Inverse Problems: Mathematical and Analytical Techniques with Applications to Engineering

INVERSE PROBLEMS MATHEMATICAL AND ANALYTICAL TECHNIQUES WITH APPLICATIONS TO ENGINEERING MATHEMATICAL AND ANALYTICAL...

Author: Alexander G. Ramm

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INVERSE PROBLEMS

MATHEMATICAL AND ANALYTICAL TECHNIQUES WITH APPLICATIONS TO ENGINEERING

MATHEMATICAL AND ANALYTICAL TECHNIQUES WITH APPLICATIONS TO ENGINEERING Alan Jeffrey, Consulting Editor

Published: Inverse Problems A. G. Ramm Singular Perturbation Theory R. S. Johnson

Forthcoming: Methods for Constructing Exact Solutions of Partial Differential Equations with Applications S. V. Meleshko The Fast Solution of Boundary Integral Equations S. Rjasanow and O. Steinbach Stochastic Differential Equations with Applications R. Situ

INVERSE PROBLEMS

MATHEMATICAL AND ANALYTICAL TECHNIQUES WITH APPLICATIONS TO ENGINEERING

ALEXANDER G. RAMM

Springer

eBook ISBN: Print ISBN:

0-387-23218-4 0-387-23195-1

©2005 Springer Science + Business Media, Inc. Print ©2005 Springer Science + Business Media, Inc. Boston All rights reserved No part of this eBook may be reproduced or transmitted in any form or by any means, electronic, mechanical, recording, or otherwise, without written consent from the Publisher Created in the United States of America Visit Springer's eBookstore at: and the Springer Global Website Online at:

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To Luba and Olga

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CONTENTS

Foreword Preface

xv

xvii

1. Introduction

1

1.1 Why are inverse problems interesting and practically important?

1

1.2 Examples of inverse problems 2 1.2.1 Inverse problems of potential theory 2 1.2.2 Inverse spectral problems 2 1.2.3 Inverse scattering problems in quantum physics; finding the potential from the impedance function 2 1.2.4 Inverse problems of interest in geophysics 3 1.2.5 Inverse problems for the heat and wave equations 3 1.2.6 Inverse obstacle scattering 4 1.2.7 Finding small subsurface inhomogeneities from the measurements of the scattered field on the surface 5 1.2.8 Inverse problem of radiomeasurements 5 1.2.9 Impedance tomography (inverse conductivity) problem 5 1.2.10 Tomography and other integral geometry problems 5 1.2.11 Inverse problems with “incomplete data” 6 1.2.12 The Pompeiu problem, Schiffer’s conjecture, and inverse problem of plasma theory 7 1.2.13 Multidimensional inverse potential scattering 8 1.2.14 Ground-penetrating radar 8 1.2.15 A geometrical inverse problem 9 1.2.16 Inverse source problems 10

viii

Contents

Identification problems for integral-differential equations 12 Inverse problem for an abstract evolution equation 12 Inverse gravimetry problem 12 Phase retrieval problem (PRP) 12 Non-overdetermined inverse problems 12 Image processing, deconvolution 13 Inverse problem of electrodynamics, recovery of layered medium from the surface scattering data 13 1.2.24 Finding ODE from a trajectory 13 1.2.17 1.2.18 1.2.19 1.2.20 1.2.21 1.2.22 1.2.23

1.3 Ill-posed problems

14

1.4 Examples of Ill-posed problems 15 1.4.1 Stable numerical differentiation of noisy data 15 1.4.2 Stable summation of the Fourier series and integrals with randomly perturbed coefficients 15 1.4.3 Solving ill-conditioned linear algebraic systems 15 1.4.4 Fredholm and Volterra integral equations of the first kind 16 1.4.5 Deconvolution problems 16 1.4.6 Minimization problems 16 1.4.7 The Cauchy problem for Laplace’s equation 16 1.4.8 The backwards heat equation 17 2. Methods of solving ill-posed problems

19

2.1 Variational regularization 19 2.1.1 Pseudoinverse. Singular values decomposition 19 2.1.2 Variational (Phillips-Tikhonov) regularization 20 2.1.3 Discrepancy principle 22 2.1.4 Nonlinear ill-posed problems 23 2.1.5 Regularization of nonlinear, possibly unbounded, operator 24 2.1.6 Regularization based on spectral theory 25 2.1.7 On the notion of ill-posedness for nonlinear equations 26 2.1.8 Discrepancy principle for nonlinear ill-posed problems with monotone operators 26 2.1.9 Regularizers for Ill-posed problems must depend on the noise level 29 2.2 Quasisolutions, quasinversion, and Backus-Gilbert method 30 2.2.1 Quasisolutions for continuous operator 30 2.2.2 Quasisolution for unbounded operators 31 2.2.3 Quasiinversion 32 2.2.4 A Backus-Gilbert-type method: Recovery of signals from discrete and noisy data 32 2.3 Iterative methods

40

2.4 Dynamical system method (DSM) 41 2.4.1 The idea of the DSM 41 2.4.2 DSM for well-posed problems 42 2.4.3 Linear ill-posed problems 45 2.4.4 Nonlinear ill-posed problems with monotone operators 49 2.4.5 Nonlinear ill-posed problems with non-monotone operators 57 2.4.6 Nonlinear ill-posed problems: avoiding inverting of operators in the Newton-type continuous schemes 59 2.4.7 Iterative schemes 62

ix

2.4.8 A spectral assumption 64 2.4.9 Nonlinear integral inequality 2.4.10 Riccati equation 70

65

2.5 Examples of solutions of ill-posed problems 71 2.5.1 Stable numerical differentiation: when is it possible? 71 2.5.2 Stable summation of the Fourier series and integrals with perturbed coefficients 85 2.5.3 Stable solution of some Volterra equations of the first kind 2.5.4 Deconvolution problems 87 2.5.5 Ill-conditioned linear algebraic systems 88 2.6 Projection methods for ill-posed problems

87

89

3. One-dimensional inverse scattering and spectral problems

91

3.1 Introduction 92 3.1.1 What is new in this chapter? 92 3.1.2 Auxiliary results 92 3.1.3 Statement of the inverse scattering and inverse spectral problems 97 3.1.4 Property C for ODE 98 3.1.5 A brief description of the basic results 99 3.2 Property C for ODE 104 3.2.1 Property 104 3.2.2 Properties and

105

3.3 Inverse problem with I-function as the data 108 3.3.1 Uniqueness theorem 108 3.3.2 Characterization of the I-functions 110 3.3.3 Inversion procedures 112 3.3.4 Properties of 112 3.4 Inverse spectral problem 122 3.4.1 Auxiliary results 122 3.4.2 Uniqueness theorem 124 3.4.3 Reconstruction procedure 126 3.4.4 Invertibility of the reconstruction steps 128 3.4.5 Characterization of the class of spectral functions of the Sturm-Liouville operators 130 3.4.6 Relation to the inverse scattering problem 130 3.5 Inverse scattering on half-line 132 3.5.1 Auxiliary material 132 3.5.2 Statement of the inverse scattering problem on the half-line. Uniqueness theorem 137 3.5.3 Reconstruction procedure 139 3.5.4 Invertibility of the steps of the reconstruction procedure 143 3.5.5 Characterization of the scattering data 145 3.5.6 A new Marchenko-type equation 147 3.5.7 Inequalities for the transformation operators and applications 148 3.6 Inverse scattering problem with fixed-energy phase shifts as the data 3.6.1 Introduction 156 3.6.2 Existence and uniqueness of the transformation operators independent of angular momentum 157

156

x

Contents

3.6.3 Uniqueness theorem 165 3.6.4 Why is the Newton-Sabatier (NS) procedure fundamentally wrong? 166 3.6.5 Formula for the radius of the support of the potential in terms of scattering data 172 3.7 Inverse scattering with “incomplete data” 176 3.7.1 Uniqueness results 176 3.7.2 Uniqueness results: compactly supported potentials 180 3.7.3 Inverse scattering on the full line by a potential vanishing on a half-line 181 3.8 Recovery of quarkonium systems 181 3.8.1 Statement of the inverse problem 3.8.2 Proof 183 3.8.3 Reconstruction method 185

181

3.9 Krein’s method in inverse scattering 186 3.9.1 Introduction and description of the method 186 3.9.2 Proofs 192 3.9.3 Numerical aspects of the Krein inversion procedure 200 3.9.4 Discussion of the ISP when the bound states are present 201 3.9.5 Relation between Krein’s and GL’s methods 201

3.10 Inverse problems for the heat and wave equations 202 3.10.1 Inverse problem for the heat equation 202 3.10.2 What are the “correct” measurements? 203 3.10.3 Inverse problem for the wave equation 204 3.11 Inverse problem for an inhomogeneous Schrödinger equation 204 3.12 An inverse problem of ocean acoustics 208 3.12.1 The problem 208 3.12.2 Introduction 209 3.12.3 Proofs: uniqueness theorem and inversion algorithm 212 3.13 Theory of ground-penetrating radars 216 3.13.1 Introduction 216 3.13.2 Derivation of the basic equations 217 3.13.3 Basic analytical results 219 3.13.4 Numerical results 221 3.13.5 The case of a source which is a loop of current 222 3.13.6 Basic analytical results 225 4. Inverse obstacle scattering 227 4.1 Statement of the problem 227 4.2 Inverse obstacle scattering problems 234 4.3 Stability estimates for the solution to IOSP 240 4.4 High-frequency asymptotics of the scattering amplitude and inverse scattering problem 243 4.5 Remarks about numerical methods for finding S from the scattering data 245 4.6 Analysis of a method for identification of obstacles 247

xi

5. Stability of the solutions to 3D Inverse scattering problems with fixed-energy data 255 5.1 Introduction 255 5.1.1 The direct potential scattering problem 5.1.2 Review of the known results 256

256

5.2 Inverse potential scattering problem with fixed-energy data 264 5.2.1 Uniqueness theorem 264 5.2.2 Reconstruction formula for exact data 264 5.2.3 Stability estimate for inversion of the exact data 267 5.2.4 Stability estimate for inversion of noisy data 270 5.2.5 Stability estimate for the scattering solutions 273 5.2.6 Spherically symmetric potentials 274 5.3 Inverse geophysical scattering with fixed-frequency data 275 5.4 Proofs 5.4.1 5.4.2 5.4.3 5.4.4 5.4.5 5.4.6 5.4.7 5.4.8

of some estimates 277 Proof of (5.1.18) 277 Proof of (5.1.20) and (5.1.21) Proof of (5.2.17) 283 Proof of (5.4.49) 285 Proof of (5.4.51) 286 Proof of (5.2.13) 287 Proof of (5.2.23) 289 Proof of (5.1.30) 292

278

5.5 Construction of the Dirichlet-to-Neumann map from the scattering data and vice versa 293 5.6 Property C

298

5.7 Necessary and sufficient condition for scatterers to be spherically symmetric 300 5.8 The Born inversion

307

5.9 Uniqueness theorems for inverse spectral problems 6. Non-uniqueness and uniqueness results

312

317

6.1 Examples of nonuniqueness for an inverse problem of geophysics 6.1.1 Statement of the problem 317 6.1.2 Example of nonuniqueness of the solution to IP 318

317

6.2 A uniqueness theorem for inverse boundary value problems for parabolic equations 319 6.3 Property C and an inverse problem for a hyperbolic equation 6.3.1 Introduction 321 6.3.2 Statement of the result. Proofs 321 6.4 Continuation of the data

321

330

7. Inverse problems of potential theory and other inverse source problems 7.1 Inverse problem of potential theory 7.2 Antenna synthesis problems

333

336

7.3 Inverse source problem for hyperbolic equations

337

333

xii

Contents

8. Non-overdetermined inverse problems 8.1 Introduction

339

8.2 Assumptions

340

8.3 The problem and the result 8.4 Finding 8.5 Appendix

from

339

340 342

347

9. Low-frequency inversion

349

9.1 Derivation of the basic equation. Uniqueness results 9.2 Analytical solution of the basic equation

353

9.3 Characterization of the low-frequency data 9.4 Problems of numerical implementation

349

355

355

9.5 Half-spaces with different properties 356 9.6 Inversion of the data given on a sphere 9.7 Inversion of the data given on a cylinder 9.8 Two-dimensional inverse problems 9.9 One-dimensional inversion

357 358

359

362

9.10 Inversion of the backscattering data and a problem of integral geometry 363 9.11 Inversion of the well-to-well data 9.12 Induction logging problems

364

366

9.13 Examples of non-uniqueness of the solution to an inverse problem of geophysics 369 9.14 Scattering in absorptive medium 9.15 A geometrical inverse problem

371 371

9.16 An inverse problem for a biharmonic equation

373

9.17 Inverse scattering when the background is variable 375 9.18 Remarks concerning the basic equation

377

10. Wave scattering by small bodies of arbitrary shapes

379

10.1 Wave scattering by small bodies 379 10.1.1 Introduction 379 10.1.2 Scalar wave scattering by a single body 380 10.1.3 Electromagnetic wave scattering by a single body 10.1.4 Many-body wave scattering 385

383

10.2 Equations for the self-consistent field in media consisting of many small particles 388 10.2.1 Introduction 388 10.2.2 Acoustic fields in random media 390 10.2.3 Electromagnetic waves in random media 394

xiii

10.3 Finding small subsurface inhomogeneities from scattering data 395 10.3.1 Introduction 396 10.3.2 Basic equations 397 10.3.3 Justification of the proposed method 398 10.4 Inverse problem of radiomeasurements 11. The Pompeiu problem

401

405

11.1 The Pompeiu problem 405 11.1.1 Introduction 405 11.1.2 Proofs 407 11.2 Necessary and sufficient condition for a domain, which fails to have Pompeiu property, to be a ball 414 11.2.1 Introduction 414 11.2.2 Proof 416 Bibliographical Notes References Index

441

425

421


FOREWORD

The importance of mathematics in the study of problems arising from the real world, and the increasing success with which it has been used to model situations ranging from the purely deterministic to the stochastic, is well established. The purpose of the set of volumes to which the present one belongs is to make available authoritative, up to date, and self-contained accounts of some of the most important and useful of these analytical approaches and techniques. Each volume provides a detailed introduction to a specific subject area of current importance that is summarized below, and then goes beyond this by reviewing recent contributions, and so serving as a valuable reference source. The progress in applicable mathematics has been brought about by the extension and development of many important analytical approaches and techniques, in areas both old and new, frequently aided by the use of computers without which the solution of realistic problems would otherwise have been impossible. A case in point is the analytical technique of singular perturbation theory which has a long history. In recent years it has been used in many different ways, and its importance has been enhanced by it having been used in various fields to derive sequences of asymptotic approximations, each with a higher order of accuracy than its predecessor. These approximations have, in turn, provided a better understanding of the subject and stimulated the development of new methods for the numerical solution of the higher order approximations. A typical example of this type is to be found in the general study of nonlinear wave propagation phenomena as typified by the study of water waves.

xvi

Foreword

Elsewhere, as with the identification and emergence of the study of inverse problems, new analytical approaches have stimulated the development of numerical techniques for the solution of this major class of practical problems. Such work divides naturally into two parts, the first being the identification and formulation of inverse problems, the theory of ill-posed problems and the class of one-dimensional inverse problems, and the second being the study and theory of multidimensional inverse problems. On occasions the development of analytical results and their implementation by computer have proceeded in parallel, as with the development of the fast boundary element methods necessary for the numerical solution of partial differential equations in several dimensions. This work has been stimulated by the study of boundary integral equations, which in turn has involved the study of boundary elements, collocation methods, Galerkin methods, iterative methods and others, and then on to their implementation in the case of the Helmholtz equation, the Lamé equations, the Stokes equations, and various other equations of physical significance. A major development in the theory of partial differential equations has been the use of group theoretic methods when seeking solutions, and in the introduction of the comparatively new method of differential constraints. In addition to the useful contributions made by such studies to the understanding of the properties of solutions, and to the identification and construction of new analytical solutions for well established equations, the approach has also been of value when seeking numerical solutions. This is mainly because of the way in many special cases, as with similarity solutions, a group theoretic approach can enable the number of dimensions occurring in a physical problem to be reduced, thereby resulting in a significant simplification when seeking a numerical solution in several dimensions. Special analytical solutions found in this way are also of value when testing the accuracy and efficiency of new numerical schemes. A different area in which significant analytical advances have been achieved is in the field of stochastic differential equations. These equations are finding an increasing number of applications in physical problems involving random phenomena, and others that are only now beginning to emerge, as is happening with the current use of stochastic models in the financial world. The methods used in the study of stochastic differential equations differ somewhat from those employed in the applications mentioned so far, since they depend for their success on the Ito calculus, martingale theory and the Doob-Meyer decomposition theorem, the details of which are developed as necessary in the volume on stochastic differential equations. There are, of course, other topics in addition to those mentioned above that are of considerable practical importance, and which have experienced significant developments in recent years, but accounts of these must wait until later. Alan Jeffrey University of Newcastle Newcastle upon Tyne United Kingdom

PREFACE

This book can be used for courses at various levels in ill-posed problems and inverse problems. The bibliography of the subject is enormous. It is not possible to compile a complete bibliography and no attempt was made to do this. The bibliography contains some books where the reader will find additional references. The author has used extensively his earlier published papers, and referenced these, as well as the papers of other authors that were used or mentioned. Let us outline some of the novel features in this book. In Chapter 1 the statement of various inverse problems is given. In Chapter 2 the presentation of the theory of ill-posed problems is shorter and sometimes simpler than that published earlier, and quite a few new results are included. Regularization for ill-posed operator equations with unbounded nonlinear operators is studied. A novel version of the discrepancy principle is formulated for nonlinear operator equations. Convergence rate estimates are given for Backus-Gilbert-type methods. The DSM (Dynamical systems method) in ill-posed problems is presented in detail. The presentation is based on the author’s papers and the joint papers of the author and his students. These results appear for the first time in book form. Papers [R216], [R217], [R218], [R220], [ARS3], [AR1] have been used in this Chapter. In Chapter 3 the presentation of one-dimensional inverse problems is based mostly on the author’s papers, especially on [R221]. It contains many novel results, which are described at the beginning of the Chapter. The presentation of the classical results, for example, Gel’fand-Levitan’s theory, and Marchenko’s theory, contains many novel points. The presentation of M. G. Krein’s inversion theory with complete proofs is

xviii

Preface

given for the first time. The Newton-Sabatier inversion theory, which has been in the literature for more than 40 years, and was presented in two monographs [CS], [N], is analyzed and shown to be fundamentally wrong in the sense that its foundations are wrong (cf. [R206]). This Chapter is based on the papers [R221], [R199], [R197], [R196], [R195], [R192], [R185]. One of the first papers on inverse spectral problems was Ambartsumian’s paper (1929) [Am], where it was proved that one spectrum determines the one-dimensional Neumann Schrödinger’s operator uniquely. This result is an exceptional one: in general one spectrum does not determine the potential uniquely (see Section 3.7 and [PT]). Only 63 years later a multidimensional analog of Ambartsumian’s result was obtained ([RSt1]). The main technical tool in this Chapter and in Chapter 5 is Property C, that is, completeness of the set of products of solutions to homogeneous differential equations. For partial differential equations this tool has been introduced in [R87] and developed in many papers and in the monograph [R139]. For ordinary differential equations completeness of the products of solutions to homegeneous ordinary equations has been used in different forms in [B], [L1]. In our book Property C for ODE is presented in the form introduced and developed by the author in [R196]. In Chapter 4 the presentation of inverse obstacle scattering problems contains many novel points. The requirements on the smoothness of the boundary are minimal, stability estimates for the inversion procedure corresponding to fixed-frequency data are given, the high-frequency inversion formulas are discussed and the error of the inversion from noisy data is estimated. Analysis of the currently used numerical methods is given. This Chapter is based on [R83], [R155], [R162], [R164], [R167], [R167], [R171], [RSa]. In Chapter 5 a presentation of the solution of the 3D inverse potential scattering problem with fixed-energy noisy data is given. This Chapter is based on the series of the author’s papers, especially on the paper [R203]. The basic concept used in the analysis of the inverse scattering problem in Chapter 5 is the concept of Property C, i.e., completeness of the set of products of solutions to homogeneous partial differential equations. This concept was introduced by the author ([R87]) and applied to many inverse problems (see [R139] and references therein). An important part of the theory consists of obtaining stability estimates for the potential, reconstructed from fixedenergy noisy data (and from exact data). Error estimates for the Born inversion are given under suitable assumptions. It is shown that the Born inversion may fail while the Born approximation works well. In other words, the Born approximation may be applicable for solving the direct scattering problem, while the Born inversion, that is, inversion based on the Born approximation, may fail. The Born inversion is still popular in applications, therefore these error estimates will hopefully be useful for practitioners. The author’s inversion method for fixed-energy scattering data, is compared with that based on the usage of the Dirichlet-to-Neumann map. The author shows why the difficulties in numerical implementation of his method are less formidable than the difficulties in implementing the inversion method based on the Dirichlet-to-Neumann map.

xix

A necessary and sufficient condition for a scatterer to be spherically symmetric is given ([R128]). In Chapter 6 an example of non-uniqueness of the solution to a 3D problem of geophysics is given. It illustrates the crucial role of the uniqueness theorems in a study of inverse problems. One may try to solve numerically such a problem, by a parameterfitting, which is very popular among practitioners. But if the uniqueness result is not established, the numerical results may be meaningless. Some uniqueness theorems for inverse boundary value problem and for an inverse problem for hyperbolic equations are established in this Chapter. In Chapter 7, inverse problems of potential theory and antenna synthesis are briefly discussed. The presentation of the theory on this topic is not complete: there are books and many papers on antenna synthesis (e.g., [MJ], [ZK], [AVST], [R21], [R26]) including nonlinear problems of antenna synthesis [R23], [R27]). Chapter 8 contains a discussion of non-overdetermined problems. These are, roughly speaking, the inverse problems in which the unknown function depends on the same number of variables as the data function. Examples of such problems are given. Most of these problems are open: even uniqueness theorems are not available. Such a problem, namely, recovery of an unknown coefficient in a Schrödinger equation in a bounded domain from the knowledge of the values of the spectral function on the boundary is discussed under the assumption that all the eigenvalues are simple, that is, the corresponding eigenspaces are one-dimensional. The presentation follows [R198]. In Chapter 9 the theory of the inversion of low-frequency data is presented. This theory is based on the series of author’s papers, starting with [R68], [R77], and uses the presentation in [R83] and [R139]. Almost all of the results in this Chapter are from the above papers and books. Chapter 10 is a summary of the author’s results regarding the theory of wave scattering by small bodies of arbitrary shapes. These results have been obtained in a series of the author’s papers and are summarized in [R65], [R50]. The solution of inverse radiomeasurements problem ([R33], [R65]) is based on these results. Also, these results are used in the solution of the problem of finding small subsurface inhomogeneities from the scattering data, measured on the surface. The solution to this problem can be used in modeling ultrasound mammography, in finding small holes in metallic objects, and in many other applied problems. In Chapter 11 the classical Pompeiu problem is presented following the papers [R177], [R186]. The author thanks several publishers of his papers, mentioned above, for the permission to use these papers in the book. There are many questions that the author did not discuss in this book: inverse scattering for periodic potentials and other periodic objects, such as gratings, periodic objects, (see, e.g., [L] for one-dimensional scattering problems for periodic potentials), the Carleman estimates and their applications to inverse problems ([Bu2], [H], [LRS]), the inverse problems for elasticity and Maxwell’s equations ([RK], [Ya]), the methods based on controllability results ([Bel]), problems of tomography and integral geometry ([RKa], [R139]), etc. Numerous parameter-fitting schemes for solving various

xx

Preface

engineering problems are not discussed. There are many papers published, which use parameter-fitting for solving inverse problems. However, in most cases there are no error estimates for parameter-fitting schemes for solving inverse problems, and one cannot guarantee any accuracy of the inversion result. In [GRS] the concept of stability index is introduced and applied to a parameter-fitting scheme for solving a one-dimensional inverse scattering problem in quantum physics. This concept allows one to get some idea about the error estimate in a parameter-fitting scheme. The applications of inverse scattering to integration of nonlinear evolution equations are not discussed as there are many books on this topic (see e.g., [M], [FT] and references therein).

1. INTRODUCTION

1.1

WHY ARE INVERSE PROBLEMS INTERESTING AND PRACTICALLY IMPORTANT?

Inverse problems are the problems that consist of finding an unknown property of an object, or a medium, from the observation of a response of this object, or medium, to a probing signal. Thus, the theory of inverse problems yields a theoretical basis for remote sensing and non-destructive evaluation. For example, if an acoustic plane wave is scattered by an obstacle, and one observes the scattered field far from the obstacle, or in some exterior region, then the inverse problem is to find the shape and material properties of the obstacle. Such problems are important in identification of flying objects (airplanes missiles, etc.), objects immersed in water (submarines, paces of fish, etc.), and in many other situations. In geophysics one sends an acoustic wave from the surface of the earth and collects the scattered field on the surface for various positions of the source of the field for a fixed frequency, or for several frequencies. The inverse problem is to find the subsurface inhomogeneities. In technology one measures the eigenfrequencies of a piece of a material, and the inverse problem is to find a defect in this material, for example, a hole in a metal. In geophysics the inhomogeneity can be an oil deposit, a cave, a mine. In medicine it may be a tumor, or some abnormality in a human body. If one is able to find inhomogeneities in a medium by processing the scattered field on the surface, then one does not have to drill a hole in a medium. This, in turn, avoids expensive and destructive evaluation. The practical advantages of remote sensing are what makes the inverse problems important.

2

1. Introduction

1.2

EXAMPLES OF INVERSE PROBLEMS

1.2.1

Inverse problems of potential theory

Suppose a body

with a density

generates gravitational potential

Is it possible to find given the potential u (x) for far away from D? A point mass m and a uniformly distributed mass m in a ball of radius a produce the same potential Thus, it is not possible to find uniquely from the knowledge of u in However, if one knows a priori that in D, then it is possible to find D from the knowledge of u (x) in provided that D is, for example, star-shaped, that is, every ray issued from some interior point intersects the boundary of D at only one point. 1.2.2

Inverse spectral problems

Let be the Sturm-Liouville operator defined by the Dirichlet boundary conditions as self-adjoint operator in and be its eigenvalues. To what extent does the set of these eigenvalues determine q (x )? Roughly speaking, one spectrum, that is, the set determines “half of q (x),” in the sense that if q (x) is known on then one spectrum determines uniquely q (x) on A classical result due to Borg [B]and Marchenko [M] says that two spectra uniquely determine the operator i.e., the potential q and the boundary conditions at x = 0 and x = 1 of the type and where and are constants, and one assumes that the two spectra correspond to the same and two distinct The author (see [R196]) asked the following question: if q (x) is known on the segment [b, 1], 0 < b < 1, then what part of the spectrum one needs to know in order to uniquely recover q (x) on [0, b]? It is assumed that q is real valued: and Let be the spectral measure of the self-adjoint operator l. This notion is defined in Chapter 3. The inverse spectral problem is: given find q (x), and the boundary conditions, characterizing Similar problems can be formulated in the multidimensional cases, when a bounded domain D plays the role of the segment [0, 1], the role of the spectral data is played by the eigenvalues and the values on the boundary S of D of the normal derivatives of the normalized eigenfunctions, One may choose other spectral data. 1.2.3

Inverse scattering problems in quantum physics; finding the potential from the impedance function

Consider

the

by function), and by

Dirichlet operator in the Jost function, by

Denote by the Jost solution, the I-function (impedance the scattering data (see Chapter 3).

3

The inverse problem of quantum scattering on the half-axis consists of finding q (x), given It was studied in [M]. The inverse problem of finding q (x), given is of interest in many applications. The I-function has the physical meaning of the impedance function, it is the ratio in the problem of electromagnetic wave falling perpendicularly onto the earth, when the dielectric permittivity and conductivity of the earth depend on the vertical coordinate only. One can prove that coincides with the Weyl function (Chapter 3). It turns out that known determines uniquely q (x), and one can explicitly calculate and given ([R196]). 1.2.4

Inverse problems of interest in geophysics

There are many inverse problems of interest in geophysics. A typical one consists of finding an unknown inhomogeneity in the velocity profile (refraction coefficient) from the scattered acoustic field measured on the surface of the earth and generated by a point source, situated on the surface of the earth at varying positions. Its mathematical formulation (in a simplified form) is:

where

is the known background refraction coefficient, where supp v(x) is the support of v, and v is an inhomogeneity in the refraction coefficient (or in the velocity profile), u is the acoustic pressure, u satisfies the radiation condition at infinity (or the limiting absorption principle). An inverse problem of geophysics consists of finding the function v from the knowledge of the scattered field on the surface of the Earth, that is from known for all at a fixed or for all where is a small number (the case of low-frequency surface data) (cf [LRS], [Ro], [R83], [R139]). Another problem is to find the conductivity of the medium from the measurements of the electromagnetic waves, scattered by a source that moves in a borehole along the vertical line. ([R83], [R139]) 1.2.5

Inverse problems for the heat and wave equations

A typical inverse problem for the heat equation

is to find q (x) from the flux measurements: The extra data (measured data), allow one to find q ( x ) . Another inverse problem is to find the unknown conductivity from boundary measurements. For example, let Can one find given a(t) and

4

1. Introduction

Consider the inverse conductivity problem: let in on where N is the outer unit normal to S, the extra data is the flux g at the boundary. Suppose that the set is known. Can one determine uniquely? Here D is a bounded domain with a sufficiently smooth boundary S, and is the Sobolev space. In applications in medicine, f is the electrostatic potential, which can be applied to a human chest, and g is the flux of the electrostatic field, which can be measured. If one can determine from these measurements then some diagnostic information is obtained ([R157], [R139], [R131], [R103], [Gro], [LRS], [Ro], [Is1]). There are many inverse problem for the wave equation. One of them is to find the velocity c (x) in the equation at t = 0, u = 0 on S, given the extra data on S for a fixed and all t > 0, or for varying on S, and where T > 0 is some number. ([LRS], [Ro], [RRa], [RSj]). 1.2.6

Inverse obstacle scattering

Let be bounded domain with a Lipschitz boundary be the exterior domain, be the unit sphere in The scattering problem consists of finding the scattering solution, i.e., the solution to the problem

The coefficient A is called the scattering amplitude. Existence and uniqueness of the solution to (1.2.2)–(1.2.3) are proved in [RSa] without any assumption on the smoothness of the boundary. If the Neumann boundary condition

is used in place of (1.2.3), then the existence and uniqueness of the solution to (1.2.2)– (1.2.3N) are proved in [RSa] under the assumption of compactness of the embedding where R > 0 is such that and is the Sobolev space. See also [GoR]. The inverse obstacle scattering problem consists of finding S and the boundary condition on S, given in the following cases: (1) either at a fixed for all and all or, (2) at a fixed for all and running through open subsets of (3) for fixed and and all

or,

Uniqueness of the solution of the first inverse problem is proved by M. Schiffer (1964), (see [R83]) of the second by A. G. Ramm (1986) (see [R83]), and the third

5

problem is still open. See also [R154], [R155], [R162], [R159], [R164], [R167], [R171], [CK]. One may consider a penetrable layered obstacle, and ask if the scattering amplitude at a fixed allows one to determine the boundaries of all the layers uniquely, and the constant velocity profiles in each of the layers. See [RPY] for an answer to this question. 1.2.7

Finding small subsurface inhomogeneities from the measurements of the scattered field on the surface

Suppose there are few small, in comparison with the wave-length, holes in the metallic body. A source of acoustic waves is on the surface of the body, and the scattered field is measured on the surface of the body for various positions of the acoustic source, at a fixed frequency. The inverse problem is to find the number of the small holes, their locations, and their volume. A similar problem is important in medicine, where the small bodies are the cancer cells to be found in the healthy tissue of a human’s body. In the ultrasound mammography modeling, one deals with the tissue of a woman’s breast ([R193], [GR1]). 1.2.8

Inverse problem of radiomeasurements

Suppose a complicated electromagnetic field distribution (E, H) exists in the aperture of a mirror antenna. For many practical reasons one wants to know this distribution. Let be the field scattered by a small probe placed at a point x in the aperture of the antenna. Given the shape and electromagnetic constants and of the probe, the inverse problem of radiomeasurements consists of finding (E(x), H(x)) from the knowledge of (See [R65] for a solution to this problem). 1.2.9

Impedance tomography (inverse conductivity) problem

This problem was briefly mentioned in Section 1.2.5. 1.2.10

Tomography and other integral geometry problems

Define

where is the unit sphere in The function is called the Radon transform of f. The function f can be assumed piecewise-continuous and absolutely integrable over every plane so that the Radon transform would be well defined in the classical sense. But in fact, one can define the Radon transform for much larger sets of functions and on distributions [R170], [RKa], [Hel]. Given one can uniquely recover f (x) provided, for example, that or where is the weighted space with the norm Practically interesting questions are: (a) How are singularities of f and related? (b) Given the noisy measurements of f at a grid, how does one find the discontinuities of f ?

6

1. Introduction

A grid is a set of points

where The noisy measurement are where are identically distributed, independent random variables with zero mean value and a finite variance See [R176], [RKa] for a detailed investigation of the above problem. An open problem is: what are the minimal assumptions on the growth of f (x) at infinity that guarantee the injectivity of the Radon transform? There is an example of a smooth function such that for all p, and [RKa]. In many applications one integrates f not over the planes but over some other family of manifolds. The problem of integral geometry is to recover f from the knowledge of its integral over a family of manifolds. For example, if the family of manifolds is a family of spheres of various radii r > 0 and centers s running over some surface S, then are the spherical means of f, and the problem is to recover f from the knowledge of and Conditions on S that guarantee the injectivity of the operator M are given in [R211], where some inversion formulas are also derived. 1.2.11

Inverse problems with “incomplete data”

Suppose that not all the scattering data in Section 1.2.3 are given, for example, is given, but and J are unknown. In general, one cannot recover a from these “incomplete” data. However, if one knows a priori that q (x) has compact support, or then the data alone determine q (x) uniquely. Such type of inverse problems we call inverse problems with “incomplete” data. The “incompleteness” of the data is remedied by the additional a priori assumption about q (x), so, in fact, the data are complete in the sense that q (x) is determined uniquely by these data. Another example of an inverse problem with “incomplete” data, is recovery of q (x) = 0 for x < 0, from the knowledge of the reflection coefficient in the full-axis (full-line) scattering problem:

The coefficients and are reflection and transmission coefficients, respectively. A general cannot be uniquely recovered from the knowledge of alone: one needs to know additionally the bound states and norming constants to recover q uniquely. However if one knows a priori that q (x) = 0 for for example, for x < 0, then q (x) is uniquely determined by alone.

7

1.2.12 The Pompeiu problem, Schiffer’s conjecture, and inverse problem of plasma theory

Let

Assume

where SO(n) is the group of rotations, and is a bounded domain. The problem (going back to Pompeiu (1929)) is to prove that (1.2.4) implies that D is a ball. Originally Pompeiu claimed that (1.2.4) implies that f = 0, but this claim is wrong. References related to this problem are given in [R186], [R177], [Z]. One can prove that (1.2.4) holds iff (= if and only if) for all and some where some

and

Iff

and

then the overdetermined problem

has a solution. The Schiffer’s conjecture is: if D is a bounded connected domain homeomorphic to a ball, and

then D is a ball. The Pompeiu problem in the form (1.2.4) is equivalent to the following conjecture: if

and D is homeomorphic to a ball, then D is a ball. An inverse problem of plasma theory consists of the following. Let

where u is a non trivial solution to (1.2.8), (i.e., if f (0) = 0 then and let the extra data (measured data) be the value Assume that f (u) is an entire function of u. The inverse problem, of interest in plasma theory, is: given can one recover f (u) uniquely. Even for j = 0, 1, the problem is open (cf [Vog]).

8

1. Introduction

1.2.13 Multidimensional inverse potential scattering

Let

where is given, and The coefficient is called the scattering amplitude, and the solution to (1.2.9)–(1.2.10) is called the scattering solution. The direct scattering problem is: given q, and find and, in particular, This problem has been studied in great detail (see, e.g., [CFKS], [[R121], Appendix]) under various assumptions on q ( x ) . We assume that for and often we assume additionally that The inverse scattering problem (ISP) consists of finding q ( x ) , given Consider several cases: (1) A is given for all and all (2) A is given for all and a fixed (3) A is given for a fixed all and all (4) is given for all and all

(back scattering data).

In case (1) uniqueness of the solution to ISP has been established long ago, and follows easily from the asymptotics of A as An inversion formula based on highenergy asymptotics of A is known (Born inversion), (cf [Sai]). In case (2) the uniqueness of the solution to ISP is proved by Ramm [R109], [R100], (see also [R105], [R112], [R114], [R115], [R120], [R125], [R130], [R133], [R140], [R142], [R143], [RSt2], [R203]), an inversion formula for the exact data is derived in [R109], [R143], an inversion formula for the noisy data is derived by in [R143], and stability estimates for the inversion formulas for the exact and noisy data are derived by in [R143], [R203]. In case (3) and (4) uniqueness of the solution to ISP is an open problem, but in the case (4) uniqueness holds if one assumes a priori that q is sufficiently small. A generic uniqueness result is given in [St1]. See also [StU]. 1.2.14

Ground-penetrating radar

Let the source of electromagnetic waves be located above the ground, and the scattered field be observed on the ground. From these data one wants to get information about the properties of the ground. Mathematical modeling of this problem is based on the Maxwell equations

where is the region of the ground,

is the vertical coordinate, is the source,

f (t)

9

describes the shape of the pulse of the current j along a wire going along the axis at the height above the ground. Assume for (in the air), for f (t) = 0 for t < 0 and t > T, and are dielectric and magnetic constants, for Differentiate the second equation (1.2.1) with respect to t, and get Let Then

where for groundpenetrating radar. The ground-penetrating radar inverse problem is: given find and One may use the source which is a current along a loop of wire, is the unit vector in cylindrical coordinates. In this case, one looks for E of the form: and from (1.2.11) one gets:

Let

where is the Bessel function. Set Then u solves (1.2.12), and the inverse problem is the same as above

([R185]). 1.2.15

A geometrical inverse problem

Let

Here D is domain homeomorphic to an annulus, is its inner boundary, outer boundary. The geometrical inverse problem is: given and find

is its

10

1. Introduction

One can interpret the data as the Cauchy data on for an electrostatic potential u, and then is the surface on which the potential is vanishing if or the charge distribution is vanishing if ([R139]). 1.2.16

Inverse source problems

(1) Inverse source problems in acoustics Let in f = 0 for uniformly in

tern A by the formula:

as

satisfies the radiation condition Define the radiation patThen

The inverse source problems are: (i) Given and find f (x). (ii) Given and a fixed find f (x). Clearly, by (1.2.15) problem (i) has at most one solution, but an a priori given function may be not of the form (1.2.15): the right-hand side of (1.2.15) is an entire function of exponential type of the vector Problem (ii), in general, may have many solutions, since may vanish for all at some (2) Inverse source problem in electrodynamics Consider Maxwell’s equations (1.2.11) in and assume j = 0 for radiation condition for (E, H) is:

The

where E and H in (1.2.11) are assumed monochromatic with time dependence, and where are the constant values of and near infinity. The inverse source problem is: given find j. Again, one should specify for what and the function is known. One can derive the relation between A and j. Namely, assuming and constants, and j smooth and compactly supported, one starts with the equations

then gets

11

then

so

Let

Thus

and

Then

where [a, b] is the vector product. So

and

It is now clear, that even if A is known for all and all vector J is not uniquely determined, but only its component orthogonal to is determined. Therefore, the solution to the inverse source problem in electrodynamics is not unique and may not exist, in general. The antenna synthesis problems are inverse source-type problems of electrodynamics. For example, if j is the current along a linear antenna (which is a wire along is the length of the antenna, then

so

is determined uniquely by the data A. Finding which produces the desired diagram is the problem of linear antenna synthesis. There is a large body of literature on this subject. Let for The inverse source problem is: given find f (x, t), The questions mentioned in this subsection were discussed in many papers and books ([AVST], [MJ], [ZK], [R11], [R21], [R26], [R27], [R28], [R73], [Is2]).

12

1. Introduction

1.2.17 Identification problems for integral-differential equations

Consider a Cauchy problem.

where is a closed, linear, densely defined in a Hilbert space H, operator, D(A) is its domain of definition, Assume that h (t) is unknown, and the extra data are given for The inverse problem is: given find h(t). See [LR]. 1.2.18 Inverse problem for an abstract evolution equation

Consider a Cauchy problem with the extra data (measured data) Here A(t) is a one-parametric family of closed, densely defined, linear, operators on a Banach space X, which generates an evolution family is a given function on [0, T] with values in X, and is an unknown scalar function. The inverse problem is: given f(t), and w, find (see [RKo]). In applications may be a control function which should be chosen so that the measured data are reproduced. 1.2.19 Inverse gravimetry problem

Let be the of the gravitational field generated by some masses, located in the region Assume that the values u(x, 0) := f ( x ) are known, in the region and u(x, –h) := g(x). Then is the equation for g. The inverse gravimetry problem is: given f, find g. See [VA], [RSm1]. 1.2.20

Phase retrieval problem (PRP)

Let The PRP consists of finding arg given Clearly, the solution to this problem is not unique: produces the same Under suitable assumptions on f ( x ) and one can get uniqueness results for PRP. See [KST], [R139]. 1.2.21

Non-overdetermined inverse problems

Formally we call an inverse problem non-overdetermined if the unknown function, which is to be found, depends on the same number of variables as the data. For example, problems in cases (1) and (2) in Section 1.2.13 are overdetermined, while in cases (3) and (4) they are not overdetermined. In multidimensional inverse scattering problems uniqueness of the solution is an open problem for most of the non-overdetermined

13

problems. For example, in Section 1.2.13 case (3) is a non-overdetermined problem, and uniqueness of its solution is an open problem. Recently Ramm ([R198]) proved that the spectral data and all determine q (x) uniquely provided that all the eigenvalues are simple. Here in is the kernel of the resolution of the identity of the selfadjoint Neumann operator in The above inverse problem is not overdetermined, because depends on three variables in and q (x) is also a function of three variables in It is an open problem to find out if this result remains valid without the assumption about the simplicity of all the eigenvalues. 1.2.22 Image processing, deconvolution

In many applications one is interested in the following inverse problem: given the properties of a linear device and the output signal, find the input signal. By the properties of a linear device one means its point-spread function (scattering function) or transfer function. For example, Given and f (x), one wants to find In practice the output signal f (x) is noisy, i.e., is given in place of f(x), where the norm depends on the problem at hand. See [BG], [RSm6], [RG]. 1.2.23

Inverse problem of electrodynamics, recovery of layered medium from the surface scattering data

There are many inverse problems arising in electrodynamics. If (1.2.11) are the governing equations,

where e is a constant vector, and are known constants: and respectively, outside a bounded domain and the data are measured on the surface of the Earth for various orientations of e then the inverse problem is to determine and from the above data. See [RSo], [RK]. 1.2.24

Finding ODE from a trajectory

Let and are constants, u = u ( t ) , Can one find and uniquely? In general, the answer is no. A trivial example is u ( t ) = 0. What trajectory allows one to find and uniquely? Suppose

14

1. Introduction

Then the system of equations

for finding and is uniquely solvable, so that (1.2.20) guarantees that u(t), determines and uniquely. More generally, consider a system where is a constant matrix, u (0) = v. If there are points such that the system is a linearly independent system of vectors, then u (t) determines the matrix uniquely. The reader can easily prove this. One can find more details in [Den]. 1.3

ILL-POSED PROBLEMS

Why are ill-posed problems important in applications? How are they related to inverse problems? Let

where X and Y are Banach spaces, or metric spaces, and A is a nonlinear operator, in general. Problem (1.3.1) is called well-posed if A is a homeomorphism of X onto Y . In other words, the solution to (1.3.1) exists for any is unique, and depends on f continuously, so that is a continuous map. If some of these conditions do not hold, then the problem is called ill-posed. Ill-posed problems are important in many application, in which one may reduce a physical problem to equation (1.3.1) where A is not boundedly invertible. For example, consider the equation

If then A is not boundedly invertible: A is injective, its range belongs to the Sobolev space is an unbounded operator in If noisy data are given, then may be not in the range of A. A practically interesting problem is: can one find an operator such that as In other words, can one estimate stably, given and Any Fredholm first-kind integral equation with linear compact operator is of the form (1.3.1). Such an operator in an infinite-dimensional space cannot have closed range and cannot be boundedly invertible. Since many inverse problems can be reduced to ill-posed equations (1.3.1), these inverse problems are ill-posed. That is how Ill-posed problems are related to inverse problems. Methods for stable solution of Ill-posed problems are developed in Chapter 2. The literature on Ill-posed problems is enormous ([IVT], [EHN], [Gro], [TLY], [VV], [VA], [R58]).

15

1.4 EXAMPLES OF ILL-POSED PROBLEMS

1.4.1

Stable numerical differentiation of noisy data

This example has been mentioned in Section 1.3. Methods for stable numerical differentiation of noisy data are given in Chapter 2. In navigation a ship receives a navigation signal which is a univalent function f (x) (that is, a smooth function which has precisely one point of maximum), and the course of the ship is determined by this point. The function f is observed in an additive noise. Given noisy data one wants to find A possible approach to this problem, is to search for a point at which One can see from a simple example that small perturbations of f can lead to large perturbations of let Then No matter how small is, one can choose so large that will take arbitrary large values at some points x. 1.4.2

Stable summation of the Fourier series and integrals with randomly perturbed coefficients

Let be an orthonormal basis of a Hilbert space H and L be a linear system (a linear operator) such that where are some numbers. Due to the inner noise in L, one observes the noisy output where is the noise. Thus, if the input signal is the output is One has noisy Fourier coefficients and one wants to recover the function If one has a Fourier integral, one can formulate a similar problem. To see that this problem is ill-posed, in general, let us take denote and assume that is playing the role of noise, Then one has the problem: given the noisy Fourier coefficients find This problem is ill-posed because the series may diverge. For example, if and then the series diverges. Similarly, if is the Fourier transform of a function and is the noisy data, then the problem is to calculate f (x) with “minimal error”, given noisy data The notion of “minimal error” should be specified. 1.4.3

Solving ill-conditioned linear algebraic systems

Let

be a linear operator such that its condition number is large. Then the linear algebraic system Au = f can be considered practically as an ill-posed problem, because small perturbation of the data f may lead to a large perturbation in the solution u. One has: so that the relative error in the data may result in relative error in the solution. In the above derivation we use the inequality Methods for stable solution of ill-conditioned algebraic systems are given in Chapter 2.

16

1. Introduction

1.4.4

Fredholm and Volterra integral equations of the first kind

If or and is a continuous kernel in D := [a, b] × [a, b], then the operators A and V are compact in these operator are not boundedly invertible in H. Therefore problems Au = f and Vu = f are ill-posed. 1.4.5

Deconvolution problems

These are problems, arising in applications: an input signal u generates an output signal f by the formula

Often one has:

The deconvolution problem consists of finding u, given f and For (1.4.2) the identification problem is of practical interest: given u (t) and f (t), find A (t). The function A(t) characterizes the linear system which generates the output f (t) gives the input u(t). Mathematically the deconvolution problems are the problems from Section 1.4.4. 1.4.6

Minimization problems

Consider the minimization problem Suppose that is the infimum of If f is perturbed, that is, then the infimum of may be not attained, or it may be attained at an element which is far away from Thus the map may be not continuous. In this case the minimization problem is ill-posed. Such problems were studied [Vas]. 1.4.7

The Cauchy problem for Laplace’s equation

Claim 1. The Cauchy problem for Laplace’s equation is an ill-posed problem. Consider the problem:

in the half-plane It is clear that solves the above problem, and this solution is unique (by the uniqueness of the solution to the Cauchy problem for elliptic equations). This example belongs to J. Hadamard, and it shows that the Cauchy data may be arbitrarily small (take while the solution tends to infinity, as at any point Thus, the claim is verified (cf. [LRS], [R139]).

17

1.4.8

The backwards heat equation

Consider the backwards heat equation problem:

Given v(x), one wants to find u (x, 0) := w (x). By separation of variables one finds Therefore, vided that this series converges, in that is, provided that

pro-

This cannot happen unless decays sufficiently fast. Therefore the backwards heat equation problem is ill-posed: it is not solvable for a given v (x) unless (1.4.3) holds, and small perturbations of the data v in may lead to arbitrary large perturbations of the function w(x), but also may lead to a function v for which the solution u(x, t) does not exist for t < T (cf. [LRS], [IVT]).


2. METHODS OF SOLVING ILL-POSED PROBLEMS

2.1

VARIATIONAL REGULARIZATION

2.1.1

Pseudoinverse. Singular values decomposition

Consider linear Equation (1.3.1). Let be a linear closed operator, D(A) and R(A) be its domain and range, and be Hilbert spaces, N(A) := {u : Au = 0}, is the adjoint operator, the bar stands for the closure, and is the orthogonal sum. If A is injective, i.e., N(A) = {0}, and surjective, i.e. and then the inverse operator is defined on I is the identity operator. A closed, linear, defined on all of operator, is bounded, so A is an isomorphism of onto if it is injective, surjective, and If P is the orthoprojector onto N(A) in and Q is the orthoprojector onto then one defines a pseudoinverse (generalized inverse) Thus, and for The operator is bounded iff If i.e., for some then the problem has a solution every element is also a solution, and there is a unique solution with minimal norm, namely the solution such that If then the infimum of is not attained. If A is bounded and then the element solves the equation and is the minimal norm solution to this equation,

20

2. Methods of solving ill-posed problems

i.e., Conversely, if

Indeed, if and and

and with Thus,

then then if where

One can prove the formula: is a regularization parameter (see Section 2.1.2) and Let us define the singular value decomposition. Let be a linear compact operator, is a compact selfadjoint operator,

are called s -values of A. If

test). If

then Thus,

and

If is arbitrary, then Thus, an element are eigenvalues of then where

(Picard’s Then

If dim then A can be written as where A is an m × n matrix, U and V are unitary matrices (n × n and m × m, respectively), whose columns are vectors and respectively, and S is an m × n matrix with the diagonal elements r is the rank of the matrix A, and other elements of S are zeros. The matrix can be calculated by the formula where is an n × m matrix with diagonal elements and other elements of are zeros. 2.1.2

Variational (Phillips-Tikhonov) regularization

Assume is linear, is not necessarily in R(A). The problem Au = f is assumed ill-posed (cf. Sec. 1.3). Consider the problem:

where

is a parameter.

Theorem 2.1.1. Assume Au = f, and (i) The minimizer (ii) If and

Then:

of (2.1.1) does exist and is unique satisfies the condition where

as

then

21

Proof. Functional (2.1.1) is quadratic. A necessary and sufficient condition for its minimizer is the Euler’s equation:

which has a unique solution has Thus Choose so that so as and that prove that

Claim (i) is proved. One if Below stands for various positive constants. as and let Then This implies and we claim This claim we prove later. Thus Let us One has

Here the estimate and the relation as were used. To prove the first estimate, one uses the formula: and the polar representation of A* yields, where V is an isometry, One has where the spectral representation for T was used. Let us prove the second relation: where P is the orthoprojector onto N(A), and is the resolution of the identity of the self-adjoint operator B (see [KA]). If then Finally, let us prove the claim used above. Lemma 2.1.2. If B is a monotone hemicontinuous operator in a Hilbert space H, D(B) = H, then

In our proof above, hemicontinuous, if

is a linear operator, so B is monotone, that is, Recall that a nonlinear operator A is called is a continuous function of for any

Proof of Lemma 2.1.2. Clearly, If then that is, monotone operator is w -closed. Indeed, if B is monotone, then for any Passing to the limit and using hemicontinuity of B, one gets: This implies so Lemma 2.1.2 is proved. The claim is proved for nonlinear monotone operators in Theorem 2.1.6 below. Theorem 2.1.1 is proved.


22

2.1.3

Discrepancy principle

Theorem 2.1.1 gives an a priori choice of which guarantees convergence An a posteriori choice of is given by Theorem 2.1.3 below. Theorem 2.1.3. (Discrepancy principle). Assume equation

If

is the root of the

then Proof. First, let us prove that equation (2.1.3) has a unique solution. Write this equation as

where is the resolution of the identity of the selfadjoint operator and the commutation formula was used. One checks this formula easily. If then If then where is the orthoprojector on Indeed, because and since and Thus, if then equation (2.1.3) has a solution. This solution is unique because is a monotone increasing function of for each fixed Now let us prove Since

One has it follows that Therefore (*) If then one can select a weakly convergent sequence as In the proof of Theorem 2.1.1 it was proved that where is the unique minimal-norm solution of the equation By the lower semicontinuity of the norm in H, one has Together with (*), one gets This and the weak convergence imply

Our proof is based on the following useful result. Theorem 2.1.4. If Proof. If Thus, as

then

and

then

Also one has and

23

2.1.4

Nonlinear ill-posed problems

Lemma 2.1.5. Assume that A in (1.3.1) is a closed, nonlinear, injective map. If K is a compactum, then the inverse operator is continuous on A(K). Proof. Since A is injective, Then closed, and A, one has

is well-defined on A(K ). Let where is a subsequence denoted again Since A is imply and, by the injectivity of Lemma 2.1.5 is proved.

Claim: Let us assume that and Then

is monotone, continuous, D(A) = H, A(u) = f,

Proof. Indeed, Multiply this equation by the monotonicity of A, to get Thus, Let a sequence, denoted again by such that as Then Since A is monotone, it is w-closed (see the proof of Theorem 2.1.1), so and is the minimal norm solution to equation (1.3.1).

and use Select

Theorem 2.1.6. If is monotone and hemicontinuous, if D(A) = H, if where is the minimal-norm solution to A(u) = f, and if then the minimal-norm solution to (1.3.1) is unique and Proof. We have Thus, Let Then and Thus, This and the weak convergence imply strong convergence as in Theorem 2.1.3. The minimal norm solution to (1.3.1) is unique if A is monotone and continuous, because in this case the set of solutions N is convex and closed. Its closedness is obvious, if A is continuous. Its convexity follows from the monotonicity of A and the following lemma: Lemma 2.1.7 (Minty). If A is monotone and continuous, then (a) is equivalent to (b) Proof. If (a) holds, then then take Take and get Lemma 2.1.7 is proved.

and (b) holds by the monotonicity of A. If (b) holds, where w is arbitrary, and get Thus, A(u) = f, and (a) holds.

To prove that N is convex, one assumes that and derives that Indeed, if then, by Lemma 2.1.7, Thus, Thus,


24

To prove uniqueness of the minimal norm element of a convex and closed set N in a Hilbert space, one assumes that there are two such elements, and Then and so that any element of the segment joining and has minimal norm m. Since Hilbert space is strictly convex, this implies Indeed, take t = 1/2. Then So Thus, Theorem 2.1.6 is proved. Consider the equation

Theorem 2.1.8. Assume that A is monotone and continuous, and equation (1.3.1) has a solution. If and as then the unique solution to (2.1.4) converges strongly to u, the unique solution to (1.3.1) of minimal norm. Proof. Because A is monotone and the equation has a solution, and this solution is unique. Let solve (2.1.4), and One has By Theorem 2.1.6, Let us prove We have Multiply this by and use the monotonicity of A to get This implies if Theorem 2.1.8 is proved. 2.1.5

Regularization of nonlinear, possibly unbounded, operator

Assume that : is a closed, injective, possibly nonlinear, map in Banach space X. is a functional such that the set is precompact in X for any constant (3) Equation (1.3.1) has a solution (4) The last assumption can be replaced in some cases when A is an unbounded operator, by the assumption.

(1) (2)

Define the functional

Let equality

where is a parameter, Consider the minimization problem:

where Denote and

is the smallest integer satisfying the inOne has By assumption (2), as one can select a

25

convergent subsequence, denoted again Thus, A(u) = f by the closedness of A, and limit of any subsequence is the same, namely We have proved:

such that by the injectivity of A. Since the it follows that

Theorem 2.1.9. If (1.3.1) has a solution, then, under the assumptions (1)–(4) (or any sequence such that converges strongly to the solution of (1.3.1) as Remark 2.1.10. In the proof of Theorem 2.1.9 we do not need existence of the minimizer of the function (2.1.5). 2.1.6

Regularization based on spectral theory

Assume that A in (1.3.1) is a linear bounded operator, i.e., is the unique minimal-norm solution.

and

Lemma 2.1.11. Solvable equation (1.3.1) with bounded linear operator A is equivalent to the equation

Proof. If u solves (1.3.1), apply A* to (1.3.1) and get (2.1.6), so u solves (2.1.6). If u solves (2.1.6) and (1.3.1) is solvable, i.e., then and This implies so Au = f. Thus u solves (1.3.1).

Equation (2.1.6) is a solvable equation with monotone, continuous operator, so Theorem 2.1.6 is applicable and yields the following theorem: Theorem 2.1.12. If the minimal-norm solution to (1.3.1), then solution of the equation

Lemma 2.1.13. Consider the elements resolution of the identity of B = A* A, on s and piecewise-continuous function. Let

as where

so that

and is is the unique

where is the does not depend and is a as Then

26


Proof. If

then

and

Thus, where

and

One has

as If the rate of decay of of can be estimated, then a quasioptimal choice of minimizing with respect to for a fixed

Thus and the rate of growth can be made by

Remark 2.1.14. We have used the spectral theorem for a selfadjoint operator B, namely is the resolution of the identity of B, the formula where

Remark 2.1.15. Similarly, one can use the theory of spectral operators in place of the spectral theory of selfadjoint operators, in particular Riesz bases formed by the root vectors. 2.1.7

On the notion of ill-posedness for nonlinear equations

If A is a linear operator, then problem (1.3.1) is ill-posed if either or or R(A) is not closed, i.e. is unbounded. If A is nonlinear and Frechet differentiable, then there are several possibilities. If is boundedly invertible at some u, then A(u) is a local homeomorphism at this point, but it may be not a global homeomorphism. If is not boundedly invertible, this does not imply, in general, that A is not a homeomorphism. For example, a homeomorphism A(u) may have a compact derivative, so its linearization yields an ill-posed problem. On the other hand, A(u) may be compact, so (1.3.1) is an ill-posed problem, but may be a finite-rank operator, so that the range of is closed. In spite of the above, we will often call a nonlinear equation problem (1.3.1) ill-posed if is not boundedly invertible, and well-posed if is boundedly invertible, deviating therefore from the usual terminology. 2.1.8

Discrepancy principle for nonlinear ill-posed problems with monotone operators

Assume that A in (1.3.1) is monotone, i.e., D(A) = H, A is continuous, is unbounded or does not exist, so (1.3.1) is an ill-posed problem, Consider the discrepancy principle for finding assuming that A is nonlinear monotone:

where C = const > 1,

is any element such that where of the regularization parameter We need three lemmas.

and

plays the role

27

Lemma 2.1.16. If A is monotone and continuous, and the set nonempty, then it is convex and closed.

is

Lemma 2.1.17. If A is monotone and continuous, then it is w-closed, that is, and imply A(u) = f, where and stand for the weak and strong convergence in H, respectively. Lemma 2.1.17 in a stronger form (hemicontinuity of A replaces continuity, and it is assumed in this case that the monotone operator A is defined on all of H ) follows from the proof of Lemma 2.1.2. Lemma 2.1.18. If

then

and

Proof of Lemma 2.1.16. If and If A is monotone, A(u) = f and A(v) = f, then and vice versa. Thus, for any

then A(u) = f, so

is closed. and the element

Lemma 2.1.18 is Theorem 2.1.4. Theorem 2.1.19. Assume: (i) A is a monotone, continuous operator, defined on all of H, (ii) equation A(u) = f is solvable, is its minimal-norm solution, and (iii) where C > 1 is a constant. Then: (j) the equation

is solvable for

for any fixed

Here

is any element satisfying inequality

where and and (jj) if

solves (2.1.8), and

then

Remark 2.1.20. The equation is uniquely solvable for any and any If is its solution, and where C = const > 1, then equation (2.1.8) with replaced by is solvable for If is its solution, then If A is injective, and if then where solves the equation If A is not injective, then it is not true, in general, that where is the minimal-norm solution to the equation A(u) = f even if one assumes that A is a linear operator.

28


Proof of Theorem 2.1.19. If A is monotone, continuous and is defined on all of H, then the set is convex and closed, so it has a unique minimal-norm element To prove the existence of a solution to (2.1.8), we prove that the function is greater than for sufficiently large and smaller than for sufficiently small If this is proved, then the continuity of with respect to on implies that the equation has a solution. Let us give the proof. As we use the inequality:

and, as

we use another inequality:

As

one gets where c > 0 is a constant depending on Thus, by the continuity of A, one obtains As one gets Thus, Therefore equation has a solution Let us now prove that if then From the estimate

and from the equation (2.1.8), it follows that Thus, one may assume that and from (2.1.8) it follows that as By w-closedness of monotone continuous operators (hemicontinuity in place of continuity would suffice), one gets A(U) = f, and from it follows that Because A is monotone, the minimal norm solution to the equation A(u) = f in H is unique. Consequently, Thus, and By Theorem 2.1.4, it follows that Note that because due to the assumption where C > 1. Theorem 2.1.19 is proved.

Proof of Remark 2.1.20. Let solve the equation let solve the equation and Then Multiply this equation by and use the monotonicity of A to get The triangle inequality yields: Note that and as Thus, Therefore Fix and let Then and where is the minimal-norm solution to the equation A(u) = f. If A is injective, then this equation has only one solution Since one gets the inequality

29

Consequently, equation (2.1.8), with w replacing has a solution We claim that in fact, as Indeed, from (2.1.8), with w replacing one gets and we prove below that where This implies We now claim that the limit does exist, that u solves the equation A(u) = f, and It is sufficient to check that where c = const does not depend on as Indeed, if then a subsequence, denoted again, converges weakly to an element and (2.1.8) implies Since A is monotone, it is w–closed, so By the injectivity of A, any subsequence converges weakly to the same element so Consequently, as claimed. The inequality follows from the assumption To prove the inequality note that where Since C > 1, this implies where Thus, where The last statement of Remark 2.1.20 is illustrated by the following example: Example 2.1.21. Let Aw = (w, p ) p , where (q, p) = 0, One has where is the minimal-norm solution to the equation Au = p. Equation has the unique solution Equation (2.1.8) is This equation yields where and we assume 1 (see the second inequality in the assumption (iii) of Theorem 2.1.19). Let Then, and Au = p. Therefore is not p, i.e., u is not the minimal-norm solution to the equation Au = p. Remark 2.1.20 is proved. Remark 2.1.22. It is easy to prove that if conditions (i) and (ii) of Theorem 2.1.19 hold and and if where then where and is the minimal-norm solution to the equation A(u) = f. In particular, if 0 < a < l, then Indeed, where is the unique solution to the equation It is well known that provided that and, clearly, one multiplies the identity by and uses the monotonicity of A and the inequality The result similar to the one in the above remark can be found in [ARy]. 2.1.9

Regularizers for ill-posed problems must depend on the noise level

In this Section we prove the following simple claim: Claim 2. There is no regularizer independent of the noise level to a linear ill-posed problem. If such a regularizer exists, then the problem is well-posed.


30

Let A be a linear operator in a Banach space X. Assume that A is injective and is unbounded, that equation

is solvable, and g is such that

where is the norm in X and is the noise level. Nothing is assumed about the statistical nature of noise. In particular, we do not assume that the noise has zero mean value or finite variance. Question: Can one find a linear operator R with the property:

for any Answer: no.

where Ran( A) is the range of A, and any

satisfying (2.1.10)?

Proof. If such an R is found, then, taking and using the fact that is arbitrary, one concludes that on the range of A. Secondly, writing where

and w is arbitrary otherwise, one concludes from (2.1.11) and from the fact that that

for any w satisfying (2.1.12). Since R is linear, this implies that R is bounded, which contradicts the equation on Ran(A) and the unboundedness of which is the necessary condition for the ill-posedness of (2.1.9). A similar result one can find in [LY]. 2.2

2.2.1

QUASISOLUTIONS, QUASINVERSION, AND BACKUS-GILBERT METHOD

Quasisolutions for continuous operator

Assume that equation (1.3.1) is solvable, its solution where K is a compactum in a Banach space X, and A is continuous. Consider the problem

31

where is the infimum of the function of and A minimizer for (2.2.1) is called a quasisolution to (1.3.1) with Let be a minimizing sequence for (2.2.1). Since K is a compactum, one may assume that as Thus so is a minimizer for the problem (2.2.1). The above argument shows that if f replaces in (2.2.1), and if equation (1.3.1) is solvable and its solution belongs to K, then any minimizer for (2.2.1) with is a solution to (1.3.1). Let us prove that where is a minimizer for (2.2.1), and u is a solution to (1.3.1), where existence of a solution (1.3.1) is assumed. Indeed, so one may assume that as By continuity of A, one has Thus, The last conclusion follows from the solvability of (1.3.1), which yields f = A(u) and from the inequality We have proved: Theorem 2.2.1. If equation (1.3.1) is solvable, K is a compactum containing all the solutions to (1.3.1), and then (2.2.1) has a minimizer and for every minimizer and some solution u to (1.3.1). Remark. Suppose that X is strictly convex, i.e., if then u = v. For example Hilbert spaces H are strictly convex, the spaces are strictly convex, but and C(D) are not. Suppose that K is a convex compactum, i.e., convex closed compact set. The metric projection of an element onto K is the element such that If X is strictly convex, then is unique, and if K is a convex compactum, then depends continuously on f. If A is injective and closed, not necessarily linear, and K is a compactum, then is continuous on the set AK. Indeed, if and then a subsequence, denoted again converges to u because K is compact, and if A is closed, then A(u) = f, which proves the claim. Therefore, if X is strictly convex and K is a convex compactum, and if A is an injective bounded linear operator, then the quasisolution depends continuously on f in the norm of X. 2.2.2

Quasisolution for unbounded operators

Assume that A is closed, possibly nonlinear, injective, unbounded operator, is possibly, unbounded, equation (1.3.1) is solvable, assumptions (1)–(4) of Section 2.1.5 hold, and K is a compactum containing all the solutions to (1.3.1). Theorem 2.2.2. Under the above assumptions, if K, then where u is a solution to (1.3.1). Proof. One can assume because

as

where

because K is a compactum. Note that Thus,

32


By closedness of A, one gets A(w) = f, so w is a solution to (1.3.1), and one can denote w by u. 2.2.3

Quasiinversion

Let A be a linear bounded operator in (1.3.1), and (1.3.1) is solvable, consider the equation is a parameter, Q is an operator chosen so that and where u is a solution to (1.3.1). If A is unbounded, a similar idea can be applied to equation (1.3.1): consider the equation where Q is chosen so that The problem is: how does one choose Q with these properties? If A is a linear bounded operator, then Q = I can be used by Theorem 2.1.1. If A is unbounded, some assumptions on its spectrum are needed. See [LL] for details. 2.2.4

A Backus-Gilbert-type method: Recovery of signals from discrete and noisy data

We discuss in this section the following Problem 2. Let D and the bar denote variance and mean value respectively,

Problem 1. Given

estimate

The idea is as follows: the estimate is sought in the form

The problem is to find such that: (1) if then (2) if then where is a certain number. We find optimal, in a certain sense. Namely, if then are found from the requirements:

Note that

If then are found from {(A) and if this problem is solvable and, if not, one increases so that this problem becomes solvable.

33

2.2.4.1. A typical problem we are concerned with is the problem of estimating the spectrum of a compactly supported function from the knowledge of the spectrum at a finite number of frequencies. More precisely, let

Suppose that the numbers:

are given. At the moment we assume that are given exactly, i.e., there is no noise. The case when the data are noisy will be considered below. Problem 2. Given

find an estimate

To be specific let us assume that

where the functions

of

such that

and that the estimate is of the form

will be chosen soon. From (2.2.5) and (2.2.2) it follows that

and

where and the star denotes complex conjugate. Property (2.2.4), convergence, holds if is a delta-sequence, i.e.,


34

Let

and let

be a sequence of functions such that

where

One can interpret (2.2.11) as the requirement that the estimate is exact for f (x) = const. Given (2.2.11), the smaller Q(x), the better is the quality of the delta-sequence Thus we are led to the optimization problem: Find such a sequence that

Note that the general problem of the type

where is a linearly independent set of functions, and is a bounded domain in can be treated in exactly the same way as before. If the problem (2.2.13) has the unique solution then (2.2.6) is the optimal estimate which, as we prove, has the convergence property (2.2.4). 2.2.4.2. If the data are noisy, that is are given in place of random vector with the covariance matrix

where

35

where the bar denotes the mean value, then the variance

where ( , ) is the inner product in

Let us fix

can be computed

and require that

The optimization problem for finding the vector formulated as follows:

can be

Here Clearly, problem (2.2.18) is not solvable for all We will discuss this important point below. If (2.2.18) is solvable, the solution is unique, and the optimal estimate is given by

This estimate has variance Our arguments so far are close to the usual ones. The new point is our convergence requirement (2.2.4). We prove the convergence property of our estimate and give the rate of convergence. The case when the data are the finite number of moments is treated, and the optimization requirements are introduced. The problem we discuss is of interest in geophysics and many other applications. 2.2.4.3. Here a solution of the estimation problems is given. We start with problem (2.2.13). Let us write Q(x) as a quadratic form

where


36

and

is a self-adjoint positive definite matrix. Let us write (2.2.13) as

Using the Lagrange multiplier condition for the minimizer

one obtains the standard necessary and sufficient

Therefore

is uniquely defined by (2.2.25), since is positive definite, and the denominator in (2.2.25) does not vanish. The minimum of Q(x) is

We assume that

If (2.2.27) holds, then (2.2.4) holds. Indeed, (2.2.13) and (2.2.27) imply that

In this argument we assume that the function f (x) satisfies the inequality

37

This inequality is satisfied if, for example, the derivative of f (x) exists except at a finite number of points and is uniformly bounded. Let us illustrate the assumption (2.2.27). Let Then

where where is the cofactor corresponding to the element

of the matrix One can show that (2.2.24) holds at any point at which f (x) is differentiable and Indeed, if we do not take the but use then the error of the estimate will be not less that On the other hand, for this choice of h, the kernel (2.2.8) is the Dirichlet kernel. From the theory of the Fourier series one knows that (2.2.9) holds in and at any point at which f (x) is differentiable. In practice it is advisable to choose the system in such a way that tends rapidly to zero. Note that depends only on the system and therefore we can control this quantity to some extent by choosing the system Let us note that one can estimate f (x) at a given point optimally using the same procedure. In this case the convergence condition (2.2.4) will hold for If is fixed, we can choose the system so that

In this case

and we can choose

so that, in addition to (2.2.29), the condition

is satisfied. For example, take

then (2.2.29) reduces to

38


and one can choose Then

which behave nearly like can be made very large, and

in a small neighborhood of as in (2.2.31).

2.2.4.4. In this section we solve problem (2.2.18). As we have already mentioned, this problem may not be solvable for every because there may be no h which satisfies both restrictions of (2.2.18). Since the set is convex and Q(h) is a strictly convex function of h, it is clear that the solution to (2.2.18) is unique when it exists. For the solution to exist it is necessary and sufficient that the set M of h, which satisfy the restrictions (2.2.18), be not empty. Let us give an analytic solution to problem (2.2.18). If for the optimal h the inequality holds, then the solution to (2.2.18) is the same as the solution to (2.2.13) and is given by formula (2.2.25). Therefore, first one checks if the function (2.2.25) satisfies the inequality

If it does, then it is the solution to (2.2.18). If it does not, then the solution satisfies the equality

By the Lagrange method the necessary condition for the optimal h, for the solution to problem (2.2.18), is

where

and

are the Lagrange multipliers. It follows from (2.2.36) that

Taking the complex conjugate in the first equation (2.2.36) we see that

From (2.2.37) and (2.2.38) one gets

Substituting (2.2.40) into

39

yields an equation for

The roots of Eq. (2.2.42), give by formula (2.2.38), and h by formula (2.2.37). Finally choose the for which (Bh, h) = min. This solves problem (2.2.18). 2.2.4.5. One can simplify the solution to problem (2.2.18) in the following way. If the problem (2.2.18) takes the form

Let us choose the coordinate system so that

and normalize

so that

In this case (2.2.43) can be written as

where

Thus, problem (2.2.43) in with two constraints is reduced to problem (2.2.46) in with one constraint. Problem (2.2.46) can be solved by the Lagrange multipliers method. One has

where

is the Lagrange multiplier. Thus,

40


Substitute (2.2.49) into the constraint equation (2.2.46) to obtain an equation for If this equation is solved then (2.2.49) gives the corresponding H. If there are several solutions then the is the one that minimizes the quadratic form (2.2.46). 2.3

ITERATIVE METHODS

There is a vast literature on iterative methods [VV], [BG], [R65]. First, we prove the following result. Theorem 2.3.1. Every solvable equation (1.3.1) with bounded linear operator A can be solved by a convergent iterative method. Proof. It was proved in Section 2.1.6 that if A is a bounded linear operator and equation (1.3.1) is solvable, then it is equivalent to the equation (2.1.6). By y we denote the minimal norm solution to (1.3.1), i.e., the solution orthogonal to N(A), the null– space of A. Note that N(B) = N(A). Without loss of generality assume (if then one can divide by equation (2.1.6)). Consider the iterations

where

is arbitrary. Denote

Thus,

and write (2.3.1) as Since

one gets:

where is the resolution of the identity of B. Thus, where P is the orthoprojector onto N(B) = N(A). If one takes and then by induction, and In particular, if then (by induction, since and where A+ is the pseudoinverse of A, defined in Section 2.1.1. Exercise 2.3.2. Prove that if

then

in (2.3.1).

Assume now that is given in place of f, Let us show that if one stops iterations (2.3.1), with in place of f and at then as if is properly chosen, and is defined by (2.3.1). Let Then and Thus, because if and we had assumed which, together with implies Therefore, if is chosen so that then Indeed, We have proved in Theorem 2.3.1

41

that rize the result. Theorem 2.3.3.

as

and

If

Let us summa-

then

where is the minimal norm solution to (1.3.1), is obtained by iterations (2.3.1), with in place of f and

and

A general approach to construction of convergent iterative methods for nonlinear problems is developed in Section 2.4. 2.4

DYNAMICAL SYSTEM METHOD (DSM)

2.4.1

The idea of the DSM

Consider the equation:

We assume in Section 2.4 that

that is,

where

is the minimal-norm solution to (2.4.1), H is a real Hilbert space. Many of our results hold in reflexive Banach spaces X and F : X X*, but we do not go into detail. The element in (2.4.2) will be specified later. In Section 2.4 we will call (2.4.1) a wellposed problem if

and ill-posed if is not boundedly invertible. We assume existence of a solution to (2.4.1) unless otherwise stated, but uniqueness of the solution is not assumed. If (2.4.3) holds, then one can construct Newton-type methods for solving (2.4.1). But if (2.4.3) fails, then it seems that there is no general approach to solving (2.4.1). One of our goals is to develop such an approach, which we call the Dynamical Systems Method (DSM). The DSM consists of finding a nonlinear locally Lipschitz operator such that the Cauchy problem:

has the following three properties:

42


that is, (2.4.4) is globally uniquely solvable, its unique solution has a limit at infinity and this limit solves (2.4.1). On first motivation is to develop a general approach to solving equation (2.4.1), especially nonlinear and ill-posed. Our second motivation is to develop a general approach to constructing convergent iterative methods for solving (2.4.1). We justify the DSM in the following cases (1) For well-posed problems, (2) For ill-posed linear problems with bounded linear operator A and also for (3) For ill-posed problem with monotone nonlinear A, for

and and also for

(4) For ill-posed problem with nonlinear A assuming some additional condition. We give a general construction of convergent iterative schemes for well-posed nonlinear problems, and also for ill-posed nonlinear problems with monotone and nonmonotone operators. These results are presented in subsections below. 2.4.2

DSM for well-posed problems

Consider (2.4.1), let (2.4.2) hold, and assume

where

is an integrable function,

where

is such that

is a constant. Assume

Remark 2.4.1. Sometimes the assumption (2.4.7) can be used in the following modified form:

where is a constant. The statement and proof of Theorem 2.4.2 can be easily adjusted to this assumption.

Our first basic result is the folowing:

43

Theorem 2.4.2. (i) If (2.4.6)–(2.4.8) hold and

then (2.4.4) has a global solution, (2.4.5) holds, (2.4.1) has a solution and

(ii) If (2.4.6)–(2.4.8) hold, 0 < a < 2, and

where T > 0 is defined by the equation

then (2.4.4) has a global solution, (2.4.5) holds, (2.4.1) has a solution and u(t) = y for (iii) If (2.4.6)–(2.4.8) hold, a > 2, and

where

then (2.4.4) has a global solution, (2.4.5) holds, (2.4.1) has a solution and

as

Let us sketch the proof.

44


Proof of Theorem 2.4.2. The assumptions about imply local existence and uniqueness of the solution u(t) to (2.4.4). To prove global existence of u, it is sufficient to prove a uniform with respect to t bound on Indeed, if the maximal interval of the existence of u(t) is finite, say [0, T), and is locally Lipschitz with respect to u, then as Assume a = 2. Let Since H is real, one uses (2.4.4) and (2.4.6) to get so and integrating this inequality one gets the second inequality (2.4.10), because Using (2.4.7), (2.4.4) and the second inequality (2.4.10), one gets:

Because it follows from (2.4.10’) that the limit exists, and by (2.4.9). From the second inequality (2.4.10) and the continuity of F one gets so solves (2.4.1). Taking and setting s = t in (2.4.10’) yields the first inequality (2.4.10). The inclusion for all follows from (2.4.9) and (2.4.10’). The first part of Theorem 2.4.2 is proved. The proof of the other parts is similar. There are many applications of this theorem. We mention just a few, and assume that and Example 2.4.3. [Continuous Newton-type method]: Assume that (2.4.3) holds, then (2.4.9) takes the form (*) and (*) implies that (2.4.4) has a global solution, (2.4.5) and (2.4.10) hold, and (2.4.1) has a solution in This result belongs to Gavurin ([Gav]). Example 2.4.4. [Continuous simple iterations method:] Let for all Then and the conclusions of Example 1 hold. Example 2.4.5. [Continuous gradient method:] Let hold, (2.4.9) is (**) conclusions of Example 2.4.3.

and assume (2.4.9) is:

(2.4.2) and (2.4.3) and (**) implies the

Example 2.4.6. [Continuous Gauss-Newton method:] Let (2.4.2) and (2.4.3) hold, (2.4.9) is (***) (***) implies the conclusions of Example 2.4.3. Example 2.4.7. [Continuous modified Newton method:] Let sume and let (2.4.2) hold. Then

and

AsChoose

45

and Thus, if

Then (2.4.9) is

that is, then the conclusions of Example 2.4.3 hold.

Example 2.4.8. [Descent methods.] Let where f = f(u(t)) is a differetiable functional and h is an element of H. From (2.4.4) one gets Thus, where Assume Then Therefore does exist, and c = const > 0. If and then and (2.4.4) is a descent method. For this one has and where is defined in (2.4.3). Condition (2.4.9) is: If this inequality holds, then the conclusions of Example 2.4.3 hold. In Example 2.4.8 we have obtained some results from [Alb]. Our approach is more general than that in [Alb], since the choices of f and h do not allow one, for example, to obtain used in Example 2.4.7. Remark 2.4.9. A method for proving the existence of a solution to equation (2.4.1) can be stated as follows. Consider (2.4.4) with and assume that (2.4.4) is locally solvable and where u(t) solves (2.4.4). Let Then so and Assume that Then does exist and as Therefore F(u(t))= 0, so solves (2.4.1). The proof of Theorem 2.4.2 is given by this method and Theorem 2.4.20 below is an example of many applications of this method. Conditions (2.4.7) and (2.4.9) are essential: if and H is the real line with the usual product of real numbers as the inner product and then condition (2.4.7) is not satisfied and equation (2.4.1), i.e. does not have a solution in H. Exercise (cf. [R222]). Use the above Remark to prove the following result: Assume F : H H, (2.4.2)–(2.4.3) hold, and lim Then the map F is surjective. This result is related to Hadamard's theorem about homeomorphisms and its generalization by Meyer (see [OR], p. 139). 2.4.3

Linear ill-posed problems

We assume that (2.4.3) fails. Consider

Let us denote by (A) the folowing assumption:

46


(A) : A is a linear, bounded operator in H, defined on all of H, the range R(A) is not closed,

so (2.4.15) is an ill-posed problem, there is a the null-space of A. Let solves

and, if function to assume that

such that

where N is

is the adjoint of A. Every solution to (2.4.15)

then every solution to (2.4.16) solves (2.4.15). Choose a continuous monotonically decaying to zero on Sometimes it is convenient

For example, the functions where and are positive constants, satisfy (2.4.17). There are many such functions. One can prove the following: Claim 3. If is a continuous monotonically decaying function on 0, and (2.4.17) holds, then

In this Section we do not use assumption (2.4.17): in the proof of Theorem 2.4.9 one uses only the monotonicity of a continuous function and (2.4.17’). One can drop assumption (2.4.17’), but then convergence is proved in Theorem 2.4.9 to some element of N, not necessarily to the normal solution that is, to the solution orthogonal to N, or, which is the same, to the minimal-norm solution to (2.4.15). However, (2.4.17) is used (in a slightly weaker form) in the next section. Consider problems (2.4.4) with

where

Without loss of generality one may assume that which we do in what follows. Our main result is Theorem 2.4.9, stated below. It yields the following: Conclusion: Given noisy data every linear ill-posed problem (2.4.15) under the assumptions (A) can be stably solved by the DSM. The result presented in Theorem 2.4.9 was essentially obtained in [R200], but the proof given here is different and much shorter. Theorem 2.4.9. Problem (2.4.4) with from (2.4.18) has a unique global solution u(t), (2.4.5) holds, and Problem (2.4.4) with from (2.4.18) has a unique global

47

solution

This

There exists

such that

can be chosen, for example, as a root of the equation

or of the equation (2.4.20’), see below.

Proof of Theorem 2.4.9. Linear equations (2.4.4) with bounded operators have unique global solutions. If then the solution u to (2.4.4) is

where as is the resolution of the identity corresponding to the selfadjoint operator B, and is a nonexpansive operator, because Actually, (2.4.21) can be used also when B is unbounded, Using L‘Hospital’s rule one checks that

provided only that and From (2.4.21), (2.4.22), and the Lebesgue dominated convergence theorem, one gets where P is the orthogonal projection operator onto the null-space of B. Under our assumptions (A), so If then In general, the rate of convergence of v to zero can be arbitrarily slow for a suitably chosen f. Under an additional a priori assumption on f (for example, the source-type assumptions), this rate can be estimated. Let us describe a method for deriving a stopping rule. One has:

Since

any choice of

gives a stopping rule: for such

one has

such that

48


To prove that (2.4.20) gives such a rule, it is sufficient to check that

Let us prove (2.4.25). Denote

Then

Integrating (2.4.26), and using the property one gets (2.4.25). Alternatively, multiply (2.4.26) by w, let use and get Thus, exp exp A more precise estimate, used at the end of the proof of Theorem 2.4.10 below, yields:

and the corresponding stopping time

can be taken as the root of the equation:

Theorem 2.4.9 is proved. If the rate of decay of v is known, then a more efficient stopping rule can be derived: is the minimizer of the problem:

For example, if then is the root of the equation that one gets from (2.4.27) with One can also use a stopping rule based on an a posteriori choice of the stopping time, for example, the choice by a discrepancy principle. A method, much more efficient numerically than Theorem 2.4.9, is given below in Theorem 2.4.12 and in Theorem 2.4.10 (see (2.4.29)). For linear equation (2.4.16) with exact data this method uses (2.4.4) with

and for noisy data it uses (2.4.4) with The linear operator is monotone, so Theorem 2.4.12 is applicable. For exact data, (2.4.4) with defined in (2.4.28), yields:

49

and (2.4.5) holds if is monotone, continuous, decreasing to 0 as Let us formulate the result: Theorem 2.4.10. Assume (A), and let B := A* A, q := A* f. Assume to be a continuous, monotonically decaying to zero function on Then, for any problem (2.4.29) has a unique global solution, and is the minimalnorm solution to (2.4.15). If is given in place of f, then (2.4.19) holds, with solving (2.4.29) with q replaced by and is chosen, for example, as the root of (3.4.20’) (or by a discrepancy principle). Proof of Theorem 2.4.10. One has is

where

and the solution to (2.4.29)

where

and

is the resolution of the identity of the selfadjoint operator B. One has

From (2.4.30)–(2.4.32) it follows that where is the minimal-norm solution to (2.4.15), N := N(B) = N(A) is the null-space of B and of A, and is the orthoprojector onto N in H. This proves the first part of Theorem 2.4.10. To prove the second part, denote where we dropped the dependence on in w and g for brevity. Then w (0) = 0. Thus so where the known estimate was used: proved. 2.4.4

Theorem 2.4.10 is

Nonlinear ill-posed problems with monotone operators

There is a large body of literature on equations (2.4.1) and (2.4.4) with monotone operators. In the result we present, the problem is nonlinear and ill-posed, the new technical tool, Theorem 2.4.11, is used, and the stopping rules are discussed. Consider (2.4.33) with monotone F under standard assumptions (2.4.2), and

50


where A* is its adjoint, is the same as in Theorem 2.4.10, and in Theorem 2.4.12 is further specified, is an element we can choose to improve the numerical performance of the method. If noisy data are given, then, as in Section 3.3, we take

where B is a monotone nonlinear operator, and solves (2.4.4) with in place of To prove that (2.4.33) with the above has a global solution and (2.4.5) holds, we use the following: Theorem 2.4.11. Let exists a positive function

for some real number

If there

such that

where is the initial condition in (2.4.35), then a nonnegative solution g to the following differential inequality:

exists for all

and satisfies the estimate:

for all

where

There are several novel features in this result. First, differential equation, that one gets from (2.4.35) by replacing the inequality sign by the equality sign, is a Riccati equation, whose solution may blow up in a finite time, in general. Conditions (2.4.34) guarantee the global existence of the solution to this Riccati equation with the initial condition (2.4.35). Secondly, this Riccati differential equation cannot be integrated analytically by separation of variables. Thirdly, the coefficient may grow to infinity as so that the quadratic term does not necessarily have a small coefficient, or the coefficient smaller than Without loss of generality one may assume in Theorem 2.4.11. This Theorem is proved in Section 2.4.9. The main result in this Section is new. It claims a global convergence in the sense that no assumptions on the choice of the initial approximation are made. Usually

51

one assumes that is sufficiently close to the solution of (2.4.1) in order to prove convergence. We take in Theorem 2.4.12 because in this theorem does not play any role. The proof is valid for any choice of but then the definition of r in Theorem 2.4.12 is changed. Theorem 2.4.12. If (2.4.2) holds, R = 3r, where and is the (unique) minimal norm solution to (2.4.1), then, for any choice of problem (2.4.4) with defined in (2.4.33), and with some positive constants and specified in the proof of Theorem 2.4.12, has a global solution, this solution stays in the ball and (2.4.5) holds. If solves (2.4.4) with in place of then there is a such that Proof of Theorem 2.4.12. Let us sketch the steps of the proof. Let V solve the equation

Under our assumptions on F, it is easy to prove that: (i) (2.4.38) has a unique solution for every t > 0, and (ii) Indeed, so This implies (ii). If F is Fréchet differentiable, then V is differentiable, and It is also known (see Section 2.1.4) that if where y is the minimal-norm solution to (2.4.1), then

We will show that the global solution u to (2.4.4), with the from (2.4.33), does exist, and This is done by deriving a differential inequality for w := u – V, and by applying Theorem 2.4.11 to Since one obtains (2.4.5). We also check that where for any choice of and a suitable choice of Let us derive the differential inequality for w. One has

and F(u) –F(V) = Aw + K, where and is the constant from (2.4.2). Multiply (2.4.39) by w, use the monotonicity of F, that is, the property and the estimate and get:

52


where

Inequality (2.4.40) is of the type (2.4.35): Choose

Clearly as Let us check three conditions (2.4.34). One has Take where are constants, and choose these constants so that for example, Then the first condition (2.4.34) is satisfied. The second condition (2.4.34) holds if

One has

Choose

Then

so (2.4.42) holds. Thus, the second condition (2.4.34) holds. The last condition (2.4.34) holds because

By Theorem 2.4.11 one concludes that

when

and

This estimate implies the global existence of the solution to (2.4.4), because if u(t) had a finite maximal interval of existence, [0, T), then u(t) could not stay bounded when which contradicts the boundedness of and from (2.4.45) it follows that We have proved the first part of Theorem 2.4.12, namely properties (2.4.5). To derive a stopping rule we argue as in Section 2.4. One has:

We have already proved that The rate of decay of v can be arbitrarily slow, in general. Additional assumptions, for example, the source-type ones, can be used to estimate the rate of decay of v (t). One derives differential inequality (2.4.35) for and estimates using (2.4.36). The analog of (2.4.40) for contains additional term on the right-hand side. If

53

then conditions (2.4.34) hold, and Let be the root of the Then and because and but the convergence can be slow. See [ARS3], [KNR] for the rate of convergence under source assumptions. If the rate of decay of v(t) is known, then one chooses as the minimizer of the problem, similar to (2.4.27), equation

where the minimum is taken over t > 0 for a fixed small timal stopping rule. Theorem 2.4.12 is proved.

This yields a quasiop-

Let us give another result: Theorem 2.4.13. Assume that F is monotone, rem 2.4.10, and (2.4.17), and (2.4.2) hold. Then (2.4.5) holds.

as in Theo-

Proof of Theorem 2.4.13. As in the proof of Theorem 2.4.12, it is sufficient to prove that where g, w, and V are the same as in Theorem 2.4.12, and u solves (2.4.4) with the defined in Theorem 2.4.13. Similarly to the derivation of (2.4.39), one gets:

Multiply (2.4.48) by w, use the monotonicity of F and the estimate which was used also in the proof of Theorem 2.4.12, and get:

This implies

From our assumptions relation (2.4.17’) follows, and (2.4.50) together with (2.4.17) imply Theorem 2.4.13 is proved. Remark 2.4.14. One can drop the smoothness of F assumption (2.4.2) in Theorem 2.4.13 and assume only that F is a monotone hemicontinuous operator defined on all of H. Claim 4. If (2.4.4) with

= const > 0, then and is any number such that

where u(t) solves

54


Proof of the claim. One has where solves (2.4.38) with = const > 0. Under our assumptions on F, equation (2.4.38) has a unique solution, and So, to prove the claim, it is sufficient to prove that provided that Let and Because one has the equation: Multiplying this equation by w, and using the monotonicity of F, one gets Therefore provided that The claim is proved. Remark 2.4.15. One can prove claims (i) and (ii), formulated below formula (2.4.38), using DSM version presented in Theorem 2.4.20 below. Claim 5. Assume that F is monotone, (2.4.2) holds, and (ii), formulated below formula (2.4.38), hold.

Then claims (i) and

Proof. First, note that (ii) follows easily from (i), because the assumptions F is monotone, and imply, after multiplying the inequality from which claim (ii) follows. Claim (i) follows from Theorem 2.4.20, proved below. Claim 6. Assume that the operator F is monotone, hemicontinuous, defined on all of H, equation F(u) = 0 has a solution, possibly non-unique, is the minimal-norm element of and where and b are constants. Then (2.4.5) holds for the solution to (2.4.4). Proof. The steps of the proof are: 1) we prove that

for the solution to (2.4.4) with where Inequality implies that from any sequence one can select a subsequence, denoted again such that where is some element. We prove that F(v) = 0. 2) we prove that the solution u (t) to the equation

satisfies the relation: arbitrary fixed number. 3) we prove that 4) passing to the limit as and

as

where h > 0 is an

in equation (*) yields F (v) = 0. We prove that

Let us give the details of the proof. 1) Let

note

where u solves (*).Then Multiply this equation by w, use the monotonicity of F, deand get Because this implies

55

This in-

SO

equality implies 2) Denote

Thus where u solves (*), and

Then Multiply this equation

by and use the monotonicity of F to get: Because one gets and

This implies

Under our assumptions about one can check that where 0 < a := 1 – b < 1. Also

as Thus

Divide (2.4.52) by h and let Then one gets 3) Denote so one has in equation (*), yields F(v) = 0. The limit 4) Passing to the limit exists because and so that Lemma 2.1.2 imSince one gets plies F(v) = 0. Let us prove that then lim If this implies strong convergence and together with the weak convergence

Let us prove that lim and

One has Since F is monotone, Thus,

it follows that because Let us prove that Replacing v by in the above argument yields so Since is the unique minimal-norm solution to (2.4.1) and v solves (2.4.1), it follows that Since the limit is the same for every subsequence for which the weak limit of exists, one concludes that the strong limit Indeed, assuming that for some sequence the limit of does not exist, one selects a subsequence, denoted again for which the weak limit of does exist, and proves as before that this limit is thus getting a contradiction. Claim 6 is proved. For convenience of the reader let us prove the global existence and uniqueness of the solution to (2.4.4) with where F is a monotone, hemicontinuous operator in H (cf [Dei]). Uniqueness of the solution is trivial: if there are two solutions, u and v, then their difference w := u – v solves the problem

56


Multiply this by w and use the monotonicity of F to get g (0) = 0, where Thus, g = 0, so w = 0, and uniqueness is proved. Let us prove the global existence of the solution to (2.4.4) with Consider the equation:

We wish to prove that

where Our assumptions (monotonicity and hemicontinuity of F) imply demicontinuity of F. Fix an arbitrary T > 0, and let be the ball centered at with radius r > 0. Let Then If then and Define

From (**) one gets:

One has:

Using the monotonicity of F, the estimate one gets:

Therefore

and the estimate

57

This implies

Therefore there exists the strong limit w(t):

This function w satisfies the integral equation:

and solves the Cauchy problem

If F is continuous, then this Cauchy problem and the preceding integral equation are equivalent. If F is demicontinuous, then they are also equivalent, but the derivative in the Cauchy problem should be understood in the weak sense. We have proved the existence of the unique local solution to(***). To prove that the solution to (***) exists for any let us assume that the solution exists on [0, T), but not on a larger interval [0, T + d), and show that this leads to a contradiction. It is sufficient to prove that the finite limit: does exist, because then one can solve locally, on the interval [T, T + d), equation (***) with the initial data and construct the solution to (***) on the interval [0, T + d), thus getting a contradiction. To prove that W exists, consider

One has

Using the monotonicity of F, one gets Thus,

The right-hand side of the above inequality tends to zero as Cauchy test imply the existence of W. The proof is complete. 2.4.5

This, and the

Nonlinear ill-posed problems with non-monotone operators

Assume that F(u) := B(u) — f, B is a non-monotone operator, where I is the identity operator,

is as

58


in Theorem 2.4.10 and

and is defined similarly, with replacing f and The main result of this Section is:

replacing u.

Theorem 2.4.16. If (2.4.2) holds, u, (that is, is sufficiently small), and R 0, intersects a punctured ball where a = const > 0. This implies the existence of an element such that where and The condition is satisfied because can be chosen arbitrarily small. 2.4.6

Nonlinear ill-posed problems: avoiding inverting of operators in the Newton-type continuous schemes

In the Newton-type methods for solving well-posed nonlinear problems, for example, in the continuous Newton method with the difficult and expensive part of the solution is inverting the operator In this section we give a method to avoid inverting of this operator. This is especially important in the ill-posed problems, where one has to invert some regularized versions of and to face more difficulties than in the well-posed problems. Consider problem (2.4.1) and assume (2.4.2), and (2.4.3). Thus, we discuss our method in the simplest well-posed case. Replace (2.4.4) by the following Cauchy problem (dynamical system):

where

T := A* A, and Q = Q(t) is a bounded operator in H.

60


First, let us state our new technical tool: an operator version of the Gronwall inequality. Theorem 2.4.17. Let

where T(t), G(t), and Q(t) are linear bounded operators on a real Hilbert space H. If there exists such that

then

A simple proof of Theorem 2.4.17 is left to the reader. It can be found in [RSm4]. Let us turn now to the proof of Theorem 2.4.18, formulated at the end of this Section. This theorem is the main result of Section 2.4.6. Applying (2.4.62) to (2.4.59), and using (2.4.2) and (2.4.3), which implies

one gets:

as long as Let K, where (2.4.58) as

Let

and

Since one has is the constant from (2.4.2). Rewrite

Multiply (2.4.65) by w and get

We prove below that

61

From (2.4.66) and (2.4.67) one gets the following differential inequality:

which implies:

provided that

Inequality (2.4.70) holds if is sufficiently close to y. From (2.4.69) it follows that Thus, (2.1.6) holds. The trajectory provided that

This inequality holds if is sufficiently close to y, that is, r is sufficiently small. To complete the argument, let us prove (2.4.67). One has:

and one has

Using (2.4.69) and Theorem 2.4.17, one gets

Thus,

If is sufficiently close to y and is sufficiently close to made arbitrarily small. We have proved:

then

can be

Theorem 2.4.18. If (2.4.2), and (2.4.3) hold, and are sufficiently close to and y, respectively, then problem (2.4.58)–(2.4.59) has a unique global solution, (2.4.5) holds, and u(t) converges to y, which solves (2.4.1), exponentially fast.

In [RSm4] a generalization of Theorem 2.4.18 is given for ill-posed problems. Exercise. 1. Let where is a constant, (2.4.2) holds in where y is a solution to (2.4.1) and R > 0 is a constant. Assume

62


that there exists a and

Then, if R,

such that

where w is some element,

Assume that:

and w are sufficiently small, then the above Cauchy problem has a

unique global solution

and

Precise meaning of the above

smallness condition is explained in [RSm4]. 2. If F is monotone, then the assumption can be dropped, and, under the assumption is sufficiently small, one derives the existence and uniqueness of the global solution to the Cauchy problem of n. 1 of this Exercise. The method of the proof is the same as in [RSm4]. 2.4.7

Iterative schemes

In this section we present a method for constructing convergent iterative schemes for a wide class of well-posed equations (2.4.1). Some methods for constructing convergent iterative schemes for a wide class of Ill-posed problems are given in [AR1]. There is an enormous literature on iterative methods ([BG], [VV]). Consider a discretization scheme for solving (2.4.4) with so that we assume no explicit time dependence in

One of the results from [AR1], concerning the well-posed equations (2.4.1) is Theorem 2.4.19, formulated below. Its proof is shorter than in [AR1]. Theorem 2.4.19. Assume (2.4.2), (2.4.3), (2.4.6)–(2.4.9) with a = 2, sufficiently small, and which

is sufficiently close to

for Then, if h > 0 is then (2.4.75) produces a sequence for

where

Proof of Theorem 2.4.19. The proof is by induction. For n = 0 estimates (2.4.76) are clear. Assuming these estimates for let us prove them for j = n + 1. Let and let solve problem (2.4.4) on the interval with By (2.4.10) (with and (2.4.76) one gets:

63

One has:

and

where we have used the formula

and the estimate:

From (2.4.77)–(2.4.80) it follows that:

provided that

Inequality (2.4.82) holds if h is sufficiently small and So, the first inequality (2.4.76), with n + 1 in place of n, is proved if h is sufficiently small and Now

Using (2.4.2) and (2.4.79), one gets:

From (2.4.83) and (2.4.84) it follows that:

provided that

Inequality (2.4.86) holds if h is sufficiently small and So, the second inequality (2.4.76) with n + 1 in place of n is proved if h is sufficiently small and Theorem 2.4.19 is proved.

64


In the well-posed case, if

the discrete Newton’s method

converges superexponentially if is sufficiently close to y. Indeed, if then where Thus, satisfies the inequality: where Therefore and if then the method converges superexponentially. If one uses the iterative method with then, in the well-posed case, assuming that this method converges, it converges exponentially, that is, slower than in the case h = 1. The continuous analog of the above method

where a = const > 0, converges at the rate so

Indeed, if Thus

then

In the continuous case one does not have superexponential convergence no matter what a > 0 is (see [R210]). 2.4.8

A spectral assumption

In this section we introduce the spectral assumption which allows one to treat some nonlinear non-monotone operators. Assumption S: The set where and are arbitrarily small, fixed numbers, consists of the regular points of the operator for all Assumption S implies the estimate:

because and our case,

where s (A) is the spectrum of a linear operator A, is the distance from a point of a complex plane to the spectrum. In and if

Theorem 2.4.20. If (2.4.2) and (2.4.87) hold, and (2.4.38), with is solvable, problem (2.4.4), with and has a unique global solution, and solution to the equation is isolated.

then problem defined in (2.4.33) Every

65

Proof of Theorem 2.4.20. Let solves locally (2.4.4), where is defined in (2.4.33) and

where u = u(t) Then:

so

Thus,

Therefore equation

has a solution in where Every solution to equation (2.4.91) is isolated. Indeed, if and then so where Thus, using (2.4.87), one gets unless Consequently, if is sufficiently small, then Theorem 2.4.20 is proved. 2.4.9

Nonlinear integral inequality

The main result of this section is Theorem 2.4.22, which is used extensively in Sections 2.4.4–2.4.8. The following lemma is a version of some known results concerning integral inequalities. Lemma 2.4.21. Let f (t, w), g (t, u) be continuous on region and if Assume that g (t, u) is such that the Cauchy problem

has a unique solution. If

then

for all t for which u (t) and w (t) are defined.

Proof of Lemma Lemma 2.4.21. Step 1. Suppose first f ( t , w ) < g ( t , u ) , if Since there exists such that u (t) > w (t) on

and Assume

66


that for some

one has

Then for some

one has

One gets

This contradiction proves that there is no point Step 2. Now consider the case

where

such that if

Define

tends monotonically to zero. Then

By Step 1

Fix an arbitrary compact set

Since g (t, u) is continuous, the sequence is uniformly bounded and equicontinuous on Therefore there exists a subsequence which converges uniformly to a continuous function u (t). By continuity of g (t, u) we can pass to the limit in (2.4.94) and get

Since is arbitrary (2.4.95) is equivalent to the initial Cauchy problem that has a unique solution. The inequality implies If the solution to the Cauchy problem (2.4.92) is not unique, the inequality holds for the maximal solution to (2.4.92). The following theorem was stated earlier as Theorem 2.4.11. For convenience of the reader, we repeat its formulation. Theorem 2.4.22. Let a positive function

for some real number such that

If there exists

67

then a non-negative solution to the following inequalities:

satisfies the estimate:

for all

where

Remark 2.4.23. Without loss of generality one can assume In [Alb1] a differential inequality was studied under some assumptions which include, among others, the positivity of for v > 0. In Theorem 2.4.22 the term (which is analogous to some extent to the term can change sign. Our Theorem 2.4.22 is not covered by the result in [Alb1]. In particular, in Theorem 2.4.22 an analog of for the case is the function This function goes to as v goes to so it does not satisfy the positivity condition imposed in [Alb1]. Lemma 1 in [Alb1] is wrong. Its corrected version is given in [ARS3], where a counterexample to Lemma 1 from [Alb1] is constructed. In [ARS3] the following result is proved: Lemma. Let

and

Assume: 1) for t > 0, 2) as 3) f (0) = 0, f (0) > 0 for u > 0, 4) there exists c > 0 such that for Then as Unlike in the case of Bihari integral inequality ([BB]) one cannot separate variables in the right hand side of the first inequality (2.4.97) and estimate v(t) by a solution of the Cauchy problem for a differential equation with separating variables. The proof below is based on a special choice of the solution to the Riccati equation majorizing a solution of inequality (2.4.97). Proof of Theorem 2.4.22. Denote:

68


then (2.4.97) implies:

where

Consider Riccati’s equation:

One can check by a direct calculation that the the solution to problem (2.4.102) is given by the following formula [Kam, eq. 1.33]:

Define f and g as follows:

and consider the Cauchy problem for equation (2.4.102) with the initial condition Then C in (2.4.103) takes the form:

From (2.4.96)) one gets

Since fg = –1 one has:

Thus

69

It follows from conditions (2.4.96) and from the second inequality in (2.4.97) that the solution to problem (2.4.102) exists for all and the following inequality holds with v(t) defined by (2.4.99):

From Lemma 2.4.21 and from formula (2.4.105) one gets:

and thus estimate (2.4.98) is proved. To illustrate conditions of Theorem 2.4.22 consider the following examples of functions and satisfying (2.4.96) for Example 2.4.24. Let

where Choose the following conditions

From (2.4.96), (2.4.97) one gets

Thus, one obtains the following conditions:

and

Therefore for such

and

and

a function

with the desired properties exists if

70


In this case one can choose as one needs the following conditions:

However in order to have

and

Example 2.4.25. If

then conditions (2.4.96), (2.4.97) are satisfied if

Example 2.4.26. Here log stands for the natural logarithm. For some

conditions (2.4.96), (2.4.97) are satisfied if

In all considered examples tends to infinity as and provide a decay of a nonnegative solution to integral inequality (2.4.97) even if tends to infinity. Moreover, in the first and the third examples v(t) tends to zero as when and 2.4.10 Riccati equation

An alternative approach to a study of Riccati equation (2.4.101) with non-negative coefficients a (t) and b (t) is based on the iterative method for solving integral equation Let Then Assume and consider the process By induction, If where T > 0 is an arbitrary number, then exists and solves (2.4.101).

71

Assume

Then

where is a constant estimated below. Inequality (2.4.117) holds for n = 0, and, by induction,

Using (2.4.116) one gets:

If

then (2.4.117) has been proved by induction. Inequality (2.4.118) holds if and 2.5

2.5.1

EXAMPLES OF SOLUTIONS OF ILL-POSED PROBLEMS

Stable numerical differentiation: when is it possible?

In many applications one has to estimate a derivative given the noisy values of the function f to be differentiated. As an example we refer to the analysis of photoelectric response data. The goal of that experiment is to determine the relationship between the intensity of light falling on certain plant cells and their rate of uptake of various substances. Rather than measuring the uptake rate directly, the experimentalists measure the amount of each substance not absorbed as a function of time, the uptake rate being defined as minus the derivative of this function. As for the other example, one can mention the problem of finding the heat capacity of a gas as a function of temperature T. Experimentally one measures the heat content and the heat capacity is determined by numerical differentiation. One can give many other examples of practical problems in which one has to differentiate noisy data. In navigation problems one selects the direction of the motion

72


of a ship by the maximum of a certain univalent curve, called the navigation characteristic. This direction can be obtained by differentiation of this curve. Since the navigation characteristic is communicated with some errors, one has to differentiate it numerically in order to find its maximum. In [R83, p. 94], the shape of a convex obstacle is found by differentiation of a support function of this obstacle. The support function is found from the experimentally measured scattering data, and by this reason the support function is noisy. In [R121, pp. 81–92], optimal estimates for the derivatives of random functions are obtained. In [RKa, p. 438], numerical differentiation of functions, contaminated by random noise is discussed. The noise has zero mean value and finite variance, and is identically distributed independently of the point x. It is proved that in this case the error of an optimal formula of numerical differentiation can be made where p is the number of observation points and is the error for a noise which is non-random (see the precise formulation of the result in [RKa]). Section 2.5.1 is essentially paper [RSm5]. The differentiation of noisy data is an ill-posed problem: small (in some norm) perturbations of a function may lead to large errors in its derivative. Indeed, if one takes then and so that small in perturbations of f result in large perturbations of in Various methods have been developed for stable numerical differentiation of f given We mention three groups of methods: being a regularization (1) regularized difference methods with a step size parameter, see [R18], where this idea was proposed for the first time, and [R156], [R58], [R168]. As an example of such a method one may consider:

If and where functions whose n-th derivative belongs to

where

is the Sobolev space of then

is an estimation constant: The error in the interval can be estimated slightly better (by a quantity In this paper The right-hand side of (2.5.2) attains the absolute we denote by while if one uses the error estimate minimum

73

for the interval then one gets the absolute minimum at When the function one can modify (2.5.1) near the ends so that it has the order of the error of approximation as and results in an algorithm of order For example one can take

The difference methods use only local values of the function which is natural when one estimates a derivative, and these methods are very simple, which is an advantage. (2) An alternative approach is first to smooth by a mollification with a certain kernel, for example with a Gaussian kernel, or to use a mollification by splines, and then to differentiate the resulting smooth approximation, see e.g. [VA]. If one applies mollification with the Gaussian kernel then

where that

stands for the convolution, with f (0) = f (1) = 0 and

and

Assume One has

From the Cauchy inequality the first term in the right-hand side of (2.5.5) can be estimated as follows:

because one gets:

By a partial integration

75

or

where

> is a regularization parameter. One proves that for a suitable choice of the above functionals have a unique minimizer and as An optimal choice of the regularization parameter in this approach is a nontrivial problem. Some methods for choosing are presented in Section 2.1.

The above methods have been discussed in the literature, and their analysis is not our goal. Our goal is to study two principally different statements of the problem of stable numerical differentiation, and to understand when it is possible in principle to get a stable approximation to given noisy data In Problem I a new notion of regularizer is introduced. Our treatment of the stable differentiation is an example of application of this new notion. In Section 2.1 a regularization method for unbounded linear and nonlinear operators is discussed. Statements of the problem of stable numerical differentiation

First, we recall some standard definitions (cf. Section 1.3). The problem of finding a solution u to the equation

where X and Y are Banach spaces, A is an operator, possibly nonlinear, is well-posed (in the sense of J. Hadamard) if the following conditions hold: (a) for every element there exists a solution (b) this solution is unique; (c) the problem is stable under small perturbations of the initial data in the sense:

If at least one of the conditions (a), (b) or (c) is violated, then the problem is called ill-posed. The problem of numerical differentiation can be written as

We study the cases p = 2 and in detail. Problem (2.5.14) is solvable only if So, condition (a) is not satisfied, condition (c) is not satisfied either, and condition (b) is satisfied. Therefore, problem (2.5.14) is ill-posed. Practically, one does not know f and the only information available for computational processing is together with an a priori information about f, for example, one

76


may know that

a = 0, a = 1, or 1
1 and mple, one can use

then there does exist an operator defined in 2.4.92 with

linear or If

such that 2.5.17 holds. For exa-

The error of the corresponding differentiation formula is

The main result on the stable numerical differentiation problem in the second formulation is stated in Theorem 2.5.2: Theorem 2.5.2. If a = 1, then there exists an operator that inequality (2.5.18) holds.

such

The principal difference is: for a = 1 one cannot differentiate stably in the sense formulated in Problem I. In the sense of Problem II Stable differentiation is possible in principle. However the approximation error, cannot be estimated, and this error may go to zero arbitrarily slowly as This is in sharp contrast with the practically computable error estimate given in (2.5.21). Moreover, no matter how small the error bound is, there exists a function such that not only (with any fixed function defined in (2.5.35), but any other operator linear or nonlinear, will satisfy the inequality where c > 0 does not depend on This follows from (2.5.19). Proofs Proof of Theorem 2.5.1. First, consider the case

Take

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Extend from (0, 2h) to (2h, 1) by zero and denote the extended function by again. Then and the norms a = 0 and a = 1 are preserved. Define Note that

Choose

so that

Then for

one has:

Let

One gets

If then If a = 0, then (2.5.23) implies that k = 1, 2, with For any fixed and the constant M in (2.5.22) can be chosen arbitrary. Therefore inequality (2.5.25) proves that (2.4.10) is false in the class and, in fact, as Suppose now that One has

Thus, for given

and

one can take

so that

k = 1, 2, holds, and then take M so that For these and M the functions k = 1, 2. One obtains from (2.5.25) the following inequality

which implies that estimate (2.4.10) is false in the class Now let p = 2. For defined in (2.5.22) one has

79

Extend from (0, 2h) to (2h, 1) by zero and denote the extended function again. Then and fine Choose to satisfy the identity

then

De-

k = 1 , 2 . Thus,

where

Therefore

If a = 0, then (2.5.29) yields k = 1 , 2 , with gets as Given constants and (in the case a = 1), one takes and then takes M so that and M the functions

Finally, consider using the Lagrange formula:

and one so that With this choice of

k = 1, 2, and one obtains from (2.5.30)

For the operator

defined by (2.4.92) one gets

Minimizing the right-hand side of (2.5.32) with respect to

yields

80


The case a = 2 is treated in the introduction (see estimate 2.4.93). So one arrives at 2.4.13–2.4.14. This completes the proof. Proof of Theorem 2.5.2. We give two proofs based on quite different methods. The first proof uses the construction of the regularizing operator defined in (2.4.92). The right-hand side of the error estimate of the type (2.4.93) is now replaced by where as provided that a = 1. Minimizing E with respect to h for a fixed one obtains a minimizer as and the error estimate as Therefore one gets (2.6). Alternatively, if one chooses as such that then The first proof is completed. Remark 2.5.3. The statement of Theorem 2.5.2 with (2.4.11) replaced by (2.4.12) is obvious: since f is fixed, one may take This is, of course, of no practical use because is unknown. The second proof is based on the DSM. The ideas of this proof have an advantage of being applicable to a wide variety of ill-posed problems, and not only to stable numerical differentiation. By this reason we give this proof in detail. In order to show that for a = 1 there exists an operator is a real Hilbert space) such that (2.5.18) holds we will use the DSM (dynamical systems method) (see Section 2.4). This approach consists of the following steps: Step 1. Solve the Cauchy problem:

where A is defined in (2.4.7), p = 2,

Step 2. Calculate as Then

where

is a number such that

and

with given by (2.5.49) below and To verify (2.5.36) consider the problem

and

and

81

Since A is monotone in

for any and admits the estimate

the solution

to (2.5.37) exists, is unique, and

Differentiate (2.5.37) with respect to t (this is possible by the implicit function theorem) and get

where (2.5.39) was used. Denote

From (2.5.37) and (2.5.33) one obtains

Multiply (2.5.42) by

Let

Since

So,

and get

then (2.5.39) and (2.5.43) imply

from (2.5.44) and (2.5.40) it follows that

82


Under assumption (2.5.34), one has

Indeed, (2.5.34) implies and Conclusion (2.5.46) follows. If one chooses so that and as then by (2.5.47’) and (2.5.46), using L’Hospital’s rule one obtains

The existence of the solution to (2.5.33) on is linear with a bounded operator. We claim that

is obvious, since equation (2.5.33)

For convenience of the reader this claim is established below. Equations (2.5.34), (2.5.45), (2.5.47), and (2.5.48) imply:

Let us now prove (2.5.48). Because and one can rewrite (2.5.37) as This and the monotonicity of A imply so, since h(t) > 0, one gets and Thus, w converges weakly in to some element as Because A is monotone, it is w-closed, that is and imply so and The inequality can be written as and because Therefore the claim (2.5.48) is proved and the second proof is completed. Numerical aspects

Figures 1–4 illustrate the impossibility to differentiate stably a function, which does a < 1. If one takes the function not have a bound on

83

and

then

and any formula of numerical differentiation will give error not going to zero as because, by (2.5.27), one has:

In Figure 1 one can see f (x) given by (2.5.50) with

and

84


Figure 2 presents

Figure 3 shows the case

and

The derivatives are given in Figure 4:

Even if the bound on in some norm is given, one can experience difficulties with stable differentiation. Namely, if is fixed and is very large, then in finite difference scheme (2.4.92) is very small, and practitioners might not have sufficiently many observation points. Another difficulty is: the estimated error in such a case is very big and does not give any information regarding the accuracy of computations. This is illustrated in Table 2.1 below for the function and Figures 5–8 show the exact and computed derivatives of The derivatives of this function were computed in the presence of the noise function

and with different step sizes. One can see in Figure 5 that for the computed derivative is very accurate. However as h grows, the accuracy decreases.

85

2.5.2

Stable summation of the Fourier series and integrals with perturbed coefficients

Assume that f (x) is a smooth

where

and

function

86


Assume that

are given such that

and are not known. The problem is: given

calculate stably

In other words, calculate

such that

The norm here can depend on the particular problem. Let us assume that this is the norm. Let us look for of the form:

One has, using Parseval’s equality,

where the constants and are defined by the last equation. Minimizing the righthand side of (2.5.62) with respect to v > 1, with being fixed, we find the optimal

and the error estimate:

If s = 2, then

Let us summarize the result.

Theorem 2.5.4. Assume that are given such that (2.5.58) and (2.5.59) hold. Define by formula (2.5.61) with where is defined in (2.5.63). Then (2.5.60) holds with defined by (2.5.64).

87

The above arguments are applicable also to Fourier integrals with perturbed Fourier transforms, which play the role of the perturbed coefficients. 2.5.3

Stable solution of some Volterra equations of the first kind

Consider equation (1.3.1) with bounded function, and is a kernel for which then (1.3.1), after a differentiation, yields

If

is a

provided that This is a Fredholm second-kind integral equation, for which the Fredholm alternative is valid in If is not differentiable with respect to x, or V(x, x) may vanish, then equation (1.3.1) with A = V can be solved by the methods discussed in Section 2.1–2.4. A different general approach to stable solution of the equation Vu = f consists of the factorization where S is a compact operator, I is the identity, the null-space N(I + S) = {0} is trivial, so that by Fredholm’s alternative the equation (I + S)u = w can be stably solved by a projection method, and the operator Q is such that w can be stably found from the equation Qw = f. If the noisy data are given in place of f, then a stable solution to the equation is given by a formula and a stable solution of the equation Vu = f with noisy data is given by the formula A numerical implementation of this scheme is given in [RSm6]. If and then the above scheme leads to the operator Q whose stable inversion is equivalent to stable differentiation, A discretization method for stable solution of Volterra integral equations of the first kind is proposed and justified in [RG], where the error estimates for the proposed method are also obtained. 2.5.4

Deconvolution problems

Equation (1.3.1) with tions 2.1–2.4, provided that

can be solved by the methods of SecIn the special case, when the kernel

where Q and P are elliptic operators, is the delta-function, one can use the theory from [R121], [R189], and prove, under suitable assumptions, that the operator is an isomorphism of onto where n = ord Q, m = ord P, is the usual Sobolev space, is its dual with respect to inner product, is, in other words, the closure of in the norm of that is, the subset of the elements of with support

88


If L is a selfadjoint elliptic operator in ord L := s and and are positive polynomials, deg Q = n, deg then Q := Q(L) and P := P(L) are elliptic operators, and if solves (2.5.1) then Equation Rh = f has a unique solution in for any and is an isomorphism. In [R121], [R189] one finds analytical formulas for the solution h. Under the above assumptions, the equation (*) Rh = f in does not have integrable solutions, in general. It has only distributional solutions of finite order of singularity, in general. Finding a solution of minimal order of singularity (mos solution) is a well posed problem ([R121]). The minimal order of singularity is equal to a. Since R is an isomorphism of onto the problem of solving equation (*) is well-posed. One does not need regularization methods for finding the solution to (*). Example 2.5.1.

In this example

One can see that generically h (x) is not an integrable function, it is a distribution, If and only if and and is Thus, equation

has a solution

which depends on where

continuously in the norm Details one can see in

[R121]. 2.5.5

Ill-conditioned linear algebraic systems

Let in (1.3.1). Then A can be represented by a matrix non-singular, define its condition number

If A is

One has (see (1.4.3)) Thus, if errors

is large then small relative errors

in the data may lead to large relative

in the solution. Solving linear algebraic ill-conditioned is an ill-posed prob-

lem practically. Note that if A is not singular (not degenerate: det are eigenvalues of A* A.

because Also,

where

89

Examples below are taken from [VA]. f = (4.1, 9.7), u = (1, 0), Au = f. Let g =

Example 2.5.2.

(4.11, 9.70). Then v = (0.34, 0.97). Here

249.5.

Example 2.5.3. Hilbert matrix

if n = 6,

if n = 10. 2.6

Let

PROJECTION METHODS FOR ILL-POSED PROBLEMS

be a linear, injective, bounded operator in a Hilbert space is unbounded, so that the problem of solving the equation Au = f is

ill-posed. Let be a sequence of finite-dimensional subspaces of H which is limit-dense in H, that is, for any one has where is the distance from f to Let be the orthoprojection operators on Assume that is given such that The problem consists of finding which solves the equation (*) and such that Equation (*) is an equation of a projection method. A general approach to finding a stable approximation is the following one: let be a regularizer in the sense and define Then where

as as and is chosen suitably. More precisely, for any fixed one has because the sequence is limit-dense. Thus, one can choose so that then and for a fixed one can choose so that as Therefore as so that u is stably approximated. There are many ways to choose the regularizer one can use a Variational Regularization, Quasisolutions, Iterative Regularization, and the DSM. One can also use the method developed in Section 2.5.2 for constructing a convergent projection method for solving (1.3.1). Assume that X = Y = H in (1.3.1), A is a linear compact operator, and (1.3.1) is solvable, i.e, f = Ay for some We assume that that is, is the minimal-norm solution. By Lemma 2.1.11, a solvable equation (1.3.1) is equivalent to equation (2.1.6). Denote := A* f, assume that is given in place of and Let where j = 1, 2, , is an orthonormal system of eigenvectors of the selfadjoint compact operator B, and are the corresponding eigenvalues. Let and The element is a regularizer if is chosen properly, as in Section 2.5.2. In this case one has See more on this topic in [IVT], [VA].


3. ONE-DIMENSIONAL INVERSE SCATTERING AND SPECTRAL PROBLEMS

Inverse scattering and spectral one-dimensional problems are discussed in this Chapter systematically in a self-contained way. Many novel results due to the author are presented. The classical results are often presented in a new way. Several highlights of the new results include: (1) Analysis of the invertibility of the steps in the Gel’fand-Levitan and Marchenko inversion procedures; (2) Theory of the inverse problem with I-function as the data and its applications; (3) Proof of the Property C for ordinary differential operators, numerous applications of Property C; (4) Inverse problems with “incomplete” data; (5) Spherically symmetric inverse scattering problem with fixed-energy phase shifts: uniqueness result is obtained for the case when part of the phase shifts is known. The Newton-Sabatier (NS) scheme for inversion of fixed-energy phase shifts is analyzed. This analysis shows that the NS scheme is fundamentally wrong in the sense that its foundations are wrong, and this scheme is not a valid inversion method; (6) Complete presentation of the Krein inverse scattering theory is given for the first time. Consistency of this theory is proved; (7) Quarkonium systems are recovered from experimental data; (8) A study of the properties of I-function; (9) Some new inverse problems for the heat and wave equations are studied;

92


(10) A study of an inverse scattering problem for an inhomogeneous Schrödinger equation; (11) A study of inverse problem of ocean acoustics; (12) Theory of ground-penetrating radars is given. 3.1

INTRODUCTION

3.1.1

What is new in this chapter?

There are excellent books [M] and [L], where inverse spectral and scattering problems are discussed in detail. Therefore let us point out the novel points in this Chapter. (1) A new approach to the uniqueness of the solutions to these problems. This approach is based on Property C for Sturm-Liouville operators; (2) the inverse problem with I-function as the data is studied and applied to many inverse problems; (3) a detailed analysis of the invertibility of the steps in Marchenko and Gel’fand-Levitan (GL) inversion procedures is given; (4) inverse problems with “incomplete” data are studied; (5) a detailed presentation of Krein’s inversion method with proofs is given apparently for the first time; (6) a number of new results for various inverse problems are presented. These include, in particular, (a) a uniqueness theorem for recovery of a compactly supported sphericallysymmetric potential from a part of the corresponding fixed-energy phase shifts; (b) a method for finding confining potential (a quarkonium system) from a few experimental data; (c) solution of several new inverse problems for the heat- and wave equations; (d) a uniqueness theorem for finding a potential q from a part of the corresponding fixed-energy phase shifts; and many other results which are taken from [R196], and especially [R221] and [R139]. (e) a solution to an inverse problem of ocean acoustics. An attempt to study a similar problem has been made in [GX]. We show that the method presented in [GX] is invalid; (f) a theory of ground-penetrating radars.

Due to the space limitations, several important questions are not discussed: inverse scattering on the full line, iterative methods for finding potential q : (a) from two spectra [R181], [R221], (b) from S-matrix alone when q is compactly supported [R184], approximate methods for finding q from fixed-energy phase shifts [RSch], [RSm2], property of resonances [R139], inverse scattering for systems of equations, etc. 3.1.2

Auxiliary results

Let

where consists of functions belonging to and overline stands for complex conjugate.

and for any

93

Consider the differential expression with domain of definition where is the set of vanishing in a neighborhood of infinity, If H is the Hilbert space then is densely defined symmetric linear operator in H, essentially self-adjoint, that is, the closure of in H is selfadjoint. It is possible to construct a selfadjoint operator assuming that Such a theory is technically more difficult, because it is not even obvious a priori that the set is dense in H (in fact, it is). Such a theory is presented in [Nai]. If one drops the assumption then is not a domain of definition of since there are functions for which In the future we mean by a self-adjoint operator generated by the differential expression and the boundary condition u(0) = 0. This operator has absolutely continuous spectrum, which fills and discrete, finite, negative spectrum where are the eigenvalues of all of them are simple,

where

are corresponding eigenfunctions, which are real-valued functions, and

The functions

and

are defined as the unique solutions to the problems:

These functions are well defined for any Their existence and uniqueness can be proved by using the Volterra equations for and If then the Jost solution f (x, k) exists and is unique. This solution is defined by the problem:

Existence and uniqueness of f is proved by means of the Volterra equation:

If

then this equation implies that f (x, k) is an analytic function of k in for k > 0. The Jost function is defined as f (k) := f (0, k). It has exactly J simple roots where are the negative eigenvalues of The number k = 0 can be a zero of f (k). If f (0) = 0, then where Existence of is a fine result under the only

94


assumption

(see Theorem 3.1.3 below, and [R139]) and an easy one if The phase shift is defined

by the formula

where the last equation in (3.1.7) follows from (3.1.6). Because for One defines the S-matrix by the formula

one has

The function S(k) is not defined for complex k if but if then f (k) is an entire function of k and S (k) is meromorphic in If q (x) = 0 for x > a, then f (k) is an entire function of exponential type (see Section 3.5.1). If then at the Jost solution is proportional to and both belong to The integral equation for is:

One has:

because the right-hand side of (3.1.10) solves equation (3.1.5) and satisfies conditions (3.1.3) at x = 0. The first condition (3.1.3) is obvious, and the second one follows from the Wronskian formula:

If then as one can derive easily from equation (3.1.6). In fact, If k > 0, then If then is a real-valued function. The function f (x, k) is analytic in but is, in general, not defined for In particular, (3.1.11), in general, is valid on the real axis only. However, if then f (k) is defined on the whole complex plane of k, as was mentioned above. Let us denote for and let be the second, linearly independent, solution to equation (3.1.5) for If then One can write a formula, similar to (3.1.10), for

95

where

For

it is necessary and sufficient that

In fact,

where To prove the second relation in (3.1.13), one differentiates (3.1.5) with respect to k and gets

Existence of the derivative with respect to k in follows easily from equation (3.1.6). Multiply (3.1.13a) by f and (3.1.5) by subtract and integrate over then by parts, put and get:

Thus,

It follows from (3.1.14) that constants:

The numbers

are called the norming

Definition 3.1.1. Scattering data is the triple:

The Jost function f (k) may vanish at k = 0. If f (0) = 0, then the point k = 0 is called a resonance. If then the zeros of f (k) in are called resonances. As we have seen above, there are finitely many zeros of f (k) in these zeros are simple, their number J is the number of negative eigenvalues of the selfadjoint Dirichlet operator If then the negative spectrum of is finite [M]. The phase shift defined in (3.1.17), is related to S(k):

96


so that S (k) and are interchangeable in the scattering data. One has because solves (3.1.3). Therefore

Thus

In Section 3.4.1 the notion of spectral function is defined. It will be proved in Section 3.5.1 for that the formula for the spectral function is:

where are defined in (3.1.18)–(3.1.19). The spectral function is defined in Section 3.4.5 for any Such a q may grow at infinity. On the other hand, the scattering theory is constructed for Let us define the index of S(k):

This definition implies that

Therefore:

because a simple zero k = 0 contributes to the index, and the index of an analytic in function f (k), such that equals to the number of zeros of f (k) in plus half of the number of its zeros on the real axis, provided that all the zeros are simple. This follows from the argument principle. In Section 3.4.2 and Section 3.5.2 the existence and uniqueness of the transformation (transmutation) operators will be proved. Namely,

97

and

and the properties of the kernels A(x, y) and K(x, y) are discussed in Section 3.5.2 and Section 3.4.2 respectively. The transformation operator I + K transforms the solution to the equation (3.1.3) with q = 0 into the solution of (3.1.3), satisfying the same as boundary conditions at x = 0. The transformation operator I + A transforms the solution to equation (3.1.5) with q = 0 into the solution f of (3.1.5) satisfying the same as “boundary conditions at infinity”. One can prove (see [M] and Sec. 5.7) the following estimates

and A(x, y) solves the equation:

By we denote Sobolev spaces The kernel A(x, y) is the unique solution to (3.1.27), and also of the problem (3.5.1)–(3.5.3). 3.1.3

Statement of the inverse scattering and inverse spectral problems

ISP: Inverse scattering problem (ISP) consists of finding tering data (see (3.1.6)). A study of ISP consists of the following:

from the corresponding scat-

(1) One proves that ISP has at most one solution. to be scattering data corre(2) One finds necessary and sufficient conditions for sponding to a (characterization of the scattering data problem). (3) One gives a reconstruction method for calculating from the corresponding

In Section 3.5 these three problems are solved. ISpP: Inverse spectral problem consists of finding q from the corresponding spectral function. A study of ISpP consists of the similar steps: (1) One proves that ISpP has at most one solution in an appropriate class of q: if and from this class generate the same then

98


(2) One finds necessary and sufficient conditions on which guarantee that a spectral function corresponding to some q from the above class. (3) One gives a reconstruction method for finding q (x) from the corresponding 3.1.4

is

Property C for ODE

Denote by operators corresponding to potentials corresponding Jost solutions, m = 1, 2. Definition 3.1.2. We say that a pair is complete (total) in This means that if

and by

has property

iff the set

then

We prove in Section 2.1 that a pair does have property and let correspond to

Definition 3.1.3. We say that a pair is complete in for any b > 0, This means that if

the

has property

if

Let

iff the set

then:

In Theorem 3.2.5 we prove that there is a

for which

for a suitable Therefore Property with hold, in general. Property is defined similarly to Property with functions In Section 3.2 we prove that properties and cations of these properties throughout this work.

does not replacing

hold, and give many appli-

99

3.1.5

A brief description of the basic results

The basic results of this work include: (1) Proof of properties and Demonstration of many applications of these properties. (2) Analysis of the invertibility of the steps in the inversion procedures of Gel’fandLevitan (GL) for solving inverse spectral problem:

where

the kernel L = L(x, y) is:

is the spectral function of Levitan equation

with q = 0, and K solves the Gel’fand-

Our basic result is a proof of the invertibility of all the steps in (3.1.31):

which holds under a weak assumption on

where

Namely, assume that

is the set of non-decreasing functions of bounded variation on every interval such that the following two assumptions, and hold. Denote the set of functions vanishing in a neighborhood of infinity. Let and Assumption is:

100


Let

and belong to Assumption is:

and

(see Section 3.4.2).

In order to insure the one-to-one correspondence between spectral functions and selfadjoint operators we assume that q is such that the corresponding is “in the limit point at infinity case”. This means that the equation has exactly one nontrivial solution in If then is “in the limit point at infinity case”. (3) Analysis of the invertibility of the Marchenko inversion procedure for solving ISP:

where

and A(x, y) solves the Marchenko equation

Our basic result is a proof of the invertibility of the steps in (3.1.40):

under the assumption

We also derive a new equation for

101

This equation is:

The function A(y) is of interest because

Therefore the knowledge of A(y) is equivalent to the knowledge of f(k). In Section 3.5.5 we give necessary and sufficient conditions for to be the scattering data corresponding to We also prove that if

and, in particular, if

then S(k) alone determines q (x) uniquely, because it determines under the assumption (3.1.47) or (3.1.48).

and J uniquely

(4) We give a very short and simple proof of the uniqueness theorem which says that the I-function,

determines

uniquely. The I-function is equal to Weyl’s m -function if

We give many applications of the above uniqueness theorem. In particular, we give short and simple proofs of the Marchenko’s uniqueness theorems which say that determines uniquely, and determines q uniquely. We prove that if (3.1.48) (or (3.1.47)) holds, then either of the four functions S(k), f (k), determines q (x) uniquely. This result is applied in Section 3.10 to the heat and wave equations. It allows one to study some new inverse problems. For example, let

102


Assume

and let the extra data (measured data) be

The inverse problem is: given these data, find q (x). Another example: Let

The extra data are

The inverse problem is: given these data, find q (x). Using the above uniqueness results, we prove that these two inverse problems have at most one solution. The proof gives also a constructive procedure for finding q . (5) We have already mentioned uniqueness theorems for some inverse problems with “incomplete data”. “Incomplete data” means the data, which are a proper subset of the classical data, but “incompleteness” of the data is compensated by the additional assumptions on q. For example, the classical scattering data are the triple (3.1.16), but if (3.1.48) or (3.1.47) is assumed, then the “incomplete data” alone, such as S(k), or or f(k), or determine q uniquely. Another general result of this nature, that we prove in Section 3.7, is the following one. Consider, for example, the problem

Other boundary conditions can also be considered. Assume that the following data are given.

103

where 0 < b < 1, and

Assume also

We prove Theorem 3.1.4. Data (3.1.60)–(3.1.61) determine uniquely q (x) on the interval if If (3.1.62) is assumed additionally, then q is uniquely determined if The gives the “part of the spectra” sufficient for the unique recovery of q on [0, b]. For example, if and (3.1.62) holds, then so “one spectrum” determines If then so “half of the spectrum” determines uniquely q on If then of the spectrum” determine uniquely q uniquely q on on If b = 1, then and “two spectra” determine q uniquely on the whole interval [0, 1]. The last result belongs to Borg [B]. By “two spectra” one means where are the eigenvalues of the problem:

In fact, two spectra determine not only q but the boundary conditions as well [M]. (6) Our basic results on the spherically symmetric inverse scattering problem with fixed-energy data are the following. The first result: If q = q (r) = 0 for r > a, a > 0 is an arbitrary large fixed number, and then the data determine q (r) uniquely. Here is the phase shift at a fixed energy is the angular momentum, and is any fixed set of positive integers such that

The second result is: If q = q(x), q = 0 for where then the knowledge of the scattering amplitude at a fixed energy and all determine q (x) uniquely, see Section 5.2. Here are arbitrary small open subsets in and is the unit sphere in The scattering amplitude is defined in Section 3.6.1.

104


The third result is: The Newton-Sabatier inversion procedure (see [CS], [N]) is fundamentally wrong. (7) Following [R197] we present, apparently for the first time, a detailed exposition

(with proofs) of the Krein inversion theory for solving inverse scattering problem and prove the consistency of this theory. (8) We give a method for recovery of a quarkonium system (a confining potential) from a few experimental measurements. (9) We study various properties of the I-function. (10) We study an inverse scattering problem for inhomogeneous Schrödinger equation. (11) We study an inverse problem of ocean acoustics. (12) We develop a theory of ground-penetrating radars. 3.2

PROPERTY C FOR ODE

3.2.1

Property

By ODE in this section, the equation

is meant. Assume Then the Jost solution f (x, k) is uniquely defined. In Section 1.3 Definition 3.1.2, property is explained. Let us prove Theorem 3.2.1. If

j = 1, 2 then property

holds.

Proof. We use (3.1.24) and (3.1.25). Denote

for some

where

Set

Let

and get

105

Thus,

where V* is the adjoint to a Volterra operator,

Using (3.2.3) and (3.2.4) one rewrites (3.2.2) as

The right-hand side is an analytic function of k in vanishing for all k > 0 Thus, it vanishes identically in and, consequently, for k < 0 Therefore

Since A(x, y) and T(x, y) are bounded continuous functions, the Volterra equation (3.2.6) has only the trivial solution h = 0. Define functions asymptotics:

and

Let us denote

and

as the solutions to equation (3.1.5) with the following

Definition 3.2.2. The pair

has property C_ iff the set

Similar definition can be given with As above, one proves: Theorem 3.2.3. If By

replacing

is complete in

j = 1, 2.

j = 1, 2, then property C_ holds for

we mean the set

3.2.2 Properties

and

We prove only property Section 3.1.4. Theorem 3.2.4. If

Property

is proved similarly. Property

j = 1, 2, then property

holds for

is defined in

106


Proof. Our proof is similar to the proof of Theorem 3.2.1. Using (3.1.23) and denoting one writes

Assume:

Then

where

Then

Let

where

and

if

if

107

Therefore

From (3.2.12) and (3.2.13), taking

one gets:

and (using completeness of the system cos(ks), equation:

in

the following

The kernels K, and are bounded and continuous functions. Therefore, if and h(x) = 0 for x > b, (3.2.14) implies:

where c > 0 is a constant which bounds the kernels 2K, 2y = s. From the above inequality one gets

and

from above and

where is sufficiently small so that and Then inequality (3.2.15) implies h(x) = 0 if Repeating this argument, one proves, in finitely many steps, that h(x) = 0, 0 < x < b. Theorem 3.2.4 is proved. The proof of Theorem 3.2.4 is not valid if Let us give a counterexample. Theorem 3.2.5. There exist

and an

The result is not valid either if

such that

Proof. Let and are two potentials in such that have one negative eigenvalue which is the same for and but that Let Let us prove that (3.2.16) holds. One has

and so

108


subtract from the first equation the second and get:


integrate over

At x = 0 we use conditions (3.2.17), and at and are the same (because (3.2.18) vanishes. Theorem 3.2.5 is proved. 3.3

and then by parts to get

the phase shifts corresponding to and therefore the right-hand side of

INVERSE PROBLEM WITH I-FUNCTION AS THE DATA

3.3.1

Uniqueness theorem

Consider equation (3.1.5) and assume the I-function:

Then f (x, k) is analytic in

Define

From (3.3.1) it follows that I(k) is meromorphic in with the finitely many simple poles Indeed, are simple zeros of f (k) and as follows from (3.1.4). Using (3.1.19), one gets

where and f (0) = 0 then a subtle result.

and exists and

iff f (0) = 0. We prove that if (Theorem 3.3.3 below). This is

Lemma 3.3.1. The I (k) equals to the Weyl function m (k). Proof. The m (k) is a function such that Clearly

where and (3.1.4).

Im k > 0. Thus,

Our basic uniqueness theorem is:

if Im k > 0.

because of (3.1.3)

109

Theorem 3.3.2. If Proof. Let Then one has

Multiply (3.3.3) by

By property

j = 1, 2, generate the same I(k), then be the Jost solution (3.1.5) corresponding to

integrate over

then by parts, using (3.3.3), and get

(Theorem 3.2.1), p(x) = 0.

Remark. If

j = 1, 2, and (*) where then for almost all This result is proved in [GS1] for Our proof is based on (3.3.3), from which, using (*), one gets (**) Note that where V* is the adjoint to a Volterra operator (see the formula below (3.2.4)). Thus, (**) can be written as (***) where Theorem 96 in [Tit] and (***) imply for almost all Since V is a Volterra operator, it follows that p = 0 for almost all as claimed. Theorem 3.3.3. If

and f (0) = 0, then

Proof. Let us prove that Let A(x, y) is defined in (3.1.24) and gets difficulty is to prove that To prove that and with x = 0: existence of is proved, one gets here the existence of the limit from (3.1.24):

and

exists and

and and where by (3.1.25). Integrating by parts, one Thus, The basic If this is done, then exists one uses the Wronskian formula (3.3.2) Divide by k and let Since We have used so The existence of it follows

110


From (3.1.26) one sees that Thus, to complete the proof, one has to prove To prove this, use (3.1.43) with x = 0 and (3.1.41). Since one has Therefore (3.1.43) with x = 0 yields:

Integrate (3.3.7) over

where

to get:

Integrating by parts yields:

Because (3.3.9) imply:

one has

From this equation and from the inclusion as follows. Choose a such that Then (3.3.10) can be written as:

Since

Therefore (3.3.8) and

one derives and let

and it follows that is bounded. Thus The operator has norm Therefore equation (3.3.3) is uniquely solvable in Theorem 3.3.3 is proved. 3.3.2

Characterization of the I-functions

One has

where the Wronskian formula was used with x = 0:

111

From (3.1.20) and (3.3.12) with

one gets:

The I (k) determines uniquely the points as the (simple) poles of I (k) on the imaginary axis, and the numbers by (3.3.2). Therefore I (k) determines uniquely the spectral function by formula (3.1.20). The characterization of the class of spectral functions given in Section 3.4.6, induces a characterization of the class of I-functions. The other characterization of the I-functions one obtains by establishing a oneto-one correspondence between the I-function and the scattering data (3.1.16). Namely, the numbers and J, are obtained from I (k) since are the only poles of I (k) in the numbers are obtained by the formula (see (3.1.15) and (3.3.2)):

if f (k) is found from I (k). Finally, f (k) can be uniquely recovered from I (k) by solving a Riemann problem. To derive this problem, define

and

Assumption (3.3.16), means that Define

Write (3.3.12) as

where in

and (3.3.17) means f (0) = 0.

or

is analytic in in has similar properties in

the closure of

112


g (k) > 0 for k > 0, g (k) is bounded in a neighborhood of k = 0 and has a finite limit at k = 0. From (3.3.20) and the properties of h, one gets:

and

In Section 3.4 we prove:

where

Taking in (3.3.24), one finds step by step all the numbers and J. If then Thus the data (3.1.16) are algorithmically recovered from I (k) known for all k > 0. A characterization of is given in Section 3.5.5, and thus an implicit characterization of I (k) is also given. 3.3.3

Inversion procedures

Both procedures in Section 3.3.2, which allow one to construct either or from I (k) can be considered as inversion procedures because in Section 3.4 and Section 3.5 reconstruction procedures are given for recovery of q (x) from either or All three data, I (k), and are equivalent. Thus, our inversion schemes are:

where (3.1.31) gives the details of the step of the step 3.3.4

and (3.1.40) gives the details

Properties of I(k)

In this section, we derive the following formula for I (k): Theorem 3.3.4. One has

113

where

if and only if f (0) = 0, are the constants defined in (3.3.2), if and if f (0) = 0 and

We prove this result in several steps which are formulated as lemmas. Using (3.1.24) one gets

One has (cf. (3.3.16))

is analytic in

the factor

in (3.3.29) is present if and only if f (0) = 0,

and

Lemma 3.3.5. If

and

Proof. It is sufficient to prove that, for any j,

Since and since (3.1.25)–(3.1.26)), it is sufficient to check that

Note that

then

the function

provided that

(see

114


One has

thus

where

From (3.3.34) one obtains (3.3.33) since

Lemma 3.3.6. If f (0) = 0 and

Lemma 3.3.5 is proved. then (3.3.31) holds.

Proof. The proof goes as above with one difference: if f (0) = 0, then in formula (3.3.27) and in formula (3.3.34) with one has

Thus, using (3.1.25), one gets

where c > 0 is a constant. Similarly one checks that Lemma 3.3.6 is proved. Lemma 3.3.7. Formula (3.3.27) holds. Proof. Write

is present

115

Clearly

By the Wiener-Levy theorem [GeRS], one has

where and

is defined in (3.3.29). Actually, the Wiener-Levy theorem yields However, since one can prove that Indeed, are related by the equation:

which implies

or

where is the convolution operation. the convolution Since and as claimed. entiating (3.3.36) one sees that From the above formulas one gets:

So, differ-

where c is a constant defined in (3.3.39) below, the constants are defined in (3.3.40) and the function is defined in (3.3.41). We will prove that c = 0 (see (3.3.43)). To derive (3.3.37), we have used the formula:

116


and made the following transformations:

where

Comparing (3.3.38) and (3.3.37) one concludes that

To complete the proof of Lemma 3.3.7 one has to prove that c = 0, where c is defined in (3.3.39). This follows from the asymptotics of I (k) as Namely, one has:

From (3.3.42) and (3.3.28) one gets:

From (3.3.43) and (3.3.37) it follows that c = 0. Lemma 3.3.7 is proved. Lemma 3.3.8. One has if f (0) = 0 Proof. From (3.3.2) one gets:

and

if

and

117

If j = 0, then

Here by we mean the right-hand side of (3.3.45) since I (k) is, in general, not analytic in a disc centered at k = 0, it is analytic in and, in general, cannot be continued analytically into By Theorem 3.3.3 the right-hand side of (3.3.45) is well defined and

From (3.1.24) one gets:

Since A(y) is a real-valued function if q (x) is real-valued (this follows from the integral equation (3.1.27), formula (3.3.47) shows that

and (3.3.46) implies

Lemma 3.3.8 is proved. One may be interested in the properties of function a (t) in (3.3.27). These can be obtained from (3.3.41) and (3.3.31) as in the proof of Lemma 3.3.5 and Lemma 3.3.6. In particular, the statements of Theorem 3.3.4 are obtained. Remark 3.3.9. Even if compactly supported.

is compactly supported, one cannot claim that a(t) is

Proof. Assume for simplicity that J = 0 and In this case, if a (t) is compactly supported then I(k) is an entire function of exponential type. It is proved in [R139], p. 278, that if is compactly supported, then f (k) has infinitely many zeros in The function if f (z) = 0. Indeed, if f (z) = 0 and then by the uniqueness of the solution of the Cauchy problem for equation (3.1.5) with k = z. Since one has a contradiction, which proves that if f(z) = 0. Thus, I (k) cannot be an entire function if and q (x) is compactly supported.

118


Let us consider the following question: What are the potentials for which a (t) = 0 in (3.3.27)? In other words, let us assume

and find q (x) corresponding to I-function (3.3.50), and describe the decay properties of q (x) as We give two approaches to this problem. The first one is as follows. By definition

Using (3.3.51) and (3.1.11) one gets [ I (k) – I (–k)] f (k) f (–k) = 2ik, or

By (3.3.44) one can write (see (3.1.20)) the spectral function corresponding to the I-function (3.3.50)

where is the delta-function. Knowing one can recover q (x) algorithmically by the scheme (3.1.31). J=1. Consider an example. Suppose

Then (3.3.53) yields:

Thus, (3.1.33) yields:

119

and, setting

and taking for simplicity

one finds:

where the known formula was used:

Thus,

Equation (3.1.34) with kernel (3.3.59) is not an integral equation with degenerate kernel:

This equation can be solved analytically [R121], but the solution is long. By this reason we do not give the theory developed in [R121], but give the second approach to a study of the properties of q (x) given I (k) of the form (3.3.54). This approach is based on the theory of the Riemann problem [G]. Equations (3.3.52) and (3.3.54) imply

The function

120


Write (3.3.61) as

Thus,

The function in in so Consider (3.3.63) as a Riemann problem. One has

is analytic in

Therefore (see [G]) problem (3.3.63) is uniquely solvable. Its solution is:

as one can check. Thus, by (3.3.62),

The corresponding S-matrix is:

Thus,

and

Equation (3.1.43) implies

so

121

Thus, if by the number

and a (t) = 0 then q (x) decays exponentially at the rate determined

If f (0) = 0, J = 0, and a (t) = 0, then

Let

and Thus, since (3.3.73). One has:

Then equation (3.3.72) implies:

in

and

and So one gets:

Equation (3.1.43) yields:

Solving (3.3.76) yields:

is uniquely determined by the Riemann problem


122

The corresponding potential (3.1.42) is

If case by (3.3.41), the function the details to the reader. 3.4

then a (t) in (3.3.27) decays exponentially. Indeed, in this decay exponentially, so g (t) decays exponentially, and, with h(t) decaying exponentially. We leave

INVERSE SPECTRAL PROBLEM

3.4.1

Auxiliary results

Transformation operators

If and are linear operators in a Banach space X , and T is a boundedly invertible linear operator such that then T is called a transformation (transmutation) operator. If then so that T sends eigenfunctions of into eigenfunctions of with the same eigenvalue. Let j = 1, 2, be selfadjoint in operators generated by the Dirichlet boundary condition at x = 0. Other selfadjoint boundary conditions can be considered also, for example,

Theorem 3.4.1. Transformation operator for a pair exists and is of the form Tf = (I + K) f, where the operator I + K is defined in (3.1.23) and the kernel K(x, y) is the unique solution to the problem:

Proof. Consider for simplicity the case If then (3.4.3) can be written as

If

and

The proof is similar in the case

then

123

Since and f is arbitrary otherwise, (3.4.5) implies (3.4.1), (3.4.2) and (3.4.4). Conversely, if K(x, y) solves (3.4.1), (3.4.2) and (3.4.4), then I + K is the transformation operator. To finish the proof of Theorem 3.4.1 we need to prove existence of the solution to (3.4.1), (3.4.2) and (3.4.4). Let Then (3.4.1), (3.4.2) and (3.4.4) can be written as

Integrate (3.4.6) to get

Integrate (3.4.7) with respect to

over

and get

This is a Volterra integral equation which has a solution, this solution is unique, and it can be obtained by iterations. Theorem 3.4.1 is proved. Spectral function

Consider the problem (3.1.1). The classical result, going back to Weyl, is: Theorem 3.4.2. There exists a monotone increasing function possibly nonunique, such that for every there exists such that

where the limit is understood in sense. If the potential q in (3.1.1) generates the Dirichlet operator in the limit point at infinity case, then is uniquely defined by q, otherwise is defined by q nonuniquely. The spectral function of has the following properties:

The remainder in (4.1.10) can be improved if additional assumptions on q are made. For example, for one can get the remainder Theorem 3.4.3. (Weyl). For any

The function

is analytic in

there exists

and in

such that

124


The function is called Weyl’s function, or m -function, and W is Weyl’s solution. Theorem 3.4.2 and Theorem 3.4.3 are proved in [M]. 3.4.2

Uniqueness theorem

Let be a non-decreasing function of bounded variation on every compact subset of the real axis. Let where is a subset of functions which vanish near infinity. Let

Our first assumption

and

on

is:

This implication should hold for any It holds, for example, if on a set which has a finite limit point: in this case the entire function of vanishes identically, and thus h = 0. Denote by a subset of and assume that if where is defined in (3.4.12), then

Our second assumption

on

and

is:

Let us start with two lemmas. Lemma 3.4.4. Spectral functions at infinity case belong to

of an operator

in the limit-point

Proof. Let

be two spectral functions corresponding to and and Let I + V and I + W be the transformation operators corresponding to and respectively, such that

where

is the regular solution (3.1.1) corresponding to

Condition

implies

125

where, for example,

It follows from (3.4.17) that

where U is a unitary operator in Indeed, U is an isometry and it is surjective because is. To finish the proof, one uses Lemma 3.4.5 below and concludes from (3.4.19) that so V = W, and Since, by assumption, q is in the limit-point at infinity case, there is only one spectral function corresponding to q, so Lemma 3.4.5. If U is unitary and V and W are Volterra operators, then (3.4.19) implies V = W. Proof. From (3.4.19) one gets where

Since U is unitary, one has (I + V) Because V is a Volterra operator, is also a Volterra (of the same type as V in (3.4.18)). Thus, or

The left-hand side in (3.4.20) is a Volterra operator of the type V in (3.4.18), while its right-hand side is a Volterra operator of the type Since they are equal, each of them must be equal to zero. Thus, or or V = W. Theorem 3.4.6. The spectral function determines Proof. If

and

uniquely.

have the same spectral function

then

where

Let I + K

be the Then

transformation

operator

and From (3.4.21) one gets

126


Thus is isometric, and, because is a Volterra operator, the range of is the whole space Therefore is unitary. This implies Indeed, (unitarity) and (Volterra property of Thus so Therefore and so This result was proved by Marchenko (see [M]). Remark 3.4.7. If c = const > 0, then the above argument is applicable and shows that c must be equal to 1, c = 1 and Indeed, the above argument yields the unitarity of the operator which implies c = 1 and Here the following lemma is useful: Lemma 3.4.8. If b I + Q = 0, where b = const and Q is a compact linear operator, then b = 0 and Q = 0. A simple proof is left to the reader. 3.4.3 Reconstruction procedure

Assume that the spectral function corresponding to is given. How can one reconstruct that is, to find q (x)? We assume for simplicity the Dirichlet boundary condition at x = 0, but the method allows one to reconstruct the boundary condition without knowing it a priori. The reconstruction procedure (the GL, i.e., the Gel’fand-Levitan procedure) is given in (3.1.31)–(3.1.34). Its basic step consists of the derivation of equation (3.1.34) and of a study of this equation. Let us derive (3.1.34). We start with the formula

and assume that L(x, y) is a continuous function of x, y in [0, b) × [0, b) for any If

one gets from (3.4.22) the relation:

Using (3.1.23), one gets (3.4.23), one gets

Applying

to

in

127

The right-hand side can be rewritten as:

From (3.4.24) and (3.4.25) one gets, using continuity at y = x, equation (3.1.34). In the above proof the integrals (3.4.23)–(3.4.25) are understood in the distributional sense. If the first inequality (3.4.10) holds, then the above integrals over are well defined in the classical sense. If one assumes that the integral in (3.4.26) converges to a function L (x) which is twice differentiable in the classical sense:

then the above proof can be understood in the classical sense, provided that for any If is a spectral function corresponding to then the sequence satisfies It is known (see [L]) that the sequence

satisfies

and converges to zero.

Lemma 3.4.9. Assume (3.4.13) and suppose that the function

Then equation (3.1.34) has a solution in

for any b > 0, and this solution is unique.

Proof. Equation (3.1.34) is of Fredholm-type: its kernel

is in for any Therefore Lemma 3.4.9 is proved if it is proved that the homogeneous version of (3.1.34) has only the trivial solution. Let

128


Because L(x, y) is a real-valued function, one may assume that h(y) is real-valued. Multiply (3.4.28) by h(y), integrate over (0, x), and use (3.4.12), (3.1.33) and Parseval’s equation to get

From (3.4.13) and (3.4.29) it follows that h = 0. If the kernel K(x, y) is found from equation (3.1.34), then q (x) is found by formula (3.4.4). 3.4.4

Invertibility of the reconstruction steps

Our basic result is: Theorem 3.4.10. Assume (3.4.13), (3.4.14), and suppose of the steps in (3.1.31) is invertible, so that (3.1.35) holds. Proof. 1. Step. is done by formula (3.1.33). Let us prove and corresponding to the same L(x, y), and then

Multiply (3.4.30) by h(x)h(y),

use (3.4.12) and get

By (3.4.14) it follows that v = 0, so

Thus

Then each

If there are

2. Step. is done by solving (3.1.34). Lemma 3.4.9 says that K is uniquely determined by L. Let us do the step Put y = x in (3.1.34), use (3.4.26) and (3.4.27) and get:

or

129

This is a Volterra integral equation for L(x) which has a solution and the solution is unique. Thus the step is done. The functions L(x) and K(x, x) are of the same smoothness. 3. Step. is done by formula (3.4.4), q (x) is one derivative less smooth than K(x, x) and therefore one derivative less smooth than L(x). Thus The step is done by solving the Goursat problem (3.4.1), (3.4.2), (3.4.4) (with or, equivalently, by solving Volterra equation (3.4.8), which is solvable and has a unique solution. The corresponding K(x, y) is in if

Theorem 3.4.10 is proved. Let us prove that the q obtained by formula (3.4.4) generates the function identical to the function K obtained in Step 2. The idea of the proof is to show that both K and solve the problem (3.4.1), (3.4.2), (3.4.4) with the same and This is clear for In order to prove it for K, it is sufficient to derive from equation (3.1.34) equations (3.4.1) and (3.4.2) with q given by (3.4.4). Let us do this. Equation (3.4.2) follows from (3.1.5) because L(x, 0) = 0. Define Apply D to (3.1.34) assuming L(x, y) twice differentiable with respect to x and y, in which case K(x, y) is also twice differentiable. (See Remark 3.4.12). By (3.4.27), DL = 0, so

Integrate by parts the last integral, (use (3.4.2)), and get

where K = K(x, x), L = L(x, y ) , and Subtract from (3.4.34) equation (3.1.34) multiplied by q (x), denote DK(x, y) – q(x)K(x, y) := v(x, y), and get:

provided that which is true because of (3.4.4). Equation (3.4.35) has only the trivial solution by Lemma 3.4.9. Thus v = 0, and equation (3.4.1) is derived. We have proved

130


Lemma 3.4.11. If L(x, y) is twice differentiable continuously or in K(x, y) of (3.1.34) solves (3.4.1), (3.4.2) with q given by (3.4.4).

then the solution

Remark 3.4.12. If a Fredholm equation

in a Banach space X depends on a parameter x continuously in the sense and at equation (3.4.36) has where N(A) = {u : Au = 0}, then the solution u(x) exists, is unique, and depends continuously on x in some neighborhood of If the data, that is, A(x) and f(x), have m derivatives with respect to x, then the solution has the same number of derivatives. Derivatives are understood in the strong sense for the elements of X and in the operator norm for the operator A(x). A simple proof of this known result is left to the reader. Hint. Use the identity which shows that if the operator is bounded, and is sufficiently small, then exists and is bounded. 3.4.5

Characterization of the class of spectral functions of the Sturm-Liouville operators

From Theorem 3.4.10 it follows that if (3.4.13) holds and then Condition (3.4.14) was used only to prove so if one starts with a then by diagram (3.1.35) one gets L(x, y) by formula (3.4.27), where If (3.4.14) holds, then one gets from L(x) a unique Recall that assumption is (3.4.13). Let be the assumption:

Theorem 3.4.13. If holds, and is a spectral function of assumption holds. Conversely, if assumptions and hold, then of

then is a spectral function

Proof. If holds and then by (4.1.4). If and hold, then by (3.1.32), because equation (3.1.34) is uniquely solvable, and (3.1.35) holds by Theorem 3.4.10.

3.4.6

Relation to the inverse scattering problem

Assume in this Section that spectral function is (3.1.20).

Then the scattering data

are (3.1.16) and the

131

Let us show how to get given If is given then finds then is recovered because

and J are known. If one

as follows from (3.1.19) and (3.1.15). To find f (k), consider the Riemann problem

which can be written as (see (3.3.29)):

Note that if The function zeros in and has similar properties in (3.4.40) have unique solutions:

is analytic in and has no Therefore problems (3.4.39) and

and

One can calculate f(x) for k > 0 by taking k = k + i0 in (3.4.43) or (3.4.44). Thus, to find given one goes through the following steps: (1) one finds J, (2) one calculates If then one calculates by formulas (3.4.41), (3.4.43), where is defined in (3.3.29), and by formula (3.4.37), and, finally, by formula (3.1.20). If

then one calculates f (k) by formulas (3.4.42) and (3.4.44), where is an arbitrary number such that If f (k) is found, one

132


calculates by formula (3.4.37), and then by formula (3.1.20). Note that in (3.4.42) depends on but f (k) in (3.4.44) does not. This completes the description of the step Let us show how to get given From formula (3.1.20) one finds J, and If then if Thus, if then is analytic in and vanishes at infinity. It can be found in from the values of its real part by Schwarz’s formula for the half-plane:

If

If

then

so

then the same formula (3.4.46) remains valid. One can see this because is analytic in has no zeroes in tends to 1 at infinity, and

if

Let us summarize the step one finds J, calculates f(k) by formula (3.4.46), and then and are calculated by formula (3.4.37). To calculate f(k) for k > 0 one takes k = k + i0 in (3.4.46) and gets:

3.5

3.5.1

INVERSE SCATTERING ON HALF-LINE

Auxiliary material

Transformation Operators

Theorem 3.5.1. If then there exists a unique operator I + A such that (3.1.24)– (3.1.27) hold, and A(x, y) solves the following Goursat problem:

133

Proof. Equations (3.5.1) and (3.5.2) are derived similarly to the derivation of the similar equations for K(x, y) in Theorem 3.4.1. Relations (3.5.3) follow from the estimates (3.1.25)–(3.1.26), which give more precise information than (3.5.3). Estimates (3.1.25)–(3.1.27) can be derived from the Volterra equation (3.1.27) which is solvable by iterations. Equation (3.1.27) can be derived, for example, similarly to the derivation of equation (3.4.8), or by substituting (3.1.24) into (3.1.6). A detailed derivation of all of the results of Theorem 3.5.1 can be found in [M]. Statement of the direct scattering problem on half-axis Existence and uniqueness of its solution

The direct scattering problem on half-line consists of finding the solution to the equation:

satisfying the boundary conditions at r = 0 and at

where is called the phase shift, and it has to be found. An equivalent formulation of (3.5.6) is:

where

where one gets

Clearly

is defined in (3.1.1), see also (3.1.10). From (3.5.8), (3.1.7) and (3.5.6)

Existence and uniqueness of the scattering solution existence and uniqueness of the regular solution (3.1.9).

follows from (3.5.8) because follows from (3.1.1) or from

Higher angular momenta

If one studies the three-dimensional scattering problem with a spherically-symmetric potential q (x) = q (r), then the scattering solution solves

134


the problem:

Here is the unit sphere in is given, is called the scattering amplitude. If q = q(r), then The converse is a theorem of Ramm [R139], p. 130. The scattering solution solves the integral equation:

It is known that

are orthonormal in spherical harmonics, summation over m in (3.5.13) is understood but not shown, and function. If q = q ( r), then

where

Relation (3.5.16) is equivalent to

and is the Bessel

135

similar to (3.5.8), which is (3.5.18) with If q = q(r), then the scattering amplitude can be written as

while in the general case q = q (x), one has

If q = q (r) then

in (3.5.18) are related to

in (3.5.19) by the formula

In the general case q = q (x), one has a relation between S -matrix and the scattering amplitude:

so that (3.5.21) is a consequence of (3.5.22) in the case q = q (r) : are the eigenvalues of S in the eigenbasis of the spherical harmonics. Since S is unitary, one has so for some real numbers which are called phase-shifts. These numbers are the same as in (3.5.17) (cf. (3.5.18)). From (3.5.21) one gets

The Green function

which solves the equation

can be written explicitly:

and the function

solves the equation:

136


The function is the Wronskian is defined in (3.5.25) and is the solution to (3.5.15) (with q = 0) with the asymptotics

Let

be the regular solution to (3.5.15) which is defined by the asymptotics as

Then

Lemma 3.5.2. One has:

where k > 0 is an arbitrary fixed number. We omit the proof of this lemma. Eigenfunction expansion

We assume that

and

let

be the resolvent kernel of this with respect to

and over

Then and divide by

The function gh is analytic with respect to except for the points because as

Integrate to get

on the complex plane with the cut which are simple poles of g h, and Therefore:

137

One has (cf. (3.1.10)):

Also

are defined in (3.1.15), and

Therefore

This implies (cf. (3.1.20), (3.1.19), (3.1.15)):

We have proved the eigenfunction expansion theorem for is dense in one gets the theorem for Theorem 3.5.3. If then (3.5.33) holds for any converge in sense. Parseval’s equality is:

3.5.2

Since this set

and the integrals

Statement of the inverse scattering problem on the half-line. Uniqueness theorem

In Section 1.2 the statement of the ISP is given. Let us prove the uniqueness theorem.

138


Theorem 3.5.4. If

generate the same data (3.1.16), then

Proof. We prove that the data (3.1.16) determine uniquely I (k), and this implies by Theorem 3.3.2. Claim 1. If (3.1.16) is given, then f(k) is uniquely determined. Assume there are and corresponding to the data (3.1.16). Then

The left-hand side of (3.5.36) is analytic in and tends to 1 as and the right-hand side of (3.5.36) is analytic in and tends to 1 as By analytic continuation is an analytic function in which tends to 1 as Thus, by Liouville theorem, so Claim 2. If (3.1.16) is given, then is uniquely defined. Assume there are and corresponding to (3.1.16). By the Wronskian relation (3.1.11), taking into account that by Claim 1, one gets

Denote

Then:

The function is analytic in and tends to zero as and has similar properties in It follows that so Let us check that is analytic in One has to check that This follows from (3.1.15): if and are given, then are uniquely determined. Let us check that as Using (3.3.5) it is sufficient to check that A(0, 0) is uniquely determined by f(k), because the integral in (3.3.5) tends to zero as by the Riemann-Lebesgue lemma. From (3.1.46), integrating by parts one gets:

Thus

Claim 2 is proved. Thus, Theorem 3.5.4 is proved.

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3.5.3

Reconstruction procedure

This procedure is described in (3.1.40). Let us derive equation (3.1.43). Our starting point is formula (3.5.34):

From (3.1.10) and (3.1.24) one gets:

Apply to (3.5.41) operator

acting on the functions of y, and get:

From (3.5.42), (3.5.43), and (3.1.24) with

one gets:

One has

From (3.1.41), (3.5.44) and (3.5.45) one gets (3.1.43). By continuity equation (3.1.43), derived for remains valid for Theorem 3.5.5. If solution in

and F is defined by (3.1.41) then equation (3.1.43) has a for any and this solution is unique.

Let us outline the steps of the proof.

140


Step 1. If

where

then F(x), defined by (3.1.41) satisfies the following estimates:

is defined in (3.1.25), and

Step 2. Equation

is of Fredholm type in and in It has only the trivial solution h = 0. Using estimates (3.5.46)–(3.5.48) and the criteria of compactness in one checks that is compact in these spaces for any The space because where We need the following lemma: Lemma 3.5.6. Let h solve (3.5.49). If If then

then

Proof. If h solves (3.5.49), then as Also as So the first claim is proved. Also as If

If

then

where

then

as

Lemma 3.5.7. If

solves (3.5.49) and

then h = 0.

Proof. By Lemma 3.5.6, It is sufficient to give a proof assuming x = 0. The function F(x) is real-valued, so one can assume that h is real-valued. Multiply (3.5.49) by h and integrate over to get

141

where

one gets

Also,

Therefore (3.5.50) implies

and

where

Since h is real valued, one has

of S implies and of (3.5.51), one has equality sign in the Cauchy inequality This means that and (3.1.16) implies

Because and vanishes as as

The unitarity Because

then and if is analytic in Also and vanishes is analytic in Therefore, by analytic continuation, is analytic in and

one has

vanishes as By Liouville theorem, then, by Theorem 3.3.3, works.

so

and If and the above argument

Because is compact in the Fredholm alternative is applicable to (3.5.49), and Lemma 3.5.7 implies that (3.1.43) has a solution in for any and this solution is unique. Note that the free term in (3.1.43) is – and this function of belongs to (cf. (3.5.48)). Because is compact in Lemma 3.5.7 and Lemma 3.5.6 imply existence and uniqueness of the solution to (3.1.43) in for any and for any Note that the solution to (3.1.43) in is the same as its solution in This is established by the argument used in the proof of Lemma 3.5.6. We give a method for the derivation of the estimates (3.5.46)–(3.5.48). Estimate (3.5.48) is an immediate consequence of the first estimate (3.5.46). Indeed,

Let us prove the first estimate (3.5.46). Put in (3.1.43)

142


Thus


where c = const > 0 stands for various constants and we have used the estimate

This estimate can be derived from (3.1.25). Write (3.1.43) as

Let us prove that equation (3.5.56) is uniquely solvable for F in p = 1 for all where N is a sufficiently large number. In fact, we prove that the operator in (3.5.56) has small norm in if N is sufficiently large. Its norm in is not more than

because

We have used estimate (3.1.25) above. The function so our claim is proved for Consider the case p = 1. One has the following upper estimate for the norm of the operator in (3.5.56) in Also Thus equation (3.5.56) is uniquely solvable in for all if N is sufficiently large. In order to finish the proof of the first estimate (3.5.46) it is sufficient to prove that This estimate is obvious for (cf. (3.1.41)). Let us prove it for Using (3.3.29), (3.3.31), (3.3.32), one gets

143

where all the Fourier transforms are taken of conclude that if one can prove that function. One has transform of

functions. Thus, one can is the Fourier and

From (3.1.25) it follows that We have proved that Differentiate (3.5.53) to get

or

One has

and

Use

Let us check that (3.1.26) and get The third estimate (3.5.46), cause mate rem 3.5.5 is proved. 3.5.4

The desired estimate is derived. follows from (3.5.60) beand The estifollows similarly from (3.5.53) and (3.1.25). Theo-

Invertibility of the steps of the reconstruction procedure

The reconstruction procedure is (3.1.40). 1. The step

is done by formula (3.1.41). To do the step one takes in (3.1.41) and finds and J . Thus is found and is found. From one finds 1 – S(k) by the inverse Fourier transform. So S(k) is found and the data (see (3.1.16)) is found

144


2. The step is done by solving equation (3.1.43). By Theorem 3.5.5 this equation is uniquely solvable in for all if that is, if F came from corresponding to

To do the step one finds the zeros of f (k) in the number J, are found by formula (3.1.15), where

Thus and by formula (3.1.41). We also give a direct way to do the step Write equation (3.1.43) with

The norm of the operator

in

and

then the numbers The numbers

as

is estimated as follows:

where and estimate (3.1.25) was used. If is sufficiently large then for because as if Therefore equation (3.5.62) is uniquely solvable in for all (by the contraction mapping principle), and so is uniquely determined for all Now rewrite (3.5.62) as

This is a Volterra equation for on the finite interval It is uniquely solvable since its kernel is a continuous function. One can put x = 0 in (3.5.64) and the kernel is a continuous function of v and and the right-hand side of (3.5.64) at x = 0 is a continuous function of Thus is uniquely recovered for all from A(x, y), Step is done.

145

is done by formula (3.1.42). The converse step is done by 3. The step solving Volterra equation (3.1.27), or, equivalently, the Goursat problem (3.5.1)– (3.5.3). We have proved: Theorem 3.5.8. If and are the corresponding data (3.1.16), then each step in (3.1.40) is invertible. In particular, the potential obtained by the procedure (3.1.40) equals to the original potential q. Remark 3.5.9. If and is the solution to (3.1.27), then satisfies equation (3.1.43) and, by the uniqueness of its solution, where A is the function obtained by the scheme (3.1.40). Therefore, the q obtained by (3.1.40) equals to the original q. Remark 3.5.10. One can verify directly that the solution A(x, y) to (3.1.43) solves the Goursat problem (3.5.1)–(3.5.3). This is done as in Section 3.4.4, Step 3. Therefore q (x), obtained by the scheme (3.1.40), generates the same A(x, y) which was obtained at the second step of this scheme, and therefore this q generates the original scattering data. Remark 3.5.11. The uniqueness Theorem 3.5.4 does not imply that if one starts with a computes the corresponding scattering data (3.1.16), and applies inversion scheme (3.1.40), then the q, obtained by this scheme, is equal to Logically it is possible that this q generates data which generate by the scheme (3.1.40) potential etc. To close this loop one has to check that This is done in Theorem 3.5.8, because 3.5.5

Characterization of the scattering data

In this Section we give a necessary and sufficient condition for the data (3.1.16) to be the scattering data corresponding to In Section 3.5.7 we give such conditions on for q to be compactly supported, or Theorem 3.5.12. If

then the following conditions hold:

(1) (3.1.22), index condition; (2) (3) (3.5.47) hold. Conversely, if

satisfies conditions (1)–(3), then

corresponds to a unique

Proof. The necessity of conditions (1)–(3) has been proved in Theorem 3.5.5. Let us prove the sufficiency. If conditions (1)–(3) hold, then the scheme (3.1.40) yields a unique potential, as was proved in Remark 3.5.9. Indeed, equation (3.1.43) is of Fredholm type in for every if F satisfies (3.5.47). Moreover, equation

146


(3.5.49) has only the trivial solution if conditions (1)–(3) hold. Every solution to (3.5.49) in is also a solution in and in and the proof of the uniqueness of the solution to (3.5.49) under the conditions (1)–(3) goes as in Theorem 3.5.5. The role of f (k) is played by the unique solution of the Riemann problem:

which consists of finding two functions and satisfying (3.5.65) such that is an analytic function in and is an analytic function in such that and if ind if ind S(k) = –2J. Existence of a solution to (3.5.65) follows from the non-negativity of ind S(–k) = –indS(k). Uniqueness of the solution to the above problem is proved as follows. Denote and Assume that and solve the above problem. Then (3.5.65) implies

The function is analytic in and tends to 1 at infinity in The function is analytic in and tends to 1 at infinity in Both functions agree on Thus is analytic in and tends to 1 at infinity. Therefore To complete the proof we need to check that q, obtained by (3.1.40), belongs to In other words, that To prove this, use (3.5.59) and (3.5.60). It is sufficient to check that and The first inclusion follows from Let us prove that One has Because and it follows that the limit exists. This limit has to be zero: if as and then Now The last inequality follows from (3.5.53): since it is sufficient to check that One has Here

Note that to

because

is monotonically decreasing and belongs

147

3.5.6

A new Marchenko-type equation

The basic result of this Section is: Theorem 3.5.13. Equation

holds, where A(y) := A(0, y), A(y) = 0 for y < 0, A(x, y) is defined in (3.1.24) and F(x) is defined in (3.1.41). Proof. Take the Fourier transform of (3.5.67) in the sense of distributions and get:

where, by (3.1.41),

Use (3.1.46), the equation

add 1 to both sides of (3.5.48), and get:

From (3.5.70) and (3.5.71) one gets:

where the equation was used. This equation holds because and the product makes sense because is analytic in Equation (3.5.72) holds obviously, and since each of our steps was invertible, equation (3.5.67) holds. Remark 3.5.14. Equation (3.5.67) has a unique solution A(y), such that and A(y) = 0 for y < 0. Proof. Equation (3.5.68) for y > 0 is identical with (3.1.43) because A ( – y ) = 0 for y > 0. Equation (3.1.43) has a solution in and this solution is unique, see Theorem 3.5.5. Thus, equation (3.5.68) cannot have more than one solution, because every solution A(y) = 0 for y < 0, of (3.5.68) solves (3.1.43), and (3.1.43) has no more than one solution. On the other hand, the solution

148


of (3.1.43) does exist, is unique, and solves (3.5.68), as was shown in the proof of Theorem 3.5.13. This proves Remark 3.5.14. 3.5.7

Inequalities for the transformation operators and applications

Inequalities for A and F

The scattering data (1.2.17) satisfy the following conditions: (A) (B) (C)

is a nonpositive integer,

If one wants to study the characteristic properties of the scattering data, that is, a necessary and sufficient condition on these data to guarantee that the corresponding potential belongs to a prescribed functional class, then conditions (A) and (B) are always necessary for a real-valued q to be in the usual class in the scattering theory, or in some other class for which the scattering theory is constructed, and a condition of the type (C) determines actually the class of potentials q. Conditions (A) and (B) are consequences of the selfadjointness of the Hamiltonian, fmiteness of its negative spectrum, and of the unitarity of the S-matrix. Our aim is to derive inequalities for F and A from equation (3.1.43). This allows one to describe the set of q, defined by (3.1.42). Let us assume:

The function is monotone decreasing, Equation (3.1.43) is of Fredholm type in and p = 1. The norm of the operator in (3.1.43) can be estimated:

Therefore (3.1.43) is uniquely solvable in

for any

if

This conclusion is valid for any F satisfying (3.5.75), and conditions (A), (B), and (C) are not used. Assuming (3.5.75) and (3.5.73) and taking let us derive inequalities for A = A(x, y). Define

149


Thus, if (3.5.75) holds, then

By c < 0 different constants depending on

are denoted. Let

Then (3.1.43) yields

Differentiate (3.1.43) with respect to x and y and get:

and

Denote

Then, using (3.5.79) and (3.5.76), one gets

and using (3.5.78) one gets:

So

150


so

Let y = x in (3.1.43), then differentiate (3.1.43) with respect to x and get:

From (3.5.76), (3.5.77), (3.5.82) and (3.5.83) one gets:

Thus,

provided that (3.5.73) implies ically, then

Assumption decreases monotonas Thus and because due to (3.5.73). Thus, (3.5.73) implies (3.5.76), (3.5.77), (3.5.80), (3.5.81), and (3.5.84), while (3.5.84) and (3.1.42) imply where and satisfies (3.5.75). Let us assume now that (3.5.76), (3.5.77), (3.5.81), and (3.5.82) hold, where and are some positive monotone decaying functions (which have nothing to do now with the function F, solving equation (3.1.43), and derive estimates for this function F. Let us rewrite (3.1.43) as: If

Let x + Y = z, s + y = v. Then,


and and

151

Thus, using (3.5.77) and (3.5.75), one obtains:

Also from (3.5.87) it follows that:


Let us summarize the results: Theorem 3.5.15. If

Conversely, if

and (3.5.73) hold, then one has:

and

then

In the next section we replace the assumption in this case is based on the Fredholm alternative.

by

The argument

Characterization of the scattering data revisited

First, let us give necessary and sufficient conditions on for q to be in These conditions are obtained in Section 3.5.5, but we give a short new argument. We assume throughout that conditions (A), (B), and (C) hold. These conditions are known to be

152


necessary for Indeed, conditions (A) and (B) are obvious, and (C) is proved in Theorem 3.5.15 and Theorem 3.5.18. Conditions (A), (B), and (C) are also sufficient for Indeed if they hold, then we prove that equation (3.1.43) has a unique solution in for all This was proved in Theorem 3.5.5, but we give another proof. Theorem 3.5.16. If (A), (B), and (C) hold, then (3.1.43) has a solution in and this solution is unique. Proof. Since to prove that

is compact in

for any

by the Fredholm alternative it is sufficient

implies h = 0. Let us prove it for x = 0. The proof is similar for x > 0. If because If then because Thus, if and solves (3.5.94), then Denote Then,

then

Since F (x) is real-valued, one can assume h real-valued. One has, using Parseval’s equation:

Thus, using (3.5.95), one gets

where we have used real-valuedness of h, i.e., one has where (A) was used. Since Thus, so the equality sign is attained in the Cauchy inequality. Therefore, By condition (B), the theory of Riemann problem guarantees existence and uniqueness of an analytic in function such that

153

and

is analytic in Here the property

in is used.

One has

The function is analytic in and is analytic in they agree on so is analytic in Since and it follows that Thus, and, consequently, h(x)=0, as claimed. Theorem 3.5.16 is proved. The unique solution to equation (3.1.44) satisfies the estimates given in Theorem 3.5.15. In the proof of Theorem 3.5.15 the estimate was established. So, by (3.1.42), The method developed in the previous Section gives accurate information about the behavior of q near infinity. An immediate consequence of Theorem 3.5.15 and Theorem 3.5.16 is: Theorem 3.5.17. If (A), (B), and (C) hold, then q, obtained by the scheme (3.1.40) belongs to Investigation of the behavior of q (x) on requires additional argument. Instead of using the contraction mapping principle and inequalities, one has to use the Fredholm theorem, which says that for any where the operator norm is taken for acting in and and the constant c does not depend on Such an analysis yields: Theorem 3.5.18. If and only if (A), (B), and (C) hold, then Proof. It is sufficient to check that Theorem 3.5.15 holds with To get (3.5.76) with one uses (3.1.44) and the estimate:

where the constant c > 0 does not depend on x. Similarly:

replacing

154




Similarly, from (3.5.83) and (3.5.96)–(3.5.99) one gets (3.5.84). Then one checks (3.5.85) as in the proof of Theorem 3.5.15. Consequently Theorem 3.5.15 holds with Theorem 3.5.18 is proved. Compactly supported potentials

In this Section necessary and sufficient conditions are given for if x > a, Recall that the Jost solution is:

Lemma 3.5.19. If F(x + y) = 0 for

then for x > a, A(x, y) = 0 for (cf (3.1.43)), and F(x) = 0 for

Thus, (3.1.43) with x = 0 yields A(0, y) := A(y) = 0 for

The Jost function

is an entire function of exponential type that is, S(k) = f ( – k ) / f ( k ) is a meromorphic function in In (3.5.102) space, and the inclusion (3.5.102) follows from Theorem 3.5.15. Let us formulate the assumption (D):

and is the Sobolev

(D) the Jost function f (k) is an entire function of exponential type Theorem 3.5.20. Assume (A), (B), (C) and (D). Then then (A), (B), (C) and (D) hold.

Conversely, if

Necessity. If then (A), (B) and (C) hold by Theorem 3.5.18, and (D) is proved in Lemma 3.5.19. The necessity is proved. Sufficiency. If (A), (B) and (C) hold, then One has to prove that q = 0 for x > a. If (D) holds, then from the proof of Lemma 3.5.19 it follows that A(y) = 0 for We claim that F (x) = 0 for

155

If this is proved, then (3.1.43) yields A(x, y) = 0 for and so q = 0 for x > a by (3.1.42). Let us prove the claim. Take x > 2a in (3.1.41). The function 1 – S(k) is analytic in except for J simple poles at the points If x > 2a then one can use the Jordan lemma and the residue theorem and get:

Since f (k) is entire, the Wronskian formula

is valid on

because

and at

it yields:

This and (3.5.103) yield

Thus, for x > 2a. The sufficiency is proved. Theorem Theorem 3.5.20 is proved. In [M] a condition on which guarantees that q = 0 for x > a, is given under the assumption that there is no discrete spectrum, that is Square integrable potentials

Let us introduce conditions (3.5.104)–(3.5.106)

Theorem 3.5.21. If (A), (B), (C), and any one of the conditions (3.5.104)–(3.5.106) hold, then Proof. We refer to [R139] for the proof.

156

3.6


INVERSE SCATTERING PROBLEM WITH FIXED-ENERGY PHASE SHIFTS AS THE DATA

3.6.1

Introduction

In Subsection 3.5.1 the scattering problem for spherically symmetric q was formulated, see (3.5.15)–(3.5.17). The are the fixed-energy (k = const > 0) phase shifts. Define

where

is a regular solution to

such that

and is the Bessel function. In (3.6.3) is the transformation kernel, I + K is the transformation operator. In (3.6.2) we assume that k = 1 without loss of generality. The is uniquely defined by its behavior near the origin:

For

we will use the known formula ([GR, 8.411.8]):

where is the gamma-function. The inverse scattering problem with fixed-energy phase shifts as the data consists of finding q (r) from these data. We assume throughout this chapter that q (r) is a real-valued function, q (r) = 0 for r > a,

Conditions (3.6.6) imply that In the literature there are books [CS] and [N] where the Newton-Sabatier (NS) theory is presented, and many papers were published on this theory, which attempts to solve the above inverse scattering problem with fixed-energy phase shifts as the data. In Section 3.6.5 it is proved that the NS theory is fundamentally wrong and is not an inversion method. The main results of this Chapter are Theorems 3.6.2–3.6.4, and the proof of the fact that the Newton-Sabatier theory is fundamentally wrong.

157

3.6.2

Existence and uniqueness of the transformation operators independent of angular momentum

The existence and uniqueness of in (3.6.3) we prove by deriving a Goursat problem for it, and investigating this problem. Substitute (3.6.3) into (3.6.2), drop index for notational simplicity and get

We assume first that is twice continuously differentiable with respect to its variables in the region This assumption requires extra smoothness of If q (r) satisfies condition (3.6.6), then equation (3.6.13) below has to be understood in the sense of distributions. Eventually we will work with an integral equation (3.6.40) (see below) for which assumption (3.6.6) suffices. Note that

provided that

We assume (3.6.9) to be valid. Denote

Then

Combining (3.6.7)–(3.6.11) and writing again

in place of u, one gets

158


Let us prove that (3.6.12) implies:

This proof requires a lemma. Lemma 3.6.1. Assume that

and

then

Proof. Equations (3.6.15) and (3.6.5) imply:

Therefore

Recall that the Legendre polynomials are defined by the formula

and they form a complete system in Therefore (3.6.17) implies

Equation (3.6.19) implies

If

159

and

Therefore A(r) = 0. Also because the left-hand side of (3.6.20) is an entire function of t, which vanishes on the interval [–1, 1] and, consequently, it vanishes identically, so that and therefore Lemma 3.6.1 is proved. We prove that the problem (3.6.13), (3.6.14), (3.6.9), which is a Goursat-type problem, has a solution and this solution is unique in the class of functions which are twice continuously differentiable with respect to and r, In this section we assume that This assumption implies that is twice continuously differentiable. If (3.6.6) holds, then the arguments in this section which deal with integral equation (3.6.40) remain valid. Specifically, existence and uniqueness of the solution to equation (3.6.40) is proved under the only assumption as far as the smoothness of q (r) is concerned. By a limiting argument one can reduce the smoothness requirements on q to the condition (3.6.6), but in this case equation (3.6.13) has to be understood in distributional sense. Let us rewrite the problem we want to study:

The difficulty in the study of this Goursat-type problem comes from the fact that the coefficients in front of the second derivatives of the kernel are variable. Let us reduce problem (3.6.22)–(3.6.24) to the one with constant coefficients. To do this, introduce the new variables:

Note that

and

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Let

A routine calculation transforms equations (3.6.22)–(3.6.24) to the following ones:

where g (r) is defined in (3.6.23). Here we have defined

and took into account that that Note that

implies

while

implies, for any fixed

for any and B > 0, where c (A, B) > 0 is a constant. To get rid of the second term on the left-hand side of (3.6.29), let us introduce the new kernel by the formula:

Then (3.6.29)–(3.6.31) can be written as:

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We want to prove existence and uniqueness of the solution to (3.6.36)–(3.6.38). In order to choose a convenient Banach space in which to work, let us transform problem (3.6.36)–(3.6.38) to an equivalent Volterra-type integral equation. Integrate (3.6.36) with respect to from 0 to and use (3.6.37) to get

Integrate (3.6.39) with respect to

from

to

and use (3.6.39) to get

where

Consider the space X of continuous functions such that for any B > 0 and any

defined in the half-plane one has

where is a number which will be chosen later so that the operator V in (3.6.40) will be a contraction mapping on the Banach space of functions with norm (3.6.42) for a fixed pair A, B. To choose let us estimate the norm of V. One has:

where c > 0 is a constant depending on A, B and

Indeed, one has:

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and, using the substitution

one gets:

From these estimates inequality (3.6.44) follows. It follows from (3.6.44) that V is a contraction mapping in the space of continuous functions in the region with the norm (3.6.42) provided that

Therefore equation (3.6.40) has a unique solution

in the region

for any real A and B > 0 if (3.6.45) holds. This means that the above solution is defined for any and any Equation (3.6.40) is equivalent to problem (3.6.36)–(3.6.38) and, by (3.6.35), one has:

Therefore we have proved the existence and uniqueness of that is, of the kernel of the transformation operator (3.6.3). Recall that and are related to and by formulas (3.6.26). Let us formulate the result: Theorem 3.6.2. The kernel of the transformation operator (3.6.3) solves problem (3.6.22)–(3.6.24). The solution to this problem does exist and is unique in the class of twice continuously differentiable functions for any potential If then has first derivatives which are bounded and equation (3.6.22) has to be understood in the sense of distributions. The following estimate holds for any r > 0 :

Proof of Theorem 3.6.2. We have already proved all the assertions of Theorem 3.6.2 except for the estimate (3.6.48). Let us prove this estimate.

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Note that

Indeed, if and

is fixed, then, by (3.6.26),

Therefore Thus:

The following estimate holds:

where j = 1 , 2 , are arbitrarily small numbers and is defined in formula (3.6.57) below, see also formula (3.6.54) for the definition of Estimate (3.6.51) is proved below, in Theorem 3.6.2. From (3.6.50) and estimate (3.6.61) (see below) estimate (3.6.48) follows. Indeed, denote by I the integral on the right-hand side of (3.6.50). Then, by (3.6.61) one gets:

Theorem 3.6.2 is proved. Theorem 3.6.3. Estimate (3.6.51) holds. Proof of Theorem 3.6.3. From (3.6.40) one gets:

where

(see (3.6.37)), and

It is sufficient to consider inequality (3.6.53) with if to (3.6.53) satisfies (3.6.51) with then the solution with any satisfies (3.6.51) with

and the solution of (3.6.53)

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Therefore, assume that

then (3.6.53) reduces to:

Inequality (3.6.51) follows from (3.6.55) by iterations. Let us give the details. Note that

Here we have used the notation

and the fact that also that Furthermore,

is a monotonically increasing function, since for any

Note

Let us prove by induction that

For n = 1 and n = 2 we have checked (3.6.59). Suppose (3.6.59) holds for some n, then

By induction, estimate (3.6.58) is proved for all n = 1, 2, 3, . . . . Therefore (3.6.55) implies

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where we have used Theorem 2 from [Lev, section 1.2], namely the order of the entire function and its type is 2. The constant c > 0 in (3.6.51) depends on j = 1, 2. Recall that the order of an entire function is the number where The type of is the number It is known [Lev], that if is an entire function, then its order and type can be calculated by the formulas:

If ved.

then the above formulas yield

3.6.3

and

Theorem 3.6.3 is pro-

Uniqueness theorem

Denote by

any fixed subset of the set

of integers {0, 1, 2, ...} with the property:

Theorem 3.6.4. ([R393]) Assume that q satisfies (3.6.6) and (3.6.63) holds. Then the data determine uniquely. The idea of the proof is based on Property C-type argument. Step 1: If and relation holds for

generate the same data

then the following orthogonality

where is the scattering solution corresponding to j = 1 , 2. Step 2: Define where is the Gamma-function. Check that is holomorphic in and are real numbers, (where N is the Nevanlinna class in that is

where and, by property

vanishes Theorem 3.6.4 is proved.

then

in

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3.6.4


Why is the Newton-Sabatier (NS) procedure fundamentally wrong?

The NS procedure is described in [N] and [CS]. A vast bibliography of this topic is given in [CS] and [N]. Below two cases are discussed. The first case deals with the inverse scattering problem with fixed-energy phase shifts as the data. This problem is understood as follows: an unknown spherically symmetric potential q from an a priori fixed class, say a l = 0, 1 , 2 , . . . . The standard scattering class, generates fixed-energy phase shifts inverse scattering problem consists of recovery of q from these data. The second case deals with a different problem: given some numbers l= 0, 1 , 2 , . . . , which are assumed to be fixed-energy phase shifts of some potential q , from a class not specified, find some potential which generates fixed-energy phase shifts equal to l = 0, 1 , 2 , . . . . This potential may have no physical interest because of its non-physical behavior at infinity or other undesirable properties. Also, in the second case it may be that We first discuss NS procedure assuming that it is intended to solve the inverse scattering problem in case 1. Then we discuss NS procedure assuming that it is intended to solve the problem in case 2. Discussion of case 1:

In [N2] and [N] a procedure was proposed by R. Newton for inverting fixed-energy l = 0, 1 , 2 , . . . , corresponding to an unknown spherically symmetric phase shifts potential q(r). R. Newton did not specify the class of potentials for which he tried to develop an inversion theory and did not formulate and proved any results which would justify the inversion procedure he proposed (NS procedure). His arguments are based on the following claim, which is implicit in his works, but crucial for the validity of NS procedure: Claim N1: The basic integral equation.

is uniquely solvable for all r > 0.

Here

are real numbers, the energy is fixed: is taken without loss of generality, are the Bessel functions. It is assumed in [CS], p. 196, that If equation (3.6.65) is uniquely solvable for all r > 0, then the potential that NS

167

procedure yields, is defined by the formula:

The R. Newton’s ansatz (3.6.65)–(3.6.66) for the transformation kernel K (r, s) of the Schrödinger operator, corresponding to some q (r ), namely, that K (r, s) is the unique solution to (3.6.65)–(3.6.66), is not correct for a generic potential, as follows from our argument below (see the justification of Conclusions). If for some r > 0 equation (3.6.65) is not uniquely solvable, then NS procedure breaks down: it leads to locally non-integrable potentials for which the scattering theory is, in general, not available.

In the original paper [N2] and in his book [N] R. Newton did not study the question fundamental for any inversion theory: does the reconstructed potential generate the data from which it was reconstructed? In [CS, p. 205], there are two claims: Claim (i): generates the original shifts “provided that are not “exceptional” ”, and Claim (ii): the NS procedure “yields one (only one) potential which decays faster than ” and generates the original phase shifts If one considers NS procedure as a solution to inverse scattering problem of finding an unknown potential q from a certain class, for example from the fixed-energy phase shifts, generated by this q , then the proof, given in [CS], of Claim (i) is not convincing: it is not clear why the potential obtained by NS procedure, has the transformation operator generated by the potential corresponding to the original data, that is, to the given fixed-energy phase shifts. In fact, as follows from Proposition 3.6.5 below, the potential cannot generate the kernel K (r, s) of the transformation operator corresponding to a generic original potential Claim (ii) is incorrect because the original generic potential generates the phase shifts and if the potential obtained by NS procedure and therefore not equal to q (r) by Proposition 3.6.5 then one has two different potentials q (r) and which both decay faster than and both generate the original phase shifts contrary to Claim (ii). Our aim is to formulate and justify the following Conclusions: Claim N1 and ansatz (3.6.65)–(3.6.66) are not proved by R. Newton and, in general, are wrong. Moreover, one cannot approximate with a prescribed accuracy in the norm a generic potential by the potentials which might possibly be obtained by the NS procedure. Therefore NS procedure cannot be justified even as an approximate inversion procedure.

Let us justify these conclusions: Claim N1 formulated above (and basic for NS procedure) is wrong, in general, for the following reason:

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Given fixed-energy phase shifts, corresponding to a generic potential either cannot carry through NS procedure because:

one

(a) the system (12.2.5a) in [CS], which should determine numbers in formula (3.6.66), given the phase shifts may be not solvable, or (b) if the above system is solvable, equation (3.6.65) may be not (uniquely) solvable for some r > 0 , and in this case NS procedure breaks down since it yields a potential which is not locally integrable. If equation (3.6.65) is solvable for all r > 0 and yields a potential by formula (3.6.67), then this potential is not equal to the original generic potential as follows from Proposition 3.6.5: Proposition 3.6.5. If equation (3.6.65) is solvable for all r > 0 and yields a potential by formula (3.6.67), then this is a restriction to (0, of a function analytic in a neighborhood of (0, Since a generic potential is not a restriction to (0, of an analytic function, one concludes that even if equation (3.6.65) is solvable for all r > 0, the potential defined by formula (3.6.67), is not equal to the original generic potential and therefore the inverse scattering problem of finding an unknown from its fixed-energy phase shifts is not solved by NS procedure. The ansatz (3.6.65)–(3.6.66) for the transformation kernel is, in general, incorrect, as follows also from Proposition 3.6.5. Indeed, if the ansatz (3.6.65)—(3.6.66) would be true and formula (3.6.67) would yield the original generic q , that is this would contradict Proposition 3.6.5 If formula (3.6.67) would yield a which is different from the original generic q , then NS procedure does not solve the inverse scattering problem formulated above. Note also that it is proved in [R192] that independent of the angular momenta l transformation operator, corresponding to a generic does exist, is unique, and is defined by a kernel K (r , s) which cannot have representation (3.6.66), since it yields by the formula similar to (3.6.67) the original generic potential q , which is not a restriction of an analytic in a neighborhood of (0, function to (0, The conclusion, concerning impossibility of approximation of a generic by potentials which can possibly be obtained by NS procedure, is proved in Claim 7, see proof of Claim 7 below. Thus, our conclusions are justified. Let us give some additional comments concerning NS procedure. Uniqueness of the solution to the inverse problem in case 1 was first proved by A. G. Ramm in 1987 (see [R100], [R109]) for a class of compactly supported potentials, while R. Newton’s procedure was published in [N2], when no uniqueness results for this inverse problem were known. It is still an open problem if for the standard in scattering theory class of potentials the uniqueness theorem for the solution of the above inverse scattering problem holds.

169

We discuss the inverse scattering problem with fixed-energy phase shifts (as the data) for potentials because only for this class of potentials a general theorem of existence and uniqueness of the transformation operators, independent of the angular momenta l , has been proved, see [R192]. In [N2], [N], and in [CS] this result was not formulated and proved, and it was not dear for what class of potentials the transformation operators, independent of l , do exist. For slowly decaying potentials the existence of the transformation operators, independent of is not established, in general, and the potentials, discussed in [CS] and [N] in connection with NS procedure, are slowly decaying. Starting with [N2], [N], and [CS] Claim N1 was not proved or the proofs given (see [CT]) were incorrect (see [R207]). This equation is uniquely solvable for sufficiently small r > 0, but, in general, it may be not solvable for some r > 0, and if it is solvable for all r > 0, then it yields by formula (3.6.67) a potential which is not equal to the original generic potential as follows from Proposition 3.6.5. Existence of “transparent” potentials is often cited in the literature. A “transparent” potential is a potential which is not equal to zero identically, but generates the fixedenergy shifts which are all equal to zero. In [CS, p. 207], there is a remark concerning the existence of “transparent” potentials. This remark is not justified because it is not proved that for the values used in [CS, p. 207], equation (3.6.65) is solvable for all r > 0. If it is not solvable even for one r > 0 , then NS procedure breaks down and the existence of transparent potentials is not established. In the proof, given for the existence of the “transparent” potentials in [CS, p. 197], formula (12.3.5), is used. This formula involves a certain infinite matrix M. It is claimed in [CS, p. 197], that this matrix M has the property MM = I, where I is the unit matrix, and on [CS, p. 198], formula (12.3.10), it is claimed that a vector exists such that Mv = 0. However, then MMv = 0 and at the same time which is a contradiction. The difficulties come from the claims about infinite matrices, which are not formulated clearly: it is not clear in what space M, as an operator, acts, what is the domain of definition of M, and on what set of vectors formula (12.3.5) in [CS] holds. The construction of the “transparent” potential in [CS] is based on the following logic: take all the fixed-energy shifts equal to zero and find the corresponding from the infinite linear algebraic system (12.2.7) in [CS]; then construct the kernel f (r , s) by formula (3.6.66) and solve equation (3.6.65) for all r > 0, finally construct the “transparent” potential by formula (3.6.67). As was noted above, it is not proved that equation (3.6.65) with the constructed above kernel f (r , s) is solvable for all r > 0. Therefore the existence of the “transparent” potentials is not established. The physicists have been using NS procedure without questioning its validity for several decades. Apparently the physicists still believe that NS procedure is “an analog of the Gel’fand-Levitan method” for inverse scattering problem with fixed-energy phase shifts as the data. In fact, the NS procedure is not a valid inversion method. Since modifications of NS procedure are still used by some physicists, who believe that this procedure is an inversion theory, the author pointed out some questions concerning this procedure. This concludes the discussion of case 1.

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Discussion of case 2: Suppose now that one wants just to construct a potential corresponding to some q .

which generates the phase shifts

This problem is actually not an inverse scattering problem because one does not recover an original potential from the scattering data, but rather wants to construct some potential which generates these data and may have no physical meaning. Therefore this problem is much less interesting practically than the inverse scattering problem. However, NS procedure does not solve this problem either: there is no guarantee that this procedure is applicable, that is, that the steps a) and b), described in the justification of the conclusions, can be done, in particular, that equation (3.6.65) is uniquely solvable for all r > 0.

If these steps can be done, then one needs to check that the potential obtained by formula (3.6.67), generates the original phase shifts. This was not done in [N2] and [N]. This concludes the discussion of case 2. The rest of the paper contains formulation and proof of Remark 3.6.6 and Claim 7. It was mentioned in [N3] that if then the numbers in formula (3.6.66) cannot satisfy the condition This observation can be obtained also from the following Remark 3.6.6. For any potential such that equation (3.6.65) is not solvable for some r > 0 and any choice of

the basic such that

Since generically, for one has this gives an additional illustration to the conclusion that equation (3.6.65), in general, is not solvable for some r > 0. Conditions and are incompatible. In [CS, p. 196], a weaker condition is used, but in the examples ([CS, pp. 189–191]), for all so that in all of these examples. Claim 7. The set of the potentials procedure, is not dense (in the norm

which can possibly be obtained by the NS in the set

Let us prove Remark 3.6.6 and Claim 7. Proof of Remark 3.6.6. Writing (3.6.67) as one gets the following relation:

and assuming

If (3.6.65) is solvable for all r > 0, then from (3.6.66) and (3.6.65) it follows that where so that I – K

171

is a transformation operator, where K is the operator with kernel K(r, s), where

where are the phase shifts at k = 1 and prove that Thus, if

and

is the Jost function at k = 1. One can then

If then (3.6.69) contradicts (3.6.68). It follows that if then equation (3.6.65) cannot be uniquely solvable for all r > 0, so that NS procedure cannot be carried through if and This proves Remark 3.6.6. Proof of Claim 7. Suppose that and because otherwise NS procedure cannot be carried through as was proved in Remark 3.6.6. If then there is also no guarantee that NS procedure can be carried through. However, we claim that if one assumes that it can be carried through, then the set of potentials, which can possibly be obtained by NS procedure, is not dense in in the norm In fact, any potential q such that and the set of such potentials is dense in cannot be approximated with a prescribed accuracy by the potentials which can be possibly obtained by the NS procedure. Let us prove this. Suppose that

where the potentials We assume verge in the norm to

are obtained by the NS procedure, so that because otherwise obviously cannot conDefine a linear bounded on functional

where The potentials which can possibly be obtained by the NS procedure, belong to the null-space of f, that is f (v) = 0. If then Since f is a linear bounded functional and one gets: So if then Therefore, no potential with can be approximated arbitrarily accurately by a potential which can possibly be obtained by the NS procedure. Claim 7 is proved.

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3.6.5


Formula for the radius of the support of the potential in terms of scattering data

The aim of this section is to prove formula (3.6.70). Let us make the following assumption. Assumption (A): the potential q (r) , is spherically symmetric, real-valued, and q (r) = 0 for r > a , but on for all sufficiently small The number a > 0 we call the radius of compactness of the potential, or simply the radius of the potential. Let denote the scattering amplitude corresponding to the potential q at a fixed energy Without loss of generality let us take k = 1 in what follows. By the unit vectors in the direction of the scattered, respectively, incident wave, are meant, is the unit sphere in Let us use formulas (3.5.19) and (3.5.20). It is of interest to obtain some information about q from the (fixed-energy) scattering data, that is, from the scattering amplitude or, equivalently, from the coefficients Very few results of such type are known. A result of such type is a necessary and sufficient condition for it was proved [R139, p. 131] (see Section 5.7, Vol. 2), that if and only if Of course, the necessity of this condition was a common knowledge, but the sufficiency, that is, the implication: is a deeper result [R139]. A (modified) conjecture from [R139, p. 356] says that if the potential q (x) is compactly supported, and a > 0 is its radius (defined for non-spherically symmetric potentials in the same way as for the spherically symmetric), then

where are the fixed-energy (k = 1) phase shifts. We prove (3.6.70) for the spherically symmetric potentials q = q (r). If q = q (r) then where depends only on l and k, but not on or Since k = 1 is fixed, depends only on l for q = q (r). Assuming q = q(r), one takes and calculates where

Here we have used

formula (14.4.46) in [RKa, p. 413], and

are the Gegenbauer polynomials (see

[RKa, p. 408]). Since where are the Legendre polynomials (see, e.g., [RKa, p. 409]), one gets: Formula (3.6.70) for q = q (r) can be written as Indeed, as is well known (see, e.g., [MPr], p. 261). Thus and formula for (3.6.70) yields:

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Note that assumption (A) implies the following assumption: Assumption the potential q (r) does not change sign in some left neighborhood of the point a. This assumption in practice is not restrictive, however, as shown in [R139, p. 282], the potentials which oscillate infinitely often in a neighborhood of the right end of their support, may have some new properties which the potentials without this property do not have. For example, it is proved in [R139], p. 282, that such infinitely oscillating potentials may have infinitely many purely imaginary resonances, while the potentials which do not change sign in a neighborhood of the right end of their support cannot have infinitely many purely imaginary resonances. Therefore it is of interest to find out if assumption is necessary for the validity of (3.6.71). The main result is: Theorem 3.6.7. Let assumption (A) hold. Then formula (3.6.71) holds with by lim.

replaced

This result can be stated equivalently in terms of the fixed-energy phase shift

Below, we prove an auxiliary result: Lemma 3.6.8. If some interval where

is real-valued and does not change sign in and a is the radius of q, then

Below we prove (3.6.72) and, therefore, (3.6.70) for spherically symmetric potentials. Proof of Lemma 3.6.8. First, we obtain a slightly different result than (3.6.73) as an immediate consequence of the Paley-Wiener theorem. Namely, we prove Lemma 3.6.8 with a continuous parameter t replacing the integer and replacing lim. This is done for and without additional assumptions about However, we are not able to prove Lemma 3.6.8 assuming only that Since q (r) is compactly supported, one can write

Let us recall that Paley-Wiener theorem implies the following claim (see [Lev]):

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If and

is the smallest interval containing the support of g (u), then

Thus, using (3.6.74) and (3.6.75), one gets:

Formula (3.6.76) is similar to (3.6.73) with m replaced by t and lim replaced by Remark 3.6.9. We have used formula (3.6.75) with theorem it is assumed that However, for one has:

Thus,

while in the Paley-Wiener on for any

and

Therefore formula (3.6.76) follows. To prove (3.6.73), we use a different approach independent of the Paley-Wiener theorem. We will use (3.6.73) below, in formula (3.6.87). In this formula the role of q (r) in (3.6.73) is played by where Let us prove (3.6.73). Assume without loss of generality that near a. Let We have where is an arbitrary small positive number. Thus, I > 0 for all sufficiently large m, and One has and as Since is arbitrary small, it follows that This completes the proof of (3.6.73). Lemma 3.6.8 is proved. Proof of formula (3.6.72). From (3.5.19) and (3.5.23) denoting

one gets where,

and

175

where and

as are the spherical Bessel functions, solves (3.5.15)–(3.5.17), and the integral

where

and is the Hankel function. It is known [RKa, p. 407] that

and [AlR, Appendix 4]:

It follows from (3.6.81) that is sufficiently large. Define

does not have zeros on any fixed interval (0, a] if l Then (3.6.78) yields


Thus

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This implies that for sufficiently large equation (3.6.83) has small kernel and therefore is uniquely solvable in C(0, a) and one has

uniformly with respect to In the book [N, formula (12.180)], which gives the asymptotic behavior of for large l, is misleading: the remainder in this formula is of order which is much greater, in general, than the order of the main term in this formula. That is why we had to find a different approach, which yielded formula (3.6.87). From (3.6.77), (3.6.80), (3.6.81), and (3.6.87) one has:

Therefore, using (3.6.73), one gets:

Theorem 3.6.7 is proved. Remark 3.6.10. Since as and formulas (3.6.89) and sin imply phase shift at a fixed positive energy. This is formula (3.6.72). 3.7

as where

is the

INVERSE SCATTERING WITH “INCOMPLETE DATA”

3.7.1 Uniqueness results

Consider equation (3.1.3) on the interval [0, 1] with boundary conditions u (0) = u(1) = 1 (or some other selfadjoint homogeneous separated boundary conditions), and Fix Assume q (x) on [b, 1] is known and a subset of the eigenvalues of the operator corresponding to the chosen boundary conditions is known. Here

177

We assume sometimes that

Theorem 3.7.1. If (3.7.1) holds and then the data determine q (x) on [0, b] uniquely. If (3.7.1) and (3.7.2) hold, the same conclusion holds also if The number is “the percentage” of the spectrum of l which is sufficient to determine q on [0, b] if and (3.7.2) holds. For example, if and then “one spectrum” determines q on the half-interval If then “half of the spectrum” determines q on Of course, q is assumed known on [b, 1]. If b = 1, then “two spectra” determines q on the whole interval. By “two spectra” one means the set where is the set of eigenvalues of corresponding to the same boundary condition u(0) = 0 at one end, say at x = 0, and some other selfadjoint boundary condition at the other end, say or The last result is a well-known theorem of Borg ([B]), which was strengthened in [M], where it is proved that not only the potential but the boundary conditions as well are uniquely determined by two spectra. A version of “one spectrum” result was mentioned in [L1, p. 81]. In [L] and [M] there is an algorithm for recovery of q from two spectra. A numerical method for solving this inverse problem is given in [RuS]. Proof of Theorem 3.7.1. First, assume the same data, then as above, one gets

If there are

where

Thus

The function is an entire function of of order and is an entire even function of k of exponential type

The indicator of g is defined by the formula

and

which produce

(see (3.1.11) with One has

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where estimate

Since

one gets from (3.7.5) and (3.7.6) the following

It is known [Lev, formula (4.16)] that for any entire function type one has:

where n(r) is the number of zeros of g(k) in the disk

of exponential

From (3.7.7) one gets

From (3.7.2) and the known asymptotics of the Dirichlet eigenvalues:

one gets for the number of zeros the estimate

From (3.7.8), (3.7.9) and (3.7.11) it follows that

Therefore, if then Theorem 3.7.1 is proved in the case Assume now that and

If

We claim that if an entire function

and (3.7.13) holds, then above.

then, by property

in (3.7.3) of order

p (x) = 0.

vanishes at the points

If this is proved, then Theorem 3.7.1 is proved as

179

Let us prove the claim. Define

and recall that

Since

is entire, of order

the function

Let us use a Phragmen-Lindelöf lemma.

Lemma 3.7.2. [Lev, Theorem 1.22] If an entire function then If, in addition

We use this lemma to prove that Theorem 3.7.1 is proved. The function is entire of order Let us check that

of order as

has the property then

If this is proved then

and that

One has, using (3.7.5), (3.7.15), (3.7.16) and taking into account that

Here we have used elementary inequalities:

and

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with and the assumption (3.7.13). We also used the relation:

Estimate (3.7.20) implies (3.7.18) and (3.7.19). An estimate similar to (3.7.20) has been used in the literature (see [GS]). Theorem 3.7.1 is proved. 3.7.2

Uniqueness results: compactly supported potentials

Consider the inverse scattering problem of Section 3.5.2 and assume

Theorem 3.7.3. If satisfies (3.7.22), then any one of the data S(k), determine q uniquely.

f (k),

Proof. We prove first Note that without assumption (3.7.22), or an assumption which implies that f (k) is an entire function on the result does not hold. If (3.7.22) holds, or even a weaker assumption:

then f (k), the Jost function (3.1.5), is an entire function of k, and S(k) is a meromorphic function on with the only poles in at the points Thus, and J are determined by S(k). Using (3.1.15) and (3.1.11), which holds for all because f (k) and f (x, k) are entire functions of k, one finds Thus all the data (3.1.16) are found from S(k) if (3.7.22) (or (3.7.23)) holds. If the data (3.1.16) are known, then q is uniquely determined, see Theorem 3.5.4. If is given, then so If f (k) is given, then so If is given, then one can uniquely find f(k) from (3.1.11). Indeed, assume there are two f (k), and corresponding to the given Subtract from (3.1.11) with equation (3.1.11) with denote and get (*) Since and one can conclude that w = 0 if one can check that is analytic in The function has at most finitely many zeros in and these zeros are simple. From (*) one concludes that if then because if then (see (3.1.11)). Thus is analytic in Similarly is analytic in These two functions agree on the real axis, so, by analytic continuation, the function is analytic in and vanishes at infinity. Thus it vanishes identically. So w (k) = 0, and f (k) is uniquely determined by Thus Theorem 3.7.3 is proved.

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In [R181], [R184] a convergent iterative method is given for finding a compactly supported (or decaying faster than exponential) potential q (x) from its S-matrix alone. 3.7.3

Inverse scattering on the full line by a potential vanishing on a half-line

The scattering problem on the full line consists of finding the solution to:

where r (k) and t (k) are, respectively, the reflection and transmission coefficients. The above scattering problem describes plane wave scattering by a potential, the plane wave is incident from in the positive direction of the x-axis. The inverse scattering problem consists of finding q(x) given the scattering data

where are norming constants, and are the negative eigenvalues of the operator It is known [M], that the data (3.7.27) determine uniquely. Assume that

Theorem 3.7.4. If

and (3.7.28) holds, then

determines q uniquely.

Proof. If (3.7.28) holds, then for x < 0, and u = t (k) f (x, k) for x > 0, where f (k, x) is the Jost solution (3.1.5). Thus

Therefore r (k) determines I (k), so by Theorem 3.3.2 q is uniquely determined. 3.8

3.8.1

RECOVERY OF QUARKONIUM SYSTEMS

Statement of the inverse problem

The problem discussed in this Section is: to what extent does the spectrum of a quarkonium system together with other experimental data determines the interquark potential? This problem was discussed in [TQR], where one can find further references. The method given in [TQR] for solving the problem is this: one has few scattering data which will be defined precisely later, one constructs, using the known results of inverse scattering theory, a Bargmann potential with the same scattering data and

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considers this a solution to the problem. This approach is wrong because the scattering theory is applicable to the potentials which tend to zero at infinity, while our confining potentials grow to infinity at infinity and no Bargmann potential can approximate a confining potential on the whole semiaxis (0, Our aim is to give an algorithm which is consistent and yields a solution to the above problem. The algorithm is based on the Gel’fand-Levitan procedure of Section 3.4.3. A Bargman potential has a rational Jost function. Such potentials decay exponentially at infinity, while the confining potentials grow (see (3.8.2)). Let us formulate the problem precisely. Consider the Schrödinger equation

where q(r) is a real-valued spherically symmetric potential,

The functions these states. We define the resulting equation for

are the bound states, are the energies of which correspond to s-waves, and consider

One can measure the energies of the bound states and the quantities experimentally. Therefore the following inverse problem (IP) is of interest: (IP): given:

can one recover p (r)?

In [TQR] this question was considered but the approach in [TQR] is inconsistent and no exact results are obtained. The inconsistency of the approach in [TQR] is the following: on the one hand [TQR] uses the inverse scattering theory which is applicable to the potentials decaying sufficiently rapidly at infinity, on the other hand, [TQR] is concerned with potentials which grow to infinity as It is nevertheless of some interest that numerical results in [TQR] seem to give some approximation of the potentials in a neighborhood of the origin. Here we present a rigorous approach to (IP) and prove the following result: Theorem 3.8.1. IP has at most one solution and the potential q (r) can be reconstructed from data (3.8.4) algorithmically. The reconstruction algorithm is based on the Gel’fand-Levitan procedure for the reconstruction of q(x) from the spectral function. We show that the data (3.8.4) allow one to write the spectral function of the selfadjoint in operator defined by the differential expression (3.8.3) and the boundary condition (3.8.3) at zero.

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In Section 3.8.2 proofs are given and the recovery procedure is described. Since in experiments one has only finitely many data the question arises: How does one use these data for the recovery of the potential?

We give the following recipe: the unknown confining potential is assumed to be of the form (3.8.2) and it is assumed that for j > J the data for this potential are the same as for the unperturbed potential In this case an easy algorithm is given for finding q (r). This algorithm is described in Section 3.8.3. 3.8.2

Proof

We prove Theorem 3.8.1 by reducing (IP) to the problem of recovery of q(r) from the spectral function (Section 3.4). Let us recall that the selfadjoint operator L has discrete spectrum since The formula for the number of eigenvalues (energies of the bound states), not exceeding is known:

This formula yields, under the assumption q (r) ~ r as totics of the eigenvalues:

The spectral function

where

Here

the following asymp-

of the operator L is defined by the formula

are the normalizing constants:

and

is the unique solution of the problem:

If then tion to the Volterra integral equation:

The function

is the unique solu-

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For any fixed r the function is an entire function of E of order that is, where c denotes various positive constants. At where are the eigenvalues of (3.8.3), one has In fact, if a > 0, then for some Let us relate and From (3.8.7) with and from (3.8.3), it follows that

Therefore

Thus data (3.8.4) define uniquely the spectral function of the operator L by the formula:

Given one can use the Gel’fand-Levitan (GL) method (Section 3.4) for recovery of q (r). According to this method, define

where is the spectral function of the unperturbed problem, which in our case is the problem with q(r) = r, then set

where are the eigenfunctions of the problem (3.8.7) with q(r) = r, and solve the second kind Fredholm integral equation for the kernel K(x, y):

The kernel L(x, y) in equation (3.8.14) is given by formula (3.8.13). If K(x, y) solves (3.8.14), then

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3.8.3

Reconstruction method

Let us describe the algorithm we propose for recovery of the function q(r) from few experimental data Denote by the data corresponding to These data are known and the corresponding eigenfunctions (3.8.3) can be expressed in terms of Airy function Ai(r), which solves the equation and decays at see [Leb]. The spectral function of the operator corresponding to is

Define

and

where

can be obtained by solving the Volterra equation (3.8.9) with and represented in the form:

where K(x, y) is the transformation kernel corresponding to the potential and are the eigenfunctions of the unperturbed problem:

Note that for the functions (3.8.20) do not belong to but We denoted in this section the eigenfunctions of the unperturbed problem by rather than for simplicity of notations, since the eigenfunctions of the perturbed problem are not used in this section. One has: where is the j –th positive root if the equation Ai (– E) = 0 and, by formula (3.8.10), one has These formulas make the calculation of and easy since the tables of Airy functions are available [Leb].

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The equation analogous to (3.8.14) is:

where

and Equation (3.8.22) has degenerate kernel and therefore can be reduced to a linear algebraic system. If K(x, y) is found from (3.8.22), then

Equation (3.8.14) and, in particular (3.8.22), is uniquely solvable by the Fredholm alternative: the homogeneous version of (3.8.14) has only the trivial solution. Indeed, if then so that, by Parseval equality, Here where are defined by (3.8.20). This implies that for all j = 1, 2, . . . . Since is an entire function of exponential type and since the density of the sequence is infinite, i.e., because as was shown in the beginning of Section 3.8.2, it follows that and consequently h(t) = 0, as claimed. In conclusion consider the case when for all and is the new eigenvalue, with the corresponding data In this case so that equation (3.8.14) takes the form

Thus, one gets:

3.9

3.9.1

KREIN’S METHOD IN INVERSE SCATTERING

Introduction and description of the method

Consider inverse scattering problem studied in Section 3.5.1 and for simplicity assume that there are no bound states. This assumption is removed in Section 3.9.4. This chapter is a commentary to Krein’s paper [K1]. It contains not only a detailed proof of the results announced in [K1] but also a proof of the new results not mentioned in [K1]. In particular, it contains an analysis of the invertibility of the steps in the inversion procedure based on Krein’s results, and a proof of the consistency of this procedure, that is, a proof of the fact that the reconstructed potential generates the scattering data from

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which it was reconstructed. A numerical scheme for solving inverse scattering problem, based on Krein’s inversion method, is proposed, and its advantages compared with the Marchenko and Gel’fand–Levitan methods are discussed. Some of the results are stated in Theorem 3.9.2–Theorem 3.9.5 below. Consider the equation for a function

Equation (3.9.1) shows that

so

in operator form, and

Let us assume that H(t) is a real-valued even function

Then (3.9.1) is uniquely solvable for any x > 0, and there exists a limit

where

solves the equation

Given H(t), one solves (3.9.1), finds

then defines

where

Formula (3.9.8) gives a one-to-one correspondence between E(x, k) and

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Remark 3.9.1. In [K1] is used in place of in the definition of By formula (3.9.46) (see Section 3.9.2 below) one has but in general. The theory presented below cannot be constructed with in place of in formula (3.9.8). Note that

where

and

Furthermore,

Note that

where

The function is called the phase shift. One has We have changed the notations from [K1] in order to show the physical meaning of the function (3.9.9): is the Jost function of the scattering theory. The function is the solution to the scattering problem: it solves equation (3.1.3), and satisfies the correct boundary conditions: and as Krein [K1] calls the S -function, and is the S -matrix used in physics. Assuming no bound states, one can solve the inverse scattering problem (ISP): Given find q (x). A solution of the ISP, based on the results of [K1], consists of four steps:

(1) Given S(k), find f (k) by solving the Riemann problem (3.9.62). (2) Given f (k), calculate H(t) using the formula

(3) Given H(t), solve (3.9.1) for

and then find

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(4) Define

where

and calculate the potential

One can also calculate q(x) by the formula:

Indeed, see (3.9.14), see (3.9.47), and see (3.9.46). There is an alternative way, based on the Wiener-Levy theorem, to do step (1). Namely, given S(k), find the phase shift, then calculate the function and finally calculate The potential generates the S -matrix S(k), with which we started, provided that the following conditions (3.9.18)–(3.9.21) hold:

the overbar stands for complex conjugation, and

where

By the index (3.9.19) one means the increment of the argument of S(k) (when k runs from to along the real axis) divided by (cf (3.1.21)). The function (3.9.7) satisfies equation (3.1.5). Recall that we have assumed that there are no bound states. In Section 3.9.2 the above method is justified and the following theorems are proved:

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Theorem 3.9.2. If (3.9.18)–(3.9.20) hold, then q (x) defined by (3.9.16) is the unique solution to ISP and this q (x) has S (k) as the scattering matrix. Theorem 3.9.3. The function f (k), defined by (3.9.10), is the Jost function corresponding to potential (3.9.16). Theorem 3.9.4. Condition (3.9.4) implies that equation (3.9.1) is solvable for all and its solution is unique. Theorem 3.9.5. If condition (3.9.4) holds, then relation (3.9.11) holds and is the unique solution to the equation

The diagram explaining the inversion method for solving ISP, based on Krein’s results, can be shown now:

In this diagram denotes step number m. Steps and are trivial. Step is almost trivial: it requires solving a Riemann problem with index zero and can be done analytically, in closed form. Step is the basic (non-trivial) step which requires solving a family of Fredholm-type linear integral equations (3.9.1). These equations are uniquely solvable if assumption (3.9.4) holds, or if assumptions (3.9.18)–(3.9.20) hold. In Section 3.9.2 we analyze the invertibility of the steps in diagram (3.9.23). Note also that, if one assumes (3.9.18)–(3.9.20), diagram (3.9.23) can be used for solving the inverse problems of finding q (x) from the following data: (a) from or (b) from (c) from the spectral function

Indeed, if (3.9.18)–(3.9.20) hold, then a) and b) are contained in diagram (3.9.23), and c) follows from the known formula Then (still assuming (3.9.19)) one has: Note that the general case of the inverse scattering problem on the half-axis, when can be reduced to the case v = 0 by the procedure, described in Section 3.9.4, provided that S(k) is the S-matrix corresponding to a potential Necessary and sufficient conditions for this are conditions (3.9.18)– (3.9.20).

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Subsection 3.9.3 contains a discussion of the numerical aspects of the inversion procedure based on Krein’s method. There are advantages in using this procedure (as compared with the Gel’fand-Levitan procedure): integral equation (3.9.1), solving of which constitutes the basic step in the Krein inversion method, is a Fredholm convolution-type equation. Solving such an equation numerically leads to inversion of Toeplitz matrices, which can be done efficiently and with much less computer time than solving the Gel’fand-Levitan equation (3.1.34). Combining Krein’s and Marchenko’s inversion methods yields an efficient way to solve inverse scattering problems. Indeed, for small x equation (3.9.1) can be solved by iterations since the norm of the integral operator in (3.9.1) is less than 1 for sufficiently small x, say Thus q (x) can be calculated for by diagram (3.9.23). For one can solve by iterations Marchenko’s equation (3.1.43) for the kernel A(x, y), where, if (3.9.19) holds, the function F(x) is defined by the (3.1.41) with Indeed, for x > 0 the norm of the operator in (3.1.41) is less than 1 and it tends to 0 as Finally let us discuss the following question: in the justification of both the Gel’fandLevitan and Marchenko methods, the eigenfunction expansion theorem and the Parseval relation play a fundamental role. In contrast, the Krein method apparently does not use the eigenfunction expansion theorem and the Parseval relation. However, implicitly, this method is also based on such relations. Namely, assumption (3.9.4) implies that the S-matrix corresponding to the potential (3.9.16), has index 0. If, in addition, this potential is in then conditions (3.9.18) and (3.9.20) are satisfied as well, and the eigenfunction expansion theorem and Parseval’s equality hold. Necessary and sufficient conditions, imposed directly on the function H(t), which guarantee that conditions (3.9.18)–(3.9.20) hold, are not known. However, it follows that conditions (3.9.18)–(3.9.20) hold if and only if H(t) is such that the diagram (3.9.23) leads to a Alternatively, conditions (3.9.18)–(3.9.20) hold (and consequently, if and only if condition (3.9.4) holds and the function f (k), which is uniquely defined as the solution to the Riemann problem

by the formula generates the S-matrix S(k) by formula (3.9.15), and this S(k) satisfies conditions (3.9.18)–(3.9.20). Although the above conditions are verifiable, they are not quite satisfactory because they are implicit, they are not formulated in terms of structural properties of the function H(t) (such as smoothness, rate of decay, etc.). In Subsection 3.9.2 Theorem 3.9.2–Theorem 3.9.5 are proved. In Subsection 3.9.3 numerical aspects of the inversion method based on Krein’s results are discussed. In Subsection 3.9.4 the ISP with bound states is discussed. In Subsection 3.9.5 a relation between Krein’s and Gel’fand–Levitan’s methods is explained.

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3.9.2


Proofs

Proof of Theorem 3.9.4. If

then

where the Parseval equality was used,

Thus is a positive definite selfadjoint operator in the Hilbert space if (3.9.4) holds. Note that, since one has as (3.9.4) implies

so

A positive definite selfadjoint operator in a Hilbert space is boundedly invertible. Theorem 3.9.4 is proved. Note that our argument shows that

Before we prove Theorem 3.9.5, let us prove a simple lemma. For results of this type, see [K2]. Lemma 3.9.6. If (3.9.4) holds, then the operator

is a bounded operator in For

Proof. Let

one has

One has

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where we have used the assumption H(t) = H ( – t ) . Similarly,

Finally, using Parseval’s equality, one gets:

where

Since

one gets from (3.9.33) the estimate:

To prove (3.9.30), one notes that

Estimate (3.9.30) is obtained. Lemma 3.9.6 is proved. Proof of Theorem 3.9.5. Define for t or s greater than x. Let Then (3.9.1) and (3.9.6) imply

If condition (3.9.4) holds, then equations (3.9.6) and (3.9.22) have solutions in and, since it is clear that this solution belongs to and consequently to because The proof of Theorem 3.9.4 shows that such a solution is unique and does exist. From (3.9.28) one gets

For any fixed s > 0 one sees that here stands for any of the three norms

as

where the norm Therefore (3.9.36) and

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(3.9.35) imply

since Also

where

for any fixed s > 0 and

are some constants. Finally, by (3.9.30), one has;

From (3.9.39) and (3.9.41) relation (3.9.11) follows. Theorem 3.9.5 is proved. Let us now prove Theorem 3.9.3. We need several lemmas. Lemma 3.9.7. The function (3.9.8) satisfies the equations

where

and a (x) is defined in (3.9.14).

Proof. Differentiate (3.9.8) and get

We will check below that

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and

Thus, by (3.9.45),

Therefore (3.9.44) can be written as

By (3.9.46) one gets

Thus

From (3.9.48) and (3.9.50) one gets (3.9.42). Equation (3.9.43) can be obtained from (3.9.42) by changing k to –k. Lemma 3.9.7 is proved if formulas (3.9.45)–(3.9.46) are checked. To check (3.9.46), use H ( – t ) = H(t) and compare the equation for

with equation (3.9.1). Let u = x – y. Then (3.9.51) can be written as

which is equation (3.9.1) for Since (3.9.1) has at most one solution, as we have proved above (Theorem 3.9.4), formula (3.9.46) is proved.

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To prove (3.9.45), differentiate (3.9.1) with respect to x and get:

Set s = x in (3.9.1), multiply (3.9.1) by compare with (3.9.53) and use again the uniqueness of the solution to (3.9.1). This yields (3.9.45). Lemma 3.9.7 is proved. Lemma 3.9.8. Equation (3.1.5) holds for

defined in (3.9.7).

Proof. From (3.9.7) and (3.9.42)–(3.9.43) one gets

Using (3.9.42)–(3.9.43) again one gets

Lemma 3.9.8 is proved. Proof of Theorem 3.9.3. The function satisfies the conditions

defined in (3.9.7) solves equation (3.1.5) and

The first condition is obvious (in [Kl] there is a misprint: it is written that and the second condition follows from (3.9.7) and (3.9.42):

Let f (x, k) be the Jost solution. Since f (x, k) and f (x, –k) are linearly independent, one has where are some constants independent of x but depending on k. From (3.9.56) one gets Indeed, the choice of and guarantees that the first condition (3.9.56) is obviously satisfied, while the second follows from the Wronskian formula: Comparing this with (3.9.12) yields the conclusion of Theorem 3.9.3. Invertibility of the steps of the inversion procedure and proof of Theorem 1.1

Let us start with a discussion of the inversion steps (1)–(4) described in the introduction.

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Then we discuss the uniqueness of the solution to ISP and the consistency of the inversion method, that is, the fact that q (x), reconstructed from S(k) by steps 1)–4), generates the original S(k). Let us go through steps l)–4) of the reconstruction method and prove their invertibility. The consistency of the inversion method follows from the invertibility of the steps of the inversion method. Step 1. Assume S(k) satisfying (3.9.18)–(3.9.20) is given. Then solve the Riemann problem

Since one has Therefore the problem (3.9.57) of finding an analytic function in in (and an analytic function in from equation (3.9.57) can be solved in closed form. Namely, define

Then f (k) solves (3.9.57),

Indeed,

by the known jump formula for the Cauchy integral. Integral (3.9.58) converges absolutely at infinity, ln S(– y) is differentiable with respect to y for and is bounded on the real axis, so the Cauchy integral in (3.9.58) is well defined. To justify the above claims, one uses the known properties of the Jost function

where estimates (3.1.25) and (3.1.26) hold and A(y) is a real-valued function. Thus

Therefore

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Also

Estimate (3.1.25) implies

so that

is bounded for all is differentiable for and ln S(– y) is bounded on the real axis, as claimed. Note that

The converse step

is trivial:

analytic in in holds. Step 2. This step is done by formula (3.9.13):

One has

If

then f (k) is

as

and (3.9.66)

Indeed, it follows from (3.9.61) that

The function

is continuous because as

by (3.1.26), and

since

Thus, Also, Indeed, integrating by parts, one gets from (3.9.67) the and relation: one uses (3.9.60), (3.1.25)– therefore To check that (3.1.26), and (3.9.64)–(3.9.65), to conclude that and, since on and it follows that The inclusion follows from (3.9.60), (3.1.25)–(3.1.26), and (3.9.64)–(3.9.65).

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By (3.9.67), the function

is the Fourier transform of

and, by (3.9.68), Thus, form is

as behaves, essentially, as

plus a function, whose Fourier trans-

Estimate (3.1.26) shows how

behaves. Equation (3.9.1) shows

that is as smooth as H(t), so that formula (3.9.17) for q (x) shows that q is essentially as smooth as The converse step

is also done by formula (3.9.13): Fourier inversion gives and factorization of yields the unique f (k), since f (k) does not vanish in and tends to 1 at infinity. Step 3. This step is done by solving equation (3.9.1). By Theorem 3.9.4 equation (3.9.1) is uniquely solvable since condition (3.9.4) is assumed. Formula (3.9.13) holds and the known properties of the Jost function are used: as for since The converse step is done by formula (3.9.3). The converse step

constitutes the essence of the inversion method. This step is done as follows:

Given a (x), system (3.9.42)–(3.9.43) is uniquely solvable for E (x, k). Note that the step can be done by solving the uniquely solvable integral equation (3.1.6): with and then calculating f (k) = f (0, k). Step 4. This step is done by formula (3.9.16). The converse step

can be done by solving the Riccati problem (3.9.16) for a (x) given q (x) and the initial condition 2H(0). Given q (x), one can find 2H(0) as follows: one finds f (x, k) by solving equation (3.1.6), which is uniquely solvable if then one gets f (k) := f (0, k), and then calculates 2H(0) using formula (3.9.67) with t = 0:

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Proof of Theorem 3.9.2. If (3.9.18)–(3.9.20) hold, then, as has been proved in Section 3.5.5, there is a unique which generates the given S-matrix S(k). It is not proved in [K1] that q (x) defined in (1.19) (and obtained as a final result of steps 1)–4)) generates the scattering matrix S(k) with which we started the inversion. Let us now prove this. We have already discussed the following diagram:

To close this diagram and therefore establish the basic one-to-one correspondence one needs to prove This is done by the scheme (3.9.72). Note that the step requires solving Riccati equation (3.9.16) with the boundary condition a (0) = 2H(0). Existence of the solution to this problem on all of is guaranteed by the assumptions (3.9.18)–(3.9.20). The fact that these assumptions imply is proved in Section 3.5.5. Theorem 3.9.2 is proved. Uniqueness theorems for the inverse scattering problem are not given in [K1]. They can be found in Section 3.5.5. Remark 3.9.9. From our analysis one gets the following result: Proposition 3.9.10. If and has no bounds states and no resonance at zero, then Riccati equation (3.9.16) with the initial condition (3.9.15) has the solution a (x) defined for all 3.9.3

Numerical aspects of the Krein inversion procedure

The main step in this procedure from the numerical viewpoint is to solve equation (3.9.1) for all x > 0 and all 0 < s < x, which are the parameters in equation (3.9.1). Since equation (3.9.1) is an equation with the convolution kernel, its numerical solution involves inversion of a Toeplitz matrix, which is a well developed area of numerical analysis. Moreover, such an inversion requires much less computer memory and time than the inversion based on the Gel’fand-Levitan or Marchenko methods. This is the main advantage of Krein’s inversion method. This method may become even more attractive if it is combined with the Marchenko method. In the Marchenko method the equation to be solved is (3.1.43) where F (x) is defined in (3.1.41) and is known if S(k) is known. The kernel A (x, y) is to be found from (3.1.41) and if A(x, y) is found then the potential is recovered by the formula: Equation (3.1.41) can be written in operator form: The operator is a contraction mapping in the Banach space for x > 0. The operator in (3.9.1) is a contraction mapping in for where is chosen to that Therefore it seems reasonable from the numerical point of view to use the following approach:

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1. Given S(k), calculate f(k) and H(t) as explained in Steps 1 and 2, and also F(x) by formula (3.1.41). 2. Solve by iterations equation (3.9.1) for where is chosen so that the iteration method for solving (3.9.6) converges rapidly. Then find q (x) as explained in Step 4. 3. Solve equation (3.1.43) for by iterations. Find q (x) for by formula (3.1.42). 3.9.4

Discussion of the ISP when the bound states are present

If the given data are (3.9.15), then one defines if and if where is arbitrary, and is chosen so that Then one defines if or if Since and one has The theory of Subsection 3.9.2 applies to and yields From one gets q (x) by adding bound states and norming constants using the known procedure (e.g. see [M]). 3.9.5

Relation between Krein’s and GL’s methods

The GL (Gel’fand-Levitan) method in the case of absence of bound states of the following steps (see Section 3.4, for example): Step 1. Given f(k), the Jost function, find

where Step 2. Solve the integral equation (3.1.34) for K(x, y). Step 3. Find Krein’s function, H(t), see (3.9.13), can be written as follows:

Thus, the relation between the two methods is given by the formula:

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In fact, the GL method deals with the inversion of the spectral foundation of the operator defined in by the Dirichlet boundary condition at x = 0. However, if (in this case there are no bound states and no resonance at k = 0), then (see (3.1.20)):

so

in this case is uniquely defined by 3.10

INVERSE PROBLEMS FOR THE HEAT AND WAVE EQUATIONS

3.10.1

Inverse problem for the heat equation

Consider problem (3.1.55)–(3.1.58). Assume

One can also take where is the delta-function. We prove that the inverse problem of finding from the conditions (3.1.55)–(3.1.58) has at most one solution. If (3.1.58) is replaced by the condition

then q (x), in general, is not uniquely defined by the conditions (3.1.55), (3.1.56), (3.1.57) and (3.10.2), but q is uniquely defined by these data if, for example, or if q (x) is known on Let us take the Laplace transform of (3.1.55)–(3.1.58) and put Then (3.1.55)–(3.1.58) can be written as

and (3.10.2) takes the form

Theorem 3.10.1. The data known on a set of finite positive limit point, determine q uniquely. Proof. Since and

and are analytic in are known for all If

and

which has a

one can assume that is defined in (3.1.3) then so

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Thus the function is meromorphic in its zeros on the axis are corresponding to the boundary conditions the eigenvalues of and its poles on the axis are the eigenvalues of corresponding to u(0) = u(1) = 0. The knowledge of two spectra determines q uniquely (Section 3.7.1). An alternative proof of Theorem 3.10.1, based on property and generate the same data, solves (3.10.3)–(3.10.4) with and get (*) Multiply (*) by integrate over [0, 1] to get

By property

is: assume that where j = 1, 2, and

it follows from (3.10.7) that p = 0. Theorem 3.10.1 is proved.

Theorem 3.10.2. Data (3.10.3), (3.10.5) do not determine q uniquely in general. They do if q (x) is known on orif Proof. Arguing as in the first proof of Theorem 3.10.1, one concluded that the data (3.10.3), (3.10.5) yields only one (Dirichlet) spectrum of since One spectrum determines q only on “a half of the interval”, see Section 3.7.1. Theorem 3.10.2 is proved. 3.10.2 What are the “correct” measurements?

From Theorem 3.10.1 and Theorem 3.10.2 it follows that the measurements are much more informative than for the problem (3.1.55)– (3.1.57). In this section we state a similar result for the problem

The extra data, that is, measurements, are

which is the flux. Assume:

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is the Sobolev space. Physically, a (x) is the conductivity, u is the temperature. We also consider in place of (3.10.10) the following data:

Our results are similar to those in Subsection 3.10.1: Data determine q(x) uniquely, while data do not, in general, determine a (x) uniquely. Therefore, the measurements are much more informative than the measurements 3.10.3

Inverse problem for the wave equation

Consider inverse problem (3.1.50)–(3.1.54). Our result is Theorem 3.10.3. The above inverse problem has at most one solution. Proof. Take the Fourier transform of (3.1.50)–(3.1.54) and get:

From (3.10.13) one gets where f(x, k) is the Jost solution, and from (3.10.14) one gets and because q = 0 for x > 1. Thus is known. By Theorem 3.7.3, q is uniquely determined. Theorem 3.10.3 is proved. Remark 3.10.4. The above method allows one to consider other boundary conditions at x = 0, such as or h = const > 0, and different data at x = 1, for example, 3.11

INVERSE PROBLEM FOR AN INHOMOGENEOUS SCHRÖDINGER EQUATION

In this Section an inverse problem is studied for an inhomogeneous Schrödinger equation. Most of the earlier studies dealt with inverse problems for homogeneous equations. Let

Assume that q (x) is a real-valued function, q (x) = 0 for Suppose that the data are given.

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The inverse problem is: (IP) Given the data, find q (x). This problem is of practical interest: think about finding the properties of an inhomogeneous slab (the governing equation is plasma equation) from the boundary measurements of the field, generated by a point source inside the slab. Assume that the self-adjoint operator in has no negative eigenvalues (this is the case when for example). The operator is the closure in of the symmetric operator defined on by the formula Our result is: Theorem 3.11.1. Under the above assumptions IP has at most one solution. Proof of Theorem 11.1: The solution to (3.11.1) is

Here f (x, k) and g(x, k) solve homogeneous version of equation (3.11.1) and have the following asymptotics:

where the prime denotes differentiation with respect to x-variable, and a (k) is defined by the equation

It is known that (see e.g. [M]):

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a (k) in analytic in b (k) in general does not admit analytic continuation from if q (x) is compactly supported, then a (k) and b (k) are analytic functions of The functions

but

are the data, they are known for all k > 0. Therefore one can assume the functions

to be known for all k > 0 because

as follows from the assumption q = 0 if and from (3.11.3). From (3.11.12), (3.11.7) and (3.11.6) it follows that

From (3.11.14) and (3.11.15) it follows:

Eliminating b(–k) from (3.11.16) and (3.11.17), one gets:

or

where

Problem (3.11.19) is a Riemann problem for the pair the function a(k) is analytic in and a ( – k ) is analytic in The

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functions a (k) and a (–k) tend to one as k tends to infinity in and, respectively, in see equation (3.11.9). The function a(k) has finitely many simple zeros at the points where are the negative eigenvalues of the operator defined by the differential expression in The zeros are the only zeros of a (k) in the upper half-plane k. Define

One has

where J is the number of negative eigenvalues of the operator (3.11.22) and (3.11.20), one gets

and, using (3.11.12),

Since has no negative eigenvalues, it follows that J = 0. In this case ind f (k) = ind g (k) = 0 (see Lemma 1 below), so ind m(k) = 0, and a(k) is uniquely recovered from the data as the solution of (3.11.19) which tends to one at infinity, see equation (3.11.9). If a(k) is found, then b(k) is uniquely determined by equation (3.11.17) and so the reflection coefficient is found. The reflection coefficient determines a compactly supported q(x) uniquely [R196], but we give a new proof. If q (x) is compactly supported, then the reflection coefficient is meromorphic. Therefore, its values for all k > 0 determine uniquely r(k) in the whole complex k-plane as a meromorphic function. The poles of this function in the upper half-plane are the numbers They determine uniquely the numbers which are a part of the standard scattering data where are the norming constants. Note that if then otherwise equation (3.11.6) would imply in contradiction to the first relation (3.11.3). If r (k) is meromorphic, then the norming where constants can be calculated by the formula the dot denotes differentiation with respect to k, and Res denotes the residue. So, for compactly supported potential the values of r(K) for all k > 0 determine uniquely the and standard scattering data, that is, the reflection coefficient, the bound states the norming constants These data determine the potential uniquely. Theorem 3.11.1 is proved.

Lemma 3.11.2. If J = 0 then ind f = ind g = 0. Proof. We prove ind f = 0. The proof of the equation ind g = 0 is similar. Since ind f(k) equals to the number of zeros of f (k) in we have to prove that f (k)

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does not vanish in If then z = i k, k > 0, and is an eigenvalue of the operator in with the boundary condition u(0) = 0. From the variational principle one can find the negative eigenvalues of the operator in with the Dirichlet condition at x = 0 as consequitive minima of the quadratic functional. The minimal eigenvalue is:

where is the Sobolev space of u(0) = 0. On the other hand, if J = 0, then

satisfying the condition

Since any element u of can be considered as an element of if one extends u to the whole axis by setting u = 0 for x < 0, it follows from the variational definitions (3.11.24) and (3.11.25) that Therefore, if J =0, then and therefore This means that the operator on with the Dirichlet condition at x = 0 has no negative eigenvalues. This means that does not have zeros in ifJ = 0. Thus J = 0 implies ind f(k) = 0. Lemma 3.11.2 is proved. Remark 3.11.3. The above argument shows that in general

so that (3.11.23) implies

Therefore the Riemann problem (3.11.19) is always solvable. It is an open problem to find out if the IP has at most one solution without assuming the absence of negative eigenvalues of the operator 3.12

AN INVERSE PROBLEM OF OCEAN ACOUSTICS

3.12.1 The problem

In this Section the result from [R199] is presented and it is shown that the approach to a somewhat similar problem in the paper [GX] is invalid.

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Let

where is the deltafunction, n(z) is the refraction coefficient, which is assumed to be a real-valued integrable function, k > 0 is a fixed wavenumber. The solution to (3.12.1)-(3.12.2) is selected by the limiting absorption principle. It is proved that if then n(z) is uniquely determined by the data known 3.12.2

Introduction

In [GX] the following inverse problem is studied:

Here k > 0 is a fixed wavenumber, n(z) > 0 is the refraction coefficient, which is assumed in [GX] to be a continuous real-valued function satisfying the condition the layer models shallow ocean, is the delta-function, is a function satisfying the following conditions (see [GX], p. 127):

The solution to (3.12.3)–(3.12.4) in [GX] is required to satisfy some radiation conditions. It is convenient to define the solution as that is by the limiting absorption principle. We do not show the dependence on k in u(x) since k > 0 is fixed throughout this Section. The function is the unique solution to problem (3.12.3)– (3.12.4) in which equation (3.12.3) is replaced by the equation with absorption:

One defines the differential operator corresponding to differential expression (3.12.3) and the boundary conditions (3.12.4) in as a selfadjoint operator (for example, as the Friedrichs extension of the symmetric operator with the domain consisting of functions vanishing near infinity and satisfying conditions

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(3.12.4)), and then the function is uniquely defined. By we mean the usual Sobolev space. One can prove that the limit of this function does exist globally in the weighted space and locally in outside a neighborhood of the set provided where are defined in (3.12.10) below. This limit defines the unique solution to problem (3.12.3)–(3.12.4) satisfying the limiting absorption principle if If where is the delta-function, then an analytical formula for can be written:

where is the modified Bessel function (the Macdonald function), and are defined in (3.12.9) below, and and are defined in formula (3.12.10) below. This formula can be checked directly. It is obtained by separation of variables. The known formula was used, and is the Fourier transform defined above formula (3.12.6). From the formula for the known asymptotics for large values of r, the boundedness of as and formula (3.12.11) below, one can see that the limit of as does exist for any and if and only if If for some then the limiting absorption principle holds if and only if If then the limiting absorption principle holds and the solution to problem (3.12.3)–(3.12.4) is well defined. If for some then we define the solution to problem (3.12.3)–(3.12.4) with by the formula:

This solution is unique in the class of functions of the form in where if then satisfies the radiation condition uniformly in directions and if then

as as

The inverse problem (IP) consists of finding n(z) given and assuming that in (3.12.3). By the cylindrical symmetry one has It is claimed in [GX], that the above inverse problem (without the assumption has not more than one solution, and a method for finding this solution is proposed. The arguments in [GX] are erroneous (see Remark 3.12.3 below, where some of the incorrect statements from [GX], which invalidate the approach in [GX], are pointed out).

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Our aim is to prove that if then n (z) can be uniquely and constructively determined from the data g (r) known for all r > 0. It is an open problem to find all such f (z) for which the IP has at most one solution. Let us outline our approach to IP. Take the Fourier transform of (1.1)–(1.2) with respect to and let

and

Then

IP: The inverse problem is: given for all q (z). The solution to (3.12.6)–(3.12.7) is:

where

and a fixed

find

are the real-valued normalized eigenfunctions of the operator

We can choose the eigenfunctions real-valued since the function is assumed real-valued. One can check that all the eigenvalues are simple, that is, there is just one eigenfunction corresponding to the eigenvalue (up to a constant factor, which for real-valued normalized eigenfunctions can be either 1 or –1). It is known (see e.g. [M], p. 71) that

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The data can be written as

where are defined in (3.12.9). The series (3.12.12) converges absolutely and uniformly on compact sets of the complex plane outside the union of small discs centered at the points Thus, is a meromorphic function on the whole complex with simple poles at the points Its residue at equals If then proof of the inequality uniquely the set

(see Section 3.12.3 for a and the data (3.12.12) determine

In Section 3.12.3 we prove the basic result: Theorem 3.12.1. If uniquely.

then the data (3.12.13) determine

An algorithm for calculation of q (z) from the data is also described in Section 3.12.3. Remark 3.12.2. The proof and the conclusion of Theorem 3.12.1 remain valid for other boundary conditions, for example, with the data known for all 3.12.3

Proofs: uniqueness theorem and inversion algorithm

Proof of Theorem 3.12.1. The data (3.12.12) with that is, with determine uniquely since are the poles of the meromorphic function which is uniquely determined for all by its values for all (in fact, by its values at any infinite sequence of which has a finite limit point on the real axis). The residues of at are also uniquely determined. Let us show that: (i) (ii) The set (3.12.13) determines

uniquely.

Let us prove (i): If then equation (3.12.10) and the Cauchy data that

which is impossible since

where

imply

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Let us prove (ii): It is sufficient to prove that the set (3.12.13) determines the norming constants

and therefore the set

where the eigenvalues

and

are defined in (3.12.10),

are the zeros of the equation

The function

is an entire function of

of order 1/2, so that (see [Lev]):

From the Hadamard factorization theorem for entire functions of order < 1 formula (3.12.16) follows but the constant factor remains undetermined. This factor is determined by the data because the main term of the asymptotics of function (3.12.16) for large positive s is cos(s), and the result in [M], p. 243, (see Claim 8 below) implies that the constant in formula (3.12.16) can be computed explicitly:

where are the roots of the equation and the infinite product in (3.12.17) converges because of (3.12.11). A simple derivation of (3.12.17), independent of the result formulated in Claim 8 below, is based on the formula:

For convenience of the reader let us formulate the result from [M], p. 243, which yields formula (3.12.17) as well:

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Claim 8. The function

where B = const,

admits the representation

and

if and only if

where

are some numbers satisfying the condition: are the roots of the even function and positive roots of

are the

The equality

where is defined in (3.12.17), is easy to prove: if w is the left-hand side and v the right-hand side of the above equality, then w and v are entire functions of the infinite products converge absolutely, and taking the infinite product and using (3.12.17), one concludes that w /v = 1, as claimed. In fact, one can establish formula (3.12.18) and prove that in (3.12.18) is defined by (3.12.17) without assuming a priori that (3.12.17) holds and without using Claim 8. The following assumption suffices for the proof of (3.12.18):

Indeed, if i) holds then both sides of (3.12.18) are entire functions with the same set of zeros and their ratio is a constant. This constant equals to 1 if there is a sequence of points at which this ratio converges to 1. Using the known formula: and the assumption i) one checks easily that the ratio of the left- and right-hand sides of (3.12.18) tends to 1 along the positive imaginary semiaxis. Thus, we have proved formulas (3.12.16)–(3.12.17) without reference to Claim 8. The above claim is used with in our paper. The fact that admits the representation required in the claim is checked by means of the formula for in terms of the transformation operator: and the properties of the kernel are studied in Section 3.4.1. Thus, This is the representation of used in Claim 8.

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Let us derive a formula for Denote (3.12.14), with replaced by with respect to and get:

by

differentiate

Since q (z) is assumed real-valued, one may assume real-valued. Multiply (3.12.19) and (3.12.14) by subtract and integrate over (0, 1) to get

where the boundary conditions From (3.12.16) with one finds the number

Claim 9. The data (3.12.20) determine uniquely

where

were used.

and equation

Indeed, the numbers are the known numbers from formula (3.12.21). Denote by the quantities known from the data (3.12.13). Then it follows from (3.12.20) that so that

Claim 9 is proved. Thus, the data (3.12.13) determine uniquely and analytically by the above formula, and consequently q (z) is uniquely determined by the following known result (see Section 3.4): The spectral function of the operator L determines q (z) uniquely. The spectral function of the operator L is defined by the formula:

The Gel’fand-Levitan algorithm (see Section 3.4) allows one to reconstruct analytically q (z) from the spectral function and therefore from the data (3.12.13), since, as we have proved already, these data determine the spectral function uniquely. Thus Theorem 1 is proved. Let us describe an algorithm for calculation of q (z) from the data

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Step 1 : Calculate the Fourier transform of Given find its poles and consequently the numbers then find its residues, and consequently the numbers Step 2 : Calculate the function (3.12.16), and the constant by formulas (3.12.16) and (3.12.17). Calculate the numbers by formula (3.12.21) and by formula (3.12.22). Calculate the spectral function by formula (3.12.23). Step 3 : Use the known Gel’fand – Levitan algorithm to calculate q (z) from This completes the description of the inversion algorithm for IP. Remark 3.12.3. There are inaccuracies, errors, and erroneous arguments in [GX] which invalidate the approach in [GX]. The authors assume (see (C2) on p. 128 in [GX]) that where is a negative decreasing sequence of numbers satisfying formula (3.1)” (a typo, made in [GX] in the above statement, is corrected: in [GX] it was written for the “negative decreasing sequence”, not There are many such sequences. Which one should one choose? The authors do not give any answer to this question. They write on the line below formula (3.1) in [GX], p. 128: “Let be the sequence generated by g(r)”. The function g(r) does not generate a uniquely defined sequence of The authors do not explain how a given function g generates such a sequence. If one chooses a sequence satisfying formula (3.1), then this sequence will not satisfy the problem on lines 2 and 3 below (3.1), unless the authors know a priori the eigenvalues of the problem (3.12.11)–(3.12.13), but these eigenvalues are unknown because is unknown. The authors use the same notation for different objects, and this causes errors in their arguments. Consequently the arguments in [GX] below formula (3.1) are without foundation, they are erroneous. Further arguments in [GX] are also erroneous. For example, the authors have to solve equation (3.13) in [GX] for If one corrects the typo in (3.13), replacing a by where are unknown numbers (since is not known), then the problem of finding from (3.13) means that one has to determine simultaneously a sequence and a function given one function g (r). This is not possible, in general. The authors do not discuss this problem and, apparently, do not see the difficulty. Instead they write that (3.13) is “some kind of H–transform”, forgetting that the numbers are unknown. One can also point out that (3.13) is not the H–transform in the standard sense (i.e., with H the Struwe function, and not a Hankel function, and with the integration over (0, and not over the set used in [GX] but not clearly defined there). 3.13

THEORY OF GROUND-PENETRATING RADARS

3.13.1 Introduction

In many physical and technical applications the problem of determining the inner structure of a material arises. In particular, such problems arise in geophysics when

217

one wants to get information about the medium from the observations of the electromagnetic fields on the surface of the Earth. Let the source of electromagnetic waves be located above the ground. These waves, radiated by the source, penetrate into the ground, interact with it and the resulting electromagnetic field is observed on the surface of the Earth. The present work shows how to get information about the inner structure of the Earth layers from these observations. The proposed algorithm might be useful, for example, in geophysical exploration, detecting the location of nuclear waste, etc. The mathematical model of the above problem is based on Maxwell’s equations for the electromagnetic fields E and H:

written in the Cartesian coordinates (x, y, z). The plane (x, y) is assumed to be parallel to the Earth’s surface, the z-axis is perpendicular to the Earth’s surface and directed into the Earth. In (3.13.1) t is time, and are conductivity, permittivity and magnetic permeability respectively, is the exterior source which is supposed to be a wire located above the Earth’s surface and going along the through the point is the delta-function, f (t) is a function of time which shows the shape of the electromagnetic pulse. It is assumed that

where

is the dielectric constant of free space while is the magnetic permeability of vacuum.

Problem 2 (The Ground-Penetrating Radar Problem (GPR):). Given E on the plane z = 0 for all t > 0, find and for 0 < z < L. 3.13.2 Derivation of the basic equations

Let us differentiate the second equation in (1) with respect to t and substitute by rot E taken from the first equation. One gets

in it

Assuming that the vector is directed along the and depends on t, x and z, one can reduce (3.13.5) to a one-dimensional inverse problem

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for a differential equation satisfied by

are the Fourier transforms of unit. The characteristic value of u is

where

and f(t) respectively, and i is the imaginary The function u solves the problem:

where the conditions at infinity are written under the assumption that is sufficiently small. In this Section we use only such k. In (3.13.7), and Note that, due to (3.13.2) and (3.13.3), B(z) = 0 outside the interval [0, L]. Equation (3.13.7) may be rewritten as follows

where

and

Equation (3.13.8) is equivalent to the equation

In (3.13.9) g (z, s, k,

is the Green function that solves the equation

219

and satisfies the conditions:

The solution to (3.13.10)–(3.13.11) is

where

and we use the function g assuming that k is sufficiently small, so that and are both positive. Let C(0, L) denote the Banach space of continuous on [0, L] functions with the usual norm. Denote

3.13.3 Basic analytical results

Theorem 3.13.1. Equation (3.13.9) is uniquely solvable in C(0, L) for sufficiently small k; its solution is analytic with respect to k in a neighborhood of the point k = 0 and can be obtained by iterations

This theorem is proved similarly to the proof of Lemma 1 in [R83], p. 219 (see also Chapter 9, Vol. 2).

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Let u = v + i w, where v and w are the real and imaginary parts of u respectively. The functions w and v solve the equations

Put z = 0. Taking into account the analyticity of v and w in the neighborhood of k = 0, one differentiates equation (3.13.13) with respect to k at k = 0 and gets:

Similarly, one differentiates twice equation (3.13.14) with respect to k and puts k = 0. This yields:

If one introduces the variable and substitutes it into in the above equations instead of then (3.13.15) and (3.13.16) will express the Laplace transforms of the unknown functions and in terms of the data on the plane z = 0. Since an can be uniquely recovered from its Laplace transform, one gets the following uniqueness theorem: Theorem 3.13.2. The functions on the plane z=0.

and

are uniquely determined by the data

Some methods are given in [R139] for recovery of and from their Laplace transforms known for only. Let us rewrite (3.13.15) as where denotes the Laplace transform operator and F is the right-hand side of (3.13.15). A possible numerical inversion method consists of minimizing the functional

with respect to in an appropriate norm (we have chosen norm), where a is a regularization parameter. If one looks for of the form of a linear combination of finitely many linearly independent functions, then (3.13.17) is equivalent to determining the coefficients of the linear combination which minimize (3.13.17). To find the unknown coefficients, one differentiates (3.13.17) with respect to these coefficients and gets a system of linear algebraic equations for them. If is determined, one can

221

use its values to recover an analogous procedure.

(and consequently

from equation (3.13.16) using

3.13.4 Numerical results

To test the proposed algorithm a number of calculations were carried out. First, the direct problem (3.13.9) was numerically solved by iterations for some known and To evaluate the integrals in (3.13.9), the Simpson’s method was used. Then, the obtained values of the function u on the plane z = 0 were used to compute the Laplace transforms (3.13.15), (3.13.16) of and respectively. Finally, the functions and considered as unknown ones, were recovered from their Laplace transforms on the basis of (3.13.17), and the computed values of and were compared with given ones to estimate the accuracy of the the obtained results. As basis functions for expansion of and on the interval [0, L], the following piecewise-constant functions were chosen:

where the points form a partition of [0, L] (here N is the number of subintervals of [0, L]). The functions form an orthonormal system in To demonstrate the efficiency of the proposed approach for the recovery of and which are continuous in the interval (0, L), we considered the following example:

The results of the calculations are shown in Figs. 1, 2 in [RSh] for and respectively. In these Figures z-coordinate is taken along the horizontal axis, the vertical axis shows the values of the functions, the solid lines denote the exact values of and respectively, and the black circles indicate the obtained numerical results. It is worth to note that the recovery of and is carried out, in fact, on the basis of the noisy data in (3.13.15) and (3.13.16): indeed, each step of the numerical scheme contributes some computational error. The accuracy of the calculation of the Laplace transforms (3.13.15), (3.13.16) in was about and respectively. The accuracy of the obtained results is about for and approximately for This accuracy is defined by the formula: where f denotes the exact function and f* is the calculated one. Here by f we mean or the Laplace transforms (3.13.15), (3.13.16). To investigate the applicability of our approach for recovery of discontinuous functions, we took

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and for L=0.5, Fig. 3 and Fig. 4 in [RSh] show the values of the recovered and respectively. To recover the discontinuous function an adaptive grid generation procedure was used. This procedure is based on the following coordinate transformation:

where f denotes the unknown function to be recovered, and c is some positive constant. The transformation (3.13.18) is one-to-one. A uniform grid i = 0, . . . , N, along the corresponds to the grid i = 0, . . . , N, along the z-axis with larger density of the nodes in places with large provided that c is sufficiently small. In practice, the procedure for the recovery of is organized as follows. First, the constant c is taken large. This corresponds to a uniform grid along the z-axis. Then minimization (3.13.17) is done. At the next step the value of c is slightly decreased. To get the new distribution of nodes along the z-axis, we integrate (3.13.18) numerically and obtain the nodes j = 0, ..., N, along the Then a uniform grid is constructed along the and the values of are calculated at these points by linear interpolation of the data from the nodes obtained after solving (3.13.18). Using the new distribution of minimization procedure (3.13.17) is done again and the new grid is generated on the basis of (3.13.18) with the smaller value of the constant c. Such steps are repeated until the minimal possible error of the recovery of is attained. In such a way, one can reduce the computational error of the recovery of to (the accuracy of the calculation of the Laplace transform of from the data was about The accuracy of the recovery of is approximately (the accuracy of the calculation of the Laplace transform (3.13.16) of was about The proposed algorithm allows one to recover two coefficients simultaneously using a reasonable a priori information about the unknown functions and 3.13.5 The case of a source which is a loop of current

Consider now the case of a source which is a loop of electric current and prove that the information about the electromagnetic field on the Earth’s surface allows one to recover uniquely both conductivity and permittivity, and analytical recovery from exact data is possible. These parameters may have polynomial growth as the vertical coordinate approaches infinity. Mathematical model of the above problem is based on Maxwell’s equations (3.13.1), written in the cylindrical coordinates (r, z). The plane z = 0 is assumed to be the Earth’s surface, the z-axis is perpendicular to the Earth’s surface and directed into the ground. In (3.13.1) t is time, and are conductivity, permittivity and magnetic permeability respectively, is the exterior source which is supposed to be a loop of a radius located above the Earth’s surface at the point is the delta-function, f (t) is a piecewise-continuos function

223

of time which shows the shape of the electromagnetic pulse. The z-axis is supposed to pass through the center of the loop and perpendicular to the plane in which that loop lies. It is assumed that

The functions functions and functions on tributional sense.

and constant are considered to be known. The are assumed to be piecewise-continuous uniformly bounded The integral transforms, used below, are understood in the dis-

Problem 3 (The Ground-Penetrating Radar Problem (GPR)). Given E on the plane z = 0 for all t > 0, find and for Our basic result is formulated in the following theorem. Theorem 3.13.3. Under the above assumption the functions determined by the surface data E(r, t) known for all t > 0, and

and

are uniquely for fixed

Let us derive the basic equations. Differentiate the second equation in (3.13.1) with respect to t and substitute in it by rot E taken from the first equation. One gets

Assume that the vector where is the unit vector of the cylindrical coordinates. In this case equation (3.13.22) takes the form of a onedimensional inverse problem for a differential equation. First one rewrites equation (3.13.22) in cylindrical coordinates taking into account the well-known relation

where

stands for Laplace operator, and One gets

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Here

and

Defining the Fourier transform of E as

and applying it to (3.13.23) one gets

where Hankel-Bessel transform

to equation (3.13.24), where

is the Fourier transform of f (t). Then one applies the

is the Bessel function. Let us denote

The function u solves the problem:

Equation (3.13.25) is equivalent to the equation

In (3.13.26),

is the Green function that solves the equation

and satisfies the condition:

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The solution to (3.13.27)–(3.13.28) is

3.13.6

Basic analytical results

One can prove the existence and uniqueness of the solution to (3.13.26) for sufficiently small k and show its analyticity with respect to k in some neighborhood of the point k = 0. Let us derive an equation from which B(z) can be obtained. Put z = 0 in (3.13.26) and get:

Passing in 3.13.30) to the limit and denoting by function in the left-hand side of (3.1) one gets

Since

the limit of he known

and the function B(z) is known for z < 0, one gets from (3.13.31)

where

is a known function computable from the surface data and the values of Therefore setting

for z < 0.

one gets

So is uniquely determined as the inverse Laplace transform of the known function b(v). The function b(v) is defined by formulas (3.13.33) and (3.13.32), and in (3.13.32) is the limit, as of the left-hand side of (3.13.30).

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If

has been found equation (3.13.26) yields

Therefore

Denote the right-hand side of (3.13.36) by equation (3.13.36) yields

Since

is known for z < 0,

where

Setting

one gets

Hence is uniquely determined as the inverse Laplace transform of the known function a(v). The function a(v) is computable from the surface data, the values of for z < 0 and from the function B(z) calculated by solving equation (3.13.34).

4. INVERSE OBSTACLE SCATTERING

4.1

STATEMENT OF THE PROBLEM

First, let us formulate the direct scattering problem. Assume that is a bounded domain (obstacle) with the boundary S. D may consist of finitely many connected components. By denote the complement of D. Assume that D has finite perimeter relative to By the perimeter of the set D we mean where is the characteristic function of D, and is the space of functions of bounded variation in This space consists of functions u such that (in the sense of distribution theory) is a charge, and the total variation of this charge is If D is a Lipschitz domain, then D has a finite perimeter. The set of domains with finite perimeter is much larger than the set of Lipschitz domains. If is the local equation of S, then Lipschitz boundary is the one for which D has finite-perimeter if S has finite (n – 1)dimensional Hausdorff measure, that is where the infimum is taken over all coverings of S by open balls of radii Let v be a unit vector, We say that v is the normal (in the sense of Federer) to a set E at the point if

where By

is n-dimensional Lebesgue measure, we denote the boundary of E. The set of

for which normals to E

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4. Inverse obstacle scattering

exist is denoted E and is called the reduced boundary of E. The following result is known (e.g. [Maz]): If then the set E is measurable with respect to and if u is a compactly supported Lipschitz function in then

More generally, let D be a bounded domain with finite perimeter, and assume that the normal to D exists everywhere on Then for which has a trace on everywhere, the Green’s formula holds. Let us consider first the direct scattering problem. Let D be a bounded domain in which may have finitely many connected components, is the unit sphere in Let

where

is one of the following boundary conditions:

N is the exterior unit normal to S, is a piecewise-continuous bounded function with finitely many discontinuity curves on S. If S is rough, so that N may be not defined on S, we have to define the meaning of the boundary condition. For the Dirichlet condition this meaning consists of the inclusion where is the closure in the norm of of the functions, and is the set of functions such that R > 0 is such that Equation (4.1.1) we replace by the identity:

where is the set of functions vanishing near infinity. We call u the weak solution of the scattering problem. Thus, problem (4.1.1)–(4.1.4) with is formulated for non-smooth D as the problem of finding of the form where v satisfies (4.1.2), and u satisfies (4.1.5).

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If

then (4.1.5) holds functions vanishing near infinity, and the conditions (4.1.2), hold. If then u satisfies the identity:

is the set of v solves

u is of the form and v satisfies (4.1.2). For (4.1.6) to make sense one has to define the traces of functions from on S. The integration in (4.1.6) over S is with respect to measure. If S is Lipschitz, or D is an extension domain, or D satisfies the cone condition, then the embedding operator from into is compact, so the traces of the elements of on S are well defined. If then we assume that the domain D is bounded; if we assume additionally that the embedding operator is compact; if we assume get additionally that the embedding operator is compact. The role of these assumptions will be made clear later, but our basic idea can be explained now: the left-hand sides of (4.1.5) (or (4.1.6)) is a symmetric, closed, densely defined in bounded from below, quadratic form

This form defines a unique self-adjoint operator L, which is denoted by When we want to emphasize the corresponding boundary condition, we write Consequently, if Im k > 0 and Re k > 0, so that Im then is a bounded operator in Its kernel G(x, y, k) is called the resolvent kernel of the Laplacian corresponding to one of the above boundary conditions. We prove the limiting absorption principle, thus defining G(x, y, k) for k > 0 as the limit Let us first prove that the quadratic forms (4.1.5) (and (4.1.6)) with domains (and corresponding to (and are closed. These forms are symmetric and, under our assumptions on D, bounded from below. Closedness means (see [Kat]) that if in and as then and If a quadratic form is bounded from below, then the form for a suitable constant c > 0. We will assume therefore, without loss of generality, that Then the norm is equivalent to the norm t[u, u]. Since is a complete Hilbert space, the form t with the domain (and with the domain is closed. This covers conditions and Consider condition If is compact, then one has [Rl] the inequality

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where

is an arbitrary small number, d s is measure on S, and and R is throughout a sufficiently large number, so that Inequality (4.1.7) guarantees that the form t defined by formula (4.1.6) is bounded from below in and that the norm it defines is equivalent to the norm of Therefore this form with the domain is closed. Actually, one could replace the assumption of compactness of the embedding operator by the assumption of its boundedness, with relative bound < 1, that is, by the inequality where const < 1, and where the constants are independent of The closed symmetric forms generate the corresponding self-adjoint operators which we denote for any of the conditions The scattering problem can be formulated as the equation Let us show how this is done. Take a cut-off function if in a neighborhood of S, and let Then w = v for r > R, Thus we have reduced the scattering problem to the problem of solving the equation with the self-adjoint operator L and Actually This equation has a solution in H if Im and the solution is unique. Define and take n = 3 in what follows. Let us take k > 0, sufficiently small, and prove the limiting absorption principle (LAP): Theorem 4.1.1. For any n = 3, there exists in the norm of The function w(x, k) is the weak solution of the problem

where j = 1, 2, or 3, and w satisfies (4.1.2). Proof. First, let us prove Lemma 4.1.2. If w is a weak solution of (4.1.8) with f = 0 and satisfies (4.1.2), then w = 0. Proof. From (4.1.2) it follows that

From (4.1.5) (or (4.1.6)), taking for 0 for then taking complex conjugate equation, subtracting

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from (4.1.5) its complex conjugate, and taking

one gets:

From (4.1.8) and (4.1.9) it follows that

Any solution to equation (4.1.1) which satisfy (4.1.11), vanishes identically everywhere in its domain of definition. Indeed, any solution w to (4.1.1) can be written as

where are the orthonormal in spherical harmonics, are constants, are spherical Hankel functions, as (4.1.11) one derives for all so w = 0. Lemma 4.1.2 is proved.

From

Let us continue with the proof of Theorem 4.1.1. The steps of the proof are: Step 1. We prove that

c does not depend on

By c we denote various constants.

Step 2. From (4.1.12) it follows that This and (4.1.5) (or (4.1.6)) imply so and in because is compact (or because and are compact, when holds). Thus, one can pass to the limit as in (4.1.5) (or (4.1.6)), and get that w is a weak solution of (4.1.7). From elliptic regularity it follows that in where is any strictly interior bounded subset of This allows one to pass to the limit in the representation

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where N is the normal to the sphere get

pointing into

and

It follows that

Let us now prove a generalization of Ramm’s lemma ([R83], p. 46) for domains with finite perimeter. Lemma 4.1.3. Let G(x, y, k) be the resolvent kernel of the Laplacian L (for any of the boundary conditions j = 1, 2, 3) in where D is a bounded domain with finite perimeter, such that and are compact. Then

where if n = 3, is uniform with respect to x running through compact sets, and u (x, k) is the scattering solution, i.e, the solution to (4.1.1)–(4.1.4). If is there solvent kernel of in is the Hankel function. Proof. For domains with finite perimeter Green’s formula holds. Under our assumption on D the existence and uniqueness of G have been proved above. Green’s formula yields the representation formula (for

Here and below integration is taken actually over the essential (reduced) boundary [VH]. One has G(x, y, k) = G(y, x, k), and

Using this formula and (4.1.15) one gets (4.1.14) with

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The function (4.1.16) solves (4.1.1)–(4.1.4) with if G is the resolvent kernel of The argument is similar for j = 2, 3. For formula (4.1.15) takes the form

and (4.1.16) becomes

and for

one gets:

while (4.1.16) becomes:

Lemma 4.1.3 is proved. We need also Lemma 4.1.4. If and

j = 1, 2, are domains with finite perimeter, then are domains with finite perimeter.

Proof. Let

be the characteristic function (= indicator) for Then Therefore it is sufficient to prove that is a charge if are charges, j = 1,2, and in our case c = 1. It is proved in [VH], p. 189, that where is the average value of Since are function of bounded variation and are charges, then is a charge. Lemma 4.1.4 is proved. For convenience of the reader let us give the differentiation formula from [VH] for B V functions: if u, B V then where prime stands for a derivative, and are charges, and is the average value defined as where x is a regular point of v(x), i.e., and exist for some unit vector a, and

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Alternatively, if

e.g.

then one defines and This limit does exist if x is a regular point of v. One gets from (4.1.13) that w satisfies (4.1.2) and so in To complete the proof of Theorem 4.1.1 one has to prove (4.1.12). Suppose (4.1.12) is false. Then there is a sequence such that Define Then

By the argument given in step 2,

it follows that in and z solves (4.1.2), so z = 0 by Lemma 4.1.2. On the other hand, This contradiction proves (4.1.12). Theorem 4.1.1 is proved. 4.2

INVERSE OBSTACLE SCATTERING PROBLEMS

Let us now turn to the inverse scattering problem. The scattered field v admits a representation (4.1.12), since w = v for Thus

The function is called the scattering amplitude. The inverse obstacle scattering problem (IOSP) consists of function S and the boundary condition on S from the knowledge of the scattering amplitude. Consider three cases: (1) (2) (3)

is known for all and all k > 0, is known for all is known for all and

is fixed; is fixed, are fixed.

In case (1) the uniqueness theorem for IOSP was proved by M. Schiffer (1964), who assumed a priori that the boundary condition is the Dirichlet one. In case (2) the uniqueness theorem for IOSP was proved by A. G. Ramm (1985) (see [R83]), who also proved that the boundary condition is uniquely determined by the scattering amplitude. In case (3) the uniqueness theorem for IOSP is still an open problem, although it is a trivial remark that this uniqueness theorem holds if one assumes a priori that the obstacle is sufficiently small (cf. [R139], [R203]). The case of penetrable obstacles had also been considered: both the direct scattering problem (e.g., see [R71], [R83]), and the inverse scattering problem (see [RPY]) have been studied. Let us prove uniqueness theorem for IOSP. Throughout we take n = 3, but the proofs and the results remain valid in By we denote an open subset of and

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Theorem 4.2.1. Assume that D is a bounded domain with finite perimeter, and is known and Then S and the boundary condition on S are uniquely determined. If we assume additionally that is compact, and if we assume that and are compact. Proof. If one proves that S is uniquely defined by the data, then the boundary condition is also uniquely defined. Indeed, we prove below that the data determine uniquely the scattering solution in If S is known, then one can check if u = 0 on S, or on S, or calculate on S, thus determining uniquely the boundary condition on S. Let us prove that S is uniquely determined by the data. Assume the contrary. Let and generate the same data. Outside of the ball B(R), which contains the scattering solutions and are identical. This follows from Lemma 4.2.2. If u solves equation (*) in and then v = 0 in and in the connected component of the domain in which v solves (*).

as containing

Proof. The last statement of Lemma 4.2.2 follows from the unique continuation theorem for elliptic equations. Let us prove the first one. Write the solution to (*) as are constants, are Hankel functions, as

as Thus

If then and v = 0. Lemma 4.2.2 is proved.

Similar lemma was proved originally by V. Kupradze (1934), then by I. Vekua (1943), and F. Rellich (1943). Our proof is taken from [R83, p. 25]. T. Kato (1959) [Kat1] proved a similar lemma for solutions of the equation where as Indeed, if and generate the same data, then so, by Lemma 4.2.2, in Let be a connected component of Without loss of generality assume that for example. Then solves in equation (*) of Lemma 4.1.2 and satisfies one of the homogeneous boundary condition On the satisfies by definition, and on the satisfies because in Since solves (*) in for any we get a contradiction, because for This is proved as usual:

where If j = 1, 2, 3, then the surface integral vanishes and holds. Thus we constructed a continuum of orthogonal, not identically vanishing,

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functions in Since proves that the assumption

is a separable space, this is a contradiction, which is wrong. Theorem 4.2.1 is proved.

Remark 4.2.3. Our proof follows the ideas of [R171]. The original proof of M. Schiffer relied on the fact that the Dirichlet Laplacian in any bounded domain has a discrete spectrum, so it is a contradiction to have every an eigenvalue of the Dirichlet Laplacian in But if the Neumann Laplacian is considered then there are bounded domains in which this Laplacian has continuous spectrum. To avoid this difficulty, we rely on the separability of

Remark 4.2.4. The scattering amplitude corresponding to a bounded Lipschitz obstacle is a meromorphic function of k, and an analytic function of and on the variety [R83]. Therefore the knowledge of on open subsets of however small, and on an open set (a, b), determines A uniquely on Consider now IOSP with the fixed-frequency data Theorem 4.2.5. Under the assumption of Theorem 4.2.1 the fixed-frequency data is fixed, j = 1, 2, are open subsets of however small, determine S and the boundary condition uniquely. Proof. As in the proof of Theorem 4.2.1, it is sufficient to prove that S is uniquely determined by the data. We need Lemma 4.2.6. If

and

Proof. The function and if now from Lemma 4.2.2.

generate the same data

then

solves the homogeneous Helmholtz equation in The conclusion of Lemma 4.2.6 follows

If

and generate the same data then by Lemma 4.2.6. If then take a point same as in the proof of Theorem 4.2.1) and as Since one gets a contradiction, which proves that

Then

(the notations are the but

Theorem 4.2.5 is proved.

Remark 4.2.7. If the data are and are fixed, then it is not known whether these data determine S uniquely. However, if one knows a priori that the obstacle is sufficiently small, so that is not an eigenvalue of the Laplacian in D, corresponding to the boundary condition then, as in the proof of Theorem 4.2.5, one gets so the obstacle is determined uniquely by under this a priori assumption.

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Alternatively, if the obstacle has any given size, say its diameter is not greater than d, then m = 1, 2, . . . , M, p = 1, 2, . . . , p, determine D sufficiently many data uniquely. One should take M + P greater than the upper bound on the number of the eigenvalues of the Laplacian in D, and then the uniqueness of the solution to IOSP with the above data holds. Remark 4.2.8. Let us discuss the case when the region does not have interior points, for example, it is an open surface in The proof of Theorem 4.2.1 and Theorem 4.2.5 covers this case as well: such a region is uniquely defined by the data of each of these theorems. Indeed, suppose that there are two regions and Then and as This contradiction shows that If then

Let

be the edge of Then as This and the relation (4.2.0) imply if If but then there is a bounded domain D with the boundary Since the region D has interior points, and the proof of Theorem 4.2.1 applies. This proof yields a contradiction, which shows that Let us consider the following inverse scattering problem. Let

in and k are positive constants, conditions hold on S:

in D,

where and the transmission boundary

where denote limiting values on S from N is the outer normal to S pointing into S is Lipschitz. In we assume that u satisfies the radiation condition: is fixed, The inverse scattering problem consists of recovery of S, and k from the scattering data where and are known. Our main result is Theorem 4.2.9. [RPY] The data

determine S,

and k uniquely.

Proof. Assume there are two sets j = 1,2, which generate the same data are the corresponding solutions to (4.2.1)–(4.2.2). Then in

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and the weak formulation of the problems (4.2.1)–(4.2.2) imply:

Define

where is the resolvent kernel corresponding to the operator problem (4.2.1)–(4.2.2),

j = 1, 2, of the

From (4.2.4) and Lemma 4.2.2 it follows that w = 0 in If then there is a point and a ball

such that

The integrals on the right-hand side of (4.2.5) are bounded as therefore so is the integral on the left-hand side of (4.2.5). Choose is a cut-off function, in a neighborhood of and vanishes outside B(R), Let be local coordinates in B = B(t, r), with the origin at the point t and axis directed along the normal to at the point t. Let If then the author has proved ([R180]) that

where and one can differentiate (4.2.6). If and

and

then

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Thus

and, as

one gets:

Therefore, as like

the left-hand side of (4.2.5) behaves

For (4.2.7) to hold it is necessary that This contradiction shows that In deriving (4.2.7), one takes into account that: (1) in B because (2) r = R if and

and

Let us prove that is uniquely determined by the scattering data. We have already proved that so If then, as the left-hand side of (4.2.5) is bounded and behaves like

Thus as claimed. Finally, let us prove that If this is not true, then Let tion one gets

and using the formulas for

one concludes that

is also uniquely determined by the scattering data. We derive a contradiction from this assumption. Subtracting from the equation equa-

This orthogonality relation follows from the formula where the boundary integrals vanish because on S. The set of products is complete in by property C for the pair (see Section 5.6). Therefore p = 0, and Theorem 4.2.9 is proved.

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Exercise (cf. [R174]). Let (4.1.1) hold and assume that the boundary S is a union of two connected sets and the normal derivative on and on The total cross-section is fixed. The function w can be treated as a control function. Prove that can be chosen so that the cross-section becomes as small as one wishes, that is, This means that one can make an obstacle (say, an aircraft) practically invisible for any given fixed direction of an incident field at any fixed wavenumber k > 0. 4.3

STABILITY ESTIMATES FOR THE SOLUTION TO IOSP

Suppose there are two bounded star-shaped obstacles j = 1,2, that is, their boundaries can be represented by the equations j = 1,2. Assume that where and are constants. If is the local equation of a surface then we assume that where The set of all such star-shaped surfaces S (or obstacles D) is denoted by Let be the Hausdorff distance between obstacles and Let be a connected component of and Suppose that where Assume that is fixed. Our basic result is (see [R162]): Theorem 4.3.1. Under the above assumptions one has constants independent of

are

Suppose is the scattering amplitude corresponding to an obstacle Let us define an algebraic variety How does one reconstruct S from We prove Proposition 4.3.2. There exists a function

where

such that

is the indicator of D,

Remark 4.3.3. It is an open problem to construct an algorithm for calculating given Such an algorithm we construct for inverse potential scattering theory in Chapter 5.

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Proof of Theorem 4.3.1. First we prove that Then we prove that in By we denote various positive constants. Next, we prove that

Here Estimate (4.3.1) implies the conclusion of Theorem 4.3.1. Let us go through the steps of the proof. By we denote the scattering solution corresponding to Step 1. Assume that corresponding boundaries. Since again by which converges Since that one gets so Step 2. Since

on

and let j = 1, 2, be the one can select a subsequence, denoted (as in and depends on continuously, it follows from By Theorem 4.2.5 of Section 4.2, in contradiction with Thus

one has

Step 3. Let From the results of Section 5.4 it follows that

for

R.Let

Let us prove the estimate:

One has

The function can be continued analytically on the complex plane to the sector if for We assume so that when Then and extends from into as function. Take a point and let K be a cone with vertex x, with opening angle Such a cone does exist since is sufficiently smooth. One has and By the two constants theorem [E], p. 296, one gets where is the harmonic measure of the set with respect to the domain is the union of the two rays which form the boundary of the sector and of the ray L. Let us prove the estimate:

as

along the real axis, c and are positive constants. Let Let us map conformally the sector onto the half-plane

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and

using the map Then L is mapped onto the ray (see [E], p. 293). By the Hopf lemma ([E], p. 34), so as Thus (4.3.2) follows, and

On the other hand,

Thus

so

stants. This implies

are con-

where

(4.3.3) in more detail. One has Also, on so Since

Finally let us explain on

on because on and in a neighborhood of

one gets

Theorem 4.3.1 is proved. Remark 4.3.4. For convenience of the reader we recall the known results about harmonic measure and two-constants theorem (see [E] for details). Let D be a domain on the complex plane, Define the harmonic measure of E relative to D at the point denoted as the value of harmonic function in D, which is equal to 1 on E and equal to zero on S\ E, If a function w := w(z) maps conformally the domain D onto a domain and the set E is mapped onto then The theorem about two constants says: if f (z) is holomorphic in the domain D with the boundary S, and then

Let us now prove Proposition 4.3.2. First we prove: Lemma 4.3.5. ([R139, p. 183]). If u(x, k), solution, then the set is complete in

k = const > 0, is the scattering

Proof of Lemma 4.3.5. If the conclusion is not true, then there is a such that This implies where G(x, y, k) is the Green’s function (resolvent kernel of the Dirichlet Laplacian in (see Lemma 4.1.3). Since as one gets p (s) = 0 on S. Here is the delta-function on S. The Lemma is proved.

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From Lemma 4.3.5 it follows that there exists a function

where is arbitrary small, and vector. One has

Multiplying this equation by gets:

such that

is an arbitrary fixed

and integrating over

with respect to

one

One has

Proposition 4.3.2 is proved. A different type of stability estimates for obstacle scattering problem are obtained in [R164], [R7]. There one solves a potential scattering problem with potential where t > 0 is a parameter, and is the characteristic function of the domain D, the obstacle. The limit, as of the scattering solution for this problem, is the scattering solution of the obstacle scattering problem with the Dirichlet boundary condition on the boundary S of D, and the scattering amplitude for the potential scattering problem tends to the scattering amplitude for the obstacle scattering problem as In [R164] one finds the estimates of the rate of convergence as For example, where c = const > 0 does not depend on t. 4.4

HIGH-FREQUENCY ASYMPTOTICS OF THE SCATTERING AMPLITUDE AND INVERSE SCATTERING PROBLEM

Let us assume that D is a bounded, strictly convex obstacle with boundary S. Let be a unit vector, be the illuminated part of S by the parallel straight lines in the direction of and be the shadowed part. Define the support function as this is the distance from the origin to the tangent plane to S, orthogonal to and closest to the origin. The equation of this plane is The origin is chosen so that is not empty.

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The surface S is the envelope of the set of these tangent planes. Therefore the equation of S is

One has to understand the derivatives in (4.4.1) as follows. The function is defined originally on because We extend this function to as a homogeneous function of order 1, so that t = const. Thus by Euler’s theorem. The differentiation in (4.4.1) is the differentiation of the extended to function For example, if then and therefore the relation is not correct. If one recovers the function from the scattering data then the surface S is uniquely and constructively recovered by formula (4.4.1). Let us show how to find from the data known for We assume the Dirichlet condition on S for definiteness. In this case

Then k ) is the scattering solution. Denote and taking the normal derivative at the boundary yields the equation for In the on S, on high-frequency Kirchhoff approximation, one takes on Thus, in this approximation,

where u (x,

Let us calculate this integral using the stationary phase method ([H]):

as Here is a smooth real-valued function which has only one nondegenerate critical point i.e., where is the number of positive (negative) eigenvalues of the matrix and f (x) is a smooth function, The smoothness requirements on f and can be relaxed. In our case The stationary point (critical point) of the function In a neighborhood of this point at which is the point one has where m = 1, 2, are local is the coordinates on S, (0, 0) are the coordinates on S of the stationary point the Gaussian second differential form of the surface S at the point curvatures of S at K > 0 since S is strictly convex. Using the stationary phase

245

formula, one gets

From (4.4.2) one can find and then S by formula (4.4.1). Although the phase is determined modulo the additive constant factor does not bring difficulties. For example, choose Then One has

and

where m is an integer. Therefore, the parametric equation of S is:

Alternatively, one can determine using various and in (4.4.2) (cf. [GR5], where numerical experiments are given). Finally, let us derive the stability estimates for the recovery of S from the scattering data by formula (4.4.1). Suppose that (the noisy data) is known in place of and Then S can be stably recovered by formulas (4.4.1) if one uses stable differentiation formulas from Section 2.5.1. One can also estimate the error of recovery of S: this error is because of the estimate which holds for a surface (See Section 2.5.1 for the error estimates of the formulas of stable differentiation.) Remark 4.4.1. In [GR5] the above results were used as a basis for an efficient numerical method for finding an obstacle from the scattering data. In [R83] an analytic method is given for finding a closed surface from the knowledge of its two principal curvatures. This is an overdetermined problem, because the Gaussian curvature above determines uniquely the surface under suitable assumptions. However, if one knows two principal curvatures one has a constructive method for finding the corresponding S [R83, p. 358], while if one knows the Gaussian curvature, there is no constructive method for finding S. 4.5

REMARKS ABOUT NUMERICAL METHODS FOR FINDING S FROM THE SCATTERING DATA

There is no numerical method for finding the surface from the scattering data similar to the method developed in Chapter 5 for finding the potential from the scattering data. There are several parameter-fitting schemes for finding S from the scattering

246


data. Let us describe one of them ([R159], [R139]) which can be also used for solving inverse potential and inverse geophysical scattering problems. Assume that is the equation of S, the data are the values is fixed, the Dirichlet condition holds on S (other conditions can be treated similarly), is the scattering solution, is its normal derivative on S, d s is the surface area element Consider the functional:

where

and

Theorem 4.5.1. Functional (4.5.1) has global minimum equal to zero, and it has unique global minimizer {f, h}. The f component of this minimizer determines the surface S by the equation Proof. Since u = 0 on S, it follows that is the equation of S and where the global minimizer of is unique. Assume that and

and if Let us prove that Then Define

Then in u = 0 on satisfies the radiation condition, and the corresponding to scattering amplitude is By the uniqueness theorem (Theorem 4.2.5) it follows that Theorem 4.5.1 is proved. Remark 4.5.2. For the inverse potential scattering problem the functional similar to (4.5.1) is

where for is fixed, a > 0 is an arbitrary large fixed number.

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A theorem similar to Theorem 4.5.1 is proved in [Rl39]: functional (4.5.2) has the global minimum equal to zero and the unique global minimizer, {q, v}, and the q defined by this minimizer, solves the inverse potential scattering problem with fixed-energy data.

In [CK] and in many papers some functionals are minimized for solving inverse obstacle scattering problem (IOSP), and these functionals do not attain their infimum. In contrast, functionals (4.5.1) and (4.5.2) do attain their infimum, this infimum equals to zero, and is attained on a unique global minimizer whis is the solution of IOSP. However, solving IOSP by numerical minimization of any functional is a parameterfitting procedure which does not allow one to get estimates of error of the solution and does not give any information about stability of the solution with respect to noise in the data. Therefore, altough parameter-fitting procedures are applied widely for solving various inverse problems in practice, they do not provide reliable solutions to these problems. 4.6

ANALYSIS OF A METHOD FOR IDENTIFICATION OF OBSTACLES

In this Section some difficulties are pointed out in the methods for identification of obstacles based on the numerical verification of the inclusion of a function in the range of an operator. Numerical examples are given to illustrate theoretical conclusions. Alternative methods of identification of obstacles are mentioned: the Support Function Method (SFM) and the Modified Rayleigh Conjecture (MRC) method. During the last decade there are many papers published, in which methods for identification of an obstacle are proposed, which are based on a numerical verification of the inclusion of some function in the range R(B) of a certain operator B. Examples of such methods include [CK], [CCM], [Kir1]. It is proved in this paper that the methods, proposed in the above papers, have essential difficulties. This also is demonstrated by numerical experiments. Although it is true that when it turns out that in any neighborhood of f, however small, there are elements from R(B). Also, although when there are elements in every neighborhood of f, however small, which do not belong to R(B) even if Therefore it is not possible to construct a stable numerical method for identification of D based on checking the inclusions and We prove below that the range R(B) is dense in the space Assumption (A): We assume throughout that is not a Dirichlet eigenvalue of the Lapladan in D. Let us introduce some notations: N(B] and R(B) are, respectively, the null-space and the range of a linear operator B, is a bounded domain (obstacle) with a smooth boundary S, is a unit vector, N is the unit normal to S pointing into where and is the scattering solution:

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where is called the scattering amplitude, corresponding to the obstacle D and the Dirichlet boundary condition. Let G = G(x, y, k) be the resolvent kernel of the Dirichlet Laplacian in

and G satisfies the outgoing radiation condition. If

and w satisfies the radiation condition, then ([R83]) one has

We write

for

and

as follows from Lemma 4.1.3. One can write the scattering amplitude as:

The following claim is proved in [Kir1]: Claim 10.

if and only if

Proof of the claim. Our proof is based on the results in [R83]. a) Let us assume that f = Bh, i.e., and prove that Define where The function p (y) solves the Helmholtz equation (4.6.4) in the region and as because of (4.6.7) and of the relation Bh = f. Therefore (see Lemma 4.1.2) p = 0 in the region Since is bounded in and as we get a contradiction unless Thus, implies b) Let us prove that implies Define and h := g (s, z). Then, by Green’s formula, one has Taking one gets f = Bh, so The claim is proved. Consider (4.6.6) and

and

Theorem 4.6.1. The ranges R(B) and R(A) are dense in

where B is defined in

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Proof. Recall that Assumption (A) holds. It is sufficient to prove that N(B*) = {0} and N(A*) = {0}. Assume where the overline stands for complex conjugate. Taking complex conjugate and denoting by q again, one gets Define Then on S, and w solves equation (4.6.1) in By the uniqueness of the solution to the Cauchy problem, w = 0 in Let us derive from this that q = 0. One has where and satisfies the radiation condition. Therefore, in as follows from the Lemma 4.6.2 proved below. By the unique continuation, in and this implies q = 0 by the injectivity of the Fourier transform. This proves the first statement of Theorem 4.6.1. Its second statement is proved below. Let us now prove the Lemma, mentioned above. We keep the notations used in the above proof. Lemma 4.6.2. If

in

Proof. The idea of the proof is simple: since and V satisfies it, one concludes that is ([R83], p.54):

then

in

does not satisfy the radiation condition, Let us give the details. The key formula

where and one assumes If then, by Lemma 4.6.2, assuming and using the relation in one gets for all Thus, Lemma 4.6.2 is proved under the additional assumption If then one uses a similar argument in a weak sense, i.e., with one considers the inner product in of and a smooth test function and applies Lemma 4.6.2 to the function Then, using arbitrariness of h, one concludes that q = 0 as an element of Lemma 4.6.2 is proved. Let us prove the second statement of Theorem 4.6.1.

Proof. Assume now that A*q = 0. Taking complex conjugate, and using the reciprocity relation: one gets an equation:

where

Define Then where and satisfies the radiation condition. Equation (4.6.9) implies that as Since function V solves equation (4.6.1) and one concludes that V = 0 in so that in Thus, Since solves equation (4.6.1) in D and one gets,

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using Assumption (A), that in D. This and the unique continuation property imply in Consequently, h = 0, so q = 0, as claimed. Theorem 4.6.1 is proved. Remark 4.6.3. In [CK] the 2D inverse obstacle scattering problem is considered. It is proposed to solve the equation (1.9) in [CK]:

where A is the scattering amplitude at a fixed k > 0, is the unit circle, and z is a point in If is found, the boundary S of the obstacle is to be found by finding those z for which is maximal. Assuming that is not a Dirichlet or Neumann eigenvalue of the Laplacian in D, that D is a smooth, bounded, simply connected domain, the authors state Theorem 2.1 [CK], p. 386, which says that for every there exists a function such that

and (see [CK], p. 386),

There are several questions concerning the proposed method. First, equation (4.6.10), in general, is not solvable. The authors propose to solve it approximately, by a regularization method. The regularization method applies for stable solution of solvable ill-posed equations (with exact or noisy data). If equation (4.6.10) is not solvable, it is not clear what numerical “solution” one seeks by a regularization method. Secondly, since the kernel of the integral operator in (4.6.11) is smooth, one can always find, for any infinitely many with arbitrary large such that (4.6.12) holds. Therefore it is not clear how and why, using (4.6.11), one can find S numerically by the proposed method. Remark 4.6.4. In [CK], p. 386, Theorem 2.1, it is claimed that for every and every there exists a function such that inequality (4.6.12) (which is (2.8) on p. 386 of [CK]) holds and as Such a is used in [CK] in a “simple method for solving inverse scattering problem”. However, there exist infinitely many such that inequality (4.6.12) holds and regardless of where is. Therefore it is not clear how one can use the method proposed in [CK] for solving the inverse scattering problem with any degree of confidence in the result. Remark 4.6.5. In [BLW] it is mentioned that the methods (called LSM-linear sampling methods) proposed in papers [CCM], [CK], [Kir1] produce numerically results which are inferior

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to these obtained by the linearized Born-type inversion. There is no guarantee of any accuracy in recovery of the obstacle by LSM. Therefore it is of interest to experiment numerically with other inversion methods. In [R83], p. 94, (see also [R139], [R75], [R76]) a method (SFM-support function method) is proposed for recovery of strictly convex obstacles from the scattering amplitude. This method allows one to recover the support function of the obstacle, and the boundary of the obstacle is obtained from this function explicitly. Error estimates for this method are obtained in the case when the data are noisy [R83], p. 104. This method is asymptotically exact for large wavenumbers, but it works numerically even for ka ~ 1, as shown in [GR5]. For the Dirichlet, Neumann and Robin boundary conditions this method allows one to recover the support function without a priori knowledge of the boundary condition. If the obstacle is not convex, the method recovers the convex hull of the obstacle. Numerically one can recover the obstacle, after its convex hull is found, by using Modified Rayleigh Conjecture (MRC) method, introduced in [R205], or by a parameter-fitting method. In Proposition 4.3.2 (see also [R162]) a formula for finding an acoustically soft obstacle from the fixed-frequency scattering data is given. It is an open problem to develop an algorithm based on this formula. The numerical implementation of the Linear Sampling Method (LSM) suggested in [CK] consists of solving a discretized version of

where A is the scattering amplitude at a fixed k > 0, is the unit circle, and z is a point on Let i = 1, . . . , N, j = 1, . . . , N be the square matrix formed by the measurements of the scattering amplitude for N incoming, and N outgoing directions. then the discretized version of (4.6.13) is

where the vector f is formed by

see [BLW] for details. Denote the Singular Value Decomposition of the far field operator by Let be the singular values of F, and Then the norm of the sought function g is given by

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Figure 9. Identification of two circles of radius 1.0 centered at ( – d , 0.0) and ( d , 0.0) for d = 2.0.

A different LSM is suggested in [Kir1]. In it one solves

instead of (4.6.14). The corresponding expression for the norm of

is

A detailed numerical comparison of the two LSMs and the linearized tomographic inverse scattering is given in [BLW]. The conclusions of [BLW], as well as of our own numerical experiments are that the method (4.6.18) gives a somewhat better, but a comparable identification, than (4.6.16). The identification is significantly deteriorating if the scattering amplitude is available only for a limited aperture, or if the data are corrupted by noise. Also, the points with the smallest values of the are the best in locating the inclusion, and not the largest one, as required by the theory in [Kir1] and in [CK]. In Figures 9 and 10 (cf [RGu2]) the implementation of the Colton-Kirsch LSM (4.6.17) is denoted by g nck, and of the Kirsch method (4.6.18) by g n k. The Figures

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Figure 10. Identification of two circles of radius 1.0 centered at ( – d , 0.0) and ( d , 0.0) for d = 1.5.

show a contour plot of the logarithm of the The original obstacle consisted of two circles of radius 1.0 centered at the points ( – d , 0.0) and ( d , 0.0). The results of the identification for d = 2.0 are shown in Figure 9, and the results for d = 1.5 are shown in Figure 10. Note that the actual radius of the circles is 1.0, but it cannot be seen from the LSM identification. Also, one cannot determine the separation between the circles, nor their shapes. Still, the methods are fast, they locate the obstacles, and do not require any knowledge of the boundary conditions on the obstacle. The Support Function Method ([GR5], [R83]) showed a better identification for the convex parts of obstacles. Its generalization for unknown boundary conditions is discussed in [RGu3]. The LSM identification was performed for the scattering amplitude of the obstacles computed by the boundary integral equations method, see [CK]. No noise was added to the synthetic data. In all the experiments we used k = 1.0, and N = 60. In [GRS] the concept of stability index is introduced and applied to a parameterfitting scheme for solving a one-dimensional inverse scattering problem in quantum physics. This concept allows one to get some idea about the error estimate in a parameter-fitting scheme. In this Section we have used [RGu2].


5. STABILITY OF THE SOLUTIONS TO 3 D INVERSE SCATTERING PROBLEMS WITH FIXED-ENERGY DATA

In this Chapter some of the author’s results on inverse potential scattering problem with fixed-energy data are presented. The presentation is based on paper [R203]. Inversion formulas and stability results for the solutions to 3D Inverse scattering problems with fixed energy data are obtained. Inversion of exact and noisy data is considered. The inverse potential scattering problem with fixed-energy scattering data is discussed in detail, inversion formulas for the exact and for noisy data are derived, error estimates for the inversion formulas are obtained. Global estimates for the scattering amplitude are given when the potential grows to infinity in a bounded domain. Inverse geophysical scattering problem is discussed briefly. An algorithm for constructing the Dirichletto-Neumann map from the scattering amplitude and vice versa is obtained.

5.1

INTRODUCTION

In this Chapter 3 D inversion scattering problems with fixed-energy data are discussed. These problems include inverse problems of potential, obstacle, and geophysical scattering (IPS, IOS, IGS). Inverse potential scattering problem is discussed in detail: uniqueness of its solution, reconstruction formulas for inversion of the exact data and for inversion of noisy data are given and error estimates for these formulas are obtained. These estimates yield the Stability estimates for the solution of the inverse scattering problem. For the inverse obstacle scattering the uniqueness theorem is proved for rough domains, stability estimates are obtained for domains, that is, for

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5. Inverse scattering problem

domains whose boundary in local coordinates is a graph of function, and reconstruction formulas are discussed in Chapter 4. For inverse geophysical scattering the inverse scattering problem is reduced to inverse scattering problem for a potential. Construction of the Dirichlet-to-Neumann map from the scattering data and vice versa is given. Analytical example of nonuniqueness of the solution of an inverse 3D problem of geophysics is given. The results discussed below were obtained mostly by the author, (see [R203], [R139] and the bibliography of the author’s papers). The presentation and some of the estimates are improved. Only some of the results from the cited papers are included. 5.1.1

The direct potential scattering problem

We want to study the inverse potential scattering problem of finding q (x) given some scattering data. Consider the direct scattering problem first and let us formulate some basic results which we need. Let

Here u(x, k) is the scattering solution, k = const > 0 is fixed. Without loss of generality we take k = 1 in what follows unless other choice is suggested explicitly. A unit vector is given, where is the unit sphere in Vector has a physical meaning of the direction of the incident plane wave, while is the direction of the scattered wave, is the fixed energy. The function is called the scattering amplitude. It describes the first term of the asymptotics of the scattered field as along the direction The function q (x) is called the potential. We assume that

where a > 0 is an arbitrary large fixed number which we call the range of q (x), and the overbar stands for complex conjugate. In many results is sufficient, but is used in the proof of a crucial estimate (5.2.17) below. 5.1.2 Review of the known results

Let us formulate some of the known results about the solution to problem (5.1.1)– (5.1.2), the scattering solution. These results can be found in many books, for example,

257

in the appendix to [R121], where a brief but self-contained presentation of the scattering theory is given. The scattering problem has a unique solution if

In fact, the above result is proved for much larger class of q ([H]), but for inverse scattering problem with noisy data it is necessary to assume q (x) compactly supported [R139]. Indeed, represent the potential q (x) as where for and for Call the tail of the potential q . If one assumes a priori that where b > 3, then the contribution of the tail of the potential to the scattering amplitude is of order and tends to 0 as At some value of a, say at this contribution becomes of the order of the noise in the scattering data. One cannot, in principle, discriminate between the noise and the contribution of the tail of the potential for Therefore the tail of q for cannot be determined from noisy data. One has

By c > 0 we denote various constants. If then u(x, k) extends as a meromorphic function to the whole complex k-plane. Let G(x, y, k) denote the resolvent kernel of the self-adjoint Schrödinger operator in

The function u (x, k) can be defined by the formula:

where is uniform with respect to x varying in compact sets and formula (5.1.7) can be differentiated with respect to x [R83] (cf. Lemma 4.1.3). The function G(x, y, k) is a meromorphic function of k on the whole complex k-plane. It has at most finitely many simple poles in and if infinitely many poles, possibly not simple, in There are no poles on the real line except, possibly, at k = 0. The functions solving (5.1.1) with are called eigenfunctions of the discrete spectrum of L, are the negative eigenvalues of L. There are at most finitely many of these if The eigenfunction expansion formulas are known:

258


where

(see e.g. [R121], [CFKS]). In Section 5.1 the results are collected whose proofs are given, e.g., in [R121], Appendix and in [R139]. If is the resolution of the identity of the selfadjoint operator L, and is its kernel, then

Properties of the scattering amplitude

The scattering amplitude has the following well-known properties:

In particular,

If

and k = 1, then the scattering amplitude is an analytic function of on the algebraic variety

This variety is non-compact, intersects (many) such that

over

and, given any

In particular, if one chooses the coordinate system in which the unit vector along the then the vectors

and

there exist

is

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satisfy (5.1.9) for any complex numbers and satisfying the last equation in (5.1.10) and such that There are infinitely many such If than the function is a meromorphic function of which has poles at the same points as G(x, y, k). One has

The S-matrix is defined by the formula

and is a unitary operator in If then

Thus, A is a normal operator in

Therefore

and

The fundamental equation

Denote

Then

that is

Completeness properties of the scattering solutions

(a) If

and

then

Let main,

in D, is the Sobolev space.

where

is a bounded do-

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(b) The set small, and any fixed

The

is total in

that is, for any however there exists such that

depends on w(x).

Special solutions

There exists

where

such that

is an arbitrary bounded domain.

Property C for the pair

Let be linear formal partial differential operators, that is, formal differential expressions. Let in D}, where is an arbitrary fixed bounded domain and the equation is understood in the sense of the distribution theory. Consider the subsets of j = 1,2, which form an algebra in the sense that the products where and If p = 1 define and if define We write meaning that run through the above subsets of Definition 5.1.1. We say that the pair of linear partial differential operators property if and only if the set is total in that is, if

then

If the above holds for p = 2, we say that property C holds for the pair

Theorem 5.1.2. ([R100], [R109]) Let is arbitrary fixed. Then the pair

has property C.

has and

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Proof. Note that j = 1, 2, where loss of generality take k = 1, let

Choose

are defined in Section 5.1.2. Without One has (see (5.1.19)):

such that (5.1.9) holds with an arbitrary fixed

Since the set is total in conclusion of Theorem 5.1.2 follows.

Then

is a bounded domain, the

Remark 5.1.3. One cannot take unbounded domain D in the above argument because in (5.1.24) holds for bounded domains. One can take the space of f (x) larger than for example, the space of distribution of finite order of singulartiy if q (x) is sufficiently smooth [R139]. Theorem 5.1.4. The set where is an arbitrary fixed bounded domain, and

is complete in is fixed.

Proof. The conclusion of Theorem 5.1.4 follows from Theorem 5.1.2 and (5.1.18). Properties of the Fourier coefficients of

We denote

and write

where summation over m is understood in (5.1.25) and similar formulas below, e.g. (5.1.31), (5.1.37), etc.

Here nomial,

is the Legendre polyare the angles corresponding to the point

Consider a subset where and

consisting of the vectors run through the whole complex plane. Clearly

but

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is a proper subset of M. Indeed, any with is an element of If then so and one gets However, there are vectors which do not belong to Such vectors one obtains choosing such that There are infinitely many such vectors. The same is true for vectors Note that in (5.1.9) one can replace M by for any Let us state two estimates ([R139]):

and

where

and is the Bessel function regular at r = 0. Note that admits a natural analytic continuation from to M by taking be arbitrary complex numbers. The resulting

defined by (5.1.26), and in (5.1.26) to

A global perturbation formula

Let

be the scattering amplitude corresponding to Then [R139]

j = 1,2. Define

Formula for the scattering solution outside the support of the potential

Let supp the data tering solution

where

The fixed-energy scattering data or, equivalently, allow one to write an analytic formula for the scatin the region



263

is the Hankel function, and the normalizing factor is chosen so that

Formula (5.1.31) follows from (5.1.2), (5.1.25), (5.1.32) and Lemma 4.1.2. Note that [GR], formula (7.1463):

where This formula implies that is a monotonically increasing function of It is known [GR, formula 8.478] that is a monotonically decreasing function of r if and

The following known estimate (e.g., [MPr]) can be useful:

where are the normalized in spherical harmonics (5.1.26). Let us give a formula for the Green function G(x, y, k) (see (5.1.5), (5.1.6)) in the region where Let and denote by the Fourier coefficients of the scattering amplitude:

Then

where

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Indeed, clearly the function (5.1.37) solves (5.1.5) in the region where q (x) = 0, it satisfies (5.1.6), and

By (5.1.36), (5.1.31), (5.1.25), and (5.1.7), it follows that the function (5.1.37) has the same main term of asymptotics (5.1.38) as the Green function of the Schrödinger operator. Therefore the function (5.1.37) is identical to the Green function (5.1.5)– (5.1.6) in the region 5.2

INVERSE POTENTIAL SCATTERING PROBLEM WITH FIXED-ENERGY DATA

The IPS problem can now be formulated: given Throughout this section k = 1. 5.2.1

find

Uniqueness theorem

The first result is the uniqueness theorem of Ramm ([R100], [R109]). Theorem 5.2.1. (Ramm) If and where j = 1, 2, are arbitrary small open subsets of

then

Proof. The function is analytic with respect to and on the variety (5.1.8). Therefore its values on extend uniquely by analyticity to M × M. In particular is uniquely determined in By (5.1.30) one gets:

By property C (formulas (5.1.22)–(5.1.23)) and by (5.1.18), the orthogonality relation (5.2.1) implies

5.2.2 Reconstruction formula for exact data

Fix an arbitrary Denote

and choose arbitrary

satisfying (5.1.9).

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Multiply (5.1.11) by over with respect to

where

will be fixed later, and integrate

If then estimates (5.1.27) and (5.1.28) imply that the series (5.1.25) converges, when is replaced by uniformly and absolutely on where is an arbitrary compact subset of M. Formula (5.1.11) implies that can be replaced by since is a compact set in Define

and rewrite (5.2.3), with

as

where

The following estimate, proved in Section 5.4.3 below, holds for a suitable choice of

From (5.2.5) and (5.2.7) one gets the reconstruction formula for inversion of exact, fixedenergy, three-dimensional scattering data:

and the error estimate:

where (5.1.9) is always assumed.

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Let us give an algorithm for computing the function (5.2.9), hold, given the scattering data

Fix arbitrarily two numbers

and define the

for which (5.2.7), and therefore

and b such that

in the annulus:

Consider the minimization problem

where the infimum is taken over all It is proved in Section 5.4.3 that

and

is given by (5.2.4).

The symbol means that is sufficiently large. The constant c > 0 in (5.2.13) depends on the norm but not on the potential q (x) itself. An algorithm for computing a function which can be used for inversion of the fixed-energy 3D scattering data by formula (5.2.9), is as follows: a) Find any approximate solution to (5.2.12) in the sense

where in place of 2 in (5.2.14) one could put any fixed constant greater than 1. (b) Any such generates an estimate of with the error This estimate is calculated by the formula

where

is any function satisfying (5.2.14).

We have obtained the following result: Theorem 5.2.2. If (5.1.3), (5.1.9) and (5.2.14) hold, then

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Proof. The proof is the same as the proof of (5.2.5)–(5.2.7) and is based on the following estimate:

The proof of (5.2.17) is given in Section 5.4.3. 5.2.3

Stability estimate for inversion of the exact data

Let the potentials Let us assume that

j = 1, 2, generate the scattering amplitudes

We want to estimate The main tool is formula (5.1.30). The result is: Theorem 5.2.3. If

and (5.2.18) holds then

where the constant c > 0 does not depend on

Proof. Multiply both sides of (5.1.30) by and integrate with respect to

Choose

where as depend on

and

where and over

j = 1, 2, to get:

such that

and note that and c > 0 in (5.2.21) does not From (5.2.18), (5.2.20) and (5.2.21) one gets

268


where

One can choose

and

such that (see Section 5.4.7):

where c > 0 stands for various constants. Thus (5.2.22) yields:

where c, and are some positive constants, means that small and means that s > 0 is large. However, our argument is valid for and One gets

is

and the minimizer is

From (5.2.24)–(5.2.26) one gets (5.2.19).

Remark 5.2.4. In the above proof the difficult part is the proof of (5.2.23). Estimate (5.2.23) can be derived for approximating the minimizer of the problem:

and

where the infimum is taken over all

In Section 5.4.7 we consider the problem of finding with minimal norm among all which satisfy the inequality:

269

The necessity to consider the with the minimal norm simple observation: there exists a sequence of

comes from the such that

To prove (5.2.29) note that

where the operator is a continuous bijection of onto itself, and is the usual space of continuous in functions equipped with the sup-norm (e.g. see Appendix in [R121]). Since T is compact in the above statement follows from the injectivity of I + T, which we now prove: If f + Tf = 0, then f is extended to by the formula f = – T f , and satisfies the following equation in and the radiation condition of the type (5.1.6) with k = 1. Therefore and the injectivity of I + T is proved. Thus I + T and are continuous bijections of into itself for any Writing one concludes that (5.2.29) is equivalent to

Existence of a normalized sequence ness of the operator

satifying (5.2.31) follows from the compact-

Of course, the same argument is applicable to the operator but the bijectivity of I + T in is of independent interest. It follows from (5.2.29) that, for a given one can find v in (5.2.29) with an arbitrary large norm By this reason we are interested in v with minimal norm. Estimate (5.2.23) gives a bound on the growth of the minimal value of the norm where satisfies (5.2.28) with

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5.2.4


Stability estimate for inversion of noisy data

Assume now that the scattering data are given with some error: a function is given such that

We emphasize that is not necessarily a scattering amplitude corresponding to some potential, it is an arbitrary function in satisfying (5.2.32). It is assumed that the unknown function is the scattering amplitude corresponding to a The problem is: Find an algorithm for calculating

and estimate the rate at which

such that

tends to zero.

An algorithm for inversion of noisy data will now be described. Let

where [x] is the integer nearest to x > 0,

Consider the variational problem with constraints:

271

the norm is defined in (5.2.11), and it is assumed that

where is an arbitrary fixed vector, c > 0 is a sufficiently large constant, and the supremum is taken over and under the constraint (5.2.41). Given one can always find and such that (5.2.42) holds. We prove that in fact

Let that

be any approximate solution to problem (5.2.40)–(5.2.41) in the sense

Calculate

Theorem 5.2.5. If (5.2.42) and (5.2.44) hold, then

Proof. One has:

The rest of the proof consists of the following steps: Step 1. We prove that

where the norm is defined in (5.2.11).

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This estimate and (5.2.43) imply (see the proof of (5.2.19) and (5.2.26)) that

Step 2. We prove that

where and the pair solves (5.2.40)–(5.2.41) approximately in the sense specified above. (See formula (5.2.44)). This estimate follows from (5.2.41) and from the inequality

Let us prove (5.2.51). One has

where

is given in (5.2.34) and

Using (5.1.28), (5.1.29) one gets

Here we have used the estimate

which follows from (5.2.32) and the Parseval equality, and which implies

We also took into account that there are with because large N one has c > 1.

spherical harmonics and

For so we write

273

To estimate

use (5.1.27)–(5.1.29) and get:

Minimizing with respect to N > 1 the function

one gets

where is given in (5.2.34) and is defined by (5.2.38). Thus, from (5.2.52)– (5.2.56) one gets (5.2.51). Theorem 5.2.5 is proved. 5.2.5

Stability estimate for the scattering solutions

Let us assume (5.2.18) and derive the following estimate: Theorem 5.2.6. If

where

and (5.2.18) holds then

is defined by (5.2.38) and (5.2.34), and c > 0 is a constant.

Proof. Using (5.1.31), one gets:

As stated below formula (5.1.33), one has


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It follows from (5.2.60) and (5.1.34) that

From (5.2.60)–(5.2.61) one gets

where we have used monotone decay of as a function of r. Using estimate (5.1.27) in order to estimate j = 1, 2, one gets:

Minimization of the right-hand side of (5.2.63) with respect to (5.2.56), the estimate similar to (5.2.57):

Since

yields, as in

solves the equation

one can use the known elliptic estimate:

where

is a strictly inner subset of

where is any annulus implies (5.2.58) in 5.2.6

If

and

and get:

By the embedding theorem, (5.2.68)

Spherically symmetric potentials

then

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In Section 5.6 the converse is proved: if and (5.2.69) holds then q(x) = q (r). The scattering data in the case of the spherically symmetric potential is equivalent to the set of the phase shifts The phase shifts are defined as follows:

From Theorem 5.2.1 it follows that if then the set determines uniquely q(r). A much stronger result is proved by the author in Section 3.6 (see also [R192]). To formulate this result, denote by any subset of positive integers such that

Theorem 5.2.7. ([R192]) If uniquely.

then the data

determine q(r)

In [GR2] and [ARS2] examples are given of quite different potentials piecewise-constant, for r > 5, for which

and

where and are of order of magnitude of 1. This result shows that the stability estimate (5.2.19) is accurate. 5.3

INVERSE GEOPHYSICAL SCATTERING WITH FIXED-FREQUENCY DATA

Consider the problem

where is a compactly supported real-valued function with support in the lower half-space. In acoustics u has the physical meaning of the pressure, v(x) is the inhomogeneity in the velocity profile. We took the fixed wavenumber k = 1 without loss of generality. The source y is on the plane i.e., on the surface of the Earth, the receiver The data are the values The inverse geophysical scattering problem is: given the above data, find v(x). The uniqueness theorem for the solution to this problem is obtained in [R100], [R105].

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Problem (5.3.1)–(5.3.2) differs from the inverse potential scattering by the source: it is a point source in (5.3.1) and a plane wave in (5.1.2). Let us show how to reduce the inverse geophysical scattering problem to inverse potential scattering problem using the “lifting” [R139]. Suppose the data w(x, y), are given. Fix y and solve the problem:

This problem has a unique solution and there is a Poisson-type analytical formula for the solution to (5.3.3)–(5.3.5), since the Green function of the Dirichlet operator in the half-space is known explicitly, analytically:

Therefore the data determine uniquely and explicitly (analytically) the data We have lifted the data from P to as far as x-dependence is concerned and similarly we can get given If w(x, y) is known for all then one uses formula (5.1.7) and calculates analytically the scattering solution corresponding to the potential q (x) := – v ( x ) and k = 1, where Given for all and one can calculate the scattering amplitude If the scattering amplitude corresponding to the compactly supported is known then the uniqueness of the solution to inverse geophysical problem follows from Theorem 5.2.1. Stability estimates obtained for the solution to inverse potential scattering problem with fixed-energy data remain valid for the inverse geophysical problem: via the lifting process one gets the scattering amplitude corresponding to the potential q (x) = –v(x), and the stability estimates for obtained in sections 2.3 and 2.4, yield stability estimates for Practically, however, there are two points to have in mind. The first point is: if the noisy data are given, where then one has to overcome the following difficulty in the lifting process: data such that may not decay as on P, and this brings the main difficulty. The second point is: if one uses the inversion algorithms presented in sections 2.2– 2.4, then one uses the data Of course, the exact data

277

determine uniquely the data but practically finding the full data from the partial data is an ill-posed problem. One can also consider two-parameter inversion, corresponding to the governing equation where is a c = const, b (x) and a (x) are known constants outside a compact domain, It is proved in [R139] that both a(x) and b(x) cannot be determined simultaneously from the fixed-energy scattering data, but they can be determined simultaneously from the scattering data known at two distinct values of k and all and running independently through two open subsets of 5.4

PROOFS OF SOME ESTIMATES

Here we prove some technical results used above: estimates (5.1.18), (5.1.19), (5.1.20), (5.1.30), (5.2.13), and (5.2.17). 5.4.1

Proof of (5.1.18)

It is sufficient to prove (5.1.18) with in place of since w(x) and solve equation (5.1.1), the elliptic estimate (see [GT])

where implies that in place of

is strictly inner subdomain of and if If (5.1.18), with k = 1 and is false then

Therefore

This implies

where G(x, y) is the Green function of the operator L. Indeed, denote the integral on the left-hand side of (5.4.4) by

Then

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The second relation (5.4.5) follows from (5.4.3) and (5.1.7). From (5.4.5) one gets (5.4.4) by Lemma 4.1.2. From (5.4.4) it follows that

Thus

Multiply (5.4.7) by integrate over D, then by parts using the boundary conditions (5.4.7), use the equation Lw = 0 and get

Thus w(x) = 0. Estimate (5.1.18) is proved. 5.4.2

Proof of (5.1.20) and (5.1.21)

From (5.1.19) one gets

Denote

and define

Note that Lw = – f (x). We will prove below that:

where is an arbitrary compact domain We will also prove that

Let us show that (5.4.11) implies existence of the special solutions (5.1.19). If (5.4.11) and (5.4.12) hold, then (5.1.20) and (5.1.21) are easily derived. Indeed, rewrite (5.4.9) as

279


Therefore the operator has the norm going to zero as Thus equation (5.4.13) is uniquely solvable in if Moreover, the following estimate holds:

Estimate (5.1.20) follows. To derive (5.1.21) from (5.4.12) one writes

Therefore (5.4.13), (5.4.15) and (5.4.16) yield (5.1.21):

Proof of (5.4.11). If then Choose the coordinate system such that are the orthonormal basis vectors. Then

a · b = 0,

This function vanishes if and only if

Equation (5.4.19) defines a circle or radius in the plane centered at Let be a toroidal neighborhood of where the section of the torus by a plane orthogonal to is a square with size and the center at Denote where is defined in (5.4.10). One has

280


where holds if Let

and we have used the Cauchy inequality and an elementary inequality

which

and Then

where we have used the relations as and took into account that if From (5.4.21) and (5.4.22) one gets

By c > 0 we denote various constants independent of and t. Let us estimate

where the Parseval equality was used and by One has

Let us estimate

the integral in (5.4.24) is denoted.

281

Let

Then the integral on the right-hand side of (5.4.26) can be written as:

If

then

Let us use the elementary inequalities:

Then

Thus, with

one gets

and

From (5.4.3), (5.4.4) and (5.4.27) one gets

so

Let us estimate

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where One has:

and

where we have used the monotonicity of arctg x, for example,

etc., and the relation By (5.4.28),

as

From (5.4.35) and (5.4.36) one gets:

Thus

From (5.4.20), (5.4.23), (5.4.25), (5.4.33) and (5.4.37) one gets:

Choose

Then (5.4.39) yields

Estimate (5.4.11) is proved. Let us prove (5.4.12). Let

283

Define

Then

In [H, vol. 2, p. 31], it is proved that

where

Therefore

is a constant and we have used the relation

Estimate (5.4.41) is identical to (5.4.12). 5.4.3

If

Proof of (5.2.17)

is defined by (2.4), where

solves (5.1.1) then

solves the equation

Let

Note that

on N, where

Define

Then

Denote

The following Hardy-type inequality will be useful:

is the differential of

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If

f (0) = 0, then

Let us sketch the basic steps of the proof of (5.2.17)

Step 1. If

and

where

is the set of

where

is a sufficiently small number.

Step 2. Let

functions with compact support in the ball

be a bounded domain with a smooth boundary and in A, A is a strictly inner subdomain of

If

then

Step 3. Write (5.4.42) as

Let

Then

Apply (5.4.49) to (5.4.53) and get

then

285

Since

in

one gets:

So

Since

in

one gets:

Using (5.4.57), one gets

From (5.4.58), (5.4.57) and (5.4.55) one obtains:

Since is arbitrarily small, the desired inequality (5.2.17) follows. To complete the proof one has to prove (5.4.49) and (5.4.51). 5.4.4

Proof of (5.4.49)

Write (5.4.49), using Parseval’s equality, as

If

then

so

If

then use the local coordinates in which the set N is defined by the equations:

and the is along vector defined by the equation on N, these local coordinates can be defined.

Since

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Put Then port, and (5.4.47) yields:

at

if

has compact sup-

Integrating (5.4.63) over the remaining variables, one gets:

Since

is compact, one has

Using Parseval’s equality, S. Bernstein’s inequality for the derivative of entire functions of exponential type, and the condition supp one gets:

From (5.4.64)–(5.4.66) it follows that

Inequality (5.4.49) is proved. 5.4.5

Proof of (5.4.51)


where One has

take the real part and integrate by parts to get:

and summation is done over the repeated indices.

287

From (5.4.69) and (5.4.68) one gets:

Thus

Inequality (5.4.51) is proved. Let us prove that

implies

Indeed, (5.4.72), (5.1.19) and (5.1.20) imply:

If (5.4.73) is false for some that

then there is a sequence

This contradicts (5.4.74) since (5.4.75) implies

Therefore estimate (5.4.73) is proved. 5.4.6

Proof of (5.2.13)

One has

such

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where

where By (5.1.18), there exists a

such that

Therefore

so that

and

This implies

Thus, inequality (5.2.13) follows. We claim that is To prove this, one uses the above inequalities and gets:

This implies the following estimate:

Recall that

as

Therefore,

289

Thus, the above claim is verified, since, as and

one has

Uniqueness class for the solution to the equation

Taking the distributional Fourier transform of (5.4.77) one gets:

Thus supp rem 7.1.27 in [H, vol. 1, p. 174], one has:

and

is the circle (5.4.19). By theo-

Using (5.4.77) we derive for

Combining (5.4.79) and (5.4.80) one gets

Thus as claimed. Estimate (5.4.80) is valid in 5.4.7

It was used in [SU] and [R139].

Proof of (5.2.23)

Let

and such that (5.4.72) holds. We wish to prove that

where

where the infimum is taken over all

and

stands for various constants.

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Let us describe the steps of the proof. Step 1. Prove the estimate

where

The choice of in (5.4.83) is justified below (see (5.4.97)) and estimate (5.4.82) is proved also below. Step 2. Minimize the right-hand side of (5.4.82) with respect to

to get

The minimizer is Step 3. Take

so, for

in (5.4.84). Then

one has:

From (5.4.84) and (5.4.85) one gets:

Estimate (5.4.81) is obtained. Proof of (5.4.82). Since onto inequality (5.4.72) with

where c = const > 0 does not depend on

where is a bijection of is equivalent to

and

We take b > a, therefore the function has the maximal values, as of the same order of magnitude as the function The function solves

291

the equation

Indeed, Therefore one can write:

where are defined in (5.1.26), some coefficients. It is known that

as claimed.

are defined in (5.1.29), and

are

so

where Choose

where is the same as in (5.4.83). Then (5.4.88) implies:

where formula (5.1.29) was used. From (5.4.92) and formula (5.4.91) with

where we have used the formula the coefficient

one gets:

we estimated by of the main term of that is, the

292


function we used estimate (5.1.28), which gives replaced by the smaller quantity. Choose r > b and use (5.1.29) to get the inequality:

and we

which implies (5.4.94). Thus (5.4.94) holds if where c stands for various constants. One has and the minimizer is Consider therefore the equation and solve it asymptotically for as where is arbitrary large but fixed. Taking logarithm, one gets Thus

and

Hence, we have justified (5.4.83). From (5.4.94), (5.4.96) and (5.1.29), one gets

Estimate (5.4.82) is established. 5.4.8

Proof of (5.1.30)

Let be the Green function corresponding to one gets

j = 1, 2. By Green’s formula

Take

and use (5.1.7) to get:

Take

use (5.1.7) and (5.1.2) and get:

Since

formula (5.4.101) is equivalent to (5.1.30).

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5.5

CONSTRUCTION OF THE DIRICHLET-TO-NEUMANN MAP FROM THE SCATTERING DATA AND VICE VERSA

Consider a ball

and assume that the problem

is uniquely sovable for any Then the D – N map is defined as

where

where

on

is the normal derivative of

is the Sobolev space.

N is the normal to

pointing into

If is known, then q (x) can be found as follows. The special solution (5.1.19)–(5.1.22) satisfies the equation:

where

so that

and

in

Thus

is the Green function of the operator L, see (5.4.10), that is

The function G(x) can be considered known. Since one can write, for

294

Here


is

for q (x) = 0, we have used Green’s formula and took into account that

where solves problem (5.5.1) with q (x) = 0 and on From (5.5.3) and (5.5.6) taking one gets a linear Fredholm- type equation for

If is known, one can find from (5.5.7) calculation. Define

and then find q (x) using the following

By Green’s formula, as in (5.5.6), one gets

From (5.5.8) one gets, using (5.1.19), (5.1.20) and (5.1.9):

Therefore the knowledge of allows one to recover by formula (5.5.10), but first one has to solve equation (5.5.7). We leave to the reader to check that the homogeneous equation (5.5.7) has only the trivial solution so that Fredholm-type equation (5.5.7) is uniquely solvable in (see a proof in [R172]). Practically, however, there are essential difficulties: a) the function G(x, y) is not known, analytically and it is difficult to solve equation (5.5.7) by this reason, b) the D – N map is not given analytically as well. Let us show how to construct

from the scattering amplitude

and vice versa. If

is given then we have shown how to find q (x) and if q (x) is found then the scattering amplitude, can be found. Conversely, suppose is known. Then the scattering solution can be calculated in by formula (5.1.31).

295

Let q (x) – 1 in

be given, g (x, y) be the Green function of the operator which satisfies the radiation condition (5.1.6), and define

such that

in

Since explicitly:

where

in

w = f on

and w satisfies (5.1.6), one can find w

are the Fourier coefficients of f :

Therefore the function

is known. By the jump formula for single-layer potentials one has:

The map is constructed as soon as we find found. To find consider the asymptotics of w(x) as and (5.5.11), one gets:

because

is already Using (5.1.7)

where we have used (5.5.13) and the asymptotics as As we have already mentioned, the function is known explicitly (see formula (5.1.31)), and equation (5.5.16) is uniquely solvable for Analytical solution of equation

296


(5.5.16) for

can be obtained as a series

Substitute (5.4.91) with into (5.1.31), take r = a in (5.1.31) and and substitute (5.1.31) into (5.5.16). By our choice of the spherical harmonics (5.1.26) both systems and form orthonormal bases of Therefore one gets:

Denote

Using (5.1.26) one gets:

Also define

by the formula:

The above definition differs from (5.1.36) and is used for convenience in this section. Equating the coefficients in front of in (5.5.18) one gets

or

297

The matrix of the linear system (5.5.23) is ill-conditioned (see [R203], where estimates of the entries of the matrix of (5.5.19) are obtained and the case of the noisy data is mentioned). Finally let us show (see [R139]) that it is impossible to get an estimate

if

where

and we assume that

Indeed, choose a

such that the problem

has a non-trival solution. Define Then

and

Because of our assumption (5.5.24), one gets:

Were (5.5.24) true, it would imply for in (5.5.31) has small norm, so

that the operator contrary to our assumption.

Let us compare the first method for solving the inverse scattering problem with fixed-energy data, described in Section 5.2.2, with the second method, described in Section 5.5.

The only difficulty in the first method is solving (5.2.12), where is defined in (5.2.4), that is, approximate minimizing a quadratic functional so that (5.2.15)

298


be satisfied. This can be done by using a necessary condition for the minimizer, which is a linear equation. If the minimizer is sought on a finite-dimensional space of the elements then the linear equation is a linear algebraic system for finding the coefficients and the functional is a function of finitely many variables The computational difficulty in this case is the ill-condition nature of the matrix of the above linear algebraic system. Alternatively, one can use one of the known methods for global minimization of convex functionals, since the quadratic functional is convex. In contrast, the second method faces several difficulties: 1) Computing the Green function (5.5.5) is very difficult because the integral in (5.5.5) is taken over the whole space and the integrand is not absolutely integrable, 2) Solving (5.5.16) is very difficult because (5.5.16) is a severely ill-posed problem: it is a first kind Fredholm-type integral equation with analytic kernel, 3) Solving (5.5.7) is difficult, because G and are very difficult to compute. 5.6

PROPERTY C

In this Section we present an outline of the theory developed by the author as a tool for solving inverse problems. The basic definition is 5.1.1. Consider the case of partial differential expressions with constant coefficients:

Define algebraic varieties:

Let tions.

where

is understood in the sense of distribu-

Definition 5.6.1. We say that is transversal to and write iff there exist at least one point and at least one point such that the corresponding tangent spaces and are transversal Here is the tangent space in to at the point Note that

iff

Definition 5.6.2. We say that an operator L has property C if the pair {L, L} has this property, (cf. definition 5.1.1). For the operators with constant coefficients the domain D in the definition 5.1.1 is an arbitrary bounded domain, and is an arbitrary fixed number. We take p = 2.

The basic result of this Section is

299

Theorem 5.6.3. A necessary and sufficient condition for the pair C is

to have property

Proof. Sufficiency: Suppose

where is a fixed number, and D is a bounded domain. We want to derive f = 0 from (5.6.3). Take so Then (5.6.3) implies:

If the set contains a ball or then the entire function vanishes in this ball and, by analyticity, in Consequently, Thus, we have to check that the set contains a ball. This follows from the assumption Indeed, take a basis in It consists of complex vectors Since there exists a vector in which has a non-zero projection onto the normal to The system forms a basis in In a neighborhoods of the points and one can find vectors on and arbitrary close in the norm to vectors and These vectors form a basis of also. Their linear span contains a ball. The sufficiency is proved. Necessity: Assume that the pair has property C. We want to prove that Assume the contrary and find such that (5.6.3) holds, so the set is not complete in The contrary means that is a union of parallel hy– perplanes in Let b be a normal vector to these hyperplanes. By choosing properly a basis in one may assume that where is an orthonormal basis. Then and so are constants, j = 1, 2. Let Every bounded domain is a subdomain of such D. Let

where Then

300


Thus

is orthogonal to all the products Any element is a limit of a linear combination of the exponential solutions if the polynomial is irreducible. Otherwise there are elements in of the form where P(j) are polynomials of some degree where depends on Our argument in (5.6.6) is given for If then g can be chosen orthogonal to all the powers Thus, the necessity is proved. Let us apply Theorem 5.6.3 to the case of one operator L using Definition 5.6.2. The condition is satisfied if and only if is not a union of parallel hyperplanes. Algebraically this condition means, that is not of the form where are constants independent of j. Consider, for example, the classical operators:

namely, Laplace, Schroedinger, heat, and wave operators with constant coefficients. The corresponding varieties are:

The condition

is satisfied. Therefore the following result holds:

Theorem 5.6.4. The operators

and

have property C. 5.7

NECESSARY AND SUFFICIENT CONDITION FOR SCATTERERS TO BE SPHERICALLY SYMMETRIC

In this section we give an easily verifiable condition on the scattering amplitude for the corresponding scatterer to be spherically symmetric. The scatterer may be a potential, an inhomogeneity, or an obstacle. We first consider some transformation laws for scattering amplitudes. Let

the unit sphere, We are interested in the following equation: if one changes in (1) to where L is a linear transformation in what is the corresponding change in For example, let

301

where O(3) is the group of rotations in

or

or

Here R, and are operators of rotation, scaling by a factor and translation by a vector – a respectively. The basic results are the following formulas for the corresponding transformation of the scattering amplitudes:

As an application, we prove that a necessary and sufficient condition for q (x) to be spherically symmetric is that the following equation holds

at a fixed k > 0, provided that

where b > 0 is an arbitrary large number and the overbar denotes complex conjugate. A similar result is proved for scattering by an obstacle. Namely let

Here is the exterior of a bounded domain D with a smooth connected boundary N is the unit normal to pointing into Let us formulate the results. Theorem 5.7.1. Formulas (5.7.5), (5.7.6), and (5.7.7) hold.

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Theorem 5.7.2. Assume that (5.7.9) holds and holds at a fixed k > 0 iff q (x) = q (r).

Then (5.7.8)

Theorem 5.7.3. Assume that (5.7.8) holds at a fixed k > 0 with Then is a sphere and The converse is trivially true. It follows from our result that if (5.7.8) holds at a single k > 0 then (5.7.8) holds for all k > 0 (provided that (5.7.9) holds). Let us give proofs. Proof. 1. Proof of formulas (5.7.5), (5.7.6), and (5.7.7). Formulas (5.7.5), (5.7.6), and (5.7.7) are direct consequences of the definition of the scattering amplitude. Formula (5.7.5) means that the scattering amplitude is the same in a coordinate system and the coordinate system in which each vector becomes Ra, where is an arbitrary rotation, and O(3) is the group of rotations in The “rotated” potential is defined in (5.7.5). A rigorous derivation of formulas (5.7.5), (5.7.6), and (5.7.7) is based on writing the asymptotics of the scattering solution in the new coordinate system. This derivation is the same in all three cases. Let us give it briefly. a) Derivation of formula (5.7.5). Consider

Let

since Note that

Note that

for any Here is the transposed operator of rotation. is invariant under rotations. Write (5.7.12) and (5.7.13) in the

303

Note that, by (5.7.13), scattering solution:

Thus, the function w in (5.7.16) is the

corresponding to the incident direction Then

Let

This is formula (5.7.5). b) Derivation of formula (5.7.6). Consider

Let

Then (5.7.18) can be written as

The scattering solution, corresponding to (5.7.20), is

while (5.7.19) can be written as

Compare (5.7.20), (5.7.21), and (5.7.22) to get

Let


This is formula (5.7.6) with p = q and

304


c) Derivation of formula (5.7.7). Consider

Let

Then (24) becomes

and

while

Note that

Thus (5.7.29) becomes

Since u solves the homgeneous linear equation (5.7.25), the expression in brackets in (5.7.31) solves equation (5.7.27). Compare (5.7.28) and (5.7.31) to get

This is formula (5.7.7). Proofs of Theorems 5.7.2 and 5.7.3. Proof. a) Proof of Theorem 5.7.2. The basic auxiliary result is the uniqueness theorem 5.2.1: Proposition 5.7.4. Let k > 0 and all then

j = 1,2. If

at a fixed

305

Theorem 5.7.2 follows immediately from formula (5.7.5) and Proposition 5.7.4. Indeed, it is well known (and follows immediately from separation of variables) that if the potential is spherically symmetric

then (5.7.8) holds for all k > 0. Assume that (5.7.9) holds and (5.7.8) holds at a fixed k > 0 with Then

Here the second equation follows from (5.7.13) and the third equation is formula (5.7.5). Since if and since and are arbitrary, one can write (5.7.34) as

where

By Proposition 5.7.4, it follows that

This is equivalent to (5.7.33). Theorem 5.7.2 is proved. Proof. b) Proof of Theorem 5.7.3. The basic auxiliary result is the following uniqueness theorem, which follows from Theorem 5.2.1: Proposition 5.7.5. Assume that a fixed k > 0. Then and

for all

and

Note that if is a sphere and then (5.7.8) holds with for all and all k > 0. This follows from the analytical solution of the scattering problem by separation of variables. Assume now that (5.7.8) holds at a single fixed k > 0. Then, by (5.7.8) and (5.7.13),

and

by formula analogous to (5.7.5). Here is the surface rotated by the element and From (5.7.37) and (5.7.38) it follows that

306


By proposition 5.7.5, equation (5.7.39) implies

Thus proved.

is a sphere and

on the sphere

Theorem 5.7.3 is

Remark

5.7.6. Consider compactly supported potentials q = q (p, z), Let denote rotations about the Then By formula (5.7.5) for one has

This symmetry property

is a necessary property of the scattering amplitude corresponding to the potential which is axially symmetric about the i.e., q(x) = q(p, z). If then (5.7.42) is also sufficient for q (x) to be axially symmetric about the This is proved as in Theorem 5.7.1. Remark 5.7.7. Consider the equation

for At a fixed k > 0 the scattering of a plane wave by the inhomogeneity v(x) is identical with the scattering of this wave by the potential Therefore, by Theorem 5.7.2, if and only if condition (5.7.8) holds at a fixed k > 0, where is the scattering amplitude corresponding to equation (5.7.43). Remark 5.7.8. The argument in this Section is based on uniqueness theorems. If Proposition 2 is no longer valid. However, for this class of potentials the following uniqueness theorem holds: if for all and all sufficiently large k > 0, then The argument used in the proof of Theorem 5.7.2 shows that if and if (5.7.8) holds for all k > 0, then The uniqueness theorem holds even for with Therefore, the following theorem holds. Theorem 5.7.9. If large k > 0.

then q (x) = q (r) iff (5.7.8) holds for all sufficiently

Remark 5.7.10. If then is not necessarily continuous in and so that (5.7.8) is understood as equality of kernels of operators on

307

5.8

THE BORN INVERSION

1. In this section we give an error estimate and a theoretical analysis of the widely used Born inversion. We study this inversion for inverse protential scattering problem, but the methodology is valid for other inverse problems. Let us first describe the Born inversion. The starting point is the formula

for the scattering amplitude. The scattering solution in (5.8.1) is substituted by the incident field according to the Born approximation. The error of this approximation can be estimated in terms of if this norm is sufficiently small and suitably chosen. The other possibility to estimate the error of the Born approximation is to assume that k is sufficiently large. The estimate of the error can be obtained from the analysis of the integral equation

If

then (5.8.2) is uniquely solvable in

by iterations and

If e.g., [R121], p. 231, cf (5.1.13))

then (see,

Define

The Born inversion consists of finding q (x) from the equation

that is:

308


Given an arbitrary

one can find (non-uniquely)

and k > 0 such that

Then (5.8.7) becomes

Therefore is known and can be found by the Fourier inversion of the righthand side of (5.8.9). This is the standard description of the Born inversion. Our aim is to analyze this procedure. First, note that equation (5.8.7) is not valid exactly: the left side of (5.8.7) differs from its right side by a quantity of order of the error of the Born approximation. Secondly, the function in (5.8.9) depends actually not only on but on and k, in spite of the fact that the left side of (5.8.9) depends on only. This is true because (5.8.9) is not an exact equation. The third observation we formulate as a lemma. Lemma 5.8.1. Assume If (5.8.7) holds for all and all k > 0, then q = 0. Assume If (5.8.7) holds for all and a fixed k > 0 then q = 0. Proof. If (5.8.7) holds then optical theorem

for all

and all k > 0. By the

Therefore for all and all k > 0. This implies, by the uniqueness theorem for IP1 that q (x) = 0. If then the argument is the same but one uses the uniqueness theorem for IP12. Lemma 1 is proved. The conclusion is: equation (5.8.7) considered as an exact equation is not solvable unless q = 0. Therefore in the Born inversion one should treat the measurements of the scattering data, however accurate, as noisy Born data. In particular, it is not advisable to measure the scattering data with the accuracy that exceeds the accuracy of the Born approximation for the solution of the direct scattering problem. The last question we wish to discuss is this: is it true that the Born inversion allows one to recover q with the accuracy which grows as q gets smaller in a suitable sense? This question should be clarified: if equation (5.8.7) considered as an exact equation is not solvable unless q = 0, then what is the meaning of Born’s inversion? What is the meaning of the smallness of q? How does the inversion result depend on the choice of in (5.8.9)? We noted that is not uniquely determined by the choice of it depends on and k.

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The result we will prove is valid for any choice of q is defined by the inequality

in (5.8.9). The smallness of

This condition can be expressed directly in terms of Thus (5.8.11) holds if

where is defined in (5.8.3). Finally, the Born inversion is understood as follows: the function in (5.8.9) is considered as the noisy values of Namely, let us assume that

Here the values are the measured values of at so that both the measurement noise and the error of the Born approximation are estimated by The Born inversion consists of finding q (x) given which satisfies (5.8.13), and some a priori information about the unknown q which is given by the estimate

Define

Let us formulate the result. Theorem 5.8.2. Assume (5.8.13) and (5.8.14). Then

where

and

are defined in (5.8.16) and (5.8.15).

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Proof. One has

Let (5.8.15), and

This minimum is attained at

where

is given in

Theorem 5.8.2 is proved. Remark 5.8.3. It follows from (5.8.5) that the Born approximation for the Schrödinger equation becomes exact as For the acoustic equation

where v (x) is a compactly supported square integrable function, the Born approximation becomes exact as and the order of the error of the Born approximation is O(l) as so that the error of the Born approximation does not decay as for equation (5.8.20). One can choose such that the scattered field on is as where N is arbitrary large integer. At the same time the measurements error is 0(1) and the error of the Born approximation is O(l). This analysis shows that it is not possible to use Born’s inversion for equation (5.8.20) at high frequencies for recovery of smooth v(x). However, one may hope to use it for recovery of discontinuities of v (x). The reason is that the Born approximation for large k is and the behavior of the Fourier transform for large frequencies is determined by the discontinuities of q . The practical conclusions which follow from the results of this section are: the Born inversion is an ill-posed problem; it needs a Regularization, given by formula (5.8.16). Even in case of an arbitrary small if the Regularization is not used (for example, the integral in (5.8.16) is taken over the whole space rather than over the ball then the error of the Born inversion may be unlimited. The last point can be explained in a different way. The original nonlinear equation which relates q and A can be written as B(q) = A, where is the mapping which associates the scattering amplitude A with the given q . The Born approximation is a linearization of this equation near

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q = 0. More generally, if one linearizes around one gets This equation may be not solvable, since may be unbounded and may not belong to the range of As Lemma 5.8.1 shows, this is exactly the situation with the Born inversion.

2. In this section we give a formula for the Born inversion of the data which are given on The mathematical formulation of the problem is: given the equation

where k > 0 is fixed, and find q (x) by integrating over We gave earlier a solution to (5.8.21) based on the fact that defines the Fourier transform of q (x) in the ball However, this solution required to choose and such that where is a given vector. Although, in principle, such a choice presents no difficulties, computationally it is desirable to avoid this step and obtain an inversion formula of the type

where is some universal kernel which does not depend on the data but depends on k, R and N. One has

where

Let us formulate the result in

Define

where

Here is the Gamma function and the braces is

is the Bessel function, the expression in

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is the volume of the ball of radius k. Define

and

where

is the surface area of the sphere

Theorem 5.8.4. If is defined by (5.8.27), and by (5.8.22), then formula (5.8.23) holds.

by (5.8.24)

by (5.8.25),

Proof of Theorem 5.8.4 and a detailed discussion of the solution to (5.8.21) given by (5.8.22) one can find in [R83], pp. 259–270, where the stability of the inversion is discussed as well. Namely, suppose that is given such that is a small given number. Then one can find such that formula (5.8.22) with in place of F and in place of N defines with the property as Estimates of are given in [R83], p. 269. An application of the sequence (5.8.25), which is a delta-sequence in of entire functions of exponential type to the problem of spectral extrapolation, that is, inversion of the Fourier transform of a compactly supported function from a compact, is done in [R83], [R139] and in [RKa]. This problem is of interest in many applications. 5.9

UNIQUENESS THEOREMS FOR INVERSE SPECTRAL PROBLEMS

Let us formulate the inverse spectral problem (ISP). Let

Here each

is the normal derivative of on Assume that is counted according to its multiplicity, that

that

and that Consider the inverse spectral problem:

If the boundary condition is the Dirichlet one, then the inverse spectral problem is:

The basic result is

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Theorem 5.9.1. The ISP has at most one solution. Outline of the Proof. The idea of the proof is to assume that there are two pairs of functions j = 1,2, which produce the same spectral data, and to derive from this assumption that the distribution is orthogonal to the set of products where

Since the pair has property C it follows that This implies that p(x) = 0 and so and This is the outline of the proof. Note that completeness of the set of products of solutions to the Schrödinger equations holds not only in but in the class of distributions with compact support. Let us turn to the proof. Let us derive the orthogonality relation. We start with a simple but important observation: Lemma 5.9.2. Let

Then

so that the coefficients

and the eigenvalues

are uniquely determined by the spectral data

Here

are counted according to their multiplicities.

Proof of Lemma 5.9.2. Multiply equation (5.9.6) by parts to get

Since the set forms an orthonormal basis of (5.9.8). Lemma 5.9.2 is proved. Let us continue the proof of Theorem 5.9.1.

integrate over D and then by

formula (5.9.7) follows from

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Suppose that the pairs on with equal to this equation with

and

produce the same spectral data:

Choose an arbitrary and solve problem (5.9.6) j = 1, 2. Subtract equation (5.9.6) with from to get

By Lemma 5.9.2,

where the coefficients are the same for and assumption, one concludes from (5.9.10) that

Since

on

by the

A detailed proof of (5.9.11) is given in Lemma 5.9.3. Multiply (5.9.9) by an arbitrary element integrate over D and then by parts to get

where we used (5.9.11). Write (5.9.12) as the orthogonality relation:

Here is the delta-function supported on The set of products in (5.9.13) is complete not only in but also in the set of distributions with support in D of order of singularity which increases with the smoothness of q (x). If then the order of singularity allowed is so that is an admissible distribution. Recall that the order of singularity s =s( f ) of a distribution f is the smallest integer such that f is representable in the form where h is a locally integrable function. Since where is the characteristic function of the domain D and v is the normal to pointing into D, one concludes that where s( f ) is the order of singularity of f. Thus for any continuous function 0. Thus, one concludes that (5.9.11) implies p = 0 and Theorem 5.9.1 is proved. We now prove

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Lemma 5.9.3. Under the assumptions of Theorem 5.9.1 equation (5.9.10) implies (5.9.11). Proof. The difficulty of the proof comes from the fact that series (5.9.10) converge in but they do not converge fast, so that it requires a proof to see that one can calculate the value of at the boundary by taking in the series For example, it is clear that one cannot put the operation under the sign of the sum in (5.9.10) (changing the order of application of the operation and the operation since while m = l, 2. In order to prove (5.9.11), write (5.9.10) as

where

is the unique solution to the problem

This problem is uniquely solvable for assumption Clearly

in a neighborhood of

because of the

where in

is the derivative in and is an integer. The series (5.9.14) converges provided that Indeed, it is well known that as d is the dimension of the space d = 3 in our case. The known elliptic estimate yields

where c = const > 0, c does not depend on m. Thus the series (5.9.14) converges in that is or if If then so one can take and the is representable by a series which converges in assumption on implies on Therefore where is a polynomial in of degree that is of degree 2, with coefficients depending on s. It is clear from (5.9.14) that, in the sense of distributions, as Therefore and Lemma 5.9.3 is proved. Remark 5.9.4. The smoothness assumption on q can be further reduced (from to in (see [H vol. III, p. 4]).

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Remark 5.9.5. If the Robin boundary condition (5.9.6) is replaced by the Dirichlet condition u = f on then the spectral data in Theorem 5.9.1 should be replaced by the spectral data

The uniqueness theorem for the ISP with these data is valid and its proof is the same as above. Theorem 5.9.1 was proved in [NSU]. In [R120], [R139], and in this Section the proof of this theorem is based on the Property C. Let us give a useful result. Lemma 5.9.6. The system integer and Proof. Assume

Let

in D,

One can assume that implies f = u = 0 on

is complete in

Here n is any fixed positive

and

Write (5.9.16) as

are real-valued since q is real-valued. Thus Thus This and the condition and the conclusion follows.

6. NON-UNIQUENESS AND UNIQUENESS RESULTS

6.1 EXAMPLES OF NONUNIQUENESS FOR AN INVERSE PROBLEM OF GEOPHYSICS

6.1.1

Statement of the problem

In this section the result from [R165] is presented. Let be a bounded domain, part S of the boundary of D is on the plane f(x, t) is a source of the wavefield u(x, t), and c (x) > 0 is a velocity profile. The wavefield, e.g., the acoustic pressure, solves the problem:

Here N is the unit outer normal to is the normal derivative of u on If is known, then the direct problem (6.1.1)–(6.1.3) is uniquely solvable. The inverse problem (IP) we are interested in is the following one: (IP) Given the data can one recover uniquely? The basic result is: the answer to the above question is no. An analytical construction is presented of two constant velocities j = 1, 2, which can be chosen arbitrary, and a source, which is constructed after are chosen, such that the solutions to problems (6.1.1)–(6.1.3) with

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6. Non-uniqueness and uniqueness results

j = 1,2, produce the same surface data on S for all times:

The domain D we use is a box: This construction is given in the next section. At the end of section 8.2 the data on S are suggested, which allow one to uniquely determine 6.1.2

Example of nonuniqueness of the solution to IP

Our construction is valid for any For simplicity we take n = 2, Let The solution to (6.1.1)–(6.1.3) with can be found analytically

where

The data are

For these data to be the same for

and

it is necessary and sufficient that

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Taking Laplace transform of (6.1.9) and using (6.1.7) one gets an equation, equivalent to (6.1.9),

Take arbitrary and find for which (6.1.10) holds. This can be done by infinitely many ways. Since (6.1.10) is equivalent to (6.1.9), the desired example of nonuniqueness of the solution to IP is constructed. Let us give a specific choice: for or Then (6.1.10) holds. Therefore, if

then the data ficients are

In (6.1.11) the values of the coef-

Remark 6.1.1. The above example brings out the question: What data on S are sufficient for the unique identifiability of The answer to this question one can find in [R139]. In particular, if one takes and allows x and y run through S, then the data determine uniquely. In fact, the low frequency surface data where is an arbitrary small fixed number, determine uniquely under mild assumptions on D and By is meant the Fourier transform of u(x, y, t) with respect to t. Remark 6.1.2. One can check that the non-uniqueness example with constant velocities is not possible to construct, as was done above, if the sources are concentrated on S, that is, if

6.2

A UNIQUENESS THEOREM FOR INVERSE BOUNDARY VALUE PROBLEMS FOR PARABOLIC EQUATIONS

Consider the problem:

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Here is the delta-function, D is a bounded domain in with a smooth boundary S, and q (x) are realvalued functions, where and are positive constants, and where is the closure of D. Let where N is the unit exterior normal to S. The IP (inverse problem) is: given the set of ordered pairs {f (s), h (s, t)} for all and all find a (x) and q (x). We prove that IP has at most one solution by reducing the uniqueness of the solution to IP to the Ramm’s uniqueness theorem for the solution to elliptic boundary value problem [R139] (cf Chapter 5). This theorem says: Let

and assume that the above problem is uniquely solvable for two distinct real values of Suppose that the set of ordered pairs {f, h} is known at these values of for all where and is the normal derivative on S of the solution to (6.2.5). Then the operator L is uniquely determined, that is, the functions a (x) and q (x) are uniquely determined.

We apply this theorem as follows. First, we claim that the data h(s, t), known for are uniquely determined for all t > 0. If is replaced by a function then the data h (s, t) known for are uniquely determined for t > T. Secondly, if this claim is established, then Laplace-transform problem (6.2.1)-(6.2.3) to get the elliptic problem studied in [R139]:

and the data where are known for all The data H(s, Thus, Ramm’s theorem yields uniqueness of the determination of L, and the proof is completed. Proof. We now sketch the proof of the claim: The solution to the time-dependent problem can be written as:

where in D, on S, The coefficients in this series can be calculated by the formula: Note that the series for u (x, t) and the series obtained by the termwise differentiation of it with respect to t converge absolutely and uniformly in each of the

321

terms is analytic with respect to t in the region and consequently so are these series. Therefore the functions u(x, t) and are analytic with respect to t in the region so the data are uniquely determined for t > T as claimed. At t = 0 the series (6.2.6) is singular: it does not converge uniformly or even in By this reason the above argument is formal. One can make it rigorous if one replaces the delta-function in (6.2.3) by a function and uses the argument similar to the one in Lemma 5.9.3. 6.3

PROPERTY C AND AN INVERSE PROBLEM FOR A HYPERBOLIC EQUATION

Let in D × [0, T], where is a bounded domain with a smooth boundary Suppose that for every the value is known, where N is the outer normal to u solves and and satisfies the initial conditions at t = 0. Then q(x, t) is uniquely determined by the data in the subset S of D × [0, T] consisting of the lines which make 45° with the t-axis and which meet the planes t = 0 and t = T outside provided that q(x, t) is known outside S. Here is the closure of D. 6.3.1

Introduction

Property C, described in Chapter 5, yields a general method for proving uniqueness theorems for multidimensional inverse problems. A number of such theorems are proved for elliptic equations in Chapter 5. The purpose of this Section is to prove the uniqueness theorem formulated in the abstract and a more general one. This theorem is another example of the application of property C. In Section 6.3.2 we formulate the result and give the proof. (cf [RRa]). 6.3.2

Statement of the result. Proofs

1. Assume that

where is a bounded domain with a smooth boundary d < a < b < T – d, and d = diamD. Let

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Assume that

and q (x, t) is a continuous in t element of

In Proposition 6.3.10 below we relax assumption (6.3.4): we assume that q(x, t) is known outside the subset S defined in this Proposition and we prove that q (x, t) is uniquely determined in S by the data (6.3.7). The set D × [a, b] is a subset of S if d < a < b < T – d. Problem (6.3.1), (6.3.2) and (6.3.3) has the unique solution (that is the solution exists and is unique). Let

where is the normal derivative of u on The set of pairs

N is the outer normal to

is our data. The inverse problem consists in finding q (x, t) given the data (6.3.7). 2. Our result is: Theorem 6.3.1. Data (6.3.7) determine q(x, t) uniquely provided that d < a < b < T – d. This means that if there are two functions are the same then

j = 1, 2, for which data (6.3.7)

Proof of Theorem 6.3.1 Assume that j = 1, 2, produce the same data (6.3.7). Subtract equation (6.3.1) with from equation (6.3.1) with to get

By the assumption

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Multiply (6.3.8) by property

and integrate over

where

is an arbitrary solution to equation (6.3.1) with the

to get

where we have integrated by parts and used (6.3.3), (6.3.10) and (6.3.9). Thus

where

in solves (6.3.10)} and in solves (6.3.3)}. Equation (6.3.11) implies that provided that the following lemma holds: Lemma 6.3.2. The set

and

is complete in

Since we assume that it is sufficient to prove completeness in namely that (6.3.11) implies if If then and Theorem 1 is proved. To complete the proof of Theorem 6.3.1 let us prove Lemma 6.3.2, but first let us outline the basic ideas. We would like to prove Lemma 6.3.2 in three steps. Proof. Step 1: If T – d/2 > b then the function takes all the values in the set in Step 2: If a > d/2 then the function takes all the values in the set in Therefore the set is the set of products of arbitrary solutions to the equations and in Step 3: The set of products of these solutions is complete in Steps 1, 2 are analogous and can be eliminated if one can prove that the set of products of solutions from some subsets of and is complete in This is done in sections 3–5 below under the assumption that a > d and b < T – d. We are not able to give a proof under the assumptions a > d/2 and b < T – (d/2), see Remark 6.3.4. 3. Let us construct the solutions, which belong to and such that the set of their products is complete in One can prove (see Lemma 6.3.5

324


below) that there exist the solutions

with

and

where

is the delta-function, and convergence is meant in the sense of distributions,

is the empty set. For q = q (x) the solutions similar to (6.3.12) were used in [RSy]. Using (6.3.12)–(6.3.16), taking and then one obtains

This set is complete in Indeed, let Then

in the sense that equation (6.3.18) below implies (6.3.20).

Thus

Therefore

Lemma 6.3.3. If for then (6.3.19) implies

where a > d, b < T – d, d = diamD,

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Proof. The integral in (6.3.19) gives the ray transform of be the X-ray transform in which the rays are the lines

Namely if

which is defined to

is the ray transform operator, then

The operator defined by formula (6.3.22) on continuous functions q (x, t) of x, t, with compact support D in x variable, and integrable in t variable can be extended to an operator on the space of temperate distributions (see Remark 6.3.4 and paper [RSj]). By the assumption, q (x, t) = 0 for in particular, for Without loss of generality one can assume that D is a ball If then the rays (6.3.21) intersect the region t > d/2. We required that This condition is satisfied and the rays (6.3.21) cover the region if

Indeed, so that the ends of these rays do not belong to D. By the assumption, q(x, t) = 0 outside Thus, (6.3.19) says that the ray transform vanishes everywhere (that is, the restriction can be dropped) provided that a > d and b < T – d. Indeed, in this case, if one has so that for and, since for one has for and Thus, if (6.3.19) holds for if T – b > d, and if a > d, then (6.3.19) holds for all This implies (6.3.20) by Lemma 6.3.6 below. Lemma 6.3.3 is proved. Remark 6.3.4. If only the assumption a > d/2, b < T – (d/2) is given, then the above simple argument is not sufficient. It would be interesting to find out if the uniqueness theorem holds under this assumption. It is dear that for a < d /2 the uniqueness theorem does not hold because the domain of dependence for equation (6.3.1) is a cone of height d/2. If then equation (6.3.19) becomes

d/2 < a < b < T – (d/2), so that T > d. It follows from (24) that if runs in of is an arbitrary small positive number, then (6.3.24) implies that the X-ray transform of vanishes so that This is the result obtained in [RSy].

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4. Let us now give the proof of the existence of the solutions (6.3.12), (6.3.13), (6.3.14). Let

where

If

then

Let

Then

where has

By Riemann-Lebesgue’s lemma one

where Multiply (6.3.30) by and integrate over D, then by parts using (6.3.29), then integrate with respect to t from 0 to t, and get

where

is the norm in


Let

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Let

Then for A,

Multiply (6.3.36) by to t and use v (0) = 0 to get

where one gets

to get

Using the inequality

Integrate from 0

It follows from (6.3.33), (6.3.34) and (6.3.37) that

Since

one has

Therefore the following lemma is proved: Lemma 6.3.5. Assume that

Then there exists a solution (6.3.25) to problem (6.3.27) such that (6.3.39) holds. One can choose with shrinking to a given point such that (6.3.15) holds. 5. Let us give a formula for finding q (x, t). Our argument is similar to the one given in [R103]. Write multiply this equation by a and integrate over to get

where v is the outer normal to Choose such that at t = T, Then the right-hand side of (6.3.41) is known if the data (6.3.7) are known.

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Choose u and to the limit

of the form (6.3.12) and (6.3.13), so that (6.3.14), (6.3.15) hold, pass to get as in (6.3.19):

The knowledge of the integral on the right in (6.3.42) for all and all q(x, t) = 0 for determines q(x, t) in a > d, b < T – d, uniquely, as we showed above. Therefore (6.3.42) can be considered as an inversion formula. Note that the left-hand side of (6.3.42) does not depend on because the right-hand side of (6.3.42) does not depend on To apply formulas (6.3.41), (6.3.42) practically one chooses on find the corresponding from (6.3.7) so that I in (6.3.41) is known without knowledge of q. Moreover this u satisfies (6.3.13), (6.3.14) so that (6.3.42) holds. 6. Finally let us sketch the proof of injectivity of the ray transform. Lemma 6.3.6. Let f(x, t) be continuous in function with values in is a bounded domain, f = 0 for Assume that

where

is an open set in

Proof. Let

where

Then be arbitrary. Then, by Parseval’s equality,

Thus

For an arbitrary fixed

one can choose such that Thus (6.3.45) says that the Fourier transform of in the variable t vanishes for Since is assumed compactly supported in t, the function is entire in Since for one concludes that Thus and Lemma 6.3.6 is proved. Remark 6.3.7. In [RSj] a stronger result is proved: (6.3.43) implies temperate distribution with support in a cylinder

for f(x, t) a A similar result is

329

established in [St]. The result of Lemma 6.3.6 can be derived from the known results on Radon’s transform. Remark 6.3.8. The line integral in (6.3.43) is defined in the sense of distributions by the formula

If (the Schwartz class) then of temperate distributions).

is defined by (6.3.46) as an element of

(the class

Remark 6.3.9. Our argument actually proves: Proposition 6.3.10. Given the data (6.3.7), q(x, t) can be uniquely reconstructed in the subset of consisting of the points of the lines which make 45° with the t-axis and which meet the planes t = 0 and t = T outside provided that q(x, t) is known outside Here is the closure of One can put in (6.3.7) a different but sufficiently rich set, for example, If in addition to (6.3.7) one knows the Cauchy data u(x, T) and for all where u(x, t) is the solution to (6.3.1), (6.3.2), (6.3.3), then one can uniquely reconstruct q (x, t) in the subset of consisting of the points of the lines which make 45° with the t-axis and which meet the plane t = 0 outside provided that q (x, t) is known outside This formulation is more general than the one in Theorem 6.3.1. Remark 6.3.11. The proof of Lemma 6.3.5 given in section II.4 is self-contained. One can give an alternative proof using an estimate (see, e.g., [H]):

where at t = 0, on and c denotes various positive constants which do not depend on (6.3.30) one obtains

is sufficiently large and w. From (6.3.47) and

330


Thus, for sufficiently large

one obtains

where 6.4

CONTINUATION OF THE DATA

In this section we discuss the following problem. Suppose that

Here is a bounded region with a smooth boundary L is a formally selfadjoint elliptic differential operator which, together with boundary condition (6.4.2) defines a selfadjoint operator in In place of (6.4.2) one can have any boundary condition for which L, together with this condition, defines a selfadjoint elliptic operator in of which the standard elliptic inequalities hold and the uniqueness of the solution to the Cauchy problem. Suppose that the data

are known, where determine the data

is an open subset of

The problem is: do the data (6.4.4)

The answer found in [Ro1], is yes. We present here the results of this paper. Lemma 6.4.1. Under the above assumptions data (6.4.4) determine data (6.4.5) uniquely. Proof. Note that data (6.4.4) are equivalent to the data

where

331

and s (L) is the spectrum of the selfadjoint operator L defined by the differential expression L and boundary condition (6.4.2): s (L) is the set of eigen-values of L since it is known that the spectrum of L is discrete with the only limit point at infinity. It is also well known that

where the bar stands for complex conjugate, eigenfunctions of L corresponding to the eigenvalue

are the orthonormal for

Claim 11. There exists a set of points is positive definite.

such that the matrix

Proof. Consider the repeated indices is understood. Thus

Therefore any one has

where j is fixed and summation over

is positive definite iff C is nonsingular, that is Suppose that there exists a such that

implies and for any

for

Then the system is linearly dependent as a system of functions on However, since for L the uniqueness of the solution to the Cauchy problem holds, one can conclude that the system is linearly dependent in which is impossible since the functions are orthonormal. Let us give more details. First, note that iff Suppose that for some

This means that

However,

332



By the uniqueness of the solution to the Cauchy problem, one concludes that in D. This implies that the system is linearly dependent, which is a contradiction. Claim 1 is proved. It is now easy to finish the proof of Lemma 1. Suppose the data (6.4.4) are given. Then the data (6.4.6) are given. By formula (6.4.8) the function (6.4.6) has a residue at the pole which is

Take matrix

Lemma 6.4.1 allows one to find as the number for which the is positive definite. By the assumption, one knows for all and for all Let us calculate r (x, z) for One has

Take for

and

Since Consider the linear system

one knows

with respect to

In the proof of Lemma 1 it is established that the determinant if the points are chosen so that the matrix is positive definite. We assume that are chosen so that this condition holds. Then are uniquely determined for any Therefore the function is uniquely determined for all Lemma 1 is proved.

Remark 6.4.2. The conclusion of Lemma 6.4.1 holds for the solution to the hyperbolic equation

u satisfies (6.4.2). The proof is the same. Remark 6.4.3. It is essential for the proof that the eigenspaces are finite-dimensional and that the spectrum of L is discrete.

7. INVERSE PROBLEMS OF POTENTIAL THEORY AND OTHER INVERSE SOURCE PROBLEMS

7.1

INVERSE PROBLEM OF POTENTIAL THEORY

The potential generated by a charge distribution

is given by the formula

where the integral is taken over the support D of If for and u (x) is known for then the inverse problem of potential theory is to find In Section 1.2.1 we have mentioned that this problem, in general, has many solutions, and gave examples of non-uniqueness. Therefore, let us assume in D, In this case the inverse problem is to find D given u (x) for where One of the basic earlier result in the Novikov’s theorem Theorem 7.1.1. ([Nov]) If j = 1,2, are two domains, star-shaped with respect to a common point D, and for then Here Proof. Let D, in D,

in one has

in

where Denote in

is the characteristic function of in

and

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7. Inverse problems of potential theory

Multiply (7.1.2) by an arbitrary to get

in

and integrate over

where the boundary integrals vanish because u (x) = 0 outside Originally u = 0 in the region but the unique continuation theorem for harmonic function implies u = 0 in Thus

One can check by a direct calculation that if

Choose a harmonic function H (x) in

then

such that

By the maximum principle, in Let h be the harmonic function defined by H (x) via formula (7.1.4), and be defined above formula (7.1.2). Let j = 1, 2, be the equations of in the spherical coordinates, Then (7.1.3) yields:

This and (7.1.5) imply Theorem 7.1.1 is proved.

on

and

must be empty. Theorem

335

Remark 7.1.2. Our proof differs from the original proof in [Nov]. The proof remains valid if is a known function, and not necessarily a constant. Exercise 7.1.3. If is a bounded starshaped domain with Lipshitz boundary, then the set of harmonic polynomials is dense in in norm. Here are the spherical harmonics. Hint. The traces on are dense in Exercise 7.1.4. If and

Exercise 7.1.5. If

of the linear combinations of the harmonic polynomials

for where

j = 1, 2, then

is a bounded domain and

then D is a ball of radius Hint. From (7.1.6) one gets, as the relation where is the volume of D. If D is a ball of radius r, then so Let us prove that (7.1.6) implies that D is a ball. From (7.1.6) as one gets Let h(0) = 1. Multiply the equation by h (x) and integrate over to get:

Here the relations and is an element of the group of rotations, then if an arbitrary unit vector, and g be the rotation for angle around by h(gy) in (7.1.7), differentiating with respect to and taking

where × stands for the cross product, is the unit sphere in gets and Gauss’ formula yields Since can be arbitrary, one gets

were used. If g Let be Replacing h (y) one gets

Since is arbitrary, one

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Finally, (7.1.8) implies that S is a sphere, so D is a ball Indeed, let s = s ( p , q) be the parametric equation of S, then and (7.1.8) implies Since the vectors and are linearly independent, one gets so s · s = const. This is an equation of a sphere. Exercise 7.1.6. [R195]. Assume

For what n and b can one conclude that

and

given (7.1.9)?

Hint. Fourier-transform (7.1.9) to get where and are entire functions of exponential type. Therefore is a meromorphic function of If (b – n)/2 is not an integer, one gets a contradiction unless 7.2

ANTENNA SYNTHESIS PROBLEMS

In Section 1.2.16 assume that in (1.2.15) is known for all k > 0 and all Then f(y) is uniquely determined by the Fourier inversion. If is known for all at a fixed then f (y), in general, is not uniquely determined because the value does not determine for all k > 0. The synthesis problem for linear antenna reduces to finding the current j (z) given the diagram (cf (1.2.18)):

Let

Then (7.2.1) can be written as:

Clearly known on the interval determines uniquely because is an entire function (of exponential type l) of If is known for all then j ( z ) is uniquely determined. The practical difficulty consists of finding j(z) from the knowledge of F on the interval [– k, k] . Also, the data are not known exactly in practice: one knows such that where the norm can be or norm. In this case the problem is to find a stable approximation j (z) to the exact current j (z) corresponding to the exact data that is

337

One can solve equation (7.2.2) with noisy data using the methods developed in Chapter 2. Alternatively, one can use the methods proposed in [R204], [R205], and find j (z) from exact data by the formula

where

where can be chosen arbitrary. If the noisy data, are given, then formula (7.2.4) with in place of with limit sign dropped, and with properly chosen, yields satisfying (7.2.3), that is, yields a stable approximation to the exact current j ( z ) . How to choose one can read in [R139], p. 205. There are books [ZK], [MJ], on antenna synthesis. 7.3

INVERSE SOURCE PROBLEM FOR HYPERBOLIC EQUATIONS

Let

The inverse problem is: given the data (7.3.2), find f (x, t). Let

Thus

and

Then

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The last two equations can be written as

where

System (7.3.7)(7.3.8) is uniquely solvable for and Therefore is determined uniquely up to an arbitrary function orthogonal in to and to for each This gives a solution of the inverse source problem stated in this Section. We followed [R73].

8. NON-OVERDETERMINED INVERSE PROBLEMS

8.1

INTRODUCTION

The inverse problems in which one wants to find a function of n variables from the data which can be considered as a function of m = n number of variables we call non-overdetermined, and if m > n then the problem is overdetermined. Among multidimensional inverse problems many of non-overdetermined ones are still open. For example, 1) the inverse obstacle scattering problem with the data (see Section 4.2, by subzero index we denote the fixed value of the parameter), 2) inverse potential scattering problems with the data or (see Section 1.2.13) (see Section 5.6), etc. 3) Inverse geophysical problems with the data

In example 1) the data are the values of a function of two variables, and the unknown is a surface in which is given also by a function of two varibles. In example 2) the unknown potential q (x) is a function of three variables, and the data are also functions of three variables. In example 3) the unknown is a velocity profile (or refraction coefficient), which depends on three variables, and the data also depends on three variables. For all of these problems currently one does not have even Uniqueness theorems which would say that the data determine the unknown object uniquely. In this Chapter we prove a uniqueness theorem for a non-overdetermined inverse problem of finding a potential from the values on the diagonal of the kernel of the

340

8. Non-overdetermined inverse problems

spectral function of a Schroedinger operator in a bounded domain. We have to assume additionally that the eigenvalues of this operator are all simple (see [R198]). 8.2

ASSUMPTIONS

Let

be a bounded domain with a connected boundary S, be a selfadjoint operator defined in by the Neumann boundary condition, be its spectral function, where j = 1, 2, . . . . The potential q (x) is a realvalued function, It is proved that q (x) is uniquely determined by the data if all the eigenvalues of L are simple. 8.3

THE PROBLEM AND THE RESULT

Let be a bounded domain with a connected boundary S, that is, locally the equation of S is given by the function We assume n = 3. This assumption is used only in the proof of Lemma 8.4.1 below and can be dropped. If n > 3 one should refer to the existence of the coordinate system in which the metric tensor, used in the proof of Lemma 8.4.1, has the property: for The basic ideas of our proof, outlined below, are valid for Let where is the Laplacian and the potential, is a real-valued function. Let j = 1, 2, . . . , be the normalized real-valued eigenfunctions of the selfadjoint operator L defined in by the Neumann boundary condition:

where N is the unit normal to S pointing into valued functions. By elliptic regularity, the functions the boundary S. The spectral function of L is defined by the formula:

We choose to be realare and on

the eigenvalues are counted according to their multiplicities. The inverse problem (IP) we are interested in is: IP: given and find q(x). We are concerned with the uniqueness of the solution to IP. This inverse problem is not overdetermined: the data is a function of n variables s, (S is an (n – 1)-dimensional manifold, and the unknown q (x) is also a function of n variables. It seems that no uniqueness theorems for multidimensional inverse scattering-type problems with non-overdetermined data have been obtained so far. The IP, studied

341

here, is a multidimensional one of the above type and with non-overdetermined data. We prove a uniqueness theorem for this problem. The inverse problem with the overdetermined data, which is the spectral function known and has been considered in [Be1]–[Be2] (see references therein). Here we follow [R198]. In Section 5.9 the following uniqueness theorem is obtained: Theorem 8.3.1. The data

determine q (x) uniquely.

The result we want to prove is: Theorem 8.3.2. The IP has at most one solution if all the eigenvalues of L in (8.2.1)–(8.2.2) are simple. It is possible that the conclusion of Theorem 8.3.2 holds without the assumption concerning the simplicity of the eigenvalues, but we do not have a proof of this currently. Generically one expects the eigenvalues to be simple, if there are no symmetries in the problem. The strategy of our proof is outlined in the following three steps: Step 1. This step is done under the additional assumption that all the eigenvalues are simple. It is an open question whether Theorem 8.3.2 holds without this assumption. Any assumption that will make Step 1 possible is sufficient for our purpose. Step 2. Step 3. Apply Theorem 8.3.1 to the data Step 1 does not require much work. Step 2 requires the basic work. Let us outline the ideas needed for the completion of Step 2. One can prove that cannot have zeros of infinite order, that is, if and where are positive constants and are arbitrary points on S, then In the Appendix it is proved that if a real-valued function does not have zeros of infinite order on S, then, up to its sign, the function f is uniquely defined on S by its square A procedure for finding, up to its sign, on S from the knowledge of on S is described below. It is known that cannot vanish on an open (in S) subset of S (uniqueness of the solution to the Cauchy problem for elliptic equations), and for almost all points of S. Take an arbitrary point at which By the continuity of on S, there is a maximal domain containing the point in which Define in Let be the boundary of We prove that is uniquely determined in The problem is to determine the sign of for in a neighborhood of where when

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The basic idea for determining this sign is to extend from across with maximal smoothness. Such an extension is unique and determines the sign of outside of For example, if one defines for x > 0, then the unique maximally smooth extension of in the region x < 0 is To determine the sign of

outside of calculate the minimal integer m for such that Here along a curve originating at a point in transversal to and passing through a point into The integer does exist if the zeros of are of finite order. That these zeros are indeed of finite order is proved in Lemma 8.4.1. Let us describe the way to continue along the curve across into Define sgn where is any point in a sufficiently small neighborhood of This algorithm determines uniquely forall given the data Since the eigenvalues are assumed simple, the above algorithm produces the trace of the eigenfunction on S. Since q (x) and S are up to the boundary, so are the eigenfunctions and the above algorithm produces a function on S. Therefore one can use the Malgrange preparation theorem ([CH], p. 43) to study the set of singular points of The above argument deals with the continuation of an eigenfunction through its “zero line” on S. There are at most finitely many points on a compact surface S at which several “zero lines” intersect and only finitely many “zero lines” can intersect at one point. Otherwise there would be a point on S which is a zero of infinite order of and this is impossible by Lemma 8.4.1 If one continues across by the above rule, the function will be uniquely determined on all of S by the choice of its sign at the initial point In the Appendix, at the end of this Chapter, we include a proof of a statement we have used in a discussion of Step 2. It is proved in the Appendix that a smooth real-valued function, which is defined on a smooth connected manifold M without boundary, and has no zeros of infinite order on M, is uniquely, up to a sign, determined on M by its square. In part of this proof some ideas communicated to me by Yu. M. Berezanskii are used. This completes the description of Step 2. The smoothness of the data is assumed for technical reasons: it guarantees the existence of an integer m in the above construction. It would be interesting to weaken this assumption and to find out if one can prove Lemma 8.4.1 assuming S is and In section 8.4 a detailed discussion of Step 2 is given. 8.4

FINDING

FROM

In Section 8.3 a method for finding discussed.

from the knowledge of

has been

343

Here a justification of this method is presented. This justification consists mainly of the proof of the following lemma: Lemma 8.4.1. The function

does not have zeros of infinite order.

Proof. It is known (see [H], p. 14, where a stronger result is obtained) that a solution to the second order elliptic inequality:

where c = const > 0, and is a strictly elliptic homogeneous second order differential expression (summation is understood over the repeated indices):

cannot have a zero of infinite order at a point

provided that:

and

By zero of infinite order of a solution u of (3.1) a point

is meant such that

where are positive constants independent of and is sufficiently small so that the ball We use this result to prove that the same is true if and Let be the local equation of S in a neighborhood of the point which we choose as the origin, and be directed along the normal to S pointed into Let us introduce the new orthogonal coordinates

so that is the equation of S in a neighborhood of the origin, and assume n = 3. For example, one can use the coordinate system in which the z-axis is directed along the outer normal to S, and the x, y-coordinates are isothermal coordinates on S, which are known to exist for two-dimensional in ( and even for in [Gu], p. 246). For an arbitrary in one can prove that locally one can introduce (non-uniquely) the coordinates in which the metric

344


tensor is diagonal in a neighborhood of S. To do this, one takes two arbitrary linearly independent vector fields tangent to S and orthogonalize them with respect to the Euclidean metric in using the Gram-Schmidt procedure, which is always possible. The third axis of the coordinate system, which we are constructing, is directed along the normal to S at each point of the patch on S. Let A and B be the resulting orthogonal vector fields tangent to S. Then one can find (non-uniquely) a function defined on the local chart, such that the Lie bracket of the vector fields and B vanishes, Then the flows of the vector fields and B commute and provide the desired orthogonal coordinate system in which the metric tensor is diagonal ([CM]). The condition can be written as the following linear partial differential equation: where x, x and y are the parameters, and z = f (x, y) is the local equation of the surface S on the chart. The above equation for has many solutions. One can find a solution by the standard method of characteristics. A unique solution is specified by prescribing some Cauchy data, which geometrically means that a noncharacteristic curve through which the surface passes, should be specified ([CG]). If n > 3 the situation is less simple if one wants to use the same idea in the argument: there are many ((n – 1)(n – 2)/2) Lie brackets to vanish in the case of n – 1 vector fields tangent to S in and it is not clear for what S these conditions can be satisfied. However, for our argument it is sufficient to have the coordinate system in which the metric tensor has zero elements for and does not vanish. In this case the even continuation (8.4.8)–(8.4.10), that is used below, still allows one to claim that the function (8.4.8) solves the same equation in the region as it solves in the original region If the elements for do not vanish, then the equation in the region will have some of the coefficients in front of the second mixed derivatives with the minus sign, while these coefficients in the region enter with the plus sign. So, in this case the principal part of the operator M, which is used in (8.4.11), will be different in the regions and Recall that the Laplace operator in the new coordinates has the form of the Laplace-Beltrami operator: where summation is understood over the repeated indices, and If is a zero of of infinite order in the sense then equations (8.2.1) and (8.2.2) imply that so that condition similar to (8.4.5) holds for the integrals over Indeed, the derivatives of in the tangential to S directions at the point vanish by the assumption. From (1.1) and (1.2) it follows that the normal derivatives of of the first and second order vanish at Differentiating equation (1.1) along the normal one concludes that all the normal derivatives of vanish at the point Thus we may assume that is a zero of of infinite order, that is, inequalities (8.4.5) hold for for the integral over In the

one writes the equation for

345

as follows:

We drop the subscript j of is defined as:

and of

in what follows. In (8.4.7) the operator M

over the repeated indices summation is understood, is the metric tensor of the new coordinate system, and the coefficients in front of the second-order derivatives in M are extended to the region as even functions of so that the extended coefficients are Lipschitz in a ball centered at with radius Let us define w in a neighborhood of the origin, by setting

and

The functions is Lipschitz in Furthermore,

defined by (8.4.10), are Lipschitz in the ball where

where c = const > 0, and

if one assumes

for all sufficiently small

if

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Note that the change of variables is a smooth diffeomorphism in a neighborhood of the origin, which maps the region onto a neighborhood of the origin in in and one can always choose a half-ball belonging to this neighborhood, so that (8.4.13) follows from (8.4.5). Therefore in and consequently in This implies that in D by the unique continuation theorem (see [H] and [W]). This is a contradiction since Lemma 8.4.1 is proved. This lemma provides a justification of the argument given for Step 2 in Section 8.3. Remark 8.4.2. In this remark we comment on the numerical recovery of the potential from the data We have explained how to get the data from Therefore the spectral function

and the resolvent kernel

are known on S. If is known, then the Neumann-to-Dirichlet (N-D) map is known. This map at every fixed associates to every a function by the formula

Here

The unique solution to (8.4.17) is given by (8.4.16). We now outline an alternative proof of Theorem 8.2.2: steps 1 and 2 are the same as in Theorem 8.2.2; step 3 consists of the construction of the N-D map by formula (8.4.16) and reference to Chapter 5. In Chapter 5 the result similar to Theorem 8.4.3 (see below) is proved for the D-N (Dirichlet-to-Neumann) map. As above, m = 1, 2, stands for the Neumann operator in D. Let us assume that is a regular point for and Theorem 8.4.3. If Proof. Take an arbitrary

and

generate the same N-D map then and consider the problems

347

Subtract from the equation with m = 1 the equation with m = 2 and get

where The condition w = 0 on S follows from the basic assumption, namely the N-D map is the same for and Let in D, be arbitrary. Multiply (8.4.18) by integrate by parts using boundary conditions (8.4.18), and get the orthogonality relation:

The last inclusion in (8.4.19) follows since is arbitrary. From (8.4.19) and property C it follows that Theorem 8.4.3 is proved. 8.5

APPENDIX

Lemma 8.5.1. Let M be a connected manifold and suppose that a real-valued function does not have zeros of infinite order, i.e., for an arbitrary point for some multiindex Then, up to sign, the function f is uniquely defined on M by its square Remark 8.5.2 The conclusion is not true, even locally, if f has zeros of infinite order. Indeed, if, for example, and x = 0 is a zero of infinite order, then the functions for for x > 0, for for x > 0, and the functions are all and Remark 8.5.3 If dim M > 1, order, then there might be no function Indeed, an example is

does not have zeros of infinite such that on M.

Proof of Lemma 8.5.1. All functions below are real-valued. Let T := {s : h(s) = 0}, where h := f – g. Then T is closed. If T = M, then g = f on M, and Lemma holds. If then is an open, non-empty set, and h = 2 f on If (the overline stands for the closure), then g = – f on M, and Lemma holds. Otherwise, the set is an open, non-empty set, and it may happen that Since Q is a proper subset of S (because T is), and since (because S is connected and cannot contain a proper subset which is simultaneously open and closed), it follows that there exists a point Let and One has for any multiindex because h = 0 on the open set U. Thus, by the continuity of On the other hand, for some multiindex because and has zeros of at most finite order. This contradiction proves that either T = S and then f = g, or and then f = –g. Lemma 8.5.1 is proved.


9. LOW-FREQUENCY INVERSION

9.1

DERIVATION OF THE BASIC EQUATION. UNIQUENESS RESULTS

This Chapter is based on [R77], [R83]. In order to give a motivation for the theory presented in this section we start with the following example. Consider the inverse problem of geophysics, see Section 1.2.4. The integral equation equivalent to this problem is:

It is easy to check that for sufficiently small k the integral operator in (9.1.1) has small norm in the Banach space C(D) of the continuous in D functions, that the scattered field

is continuous in

and that (9.1.1) can be written as

350


where

is uniform in x,


Define the low frequency data by the formula


where we took x, since the field u(x, y, k) is known on P. Uniqueness of the solution to (9.1.6) will be proved if one proves that the homogeneous equation

has only the trivial solution v = 0. Note that in D for all Lemma 9.1.1. The set

is complete in

in the set

Here and below stands for the familiar Sobolev spaces of functions whose derivatives up to the order belong to p is fixed in the interval We give two different proofs. Both can be generalized to the case when in formula (9.1.7) and in Lemma 9.1.1 the function is substituted by G(x, y), the Green function of a general second order elliptic operator in Lemma 9.1.1 the operator is substituted by it is assumed that zero is not an eigenvalue of the Dirichlet operator in D, that the Dirihlet problem in has only the trivial solution, and that the unique continuation principle for the solutions of the equation holds. Proof. First Proof Let

and assume that

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We want to prove that this implies u = 0. Define in w = 0 on P. Therefore

Then

By the unique continuation principle for the solutions to the equation (9.1.10) one concludes that

By the elliptic regularity result, bedding theorem imply

This, equation (9.1.11) and the em-

The embedding theorem says that if then w and are well defined on a smooth manifold of dimension n – 1 and if is a one-parametric family of manifolds in a neighborhood of such that is a small number, and is parallel to in the sense defined below, then

that is, w and depend continuously on t in the sense (9.1.13). The manifold is called parallel to if the normal to at each point intersects at exactly one point whose distance from the point is equal to t, From (9.1.12) one obtains

By the assumption in D. Multiply (9.1.14) by (the bar stands for complex conjugate), integrate over D, then by parts, using boundary conditions (9.1.14) for w, to get Thus u = 0 in D. Lemma 9.1.1 is proved. Remark 9.1.2. If the general second order elliptic operator replaces and G(x, y) replaces then the proof goes without any changes and one uses the symmetry of in the argument given below formula (9.1.14). Namely, This argument for the operator

for example, requires that

Second Proof of Lemma 9.1.1 The advantage of this proof is that it allows a generalization to the case of nonsymmetric operators

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It is sufficient to prove that the set of functions

is dense in Indeed, clearly in D. Therefore, if can approximate in an arbitrary element of with an arbitrary accuracy, then, by the maximum principle for harmonic functions, it can approximate an arbitrary in with an arbitrary accuracy. Thus, in order to prove Lemma 9.1.1, it is sufficient to prove that if then


Clearly

In particular, in v = 0 on P. This implies (for example, by the maximum principle) that v = 0 in By the unique continuation property for harmonic functions one concludes, using (9.1.18), that v = 0 in Therefore u = 0 on From this and from (9.1.18) it follows that v = 0 in D. Since v(x) is a simple layer potential, the jump relation for the normal derivative of v implies f = 0. This completes the second proof of Lemma 9.1.1. Remark 9.1.3. If

replaces

in this argument, then the crucial point is the estimate

which holds for in D, the coefficients of on

and

is dense in

provided that zero is not an eigenvalue of the Dirichlet operator and are sufficiently smooth (say and for The constant c in (9.1.19) does not depend on u, it depends Estimate (9.1.19) shows that if a set of functions

then it is dense in

(and in

in

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Remark 9.1.4. We use the results similar to Lemma 9.1.1 quite often. The result and proofs of Lemma 9.1.1 remain valid if P is replaced by an arbitrary closed compact Lipschitz surface which contains D inside and for some noncompact surfaces S which contain D inside and have the property that the Dirichlet problem for the operator in the exterior region (that is, the region not containing D) with the boundary S has at most one solution in the class of functions representable either in the form: or in the form:

We are now ready to prove the following uniqueness result. Lemma 9.1.5. If

and (9.1.7) holds then v = 0.

Proof. By Lemma 9.1.1, equation (9.1.7) implies

This implies that v(z) = 0 according to Lemma 9.1.6 below. Lemma 9.1.5 is proved. Lemma 9.1.6. The set is an arbitrary bounded domain.

is complete in

where

In Lemmas 9.1.5 and 9.1.6 the central idea of the Property C, a method of proving uniqueness theorems for inverse problems, based on completeness of the set of products of solutions to homogeneous PDE is presented. This method for proving uniqueness theorems for inverse scattering and other inverse problems has been introduced by the author in the study of inverse scattering problems in quantum mechanics and in geophysics and applied to many other inverse problems. It was the main tool in Chapter 5. A proof of Lemma 9.1.6 follows from the general argument given in Chapter 5. It is clear from this proof that not all of the functions are needed in order to derive from (9.1.21) that v = 0, or, which is the same, to conclude that the set of products is complete in First, we can restrict ourselves to the exponential solutions of the Laplace equation. This is not a restriction really, since the set of exponential solutions of this equation is dense in in the set of all harmonic functions in D, i.e., all elements of However, there are still many degrees of freedom left. For the general results concerning Property C see Chapter 5. 9.2

ANALYTICAL SOLUTION OF THE BASIC EQUATION

In this section we give an analytical solution to the basic equation (9.1.6).

354


Take the Fourier transform in the variables and ing that belong to P, and use the formula

where

asssum-

to get:

where

Let

and write (9.2.2) as

Here we took into account that supp

so that in (9.2.2), and defined in the coordinates (9.2.4). The Jacobian

does not vanish if and are linearly independent. The function v (z) depends on 3 variables, while F depends on 4 variables. Therefore we can use only a subset of the data for recovery of v(z). Let


355

Equation (9.2.8) defines v (z) uniquely: one should take the inverse Fourier transform in the variables p and the inverse Laplace transform in the variable s. We have proved Theorem 9.2.1. Equation (9.1.6) can be solved by inverting the integral transform (9.2.8). 9.3

CHARACTERIZATION OF THE LOW-FREQUENCY DATA

Let us give a characterization of the low-frequency data, that is, a necessary and sufficient condition on the function f(x, y) in (9.1.6) for this equation to be solvable in the class of compactly supported in square integrable v(z). Theorem 9.3.1. Equation (9.1.6) is solvable in the above class of v iff is an entire function of and s of exponential type where R is a positive number, and

Proof. The necessity of these conditions follows from formula (9.2.8) because v(z) is compactly supported in Their sufficiency follows from the Paley-Wiener theorem: i) a function

h(p)

is entire of exponential

type

R and

iff and ii) a function h(x) is an entire function of exponential type R such that iff where

Exercise. Prove that 9.4

is bounded for

PROBLEMS OF NUMERICAL IMPLEMENTATION

In this section we consider some practical questions related to inversion of (9.2.8). Let

so that

If we take then in (9.2.6) does not vanish and, given and s one finds and within the above ranges, then and and then Therefore, we can compute in (9.2.8) by formula (9.2.7). In the threedimensional space with Cartesian coordinates the points fill in the circle of radius s centered at (0, 0, s) and lying in the plane parallel to plane. As s run from 0 to the set of points Q fill in the cone in the upper half-space with vertex at the origin and a vertex angle of We invert the Laplace transform in (9.2.8) by the formula

356


where

If we use formula (9.4.2) for inversion then should be known for and Given any particular we can choose and calculate Formula (9.4.2) requires the knowledge of on the line However, one can invert the Laplace transform of a compactly supported function using only its values on the real axis (see [AR2]). If one wishes to fix in (9.4.2), then is recovered for The Fourier transform of a compactly supported function is therefore known for One can invert this Fourier transform from the compact region 9.5

HALF-SPACES WITH DIFFERENT PROPERTIES

In this section we consider the case when the lower half-space has different properties from the upper one. Let are the densities and velocities in and respectively, the assumptions about v(x) are the same as in Section 9.1. The inverse problem IP in this case can be formulated as follows. The governing equations are:

The inverse problem IP is: Given u(x, y, k) for all x, and all The solution to (9.5.1)–(9.5.2) with

find v(x). m = 1, 2, is

where

If and

then the limit of (9.5.3) is

Therefore, the integral equation analogous to (9.1.6) for IP is essentially the same:

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9.6

INVERSION OF THE DATA GIVEN ON A SPHERE

In this section we consider inversion of the data given on a sphere. In this case equation (9.1.6) is given with x, where is a sphere which contains D = supp v. Let us consider a more general equation which comes from the Born approximation of the exact equation (9.1.1). The Born approximation consists in taking u = g (z, y, k) under the sign of the integral in (9.1.1). The resulting equation is

Let us use the well known formulas (see, e.g., [GR])

where respectively,

and

Substitute (9.6.2) into (9.6.1) and get

Thus

are the spherical Bessel and Hankel functions

358


Thus

are expressed in terms of the data analytically. Multiply sum over and use (9.6.4) to get

by

Choose an arbitrary and such that The left-hand side of (9.6.7) is a known function, which we denote if and (9.6.7) gives for Since v (z) is compactly supported, we reduced the problem of solving (9.6.1) to inversion of the Fourier transform from the compact This problem is solved analytically with arbitrary accuracy for exact data in [R51] and the solution was used in [R83], [R139] and [RKa], see also Theorem 5.8.4. We have proved that equation (9.6.1) can be solved analytically if k > 0. If k = 0 then the above argument gives and this is not sufficient for recovery of v(z). However, equation (9.6.1) at k = 0 has at most one solution. The data at k = 0 determine uniquely the numbers

Finding v (z) from (9.6.8) is a moment problem which has at most one solution.

9.7

INVERSION OF THE DATA GIVEN ON A CYLINDER

In this section we consider inversion of the data given on a cylinder, so that equation (9.1.6) is given with x, The idea of inversion is the same as in section 9.6. We start with the formula

where

and

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Equation (9.6.1) can be written as

where

Here

and

is the datum on

Multiply (9.7.5) by


sum over n,

and use (9.7.2) to get

where

Let When v, run through vector runs through the ball in Thus, equation (9.7.6) reduces to finding v (z), a compactly supported function, from its Fourier transform known for If k = 0 then equation (9.6.1) with x, has at most one solution. We leave further details for the reader. 9.8

TWO-DIMENSIONAL INVERSE PROBLEMS

In this section we consider two-dimensional problems. The difficulty in twodimensional problems comes from the fact that Green’s function of the Helmholtz

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equation in

and it has no finite limit as

where

and we denote in this section where is Euler’s constant. Consider equation (9.1.6) in with g given by (9.8.1). The limit (9.1.5) does not exist now, so the argument which led to (9.1.6) has to be modified. It follows from (9.1.1) with g given by (9.8.1) that

Therefore

and

The quantities and are calculated explicitly given the lowfrequency data. The number is the intensity of the inhomogeneity. Equation (9.8.5) is analogous to (9.1.6). It follows from (9.8.4) that

This is an integral equation arising in inverse problems of potential theory. In general, it is not uniquely solvable. However, if for and is zero otherwise, then take in (9.8.6), and get

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Differentiate (9.8.7) with respect to and get:

put

Take the Fourier transform of (9.8.8) in

to get

One has ([Er])

so

Thus (9.8.9) becomes

or

Therefore the Laplace transform of is known and can be recovered analytically by the method given in [R139]. Thus, under the assumption an analytical inversion is given of the zero offset data (i.e., x = y) known at the line Consider equation (9.8.5). Let x, and and let

Take the Fourier transform in

and

Differentiate (9.8.5) with respect to Then

of (9.8.13) and use (9.8.10) to get

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Let

where is calculated in the new coordinates. The Jacobian of the transformation is if Take and and invert the Fourier transform in and the Laplace transform in t in (9.8.15) to get v uniquely and analytically. 9.9

ONE-DIMENSIONAL INVERSION

In this section we study the one-dimensional problem of finding given the values of u(x, y, k) for y = 0, is a fixed (small) number. This section is of methodological nature. The basic integral equation is

Let


Equation (9.9.4), as

is uniquely solvable by iterations. One has

Let

Substitute (9.9.6) into (9.9.4) and equate coefficients in front of similar powers of This gives some recurrence formulas for From these formulas one can find the moments and, therefore, v (z).

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9.10

INVERSION OF THE BACKSCATTERING DATA AND A PROBLEM OF INTEGRAL GEOMETRY

In this section we invert analytically the backscattering data in the Born approximatisn. Sut x = y in (9.6.1) and assume that is known for all and all The problem is to find v(z) from equation (9.6.1) with the above data:

Differentiate (9.15.1) in k, put

and get

Take the Fourier transform of (9.10.2) in

where for

Define can be written as

using the formula

one has

and get

for

Since

(9.10.4)

where is the right-hand side of (9.10.5) expressed in the variable Thus v (z) can be found uniquely from (9.10.5) by inversion of the Fourier transform. Note that is an entire function of of exponential type, that is since v(z) is compactly supported. This necessary condition on the data is also sufficient for v (z) to have compact support, as follows from the Paley-Wiener theorem. Let us give a simple example: Let where is Dirac’s function. Then Thus is an entire function and its inverse Fourier transform is Let us summarize the result. Proposition 9.10.1. Equation (9.10.2) has at most one solution in the class of compactly supported distributions. For equation (9.10.2) to have a compactly supported solution it is

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necessary and sufficient that the function of exponential type, where

be an entire function of

Equation (9.10.2) is equivalent to an Integral geometry problem of finding v(z) from its integrals over the spheres centered at and of radii Indeed, define where the bar stands for complex conjugate. Multiply (9.10.2) by and integrate in over to get

where

is the delta function and h = 0 for t < 0. Equation (9.10.6) can be written as

Thus the integrals of v (z) over the family of spheres are known, and v(z) is recovered from these data via equation (9.10.5). Necessary and sufficient conditions on the function th(x,t) for this function to be of the form (9.10.7) follow from the given above necessary and sufficient conditions on for this function to be of the form (9.10.2). 9.11

INVERSION OF THE WELL-TO-WELL DATA

In this section we study inversion of the well-to-well data. The governing equation is

where and k > 0 is fixed. Equation (9.11.1) models the well-to-well imaging problem in the Born approximation in the case when the inhomogeneity v(z) does not depend on one of the coordinates, say and the acoustic geophysical data are collected along two boreholes, lines and at a fixed k > 0. We wish to solve equation (9.11.1) for v (z). Take the Fourier transform of (9.11.1) with respect to and and get

365

If then plane with the cut (–k, k) so that k is taken to be k + i0. In particular, for and

The radical is defined on the complex for and in the calculations one obtains:

We have used the formula

Consider the case when the data are given for Let

and

and the right-hand side of (9.11.5) should be expressed in the variables p, q. The Jacobian

Equation (9.11.2) can be written as

Consider the image of S under the transformation When run through the boundary of S then (p, q) run through the boundary of Take and let run through the segment (– k, k). On (p, q) plane one gets the curve that is an upper half of the circle centered at (–k, 0) with radius k. A similar argument shows that the boundary of consists of the circles and The diagonal of S is mapped into the origin of the (p, q) plane. Therefore, given the data in S one knows F(p,q) in the disks with the boundaries Equation (9.11.7) shows that the Fourier transform of v (z) is known in Since v (z) is compactly supported, the Fourier transform can be uniquely inverted from for v (z) by the method of [R139] (see also [RKa]). Consider the case when

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and is studied. Write (9.11.3) as

are real}. In this case imaging with evanescent waves

where

where the right-hand side of (9.11.9) should be expressed in the variables p , q. As above, the Jacobian of the transformation does not vanish for The image of the region under this transformation has the boundary which consists of the hyperbolas and The knowledge of in allows one to uniquely recover v(z) by inverting the Fourier and Laplace transforms in (9.11.6). One can consider the case when k = 0, so that (9.11.8) holds with and again v (z) can be uniquely recovered from the data given in Finally, let us briefly discuss the low frequency exact inversion theory for the problem at hand. One starts with the equation

takes the Fourier transform of (9.11.10) with respect to and and let to get equation (9.11.8), where and Therefore the inversion problem is solved as above. In this argument we avoided the direct study of the asymptotic behavior of the solution to (9.11.10) as If one takes the Fourier transform of (9.11.10), then the singular term with defined below formula (9.8.1), vanishes if one considers the region as we do in (9.11.8). This makes it possible to pass to the limit in the Fourier transformed equation (9.10.2) while it is not possible to pass to the limit in (9.11.9) directly since g (x, y, k) does not have a finite limit as (see formula (9.8.1)). 9.12

INDUCTION LOGGING PROBLEMS

In this section we study induction logging problems. Consider first equation (9.1.6) with for and v = 0 otherwise. Here Let x = (0, 0, x), y = (0, 0, y), y = x – d , d = const, Then one has:

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where

Equation (9.12.1) is of convolution type and can be solved by the Fourier transform. If is known in place of the exact data is a known small number, the noise level, then a regularization is needed for a stable solution of (9.7.7), that is, for finding such that as If the measurements are taken strictly inside the borehole, then equation (9.12.1) is to be replaced by the equation

This is a Wiener-Hopf equation of the first kind which can be solved by the known factorization methods or by reducing it to a Riemann-Hilbert problem. The same method is valid for inversion of the data at a fixed k > 0 in the Born approximation. The analogue of h(x) in (9.12.3) is the function where The physical interpretation of the problem is as follows. A point source of the acoustic waves situated at the point y is moved along the borehole which is the The acoustic field is measured at the point x which is also on the The distance between x and y is d. One wishes to find the inhomogeneity v from the logging measurements, that is, from Consider now the two-dimensional case in which v = 0 for or is measured for all x, y on the that is on the axis of the borehole, k > 0. The basic equation is

Let us use the spheroidal coordinates:

The Jacobian is the angle between the plane containing x and y and the plane containing x – z and y – z. In our case x and y

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are directed along In the variables

and we take as the plane containing x and y the plane equation (9.12.4) takes the form:

It is clear from (9.12.7) that the data f should depend on Assume that f (x, y, k) is given for all k > 0, and denote

and

The function p(s , m) can be found from (9.12.7) by taking the inverse Fourier transform in the variable k, and equation (9.12.8) can be solved for given p(s, m). Let Denote by E(s, the ellipse Equation (9.12.8) can be written as

Here

is the element of the arc length of the ellipse, where expressed in the variables Equation (9.12.9) is an integral geometry problem of finding w from the knowledge of its integrals over a family of ellipses. In the literature [LRS] a similar problem was studied in the case when the family of ellipses was different: one focus was at the origin and the other one ran through a straight line. Suppose that Then (9.12.8) reduces to

Let


Differentiate (9.12.11) to get

Several other cases of analytic solvability of equation (9.12.4) are considered in [R139].

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9.13

EXAMPLES OF NON-UNIQUENESS OF THE SOLUTION TO AN INVERSE PROBLEM OF GEOPHYSICS

In this section we give examples of non-uniqueness of the solution to an inverse problem of geophysics. Consider equation (9.1.6) with x = y. It corresponds to the inverse problem in which the data are collected at the point at which the source is, that is, at the zero-offset. The uniqueness problem reduces to the homogeneous equation

A similar equation corresponding to the Born inversion is (k > 0 is fixed):

Proposition 9.13.1. Equation (9.13.1) has infinitely many non-trivial solutions. Remark 9.13.2. Equation (9.13.1) the trivial solution.

where B is an open set in

has only

Remark 9.13.3. Equation

has infinitely many non-trivial solutions. Proof of Proposition 9.13.1. Define

Take an arbitrary

if

put Then v defined by (9.13.4) is a non-trivial solution to (9.13.1). Indeed, take the Fourier transform in of (9.13.1) to get

Here

and

and ([GR], formula 6.65.4):

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where

is the modified Bessel function. Substitute (9.13.4) into (9.13.5) to get

After an integration by parts on the left, and taking into account that one obtains an identity which implies that the function (9.13.4): is a non-trivial solution to (9.13.1) for any w = 0 if and any where the cylinder The proof is completed.

Proof of Remark 9.13.2. Suppose (9.13.1) holds for all open set. The function

where

is an

is an analytic function of and in a neighborhood in of the region Therefore, by the unique continuation property, the condition h(x) = 0 in implies h(x) = 0 in Take the Fourier transform of (9.13.8) in to get

Since v(z) and h(x) are compactly supported (they vanish outside D), it follows that and are entire functions of Thus (9.13.9) yields a contradiction since is not a meromorphic function while is. This contradiction proves the Remark.

Exercise 9.13.4. Does there exist a non-trivial solution to (9.13.1) in the case when (9.13.1) holds for all and v = 0 for

Proof of Remark 9.13.3. Take any ball let be a concentric ball and take in in and are constants, chosen so that is the volume of B. Then the function where is a potential with charge density v (z). By symmetry it is clear that and for where const is proportional to the total charge which is zero by the construction. So in but

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9.14

SCATTERING IN ABSORPTIVE MEDIUM

In this section we consider inverse scattering in an absorptive medium. Let

u satisfies the radiation condition, v(x) is the same as in Section 9.1 and Ima = 0, a(x) = 0 outside D. We are interested in IP : given u(x, y, k) and find v(x) and a(x). The problem with the data given for can be studied as in Section 9.1. We want to show that the method for solving IP in Section 9.1 yields the solution to our IP as well. We start with the equation

This equation can be solved uniquely by iterations if k is small enough and

The function can be calculated given the data and therefore can be considered known. Equation (9.14.3) is uniquely solvable for a(z) as in Section 9.1. If a is found from (9.14.3), then one obtains an equation for v:

Again, f(x, y) is computable from the data and therefore can be considered known, and (9.14.4) can be uniquely solved for v(z). We have proved Proposition 9.14.1. Problem IP has at most one solution and can be solved analytically. 9.15

A GEOMETRICAL INVERSE PROBLEM

In this section we consider a geometrical inverse problem IP, Section 1.2.15, of interest in applications. One can think that u is the velocity potential of an incompressible fluid, the pressure and the normal component of its velocity are known on and one wants to find the surface on which the normal component of the velocity vanishes, the surface of a reservoir, for example. Proposition 9.15.1. IP has at most one solution. Proof. Suppose there are two solutions and and is a region bounded by or by a component of Then u solves the homogeneous Dirichlet problem

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for the Laplace equation, so u = 0 in outside the region which lies inside Proposition 9.15.1 is proved.

By the unique continuation property, u = 0 This is a contradiction since

Let us give a method for solving IP. First, without loss of generality assume that is the unit circle. If not, one first maps conformally onto the unit disk. Then is mapped onto the unit circle. So, assume that is the unit disk. Any harmonic function outside can be written as

The constants from the equations

are uniquely and constructively determined, given

and

where

Let be the equation of found from the equation u = 0 on

if

Then the unknown function

The same idea can be easily used for other boundary conditions on then, using the formulas

can be

For example,

where and are the coordinate unit vectors in polar coordinates, N is the outer normal to and is given by formula (9.15.1) with the coefficients found from (9.15.4) and (9.15.5). The equation for in the case of the boundary condition

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is

where

is given by (9.15.1) as explained above.

Example 9.15.2. Let

Then

One finds u = 1 – In r and (9.15.6) is 1 – In r = 0, r = e. Therefore

Thus is the circle

Exercise 9.15.3. Does the problem in the above Example have a solution if Example 9.15.4. Let is is no bounded Example 9.15.5. Let the problem

Then, one finds,

and (9.15.8) Therefore, there

which solves the problem. Then

and any closed curve

solves

where D is the region bounded by Remark 9.15.6. The last example shows that the inverse problem analogous to IP with the Neumann boundary condition on is not uniquely solvable. Example 2 shows that the inverse problem may have no solutions. 9.16

AN INVERSE PROBLEM FOR A BIHARMONIC EQUATION

Consider the problem

The inverse problem, IP, is: Given the data (9.16.2)–(9.16.3), find a curve

such that

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One can use other boundary conditions on such as on for example, where N is the outer normal to The basic idea is to measure the Cauchy data for elliptic equation, to determine from these data the solution, and to find the zero set of the solution from (9.16.4). Of course, the Cauchy problem for elliptic equations is ill-posed, but its solution may still be of practical use. The problems such as IP can be of interest in elasticity theory as a way to determine cracks given boundary measurements. Note that uniqueness of the solution to IP is not claimed. What we outline is a way to find some of the solutions to IP. Let us look for a solution to (9.16.1) of the form

where should be chosen so that the boundary conditions (9.16.2), (9.16.3) are satisfied. This requirement is equivalent to the equations

Here and the Fourier transform is taken in the sense of distributions. We have six functions and four equations (9.16.6)–(9.16.9). Two additional equations can be imposed. Let us choose these additional conditions as:

Then (9.16.6)–(9.16.9) can be solved uniquely and explicitly

Formulas ((9.16.5), (9.16.10) and (9.16.11) solve problem (9.16.1)–(9.16.3). The solution is smooth if the data are such that the integral (9.16.5) with and given by (9.16.10) and (9.16.11) converges rapidly.

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If the solution u(x, y) to (9.16.1)–(9.16.3) is found, then (9.16.4) leads to the equation for y = y(x),

where y = y(x) is the equation of Remark 9.16.1. The IP may have no solutions. Example: Let if and zero otherwise, (9.16.4) is exp[2(ix + y) ] = 1, or Thus, no closed curve in 9.17

y = 0,

then Equation n = 0, ±1, ... .

solves IP.

INVERSE SCATTERING WHEN THE BACKGROUND IS VARIABLE

Let

then, with

in

one has:

We assume for simplicity that the radiation condition for u and G selects the unique solutions to equation (9.17.1). Under mild assumptions on one can prove that G(x, y, k) has a limit as

If (9.17.3) holds, then one derives from (9.17.2) that

where

Therefore, the basic equation (9.17.4) is the same as (9.1.6). The crucial difference is in the low-frequency data f (x, y): the data (9.17.5) differ from the function on the

376 9. Low-frequency inversion

left-hand side of (9.1.5). Indeed, formally,

The practical conclusion is: in order to use the knowledge of the Variable background one should calculate accurately the Green function G(x, y, k) in (9.17.1) and use G(x, y, k) in (9.17.5) for calculating f(x, y) from the data u(x, y, k), x, Although (9.17.3) holds, one cannot use g(x, y, k) in place of G(x, y, k) in (9.17.5). A simple sufficient condition for (9.17.3) to hold is: where c = const > 0, and is such that for the Green function corresponding to equation (9.17.3) holds. For example, has this property, and many other examples can be given. In some cases G(x, y, k) can be calculated analytically. For example, let

This refraction coefficient is of interest in the study of layered structures. One can calculate G(x, y, k) for defined in (9.17.7) by taking the Fourier transform in and of (9.17.1). Let

Then

where If calculated analytically:

is given by (9.17.7) then

the solution to (9.17.9), can be

377

for

is defined by symmetry,

The constants and 9.18

and are uniquely determined by the requirement that are continuous at and

REMARKS CONCERNING THE BASIC EQUATION

We have:

Here

is an open set on

and L is a line on P.

Remark 9.18.1. If L is not a piecewise-analytic curve then (9.18.1) implies that v (z) = 0. Indeed, w(x, y) as a function of y is harmonic for in particular, in a neighborhood of P. Therefore, the set of zeros of this function is an analytic set whose intersection with an analytic set has to be piecewise-analytic. Therefore, if it is not, one concludes that an open set on P, containing L, has to be the set of zeros of w(x, y). If w(x, y) = 0 for all and all where and are open sets on P, then v(z) = 0.

Remark 9.18.2. If L is an analytic curve then the question whether (9.18.1) implies v(z) = 0, is open. Remark 9.18.3. If L is a union of two straight lines and their angle of intersection is where is an irrational number, then (9.18.1) implies that v(z) = 0. Indeed, the line which is the reflection of with respect to has to be the line of zeros of a harmonic function if is. This follows from the fact that is harmonic if is and Thus, the set of zeros of the harmonic function w(x, y) of y will be dense on P if belongs to the set of its zeros. Thus w(x, y) = 0 if w(x, y) = 0 for If w(x, y) = 0 for all and all then v(z) = 0.


10. WAVE SCATTERING BY SMALL BODIES OF ARBITRARY SHAPES

10.1 WAVE SCATTERING BY SMALL BODIES 10.1.1

Introduction

The theory of wave scattering by small bodies was initiated by Rayleigh (1871). Thompson (1893) was the first to understand the role of magnetic dipole radiation. Since then, many papers have been published on the subject because of its importance in applications. From a theoretical point of view there are two directions of investigation: (i) to prove that the scattering amplitude can be expanded in powers of ka, where and a is a characteristic dimension of a small body, (ii) to find the coefficients of the expansion efficiently. Among contributors to the first topic are Stevenson, Vainberg, Ramm, Senior, Dassios, Kleinman and others. To my knowledge, there were no results concerning the second topic for bodies of arbitrary shapes. Such results are of interest in geophysics, radiophysics, optics, colloidal chemistry and solid state theory. In this Section we review the results of the author on the theory of scalar and vector wave scattering by small bodies of arbitrary shapes with the emphasis on practical applicability of the formulas obtained and on the mathematical rigor of the theory.

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For the scalar wave scattering by a single body, our main results can be described as follows: (1) Analytical formulas for the scattering amplitude for a small body of an arbitrary shape are obtained; dependence of the scattering amplitude on the boundary conditions is described. (2) An analytical formula for the scattering matrix for electromagnetic wave scattering by a small body of an arbitrary shape is given. Applications of these results are outlined (calculation of the properties of a rarefied medium; inverse radio measurement problem; formulas for the polarization tensors and capacitances). (3) The multi-particle scattering problem is analyzed and interaction of the scattered waves is taken into account. For the self-consistent field in a medium consisting of many particles integral-differential equations are found. The equations depend on the boundary conditions on the particle surfaces. These equations offer a possibility of solving the inverse problem of finding the medium properties from the scattering data. For about 5 to 10 bodies the fundamental integral equations of the theory can be solved numerically to study the interaction between the bodies.

In Section 10.1.2 the results concerning the scalar wave scattering are described. In Section 10.1.3 the electromagnetic scattering is studied and the solution of the inverse problem of radio-measurements is outlined. In Section 10.2 the many-body problem is examined. We use [R65] and the author’s papers, in particular [R215]. 10.1.2

Scalar wave scattering by a single body

In this Section we denote by the boundary of the body D, by S the surface area of by f the scattering amplitude, by n the unit vector in the direction of scattering, and by v the unit vector in the direction of the incident plane wave. Consider the problem

where D is a bounded domain with a smooth boundary N is the outer normal to is the initial field which is usually taken in the form We look for a solution of the problem (10.1.1)–(10.1.2) of the form

381

and for the scattering amplitude f we have the formula

where

Putting (10.1.3) in the boundary conditions (10.1.1) we get the integral equation for

where

Expanding A(k) and T(k) in the powers k and equating the corresponding terms in (10.1.7) we obtain, for h = 0, i.e., for the Neumann boundary condition, the following equations:

etc. where

Expanding f in formula (10.1.4) we obtain, up to the terms of the second order:

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From (10.1.9) it follows that and from (10.1.10) it follows that Some calculations lead to the following final result (see [R65]):

Usually

and in this case formula (10.1.14) can be written as:

where over the repeated indices the summation is understood, V is the volume of the body D and is the magnetic polarizability tensor of D. Note that the scattering is anisotropic and is defined by the tensor Formula (10.1.27) below allows one to calculate this tensor. For (the Dirichet boundary condition) integral equation (10.1.6) takes the form:

Hence

Since the field where the origin is assumed to be inside the body D. From the above equation it follows that and

where C is the electrical capacitance of a conductor with the shape D. Hence, for the Dirichet boundary condition, f ~ a, where a is a characteristic radius of D, and the scattering is isotropic. For using the same line of arguments, it is possible to obtain the following approximate formula for the scattering amplitude:

where S = meas i.e., the surface area of and C is the electrical capacitance of the conductor D. If h is very small then the formula for f should be changed and the terms analogous to (10.1.15) should be taken into account.

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10.1.3 Electromagnetic wave scattering by a single body

If a homogeneous body D with parameters is placed into a homogeneous medium with parameters then the following formula for the scattering matrix was established by the author (see [R65]):

where

is defined by the formula

is the angle of scattering, are the components of the initial field, are the components of the scattered field in the far-field region multiplied by the plane YOZ is the plane of scattering, is the polarizability tensor, and is the magnetic polarizability tensor. Analytic formulas for the polarizability tensors are given below. If one knows one can find all the values of interest to the physicists for electromagnetic wave propagation in a rarefied medium consisting of small bodies. The tensor of refraction coefficient can be calculated by the formula where N is the number of bodies per unit volume, and are the elements of the S-matrix corresponding to the forward scattering, that is for (see [N]). The tensor can be calculated analytically by the formula (see [R65]):

where A and q are some constants depending only on the geometry of the surface, and

In (10.1.22)

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In particular

For the particles with and not very large, so that the depth of the skin layer is considerably larger than a, one can neglect the magnetic dipole radiation and in formula (10.1.19) for the scattering matrix one can omit the terms with the multipliers The vectors of the electric P and magnetic M polarizations can be found by the following formulas, respectively,

where

is the initial field, over the repeated indices one sums up, and

where is the initial field, and the second term on the right hand side of equality (10.1.27) should be omitted if the skin-layer depth The scattering amplitudes can be found from the formulas:

where [A, B] stands for the vector product A × B, and P and M can be calculated by formulas (10.1.26), (10.1.27), (10.1.21), (10.1.22), (10.1.24). If one can neglect the second term on the right-hand side of (10.1.28). It is possible to give a simple solution to the following inverse problem which can be called the inverse problem of radiomeasurements. Suppose an initial electromagnetic field is scattered by a small probe. Assume that the scattered field can be measured in the far field region. The problem is: calculate the initial field at the point where the small probe detects This problem is of interest, for example, when one wants to determine the electromagnetic field distribution in an antenna’s aperture. Let us assume for simplicity that for the probe so that

385

From (10.1.30) one can find A measurement in the direction, where Hence Thus (cf (10.1.26)):

where results in But

Since V and are known and can be calculated by formulas (10.1.21), (10.1.22) and the matrix is positive definite (because is the energy) it follows that system (10.1.31) is uniquely solvable for Its solution is the desired vector E. More details are given in Section 10.4. Let us give a formula for the capacitance of a conductor D of an arbitrary shape, which proved to be very useful in practice ([R65]):

It can be proved ([R65]) that

where A and q are constants which depend only on the geometry of It is shown in [R65] that formula (10.1.32) with n = 0 gives a relative error not more than 0.03, and with n = 1 not more than 0.01, for a wide variety of shapes. Therefore this formula can be used for developing a universal computer code for calculation of electrical capacitances for conductors of arbitrary shapes. Such a program is started in [BoR]. Remark 10.1.1. The theory is also applicable for small layered bodies (see [R65]). Remark 10.1.2. Two sided variational estimates for

and C were given in [R65].

10.1.4 Many-body wave scattering

First we describe a method for solving the scattering problem for r bodies, r ~ 5 – 10, and then we derive an integral-differential equation for the self-consistent field in a medium consisting of many small bodies. We look for a solution of the

386


scalar wave scattering problem of the form

Applying the boundary condition,

we obtain the system of r integral equations for the r unknown functions In general this system can be solved numerically. When where and is the distance between i-th and j-th body, the system of the integral equations has dominant diagonal terms and it can be easily solved by an iterative process, the zero approximation being the initial field If but not necessarily the average (self-consistent) field in the medium consisting of small particles can be found from the integral equation ([R144]):

Here q (y) is the average value of over the volume d y in a neighborhood of y for bodies with impedance boundary conditions. For (the Dirichlet boundary condition) and identical bodies, one has q (y) = N(y)C, where N(y) is the number of the bodies per unit volume and C is the capacitance of a body. For the Neumann boundary condition the corresponding equation is the integraldifferential equation (cf (10.1.14)):

where

V is the volume of a body, and is its magnetic polarizability tensor (see formula (10.1.27)). The solution to equations (10.1.35) and (10.1.36) can be considered as the self-consistent (effective) field acting in the medium. Equations (10.1.35) and (10.1.36) allow one to solve the inverse problems of the determination of the medium properties from the scattering data. For example, from

387

(10.1.35) it follows that the scattering amplitude has the form

For a rarefied medium it is reasonable to replace u by and to obtain

If

formula (10.1.39) is valid for

(the Born approximation)

with the error

if

Hence if f is known for all and all where is the unit sphere in the Fourier transform of q (y) is known, and q can be uniquely determined. If q (y) is compactly supported, i.e., q (y) = 0 outside some bounded domain, then f is an entire function of k, and knowing f in any interval for all one can find f for all uniquely by analytic continuation, and thus one can determine q (y) uniquely. Let us consider the r-body problem for a few bodies, r ~ 10 (small r). Assume that the Dirichlet boundary condition holds. Let us look for a solution of the form

The scattering amplitude is equal to

where as:

where

is some point inside the j-th body. Since

this formula can be rewritten

388


This is the same line of arguments as in Section 10.1.2. Using the Dirichlet boundary condition one gets:

With the accuracy of O(ka), this can be written as:

where (10.1.48) as:

If

is the capacitance of the m–th body one can rewrite

This is a linear system for If this system can be easily solved by an iterative process. If are known, then the scattering amplitude can be found from (10.1.45). More about the described theory the reader can find in the monograph [R65]. 10.2

EQUATIONS FOR THE SELF-CONSISTENT FIELD IN MEDIA CONSISTING OF MANY SMALL PARTICLES

In this Section an integral-differential equation is derived for the self-consistent (effective) field in media consisting of many small bodies randomly distributed in some region. Acoustic and electromagnetic fields are considered in such a medium. Each body has a characteristic dimension where is the wavelength in the free space. The minimal distance d between any of the two bodies satisfies the condition but it may also satisfy the condition in acoustic scattering. In electromagnetic scattering our assumptions are and Let us derive an integral-differential equation for the self-consistent acoustic or electromagnetic fields in the above media. 10.2.1 Introduction

We propose a general method for derivation of equations for the self-consistent (effective) field in a medium consisting of many small particles and illustrate it by the derivation of such equations for acoustic and electromagnetic waves. Equation (10.2.17) (see below) is of the type obtained in [R65], and earlier in [MK] by a different argument. It is simpler than equation (10.2.25). This can be explained from the physical point of view: scattering of an acoustic wave by small, in comparison with the wavelength in the free space, acoustically soft body is isotropic, and the

389

scattered field in the far-field zone is described by one scalar, namely the electrical capacitance of the perfect conductor of the same shape as the small body, and the scattered field is of order O(a), where a is the characteristic size of the small body. If a small body is acoustically hard, that is, condition (10.2.21) holds on its boundary, then the scattering is anisotropic, the scattered field in the far-field zone is described by a tensor, and it is of order Therefore, this field is much smaller (by a factor of order than the one for an acoustically soft body of the same size a and of the same shape. Here k is the wavenumber in the free space. In [BoW] the Lorentz-Lorenz formula is derived. This formula relates the polarizability of a uniform dielectric to the density of the distribution of molecules and the polarizability of these molecules. In this theory one assumes that the polarizabilty of a unit volume of the medium is a constant vector, the molecules are modeled as identical spheres uniformly distributed in the space. In this case the polarizability tensor is proportional to the unit matrix, and the coefficient of proportionality is the cube of the radius of the small sphere times some constant. This, together with additional assumptions, yields a relation between the dielectric constant of the medium, the polarizability of the molecule, and the number of the molecules per unit volume. The derivation of this formula in [BoW] is based on the equation of electrostatics. Our basic physical assumptions, (10.2.1), allow for rarefied medium, when but also, in acoustic wave scattering theory, for medium which is dense, when Equations (10.2.25) and (10.2.33)–(10.2.36), that we will derive, have an unusual feature: the integrand depends on the direction from y to x. This happens because of the anisotropy of the scattered field in the case of non-spherical homogeneous small bodies. A possible application of equation (10.2.17) is a method for finding the density of the distribution of small bodies from the scattering data. Namely, the function C(y) in (10.2.17)–(10.2.18) determines this distribution. On the other hand, this function can be determined from the field scattered by the region The uniqueness results and computational methods for solving this inverse scattering problem are developed in [R139]. Below we study the dynamic fields, so that the wavenumber is positive. Consider a random medium consisting of many small bodies located in a region Let be the radius of the body defined as and We assume where is the wavelength of the field in the free space (or in a homogeneous space in which the small bodies are embedded). Let Assume

We do not assume that in the acoustic wave scattering, but assume this in electromagnetic wave scattering. The difference in the physical assumptions between acoustic and electromagnetic theory is caused by the necessity to apply twice the operation to the expression of the type in the electromagnetic theory, where

390


p is a vector independent of r and r is the distance from a small body to the observation point. We consider acoustic field in the above medium, and derive in Section 10.2.2 an integral-differential equation for the self-consistent field in this medium. The notion of the self–consistent field is defined in Section 10.2.2. Roughly speaking, it is the field, acting on one of the small bodies from all other bodies, plus the incident field. In Section 10.2.3 we derive a similar equation for the self-consistent electromagnetic field in the medium. Each small body may have an arbitrary shape. The key results from [R65], that we use, are the formulas for the S-matrix for acoustic and electromagnetic wave scattering by a single homogeneous small body of an arbitrary shape. These formulas are given in Section 10.2.2 and Section 10.2.3. Wave propagation in random media is studied in [Ish]. 10.2.2 Acoustic fields in random media

Assume first that the small bodies are acoustically soft, that is where boundary of The governing equation for the acoustic pressure u is

is the

where the direction of the incident field is given, wave number, is the unit sphere in Let us look for v of the form:

is the

If (10.2.1) holds, and

then (10.2.6) can be written as

391

Define the self-consistent field

at the point

by the formula

and at any point x, such that

by the formula:

If (10.2.9) holds, then

that is, removal of one small body does not change the self-consistent field in the region which contains no immediate neighborhood of this body. On the surface S of the body the total field u = 0, so the self-consistent field on differs from u, while at a point x such that (10.2.9) holds, if (10.2.1) holds. Let us derive a formula for . By (10.2.3) one gets:

where

so that

Thus, one may neglect

in (10.2.11) and consider the resulting equation

392


as an equation for the charge distribution on the surface charged to the Then, by (10.2.7),

where is the electrostatic capacitance of the conductor (10.2.15) one gets

of a perfect conductor

From (10.2.10) and

as

Let us emphasize the physical assumptions we have used in the derivation of (10.2.16). First, the assumption allows one to claim that, uniformly with respect to all small bodies, the term o(1) in (10.2.16) tends to zero as Secondly, the assumption allows one to claim that the m-th small body is in the far zone with respect to the j-th body for any The expression under the sum in (10.2.16) is the field, scattered by j-th body and calculated at the point x, such that that is, in the far zone from the j-th body. So, physically, the equations for the self-consistent field in the medium, derived here, are valid not only for the rarified medium (that is, when and but also for not too dense medium, that is, when and but, possibly, The limiting equation for is:

where

and the summation is taken over all small bodies located in the volume d y around point y. If one assumes that the capacitances are the same for all these bodies around point y, and are equal to c(y), then C(y) = c(y)N(y), where N(y) is the number of Small bodies in the volume d y. Equation (10.2.17) is the integral equation for the self-consistent field in the medium in the region This field satisfies the Schrödinger equation:

393

Since

in (10.2.18), and the number N of the terms in the sum (10.2.18) is provided that d y is a unit cube, one concludes that so

If one had i.e., small bodies have nonzero limit of volume density, then the assumption would be violated. Let us now assume that the small bodies are acoustically hard, i.e., the Neumann boundary condition

replaces (10.2.3), is the outer normal to S. In this case the derivation of the equation for is more complicated, because the formula for is less simple. It is proved in [R65] that, for the boundary condition (10.2.21), one gets

Here and below, one sums up over the repeated indices, is the volume of is the Laplacean, the small body is located around point y, the scattered field is calculated at point defined by the formula [R65]

and

is the magnetic polarizability tensor of

is the electric polarizability tensor, defined by the formula:

Here P is the dipole moment induced on the constant placed in the electrostatic field E dielectric constant Analytical formulas for calculation of the geometry of are derived in [R65] and From (10.2.10) and (10.2.22) one gets

where V(y) and

which is

dielectric body with the dielectric in the homogeneous medium with the with an arbitrary accuracy, in terms of given in Section 10.1.

are defined by the formulas

394


where is the magnetic polarizability tensor of the j-th small body, and the summation is over all small bodies in the volume d y around point y, so that V(y) is the density of the distribution of the volumes of small bodies at a point y. The novel feature of equation (10.2.25) is the dependence of the integrand in (10.2.25) on the direction x – y. This one can understand, if one knows that the acoustic wave scattering by a small soft body is isotropic and depends on one scalar, the electrostatic capacitance of the conductor while acoustic wave scattering by a small hard obstacle is unisotropic and depends on the tensor Finally, if the third boundary condition holds:

then (see [R65]), if h is not too small, one has:

where

is the area of

so that

where

10.2.3 Electromagnetic waves in random media

In this section our basic assumptions are:

The reason for the change in the assumption compared with (10.2.2), where d is not necessarily greater than is the following: formula (10.2.35) below is valid if and where d is the distance from a small body to the point of observation. This comes from applying twice the operation of to the vector potential. In acoustic scattering the formula for the scattered field (see e.g., (10.2.8)) is valid if while formula (10.2.35) (see below) is valid for and Let Denote by the 6×6 matrix which sends into where is the scattered field, is the distance from a small body located at a point y to the observation point If is known, then the equation for the self-consistent field in the random medium situated in a region and consisting of many small bodies, satisfying

395

conditions (10.2.32), is

where

as follows from the argument given for the derivation of (10.2.25). Let us give a formula for assuming without loss of generality, that the origin is situated inside a single body which has parameters and (dielectric permittivity, magnetic permeability, and conductivity, respectively), and drop index j in the formula for In [R65] one can find the formulas

where is the unit vector in the direction of the scattered wave. Formulas (10.2.34), (10.2.35) can be written as

where the matrix is expressed in terms of the tensors because P is calculated in (10.2.24),

and

is defined in (10.2.23), and 10.3

FINDING SMALL SUBSURFACE INHOMOGENEITIES FROM SCATTERING DATA

A new method for finding small inhomogeneities from surface scattering data is proposed and mathematically justified in this Section. The method allows one to find small holes and cracks in metallic and other objects from the observation of the acoustic field scattered by the objects.

396


10.3.1 Introduction

In many applications one is interested in finding small inhomogeneities in a medium from the observation of the scattered field, acoustic or electromagnetic, on the surface of the medium. We have two typical examples of such problems in mind. The first one is in the area of material science and technology. Suppose that a piece of metal or other material is given and one wants to examine if it has small cavities (holes or cracks) inside. One irradiates the metal by acoustic waves and observes on the surface of the metal the scattered field. From these data one wants to determine: 1) are there small cavities inside the metal? 2) if there are cavities, then where are they located and what are their sizes?

Similar questions can be posed concerning localization not only of the cavities, but any small in comparison with the wavelength inhomogeneities. Our methods allow one to answer such questions. As a second example, we mention the mammography problem. Currently x-ray mammography is widely used as a method of early diagnistics of breast cancer in women. However, it is believed that the probability for a woman to get a new cancer cell in her breast as a result of an x-ray mammography test is rather high. Therefore it is quite important to introduce ultrasound mammography tests. This is being done currently. A new cancer cells can be considered as small inhomogeneities in the healthy breast tissue. The problem is to localize them from the observation on the surface of the breast of the scattered acoustic field. The purpose of this short Section is to describe a new idea of solving the problem of finding inhomogeneities, small in comparison with the wavelength, from the observation of the scattered acoustic or electromagnetic waves on the surface of the medium. For simplicity we present the basic ideas in the case of acoustic wave scattering. These ideas are based on the earlier results on wave scattering theory by small bodies, presented in [R65] (see also [R50] and references therein, and [KR1]). Our objective in solving the inverse scattering problem of finding small inhomogeneities from surface scattering data are: 1) to develop a computationally simple and stable method for a partial solution of the above inverse scattering problem. The exact inversion procedures (see Chapter 5 and [R139] and references therein) are computationally difficult and unstable. In practice it is often quite important, and sometimes sufficient for practical purposes, to get a “partial inversion”, that is, to answer questions of the type we asked above: given the scattering data, can one determine if these data correspond to some small inhomogeneities inside the body? If yes, where are these inhomogeneities located? What are their intensities? We define the notion of intensity of an inhomogeneity below formula (10.3.1). In Section 10.3.2 the basic idea of our approach is described. In Section 10.3.3 its short justification is presented. Some theoretical and numerical results based on a version of the proposed approach one can find in [R193].

397

10.3.2

Basic equations

Let the governing equation be

where u satisfies the radiation condition, k = const > 0, and v(x) is the inhomogeneity in the velocity profile. Assume that where denotes the third component of vector x in Cartesian coordinates, is a ball, centered at with radius Denote

Problem 4 (Inverse Problem (IP):). Given u(x, y, k) for all x, and a fixed k > 0, find In this Section we propose a numerical method for solving the (IP). To describe this method let us introduce the following notations:

are the points at which the data

are collected

The proposed method for solving the (IP) consists in finding the global minimizer of function (10.3.8). This minimizer gives the estimates of the positions of the small inhomogeneities and their intensities This is explained in more detail below formula (10.3.14). Numerical realization of the proposed

398


method, including a numerical procedure for estimating the number M of small inhomogeneities from the surface scattering data is described in [GR1]. Our approach with a suitable modification is valid in the situation when the Born approximation fails, for example, in the case of scattering by delta-type inhomogeneities [AlbS]. In this case the basic condition which guarantees the applicability of the Born approximation, is violated. Here and was defined below formula (10.3.1). We assume throughout that M is not very large, between 1 and 15. In the scattering by a delta–type inhomogeneity the assumption is V as so that for any fixed k > 0 one has as and clearly condition (*) is violated. In our notations this delta-type inhomogeneity is of the form The scattering theory by the delta-type potentials (see [AlbS]) requires some facts from the theory of selfadjoint extensions of symmetric operators in Hilbert spaces and in this Section we will not go into detail (see [GRa]). 10.3.3 Justification of the proposed method

We start with an exact integral equation equivalent to equation (10.3.1) with the radiation condition:

For small inhomogeneities the integral on the right-hand side of (10.3.9) can be approximately written as

where is defined by the first equation in formula (10.3.10), it is the error due to replacing u under the sign of integral in (10.3.9) by g, and is a point close to One has if and if Thus, the error term in (10.3.10) equals to if Therefore the function u(z, y, k) under the sign of the integral in (10.3.9) can be replaced by g(z, y, k) with a small relative error where and provided that:

399

where M is the number of inhomogeneities, d is the minimal distance from m = 1, 2, …, M to the surface P. A sufficient condition for the validity of the Born approximation, that is, the approximation u(x, y, k) ~ g(x, y, k) for x, is the smallness of the relative error for which holds if:

One has:

if and if is not small, so that the Born approximation may be not applicable. Note that u in (10.3.9) has dimension where L is the length, v (z) is dimensionless, and has dimension In many applications it is natural to assume If the Born approximation is not valid, for example, if as which is the case of scattering by delta-type inhomogeneities, then the error term in formula (10.3.10) can still be negligible: so if If one understands a sufficient condition for the validity of the Born approximation as the condition which guarantees the smallness of for all x, then condition (10.3.12) is such a condition. However, if one understands a sufficient condition for the validity of the Born approximation as the condition which guarantees the smallness of for x, y running only through the region where the scattered field is measured, in our case when x, then a much weaker condition (10.3.11) will suffice. In the limit and formula (10.3.10) takes the form (10.3.13), (see [GRa]). It is shown in [GRa] (see also [A1bS]) that the resolvent kernel of the Schrödinger operator with the delta-type potential supported on a finite set of points (in our case on the set of points has the form

where are some constants. These constants are determined by a selfadjoint realization of the corresponding Schrödinger operator with delta-type potential. There is an family of such realizations (see [GRa] more details). Although in general the matrix is not diagonal, under a practically reasonable assumption (10.3.11) one can neglect the off-diagonal terms of the matrix and then formula (10.3.13) reduces practically to the form (10.3.10) with the term neglected.

400 10. Wave scattering by small bodies of arbitrary shapes

We have assumed in (10.3.10) that the point

exists such that

This is an equation of the type of mean-value theorem. However, such a theorem does not hold, in general, for complex-valued functions. Therefore, if one wishes to have a rigorous derivation, one has to add to the error term in (10.3.10) the error which comes from replacing of the integral in (10.3.10) by the term The error of such an approximation can be easily estimated. We do not give such an estimate, because the basic conclusion that the error term is negligible compared with the main term remains valid under our basic assumption From (10.3.10) and (10.3.7) it follows that

Therefore, parameters and can be estimated by the least-squares method if one finds the global minimum of the function (10.3.8):

Indeed, if one neglects the error of the approximation (10.3.10), then the function (10.3.8) is a smooth function of several variables, namely, of and the global minimum of this function is zero and is attained at the actual intensities and at the values i = 1, 2, . . . , M. This follows from the simple argument: if the error of approximation is neglected, then the approximate equality in (10.3.14) becomes an exact one. Therefore so that function (10.3.8) equals to zero. Since this function is non-negative by definition, it follows that the values and are global minimizers of the function (10.3.8). Therefore we take the global minimizers of function (10.3.8) as approximate values of the positions and intensities of the small inhomogeneities. In general we do not know that the global minimizer is unique, and in practice it is often not unique. For the case of one small inhomogeneity (m = 1) uniqueness of the global minimizer is proved in [KR1] for all sufficiently small for a problem with a different functional. The problem considered in [KR1] is the (IP) with M = 1, and the functional minimized in [KR1] is specific for one inhomogeneity. In [R65] analytical formulas for the scattering matrix are derived for acoustic and electromagnetic scattering problems. An important ingredient of our approach from the numerical point of view is the solution of the global minimization problem (10.3.14). The theory of global minimization is developed extensively and the literature of this subject is quite large (see, e.g., [BarJ] and [BarJ1] and references therein).

401

10.4

INVERSE PROBLEM OF RADIOMEASUREMENTS

1. Suppose that we are interested in measuring the electromagnetic field in the aperture of a mirror antenna. A possible method for making such measurements is as follows. Let us assume that the wavelength range is and let us place at some point in the aperture of the antenna a small probe of dimension a, Let denote the electromagnetic field at the point and denote the field scattered by the probe in the far–field zone. Note that for a small probe the far-field zone, which is defined by the known condition is in fact close to the probe. For example, if a = 0.3 cm then Therefore if r = 0.2 cm then Let us assume for simplicity that the probe material is such that the magnetic dipole radiation from the probe is negligible. In this case the electric field scattered by the probe in the direction n can be calculated from formula (10.2.34) and we assume M = 0, so and

where

Here V is the volume of the probe, is its dielectric constant, is the electric polarizability tensor, is the wave number of the field in the aperture, is the dielectric field at the point where the probe was placed, and is the unit vector. Let and be two noncollinear unit vectors, and j = 1, 2, be the scattered fields corresponding to We will solve the following Problem 5. Find

from the measured

We assume that the tensor approximate formulas for follows that

Therefore

j = 1, 2

is known. In Section 10.1.3 explicit analytical are given, see (10.1.21)–(10.1.25). From (10.4.1) it

402


Let us choose for simplicity

perpendicular to

Then it follows from (10.4.4) that

Therefore

Thus one can find vector P from the knowledge of can be found from the linear system

and

If P is known then

The matrix of this system is positive definite because the tensor has this property. This follows also from the fact that is the energy of the dipole P in the field Therefore the system (10.4.8) can be uniquely solved for We have proved that the above problem has a unique solution and gave a simple algorithm for the solution of this problem. The key point in the above argument is the fact that the matrix is known explicitly. 2. In applications the problem of finding the distribution of particles according to their sizes is often of interest. Suppose that there is a medium consisting of many particles and condition is satisfied. We assume that the medium is rarefied, i.e., where a is the characteristic dimension of the particles. The scattering amplitude for a single particle can be calculated from formulas (10.4.1) and (10.4.2). The scattering amplitude is a function f(n,k,r) of the radius r of the particle. Suppose that is density of the distribution of the particles according to their sizes, so that is the number of the particles per unit volume with the radius in the interval (r, r + dr). Then the total scattered field in the direction n can be calculated by the formula

Let us assume that we can measure F(n, k) for a fixed k and all directions n. Then (10.4.9) can be considered as an integral equation of the first kind for the unknown function 3. Suppose that we can measure the electric field scattered by a small particle of an unknown shape. The initial field we denote by the scattered field by Let us assume that the magnetic dipole radiation is negligible. The problem is to find the shape of the small particle. Every small particle scatters electromagnetic wave like some ellipsoid. Indeed, the main term in the scattered field is the dipole scattering. We have seen in Section 10.4.1

403

that the knowledge of the scattered field allows one to find the dipole moment P and equation (10.4.2) holds. This equation allows one to find the corresponding to the particle. This tensor is determined if one knows its diagonal form. Let be the eigenvalues of tensor Then an ellipsoid with the semiaxes proportional to scatters as the above body. Therefore one can identify the shape of the small scatterer by giving the three numbers These numbers are eigenvalues of the tensor They can be calculated from the knowledge of the initial field and the measured scattered field For example, one can take Then We assume that the particle is homogeneous and its dielectric constant is known, so that in (10.4.2) is known. For an ellipoid the polarizability tensor in the diagonal form is where is the dielectric constant of the ellipsoid and are the depolarization coefficients. These coefficients are calculated explicitly with the help of the elliptic integrals and they are tabulated ([LaL]).


11. THE POMPEIU PROBLEM

11.1

THE POMPEIU PROBLEM

In this Section a short and self-contained presentation of some of the results known about the Pompeiu problem is given. In particular, an author’s result is proved. This result says: if has Pompeiu’s property (P-property) then has it, provided is sufficiently close to in the following sense: is sufficiently small. Here 11.1.1 Introduction

The Pompeiu problem (P-problem) can be stated as follows: Let where is the Schwartz class of distributions, and

where is a bounded domain and G is the group of all rigid motion of (G consists of all translations and rotations). Does (11.1.1) imply that f = 0? If yes, one says that D has Pompeiu’s property, P-property. Equation (11.1.1) can be written as

406

11. The Pompeiu problem

where SO(n) is the group of rotations, so g is an arbitrary orthogonal matrix with det g = 1. Since 1929 the P-problem has been open. Large literature exists on this problem. It was studied in non-euclidean spaces, its relations to harmonic analysis and inverse problems for differential equations were understood. Examples of domains D having P-property, published up to now, include: convex domains with at least one corner, or with non-analytic boundary, and ellipsoids which are not balls. Balls do not have P-property: any non-trivial function with support on the compact subset of zeros of the Fourier transform of the characteristic function of a ball B will satisfy (11.1.1) (see formula (11.1.6) below). This Fourier transform is Here is the Bessel function of the argument where a is the radius of the ball centered at the origin, and is the Fourier transform variable. The set of zeros of is a discrete set j = 1, 2, 3 , … , where the dependence on n is suppressed, and as a compact subset of zeros of one can take a sphere for some fixed positive integer j. Thus, where is the delta function supported on the sphere of radius and is a certain function. If one takes then one gets which is a solution to (11.1.1). Here is a positive constant which can be written down explicitly. If one takes where then is a solution to (11.1.1). Here is another constant which could be written explicitly, and A group of smaller than G was considered (see [Z]). In this paper we assume D to be strictly convex, homeomorphic to a ball, and its boundary S to be piecewise-smooth. We consider P-problem in The following results are proved in section 11.1.2 in a self-contained way: 1) a bounded domain and some k > 0; here is its Fourier transform, 2) a connected bounded domain

does not have P-property iff is the characteristic function of D, does not have P-property iff the problem

where N is the unit exterior normal to S, has a solution for some k > 0, or, equivalently, the problem

has a solution; here Note that the two boundary conditions (11.1.4) imply on S. (or Lipschitz), then is necessarily 2a) If (11.1.3) (or (11.1.4)) holds and simultaneously a Neumann and a Dirichlet eigenfunction of the Laplacian, and 2b) S is a real analytic hypersurface.

407

3) if a bounded, strictly convex domain D with smooth boundary is not a ball, then all the surfaces of zeros of the function in for sufficiently large are not spheres, or, equivalently, 3a) if is a bounded strictly convex domain and for all and for a sequence then D is a ball. j = 1, 2, where is a class of smooth strictly convex do4) assume that mains with uniformly bounded of the functions representing locally the boundaries of and Gaussian curvatures, uniformly bounded from below by a positive constant.

The author’s main result, Theorem 11.1.6, is: if a bounded domain has P-property and is sufficiently small, then has P-property. Here A relationship of the P-problem with an inverse problem for metaharmonic potentials is established. The following remark seems new. Consider some functional space X and let be the space of the Fourier transformed elements of X. Let us assume that the only element of supported on the set is the zero element. Then any bounded domain homeomorphic to a ball, has P-property. Therefore we assume that Examples of functional spaces X with the above property are spaces with Results 1) and 2) can be found in [BST], [W1], [W2], result 3) is from [Ber1], and result 4) is from [R177]. Bibliography on the Pompeiu problem the reader finds in [Z], [R186], and papers [Ag], [Av], et al., deal with Pompeiu’s problem. The new ideas and techniques in this Section include: a) the usage of the set defined below formula (11.1.6), and its rotational invariance, and b) the usage of formula (11.1.10) below, which is formula (4.7.1) from [RKa], and of the orthogonality condition (11.1.7).

Our proofs are often shorter and simpler than the published ones. 11.1.2 Proofs

1. In this Section we prove:

Theorem 11.1.1. Equation (11.1.1) holds for some Moreover: i) (11.1.3) is solvable iff there exists a k > 0 such that ii) if (11.1.3) is solvable then S is an analytic hypersurface; and

iff (11.1.3) is solvable.

408


iii) If (11.1.3) is solvable then is necessarily simultaneously a Neumann and a Dirichlet eigenvalue of the Laplacian in D. Proof. Let us prove claim i). Assume (11.1.1) holds. Let


In fact (11.1.2) and (11.1.5) are equivalent, and they are equivalent to

where the bar stands for complex conjugate and

is the indicator of D :

Let where Thus, is rotation invariant. Since D is compact, is an entire function of exponential type, so is an analytic set. If a point belongs to then all the points of the sphere belong to due to the rotation invariance of It follows from (11.1.6) that iff is not empty there exists an supp which satisfies (11.1.1). It follows that any bounded domain D has P -property if one restricts the class of admissible f(x) in formula (11.1.1) to the set of functions f(x) with the following property: (P): if vanishes on the complement to then The set is rotationally invariant and can be identified with the discrete set consisting of those k > 0 for which contains the spheres This set is discrete since the function is an entire function of For example, if we restrict the set of f(x) in (11.1.1) to be p = 1, 2, then the above property (P) holds, and any bounded domain D, including balls, has Pompeiu property. If then and, since is nonempty and rotation invariant, it either contains a sphere a > 0, or The last case cannot occur since Consider the first case. If contains a > 0, then for This implies where is an entire function. Taking the inverse Fourier transform, one gets equation (11.1.3) for This equation is satisfied in suppu = D, so u = 0 in Since by elliptic

409

regularity and u = 0 in one gets boundary conditions (11.1.3) on S. The necessity of (11.1.3) is proved. To prove sufficiency, assume that (11.1.3) holds. Extend u(x) by zero to and let

Then, because of the boundary conditions (11.1.3), the function U(x) solves the equation

and its Fourier transform yields where is an entire function of exponential type. Thus, vanishes on the sphere contains and there exists an for which (11.1.1) holds. We have proved claim i) of Theorem 11.1.1. Note that if (11.1.3) holds, then –1 has to be orthogonal to any solution of the homogeneous equation (11.1.3). Since are such solutions, it follows that

This yields an independent proof of the implication: existence of a solution to (3) for all with Claim ii) follows from the results on regularity of free boundary [KN]. Namely, it is proved in [KN] that if S is Im u = 0, and (11.1.3) holds, then S is a real–analytic hypersurface. In [W2] this is proved for Lipschitz S. Let us prove claim iii). If equation (11.1.3) is solvable, then so is (11.1.4). Thus is a Neumann eigenvalue of the Laplacian in D. Moreover, the solution of (11.1.4) satisfies the condition on S. Therefore, for each j = 1, 2, 3, the function is a solution to the equation

which satisfies the Dirichlet boundary condition

Claim iii) is proved. Thus, Theorem 11.1.1 is proved.

410


Let orthogonal to by

and denote the point on S at which the tangent to S plane is and is directed along inner (outer) normal to S at Denote

the width of D in the direction points

and by

the Gaussian curvature at the

Theorem 11.1.2. If there is an such that j = 1, 2, such that then the set is compact.

or there are

Proof. By formula (4.7.1) in [RKa] we have

where

Thus, the zeros of equation

in

when

can be found asymptotically from the

or

where is an integer. One sees that t > 0 cannot satisfy (11.1.11) unless Since the set is rotation invariant, it is empty for sufficiently large provided that there is an such that For equation (11.1.11) to be satisfied by t independent of it is necessary and sufficient that and These two equations imply (see Corollary 11.1.5 below) that D is a ball, in contradiction to the assumptions of Theorem 11.1.2. Therefore, the assumptions of Theorem 11.1.2 imply that, for sufficiently large t, the surface of real zeros of is not a sphere. Thus, the set is compact. It is well-known that there are convex smooth (and not smooth) bodies of constant width which are not balls. However, the following lemma holds:

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Lemma 11.1.3. If S, whose width is constant: D is a disk.

is a strictly convex connected domain with a smooth boundary and then

Proof. Let s be the length of S considered as a natural parameter on S: each point on S is uniquely defined by the value of this parameter. Since D is convex, each point of S is also uniquely defined by a unit vector Namely, given one defines as the (unique) point of S at which is the interior unit normal to S. Thus, can be considered as a function of s, and, by Frenet’s formulas,

where is the unit vector tangent to S at the point of S at the point and is the radius of curvature, equation with respect to s yields

Since

and

is the curvature Differentiating the

one gets

In this calculation we have used the assumption which allowed us to conclude that where d s is the same as in the formula Differentiating (11.1.13) and using (11.1.12), one gets

or

or

Equation (11.1.14) implies that S is a circle of radius Lemma 11.1.4 ([A1, p. 304]). If n > 2, is a strictly convex connected domain with a smooth boundary S, whose width is constant and then D is a ball. In fact, it is proved in [Al] that any S in homeomorphic to a sphere and such that there exists a with the properties j = 1, 2, ..., n, must be a sphere. Here j = 1, 2, ..., n, are the principal curvatures of S.

412


Corollary 11.1.5. Assume that boundary S. If for all

is a bounded strictly convex domain with a smooth and for a sequence then D is a ball.

Proof. It follows from the assumption and formula (11.1.10) that This implies that D is a ball. For n = 2 this is proved in Lemma 11.1.3. For n > 2 this is a result from [Al] as stated in Lemma 11.1.4. Let and assume for simplicity that and belong to a class of strictly convex smooth domains with Gaussian curvature, uniformly bounded from below by a positive constant and the of the functions representing locally the boundary of is uniformly bounded by an arbitrary large positive constant C, which characterizes the class together with the lower bound on the Gaussian curvature. Theorem 11.1.6. If has P-property, small, then has P-property.

and

Proof. If has P-property then there is no k > 0 such that Let be a connected component of the set of real zeros of Therefore if

is sufficiently

Note that

then

Here is an arbitrary large fixed number. The above argument shows that there are no spherical surfaces of zeros of in the ball of radius if there are no such surfaces for and if differs from sufficiently little (precisely, if Outside this ball, for sufficiently large the set of zeros of is given asymptotically by the equation

For t sufficiently large this equation yields equation (11.1.11) as the asymptotic equation for It is clear from (11.1.11) and from Lemma 11.1.4 that various components are, as different from spheres if is not a ball. Since the sets of zeros of is in a locally of the set of zeros of for all and the sets of zeros of in this region are not in a of any sphere centered at the origin, if is sufficiently small, it follows that does not have spherical

413

surfaces of zeros in the region if is sufficiently small, and, as we proved above, does not have spherical surfaces of zeros in the region Clearly, Thus, does not have spherical surfaces of zeros, and therefore has P-property. Thus, we have proved that the set of the domains having P-property is open if the distance between and is defined to be Let us now discuss in more detail the choice of the number above. If has Pproperty, then it is not a ball. If it is not a ball, then any domain sufficiently close to in the sense of Theorem 11.1.6, is not a ball, that is, either it is not of constant width, or there are directions such that is different from This implies that does not have spherical surfaces of zeros on any a priori fixed compact K. If this domain is smooth and strictly convex, and if one assumes a uniform positive bound on the Gaussian curvature of all these domains from below and a uniform bound in of the functions representing locally the boundaries of the domains, then outside K there are no spherical surfaces of zeros of either, and the existence of the which can be chosen simultaneously for all such domains in the proof of Theorem 11.1.6 is clear. Namely, choose such that O(1/t) in (11.1.10) is less than c, where c > 0 is a constant depending on the lower bound on Gaussian curvatures and on the of the functions representing the boundaries of the domains. Note that c can be chosen uniformly for all domains of the above class. For c sufficiently small, the expression in the brackets in (11.1.10) is not vanishing for some (depending possibly on the choice of the domain). For this find as where max is taken over and min is taken over t in the interval Now choose and let meas Then for any domain in this set does not have spherical surface of zeros. Thus has P-property. This argument does not require that the boundaries of the domains should be close in some Sobolev norm. It does require an a priori uniform positive lower bound on Gaussian curvatures and a uniform upper bound on the of the functions representing locally the boundaries of the domains in and the convexity of the domains. The last requirement is not crucial (see [RKa], Ch. 4), however a detailed discussion in the absence of convexity is longer, and one has to exclude various pathologies, e.g., existence of countably many critical points of the functions representing the surfaces locally. Theorem 11.1.6 is proved. Let us now establish a relation of the P-problem with an inverse problem for metaharmonic potentials. Define metaharmonic potential of constant unit density by the formula

Suppose (11.1.3) has a solution with compact support. Then this solution can be written as (11.1.16) and u(x) = 0 in This follows from (11.1.3), Green’s formula, and the assumption that the solution to (11.1.3) has compact support. Thus, existence of the solution to (11.1.3), which has compact support, implies that the

414


inverse problem for metaharmonic potentials, which consists of finding D, given the values of the potential near infinity, does not have a unique solution. 11.2

NECESSARY AND SUFFICIENT CONDITION FOR A DOMAIN, WHICH FAILS TO HAVE POMPEIU PROPERTY, TO BE A BALL

In this Section a necessary and sufficient condition is given for a domain, homeomorphic to a ball, which fails to have Pompeiu property, to be a ball. (cf [R186]) 11.2.1 Introduction

Let be a bounded domain with Lipschitz boundary S, SO(n) be the rotational group. Suppose that where is the Schwartz class of distributions, condition (11.1.2) holds, and so D has If f ( x ) belongs to some functional space the elements of which decay at infinity sufficiently fast, for example p = 1, 2, then any bounded domain D, including balls, has P-property. Indeed, arguing as in Section 11.1, let

Let denote the characteristic function of D. One can write (11.1.2) as a convolution of a distribution f , which is a locally integrable function, and a compactly supported function Taking the Fourier transform, one gets:

where the bar stands for complex conjugate and

Since g in (11.2.1) is arbitrary, one concludes that the support of is the discrete set of spheres of radii such that for This set is discrete since is an entire function of if the domain D is bounded. If X is any functional space such that the only element with supported on a discrete set of spheres is f ( x ) = 0, then any bounded domain D has P-property. As such X one can take spaces of functions decaying at infinity sufficiently fast, and we have mentioned two examples of such spaces above. For this reason we assumed that f belongs to a space of functions which do not decay at infinity. The long-standing conjecture, called now the Pompeiu problem, is: (C) : A ball B is the only domain, homeomorphic to a ball, which fails to have P-property.

415

In this Section we give a result related to this conjecture. We write if D has P-property, and if it fails to have P-property. We need the following result proved in Section 1.1: iff the following problem is solvable for a positive number

Here,

is the normal derivative of u,

is the exterior unit normal to S at the point

It follows immediately from (11.2.2) that is both a Dirichlet and Neumann eigenvalue of the Laplacian in D . Indeed, define Then in D, and on S, on S. Thus is a Neumann eigenvalue of the Laplacian. Moreover, since w = const on S, and on S, it follows that grad w = 0 on S, so (for any j = 1, 2, 3, ...) solves the problem in D, v = 0 on S. Thus is a Neumann and a Dirichlet eigenvalue simultaneously. Denote by the eigenspace of the Dirichlet Laplacian corresponding to the eigenvalue by an orthonormal basis of by the linear span of the functions and by the orthogonal complement of in Let be an arbitrary unit vector, is the unit sphere in Let v be the velocity corresponding to the rotation of about the axis directed along and passing through the gravity center of D. If n = 3 then where is the vector product. For simplicity let us take n = 3 in what follows, but the argument and the result hold for if one writes Gx in place of By (a, b) the inner product in is denoted. The result of this Section can now be stated. Theorem 11.2.1. Assume that a domain D, homeomorphic to a ball, fails to have P-property, that is, Then D is a ball if and only if for any one has:

The above result says that the conjecture (C) is true provided that (11.2.3) holds. In other words, equation (11.2.3) means that the function for any does not have a non-zero projection onto the finite-dimensional subspace in

If D is a ball, then

so that equation (11.2.3) is satisfied trivially. We note the following geometrically obvious lemma, an easy proof of which is left to the reader.

416


Lemma 11.2.2. Iff S, then S is a sphere.

on S for all

then S is a sphere. Iff

on

In Section 11.2.2 a proof of Theorem 11.2.1 is given. 11.2.2

Proof

Proof. 1) Necessity: Suppose only if D = B. Then, if one concludes that so (11.2.3) is trivially satisfied. 2) Sufficiency: Suppose (11.2.3) holds and We want to prove that D = B. Let be the Sobolev space, and be a ball containing D and centered at the gravity center of D. If then (11.2.2) holds. Therefore, for any

one has

This is verified by multiplying (11.2.2) by h, integrating by parts and taking into account the zero Cauchy data for u in (11.2.2). If then for any one has Fix an arbitrary and let the of the coordinate system be directed along Choose as g a rotation about which is given by the matrix

Then

Therefore the velocity field in

corresponding to this rotation, is

417

Taking h(gx) in place of h(x) in (11.2.5), differentiating with respect to afterwards, one gets:

and taking

Using Gauss formula, one obtains from (11.2.9) the following equation:

Denote


If, for any

one could choose

such that

where is arbitrarily small and the norm in (11.2.13) is (11.2.12) would imply

or (use (11.2.11) and take into account that

then

is arbitrary) the following equation:

By Lemma 11.2.2, equation (11.2.12) implies that S is a sphere, so D is a ball, D = B. To conclude the proof, we now show that (11.2.13) is possible iff We drop in what follows the for brevity. Let us first prove that the boundary-value problem (11.2.21) (see below) has a solution iff To prove this, pick an arbitrary such that

and define w by the formula:

418


Then

It is known that (11.2.18) is solvable iff the following orthogonality conditions hold:

where is a basis of the eigenspace of the Dirichlet Laplacian corresponding to the eigenvalue Integrating by parts in (11.2.19) yields

This means that Conditions (11.2.20) are necessary and sufficient for the solvability of the problem

To complete the proof of Theorem 11.2.1, it is sufficient to verify the following Claim 11.2.3, which implies inequality (11.2.13) for Claim 11.2.3. The set of restrictions of the elements of

to S is complete in

Let us prove Claim 11.2.3. Proof. We start the proof by noting that the set

is complete in

then

in

Indeed, if

where

and

419

Multiply the last equation by h, integrate over and then by parts, use equation (11.2.21) for h and the zero Cauchy data for w and get

so h = 0 as we wanted to show. Let us now finish the proof of Claim 11.2.3. Assume that

We want to prove that (11.2.19) holds iff Equation (11.2.22) implies v (x) = 0 in and v solves the homogeneous problem (11.2.21) in D. By the jump formula for the normal derivative of the single layer potential v ( x ) , one has Thus, (11.2.22) implies that The converse is easy to prove also. Therefore Claim 11.2.3 is proved. The proof of Theorem 11.2.1 is complete.


BIBLIOGRAPHICAL NOTES

In Chapter 1 statements of many inverse problems are given. In Chapter 2, Sections 2.1–2.3 contain material that can be found in [IVT], [Mor], [MG], [Gro], [VV], [TLY], [R121], [EHN]. There are some new results regarding regularization of unbounded nonlinear ill-posed operator equations, and a new version of the discrepancy principle. In the presentation of the Backus-Gilbert method the rate of convergence of the proposed version of this method is estimated. We did not discuss the optimality of the methods for solving ill-posed problems (see, e.g., [IVT], [EHN]), but argue that a stable solution of an ill-posed problem is not possible without a priori knowledge of the noise level in the data. Section 2.4 presents the DSM (Dynamical systems method) and is based on papers [R216], [R217], [R218], [R220], [R212], [AR1], [ARS3]. This method was not discussed in the books on ill-posed problems earlier. The author thinks that this method is promising from the computational point of view. Subsection 2.5.1 is based on [RSm5], where the reader can find many references regarding numerical differentiation of noisy data, see also [RSm3]. Subsection 2.5.2 is based on [R58], [R39]. Sections 3.1–3.11 are based on [R221], [R192], and other author’s papers. The author thanks Cubo Mathematical Journal for permission to use paper [R221]. Chapter 3 contains many new results, and the presentation of the classical results contains many novel points, especially in the presentation of Gel’fand-Levitan’s (GL) theory and Marchenko’s theory. Krein’s inversion theory is presented with complete proofs for the first time. Its presentation in Section 3.9, is based on [R197]. The analysis of the

422

Bibliographical notes

invertibility of the inversion steps in the GL’s theory and in Marchenko’s theory is discussed in detail, which was not done earlier. A detailed analysis of the Newton-Sabatier (NS) theory for inverting fixed-energy phase shifts is given in Section 3.6. This theory was presented in [N], [CS]. Our analysis concludes that the NS inversion theory is fundamentally wrong in the sense that its foundations are wrong. In [Sab] P. Sabatier made an attempt to disagree with the above conclusion but, in fact, failed to address the main point from [R206], namely, that there is no proof that the basic equation in the NS theory (equation (3.6.65) in Section 3.6.4) has a solution for all r > 0, and that this solution is unique. It is shown in [R206] (and in Section 3.6) that equation (3.6.65) generically does not have a unique solution for some r > 0 and in this case the NS inversion procedure breaks down. In [R207] a counterexample is given to a uniqueness theorem for a modified equation (3.6.65) proposed in [CT]. A reply to [Sab] is given in [R150]. In Section 3.6.5 a result from [RAI] is presented. In Section 3.7 some results from [R196] are presented. In [GS] a different approach to these results is given. In Section 3.8 the problem of finding a confining potential (a quarkonium system) from a few measurements is given. This problem was considered in [TQR], but the approach in [TQR] was not correct by the reasons explained in Section 3.8 (and in [R183]). Section 3.10 contains a solution of some new inverse problems for the heat and wave equations. In Section 3.11 the result from [R191] is presented. In Section 3.12 an inverse problem of ocean acoustics is solved ([R199]), and it is shown that the method used for the study of a related problem in [GX] is invalid. In Section 3.13 a theory of ground-penetrating radars is developed ([R185], [R195], [RSh]). The central role in Chapter 3 plays Property C for ordinary differential equations, a uniqueness theorem for I-function and its numerous applications. The I-function for the class of potentials, used in scattering theory, is equal to the Weyl’s m -function, which is in one-to-one correspondence with the spectral function. In [GS] and [GS1] some properties of m-functions are given and their applications to inverse problems are discussed. In [R181], [R184] a uniqueness theorem and a convergent iterative method are given for the recovery of a compactly supported (or decaying faster than any exponential) potential from S-matrix alone, without the knowledge of bound states and norming constants. Chapter 4 is based on the material in [R83], [R162], [R164], [R171], [R31], [R208], [R75], [R139], [R154], [R155], [R164], [RSa], [RPY], [GR5], [RGu2]. In [BLW], [CK], [CKi], [CCM] various numerical methods for solving inverse obstacle scattering problems are discussed. These methods and other numerical methods are analyzed in Section 4.5–4.6. The novel points in this Chapter include (but not limited to) the consideration of the scattering and inverse scattering problems for very rough boundaries, stability estimates for the solution of the obstacle inverse scattering problems with fixed-frequency data and with high-frequency data, and the analysis of the numerical methods for solving obstacle inverse scattering problems. In [Fed], [VH], [Maz], various classes of rough domains are studied, in particular, the class of domains with finite perimeter, which contains the class of Lipschitz domains as a proper subclass. The results of Chapter 5 are obtained from a series of the author’s papers, and summarized in [R203]. Chapter 5 is based on this paper. The author thanks Birkhäuser

423

for permission to use this paper. The notion of Property C was introduced in [R87] and applied to many inverse problems, for example, [R100], [R102], [R103], [R105], [R109], [R112], [R114], [R115], [R120], [R125], [R126], [RX], [R127], [RSj], [R129], [R130], [R132], [R133], [R143], [R145], [R146], [R149]. Sections 5.7, and 5.8 are based on [R128] and [R124] respectively. Section 5.5 is based on [R172]. The inverse potential scattering problem has been studied in [Fad], [N4], [No], [NK], [No1], [RW1], [RW2], and in other works. Chapter 6 is based on the papers [R165], [RRa], and uses some material from [R139]. Section 6.4 is based on [Ro1]. The example of non-uniqueness in Section 6.1 shows that the usage of numerical parameter-fitting as a method for solving inverse problems may be meaningless. This is important to have in mind, because many published papers on inverse problems are based on just such parameter-fitting procedures and neglect the analysis of the inverse problem, in particular, the analysis of the uniqueness of the solution to an inverse problem. This analysis is crucial. Chapter 7 is based on the material from [R139], [R21], [R73]. There is a large body of literature on antenna synthesis, see [MJ], [ZK], [R23], [R26], [R28], [R26], which we had not discussed. Chapter 8 is based on [R198]. Most of the uniqueness results for multidimensional inverse problems are obtained for overdetermined problems. We formulate several basic non-overdetermined inverse problems which are still open: even uniqueness theorems are not obtained for these problems. Chapter 9 is based on a series of author’s papers, starting with [R68], [R77], and summarized in the monographs [R83]. Our presentation is based on these papers and on the book [R139]. Almost all of the results in this Chapter are from these sources. However, equation (9.1.7) was derived earlier in [LRS], but not studied mathematically there. Chapter 10 is based on a series of author’s papers on wave scattering by small bodies of arbitrary shapes. This is a classical area of research, originated by Lord Rayleigh in 1871, who understood that the main term in the acoustic field, scattered by a small in comparison with the wavelength body is given by a dipole radiation. Lord Rayleigh had published many papers on this subject before his death in 1919. There are hundreds of papers, mostly dealing with applied sciences, which use wave scattering theory by small bodies. However, there were no analytic formulas for calculation of this dipole radiation for bodies of arbitrary shapes. Such formulas were found in [R22], [R25], [R32], and the theory was presented in [R50], [R51] and summarized in [R65]. Our basic results include analytic formulas, which allow one to calculate the capacitances of conductors of arbitrary shapes and the polarizability tensors of homogeneous dielectric bodies of arbitrary shapes with arbitrary accuracy in terms of the geometry of these bodies and their dielectric constant. These formulas allow one to derive an analytic formula for the scattering matrix for electromagnetic wave scattering by small bodies of arbitrary shapes. This formula allows one to solve an inverse problem of radiomeasurements ([R65], [R33]). Section 10.2 deals with waves in a media consisting of many small bodies and is based on [R215]. An interesting book [MK], deals with the wave scattering in a medium consisting of many small particles. There is

424

Bibliographical notes

a large body of literature on wave scattering in random media [Ish]. Section 10.3 is based on [R193], and numerical results obtained by the proposed method are given in [GR1]. Chapter 11 deals with the Pompeiu problem. It is based on papers [R177], [R186]. There are many papers on this problem, see [Z], [BST], [Av], [Ber1], [GaS], [Kob], and references therein. In the book of relatively small size, like this book, many problems of the theory of inverse problems have not been discussed. A short and incomplete list of these include Carleman estimates and their applications, inverse problems for systems, in particular, for elasticity and Maxwell’s equations, control theory methods, integral geometry problems, to say nothing about many concrete inverse problems.

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[RSm3] [RSm4]

[RSm5] [RSm6] [RSF] [RSo] [RSt1] [RSt2] [RSZ] [RUG] [RW1] [RW2] [RWWW]

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439

[RWeg]

[RX] [RZ] [RK] [Ro] [Ro1] [RuS] [Sab]

[Sai]

[St] [St1] [StU]

[SU] [TQR] [TLY] [Tit] [VV] [Vas]

[VA] [Vog] [VH] [W1] [W2] [W]

[Ya] [Y] [Z]

[ZK]

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INDEX

Antenna synthesis, xix, 11, 336, 337, 423 Backus-Gilbert, 30, 32 Backwards heat equation, 17 Biharmonic equation, 373 Born inversion, xviii, 8, 307–311, 369 Deconvolution, 13, 16, 87 Dynamical systems method, xvii, 421 Dynamical systems method (DSM), 41

Inverse problems for the heat and wave equations, 3, 91, 202, 422 Inverse problems of potential theory, xix, 2, 333, 360 Inverse radiomeasurements problem, xix Inverse scattering problems, 2, 12, 191, 255, 353, 422 Inverse source problem, 10, 11, 337, 338 Inverse source problems, 10, 333 Krein’s method, 186, 191

Finding ODE from a trajectory, 13 Low-frequency inversion, 349 Gel’fand-Levitan (GL) method, 184 Geometrical inverse problem, 9, 371 Ground-penetrating radars, 92, 216 I-function, 104, 108, 110, 111 Ill-posed problems, xvii, 14, 15, 19, 23, 26, 29, 45, 49, 57, 59, 62, 71, 74, 80, 89, 421 Image processing, 13 Induction logging, 366 Integral geometry, xix, 5, 6, 363, 364, 368 Inverse, 2, 97, 312 Inverse geophysical problems, 339 Inverse gravimetry problem, 12 Inverse obstacle scattering, xviii, 4, 255, 339, 422 Inverse problem of ocean acoustics, 92, 208

Marchenko inversion procedure, 91 Monotone operators, 21, 26, 49, 57, 58, 64 Non-overdetermined problems, 13 Non-uniqueness, xix, 317, 319, 333, 369 Pompeiu problem, xix, 7, 405, 407, 414, 424 Projection method, 89 Property C, xviii, 91, 92, 98, 165, 298, 353, 423 Pseudoinverse, 19, 40 Quasiinversion, 32 Quasisolution, 30, 31, 89

442

Index

Random media, 389, 390, 394 Regularization, 19, 20, 24, 25, 74, 75, 88, 89, 220, 310, 367 S-matrix, 383 Singular value decomposition, 20 Small bodies, xix, 5, 379, 383, 385, 388–390, 392–394, 396, 423 Spectral assumption, 64 Spectral problems, 2, 91, 97, 312 Stability estimates, xviii, 8, 255, 276, 422

Stable differentiation, 75, 77, 84 Stable summation, 15, 85 Uniqueness theorems, xix, 101, 102, 200, 306, 312, 321, 339, 340, 353 Variable background, 376 Variational, 19, 74, 89 Volterra equations, 87, 93 Wave scattering, xix, 181, 379, 380, 383, 385, 386, 389, 390, 394, 396, 423, 424 Well-to-well data, 364

Inverse Problems: Mathematical and Analytical Techniques with Applications to Engineering

Inverse Problems: Mathematical and Analytical Techniques with Applications to Engineering

Inverse Problems: Mathematical and Analytical Techniques with Applications to Engineering