ik (x).

(3)

2◦ . With the aid of (1), the integral equation can be rewritten in the form n Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice and taking into account (3), we find that n n σ1 y(x) – Ak λ2k Ik = fxx (x), σ1 = 2 Ak λ k . k=1

(5)

k=1

Eliminating the integral In from (4) and (5) yields n–1 σ1 y(x) + Ak (λ2n – λ2k )Ik = fxx (x) + λ2n f (x).

(6)

k=1

Differentiating (6) with respect to x twice and eliminating In–1 from the resulting equation with the aid of (6), we obtain a similar equation whose left-hand side is a second-order n–2 linear differential operator (acting on y) with constant coefficients plus the sum Bk Ik . k=1

If we successively eliminate In–2 , In–3 , . . . , with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n – 1) with constant coefficients.

13.

3◦ . The right-hand side f (x) must satisfy certain conditions. To find these conditions, one should set x = a in the integral equation and its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.) b k sin x – sink t y(t) dt = f (x), 0 < k < 1. 0

This is a special case of equation 3.8.3 with g(x) = sink x. Solution: 1 d fx (x) y(x) = . 2k dx cos x sink–1 x The right-hand side f (x) must satisfy certain conditions. As follows from item 3◦ of equation 3.8.3, the admissible general form of the right-hand side is given by f (x) = F (x) + Ax + B, A = –Fx (b), B = 12 bFx (b) – F (0) – F (b) ,

15.

where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative). b y(t) dt = f (x), 0 < k < 1. |sin(λx) – sin(λt)|k a This is a special case of equation 3.8.7 with g(x) = sin(λx)+β, where β is an arbitrary number. a k sin(λx) – t y(t) dt = f (x).

16.

This is a special case of equation 3.8.5 with g(x) = k sin(λx). a x – k sin(λt) y(t) dt = f (x).

14.

0

0

This is a special case of equation 3.8.6 with g(x) = k sin(λt).

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3.5-3. Kernels Containing Tangent 17.

b

tan(λx) – tan(λt) y(t) dt = f (x).

a

18.

This is a special case of equation 3.8.3 with g(x) = tan(λx). Solution: 1 d 2 y(x) = cos (λx)fx (x) . 2λ dx The right-hand side f (x) of the integral equation must satisfy certain relations (see item 2◦ of equation 3.8.3). a tan(βx) – tan(µt) y(t) dt = f (x), β > 0, µ > 0.

19.

This is a special case of equation 3.8.4 with g(x) = tan(βx) and λ = µ/β. b k tan x – tank t y(t) dt = f (x), 0 < k < 1.

0

0

This is a special case of equation 3.8.3 with g(x) = tank x. Solution: 1 d y(x) = cos2 x cotk–1 xfx (x) . 2k dx The right-hand side f (x) must satisfy certain conditions. As follows from item 3◦ of equation 3.8.3, the admissible general form of the right-hand side is given by f (x) = F (x) + Ax + B, A = –Fx (b), B = 12 bFx (b) – F (0) – F (b) ,

21.

where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative). b y(t) dt = f (x), 0 < k < 1. k a |tan(λx) – tan(λt)| This is a special case of equation 3.8.7 with g(x) = tan(λx)+β, where β is an arbitrary number. a k tan(λx) – t y(t) dt = f (x).

22.

This is a special case of equation 3.8.5 with g(x) = k tan(λx). a x – k tan(λt) y(t) dt = f (x).

20.

0

0

This is a special case of equation 3.8.6 with g(x) = k tan(λt). 3.5-4. Kernels Containing Cotangent 23.

b

cot(λx) – cot(λt) y(t) dt = f (x).

a

24.

This is a special case of equation 3.8.3 with g(x) = cot(λx). b k cot x – cotk t y(t) dt = f (x), 0 < k < 1. a

This is a special case of equation 3.8.3 with g(x) = cotk x.

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3.5-5. Kernels Containing a Combination of Trigonometric Functions 25.

∞

cos(xt) + sin(xt) y(t) dt = f (x).

–∞

Solution: 1 y(x) = 2π

∞

cos(xt) + sin(xt) f (t) dt.

–∞

Up to constant factors, the function f (x) and the solution y(t) are the Hartley transform pair. • Reference: D. Zwillinger (1989). 26.

∞

sin(xt) – xt cos(xt) y(t) dt = f (x).

0

This equation can be reduced to a special case of equation 3.7.1 with ν = 32 . Solution: 2 ∞ sin(xt) – xt cos(xt) y(x) = f (t) dt. π 0 x2 t2 3.5-6. Equations Containing the Unknown Function of a Complicated Argument

π/2

y(ξ) dt = f (x),

27.

ξ = x sin t.

0

Schl¨omilch equation. Solution:

π/2 2 y(x) = fξ (ξ) dt , f (0) + x π 0

ξ = x sin t.

• References: E. T. Whittaker and G. N. Watson (1958), F. D. Gakhov (1977).

π/2

y(ξ) dt = f (x),

28.

ξ = x sink t.

0

Generalized Schl¨omilch equation. This is a special case of equation 3.5.29 for n = 0 and m = 0. Solution: x 1 2k k–1 d y(x) = x k xk sin t f (ξ) dt , ξ = x sink t. π dx 0

π/2

sinλ t y(ξ) dt = f (x),

29.

ξ = x sink t.

0

This is a special case of equation 3.5.29 for m = 0. Solution: x λ+1 2k k–λ–1 d λ+1 k k y(x) = sin t f (ξ) dt , x x π dx 0

ξ = x sink t.

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π/2

sinλ t cosm t y(ξ) dt = f (x),

30.

ξ = x sink t.

0 ◦

1 . Let λ > –1, m > –1, and k > 0. The transformation 2

z = xk ,

ζ = z sin2 t,

w(ζ) = ζ

λ–1 2

k y ζ2

leads to an equation of the form 1.1.43:

z

(z – ζ)

m–1 2

w(ζ) dζ = F (z),

F (z) = 2z

λ+m 2

k f z2 .

0

2◦ . Solution with –1 < m < 1: y(x) =

π(1 – m) k–λ–1 d λ+1 π/2 2k sinλ+1 t tanm t f (ξ) dt , sin x k x k π 2 dx 0

where ξ = x sink t. 3.5-7. A Singular Equation

t – x

2π

31.

cot 0

2

y(t) dt = f (x),

0 ≤ x ≤ 2π.

Here the integral is understood in the sense of the Cauchy principal value and the right-hand 2π

side is assumed to satisfy the condition f (t) dt = 0. 0 Solution: 2π 1 t–x y(x) = – 2 f (t) dt + C, cot 4π 0 2 where C is an arbitrary constant.

2π

It follows from the solution that y(t) dt = 2πC. 0 The equation and its solution form a Hilbert transform pair (in the asymmetric form). • Reference: F. D. Gakhov (1977).

3.6. Equations Whose Kernels Contain Combinations of Elementary Functions 3.6-1. Kernels Containing Hyperbolic and Logarithmic Functions

b

1.

lncosh(λx) – cosh(λt) y(t) dt = f (x).

a

This is a special case of equation 1.8.9 with g(x) = cosh(λx).

b

2.

lnsinh(λx) – sinh(λt) y(t) dt = f (x).

a

This is a special case of equation 1.8.9 with g(x) = sinh(λx).

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a

3. –a

sinh 12 A y(t) dt = f (x), ln 2 sinh 12 |x – t|

–a ≤ x ≤ a.

Solution with 0 < a < A: a 1 d y(x) = w(t, a)f (t) dt w(x, a) 2M (a) da –a ξ 1 a d 1 d – w(x, ξ) w(t, ξ)f (t) dt dξ 2 |x| dξ M (ξ) dξ –ξ ξ a 1 d w(x, ξ) – w(t, ξ) df (t) dξ, 2 dx |x| M (ξ) –ξ where the prime stands for the derivative with respect to the argument and –1 sinh 12 A M (ξ) = ln , sinh 12 ξ

cosh 12 x M (ξ) w(x, ξ) = √ . π 2 cosh ξ – 2 cosh x

• Reference: I. C. Gohberg and M. G. Krein (1967).

b

4.

lntanh(λx) – tanh(λt) y(t) dt = f (x).

a

This is a special case of equation 1.8.9 with g(x) = tanh(λx).

a

5. –a

ln coth 14 |x – t| y(t) dt = f (x),

Solution: y(x) =

–a ≤ x ≤ a.

a 1 d w(t, a)f (t) dt w(x, a) 2M (a) da –a ξ 1 a d 1 d – w(x, ξ) w(t, ξ)f (t) dt dξ 2 |x| dξ M (ξ) dξ –ξ ξ a 1 d w(x, ξ) – w(t, ξ) df (t) dξ, 2 dx |x| M (ξ) –ξ

where the prime stands for the derivative with respect to the argument and M (ξ) =

P–1/2 (cosh ξ) , Q–1/2 (cosh ξ)

w(x, ξ) =

1 √ , πQ–1/2 (cosh ξ) 2 cosh ξ – 2 cosh x

and P–1/2 (cosh ξ) and Q–1/2 (cosh ξ) are the Legendre functions of the first and second kind, respectively. • Reference: I. C. Gohberg and M. G. Krein (1967). 3.6-2. Kernels Containing Logarithmic and Trigonometric Functions

b

6.

lncos(λx) – cos(λt) y(t) dt = f (x).

a

This is a special case of equation 1.8.9 with g(x) = cos(λx).

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lnsin(λx) – sin(λt) y(t) dt = f (x).

b

7. a

This is a special case of equation 1.8.9 with g(x) = sin(λx).

a

8. –a

sin 12 A y(t) dt = f (x), ln 2 sin 12 |x – t|

–a ≤ x ≤ a.

Solution with 0 < a < A: a 1 d y(x) = w(t, a)f (t) dt w(x, a) 2M (a) da –a a ξ 1 1 d d – w(x, ξ) w(t, ξ)f (t) dt dξ 2 |x| dξ M (ξ) dξ –ξ ξ a 1 d w(x, ξ) – w(t, ξ) df (t) dξ, 2 dx |x| M (ξ) –ξ where the prime stands for the derivative with respect to the argument and 1 –1 sin 2 A M (ξ) = ln , sin 12 ξ

cos 12 ξ M (ξ) w(x, ξ) = √ . π 2 cos x – 2 cos ξ

• Reference: I. C. Gohberg and M. G. Krein (1967).

3.7. Equations Whose Kernels Contain Special Functions 3.7-1. Kernels Containing Bessel Functions

∞

tJν (xt)y(t) dt = f (x),

1. 0

ν > – 12 .

Here Jν is the Bessel function of the first kind. Solution: ∞ y(x) =

tJν (xt)f (t) dt. 0

The function f (x) and the solution y(t) are the Hankel transform pair. • Reference: V. A. Ditkin and A. P. Prudnikov (1965). 2.

b

Jν (λx) – Jν (λt) y(t) dt = f (x).

a

This is a special case of equation 3.8.3 with g(x) = Jν (λx), where Jν is the Bessel function of the first kind. 3.

Yν (λx) – Yν (λt) y(t) dt = f (x).

b

a

This is a special case of equation 3.8.3 with g(x) = Yν (λx), where Yν is the Bessel function of the second kind.

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3.7-2. Kernels Containing Modified Bessel Functions 4.

Iν (λx) – Iν (λt) y(t) dt = f (x).

b

a

This is a special case of equation 3.8.3 with g(x) = Iν (λx), where Iν is the modified Bessel function of the first kind. 5.

Kν (λx) – Kν (λt) y(t) dt = f (x).

b

a

This is a special case of equation 3.8.3 with g(x) = Kν (λx), where Kν is the modified Bessel function of the second kind (the Macdonald function).

∞

6.

√

zt Kν (zt)y(t) dt = f (z).

0

Here Kν is the modified Bessel function of the second kind. Up to a constant factor, the left-hand side of this equation is the Meijer transform of y(t) (z is treated as a complex variable). Solution: c+i∞ √ 1 y(t) = zt Iν (zt)f (z) dz. πi c–i∞ For specific f (z), one may use tables of Meijer integral transforms to calculate the integral. • Reference: V. A. Ditkin and A. P. Prudnikov (1965).

∞

7.

K0 |x – t| y(t) dt = f (x).

–∞

Here K0 is the modified Bessel function of the second kind. Solution: 2

∞ 1 d y(x) = – 2 –1 K0 |x – t| f (t) dt. 2 π dx –∞ • Reference: D. Naylor (1986). 3.7-3. Other Kernels

a

K

8. 0

√ 2 xt y(t) dt x+t

x+t

1

Here K(z) =

= f (x). dt

(1 – t2 )(1 – z 2 t2 )

0

is the complete elliptic integral of the first kind.

Solution: 4 d y(x) = – 2 π dx

a

x

tF (t) dt √ , t2 – x2

d F (t) = dt

0

t

sf (s) ds √ . t2 – s 2

• Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

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a

F

9. 0

β 2

,

β+1 2

, µ;

4x2 t2 (x2 + t2 )2

y(t) dt (x2 + t2 )β

= f (x).

Here 0 < a ≤ ∞, 0 < β < µ < β + 1, and F (a, b, c; z) is the hypergeometric function. Solution: a x2µ–2 tg(t) dt d y(x) = , 2 Γ(1 + β – µ) dx x (t – x2 )µ–β t 2µ–1 2 Γ(β) sin[(β – µ)π] 1–2β d s f (s) ds g(t) = . t πΓ(µ) dt 0 (t2 – s 2 )µ–β If a = ∞ and f (x) is a differentiable function, then the solution can be represented in the form

∞ d (xt)2µ ft (t) β 1–β 4x2 t2 y(x) = A F µ– , µ+ , µ + 1; 2 2 2 dt, dt 0 (x2 + t2 )2µ–β 2 2 (x + t ) where A =

Γ(β) Γ(2µ – β) sin[(β – µ)π] . πΓ(µ) Γ(1 + µ)

• Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

3.8. Equations Whose Kernels Contain Arbitrary Functions 3.8-1. Equations With Degenerate Kernel 1.

g1 (x)h1 (t) + g2 (x)h2 (t) y(t) dt = f (x).

b

a

This integral equation has solutions only if its right-hand side is representable in the form f (x) = A1 g1 (x) + A2 g2 (x),

A1 = const, A2 = const .

In this case, any function y = y(x) satisfying the normalization type conditions b b h1 (t)y(t) dt = A1 , h2 (t)y(t) dt = A2 a

(1)

(2)

a

is a solution of the integral equation. Otherwise, the equation has no solutions. 2.

b n a

gk (x)hk (t) y(t) dt = f (x).

k=0

This integral equation has solutions only if its right-hand side is representable in the form f (x) =

n

Ak gk (x),

(1)

k=0

where the Ak are some constants. In this case, any function y = y(x) satisfying the normalization type conditions b hk (t)y(t) dt = Ak (k = 1, . . . , n) (2) a

is a solution of the integral equation. Otherwise, the equation has no solutions.

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3.8-2. Equations Containing Modulus

b

|g(x) – g(t)| y(t) dt = f (x).

3. a

Let a ≤ x ≤ b and a ≤ t ≤ b; it is assumed in items 1◦ and 2◦ that 0 < gx (x) < ∞. 1◦ . Let us remove the modulus in the integrand:

x

g(x) – g(t) y(t) dt +

a

b

g(t) – g(x) y(t) dt = f (x).

(1)

x

Differentiating (1) with respect to x yields gx (x)

x

y(t) dt –

gx (x)

a

b

y(t) dt = fx (x).

(2)

x

Divide both sides of (2) by gx (x) and differentiate the resulting equation to obtain the solution 1 d fx (x) y(x) = . (3) 2 dx gx (x) 2◦ . Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = a and x = b, in (1), we obtain two corollaries

b

g(t) – g(a) y(t) dt = f (a),

a

b

g(b) – g(t) y(t) dt = f (b).

(4)

a

Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f (x): f (b) g(b) – g(a) x = f (a) + f (b), gx (b) f (a) g(a) – g(b) x = f (a) + f (b). gx (a)

(5)

Let us point out a useful property of these constraints: fx (b)gx (a) + fx (a)gx (b) = 0. Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation: f (x) = F (x) + Ax + B, (6) where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative), and the coefficients A and B are given by gx (a)Fx (b) + gx (b)Fx (a) , gx (a) + gx (b) g(b) – g(a) B = – 12 A(a + b) – 12 F (a) + F (b) – A + Fx (a) . 2gx (a) A=–

3◦ . If g(x) is representable in the form g(x) = O(x – a)k with 0 < k < 1 in the vicinity of the point x = a (in particular, the derivative gx is unbounded as x → a), then the solution of the integral equation is given by formula (3) as well. In this case, the right-hand side of the integral equation must satisfy the conditions f (a) + f (b) = 0,

fx (b) = 0.

(7)

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As before, the right-hand side of the integral equation is given by (6), with A = –Fx (b), B = 12 (a + b)Fx (b) – F (a) – F (b) . 4◦ . For gx (a) = 0, the right-hand side of the integral equation must satisfy the conditions fx (a) = 0, g(b) – g(a) fx (b) = f (a) + f (b) gx (b). As before, the right-hand side of the integral equation is given by (6), with A = –Fx (a), 4.

B=

1 2

g(b) – g(a) (a + b)Fx (a) – F (a) – F (b) + Fx (b) – Fx (a) . 2gx (b)

a

g(x) – g(λt) y(t) dt = f (x),

λ > 0.

0

Assume that 0 ≤ x ≤ a, 0 ≤ t ≤ a and 0 < gx (x) < ∞. 1◦ . Let us remove the modulus in the integrand: x/λ a g(x) – g(λt) y(t) dt + g(λt) – g(x) y(t) dt = f (x).

(1)

x/λ

0

Differentiating (1) with respect to x yields x/λ gx (x) y(t) dt – gx (x) 0

a

y(t) dt = fx (x).

(2)

x/λ

Let us divide both sides of (2) by gx (x) and differentiate the resulting equation to obtain y(x/λ) = 12 λ fx (x)/gx (x) x . Substituting x by λx yields the solution λ d fz (z) y(x) = , z = λx. (3) 2 dz gz (z) 2◦ . Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = 0 in (1) and (2), we obtain two corollaries a a g(λt) – g(0) y(t) dt = f (0), gx (0) y(t) dt = –fx (0). (4) 0

0

Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f (x): fx (0)gx (λa) + fx (λa)gx (0) = 0, f (λa) g(λa) – g(0) x = f (0) + f (λa). gx (λa)

(5)

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation: f (x) = F (x) + Ax + B, (6) where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative), and the coefficients A and B are given by gx (0)Fx (λa) + gx (λa)Fx (0) , gx (0) + gx (λa) g(λa) – g(0) B = – 12 Aaλ – 12 F (0) + F (λa) – A + Fx (0) . 2gx (0) A=–

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3◦ . If g(x) is representable in the form g(x) = O(x)k with 0 < k < 1 in the vicinity of the point x = 0 (in particular, the derivative gx is unbounded as x → 0), then the solution of the integral equation is given by formula (3) as well. In this case, the right-hand side of the integral equation must satisfy the conditions fx (λa) = 0.

f (0) + f (λa) = 0,

(7)

As before, the right-hand side of the integral equation is given by (6), with A = –Fx (λa), 5.

B=

1 2

aλFx (λa) – F (0) – F (λa) .

a

g(x) – t y(t) dt = f (x).

0

Assume that 0 ≤ x ≤ a, 0 ≤ t ≤ a; g(0) = 0, and 0 < gx (x) < ∞. 1◦ . Let us remove the modulus in the integrand:

g(x)

g(x) – t y(t) dt +

a

t – g(x) y(t) dt = f (x).

(1)

g(x)

0

Differentiating (1) with respect to x yields gx (x)

g(x)

y(t) dt –

gx (x)

a

y(t) dt = fx (x).

(2)

g(x)

0

Let us divide both sides of (2) by gx (x) and differentiate the resulting equation to obtain 2gx (x)y g(x) = fx (x)/gx (x) x . Hence, we find the solution: y(x) =

1 d fz (z) , 2gz (z) dz gz (z)

z = g –1 (x),

(3)

where g –1 is the inverse of g. 2◦ . Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = 0 in (1) and (2), we obtain two corollaries

a

ty(t) dt = f (0),

gx (0)

0

a

y(t) dt = –fx (0).

(4)

0

Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f (x): fx (0)gx (xa ) + fx (xa )gx (0) = 0, g(xa )

xa = g –1 (a);

fx (xa ) = f (0) + f (xa ). gx (xa )

(5)

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation in question: f (x) = F (x) + Ax + B,

(6)

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where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative), and the coefficients A and B are given by gx (0)Fx (xa ) + gx (xa )Fx (0) , xa = g –1 (a), gx (0) + gx (xa ) g(xa ) B = – 12 Axa – 12 F (0) + F (xa ) – A + Fx (0) . 2gx (0) A=–

3◦ . If g(x) is representable in the vicinity of the point x = 0 in the form g(x) = O(x)k with 0 < k < 1 (i.e., the derivative gx is unbounded as x → 0), then the solution of the integral equation is given by formula (3) as well. In this case, the right-hand side of the integral equation must satisfy the conditions fx (xa ) = 0.

f (0) + f (xa ) = 0,

(7)

As before, the right-hand side of the integral equation is given by (6), with A = –Fx (xa ), B = 12 xa Fx (xa ) – F (0) – F (xa ) . 6.

a

x – g(t) y(t) dt = f (x).

0

Assume that 0 ≤ x ≤ a, 0 ≤ t ≤ a; g(0) = 0, and 0 < gx (x) < ∞. 1◦ . Let us remove the modulus in the integrand:

g –1 (x)

x – g(t) y(t) dt +

g(t) – x y(t) dt = f (x),

a

(1)

g –1 (x)

0

where g –1 is the inverse of g. Differentiating (1) with respect to x yields

g –1 (x)

a

y(t) dt – g –1 (x)

0

y(t) dt = fx (x).

(2)

Differentiating the resulting equation yields 2y g –1 (x) = gx (x)fxx (x). Hence, we obtain the solution y(x) = 12 gz (z)fzz (z), z = g(x). (3) 2◦ . Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy certain relations. By setting x = 0 in (1) and (2), we obtain two corollaries a a g(t)y(t) dt = f (0), y(t) dt = –fx (0). (4) 0

0

Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f (x): xa fx (xa ) = f (0) + f (xa ),

fx (0) + fx (xa ) = 0,

xa = g(a).

(5)

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation:

A=

– 12

Fx (0) + Fx (xa ) ,

f (x) = F (x) + Ax + B, B = 12 xa Fx (0) – F (xa ) – F (0) ,

xa = g(a),

where F (x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

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b

7. a

y(t) |g(x) – g(t)|k

dt = f (x),

0 < k < 1.

Let gx ≠ 0. The transformation z = g(x),

τ = g(t),

leads to an equation of the form 3.1.30: B w(τ ) dτ = F (z), k A |z – τ |

w(τ ) =

1 y(t) gt (t)

A = g(a),

B = g(b),

where F = F (z) is the function which is obtained from z = g(x) and F = f (x) by eliminating x.

1

8.

y(t) |g(x) – h(t)|k

0

dt = f (x),

0 < k < 1.

Let g(0) = 0, g(1) = 1, gx > 0; h(0) = 0, h(1) = 1, and ht > 0. The transformation z = g(x),

τ = h(t),

w(τ ) =

1 y(t) ht (t)

leads to an equation of the form 3.1.29: 1 0

w(τ ) dτ = F (z), |z – τ |k

where F = F (z) is the function which is obtained from z = g(x) and F = f (x) by eliminating x.

b

y(t) ln |g(x) – g(t)| dt = f (x).

9. a

Let gx ≠ 0. The transformation z = g(x),

τ = g(t),

leads to Carleman’s equation 3.4.2: B ln |z – τ |w(τ ) dτ = F (z),

w(τ ) =

1

y(t) gt (t)

A = g(a),

B = g(b),

A

where F = F (z) is the function which is obtained from z = g(x) and F = f (x) by eliminating x.

1

y(t) ln |g(x) – h(t)| dt = f (x).

10. 0

Let g(0) = 0, g(1) = 1, gx > 0; h(0) = 0, h(1) = 1, and ht > 0. The transformation z = g(x),

τ = h(t),

w(τ ) =

1

y(t) ht (t)

leads to an equation of the form 3.4.2: 1 ln |z – τ |w(τ ) dτ = F (z), 0

where F = F (z) is the function which is obtained from z = g(x) and F = f (x) by eliminating x.

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3.8-3. Equations With Difference Kernel: K(x, t) = K(x – t)

∞

11.

K(x – t)y(t) dt = Axn ,

n = 0, 1, 2, . . .

–∞

1◦ . Solution with n = 0: y(x) =

A , B

∞

K(x) dx.

B= –∞

2◦ . Solution with n = 1: y(x) =

A AC x+ 2 , B B

∞

B=

K(x) dx,

∞

C=

–∞

xK(x). –∞

3◦ . Solution with n ≥ 2: y(x) =

∞

12.

dn Aeλx , dλn B(λ) λ=0

∞

B(λ) =

K(x)e–λx dx.

–∞

K(x – t)y(t) dt = Aeλx .

–∞

Solution: y(x) =

∞

13.

A λx e , B

K(x – t)y(t) dt = Axn eλx ,

∞

B=

K(x)e–λx dx.

–∞

n = 1, 2, . . .

–∞

1◦ . Solution with n = 1: A λx AC λx xe + 2 e , B B ∞ ∞ –λx B= K(x)e dx, C = xK(x)e–λx dx. y(x) =

–∞

–∞

2◦ . Solution with n ≥ 2: dn Aeλx y(x) = , dλn B(λ)

∞

B(λ) =

K(x)e–λx dx.

–∞

∞

K(x – t)y(t) dt = A cos(λx) + B sin(λx).

14. –∞

Solution: AIc + BIs BIc – AIs y(x) = cos(λx) + sin(λx), 2 2 Ic + Is I2 + I2 ∞ c ∞ s Ic = K(z) cos(λz) dz, Is = K(z) sin(λz) dz. –∞

–∞

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∞

K(x – t)y(t) dt = f (x).

15. –∞

The Fourier transform is used to solve this equation. 1◦ . Solution:

∞

f˜(u) iux e du, ˜ –∞ K(u) ∞ ∞ 1 1 ˜ f˜(u) = √ f (x)e–iux dx, K(u) = √ K(x)e–iux dx. 2π –∞ 2π –∞ y(x) =

1 2π

The following statement is valid. Let f (x) ∈ L2 (–∞, ∞) and K(x) ∈ L1 (–∞, ∞). Then for a solution y(x) ∈ L2 (–∞, ∞) of the integral equation to exist, it is necessary and sufficient ˜ that f˜(u)/K(u) ∈ L2 (–∞, ∞). 2◦ . Let the function P (s) defined by the formula 1 = P (s)

∞

e–st K(t) dt

–∞

be a polynomial of degree n with real roots of the form s s s P (s) = 1 – 1– ... 1 – . a1 a2 an Then the solution of the integral equation is given by y(x) = P (D)f (x),

D=

d . dx

• References: I. I. Hirschman and D. V. Widder (1955), V. A. Ditkin and A. P. Prudnikov (1965).

∞

K(x – t)y(t) dt = f (x).

16. 0

The Wiener–Hopf equation of the first kind. This equation is discussed in Subsection 10.5-1 in detail. 3.8-4. Other Equations of the Form

∞

17.

b a

K(x, t)y(t) dt = F (x)

K(ax – t)y(t) dt = Aeλx .

–∞

Solution:

λ A y(x) = exp x , B a

∞

B= –∞

λ K(z) exp – z dz. a

∞

K(ax – t)y(t) dt = f (x).

18. –∞

The substitution z = ax leads to an equation of the form 3.8.15:

∞

K(z – t)y(t) dt = f (z/a). –∞

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∞

19.

K(ax + t)y(t) dt = Aeλx .

–∞

Solution: y(x) =

λ A exp – x , B a

∞

B= –∞

λ K(z) exp – z dz. a

∞

K(ax + t)y(t) dt = f (x).

20. –∞

The transformation τ = –t, z = ax, y(t) = Y (τ ) leads to an equation of the form 3.8.15: ∞ K(z – τ )Y (τ ) dt = f (z/a). –∞

∞

21.

[eβt K(ax + t) + eµt M (ax – t)]y(t) dt = Aeλx .

–∞

Solution: y(x) = A

Ik (q)epx – Im (p)eqx , Ik (p)Ik (q) – Im (p)Im (q)

where

∞

Ik (q) =

p=–

K(z)e(β+q)z dz,

∞

Im (q) =

–∞

λ – β, a

q=

λ – µ, a

M (z)e–(µ+q)z dz.

–∞

∞

g(xt)y(t) dt = f (x).

22. 0

By setting x = ez ,

y(t) = eτ w(τ ),

t = e–τ ,

g(ξ) = G(ln ξ),

f (ξ) = F (ln ξ),

we arrive at an integral equation with difference kernel of the form 3.8.15: ∞ G(z – τ )w(τ ) dτ = F (z). –∞

23.

∞

g

x

t By setting

y(t) dt = f (x).

0

x = ez ,

t = eτ ,

y(t) = e–τ w(τ ),

g(ξ) = G(ln ξ),

f (ξ) = F (ln ξ),

we arrive at an integral equation with difference kernel of the form 3.8.15: ∞ G(z – τ )w(τ ) dτ = F (z). –∞

24.

∞

g xβ tλ y(t) dt = f (x),

β > 0,

λ > 0.

0

By setting x = ez/β ,

t = e–τ /λ ,

y(t) = eτ /λ w(τ ),

g(ξ) = G(ln ξ),

f (ξ) =

1 λ F (β

ln ξ),

we arrive at an integral equation with difference kernel of the form 3.8.15: ∞ G(z – τ )w(τ ) dτ = F (z). –∞

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25.

∞

g

xβ

tλ By setting

y(t) dt = f (x),

β > 0,

λ > 0.

0

x = ez/β ,

t = eτ /λ ,

y(t) = e–τ /λ w(τ ),

g(ξ) = G(ln ξ),

f (ξ) =

1 λ F (β

ln ξ),

we arrive at an integral equation with difference kernel of the form 3.8.15: ∞ G(z – τ )w(τ ) dτ = F (z). –∞

3.8-5. Equations of the Form

b a

K(x, t)y(· · ·) dt = F (x)

b

f (t)y(xt) dt = Ax + B.

26. a

Solution: y(x) =

A B x+ , I1 I0

b

I0 =

f (t) dt,

b

I1 =

tf (t) dt.

a

a

b

f (t)y(xt) dt = Axβ .

27. a

Solution:

A y(x) = xβ , B

b

f (t)tβ dt.

B= a

b

f (t)y(xt) dt = A ln x + B.

28. a

Solution: y(x) = p ln x + q, where A p= , I0

B AIl q= – 2 , I0 I0

I0 =

b

f (t) dt,

b

Il =

f (t) ln t dt.

a

a

b

f (t)y(xt) dt = Axβ ln x.

29. a

Solution: y(x) = pxβ ln x + qxβ , where p=

A , I1

q=–

AI2 , I12

b

f (t)tβ dt,

I1 = a

b

f (t)tβ ln t dt.

I2 = a

b

f (t)y(xt) dt = A cos(ln x).

30. a

Solution: AIc AIs cos(ln x) + 2 sin(ln x), Ic2 + Is2 Ic + Is2 b b Ic = f (t) cos(ln t) dt, Is = f (t) sin(ln t) dt. y(x) =

a

a

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b

f (t)y(xt) dt = A sin(ln x).

31. a

Solution: AIs AIc cos(ln x) + 2 sin(ln x), Ic2 + Is2 Ic + Is2 b b Ic = f (t) cos(ln t) dt, Is = f (t) sin(ln t) dt. y(x) = –

a

a

b

f (t)y(xt) dt = Axβ cos(ln x) + Bxβ sin(ln x).

32. a

Solution: y(x) = pxβ cos(ln x) + qxβ sin(ln x), where p=

AIc – BIs , Ic2 + Is2

b

f (t)tβ cos(ln t) dt,

Ic =

AIs + BIc , Ic2 + Is2 b Is = f (t)tβ sin(ln t) dt.

q=

a

a

b

f (t)y(x – t) dt = Ax + B.

33. a

Solution: y(x) = px + q, where p=

A , I0

q=

AI1 B + , I0 I02

b

I0 =

f (t) dt,

b

I1 =

a

tf (t) dt. a

b

f (t)y(x – t) dt = Aeλx .

34. a

Solution: y(x) =

A λx e , B

B=

b

f (t) exp(–λt) dt. a

b

f (t)y(x – t) dt = A cos(λx).

35. a

Solution: AIs AIc sin(λx) + 2 cos(λx), Ic2 + Is2 Ic + Is2 b b Ic = f (t) cos(λt) dt, Is = f (t) sin(λt) dt. y(x) = –

a

a

b

f (t)y(x – t) dt = A sin(λx).

36. a

Solution: AIc AIs sin(λx) + 2 cos(λx), Ic2 + Is2 Ic + Is2 b b Ic = f (t) cos(λt) dt, Is = f (t) sin(λt) dt. y(x) =

a

a

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b

f (t)y(x – t) dt = eµx (A sin λx + B cos λx).

37. a

Solution: y(x) = eµx (p sin λx + q cos λx), where p=

AIc – BIs , Ic2 + Is2

AIs + BIc , Ic2 + Is2 b Is = f (t)e–µt sin(λt) dt. q=

b

f (t)e–µt cos(λt) dt,

Ic = a

a

b

f (t)y(x – t) dt = g(x).

38. a

1◦ . For g(x) =

n

Ak exp(λk x), the solution of the equation has the form

k=1

y(x) =

n Ak exp(λk x), Bk

b

Bk =

f (t) exp(–λk t) dt. a

k=1

2◦ . For a polynomial right-hand side, g(x) =

n

Ak xk , the solution has the form

k=0

y(x) =

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. n 3◦ . For g(x) = eλx Ak xk , the solution has the form k=0

y(x) = eλx

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. n 4◦ . For g(x) = Ak cos(λk x), the solution has the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. n 5◦ . For g(x) = Ak sin(λk x), the solution has the form k=1

y(x) =

n k=1

Bk cos(λk x) +

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients.

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6◦ . For g(x) = cos(λx)

n

Ak xk , the solution has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n 7◦ . For g(x) = sin(λx) Ak xk , the solution has the form k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n 8◦ . For g(x) = eµx Ak cos(λk x), the solution has the form k=1

y(x) = eµx

n

Bk cos(λk x) + eµx

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. n 9◦ . For g(x) = eµx Ak sin(λk x), the solution has the form k=1 µx

y(x) = e

n

µx

Bk cos(λk x) + e

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. n 10◦ . For g(x) = cos(λx) Ak exp(µk x), the solution has the form k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. n 11◦ . For g(x) = sin(λx) Ak exp(µk x), the solution has the form k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients.

b

f (t)y(x + βt) dt = Ax + B.

39. a

Solution: y(x) = px + q, where p=

A , I0

q=

B AI1 β – , I0 I02

I0 =

b

f (t) dt, a

I1 =

b

tf (t) dt. a

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b

f (t)y(x + βt) dt = Aeλx .

40. a

Solution: A λx e , B

y(x) =

b

B=

f (t) exp(λβt) dt. a

b

f (t)y(x + βt) dt = A sin λx + B cos λx.

41. a

Solution: y(x) = p sin λx + q cos λx, where p= Ic =

AIc + BIs , Ic2 + Is2

BIc – AIs , Ic2 + Is2 b Is = f (t) sin(λβt) dt. q=

b

f (t) cos(λβt) dt, a

a

1

y(ξ) dt = f (x),

42.

ξ = g(x)t.

0

Assume that g(0) = 0, g(1) = 1, and gx ≥ 0. 1

1◦ . The substitution z = g(x) leads to an equation of the form 3.1.41: y(zt) dt = F (z), 0 where the function F (z) is obtained from z = g(x) and F = f (x) by eliminating x. 2◦ . Solution y = y(z) in the parametric form: y(z) =

g(x) f (x) + f (x), gx (x) x

z = g(x).

1

tλ y(ξ) dt = f (x),

43.

ξ = g(x)t.

0

Assume that g(0) = 0, g(1) = 1, and gx ≥ 0. 1

1◦ . The substitution z = g(x) leads to an equation of the form 3.1.42: tλ y(zt) dt = F (z), 0 where the function F (z) is obtained from z = g(x) and F = f (x) by eliminating x. 2◦ . Solution y = y(z) in the parametric form: y(z) =

g(x) f (x) + (λ + 1)f (x), gx (x) x

z = g(x).

b

f (t)y(ξ) dt = Axβ ,

44.

ξ = xϕ(t).

a

Solution: A y(x) = xβ , B

B=

b

β f (t) ϕ(t) dt.

(1)

a

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b

f (t)y(ξ) dt = g(x),

45.

ξ = xϕ(t).

a

1◦ . For g(x) =

n

Ak xk , the solution of the equation has the form

k=0

y(x) =

n Ak k x , Bk

n

k f (t) ϕ(t) dt.

b

λ f (t) ϕ(t) k dt.

a

k=0

2◦ . For g(x) =

b

Bk =

Ak xλk , the solution has the form

k=0

n Ak λk y(x) = x , Bk

Bk = a

k=0

3◦ . For g(x) = ln x

n

Ak xk , the solution has the form

k=0

y(x) = ln x

n

Bk xk +

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. 4◦ . For g(x) =

n

Ak ln x)k , the solution has the form

k=0

y(x) =

n

Bk ln x)k ,

k=0

where the constants Bk are found by the method of undetermined coefficients. 5◦ . For g(x) =

n

Ak cos(λk ln x), the solution has the form

k=1

y(x) =

n

Bk cos(λk ln x) +

k=1

n

Ck sin(λk ln x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. 6◦ . For g(x) =

n

Ak sin(λk ln x), the solution has the form

k=1

y(x) =

n k=1

Bk cos(λk ln x) +

n

Ck sin(λk ln x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients.

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b

f (t)y(ξ) dt = g(x),

46.

ξ = x + ϕ(t).

a

1◦ . For g(x) =

n

Ak exp(λk x), the solution of the equation has the form

k=1

y(x) =

n Ak exp(λk x), Bk

b

Bk =

f (t) exp λk ϕ(t) dt.

a

k=1

2◦ . For a polynomial right-hand side, g(x) =

n

Ak xk , the solution has the form

k=0

y(x) =

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. n 3◦ . For g(x) = eλx Ak xk , the solution has the form k=0

y(x) = eλx

n

Bk xk ,

k=0

where the constants Bk are found by the method of undetermined coefficients. n 4◦ . For g(x) = Ak cos(λk x) the solution has the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. n 5◦ . For g(x) = Ak sin(λk x), the solution has the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. n Ak xk , the solution has the form 6◦ . For g(x) = cos(λx) k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients. n Ak xk , the solution has the form 7◦ . For g(x) = sin(λx) k=0

y(x) = cos(λx)

n k=0

Bk xk + sin(λx)

n

Ck xk ,

k=0

where the constants Bk and Ck are found by the method of undetermined coefficients.

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8◦ . For g(x) = eµx

n

Ak cos(λk x), the solution has the form

k=1

y(x) = eµx

n

Bk cos(λk x) + eµx

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. 9◦ . For g(x) = eµx

n

Ak sin(λk x), the solution has the form

k=1

y(x) = eµx

n k=1

Bk cos(λk x) + eµx

n

Ck sin(λk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. 10◦ . For g(x) = cos(λx)

n

Ak exp(µk x), the solution has the form

k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients. 11◦ . For g(x) = sin(λx)

n

Ak exp(µk x), the solution has the form

k=1

y(x) = cos(λx)

n k=1

Bk exp(µk x) + sin(λx)

n

Bk exp(µk x),

k=1

where the constants Bk and Ck are found by the method of undetermined coefficients.

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Chapter 4

Linear Equations of the Second Kind With Constant Limits of Integration Notation: f = f (x), g = g(x), h = h(x), v = v(x), w = w(x), K = K(x) are arbitrary functions; A, B, C, D, E, a, b, c, l, α, β, γ, δ, µ, and ν are arbitrary parameters; n is a nonnegative integer; and i is the imaginary unit. Preliminary remarks. A number λ is called a characteristic value of the integral equation y(x) – λ

b

K(x, t)y(t) dt = f (x) a

if there exist nontrivial solutions of the corresponding homogeneous equation (with f (x) ≡ 0). The nontrivial solutions themselves are called the eigenfunctions of the integral equation corresponding to the characteristic value λ. If λ is a characteristic value, the number 1/λ is called an eigenvalue of the integral equation. A value of the parameter λ is said to be regular if for this value the homogeneous equation has only the trivial solution. Sometimes the characteristic values and the eigenfunctions of a Fredholm integral equation are called the characteristic values and the eigenfunctions of the kernel K(x, t). In the above equation, it is usually assumed that a ≤ x ≤ b.

4.1. Equations Whose Kernels Contain Power-Law Functions 4.1-1. Kernels Linear in the Arguments x and t

1.

b y(x) – λ (x – t)y(t) dt = f (x). a

Solution: y(x) = f (x) + λ(A1 x + A2 ), where A1 =

12f1 + 6λ (f1 ∆2 – 2f2 ∆1 ) –12f2 + 2λ (3f2 ∆2 – 2f1 ∆3 ) , A2 = , λ2 ∆41 + 12 λ2 ∆41 + 12 b b f1 = f (x) dx, f2 = xf (x) dx, ∆n = bn – an . a

a

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2.

b y(x) – λ (x + t)y(t) dt = f (x). a

The characteristic values of the equation: 6(b + a) + 4 3(a2 + ab + b2 ) λ1 = , (a – b)3

6(b + a) – 4 3(a2 + ab + b2 ) λ2 = . (a – b)3

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x + A2 ), where A1 =

12f1 – 6λ(f1 ∆2 – 2f2 ∆1 ) 12f2 – 2λ(3f2 ∆2 – 2f1 ∆3 ) , A2 = , 4 2 12 – 12λ∆2 – λ ∆1 12 – 12λ∆2 – λ2 ∆41 b b f1 = f (x) dx, f2 = xf (x) dx, ∆n = bn – an . a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x), where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 : 1 b+a y1 (x) = x + – . λ1 (b – a) 2 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively.

3.

4◦ . The equation has no multiple characteristic values. b y(x) – λ (Ax + Bt)y(t) dt = f (x). a

The characteristic values of the equation: 3(A + B)(b + a) ± 9(A – B)2 (b + a)2 + 48AB(a2 + ab + b2 ) λ1,2 = . AB(a – b)3 1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x + A2 ), where the constants A1 and A2 are given by A1 =

12Af1 – 6ABλ(f1 ∆2 – 2f2 ∆1 ) 12Bf2 – 2ABλ(3f2 ∆2 – 2f1 ∆3 ) , A2 = , 12 – 6(A + B)λ∆2 – ABλ2 ∆41 12 – 6(A + B)λ∆2 – ABλ2 ∆41 b b f1 = f (x) dx, f2 = xf (x) dx, ∆n = bn – an . a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x), where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 : 1 b+a y1 (x) = x + – . λ1 A(b – a) 2 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively.

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4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = 4 is double: (A + B)(b2 – a2 ) y(x) = f (x) + Cy∗ (x),

4.

where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ : (A – B)(b + a) y∗ (x) = x – . 4A b y(x) – λ [A + B(x – t)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = 1. Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.8. 5.

b y(x) – λ (Ax + Bt + C)y(t) dt = f (x). a

This is a special case of equation 4.9.7 with g(x) = x and h(t) = 1. Solution: y(x) = f (x) + λ(A1 x + A2 ), where A1 and A2 are the constants determined by the formulas presented in 4.9.7. 6.

b

|x – t| y(t) dt = f (x).

y(x) + A a

This is a special case of equation 4.9.36 with g(t) = A. 1◦ . The function y = y(x) obeys the following second-order linear nonhomogeneous ordinary differential equation with constant coefficients: yxx + 2Ay = fxx (x).

(1)

The boundary conditions for (1) have the form (see 4.9.36) yx (a) + yx (b) = fx (a) + fx (b),

(2)

y(a) + y(b) + (b – a)yx (a) = f (a) + f (b) + (b – a)fx (a).

Equation (1) under the boundary conditions (2) determines the solution of the original integral equation. 2◦ . For A < 0, the general solution of equation (1) is given by x y(x) = C1 cosh(kx) + C2 sinh(kx) + f (x) + k sinh[k(x – t)]f (t) dt,

k=

√

–2A,

(3)

a

where C1 and C2 are arbitrary constants. For A > 0, the general solution of equation (1) is given by x y(x) = C1 cos(kx) + C2 sin(kx) + f (x) – k sin[k(x – t)]f (t) dt,

k=

√

2A.

(4)

a

The constants C1 and C2 in solutions (3) and (4) are determined by conditions (2).

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3◦ . In the special case a = 0 and A > 0, the solution of the integral equation is given by formula (4) with Is (1 + cos λ) – Ic (λ + sin λ) Is sin λ + Ic (1 + cos λ) , C2 = k , 2 + 2 cos λ + λ sin λ 2 + 2 cos λ + λ sin λ b b √ k = 2A, λ = bk, Is = sin[k(b – t)]f (t) dt, Ic = cos[k(b – t)]f (t) dt. C1 = k

0

0

4.1-2. Kernels Quadratic in the Arguments x and t

7.

b y(x) – λ (x2 + t2 )y(t) dt = f (x). a

The characteristic values of the equation: λ1 =

1 3 3 (b

– a3 ) +

1 1 5 5 (b

, – a5 )(b – a)

λ2 =

1 3 3 (b

– a3 ) –

1 1 5 5 (b

. – a5 )(b – a)

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x2 + A2 ), where the constants A1 and A2 are given by f1 – λ 13 f1 ∆3 – f2 ∆1 f2 – λ 13 f2 ∆3 – 15 f1 ∆5 A1 = 2 1 2 1 , A2 = 2 1 2 1 , λ 9 ∆3 – 5 ∆1 ∆5 – 23 λ∆3 + 1 λ 9 ∆3 – 5 ∆1 ∆5 – 23 λ∆3 + 1 b b f1 = f (x) dx, f2 = x2 f (x) dx, ∆n = bn – an . a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

b5 – a5 , 5(b – a)

y1 (x) = x2 +

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0: y(x) = f (x) + Cy2 (x),

y2 (x) = x2 –

b5 – a5 . 5(b – a)

where C is an arbitrary constant and y2 (x) is an eigenfunction of the equation corresponding to the characteristic value λ2 . 4◦ . The equation has no multiple characteristic values.

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8.

b y(x) – λ (x2 – t2 )y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 = ±

1 1 3 9 (b

–

a3 )2

.

– 15 (b5 – a5 )(b – a)

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x2 + A2 ), where the constants A1 and A2 are given by f1 + λ 13 f1 ∆3 – f2 ∆1 –f2 + λ 13 f2 ∆3 – 15 f1 ∆5 A1 = 2 1 , A2 = , λ 5 ∆1 ∆5 – 19 ∆22 + 1 λ2 15 ∆1 ∆5 – 19 ∆22 + 1 b b f1 = f (x) dx, f2 = x2 f (x) dx, ∆n = bn – an . a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = x2 +

3 – λ1 (b3 – a3 ) , 3λ1 (b – a)

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively.

9.

4◦ . The equation has no multiple characteristic values. b y(x) – λ (Ax2 + Bt2 )y(t) dt = f (x). a

The characteristic values of the equation: 1 1 4 2 2 (A + B)∆ ± 3 3 9 (A – B) ∆3 + 5 AB∆1 ∆5 λ1,2 = , 2AB 19 ∆23 – 15 ∆1 ∆5

∆n = bn – an .

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x2 + A2 ), where the constants A1 and A2 are given by

Af1 – ABλ 13 f1 ∆3 – f2 ∆1 A1 = , ABλ2 19 ∆23 – 15 ∆1 ∆5 – 13 (A + B)λ∆3 + 1 1 Bf2 – ABλ 3 f2 ∆3 – 15 f1 ∆5 A2 = , ABλ2 19 ∆23 – 15 ∆1 ∆5 – 13 (A + B)λ∆3 + 1 b b f1 = f (x) dx, f2 = x2 f (x) dx. a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = x2 +

3 – λ1 A(b3 – a3 ) , 3λ1 A(b – a)

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 .

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3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where λ∗ =

6 is the double (A + B)(b3 – a3 )

characteristic value: y(x) = f (x) + C1 y∗ (x), where C1 is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ : (A – B)(b3 – a3 ) y∗ (x) = x2 – . 6A(b – a) 10.

b y(x) – λ (xt – t2 )y(t) dt = f (x). a

This is a special case of equation 4.9.8 with A = 0, B = 1, and h(t) = t. Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.8. 11.

b y(x) – λ (x2 – xt)y(t) dt = f (x). a

This is a special case of equation 4.9.10 with A = 0, B = 1, and h(x) = x. Solution: y(x) = f (x) + λ(E1 x2 + E2 x), where E1 and E2 are the constants determined by the formulas presented in 4.9.10. 12.

b y(x) – λ (Bxt + Ct2 )y(t) dt = f (x). a

This is a special case of equation 4.9.9 with A = 0 and h(t) = t. Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.9. 13.

b y(x) – λ (Bx2 + Cxt)y(t) dt = f (x). a

This is a special case of equation 4.9.11 with A = 0 and h(x) = x. Solution: y(x) = f (x) + λ(A1 x2 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.11. 14.

b y(x) – λ (Axt + Bx2 + Cx + D)y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = Bx2 + Cx + D, h1 (t) = 1, g2 (x) = x, and h2 (t) = At. Solution: y(x) = f (x) + λ[A1 (Bx2 + Cx + D) + A2 x], where A1 and A2 are the constants determined by the formulas presented in 4.9.18.

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15.

b y(x) – λ (Ax2 + Bt2 + Cx + Dt + E)y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = Ax2 + Cx, h1 (t) = 1, g2 (x) = 1, and h2 (t) = Bt2 + Dt + E. Solution: y(x) = f (x) + λ[A1 (Ax2 + Cx) + A2 ], where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 16.

b y(x) – λ [Ax + B + (Cx + D)(x – t)]y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = Cx2 + (A + D)x + B, h1 (t) = 1, g2 (x) = Cx + D, and h2 (t) = –t. Solution: y(x) = f (x) + λ[A1 (Cx2 + Ax + Dx + B) + A2 (Cx + D)], where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 17.

b y(x) – λ [At + B + (Ct + D)(t – x)]y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = 1, h1 (t) = Ct2 + (A + D)t + B, g2 (x) = x, and h2 (t) = –(Ct + D). Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 18.

b y(x) – λ (x – t)2 y(t) dt = f (x). a

This is a special case of equation 4.9.19 with g(x) = x, h(t) = –t, and m = 2. 19.

b y(x) – λ (Ax + Bt)2 y(t) dt = f (x). a

This is a special case of equation 4.9.19 with g(x) = Ax, h(t) = Bt, and m = 2. 4.1-3. Kernels Cubic in the Arguments x and t 20.

b y(x) – λ (x3 + t3 )y(t) dt = f (x). a

The characteristic values of the equation: λ1 =

1 4 4 (b

–

a4 )

+

1 1 7 7 (b

, –

a7 )(b

– a)

λ2 =

1 4 4 (b

–

a4 )

–

1 1 7 7 (b

. – a7 )(b – a)

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x3 + A2 ),

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where the constants A1 and A2 are given by f2 – λ 14 f2 ∆4 – 17 f1 ∆7 f1 – λ 14 f1 ∆4 – f2 ∆1 A1 = 2 1 2 1 , A2 = 2 1 2 1 , λ 16 ∆4 – 7 ∆1 ∆7 – 12 λ∆4 + 1 λ 16 ∆4 – 7 ∆1 ∆7 – 12 λ∆4 + 1 b b f1 = f (x) dx, f2 = x3 f (x) dx, ∆n = bn – an . a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0:

b7 – a7 , 7(b – a)

3

y(x) = f (x) + Cy1 (x),

y1 (x) = x +

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0:

b7 – a7 , 7(b – a)

3

y(x) = f (x) + Cy2 (x),

y2 (x) = x –

where C is an arbitrary constant and y2 (x) is an eigenfunction of the equation corresponding to the characteristic value λ2 . 4◦ . The equation has no multiple characteristic values. 21.

b y(x) – λ (x3 – t3 )y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 = ±

1 1 4 4 (a

–

b4 )2

.

– 17 (a7 – b7 )(b – a)

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x3 + A2 ), where the constants A1 and A2 are given by f1 + λ 14 f1 ∆4 – f2 ∆1 A1 = 2 1 , 1 λ 7 ∆1 ∆7 – 16 ∆24 + 1 b f1 = f (x) dx, f2 = a

–f2 + λ 14 f2 ∆4 – 17 f1 ∆7 A2 = 2 1 , 1 λ 7 ∆1 ∆7 – 16 ∆24 + 1 b

x3 f (x) dx,

∆n = bn – an .

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = x3 +

4 – λ1 (b4 – a4 ) , 4λ1 (b – a)

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . The equation has no multiple characteristic values.

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22.

b y(x) – λ (Ax3 + Bt3 )y(t) dt = f (x). a

The characteristic values of the equation:

λ1,2 =

1 4 (A

+ B)∆4 ±

2AB

1 4 2 2 16 (A – B) ∆4 + 7 AB∆1 ∆7 1 1 2 16 ∆4 – 7 ∆1 ∆7

,

∆n = bn – an .

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 x3 + A2 ), where the constants A1 and A2 are given by Af1 – ABλ 14 f1 ∆4 – f2 ∆1 , 1 ABλ2 16 ∆24 – 17 ∆1 ∆7 – 14 λ(A + B)∆4 + 1 Bf2 – ABλ 14 f2 ∆4 – 17 f1 ∆7 1 2 1 A2 = , ABλ2 16 ∆4 – 7 ∆1 ∆7 – 14 λ(A + B)∆4 + 1 b b f1 = f (x) dx, f2 = x3 f (x) dx.

A1 =

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = x3 +

4 – λ1 A(b4 – a4 ) , 4λ1 A(b – a)

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where λ∗ =

8 is the double (A + B)(b4 – a4 )

characteristic value: y(x) = f (x) + Cy∗ (x),

y∗ (x) = x3 –

(A – B)(b4 – a4 ) , 8A(b – a)

where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . 23.

b y(x) – λ (xt2 – t3 )y(t) dt = f (x). a

This is a special case of equation 4.9.8 with A = 0, B = 1, and h(t) = t2 . Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.8.

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24.

b y(x) – λ (Bxt2 + Ct3 )y(t) dt = f (x). a

This is a special case of equation 4.9.9 with A = 0 and h(t) = t2 . Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.9. 25.

b y(x) – λ (Ax2 t + Bxt2 )y(t) dt = f (x). a

This is a special case of equation 4.9.17 with g(x) = x2 and h(x) = x. Solution: y(x) = f (x) + λ(A1 x2 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.17. 26.

b y(x) – λ (Ax3 + Bxt2 )y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = x3 , h1 (t) = A, g2 (x) = x, and h2 (t) = Bt2 . Solution: y(x) = f (x) + λ(A1 x3 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 27.

b y(x) – λ (Ax3 + Bx2 t + Cx2 + D)y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = Ax3 + Cx2 + D, h1 (t) = 1, g2 (x) = x2 , and h2 (t) = Bt. Solution: y(x) = f (x) + λ[A1 (Ax3 + Cx2 + D) + A2 x2 ], where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 28.

b y(x) – λ (Axt2 + Bt3 + Ct2 + D)y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = x, h1 (t) = At2 , g2 (x) = 1, and h2 (t) = Bt3 + Ct2 + D. Solution: y(x) = f (x) + λ(A1 x + A2 ), where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 29.

b y(x) – λ (x – t)3 y(t) dt = f (x). a

This is a special case of equation 4.9.19 with g(x) = x, h(t) = –t, and m = 3. 30.

b y(x) – λ (Ax + Bt)3 y(t) dt = f (x). a

This is a special case of equation 4.9.19 with g(x) = Ax, h(t) = Bt, and m = 3.

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4.1-4. Kernels Containing Higher-Order Polynomials in x and t 31.

b y(x) – λ (xn + tn )y(t) dt = f (x),

n = 1, 2, . . .

a

The characteristic values of the equation: 1 √ λ1,2 = , ∆n ± ∆0 ∆2n 1◦ . Solution with λ ≠ λ1,2 :

where

∆n =

1 (bn+1 – an+1 ). n+1

y(x) = f (x) + λ(A1 xn + A2 ), where the constants A1 and A2 are given by f1 – λ(f1 ∆n – f2 ∆0 ) f2 – λ(f2 ∆n – f1 ∆2n ) A1 = 2 2 , A2 = 2 2 , λ (∆n – ∆0 ∆2n ) – 2λ∆n + 1 λ (∆n – ∆0 ∆2n ) – 2λ∆n + 1 b b 1 f1 = (bn+1 – an+1 ). f (x) dx, f2 = xn f (x) dx, ∆n = n+1 a a 2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x), y1 (x) = xn + ∆2n /∆0 , where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0: y(x) = f (x) + Cy2 (x),

y2 (x) = xn –

∆2n /∆0 ,

where C is an arbitrary constant and y2 (x) is an eigenfunction of the equation corresponding to the characteristic value λ2 .

32.

4◦ . The equation has no multiple characteristic values. b y(x) – λ (xn – tn )y(t) dt = f (x), n = 1, 2, . . . a

The characteristic values of the equation: –1/2 1 1 n+1 n+1 2 2n+1 2n+1 λ1,2 = ± (b (b –a ) – –a )(b – a) . (n + 1)2 2n + 1 1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 xn + A2 ), where the constants A1 and A2 are given by f1 + λ(f1 ∆n – f2 ∆0 ) –f2 + λ(f2 ∆n – f1 ∆2n ) A1 = 2 , A2 = , λ (∆0 ∆2n – ∆2n ) + 1 λ2 (∆0 ∆2n – ∆2n ) + 1 b b 1 f1 = f (x) dx, f2 = xn f (x) dx, ∆n = (bn+1 – an+1 ). n+1 a a 2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 – λ 1 ∆n y(x) = f (x) + Cy1 (x), y1 (x) = xn + , λ1 ∆0 where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . The equation has no multiple characteristic values.

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33.

b y(x) – λ (Axn + Btn )y(t) dt = f (x),

n = 1, 2, . . .

a

The characteristic values of the equation: (A + B)∆n ± (A – B)2 ∆2n + 4AB∆0 ∆2n λ1,2 = , 2AB(∆2n – ∆0 ∆2n )

∆n =

1 (bn+1 – an+1 ). n+1

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 xn + A2 ), where the constants A1 and A2 are given by A1 =

Af1 – ABλ(f1 ∆n – f2 ∆0 ) 2 ABλ (∆2n – ∆0 ∆2n ) – (A + B)λ∆n

+1

,

Bf2 – ABλ(f2 ∆n – f1 ∆2n ) , ABλ2 (∆2n – ∆0 ∆2n ) – (A + B)λ∆n + 1 b b f1 = f (x) dx, f2 = xn f (x) dx.

A2 =

a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = xn +

1 – Aλ1 ∆n , Aλ1 ∆0

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = is double: y(x) = f (x) + Cy∗ (x),

y∗ (x) = xn –

2 (A + B)∆n

(A – B)∆n . 2A∆0

Here C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . 34.

b y(x) – λ (x – t)tm y(t) dt = f (x),

m = 1, 2, . . .

a

This is a special case of equation 4.9.8 with A = 0, B = 1, and h(t) = tm . Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.8. 35.

b y(x) – λ (x – t)xm y(t) dt = f (x),

m = 1, 2, . . .

a

This is a special case of equation 4.9.10 with A = 0, B = 1, and h(x) = xm . Solution: y(x) = f (x) + λ(A1 xm+1 + A2 xm ), where A1 and A2 are the constants determined by the formulas presented in 4.9.10.

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36.

b y(x) – λ (Axm+1 + Bxm t + Cxm + D)y(t) dt = f (x),

m = 1, 2, . . .

a

This is a special case of equation 4.9.18 with g1 (x) = Axm+1 +Cxm +D, h1 (t) = 1, g2 (x) = xm , and h2 (t) = Bt. Solution: y(x) = f (x) + λ[A1 (Axm+1 + Cxm + D) + A2 xm ], where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 37.

b y(x) – λ (Axtm + Btm+1 + Ctm + D)y(t) dt = f (x),

m = 1, 2, . . .

a

This is a special case of equation 4.9.18 with g1 (x) = x, h1 (t) = Atm , g2 (x) = 1, and h2 (t) = Btm+1 + Ctm + D. Solution: y(x) = f (x) + λ(A1 x + A2 ), where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 38.

b y(x) – λ (Axn tn + Bxm tm )y(t) dt = f (x),

n, m = 1, 2, . . . ,

n ≠ m.

a

This is a special case of equation 4.9.14 with g(x) = xn and h(t) = tm . Solution: y(x) = f (x) + λ(A1 xn + A2 xm ), where A1 and A2 are the constants determined by the formulas presented in 4.9.14. 39.

b y(x) – λ (Axn tm + Bxm tn )y(t) dt = f (x),

n, m = 1, 2, . . . ,

n ≠ m.

a

This is a special case of equation 4.9.17 with g(x) = xn and h(t) = tm . Solution: y(x) = f (x) + λ(A1 xn + A2 xm ), where A1 and A2 are the constants determined by the formulas presented in 4.9.17. 40.

b y(x) – λ (x – t)m y(t) dt = f (x),

m = 1, 2, . . .

a

This is a special case of equation 4.9.19 with g(x) = x and h(t) = –t. 41.

b y(x) – λ (Ax + Bt)m y(t) dt = f (x),

m = 1, 2, . . .

a

This is a special case of equation 4.9.19 with g(x) = Ax and h(t) = Bt. 42.

b

|x – t|tk y(t) dt = f (x).

y(x) + A a

This is a special case of equation 4.9.36 with g(t) = Atk . Solving the integral equation is reduced to solving the ordinary differential equation yxx + 2Axk y = fxx (x), the general solution of which can be expressed via Bessel functions or modified Bessel functions (the boundary conditions are given in 4.9.36).

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43.

b

|x – t|2n+1 y(t) dt = f (x),

y(x) + A

n = 0, 1, 2, . . .

a

Let us remove the modulus in the integrand:

x

b

(x – t)2n+1 y(t) dt + A

y(x) + A a

(t – x)2n+1 y(t) dt = f (x).

(1)

x

The k-fold differentiation of (1) with respect to x yields

x

b

(x – t)2n+1–k y(t) dt + (–1)k ABk

yx(k) (x) + ABk a

(t – x)2n+1–k y(t) dt = fx(k) (x), x

Bk = (2n + 1)(2n) . . . (2n + 2 – k),

(2)

k = 1, 2, . . . , 2n + 1.

Differentiating (2) with k = 2n + 1, we arrive at the following linear nonhomogeneous differential equation with constant coefficients for y = y(x): yx(2n+2) + 2(2n + 1)! Ay = fx(2n+2) (x).

(3)

Equation (3) must satisfy the initial conditions which can be obtained by setting x = a in (1) and (2):

b

(t – a)2n+1 y(t) dt = f (a),

y(a) + A a

yx(k) (a)

k

+ (–1) ABk

(4)

b

(t – a)

2n+1–k

y(t) dt =

fx(k) (a),

k = 1, 2, . . . , 2n + 1.

a

These conditions can be reduced to a more habitual form containing no integrals. To this end, y must be expressed from equation (3) in terms of yx(2n+2) and fx(2n+2) and substituted into (4), and then one must integrate the resulting expressions by parts (sufficiently many times).

4.1-5. Kernels Containing Rational Functions

44.

b

1 1 + y(t) dt = f (x). y(x) – λ x t a This is a special case of equation 4.9.2 with g(x) = 1/x. Solution:

A1 y(x) = f (x) + λ + A2 , x where A1 and A2 are the constants determined by the formulas presented in 4.9.2.

45.

b

1 1 y(x) – λ – y(t) dt = f (x). x t a This is a special case of equation 4.9.3 with g(x) = 1/x. Solution:

A1 y(x) = f (x) + λ + A2 , x where A1 and A2 are the constants determined by the formulas presented in 4.9.3.

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46.

b

A B y(x) – λ + y(t) dt = f (x). x t a

47.

This is a special case of equation 4.9.4 with g(x) = 1/x. Solution:

A1 y(x) = f (x) + λ + A2 , x where A1 and A2 are the constants determined by the formulas presented in 4.9.4. b

A B y(x) – λ + y(t) dt = f (x). x+α t+β a A B This is a special case of equation 4.9.5 with g(x) = and h(t) = . x+α t+β Solution:

A y(x) = f (x) + λ A1 + A2 , x+α

48.

49.

50.

51.

where A1 and A2 are the constants determined by the formulas presented in 4.9.5. b

x t y(x) – λ – y(t) dt = f (x). t x a This is a special case of equation 4.9.16 with g(x) = x and h(t) = 1/t. Solution:

A2 y(x) = f (x) + λ A1 x + , x where A1 and A2 are the constants determined by the formulas presented in 4.9.16. b

Ax Bt y(x) – λ + y(t) dt = f (x). t x a This is a special case of equation 4.9.17 with g(x) = x and h(t) = 1/t. Solution:

A2 y(x) = f (x) + λ A1 x + , x where A1 and A2 are the constants determined by the formulas presented in 4.9.17. b

x+α t+α y(x) – λ A +B y(t) dt = f (x). t+β x+β a 1 This is a special case of equation 4.9.17 with g(x) = x + α and h(t) = . t+β Solution: A2 y(x) = f (x) + λ A1 (x + α) + , x+β where A1 and A2 are the constants determined by the formulas presented in 4.9.17. b (x + α)n (t + α)n y(x) – λ A y(t) dt = f (x), n, m = 0, 1, 2, . . . + B (t + β)m (x + β)m a This is a special case of equation 4.9.17 with g(x) = (x + α)n and h(t) = (t + β)–m . Solution: A2 n y(x) = f (x) + λ A1 (x + α) + , (x + β)m where A1 and A2 are the constants determined by the formulas presented in 4.9.17.

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52.

y(x) – λ

∞

1

y(t) x+t

1 ≤ x < ∞,

dt = f (x),

Solution:

∞

y(x) = 0 ∞ F (τ ) =

–∞ < πλ < 1.

τ sinh(πτ ) F (τ ) P 1 (x) dτ , cosh(πτ ) – πλ – 2 +iτ f (x)P– 1 +iτ (x) dx, 2

1

where Pν (x) = F –ν, ν + 1, 1; 12 (1 – x) is the Legendre spherical function of the first kind, for which the integral representation P– 1 +iτ (cosh α) = 2

2 π

α

√

0

cos(τ s) ds 2(cosh α – cosh s)

(α ≥ 0)

can be used. • Reference: V. A. Ditkin and A. P. Prudnikov (1965). 53.

2

2

(x + b )y(x) =

λ

π

∞

–∞

a3 y(t)

dt.

a2 + (x – t)2

This equation is encountered in atomic and nuclear physics. We seek the solution in the form y(x) =

∞ m=0

Am x . x2 + (am + b)2

(1)

The coefficients Am obey the equations

mAm

m + 2b a

+ λAm–1 = 0,

∞

Am = 0.

(2)

m=0

Using the first equation of (2) to express all Am via A0 (A0 can be chosen arbitrarily), substituting the result into the second equation of (2), and dividing by A0 , we obtain 1+

∞ (–λ)m 1 = 0. m! (1 + 2b/a)(2 + 2b/a) . . . (m + 2b/a)

(3)

m=1

It follows from the definitions of the Bessel functions of the first kind that equation (3) can be rewritten in the form √ λ–b/a J2b/a 2 λ = 0. (4) In this sort of problem, a and λ are usually assumed to be given and b, which is proportional to the system energy, to be unknown. The quantity b can be determined by tables of zeros of Bessel functions. In some cases, b and a are given and λ is unknown. • Reference: I. Sneddon (1951).

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4.1-6. Kernels Containing Arbitrary Powers 54.

b y(x) – λ (x – t)tµ y(t) dt = f (x). a

This is a special case of equation 4.9.8 with A = 0, B = 1, and h(t) = tµ . Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.8. 55.

b y(x) – λ (x – t)xν y(t) dt = f (x). a

This is a special case of equation 4.9.10 with A = 0, B = 1, and h(x) = xν . Solution: y(x) = f (x) + λ(E1 xν+1 + E2 xν ), where E1 and E2 are the constants determined by the formulas presented in 4.9.10. 56.

b y(x) – λ (xµ – tµ )y(t) dt = f (x). a

This is a special case of equation 4.9.3 with g(x) = xµ . Solution: y(x) = f (x) + λ(A1 xµ + A2 ), where A1 and A2 are the constants determined by the formulas presented in 4.9.3. 57.

b y(x) – λ (Axν + Btν )tµ y(t) dt = f (x). a

This is a special case of equation 4.9.6 with g(x) = xν and h(t) = tµ . Solution: y(x) = f (x) + λ(A1 xν + A2 ), where A1 and A2 are the constants determined by the formulas presented in 4.9.6. 58.

b y(x) – λ (Dxν + Etµ )xγ y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = xν+γ , h1 (t) = D, g2 (x) = xγ , and h2 (t) = Etµ . Solution: y(x) = f (x) + λ(A1 xν+γ + A2 xγ ), where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 59.

b y(x) – λ (Axν tµ + Bxγ tδ )y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = xν , h1 (t) = Atµ , g2 (x) = xγ , and h2 (t) = Btδ . Solution: y(x) = f (x) + λ(A1 xν + A2 xγ ), where A1 and A2 are the constants determined by the formulas presented in 4.9.18.

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60.

b y(x) – λ (A + Bxtµ + Ctµ+1 )y(t) dt = f (x). a

This is a special case of equation 4.9.9 with h(t) = tµ . Solution: y(x) = f (x) + λ(A1 + A2 x), where A1 and A2 are the constants determined by the formulas presented in 4.9.9.

61.

b y(x) – λ (Atα + Bxβ tµ + Ctµ+γ )y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = 1, h1 (t) = Atα + Ctµ+γ , g2 (x) = xβ , and h2 (t) = Btµ . Solution: y(x) = f (x) + λ(A1 + A2 xβ ), where A1 and A2 are the constants determined by the formulas presented in 4.9.18.

62.

b y(x) – λ (Axα tγ + Bxβ tγ + Cxµ tν )y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = Axα + Bxβ , h1 (t) = tγ , g2 (x) = xµ , and h2 (t) = Ctν . Solution: y(x) = f (x) + λ[A1 (Axα + Bxβ ) + A2 xµ ], where A1 and A2 are the constants determined by the formulas presented in 4.9.18.

63.

b (x + p1 )β (x + p2 )µ y(x) – λ A y(t) dt = f (x). +B (t + q1 )γ (t + q2 )δ a This is a special case of equation 4.9.18 with g1 (x) = (x + p1 )β , h1 (t) = A(t + q1 )–γ , g2 (x) = (x + p2 )µ , and h2 (t) = B(t + q2 )–δ . Solution:

y(x) = f (x) + λ A1 (x + p1 )β + A2 (x + p2 )µ , where A1 and A2 are the constants determined by the formulas presented in 4.9.18.

64.

b

xµ + a xγ + c y(x) – λ A ν +B δ y(t) dt = f (x). t +b t +d a A This is a special case of equation 4.9.18 with g1 (x) = xµ + a, h1 (t) = ν , g2 (x) = xγ + c, t +b B and h2 (t) = δ . t +d Solution: y(x) = f (x) + λ[A1 (xµ + a) + A2 (xγ + c)], where A1 and A2 are the constants determined by the formulas presented in 4.9.18.

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4.1-7. Singular Equations In this subsection, all singular integrals are understood in the sense of the Cauchy principal value. 65.

Ay(x) +

B π

1

–1

y(t) dt t–x

= f (x),

–1 < x < 1.

Without loss of generality we may assume that A2 + B 2 = 1. 1◦ . The solution bounded at the endpoints: B π

y(x) = Af (x) –

1

–1

g(x) f (t) dt , g(t) t – x

g(x) = (1 + x)α (1 – x)1–α ,

(1)

where α is the solution of the trigonometric equation A + B cot(πα) = 0

(2)

1

on the interval 0 < α < 1. This solution y(x) exists if and only if –1

f (t) dt = 0. g(t)

◦

2 . The solution bounded at the endpoint x = 1 and unbounded at the endpoint x = –1: y(x) = Af (x) –

B π

1

–1

g(x) f (t) dt , g(t) t – x

g(x) = (1 + x)α (1 – x)–α ,

(3)

where α is the solution of the trigonometric equation (2) on the interval –1 < α < 0. 3◦ . The solution unbounded at the endpoints: y(x) = Af (x) –

B π

1

–1

g(x) f (t) dt + Cg(x), g(t) t – x

g(x) = (1 + x)α (1 – x)–1–α ,

(4)

where C is an arbitrary constant and α is the solution of the trigonometric equation (2) on the interval –1 < α < 0. • Reference: I. K. Lifanov (1996). 66.

1

1 1 – y(t) dt = f (x), y(x) – λ t – x x + t – 2xt 0

0 < x < 1.

Tricomi’s equation. Solution:

1 α 1 t (1 – x)α 1 C(1 – x)β 1 y(x) = , – f (t) dt + f (x) + α α 1 + λ2 π 2 t – x x + t – 2xt x1+β 0 x (1 – t) 2 βπ α = arctan(λπ) (–1 < α < 1), tan = λπ (–2 < β < 0), π 2 where C is an arbitrary constant. • References: P. P. Zabreyko, A. I. Koshelev, et al. (1975), F. G. Tricomi (1985).

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4.2. Equations Whose Kernels Contain Exponential Functions 4.2-1. Kernels Containing Exponential Functions

1.

b y(x) – λ (eβx + eβt )y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 =

eβb – eβa ±

β 1 2 β(b

. – a)(e2βb – e2βa )

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 eβx + A2 ), where the constants A1 and A2 are given by

f1 – λ f1 ∆β – (b – a)f2 f2 – λ(f2 ∆β – f1 ∆2β )

A1 = 2 2 , A2 = 2 2 , λ ∆β – (b – a)∆2β – 2λ∆β + 1 λ ∆β – (b – a)∆2β – 2λ∆β + 1 b b 1 f1 = f (x) dx, f2 = f (x)eβx dx, ∆β = (eβb – eβa ). β a a 2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

e2βb – e2βa , 2β(b – a)

y1 (x) = eβx +

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0: y(x) = f (x) + Cy2 (x),

y2 (x) = eβx –

e2βb – e2βa , 2β(b – a)

where C is an arbitrary constant and y2 (x) is an eigenfunction of the equation corresponding to the characteristic value λ2 . 4◦ . The equation has no multiple characteristic values. 2.

b y(x) – λ (eβx – eβt )y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 = ±

β (eβb – eβa )2 –

1 2 β(b

. – a)(e2βb – e2βa )

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1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 eβx + A2 ), where the constants A1 and A2 are given by

f1 + λ f1 ∆β – (b – a)f2 –f2 + λ(f2 ∆β – f1 ∆2β ) A1 = 2 , A2 = 2 , λ (b – a)∆2β – ∆2β + 1 λ (b – a)∆2β – ∆2β + 1 b b 1 f1 = f (x) dx, f2 = f (x)eβx dx, ∆β = (eβb – eβa ). β a a 2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 – λ1 ∆β y(x) = f (x) + Cy1 (x), y1 (x) = eβx + , λ1 (b – a) where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively.

3.

4◦ . The equation has no multiple characteristic values. b y(x) – λ (Aeβx + Beβt )y(t) dt = f (x). a

The characteristic values of the equation: (A + B)∆β ± (A – B)2 ∆2β + 4AB(b – a)∆2β

λ1,2 = , 2AB ∆2β – (b – a)∆2β

∆β =

1 βb βa (e – e ). β

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 eβx + A2 ), where the constants A1 and A2 are given by

Af1 – ABλ f1 ∆β – (b – a)f2

A1 = , ABλ2 ∆2β – (b – a)∆2β – (A + B)λ∆β + 1 A2 =

Bf – ABλ(f2 ∆β – f1 ∆2β )

22 , ∆β – (b – a)∆2β – (A + B)λ∆β + 1 b b f1 = f (x) dx, f2 = f (x)eβx dx. ABλ2

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0:

a

1 – Aλ1 ∆β , A(b – a)λ1 where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . y(x) = f (x) + Cy1 (x),

y1 (x) = eβx +

3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 2 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = (A + B)∆β is double: (A – B)∆β y∗ (x) = eβx – y(x) = f (x) + Cy∗ (x), , 2A(b – a) where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ .

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4.

b

β(x–t) y(x) – λ + B y(t) dt = f (x). Ae a

This is a special case of equation 4.9.18 with g1 (x) = eβx , h1 (t) = Ae–βt , g2 (x) = 1, and h2 (t) = B. Solution: y(x) = f (x) + λ(A1 eβx + A2 ), where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 5.

b

βx+µt y(x) – λ Ae + Be(β+µ)t y(t) dt = f (x). a

This is a special case of equation 4.9.6 with g(x) = eβx and h(t) = eµt . Solution: y(x) = f (x) + λ(A1 eβx + A2 ), where A1 and A2 are the constants determined by the formulas presented in 4.9.6. 6.

b

α(x+t) y(x) – λ Ae + Beβ(x+t) y(t) dt = f (x). a

This is a special case of equation 4.9.14 with g(x) = eαx and h(t) = eβt . Solution: y(x) = f (x) + λ(A1 eαx + A2 eβx ), where A1 and A2 are the constants determined by the formulas presented in 4.9.14. 7.

b αx+βt y(x) – λ + Beβx+αt y(t) dt = f (x). Ae a

This is a special case of equation 4.9.17 with g(x) = eαx and h(t) = eβt . Solution: y(x) = f (x) + λ(A1 eαx + A2 eβx ), where A1 and A2 are the constants determined by the formulas presented in 4.9.17. 8.

b

y(x) – λ De(γ+µ)x + Eeνt+µx y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = e(γ+µ)x , h1 (t) = D, g2 (x) = eµx , and h2 (t) = Eeνt . Solution: y(x) = f (x) + λ[A1 e(γ+µ)x + A2 eµx ], where A1 and A2 are the constants determined by the formulas presented in 4.9.18. 9.

b y(x) – λ (Aeαx+βt + Beγx+δt )y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = eαx , h1 (t) = Aeβt , g2 (x) = eγx , and h2 (t) = Beδt . Solution: y(x) = f (x) + λ(A1 eαx + A2 eγx ), where A1 and A2 are the constants determined by the formulas presented in 4.9.18.

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10.

b n γk (x–t) y(t) dt = f (x). y(x) – λ Ak e a

k=1

This is a special case of equation 4.9.20 with gk (x) = eγk x and hk (t) = Ak e–γk t . 11.

y(x) –

1

2

∞

e–|x–t| y(t) dt = Aeµx ,

0 < µ < 1.

0

Solution:

y(x) = C(1 + x) + Aµ–2 (µ2 – 1)eµx – µ + 1 ,

where C is an arbitrary constant. • Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975). 12.

y(x) + λ

∞

e–|x–t| y(t) dt = f (x).

0

∞ √ λ y(x) = f (x) – √ exp – 1 + 2λ |x – t| f (t) dt 1 + 2λ 0

∞

√ λ+1 + 1– √ exp – 1 + 2λ (x + t) f (t) dt, 1 + 2λ 0

Solution:

where λ > – 12 . • Reference: F. D. Gakhov and Yu. I. Cherskii (1978). 13.

y(x) – λ

∞

e–|x–t| y(t) dt = 0,

λ > 0.

–∞

The Lalesco–Picard equation. Solution: √ √    C1 exp x 1 – 2λ + C2 exp –x 1 – 2λ y(x) = C1 + C2 x  √ √  C1 cos x 2λ – 1 + C2 sin x 2λ – 1

for 0 < λ < 12 , for λ = 12 , for λ > 12 ,

where C1 and C2 are arbitrary constants. • Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971). 14.

y(x) + λ

∞

e–|x–t| y(t) dt = f (x).

–∞ ◦

1 . Solution with λ > – 12 : y(x) = f (x) – √

λ 1 + 2λ

∞

√ exp – 1 + 2λ |x – t| f (t) dt.

–∞

2◦ . If λ ≤ – 12 , for the equation to be solvable the conditions

∞

∞

f (x) cos(ax) dx = 0, –∞

f (x) sin(ax) dx = 0, –∞

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where a =

√

–1 – 2λ, must be satisfied. In this case, the solution has the form a2 + 1 ∞ y(x) = f (x) – sin(at)f (x + t) dt, (–∞ < x < ∞). 2a 0

In the class of solutions not belonging to L2 (–∞, ∞), the homogeneous equation (with f (x) ≡ 0) has a nontrivial solution. In this case, the general solution of the corresponding nonhomogeneous equation with λ ≤ – 12 has the form a2 + 1 ∞ y(x) = C1 sin(ax) + C2 cos(ax) + f (x) – sin(a|x – t|)f (t) dt. 4a –∞

15.

• Reference: F. D. Gakhov and Yu. I. Cherskii (1978). b y(x) + A eλ|x–t| y(t) dt = f (x). a

This is a special case of equation 4.9.37 with g(t) = A. 1◦ . The function y = y(x) obeys the following second-order linear nonhomogeneous ordinary differential equation with constant coefficients: yxx + λ(2A – λ)y = fxx (x) – λ2 f (x).

(1)

The boundary conditions for (1) have the form (see 4.9.37) yx (a) + λy(a) = fx (a) + λf (a),

(2)

yx (b) – λy(b) = fx (b) – λf (b).

Equation (1) under the boundary conditions (2) determines the solution of the original integral equation. 2◦ . For λ(2A – λ) < 0, the general solution of equation (1) is given by 2Aλ x y(x) = C1 cosh(kx) + C2 sinh(kx) + f (x) – sinh[k(x – t)] f (t) dt, k a k = λ(λ – 2A), where C1 and C2 are arbitrary constants. For λ(2A – λ) > 0, the general solution of equation (1) is given by 2Aλ x y(x) = C1 cos(kx) + C2 sin(kx) + f (x) – sin[k(x – t)] f (t) dt, k a k = λ(2A – λ). For λ = 2A, the general solution of equation (1) is given by x y(x) = C1 + C2 x + f (x) – 4A2 (x – t)f (t) dt.

(3)

(4)

(5)

a

The constants C1 and C2 in solutions (3)–(5) are determined by conditions (2). 3◦ . In the special case a = 0 and λ(2A – λ) > 0, the solution of the integral equation is given by formula (4) with A(kIc – λIs ) λ A(kIc – λIs ) , C2 = – , (λ – A) sin µ – k cos µ k (λ – A) sin µ – k cos µ b b k = λ(2A – λ), µ = bk, Is = sin[k(b – t)]f (t) dt, Ic = cos[k(b – t)]f (t) dt. C1 =

0

0

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16.

y(x) +

b n a

Ak exp(λk |x – t|) y(t) dt = f (x),

–∞ < a < b < ∞.

k=1

1◦ . Let us remove the modulus in the kth summand of the integrand: b x b Ik (x) = exp(λk |x – t|)y(t) dt = exp[λk (x – t)]y(t) dt + exp[λk (t – x)]y(t) dt. (1) a

a

x

Differentiating (1) with respect to x twice yields x Ik = λk exp[λk (x – t)]y(t) dt – λk a

Ik

= 2λk y(x) +

λ2k

b

exp[λk (t – x)]y(t) dt,

x

x

exp[λk (x – t)]y(t) dt + a

(2)

b

λ2k

exp[λk (t – x)]y(t) dt, x

where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2), we find the relation between Ik and Ik : Ik = 2λk y(x) + λ2k Ik ,

Ik = Ik (x).

(3)

2◦ . With the aid of (1), the integral equation can be rewritten in the form y(x) +

n

Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice and taking into account (3), we find that yxx (x) + σn y(x) +

n

Ak λ2k Ik = fxx (x),

σn = 2

n

Ak λ k .

(5)

Ak (λ2k – λ2n )Ik = fxx (x) – λ2n f (x).

(6)

k=1

k=1

Eliminating the integral In from (4) and (5) yields yxx (x) + (σn – λ2n )y(x) +

n–1 k=1

Differentiating (6) with respect to x twice and eliminating In–1 from the resulting equation with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear n–2 differential operator (acting on y) with constant coefficients plus the sum Bk Ik . If we k=1

successively eliminate In–2 , In–3 , . . . , with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. 3◦ . The boundary conditions for y(x) can be found by setting x = a in the integral equation and all its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.) 4.2-2. Kernels Containing Power-Law and Exponential Functions 17.

b y(x) – λ (x – t)eγt y(t) dt = f (x). a

This is a special case of equation 4.9.8 with A = 0, B = 1, and h(t) = eγt .

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18.

b y(x) – λ (x – t)eγx y(t) dt = f (x). a

This is a special case of equation 4.9.10 with A = 0, B = 1, and h(x) = eγx . 19.

b y(x) – λ (x – t)eγx+µt y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = xeγx , h1 (t) = eµt , g2 (x) = eγx , and h2 (t) = –teµt . 20.

b y(x) – λ [A + (Bx + Ct)eγx ]y(t) dt = f (x). a

This is a special case of equation 4.9.11 with h(x) = eγx . 21.

b y(x) – λ (x2 + t2 )eγ(x+t) y(t) dt = f (x). 0

This is a special case of equation 4.9.15 with g(x) = x2 eγx and h(t) = eγt . 22.

b y(x) – λ (x2 – t2 )eγ(x–t) y(t) dt = f (x). 0

This is a special case of equation 4.9.18 with g1 (x) = x2 eγx , h1 (t) = e–γt , g2 (x) = eγx , and h2 (t) = –t2 e–γt . 23.

b y(x) – λ (Axn + Btn )eαx+βt y(t) dt = f (x),

n = 1, 2, . . .

0

This is a special case of equation 4.9.18 with g1 (x) = xn eαx , h1 (t) = Aeβt , g2 (x) = eαx , and h2 (t) = Btn eβt . 24.

b n νk αk x+βk t y(x) – λ y(t) dt = f (x), Ak t e a

n = 1, 2, . . .

k=1

This is a special case of equation 4.9.20 with gk (x) = eαk x and hk (t) = Ak tνk eβk t . 25.

b n y(x) – λ Ak xνk eαk x+βk t y(t) dt = f (x), a

n = 1, 2, . . .

k=1

This is a special case of equation 4.9.20 with gk (x) = Ak xνk eαk x and hk (t) = eβk t . 26.

b y(x) – λ (x – t)n eγ(x–t) y(t) dt = f (x),

n = 1, 2, . . .

a

This is a special case of equation 4.9.20. 27.

b y(x) – λ (x – t)n eαx+βt y(t) dt = f (x),

n = 1, 2, . . .

a

This is a special case of equation 4.9.20. 28.

b y(x) – λ (Ax + Bt)n eαx+βt y(t) dt = f (x),

n = 1, 2, . . .

a

This is a special case of equation 4.9.20.

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29.

b

teλ|x–t| y(t) dt = f (x).

y(x) + A a

This is a special case of equation 4.9.37 with g(t) = At. The solution of the integral equation can be written via the Bessel functions (or modified Bessel functions) of order 1/3. 30.

∞

y(x) +

(a + b|x – t|) exp(–|x – t|)y(t) dt = f (x). 0

Let the biquadratic polynomial P (k) = k 4 + 2(a – b + 1)k 2 + 2a + 2b + 1 have no real roots and let k = α + iβ be a root of the equation P (k) = 0 such that α > 0 and β > 0. In this case, the solution has the form ∞ y(x) = f (x) + ρ exp(–β|x – t|) cos(θ + α|x – t|)f (t) dt 0 [α + (β – 1)2 ]2 ∞ + exp[–β(x + t)] cos[α(x – t)]f (t) dt 4α2 β 0 ∞ R + exp[–β(x + t)] cos[ψ + α(x + t)]f (t) dt, 4α2 0 where the parameters ρ, θ, R, and ψ are determined from the system of algebraic equations obtained by separating real and imaginary parts in the relations ρeiθ =

µ , β – iα

Reiψ =

(β – 1 – iα)4 . 8α2 (β – iα)

• Reference: F. D. Gakhov and Yu. I. Cherskii (1978).

4.3. Equations Whose Kernels Contain Hyperbolic Functions 4.3-1. Kernels Containing Hyperbolic Cosine

1.

y(x) – λ

b

cosh(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cosh(βx) and h(t) = 1. 2.

y(x) – λ

b

cosh(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = cosh(βt). 3.

y(x) – λ

b

cosh[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.13 with g(x) = cosh(βx) and h(t) = sinh(βt). Solution:

y(x) = f (x) + λ A1 cosh(βx) + A2 sinh(βx) , where A1 and A2 are the constants determined by the formulas presented in 4.9.13.

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4.

y(x) – λ

b

cosh[β(x + t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.12 with g(x) = cosh(βx) and h(t) = sinh(βt). Solution:

y(x) = f (x) + λ A1 cosh(βx) + A2 sinh(βx) ,

5.

where A1 and A2 are the constants determined by the formulas presented in 4.9.12. b n y(x) – λ Ak cosh[βk (x – t)] y(t) dt = f (x), n = 1, 2, . . . a

6.

k=1

This is a special case of equation 4.9.20. b cosh(βx) y(x) – λ y(t) dt = f (x). a cosh(βt) This is a special case of equation 4.9.1 with g(x) = cosh(βx) and h(t) =

7.

y(x) – λ

b

a

cosh(βt) cosh(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

8.

y(x) – λ

1 . cosh(βt)

1 and h(t) = cosh(βt). cosh(βx)

b

coshk (βx) coshm (µt)y(t) dt = f (x).

a

9.

This is a special case of equation 4.9.1 with g(x) = coshk (βx) and h(t) = coshm (µt). b y(x) – λ tk coshm (βx)y(t) dt = f (x). a

10.

This is a special case of equation 4.9.1 with g(x) = coshm (βx) and h(t) = tk . b y(x) – λ xk coshm (βt)y(t) dt = f (x).

11.

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = coshm (βt). b y(x) – λ [A + B(x – t) cosh(βx)]y(t) dt = f (x).

12.

This is a special case of equation 4.9.10 with h(x) = cosh(βx). b y(x) – λ [A + B(x – t) cosh(βt)]y(t) dt = f (x).

a

a

a

13.

This is a special case of equation 4.9.8 with h(t) = cosh(βt). ∞ y(t) dt y(x) + λ = f (x). –∞ cosh[b(x – t)] Solution with b > π|λ|: ∞ 2λb sinh[2k(x – t)] y(x) = f (x) – √ f (t) dt, 2 2 2 b – π λ –∞ sinh[2b(x – t)]

k=

πλ b arccos . π b

• Reference: F. D. Gakhov and Yu. I. Cherskii (1978).

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4.3-2. Kernels Containing Hyperbolic Sine 14.

y(x) – λ

b

sinh(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinh(βx) and h(t) = 1. 15.

y(x) – λ

b

sinh(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = sinh(βt). 16.

y(x) – λ

b

sinh[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.16 with g(x) = sinh(βx) and h(t) = cosh(βt). Solution:

y(x) = f (x) + λ A1 sinh(βx) + A2 cosh(βx) , where A1 and A2 are the constants determined by the formulas presented in 4.9.16. 17.

y(x) – λ

b

sinh[β(x + t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.15 with g(x) = sinh(βx) and h(t) = cosh(βt). Solution:

y(x) = f (x) + λ A1 sinh(βx) + A2 cosh(βx) , where A1 and A2 are the constants determined by the formulas presented in 4.9.15. 18.

b n y(x) – λ Ak sinh[βk (x – t)] y(t) dt = f (x), a

n = 1, 2, . . .

k=1

This is a special case of equation 4.9.20. 19.

y(x) – λ

b

a

sinh(βx) sinh(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = sinh(βx) and h(t) =

20.

y(x) – λ

b

a

sinh(βt) sinh(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

21.

y(x) – λ

1 . sinh(βt)

1 and h(t) = sinh (βt). sinh(βx)

b

sinhk (βx) sinhm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinhk (βx) and h(t) = sinhm (µt). 22.

y(x) – λ

b

tk sinhm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinhm (βx) and h(t) = tk .

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23.

y(x) – λ

b

xk sinhm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = sinhm (βt). 24.

b y(x) – λ [A + B(x – t) sinh(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = sinh(βt). 25.

b y(x) – λ [A + B(x – t) sinh(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = sinh(βx). 26.

b

y(x) + A

sinh(λ|x – t|)y(t) dt = f (x). a

This is a special case of equation 4.9.38 with g(t) = A. 1◦ . The function y = y(x) obeys the following second-order linear nonhomogeneous ordinary differential equation with constant coefficients: yxx + λ(2A – λ)y = fxx (x) – λ2 f (x).

(1)

The boundary conditions for (1) have the form (see 4.9.38) sinh[λ(b – a)]ϕx (b) – λ cosh[λ(b – a)]ϕ(b) = λϕ(a), sinh[λ(b – a)]ϕx (a) + λ cosh[λ(b – a)]ϕ(a) = –λϕ(b),

ϕ(x) = y(x) – f (x).

(2)

Equation (1) under the boundary conditions (2) determines the solution of the original integral equation. 2◦ . For λ(2A – λ) = –k 2 < 0, the general solution of equation (1) is given by 2Aλ x y(x) = C1 cosh(kx) + C2 sinh(kx) + f (x) – sinh[k(x – t)]f (t) dt, k a where C1 and C2 are arbitrary constants. For λ(2A – λ) = k 2 > 0, the general solution of equation (1) is given by 2Aλ x y(x) = C1 cos(kx) + C2 sin(kx) + f (x) – sin[k(x – t)]f (t) dt. k a For λ = 2A, the general solution of equation (1) is given by x 2 y(x) = C1 + C2 x + f (x) – 4A (x – t)f (t) dt.

(3)

(4)

(5)

a

The constants C1 and C2 in solutions (3)–(5) are determined by conditions (2). 27.

b

y(x) + A

t sinh(λ|x – t|)y(t) dt = f (x). a

This is a special case of equation 4.9.38 with g(t) = At. The solution of the integral equation can be written via the Bessel functions (or modified Bessel functions) of order 1/3.

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28.

b

sinh3 (λ|x – t|)y(t) dt = f (x).

y(x) + A a

Using the formula sinh3 β = 14 sinh 3β – 34 sinh β, we arrive at an equation of the form 4.3.29 with n = 2: b

1 3 y(x) + 4 A sinh(3λ|x – t|) – 4 A sinh(λ|x – t|) y(t) dt = f (x). a

29.

y(x) +

b n a

Ak sinh(λk |x – t|) y(t) dt = f (x),

–∞ < a < b < ∞.

k=1

1◦ . Let us remove the modulus in the kth summand of the integrand: b x b Ik (x) = sinh(λk |x – t|)y(t) dt = sinh[λk (x – t)]y(t) dt + sinh[λk (t – x)]y(t) dt. (1) a

a

x

Differentiating (1) with respect to x twice yields x Ik = λk cosh[λk (x – t)]y(t) dt – λk a

Ik

= 2λk y(x) +

λ2k

b

cosh[λk (t – x)]y(t) dt,

x

x

sinh[λk (x – t)]y(t) dt + a

(2)

b

λ2k

sinh[λk (t – x)]y(t) dt, x

where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2), we find the relation between Ik and Ik : Ik = 2λk y(x) + λ2k Ik ,

Ik = Ik (x).

(3)

2◦ . With the aid of (1), the integral equation can be rewritten in the form y(x) +

n

Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice and taking into account (3), we find that yxx (x) + σn y(x) +

n

Ak λ2k Ik = fxx (x),

n

Ak λ k .

(5)

Ak (λ2k – λ2n )Ik = fxx (x) – λ2n f (x).

(6)

k=1

σn = 2

k=1

Eliminating the integral In from (4) and (5) yields (x) + (σn – λ2n )y(x) + yxx

n–1 k=1

Differentiating (6) with respect to x twice and eliminating In–1 from the resulting equation with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear n–2 differential operator (acting on y) with constant coefficients plus the sum Bk Ik . If we k=1

successively eliminate In–2 , In–3 , . . . , with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. 3◦ . The boundary conditions for y(x) can be found by setting x = a in the integral equation and its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.)

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4.3-3. Kernels Containing Hyperbolic Tangent

30.

y(x) – λ

b

tanh(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanh(βx) and h(t) = 1. 31.

y(x) – λ

b

tanh(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = tanh(βt). 32.

b y(x) – λ [A tanh(βx) + B tanh(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.4 with g(x) = tanh(βx). 33.

y(x) – λ

b

a

tanh(βx) tanh(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = tanh(βx) and h(t) =

34.

y(x) – λ

b

a

tanh(βt) tanh(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

35.

y(x) – λ

1 . tanh(βt)

1 and h(t) = tanh(βt). tanh(βx)

b

tanhk (βx) tanhm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanhk (βx) and h(t) = tanhm (µt). 36.

y(x) – λ

b

tk tanhm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanhm (βx) and h(t) = tk . 37.

y(x) – λ

b

xk tanhm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = tanhm (βt). 38.

b y(x) – λ [A + B(x – t) tanh(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = tanh(βt). 39.

b y(x) – λ [A + B(x – t) tanh(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = tanh(βx).

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4.3-4. Kernels Containing Hyperbolic Cotangent

40.

y(x) – λ

b

coth(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = coth(βx) and h(t) = 1. 41.

y(x) – λ

b

coth(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = coth(βt). 42.

b y(x) – λ [A coth(βx) + B coth(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.4 with g(x) = coth(βx). 43.

y(x) – λ

b

a

coth(βx) coth(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = coth(βx) and h(t) =

44.

y(x) – λ

b

a

coth(βt) coth(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

45.

y(x) – λ

1 . coth(βt)

1 and h(t) = coth(βt). coth(βx)

b

cothk (βx) cothm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cothk (βx) and h(t) = cothm (µt). 46.

y(x) – λ

b

tk cothm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cothm (βx) and h(t) = tk . 47.

y(x) – λ

b

xk cothm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = cothm (βt). 48.

b y(x) – λ [A + B(x – t) coth(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = coth(βt). 49.

b y(x) – λ [A + B(x – t) coth(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = coth(βx).

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4.3-5. Kernels Containing Combination of Hyperbolic Functions 50.

y(x) – λ

b

coshk (βx) sinhm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = coshk (βx) and h(t) = sinhm (µt). 51.

b y(x) – λ [A sinh(αx) cosh(βt) + B sinh(γx) cosh(δt)]y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = sinh(αx), h1 (t) = A cosh(βt), g2 (x) = sinh(γx), and h2 (t) = B cosh(δt). 52.

y(x) – λ

b

tanhk (γx) cothm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanhk (γx) and h(t) = cothm (µt). 53.

b y(x) – λ [A tanh(αx) coth(βt) + B tanh(γx) coth(δt)]y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = tanh(αx), h1 (t) = A coth(βt), g2 (x) = tanh(γx), and h2 (t) = B coth(δt).

4.4. Equations Whose Kernels Contain Logarithmic Functions 4.4-1. Kernels Containing Logarithmic Functions 1.

y(x) – λ

b

ln(γx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = ln(γx) and h(t) = 1. 2.

y(x) – λ

b

ln(γt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = ln(γt). 3.

b y(x) – λ (ln x – ln t)y(t) dt = f (x). a

This is a special case of equation 4.9.3 with g(x) = ln x. 4.

y(x) – λ

b

a

ln(γx) ln(γt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = ln(γx) and h(t) =

5.

y(x) – λ

b

a

ln(γt) ln(γx)

1 . ln(γt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

1 and h(t) = ln(γt). ln(γx)

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6.

y(x) – λ

b

lnk (γx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnk (γx) and h(t) = lnm (µt).

4.4-2. Kernels Containing Power-Law and Logarithmic Functions

7.

y(x) – λ

b

tk lnm (γx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (γx) and h(t) = tk . 8.

y(x) – λ

b

xk lnm (γt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = lnm (γt). 9.

b y(x) – λ [A + B(x – t) ln(γt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = ln(γt). 10.

b y(x) – λ [A + B(x – t) ln(γx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = ln(γx). 11.

b y(x) – λ [A + (Bx + Ct) ln(γt)]y(t) dt = f (x). a

This is a special case of equation 4.9.9 with h(t) = ln(γt). 12.

b y(x) – λ [A + (Bx + Ct) ln(γx)]y(t) dt = f (x). a

This is a special case of equation 4.9.11 with h(x) = ln(γx). 13.

b y(x) – λ [Atn lnm (βx) + Bxk lnl (γt)]y(t) dt = f (x). a

This is a special case of equation 4.9.18 with g1 (x) = lnm (βx), h1 (t) = Atn , g2 (x) = xk , and h2 (t) = B lnl (γt).

4.5. Equations Whose Kernels Contain Trigonometric Functions 4.5-1. Kernels Containing Cosine

1.

y(x) – λ

b

cos(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cos(βx) and h(t) = 1.

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2.

y(x) – λ

b

cos(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = cos(βt). 3.

y(x) – λ

b

cos[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.12 with g(x) = cos(βx) and h(t) = sin(βt). Solution:

y(x) = f (x) + λ A1 cos(βx) + A2 sin(βx) , where A1 and A2 are the constants determined by the formulas presented in 4.9.12. 4.

y(x) – λ

b

cos[β(x + t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.13 with g(x) = cos(βx) and h(t) = sin(βt). Solution:

y(x) = f (x) + λ A1 cos(βx) + A2 sin(βx) , where A1 and A2 are the constants determined by the formulas presented in 4.9.13. 5.

y(x) – λ

∞

cos(xt)y(t) dt = 0.

0

Characteristic values: λ = ± 2/π. For the characteristic values, the integral equation has infinitely many linearly independent eigenfunctions. Eigenfunctions for λ = + 2/π have the form y+ (x) = f (x) +

2 π

∞

f (t) cos(xt) dt,

(1)

0

where f = f (x) is any continuous function absolutely integrable on the interval [0, ∞). Eigenfunctions for λ = – 2/π have the form y– (x) = f (x) –

2 π

∞

f (t) cos(xt) dt,

(2)

0

where f = f (x) is any continuous function absolutely integrable on the interval [0, ∞). In particular, from (1) and (2) with f (x) = e–ax we obtain

a 2 y+ (x) = e + π a2 + x2 2 a –ax y– (x) = e – 2 π a + x2 –ax

2 for λ = + , π 2 for λ = – , π

where a is any positive number. • Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971).

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6.

y(x) – λ

∞

cos(xt)y(t) dt = f (x).

0

Solution: y(x) = where λ ≠ ± 2/π.

f (x) λ π 2 + 1– 2λ 1 – π2 λ2

∞

cos(xt)f (t) dt, 0

• Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971). 7.

b n y(x) – λ Ak cos[βk (x – t)] y(t) dt = f (x), a

n = 1, 2, . . .

k=1

This equation can be reduced to a special case of equation 4.9.20; the formula cos[β(x – t)] = cos(βx) cos(βt) + sin(βx) sin(βt) must be used. 8.

y(x) – λ

b

a

cos(βx) cos(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = cos(βx) and h(t) =

9.

y(x) – λ

b

a

cos(βt) cos(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

10.

y(x) – λ

1 . cos(βt)

1 and h(t) = cos(βt). cos(βx)

b

cosk (βx) cosm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cosk (βx) and h(t) = cosm (µt). 11.

y(x) – λ

b

tk cosm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cosm (βx) and h(t) = tk . 12.

y(x) – λ

b

xk cosm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = cosm (βt). 13.

b y(x) – λ [A + B(x – t) cos(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = cos(βx). 14.

b y(x) – λ [A + B(x – t) cos(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = cos(βt).

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4.5-2. Kernels Containing Sine 15.

y(x) – λ

b

sin(βx)y(t) dt = f (x).

a

16.

This is a special case of equation 4.9.1 with g(x) = sin(βx) and h(t) = 1. b y(x) – λ sin(βt)y(t) dt = f (x).

17.

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = sin(βt). b y(x) – λ sin[β(x – t)]y(t) dt = f (x).

a

a

This is a special case of equation 4.9.16 with g(x) = sin(βx) and h(t) = cos(βt). Solution:

y(x) = f (x) + λ A1 sin(βx) + A2 cos(βx) ,

18.

where A1 and A2 are the constants determined by the formulas presented in 4.9.16. b y(x) – λ sin[β(x + t)]y(t) dt = f (x). a

This is a special case of equation 4.9.15 with g(x) = sin(βx) and h(t) = cos(βt). Solution:

y(x) = f (x) + λ A1 sin(βx) + A2 cos(βx) ,

19.

where A1 and A2 are the constants determined by the formulas presented in 4.9.15. ∞ y(x) – λ sin(xt)y(t) dt = 0. 0

Characteristic values: λ = ± 2/π. For the characteristic values, the integral equation has infinitely many linearly independent eigenfunctions. Eigenfunctions for λ = + 2/π have the form ∞ 2 y+ (x) = f (x) + f (t) sin(xt) dt, π 0 where f = f (x) is any continuous function absolutely integrable on the interval [0, ∞). Eigenfunctions for λ = – 2/π have the form ∞ 2 y– (x) = f (x) – f (t) sin(xt) dt, π 0 where f = f (x) is any continuous function absolutely integrable on the interval [0, ∞).

20.

• Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971). ∞ y(x) – λ sin(xt)y(t) dt = f (x). 0

Solution: y(x) = where λ ≠ ± 2/π.

f (x) λ π 2 + 1– 2λ 1 – π2 λ2

∞

sin(xt)f (t) dt, 0

• References: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971), F. D. Gakhov and Yu. I. Cherskii (1978).

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21.

b n y(x) – λ Ak sin[βk (x – t)] y(t) dt = f (x), a

22.

23.

24.

n = 1, 2, . . .

k=1

This equation can be reduced to a special case of equation 4.9.20; the formula sin[β(x – t)] = sin(βx) cos(βt) – sin(βt) cos(βx) must be used. b sin(βx) y(x) – λ y(t) dt = f (x). a sin(βt) 1 This is a special case of equation 4.9.1 with g(x) = sin(βx) and h(t) = . sin(βt) b sin(βt) y(x) – λ y(t) dt = f (x). a sin(βx) 1 This is a special case of equation 4.9.1 with g(x) = and h(t) = sin (βt). sin(βx) b y(x) – λ sink (βx) sinm (µt)y(t) dt = f (x). a

25.

This is a special case of equation 4.9.1 with g(x) = sink (βx) and h(t) = sinm (µt). b y(x) – λ tk sinm (βx)y(t) dt = f (x).

26.

This is a special case of equation 4.9.1 with g(x) = sinm (βx) and h(t) = tk . b y(x) – λ xk sinm (βt)y(t) dt = f (x).

27.

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = sinm (βt). b y(x) – λ [A + B(x – t) sin(βt)]y(t) dt = f (x).

28.

This is a special case of equation 4.9.8 with h(t) = sin(βt). b y(x) – λ [A + B(x – t) sin(βx)]y(t) dt = f (x).

29.

This is a special case of equation 4.9.10 with h(x) = sin(βx). b y(x) + A sin(λ|x – t|)y(t) dt = f (x).

a

a

a

a

a

This is a special case of equation 4.9.39 with g(t) = A. 1◦ . The function y = y(x) obeys the following second-order linear nonhomogeneous ordinary differential equation with constant coefficients: yxx + λ(2A + λ)y = fxx (x) + λ2 f (x).

(1)

The boundary conditions for (1) have the form (see 4.9.39) sin[λ(b – a)]ϕx (b) – λ cos[λ(b – a)]ϕ(b) = λϕ(a), sin[λ(b – a)]ϕx (a) + λ cos[λ(b – a)]ϕ(a) = –λϕ(b),

ϕ(x) = y(x) – f (x).

(2)

Equation (1) under the boundary conditions (2) determines the solution of the original integral equation.

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2◦ . For λ(2A + λ) = –k 2 < 0, the general solution of equation (1) is given by

2Aλ k

y(x) = C1 cosh(kx) + C2 sinh(kx) + f (x) –

x

sinh[k(x – t)] f (t) dt,

(3)

a

where C1 and C2 are arbitrary constants. For λ(2A + λ) = k 2 > 0, the general solution of equation (1) is given by 2Aλ y(x) = C1 cos(kx) + C2 sin(kx) + f (x) – k

x

sin[k(x – t)] f (t) dt.

(4)

a

For λ = 2A, the general solution of equation (1) is given by

x

y(x) = C1 + C2 x + f (x) + 4A2

(x – t)f (t) dt.

(5)

a

The constants C1 and C2 in solutions (3)–(5) are determined by conditions (2). 30.

b

y(x) + A

t sin(λ|x – t|)y(t) dt = f (x). a

This is a special case of equation 4.9.39 with g(t) = At. The solution of the integral equation can be written via the Bessel functions (or modified Bessel functions) of order 1/3. 31.

b

sin3 (λ|x – t|)y(t) dt = f (x).

y(x) + A a

Using the formula sin3 β = – 14 sin 3β + with n = 2: y(x) +

32.

y(x) +

a

sin β, we arrive at an equation of the form 4.5.32

b

a

b n

3 4

– 14 A sin(3λ|x – t|) + 34 A sin(λ|x – t|) y(t) dt = f (x).

Ak sin(λk |x – t|) y(t) dt = f (x),

–∞ < a < b < ∞.

k=1

1◦ . Let us remove the modulus in the kth summand of the integrand:

b

sin(λk |x – t|)y(t) dt =

Ik (x) = a

x

b

sin[λk (x – t)]y(t) dt + a

sin[λk (t – x)]y(t) dt. (1) x

Differentiating (1) with respect to x twice yields Ik = λk

x

a

Ik

b

cos[λk (x – t)]y(t) dt – λk

= 2λk y(x) –

λ2k

cos[λk (t – x)]y(t) dt, x

x

sin[λk (x – t)]y(t) dt – a

(2)

b

λ2k

sin[λk (t – x)]y(t) dt, x

where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2), we find the relation between Ik and Ik : Ik = 2λk y(x) – λ2k Ik ,

Ik = Ik (x).

(3)

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2◦ . With the aid of (1), the integral equation can be rewritten in the form y(x) +

n

Ak Ik = f (x).

(4)

k=1

Differentiating (4) with respect to x twice and taking into account (3), we find that yxx (x) + σn y(x) –

n

Ak λ2k Ik = fxx (x),

σn = 2

n

Ak λ k .

(5)

Ak (λ2n – λ2k )Ik = fxx (x) + λ2n f (x).

(6)

k=1

k=1

Eliminating the integral In from (4) and (5) yields yxx (x) + (σn + λ2n )y(x) +

n–1 k=1

Differentiating (6) with respect to x twice and eliminating In–1 from the resulting equation with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear n–2 differential operator (acting on y) with constant coefficients plus the sum Bk Ik . If we k=1

successively eliminate In–2 , In–3 , . . . , with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients. 3◦ . The boundary conditions for y(x) can be found by setting x = a in the integral equation and all its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.) 33.

y(x) – λ

∞

sin(x – t) x–t

–∞

y(t) dt = f (x).

Solution: λ y(x) = f (x) + √ 2π – πλ

∞

–∞

sin(x – t) f (t) dt, x–t

λ≠

2 . π

• Reference: F. D. Gakhov and Yu. I. Cherskii (1978). 4.5-3. Kernels Containing Tangent 34.

y(x) – λ

b

tan(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tan(βx) and h(t) = 1. 35.

y(x) – λ

b

tan(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = tan(βt). 36.

b y(x) – λ [A tan(βx) + B tan(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.4 with g(x) = tan(βx).

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37.

y(x) – λ

b

a

tan(βx) tan(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = tan(βx) and h(t) =

38.

y(x) – λ

b

a

tan(βt) tan(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

39.

y(x) – λ

1 . tan(βt)

1 and h(t) = tan(βt). tan(βx)

b

tank (βx) tanm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tank (βx) and h(t) = tanm (µt). 40.

y(x) – λ

b

tk tanm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanm (βx) and h(t) = tk . 41.

y(x) – λ

b

xk tanm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = tanm (βt). 42.

b y(x) – λ [A + B(x – t) tan(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = tan(βt). 43.

b y(x) – λ [A + B(x – t) tan(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = tan(βx).

4.5-4. Kernels Containing Cotangent

44.

y(x) – λ

b

cot(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cot(βx) and h(t) = 1. 45.

y(x) – λ

b

cot(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = cot(βt). 46.

b y(x) – λ [A cot(βx) + B cot(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.4 with g(x) = cot(βx).

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47.

y(x) – λ

b

a

cot(βx) cot(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = cot(βx) and h(t) = 48.

y(x) – λ

b

a

cot(βt) cot(βx)

1 . cot(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

1 and h(t) = cot(βt). cot(βx)

49.

y(x) – λ

50.

This is a special case of equation 4.9.1 with g(x) = cotk (βx) and h(t) = cotm (µt). b y(x) – λ tk cotm (βx)y(t) dt = f (x).

51.

This is a special case of equation 4.9.1 with g(x) = cotm (βx) and h(t) = tk . b y(x) – λ xk cotm (βt)y(t) dt = f (x).

b

cotk (βx) cotm (µt)y(t) dt = f (x).

a

a

a

52.

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = cotm (βt). b y(x) – λ [A + B(x – t) cot(βt)]y(t) dt = f (x).

53.

This is a special case of equation 4.9.8 with h(t) = cot(βt). b y(x) – λ [A + B(x – t) cot(βx)]y(t) dt = f (x).

a

a

This is a special case of equation 4.9.10 with h(x) = cot(βx). 4.5-5. Kernels Containing Combinations of Trigonometric Functions 54.

y(x) – λ

b

cosk (βx) sinm (µt)y(t) dt = f (x).

a

55.

This is a special case of equation 4.9.1 with g(x) = cosk (βx) and h(t) = sinm (µt). b y(x) – λ [A sin(αx) cos(βt) + B sin(γx) cos(δt)]y(t) dt = f (x).

56.

This is a special case of equation 4.9.18 with g1 (x) = sin(αx), h1 (t) = A cos(βt), g2 (x) = sin(γx), and h2 (t) = B cos(δt). b y(x) – λ tank (γx) cotm (µt)y(t) dt = f (x).

57.

This is a special case of equation 4.9.1 with g(x) = tank (γx) and h(t) = cotm (µt). b y(x) – λ [A tan(αx) cot(βt) + B tan(γx) cot(δt)]y(t) dt = f (x).

a

a

a

This is a special case of equation 4.9.18 with g1 (x) = tan(αx), h1 (t) = A cot(βt), g2 (x) = tan(γx), and h2 (t) = B cot(δt).

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4.5-6. A Singular Equation 58.

t – x cot y(t) dt = f (x), 0 ≤ x ≤ 2π. 2π 0 2 Here the integral is understood in the sense of the Cauchy principal value. Without loss of generality we may assume that A2 + B 2 = 1. Solution: 2π 2π B t–x B2 y(x) = Af (x) + cot f (t) dt. f (t) dt + 2π 0 2 2πA 0

Ay(x) –

B

2π

• Reference: I. K. Lifanov (1996).

4.6. Equations Whose Kernels Contain Inverse Trigonometric Functions 4.6-1. Kernels Containing Arccosine 1.

y(x) – λ

b

arccos(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arccos(βx) and h(t) = 1. 2.

y(x) – λ

b

arccos(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arccos(βt). 3.

y(x) – λ

b

a

arccos(βx) arccos(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = arccos(βx) and h(t) =

4.

y(x) – λ

b

a

arccos(βt) arccos(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

5.

y(x) – λ

1 . arccos(βt)

1 and h(t) = arccos(βt). arccos(βx)

b

arccosk (βx) arccosm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arccosk (βx) and h(t) = arccosm (µt). 6.

y(x) – λ

b

tk arccosm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arccosm (βx) and h(t) = tk . 7.

y(x) – λ

b

xk arccosm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = arccosm (βt).

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8.

b y(x) – λ [A + B(x – t) arccos(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = arccos(βx). 9.

b y(x) – λ [A + B(x – t) arccos(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = arccos(βt). 4.6-2. Kernels Containing Arcsine 10.

y(x) – λ

b

arcsin(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arcsin(βx) and h(t) = 1. 11.

y(x) – λ

b

arcsin(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arcsin(βt). 12.

y(x) – λ

b

a

arcsin(βx) arcsin(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = arcsin(βx) and h(t) =

13.

y(x) – λ

b

a

arcsin(βt) arcsin(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

14.

y(x) – λ

1 . arcsin(βt)

1 and h(t) = arcsin (βt). arcsin(βx)

b

arcsink (βx) arcsinm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arcsink (βx) and h(t) = arcsinm (µt). 15.

y(x) – λ

b

tk arcsinm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arcsinm (βx) and h(t) = tk . 16.

y(x) – λ

b

xk arcsinm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = arcsinm (βt). 17.

b y(x) – λ [A + B(x – t) arcsin(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = arcsin(βt). 18.

b y(x) – λ [A + B(x – t) arcsin(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = arcsin(βx).

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4.6-3. Kernels Containing Arctangent

19.

y(x) – λ

b

arctan(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arctan(βx) and h(t) = 1. 20.

y(x) – λ

b

arctan(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arctan(βt). 21.

b y(x) – λ [A arctan(βx) + B arctan(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.4 with g(x) = arctan(βx). 22.

y(x) – λ

b

a

arctan(βx) arctan(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = arctan(βx) and h(t) =

23.

y(x) – λ

b

a

arctan(βt) arctan(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

24.

y(x) – λ

1 . arctan(βt)

1 and h(t) = arctan(βt). arctan(βx)

b

arctank (βx) arctanm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arctank (βx) and h(t) = arctanm (µt). 25.

y(x) – λ

b

tk arctanm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arctanm (βx) and h(t) = tk . 26.

y(x) – λ

b

xk arctanm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = arctanm (βt). 27.

b y(x) – λ [A + B(x – t) arctan(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = arctan(βt). 28.

b y(x) – λ [A + B(x – t) arctan(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = arctan(βx).

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4.6-4. Kernels Containing Arccotangent

29.

y(x) – λ

b

arccot(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arccot(βx) and h(t) = 1. 30.

y(x) – λ

b

arccot(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arccot(βt). 31.

b y(x) – λ [A arccot(βx) + B arccot(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.4 with g(x) = arccot(βx). 32.

y(x) – λ

b

a

arccot(βx) arccot(βt)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = arccot(βx) and h(t) =

33.

y(x) – λ

b

a

arccot(βt) arccot(βx)

y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) =

34.

y(x) – λ

1 . arccot(βt)

1 and h(t) = arccot(βt). arccot(βx)

b

arccotk (βx) arccotm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arccotk (βx) and h(t) = arccotm (µt). 35.

y(x) – λ

b

tk arccotm (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = arccotm (βx) and h(t) = tk . 36.

y(x) – λ

b

xk arccotm (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = xk and h(t) = arccotm (βt). 37.

b y(x) – λ [A + B(x – t) arccot(βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = arccot(βt). 38.

b y(x) – λ [A + B(x – t) arccot(βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = arccot(βx).

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4.7. Equations Whose Kernels Contain Combinations of Elementary Functions 4.7-1. Kernels Containing Exponential and Hyperbolic Functions

1.

y(x) – λ

b

eµ(x–t) cosh[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx cosh(βx), h1 (t) = e–µt cosh(βt), g2 (x) = eµx sinh(βx), and h2 (t) = –e–µt sinh(βt). 2.

y(x) – λ

b

eµ(x–t) sinh[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx sinh(βx), h1 (t) = e–µt cosh(βt), g2 (x) = eµx cosh(βx), and h2 (t) = –e–µt sinh(βt). 3.

y(x) – λ

b

teµ(x–t) sinh[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx sinh(βx), h1 (t) = te–µt cosh(βt), g2 (x) = eµx cosh(βx), and h2 (t) = –te–µt sinh(βt). 4.7-2. Kernels Containing Exponential and Logarithmic Functions

4.

y(x) – λ

b

eµt ln(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = ln(βx) and h(t) = eµt . 5.

y(x) – λ

b

eµx ln(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = eµx and h(t) = ln(βt). 6.

y(x) – λ

b

eµ(x–t) ln(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = eµx ln(βx) and h(t) = e–µt . 7.

y(x) – λ

b

eµ(x–t) ln(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = eµx and h(t) = e–µt ln(βt). 8.

y(x) – λ

b

eµ(x–t) (ln x – ln t)y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx ln x, h1 (t) = e–µt , g2 (x) = eµx , and h2 (t) = –e–µt ln t.

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9.

y(x) +

b2 – a2

2a

0

∞

x exp –aln y(t) dt = f (x). t t

1

Solution with a > 0, b > 0, and x > 0: a2 – b2 y(x) = f (x) + 2b

∞

0

x 1 exp –bln f (t) dt. t t

• Reference: F. D. Gakhov and Yu. I. Cherskii (1978). 4.7-3. Kernels Containing Exponential and Trigonometric Functions

10.

y(x) – λ

b

eµt cos(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cos(βx) and h(t) = eµt . 11.

y(x) – λ

b

eµx cos(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = eµx and h(t) = cos(βt). 12.

y(x) – λ

∞

eµ(x–t) cos(xt)y(t) dt = f (x).

0

Solution: f (x) λ y(x) = π 2 + 1– 2λ 1 – π2 λ2 13.

y(x) – λ

∞

eµ(x–t) cos(xt)f (t) dt,

λ ≠ ± 2/π.

0

b

eµ(x–t) cos[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx cos(βx), h1 (t) = e–µt cos(βt), g2 (x) = eµx sin(βx), and h2 (t) = e–µt sin(βt). 14.

y(x) – λ

b

eµt sin(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sin(βx) and h(t) = eµt . 15.

y(x) – λ

b

eµx sin(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = eµx and h(t) = sin(βt). 16.

y(x) – λ

∞

eµ(x–t) sin(xt)y(t) dt = f (x).

0

Solution: y(x) =

f (x) λ + 1 – π2 λ2 1 – π2 λ2

∞

eµ(x–t) sin(xt)f (t) dt,

λ ≠ ± 2/π.

0

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17.

y(x) – λ

b

eµ(x–t) sin[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx sin(βx), h1 (t) = e–µt cos(βt), g2 (x) = eµx cos(βx), and h2 (t) = –e–µt sin(βt). 18.

y(x) – λ

b

eµ(x–t)

a

n

Ak sin[βk (x – t)] y(t) dt = f (x),

n = 1, 2, . . .

k=1

This is a special case of equation 4.9.20. 19.

y(x) – λ

b

teµ(x–t) sin[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx sin(βx), h1 (t) = te–µt cos(βt), g2 (x) = eµx cos(βx), and h2 (t) = –te–µt sin(βt). 20.

y(x) – λ

b

xeµ(x–t) sin[β(x – t)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = xeµx sin(βx), h1 (t) = e–µt cos(βt), g2 (x) = xeµx cos(βx), and h2 (t) = –e–µt sin(βt). 21.

y(x) – λ

b

eµt tan(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tan(βx) and h(t) = eµt . 22.

y(x) – λ

b

eµx tan(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = eµx and h(t) = tan(βt). 23.

y(x) – λ

b

eµ(x–t) [tan(βx) – tan(βt)]y(t) dt = f (x).

a

This is a special case of equation 4.9.18 with g1 (x) = eµx tan(βx), h1 (t) = e–µt , g2 (x) = eµx , and h2 (t) = –e–µt tan(βt). 24.

y(x) – λ

b

eµt cot(βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cot(βx) and h(t) = eµt . 25.

y(x) – λ

b

eµx cot(βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = eµx and h(t) = cot(βt). 4.7-4. Kernels Containing Hyperbolic and Logarithmic Functions 26.

y(x) – λ

b

coshk (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = coshk (βx) and h(t) = lnm (µt).

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27.

y(x) – λ

b

coshk (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = coshk (βt). 28.

y(x) – λ

b

sinhk (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinhk (βx) and h(t) = lnm (µt). 29.

y(x) – λ

b

sinhk (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = sinhk (βt). 30.

y(x) – λ

b

tanhk (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanhk (βx) and h(t) = lnm (µt). 31.

y(x) – λ

b

tanhk (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = tanhk (βt). 32.

y(x) – λ

b

cothk (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cothk (βx) and h(t) = lnm (µt). 33.

y(x) – λ

b

cothk (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = cothk (βt).

4.7-5. Kernels Containing Hyperbolic and Trigonometric Functions

34.

y(x) – λ

b

coshk (βx) cosm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = coshk (βx) and h(t) = cosm (µt). 35.

y(x) – λ

b

coshk (βt) cosm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cosm (µx) and h(t) = coshk (βt). 36.

y(x) – λ

b

coshk (βx) sinm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = coshk (βx) and h(t) = sinm (µt).

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37.

y(x) – λ

b

coshk (βt) sinm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinm (µx) and h(t) = coshk (βt). 38.

y(x) – λ

b

sinhk (βx) cosm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinhk (βx) and h(t) = cosm (µt). 39.

y(x) – λ

b

sinhk (βt) cosm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cosm (µx) and h(t) = sinhk (βt). 40.

y(x) – λ

b

sinhk (βx) sinm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinhk (βx) and h(t) = sinm (µt). 41.

y(x) – λ

b

sinhk (βt) sinm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinm (µx) and h(t) = sinhk (βt). 42.

y(x) – λ

b

tanhk (βx) cosm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanhk (βx) and h(t) = cosm (µt). 43.

y(x) – λ

b

tanhk (βt) cosm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cosm (µx) and h(t) = tanhk (βt). 44.

y(x) – λ

b

tanhk (βx) sinm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tanhk (βx) and h(t) = sinm (µt). 45.

y(x) – λ

b

tanhk (βt) sinm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sinm (µx) and h(t) = tanhk (βt).

4.7-6. Kernels Containing Logarithmic and Trigonometric Functions

46.

y(x) – λ

b

cosk (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cosk (βx) and h(t) = lnm (µt).

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47.

y(x) – λ

b

cosk (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = cosk (βt). 48.

y(x) – λ

b

sink (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = sink (βx) and h(t) = lnm (µt). 49.

y(x) – λ

b

sink (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = sink (βt). 50.

y(x) – λ

b

tank (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = tank (βx) and h(t) = lnm (µt). 51.

y(x) – λ

b

tank (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = tank (βt). 52.

y(x) – λ

b

cotk (βx) lnm (µt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = cotk (βx) and h(t) = lnm (µt). 53.

y(x) – λ

b

cotk (βt) lnm (µx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = lnm (µx) and h(t) = cotk (βt).

4.8. Equations Whose Kernels Contain Special Functions 4.8-1. Kernels Containing Bessel Functions 1.

y(x) – λ

b

Jν (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = Jν (βx) and h(t) = 1. 2.

y(x) – λ

b

Jν (βt)y(t) dt = f (x).

a

3.

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = Jν (βt). ∞ y(x) + λ tJν (xt)y(t) dt = f (x), ν > – 12 . 0

Solution:

f (x) λ y(x) = – 1 – λ2 1 – λ2

∞

tJν (xt)f (t) dt,

λ ≠ ±1.

0

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4.

y(x) + λ

∞

√ Jν 2 xt y(t) dt = f (x).

0

By setting x = 12 z 2 , t = 12 τ 2 , y(x) = Y (z), and f (x) = F (z), we arrive at an equation of the form 4.8.3: ∞ Y (z) + λ τ Jν (zτ )Y (τ ) dτ = F (z). 0

5.

b y(x) – λ [A + B(x – t)Jν (βt)]y(t) dt = f (x). a

6.

This is a special case of equation 4.9.8 with h(t) = Jν (βt). b y(x) – λ [A + B(x – t)Jν (βx)]y(t) dt = f (x).

7.

This is a special case of equation 4.9.10 with h(x) = Jν (βx). b y(x) – λ [AJµ (αx) + BJν (βt)]y(t) dt = f (x).

8.

This is a special case of equation 4.9.5 with g(x) = AJµ (αx) and h(t) = BJν (βt). b y(x) – λ [AJµ (x)Jν (t) + BJν (x)Jµ (t)]y(t) dt = f (x).

9.

This is a special case of equation 4.9.17 with g(x) = Jµ (x) and h(t) = Jν (t). b y(x) – λ Yν (βx)y(t) dt = f (x).

10.

This is a special case of equation 4.9.1 with g(x) = Yν (βx) and h(t) = 1. b y(x) – λ Yν (βt)y(t) dt = f (x).

11.

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = Yν (βt). b y(x) – λ [A + B(x – t)Yν (βt)]y(t) dt = f (x).

12.

This is a special case of equation 4.9.8 with h(t) = Yν (βt). b y(x) – λ [A + B(x – t)Yν (βx)]y(t) dt = f (x).

13.

This is a special case of equation 4.9.10 with h(x) = Yν (βx). b y(x) – λ [AYµ (αx) + BYν (βt)]y(t) dt = f (x).

14.

This is a special case of equation 4.9.5 with g(x) = AYµ (αx) and h(t) = BYν (βt). b y(x) – λ [AYµ (x)Yµ (t) + BYν (x)Yν (t)]y(t) dt = f (x).

15.

This is a special case of equation 4.9.14 with g(x) = Yµ (x) and h(t) = Yν (t). b y(x) – λ [AYµ (x)Yν (t) + BYν (x)Yµ (t)]y(t) dt = f (x).

a

a

a

a

a

a

a

a

a

a

This is a special case of equation 4.9.17 with g(x) = Yµ (x) and h(t) = Yν (t).

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4.8-2. Kernels Containing Modified Bessel Functions 16.

y(x) – λ

b

Iν (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = Iν (βx) and h(t) = 1. 17.

y(x) – λ

b

Iν (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = Iν (βt). 18.

b y(x) – λ [A + B(x – t)Iν (βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = Iν (βt). 19.

b y(x) – λ [A + B(x – t)Iν (βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = Iν (βx). 20.

b y(x) – λ [AIµ (αx) + BIν (βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.5 with g(x) = AIµ (αx) and h(t) = BIν (βt). 21.

b y(x) – λ [AIµ (x)Iµ (t) + BIν (x)Iν (t)]y(t) dt = f (x). a

This is a special case of equation 4.9.14 with g(x) = Iµ (x) and h(t) = Iν (t). 22.

b y(x) – λ [AIµ (x)Iν (t) + BIν (x)Iµ (t)]y(t) dt = f (x). a

This is a special case of equation 4.9.17 with g(x) = Iµ (x) and h(t) = Iν (t). 23.

y(x) – λ

b

Kν (βx)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = Kν (βx) and h(t) = 1. 24.

y(x) – λ

b

Kν (βt)y(t) dt = f (x).

a

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = Kν (βt). 25.

b y(x) – λ [A + B(x – t)Kν (βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.8 with h(t) = Kν (βt). 26.

b y(x) – λ [A + B(x – t)Kν (βx)]y(t) dt = f (x). a

This is a special case of equation 4.9.10 with h(x) = Kν (βx).

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27.

b y(x) – λ [AKµ (αx) + BKν (βt)]y(t) dt = f (x). a

This is a special case of equation 4.9.5 with g(x) = AKµ (αx) and h(t) = BKν (βt). 28.

b y(x) – λ [AKµ (x)Kµ (t) + BKν (x)Kν (t)]y(t) dt = f (x). a

This is a special case of equation 4.9.14 with g(x) = Kµ (x) and h(t) = Kν (t). 29.

b y(x) – λ [AKµ (x)Kν (t) + BKν (x)Kµ (t)]y(t) dt = f (x). a

This is a special case of equation 4.9.17 with g(x) = Kµ (x) and h(t) = Kν (t).

4.9. Equations Whose Kernels Contain Arbitrary Functions 4.9-1. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t) 1.

y(x) – λ

b

g(x)h(t)y(t) dt = f (x).

a

1◦ . Assume that λ ≠ Solution:

–1

b a

g(t)h(t) dt

.

y(x) = f (x) + λkg(x),

k=

where

–1

b

1–λ

g(t)h(t) dt a

2◦ . Assume that λ = b

For

h(t)f (t) dt. a

–1

b a

b

g(t)h(t) dt

.

h(t)f (t) dt = 0, the solution has the form

a

y = f (x) + Cg(x), where C is an arbitrary constant. b

For h(t)f (t) dt ≠ 0, there is no solution. a The limits of integration may take the values a = –∞ and/or b = ∞, provided that the corresponding improper integral converges. 2.

b y(x) – λ [g(x) + g(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1 =

g1 +

where

√

1 , (b – a)g2

a

g1 –

b

g(x) dx,

g1 =

λ2 =

√

1 , (b – a)g2

b

g 2 (x) dx.

g2 = a

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1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 ], where the constants A1 and A2 are given by A1 =

f1 – λ[f1 g1 – (b – a)f2 ] f2 – λ(f2 g1 – f1 g2 ) , A2 = 2 , 2 – (b – a)g2 ]λ – 2g1 λ + 1 [g1 – (b – a)g2 ]λ2 – 2g1 λ + 1 b b f1 = f (x) dx, f2 = f (x)g(x) dx.

[g12

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

g2 , b–a

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0: y(x) = f (x) + Cy2 (x),

y2 (x) = g(x) –

g2 , b–a

where C is an arbitrary constant and y2 (x) is an eigenfunction of the equation corresponding to the characteristic value λ2 .

3.

4◦ . The equation has no multiple characteristic values. b y(x) – λ [g(x) – g(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1 =

1 g12

where

– (b – a)g2

g1 =

,

g12

b

g(x) dx,

1

λ2 = –

– (b – a)g2

,

b

g 2 (x) dx.

g2 =

a

a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 ], where the constants A1 and A2 are given by A1 =

f1 + λ[f1 g1 – (b – a)f2 ] –f2 + λ(f2 g1 – f1 g2 ) , A2 = , 2 2 [(b – a)g2 – g1 ]λ + 1 [(b – a)g2 – g12 ]λ2 + 1 b b f1 = f (x) dx, f2 = f (x)g(x) dx. a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 – λ1 g1 , λ1 (b – a) where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . The equation has no multiple characteristic values.

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4.

b y(x) – λ [Ag(x) + Bg(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1,2

(A + B)g1 ± (A – B)2 g12 + 4AB(b – a)g2 , = 2AB[g12 – (b – a)g2 ]

where

g1 =

b

g(x) dx,

b

g 2 (x) dx.

g2 =

a

a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 ], where the constants A1 and A2 are given by A1 =

Af1 – λAB[f1 g1 – (b – a)f2 ] Bf2 – λAB(f2 g1 – f1 g2 ) , A2 = , 2 2 2 AB[g1 – (b – a)g2 ]λ – (A + B)g1 λ + 1 AB[g1 – (b – a)g2 ]λ2 – (A + B)g1 λ + 1 b b f1 = f (x) dx, f2 = f (x)g(x) dx. a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

1 – λ1 Ag1 , λ1 A(b – a)

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = is double: y(x) = f (x) + Cy∗ (x),

y∗ (x) = g(x) –

2 (A + B)g1

(A – B)g1 . 2A(b – a)

Here C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . 5.

b y(x) – λ [g(x) + h(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 where

s1 =

s1 + s3 ± (s1 – s3 )2 + 4(b – a)s2 , = 2[s1 s3 – (b – a)s2 ]

b

g(x) dx, a

s2 =

b

g(x)h(x) dx, a

s3 =

b

h(x) dx. a

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1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 ], where the constants A1 and A2 are given by A1 =

f1 – λ[f1 s3 – (b – a)f2 ] f2 – λ(f2 s1 – f1 s2 ) , A2 = , [s1 s3 – (b – a)s2 ]λ2 – (s1 + s3 )λ + 1 [s1 s3 – (b – a)s2 ]λ2 – (s1 + s3 )λ + 1 b b f1 = f (x) dx, f2 = f (x)h(x) dx. a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

1 – λ1 s 1 , λ1 (b – a)

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 .

6.

3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 2 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = s1 + s3 is double: s1 – s3 y∗ (x) = g(x) – y(x) = f (x) + Cy∗ (x), . 2(b – a) Here C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . b y(x) – λ [Ag(x) + Bg(t)]h(t) y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 = where

(A + B)s1 ± (A – B)2 s12 + 4ABs0 s2 , 2AB(s12 – s0 s2 )

b

s0 =

h(x) dx, a

b

s1 =

g(x)h(x) dx, a

b

g 2 (x)h(x) dx.

s2 = a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 ], where the constants A1 and A2 are given by A1 =

Af1 – ABλ(f1 s1 – f2 s0 ) Bf2 – ABλ(f2 s1 – f1 s2 ) , A2 = , AB(s12 – s0 s2 )λ2 – (A + B)s1 λ + 1 AB(s12 – s0 s2 )λ2 – (A + B)s1 λ + 1 b b f1 = f (x)h(x) dx, f2 = f (x)g(x)h(x) dx. a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

1 – λ1 As1 , λ1 As0

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 .

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3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = is double:

2 (A + B)s1

y(x) = f (x) + Cy∗ (x), where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . Two cases are possible. (a) If A ≠ B, then 4ABs0 s2 = –(A – B)2 s12 and y∗ (x) = g(x) –

(A – B)s1 . 2As0

(b) If A = B, then, in view of 4ABs0 s2 = –(A – B)2 s12 = 0, we have g(x) y∗ (x) =

for s0 ≠ 0 and s2 = 0, for s0 = 0 and s2 ≠ 0, for s0 = s2 = 0,

1 C1 g(x) + C2

where C1 and C2 are arbitrary constants. 7.

b y(x) – λ [Ag(x) + Bg(t) + C]h(t) y(t) dt = f (x). a

The characteristic values of the equation: (A + B)s1 + Cs0 ± (A – B)2 s12 + 2(A + B)Cs1 s0 + C 2 s02 + 4ABs0 s2 λ1,2 = , 2AB(s12 – s0 s2 ) where

b

s0 =

h(x) dx,

s1 =

a

b

g(x)h(x) dx, a

b

g 2 (x)h(x) dx.

s2 = a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 ], where the constants A1 and A2 are given by Af1 – ABλ(f1 s1 – f2 s0 ) , – s0 s2 )λ2 – [(A + B)s1 + Cs0 ]λ + 1 C1 f1 + Bf2 – ABλ(f2 s1 – f1 s2 ) A2 = , AB(s12 – s0 s2 )λ2 – [(A + B)s1 + Cs0 ]λ + 1 b b f1 = f (x)h(x) dx, f2 = f (x)g(x)h(x) dx. A1 =

AB(s12

a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 (x), y(x) = f (x) + Cy

y1 (x) = g(x) +

1 – λ1 As1 , λ1 As0

is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding where C to the characteristic value λ1 .

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3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = 2 is double: (A + B)s1 + Cs0 ∗ (x), y(x) = f (x) + Cy is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding where C to λ∗ . Two cases are possible. (a) If As1 – (Bs1 + Cs0 ) ≠ 0, then 4As0 (Bs2 + Cs1 ) = –[(A – B)s1 – Cs0 ]2 and y∗ (x) = g(x) –

(A – B)s1 – Cs0 . 2As0

(b) If (A – B)s1 = Cs0 , then, in view of 4As0 (Bs2 + Cs1 ) = –[(A – B)s1 – Cs0 ]2 = 0, we have g(x) for s0 ≠ 0 and Bs2 = –Cs1 , for s0 = 0 and Bs2 ≠ –Cs1 , y∗ (x) = 1 1 g(x) + C 2 for s0 = 0 and Bs2 = –Cs1 , C 1 and C 2 are arbitrary constants. where C 8.

b y(x) – λ [A + B(x – t)h(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1,2

A(b – a) ± [A(b – a) – 2Bh1 ]2 + 2Bh0 [A(b2 – a2 ) – 2Bh2 ] = , B A(b – a)[2h1 – (b + a)h0 ] – 2B(h21 – h0 h2 )

where

h0 =

b

h(x) dx,

h1 =

a

b

xh(x) dx,

b

x2 h(x) dx.

h2 =

a

a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 + A2 x), where the constants A1 and A2 are given by

f1 – λ B(f1 h1 + f2 h2 ) – 12 Af2 (b2 – a2 )

A1 = , B A(b – a) h1 – 12 (b + a)h0 – B(h21 – h0 h2 ) λ2 + A(b – a)λ + 1 f – λ[A(b – a)f2 – B(f1 h0 + f2 h1 )]

2 A2 = , B A(b – a) h1 – 12 (b + a)h0 – B(h21 – h0 h2 ) λ2 + A(b – a)λ + 1 b b b f (x) dx – B xf (x)h(x) dx, f2 = B f (x)h(x) dx. f1 = A a

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = 1 +

2 – 2λ1 [A(b – a) – Bh1 ] x, λ1 [A(b2 – a2 ) – 2Bh2 ]

where C is an arbitrary constant, and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 .

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3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = (A ≠ 0) is double:

2 A(b – a)

y(x) = f (x) + Cy∗ (x), where C is an arbitrary constant, and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . Two cases are possible. (a) If A(b – a) – 2Bh1 ≠ 0, then A(b – a) – 2Bh1 x. A(b2 – a2 ) – 2Bh2

y∗ (x) = 1 –

(b) If A(b – a) – 2Bh1 = 0, then, in view of h0 [A(b2 – a2 ) – 2Bh2 ] = 0, we have  1 for h0 ≠ 0 and A(b2 – a2 ) = 2Bh2 , y∗ (x) = x for h0 = 0 and A(b2 – a2 ) ≠ 2Bh2 ,  C1 + C2 x for h0 = 0 and A(b2 – a2 ) = 2Bh2 , where C1 and C2 are arbitrary constants. 9.

b y(x) – λ [A + (Bx + Ct)h(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1,2

√ A(b – a) + (C + B)h1 ± D , = B A(b – a)[2h1 – (b + a)h0 ] + 2C(h21 – h0 h2 )

D = [A(b – a) + (C – B)h1 ]2 + 2Bh0 [A(b2 – a2 ) + 2Ch2 ], where

h0 =

b

h(x) dx, a

h1 =

b

xh(x) dx, a

b

x2 h(x) dx.

h2 = a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ(A1 + A2 x), where the constants A1 and A2 are given by

A1 = ∆–1 f1 – λ Bf1 h1 – Cf2 h2 – 12 A(b2 – a2 )f2 ,

A2 = ∆–1 f2 – λ A(b – a)f2 – Bf1 h0 + Cf2 h1 ,

∆ = B A(b – a) h1 – 12 (b + a)h0 + C(h21 – h0 h2 ) λ2 + [A(b – a) + (B + C)h1 ]λ + 1, b b b f1 = A f (x) dx + C xf (x)h(x) dx, f2 = B f (x)h(x) dx. a

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 (x), y(x) = f (x) + Cy

y1 (x) = 1 +

2 – 2λ1 [A(b – a) + Ch1 ] x, λ1 [A(b2 – a2 ) + 2Ch2 ]

is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding where C to the characteristic value λ1 .

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3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = 2 is double: A(b – a) + (B + C)h1 ∗ (x), y(x) = f (x) + Cy is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding where C to λ∗ . Two cases are possible. (a) If A(b – a) + (C – B)h1 ≠ 0, then y∗ (x) = 1 –

A(b – a) + (C – B)h1 x. A(b2 – a2 ) + 2Ch2

(b) If A(b – a) + (C – B)h1 = 0, then, in view of h0 [A(b2 – a2 ) + 2Ch2 ] = 0, we have  1 for h0 ≠ 0 and A(b2 – a2 ) = –2Ch2 , y∗ (x) = x for h0 = 0 and A(b2 – a2 ) ≠ –2Ch2 ,  C1 + C2 x for h0 = 0 and A(b2 – a2 ) = –2Ch2 , 1 and C 2 are arbitrary constants. where C 10.

b y(x) – λ [A + B(x – t)h(x)]y(t) dt = f (x). a

The characteristic values of the equation: A(b – a) ± [A(b – a) + 2Bh1 ]2 – 2Bh0 [A(b2 – a2 ) + 2Bh2 ] λ1,2 = , B{h0 [A(b2 – a2 ) + 2Bh2 ] – 2h1 [A(b – a) + Bh1 ]} where

h0 =

b

h(x) dx,

h1 =

a

b

xh(x) dx, a

b

x2 h(x) dx.

h2 = a

1◦ . Solution with λ ≠ λ1,2 :

y(x) = f (x) + λ AE1 + (BE1 x + E2 )h(x) , where the constants E1 and E2 are given by

E1 = ∆–1 f1 + λB(f1 h1 – f2 h0 ) ,

E2 = B∆–1 –f2 + λf2 A(b – a) + Bh1 + λf1 12 A(b2 – a2 ) + Bh2 , ∆ = B h0 12 A(b2 – a2 ) + Bh2 – h1 [A(b – a) + Bh1 ] λ2 – A(b – a)λ + 1, b b f1 = f (x) dx, f2 = xf (x) dx. a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = A + Bxh(x) +

1 – λ1 [A(b – a) + Bh1 ] h(x), λ 1 h0

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 .

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3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = (A ≠ 0) is double:

2 A(b – a)

y(x) = f (x) + Cy∗ (x), where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . Two cases are possible. (a) If A(b – a) ≠ –2Bh1 , then y∗ (x) = A + Bxh(x) –

A(b – a) + 2Bh1 h(x). 2h0

(b) If A(b – a) = –2Bh1 , then, in view of h0 [A(b2 – a2 ) + 2Bh2 ] = 0, we have   A + Bxh(x) for h0 ≠ 0 and A(b2 – a2 ) = –2Bh2 , y∗ (x) = h(x) for h0 = 0 and A(b2 – a2 ) ≠ –2Bh2 ,  C1 [A + Bxh(x)] + C2 h(x) for h0 = 0 and A(b2 – a2 ) = –2Bh2 , where C1 and C2 are arbitrary constants. 11.

b y(x) – λ [A + (Bx + Ct)h(x)]y(t) dt = f (x). a

The characteristic values of the equation: √ A(b – a) + (B + C)h1 ± D λ1,2 = , C{2h1 [A(b – a) + Bh1 ] – h0 [A(b2 – a2 ) + 2Bh2 ]} D = [A(b – a) + (B – C)h1 ]2 + 2Ch0 [A(b2 – a2 ) + 2Bh2 ], where

h0 =

b

h(x) dx, a

b

h1 =

xh(x) dx, a

b

x2 h(x) dx.

h2 = a

1◦ . Solution with λ ≠ λ1,2 :

y(x) = f (x) + λ AE1 + (BE1 x + E2 )h(x) , where the constants E1 and E2 are given by E1 = ∆–1 [f1 – λC(f1 h1 – f2 h0 )],

E2 = C∆–1 f2 – λf2 A(b – a) + Bh1 – λf1 12 A(b2 – a2 ) + Bh2 ,

∆ = C h1 [A(b – a) + Bh1 ] – h0 12 A(b2 – a2 ) + Bh2 λ2 – [A(b – a) + (B + C)h1 ]λ + 1, b b f1 = f (x) dx, f2 = xf (x) dx. a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 (x), y(x) = f (x) + Cy

y1 (x) = A + Bxh(x) +

1 – λ1 [A(b – a) + Bh1 ] h(x), λ 1 h0

is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding where C to the characteristic value λ1 .

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3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = 2 (A ≠ 0) is double: A(b – a) + (B + C)h1 ∗ (x), y(x) = f (x) + Cy is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding where C to λ∗ . Two cases are possible. (a) If A(b – a) + Bh1 ≠ Ch1 , then y∗ (x) = A + Bxh(x) –

A(b – a) + (B – C)h1 h(x). 2h0

(b) If A(b – a) + Bh1 = Ch1 , then, in view of h0 [A(b2 – a2 ) + 2Bh2 ] = 0, we have   A + Bxh(x) for h0 ≠ 0 and A(b2 – a2 ) = –2Bh2 , y∗ (x) = h(x) for h0 = 0 and A(b2 – a2 ) ≠ –2Bh2 ,  2 h(x) for h0 = 0 and A(b2 – a2 ) = –2Bh2 , C1 [A + Bxh(x)] + C

12.

1 and C 2 are arbitrary constants. where C b y(x) – λ [g(x)g(t) + h(x)h(t)]y(t) dt = f (x). a

The characteristic values of the equation: s1 + s3 + (s1 – s3 )2 + 4s22 λ1 = , 2(s1 s3 – s22 ) where

b

g 2 (x) dx,

s1 =

s1 + s3 – (s1 – s3 )2 + 4s22 , 2(s1 s3 – s22 )

b

s2 =

a

λ2 =

g(x)h(x) dx, a

b

h2 (x) dx.

s3 = a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 h(x)], where the constants A1 and A2 are given by A1 =

f1 – λ(f1 s3 – f2 s2 ) , (s1 s3 – s22 )λ2 – (s1 + s3 )λ + 1 b f1 = f (x)g(x) dx,

f2 – λ(f2 s1 – f1 s2 ) , (s1 s3 – s22 )λ2 – (s1 + s3 )λ + 1 b f2 = f (x)h(x) dx. A2 =

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 – λ1 s1 h(x), λ1 s2 where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = 1/s1 is double: 1 g(x) + C 2 h(x), y(x) = f (x) + C 1 and C 2 are arbitrary constants. where C

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13.

b y(x) – λ [g(x)g(t) – h(x)h(t)]y(t) dt = f (x). a

The characteristic values of the equation: s1 – s3 + (s1 + s3 )2 – 4s22 λ1 = , 2(s22 – s1 s3 ) where

b

g 2 (x) dx,

s1 =

s1 – s3 – (s1 + s3 )2 – 4s22 λ2 = , 2(s22 – s1 s3 )

b

s2 =

g(x)h(x) dx,

a

b

h2 (x) dx.

s3 =

a

a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 h(x)], where the constants A1 and A2 are given by A1 =

f1 + λ(f1 s3 – f2 s2 ) , – s1 s3 )λ2 – (s1 – s3 )λ + 1 b f1 = f (x)g(x) dx,

(s22

A2 =

(s22 b

f2 =

a

–f2 + λ(f2 s1 – f1 s2 ) , – s1 s3 )λ2 – (s1 – s3 )λ + 1

f (x)h(x) dx. a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

1 – λ1 s1 h(x), λ1 s2

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = is double: y∗ (x) = g(x) –

y(x) = f (x) + Cy∗ (x),

2 s1 – s3

s1 + s3 h(x), 2s2

where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . 14.

b y(x) – λ [Ag(x)g(t) + Bh(x)h(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 where

b 2

s1 =

As1 + Bs3 ± (As1 – Bs3 )2 + 4ABs22 , = 2AB(s1 s3 – s22 )

g (x) dx, a

s2 =

b

g(x)h(x) dx, a

b

h2 (x) dx.

s3 = a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 h(x)],

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where the constants A1 and A2 are given by Af1 – λAB(f1 s3 – f2 s2 ) A1 = , AB(s1 s3 – s22 )λ2 – (As1 + Bs3 )λ + 1 b f1 = f (x)g(x) dx,

Bf2 – λAB(f2 s1 – f1 s2 ) , AB(s1 s3 – s22 )λ2 – (As1 + Bs3 )λ + 1 b f2 = f (x)h(x) dx. A2 =

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: 1 – λ1 As1 h(x), λ1 As2 where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 2 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = As1 + Bs3 is double: y(x) = f (x) + Cy∗ (x), where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . Two cases are possible. (a) If As1 ≠ Bs3 , then 4ABs22 = –(As1 – Bs3 )2 , AB < 0, and As1 – Bs3 y∗ (x) = g(x) – h(x). 2As2 (b) If As1 = Bs3 , then, in view of s2 = 0, we have y∗ (x) = C1 g(x) + C2 h(x),

15.

where C1 and C2 are arbitrary constants. b y(x) – λ [g(x)h(t) + h(x)g(t)]y(t) dt = f (x). a

The characteristic values of the equation: 1 λ1 = , √ s1 + s2 s3 where b

s1 =

h(x)g(x) dx,

λ2 =

1 √

s2 s3

,

b

h2 (x) dx,

s2 =

a

s1 –

a

1◦ . Solution with λ ≠ λ1,2 :

b

g 2 (x) dx.

s3 = a

y(x) = f (x) + λ[A1 g(x) + A2 h(x)], where the constants A1 and A2 are given by f1 – λ(f1 s1 – f2 s2 ) A1 = 2 , (s1 – s2 s3 )λ2 – 2s1 λ + 1 b f1 = f (x)h(x) dx,

f2 – λ(f2 s1 – f1 s3 ) , (s12 – s2 s3 )λ2 – 2s1 λ + 1 b f2 = f (x)g(x) dx.

A2 =

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0:

s3 h(x), s2 where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

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3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0: y(x) = f (x) + Cy2 (x),

s3 h(x), s2

y2 (x) = g(x) –

where C is an arbitrary constant and y2 (x) is an eigenfunction of the equation corresponding to the characteristic value λ2 . 4◦ . The equation has no multiple characteristic values.

16.

b y(x) – λ [g(x)h(t) – h(x)g(t)]y(t) dt = f (x). a

The characteristic values of the equation: 1

λ1 =

s12 – s2 s3

where

s12 – s2 s3

b

s1 =

h(x)g(x) dx,

,

b

h2 (x) dx,

s2 =

a

1

λ2 = –

,

a

b

g 2 (x) dx.

s3 = a

◦

1 . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 h(x)], where the constants A1 and A2 are given by f1 + λ(f1 s1 – f2 s2 ) –f2 + λ(f2 s1 – f1 s3 ) , A2 = , (s2 s3 – s12 )λ2 + 1 (s2 s3 – s12 )λ2 + 1 b b f1 = f (x)h(x) dx, f2 = f (x)g(x) dx.

A1 =

a

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

s12 – s2 s3 – s1 h(x), s2

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . Solution with λ = λ2 ≠ λ1 and f1 = f2 = 0: y(x) = f (x) + Cy2 (x),

y2 (x) = g(x) –

s12 – s2 s3 + s1 h(x), s2

where C is an arbitrary constant and y2 (x) is an eigenfunction of the equation corresponding to the characteristic value λ2 . 4◦ . The equation has no multiple characteristic values.

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17.

b y(x) – λ [Ag(x)h(t) + Bh(x)g(t)]y(t) dt = f (x). a

The characteristic values of the equation: λ1,2 = where

s1 =

(A + B)s1 ± (A – B)2 s12 + 4ABs2 s3 , 2AB(s12 – s2 s3 )

b

h(x)g(x) dx,

b 2

s2 =

h (x) dx,

a

b

g 2 (x) dx.

s3 =

a

a

1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g(x) + A2 h(x)], where the constants A1 and A2 are given by A1 =

Af1 – λAB(f1 s1 – f2 s2 ) , AB(s12 – s2 s3 )λ2 – (A + B)s1 λ + 1 b f1 = f (x)h(x) dx, a

Bf2 – λAB(f2 s1 – f1 s3 ) , AB(s12 – s2 s3 )λ2 – (A + B)s1 λ + 1 b f2 = f (x)g(x) dx.

A2 =

a

2◦ . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x),

y1 (x) = g(x) +

1 – λ1 As1 h(x), λ1 As2

where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 . 3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = is double: y(x) = f (x) + Cy∗ (x),

y∗ (x) = g(x) –

2 (A + B)s1

(A – B)s1 h(x). 2As2

Here C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . 18.

b y(x) – λ [g1 (x)h1 (t) + g2 (x)h2 (t)]y(t) dt = f (x). a

The characteristic values of the equation λ1 and λ2 are given by s11 + s22 ± (s11 – s22 )2 + 4s12 s21 λ1,2 = , 2(s11 s22 – s12 s21 ) provided that the integrals b b b b s11 = h1(x)g1(x) dx, s12 = h1(x)g2(x) dx, s21 = h2(x)g1(x) dx, s22 = h2(x)g2(x) dx a

a

a

a

are convergent.

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1◦ . Solution with λ ≠ λ1,2 : y(x) = f (x) + λ[A1 g1 (x) + A2 g2 (x)], where the constants A1 and A2 are given by A1 =

f1 – λ(f1 s22 – f2 s12 ) , (s11 s22 – s12 s21 )λ2 – (s11 + s22 )λ + 1 b f1 = f (x)h1 (x) dx,

f2 – λ(f2 s11 – f1 s21 ) , (s11 s22 – s12 s21 )λ2 – (s11 + s22 )λ + 1 b f2 = f (x)h2 (x) dx.

A2 =

a

a

◦

2 . Solution with λ = λ1 ≠ λ2 and f1 = f2 = 0: y(x) = f (x) + Cy1 (x), where C is an arbitrary constant and y1 (x) is an eigenfunction of the equation corresponding to the characteristic value λ1 : y1 (x) = g1 (x) +

1 – λ1 s11 λ1 s21 g2 (x) = g1 (x) + g2 (x). λ1 s12 1 – λ1 s22

3◦ . The solution with λ = λ2 ≠ λ1 and f1 = f2 = 0 is given by the formulas of item 2◦ in which one must replace λ1 and y1 (x) by λ2 and y2 (x), respectively. 4◦ . Solution with λ = λ1,2 = λ∗ and f1 = f2 = 0, where the characteristic value λ∗ = is double (there is no double characteristic value provided that s11 = –s22 ):

2 s11 + s22

y(x) = f (x) + Cy∗ (x), where C is an arbitrary constant and y∗ (x) is an eigenfunction of the equation corresponding to λ∗ . Two cases are possible. (a) If s11 ≠ s22 , then s12 = – 14 (s11 – s22 )2/s21 , s21 ≠ 0, and y∗ (x) = g1 (x) –

s11 – s22 g2 (x). 2s12

Note that in this case, s12 and s21 have opposite signs. (b) If s11 = s22 , then, in view of s12 s21 = – 14 (s11 – s22 )2 = 0, we have g (x) for s12 ≠ 0 and s21 = 0, 1 for s12 = 0 and s21 ≠ 0, y∗ (x) = g2 (x) C1 g1 (x) + C2 g2 (x) for s12 = 0 and s21 = 0, where C1 and C2 are arbitrary constants. 19.

b y(x) – λ [g(x) + h(t)]m y(t) dt = f (x),

m = 1, 2, . . .

a k m–k This is a special case of equation 4.9.20, with gk (x) = g k (x), hk (t) = Cm h (t), and k = 1, . . . , m. Solution: m y(x) = f (x) + λ Ak g k (x), k=0

where the Ak are constants that can be determined from 4.9.20.

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20.

b n y(x) – λ gk (x)hk (t) y(t) dt = f (x), a

n = 2, 3, . . .

k=1

The characteristic values of the integral equation (counting the multiplicity, we have exactly n of them) are the roots of the algebraic equation ∆(λ) = 0, where

1 – λs11 –λs12 –λs21 1 – λs22 ∆(λ) = .. .. . . –λsn1 –λsn2 and the integrals

s12 s11 – λ–1 s s – λ–1 21 22 = (–λ)n . .. .. . sn1 · · · 1 – λsnn sn2 ··· ··· .. .

–λs1n –λs2n .. .

··· ··· .. .

s1n s2n .. .

· · · snn – λ

, –1

b

smk =

hm (x)gk (x) dx;

m, k = 1, . . . , n,

a

are assumed to be convergent. Solution with regular λ: y(x) = f (x) + λ

n

Ak gk (x),

k=1

where the constants Ak form the solution of the following system of algebraic equations: b n Am – λ smk Ak = fm , fm = f (x)hm (x) dx, m = 1, . . . , n. a

k=1

The Ak can be calculated by Cramer’s rule: Ak = ∆k (λ)/∆(λ), –λs1n 1 – λs11 · · · –λs1k–1 f1 –λs1k+1 · · · · · · –λs2k–1 f2 –λs2k+1 · · · –λs2n –λs21 ∆k (λ) = . ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ –λsn1 · · · –λsnk–1 fn –λsnk+1 · · · 1 – λsnn For solutions of the equation in the case in which λ is a characteristic value, see Subsection 11.2-2.

where

• Reference: S. G. Mikhlin (1960). 4.9-2. Equations With Difference Kernel: K(x, t) = K(x – t) 21.

y(x) = λ

π

K(x – t)y(t) dt,

K(x) = K(–x).

–π

Characteristic values: 1 1 π λn = , an = K(x) cos(nx) dx πan π –π The corresponding eigenfunctions are y0 (x) = 1,

yn(1) (x) = cos(nx),

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

yn(2) (x) = sin(nx)

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

For each value λn with n ≠ 0, there are two corresponding linearly independent eigenfunctions yn(1) (x) and yn(2) (x). • Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971).

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22.

∞

y(x) +

K(x – t)y(t) dt = Aeλx .

–∞

Solution:

23.

A λx e , 1+q

y(x) =

∞

q=

K(x)e–λx dx.

–∞

∞

y(x) +

K(x – t)y(t) dt = A cos(λx) + B sin(λx). –∞

Solution: AIc + BIs BIc – AIs y(x) = cos(λx) + sin(λx), Ic2 + Is2 Ic2 + Is2 ∞ ∞ Ic = 1 + K(z) cos(λz) dz, Is = K(z) sin(λz) dz. –∞

24.

–∞

∞

y(x) –

K(x – t)y(t) dt = f (x). –∞

Here –∞ < x < ∞, f (x) ∈ L1 (–∞, ∞), and K(x) ∈ L1 (–∞, ∞). For the integral equation to be solvable (in L1 ), it is necessary and sufficient that √ 1 – 2π K(u) ≠ 0, –∞ < u < ∞, where K(u) =

√1 2π

∞ –∞

(1)

K(x)e–iux dx is the Fourier transform of K(x). In this case, the

equation has a unique solution, which is given by ∞ y(x) = f (x) + R(x – t)f (t) dt, 1 R(x) = √ 2π

25.

–∞ ∞

iux du, R(u)e

–∞

R(u) =

K(u) . √ 1 – 2π K(u)

• Reference: V. A. Ditkin and A. P. Prudnikov (1965). ∞ y(x) – K(x – t)y(t) dt = f (x). 0

The Wiener–Hopf equation of the second kind.* Here 0 ≤ x < ∞, K(x) ∈ L1 (–∞, ∞), f (x) ∈ L1 (0, ∞), and y(x) ∈ L1 (0, ∞). For the integral equation to be solvable, it is necessary and sufficient that ˇ Ω(u) = 1 – K(u) ≠ 0,

–∞ < u < ∞,

(1)

∞

ˇ where K(u) = K(x)eiux dx is the Fourier transform (in the asymmetric form) of K(x). –∞ In this case, the index of the equation can be introduced, ν = –ind Ω(u) = – 1◦ . Solution with ν = 0:

y(x) = f (x) +

∞ 1 arg Ω(u) –∞ . 2π

∞

R(x, t)f (t) dt, 0

* A comprehensive discussion of this equation is given in Subsection 11.9-1, Section 11.10, and Section 11.11.

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where

∞

R(x, t) = R+ (x – t) + R– (t – x) +

R+ (x – s)R– (t – s) ds, 0

and the functions R+ (x) and R– (x) satisfy the conditions R+ (x) = 0 and R– (x) = 0 for x < 0 and are uniquely defined by their Fourier transforms as follows: ∞ ∞ ln Ω(t) 1 1 ±iut 1+ dt . R± (t)e dt = exp – ln Ω(u) ∓ 2 2πi –∞ t – u 0 Alternatively, R+ (x) and R– (x) can be obtained by constructing the solutions of the equations ∞ R+ (x) + K(x – t)R+ (t) dt = K(x), 0 ≤ x ≤ ∞, 0 ∞ K(t – x)R– (t) dt = K(–x), 0 ≤ x ≤ ∞. R– (x) + 0 ◦

2 . Solution with ν > 0: y(x) = f (x) +

ν

m–1 –x

Cm x

e

∞

+

ν m–1 –t R (x, t) f (t) + Cm t e dt, ◦

0

m=1

m=1

where the Cm are arbitrary constants, R◦ (x, t) = R+(0) (x – t) + R–(1) (t – x) +

∞

R+(0) (x – s)R–(1) (t – s) ds,

0

and the functions R+(0) (x) and R–(1) (x) are uniquely defined by their Fourier transforms:

ν ∞ ∞ u–i (1) (0) ±iut ±iut 1+ 1+ R± (t)e dt = R± (t)e dt , u+i 0 0 ∞ ∞ ln Ω◦ (t) 1 1 (0) ±iut ◦ 1+ R± (t)e dt = exp – ln Ω (u) ∓ dt , 2 2πi –∞ t – u 0 Ω◦ (u)(u + i)ν = Ω(u)(u – i)ν . 3◦ . For ν < 0, the solution exists only if the conditions ∞ f (x)ψm (x) dx = 0, m = 1, 2, . . . , –ν, 0

are satisfied. Here ψ1 (x), . . . , ψν (x) is the system of linearly independent solutions of the transposed homogeneous equation ∞ ψ(x) – K(t – x)ψ(t) dt = 0. 0

Then y(x) = f (x) +

∞

R∗ (x, t)f (t) dt,

0

where ∗

R (x, t) =

R+(1) (x

– t) +

R–(0) (t

– x) +

∞

R+(1) (x – s)R–(0) (t – s) ds,

0

and the functions R+(1) (x) and R–(0) (x) are uniquely defined in item 2◦ by their Fourier transforms. • References: V. I. Smirnov (1974), F. D. Gakhov and Yu. I. Cherskii (1978), I. M. Vinogradov (1979).

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4.9-3. Other Equations of the Form y(x) + 26.

b a

K(x, t)y(t) dt = F (x)

∞

y(x) –

K(x + t)y(t) dt = f (x). –∞

The Fourier transform is used to solving this equation. Solution: √ ∞ 1 f (u) + 2π f(–u)K(u) y(x) = √ eiux du, √ K(–u) 2π –∞ 1 – 2π K(u) where

27.

1 f(u) = √ 2π

∞

f (x)e–iux dx,

–∞

1 K(u) = √ 2π

∞

K(x)e–iux dx.

–∞

• Reference: V. A. Ditkin and A. P. Prudnikov (1965). ∞ y(x) + eβt K(x + t)y(t) dt = Aeλx . –∞

Solution:

28.

eλx – k(λ)e–(β+λ)x y(x) = , 1 – k(λ)k(–β – λ) ∞

y(x) +

∞

k(λ) =

K(x)e(λ+β)x dx.

–∞

[eβt K(x + t) + M (x – t)]y(t) dt = Aeλx .

–∞

Solution: y(x) = A

Ik (λ)epx – [1 + Im (p)]eλx , Ik (λ)Ik (p) – [1 + Im (λ)][1 + Im (p)]

where

∞

Ik (λ) =

K(z)e(β+λ)z dz,

–∞

29.

y(x) –

∞

Im (λ) =

p = –λ – β,

M (z)e–λz dz.

–∞

∞

K(xt)y(t) dt = f (x). 0

The solution can be obtained with the aid of the inverse Mellin transform: c+i∞ f(1 – s) 1 f (s) + K(s) y(x) = x–s ds, K(1 – s) 2πi c–i∞ 1 – K(s) stand for the Mellin transforms of the right-hand side and of the kernel of the where f and K integral equation, ∞ ∞ s–1 f (s) = f (x)x dx, K(s) = K(x)xs–1 dx. 0

30.

0

• Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971). ∞ y(x) – K(xt)tβ y(t) dt = Axλ . 0

Solution: y(x) = A

xλ + Iβ+λ x–β–λ–1 , 1 – Iβ+λ I–λ–1

Iµ =

∞

K(ξ)ξ µ dξ.

0

It is assumed that all improper integrals are convergent.

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31.

∞

y(x) –

K(xt)tβ y(t) dt = f (x).

0

The solution can be obtained with the aid of the inverse Mellin transform as follows: 1 y(x) = 2πi

c+i∞

c–i∞

f(1 + β – s) f(s) + K(s) x–s ds, K(1 + β – s) 1 – K(s)

stand for the Mellin transforms of the right-hand side and of the kernel of the where f and K integral equation, f(s) =

∞

s–1

f (x)x

dx,

K(s) =

0

32.

∞

y(x) –

∞

K(x)xs–1 dx.

0

g(xt)xλ tµ y(t) dt = f (x).

0

This equation can be rewritten in the form of equation 4.9.31 by setting K(z) = z λ g(z) and β = µ – λ. 33.

y(x) – 0

∞

1 t

K

x t

y(t) dt = 0.

Eigenfunctions of this integral equation are determined by the roots of the following transcendental (algebraic) equation for the parameter λ:

∞

K

1

0

z

z λ–1 dz = 1.

(1)

1◦ . For a real simple root λn of equation (1), there is a corresponding eigenfunction yn (x) = xλn . 2◦ . For a real root λn of multiplicity r, there are corresponding r eigenfunctions yn1 (x) = xλn ,

yn2 (x) = xλn ln x,

...,

ynr (x) = xλn lnr–1 x.

3◦ . For a complex simple root λn = αn + iβn of equation (1), there is a corresponding pair of eigenfunctions yn(1) (x) = xαn cos(βn ln x),

yn(2) (x) = xαn sin(βn ln x).

4◦ . For a complex root λn = αn +iβn of multiplicity r, there are corresponding r eigenfunction pairs (1) (2) yn1 (x) = xαn cos(βn ln x), (x) = xαn sin(βn ln x), yn1 (1) yn2 (x) = xαn ln x cos(βn ln x),

(2) yn2 (x) = xαn ln x sin(βn ln x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

r–1

(2) (x) = xαn lnr–1 x sin(βn ln x). ynr

(1) (x) ynr

αn

=x

ln

x cos(βn ln x),

The general solution is the linear combination (with arbitrary constants) of the eigenfunctions of the homogeneous integral equation.

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34.

∞

y(x) –

1 t

0

K

x t

y(t) dt = Axb .

A solution: y(x) =

A b x , B

∞

B =1–

K

1 ξ

0

ξ b–1 dξ.

It is assumed that the improper integral is convergent and B ≠ 0. The general solution of the integral equations is the sum of the above solution and the solution of the homogeneous equation 4.9.33. 35.

∞

y(x) –

1 t

0

K

x t

y(t) dt = f (x).

The solution can be obtained with the aid of the inverse Mellin transform: 1 y(x) = 2πi

c+i∞

c–i∞

f(s) x–s ds, 1 – K(s)

stand for the Mellin transforms of the right-hand side and the kernel of the where f and K integral equation, f(s) =

∞

K(s) =

f (x)xs–1 dx,

0

∞

K(x)xs–1 dx.

0

Example. For f (x) = Ae–λx and K(x) =

1 –x e , 2

the solution of the integral equation has the form

 4A     (3 – 2C)(λx)3 y(x) = ∞ 1     –2A sk ψ(s ) (λx) k k=1

for λx > 1, for λx < 1.

Here C = 0.5772 . . . is the Euler constant, ψ(z) = [ln Γ(z)]z is the logarithmic derivative of the gamma function, and the sk are the negative roots of the transcendental equation Γ(sk ) = 2, where Γ(z) is the gamma function.

• Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971). 36.

b

|x – t|g(t)y(t) dt = f (x),

y(x) +

a ≤ x ≤ b.

a

1◦ . Let us remove the modulus in the integrand, y(x) +

x

b

(x – t)g(t)y(t) dt + a

(t – x)g(t)y(t) dt = f (x).

(1)

g(t)y(t) dt = fx (x).

(2)

x

Differentiating (1) with respect to x yields yx (x) +

x

b

g(t)y(t) dt – a

x

Differentiating (2), we arrive at a second-order ordinary differential equation for y = y(x), yxx + 2g(x)y = fxx (x).

(3)

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2◦ . Let us derive the boundary conditions for equation (3). We assume that the limits of integration satisfy the conditions –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain two consequences b y(a) + (t – a)g(t)y(t) dt = f (a), a (4) b y(b) +

(b – t)g(t)y(t) dt = f (b).

a yxx

Let us express g(x)y from (3) via and fxx and substitute the result into (4). Integrating by parts yields the desired boundary conditions for y(x), y(a) + y(b) + (b – a)[fx (b) – yx (b)] = f (a) + f (b), (5) y(a) + y(b) + (a – b)[fx (a) – yx (a)] = f (a) + f (b). Note a useful consequence of (5),

yx (a) + yx (b) = fx (a) + fx (b),

37.

(6)

which can be used together with one of conditions (5). Equation (3) under the boundary conditions (5) determines the solution of the original integral equation. Conditions (5) make it possible to calculate the constants of integration that occur in the solution of the differential equation (3). b y(x) + eλ|x–t| g(t)y(t) dt = f (x), a ≤ x ≤ b. a

1◦ . Let us remove the modulus in the integrand: x y(x) + eλ(x–t) g(t)y(t) dt + a

b

eλ(t–x) g(t)y(t) dt = f (x).

(1)

x

Differentiating (1) with respect to x twice yields x 2 λ(x–t) 2 yxx (x) + 2λg(x)y(x) + λ e g(t)y(t) dt + λ a

b

eλ(t–x) g(t)y(t) dt = fxx (x).

(2)

x

Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary differential equation for y = y(x), yxx + 2λg(x)y – λ2 y = fxx (x) – λ2 f (x).

(3)

◦

2 . Let us derive the boundary conditions for equation (3). We assume that the limits of integration satisfy the conditions –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain two consequences b –λa y(a) + e eλt g(t)y(t) dt = f (a), a (4) b y(b) + eλb

e–λt g(t)y(t) dt = f (b). a

Let us express g(x)y from (3) via yxx and fxx and substitute the result into (4). Integrating by parts yields the conditions

eλb ϕx (b) – eλa ϕx (a) = λeλa ϕ(a) + λeλb ϕ(b),

ϕ(x) = y(x) – f (x). e–λb ϕx (b) – e–λa ϕx (a) = λe–λa ϕ(a) + λe–λb ϕ(b), Finally, after some manipulations, we arrive at the desired boundary conditions for y(x): ϕx (a) + λϕ(a) = 0,

ϕx (b) – λϕ(b) = 0;

ϕ(x) = y(x) – f (x).

(5)

Equation (3) under the boundary conditions (5) determines the solution of the original integral equation. Conditions (5) make it possible to calculate the constants of integration that occur in solving the differential equation (3).

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38.

b

y(x) +

a ≤ x ≤ b.

sinh(λ|x – t|)g(t)y(t) dt = f (x), a

1◦ . Let us remove the modulus in the integrand:

x

y(x) +

b

sinh[λ(x – t)]g(t)y(t) dt +

sinh[λ(t – x)]g(t)y(t) dt = f (x).

a

(1)

x

Differentiating (1) with respect to x twice yields x yxx (x) + 2λg(x)y(x) + λ2 sinh[λ(x – t)]g(t)y(t) dt a

b

+ λ2

sinh[λ(t – x)]g(t)y(t) dt = fxx (x).

(2)

x

Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary differential equation for y = y(x), yxx + 2λg(x)y – λ2 y = fxx (x) – λ2 f (x).

(3)

2◦ . Let us derive the boundary conditions for equation (3). We assume that the limits of integration satisfy the conditions –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain two corollaries b y(a) + sinh[λ(t – a)]g(t)y(t) dt = f (a), a (4) b y(b) +

sinh[λ(b – t)]g(t)y(t) dt = f (b). a

Let us express g(x)y from (3) via yxx and fxx and substitute the result into (4). Integrating by parts yields the desired boundary conditions for y(x),

sinh[λ(b – a)]ϕx (b) – λ cosh[λ(b – a)]ϕ(b) = λϕ(a), sinh[λ(b – a)]ϕx (a) + λ cosh[λ(b – a)]ϕ(a) = –λϕ(b);

ϕ(x) = y(x) – f (x).

(5)

Equation (3) under the boundary conditions (5) determines the solution of the original integral equation. Conditions (5) make it possible to calculate the constants of integration that occur in solving the differential equation (3). 39.

b

y(x) +

a ≤ x ≤ b.

sin(λ|x – t|)g(t)y(t) dt = f (x), a

1◦ . Let us remove the modulus in the integrand: y(x) +

x

b

sin[λ(x – t)]g(t)y(t) dt + a

sin[λ(t – x)]g(t)y(t) dt = f (x).

(1)

x

Differentiating (1) with respect to x twice yields x yxx (x) + 2λg(x)y(x) – λ2 sin[λ(x – t)]g(t)y(t) dt a

b

– λ2

sin[λ(t – x)]g(t)y(t) dt = fxx (x).

(2)

x

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Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary differential equation for y = y(x), yxx + 2λg(x)y + λ2 y = fxx (x) + λ2 f (x).

(3)

◦

2 . Let us derive the boundary conditions for equation (3). We assume that the limits of integration satisfy the conditions –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain two consequences b y(a) + sin[λ(t – a)]g(t)y(t) dt = f (a), a (4) b y(b) +

sin[λ(b – t)]g(t)y(t) dt = f (b). a

Let us express g(x)y from (3) via yxx and fxx and substitute the result into (4). Integrating by parts yields the desired boundary conditions for y(x), sin[λ(b – a)]ϕx (b) – λ cos[λ(b – a)]ϕ(b) = λϕ(a), (5) sin[λ(b – a)]ϕx (a) + λ cos[λ(b – a)]ϕ(a) = –λϕ(b); ϕ(x) = y(x) – f (x). Equation (3) under the boundary conditions (5) determines the solution of the original integral equation. Conditions (5) make it possible to calculate the constants of integration that occur in solving the differential equation (3).

4.9-4. Equations of the Form y(x) + 40.

b a

K(x, t)y(· · ·) dt = F (x)

b

y(x) +

f (t)y(x – t) dt = 0. a

Eigenfunctions of this integral equation* are determined by the roots of the following characteristic (transcendental or algebraic) equation for µ: b f (t) exp(–µt) dt = –1. (1) a

1◦ . For a real (simple) root µk of equation (1), there is a corresponding eigenfunction yk (x) = exp(µk x). ◦

2 . For a real root µk of multiplicity r, there are corresponding r eigenfunctions yk1 (x) = exp(µk x),

yk2 (x) = x exp(µk x),

...,

ykr (x) = xr–1 exp(µk x).

3◦ . For a complex (simple) root µk = αk + iβk of equation (1), there is a corresponding pair of eigenfunctions yk(1) (x) = exp(αk x) cos(βk x),

yk(2) (x) = exp(αk x) sin(βk x).

4◦ . For a complex root µk = αk + iβk of multiplicity r, there are corresponding r pairs of eigenfunctions (1) yk1 (x) = exp(αk x) cos(βk x),

(2) (x) = exp(αk x) sin(βk x), yk1

(1) yk2 (x) = x exp(αk x) cos(βk x),

(2) yk2 (x) = x exp(αk x) sin(βk x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (1) ykr (x)

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

(2) = x exp(αk x) cos(βk x), ykr (x) = xr–1 exp(αk x) sin(βk x). The general solution is the linear combination (with arbitrary constants) of the eigenfunctions of the homogeneous integral equation. r–1

* In the equations below that contain y(x – t) in the integrand, the arguments can have, for example, the domain (a) –∞ < x < ∞, –∞ < t < ∞ for a = –∞ and b = ∞ or (b) a ≤ t ≤ b, –∞ ≤ x < ∞, for a and b such that –∞ < a < b < ∞. Case (b) is a special case of (a) if f (t) is nonzero only on the interval a ≤ t ≤ b.

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For equations 4.9.41–4.9.46, only particular solutions are given. To obtain the general solution, one must add the particular solution to the general solution of the corresponding homogeneous equation 4.9.40. b 41. y(x) + f (t)y(x – t) dt = Ax + B. a

A solution: y(x) = px + q, where the coefficients p and q are given by A , 1 + I0

p= 42.

AI1 B + , (1 + I0 )2 1 + I0

q=

I0 =

b

f (t) dt,

b

I1 =

a

tf (t) dt. a

b

f (t)y(x – t) dt = Aeλx .

y(x) + a

A solution:

43.

b A λx y(x) = e , B =1+ f (t) exp(–λt) dt. B a The general solution of the integral equation is the sum of the specified particular solution and the general solution of the homogeneous equation 4.9.40. b y(x) + f (t)y(x – t) dt = A sin(λx). a

A solution:

AIc AIs sin(λx) + 2 cos(λx), Ic2 + Is2 Ic + Is2 where the coefficients Ic and Is are given by b b Ic = 1 + f (t) cos(λt) dt, Is = f (t) sin(λt) dt. y(x) =

a

44.

a

b

y(x) +

f (t)y(x – t) dt = A cos(λx). a

A solution:

AIs AIc sin(λx) + 2 cos(λx), Ic2 + Is2 Ic + Is2 where the coefficients Ic and Is are given by b b Ic = 1 + f (t) cos(λt) dt, Is = f (t) sin(λt) dt. y(x) = –

a

45.

a

b

f (t)y(x – t) dt = eµx (A sin λx + B cos λx).

y(x) + a

A solution: y(x) = eµx (p sin λx + q cos λx), where the coefficients p and q are given by p=

AIc – BIs , Ic2 + Is2

b

f (t)e–µt cos(λt) dt,

Ic = 1 + a

AIs + BIc , Ic2 + Is2 b Is = f (t)e–µt sin(λt) dt.

q=

a

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46.

b

y(x) +

f (t)y(x – t) dt = g(x). a n

1◦ . For g(x) =

Ak exp(λk x), the equation has a solution

k=1

y(x) =

n Ak exp(λk x), Bk

b

Bk = 1 +

f (t) exp(–λk t) dt. a

k=1

2◦ . For polynomial right-hand side of the equation, g(x) =

n

Ak xk , a solution has the form

k=0

y(x) =

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 3◦ . For g(x) = eλx Ak xk , a solution of the equation has the form k=0

y(x) = eλx

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 4◦ . For g(x) = Ak cos(λk x), a solution of the equation has the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 5◦ . For g(x) = Ak sin(λk x), a solution of the equation has the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n Ak xk , a solution of the equation has the form 6◦ . For g(x) = cos(λx) k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. n Ak xk , a solution of the equation has the form 7◦ . For g(x) = sin(λx) k=0

y(x) = cos(λx)

n k=0

Bk xk + sin(λx)

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients.

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8◦ . For g(x) = eµx

n

Ak cos(λk x), a solution of the equation has the form

k=1

y(x) = eµx

n

Bk cos(λk x) + eµx

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 9◦ . For g(x) = eµx

n

Ak sin(λk x), a solution of the equation has the form

k=1

y(x) = eµx

n

Bk cos(λk x) + eµx

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 10◦ . For g(x) = cos(λx)

n

Ak exp(µk x), a solution of the equation has the form

k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 11◦ . For g(x) = sin(λx)

n

Ak exp(µk x), a solution of the equation has the form

k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 47.

b

y(x) +

f (t)y(x + βt) dt = Ax + B. a

A solution:* y(x) = px + q, where A p= , 1 + I0 48.

B AI1 β q= – , 1 + I0 (1 + I0 )2

I0 =

b

f (t) dt,

I1 =

a

b

tf (t) dt. a

b

f (t)y(x + βt) dt = Aeλx .

y(x) + a

A solution: y(x) =

A λx e , B

B =1+

b

f (t) exp(λβt) dt. a

* In the equations below that contain y(x + βt), β > 0, in the integrand, the arguments can have, for example, the domain (a) 0 ≤ x < ∞, 0 ≤ t < ∞ for a = 0 and b = ∞ or (b) a ≤ t ≤ b, 0 ≤ x < ∞ for a and b such that 0 ≤ a < b < ∞. Case (b) is a special case of (a) if f (t) is nonzero only on the interval a ≤ t ≤ b.

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49.

b

y(x) +

f (t)y(x + βt) dt = A sin λx + B cos λx. a

A solution: y(x) = p sin λx + q cos λx, where the coefficients p and q are given by p= Ic = 1 +

AIc + BIs , Ic2 + Is2

BIc – AIs , Ic2 + Is2 b Is = f (t) sin(λβt) dt.

q=

b

f (t) cos(λβt) dt, a

50.

a

b

y(x) +

f (t)y(x + βt) dt = g(x). a n

1◦ . For g(x) =

Ak exp(λk x), a solution of the equation has the form

k=1

n Ak y(x) = exp(λk x), Bk

b

Bk = 1 +

f (t) exp(βλk t) dt. a

k=1

2◦ . For polynomial right-hand side of the equation, g(x) =

n

Ak xk , a solution has the form

k=0

y(x) =

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 3◦ . For g(x) = eλx Ak xk , a solution of the equation has the form k=0

λx

y(x) = e

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 4◦ . For g(x) = Ak cos(λk x), a solution of the equation has the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 5◦ . For g(x) = Ak sin(λk x), a solution of the equation has the form k=1

y(x) =

n k=1

Bk cos(λk x) +

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients.

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6◦ . For g(x) = cos(λx)

n

Ak xk , a solution of the equation has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. 7◦ . For g(x) = sin(λx)

n

Ak xk , a solution of the equation has the form

k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. 8◦ . For g(x) = eµx

n

Ak cos(λk x), a solution of the equation has the form

k=1

y(x) = eµx

n

Bk cos(λk x) + eµx

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 9◦ . For g(x) = eµx

n

Ak sin(λk x), a solution of the equation has the form

k=1

y(x) = eµx

n k=1

Bk cos(λk x) + eµx

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 10◦ . For g(x) = cos(λx)

n

Ak exp(µk x), a solution of the equation has the form

k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 11◦ . For g(x) = sin(λx)

n

Ak exp(µk x), a solution of the equation has the form

k=1

y(x) = cos(λx)

n k=1

Bk exp(µk x) + sin(λx)

n

Bk exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients.

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51.

b

y(x) +

f (t)y(xt) dt = 0. a

Eigenfunctions of this integral equation* are determined by the roots of the following transcendental (or algebraic) equation for λ: b f (t)tλ dt = –1. (1) a ◦

1 . For a real (simple) root λk of equation (1), there is a corresponding eigenfunction yk (x) = xλk . 2◦ . For a real root λk of multiplicity r, there are corresponding r eigenfunctions yk1 (x) = xλk ,

yk2 (x) = xλk ln x,

ykr (x) = xλk lnr–1 x.

...,

3◦ . For a complex (simple) root λk = αk + iβk of equation (1), there is a corresponding pair of eigenfunctions yk(1) (x) = xαk cos(βk ln x),

yk(2) (x) = xαk sin(βk ln x).

4◦ . For a complex root λk = αk + iβk of multiplicity r, there are corresponding r pairs of eigenfunctions (1) yk1 (x) = xαk cos(βk ln x),

(2) (x) = xαk sin(βk ln x), yk1

(1) yk2 (x) = xαk ln x cos(βk ln x),

(2) yk2 (x) = xαk ln x sin(βk ln x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (1) ykr (x)

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

= xαk lnr–1 x cos(βk ln x),

(2) ykr (x) = xαk lnr–1 x sin(βk ln x).

The general solution is the linear combination (with arbitrary constants) of the eigenfunctions of the homogeneous integral equation. For equations 4.9.52–4.9.58, only particular solutions are given. To obtain the general solution, one must add the particular solution to the general solution of the corresponding homogeneous equation 4.9.51. 52.

b

y(x) +

f (t)y(xt) dt = Ax + B. a

A solution: y(x) = 53.

A B x+ , 1 + I1 1 + I0

I0 =

b

f (t) dt,

I1 =

a

b

tf (t) dt. a

b

f (t)y(xt) dt = Axβ .

y(x) + a

A solution: y(x) =

A β x , B

b

f (t)tβ dt.

B =1+ a

* In the equations below that contain y(xt) in the integrand, the arguments can have, for example, the domain (a) 0 ≤ x ≤ 1, 0 ≤ t ≤ 1 for a = 0 and b = 1, (b) 1 ≤ x < ∞, 1 ≤ t < ∞ for a = 1 and b = ∞, (c) 0 ≤ x < ∞, 0 ≤ t < ∞ for a = 0 and b = ∞, or (d) a ≤ t ≤ b, 0 ≤ x < ∞ for a and b such that 0 ≤ a < b ≤ ∞. Case (d) is a special case of (c) if f (t) is nonzero only on the interval a ≤ t ≤ b.

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54.

b

y(x) +

f (t)y(xt) dt = A ln x + B. a

A solution: y(x) = p ln x + q, where A , 1 + I0

p= 55.

q=

B AIl – , 1 + I0 (1 + I0 )2

b

I0 =

f (t) dt,

Il =

a

b

f (t) ln t dt. a

b

f (t)y(xt) dt = Axβ ln x.

y(x) + a

A solution: y(x) = pxβ ln x + qxβ , where p= 56.

A , 1 + I1

q=–

AI2 , (1 + I1 )2

b

f (t)tβ dt,

I1 = a

b

f (t)tβ ln t dt.

I2 = a

b

y(x) +

f (t)y(xt) dt = A cos(ln x). a

A solution: AIc AIs cos(ln x) + 2 sin(ln x), Ic2 + Is2 Ic + Is2 b b Ic = 1 + f (t) cos(ln t) dt, Is = f (t) sin(ln t) dt. y(x) =

a

57.

a

b

y(x) +

f (t)y(xt) dt = A sin(ln x). a

A solution: AIs AIc cos(ln x) + 2 sin(ln x), Ic2 + Is2 Ic + Is2 b b Ic = 1 + f (t) cos(ln t) dt, Is = f (t) sin(ln t) dt. y(x) = –

a

58.

a

b

f (t)y(xt) dt = Axβ cos(λ ln x) + Bxβ sin(λ ln x).

y(x) + a

A solution: y(x) = pxβ cos(λ ln x) + qxβ sin(λ ln x), where p=

AIc – BIs , Ic2 + Is2

b

f (t)tβ cos(λ ln t) dt,

Ic = 1 + a

AIs + BIc , Ic2 + Is2 b Is = f (t)tβ sin(λ ln t) dt.

q=

a

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59.

b

y(x) +

f (t)y(ξ) dt = 0,

ξ = xϕ(t).

a

Eigenfunctions of this integral equation are determined by the roots of the following transcendental (or algebraic) equation for λ: b f (t)[ϕ(t)]λ dt = –1. (1) a ◦

1 . For a real (simple) root λk of equation (1), there is a corresponding eigenfunction yk (x) = xλk . 2◦ . For a real root λk of multiplicity r, there are corresponding r eigenfunctions yk1 (x) = xλk ,

yk2 (x) = xλk ln x,

ykr (x) = xλk lnr–1 x.

...,

3◦ . For a complex (simple) root λk = αk + iβk of equation (1), there is a corresponding pair of eigenfunctions yk(1) (x) = xαk cos(βk ln x),

yk(2) (x) = xαk sin(βk ln x).

4◦ . For a complex root λk = αk + iβk of multiplicity r, there are corresponding r pairs of eigenfunctions (1) yk1 (x) = xαk cos(βk ln x),

(2) (x) = xαk sin(βk ln x), yk1

(1) yk2 (x) = xαk ln x cos(βk ln x),

(2) yk2 (x) = xαk ln x sin(βk ln x),

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (1) ykr (x)

60.

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

= xαk lnr–1 x cos(βk ln x),

(2) ykr (x) = xαk lnr–1 x sin(βk ln x).

The general solution is the linear combination (with arbitrary constants) of the eigenfunctions of the homogeneous integral equation. b y(x) + f (t)y(ξ) dt = Axβ , ξ = xϕ(t). a

A solution:

61.

b A β y(x) = x , B =1+ f (t)[ϕ(t)]β dt. B a It is assumed that B ≠ 0. A linear combination of eigenfunctions of the corresponding homogeneous equation (see 4.9.59) can be added to this solution. b y(x) + f (t)y(ξ) dt = g(x), ξ = xϕ(t). a

1◦ . For g(x) =

n

Ak xk , a solution of the equation has the form

k=0

y(x) =

n Ak k x , Bk

n

(1)

a

k=0

2◦ . For g(x) = ln x

b

f (t)[ϕ(t)]k dt.

Bk = 1 +

Ak xk , a solution has the form

k=0

y(x) = ln x

n k=0

Bk xk +

n

Ck xk ,

(2)

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients.

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3◦ . For g(x) =

n

Ak (ln x)k , a solution of the equation has the form

k=0 n

y(x) =

Bk (ln x)k ,

(3)

k=0

where the constants Bk can be found by the method of undetermined coefficients. 4◦ . For g(x) =

n

Ak cos(λk ln x), a solution of the equation has the form

k=1

y(x) =

n

Bk cos(λk ln x) +

k=1

n

Ck sin(λk ln x),

(4)

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 5◦ . For g(x) =

n

Ak sin(λk ln x), a solution of the equation has the form

k=1

y(x) =

n

Bk cos(λk ln x) +

k=1

n

Ck sin(λk ln x),

(5)

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. Remark. A linear combination of eigenfunctions of the corresponding homogeneous equation (see 4.9.59) can be added to solutions (1)–(5).

4.10. Some Formulas and Transformations Let the solution of the integral equation y(x) +

b

K(x, t)y(t) dt = f (x)

(1)

a

have the form

b

y(x) = f (x) +

R(x, t)f (t) dt.

(2)

a

Then the solution of the more complicated integral equation y(x) +

b

K(x, t) a

has the form

y(x) = f (x) +

g(x) y(t) dt = f (x) g(t)

(3)

g(x) f (t) dt. g(t)

(4)

b

R(x, t) a

Below are formulas for the solutions of integral equations of the form (3) for some specific functions g(x). In all cases, it is assumed that the solution of equation (1) is known and is given by (2).

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1◦ . The solution of the equation

b

K(x, t)(x/t)λ y(t) dt = f (x)

y(x) + a

has the form

b

R(x, t)(x/t)λ f (t) dt.

y(x) = f (x) + a

2◦ . The solution of the equation

b

K(x, t)eλ(x–t) y(t) dt = f (x)

y(x) + a

has the form

b

R(x, t)eλ(x–t) f (t) dt.

y(x) = f (x) + a

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Chapter 5

Nonlinear Equations With Variable Limit of Integration Notation: f , g, h, and ϕ are arbitrary functions of an argument specified in the parentheses (the argument can depend on t, x, and y); A, B, C, a, b, c, k, β, λ, and µ are arbitrary parameters.

5.1. Equations With Quadratic Nonlinearity That Contain Arbitrary Parameters 5.1-1. Equations of the Form

x 0

y(t)y(x – t) dt = F (x)

x

y(t)y(x – t) dt = Ax + B,

1.

A, B > 0.

0

A A √ A 1 y(x) = ± B √ exp – x + erf x , B B B πx z 2 where erf z = √ exp –t2 dt is the error function. π 0 x y(t)y(x – t) dt = A2 xλ .

Solutions:

2.

0

√

Solutions: y(x) = ±A

3.

where Γ(z) is the gamma function. x y(t)y(x – t) dt = Axλ–1 + Bxλ ,

Γ(λ + 1) λ–1 x 2 , Γ λ+1 2

λ > 0.

0

√

Solutions:

B λ+1 λ AΓ(λ) λ–2 B x 2 exp –λ x Φ , ;λ x , Γ(λ/2) A 2 2 A where Φ(a, c; x) is the degenerate hypergeometric function (Kummer’s function). x y(t)y(x – t) dt = A2 eλx . y(x) = ±

4.

0

A Solutions: y(x) = ± √ eλx . πx

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x

y(t)y(x – t) dt = (Ax + B)eλx ,

5.

A, B > 0.

0

Solutions:

A A √ λx 1 A √ y(x) = ± B e erf x , exp – x + B B B πx z 2 where erf z = √ exp –t2 dt is the error function. π 0

x

y(t)y(x – t) dt = A2 xµ eλx .

6. 0

Solutions: y(x) = ±

x

7.

√ A Γ(µ + 1) µ–1 λx x 2 e . Γ µ+1 2

y(t)y(x – t) dt = Axµ–1 + Bxµ eλx .

0

Solutions: √ y(x) = ±

AΓ(µ) µ–2 B µ+1 µ B x 2 exp λ – µ x Φ , ;µ x , Γ(µ/2) A 2 2 A

where Φ(a, c; x) is the degenerate hypergeometric function (Kummer’s function).

x

y(t)y(x – t) dt = A2 cosh(λx).

8. 0

A d Solutions: y(x) = ± √ π dx

x

0

I0 (λt) dt √ , where I0 is the modified Bessel function. x–t

x

y(t)y(x – t) dt = A sinh(λx).

9. 0

√ Solutions: y = ± Aλ I0 (λx), where I0 is the modified Bessel function.

x

10.

√ y(t)y(x – t) dt = A sinh(λ x ).

0

11.

√ Solutions: y = ± A π 1/4 2–7/8 λ3/4 x–1/8 I–1/4 λ 12 x , where I–1/4 is the modified Bessel function. x y(t)y(x – t) dt = A2 cos(λx). 0

A d Solutions: y(x) = ± √ π dx

0

x

J0 (λt) dt √ , where J0 is the Bessel function. x–t

x

y(t)y(x – t) dt = A sin(λx).

12. 0

√ Solutions: y = ± Aλ J0 (λx), where J0 is the Bessel function.

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x

13.

√ y(t)y(x – t) dt = A sin(λ x ).

0

√ Solutions: y = ± A π 1/4 2–7/8 λ3/4 x–1/8 J–1/4 λ 12 x , where J–1/4 is the Bessel function.

x

y(t)y(x – t) dt = A2 eµx cosh(λx).

14. 0

A d Solutions: y(x) = ± √ eµx dx π

x

0

I0 (λt) dt √ , where I0 is the modified Bessel function. x–t

x

y(t)y(x – t) dt = Aeµx sinh(λx).

15. 0

√ Solutions: y = ± Aλ eµx I0 (λx), where I0 is the modified Bessel function.

x

y(t)y(x – t) dt = A2 eµx cos(λx).

16. 0

A d Solutions: y(x) = ± √ eµx dx π

x

0

J0 (λt) dt √ , where J0 is the Bessel function. x–t

x

y(t)y(x – t) dt = Aeµx sin(λx).

17. 0

√ Solutions: y = ± Aλ eµx J0 (λx), where J0 is the Bessel function. 5.1-2. Equations of the Form

x 0

K(x, t)y(t)y(x – t) dt = F (x)

x

tk y(t)y(x – t) dt = Axλ ,

18.

A > 0.

0

Solutions:

1/2 λ–k–1 AΓ(λ + 1) y(x) = ± λ+1+k λ+1–k x 2 , Γ 2 Γ 2

where Γ(z) is the gamma function.

x

tk y(t)y(x – t) dt = Aeλx .

19. 0

Solutions:

1/2 A – k+1 2 eλx , y(x) = ± x 1–k Γ k+1 2 Γ 2

where Γ(z) is the gamma function.

x

tk y(t)y(x – t) dt = Axµ eλx .

20. 0

Solutions:

1/2 µ–k–1 AΓ(µ + 1) y(x) = ± µ+k+1 µ–k+1 x 2 eλx , Γ 2 Γ 2

where Γ(z) is the gamma function.

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x

21.

y(t)y(x – t) ax + bt

0

dt = Axλ .

Solutions:

y(x) = ±

x

22.

y(t)y(x – t) ax + bt

0

y(x) = ±

x

23.

y(t)y(x – t) ax + bt

0

y(x) = ±

x

0

y(x) = ±

x

0

A λ/2 x , I

x

0

1 b ln 1 + . b a

I=

1

z µ/2 (1 – z)µ/2

I= 0

dz . a + bz

1

I=

z λ/2 (1 – z)λ/2 √

0

dz a + bz 2

.

y(t)y(x – t) dt = Aeλx . √ ax2 + bt2 y(x) = ±

26.

A λx e , I

A µ/2 λx x e , I

Solutions:

0

y(t)y(x – t) dt = Axλ . √ 2 2 ax + bt

Solutions:

25.

dz . a + bz

z λ/2 (1 – z)λ/2

dt = Axµ eλx .

Solutions:

24.

1

I=

dt = Aeλx .

Solutions:

A λ/2 x , I

A λx e , I

1

I=

√

0

dz a + bz 2

.

y(t)y(x – t) dt = Axµ eλx . √ ax2 + bt2

Solutions: y(x) = ±

5.1-3. Equations of the Form

A µ/2 λx x e , I x 0

1

I=

z µ/2 (1 – z)µ/2 √

0

dz a + bz 2

.

G(· · ·) dt = F (x)

x

y(t)y(ax + bt) dt = Axλ .

27. 0

Solutions:

y(x) = ±

A λ–1 x 2 , I

I=

1

z

λ–1 2

(a + bz)

λ–1 2

dz.

0

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x

a ≥ 1.

y(t)y(ax – t) dt = Aeλx ,

28. 0

Solutions:

y(x) = ±

1

√

I= 0

dz . z(a – z)

x

a ≥ 1.

y(t)y(ax – t) dt = Axµ eλx ,

29.

A exp(λx/a) √ , I x

0

Solutions:

y(x) = ±

x

5.1-4. Equations of the Form y(x) + 30.

A µ–1 x 2 exp(λx/a), I

a

1

I=

z

µ–1 2

(a – z)

µ–1 2

dz.

0

K(x, t)y 2 (t) dt = F (x)

x

y 2 (t) dt = Bx + C.

y(x) + A a

By differentiation, this integral equation can be reduced to a separable ordinary differential equation. 1◦ . Solution with AB > 0:

(k + ya ) exp[2Ak(x – a)] + ya – k y(x) = k , (k + ya ) exp[2Ak(x – a)] – ya + k

k=

2◦ . Solution with AB < 0:

ya y(x) = k tan Ak(a – x) + arctan , k 3◦ . Solution with B = 0: y(x) = 31.

k=

–

B , A

B , A

ya = aB + C.

ya = aB + C.

C . AC(x – a) + 1

x

(x – t)y 2 (t) dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = ky 2 . Solution in an implicit form: y –1/2 4Au – 2kF (u) + B 2 – 4AC du = ±(x – a), y0

F (u) = 32.

1 3

u3 – y03 ,

y0 = Aa2 + Ba + C.

x

tλ y 2 (t) dt = Bxλ+1 + C.

y(x) + A a

This is a special case of equation 5.8.6 with f (y) = Ay 2 . By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form: y du (λ + 1) ya = Baλ+1 + C. + xλ+1 – aλ+1 = 0, 2 – B(λ + 1) Au ya

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33.

x

x–λ–1 y 2 (t) dt = Bxλ ,

y(x) + A 0

λ > – 12 .

Solutions: y1 (x) = β1 xλ and y2 (x) = β2 xλ , where β1,2 are the roots of the quadratic equation Aβ 2 + (2λ + 1)β – B(2λ + 1) = 0. 34.

x

y 2 (t) dt

y(x) +

= A. ax + bt Solutions: y1 (x) = λ1 and y2 (x) = λ2 , where λ1,2 are the roots of the quadratic equation b 2 ln 1 + λ + bλ – Ab = 0. a x 2 y (t) dt y(x) + A = Bx. x2 + t 2 0 Solutions: y1 (x) = λ1 x and y2 (x) = λ2 x, where λ1,2 are the roots of the quadratic equation 1 – 14 π Aλ2 + λ – B = 0. 0

35.

36.

37.

x

y 2 (t) dt = A. √ ax2 + bt2 0 Solutions: y1 (x) = λ1 and y2 (x) = λ2 , where λ1,2 are the roots of the quadratic equation 1 dz 2 √ Iλ + λ – A = 0, I= . a + bz 2 0 x n – λ+1 2 n y (t) dt = Bxλ . y(x) + A ax + btn

y(x) +

0

Solutions: y1 (x) = β1 xλ and y2 (x) = β2 xλ , where β1,2 are the roots of the quadratic equation 1 – λ+1 n dz. AIβ 2 + β – B = 0, I= z 2λ a + bz n 0

38.

x

eλt y 2 (t) dt = Beλx + C.

y(x) + A a

This is a special case of equation 5.8.11 with f (y) = Ay 2 . By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form: y du λ y0 = Beλa + C. + eλx – eλa = 0, 2 – Bλ Au y0 39.

x

eλ(x–t) y 2 (t) dt = B.

y(x) + A a

This is a special case of equation 5.8.12. By differentiation, this integral equation can be reduced to the separable ordinary differential equation yx + Ay 2 – λy + λB = 0, Solution in an implicit form: y B

y(a) = B.

du + x – a = 0. Au2 – λu + λB

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40.

x

eλ(x–t) y 2 (t) dt = Aeλx + B.

y(x) + k a

Solution in an implicit form: y y0

41.

du = x – a, λu – ku2 – λB

y0 = Aeλa + B.

x

sinh[λ(x – t)]y 2 (t) dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = ky 2 . Solution in an implicit form: y 2 2 –1/2 λ u – 2λ2 Cu – 2kλF (u) + λ2 (C 2 – 4AB) du = ±(x – a), y0

F (u) = 42.

1 3

u3 – y03 ,

y0 = Aeλa + Be–λa + C.

x

sinh[λ(x – t)]y 2 (t) dt = A cosh(λx) + B.

y(x) + k a

This is a special case of equation 5.8.15 with f (y) = ky 2 . Solution in an implicit form: y –1/2 2 2 du = ±(x – a), λ u – 2λ2 Bu – 2kλF (u) + λ2 (B 2 – A2 ) y0

F (u) = 43.

1 3

u3 – y03 ,

y0 = A cosh(λa) + B.

x

sinh[λ(x – t)]y 2 (t) dt = A sinh(λx) + B.

y(x) + k a

This is a special case of equation 5.8.16 with f (y) = ky 2 . Solution in an implicit form: y –1/2 2 2 du = ±(x – a), λ u – 2λ2 Bu – 2kλF (u) + λ2 (A2 + B 2 ) y0

F (u) = 44.

1 3

u3 – y03 ,

y0 = A sinh(λa) + B.

x

sin[λ(x – t)]y 2 (t) dt = A sin(λx) + B cos(λx) + C.

y(x) + k a

This is a special case of equation 5.8.17 with f (y) = ky 2 . Solution in an implicit form: y –1/2 2 du = ±(x – a), λ D – λ2 u2 + 2λ2 Cu – 2kλF (u) y0

y0 = A sin(λa) + B cos(λa) + C, 5.1-5. Equations of the Form y(x) + 45.

x a

D = A2 + B 2 – C 2 ,

F (u) =

1 3

u3 – y03 .

K(x, t)y(t)y(x – t) dt = F (x)

x

y(t)y(x – t) dt = AB 2 x + B.

y(x) + A 0

A solution: y(x) = B.

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46.

x

y(t)y(x – t) dt = (AB 2 x + B)eλx .

y(x) + A 0

47.

48.

49.

A solution: y(x) = Beλx . λ x y(x) + y(t)y(x – t) dt = 2β 0

1 β 2

sinh(λx).

A solution: y(x) = βI1 (λx), where I1 (x) is the modified Bessel function. λ x y(x) – y(t)y(x – t) dt = 12 β sin(λx). 2β 0 A solution: y(x) = βJ1 (λx), where J1 (x) is the Bessel function. x y(x) + A x–λ–1 y(t)y(x – t) dt = Bxλ . 0

Solutions: y1 (x) = β1 xλ and y2 (x) = β2 xλ , where β1,2 are the roots of the quadratic equation 1 Γ2 (λ + 1) 2 AIβ + β – B = 0, I= z λ (1 – z)λ dz = . Γ(2λ + 2) 0

5.2. Equations With Quadratic Nonlinearity That Contain Arbitrary Functions 5.2-1. Equations of the Form

x a

G(· · ·) dt = F (x)

x

K(x, t)y 2 (t) dt = f (x).

1. a

The substitution w(x) = y 2 (x) leads to the linear equation x K(x, t)w(t) dt = f (x). a

x

K(t)y(x)y(t) dt = f (x).

2. a

Solutions:

y(x) = ±f (x) 2

–1/2

x

K(t)f (t) dt

.

a

3.

x

f

t

y(t)y(x – t) dt = Axλ .

x Solutions: 0

y(x) = ±

4.

x

f

t

x Solutions:

A λ–1 x 2 , I

1

I=

f (z)z

λ–1 2

(1 – z)

λ–1 2

dz.

0

y(t)y(x – t) dt = Aeλx .

0

y(x) = ±

A eλx √ , I x

I= 0

1

f (z) dz √ . z(1 – z)

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x

f

5.

t

y(t)y(x – t) dt = Axµ eλx .

x

0

Solutions:

y(x) = ±

x

f

6.

y(x) = ±

x

f

x

f

8. 0

A λ–1 x 2 , I

y(x) = ±

(1 – z)

µ–1 2

dz.

0

1

I=

f (z)z

λ–1 2

(a + bz)

λ–1 2

dz.

0

a ≥ 1.

y(t)y(ax – t) dt = Aeλx ,

Solutions:

µ–1 2

t x

0

f (z)z

y(t)y(ax + bt) dt = Axλ .

Solutions:

7.

1

I=

t x

0

A µ–1 λx x 2 e , I

A exp(λx/a) √ , I x

1

I= 0

f (z) dz √ . z(a – z)

t

y(t)y(ax – t) dt = Axµ eλx ,

x

a ≥ 1.

Solutions: y(x) = ±

5.2-2. Equations of the Form y(x) + 9.

A µ–1 x 2 exp(λx/a), I x a

I=

1

f (z)z

µ–1 2

(a – z)

µ–1 2

dz.

0

K(x, t)y 2 (t) dt = F (x)

x

f (t)y 2 (t) dt = A.

y(x) + a

Solution:

y(x) = A 1 + A

–1

x

f (t) dt

.

a

10.

x

eλ(x–t) g(t)y 2 (t) dt = f (x).

y(x) + a

Differentiating the equation with respect to x yields x yx + g(x)y 2 + λ eλ(x–t) g(t)y 2 (t) dt = fx (x).

(1)

a

Eliminating the integral term from (1) with the aid of the original equation, we arrive at a Riccati ordinary differential equation, yx + g(x)y 2 – λy + λf (x) – fx (x) = 0,

(2)

under the initial condition y(a) = f (a). Equation (2) can be reduced to a second-order linear ordinary differential equation. For the exact solutions of equation (2) with various specific functions f and g, see, for example, E. Kamke (1977) and A. D. Polyanin and V. F. Zaitsev (1995).

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11.

x

g(x)h(t)y 2 (t) dt = f (x).

y(x) + a

Differentiating the equation with respect to x yields yx

2

+ g(x)h(x)y +

gx (x)

x

h(t)y 2 (t) dt = fx (x).

(1)

a

Eliminating the integral term from (1) with the aid of the original equation, we arrive at a Riccati ordinary differential equation, yx + g(x)h(x)y 2 –

gx (x) g (x) y = fx (x) – x f (x), g(x) g(x)

(2)

under the initial condition y(a) = f (a). Equation (2) can be reduced to a second-order linear ordinary differential equation. For the exact solutions of equation (2) with various specific functions f , g, and h, see, for example, E. Kamke (1977) and A. D. Polyanin and V. F. Zaitsev (1995). 12.

x

x–λ–1 f

y(x) + 0

t x

y 2 (t) dt = Axλ .

Solutions: y1 (x) = β1 xλ and y2 (x) = β2 xλ , where β1,2 are the roots of the quadratic equation Iβ 2 + β – A = 0,

1

f (z)z 2λ dz.

I= 0

13.

x

eλt+βx f (x – t)y 2 (t) dt = 0.

y(x) – –∞

This is a special case of equation 5.3.19 with k = 2. 14.

∞

y(x) –

eλt+βx f (x – t)y 2 (t) dt = 0.

x

A solution: 1 y(x) = e–(λ+β)x , A

5.2-3. Equations of the Form y(x) + 15.

y(x) + 0

x

1 x

f

t x

x a

∞

A=

e–(λ+2β)z f (–z) dz.

0

G(· · ·) dt = F (x)

y(t)y(x – t) dt = Aeλx .

Solutions: y1 (x) = B1 eλx ,

y2 (x) = B2 eλx ,

where B1 and B2 are the roots of the quadratic equation 2

IB + B – A = 0,

I=

1

f (z) dz. 0

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16.

x

x–λ–1 f

y(x) + A 0

t x

y(t)y(x – t) dt = Bxλ .

Solutions: y1 (x) = β1 xλ and y2 (x) = β2 xλ , where β1,2 are the roots of the quadratic equation AIβ 2 + β – B = 0,

1

f (z)z λ (1 – z)λ dz.

I= 0

17.

∞

f (t – x)y(t – x)y(t) dt = ae–λx .

y(x) + x

Solutions: y(x) = bk e–λx , where bk (k = 1, 2) are the roots of the quadratic equation ∞ I= f (z)e–2λz dz. b2 I + b – a = 0, 0

To calculate the integral I, it is convenient to use tables of Laplace transforms (with parameter p = 2λ).

5.3. Equations With Power-Law Nonlinearity 5.3-1. Equations Containing Arbitrary Parameters 1.

x

tλ y k (t) dt = Bxλ+1 + C.

y(x) + A a

By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form: y du (λ + 1) y0 = Baλ+1 + C. + xλ+1 – aλ+1 = 0, k – B(λ + 1) Au y0 2.

x

y(x) +

y k (t) ax + bt

0

dt = A.

A solution: y(x) = λ, where λ is a root of the algebraic (or transcendental) equation b k ln 1 + λ + bλ – Ab = 0. a 3.

x

y(x) + Ax 0

y k (t) dt x2 + t 2

= B.

A solution: y(x) = λ, where λ is a root of the algebraic (or transcendental) equation λ + 14 Aπλk = B.

4.

x

y k (t) dt = A. √ ax2 + bt2 0 A solution: y(x) = λ, where λ is a root of the algebraic (or transcendental) equation y(x) +

k

Iλ + λ – A = 0,

I= 0

1

√

dz a + bz 2

.

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5.

x

y(x) + A

axn + btn

λ–kλ–1 n

y k (t) dt = Bxλ .

a

A solution: y = βxλ , where β is a root of the algebraic (or transcendental) equation 1 λ–kλ–1 n AIβ k + β – B = 0, I= z kλ a + bz n dz. 0

6.

x

eλt y µ (t) dt = Beλx + C.

y(x) + A a

7.

By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form: y du λ y0 = Beλa + C. + eλx – eλa = 0, µ y0 Au – Bλ x y(x) + k eλ(x–t) y µ (t) dt = Aeλx + B. a

Solution in an implicit form: y y0

8.

dt = x – a, λt – ktµ – λB

y0 = Aeλa + B.

x

sinh[λ(x – t)]y µ (t) dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = ky µ . Solution in an implicit form: y 2 2 –1/2 λ u – 2λ2 Cu – 2kλF (u) + λ2 (C 2 – 4AB) du = ±(x – a), y0

F (u) = 9.

1 µ+1 – y0µ+1 , u µ+1

y0 = Aeλa + Be–λa + C.

x

sinh[λ(x – t)]y µ (t) dt = A cosh(λx) + B.

y(x) + k a

This is a special case of equation 5.8.15 with f (y) = ky µ . Solution in an implicit form: y 2 2 –1/2 λ u – 2λ2 Bu – 2kλF (u) + λ2 (B 2 – A2 ) du = ±(x – a), y0

F (u) = 10.

1 µ+1 – y0µ+1 , u µ+1

y0 = A cosh(λa) + B.

x

sinh[λ(x – t)]y µ (t) dt = A sinh(λx) + B.

y(x) + k a

This is a special case of equation 5.8.16 with f (y) = ky µ . Solution in an implicit form: y –1/2 2 2 du = ±(x – a), λ u – 2λ2 Bu – 2kλF (u) + λ2 (A2 + B 2 ) y0

F (u) =

1 µ+1 – y0µ+1 , u µ+1

y0 = A sinh(λa) + B.

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11.

x

sin[λ(x – t)]y µ (t) dt = A sin(λx) + B cos(λx) + C.

y(x) + k a

This is a special case of equation 5.8.17 with f (y) = ky µ . Solution in an implicit form:

y

λ2 D – λ2 u2 + 2λ2 Cu – 2kλF (u)

–1/2

du = ±(x – a),

y0

y0 = A sin(λa) + B cos(λa) + C,

D = A2 + B 2 – C 2 ,

F (u) =

1 µ+1 – y0µ+1 . u µ+1

5.3-2. Equations Containing Arbitrary Functions

x

K(x, t)y λ (t) dt = f (x).

12. a

The substitution w(x) = y λ (x) leads to the linear equation

x

K(x, t)w(t) dt = f (x). a

13.

x

f (t)y k (t) dt = A.

y(x) + a

Solution:

y(x) = A1–k + (k – 1)

x

f (t) dt

1 1–k

.

a

14.

x

f (x)g(t)y k (t) dt = 0.

y(x) – a

1◦ . Differentiating the equation with respect to x and eliminating the integral term (using the original equation), we obtain the Bernoulli ordinary differential equation yx – f (x)g(x)y k –

fx (x) y = 0, f (x)

y(a) = 0.

2◦ . Solution with k < 1: y(x) = f (x) (1 – k)

x k

f (t)g(t) dt

1 1–k

.

a

Additionally, for k > 0, there is the trivial solution y(x) ≡ 0. 15.

x

xλ–kλ–1 f

y(x) + 0

t x

y k (t) dt = Axλ .

A solution: y(x) = βxλ , where β is a root of the algebraic equation k

Iβ + β – A = 0,

1

f (z)z kλ dz.

I= 0

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16.

x

y(x) +

f

t

x

0

y(t) dt = Ax2 .

Solutions: yk (x) = Bk2 x2 , where Bk (k = 1, 2) are the roots of the quadratic equations B 2 ± IB – A = 0,

1

I=

zf (z) dz. 0

17.

x

ta f

y(x) – 0

t x

k ≠ 1.

y k (t) dt = 0,

A solution: y(x) =

1+a Ax 1–k

,

A

1–k

1

a+k

z 1–k f (z) dz.

= 0

18.

∞

k ≠ 1.

eλt+βx f (x – t)y k (t) dt = 0,

y(x) – x

A solution:

λ+β y(x) = A exp x , 1–k 19.

A1–k =

∞

exp 0

λ + βk z f (–z) dz. 1–k

x

k ≠ 1.

eλt+βx f (x – t)y k (t) dt = 0,

y(x) – –∞

A solution:

λ+β y(x) = A exp x , 1–k

A

1–k

= 0

∞

λ + βk exp – z f (z) dz. 1–k

5.4. Equations With Exponential Nonlinearity 5.4-1. Equations Containing Arbitrary Parameters 1.

x

y(x) + A

exp[λy(t)] dt = B. a

Solution: y(x) = – 2.

1 ln Aλ(x – a) + e–Bλ . λ

x

y(x) + A

exp[λy(t)] dt = Bx + C. a

For B = 0, see equation 5.4.1. Solution with B ≠ 0:

1 A –λy0 A λB(a–x) y(x) = – ln , – + e e λ B B

y0 = aB + C.

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3.

x

(x – t) exp[λy(t)] dt = Ax2 + Bx + C.

y(x) + k a

1◦ . This is a special case of equation 5.8.5 with f (y) = keλy . The solution of this integral equation is determined by the solution of the second-order autonomous ordinary differential equation yxx + keλy – 2A = 0 under the initial conditions yx (a) = 2Aa + B.

y(a) = Aa2 + Ba + C, 2◦ . Solution in an implicit form:

y

4Au – 2F (u) + B 2 – 4AC

y0

F (u) = 4.

k λu λy0 , e –e λ

–1/2

du = ±(x – a),

y0 = Aa2 + Ba + C.

x

tλ exp[βy(t)] dt = Bxλ+1 + C.

y(x) + A a

By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form:

y

du + xλ+1 – aλ+1 = 0, Aeβu – B(λ + 1)

(λ + 1) y0

5.

x

exp[λy(t)]

y(x) +

ax + bt

0

y0 = Baλ+1 + C.

dt = A.

A solution: y(x) = β, where β is a root of the transcendental equation b λβ ln 1 + e + bβ – Ab = 0. a

6.

x

exp[λy(t)] dt = A. √ ax2 + bt2 0 A solution: y(x) = β, where β is a root of the transcendental equation y(x) +

keλβ + β – A = 0,

k= 0

7.

x

y(x) + A

1

√

dz a + bz 2

.

exp λt + βy(t) dt = Beλx + C.

a

By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form:

y

λ y0

du + eλx – eλa = 0, Aeβu – Bλ

y0 = Beλa + C.

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8.

x

y(x) + k

exp λ(x – t) + βy(t) dt = A.

a

Solution in an implicit form:

y

A

9.

x

y(x) + k

dt = x – a. λt – keβt – λA

exp λ(x – t) + βy(t) dt = Aeλx + B.

a

Solution in an implicit form: y y0

10.

dt = x – a, λt – keβt – λB

y0 = Aeλa + B.

x

sinh[λ(x – t)] exp[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

11.

This is a special case of equation 5.8.14 with f (y) = keβy . x y(x) + k sinh[λ(x – t)] exp[βy(t)] dt = A cosh(λx) + B.

12.

This is a special case of equation 5.8.15 with f (y) = keβy . x y(x) + k sinh[λ(x – t)] exp[βy(t)] dt = A sinh(λx) + B.

a

a

13.

This is a special case of equation 5.8.16 with f (y) = keβy . x y(x) + k sin[λ(x – t)] exp[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = keβy . 5.4-2. Equations Containing Arbitrary Functions 14.

x

y(x) +

f (t) exp[λy(t)] dt = A. a

Solution: y(x) = – 15.

x 1 ln λ f (t) dt + e–Aλ . λ a

x

y(x) +

g(t) exp[λy(t)] dt = f (x). a

1◦ . By differentiation, this integral equation can be reduced to the first-order ordinary differential equation yx + g(x)eλy = fx (x) (1) under the initial condition y(a) = f (a). The substitution w = e–λy reduces (1) to the linear equation wx + λfx (x)w – λg(x) = 0, w(a) = exp –λf (a) . 2◦ . Solution:

x 1 y(x) = f (x) – ln 1 + λ g(t) exp λf (t) dt . λ a

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16.

y(x) +

1

x

f

x

0

t

x

exp[λy(t)] dt = A.

A solution: y(x) = β, where β is a root of the transcendental equation λβ

β + Ie

– A = 0,

I=

1

f (z) dz. 0

5.5. Equations With Hyperbolic Nonlinearity 5.5-1. Integrands With Nonlinearity of the Form cosh[βy(t)] 1.

x

y(x) + k

cosh[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k cosh(βy). 2.

x

y(x) + k

cosh[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k cosh(βy). 3.

x

(x – t) cosh[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k cosh(βy). 4.

x

tλ cosh[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k cosh(βy). 5.

x

y(x) +

g(t) cosh[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = cosh(βy). 6.

x

cosh[βy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = cosh(βy).

x

cosh[βy(t)] dt = A. √ ax2 + bt2 0 This is a special case of equation 5.8.9 with f (y) = cosh(βy).

7.

y(x) +

8.

y(x) + k

x

eλt cosh[βy(t)] dt = Beλx + C. a

This is a special case of equation 5.8.11 with f (y) = k cosh(βy). 9.

x

eλ(x–t) cosh[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k cosh(βy).

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10.

x

eλ(x–t) cosh[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k cosh(βy). 11.

x

sinh[λ(x – t)] cosh[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k cosh(βy). 12.

x

y(x) + k

sinh[λ(x – t)] cosh[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k cosh(βy). 13.

x

y(x) + k

sinh[λ(x – t)] cosh[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k cosh(βy). 14.

x

y(x) + k

sin[λ(x – t)] cosh[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k cosh(βy).

5.5-2. Integrands With Nonlinearity of the Form sinh[βy(t)] 15.

x

y(x) + k

sinh[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k sinh(βy). 16.

x

y(x) + k

sinh[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k sinh(βy). 17.

x

(x – t) sinh[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k sinh(βy). 18.

x

tλ sinh[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k sinh(βy). 19.

x

y(x) +

g(t) sinh[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = sinh(βy). 20.

y(x) + 0

x

sinh[βy(t)] ax + bt

dt = A.

This is a special case of equation 5.8.8 with f (y) = sinh(βy).

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21.

x

sinh[βy(t)] dt = A. √ ax2 + bt2

y(x) + 0

This is a special case of equation 5.8.9 with f (y) = sinh(βy). 22.

x

eλt sinh[βy(t)] dt = Beλx + C.

y(x) + k a

This is a special case of equation 5.8.11 with f (y) = k sinh(βy). 23.

x

eλ(x–t) sinh[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k sinh(βy). 24.

x

eλ(x–t) sinh[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k sinh(βy). 25.

x

sinh[λ(x – t)] sinh[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k sinh(βy). 26.

x

y(x) + k

sinh[λ(x – t)] sinh[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k sinh(βy). 27.

x

y(x) + k

sinh[λ(x – t)] sinh[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k sinh(βy). 28.

x

y(x) + k

sin[λ(x – t)] sinh[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k sinh(βy).

5.5-3. Integrands With Nonlinearity of the Form tanh[βy(t)] 29.

x

y(x) + k

tanh[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k tanh(βy). 30.

x

y(x) + k

tanh[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k tanh(βy). 31.

x

(x – t) tanh[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k tanh(βy).

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32.

x

tλ tanh[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k tanh(βy). 33.

x

y(x) +

g(t) tanh[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = tanh(βy). 34.

x

tanh[βy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = tanh(βy). 35.

x

tanh[βy(t)] dt = A. √ ax2 + bt2

y(x) + 0

This is a special case of equation 5.8.9 with f (y) = tanh(βy). 36.

x

eλt tanh[βy(t)] dt = Beλx + C.

y(x) + k a

This is a special case of equation 5.8.11 with f (y) = k tanh(βy). 37.

x

eλ(x–t) tanh[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k tanh(βy). 38.

x

eλ(x–t) tanh[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k tanh(βy). 39.

x

sinh[λ(x – t)] tanh[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k tanh(βy). 40.

x

y(x) + k

sinh[λ(x – t)] tanh[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k tanh(βy). 41.

x

y(x) + k

sinh[λ(x – t)] tanh[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k tanh(βy). 42.

x

y(x) + k

sin[λ(x – t)] tanh[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k tanh(βy).

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5.5-4. Integrands With Nonlinearity of the Form coth[βy(t)] 43.

x

y(x) + k

coth[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k coth(βy). 44.

x

y(x) + k

coth[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k coth(βy). 45.

x

(x – t) coth[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k coth(βy). 46.

x

tλ coth[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k coth(βy). 47.

x

y(x) +

g(t) coth[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = coth(βy). 48.

x

coth[βy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = coth(βy). 49.

x

coth[βy(t)] dt = A. √ ax2 + bt2

y(x) + 0

This is a special case of equation 5.8.9 with f (y) = coth(βy). 50.

x

eλt coth[βy(t)] dt = Beλx + C.

y(x) + k a

This is a special case of equation 5.8.11 with f (y) = k coth(βy). 51.

x

eλ(x–t) coth[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k coth(βy). 52.

x

eλ(x–t) coth[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k coth(βy). 53.

x

sinh[λ(x – t)] coth[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k coth(βy).

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54.

x

y(x) + k

sinh[λ(x – t)] coth[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k coth(βy). 55.

x

y(x) + k

sinh[λ(x – t)] coth[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k coth(βy). 56.

x

y(x) + k

sin[λ(x – t)] coth[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k coth(βy).

5.6. Equations With Logarithmic Nonlinearity 5.6-1. Integrands Containing Power-Law Functions of x and t 1.

x

y(x) + k

ln[λy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k ln(λy). 2.

x

y(x) + k

ln[λy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k ln(λy). 3.

x

(x – t) ln[λy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k ln(λy). 4.

x

tλ ln[µy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k ln(µy). 5.

x

ln[λy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = ln(λy).

6.

x

ln[λy(t)] dt = A. √ ax2 + bt2 0 This is a special case of equation 5.8.9 with f (y) = ln(λy).

y(x) +

5.6-2. Integrands Containing Exponential Functions of x and t 7.

x

eλt ln[µy(t)] dt = Beλx + C.

y(x) + k a

This is a special case of equation 5.8.11 with f (y) = k ln(µy).

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8.

x

eλ(x–t) ln[µy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k ln(µy). 9.

x

eλ(x–t) ln[µy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k ln(µy).

5.6-3. Other Integrands 10.

x

y(x) +

g(t) ln[λy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = ln(λy). 11.

x

sinh[λ(x – t)] ln[µy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k ln(µy). 12.

x

y(x) + k

sinh[λ(x – t)] ln[µy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k ln(µy). 13.

x

y(x) + k

sinh[λ(x – t)] ln[µy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k ln(µy). 14.

x

y(x) + k

sin[λ(x – t)] ln[µy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k ln(µy).

5.7. Equations With Trigonometric Nonlinearity 5.7-1. Integrands With Nonlinearity of the Form cos[βy(t)] 1.

x

y(x) + k

cos[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k cos(βy). 2.

x

y(x) + k

cos[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k cos(βy). 3.

x

(x – t) cos[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k cos(βy).

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4.

x

tλ cos[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k cos(βy). 5.

x

y(x) +

g(t) cos[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = cos(βy). 6.

x

cos[βy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = cos(βy). 7.

x

cos[βy(t)] dt = A. √ ax2 + bt2

y(x) + 0

This is a special case of equation 5.8.9 with f (y) = cos(βy). 8.

x

eλt cos[βy(t)] dt = Beλx + C.

y(x) + k a

This is a special case of equation 5.8.11 with f (y) = k cos(βy). 9.

x

eλ(x–t) cos[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k cos(βy). 10.

x

eλ(x–t) cos[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k cos(βy). 11.

x

sinh[λ(x – t)] cos[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k cos(βy). 12.

x

y(x) + k

sinh[λ(x – t)] cos[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k cos(βy). 13.

x

y(x) + k

sinh[λ(x – t)] cos[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k cos(βy). 14.

x

y(x) + k

sin[λ(x – t)] cos[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k cos(βy).

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5.7-2. Integrands With Nonlinearity of the Form sin[βy(t)] 15.

x

y(x) + k

sin[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k sin(βy). 16.

x

y(x) + k

sin[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k sin(βy). 17.

x

(x – t) sin[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k sin(βy). 18.

x

tλ sin[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k sin(βy). 19.

x

y(x) +

g(t) sin[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = sin(βy). 20.

x

sin[βy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = sin(βy). 21.

x

sin[βy(t)] dt = A. √ ax2 + bt2

y(x) + 0

This is a special case of equation 5.8.9 with f (y) = sin(βy). 22.

x

eλt sin[βy(t)] dt = Beλx + C.

y(x) + k a

This is a special case of equation 5.8.11 with f (y) = k sin(βy). 23.

x

eλ(x–t) sin[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k sin(βy). 24.

x

eλ(x–t) sin[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k sin(βy). 25.

x

sinh[λ(x – t)] sin[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k sin(βy).

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26.

x

y(x) + k

sinh[λ(x – t)] sin[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k sin(βy). 27.

x

y(x) + k

sinh[λ(x – t)] sin[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k sin(βy). 28.

x

y(x) + k

sin[λ(x – t)] sin[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k sin(βy).

5.7-3. Integrands With Nonlinearity of the Form tan[βy(t)] 29.

x

y(x) + k

tan[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k tan(βy). 30.

x

y(x) + k

tan[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k tan(βy). 31.

x

(x – t) tan[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k tan(βy). 32.

x

tλ tan[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k tan(βy). 33.

x

y(x) +

g(t) tan[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = tan(βy). 34.

x

tan[βy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = tan(βy). 35.

x

tan[βy(t)] dt = A. √ ax2 + bt2

y(x) + 0

This is a special case of equation 5.8.9 with f (y) = tan(βy). 36.

x

eλt tan[βy(t)] dt = Beλx + C.

y(x) + k a

This is a special case of equation 5.8.11 with f (y) = k tan(βy).

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37.

x

eλ(x–t) tan[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k tan(βy). 38.

x

eλ(x–t) tan[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k tan(βy). 39.

x

sinh[λ(x – t)] tan[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k tan(βy). 40.

x

y(x) + k

sinh[λ(x – t)] tan[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k tan(βy). 41.

x

y(x) + k

sinh[λ(x – t)] tan[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k tan(βy). 42.

x

y(x) + k

sin[λ(x – t)] tan[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k tan(βy).

5.7-4. Integrands With Nonlinearity of the Form cot[βy(t)] 43.

x

y(x) + k

cot[βy(t)] dt = A. a

This is a special case of equation 5.8.3 with f (y) = k cot(βy). 44.

x

y(x) + k

cot[βy(t)] dt = Ax + B. a

This is a special case of equation 5.8.4 with f (y) = k cot(βy). 45.

x

(x – t) cot[βy(t)] dt = Ax2 + Bx + C.

y(x) + k a

This is a special case of equation 5.8.5 with f (y) = k cot(βy). 46.

x

tλ cot[βy(t)] dt = Bxλ+1 + C.

y(x) + k a

This is a special case of equation 5.8.6 with f (y) = k cot(βy). 47.

x

y(x) +

g(t) cot[βy(t)] dt = A. a

This is a special case of equation 5.8.7 with f (y) = cot(βy).

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48.

x

cot[βy(t)]

y(x) +

ax + bt

0

dt = A.

This is a special case of equation 5.8.8 with f (y) = cot(βy).

x

cot[βy(t)] dt = A. √ ax2 + bt2 0 This is a special case of equation 5.8.9 with f (y) = cot(βy).

49.

y(x) +

50.

y(x) + k

x

eλt cot[βy(t)] dt = Beλx + C. a

This is a special case of equation 5.8.11 with f (y) = k cot(βy). 51.

x

eλ(x–t) cot[βy(t)] dt = A.

y(x) + k a

This is a special case of equation 5.8.12 with f (y) = k cot(βy). 52.

x

eλ(x–t) cot[βy(t)] dt = Aeλx + B.

y(x) + k a

This is a special case of equation 5.8.13 with f (y) = k cot(βy). 53.

x

sinh[λ(x – t)] cot[βy(t)] dt = Aeλx + Be–λx + C.

y(x) + k a

This is a special case of equation 5.8.14 with f (y) = k cot(βy). 54.

x

y(x) + k

sinh[λ(x – t)] cot[βy(t)] dt = A cosh(λx) + B. a

This is a special case of equation 5.8.15 with f (y) = k cot(βy). 55.

x

y(x) + k

sinh[λ(x – t)] cot[βy(t)] dt = A sinh(λx) + B. a

This is a special case of equation 5.8.16 with f (y) = k cot(βy). 56.

x

y(x) + k

sin[λ(x – t)] cot[βy(t)] dt = A sin(λx) + B cos(λx) + C. a

This is a special case of equation 5.8.17 with f (y) = k cot(βy).

5.8. Equations With Nonlinearity of General Form 5.8-1. Equations of the Form

x

1.

x a

G(· · ·) dt = F (x)

K(x, t)ϕ y(t) dt = f (x).

a

The substitution w(x) = ϕ y(x) leads to the linear equation

x

K(x, t)w(t) dt = f (x). a

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x

2.

K(x, t)ϕ t, y(t) dt = f (x).

a

The substitution w(x) = ϕ x, y(x) leads to the linear equation

x

K(x, t)w(t) dt = f (x). a

5.8-2. Equations of the Form y(x) + 3.

x

y(x) +

x a

K(x, t)G y(t) dt = F (x)

f y(t) dt = A.

a

Solution in an implicit form:

y

A

4.

x

y(x) +

du + x – a = 0. f (u)

f y(t) dt = Ax + B.

a

Solution in an implicit form: y y0

5.

x

y(x) +

du = x – a, A – f (u)

y0 = Aa + B.

(x – t)f y(t) dt = Ax2 + Bx + C.

a

1◦ . This is a special case of equation 5.8.19. The solution of this integral equation is determined by the solution of the second-order autonomous ordinary differential equation yxx + f (y) – 2A = 0

under the initial conditions y(a) = Aa2 + Ba + C,

yx (a) = 2Aa + B.

2◦ . Solutions in an implicit form: y –1/2 du = ±(x – a), 4Au – 2F (u) + B 2 – 4AC y0

u

F (u) =

f (t) dt,

y0 = Aa2 + Ba + C.

y0

6.

x

y(x) +

tλ f y(t) dt = Bxλ+1 + C.

a

By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form: y du (λ + 1) ya = Baλ+1 + C. + xλ+1 – aλ+1 = 0, f (u) – B(λ + 1) ya

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7.

x

y(x) +

g(t)f y(t) dt = A.

a

Solution in an implicit form:

y

A

8.

x

y(x) +

f y(t) ax + bt

0

du + f (u)

x

g(t) dt = 0. a

dt = A.

A solution: y(x) = λ, where λ is a root of the algebraic (or transcendental) equation b ln 1 + f (λ) + bλ – Ab = 0. a f y(t) y(x) + dt = A. √ ax2 + bt2 0 A solution: y(x) = λ, where λ is a root of the algebraic (or transcendental) equation

9.

x

kf (λ) + λ – A = 0,

k= 0

10.

y(x) + x

x

f y(t)

0

dt x2 + t 2

1

√

dz a + bz 2

.

= A.

A solution: y(x) = λ, where λ is a root of the algebraic (or transcendental) equation λ + 14 πf (λ) = A. 11.

x

y(x) +

eλt f y(t) dt = Beλx + C.

a

By differentiation, this integral equation can be reduced to a separable ordinary differential equation. Solution in an implicit form: y du λ y0 = Beλa + C. + eλx – eλa = 0, y0 f (u) – Bλ 12.

x

y(x) +

eλ(x–t) f y(t) dt = A.

a

Solution in an implicit form:

y

A

13.

x

y(x) +

du = x – a. λu – f (u) – λA

eλ(x–t) f y(t) dt = Aeλx + B.

a

Solution in an implicit form: y y0

du = x – a, λu – f (u) – λB

y0 = Aeλa + B.

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14.

x

y(x) +

sinh[λ(x – t)]f y(t) dt = Aeλx + Be–λx + C.

a

1◦ . This is a special case of equation 5.8.23. The solution of this integral equation is determined by the solution of the second-order autonomous ordinary differential equation yxx + λf (y) – λ2 y + λ2 C = 0

under the initial conditions y(a) = Aeλa + Be–λa + C,

yx (a) = Aλeλa – Bλe–λa .

2◦ . Solution in an implicit form: y –1/2 2 2 du = ±(x – a), λ u – 2λ2 Cu – 2λF (u) + λ2 (C 2 – 4AB) y0

u

F (u) =

y0 = Aeλa + Be–λa + C.

f (t) dt, y0

15.

x

y(x) +

sinh[λ(x – t)]f y(t) dt = A cosh(λx) + B.

a

This is a special case of equation 5.8.14. Solution in an implicit form: y –1/2 2 2 du = ±(x – a), λ u – 2λ2 Bu – 2λF (u) + λ2 (B 2 – A2 ) y0

u

F (u) =

f (t) dt,

y0 = A cosh(λa) + B.

y0

16.

x

y(x) +

sinh[λ(x – t)]f y(t) dt = A sinh(λx) + B.

a

This is a special case of equation 5.8.23. Solution in an implicit form: y –1/2 2 2 du = ±(x – a), λ u – 2λ2 Bu – 2λF (u) + λ2 (A2 + B 2 ) y0

u

F (u) =

f (t) dt,

y0 = A sinh(λa) + B.

y0

17.

x

y(x) +

sin[λ(x – t)]f y(t) dt = A sin(λx) + B cos(λx) + C.

a

1◦ . This is a special case of equation 5.8.25. The solution of this integral equation is determined by the solution of the second-order autonomous ordinary differential equation yxx + λf (y) + λ2 y – λ2 C = 0

under the initial conditions y(a) = A sin(λa) + B cos(λa) + C,

yx (a) = Aλ cos(λa) – Bλ sin(λa).

2◦ . Solution in an implicit form: y 2 –1/2 λ D – λ2 u2 + 2λ2 Cu – 2λF (u) du = ±(x – a), y0

y0 = A sin(λa) + B cos(λa) + C,

D = A2 + B 2 – C 2 ,

u

F (u) =

f (t) dt. y0

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5.8-3. Equations of the Form y(x) + 18.

x

y(x) +

x a

K(x, t)G t, y(t) dt = F (x)

f t, y(t) dt = g(x).

a

The solution of this integral equation is determined by the solution of the first-order ordinary differential equation yx + f (x, y) – gx (x) = 0

19.

under the initial condition y(a) = g(a). For the exact solutions of the first-order differential equations with various f (x, y) and g(x), see E. Kamke (1977), A. D. Polyanin and V. F. Zaitsev (1995), and V. F. Zaitsev and A. D. Polyanin (1994). x y(x) + (x – t)f t, y(t) dt = g(x). a

Differentiating the equation with respect to x yields x yx + f t, y(t) dt = gx (x).

(1)

a

In turn, differentiating this equation with respect to x yields the second-order nonlinear ordinary differential equation yxx + f (x, y) – gxx (x) = 0.

(2)

By setting x = a in the original equation and equation (1), we obtain the initial conditions for y = y(x): y(a) = g(a), yx (a) = gx (a). (3)

20.

Equation (2) under conditions (3) defines the solution of the original integral equation. For the exact solutions of the second-order differential equation (2) with various f (x, y) and g(x), see A. D. Polyanin and V. F. Zaitsev (1995), and V. F. Zaitsev and A. D. Polyanin (1994). x y(x) + (x – t)n f t, y(t) dt = g(x), n = 1, 2, . . . a

Differentiating the equation n+1 times with respect to x, we obtain an (n+1)st-order nonlinear ordinary differential equation for y = y(x): yx(n+1) + n! f (x, y) – gx(n+1) (x) = 0. This equation under the initial conditions y(a) = g(a),

21.

yx (a) = gx (a),

...,

yx(n) (a) = gx(n) (a),

defines the solution of the original integral equation. x y(x) + eλ(x–t) f t, y(t) dt = g(x). a

Differentiating the equation with respect to x yields x yx + f x, y(x) + λ eλ(x–t) f t, y(t) dt = gx (x). a

Eliminating the integral term herefrom with the aid of the original equation, we obtain the first-order nonlinear ordinary differential equation yx + f (x, y) – λy + λg(x) – gx (x) = 0. The unknown function y = y(x) must satisfy the initial condition y(a) = g(a). For the exact solutions of the first-order differential equations with various f (x, y) and g(x), see E. Kamke (1977), A. D. Polyanin and V. F. Zaitsev (1995), and V. F. Zaitsev and A. D. Polyanin (1994).

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22.

x

y(x) +

cosh[λ(x – t)]f t, y(t) dt = g(x).

a

Differentiating the equation with respect to x twice yields x yx (x) + f x, y(x) + λ sinh[λ(x – t)]f t, y(t) dt = gx (x), a x 2 yxx (x) + f x, y(x) x + λ cosh[λ(x – t)]f t, y(t) dt = gxx (x).

(1) (2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order nonlinear ordinary differential equation yxx + f (x, y) x – λ2 y + λ2 g(x) – gxx (x) = 0. (3) By setting x = a in the original equation and in (1), we obtain the initial conditions for y = y(x): y(a) = g(a), yx (a) = gx (a) – f a, g(a) . (4)

23.

Equation (3) under conditions (4) defines the solution of the original integral equation. x y(x) + sinh[λ(x – t)]f t, y(t) dt = g(x). a

Differentiating the equation with respect to x twice yields x yx (x) + λ cosh[λ(x – t)]f t, y(t) dt = gx (x), a x 2 yxx (x) + λf x, y(x) + λ sinh[λ(x – t)]f t, y(t) dt = gxx (x).

(1) (2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order nonlinear ordinary differential equation yxx + λf (x, y) – λ2 y + λ2 g(x) – gxx (x) = 0.

(3)

By setting x = a in the original equation and in (1), we obtain the initial conditions for y = y(x): y(a) = g(a),

24.

yx (a) = gx (a).

(4)

Equation (3) under conditions (4) defines the solution of the original integral equation. For the exact solutions of the second-order differential equation (3) with various f (x, y) and g(x), see A. D. Polyanin and V. F. Zaitsev (1995), and V. F. Zaitsev and A. D. Polyanin (1994). x y(x) + cos[λ(x – t)]f t, y(t) dt = g(x). a

Differentiating the equation with respect to x twice yields x yx (x) + f x, y(x) – λ sin[λ(x – t)]f t, y(t) dt = gx (x), a x yxx (x) + f x, y(x) x – λ2 cos[λ(x – t)]f t, y(t) dt = gxx (x).

(1) (2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order nonlinear ordinary differential equation yxx + f (x, y) x + λ2 y – λ2 g(x) – gxx (x) = 0. (3) By setting x = a in the original equation and in (1), we obtain the initial conditions for y = y(x): y(a) = g(a), yx (a) = gx (a) – f a, g(a) . (4) Equation (3) under conditions (4) defines the solution of the original integral equation.

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25.

sin[λ(x – t)]f t, y(t) dt = g(x).

x

y(x) + a

Differentiating the equation with respect to x twice yields

cos[λ(x – t)]f t, y(t) dt = gx (x), a x 2 yxx (x) + λf x, y(x) – λ sin[λ(x – t)]f t, y(t) dt = gxx (x).

yx (x) + λ

x

(1) (2)

a

Eliminating the integral term from (2) with the aid of the original equation, we arrive at the second-order nonlinear ordinary differential equation yxx + λf (x, y) + λ2 y – λ2 g(x) – gxx (x) = 0.

(3)

By setting x = a in the original equation and in (1), we obtain the initial conditions for y = y(x): y(a) = g(a),

yx (a) = gx (a).

(4)

Equation (3) under conditions (4) defines the solution of the original integral equation. For the exact solutions of the second-order differential equation (3) with various f (x, y) and g(x), see A. D. Polyanin and V. F. Zaitsev (1995), and V. F. Zaitsev and A. D. Polyanin (1994).

5.8-4. Other Equations 26.

y(x) +

1

x

f

x

t x

0

, y(t) dt = A.

A solution: y(x) = λ, where λ is a root of the algebraic (or transcendental) equation λ + F (λ) – A = 0,

1

F (λ) =

f (z, λ) dz. 0

27.

x

y(x) +

f 0

t x

,

y(t) t

dt = Ax.

A solution: y(x) = λx, where λ is a root of the algebraic (or transcendental) equation λ + F (λ) – A = 0,

F (λ) =

1

f (z, λ) dz. 0

28.

∞

y(x) +

f t – x, y(t – x) y(t) dt = ae–λx .

x

Solutions: y(x) = bk e–λx , where bk are roots of the algebraic (or transcendental) equation b + bI(b) = a,

I(b) =

∞

f (z, be–λz )e–λz dz.

0

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Chapter 6

Nonlinear Equations With Constant Limits of Integration Notation: f , g, h, and ϕ are arbitrary functions of an argument specified in the parentheses (the argument can depend on t, x, and y); and A, B, C, a, b, c, s, β, γ, λ, and µ are arbitrary parameters; and k, m, and n are nonnegative integers.

6.1. Equations With Quadratic Nonlinearity That Contain Arbitrary Parameters 6.1-1. Equations of the Form

b a

K(t)y(x)y(t) dt = F (x)

1

y(x)y(t) dt = Axλ ,

1.

A > 0,

λ > –1.

0

This is a special case of equation 6.2.1 with f (x) = Axλ , g(t) = 1, a = 0, and b = 1. √ Solutions: y(x) = ± A(λ + 1) xλ .

1

y(x)y(t) dt = Aeβx ,

2.

A > 0.

0

This is a special case of equation 6.2.1 with f (x) = Aeβx , g(t) = 1, a = 0, and b = 1. Aβ βx Solutions: y(x) = ± e . eβ – 1

1

y(x)y(t) dt = A cosh(βx),

3.

A > 0.

0

This is a special case of equation 6.2.1 with f (x) = A cosh(βx), g(t) = 1, a = 0, and b = 1. Solutions: y(x) = ±

Aβ cosh(βx). sinh β

1

y(x)y(t) dt = A sinh(βx),

4.

Aβ > 0.

0

This is a special case of equation 6.2.1 with f (x) = A sinh(βx), g(t) = 1, a = 0, and b = 1. Solutions: y(x) = ±

Aβ sinh(βx). cosh β – 1

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1

y(x)y(t) dt = A tanh(βx),

5.

Aβ > 0.

0

This is a special case of equation 6.2.1 with f (x) = A tanh(βx), g(t) = 1, a = 0, and b = 1. Solutions: y(x) = ±

Aβ tanh(βx). ln cosh β

1

y(x)y(t) dt = A ln(βx),

6.

A(ln β – 1) > 0.

0

This is a special case of equation 6.2.1 with f (x) = A ln(βx), g(t) = 1, a = 0, and b = 1. Solutions: y(x) = ±

A ln(βx). ln β – 1

1

y(x)y(t) dt = A cos(βx),

7.

A > 0.

0

This is a special case of equation 6.2.1 with f (x) = A cos(βx), g(t) = 1, a = 0, and b = 1. Solutions: y(x) = ±

Aβ cos(βx). sin β

1

y(x)y(t) dt = A sin(βx),

8.

Aβ > 0.

0

This is a special case of equation 6.2.1 with f (x) = A sin(βx), g(t) = 1, a = 0, and b = 1. Solutions: y(x) = ±

Aβ sin(βx). 1 – cos β

1

y(x)y(t) dt = A tan(βx),

9.

Aβ > 0.

0

This is a special case of equation 6.2.1 with f (x) = A tan(βx), g(t) = 1, a = 0, and b = 1. Solutions: y(x) = ±

–Aβ tan(βx). ln |cos β|

1

tµ y(x)y(t) dt = Axλ ,

10.

A > 0,

µ + λ > –1.

0

This is a special case of equation 6.2.1 with f (x) = Axλ , g(t) = tµ , a = 0, and b = 1. √ Solutions: y(x) = ± A(µ + λ + 1) xλ .

1

eµt y(x)y(t) dt = Aeβx ,

11.

A > 0.

0

This is a special case of equation 6.2.1 with f (x) = Aeβx , g(t) = eµt , a = 0, and b = 1. A(µ + β) βx Solutions: y(x) = ± e . eµ+β – 1

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6.1-2. Equations of the Form

b

G(· · ·) dt = F (x)

1

0 ≤ x ≤ 1.

y(t)y(xt) dt = A,

12.

a

0

This is a special case of equation 6.2.2 with f (t) = 1, a = 0, and b = 1. 1◦ . Solutions: y1 (x) = y3 (x) = y5 (x) =

√ √ √

A, A (3x – 2), A (10x2 – 12x + 3),

√ y2 (x) = – A, √ y4 (x) = – A (3x – 2), √ y6 (x) = – A (10x2 – 12x + 3).

2◦ . The integral equation has some other solutions; for example, √ √ A A C y7 (x) = (2C + 1)x – C – 1 , (2C + 1)xC – C – 1 , y8 (x) = – √C √C y9 (x) = A (ln x + 1), y10 (x) = – A (ln x + 1), where C is an arbitrary constant.

13.

3◦ . See 6.2.2 for some other solutions. 1 y(t)y(xtβ ) dt = A, β > 0. 0

1◦ . Solutions: y1 (x) =

√

A, √ y3 (x) = B (β + 2)x – β – 1 ,

√ y2 (x) = – A, √ y4 (x) = – B (β + 2)x – β – 1 ,

2A . β(β + 1)

where B =

2◦ . The integral equation has some other (more complicated solutions) of the polynomial n form y(x) = Bk xk , where the constants Bk can be found from the corresponding system k=0

14.

of algebraic equations. ∞ y(t)y(xt) dt = Ax–λ ,

λ > 0,

1 ≤ x < ∞.

1

This is a special case of equation 6.2.3 with f (t) = 1, a = 1, and b = ∞. 1◦ . Solutions: y1 (x) = Bx–λ , y2 (x) = –Bx–λ , –λ y3 (x) = B (2λ – 3)x – 2λ + 2 x , y4 (x) = –B (2λ – 3)x – 2λ + 2 x–λ , where B =

√

λ > 12 ; λ > 32 ;

A(2λ – 1).

◦

2 . For sufficiently large λ, the integral equation has some other (more complicated) solutions n of the polynomial form y(x) = Bk xk , where the constants Bk can be found from the k=0

corresponding system of algebraic equations. See 6.2.2 for some other solutions.

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∞

15.

e–λt y(t)y(xt) dt = A,

λ > 0,

0 ≤ x < ∞.

0

This is a special case of equation 6.2.2 with f (t) = e–λt , a = 0, and b = ∞. 1◦ . Solutions: y1 (x) =

√

Aλ, y3 (x) = 12 Aλ (λx – 2),

√ y2 (x) = – Aλ, y4 (x) = – 12 Aλ (λx – 2).

2◦ . The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = Bk xk , where the constants Bk can be found from the corresponding system k=0

of algebraic equations. See 6.2.2 for some other solutions.

1

0 ≤ x < ∞.

y(t)y(x + λt) dt = A,

16. 0

17.

This is a special case of equation 6.2.7 with f (t) ≡ 1, a = 0, and b = 1. Solutions: √ √ y1 (x) = A, y2 (x) = – A,

y3 (x) = 3A/λ (1 – 2x), y4 (x) = – 3A/λ (1 – 2x). ∞ y(t)y(x + λt) dt = Ae–βx , A, λ, β > 0, 0 ≤ x < ∞. 0

This is a special case of equation 6.2.9 with f (t) ≡ 1, a = 0, and b = ∞. Solutions:

y1 (x) = Aβ(λ + 1) e–βx , y2 (x) = – Aβ(λ + 1) e–βx , y3 (x) = B β(λ + 1)x – 1 e–βx , y4 (x) = –B β(λ + 1)x – 1 e–βx ,

where B = Aβ(λ + 1)/λ.

1

–∞ < x < ∞.

y(t)y(x – t) dt = A,

18. 0

This is a special case of equation 6.2.10 with f (t) ≡ 1, a = 0, and b = 1. 1◦ . Solutions with A > 0: √ y1 (x) = A, √ y3 (x) = 5A(6x2 – 6x + 1), 2◦ . Solutions with A < 0: y1 (x) =

√

–3A (1 – 2x),

√ y2 (x) = – A, √ y4 (x) = – 5A(6x2 – 6x + 1). √ y2 (x) = – –3A (1 – 2x).

The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = Bk xk , where the constants Bk can be found from the corresponding system k=0

19.

of algebraic equations. ∞ x e–λt y y(t) dt = Axb , t 0 √ Solutions: y(x) = ± Aλ xb .

λ > 0.

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6.1-3. Equations of the Form y(x) + 20.

b a

K(x, t)y 2 (t) dt = F (x)

b

xλ y 2 (t) dt = 0.

y(x) + A a

Solutions: y1 (x) = 0, 21.

y2 (x) = –

2λ + 1 xλ . A(b2λ+1 – a2λ+1 )

b

xλ tµ y 2 (t) dt = 0.

y(x) + A a

Solutions: y1 (x) = 0, 22.

y2 (x) = –

2λ + µ + 1 xλ . – a2λ+µ+1 )

A(b2λ+µ+1

b

e–λx y 2 (t) dt = 0.

y(x) + A a

Solutions: y1 (x) = 0, 23.

y2 (x) =

2λ e–λx . A(e–2λb – e–2λa )

b

e–λx–µt y 2 (t) dt = 0.

y(x) + A a

Solutions: y1 (x) = 0, 24.

y2 (x) =

2λ + µ e–λx . – e–(2λ+µ)a ]

A[e–(2λ+µ)b

b

xλ e–µt y 2 (t) dt = 0.

y(x) + A a

This is a special case of equation 6.2.20 with f (x) = Axλ and g(t) = e–µt . 25.

b

e–µx tλ y 2 (t) dt = 0.

y(x) + A a

This is a special case of equation 6.2.20 with f (x) = Ae–µx and g(t) = tλ . 26.

1

y 2 (t) dt = Bxµ ,

y(x) + A

µ > –1.

0

This is a special case of equation 6.2.22 with g(t) = A, f (x) = Bxµ , a = 0, and b = 1. A solution: y(x) = Bxµ + λ, where λ is determined by the quadratic equation

2AB B2 1 λ + 1+ λ+ = 0. A µ+1 2µ + 1 2

27.

b

tβ y 2 (t) dt = Bxµ .

y(x) + A a

This is a special case of equation 6.2.22 with g(t) = Atβ and f (x) = Bxµ .

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28.

b

eβt y 2 (t) dt = Beµx .

y(x) + A a

This is a special case of equation 6.2.22 with g(t) = Aeβt and f (x) = Beµx . 29.

b

xβ y 2 (t) dt = Bxµ .

y(x) + A a

This is a special case of equation 6.2.23 with g(x) = Axβ and f (x) = Bxµ . 30.

b

eβx y 2 (t) dt = Beµx .

y(x) + A a

This is a special case of equation 6.2.23 with g(x) = Aeβx and f (x) = Beµx . 6.1-4. Equations of the Form y(x) + 31.

b a

K(x, t)y(x)y(t) dt = F (x)

b

tβ y(x)y(t) dt = Bxµ .

y(x) + A a

This is a special case of equation 6.2.25 with g(t) = Atβ and f (x) = Bxµ . 32.

b

eβt y(x)y(t) dt = Beµx .

y(x) + A a

This is a special case of equation 6.2.25 with g(t) = Aeβt and f (x) = Beµx . 33.

b

xβ y(x)y(t) dt = Bxµ .

y(x) + A a

This is a special case of equation 6.2.26 with g(x) = Axβ and f (x) = Bxµ . 34.

b

eβx y(x)y(t) dt = Beµx .

y(x) + A a

This is a special case of equation 6.2.26 with g(x) = Aeβx and f (x) = Beµx . 6.1-5. Equations of the Form y(x) + 35.

y(x) + A

b a

G(· · ·) dt = F (x)

1

y(t)y(xt) dt = 0. 0

This is a special case of equation 6.2.30 with f (t) = A, a = 0, and b = 1. 1◦ . Solutions: 1 (I1 – I0 )x + I1 – I2 C x , (2C + 1)xC , y2 (x) = A I0 I2 – I12 A Im = , m = 0, 1, 2, 2C + m + 1 where C is an arbitrary nonnegative constant. n There are more complicated solutions of the form y(x) = xC Bk xk , where C is an y1 (x) = –

k=0

arbitrary constant and the coefficients Bk can be found from the corresponding system of algebraic equations.

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2◦ . A solution: y3 (x) =

(I1 – I0 )xβ + I1 – I2 C x , I0 I2 – I12

Im =

A , 2C + mβ + 1

where C and β are arbitrary constants.

n

There are more complicated solutions of the form y(x) = xC

m = 0, 1, 2,

Dk xkβ , where C and β

k=0

are arbitrary constants and the coefficients Dk can be found from the corresponding system of algebraic equations. 3◦ . A solution: y4 (x) =

xC (J1 ln x – J2 ) , J0 J2 – J12

1

t2C (ln t)m dt,

Jm =

m = 0, 1, 2,

0

where C is an arbitrary constant. There are more complicated solutions of the form y(x) = xC

n

Ek (ln x)k , where C is

k=0

an arbitrary constant and the coefficients Ek can be found from the corresponding system of algebraic equations. 36.

∞

y(x) + A

y(t)y(xt) dt = 0. 1

This is a special case of equation 6.2.30 with f (t) = A, a = 1, and b = ∞. 37.

y(x) + λ

∞

y(t)y(xt) dt = Axβ .

1

This is a special case of equation 6.2.31 with f (t) = λ, a = 0, and b = 1. 38.

y(x) + A

1

y(t)y(x + λt) dt = 0. 0

This is a special case of equation 6.2.35 with f (t) ≡ A, a = 0, and b = 1. 1◦ . A solution: y(x) =

C(λ + 1) eCx , A[1 – eC(λ+1) ]

where C is an arbitrary constant. 2◦ . There are more complicated solutions of the form y(x) = eCx

n

Bm xm , where C is an

m=0

arbitrary constant and the coefficients Bm can be found from the corresponding system of algebraic equations. 39.

y(x) + A

∞

y(t)y(x + λt) dt = 0,

λ > 0,

0 ≤ x < ∞.

0

This is a special case of equation 6.2.35 with f (t) ≡ A, a = 0, and b = ∞. A solution: C(λ + 1) –Cx y(x) = – e , A where C is an arbitrary positive constant.

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40.

∞

y(x) + A

e–λt y

x

0

t

y(t) dt = 0,

λ > 0.

λ C x , where C is an arbitrary constant. A ∞ x y(x) + A e–λt y λ > 0. y(t) dt = Bxb , t 0 Solutions: y1 (x) = β1 xb , y2 (x) = β2 xb ,

A solution: y(x) = – 41.

where β1 and β2 are the roots of the quadratic equation Aβ 2 + λβ – Bλ = 0.

6.2. Equations With Quadratic Nonlinearity That Contain Arbitrary Functions 6.2-1. Equations of the Form

b a

G(· · ·) dt = F (x)

b

g(t)y(x)y(t) dt = f (x).

1. a

Solutions:

y(x) = ±λf (x),

λ=

–1/2

b

f (t)g(t) dt

.

a

b

f (t)y(t)y(xt) dt = A.

2. a ◦

1 . Solutions* y1 (x) =

A/I0 ,

y3 (x) = q(I1 x – I2 ), where

y2 (x) = – A/I0 , y4 (x) = –q(I1 x – I2 ),

1/2 A Im = t f (t) dt, q = , m = 0, 1, 2. I0 I22 – I12 I2 a The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = Bk xk , where the constants Bk can be found from the corresponding system

b

m

k=0

of algebraic equations. 2◦ . Solutions:

q=

y5 (x) = q(I1 xC – I2 ), y6 (x) = –q(I1 xC – I2 ), 1/2 b A , I = tmC f (t) dt, m = 0, 1, 2, m I0 I22 – I12 I2 a

where C is an arbitrary constant. The equation has more complicated solutions of the form y(x) =

n

Bk xkC , where C is

k=0

an arbitrary constant and the coefficients Bk can be found from the corresponding system of algebraic equations. * The arguments of the equations containing y(xt) in the integrand can vary, for example, within the following intervals: (a) 0 ≤ t ≤ 1, 0 ≤ x ≤ 1 for a = 0 and b = 1; (b) 1 ≤ t < ∞, 1 ≤ x < ∞ for a = 1 and b = ∞; (c) 0 ≤ t < ∞, 0 ≤ x < ∞ for a = 0 and b = ∞; or (d) a ≤ t ≤ b, 0 ≤ x < ∞ for arbitrary a and b such that 0 ≤ a < b ≤ ∞. Case (d) is a special case of (c) if f (t) is nonzero only on the interval a ≤ t ≤ b.

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3◦ . Solutions: y7 (x) = p(J0 ln x – J1 ),

1/2 A p= , J02 J2 – J0 J12

y8 (x) = –p(J0 ln x – J1 ), b Jm = (ln t)m f (t) dt. a

The equation has more complicated solutions of the form y(x) =

n

Ek (ln x)k , where the

k=0

constants Ek can be found from the corresponding system of algebraic equations.

b

f (t)y(t)y(xt) dt = Axβ .

3. a ◦

1 . Solutions: y1 (x) =

A/I0 xβ ,

y2 (x) = –

y3 (x) = q(I1 x – I2 ) xβ , where

t2β+m f (t) dt,

Im =

A/I0 xβ ,

y4 (x) = –q(I1 x – I2 ) xβ ,

b

q=

a

A , I2 (I0 I2 – I12 )

m = 0, 1, 2.

2◦ . The substitution y(x) = xβ w(x) leads to an equation of the form 6.2.2:

b

g(t)w(t)w(xt) dt = A,

g(x) = f (x)x2β .

a

Therefore, the integral equation in question has more complicated solutions.

b

f (t)y(t)y(xt) dt = A ln x + B.

4. a

This equation has solutions of the form y(x) = p ln x + q. The constants p and q are determined from the following system of two second-order algebraic equations: I1 p2 + I0 pq = A, where

I2 p2 + 2I1 pq + I0 q 2 = B,

b

f (t)(ln t)m dt,

Im =

m = 0, 1, 2.

a

b

f (t)y(t)y(xt) dt = Axλ ln x + Bxλ .

5. a

The substitution y(x) = xλ w(x) leads to an equation of the form 6.2.4:

b

g(t)w(t)w(xt) dt = A ln x + B,

g(t) = f (t)t2λ .

a

∞

f (t)y(t)y

6.

x t

0

dt = Axλ .

Solutions: y1 (x) =

A λ x , I

y2 (x) = –

A λ x , I

I=

∞

f (t) dt. 0

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b

f (t)y(t)y(x + λt) dt = A,

7. a ◦

1 . Solutions* y1 (x) =

λ > 0.

y2 (x) = – A/I0 , y4 (x) = –q(I0 x – I1 ),

A/I0 , y3 (x) = q(I0 x – I1 ),

where

b m

Im =

t f (t) dt,

q=

A , – I0 I12 )

λ(I02 I2

a

m = 0, 1, 2.

2◦ . The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = Bk xk , where the constants Bk can be found from the corresponding system k=0

of algebraic equations.

b

f (t)y(t)y(x + λt) dt = Ax + B,

8.

λ > 0.

a

A solution: y(x) = βx + µ, where the constants β and µ are determined from the following system of two second-order algebraic equations: b I0 βµ + I1 β 2 = A, I0 µ2 + (λ + 1)I1 βµ + λI2 β 2 = B, Im = tm f (t) dt. (1) a

Multiplying the first equation by B and the second by –A and adding the resulting equations, we obtain the quadratic equation AI0 z 2 + (λ + 1)AI1 – BI0 z + λAI2 – BI1 = 0, z = µ/β. (2) In general, to each root of equation (2) two solutions of system (1) correspond. Therefore, the original integral equation can have at most four solutions of this form. If the discriminant of equation (2) is negative, then the integral equation has no such solutions. The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = βk xk , where the constants βk can be found from the corresponding system k=0

of algebraic equations.

b

f (t)y(t)y(x + λt) dt = Ae–βx ,

9.

λ > 0.

a ◦

1 . Solutions: y1 (x) =

A/I0 e–βx ,

y2 (x) = –

y3 (x) = q(I0 x – I1 )e–βx , where

m –β(λ+1)t

Im =

t e a

f (t) dt,

q=

A/I0 e–βx ,

y4 (x) = –q(I0 x – I1 )e–βx ,

b

A , λ(I02 I2 – I0 I12 )

m = 0, 1, 2.

2◦ . The equation has more complicated solutions of the form y(x) = e–βx

n

Bk xk , where

k=0

the constants Bk can be found from the corresponding system of algebraic equations. * The arguments of the equations containing y(x+λt) in the integrand can vary within the following intervals: (a) 0 ≤ t < ∞, 0 ≤ x < ∞ for a = 0 and b = ∞ or (b) a ≤ t ≤ b, 0 ≤ x < ∞ for arbitrary a and b such that 0 ≤ a < b < ∞. Case (b) is a special case of (a) if f (t) is nonzero only on the interval a ≤ t ≤ b.

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3◦ . The substitution y(x) = e–βx w(x) leads to an equation of the form 6.2.7: b e–β(λ+1)t f (t)w(t)w(x + λt) dt = A. a

b

f (t)y(t)y(x – t) dt = A.

10. a ◦

1 . Solutions* y1 (x) =

A/I0 ,

y3 (x) = q(I0 x – I1 ), where

b

tm f (t) dt,

Im =

y2 (x) = – A/I0 , y4 (x) = –q(I0 x – I1 ),

q=

a

A , I0 I12 – I02 I2

m = 0, 1, 2.

2◦ . The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = λk xk , where the constants λk can be found from the corresponding system k=0

11.

of algebraic equations. For n = 3, such a solution is presented in 6.1.18. b f (t)y(t)y(x – t) dt = Ax + B. a

A solution: y(x) = λx + µ, where the constants λ and µ are determined from the following system of two second-order algebraic equations: b I0 λµ + I1 λ2 = A, I0 µ2 – I2 λ2 = B, Im = tm f (t) dt, m = 0, 1, 2. (1) a

Multiplying the first equation by B and the second by –A and adding the results, we obtain the quadratic equation AI0 z 2 – BI0 z – AI2 – BI1 = 0,

z = µ/λ.

(2)

In general, to each root of equation (2) two solutions of system (1) correspond. Therefore, the original integral equation can have at most four solutions of this form. If the discriminant of equation (2) is negative, then the integral equation has no such solutions. The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = λk xk , where the constants λk can be found from the corresponding system k=0

of algebraic equations.

b

f (t)y(t)y(x – t) dt =

12. a

n

A k xk .

k=0

This equation has solutions of the form y(x) =

n

λk xk ,

(1)

k=0

where the constants λk are determined from the system of algebraic equations obtained by substituting solution (1) into the original integral equation and matching the coefficients of like powers of x. * The arguments of the equations containing y(x–t) in the integrand can vary within the following intervals: (a) –∞ < t < ∞, –∞ < x < ∞ for a = –∞ and b = ∞ or (b) a ≤ t ≤ b, –∞ ≤ x < ∞, for arbitrary a and b such that –∞ < a < b < ∞. Case (b) is a special case of (a) if f (t) is nonzero only on the interval a ≤ t ≤ b.

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b

f (t)y(x – t)y(t) dt = Aeλx .

13. a

Solutions: y1 (x) =

A/I0 eλx ,

y2 (x) = –

y3 (x) = q(I0 x – I1 )eλx , where

m

Im =

t f (t) dt,

A/I0 eλx ,

y4 (x) = –q(I0 x – I1 )eλx ,

b

q=

a

A , I0 I12 – I02 I2

m = 0, 1, 2.

The integral equation has more complicated solutions of the form y(x) = eλx

n

Bk xk , where

k=0

the constants Bk can be found from the corresponding system of algebraic equations.

b

f (t)y(t)y(x – t) dt = A sinh λx.

14. a

A solution: y(x) = p sinh λx + q cosh λx.

(1)

Here p and q are roots of the algebraic system I0 pq + Ics (p2 – q 2 ) = A,

Icc q 2 – Iss p2 = 0,

(2)

where the notation

b

I0 =

f (t) dt,

Ics =

a

f (t) cosh(λt) sinh(λt) dt, a

b

f (t) cosh2 (λt) dt,

Icc =

b

a

b

f (t) sinh2 (λt) dt

Iss = a

is used. Different solutions of system (2) generate different solutions (1) of the integral equation.

It follows from the second equation of (2) that q = ± Iss /Icc p. Using this expression to eliminate q from the first equation of (2), we obtain the following four solutions: y1,2 (x) = p sinh λx ± k cosh λx , y3,4 (x) = –p sinh λx ± k cosh λx , Iss A k= , p= . 2 Icc (1 – k )Ics ± kI0

b

f (t)y(t)y(x – t) dt = A cosh λx.

15. a

A solution: y(x) = p sinh λx + q cosh λx.

(1)

Here p and q are roots of the algebraic system I0 pq + Ics (p2 – q 2 ) = 0,

Icc q 2 – Iss p2 = A,

(2)

where we use the notation introduced in 6.2.14. Different solutions of system (2) generate different solutions (1) of the integral equation.

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b

f (t)y(t)y(x – t) dt = A sin λx.

16. a

A solution: y(x) = p sin λx + q cos λx.

(1)

Here p and q are roots of the algebraic system I0 pq + Ics (p2 + q 2 ) = A,

Icc q 2 – Iss p2 = 0,

(2)

where

b

I0 =

f (t) dt, a

b

Ics =

f (t) cos(λt) sin(λt) dt, a

b

f (t) cos2 (λt) dt,

Icc = a

b

f (t) sin2 (λt) dt.

Iss = a

It follows from the second equation of (2) that q = ± Iss /Icc p. Using this expression to eliminate q from the first equation of (2), we obtain the following four solutions: y1,2 (x) = p sin λx ± k cos λx , y3,4 (x) = –p sin λx ± k cos λx , Iss A k= , p= . Icc (1 + k 2 )Ics ± kI0

b

f (t)y(t)y(x – t) dt = A cos λx.

17. a

A solution: y(x) = p sin λx + q cos λx.

(1)

Here p and q are roots of the algebraic system I0 pq + Ics (p2 + q 2 ) = 0,

Icc q 2 – Iss p2 = A,

(2)

where we use the notation introduced in 6.2.16. Different solutions of system (2) generate different solutions (1) of the integral equation.

1

y(t)y(ξ) dt = A,

18.

ξ = f (x)t.

0 ◦

1 . Solutions: y1 (t) = y3 (t) = y5 (t) =

√ √ √

A, A (3t – 2), A (10t2 – 12t + 3),

√ y2 (t) = – A, √ y4 (t) = – A (3t – 2), √ y6 (t) = – A (10t2 – 12t + 3).

2◦ . The integral equation has some other (more complicated) solutions of the polynomial n form y(t) = Bk tk , where the constants Bk can be found from the corresponding system of k=0

algebraic equations. 3◦ . The substitution z = f (x) leads to an equation of the form 6.1.12.

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6.2-2. Equations of the Form y(x) + 19.

b a

K(x, t)y 2 (t) dt = F (x)

b

f (x)y 2 (t) dt = 0.

y(x) + a

Solutions: y1 (x) = 0 and y2 (x) = λf (x), where λ = – 20.

b a

f 2 (t) dt

–1 .

b

f (x)g(t)y 2 (t) dt = 0.

y(x) + a

This is a special case of equation 6.8.29.

Solutions: y1 (x) = 0 and y2 (x) = λf (x), where λ = – 21.

b a

f 2 (t)g(t) dt

–1 .

b

y 2 (t) dt = f (x).

y(x) + A a

This is a special case of equation 6.8.27. A solution: y(x) = f (x) + λ, where λ is determined by the quadratic equation A(b – a)λ2 + (1 + 2AI1 )λ + AI2 = 0,

where

I1 =

b

f (t) dt, a

22.

b

f 2 (t) dt.

I2 = a

b

g(t)y 2 (t) dt = f (x).

y(x) + a

This is a special case of equation 6.8.29. A solution: y(x) = f (x) + λ, where λ is determined by the quadratic equation I0 λ2 + (1 + 2I1 )λ + I2 = 0,

where

b

f m (t)g(t) dt,

Im =

m = 0, 1, 2.

a

23.

b

g(x)y 2 (t) dt = f (x).

y(x) + a

Solution: y(x) = λg(x) + f (x), where λ is determined by the quadratic equation Igg = a

24.

y(x) +

Igg λ2 + (1 + 2If g )λ + If f = 0, b b 2 g (t) dt, If g = f (t)g(t) dt, If f = a

b

f 2 (t) dt.

a

g1 (x)h1 (t) + g2 (x)h2 (t) y 2 (t) dt = f (x).

b

a

A solution: y(x) = λ1 g1 (x) + λ2 g2 (x) + f (x), where the constants λ1 and λ2 can be found from a system of two second-order algebraic equations (this system can be obtained from the more general system presented in 6.8.42).

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6.2-3. Equations of the Form y(x) + 25.

b a

Knm (x, t)y n (x)y m (t) dt = F (x), n + m ≤ 2

b

y(x) +

g(t)y(x)y(t) dt = f (x). a

Solutions: y1 (x) = λ1 f (x),

y2 (x) = λ2 f (x),

where λ1 and λ2 are the roots of the quadratic equation b Iλ2 + λ – 1 = 0, I= f (t)g(t) dt. a

26.

b

y(x) +

g(x)y(x)y(t) dt = f (x). a

A solution:

f (x) , 1 + λg(x) where λ is a root of the algebraic (or transcendental) equation b f (t) dt λ– = 0. 1 + λg(t) a Different roots generate different solutions of the integral equation. b y(x) + g1 (t)y 2 (x) + g2 (x)y(t) dt = f (x). y(x) =

27.

a

Solution in an implicit form:

y(x) + Iy 2 (x) + λg2 (x) – f (x) = 0,

b

I=

g1 (t) dt,

(1)

a

where λ is determined by the algebraic equation b λ= y(t) dt.

(2)

a

28.

Here the function y(x) = y(x, λ) obtained by solving the quadratic equation (1) must be substituted in the integrand of (2). b y(x) + g1 (t)y 2 (x) + g2 (x)y 2 (t) dt = f (x). a

Solution in an implicit form:

2

y(x) + Iy (x) + λg2 (x) – f (x) = 0,

I=

b

g1 (t) dt,

(1)

a

where λ is determined by the algebraic equation b λ= y 2 (t) dt.

(2)

a

29.

Here the function y(x) = y(x, λ) obtained by solving the quadratic equation (1) must be substituted into the integrand of (2). b y(x) + g11 (x)h11 (t)y 2 (x) + g12 (x)h12 (t)y(x)y(t) + g22 (x)h22 (t)y 2 (t) a + g1 (x)h1 (t)y(x) + g2 (x)h2 (t)y(t) dt = f (x). This is a special case of equation 6.8.44.

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b

6.2-4. Equations of the Form y(x) + 30.

a

G(· · ·) dt = F (x)

b

y(x) +

f (t)y(t)y(xt) dt = 0. a

1◦ . Solutions: 1 C (I1 – I0 )x + I1 – I2 C x , y2 (x) = x , I0 I0 I2 – I12 b Im = f (t)t2C+m dt, m = 0, 1, 2,

y1 (x) = –

a

where C is an arbitrary constant.

n

There are more complicated solutions of the form y(x) = xC

Bk xk , where C is an

k=0

arbitrary constant and the coefficients Bk can be found from the corresponding system of algebraic equations. 2◦ . A solution: y3 (x) =

(I1 – I0 )xβ + I1 – I2 C x , I0 I2 – I12

b

f (t)t2C+mβ dt,

Im =

m = 0, 1, 2,

a

where C and β are arbitrary constants. There are more complicated solutions of the form y(x) = xC

n

Dk xkβ , where C and β

k=0

are arbitrary constants and the coefficients Dk can be found from the corresponding system of algebraic equations. 3◦ . A solution: y4 (x) =

xC (J1 ln x – J2 ) , J0 J2 – J12

b

f (t)t2C (ln t)m dt,

Jm =

m = 0, 1, 2,

a

where C is an arbitrary constant. There are more complicated solutions of the form y(x) = xC

n

Ek (ln x)k , where C is

k=0

an arbitrary constant and the coefficients Ek can be found from the corresponding system of algebraic equations. 4◦ . The equation also has the trivial solution y(x) ≡ 0. 5◦ . The substitution y(x) = xβ w(x) leads to an equation of the same form, w(x) +

b

g(t)w(t)w(xt) dt = 0,

g(x) = f (x)x2β .

a

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31.

b

f (t)y(t)y(xt) dt = Axβ .

y(x) + a

1◦ . Solutions: y1 (x) = k1 xβ ,

y2 (x) = k2 xβ ,

where k1 and k2 are the roots of the quadratic equation Ik 2 + k – A = 0,

b

f (t)t2β dt.

I= a

2◦ . Solutions: y(x) = xβ (λx + µ), where λ and µ are determined from the following system of two algebraic equations (this system can be reduced to a quadratic equation): I2 λ + I1 µ + 1 = 0,

I1 λµ + I0 µ2 + µ – A = 0

b

f (t)t2β+m dt, m = 0, 1, 2.

where Im = a

3◦ . There are more complicated solutions of the form y(x) = xβ

n

Bm xm , where the Bm

m=0

can be found from the corresponding system of algebraic equations. 32.

b

y(x) +

f (t)y(t)y(xt) dt = A ln x + B. a

This equation has solutions of the form y(x) = p ln x + q, where the constants p and q can be found from a system of two second-order algebraic equations. 33.

∞

y(x) +

f (t)y(t)y

x

0

t

dt = 0.

1◦ . A solution:

y(x) = –kxC ,

–1

∞

k=

f (t) dt

,

0

where C is an arbitrary constant. 2◦ . The equation has the trivial solution y(x) ≡ 0. 3◦ . The substitution y(x) = xβ w(x) leads to an equation of the same form, ∞ x w(x) + f (t)w(t)w dt = 0. t 0 34.

y(x) +

∞

f (t)y 0

x t

y(t) dt = Axb .

Solutions: y1 (x) = λ1 xb ,

y2 (x) = λ2 xb ,

where λ1 and λ2 are the roots of the quadratic equation ∞ 2 Iλ + λ – A = 0, I= f (t) dt. 0

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35.

b

y(x) +

f (t)y(t)y(x + λt) dt = 0,

λ > 0.

a

1◦ . Solutions: 1 I2 – I1 x exp(–Cx), y2 (x) = 2 exp(–Cx), I0 I1 – I0 I2 b = tm exp –C(λ + 1)t f (t) dt, m = 0, 1, 2,

y1 (x) = – Im

a

where C is an arbitrary constant. 2◦ . There are more complicated solutions of the form y(x) = exp(–Cx)

n

Ak xk , where C

k=0

is an arbitrary constant and the coefficients Ak can be found from the corresponding system of algebraic equations. 3◦ . The equation also has the trivial solution y(x) ≡ 0. 4◦ . The substitution y(x) = eβx w(x) leads to a similar equation: b w(x) + g(t)w(t)w(x + λt) dt = 0, g(t) = eβ(λ+1)t f (t). a

36.

b

f (t)y(x + λt)y(t) dt = Ae–µx ,

y(x) +

λ > 0.

a

1◦ . Solutions: y1 (x) = k1 e–µx ,

y2 (x) = k2 e–µx ,

where k1 and k2 are the roots of the quadratic equation b Ik 2 + k – A = 0, I= e–µ(λ+1)t f (t) dt. a ◦

2 . There are more complicated solutions of the form y(x) = e–µx

n

Bm xm , where the Bm

m=0

can be found from the corresponding system of algebraic equations. 3◦ . The substitution y(x) = eβx w(x) leads to an equation of the same form, b w(x) + g(t)w(t)w(x – t) dt = Ae(λ–β)x , g(t) = f (t)eβ(λ+1)t . a

37.

b

y(x) +

f (t)y(t)y(x – t) dt = 0. a

1◦ . Solutions: 1 I2 – I1 x exp(Cx), y2 (x) = 2 exp(Cx), I0 I1 – I0 I2 where C is an arbitrary constant and m = 0, 1, 2. y1 (x) = –

b

tm f (t) dt,

Im =

2◦ . There are more complicated solutions of the form y(x) = exp(Cx)

a n

Ak xk , where C is

k=0

an arbitrary constant and the coefficients Ak can be found from the corresponding system of algebraic equations. 3◦ . The equation also has the trivial solution y(x) ≡ 0. 4◦ . The substitution y(x) = exp(Cx)w(x) leads to an equation of the same form: b w(x) + f (t)w(t)w(x – t) dt = 0. a

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38.

b

f (t)y(x – t)y(t) dt = Aeλx .

y(x) + a

1◦ . Solutions: y1 (x) = k1 eλx ,

y2 (x) = k2 eλx ,

where k1 and k2 are the roots of the quadratic equation Ik 2 + k – A = 0, I=

b

f (t) dt.

a

2◦ . The substitution y(x) = eβx w(x) leads to an equation of the same form, b w(x) + f (t)w(t)w(x – t) dt = Ae(λ–β)x . a

39.

b

y(x) +

f (t)y(t)y(x – t) dt = A sinh λx. a

A solution: y(x) = p sinh λx + q cosh λx.

(1)

Here p and q are roots of the algebraic system p + I0 pq + Ics (p2 – q 2 ) = A, where

b

I0 =

f (t) dt,

f (t) cosh(λt) sinh(λt) dt, a

b

f (t) cosh2 (λt) dt,

Icc =

b

f (t) sinh2 (λt) dt.

Iss =

a

40.

(2)

b

Ics =

a

q + Icc q 2 – Iss p2 = 0,

a

Different solutions of system (2) generate different solutions (1) of the integral equation. b y(x) + f (t)y(t)y(x – t) dt = A cosh λx. a

A solution: y(x) = p sinh λx + q cosh λx.

(1)

Here p and q are roots of the algebraic system p + I0 pq + Ics (p2 – q 2 ) = 0,

41.

q + Icc q 2 – Iss p2 = A,

(2)

where we use the notation introduced in 6.2.39. Different solutions of system (2) generate different solutions (1) of the integral equation. b y(x) + f (t)y(t)y(x – t) dt = A sin λx. a

A solution: y(x) = p sin λx + q cos λx.

(1)

Here p and q are roots of the algebraic system p + I0 pq + Ics (p2 + q 2 ) = A, where

b

I0 =

f (t) dt,

Icc =

b

f (t) cos (λt) dt, a

f (t) cos(λt) sin(λt) dt, a

2

(2)

b

Ics =

a

q + Icc q 2 – Iss p2 = 0,

b

f (t) sin2 (λt) dt.

Iss = a

Different solutions of system (2) generate different solutions (1) of the integral equation.

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42.

b

y(x) +

f (t)y(t)y(x – t) dt = A cos λx. a

A solution: y(x) = p sin λx + q cos λx.

(1)

Here p and q are roots of the algebraic system p + I0 pq + Ics (p2 + q 2 ) = 0,

q + Icc q 2 – Iss p2 = A,

(2)

where we use the notation introduced in 6.2.41. Different solutions of system (2) generate different solutions (1) of the integral equation.

6.3. Equations With Power-Law Nonlinearity 6.3-1. Equations of the Form

b a

G(· · ·) dt = F (x)

b

tλ y µ (x)y β (t) dt = f (x).

1. a

A solution:

1 y(x) = A f (x) µ ,

b

– 1 µ+β β µ t f (t) dt . λ

A= a

b

eλt y µ (x)y β (t) dt = f (x).

2. a

A solution: 1 y(x) = A f (x) µ ,

b

– 1 µ+β β e f (t) µ dt . λt

A= a

3.

∞

s f (xa t)tb y xk t y(t) dt = Axc .

0

A solution: A 1 a + c + ab s+1 λ y(x) = x , λ= , I k – a – as ∞ a + c + as + bk + cs I= . f (t)tβ dt, β = k – a – as 0 6.3-2. Equations of the Form y(x) + 4.

b a

K(x, t)y β (t) dt = F (x)

b

tλ y β (t) dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atλ y β . 5.

b

eµt y β (t) dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Aeµt y β .

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b

eλ(x–t) y β (t) dt = g(x).

6.

y(x) + A

7.

This is a special case of equation 6.8.28 with f (t, y) = Ay β . b y(x) – g(x)y β (t) dt = 0.

a

a

A solution:

y(x) = λg(x),

b β

λ=

g (t) dt

1 1–β

.

a

8.

For β > 0, the equation also has the trivial solution y(x) ≡ 0. b y(x) – g(x)y β (t) dt = h(x). a

9.

This is a special case of equation 6.8.29 with f (t, y) = –y β . b y(x) + A cosh(λx + µt)y β (t) dt = h(x).

10.

This is a special case of equation 6.8.31 with f (t, y) = Ay β . b y(x) + A sinh(λx + µt)y β (t) dt = h(x).

11.

This is a special case of equation 6.8.32 with f (t, y) = Ay β . b y(x) + A cos(λx + µt)y β (t) dt = h(x).

a

a

a

12.

This is a special case of equation 6.8.33 with f (t, y) = Ay β . b y(x) + A sin(λx + µt)y β (t) dt = h(x). a

13.

This is a special case of equation 6.8.34 with f (t, y) = Ay β . ∞

t y(x) + f y(t) dt = Ax2 . x 0 Solutions: yk (x) = βk2 x2 , where βk (k = 1, 2) are the roots of the quadratic equations ∞ β 2 ± Iβ – A = 0, I= zf (z) dz.

14.

∞

y(x) –

tλ f

0

A solution:

t x

0

β y(t) dt = 0, 1+λ

y(x) = Ax 1–β , 15.

β ≠ 1.

∞

A1–β =

λ+β

z 1–β f (z) dz. 0

∞

y(x) –

β eλt f (ax + bt) y(t) dt = 0,

b ≠ 0, aβ ≠ –b.

–∞

A solution:

aλ y(x) = A exp – x , aβ + b

A

1–β

∞

=

exp –∞

λb z f (bz) dz. aβ + b

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6.3-3. Equations of the Form y(x) + 16.

b a

G(· · ·) dt = F (x)

b

y β (x)y µ (t) dt = f (x).

y(x) + A a

Solution in an implicit form: y(x) + Aλy β (x) – f (x) = 0,

(1)

where λ is determined by the algebraic (or transcendental) equation b λ= y µ (t) dt.

(2)

a

17.

Here the function y(x) = y(x, λ) obtained by solving the quadratic equation (1) must be substituted in the integrand of (2). b y(x) + g(t)y(x)y µ (t) dt = f (x). a

A solution: y(x) = λf (x), where λ is determined from the algebraic (or transcendental) equation b Iλµ+1 + λ – 1 = 0, I= g(t)f µ (t) dt. a

18.

b

g(x)y(x)y µ (t) dt = f (x).

y(x) + a

A solution:

f (x) , 1 + λg(x) where λ is a root of the algebraic (or transcendental) equation b f µ (t) dt λ– = 0. µ a [1 + λg(t)] Different roots generate different solutions of the integral equation. b y(x) + g1 (t)y 2 (x) + g2 (x)y µ (t) dt = f (x). y(x) =

19.

a

Solution in an implicit form:

y(x) + Iy 2 (x) + λg2 (x) – f (x) = 0,

I=

b

g1 (t) dt,

(1)

a

where λ is determined by the algebraic (or transcendental) equation b λ= y µ (t) dt.

(2)

a

20.

Here the function y(x) = y(x, λ) obtained by solving the quadratic equation (1) must be substituted in the integrand of (2). b y(x) + g1 (x)h1 (t)y k (x)y s (t) + g2 (x)h2 (t)y p (x)y q (t) dt = f (x).

21.

This is a special case of equation 6.8.44. b y(x) + A y(xt)y β (t) dt = 0.

a

a

This is a special case of equation 6.8.45 with f (t, y) = Ay β .

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6.4. Equations With Exponential Nonlinearity 6.4-1. Integrands With Nonlinearity of the Form exp[βy(t)] 1.

b

y(x) + A

exp[βy(t)] dt = g(x). a

2.

This is a special case of equation 6.8.27 with f (t, y) = A exp(βy). b y(x) + A tµ exp[βy(t)] dt = g(x). a

3.

This is a special case of equation 6.8.27 with f (t, y) = Atµ exp(βy). b y(x) + A exp µt + βy(t) dt = g(x).

4.

This is a special case of equation 6.8.27 with f (t, y) = A exp(µt) exp(βy). b y(x) + A exp λ(x – t) + βy(t) dt = g(x).

5.

This is a special case of equation 6.8.28 with f (t, y) = A exp(βy). b y(x) + g(x) exp[βy(t)] dt = h(x).

6.

This is a special case of equation 6.8.29 with f (t, y) = exp(βy). b y(x) + A cosh(λx + µt) exp[βy(t)] dt = h(x).

7.

This is a special case of equation 6.8.31 with f (t, y) = A exp(βy). b y(x) + A sinh(λx + µt) exp[βy(t)] dt = h(x).

8.

This is a special case of equation 6.8.32 with f (t, y) = A exp(βy). b y(x) + A cos(λx + µt) exp[βy(t)] dt = h(x).

9.

This is a special case of equation 6.8.33 with f (t, y) = A exp(βy). b y(x) + A sin(λx + µt) exp[βy(t)] dt = h(x).

a

a

a

a

a

a

a

This is a special case of equation 6.8.34 with f (t, y) = A exp(βy). 6.4-2. Other Integrands

b

exp βy(x) + γy(t) dt = h(x).

10.

y(x) + A

11.

This is a special case of equation 6.8.43 with g(x, y) = A exp(βy) and f (t, y) = exp(γy). b y(x) + A y(xt) exp[βy(t)] dt = 0.

a

a

This is a special case of equation 6.8.45 with f (t, y) = A exp(βy).

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6.5. Equations With Hyperbolic Nonlinearity 6.5-1. Integrands With Nonlinearity of the Form cosh[βy(t)] 1.

b

y(x) + A

cosh[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A cosh(βy). 2.

b

tµ coshk [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ coshk (βy). 3.

b

y(x) + A

cosh(µt) cosh[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A cosh(µt) cosh(βy). 4.

b

eλ(x–t) cosh[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A cosh(βy). 5.

b

y(x) +

g(x) cosh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = cosh(βy). 6.

b

y(x) + A

cosh(λx + µt) cosh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.31 with f (t, y) = A cosh(βy). 7.

b

y(x) + A

sinh(λx + µt) cosh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.32 with f (t, y) = A cosh(βy). 8.

b

y(x) + A

cos(λx + µt) cosh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.33 with f (t, y) = A cosh(βy). 9.

b

y(x) + A

sin(λx + µt) cosh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.34 with f (t, y) = A cosh(βy). 6.5-2. Integrands With Nonlinearity of the Form sinh[βy(t)] 10.

b

y(x) + A

sinh[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A sinh(βy).

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11.

b

tµ sinhk [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ sinhk (βy). 12.

b

y(x) + A

sinh(µt) sinh[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A sinh(µt) sinh(βy). 13.

b

eλ(x–t) sinh[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A sinh(βy). 14.

b

y(x) +

g(x) sinh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = sinh(βy). 15.

b

y(x) + A

cosh(λx + µt) sinh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.31 with f (t, y) = A sinh(βy). 16.

b

y(x) + A

sinh(λx + µt) sinh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.32 with f (t, y) = A sinh(βy). 17.

b

y(x) + A

cos(λx + µt) sinh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.33 with f (t, y) = A sinh(βy). 18.

b

y(x) + A

sin(λx + µt) sinh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.34 with f (t, y) = A sinh(βy). 6.5-3. Integrands With Nonlinearity of the Form tanh[βy(t)] 19.

b

y(x) + A

tanh[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A tanh(βy). 20.

b

tµ tanhk [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ tanhk (βy). 21.

b

y(x) + A

tanh(µt) tanh[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A tanh(µt) tanh(βy).

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22.

b

eλ(x–t) tanh[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A tanh(βy). 23.

b

y(x) +

g(x) tanh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = tanh(βy). 24.

b

y(x) + A

cosh(λx + µt) tanh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.31 with f (t, y) = A tanh(βy). 25.

b

y(x) + A

sinh(λx + µt) tanh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.32 with f (t, y) = A tanh(βy). 26.

b

y(x) + A

cos(λx + µt) tanh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.33 with f (t, y) = A tanh(βy). 27.

b

y(x) + A

sin(λx + µt) tanh[βy(t)] dt = h(x). a

This is a special case of equation 6.8.34 with f (t, y) = A tanh(βy). 6.5-4. Integrands With Nonlinearity of the Form coth[βy(t)] 28.

b

y(x) + A

coth[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A coth(βy). 29.

b

tµ cothk [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ cothk (βy). 30.

b

y(x) + A

coth(µt) coth[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A coth(µt) coth(βy). 31.

b

eλ(x–t) coth[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A coth(βy). 32.

b

y(x) +

g(x) coth[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = coth(βy).

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33.

b

y(x) + A

cosh(λx + µt) coth[βy(t)] dt = h(x). a

34.

This is a special case of equation 6.8.31 with f (t, y) = A coth(βy). b y(x) + A sinh(λx + µt) coth[βy(t)] dt = h(x).

35.

This is a special case of equation 6.8.32 with f (t, y) = A coth(βy). b y(x) + A cos(λx + µt) coth[βy(t)] dt = h(x).

36.

This is a special case of equation 6.8.33 with f (t, y) = A coth(βy). b y(x) + A sin(λx + µt) coth[βy(t)] dt = h(x).

a

a

a

This is a special case of equation 6.8.34 with f (t, y) = A coth(βy). 6.5-5. Other Integrands 37.

b

y(x) + A

cosh[βy(x)] cosh[γy(t)] dt = h(x). a

38.

This is a special case of equation 6.8.43 with g(x, y) = A cosh(βy) and f (t, y) = cosh(γy). b y(x) + A y(xt) cosh[βy(t)] dt = 0. a

39.

This is a special case of equation 6.8.45 with f (t, y) = A cosh(βy). b y(x) + A sinh[βy(x)] sinh[γy(t)] dt = h(x).

40.

This is a special case of equation 6.8.43 with g(x, y) = A sinh(βy) and f (t, y) = sinh(γy). b y(x) + A y(xt) sinh[βy(t)] dt = 0.

41.

This is a special case of equation 6.8.45 with f (t, y) = A sinh(βy). b y(x) + A tanh[βy(x)] tanh[γy(t)] dt = h(x).

42.

This is a special case of equation 6.8.43 with g(x, y) = A tanh(βy) and f (t, y) = tanh(γy). b y(x) + A y(xt) tanh[βy(t)] dt = 0.

43.

This is a special case of equation 6.8.45 with f (t, y) = A tanh(βy). b y(x) + A coth[βy(x)] coth[γy(t)] dt = h(x).

44.

This is a special case of equation 6.8.43 with g(x, y) = A coth(βy) and f (t, y) = coth(γy). b y(x) + A y(xt) coth[βy(t)] dt = 0.

a

a

a

a

a

a

This is a special case of equation 6.8.45 with f (t, y) = A coth(βy).

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6.6. Equations With Logarithmic Nonlinearity 6.6-1. Integrands With Nonlinearity of the Form ln[βy(t)] 1.

b

y(x) + A

ln[βy(t)] dt = g(x). a

2.

This is a special case of equation 6.8.27 with f (t, y) = A ln(βy). b y(x) + A tµ lnk [βy(t)] dt = g(x). a

3.

This is a special case of equation 6.8.27 with f (t, y) = Atµ lnk (βy). b y(x) + A ln(µt) ln[βy(t)] dt = g(x).

4.

This is a special case of equation 6.8.27 with f (t, y) = A ln(µt) ln(βy). b y(x) + A eλ(x–t) ln[βy(t)] dt = g(x).

5.

This is a special case of equation 6.8.28 with f (t, y) = A ln(βy). b y(x) + g(x) ln[βy(t)] dt = h(x).

6.

This is a special case of equation 6.8.29 with f (t, y) = ln(βy). b y(x) + A cosh(λx + µt) ln[βy(t)] dt = h(x).

7.

This is a special case of equation 6.8.31 with f (t, y) = A ln(βy). b y(x) + A sinh(λx + µt) ln[βy(t)] dt = h(x).

8.

This is a special case of equation 6.8.32 with f (t, y) = A ln(βy). b y(x) + A cos(λx + µt) ln[βy(t)] dt = h(x).

a

a

a

a

a

a

9.

This is a special case of equation 6.8.33 with f (t, y) = A ln(βy). b y(x) + A sin(λx + µt) ln[βy(t)] dt = h(x). a

This is a special case of equation 6.8.34 with f (t, y) = A ln(βy). 6.6-2. Other Integrands 10.

b

y(x) + A

ln[βy(x)] ln[γy(t)] dt = h(x). a

11.

This is a special case of equation 6.8.43 with g(x, y) = A ln(βy) and f (t, y) = ln(γy). b y(x) + A y(xt) ln[βy(t)] dt = 0. a

This is a special case of equation 6.8.45 with f (t, y) = A ln(βy).

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6.7. Equations With Trigonometric Nonlinearity 6.7-1. Integrands With Nonlinearity of the Form cos[βy(t)] 1.

b

y(x) + A

cos[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A cos(βy). 2.

b

tµ cosk [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ cosk (βy). 3.

b

y(x) + A

cos(µt) cos[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A cos(µt) cos(βy). 4.

b

eλ(x–t) cos[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A cos(βy). 5.

b

y(x) +

g(x) cos[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = cos(βy). 6.

b

y(x) + A

cosh(λx + µt) cos[βy(t)] dt = h(x). a

This is a special case of equation 6.8.31 with f (t, y) = A cos(βy). 7.

b

y(x) + A

sinh(λx + µt) cos[βy(t)] dt = h(x). a

This is a special case of equation 6.8.32 with f (t, y) = A cos(βy). 8.

b

y(x) + A

cos(λx + µt) cos[βy(t)] dt = h(x). a

This is a special case of equation 6.8.33 with f (t, y) = A cos(βy). 9.

b

y(x) + A

sin(λx + µt) cos[βy(t)] dt = h(x). a

This is a special case of equation 6.8.34 with f (t, y) = A cos(βy). 6.7-2. Integrands With Nonlinearity of the Form sin[βy(t)] 10.

b

y(x) + A

sin[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A sin(βy).

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11.

b

tµ sink [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ sink (βy). 12.

b

y(x) + A

sin(µt) sin[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A sin(µt) sin(βy). 13.

b

eλ(x–t) sin[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A sin(βy). 14.

b

y(x) +

g(x) sin[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = sin(βy). 15.

b

y(x) + A

cosh(λx + µt) sin[βy(t)] dt = h(x). a

This is a special case of equation 6.8.31 with f (t, y) = A sin(βy). 16.

b

y(x) + A

sinh(λx + µt) sin[βy(t)] dt = h(x). a

This is a special case of equation 6.8.32 with f (t, y) = A sin(βy). 17.

b

y(x) + A

cos(λx + µt) sin[βy(t)] dt = h(x). a

This is a special case of equation 6.8.33 with f (t, y) = A sin(βy). 18.

b

y(x) + A

sin(λx + µt) sin[βy(t)] dt = h(x). a

This is a special case of equation 6.8.34 with f (t, y) = A sin(βy). 6.7-3. Integrands With Nonlinearity of the Form tan[βy(t)] 19.

b

y(x) + A

tan[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A tan(βy). 20.

b

tµ tank [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ tank (βy). 21.

b

tan(µt) tan[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = A tan(µt) tan(βy).

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22.

b

eλ(x–t) tan[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A tan(βy). 23.

b

y(x) +

g(x) tan[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = tan(βy). 24.

b

y(x) + A

cosh(λx + µt) tan[βy(t)] dt = h(x). a

This is a special case of equation 6.8.31 with f (t, y) = A tan(βy). 25.

b

y(x) + A

sinh(λx + µt) tan[βy(t)] dt = h(x). a

This is a special case of equation 6.8.32 with f (t, y) = A tan(βy). 26.

b

y(x) + A

cos(λx + µt) tan[βy(t)] dt = h(x). a

This is a special case of equation 6.8.33 with f (t, y) = A tan(βy). 27.

b

y(x) + A

sin(λx + µt) tan[βy(t)] dt = h(x). a

This is a special case of equation 6.8.34 with f (t, y) = A tan(βy). 6.7-4. Integrands With Nonlinearity of the Form cot[βy(t)] 28.

b

y(x) + A

cot[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A cot(βy). 29.

b

tµ cotk [βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.27 with f (t, y) = Atµ cotk (βy). 30.

b

y(x) + A

cot(µt) cot[βy(t)] dt = g(x). a

This is a special case of equation 6.8.27 with f (t, y) = A cot(µt) cot(βy). 31.

b

eλ(x–t) cot[βy(t)] dt = g(x).

y(x) + A a

This is a special case of equation 6.8.28 with f (t, y) = A cot(βy). 32.

b

y(x) +

g(x) cot[βy(t)] dt = h(x). a

This is a special case of equation 6.8.29 with f (t, y) = cot(βy).

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33.

b

y(x) + A

cosh(λx + µt) cot[βy(t)] dt = h(x). a

34.

This is a special case of equation 6.8.31 with f (t, y) = A cot(βy). b y(x) + A sinh(λx + µt) cot[βy(t)] dt = h(x).

35.

This is a special case of equation 6.8.32 with f (t, y) = A cot(βy). b y(x) + A cos(λx + µt) cot[βy(t)] dt = h(x).

36.

This is a special case of equation 6.8.33 with f (t, y) = A cot(βy). b y(x) + A sin(λx + µt) cot[βy(t)] dt = h(x).

a

a

a

This is a special case of equation 6.8.34 with f (t, y) = A cot(βy). 6.7-5. Other Integrands 37.

b

y(x) + A

cos[βy(x)] cos[γy(t)] dt = h(x). a

38.

This is a special case of equation 6.8.43 with g(x, y) = A cos(βy) and f (t, y) = cos(γy). b y(x) + A y(xt) cos[βy(t)] dt = 0. a

39.

This is a special case of equation 6.8.45 with f (t, y) = A cos(βy). b y(x) + A sin[βy(x)] sin[γy(t)] dt = h(x).

40.

This is a special case of equation 6.8.43 with g(x, y) = A sin(βy) and f (t, y) = sin(γy). b y(x) + A y(xt) sin[βy(t)] dt = 0.

41.

This is a special case of equation 6.8.45 with f (t, y) = A sin(βy). b y(x) + A tan[βy(x)] tan[γy(t)] dt = h(x).

42.

This is a special case of equation 6.8.43 with g(x, y) = A tan(βy) and f (t, y) = tan(γy). b y(x) + A y(xt) tan[βy(t)] dt = 0.

43.

This is a special case of equation 6.8.45 with f (t, y) = A tan(βy). b y(x) + A cot[βy(x)] cot[γy(t)] dt = h(x).

44.

This is a special case of equation 6.8.43 with g(x, y) = A cot(βy) and f (t, y) = cot(γy). b y(x) + A y(xt) cot[βy(t)] dt = 0.

a

a

a

a

a

a

This is a special case of equation 6.8.45 with f (t, y) = A cot(βy).

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6.8. Equations With Nonlinearity of General Form 6.8-1. Equations of the Form

b

1.

b a

G(· · ·) dt = F (x)

y(x)f t, y(t) dt = g(x).

a

A solution: y(x) = λg(x), where λ is determined by the algebraic (or transcendental) equation b λ f t, λg(t) dt = 1. a

b

2.

y k (x)f t, y(t) dt = g(x).

a

A solution: y(x) = λ[g(x)]1/k , where λ is determined from the algebraic (or transcendental) b equation λk f t, λg 1/k (t) dt = 1. a

b

3.

ϕ y(x) f t, y(t) dt = g(x).

a

A solution in an implicit form: λϕ y(x) – g(x) = 0,

(1)

where λ is determined by the algebraic (or transcendental) equation λ – F (λ) = 0,

F (λ) =

b

f t, y(t) dt.

(2)

a

Here the function y(x) = y(x, λ) obtained by solving (1) must be substituted into (2). The number of solutions of the integral equation is determined by the number of the solutions obtained from (1) and (2).

b

4.

y(xt)f t, y(t) dt = A.

a ◦

1 . Solutions: y(x) = λk , where λk are roots of the algebraic (or transcendental) equation b

λ

a

f (t, λ) dt = A.

2◦ . Solutions: y(x) = px + q, where p and q are roots of the following system of algebraic (or transcendental) equations:

b

tf (t, pt + q) dt = 0, a

q

b

f (t, pt + q) dt = A. a

In the case f t, y(t) = f¯(t)y(t), see 6.2.2 for solutions of this system. 2◦ . The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = Bk xk , where the constants Bk can be found from the corresponding system k=0

of algebraic (or transcendental) equations. 4◦ . The integral equation can have logarithmic solutions similar to those presented in item 3◦ of equation 6.2.2.

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b

5.

y(xt)f t, y(t) dt = Ax + B.

a

1◦ . A solution: y(x) = px + q,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: p

b

tf (t, pt + q) dt – A = 0,

b

q

f (t, pt + q) dt – B = 0.

a

(2)

a

Different solutions of system (2) generate different solutions (1) of the integral equation. 2◦ . The integral equation has some other (more complicated) solutions of the polynomial n form y(x) = Bk xk , where the constants Bk can be found from the corresponding system k=0

of algebraic (or transcendental) equations.

b

6.

y(xt)f t, y(t) dt = Axβ .

a

A solution: y(x) = kxβ ,

(1)

where k is a root of the algebraic (or transcendental) equation kF (k) – A = 0,

b

F (k) =

tβ f t, ktβ dt.

(2)

a

Each root of equation (2) generates a solution of the integral equation which has the form (1).

b

7.

y(xt)f t, y(t) dt = A ln x + B.

a

A solution: y(x) = p ln x + q,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: p

b

f (t, p ln t + q) dt – A = 0, a

b

(p ln t + q)f (t, p ln t + q) dt – B = 0.

(2)

a

Different solutions of system (2) generate different solutions (1) of the integral equation.

b

8.

y(xt)f t, y(t) dt = Axβ ln x.

a

This equation has solutions of the form y(x) = pxβ ln x + qxβ , where p and q are some constants.

b

9.

y(xt)f t, y(t) dt = A cos(β ln x).

a

This equation has solutions of the form y(x) = p cos(β ln x) + q sin(β ln x), where p and q are some constants.

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b

10.

y(xt)f t, y(t) dt = A sin(β ln x).

a

11.

This equation has solutions of the form y(x) = p cos(β ln x) + q sin(β ln x), where p and q are some constants. b y(xt)f t, y(t) dt = Axβ cos(β ln x) + Bxβ sin(β ln x). a

12.

This equation has solutions of the form y(x) = pxβ cos(β ln x) + qxβ sin(β ln x), where p and q are some constants. b y(x + βt)f t, y(t) dt = Ax + B, β > 0. a

A solution: y(x) = px + q,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations:

b

p

b

f (t, pt + q) dt – A = 0,

(βpt + q)f (t, pt + q) dt – B = 0.

a

(2)

a

Different solutions of system (2) generate different solutions (1) of the integral equation.

b

13.

y(x + βt)f t, y(t) dt = Ae–λx ,

β > 0.

a

Solutions: y(x) = kn e–λx , where kn are roots of the algebraic (or transcendental) equation kF (k) – A = 0,

F (k) =

b

f t, ke–λt e–βλt dt.

a

b

14.

y(x + βt)f t, y(t) dt = A cos λx,

β > 0.

a

15.

This equation has solutions of the form y(x) = p sin λx + q cos λx, where p and q are some constants. b y(x + βt)f t, y(t) dt = A sin λx, β > 0. a

16.

This equation has solutions of the form y(x) = p sin λx + q cos λx, where p and q are some constants. b y(x + βt)f t, y(t) dt = e–µx (A cos λx + B sin λx), β > 0.

17.

This equation has solutions of the form y(x) = e–µx (p sin λx + q cos λx), where p and q are some constants. b y(x – t)f t, y(t) dt = Ax + B.

a

a

This equation has solutions of the form y(x) = px + q, where p and q are some constants.

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b

18.

y(x – t)f t, y(t) dt = Aeλx .

a

This equation has solutions of the form y(x) = peλx , where p is some constant.

b

19.

y(x – t)f t, y(t) dt = A cos λx.

a

20.

This equation has solutions of the form y(x) = p sin λx + q cos λx, where p and q are some constants. b y(x – t)f t, y(t) dt = e–µx (A cos λx + B sin λx). a

This equation has solutions of the form y(x) = e–µx (p sin λx + q cos λx), where p and q are some constants. 6.8-2. Equations of the Form y(x) + 21.

b

y(x) +

b a

K(x, t)G y(t) dt = F (x)

|x – t|f y(t) dt = Ax2 + Bx + C.

a

This is a special case of equation 6.8.35 with f (t, y) = f (y) and g(x) = Ax2 + Bx + C. The function y = y(x) obeys the second-order autonomous differential equation yxx + 2f (y) = 2A,

whose solution can be represented in an implicit form: y du

= ±(x – a), 2 wa + 4A(u – ya ) – 4F (u, ya ) ya

F (u, v) =

u

f (t) dt,

(1)

v

where ya = y(a) and wa = yx (a) are constants of integration. These constants, as well as the unknowns yb = y(b) and wb = yx (b), are determined by the algebraic (or transcendental) system ya + yb – (a – b)wa = (b2 + 2ab – a2 )A + 2bB + 2C, wa + wb = 2(a + b)A + 2B, wb2 = wa2 + 4A(yb – ya ) – 4F (yb , ya ), yb du

= ±(b – a). 2 wa + 4A(u – ya ) – 4F (u, ya ) ya

(2)

Here the first equation is obtained from the second condition of (5) in 6.8.35, the second equation is obtained from condition (6) in 6.8.35, and the third and fourth equations are consequences of (1). Each solution of system (2) generates a solution of the integral equation. 22.

b

y(x) +

eλ|x–t| f y(t) dt = A + Beλx + Ce–λx .

a

This is a special case of equation 6.8.36 with f (t, y) = f (y) and g(x) = A + Beλx + Ce–λx . The function y = y(x) satisfies the second-order autonomous differential equation yxx + 2λf (y) – λ2 y = –λ2 A,

(1)

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whose solution can be written in an implicit form: y du

= ±(x – a), 2 + λ2 (u2 – y 2 ) – 2Aλ2 (u – y ) – 4λF (u, y ) w ya a a a a

u

f (t) dt, (2)

F (u, v) = v

where ya = y(a) and wa = yx (a) are constants of integration. These constants, as well as the unknowns yb = y(b) and wb = yx (b), are determined by the algebraic (or transcendental) system wa + λya = Aλ + 2Bλeλa , wb – λyb = –Aλ – 2Cλe–λb ,

23.

(3) wb2 = wa2 + λ2 (yb2 – ya2 ) – 2Aλ2 (yb – ya ) – 4λF (yb , ya ), yb du

= ±(b – a). 2 2 2 2 wa + λ (u – ya ) – 2Aλ2 (u – ya ) – 4λF (u, ya ) ya Here the first and second equations are obtained from conditions (5) in 6.8.86, and the third and fourth equations are consequences of (2). Each solution of system (3) generates a solution of the integral equation. b y(x) + eλ|x–t| f y(t) dt = β cosh(λx). a

This is a special case of equation 6.8.22 with A = 0 and B = C = 12 β. 24.

b

y(x) +

eλ|x–t| f y(t) dt = β sinh(λx).

a

This is a special case of equation 6.8.22 with A = 0, B = 12 β, and C = – 12 β. 25.

b

y(x) +

sinh λ|x – t| f y(t) dt = A + B cosh(λx) + C sinh(λx).

a

This is a special case of equation 6.8.37 with f (t, y) = f (y) and g(x) = A + B cosh(λx) + C sinh(λx). The function y = y(x) satisfies the second-order autonomous differential equation yxx + 2λf (y) – λ2 y = –λ2 A,

whose solution can be represented in an implicit form: y du

= ±(x – a), 2 + λ2 (u2 – y 2 ) – 2Aλ2 (u – y ) – 4λF (u, y ) w ya a a a a

26.

u

f (t) dt,

F (u, v) = v

where ya = y(a) and wa = yx (a) are constants of integration, which can be determined from the boundary conditions (5) in 6.8.37. b y(x) + sin λ|x – t| f y(t) dt = A + B cos(λx) + C sin(λx). a

This is a special case of equation 6.8.38 with f (t, y) = f (y) and g(x) = A+B cos(λx)+C sin(λx). The function y = y(x) satisfies the second-order autonomous differential equation yxx + 2λf (y) + λ2 y = λ2 A,

whose solution can be represented in an implicit form: y du

= ±(x – a), 2 2 2 2 wa – λ (u – ya ) + 2Aλ2 (u – ya ) – 4λF (u, ya ) ya

u

f (t) dt,

F (u, v) = v

where ya = y(a) and wa = yx (a) are constants of integration, which can be determined from the boundary conditions (5) in 6.8.38.

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6.8-3. Equations of the Form y(x) + 27.

b

y(x) +

b a

K(x, t)G t, y(t) dt = F (x)

f t, y(t) dt = g(x).

a

A solution: y(x) = g(x) + λ, where λ is determined by the algebraic (or transcendental) equation b λ + F (λ) = 0, F (λ) = f t, g(t) + λ dt. a

28.

b

y(x) +

eλ(x–t) f t, y(t) dt = g(x).

a

A solution: y(x) = βeλx + g(x), where λ is determined by the algebraic (or transcendental) equation b β + F (β) = 0, F (β) = e–λt f t, βeλt + g(t) dt. a

29.

b

y(x) +

g(x)f t, y(t) dt = h(x).

a

A solution: y(x) = λg(x) + h(x), where λ is determined by the algebraic (or transcendental) equation b λ + F (λ) = 0, F (λ) = f t, λg(t) + h(t) dt. a

30.

b

y(x) +

(Ax + Bt)f t, y(t) dt = g(x).

a

A solution: y(x) = g(x) + λx + µ, where the constants λ and µ are determined from the algebraic (or transcendental) system b b λ+A f t, g(t) + λt + µ dt = 0, µ+B tf t, g(t) + λt + µ dt = 0. a

31.

b

y(x) +

a

cosh(λx + µt)f t, y(t) dt = h(x).

a

Using the formula cosh(λx + µt) = cosh(λx) cosh(µt) + sinh(µt) sinh(λx), we arrive at an equation of the form 6.8.39: b y(x) + cosh(λx)f1 t, y(t) + sinh(λx)f2 t, y(t) dt = h(x), a f1 t, y(t) = cosh(µt)f t, y(t) , f2 t, y(t) = sinh(µt)f t, y(t) . 32.

b

y(x) +

sinh(λx + µt)f t, y(t) dt = h(x).

a

Using the formula sinh(λx + µt) = cosh(λx) sinh(µt) + cosh(µt) sinh(λx), we arrive at an equation of the form 6.8.39: b y(x) + cosh(λx)f1 t, y(t) + sinh(λx)f2 t, y(t) dt = h(x), a f1 t, y(t) = sinh(µt)f t, y(t) , f2 t, y(t) = cosh(µt)f t, y(t) .

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33.

b

y(x) +

cos(λx + µt)f t, y(t) dt = h(x).

a

Using the formula cos(λx + µt) = cos(λx) cos(µt) – sin(µt) sin(λx), we arrive at an equation of the form 6.8.39: b y(x) + cos(λx)f1 t, y(t) + sin(λx)f2 t, y(t) dt = h(x), a f1 t, y(t) = cos(µt)f t, y(t) , f2 t, y(t) = – sin(µt)f t, y(t) . 34.

b

y(x) +

sin(λx + µt)f t, y(t) dt = h(x).

a

Using the formula sin(λx + µt) = cos(λx) sin(µt) + cos(µt) sin(λx), we arrive at an equation of the form 6.8.39: b y(x) + cos(λx)f1 t, y(t) + sin(λx)f2 t, y(t) dt = h(x), a f1 t, y(t) = sin(µt)f t, y(t) , f2 t, y(t) = cos(µt)f t, y(t) . 35.

b

y(x) +

|x – t|f t, y(t) dt = g(x),

a ≤ x ≤ b.

a

1◦ . Let us remove the modulus in the integrand: x b y(x) + (x – t)f t, y(t) dt + (t – x)f t, y(t) dt = g(x). a

(1)

x

Differentiating (1) with respect to x yields x yx (x) + f t, y(t) dt – a

b

f t, y(t) dt = gx (x).

(2)

x

Differentiating (2), we arrive at a second-order ordinary differential equation for y = y(x): yxx + 2f (x, y) = gxx (x).

(3)

2◦ . Let us derive the boundary conditions for equation (3). We assume that –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain the relations b y(a) + (t – a)f t, y(t) dt = g(a), a (4) b y(b) + (b – t)f t, y(t) dt = g(b). a

Let us solve equation (3) for f (x, y) and substitute the result into (4). Integrating by parts yields the desired boundary conditions for y(x): y(a) + y(b) + (b – a) gx (b) – yx (b) = g(a) + g(b), (5) y(a) + y(b) + (a – b) gx (a) – yx (a) = g(a) + g(b). Let us point out a useful consequence of (5): yx (a) + yx (b) = gx (a) + gx (b),

(6)

which can be used together with one of conditions (5). Equation (3) under the boundary conditions (5) determines the solution of the original integral equation (there may be several solutions). Conditions (5) make it possible to calculate the constants of integration that occur in solving the differential equation (3).

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36.

b

y(x) +

eλ|x–t| f t, y(t) dt = g(x),

a ≤ x ≤ b.

a

1◦ . Let us remove the modulus in the integrand:

λ(x–t)

y(x) +

x

e

b

f t, y(t) dt +

a

eλ(t–x) f t, y(t) dt = g(x).

(1)

x

Differentiating (1) with respect to x twice yields yxx (x) + 2λf x, y(x) + λ2

x

eλ(x–t) f t, y(t) dt + λ2

a

b

eλ(t–x) f t, y(t) dt = gxx (x). (2)

x

Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary differential equation for y = y(x): yxx + 2λf (x, y) – λ2 y = gxx (x) – λ2 g(x).

(3)

2◦ . Let us derive the boundary conditions for equation (3). We assume that –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain the relations

eλt f t, y(t) dt = g(a),

b

–λa

y(a) + e

a b

λb

–λt

y(b) + e

e

(4)

f t, y(t) dt = g(b).

a

Let us solve equation (3) for f (x, y) and substitute the result into (4). Integrating by parts yields eλb ϕx (b) – eλa ϕx (a) = λeλa ϕ(a) + λeλb ϕ(b), ϕ(x) = y(x) – g(x); e–λb ϕx (b) – e–λa ϕx (a) = λe–λa ϕ(a) + λe–λb ϕ(b). Hence, we obtain the boundary conditions for y(x): ϕx (a) + λϕ(a) = 0,

ϕx (b) – λϕ(b) = 0;

ϕ(x) = y(x) – g(x).

(5)

Equation (3) under the boundary conditions (5) determines the solution of the original integral equation (there may be several solutions). Conditions (5) make it possible to calculate the constants of integration that occur in solving the differential equation (3). 37.

b

y(x) +

sinh λ|x – t| f t, y(t) dt = g(x),

a ≤ x ≤ b.

a

1◦ . Let us remove the modulus in the integrand:

x

y(x) +

sinh[λ(x – t)]f t, y(t) dt +

a

b

sinh[λ(t – x)]f t, y(t) dt = g(x).

(1)

x

Differentiating (1) with respect to x twice yields x yxx (x) + 2λf x, y(x) + λ2 sinh[λ(x – t)]f t, y(t) dt a

b

+ λ2

sinh[λ(t – x)]f t, y(t) dt = gxx (x).

(2)

x

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Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary differential equation for y = y(x): yxx + 2λf (x, y) – λ2 y = gxx (x) – λ2 g(x).

(3)

◦

2 . Let us derive the boundary conditions for equation (3). We assume that –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain the relations b y(a) + sinh[λ(t – a)]f t, y(t) dt = g(a), a (4) b y(b) + sinh[λ(b – t)]f t, y(t) dt = g(b). a

Let us solve equation (3) for f (x, y) and substitute the result into (4). Integrating by parts yields sinh[λ(b – a)]ϕx (b) – λ cosh[λ(b – a)]ϕ(b) = λϕ(a), sinh[λ(b –

38.

a)]ϕx (a)

ϕ(x) = y(x) – g(x);

+ λ cosh[λ(b – a)]ϕ(a) = –λϕ(b).

(5)

Equation (3) under the boundary conditions (5) determines the solution of the original integral equation (there may be several solutions). Conditions (5) make it possible to calculate the constants of integration that occur in solving the differential equation (3). b y(x) + sin(λ|x – t|) f t, y(t) dt = g(x), a ≤ x ≤ b. a

1◦ . Let us remove the modulus in the integrand: x y(x) + sin[λ(x – t)]f t, y(t) dt + a

b

sin[λ(t – x)]f t, y(t) dt = g(x).

(1)

x

Differentiating (1) with respect to x twice yields x yxx (x) + 2λf x, y(x) – λ2 sin[λ(x – t)]f t, y(t) dt a

–λ

b

2

sin[λ(t – x)]f t, y(t) dt = gxx (x).

(2)

x

Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary differential equation for y = y(x): yxx + 2λf (x, y) + λ2 y = gxx (x) + λ2 g(x).

(3)

◦

2 . Let us derive the boundary conditions for equation (3). We assume that –∞ < a < b < ∞. By setting x = a and x = b in (1), we obtain the relations b y(a) + sin[λ(t – a)] f t, y(t) dt = g(a), a (4) b y(b) + sin[λ(b – t)] f t, y(t) dt = g(b). a

Let us solve equation (3) for f (x, y) and substitute the result into (4). Integrating by parts yields sin[λ(b – a)] ϕx (b) – λ cos[λ(b – a)] ϕ(b) = λϕ(a), sin[λ(b –

a)] ϕx (a)

+ λ cos[λ(b – a)] ϕ(a) = –λϕ(b).

ϕ(x) = y(x) – g(x);

(5)

Equation (3) under the boundary conditions (5) determines the solution of the original integral equation (there may be several solutions). Conditions (5) make it possible to calculate the constants of integration that occur in solving the differential equation (3).

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6.8-4. Equations of the Form y(x) + 39.

y(x) +

b a

G x, t, y(t) dt = F (x)

b

g1 (x)f1 t, y(t) + g2 (x)f2 t, y(t) dt = h(x).

a

A solution: y(x) = h(x) + λ1 g1 (x) + λ2 g2 (x), where the constants λ1 and λ2 are determined from the algebraic (or transcendental) system

b

f1 t, h(t) + λ1 g1 (t) + λ2 g2 (t) dt = 0,

b

f2 t, h(t) + λ1 g1 (t) + λ2 g2 (t) dt = 0.

λ1 + a

λ2 +

a

40.

y(x) +

b n a

gk (x)fk t, y(t) dt = h(x).

k=1

A solution: y(x) = h(x) +

n

λk gk (x),

k=1

where the coefficients λk are determined from the algebraic (or transcendental) system λm + a

b

n fm t, h(t) + λk gk (t) dt = 0;

m = 1, . . . , n.

k=1

Different roots of this system generate different solutions of the integral equation. • Reference: A. F. Verlan’ and V. S. Sizikov (1986). b 6.8-5. Equations of the Form F x, y(x) + a G x, t, y(x), y(t) dt = 0 41.

b

y(x) +

y(x)f t, y(t) dt = g(x).

a

A solution: y(x) = λg(x), where λ is determined by the algebraic (or transcendental) equation λ + λF (λ) – 1 = 0,

F (λ) =

b

f t, λg(t) dt.

a

42.

b

y(x) +

g(x)y(x)f t, y(t) dt = h(x).

a

A solution: y(x) = equation

h(x) , where λ is determined from the algebraic (or transcendental) 1 + λg(x) b h(t) λ – F (λ) = 0, F (λ) = f t, dt. 1 + λg(t) a

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43.

b

y(x) +

g x, y(x) f t, y(t) dt = h(x).

a

Solution in an implicit form: y(x) + λg x, y(x) – h(x) = 0,

(1)

where λ is determined from the algebraic (or transcendental) equation λ – F (λ) = 0,

F (λ) =

b

f t, y(t) dt.

(2)

a

Here the function y(x) = y(x, λ) obtained by solving (1) must be substituted into (2). The number of solutions of the integral equation is determined by the number of the solutions obtained from (1) and (2). 44.

f x, y(x) +

b n a

gk x, y(x) hk t, y(t) dt = 0.

k=1

Solution in an implicit form: n f x, y(x) + λk gk x, y(x) = 0,

(1)

k=1

where the λk are determined from the algebraic (or transcendental) system λk – Hk (%λ) = 0, k = 1, . . . , n; b Hk (%λ) = hk t, y(t) dt, %λ = {λ1 , . . . , λn }.

(2)

a

Here the function y(x) = y(x, %λ) obtained by solving (1) must be substituted into (2). The number of solutions of the integral equation is determined by the number of the solutions obtained from (1) and (2). 6.8-6. Other Equations 45.

b

y(x) +

y(xt)f t, y(t) dt = 0.

a

1◦ . A solution: y(x) = kxC ,

(1)

where C is an arbitrary constant and the dependence k = k(C) is determined by the algebraic (or transcendental) equation 1+

b

tC f t, ktC dt = 0.

(2)

a

Each root of equation (2) generates a solution of the integral equation which has the form (1). 2◦ . The integral equation can have some other solutions similar to those indicated in items 1◦ –3◦ of equation 6.2.30.

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46.

b

y(x) +

y(xt)f t, y(t) dt = Ax + B.

a

A solution: y(x) = px + q,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: b p+p tf (t, pt + q) dt – A = 0, a (2) b q+q

f (t, pt + q) dt – B = 0. a

Different solutions of system (2) generate different solutions (1) of the integral equation. 47.

b

y(x) +

y(xt)f t, y(t) dt = Axβ .

a

A solution: y(x) = kxβ ,

(1)

where k is a root of the algebraic (or transcendental) equation b k + kF (k) – A = 0, F (k) = tβ f t, ktβ dt.

(2)

a

Each root of equation (2) generates a solution of the integral equation which has the form (1). 48.

b

y(x) +

y(xt)f t, y(t) dt = A ln x + B.

a

A solution: y(x) = p ln x + q,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: b p+p f (t, p ln t + q) dt – A = 0, a (2) b q+

(p ln t + q)f (t, p ln t + q) dt – B = 0. a

Different solutions of system (2) generate different solutions (1) of the integral equation. 49.

b

y(x) +

y(xt)f t, y(t) dt = Axβ ln x.

a

A solution: y(x) = pxβ ln x + qxβ ,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: b p+p tβ f (t, ptβ ln t + qtβ ) dt = A, a (2) b (ptβ ln t + qtβ )f (t, ptβ ln t + qtβ ) dt = 0.

q+ a

Different solutions of system (2) generate different solutions (1) of the integral equation.

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50.

b

y(x) +

y(xt)f t, y(t) dt = A cos(ln x).

a

A solution: y(x) = p cos(ln x) + q sin(ln x), where p and q are roots of the following system of algebraic (or transcendental) equations: b p+ p cos(ln t) + q sin(ln t) f t, p cos(ln t) + q sin(ln t) dt = A, q+

a b

q cos(ln t) – p sin(ln t) f t, p cos(ln t) + q sin(ln t) dt = 0.

a

51.

b

y(x) +

y(xt)f t, y(t) dt = A sin(ln x).

a

A solution: y(x) = p cos(ln x) + q sin(ln x), where p and q are roots of the following system of algebraic (or transcendental) equations: b p+ p cos(ln t) + q sin(ln t) f t, p cos(ln t) + q sin(ln t) dt = 0, q+

a b

q cos(ln t) – p sin(ln t) f t, p cos(ln t) + q sin(ln t) dt = A.

a

52.

b

y(x) +

y(xt)f t, y(t) dt = Axβ cos(ln x) + Bxβ sin(ln x).

a

A solution: y(x) = pxβ cos(ln x) + qxβ sin(ln x),

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: b p+ tβ p cos(ln t) + q sin(ln t) f t, ptβ cos(ln t) + qtβ sin(ln t) dt = A, a (2) b β β β q+ t q cos(ln t) – p sin(ln t) f t, pt cos(ln t) + qt sin(ln t) dt = B. a

Different solutions of system (2) generate different solutions (1) of the integral equation. 53.

b

y(x) +

y xtβ f t, y(t) dt = g(x),

β > 0.

a

1◦ . For g(x) =

n

Ak xk , the equation has a solution of the form

k=1

y(x) =

n

Bk xk ,

k=1

where Bk are roots of the algebraic (or transcendental) equations

b n kβ m % % dt. Fk (B) = t f t, Bm t Bk + Bk Fk (B) – Ak = 0, a

m=1

Different roots of this system generate different solutions of the integral equation.

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2◦ . For g(x) = ln x

n

Ak xk , the equation has a solution of the form

k=0

y(x) = ln x

n

Bk xk +

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. 3◦ . For g(x) =

n

Ak ln x)k , the equation has a solution of the form

k=0

y(x) =

n

Bk ln x)k ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. 4◦ . For g(x) =

n

Ak cos(λk ln x), the equation has a solution of the form

k=1

y(x) =

n

Bk cos(λk ln x) +

k=1

n

Ck sin(λk ln x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 5◦ . For g(x) =

n

Ak sin(λk ln x), the equation has a solution of the form

k=1

y(x) =

n

Bk cos(λk ln x) +

k=1

n

Ck sin(λk ln x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 54.

b

y(x) +

y(x – t)f t, y(t) dt = 0.

a ◦

1 . A solution: y(x) = keCx ,

(1)

where C is an arbitrary constant and the dependence k = k(C) is determined by the algebraic (or transcendental) equation 1+

b

f t, keCt e–Ct dt = 0.

(2)

a

Each root of equation (2) generates a solution of the integral equation which has the form (1). 2◦ . The equation has solutions of the form y(x) =

n

Em xm , where the constants Em can

m=0

be found by the method of undetermined coefficients.

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55.

b

y(x) +

y(x – t)f t, y(t) dt = Ax + B.

a

A solution: y(x) = px + q,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations:

b

p+p

f (t, pt + q) dt – A = 0,

q+

a b

(2)

(q – pt)f (t, pt + q) dt – B = 0. a

Different solutions of system (2) generate different solutions (1) of the integral equation. 56.

b

y(x) +

y(x – t)f t, y(t) dt = Aeλx .

a

Solutions: y(x) = kn eλx , where kn are roots of the algebraic (or transcendental) equation k + kF (k) – A = 0,

F (k) =

b

f t, keλt e–λt dt.

a

57.

b

y(x) +

y(x – t)f t, y(t) dt = A sinh λx.

a

A solution: y(x) = p sinh λx + q cosh λx,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations:

b

p+ a b

q+

(p cosh λt – q sinh λt)f t, p sinh λt + q cosh λt dt = A,

(2)

(q cosh λt – p sinh λt)f t, p sinh λt + q cosh λt dt = 0. a

Different solutions of system (2) generate different solutions (1) of the integral equation. 58.

b

y(x) +

y(x – t)f t, y(t) dt = A cosh λx.

a

A solution: y(x) = p sinh λx + q cosh λx, where p and q are roots of the following system of algebraic (or transcendental) equations:

b

p+

a b

q+

(p cosh λt – q sinh λt)f t, p sinh λt + q cosh λt dt = 0, (q cosh λt – p sinh λt)f t, p sinh λt + q cosh λt dt = A.

a

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59.

b

y(x) +

y(x – t)f t, y(t) dt = A sin λx.

a

A solution: y(x) = p sin λx + q cos λx,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: b p+ (p cos λt + q sin λt)f t, p sin λt + q cos λt dt = A, a (2) b q+ (q cos λt – p sin λt)f t, p sin λt + q cos λt dt = 0. a

60.

Different solutions of system (2) generate different solutions (1) of the integral equation. b y(x) + y(x – t)f t, y(t) dt = A cos λx. a

A solution: y(x) = p sin λx + q cos λx, where p and q are roots of the following system of algebraic (or transcendental) equations: b p+ (p cos λt + q sin λt)f t, p sin λt + q cos λt dt = 0,

a b

q+

(q cos λt – p sin λt)f t, p sin λt + q cos λt dt = A.

a

61.

b

y(x) +

y(x – t)f t, y(t) dt = eµx (A sin λx + B cos λx).

a

A solution: y(x) = eµx (p sin λx + q cos λx),

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: b p+ (p cos λt + q sin λt)e–µt f t, peµt sin λt + qeµt cos λt dt = A, a (2) b –µt µt µt q+ (q cos λt – p sin λt)e f t, pe sin λt + qe cos λt dt = B. a

62.

Different solutions of system (2) generate different solutions (1) of the integral equation. b y(x) + y(x – t)f t, y(t) dt = g(x). a

1◦ . For g(x) =

n

Ak exp(λk x), the equation has a solution of the form

k=1

y(x) =

n

Bk exp(λk x),

k=1

where the constants Bk are determined from the nonlinear algebraic (or transcendental) system % – Ak = 0, k = 1, . . . , n, Bk + Bk Fk (B) b n % % B = {B1 , . . . , Bn }, Fk (B) = f t, Bm exp(λm t) exp(–λk t) dt. a

m=1

Different solutions of this system generate different solutions of the integral equation.

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2◦ . For a polynomial right-hand side, g(x) =

n

Ak xk , the equation has a solution of the

k=0

form y(x) =

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 3◦ . For g(x) = eλx Ak xk , the equation has a solution of the form k=0

y(x) = eλx

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 4◦ . For g(x) = Ak cos(λk x), the equation has a solution of the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 5◦ . For g(x) = Ak sin(λk x), the equation has a solution of the form k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 6◦ . For g(x) = cos(λx) Ak xk , the equation has a solution of the form k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 7◦ . For g(x) = sin(λx) Ak xk , the equation has a solution of the form k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 8◦ . For g(x) = eµx Ak cos(λk x), the equation has a solution of the form k=1

y(x) = eµx

n k=1

Bk cos(λk x) + eµx

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients.

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9◦ . For g(x) = eµx

n

Ak sin(λk x), the equation has a solution of the form

k=1 n

y(x) = eµx

Bk cos(λk x) + eµx

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 10◦ . For g(x) = cos(λx)

n

Ak exp(µk x), the equation has a solution of the form

k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 11◦ . For g(x) = sin(λx)

n

Ak exp(µk x), the equation has a solution of the form

k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Bk exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. 63.

b

y(x) +

y(x + βt)f t, y(t) dt = Ax + B.

a

A solution: y(x) = px + q,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations:

b

p+p

f (t, pt + q) dt – A = 0,

q+

a b

(2)

(βpt + q)f (t, pt + q) dt – B = 0. a

Different solutions of system (2) generate different solutions (1) of the integral equation. 64.

b

y(x) +

y(x + βt)f t, y(t) dt = Aeλx .

a

Solutions: y(x) = kn eλx , where kn are roots of the algebraic (or transcendental) equation k + kF (k) – A = 0,

F (k) =

b

f t, keλt eβλt dt.

a

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65.

b

y(x) +

y(x + βt)f t, y(t) dt = A sin λx + B cos λx.

a

A solution: y(x) = p sin λx + q cos λx,

(1)

where p and q are roots of the following system of algebraic (or transcendental) equations: b p+ p cos(λβt) – q sin(λβt) f t, p sin λt + q cos λt dt = A, a (2) b q+ q cos(λβt) + p sin(λβt) f t, p sin λt + q cos λt dt = B. a

Different solutions of system (2) generate different solutions (1) of the integral equation. 66.

b

y(x) +

y(x + βt)f t, y(t) dt = g(x).

a

1◦ . For g(x) =

n

Ak exp(λk x), the equation has a solution of the form

k=1

y(x) =

n

Bk exp(λk x),

k=1

where the constants Bk are determined from the nonlinear algebraic (or transcendental) system % – Ak = 0, k = 1, . . . , n, Bk + Bk Fk (B) b n % = {B1 , . . . , Bn }, Fk (B) % = B f t, Bm exp(λm t) exp(λk βt) dt. a

m=1

Different solutions of this system generate different solutions of the integral equation. n 2◦ . For a polynomial right-hand side, g(x) = Ak xk , the equation has a solution of the k=0

form y(x) =

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 3◦ . For g(x) = eλx Ak xk , the equation has a solution of the form k=0

y(x) = eλx

n

Bk xk ,

k=0

where the constants Bk can be found by the method of undetermined coefficients. n 4◦ . For g(x) = Ak cos(λk x), the equation has a solution of the form k=1

y(x) =

n k=1

Bk cos(λk x) +

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients.

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Page 421

5◦ . For g(x) =

n

Ak sin(λk x), the equation has a solution of the form

k=1

y(x) =

n

Bk cos(λk x) +

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 6◦ . For g(x) = cos(λx) Ak xk , the equation has a solution of the form k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 7◦ . For g(x) = sin(λx) Ak xk , the equation has a solution of the form k=0

y(x) = cos(λx)

n

Bk xk + sin(λx)

k=0

n

Ck xk ,

k=0

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 8◦ . For g(x) = eµx Ak cos(λk x), the equation has a solution of the form k=1

y(x) = eµx

n

Bk cos(λk x) + eµx

k=1

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 9◦ . For g(x) = eµx Ak sin(λk x), the equation has a solution of the form k=1

y(x) = eµx

n k=1

Bk cos(λk x) + eµx

n

Ck sin(λk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 10◦ . For g(x) = cos(λx) Ak exp(µk x), the equation has a solution of the form k=1

y(x) = cos(λx)

n

Bk exp(µk x) + sin(λx)

k=1

n

Ck exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients. n 11◦ . For g(x) = sin(λx) Ak exp(µk x), the equation has a solution of the form k=1

y(x) = cos(λx)

n k=1

Bk exp(µk x) + sin(λx)

n

Ck exp(µk x),

k=1

where the constants Bk and Ck can be found by the method of undetermined coefficients.

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67.

b

y(x) +

y(ξ)f t, y(t) dt = 0,

ξ = xϕ(t).

a

1◦ . A solution: y(x) = kxC ,

(1)

where C is an arbitrary constant and the dependence k = k(C) is determined by the algebraic (or transcendental) equation

b

C f t, ktC dt = 0.

ϕ(t)

1+

(2)

a

Each root of equation (2) generates a solution of the integral equation which has the form (1). 2◦ . The equation has solutions of the form y(x) =

n

Em xm , where the constants Em can

m=0

be found by the method of undetermined coefficients. 68.

b

y(x) +

y(ξ)f t, y(t) dt = g(x),

ξ = xϕ(t).

a

1◦ . For g(x) =

n

Ak xk , the equation has a solution of the form

k=1

y(x) =

n

Bk xk ,

k=1

where Bk are roots of the algebraic (or transcendental) equations % – Ak = 0, Bk + Bk Fk (B) k = 1, . . . , n,

b n k m % % B = {B1 , . . . , Bn }, Fk (B) = dt. Bm t ϕ(t) f t, a

m=1

Different roots generate different solutions of the integral equation. 2◦ . For solutions with some other functions g(x), see items 2◦ –5◦ of equation 6.8.53. 69.

b

y(x) +

y(ξ)f t, y(t) dt = 0,

ξ = x + ϕ(t).

a

1◦ . A solution: y(x) = keCx ,

(1)

where C is an arbitrary constant and the dependence k = k(C) is determined by the algebraic (or transcendental) equation 1+

b

eCϕ(t) f t, keCt dt = 0.

(2)

a

Each root of equation (2) generates a solution of the integral equation which has the form (1). 2◦ . The equation has a solution of the form y(x) =

n

Em xm , where the constants Em can

m=0

be found by the method of undetermined coefficients.

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70.

b

y(x) +

y(ξ)f t, y(t) dt = g(x),

ξ = x + ϕ(t).

a

1◦ . For g(x) =

n

Ak exp(λk x) the equation has a solution of the form

k=1

y(x) =

n

Bk exp(λk x),

k=1

where the constants Bk are determined from the nonlinear algebraic (or transcendental) system % – Ak = 0, Bk + Bk Fk (B) k = 1, . . . , n, b n % = {B1 , . . . , Bn }, Fk (B) % = B f t, Bm exp(λm t) exp λk ϕ(t) dt. a

m=1

2◦ . Solutions for some other functions g(x) can be found in items 2◦ –11◦ of equation 6.8.66.

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Page 424

Part II

Methods for Solving Integral Equations

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Page 425

Chapter 7

Main Definitions and Formulas. Integral Transforms 7.1. Some Definitions, Remarks, and Formulas 7.1-1. Some Definitions A function f (x) is said to be square integrable on an interval [a, b] if f 2 (x) is integrable on [a, b]. The set of all square integrable functions is denoted by L2 (a, b) or, briefly, L2 .* Likewise, the set of all integrable functions on [a, b] is denoted by L1 (a, b) or, briefly, L1 . Let us list the main properties of functions from L2 . 1◦ . The sum of two square integrable functions is a square integrable function. 2◦ . The product of a square integrable function by a constant is a square integrable function. 3◦ . The product of two square integrable functions is an integrable function. 4◦ . If f (x) ∈ L2 and g(x) ∈ L2 , then the following Cauchy–Schwarz–Bunyakovsky inequality holds: (f , g)2 ≤ f 2 g2 , b b 2 (f , g) = f (x)g(x) dx, f = (f , f ) = f 2 (x) dx. a

a

The number (f , g) is called the inner product of the functions f (x) and g(x) and the number f is called the L2 -norm of f (x). 5◦ . For f (x) ∈ L2 and g(x) ∈ L2 , the following triangle inequality holds: f + g ≤ f + g. 6◦ . Let functions f (x) and f1 (x), f2 (x), . . . , fn (x), . . . be square integrable on an interval [a, b]. If lim

n→∞

2 fn (x) – f (x) dx = 0,

b

a

then the sequence f1 (x), f2 (x), . . . is said to be mean-square convergent to f (x). Note that if a sequence of functions {fn (x)} from L2 converges uniformly to f (x), then f (x) ∈ L2 and {fn (x)} is mean-square convergent to f (x). * In the most general case the integral is understood as the Lebesgue integral of measurable functions. As usual, two equivalent functions (i.e., equal everywhere, or distinct on a negligible set (of zero measure)) are regarded as one and the same element of L2 .

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Page 427

The notion of an integrable function of several variables is similar. For instance, a function f (x, t) is said to be square integrable in a domain S = {a ≤ x ≤ b, a ≤ t ≤ b} if f (x) is measurable and b b 2 f ≡ f 2 (x, t) dx dt < ∞. a

a

Here f denotes the norm of the function f (x, t), as above. 7.1-2. The Structure of Solutions to Linear Integral Equations A linear integral equation with variable integration limit has the form x βy(x) + K(x, t)y(t) dt = f (x),

(1)

a

where y(x) is the unknown function. A linear integral equation with constant integration limits has the form b βy(x) + K(x, t)y(t) dt = f (x).

(2)

a

For β = 0, Eqs. (1) and (2) are called linear integral equations of the first kind, and for β ≠ 0, linear integral equations of the second kind.* Equations of the form (1) and (2) with specific conditions imposed on the kernels and the right-hand sides form various classes of integral equations (Volterra equations, Fredholm equations, convolution equations, etc.), which are considered in detail in Chapters 8–12. For brevity, we shall sometimes represent the linear equations (1) and (2) in the operator form L [y] = f (x).

(3)

A linear operator L possesses the properties L [y1 + y2 ] = L [y1 ] + L [y2 ], L [σy] = σL [y],

σ = const .

A linear equation is called homogeneous if f (x) ≡ 0 and nonhomogeneous otherwise. An arbitrary homogeneous linear integral equation has the trivial solution y ≡ 0. If y1 = y1 (x) and y2 = y2 (x) are particular solutions of a linear homogeneous integral equation, then the linear combination C1 y1 + C2 y2 with arbitrary constants C1 and C2 is also a solution (in physical problems, this property is called the linear superposition principle). The general solution of a linear nonhomogeneous integral equation (3) is the sum of the general solution Y = Y (x) of the corresponding homogeneous equation L [Y ] = 0 and an arbitrary particular solution y¯ = y(x) ¯ of the nonhomogeneous equation L [y] ¯ = f (x), that is, y = Y + y. ¯

(4)

If the homogeneous integral equation has only the trivial solution Y ≡ 0, then the solution of the corresponding nonhomogeneous equation is unique (if it exists). Let y¯1 and y¯2 be solutions of nonhomogeneous linear integral equations with the same left-hand sides and different right-hand sides, L [y¯1 ] = f1 (x) and L [y¯2 ] = f2 (x). Then the function y¯ = y¯1 + y¯2 is a solution of the equation L [y] ¯ = f1 (x) + f2 (x). The transformation x = g(z),

t = g(τ ),

y(x) = ϕ(z)w(z) + ψ(z),

(5)

(gz

where g(z), ϕ(z), and ψ(z) are arbitrary continuous functions ≠ 0), reduces Eqs. (1) and (2) to linear equations of the same form for the unknown function w = w(z). Such transformations are frequently used for constructing exact solutions of linear integral equations. * In Chapters 1–4, which deal with equations with variable and constant limits of integration, we sometimes consider more general equations in which the integrand contains the unknown function y(z), where z = z(x, t), instead of y(t).

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7.1-3. Integral Transforms Integral transforms have the form

b

f˜(λ) =

ϕ(x, λ)f (x) dx. a

The function f˜(λ) is called the transform of the function f (x) and ϕ(x, λ) is called the kernel of the integral transform. The function f (x) is called the inverse transform of f˜(λ). The limits of integration a and b are real numbers (usually, a = 0, b = ∞ or a = –∞, b = ∞). In Subsections 7.2–7.6, the most popular (Laplace, Mellin, Fourier, etc.) integral transforms, applied in this book to the solution of specific integral equations, are described. These subsections also describe the corresponding inversion formulas, which have the form f (x) = ψ(x, λ)f˜(λ) dλ L

and make it possible to recover f (x) if f˜(λ) is given. The integration path L can lie either on the real axis or in the complex plane. Integral transforms are used in the solution of various differential and integral equations. Figure 1 outlines the overall scheme of solving some special classes of linear integral equations by means of integral transforms (by applying appropriate integral transforms to this sort of integral equations, one obtains first-order linear algebraic equations for f˜(λ)). In many cases, to calculate definite integrals, in particular, to find the inverse Laplace, Mellin, and Fourier transforms, methods of the theory of functions of a complex variable can be applied, including the residue theorem and the Jordan lemma, which are presented below in Subsections 7.1-4 and 7.1-5. 7.1-4. Residues. Calculation Formulas The residue of a function f (z) holomorphic in a deleted neighborhood of a point z = a (thus, a is an isolated singularity of f ) of the complex plane z is the number 1 res f (z) = f (z) dz, i2 = –1, z=a 2πi cε where cε is a circle of sufficiently small radius ε described by the equation |z – a| = ε. If the point z = a is a pole of order n* of the function f (z), then we have res f (z) =

z=a

dn–1 1 (z – a)n f (z) . lim n–1 (n – 1)! z→a dx

For a simple pole, which corresponds to n = 1, this implies res f (z) = lim (z – a)f (z) . z=a

z→a

ϕ(z) , where ϕ(a) ≠ 0 and ψ(z) has a simple zero at the point z = a, i.e., ψ(a) = 0 and ψ(z) ψz (a) ≠ 0, then ϕ(a) res f (z) = . z=a ψz (a) If f (z) =

* In a neighborhood of this point we have f (z) ≈ const (z – a)–n .

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Fig. 1. Principal scheme of applying integral transforms for solving integral equations

7.1-5. The Jordan Lemma If a function f (z) is continuous in the domain |z| ≥ R0 , Im z ≥ α, where α is a chosen real number, and if lim f (z) = 0, then z→∞ lim eiλz f (z) dz = 0 R→∞

CR

for any λ > 0, where CR is the arc of the circle |z| = R that lies in this domain. • References for Section 7.1: A. G. Sveshnikov and A. N. Tikhonov (1970), M. L. Krasnov, A. I. Kiselev, and G. I. Makarenko (1971).

7.2. The Laplace Transform 7.2-1. Definition. The Inversion Formula The Laplace transform of an arbitrary (complex-valued) function f (x) of a real variable x (x ≥ 0) is defined by ∞ f˜(p) = e–px f (x) dx, (1) 0

where p = s + iσ is a complex variable.

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The Laplace transform exists for any continuous or piecewise-continuous function satisfying the condition |f (x)| < M eσ0 x with some M > 0 and σ0 ≥ 0. In the following, σ0 often means the greatest lower bound of the possible values of σ0 in this estimate; this value is called the growth exponent of the function f (x). For any f (x), the transform f˜(p) is defined in the half-plane Re p > σ0 and is analytic there. For brevity, we shall write formula (1) as follows: f˜(p) = L f (x) ,

or

f˜(p) = L f (x), p .

Given the transform f˜(p), the function can be found by means of the inverse Laplace transform f (x) =

1 2πi

c+i∞

f˜(p)epx dp,

i2 = –1,

(2)

c–i∞

where the integration path is parallel to the imaginary axis and lies to the right of all singularities of f˜(p), which corresponds to c > σ0 . The integral in (2) is understood in the sense of the Cauchy principal value:

c+i∞

c+iω

f˜(p)epx dp = lim

ω→∞

c–i∞

f˜(p)epx dp. c–iω

In the domain x < 0, formula (2) gives f (x) ≡ 0. Formula (2) holds for continuous functions. If f (x) has a (finite) jump discontinuity at a point x = x0 > 0, then the left-hand side of (2) is equal to 12 [f (x0 – 0) + f (x0 + 0)] at this point (for x0 = 0, the first term in the square brackets must be omitted). For brevity, we write the Laplace inversion formula (2) as follows: f (x) = L–1 f˜(p) ,

or

f (x) = L–1 f˜(p), x .

7.2-2. The Inverse Transforms of Rational Functions Consider the important case in which the transform is a rational function of the form R(p) f˜(p) = , Q(p)

(3)

where Q(p) and R(p) are polynomials in the variable p and the degree of Q(p) exceeds that of R(p). Assume that the zeros of the denominator are simple, i.e., Q(p) ≡ const (p – λ1 )(p – λ2 ) . . . (p – λn ). Then the inverse transform can be determined by the formula f (x) =

n R(λk ) exp(λk x), Q (λk )

(4)

k=1

where the primes denote the derivatives.

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If Q(p) has multiple zeros, i.e., Q(p) ≡ const (p – λ1 )s1 (p – λ2 )s2 . . . (p – λm )sm , then f (x) =

m k=1

1 dsk –1 (p – λk )sk f˜(p)epx . lim s –1 k p→s (sk – 1)! k dp

7.2-3. The Convolution Theorem for the Laplace Transform x

The convolution of two functions f (x) and g(x) is defined as the integral usually denoted by f (x) ∗ g(x). The convolution theorem states that

0

f (t)g(x – t) dt, and is

L f (x) ∗ g(x) = L f (x) L g(x) , and is frequently applied to solve Volterra equations with kernels depending on the difference of the arguments. 7.2-4. Limit Theorems Let 0 ≤ x < ∞ and f˜(p) = L f (x) be the Laplace transform of f (x). If a limit of f (x) as x → 0 exists, then lim f (x) = lim pf˜(p) . x→0

p→∞

If a limit of f (x) as x → ∞ exists, then lim f (x) = lim pf˜(p) .

x→∞

p→0

7.2-5. Main Properties of the Laplace Transform The main properties of the correspondence between functions and their Laplace transforms are gathered in Table 1. There are tables of direct and inverse Laplace transforms (see Supplements 4 and 5), which are handy in solving linear integral and differential equations. 7.2-6. The Post–Widder Formula In applications, one can find f (x) if the Laplace transform f˜(t) on the real semiaxis is known for t = p ≥ 0. To this end, one uses the Post–Widder formula (–1)n n n+1 ˜(n) n

f (x) = lim . (5) ft n→∞ n! x x Approximate inversion formulas are obtained by taking sufficiently large positive integer n in (5) instead of passing to the limit. • References for Section 7.2: G. Doetsch (1950, 1956, 1958), H. Bateman and A. Erd´elyi (1954), I. I. Hirschman and D. V. Widder (1955), V. A. Ditkin and A. P. Prudnikov (1965), J. W. Miles (1971), B. Davis (1978), Yu. A. Brychkov and A. P. Prudnikov (1989), W. H. Beyer (1991).

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TABLE 1 Main properties of the Laplace transform No

Function

Laplace Transform

Operation

1

af1 (x) + bf2 (x)

Linearity

2

f (x/a), a > 0 f (x – a), f (ξ) ≡ 0 for ξ < 0

af˜1 (p) + bf˜2 (p) af˜(ap) e–ap f˜(p)

Shift of the argument

(–1)n f˜p(n) (p)

Differentiation of the transform Integration of the transform Shift in the complex plane

3

Scaling

4

xn f (x); n = 1, 2, . . .

5

1 f (x) x

6

eax f (x)

f˜(p – a)

7

fx (x)

8

fx(n) (x)

pf˜(p) – f (+0) n pn f˜(p) – pn–k fx(k–1) (+0)

∞ p

9

xm fx(n) (x), m ≥ n

10

dn m x f (x) , m ≥ n n dx

k=1 d m

–

dp

(–1)m pn

x

11 x

12

0

f˜(q) dq

Differentiation Differentiation

pn f˜(p)

Differentiation

dm ˜ f (p) dpm

Differentiation

f (t) dt

f˜(p) p

Integration

f1 (t)f2 (x – t) dt

f˜1 (p)f˜2 (p)

Convolution

0

7.3. The Mellin Transform 7.3-1. Definition. The Inversion Formula Suppose that a function f (x) is defined for positive x and satisfies the conditions 1 ∞ |f (x)|xσ1 –1 dx < ∞, |f (x)|xσ2 –1 dx < ∞ 0

1

for some real numbers σ1 and σ2 , σ1 < σ2 . The Mellin transform of f (x) is defined by ∞ fˆ(s) = f (x)xs–1 dx,

(1)

0

where s = σ + iτ is a complex variable (σ1 < σ < σ2 ). For brevity, we rewrite formula (1) as follows: fˆ(s) = M{f (x)},

or

fˆ(s) = M{f (x), s}.

Given fˆ(s), the function can be found by means of the inverse Mellin transform σ+i∞ 1 f (x) = (σ1 < σ < σ2 ) fˆ(s)x–s ds, 2πi σ–i∞

(2)

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where the integration path is parallel to the imaginary axis of the complex plane s and the integral is understood in the sense of the Cauchy principal value. Formula (2) holds for continuous functions. If f(x) has a (finite) jump discontinuity at a point x = x0 > 0, then the left-hand side of (2) is equal to 12 f (x0 – 0) + f (x0 + 0) at this point (for x0 = 0, the first term in the square brackets must be omitted). For brevity, we rewrite formula (2) in the form f (x) = M–1 {fˆ(s)}, or f (x) = M–1 {fˆ(s), x}. 7.3-2. Main Properties of the Mellin Transform The main properties of the correspondence between the functions and their Mellin transforms are gathered in Table 2. TABLE 2 Main properties of the Mellin transform No

Function

Mellin Transform

Operation

1

af1 (x) + bf2 (x)

Linearity

2

f (ax), a > 0

afˆ1 (s) + bfˆ2 (s) a–s fˆ(s)

3

xa f (x)

4

f (x2 )

fˆ(s + a)

1 ˆ 1 2f 2s

5

f (1/x)

fˆ(–s)

6

xλ f axβ , a > 0, β ≠ 0

7

fx (x)

s +λ

1 – s+λ a β fˆ β β –(s – 1)fˆ(s – 1)

8

xfx (x)

9 10

fx(n) (x) d n x f (x) dx

–s fˆ(s) Γ(s) ˆ (–1)n f (s – n) Γ(s – n) (–1)n s n fˆ(s)

∞

11 12

xα tβ f1 (xt)f2 (t) dt 0 ∞ x

α x tβ f1 f2 (t) dt t 0

Scaling Shift of the argument of the transform Squared argument Inversion of the argument of the transform Power law transform Differentiation Differentiation Multiple differentiation Multiple differentiation

fˆ1 (s + α)fˆ2 (1 – s – α + β)

Complicated integration

fˆ1 (s + α)fˆ2 (s + α + β + 1)

Complicated integration

7.3-3. The Relation Among the Mellin, Laplace, and Fourier Transforms There are tables of direct and inverse Mellin transforms (see Supplements 8 and 9), which are useful in solving specific integral and differential equations. The Mellin transform is related to the Laplace and Fourier transforms as follows: M{f (x), s} = L{f (ex ), –s} + L{f (e–x ), s} = F{f (ex ), is}, which makes it possible to apply much more common tables of direct and inverse Laplace and Fourier transforms. • References for Section 7.3: V. A. Ditkin and A. P. Prudnikov (1965), Yu. A. Brychkov and A. P. Prudnikov (1989).

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7.4. The Fourier Transform 7.4-1. Definition. The Inversion Formula The Fourier transform is defined as follows:

1 f˜(u) = √ 2π

∞

f (x)e–iux dx.

(1)

–∞

For brevity, we rewrite formula (1) as follows: f˜(u) = F{f (x)},

f˜(u) = F{f (x), u}.

or

Given f˜(u), the function f (x) can be found by means of the inverse Fourier transform 1 f (x) = √ 2π

∞

f˜(u)eiux du.

(2)

–∞

Formula (2) holds for continuous functions. If f (x) has a (finite) jump discontinuity at a point x = x0 , then the left-hand side of (2) is equal to 12 f (x0 – 0) + f (x0 + 0) at this point. For brevity, we rewrite formula (2) as follows: f (x) = F–1 {f˜(u)},

f (x) = F–1 {f˜(u), x}.

or

7.4-2. An Asymmetric Form of the Transform Sometimes it is more convenient to define the Fourier transform by fˇ(u) =

∞

f (x)e–iux dx.

(3)

–∞

For brevity, we rewrite formula (3) as follows: fˇ(u) = F{f (x)} or fˇ(u) = F{f (x), u}. In this case, the Fourier inversion formula reads 1 2π

∞

fˇ(u)eiux du,

(4)

and we use the following symbolic notation for relation (4): F–1 {fˇ(u), x}.

f (x) = F–1 {fˇ(u)}, or f (x) =

f (x) =

–∞

7.4-3. The Alternative Fourier Transform Sometimes, for instance, in the theory of boundary value problems, the alternative Fourier transform is used (and called merely the Fourier transform) in the form 1 F(u) = √ 2π

∞

f (x)eiux dx.

(5)

–∞

For brevity, we rewrite formula (5) as follows: F(u) = F{f (x)},

or

F(u) = F{f (x), u}.

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For given F(u), the function f (x) can be found by means of the inverse transform 1 f (x) = √ 2π

∞

F(u)e–iux du.

(6)

–∞

For brevity, we rewrite formula (6) as follows: f (x) = F–1 {F(u)},

f (x) = F–1 {F(u), x}.

or

The function F(u) is also called the Fourier integral of f (x). We can introduce an asymmetric form for the alternative Fourier transform similarly to that of the Fourier transform: ∞ ∞ 1 –iux ˇ ˇ F(u) = f (x)eiux dx, f (x) = du, (7) F(u)e 2π –∞ –∞ ˇ where the direct and the inverse transforms (7) are briefly denoted by F(u) = Fˇ f (x) and f (x) = ˇ ˇ ˇ Fˇ –1 F(u) , or by F(u) = Fˇ f (x), u and f (x) = Fˇ –1 F(u) x .

7.4-4. The Convolution Theorem for the Fourier Transform The convolution of two functions f (x) and g(x) is defined as 1 f (x) ∗ g(x) ≡ √ 2π

∞

f (x – t)g(t) dt. –∞

By performing substitution x – t = u, we see that the convolution is symmetric with respect to the convolved functions: f (x) ∗ g(x) = g(x) ∗ f (x). The convolution theorem states that F f (x) ∗ g(x) = F f (x) F g(x) .

(8)

For the alternative Fourier transform, the convolution theorem reads F f (x) ∗ g(x) = F f (x) F g(x) .

(9)

Formulas (8) and (9) will be used in Chapters 10 and 11 for solving linear integral equations with difference kernel. • References for Section 7.4: V. A. Ditkin and A. P. Prudnikov (1965), J. W. Miles (1971), B. Davis (1978), Yu. A. Brychkov and A. P. Prudnikov (1989), W. H. Beyer (1991).

7.5. The Fourier Sine and Cosine Transforms 7.5-1. The Fourier Cosine Transform Let a function f (x) be integrable on the semiaxis 0 ≤ x < ∞. The Fourier cosine transform is defined by ∞ 2 ˜ fc (u) = f (x) cos(xu) dx, 0 < u < ∞. (1) π 0

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For given f˜c (u), the function can be found by means of the Fourier cosine inversion formula ∞ 2 f (x) = f˜c (u) cos(xu) du, 0 < x < ∞. (2) π 0 The Fourier cosine transform (1) is denoted for brevity by f˜c (u) = Fc f (x) . It follows from formula (2) that the Fourier cosine transform has the property F2c = 1. There are tables of the Fourier cosine transform (see Supplement 7) which prove useful in the solution of specific integral equations. Sometimes the asymmetric form of the Fourier cosine transform is applied, which is given by the pair of formulas ∞ 2 ∞ ˇ ˇ fc (u) = f (x) cos(xu) dx, f (x) = (3) fc (u) cos(xu) du. π 0 0 The direct and inverse Fourier cosine transforms (3) are denoted by fˇc (u) = Fc f (x) and f (x) = ˇ F–1 c fc (u) , respectively. 7.5-2. The Fourier Sine Transform Let a function f (x) be integrable on the semiaxis 0 ≤ x < ∞. The Fourier sine transform is defined by ∞ 2 f˜s (u) = f (x) sin(xu) dx, 0 < u < ∞. (4) π 0 For given f˜s (u), the function f (x) can be found by means of the inverse Fourier sine transform ∞ 2 f (x) = 0 < x < ∞. (5) f˜s (u) sin(xu) du, π 0 The Fourier sine transform (4) is briefly denoted by f˜s (u) = Fs f (x) . It follows from formula (5) that the Fourier sine transform has the property F2s = 1. There are tables of the Fourier sine transform (see Supplement 6), which are useful in solving specific integral equations. Sometimes it is more convenient to apply the asymmetric form of the Fourier sine transform defined by the following two formulas: ∞ 2 ∞ ˇ ˇ fs (u) = fs (u) sin(xu) du. f (x) sin(xu) dx, f (x) = (6) π 0 0 The direct and inverse Fourier sine transforms (6) are denoted by fˇs (u) = Fs f (x) and f (x) = F–1 fˇs (u) , respectively. s • References for Section 7.5: V. A. Ditkin and A. P. Prudnikov (1965), J. W. Miles (1971), Yu. A. Brychkov and A. P. Prudnikov (1989), W. H. Beyer (1991).

7.6. Other Integral Transforms 7.6-1. The Hankel Transform The Hankel transform is defined as follows: ∞ f˜ν (u) = xJν (ux)f (x) dx,

0 < u < ∞,

(1)

0

where ν > – 12 and Jν (x) is the Bessel function of the first kind of order ν (see Supplement 10).

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For given f˜ν (u), the function f (x) can be found by means of the Hankel inversion formula

∞

f (x) =

0 < x < ∞.

uJν (ux)f˜ν (u) du,

(2)

0

Note that if f (x) = O(xα ) as x → 0, where α + ν + 2 > 0, and f (x) = O(xβ ) as x → ∞, where β + 32 < 0, then the integral (1) is convergent. The inversion formula (2) holds for continuous functions. If f (x) has a (finite) jump discontinuity at a point x = x0 , then the left-hand side of (2) is equal to 12 [f (x0 – 0) + f (x0 + 0)] at this point. For brevity, we denote the Hankel transform (1) by f˜ν (u) = Hν f (x) . It follows from formula (2) that the Hankel transform has the property H2ν = 1.

7.6-2. The Meijer Transform The Meijer transform is defined as follows: fˆµ (s) =

2 π

∞

√

sx Kµ (sx)f (x) dx,

0 < s < ∞,

(3)

0

where Kµ (x) is the modified Bessel function of the second kind (the Macdonald function) of order µ (see Supplement 10). For given f˜µ (s), the function f (x) can be found by means of the Meijer inversion formula 1 f (x) = √ i 2π

c+i∞

√

sx Iµ (sx)fˆµ (s) ds,

0 < x < ∞,

(4)

c–i∞

where Iµ (x) is the modified Bessel function of the first kind of order µ (see Supplement 10). For the Meijer transform, a convolution is defined and an operational calculus is developed.

7.6-3. The Kontorovich–Lebedev Transform and Other Transforms The Kontorovich–Lebedev transform is introduced as follows: F (τ ) =

∞

Kiτ (x)f (x) dx,

0 < τ < ∞,

(5)

0

where Kµ (x) is the modified Bessel function of the second kind (the Macdonald function) of order µ √ (see Supplement 10) and i = –1. For given F (τ ), the function can be found by means of the Kontorovich–Lebedev inversion formula ∞ 2 f (x) = 2 τ sinh(πτ )Kiτ (x)F (τ ) dτ , 0 < x < ∞. (6) π x 0 There are also other integral transforms, of which the most important are listed in Table 3 (for the constraints imposed on the functions and parameters occurring in the integrand, see the references given at the end of this section).

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TABLE 3 Main integral transforms Integral Transform

Definition

Laplace transform

f(p) =

Two-sided Laplace transform

f∗(p) =

∞

e–pxf (x) dx

Inversion Formula

f (x) =

0

∞

e–pxf (x) dx

f (x) =

–∞

∞ 1 f(u) = √ e–iuxf (x) dx 2π –∞ ∞ 2 fs(u) = sin(xu)f (x) dx π 0 ∞ 2 fc(u) = cos(xu)f (x) dx π 0 ∞ 1 fh(u) = √ (cos xu + sin xu)f (x) dx 2π –∞ ∞ f(s) = xs–1f (x) dx

Fourier transform Fourier sine transform Fourier cosine transform Hartley transform Mellin transform

0

Hankel transform

fν (w) =

∞

xJν (xw)f (x) dx

Fν (u) =

Y -transform

f(s) =

f(r) =

Bochner transform

2 π

∞√

sx Kν (sx)f (x) dx

0

∞

Jn/2–1(2πxr)G(x, r)f (x) dx,

G(x, r) = 2πr(x/r)n/2, n = 1, 2, . . . ∞ Fa(u) = Wν (xu, au)xf (x) dx, a

F (τ ) =

Meler–Fock transform

(τ ) = F

Kiτ (x)f (x) dx

∞

ux Hν (ux)Fν (u) du

∞ –∞

f (x) dx x–s

c+i∞√

sx Iν (sx)f(s) ds

c–i∞

Wν (xu, au) uFa(u) du Jν2 (au) + Yν2(au)

0

f (x) =

2 π 2x ∞

f (x) =

2

∞

P– 1 +iτ (x)f (x) dx

Jn/2–1(2πrx)G(r, x)f(r) dr

f (x) =

∞

(s) = 1 F π

Hilbert transform*

epxf∗(p) dp

0

0

1

∞√

f (x) =

∞

c–i∞

1 f (x) = √ i 2π

Wν (β, µ) ≡ Jν (β)Yν (µ) – Jν (µ)Yν (β)

Kontorovich– Lebedev transform

c+i∞

0

0

Weber transform

0

f (x) =

0

Meijer transform (K-transform)

epxf(p) dp

c–i∞

1 2πi

ux Yν (ux)f (x) dx

c+i∞

∞ 1 f (x) = √ eiuxf(u) du 2π –∞ ∞ 2 f (x) = sin(xu)fs(u) du π 0 ∞ 2 f (x) = cos(xu)fc(u) du π 0 ∞ 1 f (x) = √ (cos xu + sin xu)fh(u) du 2π –∞ c+i∞ 1 f (x) = x–s f(s) ds 2πi c–i∞ ∞ f (x) = wJν (xw)fν (w) dw

0 ∞√

1 2πi

0

f (x) = –

1 π

∞

τ sinh(πτ )Kiτ (x)F (τ ) dτ 0

(τ ) dτ τ tanh(πτ )P– 1 +iτ (x)F 2

∞ –∞

(s) F ds s–x

√

Notation: i = –1, Jµ(x) and Yµ(x) are the Bessel functions of the first and the second kind, respectively, Iµ(x) and Kµ(x) are the modified Bessel functions of the first and the second kind, respectively, Pµ(x) is the Leg∞ (–1)j (x/2)µ+2j+1

. endre spherical function of the second kind, and Hµ(x) is the Struve function, Hµ(x) = 3 3 j=0 Γ j + 2 Γ µ + j + 2 * REMARK. In the direct and inverse Hilbert transforms, the integrals are understood in the sense of the Cauchy principal value.

• References for Section 7.6: H. Bateman and A. Erd´elyi (1954), V. A. Ditkin and A. P. Prudnikov (1965), J. W. Miles (1971), B. Davis (1978), D. Zwillinger (1989), Yu. A. Brychkov and A. P. Prudnikov (1989), W. H. Beyer (1991).

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Chapter 8

Methods for Solving Linear Equations x of the Form a K(x, t)y(t) dt = f (x) 8.1. Volterra Equations of the First Kind 8.1-1. Equations of the First Kind. Function and Kernel Classes In this chapter we present methods for solving Volterra linear equations of the first kind. These equations have the form x K(x, t)y(t) dt = f (x), (1) a

where y(x) is the unknown function (a ≤ x ≤ b), K(x, t) is the kernel of the integral equation, and f (x) is a given function, the right-hand side of Eq. (1). The functions y(x) and f (x) are usually assumed to be continuous or square integrable on [a, b]. The kernel K(x, t) is usually assumed either to be continuous on the square S = {a ≤ x ≤ b, a ≤ t ≤ b} or to satisfy the condition b b K 2 (x, t) dx dt = B 2 < ∞, (2) a

a

where B is a constant, that is, to be square integrable on this square. It is assumed in (2) that K(x, t) ≡ 0 for t > x. The kernel K(x, t) is said to be degenerate if it can be represented in the form K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t). The kernel K(x, t) of an integral equation is called difference kernel if it depends only on the difference of the arguments, K(x, t) = K(x – t). Polar kernels L(x, t) K(x, t) = + M (x, t), 0 < β < 1, (3) (x – t)β and logarithmic kernels (kernels with logarithmic singularity) K(x, t) = L(x, t) ln(x – t) + M (x, t),

(4)

where L(x, t) and M (x, t) are continuous on S and L(x, x) ≡/ 0, are often considered as well. Polar and logarithmic kernels form a class of kernels with weak singularity. Equations containing such kernels are called equations with weak singularity. The following generalized Abel equation is a special case of Eq. (1) with the kernel of the form (3): x y(t) dt = f (x), 0 < β < 1. (x – t)β a In case the functions K(x, t) and f (x) are continuous, the right-hand side of Eq. (1) must satisfy the following conditions:

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1◦ . If K(a, a) ≠ 0, then f (x) must be constrained by f (a) = 0. 2◦ . If K(a, a) = Kx (a, a) = · · · = Kx(n–1) (a, a) = 0, 0 < Kx(n) (a, a) < ∞, then the right-hand side of the equation must satisfy the conditions f (a) = fx (a) = · · · = fx(n) (a) = 0. 3◦ . If K(a, a) = Kx (a, a) = · · · = Kx(n–1) (a, a) = 0, Kx(n) (a, a) = ∞, then the right-hand side of the equation must satisfy the conditions f (a) = fx (a) = · · · = fx(n–1) (a) = 0. For polar kernels of the form (4) and continuous f (x), no additional conditions are imposed on the right-hand side of the integral equation. Remark 1. Generally, the case in which the integration limit a is infinite is not excluded.

8.1-2. Existence and Uniqueness of a Solution Assume that in Eq. (1) the functions f (x) and K(x, t) are continuous together with their first derivatives on [a, b] and on S, respectively. If K(x, x) ≠ 0 (x ∈ [a, b]) and f (a) = 0, then there exists a unique continuous solution y(x) of Eq. (1). Remark 2. The problem of existence and uniqueness of a solution to a Volterra equation of the first kind is closely related to conditions under which this equation can be reduced to Volterra equations of the second kind (see Section 8.3). Remark 3. A Volterra equation of the first kind can be treated as a Fredholm equation of the first kind whose kernel K(x, t) vanishes for t > x (see Chapter 10).

• References for Section 8.1: E. Goursat (1923), H. M. M¨untz (1934), F. G. Tricomi (1957), V. Volterra (1959), S. G. Mikhlin (1960), M. L. Krasnov, A. I. Kiselev, and G. I. Makarenko (1971), J. A. Cochran (1972), C. Corduneanu (1973), V. I. Smirnov (1974), P. P. Zabreyko, A. I. Koshelev, et al. (1975), A. J. Jerry (1985), A. F. Verlan’ and V. S. Sizikov (1986).

8.2. Equations With Degenerate Kernel: K(x, t) = g1 (x)h1 (t) + · · · + gn (x)hn (t) 8.2-1. Equations With Kernel of the Form K(x, t) = g1 (x)h1 (t) + g2 (x)h2 (t) Any equation of this type can be rewritten in the form x g1 (x) h1 (t)y(t) dt + g2 (x) a

x

h2 (t)y(t) dt = f (x).

(1)

a

It is assumed that g1 (x) ≠ const g2 (x), h1 (t) ≠ const h2 (t), 0 < g12 (a) + g22 (a) < ∞, and f (a) = 0. The change of variables x

u(x) =

h1 (t)y(t) dt

(2)

a

followed by the integration by parts in the second integral in (1) with regard to the relation u(a) = 0 yields the following Volterra equation of the second kind: x [g1 (x)h1 (x) + g2 (x)h2 (x)]u(x) – g2 (x)h1 (x) a

h2 (t) h1 (t)

u(t) dt = h1 (x)f (x).

(3)

t

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The substitution

x w(x) = a

h2 (t) h1 (t)

u(t) dt

(4)

t

reduces Eq. (3) to the first-order linear ordinary differential equation h2 (x) h2 (x) [g1 (x)h1 (x) + g2 (x)h2 (x)]wx – g2 (x)h1 (x) w = f (x)h1 (x) . h1 (x) x h1 (x) x

(5)

1◦ . In the case g1 (x)h1 (x) + g2 (x)h2 (x) ≡/ 0, the solution of equation (5) satisfying the condition w(a) = 0 (this condition is a consequence of the substitution (4)) has the form x h2 (t) f (t)h1 (t) dt w(x) = Φ(x) , (6) h (t) Φ(t)[g (t)h 1 1 1 (t) + g2 (t)h2 (t)] a t x h2 (t) g2 (t)h1 (t) dt Φ(x) = exp . (7) h1 (t) t g1 (t)h1 (t) + g2 (t)h2 (t) a Let us differentiate relation (4) and substitute the function (6) into the resulting expression. After integrating by parts with regard to the relations f (a) = 0 and w(a) = 0, for f ≡/ const g2 we obtain x f (t) g2 (x)h1 (x)Φ(x) dt u(x) = . g1 (x)h1 (x) + g2 (x)h2 (x) a g2 (t) t Φ(t) Using formula (2), we find a solution of the original equation in the form x 1 d f (t) dt g2 (x)h1 (x)Φ(x) y(x) = , h1 (x) dx g1 (x)h1 (x) + g2 (x)h2 (x) a g2 (t) t Φ(t)

(8)

where the function Φ(x) is given by (7). If f (x) ≡ const g2 (x), the solution is given by formulas (8) and (7) in which the subscript 1 must be changed by 2 and vice versa. 2◦ . In the case g1 (x)h1 (x) + g2 (x)h2 (x) ≡ 0, the solution has the form 1 d (f /g2 )x 1 d (f /g2 )x y(x) = = – . h1 dx (g1 /g2 )x h1 dx (h2 /h1 )x 8.2-2. Equations With General Degenerate Kernel A Volterra equation of the first kind with general degenerate kernel has the form n

gm (x)

x

hm (t)y(t) dt = f (x).

(9)

a

m=1

Using the notation wm (x) =

x

hm (t)y(t) dt,

m = 1, . . . , n,

(10)

a

we can rewrite Eq. (9) as follows: n

gm (x)wm (x) = f (x).

(11)

m=1

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On differentiating formulas (10) and eliminating y(x) from the resulting equations, we arrive at the following linear differential equations for the functions wm = wm (x): h1 (x)wm = hm (x)w1 ,

m = 2, . . . , n,

(12)

(the prime stands for the derivative with respect to x) with the initial conditions wm (a) = 0,

m = 1, . . . , n.

Any solution of system (11), (12) determines a solution of the original integral equation (9) by each of the expressions w (x) y(x) = m , m = 1, . . . , n, hm (x) which can be obtained by differentiating formula (10). System (11), (12) can be reduced to a linear differential equation of order n – 1 for any function wm (x) (m = 1, . . . , n) by multiple differentiation of Eq. (11) with regard to (12). • References for Section 8.2: E. Goursat (1923), A. F. Verlan’ and V. S. Sizikov (1986).

8.3. Reduction of Volterra Equations of the First Kind to Volterra Equations of the Second Kind 8.3-1. The First Method Suppose that the kernel and the right-hand side of the equation

x

K(x, t)y(t) dt = f (x),

(1)

a

have continuous derivatives with respect to x and that the condition K(x, x) ≡/ 0 holds. In this case, after differentiating relation (1) and dividing the resulting expression by K(x, x) we arrive at the following Volterra equation of the second kind: y(x) + a

x

Kx (x, t) f (x) y(t) dt = x . K(x, x) K(x, x)

(2)

Equations of this type are considered in Chapter 9. If K(x, x) ≡ 0, then, on differentiating Eq. (1) with respect to x twice and assuming that Kx (x, t)|t=x ≡/ 0, we obtain the Volterra equation of the second kind x Kxx (x, t) f (x) y(x) + y(t) dt = xx . Kx (x, t)|t=x a Kx (x, t)|t=x If Kx (x, x) ≡ 0, we can again apply differentiation, and so on. If the first m – 2 partial derivatives of the kernel with respect to x are identically zero and the (m – 1)st derivative is nonzero, then the m-fold differentiation of the original equation gives the following Volterra equation of the second kind: x Kx(m) (x, t) fx(m) (x) y(x) + y(t) dt = (m–1) . (m–1) (x, t)|t=x Kx (x, t)|t=x a Kx

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8.3-2. The Second Method

Let us introduce the new variable Y (x) =

x

y(t) dt a

and integrate the right-hand side of Eq. (1) by parts taking into account the relation f (a) = 0. After dividing the resulting expression by K(x, x), we arrive at the Volterra equation of the second kind x Kt (x, t) f (x) Y (x) – Y (t) dt = , K(x, x) K(x, x) a for which the condition K(x, x) ≡/ 0 must hold. • References for Section 8.3: E. Goursat (1923), V. Volterra (1959).

8.4. Equations With Difference Kernel: K(x, t) = K(x – t) 8.4-1. A Solution Method Based on the Laplace Transform Volterra equations of the first kind with kernel depending on the difference of the arguments have the form x K(x – t)y(t) dt = f (x).

(1)

0

To solve these equations, the Laplace transform can be used (see Section 7.2). In what follows we need the tran