Introduction to Inverse Problems for Differential Equations [2nd ed. 2021] 3030794261, 9783030794262

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Alemdar Hasanov Hasanoğlu Vladimir G. Romanov

Introduction to Inverse Problems for Differential Equations Second Edition

Introduction to Inverse Problems for Differential Equations

Alemdar Hasanov Hasanoˇglu • Vladimir G. Romanov

Introduction to Inverse Problems for Differential Equations

Second Edition

Alemdar Hasanov Hasanoˇglu Department of Mathematics University of Kocaeli Izmit, Kocaeli, Turkey

Vladimir G. Romanov Sobolev Institute of Mathematics Novosibirsk, Russia

ISBN 978-3-030-79426-2 ISBN 978-3-030-79427-9 (eBook) https://doi.org/10.1007/978-3-030-79427-9 © Springer Nature Switzerland AG 2017, 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

This book is dedicated to our wives, S¸ afak Hasanoˇglu and Nina D. Romanova, who have been patient during all our lives together.

Preface to the Second Edition

In the past 5 years since the publication of the first edition of this book, the field of inverse problems has continued to develop, penetrating at a rapid pace in almost all areas of science and technology. In terms of inverse problems for differential equations, two important classes of problems should be distinguished here: inverse problems in wave phenomena and in vibration. The newly added Part III of the book, which consists of Chaps. 10 and 11, deals precisely with these problems. Chapter 10 includes basic inverse coefficient problems with boundary measured output, for one-dimensional damped wave equation in bounded domain. It should be emphasized that in contrast to the inverse problems related to undamped wave equation, the presence of damping factor changes the entire structure of the problem as well as its solution. Namely, the damping factor leads to a number of characteristic features in the propagation of waves through the conductor and in reflection from conducting surfaces. As a consequence, subdomains, called subdomains defined by characteristics, are formed by the wave reflected from the boundary. The analysis carried out in this chapter clearly shows that all properties of inverse coefficient problems for damped wave equation, as well as properties of Neumann-to-Dirichlet and Dirichlet-to-Neumann operators associated with them, are completely determined by properties of the direct problem solution in these subdomains and along the characteristics, which, by the way, have not been properly studied so far. Chapter 11 is devoted to the inverse problems of vibration, since in addition to classical applications in recent decades, beams and plates have also become integral elements in nanotechnology, especially in medical diagnostics and nanoscale measurement. All inverse problems included in this chapter are formulated for a general form of Euler-Bernoulli beam and Kirchhoff plate equations given in a finite domain and correspond to real physical models. Furthermore, in all these inverse problems, the observation, that is, the measured output, is given at the boundary of a finite domain, and the final time is assumed to finite and may be small enough. Along with some revisions and updates to existing sections, a supporting material related to Chap. 11 has been added to the second edition as Appendix C.

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Preface to the Second Edition

The contents of the new chapters are also self-contained and accessible to beginning graduate students. Over the past 5 years, we have received much feedback and notes from readers, and also from our colleagues. We take this opportunity to thank them for carefully reading the book and a number of suggestions for improving the second edition. We thank Onur Baysal for helping us prepare the new Sect. C.1 of Appendix C. Our special thanks go to Rahman Hasanoˇglu, who spent a lot of time to assisting us in the preparation of the revised LATEX version of the text. We hope that this book will continue to engage young researchers in inverse problems and their applications. Izmit, Turkey Novosibirsk, Russia April 2021

Alemdar Hasanov Hasanoˇglu Vladimir Romanov

Preface to the First Edition

Mathematical models of the most physical phenomena are governed by initial and boundary value problems for partial differential equations (PDEs). Inverse problems governed by these equations arise naturally in almost all branches of science and engineering. The main objective of this textbook is to introduce students and researchers to inverse problems arising in PDEs. This book presents a systematic exposition of the main ideas and methods in treating inverse problems for PDEs arising in basic mathematical models, though we make no claim to cover all of the topics. More detailed additional information related to each chapter can be found in the following books/lecture notes/monographs of Aster, Borchers and Thurber [4], Bal [7], Baunmeister [13], Beilina and Klibanov [15], Belishev and Blagovestchenskii [19], Chadan and Sabatier [28], Colton and Kress [33], Engl, Hanke and Neubauer [37], Groetsch [48], Háo [54], Hofmann [67], Isakov [70, 72], Itou and Jin [73], Kabanikhin [75], Kaipio and E. Somersalo [76], Kirsch [81], Lavrentiev [88], Lavrentiev, Romanov and Shishatski [90], Lavrentiev, Romanov and Vasiliev [89], Lebedev, Vorovich and Gladwell [91], Louis [93], Morozov [101], Mueller and Siltanen [102] Nakamura and Potthast [106], Natterer [108], Ramm [126], Romanov [132, 133], Romanov and Kabanikhin [138], Schuster, Kaltenbacher, Hofmann and Kazimierski [141], Tarantola [143], Tikhonov and Arsenin [149], Tikhonov, Concharsky, Stepanov and Yagola [151], Vogel [154]. In Introduction we discuss the nature of ill-posedness in differential and integral equations based on well-known mathematical models. Further, we pursue an indepth analysis of a reason of ill-posedness of an inverse problem governed by integral operator. We tried to answer the question “why this problem is ill-posed?”, by arriving to the physical meaning of the mathematical model, on one hand, and then explaining this in terms of compact operators, on the other hand. Main notions and tools, including best approximation, Moore-Penrose (generalized) inverse, Singular Value Decomposition, regularization strategy, Tikhonov Regularization for linear inverse problems and Morozov’s Discrepancy Principle, are given in Chap. 2. In Chap. 3, we tried to illustrate an implementation of all these notions and tools to inverse source problems with final overdetermination for evolution equations and to the backward problem for the parabolic equation, including some numerical ix

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reconstructions. The reason for the choice of this problem stems from the fact that historically, one of the first successful applications of inverse problems was Tikhonov’s work [145] on inverse problem with final overdetermination for heat equation. The second part of the book consists of almost independent six chapters. The choice of these chapters is motivated by the fact that the inverse coefficient and source problems considered here are based on the basic and commonly mathematical models governed by PDEs. These chapters describe not only these inverse problems, but also main inversion methods and techniques. Since the most distinctive features of any inverse problem related to PDEs are hidden in the properties of corresponding direct problems solutions, special attention is paid to the investigation of these properties. Chapter 4 deals with some inverse problems related to the second order hyperbolic equations. Starting with the simplest inverse source problems for the wave equation with the separated right-hand side containing spatial or time dependent unknown source functions, we use the reflection method to demonstrate a method for finding the unknown functions based on integral equations. The next and more complex problem here is the problem of recovering the potential in the string equation. The direct problem is stated for semi-infinite string with homogeneous initial data and nonhomogeneous Dirichlet data at the both ends. The Neumann output is used as an additional data for recovering an unknown potential. Using the method the successive approximations for obtained system of integral equations, we prove a local solvability for the considered inverse problem. Note that the typical situation for nonlinear inverse problems is that the only local solvability can be proved (see [134]). Nevertheless, for the considered inverse problem the uniqueness and global stability estimates of solutions are derived. As an application, inverse coefficient problems for layered media is studied in the final part of Chap. 4. In Chap. 5 the inverse problems for the electrodynamic equations are studied. These problems are motivated by geophysical applications. In the considered physical model we assume that the space R3 is divided in the two half-spaces R3− =: {x ∈ R3 | x3 < 0} and R3+ =: {x ∈ R3 | x3 > 0}. The domain R3− is filled by homogeneous nonconductive medium (air) while the domain R3+ contains a ground which is a non-homogeneous medium with variable permittivity, permeability and conductivity depending on the variable x3 only. The tangential components of the electromagnetic field are assumed to be continuous across the interface x3 = 0. The direct problem for electrodynamic equations with zero initial data and a dipole current applied at the interface is stated. The output data here is a tangential components of the electrical field given on the interface as a function of time t > 0. A method for reconstruction of one of the unknown coefficients (permittivity, permeability or conductivity) is proposed when two others are given. This method is based on the Riemannian invariants and integral equations which lead to a well convergent successive approximations. For the solutions of these inverse problems stability estimates are stated.

Preface to the First Edition

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Coefficient inverse problems for parabolic equations are studied in Chap. 6. First of all, the relationship between the solutions of direct problems corresponding to the parabolic and hyperbolic equations is derived. This relationship allows to show the similarity between the outputs corresponding inverse problems for parabolic and hyperbolic equations. Since this relationship is a special Laplace transform, it is invertible. Then it is shown that inverse problems for parabolic equations can be reduced to corresponding problems for hyperbolic equations, studied in Chap. 4. This, in particular, allows to use uniqueness theorems obtained for hyperbolic inverse problem. Further it is shown that the inverse problem of recovering the potential is closely related to the well-known inverse spectral problem for the SturmLiouville operator. Chapter 7 deals with inverse problems for the elliptic equations. Here the inverse scattering problem for stationary Schrödinger equation is considered. For the sake of simplicity, we study this problem in Born approximation using the scattering amplitude for recovering a potential, following the approach proposed by R.G. Novikov [65]. Moreover, we study also the inverse problem which output is a measured value on a closed surface S of a trace of the solution of the problem for the Schrödinger equation with point sources located at the same surface. The later problem is reduced to the X-ray tomography problem. In the final part of this chapter we define the Dirichlet-to-Neumann operator which originally has been introduced in [84]. Using this operator, we study the inverse problem of recovering the potential for an elliptic equation in the Born approximation and reduce it to the moment problem which has a unique solution. Inverse problems related to the transport equation without scattering are studied in Chap. 8. We derive here the stability estimate for the solution of X-ray tomography problem and then inversion formula. Chapter 8 is devoted to the inverse kinematic problem. On one hand, this is a problem of reconstructing the wave speed inside a domain from given travel times between arbitrary boundary points. On the other hand, this is also the problem of recovering conformal Riemannian metric via Riemannian distances between boundary points, which plays very important role in geophysics. It suffices to recall that much of our knowledge about the internal structure of the Earth are based on solutions of this problem. We derive first the equations for finding rays and fronts going from a point source. Then we consider one-dimensional and two-dimensional inverse problems and derive a stability estimates for the solutions of these problems. For the convenience of the reader a short review related to invertibility of linear, in particular compact, operators is given in Appendix A. Some necessary energy estimates for a weak and regular weak solutions of a one-dimensional parabolic equation are given in Appendix B. The presentation of the material, especially Introduction and Part I, is selfcontained and is intended to be accessible also to undergraduate and beginning graduate students, whose mathematical background include only basic courses in advanced calculus, PDEs and functional analysis. The book can be used as a backbone for a lecture on inverse and ill-posed problems for partial differential equations.

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This book project took us about two and a half years to complete, starting from May, 2014. We would not be able to finish this book without support of our family members, students and colleagues. We are deeply indebted to a number of colleagues and students who have read this material and given us many valuable suggestions for improving the presentation. We would like to thank our colleagues Onur Baysal, Andreas Neubauer, Roman Novikov, Burhan Pekta¸s, Cristiana Sebu and Marian Slodišcka who have read various parts of the manuscript and made numerous suggestions for improvement. Our special thanks anonymous reviewers who read the book proposal and the first version of the manuscript, as well as to the staff members of Springer, Dr. Martin Peters and Mrs. Ruth Allewelt, for their support and assistance with this project. We also are grateful to Mehriban Hasanoˇglu for number of corrections. We would be most grateful if you could send your suggestions and list of mistakes to the email addresses: [email protected]; [email protected]. Izmit, Turkey Novosibirsk, Russia November 2016

Alemdar Hasanov Hasanoˇglu Vladimir Romanov

Contents

1

Introduction Ill-Posedness of Inverse Problems .. . . .. . . . . . . . . . . . . . . . . . . . 1.1 Some Basic Definitions and Examples .. . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 Continuity with Respect to Coefficients and Source: Sturm-Liouville Equation .. . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3 Why a Fredholm Integral Equation of the First Kind Is an Ill-Posed Problem? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

Part I

1 1 9 14

Introduction to Inverse Problems

2

Functional Analysis Background of Ill-Posed Problems . . . . . . . . . . . . . . . 2.1 Best Approximation and Orthogonal Projection .. . . . . . . . . . . . . . . . . . . 2.2 Range and Null-Space of Adjoint Operators .. . .. . . . . . . . . . . . . . . . . . . . 2.3 Moore-Penrose Generalized Inverse.. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.4 Singular Value Decomposition . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.5 Regularization Strategy. Tikhonov Regularization . . . . . . . . . . . . . . . . . 2.6 Morozov’s Discrepancy Principle . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

23 24 31 34 39 46 59

3

Inverse Source Problems with Final Overdetermination . . . . . . . . . . . . . . 3.1 Inverse Source Problem for Heat Equation . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.1 Compactness of Input-Output Operator and Fréchet Gradient . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.2 Singular Value Decomposition of Input-Output Operator .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.3 Picard Criterion and Regularity of the Input and Output. Solvability and Stability Estimate . . . . . . . . . . . . . . . . 3.1.4 The Regularization Strategy by SVD. Truncated SVD . . . 3.2 Inverse Source Problems for Wave Equation . . .. . . . . . . . . . . . . . . . . . . . 3.2.1 Non-uniqueness and Uniqueness of a Solution . . . . . . . . . . . 3.3 Backward Parabolic Problem .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4 Computational Issues in Inverse Source Problems . . . . . . . . . . . . . . . . . 3.4.1 The Galerkin Finite Element Method... . . . . . . . . . . . . . . . . . . . 3.4.2 The Conjugate Gradient Algorithm .. . .. . . . . . . . . . . . . . . . . . . .

65 66 69 74 81 87 94 98 102 110 110 113 xiii

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3.4.3

3.4.4 Part II 4

Convergence of Gradient Algorithms for Functionals with Lipschitz Continuous Fréchet Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 118 Numerical Examples . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 122

Inverse Problems for Differential Equations

Inverse Problems for Hyperbolic Equations . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1 Inverse Source Problems .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.1 Recovering a Time Dependent Source Term . . . . . . . . . . . . . . 4.1.2 Recovering a Spacewise Dependent Source Term . . . . . . . . 4.2 Problem of Recovering the Potential in String Equation . . . . . . . . . . . 4.2.1 Some Properties of the Direct Problem .. . . . . . . . . . . . . . . . . . . 4.2.2 Existence of the Local Solution to the Inverse Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2.3 Global Stability and Uniqueness .. . . . . .. . . . . . . . . . . . . . . . . . . . 4.3 Inverse Coefficient Problems for Layered Media .. . . . . . . . . . . . . . . . . .

129 129 130 133 135 135

5

One-dimensional Inverse Problems in Electrodynamics . . . . . . . . . . . . . . . 5.1 Formulation of Inverse Electrodynamic Problems .. . . . . . . . . . . . . . . . . 5.2 The Direct Problem: Existence and Uniqueness of a Solution .. . . . 5.3 One-dimensional Inverse Problems . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3.1 Problem of Finding the Permittivity Coefficient . . . . . . . . . . 5.3.2 Problem of Finding the Conductivity Coefficient .. . . . . . . .

151 151 152 162 162 169

6

Inverse Problems for Parabolic Equations. . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.1 Relationship Between Solutions of Parabolic and Hyperbolic Direct Problems . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.2 Problem of Recovering the Potential in Heat Equation.. . . . . . . . . . . . 6.3 Uniqueness Theorems for Inverse Problems .. . .. . . . . . . . . . . . . . . . . . . . 6.4 Relationship With Inverse Spectral Problems for Sturm-Liouville Operator . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.5 Identification of a Leading Coefficient From Dirichlet Measured Output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.5.1 Some Properties of the Direct Problem Solution.. . . . . . . . . 6.5.2 Compactness and Lipschitz Continuity of the Input-Output Operator: Regularization .. . . . . . . . . . . . 6.5.3 Integral Relationship and Gradient Formula . . . . . . . . . . . . . . 6.5.4 Reconstruction of an Unknown Coefficient . . . . . . . . . . . . . . . 6.6 Identification of a Leading Coefficient From Neumann Measured Output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.6.1 Compactness of the Input-Output Operator . . . . . . . . . . . . . . . 6.6.2 Lipschitz Continuity of the Input-Output Operator and Solvability of the Inverse Problem . . . . . . . . . 6.6.3 Integral Relationship and Gradient Formula . . . . . . . . . . . . . .

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140 145 147

171 174 176 180 182 183 185 192 195 200 204 205 208

Contents

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7

Inverse Problems for Elliptic Equations . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.1 The Inverse Scattering Problem at a Fixed Energy . . . . . . . . . . . . . . . . . 7.2 The Inverse Scattering Problem with Point Sources .. . . . . . . . . . . . . . . 7.3 Dirichlet-to-Neumann Map . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

213 213 216 222

8

Inverse Problems for Stationary Transport Equations . . . . . . . . . . . . . . . . . 8.1 The Transport Equation Without Scattering . . . .. . . . . . . . . . . . . . . . . . . . 8.2 Uniqueness and a Stability Estimate in the Tomography Problem .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.3 Inversion Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

227 227

The Inverse Kinematic Problems . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 9.1 The Problem Formulation .. . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 9.2 Rays and Fronts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 9.3 The One-Dimensional Problem.. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 9.4 The Two-Dimensional Problem . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

235 235 236 240 242

9

Part III

230 231

Inverse Problems in Wave Phenomena and Vibration

10 Inverse Problems for Damped Wave Equations . . . . .. . . . . . . . . . . . . . . . . . . . 10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.1.1 Structure of Solutions of Wave Equation and Dependence on the Dirichlet Input .. . . . . . . . . . . . . . . . . . . 10.1.2 Necessary Estimates for the Direct Problem Solution . . . . 10.1.3 The Dirichlet-to-Neumann Operator.. .. . . . . . . . . . . . . . . . . . . . 10.1.4 Existence of a Quasi-Solution of the Inverse Problem .. . . 10.1.5 Uniqueness of the Solution to the Inverse Problem .. . . . . . 10.1.6 Fréchet Differentiability of the Tikhonov Functional .. . . . 10.2 Recovering a Potential from Neumann-to-Dirichlet Operator . . . . . 10.2.1 Regularity Properties of the Direct Problem Solution in Subdomains Defined by the Final Time and Characteristics . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2.2 Existence and Uniqueness of the Inverse Problem Solution . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2.3 Necessary Estimates for the Weak and Regular Weak Solutions of the Direct Problem . . . . . . . . . . . . . . . . . . . . 10.2.4 The Neumann-to-Dirichlet Operator.. .. . . . . . . . . . . . . . . . . . . . 10.2.5 Existence of a Quasi-Solution. Gradient Formula for the Tikhonov Functional . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.3 Recovering a Potential From Dirichlet-to-Neumann Operator .. . . . 10.3.1 Regularity Properties of the Direct Problem Solution. Formation of Discontinuities .. . . . . . . . . . . . . . . . . . . 10.3.2 Local Existence Theorem and Stability Estimate . . . . . . . . . 10.3.3 Necessary Estimates for the Weak and Regular Weak Solutions of the Direct Problem . . . . . . . . . . . . . . . . . . . .

249 250 252 268 275 277 279 285 291

292 303 310 318 321 325 326 339 345

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10.3.4 The Dirichlet-to-Neumann Operator.. .. . . . . . . . . . . . . . . . . . . . 349 10.3.5 Fréchet Differentiability of the Tikhonov Functional. Gradient Formula .. . . . . . . . .. . . . . . . . . . . . . . . . . . . . 353 10.3.6 Lipschitz Continuity of the Fréchet Gradient . . . . . . . . . . . . . 359 11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.1.1 The Initial Boundary Value Problem for Simply Supported Beam . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.1.2 The Initial Boundary Value Problem for a Cantilever Beam Under Free-End Shear Force.. . . . . . . . . . . 11.2 Identification of a Temporal Load in a Simply Supported Beam From Boundary Measured Slope . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.2.1 The Input-Output Operator: Compactness and Lipschitz Continuity .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.2.2 Existence of a Quasi-Solution of the Inverse Problem .. . . 11.2.3 Fréchet Differentiability of the Tikhonov Functional. Gradient Formula .. . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.2.4 Lipschitz Continuity of the Fréchet Gradient . . . . . . . . . . . . . 11.2.5 Reconstruction of an Unknown Temporal Load from Boundary Measured Slope . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.3 Determination of Unknown Shear Force in a Cantilever Beam From Boundary Measured Bending Moment .. . . . . . . . . . . . . . . 11.3.1 The Neumann-to-Neumann Operator and Existence of a Quasi-Solution.. . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.3.2 Fréchet Gradient of the Tikhonov Functional. Lipschitz Continuity of the Gradient . .. . . . . . . . . . . . . . . . . . . . 11.3.3 Numerical Reconstruction of Unknown Shear Force from Boundary Measured Moment . . . . . . . . . . . . . . . . . 11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam Equation from Final Time Output . . . . . . . . . . . . . . . 11.4.1 The Singular Value Decomposition and Sufficient Condition for the Uniqueness.. . . . . . . . . . . . . . . . . . 11.4.2 Application to Forced Vibration Under Pure Spatial Load.. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.4.3 Application to Forced Vibration Under Harmonic Load . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.4.4 Adjoint Method Based on the Quasi-Solution Approach .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.4.5 Numerical Reconstruction of Unknown Spatial Load from Final Time Output .. . . . . . . . .. . . . . . . . . . . . . . . . . . . .

363 364 364 374 379 379 381 382 385 386 391 393 396 400 405 407 416 419 423 427

Contents

11.5 Determination of the Flexural Rigidity of a Simply Supported Damped Beam from Measured Boundary Slope . . . . . . . 11.5.1 Properties of the Input-Output Operator and Existence of a Quasi-Solution.. . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.5.2 Fréchet Differentiability of the Tikhonov Functional and Gradient Formula .. . . . .. . . . . . . . . . . . . . . . . . . . 11.6 Determination of the Flexural Rigidity of a Cantilever Beam from Measured Boundary Bending Moment . . . . . . . . . . . . . . . . 11.6.1 The Neumann-to-Neumann Operator. Existence of a Quasi-Solution . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.6.2 Fréchet Differentiability of the Tikhonov Functional and Gradient Formula .. . . . .. . . . . . . . . . . . . . . . . . . . 11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured Boundary Slope .. . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.7.1 Necessary Estimates for the Direct Problem Solution . . . . 11.7.2 The Input-Output Operator. Existence of a Quasi-Solution . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 11.7.3 Fréchet Differentiability of the Tikhonov Functional. Gradient Formula .. . . . . . . . .. . . . . . . . . . . . . . . . . . . .

xvii

431 432 438 440 441 446 449 453 458 460

A

Invertibility of Linear Operators . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 465 A.1 Invertibility of Bounded Below Linear Operators . . . . . . . . . . . . . . . . . . 465 A.2 Invertibility of Linear Compact Operators . . . . . .. . . . . . . . . . . . . . . . . . . . 468

B

Necessary Estimates For One-Dimensional Parabolic Equation . . . . . . B.1 The Problem with Non-homogeneous Initial Data and Source . . . . B.2 The Problem with Neumann Input . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . B.3 The Problem with Dirichlet Input . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

473 475 478 481

C

Necessary Estimates For Euler-Bernoulli Beam Equation . . . . . . . . . . . . . C.1 Existence of Weak Solutions . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . C.1.1 Uniform Estimate in Finite Dimensional Subspace .. . . . . . C.1.2 Existence and Uniqueness Theorems ... . . . . . . . . . . . . . . . . . . . C.2 The Problem with Neumann Inputs (Bending Moments) . . . . . . . . . . C.3 The Problem with Dirichlet Input (Slope) . . . . . .. . . . . . . . . . . . . . . . . . . . C.4 The Problem with Non-homogeneous Initial Data . . . . . . . . . . . . . . . . .

485 485 487 490 493 496 499

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 505 Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 513

Chapter 1

Introduction Ill-Posedness of Inverse Problems

1.1 Some Basic Definitions and Examples Inverse problems arise in almost all areas of science and technology, in modeling of problems motivated by various physical and social processes. Most of these models are governed by differential and integral equations. If all the necessary inputs in these models are known, then the solution can be computed and behavior of the physical system under various conditions can be predicted. In terms of differential problems, the necessary inputs include such information as initial or boundary data, coefficients and force term, also shape and size of the domain. If all these data are enough to describe the system adequately, then it is possible to use the mathematical model for studying the physical system. For example, consider the case of the simplest one-dimensional model of steady-state heat transfer 

  (Lu))(x) := − k(x)u (x) + q(x)u(x) = F (x), x ∈ (0, ), u(0) = 0,

(k(x)u (x))x= = ϕ, ϕ ∈ R,

(1.1.1)

where the heat coefficients (k(x), q(x)), external heat source F (x) and the heat flux ϕ at the end x =  of the rod with length  > 0 are the given input data. If we assume that k ∈ C 1 (0, ) ∩ C 0 [0, ], q, F ∈ C 0 (0, ), then the solution u(x) of the two-point boundary value problem (1.1.1) is evidently in C 2 (0, ) ∩ C 1 [0, ]. This solution exists, is unique and continuously depends on the above given input data. If the rod is non-homogeneous and external heat source has a discontinuity, then the solution u(x) of the boundary value problem (1.1.1) needs to be considered in larger class of functions. Specifically, if k ∈ K := {k ∈ L2 (0, ) : 0 < c0 ≤ k(x) ≤ c1 }, q ∈ Q := {q ∈ L2 (0, ) : 0 ≤ q(x) ≤ c2 }, F ∈

(1.1.2)

L2 (0, ),

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_1

1

2

1 Introduction Ill-Posedness of Inverse Problems

then there exists a unique solution u(x) of the boundary value problem (1.1.1) in V 1 (0, ) := {u ∈ H 1 (0, ) : u(0) = 0}. Here and below H 1 (0, ) is the Sobolev space [1]. Moreover, this solution continuously depends on the above given input data, as we will show below. This means that the parameter-to-solution maps Ak : k ∈ K ⊂ L2 (0, ) → u ∈ V 1 (0, ), Aq : q ∈ Q ⊂ L2 (0, ) → u ∈ V 1 (0, ),

(1.1.3)

AF : F ∈ L2 (0, ) → u ∈ V 1 (0, ) are continuous. The sets K and Q are called the sets of input data. The above mentioned three conditions (existence, uniqueness and continuous dependence) related to solvability of various initial boundary value problems has been identified with the notion of a “well-posed problem”. Jacques Hadamard played a pioneering role recognizing first and developing then the idea of well-posedness at the beginning of the twentieth century. But initially, in [50], Hadamard included only existence and uniqueness in the definition of well-posedness, insisting on the fact that continuous dependence on the initial data is important only for the Cauchy problem. This latter feature of a “well-posed problem” was added to Hadamard’s definition by Hilbert and Courant in [35]. Many years later, Hadamard used this corrected definition in his book [51], referring to Hilbert and Courant. We formulate the definition of well-posedness for a linear operator equation Au = F , A ∈ L(H, H˜ ), where L(H, H˜ ) denotes the set of all bounded linear operators and H , H˜ are Hilbert spaces. Definition 1.1.1 (Well-Posedness) The linear equation/problem Au = F , A ∈ L(H, H˜ ) is defined to be well-posed in Hadamard’s sense, if (i1) (i2) (i3)

a solution of this equation exists, for all admissible data F ∈ H˜ ; this solution is unique; it depends continuously on the data.

Otherwise, the equation Au = F is defined to be ill-posed in Hadamard’s sense. According to this definition the boundary value problem (1.1.1) is well-posed. However, in practice it is not possible to measure experimentally all these inputs, since in real physical systems the inputs can be defined as measurable and unmeasurable ones. Instead, it is possible to measure experimentally certain (additional) outputs of the system and use this information together with other inputs to recover the missing, unmeasurable input. Hence, in a broad sense, an inverse problem can be defined as the problem of determining unmeasurable parameters of a system from measurable parameters, by using a mathematical/physical model. Let us consider the case where q ∈ Q, F ∈ L2 (0, ), ϕ ∈ R are the given inputs, but the coefficient k(x) in (1.1.1) is unknown and needs to be determined from some additional (physically reasonable) condition(s). Example 1.1.1 Inverse coefficient problem for the Sturm-Liouville operator Lu := −(ku ) + qu.

1.1 Some Basic Definitions and Examples

3

Let q(x) ≡ 0 in (1.1.1) and the function F (x) be given as follows:  F (x) =

0, x ∈ [0, ξ ], 1, x ∈ (ξ, ],

where ξ ∈ [0, ] is an arbitrary parameter. Then solution to problem (1.1.1) depends on the parameter ξ ∈ [0, ], that is, u = u(x, ξ ). We assume that the measured value of u(x, ξ ) at x =  is given as an additional information, i.e. as a measured output defined as f (ξ ) := u(, ξ ),

ξ ∈ [0, ].

(1.1.4)

Based on this information we are going to identify the unknown coefficient k ∈ K, i.e. the input. We define the problem of identifying the unknown coefficient k(x) in (1.1.1) from the output f ∈ F ⊂ L2 (0, ) given by (1.1.4) as a inverse coefficient (or parameter identification) problem for the Sturm-Liouville operator. In this context, for the given inputs k(x), q(x), F (x) and ϕ, the boundary value problem (1.1.1) is defined as a direct problem. We formulate the inverse problem of identifying the unknown coefficient k(x) in terms of an operator equation. To this end, introduce the set of admissible coefficients as follows: k ∈ K := {k ∈ L2 (0, ) : 0 < c0 ≤ k(x) ≤ c1 } Assume that k ∈ K is a given admissible coefficient. Since there exists a unique solution u := u(x, ξ ; k) of the boundary value problem (1.1.1) in V 1 (0, ), the trace u(x, ξ ; k)x= is uniquely determined for each k ∈ K. We define this mapping as an input-output operator  : k → f as follows: [k](ξ ) := u(x, ξ ; k)x=, u ∈ V 1 (0, ), k ∈ K,

(1.1.5)

and reformulate this inverse problem as the problem of solving the following operator equation [k] = f, k ∈ K, f ∈ F .

(1.1.6)

Let us analyze now the inverse coefficient problem. Taking into account conditions (1.1.3), we integrate Eq. (1.1.1) on (x, ) and use the boundary condition (k(x)u (x))x= = ϕ. Then we obtain: k(x)u (x, ξ ) = ϕ + ψ(x, ξ ),

(1.1.7)

4

1 Introduction Ill-Posedness of Inverse Problems

where  ψ(x, ξ ) =

 − ξ, x ∈ [0, ξ ],  − x, x ∈ (ξ, ], ξ ∈ [0, ].

(1.1.8)

Integrating (1.1.7) and using the boundary condition u(0) = 0 we get: 

x

u(x, ξ ) = 0

ϕ + ψ(τ, ξ ) dτ, k(τ )

x ∈ [0, ].

(1.1.9)

Hence, the formula for finding k(x) is  f (ξ ) = 



ϕ + ψ(τ, ξ ) dτ k(τ )

ξ

ϕ+−ξ dτ + k(τ )

0

= 0



 ξ

ϕ+−τ dτ, k(τ )

(1.1.10) ξ ∈ [0, ].

Taking the first and then the second derivatives of both sides we find: f  (ξ ) = −



ξ 0

dτ , k(τ )

1 f (ξ ) = − , k(ξ ) 

(1.1.11)

ξ ∈ [0, ].

It follows from (1.1.11) that f  (0) = 0 and f ∈ H 2 (0, ). Furthermore, k ∈ K implies: 0 < 1/c1 ≤ −f  (ξ ) ≤ 1/c0 . Then, introducing the subspace V 2 (0, ) := {f ∈ H 2 (0, ) : f  (0) = 0}, where H 2 (0, ) is the Sobolev space, we conclude that if k ∈ K, then the set of admissible outputs is F := {f ∈ V 2 (0, ) : 0 < 1/c1 ≤ −f  (ξ ) ≤ 1/c0 }.

(1.1.12)

Moreover, the unknown coefficient is defined by the inversion formula: k(ξ ) = −

1 f  (ξ )

,

ξ ∈ [0, ].

(1.1.13)

We can deduce from formula (1.1.13) the following estimate: if ki ∈ K and fi ∈ F , i = 1, 2, then

k1 − k2 L2 (0,) ≤ c12 f1 − f2 H 2 [0,] , c1 > 0. This estimate is called the stability estimate of the inverse problem.

(1.1.14)

1.1 Some Basic Definitions and Examples

5

As noted above, under conditions (1.1.2) there exists a unique solution u(x) of the boundary value problem (1.1.1) in H 1 (0, ) for each k ∈ K ⊂ L2 (0, ). Moreover, as it is shown in the next section, this solution continuously depends on the input data k(x), F (x) and ϕ. Therefore, the direct problem is well-posed from L2 (0, ) to H 1 (0, ), by Definition 1.1.1. To examine properties of the inverse problem we need the following definition. Definition 1.1.2 Let B and B˜ be normed spaces, and A : B → B˜ be a linear operator. A is called a compact operator if the set {Au : u B ≤ 1} ⊂ B˜ is a ˜ pre-compact, i.e. has compact closure in B. This definition is equivalent to the following one: for every bounded sequence {un } ⊂ B, the sequence of images {Aun } ⊂ B˜ has a subsequence which converges ˜ Compact operators are also called completely continuous to some element of B. operators. Consider now the inverse problem (1.1.6), i.e. the problem of determining the unknown k ∈ K ⊂ L2 (0, ) from the given output f ∈ L2 (0, ). We need the following definition. Assume first the input-output operator (1.1.5) acts from L2 (0, ) to L2 (0, ), that is, the range of the operator  is the entire space: R() = L2 (0, ). We may use formula (1.1.10) to get the explicit form of this operator: 



[k](ξ ) := 0

ϕ + ψ(τ, ξ ) dτ, k(τ )

ξ ∈ [0, ].

Using formula (1.1.8) in the above integral we may derive, after elementary transformations, the following explicit form of the input-output operator: 



[k](ξ ) := ϕ 0

dη + k(η)

 ξ

 η 0

1 dτ dη, k(τ )

ξ ∈ [0, ].

(1.1.15)

From the theory of integral operators [155] it follows that in this case the inputoutput operator (1.1.15) is compact. As we will prove in Lemma 1.3.1, if the operator  : L2 (0, ) → L2 (0, ) is compact, then the problem [k] = f is ill-posed. Thus, if the input-output operator is defined from L2 (0, ) to L2 (0, ), then the inverse problem (1.1.6) is ill-posed. Assume now that the input-output operator is defined from L2 (0, ) not to 2 L (0, ), but to H 1 (0, ) ⊂ L2 (0, ), that is, R() = H 1 (0, ). The right hand side of the first formula of (1.1.11) implies that the range restriction  : L2 (0, ) → H 1 (0, ) ⊂ L2 (0, ) is well-defined, since k ∈ K ⊂ L2 (0, ). Again, it follows from the theory of integral operators that in this case the input-output operator (1.1.15) is also compact. As a result, we conclude that if the input-output operator (1.1.15) is defined from L2 (0, ) to H 1 (0, ), then the inverse problem (1.1.6) is still ill-posed. We continue the range restriction, assuming finally that the input-output operator is defined from L2 (0, ) to F ⊂ H 2 (0, ), where F is the set of outputs defined by (1.1.12), that is,  : L2 (0, ) → F ⊂ H 2 (0, ). This mapping is well-defined, due

6

1 Introduction Ill-Posedness of Inverse Problems

to the second formula of (1.1.11), which also allows to derive the explicit form of the inverse operator −1 : R() = F ⊂ H 2 (0, ) → L2 (0, ): k(x) := (−1 f )(x) := −

1 f  (x)

,

x ∈ [0, ].

Thus, if the input-output operator is defined as  : L2 (0, ) → F ⊂ H 2 (0, ), i.e. if R() = F , then the inverse problem defined by (1.1.6) or by (1.1.1) and (1.1.4) is well-posed.  The above analysis of the simplest inverse coefficient problem shows some distinguished features of inverse problems. The first feature is that if even the direct problem (1.1.1) is linear, the inverse coefficient problem is nonlinear. Note that, inverse source problems corresponding to linear direct problems are linear. The second feature is that besides the linear ill-posedness arising from the differentiation operation in (1.1.13), there is a nonlinear ill-posedness due to the presence of the quotient. Specifically, in practice, the output f ∈ L2 (0, ) is obtained from measurements and always contains a random noise. As a result, errors at small values of f  (x) in (1.1.13) are amplified much stronger which can lead to instability. These two features show that inverse coefficient problems are nonlinear and most sensitive, while so-called inverse source problems are linear and less sensitive. Another consequence of the above analysis is that, depending on the choice of functional spaces, an inverse problem can be ill-posed or well-posed. To explain the cause of this change in the framework of the linear theory of ill-posed problems, ˆ given in Chap. 2, we define the new input-output operator as follows: [k] = [1/k], where  is defined by (1.1.5). Then the inverse problem becomes a linear one: ˆ [k](ξ ) := u(x, ξ ; 1/k)x= , u ∈ V 1 (0, ), k ∈ K. ˆ results in the restriction of The first statement here is that the restriction of R() ∗ ⊥ ∗ ˆ ) of the adjoint operator  ˆ , since R() ˆ = N ( ˆ ∗ )⊥ , as the null-space N ( proved in Sect. 2.2. But what does it mean in terms of differential operators? Picard’s ˆ ∗ )⊥ is one of the necessary Theorem in Sect. 2.4 says that the condition f ∈ N ( ˆ and sufficient conditions for solvability of the operator equation [k] = f . Hence the range restriction is the restriction of the class {f } of the right hand side functions, in order to get better functions to ensure a convergent Singular Value Expansion (SVE) (see Sects. 3.1.2–3.1.3). In terms of differential problems this is equivalent to increasing the smoothness of the function f (x), i.e. restricting the class of outputs, as stated in the Sect. 3.1.3. The second statement is that the range restriction, stepˆ = F of by-step from L2 (0, ) to H 2 (0, ), led us to the fact that the range R() the input-output operator became compact in L2 (0, ). Indeed, Rellich’s lemma for Sobolev spaces asserts that if ⊂ Rn is a bounded domain, then the embedding H 1 ( ) → L2 ( ) is a compact operator. This lemma implies, in particular, that every bounded set in H 1 ( ) is compact in L2 ( ). The range restriction of the inputˆ : L2 (0, ) → L2 (0, ) from L2 (0, ) to the set F ⊂ H 2 (0, ) ⊂ output operator 

1.1 Some Basic Definitions and Examples

7

ˆ −1 ) := F of the inverse operator to L2 (0, ) means that we defined the domain D( be compact in L2 (0, ) (moreover, in H 1 (0, )). In this compact set F , the stability ˆ can be treated also as the Lipschitz continuity estimate (1.1.14) for the operator  ˆ −1 : f → 1/k: of the inverse operator     ˆ −1 ˆ −1 f2    f1 − 

L2 (0,)

≤ f1 − f2 H 2 (0,), for all f1 , f2 ∈ F .

Thus, a common situation in inverse and ill-posed problems rather is that compactness plays a dual role in the ill-posedness of inverse problems. Whilst compactness, as a property of the operator, plays a negative role, making actually ˆ −1 ) := F of the problem worse, i.e. ill-posed, as a property of the domain D( the inverse operator plays an essential positive role. This positive role has first been discovered by A.N. Tikhonov [146, 149]. The fundamental Tikhonov’s lemma on the continuity of the inverse of an operator, which is injective, continuous and defined on a compact set, clearly illustrates this positive role and until now is used as an important tool in regularization of ill-posed problems. Lemma 1.1.1 (Tikhonov) Let H and H˜ be metric spaces and A : U ⊂ H → H˜ be a one-to-one continuous operator with A(U ) = V . If U ∈ H is a compact set, then the inverse operator A−1 : V ⊂ H˜ → U ⊂ H is also continuous. Proof Let v ∈ V be any element. Since the operator A : U ⊂ H → V ⊂ H˜ is one-to-one, there exists a unique element u ∈ U such that u = A−1 v. Now suppose, contrary to the assertion, that the inverse operator A−1 : V ⊂ H˜ → U ⊂ H is not continuous. This implies that there exists a positive number  > 0 and an element vδ ∈ V such that for all δ > 0 the following conditions hold: ρH˜ (v, vδ ) < δ, but ρH (A−1 v, A−1 vδ ) ≥ .

(1.1.16)

Due to the arbitrariness of δ > 0, there exists a sequence of positive numbers + {δn }∞ n=1 such that δn → 0 , as n → ∞. In view of (1.1.16) this implies that the corresponding sequence of elements {vδn } ⊂ V satisfy the conditions: ρH˜ (v, vδn ) < δn , but ρH (A−1 v, A−1 vδn ) ≥ .

(1.1.17)

Taking the limit here, as n → ∞ we conclude lim ρ ˜ (v, vδn ) n→∞ H

= 0.

(1.1.18)

−1 Being a subset of the compact U , the sequence {uδn }n=∞ n=1 ⊂ U , uδn := A vδn , ∞ ∞ has a convergent subsequence {uδm }m=1 ⊂ {uδn }n=1 . Then there exists an element u˜ ∈ U such that limm→∞ ρH (u, ˜ uδm ) = 0. By the continuity of the operator A : U ⊂ H → V ⊂ H˜ , this implies:

˜ Auδm ) lim ρ ˜ (Au, m→∞ H

= 0.

(1.1.19)

8

1 Introduction Ill-Posedness of Inverse Problems

By (1.1.18), limm→∞ ρH˜ (v, vδm ) = 0, where vδm := Auδm . With (1.1.19) this implies that Au˜ = v = Au. Taking now into account the fact that the operator A : U ⊂ H → V ⊂ H˜ is one-to-one, we find u˜ = u. On the other hand, u = A−1 v, uδm := A−1 vδm , by definition, and we deduce from the second assertion of (1.1.17) that ρH (u, uδm ) ≥ . Taking a limit here as m → ∞ we get: ρH (u, u) ˜ ≥ . This contradicts the previous assertion u˜ = u and completes the proof.   Let us return now to Definition 1.1.1. The first condition (existence) means that the operator A ∈ L(H, H˜ ) is surjective, the second condition (uniqueness) means that this operator is injective, and the last condition means that the inverse operator A−1 : H˜ → H is continuous. In terms of operator theory the first condition (existence) is equivalent to the condition R(A) = H˜ , where R(A) is the range of the operator A. The second condition (uniqueness) is equivalent to the condition N (A) = {0}, where N (A) := {u ∈ D(A) : Au = 0} is the kernel of the operator A. It is important, from applications point of view, to note that in finite dimensional spaces the conditions (i1) and (i2) of Definition 1.1.1 are equivalent, as it follows from Halmos’s theorem below. Theorem 1.1.1 Let A : H → H be a linear operator defined on the finite dimensional space H . Then the following assertions are equivalent: (H1) (H2)

N (A) = {0}, i.e. A is injective; A is surjective.

Hence, in a finite dimensional space the inverse problem Au = F has a unique solution u = A−1 F , for each F ∈ H , i.e. the inverse operator A−1 : H → H exists (but may not be continuous!), if only one of the properties (H1)–(H2) holds. However, in infinite-dimensional spaces these properties are not equivalent, as the following example shows. Example 1.1.2 The case when injectivity does not imply surjectivity. Let A : C([0, 1]) → C([0, 1]) be a linear operator defined by 

x

(Au)(x) :=

u(ξ )dξ, u ∈ C([0, 1]).

(1.1.20)

0

Evidently, (Au)(0) = 0, for all u ∈ C([0, 1]), so the range R(A) of the operator A is a proper subset of C([0, 1]), i.e. C([0, 1]) \ R(A) = ∅. This means operator A is not surjective. Obviously, this operator is injective, since differentiating the equation Au = 0, where A is defined by (1.1.20), we get u(x) = 0. So, Au = 0 implies u(x) = 0 in C([0, 1]).  In view of Hadamard’s definition, we can distinguish the following three types of Hadamard’s ill-posedness: (p1) (p2) (p3)

Non-existence (A is not surjective); Non-uniqueness (A is not injective); Instability (the inverse operator A−1 is not continuous).

1.2 Continuity with Respect to Coefficients and Source: Sturm-Liouville. . .

9

For an adequate mathematical model of a physical process it is reasonable to require an existence of a solution in an appropriate class of functions, at least for an exact data. With regard to the uniqueness of a solution, this is the most important issue in inverse problems theory and is often not easy to prove. In the case when the uniqueness can not be guaranteed by given data, one needs either to impose an additional data or to restrict the set of admissible solutions using a-priori information on the solution. In many applied problems the non-uniqueness can be used as an advantage to obtain a desired solution among several ones. Nevertheless, the main issue in inverse and ill-posed problems is usually stability, i.e. continuous dependence of the solution on measured output. Moreover, this datum always contain random noise, hence the equation Au = F cannot be satisfied exactly in general. Even if the noise level in a measured output is small, many algorithms developed for well-posed problems do not work in case of a violation of the third condition (p3) due to round-off errors, if they do not address the instability. An algorithm using differentiation may serve as an example to this, since differentiation has the properties of an ill-posed problem. This is the reason why regularization methods play a central role in the theory and applications of inverse and ill-posed problems to overcome the above mentioned instabilities [37, 81].

1.2 Continuity with Respect to Coefficients and Source: Sturm-Liouville Equation This introductory section familiarizes the reader with some aspects of continuous dependence of the solution of the boundary problem (1.1.1) on the coefficients k ∈ K, q ∈ Q and the source function F ∈ L2 (0, ). It is important to distinguish the character of continuity with respect to different coefficients and also the source function in inverse/identification problems, as well as in optimal control governed by PDEs. Let us define the weak solution of problem (1.1.1) as the solution u ∈ V 1 (0, ) of the integral identity: 

 0

[k(x)u v  + q(x)uv]dx =





F (x)vdx + ϕv(), ∀v ∈ V 1 (0, ), (1.2.1)

0

where V 1 (0, ) := {u ∈ H 1 (0, ) : u(0) = 0}. Denote by C 0,λ [0, ] the Hölder space of functions with exponent λ ∈ (0, 1], that is, there exists a positive constant M such that |u(x1 ) − u(x2 )| ≤ M|x1 − x2 |λ , for all x1 , x2 ∈ [0, ].

10

1 Introduction Ill-Posedness of Inverse Problems

Lemma 1.2.1 Let conditions (1.1.2) hold. Then there exists a unique solution u ∈ V 1 (0, ) of the boundary value problem (1.1.1). This solution is Hölder continuous with exponent λ = 1/2, i.e. for all x1 , x2 ∈ [0, ],   

F L2 (0,) + 2/ |ϕ| |x1 − x2 |1/2. |u(x1) − u(x2 )| ≤ √ 2 c0

(1.2.2)

Moreover, the following estimate holds:  2 + 2 

u V 1 (0,) ≤ √

F L2 (0,) + 2/ |ϕ| 2 c0

(1.2.3)

Proof We use the integral identity (1.2.1) to introduce the symmetric bilinear a : V 1 (0, ) × V 1 (0, ) → R and linear b : V 1 (0, ) → R functionals: 



a(u, v) :=

[k(x)u v  + q(x)uv]dx,

0





b(u) :=

F (x)vdx + ϕv(),

u, v ∈ V 1 (0, ).

0

By conditions (1.1.2), a(u, v) is a strongly positive bounded bilinear functional and b(v) is a bounded linear functional. Then, by the Main Theorem on Quadratic Variational Problems [156, Sect. 2.4] there exists a unique solution u ∈ V 1 (0, ) of the variational problem a(u, v) = b(v),

∀v ∈ V 1 (0, ).

Let us prove now the Hölder continuity of the solution. For any x1 , x2 ∈ [0, ] we have:

 x 2



|u(x1 ) − u(x2 )| =

u (ξ )dξ

x1 (1.2.4) 1/2  x2  2 1/2  1/2 (u (ξ )) dξ |x1 − x2 | ≤ u L2 (0,) |x1 − x2 | . ≤ x1

To estimate the norm u L2 (0,) through the inputs of problem (1.1.1), we use the energy identity 

 0

[k(x)(u (x))2 + q(x)(u(x))2]dx =



 0

F (x)udx + ϕu(b), u ∈ V 1 (0, )

1.2 Continuity with Respect to Coefficients and Source: Sturm-Liouville. . .

11

√  and the Poincaré inequality u L2 (0,) ≤ / 2 u L2 (0,). We have:

u 2L2 (0,)

1 ≤ c0



1 = c0



[k(x)(u (x))2 + q(x)(u(x))2]dx

0









F (x)udx + ϕ

0





u (x)dx 0

√ 1 

F L2 (0,) u L2 (0,) +  |ϕ| u L2 (0,) c0   √  1 ≤ √ F L2 (0,) +  |ϕ| u L2 (0,). c0 2

≤

After dividing both sides by u L2 (0,) = 0, we obtain 

u L2 (0,)

1 ≤ c0



 √  √ F L2 (0,) +  ϕ , 2

(1.2.5)

which with (1.2.4) leads the desired estimate (1.2.2). To prove estimate (1.2.3) one needs to use in (1.2.5) the inequality √

u L2 (0,) ≥ β0 u V 1 (0,) with β0 = / 2 + 2, which follows from the Poincaré inequality.   Inequality (1.2.2) means that the weak solution u ∈ V 1 (0, ) of the boundary value problem (1.1.1) belongs to the Hölder space C 0,λ [0, ], with exponent λ ∈ (0, 1/2]. Furthermore, estimate (1.2.3) means continuity of this weak solution with respect to the source term F ∈ L2 (0, ) and Neumann boundary output ϕ ∈ R. The following theorem shows that the nature of the continuous dependence of the weak solution u ∈ V 1 (0, ) of the boundary value problem (1.1.1), on the principal coefficient k(x) and on the second coefficient (potential) q(x) differ from each other. Theorem 1.2.1 Let conditions (1.1.2) hold. Assume that {kn } ⊂ K, {qn } ⊂ Q and {Fn } ⊂ L2 (0, ) are the sequences of coefficients and source functions. Denote by {un } ⊂ V 1 (0, ) the sequence of corresponding weak solutions, that is, for each n = 1, 2, 3, . . ., the function un (x) := u[x; kn, qn , Fn ] is the weak solution of the boundary value problem 

  − kn (x)un (x) + qn (x)un (x) = Fn (x), x ∈ (0, ), un (0) = 0,

(kn (x)un (x))x= = ϕ, ϕ ∈ R.

(1.2.6)

12

1 Introduction Ill-Posedness of Inverse Problems

If ⎧ 1 1 ⎪  weakly in L2 (0, ), ⎪ ⎪ ⎨ kn (x) k(x) qn (x)  q(x) weakly in L2 (0, ), ⎪ ⎪ ⎪ ⎩ Fn (x)  f (x) weakly in L2 (0, ),

(1.2.7) as n → ∞,

then the sequence of solutions of problem (1.2.6) converges to the solution u ∈ V 1 (0, ), u(x) := u[x; k, q, f ], of the boundary value problem (1.1.1), in the norm of C 0,λ , with 0 < λ < 1/2, as n → ∞. Proof First we derive an integral representation for the solution of the boundary value problem (1.1.1). To this end, we integrate Eq. (1.1.1) over (x, ), use the Neumann boundary condition (k(x)u (x))x= = ϕ, then integrate again over (0, x) and use the Dirichlet boundary condition u(0) = 0. This yields: 

x

u(x) = ϕ  P (ξ ) :=

0 

dξ + k(ξ )



x 0

P (ξ ) dξ, k(ξ )

(1.2.8)

[F (z) − q(z)u(z)]dz.

ξ

Now we use estimate (1.2.3) applied to the solution un of problem (1.2.6) to get  2 + 2 

Fn L2 (0,) + 2/ |ϕ| .

un V 1 (0,) ≤ √ 2c0

(1.2.9)

Since Fn  F weakly in L2 (0, ), as n → ∞, due to the weak convergence criteria [94] we have: (a) { Fn 0 } is uniformly bounded;  x  x (b) Fn (ξ )dξ → F (ξ )dξ, for all x ∈ (0, ]. 0

0

The uniform boundedness of the sequence Fn 0 with estimate (1.2.9) implies the uniform boundedness of the sequence {un } ⊂ V 1 (0, ) in H 1 -norm. Since every bounded sequence in H 1 (0, ) is a compact in C 0,λ [0, ] for 0 < λ < 1/2, the sequence {un } is a compact in C 0,λ [0, ]. Hence we can extract a subsequence {um } ⊂ V 1 (0, ) that converges to a function u ∈ C 0,λ [0, ] in C 0,λ -norm. Taking into account that the convergence in C 0,λ [0, ] is a uniform convergence, we have um (x) → u(x) uniformly, for all x ∈ [0, ], as n → ∞. Finally, we prove that the limit function u(x) is the solution of the boundary value problem (1.1.1) corresponding to the limit functions k(x), q(x) and F (x) in (1.2.7).

1.2 Continuity with Respect to Coefficients and Source: Sturm-Liouville. . .

13

The integral representation (1.2.8) for um ∈ V 1 (0, ), m = 1, 2, 3, . . ., yields: 

x

um (x) = ϕ 0 

 Pm (ξ ) :=

dξ + km (ξ )



x 0

Pm (ξ ) dξ, x ∈ [0, ], km (ξ )

(1.2.10)

[Fm (z) − qm (z)um (z)]dz.

ξ

By the above weak convergence criteria, Fm  F in L2 (0, ) implies: 







Fm (z)dz →

ξ

F (z)dz, for all ξ ∈ [0, ], as m → ∞.

(1.2.11)

ξ

Due to the fact that qm (ξ )  q(ξ ) weakly in L2 (0, ) and um (ξ ) → u(ξ ), in C 0,λ -norm, we conclude qm (ξ )um (ξ )  q(ξ )u(ξ ) weakly in L2 (0, ). Hence 







qm (z)um (z)dz →

ξ

q(z)u(z)dz, ∀z ∈ [0, ], as m → ∞,

ξ

which, with (1.2.11), implies Pm (ξ ) → P (ξ ), as m → ∞. Now we prove that 

x 0

Pm (ξ ) dξ → km (ξ )



x 0

P (ξ ) dξ, ∀x ∈ [0, ], as m → ∞. k(ξ )

(1.2.12)

Indeed,

 x

 x

Pm (ξ ) P (ξ )



dξ − dξ

0 km (ξ ) 0 k(ξ )

 

 x 



1 1 Pm (ξ ) − P (ξ )



x

dξ +

− P (ξ )dξ

≤

km (ξ ) km (ξ ) k(ξ ) 0 0

 x    x



1 1 1

|Pm (ξ ) − P (ξ )| dξ +

≤ − P (ξ )dξ

, c0 > 0. c0 0 km (ξ ) k(ξ ) 0 The first right hand side term tends to zero due to the above convergence Pm (ξ ) → P (ξ ) in (1.2.12) as m → ∞, and the second term also tends to zero due to the weak convergence criteria. Therefore, 

x

0



0

x

dξ → km (ξ )



x

dξ , k(ξ ) 0  x Pm (ξ ) P (ξ ) dξ → dξ, as n → ∞, km (ξ ) 0 k(ξ )

14

1 Introduction Ill-Posedness of Inverse Problems

for all x ∈ [0, ], and the function um (x), represented by (1.2.10) converges to the solution u(x) defined by the integral representation (1.2.8), for all x ∈ [0, ], as m → ∞. Since the problem (1.1.1) has a unique solution, the sequence {un } ⊂ V 1 (0, ) converges to the solution u(x) of problem (1.1.1), for all x ∈ [0, ], as n → ∞. This completes the proof.   Theorem 1.2.1 clearly illustrates that the nature of convergence of the leading coefficient k(x) differs from the nature of convergence of the second coefficient q(x). A necessary condition for the convergence of the sequence of solutions {un } is the convergence of the sequence 1/kn (x)  1/k(x), but not the convergence of the sequence of coefficients kn (x)  k(x), while in case of the coefficient q(x), the necessary condition is the convergence of the sequence of coefficients qn (x)  q(x). Recall that we previously dealt with the term 1/k(x) in Example 1.1.1.

1.3 Why a Fredholm Integral Equation of the First Kind Is an Ill-Posed Problem? Consider the following Fredholm integral equation of the first kind 

1

K(x, y)u(y)dy = F (x), x ∈ [0, 1].

(1.3.1)

0

We assume that u ∈ C[0, 1] is the unknown function, K : [0, 1] × [0, 1] → R is the continuous kernel and F ∈ C[0, 1] is a given function. It is known that if D := {v ∈ C[0, 1] : v C[0,1] ≤ M, M > 0}, then the linear continuous operator A : D ⊂ C[0, 1] → C[0, 1], defined as 

1

(Au) (x) :=

K(x, y)u(y)dy,

(1.3.2)

0

is a linear compact operator from C[0, 1] to C[0, 1], i.e. transforms each bounded set in C[0, 1] to a relatively compact set in C[0, 1] (see, for instance, [155]). Let us show that the problem (1.3.1) is ill-posed in sense of the third Hadamard’s condition (p3). Assume that u := u(x; F ) is a solution of (1.3.1) for given F ∈ C[0, 1]. To show that the dependence u(· ; F ) is not continuous in C[0, 1], we define the sequence of continuous functions 

1

εn (x) = 0

K(x, y) sin(nπy)dy, n = 1, 2, 3, . . . .

(1.3.3)

1.3 Why a Fredholm Integral Equation of the First Kind Is an Ill-Posed Problem?

15

By the continuity of the kernel K(x, y), we deduce that εn C[0,1] → 0, as n → ∞. Now we define the “perturbed” source functions Fn (x) = F (x) + εn (x), n = 1, 2, 3, . . . . Then the function un (x) = u(x) + sin(nπy) will be a solution of the “perturbed” problem 

1

K(x, y)un (y)dy = Fn (x), x ∈ [0, 1],

0

for each n = 1, 2, 3, . . . . Evidently, the norm F − Fn C[0,1] = εn C[0,1] tends to zero, as n → ∞, although un − u C[0,1] = sin(nπy) C[0,1] = 1, for all n = 1, 2, 3, . . . . This shows that if the Fredholm operator (1.3.2) is defined as A : C[0, 1] → C[0, 1], then problem (1.3.1) is ill-posed. Note that the same conclusion is still holds, if K ∈ L2 ((0, 1) × (0, 1)) and the Fredholm operator (1.3.2) is defined as A : L2 (0, 1) → L2 (0, 1). In this case the integral (1.3.3) tends to zero as n → ∞, by the Riemann-Lebesgue Lemma.  To answer the question “why the problem (1.3.1) is ill-posed?” we need to study this problem deeper, by arriving to the physical meaning of the mathematical model. Example 1.3.1 Relationship between the Fredholm integral equation and the differential problem. Let us analyze again problem (1.3.1) assuming that the kernel is given by the formula  (1 − x)y, 0 ≤ y ≤ x, ˚ K(x, y) = (1.3.4) x(1 − y), x ≤ y ≤ 1. It is easy to verify that 

1

˚ y) sin(nπy) dy = K(x,

0

1 sin(nπx), (nπ)2

n = 1, 2, . . . .

(1.3.5)

Equation (1.3.5) shows that the numbers {(nπ)−2 }∞ n=1 are eigenvalues of the integral operator ˚ (Au)(x) :=



1

˚ y)u(y)dy, K(x,

(1.3.6)

0

√ and { 2 sin(nπx)}∞ n=1 are the orthonormal eigenfunctions. Using (1.3.4) and (1.3.5) ˚ y): we can define the Fourier sine series representation of the kernel K(x, ˚ y) = 2 K(x,

∞  sin(nπx) sin(nπy) n=1

(nπ)2

.

(1.3.7)

16

1 Introduction Ill-Posedness of Inverse Problems

Now, to solve the Fredholm integral equation ˚ (Au)(x) :=



1

˚ y)u(y)dy = F (x), x ∈ [0, 1], K(x,

(1.3.8)

0

we use the Fourier sine series of the functions F (x) and u(x): F (x) =

∞ ∞ √  √  2 Fn sin(nπx), u(x) = 2 un sin(nπx), n=1

n=1

where Fn =

√  2

1

F (ξ ) sin(nπξ )dξ, un =

0

√  2

1

u(ξ ) sin(nπξ )dξ 0

are the Fourier coefficients. Substituting these in (1.3.8) and using (1.3.5) we deduce: ∞ ∞   un sin(nπx) = Fn sin(nπx). (nπ)2 n=1

n=1

This implies: un = (nπ)2 Fn ,

n = 1, 2, . . . .

(1.3.9)

The relationship (1.3.9) can be treated as an input-output relationship for problem (1.3.8). Thus, the Fourier series solution of the Fredholm integral equation (1.3.8) with the kernel given by (1.3.4) is the function u(x) =

∞ √  2 (nπ)2 Fn sin(nπx), x ∈ [0, 1],

(1.3.10)

n=1

if the series converges in the considered solution set C[0, 1]. However, there are very simple cases where this fails to happen, even in L2 [0, 1]. Indeed, let F (x) ≡ 1, x ∈ [0, 1]. Calculating the Fourier coefficients Fn we get: Fn =

√ [1 − (−1)n ] , n = 1, 2, 3, . . . . 2 nπ

This means the series (1.3.10) fails to converge.  The above example tells us that problem (1.3.1) or (1.3.8) may not have a solution for each function F (x) from C[0, 1].

1.3 Why a Fredholm Integral Equation of the First Kind Is an Ill-Posed Problem?

17

To understand the reason of this phenomenon, we interchange roles of u(x) and F (x) in problem (1.3.8), assuming now that F (x) is the unknown function and u(x) is the given one. Differentiating the left hand side of (1.3.8) and taking into account (1.3.4), we get d2 dx 2



1 0

  x  1 d2 ˚ − yu(y)dy + (1 − y)u(y)dy K(x, y)u(y)dy = dx 2 0 x = −u(x).

Then we obtain the following formal equation −F  (x) = u(x), x ∈ (0, 1) with respect to the unknown function F (x). Note that in terms of problem (1.3.8) this equation, in particular, implies a necessary condition for the existence of a solution in C[0, 1]: the function F (x) should belong to the space C 2 [0, 1]. Clearly, function F (x) should also satisfy the boundary conditions F (0) = F (1) = 0, as the integral equation (1.3.8) with the kernel given by formula (1.3.7) shows. Thus, if we assume in (1.3.8) that F (x) is unknown function and u(x) is the given function, then we conclude that F (x) is the solution to the boundary value problem 

−F  (x) = u(x), x ∈ (0, 1), F (0) = F (1) = 0.

(1.3.11)

˚ y) is the Green’s function for the operator −d 2 /dx 2 under the Evidently, K(x, boundary conditions (1.3.11). Let us compare now problems (1.3.8) and (1.3.11), taking into account the swapping of the functions u(x) and F (x). It follows from the above considerations that, problems (1.3.8) and (1.3.11) can be defined as inverse to each other, as stated in [80]. Then, it is natural to ask the question: which problem is the direct (i.e. original) problem, and which problem is the inverse problem? To answer this question, we need to go back to the physical model of the problem. The boundary value problem (1.3.11) is the simplest mathematical model of deflection of a string, occupying the interval [0, 1]. The Dirichlet conditions in (1.3.11) mean that the string is clamped at its end points. In this model, the function u(x), as a given right hand side of the differential equation (1.3.11), represents a given pressure, and the function F (x), as a solution of the boundary value problem (1.3.11), represents the corresponding deflection. The unique (classical) solution of the two-point boundary value problem (1.3.11) is the function 

1

F (x) =

˚ y)u(y)dy, x ∈ [0, 1], K(x,

(1.3.12)

0

˚ y) is the Green’s function defined by (1.3.4). Hence each where the kernel K(x, pressure u ∈ C[0, 1], defines uniquely the deflection function F ∈ C 2 [0, 1], F (0) =

18

1 Introduction Ill-Posedness of Inverse Problems

F (1) = 0. In other words, the boundary value problem (1.3.11) is a well-posed direct problem, with the unique solution (1.3.12). On the other hand, as we have seen above, the integral equation (1.3.8) may not have a solution for each continuous function F (x) (deflection). The physical interpretation is clear: each (admissible) pressure generates a unique deflection, but an arbitrary function cannot be regarded as a deflection. Applied to the integral equation (1.3.8) this means that in order to F (x) be a possible defection it needs, at least, to satisfy the rigid clamped boundary conditions F (0) = F (1) = 0 and to have continuous second derivative. This is a reason, in the language of the physical model, why a Fredholm integral equation of the first kind is an ill-posed problem. To finish the above analysis, now we return to the integral equation (1.3.8) with the kernel (1.3.4) and ask: what type of functions F (x) are admissible in order to get a convergent series (1.3.10)? Example 1.3.2 Existence and non-existence of a solution of the Fredholm integral equation. The solution of the integral equation (1.3.8) with the kernel (1.3.4) is the function u(x) given by series (1.3.10). First, we assume that F (x) in (1.3.8) is given by formula  x/2, 0 ≤ x ≤ 1/2, F (x) = (1.3.13) (1 − x)/2, 1/2 < x ≤ 1. Note that this function is continuous, but not continuously differentiable. Calculating the Fourier sine coefficients we get: √  √

nπ  2 2 0, n = 2k, = sin Fn = (nπ)2 2 (nπ)2 (−1)n−1 , n = 2k − 1, k = 1, 2, 3, . . . . Substituting this in (1.3.10) we obtain the series solution of problem (1.3.8): u(x) = 2

∞ 

(−1)n−1 sin [(2n − 1)πx] , x ∈ [0, 1].

(1.3.14)

n=1

This is exactly the Fourier sine series expansion of Dirac delta function δ(x − 1/2), which not only does not satisfy the above differentiability conditions, but also is not even a regular generalized function. Hence, the function (1.3.14) corresponding to F (x), given by formula (1.3.13), is not regarded as a solution of the Fredholm integral equation (1.3.8) in the above mentioned sense. Let us use in (1.3.8) more smooth input: F (x) = x(1 − x)/2, x ∈ [0, 1],

(1.3.15)

1.3 Why a Fredholm Integral Equation of the First Kind Is an Ill-Posed Problem?

19

restricting the class of admissible functions {F (x)}. Evidently, this function has continuous derivatives up to order 2 and satisfies the conditions F (0) = F (1) = 0. The Fourier sine coefficients of function (1.3.15) are Fn =

√ 2[1 − (−1)n ] 2 , n = 1, 2, 3, . . . . (nπ)3

Substituting this in (1.3.10) we obtain the series solution: u(x) =

∞ ∞ 4  sin[(2m − 1)πx] 2  [1 − (−1)n ] sin(nπx) = , x ∈ [0, 1]. π n π 2m − 1 n=1

m=1

This is the Fourier sine series expansion of the function u(x) ≡ 1, x ∈ [0, 1]. Hence, for the smooth enough function F (x) = x(1 − x)/2, the solution of the Fredholm integral equation of the first kind with the kernel (1.3.4), is the function u(x) ≡ 1, x ∈ [0, 1].  Let us return now to the last two examples and try to reformulate in terms of compact operators the above conclusion “every (admissible) pressure generates a deflection, but an arbitrary function cannot be regarded as a deflection”, obtained in terms of a physical model. The lemma below explains this conclusion, in terms of compact operators. It simply asserts that any neighborhood of an arbitrary element F ∈ R(A) from the range of a linear compact operator A might not have a preimage in H . Remember that if K(x, y) in (1.3.1) is a bounded Hilbert-Schmidt kernel, i.e. if K ∈ L2 ((0, 1) × (0, 1)), then the integral operator A : L2 (0, 1) → L2 (0, 1), defined by (1.3.2), is a compact operator. Lemma 1.3.1 Let A : D(A) ⊂ H → H˜ be a compact operator between two infinite-dimensional Hilbert spaces H and H˜ , with bounded domain D(A). Assume that f ∈ R(A) is an arbitrary element from the range R(A) ⊂ H˜ of the operator A. Then for any ε > 0 there exists an element f0 ∈ Vε (f ) := {g ∈ H˜ : g −f H˜ ≤ ε} such that f0 ∈ / R(A). Proof Let f ∈ R(A) be an arbitrary element. To prove the lemma, assume on the ˜ contrary that there exists ε0 > 0 such that Vε0 (f ) ⊂ R(A). Let {fi }∞ i=1 ⊂ H be an ∞ ˜ orthonormal basis. We define the sequence {gi }i=1 ⊂ H , with gi := f + ε0 fi /2. Evidently, {gi }∞ i=1 ⊂ Vε0 (f ), since gi − f H˜ = (ε0 /2) fi H˜ = ε0 /2 < ε0 , ˜ which means the sequence {gi }∞ i=1 is bounded in H . However, for all i, j = 1, ∞, √

gi − gj H˜ = ε0 / 2 and this sequence does not contain any Cauchy subsequence. Hence, the set Vε0 (f ) ⊂ R(A) is not a precompact. On the other hand, D(A) ⊂ H is a bounded set and A is a compact operator. Hence, as an image Vε0 (f ) := A(D0 ) of a bounded set D0 ⊂ D(A), the set Vε0 (f ) ⊂ R(A) needs to be a precompact. This contradiction completes the proof of the lemma.  

20

1 Introduction Ill-Posedness of Inverse Problems

Corollary 1.3.1 Let conditions of Lemma 1.3.1 hold. Assume, in addition, that A : D(A) ⊂ H → H˜ is a linear injective operator. Then the inverse operator A−1 : R(A) → D(A) is discontinuous, that is, the problem Au = f , u ∈ D, is ill-posed. Proof Assume on the contrary that the inverse operator A−1 : R(A) → D(A) is continuous. Then, as an image A−1 (R(A)) := D(A) (under the inverse operator A−1 ) of the compact set R(A), the set D(A) needs to be compact, while it is a bounded set.   These results give us some insights into the structure and dimension of the range R(A) of a linear compact operator. In addition to above obtained result that the problem (1.3.1) is ill-posed in sense of the third Hadamard’s condition (p3), we conclude from Corollary 1.3.1 that this problem is ill-posed also in sense of the first Hadamard’s condition (p1). The reason is that, the range R(A) of a linear compact operator is not dense everywhere, by the assertion of Lemma 1.3.1. As shown in Appendix A, the range R(A) of a linear compact operator A, defined on a Hilbert space H , is “almost finite-dimensional”, i.e. can be approximated to any given accuracy by a finite dimensional subspace in R(A). Note that, if the range R(A) of a bounded linear operator is finite dimensional, then it is a compact operator. This follows from the Heine-Borel Theorem, since the closure A(D) ⊂ H˜ of the image A(D) of a bounded set D ⊂ H is closed and bounded in the finite dimensional subspace R(A) ⊂ H˜ , so, is compact. Finally, note that Lemma 1.3.1 also asserts that if a compact operator has a bounded inverse, then H must be finite dimensional. Then it can be shown that, if H is infinite dimensional, then there is no injective compact operator from H onto H˜ . Details of these assertions are analyzed in Appendix A.

Part I

Introduction to Inverse Problems

Chapter 2

Functional Analysis Background of Ill-Posed Problems

The main objective of this chapter is to present some necessary results of functional analysis, frequently used in study of inverse problems. For simplicity, we derive these results in Hilbert spaces. Let H be a vector space over the field of real (R) or complex (C) numbers. The mapping (·, ·)H defined on H × H is called an inner product (or scalar product) of two elements of H , if the following conditions are satisfied: (i1) (u1 + u2 , v)H = (u1 , v)H + (u2 , v)H , ∀u1 , u2 , v ∈ H ; (i2) (αu, v)H = α(u, v)H , ∀α ∈ C, ∀u, v ∈ H ; (i3) (u, v)H = (v, u)H , ∀u, v ∈ H ; (i4) (u, u)H ≥ 0, ∀u ∈ H and (u, u)H = 0 iff u = 0. Through the following, we will omit the subscript H in the scalar (dot) product whenever it is clear from the text. The vector space H together with the inner product is called an inner product space or a pre-Hilbert space. The norm in a pre-Hilbert space is defined by the above introduced scalar product: u H := (u, u)1/2 , u ∈ H . Hence a pre-Hilbert space H is a normed space. If, in addition, a pre-Hilbert space is complete, it is called a Hilbert space. Thus, a Hilbert space is a Banach space, i.e. complete normed space, with a norm defined through the scalar product. A Hilbert space is called infinite dimensional (finite dimensional) if the underlying vector space is infinite dimensional (finite dimensional). The basic representative of infinite dimensional space Hilbert in the weak solution theory of PDEs is the space of square integrable functions   L2 (a, b) := u : (a, b) → R :

b

 u2 (x)dx < +∞ ,

a

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_2

23

24

2 Functional Analysis Background of Ill-Posed Problems

with the scalar product  (u, v)L2 (a,b) :=

b

u(x)v(x)dx, u, v ∈ L2 (a, b).

a

A finite dimensional analogue of this space of square-summable sequences in Rn :   n  2 n 2 l := x := (x1 , x2 , . . . , xn ) ∈ R : xk < +∞ , k=1

with the scalar product (x, y)l 2 :=

n 

xk yk , x, y ∈ Rn .

k=1

2.1 Best Approximation and Orthogonal Projection Let H be an inner product space. The elements u, v ∈ H are called orthogonal, if (u, v) = 0. This property is denoted by u ⊥ v. Evidently, 0 ⊥ v, for any v ∈ H , since (0, v) = 0, ∀v ∈ H . Let U ⊂ H be a non-empty subset and u ∈ H be an arbitrary element. If (u, v) = 0 for all v ∈ U , then the element u ∈ H is called orthogonal to the subset U and is denoted by u ⊥ U . The set of all elements u ∈ H orthogonal to U ⊂ H is called an orthogonal complement of U and is denoted by U ⊥: U ⊥ := {u ∈ H : (u, v) = 0, ∀v ∈ U }.

(2.1.1)

The subsets U, V ⊂ H are called orthogonal subsets of H if (u, v) = 0 for all u ∈ U and v ∈ V . This property is denoted by U ⊥ V . Evidently, if U ⊥ V , then U ∩ V = {0}. Theorem 2.1.1 Let U ⊂ H be a subset of an inner product space H . Then an orthogonal complement U ⊥ of U is a closed subspace of H . Moreover, the following properties are satisfied: (p1) U ∩ U ⊥ ⊂ {0} and U ∩ U ⊥ = {0} iff U is a subspace;  ⊥  ⊥ (p2) U ⊂ U ⊥ =: U ⊥⊥ , U ⊥ := {u ∈ H : (u, v) = 0, ∀v ∈ U ⊥ }; (p3) If U1 ⊂ U2 ⊂ H, then U2⊥ ⊂ U1⊥ . Proof Evidently, αu + βv ∈ U ⊥ , for all α, β ∈ C and u, v ∈ U ⊥ , by the above definition of the scalar product: (αu + βv, w) = α(u, w) + β(v, w) = 0, for all

2.1 Best Approximation and Orthogonal Projection

25

w ∈ U . This implies that U ⊥ is a subspace of H . It is easy to prove, by using continuity of the scalar product, that if {un } ⊂ U ⊥ and un → u, as n → ∞, then u ∈ U ⊥ . So, U ⊥ is closed. To prove (p1), we assume that there exists an element u ∈ U ∩ U ⊥ . Then, by definition (2.1.1) of U ⊥ , (u, u) = 0, which implies u = 0. Hence U ∩ U ⊥ ⊂ {0}. If, in addition, U is a subspace of H , then 0 ∈ U . This yields: U ∩ U ⊥ = {0}. To prove (p2) we assume in contrary, that there exists an element u ∈ U such that  ⊥ u∈ / U ⊥ . This means an existence of such an element v ∈ U ⊥ that (u, v) = 0. On the other hand, for all u ∈ U and v ∈ U ⊥ , we have (u, v) = 0, which is a  ⊥ contradiction. Hence U ⊂ U ⊥ . Finally, to prove (p3), let v ∈ U2⊥ be an arbitrary element. Then (u, v) = 0, for all u ∈ U2 . Since U1 ⊂ U2 , this holds for all u ∈ U1 as well. Hence for any element v ∈ U2⊥ , the condition (u, v) = 0 holds for all u ∈ U1 . By the definition, this implies v ∈ U1⊥ , which completes the proof.   Definition 2.1.1 (Best Approximation) Let U ⊂ H be a subset of an inner product space H and v ∈ H be a given element. If

v − u H = inf v − w H , w∈U

(2.1.2)

then the element u ∈ U is called the best approximation to the element v ∈ H with respect to the set U ⊂ H . The right hand side of (2.1.2) is a distance between a given element v ∈ H and the set U ⊂ H . Hence, the best approximation is an element with the smallest distance to the set U ⊂ H . This notion plays a crucial role in inverse problems theory and applications, since any measured output can only be given with some measurement error. As a result of this, the exact equality in any inverse problem ?u = F is never achieved. First we prove that if U ⊂ H is a closed linear subspace, then the best approximation is determined uniquely. To this end, we use the main theorem on quadratic variational problems and its consequence, called perpendicular principle (Ch. 2.4, [155]). Theorem 2.1.2 Let a : H × H → R be a symmetric, bounded, strongly positive bilinear form on a real Hilbert space H , and b : H → R be a linear bounded functional on H . Then (i) The minimum problem J (v) = min J (w), J (w) := w∈H

has a unique solution v ∈ H .

1 a(w, w) − b(w) 2

(2.1.3)

26

2 Functional Analysis Background of Ill-Posed Problems

(ii) This minimum problem is equivalent to the following variational problem: Find v ∈ H such that a(v, w) = b(w), ∀w ∈ H.

(2.1.4)

This theorem allows to prove that in a closed linear subspace U of a real or complex Hilbert space H , the best approximation is uniquely determined. Theorem 2.1.3 Let U be a closed linear subspace of a Hilbert space H and v ∈ H be a given element.Then the best approximation problem (2.1.2) has a unique solution u ∈ U . Moreover v − u ∈ U ⊥ . Proof We rewrite the norm v − u H as follows:

v − u H := (v, v) − (v, u) − (u, v) + (u, u)   1 = a(v, v) + 2 a(u, u) − b(u) , 2

(2.1.5)

where a(u, w) := Re(u, w), b(u) :=

1 [(v, u) + (u, v)] = Re(v, u). 2

The right hand side of (2.1.5) shows that the best approximation problem (2.1.2) is equivalent to the variational problem (2.1.3) with the above defined bilinear and linear forms. As a consequence, from Theorem 2.1.2 it follows that if H is a real Hilbert space, then there exists a unique best approximation v ∈ U to the element u ∈ H. If H is a complex Hilbert space, then we can introduce the new scalar product (v, w)∗ := Re(v, w), v, w ∈ H and again apply Theorem 2.1.2. We prove now that if u ∈ U is the best approximation to the element v ∈ H , then v − u ⊥ U . Indeed, it follows from (2.1.2) that

v − u 2H ≤ v − (u + λw) 2H , ∀λ ∈ C, w ∈ U. This implies, (v − u, v − u) ≤ (v − u, v − u) − λ(v − u, w) − λ(w, v − u) + |λ|2 (w, w) or 0 ≤ −λ(v − u, w) − λ(w, v − u) + |λ|2 (w, w). Assume that v − u = 0, w = 0. Then taking λ = (w, v − u)/ w 2 we obtain: 0 ≤ −|(v − u, w)|2 , which means that (v − u, w) = 0, ∀w ∈ U . Note that this remains true also if v − u = 0.  

2.1 Best Approximation and Orthogonal Projection

27

Corollary 2.1.1 (Orthogonal Decomposition) Let U be a closed linear subspace of a Hilbert space H . Then there exists a unique decomposition of a given arbitrary element v ∈ H of the form v = u + w, u ∈ U, w ∈ U ⊥ .

(2.1.6)

Existence of this orthogonal decomposition follows from Theorem 2.1.3. We prove the uniqueness. Assume, in contrary, that there exists another decomposition of v ∈ H such that v = u1 + w1 , u1 ∈ U, w1 ∈ U ⊥ . Since U is a linear subspace of H , we have u − u1 ∈ U , w − w1 ∈ U ⊥ , and (u − u1 ) + (w − w1 ) = 0. Multiplying scalarly both sides by u − u1 we get (u − u1 , u − u1 ) + (w − w1 , u − u1 ) = 0, which implies u = u1 , since the second term is zero due to w −w1 ∈ U ⊥ . Similarly, we conclude that w = w1 . This completes the proof.  The orthogonal decomposition (2.1.6) can also be rewritten in terms of the subspaces U and U ⊥ as follows: H = U ⊕ U⊥ . Example 2.1.1 Let H := L2 (−1, 1), U := {u ∈ L2 (−1, 1) : u(−x) = u(x), a.e. in (−1, 1)} be the set of even functions and U ⊥ := {u ∈ L2 (−1, 1) : u(−x) = −u(x), a.e. in (−1, 1)} be the set of odd functions. Then L2 (−1, 1) = U ⊕ U ⊥ . Remark that for any v ∈ L2 (−1, 1), v(x) = [v(x) + v(−x)]/2 + [v(x) − v(−x)]/2.  Corollary 2.1.1 shows that there exists a mapping which uniquely transforms each element v ∈ H to the element u of a closed linear subspace U of a Hilbert space H . This assertion is called Orthogonal Projection Theorem. Definition 2.1.2 (Orthogonal Projection) The operator P : H →  U , with P v = u, in the decomposition (2.1.6) which maps each element v ∈ H to the element u ∈ U is called the projection operator or orthogonal projection. Using this definition, we may rewrite the best approximation problem (2.1.2) as follows:

v − P v H = inf v − w H . w∈U

28

2 Functional Analysis Background of Ill-Posed Problems

We denote by N (P ) := {v ∈ H : P v = 0} and R(P ) := {P v : v ∈ H } the null-space and the range of the projection operator P , respectively. The theorem below shows that the orthogonal projection P : H → U is a linear continuous self-adjoint operator. Theorem 2.1.4 The orthogonal projection P : H → U defined from Hilbert space H onto the closed subspace U ⊂ H is a linear continuous self-adjoint operator with P 2 = P and P = 1, for U = {0}. Conversely, if P : H → H is a linear continuous self-adjoint operator with P 2 = P , then it defines an orthogonal projection from H onto the closed subspace R(P ) ⊂ H . Proof It follows from (2.1.6) that

v 2 := u + w 2 = u 2 + w 2 , since (u, w) = 0, by u ∈ U and w ∈ U ⊥ . This implies P v ≤ v , P v := u for all v ∈ H . In particular, for v ∈ U ⊂ H we have P v = v, which means P = 1. We prove now that P is self-adjoint. Let vk = uk + wk , uk ∈ U , wk ∈ U ⊥ , k = 1, 2. Multiplying both sides of v1 = u1 + w1 by u2 and both sides of v2 = u2 + w2 by u1 , then taking into account (uk , vm ) = 0, k, m = 1, 2, we conclude (v1 , u2 ) = (u1 , u2 ), (v2 , u1 ) = (u2 , u1 ). This implies (v1 , u2 ) = (v2 , u1 ). In view of the definition P vk = uk we deduce that (v1 , P v2 ) = (P v1 , v2 ), ∀v1 , v2 ∈ H, i.e. the projection operator P is self-adjoint. Assuming now v = u ∈ U ⊂ H in (2.1.6) we have: u = u + 0, where 0 ∈ U ⊥ . Then P u = u, and for any v ∈ H we have P 2 v = P (P v) = P u = u = P v. Hence P 2 v = P v, for all v ∈ H . We prove now the second part of the theorem. Evidently, R(P ) := {P v : v ∈ H } is a linear subspace. We prove that the range of the projection operator R(P ) is closed. Indeed, let {un } ⊂ R(P ), such that un → u, as n → ∞. Then there exists such an element vn ∈ H that un = P vn . Together with the property P 2 vn = P vn this implies: P un = P 2 vn = P vn = un . Hence, P un = un for all un ∈ R(P ). Letting to the limit and using the continuity of the operator P , we obtain: u = lim un = lim P un = P u, n→∞

n→∞

i.e. P u = u, which means u ∈ R(P ). Thus R(P ) is a linear closed subspace of H and all its elements are fixed points of the operator P , that is, P u = u for all u ∈ R(P ). On the other hand, P is a self-adjoint operator with P 2 = P , by the assumption. Then for any v ∈ H , (P v, (I − P )v) = (P v, v) − (P v, P v) = (P v, v) − (P 2 v, v) = 0.

2.1 Best Approximation and Orthogonal Projection

29

Since v ∈ H is an arbitrary element, the orthogonality (P v, (I − P )v) = 0 means that (I − P )v ∈ R(P )⊥ . Then the identity v = P v + (I − P )v, ∀v ∈ H with Corollary 2.1.1 implies that P is a projection operator, since P v ∈ R(P ) ⊂ H   and (I − P )v ∈ R(P )⊥ . Remark 2.1.1 Based on the properties of the orthogonal projection P : H → H , we conclude that the operator I − P : H → H is also an orthogonal projection. Indeed, (I − P )2 = I − 2P + P 2 = I − P . Remark 2.1.2 Let us define the set M := {u ∈ H : P u = u}, i.e. the set of fixed points of the orthogonal projection P . It follows from the proof of Theorem 2.1.4 that M = R(P ). Similarly, R(I − P ) = N (P ). Some other useful properties of the projection operator P : H → H are summarized in the following corollary. Corollary 2.1.2 Let P : H → H be an orthogonal projection defined on a Hilbert space H . Then the following assertions hold: (p1) N (P ) and R(P ) are closed linear subspaces of H . (p2) Each element v ∈ H can be written uniquely as the following decomposition: v = u + w, u ∈ R(P ), w ∈ N (P ).

(2.1.7)

v 2 = u 2 + w 2 .

(2.1.8)

Moreover,

(p3)

N (P ) = R(P )⊥ and R(P ) = N (P )⊥ .

Proof The assertions (p1) − (p2) follow from Corollary 2.1.1 and Theorem 2.1.4. We prove (p3). Evidently, N (P ) ⊥ R(P ) and N (P ) ⊂ R(P )⊥ . Hence to prove the first part of the assertion (p3) we need to show that R(P )⊥ ⊂ N (P ). Let v ∈ R(P )⊥ ⊂ H . Then, there exist such elements u ∈ R(P ), w ∈ N (P ) that v = u + w, according to (2.1.7). Multiplying both sides by an arbitrary element v˜ ∈ R(P ) we obtain: (v, v) ˜ = (u, v) ˜ + (w, v). ˜ The left hand side is zero, since v ∈ R(P )⊥ . Also, (w, v) ˜ = 0, by N (P ) ⊥ R(P ). Thus, (u, v) ˜ = 0, for all v˜ ∈ R(P ). But u ∈ R(P ). This implies u = 0, and as a result, v = 0 + w, where w ∈ N (P ). Hence v ∈ N (P ). The second part of the assertion (p3) can be proved similarly.   Let us explain the above results with the following example. Example 2.1.2 Fourier Series and Orthogonal Projection.

30

2 Functional Analysis Background of Ill-Posed Problems

Let {ϕn }∞ n=1 be an orthonormal basis of an infinite-dimensional real Hilbert space H . Then any element u ∈ H can be written uniquely as the following convergent Fourier series: u=

∞ N ∞    (u, ϕn )ϕn ≡ (u, ϕn )ϕn + (u, ϕn )ϕn . n=1

n=1

n=N+1

Let us consider the finite system {ϕn }N n=1 which forms a basis for the finitedimensional Hilbert subspace HN ⊂ H . We define the operator P : H → HN as follows: P u :=

N  (u, ϕn )ϕn , u ∈ H.

(2.1.9)

n=1

Evidently, P is a linear bounded operator. Moreover, P 2 = P . Indeed,  P u := P 2

N  (u, ϕn )ϕn

 =

n=1

N N   m=1

 (u, ϕn )ϕn , ϕm ϕm

n=1

=

N 

(u, ϕm )ϕm = P u,

m=1

by (ϕn , ϕm ) = δn,m . Thus, P : H → HN , defined by (2.1.9), is a projection operator from the infinite-dimensional Hilbert space H onto the finite-dimensional Hilbert space HN ⊂ H , with R(P ) = HN . To show the orthogonality of R(P ) and N (P ), let u ∈ R(P ) and v ∈ N (P ) be arbitrary elements. Then P u = u and 

N  (v, u) = (v, P u) = v, (u, ϕn )ϕn n=1

=



 =

N  (u, ϕn )(v, ϕn ) n=1

N 



(v, ϕn )ϕn , u = (P v, u).

n=1

But P v = 0, due to v ∈ N (P ). Hence (v, u) = 0, for all u ∈ R(P ) and v ∈ N (P ), which implies R(P ) ⊥ N (P ). Now we show the projection error, defined as u − P u :=

∞  n=N+1

(u, ϕn )ϕn ,

2.2 Range and Null-Space of Adjoint Operators

31

is orthogonal to HN . Let v ∈ HN be any element. Then  (u − P u, v) :=

∞ 

 (u, ϕn )ϕn , v

n=N+1

=

∞ 

(u, ϕn )(ϕn , v)

n=N+1

and the right hand side tends to zero as N → ∞, due to the convergent Fourier series. Thus u − P u ⊥ HN for all v ∈ HN . Finally, we use the above result to estimate the approximation error u − v HN , where v ∈ HN is an arbitrary element. We have:

u − v 2HN := (u − v, u − v) = (u − P u + P u − v, u − P u + P u − v) = u − P u 2HN + v − P u 2HN , by (2.1.8). The right hand side has minimum value when v = P u, i.e. v ∈ HN is a projection of u ∈ H . In this case we obtain: inf u − v HN = u − P u HN ,

v∈HN

which is the best approximation problem.



2.2 Range and Null-Space of Adjoint Operators Relationships between the null-spaces and ranges of a linear operator as well as its adjoint play an important role in inverse problems. The results given below show that the range of a linear bounded operator can be derived through the null-space of its adjoint. Note that for the linear bounded operator A : H → H˜ , defined between Hilbert spaces H and H˜ , the adjoint operator A∗ : H˜ → H is defined as follows: (Au, v)H˜ = (u, A∗ v)H , ∀u ∈ H, v ∈ H˜ . Theorem 2.2.1 Let A : H → H˜ be a linear bounded operator, defined between Hilbert spaces H and H˜ , and A∗ : H˜ → H be its adjoint. Then (p1) (p2)

R(A)⊥ = N (A∗ ); R(A) = N (A∗ )⊥ ,

where R(A) and N (A∗ ) are the range and null-space of the operators A and A∗ , correspondingly. Proof Let v ∈ N (A∗ ). Then A∗ v = 0, and for all u ∈ H we have: 0 = (u, A∗ v)H = (Au, v)H˜ .

32

2 Functional Analysis Background of Ill-Posed Problems

This implies that v ∈ R(A)⊥ , i.e. N (A∗ ) ⊂ R(A)⊥ . Now suppose v ∈ R(A)⊥ . Then (Au, v)H˜ = 0, for all u ∈ H . Hence 0 = (Au, v)H˜ = (u, A∗ v)H , for all u ∈ H , which means v ∈ N (A∗ ). This implies R(A)⊥ ⊂ N (A∗ ). To prove (p2) let us assume first that v ∈ R(A) is an arbitrary element. Then there exists such a sequence {vn } ∈ R(A) that vn = Aun and lim vn = v. n→∞

Assuming w ∈ N (A∗ ) we conclude that A∗ w = 0, so (vn , w)H˜ = (Aun , w)H˜ = (un , A∗ w)H = 0. Hence |(v, w)| ≤ |(v − vn , w)| + |(vn , w)| ≤ (v − vn , w) w . The right hand side tends to zero as n → ∞, which implies (v, w) = 0, for all w ∈ N (A∗ ), i.e. v ∈ N (A∗ )⊥ . Therefore R(A) ⊂ N (A∗ )⊥ . To prove N (A∗ )⊥ ⊂ R(A), we need to prove that if v ∈ / R(A), then v ∈ / N (A∗ )⊥ . Since R(A) is a closed subspace of the Hilbert space H , by Corollary 2.1.1 there exists a unique decomposition of the above defined element v ∈ H : v = v0 + w0 , v0 ∈ R(A), w0 ∈ R(A)⊥ , with v0 := P v. Then (v, w0 ) := (v0 + w0 , w0 ) = w0 2 = 0. But by (p1), w0 ∈ N (A∗ ), so (v, w0 ) = 0, which means that v ∈ / N (A∗ )⊥ . This completes the proof.   Lemma 2.2.1 Let A : H → H˜ be a bounded linear operator. Then (A∗ )∗ = A and

A 2 = A∗ 2 = AA∗ = A∗ A . Proof We can easily show that (A∗ )∗ = A. Indeed, for all u ∈ H , v ∈ H˜ ,  ∗ (v, A∗ u)H˜ = (A∗ v, u)H = (u, A∗ v)H = (Au, v)H˜ = (v, Au)H˜ . Further, it follows from the definition A∗ v 2 := (A∗ v, A∗ v)H , v ∈ H˜ , that

A∗ v 2 = (AA∗ v, v)H˜ ≤ A

A∗ v

v , and hence A∗ v ≤ A

v , which implies boundedness of the adjoint operator: A∗ ≤ A . In the same way we can deduce A ≤ A∗ , interchanging the roles of the operators A and A∗ . Therefore,

A∗ = A . To prove the second part of the lemma, we again use the definition Au 2 := (Au, Au)H˜ , u ∈ H . Then, Au 2 = (A∗ Au, u)H ≤ A∗ Au u , and we get

A 2 ≤ A∗ A . On the other hand, A∗ A ≤ A∗

A = A 2 , since A∗ =

A . Thus A 2 = A∗ A .  

2.2 Range and Null-Space of Adjoint Operators

33

Corollary 2.2.1 Let conditions of Theorem 2.2.1 hold. Then (c1) (c2)

N (A∗ ) = N (AA∗ ); R(A) = R(AA∗ ).

Proof Let v ∈ N (A∗ ). Then A∗ v = 0 and hence AA∗ v = 0, which implies v ∈ N (AA∗ ), i.e. N (A∗ ) ⊂ N (AA∗ ). Suppose now v ∈ N (AA∗ ). Then AA∗ v = 0 and A∗ v 2H := (A∗ v, A∗ v)H = (AA∗ v, v)H˜ = 0. This implies A∗ v = 0, i.e. v ∈ N (A∗ ), which completes the proof of (c1). To prove (c2) we use the formula (AA∗ )∗ := A∗∗ A∗ = AA∗ and the second relationship (p2) in Theorem 2.2.1, replacing here A by AA∗ . We have R(AA∗ ) = N (AA∗ )⊥ . Taking into account here (c1) we conclude R(AA∗ ) = N (A∗ )⊥ . With the relationship (p1) in Theorem 2.2.1, this completes the proof.   Remark that if A is a bounded linear operator defined on a Hilbert space H , then AA∗ and A∗ A are positive. Now we briefly show a crucial role of adjoint operators in studying the solvability of linear equations. Let A : H → H˜ be a bounded linear operator. Consider the linear operator equation Au = f, u ∈ H, f ∈ H˜ .

(2.2.1)

Denote by v ∈ H˜ a solution of the homogeneous adjoint equation A∗ v = 0. Then we have: (f, v)H˜ := (Au, v)H˜ = (u, A∗ v)H = 0. Hence (f, v) = 0 for all v ∈ N (A∗ ), and by H˜ = N (A∗ )⊥ ⊕ N (A∗ ), this implies: f ∈ N (A∗ )⊥ . On the other hand, Fredholm alternative asserts that Eq. (2.2.1) has a (non-unique) solution if and only if f ⊥ v for each solution v ∈ H˜ of the homogeneous adjoint equation A∗ v = 0. This leads to the following result. Proposition 2.2.1 Let A : H → H˜ be a bounded linear operator on a Hilbert space H . Then a necessary condition for existence of a solution u ∈ H of Eq. (2.2.1) is the condition f ∈ N (A∗ )⊥ .

(2.2.2)

Using Theorem 2.2.1 we can write H as the orthogonal (direct) sum H = R(A) ⊕ N (A∗ ).

(2.2.3)

If the range R(A) of A is closed in H , that is, if R(A) = R(A), then using (2.2.3) and Proposition 2.2.1, we obtain the following necessary and sufficient condition for the solvability of Eq. (2.2.1).

34

2 Functional Analysis Background of Ill-Posed Problems

Theorem 2.2.2 Let A : H → H˜ be a bounded linear operator with closed range. Then Eq. (2.2.1) has a solution u ∈ H if and only if condition (2.2.2) holds. This theorem provides a useful tool for proving existence of a solution of the closed range operator equation (2.2.1) through the null-space of the adjoint operator.

2.3 Moore-Penrose Generalized Inverse Let A : H → H˜ be a bounded linear operator between the real Hilbert spaces H and H˜ . Consider the operator equation (2.2.1). Evidently, a solution of (2.2.1) exists if and only if f ∈ R(A) ⊂ H˜ , i.e. if the first condition (i1) of Hadamard’s Definition 1.1.1 holds. Assume now that f ∈ H˜ does not belong to the range R(A) of the operator A which usually appears in applications. It is natural to extend the notion of solution for this case, looking for an approximate (or generalized) solution of (2.2.1) which satisfies this equation as well as possible. For this aim we introduce the residual f − Au H˜ and then look for an element u ∈ H , as in Definition 2.1.1, which minimizes this norm:

f − Au H˜ = inf f − Av H˜ . v∈H

(2.3.1)

The minimum problem (2.3.1) is called a least squares problem and accordingly, the best approximation u ∈ H is called a least squares solution to (2.2.1). Let us consider first the minimum problem (2.3.1) from differential calculus viewpoint. Introduce the functional J (u) =

1

Au − f 2H˜ , u ∈ H. 2

Using the identity   1 J (u + h) − J (u) = A∗ (Au − f ), h + Au 2H , ∀h ∈ H, 2 we obtain the Fréchet differential 

   J  (u), h = A∗ (Au − f ), h , h ∈ H

of this functional. Hence the least squares  u ∈ H of the operator equation  solution (2.2.1) is defined from the condition J  (v), h = 0, for all h ∈ H , as follows: (A∗ (Au − f ), h) = 0, i.e. as a solution of the equation A∗ Au = A∗ f.

(2.3.2)

2.3 Moore-Penrose Generalized Inverse

35

This shows that least squares problem (2.3.1) is equivalent to Eq. (2.3.2) with the formal solution  −1 ∗ u = A∗ A A f, f ∈ H˜ .

(2.3.3)

Equation (2.3.2) is called the normal equation. The normal equation plays an important role in studying ill-posed problems as we will see in next sections. First of all, remark that the operator A in (2.2.1) may not be injective, which means non-uniqueness in view of Hadamard’s definition. The first important property of the normal equation (2.3.2) is that the operator A∗ A is injective on the range R(A∗ ) of the adjoint operator A∗ , even if the bounded linear operator A is not injective. For this reason, the normal equation is the most appropriate one to obtain a least squares solution of an inverse problem. Lemma 2.3.1 Let A : H → H˜ be a bounded linear operator and H , H˜ Hilbert spaces. Then the operator A∗ A : R(A∗ ) ⊂ H → H is injective. Proof Note, first of all, that R(A∗ ) = R(A∗ A), as it follows from Corollary 2.2.1 (replacing A by A∗ ). Let u ∈ R(A∗ ) be such an element that A∗ Au = 0. Then Au ∈ N (A∗ ), by definition of the null-space. But N (A∗ ) = R(A)⊥ , due to Theorem 2.2.1. On the other hand, Au ∈ R(A). Thus, Au ∈ R(A) ∩ R(A)⊥ which implies Au = 0. This in turn means that u ∈ N (A) = R(A∗ )⊥ . With the above assumption u ∈ R(A∗ ), we conclude that u ∈ R(A∗ ) ∩ R(A∗ )⊥ , i.e. u = 0. Therefore A∗ Au = 0 implies u = 0, which proves the injectivity of the operator A∗ A.   Let us explain now an interpretation of the “inverse operator” (A∗ A)−1 A∗ in (2.3.3), in view of the orthogonal projection. As noted above, f ∈ H˜ may not belong to the range R(A) of the operator A. So, we assume that f ∈ H˜ \ R(A) is an arbitrary element and try to construct a unique linear extension of an “inverse operator” from R(A) to the subspace R(A) ⊕ R(A)⊥ . Since the closure R(A) of the range R(A) is a closed subspace ⊥ of H˜ , by Corollary 2.1.2, H˜ = R(A) ⊕ R(A)⊥ (note that R(A) = R(A)⊥ = R(A)⊥ ). Hence R(A)⊕R(A)⊥ is dense in H˜ . By the same corollary, the projection Pf of the arbitrary element f ∈ R(A) ⊕ R(A)⊥ is in R(A). This means that there exists such an element u ∈ H that Au = Pf, f ∈ R(A) ⊕ R(A)⊥ .

(2.3.4)

By the Orthogonal Decomposition, the element Au being the projection of f onto R(A), is an element of R(A). Furthermore, the element u ∈ H is a least squares solution to (2.2.1), i.e. is a solution of the minimum problem (2.3.1). This implies that a least squares solution u ∈ H of (2.2.1) exists if and only if f is an element of the dense in H˜ subspace R(A) ⊕ R(A)⊥ .

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2 Functional Analysis Background of Ill-Posed Problems

On the other hand, for each element f ∈ R(A) ⊕ R(A)⊥ the following (unique) decomposition holds: f = Pf + h, Pf ∈ R(A), h ∈ R(A)⊥ .

(2.3.5)

Hence for each projection Pf ∈ R(A) we have f − Pf ∈ R(A)⊥ . Taking into account (2.3.4), we conclude from (2.3.5) that f − Au ∈ R(A)⊥ .

(2.3.6)

But R(A)⊥ = N (A∗ ), by Theorem 2.2.1. Hence f − Au ∈ N (A∗ ).

(2.3.7)

By definition of the null-space, (2.3.7) implies that A∗ (f − Au) = 0. Thus, again we arrive at the same result: u ∈ H satisfies the normal equation (2.3.2). Therefore we have constructed a mapping A† from R(A)⊕R(A)⊥ into H , which associates each element f ∈ R(A) ⊕ R(A)⊥ to the least squares solution u ∈ H of the operator equation (2.2.1). Furthermore, the domain D(A† ) := R(A) ⊕ R(A)⊥ of this mapping is obtained in a natural way. This mapping is called the Moore-Penrose (generalized) inverse of the bounded linear operator A and is denoted by A† : A† : R(A) ⊕ R(A)⊥ → H.

(2.3.8)

Evidently, the generalized inverse is a densely defined linear operator, that is, D(A† ) = H˜ , since a least squares solution exists only if f is an element of the dense in H˜ subspace R(A) ⊕ R(A)⊥ . To complete this definition, let us answer the question: the operator A† is an inverse of which operator? First of all, the normal equation (2.3.2) shows that a least squares solution exists if and only if N (A∗ A) = {0}, or equivalently, N (A) = 0, due to N (A∗ A) = N (A), by Corollary 2.2.1. Since A∗ A : H → H is a self-adjoint operator, it follows from Theorem 2.2.1 and Corollary 2.2.1 that H := R(A∗ A) ⊕ R(A∗ A)⊥ = R(A∗ ) ⊕ N (A∗ A)⊥ = N (A)⊥ ⊕ N (A)

(2.3.9)

The last line of decompositions (2.3.9) shows that to ensure the existence, we need to restrict the domain of the linear operator A : H → H˜ from D(A) to N (A)⊥ . By this way, we define this operator as follows: ˚ : N (A)⊥ ⊂ H → R(A) ⊂ H˜ . ˚ := A|N (A)⊥ , A A

2.3 Moore-Penrose Generalized Inverse

37

˚ = {0} and R(A) ˚ = R(A). Therefore, It follows from this construction that N (A) the inverse operator ˚−1 : R(A) ⊂ H˜ → N (A)⊥ ⊂ H A

(2.3.10)

˚−1 ) of this inverse operator is in H˜ and does not exists. However, the range R(A contain the elements f ∈ H˜ \ R(A). For this reason, at the second stage of the above construction, we extended this range from R(A) to R(A) ⊕ R(A)⊥ , in order include those elements which may not belong to R(A). Thus, following to [37], we can define the Moore-Penrose inverse A† as follows. Definition 2.3.1 (Moore-Penrose Inverse) The Moore-Penrose (generalized) inverse of a bounded linear operator A is the unique linear extension of the inverse operator (2.3.10) from R(A) to R(A) ⊕ R(A)⊥ : A† : R(A) ⊕ R(A)⊥ → N (A)⊥ ⊂ H,

(2.3.11)

N (A† ) = R(A)⊥ .

(2.3.12)

with

The requirement (2.3.12) in this definition is due to (2.3.6) and (2.3.7). This requirement implies, in particular, that the Moore-Penrose inverse A† is a linear operator. Corollary 2.3.1 Let f ∈ D(A† ). Then u ∈ H is a least squares solution of the operator equation Au = f if and only if it is a solution of the normal equation (2.3.2). Proof It follows from (2.3.1) that u ∈ H is a least squares solution of Au = f if and only if Au is the closest element to f in R(A). The last assertion is equivalent to (2.3.6), i.e. f − Au ∈ R(A)⊥ . By Theorem 2.2.1, R(A)⊥ = N (A∗ ). Hence f − Au ∈ N (A∗ ), which means A∗ (f − Au) = 0 or A∗ Au = A∗ f .   The following theorem shows that the Moore-Penrose inverse of a bounded linear operator is a closed operator. Remark that a linear operator L : H1 → H2 is closed if and only if for any sequence {un } ⊂ D(L), satisfying lim un = u, and

n→∞

lim Lun = v,

n→∞

the conditions hold: u ∈ D(L), and v = Lu.

(2.3.13)

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2 Functional Analysis Background of Ill-Posed Problems

Theorem 2.3.1 Let A : H → H˜ be a linear bounded operator from the Hilbert spaces H into H˜ . Then the Moore-Penrose inverse A† , defined by (2.3.11) and (2.3.12), is a closed operator. Proof Let {fn } ⊂ D(A† ), n = 1, 2, 3, . . ., fn → f , and A† fn → u, as n → ∞. Denote by un := A† fn the unique solution of the normal equation (2.3.2) for each n, that is, A∗ Aun = A∗ fn , un ∈ N (A)⊥ . Since N (A)⊥ is closed, {un } ⊂ N (A)⊥ and un → u, as n → ∞, which implies u ∈ N (A⊥ ), i.e. u ∈ R(A† ), by (2.3.11). Hence, the first condition of (2.3.13) holds. Now we prove that u = A† f . Due to the continuity of the operators A∗ A and A∗ we have: A∗ Aun → A∗ Au, and A∗ Aun = A∗ fn → A∗ f, as n → ∞. The left hand sides are equal, so A∗ Au = A∗ f , i.e. u ∈ N (A)⊥ is the solution of the normal equation (2.3.3). By Corollary 2.3.1, u = A† f . This completes the proof.   Corollary 2.3.2 Let conditions of Theorem 2.3.1 hold. Then the Moore-Penrose inverse A† is continuous if and only if R(A) is closed. Proof Let A† , defined by (2.3.11) and (2.3.12), be a continuous operator. Assume that {fn } ⊂ R(A) be a convergent sequence: fn → f , as n → ∞. We need to prove that f ∈ R(A). Denote by un := A† fn . Then un ∈ N (A)⊥ , for all n = 1, 2, 3, . . .. Since A† is continuous and N (A)⊥ is closed we conclude: u := lim un = lim A† fn = A† f, and u ∈ N (A)⊥ . n→∞

n→∞

On the other hand, Aun = fn and A : H → H˜ is a continuous operator, so Au = f , where f is the above defined limit of the sequence {fn } ⊂ R(A). But the element Au, being the projection of f onto R(A), is an element of R(A), by (2.3.4). Hence, f ∈ R(A), which implies that R(A) is closed. To prove the second part of the corollary, we assume now R(A) is closed. By definition (2.3.11), D(A† ) := R(A) ⊕ R(A)⊥ , which implies D(A† ) is closed, since the orthogonal complement R(A)⊥ is a closed subspace. As a consequence, the graph GA† := {(f, A† f : f ∈ D(A† )} of the operator A† is closed. Then, as a closed graph linear operator, A† is continuous.   Remember that in the case when A is a linear compact operator, the range R(A) is closed if and only if it is finite-dimensional. Remark, finally, that the notion of generalized inverse has been introduced by E. H. Moore and R. Penrose [99, 120, 121]. For ill-posed problems this very useful concept has been developed in [37, 48].

2.4 Singular Value Decomposition

39

2.4 Singular Value Decomposition As we have seen already in the introduction, inverse problems with compact operators are a challenging case. Most inverse problems related to differential equations are represented by these operators. Indeed, all input-output operators corresponding to these inverse problems, are compact operators. Hence, compactness of the operator A is a main source of ill-posedness of the operator equation (2.2.1) and our interest will be directed towards the case when A : H → H˜ in (2.2.1) is a linear compact operator. If A is a self-adjoint compact operator, i.e. if for all u ∈ H and v ∈ H˜ , (Au, v) = (u, Av), then we may use the spectral representation [38] Au =

∞ 

λn (u, un )un , ∀u ∈ H,

(2.4.1)

n=1

where λn , n = 1, 2, 3, . . . are nonzero real eigenvalues (repeated according to its multiplicity) and {un } ⊂ H is the complete set of corresponding orthonormal eigenvectors un . The set {λn , un }, consisting of all pairs of nonzero eigenvalues and corresponding eigenvectors, is defined an eigensystem of the self-adjoint operator A. It is also known from the spectral theory of self-adjoint compact operators that λn → 0, as n → ∞. If dimR(A) = ∞ then for any ε > 0 the index set {n ∈ N : |λn | ≥ ε} is finite. Here and below N is the set of natural numbers. If the range R(A) of a compact operator is finite, λn = 0, for all n > dimR(A). However, if A is not-self-adjoint, there are no eigenvalues, hence no eigensystem. In this case, the notion singular system replaces the eigensystem. To describe this system we use the operators A∗ A and AA∗ . Both A∗ A : H → H and AA∗ : H˜ → H˜ are compact self-adjoint nonnegative operators. We denote the eigensystem of the self-adjoint operator A∗ A by {μn , un }. Then A∗ Aun = μn un , for all un ∈ H , which implies (A∗ Aun , un ) = μn (un , un ). Hence Aun 2˜ = μn un 2H ≥ 0, H which means all nonzero eigenvalues are positive: μn > 0, n ∈ N, where N := {n ∈ N : μn = 0} (at most countable) index set of positive eigenvalues. We denote the square roots of the positive eigenvalues μn of the self-adjoint operator √ A∗ A : H → H by σn := μn , n ∈ N. Below we assume that these eigenvalues are ordered as follows: μ1 ≥ μ2 ≥ . . . ≥ μn ≥ . . . > 0. Definition 2.4.1 (Singular Value) Let A : H → H˜ be a linear compact operator √ with adjoint A∗ : H˜ → H , H and H˜ be Hilbert spaces. The square root σn := μn ∗ of the eigenvalue μn > 0 of the self-adjoint operator A A : H → H is called the singular value of the operator A. Using the spectral representation (2.4.1) for the self-adjoint compact operator A∗ A we have: A∗ Au =

∞  n=1

σn2 (u, un )un , ∀u ∈ H.

(2.4.2)

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2 Functional Analysis Background of Ill-Posed Problems

Let us introduce now the orthonormal system {vn } in H˜ , through the orthonormal system {un } ⊂ H as follows: vn := Aun / Aun . Applying A∗ to both sides we get: A∗ vn = A∗ Aun / Aun . By the above definition, A∗ Aun = σn2 un and Aun = σn . This implies: A∗ vn = σn un . ∗ Act now ∞ with the adjoint operator A on both sides of the Fourier representation v = n=1 (v, vn )vn , v ∈ H˜ , where vn = Aun / Aun . Taking into account also the definition A∗ vn = σn un , we have:

A∗ v =

∞ 

σn (v, vn )un , v ∈ H˜ .

(2.4.3)

n=1

Applying A to both sides of (2.4.3) and using Aun = σn we obtain the spectral representation for the self-adjoint operator AA∗ : AA∗ v =

∞ 

σn2 (v, vn )vn , v ∈ H˜ .

(2.4.4)

n=1

It is seen from (2.4.2) and (2.4.4) that the eigenvalues σn2 > 0, n ∈ N, of the selfadjoint operators AA∗ and A∗ A are the same, as expected. The representation (2.4.3) is called singular value expansion of the adjoint operator A∗ : H˜ → H . For the operator A : H → H˜ this expansion can be derived in a similar way: Au =

∞ 

σn (u, un )vn , u ∈ H.

(2.4.5)

n=1

Substituting v = vn in (2.4.3) and u = un in (2.4.5), we obtain the following formulae: Aun = σn vn , A∗ vn = σn un .

(2.4.6)

The triple {σn , un , vn } is called the singular system for the non-self-adjoint operator A : H → H˜ . As we will see in the next chapter, some input-output operators related to inverse source problems are self-adjoint. If A : H → H˜ is a self-adjoint operator with the eigensystem {λn , un }, then Aun = |λn | and vn := Aun / Aun = λun /|λn |,

2.4 Singular Value Decomposition

41

by the above construction. Therefore, the singular system for the self-adjoint operator A : H → H˜ is defined as the triple {σn , un , vn }, with σ = |λn | and vn = λun /|λn |. Example 2.4.1 Singular values of a self-adjoint integral operator. Assuming H = L2 (0, π), we define the non-self-adjoint integral operator A : H → H as follows:  π (Au)(x) := u(ξ )dξ, x ∈ (0, π), u ∈ H. (2.4.7) x

By using the integration by parts formula and the definition (Au, v)L2 (0,π) = (u, A∗ v)L2 (0,π) , ∀u, v ∈ H , we can easily construct the adjoint operator A∗ : H → H:  x ∗ v(ξ )dξ, x ∈ (0, π), v ∈ H. (2.4.8) (A v)(x) = 0

Evidently, both integral operators (2.4.7) and (2.4.8) are compact. Indeed, let 2 {un }∞ n=1 be a bounded sequence in L (0, π) with un L2 (0,π) ≤ M, M > 0. Then for any x1 , x2 ∈ [0, π], (2.4.7) implies:



|(Aun )(x1 ) − (Aun )(x2 )| ≤

x2 x1



un (ξ )dξ

≤ M|x1 − x2 |1/2 .

This shows that {(Au)n } is an equicontinuous family of functions in C[0, π]. Hence, there exists a subsequence {(Au)m } ⊂ {(Au)n } that converges uniformly in C[0, π] to a continuous function v. Since uniform convergence implies convergence in L2 [0, π], we conclude that the subsequence {(Au)m } converges in L2 [0, π]. Therefore the integral operator defined by (2.4.7) is compact because the image of a bounded sequence always contains a convergent subsequence. Now we define the self-adjoint integral operator A∗ A: (A∗ Au)(x) =



x 0



π

u(η)dηdξ, u ∈ H.

(2.4.9)

ξ

To find the nonzero positive eigenvalues μn > 0, n ∈ N, of the self-adjoint integral operator A∗ A, defined by (2.4.9), the eigenvalue problem should be solved for the integral equation (A∗ Au)(x) = σ 2 u(x), x ∈ (0, π).

(2.4.10)

Differentiating both sides of (2.4.10) twice with respect to x ∈ [0, π] we arrive at the Sturm-Liouville equation: −u (x) = λu(x), λ = 1/σ 2 . To derive the boundary conditions, we first substitute x = 0 in (2.4.10). Then we get u(0) = 0, by (2.4.9).

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2 Functional Analysis Background of Ill-Posed Problems

Differentiating both sides of (2.4.10) and substituting x = π we conclude u (π) = 0. Hence, problem (2.4.10) is equivalent (in well-known sense) to the eigenvalue problem 

−u (x) = λu(x), a.e. x ∈ (0, π), λ = 1/σ 2 , u(0) = u (π) = 0,

(2.4.11)

for the self-adjoint positive-defined differential operator Au := −u . The solution ˚2 [0, π] := {u ∈ H 2 (0, π) : u(0) = 0}, where H 2 (0, π) is of this problem is in H the Sobolev space. 2, Solving the eigenvalue problem (2.4.11) we find the eigenvalues√λn = (n−1/2) √ n ∈ N, and the corresponding normalized eigenvectors un (x) = 2/π sin( λn x). Hence, the eigenvalues σn2 = 1/λn and the corresponding eigenvectors un (x) of the self-adjoint integral operator A∗ A are σn2 = (n − 1/2)−2 , un (x) =



2/π sin ((n − 1/2)x) .

By Definition 2.4.1, σn = (n − 1/2)−1 , n ∈ N, are the eigenvalues of the non-selfadjoint integral operator A. The corresponding eigenvectors, given in Eqs. (2.4.6) are un (x) =



2/π sin((n − 1/2)x), vn (x) =



2/π cos((n − 1/2)x).

Thus, the singular system {σn , un , vn } for the non-self-adjoint integral operator (2.4.7) is defined as follows:     (n − 1/2)−1 , 2/π sin((n − 1/2)x), 2/π cos((n − 1/2)x) .  Remark 2.4.1 The above example insights into the degree of ill-posedness of simplest integral equations Au = f and A∗ Au = A∗ f , with the operators A and A∗ A, defined by (2.4.7) and (2.4.9). In the first case one needs an operation differentiation (which is an ill-posed procedure) to find u = A−1 f . As a result, σn = O(n−1 ). In the second case the operator A∗ A, defined by (2.4.9), contains two integration and hence one needs to differentiate twice to find u = (A∗ A)−1 A∗ f , which results in the singular values as O(n−2 ). We come back to this issue in the next chapter. The above considerations lead to so-called Singular Value Decomposition (or normal form) of compact operators.

2.4 Singular Value Decomposition

43

Theorem 2.4.1 (Picard) Let H and H˜ be Hilbert spaces and A : H → H˜ be a linear compact operator with the singular system {σ, un , vn }. Then the equation Au = f has a solution if and only if f ∈ N (A∗ )⊥ and

∞  1 |(f, vn )|2 < +∞. σn2

(2.4.12)

∞  1 (f, vn )un σn

(2.4.13)

n=1

In this case u := A† f =

n=1

is the solution of the equation Au = f . Proof Let the equation Au = f has a solution. Then f must be in R(A). But by Theorem 2.2.1, R(A) = N (A∗ )⊥ . Hence f ∈ N (A∗ )⊥ and the first part of (2.4.12) holds. To prove the second part of (2.4.12) we use the relation A∗ vn = σn un in (2.4.6) to get σn (u, un ) = (u, A∗ vn ) = (Au, vn ) = (f, vn ). Hence, (u, un ) = (f, vn )/σn . Using this in u=

∞  (u, un )un , u ∈ H

(2.4.14)

n=1

we obtain: u=

∞  1 (f, vn )un . σn n=1

But the orthonormal system {un } is complete, so the Fourier series (2.4.14) is convergent. By the convergence criterion this implies the second condition of (2.4.12): ∞ ∞   1 2 |(f, v )| = |(u, un )|2 < +∞. n σn2 n=1

n=1

To prove the second part of the theorem, we assume now that conditions (2.4.12) hold. Then series (2.4.13) converges. Acting on both sides of this series by the

44

2 Functional Analysis Background of Ill-Posed Problems

operator A, using f ∈ N (A∗ )⊥ = R(A) and Aun = σn vn we get: Au =

∞  1 (f, vn )Aun σn n=1

∞  (f, vn )vn = f. n=1

 

This completes the proof.

Since μn > 0, n ∈ N are eigenvalues of the self-adjoint operator A∗ A (as well √ as AA∗ ) and σn := μn , we have: σn → 0, as n → ∞, if dimR(A) = ∞. Then it follows from formulae (2.4.12)–(2.4.13) that A† is an unbounded operator. Indeed, for any fixed eigenvector vk , with vk = 1, we have:

A† vk =

1 → ∞, as n → ∞. σk

The second condition (2.4.12), called Picard criterion, shows that the best approximate solution of the equation Au = f exists if only the Fourier coefficients (f, vn ) of f decay faster than the singular values σn . Concrete examples related to this issue will be given in the next chapter. As noted in Remark 2.4.1, singular value decomposition reflects the ill-posedness of the equation Au = f with a compact operator A between the infinite dimensional Hilbert spaces H and H˜ . Indeed, the decay rate of the non-increasing sequence {σn }∞ n=1 characterizes the degree of ill-posedness of an ill-posed problem. In particular, the amplification factors of a measured output errors in nth Fourier component of the series (2.4.13), corresponding to the integral operators (2.4.7) and (2.4.9), increase as n and n2 , respectively, due to the factor 1/σn . In terms of corresponding problems Au = f and A∗ Au = A∗ f this means that the second problem is more ill-posed than the first one. Hence, solving numerically the second ill-posed problem is more difficult than the first one. These considerations motivate the following definition of ill-posedness of problems governed by compact operators, proposed in [67]. Definition 2.4.2 (Degree of Ill-Posedness) Let A : H → H˜ be a linear compact operator between the infinite dimensional Hilbert spaces H and H˜ . If there exists a constant C > 0 and a real number s ∈ (0, ∞) such that σn ≥

C , for all n ∈ N, ns

(2.4.15)

then the equation Au = f is called moderately ill-posed of degree at most s. If for any  > 0, condition (2.4.15) does not hold with s replaced by s −  > 0, then the equation Au = f is called moderately ill-posed of degree s. If no such number s ∈ (0, ∞) exists such that condition (2.4.15) holds, then the equation Au = f is called severely ill-posed.

2.4 Singular Value Decomposition

45

Typical behavior of severe ill-posedness is exponential decay of the singular values of the compact operator A. As we will show in the next chapter, classical backward parabolic problem is severely ill-posed, whereas the final time output inverse source problems related to parabolic and hyperbolic equations are only moderately ill-posed. Remark that some authors use a more detailed classification distinguishing the mildly ill-posedness ( s ∈ (0, 1)) and the moderately ill-posedness (s ∈ (1, ∞)). In applications, to obtain an approximation of A† f one can truncate the series (2.4.13): uN :=

N  1 (f, vn )un . σn

(2.4.16)

n=1

This method of obtaining the approximate solution uN is called the truncated singular value decomposition (TSVD). To understand the role of the cutoff parameter N, we assume that the right hand side f ∈ H of the equation Au = f is given with some measurement error δ > 0, i.e. f − f δ ≤ δ, where f δ is a noisy data. Then uN,δ :=

N  1 δ (f , vn )un σn

(2.4.17)

n=1

is an approximate solution of the equation Au = f δ corresponding to the noisy measured output f δ . Let us estimate the norm uN − uN,δ , i.e. the difference between the approximate solutions corresponding to the noise free (f ) and noisy (f δ ) data. From (2.4.16)–(2.4.17) we deduce the estimate:

uN − uN,δ 2 =

N 

2 1

(f − f δ , vn )

σn2 n=1

≤

1 σN2

N 

2



(f − f δ , vn ) 2 ≤ δ . σN2 n=1

Using this estimate we find the accuracy error uN,δ − A† f , i.e. the difference between the best approximate solution A† f , corresponding to the noise free output f , and the approximate solution uN,δ , obtained by TSVD and corresponding to the noisy output f δ :

uN,δ − A† f ≤ uN − A† f + uN − uN,δ

≤ uN − A† f +

δ . σN

(2.4.18)

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2 Functional Analysis Background of Ill-Posed Problems

The first term uN − A† f on the right hand side of estimate (2.4.18) depends only on the cutoff parameter N and does not depend on the measurement error δ > 0. This term is called the regularization error. The second term uN − uN,δ on the right hand side of (2.4.18) depends not only on the cutoff parameter N, but also on the measurement error δ > 0. This term is called the data error. This term exhibits some very distinctive features of a solution of the ill-posed problems. Namely, the approximation error uN,δ − A† f decreases with δ > 0, for a fixed value of the cutoff parameter N, on one hand. On the other hand, for a given δ > 0 this error √ tends to infinity, as N → ∞, since σN := μN → 0. Hence, the parameter N = N(δ) needs to be chosen depending on δ > 0 such that δ → 0, as δ → 0. σN(δ)

(2.4.19)

We use the right hand side of (2.4.17) to introduce the operator RN(δ) : H˜ → H , RN(δ) f δ :=

N(δ)  n=1

1 δ (f , vn )un . σn

(2.4.20)

It follows from estimate (2.4.18) that if condition (2.4.19) holds, then

RN(δ) f δ − A† f → 0, δ → 0. Hence, the basic idea of TSVD in solving ill-posed problems is finding a finite dimensional approximation of the unbounded operator A† . A class of such finite dimensional operators defined by (2.4.20) and approximating the unbounded operator A† can be defined as regularization method or regularization strategy. The cutoff parameter N(δ) plays the role of the parameter of regularization [53].

2.5 Regularization Strategy. Tikhonov Regularization Let A : H → H˜ be a linear injective bounded operator between infinitedimensional real Hilbert spaces H and H˜ . Consider the linear ill-posed operator equation Au = f, u ∈ H, f ∈ R(A).

(2.5.1)

By the condition f ∈ R(A), the operator equation (2.5.1) is ill-posed in the sense that a solution u ∈ H exists, but doesn’t depend continuously on the output f ∈ R(A). In practice this output always contains a random noise. We denote by f δ ∈ H˜

2.5 Regularization Strategy. Tikhonov Regularization

47

the noisy output and assume that

f δ − f H˜ ≤ δ, f ∈ R(A), f δ ∈ H˜ , δ > 0.

(2.5.2)

Then the exact equality in the equation Au = f δ can not be satisfied due to the noisy output f δ and we may only consider the minimization problem J (u) = inf J (v) v∈H

(2.5.3)

for the Tikhonov functional J (u) =

1

Au − f δ 2H˜ , u ∈ H, f δ ∈ H˜ , 2

(2.5.4)

where Au − f δ 2˜ := (Au − f δ , Au − f δ )H˜ . H A solution u ∈ H of the minimization problem (2.5.3)–(2.5.4) is called quasisolution or least squares solution of the ill-posed problem (2.5.1). If, in addition, this solution is defined as the minimum-norm solution, i.e. if

u H = inf { w H : w ∈ H is a least squares solution of (2.5.1)}, then this solution is called best approximate solution of (2.5.1). Note that the concept of quasi-solution has been introduced in [74]. Since H is infinite-dimensional and A is compact, the minimization problem for the functional (2.5.4) is ill-posed, the functional J (u) doesn’t depend continuously on the output f δ ∈ H˜ . One of the possible ways of stabilizing the functional is to add the penalty term α u − u0 2H , as in Optimal Control Theory, and then consider the minimization problem for the regularized Tikhonov functional Jα (u) :=

1 1

Au − f δ 2H˜ + α u − u0 2H , u ∈ H, f δ ∈ H˜ . 2 2

(2.5.5)

Here α > 0 is the parameter of regularization and u0 ∈ H is an initial guess. Usually u0 ∈ H is one of the possible good approximations to the exact solution u ∈ H , but if such an initial guess is not known, we may take u0 = 0. Below we assume that u0 = 0. This approach is defined as Tikhonov regularization [147, 148] or TikhonovPhillips regularization [123]. Theorem 2.5.1 Let A : H → H˜ be a linear injective bounded operator between real Hilbert spaces H and H˜ . Then the regularized Tikhonov functional (2.5.5) has a unique minimum uδα ∈ H , for all α > 0. This minimum is the solution of the linear equation 

 A∗ A + αI uδα = A∗ f δ , uδα ∈ H, f δ ∈ H˜ , α > 0

(2.5.6)

48

2 Functional Analysis Background of Ill-Posed Problems

and has the form  −1 ∗ δ uδα = A∗ A + αI A f .

(2.5.7)

Moreover, the operator A∗ A + αI is boundedly invertible, hence the solution uδα continuously depends on f δ . Proof First of all, note that the Fréchet differentiability of the functional (2.5.5) follows from the identity:   1 1 Jα (u + v) − Jα (u) = A∗ (Au − f δ ) + αu, v + Av 2 + α v 2 , ∀u, v ∈ H, 2 2 where A∗ : H˜ → H is the adjoint operator of A. This identity implies: 

(Jα (u), v) = (Au − f δ , Av) + α(u, v), ∀v ∈ H ; Jα (u; v, v) = Av 2 + α v 2 , ∀v ∈ H.

(2.5.8)

Formula (2.5.8) for the second Fréchet derivative Jα (u; v, v) shows that the regularized Tikhonov functional Jα (u) defined on a real Hilbert H space is strictly convex, since α > 0, and lim u →+∞ Jα (u) = +∞. Then it has a unique minimizer uδα ∈ H and this minimum is characterized by the following necessary and sufficient condition (Jα (u), v) = 0, ∀v ∈ H,

(2.5.9)

where Jα (u) is the first Fréchet derivative of the regularized Tikhonov functional. Thus, condition (2.5.9) with formula (2.5.8) implies that the minimum uδα ∈ H of the regularized Tikhonov functional is the solution of the linear equation (2.5.6). This solution is defined by (2.5.7), since the operator A∗ A + αI is boundedly invertible. This follows from the Lax-Milgram Theorem [155] and the positive definiteness of the operator A∗ A + αI :  ∗  (A A + αI )v, v = Av 2 + α v 2 ≥ α v 2 , ∀v ∈ H, α > 0. Evidently, the operator A∗ A + αI is one-to-one for each positive α. Indeed, multiplication of the homogeneous equation (A∗ A + αI )v = 0 by v ∈ H implies: (A∗ Av, v) + α(v, v) = (Av, Av) + α(v, v) = 0. This holds if and only if v = 0. This completes the proof.   From (2.5.8) we deduce the gradient formula for the regularized Tikhonov functional. Corollary 2.5.1 For the Fréchet gradient Jα (u) of the regularized Tikhonov functional (2.5.4) the following formula holds:   Jα (u) = A∗ Au − f δ + αu, u ∈ H.

(2.5.10)

2.5 Regularization Strategy. Tikhonov Regularization

49

The main consequence of Picard’s Theorem 2.4.1 and Theorem 2.5.1 is that the solution uδα of the normal equation (2.5.6) corresponding to the noisy output f δ can be represented by the following series: ∞  q(α; σn )

uδα =

n=1

σn

(f δ , vn )un , α > 0,

(2.5.11)

where q(α; σ ) =

σ2 +α

σ2

(2.5.12)

is called the filter function. Corollary 2.5.2 Let conditions of Theorem 2.4.1 hold and f δ ∈ N (A∗ )⊥ . Then the unique regularized solution uδα ∈ H , given by (2.5.7), can be represented as the convergent series (2.5.11). Proof It follows from the normal equation (2.5.6) that αuδα = A∗ f δ − A∗ Auδα . By Corollary 2.2.1, R(A∗ A) = R(A∗ ), so this implies that uδα ∈ R(A∗ ). But due to Theorem 2.2.1, R(A∗ ) = N (A)⊥ and we conclude that uδα ∈ N (A)⊥ . The orthonormal system {um } spans N (A)⊥ . Hence ∞ 

uδα =

c m um .

(2.5.13)

m=1

To find the unknown parameters cm we substitute (2.5.13) into the normal equation (2.5.6): ∞ 

(σm2 + α)cm um = A∗ f δ .

m=1

Multiplying both sides by un we get: (σn2 + α)cn = (A∗ f δ , un ).   But (A∗ f δ , un ) = f δ , Aun = σn (f δ , vn ). Therefore (σn2 + α)cn = σn (f δ , vn ) and the unknown parameters are defined as follows: cm =

σm (f δ , vm ) +α

σm2

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2 Functional Analysis Background of Ill-Posed Problems

Using this in (2.5.13) we obtain: uδα =

∞ 

 σm  δ f , vm un . +α

σ2 m=1 m

With formula (2.5.12), this implies (2.5.11).

 

Remark that the filter function has the following properties:  q(α; σ ) ≤ min

 σ2 σ , ∀α > 0. √ ; 2 α α

(2.5.14)

These properties follow from the obvious inequalities: σ 2 + α ≡ (σ −

√ 2 √ √ α) + 2σ α ≥ 2σ α,

σ 2 /α σ2 σ2 ≡ ≤ , ∀α > 0, σ > 0. σ2 + α σ 2 /α + 1 α The linear equation (2.5.6) is defined as a regularized form of the normal equation A∗ Au = A∗ f δ [149]. If A∗ A is invertible, then for α = 0 formula (2.5.7) implies that uδ0 = (A∗ A)−1 A∗ f δ =: A† f δ , α = 0

(2.5.15)

is the solution of the normal equation A∗ Auδ = A∗ f δ , uδ ∈ H, f δ ∈ H˜ .

(2.5.16)

Evidently uδ0 ∈ N (A)⊥ , by Definition 2.3.1 of Moore-Penrose inverse A† . It follows from (2.5.15) that Moore-Penrose inverse A† := (A∗ A)−1 A∗ arises naturally as a result of minimization of the Tikhonov functional. Remark that the provisional replacement of the compact operators AA∗ or A∗ A by the non-singular operators AA∗ + αI or A∗ A + αI , is the main idea of Tikhonov’s regularization procedure. This procedure has originally been introduced by Tikhonov in [146] for uniform approximations of solutions of Fredholm’s equation of the first kind. However, Tikhonov does not point to the relation of his ideas to Moore-Penrose inverse. Let us assume now that α > 0. The right hand side of (2.5.7) defines family of continuous operators  −1 ∗ Rα := A∗ A + αI A : H˜ → H, α > 0,

(2.5.17)

depending on the parameter of regularization α > 0. Obviously, when the measured output is noise free, then the regularized solution Rα f should converge (in some

2.5 Regularization Strategy. Tikhonov Regularization

51

sense) to A† f , as α → 0, that is, Rα f → A† f , for all f ∈ D(A† ). In practice this output is always noisy and we may only assume that it is known up to some error δ > 0, that is, f − f δ ≤ δ. Hence the parameter of regularization α > 0 depends on the noise level δ > 0 and the noise output f δ ∈ H˜ , and should be chosen appropriately, keeping an error uδα − u as small as possible. A strategy of choosing the parameter of regularization α = α(δ, f δ ) is called a parameter choice rule. These considerations lead to the following definition of regularization strategy. Definition 2.5.1 (Regularization Strategy) Let A : H → H˜ be an injective compact operator. Assume that f, f δ ∈ H˜ be noise free and noisy output respectively, that is, f −f δ ≤ δ < f δ , δ > 0. A family {Rα(δ,f δ ) }α of bounded linear operators is called a regularization strategy or a convergent regularization method if for all f ∈ D(A† ),     lim sup Rα(δ,f δ ) f δ − A† f  δ→0+

H

   : f δ ∈ H˜ , f − f δ  ≤ δ = 0,

(2.5.18)

with α : R+ × H˜ → R+ , such that   lim sup α(δ, f δ ) : f δ ∈ H˜ , f − f δ ≤ δ = 0. δ→0+

(2.5.19)

If the parameter of regularization α = α(δ, f δ ) depends only on the noise level δ > 0, the rule α(δ, f δ ) is called a-priori parameter choice rule. Otherwise, this rule is called a-posteriori parameter choice rule. Tikhonov regularization is a typical example of a priori parameter choice rule, since the choice of the parameter of regularization α > 0 is made a priori, i.e. before computations, as we will see in Theorem 2.5.2. The Morozov’s Discrepancy Principle is a regularization strategy with a-posteriori parameter choice rule, since the choice of the parameter of regularization α > 0 is made during the process of computing. Regarding the iterative methods, the number of iterations plays here the role of the regularization parameter. As we will show in the next chapter, Landweber’s Method and Conjugate Gradient Method together with appropriate parameter choice rule are also a regularization strategy in sense of Definition 2.5.1. We remark finally that, as in the case of Tikhonov regularization, some widely used regularization operators Rα are linear. In particular, the Morozov’s Discrepancy Principle and Landweber’s Method can be formulated as linear regularization methods. However, the Conjugate Gradient Method is a nonlinear regularization method, since the right hand side of the equation Au = f does not depend linearly on the parameter of regularization α. Above definition, with Theorem 2.5.1, implies that the family of operators {Rα }α>0 , Rα : H˜ → H , defined as  −1 ∗ A , α > 0, Rα := A∗ A + αI

(2.5.20)

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2 Functional Analysis Background of Ill-Posed Problems

is the regularization strategy corresponding to Tikhonov regularization. That is, the regularization operators Rα , α > 0, approximate the unbounded inverse A† of the operator A on R(A). On the other hand, Corollary 2.5.1, implies that the singular value expansion of this operator is Rα f δ :=

∞  q(α; σn ) n=1

σn

(f δ , vn )un , α > 0.

(2.5.21)

It is easy to prove that the family of operators {Rα } are not uniformly bounded, i.e. there exists a sequence {αm }∞ m=1 , αm > 0, such that Rαm → ∞, as αm → 0. Indeed, taking fαδ = vm in (2.5.21), where vm is any fixed eigenvector, with vm = 1, we have:   Rαm vm  =

σm2

σm . + αm

For the sequence {αm } satisfying the conditions αm → 0 and αm /σm → 1, as m → ∞, we conclude:

Rαm vm =

1 → ∞, as m → ∞. σm + αm /σm

The regularization strategy Rα possesses this property in general case as well. Lemma 2.5.1 Let Rα be regularization strategy corresponding to the compact operator A : H → H˜ . Then the operators {Rα }α>0 , Rα : H˜ → H , are not uniformly bounded, i.e. there exists a sequence {αm }∞ m=1 , αm > 0, such that

Rαm → ∞, as αm → 0. Proof Assume, in contrary, that there exists a constant M > 0, independent on α > 0, such that Rα ≤ M, for all α > 0. Then for any f ∈ R(A) ⊂ H˜ we have:

A† f ≤ A† f − Rα f + Rα f ≤ A† f − Rα f + Rα

f

≤ A† f − Rα f + M f . The first norm on the right-hand side tends to zero as α → 0, by definition (2.5.18). Then passing to the limit we get: A† f ≤ M f , for all f ∈ R(A), which implies

A† ≤ M, i.e. boundedness of the generalized inverse A† . This contradiction completes the proof.   Therefore, one needs to find a bounded approximation of the unbounded operator A† . This approximation will evidently depends on both, the parameter of regularization α > 0 and the noisy output f δ . So, the main problem of regularization strategy is to find such a bounded approximation, which will be convergent to the

2.5 Regularization Strategy. Tikhonov Regularization

53

best approximate solution A† f corresponding to the exact data f ∈ N (A∗ )⊥ , as α → 0 and δ → 0. The following theorem gives an answer to this issue. Theorem 2.5.2 Let A : H → H˜ be a linear injective bounded operator between Hilbert spaces H and H˜ . Assume that conditions of Theorem 2.4.1 hold. Denote by f δ ∈ N (A∗ )⊥ the noisy data: f − f δ H˜ ≤ δ, δ > 0. Suppose that the conditions hold: α(δ) → 0 and

δ2 → 0, as δ → 0. α(δ)

(2.5.22)

Then the solution uδα := Rα(δ)f δ of the regularized form of the normal equation (2.5.6) converges to the best approximate solution u := A† f of Eq. (2.5.1) in the norm of the space H , that is,

Rα(δ) f δ − A† f H → 0, δ → 0.

(2.5.23)

Proof Let us estimate the norm Rα(δ)f δ − A† f H using (2.4.13) and (2.5.21). We have

Rα(δ) f − A δ

= ≤2

n=1

f 2H

2 ∞    q(α(δ); σn ) δ  1  := (f , vn ) − (f, vn ) un    σn σn n=1

∞   n=1

∞ 

†

 q(α(δ); σn ) δ  (f − f, vn ) +  σn



q(α(δ); σn ) 1 − σn σn



2  (f, vn ) un  

∞

 q 2 (α(δ); σn ) δ α 2 (δ) 2 |(f − f, v )| + 2 |(f, vn )|2 . (2.5.24) n σn2 σn2 (σn2 + α(δ))2 n=1

Denote by S1 and S2 the first and the second summands on the right-hand side of (2.5.24), respectively. We use from (2.5.14) the property q 2 (α; σn )/σn2 ≤ 1/(4α) of the filter function and Parseval’s identity to estimate the term S1 : ∞

S1 ≤

1  1 δ2

f δ − f 2 ≤ . |(f δ − f, vn )|2 ≤ 2α(δ) 2α(δ) 2α(δ)

(2.5.25)

n=1

For estimating the term S2 we rewrite it in the following form: S2 = 2

N  n=1

∞  α 2 (δ) α 2 (δ) 2 |(f, v )| + 2 |(f, vn )|2 n σn2 (σn2 + α(δ))2 σn2 (σn2 + α(δ))2 n=N+1

=: S2N + R2N . (2.5.26)

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2 Functional Analysis Background of Ill-Posed Problems

To estimate the Nth partial sum S2N we use the properties σ1 ≥ σ2 ≥ . . . ≥ σn ≥ . . . > 0 and σn → 0, as n → ∞, of the singular values. For any small value α = α(δ) > 0 of the parameter of regularization, depending on the noise level δ > 0, there exists such a positive integer N = N(α(δ)) ≡ N(δ) that min

1≤n≤N(δ)

2 σn2 = σN(δ) ≥



2 α(δ) > σN(δ)+1 ,

(2.5.27)

where N = N(α(δ)) → ∞, as δ → 0. For such N = N(δ) we estimate the Nth partial sum S2N in (2.5.26) as follows: S2N ≤

=

N(δ)  1 2α 2 (δ) |(f, vn )|2 2 2 (σN(δ) + α(δ)) n=1 σn2

N(δ)  1 2α(δ) |(f, vn )|2 √ √ 2 2 (σN(δ) / α(δ) + α(δ)) n=1 σn2

√ 2 / α(δ) ≥ 1. Hence By the condition (2.5.27), σN(δ) S2N ≤

N(δ)  1 2α(δ) |(f, vn )|2 . √ 2 (1 + α(δ)) n=1 σn2

(2.5.28)

Let us estimate now the series remainder term R2N defined in (2.5.26). We have R2N = 2

∞ 

1

(σn2 /α(δ) n=N(δ)+1 ≤2

+ 1)2

∞  n=N(δ)+1

1 |(f, vn )|2 σn2 1 |(f, vn )|2 . σn2

Taking into account this estimate with estimates (2.5.25) and (2.5.28) in (2.5.24) we finally deduce:

Rα(δ) f δ − A† f 2H ≤

N(δ)  1 2α(δ) δ2 + |(f, vn )|2 √ 2 2α (1 + α(δ)) n=1 σn2

+2

∞  n=N(δ)+1

1 |(f, vn )|2 . σn2

(2.5.29)

The first right hand side term tends to zero, as δ → 0, by the second condition of (2.5.22). The factor before the partial sum of the second right hand side term

2.5 Regularization Strategy. Tikhonov Regularization

55

tends to zero, since α(δ) → 0, as δ → 0, by the first condition of (2.5.22), and this partial sum is finite due to the convergence condition (2.4.12) of the Picard’s Theorem 2.4.1. The third right hand side term also tends to zero, as δ → 0, since in this case N(δ) → ∞, and, as a result, the series remainder term tends to zero, by the same convergence condition. This implies (2.5.23).   Estimate (2.5.29) clearly shows the role of all parameters δ > 0, α(δ) > 0 and N = N(δ) in the regularization strategy. As we have seen in the previous section, the number N of first terms in singular value expansion of a compact operator can also be considered as a regularization parameter. As noted above in iterative methods, the number of iterations also plays role of the regularization parameter, that is, α ∼ 1/N. For more detailed analysis of regularization methods for ill-posed problems we refer to the books [37, 81, 141]. Remark finally that besides the Tikhonov regularization, there are other regularization methods, such as Lavrentiev regularization, asymptotic regularization, local regularization, etc. Since zero is the only accumulation point of the singular values of a compact operator, the underlying idea in all these regularization techniques is modifying the smallest singular values, shifting all singular values by α > 0. In other words, the idea is to approximate the compact operator A or A∗ A by a family of operators A + αI or A∗ A + αI . The first one corresponds to Lavrentiev regularization. Specifically, while Tikhonov regularization is based on the normal equation A∗ Au = A∗ f δ , Lavrentiev regularization is based on the original equation Au = f δ . The main advantage of the Tikhonov regularization over the Lavrentiev regularization is that the operator A∗ A is always injective, due to Lemma 2.3.1, even if the operator A :→ H˜ is not injective. The second advantage of Tikhonov regularization is that the class of admissible values of the parameter of regularization α > 0 for convergent regularization method is larger than the same class in Lavrentiev regularization. Specifically, when α = δ the condition δ 2 /α(δ) → 0, as δ → 0, for convergent regularization strategy in Tikhonov regularization holds, but does not hold in Lavrentiev regularization, as we will see below. Corollary 2.5.2 can be adopted to the case of Lavrentiev regularization when the linear bounded injective operator A : H → H is self-adjoint and positive semidefinite. Corollary 2.5.3 Let conditions of Theorem 2.4.1 hold and f δ ∈ R(A). Then the unique solution uδα ∈ H of the regularized equation (A + αI ) uδα = f δ

(2.5.30)

can be represented as the series uδα =

∞  q(α; ˜ σn ) n=1

σn

(f δ , un )un , α > 0,

(2.5.31)

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2 Functional Analysis Background of Ill-Posed Problems

where q(α; ˜ σ) =

σ . σ +α

(2.5.32)

Proof We use the singular system {σn , un , un } for the non-self-adjoint operator A : H → H˜ , that is, Aun = σn un and the representation uδα =

∞ 

c m um .

(2.5.33)

m=1

Substituting this into the normal equation (2.5.30) we obtain: ∞ 

(σm + α)cm um = f δ .

m=1

Multiplying both sides by un we find the unknown parameters cm : cm =

1 (f δ , um ) σm + α

Using this in (2.5.33) we find: uδα =

∞  m=1

 δ  1 f , um um . σm + α

By (2.5.32), this is exactly the required expansion (2.5.31).

 

Now the question we seek to answer here is that under which conditions it is possible to construct a convergent regularization strategy for Lavrentiev regularization. The following theorem, which is an analogue of Theorem 2.5.2, answers this question. Theorem 2.5.3 Let the linear bounded injective operator A : H → H˜ be a selfadjoint and positive semi-definite. Assume that conditions of Theorem 2.4.1 hold. Denote by f δ ∈ R(A) the noisy data: f − f δ H˜ ≤ δ, δ > 0. Suppose that the following conditions hold: α(δ) → 0 and

δ → 0, as δ → 0. α(δ)

(2.5.34)

Then the operator Rα(δ) defined by the right hand side of (2.5.31) is a convergent regularization strategy, that is, Rα(δ)f δ :=

∞  q(α(δ); ˜ σn ) n=1

σn

(f δ , un )un , α > 0,

(2.5.35)

2.5 Regularization Strategy. Tikhonov Regularization

57

that is, the solution uδα := Rα(δ) f δ of the regularized form of the normal equation (2.5.30) converges to the best approximate solution u := A† f of Eq. (2.5.1) in the norm of the space H . Proof The proof is similar to the proof of Theorem 2.5.2. Here we first use the following property σ σ ≤ σ +α α

(2.5.36)

of the filter function (2.5.32) in (2.5.24) to obtain the estimate

Rα(δ)f δ − A† f 2H :=

2 ∞    q(α(δ);  σn ) δ 1  ˜ (f , u ) − (f, u ) un  n n   σn σn n=1

≤2

∞  q˜ 2 (α(δ); σn )

σn2

n=1

|(f δ − f, un )|2 + 2

∞  n=1

α 2 (δ) |(f, un )|2 . (2.5.37) σn2 (σn + α(δ))2

Denote by S1 and S2 the first and the second right-hand side summands of (2.5.37), respectively. By (2.5.36) we conclude q˜ 2 (α; σn )/σn2 ≤ 1/α 2 . Using this inequality in (2.5.37) we obtain the estimate S1 := 2

∞  q˜ 2 (α(δ); σn ) n=1

σn2

|(f δ − f, un )|2 ≤

1 δ2 δ 2

f . (2.5.38) − f

≤ α 2 (δ) α 2 (δ)

Second, we rewrite the term S2 in the following form: S2 = 2

N  n=1

∞  α 2 (δ) α 2 (δ) 2 |(f, u )| + 2 |(f, un )|2 n σn2 (σn + α(δ))2 σn2 (σn + α(δ))2 n=N+1

=: S2N + R2N .

(2.5.39)

For estimating the first right hand side term S2 in (2.5.39), we use the properties σ1 ≥ σ2 ≥ . . . ≥ σn ≥ . . . > 0, σn → 0, as n → ∞, of the eigenvalues λn = σn of self-adjoint positive semidefinite operator A. For any small values α = α(δ) > 0 of the parameter of regularization, depending on the noise level δ > 0, there exists such a positive integer N = N(α(δ)) ≡ N(δ) that min

1≤n≤N(δ)

2 2 σn2 = σN(δ) ≥ α(δ) > σN(δ)+1 ,

(2.5.40)

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2 Functional Analysis Background of Ill-Posed Problems

where N = N(α(δ)) → ∞, as δ → 0. For such N = N(δ) we get the following estimate for S2N in (2.5.39): S2N ≤

N(δ)  1 2α 2 (δ) |(f, vn )|2 2 (σN(δ) + α(δ)) σn2 n=1

=

N(δ)  1 2α(δ) √ √ |(f, vn )|2 2 (σN(δ) / α(δ) + α(δ)) n=1 σn2

√ By the condition (2.5.40), σN(δ) / α(δ) ≥ 1, which implies: S2N ≤ 2α(δ)

N(δ)  n=1

1 |(f, un )|2 . σn2

(2.5.41)

The factor before the above partial sum, which is finite by the convergence condition (2.4.12) of the Picard’s Theorem 2.4.1, tends to zero, as α(δ) → 0. Hence, S2N → 0, as α(δ) → 0. Let us estimate now the series remainder term R2N defined in (2.5.39). We have R2N = 2

∞  n=N(δ)+1

1 1 |(f, vn )|2 2 (σn /α(δ) + 1) σn2 ≤2

∞  n=N(δ)+1

1 |(f, vn )|2 . σn2

Substituting this estimate with (2.5.38) and (2.5.41) into (2.5.37) we conclude:  1 δ2 + 2α(δ) |(f, un )|2 2 α(δ) σn2 N(δ)

Rα(δ) f δ − A† f 2H ≤

+2

∞  n=N(δ)+1

n=1

1 |(f, vn )|2 . σn2

(2.5.42)

It is easy to verify that under the conditions (2.5.34) all three right hand side terms tend to zero, as δ → 0.   Comparing the convergence conditions in Theorems 2.5.2 and 2.5.3 we first observe that, in Tikhonov regularization the class of admissible values of the parameter of regularization α > 0 for convergent regularization method is larger than the same class in Lavrentiev regularization. √ While, for example, in Tikhonov regularization the values α = δ and α = δ of the parameter of regularization are admissible for convergent regularization method, as the second condition of

2.6 Morozov’s Discrepancy Principle

59

(2.5.22) shows, in Lavrentiev regularization they are not admissible by the condition (2.5.34). Moreover, the dependence on the parameter of regularization α > 0 of the numbers N = N(α(δ)), defined in proofs of these theorems and corresponding to these regularizations, are different. Thus, if σn = O(1/n2 ), then condition (2.5.27) of Theorem 2.5.2 implies that there exists the constants c2 > c1 > 0 such that  c2 c1 ≥ α(δ) > . N4 (N + 1)4 For the same σn = O(1/n2 ), the condition (2.5.40) of Theorem 2.5.3 implies: c˜1 c˜2 ≥ α(δ) > , c˜2 > c˜1 > 0. N4 (N + 1)4   As a result, N(α(δ)) = O α(δ)−1/8 in Tikhonov regularization and N(α(δ)) =   O α(δ)−1/4 in Lavrentiev regularization. Note, finally that the topic of Tikhonov regularization is very broad and here we described it for only linear inverse problems. We refer the reader to the books [73, 141] on the mathematical theory of regularization methods related to nonlinear inverse problems.

2.6 Morozov’s Discrepancy Principle In applications the right hand side f ∈ H˜ of the equation Au = f always contains a noise. Instead of this equation one needs to solve the equation Auδ = f δ , where f δ is a noisy data: f − f δ ≤ δ, where δ > 0. This, in particular, implies that in the ideal case the residual or discrepancy Auδ − f δ can only be at most in the order of δ. On the other hand, in order to construct a bounded approximation Rα(δ) of the unbounded operator A† one needs to choose the parameter of regularization α > 0 depending on the noise level δ > 0. Thus, the parameter of regularization needs to be chosen by a compromise between the residual Auδ − f δ and the given bound δ > 0 for the noise level. This is the main criteria of so-called Morozov’s Discrepancy Principle due to Morozov [100, 101]. This principle is now one of the simplest tools and most widely used regularization method for ill-posed problems. Definition 2.6.1 (Discrepancy Principle) Let f δ ∈ H˜ be a noisy output with an arbitrary given noise level δ > 0, that is, f δ − f ≤ δ, where f ∈ R(A) is a noise free (exact) output. If there exists such a value α = α(δ) of the parameter of regularization, depending on δ > 0, that the corresponding solution uδα(δ) ∈ H of the equation Au = f δ satisfies the condition β1 δ ≤ Auδα(δ) − f δ ≤ β2 δ, β2 ≥ β1 ≥ 1,

(2.6.1)

60

2 Functional Analysis Background of Ill-Posed Problems

then the parameter of regularization α = α(δ) is said to be chosen according to Morozov’s Discrepancy Principle. Let us assume that the size δ := δf = f δ − f > 0 of the noise is known (although we do not know the random perturbation δf ). Denote by u ∈ N (A)⊥ the solution of the equation Au = f with a noise free (exact) data f ∈ N (A∗ )⊥ = R(A). Then

Au − f δ = Au − f − δf = δf =: δ, δ > 0. Thus, if u is the exact solution, corresponding to the exact data f , and f δ is the noisy data, then

Au − f δ = δ.

(2.6.2)

Now, having the size δ = f − f δ > 0 of noise, we want to use Tikhonov regularization to find the solution uδα defined by (2.5.7) and corresponding to the noisy data f δ := f + δf . Then, as it follows from (2.6.2), the best that we can require from the parameter of regularization α > 0 is the residual (or discrepancy)

Auδα(δ) − f δ = δ. We will prove that there exists such a value of the parameter of regularization α = α(δ) which satisfies the following conditions:

Auδα(δ) − f δ = f − f δ = δ, δ > 0.

(2.6.3)

Theorem 2.6.1 Let conditions of Theorem 2.4.1 hold. Denote by f δ ∈ H˜ the noisy data with f δ > δ > 0. Then there exists a unique value α = α(δ) of the parameter of regularization satisfying conditions (2.6.3). Proof By the unique decomposition, for any f δ ∈ H˜ = R(A) ⊕ R(A)⊥ we have fδ =

∞  (f δ , vn )vn + Pf δ ,

(2.6.4)

n=1

where Pf δ ∈ R(A)⊥ is the projection of the noisy data on R(A)⊥ . Now we rewrite the solution uδα , given by (2.5.11), of the normal equation (2.5.6) in the following form: uδα =

∞  n=1

σn (f δ , vn )un , α > 0. σn2 + α

(2.6.5)

2.6 Morozov’s Discrepancy Principle

61

Acting on this solution by the operator A and using the relation Aun = σn vn we deduce: Auδα =

∞ 

σn2 (f δ , vn )vn . +α

σ2 n=1 n

With (2.6.4) this yields: f δ − Auδα =

∞ 

σ2 n=1 n

α (f δ , vn )vn + Pf δ . +α

Then we obtain the discrepancy:

Auδα − f δ 2 =

∞  n=1

α2 δ 2 δ 2 2 |(f , vn )| + Pf .  2 σn + α

(2.6.6)

It can be verified that g(α) := Auδα − f δ is a monotonically increasing function for α > 0. To complete the proof of the theorem we need to show that the equation g(α) = δ, α ∈ (0, +∞), has a unique solution. For α → 0+ we use Pf = 0 for the noise free data f ∈ R(A) to get lim g(α) = Pf δ = P (f δ − f ) ≤ f δ − f = δ,

α→0+

(2.6.7)

by (2.6.3). Note that the sum in (2.6.6) tends to zero, as α → 0+ . For α → +∞ we use the following limit lim  α→+∞

α2 σn2 + α

lim  2 = α→+∞

1 σn2 /α 2 + 1

2 = 1

and the identity (2.1.8) to get  lim g(α) =

α→+∞

∞ 

1/2 |(f , vn )| + Pf

δ

2

δ 2

=: f δ ,

n=1

by (2.6.6). By the assumption f δ > δ we conclude: lim g(α) > δ.

α→+∞

This implies with (2.6.7) that the equation g(α) = δ has a unique solution for α ∈ (0, +∞).  

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2 Functional Analysis Background of Ill-Posed Problems

Remark that f δ > δ > 0 is a natural condition in the theorem. Otherwise, i.e. if f δ < δ, then uδα(δ) = 0 can be assigned as a regularized solution. The theorem below shows that Morozov’s Discrepancy Principle provides a convergent regularization strategy. First we need the following notion. Definition 2.6.2 (Source Condition) Let A : H → H˜ be a linear injective compact operator between Hilbert spaces H and H˜ , and A∗ : H˜ → H its adjoint. Denote by u ∈ H the solution u = A† f of the equation Au = f with the noise free (exact) data f ∈ R(A). If there exists such an element v ∈ H˜ with v H˜ ≤ M such that u = A∗ v, then we say that u ∈ H satisfies the source condition. Remark that besides of the classical concept of source condition, in recent years different new concepts, including approximate source conditions, have been developed [68]. Theorem 2.6.2 Let A : H → H˜ be a linear injective compact operator between Hilbert spaces H and H˜ . Assume that the solution u = A† f of the equation Au = f with the noise free data f ∈ R(A) satisfies the source condition. Suppose that the parameter of regularization α = α(δ), δ > 0, is defined according to conditions (2.6.3). Then the regularization method Rα(δ) is convergent, that is,

Rα(δ) f δ − A† f H → 0, δ → 0,

(2.6.8)

where f δ ∈ H˜ is the noisy data and Rα(δ) f δ =: uδα(δ) is the regularized solution. Proof Since the regularized Tikhonov functional Jα (v) :=

1 1

Av − f δ 2H˜ + α v 2H , v ∈ H, f δ ∈ H˜ 2 2

(2.6.9)

attains its infimum on uδα(δ), we have Jα (uδα(δ)) ≤ Jα (v), for all v ∈ H . In particular, for the unique solution u ∈ N (A)⊥ of the equation Au = f with the noise free data f ∈ R(A) we have: Jα (uδα(δ) ) ≤ Jα (u).

(2.6.10)

According to (2.6.3), Auδα(δ) − f δ = δ and Au − f δ = δ, δ > 0. Taking into account this in (2.6.9) we conclude: 1 2 1 δ + α uδα(δ) 2H , 2 2 1 2 1 Jα (u) = δ + α u 2H . 2 2 Jα (uδα(δ)) =

2.6 Morozov’s Discrepancy Principle

63

This implies that with (2.6.10) that

uδα(δ) H ≤ u H , ∀δ > 0, α > 0,

(2.6.11)

i.e. the set {uδα(δ)}δ>0 is uniformly bounded in H by the norm of the solution u ∈ N (A)⊥ of the equation Au = f . Having the uniform boundedness of {uδα(δ)}δ>0 we estimate now the difference between the regularized and exact solutions:

uδα(δ) − u 2 = uδα(δ) 2 − 2Re(uδα(δ), u) + u 2

 ≤ 2 u 2 − Re(uδα(δ) , u) = 2Re(u − uδα(δ) , u). Since u = A∗ v, we transform this estimate as follows:

uδα(δ) − u 2 ≤ 2Re(u − uδα(δ), A∗ v) = 2Re(Au − Auδα(δ), v) = 2Re(f − Auδα(δ), v) ≤ 2Re(f − f δ , v) + 2Re(f δ − Auδα(δ), v)  ≤ 2 v f − f δ + f δ − Auδα(δ) . Using conditions (2.6.3) and the condition v H˜ ≤ M on the right hand side we finally obtain: √

uδα(δ) − u ≤ 2 Mδ. The right hand side tends to zero as δ → 0+ , which is the desired result.

(2.6.12)  

Remark 2.6.1 Estimate (2.6.12) depends on the norm of the element v ∈ H˜ which image u = A∗ v under the adjoint operator A∗ : H˜ → H is the solution of the equation Au = f with the noise free data f ∈ R(A). This element v ∈ H˜ is called a sourcewise element. In this case, we say that the exact solution u = A† f satisfies the source condition u = A∗ v, v ∈ H˜ . In applications, it is not necessary to satisfy the condition (2.6.3) exactly. Instead, the relaxed form condition β∗ δ ≤ Auδα(δ) − f δ ≤ β ∗ δ,

β ∗ > β∗ > 0, δ > 0

(2.6.13)

can be used. Therefore, if the noise level δ > 0 is known, then in the iteration algorithm one of the forms of condition (2.6.13) is used as a stopping rule, according to Morozov’s

64

2 Functional Analysis Background of Ill-Posed Problems

Discrepancy Principle. Iteration is terminated if δ,n(δ)

δ,n(δ)−1

Auα(δ) − f δ ≤ τM δ < Auα(δ)

− f δ ,

τM > 1, δ > 0,

(2.6.14)

that is, if for the first time the condition δ

Auδ,n(δ) α(δ) − f ≤ τM δ

holds. Here τM > 1 is a fixed parameter.

τM > 1, δ > 0

(2.6.15)

Chapter 3

Inverse Source Problems with Final Overdetermination

Inverse source problems for evolution PDEs ut = Au + F , t ∈ (0, T ], represent a well-known area in inverse problems theory and have extensive applications in various fields of science and technology. These problems play a key role in providing estimations of unknown and inaccessible source terms involved in the associated mathematical model, using some measured output. An inverse problem with the final overdetermination uT := u(T ), T > 0, for one-dimensional heat equation has first been considered by A.N. Tikhonov in study of geophysical problems [145]. In this work the heat equation with prescribed lateral and final data is studied in half-plane and the uniqueness of the bounded solution is proved. For parabolic equations in a bounded domain, when in addition to usual initial and boundary conditions, a solution is given at the final time, well-posedness of inverse source problem has been proved in [70, 71]. In this chapter we study inverse source problems for one-dimensional linear evolution equations with final overdetermination. The main reason of considering of this class of problems is to demonstrate an understanding of major concepts of inverse problems. First, we discuss the most widely studied and classical inverse source problem with final overdetermination for one-dimensional heat equation. Along with the other objectives of this chapter, we will show how the unique regularized solution obtained by Tikhonov regularization applied to the input-output operator  : H → H˜ is related to the Singular Value Decomposition (SDV) of the Moore-Penrose inverse † := (∗ )−1 ∗ of this operator. To this end, an adjoint problem, corresponding to each considered inverse source problem is introduced. This allows us to derive a gradient formula for the Fréchet derivative Jα (F ) of the regularized Tikhonov functional Jα (F ) through the weak solution of the adjoint problem. Using the gradient formula and solving the equation Jα (F ) = 0 with respect to the unknown source F , a quasi-solution of the inverse problem is obtained as a singular value expansion of the Moore-Penrose inverse † . Furthermore, it is shown unlike the final time output inverse source problem for parabolic equation and the backward parabolic problem, the final time output inverse source problem © Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_3

65

66

3 Inverse Source Problems with Final Overdetermination

for undamped wave equation ut t = (k(x)ux )x + F (x)G(t) is not a feasible problem in the sense that the unknown function F (x) can not be determined uniquely from the output uT (x) := u(x, T ) or νT (x) := ut (x, T ). A detailed study of this issue for damped wave and Euler-Bernoulli beam equations is given in Sect. 11.4.

3.1 Inverse Source Problem for Heat Equation In this section we demonstrate the Tikhonov regularization method and the Singular Value Decomposition (SVD) to the heat source identification problem with final data measurement. We develop the adjoint problem approach based on weak solution theory for PDEs. This approach allows not only to derive a gradient formula for the Fréchet derivative of the regularized Tikhonov functional, but also permits to prove the Lipschitz continuity of this gradient [56]. In the constant coefficient case, we prove that the solution of the nonlinear equation Jα (F ) = 0 is the singular value expansion of the unique minimizer of the regularized Tikhonov functional. Consider one dimensional heat conduction in a finite rod with the thermal conductivity k(x), occupying the interval (0, ). Assume that the source term is given in the separated form F (x)G(x, t), where the spacewise-dependent component F (x) is assumed to be unknown. Suppose that the process is governed by the following mixed initial-boundary value problem: ⎧ ⎨ ut = (k(x)ux )x + F (x)G(x, t), (x, t) ∈ T , u(x, 0) = f (x), x ∈ (0, ), ⎩ u(0, t) = 0, ux (, t) = 0, t ∈ [0, T ],

(3.1.1)

where T := {0 < x < , 0 < t ≤ T } and T > 0 is the final time and k(x) > 0 is the thermal conductivity. The initial and boundary conditions are assumed to be homogeneous without loss of generality, since the parabolic problem is linear. It is assumed that the spacewise-dependent source F (x) in (3.1.1) is unknown and needs to be identified from the measured temperature uT (x) := u(x, T ),

x ∈ (0, ),

(3.1.2)

at the final time T > 0. The final time measured output uT is assumed to be nonsmooth, that is uT ∈ L2 (0, ) and can contain a random noise. The problem of determining the pair u(x, t), F (x) in (3.1.1)–(3.1.2) will be defined as an inverse source problem with final data for heat equation. For a given function F (x) from some class of admissible sources, the problem (3.1.1) will be referred as a direct (or forward) problem. The analysis of inverse problems presented in this chapter is based on the weak solution theory for PDEs, since in practice, input and output data may not be smooth.

3.1 Inverse Source Problem for Heat Equation

67

Assuming that these data satisfy the following conditions 

k ∈ L∞ (0, ), 0 < k0 ≤ k(x) ≤ k1 < ∞, f, F ∈ L2 (0, ), G ∈ L2 (0, T ; L∞ (0, )),

(3.1.3)

where the norm of G(x, t) in the Banach space L2 (0, T ; L∞ (0, )) is defined as follows: 

G L2 (0,T ;L∞ (0,)) :=

T 0

1/2

G 2L∞ (0,)dt

.

We define the weak solution of the direct problem (3.1.1) as a function u ∈ L2 (0, T ; V(0, )), with ut ∈ L2 (0, T ; H −1 (0, )), satisfying the integral identity 







(ut v + k(x)ux vx )dx =

0

F (x)G(x, t)v(x)dx, a.e. t ∈ (0, T ), (3.1.4)

0

and the initial condition u(x, 0) = f (x), for every v ∈ V(0, ) := {w ∈ H 1 (0, ) : w(0) = 0}. Here and below H 1 (0, ) is the Sobolev space of functions which consists of all integrable real functions w : (0, ) → R such that w, w ∈ L2 (0, ) and H −1 (0, ) is the dual space of H 1 (0, ). Remark that the definition u ∈ L2 (0, T ; V(0, )) of the weak solution does not directly imply the continuity of this solution with respect to the time variable t ∈ (0, T ]. Hence, it is not clear in this definition the pointwise values u(x, 0) and u(x, T ) at the initial and final times. As it is follows from the theorem below, the conditions u ∈ L2 (0, T ; V(0, )) and ut ∈ L2 (0, T ; H −1 (0, )) imply u ∈ C([0, T ]; L2 (0, )), which means that values u(x, 0) and u(x, T ) are well defined (see [38], Ch. 5.9, Th. 3). Theorem 3.1.1 Let u ∈ L2 (0, T ; V(0, )) and ut ∈ L2 (0, T ; H −1 (0, )). (i) Then u ∈ C([0, T ]; L2 (0, )). (ii) The real valued function t → u(·, t) 2L2 (0,) is weakly differentiable and 1 d 2 dt



 0

 u2 (x, t)dx =



ut (x, t)u(x, t)dx, for a.e. t ∈ (0, T );

0

(iii) There exists a constant C0 = C0 (T ) > 0 such that the following estimate holds:   max u L2 (0,) ≤ C0 u L2 (0,T ;V (0,)) + ut L2 (0,T ;H −1 (0,)) .

t ∈[0,T ]

(3.1.5)

Theorem 3.1.1 with estimate (3.1.5) shows, in particular, that the weak solution u(x, t), as a function of t ∈ [0, T ], can be identified with its continuous

68

3 Inverse Source Problems with Final Overdetermination

representative and the initial and final time values u(x, 0), hence u(x, T ) and x ∈ [0, ] make sense. We use estimates in Lemma B.1.1 of Appendix B.1 for the weak solution of the direct problem (3.1.1), taking the inequality also into account

F G L2 (0,T ;L2 (0,)) ≤ F L2 (0,) G L2 (0,T ;L∞ (0,).

(3.1.6)

Then we have:

u 2L∞ (0,T ;L2 (0,))  ≤ Ce2 F 2L2 (0,) G 2L2 (0,T ;L∞ (0,)) + f 2L2 (0,) ,

u 2L2 (0,T ;L2 (0,))   ≤ Ce2 − 1 F 2L2 (0,) G 2L2 (0,T ;L∞ (0,)) + f 2L2 (0,) , (3.1.7)

ux 2L2 (0,T ;L2 (0,))  2 2 e2 F 2 2 , ≤C

G

+

f

2 ∞ 2 L (0,) L (0,T ;L (0,)) L (0,) e2 = Ce2 /(2k0 ). Ce2 = exp(T ), C If, in addition to conditions (3.1.3), the regularity conditions k ∈ C 1 (0, ), f ∈ H 2 (0, ), Gt ∈ L2 (0, T ; L2 (0, )),

(3.1.8)

are also satisfied, then there exists a regular weak solution of the direct problem (3.1.1), defined initially as u ∈ L2 (0, T ; H 2 (0, )) ∩ L∞ (0, T ; V(0, )), with ut ∈ L2 (0, T ; L2 (0, )) [38, Ch. 7.2]. Furthermore, it can be proved that u ∈ C([0, T ]; H 1(0, )) (see Theorem 4 [38], Ch. 5.9). This result is given by the following extension of Theorem 3.1.1. Theorem 3.1.2 Let u ∈ L2 (0, T ; H 2 (0, )) and ut ∈ L2 (0, T ; L2 (0, )). (i) Then u ∈ C([0, T ]; H 1(0, )), and hence, u ∈ C([0, T ] × (0, )). (ii) Furthermore, the following estimate holds:   max u H 1 (0,) ≤ C2 u L2 (0,T ;H 2 (0,)) + ut L2 (0,T ;L2 (0,)) ,

t ∈[0,T ]

(3.1.9)

where the constant C2 > 0 depends only on T > 0 and l > 0. This theorem with estimate (3.1.9), implies that, for a fixed t ∈ [0, T ] the function u(x, t), as a one dimensional function of x ∈ (0, ), is an element of the space H 1 (0, ). In view of estimates in Lemma B.1.2 of Appendix B.1 and inequality (3.1.6), the estimates for the regular weak solution of the direct problem (3.1.1) can be written

3.1 Inverse Source Problem for Heat Equation

69

in the following form:

ut 2L∞ (0,T ;L2 (0,))  ≤ Ce2 C˜ T2 F 2L2 (0,) G 2H 1 (0,T ;L∞ (0,)) + (kf  ) 2L2 (0,) ,

ut 2L2 (0,T ;L2 (0,)) (3.1.10)   ≤ Ce2 − 1 C˜ T2 F 2L2 (0,) G 2H 1 (0,T ;L∞ (0,)) + (kf  ) 2L2 (0,) ,

uxt 2L2 (0,T ;L2 (0,))  2  ) 2 e2 C˜ 2 F 2 2 , ≤C

G

+

(kf 1 ∞ 2 T L (0,) H (0,T ;L (0,)) L (0,) with Ce > 0 introduced in (3.1.7). Furthermore, using the identity ux (x, T ) = f  (x) +



T

uxt (x, t)dt, x ∈ (0, )

0

and the third estimate of (3.1.10) we derive the estimate for the output:

ux (·, T ) 2L2 (0,) ≤ f  2L2 (0,)  (3.1.11) 2  ) 2 e2 C˜ 2 F 2 2 . +T C

G

+

(kf T L (0,) H 1 (0,T ;L∞ (0,)) L2 (0,)

3.1.1 Compactness of Input-Output Operator and Fréchet Gradient Let us define the set of admissible spacewise-dependent sources F = {F ∈ L2 (0, ) : F L2 (0,) ≤ γF }. Evidently, F is a nonempty closed convex set in L2 (0, ). Denote by u(x, t; F ) the corresponding weak solution of (3.1.1) for a given F ∈ F . We introduce the inputoutput operator  : F → L2 (0, ), defined as (F )(x) := u(x, T ; F ), x ∈ (0, ), that is, operator  transforms each admissible input F (x) to the output u(x, T ; F ). Then the inverse source problem, defined by (3.1.1)–(3.1.2), can be reformulated as the following operator equation: F = uT , uT ∈ L2 (0, ), F ∈ F .

(3.1.12)

Let us analyze the compactness of the input-output operator  : F → L2 (0, ). Due to linearity of the problem, we may assume here that the initial data is zero: f (x) = 0.

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3 Inverse Source Problems with Final Overdetermination

Lemma 3.1.1 Let conditions (3.1.3) and (3.1.8) hold. Then the input-output operator  : F ⊂ L2 (0, ) → u(x, T ; F ) ∈ L2 (0, )

(3.1.13)

is a linear compact operator. Proof Let {Fm } ⊂ F be a bounded sequence and {u(x, t; Fm )} is the corresponding sequence of regular weak solutions of the direct problem (3.1.1). Then the sequence of outputs {u(x, T ; Fm )} is bounded in the norm of H 1 (0, ), by estimate (3.1.11), that is e2 C˜ T2 Fm 2 2

ux (·, T ; Fm ) 2L2 (0,) ≤ T C

G 2H 1 (0,T ;L∞ (0,)). L (0,) Since {Fm } ⊂ F , Fm L2 (0,) ≤ γF , by the definition of F , and the above estimate implies the uniform boundedness of the sequence of outputs {u(x, T ; Fm )} in the norm of H 1 (0, ). By the compact embedding H 1 (0, ) → L2 (0, ), the sequence of outputs {u(x, T ; Fm )} is relatively compact subset of L2 (0, ), hence  is a compact operator.   From Lemma 3.1.1 it follows that the inverse problem (3.1.12) (or (3.1.1)– (3.1.2)) is ill-posed. Then we may use Tikhonov regularization 1 Jα (F ) = J (F ) + α F 2L2 (0,) , F ∈ F , 2

(3.1.14)

introducing the Tikhonov functional J (F ) =

1

u(·, T ; F ) − uT 2L2 (0,), F ∈ F 2

(3.1.15)

and the parameter of regularization α > 0. Since  is a linear compact operator, by Theorem 2.1.1 the unique minimum Fα ∈ F of the regularized functional (3.1.14) exists and is the solution of the regularized normal equation (∗  + αI )Fα = ∗ uT ,

(3.1.16)

where ∗ : L2 (0, ) → L2 (0, ) is the adjoint operator. This unique solution can be written as follows: Fα = Rα uT , Rα := (∗  + αI )−1 ∗ : L2 (0, ) → L2 (0, ), (3.1.17) where Rα : L2 (0, 1) → L2 (0, ), α > 0, is a regularization strategy. Consider first the non-regularized case, when α = 0. The following lemma establishes the gradient formula for the Tikhonov functional (3.1.15) through the solution of (unique) adjoint problem and known source G(x, t).

3.1 Inverse Source Problem for Heat Equation

71

Lemma 3.1.2 Let conditions (3.1.3) hold. Denote by u(x, t; F ) is the weak solution of the parabolic problem (3.1.1) corresponding to a given F ∈ L2 (0, ). Then the Tikhonov functional (3.1.15) is Fréchet differentiable and for the Fréchet gradient J  (F ) the following gradient formula holds: J  (F )(x) = (ψ(x, ·; F ), G(x, ·))L2 (0,T ) :=



T

ψ(x, t; F )G(x, t)dt, (3.1.18) 0

for a.e. x ∈ (0, ), where ψ(x, t; F ) is the weak solution of the following adjoint problem: ⎧ ⎨ ψt = −(k(x)ψx )x , (x, t) ∈ T , ψ(x, T ) = u(x, T ; F ) − uT (x), x ∈ (0, ), ⎩ ψ(0, t) = 0, ψx (, t) = 0, t ∈ [0, T ).

(3.1.19)

Proof Assuming that F, F +δF ∈ F , we calculate the increment δJ (F ) := J (F + δF ) − J (F ) of the Tikhonov functional (3.1.15): 



δJ (F ) =

[u(x, T ; F ) − uT (x)]δu(x, T ; F )dx

0

1 + 2





[δu(x, T ; F )]2 dx,

(3.1.20)

0

where δu(x, t; F ) := u(x, t; F + δF ) − u(x, t; F ) is the weak solution of the following parabolic problem ⎧ ⎨ δut = (k(x)δux )x + δF (x)G(x, t), (x, t) ∈ T , δu(x, 0) = 0, x ∈ (0, ), ⎩ δu(0, t) = 0, δux (, t) = 0, t ∈ (0, T ].

(3.1.21)

We transform the first integral on the right hand side of (3.1.20), assuming that ψ(x, t; F ) and δu(x, t; F ) are the solutions of problems (3.1.19) and (3.1.21), accordingly. We have: 







[u(x, T ; F ) − uT (x)]δu(x, T ; F )dx =

0

  T

 = 0

 

ψ(x, T ; F )δu(x, t; F )dx

0

 (ψ(x, t; F )δu(x, t; F ))t dt dx

0

{ψt (x, t; F )δu(x, t; F ) + ψ(x, t; F )δut (x, t; F )} dxdt

=  

T

{−(k(x)ψx )x δu(x, t; F ) + ψ(x, t; F )(k(x)δux )x } dxdt

= T

72

3 Inverse Source Problems with Final Overdetermination

  +

δF (x)G(x, t)ψ(x, t; F )dxdt T



T

= 0

{−k(x)ψx δu + k(x)δux ψ}x= x=0 dt +

  δF (x)G(x, t)ψ(x, t; F )dxdt. T

Taking into account here the boundary conditions given in (3.1.19) and (3.1.21) we obtain the following integral identity: 



  [u(x, T ; F ) − uT (x)]δu(x, T ; F )dx =

ψ(x, t; F )δF (x)G(x, t)dxdt,

0

T

for all F, δF ∈ L2 (0, ). With formula (3.1.20) this implies:   T

 δJ (F ) = 0

ψ(x, t; F )G(x, t)dt δF (x)dx +

0

1 2



l

[δu(x, T ; F )]2 dx.

(3.1.22)

0



The last right hand side integral is of the order O δF 2L2 (0,) by estimate (3.1.7). This completes the proof of lemma.   Notice that the adjoint problem (3.1.19) is a well-posed problem as the change of the time variable τ = T − t shows. In particular, this means that the above estimates derived for the direct problem (3.1.1) can also be used for this problem. One of most important issues in numerical solution of inverse problems is the Lipschitz continuity of the Fréchet gradient. Lemma 3.4.4 in Sect. 3.4.3 shows that if {F (n) } is the a sequence of iterations defined by the gradient algorithm, then J (F (n) ) is a monotone decreasing sequence. Lemma 3.1.3 Let conditions (3.1.3) hold. Then Fréchet gradient of the Tikhonov functional (3.1.15) is Lipschitz continuous,

J  (F + δF ) − J  (F ) L2 (0,) ≤ L1 δF L2 (0,), ∀F, F + δF ∈ F ,   2 with the Lipschitz constant L21 = Ce2 Ce2 + 1 MG

G L2 (0,T ;L∞ (0,)) and Ce = exp(T ).

(3.1.23)

> 0, where MG

Proof By the definition, 

J (F + δF ) − J



 (F ) 2L2 (0,)

=

2

  T

δψ(x, t; F )G(x, t)dt 0

0

dx,

=

3.1 Inverse Source Problem for Heat Equation

73

where δψ(x, t; F ) := ψ(x, t; F + δF ) − ψ(x, t; F ) is the weak solution of the problem: ⎧ ⎨ δψt = −(k(x)δψx )x , (x, t) ∈ T ; δψ(x, T ) = δu(x, T ; F ), x ∈ (0, ); ⎩ δψ(0, t) = 0, δψx (, t) = 0, t ∈ [0, T ).

(3.1.24)

Applying the Hölder inequality we deduce that 2

δψ(·, ·; F ) 2L2 (0,T ;L2 (0,)).

J  (F + δF ) − J  (F ) 2L2 (0,) ≤ MG

(3.1.25)

We need to prove that the right hand side norm in (3.1.25) is of the order  O δF L2 (0,) . To this end we apply the second estimate of (3.1.7) to the weak solution δψ(x, t; F ) of the adjoint problem (3.1.24). We have:



δψ 2L2 (0,T ;L2 (0,)) ≤ Ce2 + 1 δ(·, T ) 2L2 (0,). Next, we apply the first estimate of (3.1.7) to the weak solution δψ(x, t; F ) of problem (3.1.21), to estimate the norm ) δ(·, T ) 2L2 (0,). This yields:

δ(·, T ) 2L2 (0,) ≤ Ce2 δF 2L2 (0,) G 2L2 (0,T ;L∞ (0,)). These show that the right-hand-side of (3.1.25) is of the order 

two estimates 2   O δF L2 (0,) This implies the desired result. An implementation of the gradient formula (3.1.18) in Conjugate Gradient Algorithm will be discussed in subsequent sections. Obviously, it is an implicit formula, since the solution ψ(x, t; F ) of the adjoint problem (3.1.19) depends on the unknown source implicitly. We may use now formula (3.1.18) in (3.1.14) to obtain the gradient formula for the regularized functional. Corollary 3.1.1 For the Fréchet gradient of the regularized functional Jα (F ), α > 0, defined by (3.1.14) the following formula holds: Jα (F )(x) =



T

ψ(x, t; F )G(x, t)dt + αF (x), a.e. x ∈ (0, ).

(3.1.26)

0

Taking into account Corollary 2.5.1 in Sect. 2.5, we may rewrite the gradient formula (3.1.26) for the regularized functional through the input-output mapping  : L2 (0, ) → L2 (0, ) as follows: Jα (F ) = ∗ (F − uT ) + αF, F ∈ L2 (0, ).

(3.1.27)

74

3 Inverse Source Problems with Final Overdetermination

With the necessary condition Jα (F ) = 0, formula (3.1.26) allows to derive the following representation for the unique regularized solution of the inverse source problem with final data. Corollary 3.1.2 The unique regularized solution Fα ∈ L2 (0, ) of the inverse problem (3.1.1)–(3.1.2) can be represented as follows: Fα (x) = −

1 α



T

ψ(x, t; Fα )G(x, t)dt, a.e. x ∈ (0, ), α > 0.

(3.1.28)

0

This representation is an analogue of the representation 1 uδα = − A∗ (Auδα − f δ ), f δ ∈ H˜ , α for the solution of the regularized normal equation (A∗ A + αI )uδα = f δ , introduced in Sect. 2.5. Remark 3.1.1 Let the source term in the heat equation (3.1.1) has the separable form F (x)G(t). As it was shown above, the regularized solution Fα ∈ L2 (0, ) of the inverse problem (3.1.1)–(3.1.2) is defined uniquely for all G ∈ L2 (0, T ), as a solution of the normal equation (3.1.16). However, the non-regularized solution F ∈ L2 (0, ) of the inverse problem (3.1.1)–(3.1.2) may not be unique for all G ∈ L2 (0, T ). Specifically, the time dependent source G ∈ L2 (0, T ) must ensure the fulfillment of the condition σn = 0, for all n ∈ N, as we will see in Sect. 3.1.2. For example, if the function G(t) satisfies the conditions G(t) > 0 and G ∈ L2 (0, T ), then σn = 0, for all n ∈ N and the non-regularized solution is unique. In this case the gradient formula (3.1.18) has the form: J  (F )(x) =



T

ψ(x, t; F )G(t)dt ≡ ∗ (F − uT ) (x),

(3.1.29)

0

a.e. for all x ∈ (0, ).

3.1.2 Singular Value Decomposition of Input-Output Operator In this subsection we assume that the source term in the heat equation (3.1.1) has the separable form F (x)G(t), the time dependent source function G(t) satisfies the conditions

G L2 (0,T ) > 0, G(t) ≥ 0, t ∈ [0, T ]

(3.1.30)

3.1 Inverse Source Problem for Heat Equation

75

and the initial data is zero: f (x) = 0. Specifically, consider the problem of determining the unknown space-wise dependent source F ∈ L2 (0, ) in ⎧⎧ ⎨ ut = (k(x)ux )x + F (x)G(t), (x, t) ∈ T ; ⎪ ⎪ ⎨ u(x, 0) = 0, x ∈ (0, ); ⎩ ⎪ u(0, t) = 0, ux (, t) = 0, t ∈ (0, T ]; ⎪ ⎩ uT (x) := u(x, T ), x ∈ (0, ).

(3.1.31)

where uT ∈ L2 (0, ) is a noise free output satisfying the following consistency conditions: uT (0) = 0, uT () = 0.

(3.1.32)

First we will prove that the input-output operator  : L2 (0, ) → L2 (0, ) corresponding to the inverse problem (3.1.31) is a self-adjoint operator. Then we will use the singular value decomposition (SVD) of compact operators on a Hilbert space given in Sect. 2.4, to obtain singular value expansion of the regularized solution Fα ∈ L2 (0, ) of this inverse problem. Lemma 3.1.4 Let conditions (3.1.3) and (3.1.30) hold. Then the input-output operator , defined by (3.1.13) and corresponding to the inverse problem (3.1.31), is self-adjoint and positive defined. Moreover, (ϕm )(x) = κm ϕm (x),

(3.1.33)

that is, {κm , ϕm } is the eigensystem of the input-output operator , where {ϕn }∞ n=1 are orthonormal eigenvectors of the Sturm-Liouville operator L : V(0, ) → L2 (0, ) defined by 

(Lϕ)(x) := −(k(x)ϕ  (x)) = λϕ(x), x ∈ (0, ); ϕ(0) = 0, ϕ  () = 0,

(3.1.34)

corresponding to eigenvalues {λn }∞ n=1 , and  κn =

T

exp(−λn (T − t)G(t)dt, n = 1, 2, 3, . . . .

(3.1.35)

0

Proof Evidently, the differential operator L : V(0, ) → L2 (0, ) defined by (3.1.34) is self-adjoint and positive defined. Hence there exists an infinite number of positive eigenvalues {λn }∞ n=1 , 0 < λ1 < λ2 < λ3 . . ., repeated according to their (finite) multiplicity that λn → ∞, as n → ∞. We assume, without loss of generality, that the eigenvectors {ϕn }∞ n=1 , corresponding to the eigenvalues λn , are normalized (dividing the both sides of (3.1.34) by ϕn L2 (0,) ). Then, these eigenvectors {ϕn }∞ n=1 form an orthonormal

76

3 Inverse Source Problems with Final Overdetermination

basis in L2 (0, ) for the operator L, defined by (3.1.34). Indeed, the identities  (x)) = λϕ (x) imply: −(k(x)ϕn (x)) = λϕn (x) and −(k(x)ϕm m (λn − λm )ϕn (x)ϕm (x) =

d   k(x)ϕn (x)ϕm (x) − k(x)ϕn (x)ϕm (x) . dx

Integrate this identity on [0, ] and use the boundary conditions (3.1.34): (λn − λm )ϕn (x)ϕm (x) = 0, ∀n, m = 1, 2, 3, . . . . Therefore, for λn = λm ,  (ϕn (x), ϕm (x))L2 (0,) :=



ϕn (x)ϕm (x)dx = 0, n = m,

0

which means that (ϕn , ϕm )L2 (0,) = δnm , n, m = 1, 2, 3, . . . . Here and below δnm is the Kronecker delta. With this orthonormal basis we use the Fourier series expansion u(x, t) =

∞ 

un (t)ϕn (x)

(3.1.36)

n=1

of the solution of the initial boundary value problem given by the first three equations of (3.1.31). Here 

t

un (t) = Fn

exp(−λn (t − τ ))G(τ )dτ, t ∈ (0, T ]

(3.1.37)

0

is the solution of the Cauchy problem 

un (t) + λn un (t) = Fn G(t), t ∈ (0, T ], un (0) = 0,

for each n = 1, 2, 3, . . . and Fn := (F, ϕn )L2 (0,) is the Fourier coefficient of the function F ∈ L2 (0, ). Now we can use (3.1.36) and (3.1.37) to obtain the Fourier series expansion of the input-output operator , defined as (F )(x) := u(x, T ; F ): (F )(x) =

∞ 

κn Fn ϕn (x),

(3.1.38)

n=1

where κn is defined by (3.1.35). Remark that, {ϕn }∞ n=1 are eigenvectors of the input-output operator , corresponding to eigenvalues {κn }∞ n=1 . To show this, we replace F (x) by ϕm (x) in

3.1 Inverse Source Problem for Heat Equation

77

(3.1.38). Then we get (3.1.33): (ϕm )(x) =

∞ 

κn (ϕm , ϕn )L2 (0,)ϕn (x) = κm ϕm (x).

(3.1.39)

n=1

Furthermore,  : L2 (0, ) → L2 (0, ) is a self-adjoint operator, that is, (F, F˜ )L2 (0,) = (F, F˜ )L2 (0,), ∀F, F˜ ∈ L2 (0, ). Indeed,  (F, F˜ )L2 (0,) :=

∞ 

κn Fn ϕn (x),

n=1

∞ 

 F˜m ϕm (x)

m=1

=

∞ 

L2 (0,)

κn Fn F˜n = (F, F˜ )L2 (0,),

n=1

where F˜n := (F˜ , ϕn )L2 (0,) is the Fourier coefficient of the function F˜ ∈ L2 (0, ).   Hence,  = ∗ , where ∗ : L2 (0, ) → L2 (0, ) is the adjoint operator. Thus, the input-output operator  : L2 (0, ) → L2 (0, ) is a self-adjoint ∞ operator with eigenvectors {ϕn }∞ n=1 , corresponding to different eigenvalues {κn }n=1 . It follows from (3.1.39) that  L2 (0,) = κ1 . Hence,

 L2 (0,) = κ1 > κ2 > κ3 > . . . .

(3.1.40)

Further, formula (3.1.38) implies that (∗ F )(x) =

∞ 

κn2 Fn ϕn (x).

(3.1.41)

n=1

By Definition 2.4.1 in Sect. 2.4, the square root of eigenvalues κn2 of the self-adjoint operator ∗  is defined as the singular values of the input-output operator , that is, σn := κn > 0, n = 1, 2, 3, . . . . Hence, the singular system {σn , un , vn } for the self-adjoint operator input-output operator  is {κn , ϕn , ϕn }, according to the definition in Sect. 1.4. Indeed, un = ϕn and vn := un / un = ϕn / ϕn . But ϕn = κn ϕn , by (3.1.33), and ϕn = κn . This implies that vn = κn ϕn /|κn |. Since κn > 0, we have vn = ϕn . Therefore, if conditions (3.1.30) hold, then κn > 0, for all n = 1, 2, 3, . . . , as formula (3.1.35) shows. In this case, the singular system {κn , ϕn , ϕn } for the selfadjoint operator input-output operator  is uniquely determined by the eigensystem {κn , ϕn } of the differential operator (3.1.34).

78

3 Inverse Source Problems with Final Overdetermination

The following theorem gives a series representation of the unique solution of the regularized form (3.1.16) of the normal equation. Theorem 3.1.3 Let conditions (3.1.3) hold. Assume that uT ∈ L2 (0, ) is a noise free measured output defined in (3.1.31). Then for the unique minimum Fα ∈ L2 (0, ) of the regularized functional (3.1.14) the following singular value expansion holds: Fα (x) =

∞  q(α; κn )

κn

n=1

uT ,n ϕn (x), x ∈ (0, ),

(3.1.42)

where q(α; κ) =

κ2 κ2 + α

(3.1.43)

is the filter function, α > 0 is the parameter of regularization, κn , n = 1, 2, 3, . . ., defined by formula (3.1.35) are the eigenvalues of the input-output operator , uT ,n := (uT , ϕn ) is the nth Fourier coefficient of uT (x) and ϕn (x) are the normalized eigenfunctions corresponding to the eigenvalues λn of the operator L : V(0, ) → L2 (0, ) defined by (3.1.34). Proof Let u(x, t; F ) be the weak solution of the direct problem defined by the first three equations of (3.1.31), for a given admissible F ∈ L2 (0, ). Then for the output u(x, T ; F ) the expansion (3.1.38) holds: u(x, T ; F ) =

∞ 

κn Fn ϕn (x),

(3.1.44)

n=1

We use this with the nth Fourier coefficients uT ,n := (uT , ϕ)L2 (0,) of the noise free measured output uT ∈ L2 (0, ) in (3.1.14)–(3.1.15). Then we have: 1  (κn Fn − uT ,n )2 + αFn2 , 2 ∞

Jα (F ) =

n=1

Transforming the nth term under the sum we get: ! "  2 ∞ 1 κ α n uT ,n + 2 u2 Jα (F ) = (κn2 + α) Fn − 2 , 2 κn + α κn + α T ,n n=1

The regularized functional achieves minimum value if Fn −

κn uT ,n = 0, κn2 + α

3.1 Inverse Source Problem for Heat Equation

79

as the right hand side shows. This defines the nth Fourier coefficient of the unique minimum Fα ∈ L2 (0, ) of the regularized functional (3.1.14): Fα,n =

κn uT ,n . κn2 + α

Substituting this into the Fourier series expansion Fα (x) =

∞ 

Fα,n ϕn (x)

(3.1.45)

n=1

of the function Fα ∈ L2 (0, ) we arrive at the required expansion (3.1.42).

 

The representation formula Fα (x) = −

1 α



T

ψ(x, t; Fα )G(t)dt, a.e. x ∈ (0, ), α > 0,

(3.1.46)

0

which follows from (3.1.28) for the solution of the inverse problem (3.1.31) (the case G(x, t) ≡ G(t)), shows the dependence of the solution Fα ∈ L2 (0, ) of this inverse problem on the solution of the adjoint problem (3.1.19). The same dependence can be observed in the gradient formulae (3.1.18) and (3.1.29). Hence it is useful to illustrate a relationship between the singular value expansion (3.1.45) and the representation formula (3.1.46). Specifically, the following example shows their equivalence [63]. Example 3.1.1 The relationship between the singular value expansion (3.1.42) and the representation formula (3.1.46). We assume here that k(x) ≡ k = const > 0. Then the above defined eigenvalues λn and normalized eigenfunctions ϕn (x) of the operator (Lϕ)(x) := −kϕ  (x), defined by (3.1.34), are   λn = [(n − 1/2)π/]2, ϕn (x) = 2/ sin( λn x), n = 1, 2, 3, . . . . (3.1.47) Replacing in formula (3.1.35) λn with kλn , we get:  κn =

T

exp(−kλn (T − t)G(t)dt, n = 1, 2, 3, . . . .

(3.1.48)

0

Let us apply the Fourier method to the adjoint problem (3.1.19) using the above defined normalized eigenfunctions ϕn (x) corresponding to the eigenvalues λn . We have: ψ(x, t; F ) =

∞  n=1

ψn (t; F )ϕn (x),

(3.1.49)

80

3 Inverse Source Problems with Final Overdetermination

where the ψn (t) is the solution of the backward Cauchy problem 

ψn (t) = λn ψn (t), t ∈ [0, T ), ψn (T ) = un (T ; Fn ) − uT ,n .

The solution of this problem is  ψn (t; Fn ) = un (T ; Fn ) − uT ,n exp(−kλn (T − t)), n = 1, 2, 3, . . . .

(3.1.50)

Now we use the Fourier series expansions (3.1.45) and (3.1.49) in the representation formula (3.1.46) to obtain the formula Fα,n

1 =− α



T

ψn (t; Fα,n )G(t)dt, a.e. x ∈ (0, ), α > 0,

(3.1.51)

0

for the nth Fourier coefficient of the unique minimum Fα ∈ L2 (0, ) of the regularized functional (3.1.14). On the other hand, assuming F (x) = Fα (x) in (3.1.50), multiplying it both sides by −G(t)/α = 0, integrating over [0, T ] and then using formula (3.1.35) for κn , we obtain: −

1 α



T

ψn (t; Fα,n )G(t)dt = −

0

κn [un (T ; Fα ) − uT ,n ]. α

Comparing this formula with (3.1.51) we deduce: Fα,n = −

κn [un (T ; Fα,n ) − uT ,n ]. α

(3.1.52)

Further, it follows from the expansions (3.1.36) and (3.1.44) that between the nth Fourier coefficient Fα,n of the unique minimum Fα ∈ L2 (0, ) and nth Fourier coefficient un (T ; Fα ) of the output data u(x, T ; F α) the following relationship holds: un (T ; Fα ) = κn Fα,n .

(3.1.53)

Using this formula in (3.1.52) we conclude: un (T ; Fα,n ) κn = − [un (T ; Fα,n ) − uT ,n ], κn α which yields: un (T ; Fα ) = q(α, κn )uT ,n , n = 0, 1, 2, 3, . . . .

(3.1.54)

3.1 Inverse Source Problem for Heat Equation

81

This is a relationship between the Fourier coefficients un (T ; Fα ) and uT ,n of the output u(x, T ; Fα ) and the measured output uT (x), through the filter function q(α, κ), defined by formula (3.1.43). Formulae (3.1.53) and (3.1.54) permit to derive the nth Fourier coefficient Fα,n of the unique minimum Fα ∈ L2 (0, ) of the regularized functional (3.1.14) and nth Fourier coefficient uT ,n of the final time measured output uT (x) through the filter function q(α, κ): Fα,n =

q(α, κn ) uT ,n , n = 0, 1, 2, 3, . . . . κn

(3.1.55)

The singular value expansion (3.1.42) for the unique minimum Fα ∈ L2 (0, ) follows from the expansion (3.1.45) and formula (3.1.55). 

3.1.3 Picard Criterion and Regularity of the Input and Output. Solvability and Stability Estimate We extract here several important conclusions from Theorem 3.1.3. Let us first assume that α = 0. Then, by formula (3.1.43), q(α, σ ) = 1, and expansion (3.1.42) becomes: F (x) =

∞  1 uT ,n ϕn (x), x ∈ (0, ), κn

(3.1.56)

n=1

This expansion exactly coincides with the singular value expansion (2.4.13) given in Picard’s Theorem 2.4.1 in Sect. 2.4. Since F0 = † uT is the solution of the normal equation (3.1.16) with α = 0 and † : L2 (0, ) → L2 (0, ) is Moore-Penrose inverse of the input-output operator, we conclude that (3.1.56) is a singular value expansion of the solution of the normal equation, when α = 0. The Picard criterion given in Theorem 2.4.1 of Sect. 2.4 and applied to the inverse problem (3.1.31) implies that the series (3.1.56) converges if and only if uT ∈ N (∗ )⊥ and the following condition holds: ∞  u2T ,n n=1

κn2

< ∞.

(3.1.57)

Remember that {ϕn (x)} forms a complete orthonormal system in L2 (0, ) and therefore for the self-adjoint input-output operator  : L2 (0, ) → L2 (0, ) we deduce that N () = N (∗ ) = {0}. On the other hand, D(† ) := R() ⊕ R()⊥ , by the definition of the Moore-Penrose inverse. Since N (∗ ) = R()⊥ , by Theorem 2.2.1, this implies that D(† ) = R() and D(† ) is dense in L2 (0, ).

82

3 Inverse Source Problems with Final Overdetermination

The fulfilment of the Picard criterion (3.1.57) depends on two factors: κn and uT ,n . As formula (3.1.35) shows, an asymptotic behavior of the first factor κn depends on a class where the input G(t) is defined. The second factor is the Fourier coefficient uT ,n := (uT , ϕn )L2 (0,) of the measured output uT (x). As a consequence, the convergence of the series (3.1.56) depends on the input data G(t) and the measured output uT (x). Based on this observation, we will analyze here the relationship between the convergence of the series (3.1.56) and the regularity of the input G ∈ L2 (0, T ) and the measured output uT ∈ L2 (0, ) data. First, we present the following result which shows the necessary condition for solvability of the inverse problem (3.1.31) with the noise free measured output uT ∈ N (∗ )⊥ ≡ R(). Corollary 3.1.3 Let conditions (3.1.3) hold. Assume that uT ∈ L2 (0, ) is a noise free measured output defined in (3.1.31). If the time dependent function G ∈ L2 (0, T ) satisfies conditions (3.1.30), then ∞ 

n2 u2T ,n < ∞, uT ,n := (uT , ϕn )L2 (0,).

(3.1.58)

n=1

is the necessary condition for solvability of the inverse problem (3.1.31). Proof Using formula (3.1.35) for κn , we estimate the singular values σn := |κn | as follows:

 T



0 < κn =

exp(−λn (T − t))G(t)dt

0



T

≤

1/2 exp(−2λn (T − t))dt

0

G L2 [0,T ]

1 = √ [1 − exp(−2λn T ]1/2 G L2 [0,] . 2λn

(3.1.59)

It is known that the eigenvalues λn of the differential operator  : V(0, ) → L2 (0, ), defined in (3.1.34), are of order λn = O(n2 ). Then we conclude from the above estimate that, κn = O(1/n). Using this property in the Picard criterion (3.1.57) we arrive at the condition (3.1.58).   To find out what means the necessary condition (3.1.58) in terms of the measured output uT ∈ L2 (0, ), we need some auxiliary results from Fourier series theory. It 2 is known that if {ϕn (x)}∞ n=1 is an orthonormal basis of L (0, ) then the Fourier series of f ∈ L2 (0, ) with respect to this basis converges in L2 -norm if and only if Parseval’s identity

f 2L2 (0,) =

∞  n=1

fn2 ,

(3.1.60)

3.1 Inverse Source Problem for Heat Equation

83

holds. Here and below, fn := (f, ϕn )L2 (0,) is the nth Fourier coefficient of 2 the element f ∈ L2 (0, ). Let now assume that {ϕn (x)}∞ n=1 ⊂ L (0, ) is the orthonormal basis defined by the eigenfunctions of the Sturm-Liouville problem (3.1.34). Then the theorem below shows that in this case Parseval’s identity holds. 2 Theorem 3.1.4 Let {ϕn (x)}∞ n=1 ⊂ L (0, ) be the orthonormal basis defined by the eigenfunctions of the Sturm-Liouville problem (3.1.34). Then Parseval’s identity (3.1.60) holds for f ∈ L2 (0, ) with respect the basis {ϕn (x)} and the Fourier series

f (x) =

∞ 

fn ϕn (x)

(3.1.61)

n=1

of an element f ∈ L2 (0, ) with respect to this basis converges in L2 -norm. Proof of this theorem can be found in [47]. Now, replacing in this theorem f ∈ L2 (0, ) by uT ∈ L2 (0, ) and reformulating the above assertion in terms of the measured output we conclude that if uT ∈ L2 (0, ), then Parseval’s identity holds:

uT 2L2 (0,) =

∞ 

u2T ,n ,

(3.1.62)

n=1

More precisely, having only the condition uT ∈ L2 (0, ) we can not guarantee the fulfilment of the Picard criterion (3.1.58). In other words, more regularity is required from the output uT (x) in order to fulfil the Picard criterion. To find out such a class of functions for which the Picard criterion (3.1.58) holds, we need the following auxiliary result. 2 Lemma 3.1.5 Let conditions (3.1.3) hold. Assume that {ϕn (x)}∞ n=1 ⊂ L (0, ) is the orthonormal basis defined by the eigenfunctions corresponding to the eigenvalues {λn }∞ n=1 of the Sturm-Liouville problem (3.1.34). Then the following assertions hold true: √ (i) The system {ϕn (x)/ λn }∞ n=1 forms an orthonormal basis of V(0, ) := {v ∈ H 1 (0, ) : v(0) = 0} with the inner product

 (Lw, v)L2 (0,) :=



k(x)w (x)v  (x)dx.

(3.1.63)

vn ϕn (x),

(3.1.64)

0

(ii) If v ∈ V(0, ), then the Fourier series v(x) =

∞  n=1

84

3 Inverse Source Problems with Final Overdetermination

with the Fourier coefficients vn = (v, ϕn )L2 (0,), converges in the norm of the space V(0, ) ⊂ H 1 (0, ). Proof Multiplying the both sides of Lϕn (x) = λn ϕn (x) by ϕm (x) we conclude: 

(Lϕn , ϕn )L2 (0,) = λn ϕn 2L2 (0,) = λn , m = n,

(Lϕn , ϕm )L2 (0,) = λn (ϕn , ϕm )L2 (0,) = 0, m = m.

(3.1.65)

Hence   (Lϕn / λn , ϕm / λm )L2 (0,) =



1, m = n, 0, m =  m.

√ This implies that {ϕn (x)/ λn }∞ n=1 forms an orthonormal subset of V(0, ). To prove that this subset is an orthonormal basis of V(0, ), it is sufficient to show that (Lϕn , v)L2 (0,) = 0 implies v ≡ 0, for any v ∈ V(0, ). Indeed, (Lϕn , v)L2 (0,) = λn (ϕn , v)L2 (0,) = 0. basis of L2 (0, ), so, λm (ϕn , u)L2 (0,) = 0 Since {ϕn (x)}∞ n=1 forms an orthonormal √ ∞ implies v ≡ 0. Therefore, {ϕn (x)/ λn }n=1 forms an orthonormal basis of V(0, ) with the inner product (3.1.63). To prove the second assertion (ii), first we remark that the norms v V (0,) and

v  L2 (0,) are equivalent, due to the Dirichlet boundary condition ϕn (0) = 0 in (3.1.34). Furthermore, conditions 0 < k0 ≤ k(x) ≤ k1 < ∞ imply an equivalence 1/2 of the energy norm v L := (Lv, v)L2 (0,) , defined by the inner product (3.1.63), and the norm v  L2 (0,). Consider now the series v(x) =

∞ 

ϕn (x) μn √ , λn n=1

(3.1.66)

√ with the Fourier coefficients μn = (Lv, ϕn (x)/ λn )L2 (0,) (with respect to the basis √ √ ∞ {ϕn (x)/ λn }). Since the system {ϕn (x)/ λn }n=1 forms an orthonormal basis of V(0, ), the series (3.1.66) converges in the norm √ of V(0, ). Comparing the series (3.1.64) and (3.1.66) we deduce that μn = λn vn . This implies that the series (3.1.64) also converges in the norm of V(0, ).   Corollary 3.1.4 Let conditions of Corollary 3.1.3 hold. Assume that the output uT (x) satisfies the consistency conditions (3.1.32). Then the necessary condition (3.1.58) for solvability of the inverse problem (3.1.31) is equivalent to the condition uT ∈ V(0, ).

3.1 Inverse Source Problem for Heat Equation

85

Proof It follows from (3.1.65) and (3.1.66), applied to the measured output uT ∈ V(0, ), that

uT 2L := (LuT , uT )L2 (0,) =

∞ 

λn u2T ,n , uT ,n := (uT , ϕn )L2 (0,) .

(3.1.67)

n=1

1/2  and uT V (0,) are equivalent Since the norms uT L := LuT , uT )L2 (0,) and λn = O(n2 ), the series (3.1.67) converges if and only if condition (3.1.58) holds.   Remark 3.1.2 From the characterization of Sobolev spaces by their Fourier coefficients (Theorem 7, Sect 5.8.4 [38]) it follows that the series ∞ 

n2 u2T ,n

n=1

converges if and only if uT ∈ V(0, ) ⊂ H 1 (0, ). It is important to note that the Picard criterion (3.1.57) implicitly includes also the requirement κn = 0, for all n = 1, 2, 3 . . . .

(3.1.68)

This condition coincides with unique solvability of the inverse problem (3.1.31). Indeed, it follows from the Fourier series expansion (3.1.38) of the input-output operator  that if κm = 0 for some m, then the mth Fourier coefficient Fm := (F, ϕm )L2 (0,) of the unknown function F ∈ L2 (0, ) can not be determined uniquely. More precisely, the solution F ∈ L2 (0, ) of the inverse problem (3.1.31) can only be determined up to the additive term Cm ϕm (x), where Cm is an arbitrary constant. In this context, (3.1.30) are the sufficient conditions ensuring the fulfilment of condition (3.1.68). Remark also that if κn = 0, for some n, then the input-output operator  is not bijective, as the Fourier series expansion (3.1.38) shows. Hence in this case the bijectivity condition of Theorem 2.5.1 in Sect. 2.5 is not satisfied. The following result shows that the conditions 0 < G0 ≤ G(t) ≤ G1 < ∞, for all t ∈ [0, T ] can be considered as the sufficient condition ensuring the fulfilment of the requirement (3.1.68). Corollary 3.1.5 Let conditions (3.1.3) hold. Assume that uT ∈ L2 (0, ) is a noise free measured output defined in (3.1.31). If the time dependent function G ∈ L2 (0, T ) satisfies the conditions 0 < G0 ≤ G(t) ≤ G1 < ∞, t ∈ [0, T ],

(3.1.69)

86

3 Inverse Source Problems with Final Overdetermination

then the inverse problem (3.1.31) is uniquely solvable if and only if ∞ 

n4 u2T ,n < ∞.

(3.1.70)

n=1

Proof By using conditions (3.1.69), we can easily derive the estimate 0
0, CST = (G0 [1 − exp(−λ1 T )])−1 > 0, G0 > 0 is the lower bound of G(t) introduced in (3.1.69) and λ1 > 0 is the smallest eigenvalue of the SturmLiouville operator introduced in Lemma 3.1.4. Proof The function δu(x, t) := u(x, t) − # u(x, t) solves the direct problem defined #(x). Then the in (3.1.31) with the input F (x) replaced by δF (x) := F (x) − F

3.1 Inverse Source Problem for Heat Equation

87

relation (3.1.72) for the Fourier coefficients δuT ,n = (δuT , ϕn )L2 (0,), δuT (x) := u(x, T ) − # u(x, T ), and δFn = (δF, ϕn )L2 (0,) reads as follows: 

T

δuT ,n = δFn

exp(−λn (T − t))G(t)dt, 1, 2, 3, . . . .

(3.1.74)

0

On the other hand, the lower estimate G(t) ≥ G0 > 0 for all t ∈ [0, T ] leads to 

T

exp(−λn (T − t))G(t)dt ≥

0

G0 [1 − exp(−λ1 T )] > 0. λn

With (3.1.74) this implies that δFn ≤ CST λn δuT ,n , 1, 2, 3, . . . . Hence, ∞ 

2 δFn2 ≤ CST

n=1

∞ 

λ2n δu2T ,n .

(3.1.75)

n=1

Notice that the eigenvalues λn of the Sturm-Liouville  operator L : V(0, ) → L2 (0, ) introduced in (3.1.34) is of the order O n2 . Furthermore, from the characterization of the Sobolev space H 2 (0, ) by Fourier transform (Theorem 7, Sect 5.8.4 [38]) it follows that the series ∞ 

n4 δu2T ,n

n=1

converges if and only if uT ∈ V(0, )∩H 2 (0, ). Then, in view of Parseval’s equality (3.1.75) implies that 2

δuT 2H 2 (0,),

δF 2L2 (0,) ≤ M02 CST

with some constant M0 > 0. This completes the proof.

 

3.1.4 The Regularization Strategy by SVD. Truncated SVD Assume now α = 0. Then, by Definition 2.5.1 in Sect. 2.5, the operator  −1 ∗ Rα := ∗  + αI  : L2 (0, ) → L2 (0, ), α > 0,

(3.1.76)

88

3 Inverse Source Problems with Final Overdetermination

is the regularization strategy. According to (3.1.42), the following singular value expansion holds: (Rα uT )(x) =

∞  q(α; κn ) n=1

κn

uT ,n ϕn (x), α ≥ 0,

(3.1.77)

where uT ∈ L2 (0, ) is the noise free measured output. Comparing the damping parameters 1/κn and q(α; κ)/κ in the singular value expansions (3.1.56) and (3.1.77), we find that Tikhonov regularization is reflected in the expansion (3.1.77) of the regularized solution Fα (x) as the factor q(α; κ). For this reason, the function q(α; κ), 0 < q(α; κ) ≤ 1, α > 0, given by (3.1.43) is also called a regularizing filter, corresponding to Tikhonov regularization. Remark that in Sect. 2.5 the factor q(α; σ ) was defined as the filter function. As noted above, the measured output uT ∈ L2 (0, ) in the inverse source problem (3.1.31) is never known as an exact datum. Instead the noisy data uδT ∈ L2 (0, ) with a given noise level δ > 0 is usually available:

uδT − uT L2 (0,) ≤ δ, δ > 0.

(3.1.78)

In this case the inverse source problem (3.1.31) with the noisy output can be reformulated as the following operator equation: F δ = uδT , uδT ∈ L2 (0, ).

(3.1.79)

Applying Theorem 3.1.3 to this inverse problem and using the expansion (3.1.77), we can define formally the regularization strategy as follows: (Rα(δ) uδT )(x) =

∞  q(α(δ); κn ) n=1

κn

uδT ,n ϕn (x),

κ2 , q(α(δ); κ) = 2 κ + α(δ)

(3.1.80)

where uδT ,n := (uδT , ϕn )L2 (0,) is the nth Fourier coefficient of the noisy output. The left hand side defines the regularized solution Fαδ (x) := (Rα(δ)uδT )(x) of the inverse problem, i.e. the unique solution of the regularized form of the normal equation with noisy output uδT ∈ L2 (0, ): δ (∗  + α(δ)I )Fα(δ) = ∗ uδT .

(3.1.81)

3.1 Inverse Source Problem for Heat Equation

89

The following theorem shows that if the parameter of regularization α(δ) > 0, depending on the noise level δ > 0, is chosen properly, then the regularizing filter q(α; κ), given in (3.1.80), generates a convergent regularization strategy for the of inverse source problem (3.1.31). Theorem 3.1.5 Let conditions (3.1.3) hold. Assume that the time dependent source G(t) in (3.1.31) satisfies (3.1.69). Suppose that uδT ∈ L2 (0, ) is the noisy data given by (3.1.78). If the parameter of regularization is chosen so that α(δ) → 0 and

δ2 → 0, as δ → 0, α(δ)

(3.1.82)

then the regularized solution Fαδ (x) := (Rα(δ)uδT )(x) given by (3.1.80) converges in L2 -norm to the unique solution F = † uT given by (3.1.56), of the operator equation (3.1.12), as δ → 0, that is,

Rα(δ) uδT − F L2 (0,) → 0, as δ → 0.

(3.1.83)

Proof Let us estimate the above norm using the singular value expansions (3.1.56) and (3.1.80). We have:

Rα(δ) uδT

− F 2L2 (0,)



∞  

δ q(κn , α(δ)) 1

2

= − uT ,n

uT ,n κ κ

n

n=1

n

 2 ∞



δ

(u − uT ,n ) q(κn , α(δ)) + uT ,n q(κn , α(δ)) − 1

=

T ,n κn κn κn

n=1

≤2

∞ 

∞

(uδT ,n − uT ,n )2

n=1

 q 2 (κn , α(δ)) α(δ)2 +2 u2T ,n 2 . 2 κn (κn + α(δ))2 κn2

(3.1.84)

n=1

By the property (q(κ, α)/κ)2 ≤ 1/(4α) of the filter function, given by estimate (2.5.14) in Sect. 2.5, the first summand on the right hand side of (3.1.84) can be estimated as follows: 2

∞ 

 (uδT ,n

n=1

− uT ,n )

2

q(κn , α(δ)) κn

2

∞

≤

1  δ δ2 . (uT ,n − uT ,n )2 = 2α(δ) 2α(δ) n=1

The right hand side tends to zero, by condition (3.1.82) of the theorem.

(3.1.85)

90

3 Inverse Source Problems with Final Overdetermination

To estimate the second right hand summand in (3.1.84), we rewrite it in the following form 2

∞ 

u2T ,n

n=1

=

N  n=1

α(δ)2 (κn2 + α(δ))2 κn2

∞  u2T ,n u2T ,n 2α(δ)2 2α(δ)2 + , (κn2 + α(δ))2 κn2 (κn2 + α(δ))2 κn2

(3.1.86)

n=N+1

and estimate each right hand side terms separately. It follows from estimate (3.1.71) that κn = O(1/n2 ). This implies that no matter how small the parameter of regularization α(δ) > 0 was selected, there exists such a natural number N = N(δ) that  2 min κn2 = κN2 ≥ α(δ) > κN+1 . (3.1.87) 1≤n≤N

Then N(δ) = O(α(δ)−1/8 ) → ∞, as δ → 0. This and (3.1.87) allow to estimate the partial sum on right hand side of (3.1.86) as follows: N  n=1

N(δ)  u2T ,n u2T ,n 2α(δ)2 2α(δ) √ ≤ . (κn2 + α(δ))2 κn2 (1 + α(δ) )2 n=1 κn2

(3.1.88)

By the solvability condition (3.1.57), the sum is finite and the right hand side tends to zero, as δ → 0, by the first condition of (3.1.82). The last right hand side term in (3.1.86) can easily be estimated by using the inequality α 2 /(κn2 + α)2 < 1, for all n > N(δ) + 1: ∞  n=N(δ)+1

u2T ,n 2α(δ)2 ≤2 (κn2 + α(δ))2 κn2

∞ 

u2T ,n

n=N(δ)+1

κn2

.

Again, by the solvability condition (3.1.57), the right hand side tends to zero, as δ → 0. This, with (3.1.84), (3.1.85), (3.1.86) and (3.1.88) implies

Rα(δ) uδT − F L2 (0,) ≤ N(δ) ∞  u2T ,n  u2T ,n 2α(δ) δ2 + + 2 . √ 2α(δ) (1 + α(δ) )2 κn2 κn2 n=1 n=N(δ)+1

(3.1.89)

This estimate yields (3.1.83), since all the right hand side terms tend to zero, as δ → 0, by conditions (3.1.82). This completes the proof.  

3.1 Inverse Source Problem for Heat Equation

91

As introduced in Sect. 1.4, the Singular Value Decomposition (SVD) can also be used to obtain an approximation F N (x) =

N  1 uT ,n ϕn (x), x ∈ (0, ), κn

(3.1.90)

n=1

of the solution (3.1.56) of the normal equation (3.1.16) without regularization (α = 0). The proof scheme of Theorem 3.1.5 shows that the in the case when α > 0 SVD can also be used to obtain a regularization strategy in finite dimensional space L2N (0, ) ⊂ L2 (0, ). This strategy is called Truncated Singular Value Decomposition (TSVD), with Tikhonov regularization. Indeed, let the cutoff parameter N = N(δ) be defined as in the proof of the theorem, by (3.1.87): N(δ) = O(α(δ)−1/8 ). We define the Nth partial sum ⎧ N(δ) 

 q(α; κn ) ⎪ κ2 ⎪ ⎪ RN(δ) uδ (x) := , uδT ,n ϕn (x), q(α; κ) = 2 ⎨ α(δ) T κn κ +α n=1  T ⎪ ⎪ ⎪ ⎩ α ≥ 0, κn = exp(λn (T − t)G(t)dt, n = 1, 2, 3, . . .

(3.1.91)

0

of the series (3.1.80) as an approximation of the regularization strategy. We first δ estimate the approximation error RN(δ) α(δ) uT − F : 2    N(δ) N(δ)

Rα(δ) uδT − F 2L2 (0,) ≤ 2 Rα(δ) uδT − F N  2

L (0,)

 2   + 2 F N − F  2

L (0,)

.

Following (3.1.84) we get: δ 2

RN(δ) α(δ) uT − F L2 (0,) ≤ 4

N(δ) 

(uδT ,n − uT ,n )2

n=1

+4

N(δ)  n=1

u2T ,n

q 2 (κn , α(δ)) κn2

∞  u2T ,n α(δ)2 + 2 (κn2 + α(δ))2 κn2 κn2 N(δ)+1

For estimating the first and the second right hand side norms we use estimates (3.1.85) and (3.1.88). Using (3.1.85) and (3.1.88) for estimating the first and the second right hand side norms, we deduce the following error estimate: N(δ)

Rα(δ) uδT − F 2L2 (0,) ≤ N(δ) ∞  u2T ,n  u2T ,n 4α(δ) δ2 + √ + 2 . α(δ) (1 + α(δ) )2 κn2 κn2 n=1 N(δ)+1

(3.1.92)

92

3 Inverse Source Problems with Final Overdetermination

Remark that the approximation error estimate (3.1.92) is almost the same as estimate (3.1.89) in Theorem 3.1.5. In the numerical implementation, it is convenient to use the special case α(δ) ∼ δ. Evidently, in this case conditions (3.1.82) of Theorem 3.1.5 hold. In the numerical example below we demonstrate the role of the above theoretical results related to the TSVD algorithm with Tikhonov regularization applied to the inverse problem of determining F ∈ L2 (0, ) in (3.1.31). Here we consider the simplest version of the heat equation assuming k(x) = 1, that is, only Simpson’s rule in the numerical integration is responsible for increase of an error. Example 3.1.2 Implementation of TSVD: identification of an unknown space-wise dependent source in (3.1.31).   The function u(x, t) = sin(πx/2) 1 − exp(−π 2 t/2) (x, t) ∈ [0, 1] × [0, 1] is the exact solution of the heat equation ut = uxx + F (x)G(t) with the source functions F (x) = sin(πx/2), G(t) = (π 2 /4) 1 + exp(−π 2 t/2) , and with homogeneous boundary and initial conditions: u(0, t) = ux (1, t) = 0, u(x, 0) = 0. The synthetic noise free output is

 uT (x) = 1 − exp(−π 2 /2) sin(πx/2), x ∈ [0, 1].

(3.1.93)

The random noisy output data uδT (x), uT − uδT L2 (0,) ≤ δ, δ > 0, is generated from (3.1.93) using the MATLAB random function rand. The TSVD regularization strategy defined by (3.1.91) is used in determination of the approximate solution δ δ Fαδ,N := RN(δ) α(δ) uT from the noisy output uT (x). Note that the smoothness of G(t) and uT (x) are enough to ensure the fulfilment of the unique solvability condition (3.1.70) in Corollary 3.1.5. For synthetic noise free output data (3.1.93) (δ = 0) the TSVD formula (3.1.90) is used to obtain the approximate solution F N . The accuracy error E(n; Fαδ,n ; δ) :=

F − Fαδ,n L2 (0,) obtained for the values N = 3 ÷ 4 of the cutoff parameter is 1.7 × 10−6 . For noisy output data this error increase drastically, when N > 3, even for the small value δ = 0.01 of the noise level. For the synthetic noisy output data uδT (x), with δ = 0.01 and δ = 0.6, the TSVD regularization strategy (3.1.91) is used to obtain the approximate solution Fαδ,N . The dependence of the accuracy error E(n; Fαδ,n ; δ) on the parameter of regularization α and the cutoff parameter N are given in Tables 3.1 and 3.2. Bold-faced values of the accuracy error correspond to the optimal values of the cutoff parameter and agree Table 3.1 Dependence of the accuracy error on the parameter of regularization α and the cutoff parameter N: δ = 0.01

N 3 5 10

α 0 2.5 × 10−2 1.5 × 10−1 3.9 × 10−1

0.1 1.2 × 10−1 1.2 × 10−1 1.2 × 10−1

0.01 1.4 × 10−2 1.5 × 10−2 1.5 × 10−2

0.001 1.5 × 10−2 3.1 × 10−2 3.2 × 10−2

3.1 Inverse Source Problem for Heat Equation Table 3.2 Dependence of the accuracy error on the parameter of regularization α and the cutoff parameter N: δ = 0.6

N 3 10 20

α 0 1.5 × 100 2.4 × 101 7.9 × 101

93

0.1 1.4 × 10−1 1.4 × 10−1 1.5 × 10−1

0.01 2.7 × 10−1 3.4 × 10−1 3.4 × 10−1

0.001 2.8 × 10−1 1.9 × 10−1 2.0 × 10−1

Fig. 3.1 Influence of cutoff (left figure) and regularization (right figure) parameters on reconstruction accuracy: δ = 0.01

with the estimate N(δ) = O(α(δ)−1/8 ). For the optimal value N = 3 the parameter of regularization α should small to ensure the accuracy, but, at the same time, should not be too small to ensure the stability. Both situations are illustrated in the third and fifth columns of the tables. Although it would seem from the above tables that the difference between the errors for N = 3 and N = 5 are small, the deviation of the approximate solution F N , corresponding to N = 5, from the exact solution is large enough, even at low-level noise δ = 0.01, as the left Fig. 3.1 shows. The reconstructed approximate solutions F N from the noisy output uδT (x), with δ = 0.01, obtained for two values α1 = 0.1 and α2 = 0.01 of the parameter of regularization are plotted in the right Fig. 3.1. Better reconstruction obviously is obtained when α2 = 0.01, since in this case both conditions of the Theorem 3.1.5 are met approximately: α  1 and δ 2 /α  1. By increasing the noise level, impact of the first condition increases. Figure 3.2 illustrates the situation: for the higher noise level δ = 0.1, better reconstruction is obtained when α2 = 0.02. Table 3.3 reflects the accuracy error depending on cutoff and regularization parameters. 

94

3 Inverse Source Problems with Final Overdetermination

Fig. 3.2 Influence of the parameter of regularization on accuracy of reconstruction: δ = 0.1 Table 3.3 The accuracy error depending on the cutoff parameter and the parameter of regularization: δ = 0.1

α\N 0 0.001 0.01 0.1

3 3.15 × 10−1 1.69 × 10−1 5.4 × 10−2 1.23 × 10−1

5 1.82 × 10−1 3.72 × 10−1 6.8 × 10−2 1.24 × 10−1

10 4.73 × 10−1 3.83 × 10−1 6.8 × 10−2 1.26 × 10−1

3.2 Inverse Source Problems for Wave Equation In this section we use the approach given in the previous section to demonstrate on a simple model that the final data inverse source problem for undamped wave equation has generic poor properties. Consider the problem of determining the unknown spatial load F (x) in ⎧ ⎨ ut t = (k(x)ux )x + F (x)G(t), (x, t) ∈ T := (0, ) × (0, T ); u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ); ⎩ u(0, t) = 0, u(, t) = 0, t ∈ [0, T ),

(3.2.1)

from the final state overdetermination uT (x) := u(x, T ),

x ∈ (0, ).

(3.2.2)

3.2 Inverse Source Problems for Wave Equation

95

The one-dimensional equation (3.2.1) describes traveling-wave phenomena, and governs, for instance, the transverse displacement u(x, t) of an oscillating elastic string [150]. k(x) > 0 is the elastic modulus, F (x) and G(t) are the a spatial and temporal load distributions along the one-dimensional elastic string, respectively. We define the problem of determining the unknown spatial load F (x), i.e. problem (3.2.1)–(3.2.2), as an inverse source problem for wave equation with final state overdetermination. For a given source F ∈ L2 (0, ), the initial-boundary value problem (3.2.1) is defined as the direct problem. In practice, instead of the final state overdetermination (3.2.2) the final velocity overdetermination νT (x) := ut (x, T ),

x ∈ (0, ).

(3.2.3)

can be given as a measured output. In this case, (3.2.1) and (3.2.3) define the problem of determining the unknown spatial load F (x), i.e. an inverse source problem for wave equation with final velocity overdetermination. Consider the direct problem (3.2.1). Theorem 3.2.1 Assume that the inputs in (3.2.1) satisfy the following basic conditions:  0 < c0 ≤ k(x) ≤ c1 < +∞, k ∈ L∞ (0, ), (3.2.4) F ∈ L2 (0, ), G ∈ L2 (0, T ). Then there exists a weak solution of the direct problem (3.2.1), defined as u ∈ L2 (0, T ; V(0, )) with ut ∈ L2 (0, T ; L2 (0, )) and ut t ∈ L2 (0, T ; H −1(0, )), where V(0, )) = {v ∈ H 1 (0, ) : v(0) = v(l) = 0}. Furthermore, for this weak solution the following a priori estimates hold: e2 F 2 2

ut 2L2 (0,T ;L2 (0,)) ≤ C

G 2L2 (0,T ) , L (0,)

ux 2L2 (0,T ;L2 (0,)) ≤

e2 C

F 2L2 (0,) G 2L2 (0,T ) , 2c0

(3.2.5)

e2 = exp(T ) − 1 and c0 > 0 is the constant defined in (3.2.4). where C Proof The first part of the theorem, i.e. the existence and uniqueness of the weak solution follows from the general theory of hyperbolic PDE [38]. To prove estimates (3.2.5) we multiply both sides of Eq. (3.2.1) by 2ut (x, t), integrate over t := (0, )×(0, t) and use the integration by parts formula. With the homogeneous initial and boundary conditions this leads to the following energy identity: 

 0

u2t dx + 2

 t 0

 0

k(x)u2x dxdτ = 2

 t



F (x)G(τ )uτ (x, τ )dxdτ, 0

0

96

3 Inverse Source Problems with Final Overdetermination

a.e. t ∈ [0, T ]. This leads to the main integral inequality: 



u dx + 2c0 2

 t

0

0



0

u2x dxdτ

≤

 t 0

 0

u2τ dxdτ + F 2L2 (0,) G 2L2 (0,T ) ,

a.e. t ∈ [0, T ]. As a first consequence of this inequality we deduce that  0



u2t dx ≤

 t 0

0



u2τ dxdτ + F 2L2 (0,) G 2L2 (0,T ) , t ∈ [0, T ],

which with Gronwall-Bellman Lemma B.0.1 implies: 

 0

u2t dx ≤ F 2L2 (0,) G 2L2 (0,T ) exp(t), t ∈ [0, T ].

(3.2.6)

The first estimate of (3.2.5) follows from this inequality. To prove the second estimate of (3.2.5) we use the next consequence  t



2c0 0

0

u2x dxdτ ≤

 t 0

 0

u2τ dxdτ + F 2L2 (0,) G 2L2 (0,T )

of the main integral inequality with estimate (3.2.6). This leads to the second estimate of (3.2.5).   Theorem 3.2.2 Assume that in addition to the basic conditions (3.2.4), the temporal load G(t) satisfies the following regularity condition G ∈ H 1 (0, T ). Then there exists a regular weak solution of the direct problem (3.2.1) defined as u ∈ L2 (0, T ; H 2 (0, )) with ut ∈ L2 (0, T ; V(0, )), ut t ∈ L2 (0, T ; L2 (0, )) and ut t t ∈ L2 (0, T ; H −1 (0, )). Furthermore, for this regular weak solution the following a priori estimates hold: 2 F 2 2 e2 C

ut t 2L2 (0,T ;L2 (0,)) ≤ C

G L2 (0,T ) , T L (0,)

uxt 2L2 (0,T ;L2 (0,)) ≤

e2 C 2 C T

F 2L2 (0,) G 2L2 (0,T ) , 2c0

(3.2.7)

e > 0 is the constant introduced in 2 = max((2T /3) + 1, 2/T ) and C where C T Theorem 3.2.1. Proof The existence and uniqueness of the regular weak solution follows from the general theory of hyperbolic PDE [38]. Differentiate Eq. (3.2.1) with respect to t ∈ (0, T ), multiply both sides by ut t (x, t), integrate over t and use the initial and boundary conditions, also the consequence 

 0

 2 u2t t (x, 0+ )dx = G(0+ )



 0

F 2 (x)dx

3.2 Inverse Source Problems for Wave Equation

97

of the limit equation ut t (x, 0+ ) = (k(x)ux (x, 0+ ))x + F (x)G(0+ ) at t = 0+ . Then we obtain the following integral identity: 

 0

u2t t

+

k(x)u2xt



dx = 2

 t 0



F (x)G (τ )ut (x, τ )dxdτ, a. e. t ∈ [0, T ].

0

a. e. t ∈ [0, T ]. This leads to the main integral inequality: 

 0

u dx + 2c0 2

 t 0

 0

u2x dxdτ

≤

 t 0

 0

T2 F 2 2 u2 dxdτ + C

G 2H 1 (0,T ) , L (0,)

a.e. t ∈ [0, T ]. The required estimates (3.2.7) are easily derived from this inequality.

 

Consider the problem (3.2.1)–(3.2.2) of determining the unknown spatial load F (x) from final state overdetermination. Denote by u := u(x, t; F ) the unique weak solution of the direct problem (3.2.1), corresponding to a given source F ∈ L2 (0, ). Introducing the input-output operator  : L2 (0, ) → L2 (0, ), defined as (F )(x) := u(x, T ; F ), we can reformulate the inverse problem as the following operator equation F = uT , F ∈ L2 (0, ), uT ∈ L2 (0, ).

(3.2.8)

Lemma 3.2.1 Let conditions (3.2.4) holds. Assume that u0 (x) = u1 (x) = 0. Then the input-output operator  : F ∈ L2 (0, ) → u(x, T ; F ) ∈ L2 (0, ), corresponding to the inverse source problem (3.2.1)–(3.2.2) is a linear compact operator. The proof is almost exactly the same as that of Lemma 3.1.1 and is left as an exercise to the reader.  Thus, the hyperbolic inverse source problem (3.2.1)–(3.2.2) is ill-posed. Hence the approach given in Sect. 2.1 can be used for a minimum of the regularized Tikhonov functional Jα (F ) =

1 2



 0

1 [(F )(x) − uT (x)]2 dx + α F L2 (0,), 2

(3.2.9)

where α > 0 is the parameter of regularization. Consider now the problem of determining the unknown spatial load F (x) from final velocity overdetermination, i.e. the inverse problem defined by (3.2.1) and

98

3 Inverse Source Problems with Final Overdetermination

(3.2.3). We define the input-output operator  : L2 (0, ) → L2 (0, ), (F )(x) := ut (x, T ; F ). In a similar way, we can prove that the problem F = νT , F ∈ L2 (0, ), νT ∈ L2 (0, )

(3.2.10)

is also ill-posed.

3.2.1 Non-uniqueness and Uniqueness of a Solution Consider first the inverse problem (3.2.1)–(3.2.2) assuming that the initial strain u0 (x) and the initial velocity u1 (x) are zero. To prove an existence of unique minimum of functional (3.2.9) we need to apply Theorem 2.5.1 in Sect. 2.5. For this aim we analyze some properties of the input-output operator  : L2 (0, ) → L2 (0, ) and the operator equation (3.2.8), adopting Lemma 3.1.4 to the considered case. Lemma 3.2.2 Let conditions (3.2.4) hold. Then the input-output operator  : L2 (0, ) → L2 (0, ), (F )(x) := u(x, T ; F ), corresponding to the inverse problem (3.2.1)–(3.2.2), is self-adjoint. Furthermore, (ϕn )(x) = κn ϕn (x),

(3.2.11)

that is, {κn , ϕn } is the eigensystem of the input-output operator , 1 κn = √ λn



T

sin



 λn (T − t) G(t)dt, n = 1, 2, 3, . . .

(3.2.12)

0

∞ and {ϕn }∞ n=1 are orthonormal eigenvectors corresponding to eigenvalues {λn }n=1 2 of the differential operator L : V(0, ) → L (0, ), V(0, ) := {v ∈ H 1 (0, ) : v(0) = v(l) = 0}, defined by



(Lϕ)(x) := −(k(x)ϕ  (x)) = λϕ(x), x ∈ (0, ); ϕ(0) = 0, ϕ() = 0,

(3.2.13)

Proof Let {ϕn }∞ n=1 be the orthonormal eigenvectors corresponding to the positive eigenvalues {λn }∞ n=1 , 0 < λ1 < λ2 < λ3 . . . of the self-adjoint positive defined differential operator L : V(0, ) → L2 (0, ) defined by (3.2.13). Then we can make use of Fourier series expansion u(x, t) =

∞  n=1

un (t)ϕn (x)

(3.2.14)

3.2 Inverse Source Problems for Wave Equation

99

of the solution of the initial boundary value problem (3.2.1) through the orthonormal 2 basis {ϕn }∞ n=1 in L (0, ). Here Fn un (t) = √ λn



t

sin



 λn (t − τ ) G(τ )dτ, t ∈ (0, T ]

(3.2.15)

0

is the solution of the Cauchy problem 

un (t) + λn un (t) = Fn G(t), t ∈ (0, T ), un (0) = 0, un (0) = 0,

for each n = 1, 2, 3, . . . and Fn := (F, ϕn )L2 (0,) is the Fourier coefficient of the function F ∈ L2 (0, ). Now we can use (3.2.14) and (3.2.15) to obtain the Fourier series expansion of the input-output operator , defined as (F )(x) := u(x, T ; F ): (F )(x) =

∞ 

(F, ϕn )L2 (0,) κn ϕn (x),

(3.2.16)

n=1

where κn is defined by (3.2.12). To show that {ϕn }∞ n=1 are eigenvectors of the input-output operator , corresponding to eigenvalues {κn }∞ n=1 , we replace F (x) by ϕm (x) in (3.2.16): (ϕm )(x) =

∞ 

(ϕm , ϕn )L2 (0,) κn ϕn (x) = κm ϕm (x).

n=1

This implies (3.2.11). The proof of the self-adjointness of the operator  : L2 (0, ) → L2 (0, ) is the same as in Lemma 3.1.4.   The assertions of Lemma 3.2.2 can be proved for the problem of determining the unknown spatial load F (x) from final velocity overdetermination. Lemma 3.2.3 Let conditions (3.2.4) hold. Then the input-output operator  : L2 (0, ) → L2 (0, ), (F )(x) := ut (x, T ; F ), corresponding to the inverse problem (3.2.1) and (3.2.3), is self-adjoint. Furthermore, (ϕn )(x) = κn ϕn (x), that is, {κn , ϕn } is the eigensystem of the input-output operator , {ϕn }∞ n=1 are orthonormal eigenvectors corresponding to eigenvalues {λn }∞ n=1 of the differential operator  : V(0, ) → L2 (0, ), defined by (3.2.13), and 

T

κn = 0

 cos( λn (T − t))G(t)dt, n = 1, 2, 3, . . . .

(3.2.17)

100

3 Inverse Source Problems with Final Overdetermination

The Fourier series expansion of the input-output operator  corresponding to the inverse problem (3.2.1) and (3.2.3) is (F )(x) =

∞  (F, ϕn )L2 (0,) κn ϕn (x),

(3.2.18)

n=1

where κn is defined by (3.2.17). The Fourier series expansions (3.2.16) and (3.2.18) of the input-output operators  and , corresponding to the above inverse problems, show that the necessary condition for bijectivity of these operators is the condition κn = 0. Remark that this is a main condition of Theorem 2.5.1 in Sect. 2.5. It follows from formulae (3.2.12) and (3.2.17) for the eigenvalues κn that even positivity of the time dependent source G ∈ L2 (0, T ) can not guarantee the condition κn = 0 for unique determination of the unknown source F (x). To show this explicitly consider the following examples. Example 3.2.1 Identification of a spatial load in constant coefficient wave equation from final state overdetermination when G(t) ≡ 1. Consider the problem of determining the unknown spatial load F (x) in ⎧ ⎨ ut t = uxx + F (x), (x, t) ∈ T := (0, 1) × (0, T ); u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, 1); ⎩ u(0, t) = 0, u(1, t) = 0, t ∈ (0, T ),

(3.2.19)

from the final state overdetermination (3.2.2). For G(t) ≡ 1 formula (3.2.12) implies: κn =

 1 [1 − cos( λn T )], λn = π 2 n2 , n = 1, 2, 3, . . . . λn

(3.2.20)

Formula (3.2.12) shows that κn = 0, if nT = 2m, where m is a natural number. Hence κn = 0 for all n = 1, 2, 3, . . . if and only if T =

2m , for all n = 1, 2, 3, . . . . n

(3.2.21)

Thus, for unique determination of the unknown source F (x) from final state overdetermination (3.2.2), the final time T > 0 must satisfy the condition (3.2.21).  Example 3.2.2 Identification of a spatial load in constant coefficient wave equation from final velocity overdetermination when G(t) ≡ 1.

3.2 Inverse Source Problems for Wave Equation

101

Consider now the problem of determining the unknown spatial load F (x) in (3.2.19) from the final velocity overdetermination (3.2.3). Formula (3.2.17) implies (G(t) ≡ 1):  1 κn = √ sin( λn T ), λn = π 2 n2 , n = 1, 2, 3, . . . . λn

(3.2.22)

Hence κn = 0, if nT = m, where m = m(T ) is a natural number. This means that the final time should satisfy the condition T =

m , f orall n = 1, 2, 3, . . . . n

(3.2.23)

Thus, for unique determination of the unknown source F (x) from the final velocity overdetermination (3.2.2) the final time T > 0 must satisfy condition (3.2.23).  As a matter of fact, both conditions (3.2.21) and (3.2.23) are equivalent and mean that the value T > 0 of the final time cannot be a rational number. Evidently, in practice for arbitrary given final time T > 0 the fulfilment of this necessary condition is impossible. Thus, both final data overspecifications (3.2.2) and (3.2.3) for the wave equation are not feasible. Although this issue will be explored in detail in Sect. 11.4, there is an important point to be emphasized here. Namely, in addition to the final time T > 0, the function G(t) also plays an important role in the problem of determining the unknown spatial load F (x) from final state or velocity overdetermination. The following example clearly illustrates the role of this function. Example 3.2.3 Identification of a spatial load in constant coefficient wave equation from final state overdetermination when G(t) = t, t ∈ (0, T ), T > 0. Consider the problem of determining the unknown spatial load F (x) in (3.2.1) from the final state overdetermination (3.2.2), assuming that G(t) = t. Formula (3.2.12) with G(t) = t implies: κn =

1 3/2

λn

  T −

T

 cos( λn (T − t)) dt



0

  1 1 = 3/2 T − √ sin( λn T ) > 0, n = 1, 2, 3, . . . , λn λn

√ for any T > 1/ λ1 , where λ1 > 0 is the smallest eigenvalue of the Sturm-Liouville operator L : V(0, ) → L2 (0, ) introduced (3.1.34). √ Thus, for all values T > 1/ λ1 of the final time all eigenvalues κn of the input-output operator  are positive, which implies the uniqueness of the solution of the inverse source problem (3.2.1)–(3.2.2) for the undamped wave equation ut t = (k(x)ux )x + F (x)t. 

102

3 Inverse Source Problems with Final Overdetermination

3.3 Backward Parabolic Problem As a next application of the above approach we consider the backward parabolic problem. Specifically, consider the problem of determining the unknown initial temperature f (x) ∈ L2 (0, ) in ⎧ ⎨ ut = (k(x)ux )x , (x, t) ∈ T := (0, ) × (0, T ], u(x, 0) = f (x), x ∈ (0, ), ⎩ u(0, t) = 0, ux (, t) = 0, t ∈ (0, T ),

(3.3.1)

from the measured temperature uT (x) at the final time t = T : uT (x) := u(x, T ),

x ∈ (0, ).

(3.3.2)

This problem is defined as a backward parabolic problem (BPP). Denote by u(x, t; f ) be the weak solution of the forward problem (3.3.1), for a given initial data f ∈ L2 (0, ), and introduce the input-output operator  : L2 (0, ) → L2 (0, ), defined by (f )(x) := u(x, 0; f ), x ∈ (0, ). Then the backward parabolic problem can be reformulated as the following operator equation: f = uT , f ∈ L2 (0, ).

(3.3.3)

Using Lemma 3.1.1, it can be easily proved that (3.3.3) is a linear compact operator. This, in particular, implies that backward parabolic problem (3.3.1)–(3.3.2) is illposed. We use Tikhonov regularization introducing the regularized functional 1 Jα (f ) := J (f ) + α f 2L2 (0,), 2

(3.3.4)

1

u(·, T ; f ) − uT 2L2 (0,), f ∈ L2 (0, ) 2

(3.3.5)

where J (f ) =

and α > 0 is the parameter of regularization. The following lemma shows that the Fréchet gradient J  (f ) of the Tikhonov functional (3.3.5) can be derived through the solution ψ(x, t) := ψ(x, t; f ) of the adjoint problem solution ⎧ ⎨ ψt = −(k(x)ψx )x , (x, t) ∈ T ; ψ(x, T ) = u(x, T ; f ) − uT (x), x ∈ (0, ); ⎩ ψ(0, t) = 0, ψx (, t) = 0, t ∈ [0, T ).

(3.3.6)

3.3 Backward Parabolic Problem

103

Note that this is the same adjoint problem (3.1.19), corresponding to the parabolic inverse problem, with u(x, T ; F ) replaced by u(x, T ; f ). Lemma 3.3.1 Let conditions (3.1.3) hold. Denote by u(x, t; f ) the weak solution of the parabolic problem (3.3.1) corresponding to a given initial data f ∈ L2 (0, ). Then the Tikhonov functional (3.3.5) is Fréchet differentiable. Furthermore, for the Fréchet gradient J  (f ) the following formula holds: J  (f )(x) = ψ(x, 0; f ),

for a.e. x ∈ (0, ),

(3.3.7)

where ψ(x, t; f ) is the weak solution of the adjoint problem (3.3.6). Proof Let f, f + δf ∈ L2 (0, ) be given initial data and u(x, t; f ), u(x, t; f + δf ) be corresponding solutions of (3.3.1). Then δu(x, t; f ) := u(x, t; f + δf ) − u(x, t; f ) is the weak solution of the following parabolic problem ⎧ ⎨ δut = (k(x)δux )x , (x, t) ∈ T , δu(x, 0) = δf (x), x ∈ (0, ), ⎩ δu(0, t) = 0, δux (, t) = 0, t ∈ (0, T ].

(3.3.8)

We calculate the increment δJ (f ) := J (f + δf ) − J (f ) of functional (3.3.5): 

l

δJ (f ) =

[u(x, T ; f ) − uT (x)]δu(x, T ; f )dx +

0

1 2



l

[δu(x, T ; f )]2 dx,

0

(3.3.9) and then transform the first right hand side integral. Assuming that ψ(x, t) := ψ(x, t; f ) and δu(x, t; f ) := δu(x, t; f ) are solutions of problems (3.3.6) and (3.3.8), accordingly, we have: 





0

  T

 = 0



T



T 0

0



 0

ψ(x, T )δu(x, T )dx 0

  (ψ(x, t)δu(x, t))t dt dx +

0

= =



[u(x, T ; f ) − uT (x)]δu(x, T ; f )dx =









ψ(x, 0)δf (x)dx 0 

[ψt δu + ψδut ] dxdt +

0

ψ(x, 0)δf (x)dx 

0 

[−(k(x)ψx )x δu + ψ(k(x)δux )x ] dxdt + 

T

= 0

 [−k(x)ψx δu + k(x)δux ψ]x= x=0 dt +

ψ(x, 0)δf (x)dx 0 

ψ(x, 0)δf (x)dx. 0

104

3 Inverse Source Problems with Final Overdetermination

Taking into account here the initial/final and boundary conditions in (3.3.6) and (3.3.8) we obtain the following integral identity: 







[u(x, T ; f ) − uT (x)]δu(x, T ; f )dx =

0

ψ(x, 0; f )δf (x)dx, 0

for all f, δf ∈ L2 (0, ). With formula (3.3.9) this implies: 



δJ (F ) =

ψ(x, 0; f )δf (x)dx +

0

1 2



l

[δu(x, T ; f )]2 dx.

(3.3.10)

0

 The last right hand side integral is of the order O δf 2L2 (0,) by estimate (3.1.7). This completes the proof of lemma.   Although the Lipschitz continuity of the Fréchet gradient (3.3.7) can be obtained from the results given in Sect. 1.1, to show an explicit form of the Lipschitz constant, we will prove it. Lemma 3.3.2 Let conditions (3.1.3) hold. Then Fréchet gradient of the Tikhonov functional (3.3.5) is Lipschitz continuous with the Lipschitz constant L3 = 1:

J  (f + δf ) − J  (f ) L2 (0,) ≤ δf L2 (0,), ∀f, δf ∈ L2 (0, ).

(3.3.11)

Proof By definition,

J  (f + δf ) − J  (f ) L2 (0,) := δψ(·, 0; f ) L2 (0,) ,

(3.3.12)

where δψ(x, t; f ) := ψ(x, t; f + δf ) − ψ(x, t; f ) is the weak solution of the problem: ⎧ ⎨ δψt = −(k(x)δψx )x , (x, t) ∈ T ; δψ(x, T ) = δu(x, T ; f ), x ∈ (0, ); ⎩ δψ(0, t) = 0, ψx (, t) = 0, t ∈ [0, T ).

(3.3.13)

and δu(x, t; f ) is the weak solution of the auxiliary problem (3.3.8). We use the energy identity 

1 d 2 dt







(δψ(x, t))2 dx =

0

k(x) (δψx (x, t))2 dx

0

for the well-posed backward problem (3.3.13), integrate it on [0, T ] and use the final condition δψ(x, T ) = δu(x, T ; f ), to deduce the estimate 

 0





(δψ(x, 0; f )) dx ≤ 2

0

(δu(x, T ; f ))2 dx.

3.3 Backward Parabolic Problem

105

Then we use the energy identity 1 d 2 dt









(δu(x, t)) dx + 2

0

k(x) (δux (x, t))2 dx = 0

0

for the auxiliary problem (3.3.8) to obtain the estimate 

 0





(δu(x, T ; f ))2 dx ≤

(δf (x))2 dx, ∀f ∈ L2 (0, ).

0

These estimates imply δψ(·, 0; f ) L2 (0,) ≤ δf L2 (0,). With (3.3.12) this yields the proof of the theorem.   We can prove analogues of Corollaries 3.1.1 and 3.1.2 for the backward parabolic problem (3.3.1)–(3.3.2) in the same way. Lemma 3.1.4 can be adopted to the backward parabolic problem to derive main properties of the input-output operator. Lemma 3.3.3 Let conditions (3.1.3) hold. Then the input-output operator  : L2 (0, ) → L2 (0, ), (f )(x) := u(x, 0; f ), corresponding to the backward problem defined by (3.3.1) and (3.3.2), is self-adjoint. Moreover, (ϕn )(x) = κn ϕn (x),

(3.3.14)

that is, {κn , ϕn } is the eigensystem of the input-output operator , where {ϕn }∞ n=1 are orthonormal eigenvectors of the differential operator  : V(0, ) → L2 (0, ), defined by (3.1.34) corresponding to the eigenvalues {λn }∞ n=1 , and σn ≡ κn = exp(−λn T ), n = 1, 2, 3, . . .

(3.3.15)

The proof follows from the proof of Lemma 3.1.4 and is left as an exercise to the reader. Thus the singular system of the (compact) input-output operator  is {κn , ϕn , ϕn }. The eigenvalues λn of the differential operator  : V(0, ) → L2 (0, ), defined in (3.1.34), are of order λn = O(n2 ) Then, as it follows from (3.3.15) that the singular values κn of the input-output operator  are of order O(exp(−n2 )). This means that the backward parabolic problem (3.3.1)–(3.3.2) is severely ill-posed. 2 Since {ϕn }∞ n=1 forms a complete orthonormal system in L (0, ) and  : 2 2 L (0, ) → L (0, ), (f )(x) := u(x, 0; f ) is a self-adjoint operator, we have N () = N (∗ ) = {0}. Moreover, D(† ) = R() is dense in L2 (0, ). Hence, for uT ∈ L2 (0, ) the first condition uT ∈ N (∗ )⊥ in Picard’s Theorem 2.4.1 of Sect. 2.4 holds. Then we can reformulate this theorem for the backward problem defined by (3.3.1) and (3.3.2). Theorem 3.3.1 Let conditions (3.1.3) hold. Assume that uT ∈ L2 (0, ) is a noise free measured output defined by (3.3.2). Then the operator equation f = uT ,

106

3 Inverse Source Problems with Final Overdetermination

corresponding to backward problem (3.3.1)–(3.3.2), has a solution if and only if uT ∈ L2 (0, ) and

∞ 

exp(λn T )u2T ,n < ∞.

(3.3.16)

exp(λn T )uT ,n ϕn (x), x ∈ (0, )

(3.3.17)

n=1

In this case f (x) =

∞  n=1

is the solution of the operator equation F = uT , where uT ,n := (uT , ϕn ) is the nth Fourier coefficient of the measured output uT (x) and ϕn (x) are the normalized eigenfunctions corresponding to the eigenvalues λn of the operator  : V(0, ) → L2 (0, ) defined by (3.1.34). The proof of this theorem is the same as proof of Theorem 3.1.3.  The following example show that neither the truncated SVD nor Tikhonov regularization can recover information about higher Fourier coefficients in (3.3.17). Example 3.3.1 The truncated SVD and Tikhonov regularization for the backward parabolic problem. Consider the simplest backward parabolic problem, i.e. the problem of determining the unknown initial temperature f (x) ∈ L2 (0, ) in ⎧⎧ ⎨ ut = uxx , (x, t) ∈ T := (0, π) × (0, 1], ⎪ ⎪ ⎨ u(x, 0) = f (x), x ∈ (0, π), ⎩ (3.3.18) ⎪ u(0, t) = 0, u(π, t) = 0, t ∈ (0, 1), ⎪ ⎩ uT (x) := u(x, 1), x ∈ (0, π), assuming k(x) ≡ 1,  = π and T = 1. In this case the eigenvalues λn and the normalized eigenfunctions ϕn (x) of the operator (Lϕ)(x) := −ϕ  (x), with the Dirichlet conditions u(0, t) = u(π, t) = 0, are λn = n2 , ϕn (x) =



2/π sin nx, n = 1, 2, 3, . . . .

The singular values of the input-output operator  are: κn = exp(−n2 ), n = 1, 2, 3, . . . . Let uδT ∈ L2 (0, π) be a noisy data and we wish to reconstruct the initial data f δ ∈ 2 L (0, π) by the truncated SVD. Then assuming 1/N a parameter of regularization we may consider the regularization strategy N(δ)

  RN(δ) uδT (x) := exp(n2 ) uδT ,n ϕn (x), n=1

(3.3.19)

3.3 Backward Parabolic Problem

107

where uδT ,n = (uδT , φn )L2 (0,π) is the nth Fourier coefficient of the noisy output data uδT (x). Let us estimate the approximation error RN(δ) uδT − † uT L2 (0,π) as in Sect. 2.4. We have:  2  N(δ) δ  uT − † uT  2 R

L (0,π)

 2 N(δ)    2 δ  ≤ 2 exp(n ) (uT ,n − uT ,n )ϕn (x)   n=1  2

L (0,π)

 2  ∞     2  +2  exp(n ) u ϕ (x) T ,n n   n=N(δ)+1  2

.

L (0,π)

Hence  2  N(δ) δ  uT − † uT  2 R

L (0,π)

≤ 2 exp(2N(δ)2 )δ 2 + 2

∞ 

exp(2n2 )u2T ,n .

(3.3.20)

n=N(δ)+1

The first term of the right hand side of (3.3.20), i.e. the data error increase dramatically by increasing N(δ), due to the term exp(2N(δ)2 ), if even the noise level δ > 0 is small enough. This means that we may use only the first few Fourier coefficients of (3.3.19) to obtain an approximate solution by the truncated SVD. For example, if even δ = 10−2 , we can use only the first two terms of this series, since for N(δ) = 2 the data error is 2 exp(2N(δ)2 )δ 2  6000×10−4 = 0.6. By adding the next term will make this error equals to 1.31×104! Moreover, one needs to take into account also the contribution of the second term of the right hand side of (3.3.20), i.e. the approximation error. The term exp(2n2 ) here shows that the approximation error also does dramatically increase, as N increases (except the case when the Fourier coefficients uT ,n of the output decay faster than exp(−N(δ)2 ), which holds only for infinitely differentiable functions). Let us apply now Tikhonov regularization to the inverse problem (3.3.18). We use estimate (3.3.21) in Theorem 3.1.5, adopting it to the considered problem. Then we get: N(δ)

Rα(δ) uδT − f 2L2 (0,π) ≤ N(δ) ∞   4α(δ) δ2 2 2 + exp(2n )u + 2 exp(2n2 )u2T ,n . √ T ,n α (1 + α(δ) )2 n=1 N(δ)+1

(3.3.21)

108

3 Inverse Source Problems with Final Overdetermination

Following the selection (3.1.87) of the cutoff parameter N(δ) and taking into account that κn = exp(−n2 ), we deduce: no matter how small the parameter of regularization α(δ) > 0 was selected, there exists such a natural number N = N(δ) that min

1≤n≤N(δ)

exp(−2n2 ) = exp(−2N(δ)2 ) ≥

√

α > exp(−2(N(δ) + 1)2 ).

√ This implies that N(δ) = O( ln α(δ)/2). Assume again that the noise level is very low: δ = 10−3 . We choose the parameter of regularization as α(δ) = 10−5 from the √ requirements (3.1.82) of Theorem 3.1.5. Then ln α(δ)/2  1.7, which means that only the first two Fourier coefficients can be used in the Nth partial sum N(δ) 

 q(α(δ); κn ) δ u (x) := uδT ,n ϕn (x), RN(δ) α(δ) T κn

q(α; κ) =

n=1

κ2 κ2 + α

for recovering the initial data. Remark that we did not take into account an influence of the last right hand side terms of (3.3.21) to the above error, although these terms need to be taken into account also.  Remark, that Duhamel’s Principle for heat equation illustrates relationship between the singular values corresponding to the inverse source problem and the backward problem for heat equation. Let v(x, t; τ ), (x, t) ∈ τ , be the solution of the problem ⎧ ⎨ vt (x, t; τ ) = (k(x)vx (x, t; τ ))x , (x, t) ∈ τ ; v(x, τ ; τ ) = F (x)G(τ ), x ∈ (0, ); ⎩ v(0, t; τ ) = 0, vx (, t) = 0, t ∈ (0, T ]

(3.3.22)

corresponding to a fixed value of the parameter τ ∈ [0, T ], where τ := {(x, t) ∈ R2 : x ∈ (0, ), t ∈ (τ, T ]}. Then, by Duhamel’s Principle, 

t

u(x, t) =

v(x, t; τ )dτ, τ ∈ (0, t], t ∈ (0, T ]

(3.3.23)

0

is the solution of the problem ⎧ ⎨ ut = (k(x)ux )x + F (x)G(t), (x, t) ∈ T ; u(x, 0) = 0, x ∈ (0, ); ⎩ u(0, t) = 0, ux (, t) = 0, t ∈ (0, T ].

(3.3.24)

For each fixed τ ∈ (0, T ], the problem of determining the initial data F (x) in (3.3.22) from the final output data uT (x) := u(x, T ) is severely ill-posed. While the problem of determining the unknown space-wise dependent source F ∈ L2 (0, ) in (3.3.24) from the same final data is moderately ill-posed. This change of degree of

3.3 Backward Parabolic Problem

109

ill-posedness is due to the integration in (3.3.23). Specifically, formula (3.1.35) is obtained from formula (3.3.15) by integration (3.3.23) and by taking into account the factor G(t). Finally, to compare the behavior of the singular values, we consider the inverse source problem (3.1.31) for heat equation (ISPH), the backward parabolic problem (BPP) defined by (3.3.1)–(3.3.2) and the inverse source problem (3.2.1)–(3.2.2) for wave equation (ISPW). Assuming, for simplicity, that G(t) ≡ 1, k(x) = 1, κ(x) = 1, l = π, T = 1, we obtain the following formulae for the singular values σn ≡ κn of these problems:  ⎧ ⎨ σn = 1 − exp(−(n − 1/2)2 ) (n + 1/2)−2 ISPH; σ = exp(−(n − 1/2)2 ), n = 1, 2, 3, . . . BPP; ⎩ n ISPW. σn = [1 − cos n]n−2 The behavior of the singular values σn , depending on n = 1, 30, is shown in Fig. 3.3. This figure shows n−2 -decay and exp(n−2 )-decay characters of the singular values corresponding to ISPH and BPP, respectively. The behavior of the singular values corresponding to ISPW is as an oscillating function of n that also decays, as n → ∞.

Fig. 3.3 Behavior of the singular values

110

3 Inverse Source Problems with Final Overdetermination

3.4 Computational Issues in Inverse Source Problems Any numerical method for inverse problems related to PDEs requires, first of all, construction of an optimal computational mesh for solving corresponding direct problem. This mesh needs to be fine enough in order to obtain an accurate numerical solution of the direct problem, i.e. to minimize the computational noise level in synthetic output data, on one hand. On the other hand, an implementation of any iterative method for numerical solution of an inverse problem requires repeated solving the direct and the corresponding adjoint problems at each iteration step. Hence, a too fine mesh for numerical solving of these problems is sometimes unnecessary and increases the computational cost. Since the approach discussed above is based on the weak solution theory, the most appropriate method for the discretization and solving these problems, is the Finite Element Method (FEM). This method with continuous piecewise quadratic basic functions is used here. The second important point is that, the approach given in Sect. 3.1 allows to derive a gradient formula. As a consequence of this, the Conjugate Gradient Algorithm (CG-algorithm) is implemented here. Furthermore, the Lipschitz continuity of the gradient of the Tikhonov functional implies monotonicity of the numerical sequence {J (F (n) }, where F (n) is the nth iteration of the CG algorithm. As a result, the CG-algorithm is an optimal algorithm in numerical solving of this class of inverse problems. The last important point in numerical solving of ill-posed problems is that employing the same mathematical model to generate a synthetic output data, as well as to invert it, may lead to trivial inversion. This situation is defined in [33] as an “inverse crime”. As it is stated there, to avoid this trivial inversion “it is crucial that the synthetic data be obtained by a forward solver which has no connection to the inverse solver”. We refer the readers to the reference [102] for computational details of inverse crimes. All numerical experiments here related to the inverse problem (3.1.31) for the heat equation, a finer mesh is used for generation of a synthetic measured output, and a coarser mesh is used in the inversion algorithm, in order to avoid an inverse crime.

3.4.1 The Galerkin Finite Element Method In this subsection we consider the following direct problem: ⎧ ⎨ ut = (k(x)ux )x + F (x)G(t), (x, t) ∈ T := (0, ) × (0, T ]; u(x, 0) = u0 (x), x ∈ (0, ); ⎩ u(0, t) = 0, ux (, t) = 0, t ∈ (0, T ].

(3.4.1)

3.4 Computational Issues in Inverse Source Problems

111

Fig. 3.4 The piecewise quadratic basis functions

To introduce the semi-discrete analogue of this problem, the continuous piecewise quadratic basic functions are used. Then applying time discretization, fully discrete analogue of the direct problem (3.1.1) is derived. Let ωh := {xi ∈ (0, ] : xi = ih, i = 1, Nx } be a uniform space mesh with the mesh parameter h = /Nx > 0. The quadratic polynomials ⎧ ⎧ ⎪ ⎨ 2(x − xi−1/2 )(x − xi−1 )/ h2 , x ∈ [xi−1 , xi ), ⎪ ⎪ ⎪ ⎪ ξi (x) = 2(x − xi+1/2 )(x − xi+1 )/ h2 , x ∈ [xi , xi+1 ), ⎪ ⎪ ⎩ ⎪ ⎪ 0, x∈ / [xi−1 , xi+1 ], ⎪ ⎪ ⎪ ⎨ i = 1, Nx − 1;  ⎪ ⎪ 0, x ∈ [0, xNx −1 ), ⎪ ⎪ ξN (x) = ⎪ 2 , x ∈ [x ⎪ )(x − x )/ h 2(x − x N−1/2 Nx −1 Nx −1 , xNx ]; ⎪ ⎪  ⎪ 2 ⎪ −4(x − xi−1 )(x − xi )/ h , x ∈ [xi−1 , xi ], ⎪ ⎪ ⎩ ηi (x) = 0, x∈ / [xi−1 , xi ], i = 1, Nx , with the compact supports are defined as 1D continuous piecewise quadratic basic functions in FEM (Fig. 3.4). Here the midpoints xi−1/2 := (i − 1/2)h, i = 1, Nx are also defined as auxiliary nodes, in addition to the nodal points xi ∈ ωh of the uniform space mesh ωh . Then the function uh (x, t) =

N  i=1

ci (t)ξi (x) +

Nx 

di (t)ηi (x)

i=0

is a piecewise-quadratic approximation of u(x, t) in the finite-dimensional subspace Vh ⊂ H1 (0, ) spanned by the set of basis functions {ξi (x)}∪{ηi (x)}, where ci (t) := uh (xi , t) and di (t) := uh (xi−1/2 , t). For convenience we treat here and below t ∈ (0, T ] as a parameter and the functions u(t) and uh (t) as mappings u : [0, T ] → V and uh : [0, T ] → Vh , defined as u(t)(x) := u(x, t) and uh (t)(x) := uh (x, t), x ∈

112

3 Inverse Source Problems with Final Overdetermination

(0, ). For the interpolation with piecewise quadratic basic functions the following estimates hold:

u − uh V (0,) ≤ Ch2 |u|H 3 (0,), C > 0, which means the second-order accuracy in space, in the norm of H 1 (0, ) [144]. Here | · |H 3 (0,) is the semi-norm in u ∈ H 3 (0, ) and V := {v ∈ H 1 (0, ) : v(0, t) = 0}. The Galerkin FEM is applied for discretization of the direct and adjoint problems (3.4.1) and (3.1.19). Find uh (t) ∈ SN such that ⎧ ∂t uh , ξi  + a(uh , ξi ) = G(t)F (x), ξi , ⎪ ⎪ ⎪ ⎪ ⎨ uh (0), ξi  := u0 , ξi , i = 1, Nx , ⎪ ⎪ ∂t uh , ηi  + a(uh , ηi ) = G(t)F (x), ηi , ⎪ ⎪ ⎩ uh (0), ηi  := u0 , ηi , i = 0, Nx ,

t ∈ (0, T ]; t ∈ (0, T ];

(3.4.2)

where $x ⎧⎧ ∂t uh , ξi  := xii+1 ut (x, t)ξi (x)dx, ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ $x ⎪ ⎪ ⎪ a(uh , ξi ) := xii+1 k(x)uh (x)ξi (x)dx, ⎪ ⎪⎪ ⎪ $x ⎪ ⎩ ⎨⎪ F (x), ξi  := xii+1 F (x)ξi (x)dx, i = 0, Nx ; ⎧ $x ⎪ ∂t uh , ηi  := xii+1 ut (x, t)ηi (x)dx, ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ $ xi+1 ⎪   ⎪ ⎪ ⎪ ⎪ a(uh , ηi ) := xi k(x)uh (x)ηi (x)dx, ⎪ ⎪ ⎪ $x ⎩⎩ F (x), ηi  := xii+1 F (x)ηi (x)dx, i = 0, Nx .

(3.4.3)

Introducing the uniform time mesh ωτ := {tj ∈ (0, T ] : tj = j τ, j = 1, Nt }, with mesh parameter τ = T /Nt , by standard Galerkin FEM procedure, we then obtain from the semi-discrete Galerkin discretization (3.4.2)–(3.4.3) the following full discretization by using Crank-Nicolson method: ⎧ % j j+1 & j j+1 uh +uh a(uh ,ξi )+a(uh ,ξi ) G(tj )+G(tj+1 ) ⎪ ⎪ + , ξ = F (x), ξi ; i ⎪ τ 2 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ uh (0), ξi  := u0 , ξi , tj ∈ ωτ , xi ∈ ωh . & % j j+1 ⎪ j j+1 ⎪ uh +uh a(uh ,ηi )+a(uh ,ηi ) G(tj )+G(tj+1 ) ⎪ ⎪ , η = F (x), ηi ; + ⎪ i ⎪ τ 2 2 ⎪ ⎪ ⎩ uh (0), ηi  := u0 , ηi , tj ∈ ωτ , xi ∈ ωh .

(3.4.4)

The FEM scheme (3.4.4) is used in subsequent computational experiments for solving the forward and backward parabolic problems.

3.4 Computational Issues in Inverse Source Problems

113

3.4.2 The Conjugate Gradient Algorithm Above derived gradient formulae for the Tikhonov functionals allow use of the classical version of the Conjugate Gradient Algorithm for the numerical reconstruction of an unknown source. We define the convergence error (or discrepancy) e(n; F ; δ) and the accuracy error E(n; F ; δ): e(n; F (n) ; δ) := F (n) − uδT L2 (0,T ) , E(n; F (n) ; δ) := F − F (n) L2 (0,) ,

(3.4.5)

where  : L2 (0, ) → L2 (0, ), (F )(x) := u(x, T ; F ) is the input-output operator. Here and below uδT (x) is the noisy data:

uT − uδT L2 (0,) ≤ δ, uδT L2 (0,) > δ, δ > 0.

(3.4.6)

The Conjugate Gradient Algorithm discussed below is applied to the Tikhonov functional J (F ) =

1

F − uδT 2L2 (0,) , 2

(3.4.7)

corresponding to the normal equation ∗ F = ∗ uδT , although the same technique remains hold for the regularized form 1 Jα (F ) = J (F ) + α F 2L2 (0,) 2

(3.4.8)

of the Tikhonov functional corresponding to the regularized form of normal equation (∗  + αI ) Fα = ∗ uδT . The iterative Conjugate Gradient Algorithm (subsequently, CG-algorithm) for the functional (3.4.7) consists of the following basic steps. All norms and scalar products below are in L2 (0, ) and will be omitted. Step 1. Step 2.

For n = 0 choose the initial iteration F (0) (x). Compute the initial descent direction p(0) (x) := J  (F (0) )(x).

Step 3.

(3.4.9)

Find the descent direction parameter    (n) J (F ), p(n) βn = .

p(n) 2

(3.4.10)

114

Step 4.

3 Inverse Source Problems with Final Overdetermination

Find next iteration F (n+1) (x) = F (n) (x) − βn p(n) (x)

(3.4.11)

and compute the convergence error e(n; F (n) ; δ). Step 5. If the stopping condition e(n; F (n) ; δ) ≤ τM δ < e(n; F (n−1); δ), τM > 1, δ > 0

(3.4.12)

holds, then go to Step 7. Step 6. Set n := n + 1 and compute 

p(n) (x) := J  (F (n) )(x) + γn p(n−1) (x), γn =

J  (F (n) ) 2

J  (F (n−1) ) 2

(3.4.13)

and go to Step 3. Step 7. Stop the iteration process. This version of the CG-algorithm is usually called in literature as CGNE, i.e. the CG-algorithm applied to the Normal Equation [37]. We will follow here the version of the CG-algorithm given in [81], adopting the results given here to the considered class of inverse problems. First of all we make some important remarks concerning formulae (3.4.9) and (3.4.10). It follows from gradient formula (3.1.29) for the Fréchet derivative of the Tikhonov functional, i.e. from formula

 J  (F (n) ) = ∗ F (n) − uT , F (n) ∈ L2 (0, ) (3.4.14) that if the CG-algorithm starts with the initial iteration F (0) (x) ≡ 0, then Jα (F (0) ) = −∗ uT .

(3.4.15)

With formula (3.4.9) this implies that p(0) = −∗ uT .

(3.4.16)

This initial descent direction is used in some versions of the CG-algorithm. Taking into account above gradient formula (3.4.14), we can rewrite formula (3.4.10) as follows: 

 J  (F (n) ), p(n) (F (n) − uT , p(n) ) = , βn :=

p(n) 2

p(n) 2

(3.4.17)

3.4 Computational Issues in Inverse Source Problems

115

Lemma 3.4.1 Formula (3.4.10) for the descent direction parameter βn is equivalent to the following formula: βn =

J  (F (n) ) 2 .

p(n) 2

(3.4.18)

Proof By (3.4.11) and (3.4.14) we have:   J  (F (n+1) ) = ∗ (F (n) − βn p(n) ) − uT   = ∗ F (n) − uT − βn ∗ p(n) . Hence J  (F (n+1) ) = J  (F (n) ) − βn ∗ p(n) .

(3.4.19)

Using this formula we show the orthogonality

 p(n) , J  F (n+1) = 0.

(3.4.20)

Indeed,  (n)  (n+1)   (n)  (n)    p , J (F ) = p , J (F ) − βn p(n) , ∗ p(n)     = p(n) , J  (F (n) ) − βn p(n) , p(n) = 0, by formula (3.4.10). Substituting now formulae (3.4.11) and (3.4.13) in (3.4.10) and using the orthogonality (3.4.20), we deduce:  J  (F (n) ), J  (F (n) )) + γn p(n−1)

J  (F (n) ) 2 βn = = .

p(n) 2

p(n) 2 

 

This implies (3.4.18).

Note that formula (3.4.18) is more convenient in computations. Some orthogonality properties of the parameters of the CG-algorithm are summarized in the following lemma. Lemma 3.4.2 Let conditions (3.1.3) and (3.1.69) hold. Denote by {J  (F (n) )} ⊂ L2 (0, ) and {p(n) } ⊂ L2 (0, ), n = 0, 1, 2, . . . the sequences of gradients and descent directions obtained by CG-algorithm. Then the gradients are orthogonal and the descent directions are -conjugate, that is,    (n) (k) ) = 0, ), J  (F J (F  p(n) , p(k) = 0, for all n = k,

(3.4.21)

116

3 Inverse Source Problems with Final Overdetermination

where  : L2 (0, ) → L2 (0, ) is the input-output mapping corresponding to the inverse source problem (3.1.1)–(3.1.2). Proof To prove both assertions of (3.4.21) we use simultaneous induction with respect to n. Let n = 1. For k = 0 we have: 

     J  (F (0) ), J  (F (1) ) := J  (F (0) ), J  (F (0) ) − β0 J  (F (0) ), ∗ p(0) = J  (F (0) ) 2 −



J  (F (0) ) 2  (0) (0) (F ), p J = 0,

p(0) 2

due to (3.4.9) and (3.4.10). Further, using this orthogonality we deduce:

   1  (0) p(0) , p(1) = ∗ p(0) , p(1) = J (F ) − J  (F (1) ), p(1) β0  

J  (F (1) ) 2  (0) 1  (1)  (0)  (1) J (F ) +  (0) 2 J (F ), J (F ) − J (F ) = 0, = β0

J (F )

by (3.4.13) and (3.4.19). Assume now that the assertions   (n)  J (F ), J  (F (k) ) = 0,  (n)  p , p(k) = 0, for all k = 1, n − 1

(3.4.22)

hold. We need to prove that 

 J  (F (n+1) ), J  (F (k) ) = 0,  (n+1)  p for all k = 1, n. , p(k) = 0,

(3.4.23)

First we prove these assertions for k = 1, n − 1. To prove the first assertion of (3.4.23), we use (3.4.19), then the first assertion of (3.4.22), and finally formula J  (F (n) )(x) = p(n) (x) − γn p(n−1) (x) obtained from (3.4.13). Then we get: 

   J  (F (n+1) ), J  (F (k) ) = J  (F (n) ) − βn ∗ p(n) , J  (F (k) )   = −βn ∗ p(n) , J  (F (k) )   = −βn ∗ p(n) , p(k) − γk p(k−1)   = −βn p(n) , p(k) − γk p(k−1) ,

for all k = 1, n − 1. By (3.4.22) the right hand side scalar product is zero.

(3.4.24)

3.4 Computational Issues in Inverse Source Problems

117

We prove the second assertion of (3.4.23) for k = 1, n − 1, in the similar way by using assertions (3.4.22) with formulae (3.4.13) and (3.4.19). We have: 

   p(n+1) , p(k) = J  (F (n+1) ) + γn p(n) , p(k)     = J  (F (n+1) ), p(k) = J  (F (n+1) ), ∗ p(k)  1  (n+1) J (F ), J  (F (k) ) − J  (F (k+1) ) = 0. = βn

To complete the induction we need to prove assertions (3.4.23) for k = n. For the first of them we use the same technique as in (3.4.24):      (n+1) ), J  (F (n) ) = J  (F (n) ) − βn ∗ p(n) , J  (F (n) ) J (F   = J  (F (n) ) 2 − βn ∗ p(n) , J  (F (n) )   = J  (F (n) ) 2 − βn p(n) , J  (F (n) )   = J  (F (n) ) 2 − βn p(n) , (p(n) − γn p(n−1) ) 

J  (F (n) ) 2 (n) p , p(n) = 0. = J  (F (n) ) 2 − (n) 2

p

The second assertion of (3.4.23) is proved in the same way.

 

The detailed results related to application of the CG-algorithm to ill-posed problems are treated in great detail in books [37, 52, 77] and references given therein. Here, we use only some of these results, concerning to the considered here class of inverse problems. The most important convergence result is given in the convergence theorem (Theorem 7.9 [37]). We give this theorem slightly modifying it. Theorem 3.4.1 Let F (n) be iterates defined by the CG-algorithm with the stopping rule (3.4.12). (a) If uδT ∈ D(† ), then the iterates F (n) converge to † uT , as n → ∞. (b) If uT ∈ / D(† ), then F (n) → ∞, as n → ∞. This theorem implies, in particular, that the iteration must be terminated appropriately when dealing with perturbed (or noisy) data uδT ∈ / D(† ), due to numerical δ 2 instabilities. Remark that here and below uT ∈ L (0, ) is the noisy data with the noise level δ > 0, by definition (3.4.6), that is, uδT − uT L2 (0,) ≤ δ. The termination index n = n(δ, uδT ) in the stopping condition (3.4.12) of the CG-algorithm is defined according to Morozov’s discrepancy principle given in Sect. 2.6. If uδT ∈ D(† ) and the CG-algorithm is stopped according to discrepancy principle, then this algorithm guarantees a finite termination index n(δ, uδT ).

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3 Inverse Source Problems with Final Overdetermination

3.4.3 Convergence of Gradient Algorithms for Functionals with Lipschitz Continuous Fréchet Gradient Let us return to the inverse source problem discussed in Sect. 2.1. Lemma 3.1.3 asserts the Lipschitz continuity J ∈ C 1,1 of Fréchet gradient of the Tikhonov functional J (F ) corresponding to the final data inverse problem for heat equation. Here we will show that an important advantage of gradient methods comes when dealing with this class of functionals. We start with a simple gradient-type iteration algorithm, usually used in solving ill-posed problems and defined also as Landweber iteration algorithm. Consider the functional J : H → R+ defined on a real Hilbert space H . Denote by {v (n) } ⊂ U the iterations defined as follows: v (n+1) = v (n) − ωn J  (v (n) ), n = 0, 1, 2, . . . ,

(3.4.25)

where v (0) ∈ U is an initial iteration and ωn > 0 is a relaxation parameter, defined by the minimum problem:

 fn (ωn ) := inf fn (ω), fn (ω) := J v (n) − ωJ  (v (n) ) , ω≥0

(3.4.26)

for each n = 0, 1, 2, . . . . We assume here and below, without loss of generality, that J  (v (n) ) = 0, for all n = 0, 1, 2, . . . . Lemma 3.4.3 Let U ⊂ H be a convex set of a real Hilbert space, J : U → R+ be a functional defined on U ⊂ H . Assume that the functional J (u) has Lipschitz continuous Fréchet gradient, i.e. J ∈ C 1,1 , that is, for all u1 , u2 ∈ U |J  (u) − J  (v)| ≤ L u − v H ,

(3.4.27)

where L > 0 is the Lipschitz constant. Then for all u1 , u2 ∈ U the following inequality hods:   1 |J (u1 ) − J (u2 ) − J  (u2 ), u1 − u2 || ≤ L u1 − u2 2H . 2 Proof Assuming u1 = u + h, u2 = u in the increment formula 

1

 J  (u + θ h)), h dθ,

J (u + h) − J (u) = 0

we get  J (u1 ) − J (u2 ) = 0

1

 J  (u2 + θ (u1 − u2 )), u1 − u2 dθ.

(3.4.28)

3.4 Computational Issues in Inverse Source Problems

119

Using the Lipschitz condition (3.4.27) we conclude that   |J (u1 ) − J (u2 ) − J  (u2 ), u1 − u2 |



1

 J  (u2 + θ (u1 − u2 )) − J  (u2 ), u1 − u2 dθ

= 0

 ≤ L u1 − u2 2H

1 0

θ dθ =

1 L u1 − u2 2H . 2  

This completes the proof.

Lemma 3.4.4 Let conditions of Lemma 3.4.3 hold. Assume that J∗ := infU J (u) > −∞. Denote by {v (n) } ⊂ U the a sequence of iterations defined by the gradient algorithm (3.4.25) and (3.4.26). Then {J (v (n) )} ⊂ R+ is a monotone decreasing sequence and lim J  (v (n) ) H = 0,

n→∞

(3.4.29)

Proof Use inequality (3.4.28), taking u1 = v (n) − ωn J  (v (n) ) and u2 = v (n) , with the relaxation parameter ωn > 0 defined in (3.4.26). Then we get:  

Lωn2    (n) 2 J v (n) − ωn J  (v (n) ) − J (v (n) ) + ωn J  (v (n) ) 2H ≤ J (v ) . H 2   Due to (3.4.25), J (v (n+1) ) = J v (n) − ωn J  (v (n) ) . Using this in the above inequality we obtain: J (v (n) ) − J (v (n+1) ) ≥ ωn (1 − Lωn /2) J  (v (n) ) 2H , ∀ω ≥ 0. The function g(ωn ) = ωn (1 − Lωn /2), ωn > 0 reaches its maximum g∗ := g(ω˜ n ) = 1/(2L) at ω˜ n = 1/L. Hence J (v (n) ) − J (v (n+1) ) ≥

1

J  (v (n) ) 2H , ∀v (n) , v (n+1) ∈ U. 2L

(3.4.30)

The right hand side is positive, since J  (v (n) ) = 0, which means that the sequence {J (v (n) )} ⊂ R+ is monotone decreasing. Since this sequence is bounded below, this assertion also implies convergence of the numerical sequence {J (v (n) )}. Then passing to the limit in the above inequality, we obtain the second assertion (3.4.29) of the lemma.   Lemma 3.4.4 shows that in the case of Lipschitz continuity of the gradient J  (v) the relaxation parameter ω > 0 can be estimated through the Lipschitz constant L > 0.

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3 Inverse Source Problems with Final Overdetermination

Let us apply this lemma to the Tikhonov functional given by (3.4.7) and find the value of the relaxation parameter ωn > 0 at nth iteration through the gradient J  (F (n) ). Using (3.4.26) we conclude that 2

 1   f (ω) := J F (n) − ωn J  (F (n) ) = F (n) − ωn J  (F (n) ) − uδT  2 2 2 

 1 1     = F (n) − uδT  − ωn F (n) − uδT , J  (F (n) ) + ωn2 J  (F (n) ) . 2 2 Since the relaxation parameter is defined as a solution of the minimum problem (3.4.26), from the condition f  (ωn ) = 0 we obtain:  ωn :=

   ∗ F (n) − uδT , J  (F (n) )  (F (n) − uδT ), J  (F (n) ) = .     J  (F (n) )2 J  (F (n) )2

  But it follows from the gradient formula (3.4.14) that J  (F (n) ) = ∗ F (n) − uT . Hence we arrive at the formula   (n) 2 J (F ) ωn =   . J  (F (n) )2

(3.4.31)

This formula with (3.4.18) illustrate the similarity and dissimilarity between the descent direction parameter βn in the CG-algorithm and the relaxation parameter ωn in the Landweber iteration algorithm. These parameter coincide only for n = 0, i.e. β0 = ω0 , due to formula (3.4.9). As a consequence, the first iterations obtained by these algorithms are also the same. Comparison of the numerical results obtained by these algorithms will be given below. It turns out that, for a functional with Lipschitz continuous Fréchet gradient the rate of convergence of the Landweber iteration algorithm can also be estimated. Lemma 3.4.5 Let, in addition to conditions of Lemma 3.4.3, J : H → R+ with J∗ := infU J (u) > −∞ is a continuous convex functional. Assume that the set M(v (0) ) := {u ∈ H : J (u) ≤ J (v (0) )}, where v (0) is an initial iteration, is bounded. Then the a sequence of iterations {v (n) } ⊂ U defined by the gradient algorithm (3.4.25) and (3.4.26) is a minimizing sequence, i.e. lim J (v (n) ) = J∗ .

n→∞

Moreover, for the rate of convergence the sequence {J (v (n) )} following estimate holds: 0 ≤ J (v (n) ) − J∗ ≤ 2L d n−1 , n = 1, 2, 3, . . . ,

(3.4.32)

3.4 Computational Issues in Inverse Source Problems

121

where L > 0 is the Lipschitz constant and d := sup u − v H , u, v ∈ M(v (0) ) is the diameter of the set M(v (0) ). Proof It is known that the minimum problem J∗ =

inf

M(u(0) )

J (u)

for a continuous convex functional in a bounded closed and convex set has a solution. Hence each minimizing sequence {v (n) } ⊂ M(v (0) ) weakly converges to an element u∗ ∈ U∗ of the solution set U∗ := {u ∈ M(v (0) ) : J (u) = J∗ }, that is J (v (n) ) → J (u∗ ), as n → ∞. To prove the rate of the convergence, we introduce the numerical sequence {an } defined as an := J (v (n) ) − J (u∗ ), n = 1, 2, 3, . . . .

(3.4.33)

For a functional with Lipschitz continuous Fréchet gradient convexity is equivalent to the following inequality: 

J (v (n) ) − J (u∗ ) ≤ J  (v (n) ), v (n) − u∗ . Applying to the right hand the Cauchy-Schwartz inequality we conclude that an ≤ J  (v (n) )

v (n) − u∗ ≤ d J  (v (n) ) . With estimate (3.4.30) and (3.4.33) this implies:  an2 ≤ 2L d 2 J (v (n) ) − J (v (n+1) ) = 2L d 2 [an − an+1 ]. Thus the numerical sequence {an } defined by (3.4.33) has the following properties: an2 > 0, an − an+1 ≥

1 a 2 , n = 1, 2, 3, . . . . 2L d 2 n

(3.4.34)

Evidently {an } is a monotone decreasing sequence with an /an+1 > 1. Using these properties we prove now that an = O(n−1 ). Indeed, for all k = 0, n − 1 we have: 1 ak+1

−

ak2 1 ak − ak+1 1 = ≥ 2 ak ak ak+1 2L d ak ak+1 ≥

ak 1 1 > , n = 1, 2, 3, . . . . 2 2L d ak+1 2L d 2

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3 Inverse Source Problems with Final Overdetermination

Summing up these inequalities and using inequalities (3.4.34) for each summand we deduce that 1 an+1

−

n   1 n 1 1 ≥ := − , n = 1, 2, 3, . . . . a1 ak+1 ak 2L d 2 k=1

This implies that an = O(n−1 ). Taking into account (3.4.33) we arrive at the estimate of the rate of the convergence (3.4.32).  

3.4.4 Numerical Examples In this subsection we present some numerical examples to show the performance analysis of the CG-algorithm. Consider the inverse problem of determining an unknown source term F (x) in ⎧ ⎨ ut = (k(x)ux )x + F (x)G(t), (x, t) ∈ T := (0, ) × (0, T ]; (3.4.35) u(x, 0) = 0, x ∈ [0, ]; ⎩ u(0, t) = 0, u(, t) = 0, t ∈ (0, T ), from the final data uT (x) := u(x, T ), x ∈ [0, ].

(3.4.36)

In all examples below the fine mesh with the mesh parameters Nx = 201 and Nt = 801 is used in the FEM scheme (3.4.4) to generate the noise free synthetic output uT ,h . Remark that the computational noise level defined as δc := uT − uT ,h L2 (0,)/ uT L2 (0,), where uT and uT ,h outputs obtained from the exact and h

h

numerical solutions of the direct problem, on this fine mesh is estimated as 10−8 . With this accuracy the synthetic output data uT ,h generated on the fine mesh will be assumed as a noise free. The coarser mesh with the mesh parameters Nx = 100 and Nt = 201 is used in numerical solution of inverse problems. The noisy output data uδT ,h , with uT ,h − uδT ,h L2 (0,) = δ, is generated by employing the “randn” function in MATLAB, that h is, uδT ,h (x) = uT ,h (x) + γ uT ,h L2 (0,) randn(N), h

(3.4.37)

where γ > 0 is the noise level. The parameter τM > 1 in the stopping condition (3.4.12) is taken as τM = 1.1. Example 3.4.1 The performance analysis of CG-algorithm (3.4.25): reconstruction of a smooth function with one critical point.

3.4 Computational Issues in Inverse Source Problems 3

123

2.5

2.5 2 2 F(x)

F(x)

1.5 1.5 F(x) F(0)(x) Fα(n)(x), n=3, α=0

1

Fα(n)(x), n=20, α=10-4

0.5

Fα(n)(x), n=20, α=10-5

0.5

Fα(n)(x), n=20, α=10-5

Fα(n)(x), n=5, α=10-6

F(n) (x), n=3, α=10-6 α

0 0

F(x) F(0)(x) Fα(n)(x), n=5, α=0

1

0.2

0.4

x

0.6

F(n) (x), n=4, α=10-7 α

0 0.8

1

0

0.2

0.4

x

0.6

0.8

1

Fig. 3.5 Reconstructions of the smooth function with one critical point: γ = 5% (left figure), and the smooth function with three critical points: γ = 3% (right figure) Table 3.4 Errors depending on the parameter of regularization: γ = 5% α e(n; α; δ) E(n; α; δ)

0 4.7 × 10−3 0.2030

1.0 × 10−4 7.3 × 10−3 0.3342

1.0 × 10−5 4.7 × 10−3 0.2019

1.0 × 10−6 4.7 × 10−3 0.2028

To generate the noise free synthetic output uT ,h the simplest concave function F (x) = 10x(1 − x), x ∈ (0, ),

(3.4.38)

with k(x) = 1 + x 2 , G(t) = exp(−t), t ∈ [0, 1], is used as an input data in the direct (3.4.35). The function uT ,h , obtained from the numerical solution of the direct problem, is assumed to be the noise free output data. Then, using (3.4.37), the noisy output data uδT ,h is generated. For the value γ = 5% of the noise level, the parameter δ > 0 is obtained as δ := uT ,h − uδT ,h L2 (0,) = 4.6 × 10−3 . h The CG-algorithm is employed with and without regularization. The left Fig. 3.5 displays the reconstructions of the unknown source F (x) from the noisy data uδT ,h , with the noise level γ = 5%. The function F (x), defined by (3.4.38), is plotted by the solid line. This figure also shows the iteration numbers (n) corresponding to the values of the parameter of regularization α > 0 are given Table 3.4 shows that the value α = 10−6 of the parameter of regularization defined from the both conditions α  1 and δ 2 /α < 1 is an optimal one. Remark that the reconstructions obtained with the optimal value α = 10−6 of the parameter of regularization and without regularization α = 0 are very close, as the left Fig. 3.5 shows. This illustrates the regularizing property of the CG-algorithm.  Example 3.4.2 The performance analysis of CG-algorithm: reconstruction of a function with three critical points.

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3 Inverse Source Problems with Final Overdetermination

Table 3.5 Errors depending on the parameter of regularization: γ = 3% α e(n; α; δ) E(n; α; δ)

1.0 × 10−5 1.6 × 10−3 0.4331

0 4.5 × 10−4 0.1443

1.0 × 10−6 4.9 × 10−4 0.1472

1.0 × 10−7 4.5 × 10−4 0.1423

Table 3.6 Errors depending on the parameter of regularization: γ = 3% α e(n; α; δ) E(n; α; δ)

0 8.0 × 10−3 0.1165

1.0 × 10−3 1.5 × 10−2 0.2754

1.0 × 10−4 8.2 × 10−3 0.1306

1.0 × 10−5 8.0 × 10−3 0.1060

1.0 × 10−6 8.0 × 10−3 0.1152

In this example, the function F (x) = 5(x − x 2 ) + sin(3πx), x ∈ [0, 1], with the inputs k(x) and G(t) from Example 3.4.1 is used to generate the noise free synthetic output uT ,h . The noise free output uT ,h is then obtained from the numerical solution of the direct problem, from which the noisy output uδT ,h , with the noise level γ = 3% is generated. This level corresponds, the value δ := uT ,h − uδT ,h L2 (0,1) = h

4.3 × 10−4 of the parameter δ > 0. Table 3.5 illustrates the convergence error e(n; F ; δ) and the accuracy error E(n; F ; δ), computed by formulae (3.4.5). The right Fig. 3.5 presents the reconstructed sources for the different values of the parameter of regularization.  Example 3.4.3 The performance analysis of CG-algorithm: mixed boundary conditions in the direct problem. In both above examples we assumed in the direct problem (3.4.35) the homogeneous Dirichlet conditions u(0, t) = u(1, t) = 0. In this example we assume that the mixed boundary conditions u(0, t) = ux (1, t) = 0 are given in the direct problem, instead of the homogeneous Dirichlet conditions. The noise free synthetic output uT ,h is generated from the inputs: F (x) = sin(πx) +

√

x, k(x) = 1 + x 2 , x ∈ [0, 1], G(t) = exp(−t), t ∈ [0, 1],

assuming in the direct problem (3.4.35) the mixed boundary conditions u(0, t) = ux (, t) = 0 instead of the homogeneous Dirichlet conditions. The noisy output data uδT ,h is generated for the noise level γ = 5%, which corresponds to δ :=

uT ,h − uδT ,h L2 (0,1) = 8.0 × 10−3 . The convergence error and the accuracy error h are give in Table 3.6 illustrates the convergence error e(n; F ; δ) and the accuracy error E(n; F ; δ), computed by formulae (3.4.5). From a computational viewpoint the change of the Dirichlet conditions in the direct problem by the mixed boundary conditions means that the order of approximation of FE-scheme at x = l will be less than the order approximation at the interior mesh points. Natural effect of this case is seen from the reconstructed

3.4 Computational Issues in Inverse Source Problems 1.8

10

125

0

1.6 1.4 1.2

10–1

1 0.8 0.6 10–2

0.4 F(x) F(0)(x)

0.2 0 0.2

Randomness #1 Randomness #2

0

0.2

0.4

0.6

0.8

1

10–3

E(n;α;δ), n=5 e(n;α;δ), n=5

1

2

3

4

5

6

Fig. 3.6 Influence of randomness: α = 10−6 (left figure); and the termination index n(δ) = 3 shows the beginning of the iteration from convergence to divergence (right figure) (γ = 3%)

sources in the left Fig. 3.6 near the point x = 1. This figure also shows the degree of influence of the randomness to the reconstructions. The right Fig. 3.6 explains that the termination index n(δ) = 3 shows the beginning of the iteration from convergence to divergence.  The above computational results demonstrate that CG-algorithm is effective, robust against a middle noise level (γ = 3 ÷ 5%), and provides satisfactory reconstructions. The accuracy in the recovery of the spacewise dependent source decreases as the noise level increases. With an appropriately chosen parameter of regularization, the number of iterations of the CG-algorithm to reach the condition (3.4.12) with τM = 1.1 was about 3 ÷ 5.

Part II

Inverse Problems for Differential Equations

Chapter 4

Inverse Problems for Hyperbolic Equations

In the first part of this chapter we study two inverse source problems related to the second order hyperbolic equations ut t − uxx = ρ(x, t)g(t) and ut t − uxx = ρ(x, t)ϕ(x) for the quarter plane R2+ = {(x, t)| x > 0, t > 0}, with Dirichlet type measured output f (t) := u(x, t)|x=0 . The time-dependent source g(t) and the spacewise-dependent source ϕ(x) are assumed to be unknown in these inverse problems. Next, we study more complex problem, namely the problem of recovering the potential q(x) in the string equation ut t − uxx − q(x)u = 0 from the Neumann type measured output f (t) := ux (x, t)|x=0 . We prove the uniqueness of the solution and then derive the global stability estimate. In the final part of the chapter, inverse coefficient problems for layered media is studied as an application.

4.1 Inverse Source Problems We begin with the simplest linear inverse source problems for wave equation. In the first case, we assume that the time dependent source term g(t) in the wave equation ut t − uxx = ρ(x, t)g(t), (x, t) ∈ R2+ , is unknown and needs to be identified from boundary information. In the second case, based on the same boundary information, the inverse source problem of determining the unknown spacewise dependent source term ϕ(x) in the wave equation ut t − uxx = ρ(x, t)ϕ(x) in studied. In both cases, the reflection method [38] is used to derive solution of the inverse problems through the solution of the Volterra equation of the second kind.

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_4

129

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4 Inverse Problems for Hyperbolic Equations

4.1.1 Recovering a Time Dependent Source Term Consider the initial-boundary value problem ⎧ 2 ∂ ∂2 ⎪ ⎪ ⎪ − 2 u(x, t) = ρ(x, t)g(t), (x, t) ∈ R2+ , ⎪ ⎨ ∂t 2 ∂x u|t =0 = 0, ut |t =0 = 0, ⎪ ⎪ ⎪ ⎪ ⎩ ux |x=0 = 0

x > 0,

(4.1.1)

t>0

for the inhomogeneous wave equation on R2+ = {(x, t)| x > 0, t > 0}. If the righthand side in the wave Eq. (4.1.1) is given, then the initial-boundary value problem (4.1.1) is a well-posed problem of mathematical physics. Let us assume that the right-hand side is known only partially, namely, the function ρ(x, t) is known, but the time dependent function g(t) is unknown. Consider the following inverse source problem: find the time dependent function g(t) for a given function ρ(x, t) and the given trace f (t) := u(x, t)|x=0 ,

t ≥ 0,

(4.1.2)

defined on the half-axis x = 0, t ≥ 0, of the solution u(x, t) of problem (4.1.1). The function f (t) in (4.1.2) is called Dirichlet type measured output. The problem (4.1.1)–(4.1.2) is called an inverse source problem of determining an unknown time dependent source in the wave equation from Dirichlet boundary data. Accordingly, for a given admissible function g(t), the initial-boundary value problem (4.1.1) is defined as the direct problem. We assume that the functions ρ(x, t) and g(t) satisfy the following conditions: ρ ∈ C(R2+ ), ρx ∈ C(R2+ ), g ∈ C[0, ∞).

(4.1.3)

The last condition in (4.1.3) suggests that we will look for a solution of the inverse problem in the space C[0, ∞). Now we use the reflection method extending the functions u(x, t) and ρ(x, t) to the domain R2− = {(x, t)| x < 0, t > 0} as even, with respect to x, functions, that is, u(x, t) = u(−x, t) and ρ(x, t) = ρ(−x, t), for x < 0. Then the condition ux |x=0 = 0 is automatically satisfied. Moreover, the extended function u(x, t) solves the Cauchy problem ⎧ 2 2 ⎪ ⎨ ∂ − ∂ u(x, t) = ρ(x, t)g(t), x ∈ R, t > 0, ∂t 2 ∂x 2 ⎪ ⎩ t > 0. u|t =0 = 0, ut |t =0 = 0,

(4.1.4)

4.1 Inverse Source Problems

131

The unique solution of this problem is given by d’Alembert’s formula u(x, t) =

1 2

 ρ(ξ, τ )g(τ ) dξ dτ, (x,t )

where (x, t) = {(ξ, τ ) ∈ R2 | 0 ≤ τ ≤ t−|x−ξ |} is the characteristic triangle with the vertex at the point (x, t). Taking into account the additional condition (4.1.2) we get: f (t) =

1 2



 ρ(ξ, τ )g(τ ) dξ dτ = (0,t )

ρ(ξ, τ )g(τ ) dξ dτ,

t ≥ 0,

+ (0,t )

where + (0, t) = {(ξ, τ ) ∈ R2 | 0 ≤ τ ≤ t, 0 ≤ ξ ≤ t − τ }. This implies the following equation with respect to the unknown function g(t): 



t

t −τ

g(τ ) 0

ρ(ξ, τ ) dξ dτ = f (t),

t ≥ 0.

(4.1.5)

0

Thus the inverse problem (4.1.1)–(4.1.2) is reformulated as the integral equation (4.1.5). We deduce from Eq. (4.1.5) that under conditions (4.1.3) the function f (t) satisfies the following conditions: f (0) = f  (0) = 0.

f ∈ C 2 [0, ∞),

(4.1.6)

Indeed, differentiating (4.1.5), we get: f  (t) =



t

g(τ )ρ(t − τ, τ ) dτ,  t  g(τ )ρx (t − τ, τ ) dτ, t ≥ 0, f (t) = g(t)ρ(0, t) + 0

(4.1.7)

0

which means, fulfilment of conditions (4.1.6). The conditions f (0) = f  (0) = 0 can be treated as consistency conditions for the output f (t). The above derived conditions (4.1.6) for the function f (t) are the necessary conditions for solvability of the inverse problem (4.1.1)–(4.1.2). We prove that under the additional condition ρ(0, t) = 0, for all t ∈ [0, T ],

(4.1.8)

these conditions are also sufficient conditions for the unique solvability of the inverse problem on the closed interval [0, T ].

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4 Inverse Problems for Hyperbolic Equations

Theorem 4.1.1 Let the source function ρ(x, t) satisfies conditions (4.1.3) and (4.1.8) in DT := {(x, t) ∈ R2+ | 0 ≤ t ≤ T − x}, T > 0. Then the inverse problem (4.1.1)–(4.1.2) is uniquely solvable in C[0, T ] if and only if the function f (t) satisfies conditions f (0) = f  (0) = 0, f (t) ∈ C 2 [0, T ]. Proof We only need to prove that the conditions on the function f (t) allows to find uniquely the function g(t). Divide the both sides of the second equation in (4.1.7) by ρ(0, t). Then we obtain the following Volterra equation of the second kind: 

t

g(t) +

K(t − τ, τ )g(τ )dτ = F (t),

t ∈ [0, T ],

(4.1.9)

0

with the kernel K(x, t) and the right hand side F (t), depending on input and output data: K(x, t) = ρx (x, t)/ρ(0, t),

F (t) = f  (t)/ρ(0, t), t ∈ [0, T ].

(4.1.10)

By conditions (4.1.3) and (4.1.6) these functions are continuous, that is, K ∈ C(DT ) and F ∈ C[0, T ]. It is well known, that the Volterra equation of the second kind with such kernel and right-hand side has the unique solution g ∈ C[0, T ] [153]. This completes the proof of the theorem.   Remark that Eq. (4.1.9) can easily be solved, for example, by Picard’s method of successive approximations: g0 (t) = F (t),  t gn (t) + K(t − τ, τ )gn−1 (τ )dτ = F (t), n = 1, 2, . . . , 0

which uniformly converges in C[0, T ]. Remark 4.1.1 As formula (4.1.10) shows, the right hand side F (t) of the integral equation (4.1.9) contains the second derivative f  (t) of the output. This implies that the inverse problem (4.1.1)–(4.1.2) is ill-posed. This linear ill-posedness arises from the fact that the measured output f (t) have to be differentiated twice. If, for example, f δ (t) = f (t) + δ sin(t/δ) is a noisy data, then f δ − f C[0,T ] → 0, as δ → 0, while F δ − F C[0,T ] → ∞.

4.1 Inverse Source Problems

133

4.1.2 Recovering a Spacewise Dependent Source Term Consider the inverse source problem of identifying the unknown spacewise dependent source ϕ(x) in ⎧ 2 ∂ ∂2 ⎪ ⎪ ⎪ − 2 u(x, t) = ρ(x, t)ϕ(x), (x, t) ∈ R2+ , ⎪ ⎨ ∂t 2 ∂x u|t =0 = 0, ut |t =0 = 0, ⎪ ⎪ ⎪ ⎪ ⎩ ux |x=0 = 0,

(4.1.11)

x > 0, t >0

from the trace of the function u(x, t) given by (4.1.2). The function ρ(x, t) is assumed to be known. Let us use again the reflection method, with the same even extensions u(x, t) = u(−x, t) and ρ(x, t) = ρ(−x, t) to the domain R2− = {(x, t)| x < 0, t > 0}. As noted above, the condition ux |x=0 = 0 is automatically satisfied. Then we can derive the solution u(x, t) of the direct problem (4.1.11) by d’Alembert’s formula: u(x, t) =

1 2

 ρ(ξ, τ )ϕ(ξ ) dξ dτ. (x,t )

With the additional condition (4.1.2) this implies: 1 f (t) = 2



 ρ(ξ, τ )ϕ(ξ ) dξ dτ = (0,t )

ρ(ξ, τ )ϕ(ξ ) dξ dτ. + (0,t )

Hence the unknown spacewise dependent source ϕ(x) satisfies the integral equation: 



t

t −ξ

ϕ(ξ ) 0

ρ(ξ, τ ) dτ dξ = f (t),

t ≥ 0.

(4.1.12)

0

As in the previous problem, we conclude again that the function f (t) satisfies conditions (4.1.6). Indeed, differentiating (4.1.12), one get f  (t)



t

=

ϕ(ξ )ρ(ξ, t − ξ ) dξ,  t  f (t) = ϕ(t)ρ(t, 0) + ϕ(τ )ρt (ξ, t − ξ ) dξ, 0

(4.1.13) t ≥ 0.

0

Thus, we arrive at almost the same necessary conditions (4.1.6) for a solvability of the inverse problem. Now we return to the inverse problem defined by (4.1.11) and (4.1.2). Assuming that ρ(t, 0) = 0, for all t > 0, we can formulate the necessary and sufficient conditions for the unique solvability of this inverse problem, similar to Theorem 4.1.1.

134

4 Inverse Problems for Hyperbolic Equations

Theorem 4.1.2 Let the source function ρ(x, t) satisfies the following conditions: ρ ∈ C(DT ), ρt ∈ C(DT ), ρ(t, 0) = 0, t ∈ [0, T ], T > 0. Then the inverse problem defined by (4.1.11) and (4.1.2) is uniquely solvable in C[0, T ] if and only if the function f (t) belongs to C 2 [0, T ] and satisfies to the conditions f (0) = f  (0) = 0. Proof Dividing the both sides of the second equation in (4.1.13) by ρ(t, 0) we obtain the Volterra equation of the second kind 

t

ϕ(t) +

K2 (ξ, t − ξ )ϕ(ξ ) dξ = F2 (t),

t ∈ [0, T ],

(4.1.14)

0

which has the unique solution ϕ ∈ C[0, T ], where K2 (x, t) = ρt (x, t)/ρ(t, 0), F2 (t) = f  (t)/ρ(t, 0). Evidently, K1 ∈ C(DT ), F1 ∈ C[0, T ]. Then we conclude that Eq. (4.1.14) has the unique solution ϕ ∈ C[0, T ].   Note that the solution of the Eq. (4.1.14) can be found by the same method of successive approximations. Remark 4.1.2 If function ρ(x, t) = 0 for 0 < t < x, then the solution to problem (4.1.11) satisfies the equality u(x, t) = 0 for 0 < t < x. In this case Eq. (4.1.12) takes the form  f (t) =



t /2

t −ξ

ϕ(ξ ) 0

ρ(ξ, τ ) dτ dξ,

t ≥ 0.

(4.1.15)

ξ

The derivatives of f (t) up to the second order are defined then by the formulae: f  (t) =



t /2 0

f  (t)

1 = ϕ 2

ϕ(ξ )ρ(ξ, t − ξ ) dξ,

   t /2 t t t ρ , + ϕ(τ )ρt (ξ, t − ξ ) dξ, 2 2 2 0

(4.1.16) t ≥ 0.

These considerations lead to the following result. Theorem 4.1.3 Let ρ(x, t) = 0 for 0 < t < x ≤ T − t and ρ ∈ C(DT ), ρt ∈ C(DT ), DT = {(x, t) ∈ R2 | 0 ≤ x ≤ t ≤ T − x}, for some positive T and ρ(t, t) = 0 for t ∈ [0, T /2]. Then the inverse problem defined by (4.1.11) and (4.1.2) is uniquely solvable in the space C[0, T /2] if and only if function f (t) belong to C 2 [0, T ] and satisfy to the consistency conditions f (0) = f  (0) = 0. Remark 4.1.3 The statement of the above theorems remain to hold if we replace the wave operator in Eqs. (4.1.1) and (4.1.11) by more general operator ∂ 2 /∂t 2 − ∂ 2 /∂x 2 + q(x), where q(x), x ≥ 0, is a given continuous function for x ≥ 0.

4.2 Problem of Recovering the Potential in String Equation

135

4.2 Problem of Recovering the Potential in String Equation In this section we consider the inverse problem of recovering the potential q(x), x ∈ R+ := {x ≥ 0} in ⎧  2 2 ⎪ ⎨ Lq u := ∂ − ∂ − q(x) u = 0, (x, t) ∈ , ∂t 2 ∂x 2 ⎪ ⎩ ut |t =0 = 0, u|x=0 = h(t), u|t =0 = 0,

(4.2.1)

from the output f (t) := ux |x=0 , t ∈ (0, T ],

(4.2.2)

given as the trace of the derivative ux of the solution u(x, t) to problem (4.2.2) at x = 0. Here = R+ × [0, T ]. It is assumed that h ∈ C 2 [0, T ], h(0) = 0.

(4.2.3)

The problem (4.2.1)–(4.2.2) of recovering the potential in the string equation can be treated as a simplest inverse coefficient problem for the differential operator Lq = ∂ 2 /∂t 2 − ∂ 2 /∂x 2 − q, with Neumann data f (t). The hyperbolic problem (4.2.1) will be defined as a direct problem. Assuming q ∈ C[0, ∞), we demonstrate that the potential q(x) can be determined uniquely and in a stable way from the output (4.2.2), for x ∈ [0, T /2]. Moreover, we will prove that the solution of the inverse problem exists for small T > 0, if the function f (t) satisfies some necessary conditions.

4.2.1 Some Properties of the Direct Problem The necessary conditions for unique solvability of the inverse problem, obtained in the next subsection, follow from properties of the solution of the direct problem (4.2.1). The lemma below shows that the solution u(x, t) of the hyperbolic problem (4.2.1) can be represented as a product of a smooth function and the Heaviside step function. Lemma 4.2.1 Let the boundary data h(t) satisfies conditions (4.2.3) and q ∈ C[0, T /2], T > 0. Then the solution of the direct problem (4.2.1) exists in the triangle T := {(x, t) ∈ R2 | 0 ≤ t ≤ T − x, x ≥ 0}. Moreover, it represented as a product u(x, t) = u(x, t)θ0 (t − x).

(4.2.4)

136

4 Inverse Problems for Hyperbolic Equations

of a smooth function u ∈ C 2 ( T ), T := {(x, t) ∈ R2 | 0 ≤ x ≤ t ≤ T − x}, and the Heaviside step function  θ0 (t) =

1, t ≥ 0, 0, t < 0.

(4.2.5)

Proof The function u(x, t) is identically zero, for all 0 < t < x, t ≤ T − x, since in this domain it solves the Cauchy problem for the homogeneous equation Lq u = 0 with homogeneous initial data. This implies the presentation (4.2.4). We introduce now the function u0 (x, t) = h(t − x)θ0(t − x), where h(t) is the Dirichlet input in the direct problem (4.2.1). We represent the function u(x, t) in the form: u(x, t) = u0 (x, t) + v(x, t).

(4.2.6)

Then v(x, t) solves the problem 

Lq v = q(x)u0(x, t), v|x=0 = 0,

0 < x < t ≤ T − x,

v|t =x = 0.

(4.2.7)

Let us explain the appearance of the condition v|t =x = 0 in (4.2.7), on the characteristic line t = x. For this aim we extend the functions v(x, t), u0 (x, t) to the domain x < 0 as an odd functions with respect to x, that is, v(x, t) = −v(−x, t), u0 (x, t) = −u0 (−x, t). We also extend the coefficient q(x) to the domain x < 0, as an even function, i.e., q(x) = q(−x). Then we may use d’Alembert’s formula for the solution of the hyperbolic problem (4.2.7) to get v(x, t) =

1 2

1 = 2

 q(x)(v(ξ, τ ) + u0 (ξ, τ )) dξ dτ 

(x,t )

♦(x,t )

q(x)(v(ξ, τ ) + u0 (ξ, τ )) dξ dτ,

t ≥ |x|.

where ♦(x, t) = {(ξ, τ ) ∈ R2 | |ξ | ≤ τ ≤ t − |x − ξ |}. The area of the domain ♦(x, t) tends to zero, as t → |x| + 0. Therefore v(x, |x| + 0) = 0, which implies the fulfilment of the second condition v|t =x = 0 in (4.2.7). Let us introduce the functions v1 (x, t) :=

∂v ∂v + , ∂t ∂x

v2 (x, t) :=

∂v ∂v − . ∂t ∂x

(4.2.8)

Then the following relations hold: Lq v =

∂v1 ∂v2 ∂v2 ∂v1 − − qv = + − qv, ∂t ∂x ∂t ∂x

∂v v1 + v2 = . ∂t 2

(4.2.9)

4.2 Problem of Recovering the Potential in String Equation

137

Note that v1 |t =x = 0,

v2 |x=0 = −v1 (0, t).

(4.2.10)

Now we are going to obtain integral relationships between the above introduced functions v1 , v2 , v. Integrating the equation Lq v = qu0 on the plane (ξ, τ ) along the line ξ + τ = x + t, then using relations (4.2.9) and the first condition of (4.2.10) we find: v1 (x, t) =  t q(x + t − τ )(u0 (x + t − τ, τ ) + v(x + t − τ, τ )) dτ.

(4.2.11)

(t +x)/2

Similarly, integrating the equation Lq v = qu0 on the plane (ξ, τ ) along the line ξ − τ = x − t, using (4.2.9) and the second condition of (4.2.10) we obtain v2 (x, t) =



−v1 (0, t − x) +

t t −x

q(x − t + τ )(u0 (x − t + τ, τ ) + v(x − t + τ, τ )) dτ.

Use (4.2.11) in the first right hand side term of this relation equation. Then we get: v2 (x, t) =  t −x − q(t − x − τ )(u0 (t − x − τ, τ ) + v(t − x − τ, τ )) dτ (4.2.12) (tt−x)/2 + q(x − t + τ )(u0 (x − t + τ, τ ) + v(x − t + τ, τ )) dτ. t −x

Finally, integrate the second relation in (4.2.9) along the line ξ = x to find v(x, t) =

1 2



t

(v1 (x, τ ) + v2 (x, τ )) dτ.

(4.2.13)

x

Equations (4.2.11)–(4.2.13) form the system of integral equations with respect to the functions v1 , v2 , v defined in the domain T := {(x, t)| 0 ≤ x ≤ t ≤ T − x}. We prove that this system has a unique solution in T . For this aim, we represent the functions vp (x, t), p = 1, 2 and v(x, t) in the form vp (x, t) =

∞  n=1

vpn (x, t), p = 1, 2,

v(x, t) =

∞  n=1

v n (x, t),

(4.2.14)

138

4 Inverse Problems for Hyperbolic Equations

and look for the series solution of this system. The first terms (n = 1) of these series are given by the formulae:  v11 (x, t) =

t (t +x)/2  t −x

q(x + t − τ )u0 (x + t − τ, τ ) dτ,

v21 (x, t) = − q(t − x − τ )u0 (t − x − τ, τ ) dτ (t −x)/2  t q(x − t + τ )u0 (x − t + τ, τ ) dτ, + t −x $t v 1 (x, t) = 12 x (v11 (x, τ ) + v21 (x, τ )) dτ.

(4.2.15)

For n ≥ 2 other terms of series (4.2.14) are defined recursively as follows:  v1n (x, t)

=

t (t+x)/2 t −x

q(x + t − τ )v n−1 (x + t − τ, τ ) dτ,

q(t − x − τ )v n−1 (t − x − τ, τ ) dτ v2n (x, t) = − (t −x)/2  t + q(x − t + τ )v n−1 (x − t + τ, τ ) dτ, t −x  1 t n v n (x, t) = (v (x, τ ) + v2n (x, τ )) dτ. 2 x 1

(4.2.16)

Obviously v1n (x, t), v2n (x, t), v n (x, t) are continuous functions in T together with the first derivatives with respect to x and t, for all n ≥ 1. We use now Mathematical Induction to prove the uniform convergence of the series (4.2.14) in T . Denote by h0 = h C[0,T ] ,

q0 = q C[0,T /2]

the norms. For n = 1 we can easily derive the estimates: (t − x) , 2 h0 q0 T (t + x) ≤ , |v21 (x, t)| ≤ h0 q0 2 2 (t − x) |v 1 (x, t)| ≤ h0 q0 T , (x, t) ∈ T , 2 |v11 (x, t)| ≤ h0 q0

by using (4.2.15).

4.2 Problem of Recovering the Potential in String Equation

139

Assume that for n = k ≥ 1 the estimates (t − x)k γk (t − x)k−1 , |v2k (x, t)| ≤ k+1 , k! 2 (k − 1)! γk (t − x)k , γk = h0 q0k T k . |v k (x, t)| ≤ k+1 2 k!

|v1k (x, t)| ≤

γk 2k+1 T

(4.2.17)

hold. Using these estimates in (4.2.16), for n = k + 1 we find: |v1k+1 (x, t)| ≤ |v2k+1 (x, t)| ≤ = |v k+1 (x, t)| ≤

q 0 γk 2k+1 k! q 0 γk 2k+1 k!



t (t +x)/2



t −x (t −x)/2

− x)k

q0 γk (t 2k+1 k!

q0 γk (t − x)k+1 , 2k+2 (k + 1)!   t k k (2τ − t + x) dτ + (t − x) dτ

(2τ − x − t)k dτ =



t −x +x 2(k + 1)

γk+1 (t − x)k+1 , 2k+2 (k + 1)!

t −x

≤

γk+1 (t − x)k , 2k+2 k!

γk = h0 q0k T k .

Therefore estimates (4.2.17) hold for all n ≥ 1. Further, due to t − x ≤ T in T := {(x, t) ∈ R2 | 0 ≤ x ≤ t ≤ T − x}, all series in (4.2.14) converge uniformly and, as a result, their sums are continuous functions in T . Furthermore, for these functions the following estimates hold:   1 h0 exp q0 T (t − x) − 1 |v1 (x, t)| ≤ 2T 2   h0 1 2 ≤ q0 T − 1 := v10 , exp 2T 2  1 h0 q0 T exp q0 T (t − x) 2 2  1 h0 q0 T exp q0 T 2 := v20 , ≤ 2 2

|v2 (x, t)| ≤

(4.2.18)

  1 h0 exp q0 T (t − x) − 1 2 2   1 h0 q0 T 2 − 1 := v0 , ≤ exp 2 2

|v(x, t)| ≤

for all (x, t) ∈ T . Note that v ∈ C 1 ( T ), by the relations vt = (v1 + v2 )/2 and vx = (v1 − v2 )/2. It follows from (4.2.11)–(4.2.12) that the functions v1 (x, t) and v2 (x, t) are continuously differentiable with respect to x and t in T . To verify this,

140

4 Inverse Problems for Hyperbolic Equations

one needs to rewrite Eqs. (4.2.11) and (4.2.12) in the following form: 

(t +x)/2

v1 (x, t) =

q(ξ )(u0 (ξ, x + t − ξ ) + v(ξ, x + t − ξ )) dξ,

(4.2.19)

x



(t −x)/2

v2 (x, t) = − q(ξ )(u0 (ξ, t − x − ξ ) + v(ξ, t − x − ξ )) dξ  x0 (4.2.20) q(ξ )(u0 (ξ, t − x + ξ ) + v(ξ, t − x + ξ )) dξ. + 0

Since v1 and v2 belong to C 1 ( T ), the function v(x, t) is twice continuously differentiable in T . Then, it follows from (4.2.6) that the function u(x, t) is also twice continuously differentiable in T . This completes the proof of the lemma.  

4.2.2 Existence of the Local Solution to the Inverse Problem First, we establish a relationship between the input h(t) and the output f (t) of the inverse problem (4.2.1)–(4.2.2) and the functions v1 (x, t), v2 (x, t) introduced in (4.2.8). It follows from (4.2.6) that ux (x, t) = −h (t − x) + vx (x, t),

0 ≤ x ≤ t,

where h (t) means the derivative of h(t) with respect to its argument. Using formulae (4.2.8) we find that vx (x, t) = (v1 (x, t) − v2 (x, t))/2. Then the additional condition (4.2.2) leads to the relation f (t) := ux (0, t) = −h (t) + (v1 (0, t) − v2 (0, t))/2,

t ∈ (0, T ].

(4.2.21)

It follows from the Lemma 4.2.1 that f ∈ C 1 (0, T ]. Moreover, since v1 (0, 0) = v2 (0, 0) = 0, we conclude from (4.2.21) that f (+0) = −h (0).

(4.2.22)

Hence, the conditions (4.2.22) and f ∈ C 1 (0, T ] are the necessary conditions for solvability of the inverse problem in the class of continuous functions q(x) on the segment [0, T /2]. Let us show now that these conditions are also sufficient for a local unique solvability of the inverse problem. Theorem 4.2.1 Let f ∈ C 1 (0, T ], T > 0, and f (+0) = −h (0). Then there exist a positive number T0 ≤ T and unique function q(x) ∈ C[0, T0 /2] such that the solution to problem (4.2.1) satisfies relation (4.2.2) for t ≤ T0 .

4.2 Problem of Recovering the Potential in String Equation

141

Proof We derive first the integral relation which follows from Eqs. (4.2.19)– (4.2.20) and (4.2.21). Taking into account that u0 (x, t) = h(t − x) for t ≥ x ≥ 0 we obtain: 

t /2

q(ξ )(h(t − 2ξ ) + v(ξ, t − ξ )) dξ = fˆ(t),

(4.2.23)

0

where fˆ(t) := f (t) + h (t) ∈ C 1 [0, T ].

(4.2.24)

Taking now the derivative with respect to t of both sides of (4.2.23) and using v(x, x) = 0, we arrive at the equations: 1 q(t/2)h(0) + 2



t /2

q(ξ )(h (t − 2ξ ) + vt (ξ, t − ξ )) dξ = fˆ (t),

t ∈ [0, T ],

0

or q(x) +

1 h(0)



x

  q(ξ ) 2h (2x − 2ξ ) + v1 (ξ, 2x − ξ ) + v2 (ξ, 2x − ξ ) dξ

0

= F (x), x ∈ [0, T /2],

(4.2.25)

with the right hand side F (x) =

2fˆ (2x) 2(f  (2x) + h (2x)) = . h(0) h(0)

(4.2.26)

Equations (4.2.11)–(4.2.13) and (4.2.25) form a system of integral equations with respect to unknown functions q(x) and v1 (x, t), v2 (x, t), v(x, t). In order to simplify further notations, it is convenient to transform, first, the Eqs. (4.2.11) and (4.2.12), then to rewrite the system of integral equations in terms of an operator equation. For this aim we use (4.2.21) and (4.2.24) to deduce that both functions v1 and v2 are known at x = 0: v1 (0, t) = fˆ(t),

v2 (0, t) = −fˆ(t),

t ∈ [0, T ].

(4.2.27)

Using these conditions we integrate the equation Lq v = qu0 on the plane (ξ, τ ) along the line ξ + τ = x + t from point (x, t) ∈ T till the axis ξ = 0 to obtain the equation v1 (x, t) = fˆ(t + x) −



x 0

q(ξ ) (u0 (ξ, x + t − ξ ) + v(ξ, x + t − ξ )) dξ.

(4.2.28)

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4 Inverse Problems for Hyperbolic Equations

Similarly, integrating the equation Lq v = qu0 on the plane (ξ, τ ) along the line ξ − τ = x − t from point (x, t) ∈ T till the axis ξ = 0, we obtain the second equation v2 (x, t) = −fˆ(t − x)  x + q(ξ ) (u0 (ξ, t − x + ξ ) + v(ξ, t − x + ξ )) dξ.

(4.2.29)

0

Now we rewrite the system formed, respectively, by Eqs. (4.2.25), (4.2.28), (4.2.29), and (4.2.13), in the form of the operator equation ϕ = A ϕ, ϕ = (ϕ1 , ϕ2 , ϕ3 , ϕ4 ) := (q(x), v1 (x, t), v2 (x, t), v(x, t)),

(4.2.30)

where (A ϕ)1 (x)

$x 1  = F (x) − h(0) 0 ϕ1 (ξ )(2h (2x − 2ξ ) + ϕ2 (ξ, 2x − ξ ) +ϕ3 (ξ, 2x − ξ )) dξ, x ∈ [0, T /2],

$x (A ϕ)2 (x, t) = fˆ(t + x) − (0 ϕ1 (ξ )(h(x + t − 2ξ ) +ϕ4 (ξ, x + t − ξ )) dξ, (A ϕ)3 (x, t) = −fˆ(t − x) + (A ϕ)4 (x, t) =

$x 0

(4.2.31)

ϕ1 (ξ )(h(t − x) + ϕ4 (ξ, t − x + ξ )) dξ,

$ 1 t 2 x (ϕ2 (x, τ ) + ϕ3 (x, τ )) dτ,

(x, t) ∈ T .

Let ϕ 0 = (ϕ10 , ϕ20 , ϕ30 , ϕ40 ) be a vector with components ϕ10 = F (x),

ϕ20 = fˆ(t + x),

ϕ30 = −fˆ(t − x),

ϕ40 = 0.

Denote by C( T ) the space of continuous vector functions, with the norm

ϕ C( T ) = max ϕk C( T ) . k=1,4

Since ϕ0 ∈ C( T ), all vector functions defined by (4.2.31) are evidently elements of C( T ). We introduce in this Banach space the closed ball BT := {ϕ ∈ C( T ) : ϕ − ϕ 0 C( T ) ≤ ϕ 0 C( T ) }.

(4.2.32)

of radius ϕ 0 C( T ) > 0 centered at ϕ 0 ∈ C( T ). Evidently,

ϕ 0 C( T ) ≤ a0 (T ) := max( F C[0,T /2] ; fˆ C[0,T ] ), where fˆ(t) and F (x) are defined by (4.2.24) and (4.2.26), respectively.

(4.2.33)

4.2 Problem of Recovering the Potential in String Equation

143

Hereafter we assume that F (x) and fˆ(t) are given fixed functions. Then their norms F C[0,T /2] , fˆ C[0,T ] depend on T only. Taking it into account we have used in (4.2.33) the notation a0 (T ) for the maximum of these two norms. The similar notations for some values we shall use and later on in order indicate on a dependence of these values on T . Now we are going to prove that the operator A, defined by (4.2.30)–(4.2.31) is a contraction on the Banach space BT , if the final time T > 0 is small enough. Recall that an operator is named contracting one on BT , if the following two conditions hold: (c1) (c2)

Aϕ ∈ BT , for all ϕ ∈ BT ; for all ϕ 1 , ϕ 2 ∈ BT , the condition

A ϕ 1 − A ϕ 2 C( T ) ≤ ρ ϕ 1 − ϕ 2 C( T )

holds with some ρ ∈ (0, 1). We verify the first condition (c1). Let ϕ ∈ BT . Then

ϕ C( T ) ≤ ϕ − ϕ 0 C( T ) + ϕ 0 C( T ) ≤ 2 ϕ 0 C[0,T ] ) ≤ 2a0 (T ), by (4.2.33). Using this in (4.2.31) we estimate the norms |(A ϕ)k − ϕk0 |, k = 1, 4 as follows: $x 1  |(A ϕ)1 − ϕ10 | ≤ |h(0)| 0 |ϕ1 (ξ )|(2 |h (2x − 2ξ )| +|ϕ2 (ξ, 2x − ξ )| + |ϕ3 (ξ, 2x − ξ )|) dξ ≤ |(A ϕ)2 − ϕ20 | ≤ ≤

$x 0

≤

+ 2a0 (T )) ϕ := a1 (T ) ϕ ,

|ϕ1 (ξ )|(|h(x + t − 2ξ )| + |ϕ4 (ξ, x + t − ξ ))| dξ

T 2 ( h C[0,T ]

|(A ϕ)3 − ϕ30 | ≤

|(A ϕ)4 − ϕ40 | ≤

T  |h(0)| ( h C[0,T ]

$x 0

+ 2a0(T )) ϕ := a2 (T ) ϕ ,

|ϕ1 (ξ )|(|h(t − x)| + |ϕ4 (ξ, t − x + ξ )|) dξ

T 2 ( h C[0,T ]

+ 2a0(T )) ϕ := a3 (T ) ϕ ,

$ 1 t 2 x (|ϕ2 (x, τ )| + |ϕ3 (x, τ )|) dτ

≤ T ϕ := a4 (T ) ϕ .

Therefore Aϕ ∈ BT , if the following condition holds: max ak (T ) ≤ 1.

k=1,2,3,4

(4.2.34)

144

4 Inverse Problems for Hyperbolic Equations

We verify the second condition (c2). Let ϕ k := (ϕ1k , ϕ2k , ϕ3k , ϕ4k ) and ϕ k ∈ BT , k = 1, 2. Then one has  x 1 |ϕ11 (ξ ) − ϕ12 (ξ )|(2|h (2x − 2ξ )| |(A ϕ 1 − A ϕ 2 )1 | ≤ |h(0)| 0 +|ϕ21 (ξ, 2x − ξ )| + |ϕ31 (ξ, 2x − ξ )|) + |ϕ12 (ξ )|(|ϕ21 (ξ, 2x − ξ ) − ϕ22 (ξ, 2x − ξ )|  +|ϕ31 (ξ, 2x − ξ ) − ϕ32 (ξ, 2x − ξ )|) dξ ≤

T ( h C[0,T ] + 4a0(T )) ϕ 1 − ϕ 2 := b1 (T ) ϕ 1 − ϕ 2 , |h(0)|

Similarly, 

x

|(A ϕ 1 − A ϕ 2 )2 | ≤ 0

|ϕ11 (ξ ) − ϕ12 (ξ )|(|h(x + t − 2ξ )| + |ϕ41 (ξ, x + t − ξ )|)  +|ϕ12 (ξ )||ϕ41 (ξ, x + t − ξ ) − ϕ42 (ξ, x + t − ξ )| dξ

T ( h C[0,T ] + 4a0 (T )) ϕ 1 − ϕ 2 := b2 (T ) ϕ 1 − ϕ 2 , 2  x |ϕ11 (ξ ) − ϕ12 (ξ )|(|h(t − x)| + |ϕ41 (ξ, t − x − ξ )|) |(A ϕ 1 − A ϕ 2)3 | ≤ ≤

0

 +|ϕ12 (ξ )||ϕ41 (ξ, t − x − ξ ) − ϕ42 (ξ, t − x − ξ )| dξ

T ( h C[0,T ] + 4a0 (T )) ϕ 1 − ϕ 2 := b3 (T ) ϕ 1 − ϕ 2 , 2  1 t 1 1 2 |(A ϕ − A ϕ )4 | ≤ (|ϕ2 (x, τ ) − ϕ22 (x, τ )| + |ϕ31 (x, τ ) − ϕ32 (x, τ )|) dτ 2 x ≤

≤ T ϕ 1 − ϕ 2 := b4 (T ) ϕ 1 − ϕ 2 . Hence, A ϕ 1 − A ϕ 2 ≤ ρ ϕ 1 − ϕ 2 with ρ < 1, if T satisfies the conditions max bk (T ) ≤ ρ < 1.

k=1,2,3,4

(4.2.35)

Thus, if the final time T > 0 is chosen so (small) that both conditions (4.2.34) and (4.2.35) hold, then the operator A is contracting on BT . Then, according to Banach Contraction Mapping Principle, there exists a unique solution of the operator equation (4.2.30) in BT . This completes the proof of the theorem.  

4.2 Problem of Recovering the Potential in String Equation

145

4.2.3 Global Stability and Uniqueness Now we state a stability estimate and a uniqueness theorem when the final time T is an arbitrary fixed positive number. Denote by Q(q0 ) the set of functions q ∈ C[0, T /2], satisfying the inequality |q(x)| ≤ q0 , for x ∈ [0, T /2], with the positive constant q0 . Let H(h0 , h1 , d) be the set of functions h(t) satisfying for some fix positive constants h0 , h1 and d the following conditions: 1) h ∈ C 2 [0, T ], 2) h(t) C[0,T ] ≤ h0 , h (t) C[0,T ] ≤ h1 , 3) |h(0)| ≥ d > 0. Evidently, for q ∈ Q(q0 ) and h ∈ H(h0 , h1 , d), estimates (4.2.18) remain true in T for the solution of the system of integral equations (4.2.11)–(4.2.13). Theorem 4.2.2 Let qk ∈ Q(q0 ) be a solution of the inverse problem (4.2.1), (4.2.2) corresponding to the data hk ∈ H(h0 , h1 , d), fk ∈ C 1 [0, T ], for each k = 1, 2. Then there exists a positive number C = C(q0 , h0 , h1 , d, T ) such that the following stability estimate holds:  

q1 − q2 C[0,T /2] ≤ C f1 − f2 C 1 [0,T ] + h1 − h2 C 2 [0,T ] .

(4.2.36)

Proof To prove this theorem we use Lemma 4.2.1. For this aim, similar to representations (4.2.4) and (4.2.6), we represent the solution of the direct problem (4.2.1) corresponding to qk (x) ∈ Q(q0 ) and hk (t) ∈ H(h0 , h1 , d), k = 1, 2, in the following form: uk (x, t) = [hk (t − x) + v k (x, t)]θ0 (t − x), k = 1, 2.

(4.2.37)

Then the functions v1k = vtk + vxk , v2k = vtk − vxk , v k , qk , k = 1, 2, satisfy in T := {(x, t)| 0 ≤ x ≤ t ≤ T − x} the integral equations: v1k (x, t) = fk (t + x) + hk (t + x) $x − 0 qk (ξ )(hk (x + t − 2ξ ) + v k (ξ, x + t − ξ )) dξ, v2k $(x, t) = −fk (t − x) − hk (t − x) x + 0 qk (ξ )(hk (t − x) + v k (ξ, t − x + ξ )) dξ, $t v k (x, t) = 12 x (v1k (x, τ ) + v2k (x, τ )) dτ, $x qk (x)hk (0) + 0 qk (ξ )(2hk (2x − 2ξ ) + v1k (ξ, 2x − ξ ) +v2k (ξ, 2x − ξ )) dξ = 2(fk (2x) + hk (2x)).

146

4 Inverse Problems for Hyperbolic Equations

Introduce the differences: v˜1 = v11 − v12 , q˜ = q1 − q2 ,

v˜2 = v21 − v22 , f˜ = f1 − f2 ,

v˜ = v1 − v2 , h˜ = h1 − h2 .

Then the above equations imply that the following relations hold in T : v˜1 (x, t) = f˜(t + x) + h˜  (t + x) $x ˜ )(h1 (x + t − 2ξ ) + v 1 (ξ, x + t − ξ )) − 0 q(ξ  ˜ + t − 2ξ ) + v(ξ, +q2 (ξ )(h(x ˜ x + t − ξ )) dξ, v˜2 (x, t) = −f˜(t − x) − h˜  (t − x) $x + 0 q(ξ ˜ )(h1 (t − x) + v 1 (ξ, t − x + ξ ))  ˜ − x) + v(ξ, +q2 (ξ )(h(t ˜ t − x + ξ )) dξ, $t v(x, ˜ t) = 12 x (v˜1 (x, τ ) + v˜2 (x, τ )) dτ, $x ˜ ˜ )(2h1 (2x − 2ξ ) q(x)h ˜ + 0 q(ξ 1 (0) + q2 (x)h(0) +v11 (ξ, 2x − ξ ) + v21 (ξ, 2x − ξ )) + q2 (ξ )(2h˜  (2x − 2ξ )  +v˜1 (ξ, 2x − ξ ) + v˜2 (ξ, 2x − ξ )) dξ = 2(f˜ (2x) + h˜  (2x)). Let  ψ(x) = max |q(x)|; ˜

max

x≤t ≤T −x

|v˜1 (x, t)|;

max

x≤t ≤T −x

|v˜2 (x, t)|;

max

x≤t ≤T −x

|v(x, ˜ t)| .

Then using estimates (4.2.18) for v11 , v21 , v 1 we conclude: ˜ C1 [0,T ] v˜1 (x, t)| ≤ f˜ C[0,T ] + h

 $x ˜ C[0,T ] + ψ(ξ )) dξ, ψ(ξ )(h0 + v0 ) + q0 ( h

+ 0

˜ C1 [0,T ] |v˜2 (x, t)| ≤ f˜ C[0,T ] + h

 $x ˜ C[0,T ] + ψ(ξ )) dξ, ψ(ξ )(h0 + v0 ) + q0 ( h

+ |v(x, ˜ t)| ≤

0 T 2 (maxx≤τ ≤t

|v˜1 (x, τ )| + maxx≤τ ≤t |v˜2 (x, τ )|)

˜ C1 [0,T ] ≤ T f˜ C[0,T ] + h

  $x ˜ C[0,T ] + ψ(ξ )) dξ , ψ(ξ )(h0 + v0 ) + q0 ( h

+ 0

4.3 Inverse Coefficient Problems for Layered Media

147

$x ˜ q0 |h(0)| + ψ(ξ )(2h1 + v01 + v02 ) 0

 ˜ C1 [0,T ] + ψ(ξ ) dξ + 2( f˜ C1 [0,T ] + h

˜ C2 [0,T ] ) . +q0 2 h

|q(x)| ˜ ≤

1 d



It follows from these estimates that ψ(x) satisfies the following integral inequality: 

 ˜ C 2 [0,T ] + C2 ψ(x) ≤ C1 f˜ C 1 [0,T ] + h|

x

ψ(ξ ) dξ, 0

for all x ∈ [0, T /2], with the constants C1 = max (1 + q0 T /2; (2 + 2q0 )/d) max(1; T ), C2 = max((h0 + v0 + q0 ) max(1; T ); (2h1 + v01 + v02 + q0 )/d). Using Gronwall’s inequality, one gets the estimate

 ˜ C 2 [0,T ] exp(C2 x), x ∈ [0, T /2]. |ψ(x)| ≤ C1 f˜ C 1 [0,T ] + h|

(4.2.38)

The required stability estimate (4.2.36) follows from (4.2.38) with C C1 exp(C2 T /2). This completes the proof.

=  

We derive now one of the most important application of this theorem. The result below shows that the uniqueness, which valid for any T > 0, of the solution of inverse problem (4.2.1)–(4.2.2) is a consequence of the above proved stability theorem. Theorem 4.2.3 Let the input h(t) in the direct problem (4.2.1) satisfies conditions (4.2.3), that is, h ∈ C 2 [0, T ] and h(0) = 0. Assume that q1 , q2 ∈ C[0, T /2] are two arbitrary solutions of the inverse problem (4.2.1)–(4.2.2). Then q1 (x) = q2 (x) for all x ∈ [0, T /2].

4.3 Inverse Coefficient Problems for Layered Media Consider the following initial-boundary value problem for the acoustic equation: 

ut t − ρ −1 divx (ρ c2 ∇x u) = 0, u|t =0 = 0,

ut |t =0 = 0,

(x, t) ∈ Rm + × R,

u|x1 =0 = g(x  , t),

t ∈ R,

(4.3.1)

m where x = (x1 , x2 . . . , xm ) := (x1 , x  ) and Rm + := {x ∈ R | x1 > 0}. The coefficients c = c(x) > 0 and ρ = ρ(x) > 0 in the acoustic equation (4.3.1) are the sound speed and the density of a medium, correspondingly. We assume here

148

4 Inverse Problems for Hyperbolic Equations

that the medium is layered, that is, the coefficients depend only on the variable x1 ∈ R: c = c(x1) and ρ = ρ(x1 ). For a given admissible coefficients c = c(x1) and ρ = ρ(x1 ) the problem of finding a function u(x, t) satisfying initial-boundary value problem (4.3.1), which is defined as a direct problem, is a well-posed problem. The inverse coefficient problem here consists of the determination of the unknown coefficients c(x1 ) and ρ(x1 ) based on the following information on the solution to problem (4.3.1) on the hyperplane x1 = 0 for all x  ∈ Rm−1 and t ∈ (0, T ], T > 0: F (x  , t) := ρ(0) c2 (0)

∂u

,

∂x1 x1 =0

x  ∈ Rm−1 ,

t ∈ (0, T ].

(4.3.2)

Physically, the condition (4.3.2) means that the acoustic pressure is given at x1 = 0. Let us show that if the acoustic rigidity ρ(x1 )c(x1 ) is known at x1 = 0, i.e. if ρ(0)c(0) is given, then the inverse problem defined by (4.3.1)–(4.3.2) can be reduced to the previously considered one. Indeed, consider the Fourier transform ϕ(λ ˜  ) of a function ϕ(x  ), x  ∈ Rm−1 ,  with respect to x , i.e., ϕ(λ ˜ ) =

 Rm−1

ϕ(x  ) exp{i(x  · λ )} dx  .

Here λ := (λ2 , . . . , λm ) is the parameter of the transform and the symbol (x  · λ ) means the scalar product of vectors x  and λ . Applying the Fourier transform to the direct problem (4.3.1) as well as to the additional condition (4.3.2), we arrive at the following transformed inverse problem: 

u˜ t t − ρ −1 (ρ c2 u˜ x1 )x1 + c2 |λ |2 u˜ = 0, u| ˜ t =0 = 0,

ρ(0)c2(0)

u˜ t |t 0, μ > 0 and σ are the permittivity, permeability and conductivity coefficients, respectively, which define electro-dynamical parameters of a medium. The function j = j (x, t) is the external current source which generates the electromagnetic waves. We consider the simplest physical model of a medium, assuming that the space R3 is divided in the two half-spaces R3− =: {x ∈ R3 | x3 < 0} and R3+ =: {x ∈ R3 | x3 > 0}. Namely, in R3− the electro-dynamical parameters of the medium are assumed to be known and are constants ε = ε− > 0, μ = μ− > 0 and σ = 0. The inverse coefficient problem consists in the determination of unknown parameters in R3+ , as functions of x, from observations of the electromagnetic field on the interface S =: {x ∈ R3 | x3 = 0}. Generally speaking, the interface S is the discontinuity boundary for these parameters. For this reason we need to introduce special notations for limiting values of the parameters ε, μ on S taking from R3+ , namely, denote by ε+ = ε|x3 =0+ , μ+ = μ|x3 =0+ . Assume that the electromagnetic field vanishes until the moment t = 0, i.e., (E, H )t 0.

By the reasons that will be clearly below, we represent these conditions in the form  + − + u− 1 = r11 u1 + r12 u2 − r11 μ j (t),  + − − u+ 2 = −r12 u1 + r22 u2 − r22 μ j (t),

t > 0,

(5.2.17)

where rij given by the formulae  2 ε − μ−  , + + ε + μ− ε− μ 2 ε + μ+  =  . ε − μ+ + ε + μ−

r11 =  r22

  ε − μ+ − ε + μ−  r12 =  , ε − μ+ + ε + μ−

(5.2.18)

156

5 One-dimensional Inverse Problems in Electrodynamics

The information for the inverse problem is now given by the formula + u+ 1 + u2 = 2

√ ε+ f (t),

t > 0.

(5.2.19)

Consider the direct problem defined by (5.2.14)–(5.2.17). Since the initial data are zero, one has u1 (z, t) = u2 (z, t) ≡ 0,

0 < t < |z|.

(5.2.20)

For arbitrary T > 0 introduce the notations − T = {(z, t)| 0 < −z ≤ t ≤ T + z} and + = {(z, t)| 0 < z ≤ t ≤ T − z}. In the domain − T T we have Eqs. (5.2.14) with the matrix A = 0 and the boundary conditions (5.2.17). The second Eq. (5.2.14) has the form ∂u2 ∂u2 + = 0, ∂t ∂z

u2 |t =0 = 0,

z < 0.

− This implies that u2 (z, t) ≡ 0 for all (z, t) ∈ − T . Hence u2 = 0. Then the second boundary condition (5.2.17) takes the form

 + − u+ 2 = −r12 u1 − r22 μ j (t),

t > 0.

(5.2.21)

It means that we can consider the direct problem in the domain + T with the boundary condition (5.2.21) and find u1 (z, t) and u2 (z, t) and then calculate u− 1 (0, t) and u1 (z, t) in − using the Eq. (5.2.14) and the first boundary condition (5.2.17). The T latter is given by the formula  u1 (z, t) = r11 u+ (0, t + z) − r μ+ j (t + z), 11 1

(z, t) ∈ − T.

(5.2.22)

Write down integral equations for u1 and u2 in the domain + T . Integrating the first component of the Eq. (5.2.14) on the plane (ζ, τ ) along the characteristic line ζ + τ = z + t from the point (z + t)/2, z + t)/2) till the point (z, t) and using that u1 (z + t)/2, z + t)/2) = 0, we obtain 

t

u1 (z, t) +

(A(z + t − τ )U (z + t − τ, τ ))1 dτ = 0,

(z+t )/2

(z, t) ∈ + T.

(5.2.23)

Now we can rewrite the condition (5.2.21) in the form u+ 2 (0, t) = r12



t t /2

 (A(t − τ )U (t − τ, τ ))1 dτ − r22 μ− j (t).

(5.2.24)

5.2 The Direct Problem: Existence and Uniqueness of a Solution

157

Integrating the second component of (5.2.14) along the characteristic line ζ − τ = z − t from the point (0, t − z) till the point (z, t) and using the boundary condition (5.2.24), we get  u2 (z, t) +

t t −z

(A(z − t + τ )U (z − t + τ, τ ))2 dτ



−r12

t −z

(t −z)/2

(A(t − z − τ )U (t − z − τ, τ ))1 dτ

 +r22 μ− j (t − z) = 0,

(5.2.25)

(z, t) ∈ + T.

Relations (5.2.23) and (5.2.25) form the closed system of integral equations in domain + T . The properties of the solution to this system are given by the following existence and uniqueness theorem. Theorem 5.2.1 Let εˆ (z) ∈ C 1 [0, T /2], μˆ ∈ C 1 [0, T /2], σˆ (z) ∈ C[0, T /2] and j (t) ∈ C 1 [0, T ], T > 0. Assume, in addition, that the following bounds hold:

1 1  , maxz∈(0,T /2] ≤ q0 , εˆ (z) μ(z) ˆ  r22 μ− j C(0,T ] ≤ j0 , max( ˆε C(0,T /2] , μˆ  C(0,T /2] , σˆ C(0,T /2]) ≤ q01 , with some constants q0 , q01 and j0 . Then there exist a unique continuously differentiable solution to Eqs. (5.2.23) and (5.2.25) in the closed domain + T and this solution satisfies the conditions u1 (z, z) = 0,

$   z u2 (z, z) = −r22 μ− j (0) exp 0 (q1 (ζ ) − q3 (ζ )) dζ , z ∈ [0, T /2],

(5.2.26)

max( u1 C( + ) , u2 C( + ) ) ≤ Cj0 , T

T

where C = C(q0 , q01 , T ). Proof To prove the existence of a continuous solution in + T one can use the method of successive approximation. For this goal, functions u1 and u2 are represented in the form of series uk (z, t) =

∞  n=0

unk (z, t),

k = 1, 2,

(5.2.27)

158

5 One-dimensional Inverse Problems in Electrodynamics

where un1 and un2 are defined by the formulae  u01 (z, t) = 0, u02 (z, t) = −r22 μ− j (t − z), U n = (un1 , un2 )T ,  un1 (z, t)

t

=−

un2 (z, t) = −

(A(z + t − τ )U n−1 (z + t − τ, τ ))1 dτ,

n ≥ 1,

(z+t )/2  t t −z



+r12

(A(z + t − τ )U n−1 (z + t − τ, τ ))2 dτ

t −z (t −z)/2

(A(t − z − τ )U n−1 (t − z − τ, τ ))1 dτ,

n ≥ 1.

Functions un1 (z, t), un2 (z, t) are continuous in + T . Introduce v n (t) =

⎧ ⎨ max max |unk (z, t)|,

t ∈ [0, T /2],

k=1,2 0≤z≤t

⎩ max

max |unk (z, t)|, t ∈ [T /2, T ],

k=1,2 0≤z≤T −t

for n = 0, 1, 2, . . .. Note that |r12 | < 1. Then one has v 0 (t)

≤ j0 , 

|un1 (z, t)|

≤ 2q0 q01

|un2 (z, t)| ≤ 2q0 q01



t

v

n−1

(z+t )/2

 t n−1

v

t −z t n−1

 ≤ 2q0 q01

v

t

(τ ) dτ ≤ 2q0q1 

(τ ) dτ +

(τ ) dτ,

v n−1 (τ ) dτ,

0 t −z

(t −z)/2

v n−1 (τ ) dτ



n ≥ 1.

0

Hence, 

t

v n (t) ≤ 2q0 q01

v n−1 (τ ) dτ,

t ∈ [0, T ],

n ≥ 1.

(5.2.28)

0

From (5.2.28) we deduce the estimates v n (t) ≤ j0 (2q0q01 )n

tn Tn ≤ j0 (2q0q01 )n , t ∈ [0, T ], n = 0, 1, 2, . . . . n! n!

(5.2.29)

5.2 The Direct Problem: Existence and Uniqueness of a Solution

159

Therefore the series (5.2.27) uniformly convergence in + T and their sums are

continuous functions in + T . This means that there exists the solution to the integral equations (5.2.23) and (5.2.25) and the solution satisfies the inequalities |uk (z, t)| ≤ j0

∞  Tn ≤ j0 exp(2q0 q01 T ), k = 1, 2, (2q0 q01 )n n!

(5.2.30)

n=0

for all (z, t) ∈ + T . Moreover, substituting in (5.2.23) and (5.2.25) t = z, we obtain u1 (z, z) = 0 and the integral relation for the function ψ(z) = u2 (z, z) in the form 

z

ψ(z) +

 (q3 (ζ ) − q1 (ζ ))ψ(ζ ) dζ = −r22 μ− j (0).

0

The latter equation is equivalent to the Cauchy problem for the ordinary differential equation ψ  (z) + (q3 (z) − q1 (z))ψ(z) = 0,

 ψ(0) = −r22 μ− j (0).

The solution of this problem is given by the formula 

z

ψ(z) = ψ(0) exp

(q1 (ζ ) − q3 (ζ )) dζ ,

0

which coincides with the second formula (5.2.26). To prove the uniqueness of the found solution, we suppose that there exist two solutions of the integral equations (5.2.23) and (5.2.25), namely, u1 , u2 and u¯ 1 , u¯ 2 . Then their difference u˜ 1 = u1 − u¯ 1 , u˜ 2 = u2 − u¯ 2 satisfies the following relations for all (z, t) ∈ + T: 

t

u˜ 1 (z, t) + u˜ 2 (z, t) +

(A(z + t − τ )U˜ (z + t − τ, τ ))1 dτ = 0,

(z+t )/2  t t −z



−r12

(A(z + t − τ )U˜ (z + t − τ, τ ))2 dτ t −z

(t −z)/2

(A(t − z − τ )U˜ (t − z − τ, τ ))1 dτ = 0,

(5.2.31)

160

5 One-dimensional Inverse Problems in Electrodynamics

where U˜ := (u˜ 1 , u˜ 2 )T . Let now v(t) =

⎧ ⎨ max max |u˜ k (z, t)|, k=1,2 0≤z≤t

⎩ max

t ∈ [0, T /2],

max |u˜ k (z, t)|, t ∈ [T /2, T ].

k=1,2 0≤z≤T −t

Then we get 

t

|u˜ 1 (z, t)| ≤ 2q0q01

v(τ ) dτ, 

0 t

|u˜ 2 (z, t)| ≤ 2q0q01 0

v(τ ) dτ, (z, t) ∈ + T.

Hence, 

t

v(t) ≤ 2q0 q01

v(τ ) dτ,

t ∈ [0, T ].

(5.2.32)

0

Inequality (5.2.32) has only trivial solution v(t) ≡ 0. Therefore u1 (z, t) = u¯ 1 (z, t) and u2 (z, t) = u¯ 2 (z, t), i.e., the solution to (5.2.23)–(5.2.25) is unique. Now we need prove that the solution is continuously differentiable in + T . If the coefficients satisfy the conditions εˆ (z) ∈ C 2 [0, T /2], μˆ ∈ C 2 [0, T /2], σˆ (z) ∈ C 1 [0, T /2], then one can directly prove this assertion taking derivatives of integral relations (5.2.23) and (5.2.25) with respect to z and t. In order to avoid the additional differentiation of the coefficients, we need first to rewrite these relations changing the variable τ under the integrands on ζ . Making this, we get  u1 (z, t) − u2 (z, t) +

z

(A(ζ )U (ζ, z + t − ζ ))1 dζ = 0,

(z+t )/2  z

(A(ζ )U (ζ, t − z + ζ ))2 dζ

0



(t −z)/2

−r12 0

(A(ζ )U (ζ, t − z − ζ ))1 dζ

 +r22 μ− j (t − z) = 0, (z, t) ∈ + T.

(5.2.33)

5.2 The Direct Problem: Existence and Uniqueness of a Solution

161

Then we differentiate equations (5.2.33) to obtain integral relations for partial derivatives u˙ 1 , u˙ 2 with respect to t of u1 and u2 . These equations are as follows: 

z

u˙ 1 (z, t) −

˙ (A(ζ )U(ζ, z + t − ζ ))1 dζ

(z+t )/2

1 + (A((z + t)/2)U ((z + t)/2, (z + t)/2))1 = 0, 2  z ˙ (A(ζ )U(ζ, t − z + ζ ))2 dζ u˙ 2 (z, t) + 0



(t −z)/2

−r12

(5.2.34)

˙ (A(ζ )U(ζ, t − z − ζ ))1 dζ

0

r12 (A((t − z)/2)U ((t − z)/2, (t − z)/2))1 2  +r22 μ− j  (t − z) = 0, (z, t) ∈ + T, −

where U˙ is the vector-column with components u˙ 1 , u˙ 2 . Now one needs again to change variable in the integrands returning back to τ . In the result the equations take the form  t u˙ 1 (z, t) + (A(z + t − τ )U˙ (z + t − τ, τ ))1 dτ (z+t )/2

1 + (A((z + t)/2)U ((z + t)/2, (z + t)/2))1 = 0, 2  t u˙ 2 (z, t) + (A(z − t + τ )U˙ (z − t + τ, τ ))2 dτ t −z



−r12

t −z

(t −z)/2

(5.2.35)

(A((t − z − τ )U˙ ((t − z − τ, τ ))1 dτ

r12 (A((t − z)/2)U ((t − z)/2, (t − z)/2))1 2  +r22 μ− j  (t − z) = 0, (z, t) ∈ + T. −

In these equations the function U (z, z) is known. The Eqs. (5.2.35) quite similar to the Eqs. (5.2.23) and (5.2.25). Therefore to prove the existence and uniqueness of a continuous solution in + T of these equations the previous methods can be used. We leave the proof of these assertions as an exercise for the reader. Since the derivatives u˙ 1 (z, t), u˙ 2 (z, t) exist and are continuous in + T , then directly for the differential equations (5.2.14) follows an existence and continuous dependence of the derivatives of u1 (z, t), u2 (z, t) with respect to z in the same domain.  

162

5 One-dimensional Inverse Problems in Electrodynamics

The following theorem is a consequence of Theorem 5.2.1. Theorem 5.2.2 Let conditions of Theorem 2.2.1 hold. Assume, in addition, that additional condition j (0) = 0 holds. Then the function f (t) in (5.2.19) belongs to the functional space C 1 [0, T ] and satisfy the condition 

μ− f (0+ ) < 0. − √ < j (0) ε−

(5.2.36)

Proof One needs prove only inequality (5.2.36). It follows directly from equations (5.2.19) and (5.2.26) that  r22 μ− f (0 ) = − √ j (0). 2 ε+ +

(5.2.37)

Then one finds  r22 μ− f (0+ ) =− √ < 0. j (0) 2 ε+ Using the formula (5.2.18) for r22 , one obtains the estimate  r22 μ+ 1 =  < √ .  √ 2 ε+ ε − μ+ + ε + μ− ε−  

This implies the inequality (5.2.36).

5.3 One-dimensional Inverse Problems Consider the inverse problem. We assume that h(t) ∈ C 1 [0, T ], j (0) = 0, and f (t) ∈ C 1 [0, T ] is a given function which satisfies to the condition (5.2.36). Since only the f (t) function is a given data, i.e. information, one can find at most one unknown parameters ε, μ or σ . In this context we consider two typical inverse problems.

5.3.1 Problem of Finding the Permittivity Coefficient In this case we should accept that μ and σ are given. We assume for simplicity that μ(x3 ) = μ+ > 0 and σ (x3 ) = 0 for x3 > 0. In this case q1 (z) = q2 (z) =

εˆ  (z) := q(z), 4ˆε(z)

q3 (z) = 0.

(5.3.1)

5.3 One-dimensional Inverse Problems

163

Using relations (5.2.36) and (5.2.18) one can find ε+ . Elementary calculations lead to the formula √ 

j (0) ε− 2 +  . (5.3.2) ε + = μ+ f (0+ ) μ− Note that the inequality (5.2.36) guaranties that ε+ is positive. Because ε+ has found, the value r12 and r22 become known. In the inverse problem both functions h(t) and f (t) are given. Therefore one can + − calculate u+ 1 and u2 using the conditions (5.2.17) and (5.2.19) (recall that u2 = 0 in (5.2.17)). Indeed, then the equations hold  + − u+ 2 = −r12 u1 − r22 μ j (t), √ + + u+ t > 0. 1 + u2 = 2 ε f (t), Solving these equations we deduce:  √ 1  − r22 μ j (t) + 2 ε+ f (t) := F1 (t), 1 − r12  √ 1  − + f (t) := F (t). = − r μ j (t) + 2r ε u+ 22 12 2 2 1 − r12

u+ 1 =

(5.3.3)

Note that, according (5.2.18), 1 − r12

 2 ε + μ− =  > 0.  ε − μ+ + ε + μ−

+ + Because u+ 1 and u2 are known, one can consider Eqs. (5.2.14) in the domain T only. These equations have the form

⎧ ∂u1 ∂u1 ⎪ ⎪ ⎨ ∂t − ∂z + q(z)(u1 + u2 ) = 0, ⎪ ∂u2 ∂u2 ⎪ ⎩ + − q(z)(u1 + u2 ) = 0, (z, t) ∈ + T. ∂t ∂z

(5.3.4)

Moreover, the following conditions hold   u1 (z, z) = 0, u2 (z, z) = −r22 μ− j (0) exp 0



z

q(ζ ) dζ

(5.3.5)

164

5 One-dimensional Inverse Problems in Electrodynamics

for z ∈ [0, T /2]. The relations (5.3.3)–(5.3.5) form the complete system of equations for the inverse problem. Deduce integral equations for this problem. At first, integrating equations (5.3.4) along the characteristic lines ζ + τ = z + t and ζ − τ = z − t from arbitrary point (z, t) ∈ + T till to intersection with axis ζ = 0 and using at the intersection points conditions (5.3.3), one gets 

z

u1 (z, t) − 

q(ζ )(u1 (ζ, z + t − ζ ) + u2 (ζ, z + t − ζ )) dζ = F1 (t + z),

0 z

u2 (z, t) −

q(ζ )(u1 (ζ, t − z + ζ ) + u2 (ζ, t − z + ζ )) dζ = F2 (t − z),

0

(z, t) ∈ + T . (5.3.6) At second, substituting in the first of these equations t = z and using condition (5.3.5), one finds the relation 

z

−

q(ζ )(u1(ζ, 2z − ζ ) + u2 (ζ, 2z − ζ )) dζ = F1 (2z),

0

that is, an additional equation for finding q(z). Differentiating it with respect to z, one gets  − q(z)(u1 (z, z) + u2 (z, z)) − 2

z

q(ζ )(u˙ 1 (ζ, 2z − ζ )

0

+ u˙ 2 (ζ, 2z − ζ )) dζ = 2F1 (2z),

(5.3.7)

Use here the relations (5.3.5). Then the first term in this equation one can rewrite as follows

 z   q(ζ ) dζ . −q(z)(u1 (z, z) + u2 (z, z)) = r22 μ− j (0)q(z) exp 0

Denote p(z) = q(z) exp



z 0

 z   d exp q(ζ ) dζ = q(ζ ) dζ . dz 0

Then



z

exp 0

  q(ζ ) dζ = 1 +

z

p(ζ ) dζ > 0, 0

(5.3.8)

5.3 One-dimensional Inverse Problems

165

and q(z) =

1+

p(z) $z . 0 p(ζ ) dζ

(5.3.9)

 Dividing both sides of (5.3.7) by r22 μ− j (0), one gets 

z

p(z) − λ0

q(ζ )(u˙ 1(ζ, 2z − ζ ) + u˙ 2 (ζ, 2z − ζ )) dζ = p0 (z),

0

z ∈ [0, T /2], (5.3.10) where λ0 =

2  , r22 μ− j (0)

p0 (z) = λ0 F1 (2z)

and q(z) is defined by the formula (5.3.9). Differentiating (5.3.6) with respect to t, one obtain equations for functions u˙ 1 and u˙ 2 . They have the form: 

z

u˙ 1 (z, t) − 

0 z

u˙ 2 (z, t) − 0

q(ζ )(u˙ 1(ζ, z + t − ζ ) + u˙ 2 (ζ, z + t − ζ )) dζ = F1 (t + z), q(ζ )(u˙ 1(ζ, t − z + ζ ) + u˙ 2 (ζ, t − z + ζ )) dζ = F2 (t − z), (z, t) ∈ + T . (5.3.11)

Now we can prove the existence and uniqueness of the local solution of the inverse problem. Theorem 5.3.1 Let μ(x3 ) = μ+ > 0, σ (x3 ) = 0, for x3 > 0, and j (t) ∈ C 1 [0, T ], j (0) = 0. Assume that f (t) ∈ C 1 [0, T ], T > 0, and the condition (5.2.36) holds. Then for sufficiently small T > 0 there exist a unique continuously differentiable positive solution of the inverse problem. Proof First, we represent the Eqs. (5.3.10) and (5.3.11) in the operator form ϕ = Aϕ,

(5.3.12)

where ϕ = (ϕ1 , ϕ2 , ϕ3 ) is the vector-function with components defined by the formulae ϕ1 = u˙ 1 (z, t),

ϕ2 = u˙ 2 (z, t),

ϕ3 = p(z)

166

5 One-dimensional Inverse Problems in Electrodynamics

and the operator A = (A1 , A2 , A3 ) is given by the relations 

z

ϕ1 (z, t) = 0



z

ϕ2 (z, t) = 0

 ϕ3 (z) = λ0

z

0

q(ζ )(ϕ1(ζ, z + t − ζ ) + ϕ2 (ζ, z + t − ζ )) dζ + ϕ10 (z, t), q(ζ )(ϕ1(ζ, z + t − ζ ) + ϕ2 (ζ, z + t − ζ )) dζ + ϕ20 (z, t), q(ζ )(ϕ1(ζ, 2z − ζ ) + ϕ2 (ζ, 2z − ζ )) dζ + ϕ30 (z),

(5.3.13)

where (z, t) ∈ + T, ϕ10 (z, t) = F1 (t + z),

ϕ20 (z, t) = F2 (t − z),

ϕ30 (z) = p0 (z)

and q(z) is defined by the formula q(z) =

1+

ϕ3 (z) $z . 0 ϕ3 (ζ ) dζ

(5.3.14)

Then we define the ball B(ϕ 0 )

ϕ − ϕ 0 ≤ ϕ 0 , centered at the element ϕ 0 = (ϕ10 , ϕ20 , ϕ30 ), in the functional space of continuous

functions ϕ in + T with the norm

ϕ = max

max |ϕk (z, t)|.

k=1,2,3 (z,t )∈ + T

Now we are going to prove that that the operator A is a contracting operator on this ball, if T is enough small. Let ϕ be arbitrary element of B(ϕ 0 ). Then ϕ ≤ 2 ϕ 0 . Making estimates one finds  z 0 |q(ζ )|(|ϕ1(ζ, z + t − ζ )| + |ϕ2 (ζ, z + t − ζ )|) dζ |ϕ1 (z, t) − ϕ1 (z, t)| ≤ 0

≤

4T ϕ0

ϕ0 , 1 − T ϕ0

(z.t) ∈ + T.

We used here that |q(ζ )| ≤

2 ϕ0

. 1 − T ϕ0

5.3 One-dimensional Inverse Problems

167

Similarly,  |ϕ2 (z, t) − ϕ20 (z, t)| ≤

z

|q(ζ )|(|ϕ1(ζ, z + t − ζ )| + |ϕ2 (ζ, z + t − ζ )|) dζ

0

≤

4T ϕ0

ϕ0 , 1 − T ϕ0



|ϕ3 (z)

− ϕ30 (z)|

≤ = |λ0 |

z

(z.t) ∈ + T,

|q(ζ )|(|ϕ1(ζ, 2z − ζ )| + |ϕ2 (ζ, 2z − ζ )|) dζ

0

≤

4T |λ0 | ϕ0

ϕ0 , 1 − T ϕ0

z ∈ T /2.

From these estimates follows that Aϕ − ϕ 0 ≤ ϕ 0 if T satisfy the relations 4T max(1, |λ0 |)

ϕ 0

< 1, 1 − T ϕ 0

T ϕ 0 < 1.

(5.3.15)

This means that Aϕ ∈ B(ϕ 0 ), i.e., the operator A maps the ball B(ϕ 0 ) into itself. Take now two arbitrary elements ϕ j , j = 1, 2, belonging to B(ϕ 0 ) and denote ϕ˜ k = ϕk1 − ϕk2 , k = 1, 2, 3, j

q j (z) =

ϕ3 (z) , j = 1, 2, $z j 1 + 0 ϕ3 (ζ ) dζ

 $z $z ϕ˜ 3 (z) 1 + 0 ϕ32 (ζ ) dζ − ϕ32 (z) 0 ϕ˜ 3 (ζ ) dζ

  q(z) ˜ = . $z $z 1 + 0 ϕ31 (ζ ) dζ 1 + 0 ϕ32 (ζ ) dζ From (5.3.13) we find 

z

ϕ˜1 (z, t) =

[q 1(ζ )(ϕ˜ 1 (ζ, z + t − ζ ) + ϕ˜2 (ζ, z + t − ζ ))

0

+q(ζ ˜ )(ϕ12 (ζ, z + t − ζ ) + ϕ22 (ζ, z + t − ζ ))] dζ, 

z

ϕ˜2 (z, t) =

[q 1(ζ )(ϕ˜ 1 (ζ, z + t − ζ ) + ϕ˜2 (ζ, z + t − ζ ))

0

+q(ζ ˜ )(ϕ12 (ζ, z + t − ζ ) + ϕ22 (ζ, z + t − ζ ))] dζ,

168

5 One-dimensional Inverse Problems in Electrodynamics



z

ϕ˜3 (z) = λ0

[q 1(ζ )(ϕ˜ 1 (ζ, 2z − ζ ) + ϕ˜2 (ζ, 2z − ζ ))

0

+q(ζ ˜ )(ϕ12 (ζ, 2z − ζ ) + ϕ22 (ζ, 2z − ζ ))] dζ. Then we can easily deduce the estimate: 

z

|ϕ˜ 1 (z, t)| ≤

|q 1(ζ )|(|ϕ˜ 1 (ζ, z + t − ζ )| + |ϕ˜2 (ζ, z + t − ζ )|)

0

+q(ζ ˜ )(|ϕ12 (ζ, z +

 t − ζ )| + |ϕ22 (ζ, z + t − ζ )|) dζ ≤ a ϕ , ˜

(z, t) ∈ + T,

where

a = 2T ϕ 0

1 1 + 2T ϕ 0  . + 1 − T ϕ 0 (1 − T ϕ 0 )2

Similarly, ˜ |ϕ˜ 2 (z, t)| ≤ a ϕ ,

|ϕ˜ 3 (z)| ≤ a|λ0 | ϕ , ˜

(z, t) ∈ + T.

Take a positive ρ < 1 and choose T such small that a max(1, |λ0 |) ≤ ρ and inequalities (5.3.15) hold. Then the operator A is contracting on the ball B(ϕ 0 ). According the Banach’s principle the Eq. (5.3.12) has a unique solution which belongs to the ball B(ϕ 0 ). Hence, function q(z) is uniquely defined by the formula (5.3.9) and it is continuous for z ∈ [0, T /2]. This means that the coefficient εˆ (z) is in C 1 [0, T /2]. Moreover, this coefficient can be derived through the function q(z) by the formula:

 εˆ (z) = ε+ exp 4

z

 q(ζ ) dζ .

0

Having εˆ (z) we can find the correspondence between z and x3 using formula (5.2.10), then calculate the unknown function ε(x3 ) = εˆ (h−1 (x3 )), where h−1 (x3 ) is the inverse of the function h(z).   Note that the problem of finding the permeability μ(x3 ), when ε(x3 ) = ε+ is a given positive constant and σ (x3 ) = 0 for x3 > 0, is completely symmetric to the previous one. So, it does not require a separate investigation. Instead we consider the following problem.

5.3 One-dimensional Inverse Problems

169

5.3.2 Problem of Finding the Conductivity Coefficient In this case we assume that ε and μ are given. For the sake of simplicity, here we assume that ε(x3 ) = ε+ > 0 and μ(x3 ) = μ+ > 0 for x3 > 0. In this case q1 (z) = q2 (z) = 0,

q3 (z) =

σˆ (z) := q(z), 2ε+

(5.3.16)

and the Eqs. (5.2.14) have the form ⎧ ∂u ∂u1 1 ⎪ − + q(z)(u1 + u2 ) = 0, ⎨ ∂t ∂z ⎪ ⎩ ∂u2 + ∂u2 + q(z)(u1 + u2 ) = 0, ∂t ∂z

(z, t) ∈ + T,

(5.3.17)

which is quite similar to (5.3.4). One needs to add to these equations the boundary conditions (5.3.3) and conditions (5.3.5) along the characteristic line t = z. Then we obtain the integral equation for functions u1 (z, t), u2 (z, t) of the form: 

z

u1 (z, t) − 

q(ζ )(u1(ζ, z + t − ζ ) + u2 (ζ, z + t − ζ )) dζ = F1 (t + z),

0 z

u2 (z, t) +

q(ζ )(u1(ζ, t − z + ζ ) + u2 (ζ, t − z + ζ )) dζ = F2 (t − z),

0

(z, t) ∈ + T . (5.3.18) Substituting in the first Eq. (5.3.18) t = z and using first condition (5.3.5), we find the relation  z q(ζ )(u1(ζ, 2z − ζ ) + u2 (ζ, 2z − ζ )) dζ = F1 (2z). 0

Differentiate the last equation with respect to t and denote again p(z) = q(z) exp



z

 q(ζ ) dζ .

0

 Then dividing both sides of the obtained relation by r22 μ− j (0), we get: 

z

p(z) + λ0

q(ζ )(u˙ 1(ζ, 2z − ζ ) + u˙ 2 (ζ, 2z − ζ )) dζ = p0 (z),

0

z ∈ [0, T /2], (5.3.19)

170

5 One-dimensional Inverse Problems in Electrodynamics

where λ0 =

2  , r22 μ− j (0)

p0 (z) = λ0 F1 (2z)

(5.3.20)

and q(z) is defined by the formula (5.3.9). Differentiating now (5.3.18) with respect to t, we obtain the following equations for functions u˙ 1 and u˙ 2 : 

z

u˙ 1 (z, t) − 

0 z

u˙ 2 (z, t) + 0

q(ζ )(u˙ 1(ζ, z + t − ζ ) + u˙ 2 (ζ, z + t − ζ )) dζ = F1 (t + z), q(ζ )(u˙ 1(ζ, t − z + ζ ) + u˙ 2 (ζ, t − z + ζ )) dζ = F2 (t − z), (z, t) ∈ + T . (5.3.21)

The Eqs. (5.3.19) and (5.3.21) form the system of integral equations for determining the unknown functions u˙ 1 (z, t), u˙ 2 (z, t) and p(z) in the domain + T . This system differs from the system (5.3.10), (5.3.11) by some signs under the integral terms. Therefore for this system the Banach Contraction Mapping Principle can be applied in the space of continuous functions to conclude that this system has a unique solution, when T > 0 is enough small. After finding p(z) we can calculate q(z), using formula (5.3.9), and then find σˆ (z) = 2q(z)ε+ .In this case we obtain the correspondence between z and x3 of the form z = x3 ε+ μ+ . So,  σ (x3 ) = σˆ (x3 ε+ μ+ ). Hence, the unique solution of the inverse problem exists and is a continuous function on [0, T /2], if T is small enough. As a result the following theorem holds. Theorem 5.3.2 Let ε(x3 ) = ε+ > 0, μ(x3 ) = μ+ > 0 for x3 > 0, and j (t) ∈ C 1 [0, T ], j (0) = 0. Assume that f (t) ∈ C 1 [0, T ], T > 0, and condition (5.2.37) holds. Then for sufficiently small T > 0 there exist a unique continuously differentiable solution to the inverse problem. For more general results related to inverse problems of electrodynamics we refer the book [138].

Chapter 6

Inverse Problems for Parabolic Equations

This chapter deals with inverse coefficient problems for linear second-order 1D parabolic equations. We establish, first, a relationship between solutions of direct problems for parabolic and hyperbolic equations. Then using the results of Chap. 3 we derive solutions of the inverse problems for parabolic equations through the corresponding solutions of inverse problems for hyperbolic equations, using the Laplace transform. In the final part of this chapter, we study the relationship between the inverse problems for parabolic equation and inverse spectral problems.

6.1 Relationship Between Solutions of Parabolic and Hyperbolic Direct Problems Consider the following initial-boundary value problem ⎧  ⎨ ∂ − L v(x, t) = 0, (x, t) ∈ R2 , + ∂t ⎩ v|t =0 = 0, v|x=0 = g(t),

(6.1.1)

for the parabolic equation in the first quadrant R2+ = {(x, t)| x > 0, t > 0} of the plane R2 , where L := c2 (x)

∂2 ∂ + d(x). + b(x) 2 ∂x ∂x

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_6

171

172

6 Inverse Problems for Parabolic Equations

Consider also the similar problem ⎧ 2  ⎨ ∂ − L u(x, t) = 0, (x, t) ∈ R2+ , ∂t 2 ⎩ u|t =0 = 0, ut |t =0 = 0, u|x=0 = h(t)

(6.1.2)

for the hyperbolic equation. It turns out that the solution of the parabolic problem (6.1.1) can be expressed through the solution of the hyperbolic problem (6.1.2), if the function g(t) is related to h(t) by a special way. To show this, let us define the Laplace transforms v(x, ˜ p) = u(x, ˜ s) =

$∞ 0

v(x, t) exp(−pt) dt,

0

u(x, t) exp(−st) dt

$∞

(6.1.3)

of the functions v(x, t) and u(x, t) given in (6.1.1) and (6.1.2). Here p = p1 + ip2 √ and s = s1 + is2 are the transform parameters and i := −1 is the imaginary unit. These transforms exist if the functions v(x, t) and u(x, t) satisfy the conditions |v(x, t)| ≤ Ceσ t ,

|u(x, t)| ≤ Ceσ t ,

x ≥ 0, t ≥ 0,

with the positive constant C and real σ . Then the integrals in (6.1.3) exist for p1 > σ and s1 > σ . Moreover, Laplace transforms are analytic functions of p and s, respectively, in the complex half-planes p1 > σ and s1 > σ . We assume that the functions v(x, t), u(x, t) with derivatives with respect to x up to the second order, as well as the functions g(t) and h(t) admit the Laplace ˜ transform with respect to t. Denote by g(p) ˜ and h(s) the Laplace transforms of the input g(t) and h(t) in (5.1.1) and (5.1.2), respectively. Then we can rewrite problems (6.1.1) and (6.1.2) as follows: p v(x, ˜ p) − Lv(x, ˜ p) = 0,

x > 0;

s 2 u(x, ˜ s) − Lu(x, ˜ s) = 0,

x > 0;

v| ˜ x=0 = g(p), ˜

(6.1.4)

˜ u| ˜ x=0 = h(s).

(6.1.5)

√ ˜ √p). Then it follows from (6.1.4) Suppose now that s = p and g(p) ˜ = h( √ and (6.1.5) that v(x, ˜ p) = u(x, ˜ p). According the table of Laplace transforms [12] the last relations correspond to the formulae: 1 g(t) = √ 2 πt 3 1 v(x, t) = √ 2 πt 3



∞

τ2

h(τ )e− 4t τ dτ,

(6.1.6)

0



∞ 0

τ2

u(x, τ )e− 4t τ dτ.

(6.1.7)

6.1 Relationship Between Solutions of Parabolic and Hyperbolic Direct Problems

173

Note that τ2 ∂ τ √ e− 4t = − G(τ, t), ∂τ 2 πt 3

τ2

e− 4t G(τ, t) = √ , πt

where the function G(τ, t) solves the heat equation ∂ 2 G(τ, t) ∂G(τ, t) = , ∂t ∂τ 2

t > 0.

Under the consistency condition h(0) = 0 and the initial condition u(x, 0) = 0 in (6.1.5), formulae (6.1.6) and (6.1.7) can be also written as follows: 

∞

g(t) =

h (τ )G(τ, t) dτ,

(6.1.8)

uτ (x, τ )G(τ, t) dτ.

(6.1.9)

0



∞

v(x, t) = 0

We use the change of variables z = τ 2 /(4t) in the integral (6.1.8) to prove that g(0) = h (0).

(6.1.10)

We have:  g(0) := lim

∞

t →+0 0

2 h (τ )G(τ, t) dτ = √ lim π t →+0 2 = h (0) √ π



∞

√ 2 h (2 tz)e−z dz

0



∞

e−z dz = h (0). 2

0

In the same way we can prove v(x, 0) = u( x, 0) = 0. It can be verified directly that function v(x, t) given by formula (6.1.7) satisfy the Eq. (6.1.1), if u(x, t) solves the problem (6.1.2). Indeed,

∂ 

∂  ∞ − L v(x, t) = − −L u(x, τ )Gτ (τ, t) dτ ∂t ∂t 0 

∞

=

Lu(x, τ )Gτ (τ, t) − u(x, τ )

0

 ∂ Gτ (τ, t) dτ ∂t

174

6 Inverse Problems for Parabolic Equations



∞

=

uτ τ (x, τ )Gτ (τ, t) − u(x, τ )

0



∞

=

u(x, τ ) 0

 ∂ Gτ (τ, t) dτ ∂t

∂2 ∂ − Gτ (τ, t) dτ = 0. 2 ∂t ∂τ

Note that the integral relation (6.1.6) is invertible. Moreover, it is the Laplace transform with parameter p = 1/(4t). Indeed, it can be represented in the form: 1 g(t) = √ 4 πt 3



∞

√ z h( z)e− 4t dz.

(6.1.11)

0

 Hence, the √ function πp−3 g(1/(4p))/2 := g(p) ˆ is the Laplace transform of the function h( t). Then g(p) ˆ is an analytic function for Re(p) > 0 and, as a result, the function h(t) can be uniquely recovered from g(t) by using the inverse Laplace transform h(t) =

1 2πi



σ0 +i∞ σ0 −i∞

2

pt g(p)e ˆ dp,

t ≥ 0,

(6.1.12)

where σ0 > 0.

6.2 Problem of Recovering the Potential in Heat Equation Let v(x, t) be the solution of the parabolic equation vt − vxx − q(x)v = 0,

x > 0, t > 0,

(6.2.1)

with the following boundary and initial conditions: v|x=0 = g(t),

t ≥ 0;

v|t =0 = 0,

x ≥ 0.

(6.2.2)

For a given function g(t) from some class of admissible coefficients, we will define problem (6.2.1)–(6.2.2) as the direct problem. Consider the inverse problem of recovering the potential q(x) from the given trace r(t) := vx |x=0 ,

t ≥0

(6.2.3)

of the derivative vx on the semi-axis x = 0, t ≥ 0 of the solution v = v(x, t; q).

6.2 Problem of Recovering the Potential in Heat Equation

175

Associate to the inverse problem (6.2.1)–(6.2.3) the similar inverse problem for the hyperbolic equation, i.e. let the direct problem is given by ut t − uxx − q(x)u = 0, u|x=0 = h(t), t ≥ 0; ux |x=0 = f (t),

x > 0, t > 0,

(6.2.4)

u|t =0 = 0, ut |t =0 = 0, x ≥ 0,

t ≥ 0,

(6.2.5) (6.2.6)

and the Neumann type measured output is given by (6.2.3). We assume here that q(x) ∈ C[0, ∞) and the solutions to direct problems (6.2.1)–(6.2.2) and (6.2.4)–(6.2.5) admit the Laplace transforms with respect to t. Furthermore, suppose that the h(0) = 0 and the relationship (6.1.6) holds between the input h(t) in the parabolic inverse problem (6.2.1)–(6.2.3) and the input g(t) in the hyperbolic inverse problem (6.2.4)–(6.2.6). Then it is necessary that f (0) = −h (0) and functions r(t) and f (t) satisfy the relation 1 r(t) = √ 2 πt 3



∞

τ2

f (τ )e− 4t τ dτ,

t ≥ 0.

(6.2.7)

0

The function f (t) can be found through r(t) by the formula 1 f (t) = 2πi



σ0 +i∞ σ0 −i∞

2

rˆ (p)ept dp,

t ≥ 0,

(6.2.8)

 similar to (6.1.12), where rˆ (p) = πp−3 r(1/(4p))/2. Thus, the functions h(t) and f (t) in the inverse problem (6.2.4)–(6.2.6) are uniquely defined by the functions g(t) and r(t). This means that instead of the inverse problem (6.2.1)–(6.2.3) for parabolic equation one can solve the inverse problem (6.2.4)–(6.2.6) for hyperbolic equation, to find unknown coefficient q(x) for x ≥ 0. The last inverse problem has been studied in Sect. 4.2 under the conditions that h(0) = 0 and h(t) ∈ C 2 ([0, ∞). But in our case we have assumed that h(0) = 0. What should be done in this case? It turn out that we need to make a small change in order to reduce the present problem to the previous one. Indeed, let we suppose that g(0) = 0. Then, it follows from formula (6.1.10) that h (0) = g(0) = 0. Differentiating relations (6.2.4)–(6.2.6) with respect to t, one gets u˙ t t − u˙ xx − q(x)u˙ = 0, x > 0, t > 0,

(6.2.9)

u| ˙ x=0 = h (t), t ≥ 0; u| ˙ t =0 = 0, u˙ t |t =0 = 0, x ≥ 0,

(6.2.10)

u˙ x |x=0 = f  (t),

(6.2.11)

t ≥ 0,

176

6 Inverse Problems for Parabolic Equations

where u˙ = ut . Note that the third relation (6.2.10) follows directly from Eq. (6.2.4) with t = 0. So, the functions h (t) and f  (t) in the inverse problem (6.2.9)– (6.2.11) play the same roles, as the functions h(t) and f (t), in the inverse problem studied in Sect. 4.2. Hence, we need suppose only that h(t) ∈ C 3 [0, ∞) in order to use the uniqueness Theorem 4.2.3 in Sect. 4.2. The following theorem is a direct consequence of this conclusion. Theorem 6.2.1 Assume that the function h(t), which is associated with the function g(t) by formula (6.1.6), satisfies the following conditions: h(t) ∈ C3 [0, ∞), h(0) = 0, h (0) = 0. Suppose, in addition, that q1 , q2 ∈ C[0, ∞) are any two solutions of the inverse problem (6.2.1)–(6.2.3) corresponding data g(t) and r(t). Then q1 (x) = q2 (x) for x ∈ [0, ∞). Remark 6.2.1 If g(t) is defined by formula (6.1.6), then g(t) is an analytic function for t > 0. Then r(t) is also an analytic function of t > 0. Hence, uniqueness theorem for the inverse problem (6.2.1)–(6.2.3) is still true, if the data g(t) and r(t) are given for t ∈ [0, T ] with arbitrary fixed T > 0.

6.3 Uniqueness Theorems for Inverse Problems Consider the following initial-boundary value problem ⎧ v − (k(x)vx )x + q(x)v = 0, x > 0, t > 0; ⎪ ⎪ ⎨ t v(x, 0) = 0, x ≥ 0; ⎪ ⎪ ⎩ v(0, t) = 0, t ≥ 0,

(6.3.1)

for the parabolic equation. Let v(x, t) solves the initial-boundary value problem (6.3.1) and, in addition, the trace of k(x)vx (x, t) is given at x = 0: r(t) := k(0)vx (x, t)|x=0 ,

t ≥ 0.

(6.3.2)

Remark that in thermal conduction k(0)vx (x, t)|x=0 is defined as the heat flux, where k(x) is the thermal conductivity. Here and below we assume that k(x) and q(x) satisfy the conditions: 0 < k0 ≤ k(x) ≤ k1 ≤ ∞, k(x) ∈ C 2 [0, ∞),

|q(x)| ≤ q0 ,

q(x) ∈ C[0, ∞),

where k0 , k1 q0 are some positive constants.

x ∈ [0, ∞);

(6.3.3) (6.3.4)

6.3 Uniqueness Theorems for Inverse Problems

177

Consider the following two inverse coefficient problems. ICP1: ICP2

Find the unknown coefficient q(x) in (6.3.1) from the given by (6.3.2) output r(t). Find the unknown coefficient k(x) in (6.3.1) from the given by (6.3.2) output r(t).

In both cases the initial-boundary value problem (6.3.1) is defined as the direct problem. Obviously the coefficient k(x) in ICP1 is assumed to be known as well as the coefficient q(x) in ICP2 assumed to be known. In both inverse problems we assume that k(0) is known. Consider first ICP1. Let us try to reduce the direct problem (6.3.1) to the simplest form. For this goal, introduce the new variable y instead of x by the formula 

x

y= 0

dξ √ . k(ξ )

(6.3.5)

Let x = H (y) be the inverse function of y = y(x) given by (6.3.5). Denote by ˆ k(H (y)) = k(y),

q(H (y)) = q(y), ˆ

v(H (y), t) = v(y, ˆ t).

Since '

ˆ

k(x)vx (x, t) = vˆy (y, t) k(y)

y=y(x)



,

(k(x)vx (x, t))x = vˆyy (y, t) + vˆy (y, t)

'

 ˆ k(y)

y y=y(x)

,

the function v(y, ˆ t) satisfies the relations vˆt (y, t) − vˆyy (y, t) − vˆy (y, t)

'  ˆ + q(y) ˆ vˆ = 0, y > 0, t > 0, k(y) y

v(y, ˆ 0) = 0, v(0, ˆ t) = g(t),



k(0)vˆy (0, t) = r(t).

(6.3.6) (6.3.7)

Introduce the new function w(y, t) by the formula v(y, ˆ t) = S(y)w(y, t), where 1/4 . Then w(y, t) solves the problem: ˆ S(y) = (k(0)/k(y)) wt − wyy (y, t) + Q(y)w = 0, y > 0, t > 0, w(y, 0) = 0, w(0, t) = g(t),

  k(0) wy + S  (0)w

y=0

= r(t),

(6.3.8) (6.3.9)

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6 Inverse Problems for Parabolic Equations

where S  (0) = −kˆ  (0)/(4k(0)) and Q(y) is defined by the formula Q(y) = q(y) −

S  (y) 2 S  (y) +2 S(y) S(y)

    2 = q(y) − ln S(z) + ln S(z) .

(6.3.10)

From the latter two relations in (6.3.9) we find r(t) − S  (0)g(t). rˆ (t) = √ k(0)

wy |y=0 = rˆ (t),

(6.3.11)

Since S  (0) is known in this inverse problem the function rˆ (t) is defined by the latter equality. Consider now the boundary value problem for the hyperbolic equation ut t − uyy (y, t) + Q(y)u = 0, u(y, 0) = 0, ut (y, 0) = 0, y > 0,

y > 0, t > 0,

(6.3.12)

u(0, t) = h(t), t ≥ 0,

(6.3.13)

where the function Q(y) is defined by (6.3.10), and the inverse problem of recovering Q(y) from the data uy |y=0 = f (t),

t ≥ 0.

(6.3.14)

Assume that h(t) ∈ C3 [0, ∞), h(0) = 0, h (0) = 0, and functions g(t) and h(t) are connected by formula (6.1.6). Then f (0) = −h (0) and 

∞

rˆ (t) =

f  (τ )G(τ, t) dτ.

(6.3.15)

0

The function f (t), as a trace of the normal derivative uy on y = 0 to the problem (6.3.12)–(6.3.13) belongs to C2 [0, ∞). Then the function u˙ = ut satisfies the relations u˙ t t − u˙ yy − Q(y)u˙ = 0, u| ˙ y=0 = h (t),

t ≥ 0;

u| ˙ t =0 = 0,

y > 0, t > 0,

u˙ t |t =0 = 0,

u˙ y |y=0 = f  (t),

(6.3.16)

x ≥ 0,

(6.3.17)

t ≥ 0.

(6.3.18)

From (6.3.16) and (6.3.17) follows that u(0, ˙ 0) = h (0). Therefore, f  (0) = −h (0)   and rˆ (0) = f (0) = −h (0). Since r(0) − S  (0)g(0), g(0) = h (0), rˆ (0) = √ k(0)

6.3 Uniqueness Theorems for Inverse Problems

179

we can find S  (0) by the formula S  (0) =

r(0) h (0) +√ . h (0) k(0)h (0)

(6.3.19)

On the over hand, S  (0) = −kˆ  (0)/(4k(0)). Thus, the formula (6.3.19) gives the necessary condition of solvability to inverse problem of finding q(x) from the data g(t) and r(t), it means, that the value h (0) must be taken from the relation (6.3.19). For the inverse problem (6.3.16)–(6.3.18) the uniqueness theorem holds. Hence, the inverse problem of finding Q(y) for Eqs. (6.3.8), (6.3.9), and (6.3.11) has at most one solution. Since k(x) is known, then y = y(x) given by (6.3.5) as well as S(y) are also known. So, we can find q(y) from given Q(y) and S(y). Therefore we can determine and q(x) = q(y(x)). Taking into account the Remark 6.2.1 we come to the following uniqueness theorem. Theorem 6.3.1 Let conditions (6.3.3) hold. Assume that the function h(t), which is associated with the function g(t) by formula (6.1.6), satisfies the following conditions: h(t) ∈ C3 [0, ∞), h(0) = 0, h (0) = 0 and h (0) satisfies the relation (6.3.19). Denote by q, q2 ∈ C[0, ∞) two solutions of ICP1, defined by (6.3.1)– (6.3.2), with data g(t) and r(t), given for t ∈ [0, T ], T > 0. Then q1 (x) = q2 (x) for x ∈ [0, ∞). Consider now ICP2. Assume now that q(x) = q0 is a given constant and k(0) > 0 is known. We reduce the ICP2 to Eqs. (6.3.8) and (6.3.11). Since in this ˆ case k(y) is unknown, S  (0) in (6.3.11) is also unknown. Let us explain how to find S  (0). Consider again the problem (6.3.12)–(6.3.13). Then the relation (6.3.19) holds and defines S  (0). So the function rˆ (t) is become known and Eq. (6.3.15) uniquely determines f  (t). As in the previous case, solving the inverse problem for the hyperbolic equation we can find the function Q(y). Then using (6.3.10) we obtain the second order differential equation for function ln S(y):     Q(y) − q0 = − ln S(z) + ln S(z)

2

(6.3.20)

with given Cauchy data S(0) and S  (0). From this equation and initial data the function ln S(y) is uniquely determined for all x ∈ [0, ∞). After this we can find ˆ ˆ k(y), y = y(x) and k(x) = k(y(x)). Hence the following theorem holds. Theorem 6.3.2 Let h(t) ∈ C3 [0, ∞), h(0) = 0, h (0) = 0, and functions g(t) and h(t) are connected by formula (6.1.6). Let, moreover, q(x) = q0 be a given constant and k(0) > 0 be known and k1 (x) and k2 (x) be two solutions to the problem (6.3.1)– (6.3.2) with data g(t) and r(t) given for t ∈ [0, T ], T > 0. Then k1 (x) = k2 (x) for x ∈ [0, ∞).

180

6 Inverse Problems for Parabolic Equations

6.4 Relationship With Inverse Spectral Problems for Sturm-Liouville Operator In this section we will analyze the relations between the inverse problems for parabolic equation and inverse spectral problems. Let v(x, t) be the solution of the parabolic equation vt − vxx + q(x)v = g(x, t),

0 < x < 1, t > 0,

(6.4.1)

with given function g(x, t) and given the initial and boundary data vx |x=0 = 0, v|x=1 = 0,

t ≥ 0;

v|t =0 = 0,

x ≥ 0.

(6.4.2)

Our aim is to show that problem (6.4.1)–(6.4.2) is related to the problem of finding eigenvalues for the ordinary differential equation − y  (x) + q(x)y(x) = λy(x),

0 < x < 1,

(6.4.3)

with boundary data y  (0) = 0,

y(1) = 0.

(6.4.4)

Remember that λ is called an eigenvalue of the differential operator −y  (x) + q(x)y(x) subject to the boundary conditions (6.4.4) provided there exists a function u(x), not identically zero, solving problem (6.4.3)–(6.4.4). The solution u(x) is called the corresponding eigenfunctions. It is well known that the eigenfunctions of the differential operator form countable set λn , n = 1, 2, . . ., with the unique concentration point at infinity and the eigenfunctions yn (x), n = 1, 2, . . ., are dense in the space L2 (0, 1). Note that all λn > 0, if q(x) ≥ 0. Applying the Fourier method to problem (6.4.1)–(6.4.2) we can construct the solution in the form v(x, t) =

∞ 

vn (t)yn (x),

(6.4.5)

n=1

where vn (t) are solutions to the Cauchy problem vn + λn vn = gn (t), vn (0) = 0,

n = 1, 2, . . . ,

(6.4.6)

and gn (t) are the Fourier coefficients of g(x, t) with respect to the eigenfunctions yn (x), n = 1, 2, . . ., i.e., gn (t) =

1

yn 2



1

g(x, t)yn (x)dx, 0

n = 1, 2, . . . .

(6.4.7)

6.4 Relationship With Inverse Spectral Problems for Sturm-Liouville Operator

181

The solutions to the problem (6.4.6) have the form  vn (t) =

t

gn (τ )e−λn (t −τ )dτ,

n = 1, 2, . . . .

(6.4.8)

0

Thus, the solution to the problem (6.4.1)–(6.4.2) is given by the formulae (6.4.5), (6.4.7), and (6.4.8). From them follows that v(x, t) =

  ∞  yn (x) t −λn (t −τ ) 1 e g(x, τ )yn (x)dxdτ.

yn 2 0 0

(6.4.9)

n=1

We can represent formula (6.4.8) in an other form. To this end, denote the solution the Eq. (6.4.3) with the Cauchy data y(0) = 1,

y  (0) = 0

(6.4.10)

n = 1, 2, . . . .

(6.4.11)

by y(x, λ). Then λn satisfy the conditions y(1, λn ) = 0,

Denote by yn (x) = y(x, λn ). Introduce the spectral function ρ(λ), λ ∈ (−∞, ∞), by the formula ⎧ ⎨ 0, k ρ(λ) =  ⎩

if λ < λ1 , 1 , if λk < λ < λk+1 . 2 n=1 yn

(6.4.12)

Then formula (6.4.9) can be written as follows  v(x, t) =



∞ −∞

t

y(x, λ)

e

−λ(t −τ )

0



1

g(ξ, τ )y(ξ, λ)dξ dτ dρ(λ),

(6.4.13)

0

where the integral with respect to λ should be understood as a Stieltjes integral or in the sense of distributions. For the direct problem (6.4.1)–(6.4.2) with the given function g(x, t), we consider the inverse problem of recovering the potential q(x) from a given trace of the solution v(x, t) on semi-axis x = 0, t ≥ 0, i.e., r(t) := v|x=0 ,

t ≥ 0.

(6.4.14)

It is easy to derive the function r(t) through the spectral function ρ(λ). Indeed, by y(0, λ) = 1, we conclude:  r(t) =

∞



−∞ 0

t

e

−λ(t −τ )



1

g(ξ, τ )y(ξ, λ)dξ dτ dρ(λ), 0

t ≥ 0. (6.4.15)

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6 Inverse Problems for Parabolic Equations

Consider now case when g(x, t) = δ(x + 0)δ(t + 0). Here δ(x + 0) is the Dirac-delta function located at the point x = +0. Then two inner integrals in formula (6.4.15) are calculated explicitly and we obtain more simple formula:  r(t) =

∞ −∞

e−λt dρ(λ),

t ≥ 0.

(6.4.16)

This equality uniquely defines the spectral function ρ(λ). Then for the direct problem of finding eigenvalues and eigenfunctions related to Eqs. (6.4.3) and (6.4.4) we can consider the inverse spectral problem: given ρ(λ) find q(x). This problem is equivalent to the inverse problem for the parabolic equation. Indeed, given r(t) we find ρ(λ) and vice versa given ρ(λ) we find r(t). For the inverse spectral problem is well known uniqueness theorem stated by V.A. Marchenko first in [95] (see also, [96]). Theorem 6.4.1 The potential q(x) ∈ C[0, 1] is uniquely recovered by the spectral function ρ(λ). As a corollary, we obtain the uniqueness theorem for the inverse problem (6.4.1), (6.4.2), and (6.4.14). Theorem 6.4.2 Let g(x, t) = δ(x + 0)δ(t + 0) and q(x) ∈ C[0, 1]. Then q(x) is uniquely determined by the given function r(t). Note that the assertions of both theorems remains true if the finite interval [0, 1] is replaced on semi-infinite interval [0, ∞).

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output In this section we consider the problem of determining the space-dependent thermal conductivity k(x) of the one-dimensional heat equation ut (x, t) = (k(x)ux (x, t))x from boundary measurements in the finite domain T = {(x, t) ∈ R2 : 0 < x < , 0 < t ≤ T }. Consider the inverse problem of identifying the leading coefficient k(x) in ⎧ ⎨ ut (x, t) = (k(x)ux (x, t))x , (x, t) ∈ T , u(x, 0) = 0, 0 < x < , ⎩ −k(0)ux (0, t) = g(t), ux (, t) = 0, 0 < t < T ,

(6.5.1)

from the measured temperature f (t) at the left boundary x = 0 of a nonhomogeneous rod: f (t) := u(0, t), t ∈ [0, T ].

(6.5.2)

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

183

Note that problem (6.5.1) is a simplest 1D model of heat conduction in a rod occupying the interval (0, ). The Neumann condition −k(0)ux (0, t) = g(t) in the direct problem (6.5.1) means that the heat flow g(t) is prescribed at the left boundary of the rod as an input for all t ∈ [0, T ]. We assume that the functions k(x) and g(t) satisfy the following basic conditions:  k ∈ H 1 (0, ), 0 < c0 ≤ k(x) ≤ c1 , (6.5.3) g ∈ L2 (0, T ), g(t) > 0, for all t ∈ (0, T ) and g(0) = 0. The second condition in (6.5.3) means heating of a rod at the left end, that is heat flux at x = 0 is positive: g(t) > 0 for all t ∈ (0, T ) and is zero at an initial time t = 0, i.e. g(0) = 0. Under conditions (6.5.3) the initial boundary value problem (6.5.1) has a unique regular weak solution [38, §7.1, Theorem 5], defined as in Appendix B.2 where the necessary estimates for the weak and regular weak solutions are also proved. In next subsection we derive some important properties of the solution u(x, t) of the direct problem (6.5.1), in particular of the output u(0, t; k). Then we introduce an input-output operator and reformulate the inverse coefficient problem (6.5.1)– (6.5.2) as a nonlinear operator equation. We prove in Sect. 6.5.2 that the input-output operator is compact and Lipschitz continuous which allows to prove an existence of a quasisolution. Then we introduce an adjoint problem and derive an integral relationship relating the change δk(x) := k1 (x) − k2 (x) in the coefficient k(x) with the change in the output δu(x, t) = u(0, t; k1) − u(0, t; k2 ). This permits to obtain explicit formula for the Fréchet gradient J  (k) of the Tikhonov functional through the solutions of the direct and adjoint problems. This approach, defined as an adjoint problem approach, has been proposed by Cannon and DuChateau [27, 36] and then developed in [55, 56, 59, 60, 64]. Furthermore, these results also constitute a theoretical framework of the numerical algorithm for recovering an unknown coefficient or source term. Some numerical examples applied to severely illposed benchmark problems, presented in the final Sect. 6.5.4 demonstrate accurate reconstructions by the CG-algorithm.

6.5.1 Some Properties of the Direct Problem Solution The theorem below shows that sign of the input g(t) has a significant impact on the solution of the direct problem (6.5.1). Theorem 6.5.1 Let conditions (6.5.3) hold. Then g(t) > 0 for all t ∈ (0, T ), implies ux (x, t) ≤ 0, for all (x, t) ∈ T . Proof Let ϕ ∈ C02,1 ( T ) ∩ C0 ( T ) be an arbitrary smooth function. Multiply both sides of the parabolic equation (6.5.1) by ϕx (x, t), integrate on T and then perform

184

6 Inverse Problems for Parabolic Equations

integration by parts multiple times. After elementary transformations we get:  ux (ϕt + k(x)ϕxx ) dxdt 

T T

= 0



(uϕt )x= x=0 dt +

T 0

 (k(x)ux ϕx )x= x=0 dt −

 0

(uϕx )tt =T =0 dx.

(6.5.4)

Now we require that the function ϕ(x, t) is chosen to be the solution of the following backward parabolic problem: ⎧ ⎨ ϕt + k(x)ϕxx = F (x, t), (x, t) ∈ T , ϕ(x, T ) = 0, x ∈ (0, ), ⎩ ϕ(0, t) = 0, ϕ(, t) = 0, t ∈ (0, T ),

(6.5.5)

where an arbitrary continuous function F (x, t) which will be defined below. Then taking into account the homogeneous initial and boundary conditions given in (6.5.1) and (6.5.5) with the flux condition −k(0)ux (0, t) = g(t), in the integral identity (6.5.4) we get: 



T

ux (x, t)F (x, t)dxdt =

g(t)ϕx (0, t)dt.

(6.5.6)

0

T

We apply the maximum principle to the backward in time parabolic problem (6.5.5). For this aim we require that the arbitrary function F (x, t) satisfies the condition F (x, t) > 0 for all (x, t) ∈ T . Then ϕ(x, t) < 0 on T . With the boundary condition ϕ(0, t) = 0, this implies: ϕx (0, t) := lim

h→0+

ϕ(h, t) − ϕ(0, t) ≤ 0. h

On the other hand, by the condition g(t) > 0 we conclude that the right hand side of (6.5.6) is non-positive, so  ux (x, t)F (x, t)dxdt ≤ 0, for all F (x, t) > 0. T

This implies that ux (x, t) ≤ 0 for all (x, t) in T .

 

Remark 6.5.1 The result of the above theorem has a precise physical meaning. By definition, g(t) := −k(0)ux (0, t) is the heat flux at x = 0. The sign minus here means that, by convention, the heat flows in the positive x-direction, i.e. from regions of higher temperature to regions of lower temperature. The condition g(t) > 0, t ∈ (0, T ) implies that the heat flux at x = 0 is positive. Theorem 6.5.1 states that in the absence of other heat sources, the positive heat flux g(t) > 0 at the left end x = 0 of a rod results the nonnegative flux at all points x ∈ (0, ) of a rod, that is −k(x)ux (x, t) ≥ 0.

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

185

Corollary 6.5.1 Assume that, in addition to conditions of Theorem 6.5.1, the solution of the direct problem belongs to C 1,0 ( T ). Then there exists such  = (t) > 0 that ux (x, t) < 0, for all (x, t) ∈ T , where T := {(x, t) ∈ T : 0 < x < (t), 0 < t ≤ T } is the triangle with two rectilinear sides 1 := {(x, t) ∈ T : x = 0, t ∈ [0, T ]}, 2 := {(x, t) ∈ T : x ∈ (0, (T )], t = T } and the curvilinear side 3 := {(x, t) ∈ T : x = (t), t ∈ (0, T )}. The parameter (t) > 0 depending on t ∈ (0, T ] satisfies the condition (0) = 0. Proof Indeed, since ux (x, t) is a continuous function of x ∈ (0, ) and ux (0, t) < 0 for all t ∈ (0, T ), it remains negative throughout the right neighborhood T of x = 0, that is ux (x, t) < 0 for all (x, t) ∈ T . Since ux (0, 0) = 0, this neighborhood is the above defined triangle T with the bottom vertex at the origin (x, t) = (0, 0).  

6.5.2 Compactness and Lipschitz Continuity of the Input-Output Operator: Regularization Let us define the set of admissible coefficients K := {k ∈ H 1 (0, ) : 0 < c0 ≤ k(x) ≤ c1 < ∞}.

(6.5.7)

For a given coefficient k ∈ K we denote by u = u(x, t; k) the solution of the direct problem (6.5.1). Then introducing the input-output operator or Neumann-toDirichlet map 

[k] := u(x, t; k)|x=0+ , [·] : K ⊂ H 1 (0, ) → L2 (0, T ),

(6.5.8)

we can reformulate the inverse coefficient problem (6.5.1)–(6.5.2) in the following nonlinear operator equation form: [k](t) = f (t), t ∈ (0, T ],

(6.5.9)

where f (t) is the noise free measured output and u(0, t; k) is the output, corresponding to the coefficient k ∈ K. Therefore the inverse coefficient problem with Dirichlet measured output can be reduced to the solution of the nonlinear equation (6.5.9) or to inverting the nonlinear input-output operator defined by (6.5.8). Lemma 6.5.1 Let conditions (6.5.3) hold. Assume in addition that g ∈ H 1 (0, T ). Then the input-output operator [·] : K ⊂ H 1 (0, ) → L2 (0, T ) defined by (6.5.8) is a compact operator.

186

6 Inverse Problems for Parabolic Equations

Proof Let {km } ⊂ K, m = 1, ∞, be a bounded in H 1 (0, )-norm sequence of coefficients. Denote by {um (x, t)}, um := u(x, t; km ), the sequence of corresponding regular weak solutions of the direct problem (6.5.1). Then {u(0, t; km } is the sequence of outputs. We need to prove that this sequence is a relatively compact subset of L2 (0, T ) or, equivalently, the sequence {um (0, t)} is bounded in the norm of the Sobolev space H 1 (0, T ). To estimate the norm um (0, ·) L2 (0,T ) first we use the trace inequality (B.2.6) in Appendix B.2, i.e. the inequality 

T 0

u2m (0, τ )dt ≤

2 



T



0

 0

 u2m (x, t)dxdt + 

T





um,x (x, t)dxdt, (6.5.10) 0

0

and then estimates in (B.2.2) for the weak solution of the direct problem (6.5.1). This leads to

um (0, ·) 2L2 (0,T ) ≤ C˜ 12 g 2L2 (0,T ) ,

(6.5.11)

C˜ 12 = 2C2 / + C12

(6.5.12)

where

and C1 , C2 > 0 are the constants introduced in Lemma B.2.1. In the same way we deduce the estimate

um,t (0, ·) 2L2 (0,T ) ≤ C˜ 12 g  2L2 (0,T )

(6.5.13)

for the regular weak solution of the direct problem (6.5.1), with the same C˜ 1 > 0, from the analogue 

T 0

u2m,t (0, τ )dt

2 ≤ 

 0

T



 0

 u2m,t (x, t)dxdt

+

T





um,xt (x, t)dxdt 0

0

of the trace inequality (6.5.10). Estimates (6.5.11) and (6.5.13) imply that the sequence of outputs {um (0, t)} is uniformly bounded in the norm of H 1 (0, T ). By the compact imbedding H 1 (0, T ) → L2 (0, T ) this implies that the sequence {um (0, t)} is a precompact in L2 (0, T ). Therefore, the input-output operator  transforms each bounded in H 1 (0, ) sequence of coefficients {km (x)} to the precompact in L2 (0, ) sequence of outputs {um (0, t)}. This means that  is a compact operator. This completes the proof of the lemma.   As a consequence of Lemma 6.5.1 we conclude that the inverse coefficient problem (6.5.1)–(6.5.2) is ill-posed.

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

187

Lemma 6.5.2 Assume that conditions (6.5.3) are satisfied. Suppose u(x, t; k1 ) and u(x, t; k2 ) are the weak solutions of the direct problem (6.5.1) corresponding to the admissible coefficients k1 , k2 ∈ K, respectively, k1 (x) = k2 (x). Then the following estimate holds: 12 g 2 2

u(0, ·; k1 ) − u(0, ·; k2) 2L2 (0,T ) ≤ C

k − k2 2H 1 (0,) . L (0,) 1

(6.5.14)

where 2 2 2 = C 2 C 2 2 C 1 1 0 , C0 = (4T c0 / + )C /c0 ,

C2 = max(2/ + 1, 2/3 + 1)

(6.5.15)

and C1 > 0 is the constant introduced in Lemma B.2.1 of Appendix B. Proof Let um (x, t) := u(x, t; km ), m = 1, 2. Then the function v(x, t) := u1 (x, t) − u2 (x, t) solves the following initial boundary value problem: ⎧ ⎪ ⎨ vt = (k1 (x)vx )x + (δk(x)u2x )x , (x, t) ∈ T , v(x, 0) = 0, 0 < x < , ⎪ ⎩ −k (0)v (0, t) = δk(0)u (0, t), v (, t) = 0, 0 < t < T , 1 x 2x x

(6.5.16)

where δk(x) = k1 (x) − k2 (x). To estimate the trace norm v(0, ·) L2 (0,T ) through the norm v L2 (0,T ;L2 (0,)) and vx L2 (0,T ;L2 (0,)), we use first the identity   v (0, t) = v(x, t) −

2

x

2

vξ (ξ, t)dξ

, v ∈ L2 (0, T ; H 1 (0, )).

0

We apply to the right hand side the inequality (a − b)2 ≤ 2(a 2 + b 2 ) and then the Hölder inequality. Integrating over [0, T ] we obtain the following inequality: 

T

 v 2 (0, t)dt ≤ 2

0

T



T

v 2 (x, t)dt + 2x

0

0



 0

vx2 (x, t)dxdt.

Integrating again over [0, ] and then dividing both sides by  > 0 we arrive at the inequality: 

T 0

v 2 (0, t)dt ≤

2 



 v 2 (x, t)dxdt +  T

T

vx2 (x, t)dxdt. (6.5.17)

This inequality shows that to prove the lemma we need to estimate the right hand side norms v L2 (0,T ;L2 (0,)) and vx L2 (0,T ;L2 (0,)) through the norm δk H 1 (0,). To do this, we use the standard L2 -energy estimates for the weak solution of the initial boundary value problem (6.5.16).

188

6 Inverse Problems for Parabolic Equations

Multiplying both sides of Eq. (6.5.16) by 2v(x, t), integrating on t = {(x, τ ) ∈ R2 : 0 < x < , 0 < τ ≤ t, t ∈ (0, T ]} and then using the formulas for integration by parts we obtain the following identity: 







v dx + 2 2

0

k1 (x)vx2 dτ dx

t

t

−2 0



(k1 (x)vx v)x= x=0 dτ 

= −2

δk(x)u2x vx dτ dx + 2

t

0

t

(δk(x)u2x v)x= x=0 dτ,

for all t ∈ [0, T ]. The terms under the last left and right hand side integrals are zero at x =  due to the homogeneous boundary conditions in (6.5.16). Taking into account the Neumann boundary condition −k1 (0)δux (0, t) = δk(0)ux (0, t; k2 ) in the last left hand side integral we deduce that this term and the term at x = 0 under the last right hand side integral are mutually exclusive. Then the above integral identity becomes 







v 2 dx + 2

0

t

k1 (x)vx2 dxdτ = −2

δk(x)u2x vx dxdτ, t

for a.e. t ∈ [0, T ]. With the Hölder inequality, this identity leads to the main integral inequality: 



 v dx + 2c0 2

0

t

vx2 dxdτ

≤ 2 max |δk(x)| u2x L2 (0,T ;L2 (0,)) vx L2 (0,T ;L2 (0,)), x∈[0,]

(6.5.18)

t ∈ [0, T ]. As a first consequence of the integral inequality (6.5.18) we deduce that c0 vx L2 (0,T ;L2 (0,)) ≤ max |δk(x)| u2x L2 (0,T ;L2 (0,)), c0 > 0. x∈[0,]

(6.5.19)

To estimate the term maxx∈[0,] |δk(x)| in (6.5.19) through the norm δk H 1 (0,) we use Agmon’s inequality max |δk(x)|2 ≤ |δk(0)|2 + 2 δk L2 (0,) δk  L2 (0,), δk ∈ H 1 (0, ),

x∈[0,]

with the auxiliary inequality |δk(0)|2 ≤ (2/) δk 2L2 (0,) + (2/3) δk  2L2 (0,),

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

189

which can be proved easily. This leads to max |δk(x)|2 ≤ C2 δk 2H 1 (0,),

x∈[0,]

(6.5.20)

with the constant C > 0 defined in (6.5.15). Inequalities (6.5.19) and (6.5.20) imply:

vx L2 (0,T ;L2 (0,)) ≤

C

δk H 1 (0,) u2x L2 (0,T ;L2 (0,)), c0 > 0. c0

(6.5.21)

To estimate the first left hand side integral in (6.5.18) we use again inequality (6.5.20) to get 

 0

v 2 (x, t)dx ≤ 2C δk H 1 (0,) u2x L2 (0,T ;L2 (0,)) vx L2 (0,T ;L2 (0,)),

t ∈ [0, T ]. Integrating this inequality over [0, T ] and using estimate (6.5.21) for the right hand side norm vx L2 (0,T ;L2 (0,)) we conclude that

v 2L2 (0,T ;L2 (0,)) ≤

2T C2

δk 2H 1 (0,) u2x 2L2 (0,T ;L2 (0,)). c0

(6.5.22)

Substituting this with estimate (6.5.21) in (6.5.17) we obtain: 02 δk 2 1

v(0, ·) 2L2 (0,) ≤ C

u 2 , H (0,) 2x L2 (0,T ;L2 (0,))

(6.5.23)

0 > 0 defined in (6.5.15). with C For the norm u2x L2 (0,T ;L2 (0,)) of the solution u(x, t; k2 ) of the direct problem (6.5.1) with k(x) = k2 (x) we use the estimate

u2x 2L2 (0,T ;L2 (0,)) ≤ C12 g 2L2 (0,T ) , C1 > 0, given in Lemma B.2.1 of Appendix B. Taking into account this in (6.5.23) we finally arrive at the required estimate (6.5.14). This completes the proof of the lemma.   The main consequence of this lemma is the Lipschitz continuity of the nonlinear input-output operator [·] : K ⊂ H 1 (0, ) → L2 (0, T ). Corollary 6.5.2 Let conditions of Lemma 6.5.2 hold. Then the input-output operator defined in (6.5.8) is Lipschitz continuous,

[k1 ] − [k2 ] L2 (0,T ) ≤ L0 k1 − k2 C[0,], k1 , k2 ∈ K,

(6.5.24)

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6 Inverse Problems for Parabolic Equations

with the Lipschitz constant 1 g L2 (0,) > 0. L0 = C

(6.5.25)

Due to measurement error in the measured output the exact equality [k](t) = f δ (t), t ∈ (0, T ]

(6.5.26)

in the operator equation is not possible in practice. Here f δ ∈ L2 (0, T ) is the noisy data, f − f δ L2 (0,T ) ≤ δ, δ > 0, and f (t) is the noise free measured output introduced in (6.5.2). Hence, one needs to introduce the Tikhonov functional J (k) =

 1 [k] − f δ 2 2 , k∈K L (0,T ) 2

(6.5.27)

and consider inverse coefficient problem (6.5.1)–(6.5.2) as the minimization problem ˜ J∗ = inf J (k) ˜ K k∈

(6.5.28)

for this functional on the set of admissible coefficients K. A solution k ∈ K of this minimization problem is called a quasi-solution of the inverse coefficient problem (6.5.1)–(6.5.2). Theorem 6.5.2 Assume that conditions of Lemma 6.5.2 are satisfied. Then the minimization problem (6.5.28) for the Tikhonov functional (6.5.27) has a solution in the set of admissible coefficients K ⊂ H 1 (0, ). Proof We prove the Tikhonov functional (6.5.27) is Lipschitz continuous. Indeed, the identities





 



|J (k1) − J (k2 )| = J (k1 ) + J (k2) J (k1) − J (k2 ) ,





1



J (k1) − J (k2 ) = √ [k1 ] − f L2 (0,T ) − [k2 ] − f L2 (0,T )

2 with the inequality | a − b | ≤ a − b imply that  1  |J (k1 ) − J (k2)| ≤ √ J (k1 ) + J (k2 ) [k1 ] − [k2 ] L2 (0,T ) . 2 Hence |J (k1) − J (k2 )| 1

[k1 ] L2 (0,T ) + [k2 ] L2 (0,T ) + 2 f L2 (0,T ) [k1 ] − [k2 ] L2 (0,T ) . ≤ 2

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

191

To estimate the norms [km ] L2 (0,T ) , m = 1, 2 in the above inequality we use the trace inequality (6.5.11). Then we get:  |J (k1 ) − J (k2 )| ≤ C˜ 1 g L2 (0,T ) + f L2 (0,T ) [k1 ] − [k2 ] L2 (0,T ) , with the constant C˜ 1 > 0 introduced in (6.5.12). In view of (6.5.24), this implies that the Tikhonov functional is Lipschitz continuous, and hence, is lower semicontinuous. But a lower semicontinuous functional defined on the nonempty closed convex set K, is weakly sequentially lower semicontinuous. Then by the generalized Weierstrass existence theorem (§2.5, Theorem 2.D, [156]), it has a minimum on K.   We introduce also the regularized form Jα (k) =

 1 α [k] − f δ 2 2 + k  2L2 (0,T ) , k ∈ K L (0,T ) 2 2

(6.5.29)

of the Tikhonov functional (6.5.27). Denote by kαδ ∈ K a solution of the minimization problem for this functional. The regularization (6.5.29) including the term k  L2 (0,T ) is referred as the Tikhonov regularization with a Sobolev norm or as higher-order Tikhonov regularization (see [4]). Let us assume now that the input-output operator is injective. In this case we may apply the regularization theory for nonlinear inverse problems given in [37, Chapter 10] to guarantee the convergence of subsequences to a minimum norm solution. Remark that assuming the injectivity of the input-output operator we assume that solution of the inverse problem (6.5.1)–(6.5.2) is unique. Theorem 6.5.3 Let conditions (6.5.3) hold. Assume that the input-output operator [·] : K ⊂ H 1 (0, ) → L2 (0, T ) is injective. Denote by {f δm } ⊂ L2 (0, T ) a sequence of noisy data satisfying the conditions f − f δm L2 (0,T ) ≤ δm and δm → 0, as m → ∞. If αm := α(δm ) → 0 and

2 δm → 0, as δm → 0, α(δm )

(6.5.30)

then the regularized solutions kαδmm ∈ K converge to the best approximate solution u† := A† f of Eq. (6.5.9), as m → ∞. For the general case this theorem with proof is given in [37, Chapter 10, Theorem 10.3]. Remark that this theorem is an analogue of Theorem 2.5.2, Chap. 2 for the case of the nonlinear input-output operator. As mentioned, Lemma 6.5.1 implies that the inverse coefficient problem (6.5.1)– (6.5.2) is ill-posed. Furthermore, the following example providing further insights

192

6 Inverse Problems for Parabolic Equations

Fig. 6.1 Coefficients ki (x) (left figure) and the corresponding outputs u(0, t; ki ), i = 1, 2 (right figure)

into the severely ill-posedness of this inverse problem shows that very small changes in the measured output u(0, t; k) can lead to unacceptably large perturbations in the coefficient k(x). The outputs u(0, t; ki ), i = 1, 2, plotted in the right Fig. 6.1, are obtained from the finite-element solution of the direct problem (6.5.1) for the following coefficients (Fig. 6.1): k1 (x) = 1 + 0.25 sin(πx),  1 + (k1 (ξ ) − 1) x/ξ, x ∈ [0, ξ ], k2 (x) = 1 + (k1 (ξ ) − 1) (1 − x)/(1 − ξ ), x ∈ (ξ, 1], ξ = 0.1, with g(t) = t, t ∈ [0, T ]. The figures illustrate high sensitivity of the inverse problem to changes in output datum. That is, almost indistinguishable outputs may correspond to quite different coefficients. This is a reason that inverse coefficient problems are extremely difficult to solve numerically.

6.5.3 Integral Relationship and Gradient Formula Now we derive an integral relationship that links the change δk(x) := k1 (x)−k2(x) in the coefficient to the change δu(x, t) = u(0, t; k1 ) − u(0, t; k2 ) in output. Lemma 6.5.3 Let conditions (6.5.3) hold. Denote by um (x, t) := u(x, t; km ) and um (0, t) = u(0, t; km ) the solutions of the direct problem (6.5.1) and outputs, corresponding to the given admissible coefficients km ∈ K, m = 1, 2. Then the following integral relationship holds:  0

T

 [u1 (0, t) − u2 (0, t)]q(t)dt = −

δk(x)u2x (x, t)ϕx (x, t)dxdt, T

(6.5.31) ,

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

193

where δk(x) = k1 (x) − k2 (x) and the function ϕ(x, t) = ϕ(x, t; q) solves the backward problem ⎧ ⎪ ⎨ ϕt + (k1 (x)ϕx )x = 0, (x, t) ∈ (0, ) × [0, T ), ϕ(x, T ) = 0, x ∈ (0, ), ⎪ ⎩ −k1 (0)ϕx (0, t) = q(t), ϕx (, t) = 0, t ∈ (0, T ),

(6.5.32)

with the input q ∈ L2 (0, T ). Proof The function δu(x, t) := u1 (x, t) − u2 (x, t) solves the following initial boundary value problem ⎧ ⎪ ⎨ δut − (k1 (x)δux )x = (δk(x)u2x (x, t))x , (x, t) ∈ T , (6.5.33) δu(x, 0) = 0, x ∈ (0, ), ⎪ ⎩ −k1 (0)δux (0, t) = δk(0)u2x (0, t), δux (, t) = 0, t ∈ (0, T ), where u2x (x, t) := ux (x, t; k2 ). Multiply both sides of Eq. (6.5.33) by an arbitrary function ϕ ∈ L2 (0, T ; H 1(0, )), integrate on T and use integration by parts formula multiple times. Then we get: 

l 0

(δu(x, t)ϕ(x, t))tt =T =0 dx



− 0

T

(k1 (x)δux (x, t)ϕ(x, t) − k1 (x)δu(x, t)ϕx (x, t))x= x=0 dt

 δu(x, t)[ϕt (x, t) + (k1 (x)ϕx (x, t))x ]dxdt

− T

 =−

δk(x)u2x (x, t)ϕx (x, t)dxdt T

  + δk(x) 0

x=

T

u2x (x, t)ϕ(x, t)dt

. x=0

The first left hand side integral is zero due to the initial and final conditions in (6.5.33) and (6.5.32). At x = l, the terms under the second left hand side and the last right hand side integrals drop out due to the homogeneous Neumann conditions. The third left hand side integral is also zero due to the adjoint equation ϕt (x, t)+(k1 (x)ϕx (x, t))x = 0. Taking into account the nonhomogeneous Neumann

194

6 Inverse Problems for Parabolic Equations

conditions in (6.5.32) and (6.5.33) we conclude that 

T

− 0



T

δk(0)u2x (0, t)ϕ(0, t)dt +

δu(0, t)q(t)dt 0



=−



T

δk(x)u2x (x, t)ϕx (x, t)dxdt −

δk(0)u2x (0, t)ϕ(0, t)dt. 0

T

Since the first left hand side and the last right side integrals are equal, we arrive at the required integral relationship.   Note that the backward problem (6.5.32) is a well-posed initial boundary value problem as the change of the variable t with τ = T − t shows. From Lemma B.2.1 in Appendix B, it follows that under the condition q ∈ L2 (0, T ) the estimates in this lemma are valid for the weak solution ϕ(x, t) also. Now we use the integral relationship (6.5.31) to derive the Fréchet gradient J  (k) of the Tikhonov functional J (k) =

1

[k] − f 2L2 (0,) , k ∈ K, 2

(6.5.34)

where f ∈ L2 (0, ) is the noise free measured output. Let u(x, t; k) and u(x, t; k + δk) be the solutions of the direct problem (6.5.1) corresponding to the coefficients k, k + δk ∈ K. Calculating the increment δJ (k) := J (k + δk) − J (k) of functional (6.5.34) we get: 

T

δJ (k) = 0

1 [u(0, t; k) − f (t)]δu(0, t; k)dt + δu(0, ·; k) 2L2 (0,T ) , 2

(6.5.35)

for all k, k + δk ∈ K. Now we use Lemma 6.5.3, taking k1 (x) = k(x) + δk(x) and k2 (x) = k(x). Then δu(0, t; k) := u(0, t; k+δk)−u(0, t; k). Choosing the arbitrary input q ∈ H 1 (0, T ) in the backward problem (6.5.31) as q(t) = −[u(0, t; k)−f (t)] we deduce from the integral relationship (6.5.31) that 

T 0

[u(0, t; k) − f (t)]δu(0, t; k)dt

 δk(x)ux (x, t; k)ϕx (x, t; k + δk)dxdt,

=

(6.5.36)

T

where the function ϕ(x, t; k1), k1 (x) := k(x) + δk(x) is the solution of the adjoint problem ⎧ ⎪ ⎨ ϕt + (k1 (x)ϕx )x = 0, (x, t) ∈ (0, ) × [0, T ), (6.5.37) ϕ(x, T ) = 0, x ∈ (0, ), ⎪ ⎩ −k1 (0)ϕx (0, t) = −[u(0, t; k) − f (t)], ϕx (, t) = 0, t ∈ (0, T ).

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

195

The backward problem (6.5.32) with the input q(t) = −[u(0, t; k) − f (t)] is called the adjoint problem corresponding to the direct problem (6.5.1). Using the integral relationship (6.5.36) in the right hand side of the increment formula (6.5.35) we deduce that  δJ (k) =

  T

0

ux (x, t; k)ϕx (x, t; k + δk)dt δk(x)dx

0

+

1 2



T

[δu(0, t; k)]2 dt,

(6.5.38)

0

Estimate (6.5.14) implies that the second right hand side integral in (6.5.38) is of  the order O δk 2C[0,] . Corollary 6.5.3 Assume that conditions (6.5.3) are satisfied. Then for the Fréchet gradient J  (k) of the Tikhonov functional (6.5.34) corresponding to the inverse coefficient problem (6.5.1)–(6.5.2) the following gradient formula holds: J  (k)(x) =



T

ux (x, t; k)ϕx (x, t; k)dt, k ∈ K,

(6.5.39)

0

where u(x, t; k) and ϕ(x, t; k) are the solutions of the direct problem (6.5.1) and the adjoint problem (6.5.37) corresponding to the coefficient k ∈ K. By definition (6.5.29), the gradient formula for the regularized form of the Tikhonov functional is as follows: Jα (k)(x) = (ux (x, ·; k), ϕx (x, ·; k))L2 (0,T ) + αk  (x), k ∈ K,

(6.5.40)

for a.e. x ∈ [0, ].

6.5.4 Reconstruction of an Unknown Coefficient Having now the gradient formulae (6.5.39) and (6.5.40) we may use the Conjugate Gradient Algorithm (CG-algorithm) to the inverse coefficient problem (6.5.1)– (6.5.2). However, the version of this algorithm described in Sect. 3.4 cannot be used in this case, since the formula  βn :=

J  (p(n) ), p(n)

p(n) 2



for the descent direction parameter βn will not work. The reason is that the inputoutput mapping corresponding to the inverse problem (6.5.1)–(6.5.2) is defined only for positive functions 0 < c0 ≤ k(x) ≤ c1 and the function p(n) (x) in p(n) 2

196

6 Inverse Problems for Parabolic Equations

may not be positive. As an equivalent alternative, the descent direction parameter βn > 0 will be defined from the minimum problem Fn (βn ) := inf Fn (β), Fn (β) := J (k (n) − βJ  (k (n) )), β>0

(6.5.41)

for each n = 0, 1, 2, . . . , as in Lemma 3.4.4 of Sect. 3.4. Thus, the following version of the CG-algorithm is used below in numerical solving of the inverse coefficient problem (6.5.1)–(6.5.2). Step 1. Step 2. Step 3. Step 4.

For n = 0 choose the initial iteration k (0) (x). Compute the initial descent direction p(0) (x) := J  (k (0) )(x). Find the descent direction parameter from (6.5.41). Find the next iteration k (n+1) (x) = k (n) (x) − βn p(n) (x) and compute the convergence error e(n; k (n); δ) := f δ − u(0, t; k (n) ) L2 (0,T )

Step 5.

If the stopping condition e(n; k (n) ; δ) ≤ τM δ < e(n; k (n−1) ; δ), τM > 1, δ > 0 (6.5.42)

Step 6.

holds, then go to Step 7. Set n := n + 1 and compute ⎧ (n)  (n) (n−1) (x), ⎪ ⎨ p (x) := J (k )(x) + γn p

J  (k (n) ) 2 ⎪ ⎩ γn =

J  (k (n−1) ) 2

Step 7.

and go to Step 3. Stop the iteration process.

In the case when the algorithm is applied to the regularized form of the Tikhonov functional (6.5.29), which gradient is defined by formula (6.5.40), one needs to replace in the above algorithm J (k (n) ) and J  (k (n) ) with Jα (k (n) ) and Jα (k (n) ), respectively. For discretization and numerical solution of the direct problem (6.5.1) as well as the adjoint problem (6.5.37) the finite element algorithm with piecewise quadratic Lagrange basis functions, introduced in Sect. 3.4.1, is used. These schemes with composite numerical integration formula are also used for approximating the spatial derivatives ux (x; t; k) and ϕx (x; t; k) in the gradient formulas (6.5.39) and (6.5.40). In the examples below the finer mesh with the mesh parameters Nx = 201 and Nt = 801 are used to generate the noise free synthetic output. For this mesh the computational noise level is estimated as 10−4 . With this accuracy the synthetic output f is assumed noise-free. Coarser mesh with the parameters Nx = 51 and

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output

197

Nt = 101 is used in the numerical solution of the inverse problem, to avoid of inverse crime. The noisy output f δ , with f − f δ L2 (0,T ) = δ, is generated by employing the h “randn” function in MATLAB, that is, uδT ,h (x) = uT ,h (x) + γ uT ,h L2 (0,1)randn(N), h

where γ > 0 is the MATLAB noise level. Remark that γ > 0 and δ > 0 are of the same order. The parameter τM > 1 in the stopping condition (6.5.42) is taken below as τM = 1.05 ÷ 1.07, where n is the iteration number. We will employ also the accuracy error defined as E(n; k (n) ; δ) := k − k (n) L2 (0,T ) , for performance analysis of the CG-algorithm. In the examples below, we represent attempts to capture performance characteristics of the CG-algorithm not only in the case when the initial datum and the Neumann boundary datum at x = l are homogeneous, but also in the case when the direct problem is given in the general form: ⎧ ⎨ ut (x, t) = (k(x)ux (x, t))x , (x, t) ∈ T , (6.5.43) u(x, 0) = h(x), 0 < x < , ⎩ −k(0)ux (0, t) = g0 (t), − k(0)ux (, t) = g1 (t), 0 < t < T . In this case the input must satisfy the following consistency conditions: 

g0 (0) = −k(0)h (0), g1 (0) = −k()h (),

due to the regularity of the solution. Since the fluxes g0 (t) and g1 (t) at the endpoints initially are assumed to be positive and since k(x) ≥ c1 > 0, the above consistency conditions imply that 

g0 (t) > 0, g1 (t) > 0, for all t ∈ [0, T ], h (0) < 0, h () < 0.

(6.5.44)

On the other hand, these conditions allow to find the values k(0) and k() of the unknown coefficients at the endpoints: 

k(0) = −g0 (0)/ h (0), h (0) < 0 k() = −g1 (0)/ h (), h () < 0.

(6.5.45)

This, in turn, permits to use the function k (0)(x) = (k(0)( − x) + k()x) /, i.e. the linear approximation of k(x), as an initial iteration in the CG-algorithm,

198

6 Inverse Problems for Parabolic Equations

also. Otherwise, i.e. in the case when g1 (t) ≡ 0, h(x) ≡ 0 and g0 (t) satisfies conditions (6.5.3), i.e. g0 (t) > 0, for all t ∈ (0, T ) and g0 (0) = 0, any constant function k (0)(x) = k0 , c0 ≤ k0 ≤ c1 can be taken as an initial iteration, where c0 > 0 and c1 > 0 are lower and upper bounds for the coefficient k(x). Note that an initial iteration has indistinguishable effect on the reconstruction quality, as computational experiments show. Only the attainability of the stopping criterion (6.5.42) becomes faster. Finally, remark that reasonable values for the parameter of regularization α > 0 in the examples below were determined by numerical experiments on carefully chosen examples, of course, taking into account the convergence conditions (6.5.30). Example 6.5.1 Performance characteristics of the CG-algorithm: noise free output The synthetic output datum f (t) in this example is generated from the numerical solution of the direct problem (6.5.1) for the given k(x) = 1 + 0.5 sin(3πx/2), x ∈ [0, 1], with the source g(t) = 20 exp(−300(t − 0.25)2 ), t ∈ [0, T ], T = 0.5, where g(0) ≈ 1.44 × 10−7 . The initial iteration here is taken as k (0) (x) = 1, x ∈ [0, 1]. The right Fig. 6.2 shows the behaviour of the convergence error (bottom curve) and the accuracy error (upper curve) depending on the iteration number n. It is clearly seen that the accuracy error E(n; k (n) ; δ) remains almost the same after 30 ÷ 35th iterations, although the convergence error e(n; k (n) ; δ) still decreased after 80 iterations. For some values of the iteration number n these errors are reported in Table 6.1. The reconstructed coefficients k (n) (x) from noise free data are plotted in the left Fig. 6.2, for the iteration numbers n = 10; 50; 100. It is seen that after 50 iterations the reconstructions are almost the same. In all these reconstructions the CG-algorithm is applied without regularization (α = 0).  100

k(x)

1.5

E(n;α;γ), α=0, γ =0

k(n)(x), n=10

1.4

e(n;α;γ), α=0, γ =0

k(n)(x), n=50 k(n)(x), n=100

1.3

10–1

1.2 1.1 10–2

1 0.9 0.8

10–3

0.7 0.6 0.5

0

0.2

0.4

0.6 x

0.8

1

10–4 0

20

40

60 iterations

80

100

Fig. 6.2 The reconstructed coefficients for different values of the iteration number n (left figure) and behaviour of the convergence and accuracy errors depending on n (right figure): noise free data, for Example 6.5.1

6.5 Identification of a Leading Coefficient From Dirichlet Measured Output Table 6.1 Errors depending on the number of iterations n: noise free output (δ = 0), for Example 6.5.1

n 25 50 75 100

e(n; α; γ ) 1.90 × 10−3 2.39 × 10−4 1.75 × 10−4 1.36 × 10−4

199

E(n; α; γ ) 5.26 × 10−2 4.94 × 10−2 4.70 × 10−2 4.71 × 10−2

Example 6.5.2 Reconstruction of an unknown coefficient from noisy data with and without regularization In this example the reconstruction of the function k(x) = 1 + 0.5 sin(3πx/2), x ∈ [0, 1], is considered in the case when the inputs in the direct problem (6.5.43) are g0 (t) = 10 exp(−300(t − 0.04)2), g0 (t) = 20 exp(−300(t − 0.02)2), t ∈ [0, T ], T = 0.5, and h(x) = Ax 2 + Bx, x ∈ [0, 1]. For the Neumann data g0 (0) ≈ 6.1878 and g1 (0) ≈ 17.7384, and the parameters A and B in initial datum h(x) is chosen from the consistency conditions: h (0) = −g0 (0)/k(0), h (1) = −g1 (0)/k(1). We have: h(x) = (1/2)[h (1) − h (0)]x 2 + h (0)x. The CG-algorithm is applied to the inverse problem defined by (6.5.43) and (6.5.2) with the initial iteration k (0)(x) = (k(0)(l − x) + k(l)x) / l. The reconstructed coefficients from noise free and noisy data are plotted in the left Fig. 6.3. In the noise-free case the reconstructed coefficient almost coincides with the exact one. Influence of the parameter of regularization α > 0 on accuracy of reconstruction is seen from the plots in the left Fig. 6.3. The values of the convergence and accuracy errors are given in Table 6.2. In the case of noisy data the iteration numbers n in this table is determined by the stopping condition (6.5.42) with τM = 1.05.  Example 6.5.3 Reconstruction of Gaussian from noisy data with and without regularization

Fig. 6.3 The reconstructed coefficient from noisy data with and without regularization: left figure, for Example 6.5.2 and the right figure, for Example 6.5.3

200

6 Inverse Problems for Parabolic Equations

Table 6.2 Errors obtained for noise free and noisy output, for Example 6.5.2 δ, α δ=0 δ = 3%, α = 0 δ = 3%, α = 5.0 × 10−2

Iteration 86 3 3

e(n; α; γ ) 1.40 × 10−3 9.75 × 10−2 1.03 × 10−1

E(n; α; γ ) 4.28 × 10−2 1.46 × 10−1 1.59 × 10−1

Table 6.3 Errors in the reconstruction of Gaussian, for Example 6.5.3 δ, α δ=0 δ = 1%, α = 0 δ = 1%, α = 5.0 × 10−2

n 41 9 55

e(n; α; γ ) 6.70 × 10−3 1.89 × 10−2 1.95 × 10−2

E(n; α; γ ) 2.05 × 10−1 1.56 × 10−1 2.98 × 10−1

In this final example, we use the input from the previous √ example for the reconstruction of the k(x) = 1 + exp(−(x − 0.5)2/(2σ 2 ))/(σ 2π), with mean σ = 0.3, in the inverse coefficient problem defined by (6.5.43) and (6.5.2). The constant k (0)(x) = 2 is used as an initial iteration. The reconstructed coefficients from noise free and noisy data are plotted in the right Fig. 6.3. Table 6.3 reports the values of the convergence and accuracy errors with the iteration numbers n.  We remark finally that, in all examples the regularization effect on the accuracy of the obtained reconstructions was negligible. That is, an accuracy of the numerical solution obtained by CG-algorithm has slightly improved due to the regularization, but at the expense of the number of iterations, as Table 6.3 shows.

6.6 Identification of a Leading Coefficient From Neumann Measured Output In the previous section we studied the inverse coefficient problem where the measured output was Dirichlet type. In applications, however, there are cases where the heat flux f (t) := −k(0)ux (0, t; k), i.e. the Neumann measured output, is specified as the most available measured output. This is an important, but more complicated case since this output contains the derivative which is an additional source of an ill-posedness. Applying the methodology given in the previous section to the inverse coefficient problem with Neumann measured output, we will also emphasize some distinctive features of this problem. Consider the inverse problem of identifying the leading coefficient k(x) in ⎧ ⎨ ut (x, t) = (k(x)ux (x, t))x + F (x, t), (x, t) ∈ T , u(x, 0) = 0, 0 < x < , ⎩ u(0, t) = 0, ux (, t) = 0, 0 < t < T ,

(6.6.1)

6.6 Identification of a Leading Coefficient From Neumann Measured Output

201

from the Neumann measured output (heat flux) f (t) at the left boundary x = 0 of a nonhomogeneous rod: f (t) := −k(0)ux (0, t), t ∈ [0, T ].

(6.6.2)

At the first stage, assume that the inputs k(x) and F (x, t) satisfy the following basic conditions:  k ∈ H 1 (0, ), 0 < c0 ≤ k(x) ≤ c1 , (6.6.3) F ∈ L2 (0, T ; L2 (0, )). Under these conditions there exists a unique weak solution of the initial boundary value problem (6.6.1), defined as u ∈ L2 (0, T ; V(0, )) with ut ∈ L2 (0, T ; H −1 (0, )), where V(0, ) := {v ∈ H 1 (0, ) : v(0) = 0} [38, §7.1, Theorem 5]. If in addition to (6.6.3) the source function F (x, t) satisfies also the condition Ft ∈ L2 (0, T ; L2 (0, )),

(6.6.4)

then there exists a unique regular weak solution of the initial boundary value problem (6.6.1), defined as u ∈ L2 (0, T ; H 2 (0, )) ∩ L∞ (0, T ; V(0, )) with ut ∈ L2 (0, T ; L2 (0, )). We need some necessary estimates for the weak and regular weak solutions. Lemma 6.6.1 Assume that the inputs in (6.6.1) satisfy the basic conditions (6.6.3). Then for the weak solution of problem (6.6.1) the following estimates hold:

u 2L∞ (0,T ;L2 (0,)) ≤ Ce2 F 2L2 (0,T ;L2 (0,)),  

u 2L2 (0,T ;L2 (0,)) ≤ Ce2 + 1 F 2L2 (0,T ;L2 (0,)),

(6.6.5)

#2 F 2 2 ,

ux 2L2 (0,T ;L2 (0,)) ≤ C 1 L (0,T ;L2 (0,)) #2 = (Ce2 + 1)/(2c0 ), and c0 > 0 is the constant introduced where Ce2 = exp(T ), C 1 in (6.6.3). Furthermore, if in addition to (6.6.3), condition (6.6.4) also holds, then for the regular weak solution of problem (6.6.1) the following estimates hold:

ut 2L∞ (0,T ;L2 (0,)) ≤ Ce2 C˜ T2 F 2H 1 (0,T ;L2 (0,)),  

ut 2L2 (0,T ;L2 (0,)) ≤ Ce2 + 1 C˜ T2 F 2H 1 (0,T ;L2 (0,)), e2 C˜ 2 F 2 1

uxt 2L2 (0,T ;L2 (0,)) ≤ C , T H (0,T ;L2 (0,)) where C˜ T2 = max(2/T , 1 + 2/(3T )).

(6.6.6)

202

6 Inverse Problems for Parabolic Equations

Estimates (6.6.5) and (6.6.6) follow from the corresponding estimates given in Lemmas B.1.1 and B.1.2 of Appendix B, with u0 (x) = 0. We also need the following estimates for the regular weak solution with improved regularity of problem (6.6.1). Lemma 6.6.2 Assume that in addition to the basic conditions (6.6.3) the inputs in (6.6.1) satisfy the following regularity conditions k ∈ H 2 (0, ), F ∈ H 2 (0, T ; H 2 (0, )).

(6.6.7)

Then for the regular weak solution with improved regularity u ∈ L∞ (0, T ; H 2 (0, )), with ut ∈ L2 (0, T ; V(0, )) and ut t ∈ L2 (0, T ; H −1 (0, )) of problem (6.6.1) the following estimates hold: #2 F 2 2

ut t 2L2 (0,T ;L2 (0,)) ≤ C , 3 H (0,T ;H 2 (0,))

(6.6.8)

#2 F 2 2

uxt t 2L2 (0,T ;L2 (0,)) ≤ C , 4 H (0,T ;H 2 (0,)) where   #2 = max(1/T , 4T /3 + 1) max(1, C #2 ) Ce2 + 1 , C 3 2

(6.6.9)

#2 = Ce2 C #2 /(2c0 ), C #2 = C 2 k H 2 (0,) + c2 , C 4 2 2  1

C > 0 is the constant defined in inequality (6.5.15), c0 , c1 > 0 are the constants introduced in (6.6.3) and Ce = exp(T ). Proof Differentiate equation (6.6.1) two times with respect to t ∈ (0, T ), multiply both sides by 2ut t (x, t) and integrate then over t := (0, ) × (0, t), t ∈ (0, T ]. Applying the integration by parts formula we obtain the following integral identity:  0



u2t t dx + 2

 t 0



0

k(x)u2xt t dxdτ = 2  + 0



 t



Ft t (x, τ )uτ τ dxdτ 0

0

u2t t (x, 0+ )dx, a. e. t ∈ [0, T ].

(6.6.10)

  We use the (formal) limit equation ut t (x, 0+ ) = k(x)ux (x, 0+ ) xt + Ft (x, 0+ ) at t = 0+ and inequality (6.5.20) to estimate the last right-hand-side integral: 

 0

u2t t (x, 0+ )dx =  ≤2 0

 



 

k(x)ux (x, 0+ )

0

k(x)ut x (x, 0+ )



2 x

 xt

+ Ft (x, 0+ ) dx 2

 dx + 2 0



Ft2 (x, 0+ )dx,

6.6 Identification of a Leading Coefficient From Neumann Measured Output

203

  a.e. t ∈ [0, T ]. Here we use again the limit equation ut (x, 0+ ) = k(x)ux (x, 0+ ) x + F (x, 0+ ) at t = 0+ and then employ the analogue max |k  (x)|2 ≤ C2 k 2H 2 (0,),

x∈[0,]

of inequality (6.5.20) with the same constant C > 0. Then we obtain: 



0

u2t t (x, 0+ )dx



 

≤2

+

k(x)Fx (x, 0 )



0

  ≤ 2 C2 δk 2H 1 (0,) + c12

 0

2

x



+2



0

Ft2 (x, 0+ )dx

0

Fx2 (x, 0+ )dx + 



dx + 2



 0

2 Fxx (x, 0+ )dx



Ft2 (x, 0+ )dx, t ∈ [0, T ].

This inequality with identity (6.6.10) leads to the following inequality: 

 0



 u2t t dx + 2c0

#22 +2C



 0

t

u2xt t dxdτ ≤

Fx2 (x, 0+ )dx





+ 0

 t

u2τ τ dxdτ +

2 Fxx (x, 0+ )dx



T





+2 0

Ft2t (x, t)dxdt

Ft2 (x, 0+ )dx, (6.6.11)

#2 > 0 is the constant introduced in (6.6.9). a.e. t ∈ [0, T ], where C We estimate now the last three right-hand-side integrals through appropriate norms of F (x, t). To this end, we use the identity 

 0

Fx2 (x, 0+ )dx =





−

0

1 T



T 0

2 ((T − t)Fx (x, t))t dt

dx,

for the first right-hand-side integral to deduce that 

 0

Fx2 (x, 0+ )dx



1 T ≤2

Fx L2 (0,T ;L2 (0,)) + Fxt L2 (0,T ;L2 (0,)) T 3

2 dx.

Estimates for the next two right-hand-side integrals are derived in the same way. Using these estimates in (6.6.11), after elementary transformations we arrive at the main integral inequality 

 0



 u2t t dx

+ 2c0 t

u2xt t dxdτ

≤ t

u2τ τ dxdτ

#32 F 2 2 +C , a.e. t ∈ [0, T ], H (0,T ;H 2 (0,)) with the constant defined in (6.6.9).

(6.6.12)

204

6 Inverse Problems for Parabolic Equations

The first consequence of (6.6.12) is the inequality 

 0

u2t t dx

≤

 t 0

 0

#32 F 2 2 u2τ τ dxdτ + C , t ∈ [0, T ]. H (0,T ;H 2 (0,))

Applying Lemma B.0.1 we obtain: 

 0

#32 F 2 2 u2t t dx ≤ C exp(t), t ∈ [0, T ]. H (0,T ;H 2 (0,))

(6.6.13)

The first estimate of (6.6.8) follows from this inequality. As the second consequence of (6.6.11) we have:  t



2c0 0

0

u2x dxdτ ≤

 t 0

0



#32 F 2 2 u2τ τ dxdτ + C , t ∈ [0, T ]. H (0,T ;H 2 (0,))

With inequality (6.6.7) this leads to the second estimate of (6.6.8).

 

6.6.1 Compactness of the Input-Output Operator We introduce the set of admissible coefficients K1 := {k ∈ H 1 (0, ) : 0 < c0 ≤ k(x) ≤ c1 }, K2 := {k ∈ H 2 (0, ) : 0 < c0 ≤ k(x) ≤ c1 },

(6.6.14)

in H 1 (0, ) and H 2 (0, ), respectively. Denote by u = u(x, t; k) the solution of the direct problem (6.6.1) for a given coefficient k ∈ Km . Introduce the input-output map as follows: 

[k](t) := − (k(x)ux (x, t; k))x=0+ , [·] : Km ⊂ H m (0, ) → L2 (0, T ), m = 1, 2.

(6.6.15)

Lemma 6.6.3 Assume that conditions of Lemma 6.6.2 are satisfied. Then the inputoutput map [·] : K2 ⊂ H 2 (0, ) → L2 (0, T ) defined in (6.6.15) is a compact operator. Proof Let {km } ⊂ K2 , m = 1, ∞, be a sequence of admissible coefficients. Denote by {um (x, t)}, um (x, t) := u(x, t; km ), the sequence of corresponding regular weak solutions with improved regularity of the direct problem (6.6.1). Then {− (k(x)ux (x, t; k))x=0+ } is the sequence of outputs. We need to prove that this sequence is a relatively compact subset of L2 (0, T ) or, equivalently, the sequence {km (0)um (0, t)} is bounded in the norm of the Sobolev space H 1 (0, T ).

6.6 Identification of a Leading Coefficient From Neumann Measured Output

205

By the homogeneous Neumann condition in (6.6.1) we have: 

T



km (0)um,x (0, t)

2



T

dt =

0



2



(km (x)um,x (x, t))x dx 0

dt.

0

Using here the equation (km (x)um,x (x, t))x = um,t (x, t) − F (x, t) we obtain:

km (0)um,x (0, ·) 2L2 (0,T ;L2 (0,))  ≤ 2l um,t 2L2 (0,T ;L2 (0,)) + F 2L2 (0,T ;L2 (0,)) . In the same way we deduce that

km (0)um,xt (0, ·) 2L2 (0,T ;L2 (0,))  ≤ 2l um,t t 2L2 (0,T ;L2 (0,)) + Ft 2L2 (0,T ;L2 (0,)) .

(6.6.16)

By estimates (6.6.6) and (6.6.8) we conclude that the sequence of norms { km (0)um,x (0, ·) L2 (0,T ;L2 (0,))} and { km (0)um,xt (0, ·) L2 (0,T ;L2 (0,))} are uniformly bounded in L2 (0, T ). Hence the sequence of outputs is bounded in the norm of H 1 (0, ). As a consequence of this, is relatively compact in L2 (0, ). This implies the compactness of the input-output operator.   Remark 6.6.1 In Sect. 6.5.2 we have proved that the input-output operator [·] : K ⊂ H 1 (0, ) → L2 (0, T ) corresponding to the inverse coefficient problem (6.5.1)–(6.5.2) with Dirichlet measured output is compact if D() := K ⊂ H 1 (0, ). Lemma 6.6.3 asserts that in the case of the Neumann measured output, the corresponding input-output operator  is compact if D() := K ⊂ H 2 (0, ), that is the admissible coefficients are defined in H 2 (0, ) ⊂ H 1 (0, ).

6.6.2 Lipschitz Continuity of the Input-Output Operator and Solvability of the Inverse Problem Based on the input-output operator, we can formulate the inverse coefficient problem (6.6.1)–(6.6.2) as the following nonlinear operator equation: [k](t) = f (t), t ∈ (0, T ].

(6.6.17)

Due to measurement errors, the exact equality in (6.6.17) is not achieved in practice. Hence we introduce the Tikhonov functional J (k) =

1

[k] − f 2L2 (0,T ) , k ∈ K1 2

(6.6.18)

206

6 Inverse Problems for Parabolic Equations

and reformulate the inverse problem (6.6.17) as the problem of funding k∗ ∈ K1 such that J (k∗ ) = J∗ , where J∗ = inf J (k). k∈K1

(6.6.19)

A solution of the minimization (6.6.19) is called a quasi-solution of the inverse problem (6.6.1)–(6.6.2). To prove the Lipschitz continuity of the input-output operator [·] : K1 ⊂ H 1 (0, ) → L2 (0, T ) we need the following auxiliary result. Lemma 6.6.4 Assume that conditions (6.6.3) and (6.6.4) which guarantee existence of the regular weak solution of the direct problem (6.6.1) are satisfied. Denote by u1 (x, t) := u(x, t; k1 ) and u2 (x, t) := u(x, t; k2 ) the solutions of this problem corresponding to the admissible coefficients k1 , k2 ∈ K1 . Then the following estimate holds: #52 k1 − k2 2 1

u1,t − u2,t 2L2 (0,T ;L2 (0,)) ≤ C , H (0,)

(6.6.20)

where e2 C˜ T2 F 2 1 #52 = 2T C2 C , C H (0,T ;L2 (0,)) c0

(6.6.21)

e , C˜ T > 0 are the constants defined in Lemma 6.5.2. and c0 , C , C Proof The function by v(x, t) := u(x, t; k1 ) − u(x, t; k2 ) solves the following initial boundary value problem: ⎧ ⎨ vt = (k1 (x)vx )x + (δk(x)u2x )x , (x, t) ∈ T , v(x, 0) = 0, x ∈ (0, ), ⎩ v(0, t) = 0, vx (, t) = 0, t ∈ (0, T ),

(6.6.22)

where δk(x) = k1 (x) − k2 (x). Differentiate equation (6.6.22) with respect to t ∈ (0, T ), multiply both sides by vt (x, t) and integrate over t . Applying the integration by parts formula and using the homogeneous initial and boundary conditions we obtain the following integral identity: 

 0

 vt2 dx



+2 t

2 k1 (x)vxτ dxdτ

= −2

δk(x)u2xτ vxτ dxdτ, t ∈ [0, T ]. t

6.6 Identification of a Leading Coefficient From Neumann Measured Output

207

This identity leads to the integral inequality 

 0

 vt2 dx

+ 2c0 t

2 vxτ dxdτ

≤ 2 max |δk(x)| u2xt L2 (0,T ;L2 (0,)) vxt L2 (0,T ;L2 (0,)),

(6.6.23)

x∈[0,]

t ∈ [0, T ]. This is the same integral inequality (6.5.18) with u2x (x, t) and vx (x, t) replaced by u2xt (x, t) and vxt (x, t), respectively. In the same way as in the derivation of inequality (6.5.22), we deduce from (6.6.23) that

vt 2L2 (0,T ;L2 (0,)) ≤

2T C2

δk 2H 1 (0,) u2xt 2L2 (0,T ;L2 (0,)). c0  

With the third estimate of (6.6.6) this yields the required estimate (6.6.20).

Theorem 6.6.1 Let conditions of Lemma 6.6.4 hold. Then the input-output operator [·] : K1 ⊂ H 1 (0, ) → L2 (0, T ) is Lipschitz continuous, that is,

[k1] − [k2] L2 (0,T ) ≤ L k1 − k2 H 1 (0,), with the Lipschitz constant L = in (6.6.21).

(6.6.24)

√ #5 , where C #5 > 0 is the constant introduced lC

Proof Use the identities 



−ki (0)um,x (0, t) =





um,t (x, t)dx −

0

F (x, t)dx, m = 1, 2

0

for the regular weak solutions um (x, t) := u(x, t; km ), m = 1, 2 of the direct problem (6.6.1) to deduce that 

[k1 ] − [k2 ] 2L2 (0,T ) :=

T

 k1 (0)u1,x (0, t) − k2 (0)u2,x (0, t)

2

dt

0



T

≤l 0





u1,t (x, t) − u2,t (x, t)

2

dxdt.

0

With estimate (6.6.20) this yields the assertion of the theorem.

 

As in Theorem 6.5.2, we can prove that the Lipschitz continuity of the inputoutput operator implies the Lipschitz continuity of the Tikhonov functional, and this, in turn, implies the existence of a quasi-solution.

208

6 Inverse Problems for Parabolic Equations

Theorem 6.6.2 Assume that conditions of Lemma 6.6.4 are satisfied. Then the minimization problem (6.6.19) for the Tikhonov functional (6.6.18) has a solution in the set of admissible coefficients K1 ⊂ H 1 (0, ).

6.6.3 Integral Relationship and Gradient Formula Having above mathematical framework we can derive now the Fréchet gradient J  (k) of the Tikhonov functional (6.6.18). Let k, k + δk ∈ K1 be admissible coefficients. For convenience, we use the notations # k(x) = k(x) + δk(x), u(x, t) := u(x, t; k), # u(x, t) := u(x, t; # k) the corresponding solutions of the direct problem (6.6.1). Then −k(0)ux (0, t) and −# k(0)# ux (0, t) are the Neumann outputs. Furthermore, the increment δJ (k) := J (# k) − J (k) of the Tikhonov functional can be written as follows: 

T

δJ (k) =

 k(0)# ux (0, t) − k(0)ux (0, t) dt [k(0)ux (0, t) + f (t)] #

0

+

1 2



T

 # k(0)# ux (0, t) − k(0)ux (0, t)

2

dt.

(6.6.25)

0

We derive an important integral identity that expresses the relationship between the change δk(x) in the coefficient and the change [k + δk](t) − [k](t) := −# k(0)# ux (0, t) + k(0)ux (0, t) ≡ −k(0)δux (0, t) − δk(0)# ux (0, t), t ∈ (0, T )

(6.6.26)

in the Neumann output. Lemma 6.6.5 Assume that basic conditions (6.6.3) are satisfied. Then the following integral relationship holds: 

T

 −# k(0)# ux (0, t) + k(0)ux (0, t) p(t)dt

0

 =

δk(x)# ux (x, t)ψx (x, t)dxdt,

(6.6.27)

T

where the function ψ(x, t) = ψ(x, t; p) solves the following backward problem ⎧ ⎪ ⎨ ψt + (k(x)ψx )x = 0, (x, t) ∈ (0, ) × [0, T ), ψ(x, T ) = 0, x ∈ (0, ), ⎪ ⎩ ψ(0, t) = p(t), ψ (, t) = 0, t ∈ (0, T ), x with an arbitrary Dirichlet input satisfying the condition p ∈ H 1 (0, T ).

(6.6.28)

6.6 Identification of a Leading Coefficient From Neumann Measured Output

209

Proof Evidently the function δu(x, t) solves the initial boundary value problem ⎧ ⎪ ux )x , (x, t) ∈ T , ⎨ δut = (k(x)δux )x + (δk(x)# δu(x, 0) = 0, x ∈ (0, ), ⎪ ⎩ δu(0, t) = 0, δu (, t) = 0, t ∈ (0, T ). x

(6.6.29)

Multiply both sides of Eq. (6.6.29) by an arbitrary function ψ(x, t), integrate on T and use integration by parts formula multiple times. Then we obtain: $$

T [−ψt (x, t) − (k(x)ψx (x, t))x ] δu(x, t)dxdt $l + 0 (δu(x, t)ψ(x, t))tt =T =0 dx $T − 0 (k(x)δux (x, t)ψ(x, t) − k(x)δu(x, t)ψx (x, t))x= x=0 dt

=−

$$

T

x= $T δk(x)# ux (x, t)ψx (x, t)dxdt + δk(x) 0 # ux (x, t)ψ(x, t)dt . x=0

We assume now the function ψ(x, t) is the weak solution of the backward problem (6.6.28). Then taking into account the boundary, initial and final conditions in (6.6.28) and (6.6.29) we deduce that  −

T

ux (0, t) + k(0)δux (0, t)] p(t)dt [δk(0)#

0

 =

δk(x)# ux (x, t)ψx (x, t)dxdt T

With the identity (6.6.26) this leads to the integral relationship (6.6.27).

 

Given the form of the first right-hand-side integral in (6.6.25), we choose the Dirichlet input in (6.6.28) as follows: p(t) = − [k(0)ux (0, t) + f (t)] , t ∈ (0, T ).

(6.6.30)

The backward problem (6.6.28) with the above defined Dirichlet input is called the adjoint problem corresponding to the inverse coefficient problem (6.6.1)–(6.6.2). Substituting (6.6.30) in (6.6.27) we obtain the following input-output relationship: 

T

 k(0)# ux (0, t) − k(0)ux (0, t) dt [k(0)ux (0, t) + f (t)] #

0

 =

δk(x)# ux (x, t)ψx (x, t)dxdt, T

(6.6.31)

210

6 Inverse Problems for Parabolic Equations

Note that substitution (6.6.30) is still formal and subject to justification. Indeed, from Lemma B.3.1 of Appendix B it follows that the weak solution ψ ∈ L2 (0, T ; H 1 (0, )) of the adjoint problem exists if p ∈ H 1 (0, T ). In view of the substitution (6.6.30) this means the Neumann output −k(0)ux (0, t) and the measured output f (t) should satisfy the following conditions ux (0, ·) ∈ H 1 (0, T ), f ∈ H 1 (0, T ).

(6.6.32)

The identity (6.6.16) shows that the first condition of (6.6.32) holds, if u(x, t) is a regular weak solution with improved regularity of the problem (6.6.1). The second condition f ∈ H 1 (0, T ) means that we have to deal with a slightly smoothed Neumann measured output. With this justification of the input-output relationship (6.6.31), we can derive the Fréchet gradient J  (k) of the Tikhonov functional (6.6.18). Theorem 6.6.3 Assume that in addition to the basic conditions (6.6.3), the regularity conditions (6.6.7) hold. Suppose also that the Neumann measured output satisfies the condition f ∈ H 1 (0, T ). Then the Tikhonov functional (6.6.18) is Fréchet differentiable. Furthermore, for the Fréchet gradient J  (k) of the Tikhonov functional the following gradient formula holds: 

J  (k)(x) =

T

ux (x, t; k)ψx (x, t; k)dt, k ∈ K2 ,

(6.6.33)

0

where ψ(x, t; k) is the solution of the adjoint problem ⎧ ⎪ ⎨ ψt + (k(x)ψx )x = 0, (x, t) ∈ (0, ) × [0, T ), ψ(x, T ) = 0, x ∈ (0, ), ⎪ ⎩ ψ(0, t) = −[k(0)ux (0, t; k) − f (t)], ψx (, t) = 0,

t ∈ (0, T ).

Proof Formula (6.6.25) for the increment of the Tikhonov functional with the inputoutput relationship (6.6.31) implies that  δJ (k) =

δk(x)# ux (x, t)ψx (x, t)dxdt T

1 + 2



T 0

 2 # k(0)# ux (0, t) − k(0)ux (0, t) dt.

6.6 Identification of a Leading Coefficient From Neumann Measured Output

211

or, considering that # ux (x, t) = ux (x, t) + δux (x, t),

  T

 δJ (k) =

ux (x, t)ψx (x, t)dt δk(x)dx 0

0



+ 1 + 2

δk(x)δux (x, t)ψx (x, t)dxdt 

T T

 2 # k(0)# ux (0, t) − k(0)ux (0, t) dt.

(6.6.34)

0

In exactly the same way as in the proof of Lemma 6.6.4, one can prove that for the weak solution of problem (6.6.29) the following estimate holds: #62 δk 2 1 #6 > 0. ,C

δux 2L2 (0,T ;L2 (0,)) ≤ C H (0,) As a consequence of this estimate, the second right-hand-side integral in (6.6.34) is of the order O( δk 2H 1 (0,) ). Furthermore, the third right-hand-side integral is of the same by the Lipschitz continuity of the input-output operator, i.e. by (6.6.24). This leads to the required result.   Comparing the gradient formulas for the inverse coefficient problems with Dirichlet and Neumann measured outputs, we see that outwardly they are the same convenient form for computational purposes, and represent the scalar products (ux , φx )L2 (0,T ) and (ux , ψx )L2 (0,T ) of the direct and corresponding adjoint problems solutions. However, if we look deeper, we will see that the inverse coefficient problem with Neumann measured output requires more regularity from the coefficient k(x), from the input F (x, t), and also from the measured output f (t) parameter, simultaneously.

Chapter 7

Inverse Problems for Elliptic Equations

This chapter is an introduction to the basic inverse problems for elliptic equations. One class of these inverse problems arises when the Born approximation is used for scattering problem in quantum mechanics, acoustics or electrodynamics. In the first part of this chapter two inverse problems, the inverse scattering problem at a fixed energy and the inverse scattering problems at a fixed energy, are studied. The last problem is reduced to the tomography problem which is studied in the next chapter. In the second part of the chapter, the Dirichlet-to-Neumann operator is introduced. It is proved that this operator uniquely defines the potential q(x) in u(x)+q(x) = 0.

7.1 The Inverse Scattering Problem at a Fixed Energy Consider the stationary Schrödinger equation − u + q(x)u = |k|2 u,

x ∈ R3

(7.1.1)

of quantum mechanics at a fixed energy E = |k|2 . We look for a solution of this equation of the form: u(x, k) = eik·x + v(x, k).

(7.1.2)

Here q(x) is the potential and the vector k = (k1 , k2 , k3 ) defines the direction of the incident plane wave eik·x . Function v(x, k) satisfies the radiation conditions

 v = O r −1 ,

 ∂v − i|k|v = o r −1 , as r = |x| → ∞. ∂r

(7.1.3)

The boundary value problem (7.1.1)–(7.1.3) will be referred below as the direct problem. © Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_7

213

214

7 Inverse Problems for Elliptic Equations

Equations similar to (7.1.1) arise also in acoustics and electrodynamics. If q(x) = 0, then v(x, k) = 0. For q(x) ≡ 0 the solution to the direct problem (7.1.1)– (7.1.3) defines a scattering wave on the potential. It is well known that the problem (7.1.1)–(7.1.3) is well posed, if q(x) ∈ L∞ (R3 ) and q(x) decreases sufficiently rapidly, as r → ∞. So, it has the unique solution. For simplicity, we will assume here that the potential q(x) is a finite function with the compact domain ⊂ R3 , having smooth boundary ∂ . Moreover, we assume that q(x) ∈ C( ) and q(x) = 0 for x ∈ ∂ . It follows from (7.1.1) and (7.1.2) that the function v(x, k) satisfies the equation − v + q(x)(eik·x + v) = |k|2 v,

x ∈ R3

(7.1.4)

and the conditions (7.1.3). On the other hand, v(x, k) is solution of the integral equation v(x, k) = −

1 4π



q(y)(eik·y + v(y, k))ei|k||x−y| dy. |x − y|

(7.1.5)

Consider the asymptotic behavior of the function v(x, k), as |x| = r → ∞, so that x/r = l/|k|, where l ∈ R3 is an arbitrary vector with |l| = |k|. We have |x − y| = r − (l · y)/|k| + o(r −1 ) uniformly for all y ∈ . Then from (7.1.5) we get  ei|k|r q(y)(eik·y + v(y, k))e−il·y dy v(x, k) = − 4πr  1 +O 2 , r → ∞. r

(7.1.6)

From (7.1.6) we see that v(x, k) can be represented in the form  1 ei|k|r f (k, l) + O 2 , r → ∞. v(x, k) = − 4πr r

(7.1.7)

The function f (k, l) is called the scattering amplitude. The inverse scattering problem here is to determine the unknown √ coefficient q(x) in (7.1.4) from given f (k, l) for all k and l, such that |k| = |l| = E. This problem has been studied in the various papers and books (see, for example, the books [28, 112, 126] and papers [39, 40, 65, 115, 116]). Following [118], consider here the problem of a reconstruction of the potential q(x) in the Born approximation using scattering amplitude at a fixed energy E. We assume that q(x) is sufficiently small and we can neglect by the second order terms. So we can use the linear approximation of the problem. It is seen form Eqs. (7.1.6) and (7.1.7) that in this case the scattering amplitude is defined the following formula: 

q(y)ei(k−l)·y dy = q(k ˆ − l),

f (k, l) =

(7.1.8)

7.1 The Inverse Scattering Problem at a Fixed Energy

215

where q(λ) ˆ means the Fourier transform of the function q(x): 

q(y)eiλ·y dy,

q(λ) ˆ =

λ ∈ R3 .



Note that the function f (k, l) is √given for all k and l so that |k|2 = |l|2 = E define q(λ) ˆ for all |λ| ≤ ρ0 = 2 E. Indeed, let λ = ρν(θ, ϕ), where ν(θ, ϕ) = (sin θ cos ϕ, sin θ sin ϕ, cos θ ). Define k, l by the formulae k= l=





1 2 , 2 ρν(θ, ϕ) + 4E − ρ νθ (θ, ϕ)   1 2 2 −ρν(θ, ϕ) + 4E − ρ νθ (θ, ϕ) ,

(7.1.9)

where νθ (θ, ϕ) = (cos θ cos ϕ, cos θ sin ϕ, − sin θ ). Then for any λ, |λ| ≥ ρ0 we have: k − l = λ and |k|2 = |l|2 = E. Hence, if the energy E is sufficiently large, the Fourier image of q(x) is defined by the scattering amplitude for the large ball Bρ0 = {λ ∈ R3 : |λ| ≤ ρ0 }. Since the Fourier transform of finite continuous function is an analytical function, values of q(λ) ˆ inside the ball Bρ0 define uniquely q(λ), ˆ for all λ ∈ R3 . This conclusion leads to the following uniqueness theorem. Theorem 7.1.1 Let the scattering amplitude f (k, l) in (7.1.7) is given at a fixed energy for all k and l. Then the inverse scattering problem in the Born approximation has at most an one solution. Moreover, we can construct the approximate solution qappr (x) of the inverse problem by following formula 1 qappr (x) = (2π)3



f (k, l)e−iρν(θ,ϕ)·x ρ 2 dρ sin θ dθ dϕ,

(7.1.10)

Bρ0

where k, l given by (7.1.9). The error of the approximation qerr (x) can be estimated under some a-priory assumption on the function q(x). Let us assume, for example, that the function q(x) with the compact support in , contained in the ball centered at the origin of radius R, belongs to C m (R3 ), m > 3. Suppose that its norm satisfies the condition: q C m (R3 ) ≤ q0 . Then the Fourier image q(λ) ˆ of q(x) satisfies the estimate: |q(λ)| ˆ ≤ Cq0 |λ|−m ,

|λ| ≥ ρ0 ,

(7.1.11)

where C = C(R, m) > 0. Indeed,  q(λ) ˆ =

R3

q(x)eiρν·x dx =



R

−R

Q(s, ν)eiρs ds.

(7.1.12)

216

7 Inverse Problems for Elliptic Equations

Here ν = ν(θ, ϕ) and Q(s, ν) is the Radon transform of q(x) given by formula  Q(s, ν) =

q(x) dσ, ν·x=s

where dσ is the square element. The variable x under the last integral can be derived by the formula 

νϕ sin φ , x = sν + r νθ cos φ + sin θ

νθ =

∂ν ∂ν , νϕ = . ∂θ ∂ϕ

Substituting this with dσ = rdrdφ into the integral (7.1.12) we conclude that q(λ) ˆ =

(−1)m (iρ)m



R

−R

∂ m Q(s, ν) iρs e ds. ∂ρ m

(7.1.13)

Evidently,

m

∂ Q(s, ν)



∂ρ m ≤ Cq0 , with the above defined constant C = C(R, m), depending only on R and m. Using this in (7.1.13) we arrive at the estimate (7.1.11). Then the error of the approximation qerr (x) can be estimate as |qerr (x)| ≤

1 (2π)3

≤

$ R3 \Bρ0

|q(λ)| ˆ ρ 2 dρ sin θ dθ dϕ Cq0

2π 2 (m − 1)ρ0m−1

=

Cq0 ≤ 2π 2



∞ ρ0

dρ ρ m−2

Cq0 . 2π 2 (m − 1)E (m−1)/2

Therefore, the error of the approximate formula (7.1.10) tends to zero, as E → ∞.

7.2 The Inverse Scattering Problem with Point Sources In the previous section we have studied the inverse scattering problem with incident plane wave. Here we discuss the case when the incident wave is produced by point sources. Let q(x) is compactly supported in the ball = {x ∈ R3 : |x| ≤ R} function and q(x) ∈ C(R3 ). Consider the equation − u + q(x)u = k 2 u + δ(x − y),

x ∈ R3 ,

(7.2.1)

7.2 The Inverse Scattering Problem with Point Sources

217

where δ(z) is a Dirac delta function. In the contrast to the previous section, here k is a positive number, k 2 = E, and the point y ∈ R3 is the variable parameter. Let u(x, k, y) be a solution of this equation of the form u(x, k, y) =

eik|x−y| + v(x, k, y), 4π|x − y|

(7.2.2)

where the function v(x, k, y) satisfies the Sommerfeld conditions v = O(r −1 ),

∂v − ikv = o(r −1 ), as r = |x| → ∞. ∂r

(7.2.3)

Then, using Eq. (7.2.1), we deduce that v(x, k, y) is the solution of the following equation 

 eik|x−y| − v + q(x) + v(x, k, y) = k 2 v, 4π|x − y|

x ∈ R3 .

(7.2.4)

Inverting the Helmholtz operator −( + k 2 ), we obtain that the function v(x, k, y) is the solution of the integral equation 1 v(x, k, y) = v0 (x, k, y) − 4π



eik|ξ −x| q(ξ )v(ξ, k, y) dξ, |ξ − x|

(7.2.5)

where 1 v0 (x, k, y) = − (4π)2



eik(|ξ −x|+|ξ −y|) q(ξ ) dξ. |ξ − x||ξ − y|

(7.2.6)

We again will restrict our analysis to the case of the Born approximation for this equation assuming that q(x) is enough small, which allows to make the linearization of the problem. Then v(x, k, y) ≈ v0 (x, k, y). Now we state the following theorem. Theorem 7.2.1 Let q(x) be a function with a compact support in the ball = {x ∈ R3 : |x| ≤ R} and q(x) ∈ C 2 (R3 ). Then for any fixed y and k, v0 (x, k, y) ∈ C(R3 ) and eik|x−y| v0 (x, k, y) = 8ikπ



 1 q(ξ ) ds + o , as k → ∞. k L(x,y)

(7.2.7)

where L(x, y) = {ξ ∈ R3 : ξ = y(1 − s) + sx, s ∈ [0, 1]} is the segment of the straight line passing through points x and y.

218

7 Inverse Problems for Elliptic Equations

Proof Denote by E(x, y, t) the ellipsoid E(x, y, t) = {ξ ∈ R3 : |ξ − x| + |ξ − y| = t}, Let us fix points x and y. Represent x as x = y + ρν(θ, ϕ), where ν(θ, ϕ) is the unite vector and θ and ϕ are its spherical coordinates. Here ρ, θ and ϕ are fixed. Then the ellipsoid E(x, y, t) is defined only for t ≥ |x − y| = ρ. If t → |x − y|, then the ellipsoid degenerates into the segment L(x, y). If t > ρ, then the main axis of this ellipsoid passes through the points x and y and the intersection points of this axis with the ellipsoid are ξ 1 = y −ν(θ, ϕ)(t −ρ)/2 and ξ 2 = x +ν(θ, ϕ)(t −ρ)/2. Using these observations, we represent the variable point ξ ∈ E(x, y, t) as follows:  1 νϕ (θ, ϕ) ξ = y + (ρ + tz)ν(θ, ϕ) + r νθ (θ, ϕ) cos ϕ + sin ψ , 2 sin θ

(7.2.8)

where r = r(z; ρ, t) = r(z, |x − y|, t) in (7.2.8) is given by the formula r(z; ρ, t) =

' 1 (t 2 − ρ 2 )(1 − z2 ) 2

(7.2.9)

and νθ (θ, ϕ) =

∂ν(θ, ϕ) , ∂θ

νϕ (θ, ϕ) =

∂ν(θ, ϕ) . ∂ϕ

In formulae (7.2.8)–(7.2.9) z and ψ are variable parameters: z ∈ [−1, 1], ψ ∈ [0, 2π). Hence, ξ = ξ(z, ψ; x, y, t). Note that the unite vectors ν, νθ , νϕ / sin θ are mutually orthogonal. Then each point ξ ∈ E(x, y, t) is uniquely defined by z ∈ [−1, 1] and ψ ∈ [0, 2π). Indeed, in this case for arbitrary z ∈ [−1, 1] and ψ ∈ [0, 2π) the following equalities hold |ξ − y| = |ξ − x| =

' '

r 2 + 14 (tz + ρ)2 = 12 (t + zρ), r 2 + 14 (tz − ρ)2 = 12 (t − zρ).

Hence, |ξ − y| + |ξ − x| = t, i.-e. the point ξ ∈ E(x, y, t). On the other hand, for any ξ given by the formula (7.2.8) we have (ξ − y) · ν = (ρ + tz)/2. It means that the coordinate z characterize the cross-section of the ellipsoid E(x, y, t) by the plane orthogonal to ν. For z = −1 this cross-sections degenerates into the point ξ 1 , for z = 1 into ξ 2 . Since the vectors νθ , νϕ / sin θ are orthogonal to ν, in any crosssection defined by z ∈ (−1, 1) the coordinate ψ in (7.2.8) is uniquely determines the position of ξ in this cross-section.

7.2 The Inverse Scattering Problem with Point Sources

219

For the sake of brevity, we denote e1 = ν, e2 = νθ , e3 = νϕ / sin θ . Then, it follows from (7.2.8) that each ξ ∈ E(x, y, t) can be presented as ' 1 1 ξ = y + (ρ + tz)e1 + (t 2 − ρ 2 )(1 − z2 )(e2 cos ψ + e3 sin ψ). 2 2

(7.2.10)

For each fixed x and y family of the ellipsoids E(x, y, t), t ≥ ρ = |x − y|, covers all space R3 . So, every point ξ ∈ R3 can be uniquely defined by z, t, ψ. The 1 ,ξ2 ,ξ3 ) Jacobian J = ∂(ξ ∂(z,t,ψ) can be easily calculated:



∂ξ1 /∂z ∂ξ2 /∂z ∂ξ3 /∂z



∂(ξ1 , ξ2 , ξ3 ) :=

∂ξ1 /∂t ∂ξ2 /∂t ∂ξ3 /∂t

J = ∂(z, t, ψ)

∂ξ /∂ψ ∂ξ /∂ψ ∂ξ /∂ψ

1 2 3



te1 /2 + rz (e2 cos ψ + e3 sin ψ)



=

ze1 /2 + rt (e2 cos ψ + e3 sin ψ)



r (−e sin ψ + e cos ψ) 2 3 =

1 1 r(trt − zrz ) = (t 2 − z2 ρ 2 ). 2 8

At the same time |ξ − x||ξ − y| =

1 2 (t − z2 ρ 2 ). 4

Therefore, dξ 1 = dzdψdt. |ξ − x||ξ − y| 2 Using the above transformation in the right hand side integral of (7.2.6) we arrive at the formula v0 (x, k, y) = −

1 32π 2



∞  2π ρ

0



1 −1

eikt q(ξ ) dzdψdt,

(7.2.11)

where ξ is defined by formula (7.2.10). Since the function q(x) has a compact support in the ball , the integral in (7.2.11) with respect to t is taken along a bounded interval depended on x and y. Further, the function v0 (x, k, y) in (7.2.11) is continuous, due to the continuity of the integrand. Using in (7.2.11) the integration

220

7 Inverse Problems for Elliptic Equations

by parts formula respect to t, we get: eikρ v0 (x, k, y) = 32ikπ 2 + =





0

1 32ikπ 2 eikρ

2π

I1 32ikπ 2



∞ ρ

+



1 −1

q(ξ ) dzdψ t =ρ+0

2π 

1 −1

0

eikt ∇q(ξ ) · ξt dzdψdt

1 I2 . 32ikπ 2

(7.2.12)

Since ξ → y + νρ(1 + z)/2 as t → ρ, we have 

2π

I1 =



−1

0

 = 2π 

1 −1 1

= 4π



1

q(ξ ) dzdψ t =ρ+0

q(y + νρ(1 + z)/2) dz  q(y + s(x − y)) ds = 4π

0

q(ξ ) ds.

(7.2.13)

L(x,y)

For the second integral I2 we use formula (7.2.10) for ξ to deduce ( z t ξt = e1 + 2 2

1 − z2 (e2 cos ψ + e3 sin ψ). t 2 − ρ2

Then we can transform the integral I2 as follows: 

∞  2π

I2 = 

ρ

= ρ

0

∞  2π 0

t + 2

 

(

1 −1

eikt ∇q(ξ ) · ξt dzdψdt

 i kt t e ∇ξ q(ξ ) · e1 2 −1 1

  1 − z2 q(ξ ) · e cos ψ + ∇ q(ξ ) · e sin ψ dzdψdt. ∇ ξ 2 ξ 3 t 2 − ρ2

7.2 The Inverse Scattering Problem with Point Sources

221

Using here the integration by parts formula with respect to ψ, we get 

∞  2π

I2 = ρ

(

0



 kt t i ∇ξ q(ξ ) · e1 e 2 −1 1

  1 − z2 ∇ξ ∇ξ q(ξ ) · e2 · ξψ sin ψ t 2 − ρ2   +∇ξ ∇ξ q(ξ ) · e3 · ξψ cos ψ dzdψdt.

t − 2

It follows from (4.2.10) that (

1 − z2 1 ξψ = (1 − z2 )(−e2 sin ψ + e3 cos ψ) 2 2 t −ρ 2

is a continuous function of z and ψ. As a consequence, the integrand I2 in (7.2.12) is also continuous function of the variables t, z and ψ. Therefore the double integral with respect to z, ψ is continuous function of t. Then integral with respect to t is the Fourier transform of continuous finite function. Therefore I2 → 0 as k → ∞. This conclusion with formulae (7.2.12) and (7.2.13) imply the proof of the theorem.   Let us now assume that the scattering field is given for every (x, y) ∈ (S × S) and all k ≥ k0 , where k0 a positive number: v(x, k, y) = f (x, k, y),

(x, y) ∈ (S × S), k ≥ k0

(7.2.14)

Then, in the Born approximation, we obtain the following asymptotic formula: f (x, k, y) =

eik|x−y| 8ikπ

 q(ξ ) ds + o L(x,y)

 1 , as k → ∞. k

(7.2.15)

where L(x, y) is the segment defined in Theorem 7.2.1. Calculating the integral of q(x) over this segment we get: 

 q(ξ ) ds = 8iπ lim e−ik|x−y| kf (x, k, y) L(x,y)

k→∞

= g(x, y), (x, y) ∈ (S × S).

(7.2.16)

Therefore we arrive at the tomography problem: find q(x) in from given integrals along arbitrary strait lines jointing the boundary of . Note that this problem can be solved separately for each cross-section of the ball by any plane. We will study this problem in the next chapter. It follows from this consideration

222

7 Inverse Problems for Elliptic Equations

that the tomography problem has, at most, an unique solution, and this solution can be derived in an explicit form.

7.3 Dirichlet-to-Neumann Map Let ⊂ Rn be a bounded domain with the smooth boundary ∂ . Consider the Dirichlet problem u + q(x)u = 0, x ∈ ,

u|∂ = f (x).

(7.3.1)

Denote by u(x; q) the unique weak solution u(·; q) ∈ H 1 ( ) of this problem, corresponding to a given coefficient q(x) from some class of admissible coefficients in L2 ( ). It is well known that if ⊂ Rn is a bounded domain with the C 1 -smooth boundary ∂ , then there exists a bounded linear operator Tr : H 1 ( ) → H 1/2 (∂ ), called the trace of u on ∂ , such that Tr u := u(·; q)|∂ , and Tr u L2 (∂ ) ≤ C0 u H 1 ( ) , C0 = C0 ( ) > 0. Now, taking (formally) the derivative of the solution u(·; q) ∈ H 1 ( ) of problem (7.3.1) along the outward normal n to the boundary ∂ at x ∈ ∂ , we can find the trace: ∂u(·; q)

, q ∈ L2 ( ). ∂ ∂n

(7.3.2)

According to [92], this trace is uniquely determined as an element of H −1/2 (∂ ). Thus, we have constructed a mappings which transforms each element (coefficient) q ∈ Q to the unique element (trace of the derivative of the solution) of H 1/2(∂ ) given by (7.3.2). In this transformation, under the assumption that q ∈ Q is given, we first used the Dirichlet data given in (7.3.1), which uniquely defines the solution, then we used the normal derivative of the solution in (7.3.2), which is usually associated with the Neumann data. Subsequently, this gave rise to the terminology “Dirichlet–Neumann operator”, which was first introduced in [84]. Definition 7.3.1 Let u(x; q) be the unique weak solution u(·; q) ∈ H 1 ( ) of problem (7.3.1), corresponding to a given coefficient q ∈ L2 ( ). Then the mapping q : H 1/2(∂ ) → H −1/2 (∂ )

(7.3.3)

defined by (7.3.2) is called the Dirichlet–Neumann operator, that is, q u :=

∂u(·; q)

, q ∈ L2 ( ). ∂ ∂n

(7.3.4)

7.3 Dirichlet-to-Neumann Map

223

It is assumed that 0 is not the Dirichlet eigenvalue for the operator  + q in . Let now f ∈ L2 (∂ ) be a given input in (7.3.1). It follows from the above conclusions that for a given coefficient q ∈ L2 ( ) the Dirichlet–Neumann operator is uniquely defined by the equality q f (x) :=

∂u(·; q)

= g(x), x ∈ ∂ . ∂ ∂n

(7.3.5)

We assume now that g ∈ L2 ( ) be a given output. Then we can naturally raise the question: does the operator q define uniquely the function q(x)? The answer to this question is positive (see [115, 142]). Theorem 7.3.1 Let ⊂ Rn , n ≥ 3, be a bounded domain with smooth boundary and qk (x) ∈ L2 ( ) for k = 1, 2. Then q1 = q2 implies q1 (x) = q2 (x) in . Proof of this theorem is based on a construction of a special solution of the equation u + q(x)u = 0, which depends on a complex parameter ζ = ξ + iη satisfying the condition ζ · ζ = 0. This solution has the form uζ (x) = (1 + vζ (x)) exp(x · ζ ), where vζ (x) satisfies the estimate vζ L2 ( ) ≤ C/|ζ | for large |ζ |. Note that this solution is unbounded for x · η < 0 and |η| → ∞. We refer the reader to [115] and [142] for details of the proof. Now we consider a linearized inverse problem. Let ∈ R2 be the unite disk |x| < 1. Denote by (r, ϕ) polar coordinates of point x ∈ . We assume that the input f (x) in (7.3.1) is given as the Fourier harmonics, that is, f (x) := exp(imϕ), for m = 0, ±1, ±2, . . .. Denote the solution of the problem (7.3.1) corresponding the such Dirichlet data by um := um (r, ϕ; q). Then given operator q is equivalent to the following information ∂um (r, ϕ; q)

= g m (ϕ),

r=1 ∂r

ϕ ∈ [0, 2π],

m = 0, ±1, ±2, . . . .

(7.3.6)

Consider the inverse problem of recovering q(r, ϕ) from the given by (7.3.6) output g m (ϕ), m = 0, ±1, ±2, . . .. Assume that q(r, ϕ) and its derivative qϕ (r, ϕ) are continues functions in . Assume also that function q(r, ϕ) is small, i.e. |q(r, ϕ)| 0}. This is a well known tomography problem which occurs in tomography when a physical body is exposed to X rays and the radiation is measured outside the body in a tomographic fashion. Note that the scalar product ν · ∇x u represents the derivative of function u(x, ν) in the direction ν. Assuming that h(x, ν) > 0 one can invert equation (8.1.3) using the boundary data (8.1.2). Then we obtain the following formula for the function u(x, ν): u(x, ν) = h(ξ ∗ (x, ν), ν)e−

$

L(x, ν) σ (ξ ) ds

,

(8.1.5)

where L(x, ν) = {ξ ∈ | ξ = x − sν} is the segment of the direct line going from point x in the direction −ν and belonging to , ξ ∗ (x, ν) is the intersection point of L(x, ν) with ∂− (ν) and ds is the arc length element. Then from (8.1.5) we obtain the relation  σ (ξ ) ds = f (x, ν), x ∈ ∂+ (ν), ν ∈ S 2 , (8.1.6) L(x, ν)

where f (x, ν) = − ln(g(x, ν)/ h(ξ ∗ (x, ν), ν)). Thus, the tomography problem is transformed to the problem of recovering a coefficient σ (x) from the given integrals along the segments L(x, ν) of arbitrary direct lines crossing domain . Note that we can consider the problem as a twodimensional one for an arbitrary cross-section domain by a plane and choosing L(x, ν) belonging to this plane only. For example, one can consider the oneparametric family of cross-sections of by the planes x3 = const and solve the

8.1 The Transport Equation Without Scattering

229

tomography problem for each of this cross-sections in order to find σ (x) for all x ∈ . Another class of inverse problems, close to the tomography problem, is related to the determination of an unknown source term F (x) in the transport equation ν · ∇x u + a(x)u = F (x), x ∈ , ν ∈ S 2 ,

(8.1.7)

when the function a(x) ≥ 0 is assumed to be known. Adding to this equation the homogeneous boundary condition u(x, ν) = 0,

x ∈ ∂− (ν), ν ∈ S 2 ,

(8.1.8)

we formulate the following the inverse source problem of recovering an unknown source in the transport equation: Find F (x) in (8.1.7)–(8.1.8) from the measured data g(x, ν) given by (8.1.4). Remark that the inverse source problem is linear, while the inverse coefficient problem defined by (8.1.3)–(8.1.4) is non-linear. Let us derive an integral representation for the solution of the inverse source problem. Along the half-line L(x, ν) = {ξ ∈ ⊂ R3 : ξ = x − sν, s ∈ (0, ∞)} the Eq. (8.1.7) can be rewritten in the form −

$s $s d      u(x − sν, ν)e− 0 a(x−s ν)ds = F (x − sν)e− 0 a(x−s ν)ds . (8.1.9) ds

Let s(x, ν) be the arc length of the segment of L(x, ν) that belong . Then integrating (8.1.9) along L(x, ν) for s ∈ [0, s(x, ν)] and using the condition (8.1.8) we get 

s(x,ν)

u(x, ν) =

F (x − sν)e−

$s 0

a(x−s  ν)ds 

ds,

(x, ν) ∈ × S 2 .

0

Substituting here x ∈ ∂+ (ν) and using the additional condition (8.1.4), we obtain we integral equation for unknown function F (x): 

s(x,ν)

F (x − sν)e−

0

$s 0

a(x−s ν)ds 

ds = g(x, ν), x ∈ ∂+ (ν), ν ∈ S 2 .

Assuming that F (x) = 0 outside , the latter equation can also be rewritten in the form  $s   g(x, ν) = F (x − sν)e− 0 a(x−s ν)ds ds, x ∈ ∂+ (ν), ν ∈ S 2 . L(x,ν)

Hence, the inverse source problem is reduced to the attenuated Radon transform: find F (x) from given integrals along arbitrary half-lines L(x, ν), x ∈ ∂+ (ν), ν ∈ S 2 , with the attenuation defined by the coefficient a(x).

230

8 Inverse Problems for Stationary Transport Equations

The last problem can also be reduced to the two-dimension one for any crosssection of by a plane and choosing L(x, ν), x ∈ ∂+ (ν), ν ∈ S 2 , belonging to this cross-section only. An inversion formula for this problem was obtained by R. Novikov (see [117]).

8.2 Uniqueness and a Stability Estimate in the Tomography Problem Consider the tomography problem (8.1.6) on the plane (x1 , x2 ) ∈ R2 . Let = {x ∈ R2 | |x| ≤ 1} be the circle and ∂ = {x ∈ R2 | |x| = 1} be its boundary. Assume here that σ ∈ C 1 ( ). Then substituting x = x(ϕ) = (cosϕ, sin ϕ) for x ∈ ∂+ (ν) and ν = ν(θ ) = (cos θ, sin θ ), we can rewrite the Eq. (8.1.6) in the following form: 

σ (ξ ) ds = fˆ(ϕ, θ ),

(ϕ, θ ) ∈ [0, 2π] × [0, 2π],

(8.2.1)

L(x, ν)

where fˆ(ϕ, θ ) = f (x(ϕ), ν(θ )). For x ∈ and ν = (cos θ, sin θ ) introduce the function  v(x, θ ) = σ (x − sν) ds. (8.2.2) L(x, ν)

Then this function satisfy the equation ν(θ ) · ∇x v(x, θ ) = σ (x), x ∈ , θ ∈ [0, 2π]

(8.2.3)

and the condition v(x(ϕ), θ ) = fˆ(ϕ, θ ),

(ϕ, θ ) ∈ [0, 2π] × [0, 2π].

(8.2.4)

Taking derivative with respect to θ , we exclude the function σ (x) from the Eq. (8.2.3) and obtain ∂ (ν · ∇x v(x, θ )) = 0, x ∈ , θ ∈ [0, 2π]. ∂θ

(8.2.5)

Use Mukhometov’s identity [103], which can be checked directly: 2(νθ · ∇x v(x, θ ))

∂ ∂ ∂ (vθ vx2 ) − (vθ vx1 ) (ν · ∇x v(x, θ )) = ∂θ ∂x1 ∂x2

+|∇x v(x, θ )|2 +

∂ [(ν(θ ) · ∇x v(x, θ ))(νθ (θ ) · ∇x v(x, θ ))] , ∂θ

(8.2.6)

8.3 Inversion Formula

231

where νθ = (− sin θ, cos θ ). Then it follows from (8.2.5) and (8.2.6) that ∂ ∂ (vθ vx2 ) − (vθ vx1 ) + |∇x v(x, θ )|2 ∂x1 ∂x2 +

∂ [(ν · ∇x v(x, θ ))(νθ · ∇x v(x, θ ))] = 0. ∂θ

(8.2.7)

Integrating the equality (8.2.7) over and [0, 2π], and then taking into account that v(x, θ )) is a periodic function with respect to θ we get: 

2π 0

  ∂ ∂ (vθ vx2 ) − (vθ vx1 ) + |∇x v(x, θ )|2 dxdθ = 0. ∂x2 ∂x1

Applying Gauss’s formula, we obtain 

2π 0





2π

|∇x v(x, θ )| dxdθ = − 2



vθ (x(ϕ), θ )vϕ (x(ϕ), θ )dθ dϕ. 0



2π

0

Using (8.2.4) and the relations σ 2 (x) = |ν · ∇x v(x, θ )|2 ≤ |∇x v(x, θ )|2 , we find the stability estimate for the tomography problem as follows:  σ 2 (x)dx ≤ −

1 2π



2π 0



2π

fˆϕ (ϕ, θ )fˆθ (ϕ, θ )dθ dϕ.

0

Thus, 1

σ L2 ( ) ≤ √ fˆ H 1 ([0,2π]×[0,2π]). 2 π

(8.2.8)

The uniqueness of the solution of the tomography problem follows from estimate (8.2.8). That is, if data fˆk (x, ν) corresponds to σk (x) for k = 1, 2, and fˆ1 (x, ν) = fˆ2 (x, ν), then σ1 (x) = σ2 (x), x ∈ .

8.3 Inversion Formula Let σ (x) ∈ C(R2 ) and σ (x) = 0 for x ∈ R2 \ . Assume ν = (cos θ, sin θ ) and νθ = (− sin θ, cos θ ), p ∈ [0, ∞) and define function f (θ, p) by the formula  f (θ, p) =

∞ −∞

σ (sνθ + pν) ds,

θ ∈ [0, 2π], p ∈ [0, ∞).

(8.3.1)

The mapping : σ → f is called the Radon transform. It is clear that the Radon transform for fixed (θ, p) is the integral from σ (x) along the direct line ν · ξ = p.

232

8 Inverse Problems for Stationary Transport Equations

So, the tomography problem consists in the inverting of the Radon transform. We derive now the inversion formula following to [43]. For a fixed x ∈ and p ∈ [0, ∞) consider the equality 1 2π

 0

2π

 ∞  2π 1 f (θ, p + x · ν)dθ = σ (sνθ + pν + (x · ν)ν) dθ ds 2π −∞ 0  ∞  2π 1 = σ ((s − x · νθ )νθ + pν + x) dθ ds 2π −∞ 0  ∞  2π 1 σ (tνθ + pν + x) dθ dt = 2π −∞ 0  ∞  ' = F x, p2 + t 2 dt, −∞

(8.3.2) where  '  2π 1 F x, p2 + t 2 = σ (tνθ + pν + x) dθ. 2π 0

(8.3.3)

Note that for a fixed x the function F (x, r) represents an averagevalue of σ (ξ ) over the circumference with thecenter at x and the radius r = p2 + t 2 since |ξ − x| = |tνθ + pν| = r = p2 + t 2 , and therefore depends on x and r only. Evidently, this is a continuous function of x and r, for all x ∈ and r ∈ [0, ∞). Furthermore, F (x, r) is a finite-valued function of r and F (x, 0) = σ (x). Thus, to obtain an inversion formula, we need to calculate F (x, 0). For this goal, we use the following identity, obtained by the change of variable: 

∞

−∞

 '  F x, p2 + t 2 dt =

∞

F (x, r) 

2rdr

.

(8.3.4)

2rdr F (x, r)  . r 2 − p2

(8.3.5)

p

r 2 − p2

Hence, 1 2π



2π

 f (θ, p + x · ν)dθ =

0

∞

p

Applying to the right hand side of the latter equality the operator 1 ∂ Lg(s) = − πs ∂s



∞ s

g(p)p dp  , p2 − s 2

8.3 Inversion Formula

233

we find 1 ∂ − πs ∂s

∞ ∞



1 ∂ =− πs ∂s

s



F (x, r) 

p



∞

r

F (x, r) s

s

1 ∂ =− s ∂s

2rdr (r 2





− p2 )(p2 − s 2 )

p dp

2p dp (r 2

∞

− p2 )(p2 − s 2 )

rdr

F (x, r)rdr = F (x, s).

(8.3.6)

s

Then from (8.3.5) we obtain the inversion formula as follows 1 ∂ σ (x) = − lim 2 s ∂s + 2π s→0

∞  2π

 s

0

pdp f (θ, p + x · ν)dθ  . p2 − s 2

(8.3.7)

This formula valid and for not a finite-valued function σ (x). It is still true if we assume that σ (x) tends to zero as |x| → ∞ together with |x|k σ (x) for any positive k. Various aspects of inverse problems for transport equations can be found in [6, 8, 31].

Chapter 9

The Inverse Kinematic Problems

The problem of recovering the sound speed (or index of refraction) from travel time measurements is an important issue in determining the substructure of the Earth. The same inverse problem can also be defined as the problem of reconstructing of a Riemannian metric inside a compact domain from given distances of geodesics joining arbitrary couples of points belonging to the boundary of . This chapter studies some basic inverse kinematic problems. Specifically, in this chapter we give an analysis of the inverse kinematic problem for one- and twodimensional cases.

9.1 The Problem Formulation Let a compact domain ∈ Rn be filled with an isotropic inhomogeneous substance in which waves propagate with the speed c(x). We shall assume that c(x) is an uniformly bounded positive smooth function in the  closed domain . Introduce the Riemannian metric dτ = |dx|/c(x), |dx| = ( ni=1 dxi2)1/2 and denote by τ (x, y) the length of the geodesic line (x, y) joining points x and y. Physically, the function τ (x, y) means travel time between points x and y. It is well known that τ (x, y) is a symmetric function of their arguments, i.-e. τ (x, y) = τ (y, x) and satisfies to the eikonal equation |∇x τ (x, y)|2 = n2 (x),

(9.1.1)

where n(x) = 1/c(x) is the refractive index. Moreover, τ (x, y) satisfy the additional condition τ (x, y) ∼ n(y)|x − y|, as x → y.

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_9

(9.1.2)

235

236

9 The Inverse Kinematic Problems

The latter condition means that the travel time τ (x, y) can be calculated as in homogeneous medium with speed equal to c(y) if x close to y. In the next subsection we consider how to construct the function τ (x, y) and the geodesic lines (x, y) for a given function c(x). Now we formulate the inverse problem: find c(x) in from τ (x, y) given for all (x, y) ∈ (∂ × ∂ ). Here ∂ is a smooth boundary of . The above problem is called the inverse kinematic problem. It has various applications in the geophysics and other sciences. It was noted for a long time ago that the inverse kinematics problem has no unique solution if the medium has inner wave guides. Therefore this problem is studied usually under assumption that field of geodesic lines is regular inside . The latter means that each couple of points x, y can be joined in by only one geodesic line and boundary ∂ of is convex with respect to these geodesics.

9.2 Rays and Fronts Here we construct a solution to problem (9.1.1) and (9.1.2) following the well known method of solving the first order partial differential equations. We assume that τ (x, y) is a twice continuously differentiable function for all x = y. Introduce the function p = p(x, y) := ∇x τ (x, y). Then Eq. (9.1.1) can be rewritten in the form n 

pj2 (x, y) = n2 (x),

(9.2.1)

j =1

where pj denote j -th component of the vector p. Taking the derivative of both sides of (9.2.1) with respect to xk , one obtains 2

n  ∂pj j =1

∂xk

pj = 2n(x)nxk (x),

k = 1, 2, . . . , n.

(9.2.2)

Since ∂pj ∂ 2τ ∂pk = = ∂xk ∂xj xk ∂xj equations (9.2.2) can be transformed to the form n  ∂pk j =1

∂xj

pj = n(x)nxk (x),

k = 1, 2, . . . , n.

(9.2.3)

9.2 Rays and Fronts

237

Introduce the new variable parameter s such that value s = 0 corresponds to point y and consider in Rn the curves determined by the equations pj dxj = 2 , ds n (x)

i = 1, 2, . . . , n.

(9.2.4)

Along these curves one has n  ∂pk j =1

∂xj

pj = n2 (x)

n  ∂pk dxj dpk = n2 (x) . ∂xj ds ds j =1

Hence, Eqs. (9.2.3) are the ordinary differential equations dpk ∂ ln n(x) = , ds ∂xk

k = 1, 2, . . . , n.

(9.2.5)

Equations (9.2.4) and (9.2.5) form the system of differential equations for x and p that determines x, p as functions of s. Let us explain the meaning of the variable s. To this end, calculate the derivative of τ with respect to s along the curves determined by Eqs. (9.2.4) and (9.2.5). We find:   pj dxj dτ = = = 1. τx j ds ds n2 (x) n

n

j =1

j =1

2

Hence, τ = s because s and τ equal zero at point y. Therefore Eqs. (9.2.4) and (9.2.5) can be written in the form dx p = 2 , dτ n (x)

dp = ∇ ln n(x). dτ

(9.2.6)

Equations (9.2.6) are called Euler’s equations. The curves determined by Euler’s equations are called the characteristic lines. In space Rn they determine the geodesic lines (x, y). In geophysical applications these lines are defined as rays. In order to find these lines consider the solution to Eqs. (9.2.6) with the following initial data x|τ =0 = y,

p|τ =0 = p0 ,

(9.2.7)

where p0 = (p10 , . . . , pn0 ) is an arbitrary vector satisfying to the condition |p0 | = n(y).

(9.2.8)

238

9 The Inverse Kinematic Problems

Then the solution to the Cauchy problem (9.2.6)–(9.2.8) gives the geodesic (x, y) which goes from point y in the direction determined by the vector

dx

p0 = 2 . ζ =

dτ τ =0 n (y) 0

(9.2.9)

Hence, ζ 0 = (ζ10 , . . . , ζn0 ) satisfies to the relation n(y)|ζ 0 | = 1.

(9.2.10)

Problem (9.2.6) and (9.2.7) has a unique solution at least in the vicinity of the point y if n(x) is a smooth function. Namely, if n(x) ∈ Ck (Rn ) with k ≥ 2 then the solution can be represented in the form x = f1 (τ, y, ζ 0 ),

p = f2 (τ, y, ζ 0 ),

where f1 , f2 are Ck−1 -smooth functions of τ, y, ζ 0 . Moreover, these functions can be represented in the following form x = f (τ ζ 0 , y),

p = fˆ(τ ζ 0 , y)/τ,

where f (τ ζ 0 , y) = f1 (1, y, τ ζ 0), fˆ(τ ζ 0 , y) = f2 (1, y, τ ζ 0 ) The latter is a consequence of the following observation: if we introduce p, ˆ τˆ and pˆ 0 , ζˆ 0 by the 0 0 0 0 ˆ relations p = λp, ˆ τ = τˆ /λ, p = λpˆ , ζ = λζ , where λ > 0, and simultaneously replace p, τ, p0 on p, ˆ τˆ , pˆ 0 in (9.2.6), (9.2.7), these equations are not changed. Hence, x, pˆ can be given in the form x = f1 (τˆ , y, ζˆ 0 ) = f1 (λτ, y, ζ 0 λ−1 ), pˆ = f2 (τˆ , y, ζˆ 0 ) = f2 (λτ, y, ζ 0 λ−1 ). Taking here λ = 1/τ , one gets the above representation x = f (ζ, y), p = fˆ(ζ, y)/τ , where ζ = τ ζ 0 . The components ζ1 , . . . , ζn of the variable ζ are called the Riemannian coordinates of the point x with respect to fixed point y. From relations (9.2.6) and the definition (9.2.9) of ζ 0 follows that in a vicinity of y one has

∂x

x = f (ζ, y) = y + τ + O(τ 2 ) = y + ζ + O(|ζ |2 ). (9.2.11) ∂τ τ =0 Hence,  det



∂x

= 1. ∂ζ ζ =0

(9.2.12)

It means that function x = f (ζ, y) has an inverse ζ = g(x, y) determined, at least, in the vicinity of point the y. Furthermore, the inverse function is smooth, that is, if

9.2 Rays and Fronts

239

n(x) ∈ Ck (Rn ), then it belongs to Ck−1 . It follows from (9.2.10) that the following representation holds for the function τ 2 (x, y): τ 2 (x, y) = n2 (x)|g(x, y)|2.

(9.2.13)

Since p = ∇τ , one also has ∇τ 2 (x, y) = 2fˆ(g(x, y), y).

(9.2.14)

Now it is obvious that τ 2 (x, y) is Ck -smooth function of x and y, at least, for x close to y. For the case, when section curvatures of the Riemannian metric are nonpositive, the geodesic lines (x, y) passing through the point y have no point of intersection, for any x, except the starting point y, and in addition, τ 2 (x, y) is a smooth function anywhere. The condition τ (x, y) = O(|x − y|) as x → y is a consequence of (9.2.11) which is equivalent to the relation ζ = x − y + O(|x − y|2 )

as x → y.

(9.2.15)

Thus, in order to find the geodesic line which passes through the point y in direction ζ 0 satisfying condition (9.2.10), one should solve the Cauchy problem (9.2.6) and (9.2.7), where p0 = ζ 0 n2 (y). Each of these geodesic lines with the common point y can be represented by the equation x = f (ζ, y), where ζ = ζ 0 τ , where f (ζ, y) is a smooth function and has the inverse one ζ = g(x, y). Then formula (9.2.13) determines τ (x, y) satisfying relations (9.1.1) and (9.1.2). Physically, the Riemannian sphere τ (x, y) = constant, presents the front of a wave propagating from a point source placed at y. Since p = ∇x τ (x, y) is the normal to this sphere, it follows from (6.2.6) that geodesic lines (x, y) are orthogonal to the front. Briefly speaking, the rays and fronts are orthogonal one to other. Note that p0 can be represented in the form p0 = −∇y τ (x, y).

(9.2.16)

Indeed, the vector ∇y τ (x, y) is directed at the point y in the direction of outward normal to the Riemannian sphere τ (x, y) = constant centered at x. On the other hand, this vector is in the tangent direction to (x, y) at y, but in opposite to p0 direction. Hence, we need to put sign (−) in (9.2.16) to obtain p0 . The formulae (9.2.9) and (9.2.16) imply the following relation: ζ =−

1 ∇y τ 2 (x, y), 2n2 (y)

(9.2.17)

which expresses the Riemannian variable ζ through the geodesic distance between the points x and y.

240

9 The Inverse Kinematic Problems

9.3 The One-Dimensional Problem Let R2+ = {(x, y) ∈ R2 | y ≥ 0} and a speed c in this half-space be a positive function of y, i.e., c = c(y). Suppose that c ∈ C2 [0, ∞) and its derivative c (y) is positive on [0, ∞). Then a perturbation produced at the origin (0, 0) reaches to a point (ξ, 0), ξ > 0, along a smooth curve, which belongs (except of its ends) to R2+ and it is symmetric with respect to the line x = ξ/2. In geophysics this curve is called the ray. From a mathematical point of view, it is a geodesic line for the Riemannian metric with an element of a length dτ determined by the formula  dτ = dx 2 + dy 2 /c(y). Denote by t (ξ ) the corresponding travel time along the ray. For ξ small enough, t (ξ ) is a single-valued monotonic increasing function. It may be false if ξ is not small. Assume here that t (ξ ) is a single-valued function on the interval (0, ξ0 ). Consider the following problem: given t (ξ ) for ξ ∈ (0, ξ0 ) find c(y) in R2+ , where it is possible. The problem is called the one-dimensional inverse kinematic problem. This problem has been solved by G. Herglotz [66] in the beginning of the last century. He obtains an explicit formula for the solution to the problem. Below we explain the main ideas, which solve the problem. Consider the function τ (x, y), which means here the travel time from the origin to point (x, y) ∈ R2+ . Then eikonal equation has the form 

∂τ (x, y) ∂x

2

 +

∂τ (x, y) ∂y

2 = n2 (y),

(9.3.1)

where n(y) = 1/c(y). Use now Euler’s equations (9.2.6). Then the equations can be rewritten as follows dx p = 2 , dτ n (y) dp = 0, dτ

dy q = 2 , dτ n (y) dq = (ln n(y)) . dτ

(9.3.2)

Here p = τx (x, y) and q = τy (x, y). It follows from these equations that p is a constant along the geodesic line (x, y) joined the origin and the point (x, y). Particularly, for (ξ, 0) the parameter p is determined by ξ , i.e., p = p(ξ ).  From Eq. (9.3.1) we deduce that q = ± n2 (y) − p2 . The sign (+) should be taken at those points of the geodesic line where τy (x, y) > 0 and sign (−) for the points in which τy (x, y) < 0. From (9.3.2) we find that p dx = ± , 2 dy n (y) − p2

n2 (y) dτ = ± . dy n2 (y) − p2

(9.3.3)

9.3 The One-Dimensional Problem

241

Suppose that the equation n(y) = p has the solution y = η > 0, η = η(p). Then the ray (ξ, 0) tangents to the straight line y = η and represents the symmetric arc passing through the origin and the point (ξ, 0). This arc has its vertex at the line y = η and belongs to the strip {(x, y)| 0 ≤ y ≤ η}. The coordinates of the vertex are (ξ/2, η). In Eqs. (9.3.3) the sign (+) corresponds to the part of the ray from the origin to point (ξ/2, η) and (−) to the remain part. Integrating (9.3.3), one finds the relations for ξ and t (ξ ) in the form 

η

ξ =2 0





p dy

η

t (ξ ) = 2

n2 (y) − p2

0

n2 (y) dy  . n2 (y) − p2

(9.3.4)

Recall that here n(η) = p. Equations (9.3.4) form the complete system of relations for the inverse problems. To give an analysis of this system, we make a change of variables in the integrals by introducing the new integration variable z = n(y). Since n(y) is a monotone decreasing function, there exists an inverse monotone function y = f (z) such that n(f (z)) ≡ z. Note that the derivative f  (z) is negative and f (p0 ) = 0, where p0 = n(0). Then the relations (9.3.4) can be rewritten in the following form  ξ =2

p

p0



p f  (z) dz  , z2 − p 2

t (ξ ) = 2

p

p0

z2 f  (z) dz  , z2 − p 2

p ≤ p0 .

(9.3.5)

The first relation in (9.3.5) presents ξ as a function of p, i.e., ξ = ξ(p), while the second relation determines the other function t = tˆ(p) := t (ξ(p)). The pair ξ(p), t (ξ(p)) gives the parametric representation of the function t = t (ξ ). Note that the function tˆ(p) can be presented in the form 

tˆ(p) = p ξ(p) + 2

p

'

z2 − p2 f  (z) dz ,

p ≤ p0 .

(9.3.6)

p0

From (9.3.6) we deduce: t  (ξ ) =

tˆ (p) = p, ξ  (p)

p ≤ p0 .

(9.3.7)

Hence, having the function t (ξ ) we can find the correspondence between ξ and p, i.e. the function ξ = ξ(p). This function is defined for p ∈ [p1 , p0 ], where p1 = t  (ξ0 ). Then we can use the first equation (9.3.5) in order to find f (z) for z ∈ [p1 , p0 ]. Namely, 1 f (s) = π



p0 s

ξ(p) dp  , p2 − s 2

s ∈ [p1 , p0 ].

(9.3.8)

242

9 The Inverse Kinematic Problems

Indeed, 1 π =

1 π



s p0

 

p0 s z

s

ξ(p) dp 2  = π p2 − s 2



p0 s

2p dp  (p2 − s 2 )(z2 − p2 )



p



f  (z) dz

p dp (p2 − s 2 )(z2 − p2 )  s f  (z) dz = f  (z) dz = f (s). p0

p0

This means that n(y) = f −1 (y) is determined for y ∈ [0, f (p1 )]. Thus, given t (ξ ) for ξ ∈ [0, ξ0 ] one can uniquely find function f (z) for z ∈ [t  (ξ0 ), p0 ] and then n(y) inside the layer y ∈ [0, f (t  (ξ0 )]. If c (y) ≤ 0 for all y, then the ray (ξ, 0) reaches the point (ξ, 0) along axis x. Hence, t (ξ ) = |ξ |/c(0) and only c(0) can be found in the inverse problem. A more interesting case is when the function c (y) is positive for [0, y0 ) and changes sign at y = y0 > 0. As was noted, in this case c(y) can be determined uniquely for y ∈ [0, y0]. Remark that it is impossible uniquely find c(y) also for y > y0 using the data of the inverse problem. A characterization of a set of all possible solutions to the inverse problem in this case was given by Gerver and Markushevich [44].

9.4 The Two-Dimensional Problem Let ⊂ R2 be a compact domain with C 1 -smooth boundary S. Assume, furthermore, that the positive function n(x) ⊂ C 2 ( ) is defined in = ∪ S. Suppose that the family of geodesic lines (x, y), x ∈ , y ∈ S, of the Riemannian metric dτ = n(x)|dx| is regular in and S is convex with respect to geodesics (x, y), x ∈ S, y ∈ S. Such functions n(x) we shall call admissible. We define the function τ (x, y) for x ∈ , y ∈ S, as the Riemannian distance (travel time, from a physical point of view) between x and y. Consider the problem: find n(x) in from τ (x, y) given for x ∈ S and y ∈ S. We will derive now a stability estimate and uniqueness theorem for the solution of this inverse problem, following to [103]. Let S be given in the parametric form S = {y ∈ R2 | y = χ(t), t ∈ [0, T ]}, where χ(t) is a C 1 -smooth periodic function with the period T . We assume that an increase of t corresponds to mowing point y = χ(t) in counterclockwise. Then the output τ (x, y), x ∈ S and y ∈ S, of the inverse problem can be written as follows: τ (χ(s), χ(t)) = g(s, t),

(s, t) ∈ [0, T ] × [0, T ],

(9.4.1)

where g(s, t) is a given function. Theorem 9.4.1 Let n1 (x) and n2 (x) be admissible functions and g1 (s, t) and g2 (s, t) be the output (9.4.1) for the Riemannian metrics dτ1 = n1 (x)|dx| and

9.4 The Two-Dimensional Problem

243

dτ2 = n2 (x)|dx|. Then the following stability estimate holds 1

n1 − n2 L2 ( ) ≤ √ g1 − g2 H1 ([0,T ]×[0,T ]) . 2 π

(9.4.2)

Proof Let nj (x) for j = 1, 2 be admissible functions and τj (x, y) corresponding them Riemannian distances between x and y. Introduce the functions τˆj (x, t) = τj (x, χ(t)). Then τˆj (χ(s), t) = gj (s, t),

(s, t) ∈ [0, T ] × [0, T ],

j = 1, 2,

(9.4.3)

and ∇x τˆj (x, t) = nj (x)ν(θj ),

ν(θj ) = (cos θj , sin θj ),

j = 1, 2,

(9.4.4)

where θj = θj (x, t) is the angle between the tangent line to the geodesic j (x, χ(t)) and axis x1 . Denote τ˜ (x, t) = τˆ1 (x, t) − τˆ2 (x, t), θ˜ (x, t) = θ1 (x, t) − θ2 (x, t),

n(x) ˜ = n1 (x) − n2 (x),

g(s, ˜ t) = g1 (s, t) − g2 (s, t).

From Eqs. (9.4.4) we get ˜ ∇x τ˜ (x, t) = n(x)ν(θ 1 ) + n2 (x)(ν(θ1 ) − ν(θ2 )).

(9.4.5)

Multiplying both sides on ν(θ1 ) we obtain ν(θ1 ) · ∇x τ˜ (x, t) = n(x) ˜ + n2 (x)(1 − cos θ˜ ).

(9.4.6)

Then taking derivative with respect to t, we exclude n(x) ˜ and obtain the equation ∂ ∂ θ˜ [ν(θ1 ) · ∇x τ˜ (x, t)] = n2 (x) sin θ˜ , ∂t ∂t

x ∈ , t ∈ [0, T ].

(9.4.7)

Let ν ⊥ (θ1 ) = (− sin θ1 , cos θ1 ). Multiply both sides of (9.4.7) on 2ν ⊥ (θ1 ) · ∇x τ˜ (x, t), we get ∂ [ν(θ1 ) · ∇x τ˜ (x, t)] ∂t ∂ θ˜ . = 2n2 (x)(ν ⊥ (θ1 ) · ∇x τ˜ (x, t)) sin θ˜ ∂t

2(ν ⊥ (θ1 ) · ∇x τ˜ (x, t))

(9.4.8)

244

9 The Inverse Kinematic Problems

Transform both sides of the latter equality separately. For the left hand side use the identity 2(ν ⊥ (θ1 ) · ∇x τ˜ (x, t))

∂ [ν(θ1 ) · ∇x τ˜ (x, t)] ∂t

∂ {[ν(θ1 ) · ∇x τ˜ (x, t)](ν ⊥ (θ1 ) · ∇x τ˜ (x, t))} ∂t ∂ ∂θ1 ∂ + , [τ˜t (x, t)τ˜x2 (x, t)] − [τ˜t (x, t)τ˜x1 (x, t)] + |∇x τ˜ (x, t)|2 ∂x1 ∂x2 ∂t =

(9.4.9)

which can be checked directly. On the other hand, we have ∂ θ˜ 2n2 (x)(ν ⊥ (θ1 ) · ∇x τ˜ (x, t)) sin θ˜ ∂t ∂ θ˜ ∂t    ∂ 2 ∂ θ˜ 1 ˜ . =− n2 (x) θ˜ − sin(2θ) = −2n22 (x) sin2 θ˜ ∂t ∂t 2 = 2n2 (x)[ν ⊥ (θ1 ) · (n1 (x)ν(θ1 ) − n2 (x)ν(θ2 )] sin θ˜

(9.4.10)

As a result of the equalities (9.4.8)–(9.4.10) we arrive at the relation   ∂ 1 ˜ [ν(θ1 ) · ∇x τ˜ (x, t)](ν ⊥ (θ1 ) · ∇x τ˜ (x, t)) + n22 (x) θ˜ − sin(2θ) ∂t 2 ∂θ1 ∂ ∂ = 0. [τ˜t (x, t)τ˜x2 (x, t)] − [τ˜t (x, t)τ˜x1 (x, t)] + |∇x τ˜ (x, t)|2 + ∂x1 ∂x2 ∂t (9.4.11) Note that in the curly brackets of the latter equality stands the T periodic function of t. Therefore, if one takes integral with respect to t over [0, T ] of the first therm, it vanishes. Integrating (9.4.11) over with respect to x and over [0, T ] with respect to t, we get 

T 0

  ∂ ∂ [τ˜t (x, t)τ˜x2 (x, t)] − [τ˜t (x, t)τ˜x1 (x, t)] ∂x2 ∂x1 ∂θ1  dxdt = 0. +|∇x τ˜ (x, t)|2 ∂t

(9.4.12)

Using Gauss’s formula and the condition (9.4.3), we obtain 

T 0



T 0

  g˜t (s, t)g˜s (s, t)dsdt + 0

T

|∇x τ˜ (x, t)|2

∂θ1 dtdx = 0. ∂t

(9.4.13)

9.4 The Two-Dimensional Problem

245

In the latter equality ∂θ1/∂t ≥ 0, due the regularity condition of geodesics and |∇x τ˜ (x, t)|2 = |n1 (x)ν(θ1 ) − n2 (x)ν(θ2 )|2 = n˜ 2 (x) + 2n1 (x)n2 (x)(1 − cos θ˜ ) ≥ n˜ 2 (x). Therefore 

 n˜ 2 (x)dx ≤ −

2π

0

T



T

g˜t (s, t)g˜s (s, t)dsdt.

0

Here H 1 ([0, T ] × [0, T ]) is the Sobolev space of square integrable functions g(s, t) in the domain {(s, t) : s ∈ [0, T ], t ∈ [0, T ]} together with first partial derivatives with respect to s and t. The required estimate (9.4.2) follows from this inequality.   As a consequence of Theorem 9.4.1 we obtain the following uniqueness theorem. Theorem 9.4.2 Let n1 (x), n2 (x) and g1 (s, t), g2 (s, t) be the functions defined in Theorem 9.4.1 and g1 (s, t) = g2 (s, t). Then n1 (x) = n2 (x) for all x ∈ . We refer to the book [133] and papers [44, 66, 103, 122] for some other aspects of the inverse kinematic problems. Recently, an effective iterative method for reconstruction of the refractive index of a medium from time-off-flight measurements has been proposed in [140].

Part III

Inverse Problems in Wave Phenomena and Vibration

Chapter 10

Inverse Problems for Damped Wave Equations

Inverse problems for wave equations have been extensively studied in the last 50 years due to a large number of engineering and technological applications. The objective of this chapter is to present an analysis of the two basic inverse coefficient problems related to 1D damped wave equations m(x)ut t + μ(x)ut = (r(x)ux )x and m(x)ut t + μ(x)ut = (r(x)ux )x + q(x)u, T := {(x, t) ∈ R2 : x ∈ (0, ), t ∈ (0, T )}, when  > 0 and T > 0 are finite. Here and below μ(x) ≥ 0 is the damping coefficient. Specifically, we study the inverse problem of identifying the unknown principal coefficient r(x) from Dirichlet-to-Neumann operator, and the inverse problem of recovering the unknown potential q(x) from either from Neumann-toDirichlet or Dirichlet-to-Neumann operators. For the first class of inverse problems we refer readers to [16, 22, 23, 30, 129, 130, 132, 135], and for the second class of inverse problems we recommend [22, 23, 125, 129]. For the inverse problems associated with the recovery of the potential in the damped wave equation we refer readers to [136] and [137]. Observe that the history of notion Dirichlet-to-Neumann map dates back at least 30 years to when for elliptic operators the terminology “Dirichlet-to-Neumann data map” was fist introduced in [84, p. 290] and then in [85, p. 644]. Earlier, in 1954 at the International Congress of Mathematicians held at Amsterdam, I.M. Gel’fand has formulated the following inverse problem for the Schrödinger equation −ψ + V (x) · ψ = k 2 ψ, x ∈ ⊂ R3 , where is a bounded domain [42]. Can the potential V (x) be determined by the operator R(k), which is defined by Gelfand as the operator which establishes a relationship between the function ψ(x) and its normal derivative ∂ψ/∂ν on the surface ∂ . Here, the terminology “operator introduced by Wigner”, instead of the terminology “Dirichlet-to-Neumann operator” has been used. This work opened a new research field in inverse problems, namely, determination of an unknown coefficient from the Dirichlet-to-Neumann operator. After 33 years, the Gel’fand’s problem in its original formulation has been solved in [119]. At present, it is also well known that

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_10

249

250

10 Inverse Problems for Damped Wave Equations

the Gel’fand’s problem at fixed energy is related with the Calderon inverse boundary value problem [26] for the electrical conductivity equation (see [115]). The study of inverse problems for wave equation were initiated in [22, 23] and [129, 130]. For the one-dimensional undamped (pure) wave equation, the problem of recovering a potential on half-line was considered in details in the books [19] and [132]. It should be emphasized that in contrast to the inverse problems for undamped wave equation studied in unbounded domain, in the inverse problems for damped wave equation one needs, first of all, to take into account the damping factor μ(x)ut , the presence of which changes the entire structure of the problem as well as its solution. Specifically, the damping factor leads to a number of characteristic features in the propagation of waves through the conductor and in reflection from conducting surfaces. The first consequence of these is the formation of various domains, called “subdomains defined by characteristics”, by the wave reflected from the boundary z = . It is precisely in these areas that all analyzes related to direct, adjoint and inverse problems should be carried out. This phenomenon has first been examined in detail in [135]. Furthermore, different from the inverse problems for parabolic and elliptic PDEs, all properties of an inverse problem solution for wave equation are very much related to behaviour of the direct problem solution in the subdomains formed by the characteristics, also along these characteristics. It turns out that it is more appropriate and suitable to consider the direct problem, and as a consequence, the inverse problem in these subdomains since, as a rule, the solution or/and its partial derivatives have singularities (discontinuities) along the characteristic lines. Remark that the analysis given in subsequent sections is mainly based on ideas proposed in [131] where the quantitative analysis of regularity of the fundamental solution to the second order hyperbolic equation was first given. Notice also that the structure of the fundamental solution of the hyperbolic equation was studied in the famous book [49] of Jacques Hadamard, where it was shown that the fundamental solution is singular not only at a point, but along a certain surface which must be characteristic line.

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator In this section we study the inverse problem of identifying the unknown principal coefficient r(x) in ⎧ ⎪ ⎨ m(x)ut t + μ(x)ut = (r(x)ux )x , (x, t) ∈ T , (10.1.1) u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ u(0, t) = s(t), u(, t) = 0, t ∈ (0, T ),

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

251

from the Neumann boundary measured output f (t) given at the left boundary x = 0 of a bar: f (t) := r(0)ux (0, t), t ∈ (0, T ].

(10.1.2)

Here and below it is assumed that the following conditions hold: ⎧ T ≥ 2z(), ⎪ ⎪ ⎪ ⎪ ⎨ 0 < m0 ≤ m(x) ≤ m1 , 0 ≤ μ(x) ≤ μ1 , 0 < r0 ≤ r(x) ≤ r1 , ⎪ m ∈ H 2 (0, ), μ ∈ H 1 (0, ), r ∈ H 2 (0, ), ⎪ ⎪ ⎪ ⎩ s ∈ H j +1 (0, T ),

(10.1.3)

where the nonnegative integer j will be defined below. z = z(x) is the new variable defined through the coefficients as follows: 

x

z = z(x) = 0

dξ , c(ξ )

 c(x) =

r(x) m(x)

1/2 .

(10.1.4)

We introduce the nonlinear input-output operator or, equivalently, the Dirichletto-Neumann operator corresponding to the inverse problem (10.1.1)–(10.1.2) as follows:  (r)(t) := r(0)ux (0, t; r), t ∈ (0, T ), r ∈ R, (10.1.5)  : R → L2 (0, T ), where u(x, t) ≡ u(x, t; r) is the solution of the direct problem (10.1.1) corresponding to a given admissible coefficient r(x) from the set of admissible coefficients R which will be defined below. Then the inverse problem can be reformulated as the nonlinear operator equation (r)(t) = f (t), t ∈ (0, T ], r ∈ R.

(10.1.6)

The problem (10.1.1)–(10.1.2) is a basic mathematical model of the inverse coefficient problem for the wave equation which includes also the damping coefficient [135]. Further, the inputs in (10.1.1) are assumed to have a minimal regularity. On the other hand, this inverse problem is itself one of the most relevant problems in inverse problems theory and applications from the point of view of the Dirichlet-toNeumann operator for the wave equation. Before studying the inverse coefficient problem we would like to make the following remark related to the Dirichlet input s(t) in (10.1.1). It is natural to suppose that s(t) starts at t = 0, that is, s(t) is not identically zero in (0, ε) for any ε > 0. Otherwise, i.e. if for some ε0 > 0, s(t) ≡ 0 for all t ∈ (0, ε0 ), the

252

10 Inverse Problems for Damped Wave Equations

solution to (10.1.1) vanishes in the domain {(x, t)| 0 ≤ x ≤ , 0 ≤ t ≤ ε0 } and we can replace t = 0 with t = ε0 .

10.1.1 Structure of Solutions of Wave Equation and Dependence on the Dirichlet Input It is well known that a solution of any inverse problem related to PDEs highly depends on how deeply the corresponding direct problem is well studied. In the context of wave propagation phenomena injectivity, compactness and Lipschitz continuity properties of the Dirichlet-to-Neumann operator is very much related to behaviour of the solution of the direct problem along the characteristic lines of the wave equation. Hence one needs to analyze first the structure of the solution of this problem, especially along these characteristics. In general, the structure of the fundamental solution of hyperbolic equation was studied in the famous book [49] where it was shown that the fundamental solution is singular not only at a point, but along a certain surface which must be characteristic line. Quantitative analysis of regularity of the fundamental solution to the second order hyperbolic equation was first given in [131]. In particular, it was proved that if the coefficients of the hyperbolic equation have enough regularity, then the fundamental solution is also regular in a conoid. The approach proposed in [131] forms the basis of the microlocal analysis of the solution of the direct problem (10.1.1) given in this subsection. To understand the structure of the solution, let us look at the simplest model, assuming that the inputs in the direct problem (10.1.1) are given as follows: m(x) = 1, μ(x) = 0, r(x) = c2 > 0, s(t) = 1 and T = 2/c. Evidently, the function u(x, t) = θ0 (t − x/c) − θ0 (t + x/c − 2/c), (x, t) ∈ T , is the solution of (10.1.1) in the generalized sense, since ut t = δ  (t − x/c) − δ  (t + x/c − 2/c) and uxx = [δ  (t − x/c) − δ  (t + x/c − 2/c)]/c2. Here and below,  θ0 (s) =

0, s < 0, 1, s ≥ 0

(10.1.7)

is the Heaviside function and δ(s) is the Dirac delta function. This example provides further insight into the structure of the solution of the hyperbolic problem (10.1.1). Namely, it shows that the Heaviside function depending on the characteristic lines t − z(x) = 2n and t + z(x) = 2(n + 1), n = 0, 1, 2, . . . , must somehow be within the structure of the solution to this problem. Theorem 10.1.1 Let N = N(T ) ≥ 1 be an integer such that 2Nz() ≤ T < 2(N + 1)z() := T1 . Suppose that the inputs satisfy the conditions (10.1.3). Denote

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

253

by u0 (x, t) the solution of problem (10.1.3) with the inputs s(t) = 1 and T = T1 . Then the following representation holds: u0 (x, t) = v0 (x, t) +

N  [α2n+1 (x)θ0 (t − z(x) + 2nz()) n=0

+α2n+2 (x)θ0(t + z(x) − (2n + 2)z() )].

(10.1.8)

Here, the function v0 ∈ L2 (0, T1 ; H01 (0, )), with v0,t (x, t) ∈ L2 (0, T1 ; L2 (0, )) is the weak solution of the following problem ⎧ ⎪ ⎨ m(x)v0,t t + μ(x)v0,t = (r(x)v0,x )x + F (x, t), (x, t) ∈ T1 , v0 (x, 0) = 0, v0,t (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ v0 (0, t) = 0, v0 (, t) = 0, t ∈ [0, T1 ],

(10.1.9)

where H01 (0, ) = {v ∈ H 1 (0, ) | v(0) = 0, v() = 0}, T1 := {(x, t) ∈ R2 : x ∈ (0, ), t ∈ (0, T1 )}, F (x, t) =

N   [(r(x)α2n+1 (x)) θ0 (t − z(x) + 2nz()) n=0  (x)) θ0 (t + z(x) − (2n + 2)z())], +(r(x)α2n+2

(10.1.10)

and '  ⎧ $ 2n 4 m(0)r(0) exp − 1 x √μ(ξ )dξ ⎪ α , x ∈ [0, ], (x) = q ⎪ 2n+1 m(x)r(x) 2 0 ⎪ m(ξ )r(ξ ) ⎨ '  $ (10.1.11) 2n+1 4 m(0)r(0) exp − 1  √μ(ξ )dξ α , x ∈ [0, ], (x) = −q ⎪ 2n+2 ⎪ m(x)r(x) 2 x m(ξ )r(ξ ) ⎪ ⎩ n = 0, 1, 2, . . . , N, with 

1 q = exp − 2

 0



μ(x)dx √ m(x)r(x)

≤ 1.

(10.1.12)

Further, for any T ≤ T1 , for the weak solution of problem (10.1.9) the following stability estimates hold:

v0,t 2L2 (0,T ;L2 (0,)) ≤ C1 ( m 2H 2 (0,) + r 2H 2 (0,) + μ 2H 1 (0,)),

(10.1.13)

v0,x 2L2 (0,T ;L2 (0,)) ≤ C2 ( m 2H 2 (0,) + r 2H 2 (0,) + μ 2H 1 (0,)),

(10.1.14)

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10 Inverse Problems for Damped Wave Equations

where C1 , C2 > 0 depend only on the constants defined in (10.1.3), and also on  > 0 and T > 0. Moreover, v0 (x, t) = 0 in the domain D0 = {(x, t) | x ∈ [0, ], 0 ≤ t < z(x)}.

(10.1.15)

Proof Represent (formally) u0 (x, t) in the form (10.1.8). Substituting this representation in (10.1.1) and equating to zero the terms under the Dirac delta function we find: ⎧  μ(x)α2n+1 (x) = −2α2n+1 (x)r(x)z (x) − α2n+1 (x)(r(x)z (x)) , ⎪ ⎪ ⎪ ⎪ ⎨ α2n+1 (0) = δ0n − α2n (0), (10.1.16)  ⎪ (x)r(x)z (x) + α2n+2 (x)(r(x)z(x)) , μ(x)α2n+2 (x) = 2α2n+2 ⎪ ⎪ ⎪ ⎩ α2n+2 () = −α2n+1 (), n = 0, 1, 2, . . . , N, where δ0n is Kronecker delta, i.e. δ0n = 1, if n = 0 and δ0n = 0 overwise. It is assumed formally that α0 (0) = 0. The above boundary conditions for α2n+1 , α2n+2 are selected to ensure that the boundary conditions in (10.1.9) for v0 (x, t) are homogeneous. Therefore, the function v0 (x, t) solves problem (10.1.9) with F (x, t) defined in (10.1.10). Let us integrate the coupled system of Eqs. (10.1.16). Note that according to (10.1.4) we have r(x)z (x) = (m(x)r(x))1/2. Hence we can rewrite equation for α2n+1 (x) as follows:  (x) + α2n+1

  1 d μ(x) 1/2 log(m(x)r(x)) + α2n+1 (x) = 0, 2 (m(x)r(x))1/2 dx

or   x   d μ(ξ )dξ 1 = 0. α2n+1 (x)(m(x)r(x))1/4 exp dx 2 0 (m(ξ )r(ξ ))1/2 In the same way we can derive the equation for α2n+2 (x):      1 d μ(ξ )dξ = 0. α2n+2 (x)(m(x)r(x))1/4 exp dx 2 x (m(ξ )r(ξ ))1/2

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

255

Integrating the above differential equations and using the boundary data in (10.1.16) we find: ⎧ ' 

$ 1 x √μ(ξ )dξ 4 m(0)r(0) ⎪ α , (x) = (δ − α (0)) exp − ⎪ 2n+1 0n 2n ⎪ m(x)r(x) 2 0 m(ξ )r(ξ ) ⎪ ⎪ ⎪ ⎪ 0 ≤ x ≤ , ⎪ ⎨ ' 

$  )dξ m()r() α2n+2 (x) = −α2n+1 () 4 m(x)r(x) , exp − 12 x √μ(ξ ⎪ m(ξ )r(ξ ) ⎪ ⎪ ⎪ ⎪ 0 ≤ x ≤ , ⎪ ⎪ ⎪ ⎩ n = 0, 1, 2, . . . , N. √ By the assumption α0 (0) = 0 we have: α1 () = q 4 m(0)r(0)/(m()r()) and then α2 (0) = −q 2 , where q > 0 is defined by (10.1.12). This leads to formulae (10.1.11). From the conditions m ∈ H 2 (0, ), r ∈ H 2 (0, ) and μ ∈ H 1 (0, ) it follows that αn ∈ H 2 (0, ), n = 1, 2, 3, . . . , 2(N + 1), and hence F ∈ L2 (0, T ; L2 (0, )) for any T ≤ T1 . By the weak solution theory of hyperbolic PDEs (see, for instance, [38]) we conclude that v0 ∈ L2 (0, T ; H01(0, )), v0,t ∈ L2 (0, T ; L2 (0, )). Moreover, we deduce from (10.1.9) that v0 (x, t) = 0 for all (x, t) ∈ D0 , since F (x, t) = 0 in D0 , by (10.1.10). The domain D0 defined in (10.1.15) is plotted in (Fig. 10.1 on the left).   To prove estimates (10.1.13) and (10.1.14) we multiply both sides of (10.1.9) by 2v0,t (x, t), integrate over t := (0, ) × (0, t), t ∈ (0, T ] and use the initial and boundary conditions in (10.1.9). Then we obtain the following energy identity: 

 0

2 m(x)v0,t

2 + r(x)v0,x







dx + 2 t

2 μ(x)v0,τ dxdτ

=2

F (x, τ )v0,τ dxdτ, t

t 4z() t = −z(x) + 4z(l)

3z()

τ

D3

2z()

t = z(x) + 2z(l)

2z()

D1

D2 t = −z(x) + 2z()

z()

z()

D1 t = z(x)

0

τ = −z(ξ) + 2z()

t − z(x)

D0 

x

(x, t)

t + z(x) = τ + z(ξ) (ξ1 , τ1 )

ξ

τ = z(ξ) D0

0



Fig. 10.1 Domains Dn , n = 0, 1, 2, 3 (left); integration scheme in the domain D1 (right)

256

10 Inverse Problems for Damped Wave Equations

t ∈ [0, T ]. Here we use the bounds in (10.1.3) and apply the ε-inequality with ε = m0 to the last right-hand-side integral to get the main integral inequality 



m0 0

 2 v0,t dx + r0



≤ m0 t



0

2 v0,τ dxdτ

 2 v0,x dx + μ0

t



1 + m0

2 v0,τ dxdτ

F 2 (x, τ )dxdτ, t ∈ [0, T ].

(10.1.17)

t

As a first consequence of this integral inequality we find: 

 0

 2 v0,t dx ≤

t

2 v0,τ dxdτ +

1 m20

 F 2 (x, τ )dxdτ, t ∈ [0, T ]. t

From Gronwall-Bellman inequality it follows that 

 0

2 v0,t dx ≤

et m20

 F 2 (x, t)dxdt, t ∈ [0, T ].

(10.1.18)

T

Using this estimate and (10.1.17) we deduce the inequality 

 0

2 v0,x dx

et ≤ 2 m0 r0

 F 2 (x, t)dxdt, t ∈ [0, T ].

(10.1.19)

T

To estimate the norm F L2 ((0,)×(0,T )) we should estimate the norm

(rαn ) L2 (0,), as it follows from formula (10.1.10). To this end, denote by a(x) = (m(x)r(x))−1/4. In view of (10.1.11) we have: ⎛ ⎞ x q 2n 1 a(x) exp ⎝− μ(ξ )a 2 (ξ )dξ ⎠ , α2n+1 (x) = a(0) 2 ⎛

0

1 q 2n+1 a(x) exp ⎝− α2n+2 (x) = − a(0) 2



⎞ μ(ξ )a 2 (ξ )dξ ⎠ ,

n = 0, 1, . . . .

x

Calculate the first and second derivatives of α2n+1 (x). We find: ⎛ ⎞   x 1 1  a  (x) − μ(x)a 3(x) exp ⎝− α2n+1 (x) = μ(ξ )a 2 (ξ )dξ ⎠ , a(0) 2 2 q 2n

0

  q 2n 1   μ (x)a 3 (x) + 3μ(x)a 2(x)(x)a (x) (x) = α2n+1 a  (x) − a(0) 2 ⎛ ⎞    x 1 1  1 μ(ξ )a 2 (ξ )dξ ⎠ . − a (x) − μ(x)a 3(x) μ(x)a 2(x)) exp ⎝− 2 2 2 0

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

257

Further,   m (x) r  (x) 1 + a  (x) = − a(x) , 4 m(x) r(x)    1  m (x) r  (x)  + a (x) = − a (x) 4 m(x) r(x)  +a(x)

m (x) − m(x)



m (x) m(x)

2 +

r  (x) − r(x)



r  (x) r(x)

2  .

Since m(x), r(x), μ(x) satisfy conditions (10.1.10) the function a(x) is bounded: (m1 r1 )−1/4 ≤ a(x) ≤ (m0 r0 )−1/4 . Then from the above equalities we obtain the following estimates:



a  2L2 (0,) ≤ C0 m 2H 1 (0,) + r 2H 1 (0,) ,



a  2L2 (0,) ≤ C0 m 2H 2 (0,) + r 2H 2 (0,) , where C0 > 0 depends only on the constants m0 , m1 , r0 , r1 > 0 and on  > 0. Then the relations for derivatives of α2n+1 imply the following estimates:

 

α2n+1

2L2 (0,) ≤ C0 m 2H 1 (0,) + r 2H 1 (0,) + μ 2L2 (0,) ,

 

α2n+1

2L2 (0,) ≤ C0 m 2H 2 (0,) + r 2H 2 (0,) + μ 2H 1 (0,) , with the constant C0 depending only on m0 , m1 , r0 , r1 , μ1 and . Evidently, the same estimates hold also for α2n+2 (x). Since (r(x)αn (x)) = r  (x)αn (x) + r(x)αn (x) we deduce that



(rαn ) 2L2 (0,) ≤ C0 m 2H 2 (0,) + r 2H 2 (0,) + μ 2H 1 (0,) , n = 1, 2, . . . , where C0 > 0 depends only on the constants defined in (3) and on  > 0. In view of formula (10.1.10) we deduce that for any T > 0 the following estimate holds for the norm F 2L2 (0,T ;L2 (0,)):



F 2L2 (0,T ;L2 (0,)) ≤ T C0 m 2H 2 (0,) + r 2H 2 (0,) + μ 2H 1 (0,) . Taking into account this estimate in (10.1.18) and (10.1.19) we arrive at the required estimates (10.1.13) and (10.1.14).  

258

10 Inverse Problems for Damped Wave Equations

Theorem 10.1.2 Suppose that the inputs satisfy conditions (10.1.3). Assume, in addition, that m(x) ∈ C 2 [0, ], r(x) ∈ C 2 [0, ], μ(x) ∈ C 1 [0, ]. Then v0,t (x, t), v0,x (x, t) are piece-wise continuous functions in T1 = (0, ) × (0, T1 ), where T1 > 0 is defined in Theorem 10.1.2. These functions can fail to be continuous only along the characteristic lines t − z(x) = 2n and t + z(x) = 2(n + 1), where n = 0, 1, 2, . . . , N. Proof Rewrite the differential equation for v0 (x, t) in the following form 

 ∂ ∂ ∂ ∂ + c(x) − c(x) v0 ∂t ∂x ∂t ∂x   ∂ ∂ ∂ ∂ − c(x) + c(x) ≡ v0 ∂t ∂x ∂t ∂x μ(x) ∂v0 =− + m(x) ∂t



∂v0  r  (x)  − c(x)c (x) + F (x, t), (10.1.20) m(x) ∂x

(x, t) = F (x, t)/m(x). Introduce the new with c(x) > 0 defined in (10.1.4) and F functions w1 (x, t) and w2 (x, t) by the formulas w1 = v0,t − c(x)v0,x ,

w2 = v0,t + c(x)v0,x .

(10.1.21)

1 (w2 − w1 ). 2c(x)

(10.1.22)

Then we have: v0,t =

1 (w1 + w2 ), 2

v0,x =

From relations (10.1.21)–(10.1.22) it follows that the pair of functions w1 , w2  solves the following coupled system of differential equations: ⎧ (x, t), (x, t) ∈ T1 , w1,t + c(x)w1,x = a(x)w1 + b(x)w2 + F ⎪ ⎪ ⎪ ⎪ ⎨ w − c(x)w = a(x)w + b(x)w + F (x, t), (x, t) ∈ T1 , 2,t 2,x 1 2 (10.1.23) ⎪ ⎪ w1 (x, 0) = 0, w2 (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎩ w1 (0, t) + w2 (0, t) = 0, w1 (, t) + w2 (, t) = 0, t ∈ (0, T1 ), where  r  (x) 1 μ(x) − + c (x) , − 2 m(x) m(x)c(x)  1 μ(x) r  (x)  b(x) = − + − c (x) . 2 m(x) m(x)c(x)

a(x) =

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

259

Notice that the left-hand-sides of the first and second equations in (10.1.23) are the total derivatives of w1 and w2 with respect to t along the characteristic lines (x, t) = 0 t − z(x) = const. and t + z(x) = const., correspondingly. Since F for all (x, t) ∈ D0 , where D0 is defined in (10.1.15), the solution (w1 , w2 ) of problem (10.1.23) vanishes in D0 , i.e, w1 (x, t) = w2 (x, t) = 0 for all (x, t) ∈ D0 . Moreover, from the second equation of (10.1.23) it follows that w2 (x, z(x) + 0) = w2 (x, z(x) − 0) = 0, x ∈ [0, ],

(10.1.24)

i.e. the component w2 (x, t) is a continuous function across the characteristic line t = z(x). Observe that this statement is also true in the general case. That is, the functions w1 (x, t) and w2 (x, t) are continuous across the characteristic lines t = −z(x) + (2n + 1)z() and t = z(x) + 2nz(), n = 0, 1, 2, . . . , correspondingly. Let x = χ(s), s ∈ [0, z()] be the inverse function to s = z(x) introduced in (10.1.4), i.e. s ≡ z(χ(s)) for s ∈ [0, z()]. Consider problem (10.1.23) in each of the domains D2n+1 = {(x, t)| x ∈ [0, ], z(x) + 2nz() < t < −z(x) + 2(n + 1)z()}, D2n+2 = {(x, t)|x ∈ [0, ], −z(x) + 2(n + 1)z() < t < z(x) + 2(n + 1)z()}, n = 0, 1, . . . , N, separately (the domains on the left in Fig. 10.1). As a first step, let us assume that (x, t) ∈ D1 . Consider the characteristic line t + z(x) = τ + z(ξ ) and its intersection point (ξ1 , τ1 ) with the characteristic line τ = z(ξ ) on the ξ τ -plane as plotted on the right in Fig. 10.1. The coordinates of this point are ξ1 (x, t) = χ((t + z(x))/2),

τ1 (x, t) = (t + z(x))/2.

At this point, w2 (ξ1 , τ1 ) = 0 according to (10.1.24). Integrating the second equation of (10.1.23) on the ξ τ -plane along the characteristic line t + z(x) = τ + z(ξ ) or, which is the same, along ξ = χ(t − τ + z(x)), from the point (ξ1 , τ1 ) to (x, t), we find:  t w2 (x, t) = [aw1 + bw2 + (rα1 ) /m](χ(t − τ + z(x)), τ )dτ, (10.1.25) τ1 (x,t )

for all (x, t) ∈ D1 , where α1 (x) is given by formula (10.1.11). The boundary condition at x = 0 in (10.1.23) we can transform to the following one:  w1 (0, t) = −

t t /2

[aw1 + bw2 + (rα1 ) /m](χ(t − τ ), τ )dτ,

(10.1.26)

for all t ∈ (0, 2z() ]. Integrating now the first equation of (10.1.23) on the ξ τ plane along the characteristic line t − z(x) = τ − z(ξ ) or, which is the same, along

260

10 Inverse Problems for Damped Wave Equations

ξ = χ(τ − t + z(x)), from the point (0, t − z(x)) to (x, t) and using the boundary condition (10.1.26), we find:  w1 (x, t) =  −

t

t −z(x)

t −z(x)

(t −z(x))/2

[aw1 + bw2 + (rα1 ) /m](χ(τ − t + z(x)), τ )dτ

[aw1 + bw2 + (rα1 ) /m](χ(t − z(x) − τ ), τ )dτ,

(10.1.27)

for all (x, t) ∈ D1 . Relations (10.1.25) and (10.1.27) form a coupled system of Volterra type equations defined in the domain D1 . By the assumptions a ∈ C 1 [0, ], b ∈ C 1 [0, ] and α ∈ C 2 [0, ], this system has a unique solution w1 , w2  which belongs to C(D1 ). Notice that the function w1 (x, t) is discontinuous across the characteristic τ = z(ξ ), in general. Indeed, if z(x) = t, then w2 (z(x), t) = 0 by (10.1.24). Using this in (10.1.27) we get:  t w1 (χ(t), t) = [aw1 + (rα1 ) /m](χ(τ ), τ )dτ, t ∈ [0, z()]. 0

This is the usual Volterra integral equation which is equivalent to the following Cauchy problem dη(t) = a(χ(t))η(t) + [(rα1 ) /m](χ(t)), t ∈ [0, z()], η(0) = 0, dt for the ordinary differential equation with respect to the function η(t) = w1 (χ(t), t), which solution can be found exactly. Obviously this solution is not zero if (rα1 ) is not identically zero, while w1 (x, z(x) − 0) = 0 by (10.1.24). This implies that w1 (x, t) is discontinuous across the characteristic line τ = z(ξ ).   Let now (x, t) ∈ D2 . Consider the characteristic line t − z(x) = τ − z(ξ ) on the ξ τ -plane, as plotted in Fig. 10.2. The coordinates (ξ2 , τ2 ) of its intersection point with the characteristic line τ + z(ξ ) = 2z() are ξ2 (x, t) = χ(z() − (t − z(x))/2),

τ2 (x, t) = z() + (t − z(x))/2.

From the first equation of (10.1.23) it follows that w1 (x, t) is continuous across the characteristic line τ +z(ξ ) = 2z(). Hence, the value w1 (ξ2 , τ2 ) of this function can be considered as a given value from the previous step. Integrating the first equation of (10.1.23) on the ξ τ -plane along the characteristic line t − z(x) = τ − z(ξ ) from point (ξ2 , τ2 ) to (x, t), we find: w1 (x, t) = w1 (ξ2 (x, t), τ2 (x, t))  t + [aw1 + bw2 + (rα2 ) /m](χ(τ − t + z(x)), τ )dτ, τ2 (x,t )

(10.1.28)

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

3z()

τ

261

D3 D2

τ = z(ξ) + 2z() (x, t)

2z()

(ξ2 , τ2 ) τ = −z(ξ) + 2z()

z()

(, t + z(x) − z())

D1

τ = z(ξ)

0

D0

ξ 

Fig. 10.2 Integration scheme in the domain D2

for all (x, t) ∈ D2 . In view of the last boundary condition at x =  in (10.1.23) this yields: w2 (, t) = −w1 (ξ2 (, t), τ2 (, t))  t − [aw1 + bw2 + (rα2 ) /m](χ(τ − t + z()), τ )dτ, τ2 (,t )

2z() ≤ t < 4z().

(10.1.29)

Integrating the second equation of (10.1.23) on the ξ τ -plane along the characteristic line t + z(x) = τ + z(ξ ) from point (, t + z(x) − z()) to (x, t) and using the boundary condition in (10.1.23), we find  w2 (x, t) =  −

t t +z(x)−z()

t +z(x)−z()

[aw1 + bw2 + (rα2 ) /m](χ(t − τ + z(x)), τ )dτ

[aw1 + bw2 + (rα2 ) /m](χ(τ − t − z(x) + 2z()), τ )dτ,

τ2 (,t +z(x)−z())

−w1 (ξ2 (, t + z(x) − z()), τ2 (, t + z(x) − z())), (x, t) ∈ D2 .

(10.1.30)

262

10 Inverse Problems for Damped Wave Equations

Relations (10.1.28) to (10.1.30) form the system of Volterra type equations in the domain D2 . This system defines a unique solution that belongs to C(D2 ). Note that the function w2 (x, t) is discontinuous across the characteristic τ + z(ξ ) = 2z(). Continuing this process step by step, we similarly prove existence of the unique continuous solution w1 , w2  of problem (10.1.23) in each domain Dn , n = 3, 4, . . . . With (10.1.21) this completes the proof of the theorem.   Theorem 10.1.1 shows that if the input s(t) is a polynomial of degree zero, i.e. s(t) = 1, then the solution v0 (x, t) of problem (10.1.9) is in L2 (0, T ; H01(0, )) with v0,t (x, t) ∈ L2 (0, T ; L2 (0, )). The next theorem shows that increase of the order of the polynomial s(t) = t k /k!, k ≥ 1 leads to the regularity of this solution. Introduce the function θk (t) = t k θ0 (t)/k!, k = 1, 2, . . . , t ≥ 0,

(10.1.31)

where θ0 (t) is defined in (10.1.7). Theorem 10.1.3 Suppose that the inputs in (10.1.1) satisfy conditions (10.1.3). Then the function 

t

uk (x, t) = 0

(t − τ )k−1 u0 (x, τ )dτ, k ≥ 1, (k − 1)!

(10.1.32)

solves problem (10.1.1) with s(t) = t k /k! and T = T1 , where T1 > 0 is defined in Theorem 10.1.1. This function has the following structure: uk (x, t) = vk (x, t) +

N 

[α2n+1 (x) θk (t − z(x) − 2nz())

n=0

+α2n+2 (x) θk (t + z(x) − (2n + 2)z())], (10.1.33) where αk (x) and θk (x) are defined in (10.1.7) and (10.1.31), correspondingly, vk (x, t) is defined by the following formula: 

t

vk (x, t) = x

(t − τ )k−1 v0 (x, τ )dτ, x ∈ [0, ], 0 ≤ t ≤ T1 . (10.1.34) (k − 1)!

Moreover, for any T ≤ T1 , vk ∈ L2 (0, T ; H 2(0, ) ∩ H01 (0, )), Dtn vk ∈ L2 (0, T ; H01 (0, )), Dtn+1 vk ∈ L2 (0, T ; L2 (0, )), n = 1, . . . , k,

(10.1.35)

with

Dtn vk H 1 ( T ) ≤

T k−n

v0 H 1 ( T ) , n = 0, 1, . . . , k. (k − n)!

(10.1.36)

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

263

Proof In a fact, Theorem 10.1.3 is a consequence of Theorem 10.1.1. Indeed, formula (10.1.32) is the result of k-times repeated integration of u0 (x, t) given by (10.1.8) with respect to t. Then the boundary condition at x = 0 in (10.1.1) is replaced by t k /k!. Further, the differential equation for uk (x, t), the initial conditions and the boundary condition at x =  preserve the form. The representation (10.1.33) and formula (10.1.34) are the result of calculating of the integral in (10.1.32) by using representation (10.1.8) for function u0 (x, t). We prove now the assertions in (10.1.35). The second and third assertions follow directly from formula (10.1.34). Further, since vk (x, t) solves problem (10.1.9) with F (x, t) replaced by  Fk (x, t) =

t

0

(t − τ )k−1 F (x, τ )dτ, (k − 1)!

from Eq. (10.1.9) for vk (x, t) it follows that vk,xx ∈ L2 (0, T ; L2 (0, )). Evidently vk (0, t) = vk (, t) = 0, by formula (10.1.34). This implies the first assertion of (10.1.35). Finally, the last assertion (10.1.36) of the theorem follows directly from formula (10.1.34).   Theorem 10.1.4 Suppose that the inputs satisfy conditions (10.1.3). Then for any j = 0, 1, 2, . . . the function  u defined by  u0 (x, t) = s(0)u0 (x, t), u0 (x, t) +  uj (x, t) = 

 t j

s (k) (0)

0 k=1

(t − τ )k−1 u0 (x, τ ) dτ (k − 1)!

(10.1.37)

solves problem (10.1.1) with s(t) =

j 

s (k) (0) t k /k!

(10.1.38)

k=0

and T = T1 , where T1 > 0 is defined in Theorem 10.1.1. The function  uj (x, t) has the following structure: vj (x, t) +  uj (x, t) = 

j  N 

s (k) (0) [α2n+1 (x)θk (t − z(x) − 2nz())

k=0 n=0

+α2n+2 (x)θk (t + z(x) − (2n + 2)z())],

(10.1.39)

264

10 Inverse Problems for Damped Wave Equations

where α2n+1 (x), α2n+2 (x) are defined by formula (10.1.12) and  vj (x, t) is given as follows: ⎧  v0 (x, t) = s(0)v0 (x, t), ⎪ ⎪ ⎪ ⎪ ⎨ j $t  −τ )k−1 (10.1.40)  vj (x, t) =  v0 (x, t) + s (k) (0) (t(k−1)! v0 (x, τ )dτ, ⎪ ⎪ x k=1 ⎪ ⎪ ⎩ 0 < x < , 0 ≤ t ≤ T1 , j = 1, 2, . . . . Moreover,  vj ∈ L2 (0, T ; H01 (0, )),  vj,t ∈ L2 (0, T ; L2 (0, )),

(10.1.41)

for any T ≤ T1 , and

 vj H 1 ( T ) ≤

j  Tk k=0

k!

|s (k) (0)| v0 H 1 ( T ) .

(10.1.42)

Proof In a fact, this theorem is a consequence of Theorem 10.1.3, since all formulas of Theorem 10.1.4 immediately follow from the corresponding analogues in Theorem 10.1.3.   Corollary 10.1.1 Let the conditions of Theorems 10.1.2 and 10.1.4 hold. Then vj,x (x, t) are piece-wise continuous functions. These functions can  vj,t (x, t) and  fail to be continuous only along the characteristic lines t − z(x) = 2nz() and t + z(x) = 2(n + 1)z(), n = 0, 1, 2, . . . , N only. Introduce now the function # uj (x, t) = u(x, t) −  uj (x, t),

j = 0, 1, . . . ,

(10.1.43)

where u(x, t) is the solution of the direct problem (10.1.1) and  uj (x, t) is the solution of the same problem with s(t) replaced by (10.1.38). Notice that the structure of  uj (x, t) is given by (10.1.39). For each j the function # uj (x, t) solves the problem ⎧ uj,t = (r(x)# uj,x )x , (x, t) ∈ T , uj,t t + μ(x)# ⎪ ⎨ m(x)# uj,t (x, 0) = 0, x ∈ (0, ), # uj (x, 0) = 0, # ⎪ ⎩ # uj (0, t) = # sj (t), # uj (, t) = 0, t ∈ (0, T ),

(10.1.44)

where # sj (t) = s(t) −

j  k=0

s (k) (0)

tk . k!

(10.1.45)

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

265

Theorem 10.1.5 Suppose that the inputs satisfy the conditions (10.1.3). Assume, in addition that # sj ∈ H j +1 (0, T ). Then the following energy estimate holds:  E(t) := 0



2 2 [# uj,t (x, t) + # uj,x (x, t)]dx ≤ C32 # sj 2L2 (0,T ) exp(C2 t), (10.1.46)

for all t ∈ [0, T ], where C3 = 2

√ m1 r1 /a0 , C2 = C1 /(m1 r1 ),

C1 = [r1 (m1 + r1 ) + μ1 ]/ + a1 ,

(10.1.47)

a0 = max (m0 , r0 ), a1 = max[0,] |(m(x)r(x)) |. Moreover, if j ≥ 1, then uj ∈ L2 (0, T ; H 2 (0, ) ∩ V 1 (0, )), Dtk+1# uj ∈ L2 (0, T ; V 1 (0, )), Dtk # uj ∈ L2 (0, T ; L2 (0, )), k = 0, 1, . . . , j − 1, Dtk+2#

(10.1.48)

where V 1 (0, )) = {v ∈ H 1 (0, ) | v() = 0}. Proof Multiply both sides of Eq. (10.1.44) by 2# uj,t , integrate over t := (0, ) × (0, t), t ∈ (0, T ] and use the initial and boundary conditions. Then we obtain the following energy identity:  0



 2 2 [m(x)# uj,t (x, t) + r(x)# uj,x (x, t)]dx + 2

+2

 t 0



0

0

t

# sj (τ )r(0)# uj,x (0, τ )dτ

2 μ(x)# uj,τ (x, τ )dxdτ = 0.

Employing the ε-inequality 2λμ ≤ (λ2 /ε + μ2 ε) and taking into account (10.1.3) here we find the following inequality for the energy functional defined in (10.1.46): 1 E(t) ≤ a0 ε



t 0

(# sj (τ ))2 dτ

ε + a0



t



2 r(0)# uj,x (0, τ ) dτ,

(10.1.49)

0

for all t ∈ [0, T ], where a0 > 0 is defined in (10.1.47) and ε > 0 will be chosen below. Let us estimate the last integral in (10.1.49). Multiply both sides of (10.1.44) by 2r(x)# uj,x ( − x). This leads to the following identity:

   2 [2m(x)r(x)# uj,x# uj,t ]( − x) t − [m(x)r(x)# uj,t + r 2 (x)# u2j,x ]( − x)

x

+[2μ(x)r(x)# uj,x# uj,t + (m(x)r(x)) 2 −m(x)r(x)# u2j,t − r 2 (x)# uj,x = 0.



2 # uj,t ]( − x)

(10.1.50)

266

10 Inverse Problems for Damped Wave Equations

Integrate this equality over t and use the initial and boundary conditions in (10.1.44). Then we obtain the following integral identity: 



[2m(x)r(x)# uj,x # uj,t ]( − x)dx

0



t

+ 0

+

 2 [m(0)r(0)(# sj (τ ))2 + r(0)# uj,x (0, τ ) ]dτ

 t 0

−

 0

 t 0

 0

2 [2μ(x)r(x)# uj,x # uj,t + (m(x)r(x)) # uj,t ]( − x)dxdτ

2 2 [m(x)r(x)# uj,t + r 2 (x)# uj,x ]dxdτ = 0, t ∈ [0, T ].

In view of the energy functional (10.1.46) we find: 

t

2  r(0)# uj,x (0, τ ) dτ ≤ m1 r1 E(t) + C1

0



t

E(τ )dτ,

(10.1.51)

0

where the constants C1 > 0 is defined in (10.1.47). Using this inequality in (10.1.49) we get:    t m1 r1 ε C1 t 1 1− ε E(t) ≤ E(τ )dτ + (# s  (τ ))2 dτ. a0 a0 0 a0 ε 0 j

(10.1.52)

Choose ε > 0 as ε = a0 /(2m1 r1 ) to get the inequality: 

t

E(t) ≤ C2 0

4m1 r1 E(τ )dτ + a02



T

(# s j (t))2 dt, t ∈ [0, T ].

0

With the Gronwall-Bellmann inequality this implies the energy estimate (10.1.46). We deduce from (10.1.46) that uj ∈ L2 (0, T ; L2 (0, )). # uj ∈ L2 (0, T ; V 1(0, )), Dt # Let us assume now j ≥ 1. Differentiating Eq. (10.1.44) k + 1 times, k ≤ j − 1, with respect to t we can prove similarly that Dtk+1# uj ∈ L2 (0, T ; V 1 (0, )), Dtk+2# uj ∈ L2 (0, T ; L2 (0, )), k = 0, 1, . . . , j ; j ≥ 1.

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

267

Moreover,

Dtk # uj H 1 ( T ) ≤ C # s (k+1) L2 (0,T ) , k = 0, 1, 2, . . . , j ; j ≥ 0, with the constant C > 0 depending only on T ,  > 0 and the constants defined in (10.1.3). From the above inclusions it follows that # u0,t t ∈ L2 (0, T ; L2 (0.)) and k 2 2 Dt # uj,t t ∈ L (0, T ; L (0.)), if k ≤ j − 1 and j ≥ 1. Then using Eq. (10.1.44) we deduce that Dtk# uj,xx ∈ L2 (0, T ; L2 (0, )) for k = 0, 1, . . . , j − 1 and j ≥ 1, i,e. uj ∈ L2 (0, T ; H 2 (0, ) ∩ V 1 (0, )), Dtk+1# uj ∈ L2 (0, T ; V 1 (0, )), Dtk # Dtk+2# uj ∈ L2 (0, T ; L2 (0, )), k = 0, 1, 2, . . . , j − 1; j ≥ 1.  

This completes the proof.

Corollary 10.1.2 If conditions of Theorems 10.1.2 and 10.1.5 hold, then uj,t (x, t), Dtk# uj,x (x, t), k = 0, 1, . . . , j are continuous functions in T . In Dtk# j j j uj,t t (x, t), Dt # uj,xt (x, t), Dt # uj,xx (x, t) are pieceaddition, if j ≥ 1, then Dt # wise continuous functions in T . These functions can fail to be continuous only along the characteristic lines t − z(x) = 2nz() and t + z(x) = 2(n + 1)z(), n = 0, 1, 2, . . . , N. The above analysis leads us to the following result. Theorem 10.1.6 Suppose that the inputs satisfy the conditions (10.1.3) and s ∈ H j +1 (0, T ), j = 1, 2, . . . . Then for any (x, t) ∈ T the following representation valid: ux (x, t) = w(x, t) − s(0)z (x)

N 

[α2n+1 (x) δ(t − z(x) − 2nz())

n=0

−α2n+2 δ(t + z(x) − (2n + 2)z())], (10.1.53) where vj,x (x, t) w(x, t) = # uj,x (x, t) +  +

j  N 

 s (k) (0) [α2n+1 (x) θk (t − z(x) − 2nz())

k=0 n=1  +α2n+2 (x) θk (t + z(x) − (2n + 2)z())]

−z (x)

j  N 

s (k) (0) [α2n+1 (x) θk−1(t − z(x) − 2nz())

k=1 n=1

−α2n+2 (x) θk−1(t + z(x) − (2n + 2)z())],

(10.1.54)

268

10 Inverse Problems for Damped Wave Equations

with w(x, t) ≡ 0 for all (x, t) ∈ D0 , while # uj,x (x, t) and  vj,x (x, t) are defined in (10.1.43) and (10.1.40), correspondingly, and (# uj +  vj ) ∈ L2 (0, T ; H 2(0, ) ∩ V 1 (0, )), (# uj +  vj )t ∈ L2 (0, T ; V 1 (0, )), (# uj +  vj )x ∈ L2 (0, T ; V 1(0, )), (# uj +  vj )t t ∈ L2 (0, T ; L2 (0, )). If, in addition, the assumptions of Theorem 10.1.2 hold and s(0) = 0, then the function ux (x, t) = w(x, t) is a piece-wise continuous function in T and it can fail to be continuous only along the characteristic lines t − z(x) = 2nz() and t + z(x) = 2(n + 1)z(), n = 0, 1, 2, . . . , N. Representations (10.1.53), (10.1.54) is the result of formulas (10.1.39) and (10.1.43), as well as the result of their differentiation with respect to x. Here we also have used the following equalities  (t) for k ≥ 1. θ0 (t) = δ(t) and θk (t) = θk−1

In representation (10.1.53), the function w(x, t) is the regular part of the function ux (x, t), while the rest of this is a singular generalized function.

10.1.2 Necessary Estimates for the Direct Problem Solution First of all, we need to make a statement about the estimation of the final time T > 0. In order to find the coefficient r(x) in some subinterval x ∈ [0, x0], where x0 ∈ [0, ], one needs to know the output f (t) := r(0)ux (0, t) for all t ∈ (0, 2z(x0)), at least, since to contain information about the coefficient r(x), a signal initiated at the starting point x = 0 must reach point x = x0 and then return back to the same point x = 0. Hence, to determine the unknown coefficient r(x) for all x ∈ [0, ] the output f (t) has to be given in the time interval (0, T ) with T ≥ 2z(). However we do not know z() in advance, since it depends on the unknown coefficient r(x), as formula (10.1.4) shows. Therefore, we can only estimate z() using assumptions (10.1.3) and formula (10.1.4). Based on these, we assume that the final time T > 0 satisfies the following condition:  T > 2 m1 /r0 .

(10.1.55)

If condition (10.1.55) holds, then one can expect that r(x) can uniquely be determined for all x ∈ [0, ] by the output (2). For a simple case, when m(x) = 1 and μ(x) = 0 this conjecture is proved in Sect. 10.1.6. We introduce now the set of admissible inputs W 1 (0, T ) := {s ∈ H 1 (0, T ) | s(0) = 0}

(10.1.56)

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

269

and consider the case j = 0 in Theorem 10.1.5. From (10.1.45) it follows that if j = 0 and s ∈ W 1 (0, T ), then # s0 (t) = s(t). Hence, # uj (x, t) ≡ u(x, t) for j . As a first direct consequence of Theorem 10.1.5 we may deduce the following estimates for the weak solution of the direct problem (10.1.1). Corollary 10.1.3 Assume that conditions (10.1.3) hold and s ∈ W 1 (0, T ). Then for the weak solution u ∈ L2 (0, T ; V 1 (0, )), with ut ∈ L2 (0, T ; L2 (0, )) of the direct problem (10.1.1) the following estimates hold:

ut L2 (0,T ;L2 (0,)) ≤ C4 s  L2 (0,T ) ,

ux L2 (0,T ;L2 (0,)) ≤ C4 s  L2 (0,T ) ,

(10.1.57)

where C4 = C3 ([exp(C2 T ) − 1]/C2)1/2 ,

(10.1.58)

and the constants C2 , C3 > 0 are defined in (10.1.47). Proof Estimates (10.1.57) follow from the energy estimate (10.1.46) with # s(t) replaced by s(t): E(t) ≤ C32 s  2L2 (0,T ) exp(C2 t), t ∈ [0, T ].

(10.1.59)

Integrating this inequality over [0, T ] we arrive at the required estimates (10.1.57).   Indeed, Theorem 10.1.5 asserts that under the conditions j = 1 and s ∈ W 2 (0, T ), where W 2 (0, T ) := {s ∈ H 2 (0, T ) | s(0) = s  (0) = 0},

(10.1.60)

is the admissible set of regular inputs, there exists a regular weak solution of the direct problem (10.1.1) and estimates similar to (10.1.57) hold for this solution also. Theorem 10.1.7 Assume that conditions (10.1.3) hold and s ∈ W 2 (0, T ), where W 2 (0, T ) is defined by (10.1.60). Then there exists the regular weak solution u ∈ L2 (0, T ; H 2 (0, ) ∩ V 1 (0, )) with ut ∈ L2 (0, T ; V 1 (0, )), ut t ∈ L2 (0, T ; L2 (0, )) of the direct problem (10.1.1). If, in addition, the conditions of Theorem 10.1.2 hold, then ut t (x, t), uxt (x, t) and uxx (x, t) are piece-wise continuous functions in T . These functions can fail to be continuous only along the characteristic lines t − z(x) = 2nz() and t + z(x) = 2(n + 1)z(), n = 0, 1, 2, . . . , N.

270

10 Inverse Problems for Damped Wave Equations

For this regular weak solution the following a priori estimates hold:

ut t L2 (0,T ;L2 (0,)) ≤ C4 s  L2 (0,T ) ,

uxt L2 (0,T ;L2 (0,)) ≤ C4 s  L2 (0,T ) ,

(10.1.61)

where C4 > 0 is the same constant defined in (10.1.58). Proof The first part, i.e. the existence of the regular weak solution follows from the general theory of hyperbolic PDE [38], while the regularity properties of ut t (x, t), uxt (x, t) and uxx (x, t) are the consequence of Theorem 10.1.5. To prove estimates (10.1.61) we differentiate Eq. (10.1.1) with respect to t ∈ (0, T ), multiply both sides by ut t (x, t), integrate over t and use the initial and boundary conditions, also the limit equation at t = 0+ . Then we obtain the integral identity 

 0

[m(x)u2t t + r(x)u2xt ]dx + 2 

= −2

t

 t 0

 0

μ(x)u2τ τ dxdτ

r(0)uxτ (0, τ ) s  (τ )dτ,

0

which yields the following inequality: 



a0 0

[u2t t (x, t) + u2xt (x, t)]dx



t

≤ε 0

(r(0)uxτ (0, τ ))2 dτ +

1 ε



t

(s  (τ ))2 dτ, t ∈ [0, T ],

(10.1.62)

0

with the constant a0 > 0 defined in (10.1.47). To estimate the first right-hand-side integral we use the analogue

 ([2m(x)r(x) uxt ut t ]( − x))t − [m(x)r(x) u2t t + r 2 (x) u2xt ]( − x)

x

+[2μ(x)r(x) uxt ut t + (m(x)r(x)) u2t t ]( − x) −m(x)r(x) u2t t − (r(x) uxt )2 = 0. of the differential identity (10.1.50) with # u(x, t) replaced by ut (x, t). Integrating this identity over t , in the same way as in the proof of Theorem 10.1.5 we obtain

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

271

the following inequality: 

t

(r(0)uxτ (0, τ ))2 dτ

0





≤ m1 r1 0

 t

[u2t t + u2xt ]dx + C1

0



0

[u2τ τ + u2xτ ]dx dτ,

(10.1.63)

where C1 > 0 is the same constant defined in (10.1.47). Substituting this in (10.1.62) after elementary transformations we obtain the integral inequality    m1 r1 ε 1− [u2t t + u2xx ]dx a0 0    t C1 ε t  2 1 2 ≤ [u + uxx ]dxdτ + (s  (τ ))2 dτ, t ∈ [0, T ], a0 0 0 τ τ a0 ε 0 i.e. the analogue of the integral inequality (10.1.52). Estimates (10.1.61) are obtained from this inequality exactly in the same way.   Corollary 10.1.4 Assume that conditions of Theorem 10.1.7 hold. Then for the regular weak solution of the direct problem (10.1.1) the following estimate holds:

uxx L2 (0,T ;L2 (0,)) ≤ C6 s H 2 (0,T ) ,

(10.1.64)

where C5 = max (m1 , μ1 ; max |r  (x)|), C6 = [0,]

√ 2 C4 C5 /r0 ,

and the constant C4 > 0 is defined in (10.1.58). Proof Use the identity r(x)uxx = (r(x)ux )x − r  (x)ux to conclude that 

T



0

= = ≤

 0

1 r02 1 r02

u2xx dxdt



T



0





T



0



(r(x)uxx )2 dxdt

0



(r(x)ux )x − r  (x)ux

2

dxdt

0 T

0

1 ≤ 2 r0





m(x)ut t + μ(x)ut − r  (x)ux

2

dxdt

0

1 [C02 + max |r  (x)|2 ] [0,] r02



T 0



 0

u2t t + u2t + u2x

2 dx

(10.1.65)

272

10 Inverse Problems for Damped Wave Equations

With estimates (10.1.57) and the first estimate of (10.1.61) this leads to the required estimate (10.1.64).   Subsequently we need estimates for the norms of the higher order derivatives of the regular weak solution with improved regularity. To this end we introduce the admissible set of regular inputs with higher regularity W 3 (0, T ) := {s ∈ H 3 (0, T )| s(0) = s  (0) = s  (0) = 0},

(10.1.66)

assuming j = 2 in Theorem 10.1.5. Then s ∈ W 3 (0, T ) and # s2 (t) = s(t). Hence, # uj (x, t) ≡ u(x, t) with j = 2 and we may deduce some consequences of Theorem 10.1.5. Theorem 10.1.8 Assume that conditions (10.1.3) hold. Suppose, in addition, that s ∈ W 3 (0, T ). Then there exists the regular weak solution with improved regularity u ∈ L2 (0, T ; H 2(0, ) ∩ V 1 (0, )) with ut ∈ L2 (0, T ; H 2(0, ) ∩ V 1 (0, )), ut t ∈ L2 (0, T ; V 1 (0, )), ut t t ∈ L2 (0, T ; L2 (0, )) of the direct problem (10.1.1). If, in addition, the conditions of Theorem 10.1.2 hold, then ut t t (x, t), uxt t (x, t) and uxxt (x, t) are piece-wise continuous functions in T . These functions can fail to be continuous only along the characteristic lines t − z(x) = 2nz() and t + z(x) = 2(n + 1)z(), n = 0, 1, 2, . . . , N. For this solution the following a priori estimates hold:

ut t t L2 (0,T ;L2 (0,)) ≤ C4 s  L2 (0,T ) ,

uxt t L2 (0,T ;L2 (0,)) ≤ C4 s  L2 (0,T ) ,

(10.1.67)

where C4 > 0 is the same constant defined in (10.1.58). These estimates can be obtained in a similar way as in the proof of Theorem 10.1.7. Corollary 10.1.5 Assume that conditions of Theorem 10.1.8 hold. Then for the regular weak solution with improved regularity of the direct problem (10.1.1) the following estimate holds:

uxxt L2 (0,T ;L2 (0,)) ≤ C6 s H 3 (0,T ) ,

(10.1.68)

where the constant C6 > 0 is defined in (10.1.65). This estimate can be obtained in the same way as in the proof of Corollary 10.1.4. Finally we estimate the solution δu(x, t) := u(x, t; r+δr)−u(x, t; r), r, r+δr ∈ R of the initial-boundary value problem: ⎧   δ ⎪ ⎨ m(x)δut t + μ(x)δut − (r(x)δux )x = δr(x)ux x , (x, t) ∈ T , (10.1.69) δu(x, 0) = 0, δut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ δu(0, t) = 0, δu (, t) = 0, t ∈ (0, T ), x

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

273

where uδ (x, t) := u(x, t; r + δr) solves the direct problem (10.1.1) with r(x) replaced by r(x) + δr(x). Theorem 10.1.9 Assume that conditions of Theorem 10.1.7 hold. Then for the weak solution of problem (10.1.69) the following estimates hold:

δut L2 (0,T ;L2 (0,)) ≤ (exp(T ) − 1) C7 s H 2 (0,T ) δr H 2 (0,) ,

δux L2 (0,T ;L2 (0,)) ≤ exp(T ) C7 s H 2 (0,T ) δr H 2 (0,),

(10.1.70) (10.1.71)

where C2 = max (1, 2/; 2/3), C7 = 2C max (C4 , C6 )/a0

(10.1.72)

and the constants C4 , C6 > 0 are defined in Corollaries 10.1.3 and 10.1.4. Proof As in the proof of Theorem 10.1.5, multiply both sides of Eq. (10.1.69) by 2δut (x, t), integrate over t and use the initial and boundary conditions. Applying then the ε-inequality to the right-hand-side integral we deduce that 



a0 0

≤ε

 t

[δu2t + δu2x ]dx + 2μ0

 t 0

0



δu2τ

1 dxdτ + ε

0

0

 t 0



δu2τ dxdτ



0

δr(x)uδx )x

2

dxdτ, t ∈ [0, T ],

with a0 > 0 defined in (10.1.47). Choose the ε > 0 as ε = a0 . Then we obtain:  0

≤



[δu2t + δu2x ]dx

 t 0

 0

δu2τ dxdτ +

1 a02

 0

T

 0



δr(x)uδx )x

2

dxdt, t ∈ [0, T ].

(10.1.73)

To estimate the last right-hand-side integral in (10.1.73) we use Agmon’s inequality max |r  (x)|2 ≤ |r  (0)|2 + 2 r  L2 (0,) r  L2 (0,) , r ∈ H 2 (0, ),

x∈[0,]

with the auxiliary inequality |r  (0)|2 ≤ (2/) r  2L2 (0,) + (2/3) r  2L2 (0,) , which can be proved easily. This yields:  2 max |r  (x)|2 ≤ C2 r  L2 (0,) + r  L2 (0,) ,

x∈[0,]

(10.1.74)

274

10 Inverse Problems for Damped Wave Equations

where C > 0 is defined in (10.1.72). In the same way,  2 max |r(x)|2 ≤ C2 r L2 (0,) + r  L2 (0,) .

x∈[0,]

(10.1.75)

In view of (10.1.74) and (10.1.75) we obtain: 

T 0





δr(x)uδx )x

0

2



T

dxdt =



0

 0



[δr  (x)uδx + δr(x)uδxx ]2 dxdt 

≤ max [(δr(x)) + (δr (x)) ] 2

2

x∈[0,]

T



0

 ≤

2C2 δr 2H 2 (0,)

 0

T

0



 0

[(uδx )2 + (uδxx )2 ]dxdt [(uδx )2 + (uδxx )2 ]dxdt

Taking into account estimates (10.1.57) and (10.1.64) we conclude that 

T 0



 0

δr(x)uδx )x

2

dxdt ≤ C72 s 2H 2 (0,T ) δr 2H 2 (0,).

Substitute this in (10.1.73) we arrive at the estimate 

 0

[δu2t + δu2x ]dx ≤

 t 0

 0

δu2τ dxdτ + C72 s 2H 2 (0,T ) δr 2H 2 (0,) , t ∈ [0, T ],

with the constant C7 > 0 defined in (10.1.72). This leads the desired estimates.

 

Theorem 10.1.10 Let conditions of Theorem 10.1.7 hold and s ∈ W 3 (0, T ). Then for the regular weak solution of problem (10.1.69) the following estimates hold: √ T C7 s H 3 (0,T ) δr H 2 (0,) , √ ≤ T C7 s H 3 (0,T ) δr H 2 (0,) ,

δut t L2 (0,T ;L2 (0,)) ≤

(10.1.76)

δuxt 2L2 (0,T ;L2 (0,))

(10.1.77)

where C7 > 0 is the same constant defined in (10.1.72). The proof of this theorem is exactly the same as the previous one.

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

275

10.1.3 The Dirichlet-to-Neumann Operator Based on conditions (10.1.3) we introduce the set of admissible coefficients as follows: R := {r ∈ H 2 (0, ) : 0 < r0 ≤ r(x) ≤ r1 },

(10.1.78)

Consider the Dirichlet-to-Neumann operator introduced in (10.1.5). Lemma 10.1.1 Assume that conditions (10.1.3) hold and s ∈ W 2 (0, T ). Suppose that the set of admissible coefficients is given by (10.1.78). Then the Dirichlet-toNeumann operator [·] : R ⊂ H 2 (0, ) → L2 (0, T ) defined in (10.1.5) is a compact operator. Proof Let {r (n) } ⊂ R, n = 1, ∞, be a sequence of coefficients bounded in H 2 -norm. Denote by {u(n) (x, t)}, u(n) (x, t) := u(x, t; r (n) ), the sequence of corresponding regular weak solutions defined in Theorem 10.1.7. Then {(r (n) )(t)}, (r (n) )(t) = r (n) (0)u(n) x (0, t) is the sequence of outputs. We need to prove that this sequence is a relatively compact subset of L2 (0, T ) or, equivalently, this sequence is bounded in the norm of H 1 (0, T ). From the trace inequality (10.1.51) applied to the solution of the direct problem and the energy estimate (10.1.59) it follows that the sequence of norms (n) { r (n) (0)ux (0, ·) L2 (0,T ) }∞ n=1 is uniformly bounded. Indeed, #2  2

r (n) (0)u(n) x (0, ·) L2 (0,T ) ≤ C s L2 (0,T ) ,

(10.1.79)

#2 = m1 r1 C 2 exp(C2 T ) + C1 C 2 and the constants C1 , C2 , C3 , C4 > 0 where C 3 4 are defined in (10.1.47) and (10.1.58). In the same way, the uniform bounded(n) ness of the sequence of norms { r (n) (0)uxt (0, ·) L2 (0,T ) }∞ n=1 follows from the trace inequality (10.1.57) and estimates (10.1.61). Thus, the sequence of norms (n) 1 { r (n) (0)ux (0, ·) L2 (0,T ) }∞ n=1 is bounded in the norm of H (0, T ) and as a result 2   is relatively compact in L (0, T ). This completes the proof. Lemma 10.1.1 in particular implies that if r ∈ R and R ⊂ H 2 (0, ) is defined in (10.1.78), then the inverse problem defined by (10.1.1) and (10.1.2) is ill-posed. Lemma 10.1.2 Assume that conditions (10.1.3) hold. Suppose also s ∈ W 3 (0, T ). Then the Dirichlet-to-Neumann operator defined in (10.1.5) is Lipschitz continuous, that is

r1 − r2 L2 (0,T ) ≤ L r1 − r2 H 2 (0,),

(10.1.80)

for all r1 , r2 ∈ R, where

1/2 #7 C72 / + 4C42 C2 / L = C

s H 3 (0,T )

(10.1.81)

276

10 Inverse Problems for Damped Wave Equations

is the Lipschitz constant depending only on the constants defined in Corollary 10.1.3, Theorem 10.1.9 and also on the constant #7 = (4r12 /) exp(T ) + (m21 + μ21 )[(exp(T ) − 1)2 + T ]. C

(10.1.82)

Proof Let r1 (x) := r(x) + δr(x), r2 := r(x) and u(x, t) := u(x, t; r), uδ (x, t) := u(x, t; r + δr) are the corresponding solutions of the direct problem (10.1.1). Then the difference δu(x, t) := uδ (x, t) − u(x, t) solves the initial-boundary value problem (10.1.69). We use the identity (r(x) + δr(x))uδx − r(x)ux = r(x)δux + δr(x)uδx

(10.1.83)

to estimate the norm r1 − r2 L2 (0,T ) . We have: 

r1 − r2 2L2 (0,T )



T

:= 0





T

≤2 0



(r(x) + δr(x))uδx − r(x)ux



T

+2

(r(0) + δr(0))uδx (0, t) − r(0)ux (0, t)

0

x 0



T

≤2



0

 +2

dt

− (r(ξ )uξ )ξ ]dξ

r(x)δux (x, t) + δr(x)uδx (x, t) 

T

dt

2 [ ((r(ξ ) + δr(ξ ))uδξ )ξ

x 0

2

2



2

dt

dt

[ m(x)δut t + μ(x)δut ]2 dx dt.

0

Integrate over (0, ) and divide both sides by  > 0. After elementary transformations we obtain:

r1 − r2 2L2 (0,T ) ≤

4r12 4

δux 2L2 (0,T ;L2 (0,)) + max (δr(x) )2 uδx 2L2 (0,T ;L2 (0,))   x∈[0,]  +(m21 + μ21 ) δut t 2L2 (0,T ;L2 (0,)) + δut 2L2 (0,T ;L2 (0,)) ,

where m1 , μ1 > 0 are the constants defined in (10.1.3). Using here the second estimate in (10.1.57) and estimates (10.1.70), (10.1.71), (10.1.76) we deduce that

r1 − r2 2L2 (0,T ) #72 C72 s 2 3 ≤C

δr 2H 2 (0,) + H (0,T )

4C42  2

s L2 (0,T ) δr 2C(0,), 

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

277

#7 > 0 is defined in (10.1.82). Now we employ the inequality (10.1.75) to where C estimate the last right-hand-side norm δr C[0,T ] . This yields:

δr 2C[0,T ] ≤ C2 δr 2H 1 (0,T ) .  

Using this in the above inequality we arrive at the required result.

10.1.4 Existence of a Quasi-Solution of the Inverse Problem For clarity of further analysis, we must keep the concepts of output and measured output apart. The output f (t) introduced in (10.1.2) is uniquely determined through the solution u(x, t; r) of the direct problem (10.1.1), corresponding to a given admissible coefficient r(x), since under appropriate conditions it has a unique solution. Evidently, the regularity of the output f (t) depends on the regularity of this solution. As far the measured output, we will come back to this issue later. Let us introduce now the Tikhonov functional J (r) :=

1 2



T

[r(0)ux (0, t) − f (t)]2 dt, r ∈ R

(10.1.84)

0

and reformulate the inverse problem (10.1.1)–(10.1.2) (or the nonlinear operator equation (10.1.6)) as the problem of funding r" ∈ R such that J (r" ) = J" , where J∗ = inf J (r). r∈R

(10.1.85)

From Lemma 10.1.2 it follows that the Tikhonov functional is Lipschitz continuous. Corollary 10.1.6 Assume that conditions of Lemma 10.1.2 hold. Then the Tikhonov functional (10.1.84) is Lipschitz continuous in set of admissible coefficients defined by (10.1.78), that is |J (r1 ) − J (r2 )| ≤ LJ r1 − r2 H 2 (0,),

(10.1.86)

for all r1 , r2 ∈ R, where 

C8 = C3 (max (m1 r1 , C1 /C2 ) [2 exp(C2 T ) − 1] )1/2 ,   LJ = C8 s  L2 (0,T ) + f L2 (0,T ) L ,

(10.1.87)

where L > 0 is defined in (10.1.81) and the constants C1 , C2 , C3 > 0 are defined in (10.1.47).

278

10 Inverse Problems for Damped Wave Equations

Proof From the identities





 



|J (r1 ) − J (r2 )| = J (r1 ) + J (r2 ) J (r1 ) − J (r2 ) ,





1



J (r1 ) − J (r2 ) = √  r1 − f L2 (0,T ) −  r2 − f L2 (0,T )

2 and the inequality | a − b | ≤ a − b it follows that  1  |J (r1 ) − J (r2 )| ≤ √ J (r1 ) + J (r2 ) r1 − r2 L2 (0,T ) . 2 Hence |J (r1 ) − J (r2 )| 1

r1 L2 (0,T ) + r2 L2 (0,T ) + 2 f L2 (0,T ) r1 − r2 L2 (0,T ) . ≤ 2 To estimate the norms rk L2 (0,T ) we use the trace inequality (10.1.51) with estimate (10.1.46), applied to the solution of the direct problem (10.1.1). We have:

rk L2 (0,T ) ≡ rk (0) ux (0, ·; rk ) L2 (0,T ) ≤ C8 s  L2 (0,T ) , k = 1, 2, where C8 > 0 is defined in (10.1.87). In view of the above inequality we conclude that   |J (r1 ) − J (r2 )| ≤ C8 s  L2 (0,T ) + f L2 (0,T ) r1 − r2 L2 (0,T ) . With (10.1.80) this leads to the desired result (10.1.86).

 

The Lipschitz continuity of the Tikhonov functional (10.1.84) allows us to prove an existence of a solution of the minimization problem (10.1.85). Theorem 10.1.11 Assume that conditions (10.1.3) hold and s ∈ W 3 (0, T ). Then the minimization problem (10.1.85) has a solution in set of admissible coefficients R ⊂ H 2 (0, ) defined by (10.1.78). Proof Indeed, the set of extended inputs R is a nonempty closed convex set of the Sobolev space H 2 (0, ). By Corollary 10.1.6 the Tikhonov functional J (r) is Lipschitz continuous, and hence, is lower semicontinuous. As a lower semicontinuous functional defined on the nonempty closed convex set R, it is weakly sequentially lower semicontinuous. Then by the generalized Weierstrass existence theorem (§2.5, Theorem 2.D, [156]), it has a minimum on R.   Theorem implies the existence of a quasi-solution of the inverse problem (10.1.1)–(10.1.2) in the set of admissible coefficients R ⊂ H 2 (0, ).

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

279

10.1.5 Uniqueness of the Solution to the Inverse Problem In this subsection we prove a uniqueness result for the inverse problem (10.1.1)– (10.1.2) for the case when m(x) = 1 and μ(x) = 0. For the case of the semi-infinite space domain x > 0 the uniqueness result is well known (see, for instance, the books [19, 132]). Specifically, in this case the problem of recovering r(x) is usually reduced to problem of finding a potential q(z), where z = z(x) is defined by formula (10.1.4), and then, after finding q(z), coefficient r(x) is recovered as a solution of the second order ordinary differential equation. But to our knowledge, for the inverse problem under consideration no relevant results have been published in the literature. Even if this result covers the simple case, the theorem below provides the best understanding of the role of the condition (10.1.55) discussed in Sect. 10.1.3. The uniqueness theorem for the case of variable, but known coefficients m(x) and μ(x), is more complicate technically and lies outside of our investigation. Let z = z(x) be new variable defined by (10.1.4) and x = χ(z) be the inverse function, i.e. x ≡ χ(z(x)) defined for all x ∈ [0, ]. Denote z∗ = z(),  u(z, t) = u(χ(z), t),  c(z) = c(χ(z)) and  r(z) = r(χ(z)) =  c 2 (z). From (10.1.4) it follows that  z χ  (z) =  c(z), x = χ(z) =  c(ζ )dζ. (10.1.88) 0

Then the function  u(z, t) is solution of the problem ⎧ ⎪ ut t =  uzz + h(z) uz , (z, t) ∈ T := (0, z∗ ) × [0, T ], ⎨  u(z, 0) = 0,  ut (z, 0) = 0, z ∈ (0, z∗ ), ⎪ ⎩ u(0, t) = s(t),  u(z∗ , t) = 0, t ∈ (0, T ),

(10.1.89)

where h(z) = (ln c(z))z . The Neumann boundary output is f (t) = c(0) uz (0, t), t ∈ (0, T ].

(10.1.90)

Thus, in this case the inverse problem is reduced to the finding h(z), z∗ and c(0) from given function f (t) for all t ∈ (0, T ]. If h(z), z∗ and c(0) are known one can find first  z  c(z) = c(0) exp h(ζ )dζ , z ∈ [0, z∗ ], (10.1.91) 0

then using (10.1.88) calculate χ(z) and χ −1 (x) ≡ z(x) as an inverse function to x = χ(z) and then find r(x) by the formula r(x) =  c 2 (χ −1 (x)), x ∈ [0, ].

(10.1.92)

280

10 Inverse Problems for Damped Wave Equations

Theorem 10.1.12 Let conditions (10.1.3), (10.1.55), and also the conditions of Theorem 10.1.2 hold. Assume, in addition, that following regularity and consistency conditions hold: s ∈ C j +1 [0, T ], s(0) = s  (0) = . . . = s (j −1) (0) = 0, s (j ) (0) = 0, for some j = 1, 2, . . .. Then the output (10.1.90) uniquely determines c(0), z∗ and h(z) for all z ∈ [0, z∗ ]. Proof Assume first that j = 1, i.e. s ∈ C 2 [0, T ] and s(0) = 0, s  (0) = 0. In the first stage, let us find c(0) defined by formula (10.1.4). Since  u(z, t) ≡ 0 in {(z, t)| z ∈ (0, z∗ ), t ∈ [0, z]} and  ut +  uz is continuous across t = z we conclude that  ut +  uz = 0, t = z ∈ [0, z∗ ].

(10.1.93)

ut (0, 0+ ) = − uz (0, 0+ ) = Moreover,  ut (0, t) = s  (t) for all t ∈ (0.T ]. Hence,   s (0) = 0. Then from (10.1.90) we find: c(0) = −

f (+0) . s  (0)

(10.1.94)

According Theorem 10.1.6, function ux (x, t) is the piece-wise continuous function in T and it can fail to be continuous only along the characteristic lines t − z(x) = 2nz() and t + z(x) = 2(n + 1)z(), n = 0, 1, . . . which are contained in T . Note that in the case under consideration ( α2n+1 (x) = −α2n+2 (x) =

4

r(0) , n = 0, 1, . . . . r(x)

Therefore f (t) is a continuous function except the points t = 2nz∗ ∈ (0, T ), at which it has the finite jumps: f (2nz∗ + 0) − f (2nz∗ − 0) = −c(0)s  (0), if 2nz∗ ∈ (0, T ],

(10.1.95)

for all n = 1, 2, . . . . It is condition (10.1.55) that guarantees T > 2z∗ . Thus, we can find z∗ using equality (10.1.95). Let now T0 > 0 and T0 < 2z∗ . Consider the domain G(T0 ) = {(z, t)| z ∈ [0, T0 /2], z ≤ t ≤ T0 − z}. For t ∈ (0, T0 ] we have the Cauchy data  u(0, t) = s(t),  uz (0, t) = f (t)/c(0), t ∈ (0, T0 ].

(10.1.96)

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

281

Then using D’Alambert formula we find integral equation for  u(z, t): 

z  t +z−ζ

 u(z, t) =  u0 (z, t) −

t −z+ζ

0

h(ζ ) uζ (ζ, τ )dτ dζ, (z, t) ∈ G(T0 ),

(10.1.97)

where  u0 (z, t) =

1 s(t − z) + s(t + z) + 2 2c(0)



t +z

t −z

f (τ )dτ.

Differentiating Eq. (10.1.97) with respect to t and z we get u0,t (z, t)  ut (z, t) =   z h(ζ )[ uζ (ζ, t + z − ζ ) −  uζ (ζ, t − z + ζ )]dζ, −

(10.1.98)

0

 uz (z, t) =  u0,z (z, t)  z h(ζ )[ uζ (ζ, t + z − ζ ) +  uζ (ζ, t − z + ζ )]dζ, −

(10.1.99)

0

for all (z, t) ∈ G(T0 ). Setting in these equations t = z + 0 and using relation (10.1.93) we get 

z

0 = g(z) − 2

h(ζ ) uζ (ζ, 2z − ζ )dζ, z ∈ [0, T0 /2],

(10.1.100)

0

where u0,z (z, t)]t =z+0 = s  (2z) + f (2z)/c(0). g(z) := [ u0,t (z, t) +  Differentiating with respect to z yields  h(z) uz (z, t)|t =z+0 + 2

z

h(ζ ) uζ τ (ζ, 2z − ζ )dζ =

0

g  (z) , 2

z ∈ [0, T0 /2]. In view of Theorem 10.1.6 we have ux (x, t)|t =z(x)+0

s  (0) =− c(x)

( 4

√ s  (0) c(0) r(0) =−  . r(x) c3 (x)

Hence  uz (z, t)|t =z+0

√ s  (0) c(0) b =− √ := √ .  c(z)  c(z)

(10.1.101)

282

10 Inverse Problems for Damped Wave Equations

Therefore we can rewrite Eq. (10.1.101) as follows: bh(z) +2 √  c(z)



z

h(ζ ) uζ τ (ζ, 2z − ζ )dζ = g  (z)/2, z ∈ [0, T0 /2]. (10.1.102)

0

Recall that  c(z) is expressed through h(z) by formula (10.1.91). Differentiating Eq. (10.1.99) with respect to t we get  uzt (z, t) =  u0,zt (z, t)  z − h(ζ )[ uζ τ (ζ, t + z − ζ ) +  uζ τ (ζ, t − z + ζ )]dζ,

(10.1.103)

0

for all (z, t) ∈ G(T0 ). For the sake of brevity, we introduce the functions ϕ(z, t) =  uzt (z, t), ϕ0 (z, t) =  u0,zt (z, t) to rewrite Eqs. (10.1.102) and (10.1.103) in the following form: bh(z) +2 √  c(z)

 −



z

h(ζ )ϕ(ζ, 2z − ζ )dζ = g  (z)/2, z ∈ [0, T0 /2], (10.1.104)

0

ϕ(z, t) = ϕ0 (z, t) z

h(ζ )[ϕ(ζ, t + z − ζ ) + ϕ(ζ, t − z + ζ )]dζ, (z, t) ∈ G(T0 ).

(10.1.105)

0

These two equations form closed system of relations for finding h(z) and ϕ(z, t) in G(T0 ). The functions g(z) and ϕ0 (z, t) are continuous in G(T0 ). Therefore one can prove the local existence and uniqueness theorem in the space of continuous functions, if T0 is small enough. Instead we prove uniqueness theorem for any T0 < 2z∗ . Suppose, on the contrary, that there exist two continuous solutions of the inverse problem in domain G(T0 ), namely hk (z), ϕk (z, t) , k = 1, 2, corresponding the same s(t) and f (t) given for all t ∈ [0, T0 ]. Let  ck corresponds to hk , i.e. 

z

 ck (z) = c(0) exp

hk (ζ )dζ , k = 1, 2.

(10.1.106)

0

Denote by # h(z) = h1 (z) − h2 (z), # ϕ (z, t) = ϕ1 (z, t) − ϕ2 (z, t). the difference of the above functions. Replace h(z), ϕ(z, t)  in (10.1.104) and (10.1.105) first with h1 (z), ϕ1 (z, t) and then with h2 (z), ϕ2 (z, t) .

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

283

Subtracting the obtained relations we get the following equations: bh1 (z) bh2 (z) − √ √  c1 (z)  c2 (z)  z +2 h2 (ζ )# ϕ (ζ, 2z − ζ ) dζ = 0, z ∈ [0, T0 /2], (10.1.107) 0



z

ϕ (z, t) = − #

# h(ζ )[ϕ1 (ζ, t + z − ζ ) + ϕ1 (ζ, t − z + ζ )] dζ

0



z

−

h2 (ζ )[# ϕ(ζ, t + z − ζ ) + # ϕ (ζ, t − z + ζ )]dζ, (z, t) ∈ G(T0 ). (10.1.108)

0

Use (10.1.106) to transform the first two terms in Eq. (10.1.107) as follows:   bh2 (z) b# h(z) 1 bh1 (z) 1 − √ = √ + bh2 (z) √ − √ √  c1 (z)  c2 (z)  c1 (z)  c1 (z)  c2 (z)       bh2 (z) 1 z 1 z b# h(z) + √ exp − h1 (ζ )dζ − exp − h2 (ζ )dζ = √ 2 0 2 0  c1 (z) c(0)  z # = P (z)# h(z) + Q(z) h(ζ )dζ, 0

where b , P (z) = √  c1 (z) bh2 (x) Q(z) = − √ 2 c(0)



1 0

    ζ1 z 1 − ζ1 z exp − h1 (ζ )dζ − h2 (ζ )dζ dζ1 . 2 0 2 0

To obtain the last equality the following identity has been used:  exp(a(z)) − exp(b(z)) :=

a(z)

exp(ζ )dζ b(z)

 = (a(z) − b(z)) a=−

1 2

 0

1

exp[ζ1 a(z) + (1 − ζ1 )b(z)]dζ1,

0 z

h1 (ζ )dζ, b = −

1 2



z

h2 (ζ )dζ. 0

284

10 Inverse Problems for Damped Wave Equations

In view of the above transformations we can represent Eqs. (10.1.104) and (10.1.105) in the following form: P (z)# h(z) + Q(z)



z

# h(ζ )dζ + 2



0



z

+2

z

# h(ζ )ϕ1(ζ, 2z − ζ )dζ

0

h2 (ζ )# ϕ1 (ζ, 2z − ζ )dζ = 0, z ∈ [0, T0 /2],

(10.1.109)

0

 # ϕ (z, t) = −

z

# h(ζ )[ϕ1 (ζ, t + z − ζ ) + ϕ1 (ζ, t − z + ζ )]dζ

0



z

−

h2 (ζ )[ϕ(ζ, t + z − ζ ) + ϕ(ζ, t − z + ζ )]dζ, (z, t) ∈ G(T0 ). (10.1.110)

0

Denote by  U (z) = max |# h(z)|,

max

z≤t ≤T0 −z

|# ϕ (z, t)| ,

|Q(z)| |ϕ1 (z, t)| + |h2 (z)| , max . K = max max 0≤z≤T0 /2 |P (z)| (z,t )∈G(T0) |P (z)| 

In view of (10.1.109) and (10.1.110) we obtain the following inequality: 

z

U (z) ≤ 3K

U (ζ )dζ, z ∈ [0, T0 /2].

(10.1.111)

0

Hence, U (z) = 0 for all z ∈ [0, T0 /2] which implies h1 (z) = h2 (z), ϕ1 (z, t) = ϕ2 (z, t) for all (z, t) ∈ G(T0 ). Thus, the output f (t) given for all t ∈ (0, T0 ) uniquely determines h(z) for all z ∈ (0, T0 /2). Substituting T0 = 2z∗ − 0 we conclude that the output f (t) given for t ∈ (0, 2z∗ ) uniquely determines h(z) for all z ∈ (0, z∗ ). Hence, the unknown coefficient r(x) is also uniquely determined by this output on the whole interval (0, ). Let now j > 1. Then differentiating j − 1 times equation (10.1.89) with respect j t we obtain the same problem for the function Dt  u(z, t), with s(t) replaced with (j −1) 2 sj −1 (t) = s (t). Since sj −1 ∈ C [0, T ] and sj −1 (0) = 0, sj −1 (0) = 0 we arrive at the considered above case. This completes the proof of the theorem.  

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

285

10.1.6 Fréchet Differentiability of the Tikhonov Functional Let r, r + δr ∈ R. Denote by δJ (r) := J (r + δr) − J (r) the increment of the Tikhonov functional (10.1.84). Then 

 [r(0)ux (0, t) − f (t)] (r(0) + δr(0))uδx (0, t) − r(0)ux (0, t) dt

T

δJ (r) = 0

+

1 2



T



0

2

(r(0) + δr(0))uδx (0, t) − r(0)ux (0, t) dt,

(10.1.112)

where uδ (x, t) := u(x, t; r + δr) solves the initial-boundary value problem (10.1.1) with r(x) replaced by r(x) + δr(x). Evidently, for difference of the outputs (r(0) + δr(0))uδx (0, t) and r(0)ux (0, t) the following formula holds: (r(0) + δr(0))uδx (0, t) − r(0)ux (0, t) = r(0)δux (0, t) + δr(0)uδx (0, t).

(10.1.113)

Lemma 10.1.3 Let conditions (10.1.3) hold and s ∈ W 2 (0, T ). Assume that r, r + δr ∈ R are arbitrary inputs. Then the following integral identity holds: 

T 0



(r(0) + δr(0))uδx (0, t) − r(0)ux (0, t) p(t) dt 

T

=− 0



 0

δr(x)uδx ψx dxdt.

(10.1.114)

where ψ(x, t) is the solution of the (well-posed) adjoint problem ⎧ ⎪ ⎨ m(x)ψt t − μ(x)ψt = (r(x)ψx )x , (x, t) ∈ T , ψ(x, T ) = 0, ψt (x, T ) = 0, x ∈ (0, ), ⎪ ⎩ ψ(0, t) = p(t), ψ(, t) = 0, t ∈ [0, T ],

(10.1.115)

with the Dirichlet input p ∈ H 1 (0, T ). Proof Multiply both sides of Eq. (10.1.69) by the arbitrary function ψ(x, t), integrate over T and apply the integration by parts formula multiple times. Then we obtain: 

T



0



[m(x)ψt t − μ(x)ψt − (r(x)ψx )x ] δu dxdt

0





+ 0

[m(x)δut ψ − m(x)δuψt + μ(x)δu ψ]tt =T =0 dx

286

10 Inverse Problems for Damped Wave Equations



T

+ 0

[r(x)δuψx − r(x)δux ψ ]x= x=0 dt 



T

=− 0



0

 δr(x)uδx ψx dxdt

T

+ 0

 δr(x)uδx ψ

x= dt. x=0

Now we require that ψ(x, t) is the solution of the adjoint problem (10.1.115). Then taking into account the initial/final and the boundary conditions given in (10.1.69) and (10.1.115) we obtain the following integral identity: 

T

r(0)δux (0, t)p(t) dt 0



T

=− 0



 0

 δr(x)uδx ψx dxdt −

T 0

δr(0)uδx (0, t)p(t)dt.

With identity (10.1.113) we arrive at the input-output relationship (10.1.114).

 

We choose now the Dirichlet input p(t) in (10.1.115) as follows: p(t) = r(0)ux (0, t) − f (t), t ∈ (0, T ).

(10.1.116)

Then the integral identity (10.1.114) becomes to the following input-output relationship: 

T 0



(r(0) + δr(0))uδx (0, t) − r(0)ux (0, t) [r(0)ux (0, t) − f (t)] dt 

T

=−



0



0

δr(x)uδx ψx dxdt.

Using this in the increment formula (10.1.112) with uδ (x, t) = u(x, t) + δu(x, t) we conclude that    T  T  δJ (r) = − ux ψx dt δr(x) dx − δr(x)δux ψx dxdt 0

+

1 2

0



T 0

0

0

 (r(0) + δr(0))uδx (0, t) − r(0)ux (0, t)

2

dt.

(10.1.117)

Therefore the following formal gradient formula holds:  ∇J (r)(x) = −

T

ux (x, t)ψx (x, t)dt, 0

(10.1.118)

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

287

Let us consider now the substitution (10.1.116). From Theorems 10.1.5 and 10.1.8 it follows that r(0)ux (0, ·) ∈ W 1 (0, T ),

(10.1.119)

f ∈ W 1 (0, T ).

(10.1.120)

Hence the input p(t) given by (10.1.116) satisfies the following condition: p ∈ W 1 (0, T ).

(10.1.121)

To understand the relationship between the degree of regularity of the weak solution to the direct problem (10.1.1) and condition (10.1.119) we use the identity 

T



T

[r(0)uxt (0, t)] dt = 2

0



 r(x)uxt (x, t) −

0

2

x

(r(ξ )uξ (ξ, t))ξ t dξ

dt

0

to deduce the inequality 

T 0

2 [r(0)uxt (0, t)] dt ≤ 



T

2



0

 + 0

T



[ r(x)uxt (x, t)]2 dxdt

0





[m(x)ut t t (x, t) + μ(x)ut t (x, t)]2 dxdt.

0

In view of the first estimate of (10.1.67) of Theorem 10.1.8 we conclude from this inequality that condition (10.1.119) holds if s ∈ W 3 (0, T ), where W 3 (0, T ) is the set of regular admissible inputs defined in (10.1.66). Theorem 10.1.13 Let conditions (10.1.3) hold and s ∈ W 3 (0, T ). Assume, in addition, that the output f (t) satisfies the regularity condition (10.1.120). Then the Tikhonov functional J (r), r ∈ R, is Fréchet differentiable in the set of admissible coefficients defined in (10.1.78). Moreover, for the Fréchet gradient the formal gradient formula (10.1.118) holds, where ψ(x, t) is the weak solution of the adjoint problem ⎧ ⎪ ⎨ m(x)ψt t − μ(x)ψt = (r(x)ψx )x , (x, t) ∈ T , ψ(x, T ) = 0, ψt (x, T ) = 0, x ∈ (0, ), ⎪ ⎩ ψ(0, t) = r(0)ux (0, t) − f (t), ψ(, t) = 0, t ∈ [0, T ],

(10.1.122)

and u(x, t) = u(x, t; r) is the regular weak solution with improved regularity of the direct problem (10.1.1) corresponding to r ∈ R.

288

10 Inverse Problems for Damped Wave Equations

Proof In view of estimates (10.1.71) and (10.1.80) the second and third right-handside integrals in (10.1.117) are of the order O( δr 2H 2 (0,)). This completes the proof.   Let us analyze now the gradient formula (10.1.118) which contains the term ψx (x, t). It should be noted that (10.1.118) still is a formal formula and needs to be justified somehow. Replacing t ∈ [0, T ] with T −t in the adjoint problem (10.1.115) we see that this is the direct problem (10.1.1) with the input s(t) replaced with p(T − t). Hence the results given in Sect. 10.1.2 remain hold also for the solution ψ(x, t) of the adjoint problem (10.1.115). In particular, a representation formula of the type (10.1.53) and (10.1.54) given in Theorem 10.1.6, must be derived for the solution ψ(x; t) of the adjoint problem (10.1.122), in order transform the formal gradient formula (10.1.110) into one that is computationally useful. To this end we consider the adjoint problem (10.1.115) not in the time-domain (0, T ), but in the slightly extended to the left time-domain (T2 , T ), where T2 < 0 will be defined below. Theorem 10.1.14 Assume that the inputs in the adjoint problem (10.1.115) satisfy the conditions (10.1.3) and p ∈ W 1 (0, T ). Denote by N = N(T ) ≥ 1 a positive integer such that 2Nz() ≤ T < 2(N + 1)z() and define T2 = T − 2(N + 1)z(). #T := {(x, t) | 0 < Consider the adjoint problem (10.1.115) in the extended domain x < , t ∈ (T2 , T )} assuming that p(t) ≡ 0 for all t ≤ 0. Then the solution of #T can be represented in the following form: (10.1.115) in #(x, t) + p(T )ψ0 (x, t) ψ(x, t) = ψ +p(T )

N 

[α2n+1 (x)θ0 (T − t − z(x) − 2nz())

n−1

+α2n+2 (x)θ0 (T − t + z(x) − (2n + 2)z())]. (10.1.123) #(x, t) is the solution Here αk (x), k = 1, 2, . . . , is defined in (10.1.11), the function ψ of the adjoint problem (10.1.115) with the input p(t) replaced by p #0 (t) = p(t) − p(T ) and ψ0 (x, t) is the solution of the following problem ⎧ (x, t), m(x)ψ0,t t − μ(x)ψ0,t = (r(x)ψ0,x )x + F ⎪ ⎪ ⎪ ⎨ 0 < x < , T2 < t < T , ⎪ ψ0 (x, T ) = 0, ψ0,t (x, T ) = 0, x ∈ (0, ), ⎪ ⎪ ⎩ ψ0 (0, t) = 0, ψ0 (, t) = 0, T2 < t < T ,

(10.1.124)

10.1 Determination of Principal Coefficient from Dirichlet-to-Neumann Operator

289

(x, t) defined by with F (x, t) = F

N 

 [(r(x)α2n+1 (x)) θ0 (T − t − z(x) − 2nz())

n=0  (x)) θ0 (T − t + z(x) − (2n + 2)z())]. +(r(x)α2n+2

(10.1.125)

# ∈ L2 (T2 , T ; V 1 (0, )), ψ #t ∈ L2 (T2 , T ; L2 (0, )) and the following Moreover, ψ energy estimate holds: 



E(t) := 0

#t2 (x, t) + ψ #x2 (x, t)]dx ≤ C32 # [ψ p0 2L2 (0,T ) exp(C2 t), (10.1.126)

for all t ∈ [0, T ], where the constants C2 , C3 > 0 are defined in (10.1.47). For the weak solution of problem (10.1.124) the following estimates hold:

ψ0,t 2L2 (0,T ;L2 (0,)) ≤ C1 ( m 2H 2 (0,) + r 2H 2 (0,) + μ 2H 1 (0,)), (10.1.127)

ψ0,x 2L2 (0,T ;L2 (0,)) ≤ C2 ( m 2H 2 (0,) + r 2H 2 (0,) + μ 2H 1 (0,)), (10.1.128) where C1 , C2 > 0 depend only on the constants defined in (10.1.3), and also on  > 0 and T > 0. In addition, ψ(x, t) = 0 in the domain D0 = {(x, t)| x ∈ [0, ], T − z(x) < t ≤ T }. Proof This theorem is the consequence of Theorem 10.1.5. As noted above, the adjoint problem (10.1.115) is the well-posed problem (10.1.1) with the input s(t) replaced with p(T − t). Therefore introducing the function j (x, t), #j (x, t) = ψ(x, t) − ψ ψ #(x, t), for #0 (x, t) = ψ similar to (10.1.43), and assuming here j = 0 and ψ the sake of brevity, we obtain representation (10.1.123), similar to that given in Theorem 10.1.6. Further, for t ∈ (0, T ) we obtain the energy estimate (10.1.126), similar to (10.1.46) in Theorem 10.1.5. The function ψ0 (x, t) is an analogue of the function v0 (x, t) introduced in Theorem 10.1.1. As a result, estimates (10.1.127) and (10.1.128) are similar to estimates (10.1.13) and (10.1.14).   Note that the domain D0 introduced in the above theorem is an analogue of the domain D0 defined by (10.1.15) in Theorem 10.1.1. A distinguished feature of Theorem 10.1.14 is that it provides the decomposition of the solution to the adjoint problem (10.1.115) separating its regular and singular parts. Due to this singular part, the formal gradient formula (10.1.118) can not be used in applications.

290

10 Inverse Problems for Damped Wave Equations

The next theorem shows that this decomposition also allows us to transform the formal gradient formula (10.1.118) into a useful formula which can be justified. Theorem 10.1.15 Suppose that the conditions of Theorems 10.1.14 and 10.1.2 hold. Then for the solution of the adjoint problem (10.1.115) the following representation is valid: ψx (x, t) = φ(x, t) − p(T )z (x)

N  [α2n+1 (x)δ(T − t − z(x) − 2nz()) n=0

−α2n+2 (x)δ(T − t + z(x) − (2n + 2)z())],

(10.1.129)

where #x (x, t) + p(T )ψ0,x (x, t) φ(x, t) = ψ +p(T )

N   [α2n+1 (x)θ0 (T − t − z(x) − 2nz()) n=0  +α2n+2 (x)θ0 (T − t + z(x) − (2n + 2)z())],

(10.1.130)

and φ(x, t) is piece-wise continuous function. This function can fail to be continu#T along the characteristic lines T − t − z(x) = 2nz() ous in the extended domain and T − t + z(x) = 2(n + 1)z(), n = 0, 1, . . . , N, only. Now, using the representation (10.1.129) we can transform the formal gradient formula (10.1.118) into the following gradient formula: 

T

∇J (r)(x) = −

ux (x, t)φ(x, t)dt 0

+p(T )z (x)

N  [α2n+1 (x)ux (x, T − z(x) − 2nz()) n=0

−α2n+2 (x)ux (x, T + z(x) − 2(n + 1)z())],

(10.1.131)

where φ(x, t) is defined in (10.1.130). Let us justify now the gradient formula (10.1.131). This formula is obtained as #(x, t) introduced in a result of substitution (10.1.116). In terms of the function ψ # Theorem 10.1.14 this means that the function ψ (x, t) is the solution of the adjoint problem (10.1.115) with the input p(t) replaced by p #0 (t) = r(0)ux (0, t) − f (t) − r(0)ux (0, T ) − f (T ), t ∈ (0, T ). (10.1.132) Evidently, the input p #0 (t) automatically satisfies the consistency condition p #0 (T ) = 0, while p(t) defined in (10.1.116) does not meet this condition.

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

291

In applications, usually instead of the output f (t) the measured output obtained from experimental measurements is given. This output, defined as f(t), contains random errors (noise) which are caused by unknown and unpredictable changes in the experiment. Moreover, different from the output f (t), the measured output f(t) may not be a smooth function and one can only assume that it is a square measurable function, i.e. f ∈ L2 (0, T ). To ensure that the above theory is applicable to this situation, one needs to satisfy only the condition p ∈ W 1 (0, T ) of Theorem 10.1.14. To this end, we may apply a smoothing algorithm to the measured output f(t) in order to enforce the condition f ∈ W 1 (0, T ) of Theorem 10.1.14. #(x, t) of the The gradient formula (10.1.131), by making use of the solution ψ adjoint problem (10.1.115) with the input p0 (t) defined in (10.1.132) and the solution ψ0 (x, t) of the auxiliary adjoint problem (10.1.124), forms the basis of an effective gradient algorithm for identifying the unknown principal coefficient r(x) in (10.1.1).

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator In Sect. 4.2 we already considered the problem of recovering the potential q(x) in the string equation ut t − uxx − q(x) = 0, (x, t) ∈ T assuming that T is the 2 half-plane: + T := {(x, t) ∈ R : x ≥ 0, t ∈ (0, T )}. This section considers similar inverse problems for the damped wave equation in a finite domain. We set out in detail all three main issues related to the inverse potential problems with Neumann-to-Dirichlet operator. Namely, local existence, global uniqueness and the gradient formula for recovering the unknown potential are discussed in details. The analysis given in this section clearly answers the basic question beyond the abovementioned issues: as a result of examining the relationship between the time and space variables, two domains are determined, one in which the solution is identically zero and the other in which the solution is not zero. As a consequence, it turns out in which domain the solution of the inverse problem needs to be determined. Consider the following inverse problem of recovering the unknown potential q(x) in ⎧ ⎪ ⎨ m(x)ut t + μ(x)ut = (r(x)ux )x + q(x)u, (x, t) ∈ T , u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ r(0)ux (0, t) = f (t), u(, t) = 0, t ∈ (0, T ),

(10.2.1)

from the Dirichlet measured output ν(t) := u(0, t), t ∈ (0, T ],

(10.2.2)

imposed at the boundary x = 0, where T := (0, ) × (0, T ) is a finite domain.

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10 Inverse Problems for Damped Wave Equations

The forward problem (10.2.1) is a mathematical model of many phenomena associated with wave processes. The term q(x)u(x, t) corresponds to a restoring force from the surrounding medium, according to Hooke’s law. The Neumann input f (t) means that there is an acting force on the boundary x = 0. The Dirichlet output ν(t) can be interpreted as a measured displacement of the medium on the same boundary. As we will prove below, in this case, it does not matter what type of homogeneous boundary condition is given on the right boundary x = . Let Q be a set of admissible coefficients which will be defined below. Denote by u(x, t; q) a solution of the forward problem (10.2.1) corresponding to a given admissible coefficient q ∈ Q. Introduce the nonlinear Neumann-to-Dirichlet operator 

(q)(t) := u(0, t; q), t ∈ (0, T ), q ∈ Q,  : Q → L2 (0, T ).

(10.2.3)

Then the inverse problem (10.2.1)–(10.2.2) can be reformulated as the following nonlinear operator equation (q)(t) = ν(t), t ∈ (0, T ], q ∈ Q.

(10.2.4)

10.2.1 Regularity Properties of the Direct Problem Solution in Subdomains Defined by the Final Time and Characteristics In this subsection we assume that the coefficients m(x), μ(x), r(x), q(x) satisfy the following conditions 

m, r ∈ C 2 [0, ], μ ∈ C 1 [0, ], q ∈ C[0, ], m(x) > 0, r(x) > 0, μ(x) ≥ 0,

(10.2.5)

and, moreover, f ∈ C 2 (0, T ), f (0) = 0.

(10.2.6)

For the convenience, we rewrite Eq. (10.2.1) in the form 1 q (x)u, (r(x)ux )x + # m(x) q(x) μ(x) ,# q (x) = , # μ(x) = m(x) m(x) μ(x)ut = ut t + #

(10.2.7)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

293

and then transform it to a more convenient form. To this end, we employ the new coefficient c(x) and variable z = z(x) introduced in (10.1.4). Denote the inverse function to z(x) as x = h(z), and let v(z, t) := u(h(z), t). Then v(z(x), t) = u(x, t) and 1 vz (z(x), t), ut (x, t) = vt (z(x), t), c(x)  r(x) r(x)  (r(x)ux )x = 2 vzz (z(x), t) + vz (z(x), t). c (x) c(x)

ux (x, t) = z (x) vz (z(x), t) =

Since r(x)/c2 (x) = m(x) and r(x)/c(x) = (m(x)r(x))1/2 we obtain μ(x)ut − ut t + #

1 q (x)u (r(x)ux )x − # m(x)

μ(x)vt − vzz − p #(x)vz − # q (x)v, = vt t + # where p #(x) = ((m(x)r(x))1/2) /m(x).

(10.2.8)

 μ(z) = # μ(h(z)),  q (z) = # q (h(z)), p (z) = p #(h(z)), z∗ = z().

(10.2.9)

Let now

Then the function v(x, t) solves the equation μ(z)vt = vzz + p (z)vz +  q (z)v, (z, t) ∈ ∗T = (0, z∗ ) × (0, T ). vt t +  Hence, we arrive at the following transformed inverse problem of recovering  q (z) in ⎧ ⎪ μ(z)vt = vzz + p (z)vz +  q (z)v, (z, t) ∈ ∗T := (0, z∗ ) × (0, T ), ⎨ vt t +  (10.2.10) v(z, 0) = 0, vt (z, 0) = 0, z ∈ (0, z∗ ), ⎪ ⎩ v (0, t) = f (t), v(z∗ , t) = 0, t ∈ (0, T ), z 1 from the Dirichlet boundary measured output ν(t) := v(0, t), t ∈ (0, T ), where f1 (t) =

 −1 r(0)m(0) f (t).

(10.2.11)

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10 Inverse Problems for Damped Wave Equations

τ

t T

T t=T −z

DT

DT

(ζ, t) (1) ΓT (t)

T /2

T /2

:= ST (t)

(4)

ΓT (t)

GT (t) (2)

ΓT (t)

D0

(3)

ΓT (t)

t=z

ζ

z 0

T /2

z∗

= z()

0

T /2

Fig. 10.3 Domains D0 , DT and GT (t) ⊂ DT

Hereafter, we will consider the case when T ≤ 2z∗ only. In this case the solution of problem (10.2.1) in the domain {t + z < T , 0 ≤ z ≤ z∗ } does not depend on a homogeneous boundary condition at the end z = z∗ . We shall demonstrate that the function ν(t) given for t ∈ (0, T ) uniquely determines the coefficient  q (z) for all z ∈ (0, T /2). Introduce the domains D0 = {(z, t) : 0 ≤ t < z, 0 < z ≤ z∗ }, DT = {(z, t) : z < t < T − z, 0 ≤ z < T /2, T ≤ 2z∗ }, defined by the characteristics of the transformed wave equation in (10.2.10) and the final time T > 0 (the left in Fig. 10.3). The following theorem shows the decisive role of characteristics in revealing the properties of the direct solution. Theorem 10.2.1 Let the coefficients of Eq. (10.2.1) and the input f (t) satisfy conditions (10.2.5) and (10.2.6), correspondingly. Then the transformed direct problem (10.2.10) has the unique solution with following properties: (i) v(z, t) ≡ 0, for all (z, t) ∈ D0 ; (ii) v, vt ∈ C 2 (DT ) and v(z, t) is a continuous function in D0 ∪ DT . Moreover, vt |t =z+0 = −vz |t =z+0 =: s(z) = 0, z ∈ [0, T /2],

(10.2.12)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

295

where   1 z s(z) = −f1 (0) exp − μ(ζ ) + p (ζ )] dζ [ 2 0

(10.2.13)

and f1 (t) is the input in (10.2.10). Proof The first property of the solution is a consequence of homogeneous initial data and the homogeneous Dirichlet condition at z = z∗ . Introduce the functions w1 (z, t) = vt (z, t) − vz (z, t), w2 (z, t) = vt (z, t) + vz (z, t).

(10.2.14)

Then vt (z, t) =

1 1 (w1 (z, t) + w2 (z, t)), vz (z, t) = (w2 (z, t) − w1 (z, t)). (10.2.15) 2 2

In view of the relations vt t − vzz = w1t + w1z = w2t − w2z , we rewrite the transformed direct problem (10.2.10) in the following equivalent form for the system of equations of the first order, in the above introduced domain DT : ⎧ ⎪ w1t + w1z = a(z)w1 + b(z)w2 +  q (z)v, (z, t) ∈ DT , ⎪ ⎪ ⎪ ⎪ q (z)v, (z, t) ∈ DT , ⎨ w2t − w2z = a(z)w1 + b(z)w2 +  vt = 12 (w1 + w2 ), (z, t) ∈ DT , (10.2.16) ⎪ ∗ ⎪ ⎪ v|t =z+0 = 0, w2 |t =z+0 = 0, z ∈ (0, z ), ⎪ ⎪ ⎩ w2 (0, t) − w1 (0, t) = 2f1 (t), t ∈ (0, T ), where 1 1 μ(z) + p (z)) , b(z) = (− μ(z) + p (z)) . a(z) = − ( 2 2

(10.2.17)

Note that the conditions at t = z + 0 in (10.2.16) are the consequence of the equality v(z, z − 0) = 0 and the smoothness of the function f1 (t) for t ≥ 0 in the neighborhood of t = 0. Then the function v(z, t) can be continued continuously across the characteristic line t = z, and 0=

d v(z, z + 0) = (vt + vz )|t =z+0 = w2 (z, z + 0). dz

296

10 Inverse Problems for Damped Wave Equations

Let (z, t) ∈ DT be an arbitrary point. Integrating the second equation of (10.2.16) on the plane ζ, τ along the characteristic line τ + ζ = t + z and using the condition w2 (ζ, ζ + 0) = 0, we obtain the following integral relation: w2 (z, t) =  t [a(ζ )w1(ζ, τ ) + b(ζ )w2(ζ, τ ) +  q (ζ )v(ζ, τ )]ζ =t +z−τ dτ. (10.2.18) (z+t )/2

Similarly, integrating the third equation (10.2.16) along ζ = z, we find: v(z, t) =

1 2



t

[w1 (ζ, τ ) + w2 (ζ, τ )]dτ.

(10.2.19)

z

Now we integrate the first equation of (10.2.16) along characteristic line ζ −τ = z−t from point (z, t) ∈ DT till the point (0, t − z) and use the Neumann boundary input at z = 0. Then we obtain the relation: w1 (z, t) = −2f1 (t − z) + w2 (0, t − z)  t + [a(ζ )w1(ζ, τ ) + b(ζ )w2(ζ, τ ) +  q (ζ )v(ζ, τ )]ζ =τ +z−t dτ. t −z

We eliminate w2 (0, t − z) from this relation, using (10.2.18), to get the new and more convenient relation: w1 (z, t) = −2f1 (t − z)  t [a(ζ )w1 (ζ, τ ) + b(ζ )w2 (ζ, τ ) +  q (ζ )v(ζ, τ )]ζ =τ +z−t dτ +  +

t −z

t −z (t −z)/2

[a(ζ )w1(ζ, τ ) + b(ζ )w2(ζ, τ ) +  q (ζ )v(ζ, τ )]ζ =t −z−τ dτ. (10.2.20)

Equations (10.2.18)–(10.2.20) form the complete system for finding w1 (z, t), w2 (z, t) and v(z, t) in DT . To obtain the solution of these equations one can use the method of successive approximations in the form w1 (z, t) = w10 (z, t) + v(z, t) = v 0 (z, t) +

∞ 

w1n (z, t), w2 (z, t) = w20 (z, t) +

n=1 ∞  v n (z, t),

n=1

∞  n=1

w2n (z, t), (10.2.21)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

297

where w20 (z, t) = v 0 (z, t) = 0, w10 (z, t) = −2f1 (t − z) and $t ⎧ n w2 (z, t) = (z+t )/2[a(ζ )w1n−1 (ζ, τ ) + b(ζ )w2n−1 (ζ, τ ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + q (ζ )v n−1 (ζ, τ )]ζ =t +z−τ dτ, ⎪ ⎪ ⎪ ⎪ $ ⎪ t ⎪ ⎪ v n (z, t) = 12 z [w1n−1 (ζ, τ ) + w2n−1 (ζ, τ )]dτ, ⎪ ⎪ ⎨ $t (10.2.22) w1n (z, t) = t −z [a(ζ )w1n−1 (ζ, τ ) + b(ζ )w2n−1 (ζ, τ ) ⎪ ⎪ ⎪ ⎪ ⎪ + q (ζ )v n−1 (ζ, τ )]ζ =τ +z−t dτ ⎪ ⎪ ⎪ ⎪ $ t −z ⎪ ⎪ ⎪ + (t −z)/2[a(ζ )w1n−1 (ζ, τ ) + b(ζ )w2n−1 (ζ, τ ) ⎪ ⎪ ⎪ ⎩ + q (ζ )v n−1 (ζ, τ )]ζ =t −z−τ dτ, n = 1, 2, . . . . It is easy to see that all the functions w1n (z, t), w2n (z, t), v n (z, t), n = 0, 1, . . . , are continuous in DT . We introduce the following notations:  ⎧ ⎪ n n n ⎪ max max |w (z, t)|, max |w (z, t)|, max |v (z, t)| , ⎪ 1 2 ⎪ z∈[0,t ] z∈[0,t ] z∈[0,t ] ⎪ ⎪ ⎨ t ∈ (0, T /2),  Mn (t) = ⎪ n n n ⎪ |w (z, t)|, max |w (z, t)|, max |v (z, t)| , max max ⎪ 1 2 ⎪ ⎪ z∈[0,T −t ] z∈[0,T −t ] z∈[0,T −t ] ⎪ ⎩ t ∈ (T /2, T ), and F = 2 maxt ∈[0,T ] |f1 (t)|,   q (z)| . K = max maxz∈[0,T /2] |a(z)|, maxz∈[0,T /2] |b(z)|, maxz∈[0,T /2] | (10.2.23) Then M0 (t) ≤ F and t M1 (t) ≤ F K , M2 (t) ≤ K 1!



t

M1 (τ )dτ ≤ F K 2

0

t2 . 2!

Using the mathematical induction method, we can easily derive the following estimates: Mn (t) ≤ F K n

tn Tn ≤ F Kn , n = 1, 2, . . . . n! n!

298

10 Inverse Problems for Damped Wave Equations

Hence, the series (10.2.21) are uniformly convergent in DT and their sums determine the continuous solution for which the following estimates hold:



 |w1 (z, t)| ≤ F eKt , |w2 (z, t)| ≤ F eKt − 1 , |v(z, t)| ≤ F eKt − 1 . This, in turn, means that v ∈ C 1 (DT ). The uniqueness of the solution to problem (10.2.16) is proven in the usual way. Suppose to the contrary that the problem has another solution w¯ 1 , w¯ 2 , v. ¯ Let us denote the difference as w #1 = w1 − w¯ 1 , w #2 = w2 − w¯ 2 , # v = v − v. ¯ #2 ,# v ) satisfies the homogeneous integral equations (10.2.18)– Then the triple (# w1 , w (10.2.20) with f1 (t) = 0 in (10.2.20): ⎧ $t w #2 (z, t) = (z+t )/2[a(ζ )# w1(ζ, τ ) + b(ζ )# w2(ζ, τ ) +  q (ζ )# v (ζ, τ )]ζ =t +z−τ dτ, ⎪ ⎪ ⎪ ⎪ $ ⎪ t ⎪ ⎨# v (z, t) = 12 z [# w1 (ζ, τ ) + w #1 (ζ, τ )]dτ, $ t ⎪ ⎪ w1(ζ, τ ) + b(ζ )# w2(ζ, τ ) +  q (ζ )# v (ζ, τ )]ζ =τ +z−t dτ w #1 (z, t) = t −z [a(ζ )# ⎪ ⎪ ⎪ ⎪ $ t −z ⎩ + (t −z)/2[a(ζ )# w1(ζ, τ ) + b(ζ )# w2(ζ, τ ) +  q (ζ )# v (ζ, τ )]ζ =t −z−τ dτ. (10.2.24) Denote  ⎧ ⎪ ⎪ max max |# w (z, t)|, max |# w (z, t)|, max |# v (z, t)| , 1 2 ⎪ ⎪ z∈[0,t ] z∈[0,t ] z∈[0,t ] ⎪ ⎪ ⎨ t ∈ (0, T /2),  M(t) = ⎪ ⎪ max max |# w (z, t)|, max |# w (z, t)|, max |# v (z, t)| , ⎪ 1 2 ⎪ ⎪ z∈[0,T −t ] z∈[0,T −t ] z∈[0,T −t ] ⎪ ⎩ t ∈ (T /2, T ). Then in view of the system (10.2.24) we deduce the following inequality:  M(t) ≤ K

t

M(τ )dτ, t ∈ (0, T ).

(10.2.25)

0

Since Eq. (10.2.25) has only the trivial solution M(t) = 0 for all t ∈ (0, T ), we find that w #1 = w #2 = # v = 0 which implies: w1 = w¯ 1 , w2 = w¯ 2 , v = v. ¯

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

299

We prove the last assertion of the theorem (10.2.12) with formula (10.2.13). To this end we use the first and fourth equations in (10.2.16). We have: d w1 (z, z + 0) = (w1z + w1t )t =z+0 = a(z)w1 (z, z + 0). dz

(10.2.26)

The boundary condition in (10.2.16) leads to equality w1 (0, 0) = −2f1 (0). With Eq. (10.2.26) this leads to the relationship 

z

w1 (z, z + 0) = −2f1 (0) exp

a(ζ )dζ .

(10.2.27)

0

On the other hand, by w2 (z, z + 0) = 0 we conclude that − vz (z, z + 0) = vt (z, z + 0) =

1 w1 (z, z + 0). 2

(10.2.28)

Relations (10.2.27) and (10.2.28) lead to formulae (10.2.12) and (10.2.13). Now, we prove that the found solution has the following regularity properties: v, vt ∈ C 2 (DT ). To prove that v ∈ C 2 (DT ), differentiate Eqs. (10.2.16) with respect to t and introduce the following new notations: w˙ 1 = w1t , w˙ 2 = w2t , v˙ = vt . These new functions satisfy the following system of equations: ⎧ ⎪ w˙ 1t + w˙ 1z = a(z)w˙ 1 + b(z)w˙ 2 +  q (z)v, ˙ (z, t) ∈ DT , ⎪ ⎪ ⎪ ⎪ q (z)v, ˙ (z, t) ∈ DT , ⎨ w˙ 2t − w˙ 2z = a(z)w˙ 1 + b(z)w˙ 2 +  v˙t = 12 (w˙ 1 + w˙ 2 ), (z, t) ∈ DT , ⎪ ⎪ ⎪ v| ˙ t =z+0 = s(z), w˙ 2 |t =z+0 = s  (z), z ∈ (0, z∗ ), ⎪ ⎪ ⎩ w˙ 2 (0, t) − w˙ 1 (0, t) = 2f˙1 (t), t ∈ (0, T ),

(10.2.29)

where s  (z) = ds(z)/dz and f˙1 (t) means the derivative of f1 (t). Let us explain the conditions on the forth line of system (10.2.29). The first of them follows from equalities (10.2.12) we have proved above. The second one is the consequence of the following relations: s  (z) =

d v(z, ˙ z + 0) = (v˙z (z, t) + v˙t (z, t))|t =z+0 = w˙ 2 (z, z + 0). dz

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10 Inverse Problems for Damped Wave Equations

The system of integral equations equivalent to problem (10.2.29) can be easily constructed, similar to the construction of Eqs. (10.2.18)–(10.2.20). These equations are as follows: ⎧ $t w˙ 2 (z, t) = s  ((z + t)/2) + (z+t )/2[a(ζ )w˙ 1(ζ, τ ) + b(ζ )w˙ 2(ζ, τ ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + q (ζ )v(ζ, ˙ τ )]ζ =t +z−τ dτ, ⎪ ⎪ ⎪ ⎪ $t ⎪ 1 ⎨ v(z, ˙ t) = s(z) + 2 z [w˙ 1 (ζ, τ ) + w˙ 2 (ζ, τ )]dτ, (10.2.30) ⎪ w˙ 1 (z, t) = −2f˙1 (t − z) + s  ((t − z)/2) ⎪ ⎪ ⎪ ⎪ $t ⎪ ⎪ + t −z [a(ζ )w ˙ 1(ζ, τ ) + b(ζ )w ˙ 2(ζ, τ ) +  q (ζ )v(ζ, ˙ τ )]ζ =τ +z−t dτ ⎪ ⎪ ⎪ ⎪ $ ⎩ t −z + (t −z)/2[a(ζ )w˙ 1(ζ, τ ) + b(ζ )w˙ 2(ζ, τ ) +  q (ζ )v(ζ, ˙ τ )]ζ =t −z−τ dτ. Relations (10.2.30) form the system of equations with respect to th functions w˙ 1 (z, t), w˙ 2 (z, t), v(z, ˙ t) defined in DT . To find the solution of this system we can use again the method of successive approximations in the form w˙ 1 (z, t) = w˙ 10 (z, t) + v(z, ˙ t) =

v˙ 0 (z, t)

+

∞  n=1

∞ 

w˙ 1n (z, t), w˙ 2 (z, t) = w˙ 20 (z, t) +

∞  n=1

w˙ 2n (z, t),

v˙ n (z, t),

n=1

(10.2.31) where w20 (z, t) = s  ((z + t)/2), v 0 (z, t) = s(z), w10 (z, t) = −2f˙1 (t − z) + s  ((t − z)/2) and ⎧ $t ⎪ w˙ 2n (z, t) = (z+t )/2[a(ζ )w ˙ 1n−1 (ζ, τ ) + b(ζ )w˙ 2n−1 (ζ, τ ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + q (ζ )v˙ n−1 (ζ, τ )]ζ =t +z−τ dτ, ⎪ ⎪ ⎪ ⎪ $t ⎪ ⎪ v˙ n (z, t) = 12 z [w˙ 1n−1 (ζ, τ ) + w˙ 2n−1 (ζ, τ )]dτ, ⎪ ⎪ ⎨ $t (10.2.32) w˙ 1n (z, t) = t −z [a(ζ )w˙ 1n−1 (ζ, τ ) + b(ζ )w˙ 2n−1 (ζ, τ ) ⎪ ⎪ ⎪ ⎪ ⎪ + q (ζ )v˙ n−1 (ζ, τ )]ζ =τ +z−t dτ ⎪ ⎪ ⎪ ⎪ $ t −z ⎪ ⎪ + (t −z)/2[a(ζ )w ˙ 1n−1 (ζ, τ ) + b(ζ )w˙ 2n−1 (ζ, τ ) ⎪ ⎪ ⎪ ⎪ ⎩ + q (ζ )v˙ n−1 (ζ, τ )]ζ =t −z−τ dτ, n = 1, 2, . . . .

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

301

Evidently, all the functions w˙ 1n (z, t), w˙ 2n (z, t), v˙ n (z, t),, n = 0, 1, . . . , are continuous in DT . We introduce the following notations:  ⎧ ⎪ n n n ⎪ max max | w ˙ (z, t)|, max | w ˙ (z, t)|, max | v ˙ (z, t)| , ⎪ 1 2 ⎪ z∈[0,t ] z∈[0,t ] z∈[0,t ] ⎪ ⎪ ⎨ t ∈ (0, T /2),  n (t) = M ⎪ n n n ⎪ | w ˙ (z, t)|, max | w ˙ (z, t)|, max | v ˙ (z, t)| , max max ⎪ 1 2 ⎪ ⎪ z∈[0,T −t ] z∈[0,T −t ] z∈[0,T −t ] ⎪ ⎩ t ∈ (T /2, T ), and  F1 = max

max

(z,t )∈DT

|w˙ 10 (z, t)|, 

K = max

|w˙ 20 (z, t)|,

max

(z,t )∈DT

max |v˙ (z, t)| , 0

(z,t )∈DT



max |a(z)|, max |b(z)|, max | q (z)| .

z∈[0,T /2]

z∈[0,T /2]

z∈[0,T /2]

0 (t) ≤ F1 and Then M 2 (t) ≤ K 1 (t) ≤ F1 K t , M M 1!

 0

t

2 1 (τ )dτ ≤ F1 K 2 t . M 2!

By making similar calculations, we arrive at the following estimates: n n n (t) ≤ F1 K n t ≤ F1 K n T , n = 1, 2, . . . . M n! n!

Hence, the series (10.2.31) are uniformly convergent in DT and their sums determine the continuous solution w˙ 1 (z, t), w˙ 2 (z, t), v(z, ˙ t), The latter means that the functions v, vt , vz , vt t , vzt belong to the space C(DT ). Then from Eq. (10.2.10) it follows that vzz belong to C(DT ) also, that is v ∈ C 2 (DT ). Calculate now the value of the second derivatives of v(z, t) at t = z + 0. Using relations (10.2.12), (10.2.13), we find 

s  (z) = −s  (z)

d dz vt (z, z + 0) = (vt z + vt t )t =z+0 , d = dz vz (z, z + 0) = (vzz + vzt )t =z+0 .

(10.2.33)

Put t = z + 0 in the first equation (10.2.29) and use that w˙ 2 (z, z + 0) = s  (z). Then we obtain d w˙ 1 (z, z + 0) = a(z)w˙ 1 (z.z + 0) + b(z)s  (z) +  q (z)s(z). dz

(10.2.34)

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10 Inverse Problems for Damped Wave Equations

Moreover, we can calculate w˙ 1 (0, +0) using the boundary data at z = 0. Making this, we get   w˙ 1 (0, +0) = −2f˙1 (t) + w˙ 2 (0, t) t =+0 = −2f˙1 (0) + s  (0) = −2f˙1 (0) − f1 (0)a(0).

(10.2.35)

From Eqs. (10.2.34) and (10.2.35) it follows that   w˙ 1 (z, z + 0) =: d(z) = −2 f˙1 (0) + f1 (0)a(0) exp  +



z



z

exp

a(ζ1)dζ1

0





z

a(ζ )dζ 0



b(ζ )s  (ζ ) +  q (ζ )s(ζ ) dζ.

(10.2.36)

ζ

Thus, for finding the second derivatives of v(z, t) we obtain the following system of the algebraic equations: ⎧  ⎪ ⎨ (vt z + vt t )t =z+0 = s (z), (vzz + vzt )t =z+0 = −s  (z), ⎪ ⎩ (vt t − vzt )t =z+0 = d(z). Solving this system, we get vt t |t =z+0 = η(z) =: 12 (s  (z) + d(z)), vzt |t =z+0 = 12 (s  (z) − d(z)), , vzz |t =z+0 =

(10.2.37)

− 12 (3s  (z) − d(z)),

where s(z) and d(z) are given by formulae (10.2.13) and (10.2.36), correspondingly. From conditions (10.2.5) it follows that s ∈ C 2 [0, z∗ ] and d, η ∈ C 1 [0, z∗ ]. Finally, to complete the proof of the theorem, we prove that vt ∈ C 2 (DT ). Let w¨ 1 (z, t) = w1t t , w¨ 2 (z, t) = w2t t , v(z, ¨ t) = vt t . From the first relation of (10.2.37) we deduce that w¨ 2 (z, z + 0) = η (z), z ∈ (0, T /2).

(10.2.38)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

303

Differentiate the equations in (10.2.16) two times with respect to t, we obtain: ⎧ w¨ 1t + w¨ 1z = a(z)w¨ 1 + b(z)w¨ 2 +  q (z)v, ¨ (z, t) ∈ DT , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ w¨ 2t − w¨ 2z = a(z)w¨ 1 + b(z)w¨ 2 +  q (z)v, ¨ (z, t) ∈ DT , ⎪ ⎪ ⎪ ⎨ 1 v¨t = (w¨ 1 + w¨ 2 ), (z, t) ∈ DT , ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ = η(z), w¨ 2 |t =z+0 = η (z), z ∈ (0, z∗ ), v| ¨ t =z+0 ⎪ ⎪ ⎪ ⎪ ⎩ w¨ 2 (0, t) − w¨ 1 (0, t) = 2f¨1 (t), t ∈ (0, T ).

(10.2.39)

The system of integral equations that equivalent to problem (10.2.39) is as follows: ⎧ w¨ 2 (z, t) = η ((z + t)/2) ⎪ ⎪ ⎪ ⎪  t ⎪ ⎪ ⎪ ⎪ ⎪ + [a(ζ )w¨ 1(ζ, τ ) + b(ζ )w¨ 2(ζ, τ ) +  q (ζ )v(ζ, ¨ τ )]ζ =t +z−τ dτ, ⎪ ⎪ ⎪ (z+t )/2 ⎪ ⎪  ⎪ ⎪ 1 t ⎪ ⎪ ¨ t) = η(z) + [w¨ 1 (ζ, τ ) + w¨ 2 (ζ, τ )]dτ, ⎪ ⎨ v(z, 2 z (10.2.40) ⎪ w¨ 1 (z, t) = −2f¨1 (t − z) + η ((t − z)/2) ⎪ ⎪ ⎪ ⎪  t ⎪ ⎪ ⎪ ⎪ ⎪ + [a(ζ )w¨ 1(ζ, τ ) + b(ζ )w¨ 2(ζ, τ ) +  q (ζ )v(ζ, ¨ τ )]ζ =τ +z−t dτ ⎪ ⎪ ⎪ t −z ⎪ ⎪  t −z ⎪ ⎪ ⎪ ⎪ ⎩ + [a(ζ )w¨ 1(ζ, τ ) + b(ζ )w¨ 2(ζ, τ ) +  q (ζ )v(ζ, ¨ τ )]ζ =t −z−τ dτ. (t −z)/2

Equations (10.2.40) form the system for finding w¨ 1 (z, t), w¨ 2 (z, t) and v(z, ¨ t) in DT . To find the solution of this system one can apply exactly the same method of successive approximations described above. Hence, one can prove that the solution of the system of Eqs. (10.2.40) exists in the domain DT , and each component of the triple w¨ 1 (z, t), w¨ 2 (z, t), v(z, ¨ t) belongs to C(DT ). This implies that vt t t , vt t z ∈ C(DT ). Differentiating the first equation of (10.2.10) with respect to t, we find that vt zz ∈ C(DT ) also. Hence, vt ∈ C 2 (DT ). This completes the proof of the theorem.  

10.2.2 Existence and Uniqueness of the Inverse Problem Solution The following corollary is an immediate consequence of the above theorem. Corollary 10.2.1 Assume that conditions (10.2.5) and (10.2.6) hold. Then for solvability of the inverse problem it is necessary that the output ν(t) satisfies the

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10 Inverse Problems for Damped Wave Equations

following regularity and consistency conditions: ν ∈ C 3 (0, T ), ν(0) = 0, ν(0) ˙ = s(0), ν¨ (0) = η(0).

(10.2.41)

Below we prove that these necessary conditions are also sufficient for unique local solvability, and for a global uniqueness of the inverse problem. Theorem 10.2.2 Let the coefficients of Eq. (10.2.1) and the input f (t) satisfy conditions (10.2.5) and (10.2.6), respectively. Assume, in addition, that and the output ν(t) satisfies conditions (10.2.41). Then there exists T ∗ ≤ 2z∗ such that the inverse problem (10.2.1)–(10.2.2) has the unique solution  q (z) for T < T ∗ , and this solution belongs to C(0, T /2). Proof Consider the system of equations ⎧ w¨ 1t + w¨ 1z = a(z)w¨ 1 + b(z)w¨ 2 +  q (z)v, ¨ (z, t) ∈ DT , ⎪ ⎪ ⎨ q (z)v, ¨ (z, t) ∈ DT , w¨ 2t − w¨ 2z = a(z)w¨ 1 + b(z)w¨ 2 +  (10.2.42) 1 ⎪ v ¨ = ( w ¨ − w ¨ ), (z, t) ∈ D , 2 1 T ⎪ 2 ⎩ z ¨ z=0 = ν¨ (t), t ∈ [0, T ], w¨ 1 |z=0 = g1 (t), w¨ 2 |z=0 = g2 (t), v| which is a consequence of system (10.2.10) and relations (10.2.14) and (10.2.15). Here g1 (t) =

d2 d2 (t)) , g (t) = ν (t) − f (˙ (˙ν (t) + f1 (t)) . 1 2 dt 2 dt 2

(10.2.43)

Evidently, the problem of finding a solution of the problem (10.2.42) is equivalent to solving of the following system of integral equations: ⎧ ⎪ ⎪ w¨ 1 (z, t) =$ g1 (t − z) ⎪ ⎪ z ⎪ q (ζ )v(ζ, ¨ τ )]τ =t −z+ζ dζ, + 0 [a(ζ )w¨ 1(ζ, τ ) + b(ζ )w¨ 2 (ζ, τ ) +  ⎪ ⎪ ⎨ w¨ 2 (z, t) = g2 (t + z) ⎪ $z ⎪ ⎪ ⎪ q (ζ )v(ζ, ¨ τ )]τ =t +z−ζ dζ, − 0 [a(ζ )w¨ 1(ζ, τ ) + b(ζ )w¨ 2 (ζ, τ ) +  ⎪ ⎪ ⎪ $ ⎩ z v(z, ¨ t) = ν¨ (t) + 12 0 [w¨ 2 (ζ, t) − w¨ 1 (ζ, t)] dζ, (z, t) ∈ DT . (10.2.44) We need to complete this system by an equation for function  q (z). To this end we use the relations w˙ 2t − w˙ 2z = a(z)w˙ 1 + b(z)w˙ 2 +  q (z)v, ˙ (z, t) ∈ DT , ˙ ¨ + f1 (t), t ∈ (0, T ), w˙ 2 (0, t) = ν(t)

(10.2.45)

which are the consequence of the first relation (10.2.29), notation (10.2.14) and the output (10.2.11). Integrating (10.2.45) along the characteristic line τ + ζ = t + z

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

305

from point (0, t + z) till point (z, t), we obtain: w˙ 2 (z, t) = ν¨ (t + z) + f˙1 (t + z)  z a(ζ )w˙ 1(ζ, τ ) + b(ζ )w˙ 2(ζ, τ ) +  q (ζ )v(ζ, ˙ τ) − 0

τ =t +z−ζ

dζ.

(10.2.46)

Put here t = z + 0 and then differentiate the obtained equation with respect to z. As the result we get: d w˙ 2 (z, z + 0) = 2g2 (2z) dz −a(z)w˙ 1 (z, z + 0) − b(z)w˙ 2 (z, z + 0) −  q (z)v(z, ˙ z + 0)  z a(ζ )w¨ 1(ζ, τ ) + b(ζ )w¨ 2(ζ, τ ) +  − q (ζ )v(ζ, ¨ τ) dζ. (10.2.47) τ =2z−ζ

0

In view of (10.2.12) and (10.2.36), we deduce that ⎧ ˙ z + 0) = s(t), w˙ 1 (z, z + 0) = d(z), w˙ 2 (z, z + 0) = s  (z), ⎨ v(z, (10.2.48) d ⎩ w˙ 2 (z, z + 0) = s  (z). dz Using formula (10.2.36), represent a(z)d(z) in the following form 

z

a(z)d(z) = n(z) +

(10.2.49)

k(z, ζ ) q (ζ )dζ, 0

where n(z) and k(z, ζ ) are known functions defined by the following formulae: ⎧  $ z  n(z) = − 2f˙1 (0) + f1 (0)a(0) a(z) exp 0 a(ζ )dζ ⎪ ⎪ ⎪  z  z ⎪ ⎪ ⎨ exp a(ζ1 )dζ1 b(ζ )s  (ζ )dζ, + 0 ζ ⎪  z ⎪ ⎪ ⎪ ⎪ a(ζ1)dζ1 . ⎩ k(z, ζ ) = a(z)s(ζ ) exp

(10.2.50)

ζ

Dividing both sides of (10.2.47) by s(z) and using formulae (10.2.48) and (10.2.49), we obtain the equation  q (z) =  q0 (z) − 1 − s(z)



z 0

1 s(z)



z

k(z, ζ ) q (ζ )dζ 0

q (ζ )v(ζ, ¨ τ) a(ζ )w¨ 1(ζ, τ ) + b(ζ )w¨ 2(ζ, τ ) + 

τ =2z−ζ

dζ,

(10.2.51)

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10 Inverse Problems for Damped Wave Equations

where  q0 (z) = −

2g2 (2z) + n(z) + b(z)s  (z) . s(z)

(10.2.52)

Equations (10.2.44) and (10.2.51) form the nonlinear system of integral equations for determining functions w¨ 1 (z, t), w¨ 2 (z, t), v(z, ¨ t) and  q (z). We rewrite this system as the operator equation ϕ = A(ϕ),

(10.2.53)

where ϕ(z, t) = (ϕ1 (z, t), ϕ2 (z, t), ϕ3 (z, t), ϕ4 (z)) and the components are defined as follows: ϕ1 (z, t) = w¨ 1 (z, t), ϕ2 (z, t) = w¨ 2 (z, t), ϕ3 (z, t) = v(z, ¨ t), ϕ4 (z) =  q1 (z). The operator A = (A1 , . . . , A4 ) is determined by the right-hand-sides of Eqs. (10.2.44) and (10.2.51), i.e. ⎧ A1 (ϕ) = ϕ10 (z, t) ⎪ ⎪ ⎪ ⎪ $z ⎪ ⎪ ⎪ + 0 [a(ζ )ϕ1(ζ, τ ) + b(ζ )ϕ2(ζ, τ ) + ϕ4 (ζ )ϕ3 (ζ, τ )]τ =t −z+ζ dζ, ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ A2 (ϕ) = ϕ2 (z, t) ⎪ $z ⎨ − 0 [a(ζ )ϕ1(ζ, τ ) + b(ζ )ϕ2(ζ, τ ) + ϕ4 (ζ )ϕ3 (ζ, τ )]τ =t +z−ζ dζ, ⎪ $z ⎪ ⎪ ⎪ A3 (ϕ) = ϕ30 (z, t) + 12 0 [ϕ2 (ζ, t) − ϕ1 (ζ, t)] dζ, ⎪ ⎪ ⎪ ⎪ $z ⎪ 1 0 ⎪ ⎪ ⎪ A4 (ϕ) = ϕ4 (z, t) − s(z) 0 k(z, ζ )ϕ4 (ζ )dζ ⎪  ⎪ ⎪ ⎪ − 1 $ z a(ζ )ϕ (ζ, τ ) + b(ζ )ϕ (ζ, τ ) + ϕ (ζ )ϕ (ζ, τ ) ⎩ dζ. 1 2 4 3 s(z) 0 τ =2z−ζ

(10.2.54) In the above formulae the functions (ϕ10 , . . . , ϕ4 ) =: ϕ 0 are defined as follows: ϕ10 (z, t) = g1 (t − z), ϕ20 (z, t) = g2 (t + z), ϕ30 (z, t) = ν¨ (t), ϕ40 (z) =  q0 (z). Consider the set   M = ϕ ∈ C(DT ) : ϕ − ϕ 0 C(DT ) ≤ ϕ 0 C(DT ) , where .

ϕ C(DT ) = max ϕj C(DT ) . j =1,4

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

307

We prove that the above defined operator A is contractive on this set if T is small enough. For this, we need prove that (1) A(ϕ) ∈ M, for all ϕ ∈ M; (2) there exists some nonnegative real number ρ ∈ [0, 1) such that A(ϕ 1 ) − A(ϕ 2 ) C(DT ) ≤ ρ ϕ 1 − ϕ 2 C(DT ) for all ϕ 1 , ϕ 2 ∈ M. Assume that min |s(z)| = s0 ,

z∈[0,T /2]

max

0≤ζ ≤z≤T /2

max( max |a(z)|, z∈[0,T /2]

|k(z, ζ )| = k0 ,

max |b(z)|) = K1 .

z∈[0,T /2]

It is obvious that s0 > 0. Assume that min |s(z)| = s0 ,

z∈[0,T /2]

max( max |a(z)|, z∈[0,T /2]

max

0≤ζ ≤z≤T /2

|k(z, ζ )| = k0 ,

max |b(z)|) = K1 .

z∈[0,T /2]

It is obvious that s0 > 0. Suppose that ϕ ∈ M. Then ϕ C(DT ) ≤ 2 ϕ 0 C(DT ) and ⎧

A1 (ϕ) − ϕ10 C(DT ) ≤ 2T (K1 ϕ 0 C(DT ) + ϕ 0 2C(DT ) ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ A2 (ϕ) − ϕ 0 C(D ) ≤ 2T (K1 ϕ 0 C(D ) + ϕ 0 2 T T 2 C(DT ) ),

(10.2.55) ⎪

A3 (ϕ) − ϕ30 C(DT ) ≤ T ϕ 0 C(DT ) , ⎪ ⎪ ⎪ ⎪ ⎩

A4 (ϕ) − ϕ40 C(DT ) ≤ T [(k0 + 2K1 ) ϕ 0 C(DT ) + 2 ϕ 0 2C(DT ) ]/s0 . Let  T1 = max 1,

1 s0 , 0 2(K1 + ϕ C(DT ) ) k0 + 2K1 + 2 ϕ 0 C(DT )



and T ≤ T1 . Then from (10.2.55) it follows that A(ϕ) − ϕ 0 C(DT ) ≤ ϕ 0 C(DT ) , i.e. A(ϕ) ∈ M. Take now arbitrary ϕ 1 , ϕ 2 ∈ M and estimate the difference A(ϕ 1 ) − A(ϕ 2 ). After elementary calculations we deduce that |ϕ41 (ζ )ϕ31 (ζ, τ ) − ϕ42 (ζ )ϕ32 (ζ, τ )| = |ϕ41 (ζ )(ϕ31 (ζ, τ ) − ϕ32 (ζ, τ )) + ϕ32 (ζ, τ )(ϕ41 (ζ ) − ϕ42 (ζ ))| ≤ 4 ϕ 0 C(DT ) ϕ 1 − ϕ 2 C(DT ) .

308

10 Inverse Problems for Damped Wave Equations

Then ⎧

A1 (ϕ 1 ) − A1 (ϕ 2 ) C(DT ) ≤ T (K1 + 2 ϕ 0 C(DT ) ) ϕ 1 − ϕ 2 C(DT ) , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ A2 (ϕ 1 ) − A2 (ϕ 2 ) C(D ) ≤ T (K1 + 2 ϕ 0 C(D ) ) ϕ 1 − ϕ 2 C(D ) , T T T ⎪

A3 (ϕ 1 ) − A3 (ϕ 2 ) C(DT ) ≤ (T /2) ϕ 1 − ϕ 2 C(DT ) , ⎪ ⎪ ⎪ ⎪ ⎩

A4 (ϕ 1 ) − A4 (ϕ 2 ) C(DT ) ≤ 2sT 0 (k0 + 2K1 + 4 ϕ 0 C(DT ) ) ϕ 1 − ϕ 2 C(DT ) . Let now 

1 1 2s0 , T2 = max , 0 2 K1 + 2 ϕ C(DT ) k0 + 2K1 + 4 ϕ 0 C(DT )



and T < T2 . Then A(ϕ 1 ) − A(ϕ 2 ) C(DT ) ≤ ρ ϕ 0 C(DT ) with ρ = T /T2 < 1. Thus, operator A is contractive on M if T < T ∗ = min(T1 , T2 , 2z∗ ). Using the Banach theorem for the contractive operator, we conclude that the system (10.2.54) has the unique solution on M, if T is small enough. This completes the proof of the theorem.   We introduce now the set of admissible coefficients (potentials) Q defined as follows: Q = {q ∈ C(0, h(T /2)) :  q C(0,T /2) ≤ q0 },

(10.2.56)

where q0 is a given positive number. Then the following local existence theorem holds. Theorem 10.2.3 Let coefficients of Eq. (10.2.1) and the input f (t) satisfy conditions (10.2.5) and (10.2.6), correspondingly. Assume in addition that the output ν(t) satisfies the condition: ν ∈ C 3 [0, T ]. Then the inverse problem (10.2.1)–(10.2.2) may have only a unique solution q(x) on the set Q, for any T ≤ 2z∗ . Proof Let q ∈ Q be a given function. In view of definitions in (10.2.7) and (10.2.9) we conclude that maxz∈[0,T /2] | q (z)| ≤ q0 . Hence, the constant K > 0 introduced in (10.2.23) depends also on the constant q0 > 0, i.e. K = K(q0 ). Further, as it is proved in Theorem 10.2.2, a solution to the system of Eqs. (10.2.44) exists in C(DT ) for any T ≤ 2z∗ and satisfies the conditions: |w¨ 1 (z, t)| ≤ N, |w¨ 2 (z, t)| ≤ N, |v(z, ¨ t)| ≤ N, (z, t) ∈ DT ,

(10.2.57)

where N is some positive constant depending on K as well as the maximum of C(0, T ) norms of the functions g1 (t), g2 (t) and ν¨ (t).

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

309

Denote by ϕ 1 and ϕ 2 the solutions of system (10.2.54) corresponding to q1 , q2 ∈ Q, that is, ϕ41 = q1 and ϕ42 = q2 . We introduce the new function consisting of the difference of these functions: ϕ = (# ϕ1 , # ϕ2 , # ϕ3 , # ϕ4 ). ϕ = ϕ1 − ϕ2, # # Then

ϕji C(DT ) ≤ N, ϕ4i C(DT ) ≤ q0 , j = 1, 2, 3, i = 1, 2,

(10.2.58)

and after elementary calculations we get: |ϕ41 (ζ )ϕ31(ζ, τ ) − ϕ42 (ζ )ϕ32 (ζ, τ )| = |ϕ41 (ζ )# ϕ3(ζ, τ ) + ϕ32 (ζ, τ )# ϕ4 (ζ )| ≤ q0 |# ϕ3 (ζ, τ )| + N|# ϕ4 (ζ )|.

(10.2.59)

Now, we use Eqs. (10.2.54) for ϕ = ϕ 1 and ϕ = ϕ 2 , to estimate the differences |# ϕk (z, t)| := |Ak (ϕ 1 ) − Ak (ϕ 2 )|, k = 1, 4, by employing inequalities (10.2.58) and (10.2.59). We have: |# ϕ1 (z, t)| = |A1 (ϕ 1 ) − A1 (ϕ 2 )|  z K(|# ϕ1 (ζ, τ )| + |# ϕ2 (ζ, τ )| + |# ϕ3 (ζ, τ )|) ≤ 0

+q0 |# ϕ3 (ζ, τ )| + N|# ϕ4 (ζ )|

τ =t +z−ζ

dζ, (z, t) ∈ DT . (10.2.60)

Similarly, |# ϕ2 (z, t)| = |A2 (ϕ 1 ) − A2 (ϕ 2 )|  z K(|# ϕ1 (ζ, τ )| + |# ϕ2 (ζ, τ )| + |# ϕ3 (ζ, τ )|) ≤ 0

+q0 |# ϕ3 (ζ, τ )| + N|# ϕ4 (ζ )|

τ =t −z+ζ

dζ, (z, t) ∈ DT . (10.2.61)

Using the third equation in (10.2.55), we find |# ϕ3 (z, t)| = |A3 (ϕ 1 ) − A3 (ϕ 2 )|  1 z (|# ϕ1 (ζ, τ )| + |# ϕ2 (ζ, τ )|)dζ, (z, t) ∈ DT . (10.2.62) ≤ 2 0

310

10 Inverse Problems for Damped Wave Equations

Finally, from the forth equation of (10.2.55) we get |# ϕ4 (z)| = |A4 (ϕ 1 ) − A4 (ϕ 2 )|   1 z ≤ ϕ4 (ζ )| + K(|# ϕ1 (ζ, τ )| + |# ϕ2 (ζ, τ )| + |# ϕ3 (ζ, τ )|) k0 |# s0 0  dζ, (z, t) ∈ DT . (10.2.63) +q0 |# ϕ3 (ζ, τ )| + N|# ϕ4 (ζ )| τ =2z−ζ

Let  (z) = max

max |# ϕj (z, t)|, j = 1, 4 .

z≤t ≤T −z

Then estimates (10.2.60)–(10.2.63) lead to the inequality z (z) ≤ C

(ζ )dζ, z ∈ (0, T /2),

(10.2.64)

0

where C = max (3K + q0 + N, 1, [k0 + 3K + q0 + N]/s0 ) . It follows from inequality (10.2.64) that (z) = 0 for all z ∈ (0, T /2). Hence, # ϕj (z, t) = 0, j = 1, 4, and ϕj1 (z, t) = ϕj2 (z, t), j = 1, 4, for all (z, t) ∈ DT . This means that there can only be a unique solution to the reverse problem.   In the conclusion we note that using the similar way it is possible to state the stability estimate of the solution to the inverse problem with respect to data of the problem.

10.2.3 Necessary Estimates for the Weak and Regular Weak Solutions of the Direct Problem In this subsection we assume that the inputs in the direct problem (10.2.1) satisfy the following bounds: 

T ≤ 2z∗ , f ∈ H 1 (0, ).

(10.2.65)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

311

Denote by ST (t) the intersection of the domain DT with the straight line τ = t, t ∈ (0, T ) (Fig. 10.3 on the right):  ST (t) =

{(ζ, τ ) : ζ ∈ (0, t), τ = t}, t ∈ (0, T /2), {(ζ, τ ) : ζ ∈ (0, T − t), τ = t}, t ∈ (T /2, T ).

(10.2.66)

Theorem 10.2.4 Assume that the inputs in (10.2.1) satisfy conditions (10.2.5) and (10.2.65). Suppose that q(x) belongs to the set of admissible coefficients Q defined in (10.2.56). Then the weak solution v(z, t) of the transformed direct problem (10.2.10) belongs to C(0, T ; H 1(ST (t)) ) with vt ∈ L2 (0, T ; L2 (ST (t)) ). Moreover, for all t ∈ (0, T ) the following estimate holds:  ST (t )

 v 2 (ζ, t) + vζ2 (ζ, t) + vt2 (ζ, t) dζ ≤ CE f 2H 1 (0,T ) ,

(10.2.67)

where ⎧ ⎪ ⎨ CE = 4C2 (C3 + 1) exp [(2C1 + 1)T ] /(m(0)r(0)),  C1 = 5 max a C(0,T /2), b C(0,T /2), 1 + q0 , ⎪ ⎩ C2 = max (T , 4/T ) , C3 = max (4/T , 6T ) .

(10.2.68)

Proof Introduce the domain GT (t) in DT (Fig. 10.3 on the right): GT (t) = {(ζ, τ ) ∈ DT : τ ≤ t, t ∈ (0, T )}. Denote by T (t) := ∂GT (t) the boundary of this domain. Then (k) ∪4k=1 T (t) and (1) T (t)

= ST (t), (3) T (t)

(2) T (t)

T (t)

=

= {(ζ, τ ) : ζ = 0, τ ∈ (0, t)},

= {(ζ, τ ) : ζ = τ, 0 ≤ τ ≤ min(t, T /2)},

(4) T (t)

= {(ζ, τ ) : ζ = T − τ, T /2 ≤ τ ≤ t}.

Note that T(4) (t) is the empty set, if t < T /2. Multiply both sides of the first equation of (10.2.16) by w1 (z, t), the second equation by w2 (z, t) and the third one by 2v(z, t). Summing up these equations and then integrating over GT (t), we finally obtain the following integral identity: 

 GT (t )

 ∂τ P (ζ, τ ) + ∂ζ Q(ζ, τ ) dζ dτ =

[a(ζ )w1(ζ, τ ) GT (t )

+b(ζ )w2(ζ, τ ) + ( q (ζ ) + 1)v(ζ, τ )] (w1 (ζ, τ ) + w2 (ζ, τ ))dζ dτ,

(10.2.69)

312

10 Inverse Problems for Damped Wave Equations

where P = v 2 + (w12 + w22 )/2, Q = (w12 − w22 )/2. We apply Green’s formula to the left-hand-side integral of (10.2.69): 



∂τ P (ζ, τ ) + ∂ζ Q(ζ, τ ) dζ dτ

GT (t )



=

[P (ζ, τ ) cos(n, τ ) + Q(ζ, τ ) cos(n, ζ )] ds,

(10.2.70)

T (t )

where cos(n, ζ ) and cos(n, τ ) are the components of the unit outward normal n to T (t) at the point (ζ, τ ) and ds is the length element. We evaluate now the righthand-side line integral over each part T(k) (t) of the domain GT (t). Let  Ik (t) =

(k) T (t )

[P (ζ, τ ) cos(n, τ ) + Q(ζ, τ ) cos(n, ζ )] ds, k = 1, 4.

Then 



I1 (t) = ST (t )

v 2 (ζ, t) + (w12 (ζ, t) + w22 (ζ, t))/2 dζ,

 t  1 t 2 w2 (0, τ ) − w12 (0, τ ) dτ = 2 v(0, ˙ τ )f1 (τ )dτ I2 (t) = 2 0 0  t = 2v(0, t)f1 (t) − 2 v(0, τ )f˙1 (τ )dτ, 0



min(t,T /2) 

I3 (t) = − 0

 I4 (t) =



t

T /2

v 2 (τ, τ ) + w22 (τ, τ ) dτ = 0,

v 2 (T − τ, τ ) + w12 (T − τ, τ ) dτ, t ≥ T /2.

Hence, 



∂τ P (ζ, τ ) + ∂ζ Q(ζ, τ ) dζ dτ

GT (t )



t

≥ I1 (t) + 2v(0, t)f1 (t) − 2 0

v(0, τ )f˙1 (τ )dτ.

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

313

On the other hand,  q (ζ ) + 1)v(ζ, τ )] [a(ζ )w1 (ζ, τ ) + b(ζ )w2 (ζ, τ ) + ( GT (t )

× [w1 (ζ, τ ) + w2 (ζ, τ )] dζ dτ 



≤ C1

2

v (ζ, τ ) GT (t )

+ (w12 (ζ, τ )

+ w22 (ζ, τ ))/2





t

dζ dτ = C1

I1 (τ )dτ, 0

where C1 > 0 is defined in (10.2.68). Taking into account these in (10.2.69) and (10.2.70) we deduce that  t  t I1 (t) ≤ C1 I1 (τ )dτ + 2 v(0, τ )f˙1 (τ )dτ − 2v(0, t)f1 (t), (10.2.71) 0

0

for all t ∈ (0, T ). Now estimate the last two right-hand-side terms in (10.2.71) through the H 1 norm of the input f1 (t) in the transformed direct problem (10.2.10). To estimate v(0, t) we use the following auxiliary formulas: ⎧ $ ⎨ − 0t vζ (ζ, t)dζ, t ∈ (0, T /2), v(0, t) = $ ⎩ 1 T −t ((ζ − T + t)v(ζ, t)) dζ, t ∈ (T /2, T ). ζ T −t 0 This yields: ⎧ $t ⎨ T2 0 vζ2 (ζ, t)dζ, t ∈ (0, T /2), 2 v (0, t) ≤ ⎩ 4 $ T −t v 2 (ζ, t)dζ + T $ T −t v 2 (ζ, t)dζ, t ∈ (T /2, T ), ζ 0 T 0 Since vz2 (z, t) ≤ (w22 (z, t) + w12 (z, t))/2, by the definition vz (z, t) = (w2 (z, t) − w1 (z, t))/2, the above relations imply that v 2 (0, t) ≤ C2 I1 (t), t ∈ (0, T ),

(10.2.72)

where C2 > 0 is defined in (10.2.68). Further, the identity f1 (t) = −

1 T



T



t

((T − τ )f1 (τ ))τ dτ +

0

(f1 (τ ))τ dτ, 0

implies the estimate f12 (t) ≤ C3 f1 2H 1 (0,T ) , t ∈ (0, T ), with the constant C3 > 0 defined in (10.2.68).

(10.2.73)

314

10 Inverse Problems for Damped Wave Equations

Using the above results with the ε-inequality we can estimate now the last two right-hand-side terms in (10.2.71) as follows: −2v(0, t)f1 (t) ≤ C2 εI1 (t) + C3 ε−1 f1 2H 1 (0,T ) ,  2

t



v(0, τ )f˙1 (τ )dτ ≤ C2 ε

0

t

I1 (τ )dτ + ε

0



t

≤ C2 ε 0

−1



t 0

f˙12 (τ )dτ

I1 (τ )dτ + ε−1 f1 2H 1 (0,T ) , t ∈ (0, T ), ε > 0.

In view of inequality (10.2.71) this implies: 

t

(1 − C2 ε)I1 (t) ≤ (C1 + C2 ε) 0

I1 (τ )dτ + (C3 + 1)ε−1 f1 2H 1 (0,T ) ,

(10.2.74)

for all t ∈ (0, T ) and ε > 0. Choosing here the arbitrary parameter ε > 0 as ε = 1/(2C2 ) we conclude that  I1 (t) ≤ (2C1 + 1) 0

t

I1 (τ )dτ + 4C2 (C3 + 1) f1 2H 1 (0,T ) , t ∈ (0, T ).

Applying the Gronwall-Bellman inequality, we arrive at the estimate I1 (t) ≤ 4C2 (C3 + 1) exp [(2C1 + 1)T ] f1 2H 1 (0,T ) , t ∈ (0, T ). This is exactly the required estimate (10.2.67), since (w12 + w22 )/2 = vt2 + vζ2 .

 

The energy estimate for the regular weak solution given in the following theorem is similar to the above one. Theorem 10.2.5 Let conditions of Theorem 10.2.4 hold. Assume, in addition, that the input f (t) satisfies the regularity condition f ∈ H 2 (0, T ). Then the regular weak solution of the transformed direct problem (10.2.10) belongs to C(0, T ; H 2(ST (t)) ) with vt ∈ C(0, T ; H 1 (ST (t)) ) and vt t ∈ L2 (0, T ; L2 (ST (t)) ). Moreover, for all t ∈ (0, T ) the following estimate holds: 

 ST (t )

E f 2 2 vt2 (ζ, t) + vζ2t (ζ, t) + vt2t (ζ, t) dζ ≤ C , H (0,T )

(10.2.75)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

315

where ⎧ E = C 2 exp(2C2 T )/(m(0)r(0)), ⎪ C ⎪ ⎪ ⎪ ⎪ ⎪ 1 + 4(C2 C3 + max (T , 1/T )), 2 = 4C2 + 2C ⎨C 1 = 2 max (T , 1/T )Ce , ⎪ C ⎪ ⎪ ⎪ 

⎪   ⎪ ⎩ Ce = T 1 + a 2 C(0,T /2) exp T a C(0,T /2) /2,

(10.2.76)

and C2 , C3 > 0 are the constants defined in (10.2.68). Proof We use system (10.2.29) as a basic system of equations corresponding to the regular weak solution of the transformed direct problem. Multiply the first equation of this system by w˙ 1 (z, t), the second equation by w˙ 2 (z, t) and the third one by 2v(z, ˙ t). In the same way as in the proof of the above theorem, we arrive at the following integral identity: 



 τ ) cos(n, ζ ) ds = P(ζ, τ ) cos(n, τ ) + Q(ζ,

T (t )

 [a(ζ )w˙ 1(ζ, τ ) GT (t )

+b(ζ )w˙ 2(ζ, τ ) + ( q (ζ ) + 1)v(ζ, ˙ τ )] (w˙ 1 (ζ, τ ) + w˙ 2 (ζ, τ ))dζ dτ,

(10.2.77)

 = (w˙ 2 − w˙ 2 )/2. We evaluate the left-hand-side where P = v˙ 2 + (w˙ 12 + w˙ 22 )/2, Q 1 2 (k) line integral in (10.2.77) along each part T (t), k = 1, 4 of the boundary T (t). Denote by  Ik (t) =



 (k) T (t )

(ζ, τ ) cos(n, τ ) + Q(ζ,  τ ) cos(n, ζ ) ds, k = 1, 4. P

Then I1 (t) =



 ST (t )

v˙ 2 (ζ, t) + (w˙ 12 (ζ, t) + w˙ 22 (ζ, t))/2 dζ,

and this coincides with the integral in (10.2.75). Further, we use the boundary conditions in (10.2.29) to calculate other integrals: 1 I2 (t) = 2

 t 0

 t w˙ 22 (0, τ ) − w˙ 12 (0, τ ) dτ = 2 v(0, ¨ τ )f˙1 (τ )dτ 0

 = −2 0

t

v(0, ˙ τ )f¨1 (τ )dτ + 2v(0, ˙ t)f˙1 (t) − 2v(0, ˙ 0)f˙1 (0),

316

10 Inverse Problems for Damped Wave Equations

I3 (t) = −



min(t,T /2) 

0



v˙ 2 (τ, τ ) + w˙ 22 (τ, τ ) dτ

min(t,T /2) 

=−

 2 s 2 (ζ ) + s  (ζ ) dζ,

0

I4 (t) =



t



T /2

v˙ 2 (T − τ, τ ) + w˙ 12 (T − τ, τ ) dτ, t ≥ T /2.

On the other hand, the upper estimate for the right-hand-side integral in (10.2.77) is obtained exactly in the same way as in the proof of the previous theorem:  q (ζ ) + 1)v(ζ, ˙ τ )] [a(ζ )w˙ 1(ζ, τ ) + b(ζ )w˙ 2(ζ, τ ) + ( GT (t )



t

× (w˙ 1 (ζ, τ ) + w˙ 2 (ζ, τ ))dζ dτ ≤ C1

I1 (τ )dτ,

0

with the same constant C1 > 0. Taking into account the above estimates in (10.2.77) we conclude that I1 (t) ≤ C1



t

I1 (τ )dτ + 2



0

t

v(0, ˙ τ )f¨1 (τ )dτ − 2v(0, ˙ t)f˙1 (t)

0

+2v(0, ˙ 0)f˙1 (0) +



T /2 

 2 s 2 (ζ ) + s  (ζ ) dζ. (10.2.78)

0

Now, we need to estimate the second to fifth right-hand-side terms in (10.2.78) through the H 2 -norm of the input f1 (t). We start with the evaluation of the fourth right-hand-side term. To this end, we use the boundary condition v(z, ˙ z + 0) = s(z) in (10.2.29) and definition of s(z) to deduce that v(0, ˙ +0) = s(0) = −f˙1 (0). Hence 2v(0, ˙ 0)f˙1 (0) = −2f1 (0)f˙1 (0).

(10.2.79)

Using the identity 1 f1 (0) = − T



T

((T − τ )f1 (τ ))τ dτ

0

and its analogue for f˙1 (0), we can derive the following estimates: (f1 (0))2 ≤

2 T

 2 2 f˙1 (0) ≤ T



T



0



2  f˙1 (t) dt,

0 T

0

T

(f1 (t))2 dt + 2T 2  f˙1 (t) dt + 2T



T 0

2  f¨1 (t) dt.

(10.2.80)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

317

With (10.2.79) this implies: 2v(0, ˙ 0)f˙1 (0) ≤ 2 max (T , 1/T ) f1 2H 2 (0,T )

(10.2.81)

We estimate the third right-hand-side term −2v(0, ˙ t)f˙1 (t) in (10.2.78). To this end, we use the inequality v˙ 2 (0, t) ≤ C2  I1 (t), t ∈ (0, T ), which is obtained in exactly the same way with the same constant C2 > 0, as estimate (10.2.72). With the ε-inequality and estimate (10.2.73), with f1 (t) replaced by f˙1 (t) and H 1 replaced by H 2 , this yields: − 2v(0, ˙ t)f˙1 (t) ≤ C2 εI1 (t) + C3 ε−1 f1 2H 2 (0,T ) .

(10.2.82)

In the same way estimate we the second right-hand-side term in (10.2.78) as follows:  2

t

v(0, ˙ τ )f¨1 (τ )dτ ≤ C2 ε



0

t

I1 (τ )dτ + ε−1



0

t



f¨1 (τ )

2

dτ. (10.2.83)

0

Finally, we should estimate the last right-hand-side integral in (10.2.78), i.e. the norm s H 1 (0,T /2) , through appropriate norms of the inputs, taking into account also formula (10.2.13). In view of this formula, we have 

z

s(z) = −f (0) exp

a(ζ )dζ , s  (z) = a(z)s(z),

0

and 

s 2H 1 (0,T /2) =

T /2

(1 + a 2 (z))s 2 (z)dz ≤ Ce (f1 (0))2 ,

0

where Ce is defined by formula (10.2.76). With estimate (10.2.80) this yields: 1 f1 2 1 .

s 2H 1 (0,T /2) ≤ C H (0,T )

(10.2.84)

1 defined in (10.2.76). with C Taking into account estimates (10.2.81), (10.2.82), (10.2.83) and (10.2.84) in (10.2.78) we finally arrive at the following inequality: (1 − C2 ε)I1 (t) ≤ C2 ε



t 0

 I1 (τ )dτ + C3 ε−1 + 2 max(T , 1/T ) f1 2H 2 (0,T ) 1 f1 2 1 +ε−1 f¨1 2L2 (0,T ) + C , H (0,T )

318

10 Inverse Problems for Damped Wave Equations

with the constants defined in (10.2.76). Choosing ε = 1/(2C2 ), after elementary transformations we obtain the following inequality I1 (t) ≤ 2C2



t 0

2 f1 2 2 I1 (τ )dτ + C . H (0,T )

Applying the Gronwall-Bellman inequality, we arrive at the required estimate (10.2.75).  

10.2.4 The Neumann-to-Dirichlet Operator Consider the nonlinear Neumann-to-Dirichlet operator introduced in (10.2.3). Lemma 10.2.1 Let conditions of Theorem 10.2.4 hold. Assume, in addition, that the input f (t) satisfies the regularity condition f ∈ H 2 (0, T ). Then the Neumann-toDirichlet operator [·] : Q → L2 (0, T ) defined in (10.2.3) is a compact operator. Proof Let {q (n)} ⊂ Q, n = 1, ∞, be a sequence of coefficients uniformly bounded in C-norm by q0 > 0. Denote by {u(n) (x, t)}, u(n) (x, t) := u(x, t; q (n) ), the sequence of corresponding regular weak solutions of the direct problem (10.2.1). Then {(q (n))(t)}, (q (n) )(t) = u(n) (0, t), is the sequence of outputs. We need to prove that this sequence is a relatively compact subset of L2 (0, T ) or, equivalently, is bounded in the norm of H 1 (0, T ). This follows from the trace inequalities

u(0, ·) 2L2 (0,T ) = v(0, ·) 2L2 (0,T ) ≤

ut (0, ·) 2L2 (0,T ) =

T 2 2 vz L2 (DT ) ,

vt (0, ·) 2L2 (0,T ) ≤ T2 vzt 2L2 (D ) , T

and estimates (10.2.67) and (10.2.75).

(10.2.85)  

Lemma 10.2.1 implies that the inverse problem (10.2.1)–(10.2.2) is ill-posed. The following lemma is a consequence of Theorem 10.2.3 and shows that if the Dirichlet measured output ν(t) is noise free and smooth enough, then Neumann-toDirichlet operator is invertible. Lemma 10.2.2 Let conditions of Theorem 10.2.4 hold. Assume, in addition, that the input f (t) satisfies the regularity condition f ∈ H 3 (0, T ) and f (0) = 0. Suppose that ν(t) defined in (10.2.2) is a noise free measured output. Then the Neumann-toDirichlet operator f [·] defined in (10.2.3) maps the set Q into the set (T ) := Q ⊂ C 3 (0, T ) and it is invertible on (T ) for any T < 2z∗ . Proof Evidently, f ∈ C 2 (0, T ) since C 2 (0, T ) ⊂ H 3 (0, T ). Then we can use Corollary 10.2.1 to conclude that ν ∈ C 3 (0, T ), since ν(t) = u(0, t). Set of all outputs ν(t) forms the image of Q defined as (T ). On the other hand, if ν ∈ (T ), then according to Theorem 10.2.3 there exists only a unique q ∈ Q.  

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

319

For further considerations we shall need the following auxiliary results. Denote by uδ (x, t) := u(x, t; q + δq) and u(x, t) := u(x, t; q) the solutions of the direct problem (10.2.1) corresponding to the potentials q, q + δq ∈ Q. Then the difference δu(x, t) := uδ (x, t) − u(x, t) solves the following initial-boundary value problem: ⎧ δ ⎪ ⎨ m(x)δut t + μ(x)δut = (r(x)δux )x + q(x)δu + δq(x)u , (x, t) ∈ T , (10.2.86) δu(x, 0) = 0, δut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ r(0)δux (0, t) = 0, δu(, t) = 0, t ∈ (0, T ). In view of the variable z = z(x), introduced in (10.1.4), and the function v(z(x), t) = u(x, t) the transformed problem corresponding to (10.1.86) is as follows: ⎧ ⎪ μ(z)δvt = δvzz + p (z)δvz +  q (z)δv + δ q (z)v δ (z, t), (z, t) ∈ ∗T , ⎨ δvt t +  (10.2.87) δv(z, 0) = 0, δvt (z, 0) = 0, z ∈ (0, z∗ ), ⎪ ⎩ δv (0, t) = 0, δv(z∗ , t) = 0, t ∈ (0, T ), z q (z) = δq(h(z))/m(h(z)) and v δ (z(x), t) = where ∗T := (0, z∗ ) × (0, T ), δ δ u (h(z), t)/m(h(z)). Lemma 10.2.3 Assume that conditions of Theorem 10.2.4 hold. Then for the weak solution δv(z, t) of problem (10.2.87) the following estimate holds: 



I δ (t) =: ST (t )

δv 2 (ζ, t) + δvζ2 (ζ, t) + δvt2 (ζ, t) dζ ≤ CEδ f 2H 1 (0,T ) δ q 2C(0,T /2),

(10.2.88)

where CEδ = T CE exp ((1 + C1 )T ) and the constants C1 , CE are defined in Theorem 10.2.4. Moreover, the following conditions are met along the characteristic line t = z: δv(z, z + 0) = 0, δvz (z, z + 0) = 0, δvt (z, z + 0) = 0, z ∈ (0, T /2). (10.2.89) Proof In the similar way as in the proof of Theorem 10.2.1, we can show that the transformed problem corresponding to (10.2.87) is equivalent to the following system: ⎧ ⎪ δw1t + δw1z = a(z)δw1 + b(z)δw2 +  q (z)δv + δ q (x)v δ , (z, t) ∈ DT , ⎪ ⎪ ⎪ ⎪ q (z)δv + δ q (x)v δ , (z, t) ∈ DT , ⎨ δw2t − δw2z = a(z)δw1 + b(z)δw2 +  1 δvt = 2 (δw1 + δw2 ), (z, t) ∈ DT , (10.2.90) ⎪ ∗ ), ⎪ ⎪ δv| = 0, δw | = 0, z ∈ (0, z t =z+0 2 t =z+0 ⎪ ⎪ ⎩ δw2 (0, t) − δw1 (0, t) = 0, t ∈ (0, T ),

320

10 Inverse Problems for Damped Wave Equations

where δv(z, t) = δu(h(x), t), δw1 (z, t) = δvt (z, t) − δvz (z, t) and δw2 (z, t) = δvt (z, t) + δvz (z, t). An analogue of inequality (10.2.71) for this problem can be established in the same way:  I1δ (t) ≤ C1

t 0

 I1δ (τ )dτ +

v δ (ζ, τ )(δw1 (ζ, τ ) + δw2 (ζ, τ ))δ q (ζ )dζ dτ, GT (t )

with constant C1 defined in (10.2.68). The second integral in the right-hand side of this inequality can be estimate as follows:  v δ (ζ, τ )(δw1 (ζ, τ ) + δw2 (ζ, τ ))δ q (ζ )dζ dτ GT (t )

 =2

v δ (ζ, τ )δvt (ζ, τ )δ q (ζ )dζ dτ GT (t )

 ≤ GT (t )

δvt2 (ζ, τ )dζ dτ + δ q 2C(0,T /2) 

t

≤ 0

 (v δ (ζ, τ ))2 dζ dτ GT (t )

I δ (τ )dτ + T CE  q 2C(0,T /2) f 2H 1 (0,T ) .

Hence,  I1δ (t)

t

≤ (1 + C1 ) 0

I1δ (τ )dτ + T CE  q 2C(0,T /2) f 2H 1 (0,T ) .

In view of the Gronwall-Bellman inequality the above inequality implies (10.2.88). We prove now the equalities (10.2.89). Since boundary conditions in (10.2.90) are homogeneous and v δ (z, t) ≡ 0 in D0 , we obtain that δv(z, t) ≡ 0 in D0 . Hence, δv(z, z + 0) = 0 and δvz (z, z + 0) + δvt (z, z + 0) = 0. Therefore δw2 (z, z + 0) = 0 and from the fifth relationship of (10.2.90) we find δw1 (0, 0) = 0. Then, if we set t = z + 0 in the first equation of (10.2.90), we find: d δw1 (z, z + 0) = a(z)δw1 (z, z + 0), z ∈ (0, T /2), dz since δw2 (z, z + 0) = δv(z, z + 0) = v δ (z, z + 0) = 0. Taking into account the condition δw1 (0, 0) = 0 we obtain δw1 (z, z + 0) = 0 for all z ∈ (0, T /2). So, δw1 (z, z+0) = 0 and δw2 (z, z+0) = 0. Then δvz (z, z+0) = 0 and δvt (z, z+0) = 0 for all z ∈ (0, T /2).   Corollary 10.2.2 Assume that conditions of Lemma 10.2.3 hold. Then for the weak solution δv(z, t) of problem (10.2.87) the following estimate holds:

δvz C(ST (t )) ≤ C1δ f H 1 (0,T ) δq C(0,h(T /2)),

δvt C(ST (t )) ≤ C2δ f H 1 (0,T ) δq C(0,h(T /2)), t ∈ (0, T ),

(10.2.91)

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator

321

where h(z) is the inverse function to z = z(x), and C1δ , C2δ > 0 depend only on the constants defined in Lemma 10.2.3. Lemma 10.2.4 Let conditions of Theorem 10.2.4 hold. Then the Dirichlet-toNeumann operator [·] : Q →  L2 (0, T ) introduced in (10.2.3) is Lipschitz continuous, that is

q1 − q2 L2 (0,T ) ≤ L q1 − q2 C(0,h(T /2)),

(10.2.92)

for all q1 , q2 ∈ Q, L2 = T 2 C1δ /2 and C1δ > 0 is the constant introduced in Corollary 10.2.2. Proof By the definition,

q1 − q2 L2 (0,T ) = δu(0, ·) L2 (0,T ) , and δu(x, t) is the weak solution of problem (10.2.86). Then the required assertion (10.2.92) follows from the first trace inequality in (10.2.85) and the first estimate of (10.2.91).  

10.2.5 Existence of a Quasi-Solution. Gradient Formula for the Tikhonov Functional Due to measurement errors in the output ν(t), exact equality in Eq. (10.2.4) can never be achieved. For this reason, one needs to introduce the Tikhonov functional 1 J ( q ) := 2



T

 q ∈ Q, q ) − ν(t)]2 dt,  [v(0, t; 

(10.2.93)

0

where  := { Q q (z) = q(h(z))/m(h(z)) : q ∈ Q}

(10.2.94)

and Q is the set of admissible coefficients introduced in (10.2.56). Then we can  such that reformulate this inverse problem as the problem of funding  q∗ ∈ Q J ( q∗ ) = J" , where q ). J∗ = min J (   q ∈Q

(10.2.95)

A solution of the minimization problem (10.2.95) is defined as a quasi-solution of the transformed inverse problem (10.2.10)–(10.2.11). Based on Lemma 10.2.4 one can prove that the Tikhonov functional is Lipschitz continuous.

322

10 Inverse Problems for Damped Wave Equations

 is the Theorem 10.2.6 Let conditions of Theorem 10.2.4 hold. Assume that Q set of admissible coefficients defined as in (10.2.94). Then the minimization  ⊂ C[0, T /2]. problem (10.2.95) has a solution in Q Proof of this theorem is made similar to the proof of Theorem 10.1.11. Denote by δJ ( q ) := J ( q + δ q ) − J ( q ), the increment of the Tikhonov  Then functional (10.2.93), assuming that  q,  q + δ q ∈ Q. 

T

δJ ( q) =

q ) − ν(t)] δv(0, t)dt + [v(0, t; 

0

1 2



T

[δv(0, t)]2 dt, (10.2.96)

0

where δv(x, t) := v δ (x, t) − u(x, t) solves the initial-boundary value problem (10.2.87). Lemma 10.2.5 Assume that the inputs in (10.2.1) satisfy conditions (10.2.5)  are arbitrary inputs. Then the following and (10.2.65). Suppose that  q,  q + δ q∈Q integral relationship holds: 

T

 p(t)δu(0, t)dt = −

0

δ q (z)v δ (z, t)φ(z, t)dzdt.

(10.2.97)

DT

where φ(z, t) is the solution of the (well-posed) adjoint problem ⎧ ⎪ μ(z)φt = φzz − ( p (z)φ)z +  q (z)φ, (z, t) ∈ ∗T , ⎨ φt t −  φ(z, T ) = 0, φt (z, T ) = 0, z ∈ (0, z∗ ), ⎪ ⎩ φ (0, t) − p (0)φ(0, t) = p(t), φ(z∗ , t) = 0, t ∈ [0, T ], z

(10.2.98)

with the Neumann input p ∈ L2 (0, T ), δv(z, t) is the weak solution of problem (10.2.87) and v δ (z(x), t) := v(x, t; q + δq) is the weak solution of the transformed direct problem (10.2.10) with  q replaced by  q + δ q. Proof First of all, note that (10.2.98) is almost the same well-posed problem, similar to the direct problem (10.2.10), with the appropriate linear boundary condition at z = 0, as the change of variable t by T − t shows. Therefore from the similar arguments given in the proof of Theorem 10.2.1 we deduce that φ(z, t) ≡ 0 for all {(z, ) : T − z < t ≤ T , 0 < z < z∗ } and φ(z, T − z − 0) = 0 for all z ∈ (0, z∗ ). Hence, φz (z, T − z − 0) − φt (z, T − z − 0) = 0. Multiply both sides of Eq. (10.2.87) by the arbitrary function φ(z, t), integrate over DT and apply the integration by parts formula multiple times. Then we obtain:  μ(z)φt − φzz + ( p(z)φ)z −  q (z)φ] δv dzdt [φt t −  DT

 + 0

T /2

−z−0 μ(z)δv φ]tt =T dz [δvt φ − δvφt +  =z+0

10.2 Recovering a Potential from Neumann-to-Dirichlet Operator



T /2

+ 0

 +

T

T /2

323

−0 (z)δv φ ]z=t dt [δvφz − δvz φ − p z=0

−t −0 (z)δv φ ]z=T dt [δvφz − δvz φ − p z=0

 =

δ q (z)v δ (z, t)φ(z, t) dzdt. DT

We now require that φ(z, t) is the solution of the adjoint problem (10.2.98). Then taking into account the boundary conditions at z = 0 given in (10.2.87) and (10.2.98) with the equalities δv(z, z + 0) = δvt (z, z + 0) = δvz (z, z + 0) = 0, (10.2.99) φ(z, T − z − 0) = 0, φz (z, T − z − 0) − φt (z, T − z − 0) = 0, we arrive at the integral identity (10.2.97). Recall that the equalities in the first line of (10.2.99) were stated in Lemma 10.2.3 (see formula (10.2.89) ). As mentioned above, the second line of (10.2.99) is a consequence of Theorem 10.2.1 applied to the adjoint problem. and the equalities in the second line were given at the beginning of this proof.   Assume that the input p(t) in the adjoint problem (10.2.98) is defined as follows: p(t) = u(0, t) − ν(t), t ∈ (0, T ),

(10.2.100)

where ν(t) is the Dirichlet measured output. Then, in view of the integral relationship (10.2.97) and v δ (z, t) = v(z, t) + δv(z, t), the increment formula (10.2.96) for the Tikhonov functional has the following form: T /2  T −z

 δJ (q) = −  − 0

0

T /2  T −z

v(z, t)φ(z, t)dt δ q (z)dz

z

δ q (z)δv(z, t)φ(z, t)dtdz +

z

1 2



T

[δv(0, t)]2 dt. (10.2.101)

0

Theorem 10.2.7 Assume that the inputs in (10.2.1) satisfy conditions (10.2.5)  is the set of admissible coefficients introduced and (10.2.65). Suppose that Q in (10.2.94). Then the Tikhonov functional (10.2.93) is Fréchet differentiable. Moreover, for the Fréchet gradient ∇J (q) the following formula holds: 

T −z

∇J ( q )(z) = − z

 v(z, t)φ(z, t)dt,  q ∈ Q,

(10.2.102)

324

10 Inverse Problems for Damped Wave Equations

where φ(z, t) is the solution of the adjoint problem (10.2.98) with the Neumann input p(t) given by (10.2.100) and v(z(x), t) = u(x, t) is the weak solution of the transformed direct problem (10.2.10). Proof From the first estimate of (10.2.91) and the trace inequality (10.2.85) it follows that the second right-hand-side integral in (10.2.101) is of the order O( δ q 2C(0,T /2)). Further, by the same argument given in the proof of Lemma 10.2.3 we conclude that the third right-hand-side integral is of the order O( δ q 2C(0,T /2)), also. This leads to the required result.   The gradient formula (10.2.102) allows us to implement the following Conjugate Gradient Algorithm, adopted for the inverse coefficient problems (see Sect. 6.5.4). Step 1. Step 2. Step 3.

For n = 0 choose the initial iteration  q (0)(z). (0) q (0))(z). Compute the initial descent direction pd (z) := ∇J ( Determine the descent direction parameter βn > 0 from the minimumization problem Fn (βn ) := inf Fn (β), Fn (β) := J ( q (n) − β∇J ( q (n)) ), β>0

Step 4.

for each n = 0, 1, 2, . . . . (n) Find the next iteration  q (n+1) (z) =  q (n) (z) − βn pd (z) and compute the convergence error q (n) ) − ν δ L2 (0,T ) . e(n;  q (n); δ) := v(0, t; 

Step 5.

If the stopping condition e(n;  q (n); δ) ≤ τM δ < e(n;  q (n−1); δ), τM > 1, δ > 0

Step 6.

holds, then go to Step 7. Set n := n + 1 and compute ⎧ (n) (n−1) ⎨ pd (z) := ∇J ( q (n))(z) + γn pd (z), ⎩ γn =

Step 7.

∇J ( q (n) ) 2

∇J ( q (n−1) ) 2

and go to Step 3. Stop the iteration process.

Here, the Dirichlet measured output ν(t) is denoted by ν δ (t) in order to show the noise level δ > 0. In the case when this algorithm is applied to the regularized form 1  Jα ( q 2L2 (0,T /2) ,  q ) := J ( q ) +  q∈Q 2

10.3 Recovering a Potential From Dirichlet-to-Neumann Operator

325

of the Tikhonov functional which gradient is defined by formula  ∇Jα ( q )(z) = ∇J ( q )(z) + α q (z),  q ∈ Q, q (n) ) by Jα ( q (n) ) and one needs to replace in the above algorithm J ( q (n) ) and ∇J ( (n) ∇Jα ( q ), respectively. Examples given in Sect. 6.5.4 show that this version is very efficient for inverse coefficient problems for parabolic PDEs.

10.3 Recovering a Potential From Dirichlet-to-Neumann Operator In this section we consider the following inverse problem of recovering the unknown potential q(x) in ⎧ ⎪ ⎨ m(x)ut t + μ(x)ut = (r(x)ux )x + q(x)u, (x, t) ∈ T , u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ u(0, t) = ν(t), u(, t) = 0, t ∈ (0, T ),

(10.3.1)

from the Neumann boundary measured output f (t) := r(0)ux (0, t), t ∈ (0, T ),

(10.3.2)

imposed at the boundary x = 0, where T := (0, ) × (0, T ). Even though this inverse potential problem appears to be a problem similar to the inverse potential problem (10.2.1)–(10.2.2) considered in the previous section, they have essential differences. First of all, the structure of the solution of the direct problem (10.3.1) is significantly different from that given in (10.2.1), as we will see below. Namely, while for a regular Neumann input f (t), the solution of the direct problem in (10.2.1) is a continuous function in the entire domain DT ∪ D0 defined by the characteristics, as it was proved in Theorem 10.2.1, the solution of the direct problem (10.3.1) is a discontinuous function along the characteristic line t = z, even for the regular Dirichlet input ν ∈ C 2 (0, T ), with ν(0) = 0. Perhaps, this is the main difference between the above mentioned inverse problems, i.e. the inverse potential problems with Neumann-to-Dirichlet and Dirichlet-to-Neumann operators. Therefore, the main question here is to find out the distinctive properties of the solution of the direct problem (10.3.1) in the domains defined by the characteristics and along these characteristics, as well. Revealing these properties requires a fundamentally new approach, which is significantly different from that given in the previous section.

326

10 Inverse Problems for Damped Wave Equations

10.3.1 Regularity Properties of the Direct Problem Solution. Formation of Discontinuities As in the previous section, here we assume that the coefficients m(x), μ(x), r(x), q(x) satisfy conditions (10.2.5). In addition, we suppose that the Dirichlet input ν(t) satisfies the following regularity condition: ν ∈ C 2 [0, T ], ν(0) = 0.

(10.3.3)

In the same way as in Sect. 10.2.1 we transform the inverse problem defined by (10.3.1) and (10.3.2) to the following problem of recovering the unknown potential  q (z) in ⎧ ⎪ μ(z)vt = vzz + p (z)vz +  q (z)v, (z, t) ∈ ∗T := (0, z∗ ) × (0, T ), ⎨ vt t +  v(z, 0) = 0, vt (z, 0) = 0, z ∈ (0, z∗ ), ⎪ ⎩ v(0, t) = ν(t), v(z∗ , t) = 0, t ∈ [0, T ], (10.3.4) from the Neumann boundary measured output  f(t) := vz (0, t), f(t) := f (t)/ r(0)m(0), t ∈ (0, T ).

(10.3.5)

We will prove that only the case T ≤ 2z∗ needs to be considered. As a consequence, the solution of problem (10.3.1) in the domain {t + z < T , 0 ≤ z ≤ z∗ } does not depend on a linear homogeneous boundary condition at z = z∗ . We will demonstrate that the function f(t), t ∈ (0, T ) introduced in (10.3.5) uniquely determines the coefficient  q (z) for all z ∈ [0, T /2]. Consider the auxiliary direct problem: ⎧ ⎪  vt t +  μ(z) vt =  vzz + p (z) vz +  q (z) v, ⎪ ⎪ ⎪ ⎨ (z, t) ∈ G∗T := (0, z∗ ) × (−∞, T ), ⎪ v (z, t)|t 0, follows Estimate (10.3.61), with the introduced in the theorem constant C from estimates (10.3.59), (10.3.62) and (10.3.63).  

10.3.4 The Dirichlet-to-Neumann Operator We introduce the transformed Dirichlet-to-Neumann operator as follows:   ν q )(t) := vx (0, t;  q ), t ∈ (0, T ), q ∈ Q, (  → L2 (0, T ), ν : Q   := { Q q = q(h(z))/m(h(z)) : q ∈ Q},

(10.3.64)

corresponding to the transformed inverse problem (10.3.4)–(10.3.5), Q is the set of admissible coefficients introduced in (10.3.45). Then this problem can be reformulated as the following nonlinear operator equation:  ν  q )(t) = f(t), t ∈ (0, T ), q ∈ Q, (

(10.3.65)

where f(t) is the transformed Neumann measured output introduced in (10.3.5). Lemma 10.3.1 Let conditions of Theorem 10.3.7 hold. Then the transformed  → L2 (0, T ) defined in (10.3.64) is a ν : Q Dirichlet-to-Neumann operator  compact operator.  n = 1, ∞, be a sequence of potentials uniformly bounded in Proof Let { q (n) } ⊂ Q, C-norm by q0 > 0. Denote by {v (n) (z, t)}, v (n) (z, t) := v(x, t; q (n) ), the sequence of corresponding regular weak solutions of the transformed direct problem (10.3.4).

350

10 Inverse Problems for Damped Wave Equations

(n) ν  Then {( q (n) )(t)}, ( ν q (n) )(t) = vz (0, t) is the sequence of outputs. We need to prove that this sequence is a relatively compact subset of L2 (0, T ) or, equivalently, is bounded in the norm of H 1 (0, T ). (n) To prove the uniform boundedness of the sequence { vz (0, ·) L2 (0,T ) }, n = 1, ∞, we use the second relation in (10.3.32) to deduce that

vz(n) (0, t) = a(0)ν(t) − ν  (t) +



t 0

ν(t − τ )wz(n) (0, τ )dτ.

This implies: 

vz(n) (0, ·) 2L2 (0,T ) ≤ 3 a 2 (0) + T w(n) 2C(DT ) ν 2L2 (0,T ) + 3 ν  2L2 (0,T ) #12 ν 2 1 ≤C , H (0,T )

(10.3.66)

where #12 = 3 max (a 2 (0) + T C12 , 1) C

(10.3.67)

and {w(n) (z, t)} is the sequence of solutions of problem (10.3.9) corresponding  to the sequence of potentials { q (n) } ⊂ Q.For each n = 1, 2, 3, . . ., the norm (n)

w C(DT ) is bounded by the constant C1 > 0 defined in Theorem 10.3.1, which depends also on the norm  q (n) C[0,T /2] . Since the sequence of norms (n) {  q C[0,T /2] }, n = 1, ∞ is uniformly bounded by definition (10.3.45), the righthand-side of the above inequality is uniformly bounded. (n) We prove now that the sequence of norms { vzt (0, ·) L2 (0,T ) } is also bounded 2 in the norm of L (0, T ). This follows from the relation vzt (0, t) = a(0)ν  (t) − ν  (t) + ν(0)wz(n) (0, τ ) + (n)

 0

t

ν(t − τ )wz(n) (0, τ )dτ,

which is obtained from the second relation of (10.3.33) for z = 0, by the same argument. This completes the proof lemma.   This lemma implies that the inverse problem (10.3.1)–(10.3.2) is ill-posed. For further considerations we shall need some preliminary results. Denote by v(z, t) := v(z, t;  q ) and v(z, ˜ t) := v(z, t;  q + δ q ) the solutions of the transformed direct problem (10.3.4) corresponding to the potentials  q,  q + δ q ∈  Then the difference δv(z, t) := v(z, Q. ˜ t) − v(z, t) solves the following initial-

10.3 Recovering a Potential From Dirichlet-to-Neumann Operator

351

boundary value problem: ⎧ ⎪ μ(z)δvt = δvzz + p (z)δvz +  q (z)δv + δ q (z)v(z, ˜ t), (z, t) ∈ ∗T , ⎨ δvt t +  δv(z, 0) = 0, δvt (z, 0) = 0, z ∈ (0, z∗ ), ⎪ ⎩ δv(0, t) = 0, δv(z∗ , t) = 0, t ∈ (0, T ), (10.3.68) where δ q (z) = δq(h(z))/m(h(z)) and # v (z, t) = # u(h(z), t), according to (10.2.7) and (10.2.9), ∗T is the domain introduced in (10.2.4). Lemma 10.3.2 Let conditions of Theorem 10.3.6 hold. Then for the weak solution δv(z, t) of problem (10.3.68) the following estimate holds:  #(t )



2 δv 2 (z, t) + δvz2 (z, t) + δvt2 (z, t) dz ≤ CE,δ

δ q 2C[0,T /2] ,

(10.3.69)

for all t ∈ (0, T ), where 

2 = T C 2 max (1/2; C 2 + 1) ν 2 CE,δ e T H 1 (0,T )

CT2 = 2 max (T , 1/T ), Ce =

√1 2

T M exp(3KC0 T ).

(10.3.70)

and the constants M, K, C0 > 0 are defined in Theorem 10.3.1. Proof From the representation formula (10.3.29) it follows that 

t

δv(z, t) =

ν(t − τ )δw(z, τ )dτ,

(10.3.71)

z

where δw(z, t) is the solution of problem (10.3.9) with the source term F (z)δ(t − z) replaced by δ q (z)w(z, ˜ t), and w(z, ˜ t) := (z, t;  q + δ q ) is the solution of problem (10.3.9) corresponding to the potential  q (z) + δ q (z). We can prove that an estimate similar to (10.3.10) in Theorem 10.3.1 holds for the solution δw(z, t) also, that is

δw C 1 (DT ) ≤ Ce δ q C[0,T /2]

(10.3.72)

where the constant Ce > 0 is defined in (10.3.70). With (10.3.71) this yields:  δv 2 (z, t)dz ≤ #(t )

1 2 2 T Ce ν 2L2 (0,T ) δ q 2C[0,T /2] . 2

(10.3.73)

352

10 Inverse Problems for Damped Wave Equations

Similarly, analogues of formulas in (10.3.32) for the solution δv(z, t) problem (10.3.68) are the following relations: 

$t

ν  (t − τ )δw(z, τ )dτ, $t δvz (z, t) = −ν(t − z)δw(z, z + 0) + z ν(t − τ )δwz (z, τ )dτ,

δvt (z, t) = ν(0)δw(z, t) +

z

(10.3.74)

As in the proof of Theorem 10.3.6, we deduce from the first relation of (10.3.74) that  t 2   δvt2 (z, t) ≤ 2Ce2 ν 2 (0) δ q 2C[0,T /2] + 2T Ce2 δ q 2C[0,T /2] ν (t − τ ) dτ. z

Using the estimate ν 2 (0) ≤ CT2 ν 2H 1 (0,T ) , with the constant CT > 0 defined in (10.3.70), in the above inequality and integrating it over #(t) we conclude that  #(t )

δvt2 (z, t)dz ≤ T (CT2 + T )Ce2 ν 2H 1 (0,T ) δ q 2C[0,T /2] .

(10.3.75)

Further, from the second relation of (10.3.74) we get the following result:  δvz2 (z, t) ≤ 2Ce2 δ q 2C[0,T /2] ν 2 (t − z) + 2T Ce2 δ q 2C[0,T /2]

t

ν 2 (t − τ )dτ.

z

This yields:  #(t )

δvz2 (z, t)dz ≤ T (CT2 + T )Ce2 ν 2H 1 (0,T ) δ q 2C[0,T /2] .

(10.3.76)

Estimates (10.3.73), (10.3.75) and (10.3.76) lead to the desired result (10.3.69).   A similar estimate holds for the regular weak solution of problem (10.3.68). Lemma 10.3.3 Let conditions of Theorem 10.3.7 hold. Then for the regular weak solution δv(z, t) of problem (10.3.68) the following estimate holds: 

 #(t )

2 2 E,δ δvt2 (z, t) + δvzt (z, t) + δvt2t (z, t) dz ≤ C

δ q 2C[0,T /2] ,

E,δ > 0 is defined through the constants for all t ∈ (0, T ), where the constant C introduced in (10.3.70) and also the norm ν H 2 (0,T ) .

10.3 Recovering a Potential From Dirichlet-to-Neumann Operator

353

Now we prove that the transformed Dirichlet-to-Neumann operator defined in (10.3.64) is Lipschitz continuous. Lemma 10.3.4 Let conditions of Theorem 10.3.6 hold. Then the Dirichlet-to → ν : Q Neumann operator   L2 (0, T ) introduced in (10.3.64) is Lipschitz continuous, that is ν  ν  q1 −  q2 L2 (0,T ) ≤  L  q1 −  q2 C[0,T /2] ,



(10.3.77)

 where  for all  q1 ,  q2 ∈ Q, L = Ce T ν L2 (0,T ) is the Lipschitz constant and Ce > 0 is the constant defined in (10.3.70). Proof By the definition, ν  ν  q1 −  q2 L2 (0,T ) = δvz (0, ·) L2 (0,T ) ,



(10.3.78)

where δvz (z, t) is the weak solution of problem (10.3.68). To estimate the righthand-side norm we use the second formula in (10.3.74) at z = 0: 

t

δvz (0, t) =

ν(t − τ )δwz (0, τ )dτ.

0

With estimate (10.3.72) this yields:  δvz2 (0, t) ≤ Ce2 T δ q 2C[0,T /2]

t

ν 2 (t − τ )dτ

0

Hence  0

T

δvz2 (0, t)dt ≤ Ce2 T 2 δ q 2C[0,T /2] ν 2L2 (0,T ) .

(10.3.79)  

In view of (10.3.78) we arrive at the required result (10.3.77).

10.3.5 Fréchet Differentiability of the Tikhonov Functional. Gradient Formula Since exact equality in Eq. (10.3.65) can never be achieved due to measurement errors in the output f ∈ L2 (0, T ), we introduce the Tikhonov functional corresponding to the transformed inverse problem (10.3.4)–(10.3.5): 1 J( q ) := 2



T 0



q ) − f(t) vz (0, t; 

2

 dt,  q ∈ Q,

(10.3.80)

354

10 Inverse Problems for Damped Wave Equations

 and f(t) are defined in (10.3.64) and (10.3.5), correspondingly. Then the where Q  transformed inverse problem can be reformulated as the problem of funding  q∗ ∈ Q such that J( q∗ ) = J" , where J∗ = min J( q ).

(10.3.81)

  q ∈Q

The following existence theorem is an immediate consequence of Lemma 10.3.3.  is the Theorem 10.3.8 Let conditions of Lemma 10.3.3 hold. Assume that Q set of admissible coefficients defined in (10.3.64). Then the minimization prob ⊂ C[0, T /2]. lem (10.3.81) has a solution in Q Proof of this theorem is made similar to the proof of Theorem 10.1.11 in Sect. 10.1. Denote by δ J( q ) := J( q +δ q )− J( q ), the increment of the Tikhonov functional,  assuming that  q,  q + δ q ∈ Q. Then δ J( q) =



T



0

1 vz (0, t;  q ) − f(t) δvz (0, t)dt + 2



T

[δvz (0, t)]2 dt, (10.3.82)

0

where δv(x, t) := v(z, ˜ t) − v(z, t) solves the initial-boundary value problem (10.3.68). To establish Fréchet differentiability of the Tikhonov functional we need to derive analogues of the representation formulas (10.3.29)–(10.3.30) for the adjoint problem solution. To this end, we need an analogue of Theorem 10.3.1 related to the following auxiliary problem: ⎧ ⎪ μ(z)v˘t = v˘zz − ( p (z)v) ˘ z + q (z)v, ˘ (z, t) ∈ ∗T , ⎨ v˘t t +  v(z, ˘ t)|t 0 defined as follows:  L∇ = C5 ν 2H 1 (0,T ) + f 2L2 (0,T ) ,

(10.3.102)

where C5 = C5 (T ,  μ C[0,T /2] ,  p C[0,T /2] ,  q C[0,T /2] ) > 0 is a constant. Proof In view of (10.3.100) we deduce that

∇ J( q1 ) − ∇ J( q2 ) L2 (0,T ) 

T

≤ 0

 [v1 (z, t)ψ1 (z, t) − v2 (z, t)ψ2 (z, t)]2 dzdt, #(t )

where vk (z, t), ψk (z, t) are the solutions of the direct and adjoint problems (10.3.4)  k = 1, 2. This yields: and (10.3.88), correspondingly, for given  qk ∈ Q, q2 ) L2 (0,T )

∇ J( q1 ) − ∇ J(  ≤2 0

T



 #(t )

δv 2 (z, t)ψ12 (z, t)dzdt + 2

0

T

 #(t )

v22 (z, t)δψ 2 (z, t)dzdt,

360

10 Inverse Problems for Damped Wave Equations

or

∇ J( q1 ) − ∇ J( q2 ) L2 (0,T ) ≤ 2 δv 2C(DT ) ψ1 2L2 (0,T ;L2 (#(t )) 2 v2 2L2 (0,T ;L2 (#(t )) δψ 2C(DT ) ,

(10.3.103)

where δv(z, t) = v1 (z, t) − v2 (z, t), δψ(z, t) = ψ1 (z, t) − ψ2 (z, t) and the function δψ(z, t) is the solution of the following problem: ⎧ μ(z)δψt = ψzz − ( p(z)δψ)z +  q (z)δψ + δ q (z)ψ2 (z, t), (z, t) ∈ ∗T , ⎪ ⎨ δψt t −  δψ(z, T ) = 0, δψt (z, T ) = 0, z ∈ (0, z∗ ), ⎪ ⎩ δψ(0, t) = 0, δψ(z∗ , t) = 0, t ∈ [0, T ], q2 (z). where δ q (z) =  q1 (z) −  To estimate the norm δψ C(DT ) in (10.3.103), we use the representation formula (10.3.90) for the solution ψ(z, t) of adjoint problem (10.3.88). We have: 

T −t

δψ(z, t) =

fadj (t + τ )δ w(z, ˘ τ )dτ, (z, t) ∈ DT ,

z

where δ w(z, ˘ t) = w˘ 1 (z, t) − w˘ 2 (z, t) and w˘ k (z, t) is the solutions of problem (10.3.85) and (10.3.86) with  q (z) replaced by  qk (z), k = 1, 2. For the norm δ w

˘ C 1 (DT ) the estimate similar to (10.3.72) holds:

δ w

˘ C 1 (DT ) ≤ C˘ e δ q C[0,T /2] , C˘ e > 0. Taking into account this estimate and formula (10.3.96) for the input fadj (t) we obtain:  q 2C[0,T /2] .

δψ 2C 1 (D ) ≤ 2C˘ e2 vz2 (0, ·) 2L2 (0,T ) + f 2L2 (0,T ) δ T

With estimate (10.3.66), this yields:

δψ 2C 1 (D

T)

  2 2 #12 ν 2 1 ≤ 2C˘ e2 C +

f

δ q 2C[0,T /2] , (10.3.104) H (0,T ) L (0,T )

#2 > 0 is the constant introduced in (10.3.67). where C 1 To estimate the norm δv C(DT ) in (10.3.103) we use formula (10.3.30) and estimate (10.3.72) to deduce that q 2C[0,T /2] ν 2L2 (0,T ) .

δv 2C(DT ) ≤ Ce2 δ

(10.3.105)

10.3 Recovering a Potential From Dirichlet-to-Neumann Operator

361

Finally we estimate the norms v2 L2 (0,T ;L2 (#(t )) and ψ1 L2 (0,T ;L2 (#(t )) in (10.3.103). To this end we use estimate (10.3.58). We have:

v2 2L2 (0,T ;L2 (#(t )) ≤ T (2C42 + C12 T 2 ) ν 2L2 (0,T ) ,

(10.3.106)

with the constants C1 , C4 > 0 defined in Theorem 10.3.6. By the same argument we estimate the second norm in (10.3.103). We have:

ψ1 2L2 (0,T ;L2 (#(t )) ≤ T (2C˘ 42 + C˘ 12 T 2 ) ϕ 2L2 (0,T ) , where C˘ 4 = exp(T a

˘ C[0,T /2] /2). The latter formula leads to the estimate 

ψ1 2L2 (0,T ;L2 (#(t )) ≤ 2T (2C˘ 42 + C˘ 12 T 2 ) vz2 (0, ·) 2L2 (0,T ) + f 2L2 (0,T )  #12 ν 2 1 + f 2L2 (0,T ) . ≤ 2T (2C˘ 42 + C˘ 12 T 2 ) C H (0,T ) This estimate with estimates (10.3.104) to (10.3.106) leads to the required result (10.3.101) with the Lipschitz constant introduced in (10.3.102).   In view of Lemma 3.4.3 the Lipschitz continuity of the Fréchet gradient implies that the sequence of iterations {J( q (n))} is a monotone decreasing sequence. This is an important advantage in implementation of gradient-based algorithms, as stated in Sect. 3.4.3.

Chapter 11

Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Beams and plates are important parts of main engineering constructions such as aircraft wings, flexible robotic manipulators, large space structures and robots, marine risers and moving strips [32]. Within the scope of the linear theory of elasticity, classical Euler-Bernoulli beam and Kirchhoff plate theories have been a cornerstone tools in engineering since the development of the Eiffel Tower and the Ferris wheel in the late nineteenth century. Recently, small size nanobeams and nanoplates are found also in nanotechnology, including emerging systems for medical diagnostics and nanoscale measurements such as the Atomic Force Microscopy (AFM) and Transverse Dynamic Force Microscope (TDFM). The direct and inverse problems for the simplest Euler-Bernoulli ρ(x)ut t + (EI (x)uxx )xx = F (x, t) and Kirchhoff ut t + D2 u = g(t)f (x), (x, t) ∈ T := × (0, T ), ⊂ R2 equations have been extensively studied in the last 50 years due to a large number of engineering and technological applications (see [2, 3, 41, 46, 59, 60, 62, 64, 78, 79, 114] and references therein). The purpose of this long chapter is the mathematical formulation, analysis and numerical solution of the basic inverse problems for the general form damped EulerBernoulli beam equation ρ(x)ut t + μ(x)ut + (EI (x)uxx )xx − (Tr (x)ux )x = F (x)G(t), (x, t) ∈ T , where T := (0, ) × (0, T ) is a finite domain, and the classical Kirchhoff plate equation     ut t + μ(x)ut + D(x)(ux1 x1 + νux2 x2 ) x x + D(x)(ux2 x2 + νux1 x1 ) x x 1 1 2 2   +2(1 − ν) D(x)ux1 x2 x x = F (x)G(t), (x, t) ∈ T := × (0, T ), 1 2

where := (0, 1 ) × (0, 2 ) ⊂ R2 . In these equations, the damping term μ(x)ut appears proportional to the velocity field, which most definitely appropriate, as beams and metallic plates are usually slightly damped structures. © Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9_11

363

364

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Notice that the decomposition Q(x, t) := F (x)G(t) of the temporal and spatial components of the acting load Q(x, t) in the above equations is characteristic of many engineering structures. On the other hand, from the mathematical point of view, the problem of unique determination of an arbitrary function of two variables using two single-variable functions is infeasible. The following three basic inverse problems for the Euler-Bernoulli and Kirchhoff equations will be considered in this chapter: (ISP) (IBVP) (ICP)

inverse source problems (determination of an unknown spatial F (x) or temporal G(t) loads; inverse boundary value problems (identification of an unknown boundary data); inverse coefficient problems (determination of unknown coefficients r(x) or D(x)).

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation As with each inverse problem governed by a partial differential equation, in order to obtain the necessary results, the corresponding direct problem needs to be studied in detail. Although some results and a priori estimates for the simplest Euler-Bernoulli beam equation ρ(x)ut t + (EI (x)uxx )xx = F (x, t) have been discussed in few studies, there is no systematic research addressing the study of weak and regular weak solutions of initial boundary value problems for the above general form of dynamic Euler-Bernoulli equation. Due to the limited scope of the chapter, here only two basic direct problems are analysed. We hope that the results of this section can motivate further study in this direction.

11.1.1 The Initial Boundary Value Problem for Simply Supported Beam Consider the following initial boundary value problem for a simply supported dynamic Euler-Bernoulli beam under a spatial-temporal load (Fig. 11.1): ⎧ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = F (x)G(t), ⎪ ⎪ ⎪ ⎪ ⎨ (x, t) ∈ T := (0, ) × (0, T ), ⎪ u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ (0, ), ⎪ ⎪ ⎪ ⎩ u(0, t) = uxx (0, t) = 0, u(, t) = uxx (, t) = 0, t ∈ [0, T ].

(11.1.1)

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation

u

365

F (x) G(t)

x u(0, t) = 0 −r(0)uxx (0, t) = 0

u(, t) = 0 r()uxx (, t) = 0

Fig. 11.1 Geometry of the problem: Simply supported Euler-Bernoulli beam

The function u(x, t) represents the transverse deflection in position x ∈ (0, ) and at time t ∈ (0, T ), where T > 0 is the final time and  > 0 is the length of a beam. Further, r(x) = E(x)I (x) is the flexural rigidity of the beam, while E(x) > 0, I (x) > 0, ρ(x) > 0, μ(x) ≥ 0 and Tr (x) ≥ 0 are the elasticity modulus, moment of inertia of the cross-section, mass density, damping coefficient (or an external damping force per unit length) and axial tensile force, correspondingly. F (x) and G(t) are the spatial and temporal components of the acting load. Here and below, ux (0, t), ux (, t), −r(x)uxx (x, t)|x=0 , r(x)uxx (x, t)|x= and (r(x)uxx (x, t))x + Tr (x)ux (x, t)|x=0 , −(r(x)uxx (x, t))x + Tr (x)ux (x, t)|x= are the slope of deflection (or angle of rotation), the bending moment and shear force at the boundaries x = 0 and x =  of a beam, correspondingly, according to the relevant definitions of these physical concepts [127]. The geometry of the problem is given in Fig. 11.1. Problem (11.1.1) is a mathematical model of the response of a simply supported Euler-Bernoulli beam to the spatial-temporal load that includes all essential physical parameters defined above. Bridges, guide ways, overhead cranes, cableways, rails, tunnels, pipelines and nano-beams are some examples where this model is applied [111]. In the bridge model governed by a simply supported Euler-Bernoulli beam, F (x)G(t) can be treated as a dynamic force transmitted by the vehicle on the bridge. We assume that the inputs in (11.1.1) satisfy the following basic conditions: ⎧ ρ, r, μ, Tr ∈ L∞ (0, ), ⎪ ⎪ ⎪ ⎪ ⎨ u ∈ H 2 (0, ), u ∈ L2 (0, ), F ∈ L2 (0, ), G ∈ L2 (0, T ), 0 1 ⎪ 0 < ρ0 ≤ ρ(x) ≤ ρ1 , 0 < r0 ≤ r(x) ≤ r1 , ⎪ ⎪ ⎪ ⎩ 0 ≤ μ0 ≤ μ(x) ≤ μ1 , 0 ≤ Tr0 ≤ Tr (x) ≤ Tr1 , x ∈ (0, ).

(11.1.2)

First, we notice that in the classical theory for PDEs for parabolic and hyperbolic equations, the source term is usually assumed to be in L2 (0, T ; L2 (0, )) (see §7.1 and §7.2 of [38]), that does not contain concentrated load. However, the existence theorem given in [14] allows us to employ the source term Q(x, t) := F (x)G(t) of Eq. (11.1.1) in the Hilbert space L2 (0, T ; H −2 (0, )) which includes the negative order Sobolev space H −2 (0, ). This is an important issue in terms of applications, since the equation ρ(x)ut t + η(x)ut + (r(x)uxx )xx = δ(x − ct) P (t) describes vibration of a beam under a moving and arbitrarily varying in time load P (t) with

366

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

constant speed c > 0 [41]. Here δ(y) is the Dirac delta which is not a regular distribution. To derive the theorem ensuring the existence of a weak solution for the initial boundary value problem (11.1.1) when F ∈ L2 (0, T ; H −2 (0, )), we employ the notations introduced in [38], assuming that [U (t)](x) := u(x, t) and [Q(t)](x) := F (x)G(t) are the abstract functions defined by the mappings U : [0, T ] → V 2 (0, ) and G : [0, T ] → L2 (0, ), correspondingly, and V 2 (0, ) := {v ∈ H 2 (0, ) : v(0) = v() = 0}. Further, we introduce the symmetric bilinear functional a : H 2 (0, ) × H 2 (0, ) → R: a(u, v) := (ruxx , vxx ) + (Tr ux , vx ), and rewrite problem (11.1.1) as the following variational problem: ⎧   ⎪ ⎨ (U (t), v) + (ηU (t), v) + a(U (t), v) = (Q(t), v) ∀v ∈ V 2 (0, ), 0 < t ≤ T , ⎪ ⎩ U (0) = u0 , U  (0) = u1 . Definition 11.1.1 A function U : [0, T ] → V 2 (0, ) with U  ∈ L2 (0, T ; L2 (0, )) and U  ∈ L2 (0, T ; H −2 (0, )) is a weak solution of problem (11.1.1) if (a) U  (t), v + (ηU  (t), v) + a(U (t), v) = G(t), v, ∀v ∈ V 2 (0, ), a.e. t ∈ [0, T ]; (b) U (0) = u0 and U  (0) = u1 , where the mapping ·, · : H −2 (0, ) × V 2 (0, ) → R is used for the duality pairing between H −2 (0, ) and V 2 (0, ) ⊂ H 2 (0, ). According to Theorem 2 in §5.9 of [38], U ∈ C([0, T ]; L2 (0, )) and U  ∈ C([0, T ]; H −2(0, )). This implies that the initial conditions U (0) = u0 and U  (0) = u1 are well-defined. Theorem 11.1.1 Let conditions (11.1.2), with F ∈ L2 (0, T ; L2 (0, )) replaced by F ∈ H −2 (0, ) and G ∈ L2 (0, T ), be satisfied. Then there exists a unique weak solution to problem (11.1.1). The proof of this theorem is similar to the proof of the theorem given in [14] for the case of a clamped beam, i.e. for the boundary conditions u(0, t) = ux (0, t) = 0, u(, t) = ux (, t) = 0, t ∈ [0, T ], and left as an exercise for the reader. In order to find out some important properties of the inverse problems examined in next sections, we will need some a priori estimates for the solution of problem (11.1.1). In these a priori estimates, we assume that the source function F (x) and G(t) are of the classes L2 (0, ) and L2 (0, T ), respectively. Furthermore, since the nonhomogeneous initial conditions in problem (11.1.1) do not play an essential role in the study of subsequent inverse problems, we will consider these data to be equal to zero.

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation

367

Theorem 11.1.2 Let conditions (11.1.2) hold. Then for the weak solution u ∈ L2 (0, T ; V 2 (0, )) with ut ∈ L2 (0, T ; L2 (0, )) and ut t ∈ L2 (0, T ; H −2 (0, )) of problem ⎧ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = F (x)G(t), ⎪ ⎪ ⎪ ⎨ (x, t) ∈ T , ⎪ ⎪ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ u(0, t) = uxx (0, t) = 0, u(, t) = uxx (, t) = 0, t ∈ [0, T ],

(11.1.3)

the following estimates hold:  1 2 C1 + 1 F 2L2 (0,) G 2L2 (0,T ) , ρ0

ut 2L2 (0,T ;L2 (0,)) ≤ C12 F 2L2 (0,) G 2L2 (0,T ) ,  1 2

uxx 2L∞ (0,T ;L2 (0,)) ≤ C1 + 1 F 2L2 (0,) G 2L2 (0,T ) , r0 r 0

uxx 2L2 (0,T ;L2 (0,)) ≤ C12 F 2L2 (0,) G 2L2 (0,T ) , ρ0

ut 2L∞ (0,T ;L2 (0,)) ≤

(11.1.4)

where V 2 (0, ) := {v ∈ H 2 (0, ) : v(0) = v() = 0}, C12 = exp(T /ρ0 ) − 1 and ρ0 , r0 > 0 are the constants introduced in (11.1.2). Proof Multiply both sides of Eq. (11.1.3) by 2ut (x, t), integrate it over t := (0, ) × (0, t), use the identity

 2(r(x)uxx )xx ut ≡ 2[(r(x)uxx )x ut − r(x)uxx uxt ]x + r(x)u2xx , t

(11.1.5)

and then apply the integration by parts formula. Taking into account the initial and boundary conditions in (11.1.3) we obtain the following energy identity: 

 0

ρ(x)u2t

=2

+ r(x)u2xx

 t 0



+

Tr (x)u2x



dx + 2

 t 0

 0

μ(x)u2τ dxdτ

F (x)G(τ )uτ (x, τ )dxdτ, a.e. t ∈ [0, T ].

(11.1.6)

0

In view of the bounds in (11.1.2) this implies the main integral inequality: 



ρ0 0

 u2t dx + r0 ≤

 0

 u2xx dx +

 t 0

 0

 0

Tr (x)u2x dx + 2

 t 0

 0

μ(x)u2τ dxdτ

u2τ (x, τ )dxdτ + F 2L2 (0,) G 2L2 (0,T ) ,

(11.1.7)

368

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

t ∈ [0, T ]. As the first consequence of this inequality we deduce that 



ρ0 0

u2t dx

≤

 t 0

 0

u2τ (x, τ )dxdτ + F 2L2 (0,) G 2L2 (0,T ) ,

a.e. t ∈ [0, T ]. Applying the Gronwall-Bellman inequality we obtain: 

 0

u2t dx ≤ F 2L2 (0,) G 2L2 (0,T ) exp(t/ρ0 ), t ∈ [0, T ].

(11.1.8)

The required first and second estimates in (11.1.4) are obtained from (11.1.8). Integrate now (11.1.8) to get  t 0

 0

u2τ dxdτ ≤ F 2L2 (0,) G 2L2 (0,T ) [exp(t/ρ0 ) − 1].

Using this in the second consequence  r0 0



u2xx dx ≤

 t 0

 0

u2τ (x, τ )dxdτ + F 2L2 (0,) G 2L2 (0,T )

of the integral inequality (11.1.7), we arrive at the third and fourth estimates in (11.1.4).   Corollary 11.1.1 Let conditions of Theorem 11.1.2 hold. Then for the weak solution of problem (11.1.3) the following estimate holds:

ux 2L∞ (0,T ;L2 (0,)) ≤

ux 2L2 (0,T ;L2 (0,))

 2 2 C1 + 1 F 2L2 (0,) G 2L2 (0,T ) , 2r0

ρ0 2 2 ≤ C F 2L2 (0,) G 2L2 (0,T ) , 2r0 1

(11.1.9)

where C1 > 0 is the constant introduced in Theorem 11.1.2. Proof Let t ∈ (0, T ) be given. By Rolle’s Theorem, there exists x0 ∈ (0, ) such that ux (x0 , t) = 0. If x ∈ (0, x0 ), then  u2x (x, t) ≡

2

x0

uxx dx



x0

≤ (x − x0 )

x

x

u2ξ ξ dξ,

and 

x0 0

u2x (x, t)dx ≤

1 2 x 2 0



x0 0

u2xx (x, t)dx.

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation

369

If x ∈ (x0 , ), in the same way we deduce that 

 x0

u2x (x, t)dx

1 ≤ ( − x0 )2 2



 x0

u2xx (x, t)dx.

Hence 

 0

u2x (x, t)dx ≤

 1 max x02 , ( − x0 )2 2



 0

u2xx (x, t)dx,

and 1 2  uxx 2L∞ (0,T ;L2 (0,)), 2 1 ≤ 2 uxx 2L2 (0,T ;L2 (0,)). 2

ux 2L∞ (0,T ;L2 (0,)) ≤

ux 2L2 (0,T ;L2 (0,))

(11.1.10)

 

In view of estimates in (11.1.4), this leads to (11.1.9).

Estimates (11.1.4) and (11.1.9) allow us to derive the following trace estimates for the L2 -norms of the Dirichlet outputs ux (0, t) and ux (, t). Corollary 11.1.2 Let conditions of Theorem 11.1.2 hold. Then for the weak solution of problem (11.1.3) the following trace estimates holds:

ux (0, ·) 2L2 (0,T ) ≤ C22 F 2L2 (0,) G 2L2 (0,T ) ,

(11.1.11)

ux (, ·) 2L2 (0,T ) ≤ C22 F 2L2 (0,) G 2L2 (0,T ) ,

where C22 = 3 ρ0 C12 /r0 and C1 > 0 is the constant introduced in Theorem 11.1.2. Proof Indeed, from the identities  1  ux (0, t) = − (( − x)ux (x, t))x dx,  0   1 ux (, t) = (xux (x, t))x dx  0 it follows that  

 0  0

u2x (0, t)dx ≤ u2x (, t)dx

2 

2 ≤ 

 

T 0 T

0

 

 0 

0

 u2x (x, t)dxdt + 2  u2x (x, t)dxdt

T

0 T

+ 2 0

 



0  0

u2xx (x, t)dxdt, u2xx (x, t)dxdt,

370

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

or, taking into account inequality (11.1.10),  

 0  0

 u2x (0, t)dx

≤ 3 

u2x (, t)dx

T



T



0

≤ 3 0

 0 

0

u2xx (x, t)dxdt (11.1.12) u2xx (x, t)dxdt.

With the second estimate in (11.1.4) these inequalities lead to the required result.   Theorem 11.1.3 Assume that in addition to basic conditions (11.1.2), the following regularity conditions are satisfied: r ∈ H 2 (0, ), Tr ∈ H 1 (0, ), G ∈ H 1 (0, T ).

(11.1.13)

Then for the regular weak solution u ∈ L2 (0, T ; H 4 (0, )), with ut ∈ L2 (0, T ; V 2 (0, )), ut t ∈ L2 (0, T ; L2 (0, )) and ut t t ∈ L2 (0, T ; H −2 (0, )) of problem (11.1.3) the following estimates hold:

ut t 2L2 (0,T ;L2 (0,)) ≤ C12 C32 F 2L2 (0,) G 2H 1 (0,T ) , ρ0 2 2

uxxt 2L2 (0,T ;L2 (0,)) ≤ C C F 2L2 (0,) G 2H 1 (0,T ) , r0 1 3

(11.1.14)

where C32 = 1 + CT2 /ρ0 , CT2 = 2 max(T /3, 1/T ) and C1 > 0 is the constant introduced in Theorem 11.1.2. Proof Differentiating Eq. (11.1.3) with respect to t ∈ (0, T ), multiplying both sides by 2ut t (x, t), integrating it over t := (0, ) × (0, t) and using then the identity (11.1.5) with u(x, t) replaced by ut (x, t), in the same way as in proof of Theorem 11.1.2, we obtain the following integral identity: 

 0

 t ρ(x)u2t t + r(x)u2xxt + Tr (x)u2xt dx + 2 =2

 t 0

a.e. t ∈ [0, T ].

0



F (x)G (τ )uτ τ dxdτ +

0



 0

0



μ(x)u2τ τ dxdτ

ρ(x)u2t t (x, 0+ )dx,

(11.1.15)

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation

371

ρ(x)u2t t (x, 0+ ) + μ(x)ut (x, 0+ ) +  We use the (formal) the limit equation + + r(x)uxx (x, 0 ) xx − (Tr (x)ux (x, 0 ))x = F (x)G(0+ ) at t = 0+ to evaluate the last right-hand-side integral. This yields: 

 0



1  G(0+ )F (x) − μ(x)ut (x, 0+ ) 0 ρ(x)   2 − r(x)uxx (x, 0+ ) xx + (Tr (x)ux (x, 0+ ))x dx.

ρ(x)u2t t (x, 0+ )dx =



In view of the homogeneous initial conditions in (11.1.3), this yields: 

 0

 2 ρ(x)u2t t (x, 0+ )dx = G(0+ )



 0

1 F 2 (x)dx. ρ(x)

(11.1.16)

Hence, 

 0

ρ(x)u2t t (x, 0+ )dx

2 1  G(0+ ) ≤ ρ0





F 2 (x)dx,

0

Here we employ the identity 1 G(0 ) = − T +



T 0

((T − t)G(t))t dt

to conclude that 2  G(0+ ) ≤ CT2 G 2H 1 (0,T ) , CT2 = 2 max(T /3, 1/T ).

(11.1.17)

Then 

 0

ρ(x)u2t t (x, 0+ )dx ≤

CT2

F L2 (0,) G 2H 1 (0,T ) . ρ0

In view of the identity (11.1.15) this leads to the following integral inequality:  ρ0 0



 u2t t dx + r0 ≤

 t 0

 0

 0

 u2xxt dx +

 0

Tr (x)u2xt dx + 2

 t 0

 0

μ(x)u2τ τ dxdτ

u2τ τ (x, τ )dxdτ + C32 F L2 (0,) G 2H 1 (0,T ) , t ∈ [0, T ],

with the constant C3 > 0 defined in the theorem. The required estimates (11.1.14) can be derived from this inequality in the same manner as in the proof of Theorem 11.1.2.  

372

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Corollary 11.1.3 Let conditions of Theorem 11.1.3 hold. Then for the regular weak solution of problem (11.1.3) the following trace estimates hold:

uxt (0, ·) 2L2 (0,T ) ≤ C22 C32 F 2L2 (0,) G 2H 1 (0,T ) ,

uxt (, ·) 2H 1 (0,T ) ≤ C22 C32 F 2L2 (0,) G 2H 1 (0,T ) ,

(11.1.18)

with the constants C2 , C3 > 0 introduced in Corollary 11.1.2 and Theorem 11.1.3. Theorem 11.1.4 Assume that conditions of Theorem 11.1.3 hold. Suppose, in addition, ρ ∈ H 2 (0, ), F ∈ H 2 (0, ), G ∈ H 2 (0, T ).

(11.1.19)

Then for the regular weak solution with improved regularity u ∈ L2 (0, T ; H 6 (0, )), with ut ∈ L2 (0, T ; H 4 (0, )), ut t ∈ L2 (0, T ; V 2(0, )), ut t t ∈ L2 (0, T ; L2 (0, )) and ut t t t ∈ L2 (0, T ; H −2(0, )), of problem (11.1.3) the following estimates hold:

ut t t 2L2 (0,T ;L2 (0,)) ≤ C12 Cr2 F 2H 2 (0,) G 2H 2 (0,T ) ,

uxxt t 2L2 (0,T ;L2 (0,)) ≤

ρ0 2 2 C C F 2H 2 (0,) G 2H 2 (0,T ) , r0 1 r

(11.1.20)

where 4 Cr2 = 1 + CT2 max(Cr21 , Cr22 , Cr23 , ), Cr21 = (1 + μ21 /ρ02 ), ρ0

 1 Cr22 = 4 Tr1 ρ12 + max |ρ  (x)|2 , (11.1.21) ρ0  

r1 Cr23 = 3 ρ14 + max 4ρ12 |ρ  (x)|2 + 4|ρ  (x)|4 + 2ρ12 |ρ  (x)|2 , ρ0 and C1 , CT > 0 are the constants introduced in Theorems 11.1.2 and 11.1.3. Proof Differentiating Eq. (11.1.3) two times with respect to t ∈ (0, T ), multiplying both sides by 2ut t t (x, t), integrating it over t := (0, ) × (0, t) and using then the identity (11.1.5) with u(x, t) replaced by ut t (x, t), we obtain the following integral identity: 

 0

 t ρ(x)u2t t t + r(x)u2xxt t + Tr (x)u2xt t dx + 2

=2

 t 0





+ 0



F (x)G (τ )uτ τ τ dxdτ +

0

Tr (x)u2xt t (x, 0+ )dx +

a.e. t ∈ [0, T ].

0



 0



 0

 0

μ(x)u2τ τ τ dxdτ

ρ(x)u2t t t (x, 0+ )dx

r(x)u2xxt t (x, 0+ )dx, t ∈ [0, T ]. (11.1.22)

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation

373

As in Theorem 11.1.3, we use the limit equation at t = 0+ to evaluate the last three right-hand-side integrals in (11.1.22), denoting each of these integrals by I1 , I2 and I3 , respectively. We have: 



I1 = 0

1   + G (0 )F (x) − μ(x)ut t (x, 0+ ) ρ(x)     − r(x)uxx (x, 0+ ) xxt + Tr (x)ux (x, 0+ ) xt

2

dx.

Taking into account (11.1.16) and the homogeneous initial conditions u(x, 0) = ut (x, 0) = 0 we get: 



I1 = 0

1   + 2 G (0 )F (x) − G(0+ )μ(x)F (x)/ρ(x) dx. ρ(x)

 2 With inequality (11.1.17) and its analogue G (0+ ) ≤ CT2 G 2H 2 (0,T ) , CT2 = 2 max(T /3, 1/T ), this leads to I1 ≤ Cr21 F 2L2 (0,) G 2H 2 (0,T ) ,

(11.1.23)

with the constant Cr1 > 0 defined in (11.1.21). For I2 we have:  2 I2 = G(0+ )





 Tr (x) (F (x)/ρ(x))

2

dx.

0

Consequently, I2 ≤ Cr22 F 2H 1 (0,) G 2H 1 (0,T ) .

(11.1.24)

Finally, to evaluate the last integral I3 in (11.1.22), firstly we rewrite it as follows: 



I3 =

r(x) 0



ut t (x, 0+ )

 xx

2

dx

Then we use (11.1.16) to conclude  2 I3 = G(0+ )





 r(x) (F (x)/ρ(x))

2

dx

0

≤ Cr23 F 2H 2 (0,) G 2H 1 (0,T ) , where Cr2 , Cr3 > 0 are the constants defined in (11.1.21).

(11.1.25)

374

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Inserting (11.1.23) to (11.1.25) into the integral identity (11.1.22) we deduce that 



ρ0 0

 u2t t t dx ≤



+ r0 0

 t 0

 0

 u2xxt t dx



+ 0

Tr (x)u2xt t dx

+2

 t 0

 0

μ(x)u2τ τ τ dxdτ

u2τ τ τ dxdτ + Cr2 F 2H 2 (0,) G 2H 2 (0,T ) , t ∈ [0, T ], (11.1.26)

with Cr > 0 defined in (11.1.21). Estimates (11.1.20) are derived from the integral inequality (11.1.26) in the same way as in the proof of Theorem 11.1.2.  

11.1.2 The Initial Boundary Value Problem for a Cantilever Beam Under Free-End Shear Force Transverse vibration of a nonuniform beam under free-end shear force is governed by the following initial boundary value problem: ⎧ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = 0, in T , ⎪ ⎪ ⎪ ⎪ ⎨ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ u(0, t) = ux (0, t) = 0, r()uxx (, t) = 0, ⎪ ⎪ ⎪ ⎩ −(r()uxx (, t))x + Tr ()ux (, t) = g(t), t ∈ [0, T ].

(11.1.27)

If the left end x = 0 of a beam extending over the interval (0, ) is clamped and a load distribution along the beam is zero, then there can only be two types of force effects acting on the free-end x =  of a beam: the bending moment r()uxx (, t) and the shear force −(r()uxx (, t))x + Tr ()ux (, t). Here, we consider the case where the vibration occurs due to the transverse shear force g(t) := −(r()uxx (, t))x + Tr ()ux (, t), applied at x =  (Fig. 11.2). Theorem 11.1.5 Let conditions (11.1.2) hold. Assume that the function g(t) in (11.1.27) satisfies the following conditions: g ∈ H 1 (0, T ), g(0) = 0.

u

r()uxx (, t) = 0 −(r()uxx (, t))x + Tr ()ux (, t) = g(t) x= u(0, t) = 0 ux (0, t) = 0

Fig. 11.2 Geometry of the problem for the cantilever beam under free-end shear force

x

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation

375

Then for the weak solution u ∈ L2 (0, T ; V12 (0, )) with ut ∈ L2 (0, T ; L2 (0, )) and ut t ∈ L2 (0, T ; H −2 (0, )) of problem (11.1.27) the following estimates hold:

uxx 2L2 (0,T ;L2 (0,)) ≤ C42 g  2L2 (0,T ) ,

(11.1.28)

ut 2L2 (0,T ;L2 (0,)) ≤ C52 g  2L2 (0,T ) , where V12 (0, ) = {v ∈ H 2 (0, ) : v(0) = v  (0) = 0}, 43 (1 + T ) r0 2 [exp(T ) − 1], C52 = C 2 2ρ0 4 3r0

C42 =

and ρ0 , r0 > 0 are the constants introduced in (11.1.2). Proof Multiply both sides of Eq. (11.1.27) by 2ut (x, t), integrate it over t := (0, ) × (0, t), use the identity (11.1.5) and then apply the integration by parts formula. As in the proof of Theorem 11.1.2 we obtain the following energy identity: 

 0

 t ρ(x)u2t + r(x)u2xx + Tr (x)u2x dx + 2 0

 = 2g(t)u(, t) − 2

t

 0

μ(x)u2τ dxdτ

g  (τ )u(, τ )dτ,

(11.1.29)

0

for a.e. t ∈ [0, T ]. We employ the ε-inequality 2ab ≤ (1/ε) a 2 + ε b2 in the righthand-side integrals of (11.1.29), and then use the trace inequality u2 (, t) ≤



3 3

 0

u2xx (x, t)dx, a.e. t ∈ [0, T ],

(11.1.30)

which is a consequence of the identity  x

 u(, t) ≡ 0

uξ ξ (ξ, t)dξ dx, for all t ∈ [0, T ].

0

Next, we choose the arbitrary constant ε > 0 from the condition r0 − 3 ε/3 > 0 as follows: ε = 3r0 /(23 ). After elementary transformations we obtain the main integral inequality: 

 0

ρ0 u2t +

r0 2 dx + u 2 xx

≤ for a.e. t ∈ [0, T ].

r0 2

 t 0

 0

 0



Tr (x)u2x dx + 2

u2xx dxdτ +

 t 0

0



μ(x)u2τ dxdτ

2(1 + T )3  2

g L2 (0,T ) , 3r0

(11.1.31)

376

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

The first consequence of the integral inequality (11.1.31) is 

 0

u2xx dx

≤

 t 0



0

u2xx dxdτ +

4(1 + T )3  2

g L2 (0,T ) , t ∈ [0, T ]. 3r02

In view of the Gronwall-Bellmann inequality this implies: 

 0

u2xx dx ≤

4(1 + T )3  2

g L2 (0,T ) exp(t), t ∈ [0, T ]. 3r02

(11.1.32)

Integrating this inequality over [0, T ] we arrive at the first required estimate (11.1.28). The second consequence of (11.1.31) is the inequality 

 0

u2t dx ≤

 t

r0 2ρ0

0

 0

u2xx dxdτ +

2(1 + T )3  2

g L2 (0,T ) , 3r0 ρ0

(11.1.33)

for a.e. t ∈ [0, T ]. With estimate (11.1.32) this yields the second required estimate of (11.1.28).   Theorem 11.1.6 Assume that in addition to conditions of Theorem 11.1.5, the input g(t) satisfies the following regularity and consistency conditions: r ∈ H 2 (0, ), Tr ∈ H 1 (0, ),

(11.1.34)

g ∈ H 2 (0, T ), g  (0) = 0.

Then for the regular weak solution of problem (11.1.27) u ∈ L2 (0, T ; H 4(0, )) with ut ∈ L2 (0, T ; V12 (0, )), ut t ∈ L2 (0, T ; L2 (0, )), ut t t ∈ L2 (0, T ; H −2 (0, )) the following estimates hold:

uxxt 2L2 (0,T ;L2 (0,)) ≤ C42 g  2L2 (0,T ) ,

(11.1.35)

ut t 2L2 (0,T ;L2 (0,)) ≤ C52 g  2L2 (0,T ) , where C4 , C5 > 0 are the same constant defined in Theorem 11.1.5.

Proof Differentiating Eq. (11.1.27) with respect to the time variable t ∈ (0, T ), multiplying both sides by 2ut t (x, t), integrating then it over t and using identity (11.1.5) with u(x, t) replaced by ut (x, t), after elementary transformations we arrive at the integral identity: 

 0

ρ(x)u2t t

+ r(x)u2xxt 

+ Tr (x)u2xt 

= 2g (t)uτ (, t) − 2 0

t



dx + 2

 t 0

 0

μ(x)u2τ τ dxdτ

g  (τ )uτ (, τ )dτ, a.e. t ∈ [0, T ].

11.1 Initial Boundary Value Problems for Dynamic Euler-Bernoulli Equation

377

Here the equality 

 0

ρ(x)u2t t (x, 0+ )dx = 0,

(11.1.36)

which is a consequence of the homogeneous initial conditions and the limit equation at t = 0+ , has been taken into account. This is exactly the same integral identity as (11.1.29) with u(x, t) replaced by ut (x, t) and g(t) replaced by g  (t). Hence estimates (11.1.35) can be derived in the same way as in the proof of Theorem 11.1.5.   Theorem 11.1.7 Let conditions (11.1.2) hold. Assume, in addition, that the following regularity and consistency conditions hold: 

r ∈ H 4 (0, ), Tr ∈ H 3 (0, ), ρ, μ ∈ H 2 (0, ), g ∈ H 3 (0, T ), g  (0) = g  (0) = 0.

(11.1.37)

Then for the regular weak solution with improved regularity of problem (11.1.27) defined as u ∈ L2 (0, T ; H 6 (0, )), with ut ∈ L2 (0, T ; H 4(0, )), ut t ∈ L2 (0, T ; V12 (0, )), ut t t ∈ L2 (0, T ; L2 (0, )) and ut t t t ∈ L2 (0, T ; H −2(0, )), the following estimates hold:

uxxt t 2L2 (0,T ;L2 (0,)) ≤ C42 g  2L2 (0,T ) ,

ut t t 2L2 (0,T ;L2 (0,)) ≤ C52 g  2L2 (0,T ) .

(11.1.38)

Furthermore, if in addition, the input g(t) satisfies the regularity and consistency conditions g ∈ H 4 (0, T ), g  (0) = 0,

(11.1.39)

then for the regular weak solution with higher regularity of problem (11.1.38) defined as u ∈ L2 (0, T ; H 8 (0, )), with ut ∈ L2 (0, T ; H 6(0, )), ut t ∈ L2 (0, T ; H 4 (0, )), ut t t ∈ L2 (0, T ; V12 (0, )), ut t t t ∈ L2 (0, T ; L2 (0, )), ut t t t t ∈ L2 (0, T ; H −2(0, )), the following estimates hold

uxxt t t 2L2 (0,T ;L2 (0,)) ≤ C42 g (I V ) 2L2 (0,T ) ,

ut t t t 2L2 (0,T ;L2 (0,)) ≤ C52 g (I V ) 2L2 (0,T ) , where C4 , C5 > 0 are the same constant introduced in Theorem 11.1.5.

(11.1.40)

378

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Proof Differentiating Eq. (11.1.38) two times with respect to t ∈ (0, T ), multiplying both sides by ut t t (x, t), integrating then it over t and using identity (11.1.5) with u(x, t) replaced by ut t (x, t) we obtain the following integral identity: 

 0

ρ(x)u2t t t

+ r(x)u2xxt t  =



0

 +



0

+ Tr (x)u2xt t



dx + 2

ρ(x)u2t t t (x, 0+ )dx +

Tr (x)u2xt t (x, 0+ )dx + 2

 t 0



0



0



t



μ(x)u2τ τ τ dxdτ

r(x)u2xxt t (x, 0+ )dx g  (τ )uτ τ τ (, τ )dτ,

0

a.e. t ∈ [0, T ]. The third right-hand-side integral is zero due to (11.1.36). To evaluate the first two right-hand-side integrals in above integral identity we use the limit equation (11.1.27) at t = 0+ . We have: 



0

ρ(x)u2t t t (x, 0+ )dx





= 0



 0

1  2 (Tr (x)ux (x, 0+ ))xt − (r(x)uxx (x, 0+ ))xxt − μ(x)ut t (x, 0+ ) dx, ρ(x)

r(x)u2xxt t (x, 0+ )dx







=

r(x) 0

−

1 (Tr (x)ux (x, 0+ ))x ρ(x)

1 μ(x) (r(x)uxx (x, 0+ ))xx − ut (x, 0+ ) ρ(x) ρ(x)

2 dx. xx

In both identities above, the right-hand-side integrals are well-defined due to conditions (11.1.37), and are zero by the homogeneous initial conditions and condition (11.1.36). Then the integral identity corresponding to the regular weak solution with improved regularity is: 

 0

 t ρ(x)u2t t t + r(x)u2xxt t + Tr (x)u2xt t dx + 2 0

= 2g  (t)ut t (, τ ) + 2



t

0



μ(x)u2τ τ τ dxdτ

g  (τ )uτ τ (, τ )dτ, a.e. t ∈ [0, T ].

0

This is exactly the same integral identity as (11.1.29) with u(x, t) and g(t) replaced by ut t (x, t) and g  (t), respectively. Hence estimates (11.1.38) can be derived in the same way as in the proof of Theorem 11.1.5. Next estimates in (11.1.40) are also proved in the same way.  

11.2 Identification of a Temporal Load in a Simply Supported Beam From. . .

379

11.2 Identification of a Temporal Load in a Simply Supported Beam From Boundary Measured Slope In this section we study the inverse source problem of identifying the unknown temporal load G(t) in ⎧ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = F (x) G(t), ⎪ ⎪ ⎪ ⎨ (x, t) ∈ T , ⎪ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎩ u(0, t) = uxx (0, t) = 0, u(, t) = uxx (, t) = 0, t ∈ [0, T ],

(11.2.1)

from the measured boundary slope θ (t) := ux (0, t), t ∈ (0, T )

(11.2.2)

at the left end x = 0 of the beam. F (x) in (11.2.1) is considered a given function. Alternatively, the measured boundary slope θ (t) := ux (, t), t ∈ (0, T ), can be set at the right end x =  of the beam. Geometry of the inverse problem (11.2.1)–(11.2.2) is shown in Fig. 11.3.

11.2.1 The Input-Output Operator: Compactness and Lipschitz Continuity Let us define the set of admissible inputs (temporal loads) G = {G ∈ L2 (0, T ) : G L2 (0,T ) ≤ γG }.

(11.2.3)

u G(t) =? θ(t) u(0, t) = 0 −r(0)uxx (0, t) = 0

x u(, t) = 0 r()uxx (, t) = 0

Fig. 11.3 Simply supported Euler-Bernoulli beam bridge subjected to interface forces with measured boundary slope

380

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Evidently, G is a nonempty closed convex set in L2 (0, T ). We will need also the following set of admissible regular inputs Gr = {G ∈ H 1 (0, T ) : G H 1 (0,T ) ≤ γG },

(11.2.4)

where γG > 0 is a constant. Denote by u = u(x, t; G) the weak solution of the direct problem (11.2.1) for a given input G ∈ G. Introduce the input-output map  defined on the set of admissible inputs G and corresponding to the inverse problem (11.2.1)–(11.2.2), as follows: (G)(t) := ux (0, t; G), t ∈ (0, T ),  : G ⊂ L2 (0, T ) → L2 (0, T ).

(11.2.5)

Having the input-output map, we can reformulate the inverse problem as the linear operator equation (G)(t) = θ (t), G ∈ G, t ∈ (0, T ).

(11.2.6)

Lemma 11.2.1 Assume that conditions of Theorem 11.1.3 hold. Suppose that G ∈ Gr , where Gr is the set of admissible regular inputs introduced in (11.2.4). Then the input-output map [·] : Gr → L2 (0, T ) is a linear compact operator. Proof Let {G(m) } ⊂ Gr , m = 1, ∞ be a sequence of inputs, uniformly bounded, by the definition (11.2.4). Denote by {u(m) (x, t)}, u(m) (x, t) := u(x, t; G(m) ), the sequence of corresponding regular weak solutions of the direct problem (11.2.1). Then {(G(m) )(t)}, (G(m) )(t) = u(m) x (0, t), m = 1, ∞ is the sequence of outputs. We need to prove that this sequence is a relatively (m) compact subset of L2 (0, T ) or, equivalently, the sequence {uxt (0, t)} is bounded in the norm of the space L2 (0, T ). The last assertion follows from the first trace estimate in (11.1.18) applied to the above sequence. Namely, 2 2 2 2 (m) 2

u(m)

H 1 (0,T ) xt (0, ·) L2 (0,T ) ≤ C2 C3 F L2 (0,) G

and the norms G(m) H 1 (0,T ) , m = 1, ∞ are uniformly bounded by definition (m)

(11.2.4). Thus, the sequence {uxt (0, t)} is uniformly bounded in the norm of L2 (0, T ) which implies that the input-output map [·] : Gr → L2 (0, T ) transforms a bounded in Gr set to any compact in L2 (0, T ) set, hence is a compact operator.   From Lemma 11.2.1 it follows that the inverse source problem (11.2.1) and (11.2.2) is ill-posed.

11.2 Identification of a Temporal Load in a Simply Supported Beam From. . .

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Lemma 11.2.2 Assume that conditions (11.1.2) hold. Suppose that [·] : G ⊂ L2 (0, T ) → L2 (0, T ) is the input-output operator introduced in (11.2.5). Then this operator is Lipschitz continuous, that is

G1 − G2 L2 (0,T ) ≤ L G1 − G2 L2 (0,T ) , ∀G1 , G2 ∈ G,

(11.2.7)

where L = C2 is the Lipschitz constant and C2 > 0 is the constant introduced in Corollary 11.1.2. Proof Let G1 , G2 ∈ G and δG(t) = G1 (t) − G2 (t). Then δu(x, t) = u(x, t; G + δG) − u(x, t; G) solves the following problem: ⎧ ⎪ ⎪ ρ(x)δut t + μ(x)δut + (r(x)δuxx )xx − (Tr (x)δux )x ⎪ ⎨ = F (x)δG(t), (x, t) ∈ T , ⎪ δu(x, 0) = 0, δut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎩ δu(0, t) = δuxx (0, t) = δu(, t) = δuxx (, t) = 0, t ∈ [0, T ].

(11.2.8)

From definition (11.2.5) it follows that:

G1 − G2 L2 (0,T ) = δux (0, ·) L2 (0,T ) .

(11.2.9)

In view of Corollary 11.1.2 applied to the output δux (0, t) of problem (11.2.8), we estimate the right-hand-side norm in (11.2.9) to deduce that

G1 − G2 L2 (0,T ) ≤ C2 δG L2 (0,T ) .

(11.2.10)  

This leads to (11.2.7).

11.2.2 Existence of a Quasi-Solution of the Inverse Problem As stated earlier, the function θ (t) in (11.2.2) which is the result of experimental measurements, always contains a random noise and, as a consequence, exact equality in Eq. (11.2.7) is not possible in practice. Hence one needs to introduce the Tikhonov functional J (G) =

1

G − θ 2L2 (0,T ) , G ∈ G, θ ∈ L2 (0, T ) 2

(11.2.11)

and reformulate the inverse problem (11.2.1)–(11.2.2) as the minimization problem J (G∗ ) = inf J (G) G∈G

(11.2.12)

for this functional. A solution to this problem is defined as a quasi-solution to the inverse problem.

382

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Lemma 11.2.3 Assume that the conditions of Theorem 11.1.2 hold and θ ∈ L2 (0, T ). Then the Tikhonov functional is Lipschitz continuous, that is |J (G1 ) − J (G2 )| ≤ LJ G1 − G2 L2 (0,T ) , ∀G1 , G2 ∈ G,

(11.2.13)

where   LJ = 2C2 γG F L2 (0,) + θ L2 (0,T ) C2 C2 > 0 is the constant introduced in Corollary 11.1.2. Proof As in proof of Corollary 10.1.6 of Sect. 10.1, we can show that |J (G1 ) − J (G2 )| ≤

1

G1 L2 (0,T ) + G2 L2 (0,T ) 2 +2 θ L2 (0,T ) G1 − G2 L2 (0,T ) .

With estimates (11.1.11) and inequality (11.2.10) this implies:  |J (G1 ) − J (G2 )| ≤ C2 C2 γG F L2 (0,) + θ L2 (0,T ) G1 − G2 L2 (0,T ) , since Gk L2 (0,T ) ≤ γG , k = 1, 2. This inequality implies the desired result (11.2.13) with the Lipschitz constant LJ > 0 introduced in the theorem.   The Lipschitz continuity of the Tikhonov functional allows us to prove an existence of a quasi-solution to the inverse problem (11.2.1)–(11.2.2). Theorem 11.2.1 Suppose conditions (11.1.2) hold. Then the minimization problem (11.2.12) has a solution in the set of admissible temporal loads G, defined in (11.2.3). This theorem is proved exactly in the same way as Theorem 10.1.11 in Sect. 10.1.

11.2.3 Fréchet Differentiability of the Tikhonov Functional. Gradient Formula Let G, G + δG ∈ G. Denote by δJ (G) := J (G + δG) − J (G) the increment of the Tikhonov functional (11.2.11). Then 

T

δJ (G) =

[ux (0, t) − θ (t)] δux (0, t)dt

0

1 + 2 where ux (0, t) := ux (0, t; G).



T 0

(δux (0, t))2 dt,

(11.2.14)

11.2 Identification of a Temporal Load in a Simply Supported Beam From. . .

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Lemma 11.2.4 Assume that conditions (11.1.2) hold. Suppose G, G + δG ∈ G are two arbitrary inputs. Then the following integral relationship holds: 

T

−



T

p(t)δux (0, t) dt =

0





δG(x)φ(x, t)dxdt, 0

(11.2.15)

0

where δu(x, t) solves the problem (11.2.8) and φ(x, t) is the solution of the following backward problem: ⎧ ρ(x)φt t − μ(x)φt + (r(x)φxx )xx − (Tr (x)φx )x = 0, (x, t) ∈ T , ⎪ ⎪ ⎪ ⎨ φ(x, T ) = 0, φ (x, T ) = 0, x ∈ (0, ), t (11.2.16) ⎪ φ(0, t) = 0, − r(0)φxx (0, t) = p(t), ⎪ ⎪ ⎩ φ(, t) = 0, r()φxx (, t) = 0, t ∈ [0, T ], with the arbitrary input p ∈ L2 (0, T ). Proof Multiply both sides of Eq. (11.2.8) by arbitrary function φ(x, t), integrate over (0, T ) and apply the integration by parts formula multiple times. Using the homogeneous initial and final conditions in (11.2.8) and (11.2.16) we obtain: 



T 0



0



T

+



0



T

+ 0

ρ(x)φt t − μ(x)φt + (r(x)φxx )xx − (Tr (x)φx )x δu dxdt

(r(x)δuxx )x φ − r(x)δuxx φx + r(x)δux φxx − δu (r(x)φxx )x  [Tr (x)φx δu − Tr (x)φδux ]x= x=0 dt =

T



x= x=0

dt



F (x)δG(t)φ(x, t)dxdt. 0

0

In view of the homogeneous equation (11.2.16) and the boundary conditions in (11.2.8) and (11.2.16) this identity implies the required integral relationship.   We choose now the arbitrary input p ∈ L2 (0, T ) in the backward (adjoint) problem (11.2.16) as follows: p(t) = ux (0, t; G) − θ (t), t ∈ (0, T ).

(11.2.17)

Then the integral relationship (11.2.15) yields: 

T 0

 [ux (0, t; G) − θ (t)] δux (0, t) dt =

T





F (x) δG(t)φ(x, t)dxdt. 0

0

384

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

With this relationship, the increment formula (11.2.14) for the Tikhonov functional can be rewritten in the following form: 

T

δJ (G) =







φ(x, t)F (x)dx δG(t)dt 0

0

+

1 2



T

(δux (0, t))2 dt.

(11.2.18)

0

Theorem 11.2.2 Assume conditions of Theorem 11.1.2 hold and G ∈ G. Then the Tikhonov functional J (G) defined on set of admissible inputs G ⊂ L2 (0, T ) is Fréchet differentiable. Moreover, for the Fréchet gradient of this functional the following gradient formula holds: 



∇J (G)(t) =

φ(x, t)F (x)dx, t ∈ (0, T ),

(11.2.19)

0

where φ(x, t) = φ(x, t; G) is the weak solution solutions of the adjoint problem (11.2.15) with the input defined in (11.2.17). Proof From the estimate (11.1.11) applied to the solution of problem (11.2.8) we deduce that integral in the increment formula (11.2.18) is of the 

the right-hand-side   order O δG 2L2 (0,T ) . This leads to the gradient formula (11.2.19). Remark 11.2.1 Let us examine now the validity of the formal substitution (11.2.17), i.e. under which conditions this substitution is correct. In practice the Dirichlet measured output θ (t) usually belongs to L2 (0, T ), and from this point of view the condition p ∈ L2 (0, T ) in Lemma 11.2.4 is reasonable. However, the weak solution φ ∈ L2 (0, T ; V(0, )) of the adjoint problem (11.2.16) exists if p ∈ H 1 (0, T ), as it follows from Theorem C.2.1 of Appendix C, and hence, for the substitution (11.2.17) to be valid, the following regularity conditions must be satisfied: ux (0, ·) ∈ H 1 (0, T ), θ ∈ H 1 (0, T ).

(11.2.20)

Although this situation seems contradictory, there is actually no contradiction here, as the increment formula (11.2.18) provides further insight into this situation. In fact, this formula suggests that we don’t even need such a “smooth” weak solution as φ ∈ L2 (0, T ; V 2 (0, )) to the adjoint problem (11.2.16). From the increment formula (11.2.18) it follows that we need the solution to satisfy only the condition φ ∈ L2 (0, T ; L2 (0, )). For the adjoint problem (11.2.16) with the input −r(0)φxx (0, t) = p(t) replaced with φx (0, t) = θ (t), such a weaker solution was introduced in [60]. The same approach applies to the adjoint problem (11.2.16). The gradient formula (11.2.19) allows us use of the classical version of the Conjugate Gradient Algorithm for the numerical reconstruction of the temporal load G(t) (see Sect. 3.4.2).

11.2 Identification of a Temporal Load in a Simply Supported Beam From. . .

385

11.2.4 Lipschitz Continuity of the Fréchet Gradient As noted in the previous sections, the Lipschitz continuity of the Fréchet gradient of the Tikhonov functional is an essential property which provides the monotonicity of the iterations {J (G(n) )} in gradient-based algorithms. We prove under the regularity conditions (11.2.20) the Fréchet gradient is Lipschitz continuous. Theorem 11.2.3 Assume that conditions of Theorem 11.2.3 hold. Suppose, in addition, that the output ux (0, t) and the measured output θ (t) satisfy the regularity conditions (11.2.20). Then the Fréchet gradient of the Tikhonov functional is Lipschitz continuous in the set of admissible regular inputs Gr ⊂ H 1 (0, T ), that is,

∇J (G1 ) − ∇J (G2 ) L2 (0,T ) ≤ Lg G1 − G2 H 1 (0,T ) ,

(11.2.21)

with the Lipschitz constant 1 2 (1 + C3 )1/2 F L2 (0,) L∇ = √ 3/2 C2 C 2 depending on the constants introduced in Corollary 11.1.2, Theorems 11.1.3 and C.2.1. Proof Let G1 , G2 ∈ Gr and δG = G1 − G2 . Denote by φ(x, t; Gk ), k = 1, 2, the corresponding solutions of the adjoint problem (11.2.16) with the inputs pk (t) = ux (0, t; Gk ) − θ (t), t ∈ (0, T ). Then the function δφ(x, t) := φ(x, t; G1 ) − φ(x, t; G2 ) solves the following backward problem ⎧ ρ(x)δφt t − μ(x)δφt + (r(x)δφxx )xx − (Tr (x)δφx )x = 0, (x, t) ∈ T , ⎪ ⎪ ⎪ ⎨ δφ(x, T ) = 0, δφt (x, T ) = 0, x ∈ (0, ), (11.2.22) ⎪ ⎪ δφ(0, t) = 0, δφxx (0, t) = δux (0, t), ⎪ ⎩ δφ(, t) = 0, δφxx (, t) = 0, t ∈ [0, T ], where δux (0, t) = ux (0, t; G1 ) − ux (0, t; G2 ), while ux (0, t; G1 ) and ux (0, t; G2 ) are the outputs corresponding to the inputs G1 , G2 ∈ Gr . By the gradient formula (11.2.19) we have: 

∇J (G1 ) − ∇J (G2 ) 2L2 (0,T )

T

:=



2



δφ(x, t)F (x)dx 0

dt

0

≤ F 2L2 (0,) δφ 2L2 (0,T ;L2 (0,))

(11.2.23)

386

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

and we need to estimate the last right-hand-side norm. To this end, we employ first the inequality (11.2.10) applied to the weak solution of (11.2.22):

δφ 2L2 (0,T ;L2 (0,)) ≤

3

δφxx 2L2 (0,T ;L2 (0,)). 2

Then, we use estimate (C.2.4) in Theorem C.2.1 of Appendix C for the norm

δφxx 2L2 (0,T ;L2 (0,)). Note that (11.2.22) is the same well-posed problem (C.2.1) with the Neumann input, as the change of variable t by T − t shows. We have:

δφ 2L2 (0,T ;L2 (0,)) ≤

3 2 C δux (0, ·) 2H 1 (0,T ) , 2 1

1 > 0 is the constant introduced in Theorem C.2.1. In view of the trace where C estimates (11.1.11) and (11.1.18) applied to the regular weak solution δu(x, t) of problem (11.2.8) with the input δG(t), this implies:

δφ 2L2 (0,T ;L2 (0,)) ≤

 3 2 2 C1 C2 δF 2L2 (0,) δG 2L2 (0,T ) + C32 δG 2H 1 (0,T ) . 2

With (11.2.23) this leads to the required result.

 

11.2.5 Reconstruction of an Unknown Temporal Load from Boundary Measured Slope Wide range of basic engineering problems are related to harmonic force moving along a simply supported beam which is a commonly used model of single span bridges [41]. In the simplest case, the (harmonic) load is of the form P sin ωc t, where P is the amplitude and ωc is the circular frequency of the harmonic load. Then the vibration of the Euler-Bernoulli beam is under the moving load with constant speed c > 0 is governed by the equation ρut t + 2μut + EI uxxxx = P δ(x − ct) sin(ωc t). The two assumptions that P is the amplitude of the concentrated (point) load and the speed is a constant, guarantees the presence of the Dirac function in the above equation which simplifies the model and hence provides the use of analytical approximate methods [41]. However, in real engineering models the concentrated force is usually a pointwise load with a finite support, and the speed of the object may not be a constant. This means that the lateral source should be given at least in the form F (x) G(t). Notethat this is a special case of the so-called asynchronous loading when F (x, t) := K k=1 Fk (x) Gk (t), and is defined as synchronous loading [79]. The numerical examples below illustrate the applicability of the above theory, as well as the efficiency of the Conjugate Gradient Algorithm (CG-algorithm) with the gradient formula (11.2.19). The following version of the CG-algorithm, adopted

11.2 Identification of a Temporal Load in a Simply Supported Beam From. . .

387

from that given in Sect. 6.5.4, is used in all examples related to the inverse source problem of identifying the temporal load G(t) governed by (11.2.1)–(11.2.2). Step 1. For n = 0 choose the initial iteration G(0) (t) ≡ 0 and find the initial descent direction p(0) ≡ −∇J (G(0)). Set n = 0; Step 2. Find the descent direction parameter β∗(n)

=

||∇J (G(n) )||2L2 (0,T ) ||ux (0, ·; p(n) )||2L2 (0,T )

, (n)

and then define the next iteration G(n+1) (t) = G(n) (t) + β∗ p(n) (t). Step 3. Set n = n + 1, compute ∇J (G(n) ) and the descent direction p(n) (t) = Step 4.

∇J (G(n) ) 2L2 (0,T )

∇J (G(n−1) ) 2L2 (0,T )

p(n−1) (t) − ∇J (G(n) )(t).

If the stopping condition

||ux (0, ·; G(n) ) − θ γ ||L2 (0,T ) ≤ γ τM < ||ux (0, ·; G(n−1) ) − θ γ ||L2 (0,T ) holds, then stop the iteration process; otherwise, return to Step 2. Here, θ γ (t) is the random noisy measured output with the noise level γ > 0. At each step of the CG-algorithm, both the direct and adjoint problems are solved on a sufficiently fine mesh with by using the method of lines (MOL) [139] which provides accurate approximate solutions for many types of dynamical problems, especially linear PDEs. This method is applied to the following weak formulation of the direct problem (11.2.1). Find U : (0, T ] → V(0, ) such that ∀v ∈ V(0, ) 

(ρ(·)U  (t), v) + (μ(·)U  (t), v) + a(U (t), v) = G(t)(F (·), v), U (0) = 0, U  (0) = 0,

where V 2 (0, ) := {v ∈ H 2 (0, ) : v(0) = v() = 0}, U (t) := u(·, t) and a(u, v) := (ruxx , vxx ) + (Tr ux , vx ), a : V 2 (0, ) × V 2 (0, ) → R is the symmetric bilinear functional. We use the Finite Element Method (FEM) with cubic Hermite basis functions and uniform elements with the length h = l/Nx to get the semi-discretization of the above weak form of the direct problem. The corresponding finite dimensional space, defined as Vh ⊂ V12 (0, ) ⊂ H 2 (0, ), where V12 (0, ) is the subspace introduced in Theorem 11.1.5, is spanned by piecewise polynomials which are called the Hermite

388

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

shape functions. These cubic basis functions provide continuity of both deflection and slope along the beam which is a required property for the physical model. Then the finite difference techniques are independently performed for the time integrations to the following semi-discrete problem. Find Uh : (0, T ] → Vh such that ∀vh ∈ Vh 

(ρ(·)Uh (t), vh ) + (μ(·)Uh (t), vh ) + a(Uh (t), vh ) = G(t)(F (·), vh ), Uh (0) = 0, Uh (0) = 0.

The following time integrations are used on a uniform partition of the time interval [0, T ] with the time step size τ = T /Nt . f  (tj ) ≈ ∂τ τ fh = j

f  (tj ) ≈ ∂τ fh = j

j

j −1

2fh − 5fh j

j −2

+ 4fh τ2

j −1

j −3

− fh

,

(11.2.24)

j −2

3fh − 4fh + fh 2τ

.

(11.2.25)

The above semi-discrete problem leads to the following fully discretized problem: j Find Uh ∈ Vh such that ∀vh ∈ Vh and for each j = 1, 2, . . . , Nt j

j

j

(ρ(·)∂τ τ Uh , vh ) + (μ(·)∂τ Uh (t), vh ) + a(Uh , vh ) = G(t j )(F (·), vh ). The solution to this algebraic problem is an approximate solution to the direct problem (11.2.1) at the temporal mesh points t = tj , for j = 0, · · · , Nt , which is j denoted by Uh ≈ u(·, tj ). Due to the need for a series of numerical experiments, the mesh parameters Nx and Nt should be optimized in the sense of a balance between the CPU time and computational noise level. As a result of the computational experiments, these optimal values were found as Nx = 501 and Nt = 41. The computational inversion has been performed on a coarser mesh, to avoid inverse crime. In the numerical examples below,  = T = 1. The noise free synthetic measured output θh (tj ) := ux (0, tj ; G) is generated from the numerical solution of the direct problem for a given temporal load G(t). γ Then the random noisy data θh (tj ) := θh (tj ) + γ θh L2 (0,T ) randn(Nt ) is obtained by using the MATLAB randn function, which generates Nt arrays of random numbers whose elements are normally distributed with mean 0 and standard deviation σ = 1. Finally, Simpson’s quadrature formula is used to approximate the Tikhonov functional J (G). For all other integrations in the above discretization a three-point Gauss quadrature formula is used. Example 11.2.1 Behavior of the convergence error e(n; G; γ ) and the accuracy error E(n; G; γ ).

11.2 Identification of a Temporal Load in a Simply Supported Beam From. . .

389

The aim of this example is to examine efficiency of the algorithm applied to the vibrations of a beam under a point-mass moving load. Such type of problems have wide applications in engineering and technology. To simulate the vibration of a beam subjected to a point-mass moving with the variable velocity, the noise free output θh (tj ) is generated from the pointwise spatial load F (x) = δx0 (x) with the √ temporal component G(t) = t (1 − t). Here δx0 (x) = exp(−(x − x0 )2 /)/ π, with small enough parameter  > 0, is the approximation of Dirac delta centered at γ x0 = 0.9. The noisy outputs θh with two noise levels γ = 0.03; 0.05 are then used with the noise free output in the reconstruction algorithm. To examine the convergence and accuracy of the numerical algorithm the behavior of the convergence error e(n; G; γ ) = ux (0, ·; G(n) ) − θhδ L2 (0,T ) and the accuracy error E(n; G; γ ) = G − G(n) L2 (0,T ) are analyzed depending on the iteration number n. The results are illustrated in Fig. 11.4. The efficiency of the algorithm in terms of accuracy as well as the number of iterations can be seen from the graph on the right figure which shows the behavior of the accuracy error. The reconstructed temporal load G(t) from these noise free and noisy outputs is plotted in Fig. 11.5.  Example 11.2.2 Reconstruction of oscillatory and non-smooth temporal loads. The purpose of the numerical examples here is to demonstrate the ability of the algorithm to recover temporal loads governed by oscillating and non-smooth functions. In the first test problem, the noise free output θh (tj ) obtained from the solution of the direct problem, assuming that the temporal load is given by the are oscillating function G(t) = t sin(4πt). The spatial component is assumed to be P (x) = x. γ This output is then used to generate the random noisy outputs θh with two noise levels γ = 0.03, 0.05.

Fig. 11.4 Convergence error (left figure) and accuracy error (right figure) for Example 11.2.1

390

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Fig. 11.5 Reconstructed temporal load corresponding to Example 11.2.1

In the second test problem, the reconstruction of the non-smooth temporal load given by the function G(t) = κ(1/2 − t) + κ(t − 1/2) sin(3πt) is examined taking the spatial load as P (x) = ex . Here, κ is the Heaviside step function. As in the previous problem, the random noisy outputs with two noise levels γ = 0.03; 0.05 are generated from the synthetic noise free output θh (tj ) found from the numerical solution of the direct problem. The results of the computational experiments are plotted in Fig. 11.6 on the left and right. The reconstructed temporal loads from the noise free as well as noisy outputs are very close to the oscillating function G(t) = t sin(4πt), as it can be seen from the graph in Fig. 11.6 on the left. The reconstruction results in Fig. 11.6 on the right, related to the non-smooth temporal load, are also acceptable. However, in this case the accuracy error is very sensitive with respect to the noise level that shows the average quality of the algorithm for non-smooth temporal loads.

11.3 Determination of Unknown Shear Force in a Cantilever Beam From. . .

391

Fig. 11.6 Reconstructed temporal loads: oscillatory (left) and non-smooth (right) corresponding to Example 11.2.2

11.3 Determination of Unknown Shear Force in a Cantilever Beam From Boundary Measured Bending Moment Consider the following inverse boundary value problem of determining the unknown transverse shear force g(t) in ⎧ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = 0, in T , ⎪ ⎪ ⎪ ⎪ ⎨ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ u(0, t) = ux (0, t) = 0, r()uxx (, t) = 0, ⎪ ⎪ ⎪ ⎩ −(r()uxx (, t))x + Tr ()ux (, t) = g(t), t ∈ [0, T ],

(11.3.1)

from the measured boundary bending moment M(t) := −r(0)uxx (0, t), t ∈ (0, T )

(11.3.2)

at the left end x = 0 of the beam. Note that if the accessible boundary x = 0 of a beam extending over the interval (0, ) is clamped and a load distribution along the beam is zero, then there can only be two types of force effects acting on the inaccessible boundary x =  of a beam: the bending moment r()uxx (, t) and the shear force −(r()uxx (, t))x + Tr ()ux (, t). In this section we consider the case where the vibration occurs due to the transverse shear force g(t) := −(r()uxx (, t))x + Tr ()ux (, t), applied at the inaccessible boundary x = , which is assumed to be unknown and needs to be

392

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

u

r()uxx (, t) = 0

M (t) u(0, t) = 0 ux (0, t) = 0

−(r()uxx (, t))x + Tr ()ux (, t) = g(t)

g(t) =?

x

Fig. 11.7 Geometry of the inverse BVP with measured boundary bending moment for the cantilever beam under free-end shear force

identified from the measured bending moment M(t) at the accessible boundary x = 0 (Fig. 11.7). As noted in the previous sections, cantilever beams are an integral part of many engineering structures. More recently, they are also found in emerging systems for medical diagnostics and nanoscale measurements such as the Transverse Dynamic Force Microscope (TDFM) and the Atomic Force Microscope (AFM), since its invention [21]. Various simplified versions of the inverse problem (11.3.1)–(11.3.2) have been used for estimation of the shear force affecting the tip of the cantilever in the TDFM (and in AFMs in general) using a real-time implementable sliding mode observer (see [3, 113] and references therein). Such estimates are allow better interpretation and understanding of the scan results. The proposed formulation of the inverse problem given by (11.3.1)–(11.3.2) is important not only from the point of view of generalizing existing parameter estimation models in the sense that the Euler-Bernoulli equation includes all the physically possible variable terms, and the time interval is finite. The inverse problem (11.3.1)–(11.3.2) is itself one of the most important problems in vibration theory and applications from the view point of the Neumann-to-Neumann map. Indeed, for each admissible input g(t) there exists (in appropriate sense) a unique solution u(x, t; g) of the initial boundary value problem (11.3.1), functionally depending on the admissible input g(t). If −r(0)uxx (0, t; g) = M(t), where M(t) is the measured boundary bending moment, we say that g(t) is a solution of the inverse problem (11.3.1)–(11.3.2). Since g(t) is the Neumann input and −r(0)uxx (0, t; g) is the Neumann output, the operator (g)(t) := −r(0)uxx (0, t; g) has a meaning of the Neumann-to-Neumann map corresponding to the inverse problem (11.3.1)– (11.3.2), according to generally accepted terminology. Having this operator one can reformulate the inverse problem as the operator equation (g)(t) = M(t), and then employ the theory developed in the previous section, introducing the Tikhonov functional J (g) := (1/2) g − M 2L2 (0,T ) . Another important point, as in all inverse problems with Neumann measured output, is that the Tikhonov functional contains the second order derivative due to the Neumann output −r(0)uxx (0, t; g), which, in turn, introduces an additional source of ill-posedness. In addition, one needs to impose an additional regularity condition on the input g(t) and the measured output M(t), as we will see in the following subsections. And this is one of the main features of an inverse problem with Neumann output.

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11.3.1 The Neumann-to-Neumann Operator and Existence of a Quasi-Solution We define the following sets of admissible inputs (i.e. transverse shear forces) G m = {g ∈ H m (0, T ) : g (k) (0) = 0, k = 0, m − 1, g H m (0,T ) ≤ γ },

(11.3.3)

m = 2, 3, 4, γ > 0. Note that the sets of admissible inputs G 2 , G 3 and G 4 correspond to the regular weak solution, the regular weak solution with improved regularity, and the regular weak solution with higher regularity of the direct problem (11.3.1), respectively, examined in Sect. 11.1.2. From Theorems 11.1.6 and 11.1.7 it follows that the requirements for the regularity of the Neumann input g(t) for these solutions are g ∈ H 2 (0, T ), g ∈ H 3 (0, T ) and g ∈ H 4 (0, T ), in turn. Let g ∈ G m be a given admissible input and u = u(x, t; g) is the solution of the direct problem (11.3.1). Then, as noted above, (g)(t) := −r(0)uxx (0, t; g),  : G m → L2 (0, T ),

(11.3.4)

is the Neumann-to-Neumann operator corresponding to the inverse problem (11.3.1)–(11.3.2). Properties of this operator will be examined below on each set of admissible inputs G m , m = 2, 3, 4, separately. Lemma 11.3.1 Assume that conditions of Theorem 11.1.6 hold. Then the Neumann output −r(0)uxx (0, t) belongs to L2 (0, T ). Furthermore, the following trace estimate holds: #12 g 2 2

r(0)uxx (0, ·) 2L2 (0,T ) ≤ C , H (0,T )

(11.3.5)

  #12 = 22 max 2,  Tr2 C42 + μ21 C52 , ρ12 C52 C 1

(11.3.6)

where

and C4 , C5 > 0 are the constants introduced in Theorem 11.1.5. If, in addition, conditions (11.1.37) of Theorem 11.1.7 also hold, then #12 g 2 3

r(0)uxxt (0, ·) 2L2 (0,T ) ≤ C , H (0,T )

(11.3.7)

#1 > 0 defined in (11.3.6). with the same constant C Furthermore, if in addition to the above conditions, the regularity and consistency conditions (11.1.39) of Theorem 11.1.7 also hold, then #12 g 2 4 .

r(0)uxxt t (0, ·) 2L2 (0,T ) ≤ C H (0,T )

(11.3.8)

394

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Proof Integrating Eq. (11.3.1) over (x, ) we get: 



(r(x)uxx )x = Tr (x)ux +

[ρ(ξ )ut t + μ(ξ )ut ] dξ − g(t),

x

(x, t) ∈ T . Squaring and then integration this identity over T , after some calculations we deduce that    u2x dxdt + 22 ρ12 u2t t dxdt [(r(x)uxx )x ]2 dxdt ≤ 4Tr21 T

T

T



 +22 μ21

T

u2t dxdt + 4

T

g(t)dt. 0

Using this inequality in 

T

 (r(0)uxx (0, t))2 dt ≤ 

0

0

T





[(r(x)uxx (x, t))x ]2 dxdt

0

and taking into account the inequality

ux 2L2 (0,T ;L2 (0,)) ≤

2

uxx 2L2 (0,T ;L2 (0,)), 2

we arrive at 

r(0)uxx (0, ·) 2L2 (0,T ) ≤ 23 Tr1 uxx 2L2 (0,T ;L2 (0,)) + ρ12 ut t 2L2 (0,T ;L2 (0,))

+μ21 ut 2L2 (0,T ;L2 (0,)) + 42 g 2L2 (0,T ) .

With the estimates in Theorems 11.1.5 and 11.1.6, this leads to the required estimate #1 > 0 introduced in (11.3.6). (11.3.5), with the constant C Assume now that conditions (11.1.37) of Theorem 11.1.7 also hold. For this case, the analogous 

r(0)uxxt (0, ·) 2L2 (0,T ) ≤ 23 Tr1 uxxt 2L2 (0,T ;L2 (0,)) + ρ12 ut t t 2L2 (0,T ;L2 (0,))

+μ21 ut t 2L2 (0,T ;L2 (0,)) + 42 g  2L2 (0,T )

to the above inequality is derived in exactly the same. The required estimate (11.3.7) follows from this inequality. The last estimate (11.3.8) of the theorem can be proven similarly.   Lemma 11.3.2 Assume that in addition to the basic conditions (11.1.2), the regularity and consistency conditions (11.1.37) of Theorem 11.1.6 also hold. and

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g ∈ G 3 . Then the Neumann-to-Neumann map [·] : G 3 → L2 (0, T ) introduced in (11.3.4) is a linear compact operator. Proof Let {g (n) } ⊂ G 3 be a sequence of inputs, uniformly bounded in the norm of H 3 (0, T ). Denote by {u(n) (x, t)}, u(n) (x, t) := u(x, t; g (n) ) the sequence of corresponding regular weak solutions of the direct problem (11.3.1). Then {(g (n) )(t)}, (g (n) )(t) = −r(0)u(n) xx (0, t), n = 1, ∞ is the sequence of outputs. From estimates (11.3.5) and (11.3.7) we deduce that (n) #

r(0)u(n) xx (0, ·) L2 (0,T ) ≤ C1 g H 2 (0,T ) , (n) #

r(0)u(n) xxt (0, ·) L2 (0,T ) ≤ C1 g H 3 (0,T ) .

In addition, g (n) 2H 3 (0,T ) ≤ γ , for all n, by the definition (11.3.3) of the set of admissible inputs G 3 ⊂ H 3 (0, T ). This means the uniform boundedness of the (n) sequence of norms { r(0)uxx (0, ·) L2 (0,T ) } in H 1 (0, T ), that is, it compactness in L2 (0, T ). By definition, this implies that the Neumann-to-Neumann map [·] : G 3 → L2 (0, T ) transforms a bounded in G 3 set to a compact in L2 (0, T ) set, hence is a compact operator.   Lemma 11.3.2 also implies that the inverse boundary value problem (11.3.1)– (11.3.2) is ill-posed. Lemma 11.3.3 Assume that conditions of Theorem 11.1.5 hold, and g ∈ G 2 . Then the Neumann-to-Neumann map [·] : G 2 → L2 (0, T ) is Lipschitz continuous, that is

g1 − g2 L2 (0,T ) ≤ L0 g1 − g2 H 2 (0,T ) , g1 , g2 ∈ G 2 ,

(11.3.9)

#1 , where C #1 > 0 is the same constant introduced in (11.3.6). with L0 = C Proof Let um (x, t) := u(x, t; gm ) be the regular weak solutions of the direct problem (11.3.1) corresponding to gm ∈ G 2 , m = 1, 2. Then the function δu(x, t) = u1 (x, t) − u2 (x, t) solves the problem ⎧ ρ(x)δut t + μ(xδut + (r(x)δuxx )xx − (Tr (x)δux )x = 0, ⎪ ⎪ ⎪ ⎪ ⎪ (x, t) ∈ T , ⎪ ⎪ ⎨ δu(x, 0) = 0, δut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ ⎪ δu(0, t) = δux (0, t) = 0, r()δuxx (, t) = 0, ⎪ ⎪ ⎪ ⎩ −(r()δuxx (, t))x + Tr ()δux (, t) = δg(t), t ∈ [0, T ], with the input δg(t) = g1 (t) − g2 (t).

(11.3.10)

396

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

By the definition the input-output operator we have:

g1 − g2 L2 (0,T ) = r(0)δuxx (0, ·) L2 (0,T ) . We use trace estimate (11.3.5) for the regular weak solution of the problem (11.3.10): #12 δg 2 2 .

r(0)δuxx (0, ·) 2L2 (0,T ) ≤ C H (0,T )  

This implies the required result (11.3.9). We introduce now the Tikhonov functional J (g) :=

1

g − M 2L2 (0,T ) , g ∈ G 2 , M ∈ L2 (0, T ), 2

(11.3.11)

and define a quasi-solution to the inverse problem (11.3.1)–(11.3.2) as a solution of the minimization problem J (g∗ ) = inf J (g) g∈G 2

for this functional. As with the inverse problem (11.2.1)–(11.2.2) in previous Sect. 11.2.2, the Lipschitz continuity of the Tikhonov functional implies existence of a quasi-solution to the inverse problem (11.3.1)–(11.3.2) in G 2 . Theorem 11.3.1 Assume that conditions of Theorem 11.1.6 hold. Suppose that g ∈ G 2 and M ∈ L2 (0, T ). Then there exists a quasi-solution to the inverse problem (11.3.1)–(11.3.2) in the set of admissible inputs G 2 ⊂ H 2 (0, T ).

11.3.2 Fréchet Gradient of the Tikhonov Functional. Lipschitz Continuity of the Gradient Let g, g + δg ∈ G 2 be arbitrary inputs. Denote by δJ (g) := J (g + δg) − J (g) the increment of the Tikhonov functional defined in (11.3.11). Then 

T

δJ (g) =

[r(0)uxx (0, t) + M(t)] r(0)δuxx (0, t)dt

0

+

1 2



T

[r(0)δuxx (0, t)]2 dt,

(11.3.12)

0

where δu(x, t) := u(x, t; g + δg) − u(x, t; g) is the regular weak solution of the initial-boundary value problem (11.3.10).

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Lemma 11.3.4 Assume that conditions of Theorem 11.1.5 hold. Suppose that g, g+ δg ∈ G 2 are arbitrary inputs. Then the following integral relationship holds: 

T



T

r(0)δuxx (0, t) θ (t)dt =

0

φ(, t)δg(t)dt,

(11.3.13)

0

where δu(x, t) := u(x, t; g+δg)−u(x, t; g) is the regular weak solution of problem (11.3.10) and φ(x, t) is the solution of the following backward problem ⎧ ⎪ ρ(x)φt t − μ(x)φt + (r(x)φxx )xx − (Tr (x)φx )x = 0, ⎪ ⎪ ⎪ ⎪ ⎪ (x, t) ∈ T , ⎪ ⎨ (11.3.14) φ(x, T ) = 0, φt (x, T ) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ ⎪ φ(0, t) = 0, φx (0, t) = θ (t), φxx (, t) = 0, ⎪ ⎪ ⎪ ⎩ − (r()φxx (, t))x + Tr ()φx (, t) = 0, t ∈ [0, T ]. with the Dirichlet input θ (t). Proof Multiply both sides of Eq. (11.3.10) by arbitrary function φ(x, t), integrate over T and apply the integration by parts formula multiple times. Then we obtain: 

 T

 +

ρ(x)φt t − μ(x)φt + (r(x)φxx )xx − (Tr (x)φx )x δu dtdx



0

 + 0

T



[ρ(x)δut φ − ρ(x)δuφt + μ(x)δu φ]tt =T =0 dx (r(x)δuxx )x φ − r(x)δuxx φx + r(x)δux φxx − δu (r(x)φxx )x 

T

− 0

x= dt x=0

[Tr (x)δux φ − Tr (x)δuφx ]x= x=0 dt = 0.

Now, we require that φ(x, t) is the solution of the backward problem (11.3.14). Then taking into account the initial/final and the boundary conditions given in (11.3.1) and (11.3.14) we arrive at the integral relationship (11.3.13).   Choose formally the input θ (t) in the backward problem (11.3.14) as follows: θ (t) = r(0)uxx (0, t; g) + M(t), t ∈ [0, T ].

(11.3.15)

The backward problem (11.3.14) with the input (11.3.15) is called the adjoint problem corresponding to the direct problem (11.3.1).

398

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

With the input given in (11.3.15), the integral relationship becomes the following input-output relationship: 

T



T

[r(0)uxx (0, t; g) + M(t)] r(0)uxx (0, t)dt =

0

φ(, t)δg(t)dt, (11.3.16) 0

between the Neumann input g(t), the Neumann output −r(0)uxx (0, t) and also the measured output M(t). Taking into account (11.3.16) in (11.3.12) we arrive at the following convenient form of the increment formula to obtain a gradient formula: 

T

δJ (g) = 0

φ(, t)δg(t)dt +

1 2



T

[r(0)δuxx (0, t)]2 dt, g ∈ G 2 .

(11.3.17)

0

We should examine now the validity of the formal substitution (11.3.15). In other words, we need to find out under what conditions this replacement makes sense. Considering that backward problem (11.3.14) is transformed to the well-posed problem (C.3.1) formulated in Appendix C, by the change of variable τ = T − t, this question is answered in the positive in Theorem C.3.1. Specifically, if the input θ (t) in the backward problem (11.3.14) satisfies the condition θ ∈ H 2 (0, T ), then Theorem C.3.1 is applicable to this problem as well. And this requirement, in turn, entails the following conditions: uxx (0, ·) ∈ H 2 (0, T ), M ∈ H 2 (0, T ), t ∈ [0, T ],

(11.3.18)

as it follows from relation (11.3.15). The first condition uxx (0, ·) ∈ H 2 (0, T ) means that we are dealing with a regular weak solution with higher regularity of the direct problem (11.3.1), as it follows from Lemma 11.3.1 and Theorem 11.1.7. The second condition M ∈ H 2 (0, T ) in (11.3.18) precisely shows an essential feature of the Neumann measured output in the inverse problem (11.3.1)–(11.3.2). To explain this property more precisely, let us compare this inverse problem with the inverse source problem (11.2.1)–(11.2.2) with the Dirichlet measured output θ (t) := ux (0, t), discussed in Sect. 11.2. While for the existence of the weak solution of corresponding adjoint problem, the Dirichlet measured output θ (t), which is originally from L2 (0, T ), needs to be smoothed (or mollified) up to H 1 (0, T ) (see Remark 11.2.1), the Neumann measured output M(t), which is also originally from L2 (0, T ), should be smoothed up to H 2 (0, T ). This shows that the Neumann measured output is more expensive than the Dirichlet one, in terms of the requirements for the regularity.

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Theorem 11.3.2 Assume that conditions of Theorem 11.1.7 hold. Suppose that g ∈ G 4 and the measured output M(t) satisfies the regularity condition M ∈ H 2 (0, T ). Then the Tikhonov functional J (g) defined in (11.3.11) is Fréchet differentiable. Furthermore, for the Fréchet gradient of this functional the following gradient formula holds: ∇J (g)(t) = φ(, t; g), t ∈ (0, T ),

(11.3.19)

where φ(x, t) = φ(x, t; g) is the weak solution solutions of the adjoint problem (11.3.14) with the input (11.3.15). Proof From estimate (11.3.7) applied to the regular weak solution with improved regularity of problem (11.3.10) we deduce that #12 δg 2 3 .

r(0)δuxxt (0, ·) 2L2 (0,T ) ≤ C H (0,T ) Hence integral in the increment formula (11.3.17) is of the order 

the right-hand-side 2 O δg H 3 (0,T ) . On the other hand, applying Theorem C.3.1 of Appendix C to the adjoint problem (11.3.14) with the input (11.3.15) we deduce that the trace φ(, t; g) in (11.3.19) is well-defined since the weak solution of this problem is in L2 (0, T ; V0 (0, )), where V0 (0, ) := {v ∈ H 2 (0, ) : v(0) = 0}. This leads to the gradient formula (11.2.19).   The next important issue is the Lipschitz continuity of the Fréchet gradient of the Tikhonov functional. It turns out that the first condition in (11.3.18) is also necessary to preserve this property. Theorem 11.3.3 Under the conditions of Theorem 11.3.2, the Fréchet gradient of the Tikhonov functional is Lipschitz continuous, that is

∇J (g1 ) − ∇J (g2 ) 2L2 (0,T ) ≤ L1 g1 − g2 2H 4 (0,T ) ,

(11.3.20)

√ 2 √  #C for all g1 , g2 ∈ G 4 , with the Lipschitz constant L1 =   C 4 1 / 3 > 0, where #4 > 0 is the constant introduced in Theorem C.3.1 of Appendix C and C 1 > 0 C depends on the constants defined in Lemma 11.3.1. Proof From the gradient formula (11.3.19) we deduce that

∇J (g1 ) − ∇J (g2 ) 2L2 (0,T ) = δφ(, ·) 2L2 (0,T ) ,

(11.3.21)

400

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

where δφ(x, t) := φ(x, t; g1 ) − φ(x, t; g2 ) is the weak solution of the following backward problem ⎧ ⎪ ρ(x)δφt t − μ(x)δφt + (r(x)δφxx )xx − (Tr (x)δφx )x = 0, ⎪ ⎪ ⎪ ⎪ ⎪ (x, t) ∈ T , ⎪ ⎨ δφ(x, T ) = 0, δφt (x, T ) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ ⎪ δφ(0, t) = 0, δφx (0, t) = −r(0)δuxx (0, t), δφxx (, t) = 0, ⎪ ⎪ ⎪ ⎩ − (r()δφxx (, t))x + Tr ()δφx (, t) = 0, t ∈ [0, T ].

(11.3.22)

with the Dirichlet input −r(0)δuxx (0, t), as it follows from (11.3.14) and (11.3.15). We employ first the trace inequality (11.1.30) to estimate the trace norm

δφ(, ·) L2 (0,T ) through the norm δφxx L2 (0,T ;L2 (0,)) as

δφ(, ·) 2L2 (0,T ) ≤

3

δφxx 2L2 (0,T ;L2 (0,)), 3

and then apply Theorem C.3.1 of Appendix C to the weak solution of the problem (11.1.33). We have:

δφ(, ·) 2L2 (0,T ) ≤

3 #2 C r(0)δuxx (0, ·) 2H 2 (0,T ) . 3 4

(11.3.23)

To estimate the norm r(0)δuxx (0, ·) H 2 (0,T ) we apply the trace estimates in Lemma 11.3.1 to the regular weak solution with higher regularity of problem (11.3.10). This yields: 12 δg 2 4

r(0)δuxx (0, ·) 2H 2 (0,T ) ≤ C . H (0,T ) Inserting this with (11.3.23) into (11.3.21) we arrive at the desired result (11.1.31).  

11.3.3 Numerical Reconstruction of Unknown Shear Force from Boundary Measured Moment In this subsection, we demonstrate the effectiveness of numerical algorithms discussed in Sect. 11.2.5, on new benchmark examples related to the inverse boundary value problem (11.3.1)–(11.3.2). The modification of the Conjugate Gradient Algorithm (CG-algorithm) for the case of this inverse problem is given in the Table 11.1.

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Table 11.1 Conjugate Gradient Algorithm for the inverse problem (11.3.1)–(11.3.2) J (g) = 12 r(0)uxx (0, ·; g) + M 2

Steps 1

2 3

Choose the initial

g(0) ∈ H 4 (0, T ),

iteration (n = 0)

g(n) (0) = 0

Compute the initial descent direction

p (n) (t) = −φ(, ·; g(n) )

Find the descent direction parameter

βn =

φ(, ·; g(n) ) 2

r(0)uxx (0, ·; g(n) ) 2

4

Find the next iteration

g(n+1) (t) = g(n) (t) + βn p (n) (t)

5

Compute the convergence error

e(n; g; γ ) = r(0)uxx (0, ·; g(n) ) + M

6

If the stopping condition holds, then go to the Step 8

eγ (n)e(n; g; γ ) ≤ τM γ ≤ e(n − 1; g; γ )

7

Compute p (n+1) (t) Set n := n + 1 and return to Step 3

p (n+1) (t) = βn p (n) (t) − ∇J (g(n+1) )(t)

8

βn =

φ(, ·; g(n+1) ) 2

φ(, ·; g(n) ) 2

Stop the iteration process

Remark 11.3.1 From the formulas p(n) (t) = −φ(, ·; g (n) ) and g (n+1) (t) = g (n) (t) + αn p(n) (t) in Table 11.1 it follows that g (n+1) (t) = g (n) (t) − αn φ(, ·; g (n) ), t ∈ (0, T ), n = 0, 1, 2, . . . ,

(11.3.24)

where φ ∈ L2 (0, T ; V0 (0, )), with V0 (0, ) := {v ∈ H 2 (0, ) : v(0) = 0} is the weak solution of the backward problem (11.3.14) with the input (11.3.15), i.e. the solution of the adjoint problem corresponding to the direct problem (11.3.1). On the other hand, this weak solution φ(x, t) belongs to C([0, T ]; H 1(0, )), and satisfies the consistency condition φ(− , T ) = 0 since φ(x, T ) = 0 for all x ∈ (0, ), by the final condition in the backward problem (11.3.14). In view of the iteration scheme (11.2.24) this implies that the CG-algorithm forces the function g(t) to satisfy the consistency condition g (n+1) (T ) = 0, at each iteration step. This is the reason that in the left neighborhood (T − ε, T ), ε > 0 of the final time T > 0, there is some discrepancy between the exact functions g(t) and the function g (n) (t) found by iteration scheme (11.2.24), as we will see firsthand in the next example. Although it is possible to solve this problem by modifying the gradient formula for the Tikhonov functional, as well as the Dirichlet input in the adjoint problem, similar to what was done in [135], we will not go into the details of this issue here.  As mentioned above, the same MOL approach outlined in Sect. 11.2.5 is used for the numerical solution of the inverse problem (11.3.1)–(11.3.2). Then semi-discrete problem corresponding to the direct problem (11.3.1) can be formulated as follows.

402

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Find Uh : (0, T ] → Vh such that ∀vh ∈ Vh 

(ρ(·)Uh (t), vh ) + (μ(·)Uh (t), vh ) + a(Uh (t), vh ) = g(t)vh (), Uh (0) = 0, Uh (0) = 0.

With the time integrations defined in (11.2.24) this leads to the following fully discretized problem: j Find Uh ∈ Vh such that ∀vh ∈ Vh and for each j = 1, 2, . . . , Nt j

j

j

(ρ(·)∂τ τ Uh , vh ) + (μ(·)∂τ Uh (t), vh ) + a(Uh , vh ) = g(t j )vh (). The mesh parameters used for this problem are the same as in Sect. 11.2.5. In the numerical examples below the inverse problem is analyzed for prismatic beams with real geometric (the length  > 0, cross-sectional dimensions S := b×h) and physical parameters (with the mass density ρd and Young’s modulus E) given in [128]. Specifically, for the homogeneous prismatic cantilever beam these parameters are assumed to be as follows:  = 0.502m, b × h = 1.7 · 10−3 m × 0.89 · 10−3 m. With these values, the value of the moment of inertia of the cross-section I := bh3 /12 turned out to be = 1.2 · 10−12. As a consequence, E I = 3.1 · 109 · 1.2 · 10−12 = 3.72 · 10−3 . Furthermore, with the above parameters the following value of the mass density ρ := ρd S is obtained: ρ = 2.148 · 10−3 . The value of the damping coefficient μ > 0, which is usually in the interval 0.01 ÷ 10 s −1 , is taken to be μ = 0.05. In all examples the final time is assumed to be T = 1. γ Below, the random noisy data Mh (tj ) := Mh (tj ) + γ Mh L2 (0,T ) randn(Nt ) is obtained by using the MATLAB randn function, where the noise free synthetic measured output M(tj ) := −r(0)uxx (0, tj ; g) is generated from the numerical solution of the direct problem for a given Neumann input g(t). γ Then the random noisy data Mh (tj ) := Mh (tj ) + γ Mh L2 (0,T ) randn(Nt ) is obtained by using the MATLAB randn function. Example 11.3.1 Shear force reconstruction without the consistency condition g(T ) = 0. The function g(t) = 2t 2 + sin2 (ωg π t/2), ωg = 4, T = 1

(11.3.25)

is taken as the Neumann input in the direct problem (11.3.1) to generate the noise free synthetic measured output M(tj ) := −r(0)uxx (0, tj ; g) from the solution of the fully discretized problem. The plots in Fig. 11.8 on the left show the noise free and noisy outputs, with the noise levels γ = 0.03; 0.06. The reconstructed shear forces are plotted in Fig. 11.8 on the right. The average accuracy of the reconstructions, especially from the noisy Neumann outputs, are high enough. Some divergence between the function g(t) and the found by the CG-algorithm,

11.3 Determination of Unknown Shear Force in a Cantilever Beam From. . .

403

Fig. 11.8 Transverse shear force reconstruction: synthetic measured output (left) and reconstructed shear force (right) corresponding to Example 11.3.1

Fig. 11.9 Convergence error (left) and accuracy error (right) corresponding to Example 11.3.1

is observed in the interval (T − ε, T ) with ε = 0.1, for the reason stated in Remark 11.3.1. The behavior of the convergence error e(n; g; γ ) = r(0)uxx (0, ·; g (n) ) + γ Mh L2 (0,T ) and the accuracy error E(n; g; γ ) = g − g (n) L2 (0,T ) is analyzed depending on the iteration number n. The results are given in Fig. 11.9 on the left and right, respectively. The efficiency of the algorithm in terms of accuracy as well as the number of iterations can be seen from these graphs. 

404

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Fig. 11.10 Transverse shear force reconstruction: synthetic measured output (left) and reconstructed shear force (right) corresponding to Example 11.3.2

Example 11.3.2 Shear force reconstruction with the consistency condition g(T ) = 0. In this numerical examples the input function has been modified so that the consistency condition g(T ) = 0 is preserved: g(t) = (2t 2 + sin2 (ωg π t/2)) (1 − t), ωg = 4, T = 1.

(11.3.26)

The noise free and random noisy synthetic Neumann outputs, generated from the numerical solution of the direct problem, are plotted in Fig. 11.10 on the left. The reconstructed from these outputs transverse shear forces are shown on the right in Fig. 11.10. As can be clearly seen from this figure, the reconstruction quality in this figure is slightly better, especially around the point tNt = T corresponding to the final time, than one in the previous Fig. 11.8 on the right. The behavior of the convergence error and the accuracy error, depending on the iteration number n, is the same as in the previous example. The results of numerical experiments carried out for these inverse problems discussed here and in Sect. 11.2.5 indicate that the proposed approach is also an efficient computational tool with a high reconstruction accuracy, and is applicable for a wide class of inverse problems for Euler-Bernoulli equation.

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

405

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam Equation from Final Time Output Consider the following inverse problem of recovering the unknown spatial load F (x) in ⎧ ⎪ ⎨ ρut t + μut + (r(x)uxx )xx = F (x)G(t), (x, t) ∈ T , u(x, 0) = ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ u(0, t) = uxx (0, t) = 0, u(, t) = uxx (, t) = 0, t ∈ [0, T ],

(11.4.1)

from the final time measured output (displacement) uT (x) := u(x, T ), x ∈ [0, ].

(11.4.2)

Here, the mass density ρ > 0 and the damping coefficient μ > 0 are the assumed to be constant without loss of generality, only for the convenience of the Singular Value Decomposition (SVD) analysis. In Example 3.2.1 of Sect. 3.2, we analyzed the problem of determining the spatial load F (x) in ⎧⎧ ⎪ ⎪ ⎪ ut t = uxx + F (x), (x, t) ∈ T := (0, 1) × (0, T ), ⎪ ⎨ ⎪ ⎪ ⎨ u(x, 0) = u (x, 0) = 0, x ∈ (0, 1), t ⎪ (11.4.3) ⎪ ⎩ ⎪ u(0, t) = u(1, t) = 0, t ∈ [0, T ], ⎪ ⎪ ⎪ ⎩ uT (x) := u(x, T ), x ∈ (0, 1), and found that for unique determination of the unknown source F (x) the final time must satisfy the following condition T =

2m , for all m, n = 1, 2, 3, . . . . n

(11.4.4)

Otherwise, i.e. when T = 2m/n, an infinite number of singular values σn defined as (see formula (3.2.12) in Sect. 3.2) σn =

  1  1 − cos λn T , for all n = 1, 2, 3, . . . , λn

(11.4.5)

in the Singular Value Expansion (SVE) ∞  1 uT ,n ψn (x), x ∈ (0, ) F (x) = σn n=1

(11.4.6)

406

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

vanish, where uT ,n := (uT , ψm )L2 (0,) is the nth Fourier coefficient of the final time  output uT (x) and {λn , ψn (x)}∞ n=1 is the eigensystem of the operator −u (x) subject to the boundary conditions given in (11.4.3). As a consequence of this, the Picard criterion ∞  u2T ,n n=1

σn2

0 and the other two main parameters: the final time T > 0 and the temporal load G(t). To summarize very briefly, the damping term μut plays almost the same role as the regularization term α F 2 in the Tikhonov functional J (u) = (1/2) F − uT 2 + α F 2 , in terms of ensuring the uniqueness of the solution, as the detailed analysis carried out in the following sections show.

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

407

11.4.1 The Singular Value Decomposition and Sufficient Condition for the Uniqueness Let us define the set of admissible spatial loads F = {F ∈ L2 (0, ) : F L2 (0,) ≤ γF , γF > 0}.

(11.4.9)

Denote by u(x, t; F ) the weak solution of the direct problem (11.4.1) corresponding to given F ∈ F and introduce the input-output operator: (F )(x) := u(x, T ; F ),  : F ⊂ L2 (0, ) → L2 (0, ).

(11.4.10)

The we can reformulate the inverse problem (11.4.1)–(11.4.2) in terms of the operator equation: F = uT , F ∈ F , uT ∈ L2 (0, ).

(11.4.11)

Lemma 11.4.1 Assume that the inputs in (11.4.1) satisfy the basic conditions: 

r ∈ L∞ (0, ), 0 < r0 ≤ r(x) ≤ r1 , ρ, μ > 0, F ∈ L2 (0, ), F (x) ≡ 0, G ∈ L2 (0, T ), G(t) > 0.

(11.4.12)

Then the input-output operator  : F ⊂ L2 (0, ) → L2 (0, ) introduced in (11.4.10) is a linear compact operator. Proof Let {F (m) } ⊂ F , m = 1, 2, 3, . . . , be a sequence of inputs. Denote by {u(m) (x, t)}, u(m) (x, t) := u(x, t; F (m) ), the sequence of corresponding weak solutions u(m) ∈ L2 (0, T ; V 2 (0, )), where V 2 (0, ) := {v ∈ H 2 (0, ) : v(0) = v() = 0}, of the direct problem (11.4.1). Then {(F (m) )(x)}, (F (m) )(t) = u(m) (x, T ), is the sequence of outputs. From the second inequality of (11.1.10) it follows that

ux (·, T ) 2L2 (0,) ≤

1 2  uxx (·, T ) 2L2 (0,), 2

which in turns, means that the norms u(·, T ) V 2 (0,) and uxx (·, T ) L2 (0,) are equivalent. The third estimate (C.4.3) in Corollary C.4.1 of Appendix C applied to the solution of the direct problem (11.4.1) with the input F (m) (x) implies that

u(m) (·, T ) 2V 2 (0,) ≤ C˜ 2 F (m) 2L2 (0,) G 2L2 (0,T ) .

(11.4.13)

408

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

The norms F (m) 2L2 (0,), m = 1, 2, 3, . . . , are uniformly bounded by the definition (11.4.9), which means that the sequence of norms {u(m) (x, T )} is uniformly bounded in the norm of the subspace V 2 (0, ) ⊂ H 2 (0, ), hence compact in L2 (0, ). Thus, the input-output map [·] : F → L2 (0, ) transforms a bounded in F ⊂ L2 (0, ) set to a relatively compact set in L2 (0, ). This implies the desired assertion.   Compactness of the input-output operator implies that the inverse problem (11.4.1)–(11.4.2) is ill-posed. In fact, the input-output operator is compact not only as an operator from L2 (0, ) to L2 (0, ), but also as an operator from L2 (0, ) to H 1 (0, ), as the inequality (11.4.13) shows. In particular, this means that if we deal with more regular measured output uT (x) from H 1 (0, ), the problem will still be ill-posed. Consider now the following eigenvalue problem 

(Bw)(x) := (r(x)w (x)) = λw(x), x ∈ (0, ), w(0) = wxx (0) = 0, w() = wxx () = 0,

(11.4.14)

associated with the Euler-Bernoulli operator (Bw)(x) defined on D(B) = {v ∈ V 2 (0, ) ∩ H 4 (0, ) : vxx (0) = vxx (l) = 0}, that is B : D(B) → L2 (0, ). Evidently, this operator is self-adjoint, as it follows from the identity (Bw, v)L2 (0,) − (w, Bv)L2 (0,) =

   x=  r(x)w v − r(x)w v  + r(x)w v  − w r(x)v  , x=0

for all w, v ∈ D(B). Furthermore, there exist eigenfunctions {ψn }∞ n=1 , 

(Bψn )(x) = λn ψn (x), x ∈ (0, ), ψn (0) = ψn (0) = ψn () = ψn () = 0,

(11.4.15)

corresponding to the eigenvalues {λn }∞ n=1 , 0 < λ1 < λ2 . . . , with the asymptotic property O(n4 ) [10, 11]. In addition, the system {ψn }∞ n=1 forms an orthonormal basis for L2 (0, ). Using this basis, we can write the Fourier series expansion w(x) =

∞ 

wn ψn (x) , wn := (w, ψn )L2 (0,)

(11.4.16)

n=1

for the weak solution w ∈ V 2 (0, ) of problem (11.4.14). We prove that in fact the series (11.4.16) converges not only in L2 (0, ), but also in V 2 (0, ) ⊂ H 2 (0, ).

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

409

Lemma 11.4.2 Let the basic conditions (11.4.12) hold. Suppose that {λn , ψn }∞ n=1 is the eigensystem defined Then for any w ∈ V 2 (0, ), there exists an √ above. 2 orthonormal basis {ψn / λn }∞ n=1 for V (0, ). Furthermore, the series (11.4.16) converges in V 2 (0, ). In addition, the principal eigenvalue λ1 > 0 of the Euler-Bernoulli operator can be defined through the norm (B[w, v])1/2 defined the symmetric bilinear form 



B[w, v] :=

r(x)w (x)v  (x)dx, w, v ∈ V 2 (0, ),

(11.4.17)

0

associated with this operator, as follows: λ1 = min{B[v, v] : v ∈ V 2 (0, ), v L2 (0,) = 1}.

(11.4.18)

Proof Let w ∈ V 2 (0, ). Then we can write the Fourier series expansion (11.4.16) 2 as {ψn }∞ n=1 is an orthonormal basis for L (0, ). Furthermore, from (11.4.14) and (11.4.17) it follows that B[ψn , ψm ] = λn (ψn , ψm )L2 (0,) ≡ λn δn,m , n, m = 1, ∞,

(11.4.19)

where δn,m is the Kronecker symbol. Rewriting this as   ψn ψm B √ ,√ = δn,m , n, m = 1, ∞, λn λn √ 2 we find that {ψn / λn }∞ n=0 is an orthonormal subset of V (0, ) endowed with the new inner product defined in√(11.4.17). Next, we prove that {ψn / λn }∞ is in fact an orthonormal basis of V 2 (0, ). To n=0√ this end, we need to show that B[ψn / λn , w] = 0, for all n = 1, 2, 3, . . . , implies w ≡ 0. But this assertion is evidently holds since    ψn B √ , w = λn (ψn , w)L2 (0,), λn by (11.4.19), and the conditions (ψn , w)L2 (0,) = 0, for all n = 1, 2, 3, . . . √ ∞ 2 imply w(x) ≡ 0, as {ψn }∞ n=0 is a basis for L (0, ). Thus, {ψn / λn }n=0 is an 2 orthonormal basis of V (0, ) and, as a consequence, the series   ψn ψn w n √ , w n := B w, √ λn λn n=1

∞ 

410

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

2 converges √ in V (0, ). Comparing this series with the series (11.4.16) we 2find that w n = λn wn . This means that series (11.4.16) in fact converges also in V (0, ). To prove the last assertion of the theorem, we employ the Fourier series

˚= w

∞ 

˚n ψn , w ˚n = (˚ w , ψn )L2 (0,) w

n=1

˚ ∈ V 2 (0, ) with unit norm ˚ of the element w w L2 (0,) = 1. Then, by Parseval’s equality, ∞ 

˚n2 = ˚ w w 2L2 (0,) = 1.

n=1

˚] = λn ˚ In view of B[˚ w,w w 2L2 (0,) = λn we deduce that ˚] = B[˚ w,w

∞  n=1

˚n2 λn ≥ λ1 w

∞ 

˚n2 = λ1 . w

n=1

 

This leads to (11.4.18).

Remark 11.4.1 Consider the case when the flexural rigidity in (11.4.12) is a constant: r(x) ≡ r = constant > 0. One can prove that the eigensystem {λn , ψn }∞ n=1 of the Euler-Bernoulli operator (Bw)(x) defined on D(B) is given by (see, for instance [127]) λn =

 π4 r 4 πnx n , ψn (x) = 2/ sin , x ∈ (0, ), n = 1, 2, , . . . . 4  ρ 

The left formula also shows the dependence of the eigenvalues λn on the physical and geometrical parameters of the Euler-Bernoulli beam. Thus, λn (r1 , ρ) < λn (r2 , ρ), if r1 < r2 , and λn (r, ρ1 ) > λn (r, ρ2 ), if ρ1 < ρ2 , that is increase of the flexural rigidity r > 0 of a homogeneous beam leads to increase of the values of the eigenvalues, and, conversely, increase of the density ρ > 0 of the beam leads to decrease of the values of the eigenvalues.  Theorem 11.4.1 Let the basic conditions (11.4.12) hold. Then the input-output operator  introduced in (11.4.10) is self-adjoint. Furthermore, (ψn )(x) = σn ψn (x),

(11.4.20)

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

411

that is, {σn , ψn }∞ n=1 is the eigensystem of this operator, where 1 σn = ωn



T

e−μ(T −t )/2 sin(ωn (T − t)) G(t)dt,

0

ωn =  σn∗ =

T 0

1 2



4λn

− μ2 ,

√ 0 < μ < 2 λn ,

 (T − t)e−μ(T −t )/2G(t)dt, μ = 2 λn∗ ,

σn =

1 2 ωn



T

0

(11.4.22)

 ωn (T −t ) ωn (T −t ) G(t)dt, e−μ(T −t )/2 e − e− √ μ > 2 λn ,  ωn =

1 2



(11.4.21)

μ2

(11.4.23)

− 4λn ,

while {λn , ψn }∞ n=1 is the eigensystem of the Euler-Bernoulli operator (Bw)(x) defined on D(B). Proof We use the Fourier series expansion u(x, t) =

∞ 

un (t)ψn (x), un (t) := (u(·, t), ψn )L2 (0,) ,

(11.4.24)

n=1

for the weak solution u ∈ L2 (0, T ; V 2(0, )) of the initial boundary value problem (11.4.1) and then take the L2 -product between Eq. (11.4.1) and ψn (x). In view of (11.4.14) we arrive at the problem: 

(ut t (t), ψn ) + μ(ut (t), ψn ) + λn (u(t), ψn ) = (F, ψn ) G(t), (u(0), ψn ) = (ut (0), ψn ) = 0.

This implies that 

un (t) + μun (t) + λn un (t) = Fn G(t), un (0) = un (0) = 0,

(11.4.25)

for each n = 1, 2, 3, . . . . The roots of the characteristic equation associated with (11.4.25) are β1,2 = (−μ ±

' μ2 − 4λn )/2.

412

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

There are three possible cases depending on the discriminant √ μ2 − 4λn √ the sign of√ and determined by the conditions μ < 2 λn , μ = 2 λn and μ > 2 λn . The solutions of the Cauchy problem (11.4.25) corresponding to these cases are Fn un (t) = ωn



t

e−μ(t −τ )/2 sin(ωn (t − τ )) G(τ )dτ, t ∈ [0, T ],  √ ωn = 12 4λn − μ2 , μ < 2 λn , $t √ un (t) = Fn 0 (t − τ ) e−μ(t −τ )/2G(τ )dτ, t ∈ [0, T ], μ = 2 λn ,  t  Fn ωn (t −τ ) ωn (t −τ ) e−μ(t −τ )/2 e − e− G(τ )dτ, un (t) = 2 ωn 0  √ μ > 2 λn ,  ωn = 12 μ2 − 4λn . 0

(11.4.26)

Substituting t = T in (11.4.24) and (11.4.26) we arrive at the (formal) Fourier series expansion of the input-output operator: (F )(x) =

∞ 

σn Fn ψn (x), Fn := (F, ψn )L2 (0,) ,

(11.4.27)

n=1

where σn , n = 1, 2, 3, . . . are defined by expressions (11.4.21) to (11.4.23) Evidently ψn are eigenfunctions of the operator  corresponding to the eigenvalues σn , n = 1, ∞. Indeed, substituting F (x) = ψm (x) in (11.4.27) we get: (ψm )(x) =

∞ 

σn (ψm (x), ψn (x)) ψn (x) = σm ψm (x).

n=1

Furthermore,  (F, F˜ )L2 (0,) :=

∞  n=1

=

∞ 

σn Fn ψn (x),

∞ 

 F˜m ψm (x)

m=1

σn Fn F˜n = (F, F˜ ) ∀ F, F˜ ∈ L2 (0, ),

n=1

where F˜n := (F˜ , ψn )L2 (0,). Hence, the input-output operator is self-adjoint, that is  = ∗ , where ∗ : L2 (0, ) → L2 (0, ) is the adjoint operator. This completes the proof of the theorem.   √ √ √ Remark 11.4.2 The cases μ < 2 λn , μ = 2 λn and μ > 2 λn , specified in the above theorem, correspond to underdamped, critically damped and overdamped vibrating systems, according to the commonly accepted classification [127].

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

413

μ = It should be emphasized that only one term σn∗ associated with the case  2 λn∗ , specified in (11.4.22), can appear in the expansion (11.4.27). If μ = 2 λn∗ and √ n∗ > 1, then the terms σn , n = 1, 2, , .. n∗ − 1 associated with the case μ > 2 λn and defined by (11.4.23), appear in the expansion (11.4.27), due to the fact ∞ that the sequence of eigenvalues n }n=1 increases monotonically as n → ∞. √ {λ√ Finally, the case μ ∈ (2 λm , 2 λm+1 ) means that the terms σ1 , . . . , σm in the expansion (11.4.27) are defined by (11.4.23).  Formula (11.4.21) shows that even the positivity G(t) > 0 of the temporal load can not guarantee the positivity σn > 0 for all n = 1, 2, , . . . of the singular values, which means that the singular value expansion (11.4.27) is still formal. Recall that Picard criterion (11.4.7) also requires the positivity of the singular values. These considerations suggest that before proceeding with the solution of the inverse problem (11.4.1)–(11.4.2), by any method, one needs, first of all, to obtain a sufficient condition for the positivity of the singular values defined in (11.4.21) to (11.4.23). √ Let us consider first the most common situation μ √ < 2 λ1 . This condition guarantees the fulfilment of the conditions μ < 2 λn for all n > 1, since 0 < λ1 ≤ λ2 ≤ λ3 < . . . . Theorem 11.4.2 Let the conditions (11.4.12) and √ (11.4.13) hold. Assume that the damping coefficient satisfies the condition μ < 2 λ1 , where λ1 > 0 is the principal eigenvalue of the Euler-Bernoulli operator. Suppose that the damping coefficient, final time and the temporal load satisfy the following inequality: 

1/2 G(T ) > G(0)e−μT /2 + (1 − e−μT )/μ

G L2 (0,T ) 

 2 −1/2 . × 1 − μ/(2 λ1 )

(11.4.28)

Then the singular value expansion (11.4.27) of the solution to the inverse problem (11.4.1)–(11.4.2) is unique. Proof We transform the integral in (11.4.21) as follows. First we use the integration by parts formula to get 1 σn = ωn =



T

e−μt /2 sin(ωn t)G(T − t) dt

0

1  G(T ) − G(0)e−μT /2 cos(ωn T ) ωn2 

T

− 0

e−μt /2 cos(ωn t)

μ 2

  G(T − t) + G (T − t) dt .

414

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Then by grouping terms we obtain: σn =

1 ωn2 

   μ G(T ) − G(0)e−μT /2 cos(ωn T ) + sin(ωn T ) 2ωn T

− 0

 2  μ μ  e−μt /2 sin(ωn t) G(T − t) + G (T − t) 4ωn 2ωn . + cos(ωn t)G (T − t) dt .

After rearranging the terms, we deduce that    1 μ −μT /2 σn = 2 G(0) cos(ωn T ) + sin(ωn T ) G(T ) − e ωn + (μ/2)2 2ωn 

T

−

e−μt /2

0



  μ sin(ωn t) + cos(ωn t) G (T − t) dt . 2ωn

Use here the inequality |a cos α + b sin α| ≤







T

e 0

−μτ/2



√ a 2 + b 2 with the estimate



μ  sin(ωn τ ) + cos(ωn τ ) G (T − τ ) dτ

2ωn  ≤

1 − e−μT μ

1/2

! 



G L2 (0,T ) 1 +

μ 2ωn

2 "1/2 .

Taking into account also the relations ωn2 + (μ/2)2 = λn ,  

 2 −1 μ 2 1+ = 1 − μ/(2 λn ) , 2ωn we obtain σn ≥

 

1/2 1 G(T ) − G(0) e−μT /2 + (1 − e−μT )/μ

G L2 (0,T ) λn  

 2 −1/2 × 1 − μ/(2 λ1 ) .

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

415

This inequality with the condition (11.4.28) shows that all singular values σn defined in (11.4.21) are positive. This proves the uniqueness of the SVE given by (11.4.6), as desired.   Remark 11.4.3 For G(t) > 0, the positivity of √ the singular values in the critically damped μ = 2 λn∗ , and overdamped μ > 2 λn cases is obvious for any T > 0, as it follows from expressions (11.4.22) and (11.4.23).  Note that Theorem 11.4.2 not only proves the uniqueness of the solution, but also gives a specific relationship between the main parameters of the problem, at which the uniqueness of the solution to the inverse problem is achieved. Remark 11.4.4 Consider the case of pure spatial load, that is, G(t) = 1 in (11.4.1). Then, inequality (11.4.28) holds for all large enough values of the final time T > 0.  √ Remark 11.4.5 Consider the case μ = 2λ1 . Then 

μ2 1+ 4λ1 − μ2



1 − e−μT μ

1/2 =

1/2

√ √ 2, e−μT /2 = e−T λ1 /2 ,

1 < √ . μ

In this case the inequality (11.4.28) is valid for all large enough values of T > 0, and G(T ) > G L2 (0,T ) (2/λ1 )1/4 . The latter inequality holds if, for example, G(t) = exp(αt) with large enough α > 0.  It should be emphasized that, having only the positivity of the singular values, we can not guarantee the convergence of the SVE given by (11.4.6), as the Picard criterion (11.4.7) shows. Indeed, from formula (11.4.21) it follows that 1 0 < σn < ωn



T

1/2  2

T

sin (ωn t)dt 0

1/2 2

G (T − t) dt

0

≤ 

√ T λn − μ2 /(2λn )

G L2 (0,T ) .

416

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

With the asymptotic property λn ∼ O(n4 ) this implies that the singular values σn , n = 1, 2, 3, . . . have the asymptotic property O(n−2 ). As a consequence of this and the Picard criterion (11.4.7), we deduce that the series (11.4.6) converges if and only if ∞ 

n4 u2T ,n < ∞.

(11.4.29)

n=1

Based on characterization of Sobolev spaces by Fourier transform [38], we conclude that (11.4.29) holds, if the measured output uT (x) satisfies the following regularity and consistency conditions: uT ∈ H 2 (0, ), uT (0) = uT () = 0.

(11.4.30)

Theorem 11.4.3 Let the conditions of Theorem 11.4.2 hold. Assume, in addition, that the measured output uT (x) satisfies the conditions (11.4.30). Then the inverse problem (11.4.1)–(11.4.2) has a unique solution. Furthermore, this solution possesses the convergent SVE given by (11.4.6), with the singular values defined in (11.4.21).

11.4.2 Application to Forced Vibration Under Pure Spatial Load Assume that G(t) ≡ 1 in (11.4.1), which corresponds to pure spatial loading case. Using the analysis given in the previous subsection, we will derive here the SVE of the solution to the inverse problem of recovering the unknown spatial load F (x) in ⎧⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ρut t + μut + (r(x)uxx )xx = F (x), (x, t) ∈ T , ⎪ ⎪ ⎨ u(x, 0) = u (x, 0) = 0, x ∈ (0, ), t ⎪ (11.4.31) ⎪ ⎩ ⎪ (0, t) = u(, t) = u (, t) = 0, t ∈ [0, T ], [2pt]u(0, t) = u ⎪ xx xx ⎪ ⎪ ⎩ uT (x) := u(x, T ), x ∈ (0, ). Calculating the integral (11.4.21) with G(t) ≡ 1, we find the singular values corresponding to the underdamped case as follows: σn =

1 λn

    μ 1 − cos(T ωn ) + sin(T ωn ) e−μT /2 , 2ωn  √ ωn = 4λn − μ2 /2, μ < 2 λ1 .

(11.4.32)

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

417

In the same way, the formulae σn∗ =

    1  1 − 1 + T λn∗ e−μT /2 , μ = 2 λn∗ . λn∗

1 σn = λn

  μ/2 +  ωn −(μ/2− ωn )T e 1− 2 ωn  μ/2 −  ωn −(μ/2+ ωn )T − e , 2 ωn  √  ωn = μ2 − 4λn /2, μ > 2 λn ,

(11.4.33)

(11.4.34)

for the singular values corresponding to critically damped and overdamped vibrating systems are obtained from (11.4.22) and (11.4.23), with G(t) ≡ 1. As noted in Remark 11.4.3, for G(t) > 0, the integrals in (11.4.22) and (11.4.23) are positive, which means the singular values corresponding to the critically damped and overdamped vibrating systems and defined by formulas (11.4.33) and (11.4.34) are positive. Therefore, we will investigate the issues related to the positivity of the singular values √ and the uniqueness of the solution only for the underdamped case, when μ < 2 λ1 . Lemma 11.4.3 Let conditions (11.4.12) hold. Assume that the damping coefficient μ > 0 and the final time T > 0 satisfy the following conditions  μ < 2 λ1 , T >

μ =: T∗ , 4λ1 − μ2

(11.4.35)

where λ1 > 0 is the principal eigenvalue of the Euler-Bernoulli operator. Then the singular values defined by the formula (11.4.32) are positive. Proof Formula (11.4.32) implies that if 

 μ sin(T ωn ) e−μT /2 < 1, gσ (n; μ, T ) := cos(T ωn ) + 2ωn

(11.4.36)

for all n = 1, 2, 3, . . . , then √ the singular values are positive. In view of the inequality |a cos α + b sin α| ≤ a 2 + b2 and the relation 4ωn2 = 4λn − μ2 , we deduce that the above conditions hold, if eμT > 1 +

μ2 := σ (λn ), n = 1, 2, 3, . . . . 4λn − μ2

418

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

The function σ (λn ) decreases monotonically as λn > 0 increases, hence all of the above conditions are met under the following condition:   μ2 1 ln 1 + λ1 . T > , μ < 2 μ 4λ1 − μ2 With the inequality ln(1 + x) < x, x > 0, this leads to the required sufficient conditions in (11.4.35).   Thus, Theorem 11.4.3 can be formulated for this special case as follows. Theorem 11.4.4 Let the conditions (11.4.13) and (11.4.35) hold. Assume that the measured output uT (x) satisfies the regularity and consistency conditions (11.4.30). Then the inverse problem (11.4.31) has a unique solution F ∈ L2 (0, ). Furthermore, this solution possesses the convergent singular value expansion given by (11.4.6), with the singular values defined in (11.4.32). Let us analyze what does the second condition of (11.4.35) means in sense of admissible lower limit T∗ > 0 of the final time T > 0. The first conclusion is that, for near zero values of the damping coefficient μ > 0, the corresponding value of the lower limit is of the same order. Namely, in the case r(x) ≡ 1 and  = π considered in Remark 11.4.1, λn = n4 and the principal eigenvalue is λ1 = 1. With formula T∗ = μ/(4λ1 − μ2 ), this implies that for μ = 10−2 , the admissible lower limit is T∗ = 0.25 × 10−2. Hence, in this case, no condition other than the positivity is imposed in (11.4.35) on the final time T > 0. Furthermore, for the most critical √ case, when the difference between the values μ and 2 λ1 is of the order 10−1 , the corresponding value of the admissible lower limit T∗ is reasonable: T∗ ≈ 4.9. This means that any reasonable value of the final time satisfying the condition T ≥ 5 can be taken in order to generate the final time measured output uT (x). Behaviour of the function gσ (n; μ, T ), introducing in (11.4.36), and the singular values σn , defined by formula (11.4.32), with λn = n2 , n = 1, 2, 3, . . . , are plotted in Fig. 11.11 on the left and right. To illustrate the effect of the damping coefficient, the behavior of these functions is considered for two values μ = 0.1 (blue, high amplitude line) and μ = 1 (purple, √ low amplitude line) of the damping coefficient satisfying the conditions 0 < μ < 2 λ1 , with λ1 = 1. The figure on the left clearly shows that by increasing the value of the damping coefficient, the amplitude of the function gσ (n; μ, T ) decreases, ensuring that the maximum value of gσ (n; μ, T ) is even less than one. Note that the case gσ (n; μ, T ) = 1 corresponds to the zero singular value σn = 0, which, in turn, leads to the non-uniqueness of the solution. Furthermore, plots in Fig. 11.11 on the right show how the oscillation is drastically reduced when the effect of the damping coefficient increases. 

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

419

Fig. 11.11 Behaviour of the function gσ (n; μ, T ) introduced in (11.4.36) (the figure on the left) and the singular values σn defined by formula (11.4.32) (the figure on the right), depending on n = 1, 2, 3, . . . , for the values μ = 0.1 (blue, high amplitude line) and μ = 1 (purple, low amplitude line) of the damping parameter, with T = 4

 Consider now the critically damped case, when μ = 2 λn∗ . From formula (11.4.33) it follows that the sufficient condition for the positivity of the singular value σn∗ is eT

√

λn∗

>1+T



λn∗ .

Since the inequality exp(y) > 1 + y is true for all y > 0, the above inequality holds for all T > 0.  Therefore, the singular value σn∗ corresponding to the critically damped case μ = 2 λn∗ is positive for all T > 0. The positivity of the singular values given by formula (11.4.34) and corresponding to the overdamped case, is obvious.

11.4.3 Application to Forced Vibration Under Harmonic Load This subsection will focus on the most important class of temporal load, called harmonic load, when G(t) = cos(ωt) in (11.4.1). The parameter ω > 0 is the frequency of the applied temporal load. This leads to the problem of recovering the unknown spatial load F (x) in ⎧⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ρut t + μut + (r(x)uxx )xx = F (x) cos(ωt), (x, t) ∈ T , ⎪ ⎪ ⎨ u(x, 0) = u (x, 0) = 0, x ∈ (0, ), t ⎪ ⎪ ⎩ ⎪ u(0, t) = uxx (0, t) = u(, t) = uxx (, t) = 0, t ∈ [0, T ], ⎪ ⎪ ⎪ ⎩ uT (x) := u(x, T ), x ∈ (0, ).

(11.4.37)

420

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

√ Consider the underdamped case: μ < 2 λ1 . Substituting G(t) = cos(ωt) in (11.4.21) and calculating this integral, we find the following formula for the singular values:  1 2 2μω sin(ωT ) + (λ σn = − ω ) cos(ωT ) n (λn − ω2 )2 + μ2 ω2    μ(λn + ω2 ) sin(ωn T ) + (λn − ω2 ) cos(ωn T ) e−μT /2 , (11.4.38) − 2ωn ' √ 1 4λn − μ2 , n = 1, 2, 3, . . . . μ < 2 λ1 , ωn = 2 Lemma 11.4.4 Assume that the damping coefficient μ > 0, the frequency ω > 0 of the applied harmonic load G(t) = cos(ωt) and the final time T > 0 satisfy the following conditions   μ < 2 λ1 , 0 < ω < λ1 , ωT < π/4, (11.4.39) where λ1 > 0 is the principal eigenvalue of the Euler-Bernoulli operator. Denote by ω∗ = ω∗ (λ1 , μ) the root of the equation  eμπ/(4ω) = 2 1 +

 μ2 (λ1 + ω2 )2 , (4λ1 − μ2 )(λ1 − ω2 )2

(11.4.40)

and let T∗ =

π π , T∗ = . 4ω∗ (λ1 , μ) 4ω

(11.4.41)

Then for all of ω ∈ (0, ω∗ ) and T ∈ (T∗ , T ∗ ), the singular values defined by formula (11.4.38) are positive. Proof Rewrite formula (11.4.38) as follows: σn =

  λn − ω2 2μω sin(ωT ) + cos(ωT ) (λn − ω2 )2 + μ2 ω2 λn − ω2 ! "  μ(λn + ω2 ) −μT /2 sin(ωn T ) + cos(ωn T ) e −  . 4λn − μ2 (λn − ω2 )

(11.4.42)

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

421

By the third condition of (11.4.39) the sum in the first right-hand-side square bracket in √ formula (11.4.42) is positive. Then, using the inequality |a sin α + b cos α| ≤ a 2 + b2 for the sum in the second right-hand-side square bracket we deduce that  λn − ω2 σn ≥ cos(ωT ) (λn − ω2 )2 + μ2 ω2 1/2  (11.4.43)  μ2 (λn + ω2 )2 −μT /2 e . − 1+ (4λn − μ2 )(λn − ω2 )2 √ Since cos(ωT ) > 1/ 2, by the condition ωT < π/4, the right-hand-side of the inequality (11.4.43) is positive for all n = 1, 2, 3, . . . , if 1/2  μ2 (λn + ω2 )2 1 e−μT /2 , n = 1, 2, 3, . . . . √ > 1+ (4λn − μ2 )(λn − ω2 )2 2 This yields:  eμT > 2 1 +

 μ2 (λn + ω2 )2 =: (λn , ω), (4λn − μ2 )(λn − ω2 )2

(11.4.44)

for all n = 1, 2, 3, . . . . Evidently, (λn , ω) is a monotone decreasing function of λn . This implies that the condition  eμT > 2 1 +

 μ2 (λ1 + ω2 )2 =: (λ1 , ω) (4λ1 − μ2 )(λ1 − ω2 )2

(11.4.45)

ensures the fulfillment of the condition (11.4.44) for all n = 1, 2, 3, . . . . Consider now √ Eq. (11.4.40) which right-hand-side√is the function (λ1 , ω), defined in (0, λ1 ). We prove that for all μ ∈ (0, 2 √ λ1 ) there exists a unique root ω∗ = ω∗ (λ1 , μ) of this equation in the interval (0, λ1 ). Indeed, for a given √ λ1 > 0, (λ1 , ω)√is a monotone increasing function of ω ∈ (0, λ1 ), and tends to infinity as ω → λ1 , as (11.4.45) shows. On the other hand, the left-hand-side of equation√(11.4.40), i.e. the function eμπ/(4ω) , is a monotone decreasing function of ω ∈ (0, λ1 ) and tends to infinity √ as ω → 0. This implies that Eq. (11.4.40) has a unique root ω∗ = ω∗ (λ1 , μ) in (0, λ1 ). The above considerations lead to the following relations: ∗

eμT ≡ eμπ/(4ω) > eμπ/(4ω ) ≡ (λ1 , ω∗ )  >2 1+

 μ2 (λ1 + ω2 )2 , (4λ1 − μ2 )(λ1 − ω2 )2

422

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

for all ω ∈ (0, ω∗ (λ1 , μ)) and T ∈ (T∗ , T ∗ ), with T∗ , T ∗ > 0 defined in (11.4.41). In particular, this means the fulfillment of the condition (11.4.45). With the inequality (11.4.43) this leads to the positivity of the singular values for all n = 1, 2, 3, . . . .   The version Theorem 11.4.3 for the inverse problem (11.4.37) with time harmonic temporal load can be formulated as follows. Theorem 11.4.5 Assume that conditions (11.4.39) hold and the measured output uT (x) satisfies the regularity and consistency conditions (11.4.30). Then the inverse problem (11.4.37) has a unique solution F ∈ L2 (0, ). Furthermore, this solution possesses the convergent singular value expansion given by (11.4.6), with the singular values defined in (11.4.38). Example 11.4.1 The admissible intervals for the values of the frequency of the applied temporal load ω > 0 and the final time T > 0 for the inverse problem (11.4.37) related to simply supported Euler-Bernoulli beam with unit flexural rigidity and mass density r = ρ = 1, and with the length  = π. The formula λn = π 4 rn4 /(4 ρ) given in Remark 11.4.1 suggests that in the considered case λn = n4 which in√particular means the principal eigenvalue is λ1 = 1, and hence the condition μ < 2 λ1 for the underdamped case means μ ∈ (0, 2). Table 11.2 illustrates the admissible upper and lower limits of ω > 0 and T > 0 corresponding to some reasonable values of the damping parameter μ ∈ (0, 2) given on the first row of the table. The upper limits of ω > 0, found as a root of Eq. (11.4.40), and the final time are given on the second and third rows of the table, respectively. The lower limits of ω > 0 on the fourth row of the table are set, for ease of calculations, just half 0.5ω∗ of the upper limits. The values of the lower, T∗ > 0 and the upper, T ∗ > T∗ limits of the final time, given on the third and fifth rows of the table show that they also differ by about two times. This allows us to select any value of the final time from the interval (T∗ , T ∗ ] with sufficient length, to generate the measured output uT (x) := u(x, T ).  Table 11.2 Lower and upper limits T∗ √and T ∗ of admissible values T ∈ (T∗ , T ∗ ] of the final time corresponding to the root ω∗ ∈ (0, λ1 ) of Eq. (11.4.40) with λ1 = 1 in the inverse problem (11.4.37) with r = ρ = 1 and  = π μ ω∗ = ω∗ (λ1 , μ) T∗ = π/(4ω∗ ) ω∗ = ω∗ /2 T ∗ = π/(4ω∗ )

0.1 0.113 6.950 0.057 13.780

0.5 0.465 1.689 0.233 3.371

1.0 0.543 1.446 0.271 2.910

1.2 0.539 1.457 0.268 3.931

1.5 0.517 1.519 0.208 3.776

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

423

11.4.4 Adjoint Method Based on the Quasi-Solution Approach From the above analysis given in the previous subsections, it follows that the main merit of the approach based on the singular value decomposition of the input-output operator introduced in (11.4.10), is that it allows us to find the relationship between the input parameters: the final time, the damping coefficient and the temporal load, in which the inverse problem (11.4.1)–(11.4.2) has a unique solution. Of course, it provides a simple computational algorithm for recovering the unknown spatial load according to the SVE (11.4.6) and formulas (11.4.21) to (11.4.23) for the singular values. However, due to the rapid decay O(n−2 ) of the singular values σn , in fact, only a few Fourier coefficients uT ,n of the final time output uT (x) can be used in the series (11.4.6) to recover F (x). Moreover, if the output uT (x) contains a low random noise of, say, 10−2 , in the 3th or 4th terms of the series (11.4.6) this error is amplified by the factor of the order 10! The second drawback of this method is that it requires very smooth final time output as the conditions (11.4.30) show, which is not always the case in applications. For non-smooth final time measured output uT ∈ L2 (0, T ), of course, the convergence of the functional series (11.4.6) cannot be guaranteed. The theory based on the quasi-solution approach combined with the adjoint problem method and developed in previous sections does not contain these shortcomings and restrictions. In this subsection, this approach will be employed for the inverse problem (11.4.1)–(11.4.2), in particular, to obtain the existence of a solution, as well as the Fréchet gradient of the corresponding Tikhonov functional J (F ) =

1

F − uT 2L2 (0,), F ∈ F . 2

(11.4.46)

We reformulate the inverse problem (11.4.1)–(11.4.2) as the minimization problem J (F∗ ) = inf J (F ) F ∈F

(11.4.47)

for the Tikhonov functional. A solution of this problem is a quasi-solution of the inverse problem. Lemma 11.4.5 Let conditions (11.4.12) hold and uT ∈ L2 (0, ). Then the Tikhonov functional (11.4.46) is Lipschitz continuous, that is |J (F1 ) − J (F2 )| ≤ LJ F1 − F2 L2 (0,), ∀F1 , F2 ∈ F , where LJ =

√ √ 2T C1 2T C1 γF + uT L2 (0,) G L2 (0,T ) ,

and C1 > 0 is the constant defined in Corollary C.4.1.

(11.4.48)

424

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Proof We employ the inequality |J (F1 ) − J (F2 )| ≤

1

F1 L2 (0,) + F2 L2 (0,) + 2 uT L2 (0,) 2 × F1 − F2 L2 (0,), (11.4.49)

as in proof of Corollary 10.1.6 of Sect. 10.1, to estimate the difference |J (F1 ) − J (F2 )|. By the definition,

F1 − F2 L2 (0,) = δu(·, T ) L2 (0,),

(11.4.50)

where δu(x, t) := u(x, t; F1 ) − u(x, t; F2 ) solves the following problem ⎧ ⎪ ⎨ ρδut t + μδut + (r(x)δuxx )xx = δF (x)G(t), (x, t) ∈ T , (11.4.51) δu(x, 0) = δut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ δu(0, t) = δuxx (0, t) = 0, δu(, t) = δuxx (, t) = 0, t ∈ [0, T ]. To estimate the norm δu(·, T ) L2 (0,) in (11.4.50), we use the estimate (C.4.7) in Corollary C.4.1 applied to the weak solution of problem (11.4.51). This yields:

δu(·, T ) L2 (0,) ≤ C3 G L2 (0,T ) δF L2 (0,), C3 =

√ 2T C1 ,

(11.4.52)

that is, the input-output operator defined in (11.4.10) is Lipschitz continuous. Substituting this estimate in (11.4.49) with the estimates

Fi L2 (0,) ≤ C3 Fi L2 (0,) G L2 (0,T ) , ∀Fi ∈ F , i = 1, 2, and taking into account that Fi L2 (0,) ≤ γF , we obtain: |J (F1 ) − J (F2 )|  ≤ C3 C3 γF G L2 (0,T ) + uT L2 (0,) G L2 (0,T ) δF L2 (0,). This leads to (11.4.48).

 

Theorem 11.4.6 Assume that conditions (11.4.12) hold and uT ∈ L2 (0, ). Then the minimization problem (11.4.57) has a solution in the set of admissible spatial loads F , defined in (11.4.9). This theorem is proved exactly in the same way as Theorem 10.1.11 in Sect. 10.1.  The next step is to establish the Fréchet differentiability of the Tikhonov functional, and then to derive the gradient formula.

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

425

Let F, F + δF ∈ F . Denote by δJ (F ) := J (F + δF ) − J (F ) the increment of the Tikhonov functional (11.4.46). Then 



δJ (F ) =

[u(x, T ; F ) − uT (x)] δu(x, T )dx

0





+

(δu(x, T ))2 dx.

(11.4.53)

0

The following lemma establishes the basic relationship between the inputs and outputs of the inverse problem (11.4.1)–(11.4.2). Lemma 11.4.6 Assume that conditions (11.4.12) hold. Suppose F, F + δF ∈ F are two arbitrary inputs. Then the following integral relationship holds: 



−ρ

  T

 q(x)δu(x, T )dx =

0

G(t)φ(x, t)dt δF (x)dx,

0

(11.4.54)

0

where δu ∈ L2 (0, T ; V 2 (0, )) is the weak solution of problem (11.4.51) and φ(x, t) is the solution of the following backward problem: ⎧ ⎪ ⎨ ρφt t − μ(x)φt + (r(x)φxx )xx = 0, (x, t) ∈ T , φ(x, T ) = 0, φt (x, T ) = q(x), x ∈ (0, ), ⎪ ⎩ φ(0, t) = φxx (0, t) = 0, φ(, t) = φxx (, t) = 0,

(11.4.55)

with the arbitrary input q ∈ L2 (0, ). This lemma is proved in the same way as Lemma 11.2.4, and we leave the proof as an exercise for the reader.  Note that the backward problem (11.4.55) is a well-posed problem as the change of the variable t with τ = T − t shows. With the input q(x) defined as q(x) = −

1 [u(x, T ; F ) − uT (x)], x ∈ (0, ), ρ

(11.4.56)

the backward problem (11.4.55) is called the adjoint problem corresponding to the inverse problem (11.4.1)–(11.4.2). Substituting (11.4.56) in the integral relationship (11.4.54) we get: 

 0



T

[u(x, T ; F ) − uT (x)]δu(x, T )dx =





δF (x)G(t)φ(x, t; F )dxdt, 0

0

where φ(x, t; F ) is the solution of the adjoint problem (11.4.55) with the input (11.4.56).

426

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

With the formula (11.4.53) for the first variation of the Tikhonov functional leads to

  T

 δJ (F ) =

G(t)φ(x, t; F )dt δF (x)dx 0

0

+

1 2





(δu(x, T ))2 dx, δF ∈ F .

(11.4.57)

0

Theorem 11.4.7 Assume that conditions (11.4.12) hold. Suppose that F ∈ F and uT ∈ L2 (0, ). Then the Tikhonov functional J (F ) defined on the set of admissible inputs F ⊂ L2 (0, ) is Fréchet differentiable. Furthermore, for the Fréchet gradient of this functional the following gradient formula holds:  ∇J (F )(x) =

T

φ(x, t; F )G(t)dt, F ∈ F , x ∈ (0, ),

(11.4.58)

0

where φ(x, t) = φ(x, t; F ) is the weak solution solutions of the adjoint problem (11.4.55) with the input defined in (11.4.56). Proof

The last right-hand-side integral in formula (11.4.57) is of the order O δF 2L2 (0,) , as estimate (11.4.52) shows. Then the proof follows from the definition of the gradient.   Theorem 11.4.8 Assume that conditions of Theorem 11.4.8 hold. Then the Fréchet gradient defined by formula (11.4.58) is Lipschitz continuous, that is

∇J (F1 ) − ∇J (F2 ) L2 (0,) ≤ L∇ F1 − F2 L2 (0,), F1 , F2 ∈ F ,

(11.4.59)

where L∇ =

1 3/2 2 T C1 Cu G 2L2 (0,T ) ρ

is the Lipschitz constant and C1 , Cu > 0 are the constants introduced in (C.4.3) of Appendix C. Proof From the gradient formula (11.4.58) it follows that   T



∇J (F1 ) −

∇J (F2 ) 2L2 (0,)

=

2 δφ(x, t)G(t)dt

0

dx,

0

≤ G 2L2 (0,T ) δφ 2L2 (0,T ;L2 (0,)), (11.4.60)

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

427

where δφ(x, t) := φ(x, t; F1 ) − φ(x, t; F2 ) is the weak solution of the problem ⎧ ⎪ ⎨ ρδφt t − μ(x)δφt t + (r(x)δφxx )xx = 0, (x, t) ∈ T , (11.4.61) δφ(x, T ) = 0, δφt (x, T ) = δu(x, T )/ρ, x ∈ (0, ), ⎪ ⎩ δφ(0, t) = δφxx (0, t) = 0, δφ(, t) = δφxx (, t) = 0, t ∈ (0, T ), with the input δu(x, T )/ρ. To estimate the norm δφ 2L2 (0,T ;L2 (0,)) in (11.4.60) we use the inequality

δφ 2L2 (0,T ;L2 (0,)) ≤

1 2 T δφt 2L2 (0,T ;L2 (0,)) 2

and then apply estimate (C.4.3) in Theorem C.4.1 of Appendix C to the solution of problem (11.4.61). We have:

δφ 2L2 (0,T ;L2 (0,)) ≤

1 T 2 C12 Cu2 δu(·, T ) 2L2 (0,) . 2ρ 2

Now, we use the trace estimate (C.4.7) in Appendix C, applied to the weak solution of problem (11.4.51). This yields:

δu(·, T ) 2L2 (0,) ≤ 2T C12 δF 2L2 (0,) G 2L2 (0,T ) . Substituting this in the above inequality we deduce that

δφ 2L2 (0,T ;L2 (0,)) ≤

1 3 4 2 T C1 Cu δF 2L2 (0,) G 2L2 (0,T ) . ρ2

With inequality (11.4.60) this leads to the required assertion (11.4.59).

 

11.4.5 Numerical Reconstruction of Unknown Spatial Load from Final Time Output In this subsection, we illustrate the application of the truncated SVD (TSVD) and the conjugate gradient algorithm (CG-algorithm) to the inverse problem (11.4.31) related to the forced vibration under the harmonic load G(t) = cos(ωt). The same version of the CG-algorithm as in Sect. 3.4.2 is used here. In the first part of this subsection, we provide numerical examples to demonstrate the ability of the TSVD algorithm to recover the unknown spatial load from noise free as well as noisy final time output. This algorithm is especially convenient when the eigenvalues of the Euler-Bernoulli operator are known. As noted in Remark 11.4.1, if the flexural rigidity r(x) is constant, then the eigensystem is given

428

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

by formulas defined in Remark 11.4.1. In the numerical examples below, we assume that ρ(x) = 1, μ = 1/2,  = π, T = 3/2, ω = 1/2.

(11.4.62)

Notice that the values of the frequency of the applied temporal load ω and the final time T > 0 given in (11.4.62) are defined according to the Table 11.2. The eigensystem of the Euler-Bernoulli operator with r(x) = 1 subject to the clamped boundary conditions is / λn = n , ψn (x) = 4

2 sin(nx), x ∈ [0, π]. π

(11.4.63)

Since √ λ1 = 1, for the above value of the damping coefficient the condition μ < 2 λn for the underdamped case holds for all n ∈ N+ . Substituting λn = n4 in (11.4.38) we obtain the following formula for the singular values corresponding to the underdamped case: σn =

 1 4 2 − ω ) cos(ωT ) 2μω sin(ωT ) + (n (n4 − ω2 )2 + μ2 ω2    μ(n4 + ω2 ) sin(ωn T ) + (n4 − ω2 ) cos(ωn T ) e−μT /2 , (11.4.64) − 2ωn ' 1 4n4 − μ2 , n = 1, 2, 3, . . . . μ < 2, ωn = 2

Then the TSVD solution of the inverse problem (11.4.31) is Fα,N (x) =

N  q(α; σn ) n=1

σn

uT ,n ψn (x), x ∈ (0, π),

(11.4.65)

where N > 1 is the truncation parameter, / uT ,n =

2 π



π

uT (x) sin(nx)dx, n = 1, 2, 3, . . .

(11.4.66)

0

in the Fourier coefficients of the measured output uT (x), q(α; σ ) =

σ2 σ2 + α

(11.4.67)

is the filter function and α > 0 is the parameter of regularization. Therefore, if the eigensystem is given by (11.4.63), then the TSVD solution of the inverse problem (11.4.31), given by (11.4.65) and also by relations (11.4.64),

11.4 Unique Recovery of an Unknown Spatial Load in Damped Beam. . .

429

Fig. 11.12 Reconstruction of spatial load: by TSVD (left) and by CG-algorithm (right): r(x) = 1

(11.4.66) and (11.4.67), can be computed without the large computational effort involved in a complete TSVD computation. The synthetic noise free output uT (x) := u(x, T ) for the considered inverse problem is generated in the same way as in Sect. 11.2.5, from the numerical solution of the direct problem defined in (11.4.31), assuming that F (x) = x sin(5x) and G(t) = cos(t/2). Having the noise free output, the synthetic noisy output uδT (x), with noise level δ > 0 is generated by using standard MATLAB rand function. The reconstructed spatial loads from noise free and noisy outputs with the noise level δ = 0.05 are shown in Fig. 11.12 on the left, together with the function F (x) = x sin(5x) (exact solution). The worst numerical result is obtained when the regularization parameter is zero: α = 0 (thin dotted line), as expected. In both of the remaining two cases, δ = 0, α = 3 · 10−9  and δ = 0.05, α = 3 · 10−9 , the results of reconstructions corresponding to noise free and noisy outputs, are fairly accurate. Thus, if the eigensystem of the Euler-Bernoulli operator is known, then with a very small values α ∈ (10−9 , 10−8 ) of the regularization parameter, the TSVD solution of the inverse problem can be obtained by formula (11.4.65), taking the values of the truncation parameter N between 6 and 11. This means that the TSVD solution FN (x), corresponding to a sufficiently high accuracy of reconstruction, can be obtained by using only a few terms of the sum (11.4.65), if the optimal value of the regularization parameter α > 0 is found. Thus, if the coefficients of the Euler-Bernoulli equation are constant, then the TSVD algorithm is a simple computational tool. However, in the case of variable coefficients, additional computational work is required to find the eigenvalues numerically, which increases the overall cost of the entire computational algorithm. Furthermore, the numerically found eigenvalues introduce an additional error in the formula (11.4.65). To compare the TSVD and with the CG-algorithm, the above example with the constant flexural rigidity r ≡ 1, was solved by the CG-algorithm. The

430

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

reconstructions from noisy free and noisy, with δ = 0.05, final time output are shown in Fig. 11.12 on the right, together with the function F (x) = x sin(5x) (solid line). Comparison of the results presented in Fig. 11.12 on the left and right shows that the accuracy of reconstruction in both cases is approximately the same. However, it should be noted that in the TSVD, the optimal value α = 3 · 10−9  of the regularization parameter was found as a result of computational experiments, i.e. empirically, while in the CG-algorithm this parameter is not even needed, since sufficiently accurate reconstruction result presented in Fig. 11.12 on the right was obtained without regularization (α = 0), also. Finally, let us assume in √the above example that the flexural rigidity is not constant and r(x) = exp(− x ), in order to find out how the variability of the coefficient r(x) affects the accuracy of the reconstruction. The rest of the parameters are the same. The synthetic noise free output uT (x) := u(x, T ) here is generated, in the same way, assuming F (x) = x sin(5x) and G(t) = cos(t/2) in the direct problem. The CG-algorithm is then applied to the inverse problem (11.4.31), for reconstruction of the spatial load F (x). The reconstructed spatial loads from noise free and noisy outputs with the noise level δ = 0.05 are shown in Fig. 11.13 on the right. The behavior of the convergence error e(n; F ; δ) = u(·, T ; F (n) ) − uδT L2 (0,) depending on the iteration number n is shown in Fig. 11.13 on the left. The reconstruction results with and without regularization were obtained in about 150 iterations. Note that for very small value α = 3 · 10−9  of the regularization parameter, almost the same results are obtained. A further increase in the value of the regularization parameter, degrades the accuracy of the solution. Let us compare the results presented on the right,√in Figs. 11.12 and 11.13, corresponding to the cases r(x) = 1 and r(x) = exp(− x ), respectively. In terms

Fig. 11.13 Convergence error (left); reconstruction of spatial load by CG-algorithm (right): √ r(x) = exp(− x )

11.5 Determination of the Flexural Rigidity of a Simply Supported Damped. . .

431

of accuracy of the reconstructions, they are qualitatively the same, although the results corresponding to the second case are slightly inferior to that corresponding to the first one. This shows that the effect of the coefficients in the Euler-Bernoulli equation being constant or variable on the numerical results obtained by the CGalgorithm is very low.

11.5 Determination of the Flexural Rigidity of a Simply Supported Damped Beam from Measured Boundary Slope In this section we study the inverse problem of determining the unknown principal coefficient, i.e. the flexural rigidity r(x) in ⎧ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = F (x) G(t), ⎪ ⎪ ⎪ ⎨ (x, t) ∈ T , ⎪ ⎪ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎩ u(0, t) = uxx (0, t) = 0, u(, t) = uxx (, t) = 0, t ∈ [0, T ],

(11.5.1)

from the measured boundary slope θ (t) := ux (0, t), t ∈ (0, T )

(11.5.2)

at the left end x = 0 of the simply supported beam. Inverse coefficient problems for Euler-Bernoulli beam equation are still of great interest because of the practical significance of these problems, starting from the first studies [9–11, 45, 97, 98] on this subject until today. From the point of view of the methods and approaches used, the studies on the inverse coefficient problems for the Euler-Bernoulli beam equation can be broadly divided into two categories: methods based on spectral theory and methods based on boundary observations, i.e input-output mappings. In the above pioneering studies, these problems have been studied using the first category of methods, as inverse spectral problems. Thus, in [9], the problem of reconstruction of the coefficients p(x) and q(x) associated with the eigenvalue problem u(4) − (p(x)u ) + q(x)u = 0 from a knowledge of three spectra was studied. The results obtained here indicated that n+1 spectra associated with n + 1 distinct sets of boundary conditions are required to uniquely determine the unknown coefficients. For the simplest Euler-Bernoulli beam equation ρ(x)ut t + (r(x)uxx )xx = 0, various inverse coefficient problems related to determining the mass density ρ(x) and the flexural rigidity r(x) were studied in subsequent studies [10, 11]. The main effort in all the above cited works was the determination of the coefficients from the spectral data. However, in practice, it is difficult to acquire such a spectral data as measured output.

432

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

An identification of unknown coefficients ρ(x) and r(x) in the cantilever EulerBernoulli beam equation from two boundary observations was considered in [29], assuming that the coefficients are smooth enough, ρ, r ∈ C 4 (0, ), and the time interval is infinite, i.e. t ∈ (0, +∞), to allow the use of the Laplace transform. However, the drawback of this approach is that in real engineering models the final time T > 0 is finite. Moreover, it may be small enough, since the information propagation speed in elastic beams is not finite [87]. The resulting algorithm for constructing spectral data is practically cannot be used, since the construction procedure requires calculating the Laplace transform in the complex plane and finding an infinite number of poles in it. Therefore, a detailed study of problem (11.5.1)–(11.5.2), as well as the development of an algorithm for the numerical solution of the inverse problem of recovering ρ(x) or/and r(x) from boundary measured output on a finite time interval is important both in theoretical and applied terms. The importance of the proposed model governed by (11.5.1)–(11.5.2) is that, firstly, it is a generalization of existing mathematical models in the sense that, the Euler-Bernoulli equation (11.5.1) includes all the main physical coefficients. Secondly, the time interval is finite, and finally, the inputs in (11.5.1) may not be smooth enough. An analogue of this model for the cantilever beam, but with Neumann measured output, is discussed in the next section.

11.5.1 Properties of the Input-Output Operator and Existence of a Quasi-Solution We define the set of admissible coefficients Rm = {r ∈ H m (0, ), : 0 < r0 ≤ r(x) ≤ r1 },

(11.5.3)

which is a nonempty closed convex set in H m (0, ), m ≥ 1. The order m in the above definition will be determined below accordingly. Let r ∈ Rm . Denote by u(x, t) = u(x, t; r) the corresponding (unique) weak solution of the direct problem (11.5.1). Introduce the nonlinear input-output operator associated with the inverse problem (11.5.1)–(11.5.2) as follows: (r)(t) := ux (0, t; r), t ∈ (0, T ),  : Rm ⊂ H m (0, ) → L2 (0, T ).

(11.5.4)

11.5 Determination of the Flexural Rigidity of a Simply Supported Damped. . .

433

Then the inverse coefficient problem (11.5.1)–(11.5.2) is reduced to (r)(t) = θ (t), t ∈ (0, T ), r ∈ Rm , θ ∈ L2 (0, T ), i.e. to the problem of the invertibility of the input-output operator (11.5.4). Lemma 11.5.1 Assume that conditions of Theorem 11.1.3 ensuring the existence of the regular weak solution of the direct problem (11.5.1) are satisfied. Then the input-output operator [·] : R2 ⊂ H 2 (0, ) → L2 (0, T ) defined in (11.5.4) is a compact operator. Proof Let {r (n) } ⊂ R2 , n = 1, ∞, be a sequence of admissible coefficients uniformly bounded in the norm of H 2 (0, ). We need to prove that corresponding the sequence of outputs {(r (n) )(t)} is bounded in the norm of H 1 (0, T ), (n) (n) where (r (n) )(t) = u(n) x (0, t), while u (x, t) := u(x, t; r ) is the sequence of corresponding regular weak solutions of the direct problem (11.5.1). This statement follows from estimates (11.1.10) of Corollary 11.1.2 and (11.1.18) of Corollary 11.1.3. Thus, the input-output operator transforms any bounded H 2 (0, ) set {r (n) } ⊂ R2 to a compact in L2 (0, T ) set {(r (n) )(t)}, hence it is a compact operator.   As a consequence of this lemma, the inverse coefficient problem (11.5.1)– (11.5.2) is ill-posed. As has been proven in previous sections, the Lipschitz continuity of the inputoutput operator plays a key role for existence of a quasi-solution. To establish this property we need some auxiliary results. Let r, r + δr ∈ Rm . Then δu(x, t) := u(x, t; r + δr) − u(x, t; r) solves the following problem ⎧ ρ(x)δut t + μ(x)δut + (r(x)uxx )xx − (Tr (x)δux )x ⎪ ⎪ ⎪ ⎪ ⎨ = − (δr(x)# uxx (x, t))xx , (x, t) ∈ T , (11.5.5) ⎪ δu(x, 0) = 0, δut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ ⎩ δu(0, t) = δuxx (0, t) = 0, δu(, t) = δuxx (, t) = 0, t ∈ [0, T ]. Here and below, the following notations are used: # r(x) := r(x) + δr(x), and # u(x, t) := u(x, t;# r) is the solution of the direct problem (11.5.1) corresponding to the coefficient # r (x) := r(x) + δr(x). Theorem 11.5.1 Assume that conditions of Theorem 11.1.3 hold. Then for the solution δu(x, t) of problem (11.5.5) the following estimates hold: 2 F 2 2

δuxx 2L2 (0,T ;L2 (0,)) ≤ C

G 2H 1 (0,T ) δr 2H 1 (0,) , 2 L (0,) 2 F 2 2

δut 2L2 (0,T ;L2 (0,)) ≤ C

G 2H 1 (0,T ) δr 2H 1 (0,) , 3 L (0,)

(11.5.6)

434

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

where 12 , C 32 = 1 [exp(T ) − 1]C 12 , 2 = 2 [exp(T ) − 1]C C 2 r0 ρ0  2 = 2 1 + C12 + ρ0 C32 C2 , C2 = max(1, 2/3, 2/), C 1 r02

(11.5.7)

and C1 , C3 > 0 are the constants introduced in Theorems 11.1.2 and 11.1.3. Proof Multiply both sides of Eq. (11.5.5) by 2δut (x, t), integrate it over t := (0, ) × (0, t), use the identity 2(r(x)δuxx )xx δut ≡ 2[(r(x)δuxx )x δut − r(x)δuxx δuxt ]x

 + r(x)δu2xx t

(11.5.8)

and then apply the integration by parts formula. After elementary calculations we obtain the following integral identity: 

 0

+ r(x)δu2xx

ρ(x)δu2t

= −2

+ Tr (x)δu2x

 t 0

0





dx + 2

 t 0

 0

μ(x)δu2τ dxdτ

uxx )xx δuτ dxdτ, t ∈ [0, T ]. (δr(x)#

Taking into account the boundary conditions in (11.5.1) and (11.5.5) we transform the right-hand-side integral to the following form: −2

 t 0

 0

uxx )xx δuτ dxdτ = 2 (δr(x)#  −2



 t



δr(x)# uxxτ δuxx dxdτ 0

0

δr(x)# uxx δuxx dxdτ, t ∈ [0, T ].

0

Then we get:  0



ρ(x)δu2t =2

+ r(x)δu2xx

 t 0

 0

+ Tr (x)δu2x



dx + 2

 t

 δr(x)# uxxτ δuxx dxdτ − 2

0

 0

μ(x)δu2τ dxdτ



δr(x)# uxx δuxx dx, 0

(11.5.9)

11.5 Determination of the Flexural Rigidity of a Simply Supported Damped. . .

435

t ∈ [0, T ]. In view of the ε-inequality with ε = r0 /2 this identity yields the following integral inequality: 



ρ0 0

+2

 t 0

+

 0



2 r0



r0 + 2

δu2t dx

T 0

 0

 δu2xx dx

0

μ(x)δu2τ dxdτ ≤ 

 0



+ r0 2

Tr (x)δu2x dx

 t 0



δu2xx dxdτ

0

 |δr(x)|2 # u2xxt dxdt +



0

 |δr(x)|2 # u2xx dx , t ∈ [0, T ].

We evaluate the sum in the right-hand-side square brackets using estimates in (11.1.4) and (11.1.14). This yields:  0

T



 0

 δr(x)# u2xxt dxdt +

≤ max |δr(x)|

2

[0,]

≤



 0

δr(x)# u2xx dx

# uxxt 2L2 (0,T ;L2 (0,)) + # uxx 2L∞ (0,T ;L2 (0,))



1  1 + C12 + ρ0 C32 C2 F 2L2 (0,) G 2H 1 (0,T ) δr 2H 1 (0,). r0

Here we also used the consequence max |δr(x)|2 ≤ C2 δr 2H 1 (0,), C2 = max(1, 2/3, 2/).

(11.5.10)

of Agmon’s inequality. This leads to the main integral inequality 



ρ0 0

+2

δu2t dx +

 t 0

 0

r0 2



 0

 δu2xx dx +

μ(x)δu2τ dxdτ ≤

r0 2

 0

Tr (x)δu2x dx

 t 0

 0

δu2xx dxdτ

12 F 2 2

G 2H 1 (0,T ) δr 2H 1 (0,), t ∈ [0, T ], +C L (0,)

(11.5.11)

1 > 0 introduced in (11.5.7). with C The required estimates (11.5.6) are easily derived from this inequality in the same way as in the proof of Theorem 11.1.2.   Under more regularity conditions we can derive analogues of estimates (11.5.6). Theorem 11.5.2 Assume that conditions of Theorem 11.1.4 ensuring the existence of a regular solution with improved regularity of the direct problem (11.5.1) are

436

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

satisfied. Then for the solution δu(x, t) of problem (11.5.5) the following estimates hold: 2 F 2 2

δuxxt 2L2 (0,T ;L2 (0,)) ≤ C

G 2H 2 (0,T ) δr 2H 1 (0,) , 5 H (0,) 2 F 2 2

G 2H 2 (0,T ) δr 2H 1 (0,) ,

δut t 2L2 (0,T ;L2 (0,)) ≤ C 6 H (0,)

(11.5.12)

where 42 , C 62 = 1 [exp(T ) − 1]C 42 , 2 = 2 [exp(T ) − 1]C C 5 r0 ρ0  2 = 2 ρ0 C12 Cr2 + (C12 + 1)C32 C2 , C 4 r02

(11.5.13)

and C1 , C3 , Cr > 0 are the constants introduced in Theorems 11.1.2, 11.1.3 and 11.1.4, respectively. Proof The main integral inequality 



ρ0 0

+2

δu2t dx +

 t 0

 0

r0 2



 0

 δu2xx dx +

μ(x)δu2τ dxdτ ≤

r0 2

 0

Tr (x)δu2x dx

 t 0

 0

δu2xx dxdτ

42 F 2 2

G 2H 2 (0,T ) δr 2H 1 (0,), t ∈ [0, T ], +C H (0,) 4 > 0 defined in (11.5.13), can be derived in the same manner as in the with C proof of the previous theorem. This is the same integral inequality (11.5.11) with 1 , F 2 2 4 , F 2 2 u(x, t) C and G 2H 1 (0,T ) replaced by ut (x, t), C and L (0,) H (0,)

G 2H 2 (0,T ) , respectively. Further estimates (11.5.12) are derived from this inequality.

 

Lemma 11.5.2 Assume that conditions of Theorem 11.1.3 ensuring the existence of the regular weak solution of the direct problem (11.5.1) are satisfied. Suppose that [·] : R2 ⊂ H 2 (0, ) → L2 (0, T ) is the input-output operator introduced in (11.5.4). Then this operator is Lipschitz continuous, that is

r1 − r2 L2 (0,T ) ≤ L r1 − r2 H 1 (0,), ∀r1 , r2 ∈ R2 ,

(11.5.14)

√ 2 F L2 (0,T ) G H 1 (0,T ) is the Lipschitz constant and C 2 > 0 where L = 3 C is the constant introduced in Theorem 11.5.1.

11.5 Determination of the Flexural Rigidity of a Simply Supported Damped. . .

437

Proof By the definition (11.5.4), we have:

r1 − r2 L2 (0,T ) = δux (0, ·) L2 (0,T ) , where δux (0, t) = ux (0, t; r1 ) − ux (0, t; r2 ). With inequality (11.1.12) this yields:

r1 − r2 2L2 (0,T ) ≤ 3 δuxx 2L2 (0,T ;L2 (0,)). Taking into account inequality the first estimate in (11.5.6), this implies the required statement (11.5.14).   Remark 11.5.1 In fact, one should write the right-hand-side norm in (11.5.14) as

r1 − r2 H 2 (0,) , i.e. in the norm of H 2 (0, ), as r1 , r2 ∈ R2 . Since this inequality is obtained in the norm of H 1 (0, ) we decided to leave it as it is. Obviously, inequality (11.5.14) remains valid when the norm r1 − r2 H 1 (0,) is replaced with the norm

r1 − r2 H 2 (0,) . Since exact equality cannot be achieved in the operator equation (r)(t) = θ (t) we introduce the Tikhonov functional J (r) =

1

r − θ 2L2 (0,), r ∈ R2 , θ ∈ L2 (0, T ), 2

(11.5.15)

and reformulate the inverse problem (11.5.1)–(11.5.2) as the following minimization problem: J (r∗ ) = inf J (r). r∈R2

(11.5.16)

A solution of the minimization problem (11.5.16) is defined as a quasi-solution of the inverse problem. As stated in the previous sections, the Lipschitz continuity of the input-output operator leads to the Lipschitz continuity of the Tikhonov functional, and the latter property, in turn, to the existence of a quasi-solution of the inverse problem. Theorem 11.5.3 Assume that conditions of Theorem 11.1.3 ensuring the existence of the regular weak solution of the direct problem (11.5.1) are satisfied. Suppose that the measured output θ (t) belongs to L2 (0, T ). Then there exists a quasi-solution to the inverse coefficient problem (11.5.1)–(11.5.2) in the set of admissible coefficients R2 ⊂ H 2 (0, ). This theorem is proved in the same way as Theorem 10.1.11 in Sect. 10.1.

438

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

11.5.2 Fréchet Differentiability of the Tikhonov Functional and Gradient Formula Let r, r + δr ∈ R2 . Then the increment δJ (r) := J (r + δr) − J (r) of the Tikhonov functional (11.5.15) is 

T

δJ (r) =

1 2

[ux (0, t) − θ (t)] δux (0, t)dt +

0



T

(δux (0, t))2 dt,

(11.5.17)

0

where ux (0, t) := ux (0, t; r). Lemma 11.5.3 Assume that conditions of Theorem 11.1.3 hold. Suppose that r, r + δr ∈ R2 are two arbitrary coefficients. Then the following integral relationship holds: 

T

# M(t)δu x (0, t)dt = −

0



T





δr(x)# uxx φxx dxdt, 0

(11.5.18)

0

where δu ∈ L2 (0, T ; V 2 (0, )) is the weak solution of problem (11.5.5) and φ(x, t) is the weak solution of the following backward problem: ⎧ ρ(x)φt t − μ(x)φt + (r(x)φxx )xx − (Tr (x)φx )x = 0, (x, t) ∈ T , ⎪ ⎪ ⎪ ⎨ φ(x, T ) = 0, φ (x, T ) = 0, x ∈ (0, ), t (11.5.19) # ⎪ φ(0, t) = 0, −r(0)φ (0, t) = M(t), xx ⎪ ⎪ ⎩ φ(, t) = φxx (, t) = 0, t ∈ [0, T ], # ∈ H 1 (0, T ). with the arbitrary Neumann input M This lemma is proved in the similar way as Lemma 11.2.4, and we leave the proof as an exercise for the reader.  # ∈ H 1 (0, T ) in above lemma is imposed on the input Remark 11.5.2 Condition M based on Theorem C.2.1 in Appendix C that guarantees the existence of a weak solution of the backward problem (11.5.19). # in (11.5.19) as follows: Choose the Neumann input M(t) # = ux (0, t; r) − θ (t), t ∈ (0, T ). M(t)

(11.5.20)

Then (11.5.18) implies: 

T 0



T

[ux (0, t; r) − θ (t)]δux (0, t)dt = −





δr(x)# uxx φxx dxdt. 0

0

(11.5.21)

11.5 Determination of the Flexural Rigidity of a Simply Supported Damped. . .

439

The integral identity (11.5.21) establishes the basic relationship between the main inputs and outputs of inverse coefficient problem (11.5.1)–(11.5.2): the input r(x), the output ux (0, t; r) and also the measured output θ (t). In view of the input-output relationship (11.5.21), the increment formula (11.5.17) has the following form:  δJ (r) = −

  T

0

# uxx (x, t; r)φxx (x, t; r)dt δr(x)dx

0

+

1 2



T

(δux (0, t))2 dt,

(11.5.22)

0

Now, we must justify the validity of the substitution (11.5.20) in the backward problem (11.5.19). From Theorem C.2.1 in Appendix C, it follows that in order for this problem to have a weak solution the Neumann input M(t) must belong to H 1 (0, T ). Therefore, it is necessary that the following conditions be satisfied: ux (0, ·) ∈ H 1 (0, T ), θ ∈ H 1 (0, T ).

(11.5.23)

The first condition of (11.5.23) is automatically met, by Theorem 11.1.3 and Corollary 11.1.3. As for the second condition θ ∈ H 1 (0, T ), this is an additional condition that must be imposed on the Dirichlet measured output θ (t) which originally belongs to L2 (0, T ). Looking ahead, we note that if this were the Neumann measured output, say the measured bending moment M(t), then more regularity condition M ∈ H 2 (0, T ) would be required (see the second condition in (11.6.22)). Theorem 11.5.4 Assume that conditions of Theorem 11.1.3 ensuring the existence of the regular weak solution of the direct problem (11.5.1) are satisfied. Suppose in addition the Dirichlet measured output satisfies the condition θ ∈ H 1 (0, T ). Then the Tikhonov functional J (r), r ∈ R2 ⊂ H 2 (0, ), corresponding to the inverse coefficient problem (11.5.1)–(11.5.2) is Fréchet differentiable. Moreover, for the Fréchet gradient of this functional the following gradient formula holds:  ∇J (r)(x) = −

T

uxx (x, t; r)φxx (x, t; r)dt, x ∈ (0, ),

(11.5.24)

0

where φ(x, t; r) is the weak solution solutions of the adjoint problem (11.5.19) with the input (11.5.20).

440

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Proof Taking into account # u(x, t) = u(x, t) + δu(x, t), we rewrite formula (11.5.22) as follows: 

T

δJ (r) = − − 0

T



δr(x)uxx (x, t; r)φxx (x, t; r)dxdt 0







0 

0

1 δr(x)δuxx φxx dxdt + 2



T

(δux (0, t))2 dt.

(11.5.25)

0

The estimates 

T





δr(x)δuxx φxx dxdt 0

0

≤ max |δr(x)| δuxx L2 (0,T ;L2 (0,)) φxx L2 (0,T ;L2 (0,)), [0,]



T 0

(δux (0, t))2 dt ≤ 3 δuxx L2 (0,T ;L2 (0,)).

for second and third right-hand-side integrals in (11.5.25), with inequality (11.5.10)  and estimate (11.5.6) show that both integrals are of the order O δr 2H 1 (0,) . This completes the proof.   The gradient formula (11.5.24) is a powerful tool in the numerical reconstruction of the unknown coefficient. Furthermore, the numerical algorithm discussed in Sect. 6.5 can also be applied to the inverse problem (11.5.1)–(11.5.2).

11.6 Determination of the Flexural Rigidity of a Cantilever Beam from Measured Boundary Bending Moment In this section we study the inverse problem of determining the unknown principal coefficient, i.e. the flexural rigidity r(x) in ⎧ ⎪ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = 0, ⎪ ⎪ ⎪ ⎪ (x, t) ∈ T , ⎪ ⎨ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ u(0, t) = ux (0, t) = 0, uxx (, t) = 0, ⎪ ⎪ ⎪ ⎩ − (r()uxx (, t))x + Tr ()ux (, t) = g(t), t ∈ [0, T ],

(11.6.1)

from the measured boundary bending moment M(t) := −r(0)uxx (0, t), t ∈ (0, T )

(11.6.2)

11.6 Determination of the Flexural Rigidity of a Cantilever Beam from. . .

441

beam. Here and below, the traces at the left end x = 0 of the cantilever  − (r(x)uxx (x, t))x + Tr (x)ux (x, t) x= and (−r(x)uxx (x, t))x=0 are indicated as − (r()uxx (, t))x + Tr ()ux (, t) and −r(0)uxx (0, t), respectively. Although this problems and the inverse coefficient problem (11.5.1)–(11.5.2) considered in the previous section, may seem similar to each other, they actually have different characteristics, as we will see below. In addition, the inverse problem (11.5.1)–(11.5.2) is of particular interest from the point of view of the Neumann-toNeumann operator related to the inverse coefficient problem for the Euler-Bernoulli beam equation.

11.6.1 The Neumann-to-Neumann Operator. Existence of a Quasi-Solution We assume that the inputs in (11.6.1) satisfy the following basic conditions: ⎧ ρ, r, μ, Tr ∈ L∞ (0, ), ⎪ ⎪ ⎪ ⎪ ⎨ g ∈ H 1 (0, T ), g(0) = 0, ⎪ ⎪ 0 < ρ0 ≤ ρ(x) ≤ ρ1 , 0 < r0 ≤ r(x) ≤ r1 , ⎪ ⎪ ⎩ 0 < μ0 ≤ μ(x) ≤ μ1 , 0 ≤ Tr0 ≤ Tr (x) ≤ Tr1 , x ∈ (0, ).

(11.6.3)

Let r ∈ Rm , where Rm ⊂ H m (0, ) is the set of admissible principal coefficients r(x), defined in (11.5.3). Denote by u = u(x, t; r) the corresponding (unique) weak solution of the direct problem (11.6.1). Introduce the nonlinear Neumannto-Neumann map associated with the inverse problem (11.6.1)–(11.6.2) as follows: (r)(t) := −r(0)uxx (0, t; r), t ∈ (0, T ),  : Rm ⊂ H m (0, ) → L2 (0, T ).

(11.6.4)

Lemma 11.6.1 Assume that the basic conditions (11.6.3) hold. Suppose in addition that the conditions of Theorem 11.1.7 that ensure the existence of a regular weak solution with improved regularity are satisfied and, in addition, r ∈ R4 . Then the Neumann-to-Neumann map [·] : R4 ⊂ H 4 (0, ) → L2 (0, T ) defined in (11.6.4) is a compact operator. Proof Let {r (n) } ⊂ R4 , n = 1, ∞, be a sequence of admissible coefficients uniformly bounded in H 4 -norm. We need to prove that the sequence of corresponding outputs {(r (n) )(t)}, where (r (n) )(t) = −r (n) (0)u(n) xx (0, t), is bounded in the norm of H 1 (0, T ). To this end we need some auxiliary estimates.

442

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Let u(x, t) be a regular weak solution with improved regularity of the direct problem (11.6.1). Then using the identities 



(r(x)uxx (x, t))x =

[ρ(ξ )ut t + μ(ξ )ut ] dξ + Tr (x)ux − g(t),

x





−r(0)uxx (0, t) = 0

(r(x)uxx )x dx, t ∈ (0, T )

we deduce that 

 

−r(0)uxx (0, t) = 0





+

[ρ(ξ )ut t + μ(ξ )ut ] dξ dx

x

Tr (x)ux dx −  g(t), t ∈ (0, T ).

(11.6.5)

0

This yields: 

r(0)uxx (0, ·) 2L2 (0,T ) ≤ 4

1 3 2  ρ1 ut t 2L2 (0,T ;L2 (0,)) 3

 1 3 2 1 3 2 2 2 2 2 +  μ1 ut L2 (0,T ;L2 (0,)) +  Tr1 uxx L2 (0,T ;L2 (0,)) +  g L2 (0,T ) . 3 2 Using here the estimates in Theorems 11.1.5 and 11.1.6 we find: 2 g 2 2

r(0)uxx (0, ·) 2L2 (0,T ) ≤ C , 7 H (0,T )   2 = 43 max ρ 2 C 2 /3, μ2 C 2 /3 + Tr2 C 2 /2, 1/ . C 7 1 5 1 5 1 4 In the same way, we can derive the estimate #72 g 2 3

r(0)uxxt (0, ·) 2L2 (0,T ) ≤ C , H (0,T ) #7 > 0. with the above defined constant C These estimates imply that the sequence {(r (n))(t)} of outputs is bounded in the norm of H 1 (0, T ), and hence is relatively compact subset in L2 (0, T ). This means the compactness of the Neumann-to-Neumann operator.   As a consequence of Lemma 11.6.1, we conclude that the inverse coefficient problem (11.6.1)–(11.6.2) is ill-posed. To investigate further properties of the Neumann-to-Neumann operator, we need some auxiliary results.

11.6 Determination of the Flexural Rigidity of a Cantilever Beam from. . .

443

Let r, r + δr ∈ Rm . Then δu(x, t) = u(x, t; r + δr) − u(x, t; r) solves the following problem ⎧ ρ(x)δut t + μ(x)δut + (r(x)uxx )xx − (Tr (x)δux )x ⎪ ⎪ ⎪ ⎪ ⎪ = − (δr(x)# uxx (x, t))xx , (x, t) ∈ T , ⎪ ⎨ (11.6.6) δu(x, 0) = 0, δut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ δu(0, t) = δu (0, t) = 0, δu (, t) = 0, ⎪ x xx ⎪ ⎪ ⎩ − (r()δuxx (, t))x + Tr ()δux (, t) = (δr()# uxx (, t))x , t ∈ [0, T ], where # u(x, t) := u(x, t; r + δr) is the solution of the direct problem (11.6.1) corresponding to the coefficient r(x) + δr(x) =: # r(x). Theorem 11.6.1 Assume that conditions of Theorem 11.1.6 that ensure the existence of a regular weak solution of the direct problem (11.6.1) are satisfied. Then for the solution of problem (11.6.6) the following estimates hold: #2 g 2 2

δr 2H 1 (0,),

δuxx 2L2 (0,T ;L2 (0,)) ≤ C 9 H (0,T ) #2 g 2 2

δr 2H 1 (0,),

δut 2L2 (0,T ;L2 (0,)) ≤ C 10 H (0,T )

(11.6.7)

where 2 2 #82 , C #10 #82 , #2 = 2 [exp(T ) − 1]C = [exp(T ) − 1]C C 9 r0 r0 3 #2 = 8(1 + T ) exp(T )C2 , C 8 3r03

(11.6.8)

and C > 0 is the constant introduced in inequality (11.5.10). Proof Multiply both sides of Eq. (11.6.6) by 2δut (x, t), integrate it over t := (0, ) × (0, t), use the identity (11.5.8) and then apply the integration by parts formula. After elementary calculations we arrive at the following integral identity: 

 0

 t ρ(x)δu2t + r(x)δu2xx + Tr (x)δu2x dx + 2

= −2

 t 0

 0

 uxx )xx δuτ dxdτ + 2 (δr(x)#

0

0

 0

μ(x)δu2τ dxdτ

t

uxx (, τ ))x δuτ (, τ )dτ, (δr()#

444

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

t ∈ [0, T ]. Using integration by parts multiple times to transform the right-hand-side integrals. Taking into account the boundary conditions we obtain: 

 0

ρ(x)δu2t =2

+ r(x)δu2xx

 t 0



+ Tr (x)δu2x



dx + 2 

δr(x)# uxxτ δuxx dxdτ − 2

0

 t 0

 0

μ(x)δu2τ dxdτ



δr(x)# uxx δuxx dx,

(11.6.9)

0

t ∈ [0, T ]. It should be noted that despite the fact that the integral identities (11.5.9) and (11.6.9) are identical in form, the functions δu(x, t) and # u(x, t) in these identities are solutions to different problems. As in the derivation of the integral inequality (11.5.11), the integral identity (11.6.9) leads to the following integral inequality 



ρ0 0

δu2t dx

r0 + 2



 0

 δu2xx dx r0 ≤ 2



+

 t 0

0  0

Tr (x)δu2x dx

+2

 t 0

 0

μ(x)δu2τ dxdτ

#82 g 2 2 δu2xx dxdτ + C

δr 2H 1 (0,), H (0,)

#8 > 0 introduced in (11.6.8). for all t ∈ [0, T ], with C The rest of the proof of the theorem is the same as in Theorem 11.5.1.

 

Theorem 11.6.2 Assume that the conditions (11.1.37) of the Theorem 11.1.7 that ensure the existence of a regular weak solution with improved regularity of the direct problem (11.6.1) are satisfied. Then for the solution of problem (11.6.6) the following estimates hold: #2 g 2 3

δuxxt 2L2 (0,T ;L2 (0,)) ≤ C

δr 2H 1 (0,), 9 H (0,T ) #2 g 2 3

δut t 2L2 (0,T ;L2 (0,)) ≤ C

δr 2H 1 (0,), 10 H (0,T )

(11.6.10)

#10 > 0 are the same introduced in Theorem 11.6.1. #9 , C where C Estimates (11.6.10) are derived in exactly the same way as similar estimates in Theorem 11.5.2.  We will use the above results to prove the Lipschitz continuity of the Neumannto-Neumann operator. Lemma 11.6.2 Under the conditions of the Theorem 11.1.6 providing existence of a regular weak solution of the direct problem (11.6.1), the Neumann-to-Neumann operator introduced in (11.6.4) is Lipschitz continuous, that is

r1 − r2 L2 (0,T ) ≤ L r1 − r2 H 1 (0,), ∀r1 , r2 ∈ R2 ,

(11.6.11)

11.6 Determination of the Flexural Rigidity of a Cantilever Beam from. . .

445

where    1/2 2 #10 #92 /2 L = 3 ρ12 + μ21 C + 3Tr21 C

g H 3 (0,T )

(11.6.12)

#9 , C #10 > 0 are the constants introduced in (11.6.8). is the Lipschitz constant and C Proof Let r1 , r2 ∈ R2 . Denote by (r1 )(t) := −r1 (0)# uxx (0, t) and (r2 )(t) := −r2 (0)uxx (0, t) the outputs, where # u(x, t) := u(x, t; r1 ) and u(x, t) := u(x, t; r2 ). We employ the identity (11.6.5) for these outputs:  (r1 )(t) =



ut ) + Tr (x)# ux ] dx − g(t), ut t + μ(x)# [x (ρ(x)#

0

 (r2 )(t) =



[x (ρ(x)ut t + μ(x)ut ) + Tr (x)ux ] dx − g(t), t ∈ (0, T ),

0

Subtracting these identities from each other, squaring them, and then integrating over (0, T ) we obtain: 

r1 − r2 2L2 (0,T ) ≤ 3 ρ12 δut t 2L2 (0,T ;L2 (0,))

+μ21 δut 2L2 (0,T ;L2 (0,)) + 3Tr21 δuxx 2L2 (0,T ;L2 (0,))/2 .

With estimates in Theorem 11.6.1 and Theorem 11.6.2 this leads to estimate (11.6.11) with the Lipschitz constant defined in (11.6.12).   Remark 11.5.1 applies also to this lemma. As in the case of the previous inverse problems, the Lipschitz continuity of inputoutput operator leads to the Lipschitz continuity of Tikhonov functional. Here we also introduce this functional J (r) =

1

r − M 2L2 (0,T ) , r ∈ R2 , M ∈ L2 (0, T ), 2

(11.6.13)

corresponding to the inverse problem (11.6.1)–(11.6.2), and reformulate this problem as the minimization problem J (r∗ ) = inf J (r). r∈R2

(11.6.14)

Then we can formulate the analogue of the existence Theorem 11.5.3. Theorem 11.6.3 Assume that conditions of Theorem 11.1.6 ensuring the existence of the regular weak solution of the direct problem (11.6.1) are satisfied. Suppose that the measured output M(t) belongs to L2 (0, T ). Then there exists a quasi-solution to the inverse coefficient problem (11.6.1)–(11.6.2) in the set of admissible coefficients R2 ⊂ H 2 (0, ).

446

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

11.6.2 Fréchet Differentiability of the Tikhonov Functional and Gradient Formula Let r, r + δr ∈ R2 . For convenience, we use the notations # r(x) := r(x) + δr(x) and # u(x, t) := u(x, t; r + δr), as above. Then the increment δJ (r) := J (# r) − J (r) of the Tikhonov functional (11.6.13) can be written as follows: 

T

δJ (r) =

r(0)# uxx (0, t) − r(0)uxx (0, t)] dt [r(0)uxx (0, t) + M(t)] [#

0

1 + 2



T

r(0)# uxx (0, t) − r(0)uxx (0, t)]2 dt, [#

(11.6.15)

0

where uxx (0, t) := uxx (0, t; r). Lemma 11.6.3 Assume that conditions of Theorem 11.1.6 hold. Suppose that r, r + δr ∈ R2 are two arbitrary coefficients. Then the following integral relationship holds: 

T

# θ (t) [# r(0)# uxx (0, t) − r(0)uxx (0, t)] dt

0

 =−

T





δr(x)# uxx ψxx dxdt, 0

(11.6.16)

0

where δu ∈ L2 (0, T ; V 2 (0, )) is the regular weak solution of problem (11.6.5) and ψ(x, t) is the solution of the following backward problem: ⎧ ρ(x)ψt t − μ(x)ψt + (r(x)ψxx )xx − (Tr (x)ψx )x = 0, (x, t) ∈ T , ⎪ ⎪ ⎪ ⎪ ⎨ ψ(x, T ) = 0, ψ (x, T ) = 0, x ∈ (0, ), t (11.6.17) ⎪ (0, t) = # θ (t), ψxx (, t) = 0 ψ(0, t) = 0, ψ ⎪ x ⎪ ⎪ ⎩ −(r()ψxx (, t))x + Tr ()ψx (, t) = 0, t ∈ [0, T ], with the arbitrary Dirichlet input # θ ∈ H 2 (0, T ). Proof Multiply both sides of Eq. (11.6.6) by arbitrary function ψ(x, t), integrate over (0, T ) and perform integration by parts multiple times. Then we obtain: 

T 0



 0

ρ(x)ψt t − μ(x)ψt + (r(x)ψxx )xx − (Tr (x)ψx )x δu dxdt 



+ 0

[ρ(x)δut ψ − ρ(x)δuψt + μ(x)δuψ]tt =T =0 dt

11.6 Determination of the Flexural Rigidity of a Cantilever Beam from. . .





T

+ 0

447

(r(x)δuxx )x ψ − r(x)δuxx ψx + r(x)δux ψxx − δu (r(x)ψxx )x



T

+ 0

 [Tr (x)ψx δu − Tr (x)ψδux ]x= x=0 dt = − 

T

−



0

T



x= x=0

dt



δr(x)# uxx ψxx dxdt 0

0

uxx ψx uxx )x ψ − δr(x)# (δr(x)#

x= x=0

dt.

(11.6.18)

Now, we require that the function ψ(x, t) is a solution to problem backward problem (11.6.17). Using the homogeneous initial and final conditions in (11.6.6) and (11.6.17), as well as the non-homogeneous boundary conditions − (r()δuxx (, t))x + Tr ()δux (, t) = (δr()# uxx (, t))x and ψx (0, t) = # θ(t), we conclude from (11.6.18) that 

T

r(0)δuxx (0, t)# θ (t)dt

0





T

=− 0





T

δr(x)# uxx ψxx dxdt +

0

δr(0)# uxx (0, t)# θ (t)dt.

0

In view of the identity uxx (0, t), # r(0)# uxx (0, t) − r(0)uxx (0, t) = r(0)δuxx (0, t) − δr(0)# this implies to the required relationship (11.6.16).

 

Remark 11.6.1 Condition # θ ∈ H 2 (0, T ) in above lemma is imposed on the input based on Theorem C.3.1 in Appendix C that guarantees the existence of a weak solution of the backward problem (11.5.17). Choose the Dirichlet input # θ (t) in (11.6.17) as follows: # θ (t) = r(0)uxx (0, t) + M(t), t ∈ (0, T ).

(11.6.19)

This leads to the basic relationship 

T

r(0)# uxx (0, t) − r(0)uxx (0, t)] dx [r(0)uxx (0, t) + M(t)] [#

0

 =−

T





δr(x)# uxx ψxx dxdt 0

(11.6.20)

0

between the input, r(x), the Neumann output, r(0)uxx (0, t; r), and also the measured output M(t) in the inverse coefficient problem (11.6.1)–(11.6.2).

448

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

In view of the input-output relationship (11.6.20) and the relation # u(x, t) = u(x, t) + δu(x, t), the increment formula (11.6.15) has the following form: δJ (r) = −  0

uxx (x, t; r)ψxx (x, t; r)dt δr(x)dx 0

T



  T

 

0 

0

1 δr(x)δuxx ψxx dxdt + 2



T

(δux (0, t))2 dt,

(11.6.21)

0

Let us justify now the validity of the substitution (11.6.19) in the backward problem (11.6.17). According to Theorem C.3.1 in Appendix C, for existence of a weak solution of (11.6.17) the Dirichlet input # θ (t) should satisfy the regularity condition # θ ∈ H 2 (0, T ). In view of substitution (11.6.19) this means that the output δuxx (0, t) and the measured output M(t) should satisfy the following conditions: uxx (0, ·) ∈ H 2 (0, T ), M ∈ H 2 (0, T ).

(11.6.22)

In order to identify for which regular weak solution of the direct problem (11.6.1) the first condition of (11.6.22) is satisfied, we use the following analogue 

 

−r(0)uxxt t (0, t) =  +

0 

[ρ(ξ )ut t t t + μ(ξ )ut t t ] dξ dx

x

Tr (x)uxt t dx −  g  (t), t ∈ (0, T ).

0

of identity (11.6.5) to deduce the inequality 

r(0)uxxt t (0, ·) 2L2 (0,T ) ≤ 4

1 3 2  ρ1 ut t t t 2L2 (0,T ;L2 (0,)) 3

 1 1 + 3 μ21 ut t t 2L2 (0,T ;L2 (0,)) + 3 Tr21 uxxt t 2L2 (0,T ;L2 (0,)) + 2 g  2L2 (0,T ) . 3 2 The right-hand-side of this inequality shows that for the condition uxxt t (0, ·) ∈ L2 (0, T ) to be satisfied, the function u(x, t) must be a regular weak solution with higher regularity, which is ensured by the first part of Theorem 11.1.7. The second condition M ∈ H 2 (0, T ) of (11.6.22) suggests that the Neumann measured output M(t), which is in L2 (0, T ), as the result of the measurement, should be more regular function. Theorem 11.6.4 Assume that conditions of Theorem 11.1.7 ensuring the existence of the regular weak solution with higher regularity of the direct problem (11.6.1) are satisfied. Suppose in addition the Neumann measured output satisfies the regularity condition M ∈ H 2 (0, T ). Then the Tikhonov functional J (r), r ∈ R4 ⊂ H 4 (0, ), corresponding to the inverse coefficient problem (11.5.1)–(11.5.2)

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

449

is Fréchet differentiable. Moreover, for the Fréchet gradient of this functional the following gradient formula holds: 

T

∇J (r)(x) = −

uxx (x, t; r)ψxx (x, t; r)dt, x ∈ (0, ),

(11.6.23)

0

where φ(x, t; r) is the weak solution solutions of the adjoint problem (11.6.17) with the input (11.6.19). Examining the above justifications and comments together with Theorems 11.5.4 and 11.6.4, we can compare the inverse coefficient problem (11.5.1)–(11.5.2), considered in previous section, and the inverse coefficient problem (11.6.1)– (11.6.2), in order to identify the similarities of these problems, as well as their differences. The first important conclusion is that both gradient formulas (11.5.24) and (11.6.23) for the Tikhonov functionals corresponding to these problems are equally expressed through the dot product (uxx (x, ·), φxx (x, ·)L2 (0,T ) (or (uxx (x, ·), ψxx (x, ·)L2 (0,T ) ). Although, naturally, the functions in these formulas are solutions to different direct and adjoint problems. The main difference between these two problems is that the Neumann measured output (bending moment) requires more regularity than the Dirichlet measured output (slope) as conditions (11.5.21) and (11.6.22) show. As a result of all this, in the inverse coefficient problem (11.5.1)–(11.5.2), with Dirichlet measured output, we are dealing with a regular weak solution u ∈ L2 (0, T ; H 4(0, )∩V 2 (0, )) of the direct problem (11.5.1), while in the inverse coefficient problem (11.6.1)–(11.6.2), with Neumann measured output, we are dealing with a regular weak solution with higher regularity u ∈ L2 (0, T ; H 8(0, ) ∩ V12 (0, )) of the direct problem (11.6.1).

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured Boundary Slope In this section, the methodology discussed in the previous sections for the inverse problems related to Euler-Bernoulli beam equation, is developed for twodimensional analogue of these problems, governed by Kirchhoff plate equation. The governing equation of forced damped motion of a non-homogeneous isotropic rectangular plate has the following form [127, 152]:   ρ(x)h(x)ut t + μ(x)ut + D(x)(ux1 ,x1 + νux2 ,x2 ) x   + D(x)(ux2 ,x2 + νux1 ,x1 ) x



2 ,x2

1 ,x1

+ 2(1 − ν) D(x)ux1 ,x2

 x1 ,x2

= F (x)G(t), (x, t) ∈ T := × (0, T ).

(11.7.1)

450

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations x2

Fig. 11.14 A thin rectangular plate under transverse loading: boundary conditions and measured output (slope)

u = 0, ux2 = 0

2 Γ4 u=0 ux1 = 0

Q(x, t)

Γ3

u=0 ux1 = 0

Ω

x1

θ(x1 , t) := ux2 (x, t), on Γ1 × [0, T ] 0

u = 0, νux1 x1 + ux2 x2 = 0

Γ2

1

Here and below, := {x = (x1 , x2 ) ∈ R2 : x1 ∈ (0, 1 ), x2 ∈ (0, 2 )}, 1 , 2 > 0, is the rectangular domain, with the boundary := ∂ , occupied by the plate (Fig. 11.14). The functions F (x) and G(t) represent the spatial and temporal components of the transverse load Q(x, t) := F (x)G(t), and u(x, t) is the normal component of the displacement vector in position x ∈ and at time t ∈ (0, T ), called the deflection. Further, ρ(x) > 0 is the density of the plate, and D(x) =

E(x)h3 (x) 12(1 − ν 2 )

is the flexural rigidity of the plate, while E(x) is the Young’s modulus, ν ∈ (0, 1) is the Poisson’s ratio and h(x) > 0 is the thickness of a plate. The parameter D(x) plays the same role as the flexural rigidity r(x) = E(x)I (x) in beam bending. However, it should be noted at this point that D(x) > r(x), which means a plate is always stiffer than a beam of the same span and thickness [152]. The term μ(x)ut in the Kirchhoff plate equation (11.7.1) represents viscous-type damping which proportional to the velocity field. Considering in this section the inverse source problem of identifying the unknown spatial load F (x) in Eq. (11.7.1), we simultaneously aim to demonstrate how the ideas and methods developed in the previous sections for the inverse problems related to Euler-Bernoulli beam, can be applied to similar inverse source problem for the Kirchhoff plate as well. We hope that the reader will be able to successfully use these ideas to solve other inverse problems related to the Kirchhoff plate equation. Consider the following inverse problem of identifying the unknown spatial load distribution F (x) acting on a clamped-supported non-homogeneous isotropic plate.

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

451

Find an unknown spatial load F (x) in   ⎧ ρ(x)h(x)ut t + μ(x)ut + D(x)(ux1 ,x1 + νux2 ,x2 ) x ,x ⎪ ⎪ ⎪     1 1 ⎪ ⎪ ⎪ + D(x)(u + νu ) + 2(1 − ν) D(x)ux1 ,x2 x ,x ⎪ x2 ,x2 x1 ,x1 x2 ,x2 ⎪ 1 2 ⎪ ⎪ ⎪ ⎨ = F (x)G(t), (x, t) ∈ T := × (0, T ), (11.7.2) ⎪ ⎪ ⎪ u(x, 0) = 0, u (x, 0) = 0, x ∈ , t ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u(x, t) = 0, ∂n u(x, t) = 0, (x, t) ∈ i × [0, T ], i = 2, 3, 4, ⎪ ⎩ u(x, t) = 0, D(x)(νux1 x1 (x, t) + ux2 x2 (x, t)) = 0, (x, t) ∈ 1 × [0, T ], from the measured slope of deflection normal to the boundary θ (x1 , t) := ux2 (x, t), (x, t) ∈

1

1:

× [0, T ],

(11.7.3)

where ∂n u denotes the normal derivative (i.e. slope of deflection [127]) of u(x, t) at the boundary := ∪4i=1 i , with 1 = (0, 1 ) × {0}, 2 = {1 } × (0, 2 ), 3 = (0, 1 ) × {2 }, 4 = {0} × (0, 2 ). The function (measured output) θ (x1 , t) in (11.7.3) is the slope of the tangent to the trace of the middle surface of the plate in the plane Oxz or, which is the same, is the angle of rotation of the normal to the middle surface in the same plane Oxz. In this model, the plate is clamped (or fixed) along the three edges denoted by 2 , 3 and 4 , that is, the deflection and slope are zero along these edges. The edge of the plate denoted by 1 , is simply supported, that is, the deflection and the bending moment are both zero: u(x, t) = 0,

  M2 := D(x) νux1 x1 + ux2 x2 = 0, (x, t) ∈

1

× [0, T ].

(11.7.4)

From the first condition of (11.7.4) it follows that along the edge 1 , all the partial derivatives of u(x, t) with respect to x1 are zero. In particular, ux1 (x, t) = 0 and ux1 x1 (x, t) = 0, (x, t) ∈ 1 × [0, T ]. Taking into account the second condition of (11.7.4), this implies that ux2 x2 (x, t) = 0, (x, t) ∈ 1 × [0, T ]. Therefore, the simply supported boundary condition can also appear in the following equivalent form: u(x, t) = 0, ux2 x2 (x, t) = 0, (x, t) ∈

1

× [0, T ].

452

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Here and below, we assume that the inputs in (11.7.2) satisfy the following basic conditions: ⎧ ρ, D, μ ∈ L∞ ( ), h ∈ C( ), ⎪ ⎪ ⎪ ⎪ ⎨ 0 < D ≤ D(x) ≤ D , 0 < ρ ≤ ρ(x) ≤ ρ , 0 1 0 1 (11.7.5) ⎪ ≤ h(x) ≤ h , 0 ≤ μ ≤ μ(x) ≤ μ , 0 < h ⎪ 0 1 0 1 ⎪ ⎪ ⎩ F ∈ L2 ( ), G ∈ L2 (0, T ), G(t) ≡ 0. Introduce the set of admissible spatial loads F = {F ∈ L2 ( ) : F L2 ( ) ≤ γF },

(11.7.6)

where γF > 0 is an independent on F (x) constant. Let F ∈ F be given spatial source and u = u(x, t; F ) be the corresponding solution of the initial boundary value problem (11.7.2), defined as a direct problem. We can adopt the results given in [14] and [38] to prove that under conditions (11.7.5) there exists a unique weak solution of (11.7.2) that belongs to L2 (0, T ; V 2 ( )) with ut ∈ L2 (0, T ; L2 ( )) and ut t ∈ L2 (0, T ; H −2 ( )), where V 2 (0, ) := {v ∈ H 2 ( ) : v| = ∂n v|

\

1

= 0}.

From the above definition of the weak solution it follows that the trace ux2 | 1 := ux2 (x1 , 0, t; F ) on the boundary 1 = (0, 1 ) × {0} is well-defined. This trace is defined as an output of the inverse problem (11.7.1)–(11.7.2), corresponding to the input F ∈ F . Introduce now the input-output map corresponding to the inverse problem (11.7.2)–(11.7.3) as follows: (F )(x, t) := ux2 (x, t; F )| 1 , F ∈ F , (x, t) ∈  : F → L2 (0, T ; L2 (

1

× (0, T ).

1 )),

(11.7.7)

Then we can reformulate the inverse problem as the following linear operator equation: (F )(x, t) = θ (x1 , t), F ∈ F , θ ∈ L2 (0, T ; L2 (0, 1 )), (x, t) ∈

1

× (0, T ).

(11.7.8)

However, as noted earlier, this equality cannot hold due to measurement errors in the measured output θ (x1 , t), and therefore one should introduce the Tikhonov functional J (F ) =

1

F − θ 2L2 (0,T ;L2 ( 2

1 ))

, F ∈ F , θ ∈ L2 (0, T ; L2 (0, 1 )) (11.7.9)

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

453

and look for a quasi-solution of the inverse problem (11.7.2)–(11.7.3) as a solution of the following minimization problem J (f∗ ) = inf J (F ). F ∈F

(11.7.10)

11.7.1 Necessary Estimates for the Direct Problem Solution In this subsection, we present some estimates necessary for studying the inverse problem. Multiply both sides of Eq. (11.7.2) by ut (x, t), integrate it over t := × (0, t) and then apply the integration by parts formula. Taking into account the homogeneous initial and boundary conditions in (11.7.2), after elementary transformations we obtain the following energy identity: 1 2



1 + 2

ρ(x)h(x)u2t dx + 

 t 0



μ(x)u2τ (x, τ )dxdτ

 D(x) u2x1 x1 + 2u2x1 x2 + u2x2 x2 + 2ν(ux1 x1 ux2 x2 − u2x1 x2 ) dx =

 t 0

G(τ )F (x)uτ (x, τ )dxdτ, t ∈ [0, T ].

(11.7.11)



The sum of the first and second left-hand-side integrals in (11.7.11) represents the kinetic energy of the plate, and the third integral represent the strain energy of bending. Theorem 11.7.1 Assume that the inputs in (11.7.2) satisfy the basic conditions (11.7.5). Then for the weak solution u ∈ L2 (0, T ; V 2 ( )) of the direct problem (11.7.2) the following estimates hold:

ut 2L2 (0,T ;L2 ( ) ≤ C12 F 2L2 ( ) G 2L2 (0,T ) , 2

2 i,j =1 uxi xj L2 (0,T ;L2 ( )) ≤

C0 C12

F 2L2 ( ) G 2L2 (0,T ) , D0

(11.7.12)

where C0 = ρ0 h0 , C12 = [exp(T /C0 ) − 1] and ρ0 , D0 , h0 > 0 are the constants introduced in (11.7.5).

454

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Proof We use the inequality 2ab ≤ a 2 +b2 in the right-hand-side integral of identity (11.7.11) and then use the bounds in (11.7.5) to deduce the following integral inequality:  C0

u2t dx

+D0

+2

 

≤

 t 0



 t 0



μ(x)u2τ (x, τ )dxdτ

u2x1 x1 + 2u2x1 x2 + u2x2 x2 + 2ν(ux1 x1 ux2 x2 − u2x1 x2 ) dx

u2τ (x, τ )dxdτ + F 2L2 ( ) G 2L2 (0,T ) , t ∈ [0, T ],

(11.7.13)

with C0 = ρ0 h0 . To evaluate the third left-hand-side integral in (11.7.13) we use the identity  

ux1 x1 ux2 x2 − u2x1 x2 dx = 0, ∀u ∈ L2 (0, T ; V 2 ( )),

(11.7.14)

which holds for clamped and simply supported plates [152], and can be easily proven by transforming the integral in (11.7.14). Then we obtain the main integral inequality:  C0

u2t dx

+2

 t 0

+D0

 

≤

 t 0





μ(x)u2τ (x, τ )dxdτ

u2x1 x1 + 2u2x1 x2 + u2x2 x2 dx

u2τ (x, τ )dxdτ + F 2L2 ( ) G 2L2 (0,T ) , t ∈ [0, T ].

(11.7.15)

As a first consequence of inequality (11.7.15) we deduce that 

u2t dx

1 ≤ C0

 t 0



u2τ (x, τ )dxdτ +

1

F 2L2 ( ) G 2L2 (0,T ) , t ∈ [0, T ]. C0

Applying the Gronwall-Bellman inequality we obtain: 

u2t dx ≤

1

F 2L2 (L2 ( ) G 2L2 (0,T ) exp(t/C0 ), t ∈ [0, T ]. (11.7.16) C0

Integrating this inequality over (0, T ) we arrive at the first estimate of (11.7.12).

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

455

Use now (11.7.16) in the second consequence  

u2x1 x1 + 2u2x1 x2 + u2x2 x2 dx 1 ≤ D0

 t 0



u2τ (x, τ )dxdτ +

1

F 2L2 ( ) G 2L2 (0,T ) , t ∈ [0, T ] D0

of (11.7.15) and integrate it over (0, T ). Then we arrive at the second required estimate of (11.7.12).   Corollary 11.7.1 Under conditions of Theorem 11.7.1 the following trace estimate holds:

ux2 2L2 (0,T ;L2 (

1 ))

≤ C22 F 2L2 ( ) G 2L2 (0,T ) ,

(11.7.17)

where C22 = 2 C0 C12 /D0 , and C1 > 0 is the constant introduced in Theorem 11.7.1. Theorem 11.7.2 Assume that conditions (11.7.5) are satisfied. Suppose, in addition that the temporal load satisfies the regularity condition G ∈ H 1 (0, T ). Then for the regular weak solution u ∈ L2 (0, T ; H 4 ( )), with ut ∈ L2 (0, T ; V 2 ( )), ut t ∈ L2 (0, T ; L2 ( )) and ut t t ∈ L2 (0, T ; H −2 ( )) of the direct problem (11.7.2) the following estimates hold:

ut t 2L2 (0,T ;L2 ( )) ≤ C12 F 2L2 ( ) G 2L2 (0,T ) , 2

2 i,j =1 uxi xi t L2 (0,T ;L2 ( )) ≤

(11.7.18) C0 C12

G 2L2 ( ) G 2L2 (0,T ) , D0

where C1 > 0 is the constant introduced in Theorem 11.7.1. Proof Differentiate Eq. (11.7.2) with respect to t ∈ (0, T ), multiply both sides by 2ut t (x, t), integrate it over t := × (0, t) and use the initial and boundary conditions in (11.7.2). Then obtain the following integral identity: 

ρ(x)h(x)u2t t dx + 2  +

 t 0



μ(x)u2τ τ (x, τ )dxdτ

 D(x) u2x1 x1 t + 2u2x1 x2 t + u2x2 x2 t + 2ν(ux1 x1 t ux2 x2 t − u2x1 x2 t ) dx =2

 t 0

F (x)G (τ )uτ τ (x, τ )dxdτ, t ∈ [0, T ].

The rest of the proof is carried out in the same way as the proof of the previous theorem.  

456

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Corollary 11.7.2 Assume that conditions of Theorem 11.7.2 are satisfied. Then for the regular weak solution of the direct problem (11.7.2) the following trace estimate holds:

ux2 t 2L2 (0,T ;L2 (

1 ))

≤ C22 F 2L2 ( ) G 2L2 (0,T ) ,

(11.7.19)

where C2 > 0 is the same constant introduced in Corollary 11.7.1. Consider now the following auxiliary problem for homogeneous Kirchhoff plate equation, with nonhomogeneous Neumann boundary input (bending moment) at the boundary 1 × [0, T ]: ⎧   ρ(x)h(x)vt t + μ(x)vt + D(x)(vx1 ,x1 + νvx2 ,x2 ) x ,x ⎪ ⎪ ⎪ ⎪     1 1 ⎪ ⎪ + D(x)(v + νv ) + 2(1 − ν) D(x)vx1 ,x2 x ,x ⎪ x ,x x ,x 2 2 1 1 x2 ,x2 ⎪ 1 2 ⎪ ⎪ ⎪ ⎪ ⎪ = 0, (x, t) ∈ T := × (0, T ), ⎪ ⎨ (11.7.20) v(x, 0) = 0, vt (x, 0) = 0, x ∈ , ⎪ ⎪ ⎪ ⎪ ⎪ v(x, t) = 0, ∂n v(x, t) = 0, (x, t) ∈ i × [0, T ], i = 2, 3, 4, ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ v(x, t) = 0, D(x)(νvx1 x1 (x, t) + vx2 x2 (x, t)) = M(t), ⎪ ⎪ ⎪ ⎩ (x, t) ∈ 1 × [0, T ]. Theorem 11.7.3 Assume that the inputs in (11.7.20) satisfy conditions (11.7.5).  ∈ H 1 (0, T ; L2 ( 1 )). Then for the weak solution v ∈ Suppose in addition M L2 (0, T ; V 2 ( )), with vt ∈ L2 (0, T ; L2 ( )) and vt t ∈ L2 (0, T ; H −2 ( )) of problem (11.7.20) the following estimates hold: 2 M

 21

vx1 x2 2L2 (0,T ;L2 ( )) ≤ C 1 H (0,T ;L2 (

1 ))

,

vx1 x1 2L2 (0,T ;L2 ( )) + vx2 x2 2L2 (0,T ;L2 ( )) 2 M

 21 ≤C 1 H (0,T ;L2 ( 2 M

 21

vt 2L2 (0,T ;L2 ( )) ≤ C 2 H (0,T ;L2 (

(11.7.21) 1 ))

,

1 ))

where 2 = [exp(T ) − 1]C 2 , C 2 = D0 [exp(T ) − 1], C 1 0 2 C0 2 # C  2 T #2 2 = , CT = max(4T , 1 + 3/T ) C 0 D02 and ρ0 , D0 , h0 > 0 are the constants introduced in (11.7.5).

(11.7.22)

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

457

Proof As in the proof of Theorem 11.7.1, we derive the following integral identity corresponding to problem (11.7.20): 

ρ(x)h(x)vt2 dx + 2 

 t 0



μ(x)u2τ (x, τ )dxdτ



+

D(x) vx21 x1 + 2vx21 x2 + vx22 x2 + 2ν(vx1 x1 vx2 x2 − vx21 x2 ) dx

= −2

 t 0

1

0

 1 , τ )vx2 τ (x1 , 0, τ )dx1 dτ, t ∈ [0, T ]. M(x

(11.7.23)

We transform the right-hand-side integral in (11.7.23) by applying the integration by parts formula: −2

 t 0

1 0

 1 , τ )vx2 τ (x1 , 0, τ )dx1 dτ M(x

=2

 t 0

1 0

τ (x1 , τ )(x)vx2 (x1 , 0, τ )dx1 dτ M 

−2 0

1

 1 , t)vx2 (x1 , 0, t)dx1 dτ. M(x

Using here the ε-inequality we deduce that −2

 t 0

1 0

≤ε

 1 , τ )(x)vx2 τ (x1 , 0, τ )dx1 dτ M(x

 t  0

+

1 ε

1 0



 vx22 (x1 , 0, τ )dx1 dτ +



T 0

1 0

τ2 (x1 , t)dx1 dt + M

1 0



 vx22 (x1 , 0, t)dx1

1

  2 (x1 , t)dx1 . M

(11.7.24)

0

For the first two right-hand-side integrals in (11.7.24) we employ trace inequality to conclude that  t  1   1 2 2 ε vx2 (x1 , 0, τ )dx1 dτ + vx2 (x1 , 0, t)dx1 0

0

≤ 2 ε

0

 t  0



vx21 x2 (x, τ )dxdτ +



 vx21 x2 (x, t)dx .

(11.7.25)

458

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

Further, we apply to the last two right-hand-side integrals in (11.7.24) the first inequality in (B.3.7) of Appendix B, to get #T2 w 2 1

w 2L2 (0,T ) + w2 (t) ≤ C , w ∈ H 1 (0, T ), ∀t ∈ [0, T ], H (0,T ) #T > 0 defined in (11.7.22). Then we obtain: with the constant C 1 ε



T



0

1 0

τ2 (x1 , t)dx1 dt + M ≤



1

 2 (x1 , t)dx1 M



0

1 #2  2 C M H 1 (0,T ;L2 ( ε T

1 ))

.

(11.7.26)

Use now (11.7.25) and (11.7.26), with identity (11.7.14), in (11.7.23). This leads to  C0

vt2 dx

≤ 2 ε

+ D0

 

 t 0



vx21 x1

+ vx22 x2

vx21 x2 dxdτ +



 dx + (2D0 − 2 ε)

1 #2  2 C M H 1 (0,T ;L2 ( ε T

1 ))

vx21 x2 dx

, t ∈ [0, T ].

Choosing the parameter ε > 0 from the condition 2D0 − 2 ε > 0 as ε = D0 /2 , finally we obtain the main integral inequality corresponding to problem (11.7.20):  C0

≤ D0

 t 0



vt2 dx

+ D0

 

vx21 x2 dxdτ +



vx21 x1

+

vx22 x2



 dx + D0

#2 2 C T  21

M

H (0,T ;L2 ( D0



1 ))

u2x1 x2 dx

, t ∈ [0, T ].

Estimates (11.7.21) are derived from this inequality in the same way as in the proof of Theorem 11.7.1.  

11.7.2 The Input-Output Operator. Existence of a Quasi-Solution Lemma 11.7.1 Assume that conditions of Theorem 11.7.2 are satisfied. Then the input-output operator  : F → L2 (0, T ; L2 ( 1 )) defined in the set of admissible spatial loads F is a linear compact operator. Proof Let {F (m) } ⊂ F , m = 1, ∞ be a sequence of spatial loads, uniformly bounded by definition (11.7.6). Denote by {u(m) (x, t)}, u(m) (x, t) := u(x, t; F (m) ) the sequence of corresponding regular weak solutions of the direct problem (11.7.2).

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

459

(m)

Then {ux2 (x, t)| 1 } is the sequence of outputs. We need to prove that this sequence is relatively compact subset of L2 (0, T ; L2 ( 1 )) or, equivalently, the sequences (m) (m) {ux | 1 } and {uxt | 1 } are bounded in the norm of L2 (0, T ; L2 ( 1 )). The last assertions follow from the trace estimates (11.7.17) and (11.7.19). Namely, (m)

ux2 2L2 (0,T ;L2 (

1 ))

≤ C22 F (m) 2L2 ( ) G 2L2 (0,T ) ,

1 ))

≤ C22 F (m) 2L2 ( ) G 2L2 (0,T ) ,

(m)

ux2 t 2L2 (0,T ;L2 (

and the right-hand sides of both inequalities are bounded by the constant Cc > 0, where Cc2 = C22 γF G 2H 1 (0,T ) . Since H 1 ( 1 ) is compactly embedded into L2 ( 1 ) the above inequalities imply (m) the compactness of the sequence {ux2 (x, t)| 1 } in L2 ( 1 ).   As a consequence of this lemma, the inverse source problem (11.7.2)–(11.7.3) is ill-posed. Lemma 11.7.2 Let conditions of Theorem 11.7.1 hold. Then the input-output operator  : F → L2 (0, T ; L2 ( 1 )) is Lipschitz continuous, that is

F1 − F2 L2 (0,T ;L2 (

1 ))

≤ L F1 − F2 L2 ( ) , F1 , F2 ∈ F ,

(11.7.27)

where L = C2 G 2L2 (0,T ) is the Lipschitz constant and C2 > 0 is the constant introduced in Corollary 11.7.1. Proof Let F1 , F2 ∈ F and δF (x) = F1 (x) − F2 (x). Then δu(x, t) = u(x, t; F1 ) − u(x, t; F2 ) solves the following problem: ⎧   ⎪ ρ(x)h(x)δut t + μ(x)δut + D(x)(δux1 x1 + νδux2 x2 ) x x ⎪ 1 1 ⎪ ⎪     ⎪ ⎪ ⎪ + D(x)(δux2 x2 + νδux1 x1 ) x x + 2(1 − ν) D(x)δux1 x2 x x ⎪ 2 2 1 2 ⎪ ⎪ ⎪ ⎪ ⎪ = δF (x)G(t), (x, t) ∈ T, ⎪ ⎨ (11.7.28) δu(x, 0) = 0, δut (x, 0) = 0, x ∈ , ⎪ ⎪ ⎪ ⎪ ⎪ δu(x, t) = 0, ∂n δu(x, t) = 0, (x, t) ∈ i × [0, T ], i = 2, 3, 4, ⎪ ⎪ ⎪ ⎪ ⎪ δu(x, t) = 0, − D(x)(νδux1 x1 (x, t) + δux2 x2 (x, t)) = 0, ⎪ ⎪ ⎪ ⎪ ⎩ (x, t) ∈ 1 × [0, T ]. To estimate the trace norm

δux2 2L2 (0,T ;L2 (

1 ))

=: F1 − F2 L2 (0,T ;L2 (

1 ))

460

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

we employ Corollary 11.7.1 applied to the weak solution of problem (11.7.28). In view of estimate (11.7.17) we get:

δux2 2L2 (0,T ;L2 (

1 ))

≤ C22 δF 2L2 ( ) G 2L2 (0,T ) , F1 , F2 ∈ F ,  

which implies the required result (11.7.27).

Based on this lemma, we can prove the Lipschitz continuity of the Tikhonov functional in the same way as in Lemma 11.2.3. Theorem 11.7.4 Let conditions of Theorem 11.7.1 hold. Then the minimization problem (11.7.10) for the Tikhonov functional (11.7.9) has a solution in the set of admissible spatial loads F , defined in (11.7.6). This theorem is proved exactly in the same way as in Theorem 10.1.11 in Sect. 10.1. From Theorem 11.7.4 it follows that the inverse source problem (11.7.2)–(11.7.3) has a quasi-solution.

11.7.3 Fréchet Differentiability of the Tikhonov Functional. Gradient Formula Let F, F + δF ∈ F . Denote by δJ (F ) := J (F + δF ) − J (F ) the increment of the Tikhonov functional (11.7.9). Then 

T

δJ (F ) =



0

1 0



ux2 (x1 , 0, t; F ) − θ (x1 , t) δux2 (x1 , 0, t)dx1 dt +

1 2



T 0



1 0



δux2 (x1 , 0, t)

2

dx1 dt,

(11.7.29)

where δux2 (x1 , 0, t) = ux2 (x1 , 0, t; F + δF ) − ux2 (x1 , 0, t; F ), while ux2 (x1 , 0, t; F + δF ) and ux (x1 , 0, t; F ) are the outputs corresponding to the inputs F, F + δF ∈ F . Lemma 11.7.3 Assume that the inputs in (11.7.2) satisfy the basic conditions (11.7.5). Then the following integral identity holds: 

T 0

 0

1

M(x1 , t)δux2 (x1 , 0, t)dx1 dt  



T

=

G(t)φ(x, t)dt

0

δF (x)dx,

(11.7.30)

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

461

where δu(x, y, t) is the week solution of problem (11.7.28) and φ(x, t) is the weak solution of the following backward problem   ⎧ ρ(x)h(x)φt t − μ(x)φt + D(x)(φx1 x1 + νφx2 x2 ) x x ⎪ ⎪ ⎪     1 1 ⎪ ⎪ ⎪ + D(x)(νφ + φ ) + 2(1 − ν) D(x)φ x x x x x x ⎪ 1 1 2 2 1 2 x x x1 x2 ⎪ 2 2 ⎪ ⎪ ⎪ ⎪ = 0, (x, t) ∈ T , ⎪ ⎨ (11.7.31) φ(x, T ) = 0, φt (x, T ) = 0, x ∈ , ⎪ ⎪ ⎪ ⎪ ⎪ φ(x, t) = 0, ∂n φ(x, t) = 0, (x, t) ∈ i × [0, T ], i = 2, 3, 4, ⎪ ⎪ ⎪ ⎪ ⎪ φ(x, t) = 0, − D(x)(νφx1 x1 (x, t) + φx2 x2 (x, t)) = M(x1 , t), ⎪ ⎪ ⎪ ⎩ (x, t) ∈ 1 × [0, T ], with the input M ∈ H 1 (0, T ; L2 (0, 1 )). Proof Multiply both sides of Eq. (11.7.28) for δu(x, t) by arbitrary function φ(x, t), integrate over T and apply the integration by parts formula multiple times. Then assume that φ(x, t) solves the backward problem (11.7.31). Taking into account the homogeneous initial, final and boundary conditions in (11.7.28) and (11.7.31) we arrive at the following integral identity: 

T 0



1 0



−D(x)(νφx1 x1 + φx2 x2 )δux2 

T

=

x2 =0

dx1dt

 G(t)φ(x, t)δF (x) dxdt.

0



With the non-homogeneous Neumann boundary condition in the backward problem (11.7.31) this leads to the required integral relationship (11.7.30).   Choose now formally the arbitrary input M(x1 , t) in (11.7.31) as follows: M(x1 , t) = ux2 (x1 , 0, t; F ) − θ (x1 , t), t ∈ [0, T ].

(11.7.32)

The backward problem (11.7.31) with the input defined in (11.7.32) is the adjoint problem corresponding to the direct problem (11.7.2). Evidently, this is a well-posed problem as the change of variable t by T −t shows. Furthermore, under the condition M ∈ H 1 (0, T ; L2 (0, 1 )) for the weak solution of this adjoint problem estimates given in Theorem 11.7.3 hold.

462

11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

With the above defined input (11.7.32), identity (11.7.30) takes the following form: 

T 0



1 0

 ux2 (x1 , 0, t; F ) − θ (x1 , t) δux2 (x1 , 0, t)dx1 dt  



T

=

G(t)φ(x, t; F )dt δF (x)dx, (11.7.33) 0



The integral identity (11.7.33) reveals not only a relationship between the inputs F (t), G(t), output ux2 (x1 , 0, t; F ) and the measured output θ (x1 , t), through the solution φ(x, t) of the adjoint problem (11.7.31) with input (11.7.32). It also clearly shows how the change δF (x) in input reflects on the change δux2 (x1 , 0, t; F ) in output during the time interval [0, T ]. In view of relationship (11.7.33) the increment formula (11.7.29) for the Tikhonov functional takes the form:   T δJ (F ) = G(t)φ(x, t; F )dt δF (x)dx

0

+

1 2



T 0



1 0



δux2 (x1 , 0, t)

2

dx1 dt.

(11.7.34)

Theorem 11.7.5 Assume that the inputs in the direct problem (11.7.2) satisfy the basic conditions (11.7.5). Suppose, in addition, the measured output θ (x1 , t) satisfies the regularity condition θ ∈ H 1 (0, T ; L2 (0, 1 )). Then the Tikhonov functional J (F ) defined on the set of admissible inputs F is Fréchet differentiable. Moreover, for the Fréchet gradient of this functional the following gradient formula holds: 

T

∇J (F )(x) =

G(t)φ(x, t; F )dt, x ∈ ,

(11.7.35)

0

where φ(x, t; F ) is the weak solution of the adjoint problem (11.7.31) with the input defined in (11.7.32). Proof From the trace estimate (11.7.17) applied to the solution δu(x, t) of problem (11.7.28) it follows that

δux2 2L2 (0,T ;L2 (

1 ))

≤ C22 G 2L2 (0,T ) δF 2L2 ( ) .

This means that the last right-hand-side integral in (11.7.34) is of the order   O( δf 2L2 ( ) ). This completes the proof. Remark 11.7.1 Remark 11.2.1 is also valid for this inverse problem. Namely, one can weaken the conditions of the Theorem 11.7.5, assuming that the measured

11.7 Spatial Load Identification in a Vibrating Kirchhoff Plate from Measured. . .

463

output θ (x1 , t) belongs to L2 (0, T ; L2 (0, 1 )). This can be seen from the gradient formula (11.7.35), which indicates that we need only a weaker solution φ ∈ L2 (0, T ; L2 (0, )) to the adjoint problem (11.7.31), but not the weak solution φ ∈ L2 (0, T ; V 2(0, )). The more regular measured output θ ∈ H 1 (0, T ; L2 (0, 1 )) ensures the Lipschitz continuity of the Fréchet gradient, which in turn entails a more important property of the Tikhonov functional. Namely, in this case, the sequence of iterations {J (F (n) )} is monotonically decreasing, according to Lemma 3.4.4. Theorem 11.7.6 Assume that the inputs in the direct problem (11.7.2) satisfy the basic conditions (11.7.5), and the measured output θ (x1 , t) satisfies the regularity condition θ ∈ H 1 (0, T ; L2 (0, 1 )). Then the Fréchet gradient (11.7.35) of the Tikhonov functional J (F ) is Lipschitz continuous, that is

∇J (F1 ) − ∇J (F2 ) L2 ( ) ≤ L∇ F1 − F2 L2 ( ) , F1 , F2 ∈ F , (11.7.36) where L∇ =

√ 1 1 C2 G L2 (0,T ) G H 1 (0,T ) 1 2 T C 2

is the Lipschitz constant depending on the geometric parameters, final time, norms of the temporal load, and also the constants defined in Theorem 11.7.1 and Corollary 11.7.1. Proof Let δF (x) = F1 (x) − F2 (x). Then by the gradient formula (11.7.35) we have:  

∇J (F1 ) − ∇J (F2 ) 2L2 ( )

=

2

T

G(t)δφ(x, t)dt

dx. (11.7.37)

0

where δφ(x, t) = φ(x, t; F1 ) − φ(x, t; F2 ) solves the following backward problem: ⎧   ρ(x)h(x)δφt t + μ(x)δφt + D(x)(δφx1 x1 + νδφx2 x2 ) x x ⎪ ⎪ 1 1 ⎪ ⎪     ⎪ ⎪ ⎪ + D(x)(δφ + νδφ ) + 2(1 − ν) D(x)δφ x x x x x1 x2 x1 x2 ⎪ 2 2 1 1 x2 x2 ⎪ ⎪ ⎪ ⎪ ⎪ = 0, (x, t) ∈ T , ⎪ ⎪ ⎨ (11.7.38) δφ(x, T ) = 0, δφt (x, T ) = 0, x ∈ , ⎪ ⎪ ⎪ ⎪ ⎪ δφ(x, t) = 0, ∂n δφ(x, t) = 0, (x, t) ∈ i × [0, T ], i = 2, 3, 4, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ δφ(x, t) = 0, − D(x)(νδφx1 x1 (x, t) + δφx2 x2 (x, t)) ⎪ ⎪ ⎪ ⎪ ⎩ = δux2 (x1 , 0, t; F ), (x, t) ∈ 1 × [0, T ], where δux2 (x1 , 0, t) = ux2 (x1 , 0, t; F1 ) − ux2 (x1 , 0, t; F2 ).

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11 Inverse Problems for Euler-Bernoulli Beam and Kirchhoff Plate Equations

From (11.7.37) it follows that

∇J (F1 ) − ∇J (F2 ) 2L2 ( ) ≤ T G 2L2 (0,T ) δφ 2L2 (0,T ;L2 ( )) . To estimate the last right-hand-side norm we use the Poincaré inequality:

δφ 2L2 (0,T ;L2 ( )) ≤

21 22

δφx1 x2 2L2 (0,T ;L2 ( )) 4

to deduce that

∇J (F1 ) − ∇J (F2 ) 2L2 ( ) ≤

21 22 T

G 2L2 (0,T ) δφx1 x2 2L2 (0,T ;L2 ( )) . 4

(11.7.39)

Now, we employ the first estimate of (11.7.21) applied to the weak solution of  1 , t) = δux2 (x1 , 0, t). We have: problem (11.7.38) with the input M(x 12 δux2 2 1

δφx1 x2 2L2 (0,T ;L2 ( )) ≤ C H (0,T ;L2 ( To estimate the H 1 -trace norm δux2 2H 1 (0,T ;L2 ( (11.7.17) and (11.7.19). Then we get:

1 ))

1 ))

.

we use the trace estimates

12 C22 G 2 1

δφx1 x2 2L2 (0,T ;L2 ( )) ≤ C

δF 2L2 ( ) . H (0,T ) With estimate (11.7.39) this yields the required statement (11.7.36).

 

Appendix A

Invertibility of Linear Operators

A.1 Invertibility of Bounded Below Linear Operators The analysis given in Introduction allows us to conclude that most inverse problems are governed by compact operators. More specifically, all inverse problems related to PDEs are governed by corresponding input-output operators which are compact operator. The last property mainly resulted from the Sobolev embedding theorems. Therefore compactness of an operator, being an integral part of inverse problems, is a main source of ill-posedness of inverse problems. We begin some well-known fundamental results related to invertibility of linear operators in Banach and, in particular, in Hilbert spaces, since compact operators on a Banach space are always completely continuous. Details of these results can be found in textbooks on functional analysis (see, for example, [5, 34, 155, 156]). Let A : D ⊂ B → B˜ be a linear continuous operator on the Banach space B ˜ Consider the operator equation Au = F , F ∈ R(A) ⊂ B, ˜ into the Banach space B. where R(A) is the range of the operator A. We would like to ask whether the linear continuous operator A has a bounded inverse, because this is a first step for solving the operator equation Au = F . By definition of an inverse operator, if for any F ∈ R(A) there is at most one u ∈ D(A) such that Au = F , i.e. if the operator A is injective, then the correspondence from B˜ to B defines an operator, which is called an inverse operator. This definition directly implies that A−1 exists if and only if the equation Au = 0 has only the unique solution u = 0, i.e. the null space N (A) := {u ∈ B : Au = 0} of the operator A consists of only zero element: N (A) = {0}. Evidently, if A−1 exists, it is also a linear operator. Note that if H0 is a Hilbert space, then by0the orthogonal decomposition theorem, H = N (A) N (A)⊥ , where N (A) N (A)⊥ := {z ∈ H : z = u + v, u ∈ N (A), v ∈ N (A)⊥ } and N (A)⊥ is the orthogonal complement of N (A), i.e. (u, v)H = 0 for all u ∈ N (A) and for all v ∈ N (A)⊥ . The first group of results in this direction is related to invertibility of bounded below linear operators defined on Banach space. © Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9

465

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A Invertibility of Linear Operators

Definition A.1.1 A linear operator A defined on the Banach space B into the Banach space B˜ is called bounded below if there exists such a positive constant C > 0 that

Au B˜ ≥ C u B , ∀u ∈ D(A).

(A.1.1)

The theorem below shows an important feature of bounded below linear operators. Specifically, a bounded below linear operator A defined between the Banach spaces B and B˜ has always a continuous inverse defined on the range R(A) of A, if even A is not continuous. Theorem A.1.1 Let B and B˜ be Banach spaces and A be a linear continuous ˜ Then the inverse operator A−1 , defined on R(A), operator defined on B into B. exists and bounded if and only if it is bounded below. Proof Assume, first, that the inverse operator A−1 exists and bounded on R(A). Then there exists a constant M > 0 such that A−1 F B ≤ M F B˜ , for all F ∈ R(A). Substituting here u := A−1 F we get: u B ≤ M Au B˜ . This implies (A.1.1) with C = 1/M, i.e. A is bounded below. Assume now condition (A.1.1) holds. This, first of all, implies injectivity of the operator A. Indeed, let u1 , u2 ∈ B be arbitrary elements and Au1 = Au2 = v ∈ R(A). Then, by condition (A.1.1), 0 = Au1 − Au2 B˜ = A(u1 − u2 ) B˜ ≥ C u1 − u2 B . This u1 −u2 B = 0, i.e. implies u1 = u2 . Hence for any F ∈ R(A) there is at most one u ∈ D(A) such that Au = F , i.e. the inverse operator A−1 exists. To prove the boundedness of this operator, we substitute u = A−1 F in (A.1.1). Then we have:

F B˜ ≥ C A−1 F B . This means the boundedness of the inverse operator.   Thus, this theorem gives a necessary and sufficient condition for solvability of the operator equation Au = F with bounded below linear operator A. A simple illustration of this theorem can be given by the following example. Example A.1.1 An existence of a bounded inverse. Consider the following boundary value problem. 

(Au)(x) := −u (x) = F (x), x ∈ (0, 1), u(0) = u (1) = 0.

We rewrite problem (A.1.2) in the operator equation form onto Au = F, A : D(A) −−−→ C[0, 1],

(A.1.2)

A Invertibility of Linear Operators

467

which D(A) := C 2 (0, 1) ∩ C0 [0, 1], C0 [0, 1] := {u ∈ C[0, 1] : u(0) = u (1) = 0} and R(A) := C[0, 1]. We can show easily that the operator A is bounded below. Indeed, the identities 

x

u(x) ≡







1

u (ξ )dξ, u (x) ≡ −

0

u (ξ )dξ, ∀x ∈ [0, 1]

x

imply |u(x)| ≤ sup |u (x)|, |u (x)| ≤ sup |u (x)|. [0,1]

[0,1]

Hence,

Au C[0,1] := sup |u (x)| ≥ u C[0,1] , [0,1]

which implies that the operator A is bounded below. Integrating equation (A.1.2) and using boundary conditions, we find 

x

u(x) = 0



1

F (η)dηdξ =: (A−1 F )(x), x ∈ [0, 1].

(A.1.3)

ξ

By definition of an inverse operator, the inverse operator A−1 : R(A) → D(A), defined by (A.1.3), exists. Evidently, A−1 is bounded, although the operator A := d 2 /dx 2 is unbounded.  As the second part of the proof of the above theorem shows, condition (A.1.1) contains within itself two conditions: injectivity of the operator A and continuity of the inverse operator A−1 . So, injectivity is a part of conditions for invertibility of ˜ i.e. the operator linear operators. Further, if, in addition to injectivity, R(A) ≡ B, A is defined on B onto B˜ or, equivalently, operator A is surjective, then the linear continuous operator A : D ⊂ B → B˜ is continuously invertible for all F ∈ B˜ and ˜ In this case the operator equation Au = F has a unique solution for D(A−1 ) = B. ˜ while Theorem A.1.1 asserts that the operator equation Au = F solvable all F ∈ B, only for F ∈ R(A). In other words, under the conditions of only Theorem A.1.1, ˜ As a result the operator equation Au = F may not have a solution for all F ∈ B. of these considerations, we conclude that the best case for unique solvability of the operator equation Au = F is the case when the linear continuous operator A is bijective, i.e. injective and surjective. The following result, which follows from Open Mapping Theorem, is exactly in this direction. Theorem A.1.2 Let A : B → B˜ be a linear continuous and bijective operator from ˜ Then the inverse operator A−1 : B˜ → B a Banach space B to a Banach space B. is also continuous (and hence bounded).

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A Invertibility of Linear Operators

A.2 Invertibility of Linear Compact Operators Let us return again to the surjectivity property of the linear continuous operator A, ˜ This condition means, first of all, the range R(A) i.e. to the condition R(A) ≡ B. needs to be a closed set, since B˜ being a Banach space is a closed set. In this case, i.e. if R(A) is a closed set, we can replace B˜ by R(A), and define the operator as A : D ⊂ B → R(A). However, as we will see below, the condition R(A) ≡ B˜ is a very strong restriction, and is more difficult to satisfy for compact operators. Even for non-compact continuous operators, this condition may not be fulfilled, as the following simple example shows. Example A.2.1 A linear bounded operator with non closed range. Consider the bounded linear operator A : L2 (0, 1) → L2 (0, 1), defined as 

x

(Au)(x) :=

u(y)dy, u ∈ H := L2 (0, 1).

0

˚1 (0, 1) := {v ∈ H 1 (0, 1) : u(0) = 0} The range of this operator is the subspace H 1 ˚1 (0, 1) ⊂ L2 (0, 1) is not closed of the Sobolev space H (0, 1). Since the space H 2 1 ˚ (0, 1) of the above bounded linear operator is not in L (0, 1), the range R(A) = H closed, i.e. R(A) = R(A).  ˜ but even Thus, this example shows that not only the condition R(A) ≡ B, more weaker condition R(A) ≡ R(A) may not be satisfied. If the condition R(A) ≡ R(A) holds, then the linear bounded operator A is defined as a closed range operator. Otherwise, i.e. if R(A) = R(A), A is defined as a non-closed range operator. Note that the operator equation Au = F with non-closed range operator A is defined ill-posed, according to the concept of M.Z. Nashed [107]. The following fundamental theorem shows what means the condition R(A) ≡ B˜ in terms of invertibility of linear compact operators. In particular, this theorem implies that a compact linear operator A : B → B˜ defined in infinite-dimensional Banach space B is not invertible. Theorem A.2.1 Let A : B → B˜ be a linear compact operator defined between the ˜ Assume that R(A) ≡ B, ˜ i.e. A is surjective. If A has a Banach spaces B and B. −1 ˜ bounded inverse A on B onto B, then B is finite dimensional. Note that a compact operator A : H → H˜ is called finite dimensional (or finite-rank) operator, if its range R(A) is a finite dimensional set in H˜ . As we noted above, if a bounded linear operator is of finite dimensional, then it is a compact operator. From this theorem we can deduce some important properties related to invertibility of linear compact operators. Corollary A.2.1 A surjective linear compact operator A : B → B˜ defined on a infinite-dimensional Banach space B can not have a bounded inverse.

A Invertibility of Linear Operators

469

Let us compare now Theorem A.1.2 with Theorem A.2.1, where in both the linear ˜ Theorem A.1.2 asserts that the continuous operator A is surjective, i.e. is onto B. −1 ˜ inverse operator A : B → B is continuous, while Theorem A.2.1 asserts that if the inverse operator A−1 : B˜ → B is continuous, then B˜ must be finite dimensional. This leads to the following conclusion. Corollary A.2.2 If B˜ is infinite-dimensional, there is no injective compact linear ˜ defined from B onto B. ˜ operator A : B → B, Let us analyze now what does mean surjectivity condition for linear compact operators, imposed in Theorem A.2.1. As it was noted above, this condition ˜ is a very strong restriction and may not be satisfied for some classes (R(A) ≡ B) of compact operators. What happens if the this condition does not hold, i.e. A is not surjective, and is only a closed range compact operator, i.e. R(A) ≡ R(A)? Next theorem shows that, at least in Hilbert spaces, the closedness and finite dimensionality of the range of a linear compact operators are equivalent properties. Theorem A.2.2 Let A : H → H˜ be a linear compact operator defined between two infinite-dimensional Hilbert spaces H and H˜ . Then the range R(A) is closed if and only if it is finite dimensional. Proof The proof of this theorem is based on the well-known elementary result that the identity operator I : R(A) → R(A) is compact if and only if its range R(A) is finite dimensional. Indeed, if R(A) is closed, then it is complete in 0 H˜ , i.e. is also a Hilbert space. Then we can use decomposition theorem, H = N (A) N (A)⊥ , and define the restriction A|N (A)⊥ of the compact operator A on N (A)⊥ , according the construction given in Sect. 1.3. Evidently, this restriction is continuous and bijective onto the Hilbert space R(A). Hence we may apply Theorem A.1.2 to conclude that the restricted operator A|N (A)⊥ has a bounded inverse. Then the identity operator −1  I := A A|N (A)⊥ : R(A) → R(A), defined as a superposition of compact and continuous operators, is also compact. This completes the proof.   Besides the above mentioned importance, the above theorems are directly related to ill-posed problems. Indeed, if A : H → H˜ be a linear compact operator between the infinite-dimensional Hilbert spaces H and H˜ , and its range R(A) is also infinite-dimensional, then R(A) is not closed, by the assertion of the theorem. This means that the equation Au = F is ill-posed in the sense of the condition (p1) of Hadamard’s ill-posedness, defined in Introduction. Note that the range of almost all integral (compact) operators related to integral equations with non-degenerate kernel, is infinite-dimensional. Below we will illustrate how an infinite-dimensional range compact operator, defined by the integral equation with Hilbert-Schmidt kernel function, can uniformly be approximated by a sequence of finite rank compact operators. In inverse problems this approach, i.e. the restriction of infinite-dimensional range compact operator A to a finite dimensional subspace

470

A Invertibility of Linear Operators

of H˜ , is defined as regularization by discretization. Although this approach yields a well-posed problem, since linear operators on finite dimensional spaces are always bounded (i.e. continuous), the condition number of the finite dimensional problem may be very large unless the finite dimensional subspace is chosen properly. Thus, a finite dimensionality of a range of a linear compact operator A may “remove” the ill-posedness of the equation Au = F . On the other hand, many compact operators, associated with function spaces that occur in differential and integral equations, can be characterized as being the limit of a sequence of bounded finite dimensional operators. The following theorem gives a construction of a sequence of bounded finite dimensional operators in separable Hilbert space [91]. Theorem A.2.3 Let A : H → H˜ be an infinite-dimensional linear compact operator defined on a separable Hilbert space H . Then there exists a sequence of finite dimensional linear operators {AN } such that A is a norm limit of this sequence, i.e. A − AN H˜ → 0, as n → ∞. Proof As a separable Hilbert space, H has an orthonormal basis, defined as {ψn ∈ H : n ∈ N }, and for any u ∈ H we can write u=

∞ 

u, ψn H ψn ; Au =

n=1

∞  u, ψn H Aψn . n=1

We use this orthonormal basis to define the finite dimensional linear AN : H → H˜ operator as follows: AN u :=

N  u, ψn H Aψn .

(A.2.1)

n=1

Since both operators A and AN are compact, the operator RN := A − AN is also compact. To prove the theorem, we need to show that RN → 0 or equivalently, rN := sup RN u H˜ → 0, as N → ∞.

u H ≤1

(A.2.2)

By definition of the supremum, there exists a sequence {u(n) } ⊂ H , such that

u(n) H ≤ 1, for all n ∈ N , and RN u(n) H˜ → rN , as n → ∞. Further, by Banach-Alaoglu theorem, there is a weakly convergent subsequence {u(nm ) } ⊂ {u(n) } such that u(nm )  u∗N , and u∗N H ≤ 1. Then the compact operator RN transforms this weakly convergent subsequence to the strongly convergent one: RN u(nm ) − RN u∗N H˜ → 0, as nm → ∞. In particular, RN u(nm ) H˜ →

RN u∗N H˜ . On the other hand, RN u(n) H˜ → rN , as n → ∞. This implies that rN = RN u∗N H˜ .

A Invertibility of Linear Operators

471

Let us define now the difference RN u∗N := (A − AN )u∗N using (A.2.1) and (A.2.2): RN u∗N = A

∞ 

∗ ∗ u∗N , ψn H Aψn =: ASN , SN :=

n=N+1

∞ 

u∗N , ψn H Aψn .

i=N+1

This show that in order to prove rN → 0, as N → ∞, we need to prove the weak convergence SN  0, as N → ∞, due to the compactness of the operator RN . Using the Hölder inequality and the Parseval’s equality, we have: 1 |SN , uH | =

∞ 

u∗N , ψn H ψn ,

n=N+1

1

∞ 

=

≤

∞ 

2 u, ψn H ψn

n=1 ∞ 

u∗N , ψn H ψn ,

n=N+1



∞ 

2 u, ψn H ψn

n=N+1

1/2  |u∗N , ψn H |

n=N+1

H

H

1/2

∞ 

|u, ψn H |

n=N+1

 ≤ u∗N H

∞ 

1/2 |u, ψn H |

.

n=N+1

The u∗N H ≤ 1 and the second multiplier in the last inequality tends to zero, as N → 0. Hence, for any u ∈ H , |SN , uH | → 0, as N → 0. This completes the proof.   This theorem implies that the set of finite dimensional linear operators is dense in the space of compact operators defined on a separable Hilbert space. Example A.2.2 An approximation of a compact operator associated with an integral equation. Consider the linear bounded operator defined by (1.3.2): 

b

(Au) (x) :=

K(x, y)u(y)dy, x ∈ [a, b],

a

assuming H = H˜ = L2 (a, b), K ∈ L2 ((a, b)×(a, b)) and K L2 ((a,b)×(a,b)) ≤ M. Evidently, under these assumptions the integral operator A : H → H˜ , defined as a Hilbert-Schmidt operator with non-degenerate (or non-separable) kernel function, is a compact operator. Let {φn ∈ L2 (a, b) : n ∈ N } be an orthonormal basis for L2 (a, b). Then {ψm,n ∈ L2 ((a, b) × (a, b)) : ψm,n (x, y) := φn (x)φn (y), n, m ∈ R}

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A Invertibility of Linear Operators

forms an orthonormal basis for L2 ((a, b) × (a, b)). Hence KN (x, y) :=

N 

K, ψm,n L2 ((a,b)×(a,b))ψm,n (x, y), x, y ∈ [a, b].

n,m=1

Now we define the operator 

b

(AN u) (x) :=

KN (x, y)u(y)dy, x ∈ [a, b].

a

Evidently, AN : H → H˜ N ⊂ H˜ is a finite dimensional compact operator with the range RN (AN ) ⊂ span{ψm,n (x, y)}N m,n=1 and A − AN H˜ → 0, as n → ∞, as follows from Theorem A.2.3 in Introduction.  As was noted above, the range of a compact operator is “almost finite dimensional”. Specifically, not only a compact operator itself can be approximated by finite dimensional linear (i.e. compact) operators, but also the range of a linear compact operator can be approximated by a finite dimensional subspace. This gives a further justification for the presentation of the following theorem. Theorem A.2.4 Let A : B → B˜ be a linear operator between two infinite˜ Then for any ε > 0 there exists a finite dimensional Banach spaces B and B. dimensional subspace RN (A) ⊂ R(A) such that, for any u ∈ B inf

v∈RN (A)

Au − v B ≤ ε u B .

This well-known result can be found in the above mentioned books on functional analysis. Remark finally that the approaches proposed in Theorems A.2.3 and A.2.4 are widely used within the framework of regularization methods based on discretization.

Appendix B

Necessary Estimates For One-Dimensional Parabolic Equation

We prove here some basic estimates for the weak and regular weak solutions of the following initial boundary value problems ⎧ ⎪ ⎨ ut (x, t) = (k(x)ux (x, t))x + F (x, t), (x, t) ∈ T := (0, ) × (0, T ], (B.0.1) u(x, 0) = u0 (x), x ∈ (0, ), ⎪ ⎩ u(0, t) = 0, ux (, t) = 0, t ∈ [0, T ], ⎧ ⎪ ⎨ ut (x, t) = (k(x)ux (x, t))x , (x, t) ∈ T , u(x, 0) = 0, x ∈ (0, ), ⎪ ⎩ −k(0)ux (0, t) = f (t), ux (, t) = 0, t ∈ [0, T ]

(B.0.2)

⎧ ⎪ ⎨ ut = (k(x)ux )x − q(x)u, (x, t) ∈ T , u(x, 0) = 0, x ∈ (0, ), ⎪ ⎩ u(0, t) = ν(t), ux (, t) = 0, t ∈ [0, T ],

(B.0.3)

and

discussed in Sects. 3.1, 6.5 and Sect. 6.6. Note first of all, that energy estimates for the second order parabolic equations are given in general form in the books [38, 86]. A priori estimates given below for cannot be obtained directly from those given in general form. Our aim here is to derive all necessary a priori estimates for the weak and regular weak solutions and for their derivatives through the Neumann and Dirichlet inputs f (t) and ν(t) in B.0.2 and B.0.3. In fact, one needs indeed to proceed the following three steps to derive the energy estimates as well as to prove an existence of weak solutions by the

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9

473

474

B Necessary Estimates For One-Dimensional Parabolic Equation

Galerkin approximation: 1. Construction of the approximate solution, un (x, t) =

n 

dn,i (t)ϕi (x), (x, t) ∈ T .

i=1

where {ϕi }∞ i=1 are orthonormal eigenvectors of the differential operator L : H 2 (0, ) → L2 (0, ) defined by 

(Lϕ)(x) := −(k(x)ϕ  (x)) = λϕ(x), x ∈ (0, ); ϕ  (0) = 0, ϕ  () = 0,

corresponding to eigenvalues {λi }∞ i=1 . 2. Derivation of energy estimates for the approximate solution un (x, t); 3. Convergence of the approximate solution to a weak solution in appropriate sense. To derive necessary estimates for the weak and regular weak solutions we need the following Gronwall-Bellman inequality. Lemma B.0.1 Assume that v(t) and f (t) are the nonnegative continuous functions defined on [α, β], and r ≥ 0. Then the inequality 

t

v(t) ≤

f (τ )v(τ )dτ + r, t ∈ [α, β]

(B.0.4)

α

implies that 

t

v(t) ≤ r exp

f (τ )dτ , t ∈ [α, β].

α

Proof By the assumption, $t α

v(t) f (τ )v(τ )dτ + r

≤ 1, t ∈ [α, β].

Multiply both sides by f (t) > 0 to get d ln dt



t

f (τ )v(τ )dτ + r

≤ f (t), t ∈ [α, β].

α

Integrating this inequality over (α, t) we deduce that 

t α



t

f (τ )v(τ )dτ + r ≤ r exp α

f (τ )dτ , t ∈ [α, β].

(B.0.5)

B Necessary Estimates For One-Dimensional Parabolic Equation

475

The left-hand-side of this inequality is greater than or equal to the function v(t), by condition (B.0.4). This leads to the desired result.  

B.1 The Problem with Non-homogeneous Initial Data and Source Consider the following initial-boundary value problem (B.0.1) with the nonhomogeneous initial data u0 (x) and the source term F (x, t). Lemma B.1.1 Assume that the following conditions are satisfied:  k ∈ L∞ (0, ), 0 < c0 ≤ k(x) ≤ c1 , F ∈ L2 (0, T ; L2 (0, )), u0 ∈ L2 (0, ).

(B.1.1)

Then for the weak solution u ∈ L2 (0, T ; V(0, )) with ut ∈ L2 (0, T ; H −1(0, )) of problem (B.0.1) the following estimates hold: 

u 2L∞ (0,T ;L2 (0,)) ≤ Ce2 F 2L2 (0,T ;L2 (0,)) + u0 2L2 (0,) ,  

u 2L2 (0,T ;L2 (0,)) ≤ Ce2 − 1 F 2L2 (0,T ;L2 (0,)) + u0 2L2 (0,) , (B.1.2)  e2 F 2 2

ux 2L2 (0,T ;L2 (0,)) ≤ C

+ u0 2L2 (0,) , L (0,T ;L2 (0,)) e2 = Ce2 /(2c0 ), and where V(0, ) := {v ∈ H 1 (0, ) : v(0) = 0}, Ce2 = exp(T ), C c0 > 0 is the constant introduced in (B.1.1). Proof Multiply both sides of Eq. (B.0.1) by 2u(x, t), integrate over t := (0, ) × (0, t) and use the integration by parts formula. In view of the initial and boundary conditions in (B.0.1) we obtain the following energy identity: 



u2 (x, t)dx + 2

0

=2

 t 0

 t 0



 0

k(x)u2x (x, τ )dxdτ 



F (x, τ )u(x, τ )dxdτ +

0

0

u20 (x)dx, a. e. t ∈ [0, T ]. (B.1.3)

This leads to the main integral inequality: 



u2 (x, t)dx + 2c0

0

≤ a.e. t ∈ [0, T ].

0

 t 0

 t

 0

 0

u2x (x, τ )dxdτ

u2 (x, τ )dxdτ + F 2L2 (0,T ;L2 (0,)) + u0 2L2 (0,),

(B.1.4)

476

B

Necessary Estimates For One-Dimensional Parabolic Equation

The first consequence of (B.1.4) is the inequality 



u dx ≤ 2

0

 t 0

 0

u2 dxdτ + + F 2L2 (0,T ;L2 (0,)) + u0 2L2 (0,),

for a.e. t ∈ [0, T ]. Applying Lemma B.0.1 we obtain: 

 0

 u2 (x, t)dx ≤ F 2L2 (0,T ;L2 (0,)) + u0 2L2 (0,) exp(t),

(B.1.5)

for a.e. t ∈ [0, T ]. The first two estimates of (B.1.2) directly follows from this inequality. The third estimate of (B.1.2) follows from the second consequence  t



2c0 0

0

u2x dxdτ ≤

 t 0

 0

u2 dxdτ + F 2L2 (0,T ;L2 (0,)) + u0 2L2 (0,) ,

t ∈ [0, T ], taking into account also inequality (B.1.4) is the above right-hand-side integral.   Notice that the above weak solution is in C([0, T ]; L2 (0, )), according to Theorem 3.1.1 of Sect 3.1. Similar estimates hold also for the regular weak solution of problem B.0.1 with smoother input data. Lemma B.1.2 Let conditions (B.1.1) hold. Assume, in addition that the inputs satisfy the following regularity conditions: k ∈ H 1 (0, ), Ft ∈ L2 (0, T ; L2 (0, )), u0 ∈ H 2 (0, ).

(B.1.6)

Then for the regular weak solution u ∈ L2 (0, T ; H 2 (0, )) ∩ L∞ (0, T ; V(0, )) with ut ∈ L2 (0, T ; L2 (0, )) of problem (B.0.1) the following estimates hold: 

ut 2L∞ (0,T ;L2 (0,)) ≤ Ce2 C˜ T2 F 2H 1 (0,T ;L2 (0,)) + (ku0 ) 2L2 (0,) ,  

ut 2L2 (0,T ;L2 (0,)) ≤ Ce2 − 1 C˜ T2 F 2H 1 (0,T ;L2 (0,)) + (ku0 ) 2L2 (0,) ,   ) 2 e2 C˜ 2 F 2 1 +

(ku ,

uxt 2L2 (0,T ;L2 (0,)) ≤ C 2 2 T 0 H (0,T ;L (0,)) L (0,)

(B.1.7)

e > 0 is the constant introduced in where C˜ T2 = max(2/T , 1 + 2T /3) and C Lemma B.1.1.

B Necessary Estimates For One-Dimensional Parabolic Equation

477

Proof Differentiate formally Eq. (B.0.1) with respect to the time variable t ∈ (0, T ), multiply both sides of by 2ut (x, t) and then integrate over t . Using the initial and boundary conditions we obtain the following integral identity: 

 0

+2

u2t (x, t)dx =2

 t 0

 t 0



k(x)u2xτ (x, τ )dxdτ

0







Fτ (x, τ )uτ (x, τ )dxdτ +

0

u2t (x, 0+ )dx,

0

which implies the main integral inequality 

 0

u2t (x, t)dx

 t

+ 2c0

0

 0

u2xτ (x, τ )dxdτ

 t

+

0



0

≤

 t 0

0

 Fτ2 (x, τ )dxdτ +



 0

u2τ (x, τ )dxdτ

u2t (x, 0+ )dx,

(B.1.8)

a.e. t ∈ [0, T ].   We use the (formal) limit equation ut (x, 0+ ) = k(x)ux (x, 0+ ) x + F (x, 0+ ) at t = 0+ to estimate the last right-hand-side integral in (B.1.8). We have: 

 0

u2t (x, 0+ )dx ≤ 2





0

(k(x)u0 (x))x

2

 dx + 2



F (x, 0+ ) dx. 2

(B.1.9)

0

Now we use the identity F (x, 0+ ) = −



1 T

T



0

 0

((T − t)F (x, t))t dt, x ∈ (0, ).

to estimate the last right-hand-side integral in (B.1.9) as follows: 



F (x, 0+ ) dx ≤ 2

0

2 T



T 0



 0

F 2 (x, t)dxdt +

2T 3



T 0

 0



Ft2 (x, t)dxdt.

Hence

F (·, 0+ ) 2L2 (0,) ≤

2 2T

F 2L2 (0,T ;L2 (0,)) +

Ft 2L2 (0,T ;L2 (0,)). T 3

(B.1.10)

478

B

Necessary Estimates For One-Dimensional Parabolic Equation

Substituting this with inequality (B.1.9) in (B.1.8) we deduce that 

 0

u2t (x, t)dx

+ 2c0

 t 0

 0

u2xτ (x, τ )dxdτ

≤

 t 0

 0

u2τ (x, τ )dxdτ

+C˜ T2 F 2H 1 (0,T ;L2 (0,)) + (ku0 ) 2L2 (0,), t ∈ [0, T ], with the constant C˜ T2 defined in the theorem. This leads to the required estimates (B.1.7) are derived from this inequality.

 

B.2 The Problem with Neumann Input Lemma B.2.1 Let the following conditions hold: 

k ∈ L∞ (0, ), 0 < c0 ≤ k(x) ≤ c1 , f ∈ L2 (0, T ).

(B.2.1)

Then for the weak solution u ∈ L2 (0, T ; H 1(0, )) with ut ∈ L2 (0, T ; H −1 (0, )) of problem (B.0.2) the following estimates hold:

u 2L∞ (0,T ;L2 (0,)) ≤ C12 f 2L2 (0,T ) ,

u 2L2 (0,T ;L2 (0,)) ≤ C22 f 2L2 (0,T ) ,

ux 2L2 (0,T ;L2 (0,))

≤

(B.2.2)

C12 f 2L2 (0,T ) ,

where C12 = (/c0 ) exp(C T ), C = 2c0 /, C22 = (/(c0 C )) [exp(C T ) − 1]

(B.2.3)

and c0 > 0 is the constant introduced in (B.2.1). Proof Multiply both sides of Eq. (B.0.2) by 2u(x, t), integrate over t := (0, ) × (0, t) and use the integration by parts formula. Taking into account the boundary conditions in (B.0.2) we obtain the following energy identity: 

 0

u2 (x, t)dx + 2

 t 0

 0

 k(x)u2x (x, τ )dxdτ = 2

t

f (τ )u(0, τ )dτ, 0

(B.2.4)

B Necessary Estimates For One-Dimensional Parabolic Equation

479

for a.e. t ∈ [0, T ]. Employ now the ε-inequality 2ab ≤ (1/ε) a 2 + ε b2 in the right hand side integral to get: 



u2 (x, t)dx + 2c0

0

 t 0



t

≤ε 0



0

u2x (x, τ )dxdτ 

1 u (0, τ )dτ + ε

t

2

f 2 (τ )dτ,

(B.2.5)

0

for a.e. t ∈ [0, T ]. To estimate the first right-hand-side integral in (B.2.5) we use the identity   u (0, t) = u(x, t) −

2

x

2

uξ (ξ, t)dξ 0

with the inequality (a − b)2 ≤ 2(a 2 + b2 ) and the Hölder inequality. Integrating then over [0, t] we obtain: 

t



t

u2 (0, τ )dτ ≤ 2

0

u2 (x, τ )dτ + 2x

 t

0

0

 0

u2x (x, τ )dxdτ, t ∈ [0, T ].

Integrating now both sides over [0, ] and dividing then by  > 0 we arrive at the following trace inequality: 

t

u2 (0, τ )dτ ≤

0

2 

 t

 0

u2 (x, τ )dτ dx + 

0

 t 0

 0

u2x (x, τ )dxdτ,

(B.2.6)

for all t ∈ [0, T ]. In view of (B.2.5) we deduce from (B.2.6) that 



u (x, t)dx + (2c0 − ε) 2

0

2ε ≤ 

0

 t 0

 t

 0

0

1 u (x, τ )dxdτ + ε



u2x (x, τ )dxdτ



t

2

f 2 (τ )dτ, t ∈ [0, T ].

0

Choosing the arbitrary parameter ε > 0 from the condition 0 < ε < 2c0 / as ε = c0 /, we finally obtain the main integral inequality: 



2

u (x, t)dx + c0

0

2c0 ≤ 2 

0

 t 0

 t

 0

 0

u2x (x, τ )dxdτ

 u (x, τ )dxdτ + c0



t

2

0

f 2 (τ )dτ, t ∈ [0, T ].

(B.2.7)

480

B

Necessary Estimates For One-Dimensional Parabolic Equation

The first consequence of (B.2.7) is the inequality 

 0

2c0 u (x, t)dx ≤ 2 l 2

 t 0

 0

 u (x, τ )dxdτ + c0



T

2

f 2 (t)dt,

0

for a.e. t ∈ [0, T ]. Applying Lemma B.0.1 we deduce that 





f 2L2 (0,T ) exp(C t), t ∈ [0, T ], c0

u2 (x, t)dx ≤

0

(B.2.8)

where C > 0 is the constant defined by (B.2.3). The first estimate of (B.2.2) directly follows from this inequality. Notice that the weak solution is in C([0, T ]; L2 (0, )), according to Theorem 3.1.1 of Sect 3.1. Further, integrating (B.2.8) over (0, T ) we obtain the second estimate of (B.2.2). The next consequence of (B.2.7) is the inequality 

T





c0 0

0

u2x (x, τ )dxdt ≤

2c0 2



T



0



u2 (x, t)dxdt +

0

 c0



T

f 2 (t)dt.

0

Using here (B.2.8) we arrive at the required third estimate of (B.2.2).

 

The lemma below shows that the similar estimates hold for the regular weak solution of problem (B.0.2), if the input f (t) belongs to H 1 (0, T ). Lemma B.2.2 Let conditions (B.2.1) hold. Assume, in addition that k ∈ H 1 (0, ) and f ∈ H 1 (0, T ). Then for the regular weak solution u ∈ L2 (0, T ; H 2(0, )) ∩ L∞ (0, T ; H 1 (0, )) with ut ∈ L2 (0, T ; L2 (0, )) of problem (B.0.2) the following estimates hold: ⎧

ut 2L∞ (0,T ;L2 (0,)) ≤ C12 f  2L2 (0,T ) , ⎪ ⎪ ⎨

ut 2L2 (0,T ;L2 (0,)) ≤ C22 f  2L2 (0,T ) (B.2.9) ⎪ ⎪ ⎩ u 2 2 2  ≤ C f

xt L2 (0,T ;L2 (0,))

1

L2 (0,T )

with the constants introduced in Lemma B.2.1. Proof Differentiate formally Eq. (B.0.2) with respect to t ∈ (0, T ), multiply both sides by 2ut (x, t), integrate over [0, ] and use the integration by parts formula. Taking then into account the initial and boundary conditions we obtain the following integral identity: d dt



 0

 u2t (x, t)dx



+2 0

k(x)u2xt (x, t)dx = 2f  (t)ut (0, t), t ∈ [0, T ].

B Necessary Estimates For One-Dimensional Parabolic Equation

481

Integrating this identity over [0, t] and using the limit equation  0



u2t (x, 0+ )dx





(k(x)ux (x, 0+ ))x

=

2

dx = 0,

0

to deduce that 

 0

u2t (x, t)dx

+2

 t 0

 0

 k(x)u2xτ (x, τ )dxdτ

=2

t

f  (τ )uτ (0, τ )dτ,

0

for all t ∈ [0, T ]. This is the same integral identity (B.2.4) with u(x, t) and f (t) replaced by ut (x, t) and f  (t), correspondingly. Hence estimates (B.2.9) are derived in the same way in the proof of Lemma B.2.1.   Lemma B.2.3 Let conditions of Lemma B.2.2 hold. Assume, in addition that f ∈ H 2 (0, T ). Then for the regular weak solution with improved regularity of problem (B.0.2), defined as u ∈ L∞ (0, T ; H 2(0, )) with ut ∈ L2 (0, T ; H 1 (0, )) ∩ L∞ (0, T ; L2 (0, )) and ut t ∈ L2 (0, T ; H −1 (0, )), the following estimates hold: ⎧ 2 2  2 ⎪ ⎪ ut t L∞ (0,T ;L2 (0,)) ≤ C1 f L2 (0,T ) , ⎨

ut t 2L2 (0,T ;L2 (0,)) ≤ C22 f  2L2 (0,T ) ⎪ ⎪ ⎩ u 2 ≤ C 2 f  2 , xt t L2 (0,T ;L2 (0,))

1

(B.2.10)

L2 (0,T )

where C1 , C2 > 0 are the constants introduced in Lemma B.2.1. This lemma is proved in the same way as the previous one. Remark B.2.1 With slightly different constants, estimates (B.2.2), (B.2.9) and (B.2.10) are also valid for the case if the Neumann boundary conditions −k(0)ux (0, t) = f (t), ux (, t) = 0 in problem (B.0.2) are replaced by the mixed boundary conditions u(0, t) = 0, k()ux (, t) = f (t).

B.3 The Problem with Dirichlet Input Lemma B.3.1 Assume that the inputs in problem (B.0.3) satisfy the following basic conditions:  k, q ∈ L∞ (0, ), 0 < c0 ≤ k(x) ≤ c1 , |q(x)| ≤ c3 , (B.3.1) ν ∈ H 1 (0, T ).

482

B

Necessary Estimates For One-Dimensional Parabolic Equation

Then for the weak solution u ∈ L2 (0, T ; H 1(0, )) with ut ∈ L2 (0, T ; H −1 (0, )) of problem (B.0.3) the following estimates hold: 2 ν 2 1

u 2L∞ (0,T ;L2 (0,)) ≤ C , 1 H (0,T ) 2 ν 2 1 ,

u 2L2 (0,T ;L2 (0,)) ≤ C 2 H (0,T )

(B.3.2)

2 C ≤ 1 ν 2H 1 (0,T ) , 4c0

ux 2L2 (0,T ;L2 (0,)) where

12 = 4 C T2 exp(C02 T ), C 22 = C

2 4 C T

[exp(C02 T ) − 1], C02 T2 = max(4T + 1, 1 + 3/T ) 02 = 1 + 4c3 + c32 , C C

and c0 , c3 > 0 is the constant introduced in (B.3.1). Proof Multiply both sides of Eq. (B.0.3) by 2u(x, t), integrate over t = (0, ) × (0, t) and use the integration by parts formula. Then we obtain the following energy identity: 



 u (x, t)dx + 2 2

0

 = −2

t

k(x)u2x (x, τ )dxdτ 

t

q(x)u (x, τ )dxdτ − 2 2

k(0)ux (0, τ )ν(τ )dτ,

(B.3.3)

0

t

t ∈ [0, T ]. We use the condition |q(x)| ≤ c3 to estimate the first right-hand-side integral: 

 2

u2 (x, τ )dxdτ, t ∈ [0, T ].

q(x)u (x, τ )dxdτ ≤ 2c3

−2 t

(B.3.4)

t

Transform now the second right-hand-side integral in (B.3.3) as follows:  −2

t

k(0)ux (0, τ )ν(τ )dτ = 2

0

 t 

(k(x)ux (x, τ ))x dx ν(τ )dτ 0

=2 = −2

0



0

 t 0

 t







[uτ + q(x)u(x, τ )] ν(τ )dxdτ

0





ν (τ )u(x, τ )dxdτ + 2

0

+2



ν(t)u(x, t)dx 0

 t



q(x)ν(τ )u(x, τ )dxdτ. 0

0

B Necessary Estimates For One-Dimensional Parabolic Equation

483

Applying the ε-inequality to the right-hand-side integrals we obtain: 

t

−2 0

k(0)ux (0, τ )ν(τ )dτ ≤





+ε

u2 (x, t)dx +

0



 ε

(1 + c32 ) ε T

 t 0

 0

(ν  (t))2 dt +



0

T

u2 (x, τ )dxdτ  ν 2 (t)dt + ν 2 (t) .

(B.3.5)

0

We estimate now the last term ν 2 (t) in the square bracket through the H 1 -norm of the input ν(t). To this end we use the identities 

t

ν(t) =

ν  (τ )dτ + ν(0),

0

1 ν(0) = − T



T

(B.3.6) 

((T − t)ν(t)) dt, t ∈ [0, T ]

0

to deduce that 3 ν (t) ≤ T





T

2

T

2

ν (t)dt + 4T

0

(ν  (t))2 dt, t ∈ [0, T ].

0

Hence,

ν 2L2 (0,T ) + ν 2 (t) ≤ max(4T , 1 + 3/T ) ν 2H 1 (0,T ) , 2 ν 2 1

ν  2L2 (0,T ) + ν 2L2 (0,T ) + ν 2 (t) ≤ C , T H (0,T )

(B.3.7)

T > 0 defined in the theorem. Use this in (B.3.5) and for all t ∈ [0, T ], with C then in the energy identity (B.3.3), together with the estimate (B.3.4), choosing the parameter ε > 0 as ε = 1/2. Then we obtain the main integral inequality: 



 2

u (x, t)dx + 4c0

0

02 =C

t

 t

u2x (x, τ )dxdτ

T2 ν 2 1 u2 (x, τ )dxdτ + 4 C , t ∈ [0, T ], H (0,T )

T > 0 are the constants introduced in the theorem. 0 , C where C The required estimates (B.3.2) are easily obtained from this inequality estimates the same way as in the proof of Lemma B.2.1.   Remark B.3.1 From identities (B.3.6) it follows that for any function v ∈ H 1 (0, T ) the following inequality holds:

v 2C[0,T ] ≤ Cc2 v 2H 1 (0,T ) , Cc2 = max(3/T , 4T }.

484

B

Necessary Estimates For One-Dimensional Parabolic Equation

Next lemma below shows that similar estimates hold for the regular weak solution of problem (B.0.3) under slightly higher regularity condition with respect on the Dirichlet input ν(t). Lemma B.3.2 Assume that conditions (B.3.1) hold. Suppose, in addition that ν ∈ H 2 (0, T ) and k ∈ H 1 (0, ). Then for the regular weak solution u ∈ L2 (0, T ; H 2 (0, )) with ut ∈ L2 (0, T ; L2 (0, )) of problem (B.0.3) the following estimates hold: 2 ν 2 2

ut 2L∞ (0,T ;L2 (0,)) ≤ C , 1 H (0,T ) 2 ν 2 2 ,

ut 2L2 (0,T ;L2 (0,)) ≤ C 2 H (0,T )

uxt 2L2 (0,T ;L2 (0,))

(B.3.8)

2 ν 2 2 ≤C , 1 H (0,T )

with the same constants introduced in Lemma B.2.2. This lemma can be proved in much the same way as the previous one and left to the reader as an exercise.

Appendix C

Necessary Estimates For Euler-Bernoulli Beam Equation

In this chapter we consider some typical initial-boundary value problems for dynamic Euler-Bernoulli beam equation. We continue the manner of Appendix B by employing the Galerkin approximation and energy method to obtain necessary a priori estimates for the weak and regular weak solutions of these problems.

C.1 Existence of Weak Solutions Consider the following vibration model governed by the general form dynamic Euler-Bernoulli beam equation subject to the initial and homogeneous boundary conditions: ⎧ ⎪ ⎨ ρ(x)ut t + μ(x)ut + Lu = g(x, t), (x, t) ∈ (0, ) × (0, T ], (C.1.1) u(x, 0) = p(x), ut (x, 0) = q(x), x ∈ (0, ), ⎪ ⎩ u(0, t) = ux (0, t) = 0, u(, t) = ux (, t) = 0, t ∈ (0, T ), where Lu := (r(x)uxx )xx − (Tr (x)ux )x . The functions u(x, t) and g(x, t) express the beam deflection and the load distribution, respectively. Furthermore, r(x) = E(x)I (x), while E(x) > 0 is the elasticity modulus, I (x) > 0 is the moment of inertia of the cross-section, ρ(x) is the mass density of the beam, μ(x) ≥ 0 and Tr (x) ≥ 0 denote the damping coefficient and axial tensile (or traction) force, respectively.

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9

485

486

C

Necessary Estimates For Euler-Bernoulli Beam Equation

u u(0, t) = 0

g(x, t)

ux (0, t) = 0

u(, t) = 0

x

ux (, t) = 0

Fig. C.1 Geometry of the problem for rigid clamped Euler-Bernoulli beam under spatial-temporal load

The geometry of problem (C.1.1) is given in Fig. C.1. We assume that the following basic conditions hold: ⎧ ρ, r, μ, Tr ∈ L∞ (0, ), G ∈ L2 (0, T ; H −2(0, )), ⎪ ⎪ ⎪ ⎨ p ∈ H 2 (0, ), q ∈ L2 (0, ), 0 ⎪ ≤ ρ(x) ≤ ρ1 , 0 < r0 ≤ r(x) ≤ r1 , x ∈ (0, ), 0 < ρ 0 ⎪ ⎪ ⎩ 0 ≤ μ0 ≤ μ(x) ≤ μ1 , 0 ≤ Tr0 ≤ Tr (x) ≤ r1 , x ∈ (0, ).

(C.1.2)

Below, we assume, without loss of generality, that ρ(x) = 1. Assumption G ∈ L2 (0, T ; H −2 (0, )) in (C.1.2) deserves a comment. In the classical theory for parabolic and hyperbolic PDEs, the source term G(t), [G(t)](x) := g(x, t) in Eq. (C.1.1) is usually assumed to be in L2 (0, T ; L2 (0, )) (see §7.1 and §7.2 of [38]). The approach proposed here allows us to employ the loading term in the larger space L2 (0, T ; H −2(0, )). A specific motivation for the general form of the beam equation with such a loading term from the negative order Sobolev space H −2 (0, ), which includes also a point load is, for instance, the following application. The equation ρ(x)ut t +η(x)ut +(r(x)uxx )xx −(Tr (x)ux )x = δ(x − ct)P (t) describes vibration of a beam under a moving and arbitrarily varying in time load P (t) with constant speed c > 0 [41]. Here δ(z) is the Dirac delta (generalized) function which is not a regular distribution. The following notations are used here. H k (0, ) is the Hilbert-Sobolev space with order k and H0k (0, ) denotes the closure of C0∞ in H k (0, ) space. a(u, v) := (ruxx , vxx ) + (Tr ux , vx )

(C.1.3)

defines the symmetric bilinear functional a : H 2 (0, ) × H 2(0, ) → R, where (·, ·) denotes the usual inner product in L2 (0, ). Then we transform problem (C.1.1) to the following variational problem: ⎧  (U (t), v) + (ηU  (t), v) + a(U (t), v) = (G(t), v), ⎪ ⎪ ⎨ ∀v ∈ H02 (0, ), t ∈ (0, T ], ⎪ ⎪ ⎩ U (0) = p, U  (0) = q,

(C.1.4)

C Necessary Estimates For Euler-Bernoulli Beam Equation

487

for the abstract functions [U (t)](x) := u(x, t) and [G(t)](x) := g(x, t) which are defined by the mappings U : [0, T ] → H02 (0, ) and G : [0, T ] → L2 (0, ), respectively. For the convenience we rewrite Eq. (C.1.1) in the form ut t = z + z˘ , where z := −ηut + (Tr ux )x and z˘ = g − (ruxx )xx . Then U  (t) = z(t, ·) + z˘ (t, ·) belongs to H −2 (0, ) for all t ∈ (0, T ]. The mapping ·, · : H −2 (0, ) × H02 (0, ) → R is used for the duality pairing between H −2 (0, ) and H02 (0, ). Definition C.1.1 A function U ∈ L2 (0, T ; H02 (0, )) with U  ∈ L2 (0, T ; L2 (0, )) and U  ∈ L2 (0, T ; H −2 (0, )) is called a weak solution of (C.1.1) if the followings hold: (a) U  (t), v + (ηU  (t), v) + a(U (t), v) = G(t), v, ∀v ∈ H02 (0, ), a.e. t ∈ [0, T ], (b) U (0) = p and U  (0) = q. Notice that the initial conditions U (0) = p and U  (0) = q in Definition C.1.1 are well-defined due to Theorem 2 in §5.9 of [38].

C.1.1 Uniform Estimate in Finite Dimensional Subspace For a fixed n, we define n dimensional subspace Vn := span{φ1 , φ2 , . . . , φn } of H02 (0, l) and look for the Galerkin approximation Un (t) := un (x, t) of the form Un (t) =

n 

un,i (t)φi , for t ∈ [0, T ]

(C.1.5)

i=1

where φi ∈ H02 (0, l) is the solution (normalized in L2 -norm) of the eigenvalue problem φi (x) = λi φi (x). Actually, {φi : i ∈ N} is an orthogonal basis of H02 (0, l) space. The coefficients un,i (t) are chosen so that Un (t) satisfies an analogue of (C.1.4): ⎧ ⎪ (U  (t), v) + (ηUn (t), v) + a(Un (t), v) = G(t), v, ⎪ ⎨ n ∀ v ∈ Vn , t ∈ [0, T ], (C.1.6) ⎪ ⎪ ⎩  Un (0) = $n p, Un (0) = $n q,  where $n : L2 (0, ) → Vn is the orthogonal projection, i.e. $n h = ni=1 hi φi where hi = (h, φi ) for any h ∈ L2 (0, ). Hence we are looking for an approximate solution Un (t) in Vn . Theorem C.1.1 Assume that the basic conditions (C.1.2) are satisfied. Then for each n ∈ N there exists a unique solution to problem (C.1.6) of the form (C.1.5).

488

C

Necessary Estimates For Euler-Bernoulli Beam Equation

Proof Since {φi : i ∈ N} forms an orthonormal basis for L2 (0, ) and Un (t) is defined by (C.1.5), we have (Un , φj ) =

n

(ηUn , φj ) = a(Un , φj ) =

  i=1 un,i (t)(φi , φj ) = un,j (t), n  i=1 un,i (t)(ηφi , φj ), n i=1 un,i (t)a(φi , φj ).

(C.1.7)

Denote by ηi,j = (ηφi , φj ), ai,j := a(φi , φj ) and gj (t) := G(t), φj . Then choosing v = φj and taking into account (C.1.7), we can reformulate problem (C.1.6) as the following Cauchy problem for the constant coefficient system of linear ordinary differential equations: ⎧ n   ⎪ ⎨ u (t) + ηi,j un,i (t) + ai,j un,i (t) = gj (t), t ∈ (0, T ], n,j i=1 ⎪ ⎩ un,j (0) = (p, φj ), un,j (0) = (q, φj ).

(C.1.8)

This system can be written in the vector form # n (t) + M u # n (t) + N u# n = g# (t), t ∈ (0, T ], u # n = [un,1 , · · ·, un,n ]T , u M = [ηi,j ]n×n , N = [ai,j ]n×n , with the initial conditions u# n (0) = [(p, φ1 ), · · ·, (p, φn )]T and u# n (0) = [(q, φ1 ), · · ·, (q, φn )]T . Since M and N are constant matrices and g# ∈ L2 (0, T ; Rn ), it follows from standard theory of ODE that there exists a unique solution u# n ∈ C 1 ([0, T ]; Rn ) # n ∈ C([0, T ]; Rn ), u # n ∈ L2 (0, T ; Rn ). Hence the solution Un (t) of (C.1.6) with u exists, unique and belongs to C 1 ([0, T ]; Vn ) with Un ∈ C([0, T ]; Vn ) and Un ∈ L2 (0, T ; Vn ).   Theorem C.1.2 Assume that the basic conditions (C.1.2) are satisfied. Then there exists a constant C > 0 depending only on the constants introduced in the basic conditions (C.1.2), such that for the solution Un (t) of problem (C.1.6) the following estimate holds

 max ||Un (t)||2L2 (0,) + ||Un (t)||2H 2 (0,) + ||Un ||2L2 (0,T ;H −2 (0,)) t ∈[0,T ]

for all n ∈ N.



≤ C ||G||2L2 (0,T ;H −2 (0,)) + ||p||2H 2 (0,) + ||q||2L2(0,)

(C.1.9)

C Necessary Estimates For Euler-Bernoulli Beam Equation

489

Proof Consider Eq. (C.1.6) for v = φj and multiply both sides by un,j . Summing up the terms from j = 1 to j = n we obtain: (Un (t), Un (t)) + (ηUn (t), Un (t)) + a(Un (t), Un (t)) = G(t), Un (t), t ∈ (0, T ].

(C.1.10)

By the Riesz theorem, there exists νt ∈ H02 (0, l) such that G(t), Un (t) = (νt , Un (t)) and ||νt ||H 2 (0,l) = ||G(t)||H −2 (0,l), for all t ∈ (0, T ]. Furthermore, using the identities (Un (t), Un (t)) = [||Un (t)||2L2 (0,l) ] /2, a(Un (t), Un (t)) = [a(Un (t), Un (t))] /2 on the left-hand side and also the Cauchy-Schwarz inequality on the right-hand side of (C.1.10) we obtain: 

 √ ||Un (t)||2L2 (0,l) + a(Un (t), Un (t)) + || ηUn (t)||2L2 (0,l) ≤ 2||Un (t)||L2 (0,l)||νt ||H 2 (0,l).

With Young’s inequality this implies: 

||Un (t)||2L2 (0,l) + a(Un (t), Un (t))



≤ ||Un (t)||2L2 (0,l) + ||G(t)||2H −2 (0,l).

Denote by || · ||a the norm defined by the bilinear form a(., .). Then we can write  d  ||Un (t)||2L2 (0,l) + ||Un (t)||2a dt ≤ ||Un (t)||2L2 (0,l) + ||Un (t)||2a + ||G(t)||2H −2 (0,l), adding the term ||Un (t)||2a to the right-hand, to apply the Gronwall inequality [38]. This implies: ||Un (t)||2L2 (0,l) + ||Un (t)||2a ≤e

t



||Un (0)||2L2 (0,l)

 +

||Un (0)||2a

t

+ 0

||G(t)||2H −2 (0,l) dt



.

490

C

Necessary Estimates For Euler-Bernoulli Beam Equation

Thanks to the conditions in (C.1.2), the norms ||Un (t)||2a and ||Un (0)||2a can be estimated below and above with C||Un (t)||2H 2 (0,l) and C||Un (0)||2H 2 (0,l) for some C, C ∈ R+ , respectively. Then

||Un (t)||2L2 (0,l) + ||Un (t)||2H 2 (0,l)   t  2 2 2 ≤ C1 ||Un (0)||L2 (0,l) + ||Un (0)||H 2 (0,l) + ||G(t)||H −2 (0,l)dt , 0

for some C1 > 0. Using the projection inequalities for ||Un (0)||L2(0,l) and ||Un (0)||H 2 (0,l) and taking maximum for t ∈ [0, T ] we conclude that max

t ∈[0,T ]



||Un (t)||2L2 (0,l) + ||Un (t)||2H 2 (0,l)

≤ C1





0

 ||q||2L2(0,l) + ||p||2H 2 (0,l) + ||G||2L2 (0,T ;H −2 (0,l)) .

(C.1.11)

On the other hand, for any w ∈ H02 (0, l) with ||w||H 2 (0,l) = 1 we can write the representation w = φ + ψ, where φ ∈ Vn := span{φ1 , · · ·, φn } and ψ ∈ Vn⊥ . Hence (Un (t), w) = (Un (t), φ) = G(t), φ − (ηUn (t), φ) − a(Un (t), φ).

(C.1.12)

By the Pythagorean Theorem, we have: ||φ||H 2 (0,l) ≤ 1. With Cauchy-Schwarz inequality, (C.1.12) becomes to |(Un (t), w)| ≤ C2



 ||G(t)||H −2 (0,l) + ||Un (t)||L2 (0,l) + ||Un (t)||H 2 (0,l) .

Then by taking the supremum over w ∈ H02 (0, l) with ||w||H 2 (0,l) = 1 in the above inequality, integrating the square of both side from 0 to T and then using inequality (C.1.11), we obtain ||Un ||2L2 (0,T ;H −2 (0,l)) ≤ C2



||G||2L2(0,T ;H −2 (0,l)) + ||q||2L2(0,l) + ||p||2H 2 (0,l)



This inequality with (C.1.11) implies the desired result (C.1.9) with C = C1 + C2 .  

C.1.2 Existence and Uniqueness Theorems The uniform estimate derived in the previous section for the approximate solution Un and its derivatives, allows us to obtain the boundedness properties for the corresponding sequences {Un }, {Un } and {Un } in L∞ (0, T ; H02 (0, l)), L∞ (0, T ; L2 (0, l)) and L2 (0, T ; H −2(0, l)), respectively. Thus, the sequence of

C Necessary Estimates For Euler-Bernoulli Beam Equation

491

approximate solutions is weakly pre-compact as a bounded in a reflexive Banach space sequence. Furthermore, there exists a weakly convergent subsequence of this sequence which converges to the weak solution of problem (C.1.1). Although, only the sequence {Un } is bounded in the Hilbert space which is the special form of reflexive Banach space. The other two sequences {Un } and {Un } are bounded in L∞ in time which is not a reflexive Banach space. On the other hand, the weak*topology is a very useful tool for desired convergence property of sequences in dual of Banach spaces. In detail, if V is a Banach space, any closed unit ball in the dual space V  is weak*-compact (which means compact in the weak* topology) [24]. This result is known as Banach-Alaoglu theorem and it is the most crucial property of the weak*-topology. The sequential version of this theorem states the existence of a convergent subsequence of {Un } and {Un } in weak*-limit due to the uniform estimate (C.1.9). In our cases, corresponding dual spaces for {Un } and {Un } are isomorphic to L∞ (0, T ; H02(0, l)) and L∞ (0, T ; L2 (0, l)), respectively. Theorem C.1.3 Under the basic conditions (C.1.2) there exists a weak solution of problem (C.1.1). Proof According to the discussion above and estimate (C.1.9), there exists subsequences {Unm }, {Un m } and {Unm } and limit functions U , U  and U  such that, Unm  U in L2 (0, T ; H02(0, l)), Un m  U  in L2 (0, T ; L2 (0, l)) and Unm  U  in L2 (0, T ; H −2 (0, l)). We begin with choosing a test function ξ ∈ C ∞ (0, T ) and constructing a function w : [0, T ] → Vk≤n ⊂ H02 (0, l) such that w(t) = ξ(t)v. Then multiply (C.1.8) by ξ(t), we have Un (t), w(t) + (ηUn (t), w(t)) + a(Un (t), w(t)) = G(t), w(t).

(C.1.13)

We substitute n = nm into (C.1.13) and integrate it on [0, T ] then taking limit as m → 0 to get 

T

 ξ(t) U  (t), v + (ηU  (t), v) + a(U (t), v) dt

0

 =

T

ξ(t)G(t), vdt.

(C.1.14)

0

Hence U ∈ L∞ (0, T ; H02 (0, l)) satisfies for a.e. t ∈ (0, T ), U  (t), v + (ηU  (t), v) + a(U (t), v) = G(t), v, ∀v ∈ H02 (0, l).

(C.1.15)

Moreover, it is known that U  ∈ C([0, T ]; H −2(0, l)) by Theorem 2 in §5.9 of [38]. We use this result to show U (0) = p and U  (0) = q. Firstly, we choose ξ ∈ C ∞ [0, T ] with ξ(0) = ξ(T ) = ξ  (T ) = 0 and ξ  (0) = 1. Then using (C.1.14)

492

C

Necessary Estimates For Euler-Bernoulli Beam Equation

and applying integration by parts yield: For all v ∈ H02 (0, l), (U (0), v)  T . = ξ(t) [G(t), v − a(U (t), v)] + ξ  (t)(ηU, v) − ξ  (t)(U (t), v) dt. 0

Similarly, this equality holds for Unm and 

T

(p, v) =

ξ(t) [G(t), v − a(U (t), v)dt] + ξ  (t)(ηU, v) − ξ  (t)(U (t), v)dt,

0

as m → ∞, for every ξ ∈ C ∞ ([0, T ]) and v ∈ H02 (0, l). Thus (U (0), v) = (p, v), for all v ∈ H02 (0, l), which means U (0) = p. Similar calculation can be done to show U  (0) = q. In this case, one can choose ξ ∈ C ∞ ([0, T ]) with ξ  (0) = ξ(T ) = ξ  (T ) = 0 and ξ(0) = 1. Then the resulting form is (U  (0), v) = (q, v), for all v ∈ H02 (0, l) which yields U  (0) = q. Hence, Eq. (C.1.15) with U (0) = p and U  (0) = q verifies that U is a weak solution of (C.1.1).   Theorem C.1.4 Under the basic conditions (C.1.2) the weak solution of problem (C.1.1) is unique. Proof We assume there are two functions U1 and U2 satisfy Eq. (C.1.4) for a.e. #(t) = U1 (t) − U2 (t). Then it is obvious that t ∈ [0, T ]. Let us define the function U #(t) satisfies U # (t), w(t) + (ηU # (t), w(t)) + a(U(t), # U w(t)) = 0, ∀w ∈ H02 (0, l),

(C.1.16)

#(0) = U # (0) = 0. In order to prove the theorem, it suffices for a.e. t ∈ [0, T ], with U # ≡ 0. to show that U We first choose the arbitrary t0 ∈ [0, T ] and define w(t) ∈ H02 (0, l) such that  w(t) =

t

#(s)ds, for t ∈ [0, t0 ] and w(t) = 0 for t ∈ (t0 , T ]. U

t0

# ∈ L∞ (0, T ; H 2 (0, l)) and w(t0 ) = 0. Evidently, w ∈ C([0, T ]; H02(0, l)), since U 0 Integrate (C.1.16) over [0, t0 ] and then applying the integration by parts formula we deduce that  t0 # (t), w (t)) + (ηU #(t), w (t)) − a(U #(t), w(t))dt = 0. (U 0

#(t) to obtain On the other hand, from the definition of w(t), we substitute w (t) = U 1 # #(t0 )) + a(w(0), w(0)) + (U (t0 ), U 2



t0 0

#(t), U #(t))dt = 0. (ηU

(C.1.17)

C Necessary Estimates For Euler-Bernoulli Beam Equation

493

#(t0 ) = 0. Actually this Since each term on the left side of (C.1.17) nonnegative, U is the result we wish to prove because t0 is arbitrary point in [0, T ]. Hence for a.e. t ∈ [0, T ], U1 (t) = U2 (t) which indicates the uniqueness of the weak solution of (C.1.1).   The detailed presentation given above for problem (C.1.1) allows one to obtain the same solvability results for other direct problems considered in Chap. 11, in a similar way.

C.2 The Problem with Neumann Inputs (Bending Moments) Consider the following initial-boundary value problem: ⎧ ⎪ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = 0, ⎪ ⎪ ⎪ ⎪ ⎪ (x, t) ∈ T := (0, ) × (0, T ), ⎪ ⎨ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ ⎪ u(0, t) = 0, −r(0)uxx (0, t) = M0 (t), ⎪ ⎪ ⎪ ⎩ u(, t) = 0, r()uxx (, t) = M1 (t), t ∈ [0, T ].

(C.2.1)

This problem is a mathematical model of dynamic vibration of a beam that is pinned at both ends x = 0;  and subjected to bending moments M0 (t) and M1 (t) applied at its pinned ends. Although this type of beam is not commonly used in practice, it is useful for analysis purposes. In particular, this problem is an analogue of the adjoint problem corresponding to the inverse source problem discussed in Sect. 11.2. Theorem C.2.1 Assume that the inputs in (C.2.1) satisfy the following basic conditions: ⎧ ∞ ⎪ ⎨ ρ, r, μ, Tr ∈ L (0, ), (C.2.2) 0 < ρ0 ≤ ρ(x) ≤ ρ1 , 0 < r0 ≤ r(x) ≤ r1 , ⎪ ⎩ 0 ≤ μ0 ≤ μ(x) ≤ μ1 , 0 ≤ Tr0 ≤ Tr (x) ≤ Tr1 , x ∈ (0, ). Suppose that the Neumann inputs M0 (t) and M1 (t) satisfy the conditions: M0 , M1 ∈ H 1 (0, T ), M02 (t) + M12 (t) ≡ 0, M0 (0) = M1 (0) = 0.

(C.2.3)

494

C

Necessary Estimates For Euler-Bernoulli Beam Equation

Then for the weak solution u ∈ L2 (0, T ; V(0, )), with ut ∈ L2 (0, T ; L2 (0, )), ut t ∈ L2 (0, T ; H −2 (0, )) and V(0, ) := {v ∈ H 2 (0, ) : v(0) = v() = 0}, of problem (C.2.1) the following estimates hold: 2 M  2 2

uxx 2L2 (0,T ;L2 (0,)) ≤ C , 1 L (0,T )

(C.2.4)

2 M  2 2

ut 2L2 (0,T ;L2 (0,)) ≤ C , 2 L (0,T ) where M  2L2 (0,T ) := M0 2L2 (0,T ) + M1 2L2 (0,T ) and 12 = C 02 [exp(T ) − 1], C 22 = r0 C 02 = 24(1 + T ) . 12 , C C 2ρ0 r02

(C.2.5)

Proof Multiply both sides of Eq. (C.2.1) by 2ut (x, t), integrate it over t := (0, ) × (0, t), use the identity

 2(r(x)uxx )xx ut ≡ 2[(r(x)uxx )x ut − r(x)uxx uxt ]x + r(x) u2xx . t

(C.2.6)

Using the boundary conditions, after elementary transformations we obtain the following energy identity: 

 0

ρ(x)u2t 

t

=2

+ r(x)u2xx

+

Tr (x)u2x



dx + 2

 t 0

 M0 (τ )uxτ (0, τ ) dτ + 2

0

t

 0

μ(x)u2τ dxdτ

M1 (τ )uxτ (, τ ) dτ, t ∈ [0, T ].

(C.2.7)

0

We apply the integration by parts formula and then employ the ε-inequality to estimate the first right-hand-side integral as follows: 

t

2

 M0 (τ )uxτ (0, τ ) dτ = −2

0

0



t

≤ε 0

t

M0 (τ )ux (0, τ ) dτ + 2M0 (t)ux (0, t)

 T     2 1 u2x (0, τ ) dτ + u2x (0, t) + M0 (t) dt + M02 (t) , ε 0

(C.2.8)

t ∈ [0, T ]. To estimate the terms in the first right-hand-side square bracket we use the trace inequality (11.1.12) (see proof of Corollary 11.1.2 in Sect 11.1). We have: 

t 0

t ∈ [0, T ].

u2x (0, τ ) dτ + u2x (0, t) ≤ 3

 t 0

0



 u2xx dxdτ + 3

0



u2xx dx,

(C.2.9)

C Necessary Estimates For Euler-Bernoulli Beam Equation

495

Now we use the condition M0 (0) = 0 in the second square bracket of (C.2.8) to deduce that 

T



0

M0 (t)

2

dt + M02 (t) ≤ (1 + T ) M0 2L2 (0,T ) .

Substituting this in (C.2.8) with (C.2.9) we obtain: 

t

2

 t

M0 (τ )uxτ (0, τ ) dτ ≤ 3ε

0

0

+

 0

 u2xx dxdτ

+ 3ε



0

u2xx dx

1+T

M0 2L2 (0,T ) , t ∈ [0, T ]. ε

(C.2.10)

Exactly in the same way we estimate the second right-hand-side integral in (C.2.7): 

t

2

M1 (τ )uxτ (, τ ) dτ ≤ 3ε

 t

0

0

 0

+

 u2xx dxdτ + 3ε

 0

u2xx dx

1+T

M1 2L2 (0,T ) , t ∈ [0, T ]. ε

Taking this inequality with (C.2.10) into account in (C.2.7) we conclude that 



ρ0 0

+2

 u2t dx

 t 0

 0



+ (r0 − 6 ε) 0

 u2xx dx

μ(x)u2τ dxdτ ≤ 6 ε

0

 t 0



+  0

Tr (x)u2x dx

u2xx dxdτ +

1+T

M  2L2 (0,T ) , ε

for all t ∈ [0, T ]. Choosing the parameter ε > 0 from the positivity condition r0 − 6 ε > 0 as ε = r0 /(12), we finally arrive at the main integral inequality: 



ρ0 0

u2t dx

r0 + 2 ≤

r0 2



 0

 u2xx dx

 t 0

 0



+ 0

Tr (x)u2x dx

u2xx dxdτ +

+2

 t 0

 0

μ(x)u2τ dxdτ

r0 2 C M  2L2 (0,T ) , t ∈ [0, T ], 2 0

0 > 0 introduced in (C.2.5). with the constant C The first consequence of (C.2.11) is the inequality 

 0

u2xx dx

≤

 t 0

 0

#02 M  2 2 u2xx dxdτ + C , t ∈ [0, T ]. L (0,T )

(C.2.11)

496

C

Necessary Estimates For Euler-Bernoulli Beam Equation

In view of the Gronwall-Bellmann inequality this implies:  0



02 M  2 2 u2xx dx ≤ C exp(t), t ∈ [0, T ]. L (0,T )

This yields the first estimate of (C.2.4). The second estimate of (C.2.4) is derived from the next consequence of (C.2.11): 

 0

u2t dx ≤

r0 2ρ0

 t 0

 0

u2xx dxdτ +

r0

M 2L2 (0,T ) , t ∈ [0, T ]. 2ρ0  

C.3 The Problem with Dirichlet Input (Slope) Consider the following initial-boundary value problem: ⎧ ⎪ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x = 0, ⎪ ⎪ ⎪ ⎪ ⎪ (x, t) ∈ T := (0, ) × (0, T ), ⎪ ⎨ u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, ), ⎪ ⎪ ⎪ ⎪ u(0, t) = 0, ux (0, t) = θ (t), uxx (, t) = 0, ⎪ ⎪ ⎪ ⎩ − (r()uxx (, t))x + Tr ()ux (, t) = 0, t ∈ [0, T ].

(C.3.1)

The problem (C.3.1) is a mathematical model of dynamic vibration of a cantilever beam subjected to the non-zero slope θ (t) at the end x = 0. This problem arises also as an adjoint problem corresponding to the inverse problem discussed in Sect. 11.3. Let us introduce the new function v(x, t) = u(x, t) − xθ (t), (x, t) ∈ T .

(C.3.2)

Then the function v(x, t) solves the following initial-boundary value problem: ⎧ ⎪ ρ(x)vt t + μ(x)vt + (r(x)vxx )xx − (Tr (x)vx )x , ⎪ ⎪ ⎪ ⎪ ⎪ = −xμ(x)θ  (t) − xρ(x)θ  (t), (x, t) ∈ T , ⎪ ⎨ v(x, 0) = −θ (0)x, vt (x, 0) = −θ  (0)x, x ∈ (0, ), ⎪ ⎪ ⎪ v(0, t) = 0, v (0, t) = 0, u (, t) = 0, ⎪ ⎪ x xx ⎪ ⎪ ⎩ − (r()vxx (, t))x + Tr ()vx (, t) = 0, t ∈ [0, T ].

(C.3.3)

C Necessary Estimates For Euler-Bernoulli Beam Equation

497

Lemma C.3.1 Assume that the conditions of Theorem C.2.1 hold. Suppose, in addition, the Dirichlet input θ (t) in (C.3.1) satisfies the regularity condition: θ ∈ H 2 (0, T ).

(C.3.4)

Then for the weak solution v ∈ L2 (0, T ; V(0, )), with ut ∈ L2 (0, T ; L2 (0, )), ut t ∈ L2 (0, T ; H −2 (0, )) and V(0, ) := {v ∈ H 2 (0, ) : v(0) = v  (0) = 0}, of problem (C.3.1) the following estimates hold: #2 θ 2 2

vt 2L2 (0,T ;L2 (0,)) ≤ C , 3 H (0,T ) #2 θ 2 2 ,

vxx 2L2 (0,T ;L2 (0,)) ≤ C 4 H (0,T )

(C.3.5)

where #2 [exp(T /ρ0 ) − 1], C #2 = r0 C #2 , #2 = C C 3 2 4 ρ0 3 #0 , C #1 C #2 = 2 3 max(μ21 , ρ12 ), #T ), C #2 = max(C C 0 3 2 2 2 2 2 # #2 = 2(T /3 + 1/T ). C1 =  ( ρ1 /3 + Tr1 ), C T

(C.3.6)

Proof Multiply both sides of Eq. (C.3.3) by 2vt (x, t), integrate it over t := (0, ) × (0, t), use the identity (C.2.6) then apply the integration by parts formula. Taking into account the initial and boundary conditions in (C.3.3) we obtain the following energy identity: 

 0

 t 2 ρ(x)vt2 + r(x)vxx + Tr (x)vx2 dx + 2 0

= −2

 t 0



 0

μ(x)vτ2 dxdτ

xμ(x)θ (τ ) + xρ(x)θ  (τ ) vτ dxdτ

0





−θ (0)



 xρ(x)dx − θ (0)

0



Tr (x)dx, t ∈ [0, T ].

(C.3.7)

0

Denote by I1 and I2 the first term and the algebraic sum of the third and fourth terms on right-hand-side of (C.3.7), respectively. Then these terms can be estimated as follows: I1 ≤

 t 0

0



vτ2 dxdτ + ≤

 t 0

 t 0

  2  2 x 2 μ2 (x) θ  (τ ) + ρ 2 (x) θ  (τ ) dxdτ

0 

0



  2 #02 θ  2 2 , (C.3.8) vτ2 dxdτ + C +

θ

2 L (0,T ) L (0,T )

498

C

Necessary Estimates For Euler-Bernoulli Beam Equation

for all t ∈ [0, T ], and 



I2 ≤ 

 x 2 ρ 2 (x)dx +

0



0

Tr2 (x)dx



 2 (θ (0))2 + θ  (0)

   #12 (θ (0))2 + θ  (0) 2 . ≤C

(C.3.9)

#0 , C #1 > 0 are the constants introduced in (C.3.6). We employ now in (C.3.9) where C the inequalities 

 1 T 2  2

θ L2 (0,T ) + θ L2 (0,T ) (θ (0)) ≤ 2 T 3     2 1  2 T  2 θ (0) ≤ 2

θ L2 (0,T ) + θ L2 (0,T ) , T 3 2

which are proved in the same way as inequality (11.1.17). This leads to #12 C #T2 θ 2 2 , I2 ≤ C H (0,T ) #T > 0 introduced in (C.3.6). Substituting this inequality with with the constant C (C.3.8) in (C.3.7) we obtain the following integral inequality: 



ρ0 0

 vt2 dx

+ r0 0



 2 vxx dx

≤

+ 0

 t 0



 0

Tr (x)vx2 dx

+2

 t 0

 0

μ(x)vτ2 dxdτ

2 #22 θ 2 2 vxx dxdτ + C , t ∈ [0, T ], H (0,T )

#2 > 0 defined in (C.3.6). with C The required estimates (C.3.5) are derived from this integral inequality in the same way as in the proof of Theorem C.2.1.   In view of estimates (C.3.2) we can derive similar estimates for the weak solution of the problem (C.3.1) with the Dirichlet input θ (t). Theorem C.3.1 Assume that the conditions of Lemma C.3.1 hold. Then for the weak solution u ∈ L2 (0, T ; V0 (0, )), with ut ∈ L2 (0, T ; L2 (0, )), ut t ∈ L2 (0, T ; H −2 (0, )), V0 (0, ) := {v ∈ H 2 (0, ) : v(0) = 0} and ux (0, t) = θ (t), of problem (C.3.1) the following estimates hold: #2 θ 2 2

ut 2L2 (0,T ;L2 (0,)) ≤ C , 5 H (0,T ) #2 θ 2 2

uxx 2L2 (0,T ;L2 (0,)) ≤ C , 4 H (0,T )

C Necessary Estimates For Euler-Bernoulli Beam Equation

499

#2 = 2(C #2 + 3 /3) and C #3 , C #4 > 0 are the constants introduced in where C 5 3 Lemma C.3.1.

C.4 The Problem with Non-homogeneous Initial Data Consider the following problem with non-homogeneous initial conditions: ⎧ ⎪ ρ(x)ut t + μ(x)ut + (r(x)uxx )xx − (Tr (x)ux )x ⎪ ⎪ ⎪ ⎨ = F (x)G(t), (x, t) ∈ T ,

(C.4.1)

⎪ u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ (0, ), ⎪ ⎪ ⎪ ⎩ u(0, t) = uxx (0, t) = 0, u(, t) = uxx (, t) = 0, t ∈ [0, T ], We assume that the inputs in (C.4.1) satisfy the following conditions: ⎧ ρ, r, μ, Tr ∈ L∞ (0, ), ⎪ ⎪ ⎪ ⎪ ⎨ u ∈ H 2 (0, ), u ∈ L2 (0, ), F ∈ L2 (0, ), G ∈ L2 (0, T ), 0 1 ⎪ 0 < ρ0 ≤ ρ(x) ≤ ρ1 , 0 < r0 ≤ r(x) ≤ r1 , ⎪ ⎪ ⎪ ⎩ 0 ≤ μ0 ≤ μ(x) ≤ μ1 , 0 ≤ Tr0 ≤ Tr (x) ≤ Tr1 , x ∈ (0, ).

(C.4.2)

Theorem C.4.1 Assume that the conditions (C.4.2) hold. Then for the weak solution u ∈ L2 (0, T ; V 2 (0, )) with ut ∈ L2 (0, T ; L2 (0, )) and ut t ∈ L2 (0, T ; H −2 (0, )) of the initial boundary value problem (C.4.1) the following estimates hold:

ut 2L∞ (0,T ;L2 (0,)) ≤

 1 2 C1 + 1 F 2L2 (0,) G 2L2 (0,T ) ρ0



+Cu2 u0 2H 2 (0,) + u1 2L2 (0,)

 ,

ut 2L2 (0,T ;L2 (0,)) ≤ C12 F 2L2 (0,) G 2L2 (0,T )

 +Cu2 u0 2H 2 (0,) + u1 2L2 (0,) ,

uxx 2L∞ (0,T ;L2 (0,)) ≤

uxx 2L2 (0,T ;L2 (0,))

 1 2 C1 + 1 F 2L2 (0,) G 2L2 (0,T ) r0

 +Cu2 u0 2H 2 (0,) + u1 2L2 (0,) ,  r0 ≤ C12 F 2L2 (0,) G 2L2 (0,T ) ρ0

 +Cu2 u0 2H 2 (0,) + u1 2L2 (0,) ,

(C.4.3)

500

C

Necessary Estimates For Euler-Bernoulli Beam Equation

where V 2 (0, ) := {v ∈ H 2 (0, ) : v(0) = v() = 0}, C12 = exp(T /ρ0 ) − 1, Cu2 = max(r1 , ρ1 , Tr1 ),

(C.4.4)

and ρ0 , r0 > 0 are the constants introduced in (C.4.2). Proof Multiply both sides of Eq. (C.4.1) by 2ut (x, t), integrate it over t := (0, ) × (0, t), use the identity 

2(r(x)uxx )xx ut ≡ 2[(r(x)uxx )x ut − r(x)uxx uxt ]x + r(x)u2xx , t

(C.4.5)

and then apply the integration by parts formula. Taking into account the initial and boundary conditions in (C.4.1) we obtain the following energy identity: 



0

ρ(x)u2t

=2

+

r(x)u2xx

 t

+ Tr (x)u2x



dx + 2μ

 t 0

 0

u2τ dxdτ



F (x)G(τ )uτ (x, τ )dxdτ 0

0





+ 0

ρ(x)u2t (x, 0+ ) + r(x)u2xx (x, 0+ ) + Tr (x)u2x (x, 0+ ) dx,

t ∈ [0, T ]. We deduce from this identity the main integral inequality:  ρ0 0



 u2t dx + r0

≤

0

 t 0



0



u2xx dx + 2

 t 0

0



Tr (x)u2x dxdτ + 2

 t 0

 0

μ(x)u2τ dxdτ

u2τ dx + F 2L2 (0,) G 2L2 (0,T )

+ρ1 u1 2L2 (0,) + r1 u0 2L2 (0,) + Tr1 u0 2L2 (0,), for all t ∈ [0, T ]. Estimates (C.4.3) are easily derived from this inequality.

(C.4.6)  

C Necessary Estimates For Euler-Bernoulli Beam Equation

501

In the particular case u0 (x) = u1 (x) = 0, estimates (C.4.3) and (11.1.4) coincide. Corollary C.4.1 Assume that conditions of Theorem C.4.1 hold. Then for the final time output u(x, T ) the following estimate holds:

u(·, T ) 2L2 (0,) ≤ 2T C12 F 2L2 (0,) G 2L2 (0,T )   +2 1 + T C12 Cu2 u0 2H 2 (0,) + 2T C12 Cu2 u1 2L2 (0,),

ut (·, T ) 2L2 (0,) ≤

 1 2 C1 + 1 F 2L2 (0,) G 2L2 (0,T ) ρ0

+Cu2 u0 2H 2 (0,) + u1 2L2 (0,)

(C.4.7)  ,

where C1 , Cu > 0 are the constants introduced in (C.4.4). Proof The first estimate of (C.4.7) follows from the inequality

u(·, T ) 2L2 (0,) ≤ 2 u0 2L2 (0,) + 2T u 2L2 (0,T ;L2 (0,)), which is a consequence of the identity  u(x, T ) = u0 (x) +

T

ut (x, t)dt, |x ∈ [0, ],

0

and the second estimate of (C.4.3). Next estimate of (C.4.7) is a direct result of the first estimate of (C.4.3).

 

The second estimate in (C.4.7) is required if the final time measured output in the inverse problem is provided as the measured velocity νT (x) := ut (x, T ) at the final time instance t = T . Corollary C.4.2 Assume that conditions (C.4.2) are satisfied. Then the following estimate holds:

ux (·, T ) 2L2 (0,) ≤

 1 2 2  C1 + 1 F 2L2 (0,) G 2L2 (0,T ) 2

 +Cu2 u0 2H 2 (0,) + u1 2L2 (0,) ,

(C.4.8)

where C1 , Cu > 0 are the constants introduced in Theorem C.4.1. Proof follows from the first inequality in (11.1.10) and the third estimate of (C.4.3). 

502

C

Necessary Estimates For Euler-Bernoulli Beam Equation

Theorem C.4.2 Assume that in addition to the basic conditions (C.4.2), the inputs in (C.4.1) satisfy the following regularity conditions: 

r ∈ H 2 (0, ), G ∈ H 1 (0, T ),

(C.4.9)

u0 ∈ H 4 (0, ), u1 ∈ H 2 (0, ).

Then for the regular weak solution u ∈ L2 (0, T ; H 4 (0, )), with ut ∈ L2 (0, T ; V 2 (0, )), ut t ∈ L2 (0, T ; L2 (0, )) and ut t t ∈ L2 (0, T ; H −2 (0, )) of the initial boundary value problem (C.4.1) the following estimates hold: 

ut t 2L2 (0,T ;L2 (0,)) ≤ C12 C32 F 2L2 (0,) G 2H 1 (0,T )

 +C42 u0 2H 4 (0,) + u1 2H 2 (0,) , 

uxxt 2L2 (0,T ;L2 (0,)) ≤ C22 C32 F 2L2 (0,) G 2H 1 (0,T )

 +C42 u0 2H 4 (0,) + u1 2H 2 (0,) ,

(C.4.10)

where C32 = 1 + 4CT2 /ρ0 , Cr = Cr ( r H 2 (0,)) > 0, C4 = max(C5 , C6 ), C52 = max(r1 , Tr1 , 4μ2 /ρ0 ), C62 = 2 max(Cr2 , Tr21 )/ρ0 ,

(C.4.11)

and C1 , CT > 0 are the constants introduced in (C.4.4) and (11.1.17), respectively. Proof Differentiate Eq. (C.4.1) with respect to the time variable t ∈ (0, T ) and multiply both sides by 2ut t (x, t). Integrating then it over t := (0, ) × (0, t) and using the identity (C.4.5) with u(x, t), replaced by ut (x, t), we obtain the following integral identity: 

 0

ρ(x)u2t t =2

 t 0





+ 0

+ r(x)u2xxt 

+ Tr (x)u2xt



dx + 2μ

 t 0

 0

u2τ τ dxdτ

F (x)G (τ )uτ τ (x, τ )dxdτ

0

ρ(x)u2t t (x, 0+ ) + r(x)u2xxt (x, 0+ ) + Tr (x)u2xt (x, 0+ ) dx,

t ∈ [0, T ]. We use the limit equation ρ(x)ut t (x, 0+ ) + μut (x, 0+ ) + (r(x)uxx (x, 0+ ))xx = F (x)G(0+ )

(C.4.12)

C Necessary Estimates For Euler-Bernoulli Beam Equation

503

at t = 0+ to estimate the first term in the second right-hand-side integral in the above identity. This yields: 

 0

ρ(x)u2t t (x, 0+ )dx ≤

2 4  G(0+ ) F 2L2 (0,) ρ0

+ (r(·)u0 ) 2L2 (0,) + μ21 u1 2L2 (0,) + Tr21 u0 2L2 (0,) . Substituting this in (C.4.12) with the estimates

(r(·)u0 ) 2L2 (0,) ≤ Cr2 u0 2H 4 (0,), 

G(0+ )

2

≤ CT2 G 2H 1 (0,),

with the constants Cr , CT > 0 introduced in (C.4.11) and (11.1.17), we arrive at the following integral inequality: 



ρ0 0

 u2t t dx

+ r0 0



u2xxt dx

+ 2μ

 t 0

 0

u2τ τ dxdτ

≤

 t 0

 0

u2τ τ dx

 4μ21 4 2  2 2 CT G H 1 (0,T ) F 2L2 (0,) +

u1 2L2 (0,) + G L2 (0,T ) + ρ0 ρ0 + r1 u1 2L2 (0,) + Tr1 u1 2L2 (0,) +

4Tr21  2 4Cr2

u0 2H 4 (0,) +

u0 L2 (0,) , ρ0 ρ0

for all t ∈ [0, T ]. This inequality leads to the required estimates (C.4.10) with constants introduced in (C.4.11).  

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Index

A Accuracy error, 45, 113, 388 Adjoint operator, 31, 40 a-posteriori parameter choice rule, 51 Approximate solution, 45, 92, 215, 387, 474 a-priori parameter choice rule, 51 Attenuated Radon transform, 229 Axial tensile force, 365

B Backward parabolic problem, 102 Banach Contraction Mapping Principle, 144, 170 Bending moment, 365, 374, 391, 439, 451 Best approximation, 25 Born approximation, 214

C Characteristic lines, 164, 237, 252 Compactness, 7, 39, 185, 205, 395, 408, 442, 459, 465 dual role in ill-posedness, 7 Computational noise level, 110, 122 Conjugate Gradient Algorithm, 113, 195, 324, 386, 400 Consistency conditions for output, 75, 131, 173, 197, 280, 304, 338, 356, 377 Convergence error, 113, 196, 324, 389, 401 Convergent regularization strategy, 55, 89

D Damped wave equation, 450 Damping coefficient, 249, 365, 402, 450 Data error, 46 Descent direction parameter, 115, 195, 324 Direct problem, 18, 67, 95, 130, 133, 147, 152, 174, 228, 364 Dirichlet-to-Neumann operator, 222, 251, 349 Duhamel’s Principle, 108

E Eigensystem self-adjoint operator, 39, 406, 408 input-output operator, 75, 97, 410 Error estimate, 91 Euler-Bernoulli beam equation, 363, 431 Euler’s equations, 237

F Filter function, 49, 78, 428 Flexural rigidity, 365, 431, 440, 450 Fredholm integral equation, 14 Fredholm operator, 15

G Galerkin Finite Element Method, 110 Global stability, 145, 339 Gradient formula, 48, 70, 73, 195, 210, 287, 384, 399, 426, 439, 449, 462

© Springer Nature Switzerland AG 2021 A. Hasanov Hasano˘glu, V. G. Romanov, Introduction to Inverse Problems for Differential Equations, https://doi.org/10.1007/978-3-030-79427-9

513

514 I Ill-posedness degree, 42, 44 Hadamard’s definition, 8 Ill-posed problem, 6, 9, 14 mildly ill-posed, 45 moderately ill-posed, 45 severely ill-posed, 45, 105 Input-output operator, 69, 97, 101, 185, 206, 251, 381, 407, 432, 458 Input-output relationship, 16, 209, 286, 398, 439 Inverse coefficient problem, 3, 134, 147, 151, 177, 182, 200, 250, 326 Inverse conductivity problem, 169 Inverse crime, 110 Inverse kinematic problem, 236, 240 Inverse Laplace transform, 174 Inverse scattering problem, 214 Inverse source problem, 66, 94, 130, 133, 229, 379, 450 Inverse spectral problem, 180 Inversion formula, 4, 230, 231

K Kirchhoff plate equation, 363, 449

L Landweber iteration algorithm, 118 Laplace transform, 172 Lavrentiev regularization, 55

M Measured output Dirichlet type, 3, 130, 182, 291, 430, 450 final time, 65, 95, 102, 405 Neumann type, 175, 201, 251, 326 Moore-Penrose inverse, 36 Morozov’s Discrepancy Principle, 51, 59 Moving load, 386, 389

N Neumann-to-Dirichlet operator, 185, 291, 318 Noisy data, 45, 59, 89, 113 Normal equation, 35, 49, 50, 70

O Orthogonal complement, 24 Orthogonal decomposition, 27

Index Orthogonal projection, 27 Orthogonal Projection Theorem, 27

P Parameter choice rule, 51 Parameter identification problem, 3 Picard criterion, 44, 81 Picard’s Theorem, 43 Point source, 216 Projection operator, 27

Q Quasi-solution, 47

R Radon transform, 216, 231 Recovering the potential, 135, 174, 291 Reflection method, 129 Regularization error, 46 Regularization strategy, 46, 51 Regularized Tikhonov functional, 47 Regularizing filter, 88 Regular weak solution, 201, 269, 310, 345, 370, 476, 478, 484, 494 Residual, 34 Riemannian coordinates, 238 Riemannian distance, 242 Riemannian invariants, 154 Riemannian metric, 235

S Scattering amplitude, 215 Schrödinger equation, 213 Set of admissible coefficients, 3, 185, 204, 275, 308, 349, 432 Singular system, 39, 40, 42, 77 non-self-adjoint operator, 40, 42 self-adjoint operator, 40, 77, 105 Singular value, 39 Singular value decomposition, 42, 45 Singular value expansion (SVE), 40, 52, 78, 405 Slope of deflection, 365 Source condition, 62 Sourcewise element, 63 Spatial load, 94, 405, 449 Spectral function, 181 Spectral representation, 39 Stability estimate, 4, 86, 145, 231, 242, 253, 343

Index

515

Stopping condition, 114, 196, 324, 401 Stopping rule, 63, 117

Truncated singular value decomposition, 45, 91

T Temporal load, 95, 379 Termination index, 117 Tikhonov functional, 47, 70, 97, 113, 190, 194, 205, 277, 353, 396, 423, 437, 446 Tikhonov-Phillips regularization, 47 Tikhonov regularization, 47, 55, 70 Tikhonov regularization with a Sobolev norm, 191 Tomography problem, 221, 228 Transformed inverse problem, 148, 293, 338 Transport equation, 227, 228 Transverse deflection, 365 Transverse shear force, 374, 391

V Volterra equation, 132, 134, 338

W Weak solution Euler-Bernoulli beam equation, 486 heat equation, 475, 478, 482 Kirchhoff plate equation, 452 Sturm-Liouville problem, 83 wave equation, 94 Well-posedness, 2 Well-posed problem, 2, 72, 130, 148, 289, 322, 356, 386, 425, 461