Inverse Problems - Kansas State Mathematics Departmentramm/papers/470.pdf · · 2008-01-23Inverse...

Inverse Problems

Alexander G. RAMM

To Luba and Olga

Professor Alexander G.Ramm

Mathematics Department,Kansas State University,Manhattan, KS 66506-2602,USA

email: [email protected]: http://www.math.ksu.edu/˜ramm

Springer, New York, 2004isbn 0-387-23195-1

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Preface

This book can be used for courses at various levels in ill-posed problems and inverse problems. Thebibliography of the subject is enormous. It is not possible to compile a complete bibliography and noattempt was made to do this. The bibliography contains some books where the reader will find additionalreferences. The author has used extensively his earlier published papers, and referenced these, as wellas 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 simplerthan that published earlier, and quite a few new results are included. Regularization for ill-posed operatorequations with unbounded nonlinear operators is studied. A novel version of the discrepancy principle isformulated 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. Thepresentation is based on the author’s papers and the joint papers of the author and his students. Theseresults appear for the first time in book form. Papers [R216], [R217], [R218], [R220], [ARS3], [AR1] havebeen used in this Chapter.

In Chapter 3 the presentation of one-dimensional inverse problems is based mostly on the author’spapers, especially on [R221]. It contains many novel results, which are described at the beginningof the Chapter. The presentation of the classical results, for example, Gel’fand-Levitan’s theory, andMarchenko’s theory, contains many novel points. The presentation of M. G. Krein’s inversion theorywith complete proofs is given for the first time. The Newton-Sabatier inversion theory, which has beenin the literature for more than 40 years, and was presented in two monographs [CS],[N], is analyzed andshown to be fundamentally wrong in the sense that its foundations are wrong (cf [R206]). This Chapteris based on the papers [R221], [R199],[R197], [R196], [R195],[R192], [R185]. One of the first paperson inverse spectral problems was Ambartsumian’s paper (1929) [Am], where it was proved that onespectrum determines the one-dimensional Neumann Schrodinger’s operator uniquely. This result is anexceptional 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 setof products of solutions to homogeneous differential equations. For partial differential equations this toolhas been introduced in [R87] and developed in many papers and in the monograph [R139]. For ordinarydifferential equations completeness of the products of solutions to homegeneous ordinary equations hasbeen used in different forms in [L1], [Hr]. In our book Property C for ODE is presented in the formintroduced 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 inversionprocedure corresponding to fixed-frequency data are given, the high-frequency inversion formulas arediscussed and the error of the inversion from noisy data is estimated. Analysis of the currently usednumerical 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 thepaper [R203]. The basic concept used in the analysis of the inverse scattering problem in Chapter 5 isthe concept of Property C, i.e., completeness of the set of products of solutions to homogeneous partialdifferential equations. This concept was introduced by the author ([R87]) and applied to many inverseproblems (see [R139] and references therein). An important part of the theory consists of obtainingstability estimates for the potential, reconstructed from fixed-energy noisy data (and from exact data).Error estimates for the Born inversion are given under suitable assumptions. It is shown that the Borninversion may fail while the Born approximation works well. In other words, the Born approximationmay be applicable for solving the direct scattering problem, while the Born inversion, that is, inversion

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based on the Born approximation, may fail. The Born inversion is still popular in applications, thereforethese error estimates will hopefully be useful for practitioners.

The author’s inversion method for fixed-energy scattering data, is compared with that based on theusage of the Dirichlet-to-Neumann map. The author shows why the difficulties in numerical implemen-tation of his method are less formidable than the difficulties in implementing the inversion method basedon the Dirichlet-to-Neumann map.

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 tosolve numerically such a problem, by a parameter-fitting, which is very popular among practitioners. Butif the uniqueness result is not established, the numerical results may be meaningless. Some uniquenesstheorems for inverse boundary value problem and for an inverse problem for hyperbolic equations areestablished in this Chapter.

In Chapter 7, inverse problems of potential theory and antenna synthesis are briefly discussed. Thepresentation of the theory on this topic is not complete: there are books and many papers on antennasynthesis (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, theinverse problems in which the unknown function depends on the same number of variables as the datafunction. Examples of such problems are given. Most of these problems are open: even uniquenesstheorems are not available. Such a problem, namely, recovery of an unknown coefficient in a Schrodingerequation in a bounded domain from the knowledge of the values of the spectral function ρ(s, s, λ) on theboundary is discussed under the assumption that all the eigenvalues are simple, that is, the correspondingeigenspaces 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 onthe 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 the summary of the author’s results regarding the theory of wave scattering by smallbodies of arbitrary shapes. These results have been obtained in a series of the author’s papers and aresummarized in [R65], [R50]. The solution of inverse radiomeasurements problem ([R33], [R65]) is basedon these results. Also, these results are used in the solution of the problem of finding small subsurfaceinhomogeneities from the scattering data, measured on the surface. The solution to this problem canbe used in modeling ultrasound mammography, in finding small holes in metallic objects, and in manyother applied problems.

In Chapter 11 the classical Pompeiu problem is presented following the papers [R177], [R186].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-dimensionalscattering problems for periodic potentials), the Carleman estimates and their applications to inverseproblems ([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 engineering problemsare not discussed. There are many papers published, which use parameter-fitting for solving inverseproblems. However, in most cases there are no error estimates for parameter-fitting schemes for solvinginverse problems, and one cannot guarantee any accuracy of the inversion result. In [GRS] the conceptof stability index is introduced and applied to a parameter-fitting scheme for solving a one-dimensionalinverse scattering problem in quantum physics. This concept allows one to get some idea about the errorestimate in a parameter-fitting scheme.

The applications of inverse scattering to integration of nonlinear evolution equations are not discussedas there are many books on this topic (see e.g., [M], [FT] and references therein).

Contents

1 Introduction 101.1 Why are inverse problems interesting

and practically important? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2 Examples of inverse problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2.1 Inverse problems of potential theory . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.2 Inverse spectral problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.3 Inverse scattering problems in quantum physics; finding the potential from the

impedance function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.4 Inverse problems of interest in geophysics . . . . . . . . . . . . . . . . . . . . . . . 111.2.5 Inverse problems for the heat and wave equations . . . . . . . . . . . . . . . . . . . 121.2.6 Inverse obstacle scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.7 Finding small subsurface inhomogeneities from the measurements

of the scattered field on the surface . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.2.8 Inverse problem of radiomeasurements . . . . . . . . . . . . . . . . . . . . . . . . . 131.2.9 Impedance tomography (inverse conductivity) problem . . . . . . . . . . . . . . . . 131.2.10 Tomography and other integral geometry problems . . . . . . . . . . . . . . . . . . 131.2.11 Inverse problems with “incomplete data” . . . . . . . . . . . . . . . . . . . . . . . 141.2.12 The Pompeiu problem, Schiffer’s conjecture, and inverse problem

of plasma theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2.13 Multidimensional inverse potential scattering . . . . . . . . . . . . . . . . . . . . . 151.2.14 Ground-penetrating radar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.15 A geometrical inverse problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.16 Inverse source problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.2.17 Identification problems for integral-differential equations . . . . . . . . . . . . . . . 181.2.18 Inverse problem for an abstract evolution equation . . . . . . . . . . . . . . . . . . 181.2.19 Inverse gravimetry problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.20 Phase retrieval problem (PRP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.2.21 Non-overdetermined inverse problems . . . . . . . . . . . . . . . . . . . . . . . . . 191.2.22 Image processing, deconvolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.2.23 Inverse problem of electrodynamics, recovery of layered medium

from the surface scattering data . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.2.24 Finding ODE from a trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.3 Ill-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.4 Examples of Ill-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.4.1 Stable numerical differentiation of noisy data . . . . . . . . . . . . . . . . . . . . . 201.4.2 Stable summation of the Fourier series and integrals with

randomly perturbed coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.4.3 Solving ill-conditioned linear algebraic systems . . . . . . . . . . . . . . . . . . . . 211.4.4 Fredholm and Volterra integral equations of the first kind . . . . . . . . . . . . . . 21

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1.4.5 Deconvolution problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.4.6 Minimization problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.4.7 The Cauchy problem for Laplace’s equation . . . . . . . . . . . . . . . . . . . . . . 221.4.8 The backwards heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Methods of solving ill-posed problems 232.1 Variational regularization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.1.1 Pseudoinverse. Singular values decomposition . . . . . . . . . . . . . . . . . . . . . 232.1.2 Variational (Phillips-Tikhonov) regularization . . . . . . . . . . . . . . . . . . . . . 242.1.3 Discrepancy principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.1.4 Nonlinear ill-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.1.5 Regularization of nonlinear, possibly unbounded, operator . . . . . . . . . . . . . . 262.1.6 Regularization based on spectral theory . . . . . . . . . . . . . . . . . . . . . . . . 272.1.7 On the notion of ill-posedness for nonlinear equations . . . . . . . . . . . . . . . . 282.1.8 Discrepancy principle for nonlinear ill-posed problems with

monotone operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.1.9 Regularizers for ill-posed problems must depend on the noise level . . . . . . . . . 30

2.2 Quasisolutions, quasiinversion, and Backus-Gilbert method . . . . . . . . . . . . . . . . . 312.2.1 Quasisolutions for continuous operator . . . . . . . . . . . . . . . . . . . . . . . . . 312.2.2 Quasisolution for unbounded operators . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.3 Quasiinversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.4 Backus-Gilbert method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.3 Iterative methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.4 Dynamical system method (DSM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.4.1 The idea of the DSM. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.4.2 DSM for well-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.4.3 Linear ill-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.4.4 Nonlinear ill-posed problems with monotone operators . . . . . . . . . . . . . . . . 452.4.5 Nonlinear ill-posed problems with non-monotone operators . . . . . . . . . . . . . 502.4.6 Nonlinear ill-posed problems: avoiding inverting of operators in

the Newton-type continuous schemes . . . . . . . . . . . . . . . . . . . . . . . . . . 522.4.7 Iterative schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542.4.8 A spectral assumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.4.9 Nonlinear integral inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.4.10 Riccati equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2.5 Examples of solutions of ill-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . 602.5.1 Stable numerical differentiation: when is it possible? . . . . . . . . . . . . . . . . . 602.5.2 Stable summation of the Fourier series and integrals with perturbed coefficients . . 712.5.3 Stable solution of some Volterra equations of the first kind. . . . . . . . . . . . . . 722.5.4 Deconvolution problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 732.5.5 Ill-conditioned linear algebraic systems . . . . . . . . . . . . . . . . . . . . . . . . . 73

2.6 Projection methods for ill-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3 One-dimensional inverse scattering and spectral problems 753.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3.1.1 What is new in this chapter? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 763.1.2 Auxiliary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 763.1.3 Statement of the inverse scattering and inverse spectral problems . . . . . . . . . . 803.1.4 Property C for ODE. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.1.5 A brief description of the basic results. . . . . . . . . . . . . . . . . . . . . . . . . . 81

3.2 Property C for ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

6 CONTENTS

3.2.1 Property C+ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 843.2.2 Properties Cϕ and Cθ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

3.3 Inverse problem with I-function as the data . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3.3.1 Uniqueness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873.3.2 Characterization of the I-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 893.3.3 Inversion procedures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3.3.4 Properties of I(k) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 913.4 Inverse spectral problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

3.4.1 Auxiliary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 973.4.2 Uniqueness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

3.4.3 Reconstruction procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1013.4.4 Invertibility of the reconstruction steps . . . . . . . . . . . . . . . . . . . . . . . . . 1023.4.5 Characterization of the class of spectral functions of

the Sturm-Liouville operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

3.4.6 Relation to the inverse scattering problem . . . . . . . . . . . . . . . . . . . . . . . 1043.5 Inverse scattering on half-line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

3.5.1 Auxiliary material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

3.5.2 Statement of the inverse scattering problem onthe half-line. Uniqueness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

3.5.3 Reconstruction procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.5.4 Invertibility of the steps of the reconstruction procedure . . . . . . . . . . . . . . . 1133.5.5 Characterization of the scattering data . . . . . . . . . . . . . . . . . . . . . . . . . 115

3.5.6 A new Marchenko-type equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1163.5.7 Inequalities for the transformation operators and applications . . . . . . . . . . . 116

3.6 Inverse scattering problem with fixed-energy . . . . . . . . . . . . . . . . . . . . . . . . . . 122

3.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1223.6.2 Existence and uniqueness of the transformation operators

independent of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 1233.6.3 Uniqueness theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1283.6.4 Why is the Newton-Sabatier (NS) procedure fundamentally wrong? . . . . . . . . 129

3.6.5 Formula for the radius of the support of the potentialin terms of scattering data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

3.7 Inverse scattering with “incomplete data” . . . . . . . . . . . . . . . . . . . . . . . . . . . 1373.7.1 Uniqueness results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1373.7.2 Uniqueness results: compactly supported potentials . . . . . . . . . . . . . . . . . 140

3.7.3 Inverse scattering on the full line by a potential vanishing on a half-line . . . . . . 1403.8 Recovery of quarkonium systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

3.8.1 Statement of the inverse problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

3.8.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1423.8.3 Reconstruction method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

3.9 Krein’s method in inverse scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

3.9.1 Introduction and description of the method . . . . . . . . . . . . . . . . . . . . . . 1443.9.2 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1483.9.3 Numerical aspects of the Krein inversion procedure. . . . . . . . . . . . . . . . . . 154

3.9.4 Discussion of the ISP when the bound states are present. . . . . . . . . . . . . . . 1553.9.5 Relation between Krein’s and GL’s methods. . . . . . . . . . . . . . . . . . . . . . 155

3.10 Inverse problems for the heat and wave equations . . . . . . . . . . . . . . . . . . . . . . . 155

3.10.1 Inverse problem for the heat equation . . . . . . . . . . . . . . . . . . . . . . . . . 1553.10.2 What are the “correct” measurements? . . . . . . . . . . . . . . . . . . . . . . . . 1563.10.3 Inverse problem for the wave equation . . . . . . . . . . . . . . . . . . . . . . . . . 157

CONTENTS 7

3.11 Inverse problem for an inhomogeneous Schrodingerequation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

3.12 An inverse problem of ocean acoustics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

3.12.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

3.12.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

3.12.3 Proofs: uniqueness theorem and inversion algorithm . . . . . . . . . . . . . . . . . 163

3.13 Theory of ground-penetrating radars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

3.13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

3.13.2 Derivation of the basic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

3.13.3 Basic analytical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

3.13.4 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

3.13.5 The case of a source which is a loop of current . . . . . . . . . . . . . . . . . . . . 170

3.13.6 Basic analytical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

4 Inverse obstacle scattering 174

4.1 Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

4.2 Inverse obstacle scattering problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

4.3 Stability estimates for the solution to IOSP . . . . . . . . . . . . . . . . . . . . . . . . . . 183

4.4 High-frequency asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

4.5 Remarks about numerical methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

4.6 Analysis of a method for identification of obstacles . . . . . . . . . . . . . . . . . . . . . . 188

5 Inverse scattering problem 195

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

5.1.1 The direct potential scattering problem . . . . . . . . . . . . . . . . . . . . . . . . 196

5.1.2 Review of the known results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

5.2 Inverse potential scattering problem with fixed-energy data . . . . . . . . . . . . . . . . . 202

5.2.1 Uniqueness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

5.2.2 Reconstruction formula for exact data . . . . . . . . . . . . . . . . . . . . . . . . . 202

5.2.3 Stability estimate for inversion of the exact data . . . . . . . . . . . . . . . . . . . 204

5.2.4 Stability estimate for inversion of noisy data . . . . . . . . . . . . . . . . . . . . . 206

5.2.5 Stability estimate for the scattering solutions . . . . . . . . . . . . . . . . . . . . . 208

5.2.6 Spherically symmetric potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

5.3 Inverse geophysical scattering with fixed-frequency data . . . . . . . . . . . . . . . . . . . 210

5.4 Proofs of some estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

5.4.1 Proof of (5.1.18) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

5.4.2 Proof of (5.1.20) and (5.1.21) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

5.4.3 Proof of (5.2.17) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

5.4.4 Proof of (5.4.49) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

5.4.5 Proof of (5.4.51) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

5.4.6 Proof of (5.2.13) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

5.4.7 Proof of (5.2.23) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

5.4.8 Proof of (5.1.30) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

5.5 Construction of the Dirichlet-to-Neumann map . . . . . . . . . . . . . . . . . . . . . . . . 223

5.6 Property C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

5.7 Necessary and sufficient condition for scatterers . . . . . . . . . . . . . . . . . . . . . . . . 228

5.8 The Born inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

5.9 Uniqueness theorems for inverse spectral problems . . . . . . . . . . . . . . . . . . . . . . 236

8 CONTENTS

6 Non-uniqueness and uniqueness results 2406.1 Examples of nonuniqueness for an inverse problem of geophysics . . . . . . . . . . . . . . 240

6.1.1 Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2406.1.2 Example of nonuniqueness of the solution to IP . . . . . . . . . . . . . . . . . . . . 240

6.2 A uniqueness theorem for an inverse problem . . . . . . . . . . . . . . . . . . . . . . . . . 2426.3 Property C and an inverse problem for a hyperbolic equation . . . . . . . . . . . . . . . . 243

6.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2436.3.2 Statement of the result. Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

6.4 Continuation of the data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

7 Inverse problems of potential theory 2527.1 Inverse problem of potential theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2527.2 Antenna synthesis problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2547.3 Inverse source problem for hyperbolic equations . . . . . . . . . . . . . . . . . . . . . . . . 255

8 Non-overdetermined inverse problems 2568.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2568.2 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2568.3 The problem and the result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2578.4 Finding ϕj(s) from ϕ2

j (s) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2598.5 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

9 Low-frequency inversion 2639.1 Derivation of the basic equation. Uniqueness results . . . . . . . . . . . . . . . . . . . . . 2639.2 Analytical solution of the basic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2669.3 Characterization of the low-frequency data . . . . . . . . . . . . . . . . . . . . . . . . . . 2679.4 Problems of numerical implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2679.5 Half-spaces with different properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2689.6 Inversion of the data given on a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2689.7 Inversion of the data given on a cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2709.8 Two-dimensional inverse problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2719.9 One-dimensional inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2729.10 Inversion of the backscattering data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2739.11 Inversion of the well-to-well data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2749.12 Induction logging problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2769.13 Examples of non-uniqueness of the solution . . . . . . . . . . . . . . . . . . . . . . . . . . 2789.14 Scattering in absorptive medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2799.15 A geometrical inverse problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2809.16 An inverse problem for a biharmonic equation . . . . . . . . . . . . . . . . . . . . . . . . . 2819.17 Inverse scattering when the background is variable . . . . . . . . . . . . . . . . . . . . . . 2829.18 Remarks concerning the basic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

10 Wave scattering by small bodies of arbitrary shapes 28510.1 Wave scattering by small bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285

10.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28510.1.2 Scalar wave scattering by a single body . . . . . . . . . . . . . . . . . . . . . . . . 28610.1.3 Electromagnetic wave scattering by a single body . . . . . . . . . . . . . . . . . . . 28710.1.4 Many-body wave scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

10.2 Equations for the self-consistent field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29110.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29210.2.2 Acoustic fields in random media . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29310.2.3 Electromagnetic waves in random media . . . . . . . . . . . . . . . . . . . . . . . . 296

CONTENTS 9

10.3 Finding small subsurface inhomogeneities from scattering data . . . . . . . . . . . . . . . 29710.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29710.3.2 Basic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29810.3.3 Justification of the proposed method . . . . . . . . . . . . . . . . . . . . . . . . . . 299

10.4 Inverse problem of radiomeasurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

11 The Pompeiu problem 30311.1 The Pompeiu problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

11.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30311.1.2 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304

11.2 Necessary and sufficient condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30911.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30911.2.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

Bibliographical notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316

Chapter 1

Introduction

1.1 Why are inverse problems interestingand practically important?

Inverse problems are the problems that consist of finding an unknown property of an object, or amedium, from the observation of a response of this object, or medium, to a probing signal. Thus, thetheory 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 farfrom the obstacle, or in some exterior region, then the inverse problem is to find the shape and materialproperties of the obstacle. Such problems are important in identification of flying objects (airplanesmissiles, 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 scatteredfield on the surface for various positions of the source of the field for a fixed frequency, or for severalfrequencies. The inverse problem is to find the subsurface inhomogeneities. In technology one measuresthe 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 destructiveevaluation. The practical advantages of remote sensing are what makes the inverse problems important.

1.2 Examples of inverse problems

1.2.1 Inverse problems of potential theory

Suppose a body D ⊂ R3 with a density ρ(x), x ∈ R3, generates gravitational potential

u(x) =

∫

D

ρ(y)

4π|x− y|dy.

Is it possible to find ρ, given the potential u(x) for x ∈ B′R := x : |x| ≥ R, far away from D? A point

mass m and a uniformly distributed mass m in a ball of radius a produce the same potential u(x) = m|x|

for |x| ≥ R > a. Thus, it is not possible to find ρ(y) uniquely from the knowledge of u in B ′R. However,

if one knows a priori that ρ(x) = 1 in D, then it is possible to find D from the knowledge of u(x) in B ′R,

provided that D is, for example, star-shaped, that is, every ray issued from some interior point O ∈ D,intersects the boundary S = ∂D of D at only one point.

10

1.2. EXAMPLES OF INVERSE PROBLEMS 11

1.2.2 Inverse spectral problems

Let ` = − d2

dx2 + q(x) be the Sturm-Liouville operator defined by the Dirichlet boundary conditions asself-adjoint operator in L2[0, 1], and 0 < λ1 < λ2 ≤ · · · be its eigenvalues. To what extent does the set ofthese eigenvalues determine q(x)? Roughly speaking, one spectrum, that is, the set λj ∀j, j = 1, 2, . . .,determines “half of q(x),” in the sense that if q(x) is known on [ 1

2 , 1], then one spectrum determinesuniquely q(x) on [0, 1

2 ]. A classical result due to Borg [B]and Marchenko [M] says that two spectrauniquely determine the operator `, i.e., the potential q and the boundary conditions at x = 0 and x = 1of the type u′(1) + h1u(1) = 0 and u′(0) = h0u(0), where h0 and h1 are constants, and one assumesthat the two spectra correspond to the same h0 and two distinct h1. The author (see [R196]) asked thefollowing question: if q(x) is known on the segment [b, 1], 0 < b < 1, then what part of the spectrumone needs to know in order to uniquely recover q(x) on [0, b]? It is assumed that q is real valued: q = q,and q ∈ L(0, 1).

Let ρ(λ) be the spectral measure of the self-adjoint operator l. This notion is defined in Chapter 3.The inverse spectral problem is: given ρ(λ), find q(x), and the boundary conditions, characterizing `.

Similar problems can be formulated in the multidimensional cases, when a bounded domain D playsthe 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,∂φj

∂N |S , ∀j. One maychoose other spectral data.

1.2.3 Inverse scattering problems in quantum physics; finding the potentialfrom the impedance function

Consider the Dirichlet operator ` = − d2

dx2 + q(x), q ∈ L1,1 := q : q = q,∫∞0 x|q(x)|dx < ∞ in

L2(R+),R+ := [0,∞). Denote by f(x, k) the Jost solution, by f(k) := f(0, k) the Jost function, by

I(k) := f ′(0,k)f(k)

the I-function (impedance function), and by S := S(k), kj, sj , 1 ≤ j ≤ J the

scattering data (see Chapter 3). The inverse problem of quantum scattering on the half-axis consists offinding q(x), given S . It was studied in [M].

The inverse problem of finding q(x), given I(k) ∀k > 0, is of interest in many applications. The

I-function has the physical meaning of the impedance function, it is the ratioHy

Exin the problem of

electromagnetic wave falling perpendicularly onto the earth, when the dielectric permittivity and con-ductivity of the earth depend on the vertical coordinate only. One can prove that I(k) coincides withthe Weyl function m(k) (Chapter 3). It turns out that I(k) known ∀k > 0 determines uniquely q(x),and one can explicitly calculate S and ρ(λ), given I(k) ([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 unknowninhomogeneity in the velocity profile (refraction coefficient) from the scattered acoustic field measured onthe surface of the earth and generated by a point source, situated on the surface of the earth at varyingpositions.

Its mathematical formulation (in a simplified form) is:

[∆ + k2n0(x) + k2v(x)]u(x, k) = −δ(x− y) in R3, x ∈ R3,

where k = const > 0, n0(x) is the known background refraction coefficient, D := supp v(x) ⊂ R3− :=

x : x3 < 0, where supp v(x) is the support of v, v ∈ L2(D), and v is an inhomogeneity in the refractioncoefficient (or in the velocity profile), u is the acoustic pressure, u satisfies the radiation condition atinfinity (or the limiting absorption principle). An inverse problem of geophysics consists of finding thefunction v from the knowledge of the scattered field on the surface of the Earth, that is from u(x, y, k)known for all x, y ∈ P := x : x3 = 0 at a fixed k > 0, or for all k ∈ (0, k0), where k0 > 0 is a smallnumber (the case of low-frequency surface data) (cf [LRS], [Ro], [R83], [R139]).

12 CHAPTER 1. INTRODUCTION

Another problem is to find the conductivity of the medium from the measurements of the electro-magnetic 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

ut = uxx − q(x)u, 0 ≤ x ≤ 1, t > 0, u(x, 0) = u0(x), u(0, t) = 0, u(1, t) = a(t), (1.2.1)

is to find q(x) from the flux measurements: ux(1, t) = b(t). The extra data (measured data), b(t) ∀t > 0,allow one to find q(x). Another inverse problem is to find the unknown conductivity σ(x) from boundarymeasurements. For example, let ut = (σ(x)ux)x, u |t=0= u0(x), u(0, t) = 0, u(1, t) = a(t), ux(1, t) = b(t).Can one find σ(x), given u0(x), a(t) and b(t) ∀t > 0?

Consider the inverse conductivity problem: let ∇(σ(x)∇u) = 0 in D ⊂ R3, u = f , σ uN = g onS := ∂D, where N is the outer unit normal to S, the extra data is the flux g at the boundary. Supposethat the set f, g

∀f∈H32 (S)

is known. Can one determine σ(x) uniquely? Here D is a bounded domain

with a sufficiently smooth boundary S, and H l(S) is the Sobolev space. In applications in medicine, f isthe electrostatic potential, which can be applied to a human chest, and g is the flux of the electrostaticfield, which can be measured. If one can determine σ(x) from these measurements then some diagnosticinformation 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) inthe equation utt

c2(x)= ∆u − δ(x − y)δ(t), u = ut = 0 at t = 0, u = 0 on S, given the extra data uN

on S for a fixed y and all t > 0, or for y, varying on S, and t ∈ [0, T ], where T > 0 is some number.([LRS],[Ro], [RRa], [RSj]).

1.2.6 Inverse obstacle scattering

Let D ⊂ R3 be bounded domain with a Lipschitz boundary S,D′ := R3 \ D be the exterior domain,S2 be the unit sphere in R3. The scattering problem consists of finding the scattering solution, i.e., thesolution to the problem

(∇2 + k2

)u = 0 in D′,

u = u0 + v, u0 := eikα·x, α ∈ S2,(1.2.2)

u|S = 0, v = A(α′, α, k)eikr

r+ o

(1

r

), r := |x| → ∞, α′ :=

x

r. (1.2.3)

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

uN |S = 0 (1.2.3N)

is used in place of (1.2.3), then the existence and uniqueness of the solution to (1.2.2)- (1.2.3N) areproved in [RSa] under the assumption of compactness of the embedding H1(DR) → L2(DR), whereDR := BR\D, BR := x : |x| ≤ R, R > 0 is such that BR ⊃ D, and H1(DR) is the Sobolev space. Seealso [GoR]. The inverse obstacle scattering problem consists of finding S and the boundary conditionon S, given A(α′, α, k) in the following cases:

(1) either at a fixed α = α0 for all α′ ∈ S2 and all k > 0, or,

(2) at a fixed k = k0 > 0 for all α′ and α running through open subsets of S2, or,


(3) for fixed α = α0 and k = k0 and all α′ ∈ S2.

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 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 fixedk = k0 > 0 allows one to determine the boundaries of all the layers uniquely, and the constant velocityprofiles in each of the layers. See [RPY] for an answer to this question.

1.2.7 Finding small subsurface inhomogeneities from the measurementsof the scattered field on the surface

Suppose there are few small, in comparison with the wave-length, holes in the metallic body. A sourceof acoustic waves is on the surface of the body, and the scattered field is measured on the surface of thebody 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. Asimilar problem is important in medicine, where the small bodies are the cancer cells to be found in thehealthy tissue of a human’s body. In the ultrasound mammography modeling, one deals with the tissueof 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 an-tenna. For many practical reasons one wants to know this distribution. Let (E ′,H ′) be the field scatteredby a small probe placed at a point x in the aperture of the antenna. Given the shape and electromag-netic constants ε, µ, and σ, of the probe, the inverse problem of radiomeasurements consists of finding(E(x),H(x)) from the knowledge of E′H ′. (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 f (α, p) =∫lαp

f(x)ds, where lαp := x : α · x = p, x ∈ Rn, n ≥ 2, α ∈ Sn−1, Sn−1 is the

unit sphere in Rn, p ∈ R. The function f(α, p) is called the Radon transform of f . The function fcan be assumed piecewise-continuous and absolutely integrable over every plane lαp, so that the Radontransform would be well defined in the classical sense. But in fact, one can define the Radon transformfor much larger sets of functions and on distributions [R170], [RKa], [Hel].

Given f(α, p), one can uniquely recover f(x) provided, for example, that f(x) ∈ L1(Rn), or f(x) ∈L1(Rn, 1

1+|x| ), where L1(R, w) is the weighted space with the norm∫

Rn |f(x)|w(x)dx := ‖f‖L1(Rn,w).

Practically interesting questions are:

(a) How are singularities of f and f related?

(b) Given the noisy measurements of f at a grid, how does one find the discontinuities of f?

A grid is a set of points xj := (j1 h1, j2 h2, jn hn), where hi > 0, 1 ≤ i ≤ n, jm = 0,±1,±2,±3, . . ., 1 ≤m ≤ n. The noisy measurement are uj = f(xj) + nj , where nj are identically distributed, independentrandom variables with zero mean value and a finite variance σ2 < ∞. See [R176], [RKa] for a detailedinvestigation 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 f 6≡ 0, such that∫lαp

|f |ds < ∞ for all α, p, and f (α, p) ≡ 0, [RKa].

In many applications one integrates f not over the planes lαp, 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 ofmanifolds.

For example, if the family of manifolds is a family of spheres of various radii r > 0 and centers srunning over some surface S, then Mf := m(s, r) :=

∫|s−x|=r f(x)dx are the spherical means of f , s ∈ S,

and the problem is to recover f from the knowledge of m(s, z) ∀s ∈ S and ∀z > 0.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, r(k) ∀k > 0 is given,but kj, sj and J are unknown.

In general, one cannot recover a q ∈ L1,1 from these “incomplete” data. However, if one knows a

priori that q(x) has compact support, or |q(x)| ≤ ce−c2|x|δ

, δ > 1, then the data r(k) ∀k > 0 alonedetermine 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) ∈ L1,1(R) :=q :

∫∞−∞(1 + |x|)|q| dx, q = q, q(x) = 0 for x < 0, from the knowledge of the reflection coefficient

r(k) ∀k > 0, in the full-axis (full-line) scattering problem:

u′′ + k2u− q(x)u = 0, −∞ < x < ∞,

u = eikx + r(k)e−ikx + o(1), x −→ −∞,

u = t(k)eikx + o(1), x→ +∞.

The coefficients r(k) and t(k) are reflection and transmission coefficients, respectively.A general q ∈ L1,1(R) cannot be uniquely recovered from the knowledge of r(k) alone: one needs to

know additionally the bound states and norming constants to recover q uniquely. However if one knowsa priori that q(x) = 0 for x < x0, for example, for x < 0, then q(x) is uniquely determined by r(k)∀k>0

alone.

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

Let 0 6≡ f ∈ L1loc(R

n). Assume

∫

D

f(gx + y)dx = 0 ∀g ∈ S O(n) ∀y ∈ Rn, n ≥ 2, (1.2.4)

where SO(n) is the group of rotations, and D ⊂ Rn is a bounded domain. The problem (going back toPompeiu (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) χ(kα) = 0 for all α ∈ Sn−1 and some k > 0,

where f (ξ) :=∫D

f(x)eiζ·xdx, and χ :=

1 in D,

0 D. Iff f(k α) = 0 ∀α ∈ Sn−1 and some k > 0, then the

overdetermined problem (∇2 + k2

)u = 1 in D, u|S = uN |S = 0, (1.2.5)


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

(∇2 + k2

)u = 0 in D, u|S = 0, uN |S = 1, k2 = cons t > 0, (1.2.6)

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

(∇2 + k2

)u = 0 in D, u|S = 1, uN |S = 0, k2 = cons t > 0, (1.2.7)

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

∇2u+ f(u) = 0 in D, u|S = 0, (1.2.8)

where u is a non trivial solution to (1.2.8), (i.e., if f(0) = 0 then u 6≡ 0), and let the extra data (measureddata) be the value uN |S := h(s), ∀s ∈ S. Assume that f(u) is an entire function of u. The inverseproblem, of interest in plasma theory, is: given h(s) ∀s ∈ S, can one recover f(u) uniquely. Even forf(u) = c0 + c1u, cj = cons t, j = 0, 1, the problem is open (cf [Vog]).

1.2.13 Multidimensional inverse potential scattering

Let [∇2 + k2 − q(x)

]u = 0 in Rn, n > 2, (1.2.9)

u = eikα·x + A(α′, α, k)eikr

r+ o

(1

r

),

r := |x| → ∞,x

r= α, α ∈ S2,

(1.2.10)

where α is given, and k = cons t > 0. The coefficient A(α′, α, k) is called the scattering amplitude, andthe solution to (1.2.9)-(1.2.10) is called the scattering solution. The direct scattering problem is: givenq, k > 0 and α ∈ Sn−1, find u(x, α, k), and, in particular, A(α, α, k). This problem has been studiedin great detail (see, e.g.,[CFKS], [[R121], Appendix]) under various assumptions on q(x). We assumethat q ∈ Q := q = q, q(x) = 0 for |x| > a, q(x) ∈ L2(Ba) and often we assume additionally thatq ∈ Q := Qa ∩ L∞(Rn).

The inverse scattering problem (ISP) consists of finding q(x), given A(α, α, k). Consider severalcases:

(1) A is given for all α′, α ∈ S2 and all k > 0,

(2) A is given for all α′, α and a fixed k = k0 > 0,

(3) A is given for a fixed α = α0 all α′ ∈ S2 and all k > 0,

(4) A (−α, α, k) is given for all α ∈ S2 and all k > 0 (back scattering data).

In case (1) uniqueness of the solution to ISP has been established long ago, and follows easily from theasymptotics of A as k → ∞. An inversion formula based on high-energy 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 formulafor the noisy data is derived by in [R143], and stability estimates for the inversion formulas for the exactand noisy data are derived by in [R143], [R203]. In case (3) and (4) uniqueness of the solution to ISP isan 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 observedon 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

∇×E = −µ∂H∂t

, ∇×H = ε∂E

∂t+ σE + j, (1.2.11)

where µ = cons t > 0, ε = ε(z), σ = σ(z), z = x3 is the vertical coordinate, z > 0 is the region ofthe ground, j = f(t) δ(x) δ(z − z0)ey is the source, z0 < 0, f(t) describes the shape of the pulse of thecurrent j along a wire going along the y = x2 axis at the height |z0| above the ground. Assume ε = ε0,σ = 0, µ = µ0 for z < 0 (in the air), ε = ε1 = const, σ = 0, µ = µ0 for z > L, f(t) = 0 for t < 0 andt > T , ε0 and µ0 are dielectric and magnetic constants, ε = ε(z), σ = σ(z) for 0 < z < L. Differentiatethe second equation (1.2.1) with respect to t, and get −∇ ×∇ × E = εµEtt + σµEt + µ jt. Let E =E (x, z, t)ey, u(z, k, λ) := E (z, k, λ)/(ikµh(k)), E :=

∫∞0 dt

∫∞−∞ dx E (x, z, t)eikt, f (k) :=

∫∞0 f(t)eiktdt.

Thenu′′ − λ2u+ k2A2(z)u+ ik B(z)u = −δ

(z − z0

), u(±∞, k, λ) = 0, (1.2.12)

where u′ = dudz , A2(z) := ε(z)µ, B(z) := σ(z)µ, B(z) = 0 for z /∈ [0, 1] ground-penetrating radar. The

ground-penetrating radar inverse problem is: given E (x, 0, t) ∀x ∈ R, ∀t > 0, find ε(z) and σ(z).One may use the source j = f(t)δ(r − r0)δ(z − z0)eφ, which is a current along a loop of wire, eφ is

the unit vector in cylindrical coordinates. In this case, one looks for E of the form: E = E (r, z)eφ, andfrom (1.2.11) one gets:

A2(z)Ett +B(z)Et − Ezz − Err −1

rEr +

E

r2= −µ ft δ(r − r0) δ(z − z0), (1.2.13)

A2(z) := ε(z)µ, B(z) = σ(z)µ. If E :=∫∞0

E (z, z, t)eiktdt, then

E ′′ + k2A2(z) E + ik B(z) E + Err +1

rEr −

E

r2= −ik µ h(k) δ(r − r0)δ(z − z0)

E ′ :=d E

dz, h(k) :=

∫ ∞

0

f(t)eiktdt.

(1.2.14)

Let w :=∫∞0

E (r, z, k)J1(λr)rdr, where J1(r) is the Bessel function. Set u := wikh(k)roJ1(λr0) . Then u

solves (1.2.12), and the inverse problem is the same as above ([R185]).

1.2.15 A geometrical inverse problem

Let

∆u = 0 in D ⊂ Rn, n ≥ 2,

u|S0 = u0,

uN |S0 = u1,∣∣u0

∣∣ +∣∣u1

∣∣ 6≡ 0,

u|S1 = 0 or uN |S1 = 0.

Here D is domain homeomorphic to an annulus, S0 is its inner boundary, S1 is its outer boundary. Thegeometrical inverse problem is: given u0, u1, and S0, find S1.

One can interpret the data as the Cauchy data on S0 for an electrostatic potential u, and then S1

is the surface on which the potential is vanishing if u|S1 = 0, or the charge distribution is vanishing ifuN |S1 = 0 ([R139]).


1.2.16 Inverse source problems

(1) Inverse source problems in acoustics.

Let (∇2+k2)u = f in R3, f = 0 for |x| ≥ a, u satisfies the radiation condition limr→∞ r(∂u∂r

−iku) =

0 uniformly in α′ := xr , r = |x|. Define the radiation pattern A by the formula: u = A eikr

r + o(1r )

as r → ∞ and xr = α′. Then

A(α′, k) := − 1

4π

∫

Ba

f(y)e−ik α′·ydy, Ba :=

x : |x| ≤ a

. (1.2.15)

The inverse source problems are:

(i) Given A(α′, k) ∀α′ ∈ S2 and ∀k > 0, find f(x).

(ii) Given A(α′, k) ∀α′ ∈ S2 and a fixed k = k0 > 0, find f(x).

Clearly, by (1.2.15) problem (i) has at most one solution, but an a priori given functionA(α′, k) may be not of the form (1.2.15): the right-hand side of (1.2.15) is an entire functionof exponential type of the vector ξ := kα. Problem (ii), in general, may have many solutions,since A(α′, k0) may vanish for all α′ ∈ S2 at some k0 > 0.

(2) Inverse source problem in electrodynamics.

Consider Maxwell’s equations (1.2.11) in R3, and assume j = 0 for |x| ≥ a, The radiation conditionfor (E,H) is:

E =eikr

rA(α′, k) + o

(1

r

), r := |x| −→ ∞, α′ :=

x

r, k = ω

√εµ, (1.2.16)

where E and H in (1.2.11) are assumed monochromatic with e−iωt time dependence, σ(x) = 0,

and H =√

ε0µ0

[α′, E] + o( 1r ), where εo, µo are the constant values of ε and µ near infinity.

The inverse source problem is: given A(α′, k), find j. Again, one should specify for what α′ and kthe function A(α′, k) is known. One can derive the relation between A and j. Namely, assuming ε andµ constants, and j smooth and compactly supported, one starts with the equations

∇×E = iω µH, ∇×H = −iω εE + j(x),

then gets∇×∇× E = k2E + ω µ j, k2 = ω2ε µ,

then (− ∆ − k2

)E = iω µ j −∇∇ ·E, ∇ ·E = (iω ε)−1∇ · j,

so

E = iωµ

∫g(x, y)j dy − (iω ε)−1

∫g∇y

(∇y · j

)dy, g :=

eik|x−y|

4π|x− y| ,∫

:=

∫

R3

.

Let γ := eikr

r , and J := 14π

∫e−ik α

′·yj dy. Then

E = γ i ω µ J − 1

i ω ε∇x∇x · γ J = γ i ω µ(J − α′, α′ · J) + o

(1

r

).

Thus E = −γ i ω µ[α′[α′, J ]], where [a, b] is the vector product. So

A(α′, k) = −ik√µ

ε

[α′, [α′, J ]

], J :=

1

4π

∫e−ikα

′·yj(y)dy. (1.2.17)


and J = α′(α′, J) +√

εµ

1ikA.

It is now clear, that even if A is known for all k > 0 and all α′ ∈ S2, vector J is not uniquelydetermined, but only its component orthogonal to α′ is determined. Therefore, the solution to theinverse source problem in electrodynamics is not unique and may not exist, in general. The antennasynthesis problems are inverse source-type problems of electrodynamics. For example, if j is the currentalong a linear antenna (which is a wire along x3 := z axis, −` < z ≤ `, 2` is the length of the antenna,j = j(z) δ(x) δ(y) e3), then

J = e3f :=e34π

∫ `

−è−ikz cos θj(z)dz, cos θ = e3 · α′ and

A = −ik√µ

ε

[α′, [α′, e3]

]f,

(1.2.18)

so f = f(θ) = 14π

∫ `−` e

−ikz cos θj(z)dz is determined uniquely by the data A. Finding j(z), whichproduces the desired diagram f(θ), is the problem of linear antenna synthesis. There is a large body ofliterature on this subject.

Let utt − ∆u = f(x, t), t ∈ R, x ∈ R3, u|t=0 = u0(x), ut|t=0 = u1(x), u|t=T = v0(x), ut|t=T = 0 fort ∈ [0, T ].

The Inverse source problem is: given u0, u1, v0, v1 ∀x ∈ R3 ∀t ∈ R, find f(x, t), x ∈ R3, t ∈ [0, T ].The questions mentioned in this subsection were discussed in many papers and books ([AVST], [MJ],

[ZK], [R11], [R21], [R26], [R27], [R28], [R73], [Is2]).

1.2.17 Identification problems for integral-differential equations

Consider a Cauchy problem.

u′′ =

∫ t

0

h(t− s)Au(s)ds + f(t), 0 ≤ t ≤ T, u′(0) = u0, u′(0) = u1, (1.2.19)

where T > 0, h ∈ C([0, T ]; R), f ∈ C1([0, T ],H), u0 ∈ D(A), u1 ∈ H, (u0, ϕ) 6= 0, A : D(A) → H is aclosed, linear, densely defined in a Hilbert spaceH, operator, D(A) is its domain of definition, ϕ ∈ D(A∗),A∗ϕ = λ0ϕ, λ0 6= 0, ‖ϕ‖ = 1. Assume that h(t) is unknown, and the extra data g(t) = (u(t), ϕ) aregiven for 0 ≤ t ≤ T .

The inverse problem is: given (A,ϕ, u0, u1, f, g), find h(t).See [LR].

1.2.18 Inverse problem for an abstract evolution equation

Consider a Cauchy problem u = A(t)u + γ(t)u + f(t), u(0) = u0, with the extra data (measured data)φ(t) := (u(t), w(t)). Here A(t) is a one-parametric family of closed, densely defined, linear, operatorson a Banach space X, which generates an evolution family U (t, s), u0 ∈ X, w ∈ X∗, f(t) is a givenfunction on [0, T ] with values in X, and γ(t) is an unknown scalar function. The inverse problem is:given φ(t), f(t), u0 and w, find γ(t) (see [RKo]).

In applications γ(t) may be a control function which should be chosen so that the measured dataφ(t) are reproduced.

1.2.19 Inverse gravimetry problem

Let u(x, y) be the y-component of the gravitational field generated by some masses, located in the region−y < −h. Assume that the values u(x, 0) := f(x) are known, ∆u = 0 in the region −h < y, and


u(x,−h) := g(x). Then∫∞−∞

hg(s)ds(x−s)2+h2 = f(x), −∞ < x < ∞ 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 F (ξ) =∫Df(x)eiξ·xdx. The PRP consists of finding argF (ξ), given |F (ξ)|. Clearly, the solution to

this problem is not unique: f(x)eiϕ, ϕ ∈ R, produces the same |F (ξ)|. Under suitable assumptions onf(x) and |F (ξ)|, 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 multi-dimensional inverse scattering problems uniqueness of the solution is an open problem for most of thenon-overdetermined 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 E(s, s, λ) ∀s ∈ S and all λ ∈ R determineq(x) uniquely provided that all the eigenvalues are simple. Here [−∆+q(x)−λn]ϕn = 0 in D, ∂ϕn

∂N |S = 0,‖ϕn‖L2(D) = 1, q ∈ Qa, E(x, y, λ) is the kernel of the resolution of the identity of the selfadjoint Neumannoperator L = −∆ + q(x) in H = L2(D).

The above inverse problem is not overdetermined, because E(s, s, λ) depends on three variables inR3 and q(x) is also a function of three variables in R3. It is an open problem to find out if this resultremains 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 lineardevice and the output signal, find the input signal. By the properties of a linear device one means itspoint-spread function (scattering function) or transfer function. For example,

∫Dk(x, y)u(y)dy = f(x).

Given k(x, y) and f(x), one wants to find u(y).

In practice the output signal f(x) is noisy, i.e., fδ(x) is given in place of f(x), ‖fδ − f‖ ≤ δ, wherethe norm depends on the problem at hand.

See [BG], [RSm6], [RG].

1.2.23 Inverse problem of electrodynamics, recovery of layered mediumfrom the surface scattering data

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

j = δ(t− t0

)δ(x− x0

)δ(y − y0

)δ(z − z0

)e,

where e is a constant vector, ε(x, y, z), µ(x, y, z) and σ(x, y, z) are known constants: ε0, µ0, and σ0,respectively, outside a bounded domain D ⊂ R3 := x : z < 0, and the data E(x, x0, t), H(x, x0, t) aremeasured on the surface of the Earth z = 0 for various orientations of e then the inverse problem is todetermine ε(x), µ(x), and σ(x) from the above data.

See [RSo], [RK].


1.2.24 Finding ODE from a trajectory

Let u+ a1u+ a0 u = 0, a1 and a0 are constants, u = u(t), t ∈ (0, 1). Can one find a1 and a2 uniquely?In general, the answer is no. A trivial example is u(t) = 0. What trajectory allows one to find a1 anda0 uniquely? Suppose

u(s)u(t) − u(t)u(s) 6= 0 for some 0 < t < s < T. (1.2.20)

Then the system of equations

a1 u(s) + a0 u(s) = − ü(s), a1 u(t) + a0 u(t) = −u(t) (1.2.21)

for finding a1 and a0 is uniquely solvable, so that (1.2.20) guarantees that u(t), t ∈ [0, t], determinesa0 and a1 uniquely. More generally, consider a system u = Au, u = (u1(t)), . . . , un(t), where A =(aij)1≤i,j≤n is a constant matrix, u(0) = v. If there are points t1, t2, ......, tn ∈ [0, T ], such that thesystem u(t1), u(t2), ........, u(tn) is a linearly independent system of vectors, then u(t) determines thematrix A(ai,j)1≤i,j≤n 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

A(u) = f, A : X → Y, (1.3.1)

where X and Y are Banach spaces, or metric spaces, and A is a nonlinear operator, in general. Prob-lem (1.3.1) is called well-posed if A is a homeomorphism of X onto Y . In other words, the solutionto (1.3.1) exists for any f ∈ Y , is unique, and depends on f continuously, so that A−1 is a continuousmap. 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 problemto equation (1.3.1) where A is not boundedly invertible. For example, consider the equation

Au :=

∫ x

a

u dt = f(x), f(a) = 0. (1.3.2)

If X = Y = L2(a, b), then A is not boundedly invertible: A is injective, its range belongs to theSobolev space H1(a, b), A−1f = f ′ is an unbounded operator in L2(a, b). If noisy data fδ are given,‖fδ − f‖L2(a,b) ≤ δ, then fδ may be not in the range of A. A practically interesting problem is: can onefind an operator Rδ such that ‖Rδfδ − f ′‖ ≤ η(δ) → 0 as δ → 0? In other words, can one estimate f ′

stably, given δ and fδ?Any Fredholm first-kind integral equation with linear compact operator A : X → Y is of the

form (1.3.1). Such an operator in an infinite-dimensional space cannot have closed range and can-not 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-posedproblems is enormous ([IVT], [EHN], [Gro], [TLY], [VV], [VA], [R58]).

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 noisydata are given in Chapter 2.

1.4. EXAMPLES OF ILL-POSED PROBLEMS 21

In navigation a ship receives a navigation signal which is a univalent function f(x), that is, a smoothfunction which has precisely one point xm of maximum, and the course of the ship is determined bythis point. The function f is observed in an additive noise. Given noisy data fδ , ‖fδ − f‖ ≤ δ, onewants to find xm. A possible approach to this problem, is to search for a point at which f ′ = 0. Onecan see from a simple example that small perturbations of f can lead to large perturbations of f ′: letfδ = f + δ sin(ωx). Then f ′δ − f ′ = δω cos(ωx). No matter how small δ > 0 is, one can choose ω so largethat δω cos(ωx) will take arbitrary large values at some points x.

1.4.2 Stable summation of the Fourier series and integrals withrandomly perturbed coefficients

Let ϕn(x) be an orthonormal basis of a Hilbert space H and L be a linear system (a linear operator)such that Lϕn = cnϕn, where cn are some numbers. Due to the inner noise in L, one observes the noisyoutput (cn + εn)ϕn, where |εn| ≤ ε is the noise. Thus, if the input signal is

∑∞n=1 fnϕn, the output is∑∞

n=1 fn(cn + εn)ϕn(x). One has noisy Fourier coefficients fncn + fnεn and one wants to recover thefunction

∑∞n=1 fncnϕn(x).

If one has a Fourier integral, one can formulate a similar problem. To see that this problem isill-posed, in general, let us take cn = 1, denote fnεn = bn, and assume that bn is playing the role ofnoise, |bn| ≤ δ. Then one has the problem: given the noisy Fourier coefficients gn := fn + bn, |bn| ≤ δ,find f :=

∑∞n=1 fnϕn(x). This problem is ill-posed because the series

∑∞n=1 bnϕn(x) may diverge. For

example, if bn = δ and ϕn(x) = sin(nx), then the series∑∞

n=1 δ sin(nx) diverges. Similarly, if f (λ)

is the Fourier transform of a function f ∈ L2(Rn) and g(λ) is the noisy data, g(λ) = f (λ) + b(λ),|b(λ)| ≤ δ, then the problem is to calculate f(x) with “minimal error”, given noisy data g(λ). Thenotion of “minimal error” should be specified.

1.4.3 Solving ill-conditioned linear algebraic systems

Let A : Rn → Rn be a linear operator such that its condition number ν(A) := ‖A‖‖A−1‖ is large.Then the linear algebraic system Au = f can be considered practically as an ill-posed problem, becausesmall perturbation ∆f of the data f may lead to a large perturbation ∆u in the solution u. One has:‖∆u‖‖u‖ = ‖A−1∆f‖

‖A−1f‖ ≤ ‖A−1‖‖∆f‖‖A‖−1‖f‖ = ν(A) ‖∆f‖‖f‖ , so that the relative error ‖∆f‖

‖f‖ in the data may result

in ν(A) ‖∆f‖‖f‖ relative error in the solution. In the above derivation we use the inequality ‖A−1f‖ ≥‖A‖−1‖f‖. Methods for stable solution of ill-conditioned algebraic systems are given in Chapter 2.

1.4.4 Fredholm and Volterra integral equations of the first kind

If Au =∫ baA(x, y)u(y)dy or V u =

∫ xaA(x, y)u(y)dy, and A(x, y) is a continuous kernel in D := [a, b]×

[a, b], then the operators A and V are compact in H = L2(a, b), these operator are not boundedlyinvertible in H. Therefore problems Au = f and V u = 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

Au =

∫

D

A(x, y)u(y)dy = f(x), x ∈ D. (1.4.1)

Often one has:

Au =

∫ t

0

A(t− s)u(s)ds = f(t), t ≥ 0. (1.4.2)


The deconvolution problem consists of finding u, given f and A(x, y). For (1.4.2) the identificationproblem is of practical interest: given u(t) and f(t), find A(t). The function A(t) characterizes the linearsystem 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 ϕ(u) := ‖A(u) − f‖ = inf. Suppose that ui is the infimum of ϕ(u):ϕ(ui) ≤ ϕ(u). If f is perturbed, that is, ϕδ(u) := ‖A(u) − fδ‖, ‖fδ − f‖ ≤ δ, then the infimum of ϕδ(u)may be not attained, or it may be attained at an element uδ which is far away from ui. Thus the mapf → ui may be not continuous. In this case the minimization problem is ill-posed. Such problems werestudied [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: ∆u = 0 in the half-plane y > 0, u|y=0 = 0 uy|y=0 = vn(x) := sinn xn ,

u = u(x, y), x ∈ R, y ≥ 0. It is clear that u(x, y) = sin(n x)2n2 (en y − e−ny) 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 (taken → ∞), while the solution tends to infinity, as n → ∞, at any point (x, y), y > 0, x 6= nπ. Thus, theclaim is verified (cf [LRS] [R139]).

1.4.8 The backwards heat equation

Consider the backwards heat equation problem:

ut = uxx, t ≥ 0, x ∈ [0, π]; u(0, t) = u(π, t) = 0, u(x, T ) = v(x).

Given v(x), one wants to find u(x, 0) := w(x).

By separation of variables one finds u(x, t) =∑∞

n=1 un(t) sin(nx), un(t) = e−n2(t−T )vn, vn =

2π

∫ π0v(x) sin(nx)dx. Therefore, w(x) =

∑∞n=1 e

n2T vn sinnx, provided that this series converges, inL2(0, π), that is, provided that

∞∑

n=1

e2n2T∣∣vn∣∣2 < ∞. (1.4.3)

This cannot happen unless vn decays sufficiently fast. Therefore the backwards heat equation problemis ill-posed: it is not solvable for a given v(x) unless (1.4.3) holds, and small perturbations of the dataυ in L2(0, π)-norm may lead to arbitrary large perturbations of the function w(x), but also may lead toa function v for which the solution u(x, t) does not exist for t < T (cf [LRS], [IVT]).

Chapter 2

Methods of solving ill-posedproblems

2.1 Variational regularization

2.1.1 Pseudoinverse. Singular values decomposition

Consider linear Equation (1.3.1). Let A : X → Y be a linear closed operator, D(A) and R(A) be itsdomain and range, X = H1 and Y = H2 be Hilbert spaces, N (A) := u : Au = 0, A∗ is the adjointoperator, R(A) ⊕ N (A∗) = H2, the bar stands for the closure, and ⊕ is the orthogonal sum. If A isinjective, i.e., N (A) = 0, and surjective, i.e. R(A) = H2, and D(A) = H1, then the inverse operatorA−1 is defined on H2, A

−1A = AA−1 = I, I is the identity operator. A closed, linear, defined on allof H1 operator, is bounded, so A is an isomorphism of H1 onto H2 if it is injective, surjective, andD(A) = H1.

If N (A) 6= 0, P is the orthoprojector onto N (A) in H1, and Q is the orthoprojector onto R(A),then one defines a pseudoinverse (generalized inverse) A+: D(A+) := R(A)⊕N (A∗), A+(N (A∗)) := 0,A+Au := u−Pu. Thus, AA+A = A, A+AA+ = A+, and AA+u = Qu for u ∈ D(A+). The operator A+

is bounded iff R(A) = R(A). If f ∈ R(A), i.e., f = Au0 for some u0, then the problem ‖Au − f‖ = infhas a solution u0, every element u0 + v, ∀v ∈ N (A), is also a solution, and there is a unique solutionwith minimal norm, namely the solution u0 such that u0 ⊥ N (A), u0 = A+f . If f /∈ R(A), then theinfimum of ‖Au − f‖ is not attained. If A is bounded and f ∈ R(A), then the element u0 = A+fsolves the equation A∗Au = A∗f and is the minimal norm solution to this equation, i.e., u0 ⊥ N (A).Indeed, if Au0 = f and u0 ⊥ N (A), then A∗Au0 = A∗f . Conversely, if A∗Au = A∗f and f = Au0, withu0 ⊥ N (A), then A∗A(u−u0) = 0, (A∗A(u−u0), u−u0) = 0, and Au = Au0. Thus, u = u0 if u ⊥ N (A).One can prove the formula: A+f = limα→0(αI+A∗A)−1A∗f, where α > 0 is a regularization parameter(see Section 2.1.2) and f ∈ R(A).

Let us define the singular value decomposition. Let A : H1 → H2 be a linear compact operator,B := A∗A : H1 → H1 is a compact selfadjoint operator, Bϕj = s2j ϕj , ‖ϕj‖ = 1,

(ϕj, ϕm) = δjm :=

1 j = m,

0 j 6= m,s1 ≥ s2 ≥ · · · ≥ 0,

sj are called s-values of A. If AA∗ := T , and Aϕj/‖Aϕj‖ := ψj , then Tψj = s2j ψj , (ψj, ψm) =

(Aϕj , Aϕm)/(‖Aϕj‖‖Aϕm‖) = s2j δjm/‖Aϕj‖2 = δjm. Thus, ‖Aϕj‖ = sj , Aϕj = sj ψj , A∗ ψj = sj ϕj .

If u ∈ H1 is arbitrary, then Au =∑∞

j=1 sj(u, ϕj)ψj (∗), limj→∞ sj = 0. Thus, an element f ∈ R(A) iff

23

24 CHAPTER 2. METHODS OF SOLVING ILL-POSED PROBLEMS

∑∞j=1 |(f, ϕj)|2/s2j < ∞ (Picard’s test). If A = A∗, then s2j = λ2

j(A2), where λ2

j are eigenvalues of A2.Then ψj = ϕj , Aϕj = λj ϕj .

If dimH1 = n < ∞, dimH2 = m < ∞, then A can be written as A = V SU ∗, where A is an m × nmatrix, U and V are unitary matrices (n × n and m ×m, respectively), whose columns are vectors ϕjand ψj respectively, and S is an m × n matrix with the diagonal elements sj , 1 ≤ j ≤ r, r is the rankof the matrix A, and other elements of S are zeros. The matrix A+ can be calculated by the formulaA+ = US+V ∗, where S+ is an n×m matrix with diagonal elements s−1

j , 1 ≤ j ≤ r, and other elements

of S+ are zeros.

2.1.2 Variational (Phillips-Tikhonov) regularization

Assume A : H1 → H2, ‖A‖ < ∞ is linear, f ∈ R(A), ‖fδ − f‖ ≤ δ, fδ is not necessarily in R(A). Theproblem Au = f is assumed to be ill-posed (see Section 1.3).

Consider the problem:

F (v) :=∥∥Av − fδ

∥∥2+ α‖v‖2 = inf, (2.1.1)

where α > 0 is a parameter.

Theorem 2.1.1. Assume Au = f , and u⊥N (A). Then:

(i) The minimizer uα,δ of (2.1.1) does exist and is unique

(ii) If δ → 0 and α = α(δ) satisfies the condition δ2/α(δ) → 0 as δ → 0, then limδ→0 ‖uδ − u‖ = 0,where uδ := uα(δ),δ.

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

Bv + αv = A∗fδ , B := A∗A ≥ 0, (2.1.2)

which has a unique solution uα,δ = (B + αI)−1A∗fδ. Claim (i) is proved. One has F (uαδ) ≤ F (u) =δ2 + α‖u‖2 = α(δ2/α + ‖u‖2) ≤ cα, c = const > 0, if δ2/α ≤ c1, c1 = cons t. Thus ‖uα,δ‖ ≤ c.Below c stands for various positive constants. Choose α = α(δ) so that δ2/α(δ) → 0 as δ → 0, andlet uδ := uα(δ),δ. Then ‖uδ‖ ≤ c, so uδ u0 as δ → 0, and Buδ → A∗f . This implies Bu0 = A∗f ,and we claim that ‖u0‖ ≤ ‖w‖ ∀w : Bw = A∗f . This claim we prove later. Thus, u0 = u. Let usprove that limδ→0 ‖uδ − u‖ = 0. One has ‖(B + α)−1A∗fδ − u‖ ≤ ‖(B + α)−1A∗(fδ − f)‖ + ‖(B +α)−1A∗f − u‖ ≤ ‖(B + α)−1A∗‖δ + ‖(B + α)−1Bu − u‖ ≤ δ/(2

√α) + η(α) → 0, α → 0. Here the

estimate ‖(B +α)−1A∗‖ ≤ 1/(2√α) and the relation ‖[(B+α)−1B− I]u‖ → 0 as α→ 0, were used. To

prove the first estimate, one uses the formula: (B + α)−1A∗ = A∗(T +α)−1, T := AA∗, B := A∗A, andthe polar representation of A∗ yields the formula A∗ = V T 1/2, where V is an isometry, ‖V ‖ ≤ 1. Onehas ‖T 1/2(T + α)−1‖ = maxλ≥0 λ

1/2(λ + α)−1 = 1/(2√α), where the spectral representation for T was

used.

Let us prove the second relation: ‖[(B + α)−1B − I]u‖2 =∫ ‖B‖0

| λλ+α − 1|2d(Eλu, u)

=∫ ‖B‖0

α2

(λ+α)2 d(Eλu, u) → ‖Pu‖2 as α → 0, where P is the orthoprojector onto N (A), and Eλ is the

resolution of the identity of the self-adjoint operator B (see [KA]).If u⊥N (A), then limα→0 ‖(B + α)−1Bu− u‖ = 0.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, Bv +α(δ)v = gδ, Bu0 = g, v u0, gδ → g, α(δ) → 0 as δ → 0, then Bv → Bu0 = g.

In our proof above, B ≥ 0 is a linear operator, so B is monotone, that is, (B(u) − B(v), u − v) ≥ 0∀u, v ∈ D(B). Recall that a nonlinear operator A is called hemicontinuous, if (A(u + tv), w) is acontinuous function of t ∈ R for any u, v, w ∈ H.

2.1. VARIATIONAL REGULARIZATION 25

Proof of Lemma 2.1.2. Clearly, Bv → g. If Bv → g, v u0, then B u0 = g, that is, monotone operatoris w-closed. Indeed, if B is monotone, then (Bv −B(u0 − tw), v− u0 + tw) ≥ 0 for any w ∈ H. Passingto the limit v u0, Bv → g, t→ 0, and using hemicontinuity of B, one gets: (g−Bu0, w) ≥ 0 ∀w ∈ H.This implies Bu0 = g, so Lemma 2.1.2 is proved. 2

The claim ‖u0‖ ≤ ‖w‖ ∀w : Bw = A∗f is proved for nonlinear monotone operators in Theorem 2.1.6below. Theorem 2.1.1 is proved. 2

2.1.3 Discrepancy principle

Theorem 2.1.1 gives an a priori choice of α = α(δ) which guarantees convergence limδ→0 uδ = u. An aposteriori choice of α is given by Theorem 2.1.3 below.

Theorem 2.1.3. (Discrepancy principle). Assume ‖fδ‖ > δ. If α = α(δ) is the root of the equation

∥∥A(B + α)−1A∗fδ − fδ∥∥ = Cδ, C = const > 1, (2.1.3)

then limδ→0 ‖uα − u‖ = 0, uδ := uα(δ),δ.

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

C2δ2 =

∫ ‖B‖

0

∣∣∣∣λ

λ+ α− 1

∣∣∣∣2

d(Fλfδ , fδ

)= α2

∫ ‖B‖

0

d(Fλfδ, fδ)

(λ + α)2:= I(α, δ),

where Fλ is the resolution of the identity of the selfadjoint operator T := AA∗, and the commutationformula (B + α)−1A∗ = A∗(T + α)−1 was used. One checks this formula easily. If α → +∞, thenI(∞, δ) = ‖fδ‖2 > δ2. If α → +0, then I(+0, δ) = ‖P1 fδ‖2 ≤ δ2, where P1 is the orthoprojector onN (T ) = N (A∗). Indeed, ||P1fδ|| ≤ ||P1(fδ − f)|| + ||P1f || ≤ δ, because ||P1|| ≤ 1, and P1f = 0 sincef ∈ R(A) and R(A)⊥N (A∗). Thus, if C2 < ‖fδ‖2 then equation (2.1.3) has a solution. This solution isunique because I(α, δ) is a monotone increasing function of α for each fixed δ > 0.

Now let us prove limδ→0 ‖uδ − u‖ = 0. One has ‖Auδ − fδ‖2 + α(δ)‖uδ‖2 ≤ δ2 + α(δ)‖u‖2. Since‖Auδ − fδ‖2 > δ2, it follows that ‖uδ‖2 ≤ ‖u‖2. Therefore (*) lim supδ→0 ‖uδ‖ ≤ ‖u‖. If ‖uδ‖ ≤ ‖u‖,then one can select a weakly convergent sequence uδ u0 as δ → 0. In the proof of Theorem 2.1.1 itwas proved that u0 = u, where u is the unique minimal-norm solution of the equation Bu = A∗f . Bythe lower semicontinuity of the norm in H, one has ‖u‖ ≤ lim infδ→0 ‖uδ‖. Together with (*), one getslimδ→0 ‖uδ‖ = ‖u‖. This and the weak convergence uδ u, imply limδ→0 ‖uδ − u‖ = 0. 2

Our proof is based on the following useful result.

Theorem 2.1.4. If un y and ‖un‖ ≤ ‖y‖, then limn→∞ ‖un − y‖ = 0.

Proof. If un y then lim infn→∞ ‖un‖ ≥ ‖y‖. Also one has lim supn→∞ ‖un‖ ≤ ‖y‖. Thus, limn→∞ ‖un‖= ‖y‖, and ‖un − y‖2 = ‖un‖2 + ‖y‖2 −2<(un, y) → 0 as n → ∞. 2

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 A−1 is continuous on A(K).

Proof. Since A is injective, A−1 is well-defined on A(K). Let A(un) := fn → f , un ∈ K. Thenun → u ∈ K, where un is a subsequence denoted again un. Since A is closed, un → u and A(un) → fimply A(u) = f ∈ A(K), and, by the injectivity of A, one has A−1(fn) → A−1(f). Lemma 2.1.5 isproved. 2


Claim: Let us assume that A : H → H is monotone, continuous, D(A) = H, A(u) = f , andA(uα) + αuα = f . Then ‖uα‖ ≤ ‖u‖.

Proof. Indeed, A(uα)−A(u) +αuα = 0. Multiply this equation by uα−u and use the monotonicityof A, to get (uα, uα − u) ≤ 0. Thus, ‖uα‖ ≤ ‖u‖. Let α ↓ 0. Select a sequence, denoted again by uα,such that uα u0 as α → 0. Then A(uα) → f . Since A is monotone, it is w-closed (see the proof ofTheorem 2.1.1), so A(u0) = f , and u0 is the minimal norm solution to equation (1.3.1). 2

Theorem 2.1.6. If A : H → H is monotone and hemicontinuous, if D(A) = H, if A(u0) = f , whereu0 is the minimal-norm solution to A(u) = f , and if A(uα) +αuα = f , then the minimal-norm solutionto (1.3.1) is unique and limα→0 ‖uα − u0‖ = 0.

Proof. We have ‖uα‖ ≤ ‖u‖, ∀u ∈ u : A(u) = f := N . Thus, lim supα→0 ‖uα‖ ≤ ‖u‖ ∀u ∈ N . Letuα u0. Then ‖u0‖ ≤ lim infα→0 ‖uα‖ and lim sup‖un‖ ≤ ‖u0‖. Thus, limα→0 ‖uα‖ = ‖u0‖. This andthe weak convergence uα u imply strong convergence uα → u0 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 casethe set of solutions N is convex and closed. Its closedness is obvious, if A is continuous. Its convexityfollows from the monotonicity of A and the following lemma:

Lemma 2.1.7 (Minty). If A is monotone and continuous, then (a) (A(u)−f, v−u) ≥ 0 ∀v is equivalentto (b) (A(v) − f, v − u) ≥ 0 ∀v.

Proof. If (a) holds, then A(u) = f , and (b) holds by the monotonicity of A. If (b) holds, then takev = u + tw, t ≥ 0, where w is arbitrary, and get (A(u + tw) − f, w) ≥ 0 ∀w. Take t → 0 and get(Au− f, w) > 0 ∀w. Thus, A(u) = f , and (a) holds. Lemma 2.1.7 is proved. 2

To prove that N is convex, one assumes that u1, u2 ∈ N and derives that tu1+(1−t)u2 ∈ N ∀t ∈ (0, 1).Indeed, if uj ∈ N , then, by Lemma 2.1.7, (Av − f, v − uj) ≥ 0 ∀v. Thus, (Av − f, v − tu1 − (1− t)u2) =t(Av − f, v − u1) + (1 − t)(Av − f, v − u2) ≥ 0 ∀v. Thus, tu1 + (1 − t)u2 ∈ N ∀t ∈ (0, 1).

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, u1 and u2. Then ‖u1‖ = ‖u2‖ := m, and ‖tu1 +(1 − t)u2‖ ≤ t‖u1‖ + (1 − t)‖u2‖ = m, so that any element of the segment joining u1 and u2 hasminimal norm m. Since Hilbert space is strictly convex, this implies u1 = u2. Indeed, take t = 1/2.Then ‖(u1 + u2)/2‖2 = ‖u1‖2 = ‖u2‖2. So <(u1, u2) = ‖u1‖‖u2‖ = ‖u1‖2 = ‖u2‖2. Thus, u1 = u2.Theorem 2.1.6 is proved. 2

Consider the equation

A(uα)

+ αuα = fδ . (2.1.4)

Theorem 2.1.8. Assume that A is monotone and continuous, and equation (1.3.1) has a solution. Ifα = α(δ) → 0 and δ/α(δ) → 0 as δ → 0, then the unique solution to (2.1.4) converges strongly to u, theunique solution to (1.3.1) of minimal norm.

Proof. Because A is monotone and α > 0, the equation A(vα) + αvα = f has a solution, and thissolution is unique. Let uδ := uα(δ) solve (2.1.4), and vδ := vα(δ). One has ‖u − uδ‖ ≤ ‖u − vδ‖ +‖vδ − uδ‖. By Theorem 2.1.6, limδ→0 ‖u − vδ‖ = 0. Let us prove limδ→0 ‖vδ − uδ‖ = 0. We haveA(uδ) − A(vδ) + α(uδ − vδ) = fδ − f . Multiply this by uδ − vδ and use the monotonicity of A to getα‖uδ − vδ‖2 ≤ δ‖uδ − vδ‖. This implies limδ→0 ‖uδ − vδ‖ = 0, if limδ→0 δ/α(δ) = 0. Theorem 2.1.8 isproved. 2

2.1.5 Regularization of nonlinear, possibly unbounded, operator

Assume that :

(1) A : D(A) → X is a closed, injective, possibly nonlinear, map in Banach space X.


(2) φ ≥ 0 is a functional such that the set v : φ(v) ≤ c is precompact in X for any constant c > 0,

(3) Equation (1.3.1) has a solution y ∈ D(φ), A(y) = f ,

(4) D(A) ⊆ D(φ).

The last assumption can be replaced in some cases when A is an unbounded operator, by theassumption.

(4′) D(φ) ⊆ D(A).

Define the functional F (u) = ‖A(u) − fδ‖ + δ φ(u), where δ > 0 is a parameter, ‖fδ − f‖ ≤ δ, D(F ) =D(A) ∩D(φ). Consider the minimization problem:

F (u) = inf, infu∈D(F )

F (u) := m = m(δ). (2.1.5)

Let F (uj) ≤ m + 1j≤ m + δ, where j = j(δ) is the smallest integer satisfying the inequality 1

j(δ)≤ δ.

Denote uδ := uj(δ). One has m ≤ F (uδ) ≤ m + δ ≤ F (y) + δ = δ(2 + φ(y)) := cδ, and φ(uδ) ≤ c. Byassumption (2), as δ → 0 one can select a convergent subsequence, denoted again uδ, uδ → u, such thatlimδ→0 ‖A(uδ) − f‖ = 0. Thus, A(u) = f by the closedness of A, and u = y by the injectivity of A.Since the limit of any subsequence uδ is the same, namely y, it follows that limδ→0 ‖uδ − y‖ = 0. Wehave proved:

Theorem 2.1.9. If (1.3.1) has a solution, then, under the assumptions (1)–(4) (or (4′)), any sequenceuδ, such that F (uδ) ≤ m(δ) + δ, converges strongly to the solution y of (1.3.1) as δ → 0.

Remark 2.1.10. In the proof of Theorem 2.1.9 we do not need existence of the minimizer of thefunction (2.1.5).

2.1.6 Regularization based on spectral theory

Assume that A in (1.3.1) is a linear bounded operator, ‖A‖ < ∞, Ay = f , and y ⊥ N (A), i.e., y is theunique minimal-norm solution.

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

Bu = f1, B := A∗A ≥ 0, f1 := A∗f. (2.1.6)

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 f = Ay, then f1 = By, B(u− y) = 0, and (B(u− y), u− y) = 0. This implies‖A(u− y)‖ = 0, so Au = f . Thus u solves (1.3.1). 2

Equation (2.1.6) is a solvable equation with monotone, continuous operator, so Theorem 2.1.6 isapplicable and yields the following theorem:

Theorem 2.1.12. If 0 < α(δ) → 0, δ/α(δ) → 0 as δ → 0, ‖fδ − f‖ ≤ δ, and y is the minimal-normsolution to (1.3.1), then limδ→0 ‖uδ − y‖ = 0, where uδ is the unique solution of the equation

Buδ + α(δ)uδ = A∗fδ .

Lemma 2.1.13. Consider the elements wδ :=∫ ‖B‖0

g(s, α)dEsA∗fδ, where Es is the resolution of the

identity of B = A∗A, |sg(s, α)| ≤ c, c = const > 0 does not depend on s and α, limα↓0 g(s, α) = 1/s ∀s >0, sups

√sg(s, α) := g(α), and g(s, α) is a piecewise-continuous function. Let α = α(δ) → 0 so that

δg(α(δ)) → 0 as δ → 0. Then limδ→0 ‖wδ − y‖ = 0.


Proof. If f = Ay, then A∗f = By, and y =∫ ‖B‖0

s−1dEsA∗f . Thus, ‖wδ−y‖ ≤ ‖

∫ ‖B‖0

g(s, α)dEsA∗(fδ−

f)‖+ ‖∫ ‖B‖0

s(g(s, α)− s−1)dEs y‖ := I + η(α), where limα↓0 η(α) = 0, and I ≤ δ‖∫ ‖B‖0

g(s, α)dEsA∗‖.

One has ‖∫ ‖B‖0

g(s, α)dEsA∗‖ = ‖g(A∗A,α)A∗‖ = ‖A∗g(AA∗, α)‖ ≤ sups |

√sg(s, α)| = g(α). Thus

‖wδ − y‖ ≤ δ g(α) + η(α) → 0 as δ → 0. If the rate of decay of η(α) and the rate of growth of g(α) canbe estimated, then a quasioptimal choice of α = α(δ) can be made by minimizing δ g(α) + η(α) withrespect to α for a fixed δ. 2

Remark 2.1.14. We have used the spectral theorem for a selfadjoint operator B, namely the formulag(B) =

∫∞−∞ g(s)dEs, where Es is the resolution of the identity of B, D(g(B)) = u :

∫∞−∞ |g(s)|2d(Esu, u)

< ∞, ‖g(B)‖ = sup|s|≤||B|| |g(s)|.Remark 2.1.15. Similarly, one can use the theory of spectral operators in place of the spectral theoryof 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 N (A) 6= 0, or f /∈ R(A), or R(A)is not closed, i.e. A−1 is unbounded. If A is nonlinear and Frechet differentiable, then there are severalpossibilities. If A′(u) is boundedly invertible at some u, then A(u) is a local homeomorphism at thispoint, but it may be not a global homeomorphism. If A′(u) is not boundedly invertible, this does notimply, in general, that A is not a homeomorphism. For example, a homeomorphism A(u) may have acompact derivative, so its linearization yields an ill-posed problem. On the other hand, A(u) may becompact, so (1.3.1) is an ill-posed problem, but A′(u) may be a finite-rank operator, so that the rangeof A′(u) is closed. In spite of the above, we will often call a nonlinear equation problem (1.3.1) ill-posedif A′(u) is not boundedly invertible, and well-posed if A′(u) is boundedly invertible, deviating thereforefrom the usual terminology.

2.1.8 Discrepancy principle for nonlinear ill-posed problems withmonotone operators

Assume that A in (1.3.1) is monotone, i.e., (A(u) − A(v), u − v) ≥ 0, ∀u, v ∈ D(A), D(A) = H, A iscontinuous, A−1 is unbounded or does not exist, so (1.3.1) is an ill-posed problem, f ∈ R(A), ‖fδ−f‖ ≤ δ.Consider the discrepancy principle for finding ε = ε(δ) assuming that A is nonlinear monotone:

∥∥A(uδ,ε) − fδ∥∥ = Cδ, (2.1.7)

where C = const > 1, uδ,ε is any element such that F (uδ,ε) := ‖A(uδ,ε) − fδ‖2 + ε‖uδ,ε‖2 ≤ m(δ, ε) +δ2(C − 1 − b), where m(δ, ε) := infuF (u), b = const > 0, C > 1 + b, and ε plays the role of theregularization parameter α. We need three lemmas.

Lemma 2.1.16. If A is monotone and continuous, and the set Nf := u : A(u) = f is nonempty, thenit is convex and closed.

Lemma 2.1.17. If A is monotone and continuous, then it is w-closed, that is, un u and A(un) → fimply 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 thiscase 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 un u and ‖un‖ ≤ ‖u‖, then un → u.

Proof of Lemma 2.1.16. If A(un) = f and un → u, then A(u) = f , so Nf is closed. If A is monotone,A(u) = f and A(v) = f , then (A(z) − f, z − u) ≥ 0 and (A(z) − f, z − v) ≥ 0 ∀z and vice versa. Thus,for any λ, η ≥ 0, λ + η = 1, the element λu+ ηv ∈ Nf . 2 2


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, y is its minimal-norm solution, and

(iii) ‖fδ − f‖ ≤ δ, ‖A(0) − fδ‖ > Cδ, where C > 1 is a constant. Then:

(j) the equation ∥∥∥A(uδ,ε

)− fδ

∥∥∥ = Cδ, (2.1.8)

is solvable for ε for any fixed δ > 0. Here uδ,ε is any element satisfying inequality F (uδ,ε) ≤m+(C2 −1− b)δ2, where F (u) := ‖A(u)− fδ‖2 + ε‖u‖2, m = m(δ, ε) := infuF (u), b = const > 0,and C2 > 1 + b,

and

(jj) if ε = ε(δ) solves (2.1.8), and uδ := uδ,ε(δ), then limδ→0 ‖uδ − y‖ = 0.

Remark 2.1.20. The equation A(v) + εv = fδ is uniquely solvable for any ε > 0 and any fδ ∈ H. Ifv := vδ,ε is its solution, and ‖A(0) − fδ‖ > Cδ, where C = const > 1, then equation (2.1.8) with uδ,εreplaced by vδ,ε, is solvable for ε > 0. If ε := ε(δ) is its solution, then limδ→0 ε(δ) = 0. If A is injective,and if vδ := vδ,ε(δ), then limδ→0 ‖vδ − y‖ = 0, where y solves the equation A(y) = f .

If A is not injective, then it is not true, in general, that limδ→0 vδ = y, where y is the minimal-normsolution to the equation A(u) = f even if one assumes that A is a linear operator.

Proof of Theorem 2.1.19. If A is monotone, continuous and is defined on all of H, then the set Nf :=u : A(u) = f is convex and closed, so it has a unique minimal-norm element y. To prove the existenceof a solution to (2.1.8), we prove that the function h(δ, ε) := ‖A(uδ,ε) − fδ‖ is greater than Cδ forsufficiently large ε, and smaller than Cδ for sufficiently small ε. If this is proved, then the continuity ofh(δ, ε) with respect to ε on (0,∞) implies that the equation h(δ, ε) = Cδ has a solution.

Let us give the proof. As ε → ∞, we use the inequality:

ε∥∥uδ,ε

∥∥2 ≤ F(uδ,ε

)≤ m +

(C2 − 1 − b

)δ2 ≤ F (0) +

(C2 − 1 − b

)δ2,

and, as ε→ 0, we use another inequality:∥∥∥A(uδ,ε

)− fδ

∥∥2 ≤ F(uδ,ε

)≤ m +

(C2 − 1 − b

)δ2 ≤ F (y) +

(C2 − 1 − b

)δ2 = ε‖y‖2 +

(C2 − b

)δ2.

As ε → ∞, one gets ‖uδ,ε‖ ≤ c/√ε → 0, where c > 0 is a constant depending on δ. Thus, by the

continuity of A, one obtains limε→∞ h(δ, ε) = ‖A(0) − fδ‖ > Cδ.As ε → 0, one gets h2(δ, ε) ≤ ε‖y‖2 + (C2 − b)δ2. Thus, lim infε→0 h(δ, ε) < Cδ. Therefore equation

h(δ, ε) = Cδ has a solution ε(δ) > 0.Let us now prove that if uδ := uδ,ε(δ), then limδ→0 ‖uδ − y‖ = 0. From the estimate

∥∥∥A(uδ)− fδ

∥∥∥2

+ ε∥∥uδ

∥∥2 ≤ C2δ2 + ε‖y‖2,

and from the equation (2.1.8), it follows that ‖uδ‖ ≤ ‖y‖. Thus, one may assume that uδ U , andfrom (2.1.8) it follows that A(uδ) → f as δ → 0. By w−closedness of monotone continuous operators(hemicontinuity in place of continuity would suffice), one gets A(U ) = f , and from ‖uδ‖ ≤ ‖y‖ it followsthat ‖U‖ ≤ ‖y‖. Because A is monotone, the minimal norm solution to the equation A(u) = f in His unique. Consequently, U = y. Thus, uδ y, and ‖uδ‖ ≤ ‖y‖. By Theorem 2.1.4, it follows thatlimδ→0 ‖uδ − y‖ = 0.

Note that ‖y‖ > 0, because ‖A(0) − f‖ > 0 due to the assumption ‖A(0) − fδ‖ > Cδ, where C > 1.Theorem 2.1.19 is proved. 2


Proof of Remark 2.1.20. Let v = vε solve the equation A(v) + εv = f , let w := vδ,ε solve the equationA(w) + εw = fδ, h := fδ − f , ‖h‖ ≤ δ, and w − v := z. Then A(w) − A(v) + εz = h. Multiplythis equation by z and use the monotonicity of A to get ε‖z‖ ≤ δ. The triangle inequality yields:ε‖v‖ − δ ≤ ε‖w‖ ≤ ε‖v‖ + δ. Note that limε→∞ v = 0, and δ/ε → 0 as ε → ∞. Thus, limε→∞ ‖w‖ = 0.Therefore limε→∞ ‖A(w) − fδ‖ = ‖A(0) − fδ‖ > Cδ.

Fix δ > 0 and let ε → 0. Then limε→0 ‖v − y‖ = 0 and ‖v‖ ≤ ‖y‖, where y is the minimal-normsolution to the equation A(u) = f . If A is injective, then this equation has only one solution y. Sinceε‖w‖ ≤ ε‖v‖ + δ, one gets the inequality limsupε→0 ε||w|| = limsupε→0 ||Aw − fδ|| ≤ δ. Consequently,equation (2.1.8), with w replacing uδ,ε, has a solution ε(δ) > 0. We claim that limδ→0 ε(δ) = 0, in fact,ε(δ) = O(δ) as δ → 0. Indeed, from (2.1.8), with w replacing uδ,ε, one gets ε(δ)‖w‖ = Cδ, and we provebelow that liminfδ→0 ‖wδ‖ > 0, where wδ := vδ,ε(δ). This implies ε(δ) = O(δ).

We now claim that the limit limδ→0 wδ := u does exist, that u solves the equation A(u) = f ,and ‖u‖ > 0. It is sufficient to check that ‖wδ‖ < c where c = const does not depend on δ asδ → 0. Indeed, if ‖wδ‖ < c, then a subsequence, denoted wδ again, converges weakly to an elementy, wδ y, and (2.1.8) implies limδ→0A(wδ) = f . Since A is monotone, it is w−closed, so A(y) = f .By the injectivity of A, any subsequence wδ converges weakly to the same element y, so wδ y.Consequently, liminfδ→0 ‖wδ‖ ≥ ‖y‖ > 0 as claimed. The inequality ‖y‖ > 0 follows from the assumption‖A(0) − f‖ > 0.

To prove the inequality ‖wδ‖ < c, note that Cδ = ε‖wδ‖ ≤ ε‖wδ − vε(δ)‖ + ε‖vε(δ)‖ ≤ δ + ε‖y‖,where ε := ε(δ). Since C > 1, this implies δ/ε(δ) ≤ c1, where c1 := ‖y‖/(C − 1). Thus, ‖wδ‖ ≤ c, wherec := c1 + ‖y‖.

The last statement of Remark 2.1.20 is illustrated by the following example:

Example 2.1.21. Let Aw = (w, p)p, ‖p‖ = 1, p ⊥ N (A), f = p, fδ = p+ qδ, where (q, p) = 0, ‖q‖ = 1,Aq = 0, ‖qδ‖ = δ. One has Ay = p, where y = p is the minimal-norm solution to the equationAu = p. Equation Aw+ εw = p+ qδ, has the unique solution w = qδ/ε+ p/(1 + ε). Equation (2.1.8) isCδ = ‖qδ + (εp)/(1 + ε)‖. This equation yields ε = ε(δ) = cδ/(1 − cδ), where c := (C2 − 1)1/2, and weassume cδ < 1 (see the second inequality in the assumption (iii) of Theorem 2.1.19). Let wδ = w(δ, ε(δ)).Then, limδ→0wδ = p + c−1q := u, and Au = p. Therefore u = limδ→0 wδ is not p, i.e., u is not theminimal-norm solution to the equation Au = p.

Remark 2.1.20 is proved. 2

Remark 2.1.22. It is easy to prove that if conditions (i) and (ii) of Theorem 2.1.19 hold and A(uε,δ)+εuε,δ = fδ, and if limδ→0 δ/ε(δ) = 0, where limδ→0 ε(δ) = 0, ε := ε(δ), then limδ→0 ‖uδ − y‖ = 0, whereuδ := uε(δ),δ, and y is the minimal-norm solution to the equation A(u) = f . In particular, if ε = δa, 0 <a < 1, then limδ→0 ‖uδ−y‖ = 0. Indeed, ‖uδ−u‖ ≤ ‖uδ−vδ‖+‖vδ−u‖, where vδ is the unique solution tothe equation A(vδ)+ε(δ)vδ = f. It is well known that limδ→0 ‖vδ−y‖ = 0, provided that limδ→0 ε(δ) = 0,and, clearly, ‖uδ − vδ‖ ≤ δ/ε(δ): one multiplies the identity A(uδ) − A(vδ) + ε(δ)(uδ − vδ) = fδ − f byuδ − vδ and uses the monotonicity of A and the inequality ‖fδ − f‖ ≤ δ.

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 aregularizer exists, then the problem is well-posed.

Let A be a linear operator in a Banach space X. Assume that A is injective and A−1 is unbounded,that equation

Au = f (2.1.9)

2.2. QUASISOLUTIONS, QUASIINVERSION, AND BACKUS-GILBERT METHOD 31

is solvable, and g is such that

‖g − f‖ < δ, (2.1.10)

where ‖g‖ is the norm in X and δ > 0 is the noise level. Nothing is assumed about the statistical natureof 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:

‖Rg − u‖ −→ 0 as δ −→ 0 (2.1.11)

for any f ∈ Ran(A), where Ran(A) is the range of A, and any g ∈ X satisfying (2.1.10)?

Answer: no.

Proof. If such an R is found, then, taking g = f and using the fact that f ∈ Ran(A) is arbitrary, oneconcludes that R = A−1 on the range of A. Secondly, writing g = f + w, where

‖w‖ < δ, (2.1.12)

and w is arbitrary otherwise, one concludes from (2.1.11) and from the fact that Rf = A−1f = u, that

‖Rg − u‖ = ‖Rw‖ −→ 0 as δ → 0 (2.1.13)

for any w satisfying (2.1.12). Since R is linear, this implies that R is bounded, which contradicts theequation R = A−1 on Ran(A) and the unboundedness of A−1, which is the necessary condition for theill-posedness of (2.1.9). 2

A similar result one can find in [LY].

2.2 Quasisolutions, quasiinversion, and Backus-Gilbert method

2.2.1 Quasisolutions for continuous operator

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

∥∥A(u) − fδ∥∥ = inf := m(δ), u ∈ K, (2.2.1)

where m(δ) is the infimum of the function of ‖A(u) − fδ‖ and ‖fδ − f‖ ≤ δ. A minimizer for (2.2.1) iscalled a quasisolution to (1.3.1) with f = fδ . Let uj be a minimizing sequence for (2.2.1). Since K is acompactum, one may assume that uj → uδ as j → ∞, ‖A(uj) − fδ‖ → m(δ), A(uj) → A(uδ).

Thus ‖A(uδ)− fδ‖ = m(δ), so uδ is a minimizer for the problem (2.2.1). The above argument showsthat if f replaces fδ in (2.2.1), and if equation (1.3.1) is solvable and its solution belongs to K, thenany minimizer for (2.2.1) with fδ = f , is a solution to (1.3.1). Let us prove that limδ→0 ‖uδ − u‖ = 0,where uδ is a minimizer for (2.2.1), and u is a solution to (1.3.1), where existence of a solution (1.3.1) isassumed.

Indeed, uδ ∈ K, so one may assume that uδ → u as δ → 0. By continuity ofA, one has limδ→0A(uδ) =A(u). Thus, ‖A(u) − f‖ = limδ→0m(δ) = 0. The last conclusion follows from the solvability of (1.3.1),which yields f = A(u) and from the inequality m(δ) ≤ ‖A(uδ) − A(u)‖ + δ. 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 ‖fδ− f‖ ≤ δ, then (2.2.1) has a minimizer uδ, and limδ→0 ‖uδ−u‖ = 0 for every minimizer uδ andsome solution u to (1.3.1).


Remark. Suppose that X is strictly convex, i.e., if ||u|| = ||v|| = ||(u + v)/2||, then u = v. Forexample Hilbert spaces H are strictly convex, the spaces Lp(D), p > 1, are strictly convex, but L1(D)and C(D) are not. Suppose that K is a convex compactum, i.e., convex closed compact set. The metricprojection of an element f ∈ X onto K is the element PKf ∈ K such that ||PKf − f || = infu∈K ||u− f ||.If X is strictly convex, then PKf is unique, and if K is a convex compactum, then PKf dependscontinuously on f . If A is injective and closed, not necessarily linear, and K is a compactum, then A−1

is continuous on the set AK. Indeed, if fn = A(un), un ∈ K, and fn → f , then a subsequence, denotedagain un, converges to u because K is compact, and if A is closed, then A(u) = f , which proves theclaim. Therefore, if X is strictly convex and K is a convex compactum, and if A is an injective boundedlinear operator, then the Quasisolution u(f) = A−1PAKf 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, A−1 is possibly, unbounded,equation (1.3.1) is solvable, assumptions (1)–(4) of Section 2.1.5 hold, and K is a compactum containingall the solutions to (1.3.1).

Theorem 2.2.2. Under the above assumptions, if ‖A(wδ) − fδ‖ ≤ m(δ) + δ, where wδ ∈ K, thenlimδ→0 ‖wδ − u‖ = 0, where u is a solution to (1.3.1).

Proof. One can assume wδ → w as δ → 0, because K is a compactum. Note that limδ→0m(δ) = 0,because m(δ) ≤ ‖A(u)−fδ‖ ≤ δ. Thus, limδ→0 ‖A(wδ)−f‖ = 0. 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

2.2.3 Quasiinversion

Let A be a linear bounded operator in (1.3.1), and (1.3.1) is solvable, consider the equation (εQ+B)uε =f1, B := A∗A, f1 := A∗f , ε > 0 is a parameter, Q is an operator chosen so that ‖(εQ+B)−1‖ ≤ c(ε), andlimε→0 ‖uε−u‖ = 0, where u is a solution to (1.3.1). If A is unbounded, a similar idea can be applied toequation (1.3.1): consider the equation (εQ+B)vε = f , where Q is chosen so that ‖(εQ+B)−1‖ ≤ c1(ε),limε→0 ‖vε − u‖ = 0. The problem is: how does one choose Q with these properties? If A is a linearbounded operator, then Q = I can be used by Theorem 2.1.1. If A is unbounded, some assumptions onits spectrum are needed. See [LL] for details.

2.2.4 A Backus-Gilbert-type method: Recovery of signals from discrete andnoisy data

We discuss in this section the following Problem 1. Let D and the bar denote variance and mean valuerespectively, ∫ b

a

fψjdy = fj + εj , 1 ≤ j ≤ n, ε∗jεp = Cjp

Problem 1: Given fj + εj , 1 ≤ j ≤ n, estimate f(x).

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

fn(x) =

n∑

j=1

ϕj(x)(fj + εj).

The problem is to find ϕj(x) such that:

(1) if εj = 0 then ‖fn − f‖ → 0, n→ ∞;


(2) if ε∗jεp = Cjp then D[fn − f ] ≤ σ2; where σ > 0 is a certain number. We find ϕj, 1 ≤ j ≤ n,optimal, in a certain sense. Namely, if εj = 0, then ϕj are found from the requirements:

(A)

n∑

j=1

ϕj(x)

∫ b

a

ψjdy = 1,

∫ b

a

(

n∑

j=1

ϕj(x)ψj(y))2(x− y)2dy = min .

Note that

(A) ⇒n∑

j=1

ϕj(x)ψjdy ≡ δn(x, y) −→ δ(x− y), in an optimal way, as n→ ∞.

If εj 6= 0 then ϕj are found from (A) and (Cϕ,ϕ) ≤ σ2 if this problem is solvable and, if not, oneincreases σ2 so that this problem becomes solvable.

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

(2π)−1

∫ 1

−1

f(x) exp(−iωx)dx = F (ω). (2.2.2)

Suppose that the numbers:

Fj = F (ωj), 1 ≤ j ≤ n (2.2.3)

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

Problem 1. Given Fj, 1 ≤ j ≤ n, find an estimate fn(x) of f(x) such that

fn(x) → f(x) as n→ ∞ (convergence) (2.2.4)

The estimate fn is to be optimal in the sense specified in (2.2.13). (2.2.5)

To be specific let us assume that f(x) ∈ L2[−1, 1] and that the estimate is of the form

fn(x) =

n∑

j=1

Fjhj(x), (2.2.6)

where the functions hj(x) will be chosen soon. From (2.2.5) and (2.2.2) it follows that

fn(x) =

∫ 1

−1

An(x, y)f(y)dy, (2.2.7)

and

An(x, y) =

n∑

j=1

hj(x)ψ∗j (y), (2.2.8)

where ψj(y) = (2π)−1 exp(iωjy) and the star denotes complex conjugate.Property (2.2.4), convergence, holds if An(x, y) is a delta-sequence, i.e.,

‖∫ 1

−1

An(x, y)f(y)dy − f(x)‖ → 0, n→ ∞. (2.2.9)


Let

Q(x) = Q(h(x)) =

∫ 1

−1

|An(x, y)|2(x− y)2dy, (2.2.10)

and leth(x) = (h1(x), . . . , hn(x))

be a sequence of functions such that

∫ 1

−1

An(x, y)dy =

n∑

j=1

hj(x)aj = 1, (2.2.11)

where

aj =

∫ 1

−1

ψ∗j (y)dy. (2.2.12)

One can interpret (2.2.11) as the requirement that the estimate fn is exact for f(x) = const. Given(2.2.11), the smaller Q(x), the better is the quality of the delta-sequence An(x, y). Thus we are led tothe optimization problem: Find such a sequence hj(x), 1 ≤ j ≤ n that

Q(h(x)) = min and (2.2.11) holds. (2.2.13)

Note that the general problem of the type∫

Ω

f(x)ψ∗j (x)dx = bj, 1 ≤ j ≤ n, (2.2.14)

where ψj, 1 ≤ j ≤ n, is a linearly independent set of functions, and Ω is a bounded domain in Rd, canbe treated in exactly the same way as before.

If the problem (2.2.13) has the unique solution h(x) = h1(x), . . . , hn(x) then (2.2.6) is the optimalestimate which, as we prove, has the convergence property (2.2.4).

2.2.4.2. If the data are noisy, that is Fj + εj are given in place of Fj, where ε = ε1, . . . , εn randomvector with the covariance matrix

Cij = ε∗i εj , εj = 0, (2.2.15)

where the bar denotes the mean value, then the variance D(f − fn) can be computed

D(f − fn) = D

f(x) −

n∑

j=1

Fjhj(x) −n∑

j=1

εjhj(x)

=

n∑

i,j=1

Cijhj(x)h∗i (x) = (Ch, h), C = (cij),

(2.2.16)

where ( , ) is the inner product in Cn. Let us fix σ2 > 0 and require that

(Ch, h) ≤ σ2. (2.2.17)

The optimization problem for finding the vector h(x) = (h1(x), . . . , hn(x)) can be formulated as follows:

Minimize Q(h(x)) under the restrictions (h, a) = 1 and (Ch, h) ≤ σ2. (2.2.18)

Here a = (a∗1, a∗2, . . . , a

∗n). Clearly, problem (2.2.18) is not solvable for all σ2 > 0. We will discuss this

important point below. If (2.2.18) is solvable, the solution is unique, and the optimal estimate is givenby

fn =

n∑

j=1

(Fj + εj)hj(x), (2.2.19)


This estimate has variance ≤ σ2.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. Thecase when the data are the finite number of moments is treated, and the optimization requirements areintroduced.

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

Q(x) = (Bh, h), (2.2.20)

where

(Bh, h) =

n∑

i,j=1

bij(x)hj(x)h∗i (x) (2.2.21)

and

bij(x) = bij =

∫ 1

−1

ψ∗i (y)ψj(y)(x − y)2dy (2.2.22)

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

(Bh, h) = min, (h, a) = 1. (2.2.23)

Using the Lagrange multiplier λ, one obtains the standard necessary and sufficient condition for theminimizer hopt

Bh = λa, (h, a) = 1. (2.2.24)

Therefore h = λB−1a, λ = (B−1a, a)−1,

hopt = B−1a/(B−1a, a) (2.2.25)

is uniquely defined by (2.2.25), since B−1 is positive definite, and the denominator in (2.2.25) does notvanish. The minimum of Q(x) is

Qmin(x) = (Bhopt, hopt) = (B−1a, a)−1 := αn(x). (2.2.26)

We assume that ∫ 1

−1

αn(x)dx→ 0 as n→ ∞. (2.2.27)

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

‖fn − f‖2 ≤ ‖∫ 1

−1

n∑

j=1

hj(x)ψj(y)[f(y) − f(x)]dy‖2

≤∫ 1

−1

dx

∫ 1

−1

|An(x, y)|2(x− y)2dy

∫ 1

−1

|f(y) − f(x)|2(x− y)−2dy

≤∫ 1

−1

αn(x)dx sup−1≤x≤1

∫ 1

−1

∣∣∣∣f(y) − f(x)

x− y

∣∣∣∣2

dy → 0 as n → ∞.

(2.2.28)

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

sup−1≤x≤1

∫ 1

−1

∣∣∣∣f(x) − f(y)

x− y

∣∣∣∣2

dy < ∞. (2.2.29)


This inequality is satisfied if, for example, the derivative of f(x) exists except at a finite number ofpoints and is uniformly bounded.

Let us illustrate the assumption (2.2.27). Let ψp = (2π)−1 exp(ipπx), p = 0,±1, . . . ,±n. Then

ap =

∫ 1

−1

ψ∗p(x)dx =

sin(pπ)

pπ2=

0 p 6= 0,

π−1, p = 0

(B−1a, a) = π−2b(−1)00 , where bpj =

∫ 1

−1(x − y)2 expi(j − p)πydy, B−1 = (b(−1)pj ) =

(Bjp(det bpj)−1), where Bjp is the cofactor corresponding to the element bjp of the matrix (bpj).

One can show that (2.2.4) holds at any point at which f(x) is differentiable and αn(x) → 0 as n → ∞.Indeed, if we do not take the hopt but use h = (ψ1, . . . , ψn), then the error of the estimate will be notless that αn(x). On the other hand, for this choice of h, the kernel (2.2.8) is the Dirichlet kernel. Fromthe theory of the Fourier series one knows that (2.2.9) holds in L2 and fn(x) → f(x), n → ∞, at anypoint at which f(x) is differentiable.

In practice it is advisable to choose the system ψj in such a way that αn(x) tends rapidly to zero.Note that αn(x) depends only on the system ψj, and therefore we can control this quantity to someextent by choosing the system ψj.

Let us note that one can estimate f(x) at a given point x0 optimally using the same procedure. Inthis case the convergence condition (2.2.4) will hold for x0. If x0 is fixed, we can choose the system ψjso that

bpj = δpj =

0, p 6= j,

1, p = j.(2.2.30)

In this case

αn(x0) := αn = ‖a‖−2 =

n∑

j=1

∣∣∣∣∫ 1

−1

ψ∗j dy

∣∣∣∣2

−1

, (2.2.31)

and we can choose ψj so that, in addition to (2.2.29), the condition

‖a‖ → ∞ as n→ ∞ (2.2.32)

is satisfied. For example, take x0 = 0, then (2.2.29) reduces to

∫ 1

−1

y2ψp(y)ψ∗j (y)dy = δpj , (2.2.33)

and one can choose ψp(y) which behave nearly like |y|−1 in a small neighborhood of y = 0. Then∫ 1

−1ψj(y)dy can be made very large, and αn → 0 as n→ ∞ 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 σ2 > 0, because there may be no h which satisfies both restrictions of (2.2.18).Since the set h : (h, a) = 1, (Ch, h) ≤ σ2 is convex and Q(h) is a strictly convex function of h, it isclear that the solution to (2.2.18) is unique when it exists. For the solution to exist it is necessary andsufficient 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 (Ch, h) < σ2

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

(CB−1a,B−1a) · (B−1a, a)−2 < σ2. (2.2.34)

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

(Ch, h) = σ2 (2.2.35)


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

Bh − λCh− µa = 0, (h, a) = 1, (Ch, h) = σ2, (2.2.36)

where λ and µ are the Lagrange multipliers. It follows from (2.2.36) that

(Bh, h) = λ∗σ2 + µ∗, h = µ(B − λC)−1a, (2.2.37)

µ = ((B − λC)−1a, a)−1. (2.2.38)

Taking the complex conjugate in the first equation (36) we see that

(Bh, h) = λσ2 + µ (2.2.39)

From (2.2.37) and (2.2.38) one gets

h = (B − λC)−1a/((B − λC)−1a, a). (2.2.40)

Substituting (2.2.40) into(Ch, h) = σ2 (2.2.41)

yields an equation for λ:

(C(B − λC)−1a, (B − λC−1a) = σ2((B − λC−1a, a)2. (2.2.42)

The roots of Eq. (2.2.42), give µ by formula (2.2.38), and h by formula (2.2.37). Finally choose the hopt

for which (Bh, h) = min. This hopt solves problem (2.2.18).2.2.4.5. One can simplify the solution to problem (2.2.18) in the following way. If (Ch, h) = σ2 the

problem (2.2.18) takes the form

(Bh, h) = min, (Ch, h) = σ2, (h, a) = 1. (2.2.43)

Let us choose the coordinate system so that

aj = δjnan, δjn =

0, j 6= n,

1, j = n,(2.2.44)

and normalize ψ so thatan = 1. (2.2.45)

In this case (2.2.43) can be written as

(βH,H) + 2Re(H, βn) + bnn = min,

(γH,H) + 2Re(H, γn) + Cnn = σ2,(2.2.46)

whereH = (h1, . . . , hn−1), hn = 1;

βn = (bn1, . . . , bnn−1);

γn = (Cn1, . . . , Cnn−1).

βjp = bjp,

γpj = Cpj,

1 ≤ j, p ≤ n− 1,

1 ≤ j, p ≤ n− 1, (2.2.47)

Thus, problem (2.2.43) in Cn with two constraints is reduced to problem (2.2.46) in Cn−1 with oneconstraint. Problem (2.2.46) can be solved by the Lagrange multipliers method. One has

βH + βn − νγH − νγn = 0, (2.2.48)


where ν is the Lagrange multiplier. Thus,

H = −(β − νγ)−1(βn − νγn). (2.2.49)

Substitute (2.2.49) into the constraint equation (2.2.46) to obtain an equation for ν. If this equation issolved then (2.2.49) gives the corresponding H. If there are several solutions then the Hopt is the onethat 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 aconvergent iterative method.

Proof. It was proved in Section 2.1.6 that if A is a bounded linear operator and equation (1.3.1) issolvable, then it is equivalent to the equation (2.1.6). By y we denote the minimal norm solutionto (1.3.1), i.e., the solution orthogonal to N (A), the null-space of A. Note that N (B) = N (A). Withoutloss of generality assume ‖B‖ ≤ 1 (if ‖B‖ > 1, then one can divide by ‖B‖ equation (2.1.6)). Considerthe iterations

un+1 = un − (Bun − f1), u0 = u0, f1 := A∗f, B := A∗A, (2.3.1)

where u0 is arbitrary. Denote un − u := wn, and write (2.3.1) as wn+1 = (I − B)wn, w0 := u0 − u.

Thus, wn+1 = (I −B)n+1w0. Since 0 ≤ B ≤ I, one gets:

∥∥wn+1

∥∥2=

∫ 1

0

(1 − λ)2n+2d(Eλw0, w0

),

where Eλ is the resolution of the identity of B. Thus, limn→∞ ‖wn+1‖2 = ‖Pw0‖2, where P is theorthoprojector onto N (B) = N (A). If one takes u0 ⊥ N (A), and u = y, y ⊥ N (A), then Pwn+1 = 0 ∀nby induction, and limn→∞ ‖wn‖ = 0. In particular, if u0 = 0, then un ⊥ N (A) ∀n (by induction, sinceu1 = f1 = A∗f ⊥ N (A)), and limn→∞ ‖un − y‖ = 0, y = A+f , where A+ is the pseudoinverse of A,defined in Section 2.1.1. 2

Exercise 2.3.2. Prove that if f /∈ D(A+) in (1.3.1), then ‖un‖ → ∞ in (2.3.1).

Assume now that fδ is given in place of f, ‖fδ−f‖ ≤ δ. Let us show that if one stops iterations (2.3.1),with fδ in place of f and u0 = 0, at n = n(δ), then uδ := un(δ),δ → y as δ → 0, if n(δ) is properlychosen, and un,δ is defined by (2.3.1). Let un − un,δ := wn. Then wn+1 = wn − (Bwn − g), g := f − fδ,‖g‖ ≤ δ, and w0 = 0. Thus, wn+1 =

∑nj=0(I − B)jg, ‖wn+1‖ ≤ δn, because ‖I − B‖ ≤ 1 if 0 ≤ B ≤ I,

and we had assumed ‖B‖ ≤ 1, which, together with B ≥ 0, implies 0 ≤ B ≤ I.

Therefore, if n(δ) → ∞ is chosen so that limδ→0 δn(δ) = 0, then limδ→0 ‖uδ − y‖ = 0. Indeed,‖uδ − y‖ ≤ ‖uδ − un(δ)‖ + ‖un(δ) − y‖. We have proved in Theorem 2.3.1 that limδ→0 ‖un(δ) − y‖ = 0,and ‖uδ − un(δ)‖ ≤ δn(δ) → 0 as δ → 0. Let us summarize the result.

Theorem 2.3.3. If ‖fδ − f‖ ≤ δ, limδ→0 n(δ) = ∞, limδ→0 δn(δ) = 0, then limδ→0 ‖uδ − y‖ = 0, wherey is the minimal norm solution to (1.3.1), uδ := un(δ),δ, and un,δ is obtained by iterations (2.3.1), withfδ in place of f and u0 = 0.

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

2.4. DYNAMICAL SYSTEM METHOD (DSM) 39

2.4 Dynamical system method (DSM)

2.4.1 The idea of the DSM.

Consider the equation:

F (u) = 0, F (u) := B(u) − f. (2.4.1)

We assume in Section 2.4 that F ∈ C2loc, that is,

supu∈B(u0,R)

∥∥F (j)(u)∥∥ ≤ Mj(R), j = 1, 2, (2.4.2)

where y is the minimal-norm solution to (2.4.1), B(u0, R) := u : ‖u−u0‖ ≤ R, F (y) = 0, y ∈ B(u0, R),F : H → H, H is a real Hilbert space. Many of our results hold in reflexive Banach spaces X andF : X → X∗, but we do not go into detail. The element u0 in (2.4.2) will be specified later. InSection 2.4 we will call (2.4.1) a well-posed problem if

supu∈B(u0 ,R)

∥∥∥[F ′(u)

]−1∥∥∥ ≤ m(R), (2.4.3)

and ill-posed if F ′(u) is not boundedly invertible. We assume existence of a solution to (2.4.1) unlessotherwise 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 suchan approach, which we call the Dynamical Systems Method (DSM). The DSM consists of finding anonlinear locally Lipschitz operator Φ(t, u), such that the Cauchy problem:

u = Φ(t, u), u(0) = u0, (2.4.4)

has the following three properties:

∃!u(t) ∀t ≥ 0, ∃u(∞), F (u(∞)) = 0, (2.4.5)

that is, (2.4.4) is globally uniquely solvable, its unique solution has a limit at infinity u(∞), and thislimit solves (2.4.1).

Our first motivation is to develop a general approach to solving equation (2.4.1), especially nonlinearand ill-posed. Our second motivation is to develop a general approach to constructing convergent iterativemethods 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 f ∈ R(A), and also for fδ /∈ R(A),‖fδ − f‖ ≤ δ.

(3) For ill-posed problem with monotone nonlinear A, for f ∈ R(A), and also for fδ /∈ R(A), ‖fδ−f‖ ≤δ,

(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 arepresented in subsections below.


2.4.2 DSM for well-posed problems

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

(F ′(u)Φ(t, u), F (u)

)≤ −g1(t)

∥∥F (u)∥∥a ∀u ∈ B

(u0, R

),

∫ ∞

0

g1dt = ∞, (2.4.6)

where g1 > 0 is an integrable function, a > 0 is a constant. Assume∥∥Φ(t, u)

∥∥ ≤ g2(t)∥∥F (u)

∥∥, ∀u ∈ B(u0, R

), (2.4.7)

where g2 > 0 is such that

G(t) := g2(t) exp

(−∫ t

0

g1ds

)∈ L1

(R+

). (2.4.8)

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

∥∥Φ(t, u)∥∥ ≤ g2(t)

∥∥F (u)∥∥b ∀u ∈ B, (2.4.7’)

where b > 0 is a constant. The statement and proof of Theorem 2.4.2 can be easily adjusted to thisassumption.

Our first basic result is the folowing:

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

∥∥∥F(u0

)∥∥∫ ∞

0

G(t)dt ≤ R, a = 2, (2.4.9)

then (2.4.4) has a global solution, (2.4.5) holds, (2.4.1) has a solution y = u(∞) ∈ B(u0, R), and

∥∥u(t) − y∥∥ ≤

∥∥F (u0)∥∥∫ ∞

t

G(x)dx,∥∥F (u(t))

∥∥ ≤∥∥F (u0)

∥∥ exp

(−∫ t

0

g1(x)dx

). (2.4.10)

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

∥∥∥F(u0

)∥∥∥∫ T

0

g2ds ≤ R, (2.4.11)

where T > 0 is defined by the equation∫ T

0

g1(s)ds =∥∥∥F(u0

)∥∥∥2−a

/(2 − a), (2.4.12)

then (2.4.4) has a global solution, (2.4.5) holds, (2.4.1) has a solution y = u(∞) ∈ B(u0, R), andu(t) = y for t ≥ T .

(iii) If (2.4.6) – (2.4.8) hold, a > 2, and∫ ∞

0

g2(s)h(s)ds ≤ R, (2.4.13)

where [∥∥∥F(u0

)∥∥∥2−a

+ (a− 2)

∫ t

0

g1(s)ds

] 12−a

:= h(t), limt→∞

h(t) = 0, (2.4.14)

then (2.4.4) has a global solution, (2.4.5) holds, (2.4.1) has a solution y = u(∞) ∈ B(u0, R), and

∥∥u(t) − u(∞)∥∥ ≤

∫ ∞

t

g2(s)h(s)ds −→ 0

as t→ ∞.


Let us sketch the proof.

Proof of Theorem 2.4.2. The assumptions about Φ imply local existence and uniqueness of the solutionu(t) to (2.4.4). To prove global existence of u, it is sufficient to prove a uniform with respect to t boundon ‖u(t)‖. Indeed, if the maximal interval of the existence of u(t) is finite, say [0, T ), and Φ(t, u) islocally Lipschitz with respect to u, then ‖u(t)‖ → ∞ as t→ T .

Assume a = 2. Let g(t) := ‖F (u(t))‖. Since H is real, one uses (2.4.4) and (2.4.6) to getgg = (F ′(u)u, F ) ≤ −g1(t)g2, so g ≤ −g1(t)g, and integrating this inequality one gets the secondinequality (2.4.10), because g(0) = ‖F (u0)‖. Using (2.4.7), (2.4.4) and the second inequality (2.4.10),one gets:

∥∥u(t) − u(s)∥∥ ≤ g(0)

∫ t

s

G(x)dx, G(x) := g2(x) exp

(−∫ x

0

g1(z)dz

). (2.4.10’)

Because G ∈ L1(R+), it follows from (2.4.10’) that the limit y := limt→∞ u(t) = u(∞) exists, andy ∈ B by (2.4.9). From the second inequality (2.4.10) and the continuity of F one gets F (y) = 0, soy solves (2.4.1). Taking t → ∞ and setting s = t in (2.4.10’) yields the first inequality (2.4.10). Theinclusion u(t) ∈ B for all t ≥ 0 follows from (2.4.9) and (2.4.10’). The first part of Theorem 2.4.2 isproved. The proof of the other parts is similar. 2

There are many applications of this theorem. We mention just a few, and assume that g1 = c1 =const > 0 and g2 = c2 = const > 0.

Example 2.4.3. [Continuous Newton-type method]: Φ = −[F ′(u)]−1F (u). Assume that (2.4.3) holds,then c1 = 1, c2 = m1, (2.4.9) takes the form (*) m1(R)‖F (u0)‖ ≤ R, and (*) implies that (2.4.4) has aglobal solution, (2.4.5) and (2.4.10) hold, and (2.4.1) has a solution in B(u0, R). This result belongs toGavurin ([Gav]).

Example 2.4.4. [Continuous simple iterations method:] Let Φ = −F , and assume F ′(u) ≥ c1(R) > 0for all u ∈ B(u0, R). Then c2 = 1, c1 = c1(R), (2.4.9) is: [c1(R)]−1‖F (u0)‖ ≤ R, and the conclusionsof Example 1 hold.

Example 2.4.5. [Continuous gradient method:] Let Φ = −[F ′]∗F , (2.4.2) and (2.4.3) hold, c1 = m−21 ,

c2 = M1(R), (2.4.9) is (**) M1m21‖F (u0)‖ ≤ R, and (**) implies the conclusions of Example 2.4.3.

Example 2.4.6. [Continuous Gauss-Newton method:] Let Φ = −([F ′]∗F ′)−1[F ′]∗F , (2.4.2) and (2.4.3)hold, c1 = 1, c2 = m2

1M1, (2.4.9) is (***) M1m21‖F (u0)‖ ≤ R, and (***) implies the conclusions

of Example 2.4.3.

Example 2.4.7. [Continuous modified Newton method:] Let Φ = −[F ′(u0)]−1F (u). Assume ‖[F ′(u0)]

−1‖ ≤m0, and let (2.4.2) hold. Then c2 = m0. Choose R = (2M2m0)

−1, and c1 = 0.5. Then (2.4.9)is 2m0‖F (u0)‖ ≤ (2M2m0)

−1, that is, 4m20M2‖F (u0)‖ ≤ 1. Thus, if 4m2

0M2‖F (u0)‖ ≤ 1, then theconclusions of Example 2.4.3 hold.

Example 2.4.8. [Descent methods.] Let Φ = − f(f ′ ,h)h, where f = f(u(t)) is a differentiable functional

f : H → [0,∞), and h is an element of H. Assume that |(f ′, h)| 6= 0. Then (f ′,Φ)f ≤ −f2, sothat c1 = 1. From (2.4.4) one gets f = (f ′, u) = −f . Thus, f = f0e

−t, where f0 := f(u0). Assume‖Φ‖ ≤ c2|f |b, b > 0. Then ‖u‖ ≤ c2|f0|be−bt. Therefore u(∞) does exist, f(u(∞)) = 0, and ‖u(∞) −u(t)‖ ≤ ce−bt, c = const > 0.

If h = f ′, and f = ‖F (u)‖2, then f ′(u) = 2[F ′]∗(u)F (u), Φ = − f‖f ′‖2 f

′, and (2.4.4) is a descent

method. For this Φ one has c1 = 1, and c2 = 1inft≥0 ||f ′(u(t))|| := m1. Condition (2.4.9) is: m1‖F (u0)‖ ≤ R.

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 Φ 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 asfollows. Consider (2.4.4) with Φ = −[F ′(u)]−1F (u), and assume that (2.4.4) is locally solvable and||[F ′(u(t))]−1|| ≤ a(t), where u(t) solves (2.4.4). Let g(t) := ||F (u(t))||. Then gg = (F ′(u(t))u, F ) =−g2, so g(t) = g0e

−t, and ||u|| ≤ g0a(t)e−t. Assume that a(t)e−t ∈ L1(0,∞). Then u(∞) does exist and

||u(t) − u(∞)|| ≤ g0∫∞ta(s)e−sds → 0 as t → ∞. Therefore F (u(∞)) = limt→∞ F (u(t)) = 0, so u(∞)

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 F (u) = eu and H is the real line with the usual

product of real numbers as the inner product and |u| := ||u||, then condition (2.4.7) is not satisfied andequation (2.4.1), i.e. eu = 0, 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 limsupR→∞

Rm(R) = ∞. 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

Au = f. (2.4.15)

Let us denote by (A) the folowing assumption:

(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 y such that Ay = f , y ⊥ N , where N is the null-spaceof A.

Let B := A∗A, q := By = A∗f , A∗ is the adjoint of A. Every solution to (2.4.15) solves

Bu = q, (2.4.16)

and, if f = Ay, then every solution to (2.4.16) solves (2.4.15). Choose a continuous function ε(t) > 0,monotonically decaying to zero on R+. Sometimes it is convenient to assume that

limt→∞

(εε−2

)= 0. (2.4.17)

For example, the functions ε = c1(c0 + t)−b, 0 < b < 1, where c0 and c1 are positive constants, satisfy(2.4.17). There are many such functions. One can prove the following:

Claim 3. If ε(t) > 0 is a continuous monotonically decaying function on R+, limt→∞ ε(t) = 0, and(2.4.17) holds, then ∫ ∞

0

εds = ∞. (2.4.17’)

In this Section we do not use assumption (2.4.17): in the proof of Theorem 2.4.9 one uses only themonotonicity of a continuous function ε > 0 and (2.4.17’). One can drop assumption (2.4.17’), but thenconvergence is proved in Theorem 2.4.9 to some element of N , not necessarily to the normal solution y,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

Φ := −[Bu + ε(t)u − q

], Φδ = −

[Buδ + ε(t)uδ − qδ

], (2.4.18)


where ‖q − qδ‖ ≤ ‖A∗‖δ := Cδ. Without loss of generality one may assume that C = ‖A∗‖ = 1, whichwe do in what follows. Our main result is Theorem 2.4.9, stated below. It yields the following:

Conclusion: Given noisy data fδ , 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 hereis different and much shorter.

Theorem 2.4.9. Problem (2.4.4) with Φ from (2.4.18) has a unique global solution u(t), (2.4.5) holds,and u(∞) = y. Problem (2.4.4) with Φδ from (2.4.18) has a unique global solution uδ(t). There existstδ, such that

limδ→0

∥∥uδ(tδ) − y∥∥ = 0. (2.4.19)

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

ε(t) = δb, b ∈ (0, 1), (2.4.20)

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 Φ = −[Bu + ε(t)u − q], then the solution u to (2.4.4) is

u(t) = h−1(t)U (t)u0 + h−1(t)

∫ ‖B‖

0

exp(−tλ)∫ t

0

esλh(s)dsλdEλy, (2.4.21)

where h(t) := exp(∫ t0ε(s)ds) → ∞ as t → ∞, Eλ is the resolution of the identity corresponding to the

selfadjoint operator B, and U (t) := e−tB is a nonexpansive operator, because B ≥ 0. Actually, (2.4.21)can be used also when B is unbounded, ‖B‖ = ∞.

Using L’Hospital’s rule one checks that

limt→∞

λ∫ t0esλh(s)ds

etλh(t)= lim

t→∞λetλh(t)

λetλh(t) + etλh(t)ε(t)= 1 ∀λ > 0, (2.4.22)

provided only that ε(t) > 0 and limt→∞ ε(t) = 0. From (2.4.21), (2.4.22), and the Lebesgue dominatedconvergence theorem, one gets u(∞) = y − Py, where P is the orthogonal projection operator ontothe null-space of B. Under our assumptions (A), Py = 0, so u(∞) = y. If v(t) := ‖u(t) − y‖, thenlimt→∞ v(t) = 0. In general, the rate of convergence of v to zero can be arbitrarily slow for a suitablychosen f . Under an additional a priori assumption on f (for example, the source-type assumptions whichsay that u ∈ R(B)), this rate can be estimated.

Let us describe a method for deriving a stopping rule. One has:

‖uδ(t) − y‖ ≤ ‖uδ(t) − u(t)‖ + v(t). (2.4.23)

Since limt→∞ v(t) = 0, any choice of tδ such that

limtδ→∞

‖uδ(tδ) − u(tδ)‖ = 0, (2.4.24)

gives a stopping rule: for such tδ one has limδ→0 ‖uδ(t) − y‖ = 0.To prove that (2.4.20) gives such a rule, it is sufficient to check that

‖uδ(t) − u(t)‖ ≤ δ

ε(t). (2.4.25)

Let us prove (2.4.25). Denote w := uδ − u, p := qδ − q. Then

w = −[Bw + εw − p], w(0) = 0, ‖p‖ ≤ δ. (2.4.26)


Integrating (2.4.26), and using the property B ≥ 0, one gets (2.4.25).Alternatively, multiply (2.4.26) by w, let ‖w‖ := g, use B ≥ 0, and get g ≤ −ε(t)g + δ, g(0) = 0.

Thus, g(t) ≤ δ exp(−∫ t0εds)

∫ t0

exp(∫ s0εdτ )ds ≤ δ

ε(t) . A more precise estimate, used at the end of the

proof of Theorem 2.4.10 below, yields:

‖uδ(t) − u(t)‖ ≤ δ

2√ε(t)

,

and the corresponding stopping time tδ can be taken as the root of the equation:

2√ε(t) = δb, b ∈ (0, 1). (2.4.20’)

Theorem 2.4.9 is proved. 2

If the rate of decay of v is known, then a more efficient stopping rule can be derived: tδ is theminimizer of the problem:

v(t) + δ[ε(t)]−1 = min . (2.4.27)

For example, if v(t) ≤ cεa(t), then tδ is the root of the equation ε(t) = ( δca

)1

1+a , that one gets from(2.4.27) with v = cεa.

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 2.4.12and 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

Φ = −(B + ε(t))−1[Bu+ ε(t)u − q] = −u+ (B + ε(t))−1q, (2.4.28)

and for noisy data it uses (2.4.4) with Φδ = −uδ+(B+ε(t))−1qδ. The linear operator B ≥ 0 is monotone,so Theorem 2.4.12 is applicable. For exact data, (2.4.4) with Φ, defined in (2.4.28), yields:

u = −u+ (B + ε(t))−1q, u(0) = u0, (2.4.29)

and (2.4.5) holds if ε(t) > 0 is monotone, continuous, decreasing to 0 as t→ ∞.Let us formulate the result:

Theorem 2.4.10. Assume (A), and let B := A∗A, q := A∗f . Assume ε(t) > 0 to be a continuous,monotonically decaying to zero function on [0,∞). Then, for any u0 ∈ H, problem (2.4.29) has a uniqueglobal solution, ∃u(∞) = y, Ay = f , and y is the minimal-norm solution to (2.4.15). If fδ is given inplace of f , ‖f − fδ‖ ≤ δ, then (2.4.19) holds, with uδ(t) solving (2.4.29) with q replaced by qδ := A∗fδ,and tδ is chosen, for example, as the root of (2.4.20’) (or by a discrepancy principle).

Proof of Theorem 2.4.10. One has q = Bz, where Az = f , and the solution to (2.4.29) is

u(t) = u0e−t + e−t

∫ t

0

es(B + ε(s))−1Bzds := u0e−t +

∫ ‖B‖

0

j(λ, t)dEλz (2.4.30)

where

j(λ, t) :=

∫ t

0

λes

[λ+ ε(s)]etds, (2.4.31)

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

0 ≤ j(λ, t) ≤ 1, limt→∞

j(λ, t) = 1 λ > 0, j(0, t) = 0. (2.4.32)


From (2.4.30)–(2.4.32) it follows that ∃u(∞), u(∞) = z−PNz = y, where y is the minimal-norm solutionto (2.4.15), N := N (B) = N (A) is the null-space of B and of A, and PN is the orthoprojector onto Nin H. This proves the first part of Theorem 2.4.10.

To prove the second part, denote w := uδ−u, g := fδ − f , where we dropped the dependence on δ inw and g for brevity. Then w = −w+ (B + ε(t))−1A∗g, w(0) = 0. Thus w = e−t

∫ t0es(B+ ε(s))−1A∗gds,

so ‖w‖ ≤ δ e−t∫ t0

es

2√ε(s)

ds ≤ δ

2√ε(t)

, where the known estimate was used: ‖(B + ε)−1A∗‖ ≤ 12√ε.


2.4.4 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 theresult 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

Φ = −A−1ε(t)

(u)[F (u(t)) + ε(t)(u(t) − u0)

], (2.4.33)

where A = A(u) := F ′(u), A∗ is its adjoint, ε(t) is the same as in Theorem 2.4.10, and in Theorem 2.4.12ε(t) is further specified, u0 ∈ B(u0, R) is an element we can choose to improve the numerical performanceof the method. If noisy data are given, then, as in Section 3.3, we take

F (u) := B(u) − f, Φδ = −A−1ε(t)(uδ)

[B(uδ(t)) − fδ + ε(t)(uδ(t) − u0)

],

where ‖fδ − f‖ ≤ δ, B is a monotone nonlinear operator, B(y) = f , and uδ solves (2.4.4) with Φδ inplace of Φ.

To prove that (2.4.33) with the above Φ has a global solution and (2.4.5) holds, we use the following:

Theorem 2.4.11. Let γ(t), σ(t), β(t) ∈ C[t0,∞) for some real number t0 ≥ 0. If there exists a positivefunction µ(t) ∈ C1[t0,∞) such that

0 ≤ σ(t) ≤ µ(t)

2[γ(t) − µ(t)

µ(t)], β(t) ≤ 1

2µ(t)[γ(t) − µ(t)

µ(t)], g0µ(t0) < 1, (2.4.34)

where g0 is the initial condition in (2.4.35), then a nonnegative solution g to the following differentialinequality:

g(t) ≤ −γ(t)g(t) + σ(t)g2(t) + β(t), g(t0) = g0 ≥ 0, (2.4.35)

exists for all t ≥ t0 and satisfies the estimate:

0 ≤ g(t) ≤ 1 − ν(t)

µ(t)<

1

µ(t), (2.4.36)

for all t ∈ [t0,∞), where

0 < ν(t) =

(1

1 − µ(t0)g(t0)+

1

2

∫ t

t0

(γ(s) − µ(s)

µ(s)

)ds

)−1

. (2.4.37)

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 upin a finite time, in general. Conditions (2.4.34) guarantee the global existence of the solution to thisRiccati equation with the initial condition (2.4.35). Secondly, this Riccati differential equation cannotbe integrated analytically by separation of variables. Thirdly, the coefficient σ(t) may grow to infinity ast → ∞, so that the quadratic term does not necessarily have a small coefficient, or the coefficient smaller


than γ(t). Without loss of generality one may assume β(t) ≥ 0 in Theorem 2.4.11. This Theorem isproved in Section 2.4.9 below.

The main result in this Section is new. It claims a global convergence in the sense that no assumptionson the choice of the initial approximation u0 are made. Usually one assumes that u0 is sufficiently closeto the solution of (2.4.1) in order to prove convergence. We take in Theorem 2.4.12 u0 = 0 because inthis theorem u0 does not play any role. The proof is valid for any choice of u0, but then the definitionof r in Theorem 2.4.12 is changed.

Theorem 2.4.12. If (2.4.2) holds, u0 = 0, R = 3r, where r := ‖y‖ + ‖u0‖, and y ∈ N := z :F (z) = 0 is the (unique) minimal norm solution to (2.4.1), then, for any choice of u0, problem (2.4.4)with Φ defined in (2.4.33), u0 = 0, and ε(t) = c1(c0 + t)−b with some positive constants c1, c0, andb ∈ (0, 1), specified in the proof of Theorem 2.4.12, has a global solution, this solution stays in the ballB(u0, R) and (2.4.5) holds. If uδ(t) solves (2.4.4) with Φδ in place of Φ, then there is a tδ such thatlimδ→0 ‖uδ(tδ) − y‖ = 0.

Proof of Theorem 2.4.12. Let us sketch the steps of the proof. Let V solve the equation

F (V ) + ε(t)V = 0. (2.4.38)

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) supt≥0 ‖V ‖ ≤ ‖y‖.Indeed, (F (V ) − F (y), V − y) ≥ 0, F (y) = 0, ε > 0, so (V, V − y) ≤ 0. This implies (ii).

If F is Frechet differentiable, then V is differentiable, and ‖V (t)‖ ≤ ‖y‖|ε(t)|/ε(t). It is alsoknown (see Section 2.1.4) that if F (y)=0, where y is the minimal-norm solution to (2.4.1), thenlimt→∞ ‖V (t) − y‖ = 0.

We will show that the global solution u to (2.4.4), with the Φ from (2.4.33), does exist, and limt→∞ ‖u(t)−V (t)‖ = 0. This is done by deriving a differential inequality for w := u − V , and by applying Theo-rem 2.4.11 to g = ‖w‖. Since ‖u(t)− y‖ ≤ ‖u(t)−V (t)‖+‖V (t)− y‖, one obtains (2.4.5). We also checkthat u(t) ∈ B(u0, R), where R := 3(‖y‖ + ‖u0‖), for any choice of u0 and a suitable choice of ε.

Let us derive the differential inequality for w. One has

w = −V − A−1ε(t)(u)

[F(u(t)

)− F

(V (t)

)+ ε(t)w

], (2.4.39)

and F (u) − F (V ) = Aw + K, where ‖K‖ ≤ M2g2/2, g := ‖w‖ and M2 is the constant from (2.4.2).

Multiply (2.4.39) by w, use the monotonicity of F , that is, the property A ≥ 0, and the estimate‖V ‖ ≤ ‖y‖|ε|/ε, and get:

g ≤ −g +0.5Mg2

ε+ ‖y‖ |ε|

ε, (2.4.40)

where M := M2. Inequality (2.4.40) is of the type (2.4.35): γ = 1, σ = 0.5M/ε, β = ‖y‖ |ε|ε

. Choose

µ(t) =2M

ε(t). (2.4.41)

Clearly µ → ∞ as t → ∞. Let us check three conditions (2.4.34). One has µ(t)µ(t) = |ε|

ε . Take ε =

c1(c0 + t)−b, where cj > 0 are constants, 0 < b < 1, and choose these constants so that |ε|ε < 1

2 , for

example, bc0

= 14 . Then the first condition (2.4.34) is satisfied. The second condition (2.4.34) holds if

8M‖y‖ |ε|ε−2 ≤ 1. (2.4.42)


One has ε(0) = c1c−b0 . Choose

ε(0) = 4Mr. (2.4.43)

Then

|ε|ε−2 = bc−11 (c0 + t)b−1 ≤ bc−1

0 c−11 cb0 =

1

4ε(0)=

1

16Mr, (2.4.44)

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

2M‖u0 − V0‖ε(0)

≤ 2Mr

4Mr=

1

2< 1.

By Theorem 2.4.11 one concludes that g = ‖w(t)‖ < ε(t)2M → 0 when t → ∞, and

‖u(t) − u0‖ ≤ g + ‖V − u0‖ ≤ g(0) + r ≤ 3r. (2.4.45)

This estimate implies the global existence of the solution to (2.4.4), because if u(t) had a finite maximalinterval of existence, [0, T ), then u(t) could not stay bounded when t → T , which contradicts theboundedness of ‖u(t)‖, and from (2.4.45) it follows that ‖u(t)‖ ≤ 4r. We have proved the first part ofTheorem 2.4.12, namely properties (2.4.5).

To derive a stopping rule we argue as in Section 2.4. One has:

∥∥uδ(t) − y∥∥ ≤

∥∥uδ(t) − V (t)∥∥+

∥∥V (t) − y∥∥. (2.4.46)

We have already proved that limt→∞ v(t) := limt→∞ ‖V (t) − y‖ = 0. The rate of decay of v can bearbitrarily slow, in general. Additional assumptions, for example, the source-type ones, can be used toestimate the rate of decay of v(t). One derives differential inequality (2.4.35) for gδ := ‖uδ(t) − V (t)‖,and estimates gδ using (2.4.36). The analog of (2.4.40) for gδ contains additional term δ

εon the right-

hand side. If δε2 ≤ 1

16M , then conditions (2.4.34) hold, and gδ <ε(t)2M . Let tδ be the root of the equation

ε2(t) = 16Mδ. Then limδ→0 tδ = ∞, and limδ→0 ‖uδ(tδ) − y‖ = 0 because ‖uδ(tδ) − y‖ ≤ v(tδ) + gδ,limtδ→∞ gδ(tδ) = 0 and limtδ→∞ v(tδ) = 0, but the convergence can be slow. See [ARS3], [KNR] for therate of convergence under source assumptions. If the rate of decay of v(t) is known, then one chooses tδas the minimizer of the problem, similar to (2.4.27),

v(t) + gδ(t) = min, (2.4.47)

where the minimum is taken over t > 0 for a fixed small δ > 0. This yields a quasioptimal stopping rule.Theorem 2.4.12 is proved. 2

Let us give another result:

Theorem 2.4.13. Assume that Φ = −F (u) − ε(t)u, F is monotone, ε(t) as in Theorem 2.4.10, and(2.4.17), and (2.4.2) hold. Then (2.4.5) holds.

Proof of Theorem 2.4.13. As in the proof of Theorem 2.4.12, it is sufficient to prove that limt→∞ g(t) = 0,where g, w, and V are the same as in Theorem 2.4.12, and u solves (2.4.4) with the Φ defined inTheorem 2.4.13. Similarly to the derivation of (2.4.39), one gets:

w = −V −[F (u)− F (V ) + ε(t)w

]. (2.4.48)

Multiply (2.4.48) by w, use the monotonicity of F and the estimate ‖V ‖ ≤ |ε(t)|ε(t) ‖y‖, which was used

also in the proof of Theorem 2.4.12, and get:

g ≤ −ε(t)g +|ε(t)|ε(t)

‖y‖. (2.4.49)


This implies

g(t) ≤ e−∫ t0ε(s)ds

[g(0) +

∫ t

0

e∫ s0ε(x)dx |ε(s)|

ε(s)‖y‖ds

]. (2.4.50)

From our assumptions relation (2.4.17’) follows, and (2.4.50) together with (2.4.17) imply limt→∞ g(t) =0. Theorem 2.4.13 is proved. 2

Remark 2.4.14. One can drop the smoothness of F assumption (2.4.2) in Theorem 2.4.13 and assumeonly that F is a monotone hemicontinuous operator defined on all of H.

Claim 4. If ε(t) = ε = const > 0, then limε→0 ‖u(tε) − y‖ = 0, where u(t) solves (2.4.4) with Φ :=−F (u) − εu, and tε is any number such that limε→0 εtε = ∞.

Proof of the claim. One has ‖u(t)−y‖ ≤ ‖u(t)−Vε‖+‖Vε−y‖, where Vε solves (2.4.38) with ε(t) = ε =const > 0. Under our assumptions on F , equation (2.4.38) has a unique solution, and limε→0 ‖Vε−y‖ = 0.So, to prove the claim, it is sufficient to prove that limε→0 ‖u(tε)−Vε‖ = 0, provided that limε→0 εtε = ∞.Let g := ‖u(t)−Vε‖, and w := u(t)−Vε. Because Vε = 0, one has the equation: w = −[F (u)−F (Vε)+εw].Multiplying this equation by w, and using the monotonicity of F , one gets g ≤ −εg, so g(t) ≤ g(0)e−εt.Therefore limε→0 g(tε) = 0, provided that limε→0 εtε = ∞. The claim is proved. 2

Remark 2.4.15. One can prove claims (i) and (ii), formulated below formula (2.4.38), using DSMversion presented in Theorem 2.4.20 below.

Claim 5. Assume that F is monotone, (2.4.2) holds, and F (y) = 0. Then claims (i) and (ii), formulatedbelow formula (2.4.38), hold.

Proof. First, note that (ii) follows easily from (i), because the assumptions F (y) = 0, F is monotone,and ε > 0, imply, after multiplying F (V )−F (y) + εV = 0 by V − y, the inequality (V, V − y) ≤ 0, fromwhich claim (ii) follows. Claim (i) follows from Theorem 2.4.20, proved below. 2

Claim 6. Assume that the operator F is monotone, hemicontinuous, defined on all of H, equationF (u) = 0 has a solution, possibly non-unique, y is the minimal-norm element of NF := z : F (z) = 0,Φ = −F (u) − ε(t)u, and ε = c1(c0 + t)−b, 0 < b < 1, where c0 > 0, c1 > 0 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 supt≥0 ||u(t)|| < ∞ for the solution to (2.4.4) with Φ(t, u) := −F (u) − ε(t)u, where

0 < ε(t) 0,∫∞0 ε(s)ds = ∞. Inequality supt≥0 ||u(t)|| < ∞ implies that from any sequence tn → ∞

one can select a subsequence, denoted again tn, such that u(tn) := un v, where v ∈ H is some element.We prove that F (v) = 0.

2) we prove that the solution u(t) to the equation

u = −F (u) − ε(t)u, u(0) = u0, (∗)

satisfies the relation: g(t) := ||u(t+h)−u(t)|| → 0 as t → ∞, where h > 0 is an arbitrary fixed number.3) we prove that ||u|| = O( 1

t ) as t→ ∞.4) passing to the limit tn → ∞ in equation (*) yields F (v) = 0. We prove that u(t) → v as t → ∞,

and v = y.Let us give the details of the proof.

1) Let F (y) = 0, u− y := w, where u solves (*).Then w = −[F (u)− F (w)]− ε(t)w − ε(t)y. Multiplythis equation by w, use the monotonicity of F , denote ||w|| := z(t), and get zz ≤ −ε(t)z2 + ε(t)||y||z.Because z ≥ 0, this implies z ≤ −ε(t)z + ε(t)||y||, so z(t) ≤ e−

∫t0ε(s)ds[z(0) +

∫ t0e∫

s0ε(s′)ds′ε(s)ds||y||].

This inequality implies supt≥0 ||u(t)||< c < ∞. Thus, u(tn) v as tn → ∞.


2) Denote v := u(t + h) − u(t), where u solves (*), and ||v|| := g(t). Then v = −[F (u(t + h)) −F (u(t))] − [ε(t + h)u(t + h) − ε(t)u(t)]. Multiply this equation by v and use the monotonicity of F toget: gg ≤ −ε(t)g2 + |ε(t)|h||u(t+ h)||g. Because g ≥ 0 and ||u(t+ h)|| < c, one gets

g ≤ −ε(t)g + |ε(t)|hc. (2.4.51)

This implies

g(t) ≤ e−∫

t0ε(s)ds[g(0) + hc

∫ t

0

e∫

s0ε(s′)ds′ |ε(s)|ds]. (2.4.52)

Under our assumptions about ε(t), one can check that∫ t0ε(s)ds = O(ta), as t → ∞, where 0 < a :=

1 − b < 1. Also e−∫ t0ε(s)ds

∫ t0 e

∫ s0ε(s′)ds′ |ε(s)|ds = O( 1

t ). Thus g(t) = O( 1t ) as t → ∞.

3) Denote G(t) := ||u||. Divide (2.4.52) by h and let h → 0. Then one gets G(t) = O( 1t ), so, as

t → ∞ one has ||u(t)|| = O( 1t).

4) Passing to the limit t = tn → ∞ in equation (*), yields F (v) = 0. The limit limtn→∞ F (u(tn)) =F (v) exists because ε(tn) → 0, supn ||u(tn)|| < ∞, limtn→∞ ||u(tn)|| = 0, and u(tn) v as tn → ∞,so that Lemma 2.1.2 implies F (v) = 0. Let us prove that u(tn) → v. Since u(tn) v, one gets||v|| ≤ lim infn→∞ ||u(tn)||. If lim supn→∞ ||u(tn)|| ≤ ||v||, then limn→∞ ||u(tn)|| = ||v||, and togetherwith the weak convergence u(tn) v this implies strong convergence u(tn) → v.

Let us prove that lim supn→∞ ||u(tn)|| ≤ ||v||. One has (F (u(tn))−F (v), u(tn)−v)+ε(tn)(u(tn), u(tn)−v) = −(u(tn), u(tn) − v). Since F is monotone, ||u(t)|| = O( 1

t ) and ||u(tn) − v|| ≤ c, it follows that(u(tn), u(tn) − v) ≤ c

tnε(tn) . Thus, lim supn→∞ ||u(tn)|| ≤ ||v||, because limn→∞ tnε(tn) = ∞.

Let us prove that v = y. Replacing v by y in the above argument yields (u(tn), u(tn) − y) ≤ ctnε(tn) ,

so ||v|| = limn→∞ ||u(tn)|| ≤ ||y||. Since y is the unique minimal-norm solution to (2.4.1) and v solves(2.4.1), it follows that v = y.

Since the limit limn→∞ u(tn) = v = y is the same for every subsequence tn → ∞, for which theweak limit of u(tn) exists, one concludes that the strong limit limt→∞ u(t) = y. Indeed, assuming thatfor some sequence tn → ∞ the limit of u(tn) does not exist, one selects a subsequence, denoted againtn, for which the weak limit of u(tn) does exist, and proves as before that this limit is y, thus getting acontradiction. Claim 6 is proved. 2

For convenience of the reader let us prove the global existence and uniqueness of the solution to(2.4.4) with Φ = −F , where F is a monotone, hemicontinuous operator in H (cf [Dei]). Uniqueness ofthe solution is trivial: if there are two solutions, u and v, then their difference w := u − v solves theproblem w = −[F (u) − F (v)], w(0) = 0. Multiply this by w and use the monotonicity of F to getg ≤ 0, g(0) = 0, where g := ||w(t)||. Thus, g = 0, so w = 0, and uniqueness is proved.

Let us prove the global existence of the solution to (2.4.4) with Φ = −F . Consider the equation:

wn(t) = w0 −∫ t

0

F (wn(s −1

n))ds, t > 0, wn(t) = w0, t ≤ 0. (∗∗)

We wish to prove thatlimn→∞

wn(t) = w(t), ∀t > 0,

where w(t) = −F (w). Our assumptions (monotonicity and hemicontinuity of F ) imply demicontinuityof F . Fix an arbitrary T > 0, and let B(w0, r) be the ball centered at w0 with radius r > 0. Letsupu∈B(w0,r) ||F (u)|| := c. Then ||wn(t)− w0|| ≤ ct. If t ≤ r/c, then wn(t) ∈ B(w0, r), and ||wn(t)|| ≤ c.Define

znm(t) := wn(t) − wm(t), ||znm(t)|| := gnm(t).

From (**) one gets:

gnmgnm = −(F (wn(t −1

n)) − F (wm(t− 1

m)), wn(t) − wm(t)) := I.


One has:

I = −(F (wn(t −1

n)) − F (wm(t− 1

m)), wn(t −

1

n) − wm(t − 1

m))

−(F (wn(t −1

n)) − F (wm(t− 1

m)), wn(t) − wn(t− 1

n) − (wm(t) − wm(t− 1

m))).

Using the monotonicity of F , the estimate supw∈B(w0 ,r) ||F (w)|| ≤ c, and the estimate ||wn(t)|| ≤ c, onegets:

I ≤ 2c2(1

n+

1

m).

Therefore

gnmgnm ≤ 2c2(1

n+

1

m) → 0 as n,m→ ∞.

This implies

limn,m→∞

gnm(t) = 0, 0 ≤ t ≤ r

c.

Therefore there exists the strong limit w(t):

limn→∞

wn(t) = w(t), 0 ≤ t ≤ r

c.

This function w satisfies the integral equation:

w(t) = w0 −∫ t

0

F (w(s))ds,

and solves the Cauchy problem

w = −F (w), w(0) = w0. (∗ ∗ ∗)If F is continuous, then this Cauchy problem and the preceding integral equation are equivalent. If Fis demicontinuous, then they are also equivalent, but the derivative in the Cauchy problem should beunderstood in the weak sense. We have proved the existence of the unique local solution to(***). Toprove that the solution to (***) exists for any t ∈ [0,∞), let us assume that the solution exists on [0, T ),but not on a larger interval [0, T + d), d > 0, and show that this leads to a contradiction. It is sufficientto prove that the finite limit: limt→T w(t) := W does exist, because then one can solve locally, on theinterval [T, T + d), equation (***) with the initial data w(T ) = limt→T w(t), and construct the solutionto (***) on the interval [0, T + d), thus getting a contradiction.

To prove that W exists, consider

w(t+ h) − w(t) := z(t), ||z|| := g.

One has z = −[F (w(t+ h)) − F (w(t))]. Using the monotonicity of F , one gets (z, z) ≤ 0. Thus,

||w(t+ h) − w(t)|| ≤ ||w(h) − w(0)||.The right-hand side of the above inequality tends to zero as h → 0. This, and the Cauchy test implythe existence of W . The proof is complete. 2

2.4.5 Nonlinear ill-posed problems with non-monotone operators

Assume that F (u) := B(u) − f , B is a non-monotone operator, A := F ′(u), A := F ′(y), T := A∗A,

T := A∗A, Tε := T + εI, where I is the identity operator, ε is as in Theorem 2.4.10 and |ε(t)|ε(t)

< 1,

Φ := −T−1ε (u)[A∗(B(u) − f) + ε(u− u0)], ε = ε(t) > 0, (2.4.53)

and Φδ is defined similarly, with fδ replacing f and uδ replacing u.The main result of this Section is:


Theorem 2.4.16. If (2.4.2) holds, u, u0 ∈ B(y,R), y − u0 = T z, ‖z‖ << 1, that is, ‖z‖ is sufficientlysmall, and R << 1, then problem (2.4.4) has a unique global solution and (2.4.5) holds. If uδ(t) solves(2.4.4) (with Φδ in place of Φ), then there exists a tδ such that limδ→0 ‖uδ(tδ) − y‖ = 0.

The derivation of the stopping rule, that is, the choice of tδ, is based on the ideas presented inSection 2.4.4.

Sketch of proof of Theorem 2.4.16. Proof of Theorem 2.4.16 consists of the following steps.First, we prove that g := ‖w‖ := ‖u(t) − y‖ satisfies a differential inequality (2.4.35), and, applying

(4.4), conclude that g(t) < µ−1(t) → 0 as t → ∞. A new point in this derivation (compared with thatfor monotone operators) is the usage of the source assumption y − u0 = T z.

Secondly, we derive the stopping rule. The source assumption allows one to get a rate of convergence[KNR]. Details of the proof are technical and are not included. One can see [KNR] for some proofs.

Let us sketch the derivation of the differential inequality for g. Write B(u) − f = B(u) − B(y) =Aw+K, where ‖K‖ ≤ M2

2g2, and ε(u− u0) = εw+ ε(y− u0) = εw+ εT z. Then (2.4.53) can be written

asΦ = −w − T−1

ε A∗K − εT−1ε T z, ε := ε(t). (2.4.54)

Multiplying (2.4.4), with Φ defined in (2.4.54), by w, one gets:

gg ≤ −g2 +M2

2

∥∥T−1ε(t)A

∗∥∥g3 + ε(t)∥∥T−1

ε(t)T∥∥ ‖z‖g. (2.4.55)

Since g ≥ 0, one obtains:

g ≤ −g +M2

4√ε(t)

g2 + ε(t)‖T−1ε T‖‖z‖, (2.4.56)

where the estimate ‖T−1ε A∗‖ ≤ 1

2√ε

was used. Clearly,

‖T−1ε T ‖ ≤ ‖(T−1

ε − T−1ε )T ‖ + ‖T−1

ε T‖, ‖T−1ε T‖ ≤ 1, ε‖T−1

ε ‖ ≤ 1,

andT−1ε − T−1

ε = T−1ε (A∗A − A∗A)T−1

ε .

One has:‖A∗A − A∗A‖ ≤ 2M2M1g, ‖z‖ << 1.

Let 2M1M2‖z‖ ≤ 12 . This is possible since ‖z‖ << 1. Using the above estimates, one transforms (2.4.56)

into the following inequality:

g ≤ −1

2g +

M2

4√ε(t)

g2 + ‖z‖ε. (2.4.57)

Now, apply Theorem 2.4.11 to (2.4.57), choosing

µ =2M2√ε,

|ε|ε<

1

2, 16M2‖z‖

√ε(0) < 1, and

2M2‖u0 − y‖√ε(0)

< 1.

Then conditions (2.4.34) are satisfied, and Theorem 2.4.11 yields the estimate:

g(t) <

√ε(t)

2M2.

This is the main part of the proof of Theorem 2.4.16. 2

Remark. The assumption y − u0 = T z, ||z|| << 1, is satisfied under very weak assumption on T ,namely, TBr ∩ (Ba \ 0) 6= ∅, i.e., the image of a ball Br := u : ||u|| ≤ r∀r ∈ (0, r0), r0 = const > 0,intersects a punctured ball Ba \ 0, where a = const > 0. This implies the existence of an element u0

such that y− u0 = T z, where z ∈ Br and y− u0 ∈ Ba \0. The condition ||z|| << 1 is satisfied becauser ∈ (0, r0) can be chosen small.


2.4.6 Nonlinear ill-posed problems: avoiding inverting of operators inthe Newton-type continuous schemes

In the Newton-type methods for solving well-posed nonlinear problems, for example, in the continuousNewton method with Φ = −[F ′(u)]−1F (u), the difficult and expensive part of the solution is invertingthe operator F ′(u). In this section we give a method to avoid inverting of this operator. This is especiallyimportant in the ill-posed problems, where one has to invert some regularized versions of F ′, and to facemore 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 thesimplest well-posed case.

Replace (2.4.4) by the following Cauchy problem (dynamical system):

u = −QF, u(0) = u0, (2.4.58)

Q = −TQ +A∗, Q(0) = Q0, (2.4.59)

where A := F ′(u), T := A∗A, and Q = Q(t) is a bounded operator in H.First, let us state our new technical tool: an operator version of the Gronwall inequality:

Theorem 2.4.17. LetQ = −T (t)Q(t) +G(t), Q(0) = Q0, (2.4.60)

where T (t), G(t), and Q(t) are linear bounded operators on a real Hilbert space H. If there exists ε(t) > 0such that (

T (t)h, h)≥ ε(t)‖h‖2 ∀h ∈ H, (2.4.61)

then∥∥Q(t)

∥∥ ≤ e−∫

t0ε(s)ds

[∥∥Q(0)∥∥+

∫ t

0

∥∥G(s)∥∥e∫

s0ε(x)dx ds

]. (2.4.62)

A proof of this theorem is left to the reader (see [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

(T (t)h, h

)≥ c‖h‖2 ∀h ∈ H, c = const > 0, (2.4.63)

one gets:∥∥Q(t)

∥∥ ≤ e−ct[∥∥Q(0)

∥∥+

∫ t

0

M1ecs ds

]≤[∥∥Q0

∥∥+M1c−1]

:= c1, (2.4.64)

as long as u(t) ∈ B(u0, R).Let u(t) − y := w, ‖w‖ := g, A := F ′(y). Since F (y) = 0, one has F (u) = Aw + K, where

‖K‖ ≤ 0.5M2g2 := c0g

2, and M2 is the constant from (2.4.2). Rewrite (2.4.58) as

w = −Q[Aw +K]. (2.4.65)

Let Λ := I −QA. Multiply (2.4.65) by w and get

gg ≤ −g2 + (Λw,w) + c0g3, c0 = const > 0. (2.4.66)

We prove below thatsupt≥0

‖Λ‖ ≤ λ < 1. (2.4.67)

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

g ≤ −γg + c0g2, 0 < γ < 1, γ := 1 − λ, (2.4.68)


which implies:

g(t) ≤ re−γt, r := g(0)[1− g(0)c0]−1, (2.4.69)

provided that

g(0)c0 < 1. (2.4.70)

Inequality (2.4.70) holds if u0 is sufficiently close to y.

From (2.4.69) it folows that u(∞) = y. Thus, (2.1.6) holds.

The trajectory u(t) ∈ B(u0, R), ∀t > 0, provided that

∫ ∞

0

‖u‖dt =

∫ ∞

0

‖w‖dt ≤ r +c0r

2

2γ≤ R. (2.4.71)

This inequality holds if u0 is sufficiently close to y, that is, r is sufficiently small.

To complete the argument, let us prove (2.4.67). One has:

Λ = −QA = −TΛ +A∗(A − A), (2.4.72)

and ‖A− A‖ ≤ M2g. Using (2.4.69) and Theorem 2.4.17, one gets

‖Λ‖ ≤ e−ct[‖Λ0‖ + rM1M2

∫ t

0

e(c−γ)sds]. (2.4.73)

Thus,

‖Λ‖ ≤ ‖Λ0‖ + Cr := λ, C := M1M2 supt>0

e−γt − e−ct

c− γ. (2.4.74)

If u0 is sufficiently close to y and Q0 is sufficiently close to A−1, then λ > 0 can be made arbitrarilysmall. We have proved:

Theorem 2.4.18. If (2.4.2), and (2.4.3) hold, Q0 and u0 are sufficiently close to A−1 and y, respec-tively, then problem (2.4.58)-(2.4.59) has a unique global solution, (2.4.5) holds, and u(t) converges toy, 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 ε(t) > 0, ε < 0, |ε|ε−1 < γ, where γ > 0 is a constant, (2.4.2) holds in B(y,Rε(0)),where y is a solution to (2.4.1) and R > 0 is a constant. Assume that there exists a z ∈ B(y,Rε(0)),such that y − z = Tw, where w is some element, and T := A∗A. Assume that:

u = −Q(t)[A∗F (u(t)) + ε(t)(u(t) − u0)], A := F ′(u(t)),

˙Q(t) = −Tε(t)Q(t) + I, u(0) = u0, Q(0) = Q0, Tε := T + εI.

Then, if R, γ and w are sufficiently small, then the above Cauchy problem has a unique global solution(u(t)Q(t)

), and ||u(t) − y|| ≤ Rε(t). Precise meaning of the above smallness condition is explained in

[RSm4].

2. If F is monotone, then the assumption y − z = Tw can be dropped, and, under the assumption|ε|ε−3 is sufficiently small, one derives the existence and uniqueness of the global solution to the Cauchyproblem 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 ofwell-posed equations (2.4.1). Some methods for constructing convergent iterative schemes for a wideclass 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 Φ = Φ(u), so that we assume no explicittime dependence in Φ:

un+1 = un + hΦ(un), u0 = u0, h = const > 0. (2.4.75)

One of the results from [AR1], concerning the well-posed equations (2.4.1) is Theorem 2.4.19, formu-lated 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, g1 = c1 = const > 0, g2 = c2 =const > 0, ‖Φ′(u)‖ ≤ L1, for u ∈ B(y,R). Then, if h > 0 is sufficiently small, and u0 is sufficientlyclose to y, then (2.4.75) produces a sequence un for which

‖un − y‖ ≤ Re−chn, ‖F (un)‖ ≤ ‖F0‖e−chn, (2.4.76)

where R := c2‖F0‖c1

, F0 = F (u0), c = const > 0, c < c1.

Proof of Theorem 2.4.19. The proof is by induction. For n = 0 estimates (2.4.76) are clear. Assum-ing these estimates for j ≤ n, let us prove them for j = n + 1. Let Fn := F (un), and let wn+1(t) solveproblem (2.4.4) on the interval (tn, tn+1), tn := nh, with wn+1(tn) = un. By (2.4.10) (with G = c2e

−c1t)and (2.4.76) one gets:

‖wn+1(t) − y‖ ≤ c2c1

‖Fn‖e−c1t ≤ Re−cnh−c1t, tn ≤ t ≤ tn+1. (2.4.77)

One has:

‖un+1 − y‖ ≤ ‖un+1 − wn+1(tn+1)‖ + ‖wn+1(tn+1) − y‖, (2.4.78)

and

‖un+1 − wn+1(tn+1)‖ ≤∫ tn+1

tn

‖Φ(un) − Φ(wn+1(s))‖ds

≤ L1c2h

∫ tn+1

tn

‖F (wn+1(t))‖dt ≤ L1c1h2Re−cnh, (2.4.79)

where we have used the formula R := c2‖F0‖c1

, and the estimate:

‖F (wn+1(t))‖ ≤ ‖Fn‖e−c1(t−tn) ≤ ‖F0‖e−cnh−c1(t−tn). (2.4.80)

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

‖un+1 − y‖ ≤ Re−cnh(e−c1h + c1L1h2) ≤ Re−c(n+1)h, (2.4.81)

provided that

e−c1h + c1L1h2 ≤ e−ch. (2.4.82)

Inequality (2.4.82) holds if h is sufficiently small and c < c1. So, the first inequality (2.4.76), with n+ 1in place of n, is proved if h is sufficiently small and c < c1.

Now

‖F (un+1)‖ ≤ ‖F (un+1) − F (wn+1(t))‖ + ‖F (wn+1(t))‖, tn ≤ t ≤ tn+1. (2.4.83)


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

‖F (un+1) − F (wn+1(tn+1))‖ ≤ M1‖un+1 − wn+1(tn+1)‖ ≤M1c2L1h2‖F0‖e−cnh. (2.4.84)

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

‖F (un+1)‖ ≤ ‖F0‖e−cnh(e−c1h +M1c2L1h2) ≤ ‖F0‖e−c(n+1)h, (2.4.85)

provided thate−c1h +M1c2L1h

2 ≤ e−ch. (2.4.86)

Inequality (2.4.86) holds if h is sufficiently small and c < c1. So, the second inequality (2.4.76) withn+ 1 in place of n is proved if h is sufficiently small and c < c1. Theorem 2.4.19 is proved. 2

In the well-posed case, if F (y) = 0, the discrete Newton’s method

un+1 = un − [F ′(un)]−1F (un), u0 = u(0),

converges superexponentially if u0 is sufficiently close to y. Indeed, if vn := un − y, then vn+1 =vn− [F ′(un)]−1[F ′(un)vn+K] where ‖K‖ ≤ M2

2 ‖vn‖2. Thus, gn := ‖vn‖ satisfies the inequality: gn+1 ≤qg2n, where q := m1M2

2. Therefore gn ≤ q2

n−1g2n

0 , and if 0 < qg0 < 1, then the method convergessuperexponentially.

If one uses the iterative method un+1 = un − h[F ′(un)]−1F (un), with h 6= 1, then, in the well-posedcase, assuming that this method converges, it converges exponentially, that is, slower than in the caseh = 1.

The continuous analog of the above method

u = −a[F ′(u)]−1F (u), u(0) = u0,

where a = const > 0, converges at the rate O(e−at). Indeed, if g(t) := ‖F (u(t))‖, then gg = −ag2, sog(t) = g0e

−at, ‖u‖ ≤ am1g0e−at. Thus

‖u(t) − u(∞)‖ ≤ m1g0e−at, and F (u(∞)) = 0.

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 r, ϕ : π − ϕ0 < ϕ < π + ϕ0, ϕ0 > 0, 0 < r < r0, where ϕ0 and r0are arbitrarily small, fixed numbers, consists of the regular points of the operator A := F ′(u) for allu ∈ B(u0, R).

Assumption S implies the estimate:

‖(F ′(u) + ε)−1‖ ≤ 1

ε sinϕ0, 0 < ε < r0(1 − sinϕ0), (2.4.87)

because ‖(A−z)−1‖ ≤ 1dist(z,s(A)) , where s(A) is the spectrum of a linear operator A, and dist(z, s(A)) is

the distance from a point z of a complex plane to the spectrum. In our case, z = −ε, and dist(z, s(A)) =ε sinϕ0, if ε < r0(1 − sinϕ0).

Theorem 2.4.20. If (2.4.2) and (2.4.87) hold, and 0 < ε < r0(1 − sinϕ0), then problem (2.4.38), withε(t) = ε = const > 0, is solvable, problem (2.4.4), with Φ defined in (2.4.33) and u0 = 0, has a uniqueglobal solution, ∃u(∞), and F (u(∞)) + εu(∞) = 0. Every solution to the equation F (V ) + εV = 0 isisolated.


Proof of Theorem 2.4.20. Let g = g(t) := ‖F (u(t)) + εu(t)‖, where u = u(t) solves locally (2.4.4),where Φ is defined in (2.4.33) and u0 = 0. Then:

gg = −((F ′(u) + ε)(F ′(u) + ε)−1(F (u) + εu), F (u) + εu) = −g2, (2.4.88)

so

g = g0e−t, g0 := g(0); ‖u‖ ≤ g0

ε sinϕ0e−t. (2.4.89)

Thus,

‖u(t) − u(∞)‖ ≤ g0ε sinϕ0

e−t, ‖u(t) − u0‖ ≤ g0ε sinϕ0

, F (u(∞)) + εu(∞) = 0. (2.4.90)

Therefore equation

F (V ) + εV = 0, ε = const > 0, (2.4.91)

has a solution in B(u0, R), where R = g0ε sinϕ0

.

Every solution to equation (2.4.91) is isolated. Indeed, if F (W ) + εW = 0, and ψ := V −W , thenF (V )−F (W )+ εψ = 0, so [F ′(V ) + ε]ψ+K = 0, where ‖K‖ ≤ M2

2 ‖ψ‖2. Thus, using (2.4.87), one gets

‖ψ‖ ≥ 2ε sin ϕ0

M2, unless ψ = 0. Consequently, if ‖ψ‖ is sufficiently small, then ψ = 0. Theorem 2.4.20 is

proved. 2

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 [0, T )×D (D ⊂ R, T ≤ ∞) and f(t, w) ≤g(t, u) if w ≤ u, t ∈ (0, T ), w, u ∈ D. Assume that g(t, u) is such that the Cauchy problem

u = g(t, u), u(0) = u0, u0 ∈ D (2.4.92)

has a unique solution. If

w ≤ f(t, w), w(0) = w0 ≤ u0, w0 ∈ D, (2.4.93)

then u(t) ≥ w(t) 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 w ≤ u. Since w0 ≤ u0 andw(0) ≤ f(0, w0) < g(0, u0) = u(0), there exists δ > 0 such that u(t) > w(t) on (0, δ]. Assume that forsome t1 > δ one has u(t1) < w(t1). Then for some t2 < t1 one has

u(t2)

= w(t2)

and u(t) < w(t) for t ∈(t2, t1

].

One gets

w(t2)≥ u

(t2)

= g(t, u(t2))> f

(t, w(t2))

≥ w(t2).

This contradiction proves that there is no point t2 such that u(t2) = w(t2).

Step 2. Now consider the case f(t, w) ≤ g(t, u), if w ≤ u. Define

un = g(t, un

)+ εn, un(0) = u0, εn > 0, n = 0, 1, . . . ,

where εn tends monotonically to zero. Then

w ≤ f(t, w) ≤ g(t, u) < g(t, u) + εn, w ≤ u.


By Step 1 un(t) ≥ w(t), n = 0, 1, . . . . Fix an arbitrary compact set [0, T1], 0 < T1 < T .

un(t) = u0 +

∫ t

0

g(τ, un(τ )

)dτ + εnt. (2.4.94)

Since g(t, u) is continuous, the sequence un is uniformly bounded and equicontinuous on [0, T1]. There-fore there exists a subsequence unk which converges uniformly to a continuous function u(t). Bycontinuity of g(t, u) we can pass to the limit in (2.4.94) and get

u(t) = u0 +

∫ t

0

g(τ, u(τ )

)dτ, t ∈

[0, T1

]. (2.4.95)

Since T1 is arbitrary (2.4.95) is equivalent to the initial Cauchy problem that has a unique solution. Theinequality unk(t) ≥ w(t), k = 0, 1, . . . implies u(t) ≥ w(t). If the solution to the Cauchy problem (2.4.92)is not unique, the inequality w(t) ≤ u(t) holds for the maximal solution to (2.4.92). 2

The following theorem has been stated earlier as Theorem 2.4.11 and its statement is repeated forconvenience of the reader.

Theorem 2.4.22. Let γ(t), σ(t), β(t) ∈ C[t0,∞) for some real number t0. If there exists a positivefunction µ(t) ∈ C1[t0,∞) such that

0 ≤ σ(t) ≤ µ(t)

2

(γ(t) − µ(t)

µ(t)

), β(t) ≤ 1

2µ(t)

(γ(t) − µ(t)

µ(t)

), (2.4.96)

then a non-negative solution to the following inequalities:

v(t) ≤ −γ(t)v(t) + σ(t)v2(t) + β(t), v(t0) <1

µ(t0), (2.4.97)

satisfies the estimate:

v(t) ≤ 1 − ν(t)

µ(t)<

1

µ(t), (2.4.98)

for all t ∈ [t0,∞), where

ν(t) =

(1

1 − µ(t0)v(t0)+

1

2

∫ t

t0

(γ(s) − µ(s)

µ(s)

)ds

)−1

. (2.4.99)

Remark 2.4.23. Without loss of generality one can assume β(t) ≥ 0.

In [Alb1] a differential inequality v ≤ −A(t)ψ(v(t))+β(t) was studied under some assumptions whichinclude, among others, the positivity of ψ(v) for v > 0. In Theorem 2.4.22 the term −γ(t)v(t)+σ(t)v2 (t)(which is analogous to some extent to the term −A(t)ψ(v(t))) can change sign. Our Theorem 2.4.22is not covered by the result in [Alb1]. In particular, in Theorem 2.4.22 an analog of ψ(v), for the caseγ(t) = σ(t) = A(t), is the function ψ(v) := v − v2. This function goes to −∞ as v goes to +∞, so itdoes not satisfy the positivity condition imposed in [Alb1]. Lemma 1 in [Alb1] is wrong. Its correctedversion is given in [ARS3], where a counterexample to Lemma 1 from [Alb1] is constructed. In [ARS3]the following result is proved:

Lemma.Let u ∈ C1[0,∞),u≥ 0, andu ≤ −a(t)f(u(t)) + b(t), u(0) = u0.Assume:1) a(t), b(t) ∈ C[0,+∞), a(t) > 0, b(t) ≥ 0 for t > 0,

2)∫+∞

a(t)dt = +∞, b(t)a(t)

→ 0 as t → +∞,


3) f ∈ C[0,+∞), f(0) = 0, f(u) > 0 for u > 0,4) there exists c > 0 such that f(u) ≥ c for u ≥ 1.Then u(t) → 0 as t → +∞.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 differentialequation with separating variables. The proof below is based on a special choice of the solution to theRiccati equation majorizing a solution of inequality (2.4.97).

Proof of Theorem 2.4.22. Denote:

w(t) := v(t)e∫ t

t0γ(s)ds

, (2.4.100)

then (2.4.97) implies:w(t) ≤ a(t)w2(t) + b(t), w

(t0)

= v(t0), (2.4.101)

wherea(t) = σ(t)e

−∫ t

t0γ(s)ds

, b(t) = β(t)e∫ t

t0γ(s)ds

.

Consider Riccati’s equation:

u(t) =f(t)

g(t)u2(t) − g(t)

f(t). (2.4.102)

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

u(t) = − g(t)

f(t)+

[f2(t)

(C −

∫ t

t0

f (s)

g(s)f2(s)ds

)]−1

. (2.4.103)

Define f and g as follows:

f(t) := µ12 (t)e

− 12

∫ tt0γ(s)ds

, g(t) := −µ− 12 (t)e

12

∫ tt0γ(s)ds

, (2.4.104)

and consider the Cauchy problem for equation (2.4.102) with the initial condition u(t0) = v(t0). ThenC in (2.4.103) takes the form:

C =1

µ(t0)v(t0) − 1.

From (2.4.96)) one gets

a(t) ≤ f(t)

g(t), b(t) ≤ − g(t)

f(t).

Since fg = −1 one has:∫ t

t0

f (s)

g(s)f2(s)ds = −

∫ t

t

f (s)

f(s)ds =

1

2

∫ t

t0

(γ(s) − µ(s)

µ(s)

)ds.

Thus

u(t) =e∫ t

t0γ(s)ds

µ(t)

[1 −

(1

1 − µ(t0)v(t0)+

1

2

∫ t

t0

(γ(s) − µ(s)

µ(s)

)ds

)−1]. (2.4.105)

It follows from conditions (2.4.96) and from the second inequality in (2.4.97) that the solution to prob-lem (2.4.102) exists for all t ∈ [0,∞) and the following inequality holds with ν(t) defined by (2.4.99):

1 > 1 − ν(t) ≥ µ(t0)v(t0). (2.4.106)

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

v(t)e∫

tt0γ(s)ds

:= w(t) ≤ u(t) =1 − ν(t)

µ(t)e∫

tt0γ(s)ds

<1

µ(t)e∫

tt0γ(s)ds

, (2.4.107)

and thus estimate (2.4.98) is proved. 2


To illustrate conditions of Theorem 2.4.22 consider the following examples of functions γ, σ, and β,satisfying (2.4.96) for t0 = 0.

Example 2.4.24. Let

γ(t) = c1(1 + t)ν1 , σ(t) = c2(1 + t)ν2 , β(t) = c3(1 + t)ν3 , (2.4.108)

where c2 > 0, c3 > 0. Choose µ(t) := c(1 + t)ν, c > 0. From (2.4.96), (2.4.97) one gets the followingconditions

c2 ≤ cc12

(1 + t)ν+ν1−ν2 − cν

2(1 + t)ν−1−ν2 ,

c3 ≤ c12c

(1 + t)ν1−ν−ν3 − ν

2c(1 + t)−ν−1−ν3, cv(0) < 1. (2.4.109)

Thus, one obtains the following conditions:

ν1 ≥ −1, ν2 − ν1 ≤ ν ≤ ν1 − ν3, (2.4.110)

and

c1 > ν,2c2

c1 − ν≤ c ≤ c1 − ν

2c3, cv(0) < 1. (2.4.111)

Therefore for such γ, σ and β a function µ with the desired properties exists if

ν1 ≥ −1, ν2 + ν3 ≤ 2ν1, (2.4.112)

andc1 > ν2 − ν1, 2

√c2 c3 ≤ c1 + ν1 − ν2, 2c2 v(0) < c1 + ν1 − ν2. (2.4.113)

In this case one can choose ν = ν2 − ν1, c = 2c2c1+ν1−ν2 . However in order to have v(t) → 0 as t → +∞

one needs the following conditions:

ν1 ≥ −1, ν2 + ν3 ≤ 2ν1, ν1 > ν3, (2.4.114)

andc1 > ν2 − ν1, 2

√c2 c3 ≤ c1, 2c2 v(0) < c1. (2.4.115)

Example 2.4.25. If

γ(t) = γ0, σ(t) = σ0 eν t, β(t) = β0e

−ν t, µ(t) = µ0 eν t,

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

0 ≤ σ0 ≤ µ0

2

(γ0 − ν

), β0 ≤ 1

2µ0

(γ0 − ν

), µ0 v(0) < 1.

Example 2.4.26. Here log stands for the natural logarithm. For some t1 > 0

γ(t) =1√

log(t + t1), µ(t) = c log

(t+ t1

),

conditions (2.4.96), (2.4.97) are satisfied if

0 ≤ σ(t) ≤ c

2

(√log(t+ t1) −

1

t + t1

),

β(t) ≤ 1

2c log2(t+ t1)

(√log(t + t1) −

1

t+ t1

), v(0)c log t1 < 1.

In all considered examples µ(t) tends to infinity as t → +∞ and provides a decay of a nonnegativesolution to integral inequality (2.4.97) even if σ(t) tends to infinity. Moreover, in the first and the thirdexamples v(t) tends to zero as t → +∞ when γ(t) → 0 and σ(t) → +∞.


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 w(t) = w0+∫ t0b(s)ds+

∫ t0a(s)w2(s)ds,

w0 = const ≥ 0. Let B(t) := w0+∫ t0b ds, A(t) :=

∫ t0a(s)ds. Then w(t) = B(t)+

∫ t0w2(s) dA(s). Assume

a(t) ≥ 0, B ≥ 0, and consider the process wn+1(t) = B(t) +∫ t0w2n dA, w0(t) = B(t), w1(t) ≥ w0(t). By

induction, wn+1(t) ≥ wn(t) ∀n. If wn(t) ≤ c(T ) ∀t ∈ [0, T ], where T > 0 is an arbitrary number, thenlimn→∞wn(t) := w(t) exists and solves (2.4.101).

Assume ∫ t

0

dA

(∫ t

0

B2(s) dA

)2

≤ c0

∫ t

0

B2(s) dA. (2.4.116)

Then

wn(t) ≤ B(t) + c1

∫ t

0

B2(s) dA, (2.4.117)

where c1 is a constant estimated below. Inequality (2.4.117) holds for n = 0, and, by induction,

wn+1(t) ≤ B(t) +

∫ t

0

dA

(B + c1

∫ t

0

B2 dA

)2

≤ B(t) + 2

∫ t

0

B2 dA+

∫ t

0

c21

(∫ t

0

B2 dA

)2

dA

.

Using (2.4.116) one gets:

wn+1(t) ≤ B + 2

∫ t

0

B2 dA+ 2c21c0

∫ t

0

B2 dA = B + 2(1 + c21c0

)∫ t

0

B2 dA.

If2(1 + c21c0

)≤ c1, (2.4.118)

then (2.4.117) has been proved by induction. Inequality (2.4.118) holds if 14c0

−√

1b c20

− 1c0

< c1 <

14c0

+√

116c20

− 1c0

, and 0 < c0 <116 .

2.5 Examples of solutions of ill-posed problems

2.5.1 Stable numerical differentiation: when is it possible?

In many applications one has to estimate a derivative f ′, given the noisy values of the function f tobe differentiated. As an example we refer to the analysis of photo-electric response data. The goal ofthat experiment is to determine the relationship between the intensity of light falling on certain plantcells and their rate of uptake of various substances. Rather than measuring the uptake rate directly, theexperimentalists measure the amount of each substance not absorbed as a function of time, the uptakerate being defined as minus the derivative of this function. As for the other example, one can mentionthe problem of finding the heat capacity cp of a gas as a function of temperature T . Experimentally

one measures the heat content q(T ) =∫ TT0cp(τ )d τ, 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 of a ship by the maximum of a certainunivalent curve, called the navigation characteristic. This direction can be obtained by differentiationof this curve. Since the navigation characteristic is communicated with some errors, one has to differ-entiate it numerically in order to find its maximum. In [R83, p.94], the shape of a convex obstacle isfound by differentiation of a support function of this obstacle. The support function is found from the

2.5. EXAMPLES OF SOLUTIONS OF ILL-POSED PROBLEMS 61

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 zeromean value and finite variance, and is identically distributed independently of the point x. It is provedthat in this case the error of an optimal formula of numerical differentiation can be made O(p−0.25ε),where p is the number of observation points and ε is the error for a noise which is non-random (see theprecise 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 afunction may lead to large errors in its derivative. Indeed, if one takes fδ = f + δ sin( t

δ2 ), f ′ ∈ L∞(0, 1),then ‖fδ − f‖∞ = δ and ‖f ′δ − f ′‖∞ = 1

δ , so that small in L∞(0, 1)-norm perturbations of f result inlarge perturbations of f ′ in L∞(0, 1)-norm.

Various methods have been developed for stable numerical differentiation of f given fδ, ‖fδ−f‖ ≤ δ.We mention three groups of methods:

(1) regularized difference methods with a step size h = h(δ) being a regularization parameter, see [R18],where this idea was proposed for the first time, and [R156], [R58], [R168]. As an example of sucha method one may consider:

Rh(δ)fδ(x) :=

1h (fδ(x+ h) − fδ(x)), 0 < x < h,

12h (fδ(x+ h) − fδ(x− h)), h ≤ x ≤ 1 − h,

1h (fδ(x) − fδ(x− h)), 1 − h < x < 1, h > 0.

(2.5.1)

If fδ ∈ L∞(0, 1), and f ∈ W 2,p(0, 1), where Wn,p(0, 1) is the Sobolev space of functions whosen-th derivative belongs to Lp(0, 1), ‖fδ − f‖p ≤ δ, then

∥∥Rh(δ)fδ − f ′∥∥p≤∥∥Rh(δ)

(fδ − f

)∥∥p

+∥∥Rh(δ)f − f ′

∥∥p

≤ 2δ

h+N2,ph

2,

(2.5.2)

where N2,p is an estimation constant: ‖f ′′‖p ≤ N2,p. The error in the interval h ≤ x ≤ 1 − h can

be estimated slightly better (by a quantity δh

+N2,ph

2). In this paper by ‖ · ‖p we denote ‖ · ‖Lp(0,1).

The right-hand side of (2.5.2) attains the absolute minimum 2√δN2,p at h = h2,p(δ) := 2( δ

N2,p)

12 ,

while if one uses the error estimate for the interval h ≤ x ≤ 1 − h, then one gets the absoluteminimum

√2δN2,p at h = ( 2δ

N2,p)

12 . When the function f ∈ W 3,p(0, 1), one can modify (2.5.1)

near the ends so that it has the order O(h2) of the error of approximation as h → 0, and resultsin an algorithm of order δ2/3. For example one can take

Rh(δ)fδ(x) :=

12h (4fδ(x+ h) − fδ(x+ 2h) − 3fδ(x)), 0 < x < 2h,

12h (fδ(x+ h) − fδ(x− h)), 2h < x < 1 − 2h,

12h

(3fδ(x) + fδ(x− 2h) − 4fδ(x− h)), 1 − 2h < x < 1.

(2.5.3)

The difference methods use only local values of the function fδ, which is natural when one estimatesa derivative, and these methods are very simple, which is an advantage.

(2) An alternative approach is first to smooth fδ by a mollification with a certain kernel, for examplewith a Gaussian kernel, or to use a mollification by splines, and then to differentiate the resultingsmooth approximation, see e.g. [VA]. If one applies mollification with the Gaussian kernel wh(x) :=

1h√π

exp(−x2

h2 ), x ∈ R, h > 0, then (Mh(δ))′ : L2(0, 1) → L2(0, 1),

(Mh(δ)

)′fδ(x) :=

(w′h ? fδ

)(x) :=

∫ 1

0

w′h(x− s)fδ(s)ds, (2.5.4)


where ? stands for the convolution, fδ ∈ L2(0, 1), and ‖fδ − f‖2 ≤ δ. Assume that f ∈ H1(0, 1)with f(0) = f(1) = 0 and ‖f ′′‖2 ≤ N2,2. One has

∥∥∥(Mh(δ)

)′fδ − f ′

∥∥2≤∥∥∥(Mh(δ)

)′(fδ − f)

∥∥∥2

+∥∥(Mh(δ)

)′f − f ′

∥∥∥2

(2.5.5)

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

∥∥∥(Mh(δ)

)′(fδ − f

)∥∥∥L2(0,1)

=∥∥∥w′

h ?(fδ − f

)∥∥∥L2(0,1)

≤∥∥∥w′

h ?(fδ − f

)∥∥∥L2(R)

≤∥∥w′

h

∥∥L1(R)

∥∥fδ − f∥∥L2(0,1)

≤ 2δ

h√π,

(2.5.6)

because ‖w′h‖L1(R) = −2

∫∞0w′h(s)ds = 2wh(0) = 2

h√π. By a partial integration one gets:

(w′h ? f

)(x) =

∫ 1

0

w′h(x− s)f(s)ds =

∫ 1

0

wh(x− s)f ′(s)ds =(wh ? f

′)(x). (2.5.7)

To complete the argument one uses the inequality

∥∥wh ? z − z∥∥L2(R)

≤ h‖z′‖L2(0,1) (2.5.8)

for every z ∈ H1(0, 1) with z(0) = z(1) = 0. Here the above functions z are extended from [0, 1]to R by zero.

To verify (2.5.8) define the Fourier transform by

(F z)(t) :=1√2π

∫ ∞

−∞z(s)eistds, t ∈ R.

Using Parseval’s equation, one gets:

∥∥wh ? z − z∥∥L2(R)

=∥∥∥F

(wh ? z − z

)∥∥∥L2(R)

=∥∥∥(√

2πFwh − 1)F z∥∥∥L2(R)

=

∥∥∥∥1

−it(√

2πFwh − 1)F z′

∥∥∥∥L2(R)

≤∥∥∥∥

1

−it(√

2πFwh − 1)∥∥∥∥

L∞(R)

‖z′‖L2(0,1).

(2.5.9)

Since

ϕh(t) :=1

it

(1 −

√2πFwh

)=

1

it

[1 − e−h

2t2/4], t ∈ R,

and 1−e−τ2

τ ≤ 2, for all τ > 0, estimate (2.5.9) yields inequality (2.5.8). Thus, from (2.5.7)and (2.5.9) one obtains

∥∥(Mh(δ))′f − f ′

∥∥2≤∥∥wh ? f ′ − f ′

∥∥2≤ hN2,2. (2.5.10)

Finally, combining (2.5.5), (2.5.6) and (2.5.10) one gets

∥∥(Mh(δ))′fδ − f ′

∥∥2≤ 2δ

h√π

+ hN2,2 := ε2.

The choice h = h2,2(δ) =√

2δN2,2

√π

leads to the estimate ε2 ≤ 2√

2/√π√δN2,2.


(3) The third group of methods uses variational regularization for solving ill-posed problems ([Phi], [IVT]).One applies variational regularization to a Volterra integral equation

Au(x) :=

∫ x

0

u(s)ds = f(x). (2.5.11)

For example, if the noisy data fδ are given, ‖fδ − f‖2 ≤ δ, then one minimizes the functional

F0(u) :=∥∥Au− fδ

∥∥2

2+ α‖u‖2

2

orFm(u) :=

∥∥Au− fδ∥∥2

2+ α

∥∥u(m)∥∥2

2, m > 0,

where α > is a regularization parameter. One proves that for a suitable choice of α, α = α(δ),the above functionals have a unique minimizer uδ and ‖uδ − f ′‖2 → 0 as δ → 0. An optimalchoice of the regularization parameter α in this approach is a nontrivial problem. Some methodsfor choosing α = α(δ) are presented in Section 2.1.

The above methods have been discussed in the literature, and their analysis is not our goal. Ourgoal 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 f ′ given noisy datafδ . In Problem I a new notion of regularizer is introduced. Our treatment of the stable differentiationis an example of application of this new notion. In Section 2.1 a regularization method for unboundedlinear and nonlinear operators is discussed.

Statements of the problem of stable numerical differentiation

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

A(u) = f, A : X −→ Y, (2.5.12)

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 f ∈ Y there exists a solution u ∈ X;

(b) this solution is unique;

(c) the problem is stable under small perturbations of the initial data in the sense:

‖uδ − u‖X −→ 0 if ‖fδ − f‖Y −→ 0, where A(uδ) := fδ. (2.5.13)

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

A(u) :=

x∫

0

u(s)ds = f, A : X = Lp(0, 1) −→ Lp(0, 1), f(0) = 0. (2.5.14)

We study the cases p = 2 and p = ∞ in detail. Problem (2.5.14) is solvable only if f ′ ∈ X. 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 isfδ together with an a priori information about f , for example, one may know that f ∈ K(p, δ, a), where

K(p, δ, a) := K := f : f ∈W a,p(0, 1), ‖f (a)‖p ≤ Na,p <∞, ‖fδ − f‖p ≤ δ, (2.5.15)


a = 0, a = 1, or 1 < a ≤ 2. For 1 < a < 2

∥∥f (a)∥∥p

:= ‖f ′‖p + supx,y∈(0,1),x6=y

‖f ′(x) − f ′(y)‖p|x− y|a0

, a = 1 + a0, 0 < a0 < 1. (2.5.16)

Therefore given δ > 0 and fδ one has to estimate f ′ for any f ∈ K(p, δ, a) and the problem of stablenumerical differentiation has to be understood in the following sense:

Problem I:Find a linear or nonlinear operator Rh(δ) such that

supf∈K(p,δ,a)

‖Rh(δ)fδ − f ′‖p ≤ η(δ) −→ 0 as δ −→ 0, (2.5.17)

where η(δ) is some positive continuous function of δ ∈ (0, δ0), and δ0 > 0 is some number.The traditional formulation of the problem of stable numerical differentiation is different from the

above:Problem II:Find a linear or nonlinear operator Rh(δ) such that

supfδ∈B

pδ,f

‖Rh(δ)fδ − f ′‖p ≤ η(δ, f) −→ 0 as δ −→ 0, (2.5.18)

where Bpδ,f := fδ : ‖fδ − f‖p ≤ δ and f ∈ K(p, δ, a) is fixed,

or even in a weaker form:Find Rh(δ) such that ∥∥Rh(δ)fδ − f ′

∥∥p≤ η(δ, f) −→ 0 as δ −→ 0 (2.5.20’)

for a fixed f ∈ K(p, δ, a) and fixed family fδ ∈ Bpδ,f .

Note the principal difference in the statements of Problems I and II of stable numerical differen-tiation: in Problem I the data are fδ, Na,p, f is arbitrary in the set K(p, δ, a), and we wish to find astable approximation of f ′, which is valid uniformly with respect to f ∈ K(p, δ, a). On the other hand,in Problem II it is assumed that f ∈ K(p, δ, a) is fixed and the approximation of f ′ is either uniform withrespect to fδ ∈ Bp

δ,f or holds for a particular family fδ ∈ Bpδ,f . Since in practice we do not know f and

we do know only one family fδ, Problem I is much more important practically than Problem II. In thisSection we show when one can obtain, in principle, a stable approximation of f ′ in the sense formulatedin Problems I and II, and when it is not possible, in principle, to obtain such an approximation of f ′

from noisy data.The main result on the stable numerical differentiation problem in the first formulation is stated in

Theorem 2.5.1:

Theorem 2.5.1. There does not exist an operator Rh(δ) : Lp(0, 1) → Lp(0, 1), linear or nonlinear, forp = 2 and p = ∞, such that inequality (2.5.17) holds for a ≤ 1. If a ≤ 1, then

γ∞δ,a := infRh(δ):Lp(0,1)−→Lp(0,1)

supf∈K(p,δ,a)

‖Rh(δ)fδ − f ′‖p ≥ c > 0, p = 2,∞. (2.5.19)

If a > 1 and p ≥ 1, then there does exist an operator Rh(δ) such that 2.5.17 holds. For example, one canuse Rh(δ) defined in 2.4.92 with

h = ha(δ) :=

( 2δa0Na,p

)1a , a = 1 + a0, 0 < a0 < 1,

2( δN2,p

)12 , a = 2.

(2.5.20)

The error of the corresponding differentiation formula is

η(δ) :=

a(Na,p)

1a ( 2δ

a0)

a0a , a = 1 + a0, 0 < a0 < 1,

2(δN2,p)12 , a = 2.

(2.5.21)


The main result on the stable numerical differentiation problem in the second formulation is statedin Theorem 2.5.2:

Theorem 2.5.2. If a = 1, then there exists an operator Rh(δ) : L2(0, 1) → L2(0, 1), such that inequality(2.5.18) holds.

The principal difference is: for a = 1 one cannot differentiate stably in the sense formulated inProblem I. In the sense of Problem II stable differentiation is possible in principle. However the approxi-mation error, ‖Rh(δ)f−f ′‖2, cannot be estimated, and this error η(δ, f) may go to zero arbitrarily slowlyas δ → 0. This is in sharp contrast with the practically computable error estimate given in (2.5.21).Moreover, no matter how small the error bound δ > 0 is, there exists a function f ∈ K 2

δ,1, such thatnot only Rh(δ) (with any fixed function h(δ)), defined in (2.5.35), but any other operator Rh(δ), linearor nonlinear, will satisfy the inequality ‖Rh(δ)f − f ′‖2 ≥ c > 0, where c > 0 does not depend on δ. Thisfollows from (2.5.19).

Proofs

Proof of Theorem 2.5.1. First, consider the case p = ∞. Take

f1(x) := −M2x(x− 2h), 0 ≤ x ≤ 2h. (2.5.22)

Extend f1(x) from (0, 2h) to (2h, 1) by zero and denote the extended function by f1(x) again. Thenf1(x) ∈ W 1,∞(0, 1) and the norms ‖f (a)‖∞, a = 0 and a = 1 are preserved. Define f2(x) = −f1(x),x ∈ (0, 1). Note that

supx∈(0,1)

|fk(x)| =Mh2

2, k = 1, 2. (2.5.23)

Choose h = h∞ := h∞(δ) :=√

2δM, so that

Mh2∞

2= δ, (2.5.24)

Then for fδ(x) ≡ 0 one has: ‖fk − fδ‖∞ = δ, k = 1, 2. Let (Rh(δ)fδ)(0) = (Rh(δ)0)(0) := b. One gets

γ∞δ,a := infRh(δ)

supf∈K ∞

δ,a

‖Rh(δ)fδ − f ′‖∞ ≥ infRh(δ)

maxk=1,2

‖Rh(δ)fδ − f ′k‖∞

≥ infRh(δ)

maxk=1,2

‖Rh(δ)fδ(0) − f ′k(0)‖∞

= infb∈R

max|b−Mh∞| , |b+Mh∞| = Mh∞. (2.5.25)

If h∞ =√

2δM , then Mh∞ =

√2δM. If a = 0, then (2.5.23) implies that fk ∈ K ∞

δ,0 , k = 1, 2, with

N0,∞ :=Mh2

∞

2= δ. For any fixed δ > 0 and N0,∞ = δ 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 K ∞δ,0 and, in fact, γ∞δ,0 → ∞ as

M → ∞.

Suppose now that f ∈ K ∞δ,1 . One has

‖f ′1‖∞ = ‖f ′2‖∞ = sup0≤x≤2h∞

|M (x− h∞)| = Mh∞. (2.5.26)


Thus, for given δ and N1,∞ one can take h = h∞ :=√

2δM , so that ‖fk − fδ‖∞ = δ, k = 1, 2, holds,

and then take M so that N1,∞ =√

2δM. For these h∞ and M the functions fk ∈ K ∞δ,1 , k = 1, 2. One

obtains from (2.5.25) the following inequality

γ∞δ,1 ≥ N1,∞ > 0 as δ −→ 0, (2.5.27)

which implies that estimate (2.4.10) is false in the class K ∞δ,1 .

Now let p = 2. For f1 defined in (2.5.22) one has

‖f1‖L2(0,2h) =2√15Mh

52 , ‖f ′1‖L2(0,2h) =

√2

3Mh

32 . (2.5.28)

Extend f1(x) from (0, 2h) to (2h, 1) by zero and denote the extended function f1(x) again. Then

f1 ∈ W 1,2(0, 1), ‖f1‖L2(0,1) = c1Mh52 , and ‖f ′1‖L2(0,1) = c2Mh

32 . Define f2(x) = −f1(x), fδ(x) ≡ 0,

x ∈ (0, 1).

Choose h = h2 := ( δc1M

)25 = δ to satisfy the identity

c1Mh52

2 = δ, (2.5.29)

then ‖fk − fδ‖L2(0,1) = δ, k = 1, 2. Thus,

γ2δ,a := inf

Rh(δ)

supf∈K 2

δ,a

‖Rh(δ)fδ − f ′‖2 ≥ infRh(δ)

maxk=1,2

‖Rh(δ)fδ − f ′k‖2

= infϕ∈L

max‖ϕ− f ′1‖2, ‖ϕ+ f ′1‖2 ,

where L := ϕ : ϕ = cf ′1 + ψ, ψ ⊥ f ′. Therefore

γ2δ,a ≥ inf

c∈R, ψ⊥f ′max

√(1 − c)2‖f ′1‖2

2 + ‖ψ‖22,√

(1 + c)2‖f ′1‖22 + ‖ψ‖2

2

= infc∈R

max|1− c|‖f ′1‖2, |1 + c|‖f ′1‖2 = ‖f ′1‖2

= c2Mh322 = c2M

25

(δ

c1

) 35

. (2.5.30)

If a = 0, then (2.5.29) yields fk ∈ K 2δ,0, k = 1, 2, with N0,2 := c1Mh

52 = δ, and one gets γ2

δ,0 → ∞ asM → ∞.

Given constants δ and N1,2 (in the case a = 1), one takes h = h2 :=(

δc1M

)25

so that ‖fk− fδ‖2 = δ,

and then takes M so that C2M25

(δc1

)35

= N1,2. With this choice of h2 and M the functions fk ∈ K 2δ,1,

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

γ2δ,1 ≥ N1,2 > 0 as δ −→ 0. (2.5.31)

Finally, consider a ∈ (1, 2), p ≥ 1. For the operator Rh(δ) defined by (2.4.92) one gets using the Lagrangeformula:

‖Rh(δ)fδ − f ′‖p ≤ ‖Rh(δ)(fδ − f)‖p + ‖Rh(δ)f − f ′‖p

≤ 2δ

h+ Na,ph

a0 . (2.5.32)


Minimizing the right-hand side of (2.5.32) with respect to h ∈ (0,∞) yields

ha(δ) =

(2δ

a0Na,p

) 1a

, η(δ) = a(Na,p)1a

(2δ

a0

)a0a

, a = 1 + a0, 0 < a0 < 1.

The case a = 2 is treated in the introduction (see estimate 2.4.93). So one arrives at 2.4.13-2.4.14. Thiscompletes the proof. 2

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 Rh(δ) defined in (2.4.92). The

right-hand side of the error estimate of the type (2.4.93) is now replaced by E(h) := 2δh

+ w(h), wherew → 0 as h → 0, provided that a = 1. Minimizing E with respect to h for a fixed δ, one obtains aminimizer h(δ) → 0 as δ → 0 and the error estimate E(h(δ)) → 0 as δ → 0. Therefore one gets (2.6).Alternatively, if one chooses h1(δ) → 0 as δ → 0, such that 2δ

h1= w(h1), then E(h1(δ)) ≤ 2w(h1(δ)) → 0.

The first proof is completed. 2

Remark 2.5.3. The statement of Theorem 2.5.2 with (2.4.11) replaced by (2.4.12) is obvious: since fis fixed, one may take Rh(δ)fδ = f ′. This is, of course, of no practical use because f ′ is unknown.

The second proof is based on DSM. The ideas of this proof have an advantage of being applicable to awide variety of ill-posed problems, and not only to stable numerical differentiation. By this reason we givethis proof in detail. In order to show that for a = 1 there exists an operator Rh(δ) : L2(0, 1) → L2(0, 1),( L2(0, 1) is a real Hilbert space) such that (2.5.18) holds we use the DSM ( dynamical systems method)(see Section 2.4). This approach consists of the following steps:Step 1. Solve the Cauchy problem:

v = −[Av + h(t)v − fδ], v(0) = v0 ∈ L2(0, 1), (2.5.33)

where A is defined in (2.4.7), p = 2, v := dvdt, ‖fδ − f‖ ≤ δ and

h(t) ∈ C1[0,+∞), h(t) > 0, h(t) 0,h(t)

h2(t)−→ 0 as t −→ +∞. (2.5.34)

Step 2. Calculate v(tδ), where tδ > 0 is a number such that tδ → +∞ and δh(tδ) → 0 as δ → 0. Then

Rh(δ)fδ := v(tδ) (2.5.35)

andsup

fδ∈B2δ,f

‖Rh(δ)fδ − f ′‖2 ≤ η(δ, f) −→ 0 as δ −→ 0 (2.5.36)

with η(δ, f) given by (2.5.49) below and f ∈ K(2, δ, 1).To verify (2.5.36) consider the problem

Aw + h(t)w − f = 0. (2.5.37)

Since A is monotone in L2(0, 1):

(Aφ, φ) =

1∫

0

x∫

0

φ(τ )dτ

φ(x)dx =

1

2

1∫

0

x∫

0

φ(τ )dτ

2

′

dx ≥ 0, (2.5.38)

for any φ ∈ L2(0, 1) and h(t) > 0, the solution w(t) to (2.5.37) exists, is unique, and admits the estimate

(A(w − f ′), w − f ′) + h(t)‖w‖22 = h(t)(w, f ′), ‖w‖2 ≤ N1,2. (2.5.39)


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

[A+ h(t)I]w = −h(t)w, ‖w‖2 ≤ |h(t)|h(t)

‖w‖2 ≤ |h(t)|h(t)

N1,2, (2.5.40)

where (2.5.39) was used. Denotez(t) := v(t) − w(t). (2.5.41)

From (2.5.37) and (2.5.33) one obtains

z(t) = −w − [A+ h(t)I]z + fδ − f, z(0) = v0 − w(0). (2.5.42)

Multiply (2.5.42) by z(t) and get

(z, z) = −(w, z) − (Az, z) − h(t)(z, z) + (fδ − f, z). (2.5.43)

Let ‖z(t)‖2 := g(t), then (2.5.39) and (2.5.43) imply

gg ≤ (‖w‖2 + δ)g − h(t)g2. (2.5.44)

Since g ≥ 0, from (2.5.44) and (2.5.40) it follows that

g ≤ N1,2|h(t)|h(t)

+ δ − h(t)g(t), g(0) = ‖v0 − w(0)‖. (2.5.45)

So,

g(t) ≤ e−

t∫0

h(s)ds

g(0) +

t∫

0

e

τ∫0

h(s)ds(

N1,2|h(τ )|h(τ )

+ δ

)dτ

. (2.5.47’)

Under assumption (2.5.34), one has∞∫

0

h(t)dt = ∞. (2.5.46)

Indeed, (2.5.34) implies |h|h2 ≤ c, c = const > 0, so d

dt1h ≤ c, 1

h(t) − 1h(0) ≤ ct, 1

h(t) ≤ c0 + ct,

c0 := [h(0)]−1 > 0, and h(t) ≥ 1c0+ct

. Conclusion (2.5.46) follows.

If one chooses t = tδ so that tδ → +∞ and δh(tδ) → 0 as δ → 0, then by (2.5.47’) and (2.5.46), using

L’Hospital’s rule one obtains

‖v(tδ) − w(tδ)‖2 := g(tδ) −→ 0 as δ −→ 0. (2.5.47)

The existence of the solution to (2.5.33) on [0,+∞) is obvious, since equation (2.5.33) is linear with abounded operator.

We claim that‖w(tδ) − f ′‖2 −→ 0 as δ −→ 0 (2.5.48)

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:

supfδ∈B2

δ,f

‖v(tδ) − f ′‖2 ≤ ‖w(tδ) − f ′‖2 + e−

tδ∫0

h(s)ds[g(0)+

+

tδ∫

0

e

τ∫0

h(s)ds(

N1,2|h(τ )|h(τ )

+ δ

)dτ

:= η(δ, f) −→ 0 as δ −→ 0. (2.5.49)


0 0.2 0.4 0.6 0.8 1−1

−0.5

0

0.5

1Figure 1

f(x)=2.5x(x−0.4), 0<x<0.4, fdelta(x)=00 0.2 0.4 0.6 0.8 1

−1

−0.5

0

0.5

1Figure 2

fl(x)=5x−1, 0<x<0.4

0 0.2 0.4 0.6 0.8 1−0.5

0

0.5Figure 3

f(x)=6.25x(x−0.08), 0<x<0.08, fdelta(x)=00 0.2 0.4 0.6 0.8 1

−0.5

0

0.5Figure 4

fl(x)=12.5x−0.5, 0<x<0.08

Let us now prove (2.5.48). Because f ′ ∈ L2(0, 1) and f(0) = 0, one can rewrite (2.5.37) as A(w − f ′) +h(t)w = 0. This and the monotonicity of A imply h(t)(w,w − f ′) ≤ 0, so, since h(t) > 0, one gets(w,w− f ′) ≤ 0, and ‖w‖2 ≤ ‖f ′‖2. Thus, w converges weakly in L2(0, 1) to some element ψ, w ψ ast → ∞. Because A is monotone, it is w−closed, that is w ψ and A(w− f ′) → 0 imply A(ψ− f ′) = 0,so ψ = f ′ and w f ′. The inequality (w,w − f ′) ≤ 0 can be written as ‖w − f ′‖2

2 ≤ (f ′, f ′ − w), and(f ′, w − f ′) → 0 because w − f ′ 0. Therefore the claim (2.5.48) is proved and the second proof iscompleted.

Numerical aspects

Figures 1-4 illustrate the impossibility to differentiate stably a function, which does not have a boundon f (a), a > 1. If one takes the function

f(x) :=

N

21,∞

4δ x(x− 4δN1,∞

), 0 ≤ x ≤ 4δ

0, 4δ < x ≤ 1,(2.5.50)

and fδ ≡ 0, then

f(x) ∈ f : f ∈ W 1,∞(0, 1), ‖f ′‖∞ < N1,∞, ‖f − fδ‖∞ ≤ δ,

and any formula of numerical differentiation will give error not going to zero as δ → 0, because, by(2.5.27), one has:

infRh(δ)

‖Rh(δ)fδ − f ′‖∞ ≥ N1,∞.


In Figure 1 one can see f(x) given by (2.5.50) with δ = 0.1 and N1,∞ = 1:

f(x) :=

2.5x(x− 0.4), 0 ≤ x ≤ 0.4

0, 0.4 < x ≤ 1.(2.5.51)

Figure 2 presents

f ′(x) :=

5x− 1, 0 ≤ x ≤ 0.4

0, 0.4 < x ≤ 1.(2.5.52)

Figure 3 shows the case δ = 0.01 and N1,∞ = 0.5:

f(x) :=

6.25x(x− 0.08), 0 ≤ x ≤ 0.08

0, 0.08 < x ≤ 1.(2.5.53)

The derivatives are given in Figure 4:

f ′(x) :=

12.5x− 0.5, 0 ≤ x ≤ 0.4

0, 0.4 < x ≤ 1.(2.5.54)

Even if the bound on f (a) in some norm is given, one can experience difficulties with stable differenti-ation. Namely, if δ is fixed and Na,p is very large, then hopt in finite difference scheme (2.4.92) is verysmall, and practitioners might not have sufficiently many observation points. Another difficulty is: theestimated error εa,p in such a case is very big and does not give any information regarding the accuracyof computations. This is illustrated in Table 2.1 below for the function f(x) = sin((πx)n) and δ = 0.1.

Table 2.1:

n N2,p hopt εa,p

5 1.24 · 106 5.67 · 10−4 7.04 · 102

10 3.27 · 1011 1.11 · 10−6 3.62 · 105

15 4.41 · 1016 3.01 · 10−9 1.33 · 108

20 7.94 · 1021 7.10 · 10−12 5.63 · 1010

25 2.62 · 1027 1.24 · 10−14 3.24 · 1013

Figures 5 – 8 show the exact and computed derivatives of f(x) = sin((πx)5). The derivatives of thisfunction were computed in the presence of the noise function

e(x) = δ(cos(2x) + cos(3x2)

)/2, (2.5.55)

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


0 0.2 0.4 0.6 0.8 1−1500

−1000

−500

0

500

1000

1500Figure 5

derivative computed with hopt=0.00056778

delta

=0.

1

0 0.2 0.4 0.6 0.8 1−1500

−1000

−500

0

500

1000

1500Figure 6

exact derivative

0 0.2 0.4 0.6 0.8 1−200

−100

0

100

200Figure 7

derivative computed with h=0.005

delta

=0.1

0 0.2 0.4 0.6 0.8 1−100

−50

0

50

100Figure 8

derivative computed with h=0.01

delta

=0.1

2.5.2 Stable summation of the Fourier series and integrals with perturbedcoefficients

Assume that f(x) is a smooth 2π-periodic function

f(x) = (2π)−12

∞∑

l=−∞fl exp(ilx), (2.5.56)

where

fl = (2π)−12

∫ π

−πf(x) exp(−ilx)dx, (2.5.57)

and

(1 + l2)s2 |fl| ≤ Ms, s >

3

2. (2.5.58)

Assume that fδl are given such that

|fδl − fl| < δ, l = 0,±1,±2, . . . , (2.5.59)

and fl are not known.

The problem is: given fδl, calculate stably f ′(x). In other words, calculate Fδ such that

||Fδ − f ′|| < η(δ) −→ 0 as δ −→ 0. (2.5.60)


The norm here can depend on the particular problem. Let us assume that this is the L2[(−π, π)] norm.Let us look for Fδ of the form:

Fδ(x) = (2π)−1/2N∑

l=−Nilfδl exp(ilx). (2.5.61)

One has, using Parseval’s equality,

||Fδ − f ′||2 =

N∑

l=−Nl2|fδl − f |2 +

∑

|l|>Nl2|fl|2

≤ δ2N (N + 1)(2N + 1)

3+ 2M2

s

∞∑

l=N+1

l2

(1 + l2)s

≤ 2

3(N + 1)3δ2 +

2M2s

2s− 3(N + 1)−2s+3

:= c0ν3δ2 + c1ν

−2s+3, ν := N + 1,

(2.5.62)

where the constants c0 and c1 are defined by the last equation. Minimizing the right-hand side of (2.5.62)with respect to ν > 1, with δ > 0 being fixed, we find the optimal ν := ν(δ):

ν(δ) =

(2s − 3)c1

3c0

12s

δ−1s := c2δ

− 1s , (2.5.63)

and the error estimate:||Fδ − f ′|| ≤ c3δ

2s−3s := η(δ). (2.5.64)

If s = 2, then η(δ) = c3δ12 . Let us summarize the result.

Theorem 2.5.4. Assume that fδl are given such that (2.5.58) and (2.5.59) hold. Define Fδ by formula(2.5.61) with N = N (δ) = ν(δ) − 1, where ν(δ) is defined in (2.5.63). Then (2.5.60) holds with η(δ)defined by (2.5.64).

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 Au := V u =∫ x0 V (x, y)u(y)dy. If V (x, y) ≥ c > 0 is a bounded function,

and Vx(x, y) is a kernel for which max0≤x≤B∫ x0|Vx(x, y)|2dy ≤ c1, then (1.3.1), after a differentiation,

yields

u(x) +

∫ x

0

Vx(x, y)u dy

V (x, x)=

f ′(x)

V (x, x), 0 ≤ x ≤ l, (2.5.12)

provided that f ′ ∈ L2(0, b). This is a Fredholm second-kind integral equation, for which the Fredholmalternative is valid in L2(0, b).

If V (x, y) is not differentiable with respect to x, or V (x, x) may vanish, then equation (1.3.1) withA = V can be solved by the methods discussed in Section 2.1–2.4.

Another general approach to stable solution of the equation V u = f consists of the factorizationV u = Q(I + S)u = f , where S is a compact operator, I is the identity, the null-space N (I + S) = 0 istrivial, so that by Fredholm’s alternative the equation (I+S)u = w can be stably solved by a projectionmethod, and the operator Q is such that w can be stably found from the equation Qw = f . If the noisydata fδ are given in place of f , then a stable solution to the equation Qw = fδ is given by a formula


wδ = Rδfδ, and a stable solution of the equation V u = f with noisy data fδ is given by the formulauδ = (I+S)−1Rδfδ. A numerical implementation of this scheme is given in [RSm6]. If V (x, y) = V (x−y)and V (0) 6= 0, then the above scheme leads to the operator Q whose stable inversion is equivalent tostable differentiation, Qw =

∫ x0w(y)dy.

A discretization method for stable solution of Volterra integral equations of the first kind is proposedand justified in [RG], where the error estimates for the proposed method are also obtained.

2.5.4 Deconvolution problems

Equation (1.3.1) with Au =∫DA(x, y)u(y)dy can be solved by the methods of Sections 2.1 – 2.4, provided

that u ∈ H. In the special case, when the kernel

A(x, y) = R(x, y), and QR(x, y) = Pδ(x− y) in Rr, r ≥ 1, (2.5.13)

where Q and P are elliptic operators, δ(x − y) is the delta-function, <Q ≥ c > 0, one can use thetheory from [R121], [R189], and prove, under suitable assumptions, that the operator R, where Rh :=∫D R(x, y)h(y)dy, is an isomorphism of H−a(D) onto Ha(D), where a = n−m

2 , n = ordQ, m = ordP ,n ≥ m, Ha(D) is the usual Sobolev space, H−a(D) is its dual with respect to L2(D) inner product,H−a(D) is, in other words, the closure of C∞

0 (D) in the norm of H−a(Rr), that is, the subset of theelements of H−a(Rr) with support D,D := D ∪ S, S := ∂D.

If L is a selfadjoint elliptic operator in L2(R2), ordL := s and Q(λ) and P (λ) are positive polynomials,degQ = n, degP = m ≤ n, then Q := Q(L) and P := P (L) are elliptic operators, and if R(x, y)

solves (2.5.1) then a = s(n−m)2 . Equation Rh = f has a unique solution in H−a(D) for any f ∈ Ha(D),

and R : H−a(D) → Ha(D) is an isomorphism. In [R121], [R189] one finds analytical formulas for thesolution h. Under the above assumptions, the equation (*) Rh = f in D does not have integrablesolutions, in general. It has only distributional solutions of finite order of singularity, in general. Findinga solution of minimal order of singularity (mos solution) is a well posed problem ([R121]). The minimalorder of singularity is equal to a. Since R is an isomorphism of H−a(D) onto Ha(D) the problem ofsolving equation (*) is well-posed. One does not need regularization methods for finding the solution to(*).

Example 2.5.1.

Rh =

∫ 1

−1

e−|x−y|h(y)dy = f(x), −1 ≤ x ≤ 1.

In this example L = 1iddx

, P (λ) = 1, Q(λ) = λ2+12

, a = 1, e−|x| = 1π

∫∞−∞

eiλxdλλ2+1

, h(x) = −f ′′(x)+f(x)2

+

δ(x− 1) f′(1)+f(1)

2 + −f ′(−1)+f(−1)2 δ(x + 1).

One can see that generically h(x) is not an integrable function, it is a distribution, h ∈ H−1(−1, 1).If and only if f ′(1) + f(1) = 0 and f ′(−1) = f ′(1) and f ′′ ∈ L1(−1, 1), is h ∈ L1(−1, 1) Thus, equation

Rhδ = fδ, ‖ fδ − f ‖H1≤ δ, f ∈ H1,

has a solution hδ which depends on fδ continuously in the norm H−1(−1, 1) : ‖hδ − h‖H−1(−1,1) ≤ cδwhere c =‖ R−1 ‖H1→H−1< ∞. Details one can see in [R121].

2.5.5 Ill-conditioned linear algebraic systems

Let A : Rn → Rn in (1.3.1). Then A can be represented by a matrix (aij). If A is non-singular, defineits condition number

γ(A) := ‖A‖∥∥A−1

∥∥.One has (see (1.4.3)) ‖∆u‖

‖u‖ /‖∆f‖‖f‖ ≤ ν(A). Thus, if ν(A) is large then small relative errors ‖∆f‖

‖f‖ in the

data may lead to large relative errors ‖∆u‖‖u‖ in the solution. Solving linear algebraic ill-conditioned is


an ill-posed problem practically. Note that ν(A) = supu6=0(‖Au‖‖u‖ )/ infv 6=0(

‖Av‖‖v‖ ), because infv 6=0

‖Av‖‖v‖ =

‖A−1‖−1, if A is not singular (not degenerate: detA 6= 0) Also, ν(A) = smax

smin, where s2j are eigenvalues

of A∗A.Examples below are taken from [VA].

Example 2.5.2. A =

(4.1 2.89.7 6.6

), f = (4.1, 9.7),u = (1, 0), Au = f . Let g = (4.11, 9.70). Then

u = (0.34, 0.97). Here µ(A) = 2, 249.5.

Example 2.5.3. Hilbert matrix aij = (i+j−1)−1 , 1 ≤ i, j ≤ n, µ(A) = 1.5·107 if n = 6, µ(A) = 1.6·1013

if n=10

2.6 Projection methods for ill-posed problems

Let A : H → H be a linear, injective, bounded operator in a Hilbert space H, A−1 : R(A) → H isunbounded, so that the problem of solving the equation Au = f is ill-posed.

Let Hn ⊂ Hn+1 ⊂ . . . be a sequence of finite-dimensional subspaces of H which is limit-dense inH, that is, for any f ∈ H one has limn→∞ ρ(f,Hn) = 0, where ρ(f,Hn) = infv∈Hn ‖f − v‖ is thedistance from f to Hn. Let Pn be the orthoprojection operators on Hn. Assume that fδ is givensuch that ‖fδ − f‖ ≤ δ. The problem consists of finding uδ := um(δ), which solves the equation(∗)PmAum(δ) = Pmfδ , um(δ) ∈ Hm, and such that limδ→0 ‖um(δ) − u‖ = 0. Equation (*) is an equationof a projection method. A general approach to finding a stable approximation um(δ) is the following one:let Rh be a regularizer in the sense limh→0RhAu = u ∀u ∈ H, and define uδ = uδ(h,m) := RhPmAPmfδ.Then ‖uδ − u‖ ≤ ‖RhPmfδ −Rhfδ‖ + ‖Rhfδ −Rhf‖ + ‖Rhf − u‖ ≤ ‖Rh‖(‖(Pm − I)fδ‖+ ‖fδ − f‖) +‖RhAu − u‖ := a(h)l(m, δ) + p(h), where p(h) := ‖RhAu − u‖ → 0 as h → 0, a(h) := ‖Rh‖ → ∞ ash→ 0, l(m, δ) := ‖fδ−f‖+‖Pmfδ−fδ‖ → 0 as δ → 0 and m = m(δ) is chosen suitably. More precisely,for any fixed fδ one has limm→∞ ‖Pmfδ−fδ‖ = 0, because the sequence Hn is limit-dense. Thus, one canchoose m = m(δ) so that ‖Pmfδ − fδ‖ ≤ δ, then l(m(δ), δ) ≤ 2δ, and for a fixed δ >= 0 one can chooseh = h(δ) so that 2a(h)δ+ p(h) = min := m(δ) → 0 as δ → 0. Therefore ‖Rh(δ)Pm(δ)fδ −u‖ ≤ m(δ) → 0as δ → 0, so that u is stably approximated.

There are many ways to choose the regularizer Rh: one can use a Variational Regularization, Quasiso-lutions, Iterative Regularization, and the DSM. One can also use the method developed in Section 2.5.2for 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, ||A|| ≤ 1, and (1.3.1) is solvable, i.e, f = Ay for some y. We assumethat y ⊥ N (A), that is, y is the minimal-norm solution. By Lemma 2.1.11, a solvable equation (1.3.1)is equivalent to equation (2.1.6). Denote g := f1 := A∗f , assume that gδ is given in place of g, and||gδ − g|| ≤ δ. Let Bϕj = s2jϕj , where ϕj, j = 1, 2, ......., is an orthonormal system of eigenvectors of

the selfadjoint compact operator B, and s2j > 0 are the corresponding eigenvalues. Let cj :=gδ,j

s2j, and

uδ :=∑N

j=1 cjϕj. The element uδ is a regularizer if N = N (δ) is chosen properly, as in Section 2.5.2. Inthis case one has limδ→0 ||uδ − y|| = 0.

See more on this topic in [IVT], [VA].

Chapter 3

One-dimensional inverse scatteringand spectral problems

Inverse scattering and spectral one-dimensional problems are discussed in this Chapter systematicallyin a self-contained way. Many novel results due to the author are presented. The classical results areoften 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 proce-dures ;

(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 resultis obtained for the case when part of the phase shifts is known. The Newton-Sabatier (NS) schemefor inversion of fixed-energy phase shifts is analyzed. This analysis shows that the NS scheme isfundamentally wrong in the sense that its foundations are wrong, and this scheme is not a validinversion method;

(6) Complete presentation of the Krein inverse scattering theory is given apparently 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;

(10) A study of an inverse scattering problem for an inhomogeneous Schrodinger equation;

(11) A study of inverse problem of ocean acoustics;

(12) Theory of ground-penetrating radars.

75

76 CHAPTER 3. ONE-DIMENSIONAL INVERSE SCATTERING AND SPECTRAL PROBLEMS

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 indetail. 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 onProperty 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) inver-sion 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 firsttime;

(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 spherically-symmetric potentialfrom 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-energyphase 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 problemhas 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 thefull line, iterative methods for finding potential q: (a) from two spectra [R181],[R221], (b) from S-matrixalone when q is compactly supported [R184], approximate methods for finding q from fixed-energy phaseshifts [RSch],[RSm2], property of resonances [R139], inverse scattering for systems of equations, etc.

3.1.2 Auxiliary results

Let q(x) ∈ L1,1, L1,m = q : q(x) = q(x),∫∞0

(1+x)m|q(x)|dx < ∞, and q ∈ L2loc(R+), where L2

loc (R+)consists of functions belonging to L2(0, a) for any a < ∞, and overline stands for complex conjugate.

Consider the differential expression ù = −u′′ + q(x)u with domain of definition D(l0) = u : u(0) =0, u ∈ C2

0(0,∞), where C20 (0,∞) is the set of C2(R+)-functions vanishing in a neighborhood of infinity,

R+ := [0,∞). If H is the Hilbert space L2(R+), then `0 is densely defined symmetric linear operatorin H, essentially self-adjoint, that is, the closure ` of `0 in H is selfadjoint. It is possible to construct aselfadjoint operator ` assuming that q ∈ L1

loc(R+). Such a theory is technically more difficult, becauseit is not even obvious a priori that the set D(`0) := u : u ∈ C2

0 (R+), ù ∈ L2(R+) is dense in H (infact, it is). Such a theory is presented in [Nai]. If one drops the assumption q ∈ L2

loc, then D(`0) is nota domain of definition of ` since there are functions u ∈ D(`0) for which ù /∈ L2(R+). In the future wemean by ` a self-adjoint operator generated by the differential expression ` and the boundary conditionu(0) = 0.

3.1. INTRODUCTION 77

This operator has absolutely continuous spectrum, which fills (0,∞), and discrete, finite, negativespectrum −k2

j1≤j≤J , where −k2j are the eigenvalues of `, all of them are simple,

`ϕj := −ϕ′′j + qϕj = −k2

jϕj , ϕj(0) = 0, ϕ′j(0) = 1, (3.1.1)

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

1

cj:=

∫ ∞

0

ϕ2j dx. (3.1.2)

The functions ϕ(x, k) and θ(x, k) are defined as the unique solutions to the problems:

`ϕ = k2ϕ, x > 0; ϕ(0, k) = 0, ϕ′(0, k) = 1, (3.1.3)

`θ = k2θ, x > 0; θ(0, k) = 1, θ′(0, k) = 0. (3.1.4)

These functions are well defined for any q(x) ∈ L1loc(R+). Their existence and uniqueness can be proved

by using the Volterra equations for ϕ and θ. If q ∈ L1,1, then the Jost solution f(x, k) exists and isunique. This solution is defined by the problem:

`f := −f ′′ + qf = k2f, f(x, k) = exp(ikx) + o(1) as x −→ +∞; f(0, k) := f(k). (3.1.5)

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

f(x, k) = exp(ikx) +

∫ ∞

x

sin[k(t− x)]

kq(t)f(t, k)dt. (3.1.6)

If q ∈ L1,1, then this equation implies that f(x, k) is an analytic function of k in C+ = k : Imk > 0,f(x, k) = f(x,−k) for k > 0. The Jost function is defined as f(k) := f(0, k). It has exactly J simpleroots ikj , kj > 0, where −k2

j , 1 ≤ j ≤ J , are the negative eigenvalues of `. The number k = 0

can be a zero of f(k). If f(0) = 0, then f(0) 6= 0, where f (k) := dfdk . Existence of f(0) is a fine

result under the only assumption q ∈ L1,1 ( see Theorem 3.1.3 below, and [R139]) and an easy one ifq ∈ L1,2 := q : q = q,

∫∞0

(1 + x2)|q(x)|dx < ∞. The phase shift δ(k) is defined by the formula

f(k) = |f(k)|exp(−iδ(k)), δ(∞) = 0, f(∞) = 1, (3.1.7)

where the last equation in (3.1.7) follows from (3.1.6). Because q(x) = q(x), one has δ(−k) = −δ(k) fork ∈ R. One defines the S-matrix by the formula

S(k) :=f(−k)f(k)

, k ∈ R. (3.1.8)

The function S(k) is not defined for complex k if q ∈ L1,1, but if |q(x)| ≤ c1exp(−c2|x|γ), γ > 1, thenf(k) is an entire function of k and S(k) is meromorphic in C. If q(x) = 0 for x > a, then f(k) is anentire function of exponential type ≤ 2a (see Section 3.5.1).

If q ∈ L1,1, then at k2 = −k2j , kj > 0, the Jost solution fj(x) := f(x, ikj) is proportional to

ϕj(x) := ϕ(x, ikj), fj and ϕj both belong to L2(R+). The integral equation for ϕ is:

ϕ(x, k) =sin(kx)

k+

∫ x

0

sin k(x− s)

kq(s)ϕ(s, k)ds. (3.1.9)

One has:

ϕ(x, k) =f(x, k)f(−k) − f(x,−k)f(k)

2ik, (3.1.10)


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:

f ′(0, k)f(−k) − f ′(0,−k)f(k) = 2ik. (3.1.11)

If k = ikj then fj(x) ∈ L2(R+), as one can derive easily from equation (3.1.6). In fact, |fj(x)| ≤ ce−kjx,x ≥ 0. If k > 0, then f(x,−k) = f(x, k). If q = q then fj(x) is a real-valued function. The functionf(x, k) is analytic in C+ but is, in general, not defined for k ∈ C− := k : Imk < 0. In particular,(3.1.11), in general, is valid on the real axis only. However, if |q(x)| ≤ c1exp(−c2|x|γ), γ > 1, then f(k)is defined on the whole complex plane of k, as was mentioned above. Let us denote f(x, k) := f+(x, k)for k ∈ C+ and let f−(x, k) be the second, linearly independent, solution to equation (3.1.5) for k ∈ C+.If f+ ∈ L2(R+), then f− /∈ L2(R+). One can write a formula, similar to (3.1.10), for k ∈ C+:

ϕ(x, k) = c(k)[f−(0, k)f(x, k) − f(0, k)f−(x, k)], (3.1.12)

where c(k) = const 6= 0. For ϕ(x, ikj) ∈ L2(R+), it is necessary and sufficient that f(ikj) = 0. In fact,

f(ikj) = 0, f (ikj) 6= 0, 1 ≤ j ≤ J, (3.1.13)

where f = dfdk . To prove the second relation in (3.1.13), one differentiates (3.1.5) with respect to k and

getsf ′′ + k2f − qf = −2kf. (3.1.13a)

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

−2ikj

∫ ∞

0

f2j dx = (ff ′ − f ′f )|∞0 = f ′(0, ikj)f (ikj).

Thus, ∫ ∞

0

f2j dx =

f ′(0, ikj)f (ikj)

−2ikj:=

1

sj> 0. (3.1.14)

It follows from (3.1.14) that f(ikj) 6= 0. The numbers sj > 0 are called the norming constants:

sj = − 2ikj

f ′(0, ikj)f (ikj), 1 ≤ j ≤ J. (3.1.15)

Definition 3.1.1. Scattering data is the triple:

S := S(k), kj , sj, 1 ≤ j ≤ J, S(k) :=f(−k)f(k)

, kj > 0, sj > 0. (3.1.16)

The Jost function f(k) may vanish at k = 0. If f(0) = 0, then the point k = 0 is called a resonance.If |q(x)| ≤ c1exp(−c2|x|γ), γ > 1, then the zeros of f(k) in C− are called resonances. As we have seenabove, there are finitely many zeros of f(k) in C+, these zeros are simple, their number J is the numberof negative eigenvalues −k2

j , 1 ≤ j ≤ J , of the selfadjoint Dirichlet operator `. If q ∈ L1,1 then thenegative spectrum of ` is finite [M].

The phase shift δ(k), defined in (3.1.17), is related to S(k):

S(k) = e2iδ(k), (3.1.17)

so that S(k) and δ(k) are interchangeable in the scattering data. One has fj(x) = f ′(0, ikj)ϕj(x),

becausef(x,ikj)f ′(0,ikj)

solves (3.1.3). Therefore

∫ ∞

0

ϕ2jdx =

1

sj [f ′(0, ikj)]2:=

1

cj. (3.1.18)


Thus

cj = −2ikjf′(0, ikj)

f(ikj), 1 ≤ j ≤ J. (3.1.19)

In Section 3.4.1 the notion of spectral function ρ(λ) is defined. It will be proved in Section 3.5.1 forq ∈ L1,1 that the formula for the spectral function is:

dρ(λ) =

√λdλ

π|f(√λ)|2

, λ ≥ 0,

J∑

j=1

cjδ(λ + k2j )dλ, λ < 0,

(3.1.20)

where cj are defined in (3.1.18)-(3.1.19). The spectral function is defined in Section 3.4.5 for anyq ∈ L1

loc(R+), q = q. Such a q may grow at infinity. On the other hand, the scattering theory isconstructed for q ∈ L1,1.

Let us define the index of S(k):

J := indS(k) :=1

2π∆R argS(k) =

1

2πi

∫ ∞

−∞d lnS(k). (3.1.21)

This definition implies that indS(k) = indf(−k) − indf(k) = −2indf(k). Therefore:

indS(k) =

−2J if f(0) 6= 0,

−2J − 1 if f(0) = 0,(3.1.22)

because a simple zero k = 0 contributes 12

to the index, and the index of an analytic in C+ functionf(k), such that f(∞) = 1, equals to the number of zeros of f(k) in C+ plus half of the number of itszeros 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,

ϕ(x, k) = ϕ0(x, k) +

∫ x

0

K(x, y)ϕ0(y)dy := (I +K)ϕ0, ϕ0 :=sin(kx)

k, (3.1.23)

and

f(x, k) = eikx +

∫ ∞

x

A(x, y)eikydy := (I + A)f0, f0 := eikx, (3.1.24)

and the properties of the kernels A(x, y) and K(x, y) are discussed in Section 3.5.2 and Section 3.4.2respectively. The transformation operator I +K transforms the solution ϕ0 to the equation (3.1.3) withq = 0 into the solution ϕ of (3.1.3), satisfying the same as ϕ0 boundary conditions at x = 0. Thetransformation operator I+A transforms the solution f0 to equation (3.1.5) with q = 0 into the solutionf of (3.1.5) satisfying the same as f0 “boundary conditions at infinity”.

One can prove (see [M] and Sec. 5.7) the following estimates

|A(x, y)| ≤ cσ

(x+ y

2

), c = const > 0, σ(x) :=

∫ ∞

x

|q(t)|dt, (3.1.25)

∣∣∣∣Ay(x, y) +1

4q

(x+ y

2

) ∣∣∣∣+∣∣∣∣Ax(x, y) +

1

4q

(x+ y

2

)∣∣∣∣ ≤ cσ(x)σ

(x+ y

2

), (3.1.26)

and A(x, y) solves the equation:

A(x, y) =1

2

∫ ∞

x+y2

qds+

∫ ∞

x+y2

ds

∫ y−x2

0

dtq(s − t)A(s − t, s+ t). (3.1.27)

By Hm = Hm(R+) we denote Sobolev spaces Wm,2. 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 q ∈ L1,1 from the corresponding scattering dataS (see (3.1.6)).

A study of ISP consists of the following:

(1) One proves that ISP has at most one solution.

(2) One finds necessary and sufficient conditions for S to be scattering data corresponding to a q ∈ L1,1

(characterization of the scattering data problem).

(3) One gives a reconstruction method for calculating q ∈ L1,1 from the corresponding S .

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 q1 and q2 from thisclass generate the same ρ(λ), then q1 = q2.

(2) One finds necessary and sufficient conditions on ρ(λ) which guarantee that ρ(λ) is a spectralfunction corresponding to some q from the above class.

(3) One gives a reconstruction method for finding q(x) from the corresponding ρ(λ).

3.1.4 Property C for ODE.

Denote by `m operators ` corresponding to potentials qm ∈ L1,1, and by fm(x, k) the corresponding Jostsolutions, m = 1, 2.

Definition 3.1.2. We say that a pair `1, `2 has property C+ iff the set f1(x, k)f2(x, k)∀k>0 iscomplete (total) in L1(R+).

This means that if h ∈ L1(R+) then∫ ∞

0

h(x)f1(x, k)f2(x, k)dx = 0 ∀k > 0

⇒ h = 0. (3.1.28)

We prove in Section 2.1 that a pair `1, `2 does have property C+ if qm ∈ L1,1. Let `ϕ := −ϕ′′ +q(x)ϕ, and let ϕj correspond to q = qj,

`ϕ− k2ϕ = 0, ϕ(0, k) = 0; ϕ′(0, k) = 1; `θ − k2θ = 0, θ(0, k) = 1, θ′(0, k) = 0. (3.1.29)

Definition 3.1.3. We say that a pair `1, `2 has property Cϕ iff the set ϕ1(·, k)ϕ2(·, k) is completein L1(0, b) for any b > 0, b <∞.

This means that if h ∈ L1(0, b), then:∫ b

0

h(x)ϕ1(x, k)ϕ2(x, k)dx

∀k>0

=⇒ h = 0. (3.1.30)

In Theorem 3.2.5 we prove that there is a h 6= 0 for which∫ ∞

0

h(x)ϕ1(x, k)ϕ2(x, k)dx = 0 ∀k > 0

for a suitable q1 6= q2, q1, q2 ∈ L1,1. Therefore Property Cϕ with b = ∞ does not hold, in general.Property Cθ is defined similarly to Property Cϕ, with functions θj(x, k) replacing ϕj(x, k).In Section 3.2 we prove that properties C+, Cϕ and Cθ hold, and give many applications of these

properties throughout this work.


3.1.5 A brief description of the basic results.

The basic results of this work include:

(1) Proof of properties C+, Cϕ and Cθ. Demonstration of many applications of these properties.

(2) Analysis of the invertibility of the steps in the inversion procedures of Gel’fand-Levitan (GL) forsolving inverse spectral problem:

ρ ⇒ L ⇒ K ⇒ q, (3.1.31)

where

q = 2dK(x, x)

dx, (3.1.32)

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

L(x, y) =

∫ ∞

−∞ϕ0(x, λ)ϕ0(y, λ)dσ(λ), dσ(λ) := d[ρ(λ) − ρ0(λ)],

dρ0 =

√λdλπ , λ ≥ 0,

0, λ < 0.

(3.1.33)

ρ0 = 2λ3/2

3π , ρ0 is the spectral function of ` with q = 0, and K solves the Gel’fand-Levitan equation

K(x, y) +

∫ x

0

K(x, s)L(s, y)ds + L(x, y) = 0, 0 ≤ y ≤ x. (3.1.34)

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

ρ ⇔ L ⇔ K ⇔ q, (3.1.35)

which holds under a weak assumption on ρ. Namely, assume that

ρ ∈ G , (3.1.36)

where G is the set of non-decreasing functions ρ of bounded variation on every interval (−∞, b), b < ∞,such that the following two assumptions, A1) and A2) hold.

Denote L20(R+) the set of L2(R+) functions vanishing in a neighborhood of infinity. Let h ∈ L2

0(R+)and H(λ) :=

∫∞0h(x)ϕ0(x, λ)dx.

Assumption A1) is:

If h ∈ L20(R+) and

∫ ∞

−∞H2(λ)dρ(λ) = 0, then h = 0. (3.1.37)

Let

H := H(λ) : h ∈ C∞0 (R+), H(λ) :=

∫ ∞

0

h(x)ϕ0(x, λ)dx, (3.1.38)

ρ1 and ρ2 belong to P and ν := ρ1 − ρ2 (see Section 3.4.2).Assumption A2) is:

If

∫ ∞

−∞H2(λ)dν = 0 ∀H ∈ H , then ν = 0. (3.1.39)

In order to insure the one-to-one correspondence between spectral functions ρ and selfadjoint op-erators `, we assume that q is such that the corresponding ` is “in the limit point at infinity case”.This means that the equation (` − z)u = 0, Im z > 0 has exactly one nontrivial solution in L2(R+),ù = −u′′ + q(x)u. If q ∈ L1,1 then ` is “in the limit point at infinity case”.


(3) Analysis of the invertibility of the Marchenko inversion procedure for solving ISP:

S ⇒ F ⇒ A⇒ q, (3.1.40)

where

F (x) :=1

2π

∫ ∞

−∞[1 − S(k)]eikxdx+

J∑

j=1

sje−kjx := Fs(x) + Fd(x), (3.1.41)

q(x) = −2dA(x, x)

dx, (3.1.42)

and A(x, y) solves the Marchenko equation

A(x, y) +

∫ ∞

x

A(x, s)F (s + y)ds + F (x+ y) = 0, 0 ≤ x ≤ y <∞. (3.1.43)

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

S ⇔ F ⇔ A ⇔ q (3.1.44)

under the assumption q ∈ L1,1. We also derive a new equation for

A(y) :=

A(0, y), y ≥ 0,

0, y < 0.

This equation is:

F (y) + A(y) +

∫ ∞

0

A(s)F (s + y)ds = A(−y), −∞ < y <∞. (3.1.45)

The function A(y) is of interest because

f(k) = 1 +

∫ ∞

0

A(y)eikydy := 1 + A(k). (3.1.46)

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 S to be the scattering data corre-sponding to q ∈ L1,1. We also prove that if

|q(x)| ≤ c1 exp(−c2|x|γ), γ > 1, (3.1.47)

and, in particular, if

q(x) = 0 for x > a, (3.1.48)

then S(k) alone determines q(x) uniquely, because it determines kj, sj and J uniquely under the as-sumption (3.1.47) or (3.1.48).

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

I(k) :=f ′(0, k)

f(k), ∀k > 0, (3.1.49)

determines q ∈ L1,1 uniquely. The I-function is equal to Weyl’s m-function if q ∈ L1,1.


We give many applications of the above uniqueness theorem. In particular, we give short and simpleproofs of the Marchenko’s uniqueness theorems which say that S determines q ∈ L1,1 uniquely, and ρ(λ)determines q uniquely. We prove that if (3.1.48) (or (3.1.47)) holds, then either of the four functionsS(k), δ(k), f(k), f ′(0, k), determines q(x) uniquely. This result is applied in Section 3.10 to the heatand wave equations. It allows one to study some new inverse problems. For example, let

utt = uxx − q(x)u, x > 0, t > 0, (3.1.50)

u = ut = 0 at t = 0. (3.1.51)

u(0, t) = δ(t) or u′(0, t) = δ(t). (3.1.52)

Assumeq = 0 for x > 1, q = q, q ∈ L1(0, 1), (3.1.53)

and let the extra data (measured data) be

u(1, t) = a(t) ∀t > 0. (3.1.54)

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

ut = uxx − q(x)u, 0 ≤ x ≤ 1, t > 0, q ∈ L1[0, 1], (3.1.55)

u(x, 0) = 0 (3.1.56)

u(0, t) = 0, u(1, t) = a(t), a(t) ∈ L1(R+), a 6= 0. (3.1.57)

The extra data areux(1, t) = b(t) ∀t > 0. (3.1.58)

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 “incompletedata”. “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 δ(k), or f(k), or f ′(0, k), ∀k > 0, 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

`ϕj = λjϕj , 0 ≤ x ≤ 1; ϕj(0) = ϕj(1) = 0. (3.1.59)

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

λm(j)∀j; q(x), b ≤ x ≤ 1, q(x) ∈ L1[0, 1], q = q, (3.1.60)

where 0 < b < 1, and

m(j) =j

σ(1 + εj), |εj| < 1, εj → 0 as j −→ ∞, σ = const, 0 < σ ≤ 2. (3.1.61)

Assume also ∞∑

j=1

|εj| < ∞. (3.1.62)

We prove


Theorem 3.1.4. Data (3.1.60)-(3.1.61) determine uniquely q(x) on the interval 0 ≤ x ≤ b if σ > 2b. If(3.1.62) is assumed additionally, then q is uniquely determined if σ ≥ 2b.

The σ gives the “part of the spectra” sufficient for the unique recovery of q on [0, b]. For example, ifb = 1

2and (3.1.62) holds, then σ = 1, so “one spectrum” determines uniquely q on [0, 1

2]. If b = 1

4, then

σ = 12, so “half of the spectrum” determines uniquely q on [0, 1

4]. If b = 1

5, then “2

5of the spectrum”

determine uniquely q on [0, 15 ]. If b = 1, then σ = 2, and “two spectra” determine q uniquely on the

whole interval [0, 1]. The last result belongs to Borg [B]. By “two spectra” one means λj⋃µj,

where µj are the eigenvalues of the problem:

ùj = µjuj , uj(0) = 0, u′j(1) + huj(1) = 0. (3.1.63)

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 dataare the following.

The first result: If q = q(r) = 0 for r > a, a > 0 is an arbitrary large fixed number, r := |x|, x ∈ R3,q = q, and

∫ a0r2|q(r)|2dr < ∞, then the data δ`∀`∈L determine q(r) uniquely. Here δ` is the phase

shift at a fixed energy k2 > 0, ` is the angular momentum, and L is any fixed set of positive integerssuch that ∑

`∈L

1

`= ∞. (3.1.64)

The second result is: If q = q(x), x ∈ R3, q = 0 for |x| > a, q ∈ L2(Ba), where Ba := x : |x| ≤ a,then the knowledge of the scattering amplitude A(α′, α) at a fixed energy k2 > 0 and all α′ ∈ S2

j

determine q(x) uniquely, see Section 5.2. Here S2j , j = 1, 2, are arbitrary small open subsets in S2 and

S2 is the unit sphere in R3. The scattering amplitude is defined in Section 3.6.1.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) ofthe Krein inversion theory for solving inverse scattering problem and prove the consistency of thistheory.

(8) We give a method for recovery of a quarkonium system (a confining potential) from a few experi-mental measurements.

(9) We study various properties of the I-function.

(10) We study an inverse scattering problem for inhomogeneous Schrodinger 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 C+

By ODE in this section, the equation

(` − k2)u := −u′′ + q(x)u− k2u = 0 (3.2.1)

is meant. Assume q ∈ L1,1. Then the Jost solution f(x, k) is uniquely defined. In Section 1.3 Defini-tion 3.1.2, property C+ is explained. Let us prove

3.2. PROPERTY C FOR ODE 85

Theorem 3.2.1. If q ∈ L1,1, j = 1, 2 then property C+ holds.

Proof. We use (3.1.24) and (3.1.25). Denote A(x, y) := A1(x, y) +A2(x, s). Let

0 =

∫ ∞

0

dxh(x)f1(x, k)f2(x, k)

=

∫ ∞

0

dxh(x)

[e2ikx +

∫ ∞

x

A(x, y)eikydy

+

∫ ∞

x

∫ ∞

x

dydzA1(x, y)A2(x, z)eik(y+z)

](3.2.2)

for some h ∈ L1(R+). Set y + z = s, y − z = σ and get

∫ ∞

x

∫ ∞

x

A1(x, y)A2(x, z)eik(y+z)dydz =

∫ ∞

2x

T (x, s)eiksds, (3.2.3)

where

T (x, s) =1

2

∫ s−2x

−(s−2x)

A1

(x,s + σ

2

)A2

(x,s − σ

2

)dσ. (3.2.4)

Thus, f1f2 = (I + V ∗)e2ikx, where V ∗ is the adjoint to a Volterra operator, V ∗f := 2∫∞x/2A(x, 2s)

f(s)ds + 2∫∞x T (x, 2s)f(s)ds.

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

0 =

∫ ∞

0

dse2iks[h(s) + 2

∫ s

0

A(x, 2s− x)h(x)dx+ 2

∫ s

0

T (x, 2s)h(x)dx

], ∀k > 0. (3.2.5)

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

h(s) + 2

∫ s

0

A(x, 2s− x)h(x)dx+ 2

∫ s

0

T (x, 2s)h(x)dx = 0, ∀s > 0. (3.2.6)

Since A(x, y) and T (x, y) are bounded continuous functions, the Volterra equation (3.2.6) has only thetrivial solution h = 0. 2

Define functions g± and f± as the solutions to equation (3.1.5) with the following asymptotics:

g± = exp(±ikx) + o(1), x −→ −∞, (3.2.7)

f± = exp(±ikx) + o(1), x −→ +∞, (3.2.8)

Let us denote f+ = f and g+ = g.

Definition 3.2.2. The pair `1, `2 has property C− iff the set g1g2∀k>0 is complete in L1(R−).

Similar definition can be given with (g−,j) replacing gj, j = 1, 2.As above, one proves:

Theorem 3.2.3. If qj ∈ L1,1(R−), j = 1, 2, then property C− holds for `1, `2.

By L1,1(R) we mean the set

L1,1(R) := q : q = q,

∫ ∞

−∞(1 + |x|)|q(x)|dx < ∞. (3.2.9)


3.2.2 Properties Cϕ and Cθ.

We prove only property Cϕ. Property Cθ is proved similarly. Property Cϕ is defined in Section 3.1.4.

Theorem 3.2.4. If qj ∈ L1,1, j = 1, 2, then property Cϕ holds for `1, `2.Proof. Our proof is similar to the proof of Theorem 3.2.1. Using (3.1.23) and denoting φ = Kϕ,K := K1 +K2, one writes

φ1φ2 = sin2(kx) +

∫ x

0

K(x, y) sin(kx) sin(ky)dy

+1

2

∫ x

0

∫ x

0

K1(x, y)K2(x, s)cos[k(y − s)] − cos[k(y + s)]dyds.(3.2.10)

Assume:

0 =

∫ b

0

h(x)φ1(x, k)φ2(x, k)dx ∀k > 0. (3.2.11)

Then

0 =

∫ b

0

dxh(x) −∫ b

0

dxh(x) cos(2kx)

+

∫ b

0

ds cos(ks)

∫ b

s

dxh(x)K(x, x− s)

−∫ 2b

0

ds cos(ks)

∫ min(b,s)

s2

dxh(x)K(x, s− x) + I,

(3.2.12)

where

I :=

∫ b

0

dxh(x)

∫ x

0

∫ x

0

K1(x, y)K2(x, s)cos[k(y − s)] − cos[k(y + s)]dyds.

Let y − s := t, y + s := v. Then∫ x

0

∫ x

0

K1K2 cos[k(y − s)]dyds =

∫ x

0

ds cos(ks)B1(x, s),

where

B1(x, s) :=1

2

∫ 2x−|s|

|s|

[K1(x,

s+ v

2)K2

(x,v − s

2

)+K1

(x,v − s

2

)K2

(x,v + s

2

)]dv,

∫ x

0

∫ x

0

K1K2 cos[k(y + s)]dyds =

∫ 2x

0

B2(x, s) cos(ks)ds,

B2(x, s) :=1

2

∫ ω(s)

−ω(s)

K1

(x,t+ s

2

)K2

(x,s − t

2

)dt,

and ω = s if 0 ≤ s ≤ x; ω = 2x− s if x ≤ s ≤ 2x.Therefore

I =

∫ b

0

ds cos(ks)

∫ b

s

dxh(x)B1(x, s) −∫ 2b

0

ds cos(ks)

∫ b

s2

dxh(x)B2(x, s). (3.2.13)

From (3.2.12) and (3.2.13), taking k → ∞, one gets:

∫ b

0

h(x)dx = 0,

3.3. INVERSE PROBLEM WITH I-FUNCTION AS THE DATA 87

and (using completeness of the system cos(ks), 0 < k < ∞, in L2(0, b)) the following equation:

0 = −h(s2)

2+

∫ b

s

K(x, x− s)h(x)dx−∫ min(b,s)

s/2

dxh(x)K(x, s− x)

+

∫ b

s

dxh(x)B1(x, s) −∫ b

s/2

dxh(x)B2(x, s).

(3.2.14)

The kernels K,B1, and B2 are bounded and continuous functions. Therefore, if b <∞ and h(x) = 0 forx > b, (3.2.14) implies:

|h(y)| ≤ c

∫ b

2y

|h(x)|dx+ c

∫ b

y

|h(x)|dx,

where c > 0 is a constant which bounds the kernels 2K, 2B1 and 2B2 from above and 2y = s. From theabove inequality one gets

maxb−ε≤y≤b

|h(y)| ≤ cε maxb−ε≤y≤b

|h(y)|, (3.2.15)

where ε, 0 < ε < b, is sufficiently small so that cε < 1 and b − ε < 2b − 2ε. Then inequality (3.2.15)implies h(x) = 0 if b − ε < x < b. Repeating this argument, one proves, in finitely many steps, thath(x) = 0, 0 < x < b. Theorem 3.2.4 is proved. 2

The proof of Theorem 3.2.4 is not valid if b = ∞. The result is not valid either if b = ∞. Let us givea counterexample.

Theorem 3.2.5. There exist q1, q2 ∈ L1,1 and an h 6= 0, such that

∫ ∞

0

h(x)ϕ1(x, k)ϕ2(x, k)dx = 0 ∀k > 0. (3.2.16)

Proof. Let q1 and q2 are two potentials in L1,1 such that S1(k) = S2(k) ∀k > 0, `1 and `2 have onenegative eigenvalue −k2

1, which is the same for `1 and `2, but s1 6= s2, so that q1 6= q2. Let h := q2 − q1.Let us prove that (3.2.16) holds. One has `1ϕ1 = k2ϕ1, `2ϕ2 = k2ϕ2. subtract from the first equationthe second and get:

−ϕ′′ − k2ϕ+ q1ϕ = hϕ2, ϕ := ϕ1 − ϕ2, ϕ(0, k) = ϕ′(0, k) = 0. (3.2.17)

Multiply (3.2.17) by ϕ1, integrate over (0,∞) and then by parts to get

∫ ∞

0

hϕ2ϕ1dx = (ϕϕ′1 − ϕ′ϕ1)

∣∣∞0

= 0, ∀k > 0. (3.2.18)

At x = 0 we use conditions (3.2.17), and at x = ∞ the phase shifts corresponding to q1 and q2 are thesame (because S1(k) = S2(k)) and therefore the right-hand side of (3.2.18) vanishes. Theorem 3.2.5 isproved. 2

3.3 Inverse problem with I-function as the data

3.3.1 Uniqueness theorem

Consider equation (3.1.5) and assume q ∈ L1,1 Then f(x, k) is analytic in C+. Define the I-function:

I(k) =f ′(0, k)

f(k). (3.3.1)


From (3.3.1) it follows that I(k) is meromorphic in C+ with the finitely many simple poles ikj, 1 ≤ j ≤ J .Indeed, ikj are simple zeros of f(k) and f ′(0, ikj) 6= 0 as follows from (3.1.4). Using (3.1.19), one gets

aj := Resk=ikj I(k) =f ′(0, ikj)

f(ikj)= − cj

2ikj, kj > 0; a0 =

f ′(0, 0)

f(0), (3.3.2)

where Im aj > 0, 1 ≤ j ≤ J , and Im a0 ≥ 0, a0 6= 0 iff f(0) = 0. We prove that if q ∈ L1,1 and f(0) = 0

then f(0) exists and f(0) 6= 0 (Theorem 3.3.3 below). This is a subtle result.

Lemma 3.3.1. The I(k) equals to the Weyl function m(k).

Proof. The m(k) is a function such that θ(x, k) +m(k)ϕ(x, k) ∈ L2(R+) if Imk > 0. Clearly

f(x, k) = c(k)[θ(x, k) +m(k)ϕ(x, k)],

where c(k) 6= 0 , Im k > 0. Thus, I(k) = θ′(0,k)+m(k)ϕ′(0,k)θ(0,k)+m(k)ϕ(0,k)

= m(k) because of (3.1.3) and (3.1.4). 2

Our basic uniqueness theorem is:

Theorem 3.3.2. If qj ∈ L1,1, j = 1, 2, generate the same I(k), then q1 = q2.

Proof. Let p := q2 − q1, fj be the Jost solution (3.1.5) corresponding to qj, w := f1 − f2. Then one has

−w′′ + q1w − k2w = pf2, |w|+ |w′| = o(1), x −→ +∞. (3.3.3)

Multiply (3.3.3) by f1, integrate over R+, then by parts, using (3.3.3), and get

∫ ∞

0

pf2f1dx = w′(0)f1(0) − w(0)f ′1(0)

= f ′1(0, k)f2(k) − f ′2(0, k)f1(k) = f1(k)f2(k)[I1(k) − I2(k)] = 0. (3.3.4)

By property C+ (Theorem 3.2.1), p(x) = 0. 2

Remark. If qj ∈ L1,1, j = 1, 2, and (*) |I1(k) − I2(k)| ≤ ce−2a| Im k|, where k = |k|eiarg k, ∀|k| >0, 0 < arg k < π, then q1(x) = q2(x) for almost all x ∈ (0, a). This result is proved in [GS1] for qj ∈ L1

loc.

Our proof is based on (3.3.3), from which, using (*), one gets (**)∫∞0pf2f1dx = O(e−2a| Im k|). Note

that f1f2 = (I+V ∗)e2ikx, where V ∗ is the adjoint to a Volterra operator (see the formula below (3.2.4)).

Thus, (**) can be written as (***)∫∞0p1e

2ikxdx = O(e−2a| Im k|), where p1 := (I + V )p. Theorem 96in [Tit] and (***) imply p1 = 0 for almost all x ∈ (0, a). Since V is a Volterra operator, it follows thatp = 0 for almost all x ∈ (0, a), as claimed.

Theorem 3.3.3. If q ∈ L1,1 and f(0) = 0, then f (0) exists and f (0) 6= 0.

Proof. Let us prove that f(k) = ikA1(k), A1(0) 6= 0, A1 :=∫∞0

eiktA1(t)dt, and A1 ∈ L1(R+). Let

A1(t) :=∫∞t A(s)ds, A := f(k) − 1, and A(y) = A(0, y), where A(x, y) is defined in (3.1.24) and A(y) ∈

L1(R+) by (3.1.25). Integrating by parts, one gets A(k) = −exp(ikt)A1(t)|∞0 + ikA1 = ikA1 − 1. Thus,

f(k) = ikA1. The basic difficulty is to prove that A1 ∈ L1(R+). If this is done, then limk→0f(k)k = f (0)

exists and f (0) = iA1(0). To prove that f (0) 6= 0, one uses the Wronskian formula (3.3.2) with x = 0:f(−k)f ′(0, k)−f(k)f ′(0,−k) = 2ik. Divide by k and let k → 0. Since existence of f(0) is proved, one gets−f (0)f ′(0, 0) = i, so f (0) 6= 0. We have used here the existence of the limit limk→0 f

′(0, k) = f ′(0, 0).The existence of it follows from (3.1.24):

f ′(0, k) = ik − A(0, 0) +

∫ ∞

0

Ax(0, y)eikydy, (3.3.5)


and

f ′(0, 0) = −A(0, 0) +

∫ ∞

0

Ax(0, y)dy. (3.3.6)

From (3.1.26) one sees that Ax(0, y) ∈ L1(R+). Thus, to complete the proof, one has to prove A1 ∈L1(R+). To prove this, use (3.1.43) with x = 0 and (3.1.41). Since f(ikj) = 0, one has A(ikj) = −1.Therefore (3.1.43) with x = 0 yields:

A(y) +

∫ ∞

0

A(t)Fs(t + y)dt + Fs(y) = 0, y ≥ 0. (3.3.7)

Integrate (3.3.7) over (x,∞) to get:

A1(x) +

∫ ∞

0

A(t)

∫ ∞

x

Fs(t + y)dy dt+

∫ ∞

x

Fs(y)dy = 0, x ≥ 0, (3.3.8)

where Fs(y) ∈ L1(R+). Integrating by parts yields:

∫ ∞

0

A(t)

∫ ∞

x

Fs(t+ y)dy dt = A1(0)

∫ ∞

x

Fs(y)dy −∫ ∞

x

A1(t)Fs(x+ t)dt. (3.3.9)

Because 0 = f(0) = 1 +∫∞0A(y)dy, one has A1(0) = −1. Therefore (3.3.8) and (3.3.9) imply:

A1(x) −∫ ∞

x

A1(t)Fs(x+ t)dt = 0, x ≥ 0. (3.3.10)

From this equation and from the inclusion Fs(t) ∈ L1(R+), one derives A1 ∈ L1(R+) as follows. Choosea T (t) ∈ C∞

0 (R+) such that ‖Fs − T‖L1(R+) ≤ 0.5, and let Q := Fs − T . Then (3.3.10) can be writtenas:

A1(x) −∫ ∞

x

Q(x+ t)A1(t)dt = a(x) :=

∫ ∞

x

T (x + t)A1(t)dt, x ≥ 0. (3.3.11)

Since T ∈ C∞0 (R+) and A ∈ L1(R+), it follows that A1 is bounded. Thus a ∈ L1(R+). The operator

QA1 :=∫∞xQ(x+ t)A1(t)dt has norm ‖Q‖L1(R+)→L1(R+) ≤ 0.5. Therefore equation (3.3.3) is uniquely

solvable in L1(R+) and A1 ∈ L1(R+). Theorem 3.3.3 is proved. 2

3.3.2 Characterization of the I-functions

One has

Im I(k) =1

2i

(f ′(0, k)

f(k)− f ′(0, k)

f(k)

)=

k

|f(k)|2 , (3.3.12)

where the Wronskian formula was used with x = 0:

f(x, k)f ′(x, k)− f(x, k)f ′(x, k) = 2ik. (3.3.13)

From (3.1.20) and (3.3.12) with k =√λ, one gets:

1

πIm I(

√λ)dλ = dρ, λ ≥ 0. (3.3.14)

The I(k) determines uniquely the points ikj , 1 ≤ j ≤ J , as the (simple) poles of I(k) on the imaginaryaxis, and the numbers cj by (3.3.2). Therefore I(k) determines uniquely the spectral function ρ(λ) byformula (3.1.20). The characterization of the class of spectral functions ρ(λ), 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 one-to-one correspondencebetween the I-function and the scattering data S (3.1.16). Namely, the numbers kj and J , 1 ≤ j ≤ J ,are obtained from I(k) since ikj are the only poles of I(k) in C+, the numbers sj are obtained by theformula (see (3.1.15) and (3.3.2)):

sj = − 2ikj

aj [f(ikj)]2, (3.3.15)

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

w(k) :=

J∏

j=1

k − ikjk + ikj

if I(0) < ∞, (3.3.16)

and

w0(k) :=k

k + iκw(k), if I(0) = ∞, κ 6= kj ∀j, κ > 0. (3.3.17)

Assumption (3.3.16), means that f(0) 6= 0, and (3.3.17) means f(0) = 0.

Define

h(k) := w−1(k)f(k), I(0) < ∞ (3.3.18)

h0(k) := w−10 (k)f(k), I(0) = ∞. (3.3.19)

Write (3.3.12) as f(k) = k

Im I(k)

1f(−k) , or

h+(k) = g(k)h−(k), −∞ < k <∞, (3.3.20)

where h+(k) := h(k) is analytic in C+, h+(k) 6= 0 in C+, the closure of C+, h(∞) = 1 in C+, h−(k) :=h(−k) has similar properties in C−,

g(k) =k

Im I(k)if I(0) < ∞, g(k) =

k

Im I(k)

k2 + 1

k2if I(0) = ∞, (3.3.21)

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:

h(k) = exp

(1

2πi

∫ ∞

−∞

ln g(t)

t− kdt

), (3.3.22)

and

f(k) = w(k)h(k), Imk ≥ 0. (3.3.23)

In Section 3.4 we prove:

1

2π

∫ ∞

−∞[I(k) − ik]e−iktdk = −r0

2−

J∑

j=1

rjekjt, t < 0, (3.3.24)

where rj = −iaj . Taking t → −∞ in (3.3.24), one finds step by step all the numbers rj, kj and J . IfI(0) < ∞, then r0 = 0. Thus the data (3.1.16) are algorithmically recovered from I(k) known for allk > 0.

A characterization of S is given in Section 3.5.5, and thus an implicit characterization of I(k) is alsogiven.


3.3.3 Inversion procedures.

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

I(k) ⇒ ρ(λ) ⇒ q(x), (3.3.25)

I(k) ⇒ S ⇒ q(x), (3.3.26)

where (3.1.31) gives the details of the step ρ(λ) ⇒ q(x), and (3.1.40) gives the details of the stepS ⇒ q(x).

3.3.4 Properties of I(k)

In this section, we derive the following formula for I(k):

Theorem 3.3.4. One has

I(k) = ik +

J∑

j=0

ajk − ikj

+ a(k), a(k) =

∫ ∞

0

a(t)eiktdt, (3.3.27)

where k0, Im a0 > 0 if and only if f(0) = 0, aj are the constants defined in (3.3.2), Im aj > 0, 1 ≤ j ≤ J ,a(t) ∈ L1(R+) if f(0) 6= 0 and q ∈ L1,1, a(t) ∈ L1(R+) if f(0) = 0 and q ∈ L1,3(R+).

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

I(k) =ik − A(0, 0) +

∫∞0 Ax(0, y)e

ikydy

1 + A(k),

A(y) : = A(0, y), A(k) :=

∫ ∞

0

A(y)eikydy.

(3.3.28)

One has (cf. (3.3.16))

f(k) = 1 + A(k) := f0(k)w(k)k

k + iκ, w(k) :=

J∏

j=1

k − ikjk + ikj

, κ 6= kj ∀j (3.3.29)

f0(k) 6= 0 in C+, f0(∞) = 1, (3.3.30)

f0(k) is analytic in C+, the factor kk+i

in (3.3.29) is present if and only if f(0) = 0, and w(k) kk+iκ

:=w0(k).

Lemma 3.3.5. If f(0) 6= 0 and q ∈ L1,1(R+), then

f0(k) = 1 + b0(k), b0(x) ∈W 1,1(R+), ‖b0‖W1,1(R+) :=

∫ ∞

0

(|b0| + |b′0|)dx < ∞. (3.3.31)

Proof. It is sufficient to prove that, for any j, 1 ≤ j ≤ J , the function

k + ikjk − ikj

f(k) = 1 +

∫ ∞

0

gj(t)eikt dt, gj ∈W 1,1(R+). (3.3.32)

Sincek+ikj

k−ikj= 1+

2ikj

k−ikj, and since A(y) ∈ W 1,1(R+) provided that q ∈ L1,1(R+) (see (3.1.25)– (3.1.26)),

it is sufficient to check that

f(k)

k − ikj=

∫ ∞

0

g(t)eikt dt, g ∈W 1,1(R+). (3.3.33)


Note thatk − ikjk + ikj

=

∫ ∞

−∞eikt

[δ(t) − 2kje

−kjtθ(t)]dt, θ(t) :=

1, t ≥ 0

0, t < 0.(3.3.33’)

One has f(ikj) = 0, thus

f(k)

k − ikj=f(k) − f(ik)

k − ikj=

∫ ∞

0

dyA(y)ei(k−ikj )y − 1

k − ikje−kjy dy

=

∫ ∞

0

dyA(y)e−kjyi

∫ y

0

ei(k−ikj)s ds =

∫ ∞

0

eikshj(s) ds,

where

hj(s) := i

∫ ∞

s

A(y)e−kj (y−s) dy = i

∫ ∞

0

A(t + s)e−kj t dt. (3.3.34)

From (3.3.34) one obtains (3.3.33) since A(y) ∈W 1,1(R+). Lemma 3.3.5 is proved. 2

Lemma 3.3.6. If f(0) = 0 and q ∈ L1,2(R+), then (3.3.31) holds.

Proof. The proof goes as above with one difference: if f(0) = 0, then k0 = 0 is present in formula(3.3.27) and in formula (3.3.34) with k0 = 0 one has

h0(s) = i

∫ ∞

0

A(t+ s) dt. (3.3.35)

Thus, using (3.1.25), one gets∫ ∞

0

|h0(s)| ds ≤ c

∫ ∞

0

ds

∫ ∞

0

dt

∫ ∞

t+s2

|q(u)| du

= 2c

∫ ∞

0

ds

∫ ∞

s2

dv

∫ ∞

v

|q(u)| du ≤ 2c

∫ ∞

0

ds

∫ ∞

s2

|q(u)|u du

= 4c

∫ ∞

0

u2|q(u)| du < ∞ if q ∈ L1,2(R+),

where c > 0 is a constant. Similarly one checks that h′0(s) ∈ L1(R+) if q ∈ L1,2(R+). Lemma 3.3.6 isproved. 2

Lemma 3.3.7. Formula (3.3.27) holds.

Proof. Write

1

f(k)=

k+ik

∏Jj=1

k+ikj

k−ikj

f0(k).

Clearly

k + i

k

J∏

j=1

k + ikjk − ikj

= 1 +

J∑

j=0

cjk − ikj

, k0 := 0, kj > 0.

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

1

f0(k)= 1 +

∫ ∞

0

b(t)eikt dt, b(t) ∈W 1,1(R+), (3.3.31’)

where f0(k) is defined in (3.3.29). Actually, the Wiener-Levy theorem yields b(t) ∈ L1(R+). However,

since b0 ∈W 1,1(R+), one can prove that b(t) ∈W 1,1(R+). Indeed, b and b0 are related by the equation:

(1 + b0)(1 + b) = 1, ∀k ∈ R,


which impliesb = −b0 − b0b,

or

b(t) = −b0(t) −∫ t

0

b0(t− s)b(s) ds := −b0 − b0 ∗ b, (3.3.36)

where ∗ is the convolution operation.Since b′0 ∈ L1(R+) and b ∈ L1(R+) the convolution b′0 ∗ b ∈ L1(R+). So, differentiating (3.3.36) one

sees that b′ ∈ L1(R+), as claimed.From the above formulas one gets:

I(k) = (ik − A(0) + A1)(1 + b)

1 +

J∑

j=0

cjk − ikj

= ik + c+

J∑

j=0

ajk − ikj

+ a, (3.3.37)

where c is a constant defined in (3.3.39) below, the constants aj are defined in (3.3.40) and the functiona 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:

ikb = ik

[eikt

ikb(t)

∣∣∣∣∞

0

− 1

ik

∫ ∞

0

eiktb′(t)dt

]= −b(0) − b′,

and made the following transformations:

I(k) = ik −A(0) − b(0) − b′ + A1 − A(0)b+ A1b

J∑

j=0

cjik

k − ikj

−J∑

j=0

cj [A(0) + b(0)]

k − ikj+

J∑

j=0

g(k) − g(ikj)

k − ikjcj +

J∑

j=0

g(ikj)cjk − ikj

,

(3.3.38)

whereg(k) := −b′ + A1 −A(0)b+ A1b.

Comparing (3.3.38) and (3.3.37) one concludes that

c : = −A(0) − b(0) + i

J∑

j=0

cj, (3.3.39)

aj : = −cj [kj + A(0) + b(0) − g(ikj)] , (3.3.40)

a(k) : = g(k) +

J∑

j=0

g(k) − g(ikj)

k − ikjcj . (3.3.41)

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 k → ∞. Namely, one has:

A(k) = −A(0)

ik− 1

ikA′ (3.3.42)

From (3.3.42) and (3.3.28) one gets:

I(k) = (ik −A(0) + A1)

[1 − A(0)

ik+ o

(1

k

)]−1

= (ik −A(0) + A1)

(1 +

A(0)

ik+ o

(1

k

))= ik + o(1), k −→ +∞.

(3.3.43)

From (3.3.43) and (3.3.37) it follows that c = 0. Lemma 3.3.7 is proved. 2


Lemma 3.3.8. One has aj = irj, rj > 0, 1 ≤ j ≤ J , and r0 = 0 if f(0) 6= 0, and r0 > 0 if f(0) = 0.

Proof. From (3.3.2) one gets:

aj = − cj2ikj

= icj2kj

:= irj , rj :=cj2kj

> 0, j > 0. (3.3.44)

If j = 0, then

a0 = Resk=0

I(k) :=f ′(0, 0)

f (0). (3.3.45)

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

a0 = − i[f (0)

]2 = ir0, r0 := − 1

[f (0)]2. (3.3.46)

From (3.1.24) one gets:

f(0) = i

∫ ∞

0

A(y) y dy. (3.3.47)

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 [

f (0)]2< 0, (3.3.48)

and (3.3.46) impliesr0 > 0. (3.3.49)

Lemma 3.3.8 is proved. 2

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 q(x) 6≡ 0 is compactly supported, one cannot claim that a(t) is compactlysupported.

Proof. Assume for simplicity that J = 0 and f(0) 6= 0. In this case, if a(t) is compactly supported thenI(k) is an entire function of exponential type. It is proved in [R139], p.278, that if q(x) 6≡ 0 is compactlysupported, q ∈ L1(R+), then f(k) has infinitely many zeros in C. The function f ′(0, z) 6= 0 if f(z) = 0.Indeed, if f(z) = 0 and f ′(0, z) = 0 then f(x, z) ≡ 0 by the uniqueness of the solution of the Cauchyproblem for equation (3.1.5) with k = z. Since f(x, z) 6≡ 0, one has a contradiction, which proves thatf ′(0, z) 6= 0 if f(z) = 0. Thus, I(k) cannot be an entire function if q(x) 6≡ 0, q(x) ∈ L1(R+) and q(x) iscompactly supported. 2

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

I(k) = ik +

J∑

j=0

irjk − ikj

, (3.3.50)

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


By definitionf ′(0, k) = I(k)f(k), f ′(0,−k) = I(−k)f(−k), k ∈ R. (3.3.51)

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

f(k)f(−k) =k

ImI(k), ∀k ∈ R. (3.3.52)

By (3.3.44) one can write (see (3.1.20)) the spectral function corresponding to the I-function (3.3.50)(√λ = k):

dρ(λ) =

Im I(λ)

π dλ, λ ≥ 0,∑Jj=1 2kjrjδ(λ + k2

j ) dλ, λ < 0,(3.3.53)

where δ(λ) is the delta-function.Knowing dρ(λ) one can recover q(x) algorithmically by the scheme (3.1.31).Consider an example. Suppose f(0) 6= 0, J = 1,

I(k) = ik +ir1

k − ik1= ik +

ir1(k + ik1)

k2 + k21

= i

(k +

r1k

k2 + k21

)− r1k1

k2 + k21

. (3.3.54)

Then (3.3.53) yields:

dρ(λ) =

dλπ (

√λ+ r1

√λ

λ+k21), λ > 0,

2k1r1δ(λ + k21) dλ, λ < 0.

(3.3.55)

Thus, (3.1.33) yields:

L(x, y) =1

π

∫ ∞

0

dλr1√λ

λ+ k21

sin√λx√λ

sin√λy√λ

+ 2k1r1sh(k1x)

k1

sh(k1y)

k1, (3.3.56)

and, setting λ = k2 and taking for simplicity 2k1r1 = 1, one finds:

L0(x, y) : =2r1π

∫ ∞

0

dkk2

k2 + k21

sin(kx) sin(ky)

k2

=2r1π

∫ ∞

0

dk sin(kx) sin(ky)

k2 + k21

=r1π

∫ ∞

0

dk[cos k(x− y) − cos k(x+ y)]

k2 + k21

=r12k1

(e−k1|x−y| − e−k1(x+y)

), k1 > 0,

(3.3.57)

where the known formula was used:

1

π

∫ ∞

0

cos kx

k2 + a2dk =

1

2ae−a|x|, a > 0, x ∈ R. (3.3.58)

Thus,

L(x, y) =r12k1

[e−k1|x−y| − e−k1(x+y)

]+sh(k1x)

k1

sh(k1y)

k1. (3.3.59)

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

K(x, y) +

∫ x

0

K(x, t)

[e−k1|t−y| − e−k1(t+y)

2k1/r1+sh(k1t)

k1

sh(k1y)

k1

]dt

= −e−k1|x−y| − e−k1(x+y)

2k1/r1− sh(k1x)

k1

sh(k1y)

k1.

(3.3.60)


This equation can be solved analytically [R121], but the solution is long. By this reason we do notgive 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

f(k)f(−k) =k2 + k2

1

k2 + ν21

, ν21 := k2

1 + r1. (3.3.61)

The function

f0(k) := f(k)k + ik1

k − ik16= 0 in C+. (3.3.62)

Write (3.3.61) as

f0(k)k − ik1

k + ik1f0(−k)

k + ik1

k − ik1=k2 + k2

1

k2 + ν21

.

Thus,

f0(k) =k2 + k2

1

k2 + ν21

h, h(k) :=1

f0(−k). (3.3.63)

The function f0(−k) 6= 0 in C−, f0(∞) = 1 in C−, so h := 1f0(−k) is analytic in C−.

Consider (3.3.63) as a Riemann problem. One has

indR

k2 + k21

k2 + ν21

:=1

2πi

∫ ∞

−∞d ln

k2 + k21

k2 + ν21

= 0. (3.3.64)

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

f0(k) =k + ik1

k + iν1, h(k) =

k − iν1

k − ik1, (3.3.65)

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

f(k) =k − ik1

k + iν1. (3.3.66)

The corresponding S-matrix is:

S(k) =f(−k)f(k)

=(k + ik1)(k + iν1)

(k − ik1)(k − iν1)(3.3.67)

Thus,

Fs(x) :=1

2π

∫ ∞

−∞[1 − S(k)]eikxdk = O

(e−k1x

)for x > 0, (3.3.68)

Fd(x) = s1 e−k1x,

and

F (x) = Fs(x) + Fd(x) = O(e−k1x

). (3.3.69)

Equation (3.1.43) implies A(x, x) = O(e−2k1x), so

q(x) = O(e−2k1x

), x −→ +∞. (3.3.70)

Thus, if f(0) 6= 0 and a(t) = 0 then q(x) decays exponentially at the rate determined by the numberk1, k1 = min

1≤j≤Jkj.

3.4. INVERSE SPECTRAL PROBLEM 97

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

I(k) = ik +ir0k, (3.3.71)

f(k)f(−k) =k2

k2 + r0, r0 > 0. (3.3.72)

Let f0(k) = (k+i)f(k)k

. Then equation (3.3.72) implies:

f0(k)f0(−k) =k2 + 1

k2 + ν20

, ν20 := r0, (3.3.73)

and f0(k) 6= 0 in C+.

Thus, since indRk2+1k2+ν2

0= 0, f0(k) is uniquely determined by the Riemann problem (3.3.73).

One has:

f0(k) =k + i

k + iν0, f0(−k) =

k − i

k − iν0,

and

f(k) =k

k + iν0, S(k) =

f(−k)f(k)

=k + iν0

k − iν0,

Fs(x) =1

2π

∫ ∞

−∞

(1 − k + iν0

k − iν0

)eikxdk

=−2iν0

2π

∫ ∞

−∞

eikxdk

k − iν0= 2ν0e

−ν0x, x > 0,

(3.3.74)

and Fd(x) = 0.So one gets:

F (x) = Fs(x) = 2ν0e−ν0x, x > 0. (3.3.75)

Equation (3.1.43) yields:

A(x, y) + 2ν0

∫ ∞

x

A(x, t)e−ν0(t+y)dt = −2ν0e−ν0(x+y), y ≥ x ≥ 0. (3.3.76)

Solving (3.3.76) yields:

A(x, y) = −2ν0e−ν0(x+y) 1

1 + e−2ν0x. (3.3.77)

The corresponding potential (3.1.42) is

q(x) = O(e−2ν0x

), x −→ ∞. (3.3.78)

If q(x) = O(e−kx

), k > 0, then a(t) in (3.3.27) decays exponentially. Indeed, in this case b′(t), A1(y),

b(t), A1∗b decay exponentially, so g(t) decays exponentially, and, by (3.3.41), the functiong(k)−g(ikj)k−ikj

:= h

with h(t) decaying exponentially. We leave the details to the reader.

3.4 Inverse spectral problem

3.4.1 Auxiliary results

Transformation operators

If A1 and A2 are linear operators in a Banach space X, and T is a boundedly invertible linear operatorsuch that A1T = TA2, then T is called a transformation (transmutation) operator. If A2f = λf then


A1Tf = λTf , so that T sends eigenfunctions of A2 into eigenfunctions of A1 with the same eigenvalue.

Let `j = − d2

dx2 + qj(x), j = 1, 2, be selfadjoint in H := L2(0,∞) operators generated by the Dirichletboundary condition at x = 0. Other selfadjoint boundary conditions can be considered also, for example,u′(0) − hu(0) = 0, h = const ≥ 0.

Theorem 3.4.1. Transformation operator for a pair `1, `2 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 theproblem:

Kxx(x, y) − q1(x)K(x, y) = Kyy − q2(y)K, (3.4.1)

K(x, 0) = 0 (3.4.2)

K(x, x) =1

2

∫ x

0

(q1 − q2)dy. (3.4.3)

Proof. Consider for simplicity the case q2 = 0, q1 = q. The proof is similar in the case q2 6= 0. If q2 = 0,then (3.4.3) can be written as

q(x) = 2dK(x, x)

dx, K(0, 0) = 0. (3.4.4)

If `1Tf = T`2f and Tf = f +∫ x0 K(x, y)f dy, then

− f ′′ + q(x)f + qTf − [K(x, x)f ]′ − ∂K(x, x)

∂xf −

∫ x

0

Kxxf dy

= − f ′′ −∫ x

0

K(x, y)fyydy = −∫ x

0

Kyyf dy −K(x, y)f ′∣∣∣x

0+Kyf

∣∣∣x

0.

(3.4.5)

Since f ∈ D(`1), f(0) = 0, 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. Tofinish 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 ξ = x+ y, η = x− y, K(x, y) := B(ξ, η). Then (3.4.1), (3.4.2) and (3.4.4) can be written as

Bξη =1

4q

(ξ + η

2

)B, B(ξ, 0) =

1

2

∫ ξ/2

0

q(s)ds, B(ξ, ξ) = 0. (3.4.6)

Integrate (3.4.6) to get

Bξ(ξ, η) =1

4q

(ξ

2

)+

1

4

∫ η

0

q

(ξ + η

x

)B(ξ, τ )dτ. (3.4.7)

Integrate (3.4.7) with respect to ξ over (η, ξ) and get

B(ξ, η) =1

4

∫ ξ

η

q( s

2

)ds +

1

4

∫ ξ

η

∫ η

0

q

(s+ τ

2

)B(s, τ )dτds. (3.4.8)

This is a Volterra integral equation which has a solution, this solution is unique, and it can be obtainedby iterations. Theorem 3.4.1 is proved. 2

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 ρ(λ), possibly nonunique, such that for

every h ∈ L2(0,∞), there exists h(λ) ∈ L2(R; dρ) such that

∫ ∞

0

|h|2dx =

∫ ∞

−∞|h|2dρ(λ), h(λ) := lim

n→∞

∫ n

0

f(x)ϕ(x,√λ)dx, (3.4.9)

where the limit is understood in L2(R, dρ) sense. If the potential q in (3.1.1) generates the Dirichletoperator ` in the limit point at infinity case, then ρ(λ) is uniquely defined by q, otherwise ρ(λ) is definedby q nonuniquely. The spectral function of ` has the following properties:

∫ 0

−∞ex|λ|

1/2

dρ(λ) < ∞, ∀x > 0, ρ(λ) =2λ3/2

3π+ o(λ3/2), λ −→ +∞. (3.4.10)

The remainder o(λ3/2) in (4.1.10) can be improved if additional assumptions on q are made. Forexample, for q ∈ L1,1(0,∞), one can get the remainder O(λ).

Theorem 3.4.3. (Weyl). For any λ, Imλ 6= 0, there exists m(λ) such that

W (x, λ) := θ(x, λ) +m(λ)ϕ(x, λ) ∈ L2(R+). (3.4.11)

The function m(λ) is analytic in C+ and in C−.

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


Let ρ(λ) be a non-decreasing function of bounded variation on every compact subset of the real axis. Let

h ∈ L20(R+), where L2

0(R+) is a subset of L2(R+) functions which vanish near infinity. Let ϕ0 := sin(x√λ)√

λand

H(λ) =

∫ ∞

0

h(x)ϕ0(x, λ)dx. (3.4.12)

Our first assumption A1) on ρ(λ) is:

∫ ∞

−∞H2(λ)dρ(λ) = 0, ⇒ h(x) = 0. (3.4.13)

This implication should hold for any h ∈ L20(R+). It holds, for example, if dρ(λ) 6= 0 on a set which has

a finite limit point: in this case the entire function of λ, H(λ), vanishes identically, and thus h = 0.Denote by P a subset of ρ(λ) and assume that if ρ1, ρ2 ∈ P, ν := ρ1 − ρ2, and H := H (λ) : ∈

C∞(R+), where H(λ) is defined in (3.4.12), then

∫ ∞

−∞H2(λ)dν(λ) = 0 ∀H ∈ H

⇒ ν(λ) = 0. (3.4.14)

Our second assumption A2) on ρ(λ) is:

ρ ∈ P. (3.4.15)

Let us start with two lemmas.

Lemma 3.4.4. Spectral functions ρ(λ) of an operator `q = − d2

dx2 + q(x) in the limit-point at infinitycase belong to P.


Proof. Let ρ1, ρ2 be two spectral functions corresponding to `1 and `2, `j = `qj , j = 1, 2, ν = ρ1−ρ2 and

(∗)∫∞−∞H2(λ)dν = 0 ∀h ∈ L2

0(R+). Let I+V and I+W be the transformation operators correspondingto `1 and `2 respectively, such that

ϕ0 = (I + V )ϕ1 = (I +W )ϕ2, (3.4.16)

where ϕj is the regular solution (3.1.1) corresponding to qj. Condition (∗) implies

‖(I + V ∗)h‖ = ‖(I +W ∗)h‖ ∀h ∈ L2(0, b), (3.4.17)

where, for example,

V h =

∫ x

a

V (x, y)h(y)dy, V ∗h =

∫ b

x

V (y, x)h(y)dy. (3.4.18)

It follows from (3.4.17) thatI + V ∗ = U (I +W ∗), (3.4.19)

where U is a unitary operator in L2(0, b). Indeed, U is an isometry and it is surjective because I + V ∗

is.To finish the proof, one uses Lemma 3.4.5 below and concludes from (3.4.19) that V ∗ = W ∗, so

V = W , ϕ1 = ϕ2, and q1 = q2 := q. Since, by assumption, q is in the limit-point at infinity case, thereis only one spectral function ρ corresponding to q, so ρ1 = ρ2 = ρ. 2

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 I + V = (I + W )U ∗. Since U is unitary, one has (I + V )(I + V ∗) =(I +W )(I +W ∗). Because V is a Volterra operator, (I + V )−1 = I + V1, where V1 is also a Volterra (ofthe same type as V in (3.4.18)). Thus, (I + V1)(I +W ) = (I + V ∗)(I +W ∗

1 ), or

V1 +W + V1W = V ∗ +W ∗1 + V ∗W ∗

1 (3.4.20)

The left-hand side in (3.4.20) is a Volterra operator of the type V in (3.4.18), while its right-hand sideis a Volterra operator of the type V ∗. Since they are equal, each of them must be equal to zero. Thus,V1(I +W ) = −W , or (I + V )−1(I +W ) = I, or V = W . 2

Theorem 3.4.6. The spectral function determines `q uniquely.

Proof. If `q1 and `q2 have the same spectral function ρ(λ), then

‖h‖2 =

∫ ∞

−∞|H1(λ)|2dρ =

∫ ∞

−∞|H2(λ)|2dρ ∀h ∈ L2

0(0, b), (3.4.21)

where

Hj(λ) :=

∫ b

0

h(x)ϕj(x, k)dx, k =√λ, j = 1, 2.

Let I +K be the transformation operator ϕ2 = (I +K)ϕ1, and g := (I +K∗)h. Then H2 = (h, ϕ2) =(h, (I + K)ϕ1) = (g, ϕ1). From (3.4.21) one gets ‖h‖ = ‖(I + K∗)h‖. Thus I + K∗ is isometric, and,because K∗ is a Volterra operator, the range of I +K∗ is the whole space L2(0, b). Therefore I + K∗

is unitary. This implies K∗ = 0. Indeed, (I + K∗)−1 = I + K (unitarity) and (I + K∗)−1 = I + V ∗

(Volterra property of K∗). Thus K = V ∗, so K = V ∗ = 0. Therefore ϕ2 = ϕ1 and q1 = q2, so `q1 = `q2 .2

This result was proved by Marchenko (see [M]).

Remark 3.4.7. If ρ1 = cρ2, c = const > 0, then the above argument is applicable and shows that cmust be equal to 1, c = 1 and q1 = q2. Indeed, the above argument yields the unitarity of the operator√c(I +K∗), which implies c = 1 and K∗ = 0.


Here the following lemma is useful:

Lemma 3.4.8. If bI + Q = 0, where b = const and Q is a compact linear operator, then b = 0 andQ = 0.

A simple proof is left to the reader.

3.4.3 Reconstruction procedure

Assume that ρ(λ), the spectral function corresponding to `q, is given. How can one reconstruct `q, thatis, to find q(x)? We assume for simplicity the Dirichlet boundary condition at x = 0, but the methodallows 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

∫ ∞

−∞ϕ(x,

√λ)ϕ(y,

√λ)dρ(λ) = δ(x− y), (3.4.22)

and assume that L(x, y) is a continuous function of x, y in [0, b)× [0, b) for any b ∈ (0,∞).If 0 ≤ y < x, one gets from (3.4.22) the relation:

∫ ∞

−∞ϕ(x,

√λ)ϕ(y,

√λ)dρ(λ) = 0, 0 ≤ y < x. (3.4.23)

Using (3.1.23), one gets ϕ0 = (I +K)−1ϕ. Applying (I +K)−1 to ϕ(y,√λ) in (3.4.23), one gets

0 =

∫ ∞

−∞ϕ(x,

√λ)ϕ0(y,

√λ)dρ := I(x, y), 0 ≤ y < x. (3.4.24)

The right-hand side can be rewritten as:

I(x, y) =

∫ ∞

−∞(ϕ0 +Kϕ0(x)ϕ0(y,

√λ)d(ρ− ρ0) +

∫ ∞

−∞(ϕ0 +Kϕ0)(x)ϕ0(y,

√λ)dρ0

= L(x, y) +

∫ x

0

K(x, s)L(s, y)ds + δ(x− y) +

∫ x

0

K(x, s)δ(s − y)ds

= L(x, y) +

∫ x

0

K(x, s)L(s, y)ds +K(x, y), 0 ≤ y < x.

(3.4.25)

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 (−∞, n) 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 inthe classical sense:

L(x) := limn→∞

Ln(x) := limn→∞

∫ n

−∞

1 − cos(x√λ)

2λdσ(λ),

then the above proof can be understood in the classical sense, provided that (∗) supn,x∈(a,b) |Ln(x)| ≤c(a, b) for any −∞ < a < b < ∞. If ρ(λ) is a spectral function corresponding to `, then the sequenceLn(x) satisfies (∗). It is known (see [L]) that the sequence

Φn(x, y) =

∫ n

−∞ϕ(x,

√λ)ϕ(y,

√λ)dρ(λ) −

∫ n

−∞

sin(x√λ) sin(y

√λ)

λdρ0(λ)

satisfies (∗) and converges to zero.


Lemma 3.4.9. Assume (3.4.13) and suppose that the function L(x) ∈ H1loc(R+),

L(x) :=

∫ ∞

−∞

1 − cos(x√λ)

2λdσ(λ). (3.4.26)

Then equation (3.1.34) has a solution in L2(0, b) for any b > 0, and this solution is unique.

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

L(x, y) = L(x + y) − L(x− y), L(x, x) = L(2x), L(0) = 0, (3.4.27)

is in H1(0, b) × H1(0, b) for any b ∈ (0,∞). Therefore Lemma 3.4.9 is proved if it is proved that thehomogeneous version of (3.1.34) has only the trivial solution. Let

h(y) +

∫ x

0

L(s, y)h(s)ds = 0, 0 ≤ y ≤ x, h ∈ L2(0, x). (3.4.28)

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

0 = ‖h‖2 +

∫ ∞

−∞H2(λ)dσ = ‖h‖2 +

∫ ∞

−∞H2(λ)dρ − ‖h‖2 =

∫ ∞

−∞H2(λ)dρ. (3.4.29)

From (3.4.13) and (3.4.29) it follows that h = 0. 2

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 L(x) ∈ H1loc(R+). Then each of the steps in

(3.1.31) is invertible, so that (3.1.35) holds.

Proof. 1. Step. ρ ⇒ L is done by formula (3.1.33). Let us prove L ⇒ ρ. If there are ρ1 and ρ2

corresponding to the same L(x, y), and ν := ρ1 − ρ2, then

0 =

∫ ∞

−∞ϕ0(x,

√λ)ϕ0(y,

√λ)dν. (3.4.30)

Multiply (3.4.30) by h(x)h(y), h ∈ C∞0 (R+), use (3.4.12) and get

0 =

∫ ∞

−∞H2(λ)dν(λ) ∀H ∈ H . (3.4.31)

By (3.4.14) it follows that ν = 0, so ρ1 = ρ2. Thus L ⇒ ρ. 2

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

L(2x) +

∫ x

0

K(x, s)[L(x+ s) − L(x − s)]ds = −K(x, x), (3.4.32)

or

L(2x) +

∫ 2x

x

K(x, y − x)L(y)dy −∫ x

0

K(x, x− y)L(y)dy = −K(x, x). (3.4.33)

This is a Volterra integral equation for L(x) which has a solution and the solution is unique. Thus thestep K ⇒ L is done. The functions L(x) and K(x, x) are of the same smoothness. 2


3. Step. K ⇒ q is done by formula (3.4.4), q(x) is one derivative less smooth than K(x, x) and thereforeone derivative less smooth than L(x). Thus q ∈ L2

loc(R+). The step q ⇒ K is done by solving theGoursat problem (3.4.1), (3.4.2), (3.4.4) (with q2 = 0), or, equivalently, by solving Volterra equation(3.4.8), which is solvable and has a unique solution. The corresponding K(x, y) is in H1

loc(R+ × R+) ifq ∈ L2

loc(R+). 2


Let us prove that the q obtained by formula (3.4.4) generates the function K1(x, y) identical to thefunction K obtained in Step 2. The idea of the proof is to show that both K and K1 solve the problem(3.4.1), (3.4.2), (3.4.4) with the same q1 = q and q2 = 0. This is clear for K1. 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 D := ∂2

∂x2 − ∂2

∂y2 := ∂2x− ∂2

y .

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

DK +d

dx[K(x, x)L(x, y)] +Kx(x, x)L(x, y)

+

∫ x

0

Kxx(x, s)L(s, y)ds −∫ x

0

K(x, s)Lyy(s, y)ds = 0.

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

(DK)(x, y) +

∫ x

0

(DK)(x, s)L(s, y)ds

+ KL + (Kx +Ky)L(x, y) +K(Lx(x, y) − Ls(s, y)|s=x) = 0, 0 ≤ y ≤ x,

(3.4.34)

where K = K(x, x), L = L(x, y), K = dK(x,x)dx , Kx +Ky = K , and Lx(x, y)−Ls(s, y)|s=x = 0. 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:

v(x, y) +

∫ x

0

L(s, y)v(x, s)ds = 0, 0 ≤ y ≤ x, (3.4.35)

provided that −q(x)L(x, y) + 2KL(x, y) = 0, which is true because of (3.4.4). Equation (3.4.35) hasonly the trivial solution by Lemma 3.4.9. Thus v = 0, and equation (3.4.1) is derived.

We have proved

Lemma 3.4.11. If L(x, y) is twice differentiable continuously or in L2-sense then the solution K(x, y)of (3.1.34) solves (3.4.1), (3.4.2) with q given by (3.4.4).

Remark 3.4.12. If a Fredholm equation

(I +A(x))u = f(x) (3.4.36)

in a Banach space X depends on a parameter x continuously in the sense limh→0 ‖A(x+h)−A(x)‖ = 0,limh→0 ‖f(x + h) − f(x)‖ = 0, and at x = x0 equation (3.4.36) has N (I + A(x0)) = 0, whereN (A) = u : Au = 0, then the solution u(x) exists, is unique, and depends continuously on x in someneighborhood of x0, |x− x0| < r. If the data, that is, A(x) and f(x), have m derivatives with respect tox, 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 forthe operator A(x). A simple proof of this known result is left to the reader.

Hint. Use the identity B−1 − A−1 = A−1(A − B)B−1, which shows that if the operator A−1 isbounded, and ||B −A|| is sufficiently small, then B−1 exists and is bounded.


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

From Theorem 3.4.10 it follows that if (3.4.13) holds and L(x) ∈ H1loc(R+), then q ∈ L2

loc(R+). Condition(3.4.14) was used only to prove L ⇒ ρ, so if one starts with a q ∈ L2

loc(R+), then by diagram (3.1.35)one gets L(x, y) by formula (3.4.27), where L(x) ∈ H1

loc(R+). If (3.4.14) holds, then one gets from L(x)a unique ρ(λ).

Recall that assumption A1) is (3.4.13). Let A3) be the assumption: L(x) ∈ Hm+1loc (R+).

Theorem 3.4.13. If A1) holds, and ρ is a spectral function of `q , q ∈ Hmloc(R+), then assumption A3)

holds. Conversely, if assumptions A1) and A3) hold, then ρ is a spectral function of `q, q ∈ Hmloc(R+).

Proof. If A1) holds and q ∈ Hmloc(R+), then L(x) ∈ Hm+1

loc (R+) by (4.1.4). If A1) and A3) hold,then q ∈ Hm

loc(R+) by (3.1.32), because equation (3.1.34) is uniquely solvable, and (3.1.35) holds byTheorem 3.4.10. 2

3.4.6 Relation to the inverse scattering problem

Assume in this section that q ∈ L1,1. Then the scattering data S are (3.1.16) and the spectral functionis (3.1.20).

Let us show how to get dρ, given S . If S is given then sj , kj and J are known. If one finds f(k)then dρ is recovered because

cj = −4k2j

[f(ikj)]21

sj, (3.4.37)

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

f(k) = S(−k)f(−k), k ∈ R, f(∞) = 1, (3.4.38)

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

f0(k) = S(−k)w(−k)w(k)

f0(−k) if indS(k) = −2J, (3.4.39)

f0(k) = S(−k)w(−k)w(k)

k + iκ

k − iκf0(−k) if indS(k) = −2J − 1. (3.4.40)

Note that w(−k) = 1w(k) if k ∈ R. The function f0(k) is analytic in C+ and has no zeros in C+, and

f0(−k) has similar properties in C−. Therefore problems (3.4.39) and (3.4.40) have unique solutions:

f0(k) = exp

1

2πi

∫ ∞

−∞

log[S(−t)w−2(t)]dt

t− k

if indS(k) = −2J, Im k > 0, (3.4.41)

f0(k) = exp

1

2πi

∫ ∞

−∞

log[S(−t)w−2(t) t+iκt−iκ ]

t− k

dt if indS(k) = −2J − 1, Im k > 0, (3.4.42)

andf(k) = f0(k)w(k) if indS(k) = −2J, Imk > 0, (3.4.43)

f(k) = f0(k)w(k)k

k + iκif indS(k) = −2J − 1, Imk > 0. (3.4.44)

One can calculate f(x) for k > 0 by taking k = k+ i0 in (3.4.43) or (3.4.44). Thus, to find dρ, given S ,one goes through the following steps:

(1) one finds J , sj , kj, 1 ≤ j ≤ J ;

3.5. INVERSE SCATTERING ON HALF-LINE 105

(2) one calculates indS(k) := J . If J = −2J , then one calculates f(k) by formulas (3.4.41), (3.4.43),where w(k) is defined in (3.3.29), and cj by formula (3.4.37), and, finally, dρ by formula (3.1.20).

If J = −2J − 1, then one calculates f(k) by formulas (3.4.42) and (3.4.44), where κ > 0 is anarbitrary number such that κ 6= kj, 1 ≤ j ≤ J . If f(k) is found, one calculates cj by formula (3.4.37),and then dρ by formula (3.1.20). Note that f0(k) in (3.4.42) depends on κ, but f(k) in (3.4.44) doesnot.

This completes the description of the step S ⇒ ρ.Let us show how to get S given dρ(λ).From formula (3.1.20) one finds J , kj, cj and |f(k)|. If |f(0)| 6= 0, then |f0(k)| = |f(k)| if k ∈ R.

Thus, if |f(0)| 6= 0, then log f0(k) is analytic in C+ and vanishes at infinity. It can be found in C+ fromthe values of its real part log |f0(k)| by Schwarz’s formula for the half-plane:

log f0(k) =1

iπ

∫ ∞

−∞

log |f0(t)|t− k

dt, Imk > 0. (3.4.45)

If f(0) 6= 0, then f = f0w, so

f(k) = exp

1

iπ

∫ ∞

−∞

log |f0(t)|dtt− k

w(k), Im k > 0. (3.4.46)

If |f(0)| = 0, then the same formula (3.4.46) remains valid. One can see this because f(k)w(k)

is analytic in

C+, has no zeroes in C+, tends to 1 at infinity, and | f(k)w(k) | = |f(k)| if k ∈ R.

Let us summarize the step dρ ⇒ S : one finds J , kj, cj , calculates f(k) by formula (3.4.46), and

then S(k) = f(−k)f(k) , and sj 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:

f(k) = exp

1

iπ

∫ ∞

−∞

log |f(t)|dtt− k

+ log |f(k)|w(k)

= |f(k)|w(k) exp

1

iπP

∫ ∞

−∞

log |f(t)|dtt− k

, k > 0.

(3.4.47)

3.5 Inverse scattering on half-line

3.5.1 Auxiliary material

Transformation operators

Theorem 3.5.1. If q ∈ L1,1, 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:

Axx − q(x)A = Ayy, 0 ≤ x ≤ y ≤ ∞, (3.5.1)

A(x, x) =1

2

∫ ∞

x

q(s)ds, (3.5.2)

limx+y→∞

A(x, y) = limx+y→∞

Ax(x, y) = limx+y→∞

Ay(x, y) = 0. (3.5.3)

Proof. Equations (3.5.1) and (3.5.2) are derived similarly to the derivation of the similar equations forK(x, y) in Theorem 3.4.1. Relations (3.5.3) follow from the estimates (3.1.25)– (3.1.26), which give moreprecise 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 thederivation 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]. 2


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 ψ = ψ(r, k) to the equation:

ψ′′ + k2ψ − q(r)ψ = 0, r > 0, (3.5.4)

satisfying the boundary conditions at r = 0 and at r = ∞:

ψ(0) = 0, (3.5.5)

ψ(r) = eiδ sin(kr + δ) + o(1), r −→ +∞, (3.5.6)

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

ψ =i

2[e−ikr − S(k)eikr ] + o(1), r −→ ∞, (3.5.7)

where S(k) = f(−k)f(k) = e2iδ(k). Clearly

ψ(r, k) =i

2[f(r,−k) − S(k)f(r, k)] = a(k)ϕ(r, k), a(k) :=

k

f(k), (3.5.8)

where ϕ(r, k) is defined in (3.1.1), see also (3.1.10). From (3.5.8), (3.1.7) and (3.5.6) one gets

ϕ(r, k) =|f(k)|k

sin(kr + δ(k)) + o(1), r −→ ∞. (3.5.9)

Existence and uniqueness of the scattering solution ψ(r, k) follows from (3.5.8) because existence anduniqueness of the regular solution ϕ(r, k) follows from (3.1.1) or from (3.1.9).

Higher angular momenta.

If one studies the three-dimensional scattering problem with a spherically-symmetric potential q(x) =q(r), x ∈ R3, |x| = r, then the scattering solution ψ(r, α, k) solves the problem:

[∇2 + k2 − q(r)]ψ = 0 in R3 (3.5.10)

ψ = eikα·x +A(α′, α, k)eikr

r+ o

(1

r

), r := |x| −→ ∞, α′ :=

x

r, α ∈ S2. (3.5.11)

Here S2 is the unit sphere in R3 α ∈ S2 is given, A(α′, α, k) is called the scattering amplitude. Ifq = q(r), then A(α′, α, k) = A(α′ · α, k). The converse is a theorem of Ramm [R139], p.130. Thescattering solution solves the integral equation:

ψ = eikα·x −∫

R3

g(x, y, k)q(y)ψ(y, α, k)dy, g :=eik|x−y|

4π|x− y| . (3.5.12)

It is known that

eikα·x =

∞∑

`=0

4π

kiù`(kr)

rY`(α

′)Y`(α), α′ :=x

r, u`(r) :=

√πr

2J`+ 1

2(r), (3.5.13)

Y`(α) are orthonormal in L2(S2) spherical harmonics, Yl = Y`m, −` ≤ m ≤ `, and summation over m in(3.5.13) is understood but not shown, and J`(r) is the Bessel function.


If q = q(r), then

ψ =

∞∑

`=0

4π

ki`ψ`(r, k)

rY`(α

′)Y`(α), (3.5.14)

where

ψ′′` + k2ψ` − q(r)ψ` −

`(` + 1)

r2ψ` = 0, (3.5.15)

ψ` = eiδ` sin

(kr − `π

2+ δ`

)+ o(1), r → ∞, (3.5.16)

ψ` = O(r`+1), r → 0. (3.5.17)

Relation (3.5.16) is equivalent to

ψ` =ei

π2 (`+1)

2

[e−ikr − eiπ`Sè

ikr]+ o(1), r → ∞, (3.5.18)

similar to (3.5.8), which is (3.5.18) with ` = 0. If q = q(r), then the scattering amplitude A(α′, α, ) =A(α′ · α, k) can be written as

A(α′ · α, k) =

∞∑

`=0

A`(k)Y`(α′)Y`(α), (3.5.19)

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

A(α′, α, k) =

∞∑

`=0

Al(α, k)Y`(α′). (3.5.20)

If q = q(r) then S` in (3.5.18) are related to A` in (3.5.19) by the formula

S` = 1 − k

2πiA`. (3.5.21)

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

S = I − k

2πiA, (3.5.22)

so that (3.5.21) is a consequence of (3.5.22) in the case q = q(r) : S` are the eigenvalues of S in theeigenbasis of the spherical harmonics. Since S is unitary, one has |S`| = 1, so S` = e2iδ` for some realnumbers δ`, 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

A`(k) =4π

keiδ` sin(δ`). (3.5.23)

The Green function g`(r, ρ), which solves the equation

(d2

dr2+ k2 − `(` + 1)

r2

)g` = −δ(r − ρ),

∂gl∂r

− ikg` →r→+∞

0, (3.5.24)

can be written explicitly:

g`(r, ρ) =

F−1o` ϕo`(kρ)fo`(kr), r ≥ ρ, Fo` := e

i`π2

k` ,

F−1o` ϕo`(kr)fo`(kρ), r < ρ, ϕo`(kr) = u`(kr)

k`+1 ,(3.5.25)


and the function ψ`(r, k) solves the equation:

ψ`(r, k) = u`(kr) −∫ ∞

0

g`(r, ρ)q(ρ)ψ`(ρ, k)dρ. (3.5.26)

The function Fo`(k) is the Wronskian W [fo`, ϕo,`], ϕo`(kr) is defined in (3.5.25) and fo` is the solutionto (3.5.15) (with q = 0) with the asymptotics

fo` = eikr + o(1), r −→ +∞, fo`(kr) = ei(`+1)π

2 (u`(kr) + iv`(kr)), v` :=

√πr

2N`+ 1

2(kr). (3.5.27)

Let ϕ`(r, k) be the regular solution to (3.5.15) which is defined by the asymptotics as r → 0:

ϕ`(r, k) =r`+1

(2` + 1)!!+ o(r`+1), r −→ 0. (3.5.28)

Then

ψ`(r, k) = a`(k)ϕ`(r, k), ϕ`(r, k) =|f`(0, k)|k`+1

sin

(kr − `π

2+ δ`

)+ o(1), r −→ ∞. (3.5.29)

Lemma 3.5.2. One has:sup

`=0,1,2,...|a`(k)| < ∞, (3.5.30)

where k > 0 is an arbitrary fixed number.

We omit the proof of this lemma.

Eigenfunction expansion

We assume that q ∈ L1,1 and h ∈ C∞0 (R+), `h = −h′′ + q(x)h, λ = k2, letg = ϕ(x,

√λ)f(y,

√λ)

f(√λ)

, y ≥ x ≥ 0,

be the resolvent kernel of ` : (` − λ)g = δ(x − y), gh := g(λ)h :=∫∞0g(x, y, λ)hdy, and fj = f(y, ikj).

Then hλ

= −gh + 1λ`gh. Integrate this with respect to λ ∈ C over |λ| = N and divide by 2πi to get

h = − 1

2πi

∫

|λ|=Nghdλ+

1

2πi

∫

|λ|=N

`gh

λdλ := I1 + I2.

The function gh is analytic with respect to λ on the complex plane with the cut (0,∞) except for thepoints λ = −k2

j , 1 ≤ j ≤ J , which are simple poles of gh, and limN→∞ I2 = 0, because |`gh| = o(1) asN → ∞. Therefore:

h =1

2πi

∫ ∞

0

[g(λ+ i0)h − g(λ − i0)h]dλ+

J∑

j=1

−1

2πi

∮

|λ+k2j |=δ

ghdλ. (3.5.31)

One has (cf. (3.1.10)):

g(λ + i0) − g(λ − i0)

2i= ϕ(x, k)

f(−k)f(y, k) − f(y,−k)f(k)2i|f(k)|2

=k

|f(k)|2ϕ(x, k)ϕ(y, k), k =√λ > 0.

Also

− 1

2πi

∮

|λ+k2j |=δ

ghdλ = −Resλ=−k2jgh = −

∫ ∞

0

fj(y)h(y)dy · ϕ(x, ikj)

f(ikj)2ikj

= sjfj(x)hj , hj :=

∫ ∞

0

fjhdy,


sj are defined in (3.1.15), and

ϕ(x, ikj) =f(x, ikj)

f ′(0, ikj)=

fj(x)

f ′(0, ikj). (3.5.32)

Therefore

h(x) =

∫ ∞

0

(∫ ∞

0

ϕ(y, k)h(y)dy

)ϕ(x, k)

2k2dk

π|f(k)|2 +

J∑

j=1

sjfj(x)hj . (3.5.33)

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

δ(x − y) =2

π

∫ ∞

0

ϕ(x, k)ϕ(y, k)k2dk

|f(k)|2 +

J∑

j=1

sjfj(x)fj(y)

=

∫ ∞

−∞ϕ(x,

√λ)ϕ(y,√λ)dρ(λ).

(3.5.34)

We have proved the eigenfunction expansion theorem for h ∈ C∞0 (R+). Since this set is dense in L2(R+),

one gets the theorem for h ∈ L2(R+).

Theorem 3.5.3. If q ∈ L1,1, then (3.5.33) holds for any h ∈ L2(R+) and the integrals converge inL2(R+) sense. Parseval’s equality is:

‖h‖2L2(R+) =

J∑

j=1

sj |hj|2 +2

π

∫ ∞

0

|h(k)|2 k2dk

|f(k)|2 , h :=

∫ ∞

0

h(y)ϕ(y, k)dy. (3.5.35)

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

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

Theorem 3.5.4. If q1, q2 ∈ L1,1 generate the same data (3.1.16), then q1 = q2.

Proof. We prove that the data (3.1.16) determine uniquely I(k), and this implies q1 = q2 by Theo-rem 3.3.2.

Claim 1. If (3.1.16) is given, then f(k) is uniquely determined.Assume there are f1(k) and f2(k) corresponding to the data (3.1.16). Then

f1(k)

f2(k)=f1(−k)f2(−k)

, −∞ < k < ∞. (3.5.36)

The left-hand side of (3.5.36) is analytic in C+ and tends to 1 as |k| → ∞, k ∈ C+, and the right-hand

side of (3.5.36) is analytic in C−, and tends to 1 as |k| → ∞, k ∈ C−. By analytic continuation f1(k)f2(k)

is

an analytic function in C, which tends to 1 as |k| → ∞, k ∈ C. Thus, by Liouville theorem,f1(k)f2(k) = 1,

so f1 = f2. 2

Claim 2. If (3.1.16) is given, then f ′(0, k) is uniquely defined.Assume there are f ′1(0, k) and f ′2(0, k) corresponding to (3.1.16). By the Wronskian relation (3.1.11),

taking into account that f1(k) = f2(k) := f(k) by Claim 1, one gets[f ′1(0, k) − f ′2(0, k)

]f(−k) −

[f ′1(0,−k) − f ′2(0,−k)

]f(k) = 0. (3.5.37)

Denote w(k) := f ′1(0, k)− f ′2(0, k). Then:

w(k)

f(k)=w(−k)f(−k) , k ∈ R. (3.5.38)


The function w(k)f(k) is analytic in C+ and tends to zero as |k| → ∞, k ∈ C+, and w(−k)

f(−k) has similar

properties in C−. It follows that w(k)f(k) = 0, so f ′1(0, k) = f ′2(0, k). Let us check that w(k)

f(k) is analytic in

C+. One has to check that w(ikj) = 0. This follows from (3.1.15): if f(k), sj and kj are given, thenf ′(0, ikj) are uniquely determined.

Let us check that w(k) → 0 as |k| → ∞, k ∈ C+. 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 |k| → ∞, k ∈ C+ by theRiemann-Lebesgue lemma. From (3.1.46), integrating by parts one gets:

f(k) = 1 − A(0, 0)

ik− 1

ik

∫ ∞

0

eikyAy(0, y)dy. (3.5.39)

ThusA(0, 0) = − lim

k→∞

[i kf(k) − 1)

]. (3.5.40)

Claim 2 is proved. 2

Thus, Theorem 3.5.4 is proved. 2

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):

∫ ∞

0

ϕ(x, k)ϕ(y, k)2k2dk

π|f(k)|2 +

J∑

j=1

sjfj(x)fj(y) = 0 y > x ≥ 0. (3.5.41)

From (3.1.10) and (3.1.24) one gets:

kϕ(x, k)

|f(k)| = sin(kx+ δ) +

∫ ∞

x

A(x, y) sin(ky + δ)dy

= (I +A) sin(kx+ δ), δ = δ(k).

(3.5.42)

Apply to (3.5.41) operator (I +A)−1, acting on the functions of y, and get:

2

π

∫ ∞

0

kϕ(x, k)

|f(k)| sin(ky + δ)dk +

J∑

j=1

sjfj(x)e−kjy = 0, y > x ≥ 0. (3.5.43)

From (3.5.42), (3.5.43), and (3.1.24) with k = ikj , one gets:

(I +A)

(2

π

∫ ∞

0

sin(kx+ δ) sin(ky + δ)dk

)+ (I +A)

J∑

j=1

sje−kj (x+y) = 0, y > x ≥ 0. (3.5.44)

One has

2

π

∫ ∞

0

sin(kx+ δ) sin(ky + δ)dk =1

π

∫ ∞

0

cos[k(x− y)]dk

− 1

π

∫ ∞

0

cos[k(x+ y) + 2δ(k)

]dk = δ(x, y) − 1

2π

∫ ∞

−∞

(e2iδ(k) − 1

)eik(x+y)dk

= δ(x− y) +1

2π

∫ ∞

−∞

[1 − S(k)

]eik(x+y)dk.

(3.5.45)

From (3.1.41), (3.5.44) and (3.5.45) one gets (3.1.43). By continuity equation (3.1.43), derived fory > x ≥ 0, remains valid for y ≥ x ≥ 0. 2


Theorem 3.5.5. If q ∈ L1,1 and F is defined by (3.1.41) then equation (3.1.43) has a solution inL1(Rx) ∩L∞(Rx), Rx := [x,∞), for any x ≥ 0, and this solution is unique.

Let us outline the steps of the proof.Step 1. If q ∈ L1,1, then F (x), defined by (3.1.41) satisfies the following estimates:

∣∣F (2x)∣∣ ≤ cσ(x)

∣∣,∣∣F (2x) +A(x, x)

∣∣ ≤ cσ(x),

∣∣∣∣F ′(2x) − q(x)

4

∣∣∣∣ ≤ cσ2(x), (3.5.46)

where σ(x) is defined in (3.1.25), and

‖F‖L2(R+) + ‖F‖L1(R+) + ‖F‖L∞(R+) + ‖xF ′(x)‖L1(R+) < ∞, (3.5.47)

∫ ∞

0

∫ ∞

0

|F (s+ y)|2dsdy < ∞. (3.5.48)

Step 2. Equation

(I + Fx)h := h(y) +

∫ ∞

x

h(s)F (s + y)ds = 0, y ≥ x ≥ 0 (3.5.49)

is of Fredholm type in L1(Rx), L2(Rx) and in L∞(Rx). It has only the trivial solution h = 0.

Using estimates (3.5.46) – (3.5.48) and the criteria of compactness in Lp(Rx), p = 1, 2,∞, one checksthat Fx is compact in these spaces for any x ≥ 0. The space L1 ∩ L∞ ⊂ L2 because ‖h‖2 ≤ ‖h‖1‖h‖∞,where ‖h‖p := ‖h‖Lp(Rx). We need the following lemma:

Lemma 3.5.6. Let h solve (3.5.49). If h ∈ L1 := L1(Rx), then h ∈ L2. If h ∈ L1, then h ∈ L∞. Ifh ∈ L2, then h ∈ L∞.

Proof. If h solves (3.5.49), then ‖h‖∞ ≤ ‖h‖1 supy≥2x |F (y)| ≤ c(x)‖h‖1 < ∞, where c(x) → 0 as

x → ∞. Also ‖h‖22 ≤

∫∞x dyσ2(x+y2 )‖h‖2

1 ≤ c1(x)‖h‖21 < ∞, c1(x) → 0 as x → ∞. So the first claim is

proved. Furthermore, ‖h‖1 ≤ ‖h‖1 sups≥x∫∞x

|F (s+y)|dy = c2(x)‖h‖1, c2(x) → 0 as x→ ∞. If h ∈ L2,

then ‖h‖∞ ≤ ‖h‖2 supy≥x(∫∞x

|F (s+ y)|2ds) 12 = c3(x)‖h‖2, c3(x) → 0 as x→ ∞. 2

Lemma 3.5.7. If h ∈ L1 solves (3.5.49) and x ≥ 0, then h = 0.

Proof. By Lemma 3.5.6, h ∈ L2 ∩L∞. 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 (x,∞)to get

‖h‖2 +1

2π

∫ ∞

−∞[1 − S(k)]h2(k)dk +

J∑

j=1

sj

(∫ ∞

x

e−kjsh(s)ds

)2

= 0, h :=

∫ ∞

x

eiksh(s)ds, (3.5.50)

where ‖h‖ = ‖h‖L2(R+) one gets∫∞−∞ h2(k)dk = 0. Also, | 1

2π

∫∞−∞ S(k) h2(k)dk| ≤ 1

2π

∫∞−∞ |h(k)|2dk =

‖h‖2. Therefore (3.5.50) implies 0 = hj :=∫∞xhe−kjsds, 1 ≤ j ≤ J , and

(h, h) = (h, S(−k)h(−k)), (3.5.51)

where (h, g) :=∫∞−∞ h(k)g(k)dk. Since h is real valued, one has h(−k). The unitarity of S implies

S−1(k) = S(−k) = S(k), k ∈ R, and ‖S(−k)h(−k)‖ = ‖h(−k)‖. Because of (3.5.51), one has equality

sign in the Cauchy inequality (h, S(−k)h(−k) ≤ ‖h‖2. This means that h(k) = S(−k)h(−k), and(3.1.16) implies

h(k)

f(k)=h(−k)f(−k) , k ∈ R. (3.5.52)


Because hj = 0, one has h(ikj) = 0, and if f(0) 6= 0, then h(k)f(k) is analytic in C+ and vanishes as |k| → ∞,

k ∈ ∞, k ∈ C+. Also h(−k)f(−k) is analytic in C− and vanishes as |k| → ∞, k ∈ C−. Therefore, by analytic

continuation, h(k)f(k) is analytic in C and vanishes as |k| → ∞. By Liouville theorem, h(k)f(k) = 0, so h(k) = 0

and h = 0. If f(0) = 0, then, by Theorem 3.3.3, f(k) = ikA1(k), A1(0) 6= 0, and the above argumentworks. 2

Because Fx is compact in L2(Rx), the Fredholm alternative is applicable to (3.5.49), and Lemma 3.5.7implies that (3.1.43) has a solution in L2(Rx) for any x ≥ 0, and this solution is unique. Note that thefree term in (3.1.43) is −F (x+ y), and this function of y belongs to L2(Rx) (cf. (3.5.48)). Because Fxis compact in L1(Rx), Lemma 3.5.7 and Lemma 3.5.6 imply existence and uniqueness of the solutionto (3.1.43) in L1(Rx) for any x ≥ 0, and F (x + y) ∈ L1(Rx) for any x ≥ 0. Note that the solution to(3.1.43) in L1(Rx) is the same as its solution in L2(Rx). This is established by the argument used in theproof 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 animmediate consequence of the first estimate (3.5.46). Indeed,

∫ ∞

0

∫ ∞

0

∣∣F (s+ y)∣∣2ds dy ≤

(∫ ∞

0

maxs≥0

∣∣F (s+ y)∣∣dy)2

≤ c

(∫ ∞

0

∫ ∞

y2

|q|dtdy)2

≤ c

(∫ ∞

0

t∣∣q(t)

∣∣dt)2

< ∞.

Let us prove the first estimate (3.5.46). Put in (3.1.43) x = y:

A(x, x) +

∫ ∞

x

A(x, s)F (s+ x)dx+ F (2x) = 0. (3.5.53)

Thus

|F (2x)| ≤ |A(x, x)|+∫ ∞

x

|A(x, s)F (s+ x)|dx. (3.5.54)


|F (2x)| ≤ cσ(x) + c

∫ ∞

x

σ

(x+ s

2

)|F (s+ x)|ds

≤ cσ(x) + cσ(x)

∫ ∞

x

|F (s+ x)|ds ≤ cσ(x)

(3.5.55)

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

supx≥0

∫ ∞

x

|F (s+ x)|ds ≤∫ ∞

0

|F (t)|dt = c <∞.

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

A(x, z − x) +

∫ ∞

z

A(x, t+ x− z)F (t)dt+ F (z) = 0, z ≥ 2x ≥ 0. (3.5.56)

Let us prove that equation (3.5.56) is uniquely solvable for F in Lp(RN ), p = ∞, p = 1 for all x ≥ N2 ,

where N is a sufficiently large number. In fact, we prove that the operator in (3.5.56) has small normin Lp(RN ) if N is sufficiently large. Its norm in L∞(RN ) is not more than

supz≥N

∫ ∞

N

|A(x, t+ x− z)|dt ≤ c

∫ ∞

x+ t−N2

|q(s)|ds

≤ c

∫ ∞

N

dt

∫ ∞

t

|q(s)|ds = c

∫ ∞

N

(s −N )|q(s)|ds < 1


because q ∈ L1,1. We have used estimate (3.1.25) above. The function A(x, y) ∈ L∞(RN ), so our claimis proved for p = ∞. Consider the case p = 1. One has the following upper estimate for the norm

of the operator in (3.5.56) in L1(RN ): supt≥N∫ tN |A(x, t+ x − z)|dz ≤ supt≥N

∫ t−N0 |A(x, x+ v)|dv ≤∫∞

0dv∫∞x+ v

2|q|ds = 2

∫∞x

(s − x)|q|ds → 0 as x → ∞. Also∫∞N

|A(x, z − x)|dz < ∞. Thus equation

(3.5.56) is uniquely solvable in L1(RN ) for all x ≥ N2

if N is sufficiently large. In order to finish theproof of the first estimate (3.5.46) it is sufficient to prove that ‖F‖L∞(0,N) ≤ c < ∞. This estimate isobvious for Fd(x) (cf. (3.1.41)). Let us prove it for Fs(x). Using (3.3.29), (3.3.31), (3.3.32), one gets

1− S(k) =[f(k) − f(−k)](k + iκ)

f0(k)w(k)k= [A(k) − A(−k)](1 + b(k))(1 + g)

(1 +

iκ

k

), (3.5.57)

where all the Fourier transforms are taken of W 1,1(R+) functions. Thus, one can conclude that Fs(x) ∈L∞(R+) if one can prove that I := A(k)−A(−k)

k is the Fourier transform of L∞(R+) function. One has

I =∫∞0dyA(y) eiky−e−iky

k and

∫ ∞

−∞eikxI(k)dk =

∫ ∞

0

dyA(y)

∫ ∞

−∞

eik(x+y) − eik(x−y)

k

=

∫ ∞

0

dyA(y)iπ[1 − sgn(x − y)] = 2iπ

∫ ∞

x

dyA(y).

(3.5.58)

From (3.1.25) it follows that∫∞xA(y)dy ∈ L∞(R+). We have proved that ‖F‖L∞(R+) +‖F‖L1(R+) < ∞.

Differentiate (3.5.53) to get

2F ′(2x) + A(x, x)− A(x, x)F (2x) +

∫ ∞

x

Ax(x, s)F (s+ x)ds

+

∫ ∞

x

A(x, s)F ′(s + x)ds = 0, A :=dA(x, x)

dx

(3.5.59)

or

F ′(2x) =q(x)

4+ A(x, x)F (2x)− 1

2

∫ ∞

x

[Ax(x, s) −As(x, s)]F (s+ x)ds. (3.5.60)

One has∫∞0 x|q|dx < ∞, and

∫ ∞

0

x|A(x, x)||F (2x)|dx≤ supx≥0

(x|A(x, x)|)∫ ∞

0

|F (2x)|dx≤ c.

Let us check that I :=∫∞0x|∫∞x

[Ax(x, s) − As(x, s)]F (s + x)ds| dx < ∞. Use (3.1.26) and get I ≤c∫∞0xσ(x)

∫∞xσ (x+s

2)|F (s + x)|ds dx ≤ c

∫∞0σ(x)dx

∫∞0

|F (y)|dy · supx≥0,s≥x xσ(x+s2

) ≤ c < ∞. Thedesired estimate is derived.

The third estimate (3.5.46), |F ′(2x) − q(x)4 | ≤ cσ2(x) follows from (3.5.60) because |A(x, x)| ≤

cσ(x), |F (2x)| ≤ cσ(x), and∫∞x

|Ax(x, s) − As(x, s)||F (s + x)ds ≤ cσ(x)∫∞xσ(x+s2 )|F (s + x)|ds ≤

cσ2(x)∫∞0

|F (s+x)|ds ≤ cσ2(x). The estimate |F (2x)+A(x, x)| ≤ cσ(x) follows similarly from (3.5.53)and (3.1.25). Theorem 3.5.5 is proved.

3.5.4 Invertibility of the steps of the reconstruction procedure

The reconstruction procedure is (3.1.40).

1. The step S ⇒ F is done by formula (3.1.41). To do the step F ⇒ S , one takes x → −∞ in(3.1.41) and finds sj , kj, and J . Thus Fd(x) is found and Fs = F − Fd is found. From Fs(x) onefinds 1− S(k) by the inverse Fourier transform. So S(k) is found and the data S (see (3.1.16)) isfound


2. The step F ⇒ A is done by solving equation (3.1.43). By Theorem 3.5.5 this equation is uniquelysolvable in L1(Rx) ∩ L∞(Rx) for all x ≥ 0 if q ∈ L1,1, that is, if F came from S corresponding toq ∈ L1,1.

To do the step A ⇒ F , one finds f(k) = 1+∫∞0A(0, y)eikydy, then the numbers ikj, the zeros of f(k) in

C+, the number J , 1 ≤ j ≤ J , and S(k) = f(−k)f(k) . The numbers sj are found by formula (3.1.15), where

f ′(0, ikj) = −kj − A(0, 0) +

∫ ∞

0

Ax(0, y)e−kjydy. (3.5.61)

Thus A ⇒ S and S ⇒ F by formula (3.1.41).We also give a direct way to do the step A⇒ F .Write equation (3.1.43) with z = x+ y, v = s + y, as

(I +Bx)F := F (z) +

∫ ∞

z

A(x, v + x− z)F (v)dv = −A(x, z − x), z ≥ 2x ≥ 0. (3.5.62)

The norm of the operator Bx in L12x is estimated as follows:

‖Bx‖ ≤ supv>0

∫ v

0

|A(x, v+ x− z)|dz ≤ c supv>0

∫ v

0

σ

(x+

v − z

2

)dz ≤ c

∫ ∞

x

σ(t)dt, (3.5.63)

where σ(x) =∫∞x |q(t)|dt and estimate (3.1.25) was used. If x0 is sufficiently large then ‖Bx‖ < 1 for

x ≥ x0 because∫∞xσ(t)dt → 0 as x → ∞ if q ∈ L1,1. Therefore equation (3.5.62) is uniquely solvable

in L12x for all x ≥ x0 (by the contraction mapping principle), and so F (z) is uniquely determined for all

z ≥ 2x0.Now rewrite (3.5.62) as

F (z) +

∫ 2x0

z

A(x, v + x− z)F (v)dv = −A(x, z − x) −∫ ∞

2x0

A(x, v + x− z)F (v)dv. (3.5.64)

This is a Volterra equation for F (z) on the finite interval (0, 2x0). It is uniquely solvable since itskernel is a continuous function. One can put x = 0 in (3.5.64) and the kernel A(0, v− z) is a continuousfunction of v and z, and the right-hand side of (3.5.64) at x = 0 is a continuous function of z. ThusF (z) is uniquely recovered for all z ≥ 0 from A(x, y), y ≥ x ≥ 0. Step S ⇒ F is done.

3. The step A ⇒ q is done by formula (3.1.42). The converse step q ⇒ A is done by solving Volterraequation (3.1.27), or, equivalently, the Goursat problem (3.5.1) – (3.5.3).

We have proved:

Theorem 3.5.8. If q ∈ L1,1 and S are the corresponding data (3.1.16), then each step in (3.1.40) isinvertible. In particular, the potential obtained by the procedure (3.1.40) equals to the original potentialq.

Remark 3.5.9. If q ∈ L1,1 and Aq := Aq(x, y) is the solution to (3.1.27), then Aq satisfies equation(3.1.43) and, by the uniqueness of its solution, Aq = A, 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 thereforethis 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 q0 ∈ L1,1,computes the corresponding scattering data (3.1.16), and applies inversion scheme (3.1.40), then the qis obtained by this scheme is equal to q0. Logically it is possible that this q generates data S1 whichgenerate by the scheme(3.1.40) potential q1, etc. To close this loop one has to check that q = q0. This is

done in Theorem 3.5.8, because q0 = −2 dA(x,x)dx = q(x).


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 scatteringdata corresponding to q ∈ L1,1. In Section 3.5.7 we give such conditions on S for q to be compactlysupported, or q ∈ L2(R+).

Theorem 3.5.12. If q ∈ L1,1, then the following conditions hold:

(1) (3.1.22), the index condition;

(2) kj > 0, sj > 0, 1 ≤ j ≤ J , S(k) = S(−k) = S−1(k), k ≥ 0, S(∞) = 1;

(3) (3.5.47) hold.

Conversely, if S satisfies conditions (1) – (3), then S corresponds to a unique q ∈ L1,1.

Proof. The necessity of conditions (1) – (3) has been proved in Theorem 3.5.5. Let us prove the suffi-ciency. If conditions (1) – (3) hold, then the scheme (3.1.40) yields a unique potential, as was provedin Remark 3.5.9. Indeed, equation (3.1.43) is of Fredholm type in L1(Rx) for every x ≥ 0 if F satisfies(3.5.47). Moreover, equation (3.5.49) has only the trivial solution if conditions (1) – (3) hold. Everysolution to (3.5.49) in L1(Rx) is also a solution in L2(Rx) and in L∞(Rx), and the proof of the uniquenessof the solution to (3.5.49) under the conditions (1) – (3) goes as in Theorem 3.5.5. The role of f(k) isplayed by the unique solution of the Riemann problem:

f+(k) = S(−k)f−(k), k ∈ R, (3.5.65)

which consists of finding two functions f+(k) and f−(k) satisfying (3.5.65) such that f+ is an analyticfunction in C+, f+(ikj) = 0, f+(ikj) 6= 0, 1 ≤ j ≤ J , f+(∞) = 1, and f−(k) is an analytic function in

C− such that f−(−ikj) = 0, f−(−ikj) 6= 0, 1 ≤ j ≤ J , f−(∞) = 1, and f+(0) = 0 if indS(k) = −2J − 1,f+(0) 6= 0 if indS(k) = −2J . Existence of a solution to (3.5.65) follows from the non-negativity ofindS(−k) = −indS(k). Uniqueness of the solution to the above problem is proved as follows. Denotef+(k) := f(k) and f−(k) = f(−k). Assume that f1 and f2 solve the above problem. Then (3.5.65)implies

f1(k)

f2(k)=f1(−k)f2(−k)

, k ∈ R, f1(ikj) = f2(ikj) = 0, f1(ikj) 6= 0, f2(ikj) 6= 0, f1(∞) = f2(∞) = 1.

(3.5.66)

The function f1(k)f2(k)

is analytic in C+ and tends to 1 at infinity in C+, The function f1(−k)f2(−k) is analytic in

C− and tends to 1 at infinity in C−. Both functions agree on R. Thusf1(k)f2(k)

is analytic in C and tends to 1

at infinity. Therefore f1(k) = f2(k). To complete the proof we need to check that q, obtained by (3.1.40),

belongs to L1,1. In other words, that q = −2 dA(x,x)dx ∈ L1,1. To prove this, use (3.5.59) and (3.5.60). It

is sufficient to check that F ′(2x) ∈ L1,1, A(x, x)F (2x) ∈ L1,1 and∫∞x

[Ax(x, s) − Ax(x, s)]F (s+ x)ds ∈L1,1. The first inclusion follows from ‖xF ′‖L1(R+) < ∞. Let us prove that limx→∞[xF (x)] = 0. One

has∫ x0sF ′ds = xF (x) −

∫ x0Fds. Because xF ′ ∈ L1(R+) and F ∈ L1(R+) it follows that the limit

c0 := limx→∞xF exists. This limit has to be zero: if F = c0x + o( 1

x ) as x → ∞ and c0 6= 0, then

F 6∈ L1(R+). Now∫∞0x|F (2x)A(x, x)|dx ≤ c

∫∞0

|A(x, x)|dx < ∞. The last inequality follows from

(3.5.53): since F (2x) ∈ L1(R+) it is sufficient to check that∫∞x A(x, s)F (s+ x)ds ∈ L1(R+). One has∫∞

0dx∫∞x

|A(x, s)||F (s+ x)|ds ≤∫∞0dxσF (2x)

∫∞x

|A(x, s)|ds ≤ c. Here

σF (x) := supy≥x

|F (y)|, σF ∈ L1(R+). (3.5.67)

Note that limx→∞ xσF (x) = 0 because σF (x) is monotonically decreasing and belongs to L1(R+). 2


3.5.6 A new Marchenko-type equation

The basic result of this Section is:

Theorem 3.5.13. Equation

F (y) +A(y) +

∫ ∞

−∞A(t)F (t + y)dt = A(−y), −∞ < y < ∞, (3.5.68)

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:

F (ξ) + A(ξ) + A(−ξ)F (ξ) = A(−ξ), (3.5.69)

where, by (3.1.41),

F (ξ) = 1 − S(−ξ) + 2πJ∑

j=1

sjδ(ξi + ikj). (3.5.70)

Use (3.1.46), the equation S(ξ)f(ξ) = f(−ξ), add 1 to both sides of (3.5.48), and get:

f(ξ) + f(−ξ)F (ξ) = f(−ξ). (3.5.71)

From (3.5.70) and (3.5.71) one gets:

f(ξ) = f(−ξ)[S(−ξ) − 2π

J∑

j=1

sjδ(ξ + ikj)] = f(ξ) − 2π

J∑

j=1

sjδ(ξ + ikj)f(−ξ) = f(ξ), (3.5.72)

where the equation δ(ξ + ikj)f(−ξ) = 0 was used. This equation holds because f(ikj) = 0, and theproduct δ(ξ + ikj)f(−ξ) makes sense because f(ξ) is analytic in C+. Equation (3.5.72) holds obviously,and since each of our steps was invertible, equation (3.5.67) holds. 2

Remark 3.5.14. Equation (3.5.67) has a unique solution A(y), such that A(y) ∈ L1(R+) and A(y) = 0for y < 0.

Proof. Equation (3.5.69) for y > 0 is identical with (3.1.43) because A(−y) = 0 for y > 0. Equation(3.1.43) has a solution in L1(R+) and this solution is unique, see Theorem 3.5.5. Thus, equation (3.5.69)cannot have more than one solution, because every solution A(y) ∈ L1(R+), A(y) = 0 for y < 0, of(3.5.69) solves (3.1.43), and (3.1.43) has no more than one solution. On the other hand, the solutionA(y) ∈ L1(R+) of (3.1.43) does exist, is unique, and solves (3.5.69), as was shown in the proof ofTheorem 3.5.13. This proves Remark 3.5.14. 2

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) kj, sj > 0, S(−k) = S(k) = S−1(k), k ∈ R, S(∞) = 1,

(B) J := indS(k) := 12π

∫∞−∞ dlogS(k) is a nonpositive integer,

(C) F ∈ Lp, p = 1 and p = ∞, xF ′ ∈ L1, Lp := Lp(0,∞).


If one wants to study the characteristic properties of the scattering data, that is, a necessary andsufficient condition on these data to guarantee that the corresponding potential belongs to a prescribedfunctional class, then conditions (A) and (B) are always necessary for a real-valued q to be in L1,1, theusual 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, finiteness of its negative spectrum, and ofthe 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:

supy≥x

|F (y)| := σF (x) ∈ L1, F ′ ∈ L1,1. (3.5.73)

The function σF is monotone decreasing, |F (x)| ≤ σF (x). Equation (3.1.43) is of Fredholm type inLpx := Lp(x,∞) ∀x ≥ 0 and p = 1. The norm of the operator F := Fx in (3.1.43) can be estimated:

‖Fx‖ ≤∫ ∞

x

σF (x+ y)dy ≤ σ1F (2x), σ1F (x) :=

∫ ∞

x

σF (y)dy. (3.5.74)

Therefore (3.1.43) is uniquely solvable in L1x for any x ≥ x0 if

σ1F (2x0) < 1. (3.5.75)

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 x ≥ x0, let us derive inequalities for A = A(x, y). Define

σA(x) := supy≥x

|A(x, y)| := ‖A‖.


σA(x) ≤ σF (2x) + σA(x) supy≥x

∫ ∞

x

σF (s + y)ds ≤ σF (2x) + σA(x)σ1F (2x).

Thus, if (3.5.75) holds, then

σA(x) ≤ cσF (2x), x ≥ x0. (3.5.76)

By c > 0 different constants depending on x0 are denoted. Let

σ1A(x) := ‖A‖1 :=

∫ ∞

x

|A(x, s)|ds.

Then (3.1.43) yields σ1A(x) ≤ σ1F (2x) + σ1A(x)σ1F (2x). So

σ1A(x) ≤ cσ1F (2x), x ≥ x0. (3.5.77)

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

(I + Fx)Ax(x, y) = A(x, x)F (x+ y) − F ′(x+ y), y ≥ x ≥ 0, (3.5.78)

and

Ay(x, y) +

∫ ∞

x

A(x, s)F ′(s + y)ds = −F ′(x+ y), y ≥ x ≥ 0. (3.5.79)

Denote

σ2F (x) :=

∫ ∞

x

|F ′(y)|dy, σ2F (x) ∈ L1. (3.5.80)


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

‖Ay‖1 ≤∫ ∞

x

|F ′(x+y)|dy+σ1A(x) sups≥x

∫ ∞

x

|F ′(s+y)|dy ≤ σ2F (2x)[1+cσ1F (2x)] ≤ cσ2F (2x), (3.5.81)

and using (3.5.78) one gets:

‖Ax‖1 ≤ A(x, x)σ1F (2x) + σ2F (2x) + ‖Ax‖1σ1F (2x),

so‖Ax‖1 ≤ c[σ2F (2x) + σ1F (2x)σF (2x)]. (3.5.82)

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

A(x, x) = −2F ′(2x) + A(x, x)F (2x)−∫ ∞

x

Ax(x, s)F (x+ s)ds −∫ ∞

x

A(x, s)F ′(s+ x)ds. (3.5.83)

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

|A(x, x)| ≤ 2|F ′(2x)| + cσ2F (2x) + cσF (2x)[σ2F (2x) + σ1F (2x)σF (2x)] + cσF (2x)σ2F (2x). (3.5.84)

Thus,x|A(x, x)| ∈ L1, (3.5.85)

provided that xF ′(2x) ∈ L1, xσ2F (2x) ∈ L1, and xσF (2x)σ2F (2x) ∈ L1. Assumption (3.5.73) implies

xF ′(2x) ∈ L1. If σF (2x) ∈ L1, and σF (2x) > 0 decreases monotonically, then xσF (x) → 0 as x → ∞.Thus xσ2

F (2x) ∈ L1, and σ2F (2x) ∈ L1 because∫∞0dx∫∞x

|F ′(y)|dy =∫∞0

|F ′(y)|ydy < ∞, 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 q ∈ L1,1 where L1,1 = q : q = q,∫∞x0x|q(x)|dx < ∞, and x0 ≥ 0 satisfies (3.5.75).

Let us assume now that (3.5.76), (3.5.77), (3.5.81), and (3.5.82) hold, where σF ∈ L1 and σ2F ∈ L1

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:

F (x+ y) +

∫ ∞

x

A(x, s)F (s + y)ds = −A(x, y), y ≥ x ≥ 0. (3.5.86)

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

F (z) +

∫ ∞

z

A(x, v + x− z)F (v)dv = −A(x, z − x), z ≥ 2x. (3.5.87)


σF (2x) ≤ σA(x) + σF (2x) supz≥2x

∫ ∞

z

|A(x, v+ x− z)|dv ≤ σA(x) + σF (2x) ‖A‖1.

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

σF (2x) ≤ cσA(x). (3.5.88)

Also from (3.5.87) it follows that:

σ1F (2x) := ‖F‖1 :=

∫ ∞

2x

|F (v)|dv

≤∫ ∞

2x

|A(x, z − x)|dz +

∫ ∞

2x

∫ ∞

z

|A(x, v + x− z)||F (v)|dv dz

≤ ‖A‖1 + ‖F‖1‖A‖1, so σ1F (2x) ≤ cσ1A(x).

(3.5.89)



∫ ∞

x

|F ′(x+ y)|dy = σ2F (2x) ≤ cσA(x)σ1A(x) + ‖Ax‖ + c‖Ax‖1σ1A(x). (3.5.90)

Let us summarize the results:

Theorem 3.5.15. If x ≥ x0 and (3.5.73) hold, then one has:

σA(x) ≤ cσF (2x), σ1A(x) ≤ cσ1F (2x),

‖Ay‖1 ≤ σ2F (2x)(1 + cσ1F (2x)),

‖Ax‖1 ≤ c[σ2F (2x) + σ1F (2x)σF (2x)].

(3.5.91)

Conversely, if x ≥ x0 andσA(x) + σ1A(x) + ‖Ax‖1 + ‖Ay‖1 < ∞, (3.5.92)

then

σF (2x) ≤ cσA(x),

σ1F (2x) ≤ cσ1A(x),

σ2F (x) ≤ c[σA(x)σ1A(x) + ‖Ax‖1(1 + σ1A(x))].

(3.5.93)

In the next section we replace the assumption x ≥ x0 > 0 by x ≥ 0. The argument in this case isbased on the Fredholm alternative.

Characterization of the scattering data revisited

First, let us give necessary and sufficient conditions on S for q to be in L1,1. These conditions areobtained 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 necessary for q ∈ L1,1. 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 q ∈ L1,1. Indeed if they hold, then we prove that equation (3.1.43) has aunique solution in L1

x for all x ≥ 0. 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 L1x for any x ≥ 0 and this

solution is unique.

Proof. Since Fx is compact in L1x, ∀x ≥ 0, by the Fredholm alternative it is sufficient to prove that

(I + Fx)h = 0, h ∈ L1x, (3.5.94)

implies h = 0. Let us prove it for x = 0. The proof is similar for x > 0. If h ∈ L1, then h ∈ L∞ because‖h‖∞ ≤ ‖h‖L1σF (0). If h ∈ L1 ∩ L∞, then h ∈ L2 because‖h‖2

L2 ≤ ‖h‖L∞‖h‖L1 . Thus, if h ∈ L1 andsolves (3.5.94), then h ∈ L2 ∩ L1 ∩ L∞.

Denote h =∫∞0 h(x)eikxdx, h ∈ L2. Then,

∫ ∞

−∞h2dk = 0. (3.5.95)

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

0 = ((I + F0)h, h) =1

2π‖h‖2 +

1

2π

∫ ∞

−∞[1 − S(k)]h2(k)dk +

J∑

j=1

sjh2j , hj :=

∫ ∞

0

e−kjxh(x)dx.


Thus, using (3.5.95), one gets

hj = 0, 1 ≤ j ≤ J, (h, h) = (S(k)h, h(−k)),where we have used real-valuedness of h, i.e., h(−k) = h(k), ∀k ∈ R.

Thus, (h, h) = (h, S(−k)h(−k)), where (A) was used. Since ‖S(−k)‖ = 1, one has ‖h‖2 = |(h,S(−k)h(−k))| ≤ ‖h‖2, so the equality sign is attained in the Cauchy inequality. Therefore, h(k) =

S(−k)h(−k).By condition (B), the theory of Riemann problem guarantees existence and uniqueness of an analytic

in C+ := k : Im k > 0 function f(k) := f+(k), f(ikj) = 0, f (ikj) 6= 0, 1 ≤ j ≤ J, f(∞) = 1, such that

f+(k) = S(−k)f−(k), k ∈ R, (3.5.96)

and f−(k) = f(−k) is analytic in C− := k : Imk < 0, f−(∞) = 1 in C−, f−(−ikj) = 0, f−(−ikj) 6= 0.Here the property S(−k) = S−1(k), ∀k ∈ R is used.

One has

ψ(k) :=h(k)

f(k)=h(−k)f(−k , k ∈ R, hj := h(ikj) = 0, 1 ≤ j ≤ J.

The function ψ(k) is analytic in C+ and ψ(−k) is analytic in C−, they agree on R, so ψ(k) is analytic

in C. Since f(∞) = 1 and h(∞) = 0, it follows that ψ ≡ 0.

Thus, h = 0 and, consequently, h(x) = 0, as claimed. Theorem 3.5.16 is proved. 2

The unique solution to equation (3.1.44) satisfies the estimates given in Theorem 3.5.15. In the proofof Theorem 3.5.15 the estimate x|A(x, x)| ∈ L1(x0,∞) was established. So, by (3.1.42), xq ∈ L1(x0,∞).

The method developed in the previous Section gives accurate information about the behavior of qnear 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 toL1,1(x0,∞).

Investigation of the behavior of q(x) on (0, x0) requires additional argument. Instead of using thecontraction mapping principle and inequalities, one has to use the Fredholm theorem, which says that‖(I+Fx)

−1‖ ≤ c for any x ≥ 0, where the operator norm is taken for Fx acting in Lpx, p = 1 and p = ∞,and the constant c does not depend on x ≥ 0.

Such an analysis yields:

Theorem 3.5.18. If and only if (A), (B), and (C) hold, then q ∈ L1,1.

Proof. It is sufficient to check that Theorem 3.5.15 holds with x ≥ 0 replacing x ≥ x0. To get (3.5.76)with x0 = 0, one uses (3.1.44) and the estimate:

‖A(x, y)‖ ≤ ‖(I + Fx)−1‖‖F (x+ y)‖ ≤ cσF (2x), ‖ · ‖ = sup

y≥x| · |, x ≥ 0, (3.5.97)

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

‖A(x, y)‖1 ≤ c sups≥x

∫ ∞

x

|F (s+ y)|dy ≤ cσ1F (2x), x ≥ 0. (3.5.98)


‖Ax(x, y)‖1 ≤ c[‖F ′(x+ y)‖1 + A(x, x)‖F (x+ y)‖1]

≤ cσ2F (2x) + cσF (2x)σ1F (2x), x ≥ 0.(3.5.99)


‖Ay(x, y)‖1 ≤ c[σ2F (2x) + σ1F (2x)σ2F (2x)] ≤ σ2F (2x). (3.5.100)

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 proofof Theorem 3.5.15. Consequently Theorem 3.5.15 holds with x0 = 0. Theorem 3.5.18 is proved. 2


Compactly supported potentials

In this Section necessary and sufficient conditions are given for q ∈ La1,1 := q : q = q, q = 0 if x >

a,∫ a0x|q|dx < ∞. Recall that the Jost solution is:

f(x, k) = eikx +

∫ ∞

x

A(x, y)eikydy, f(0, k) := f(k). (3.5.101)

Lemma 3.5.19. If q ∈ La1,1, then f(x, k) = eikx for x > a, A(x, y) = 0 for y ≥ x ≥ a, F (x+ y) = 0 fory ≥ x ≥ a (cf (3.1.43)), and F (x) = 0 for x ≥ 2a.

Thus, (3.1.43) with x = 0 yields A(0, y) := A(y) = 0 for x ≥ 2a. The Jost function

f(k) = 1 +

∫ 2a

0

A(y)eikydy, A(y) ∈W 1,1(0, a), (3.5.102)

is an entire function of exponential type ≤ 2a, that is, |f(k)| ≤ ce2a|k|, k ∈ C, and S(k) = f(−k)/f(k) isa meromorphic function in C. In (3.5.102)W l,p is the Sobolev space, and the inclusion (3.5.102) followsfrom Theorem 3.5.15.

Let us formulate the assumption (D):(D) the Jost function f(k) is an entire function of exponential type ≤ 2a.

Theorem 3.5.20. Assume (A), (B), (C) and (D). Then q ∈ La1,1. Conversely, if q ∈ La1,1, then (A),(B), (C) and (D) hold.

Necessity. If q ∈ L1,1, 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 q ∈ L1,1. 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 y ≥ 2a.We claim that F (x) = 0 for x ≥ 2a.

If this is proved, then (3.1.43) yields A(x, y) = 0 for y ≥ x ≥ a, 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 C+ except for J simple poles at the

points ikj. If x > 2a then one can use the Jordan lemma and the residue theorem and get:

Fs(x) =1

2π

∫ ∞

−∞[1 − S(k)]eikxdk = −i

J∑

j=1

f(−ikj)f (ikj)

e−kjx, x > 2a. (3.5.103)

Since f(k) is entire, the Wronskian formula

f ′(0, k)f(−k) − f ′(0,−k)f(k) = 2ik

is valid on C, and at k = ikj it yields:

f ′(0, ikj)f(−ikj ) = −2kj,

because f(ikj) = 0. This and (3.5.103) yield

Fs(x) =

J∑

j=1

2ikj

f ′(0, ikj)f (ikj)e−kjx = −

J∑

j=1

sje−kjx = −Fd(x), x > 2a.

Thus, F (x) = Fs(x) + Fd(x) = 0 for x > 2a. The sufficiency is proved.Theorem Theorem 3.5.20 is proved. 2

In [M] a condition on S , which guarantees that q = 0 for x > a, is given under the assumption thatthere is no discrete spectrum, that is F = Fs.


Square integrable potentials

Let us introduce conditions (3.5.104) – (3.5.106)

2ik[f(k) − 1 +Q

2ik] ∈ L2(R) := L2, Q :=

∫ ∞

0

qds, (3.5.104)

k[1− S(k) +Q

ik] ∈ L2, (3.5.105)

k[|f(k)|2 − 1] ∈ L2. (3.5.106)

Theorem 3.5.21. If (A), (B), (C), and any one of the conditions (3.5.104) – (3.5.106) hold, thenq ∈ L2(R+).

Proof. We refer to [R139] for the proof. 2

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 δ` are the fixed-energy (k = const > 0) phase shifts. Define

Lrϕ :=

[r2

∂2

∂r2+ r2 − r2q(r)

]ϕ := L0rϕ − r2q(r)ϕ, (3.6.1)

where ϕ = ϕ`(r) is a regular solution to

Lrϕ` = `(` + 1)ϕ`, (3.6.2)

such that

ϕ` = u` +

∫ r

0

K(r, ρ)u`(ρ)ρ−2dρ, K(r, 0) = 0, (3.6.3)

and ul =√

πr2 J`+ 1

2(r), J`(r) is the Bessel function. In (3.6.3)K(r, ρ) is the transformation kernel, I+K

is the transformation operator. In (3.6.2) we assume that k = 1 without loss of generality. The ϕ` isuniquely defined by its behavior near the origin:

ϕ`(r) =r`+1

(2`+ 1)!!+ o(r`+1), r −→ 0. (3.6.4)

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

γù` := 2`Γ(`+ 1)u`(r) = r`+1

∫ 1

−1

(1 − t2)èirtdt, (3.6.5)

where Γ(z) is the gamma-function.The inverse scattering problem with fixed-energy phase shifts δ``=0,1,2... 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, ∫ a

0

r2|q(r)|2dr < ∞. (3.6.6)

3.6. INVERSE SCATTERING PROBLEM WITH FIXED-ENERGY 123

Conditions (3.6.6) imply that q ∈ L2(Ba), Ba := x : x ∈ R3, |x| ≤ a.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 scatteringproblem with fixed-energy phase shifts as the data. In Section 3.6.5 it is proved that the NS theory isfundamentally wrong and is not an inversion method. The main results of this Section are Theorems3.6.2–3.6.4, and the proof of the fact that the Newton-Sabatier theory is fundamentally wrong.

3.6.2 Existence and uniqueness of the transformation operatorsindependent of angular momentum

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

0 = − r2q(r)u+ (r2 − r2q(r))

∫ r

0

K(r, ρ)uρ−2dρ

−∫ r

0

K(r, ρ)ρ−2L0ρudρ+ r2∂2r

∫ r

0

K(r, ρ)uρ−2dρ.

(3.6.7)

We assume first that K(r, ρ) is twice continuously differentiable with respect to its variables inthe region 0 < r < ∞, 0 < ρ ≤ r. This assumption requires extra smoothness of q(r), q(r) ∈C1(0, a). If q(r) satisfies condition (3.6.6), then equation (3.6.13) below has to be understood in thesense of distributions. Eventually we will work with an integral equation (3.6.40) (see below) for whichassumption (3.6.6) suffices.

Note that∫ r

0

K(r, ρ)ρ−2L0ρudρ =

∫ r

0

L0ρK(r, ρ)uρ−2dρ+K(r, r)ur −Kρ(r, r)u, (3.6.8)

provided thatK(r, 0) = 0. (3.6.9)

We assume (3.6.9) to be valid. Denote

K :=dK(r, r)

dr. (3.6.10)

Then

r2∂2r

∫ r

0

K(r, ρ)uρ−2dρ = Ku+K(r, r)ur −2

rK(r, r)u+Kr(r, r)u+ r2

∫ r

0

Krr(r, ρ)uρ−2dρ. (3.6.11)

Combining (3.6.7) – (3.6.11) and writing again u` in place of u, one gets

0 =

∫ r

0

[LrK(r, ρ) − L0ρK(r, ρ)

]u`(ρ)ρ

−2dρ + u`(r)

×[−r2q(r) + K − 2Kr(r, r)

r+Kr(r, r) +Kρ(r, r)

], ∀r > 0, ` = 0, 1, 2, . . . .

(3.6.12)

Let us prove that (3.6.12) implies:

LrK(r, ρ) = L0ρK(r, ρ), 0 < ρ ≤ r, (3.6.13)

q(r) =2K

r2− 2K(r, r)

r=

2

r

d

dr

K(r, r)

r. (3.6.14)

This proof requires a lemma.


Lemma 3.6.1. Assume that ρf(ρ) ∈ L1(0, r) and ρA(ρ) ∈ L1(0, r). If

0 =

∫ r

0

f(ρ)u`(ρ)dρ + u`(r)A(r) ∀` = 0, 1, 2, . . ., (3.6.15)

thenf(ρ) ≡ 0 and A(r) = 0. (3.6.16)

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

0 =

∫ 1

−1

dt(1− t2)`(d

idt

)` ∫ r

0

dρρf(ρ)eiρt + rA(r)

∫ 1

−1

(1 − t2)`(d

idt

)èirtdt.

Therefore

0 =

∫ 1

−1

dtd`(t2 − 1)`

dt`[

∫ r

0

dρρf(ρ)eiρt + rA(r)eirt], l = 0, 1, 2, . . . (3.6.17)

Recall that the Legendre polynomials are defined by the formula

P`(t) =1

2``!

d`

dt`(t2 − 1)`, (3.6.18)

and they form a complete system in L2(−1, 1).Therefore (3.6.17) implies

∫ r

0

dρρf(ρ)eiρt + rA(r)eirt = 0 ∀t ∈ [−1, 1]. (3.6.19)

Equation (3.6.19) implies ∫ r

0

dρρf(ρ)eiρt = 0, ∀t ∈ [−1, 1], (3.6.20)

andrA(r) = 0. (3.6.21)

Therefore A(r) = 0. Also f(ρ) = 0 because the left-hand side of (3.6.20) is an entire function of t, whichvanishes on the interval [−1, 1] and, consequently, it vanishes identically, so that ρf(ρ) = 0 and thereforef(ρ) ≡ 0.


We prove that the problem (3.6.13), (3.6.14), (3.6.9), which is a Goursat-type problem, has a solutionand this solution is unique in the class of functions K(r, ρ), which are twice continuously differentiablewith respect to ρ and r, 0 < r < ∞, 0 < ρ ≤ r.

In this section we assume that q(r) ∈ C1(0, a). This assumption implies that K(r, ρ) is twicecontinuously differentiable. If (3.6.6) holds, then the arguments in this section which deal with integralequation (3.6.40) remain valid. Specifically, existence and uniqueness of the solution to equation (3.6.40)is proved under the only assumption

∫ a0r|q(r)|dr < ∞ 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:

r2Krr − ρ2Kρρ + [r2 − r2q(r) − ρ2]K(r, ρ) = 0, 0 < ρ ≤ r, (3.6.22)

K(r, r) =r

2

∫ r

0

sq(s)ds := g(r), (3.6.23)

K(r, 0) = 0. (3.6.24)


The difficulty in the study of this Goursat-type problem comes from the fact that the coefficients in frontof the second derivatives of the kernel K(r, ρ) are variable.

Let us reduce problem (3.6.22) – (3.6.24) to the one with constant coefficients. To do this, introducethe new variables:

ξ = ln r + ln ρ, η = ln r − lnρ. (3.6.25)

Note that

r = eξ+η2 , ρ = e

ξ−η2 , (3.6.26)

η ≥ 0, −∞ < ξ < ∞, (3.6.27)

and

∂r =1

r(∂ξ + ∂η), ∂ρ =

1

ρ(∂ξ − ∂η). (3.6.28)

Let

K(r, ρ) := B(ξ, η).

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

Bξη(ξ, η) −1

2Bη(ξ, η) + Q(ξ, η)B = 0, η ≥ 0, −∞ < ξ < ∞, (3.6.29)

B(ξ, 0) = g(e

ξ2

):= G(ξ), −∞ < ξ < ∞ (3.6.30)

B(−∞, η) = 0, η ≥ 0, (3.6.31)

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

Q(ξ, η) :=1

4

[eξ+η − eξ+ηq

(e

ξ+η2

)− eξ−η

], (3.6.32)

and took into account that ρ = r implies η = 0, while ρ = 0 implies, for any fixed η ≥ 0, that ξ = −∞.Note that

sup−∞<ξ<∞

e−ξ2G(ξ) < c, (3.6.33)

sup0≤η≤B

∫ A

−∞|Q(s, η)|ds ≤ c(A,B), (3.6.34)

for any A ∈ R 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 L(ξ, η)

by the formula:

L(ξ, η) := B(ξ, η)e−ξ2 . (3.6.35)

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

Lηξ(ξ, η) + Q(ξ, η)L(ξ, η) = 0, η ≥ 0, −∞ < ξ < ∞, (3.6.36)

L(ξ, 0) = e−ξ2G(ξ) := b(ξ) :=

1

2

∫ eξ2

0

sq(s)ds, −∞ < ξ < ∞, (3.6.37)

L(−∞, η) = 0, η ≥ 0. (3.6.38)

We want to prove existence and uniqueness of the solution to (3.6.36) – (3.6.38). In order to choose aconvenient Banach space in which to work, let us transform problem (3.6.36) – (3.6.38) to an equivalentVolterra-type integral equation.


Integrate (3.6.36) with respect to η from 0 to η and use (3.6.37) to get

Lξ(ξ, η) − b′(ξ) +

∫ η

0

Q(ξ, t)L(ξ, t)dt = 0. (3.6.39)

Integrate (3.6.39) with respect to ξ from −∞ to ξ and use (3.6.39) to get

L(ξ, η) = −∫ ξ

−∞ds

∫ η

0

dtQ(s, t)L(s, t) + b(ξ) := V L+ b, (3.6.40)

where

V L := −∫ ξ

−∞ds

∫ η

0

dtQ(s, t)L(s, t). (3.6.41)

Consider the space X of continuous functions L(ξ, η), defined in the half-plane η ≥ 0, −∞ < ξ < ∞,such that for any B > 0 and any −∞ < A < ∞ one has

‖L‖ := ‖L‖AB := sup−∞<s≤A0≤t≤B

(e−γt|L(s, t)|

)< ∞, (3.6.42)

where γ > 0 is a number which will be chosen later so that the operator V in (3.6.40) will be a contractionmapping on the Banach space of functions with norm (3.6.42) for a fixed pair A,B. To choose γ > 0,let us estimate the norm of V . One has:

‖V L‖ ≤ sup−∞<ξ≤A,0≤η≤B

(∫ ξ

−∞ds

∫ η

0

dt∣∣Q(s, t)

∣∣e−γ(η−t)e−γt∣∣L(s, t)

∣∣)

≤ ‖L‖ sup−∞<ξ≤A,0≤η≤B

∫ ξ

−∞ds

∫ η

0

dt

(2es+t + es+t

∣∣∣q(e

s+t2

)∣∣∣)e−γ(η−t) ≤ c

γ‖L‖,

(3.6.43)

where c > 0 is a constant depending on A,B and∫ a0r|q(r)|dr. Indeed, one has:

2

∫ A

−∞ds

∫ η

0

dtes+t−γ(η−t) = 2eA∫ η

0

dtet−γ(η−t)dt ≤ 2eA+B 1 − e−γB

γ=c1γ, (3.6.44)

and, using the substitution σ = es+t2 , one gets:

∫ A

−∞ds

∫ η

0

dtes+t∣∣∣q(e

s+t2

)∣∣∣e−γ(η−t) =

∫ η

0

dte−γ(η−t)∫ A

−∞dses+t|q

(e

s+t2

)|

= 2

∫ η

0

dte−γ(η−t)∫ e

A+t2

0

dσσ|q(σ)|

=2(1 − e−γB)

γ

∫ a

0

dσσ|q(σ)| :=c2γ.

From these estimates inequality (3.6.44) follows.It follows from (3.6.44) that V is a contraction mapping in the space XAB of continuous functions

in the region −∞ < ξ ≤ A, 0 ≤ η ≤ B, with the norm (3.6.42) provided that

γ > c. (3.6.45)

Therefore equation (3.6.40) has a unique solution L(ξ, η) in the region

−∞ < ξ < A, 0 ≤ η ≤ B (3.6.46)


for any real A and B > 0 if (3.6.45) holds. This means that the above solution is defined for any ξ ∈ Rand any η ≥ 0.

Equation (3.6.40) is equivalent to problem (3.6.36) – (3.6.38) and, by (3.6.35), one has:

B(ξ, η) = L(ξ, η)eξ2 . (3.6.47)

Therefore we have proved the existence and uniqueness of B(ξ, η), that is, of the kernel K(r, ρ) = B(ξ, η)of the transformation operator (3.6.3). Recall that r and ρ are related to ξ and η by formulas (3.6.26).

Let us formulate the result:

Theorem 3.6.2. The kernel K(r, ρ) 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 continuouslydifferentiable functions for any potential q(r) ∈ C1(0, a). If q(r) ∈ L∞(0, a), then K(r, ρ) has firstderivatives 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:

∫ r

0

|K(r, ρ)|ρ−1dρ <∞. (3.6.48)

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

Note that ∫ r

0

|K(r, ρ)|ρ−1dρ = r

∫ ∞

0

|L(2 ln r − η, η)|e−η2 dη < ∞ (3.6.49)

Indeed, if r > 0 is fixed, then, by (3.6.26), ξ + η = 2 ln r = const. Therefore dξ = −dη, and ρ−1dρ =12 (dξ − dη) = −dη, ξ = 2 ln r − η. Thus:

∫ r

0

|K(r, ρ)|ρ−1dρ =

∫ ∞

0

|L(2 ln r − η, η)|e 2 ln r−η2 dη = r

∫ ∞

0

|L(2 ln r − η, η)|e− η2 dη. (3.6.50)

The following estimate holds:

|L(ξ, η)| ≤ ce(2+ε1)[ηµ1(ξ+η)]12+ε2

, (3.6.51)

where εj > 0, j = 1, 2, are arbitrarily small numbers and µ1 is defined in formula (3.6.57) below, seealso formula (3.6.54) for the definition of µ.

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:

I ≤ 2 + 2

∞∑

1

[2µ1(2 log r)]n

n!= 2 exp[2µ1(2 log r)] < ∞. (3.6.52)


Theorem 3.6.3. Estimate (3.6.51) holds.

Proof of Theorem 3.6.3. From (3.6.40) one gets:

m(ξ, η) ≤ c0 + (Wm)(ξ, η), m(ξ, η) := |L(ξ, η)|, (3.6.53)

where c0 = sup−∞<ξ<∞ |b(ξ)| ≤ 12

∫ a0s|q(s)|ds (see (3.6.37)), and

Wm :=

∫ ξ

−∞ds

∫ η

0

dtµ(s+ t)m(s, t), µ(s) :=1

2es(1 + |q(e s

2 )|). (3.6.54)


It is sufficient to consider inequality (3.6.53) with c0 = 1: if c0 = 1 and the solution m0(ξ, η) to (3.6.53)satisfies (3.6.51) with c = c1, then the solution m(ξ, η) of (3.6.53) with any c0 > 0 satisfies (3.6.51) withc = c0c1.

Therefore, assume that c0 = 1, then (3.6.53) reduces to:

m(ξ, η) ≤ 1 + (Wm)(ξ, η). (3.6.55)

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

W1 =

∫ ξ

−∞ds

∫ η

0

dtµ(s+ t) =

∫ η

0

dt

∫ ξ

−∞dsµ(s + t) =

∫ η

0

dtµ1(ξ + t) ≤ ηµ1(ξ + η). (3.6.56)

Here we have used the notation

µ1(ξ) =

∫ ξ

−∞µ(s)ds, (3.6.57)

and the fact that µ1(s) is a monotonically increasing function, since µ(s) > 0. Note also that µ1(s) <∞for any s, −∞ < s < ∞.

Furthermore,

W 21 ≤∫ ξ

−∞ds

∫ η

0

dtµ(s + t)tµ1(s + t) ≤∫ η

0

dtt

∫ ξ

−∞dsµ(s + t)µ1(s + t) =

η2

2!

µ21(ξ + η)

2!. (3.6.58)

Let us prove by induction that

Wn1 ≤ ηn

n!

µn1 (ξ + η)

n!. (3.6.59)

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

Wn+11 ≤ W

(ηn

n!

µn1 (ξ + η)

n!

)=

∫ η

0

dttn

n!

∫ ξ

−∞dsµ(s + t)

µn1 (s + t)

n!≤ ηn+1

(n+ 1)!

µn+11 (ξ + η)

(n+ 1)!. (3.6.60)

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

m(ξ, η) ≤ 1 +

∞∑

n=1

ηn

n!

µn1 (ξ + η)

n!≤ ce(2+ε1)[ηµ1(η+ξ)]

12+ε2

, (3.6.61)

where we have used Theorem 2 from [Lev, section 1.2], namely the order of the entire function F (z) :=1 +

∑∞n=1

zn

(n!)2is 1

2and its type is 2. The constant c > 0 in (3.6.51) depends on εj, j = 1, 2.

Recall that the order of an entire function F (z) is the number ρ := lim supr→∞ln lnMF (r)

ln r , where

MF (r) := max|z|=r |F (z)|. The type of F (z) is the number σ := lim supr→∞lnMF (r)

rρ . It is known [Lev],that if F (z) =

∑∞n=0 cnz

n is an entire function, then its order ρ and type σ can be calculated by theformulas:

ρ = lim supn→∞

n lnn

ln 1|cn |

, σ =lim supn→∞(n|cn|

ρn )

eρ. (3.6.62)

If cn = 1(n!)2 , then the above formulas yield ρ = 1

2 and σ = 2. Theorem 3.6.3 is proved. 2

3.6.3 Uniqueness theorem.

Denote by L any fixed subset of the set N of integers 0, 1, 2, . . . with the property:

∑

`∈L`6=0

1

`= ∞ (3.6.63)


Theorem 3.6.4. ([R393]) Assume that q satisfies (3.6.6) and (3.6.63) holds. Then the data δ`∀`∈L

determine q uniquely.

The idea of the proof is based on Property C-type argument.

Step 1: If q1 and q2 generate the same data δ`∀`∈L , then the following orthogonality relation holdsfor p := q1 − q2:

h(`) :=

∫ a

0

p(r)φ1`(r)φ2`(r)dr = 0 ∀` ∈ L , (3.6.64)

where φj` is the scattering solution corresponding to qj, j = 1, 2.

Step 2: Define h1(`) := 22`[Γ(` + 1)]2h(`), where Γ is the Gamma-function. Check that h1(`) isholomorphic in Π+ := ` : Re` > 0, ` = σ + iτ, σ ≥ 0, and τ are real numbers, h1(`) ∈ N (whereN is the Nevanlinna class in Π+), that is

sup0<r<1

∫ π

−πlog+ |h1(

1 − reiϕ

1 + reiϕ)|dϕ < ∞,

where log+ x =

logx if logx > 0,

0 if logx ≤ 0.If h1 ∈ N vanishes ∀` ∈ L , then h1 = 0 in Π+, and, by property

Cϕ, p(r) = 0. Theorem 3.6.4 is proved. 2

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 symmet-ric potential q from an a priori fixed class, say L1,1, a standard scattering class, generates fixed-energyphase shifts δl, l = 0, 1, 2, . . . ,. The inverse scattering problem consists of recovery of q from these data.

The second case deals with a different problem: given some numbers δl , l = 0, 1, 2, . . ., which areassumed to be fixed-energy phase shifts of some potential q, from a class not specified, find some potentialq1, which generates fixed-energy phase shifts equal to δl, l = 0, 1, 2, . . .. This potential q1 may have nophysical interest because of its non-physical” behavior at infinity or other undesirable properties. Also,in the second case it may be that q1 6= q.

We first discuss NS procedure assuming that it is intended to solve the inverse scattering problem incase 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 phase shiftsδl, l = 0, 1, 2, . . ., corresponding to an unknown spherically symmetric potential q(r). R. Newton did notspecify the class of potentials for which he tried to develop an inversion theory and did not formulateand proved any results which would justify the inversion procedure he proposed (NS procedure). Hisarguments are based on the following claim, which is implicit in his works, but crucial for the validityof NS procedure:

Claim N1: The basic integral equation.

K(r, s) = f(r, s) −∫ r

0

K(r, t)f(t, s)dt

t2, 0 ≤ s ≤ r < ∞, (3.6.65)

is uniquely solvable for all r > 0.


Here

f(r, s) :=

∞∑

l=0

clul(r)ul(s), ul :=

√πr

2Jl+ 1

2(r), (3.6.66)

cl are real numbers, the energy k2 is fixed: k = 1 is taken without loss of generality, Jl+ 12(r) are the

Bessel functions. It is assumed in [CS], p.196, that∑∞

l=1 l−2cl < ∞. If equation (3.6.65) is uniquely

solvable for all r > 0, then the potential q1, that NS procedure yields, is defined by the formula:

q1(r) = −2

r

d

dr

K(r, r)

r. (3.6.67)

The R. Newton’s ansatz (3.6.65) - (3.6.66) for the transformation kernel K(r, s) of the Schrodinger oper-ator, corresponding to some q(r), namely, that K(r, s) is the unique solution to (3.6.65) - (3.6.66), is notcorrect 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 leadsto 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 fundamentalfor any inversion theory: does the reconstructed potential q1 generate the data from which it wasreconstructed?

In [CS, p.205], there are two claims:Claim (i): q1(r) generates the original shifts δl ”provided that δl are not ”exceptional””, and

Claim (ii): the NS procedure ”yields one (only one) potential which decays faster than r−32 ” and

generates the original phase shifts δl.If one considers NS procedure as a solution to inverse scattering problem of finding an unknown

potential q from a certain class, for example q(r) ∈ L1,1 := q : q = q,∫∞0r|q(r)|dr < ∞, from

the fixed-energy phase shifts, generated by this q, then the proof, given in [CS], of Claim (i) is notconvincing: it is not clear why the potential q1, obtained by NS procedure, has the transformationoperator generated by the potential corresponding to the original data, that is, to the given fixed-energyphase shifts. In fact, as follows from Proposition 3.6.5 below, the potential q1 cannot generate the kernelK(r, s) of the transformation operator corresponding to a generic original potential q(r) ∈ L1,1 := q :q = q,

∫∞0r|q(r)|dr < ∞.

Claim (ii) is incorrect because the original generic potential q(r) ∈ L1,1 generates the phase shiftsδl, and if q1(r), the potential obtained by NS procedure and therefore not equal to q(r) by Propo-sition 3.6.5, generates the same phase shifts δl, then one has two different potentials q(r) and q1(r),

which both decay faster than r−32 and both generate the original phase shifts δl, contrary to Claim

(ii).Our aim is to formulate and justify the followingConclusions: Claim N1 and ansatz (3.6.65) - (3.6.66) are not proved by R. Newton and, in gen-

eral, are wrong. Moreover, one cannot approximate with a prescribed accuracy in the norm ‖q‖ :=∫∞0 r|q(r)|dr a generic potential q(r) ∈ L1,1 by the potentials which might possibly be obtained by

the NS procedure. Therefore NS procedure cannot be justified even as an approximate inversion proce-dure.

Let us justify these conclusions:

Claim N1 formulated above (and basic for NS procedure) is wrong, in general, for the following reason:Given fixed-energy phase shifts, corresponding to a generic potential q ∈ L1,1, one either cannot

carry through NS procedure because:

(a) the system (12.2.5a) in [CS], which should determine numbers cl in formula (3.6.66), given thephase shifts δl , 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 q1 by formula (3.6.67), then thispotential is not equal to the original generic potential q ∈ L1,1, 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 q1 by formula(3.6.67), then this q1 is a restriction to (0,∞) of a function analytic in a neighborhood of (0,∞).

Since a generic potential q ∈ L1,1 is not a restriction to (0,∞) of an analytic function, one concludesthat even if equation (3.6.65) is solvable for all r > 0, the potential q1, defined by formula (3.6.67), is notequal to the original generic potential q ∈ L1,1 and therefore the inverse scattering problem of findingan unknown q ∈ L1,1 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 alsofrom 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 originalgeneric q, that is q1 = q, this would contradict Proposition 3.6.5 If formula (3.6.67) would yield a q1 whichis different from the original generic q, then NS procedure does not solve the inverse scattering problemformulated above. Note also that it is proved in [R192] that independent of the angular momenta ltransformation operator, corresponding to a generic q ∈ L1,1 does exist, is unique, and is defined by akernel 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 q ∈ L1,1 by potentials q1,which can possibly be obtained by NS procedure, is proved in Claim 7, see proof of Claim 7 below.

Thus, our conclusions are justified. 2

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 waspublished in [N2], when no uniqueness results for this inverse problem were known. It is still an openproblem if for the standard in scattering theory class of L1,1 potentials the uniqueness theorem for thesolution of the above inverse scattering problem holds.

We discuss the inverse scattering problem with fixed-energy phase shifts (as the data) for potentialsq ∈ L1,1, because only for this class of potentials a general theorem of existence and uniqueness of thetransformation 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 clear for what class of potentialsthe transformation operators, independent of l, do exist. For slowly decaying potentials the existenceof 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]) wereincorrect (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 q1, which is not equal to the original generic potential q ∈ L1,1, as follows fromProposition 3.6.5

Existence of ”transparent” potentials is often cited in the literature. A ”transparent” potential is apotential which is not equal to zero identically, but generates the fixed-energy shifts which are all equalto zero.

In [CS, p.207], there is a remark concerning the existence of ”transparent” potentials. This remarkis not justified because it is not proved that for the values cl, used in [CS, p.207], equation (3.6.65) issolvable for all r > 0. If it is not solvable even for one r > 0, then NS procedure breaks down and theexistence 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 matrixM has the property MM = I, where I is the unit matrix, and on [CS, p.198], formula (12.3.10), it isclaimed that a vector v 6= 0 exists such that Mv = 0. However, then MMv = 0 and at the same time


MMv = v 6= 0, 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 thedomain 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 thefixed-energy shifts equal to zero and find the corresponding cl 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) forall r > 0; finally construct the ”transparent” potential by formula (3.6.67). As was noted above, it isnot 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 isnot a valid inversion method. Since modifications of NS procedure are still used by some physicists, whobelieve that this procedure is an inversion theory, the author pointed out some questions concerning thisprocedure.

This concludes the discussion of case 1. 2

Discussion of case 2:

Suppose now that one wants just to construct a potential q1, which generates the phase shifts corre-sponding to some q.

This problem is actually not an inverse scattering problem because one does not recover an originalpotential from the scattering data, but rather wants to construct some potential which generates thesedata and may have no physical meaning. Therefore this problem is much less interesting practicallythan the inverse scattering problem.

However, NS procedure does not solve this problem either: there is no guarantee that this procedureis applicable, that is, that the steps a) and b), described in the justification of the conclusions, can bedone, 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 q1, 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. 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 Q :=

∫∞0rq(r)dr 6= 0, then the numbers cl in formula (3.6.66) cannot

satisfy the condition∑∞

0 |cl| < ∞. This observation can be obtained also from the following

Remark 3.6.6. For any potential q(r) ∈ L1,1 such that Q :=∫∞0rq(r)dr 6= 0 the basic equation (3.6.65)

is not solvable for some r > 0 and any choice of cl such that∑∞

l=0 |cl| <∞.

Since generically, for q ∈ L1,1, one has Q 6= 0, this gives an additional illustration to the conclusionthat equation (3.6.65), in general, is not solvable for some r > 0. Conditions

∑∞l=0 |cl| < ∞ and Q 6= 0

are incompatible.In [CS, p.196], a weaker condition

∑∞l=0 l

−2|cl| < ∞ is used, but in the examples ([CS, pp.189-191]),cl = 0 for all l ≥ l0 > 0, so that

∑∞l=0 |cl| <∞ in all of these examples.

Claim 7. The set of the potentials v(r) ∈ L1,1, which can possibly be obtained by the NS procedure, isnot dense (in the norm ‖q‖ :=

∫∞0r|q(r)|dr) in the set L1,1.

Let us prove Remark 3.6.6 and Claim 7.

Proof of Remark 3.6.6. Writing (3.6.67) as K(r, r) = − r2

∫ r0sq1(s)ds and assuming Q 6= 0, one gets the

following relation:

K(r, r) = −Qr2

[1 + o(1)

]−→ ∞ as r −→ ∞. (3.6.68)


If (3.6.65) is solvable for all r > 0, then from (3.6.66) and (3.6.65) it follows that K(r, s) =∑∞l=0 clϕl(r) ul(s), where ϕl(r) := ul(r)−

∫ r0K(r, t)ul(t)

dtt2

, so that I −K is a transformation operator,

where K is the operator with kernel K(r, s), ϕ′′l +ϕl − l(l+1)

r2ϕl − q1(r)ϕl = 0, where q1(r) = − 2

rddr

K(r,r)r

, ϕl = O(rl+1), as r → 0, and

ul(r) ∼ sin

(r − lπ

2

), ϕl(r) ∼ |Fl| sin

(r − lπ

2+ δl

)as r −→ ∞,

where δl are the phase shifts at k = 1 and Fl is the Jost function at k = 1. One can prove thatsupl |Fl| < ∞. Thus, if

∑∞l=0 |cl| < ∞, then

K(r, r) = O(1) as r −→ ∞. (3.6.69)

If Q 6= 0 then (3.6.69) contradicts (3.6.68). It follows that if Q 6= 0 then equation (3.6.65) cannotbe uniquely solvable for all r > 0, so that NS procedure cannot be carried through if Q 6= 0 and∑∞

l=0 |cl| < ∞. This proves Remark 3.6.6. 2

Proof of Claim 7. Suppose that v(r) ∈ L1,1 and Qv :=∫∞0rv(r)dr = 0, because otherwise NS procedure

cannot be carried through as was proved in Remark 3.6.6.If Qv = 0, 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 beobtained by NS procedure, is not dense in L1,1 in the norm ‖q‖ :=

∫∞0 r|q(r)|dr. In fact, any potential q

such that Q :=∫∞0rq(r)dr 6= 0, and the set of such potentials is dense in L1,1, 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 q ∈ L1,1,

Qq :=

∫ ∞

0

rq(r)dr 6= 0, and ‖vn − q‖ −→ 0 as n −→ ∞,

where the potentials vn ∈ L1,1 are obtained by the NS procedure, so that Qn :=∫∞0rvn(r)dr = 0. We

assume vn ∈ L1,1 because otherwise vn obviously cannot converge in the norm ‖ · ‖ to q ∈ L1,1. Definea linear bounded on L1,1 functional

f(q) :=

∫ ∞

0

rq(r)dr, |f(q)| ≤ ‖q‖,

where ‖q‖ :=∫∞0 r|q(r)|dr. The potentials v ∈ L1,1, which can possibly be obtained by the NS procedure,

belong to the null-space of f , that is f(v) = 0.If limn→∞ ‖vn − q‖ = 0, then limn→∞ |f(q − vn)| ≤ limn→∞ ‖q − vn‖ = 0. Since f is a linear

bounded functional and f(vn) = 0, one gets: f(q − vn) = f(q) − f(vn) = f(q). So if f(q) 6= 0 thenlimn→∞ |f(q − vn)| = |f(q)| 6= 0. Therefore, no potential q ∈ L1,1 with Qq 6= 0 can be approximatedarbitrarily accurately by a potential v(r) ∈ L1,1 which can possibly be obtained by the NS procedure.Claim 7 is proved. 2

3.6.5 Formula for the radius of the support of the potentialin 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), r = |x|, is spherically symmetric, real-valued,

∫ a0|q|2dr < ∞, and

q(r) = 0 for r > a, but q(r) 6= 0 on (a− ε, a) for all sufficiently small ε > 0.The number a > 0 we call the radius of compactness of the potential, or simply the radius of the

potential. Let A(α′, α) denote the scattering amplitude corresponding to the potential q at a fixed energyk2 > 0. Without loss of generality let us take k = 1 in what follows. By α′, α ∈ S2 the unit vectors in


the direction of the scattered, respectively, incident wave, are meant, S2 is the unit sphere in R3. Letus 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 A(α′, α), or, equivalently, from the coefficients A`(α). Very few results ofsuch type are known.

A result of such type is a necessary and sufficient condition for q(x) = q(|x|): it was proved [R139,p.131] (see Section 5.7), that q(x) = q(|x|) if and only if A(α′, α) = A(α′ · α). Of course, the necessityof this condition was a common knowledge, but the sufficiency, that is, the implication: A(α′, α) =A(α′ · α) ⇒ q(x) = q(|x|), 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 thespherically symmetric), then

a = lim`→∞

(2`

e

[sup

α∈S2−`≤m≤`

∣∣A`m(α)∣∣] 1

2`

)= lim`→∞

(2`

e

∣∣δ`∣∣ 12`

), (3.6.70)

where δ` are the fixed-energy (k = 1) phase shifts. We prove (3.6.70) for the spherically symmetricpotentials q = q(r).

If q = q(r) then A`m(α) = a`Y`m(α) where a` depends only on ` and k, but not on α or α′. Sincek = 1 is fixed, a` depends only on ` for q = q(r). Assuming q = q(r), one takes A(α′, α) = A(α′ ·α) and

calculates A`m(α) =∫S2 A(α′ · α)Y`m(α′) dα′ = a`Y`m(α), where a` := 2π

C( 12)

` (1)

∫ 1

−1A(t)C

( 12 )

` (t) dt, ` =

0, 1, 2, . . . Here we have used formula (14.4.46) in [RKa, p.413], and C(p)` (t) are the Gegenbauer polyno-

mials (see [RKa, p.408]). Since C(12 )` = P`(t), P`(1) = 1, where P`(t) are the Legendre polynomials (see,

e.g., [RKa, p.409]), one gets: a` = 2π∫ 1

−1A(t)P`(t) dt.

Formula (3.6.70) for q = q(r) can be written as a = lim`→∞(2`+1e

|a`| 12` ).

Indeed, sup α∈S2

−`≤m≤`|Y`m| = O (`

12 ), as is well known (see, e.g., [MPr], p.261). Thus lim`→∞

(sup α∈S2

−`≤m≤`|Y`m(α)|) 1

` = 1, and formula for (3.6.70) yields:

a =2

elim`→∞

(`|a`|

12`

). (3.6.71)

Note that assumption (A) implies the following assumption:Assumption (A′): 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 newproperties 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 cannothave infinitely many purely imaginary resonances. Therefore it is of interest to find out if assumptionA′ 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 lim replaced by lim.

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

lim`→∞

(2`+ 1

e|δ`|

12`

)= a. (3.6.72)

Below, we prove an auxiliary result:


Lemma 3.6.8. If q = q(r) ∈ L2(0,∞), q(r) is real-valued and does not change sign in some interval(a1, a] where a1 < a, and a is the radius of q, then

a = limm→∞

∣∣∣∣∫ ∞

0

q(r)rm dr

∣∣∣∣1m

,m = 1, 2, . . . . (3.6.73)

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 conse-quence of the Paley-Wiener theorem. Namely, we prove Lemma 3.6.8 with a continuous parameter treplacing the integer m and lim replacing lim. This is done for q(r) ∈ L2(0, a) and without additionalassumptions about q. However, we are not able to prove Lemma 3.6.8 assuming only that q(r) ∈ L2(0, a).

Since q(r) is compactly supported, one can write

I(t) :=

∫ ∞

0

q(r)rt dr =

∫ a

0

q(r)et ln rdr =

∫ ln a

−∞q(eu)euetudu. (3.6.74)

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

If f(z) =∫ b2b1g(u)e−iuzdu, [b1, b2] is the smallest interval containing the support of g(u), and g(u) ∈

L2(b1, b2), then

b2 = limt→+∞(t−1 ln |f(it)|

)= limt→+∞

ln |∫ b2b1g(u)etudu|t

. (3.6.75)

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

ln a = limt→+∞

(t−1 ln

∣∣∣∣∣

∫ ln a

−∞q(eu)euetudu

∣∣∣∣∣

). (3.6.76)

Formula (3.6.76) is similar to (3.6.73) with m replaced by t and lim replaced by lim.

Remark 3.6.9. We have used formula (3.6.75) with b1 = −∞, while in the Paley-Wiener theorem it isassumed that b1 > −∞. However, for b1 < b2, g 6≡ 0 on [b2 − ε, b2] for any ε > 0, one has:

∫ b2

−∞g(u)etudu =

∫ b1

−∞g(u)etudu+

∫ b2

b1

g(u)etudu := h1(t) + h2(t).

Thus, limt→∞h1(t)h2(t)

= 0, and

limt→∞ln |h1(t) + h2(t)|

t= limt→∞

ln |h2(t)|t

+ limt→∞

ln |1 + o(1)|t

= limt→∞ln |h2(t)|

t= lna.

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 rq(r)[1+ε(r, `)],where ε = O( 1

` ). Let us prove (3.6.73).

Assume without loss of generality that q ≥ 0 near a. Let I :=∫ a0q(r)rmdr =

∫ a1

0q(r)rmdr +∫ a

a1q(r)rmdr := I1 + I2. We have |I1| < cam1 , c1(a − η)m < I2 < c2a

m, where η is an arbitrary small

positive number. Thus, I > 0 for all sufficiently large m, and I1/m = I1/m2 (1 + I1

I2)1/m. One has

a − η ≤ I1/m2 ≤ a and I1

I2→ 0 as m → ∞. Since η is arbitrary small, it follows that limm→∞ I1/m = a.

This completes the proof of (3.6.73). Lemma 3.6.8 is proved. 2


Proof of formula (3.6.72). From (3.5.19) and (3.5.23) denoting a` := eiδ` sin δ`, one gets A(α′ · α) =∑∞`=0 a`Y`(α)Y`(α

′) := 4π∑∞

`=0 a`Y`(α)Y`(α′), where, a` := a`

4π, k = 1, and a` = e2iδ`−1

2i= eiδ` sin δ`,

a` = −∫ ∞

0

dru`(r)q(r)ψ`(r), (3.6.77)

where u`(r) = rj`(r) ∼ sin(r − `π2 ) as r → ∞, j`(r) are the spherical Bessel functions, j`(r) :=√

π2rJ`+ 1

2(r), and ψ`(r) solves (3.5.15) - (3.5.17), and the integral

ψ`(r) = u`(r) +

∫ ∞

0

g`(r, s)q(s)ψ`(s)ds, k = 1, (3.6.78)

whereg`(r, s) = −u`(r)w`(s), r < s; g`(r, s) = g`(s, r), (3.6.79)

w`(s) := i

√πs

2H

(1)

`+ 12

(s), u`(r) =

√πr

2J`+ 1

2(r), (3.6.80)

and H(1)` is the Hankel function.

It is known [RKa, p.407] that

Jν(r) ∼( er

2ν

)ν 1√2πν

, H(1)ν (r) ∼ −i

√2

πν

( er2ν

)−ν, Jν(r)H

(1)ν (r) ∼ − i

πν, ν −→ +∞, (3.6.81)

and [AlR, Appendix 4]:∣∣∣Jν(r)H(1)

ν (r)∣∣∣ <

(ν2 − 1

16

)− 14

, ν >1

4. (3.6.82)

It follows from (3.6.81) that u`(r) does not have zeros on any fixed interval (0, a] if ` is sufficiently large.

Define v`(r) := ψ`(r)u`(r)

. Then (3.6.78) yields

v`(r) = 1 +

∫ a

0

g`(r, s)u`(s)

u`(r)q(s)v`(s)ds. (3.6.83)


g`(r, s) ∼r

2` + 1

(rs

)`, r < s, ` −→ +∞, (3.6.84)

u`(s)

u`(r)∼(sr

)`+1

, ` −→ +∞. (3.6.85)

Thus

g`(r, s)u`(s)

u`(r)∼ s

2` + 1. (3.6.86)

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

ψ`(r) = u`(r)

[1 +O

(1

`

)]as ` −→ +∞, 0 ≤ r ≤ a, (3.6.87)

uniformly with respect to r ∈ [0, a].In the book [N, formula (12.180)], which gives the asymptotic behavior of S` for large `, is misleading:

the remainder in this formula is of order which is much greater, in general, than the order of the mainterm in this formula. That is why we had to find a different approach, which yielded formula (3.6.87).

3.7. INVERSE SCATTERING WITH “INCOMPLETE DATA” 137

From (3.6.77), (3.6.80), (3.6.81), and (3.6.87) one has:

a` = −∫ ∞

0

dr q(r)u2` (r)

[1 +O

(1

`

)]= −

∫ a

0

dr q(r)r2r2`[1 + O

(1

`

)]1

4`+ 2

(e

2`+ 1

)2`+1

.

(3.6.88)Therefore, using (3.6.73), one gets:

lim`→∞

(2`+ 1

e|a`|

12`

)= lim

`→∞

∣∣∣∣∫ a

0

dr q(r)r2r2`∣∣∣∣

12`

= a. (3.6.89)


Remark 3.6.10. Since δ` → 0 as ` → +∞, and sin δ` ∼ δ`, eiδ` ∼ 1, as δ` → 0, formulas (3.6.89)

and a` = eiδ` sin δ` imply lim`→∞(

2`+1e

|δ`| 12`

)= a, where δ` is the phase shift at a fixed positive energy.

This is formula (3.6.72).

3.7 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) = 0 (or someother selfadjoint homogeneous separated boundary conditions), and q = q, q ∈ L1[0, 1]. Fix 0 < b ≤ 1.Assume q(x) on [b, 1] is known and a subset λm(n)∀n=1,2,3,... of the eigenvalues λn = k2

n of the operator` corresponding to the chosen boundary conditions is known. Here

m(n)

n=

1

σ(1 + εn), σ = const > 0, |εn| < 1, εn −→ 0. (3.7.1)

We assume sometimes that ∞∑

n=1

|εn| < ∞. (3.7.2)

Theorem 3.7.1. If (3.7.1) holds and σ > 2b, then the data q(x), b ≤ x ≤ 1; λm(n)∀n determineq(x) on [0, b] uniquely. If (3.7.1) and (3.7.2) hold, the same conclusion holds also if σ = 2b.

The number σ is “the percentage” of the spectrum of ` which is sufficient to determine q on [0, b] ifσ ≥ 2b and (3.7.2) holds. For example, if σ = 1 and b = 1

2, then “one spectrum” determines q on the

half-interval [0, 12]. If b = 1

4, σ = 1

2, then “half of the spectrum” determines q on [0, 1

4]. Of course, q is

assumed known on [b, 1]. If b = 1, σ = 2, then “two spectra” determines q on the whole interval. By“two spectra” one means the set λn∪ µn, where µn is the set of eigenvalues of ` corresponding tothe same boundary condition u(0) = 0 at one end, say at x = 0, and some other selfadjoint boundarycondition at the other end, say u′(1) = 0 or u′(1) + hu(1) = 0, h = const > 0. The last result is awell-known theorem of Borg ([B]), which was strengthened in [M], where it is proved that not only thepotential but the boundary conditions as well are uniquely determined by two spectra. A version of “onespectrum” 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 forsolving this inverse problem is given in [RuS].

Proof of Theorem 3.7.1. First, assume σ > 2b. If there are q1 and q2 which produce the same data, thenas above, one gets

G(λ) := g(k) :=

∫ b

0

p(x)ϕ1(x, k)ϕ2(x, k) dx = (ϕ1w′ − ϕ′

1w)∣∣∣b

0= (ϕ1w

′ − ϕ′1w)

∣∣∣x=b

, (3.7.3)


where w := ϕ1 − ϕ2, p := q1 − q2, k =√λ. Thus

g(k) = 0 at k = ±√λm(n) := ±kn. (3.7.4)

The function G(λ) is an entire function of λ of order 12 (see (3.1.11) with k =

√λ), and is an entire

even function of k of exponential type ≤ 2b. One has

|g(k)| ≤ ce2b|Imk|

1 + |k|2 . (3.7.5)

The indicator of g is defined by the formula

h(θ) := hg(θ) := limr→∞

ln |g(reiθ)|r

, (3.7.6)

where k = reiθ. Since |Imk| = r| sin θ|, one gets from (3.7.5) and (3.7.6) the following estimate

h(θ) ≤ 2b| sin θ|. (3.7.7)

It is known [Lev, formula (4.16)] that for any entire function g(k) 6≡ 0 of exponential type one has:

limr−→∞

n(r)

r≤ 1

2π

∫ 2π

0

hg(θ) dθ, (3.7.8)

where n(r) is the number of zeros of g(k) in the disk |k| ≤ r. From (3.7.7) one gets

1

2π

∫ 2π

0

hg(θ) dθ ≤2b

2π

∫ 2π

0

| sin θ| dθ =4b

π(3.7.9)

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

λn = (πn)2 + c+ o(1), n→ ∞, c = const, (3.7.10)

one gets for the number of zeros the estimate

n(r) ≥ 2∑

nπσ [1+0( 1

n2 )]<r

1 = 2σr

π[1 + o(1)], r → ∞. (3.7.11)

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

σ ≤ 2b. (3.7.12)

Therefore, if σ > 2b, then g(k) ≡ 0. If g(k) ≡ 0 then, by property Cϕ, p(x) = 0. Theorem 3.7.1 isproved in the case σ > 2b.

Assume now that σ = 2b and ∞∑

n=1

|εn| < ∞. (3.7.13)

We claim that if an entire function G(λ) in (3.7.3) of order 12

vanishes at the points

λn =n2π2

σ2(1 + εn), (3.7.14)

and (3.7.13) holds, then G(λ) ≡ 0. If this is proved, then Theorem 3.7.1 is proved as above.

3.7. INVERSE SCATTERING WITH “INCOMPLETE DATA” 139

Let us prove the claim. Define

Φ(λ) :=

∞∏

n=1

(1 − λ

λn

)(3.7.15)

and recall that

Φ0(λ) :=sin(σ

√λ)

σ√λ

=

∞∏

n=1

(1 − λ

µn

), µn :=

n2π2

σ2. (3.7.16)

Since G(λn) = 0, the function

w(λ) :=G(λ)

Φ(λ)(3.7.17)

is entire, of order ≤ 12. Let us use a Phragmen-Lindelof lemma.

Lemma 3.7.2. [Lev, Theorem 1.22] If an entire function w(λ) of order < 1 has the property sup−∞<y<∞|w(iy)| ≤ c, then w(λ) ≡ c. If, in addition w(iy) → 0 as y → +∞, then w(λ) ≡ 0.

We use this lemma to prove that w(λ) ≡ 0. If this is proved then G(λ) ≡ 0 and Theorem 3.7.1 isproved.

The function w(λ) is entire of order 12 < 1.

Let us check that

sup−∞<y<∞

|w(iy)| < ∞, (3.7.18)

and that

|w(iy)| → 0 as y → +∞. (3.7.19)

One has, using (3.7.5), (3.7.15), (3.7.16) and taking into account that σ = 2b:

∣∣w(iy)∣∣ =

∣∣∣∣G(iy)

Φ(iy)

Φ0(iy)

Φ0(iy)

∣∣∣∣ ≤e2b|Im

√iy|

(1 + |y|)

(eσ|Im

√iy|

1 + |y| 12

)−1

∞∏

h=1

1 + y2

µ2n

1 + y2

λ2n

12

≤ c

1 + |y| 12

∏

n:µn≤λn

λ2n

µ2n

12

≤ c

1 + |y| 12∏

n:µn≤λn(1 + |εn|) ≤

c1

1 + |y| 12.

(3.7.20)

Here we have used elementary inequalities:

1 + a

1 + d≤ a

dif a ≥ d > 0;

1 + a

1 + d≤ 1 if 0 ≤ a ≤ d, (3.7.21)

with a := y2

µ2n, d := y2

λ2n, and the assumption (3.7.13).

We also used the relation:

∣∣∣∣sin(σ

√iy)

σ√iy

∣∣∣∣ ∼eσ|Im

√iy|

2σ|√iy| as y → +∞.

Estimate (3.7.20) implies (3.7.18) and (3.7.19). An estimate similar to (3.7.20) has been used in theliterature (see [GS]).



3.7.2 Uniqueness results: compactly supported potentials

Consider the inverse scattering problem of Section 3.5.2 and assume

q = 0 for x ≥ a > 0. (3.7.22)

Theorem 3.7.3. If q ∈ L1,1 satisfies (3.7.22), then any one of the data S(k), δ(k), f(k), f ′(k), deter-mine q uniquely.

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

|q(x)| ≤ c1e−c2|x|γ , γ > 1, c1, c2 > 0, (3.7.23)

then f(k), the Jost function (3.1.5), is an entire function of k, and S(k) is a meromorphic function on Cwith the only poles in C+ at the points ikj, 1 ≤ j ≤ J . Thus, kj and J are determined by S(k). Using(3.1.15) and (3.1.11), which holds for all k ∈ C, because f(k) and f(x, k) are entire functions of k, onefinds sj = iResk=ikj S(k). 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 δ(k) is given, then S(k) = e2iδ(k), so δ(k) ⇒ q. If f(k) is given, then S(k) = f(−k)f(k) , so f(k) ⇒ q. If

f ′(0, k) is given, then one can uniquely find f(k) from (3.1.11). Indeed, assume there are two f(k), f1

and f2, corresponding to the given f ′(0, k). Subtract from (3.1.11) with f = f1 equation (3.1.11) with

f = f2, denote f1 − f2 := w, and get (∗) f ′(0, k)w(−k) = f ′(0,−k)w(k) or w(k)f ′(0,k)

= w(−k)f ′(0,−k). Since

w(∞) = 0, and f ′(0, k) = ik − A(0, 0) +∫∞0 Ax(0, y)e

ikydy, one can conclude that w = 0 if one can

check that w(k)f ′(0,k)

is analytic in C+. The function f ′(0, k) has at most finitely many zeros in C+, and

these zeros are simple. From (∗) one concludes that if f ′(0, κ) = 0, κ ∈ C+, then w(κ) = 0, because

if f ′(0, κ) = 0 then f ′(0,−κ) 6= 0 (see (3.1.11)). Thus w(k)f ′(0,k) is analytic in C+. Similarly w(−k)

f ′(0,−k) is

analytic in C+. These two functions agree on the real axis, so, by analytic continuation, the functionw(k)f ′(0,k) is analytic in C+ and vanishes at infinity. Thus it vanishes identically. So w(k) = 0, f1 = f2, and

f(k) is uniquely determined by f ′(0, k). Thus Theorem 3.7.3 is proved. 2

In [R181], [R184] a convergent iterative method is given for finding a compactly supported (ordecaying 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:

ù− k2u = 0, x ∈ R, (3.7.24)

u = eikx + r(k)e−ikx + o(1), x −→ −∞, (3.7.25)

u = t(k)eikx + o(1), x −→ +∞, (3.7.26)

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

r(k), kj, sj , 1 ≤ j ≤ J, (3.7.27)

where sj > 0 are norming constants, kj > 0, and −k2j are the negative eigenvalues of the operator ò.

It is known [M], that the data (3.7.27) determine q ∈ L1,1(R) := q : q = q,∫∞−∞(1 + |x|)|q|dx < ∞

uniquely. Assume thatq(x) = 0, x < 0. (3.7.28)

3.8. RECOVERY OF QUARKONIUM SYSTEMS 141

Theorem 3.7.4. If q ∈ L1,1(R) and (3.7.28) holds, then r(k)∀k>0 determines q uniquely.

Proof. If (3.7.28) holds, then u = eikx+r(k)e−ikx 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

ik(1 − r(k))

1 + r(k)=u′(−0, k)

u(−0, k)=u′(+0, k)

u(+0, k)=f ′(0, k)

f(k):= I(k). (3.7.29)

Therefore r(k) determines I(k), so by Theorem 3.3.2 q is uniquely determined. 2

3.8 Recovery of quarkonium systems

3.8.1 Statement of the inverse problem

The problem discussed in this Section is: to what extent does the spectrum of a quarkonium systemtogether with other experimental data determines the interquark potential? This problem was discussedin [TQR], where one can find further references. The method given in [TQR] for solving the aboveproblem is this: one has few scattering data Ej , sj , which will be defined precisely later, one constructs,using the known results of inverse scattering theory, a Bargmann potential (i.e., a potential which hasa rational Jost function) with the same scattering data and considers this a solution to the problem.This approach is wrong because the scattering theory is applicable to the potentials which tend to zeroat infinity, while our confining potentials grow to infinity at infinity and no Bargmann potential canapproximate a confining potential on the whole semiaxis (0,∞). Our aim is to give an algorithm whichis consistent and yields a solution to the above problem. The algorithm is based on the Gel’fand-Levitanprocedure of Section 3.4.3.

Let us formulate the problem precisely. Consider the Schrodinger equation

−∇2ψj + q(r)ψj = Ejψj in R3, (3.8.1)

where q(r) is a real-valued spherically symmetric potential, r := |x|, x ∈ R3,

q(r) = r + p(r), p(r) = o(1) as r −→ ∞. (3.8.2)

The functions ψj(x), ‖ψj‖L2(R3) = 1, are the bound states, Ej are the energies of these states. We defineuj(r) := rψj(r), which correspond to s-waves, and consider the resulting equation for uj:

ùj := −u′′j + q(r)uj = Ejuj, r > 0, uj(0) = 0, ‖uj‖L2(0,∞) = 1. (3.8.3)

One can measure the energies Ej of the bound states and the quantities sj = u′j(0) experimentally.Therefore the following inverse problem (IP) is of interest:(IP): given:

Ej, sj∀j=1,2,... (3.8.4)

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 atinfinity, on the other hand, [TQR] is concerned with potentials which grow to infinity as r → +∞. Itis nevertheless of some interest that numerical results in [TQR] seem to give some approximation of thepotentials 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 ofq(x) from the spectral function. We show that the data (3.8.4) allow one to write the spectral functionof the selfadjoint in L2(0,∞) operator `, defined by the differential expression (3.8.3) and the boundarycondition (3.8.3) at zero.

In Section 3.8.2 proofs are given and the recovery procedure is described.Since in experiments one has only finitely many data Ej, sj1≤j≤J , 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 Ej, sjj>J for this potential are the same as for the unperturbedpotential q0(r) = r. 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 q(r) → +∞. The formulafor the number of eigenvalues (energies of the bound states), not exceeding λ, is known:

∑

Ej<λ

1 := N (λ) ∼ 1

π

∫

q(r)<λ

[λ− q(r)]12 dr.

This formula yields, under the assumption q(r) ∼ r as r → ∞, the following asymptotics of the eigen-values:

Ej ∼ (3π

2j)

23 as j → +∞.

The spectral function ρ(λ) of the operator L is defined by the formula

ρ(λ) =∑

Ej<λ

1

αj, (3.8.5)

where αj are the normalizing constants:

αj :=

∫ ∞

0

φ2j(r)dr. (3.8.6)

Here φj(r) := φ(r, Ej) and φ(r, E) is the unique solution of the problem:

Lφ := −φ′′ + q(r)φ = Eφ, r > 0, φ(0, E) = 0, φ′(0, E) = 1. (3.8.7)

If E = Ej, then φj = φ(r, Ej) ∈ L2(0,∞). The function φ(r, E) is the unique solution to the Volterraintegral equation:

φ(r, E) =sin(

√Er)√E

+

∫ r

0

sin[√E(r − y)]√E

q(y)φ(y,E)dy. (3.8.8)

For any fixed r the function φ is an entire function of E of order 12 , that is, |φ| < c exp(c|E|1/2), where

c denotes various positive constants. At E = Ej, where Ej are the eigenvalues of (3.8.3), one hasφ(r, Ej) := φj ∈ L2(0,∞). In fact, if q(r) ∼ cra, a > 0, then |φj| < c exp(−γr) for some γ > 0.

Let us relate αj and sj . From (3.8.7) with E = Ej and from (3.8.3), it follows that

φj =ujsj. (3.8.9)

3.8. RECOVERY OF QUARKONIUM SYSTEMS 143

Therefore

αj := ‖φj‖2L2(0,∞) =

1

s2j. (3.8.10)

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

ρ(λ) :=∑

Ej<λ

s2j . (3.8.11)

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

σ(λ) := ρ(λ) − ρ0(λ), (3.8.12)

where ρ0(λ) is the spectral function of the unperturbed problem, which in our case is the problem withq(r) = r, then set

L(x, y) :=

∫ ∞

−∞φ0(x, λ)φ0(y, λ)dσ(λ), (3.8.13)

where φ0(x, λ) are the eigenfunctions of the problem (3.8.7) with q(r) = r, and solve the second kindFredholm integral equation for the kernel K(x, y):

K(x, y) +

∫ x

0

K(x, t)L(t, y)dt = −L(x, y), 0 ≤ y ≤ x. (3.8.14)

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

p(r) = 2dK(r, r)

dr, r > 0. (3.8.15)

3.8.3 Reconstruction method

Let us describe the algorithm we propose for recovery of the function q(x) from few experimental dataEj, sj1≤j≤J . Denote by E0

j , s0j1≤j≤J the data corresponding to q0 := r. These data are known

and the corresponding eigenfunctions (3.8.3) can be expressed in terms of Airy function Ai(r), whichsolves the equation w′′ − rw = 0 and decays at +∞, see [Leb]. The spectral function of the operator L0

corresponding to q = q0 := r is

ρ0(λ) :=∑

E0j<λ

(s0j )2. (3.8.16)

Define

ρ(λ) := ρ0(λ) + σ(λ), (3.8.17)

σ(λ) :=∑

Ej<λ

s2j −∑

E0j<λ

(s0j )2, (3.8.18)

and

L(x, y) :=

J∑

j=1

s2jφ(x,Ej)φ(y,Ej) −J∑

j=1

(s0j )2φj(x)φj(y), (3.8.19)

where φ(x,E) can be obtained by solving the Volterra equation (3.8.9) with q(r) = q0(r) := r andrepresented in the form:

φ(x,E) =sin(E1/2x)

E1/2+

∫ x

0

K(x, y)sin(E1/2y)

E1/2dy, (3.8.20)


where K(x, y) is the transformation kernel corresponding to the potential q(r) = q0(r) := r, and φj arethe eigenfunctions of the unperturbed problem:

−φ′′j + rφj = Ejφj r > 0, φj(0) = 0, φ′

j(0) = 1. (3.8.21)

Note that for E 6= E0j the functions (3.8.20) do not belong to L2(0,∞), but φ(0, E) = 0. We denoted

in this section the eigenfunctions of the unperturbed problem by φj rather than φ0j for simplicity ofnotations, since the eigenfunctions of the perturbed problem are not used in this section. One has:φj(r) = cjAi(r − E0

j ), where cj = [Ai′(−E0j )]

−1, E0j > 0 is the j−th positive root if the equation

Ai(−E) = 0 and, by formula (3.8.10), one has s0j = [c2j∫∞0Ai2(r − E0

j )dr]−1/2. These formulas make

the calculation of φj(x), E0j and s0j easy since the tables of Airy functions are available [Leb].

The equation analogous to (3.8.14) is:

K(x, y) +

2J∑

j=1

cjΨj(y)

∫ x

0

K(x, t)Ψj(t)dt = −2J∑

j=1

cjΨj(x)Ψj(y), (3.8.22)

where Ψj(t) := φ(t, Ej), cj = s2j , 1 ≤ j ≤ J , and Ψj(t) = φj−J(t), cj = (s0j−J )2, J+1 ≤ j ≤ 2J . 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

p(r) = 2d

drK(r, r), q(r) = r + p(r). (3.8.23)

Equation (3.8.14) and, in particular (3.8.22), is uniquely solvable by the Fredholm alternative: thehomogeneous version of (3.8.14) has only the trivial solution. Indeed, if h +

∫ x0L(t, y)h(t)dt = 0, 0 ≤

y ≤ x, then ‖h‖2 +∫∞−∞ |h|2[dρ(λ) − dρ0(λ)] = 0, so that, by Parseval equality,

∫∞−∞ |h|2dρ(λ) = 0.

Here h :=∫ x0h(t)φ(t, λ)dt, where φ(t, λ) are defined by (3.8.20). This implies that h(Ej) = 0 for all

j = 1, 2, . . .. Since h(λ) is an entire function of exponential type ≤ x, and since the density of the

sequence Ej is infinite, i.e., limλ→∞N(λ)λ = ∞, because Ej = O(j2/3), as was shown in the beginning of

Section 3.8.2, it follows that h = 0 and consequently h(t) = 0, as claimed.In conclusion consider the case when Ej = E0

j , sj = s0j for all j ≥ 1, and E0, s0 is the new

eigenvalue, E0 < E01 , with the corresponding data s0. In this case L(t, y) = s20φ0(t, E0)φ0(y,E0), so that

equation (3.8.14) takes the form

K(x, y) + s20φ0(y)

∫ x

0

K(x, t)φ0(t, E0)dt = −s20φ0(x,E0)φ0(y,E0).

Thus, one gets:

p(r) = −2d

dr

s20φ20(x,E0)

1 + s20∫ x0 φ

20(t, E0)dt

.

3.9 Krein’s method in inverse scattering

3.9.1 Introduction and description of the method

Consider inverse scattering problem studied in Section 3.5.1 and for simplicity assume that there are nobound states. This assumption is removed in Section 3.9.4.

This Section is a commentary to Krein’s paper [K1]. It contains not only a detailed proof of theresults announced in [K1] but also a proof of the new results not mentioned in [K1]. In particular, itcontains 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 po-tential generates the scattering data from which it was reconstructed. A numerical scheme for solving

3.9. KREIN’S METHOD IN INVERSE SCATTERING 145

inverse scattering problem, based on Krein’s inversion method, is proposed, and its advantages com-pared with the Marchenko and Gel’fand-Levitan methods are discussed. Some of the results are statedin Theorem 3.9.2 – Theorem 3.9.5 below.

Consider the equation for a function Γx(t, s):

(I +Hx)Γx := Γx(t, s) +

∫ x

0

H(t− u)Γx(u, s)du = H(t− s), 0 ≤ t, s ≤ x. (3.9.1)

Equation (3.9.1) shows that Γx = (I +Hx)−1H = I − (I +Hx)

−1, so

(I +Hx)−1 = I − Γx (3.9.2)

in operator form, andHx = (I − Γx)

−1 − I. (3.9.3)

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

H(−t) = H(t), H(t) ∈ L1(R) ∩ L2(R),

1 + H(k) > 0, H(k) :=

∫ ∞

−∞H(t)eiktdt = 2

∫ ∞

0

cos(kt)H(t)dt. (3.9.4)

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

Γ(t, s) = limx→∞

Γx(t, s) := Γ∞(t, s), t, s ≥ 0, (3.9.5)

where Γ(t, s) solves the equation

Γ(t, s) +

∫ ∞

0

H(t− u)Γ(u, s)du = H(t − s), 0 ≤ t, s < ∞. (3.9.6)

Given H(t), one solves (3.9.1), finds Γ2x(s, 0), then defines

ψ(x, k) :=E(x, k) − E(x,−k)

2i, (3.9.7)

where

E(x, k) := eikx[1 −

∫ 2x

0

Γ2x(s, 0)e−iksds

]. (3.9.8)

Formula (3.9.8) gives a one-to-one correspondence between E(x, k) and Γ2x(s, 0).

Remark 3.9.1. In [K1] Γ2x(0, s) is used in place of Γ2x(s, 0) in the definition of E(x, k). By formula(3.9.46) (see Section 3.9.2 below) one has Γx(0, x) = Γx(x, 0), but Γx(0, s) 6= Γx(s, 0) in general. Thetheory presented below cannot be constructed with Γ2x(0, s) in place of Γ2x(s, 0) in formula (3.9.8).

Note thatE(x, k) = eikxf(−k) + o(1), x −→ +∞, (3.9.9)

where

f(k) := 1 −∫ ∞

0

Γ(s)eiksds, (3.9.10)

andΓ(s) := lim

x→+∞Γx(s, 0) := Γ∞(s, 0). (3.9.11)

Furthermore,

ψ(x, k) =eikxf(−k) − e−ikxf(k)

2i+ o(1), x −→ +∞. (3.9.12)


Note that ψ(x, k) = |f(k)| sin(kx + δ(k)) + o(1), x → +∞, where f(k) = |f(k)|e−iδ(k), δ(k) =−δ(−k), k ∈ R.

The function δ(k) is called the phase shift. One has S(k) = e2iδ(k).We have changed the notations from [K1] in order to show the physical meaning of the function

(3.9.9): f(k) is the Jost function of the scattering theory. The functionψ(x,k)f(k) is the solution to the

scattering problem: it solves equation (3.1.3), and satisfies the correct boundary conditions: ψ(0,k)f(k)

= 0,

and ψ(x,k)f(k) = eiδ(k) sin(kx+ δ(k)) + o(1) as x→ ∞.

Krein [K1] calls S(k) :=f(−k)f(k) the S-function, and S(k) is the S-matrix used in physics.

Assuming no bound states, one can solve the inverse scattering problem (ISP): Given S(k) ∀k > 0,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

1 + H = 1 +

∫ ∞

−∞H(t)eiktdt =

1

|f(k)|2 . (3.9.13)

(3) Given H(t), solve (3.9.1) for Γx(t, s) and then find Γ2x(2x, 0), 0 ≤ x < ∞.

(4) Definea(x) = 2Γ2x(2x, 0), (3.9.14)

wherea(0) = 2H(0), (3.9.15)

and calculate the potential

q(x) = a2(x) + a′(x), a(0) = 2H(0). (3.9.16)

One can also calculate q(x) by the formula:

q(x) = 2d

dx[Γ2x(2x, 0)− Γ2x(0, 0)]. (3.9.17)

Indeed, 2Γ2x(2x, 0) = a(x), see (3.9.14), ddxΓ2x(0, 0) = −2Γ2x(2x, 0)Γ2x(0, 2x), see (3.9.47), and Γ2x(2x, 0)

= Γ2x(0, 2x), see (3.9.46).There is an alternative way, based on the Wiener-Levy theorem, to do step (1). Namely, given S(k),

find δ(k), the phase shift, then calculate the function g(t) := − 2π

∫∞0 δ(k) sin(kt)dk, and finally calculate

f(k) = exp(∫∞0g(t)eiktdk).

The potential q ∈ L1,1 generates the S-matrix S(k), with which we started, provided that thefollowing conditions (3.9.18) – (3.9.21) hold:

S(k) = S(−k) = S−1(k), k ∈ R, (3.9.18)

the overbar stands for complex conjugation, and

indRS(k) = 0, (3.9.19)

‖F (x)‖L∞(R+) + ‖F (x)‖L1(R+) + ‖xF ′(x)‖L1(R+) < ∞, (3.9.20)

where

F (x) :=1

2π

∫ ∞

−∞[1− S(k)]eikxdk. (3.9.21)


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 2π. The function (3.9.7) satisfies equation (3.1.5). Recall that we haveassumed that there are no bound states.

In Section 3.9.2 the above method is justified and the following theorems are proved:

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 ISPand 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 x ≥ 0 and its solutionis unique.

Theorem 3.9.5. If condition (3.9.4) holds, then relation (3.9.11) holds and Γ(s) := Γ∞(s, 0) is theunique solution to the equation

Γ(s) +

∫ ∞

0

H(s− u)Γ(u)du = H(s), s ≥ 0. (3.9.22)

The diagram explaining the inversion method for solving ISP, based on Krein’s results, can be shownnow:

S(k)(3.9.58)⇒s1

f(k)(3.9.13)⇒s2

H(t)(3.9.1)⇒s3

Γx(t, s)(trivial)⇒s4

Γ2x(2x, 0)(3.9.14)⇒s5

a(x)(3.9.16)⇒s6

q(x). (3.9.23)

In this diagram sm denotes step number m. Steps s2, s4, s5 and s6 are trivial. Step s1 is almost trivial:it requires solving a Riemann problem with index zero and can be done analytically, in closed form.Step s3 is the basic (non-trivial) step which requires solving a family of Fredholm-type linear integralequations (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 oneassumes (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 f(k), ∀k > 0,

(b) from |f(k)|2, ∀k > 0, or

(c) from the spectral function dρ(λ).

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 dρ(λ) =

√λπ

dλ|f(

√λ)|2 , λ > 0,

0, λ < 0. Let λ = k2. Then (still assuming (3.9.19)) one

has: dρ = 2k2

π1

|f(k)|2 dk, k > 0.

Note that the general case of the inverse scattering problem on the half-axis, when indRS(k) := ν 6= 0,can be reduced to the case ν = 0 by the procedure, described in Section 3.9.4, provided that S(k) is theS−matrix corresponding to a potential q ∈ L1,1(R+). Necessary and sufficient conditions for this areconditions (3.9.18) – (3.9.20).

Subsection 3.9.3 contains a discussion of the numerical aspects of the inversion procedure basedon 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 Kreininversion method, is a Fredholm convolution-type equation. Solving such an equation numerically leadsto 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 methodsyields 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 operatorin (3.9.1) is less than 1 for sufficiently small x, say 0 < x < x0. Thus q(x) can be calculated for 0 ≤ x ≤ x0

2by diagram (3.9.23).

For x ≥ x0 > 0 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 Fd = 0.

Indeed, for x > 0 the norm of the operator in (3.1.41) is less than 1 and it tends to 0 as x→ +∞.Finally let us discuss the following question: in the justification of both the Gel’fand-Levitan and

Marchenko methods, the eigenfunction expansion theorem and the Parseval relation play a fundamentalrole. In contrast, the Krein method apparently does not use the eigenfunction expansion theorem and theParseval 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, thispotential is in L1,1(R+), then conditions (3.9.18) and (3.9.20) are satisfied as well, and the eigenfunctionexpansion theorem and Parseval’s equality hold. Necessary and sufficient conditions, imposed directlyon the function H(t), which guarantee that conditions (3.9.18)–(3.9.20) hold, are not known. However, itfollows that conditions (3.9.18)–(3.9.20) hold if and only ifH(t) is such that the diagram (3.9.23) leads toa q(x) ∈ L1,1(R+). Alternatively, conditions (3.9.18)–(3.9.20) hold (and consequently, q(x) ∈ L1,1(R+))if and only if condition (3.9.4) holds and the function f(k), which is uniquely defined as the solution tothe Riemann problem

Φ+(k) = [1 + H(k)]−1Φ−(k), k ∈ R, (3.9.24)

by the formula f(k) = Φ+(k), generates the S-matrix S(k) by formula (3.9.15), and this S(k) satisfiesconditions (3.9.18)–(3.9.20). Although the above conditions are verifiable, they are not quite satisfactorybecause 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 aspectsof the inversion method based on Krein’s results are discussed. In Subsection 3.9.4 the ISP with boundstates is discussed. In Subsection 3.9.5 a relation between Krein’s and Gel’fand-Levitan’s methods isexplained.

3.9.2 Proofs

Proof of Theorem 3.9.4. If v ∈ L2(0, x), then

(v +Hxv, v) =1

2π[(v, v)L2(R) + (Hv, v)L2(R)], (3.9.25)

where the Parseval equality was used, v :=∫ x0 v(s)e

iksds,

(v, v) =

∫ x

0

|v|2ds = (v, v)L2(R). (3.9.26)

Thus I+Hx is a positive definite selfadjoint operator in the Hilbert space L2(0, x) if (3.9.4) holds. Note

that, since H(t) ∈ L1(R), one has H(k) → 0 as |k| → ∞, so (3.9.4) implies

1 + H(k) ≥ c > 0. (3.9.27)

A positive definite selfadjoint operator in a Hilbert space is boundedly invertible. Theorem 3.9.4 isproved. 2

Note that our argument shows that

supx≥0

‖(I +Hx)−1‖L2(R) ≤ c−1. (3.9.28)

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

Hϕ :=

∫ ∞

0

H(t − u)ϕ(u)du (3.9.29)

is a bounded operator in Lp(R+), p = 1, 2,∞.For Γx(u, s) ∈ L1(R+) one has

‖∫ ∞

x

duH(t− u)Γx(u, s)‖L2(0,x) ≤ c1

∫ ∞

x

du|Γx(u, s)|. (3.9.30)

Proof. Let ‖ϕ‖p := ‖ϕ‖Lp(R+). One has

‖Hϕ‖1 ≤ supu∈R+

∫ ∞

0

dt|H(t− u)|∫ ∞

0

|ϕ(u)|du ≤∫ ∞

−∞|H(s)|ds‖ϕ‖1 = 2‖H‖1 ‖ϕ‖1, (3.9.31)

where we have used the assumption H(t) = H(−t). Similarly,

‖Hϕ‖∞ ≤ 2‖H‖1 ‖ϕ‖∞. (3.9.32)

Finally, using Parseval’s equality, one gets:

2π‖Hϕ‖22 = ‖Hϕ+‖2

L2(R) ≤ supk∈R

|H(k)|2‖ϕ‖22, (3.9.33)

where

ϕ+(x) :=

ϕ(x), x ≥ 0,

0, x < 0.(3.9.34)

Since |H(k)| ≤ 2‖H‖1 one gets from (3.9.33) the estimate:

‖Hϕ‖2 ≤√

2/π‖H‖1 ‖ϕ‖2. (3.9.35)

To prove (3.9.30), one notes that

∫ x

0

dt

∣∣∣∣∫ ∞

x

duH(t− u)Γx(u, s)

∣∣∣∣2

≤ supu,v≥x

∫ x

0

dt∣∣H(t − u)H(t− v)

∣∣( ∫ ∞

x

∣∣Γx(u, s)∣∣du)2

≤ c1

(∫ ∞

x

du∣∣Γx(u, s)

∣∣)2

.

Estimate (3.9.30) is obtained. Lemma 3.9.6 is proved. 2

Proof of Theorem 3.9.5. Define Γx(t, s) = 0 for t or s greater than x. Let w := Γx(t, s) − Γ(t, s). Then(3.9.1) and (3.9.6) imply

(I +Hx)w =

∫ ∞

x

H(t− u)Γ(u, s)du := hx(t, s). (3.9.36)

If condition (3.9.4) holds, then equations (3.9.6) and (3.9.22) have solutions in L1(R+), and, sincesupt∈R

|H(t)| < ∞, it is clear that this solution belongs to L∞(R+) and consequently to L2(R+), because‖ϕ‖2 ≤ ‖ϕ‖∞‖ϕ‖1. The proof of Theorem 3.9.4 shows that such a solution is unique and does exist.From (3.9.28) one gets

supx≥0

‖(I +Hx)−1‖L2(0,x) ≤ c−1. (3.9.37)


For any fixed s > 0 one sees that supx≥y ‖hx(t, s)‖ → 0 as y → ∞, where the norm here stands for anyof the three norms Lp(0, x), p = 1, 2,∞. Therefore (3.9.36) and (3.9.35) imply

‖w‖2L2(0,x) ≤ c−2‖hx‖2

L2(0,x)

≤ c−2

∥∥∥∥∫ ∞

x

H(t − u)Γ(u, s)du

∥∥∥∥L1(0,x)

∥∥∥∥∫ ∞

x

H(t− u)Γ(u, s)du

∥∥∥∥L∞(0,x)

≤ const ‖Γ(u, s)‖2L1(x,∞) −→ 0 as x −→ ∞,

(3.9.38)

since Γ(u, s) ∈ L1(R+) for any fixed s > 0 and H(t) ∈ L1(R).Also

‖w(t, s)‖2L∞(0,x) ≤ 2(‖hx‖2

L∞(0,x) + ‖Hxw‖2L∞(0,x)) (3.9.39)

≤ c1‖Γ(u, s)‖2L1(x,∞) + c2 sup

t∈R

‖H(t− u)‖2L2(0,x)‖w‖2

L2(0,x), (3.9.40)

where cj > 0 are some constants. Finally, by (3.9.30), one has;

‖w(t, s)‖2L2(0,x) ≤ c3(

∫ ∞

x

|Γ(u, s)|du)2 −→ 0 as x −→ +∞. (3.9.41)

From (3.9.39) and (3.9.41) relation (3.9.11) follows. Theorem 3.9.5 is proved. 2

Let us now prove Theorem 3.9.3. We need several lemmas.

Lemma 3.9.7. The function (3.9.8) satisfies the equations

E′ = ikE − a(x)E−, E(0, k) = 1, E− := E(x,−k), (3.9.42)

E′− = −ikE− − a(x)E, E−(0, k) = 1, (3.9.43)

where E′ = dEdx , and a(x) is defined in (3.9.14).

Proof. Differentiate (3.9.8) and get

E′ = ikE − eikx(

2Γ2x(2x, 0)e−ik2x + 2

∫ 2x

0

∂Γ2x(s, 0)

∂(2x)e−iksds

). (3.9.44)

We will check below that∂ Γx(t, s)

∂x= −Γx(t, x)Γx(x, s), (3.9.45)

andΓx(t, s) = Γx(x− t, x− s). (3.9.46)

Thus, by (3.9.45),∂Γ2x(s, 0)

∂(2x)= −Γ2x(s, 2x)Γ2x(2x, 0). (3.9.47)

Therefore (3.9.44) can be written as

E′ = ikE − e−ikxa(x) + a(x)eikx∫ 2x

0

Γ2x(s, 2x)e−iksds. (3.9.48)

By (3.9.46) one getsΓ2x(s, 2x) = Γ2x(2x− s, 0). (3.9.49)


Thus

eikx∫ 2x

0

Γ2x(s, 2x)e−iksds =

∫ 2x

0

Γ2x(2x− s, 0)eik(x−s)ds

= e−ikx∫ 2x

0

Γ2x(y, 0)eikydy. (3.9.50)

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 Γx(x− t, x− s) := ϕ,

Γx(x− t, x− s) +

∫ x

0

H(x− t − u)Γx(u, x− s)du = H(x− t − x+ s) = H(t− s), (3.9.51)

with equation (3.9.1). Let u = x− y. Then (3.9.51) can be written as

ϕ +

∫ x

0

H(t− y)ϕdy = H(t− s), (3.9.52)

which is equation (3.9.1) for ϕ. Since (3.9.1) has at most one solution, as we have proved above (Theo-rem 3.9.4), formula (3.9.46) is proved.

To prove (3.9.45), differentiate (3.9.1) with respect to x and get:

Γ′x(t, s) +

∫ x

0

H(t− u)Γ′x(u, s)du = −H(t− x)Γx(x, s), Γ′

x :=∂Γx∂x

. (3.9.53)

Set s = x in (3.9.1), multiply (3.9.1) by −Γx(x, s), compare with (3.9.53) and use again the uniquenessof the solution to (3.9.1). This yields (3.9.45).


Lemma 3.9.8. Equation (3.1.5) holds for ψ defined in (3.9.7).

Proof. From (3.9.7) and (3.9.42) – (3.9.43) one gets

ψ′′ =E′′ −E′′

−2i

=(ikE − a(x)E−)′ − (−ikE− − a(x)E)′

2i. (3.9.54)

Using (3.9.42) – (3.9.43) again one gets

ψ′′ = −k2ψ + q(x)ψ, q(x) := a2(x) + a′(x). (3.9.55)


Proof of Theorem 3.9.3. The function ψ defined in (3.9.7) solves equation (3.1.5) and satisfies the con-ditions

ψ(0, k) = 0, ψ′(0, k) = k. (3.9.56)

The first condition is obvious (in [K1] there is a misprint: it is written that ψ(0, k) = 1), and the secondcondition follows from (3.9.7) and Lemma 3.9.7:

ψ′(0, k) =E′(0, k) −E′

−(0, k)

2i=ikE − aE− − (ikE− − aE)

2i

∣∣∣∣x=0

=2ik

2i= k.

Let f(x, k) be the Jost solution. Since f(x, k) and f(x,−k) are linearly independent, one has ψ =c1f(x, k) + c2f(x,−k), where c1, c2 are some constants independent of x but depending on k.

From (3.9.56) one gets c1 = f(−k)2i

, c2 = −f(k)2i

; f(k) := f(0, k). Indeed, the choice of c1 andc2 guarantees that the first condition (3.9.56) is obviously satisfied, while the second follows from theWronskian formula: f ′(0, k)f(−k) − f(k)f ′(0,−k) = 2ik.

Comparing this with (3.9.12) yields the conclusion of Theorem 3.9.3. 2


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.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 1) – 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. S(k) ⇒ f(k).Assume S(k) satisfying (3.9.18) – (3.9.20) is given. Then solve the Riemann problem

f(k) = S(−k)f(−k), k ∈ R. (3.9.57)

Since indRS(k) = 0, one has indRS(−k) = 0. Therefore the problem (3.9.57) of finding an analyticfunction f+(k) in C+ := k : Imk > 0, f(k) := f+(k) in C+, (and an analytic function f−(k) := f(−k)in C− := k : Im k < 0) from equation (3.9.57) can be solved in closed form. Namely, define

f(k) = exp

1

2πi

∫ ∞

−∞

lnS(−y)dyy − k

, Imk > 0. (3.9.58)

Then f(k) solves (3.9.57), f+(k) = f(k), f−(k) = f(−k). Indeed,

ln f+(k) − ln f−(k) = ln S(−k), k ∈ R (3.9.59)

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 y 6= 0, and is bounded on the real axis, so the Cauchyintegral in (3.9.58) is well defined.

To justify the above claims, one uses the known properties of the Jost function

f(k) = 1 +

∫ ∞

0

A(0, y)eikydy := 1 +

∫ ∞

0

A(y)eikydy, (3.9.60)

where estimates (3.1.25) and (3.1.26) hold and A(y) is a real-valued function. Thus

f(k) = 1 − A(0)

ik− 1

ik

∫ ∞

0

A′(t)eiktdt, (3.9.61)

S(−k) =f(k)

f(−k) =1 − A(0)

ik − 1ik A

′(k)

1 + A(0)ik + 1

ik A′(−k)

= 1 + O

(1

k

). (3.9.62)

Therefore

lnS(−k) = O

(1

k

)as |k| −→ ∞, k ∈ R. (3.9.63)

Also

f (k) = i

∫ ∞

0

A(y)yeikydy, f :=∂f

∂k. (3.9.64)

Estimate (3.1.25) implies

∫ ∞

0

y|A(y)|dy ≤ 2

∫ ∞

0

t|q(t)|dt < ∞, A(y) ∈ L2(R+), (3.9.65)

so that f (k) is bounded for all k ∈ R, f(k) − 1 ∈ L2(R), S(−k) is differentiable for k 6= 0, and ln S(−y)is bounded on the real axis, as claimed. Note that

f(−k) = f(k), k ∈ R. (3.9.66)


The converse step f(k) ⇒ S(k) is trivial: S(k) = f(−k)f(k) . If indRS = 0 then f(k) is analytic in C+,

f(k) 6= 0 in C+, f(k) = 1 +O( 1k ) as |k| → ∞, k ∈ C+, and (3.9.66) holds.

Step 2. f(k) ⇒ H(t).This step is done by formula (3.9.13):

H(t) =1

2π

∫ ∞

−∞e−ikt

(1

|f(k)|2 − 1

)dk. (3.9.67)

One has H ∈ L2(R). Indeed, it follows from (3.9.61) that

|f(k)|2 − 1 = −2

k

∫ ∞

0

A′(t) sin(kt)dt+O

(1

|k|2), |k| −→ ∞, k ∈ R. (3.9.68)

The function

w(k) :=1

k

∫ ∞

0

A′(t) sin(kt)dt (3.9.69)

is continuous because A′(t) ∈ L1(R+) by (3.1.26), and w ∈ L2(R) since w = o(

1|k|

)as |k| → ∞, k ∈ R.

Thus, H ∈ L2(R).Also, H ∈ L1(R). Indeed, integrating by parts, one gets from (3.9.67) the relation: 2πH(t) =

it

∫∞−∞ e−ikt[f (k)f(−k)− f (−k)f(k)] dk

|f(k)|4 := itg(t), and g ∈ L2(R), therefore H ∈ L1(R). To check that

g ∈ L2(R), one uses (3.9.60), (3.1.25) – (3.1.26), and (3.9.64) – (3.9.65), to conclude that [f(k)f(−k) −f (−k)f(k)] ∈ L2(R), and, since f(k) 6= 0 on R and f(∞) = 1, it follows that g ∈ L2(R). The inclusion[f (k)f(−k) − f (−k)f(k)] ∈ L2(R) follows from (3.9.60), (3.1.25) – (3.1.26), and (3.9.64) – (3.9.65).

By (9.2.42), the function H ′(t) is the Fourier transform of −ik(1−|f(k)|2)|f(k)|−2, and, by (9.2.43),

k(|f(k)|2 − 1) = −2∫∞0A′(t) sin(kt)dt+O

(1|k|

), as |k| → ∞, k ∈ R. Thus, H ′(t) behaves, essentially,

as A′(t) plus a function, whose Fourier transform is O(

1|k|

). Estimate (3.1.26) shows how A′(t) behaves.

Equation (3.9.1) shows that Γx(t, 0) is as smooth as H(t), and formula (9.1.17) for q(x) shows that q isessentially is as smooth as A′(t).

The converse stepH(t) ⇒ f(k) (3.9.70)

is also done by formula (3.9.13): Fourier inversion gives |f(k)|2 = f(k)f(−k), and factorization of |f(k)|2yields the unique f(k), since f(k) does not vanish in C+ and tends to 1 at infinity.

Step 3. H ⇒ Γx(s, 0) ⇒ Γ2x(2x, 0).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 functionare used: f(k) → 1 as k → ±∞, f(k) 6= 0 for k 6= 0, k ∈ R, f(0) 6= 0 since indRS(k) = 0.

The converse step Γx(s, 0) ⇒ H(t) is done by formula (3.9.3). The converse step

Γ2x(2x, 0) ⇒ Γx(s, 0) (3.9.71)

constitutes the essence of the inversion method.This step is done as follows:

Γ2x(2x, 0)(3.9.14)⇒ a(x)

(3.9.42)−−(3.9.43)⇒ E(x, k)(3.9.8)⇒ Γx(s, 0). (3.9.72)

Given a(x), system (3.9.42) – (3.9.43) is uniquely solvable for E(x, k).Note that the step q(x) ⇒ f(k) can be done by solving the uniquely solvable integral equation (3.1.6):

with q ∈ L1,1(R+), and then calculating f(k) = f(0, k).


Step 4. a(x) := 2Γ2x(2x, 0) ⇒ q(x).This step is done by formula (3.9.16). The converse step

q(x) ⇒ a(x)

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 uniquelysolvable if q ∈ L1,1(R+), then one gets f(k) := f(0, k), and then calculates 2H(0) using formula (3.9.67)with t = 0:

2H(0) =1

π

∫ ∞

−∞

(1

|f(k)|2 − 1

)dk.

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 aunique q(x) ∈ L1,1(R+) 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:

S(k)(3.9.58)⇔ f(k)

(3.9.13)⇔ H(t)(3.9.1)⇔ Γx(s, 0)⇒Γ2x(2x, 0)

(3.9.14)⇔ a(x)(3.9.16)⇔ q(x). (3.9.73)

To close this diagram and therefore establish the basic one-to-one correspondence S(k) ⇔ q(x), oneneeds to prove Γ2x(2x, 0) ⇒ Γx(s, 0). This is done by the scheme (3.9.72).

Note that the step q(x) ⇒ a(x) requires solving Riccati equation (3.9.16) with the boundary conditiona(0) = 2H(0). Existence of the solution to this problem on all of R+ is guaranteed by the assumptions(3.9.18) – (3.9.20). The fact that these assumptions imply q(x) ∈ L1,1(R+) is proved in Section 3.5.5.Theorem 3.9.2 is proved. 2

Uniqueness theorems for the inverse scattering problem are not given in [K1]. They can be found inSection 3.5.5

Remark 3.9.9. From our analysis one gets the following result:

Proposition 3.9.10. If q(x) ∈ L1,1(R+) and has no bounds states and no resonance at zero, thenRiccati equation (3.9.16) with the initial condition (3.9.15) has the solution a(x) defined for all x ∈ R+.

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 > 0and 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 involvesinversion of a Toeplitz matrix, which is a well developed area of numerical analysis. Moreover, such aninversion requires much less computer memory and time than the inversion based on the Gel’fand-Levitanor 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 theMarchenko method the equation to be solved is (3.1.43) where F (x) is defined in (3.1.41) and is known ifS(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 potentialis recovered by the formula: Equation (3.1.41) can be written in operator form: (I + Fx)A = −F . Theoperator Fx is a contraction mapping in the Banach space L1(x,∞) for x > 0. The operator Hx in (3.9.1)is a contraction mapping in L∞(0, x) for 0 < x < x0, where x0 is chosen to that

∫ x0

0|H(t− u)|du < 1.

Therefore it seems reasonable from the numerical point of view to use the following approach:

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).

3.10. INVERSE PROBLEMS FOR THE HEAT AND WAVE EQUATIONS 155

2. Solve by iterations equation (3.9.1) for 0 < x < x0, where x0 is chosen so that the iteration methodfor solving (3.9.6) converges rapidly. Then find q(x) as explained in Step 4.

3. Solve equation (3.1.43) for x > x0 by iterations. Find q(x) for x > x0 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 w(k) =∏Jj=1

k−ikj

k+ikjif indRS(x) = −2J and W (k) =

kk+iγ

w(k) if indRS(k) = −2J − 1, where γ > 0 is arbitrary, and is chosen so that γ 6= kj, 1 ≤ j ≤ J .

Then one defines S1(k) := S(k)w2(k) if indRS = −2J or S1(k) := S(k)W 2(k) if indRS =−2J − 1. Since indRw

2(k) = 2J and indRW2(k) = 2J + 1, one has indRS1(k) = 0. The theory of

Subsection 3.9.2 applies to S1(k) and yields q1(x). From q1(x) one gets q(x) by adding bound states−k2

j and norming constants sj 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 (seeSection 3.4, for example):

Step 1. Given f(k), the Jost function, find

L(x, y) :=2

π

∫ ∞

0

dk k2

(1

|f(k)|2 − 1

)sin kx

k

sin ky

k

=1

π

∫ ∞

0

dk(|f(k)|−2 − 1

)(cos[k(x− y)] − cos[k(x+ y)])

:= M (x− y) −M (x+ y),

where M (x) := 1π

∫∞0dk(|f(k)|−2 − 1

)cos(kx).

Step 2. Solve the integral equation (3.1.34) for K(x, y).

Step 3. Find q(x) = 2dK(x,x)dx . Krein’s function, H(t), see (3.9.13), can be written as follows:

H(t) =1

2π

∫ ∞

−∞

(|f(k)|−2 − 1

)e−iktdk =

1

π

∫ ∞

0

(|f(k)|−2 − 1

)cos(kt)dk. (3.9.74)

Thus, the relation between the two methods is given by the formula:

M (x) = H(x). (3.9.75)

In fact, the GL method deals with the inversion of the spectral foundation dρ of the operator − d2

dx2 +q(x) defined in L2(R+) by the Dirichlet boundary condition at x = 0. However, if indRS(k) = 0(in this case there are no bound states and no resonance at k = 0), then (see (3.1.20)): dρ(λ) =

2k2dkπ|f(k)|2 , λ > 0, λ = k2,

0, λ < 0,so dρ(λ) in this case is uniquely defined by f(k), k ≥ 0.

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

a(t) = 0 for t > T,

∫ T

0

a(t)dt < ∞, a(t) 6≡ 0. (3.10.1)


One can also take a(t) = δ(t) where δ(t) is the delta-function. We prove that the inverse problem offinding q(x) ∈ L1[0, 1], q = q, from the conditions (3.1.55) – (3.1.58) has at most one solution. If (3.1.58)is replaced by the condition

ux(0, t) = b0(t), (3.10.2)

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, q( 1

2 −x) = q( 12 +x), or if q(x) is known on [ 12 , 1].

Let us take the Laplace transform of (3.1.55) – (3.1.58) and put v(x, λ) :=∫∞0u(x, t)e−λtdt, A(λ) :=

v(1, λ), B(λ) := vx(1, λ), B0(λ) := vx(0, λ). Then (3.1.55) – (3.1.58) can be written as

`v + λv := −v′′ + q(x)v + λv = 0, 0 ≤ x ≤ 1, v(0, λ) = 0, v(1, λ) = A(λ) (3.10.3)

v′(1, λ) = B(λ) (3.10.4)

and (3.10.2) takes the formv′(0, λ) = B0(λ). (3.10.5)

Theorem 3.10.1. The data A(λ), B(λ), known on a set of λ ∈ (0,∞), which has a finite positivelimit point, determine q uniquely.

Proof. Since A(λ) and B(λ) are analytic in∏

+ := λ : <λ > 0, one can assume that A(λ) and B(λ)

are known for all λ > 0. If k = iλ12 and ϕ is defined in (3.1.3) then v(x, λ) = c(k)ϕ(x, k), c(k) 6= 0,

A(λ) = c(k)ϕ(1, k), B(λ) = c(k)ϕ′(1, k), so

B(λ)

A(λ)=ϕ′(1, k)

ϕ(1, k). (3.10.6)

Thus the function B(λ)A(λ) is meromorphic in C, its zeros on the axis k ≥ 0 are the eigenvalues of ` =

− d2

dx2 + q(x), corresponding to the boundary conditions u(0) = u′(1) = 0 and its poles on the axis k ≥ 0are the eigenvalues of ` corresponding to u(0) = u(1) = 0. The knowledge of two spectra determines quniquely (Section 3.7.1). 2

An alternative proof of Theorem 3.10.1, based on property Cϕ, is: assume that q1 and q2 generatethe same data, p := q1 − q2, w := v1 − v2, where vj , j = 1, 2, solves (3.10.3) – (3.10.4) with q = qj, andget (∗) `1w = pv2, w(0, λ) = w(1, λ) = w′(1, λ) = 0. Multiply (∗) by ϕ1, `1ϕ1 + λϕ1 = 0, ϕ1(0, λ) = 0,ϕ′(1(0, λ) = 1, and integrate over [0, 1] to get

∫ 1

0

pv2ϕ1dx = 0 ∀λ > 0. (3.10.7)

By property Cϕ it follows from (3.10.7) that p = 0. Theorem 3.10.1 is proved. 2

Theorem 3.10.2. Data (3.10.3), (3.10.5) do not determine q uniquely in general. They do if q(x) isknown on [ 12 , 1], or if q(x+ 1

2) = q( 12 − x).

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 ϕ′(0, k) = 1. One spectrum determines q only on “a halfof the interval”, b = 1

2, see Section 3.7.1. Theorem 3.10.2 is proved. 2

3.10.2 What are the “correct” measurements?

From Theorem 3.10.1 and Theorem 3.10.2 it follows that the measurements ux(1, t)∀t>0 are much moreinformative than ux(0, t)∀t>0 for the problem (3.1.55) – (3.1.57). In this section we state a similarresult for the problem

ut = (a(x)u′)′, 0 ≤ x ≤ 1, t > 0; u(x, 0) = 0, u(0, t) = 0, (3.10.8)

3.11. INVERSE PROBLEM FOR AN INHOMOGENEOUS SCHRODINGER EQUATION 157

u(1, t) = f(t). (3.10.9)

The extra data, that is, measurements, are

a(1)u′(1, t) = g(t), (3.10.10)

which is the flux. Assume:

f 6≡ 0, f ∈ L1(0, 1), a(x) ∈W 2,1(0, 1), a(x) ≥ c > 0, (3.10.11)

W `,p is the Sobolev space. Physically, a(x) is the conductivity, u is the temperature. We also considerin place of (3.10.10) the following data:

a(0)u′(0, t) = h(t). (3.10.12)

Our results are similar to those in Subsection 3.10.1:Data f(t), g(t)∀t>0 determine q(x) uniquely, while data f(t), h(t)∀t>0 do not, in general, deter-

mine a(x) uniquely.Therefore, the measurements g(t)∀t>0 are much more informative than the measurements h(t)∀t>0.

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:

`v − k2v = 0, x ≥ 0, v(x, k) =

∫ ∞

0

eiktu(x, t)dt, (3.10.13)

v(0, k) = 1, v(1, k) = A(k) =

∫ ∞

0

a(t)eiktdt. (3.10.14)

From (3.10.13) one gets v(x, k) = c(k)f(x, k), where f(x, k) is the Jost solution, and from (3.10.14) one

gets v(x, k) =f(x,k)f(k) and A(k) =

f(1,k)f(k) = eik

f(k) , because q = 0 for x > 1. Thus f(k) = eik

A(k) is known. By

Theorem 3.7.3 q is uniquely determined. Theorem 3.10.3 is proved. 2

Remark 3.10.4. The above method allows one to consider other boundary conditions at x = 0, such asu′(0, t) = 0 or u′(0, t) = hu(0, t), h = const > 0, and different data at x = 1, for example, u′(1, t) = b(t).

3.11 Inverse problem for an inhomogeneous Schrodingerequation

In thisSection an inverse problem is studied for an inhomogeneous Schrodinger equation. Most of theearlier studies dealt with inverse problems for homogeneous equations. Let

ù− k2u := −u′′ + q(x)u− k2u = δ(x), x ∈ R1,∂u

∂|x| − iku −→ 0, |x| −→ ∞. (3.11.1)

Assume that q(x) is a real-valued function, q(x) = 0 for |x| > 1, q ∈ L∞[−1, 1]. Suppose that the datau(−1, k), u(1, k), ∀k > 0 are given.

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 ` = − d2

dx2 + q(x) in L2(R) hasno negative eigenvalues (this is the case when q(x) ≥ 0, for example). The operator ` is the closure inL2(R) of the symmetric operator `0 defined on C∞

0 (R1) by the formula `0u = −u′′ + q(x)u. Our resultis:

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

u =

g(k)[f,g] f(x, k), x > 0,f(k)[f,g]

g(x, k), x < 0.(3.11.2)

Here f(x, k) and g(x, k) solve homogeneous version of equation (3.11.1) and have the following asymp-totics:

f(x, k) ∼ eikx, x −→ +∞, g(x, k) ∼ e−ikx, x −→ −∞, (3.11.3)

f(k) := f(0, k), g(k) := g(0, k), (3.11.4)

[f, g] := fg′ − f ′g = −2ika(k), (3.11.5)

where the prime denotes differentiation with respect to x-variable, and a(k) is defined by the equation

f(x, k) = b(k)g(x, k) + a(k)g(x,−k). (3.11.6)

It is known that (see e.g.[M]):

g(x, k) = −b(−k)f(x, k) + a(k)f(x,−k), (3.11.7)

a(−k) = a(k), b(−k) = b(k), |a(k)|2 = 1 + |b(k)|2, k ∈ R, (3.11.8)

a(k) = 1 +O(1

k), k −→ ∞, k ∈ C+; b(k) = O(

1

k), |k| −→ ∞, k ∈ R, (3.11.9)

[f(x, k), g(x,−k)] = 2ikb(k), [f(x, k), g(x, k)] = −2ika(k), (3.11.10)

a(k) in analytic in C+, b(k) in general does not admit analytic continuation from R, but if q(x) iscompactly supported, then a(k) and b(k) are analytic functions of k ∈ C \ 0.

The functions

A1(k) :=g(k)f(1, k)

−2ika(k), A2(k) :=

f(k)g(−1, k)

−2ika(k)(3.11.11)

are the data, they are known for all k > 0. Therefore one can assume the functions

h1(k) :=g(k)

a(k), h2(k) :=

f(k)

a(k)(3.11.12)

to be known for all k > 0 because

f(1, k) = eik, g(−1, k) = eik, (3.11.13)

as follows from the assumption q = 0 if |x| > 1, and from (3.11.3).From (3.11.12), (3.11.7) and (3.11.6) it follows that

a(k)h1(k) = −b(−k)f(k) + a(k)f(−k) = −b(−k)h2(k)a(k) + h2(−k)a(−k)a(k), (3.11.14)

a(k)h2(k) = b(k)a(k)h1(k) + a(k)h1(−k)a(−k). (3.11.15)

From (3.11.14) and (3.11.15) it follows:

−b(−k)h2(k) + h2(−k)a(−k) = h1(k), (3.11.16)

3.11. INVERSE PROBLEM FOR AN INHOMOGENEOUS SCHRODINGER EQUATION 159

b(k)h1(k) + a(−k)h1(−k) = h2(k). (3.11.17)

Eliminating b(−k) from (3.11.16) and (3.11.17), one gets:

a(k)h1(k)h2(k) + a(−k)h1(−k)h2(−k) = h1(k)h1(−k) + h2(−k)h2(k), (3.11.18)

or

a(k) = m(k)a(−k) + n(k), k ∈ R (3.11.19)

where

m(k) := −h1(−k)h2(−k)h1(k)h2(k)

, n(k) :=h1(−k)h2(k)

+h2(−k)h1(k)

. (3.11.20)

Problem (3.11.19) is a Riemann problem for the pair a(k), a(−k), the function a(k) is analytic inC+ := k : k ∈ C, Imk > 0 and a(−k) is analytic in C−. The functions a(k) and a(−k) tend to one ask tends to infinity in C+ and, respectively, in C−, see equation (3.11.9).

The function a(k) has finitely many simple zeros at the points ikj , 1 ≤ j ≤ J , kj > 0, where −k2j

are the negative eigenvalues of the operator ` defined by the differential expression ù = −u′′ + q(x)u inL2(R).

The zeros ikj are the only zeros of a(k) in the upper half-plane k.Define

ind a(k) :=1

2πi

∫ ∞

−∞d ln a(k). (3.11.21)

One has

ind a = J, (3.11.22)

where J is the number of negative eigenvalues of the operator `, and, using (3.11.12), (3.11.22) and(3.11.20), one gets

indm(k) = −2[ind h1(k) + ind h2(k)] = −2[ind g(k) + ind f(k) − 2J ]. (3.11.23)

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 indm(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

r(k) := b(k)a(k)

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 r(k) := b(k)a(k)

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 numbersikj , j = 1, 2, . . . , J . They determine uniquely the numbers kj, 1 ≤ j ≤ J , which are a part of thestandard scattering data r(k), kj, sj, 1 ≤ j ≤ J, where sj are the norming constants. Note that ifa(ikj) = 0 then b(ikj) 6= 0: otherwise equation (3.11.6) would imply f(x, ikj) ≡ 0 in contradiction tothe first relation (3.11.3). If r(k) is meromorphic, then the norming constants can be calculated by the

formula sj = −i b(ikj)a(ikj)

= −iResk=ikj r(k), where 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 determineuniquely the standard scattering data, that is, the reflection coefficient, the bound states −k2

j and thenorming constants sj , 1 ≤ j ≤ J . These data determine the potential uniquely. Theorem 3.11.1 isproved. 2

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 thenumber of zeros of f(k) in C+, we have to prove that f(k) does not vanish in C+. If f(z) = 0, z ∈ C+,


then z = ik, k > 0, and −k2 is an eigenvalue of the operator ` in L2(0,∞) with the boundary conditionu(0) = 0.

From the variational principle one can find the negative eigenvalues of the operator ` in L2(R+)with the Dirichlet condition at x = 0 as consequitive minima of the quadratic functional. The minimaleigenvalue is:

−k2 = inf

∫ ∞

0

[u′2 + q(x)u2

]dx := κ0, u ∈

H 1(R+), ‖u‖L2(R+) = 1, (3.11.24)

whereH 1(R+) is the Sobolev space of H1(R+)-functions satisfying the condition u(0) = 0.

On the other hand, if J = 0, then

0 ≤ inf

∫ ∞

−∞[u′2 + q(x)u2] dx := κ1, u ∈ H1(R), ‖u‖L2(R) = 1. (3.11.25)

Since any element u ofH 1(R+) can be considered as an element of H1(R) 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) thatκ1 ≤ κ0. Therefore, if J = 0, then κ1 ≥ 0 and therefore κ0 ≥ 0. This means that the operator ` onL2(R+) with the Dirichlet condition at x = 0 has no negative eigenvalues. This means that f(k) doesnot have zeros in C+, if J = 0. Thus J = 0 implies ind f(k) = 0.


Remark 3.11.3. The above argument shows that in general

ind f ≤ J and ind g ≤ J, (3.11.26)

so that (3.11.23) implies

indm(k) ≥ 0. (3.11.27)

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 ofnegative 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 somewhatsimilar problem in the paper [GX] is invalid.

Let

∆u+ k2n(z)u = −δ(r)2πr

f(z) in R2 × [0, 1], (3.12.1)

u(x1, 0) = 0, u′(x1, 1) = 0, (3.12.2)

where u = u(x1, z), x1 := (x1, x2), r := |x1|, x3 := z, u′ = ∂u/∂z, δ(r) is the delta-function, n(z)is the refraction coefficient, which is assumed to be a real-valued integrable function, k > 0 is a fixedwavenumber. The solution to (3.12.1)–(3.12.2) is selected by the limiting absorption principle.

It is proved that if f(z) = δ(z − 1), then n(z) is uniquely determined by the data u(x1, 1) known∀x1 ∈ R2.

3.12. AN INVERSE PROBLEM OF OCEAN ACOUSTICS 161

3.12.2 Introduction

In [GX] the following inverse problem is studied:

[∆ + k2n(z)]u = −δ(r)2πr

f(z) in R2 × [0, 1], (3.12.3)

u(x1, 0) = u′(x1, 1) = 0, x1 := (x1, x2), x3 := z, u′ :=∂u

∂z. (3.12.4)

Here k > 0 is a fixed wavenumber, n(z) > 0 is the refraction coefficient, which is assumed in [GX] tobe a continuous real-valued function satisfying the condition 0 ≤ n(z) < 1, the layer R2 × [0, 1] modelsshallow ocean, r := |x1| =

√x2

1 + x22, δ(r) is the delta-function, δ(r)/2πr = δ(x1), f(z) ∈ C2[0, 1] is a

function satisfying the following conditions (see [GX], p. 127):

f(0) = f ′′(0) = f ′(1) = 0, f ′(0) 6= 0, f(1) 6= 0, f(z) > 0 in (0, 1). (3.12.5)

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 u(x) = limε↓0 uε(x), 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. Thefunction uε(x) is the unique solution to problem (3.12.3)–(3.12.4) in which equation (3.12.3) is replacedby the equation with absorption:

[∆ + k2n(z) − iε]uε(x) = −δ(r)2πr

f(z) in R2 × [0, 1], ε > 0.

One defines the differential operator corresponding to differential expression (3.12.3) and the boundaryconditions (3.12.4) in L2(R2× [0, 1]) as a selfadjoint operator (for example, as the Friedrichs extension ofthe symmetric operator with the domain consisting of H2(R2 × [0, 1]) functions vanishing near infinityand satisfying conditions (3.12.4)), and then the function uε(x) is uniquely defined. By Hm we meanthe usual Sobolev space. One can prove that the limit of this function u(x) = limε↓0 uε(x) does existglobally in the weighted space L2(R2 × [0, 1], 1/(1+ r)a), a > 1, and locally in H2(R2 × [0, 1]) outside aneighborhood of the set r = 0, 0 ≤ z ≤ 1, provided λj 6= 0 ∀j, where λj are defined in (3.12.10) below.This limit defines the unique solution to problem (3.12.3)–(3.12.4) satisfying the limiting absorptionprinciple if λj 6= 0 ∀j. If f(z) = δ(z−1), where δ(z−1) is the delta-function, then an analytical formulafor uε(x) can be written:

uε(x) =

∞∑

j=1

ψj(z)fj1

2πK0

(r√λ2j + iε

),

where K0(r) is the modified Bessel function (the Macdonald function), and fj = ψj(1) are defined in(3.12.9) below, and ψj(z) and λ2

j are defined in formula (3.12.10) below. This formula can be checked

directly. It is obtained by separation of variables. The known formula F −1 (λ2 + a2)−1 = (1/2π)K0(ar)was used, and Fu := u is the Fourier transform defined above formula (3.12.6).

From the formula for uε(x), the known asymptotics K0(r) =√π/2r e−r[1 +O(r−1)] for large values

of r, the boundedness of |ψj(z)| as j → ∞ and formula (3.12.11) below, one can see that the limit ofuε(x) as ε → 0 does exist for any r > 0 and z ∈ [0, 1], if and only if λj 6= 0. If λj = 0 for some j = j0, thenthe limiting absorption principle holds if and only if fj0 = 0. If λj 6= 0 ∀j, then the limiting absorptionprinciple holds and the solution to problem (3.12.3)–(3.12.4) is well defined. If λj = 0 for some j = j0,then we define the solution to problem (3.12.3)–(3.12.4) with f(z) = δ(z − 1) by the formula:

u(x) = ψj0(z)ψj0 (1)1

2πlog(1

r

)+

∞∑

j=1, j 6=j0ψj(z)ψj(1)

1

2πK0(rλj), r := |x1|.

This solution is unique in the class of functions of the form u(x) =∑∞j=1 uj(x

1)ψj(z), where ∆1uj −λ2juj = −δ(x1) in R2, ∆1w := wx1x1 + wx2x2 , uj ∈ L2(R2) if λ2

j > 0; if λ2j < 0 then uj satisfies the


radiation condition r1/2(∂uj/∂r − i|λj |uj) → 0 as r → ∞, uniformly in directions x1/r; and if λ2j = 0

then uj = (1/2π) log(1/r) + o(1) as r → ∞.The inverse problem (IP) consists of finding n(z) given g(x1) := u(x1, 1) and assuming that f(z) =

δ(z − 1) in (3.12.3).By the cylindrical symmetry one has g(x1) = g(r).It is claimed in [GX], that the above inverse problem, without the assumption f(z) = δ(z−1), has not

more than one solution, and a method for finding this solution is proposed. The arguments in [GX] areerroneous (see Remark 3.12.3 below, where some of the incorrect statements from [GX], which invalidatethe approach in [GX], are pointed out).

Our aim is to prove that if f(z) = δ(z − 1), then n(z) can be uniquely and constructively determinedfrom the data g(r) known for all r > 0. It is an open problem to find all such f(z) for which the IP hasat most one solution.

Let us outline our approach to IP.Take the Fourier transform of (1.1)–(1.2) with respect to x1 and let

v := v(z, λ) := u :=

∫

R2

u(x1, z)eix1·ζdx1, |ζ| := λ, ζ ∈ R2,

andG(λ) := g.

Then

`v := v′′ − λ2v + q(z)v = −f(z), q(z) := k2n(z), v = v(z, λ), (3.12.6)

v(0, λ) = v′(1, λ) = 0, (3.12.7)

v(1, λ) = G(λ). (3.12.8)

IP: The inverse problem is: given G(λ), for all λ > 0 and a fixed f(z) = δ(z − 1), find q(z).The solution to (3.12.6)–(3.12.7) is:

v(z, λ) =

∞∑

j=1

ψj(z)fjλ2 + λ2

j

, fj := (f, ψj) :=

∫ 1

0

f(z)ψj(z)dz, (3.12.9)

where ψj(z) are the real-valued normalized eigenfunctions of the operator L := −d2/dz2 − q(z):

Lψj = λ2jψj, ψj(0) = ψ′

j(1) = 0, ‖ψj(z)‖ = 1. (3.12.10)

We can choose the eigenfunctions ψj(z) real-valued since the function q(z) = k2n(z) is assumed real-valued. One can check that all the eigenvalues are simple, that is, there is just one eigenfunction ψjcorresponding to the eigenvalue λ2

j (up to a constant factor, which for real-valued normalized eigenfunc-tions can be either 1 or −1).

It is known (see e. g. [M], p. 71) that

λ2j = π2

(j − 1

2

)2[1 + O

( 1

j2

)]as j → +∞. (3.12.11)

The data can be written as

G(λ) =

∞∑

j=1

ψj(1)fjλ2 + λ2

j

, (3.12.12)

where fj are defined in (3.12.9). The series (3.12.12) converges absolutely and uniformly on compactsets of the complex plane λ outside the union of small discs centered at the points ±iλj . Thus, G(λ) is


a meromorphic function on the whole complex λ-plane with simple poles at the points ±iλj . Its residueat λ = iλj equals ψj(1)fj/(2iλj).

If f(z) = δ(z− 1), then fj = ψj(1) 6= 0 ∀j = 1, 2, . . . , (see Section 3.12.3 for a proof of the inequalityψj(1) 6= 0 ∀j = 1, 2, . . .) and the data (3.12.12) determine uniquely the set

λ2j , ψ

2j (1)j=1,2,... (3.12.13)

In Section 3.12.3 we prove the basic result:

Theorem 3.12.1. If f(z) = δ(z − 1) then the data (3.12.13) determine q(z) ∈ L1(0, 1) uniquely.

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 boundaryconditions, for example, u′(x1, 0) = u(x1, 1) = 0 with the data u(x1, 0) known for all x1 ∈ R2.

3.12.3 Proofs: uniqueness theorem and inversion algorithm

Proof of Theorem 3.12.1. The data (3.12.12) with f(z) = δ(z − 1), that is, with fj = ψj(1), determineuniquely λ2

jj=1,2,... since ±iλj are the poles of the meromorphic function G(λ) which is uniquelydetermined for all λ ∈ C by its values for all λ > 0 (in fact, by its values at any infinite sequence ofλ > 0 which has a finite limit point on the real axis). The residues ψ2

j (1) of G(λ) at λ = iλj are alsouniquely determined.

Let us show that:

(i) ψj(1) 6= 0 ∀j = 1, 2, . . . .

(ii) The set (3.12.13) determines q(z) ∈ L1(0, 1) uniquely.

Let us prove (i):If ψj(1) = 0 then equation (3.12.10) and the Cauchy data ψj(1) = ψ′

j(1) = 0 imply that ψj(z) ≡ 0

which is impossible since ‖ψj(z)‖ = 1, where ‖u‖2 :=∫ 1

0 |u|2dz.Let us prove (ii):It is sufficient to prove that the set (3.12.13) determines the norming constants

αj := ‖Ψj(z)‖2

and therefore the setλ2

j , αjj=1,2,...,

where the eigenvalues λ2j are defined in (3.12.10), Ψj = Ψ(z, λj), ψj(z) := Ψ(z, λj)/‖Ψj‖,

−Ψ′′ − s2Ψ − q(z)Ψ = 0, Ψ(0, s) = 0, Ψ′(0, s) = 1, (3.12.14)

and λj are the zeros of the equation

Ψ′(1, s) = 0, s = λj, j = 1, 2, . . . . (3.12.15)

The function Ψ′(1, s) is an entire function of ν = s2 of order 1/2, so that (see [Lev]):

Ψ′(1, s) = γ

∞∏

j=1

(1 − s2

λ2j

), γ = const . (3.12.16)

From the Hadamard factorization theorem for entire functions of order < 1 formula (3.12.16) follows butthe constant factor γ remains undetermined. This factor is determined by the data λ2

j∀j 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 γ in formula (3.12.16) can be computed explicitly:

γ =

∞∏

j=1

λ2j

(λ0j )

2, (3.12.17)

where λ0j are the roots of the equation cos(s) = 0, λ0

j = (2j−1)π/2, j = 1, 2, . . . , and the infinite productin (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 onthe formula:

1 = limy→+∞

Ψ′(1, iy)

cos(iy)= γ

∞∏

j=1

(λ0j )

2

λ2j

.

For convenience of the reader let us formulate the result from [M], p. 243, which yields formula(3.12.17) as well:

Claim 8. The function w(λ) admits the representation

w(λ) = cos(λ) − Bsin(λ)

λ+h(λ)

λ,

where B = const, h(λ) =∫ 1

0H(t) sin(λt)δt, and H(t) ∈ L2(0, 1) if and only if

w(λ) =

∞∏

j=1

λ2j − λ2

(λ0j )

2,

where λj = λ0j −B/j + βj/j, βj are some numbers satisfying the condition:

∑∞j=1 |βj|2 < ∞, λj are the

roots of the even function w(λ) and λ0j = (j − 1/2)π, j = 1, 2, . . . , are the positive roots of cos(λ).

The equality∞∏

j=1

λ2j − λ2

(λ0j )

2= γ

∞∏

j=1

(1 − λ2

λ2j

), (3.12.18)

where γ is defined in (3.12.17), is easy to prove: if w is the left-hand side and v the right-hand sideof the above equality, then w and v are entire functions of λ, the infinite products converge absolutely,(λ2j − λ2)/(λ0

j )2 = (λ2

j/(λ0j )

2)(1 − λ2/λ2j), 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 γ 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 assumptionsuffices for the proof of (3.12.18):

i) λ2j = (λ0

j )2 +O(1), (λ0

j )2 = π2(j − 1/2)2.

Indeed, if i) holds then both sides of (3.12.18) are entire functions with the same set of zeros andtheir ratio is a constant. This constant equals to 1 if there is a sequence of points at which this ratioconverges to 1. Using the known formula: cos(λ) =

∏∞j=1((λ

0j )

2 − λ2)/(λ0j )

2, and the assumption i) onechecks easily that the ratio of the left- and right-hand sides of (3.12.18) tends to 1 along the positiveimaginary semiaxis. Thus, we have proved formulas (3.12.16)–(3.12.17) without reference to Claim 8.

The above claim is used with w(s) = Ψ′(1, s) in our paper. The fact that Ψ′(1, s) admits therepresentation required in the claim is checked by means of the formula for Ψ′(1, s) in terms of the

transformation operator: Ψ(z, s) = sin(sz)s

+∫ z0K(z, t) sin(st)

sdt, and the properties of the kernel K(z, t)

are studied in Section 3.4.1. Thus, Ψ′(1, s) = cos(s) + K(1, 1) sin(s)s

+∫ 1

0Kz(1, t)

sin(st)s

dt. This is therepresentation of Ψ′(1, s) := w(s) used in Claim 8.


Let us derive a formula for αj := ‖Ψj‖2. Denote Ψ := dΨ/dν, differentiate (3.12.14), with s2 replacedby ν, with respect to ν and get:

−Ψ′′ − νΨ − qΨ = Ψ. (3.12.19)

Since q(z) is assumed real-valued, one may assume ψ real-valued. Multiply (3.12.19) by Ψ and(3.12.14) by Ψ, subtract and integrate over (0, 1) to get

0 < αj :=

∫ 1

0

Ψ2jδz =

(Ψ′jΨj − ΨjΨ

′j

)∣∣10

= −Ψj(1)Ψ′j(1), (3.12.20)

where the boundary conditions Ψj(0) = Ψ′j(1) = Ψj(0) = 0 were used.

From (3.12.16) with s2 = ν one finds the numbers bj := Ψ′j(1):

bj = γd

dν

∞∏

j′=1

(1 − ν

λ2j′

)∣∣∣ν=λ2

j

= − γ

λ2j

∏

j′ 6=j

(1 − λ2

j

λ2j′

). (3.12.21)

Claim 9. The data ψ2j (1) = Ψ2

j (1)/αj := tj , where αj := ‖Ψj(z)‖2, and equation (3.12.20) determineuniquely αj.

Indeed, the numbers bj are the known numbers from formula (3.12.21). Denote by tj := ψ2j (1) the

quantities known from the data (3.12.13). Then it follows from (3.12.20) that α2j = tjαjb

2j , so that

αj = tjb2j . (3.12.22)

Claim 9 is proved. 2

Thus, the data (3.12.13) determine αj = ‖Ψj‖2 uniquely and analytically by the above formula, andconsequently 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 ρ(λ) of the operator L is defined by the formula:

ρ(λ) =∑

λ2j<λ

1

αj. (3.12.23)

The Gel’fand– Levitan algorithm (see Section 3.4) allows one to reconstruct analytically q(z) fromthe spectral function ρ(λ) and therefore from the data (3.12.13), since, as we have proved already, thesedata determine the spectral function ρ(λ) uniquely. Thus Theorem 1 is proved. 2

Let us describe an algorithm for calculation of q(z) from the data g(x1):

Step 1 : Calculate G(λ), the Fourier transform of g(x1). Given G(λ), find its poles ±iλj , and consequentlythe numbers λj ; then find its residues, and consequently the numbers ψj(1)fj .

Step 2 : Calculate the function (3.12.16), and the constant γ by formulas (3.12.16) and (3.12.17). Cal-culate the numbers bj by formula (3.12.21) and αj by formula (3.12.22). Calculate the spectralfunction ρ(λ) by formula (3.12.23).

Step 3 : Use the known Gel’fand– Levitan algorithm to calculate q(z) from ρ(λ).

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 theapproach in [GX].


The authors assume (see (C2) on p.128 in [GX]) that ”g(r) =∑∞

0 αnH(1)0 (kanr), where a2

n isa negative decreasing sequence of numbers satisfying formula (3.1)” ( a typo, made in [GX] in theabove statement, is corrected: in [GX] it was written an for the ”negative decreasing sequence”, nota2n). 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 an be thesequence generated by g(r)”. The function g(r)does not generate a uniquely defined sequence of an.The authors do not explain how a given function g generates such a sequence. If one chooses a sequencesatisfying 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 eigenvaluesare unknown because n2(z) is unknown. The authors use the same notation an for different objects,and this causes errors in their arguments. Consequently the arguments in [GX] below formula (3.1) arewithout foundation, they are erroneous.

Further arguments in [GX] are also erroneous. For example, the authors have to solve equation(3.13) in [GX] for C(t2). If one corrects the typo in (3.13), replacing a by an, where an are unknownnumbers (since n2(z) is not known), then the problem of finding C(t2) from (3.13) means that onehas to determine simultaneously a sequence an and a function C(t2), given one function g(r). Thisis not possible, in general. The authors do not discuss this problem and, apparently, do not see thedifficulty. Instead they write that (3.13) is ”some kind of H-transform”, forgetting that the numbers anare unknown. One can also point out that (3.13) is not the H-transform in the standard sense (i.e, withH the Struwe function, and not a Hankel function, and with the integration over (0,∞) and not overthe 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 materialarises. In particular, such problems arise in geophysics when one wants to get information about themedium from the observations of the electromagnetic fields on the surface of the Earth. Let the sourceof electromagnetic waves be located above the ground. These waves, radiated by the source, penetrateinto the ground, interact with it and the resulting electromagnetic field is observed on the surface of theEarth. The present work shows how to get information about the inner structure of the Earth layers fromthese 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 electromag-netic fields E and H:

rot E = −µ∂H∂t

, rotH = ε∂E

∂t+ σE + j, (3.13.1)

written in the Cartesian coordinates (x, y, z). The plane (x, y) is assumed to be parallel to the Earth’ssurface, the z-axis is perpendicular to the Earth’s surface and directed into the Earth. In (3.13.1) t istime, σ(z), ε(z) and µ = const are conductivity, permittivity and magnetic permeability respectively,j = f(t)δ(x)δ(z − z0)ey is the exterior source which is supposed to be a wire located above the Earth’ssurface and going along the y-axis through the point z = z0 < 0, δ is the delta-function, f(t) is a functionof time which shows the shape of the electromagnetic pulse.

It is assumed that

ε = ε0, σ = 0, µ = µ0 for z < 0 in the air, (3.13.2)

ε = ε1 = const, σ = 0, µ = µ0 for z > L, (3.13.3)

f(t) = 0 for t < 0 and t > T, (3.13.4)

3.13. THEORY OF GROUND-PENETRATING RADARS 167

where ε0 = 8.854 · 10−12 f/m is the dielectric constant of free space while µ0 = 1.257 · 10−6 h/m is themagnetic permeability of vacuum.

Problem 2 (The Ground-Penetrating Radar Problem (GPR):). Given E on the plane z = 0for all t > 0, find ε(z) and σ(z) 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 ∂H∂t in it by −µ−1rot E

taken from the first equation. One gets

−µ−1rot rot E = ε∂2E

∂t2+ σ

∂E

∂t+∂j

∂t. (3.13.5)

Assuming that the vector E = (0, E2(x, z, t), 0) is directed along the y-axis and depends on t, x andz, one can reduce (3.13.5) to a one-dimensional inverse problem for a differential equation satisfied by

u(z, k, λ) = E(z, k, λ)/(ikµh(k)), where

E(z, k, λ) =

∫ ∞

0

dt

∫ ∞

−∞dx E2(x, z, t)e

ikt+iλx and h(k) =

∫ ∞

0

f(t)eiktdt (3.13.6)

are the Fourier transforms of E2(x, z, t) and δ(x)f(t) respectively, and i is the imaginary unit. Thecharacteristic value of u is z∗. The function u solves the problem:

uzz − λ2u+ k2A2(z)u+ ikB(z)u = −δ(z − z0), u(±∞, k, λ) = 0, (3.13.7)

where the conditions at infinity are written under the assumption that k2 is sufficiently small. In thisSection we use only such k. In (3.13.7), A2(z) = ε(z)µ and B(z) = σ(z)µ. 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

uzz − λ2u+ k2A20u+ k2(A2

1 −A20)θ(z − L)u+ k2a2(z)u+ ikB(z)u = −δ(z − z0),

u(±∞, k, λ) = 0,(3.13.8)

where A20 = µ, A2

1 = ε1µ,

a2(z) =

A2(z) − A2

0, if 0 ≤ z ≤ L

0, otherwise,

and

θ(z) =

1, if z ≥ 0

0, otherwise.

Equation (3.13.8) is equivalent to the equation

u(z, k, λ) = g(z, z0, k, λ) + k2

∫ L

0

g(z, s, k, λ)a2(s)u(s, k, λ)ds

+ ik

∫ L

0

g(z, s, k, λ)B(s)u(s, k, λ)ds.

(3.13.9)

In (3.13.9) g(z, s, k, λ) is the Green function that solves the equation

gzz − λ2g + k2A20g + k2(A2

1 − A20)θ(z − L)g = −δ(z − s) (3.13.10)


and satisfies the conditions:

g and ∂g/∂z are continuous at z = L, limz→±∞

g(z, s, k, λ) = 0. (3.13.11)

The solution to (3.13.10)-(3.13.11) is

g(z, s, k, λ) =

e−ν|z−s|

2ν+ ν−ν1

ν+ν1eνz+νs

2νe−2νL, if z < L, s < L

e−ν1z+νs

ν+ν1e(ν1−ν)L, if z > L, s < L

eνz−ν1s

ν+ν1e(ν1−ν)L, if z < L, s > L

e−ν1|z−s|

2ν1+ ν1−ν

ν1+νe−ν1z−ν1s

2ν1e2ν1L, if z > L, s > L,

(3.13.12)

whereν2 = λ2 − k2A2

0 > 0, ν21 = λ2 − k2A2

1 > 0,

and we use the function g assuming that k is sufficiently small, so that ν2 and ν21 are both positive.

Let C(0, L) denote the Banach space of continuous on [0, L] functions with the usual norm. Denote

(Q1u)(z, k, λ) =

∫ L

0

g(z, s, k, λ)a2(s)u(s, k, λ)ds,

(Q2u)(z, k, λ) =

∫ L

0

g(z, s, k, λ)B(s)u(s, k, λ)ds.

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 solutionis analytic with respect to k in a neighborhood of the point k = 0 and can be obtained by iterations

un+1(z, k, λ) = g(z, z0, k, λ) + k2(Q1un)(z, k, λ) + ik(Q2un)(z, k, λ), u0 = g; n = 0, 1, 2, . . . .

This theorem is proved similarly to the proof of Lemma 1 in [R83], p.219, and to the proof in Section9.1.

Let u = v + iw, where v and w are the real and imaginary parts of u respectively. The functions wand v solve the equations

w(z, k, λ) = k2(Q1w)(z, k, λ) + k(Q2v)(z, k, λ), (3.13.13)

v(z, k, λ) = g(z, z0, k, λ) + k2(Q1v)(z, k, λ) − k(Q2w)(z, k, λ). (3.13.14)

Put z = 0. Taking into account the analyticity of v and w in the neighborhood of k = 0, onedifferentiates equation (3.13.13) with respect to k at k = 0 and gets:

∫ L

0

e−2λsσ(s)ds = 4λ2eλ|z0|µ−1 ∂w(0, 0, λ)

∂k. (3.13.15)

Similarly, one differentiates twice equation (3.13.14) with respect to k and puts k = 0. This yields:

∫ L

0

e−2λsa2(s)ds = 4λ2eλ|z0|

1

2

[∂2v(0, 0, λ)

∂k2− ∂2g(0, z0, 0, λ)

∂k2

]

+

∫ L

0

dsB(s)g(0, s, 0, λ)

∫ L

0

g(s, t, 0, λ)B(t)g(t, z0, 0, λ)dt

.

(3.13.16)


If one introduces the variable p = 2λ and substitutes it into in the above equations instead of λ then(3.13.15) and (3.13.16) will express the Laplace transforms of the unknown functions σ(z) and a2(z) interms of the data on the plane z = 0. Since an L2-function can be uniquely recovered from its Laplacetransform, one gets the following uniqueness theorem:

Theorem 3.13.2. The functions ε(z) and σ(z) are uniquely determined by the data E2(x, t) on theplane z = 0.

Some methods are given in [R139] for recovery of σ(z) and a2(z) from their Laplace transforms knownfor p ∈ (0,∞) only. Let us rewrite (3.13.15) as L σ(z) = F (p) where L denotes the Laplace transformoperator and F is the right-hand side of (3.13.15). A possible numerical inversion method consists ofminimizing the functional

F (σ) = ||Lσ(z) − F (p)||2 + α||σ(z)||2 −→ minimum (3.13.17)

with respect to σ(z) in an appropriate norm (we have chosen L2 norm), where α is a regularizationparameter. If one looks for σ(z) of the form of a linear combination of finitely many linearly independentfunctions, then (3.13.17) is equivalent to determining the coefficients of the linear combination whichminimize (3.13.17). To find the unknown coefficients, one differentiates (3.13.17) with respect to thesecoefficients and gets a system of linear algebraic equations for them. If σ(z) is determined, one can useits values to recover a2(z) (and consequently ε(z)) from equation (3.13.16) using an analogous procedure.

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 σ(z) and ε(z).

To evaluate the integrals in (3.13.9), the Simpson’s method was used. Then, the obtained values ofthe function u on the plane z = 0 were used to compute the Laplace transforms (3.13.15), (3.13.16) ofσ(z) and a2(z) respectively. Finally, the functions σ(z) and a2(z), considered as unknown ones, wererecovered from their Laplace transforms on the basis of (3.13.17), and the computed values of σ(z) andε(z) were compared with given ones to estimate the accuracy of the the obtained results.

As basis functions for expansion of σ(z) and a2(z) on the interval [0, L], the following piecewise-constant functions were chosen:

B0i (z) =

(zi+1 − zi)−1/2, if zi ≤ z < zi+1

0, otherwise,

where the points ziNi=0 form a partition of [0, L] (here N is the number of subintervals of [0, L]). Thefunctions B0

i (z) form an orthonormal system in L2.To demonstrate the efficiency of the proposed approach for the recovery of σ(z) and ε(z) which are

continuous in the interval (0, L), we considered the following example:

σ(z) = e−6z, ε(z) = 1 + (ε1 − 1) sin(πz/L), L = 0.5, z0 = −0.5, ε1 = 1.5.

The results of the calculations are shown in Fig.1, 2 for σ and ε respectively. In these Figures z-coordinate is taken along the horizontal axis, the vertical axis shows the values of the functions, thesolid lines denote the exact values of σ(z) and ε(z) respectively, and the black circles indicate theobtained numerical results. It is worth to note that the recovery of σ and ε is carried out, in fact, on thebasis of the noisy data in (3.13.15) and (3.13.16): indeed, each step of the numerical scheme contributessome computational error. The accuracy of the calculation of the Laplace transforms (3.13.15), (3.13.16)in L2-norm was about 5 · 10−7 and 2 · 10−4 respectively. The accuracy of the obtained results is about3 ·10−4 for σ and approximately 3.5 ·10−2 for ε. This accuracy is defined by the formula: ||f − f∗||/||f ||,


where f denotes the exact function and f∗ is the calculated one. Here by f we mean σ(z), ε(z) or theLaplace transforms (3.13.15), (3.13.16).

To investigate the applicability of our approach for recovery of discontinuous functions, we took

σ(z) =

4, if 0 ≤ z ≤ 0.25,

1, if 0.25 < z ≤ 0.5,

and ε(z) = 1 + ε1e−6z for L = 0.5, z0 = −0.5, ε1 = 1.5. Fig.3 and Fig.4 in [RSh] show the values of the

recovered σ and ε respectively.

To recover the discontinuous function σ(z), an adaptive grid generation procedure was used. Thisprocedure is based on the following coordinate transformation:

ω(z) =

∫ z

0

[c2 + (df/dp)2]1/2dp, (3.13.18)

where f denotes the unknown function to be recovered, and c is some positive constant. The transfor-mation (3.13.18) is one-to-one. A uniform grid ωi = i · 4ω, i = 0, ..., N, 4ω = ω(L)/N along theω-axis corresponds to the grid zi, i = 0, ..., N, z0 = 0, zN = L along the z-axis with larger density ofthe nodes in places with large |df/dz|, provided that c is sufficiently small.

In practice, the procedure for the recovery of σ is organized as follows. First, the constant c is takenlarge. This corresponds to a uniform grid zi 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 thez-axis, we integrate (3.13.18) numerically and obtain the nodes Ωj, j = 0, ..., N, Ω0 = 0, ΩN = ω(L)along the ω-axis. Then a uniform grid ωi is constructed along the ω-axis and the values of zi arecalculated at these points ωi by linear interpolation of the data from the nodes Ωj obtained aftersolving (3.13.18). Using the new distribution of zi, minimization procedure (3.13.17) is done againand the new grid zi 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 σ is attained. In sucha way, one can reduce the computational error of the recovery of σ to 9 · 10−3 (the accuracy of thecalculation of the Laplace transform of σ from the data was about 10−6). The accuracy of the recoveryof ε is approximately 2.5 · 10−2 (the accuracy of the calculation of the Laplace transform (3.13.16) ofa2(s) was about 3 · 10−4).

The proposed algorithm allows one to recover two coefficients simultaneously using a reasonable apriori information about the unknown functions ε(z) and σ(z).

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 aboutthe electromagnetic field on the Earth’s surface allows one to recover uniquely both conductivity andpermittivity, and analytical recovery from exact data is possible. These parameters may have polynomialgrowth as the vertical coordinate approaches infinity.

Mathematical model of the above problem is based on Maxwell’s equations (3.13.1), written in thecylindrical coordinates (r, φ, z). The plane z = 0 is assumed to be the Earth’s surface, the z-axis isperpendicular to the Earth’s surface and directed into the ground. In (3.13.1) t is time, σ(z), ε(z) and µ =const are conductivity, permittivity and magnetic permeability respectively, j = f(t)δ(r− r0)δ(z− z0)eφis the exterior source which is supposed to be a loop of a radius r0 located above the Earth’s surface atthe point z = z0 < 0, δ is the delta-function, f(t) is a piecewise-continuos function of time which showsthe shape of the electromagnetic pulse. The z-axis is supposed to pass through the center of the loopand perpendicular to the plane in which that loop lies.


It is assumed that

ε = ε0(z), σ = σ0(z), µ = µ0 for z < 0 in the air, (3.13.19)

ε = ε(z), σ = σ(z), µ = µ0 for z > 0, (3.13.20)

f(t) = 0 for t < 0 and t > T. (3.13.21)

The functions ε0(z), σ0(z) and constant µ0 > 0 are considered to be known. The functions σ(z) and ε(z)are assumed to be piecewise-continuous uniformly bounded functions on [0,∞). The integral transforms,used below, are understood in the distributional sense.

Problem 3 (The Ground-Penetrating Radar Problem (GPR)). : Given E on the plane z = 0for all t > 0, find ε(z) and σ(z) for 0 < z < ∞.

Our basic result is formulated in the following theorem.

Theorem 3.13.3. Under the above assumption the functions σ(z) and ε(z) are uniquely determined bythe surface data E(r, φ, t) known for all t > 0, 0 ≤ φ < 2π for fixed z0 < 0 and r0 > 0.

Let us derive the basic equations. Differentiate the second equation in (3.13.1) with respect to t andsubstitute ∂H

∂t in it by −µ−1rot E taken from the first equation.One gets

−µ−1rot rot E = ε∂2E

∂t2+ σ

∂E

∂t+∂j

∂t. (3.13.22)

Assume that the vector E = (Er, Eφ, Ez) = E(r, z)eφ, where eφ is the unit vector of the cylindricalcoordinates. In this case equation (3.13.22) takes the form of a one-dimensional inverse problem for adifferential equation. First one rewrites equation (3.13.22) in cylindrical coordinates taking into accountthe well-known relation

rot rot E = grad div E −4E,

where 4 stands for Laplace operator, and (rot E)r = 1/r(∂Ez)/(∂φ) − (∂Eφ)/(∂z), (rot E)φ =(∂Er)/(∂z)− (∂Ez)/(∂r), (rot E)z = 1/r((∂(rEφ))/(∂r)− (∂Er)/(∂φ)), 4(E(r, z)eφ) = ((∂2E)/(∂z2)+(∂2E)/(∂r2) + 1/r(∂E)/(∂r) −E/r2)eφ. One gets

A2(z)∂2E

∂t2+B(z)

∂E

∂t− ∂2E

∂z2− ∂2E

∂r2− 1

r

∂E

∂r+E

r2= −µ∂f

∂tδ(r − r0)δ(z − z0). (3.13.23)

Here A2 = εµ, B = σµ and i = exp(iπ/2). Defining the Fourier transform of E as

E(r, z, k) :=

∫ ∞

0

E(r, z, t)eiktdt

and applying it to (3.13.23) one gets

∂2E

∂z2+ k2A2E + ikBE +

∂2E

∂r2+

1

r

∂E

∂r− E

r2= −ikµh(k)δ(r − r0)δ(z − z0), (3.13.24)

where h(k) =∫∞0f(t)eiktdt is the Fourier transform of f(t). Then one applies the Hankel-Bessel trans-

form

w(z, k, λ) :=

∫ ∞

0

E(r, z, k)J1(λr)rdr

to equation (3.13.24), where J1 is the Bessel function. Let us denote

u(z, k, λ) :=w(z, k, λ)

ikµh(k)r0J1(λr0).


The function u solves the problem:

∂2u

∂z2− λ2u+ k2A2(z)u+ ikB(z)u = −δ(z − z0), u(±∞, k, λ) = 0. (3.13.25)

Equation (3.13.25) is equivalent to the equation

u(z, k, λ) = g(z, z0, λ) + k2

∫ ∞

−∞g(z, s, λ)A2(s)u(s, k, λ)ds

+ ik

∫ ∞

−∞g(z, s, λ)B(s)u(s, k, λ)ds.

(3.13.26)

In (3.13.26), g(z, s, λ) is the Green function that solves the equation

∂2g

∂z2− λ2g = −δ(z − s) (3.13.27)

and satisfies the condition:lim

z→±∞g(z, s, λ) = 0. (3.13.28)

The solution to (3.13.27)- (3.13.28) is

g(z, s, λ) =e−λ|z−s|

2λ. (3.13.29)

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 showits 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:

u(0, k, λ)− g(0, z0, λ)

ik=

∫ ∞

−∞

e−λ|s|

2λB(s)

e−λ|s−z0 |

2λds+ O(k), k → 0. (3.13.30)

Passing in (3.13.30) to the limit k → 0 and denoting by β(λ) the limit of the known function in theleft-hand side of (3.1) one gets

∫ ∞

−∞e−λ(|s|+|s−z0|)B(s)ds = 4λ2β(λ). (3.13.31)

Since z0 < 0 and the function B(z) is known for z < 0, one gets from (3.13.31)

∫ ∞

0

e−2λsB(s)ds = b1(λ)

where

b1(λ) = eλ|z0|[4λ2β(λ) − µ0

∫ 0

−∞e−λ(|s|+|s−z0|)σ0(s)ds] (3.13.32)

is a known function computable from the surface data and the values of σ(z) for z < 0.Therefore setting

ν = 2λ, b(ν) = µ−10 b1(ν/2) (3.13.33)

one gets ∫ ∞

0

e−νsσ(s)ds = b(ν), ν > 0. (3.13.34)


So σ(z) is uniquely determined as the inverse Laplace transform of the known function b(ν). The functionb(ν) is defined by formulas (3.13.33) and (3.13.32), and β(λ) in (3.13.32) is the limit, as k → 0, of theleft-hand side of (3.13.30).

If B(z) has been found equation (3.13.26) yields

u(0, k, λ) = g(0, z0, λ) + k2

∫ ∞

−∞g(0, s, λ)A2(s)g(s, z0, λ)ds

+ ik

∫ ∞

−∞dsg(0, s, λ)B(s)

[g(s, z0, λ) + ik

∫ ∞

−∞g(s, p, λ)B(p)g(p, z0, λ)dp

]+O(k3), k → 0.

(3.13.35)

Therefore∫ ∞

−∞g(0, s, λ)A2(s)g(s, z0, λ)ds =

limk→0

u(0, k, λ)− g(0, z0, λ) − ik∫∞−∞ g(0, s, λ)B(s)g(s, z0, λ)ds

k2

+

∫ ∞

−∞ds g(0, s, λ)B(s)

∫ ∞

−∞g(s, p, λ)B(p)g(p, z0, λ)dp.

(3.13.36)

Denote the right-hand side of (3.13.36) by α(λ). Since A2(z) is known for z < 0, equation (3.13.36)yields ∫ ∞

0

e−2λsA2(s)ds = a1(λ), (3.13.37)

where

a1(λ) = eλ|z0|[4λ2α(λ) − µ0

∫ 0

−∞e−λ(|s|+|s−z0|)ε0(s)ds

]. (3.13.38)

Settingν = 2λ, a(ν) = µ−1

0 a1(ν/2)

one gets ∫ ∞

0

e−νsε(s)ds = a(ν), ν > 0. (3.13.39)

Hence ε(z) is uniquely determined as the inverse Laplace transform of the known function a(ν). Thefunction a(ν) is computable from the surface data, the values of ε(z) for z < 0 and from the functionB(z) calculated by solving equation (3.13.34).

Chapter 4

Inverse obstacle scattering

4.1 Statement of the problem

First, let us formulate the direct scattering problem. Assume that D ⊂ Rn is a bounded domain(obstacle) with the boundary S. D may consist of finitely many connected components. By D′ denotethe complement of D. Assume that D has finite perimeter relative to Rn. By the perimeter of the setD we mean P (D) := ‖χD‖BV (Rn), where χD is the characteristic function of D, and BV (Rn) is thespace of functions of bounded variation in Rn. This space consists of functions u such that ∇u (in thesense of distribution theory) is a charge, and the total variation of this charge is ‖u‖BV (Rn). If D is aLipschitz domain, then D has a finite perimeter. The set of domains with finite perimeter is much largerthan the set of Lipschitz domains. If xn = f(x′), x′ := (x1, . . . , xn−1) is the local equation of S, thenLipschitz boundary is the one for which |f(x′) − f(x′′)| ≤ M |x′ − x′′|. D has finite-perimeter if S hasfinite (n− 1)-dimensional Hausdorff measure, that is Hn−1(S) = limε→0 infBi

∑i rn−1i < ∞, where the

infimum is taken over all coverings of S by open balls of radii ri < ε.Let ν be a unit vector, x, ν ∈ Rn, A+ := y : (y − x) · ν > 0 A− := y : (y − x) · ν < 0,

A := y : (y −x) · ν = 0. We say that ν is the normal (in the sense of Federer) to a set E at the pointx ∈ E if

limr→0

r−nmn

(E ∩B(x, r) ∩A+

)= 0 = lim

r→0r−nmn

(E′ ∩B(x, r) ∩A−) = 0,

where mn is n-dimensional Lebesgue measure, E′ : Rn \E, B(x, r) = y : |x− y| ≤ r. By ∂E we denotethe boundary of E. The set of x ∈ ∂E for which normals to E exist is denoted ∂∗E and is called thereduced boundary of E. The following result is known (e.g.[Maz]):

If P (E) < ∞ then the set ∂∗E is measurable with respect to Hn−1 and if u is a compactly supportedLipschitz function in Rn then

∫

E

∇u dx =

∫

∂∗E

u(s)ν(s)Hn−1(ds) (Green′s formula).

More generally, let D be a bounded domain with finite perimeter, and assume that the normal to Dexists Hn−1-almost everywhere on ∂D. Then for u ∈ BV (D), which has a Hn−1-summable trace u on∂D Hn−1-almost everywhere, the Green’s formula holds.

Let us consider first the direct scattering problem. Let D be a bounded domain in Rn, n ≥ 2, whichmay have finitely many connected components, D′ := Rn \D, S := ∂D, Sn−1 is the unit sphere in Rn.Let

(∇2 + k2

)u = 0 in D′, u = u0 + v, u0 = ei k α·x, α ∈ Sn−1, (4.1.1)

limR→∞

∫

|s|=R

∣∣∣∣∂v

∂|x| − i k v

∣∣∣∣2

ds = 0 (4.1.2)

174

4.1. STATEMENT OF THE PROBLEM 175

Γu = 0 on S, (4.1.3)

where Γ is one of the following boundary conditions:

Γ1u = u, Γ2u := uN , Γ3u := uN + h(s)u, (4.1.4)

N is the exterior unit normal to S, h(s) ≥ 0 is a piecewise-continuous bounded function with finitelymany 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 Γ1, this meaning consists of the inclusion u ∈H1

loc(D′), where

H1(D′) is the closure in the norm ofH1(D′) of the C∞

0 (D′) functions, and H1loc(D

′) is the set of functions

such that f ∈H1(D′

R), R > 0 is such that D ⊂ BR := x : |x| ≤ R.Equation (4.1.1) we replace by the identity:

∫

D′

(∇u · ∇ϕ− k2uϕ

)dx = 0 ∀ϕ ∈

H1

0(D′), (4.1.5)

whereH1

0 (D′) is the set ofH1(D′) functions vanishing near infinity. We call u the weak solution of the

scattering problem. Thus, problem (4.1.1)–(4.1.4) with Γ = Γ1 is formulated for non-smooth D as the

problem of finding u ∈

H1loc(D

′), of the form u = u0 + v, where v satisfies (4.1.2), and u satisfies (4.1.5).If Γ = Γ2, then u ∈ H1

loc(D′), (4.1.5) holds ∀ϕ ∈ H1

0(D′), H10 (D′) is the set of H1(D′) functions

vanishing near infinity, and the conditions u = u0 + v, v solves (4.1.2), hold.If Γ = Γ3, then u satisfies the identity:

∫

D′

(∇u∇ϕ− k2uϕ

)dx+

∫

S

h(s)u(s)ϕ(s)ds = 0 ∀ϕ ∈ H10 (D′), (4.1.6)

u is of the form u = u0 + v, and v satisfies (4.1.2). For (4.1.6) to make sense one has to define the tracesof functions from H1

loc(D′) on S. The integration in (4.1.6) over S is with respect to Hn−1 measure.

If S is Lipschitz, or D is an extension domain, or D satisfies the cone condition, then the embeddingoperator from H1

loc(D′) into L2(S) is compact, so the traces of the elements of H1

loc(D′) on S are well

defined. If Γ = Γ1, then we assume that the domain D is bounded; if Γ = Γ2 we assume additionallythat the embedding operator i1 : H1

loc(D′) → L2

loc(D′) is compact; if Γ = Γ3 we assume get additionally

that the embedding operator i2 : H1loc(D

′) → L2(S) 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 H = L2(D′), boundedfrom below, quadratic form

D[t] =H1(D′) if Γ = Γ1,

(D[t] = H1(D′)

)if Γ = Γ2 or Γ = Γ3.

This form defines a unique self-adjoint operator L, which is denoted by −∆. When we want to emphasizethe corresponding boundary condition, we write Lj, j = 1, 2, 3. Consequently, if Im k > 0 and Re k > 0,so that Im k2 > 0, then (−∆ − k2)−1 is a bounded operator in H = L2(D′). Its kernel G(x, y, k) iscalled 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 limitlimε→0G(x, y, k + i ε).

Let us first prove that the quadratic forms (4.1.5) (and (4.1.6)) with domainsD[t] =H1(D′), H1(D′)

(and H1(D′)), corresponding to Γ1, Γ2 (and Γ3) are closed. These forms are symmetric and, under ourassumptions on D, bounded from below.

176 CHAPTER 4. INVERSE OBSTACLE SCATTERING

Closedness means (see [Kat]) that if un → u in L2(D′) and t[un − um, un − um] → 0 as n,m → ∞,then u ∈ D[t], and t[un − u, un − u] → 0. If a quadratic form is bounded from below, then the formt[u, u] + c(u, u) ≥ (u, u), for a suitable constant c > 0. We will assume therefore, without loss ofgenerality, that t[u, u] ≥ (u, u). Then the norm H1(D′) is equivalent to the norm t[u, u]. Since H1(D′)

is a complete Hilbert space, the form t with the domain H1(D′) (and with the domainH1(D′)), is

closed. This covers conditions Γ1 and Γ2. Consider condition Γ3. If i2 is compact, then one has [R1] theinequality ∫

S

|u|2ds ≤ ε

∫

D′R

|∇u|2dx+ C(ε)

∫

D′R

|u|2dx, (4.1.7)

where ε > 0 is an arbitrary small number, ds is Hn−1 measure on S, and D′R = D′ ∩ B(R), B(R) :=

B(0, R), andR is throughout a sufficiently large number, so thatD ⊂ B(R). Inequality (4.1.7) guaranteesthat the form t defined by formula (4.1.6) is bounded from below in L2(D′) and that the norm it definesis equivalent to the norm of H1(D′). Therefore this form with the domain H1(D′) is closed. Actually,one could replace the assumption of compactness of the embedding operator i2 by the assumption ofits boundedness, with relative bound < 1, that is, by the inequality C0

∫S|u|2ds ≤ a

∫D′

R|∇u|2dx +

C1

∫D′

R|u|2dx, where C0 = sups∈S |h(s)|, 0 < a = const < 1, and C1 = const > 0, where the constants

are independent of u ∈ H1(D′).The closed symmetric forms generate the corresponding self-adjoint operators which we denote L =

Lj = −∆ for any of the conditions Γj , j = 1, 2, 3.

The scattering problem can be formulated as the equation (L − k2)w = f . Let us show how thisis done. Take a cut-off function η ∈ C∞(Rn), η(r) = 1 if |x| := r > R, η = 0 in a neighborhoodof S, 0 ≤ η ≤ 1, and let u = η u0 + w. Then w = v for r > R, Γ(w) = Γ(u), (∇2 + k2)w = −f ,f := (∆ + k2)(η u0) ∈ C∞

0 (D′). Thus we have reduced the scattering problem to the problem of solvingthe equation (L−k2)w = f with the self-adjoint operator L and f ∈ H := L2(D′). Actually f ∈ C∞

0 (D′).This equation has a solution in H if Im k2 > 0, and the solution is unique. Define L2

γ := L2(D′, (1 +

|x|2)γ/2), and take n = 3 in what follows.

Let us take k > 0, ε > 0 sufficiently small, and prove the limiting absorption principle (LAP):

Theorem 4.1.1. For any f ∈ L2γ , γ > 1, n = 3, there exists limε↓0w(x, k + iε) = w(x, k) in the norm

of L2−γ . The function w(x, k) is the weak solution of the problem

Lw − k2w = −f, Γj(w) = 0, (4.1.8)

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

limR→∞

∫

|x|=R

[∣∣wr∣∣2 + k2|w|2 + i k

(wr w − w wr

)]ds = 0, r := |x|. (4.1.9)

From (4.1.5) (or (4.1.6)), taking ϕ = uψ( r−Rδ ) ∈ C∞(R), ψ(r) = 1 for r ≤ 1, ψ(r) = 0 for r ≥ 2,0 ≤ ψ ≤ 1, δ > 0, then taking complex conjugate equation, subtracting from (4.1.5) its complexconjugate, and taking δ → 0, one gets:

0 = limδ→0

∫

D′

(w∇w · ∇ψ − w∇ w · ∇ψ

)dx =

∫

|x|=R

(w wr − w wr

)ds, D ⊂ B(R). (4.1.10)

4.1. STATEMENT OF THE PROBLEM 177

From (4.1.8) and (4.1.9) it follows that

limR→∞

∫

|x|=R

[∣∣wr∣∣2 + k2|w|2

]ds = 0. (4.1.11)

Any solution to equation (4.1.1) which satisfy (4.1.11), vanishes identically everywhere in its domain ofdefinition.

Indeed, any solution w to (4.1.1) can be written as

w(x) =

∞∑

l=0

Y`(x0)[C+` h

+` (k r) + C−

` h−` (k r)

], r > R, x0 :=

x

r, r := |x|,

where Y` = Y`m are the orthonormal in L2(S2) spherical harmonics, −` ≤ m ≤ `, C±` are constants, h±`

are spherical Hankel functions, h` ∼ e±i k r

r as r → ∞. From (4.1.11) one derives C±` = 0 for all `, so

w = 0. Lemma 4.1.2 is proved. 2

Let us continue with the proof of Theorem 4.1.1.The steps of the proof are:

Step 1. We prove that ∥∥wε∥∥L2

−γ

≤ c, γ > 1, (4.1.12)

c does not depend on ε ∈ (0, 1). By c we denote various constants.

Step 2. From (4.1.12) it follows that wε → w in L2loc(D

′). This and (4.1.5) (or (4.1.6)) imply‖∇wε‖L2

loc(D′) ≤ c, so wε H1

loc(D′), and wε → w in L2

loc(D′) because i2 is compact (or because

i1 and i2 are compact, when Γ3 holds). Thus, one can pass to the limit as ε ↓ 0 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 wε → w in Hm(D′),where D′ is any strictly interior bounded subset of D′. This allows one to pass to the limit ε→ 0 in therepresentation

wε(x) =

∫

|x|=R

(wε gεN − wεN gε

)ds,

gε : =ei(k+i ε)|x−y|

4π|x− y| , x ∈ B′(R),

where N is the normal to the sphere |x| = R, pointing into B′(R) = R3 \B(R) and get

w(x) =

∫

|x|=R

[w gN (x, s) − wN g(x, s)

]ds, x ∈ B′(R). (4.1.13)

It follows thatsup

0<ε<1|wε(x)| ≤

c

1 + |x| , x ∈ B′(R).

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 condi-tions Γj , j = 1, 2, 3) in D′, where D is a bounded domain with finite perimeter, such that i1 and i2 arecompact. Then

G(x, y, k) = g(|y|)u(x, α, k) + O

(1

|y|2), |y| → ∞,

y

|y| = −α, (4.1.14)

where g(|y|) := ei k|y|

4π|y| if n = 3, O( 1|y|2 ) is uniform with respect to x running through compact sets, and

u(x, α, k) is the scattering solution, i.e, the solution to (4.1.1)–(4.1.4). If n ≥ 2, g(|y|) is the resolvent

kernel of −∆ in Rn, g = i4( k2π|y|)

n2 −1H

(1)n2 −1(k|y|), H

(1)p is the Hankel function.


Proof. For domains with finite perimeter Green’s formula holds. Under our assumption on D the exis-tence and uniqueness of G have been proved above. Green’s formula yields the representation formula(for Γ = Γ1):

G(y, x, k) = g(y, x, k) −∫

S

g(y, s, k)∂G(s, x, k)

∂Nsds. (4.1.15)

Here and below the integration over S is actually taken over the essential boundary (see [VH], p. 198).One has G(x, y, k) = G(y, x, k), and

g(y, x, k) = g(|y|)ei k α·x + O

(1

|y|2), |y| → ∞,

y

|y| = −α.

Using this formula and (4.1.15) one gets (4.1.14) with

u = ei k α·x − 1

4π

∫

S

ei k α·s∂G(s, x, k)

∂Nds. (4.1.16)

The function (4.1.16) solves (4.1.1)–(4.1.4) with Γ = Γ1 if G is the resolvent kernel of L1. The argumentis similar for Γ = Γj, j = 2, 3. For Γ = Γ2 formula (4.1.15) takes the form

G(y, x, k) = g(y, x, k) +

∫

S

∂g(y, s, k)

∂NsG(s, x, k)ds, Γ = Γ2,

and (4.1.16) becomes

u = ei k α·x +1

4π

∫

S

∂ ei k α·s

∂NsG(s, x, k)ds, Γ = Γ2,

and for Γ = Γ3 one gets:

G(y, x, k) = g(y, x, k) +

∫

S

[gNs(y, s, k) + h(s)g(s, y, k)

]G(s, x, k)ds, Γ = Γ3,

while (4.1.16) becomes:

u = ei k α·x +1

4π

∫

S

(∂

∂Ns+ h(s)

)ei k α·sG(s, x, k)ds, Γ = Γ3.


We need also

Lemma 4.1.4. If Dj , j = 1, 2, are domains with finite perimeter, then D12 := D1∪D2, D12 := D1∩D2,

and D12 \D12 := ∆12 are domains with finite perimeter.

Proof. Let χj be the characteristic function (=indicator) for Dj . Then χ12 = χ1 + χ2 − χ1 χ2, χ12 =

χ1 χ2, χ∆12 = χ12 − χ12. Therefore it is sufficient to prove that ∇(χ1 χ2) is a charge if ∇χj arecharges, j = 1, 2, and |χj| ≤ c = const > 0, in our case c = 1. It is proved in [VH], p.189, that∇(χ1 χ2) = χ1 ∇χ2 + χ2 ∇χ1, where χ is the average value of χ. Since χj are function of boundedvariation and ∇χj are charges, then χj∇χi is a charge. Lemma 4.1.4 is proved. 2

For convenience of the reader let us give the differentiation formula from [VH] for B V functions: ifu, υ ∈ B V then (u υ)′ = u υ′ + υ u′, where prime stands for a derivative, u′ and υ′ are charges, and υis the average value defined as v(x) := 1

2 [va(x) + v−a(x)], where x is a regular point of v(x), i.e., va(x)and v−a(x) exist for some unit vector a, and

va(x) := limy→x,(y−x,a)>0

v(y).

4.2. INVERSE OBSTACLE SCATTERING PROBLEMS 179

Alternatively, if ϕ ≥ 0, ϕ ∈ C∞0 (Rn),

∫Rn ϕ(x)dx = 1, e.g.

ϕ(x) =

cn e

−1/(1−|x|2), |x| ≤ 1

0, |x| ≥ 1,cn :=

(∫

Rn

e−1/(1−|x|2)dx

)−1

,

then one defines ϕr := 1rnϕ(x

r) and v(x) := limr→0

∫Rn v(y)ϕr (x − y)dy. This limit does exist if x is a

regular point of v.

One gets from (4.1.13) that w satisfies (4.1.2) |w(x)| ≤ c1+|x| so w ∈ L2

−γ (D′) and wε → w in

L2−γ (D

′).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 ε ↓ 0 such that ‖wε‖L2−γ

→ ∞. Define zε := w‖wε‖L2

−γ

. Then Lzε − k2ε zε =

fε := f‖wε‖L2

−γ

, kε := k + i ε. By the argument given in step 2, it follows that zε → z in L2−γ(D

′),

Lz − k2 z = 0, Γj(z) = 0, and z solves (4.1.2), so z = 0 by Lemma 4.1.2. On the other hand,1 = limε→0 ‖zε‖L2

−γ(D′) = ‖z‖L2−γ(D′) = 0. This contradiction proves (4.1.12). Theorem 4.1.1 is proved.

2

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 |x| ≥ R. Thus

v =ei k|x|

|x| A(α′, α, k) + o

(1

|x|

), |x| → ∞,

x

|x| = α′.

The function A(α′, α, k) 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 scatteringamplitude. Consider three cases:

(1) A(α′, α, k) is known for all α′ ∈ S2 and all k > 0, α = α0 is fixed;

(2) A(α′, α, k0) := A(α′, α) is known for all α′, α ∈ S2, k = k0 > 0 is fixed,

(3) A(α′, α0, k0) is known for all α′ ∈ S2, α = α0 and k = k0 are fixed.

In case (1) the uniqueness theorem for IOSP was proved by M. Schiffer (1964), who assumed a priorithat the boundary condition is the Dirichlet one. In case (2) the uniqueness theorem for IOSP wasproved by A. G. Ramm (1985) (see [R83]), who also proved that the boundary condition is uniquelydetermined by the scattering amplitude. In case (3) the uniqueness theorem for IOSP is still an openproblem, although it is a trivial remark that this uniqueness theorem holds if one assumes a priorithat the obstacle is sufficiently small ( cf [R139], [R203]). The case of penetrable obstacles had alsobeen considered: both the direct scattering problem ( e.g., see [R71], [R83]), and the inverse scatteringproblem (see [RPY]) have been studied.

Let us prove uniqueness theorem for IOSP. Throughout we take n = 3, but the proofs and the resultsremain valid in Rn, n ≥ 2. By S2 we denote an open subset of S2, and S := ∂D.

Theorem 4.2.1. Assume that D is a bounded domain with finite perimeter, and A(α′, α0, k) := A(α′, k)is known ∀α′ ∈ S2 and ∀k ∈ (a, b), 0 ≤ a ≤ b < ∞. Then S and the boundary condition on S are uniquelydetermined. If Γ = Γ2 we assume additionally that i1 is compact, and if Γ = Γ3 we assume that i1 andi2 are compact.


Proof. If one proves that S is uniquely defined by the data, then the boundary condition is also uniquelydefined. Indeed, we prove below that the data determine uniquely the scattering solution u(x, α0, k) inD′. If S is known, then one can check if u = 0 on S, or uN = 0 on S, or calculate uN

u= −h(s) on S,

thus determining uniquely the boundary condition (Γj , j = 1, 2, 3) on S. Let us prove that S is uniquelydetermined by the data.

Assume the contrary. Let S1 and S2 generate the same data. Outside of the ball B(R), whichcontains D12 = D1 ∪D2, the scattering solutions u1 and u2 are identical. This follows from

Lemma 4.2.2. If u solves equation (*) (∆ + k2) v = 0 in B′(R) and v = o( 1|x| ) as |x| → ∞, then v = 0

in B′(R), and in the connected component of the domain D′, containing B′(R), in which v solves (*).

Proof. The last statement of Lemma 4.2.2 follows from the unique continuation theorem for elliptic equa-tions. Let us prove the first one. Write the solution to (*) as v =

∑∞`=0[C

+` h

+` (k r) + C2

`h−` (k r)]Y`(x

0),

x0 := xr , r = |x|, C±

` are constants, h±` are Hankel functions, h±` (k r) ∼ e±i k r

r as r → ∞. If v = o( 1r ),

then |C+` e

i k r + C−` e

−i k r|2 = o(1) as r → ∞. Thus C±` = 0, and v = 0. Lemma 4.2.2 is proved. 2

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 solutionsof the equation [∆ + k2 − q(x)]v = 0 where q = o( 1

|x| ) as |x| → ∞.

Indeed, if S1 and S2 generate the same data, then v = u1 −u2 = o( 1|x|), so, by Lemma 4.2.2, u1 = u2

inD′12. Let D3 be a connected component ofD12\D12. Without loss of generality assume that D3 ⊂ D1,

for example. Then u2 solves in D3 equation (*) of Lemma 4.1.2 and u2 satisfies one of the homogeneousboundary condition Γj(j = 1, 2, 3). On S2 ∩ D3 the u2 satisfies Γj by definition, and on S1 ∩ D3 the u2

satisfies Γj because u2 = u1 in D′12. Since u2 solves (*) in D3 for any k ∈ (a, b), we get a contradiction,

because∫D3u2(x, k)u2(x, κ)dx = 0 (†) for k 6= κ. This is proved as usual:

0 =

∫

D3

[u2(x, κ)

(∆ + k2

)u2(x, k) − u2(x, k)

(∆ + κ2

)u2(x, κ)

]dx

=(k2 − κ2

) ∫

D3

u2(x, κ)u2(x, k)dx+

∫

S3

[u2(s, κ)u2N (s, k) − u2N (s, κ)u2(s, k)

]ds,

where S3 = ∂D3. If Γ = Γj, j = 1, 2, 3, then the surface integral vanishes and (†) holds. Thus weconstructed a continuum of orthogonal, not identically vanishing, functions in L2(D3). Since L2(D3)is a separable space, this is a contradiction, which proves that the assumption S1 6= S2 is wrong.Theorem 4.2.1 is proved. 2

Remark 4.2.3. Our proof follows the ideas of [R171]. The original proof of M. Schiffer relied on thefact that the Dirichlet Laplacian in any bounded domain has a discrete spectrum, so it is a contradictionto have every k2 ∈ (a, b) an eigenvalue of the Dirichlet Laplacian in D3. But if the Neumann Laplacianis considered then there are bounded domains in which this Laplacian has continuous spectrum. To avoidthis difficulty, we rely on the separability of L2(D3).

Remark 4.2.4. The scattering amplitude A(α′, α, k) corresponding to a bounded Lipschitz obstacle is ameromorphic function of k, and an analytic function of α′ and α on the variety M := z : z ∈ C3, z ·z =

1, z · ζ :=∑3j=1 zjζj [R83]. Therefore the knowledge of A(α′, α, k) on open subsets of S2 ×S2 , however

small, and on an open set (a, b), determines A uniquely on S2 × S2 × (0,∞).

Consider now IOSP with the fixed-frequency data A(α′, α, k0) := A(α′, α).

Theorem 4.2.5. Under the assumption of Theorem 4.2.1 the fixed-frequency data A(α′, α, k0) :=A(α′, α)∀α′ ∈ S2

1 , ∀α ∈ S22 , k0 > 0 is fixed, S2

j , j = 1, 2, are open subsets of S2, however small,determine S and the boundary condition Γj uniquely.

4.2. INVERSE OBSTACLE SCATTERING PROBLEMS 181

Proof. As in the proof of Theorem 4.2.1, it is sufficient to prove that S is uniquely determined by thedata. We need

Lemma 4.2.6. If S1 and S2 generate the same data A(α′, α), then G1(x, y, k0) = G2(x, y, k0) ∀x, y ∈D′

12.

Proof. The function v := G1 −G2 solves the homogeneous Helmholtz equation in D′12, and v = o( 1

|x| ) if

A1 = A2 := A ∀α′, α. The conclusion of Lemma 4.2.6 follows now from Lemma 4.2.2. 2

If S1 and S2 generate the same data A(α′, α), then G1(x, y, k0) = G2(x, y, k0) ∀x, y ∈ D′12 by

Lemma 4.2.6. If S1 6= S2, then take a point x ∈ S3 ∩ S1 (the notations are the same as in the proofof Theorem 4.2.1) and x 6∈ S2. Then Γj G1(x, y, k0) = 0, but Γj G2(x, y, k0) → ∞ as y → x, y ∈ D′

12.Since G1(x, y, k0) = G2(x, y, k0) ∀x, y ∈ D′

12, one gets a contradiction, which proves that S1 = S2.Theorem 4.2.5 is proved. 2

Remark 4.2.7. If the data are A(α′) := A(α′, α0, k0) ∀α′ ∈ S2, α = α0 and k = k0 > 0 are fixed, then itis not known whether these data determine S uniquely. However, if one knows a priori that the obstacleis sufficiently small, so that k2

0 is not an eigenvalue of the Laplacian in D, corresponding to the boundarycondition Γj, then, as in the proof of Theorem 4.2.5, one gets S1 = S2, so the obstacle is determineduniquely by A(α′) under this a priori assumption. Alternatively, if the obstacle has any given size,say its diameter is not greater than d, then sufficiently many data A(α′, αm, kp), m = 1, 2, . . . ,M, p =1, 2, . . . , P , determine D uniquely. One should take M +P greater than the upper bound on the numberNk2 of the eigenvalues Nk2 = #λj ≤ k2 of the Laplacian in D, and then the uniqueness of the solutionto 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, itis an open surface in R3. The proof of Theorem 4.2.1 and Theorem 4.2.5 covers this case as well: sucha region is uniquely defined by the data of each of these theorems.

Indeed, suppose that there are two regions D1 6= D2, and A1(α′, α) = A2(α

′, α) ∀α′, α. ThenG1(x, y) = 0, x ∈ D1 \D2, G1 = G2 ∀x, y 6∈ D12, and 0 = G1(x, y) = G2(x, y) → ∞ as y → x ∈ D1 \D2.This contradiction shows that D1 = D2.

If A1(α, k) = A2(α, k) ∀α ∈ S2, ∀k > 0, then

u1

(x, α0, k

):= u1(x, k) = u2(x, k) ∀x ∈ D′

12, ∀k > 0. (4.2.0)

Let γ1 be the edge of D1. Then |∇u| → ∞ as x → γ. This and the relation (4.2.0) imply γ1 = γ2,if A1(α, k) = A2(α, k)∀α ∈ S2, ∀k > 0. If γ1 = γ2 but D1 6= D2, then there is a bounded domain Dwith the boundary γ ⊂ (D1 ∪D2). Since D1 6= D2, the region D has interior points, and the proof ofTheorem 4.2.1 applies. This proof yields a contradiction, which shows that D1 = D2.

Let us consider the following inverse scattering problem. Let

Lu := ∇(a(x)∇u) + q(x)u = 0 in R3, a(x) = a+ in D, (4.2.1)

a(x) = a− := a0 in D′, q(x) = k2a+ := q+ in D, q(x) = k20a0 := q− in D′, where a±, q±, k0 and k are

positive constants, a+ 6= a−, and the transmission boundary conditions hold on S:

a+ u+N = a0 u

−N , u+ = u− on S, (4.2.2)

where u+, u+N (u−, u−N ) denote limiting values on S from D(D′), N is the outer normal to S pointing into

D′, S is Lipschitz. In D′ we assume that u satisfies the radiation condition: u = u0 +A(α′, α, k0)eik0r

r +o(1r), r = |x| → ∞, α′ = x

r, α ∈ S2, u0 := ei k0 α·x, k0 > 0 is fixed, A(α′, α, k0) := A(α′, α).

The inverse scattering problem consists of recovery of S, a+ and k from the scattering data A(α′, α),where a0 and k0 are known. Our main result is


Theorem 4.2.9. [RPY] The data A(α′, α) ∀α′, α ∈ S2, determine S, a+ and k uniquely.

Proof. Assume there are two sets (Sj , a+j , k

2j ), j = 1, 2, which generate the same data A(α′, α), uj are

the corresponding solutions to (4.2.1)- (4.2.2). Then u1 = u2 in D′12, and the weak formulation of the

problems (4.2.1)-(4.2.2) imply:∫

D12

(a1∇u1 · ∇φ− q1 u1 φ

)dx−

∫

D12

(a2∇u2 · ∇u− q2 u2 φ

)dx = 0 ∀φ ∈ C∞

0 (R). (4.2.3)

Define

w(y) := −∫

D12

(a1∇G1(x, y) · ∇φ− q1G1(x, y)φ)dx−∫

D

(a2∇G2(x, y) · ∇φ− q2G2 φ)dx,

where Gj is the resolvent kernel corresponding to the operator Lj , j = 1, 2, of the problem (4.2.1)-(4.2.2),

Gj(x, y) =ei k0|y|

4π|y| uj(x, α′) + o

(1

|y|

)as |y| → ∞,

y

|y| = −α′. (4.2.4)

From (4.2.4) and Lemma 4.2.2 it follows that w = 0 in D′12.

If S1 6= S2 then there is a point t ∈ S1 ∩D′2 and a ball B := B(t, r) ⊂ D′

2 such that∫

D12∩B

[a1(x)∇xG1(x, y) · ∇xφ− a2(x)∇xG2(x, y) · ∇xφ

]dx

= −∫

D12\B

[a1(x)∇xG1 · ∇φ− a2(x)∇xG2 · ∇φ

]dx

+

∫

D12

(q1G1 − q2G2

)φdx, ∀y ∈ D′

12 ∩B.

(4.2.5)

The integrals on the right-hand side of (4.2.5) are bounded as y → t, therefore so is the integral on theleft-hand side of (4.2.5). Choose φ := η(x)|x− y|−1, η ∈ C∞

0 (R3) is a cut-off function, 0 ≤ η ≤ 1, η = 1in a neighborhood of D12 and vanishes outside B(R), D12 ⊂ B(R). Let (x1, x2, x3) be local coordinatesin B = B(t, r), with the origin at the point t and x3 axis directed along the normal to S1 at the pointt. Let y = (0, 0, y3). If y → x then the author has proved ([R180]) that

G1(x, y) =1

4π a0

[1

r+b

R

] (1 + o(1)

)as y → x, y ∈ D′

12 ∩B, (4.2.6)

where b := (a0 − a+1 )(a0 + a+

1 )−1, r := |x− y|, R := [(x1 − y1)2 + (x2 − y2)

2 + (|x3|+ |y3|)2]1/2, and onecan differentiate (4.2.6). If x ∈ D12 and y ∈ D′

12, then y3 > 0, x3 < 0, and

∇xG1(x, y) ∼1

4π a0

[− x− y

|x− y|3 − bA

R3

], A :=

(x1 − y1, x2 − y2,−

∣∣x3

∣∣− y3

).

Thus

∇xG1(x, y) · ∇φ ∼ 1

4π a0r4[1 +

b r

R] as y → x, y3 > 0, x3 < 0,

and, as r → 0, one gets:

G2(x, y) ∼1

4π a0r, so a2∇xG2 · ∇xφ ∼ a2

4π a0r4.

Therefore, as y → t, x→ t, x3 < 0, y3 > 0, the left-hand side of (4.2.5) behaves like

1

4π a0

∫

D12∩B

[a+1 + a+

1 b− a0

]dxr4

=1

4π

∫

D12∩B

(a+1 − a0

)dxr4

= O(1), (4.2.7)

For (4.2.7) to hold it is necessary that a+1 = a0. This contradiction shows that S1 = S2. In deriving

(4.2.7), one takes into account that:

4.3. STABILITY ESTIMATES FOR THE SOLUTION TO IOSP 183

(1) a2 = a0 in B because B ⊂ D′2, and

(2) r = R if x3 < 0 and y3 > 0.

Let us prove that a+ is uniquely determined by the scattering data. We have already proved thatS1 = S2, so D12 = D. If a+

1 6= a+2 , then, as y → t, the left-hand side of (4.2.5) is bounded and behaves

like1

4πa0

∫

D∩B

[a+1

(1 + b1

)− a+

2

(1 + b2

)]dxr4

= O(1), where bj =a0 − a+

j

a0 + a+j

.

Thus a+1 (1 + b1) = a+

2 (1 + b2), and using the formulas for bj one concludes that a+1 = a+

2 , as claimed.Finally, let us prove that k2 is also uniquely determined by the scattering data.If this is not true, then q1 6= q2. We derive a contradiction from this assumption.Let p := q2 − q1, w := u1 − u2. Subtracting from the equation L1 u1 = 0, equation L2 u2 = 0, one

gets

L1w = p u2 in D, w = wN = 0 on S := ∂D, and∫

D

p φu2dx = 0 ∀φ ∈ N(L1

):=φ : φ ∈ H2(D), L1φ = 0

.

This orthogonality relation follows from the formula∫D L1 wφdx =

∫D wL1 φdx = 0, where the bound-

ary integrals vanish because w = wN = 0 on S. The set of products φu2(x, α)∀φ∈N(L1)∀α∈S2 iscomplete in L2(D) by property C for the pair L1, L2 (see Section 5.6). Therefore p = 0, and Theo-rem 4.2.9 is proved. 2

Exercise (cf [R174]). Let (4.1.1) hold and assume that the boundary S is a union of two connectedsets S1 and S2, the normal derivative uN = 0 on S1 and vN = w on S2. The total cross-sectionσ(w) := σ :=

∫S2 |A(α′, α, k)|2dα′, α ∈ S2 is fixed. The function w can be treated as a control function.

Prove that w ∈ L2(S2) can be chosen so that the cross-section σ becomes as small as one wishes,that is, infw∈L2(S2) σ(w) = 0.

This means that one can make an obstacle (say, an aircraft) practically invisible for any given fixeddirection α 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 Dj , j = 1, 2, that is, their boundaries Sj can berepresented by the equations r = fj(α), r = |x|, α = x

|x| , j = 1, 2. Assume that 0 < c0 ≤ fj ≤ C0 ∀α ∈S2, where c0 and C0 are constants. If x3 = φ(x1, x2) is the local equation of a surface S := ∂D, thenwe assume that ‖φ‖C2,λ ≤ mλ, where λ ∈ (0, 1]. The set of all such star-shaped surfaces S (or obstaclesD) is denoted by γλ. Let ρ := maxsupx∈S1

infy∈S2 |x − y|, supy∈S2infx∈S1 |x − y| be the Hausdorff

distance between obstacles D1 and D2, Sj := ∂Dj , S12 := ∂D12. Let D1 be a connected component of

D1 \D2, S1 = ∂D1 = S′1 ∪ S2, S

′1 ⊂ S1, S2 ⊂ S2, and S′

1 ⊂ (S1 ∩D′2). Suppose that ρ = |x0 − y0|, where

x0 ∈ S1, y0 ∈ S2. Assume that supα′,α∈S2 |A1(α′, α) − A2(α

′, α)| ≤ δ, k = k0 > 0 is fixed. Our basicresult is (see [R162]):

Theorem 4.3.1. Under the above assumptions one has ρ ≤ c1( log | log δ|

| log δ|)c2

, c1, c2 > 0 are constants

independent of δ.

Suppose A(α′, α) is the scattering amplitude corresponding to an obstacle D ∈ γλ. Let us define an

algebraic variety M := Mk :=θ : θ ∈ C3, θ · θ = k2

, θ · w :=

∑3j=1 θjwj. How does one reconstruct

S from A(α′, α)? We prove


Proposition 4.3.2. There exists a function vε(α, θ) ∈ L2(S2), such that

− 4π limε→0

∫

S2

A(θ′, α)vε(α, θ)dα = −ξ2

2χD(ξ), where

χD(ξ) :=

∫

D

e−i ξ·xd ξ, χD(x) is the indicator of D,

θ′, θ ∈M, θ′ − θ = ξ, A(θ′, α) :=

∞∑

`=0

A`(α)Y`(θ′), A`(α) :=

∫

S2

A(α′, α)Y`(α′)dα′.

Remark 4.3.3. It is an open problem to construct an algorithm for calculating vε(α, θ) given A(α′, α).Such an algorithm we construct for inverse potential scattering theory in Chapter 5.

Proof of Theorem 4.3.1. First we prove that limδ→0 ρ(δ) = 0. Then we prove that |u2| ≤ c ρ in D1. Byc we denote various positive constants. Next, we prove that

∣∣u2(x)∣∣ ≤ cερ

c′

if dist(x, S′

1

)= O(ρ). (4.3.1)

Here ε = e−cN(δ), N (δ) := log δlog | log δ| .

Estimate (4.3.1) implies the conclusion of Theorem 4.3.1Let us go through the steps of the proof. By uj we denote the scattering solution corresponding to

Dj .

Step 1. Assume that ρn := ρ(δn) ≥ c > 0, δn → 0, and let Sj n, j = 1, 2, be the corresponding boundaries.Since Sj n ∈ γλ, one can select a subsequence, denoted again by Sj n, which converges Sj n → Sj (asn → ∞) in C2,µ, 0 < µ < λ, and ρ(D1, D2) ≥ c > 0. Since Aj(α

′, α) depends on Sj continuously, itfollows from |A1n − A2n| ≤ δn → 0 that A1(α

′, α) = A2(α′, α). By Theorem 4.2.5 of Section 4.2, one

gets S1 = S2, so ρ(D1, D2) = 0 in contradiction with ρ(D1, D2) ≥ c > 0. Thus limδ→0 ρ(δ) = 0.

Step 2. Since u2 = 0 on S2, one has |u2(x, k)| ≤ max |∇u2| dist(x, S2) ≤ c ρ for ∈ D1.

Step 3. Let v(x) = v := u1 − u2, d := dist(x, S′1), ε := c e−cN(δ), N (δ) :=

| log δ|log | log δ| . From the results of

Section 5.4 it follows that |v(x)| ≤ ε for |x| ≥ R. Let c3ρ ≤ d ≤ c4ρ, c3 > 0, x ∈ D′12.

Let us prove the estimate: ∣∣v(x)∣∣ ≤ c1 ε

c dc′

, x ∈ D′12.

One has v(x) =∫D12

g(x − y)Hdy, g = ei k|x|

4π|x| , H ∈ C2,µ(D12). The function |x− y| = [r2 − 2r|y| cos θ +

|y|2]1/2 := p = p(r), r = |x|, can be continued analytically on the complex plane z = r ei ψ to the sectorsφ := z : |arg z| < φ, if r2 − 2z|y| cos θ + |y|2 6= 0 for z ∈ sφ. We assume Re p|Im r=0 ≥ 0, Im p ≥ 0,so that |g(x− y)| ≤ c when z ∈ sφ. Then ‖v‖C1(D′

12) ≤ c, and v extends from D′12 into D12 as C1(R3)

function. Take a point x = r x0, r = |x|, x ∈ S′1, and let K be a cone with vertex x, K ∈ D′

12, withopening angle 2φ. Such a cone does exist since S1 is sufficiently smooth. One has supr≥R |v(r)| ≤ ε and

supz∈sφ|v(z)| ≤ c. By the two constants theorem [E], p.296, one gets |v(z)| ≤ c εh(z), where h(z) is the

harmonic measure of the set ∂sφ \ L with respect to the domain sφ \ L, L := [R,∞), ∂sφ is the unionof the two rays which form the boundary of the sector sφ and of the ray L.

Let us prove the estimate:h(z) = c dc

′

[1 + o(1)], (4.3.2)

as z → r(x0) along the real axis, d := |z − r(x0)|, c and c′ are positive constants. Let ζ := z − r(x0),|ζ| = d. Let us map conformally the sector sφ onto the half-plane Re ζ ≥ 0 using the map w = ζc

′

,

c′ := π2φ . Then L is mapped onto the ray [Rc

′

,∞) and h(z) = h(ζc′

) ( see [E], p.293). By the Hopf

lemma ([E], p.34), c := ∂h∂w |w=0 > 0, h(0) = 0, so h(ζc

′

) = c ζc′

[1 + o(1)] as ζ → 0. Thus (4.3.2) follows,

and |v(x)| ≤ c1 εc dc′

, x ∈ D′12, c3ρ ≤ d ≤ c4ρ.

4.3. STABILITY ESTIMATES FOR THE SOLUTION TO IOSP 185

On the other hand,

∣∣v(x)∣∣ =

∣∣∣∣v(r(x0))

+ ∇v ·(x− r

(x0))∣∣∣∣ = O(ρ) ≤ c1ε

c5ρc′

. (4.3.3)

Thus log ρ ≤ c ρc′

log ε, so ρc′

log( 1ρ )

≤ c6(log 1ε )

−1, 0 < ρ, ε < 1, c6, c′ > 0 are constants. This implies

ρ ≤ ( cln 1

ε

)(1+w)/c′ , where limε→0w = 0. Finally let us explain (4.3.3) in more detail. One has v =

u2 − u1 = u2 on S′1, because u1 = 0 on S1. Also, u2 = 0 on S2, |∇u2| ≤ c, so |u2| = O(ρ) on S′

1 and ina O(ρ) neighborhood of S′

1.Since |∇v| ≤ |∇u1| + |∇u2| ≤ c, one gets

∣∣v(x)∣∣ = O(ρ), for x ∈ D′

12, dist(x, S′

1

)= O(ρ).


Remark 4.3.4. For convenience of the reader we recall the known results about harmonic measure andtwo-constants theorem (see [E] for details). Let D be a domain on the complex plane, S = ∂D, E ⊂ S,z ∈ D. Define the harmonic measure of E relative to D at the point z, denoted h(z, E,D), as the valueof harmonic function in D, which is equal to 1 on E and equal to zero on S \E, 0 ≤ h(z, E,D) ≤ 1. Ifa function w := w(z) maps conformally the domain D onto a domain D′, and the set E is mapped ontoE′, then h(z, E,D) = h(w(z), E′, D′). The theorem about two constants says: if f(z) is holomorphic inthe domain D with the boundary S, E ⊂ S, and lim supz→ζ |f(z)| < m, ζ ∈ E, lim supz→ζ |f(z)| < M ,

ζ ∈ S \E then |f(z)| < mh(z,E,D)M1−h(z,E,D).

Let us now prove Proposition 4.3.2. First we prove:

Lemma 4.3.5. ([R139, p. 183]). If u(x, α, k), α ∈ S2, k = const > 0, is the scattering solution, thenthe set uN (s, α, k)∀α∈S2 is complete in L2(S).

Proof of Lemma 4.3.5. If the conclusion is not true, then there is a p(s) ∈ L2(S) such that 0 =∫Sp(s)uN (s, α, k)ds ∀α ∈ S2. This implies 0 =

∫Sp(s)GN (s, y, k)ds ∀y ∈ D′, where G(x, y, k) is

the Green’s function (resolvent kernel of the Dirichlet Laplacian in D′) see Lemma 4.1.3). SinceGN (s, y, k) → δS (s − σ) as y → σ ∈ S, one gets p(s) = 0 on S. Here δS(s − σ) is the delta-function onS. The Lemma is proved. 2

From Lemma 4.3.5 it follows that there exists a function νε(α, θ) ∈ L2(S2) such that∥∥∥∥∫

S2

uN (s, α, k)νε(α, θ)dα− ∂ei k θ·s

∂Ns

∥∥∥∥L2(S)

< ε,

where ε > 0 is arbitrary small, and θ ∈M := θ ∈ C3, θ · θ = 1 is an arbitrary fixed vector. One has

−4πA(θ′, α) =

∫

S

e−i k θ′·suN (s, α, k)ds.

Multiplying this equation by νε(α, θ) and integrating over S2 with respect to α, one gets:

4π limε→0

∫

S2

A(θ′, α)νε(α, θ)dα =

∫

S

e−i k θ′·s∂e

i k θ·s

∂Nsds.

One has∫

S

e−i k θ′·s ∂e

i k θ·s

∂Nsds =

1

2

∫

S

∂

∂Nse−i k(θ

′−θ)·sds =1

2

∫

D

∆e−i ξ·xdx

= −ξ2

2χD(ξ), ξ := k(θ′ − θ), χD :=

∫

D

e−i ξ·xdx.

Proposition 4.3.2 is proved. 2


A different type of stability estimates for obstacle scattering problem are obtained in [R164], [R7].There one solves a potential scattering problem with potential tχD(x) where t > 0 is a parameter,t → ∞, and χD(x) is the characteristic function of the domain D, the obstacle. The limit, as t→ ∞, ofthe scattering solution for this problem, is the scattering solution of the obstacle scattering problem withthe Dirichlet boundary condition on the boundary S of D, and the scattering amplitude A(α′, α, t) forthe potential scattering problem tends to the scattering amplitude AS(α′, α) for the obstacle scatteringproblem as t → ∞. In [R164] one finds the estimates of the rate of convergence as t→ ∞. For example,

supα,α′∈S2 |A(α′, α, t)− AS(α′, α)| ≤ ct−14 , where c = const > 0 does not depend on t.

4.4 High-frequency asymptotics of the scattering amplitudeand inverse scattering problem

Let us assume that D is a bounded, strictly convex obstacle with C2 boundary S. Let β be a unit vector,S+ = s : β · s ≥ 0, s ∈ S be the illuminated part of S by the parallel straight lines in the direction ofβ, and S− = S \ S+ be the shadowed part. Define the support function as a(β) = mins∈S+ (β, s), this isthe distance from the origin to the tangent plane to S, orthogonal to β, and closest to the origin. Theequation of this plane is β · x = a(β). The origin is chosen so that S+ is not empty.

The surface S is the envelope of the set of these tangent planes. Therefore the equation of S is

xj =∂a(β)

∂βj, j = 1, 2, 3. (4.4.1)

One has to understand the derivatives in (4.4.1) as follows. The function a(β) is defined originally on S2,because |β| = 1. We extend this function to R3 as a homogeneous function of order 1, so that a(t β) =

t a(β) ∀t > 0, t = const. Thus∑3j=1

∂a∂βj

βj = a(β), by Euler’s theorem. The differentiation in (4.4.1) is

the differentiation of the extended to R3 function a(β). For example, if a(β) = 1 = (β12 + β2

2 + β32)1/2,

then ∂a∂βj

=βj√

β21+β2

2+β23

, and therefore the relation ∂a(β)∂βj

= 0 is not correct.

If one recovers the function a(β) from the scattering data then the surface S is uniquely and con-structively recovered by formula (4.4.1).

Let us show how to find a(β) from the data A(α′, α, k) known for k → ∞.We assume the Dirichlet condition on S for definiteness. In this case

−4π A(α′, α, k) =

∫

S

e−i k α′·suN (s, α, k)ds,

where u(x, α, k) is the scattering solution. Denote uN := µ, u0 := ei k α·x. Then u = u0 −∫Sg(x, s)µds,

g := ei k|x−y|

4π|x−y| , and taking the normal derivative at the boundary yields the equation for µ : µ+Aµ = 2 ∂u0

∂N

on S, Aµ :=∫S∂g(s,s′)∂Ns

µds′. In the high-frequency Kirchhoff approximation, one takes µ = 0 on S−,µ = 2u0N on S+. Thus, in this approximation,

A(α′, α, k) = − i k2π

∫

S+

α ·Nsei k(α−α′)·sds.

Let us calculate this integral using the stationary phase method ([H]):

∫

D

f(x)ei kΦ(x)dx =

(2π

k

)n/2ei kΦ(x0)+

iπ4 sgnΦ′′(x0)

∣∣det Φ′′(x0)∣∣−1

2

[f(x0

)+O

(1

k

)]

as k → ∞. Here Φ(x) is a smooth real-valued function which has only one non-degenerate critical point

x0 ∈ D, i.e., ∇Φ(x0) = 0, det Φ′′(x0) 6= 0, Φ′′ := ( ∂2Φ(x)

∂xi ∂xj), sgn Φ′′(x0) := n+ − n−, where n+(n−)

4.5. REMARKS ABOUT NUMERICAL METHODS 187

is the number of positive (negative) eigenvalues of the matrix Φ′′(x0), and f(x) is a smooth function,f ∈ C∞

0 (D), x ∈ Rn.The smoothness requirements on f and Φ can be relaxed. In our case Φ(s) = (α − α′) · s. The

stationary point (critical point) of the function Φ(s) = β · s|α′ − α|, β = α−α′

|α−α′| is the point s0 ∈ S at

which β = Ns0 . In a neighborhood of this point one has Φ(s) = |α′−α|(a(β)+ 12bmjzm zj+o(|z|2) where

zm, m = 1, 2, are local coordinates on S, (0, 0) are the coordinates on S of the stationary point s0, bmjis the second differential form of the surface S at the point s0, det bmj = K, the Gaussian curvatures ofS at s0, K > 0 since S is strictly convex. Using the stationary phase formula, one gets

A(α′, α, k) = −e2i k a(β)β·α

2√K(s0)

[1 +O

(1

k β · α

)], β :=

α− α′

|α− α′| . (4.4.2)

From (4.4.2) one can find a(β) and then S by formula (4.4.1). Although the phase is determined modulo2π, the additive constant factor does not bring difficulties. For example, choose α′ = −α. Then β = α,β · α = 1, Φ(s0) = 2a(β). One has

|A(−α′, α, k)| = 1

2√K(s0)

,

andlog[− 2√K(s0)A(−β, β, k)

]= 2i k a(β) + 2mπ + o(1) as k → ∞,

where m is an integer. Therefore, the parametric equation of S is:

xj =1

2ik

∂

∂ βjlog

[− A(−β, β, k)|A(−β, β, k)|

]. (4.4.3)

Alternatively, one can determine a(β) using various α′ and α in (4.4.2) (cf [GR5], where numericalexperiments 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 aδ(β), (the noisy data) is known in place of a(β), and supβ∈S2 |aδ(β) − a(β)| ≤ δ.Then S can be stably recovered by formulas (4.4.1) if one uses stable differentiation formulas fromSection 2.5.1.

One can also estimate the error of recovery of S: this error is O(δ1/2), because of the estimatesupβ |D2a(β)| ≤ M2, which holds for a surface S ⊂ C2,λ, λ > 0. (See Section 2.5.1 for the errorestimates 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 forfinding an obstacle from the scattering data. In [R83] an analytic method is given for finding a closedsurface from the knowledge of its two principal curvatures. This is an overdetermined problem, becausethe Gaussian curvature above determines uniquely the surface under suitable assumptions. However, ifone 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 methoddeveloped in Chapter 5 for finding the potential from the scattering data. There are several parameter-fitting schemes for finding S from the scattering 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 r = f(α) is the equation of S, α ∈ S2, 0 < c1 ≤ f ≤ c2, f ∈ C1,λ, λ > 0, the data arethe values A(α′, α) ∀α′, α ∈ S2, k > 0 is fixed, the Dirichlet condition holds on S (other conditions canbe treated similarly), h := uN u is the scattering solution, uN is its normal derivative on S, ds is thesurface area element ds = ds

d αdα := w(α)dα. Consider the functional:

F (f, h) : =

∥∥∥∥4πA(α′, α) +

∫

S2

e−i k α′·β f(β)h(β, α)d β

∥∥∥∥2

+

∥∥∥∥ei k α′·β f(β) −

∫

S2

g(β f(β), θ f(θ)

)h(β, θ)d θ

∥∥∥∥2

, ∀α′, α ∈ S2,

(4.5.1)

where ‖ · ‖ = ‖ · ‖L2(S2) and g := ei k|x−y|

4π|x−y| .

Theorem 4.5.1. Functional (4.5.1) has global minimum equal to zero, and it has unique global minimizerf, h. The f component of this minimizer determines the surface S by the equation r = f(α).

Proof. Since A(α′, α) = − 14π

∫Se−i k α

′·suNds = − 14π

∫S2 e

−i k α′·βf(β)h(α, β)dβ, and u = ei k α·x −∫Sg(x, s)uN (s, α)ds, u = 0 on S, it follows that F (f, h) = 0 if r = f(α) is the equation of S and

h = uNw, where ds = w dβ. Let us prove that the global minimizer of F is unique. Assume that

F (f , h) = 0. Then A(α′, α) = − 14π

∫S2 e

−i k α′·βf (β)h d β and ei k α·βf(β) =∫S2 g(βf (β), θf (θ))h dθ.

Define

u(x, α) := ei k α·x −∫

S2

g(x, s)h d θ, s := θf (θ), θ ∈ S2, D := r, θ : r ≤ f(θ), θ ∈ S.

Then (∆ + k2)u = 0 in D′, u = 0 on ∂D, v := u − ei k α·x satisfies the radiation condition, and thecorresponding to ∂D scattering amplitude is A(α′, α) = A(α′, α) ∀α′, α. By the uniqueness theorem(Theorem 4.2.5) it follows that ∂D = S. Theorem 4.5.1 is proved. 2

Remark 4.5.2. For the inverse potential scattering problem the functional similar to (4.5.1) is

F (q, v) =∥∥4πA(α′, α) +

∫

R3

e−i k α′·yq(y)[u0 + v]dy

∥∥2

1+∥∥q(x)v(x) + q(x

∫

R3

g(x, y)q(y)(u0 + v)dy∥∥

2

(4.5.2)where ‖ · ‖1 = ‖ · ‖L2(S2×S2), ‖ · ‖2 = ‖ · ‖L2(R3×S2), q ∈ Q := q : q = q, q = 0 for |x| > a, q ∈ L2(B),k > 0 is fixed, α′, α ∈ S2, v ∈ L2(Ba), Ba := x : |x| ≤ a, a > 0 is an arbitrary large fixed number.

A theorem similar to Theorem 4.5.1 is proved in [R139]: functional (4.5.2) has the global minimumequal to zero and the unique global minimizer, q, v, and the q defined by this minimizer, solves theinverse potential scattering problem with fixed-energy data.

In [CK] and in many papers some functionals are minimized for solving inverse obstacle scatteringproblem (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 minimizerwhis is the solution of IOSP. However, solving IOSP by numerical minimization of any functional is aparameter-fitting procedure which does not allow one to get estimates of error of the solution and doesnot 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 onthe numerical verification of the inclusion of a function in the range of an operator. Numerical examples

4.6. ANALYSIS OF A METHOD FOR IDENTIFICATION OF OBSTACLES 189

are given to illustrate theoretical conclusions. Alternative methods of identification of obstacles arementioned: 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 anobstacle are proposed, which are based on a numerical verification of the inclusion of some functionf := f(α, z), z ∈ R3, α ∈ S2, in the range R(B) of a certain operator B. Examples of such methodsinclude [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 thatf 6∈ R(B) when z 6∈ D, it turns out that in any neighborhood of f , however small, there are elements fromR(B). Also, although f ∈ R(B) when z ∈ D, there are elements in every neighborhood of f , howeversmall, which do not belong to R(B) even if z ∈ D. Therefore it is not possible to construct a stablenumerical method for identification of D based on checking the inclusions f 6∈ R(B) and f ∈ R(B).

We prove below that the range R(B) is dense in the space L2(S2).

Assumption (A): We assume throughout that k2 is not a Dirichlet eigenvalue of the Laplacian inD.

Let us introduce some notations: N (B) and R(B) are, respectively, the null-space and the range ofa linear operator B, D ∈ R3 is a bounded domain (obstacle) with a smooth boundary S, D′ = R3 \D,u0 = eikα·x, k = const > 0, α ∈ S2 is a unit vector, N is the unit normal to S pointing into D′,

g = g(x, y, k) := g(|x − y|) := eik|x−y|

4π|x−y| , f := e−ikα′·z, where z ∈ R3 and α′ ∈ S2, α′ := xr−1, r = |x|,

u = u(x, α, k) is the scattering solution:

(∆ + k2)u = 0 in D′, u|S = 0, (4.6.1)

u = u0 + v, v = A(α′, α, k)eikrr−1 + o(r−1), as r → ∞, xr−1 = α′, (4.6.2)

where A := A(α′, α, k) is called the scattering amplitude, corresponding to the obstacle D and theDirichlet boundary condition. Let G = G(x, y, k) be the resolvent kernel of the Dirichlet Laplacian inD′:

(∆ + k2)G = −δ(x − y) in D′, G|S = 0, (4.6.3)

and G satisfies the outgoing radiation condition.

If

(∆ + k2)w = 0 in D′, w|S = h, (4.6.4)

and w satisfies the radiation condition, then ([R83]) one has

w(x) =

∫

S

GN (x, s)h(s)ds, w = a(α′, k)eikrr−1 + o(r−1), as r → ∞, xr−1 = α′. (4.6.5)

We write a(α′) for a(α′, k), and

a(α′) := Bh :=1

4π

∫

S

uN (s,−α′)h(s)ds, (4.6.6)

as follows from Lemma 4.1.3.

One can write the scattering amplitude as:

A(α′, α, k) = − 1

4π

∫

S

uN (s,−α′)eikα·sds. (4.6.7)

The following claim is proved in [Kir1]:

Claim 10. f := e−ikα′·z ∈ R(B) if and only if z ∈ D.


Proof of the claim. Our proof is based on the results in [R83].a) Let us assume that f = Bh, i.e., f ∈ R(B), and prove that z ∈ D. Define p(y) := g(y, z) − ψ(y),

where ψ(y) :=∫SGN (s, y)h(s)ds. The function p(y) solves the Helmholtz equation (4.6.4) in the region

|y| > |z| and p(y) = o( 1|y| ) as |y| → ∞ because of (4.6.7) and of the relation Bh = f . Therefore (see

Lemma 4.1.2) p = 0 in the region |y| > |z|. Since ψ is bounded in D′ and g(y, z) → ∞ as y → z, we geta contradiction unless z ∈ D. Thus, f ∈ R(B) implies z ∈ D.

b) Let us prove that z ∈ D implies f ∈ R(B). Define ψ(y) :=∫SGN (s, y)g(s, z)ds, and h := g(s, z).

Then, by Green’s formula, one has ψ(y) = g(y, z). Taking |y| → ∞, y|y| = α′, one gets f = Bh, so

f ∈ R(B). The claim is proved. 2

Consider B : L2(S) → L2(S2), and A : L2(S2) → L2(S2), where B is defined in (4.6.6) andAq :=

∫S2 A(α′, α)q(α)dα.

Theorem 4.6.1. The ranges R(B) and R(A) are dense in L2(S2).

Proof. Recall that Assumption (A) holds. It is sufficient to prove that N (B∗) = 0 and N (A∗) = 0.Assume 0 = B∗q =

∫S2 uN (s,−α′)qdα′, where the overline stands for complex conjugate. Taking

complex conjugate and denoting q by q again, one gets 0 =∫S2 uN (s,−α′)qdα′. Define w(x) :=∫

S2 u(x,−α′)qdα′. Then w = wN = 0 on S, and w solves equation (4.6.1) in D′. By the unique-ness of the solution to the Cauchy problem, w = 0 in D′. Let us derive from this that q = 0. Onehas w = w0 + V , where w0 :=

∫S2 e

−ikα′·xqdα′, and V :=∫S2 v(x,−α′, k)qdα′ satisfies the radiation

condition. Therefore, w0(x) = 0 in D′, as follows from the Lemma 4.6.2 proved below. By the uniquecontinuation, w0(x) = 0 in R3, and this implies q = 0 by the injectivity of the Fourier transform. Thisproves the first statement of Theorem 4.6.1. Its second statement is proved below. 2

Let us now prove the Lemma, mentioned above. We keep the notations used in the above proof.

Lemma 4.6.2. If w = w0 + V = 0 in D′, then w0 = 0 in D′.

Proof. The idea of the proof is simple: since w0 does not satisfy the radiation condition, and V satisfiesit, one concludes that w0 = 0. Let us give the details. The key formula is ([R83], p.54):

∫

S2

eikα·βrq(β)dβ =2πi

k[gq(−α) − gq(α)] + o(

1

r), r → ∞, (4.6.8)

where g := eikr/r, and one assumes q ∈ C1(S2).If r := |x| → ∞, then, by Lemma 4.6.2, assuming q ∈ C1(S2), and using the relation w = w0 +V = 0

in D′, one gets q(α) = 0 for all α ∈ S2. Thus, Lemma 4.6.2 is proved under the additional assumptionq ∈ C1(S2). If q ∈ L2(S2), then one uses a similar argument in a weak sense, i.e., with x := rβ, oneconsiders the inner product in L2(S2) of w0(rβ) and a smooth test function h ∈ C∞(S2), and appliesLemma 4.6.2 to the function

∫S2 e

−ikα′·βrhdβ. Then, using arbitrariness of h, one concludes that q = 0as an element of L2(S2). Lemma 4.6.2 is proved. 2

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:A(α, β) = A(−β,−α), one gets an equation:

∫

S2

A(α, β)hdβ = 0, (4.6.9)

where h = q(−β). Define w(x) :=∫S2 u(x, β)hdβ. Then w = w0 + V , where w0 :=

∫S2 e

ikβ·xhdβ, and

V :=∫S2 v(x, β)hdβ satisfies the radiation condition. Equation (4.6.9) implies that V = o( 1

r ) as r → ∞.

Since function V solves equation (4.6.1) and V = o( 1r ), one concludes that V = 0 in D′, so that w = w0


in D′. Thus, w0|S = w|S = 0. Since w0 solves equation (4.6.1) in D and w0|S = 0, one gets, usingAssumption (A), that w0 = 0 in D. This and the unique continuation property imply w0 = 0 in R3.Consequently, h = 0, so q = 0, as claimed. Theorem 4.6.1 is proved. 2

Remark 4.6.3. In [CK] the 2D inverse obstacle scattering problem is considered. It is proposed to solvethe equation (1.9) in [CK]: ∫

S1

A(α, β)γdβ = e−ikα·z, (4.6.10)

where A is the scattering amplitude at a fixed k > 0, S1 is the unit circle, and z is a point in R2.If γ = γ(β, z) is found, the boundary S of the obstacle is to be found by finding those z for which||γ|| := ||γ(β, z)||L2(S1) is maximal. Assuming that k2 is not a Dirichlet or Neumann eigenvalue of theLaplacian 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 ε > 0 there exists a function γ ∈ L2(S1), such that

limz→S

||γ(β, z)|| = ∞, (4.6.11)

and ( see [CK], p.386),

||∫

S1

A(α, β)γdβ − e−ikα·z|| < ε. (4.6.12)

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-posedequations (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 anyz ∈ R2, infinitely many γ with arbitrary large ||γ||, such that (4.6.12) holds. Therefore it is not clearhow 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 ε > 0 and every y0 ∈ D thereexists a function γ such that inequality (4.6.12) (which is (2.8) on p.386 of [CK]) holds and ||γ|| → ∞as y0 → ∂D. Such a γ is used in [CK] in a ”simple method for solving inverse scattering problem”.However, there exist infinitely many γ such that inequality (4.6.12) holds and ||γ|| → ∞, regardless ofwhere y0 is. Therefore it is not clear how one can use the method proposed in [CK] for solving the inversescattering 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) pro-posed in papers [CCM], [CK], [Kir1] produce numerically results which are inferior to these obtained bythe linearized Born-type inversion. There is no guarantee of any accuracy in recovery of the obstacleby 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 ofstrictly convex obstacles from the scattering amplitude. This method allows one to recover the supportfunction of the obstacle, and the boundary of the obstacle is obtained from this function explicitly. Errorestimates for this method are obtained in the case when the data are noisy [R83], p.104. This methodis 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 recoverthe 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 itsconvex hull is found, by using Modified Rayleigh Conjecture (MRC) method, introduced in [R205], or bya parameter-fitting method.

In Proposition 4.3.2 (see also [R162]) a formula for finding an acoustically soft obstacle from thefixed-frequency scattering data is given. It is an open problem to develop an algorithm based on thisformula.


The numerical implementation of the Linear Sampling Method (LSM) suggested in [CK] consists ofsolving a discretized version of ∫

S1

A(α, β)γdβ = e−ikα·z, (4.6.13)

where A is the scattering amplitude at a fixed k > 0, S1 is the unit circle, α ∈ S1, and z is a point onR2.

Let F = Aαi, βj, i = 1, . . . , N , j = 1, . . . , N be the square matrix formed by the measurementsof the scattering amplitude for N incoming, and N outgoing directions. then the discretized version of(4.6.13) is

Fg = f , (4.6.14)

where the vector f is formed by

fn =ei

π4√

8πke−ikαn·z, n = 1, . . . , N, (4.6.15)

see [BLW] for details.Denote the Singular Value Decomposition of the far field operator by F = USV H . Let sn be the

singular values of F , ρ = UHf , and µ = V H f . Then the norm of the sought function g is given by

‖γ‖2 =

N∑

n=1

|ρn|2s2n

. (4.6.16)

A different LSM is suggested in [Kir1]. In it one solves

(F ∗F )1/4g = f (4.6.17)

instead of (4.6.14). The corresponding expression for the norm of γ is

‖γ‖2 =

N∑

n=1

|µn|2sn

. (4.6.18)

A detailed numerical comparison of the two LSMs and the linearized tomographic inverse scattering isgiven 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 significantlydeteriorating if the scattering amplitude is available only for a limited aperture, or if the data arecorrupted by noise. Also, the points with the smallest values of the ‖γ‖ are the best in locating theinclusion, and not the largest one, as required by the theory in [Kir1] and in [CK].

In Figures 1 and 2 (cf [RGu2]) the implementation of the Colton-Kirsch LSM (4.6.17) is denoted bygnck, and of the Kirsch method (4.6.18) by gnk. The Figures show a contour plot of the logarithm ofthe ‖γ‖. 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 1, and the results for d = 1.5are shown in Figure 2. Note that the actual radius of the circles is 1.0, but it cannot be seen from theLSM 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 boundaryconditions on the obstacle. The Support Function Method ([GR5], [R83]) showed a better identificationfor 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 computedby 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 parameter-fitting scheme for

solving a one-dimensional inverse scattering problem in quantum physics. This concept allows one to getsome idea about the error estimate in a parameter-fitting scheme. In this Section we have used [RGu2].


gnck, k=1.0, d=2.0

-4 -2 0 2 4

-4

-2

0

2

4

2.34478

20.7297

gnk, k=1.0, d=2.0

-4 -2 0 2 4

-4

-2

0

2

4

-2.89727

8.60753

Figure 1: Identification of two circles of radius 1.0 centered at (−d, 0.0) and (d, 0.0) for d = 2.0.


gnck, k=1.0, d=1.5

-4 -2 0 2 4

-4

-2

0

2

4

5.73788

24.8328

gnk, k=1.0, d=1.5

-4 -2 0 2 4

-4

-2

0

2

4

-2.87834

9.25278

Figure 2: Identification of two circles of radius 1.0 centered at (−d, 0.0) and (d, 0.0) for d = 1.5.

Chapter 5

Stability of the solutions to 3D

Inverse scattering problems withfixed-energy data

In this Chapter some of the author’s results on inverse potential scattering problem with fixed-energydata are presented. The presentation is based on paper [R203]. Inversion formulas and stability resultsfor the solutions to 3D inverse scattering problems with fixed energy data are obtained. Inversion ofexact and noisy data is considered. The inverse potential scattering problem with fixed-energy scatteringdata is discussed in detail, inversion formulas for the exact and for noisy data are derived, error estimatesfor the inversion formulas are obtained. Global estimates for the scattering amplitude are given whenthe potential grows to infinity in a bounded domain. Inverse geophysical scattering problem is discussedbriefly. An algorithm for constructing the Dirichlet-to-Neumann map from the scattering amplitude andvice versa is obtained.

5.1 Introduction

In this Chapter 3D inversion scattering problems with fixed-energy data are discussed. These problemsinclude inverse problems of potential, obstacle, and geophysical scattering (IPS, IOS, IGS).

Inverse potential scattering problem is discussed in detail: uniqueness of its solution, reconstructionformulas for inversion of the exact data and for inversion of noisy data are given and error estimates forthese formulas are obtained. These estimates yield the stability estimates for the solution of the inversescattering problem.

For the inverse obstacle scattering the uniqueness theorem is proved for rough domains, stabilityestimates are obtained for C2,λ domains, 0 < λ < 1, that is, for domains whose boundary in localcoordinates is a graph of C2,λ function, and reconstruction formulas are discussed in Chapter 4.

For inverse geophysical scattering the inverse scattering problem is reduced to inverse scatteringproblem 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 bibli-ography of the author’s papers). The presentation and some of the estimates are improved. Only someof the results from the cited papers are included.

195

196 CHAPTER 5. INVERSE SCATTERING PROBLEM

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 [

∇2 + k2 − q(x)]u(x, α, k) = 0 in R3, x ∈ R3, (5.1.1)

u = eikα·x +A(α′, α, k)eikr

r+ o

(1

r

), r := |x| −→ ∞, α′ :=

x

r(5.1.2)

Here u(x, k) is the scattering solution, k = const > 0 is fixed. Without loss of generality we takek = 1 in what follows unless other choice is suggested explicitly. A unit vector α ∈ S2 is given, whereS2 is the unit sphere in R3. Vector α has a physical meaning of the direction of the incident plane wave,while α′ ∈ S2 is the direction of the scattered wave, k2 is the fixed energy. The function A(α′, α, k) iscalled the scattering amplitude. It describes the first term of the asymptotics of the scattered field asr → ∞ along the direction α′ = x

r .The function q(x) is called the potential. We assume that

q ∈ Q := Qa ∩ L∞(R3),

Qa :=q : q(x) = q(x), q(x) ∈ L2

(Ba), q(x) = 0 if |x| > a

,

(5.1.3)

where a > 0 is an arbitrary large fixed number which we call the range of q(x), and the overbar standsfor complex conjugate.

In many results q ∈ Qa is sufficient, but q ∈ Q 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 scatteringsolution. These results can be found in many books, for example, in the appendix to [R121], where abrief but self-contained presentation of the scattering theory is given.

The scattering problem has a unique solution if q ∈ Qa

In fact, the above result is proved for much larger class of q ( [H]), but for inverse scattering problem withnoisy data it is necessary to assume q(x) compactly supported [R139]. Indeed, represent the potentialq(x) as q = q1 + q2, where q1 = 0 for |x| > a and q1 = q for |x| ≤ a. Call q2 the tail of the potential q.If one assumes a priori that q = O(|x|−b), where b > 3, then the contribution of the tail of the potentialto the scattering amplitude is of order O(|a|3−b) and tends to 0 as a → ∞. At some value of a, sayat a = a0, this contribution becomes of the order of the noise in the scattering data. One cannot, inprinciple, discriminate between the noise and the contribution of the tail of the potential for a > a0.Therefore the tail of q for a > a0 cannot be determined from noisy data.

One hassupx∈R3

∣∣u(x, k)∣∣ ≤ c, k = const > 0. (5.1.4)

By c > 0 we denote various constants. If q ∈ Qa then u(x, k) extends as a meromorphic function tothe whole complex k-plane. Let G(x, y, k) denote the resolvent kernel of the self-adjoint Schrodingeroperator Lu = −∇2 + q(x) in L2(R3):

(L− k2

)G(x, y, k) = δ(x− y) in R3, (5.1.5)

limr→∞

∫

|x|=r

∣∣∣∣∂G

∂|x| − ikG

∣∣∣∣2

ds = 0, y is fixed, k > 0. (5.1.6)


The function u(x, k) can be defined by the formula:

G(x, y, k) =eik|y|

4π|y|u(x, α, k) + o

(1

|y|

),

y

|y| = −α, (5.1.7)

where o( 1|y| ) = O( 1

|y|2 ) 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 ikj, kj > 0, 1 ≤ j ≤ J in C+ := k : Imk > 0 and if q(x) 6≡ 0, q ∈ Qa,infinitely many poles, possibly not simple, in C− = C \ C+. There are no poles on the real line except,possibly, at k = 0.

The functions uj(x) ∈ L2(R3), solving (5.1.1) with k = ikj , are called eigenfunctions of the discretespectrum of L, −k2

j are the negative eigenvalues of L. There are at most finitely many of these if q ∈ Qa.The eigenfunction expansion formulas are known:

f(x) =∑

j

fjuj(x) +

∫

R3

f (ξ)u(x, ξ)dξ, |ξ| = k, ξ = kα,

where

fj :=(f, uj

)L2(R3)

, f(ξ) =1

(2π)3

∫

R3

f(x)u(x, α, k)dx,

(see e.g. [R121], [CFKS]). The proofs of the results in this Section can be found in [R121], Appendix,and in [R139].

If Eλ is the resolution of the identity of the selfadjoint operator L, and Eλ(x, y) is its kernel, then

dEλ(x, y)

dλ=

1

πImG

(x, y,

√λ)

=

√λ

16π3

∫

S2

u(x, α,

√λ)u(y, α,

√λ)dα, λ > 0.

Properties of the scattering amplitude

The scattering amplitude has the following well-known properties:

A(α′, α, k) = (−α,−α′, k) (reciprocity),

A(α′, α, k) = A(α′, α,−k), k > 0 (reality),

ImA(α′, α, k) =k

4π

∫

S2

A(α′, β, k)A(α, β, k)dβ, k > 0 (unitarity).

In particular,

ImA(α, α, k) =k

4π

∫

S2

∣∣A(α, β, k)∣∣2dβ (optical theorem).

If q ∈ Qa, and k = 1, then the scattering amplitude is an analytic function of α′ and α on the algebraicvariety

M :=θ : θ ∈ C3, θ · θ = 1

, θ ·w :=

3∑

j=1

θjwj. (5.1.8)

This variety is non-compact, intersects R3 over S2, and, given any ξ ∈ R3, there exist (many) θ, θ′ ∈Msuch that

θ′ − θ = ξ, |θ| −→ ∞, θ, θ′ ∈M. (5.1.9)

In particular, if one chooses the coordinate system in which ξ = te3, t > 0, e3 is the unit vector alongthe x3-axis, then the vectors

θ′ =t

2e3 + ζ2e2 + ζ1e1, θ = − t

2e3 + ζ2e2 + ζ1e1, ζ2

1 + ζ22 = 1 − t2

4, (5.1.10)


satisfy (5.1.9) for any complex numbers ζ1 and ζ2 satisfying the last equation in (5.1.10) and such that|ζ1|2 + |ζ2|2 → ∞. There are infinitely many such ζ1, ζ2 ∈ C. If q ∈ Qa than the function A(α′, α, k) isa meromorphic function of k ∈ C which has poles at the same points as G(x, y, k).

One has

−4πA(α′, α, k) =

∫

Ba

e−ikα′·xq(x)u(x, α, k)dx. (5.1.11)

The S-matrix is defined by the formula

S = I +ik

2πA, S∗S = SS∗ = I, (5.1.12)

and is a unitary operator in L2(S2). Thus, A is a normal operator in L2(S2).

If ∇q ∈ Qa then

φ(x, α, k) := e−ikα·xu(x, α, k) = 1 +1

2ik

∫ ∞

0

q(x− rα)dr+ o

(1

k

), k −→ ∞. (5.1.13)

Therefore

q(x) = α · ∇x limk→∞

2ik[φ(x, α, k

)− 1], (5.1.14)

and

A(α, α, k) = − 1

4π

∫

R3

q(x)dx+ o(1), k −→ +∞. (5.1.15)

The fundamental equation

Denote u+ := u(x, α, k), u− := u(x,−α,−k), k > 0. Then u+ = Su−, that is

u+ = u− +ik

2π

∫

S2

A(α′, α, k)u−(x, α′, k)dα′. (5.1.16)

Completeness properties of the scattering solutions

(a) If h(α) ∈ L2(S2) and

∫

S2

h(α)u(x, α, k)dα= 0 ∀x ∈ B′R :=

x : |x| > R

, k > 0 is fixed (5.1.17)

then h(α) = 0.

Let ND(L) = w : Lw = 0 in D, w ∈ H2(D), where D ⊂ R3 is a bounded domain, H2(D) is theSobolev space.

(b) The set u(x, α, k)∀α∈S2 is total in ND(L − k2), that is, for any ε > 0, however small, and anyfixed w ∈ ND(L − k2), there exists νε(α) ∈ L2(S2) such that

∥∥∥∥w(x) −∫

S2

u(x, α, k)νε(α)dα

∥∥∥∥H2(D)

< ε. (5.1.18)

The νε(α) depends on w(x).


Special solutions

There exists ψ(x, θ, k), ψ ∈ ND(L − k2), such that[∇2 + k2 − q(x)

]ψ = 0 in R3, ψ = eikθ·x

[1 +R(x, θ, k)

], θ ∈M, (5.1.19)

‖R‖L∞(D) ≤ c(ln |θ|) 1

2

|θ| 12, |θ| −→ ∞, θ ∈M, (5.1.20)

‖R‖L2(D) ≤c

|θ| , |θ| −→ ∞, θ ∈ M, (5.1.21)

where D ⊂ R3 is an arbitrary bounded domain.

Property C for the pair L1 − k2, L2 − k2

Let Ljw :=∑Mj

|m|=0 ajm(x)∂|m|w(x), x ∈ Rn, n ≥ 2, j = 1, 2, be linear formal partial differential

operators, that is, formal differential expressions.Let Nj = NjD(Lj) := w : Ljw = 0 in D, where D ⊂ Rn is an arbitrary fixed bounded domain and

the equation is understood in the sense of the distribution theory. Consider the subsets of Nj , j = 1, 2,

which form an algebra in the sense that the products w1w2 ∈ Lp′

(D), where wj ∈ Nj , p′ = p

p−1, and

1 ≤ p ≤ ∞. If p = 1 define p′ = ∞, and if p = ∞ define p′ = 1. We write ∀wj ∈ Nj meaning that wjrun through the above subsets of Nj .

Definition 5.1.1. We say that the pair of linear partial differential operators L1, L2 has property Cpif and only if the set w1w2 is total in Lp(D), that is, if f(x) ∈ Lp(D) and

∫

D

f(x)w1(x)w2(x)dx = 0 ∀wj ∈ Nj , j = 1, 2, (5.1.22)

thenf(x) = 0. (5.1.23)

If the above holds for p = 2 we say that property C holds for the pair L1, L2.Theorem 5.1.2. ([R100],[R109]) Let Lj = −∇2 + qj(x), qj(x) ∈ Qa, k = const ≥ 0 is arbitrary fixed.Then the pair L1 − k2, L2 − k2 has property C.

Proof. Note that ψj ∈ Nj , j = 1, 2, where ψj are defined in Section 5.1.2. Without loss of generalitytake k = 1, let ψ(x, θ, 1) := ψ(x, θ). One has (see (5.1.19)):

ψ1(x, θ′)ψ2(x,−θ) = ei(θ

′−θ)·x(1 + R1

)(1 + R2

).

Choose θ′, θ ∈M such that (5.1.9) holds with an arbitrary fixed ξ ∈ R3. Then

ψ1(x, θ′)ψ2(x,−θ) = eiξ·x

(1 + o(1)

)as |θ| −→ ∞. (5.1.24)

Since the set eiξ·x∀ξ∈R3 is total in Lp(D), p ≥ 1, D ⊂ Rn is a bounded domain, the conclusion ofTheorem 5.1.2 follows. 2

Remark 5.1.3. One cannot take unbounded domain D in the above argument because o(1) in (5.1.24)holds for bounded domains.

One can take the space of f(x) larger than L1(D), for example, the space of distribution of finiteorder of singulartiy if q(x) is sufficiently smooth [R139].

Theorem 5.1.4. The set u1(x, α, k)u2(x, β, k)∀α,β∈S2, at a fixed k > 0, is complete in Lp(D), whereD ⊂ R3 is an arbitrary fixed bounded domain, and p ≥ 1 is fixed.

Proof. The conclusion of Theorem 5.1.4 follows from Theorem 5.1.2 and (5.1.18). 2


Properties of the Fourier coefficients of A(α′, α)

We denote A(α′, α, k)|k=1 := A(α′, α), and write

A(α′, α) =

∞∑

`=0

A`(α)Y`(α′), A`(α) :=

∫

S2

A(α′, α)Yl(α′)dα′, (5.1.25)

where Y`(α′) = Y`,m(α′),−` ≤ m ≤ `, summation over m is understood in (5.1.25) and similar formulas

below, e.g. (5.1.31), (5.1.37), etc,

Y`,m(α) =(−1)mi`√

4π

[(2` + 1)(` −m)!

(` +m)!

] 12

eimϕP`,m(cos ϑ),

Y`,m(α) = (−1)`+mY`,−m(α), Y`,m(−α) = (−1)`Y`,m(α).

(5.1.26)

Here P`,m(cosϑ) = (sinϑ)mdmP`(cosϑ)(d cosϑ)m , 0 ≤ m ≤ `, P`(x) is the Legendre polynomial, (ϑ, ϕ) are the

angles corresponding to the point α ∈ S2, P`,−m(cos ϑ) = (−1)m (`−m)!(`+m)!P`,m(cos ϑ), 0 ≤ m ≤ `.

Consider a subset M ′ ⊂ M consisting of the vectors θ = (sinϑ cosϕ, sinϑ sinϕ, cosϑ) where ϑ andϕ run through the whole complex plane. Clearly θ ∈ M , but M ′ is a proper subset of M . Indeed, anyθ ∈ M with θ3 6= ±1 is an element of M ′. If θ3 = ±1, then cosϑ = ±1, so sinϑ = 0 and one getsθ = (0, 0,±1) ∈ M ′. However, there are vectors θ = (θ1, θ2, 1) ∈ M which do not belong to M ′. Suchvectors one obtains choosing θ1, θ2 ∈ C such that θ21 + θ22 = 0. There are infinitely many such vectors.The same is true for vectors (θ1, θ2,−1). Note that in (5.1.9) one can replace M by M ′ for any ξ ∈ R3,ξ 6= 2e3.

Let us state two estimates ([R139]):

maxα∈S2

∣∣A`(α)∣∣ ≤ c

(a`

) 12(ae

2`

)`+1

, (5.1.27)

and∣∣Y`(θ)

∣∣ ≤ 1√4π

er|Imθ|

|j`(r)|, ∀r > 0, θ ∈M ′, (5.1.28)

where

j`(r) :=( π

2r

) 12

J`+ 12(r) =

1

2√

2

1

`

(er2`

)` [1 + o(1)

]as ` −→ ∞, (5.1.29)

and J`(r) is the Bessel function regular at r = 0. Note that Y`(α′), defined by (5.1.26), admits a natural

analytic continuation from S2 to M by taking ϑ and ϕ in (5.1.26) to be arbitrary complex numbers.The resulting θ′ ∈M ′ ⊂ M .

A global perturbation formula

Let Aj(α′, α) be the scattering amplitude corresponding to qj ∈ Qa, j = 1, 2. Define A := A1 − A2,

p := q1(x) − q2(x). Then [R139]

−4πA(α′, α) =

∫

Ba

p(x)u1(x, α)u2(x,−α′)dx. (5.1.30)

Formula for the scattering solution outside the support of the potential

Let supp (q) ⊂ Ba. The fixed-energy scattering dataA(α′, α)∀α′, α, or, equivalently, the data A`(α)`=0,1,2,...∀α ∈S2, allow one to write an analytic formula for the scattering solution u(x, α) in the region B ′

a := R3\Ba:

u(x, α) = eiα·x +

∞∑

`=0

A`(α)Y`(α′)h`(r), r := |x| > a, α′ :=

x

r, (5.1.31)


where A`(α) are defined in (5.1.25), Y`(α′) are defined in (5.1.26),

h`(r) := eiπ2 (`+1)

√π

2rH

(1)

`+ 12

(r),

H(1)` (r) is the Hankel function, and the normalizing factor is chosen so that

h`(r) =eir

r

[1 + o(1)

]as r −→ ∞. (5.1.32)

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):

∣∣H(1)` (r)

∣∣2 =4

π2

∫ ∞

0

K0(2rsht)(e2`t + e−2`t

)dt, (5.1.33)

where sht := et−e−t

2. This formula implies that |h`(r)| is a monotonically increasing function of `.

It is known [GR, formula 8.478] that r|h`(r)|2 is a monotonically decreasing function of r if ` > 0,and

h`(r) = − i`√

2

r

(2`

er

)` [1 + o(1)

], ` −→ +∞, r > 0. (5.1.34)

The following known estimate (e.g., [MPr]) can be useful:

∣∣Y`,m(α)∣∣ ≤ c`

m2 −1, α ∈ S2, (5.1.35)

where Y`,m(α) are the normalized in L2(S2) 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 |x| > a,

|y| > a, where supp q(x) ⊂ Ba. Let g(x, y, k) := eik|x−y|

4π|x−y| and denote by A`′` the Fourier coefficients of

the scattering amplitude:

A(α′, α) =∞∑

`=0

A`(α)Y`(α′) =

∞∑

`′,`=0

A`′`Y`′ (α)Y`(α′). (5.1.36)

Then

G(x, y, k) = g(x, y, k) +k2

4π

∞∑

`′,`=0

A`′`Y`′(α)Y`(α′)h`

(k|x|

)h`′(k|y|

), |x| > a, |y| > a, (5.1.37)

where α′ := x|x| , α := − y

|y| .

Indeed, clearly the function (5.1.37) solves (5.1.5) in the region |x| > a, |y| > a, where q(x) = 0, itsatisfies (5.1.6), and

G(x, y, k) =eik|y|

4π|y|

[eikα·x +

∞∑

`′,`=0

A`′`Y`′(α)Y`(α′)kh`

(k|x|

)]

+ o

(1

|y|

), as |y| −→ ∞,

y

|y| = −α.(5.1.38)

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 mainterm of asymptotics (5.1.38) as the Green function of the Schrodinger operator. Therefore the function(5.1.37) is identical to the Green function (5.1.5)-(5.1.6) in the region |x| > a, |y| > a.


5.2 Inverse potential scattering problem with fixed-energy data

The IPS problem can now be formulated: given A(α′, α) ∀α′, α ∈ S2, find q(x) ∈ Qa. Throughout thissection k = 1.


The first result is the uniqueness theorem of Ramm ([R100], [R109]).

Theorem 5.2.1. (Ramm) If q1, q2 ∈ Qa and A1(α′, α) = A2(α

′, α) ∀α′ ∈ S21 , ∀α ∈ S2

2 , where S2j

j = 1, 2, are arbitrary small open subsets of S2, then q1(x) = q2(x).

Proof. The function A(α′, α) is analytic with respect to α′ and α on the variety (5.1.8). Therefore itsvalues on S2

1×S22 extend uniquely by analyticity to M×M . In particular A(α′, α) is uniquely determined

in S2 × S2. By (5.1.30) one gets:

∫

Ba

p(x)u1(x, α)u2(x,−α′)dx = 0 ∀α, α′ ∈ S2. (5.2.1)

By property C ( formulas (5.1.22) -(5.1.23)) and by (5.1.18), the orthogonality relation (5.2.1) impliesp(x) ≡ 0. 2

5.2.2 Reconstruction formula for exact data

Fix an arbitrary ξ ∈ R3 and choose arbitrary θ′, θ satisfying (5.1.9).Denote

q(ξ) :=

∫

Ba

e−iξ·xq(x)dx. (5.2.2)

Multiply (5.1.11) by ν(α, θ) ∈ L2(S2), where ν(α, θ) will be fixed later, and integrate over S2 withrespect to α:

−4π

∫

S2

A(α′, α)ν(α, θ)dα=

∫

Ba

e−iα′·x∫

S2

u(x, α)ν(α, θ)dαq(x)dx. (5.2.3)

If q ∈ Qa, then estimates (5.1.27) and (5.1.28) imply that the series (5.1.25) converges, when α′ isreplaced by θ′ ∈ M , uniformly and absolutely on S2 × Mc, where Mc ⊂ M is an arbitrary compactsubset of M . Formula (5.1.11) implies that α′ can be replaced by θ′ ∈ M , since Ba is a compact set inR3.

Define

ρ(x) := ρ(x; ν) := e−iθ·x∫

S2

u(x, α)ν(α, θ)dα− 1, (5.2.4)

and rewrite (5.2.3), with α′ = θ′, as

−4π

∫

S2

A(θ′, α)ν(α, θ)dα=

∫

Ba

e−iθ′·x+iθ·x[ρ(x) + 1

]q(x)dx

:= q(ξ) +

∫

Ba

e−iξ·xρ(x)q(x)dx = q + ε,

(5.2.5)

where|ε| ≤ ‖q‖a‖ρ‖a, ‖q‖a := ‖q‖L2(Ba). (5.2.6)

The following estimate, proved in Section 5.4.3 below, holds for a suitable choice of ν(α, θ):

‖ρ‖a ≤ c|θ|−1, |θ| −→ ∞, θ ∈M. (5.2.7)

5.2. INVERSE POTENTIAL SCATTERING PROBLEM WITH FIXED-ENERGY DATA 203

From (5.2.5) and (5.2.7) one gets the reconstruction formula for inversion of exact, fixed-energy,three-dimensional scattering data:

lim|θ|→∞θ′−θ=ξ,θ,θ′∈M.

− 4π

∫

S2

A(θ′, α)ν(α, θ)dα

= q(ξ), (5.2.8)

and the error estimate:

−4π

∫

S2

A(θ′, α)ν(α, θ)dα= q(ξ) +O

(1

|θ|

), |θ| −→ ∞, θ ∈M, (5.2.9)

where (5.1.9) is always assumed.Let us give an algorithm for computing the function ν(α, θ) for which (5.2.7), and therefore (5.2.9),

hold, given the scattering data A(α′, α) ∀α′, α ∈ S2.Fix arbitrarily two numbers a1 and b such that

a < a1 < b, (5.2.10)

and define the L2-norm in the annulus:

‖ρ‖2 :=

∫

a1≤|x|≤b|ρ|2dx. (5.2.11)

Consider the minimization problem

‖ρ‖ = inf := d(θ), (5.2.12)

where the infimum is taken over all ν ∈ L2(S2), and ρ is defined in (5.2.4).It is proved in Section 5.4.3 that

d(θ) ≤ c|θ|−1 if θ ∈M, |θ| 1. (5.2.13)

The symbol |θ| 1 means that |θ| is sufficiently large. The constant c > 0 in (5.2.13) depends onthe norm ‖q‖a but not on the potential q(x) itself. An algorithm for computing a function ν(α, θ), whichcan 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

‖ρ(x, ν)‖ < 2d(θ), (5.2.14)

where in place of 2 in (5.2.14) one could put any fixed constant greater than 1.(b) Any such ν(α, θ) generates an estimate of q(ξ) with the error O( 1

|θ|), |θ| → ∞. This estimate is

calculated by the formula

q := −4π

∫

S2

A(θ′, α)ν(α, θ)dα, (5.2.15)

where ν(α, θ) ∈ L2(S2) 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

supξ∈R3

|q − q(ξ)| ≤ c

|θ| , |θ| −→ ∞, θ ∈M. (5.2.16)

Proof. The proof is the same as the proof of (5.2.5) - (5.2.7) and is based on the following estimate:

‖ρ‖a ≤ c(‖ρ‖ + |θ|−1

), |θ| 1, θ ∈M. (5.2.17)

The proof of (5.2.17) is given in Section 5.4.3. 2


5.2.3 Stability estimate for inversion of the exact data

Let the potentials q ∈ Q, j = 1, 2, generate the scattering amplitudes Aj(α′, α).

Let us assume thatsup

α′,α∈S2

∣∣A1(α′, α) −A2(α

′, α)∣∣ < δ. (5.2.18)

We want to estimate p(x) := q1(x) − q2(x).The main tool is formula (5.1.30).The result is:

Theorem 5.2.3. If qj ∈ Q and (5.2.18) holds then

supξ∈R3

∣∣q1(ξ) − q2(ξ)∣∣ ≤ c

ln | ln δ|| ln δ| , δ −→ 0, (5.2.19)

where the constant c > 0 does not depend on δ > 0, δ → 0.

Proof. Multiply both sides of (5.1.30) by ν1(α, θ)ν2(−α′, θ2), where θj ∈ M , j = 1, 2, θ1 + θ2 = ξ,|θ1| → ∞, and integrate with respect to α and α′ over S2, to get:

− 4π

∫

S2

∫

S2

A(α′, α)ν1

(α, θ1

)ν2

(− α′, θ2

)dαdα′

=

∫

Ba

dxp(x)

∫

S2

u1(x, α)ν1

(α, θ1

)dα

∫

S2

u2(x, β)ν2

(β, θ2

)dβ, β = −α′.

(5.2.20)

Choose ν1 and ν2 such that

‖ρj(νj)‖ ≤ c

|θj|, |θj| −→ ∞, θj ∈M, (5.2.21)

where ρj(νj) := ρj(x, νj) := e−iθjx∫S2 uj(x, α)νj(α, θj)dα − 1, and note that |θ1|

|θ2| → 1 as |θ1| → ∞,

θ1, θ2 ∈ M , θ1 + θ2 = ξ, |ξ| ≤ ξ0, and c > 0 in (5.2.21) does not depend on ξ. From (5.2.18), (5.2.20)and (5.2.21) one gets

|p(ξ)| ≤ c(|θ|−1 + cδ‖ν1‖L2(S2)‖ν2‖L2(S2)

), (5.2.22)

where θ := θ1. One can choose ν1 and ν2 such that ( see Section 5.4.7):

∥∥∥νj(α, θj

)‖L2(S2) ≤ cec|θ| ln |θ|, θ = θ1, |θ| −→ ∞, θ2 = ξ − θ, (5.2.23)

where c > 0 stands for various constants.Thus (5.2.22) yields:

supξ∈R3

∣∣p(ξ)∣∣ ≤ cmin

s1

[s−1 + c1δe

c2s ln |s|], 0 < δ << 1, (5.2.24)

where c, c1 and c2 are some positive constants, s := |θ|, δ 1 means that δ > 0 is small and s 1means that s > 0 is large. However, our argument is valid for s ≥ 1 and 0 < δ ≤ 1

2.

One gets

mins>0

[s−1 + c1δe

c2s ln |s|] := η(δ) ≤ c3ln | ln δ|| ln δ| , δ −→ 0, (5.2.25)

and the minimizer is

s = s(δ) = c−12

|lnδ|ln | ln δ|

[1 + o(1)

], δ −→ 0. (5.2.26)

From (5.2.24)–(5.2.26) one gets (5.2.19). 2


Remark 5.2.4. In the above proof the difficult part is the proof of (5.2.23). Estimate (5.2.23) can bederived for ν which approximately solve the minimization problem:

∥∥∥∥∥ψ(x, θ) −∫

S2

u(x, α)ν(α, θ)dα

∥∥∥∥∥L2(Bb)

≤ ce−bκ

κ, κ := |Imθ|, θ ∈ M, θ 1, (5.2.27)

and ‖ν(α, θ)‖L2(S2) = inf, where the infimum is taken over all ν ∈ L2(S2).

In Section 5.4.7 we consider the problem of finding ν ∈ L2(S2) with minimal norm ‖ν‖L2(S2) := a(ν)among all ν(α, θ) which satisfy the inequality:

∥∥∥∥∥ψ(x) −∫

S2

u(x, α)ν(α, θ)dα

∥∥∥∥∥L2(Bb)

≤ ε. (5.2.28)

The necessity to consider the ν with the minimal norm ‖ν‖L2(S2) comes from the simple observation:there exists a sequence of νn ∈ L2(S2), ‖νn‖L2(S2) = 1, such that

∥∥∥∥∥

∫

S2

u(x, α)νn(α)dα

∥∥∥∥∥L2(Bb)

−→ 0 as n −→ ∞. (5.2.29)

To prove (5.2.29) note that

u(x, α) = eiα·x −∫

Bb

ei|x−y|

4π|x− y|q(y)u(y, α)dy := eiα·x − Tu, (5.2.30)

where the operator (I + T )−1 := I + T1 is a continuous bijection of C(Bb) onto itself, and C(Bb) is theusual space of continuous in Bb, b ≥ a, functions equipped with the sup-norm (e.g. see Appendix in[R121]). Since T is compact in C(Bb), the above statement follows from the injectivity of I + T , whichwe now prove:

If f + Tf = 0, then f is extended to C(R3) by the formula f = −Tf , and satisfies the followingequation (∇2+1−q(x))f = 0 in R3 and the radiation condition of the type (5.1.6) with k = 1. Thereforef(x) ≡ 0 and the injectivity of I + T is proved.

Thus I + T and I + T1 are continuous bijections of C(Bb) into itself for any b ≥ a.

Writing u(x, α) = (I + T1)eiα·x, one concludes that (5.2.29) is equivalent to

∥∥∥∥∥

∫

S2

eiα·xνn(α)dα

∥∥∥∥∥L2(Bb)

−→ 0 as n −→ ∞. (5.2.31)

Existence of a normalized sequence νn(α) satifying (5.2.31) follows from the compactness of the operator

Q : L2(S2)−→ L2

(Bb), Qν :=

∫

S2

eiα·xν(α)dα.

Of course, the same argument is applicable to the operator Q1ν :=∫S2 u(x, α)ν(α)dα, but the bijectivity

of I + T in C(Bb), b ≥ a, is of independent interest.

It follows from (5.2.29) that, for a given ε > 0, one can find ν in (5.2.29) with an arbitrary largenorm ‖ν‖L2(S2). By this reason we are interested in ν with minimal norm. Estimate (5.2.23) gives abound on the growth of the minimal value of the norm ‖ν‖L2(S2), where ν = ν(α, θ) satisfies (5.2.28)

with ε = e−bκ

κ , κ := |Imθ|, |θ| → ∞, θ ∈M .


5.2.4 Stability estimate for inversion of noisy data

Assume now that the scattering data are given with some error: a function Aδ(α′, α) is given such that

supα′,α∈S2

∣∣A(α′, α) −Aδ(α′, α)

∣∣ ≤ δ. (5.2.32)

We emphasize that Aδ(α′, α) is not necessarily a scattering amplitude corresponding to some po-

tential, it is an arbitrary function in L∞(S2 × S2) satisfying (5.2.32). It is assumed that the unknownfunction A(α′, α) is the scattering amplitude corresponding to a q ∈ Q.

The problem is: Find an algorithm for calculating q such that

supξ∈R3

∣∣q − q(ξ)∣∣ ≤ η(δ), η(δ) −→ 0 as δ −→ 0, (5.2.33)

and estimate the rate at which η(δ) tends to zero.An algorithm for inversion of noisy data will now be described.Let

N (δ) :=

[ | ln δ|ln | ln δ|

], (5.2.34)

where [x] is the integer nearest to x > 0,

Aδ(θ′, α) :=

N(δ)∑

`=0

Aδ`(α)Y`(θ′), Aδ`(α) :=

∫

S2

Aδ(α′, α)Y`(α′)dα′, (5.2.35)

uδ(x, α) := eiα·x +

N(δ)∑

`=0

Aδ`(α)Y`(α′)h`(r), (5.2.36)

ρδ(x; ν) := e−iθ·x∫

S2

uδ(x, α)ν(α)dα− 1, θ ∈ M, (5.2.37)

µ(δ) := e−γN(δ), γ = const > 0, γ = lna1

a> 0, (5.2.38)

a(ν) := ‖ν‖L2(S2), κ := |Imθ|. (5.2.39)

Consider the variational problem with constraints:

|θ| = sup := ϑ(δ), (5.2.40)

|θ|[∥∥ρδ(ν)

∥∥+ a(ν)eκbµ(δ)]≤ c, θ ∈M, (5.2.41)

the norm is defined in (5.2.11), and it is assumed that

θ′ − θ = ξ, θ, θ′ ∈M, (5.2.42)

where ξ ∈ R3 is an arbitrary fixed vector, c > 0 is a sufficiently large constant, and the supremum istaken over θ ∈M and ν ∈ L2(S2) under the constraint (5.2.41).

Given ξ ∈ R3 one can always find θ and θ′ such that (5.2.42) holds.We prove that ϑ(δ) → ∞, in fact

ϑ(δ) ≥ c| ln δ|

(ln | ln δ|)2 , δ −→ 0. (5.2.43)

Let θ(δ), νδ(α) be any approximate solution to problem (5.2.40)-(5.2.41) in the sense that

|θ(δ)| ≥ ϑ(δ)

2. (5.2.44)


Calculate

qδ := −4π

∫

S2

Aδ(θ′, α)νδ(α)dα. (5.2.45)

Theorem 5.2.5. If (5.2.42) and (5.2.44) hold, then

supξ∈R3

∣∣qδ − q(ξ)∣∣ ≤ c

(ln | ln δ|)2| ln δ| as δ −→ 0. (5.2.46)

Proof. One has:

qδ − q(ξ) = − 4π

∫

S2

Aδ(θ′, α)νδ(α)dα− q(ξ)

= − 4π

∫

S2

A(θ′, α)νδ(α)dα− q(ξ)

+ 4π

∫

S2

[A(θ′, α)− Aδ(θ′, α)]νδ(α)dα.

(5.2.47)

The rest of the proof consists of the following steps:Step 1. We prove that

∥∥ρ(x, νδ)∥∥ ≤ c

[∥∥ρδ(x, νδ

)∥∥+ a(νδ)eκbµ(δ)

]≤ c

ϑ(δ), (5.2.48)

where the norm is defined in (5.2.11).This estimate and (5.2.43) imply (see the proof of (5.2.19) and (5.2.26)) that

∣∣∣∣∣− 4π

∫

S2

A(θ′, α)νδ(α)dα− q(ξ)

∣∣∣∣∣ ≤ c(ln | ln δ|)2

| ln δ| . (5.2.49)

Step 2. We prove that

∣∣∣∣∣

∫

S2

[A(θ′, α) − Aδ(θ

′, α)

]νδ(α)dα

∣∣∣∣∣ ≤ ca(νδ)eκbµ(δ) ≤ c

ϑ(δ)≤ c

(ln | ln δ|)2| ln δ| , (5.2.50)

where θ′ = θ′(δ) = ξ+θ(δ), and the pair θ(δ), νδ(α) solves (5.2.40)-(5.2.41) approximately in the sensespecified above. (See formula (5.2.44)).

This estimate follows from (5.2.41) and from the inequality

∥∥A(θ′, α)− Aδ(θ′, α)

∥∥L2(S2)

≤ ceκbµ(δ). (5.2.51)

Let us prove (5.2.51). One has

∥∥Aδ(θ′(δ), α) − A(θ′(δ), α)∥∥L2(S2)

≤∥∥∥∥∥

N(δ)∑

`=0

[Aδ`(α) − Al(α)

]Y`(θ

′)

∥∥∥∥∥L2(S2)

+

∥∥∥∥∥∞∑

`=N(δ)+1

A`(α)Y`(θ′)

∥∥∥∥∥L2(S2)

:= I1 + I2,

(5.2.52)

where N (δ) is given in (5.2.34) and θ′ := θ′(δ). Using (5.1.28), (5.1.29) one gets

I1 ≤ cδN 2(δ)eκb(2N )N(δ)+1

(eb)N(δ). (5.2.53)


Here we have used the estimate

supα∈S2

∞∑

`=0

∣∣A`(α) − Aδ`(α)∣∣2 ≤ 4πδ2,

which follows from (5.2.32) and the Parseval equality, and which implies

supα∈S2

∣∣A`(α) − Aδ`(α)∣∣ ≤

√4πδ. (5.2.54)

We also took into account that there are (N + 1)2 spherical harmonics Y` = Y`,m with 0 ≤ ` ≤ N ,

because∑`

m=−` 1 = 2`+ 1, and∑N

`=0(2`+ 1) = (N + 1)2. For large N one has (N + 1)2 = N2[1 + o(1)],N → ∞, so we write (N + 1)2 ≤ cN 2, c > 1.

To estimate I2, use (5.1.27) -(5.1.29) and get:

I2 ≤ c

∞∑

`=N+1

(ea2`

)`+1 eκa1

ea1

2` )`+1≤ ceκa1

(a

a1

)N+1

. (5.2.55)

Minimizing with respect to N > 1 the function

δN2 (2N )N+1

(eb)N+ sN+1, 0 < s :=

a

a1< 1. (5.2.56)

one gets

minN>1

[δN2 (2N )N+1

(eb)N+ sN+1

]≤ ce−γN(δ) = cµ(δ), γ = ln

a1

a> 0, (5.2.57)

where N (δ) is given in (5.2.34) and µ(δ) 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. 2

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 q1, q2 ∈ Qa and (5.2.18) holds then

supx∈B′

a

α∈S2

∣∣u1(x, α)− u2(x, α)∣∣ ≤ cµ(δ), |x| > a, (5.2.58)

where µ(δ) is defined by (5.2.38) and (5.2.34), and c > 0 is a constant.

Proof. Using (5.1.31), one gets:

u1 − u2 =

∞∑

`=0

[A`1(α) −A`2(α)

]Y`(α

′)h`(r), r > a. (5.2.59)

As stated below formula (5.1.33), one has

∣∣h`(r)∣∣ ≤

∣∣h`+1(r)∣∣, r > 0. (5.2.60)


∥∥u1 − u2

∥∥2

L2(Bb\Ba1 )=

∞∑

`=0

∣∣A`1(α) −A`2(α)∣∣2∫ b

a1

r2∣∣h`(r)

∣∣2dr, a < a1 < b. (5.2.61)


It follows from (5.2.60) and (5.1.34) that

sup0≤`≤N

∫ b

a1

r2∣∣h`(r)

∣∣2dr =

∫ b

a1

r2∣∣hN (r)

∣∣2dr

≤ c

(2N

e

)2N ∫ b

a1

dr

r2N≤ c

(2N

e

)2N

a−2N1 , b > a1.

(5.2.62)

From (5.2.60)-(5.2.61) one gets

∥∥u1 − u2

∥∥2

L2(Bb\Ba)≤ cδ2

(2N

ea1

)2N

+ c

∞∑

`=N+1

(∣∣A`1∣∣2 +

∣∣A`2∣∣2)( 2`

ea1

)2`

, (5.2.63)

where we have used monotone decay of r|H`(r)|2 as a function of r. Using estimate (5.1.27) in order toestimate |A`j(α)|, j = 1, 2, one gets:

∥∥u1 − u2

∥∥2

L2(Bb\Ba1 )≤ cδ2

(2N

ea1

)2N

+ c

(a

a1

)2N

, a < a1. (5.2.64)

Minimization of the right-hand side of (5.2.63) with respect to N ≥ 1 yields, as in (5.2.56), the estimatesimilar to (5.2.57): ∥∥u1 − u2

∥∥2

L2(Bb\Ba1 )≤ ce−γN(δ), γ = ln

a1

a> 0. (5.2.65)

Since u1 − u2 := w solves the equation(∇2 + 1

)w = 0 in B′

a := R3\Ba, (5.2.66)

one can use the known elliptic estimate:

‖w‖H2(D1) ≤ c

(∥∥∥(∇2 + 1

)w∥∥∥L2(D2)

+ ‖w‖L2(D2)

), D1 ⊂ D2, (5.2.67)

where D1 is a strictly inner subset of D2 and c = c(D1, D2), and get:

‖w‖H2(D1) ≤ ce−γN(δ), (5.2.68)

where D1 is any annulus a1 < a2 ≤ |x| ≤ a3 < b. By the embedding theorem, (5.2.68) implies (5.2.58)in R3. 2

5.2.6 Spherically symmetric potentials

If q(x) = q(r), r := |x|, then

A(α′, α) = A(α′ · α), A`(α) = A`Y`(α). (5.2.69)

In Section 5.6 the converse is proved: if q ∈ Qa and (5.2.69) holds then q(x) = q(r).The scattering data A(α′, α) in the case of the spherically symmetric potential is equivalent to the

set of the phase shifts δ`. The phase shifts are defined as follows:

1 +i

2πA` = e2iδ` , A` = 4πeiδ` sin δ`, k = 1. (5.2.70)

From Theorem 5.2.1 it follows that if q = q(r) ∈ Qa then the set δ``=0,1,2,... 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 L any subset of positive integers such that

∑

`∈L

1

`= ∞. (5.2.71)


Theorem 5.2.7. ([R192]) If q(x) = q(r) ∈ Qa then the data δ`∀`∈L determine q(r) uniquely.

In [GR2] and [ARS2] examples are given of quite different potentials q1(r) and q2(r), piecewise-constant, qj(r) = 0 for r > 5, for which

sup`≥0

∣∣δ(1)` − δ

(2)`

∣∣ < 10−5 and supr≥0

∣∣q1(r) − q2(r)∣∣ ≥ 1, where q1 and q2

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 [∇2 + 1 + v(x)

]w = −δ(x− y) in R3, (5.3.1)

limr→∞

∫

|S|=r

∣∣∣∣∂w

∂|x| − iw

∣∣∣∣2

ds = 0, (5.3.2)

where v(x) ∈ L2(R3) is a compactly supported real-valued function with support in R3−, the lower half-

space. In acoustics u has the physical meaning of the pressure, v(x) is the inhomogeneity in the velocityprofile. We took the fixed wavenumber k = 1 without loss of generality. The source y is on the planeP := x : x3 = 0, i.e., on the surface of the Earth, the receiver x ∈ P .

The data are the values w(x, y)∀x,y∈P .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].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 scatteringproblem to inverse potential scattering problem using the “lifting” [R139]. Suppose the data w(x, y),x ∈ P , y ∈ P , are given. Fix y and solve the problem:

(∇2 + 1

)w = 0 in R3

+ =x : x3 > 0

, (5.3.3)

w = w(x, y), x ∈ P, (5.3.4)

w satifies (5.3.2). (5.3.5)

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 ∇2 + 1 in the half-space R3

+ is knownexplicitly, analytically:

G1(x, y) =ei|x−y|

4π|x− y| −eik|x−y|

4π|x− y| , y :=(y1, y2,−y3

). (5.3.6)

Therefore the data w(x, y)∀x ∈ P determine uniquely and explicitly (analytically) the dataw(x, y)∀x ∈R3

+, y ∈ P . We have lifted the data from P to R3+ as far as x-dependence is concerned and similarly we

can get w(x, y)∀x, y ∈ R3+ given w(x, y)∀x, y ∈ P .

If w(x, y) is known for all x, y ∈ R3+, then one uses formula (5.1.7) and calculates analytically the

scattering solution u(x, α) corresponding to the potential q(x) := −v(x) and k = 1, where α ∈ S2− :=

α : α ∈ S2, α3 ≤ 0. Given u(x, α) for all x ∈ R3+ and α ∈ S2

−, one can calculate the scatteringamplitude A(α′, α) ∀α′ ∈ S2

+ := α : α ∈ S2, α3 ≥ 0.If the scattering amplitude A(α′, α), corresponding to the compactly supported q(x) = −v(x) ∈

L2(R3) is known ∀α′ ∈ S2+, ∀α ∈ S2

−, then the uniqueness of the solution to inverse geophysical problemfollows from Theorem 5.2.1.

5.4. PROOFS OF SOME ESTIMATES 211

Stability estimates obtained for the solution to inverse potential scattering problem with fixed-energydata remain valid for the inverse geophysical problem: via the lifting process one gets the scatteringamplitude A(α′, α) corresponding to the potential q(x) = −v(x), and the stability estimates for q(ξ),obtained in sections 2.3 and 2.4, yield stability estimates for v(ξ) = −q(ξ).

Practically, however, there are two points to have in mind. The first point is: if the noisy data uδ(x, y)are given, where supx,y∈P |uδ(x, y) − u(x, y)| < δ, then one has to overcome the following difficulty inthe lifting process: data ϕ(x, y), x, y ∈ P , such that |ϕ(x, y)| < δ, may not decay as |x| → ∞, |y| → ∞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 usesthe data A(α′, α)∀α′, α ∈ S2. Of course, the exact data A(α′, α)∀α′ ∈ S2

+, α ∈ S2, determine uniquelythe data A(α′, α) ∀α′, α ∈ S2, but practically finding the full data from the partial data is an ill-posedproblem.

One can also consider two-parameter inversion, corresponding to the governing equation ∇(a(x)∇u)+k2b(x)u = 0, where a(x) ≥ c > 0 is a C2−function, c = const, b(x) and a(x) are known constants outsidea compact domain, k2 = const > 0. It is proved in [R139] that both a(x) and b(x) cannot be determinedsimultaneously from the fixed-energy scattering data, but they can be determined simultaneously fromthe scattering data known at two distinct values of k and all α and α′ running independently throughtwo open subsets of S2.

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 L2(D) in place of H2(D): since w(x) and∫S2 u(x, α)νε(α)dα solve

equation (5.1.1), the elliptic estimate (see [GT])

‖ϕ‖H2(D1) ≤ c[‖Lϕ‖L2(D2) + ‖ϕ‖L2(D2)

], D1 ⊂ D2, (5.4.1)

where L = ∇2 + 1 − q(x), D1 is strictly inner subdomain of D2 and c = c(D1, D2) = const > 0, impliesthat ‖ϕ‖H2(D1) ≤ c‖ϕ‖L2(D2) if Lϕ = 0. If (5.1.18), with k = 1 and L2(D) in place of H2(D), is falsethen ∫

D

w(x)

∫

S2

u(x, α)ν(α)dα= 0 ∀ν(α) ∈ L2(S2). (5.4.2)

Therefore ∫

D

w(x)u(x, α)dx = 0 ∀α ∈ S2. (5.4.3)

This implies ∫

D

w(x)G(x, y)dx = 0 ∀y ∈ D′, (5.4.4)

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 ϕ(y). Then

Lϕ = 0 in D′; ϕ = o

(1

|y|

)as |y| −→ ∞. (5.4.5)

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

Lϕ = −w(x) in D, ϕ = 0 in D′, ϕ ∈ H2loc(R

3). (5.4.6)


ThusLϕ = −w in D, ϕ = ϕN = 0 on S. (5.4.7)

Multiply (5.4.7) by w, integrate over D, then by parts using the boundary conditions (5.4.7), use theequation Lw = 0 and get ∫

D

|w(x)|2dx = 0. (5.4.8)

Thus w(x) = 0. Estimate (5.1.18) is proved. 2

5.4.2 Proof of (5.1.20) and (5.1.21)

From (5.1.19) one gets∇2R+ 2iθ · ∇R− q(x)R = q(x) in R3. (5.4.9)

Denote L = ∇2 + 2iθ · ∇, and define

w(x) := L−1f =1

(2π)3

∫

R3

f (ξ)eiξ·x

ξ2 + 2ξ · θdξ. (5.4.10)

Note that Lw = −f(x). We will prove below that:

∥∥L−1f∥∥L∞(D1)

≤ c

(ln |θ||θ|

)12

‖f‖L2(D), θ ∈M, |θ| −→ ∞, (5.4.11)

where D1 is an arbitrary compact domain c = c(D1, ‖q‖L2(Ba)), D ⊂ Ba.We will also prove that

∥∥L−1f∥∥L2(D1)

≤ c

|θ| ‖f‖L2(D), |θ| −→ ∞, θ ∈M. (5.4.12)

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

R = L−1qR+ L−1q. (5.4.13)


∥∥L−1qR∥∥L∞(D)

≤ c

(log |θ||θ|

) 12

‖qR‖L2(D) ≤ c

(log |θ||θ|

) 12

‖q‖L2(D)‖R‖L∞(D). (5.4.14)

Therefore the operator L−1q : L∞(D) → L∞(D) has the norm going to zero as |θ| → ∞, θ ∈ M . Thusequation (5.4.13) is uniquely solvable in L∞(D) if |θ| 1, θ ∈ M . Moreover, the following estimateholds:

‖R‖L∞(D) ≤ c∥∥L−1q

∥∥L∞(D)

≤ c

(ln |θ||θ|

)12

‖q‖L2(D). (5.4.15)

Estimate (5.1.20) follows.To derive (5.1.21) from (5.4.12) one writes

∥∥L−1qR∥∥L2(D)

≤ c

|θ| ‖qR‖L2(D) ≤c

|θ|‖q‖L2(D)‖R‖L∞(D). (5.4.16)

Therefore (5.4.13), (5.4.15) and (5.4.16) yield (5.1.21):

‖R‖L2(D) ≤c

|θ|‖q‖2L2(D)

(log |θ||θ|

)12

+ ‖L−1q‖L2(D) ≤c

|θ| . (5.4.17)


Proof of (5.4.11). If θ ∈ M then θ = a + ib, a, b ∈ R3, a · b = 0, a2 − b2 = 1. Choose the coordinate

system such that a = τe2, b = te1, τ = (1 + t2)12 , ej , 1 ≤ j ≤ 3, are the orthonormal basis vectors. Then

ξ2 + 2θ · ξ = ξ21 + ξ22 + ξ23 + 2τξ2 + 2itξ1 = ξ21 +(ξ2 + τ

)2+ ξ23 − τ2 + 2itξ1. (5.4.18)

This function vanishes if and only if

ξ1 = 0,(ξ2 + τ

)2+ ξ23 = τ2. (5.4.19)

Equation (5.4.19) defines a circle Cτ or radius τ in the plane ξ1 = 0 centered at (0,−τ, 0). Let Tδ bea toroidal neighborhood of Cτ , where the section of the torus by a plane orthogonal to Cτ is a squarewith size 2δ and the center at Cτ .

Denote u(x) := L−1f , where L−1f is defined in (5.4.10). One has

∣∣u(x)∣∣ ≤ 1

(2π)3

∣∣∣∣∣

∫

Tδ

f(ξ)eiξ·xdξ

ξ2 + 2ξ · θ

∣∣∣∣∣ +1

(2π)3

∣∣∣∣∣

∫

R3\Tδ

f (ξ)eiξ·xdξ

ξ2 + 2ξ · θ

∣∣∣∣∣ := I1 + I2, (5.4.20)

I1 ≤ c‖f‖L∞(R3)

∫

Tδ

dξ

|ξ2 + 2ξ · θ|

≤ c‖f‖L1(R3)

∫ δ

−δdξ1

∫ 2π

0

dϕ

∫ τ+δ

τ−δ

ρdρ√4t2ξ21 + (ξ21 + ρ2 − τ2)2

= c‖f‖L1(R3)

∫ δ

0

∫ xi21+2τδ+δ2

ξ21−2τδ+δ2

dµ√4t2ξ21 + µ2

≤ c(D)‖f‖L2(D)

∫ δ

0

dξ1

∫ 3τδ

0

dµ√4t2ξ21 + µ2

, 0 < δ <τ

2,

(5.4.21)

where ρ2 = (ξ2 +τ )2+ξ23 and we have used the Cauchy inequality ‖f‖L1(R3) = ‖f‖L1(D) ≤ c(D)‖f‖L2(D)

and an elementary inequality ξ21 + 2τδ + δ2 ≤ 3τδ, which holds if ξ21 ≤ δ2 and τ > 2δ.Let β := 2tξ1. Then

1

2t

∫ 2tδ

0

dβ

∫ 3τδ

0

dµ√β2 + µ2

≤ 1

2t

∫ 3(t+τ)δ

0

dρρ

∫ π2

0

dϕ1

ρ=

π

4t3(t+ τ )δ ≤ cδ, (5.4.22)

where we have used the relations τt → 1 as t → ∞ and took into account that t → ∞ if |θ| → ∞. From

(5.4.21) and (5.4.22) one getsI1 ≤ c‖f‖L2(D)δ. (5.4.23)

By c > 0 we denote various constants independent of δ and t.Let us estimate I2:

I22 ≤ c‖f‖L2(R3)

∫

R3\Tδ

dξ

|ξ2 + 2θ · ξ|2 = c‖f‖2L2(D)J , (5.4.24)

where the Parseval equality was used and by J the integral in (5.4.24) is denoted. One has

J ≤∫

|ξ1|>δ

dξ

|ξ2 + 2θ · ξ|2 +

∫

|ξ1<δ,|ρ−τ |≥δ

dξ

|ξ2 + 2θ · ξ|2 := j1 + j2. (5.4.25)

Let us estimate j1:

j1 ≤ c

∫ ∞

δ

dξ1

∫ ∞

0

ρdρ

4ξ21t2 + (ξ21 + ρ2 − τ2)2

≤ c

∫ ∞

δ

dξ1

∫ ∞

ξ21−τ2

dµ

4ξ21t2 + µ2

≤ c

∫ ∞

δ

dξ1ξ1t

(π

2− arctg

ξ21 − τ2

2ξ1t

) (5.4.26)


Let ξ12t = x. Then the integral on the right-hand side of (5.4.26) can be written as:

j1 ≤ c

t

∫ ∞

δ2t

dx

x

[π

2− arctg

(x− τ2

4t21

x

)]. (5.4.27)

If t→ ∞, then τ2

4t2 → 14 . Let us use the elementary inequalities:

π

2− x ≤ arctg

1

x, 0 < x ≤ π

2; (5.4.28)

arctg1

x≤ π

2− x

2, x −→ +0. (5.4.29)

Thenπ

2− 1

y≤ arctg y ≤ π

2− 1

2y, y −→ +∞. (5.4.30)

Thus, with A := τ2

4t2, one gets

1

x

[π

2− arctg

(x− A

x

)]≤ 1

x

(x− A

x

)−1

≤ c

x2, x −→ +∞, (5.4.31)

and1

x

[π

2− arctg

(x− A

x

)]≤ c

x, x −→ +0. (5.4.32)

From (5.4.3), (5.4.4) and (5.4.27) one gets

j1 ≤ c

t

(∫ 1

δ2t

dx

x+ c

)≤ c

tln

δ

2t, t −→ +∞,

so

j1 ≤ cln δ

2t

t, t −→ +∞. (5.4.33)

Let us estimate j2:

j2 =

∫ δ

−δdξ1

∫ρ>τ+δ

0<ρ≤τ−δdρρ

∫ 2π

0

dϕ1

4ξ21t2 + (ξ21 + ρ2 − ξ2)2

≤ c

∫ δ

0

dξ1

∫ ∞

ξ21+(τ+δ)2−τ2

dµ

4ξ21t2 + µ2

+

∫ τ2−ξ21

2τδ−δ2−ξ21

dν

4ξ21t2 + ν2

(5.4.34)

where µ = ξ21 + ρ2 − τ2 and ν = τ 2 − ρ2 − ξ21 .One has:

j2 ≤ c

t

∫ δ

0

dξ1ξ1

(π

2− arctg

ξ21 + 2δτ + δ2

2ξ1t+ arctg

τ2 − ξ212ξ1t

− arctg2τδ − δ2 − ξ21

2ξ1t

)

≤ c

t

∫ δ

0

dξ1ξ1

(π

2− arctg

δ

ξ1+ arctg

t

ξ1− arctg

δ

2ξ1

):=

c

tj3,

(5.4.35)

where we have used the monotonicity of arctgx, for example,

arctgξ21 + 2δτ + δ2

2ξ1t≥ arctg

δ

ξ1,


etc., and the relation τt → 1 as t→ +∞, τ > t.

By (5.4.28),

π

2− arctg

δ

ξ1≤ ξ1

δ, arctg

t

ξ1≤ π

2− ξ1

2t, arctg

δ

2ξ1≥ π

2− 2ξ1

δ. (5.4.36)

From (5.4.35) and (5.4.36) one gets:

j2 ≤ c

t

∫ δ

0

dξ1ξ1

(ξ1δ

− ξ12t

+2ξ1t

)=c

t

(1 − δ

2t+

2δ

t

)=c

t

(1 +

3δ

t

). (5.4.37)

Thusj2 ≤ c

t, t −→ +∞. (5.4.38)

From (5.4.20), (5.4.23), (5.4.25), (5.4.33) and (5.4.37) one gets:

|u(x)| ≤ c‖f‖L2(D)

δ +

(| ln δ

2t |t

) 12

+1

t12

. (5.4.39)

Choose δ = 1t. Then (5.4.39) yields

|u(x)| ≤ c‖f‖L2(D)

(ln |θ||θ|

)12

, |θ| −→ ∞, θ ∈M. (5.4.40)

Estimate (5.4.11) is proved. 2

Let us prove (5.4.12).Let L(ξ) = ξ2 + 2θ · ξ, ∂ = −i∇.Define L (ξ) := (

∑|j|≥0 |L(j)(ξ)|2)1/2. Then

L (ξ) = (|ξ2 + 2θ · ξ|2 + 4|ξ + θ|2 + 36)12 ≥ |Imθ| + 3.

In [H, vol. 2, p.31], it is proved that ‖L−1f‖L2(D1) ≤ 1minξ |L (ξ)|‖f‖L2(D). Therefore

∥∥L−1f∥∥L2(D1)

≤ c

minξ∈R3 |L (ξ)| ‖f‖L2(D)

≤ c

|θ|‖f‖L2(D), D ⊂ D1, θ ∈M, |θ| −→ ∞,(5.4.41)

where c = c(D1, D) > 0 is a constant and we have used the relation

c1|θ| ≤ |Imθ| ≤ |θ|, c1 > 0, if θ ∈ M, |θ| −→ ∞.

Estimate (5.4.41) is identical to (5.4.12). 2

5.4.3 Proof of (5.2.17)

If ρ is defined by (2.4), where u(x, α) solves (5.1.1) then ρ solves the equation

∇2ρ+ 2iθ · ∇ρ − q(x)ρ = q(x) in R3, θ ∈M. (5.4.42)

Let h = |θ|−1, h→ 0, ρ(ξ) := ξ2 + 2β · ξ, β = hθ, β · β = h2, |β| = 1,

N := ξ : ρ(ξ) = 0, ξ ∈ R3, Nh := ξ : dist(ξ,N ) ≤ h, ξ ∈ R3, N ′h := R3\Nh,


P = P1 + iP2, P1 = ReP . Note that dP1 6= 0 on N , where dP1 is the differential of P1.Define

Fhu := u =1

(2π)32

∫

R3

u(x)e−iξ·xh−1

dx. (5.4.43)

ThenFh(− i∂ju(x)

)= ξj u(ξ); ih∂ξj u(ξ) = xju. (5.4.44)

Denote‖ρ‖a := ‖ρ‖L2(Ba), ‖ρ‖ := ‖ρ‖L2(R3), ‖ρ‖a,b = ‖ρ‖L2(Ba\Ba), b > a, (5.4.45)

∥∥g(< hD >)ρ∥∥ :=

∥∥∥(√

1 + ξ2)ρ(ξ)

∥∥∥, D = −i∇. (5.4.46)

The following Hardy-type inequality will be useful:If f(t) ∈ C1(−h, h), f(0) = 0, then

∫ h

−ht−2∣∣f(t)

∣∣2dt ≤ 4

∫ h

−h

∣∣f ′(t)∣∣2dt, h > 0. (5.4.47)

Let us sketch the basic steps of the proof of (5.2.17)Step 1. If ρ ∈ C2

0(Br) and

P (hD)ρ := (hD)2ρ + 2β · hDρ = −h2v, v ∈ L20

(Br), (5.4.48)

where L20(Br) is the set of L2(Br) functions with compact support in the ball Br , then

h∥∥ < hD >2 ρ

∥∥ ≤ c∥∥P (hD)ρ

∥∥ ∀h ∈(0, h0

), (5.4.49)

where h0 > 0 is a sufficiently small number.Step 2. Let A1 be a bounded domain with a smooth boundary and A ⊂ A1, η ∈ C∞

0 (A1), 0 ≤ η ≤ 1,η(x) = 1 in A, A is a strictly inner subdomain of A1.

IfP (hD)ρ = 0 in A1, (5.4.50)

thenh∥∥Dρ

∥∥L2(A)

≤ c‖ρ‖L2(A1). (5.4.51)

Step 3. Write (5.4.42) as

P (hD)ρ = −h2(qρ + q). (5.4.52)

Letη ∈ C∞

0 (Bb), 0 ≤ η(x) ≤ 1,

η(x) = 1 x ∈ Ba1 , a < a1 < b.

ThenP (ηρ) = (Pη − ηP )ρ − h2η(qρ + q), P = P (hD). (5.4.53)

Apply (5.4.49) to (5.4.53) and get

h∥∥ < hD >2 (ρη)

∥∥ ≤ c∥∥(Pη − ηP )ρ

∥∥+ ch2‖q‖L∞(Ba)‖ρ‖L2(Ba) + ch2‖q‖L2(Ba).

(5.4.54)

Since η = 1 in Ba, one gets:

h‖ρ‖a ≤ h∥∥ < hD >2 (ηρ)

∥∥ ≤ ch2‖ρ‖a + ch2 + c∥∥(Pη − ηP )ρ

∥∥. (5.4.55)


So

‖ρ‖a ≤ ch+ ch−1∥∥(Pη − ηP )ρ

∥∥. (5.4.56)

Since Dη = 0 in Ba one gets:

∥∥(Pη − ηP )ρ∥∥ =

∥∥ρ(hD)2η + 2h2Dη ·Dρ + 2hρβ ·Dη∥∥

≤ c(h2 + h

)‖ρ‖a1,b + ch2‖Dρ‖a1 ,b.

(5.4.57)

Using (5.4.57), one gets

h‖Dρ‖a1,b ≤ c‖ρ‖a1−ε,b+ε. (5.4.58)

From (5.4.58), (5.4.57) and (5.4.55) one obtains:

‖ρ‖a ≤ c(h+ ‖ρ‖a1−ε,b+ε

). (5.4.59)

Since ε > 0 is arbitrarily small, the desired inequality (5.2.17) follows. 2

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

h∥∥∥(1 + |ξ|2

)ρ∥∥∥ ≤ c

∥∥P (ξ)ρ∥∥. (5.4.60)

If ξ ∈ N ′h, then h(1 + |ξ|2) ≤ c|P (ξ)|, so

h2

∫

N ′h

(1 + |ξ|2

)2∣∣ρ(ξ)∣∣2dξ ≤ c2

∫

N ′h

|P (ξ)|2∣∣ρ(ξ)

∣∣2dξ

≤ c2∫

R3

∣∣P (ξ)ρ(ξ)∣∣2dξ = c2

∫

R3

∣∣P (hD)ρ∣∣2dx.

(5.4.61)

If ξ ∈ Nh, then use the local coordinates in which the set N is defined by the equations:

t = 0, ξ1 = 0, t = P1(ξ), (5.4.62)

and the ξ1-axis is along vector µ defined by the equation β = m + iµ. Since dP1 6= 0 on N , these localcoordinates can be defined.

Put f := P1(ξ)ρ(ξ). Then f = 0 at t = 0, f ∈ C∞(R3) if ρ(x) has compact support, and (5.4.47)yields: ∫ h

−h

∣∣ρ(ξ)∣∣2dt ≤ 4

∫ h

−h

∣∣f ′t∣∣2dt. (5.4.63)

Integrating (5.4.63) over the remaining variables, one gets:

∫

Nh

∣∣ρ(ξ)∣∣2dξ ≤ c

∫

Nh

∣∣∣∇ξ

(P1(ξ)ρ(ξ)

)∣∣∣2

dξ ≤ c

∫

R3

∣∣∇ξ

(P1(ξ)ρ(ξ)

)∣∣∣2

dξ. (5.4.64)

Since Nh is compact, one has

h2

∫

Nh

(1 + |ξ|2

)2∣∣ρ(ξ)∣∣2dξ ≤ ch2

∫

Nh

∣∣ρ(ξ)∣∣2dξ. (5.4.65)


Using Parseval’s equality, S. Bernstein’s inequality for the derivative of entire functions of exponentialtype, and the condition supp ρ(x) ⊂ Br, one gets:

h2

∫

R3

∣∣∣∇ξ

(P1(ξ)ρ(ξ)

)∣∣∣2

dξ =

∫

R3

|x|2∣∣P1(hD)ρ(x)

∣∣2dx = r2∫

Br

∣∣P1(hD)ρ(x)∣∣∣2

dx

≤ r2∫

R3

∣∣P (hD)ρ(x)∣∣2dx.

(5.4.66)

From (5.4.64)-(5.4.66) it follows that

h2

∫

Nh

(1 + |ξ|2

)2∣∣ρ(ξ)∣∣2dξ ≤ c

∫

R3

∣∣P (hD)ρ(x)∣∣2dx. (5.4.67)

Inequality (5.4.49) is proved. 2

5.4.5 Proof of (5.4.51)

Multiply (5.4.50) by ηρ, take the real part and integrate by parts to get:

h

∫

A1

η|∇ρ|2dx = −h2

∫

A1

(ρ∇ρ+ ρ∇ρ

)∇ηdx+ 2Re

(iβj

∫

A1

ρjρηdx

)

=h

2

∫

A1

|ρ|2∇2ηdx+ 2Re(iβj

∫

A1

ρjρηdx),

(5.4.68)

where ρj := ∂ρ∂xj

and summation is done over the repeated indices.

One has ∣∣∇2η∣∣ ≤ c,

∣∣βj∣∣ ≤ 1,

∣∣2ρjρ∣∣ ≤ h

2

∣∣ρj∣∣2 +

2

h|ρ|2. (5.4.69)

From (5.4.69) and (5.4.68) one gets:

h

∫

A1

η|∇ρ|2dx ≤ ch

∫

A1

|ρ|2dx+h

2

∫

A1

η|∇ρ|2dx+2

h

∫

A1

η|ρ|2dx. (5.4.70)

Thus

h2

∫

A

|∇ρ|2dx ≤ h2

∫

A1

η|∇ρ|2dx ≤ c

∫

A1

|ρ|2dx. (5.4.71)

Inequality (5.4.51) is proved. 2

Let us prove that

‖ψ(x, θ) −∫

S2

u(x, α)νε(α)dα‖L2(D) ≤ ε, θ ∈M, |θ| −→ ∞, (5.4.72)

implies

‖νε‖L2(S2) ≥ ceκd2 , κ = |Imθ|, d = diamD, |θ| −→ +∞. (5.4.73)

Indeed, (5.4.72), (5.1.19) and (5.1.20) imply:

‖∫

S2

u(x, α)νε(α)dα‖L2(D) ≥ ‖ψ(x, θ)‖L2(D) − ε ≥ ceκd2 , θ ∈M, |θ| 1. (5.4.74)

If (5.4.73) is false for some ε > 0, then there is a sequence θn ∈ M , |θn| → ∞, such that

‖νε‖L2(S2)e−κnd

2 −→ 0, n −→ ∞. (5.4.75)


This contradicts (5.4.74) since (5.4.75) implies

∥∥∥∥∥

∫

S2

u(x, α)νε(α)dα

∥∥∥∥∥L2(D)

≤ c∥∥νε(α)

∥∥L2(S2)

= o(e

κnd2

)as n −→ ∞. (5.4.76)

Therefore estimate (5.4.73) is proved. 2

5.4.6 Proof of (5.2.13)

One has ∫

S2

u(x, α)ν(α)dα= eiθ·x(1 + ρ),

where

ρ := e−iθ·x∫

S2

u(x, α)ν(α)dα− 1,

ψ(x, θ) = eiθ·x(1 +R), ‖R‖L2(Bb1 ) ≤c

|θ| , θ ∈M, |θ| 1,

where b1 > b.By (5.1.18), there exists a ν(α) such that

‖eiθ·x(1 + ρ) − eiθ·x(1 + R)‖L2(Bb1 ) ≤e−κb1

κ, κ = |Imθ|.

Therefore

‖(ρ− R)eiθ·x‖L2(Bb1 ) ≤e−κb1

κ,

so that

e−κb1‖ρ− R‖L(Bb1 ) ≤e−κb1

κ,

and

‖ρ− R‖L2(Bb1 ) ≤1

κ.

This implies

‖ρ‖L2(Bb1 ) ≤ ‖ρ −R‖L2(Bb1 ) + ‖R‖L2(Bb1 ) ≤c

|θ| .

Thus, inequality (5.2.13) follows.We claim that ‖ρ‖L2(Bb) is O( 1

|θ| ).

To prove this, one uses the above inequalities and gets:

e−bκ‖ρ− R‖L2(Bb) ≤ ‖(ρ −R)eiθ·x‖L2(Bb) ≤ ‖(ρ −R)eiθ·x‖L2(Bb1 ) ≤e−κb1

κ.

This implies the following estimate:

‖ρ− R‖L2(Bb) ≤e−(b1−b)κ

κ.

Recall that κ := | Im θ|, c1|θ| ≤ κ ≤ |θ|, 0 < c1 <12 , as |θ| → ∞, θ ∈M . Therefore,

‖ρ‖L2(Bb) ≥ ‖R‖L2(Bb) − ‖ρ −R‖L2(Bb) ≥c

|θ| −e−γκ

κ, γ = b1 − b > 0.


Thus, the above claim is verified, since, as |θ| → ∞, θ ∈M , one has |θ|κ →

√2 and e−γκ

κ = o(

1|θ|

). 2

Uniqueness class for the solution to the equation Lρ = 0.

Lρ := (∇2 + 2iθ · ∇)ρ = 0 in R3,

∫

R3

|ρ(x)|2(1 + |x|2)`dx < ∞, −1 < ` < 0. (5.4.77)

Taking the distributional Fourier transform of (5.4.77) one gets:

L(ξ)ρ = (ξ2 + 2θ · ξ)ρ = 0. (5.4.78)

Thus supp ρ = Cτ := ξ : ξ ∈ R3, L(ξ) = 0, and Cτ is the circle (5.4.19). By theorem 7.1.27 in [H,vol 1, p.174], one has: ∫

Cτ

|ρ|2ds ≤ c limr→∞

sup

(1

R2

∫

|x|≤R|ρ(x)|2dx

). (5.4.79)

Using (5.4.77) we derive for −1 < ` < 0:

∞ > c >

∫

R3

|u|2(1 + |x|2)`dx ≥∫

|x|≤R

|u|2dx(1 + |x|2)|`|

≥ 1

(1 + R2)|`|

∫

|x|≤R|u|2dx ≥ c

R2|`|

∫

|x|≤R|u|2dx.

(5.4.80)

Combining (5.4.79) and (5.4.80) one gets

∫

Cτ

|ρ|2ds ≤ c limR→∞

supR2|`|

R2= 0, |`| < 1.

Thus ρ(ξ) = 0, as claimed. 2

Estimate (5.4.80) is valid in Rn, n ≥ 2. It was used in [SU] and [R139].

5.4.7 Proof of (5.2.23)

Let ‖νε‖ := ‖νε‖L2(S2) and m(ε, θ) := inf ‖ν‖ where the infimum is taken over all ν ∈ L2(S2) such that(5.4.72) holds.

We wish to prove that

m(ε, θ) ≤ cec|θ| ln |θ| as |θ| −→ ∞, θ ∈M, ε =e−bκ

κ, b > a, (5.4.81)

where θ ∈M , |θ| → ∞, κ = |Imθ|, and c > 0 stands for various constants.Let us describe the steps of the proof.Step 1. Prove the estimate

m(ε, θ) ≤ ceκr(

2n(ε)

er

)n(ε)

n2(ε), r ≥ b, θ ∈ M, ε > 0, (5.4.82)

whereln(n(ε)) = ln(| ln ε|)[1 + o(1)], ε −→ +0. (5.4.83)

The choice of n(ε) 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 r ≥ b to get

m(ε, θ) ≤ c(2κ)n(ε)n2(ε). (5.4.84)


The minimizer is r = n(ε)κ .

Step 3. Take ε = ε(θ) = e−κ

κ, κ→ ∞, in (5.4.84). Then

lnn = ln(κb+ lnκ)[1 + o(1)] = (lnκ)

[1 + O

(1

lnκ

)], κ −→ +∞, (5.4.85)

so, for ε = e−κb

κ one has:c1κ ≤ n ≤ c2κ, κ −→ +∞, c1 > 0. (5.4.86)

From (5.4.84) and (5.4.85) one gets:

m(θ) = m(ε(θ), θ) ≤ cec|θ| ln |θ|, |θ| −→ ∞, θ ∈M. (5.4.87)

Estimate (5.4.81) is obtained.Proof of (5.4.82). Since u(x, α) = (I + T1)e

iα·x where I + T1 is a bijection of C(Bb) onto C(Bb),inequality (5.4.72) with D = Bb is equivalent to

‖(I + T1)−1ψ −

∫

S2

eiα·xνε(α)dα‖L2(Bb) ≤ cε, (5.4.88)

where c = const > 0 does not depend on ε and θ, (I + T1)−1ψ = (I + T )ψ,

Tψ =

∫

Ba

ei|x−y|

4π|x− y|q(y)ψ(y)dy.

We take b > a, therefore the function ϕ := ψ + Tψ, has the maximal values, as |θ| → ∞, of the sameorder of magnitude as the function ψ. The function ϕ solves the equation

(∇2 + 1

)ϕ = 0 in R3. (5.4.89)

Indeed, (∇2 + 1)ϕ = (∇2 + 1)ψ − qψ = qψ − qψ = 0, as claimed.Therefore one can write:

ϕ := ϕ(x, θ) =

∞∑

`=0

4πi`ϕ`Y`(α′)j`(r), r = |x|, α′ =

x

|x| , (5.4.90)

where Y` are defined in (5.1.26), j`(r) are defined in (5.1.29), and ϕ` = ϕ`(θ) are some coefficients.It is known that

eiα·x =

∞∑

`=o

4πi`Y`(α)Y`(α′)j`(r), (5.4.91)

so ∫

S2

eiα·xνε(α)dα =

∞∑

`=0

4πi`νε`Y`(α′)j`(r), (5.4.92)

where νε` = (νε, Y`)L2(S2).Choose

νε` = ϕ` for ` ≤ n(ε), νε` = 0 for ` > n(ε), (5.4.93)

where n(ε) is the same as in (5.4.83).Then (5.4.88) implies:

‖ϕ−∫

S2

eiα·xνε(α)dα‖2L2(Bb)

=

∞∑

`=n(ε)+1

16π2

∫ b

0

r2|j`(r)|2dr|ϕ`|2 ≤ c

∞∑

`=n(ε)+1

1

`2|ϕ`|2

(eb

2`

)2`

< cε,(5.4.94)


where formula (5.1.29) was used.

From (5.4.92) and formula (5.4.91) with α = θ ∈M , one gets:

‖νε‖2 =

n(ε)∑

`=0

|ϕ`|2 ≤ c

n(ε)∑

`=0

∑

m=−`|Y`(θ)|2 ≤ cn2(ε)

e2κr

|jn(ε)(r)|2, ∀r > 0, (5.4.95)

where we have used the formula∑n`=0

∑`m=−` = (n + 1)2, we estimated |ϕ`| by the coefficient

|(eiθ·x, Y`)L2(S2)|2 = 16π2|Y`(θ)|2 of the main term of ϕ, that is, the function eiθ·x, we used estimate

(5.1.28), which gives |ϕ`|2 ≤ c e2κr

|j`(r)|2 ∀r > 0, and we replaced |j`(r)| by |jn(ε)(r)|, the smaller quantity.

Choose r > b and use (5.1.29) to get the inequality:

∞∑

`=n(ε)+1

|ϕ`|2(eb

2`

)2`

≤∞∑

`=n(ε)+1

e2κr(b

r

)2`

≤ c1e2κr

(b

r

)2n(ε)

< cε, r > b, (5.4.96)

which implies (5.4.94). Thus (5.4.94) holds if eκr(br

)n(ε) ≤ c√ε, where c stands for various constants.

One has minr>b eκr(br

)n= en

(bκn

)n, and the minimizer is r = n

κ. Consider therefore the equation

en(bκn

)n= c

√ε and solve it asymptotically for n = n(ε) as ε → 0, where κ > 1 is arbitrary large but

fixed. Taking logarithm, one gets ln c− 12 ln 1

ε = n− n lnn+ n ln(bκ).

Thus | ln ε| = ln 1ε

= 2n lnn[1 + o(1)], and

ln | ln ε| = (lnn)(1 + o(1)), ε −→ +0. (5.4.97)

Hence, we have justified (5.4.83).

From (5.4.94), (5.4.96) and (5.1.29), one gets

‖νε‖ ≤ c n2(ε)eκr (2n(ε))n(ε)

(er)n(ε)∀r > b, κ = |Imθ|, θ ∈M. (5.4.98)

Estimate (5.4.82) is established. 2

5.4.8 Proof of (5.1.30)

Let Gj be the Green function corresponding to qj(x), j = 1, 2. By Green’s formula one gets

G2(x, y) − G1(x, y) =

∫

Ba

p(z)G1(x, z)G2(z, y)dz, p := q1(x) − q2(x). (5.4.99)

Take |y| → ∞, y|y| = −α and use (5.1.7) to get:

u2(x, α) − u1(x, α) =

∫

Ba

p(z)G1(x, z)u2(z, α)dz. (5.4.100)

Take |x| → ∞, x|x| = α′, use (5.1.7) and (5.1.2) and get:

A2(α′, α)− A1(α

′, α) =1

4π

∫

Ba

u1(z,−α′)u2(z, α)p(z)dz. (5.4.101)

Since A(α′, α) = A(−α,−α′), formula (5.4.101) is equivalent to (5.1.30). 2

5.5. CONSTRUCTION OF THE DIRICHLET-TO-NEUMANN MAP 223

5.5 Construction of the Dirichlet-to-Neumann map from thescattering data and vice versa

Consider a ball Ba ⊃ D = supp q(x) and assume that the problem

[∇2 + 1 − q(x)

]w = 0 in Ba, w = f on Sa := ∂Ba, (5.5.1)

is uniquely sovable for any f ∈ H32 (Sa), where H`(Sa) is the Sobolev space.

Then the D − N map is defined as

Λ : f −→ wN (5.5.2)

where wN is the normal derivative of w on Sa, N is the normal to Sa pointing into B′a := R3\Ba.

If Λ is known, then q(x) can be found as follows.The special solution (5.1.19)-(5.1.22) satisfies the equation:

ψ(x) = eiθ·x −∫

Ba

G(x− y)q(y)ψ(y)dy, (5.5.3)

where G(x) = eiθ·xG0(x) and ∇2G(x) +G(x) = −δ(x) in R3. Thus

∇2G0 + 2iθ · ∇G0 = −δ(x), (5.5.4)

so that G0(x − y) is the Green function of the operator L, see (5.4.10), that is

G0(x) =1

(2π)3

∫

R3

eiξ·xdξ

ξ2 + 2ξ · θ . (5.5.5)

The function G(x) can be considered known.

Since qψ = (∇2 + 1)ψ, one can write, for x ∈ B′a,

∫

Ba

G(x− s)q(y)ψ(y)dy

∫

Ba

G(∇2 + 1

)ψdy

=

∫

Sa

[G(x− s)ψN (s) − GN (x− s)ψ(s)] ds =

∫

Sa

G(x− s)(Λ − Λ0)ψ(s)ds

+

∫

Sa

[G(x− s)Λ0ψ − GN (x− s)ψ] ds =

∫

Sa

G(x− s)(Λ − Λ0)ψ(s)ds.

(5.5.6)

Here Λ0 is Λ for q(x) = 0, we have used Green’s formula and took into account that

∫

Sa

[G(x− s)Λ0ψ −GN (x− s)ψ] ds =

∫

Ba

[G(∆ + 1)ϕ− ϕ(∆ + 1)G]dy = 0,

where ϕ solves problem (5.5.1) with q(x) = 0 and ϕ = f on Sa.From (5.5.3) and (5.5.6) taking x → s ∈ Sa one gets a linear Fredholm- type equation for ψ|Sa :

ψ(s) = eiθ·s −∫

Sa

G(s − s′)(Λ − Λ0)ψ(s′)ds′. (5.5.7)

If Λ is known, one can find from (5.5.7) ψ|Sa and then find q(x) using the following calculation.

Define

t(θ′, θ) :=

∫

Ba

e−iθ′·yq(y)ψ(y, θ)dy. (5.5.8)


By Green’s formula, as in (5.5.6), one gets

t(θ′, θ) =

∫

Sa

e−iθ′·s(Λ − Λ0)ψ(s)ds. (5.5.9)

From (5.5.8) one gets, using (5.1.19), (5.1.20) and (5.1.9):

lim|θ| −→ ∞θ′ − θ = ξθ ∈M

t(θ′, θ) =

∫

Ba

e−iξ·xq(x)dx := q(ξ). (5.5.10)

Therefore the knowledge of Λ allows one to recover q(ξ) by formula (5.5.10), but first one has tosolve equation (5.5.7). We leave to the reader to check that the homogeneous equation (5.5.7) has onlythe trivial solution so that Fredholm-type equation (5.5.7) is uniquely solvable in L2(Sa) (see a proof in[R172]).

Practically, however, there are essential difficulties: a) the function G(x, y) is not known, analyticallyand it is difficult to solve equation (5.5.7) by this reason, b) the D −N map is not given analytically aswell.

Let us show how to construct Λ from the scattering amplitude A(α′, α) and vice versa. If Λ is giventhen we have shown how to find q(x) and if q(x) is found then A(α′, α), the scattering amplitude, canbe found.

Conversely, suppose A(α′, α) is known. Then the scattering solution can be calculated in B′a by

formula (5.1.31).

Let f ∈ H32 (Sa) be given, g(x, y) be the Green function of the operator −∇2 + q(x)− 1 in R3 which

satisfies the radiation condition (5.1.6), and define

w(x) =

∫

Sa

g(x, s)σ(s)ds, (5.5.11)

such that

w = f on Sa. (5.5.12)

Since (∇2 + 1)w = 0 in B′a, w = f on Sa and w satisfies (5.1.6), one can find w in B′

a explicitly:

w(x) =

∞∑

`=0

f`h`(a)

Y`(α′)h`(r), r ≥ a, z = |x|, α′ =

x

r, (5.5.13)

where f` are the Fourier coefficients of f :

f(s) =

∞∑

`=0

f`Y`(α′), s ∈ Sa. (5.5.14)

Therefore the function

w−N = lim

|x|→a,x∈B′a

∂w(x)

∂r

is known. By the jump formula for single-layer potentials one has:

w+N = w−

N + σ. (5.5.15)

The map Λ : f → w+N is constructed as soon as we find σ(s), because w−

N is already found.

5.5. CONSTRUCTION OF THE DIRICHLET-TO-NEUMANN MAP 225

To find σ, consider the asymptotics of w(x) as |x| → ∞, x|x| = β. Using (5.1.7) and (5.5.11), one

gets:

1

4π

∫

Sa

u(s,−β)σ(s)ds = η(β) :=

∞∑

`=0

f`Y`(β)

h`(a), (5.5.16)

where we have used (5.5.13) and the asymptotics h`(r) ∼ eir

r as r → +∞. As we have already mentioned,the function u(s, α′) is known explicitly (see formula (5.1.31)), and equation (5.5.16) is uniquely solvablefor σ(s). Analytical solution of equation (5.5.16) for σ(s) can be obtained as a series

σ(s) =∞∑

`=0

σ`Y`(α′), α′ =

s

|s| . (5.5.17)

Substitute (5.4.91) with α = −β into (5.1.31), take r = a in (5.1.31) and α′ = sa , and substitute

(5.1.31) into (5.5.16). By our choice of the spherical harmonics (5.1.26) both systems Y``=0,1,2,... andY``=0,1,2,... form orthonormal bases of L2(S2). Therefore one gets:

1

4π

∞∑

`=0

4πi`Y`(−β)j`(a)a2

∫

S2

Y`(α′)σ(aα′)dα′

+1

4π

∞∑

`=0

A`(−β)h`(a)a2

∫

S2

Y`(α′)σ(aα′)dα′

=

∞∑

`=0

f`Y`(β)

h`(a).

(5.5.18)

Denote ∫

S2

σ(aα′)Y`,m(α′)dα′ := σ`m. (5.5.19)

Using (5.1.26) one gets:

Y`,m(−β) = (−1)`Y`,m(β), Y`,m(−β) = (−1)`Y`,m(β)

= (−1)`+`+mY`,−m(β) = (−1)mY`,−m(β).(5.5.20)

Also define A`m,`′m′ by the formula:

A`,m(−β) =∑

`′,m′

A`m,`′m′Y`′,−m′(β). (5.5.21)

The above definition differs from (5.1.36) and is used for convenience in this section.Equating the coefficients in front of Y`,−m(β) in (5.5.18) one gets

i`(−1)mj`(a)a2σ`m +

a2

4π

∞∑

`′=0

`′∑

m′=−`′A`′m′ ,`mh`′(a)σ`′m′ =

f`,−mh`(a)

, (5.5.22)

or

σ`m +(−1)m(−i)`

4πj`(a)

∞∑

`′=0

`′∑

m′=−`′A`′m′,`mh`′(a)σ`′m′ =

f`,−m(−1)m(−i)à2j`(a)h`(a)

. (5.5.23)

The matrix of the linear system (5.5.23) is ill-conditioned (see [R203], where estimates of the entries ofthe matrix of (5.5.19) are obtained and the case of the noisy data is mentioned). 2

Finally let us show (see [R139]) that it is impossible to get an estimate

‖Qf‖ ≤ ε(|θ|)‖f‖, θ ∈ M, ‖f‖ := ‖f‖L2(D), ε(t) −→ 0 as t −→ +∞, (5.5.24)


if

Qf =

∫

D

Γ(x, y, θ)f(y)dy, (5.5.25)

whereLΓ :=

(∇2 + 2iθ · ∇

)Γ = −δ(x− y) in D, θ ∈M, (5.5.26)

Γ = 0 on S := ∂D, (5.5.27)

and we assume that

the problem Lρ = 0, ρ = 0 on S has only the trivial solution. (5.5.28)

Indeed, choose a q(x) ∈ L∞(D) such that the problem[∇2 + 1 − q(x)

]w = 0 in D, w = 0 on S, (5.5.29)

has a non-trival solution.Define ρ = e−iθ·xw. Then ρ 6≡ 0, and

Lρ− qρ = 0 in D, ρ = 0 on S. (5.5.30)

Because of our assumption (5.5.24), one gets:

ρ =

∫

D

Γ(x, y)q(y)ρ(y)dy := Tρ. (5.5.31)

Were (5.5.24) true, it would imply for |θ| 1, θ ∈M , that the operator T : L2(D) → L2(D) in (5.5.31)has small norm, so ρ = 0, contrary to our assumption. 2

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 ρ is defined in (5.2.4), that is,approximate minimizing a quadratic functional ||ρ||2, so that (5.2.15) be satisfied. This can be done byusing a necessary condition for the minimizer, which is a linear equation. If the minimizer is sought ona finite-dimensional space of the elements ν =

∑L`=0 c`Y`, then the linear equation is a linear algebraic

system for finding the coefficients c`, and the functional is a function of finitely many variables c`.The computational difficulty in this case is the ill-condition nature of the matrix of the above linearalgebraic system. Alternatively, one can use one of the known methods for global minimization ofconvex 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 Λ 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 inverseproblems.

The basic definition is 5.1.1. Consider the case of partial differential expressions with constantcoefficients:

Ljw =

Mj∑

|m|=0

ajm∂|m|w(x), ajm = const , Mj ≥ 1, x ∈ Rn, n ≥ 2, j = 1, 2. (5.6.1)

5.6. PROPERTY C 227

Define algebraic varieties:

Lj :=

z : z ∈ Cn,

Mj∑

|m|=0

ajm z|m| = 0

, j = 1, 2. (5.6.2)

Let Nj := w : Ljw = 0, where Ljw = 0 is understood in the sense of distributions.

Definition 5.6.1. We say that L1 is transversal to L2 and write L1 ∦ L2 iff there exist at least onepoint z1 ∈ L1 and at least one point z2 ∈ L2 such that the corresponding tangent spaces T1 and T2 aretransversal T1 ∦ T2. Here Tj is the tangent space in Cn to Lj at the point z(j).

Note that ex·z ∈ Nj iff z ∈ Lj.

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 arbitrarybounded domain, and p ≥ 1 is an arbitrary fixed number. We take p = 2.

The basic result of this Section is

Theorem 5.6.3. A necessary and sufficient condition for the pair L1, L2 to have property C is L1 ∦L2.

Proof. Sufficiency: Suppose

∫

D

f(x)w1w2dx = 0 ∀wj ∈ Nj , j = 1, 2, D ⊂ Rn, n ≥ 2, (5.6.3)

where f ∈ Lp(D), p ≥ 1 is a fixed number, and D is a bounded domain. We want to derive f = 0 from(5.6.3). Take wj = ex·zj , zj ∈ Lj , so wj ∈ Nj . Then (5.6.3) implies:

F (ζ) :=

∫

D

f(x)ex·ζdx = 0, ζ = z1 + z2, ∀zj ∈ Lj (5.6.4)

If the set ζ = z1 + z2 contains a ball B ⊂ Cn or B ⊂ Rn, then the entire function F (ζ) vanishesin this ball and, by analyticity, F (ζ) = 0 in Cn. Consequently, f = 0. Thus, we have to check that theset z1 + z2∀zj∈Lj contains a ball. This follows from the assumption L1 ∦ L2. Indeed, take a basis inT1. It consists of n− 1 complex vectors ξ1, . . . , ξn−1. Since T1 ∦ T2, there exists a vector ξn in T2 whichhas a non-zero projection onto the normal to T1. The system ξ1, ξ2, . . . , ξn forms a basis in Cn.

In a neighborhoods of the points z(1)and z(2) one can find vectors on L1 and L2 arbitrary close inthe norm to vectors ξj , 1 ≤ j ≤ n− 1, ξj ∈ T1, and ξn ∈ T2.

These vectors form a basis of Cn also. Their linear span contains a ball. The sufficiency is proved. 2

Necessity: Assume that the pair L1, L2 has property C. We want to prove that L1 ∦ L2. Assume thecontrary and find f(x) 6≡ 0, such that (5.6.3) holds, so the set w1w2∀wj ∈ Nj is not complete in Lp(D).The contrary means that L1 ∪ L2 is a union of parallel hyperplanes in Cn. Let b be a normal vectorto these hyperplanes. By choosing properly a basis in Cn one may assume that b = bnen, where is anorthonormal basis. Then z1 · en = 0 and z2 · en = 0, so z1 + z2 =

∑n−1m=1 pmem, pj are constants, zj ∈ Lj ,

j = 1, 2. Let D = Dn−1 × (−a, a). Every bounded domain is a subdomain of such D. Let

f(x) :=

g(xn) in D,

0 in D′ := Rn/D,(5.6.5)

where g 6≡ 0, g ∈ C∞0 (−a, a),

∫ a−a gdxn = 0.


Then ∫

D

f(x)e∑n−1

m=1 pmxmdx =

∫ a

−adxng(xn)

∫

Dn−1

e∑n−1

m=1 pmxmdx1 . . .dxn−1 = 0. (5.6.6)

Thus f(x) 6≡ 0 is orthogonal to all the products ex·z1ex·z2 , zj ∈ Lj . Any element wj ∈ Nj is a limit of alinear combination of the exponential solutions if the polynomial Lj(z) is irreducible. Otherwise thereare elements in Nj of the form wj = P (x)ezj ·x, where P (j) are polynomials of some degree γ, where0 ≤ γ < ∞ depends on Lj(z). Our argument in (5.6.6) is given for γ = 0. If γ > 0, then g can be chosenorthogonal to all the powers xjn, 0 ≤ j ≤ γ. Thus, the necessity is proved. 2

Let us apply Theorem 5.6.3 to the case of one operator L using Definition 5.6.2. The condition L ∦ Lis satisfied if and only if L is not a union of parallel hyperplanes. Algebraically this condition means,that L(z) is not of the form L(z) =

∏jj=1(b0j +

∑nm=1 bmzm)νj , where bm are constants independent of

j.Consider, for example, the classical operators:

L = ∇2, L = i∂t − ∆, L = ∂t − ∆, L = ∂2tt − ∆,

namely, Laplace, Schroedinger, heat, and wave operators with constant coefficients. The correspondingvarieties L are:

z21 + z2

2 + . . .+ z2n = 0, iz0 − z2

1 − . . .− z2n = 0, z0 − z2

1 − . . .− z2n = 0, z2

0 − z21 − . . .− z2

n = 0

The condition L ∦ L is satisfied. Therefore the following result holds:

Theorem 5.6.4. The operators L = ∆, L = i∂t − ∆, L = ∂t − ∆, and L = ∂2t − ∆ have property C.

5.7 Necessary and sufficient condition for scatterers to be spher-

ically symmetric

In this section we give an easily verifiable condition on the scattering amplitude for the correspondingscatterer 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[∇2 + k2 − q(x)

]u = 0 in R3, k > 0,

u = u0(x, θ, k) + Aq(θ′, θ, k)r−1 exp(ikr) + o(r−1), r = |x| → ∞, xr−1 = θ′,

(5.7.1)

θ, θ′ ∈ S2, the unit sphere, u0(x, θ, k) := exp(ikθ · x).We are interested in the following equation: if one changes q(x) in (1) to q L, where L is a linear

transformation in R3, what is the corresponding change in Aq(θ′, θ, k)? For example, let

(q L)(x) = (q R)(x) := q(R−1x), R ∈ O(3) (5.7.2)

where O(3) is the group of rotations in R3, or

(q L)(x) = (q λ)(x) := q(λx), λ > 0 (5.7.3)

or(q L)(x) = (q T−a)(x) := q(x− a), a ∈ R3. (5.7.4)

Here R, λ and T−a are operators of rotation, scaling by a factor λ > 0 and translation by a vector −arespectively.

5.7. NECESSARY AND SUFFICIENT CONDITION FOR SCATTERERS 229

The basic results are the following formulas for the corresponding transformation of the scatteringamplitudes:

Aq(θ′, θ) = AqR(Rθ′, Rθ), (q R)(x) := q(R−1x) (5.7.5)

Aq(θ′, θ, k) = λAλ2qλ(θ

′, θ, λk), (λ2q λ)(x) := λ2q(λx) (5.7.6)

Aq(θ′, θ, k) = AqT−a(θ′, θ, k) exp ik(θ′ − θ) · a , (q T−a)(x) := q(x− a) (5.7.7)

As an application, we prove that a necessary and sufficient condition for q(x) to be spherically symmetricis that the following equation holds:

A(θ′, θ, k) = A(θ′ · θ, k), ∀θ′, θ ∈ S2 (5.7.8)

at a fixed k > 0, provided that

q ∈ Qb :=q : q = q, q ∈ L2(Bb), q = 0 for |x| > b

(5.7.9)

where Bb := x : x ∈ R3, |x| ≤ b, b > 0 is an arbitrary large number and the overbar denotes complexconjugate. A similar result is proved for scattering by an obstacle. Namely let

(∇2 + k2)u = 0 in Ω ⊂ R3, k > 0 (5.7.10)

uN + ζ(s)u = 0 on Γ, Imζ(s) ≥ 0 (5.7.11)

Here Ω is the exterior of a bounded domain D with a smooth connected boundary Γ, N is the unitnormal to Γ pointing into Ω.

Let us formulate the results.

Theorem 5.7.1. Formulas (5.7.5), (5.7.6), and (5.7.7) hold.

Theorem 5.7.2. Assume that (5.7.9) holds and A(θ′, θ, k) := Aq(θ′, θ, k). Then (5.7.8) holds at a fixed

k > 0 iff q(x) = q(r).

Theorem 5.7.3. Assume that (5.7.8) holds at a fixed k > 0 with A(θ′, θ, k) := AΓ,ζ(θ′, θ, k). Then Γ is

a sphere and ζ(s) = const. 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 directconsequences of the definition of the scattering amplitude.

Formula (5.7.5) means that the scattering amplitude is the same in a coordinate system τ and thecoordinate system Rτ in which each vector becomes Ra, where R ∈ O(3) is an arbitrary rotation, andO(3) is the group of rotations in R3. The ”rotated ” potential q R is defined in (5.7.5). A rigorousderivation of formulas (5.7.5), (5.7.6), and (5.7.7) is based on writing the asymptotics of the scatteringsolution u(x, θ, k) in the new coordinate system. This derivation is the same in all three cases. Let usgive it briefly.a) Derivation of formula (5.7.5). Consider

[∇2x + k2 − q(R−1x)

]u = 0 in R3 (5.7.12)

u(x, θ, k) = exp(ikθ · x) +AqR(θ′, θ, )|x|−1 exp(ik|x|) + o(|x|−1), θ′ =x

|x| := x0. (5.7.13)


Let ξ := R−1x, ξ0 := ξ|ξ| = R−1θ′, |ξ| = |x|. Note that

Rθ′ ·Rθ = θ′ · θ, Rx · a = x ·R−1a (5.7.14)

since R′ = R−1 for any R ∈ O(3). Here R′ is the transposed operator of rotation. Note that ∇2x = ∇2

ξ

is invariant under rotations. Write (5.7.12) and (5.7.13) in the ξ-coordinates:

[∇2ξ + k2 − q(ξ)

]w = 0, (5.7.15)

w = exp(ikθ ·Rξ) + AqR(θ′, θ, k)|ξ|−1 exp(ik|ξ|) + o(|ξ|−1), θ′ = Rξ0 (5.7.16)

Note that, by (5.7.13), θ ·Rξ = R−1θ · ξ. Thus, the function w in (5.7.16) is the scattering solution:

w = exp(ikR−1θ · ξ) +Aq(α′, R−1θ)|ξ|−1 exp(ik|ξ|) + o(|ξ|−1) (5.7.17)

corresponding to the incident direction R−1θ. Let R−1θ := α, R−1θ′ = α′ := ξ0. Then

Aq(α′, α) = AqR(Rα′, Rα) ∀α, α′ ∈ S2.

This is formula (5.7.5).b) Derivation of formula (5.7.6). Consider

[∇2x + k2 − q(λx)

]u = 0 in R3, (5.7.18)

u = exp(ikθ · x) +Aqλ(θ′, θ, k)|ξ|−1 exp(ik|x|) + o(|x|−1), (5.7.19)

Let ξ = λx. Then (5.7.18) can be written as

[∇2ξ +

k2

λ2− q(ξ)

λ2

]w = 0. (5.7.20)

The scattering solution, corresponding to (5.7.20), is

w = exp(ik

λθ · ξ) +Aλ−2q(θ

′, θ,k

λ)exp(i k

λ|ξ|)

|ξ| + o(1

|ξ|), (5.7.21)

while (5.7.19) can be written as

u = exp(ik

λθ · ξ) + λAqλ(θ′, θ, k)|ξ|−1 exp(i

k

λ|ξ|) + o(

1

|ξ| ). (5.7.22)

Compare (5.7.20), (5.7.21), and (5.7.22) to get

Aλ−2q(θ′, θ

k

λ) = λAqλ(θ′, θ, k) (5.7.23)

Let kλ := µ, λ−2q = p. Then (5.7.23) can be written as

Ap(θ′, θ, µ) = λAλ2pλ(θ

′, θ, λµ) (5.7.24)

This is formula (5.7.6) with p = q and µ = k.c) Derivation of formula (5.7.7). Consider

[∇2x + k2 − q(x− a)

]u = 0 in R3 (5.7.25)

u = exp(ikθ · x) +AqT−a (θ′, θ, k)|ξ|−1 exp(ik|ξ|) + o(|ξ|−1) (5.7.26)

5.7. NECESSARY AND SUFFICIENT CONDITION FOR SCATTERERS 231

Let x− a = ξ. Then (24) becomes [∇2x + k2 − q(ξ)

]w = 0, (5.7.27)

and

w = exp(ikθ · ξ) + Aq(θ′, θ, k)

exp(ik|ξ|)|ξ| + o(

1

|ξ| ), (5.7.28)

while

u = exp(ikθ · ξ) exp(ikθ · a) + AqT−a(θ′, θ, k)|ξ + a|−1 exp(ik|ξ + a|) + o(1

|ξ| ). (5.7.29)

Note that

|ξ + a| = |ξ|+ ξ0 · a +O(|ξ|−1), ξ0 :=ξ

|ξ| → x0 = θ′, as |ξ| → ∞. (5.7.30)

Thus (5.7.29) becomes

u = exp(ikθ · a)[exp(ikθ · ξ) +AqT−a (θ′, θ, k) expik(θ′ − θ) · aexp(ik|ξ|)

|ξ| + o(1

|ξ|)]

(5.7.31)

Since u solves the homgeneous linear equation (5.7.25), the expression in brackets in (5.7.31) solvesequation (5.7.27). Compare (5.7.28) and (5.7.31) to get

Aq(θ′, θ, k) = AqT−a (θ′, θ, k) expik(θ′ − θ) · a (5.7.32)

This is formula (5.7.7). 2

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 qj ∈ Qa, j = 1, 2. If Aq1(θ′, θ, k) = Aq2(θ

′, θ, k) at a fixed k > 0 and all θ′,θ ∈ S2, then q1 = q2.

Theorem 5.7.2 follows immediately from formula (5.7.5) and Proposition 5.7.4. Indeed, it is wellknown (and follows immediately from separation of variables) that if the potential is spherically sym-metric

q(x) = q(|x|) = q(r), r = |x| (5.7.33)

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 A = Aq .Then

Aq(θ′, θ, k) = Aq(θ

′ · θ, k) = Aq(Rθ′ ·Rθ, k) = AqR(Rθ′, Rθ, k), ∀R ∈ O(3), (5.7.34)

Here the second equation follows from (5.7.13) and the third equation is formula (5.7.5). Since qR ∈ Qa

if q ∈ Qa, and since θ′ and θ are arbitrary, one can write (5.7.34) as

Aq(α′, α, k) = AqR(α′, α, k) ∀R ∈ O(3), ∀α′, α ∈ S2 (5.7.35)

where α′ = Rθ′, α = Rθ. By Proposition 5.7.4, it follows that

q = q R ∀R ∈ O(3) (5.7.36)

This is equivalent to (5.7.33). Theorem 5.7.2 is proved. 2

Proof. b) Proof of Theorem 5.7.3. The basic auxiliary result is the following uniqueness theorem, whichfollows from Theorem 5.2.1:


Proposition 5.7.5. Assume that AΓ1,ζ1(θ′, θ, k) = AΓ2,ζ2

(θ′, θ, k) for all θ′, θ ∈ S2 and a fixed k > 0.Then Γ1 = Γ2 and ζ1(s) = ζ2(s).

Note that if Γ is a sphere and ζ = const, then (5.7.8) holds with A(θ′, θ, k) = AΓ,ζ(θ′, θ, k) for all

θ′, θ ∈ S2 and all k > 0. This follows from the analytical solution of the scattering problem by separationof variables. Assume now that (5.7.8) holds at a single fixed k > 0. Then, by (5.7.8) and (5.7.13),

AΓ,ζ(θ′, θ, k) = AΓ,ζ(θ

′ · θ, k) = AΓ,ζ(Rθ′ ·Rθ, k) (5.7.37)

and

AΓ,ζ(θ′, θ, k) = AΓR,ζR(Rθ′ ·Rθ, k) (5.7.38)

by formula analogous to (5.7.5). Here Γ R is the surface Γ rotated by the element R ∈ O(3), and(ζ R)(s) = ζ(R−1s). From (5.7.37) and (5.7.38) it follows that

AΓ,ζ(α′, α, k) = AΓR,ζR(α′, α, k) ∀R ∈ O(3), ∀α′, α ∈ S2 (5.7.39)

By proposition 5.7.5, equation (5.7.39) implies

Γ = Γ R, ζ = ζ R ∀R ∈ O(3) (5.7.40)

Thus Γ is a sphere and ζ(s) = ζ(|s|) = const on the sphere Γ. Theorem 5.7.3 is proved. 2

Remark 5.7.6. Consider compactly supported potentials q = q(p, z), z = x3, p = (x21 + x2

2)1/2. Let Rϕ,

0 ≤ ϕ ≤ 2π, denote rotations about the x3-axis. Then q Rϕ = q. By formula (5.7.5) for R = Rϕ onehas

Aq(θ′, θ, k) = AqRϕ(Rϕθ

′, Rϕθ, k) = Aq(Rϕθ′, Rϕθ, k) (5.7.41)

This symmetry property

Aq(θ′, θ, k) = Aq(Rϕθ

′, Rϕθ, k) (5.7.42)

is a necessary property of the scattering amplitude corresponding to the potential which is axially sym-metric about the x3-axis, i.e., q(x) = q(p, z). If q ∈ Qb, then (5.7.42) is also sufficient for q(x) to beaxially symmetric about the x3-axis. This is proved as in Theorem 5.7.1.

Remark 5.7.7. Consider the equation

[∇2 + k2 + k2v(x)

]u = 0 in R3, k > 0 (5.7.43)

for v ∈ Qa. At a fixed k > 0 the scattering of a plane wave by the inhomogeneity v(x) is identical withthe scattering of this wave by the potential q(x) = −k2v(x). Therefore, by Theorem 5.7.2, v(x) = v(|x|)if and only if condition (5.7.8) holds at a fixed k > 0, where A(θ′, θ, k) is the scattering amplitudecorresponding to equation (5.7.43).

Remark 5.7.8. The argument in this Section is based on uniqueness theorems. If q ∈ Q(β) := q :|q(x)| ≤ c(1 + |x|)−β, q = q, β > 3, Proposition 2 is no longer valid. However, for this class ofpotentials the following uniqueness theorem holds: if Aq1(θ

′, θ, k) = Aq2(θ′, θ, k) for all θ′, θ ∈ S2 and all

sufficiently large k > 0, then q1 = q2. The argument used in the proof of Theorem 5.7.2 shows that ifq ∈ Q(β), β > 3, and if (5.7.8) holds for all k > 0, then q(x) = q(|x|). The uniqueness theorem holdseven for q ∈ Q(β) with β > 1. Therefore, the following theorem holds.

Theorem 5.7.9. If q ∈ Q(β), β > 1, then q(x) = q(r) iff (5.7.8) holds for all sufficiently large k > 0.

Remark 5.7.10. If q ∈ Q(β), 1 < β ≤ 3, then A(θ′, θ, k) is not necessarily continuous in θ′ and θ sothat (5.7.8) is understood as equality of kernels of operators on L2(S2).

5.8. THE BORN INVERSION 233

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 otherinverse problems. Let us first describe the Born inversion. The starting point is the formula

A(θ′, θ, k) = − 1

4π

∫exp(−ikθ′ · x)u(x, θ, k)q(x)dx,

∫=

∫

R3

(5.8.1)

for the scattering amplitude. The scattering solution in (5.8.1) is substituted by the incident fieldu → u0 = exp(ikθ · x) according to the Born approximation. The error of this approximation can beestimated in terms of ‖q‖ if this norm is sufficiently small and suitably chosen. The other possibility toestimate the error of the Born approximation is to assume that k is sufficiently large. The estimate ofthe error can be obtained from the analysis of the integral equation

u(x, θ, k) = u0(x, θ, k) −∫g(x, y, k)q(y)u(y, θ, k)dy, u0 = exp(ikθ · x) (5.8.2)

g(x, y) = (4π|x− y|)−1 exp(ik|x− y|). If

maxx∈R3

∫ |q(y)|dy4π|x− y| := ‖q‖ < 1 (5.8.3)

then (5.8.2) is uniquely solvable in C(R3) by iterations and

u = u0 + ε, |ε| ≤ (1 − ‖q‖)−1‖q‖. (5.8.4)

If |q| + |∇q| ∈ Q(β) := q : q = q, |q| + |∇q| < c(1 + |x|)−β, β > 3, then (see, e.g., [R121], p.231, cf(5.1.13))

u = u0 +u0

2ik

∫ ∞

0

q(x− rθ)dr + o

(1

k

)as k → +∞. (5.8.5)

Define

AB(θ′, θ, k) := − 1

4π

∫expik(θ − θ′) · xq(x)dx. (5.8.6)

The Born inversion consists of finding q(x) from the equation AB = A, that is:

∫expik(θ − θ′) · xq(x)dx = −4πA(θ′, θ, k). (5.8.7)

Given an arbitrary λ ∈ R3 one can find (non-uniquely) θ, θ′ ∈ S2 and k > 0 such that

k(θ − θ′) = −λ. (5.8.8)

Then (5.8.7) becomes

q(λ) :=

∫exp(−iλ · x)q(x)dx = −4πA(θ′, θ, k)|k(θ′−θ)=λ := a(λ). (5.8.9)

Therefore q(λ) is known and q(λ) can be found by the Fourier inversion of the right-hand side of (5.8.9).This is the standard description of the Born inversion. Our aim is to analyze this procedure. First, notethat equation (5.8.7) is not valid exactly: the left side of (5.8.7) differs from its right side by a quantityof order of the error of the Born approximation. Secondly, the function a(λ) in (5.8.9) depends actuallynot only on λ but on θ, θ′, and k, in spite of the fact that the left side of (5.8.9) depends on λ 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 q ∈ Q(β). If (5.8.7) holds for all θ, θ′ ∈ S2 and all k > 0, then q = 0. Assumeq ∈ Qa. If (5.8.7) holds for all θ, θ′ ∈ S2 and a fixed k > 0 then q = 0.

Proof. If (5.8.7) holds then ImA(θ, θ, k) = 0 for all θ ∈ S2 and all k > 0. By the optical theorem

ImA(θ, θ, k) =k

4π

∫

S2

|A(θ, α, k)|2dα. (5.8.10)

Therefore A(θ, α, k) = 0 for all θ, α ∈ S2 and all k > 0. This implies, by the uniqueness theorem for IP1that q(x) = 0. If q ∈ Qa then the argument is the same but one uses the uniqueness theorem for IP12.Lemma 1 is proved. 2

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, howeveraccurate, as noisy Born data. In particular, it is not advisable to measure the scattering data with theaccuracy that exceeds the accuracy of the Born approximation for the solution of the direct scatteringproblem. The last question we wish to discuss is this: is it true that the Born inversion allows one torecover q with the accuracy which grows as q gets smaller in a suitable sense? This question should beclarified: if equation (5.8.7) considered as an exact equation is not solvable unless q = 0, then what isthe meaning of Born’s inversion? What is the meaning of the smallness of q? How does the inversionresult depend on the choice of a(λ) in (5.8.9)? We noted that a(λ) is not uniquely determined by thechoice of λ: it depends on θ′, θ and k.

The result we will prove is valid for any choice of a(λ) in (5.8.9). The smallness of q is defined bythe inequality

supk>0,θ,θ′∈S2

|A− AB| < δ. (5.8.11)

This condition can be expressed directly in terms of q(x) : |A − AB | ≤ 1/(4π)∫|q|dx‖q‖(1 − ‖q‖)−1.

Thus (5.8.11) holds if1

4π

∫|q|dx‖q‖(1− ‖q‖)−1 < δ, ‖q‖ < 1 (5.8.12)

where ‖q‖ is defined in (5.8.3). Finally, the Born inversion is understood as follows: the function a(λ)in (5.8.9) is considered as the noisy values of q(λ). Namely, let us assume that

|a(λ) − q(λ)| < δ. (5.8.13)

Here the values a(λ) are the measured values of A(θ′, θ, k) at (θ′−θ)k = λ, so that both the measurementnoise and the error of the Born approximation are estimated by δ > 0. The Born inversion consists offinding q(x) given δ > 0, a(λ) which satisfies (5.8.13), and some a priori information about the unknownq which is given by the estimate

|q(λ)| ≤ c0(1 + |λ|2)−b, b >3

2, c0 = const > 0. (5.8.14)

Define

R(δ) = (c0/δ)1/(2b), c1 = c

3/(2b)0

b

3π2(2b− 3)(5.8.15)

qδ(x) := (2π)−3

∫

|λ|≤R(δ)

exp(iλ · x)a(λ)dλ. (5.8.16)

Let us formulate the result.

5.8. THE BORN INVERSION 235

Theorem 5.8.2. Assume (5.8.13) and (5.8.14). Then

|qδ(x) − q(x)| ≤ c1δ1−3/(2b), (5.8.17)

where qδ(x) and c1 are defined in (5.8.16) and (5.8.15).

Proof. One has

|qδ(x) − q(x)| ≤ (2π)−3

∣∣∣∣∣

∫

|λ|≤R(δ)

exp(iλ · x)[a(λ) − q(λ)]dλ

∣∣∣∣∣+∫

|λ|≥R(δ)

|q(λ)|dλ

≤ (2π)−3

4

3πR3(δ)δ + c04π

∫ ∞

R(δ)

r2dr

(1 + r2)b

≤ 1

6π2R3δ +

c02π2

R−2b+3

2b− 3:= φ.

(5.8.18)

Let minR>0 φ := m(δ). This minimum is attained at R = R(δ) where R(δ) is given in (5.8.15), and

m(δ) = δ1−3/2bc3/2b0

b

3π2(2b− 3):= c1δ

1−3/2b. (5.8.19)


Remark 5.8.3. It follows from (5.8.5) that the Born approximation for the Schrodinger equation be-comes exact as k → +∞. For the acoustic equation

u = 0 in R3

u = exp(ikθ · x) +A(θ′, θ, k)r−1 exp(ikr) + o(r−1), r = |x| → ∞,x

r= θ′

(5.8.20)

θ′, θ ∈ S2, k = const > 0, where v(x) is a compactly supported square integrable function, the Bornapproximation becomes exact as k → 0, and the order of the error of the Born approximation is O(1)as k → ∞, so that the error of the Born approximation does not decay as k → ∞ for equation (5.8.20).One can choose v(x) ∈ C∞

0 (R3−) such that the scattered field on P = x : x3 = 0 is O(k−N ) as k → ∞,

where N is arbitrary large integer. At the same time the measurements error is 0(1) and the error ofthe Born approximation is O(1). This analysis shows that it is not possible to use Born’s inversion forequation (5.8.20) at high frequencies for recovery of smooth v(x). However, one may hope to use it forrecovery of discontinuities of v(x). The reason is that the Born approximation for large k is q(λ) andthe behavior of the Fourier transform for large frequencies is determined by the discontinuities of q. Thepractical conclusions which follow from the results of this section are: the Born inversion is an ill-posedproblem; it needs a regularization, given by formula (5.8.16). Even in case of an arbitrary small q 6≡ 0 ifthe regularization is not used (for example, the integral in (5.8.16) is taken over the whole space ratherthan over the ball B(R(δ)) then the error of the Born inversion may be unlimited. The last point can beexplained in a different way. The original nonlinear equation which relates q and A can be written asB(q) = A, where B : q → A is the mapping which associates the scattering amplitude A with the given q.The Born approximation is a linearization of this equation near q = 0. More generally, if one linearizesaround q0, one gets B′(q0)(q− q0) = A−B(q0). This equation may be not solvable, since B′(q0) may beunbounded and A−B(q0) may not belong to the range of B′(q0). As Lemma 5.8.1 shows, this is exactlythe situation with the Born inversion.

2. In this section we give a formula for the Born inversion of the data which are given on S2 × S2.The mathematical formulation of the problem is: given the equation

∫

BR

expik(θ − θ′) · xq(x)dx = F (θ, θ′), ∀θ, θ′ ∈ S2 (5.8.21)


where k > 0 is fixed, and BR = x : |x| ≤ R find q(x) by integrating F (θ, θ′) over S2 × S2. We gaveearlier a solution to (5.8.21) based on the fact that F (θ, θ′) defines the Fourier transform of q(x) in theball |λ| < 2k. However, this solution required to choose θ and θ′ such that k(θ′ − θ) = λ, where λ ∈ R3,|λ| < 2k, is a given vector. Although, in principle, such a choice presents no difficulties, computationallyit is desirable to avoid this step and obtain an inversion formula of the type

qN (x) =

∫

S2

∫

S2

F (θ, θ′)hN [k(θ′ − θ)] exp[ik(θ′ − θ) · x]dθdθ′ (5.8.22)

where hN (kθ) is some universal kernel which does not depend on the data F (θ, θ′), but depends on k,R and N . One has

‖qN (x) − q(x)‖R := ε(N ) → 0 as N → ∞ (5.8.23)

where ‖f‖a := ‖f‖L2(Ba). Let us formulate the result in Rd, d ≥ 2. Define

AN (λ) :=

∫

Rd

δN (x) exp(−iλ · x)dx (5.8.24)

where

δN (x) =

(N

4πR2

)d/2(1 − |x|2

4R2

)N 2d/2Γ

(d+ 2

2

)Jd/2(ξ|x|)(ξ|x|)d/2

2N+d

, ξ :=2k

2N + d. (5.8.25)

Here Γ(z) is the Gamma function and Jν(z) is the Bessel function, the expression in the braces is

1

|B2k|

∫

|λ|<2k

dλ exp

(iλ · x

2N + d

)=

(2k)d/2(2N + d)d/2(2π)d/2

|x|d/2|B2k|J d

2(ξ|x|), (5.8.26)

and |Bk| = kdπd/2

Γ(1+ d2 )

is the volume of the ball of radius k. Define

hN (λ) = |λ|(4k2 − |λ|2)−(d−3)/2k2d−4(8πdωd−1)−1AN (λ), (5.8.27)

where ωd is the surface area of the sphere Sd−1, ωd = 2πd/2

Γ(d/2) .

Theorem 5.8.4. If hN (λ) is defined by (5.8.27), AN (λ) by (5.8.24) δN (x) by (5.8.25), and qN (x) by(5.8.22), then formula (5.8.23) holds.

Proof of Theorem 5.8.4 and a detailed discussion of the solution to (5.8.21) given by (5.8.22) one canfind in [R83], pp. 259-270, where the stability of the inversion is discussed as well. Namely, suppose thatFδ(θ

′, θ) is given such that maxθ,θ′∈S2 |Fδ(θ′, θ) − F (θ′, θ)| < δ, δ > 0 is a small given number. Thenone can find N = N (δ) such that formula (5.8.22) with Fδ in place of F and N (δ) in place of N definesqN(δ)(x) with the property ‖qN(δ)(x)−q(x)‖a ≤ η(δ) → 0 as δ → 0. Estimates of η(δ) are given in [R83],p.269.

An application of the sequence (5.8.25), which is a delta-sequence in BR of entire functions of expo-nential type ≤ 2k, to the problem of spectral extrapolation, that is, inversion of the Fourier transformof a compactly supported function from a compact, is done in [R83], [R139] and in [RKa]. This problemis of interest in many applications.

5.9 Uniqueness theorems for inverse spectral problems

Let us formulate the inverse spectral problem (ISP). Let

−∆φn + q(x)φn = λnφn, x ∈ D ⊂ Rr, n = 1, 2, . . ., r ≥ 2, (5.9.1)

φnN + σ(s)φn = 0 on Γ = ∂D, (5.9.2)

5.9. UNIQUENESS THEOREMS FOR INVERSE SPECTRAL PROBLEMS 237

Here φnN is the normal derivative of φn on Γ. Assume that q(x) = q(x) ∈ L2(D), that each λn iscounted according to its multiplicity, that

λn 6= 0 for all n, (5.9.3)

and that σ(s) ∈ C(Γ), 0 ≤ σ(s).Consider the inverse spectral problem:

Given the data λn, φn|Γ for all n, find σ(s) and q(x). (5.9.4)

If the boundary condition is the Dirichlet one, then the inverse spectral problem is:

Given λn, φnN |Γ for all n, find q(x). (5.9.5)

The basic result is

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 functionsqj, σj, j = 1, 2, which produce the same spectral data, and to derive from this assumption that thedistribution p(x) + σ(s)δΓ, p := q2 − q1, σ := σ2 − σ1, is orthogonal to the set of products u1u2∀u1 ∈ ND(`1) ∀u2 ∈ ND(`2) where

ND(`j) = u∣∣ `ju := (−∇2 + qj)u = 0 in D, uN + σju = 0 on Γ, u ∈ H2(D).

Since the pair `1, `2 has property C it follows that p(x) + σ(s)δΓ = 0. This implies that p(x) = 0and σ(s) = 0, so q1 = q2 and σ1 = σ2. This is the outline of the proof. Note that completeness of theset of products of solutions to the Schrodinger equations holds not only in L2(D) but in the class ofdistributions 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 f ∈ H3/2(Γ),

ù := −∇2u+ q(x)u = 0 in D, uN + σu|Γ = f. (5.9.6)

Then

u(x) =

∞∑

j=1

cjψj(x), cj = µ−1j

∫

Γ

fψjds (5.9.7)

so that the coefficients cj are uniquely determined by the spectral data µj, ψj|Γ. Here

`ψj = µjψj in D, ψjN + σψj = 0 on Γ

and the eigenvalues µj are counted according to their multiplicities.

Proof of Lemma 5.9.2. Multiply equation (5.9.6) by ψj integrate over D and then by parts to get

0 = µjcj −∫

Γ

fψjds, cj :=

∫

D

uψjdx. (5.9.8)

Since the set ψj forms an orthonormal basis of L2(D), formula (5.9.7) follows from (5.9.8). Lemma 5.9.2is proved. 2

Let us continue the proof of Theorem 5.9.1.


Suppose that the pairs q1, σ1 and q2, σ2 produce the same spectral data: µ(1)j = µ

(2)j , ψ

(1)j = ψ

(2)j

on Γ ∀j. Choose an arbitrary f ∈ H3/2(Γ) and solve problem (5.9.6) with q, σ equal to qj, σj,j = 1, 2. Subtract equation (5.9.6) with q2, σ2 from this equation with q1, σ1 to get

`1w = pu2, w := u1 − u2, p := q2 − q1

wN + σ1w = σu2 on Γ, σ := σ2 − σ1.(5.9.9)

By Lemma 5.9.2,

u1 =

∞∑

j=1

cjψ(1)j , u2 =

∞∑

j=1

cjψ(2)j (5.9.10)

where the coefficients cj are the same for u1 and u2. Since ψ(1)j = ψ

(2)j on Γ by the assumption, one

concludes from (5.9.10) thatw = 0 on Γ. (5.9.11)

A detailed proof of (5.9.11) is given in Lemma 5.9.3. Multiply (5.9.9) by an arbitrary element v1 ∈ND(`1), integrate over D and then by parts to get

∫

D

pu2v1dx = −∫

Γ

(wNv1 − wv1N )ds = −∫

Γ

σu2v1ds (5.9.12)

where we used (5.9.11). Write (5.9.12) as the orthogonality relation:

∫

D

[p(x) + σδΓ] v1(x)u2(x)dx = 0 ∀v1 ∈ ND(`1), ∀u2 ∈ ND(`2). (5.9.13)

Here δΓ is the delta-function supported on Γ. The set of products v1u2 in (5.9.13) is complete not onlyin L2(D) but also in the set of distributions with support in D of order of singularity which increaseswith the smoothness of q(x). If q ∈ L2(D) then the order of singularity allowed is ≤ 1, so that σ(s)δΓ isan admissible distribution. Recall that the order of singularity s = s(f) of a distribution f is the smallestinteger such that f is representable in the form f = Dsh where h is a locally integrable function. SinceδΓ = DνχD , where χD is the characteristic function of the domain D and ν is the normal to Γ pointinginto D, one concludes that s(δΓ) = 1, where s(f) is the order of singularity of f . Thus s(σ(s)δΓ) = 1for any continuous function σ(s) 6≡ 0. Thus, one concludes that (5.9.11) implies p = 0 and σ = 0.Theorem 5.9.1 is proved. 2

We now prove

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 L2(D) but they donot converge fast, so that it requires a proof to see that one can calculate the value of uj at the boundaryby taking x = s ∈ Γ in the series u1 − u2. For example, it is clear that one cannot put the operation∂/∂N + σ(s) under the sign of the sum in (5.9.10) (changing the order of application of the operation∂∂N + σ(s) and the operation

∑∞j=1) since ψjN + σψj = 0 ∀j while umN + σum = f 6≡ 0, m = 1, 2. In

order to prove (5.9.11), write (5.9.10) as

uj = uj(x, λ)|λ=0 :=

∫

Γ

Gj(x, s, λ)f(s)ds|λ=0

where Gj is the unique solution to the problem

(`j − λ)Gj = δ(x− y) in D, GjN + σGj = 0 on Γ.

5.9. UNIQUENESS THEOREMS FOR INVERSE SPECTRAL PROBLEMS 239

This problem is uniquely solvable for λ in a neighborhood of λ = 0 because of the assumption µj 6= 0∀j. Clearly

D`λGj = `!

∞∑

m=1

ψm(x)ψm(y)

(µm − λ)`+1(5.9.14)

where Dλ is the derivative in λ and ` ≥ 0 is an integer. The series (5.9.14) converges in L2(Γ) × L2(Γ)provided that ` > 5

2 . Indeed, it is well known that µm ∼ m2/d as m → ∞, d is the dimension of thespace d = 3 in our case. The known elliptic estimate yields

‖ψ(j)m ‖L2(Γ) ≤ c

(‖`jψ(j)

m ‖L2(D) + ‖ψ(j)m ‖L2(D)

)≤ cµ(j)

m ≤ cm2/d

where c = const > 0, c does not depend on m. Thus the series (5.9.14) converges in L2(Γ) × L2(Γ) if∑∞m=1 m

4/d/m2(`+1)/d < ∞, that is (2`+ 2)/d− 4/d > 1, or ` > (d+ 2)/2 = 5/2, so one can take ` = 3.If w(x, λ) := u1(x, λ) − u2(x, λ), then D`

λw, ` > 5/2, is representable by a series which converges in

L2(Γ)×L2(Γ) and the assumption ψ(1)m = ψ

(2)m on Γ implies D`

λw = 0 on Γ. Therefore w(s, λ) = P (s, λ)where P (s, λ) is a polynomial in λ of degree `−1, that is of degree 2, with coefficients depending on s. Itis clear from (5.9.14) that, in the sense of distributions, w(s, λ) → 0 as λ → −∞. Therefore P (s, λ) = 0,w(s, λ) = 0 and Lemma 5.9.3 is proved. 2

Remark 5.9.4. The smoothness assumption on q can be further reduced (from q ∈ L2(D) to q ∈ Lγ (D),γ > 3/2 in R3 (see [H vol. III, p. 4]).

Remark 5.9.5. If the Robin boundary condition (5.9.6) is replaced by the Dirichlet condition u =f on Γ then the spectral data µm, ψm|Γ in Theorem 5.9.1 should be replaced by the spectral data

λm, φmN |Γ ∀m; `φm = λmφm in D, φm = 0 on Γ. (5.9.15)

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 theoremis based on the Property C.

Let us give a useful result.

Lemma 5.9.6. The system φmN |Γm≥n is complete in L2(Γ). Here n is any fixed positive integer andq = q ∈ L2(D).

Proof. Assume f ∈ L2(Γ) and ∫

Γ

fφmNds = 0 ∀m ≥ n. (5.9.16)

Let ù = 0 in D, u|Γ = f . Write (5.9.16) as

0 =

∫

Γ

(uφmN − uNφm)ds =

∫

D

(u∆φm − φm∆u)dx = −λm∫

D

uφmdx, m ≥ n.

One can assume that φm are real-valued since q is real-valued. Thus um :=∫Duφmdx = 0 ∀m ≥ n.

Thus u =∑n−1m=1 umφm(x). This and the condition φm|Γ = 0 implies f = u = 0 on Γ, and the conclusion

follows. 2

Chapter 6

Non-uniqueness and uniquenessresults

6.1 Examples of nonuniqueness for an inverse problem of geo-

physics

6.1.1 Statement of the problem

In this section the result from [R165] is presented.Let D ⊂ Rn+ := x : x ∈ Rn, xn ≥ 0 be a bounded domain, part S of the boundary Γ of D is on the

plane xn = 0, 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:

c−2(x)utt − ∆u = f(x, t) in D × [0,∞), f(x, t) 6≡ 0, (6.1.1)

uN = 0 on Γ (6.1.2)

u = ut = 0 at t = 0. (6.1.3)

Here N is the unit outer normal to Γ, uN is the normal derivative of u on Γ. If c2(x) is known, then thedirect problem (6.1.1)-(6.1.3) is uniquely solvable. The inverse problem (IP) we are interested in is thefollowing one:(IP) Given the data u(x, t) ∀x ∈ S, ∀t > 0, can one recover c2(x) uniquely?

The basic result is: the answer to the above question is no.An analytical construction is presented of two constant velocities cj > 0, j = 1, 2, c1 6= c2, which can

be chosen arbitrary, and a source, which is constructed after cj > 0 are chosen, such that the solutionsto problems (6.1.1)-(6.1.3) with c2(x) = c2j , j = 1, 2, produce the same surface data on S for all times:

u1(x, t) = u2(x, t) ∀x ∈ S, ∀t > 0. (6.1.4)

The domain D we use is a box: D = x : aj ≤ xj ≤ bj , 1 ≤ j ≤ n.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 c2(x).

6.1.2 Example of nonuniqueness of the solution to IP

Our construction is valid for any n ≥ 2. For simplicity we take n = 2, D = x : 0 ≤ x1 ≤ π, 0 ≤ x2 ≤ π.Let c2(x) = c2 = const > 0. The solution to (6.1.1)-(6.1.3) with c2(x) = c2 = const can be found

240

6.1. EXAMPLES OF NONUNIQUENESS FOR AN INVERSE PROBLEM OF GEOPHYSICS 241

analytically

u(x, t) =∞∑

m=0

um(t)φm(x), m = (m1,m2) (6.1.5)

where

φm(x) = γm1m2 cos(m1x1) cos(m2x2),

∫

D

φ2m(x)dx = 1, ∆φm + λmφm = 0,

φmN = 0 on Γ, λm := m21 +m2

2,

γ00 =1

π, γm10 = γ0m2 =

√2

π,

(6.1.6)

γm1m2 = 2/π if m1 > 0 and m2 > 0,

um(t) := um(t, c) =c√λm

∫ t

0

sin[c√λm(t − τ )]fm(τ )dτ,

fm(t) :=

∫

D

f(x, t)φm(x)dx.

(6.1.7)

The data are

u(x1, 0, t) =

∞∑

m=0

um(t, c)γm1m2 cos(m1x1). (6.1.8)

For these data to be the same for c = c1 and c = c2, it is necessary and sufficient that

∞∑

m2=0

γm1m2um(t, c1) =

∞∑

m2=0

γm1m2um(t, c2), ∀t > 0, ∀m1. (6.1.9)

Taking Laplace transform of (6.1.9) and using (6.1.7) one gets an equation, equivalent to (6.1.9),

∞∑

m2=0

γm1m2fm(p)

[c21

p2 + c21λm− c22p2 + c22λm

]= 0, ∀p > 0, ∀m1. (6.1.10)

Take c1 6= c2, c1, c2 > 0, arbitrary and find fm(p) for which (6.1.10) holds. This can be done byinfinitely many ways. Since (6.1.10) is equivalent to (6.1.9), the desired example of nonuniqueness of thesolution to IP is constructed.

Let us give a specific choice: c1 = 1, c2 = 2, fm1m2= 0 for m1 6= 0, m2 6= 1 or m2 6= 2, f 02(p) = 1

p+1,

f01(p) = − p2+1(p+1)(p2+16) . Then (6.1.10) holds. Therefore, if

f(x, t) =

√2

π[f01(t) cos(x2) + f02(t) cos(2x2)] , c1 = 1, c2 = 2, (6.1.11)

then the data u1(x, t) = u2(x, t) ∀x ∈ S, ∀t > 0. In (6.1.11) the values of the coefficients are

f01(t) = − 2

17exp(−t) − 15

17

[cos(4t) − 1

4sin(4t)

], f02(t) = exp(−t). (6.1.12)

242 CHAPTER 6. NON-UNIQUENESS AND UNIQUENESS RESULTS

Remark 6.1.1. The above example brings out the question:What data on S are sufficient for the unique identifiability of c2(x)?The answer to this question one can find in [R139].In particular, if one takes f(x, t) = δ(t)δ(x − y), and allows x and y run through S, then the data

u(x, y, t) ∀x, y ∈ S, ∀t > 0, determine c2(x) uniquely. In fact, the low frequency surface data u(x, y, k),∀x, y ∈ S ∀k ∈ (0, k0), where k0 > 0 is an arbitrary small fixed number, determine c2(x) uniquelyunder mild assumptions on D and c2(x). By u(x, y, k) is meant the Fourier transform of u(x, y, t) withrespect to t.

Remark 6.1.2. One can check that the non-uniqueness example with constant velocities is not possible toconstruct, as was done above, if the sources are concentrated on S, that is, if f(x1, x2, t) = δ(x2)f1(x1, t).

6.2 A uniqueness theorem for inverse boundary value problems

for parabolic equations

Consider the problem:

ut + Lu = 0, x ∈ D, t ∈ [0, T ], (6.2.1)

u = 0 at t = 0 (6.2.2)

u = f(s)δ(t) on S. (6.2.3)

Here δ(t) is the delta-function, D is a bounded domain in Rn, n ≥ 3, with a smooth boundary S,f ∈ H3/2(S), Lu := −div[a(x)gradu] + q(x)u, a(x) and q(x) are real-valued functions, q ∈ L2(D),0 < a0 ≤ a(x) ≤ a1, where a0 and a1 are positive constants, and a(x) ∈ C2(D), where D is the closureof D. Let h(s, t) := a(s)uN , 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 t ∈ [0, T ] and allf ∈ H3/2(S), 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 theRamm’s uniqueness theorem for the solution to elliptic boundary value problem [R139] (cf Chapter 5).

This theorem says:Let

Lu− λu = 0 in D, u = f(s) on S, (6.2.4)

and assume that the above problem is uniquely solvable for two distinct real values of λ. Suppose thatthe set of ordered pairs f, h is known at these values of λ for all f ∈ H3/2(S), where h := a(s)uN , anduN 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 t ∈ [0, T ] are uniquely determined for all t > 0. If

δ(t) is replaced by a function η(t) ∈ C∞0 (0, T ),

∫ T0 η(t)dt = 1, then the data h(s, t) known for t ∈ [0, T ]

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]:

Lv − λv = 0 in D, u = f(s) on S, (6.2.5)

and the data H(s, λ), where v :=∫∞0 e−λtu(x, t)dt.

The data H(s, λ) :=∫∞0e−λth(s, t)dt are known for all λ > 0.

Thus, Ramm’s theorem yields uniqueness of the determination of L, and the proof is completed.

6.3. PROPERTY C AND AN INVERSE PROBLEM FOR A HYPERBOLIC EQUATION 243

Proof. We now sketch the proof of the claim:The solution to the time-dependent problem can be written as:

u(x, t) =

∞∑

j=0

e−λj tcjφj(x), (6.2.6)

where Lφj(x) = λjφj(x) in D,φj(x) = 0 on S, ||φj(x)||L2(D) = 1. The coefficients in this series can becalculated by the formula: cj := −

∫S f(s)a(s)φjN (s)ds.

Note that the series for u(x, t) and the series obtained by the termwise differentiation of it withrespect to t converge absolutely and uniformly in D × (0,∞), each of the terms is analytic with respectto t in the region <t > 0, and consequently so are these series.

Therefore the functions u(x, t) and h(s, t) := uN (s, t) are analytic with respect to t in the region<t > 0, so the data are uniquely determined for t > T as claimed. 2

At t = 0 the series (6.2.6) is singular: it does not converge uniformly or even in L2(D). By thisreason the above argument is formal. One can make it rigorous if one replaces the delta-function in

(6.2.3) by a C∞0 (0, T ) function η(t),

∫ T0 η(t)dt = 1, and uses the argument similar to the one in Lemma

5.9.3 (cf [R139]).

6.3 Property C and an inverse problem for a hyperbolic equa-

tion

Let (∗) utt − ∆u + q(x, t)u = 0 in D × [0, T ], where D ⊂ R3 is a bounded domain with a smoothboundary ∂D, T > d, d := diamD, q(x, t) ∈ C([0, T ], L∞(D)). Suppose that for every (∗∗) u|∂D =f(x, t) ∈ C1(∂D × [0, T ]), the value uN |∂D := h(s, t) is known, where N is the outer normal to ∂D, usolves (∗) and (∗∗) and satisfies the initial conditions u = ut = 0 at t = 0. Then q(x, t) is uniquelydetermined by the data f, h ∀f ∈ C1(∂D × [0, T ]) in the subset S of D × [0, T ] consisting of the lineswhich make 45 with the t-axis and which meet the planes t = 0 and t = T outside D× [0, T ], providedthat q(x, t) is known outside S. Here D is the closure of D.

6.3.1 Introduction

Property C, described in Chapter 5, yields a general method for proving uniqueness theorems for multi-dimensional 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 amore general one. This theorem is another example of the application of property C. In Section 6.3.2we formulate the result and give the proof.

6.3.2 Statement of the result. Proofs

1. Assume that

(2 + q)u := utt − ∆u+ q(x, t)u = 0 in DT = D × [0, T ] := D0T , (6.3.1)

where D ⊂ R3 is a bounded domain with a smooth boundary Γ, Dab := D × [a, b], d < a < b < T − d,and d = diamD. Let

u = f(s, t) on ΓT , f ∈ C1T := C1(ΓT ), ΓT := Γ × [0, T ], (6.3.2)

u = ut = 0 at t = 0. (6.3.3)


Assume thatq(x, t) = 0 for x 6∈ Dab (6.3.4)

and q(x, t) is a continuous in t element of L∞(D):

q(x, t) ∈ C([a, b], L∞(D)) := Cab. (6.3.5)

In Proposition 6.3.10 below we relax assumption (6.3.4): we assume that q(x, t) is known outside thesubset 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

uN = h(s, t) on ΓT (6.3.6)

where uN is the normal derivative of u on Γ, N is the outer normal to Γ.The set of pairs

f, h ∀f ∈ C1(ΓT ) (6.3.7)

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 q(x, t) ∈ Cab, d < a < b < T −d.This means that if there are two functions qj(x, t), j = 1, 2, for which data (6.3.7) are the same then

q1 = q2.

Proof. of Theorem 6.3.1 Assume that qj, j = 1, 2, produce the same data (6.3.7). Subtract equation(6.3.1) with q = q2 from equation (6.3.1) with q = q1 to get

`q1u := `1u := utt − ∆u+ q1u = qu2, u := u1 − u2, q := q2 − q1. (6.3.8)

By the assumptionu = uN = 0 on ΓT . (6.3.9)

Multiply (6.3.8) by w1, where w1 is an arbitrary solution to equation (6.3.1) with the property

w1 = w1t = 0 at t = T (6.3.10)

and integrate over DT to get

∫

Dab

qu2w1dxdt =

∫ T

0

dt

∫

Γ

ds(w1Nu− w1uN ) +

∫

D

(utw1 − uw1t)|T0 dx = 0

where we have integrated by parts and used (6.3.3), (6.3.10) and (6.3.9). Thus

∫

Dab

qu2w1dxdt = 0 ∀u2 ∈ N2,0, ∀w1 ∈ N1,T , (6.3.11)

where Dab := D × [a, b], N1,T := w : `1w = 0 in DT , w solves (6.3.10) and N2,0 := u : `2u = 0 in DT ,u solves (6.3.3). Equation (6.3.11) implies that q = 0 provided that the following lemma holds: 2

Lemma 6.3.2. The set w1u2 ∀w1 ∈ N1,T and ∀u2 ∈ N2,0 is complete in L2(Dab).

Since we assume that q ∈ Cab it is sufficient to prove completeness in Cab, namely that (6.3.11)implies q = 0 if q ∈ Cab ⊂ L2(Dab). If q = 0 then q1 = q2 and Theorem 1 is proved. To complete theproof of Theorem 6.3.1 let us prove Lemma 6.3.2, but first let us outline the basic ideas. We would liketo prove Lemma 6.3.2 in three steps.


Proof. Step 1: If T − d/2 > b then the function w1 takes all the values in the set w : `1w = 0 inDab := D × [a, b], w ∈ C2.

Step 2: If a > d/2 then the function u2(x, t) takes all the values in the set u : `2u = 0 inDab, u ∈ C2. Therefore the set w1u2 is the set of products of arbitrary C2 solutions to the equations`1w = 0 and `2u = 0 in Dab.

Step 3: The set of products of these solutions is complete in Cab. 2

Steps 1, 2 are analogous and can be eliminated if one can prove that the set of products of solutionsfrom some subsets of N1,T and N2,0 is complete in Dab. This is done in sections 3-5 below under theassumption that a > d and b < T − d. We are not able to give a proof under the assumptions a > d/2and b < T − (d/2), see remark 6.3.4.

3. Let us construct the solutions, which belong to N2,0 and N1,T , such that the set of their productsis complete in C([a, b], L∞(D)). One can prove (see Lemma 6.3.5 below) that there exist the solutions

w1 = φε(x+ tθ) exp[iσ(t+ θ · x)] + R1ε, θ ∈ S2, σ > 0, R1ε = R1ε(x, t, θ, σ), (6.3.12)

u2 = φε(x+ tθ) exp[−iσ(t + θ · x)] + R2ε, θ ∈ S2, σ > 0 (6.3.13)

with φε(x) ∈ C∞0 (R3),

‖ Rjε ‖L2(Dab)→ 0 as σ → ∞, j = 1, 2, (6.3.14)

and

φ2ε(x) −→

ε→0δ(x− ξ), ξ 6∈ D, (6.3.15)

where δ(x) is the delta-function, and convergence is meant in the sense of distributions,

suppφε(x) ∩D = ∅, suppφε(x) + Tθ ∩D = ∅, ∀θ ∈ S2, (6.3.16)

∅ 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 σ → +∞ and then ε→ 0, one obtains

w1u2 = δ(x + tθ − ξ) + o(1), ε→ 0. (6.3.17)

This set is complete in Cab in the sense that equation (6.3.18) below implies (6.3.20). Indeed, let ε→ 0.Then ∫

DT

q(x, t)δ(x+ tθ − ξ)dxdt = 0 ∀θ ∈ S2, ∀ξ 6∈ D. (6.3.18)

Thus ∫ T

0

q(ξ − tθ, t)dt = 0 ∀θ ∈ S2, ∀ξ 6∈ D.

Therefore ∫ T

0

q(ξ − tθ, t)dt = 0 ∀θ ∈ S2, ∀ξ 6∈ D. (6.3.19)

Lemma 6.3.3. If q(x, t) = 0 for (x, t) 6∈ Dab where a > d, b < T − d, d = diamD, q(x, t) ∈ Cab, then(6.3.19) implies

q(x, t) = 0 ∀t ∈ [a, b], ∀x ∈ D. (6.3.20)

Proof. The integral in (6.3.19) gives the ray transform of q(x, t) which is defined to be the X-ray transformin which the rays are the lines

t = t, x = ξ + tθ, 0 < t < T, ξ 6∈ D, (6.3.21)


Namely if R is the ray transform operator, then

(Rq)(ξ, θ) :=

∫ ∞

−∞q(ξ + θt, t)dt, θ ∈ S2, ξ ∈ R3. (6.3.22)

The operator R, defined by formula (6.3.22) on continuous functions q(x, t) of x, t, with compact supportD in x variable, and integrable in t variable can be extended to an operator on q : suppq ∈ D ×R, q ∈S′(R3×R), the space of temperate distributions (see Remark 6.3.4 and paper [RSj]). By the assumption,q(x, t) = 0 for x 6∈ D, in particular, for |x| > d/2. Without loss of generality one can assume that D isa ball |x| ≤ d/2. If ξ 6∈ D, then the rays (6.3.21) intersect the region x ∈ D, t > d/2. We required thatξ − θT 6∈ D. This condition is satisfied and the rays (6.3.21) cover the region Dab if

d

2< |ξ| < d

2+ b, and T − b > d. (6.3.23)

Indeed, |ξ − Tθ| ≥ T − (d/2) − b > d/2, so that the ends of these rays do not belong to D. By theassumption, q(x, t) = 0 outside Dab. Thus, (6.3.19) says that the ray transform vanishes everywhere(that is, the restriction ξ 6∈ D can be dropped) provided that a > d and b < T − d. Indeed, in this case,if ξ ∈ D one has |ξ − tθ| ≥ |t| − |ξ| > d − (d/2) = d/2, so that ξ − tp 6∈ D for t ≥ a > d and, sinceq(x, t) = 0 for x 6∈ D, one has q(ξ − tθ, t) = 0 for ξ ∈ D and t ≥ a > d. Thus, if (6.3.19) holds for ξ 6∈ D,if T − b > d, and if a > d, then (6.3.19) holds for all ξ ∈ R3. This implies (6.3.20) by Lemma 6.3.6below. Lemma 6.3.3 is proved. 2

Remark 6.3.4. If only the assumption a > d/2, b < T − (d/2) is given, then the above simple argumentis not sufficient. It would be interesting to find out if the uniqueness theorem holds under this assumption.It is clear that for a < d/2 the uniqueness theorem does not hold because the domain of dependence forequation (6.3.1) is a cone of height d/2. If q(x, t) = q(x) then equation (6.3.19) becomes

∫ T

0

q(ξ − tθ)dt = 0, ∀θ ∈ S2, ∀ξ 6∈ D, (6.3.24)

d/2 < a < b < T − (d/2), so that T > d. It follows from (24) that if ξ runs in ε-neighborhood of Γ,ξ 6∈ D, ε > 0 is an arbitrary small positive number, then (6.3.24) implies that the X-ray transform ofq(x) vanishes so that q(x) = 0. This is the result obtained in [RSy].

4. Let us now give the proof of the existence of the solutions (6.3.12), (6.3.13), (6.3.14).Let

u = φ(x+ tθ) exp[iσ(t+ x · θ)] +R, θ ∈ S2, σ > 0, (6.3.25)

φ ∈ C∞0 (R3), suppφ ∩D = ∅, supp φ+ Tθ ∩D = ∅, (6.3.26)

where R = R(x, t, σ, θ). If

(2 + q)u = 0 in DT , u = ut = 0 in D at t = 0, (6.3.27)

then

(2 + q)R = exp(iσt)g, g := − exp(iσx · θ)(2 + q)φ (6.3.28)

RN = 0 on ΓT , R = Rt = 0 in D at t = 0. (6.3.29)

Let∫ t0Rdτ := w. Then

2w =

∫ t

0

2Rdτ =

∫ t

0

[−qR+ exp(iστ )g]dτ = −∫ t

0

qwτdτ + Ω (6.3.30)


where Ω :=∫ t0 exp(iστ )gdτ , Ω = Ω(x, t, σ, θ). By Riemann-Lebesgue’s lemma one has

η :=1

2max

0≤t≤T,θ∈S2‖ Ω ‖∞→ 0 as σ → +∞ (6.3.31)

where ‖ Ω ‖∞:=‖ Ω ‖L∞(D). Multiply (6.3.30) by wt and integrate over D, then by parts using (6.3.29),then integrate with respect to t from 0 to t, and get

‖ wt ‖2 + ‖ ∇w ‖2≤ 2

∫ t

0

ds ‖ ws ‖∫ s

0

‖ q(x, τ ) ‖∞‖ wτ ‖ dτ + η

∫ t

0

‖ wτ ‖ dτ (6.3.32)

where ‖ · ‖ is the norm in L2(D). Let

sup0≤τ≤t

‖ q(x, τ ) ‖∞= Q(t),

∫ t

0

‖ wτ ‖ dτ := v(t). (6.3.33)

Then (6.3.32) can be written as

v′2 ≤‖ wt ‖2 + ‖ ∇w ‖2≤ Q(t)v2(t) + ηv. (6.3.34)

Letsup

0≤t≤TQ(t) = c. (6.3.35)

Then v′2 ≤ cv2 + ηv ≤ c1v2 + η2, where c1 = c + 1/2. Using the inequality (A + B)1/2 ≤ A1/2 + B1/2

for A,B ≥ 0, one gets

v′ ≤ γv + η, γ = c1/21 . (6.3.36)

Multiply (6.3.36) by exp(−γt) to get [v exp(−γt)]′ ≤ η exp(−γt). Integrate from 0 to t and use v(0) = 0to get

v(t) ≤ ηγ−1[exp(γt) − 1]. (6.3.37)

It follows from (6.3.33), (6.3.34) and (6.3.37) that

‖ ∇w ‖ + ‖ wt ‖≤ c2η, 0 ≤ t ≤ T, c2 = const. (6.3.38)

Since R = wt one has‖ R ‖≤ cη → 0 as σ → ∞. (6.3.39)

Therefore the following lemma is proved:

Lemma 6.3.5. Assume thatsup

0≤t≤T‖ q(x, t) ‖∞≤ c. (6.3.40)

Then there exists a solution (6.3.25) to problem (6.3.27) such that (6.3.39) holds. One can chooseφ(x) = φε(x), with suppφε(x) shrinking to a given point ξ 6∈ D, 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−2u = qu, multiply this equation by a ψ ∈ NDT (2) and integrate over DT to get

∫

DT

q(x, t)u(x, t)ψ(x, t)dxdt=

∫

∂DT

(ψνu− ψuν)ds := I (6.3.41)

where ν is the outer normal to ∂DT . Choose ψ such that ψ = ψt = 0 at t = T , x ∈ D. Then theright-hand side of (6.3.41) is known if the data (6.3.7) are known. Choose u and ψ of the form (6.3.12)and (6.3.13), so that (6.3.14), (6.3.15) hold, pass to the limit ε → 0 to get as in (6.3.19):

f(ξ, θ) := limε→0

limσ→+∞

I =

∫ T

0

q(ξ − tθ, t)dt. (6.3.42)


The knowledge of the integral on the right in (6.3.42) for all θ ∈ S2 and all ξ 6∈ D, |ξ| < b + (d/2),q(x, t) = 0 for (x, t) 6∈ Dab determines q(x, t) in Dab, 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 λ because the right-hand side of (6.3.42) does not depend on λ. To apply formulas(6.3.41), (6.3.42) practically one chooses u = pε(x+ tθ) exp[−iσ(t+ θ ·x)] on ΓT , find the correspondinguN |ΓT 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 t ∈ R function with values in L2(D), D ⊂ R3 is a boundeddomain, f = 0 for (x, t) 6∈ Dab. Assume that

(Rf)(x, θ) :=

∫ ∞

−∞f(x + tθ, t)dt = 0 ∀θ ∈ S2, ∀x ∈ R3 (6.3.43)

where S2 is an open set in S2. Then f(x, t) ≡ 0.

Proof. Let φ(x) ∈ C∞0 (R3) be arbitrary. Then, by Parseval’s equality,

0 =

∫ ∞

−∞dt

∫

R3

f(x + tθ, t)φ(x)dx =

∫ ∞

−∞dt

∫

R3

dλφ(−λ) · f (λ, t) exp(−itλ · θ) (6.3.44)

where φ = (2π)−3/2∫

R3 φ exp(iλ · x)dx. Thus

0 =

∫ ∞

−∞dtf(λ, t) exp(−itλ · θ) ∀λ ∈ R3, ∀θ ∈ S2. (6.3.45)

For an arbitrary fixed λ ∈ R3, |λ| > 0, one can choose θ ∈ S2 such that θ · λ := τ , −τ0 ≤ τ ≤ τ0,τ0 > 0. Thus (6.3.45) says that the Fourier transform F (λ,−τ ) of f (λ, t) in the variable t vanishes for−τ0 ≤ τ ≤ τ0. Since f (λ, t) is assumed compactly supported in t, the function F (λ, τ ) is entire in τ .Since F (λ, τ ) = 0 for −τ0 < τ ≤ τ0, τ0 > 0, one concludes that F (λ, τ ) ≡ 0. Thus f (λ, t) ≡ 0 andf(x, t) ≡ 0. Lemma 6.3.6 is proved. 2

Remark 6.3.7. In [RSj] a stronger result is proved: (6.3.43) implies f ≡ 0 for f(x, t) a temperatedistribution with support in a cylinder |x| < R, −∞ < t < ∞. A similar result is 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

∫

R3

Rfφdx =

∫ ∞

−∞

∫

R3

f (λ, t)φ(−λ) exp(−itλ · θ)dλdt at ∀φ ∈ C∞0 (R3). (6.3.46)

If φ ∈ S (the Schwartz class) then Rf is defined by (6.3.46) as an element of S ′ (the class of temperatedistributions).

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 S ofDT consisting of the points of the lines which make 45 with the t-axis and which meet the planes t = 0and t = T outside DT provided that q(x, t) is known outside S . Here DT is the closure of DT .

One can put in (6.3.7) a different but sufficiently rich set, for example, C∞(ΓT ). If in addition to(6.3.7) one knows the Cauchy data u(x, T ) and ut(x, T ) for all x ∈ D, 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 S1 of DT consisting of thepoints of the lines which make 45 with the t-axis and which meet the plane t = 0 outside D providedthat q(x, t) is known outside S1. This formulation is more general than the one in Theorem 6.3.1.

6.4. CONTINUATION OF THE DATA 249

Remark 6.3.11. The proof of Lemma 6.3.5 given in section II.4 is self-contained. One can give analternative proof using an estimate (see, e.g., [H]):

λ2

∫

DT

|wt|2 exp(−2λt)dxdt ≤ c

∫

DT

|w|2 exp(−2λt)dxdt (6.3.47)

where w ∈ H2(DT ), w = wt = 0 at t = 0, wN = 0 on ΓT , λ > 0 is sufficiently large and c denotesvarious positive constants which do not depend on λ and w. From (6.3.47) and (6.3.30) one obtains

λ2

∫

DT

|R|2 exp(−2λt)dxdt ≤ c

∫

DT

exp(−2λt)δ2 +

∫ t

0

|qR|2dτdx dt

≤ cη2 + c ‖ q ‖2∞

∫

DT

dx dt

∫ T

0

dτ |R|2 exp(−2λτ )

≤ cη2 + c ‖ q ‖2∞

∫

DT

dx dt|R|2 exp(−2λt).

Thus, for sufficiently large λ, one obtains

∫

DT

|R|2 exp(−2λt)dxdt ≤ c1η2, or

∫

DT

|R|2dxdt ≤ c2η2 → 0 as σ → ∞ (6.3.48)

where c2 = λ−1 exp(2λT )c1.

6.4 Continuation of the data

In this section we discuss the following problem. Suppose that

ut + Lu = δ(x− y)δ(t) in D × [0,∞] (6.4.1)

uN + σu = 0 on Γ, 0 < σ(s) ∈ C(Γ) (6.4.2)

u(x, 0) = 0 (6.4.3)

Here D ⊂ R3 is a bounded region with a smooth boundary Γ, L is a formally selfadjoint elliptic differentialoperator which, together with boundary condition (6.4.2) defines a selfadjoint operator in L2(D). Inplace of (6.4.2) one can have any boundary condition for which L, together with this condition, defines aselfadjoint elliptic operator in L2(D) of which the standard elliptic inequalities hold and the uniquenessof the solution to the Cauchy problem. Suppose that the data

u(x, y, t) ∀x ∈ Γ, ∀y ∈ Γ0, ∀t > 0 (6.4.4)

are known, where Γ0 ⊂ Γ is an open subset of Γ. The problem is: do the data (6.4.4) determine the data

u(x, y, t) ∀x, ∀y ∈ Γ, ∀t > 0 (6.4.5)

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

u(x, y, λ) ∀x ∈ Γ, ∀y ∈ Γ0 ∀λ ∈ C \ s(L) (6.4.6)


where

u(x, y, λ) :=

∫ ∞

0

exp(−λt)u(x, y, t)dt (6.4.7)

and s(L) is the spectrum of the selfadjoint operator L defined by the differential expression L andboundary condition (6.4.2): s(L) is the set of eigen-values of L since it is known that the spectrum of Lis discrete with the only limit point at infinity.

It is also well known that

u(x, y, λ) =

∞∑

j=1

∑mj

k=1 φj,k(x)φj,k(y)

λj + λ(6.4.8)

where the bar stands for complex conjugate, φj,k, 1 ≤ k ≤ mj , are the orthonormal eigenfunctions of Lcorresponding to the eigenvalue λj , λj 6= λp for j 6= p.

Claim 11. There exists a set of points xα ∈ Γ0, 1 ≤ α ≤ mj , such that the matrix∑mj

k=1 φj,k(xα)φj,k(xβ) :=Φαβ := Φ is positive definite.

Proof. Consider Φαβξαξβ =

∑mj

k=1 |φj,k(xα)ξα|2, where j is fixed and summation over the repeatedindices is understood. Thus

(Φξ, ξ) = (Cξ,Cξ), Cξ := φj,k(xα)ξα (6.4.9)

Therefore Φ is positive definite iff C is nonsingular, that is Cξ = 0 implies ξ = 0 for any ξ ∈ Cmj .Supose that there exists a ξ ∈ Cmj such that ξ 6= 0 and for any x ∈ Γ0 one has

φj,k(xα)ξα = 0 1 ≤ k ≤ mj (6.4.10)

Then the system φj,k(x) is linearly dependent as a system of functions on Γ0. However, sincefor L the uniqueness of the solution to the Cauchy problem holds, one can conclude that the systemφj,k(x) is linearly dependent in L2(D) which is impossible since the functions φj,k(x), 1 ≤ k ≤ mj , areorthonormal. Let us give more details. First, note that det φj,k(xα) 6= 0 iff detφj,α(xk) 6= 0. Supposethat for some ξ ∈ Cmj , ξ 6= 0,

φj,α(xk)ξα = 0 ∀xk ∈ Γ0 (6.4.11)

This means that

w(x) := φj,α(x)ξα = 0 ∀x ∈ Γ0 (6.4.12)

However,

Lw = λjw, wN + σw = 0 on Γ (6.4.13)


w = wN = 0 on Γ0 (6.4.14)

By the uniqueness of the solution to the Cauchy problem, one concludes that w(x) ≡ 0 in D. Thisimplies that the system φj,k(x), 1 ≤ k ≤ mj , is linearly dependent, which is a contradiction. Claim 1is proved. 2

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 −λj which is

rj(x, y) =

mj∑

k=1

φj,k(x)φj,k(y) (6.4.15)

6.4. CONTINUATION OF THE DATA 251

Take x, y ∈ Γ0. Lemma 6.4.1 allows one to find mj as the number for which the matrix r(xα, xβ) ispositive definite. By the assumption, one knows r(x, y) for all x ∈ Γ and for all y ∈ Γ0. Let us calculater(x, z) for z ∈ Γ, z 6∈ Γ0. One has

rj(ζα, z) =

mj∑

k=1

φj,k(ζα)φj,k(z) (6.4.16)

Take ζα ∈ Γ0, 1 ≤ α ≤ mj , z ∈ Γ, z 6∈ Γ0. Since rj(ζα, z) = rj(z, ζα) one knows rj(ζα, z) for ζα ∈ Γ0

and z ∈ Γ. Consider the linear system

mj∑

k=1

φj,k(ζα)φj,k(z) = rj(ζα, z), 1 ≤ α ≤ mj (6.4.17)

with respect to φj,k(z). In the proof of Lemma 1 it is established that the determinant detφj,k(ζα) 6= 0if the points ζα ∈ Γ0 are chosen so that the matrix Φαβ is positive definite. We assume that ζα ∈ Γ0 are

chosen so that this condition holds. Then φj,k(z) are uniquely determined for any z ∈ Γ. Therefore thefunction rj(x, y) is uniquely determined for all x, y ∈ Γ. Lemma 1 is proved. 2

Remark 6.4.2. The conclusion of Lemma 6.4.1 holds for the solution to the hyperbolic equation

utt + Lu = δ(x− y)δ(t), u = ut = 0 at t = 0, (6.4.18)

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 thespectrum of L is discrete.

Chapter 7

Inverse problems of potential theoryand other inverse source problems

7.1 Inverse problem of potential theory

The potential generated by a charge distribution ρ(y), is given by the formula

u(x) =

∫

D

ρ(y)

4πrxydy, x ∈ R3, rxy := |x− y|, (7.1.1)

where the integral is taken over the support D of ρ. If ρ = 0 for |y| > a, and u(x) is known for|x| > R ≥ a, then the inverse problem of potential theory is to find ρ. In Section 1.2.1 we have mentionedthat this problem, in general, has many solutions, and gave examples of non-uniqueness. Therefore, letus assume ρ = 1 in D, In this case the inverse problem is to find D given u(x) for |x| > R ≥ a, whereD ⊂ Ba := x : |x| ≤ a. One of the basic earlier result in the Novikov’s theorem

Theorem 7.1.1. ([Nov]) If Dj , j = 1, 2, are two domains, star-shaped with respect to a common pointD, Dj ⊂ Ba, and u1(x) = u2(x) for |x| > R ≥ a, then D1 = D2 Here uj(x) :=

∫Dj

(4πrxy)−1dy.

Proof. Let u1 − u2 := u, ρ := χD1 − χD2 , where χD is the characteristic function of D,χD = 1 inD,χD = 0 in D′ := R3\D. Denote D12 := D1 ∪D2, D

12 := D1 ∩D2, Dj := Dj\D12, one has ρ = 0 in

D12, ρ = 1 in D1, ρ = −1 in D2, and

u(x) :=

∫

D12

ρ(y)

4πrxydy = 0 |x| > R, 4u = −ρ in R3. (7.1.2)

Multiply (7.1.2) by an arbitrary h ∈ N (4) := h : 4h = 0 in Ba, and integrate over Ba to get∫

Ba

ρhdx = −∫

Ba

4uhdx = −∫

Ba

u4hdx = 0,

where the boundary integrals vanish because u(x) = 0 outside D12. Originally u = 0 in the region B′R :=

R3\BR, but the unique continuation theorem for harmonic function implies u = 0 in D′12 := R3\D12.

Thus ∫

D12

ρhdx = 0 ∀h ∈ N (4). (7.1.3)

One can check by a direct calculation that if h ∈ N (4) then

H ∈ N (4),H(x) :=1

r3

∫ r

0

h(s, x)s2ds, x :=x

r, r := |x|. (7.1.4)

252

7.1. INVERSE PROBLEM OF POTENTIAL THEORY 253

Choose a harmonic function H(x) in D12 such that

H(x) =

1 on S1\(S1 ∩D2),

0 on S2\(S2 ∩D1).(7.1.5)

By the maximum principle, 0 ≤ H(x) ≤ 1 in D12. Let h be the harmonic function defined by H(x) viaformula (7.1.4), and ρ be defined above formula (7.1.2). Let rj = rj(x

), j = 1, 2, be the equations of Sjin the spherical coordinates, S0

+ := x : x ∈ S2, r1(x) ≥ r2(x

), S2− = S2\S2

+. Then (7.1.3) yields:

0 =

∫

S2+

dx∫ r1(x

)

r2(x)

h(r, x)r2dr −∫

S2−

dx∫ r2(x)

r1(x)

h(r, x)r2dr

=

∫

S2+

dx[r31(x

)H(r1(x), x) − r32(x

)H(r2(x), x)

]

−∫

S2−

dx)[r32(x)H(r2(x

), x) − r31(x)H(r1(x

), x)]

≥∫

S2+

dx[r31(x) − r32(x

)] +

∫

S2−

dxr31(x)H1(r1(x

), x) ≥ 0.

This and (7.1.5) imply r1(x) = r2(x

) on S+ and S2− must be empty. Theorem Theorem 7.1.1 is proved.

2

Remark 7.1.2. Our proof differs from the original proof in [Nov]. The proof remains valid if ρ =ρ(x) ≥ 0 is a known function, and not necessarily a constant.

Exercise 7.1.3. If D ⊂ R3 is a bounded starshaped domain with Lipshitz boundary, then the set ofharmonic polynomials r`Y`(x)`≥0 is dense in ND(4) in L2(D) norm. Here Y` are the sphericalharmonics.

Hint : The traces on S = ∂D of the linear combinations of the harmonic polynomials are dense inL2(S).

Exercise 7.1.4. If uj(x) :=∫D

(ρj(y)dy)/(4πrxy), u1(x) = u2(x) for |x| > R ≥ a, and ρj(y) =f(y)vj (y), where f ≥ 0, ∂f/∂|y| ≥ 0, ∂vj/∂|y| = 0, j = 1, 2, then ρ1(x) = ρ2(x).

Exercise 7.1.5. If D ⊂ R3 is a bounded domain and

u(x) :=

∫

D

dy

4πrxy=

c

|x| for |x| > a, c = const, D ⊂ Ba, (7.1.6)

then D is a ball of radius r := (3c)1/3.

Hint : From (7.1.6) one gets, as |x| → ∞, the relation c = |D|/(4π), where |D| is the volume of D. IfD is a ball of radius r, then c = r3/3, so r = (3c)1/3. Let us prove that (7.1.6) implies that D is a ball.From (7.1.6) as |x| → ∞, one gets

∫D|y|`Y`(y)dy = 0 ∀` > 0. Let h ∈ NBR (4), h(0) = 1. Multiply the

equation 4u = −χD(x) by h(x) and integrate over BR to get:

∫

D

h(y)dy = −∫

BR

4uhdy =

∫

SR

(uhN − uNh)ds =c

R

∫

SR

hNds+c

R2

∫

SR

hds = 4πc, SR := ∂BR.

(7.1.7)

254 CHAPTER 7. INVERSE PROBLEMS OF POTENTIAL THEORY

Here the relations∫SRhNds = 0 and 1/(4πR2)

∫SRhds = h(0) = 1 were used. If g is an element of

the group of rotations, then h(gy) ∈ NBa (4) if h ∈ NBa (4). Let α be an arbitrary unit vector, and gbe the rotation for angle ϕ around α. Replacing h(y) by h(gy) in (7.1.7), differentiating with respect toϕ and taking ϕ = 0, one gets ∫

D

∇h(y)α× ydy = 0 ∀α ∈ S2,

where × stands for the cross product, S2 is the unit sphere in R3. Since α is arbitrary, one gets∫D∇h × ydy = 0, and Gauss’ formula yields

∫Sh(s)N × sds = 0 ∀h ∈ NBa(4). Since h|S can be

arbitrary, one gets

N × s = 0 on S. (7.1.8)

Finally, (7.1.8) implies that S is a sphere, so D is a ball Indeed, let s = s(p, q) be the parametric equationof S, then N = sp× sq , and (7.1.8) implies sps · sq − sqs · sp = 0. Since the vectors sp and sq are linearlyindependent, one gets s · sq = s · sp = 0, so s · s = const. This is an equation of a sphere.

Exercise 7.1.6. [R195]. Assume ρ ∈ L2(D) and

u(x) =

∫

D

ρ(y)

|x− y|b dy = 0 for |x| > a, D ⊂ Ba ⊂ Rn, n > 0. (7.1.9)

For what n and b can one conclude that ρ = 0 given (7.1.9)?

Hint. Fourier-transform (7.1.9) to get u(ξ) = constρ(ξ)|ξ|b−n, where u and ρ are entire functions ofexponential type. Therefore |ξ|b−n is a meromorphic function of ξ ∈ Cn. If (b − n)/2 is not an integer,one gets a contradiction unless ρ = 0

7.2 Antenna synthesis problems

In Section 1.2.16 assume that A(α′, α) in (1.2.15) is known for all k > 0 and all α′ ∈ S2. Then f(y) isuniquely determined by the Fourier inversion. If A(α′, k) is known for all α′ ∈ S2 at a fixed k0 > 0, thenf(y), in general, is not uniquely determined because the value A(α′, k) does not determine A(α′, k) forall k > 0.

The synthesis problem for linear antenna reduces to finding the current j(z) given the diagram f(θ),(cf (1.2.18)):

1

4π

∫ l

−le−ikz cos θj(z)dz = f(θ), 0 ≤ θ ≤ π, k = const > 0. (7.2.1)

Let cosθ := ξ, −k ≤ ξ ≤ k, 2f(θ) := F (ξ). Then (7.2.1) can be written as:

1

2π

∫ l

−le−iξzj(z)dz = F (ξ), −k ≤ ξ ≤ k. (7.2.2)

Clearly F (ξ), known on the interval ξ ∈ [−k, k], determines uniquely F (ξ) ∀ξ ∈ C, because F (ξ) isan entire function (of exponential type l) of ξ. If F (ξ) is known for all ξ ∈ (−∞,∞), then j(z) isuniquely determined. The practical difficulty consists of finding j(z) from the knowledge of F on theinterval [−k, k]. Also, the data F (ξ) are not known exactly in practice: one knows Fδ(ξ) such that‖Fδ(ξ) − F (ξ)‖ ≤ δ, where the norm can be L2(−k, k) or L∞(−k, k) norm. In this case the problem isto find a stable approximation j(z) to the exact current j(z) corresponding to the exact data F (ξ), thatis

‖jδ(z) − j(z)‖L2(−l,l) ≤ η(δ) → 0 as δ → 0. (7.2.3)

One can solve equation (7.2.2) with noisy data using the methods developed in Chapter 2.

7.3. INVERSE SOURCE PROBLEM FOR HYPERBOLIC EQUATIONS 255

Alternatively, one can use the methods proposed in [R204],[R205], and find j(z) from exact dataF (ξ) by the formula

j(z) = limN→∞

∫ k

−kF (ξ)hN (ξ)eiξzdξ, (7.2.4)

where

hN (ξ) :=

∫ ∞

−∞δN (t)e−iξtdt, δN (t) :=

(N

4πl2

)1/2(1 − t2

4l2

)N ( sin kt2N+vkt

2N+v

)2N+v

, (7.2.5)

where v ≥ 1 can be chosen arbitrary. If Fδ(ξ), the noisy data, are given, then formula (7.2.4) with Fδ(ξ)in place of F (ξ), with limit sign dropped, and with N = N (δ) properly chosen, yields jδ(z) satisfying(7.2.3), that is, yields a stable approximation to the exact current j(z). How to choose N (δ) one canread in [R139], p.205. There are books [ZK],[MJ], on antenna synthesis.

7.3 Inverse source problem for hyperbolic equations

Let

utt −4u = f(x, t), x, t ∈ R3 × R; f(x, t) = 0, t /∈ [0, T ] (7.3.1)

u(x, 0) = u0(x), ut(x, 0) = u1(x); u(x, T ) = v0(x), ut(x, T ) = v1(x). (7.3.2)

The inverse problem is: given the data (7.3.2), find f(x, t).Let w(ξ) := 1

(2π)3

∫R3 ue

−iξ·xdx := u, g = f , w := dwdt . Then

w + ξ2w = g, w(ξ, 0) = u0, w(ξ, 0) = u1, w(ξ, T ) = v0, w(ξ, T ) = v1. (7.3.3)

Thus

w(ξ, t) =

∫ t

0

sin(|ξ|(t− s))

|ξ| g(ξ, s)ds + u0 cos(|ξ|t) + u1sin(|ξ|t)

|ξ| , (7.3.4)

and

v0 =

∫ T

0

sin(|ξ|(T − s))

|ξ| gds + u0 cos(|ξ|T ) + u1sin(|ξ|T )

|ξ| (7.3.5)

v1 =

∫ T

0

cos(|ξ|(T − s))gds − |ξ|u0 sin(|ξ|T ) + |ξ|u1 cos(|ξ|T ) (7.3.6)

The last two equations can be written as

sin(|ξ|T )

∫ T

0

cos(|ξ|s)gds− cos(|ξ|s)gds∫ T

0

sin(|ξ|s)gds = b0, (7.3.7)

cos(|ξ|T )

∫ T

0

cos(|ξ|s)gds + sin(|ξ|T )

∫ T

0

sin(|ξ|s)gds = b1, (7.3.8)

where

b0 := |ξ|[v0 − u0 cos(|ξ|T ) − u1sin(|ξ|T )

|ξ| ], b1 = v1 + |ξ|u0 sin(|ξ|T ) − |ξ|ucos(|ξ|T ). (7.3.9)

System (7.3.7)(7.3.8) is uniquely solvable for∫ T0

cos(|ξ|s)gds and∫ T0

sin(|ξ|s)g(s)ds. Therefore g(ξ, s)is determined uniquely up to an arbitrary function g1(ξ, s), orthogonal in L2[0, T ] to sin(|ξ|s) and tocos(|ξ|s) for each ξ ∈ R3. This gives a solution of the inverse source problem stated in this Section. Wefollowed [R73].

Chapter 8

Non-overdetermined inverseproblems

8.1 Introduction

The inverse problems in which one wants to find a function of n variables from the data which can beconsidered as a function of m = n number of variables we call non-overdetermined, and if m > n thenthe problem is overdetermined.

Among multidimensional inverse problems many of non-overdetermined ones are still open. Forexample,

1) the inverse obstacle scattering problem with the data A(α′, α0, k0), (see Section 4.2, by subzeroindex we denote the fixed value of the parameter),

2) inverse potential scattering problems with the data A(α′, α0, k) or A(−α, α, k) (see Section 1.2.13)

3) Inverse geophysical problems with the data u(x1, y0, k) (see Section 5.6), etc.

In example 1) the data are the values of a function of two variables, and the unknown is a surfacein R3, 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) theunknown is a velocity profile (or refraction coefficient), which depends on three variables, and the dataalso depends on three variables. For all of these problems currently one does not have even Uniquenesstheorems 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 findinga potential from the values on the diagonal of the kernel of the spectral function of a Schroedingeroperator in a bounded domain. We have to assume additionally that the eigenvalues of this operator areall simple (see [R198]).

8.2 Assumptions

Let D ⊂ R3, be a bounded domain with a C∞ connected boundary S, L = −∇2 + q(x) be a selfadjointoperator defined in H = L2(D) by the Neumann boundary condition, θ(x, y, λ) be its spectral function,

θ(x, y, λ) :=∑

λj<λ

ϕj(x)ϕj(y), where Lϕj = λjϕj , ϕjN |S = 0, ‖ϕj‖L2(D) = 1, j = 1, 2, . . . . The potential

q(x) is a real-valued function, q ∈ C∞(D). It is proved that q(x) is uniquely determined by the dataθ(s, s, λ) ∀s ∈ S, ∀λ ∈ R+ if all the eigenvalues of L are simple.

256

8.3. THE PROBLEM AND THE RESULT 257

8.3 The problem and the result

Let D ⊂ Rn, be a bounded domain with a C∞ connected boundary S, that is, locally the equationxn = f(x′), x′ := (x1 . . .xn−1), of S is given by the function f(x′) ∈ C∞. We assume n = 3. Thisassumption is used only in the proof of Lemma 8.4.1 below and can be dropped. If n > 3 one shouldrefer to the existence of the coordinate system in which the metric tensor, used in the proof of Lemma8.4.1, has the property: gin = 0 for 1 ≤ i < n. The basic ideas of our proof, outlined below, are validfor n ≥ 3.

Let L = −∆ + q(x), where ∆ is the Laplacian and q(x) ∈ C∞(D), the potential, is a real-valuedfunction. Let ϕj(x), j = 1, 2, . . ., be the normalized real-valued eigenfunctions of the selfadjointoperator L defined in H = L2(D) by the Neumann boundary condition:

Lϕj = λjϕj in D, ‖ϕj‖ = 1, ‖ϕ‖ := ‖ϕ‖L2(D), (8.3.1)

ϕjN = 0 on S, (8.3.2)

where N is the unit normal to S pointing into D′ := Rn\D. We choose ϕj to be real-valued functions.By elliptic regularity, the functions ϕj(x) are C∞(D) and C∞(S) on the boundary S.

The spectral function of L is defined by the formula:

θ(x, y, λ) :=∑

λj<λ

ϕj(x)ϕj(y), (8.3.3)

the eigenvalues λj are counted according to their multiplicities.The inverse problem (IP) we are interested in is:IP: given θ(s, s, λ) ∀s ∈ S and ∀λ ∈ R+, find q(x).We are concerned with the uniqueness of the solution to IP. This inverse problem is not overdeter-

mined: the data is a function of n variables s, λ ( S is an (n− 1)–dimensional manifold, λ ∈ R) and theunknown q(x) is also a function of n variables.

It seems that no uniqueness theorems for multidimensional inverse scattering-type problems withnon-overdetermined data have been obtained so far. The IP, studied here, is a multidimensional one ofthe 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 θ(x, y, λ), known∀x, y ∈ S and ∀λ ∈ R has been considered in [Be1]–[Be2] and references therein. In this Section wefollow [R198].

In Section 5.9 the following uniqueness theorem is obtained:

Theorem 8.3.1. The data λj , ϕj(s)∀s∈S,∀j 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) aresimple.

It is possible that the conclusion of Theorem 8.3.2 holds without the assumption concerning thesimplicity of the eigenvalues, but we do not have a proof of this currently. Generically one expects theeigenvalues 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. θ(s, s, λ) ⇒ λj , ϕ2

j(s)∀s∈S∀j

This step is done under the additional assumption that all the eigenvalues λj 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. ϕ2

j (s) ⇒ ϕj(s)

258 CHAPTER 8. NON-OVERDETERMINED INVERSE PROBLEMS

Step 3. Apply Theorem 8.3.1 to the data λj, ϕj(s)∀s∈S,∀j .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 ϕj(s) cannot have zeros of infinite order, that is, if y ∈ S and |ϕj(s)| ≤ Cm|s −y|m ∀m = 1, 2, . . . , where Cm are positive constants and s ∈ S are arbitrary points on S, thenϕj(x) ≡ 0. In the Appendix it is proved that if a real-valued function f ∈ C∞(S) does not havezeros of infinite order on S, then, up to its sign, the function f is uniquely defined on S by its squaref2. A procedure for finding, up to its sign, ϕj(s) on S from the knowledge of ϕj(s)

2 on S is describedbelow.

It is known that ϕj(s) cannot vanish on an open (in S) subset of S by the uniqueness of the solutionto the Cauchy problem for elliptic equations, and ϕj(s) 6= 0 for almost all points of S. Take an arbitrarypoint s0 at which ϕ2

j (s0) 6= 0. By the continuity of ϕ2j(s) on S, there is a maximal domain F ⊂ S

containing the point s0 in which ϕ2j (s) 6= 0. Define ϕj(s) :=

√ϕ2j (s) > 0 in F . Let L ⊂ S be the

boundary of F . We prove that ϕj(s) is uniquely determined in F ′ := S\F . The problem is to determinethe sign of ϕj(s) for s ∈ F ′ in a neighborhood of L , where ϕj(s) = 0 when s ∈ L .

The basic idea for determining this sign is to extend ϕj(s) =√ϕ2j (s) from F across L with maximal

smoothness. Such an extension is unique and determines the sign of ϕj(s) outside of F . For example,

if one defines√x2 = x > 0 for x > 0, then the unique maximally smooth extension of

√x2 in the region

x < 0 is√x2 = x.

To determine the sign ofϕj(s) outside of F , calculate the minimal integerm for ϕj(s) =√ϕ2j (s), s ∈

F ⊂ S, such that γ(s) := lims→sϕj(s)|s−s|m 6= 0. Here s → s ∈ L along a curve ` originating at a point in

F , transversal to L , and passing through a point s ∈ L into F ′.The integer m < ∞ does exist if the zeros of ϕj(s) 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 ϕj(s) along the curve ` across L into F ′.Define sgn ϕj(s) = (−1)m, where s ∈ F ′ ⊂ S is any point in a sufficiently small neighborhood of s.

This algorithm determines uniquely ϕj(s) for all s ∈ S, given the data ϕ2j (s). Since the eigenvalues are

assumed simple, the above algorithm produces the trace ϕj(s) of the eigenfunction ϕj(x) on S.

Since q(x) and S are C∞-smooth up to the boundary, so are the eigenfunctions ϕj(x), and the abovealgorithm produces a C∞-smooth function on S. Therefore one can use the Malgrange preparationtheorem ([CH], p.43) to study the set of singular points of L .

The above argument deals with the continuation of an eigenfunction through its ”zero line” L on S.There are at most finitely many points on a compact surface S at which several ”zero lines” intersectand only finitely many ”zero lines” can intersect at one point. Otherwise there would be a point on Swhich is a zero of infinite order of ϕj(s), and this is impossible by Lemma 8.4.1

If one continues ϕj(s) across L by the above rule, the function ϕj(s) will be uniquely determinedon all of S by the choice of its sign at the initial point s0.

In the Appendix, at the end of this Chapter, we include a proof of a statement we have used in adiscussion of Step 2. It is proved in the Appendix that a smooth real-valued function, which is defined ona 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 byYu. M. Berezanskii are used.

This completes the description of Step 2.

The C∞ smoothness of the data is assumed for technical reasons: it guarantees the existence of aninteger m in the above construction. It would be interesting to weaken this assumption and to find outif one can prove Lemma 8.4.1 assuming S is C1,1−smooth and q ∈ L∞(D).

In section 8.4 a detailed discussion of Step 2 is given.

8.4. FINDING ϕJ (S) FROM ϕ2J (S) 259

8.4 Finding ϕj(s) from ϕ2j(s)

In Section 8.3 a method for finding ϕj(s) from the knowledge of ϕ2j(s) has been discussed.

Here a justification of this method is presented. This justification consists mainly of the proof of thefollowing lemma:

Lemma 8.4.1. The function ϕj(s) 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 orderelliptic inequality:

|M0u| ≤ c(|u|+ |gradu|) in D, (8.4.1)

where c = const > 0, and M0 is a strictly elliptic homogeneous second order differential expression(summation is understood over the repeated indices):

M0u := −gmjumj , umj :=∂2u

∂xm ∂xj, (8.4.2)

cannot have a zero of infinite order at a point y ∈ D provided that:

gmj(x) are Lipschitz-continuous and strictly elliptic, (8.4.3)

andgmj(0) are real numbers. (8.4.4)

By zero of infinite order of a solution u of (3.1) a point y ∈ D is meant such that

∫

|y−x|<ε|u(x)|2dx ≤ cmε

m ∀ε > 0, ∀m = 1, 2, ...., (8.4.5)

where cm are positive constants independent of ε, and ε > 0 is sufficiently small so that the ballB(y, ε) := x : |x− y| ≤ ε ⊂ D.

We use this result to prove that the same is true if S ∈ C∞ and y ∈ S. Let xn = f(x′), x′ =(x1, . . . , xn−1), be the local equation of S in a neighborhood of the point y ∈ S, which we choose as theorigin, and xn-axis be directed along the normal to S pointed into D′ := Rn\D.

Let us introduce the new orthogonal coordinates

ξj = ξj(xi), 1 ≤ j, i ≤ n,

so that ξn = 0 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, andthe x, y−coordinates are isothermal coordinates on S, which are known to exist for two-dimensionalC∞−surfaces in R3 ( and even for C3−surfaces in R3, [Gu], p. 246). For an arbitrary C∞−surfacein R3 one can prove that locally one can introduce (non-uniquely) the coordinates in which the metrictensor is diagonal in a neighborhood of S. To do this, one takes two arbitrary linearly independentvector fields tangent to S and orthogonalize them with respect to the Euclidean metric in R3 using theGram-Schmidt procedure, which is always possible. The third axis of the coordinate system, which weare constructing, is directed along the normal to S at each point of the patch on S. Let A and B be theresulting orthogonal vector fields tangent to S. Then one can find (non-uniquely) a function χ, definedon the local chart, such that the Lie bracket of the vector fields χA and B vanishes, [χA,B] = 0. Thenthe flows of the vector fields χA and B commute and provide the desired orthogonal coordinate systemin which the metric tensor is diagonal ([CM]). The condition [χA,B] = 0 can be written as the following

linear partial differential equation: χhx−hχx−χy = 0, where h := − fxfy

1+f2x, 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 χ has many


solutions. One can find a solution χ(x, y) by the standard method of characteristics. A unique solutionis specified by prescribing some Cauchy data, which geometrically means that a noncharacteristic curvethrough which the surface χ = χ(x, y) 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 aremany ((n− 1)(n− 2)/2) Lie brackets to vanish in the case of n− 1 vector fields tangent to S in Rn, andit is not clear for what S these conditions can be satisfied. However, for our argument it is sufficient tohave the coordinate system in which the metric tensor has zero elements gjn for 1 ≤ j ≤ n− 1, and gnn

does not vanish. In this case the even continuation (8.4.8)-(8.4.10), that is used below, still allows one toclaim that the function (8.4.8) solves the same equation in the region ξn > 0 as it solves in the originalregion ξn < 0. If the elements gjn for 1 ≤ j ≤ n − 1 do not vanish, then the equation in the regionξn > 0 will have some of the coefficients in front of the second mixed derivatives gjnwjn with the minussign, while these coefficients in the region ξn < 0 enter with the plus sign. So, in this case the principalpart M0 of the operator M , which is used in (8.4.11), will be different in the regions ξn > 0 and ξn < 0.Recall that the Laplace operator in the new coordinates has the form of the Laplace-Beltrami operator:∆w = g−1/2(g1/2gijwj)i, where g := det(gij), summation is understood over the repeated indices, andwj := ∂w

∂ξj.

If y ∈ S is a zero of ϕj(s) of infinite order in the sense |ϕj(s)| ≤ cm|s − y|m ∀m = 1, 2, ...., s ∈ S,then equations (8.2.1) and (8.2.2) imply that |ϕj(x)| ≤ cm|x− y|m ∀m = 1, 2, ...., x ∈ D ∩ B(y, ε), sothat condition similar to (8.4.5) holds for the integrals over D∩B(y, ε). Indeed, the derivatives of ϕj(x)in the tangential to S directions at the point y ∈ S vanish by the assumption. From (1.1) and (1.2) itfollows that the normal derivatives of ϕj(x) of the first and second order vanish at y. Differentiatingequation (1.1) along the normal one concludes that all the normal derivatives of ϕj(x) vanish at thepoint y. Thus we may assume that y ∈ S is a zero of ϕj(x) of infinite order, that is, inequalities (8.4.5)hold for u = ϕj(x) for the integral over D ∩B(y, ε).

In the ξ- coordinates one writes the equation for ϕj :

−∆ϕj + q(x)ϕj − λjϕj(x) = 0 in D, ϕjN = 0 on S, (8.4.6)

as follows:Mϕj = 0 for ξn < 0 in D, ϕjξn = 0 at ξn = 0. (8.4.7)

We drop the subscript j of ϕj and of λj in what follows. In (8.4.7) the operator M is defined as:

Mϕ = − 1

gjj

∂2ϕ

∂ξ2j− 1

g1/2

∂(g−1jj g

1/2)

∂ξj

∂ϕ

∂ξj+ Q(ξ)ϕ− λϕ, Q(ξ) := q(x(ξ)),

over the repeated indices summation is understood, gij is the metric tensor of the new coordinate system,g := det(gij), and the coefficients in front of the second-order derivatives in M are extended to the regionξn > 0 as even functions of ξn, so that the extended coefficients are Lipschitz in a ball centered at y ∈ Swith radius ε > 0.

Let us define w in a neighborhood of the origin, |ξ| < ε, by setting

w =

ϕ(ξ′, ξn) if ξn ≤ 0,ϕ(ξ′,−ξn) if ξn > 0,

(8.4.8)

and

Q(ξ) =

Q(ξ′, ξn) if ξn ≤ 0,Q(ξ′,−ξn) if ξn > 0,

(8.4.9)

gjj(ξ) =

gjj(ξ

′, ξn) if ξn ≤ 0,gjj(ξ

′,−ξn) if ξn > 0.(8.4.10)

The functions gjj(ξ), defined by (8.4.10), are Lipschitz in the ball B(0, ε), if gjj(ξ) is Lipschitz inB(0, ε) ∩ Rn−, where Rn− := ξ : ξn ≤ 0.

8.4. FINDING ϕJ (S) FROM ϕ2J (S) 261

Furthermore,|M0w| ≤ c(|w|+ |gradw|) in B(0, ε), (8.4.11)

where c = const > 0, and ∫

B(0,ε)

|w|2dξ ≤ cmεm ∀m = 1, 2, ....., (8.4.12)

if one assumes ∫

B(0,ε)∩Rn−

|ϕ|2dξ ≤ cmεm ∀m = 1, 2, ......, (8.4.13)

for all sufficiently small ε > 0.Note that the change of variables x→ ξ is a smooth diffeomorphism in a neighborhood of the origin,

which maps the region D ∩ B(0, ε) onto a neighborhood of the origin in Rn− in ξ− space, and one can

always choose a half-ball B(0, ε) belonging to this neighborhood, so that (8.4.13) follows from (8.4.5).Therefore w ≡ 0 in B(0, ε) and consequently ϕ ≡ 0 in B(0, ε)∩Rn

−. This implies that ϕ ≡ 0 in D bythe unique continuation theorem (see [H] and [W]). This is a contradiction since ||ϕ|| = 1.


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 dataθ(s, s, λ). We have explained how to get the data λj , ϕj(s)∀j from θ(s, s, λ). Therefore the spectralfunction

θ(s, t, λ) =∑

λj<λ

ϕj(s)ϕj (t), ∀s, t ∈ S (8.4.14)

and the resolvent kernel

G(s, t, λ) =

∫ ∞

−∞

dµθ(s, t, µ)

µ − λ(8.4.15)

are known on S. If G(s, t, λ) is known, then the Neumann-to-Dirichlet (N-D) map uN := f → h := u|Sis known. This map Λ : f → h at every fixed λ 6= λj ∀j, associates to every f ∈ H

12 (S) a function

h ∈ H32 (S) by the formula

h(s) =

∫

S

G(s, t, µ)f(t)dt. (8.4.16)

HereLu− λu = 0 in D, uN = f on S. (8.4.17)

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 Theorem8.2.2; step 3 consists of the construction of the N-D map by formula (8.4.16) and reference to Chapter5.

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, Lm,m = 1, 2, stands for the Neumann operator −∆ + qm in D. Let usassume that λ ∈ R is a regular point for L1 and L2.

Theorem 8.4.3. If L1 − λ and L2 − λ generate the same N-D map then q1 = q2.

Proof. Take an arbitrary f ∈ H12 (S) and consider the problems

(Lm − λ)um = 0 in D,umN = f on S, m = 1, 2.

Subtract from the equation with m = 1 the equation with m = 2 and get

(L1 − λ)w = pu2 in D, wN = w = 0 on S, (8.4.18)


where w := u1 − u2, p := q2 − q1. The condition w = 0 on S follows from the basic assumption, namelythe N-D map is the same for L1 and L2. Let ψ1 ∈ N (L1) := ψ : L1ψ = 0 in D,ψ ∈ H2(D) bearbitrary. Multiply (8.4.18) by ψ1, integrate by parts using boundary conditions (8.4.18), and get theorthogonality relation: ∫

D

pu2ψ1dx = 0 ∀ψ1 ∈ N (L1) ∀u2 ∈ N (L2). (8.4.19)

The last inclusion in (8.4.19) follows since f ∈ H12 (S) is arbitrary. From (8.4.19) and Property C it

follows that p(x) ≡ 0. Theorem 8.4.3 is proved. 2

8.5 Appendix

Lemma 8.5.1. Let M be a connected C∞ manifold and suppose that a real-valued function f ∈ C∞(M )does not have zeros of infinite order, i.e., for an arbitrary point x0 ∈M , f (α)(x0) 6= 0 for some multiindexα. Then, up to sign, the function f is uniquely defined on M by its square f 2.

Remark 8.5.2 The conclusion is not true, even locally, if f has zeros of infinite order.Indeed, if, for example, M = R, and x = 0 is a zero of infinite order, then the functions g1 = f for

x ≤ 0, g1 = f for x > 0, g2 = −f for x ≤ 0, g2 = f for x > 0, and the functions g3 = −g1, g4 = −g2, areall C∞(M ) and g2

j = f2. 2

Remark 8.5.3 If dimM > 1, F ≥ 0, F ∈ C∞(M ), does not have zeros of infinite order, then theremight be no function f ∈ C∞(M ), such that f2 = F on M .

Indeed, an example is F = x2 + y2, M = R2. 2

Proof of Lemma 8.5.1. All functions below are real-valued. Let g ∈ C∞(M ), g2 = f2, T := s :h(s) = 0, where h := f − g. Then T is closed. If T = M , then g = f on M , and Lemma holds. IfT 6= M , then T ′ := M \ T is an open, non-empty set, and h = 2f on T ′. If T ′ = S (the overlinestands for the closure), then g = −f on M , and Lemma holds. Otherwise, the set U := S \ T ′ ⊂ Tis an open, non-empty set, U := Q ⊂ T , and it may happen that Q 6= T . Since Q is a proper subsetof S (because T is), and since Q 6= U (because S is connected and cannot contain a proper subsetwhich is simultaneously open and closed), it follows that there exists a point s ∈ Q ∩ T ′. Let pj ∈ U ,pj → s, and qm ∈ T ′, qm → s. One has h(α)(pj) = 0 for any multiindex α, because h = 0 on theopen set U . Thus, limj→∞ h(α)(pj) = h(α)(s) = 0, by the continuity of h(α). On the other hand,limm→∞ h(α)(qm) = 2 limm→∞ f (α)(qm) = 2f (α)(s) 6= 0 for some multiindex α, because f ∈ C∞(M )and has zeros of at most finite order. This contradiction proves that either T = S and then f = g, orT ′ = S and then f = −g. Lemma 8.5.1 is proved. 2

2

Chapter 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

exampleConsider the inverse problem of geophysics, see Section 1.2.4. The integral equation equivalent to

this problem is:

u(x, y, k) = g(x, y, k) + k2

∫

D

g(x, z, k)v(z)u(z, y, k)dz, g(x, y, k) =exp(ik|x− y|)

4π|x− y| . (9.1.1)

It is easy to check that for sufficiently small k the integral operator in (9.1.1) has small norm in theBanach space C(D) of the continuous in D functions, that the scattered field

usc := u(x, y, k)− g(x, y, k) (9.1.2)

is continuous in x, y ∈ D ×D, and that (9.1.1) can be written as

usc(x, y, k) = k2

∫

D

g(x, z, k)v(z)g(z, y, k)dz + O(k4) as k → 0 (9.1.3)

where O(k4) is uniform in x, y ∈ D. It follows from (9.1.3) that

limk→0

k−2usc(x, y, k) =

∫

D

g(x, z, 0)v(z)g(z, y, 0)dz =1

16π2

∫

D

v(z)dz

|x− z||z − y| . (9.1.4)

Define the low frequency data by the formula

f(x, y) := 16π2 limk→0

k−2usc(x, y, k). (9.1.5)

Then (9.1.4) can be written as∫

D

v(z)dz

|x− z||y − z| = f(x, y), ∀x, y ∈ P := x : x3 = 0 (9.1.6)

where we took x, y ∈ P 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

∫

D

v(z)dz

|x− z||y − z| = 0, ∀x, y ∈ P (9.1.7)

has only the trivial solution v = 0.Note that ∆z1/(|x− z|) = 0 in D for all x ∈ P .

263

264 CHAPTER 9. LOW-FREQUENCY INVERSION

Lemma 9.1.1. The set |x− z|−1∀x∈P is complete in L2(D) in the set

ND(∆) = u : ∆u = 0 in D, u ∈ H2(D). (9.1.8)

Here and below H`(D) = W `,2(D), W `,p(D) stands for the familiar Sobolev spaces of functionswhose derivatives up to the order ` belong to Lp(D), p is fixed in the interval [1,∞].

We give two different proofs. Both can be generalized to the case when in formula (9.1.7) and inLemma 9.1.1 the function |x − z|−1 is substituted by G(x, y), the Green function of a general secondorder elliptic operator L u = ∂i(aij(x)∂

ju)+q(x)u, in Lemma 9.1.1 the operator ∆ is substituted by L ,it is assumed that zero is not an eigenvalue of the Dirichlet operator L in D, that the Dirihlet problemin R3

+ has only the trivial solution, and that the unique continuation principle for the solutions of theequation L u = 0 holds.

Proof. First Proof Let u ∈ ND(∆) and assume that

∫

D

udz

|x− z| = 0 ∀x ∈ P. (9.1.9)

We want to prove that this implies u = 0. Define w(x) :=∫D

udz|x−z| , x ∈ R3. Then ∆w = 0 in R3,

w = 0 on P . Therefore

∆w = 0 in D′ := R3 \D, w = 0 in R3+. (9.1.10)

By the unique continuation principle for the solutions to the equation (9.1.10) one concludes that

w = 0 in D′ (9.1.11)

By the elliptic regularity result, w ∈ H2loc(R

3). This, equation (9.1.11) and the embedding theoremimply

w = wN = 0 on Γ. (9.1.12)

The embedding theorem says that if w ∈ H2loc then w and wN are well defined on a smooth manifold

of dimension n− 1 and if Γt is a one-parametric family of manifolds in a neighborhood of Γ, such thatΓ0 = Γ, 0 ≤ t ≤ δ, δ > 0 is a small number, and Γt is parallel to Γ in the sense defined below, then

limt→0

(‖ w ‖2

L2(Γt)+ ‖ wN ‖2

L2(Γt)

)=‖ w ‖2

L2(Γ) + ‖ wN ‖L2(Γ), (9.1.13)

that is, w and wN depend continuously on t in the sense (9.1.13).The manifold Γt is called parallel to Γ0 if the normal to Γ0 at each point s0 ∈ Γ0 intersects Γt at

exactly one point st whose distance from the point s0 is equal to t, 0 ≤ t ≤ δ. From (9.1.12) one obtains

∆w = −4πu in D, w = wN = 0 on Γ. (9.1.14)

By the assumption ∆u = 0 in D. Multiply (9.1.14) by u, (the bar stands for complex conjugate),integrate over D, then by parts, using boundary conditions (9.1.14) for w, to get

∫D|u|2dx = 0. Thus

u = 0 in D. Lemma 9.1.1 is proved. 2

Remark 9.1.2. If the general second order elliptic operator L replaces ∆ and G(x, y) replaces |x−y|−1,LG = −δ(x − y), then the proof goes without any changes and one uses the symmetry of L in theargument given below formula (9.1.14). Namely, −

∫D |u|2dx =

∫D L uudx =

∫D uL udx = 0.

This argument for the operator −∆ + q(x), for example, requires that q = q.

Proof. Second Proof of Lemma 9.1.1. The advantage of this proof is that it allows a generalization tothe case of nonsymmetric operators L .

9.1. DERIVATION OF THE BASIC EQUATION. UNIQUENESS RESULTS 265

It is sufficient to prove that the set of functions

η(z) :=

∫

P

φ(x)dx

|x− z| ∀φ ∈ C∞0 (P ) (9.1.15)

is dense in L2(Γ). Indeed, clearly ∆zη = 0 in D. Therefore, if η can approximate in L2(Γ) an arbitraryelement of L2(Γ) with an arbitrary accuracy, then, by the maximum principle for harmonic functions, itcan approximate an arbitrary u ∈ ND(∆) in L2(D) with an arbitrary accuracy. Thus, in order to proveLemma 9.1.1, it is sufficient to prove that if f ∈ L2(Γ) then

∫

Γ

fηds = 0 for all ηof the form (9.1.15) implies f = 0. (9.1.16)


v(x) :=

∫

Γ

f(s)ds

|x− s| = 0 ∀x ∈ P. (9.1.17)

Clearly∆v = 0 in D ∪ Ω, v(∞) = 0. (9.1.18)

In particular, ∆v = 0 in R3+, v(∞) = 0, v = 0 on P . This implies (for example, by the maximum

principle) that v = 0 in R3+. By the unique continuation property for harmonic functions one concludes,

using (9.1.18), that v = 0 inD′. Therefore u = 0 on Γ.From this and from (9.1.18) it follows that v = 0 inD. 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 9.1.1.2

Remark 9.1.3. If L replaces ∆ in this argument, then the crucial point is the estimate

‖ u ‖H

`+ 12 (D)

≤ c ‖ u ‖H`(Γ), u ∈ ND(L ) (9.1.19)

which holds for u ∈ ND(L ) provided that zero is not an eigenvalue of the Dirichlet operator L in D,

the coefficients of L and Γ are sufficiently smooth (say Γ ∈ C2, and aij(x) ∈ C1(D), q ∈ L32+ε

loc(D),

ε > 0 for ` = 0). The constant c in (9.1.19) does not depend on u, it depends on L and Γ. Estimate(9.1.19) shows that if a set of functions η

η(z) :=

∫

P

G(x, z)φ(x)dx, φ ∈ C∞0 (P ), LG = −δ(x− y) (9.1.20)

is dense in L2(Γ) then it is dense in H12 (D) (and in L2(D)) in ND(L ).

Remark 9.1.4. We use the results similar to Lemma 9.1.1 quite often. The result and proofs ofLemma 9.1.1 remain valid if P is replaced by an arbitrary closed compact Lipschitz surface which con-tains D inside and for some noncompact surfaces S which contain D inside and have the property thatthe Dirichlet problem for the operator L 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:w(x) =

∫D G(x, z)u(z)dz, or in the form: η(z) =

∫S G(x, z)φ(x)dx, φ ∈ C∞

0 (S).

We are now ready to prove the following uniqueness result.

Lemma 9.1.5. If v ∈ L2(D) and (9.1.7) holds then v = 0.

Proof. By Lemma 9.1.1, equation (9.1.7) implies∫

D

v(z)u1(z)u2(z)dz = 0 ∀u1, u2 ∈ ND(∆). (9.1.21)

This implies that v(z) = 0 according to Lemma 9.1.6 below. Lemma 9.1.5 is proved. 2


Lemma 9.1.6. The set u1u2 ∀u1, u2 ∈ ND(∆) is complete in L2(D), where D ⊂ R3 is an arbitrarybounded domain.

In Lemmas 9.1.5 and 9.1.6 the central idea of the Property C, which yields a method of provinguniqueness theorems for inverse problems, based on completeness of the set of products of solutions tohomogeneous PDE is presented. This method for proving Uniqueness theorems for inverse scattering andother inverse problems has been introduced by the author in the study of Inverse scattering problemsin quantum mechanics and in geophysics and applied to many other inverse problems. It was the maintool in Chapter 5. A proof of Lemma 9.1.6 follows from the general argument given in Chapter 5. Itis clear from this proof that not all of the functions u1, u2 ∈ ND(∆) 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 u1u2 is complete inL2(D). First, we can restrict ourselves to the exponential solutions of the Laplace equation. This is nota restriction really, since the set of exponential solutions of this equation is dense in L2(D) in the setof all harmonic functions in D, i.e., all elements of ND(∆). However, there are still many degrees offreedom 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).Take the Fourier transform in the variables x1 := (x1, x2) and y1 = (y1, y2) asssuming that (x =

x1, x2, 0), y = (y1, y2, 0) belong to P , and use the formula

∫∫ ∞

−∞|x1 − z|−1 exp(iλ · x1)dx1 = 2π|λ|−1 expiλ · z1 − |λ||z3|, (9.2.1)

where dx1 = dx1dx2, λ · x1 = λ1x1 + λ2x2, |λ| = (λ21 + λ2

2)1/2, to get:

(2π)−2

∫dzv(z) expi(λ + µ) · z1 − (|λ| + |µ|)|z3| = |λ||µ|f(λ, µ), (9.2.2)

where∫

=∫

R3 ,

f (λ, µ) :=

∫∫∫∫ ∞

−∞f(x1, y1) expi(λ · x1 + µ · y1)dx1dy1. (9.2.3)

Letp := λ + µ, pj = λj + µj, j = 1, 2; p3 = |λ|, p4 = |µ|, (9.2.4)

and write (9.2.2) as ∫dzv(z) exp(ip · z1 + (p3 + p4)z3) = F (p1, p2, p3, p4). (9.2.5)

Here we took into account that supp v ⊂ R3− so that |z3| = −z3 in (9.2.2), and defined F (p1, p2, p3, p4) :=

4π2|λ||µ|f(λ, µ) in the coordinates (9.2.4). The Jacobian

J :=∂(p1, p2, p3, p4)

∂(λ1, λ2, µ1, µ2)=

∣∣∣∣∣∣∣∣∣∣

1 0 1 0

0 1 0 1

λ1|λ|−1 λ2|λ|−1 0 0

0 0 µ1|µ|−1 µ2|µ|−1

∣∣∣∣∣∣∣∣∣∣

=µ1λ2 − µ2λ1

|µ||λ| (9.2.6)

does not vanish if µ and λ 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).

9.3. CHARACTERIZATION OF THE LOW-FREQUENCY DATA 267

Letp3 = p4 :=

s

2, F (p1, p2,

s

2,s

2) := φ(p1, p2, s) (9.2.7)

and write (9.2.5) as

∫ ∞

0

dζ

∫∫ ∞

−∞dz1dz2v(z1, z2,−ζ) exp(ip · z1 − sζ) = φ(p1, p2, s), ζ := −z3 (9.2.8)

Equation (9.2.8) defines v(z) uniquely: one should take the inverse Fourier transform in the variablesp 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 onthe function f(x, y) in (9.1.6) for this equation to be solvable in the class of compactly supported in R3

−square integrable v(z).

Theorem 9.3.1. Equation (9.1.6) is solvable in the above class of v iff φ(p1, p2, s) is an entire functionof p1, p2 and s of exponential type ≤ R, where R is a positive number, and

supσ≥0

∫∫∫ ∞

−∞|φ(p1, p2, σ + iτ )|2dp1dp2dτ < ∞. (9.3.1)

Proof. The necessity of these conditions follows from formula (9.2.8) because v(z) is compactly supportedin R3

−. Their sufficiency follows from the Paley-Wiener theorem:i) a function h(p), p = (p1, p2), is entire of exponential type ≤ R and

∫∫∞−∞ |h(p)|2dp < ∞ iff

h(p) =∫|x1|≤R w(x1) exp(−ip · x1)dx1,

∫|x1|≤R |w|2dx1 < ∞,

andii) a function h(x) is an entire function of exponential type ≤ R such that supσ≥0

∫∞−∞ |h(σ+iτ )|2dτ <

∞ iff h(s) =∫ R0 exp(−st)f(t)dt where f ∈ L2(0, R). 2

Exercise. Prove that φ(p1, p2, s) is bounded for p ∈ R2, s ≥ 0.

9.4 Problems of numerical implementation

In this section we consider some practical questions related to inversion of (9.2.8). Let

(λ1, λ2, µ1, µ2) =s

2(cos θ1, sin θ1, cos θ2, sin θ2) (9.4.1)

so that p1 = s cosα cos β, p2 = s cosα sin β, α := θ1−θ22 , β := θ1+θ2

2 . If we take 0 < α < π2 , 0 ≤

β < 2π, then J in (9.2.6) does not vanish and , given p1, p2 and s one finds α and β within theabove ranges, then θ1 and θ2, and then (λ1, λ2, µ1, µ2). Therefore, we can compute φ(p1, p2, s) in (9.2.8)by formula (9.2.7). In the three-dimensional space with Cartesian coordinates (p1, p2, s) the pointsQ = (s cosα cos β, s cosα sin β, s), 0 < α < π

2 , 0 ≤ β < 2π, fill in the circle of radius s centered at (0, 0, s)and lying in the plane parallel to (p1, p2) plane. As s run from 0 to +∞ the set of points Q fill in thecone C in the upper half-space with vertex at the origin and a vertex angle of π

2 . We invert the Laplacetransform in (9.2.8) by the formula

v(p1, p2,−ζ) =1

2πi

∫ σ+i∞

σ−i∞exp(sζ)φ(p1, p2, s)ds, σ > 0 (9.4.2)


where

v(p1, p2,−ζ) :=

∫∫ ∞

−∞expi(p1z1 + p2z2)v(z1, z2,−ζ)dz1dz2. (9.4.3)

If we use formula (9.4.2) for inversion then φ(p1, p2, s) should be known for s = σ + iτ , −∞ < τ < ∞and p2

1 + p22 ≤ σ2. Given any particular p1, p2 we can choose σ2 ≥ p2

1 + p22 and calculate v(p1, p2, ζ).

Formula (9.4.2) requires the knowledge of φ(p1, p2, s) on the line s = σ + iτ , −∞ < τ < ∞. However,one can invert the Laplace transform of a compactly supported function using only its values on the realaxis (cf [AR2]). If one wishes to fix σ in (9.4.2), then v(p1, p2,−ζ) is recovered for p2

1 + p22 ≤ σ2. The

Fourier transform v(p1, p2,−ζ) of a compactly supported function v(z1, z2,−ζ), ζ = −z3, is thereforeknown for p2

1 + p22 ≤ σ2. One can invert this Fourier transform from the compact region p2

1 + p22 ≤ σ2.

9.5 Half-spaces with different properties

In this section we consider the case when the lower half-space has different properties from the upperone. Let ρ1, α1, ρ2, α2(x) are the densities and velocities in R3

+ and R3− respectively, ρ1 = const, ρ2 =

const, α1 = const, α2(x) = α2+v(x), α2 = const, the assumptions about v(x) are the same as in Section9.1. The inverse problem IP in this case can be formulated as follows. The governing equations are:

∇2um +k2umα2m

= −δ(x− y), m = 1, 2 (9.5.1)

∂um∂x3

=∂um∂x3

, ρ1u1 = ρ2u2 at x3 = 0. (9.5.2)

The inverse problem IP is:Given u(x, y, k) for all x, y ∈ P and all k ∈ (0, k0), find v(x).The solution to (9.5.1)-(9.5.2) with αm = const, m = 1, 2, is

u2 =1

2π

∫ ∞

0

exp(ν2x3 − ν1y3)J0(λρ)λdλ

ν1δ + ν2, x3 < 0, y3 ≥ 0 (9.5.3)

where νm := (λ2 − k2α−2m )1/2, ρ = (x1 − y1)

2 + (x2 − y2)1/21/2, δ = p2p

−11 . If y3 = 0 and k → 0 then

the limit of (9.5.3) is

u2 =1

2π(δ + 1)

∫ ∞

0

J0(λp) exp(−λ|x3|)dλ =2ρ1

ρ1 + ρ2

1

4π|x− y| , y = (y1, y2, 0) (9.5.4)

Therefore, the integral equation analogous to (9.1.6) for IP is essentially the same:

∫

D

v(z)dz

|x− z||y − z| =

(ρ1 + ρ2

2ρ1

)2

f(x, y), x, y ∈ P. (9.5.5)

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) isgiven with x, y ∈ SR, where SR is a sphere x : |x| = R, x ∈ R3 which contains D = supp v. Letus 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

∫

D

g(x, z, k)g(z, y, k)v(z)dz = f(x, y, k) := k−2usc(x, y, k), x, y ∈ SR. (9.6.1)

9.6. INVERSION OF THE DATA GIVEN ON A SPHERE 269

Let us use the well known formulas (see, e.g., [GR])

g(x, z, k) = ik

∞∑

`=0

∑

m=−`j`(k|z|)h(1)

` (k|x|)Y`m(x0)Y`m(z0), |x| > |z|, (9.6.2)

where x0 = x/|x|, z0 = z/|z|, j` and h(1)` are the spherical Bessel and Hankel functions respectively,

j`(r) :=( π

2r

)1/2

J`+ 12(r), h

(1)` (r) =

( π2r

)1/2

H(1)

`+12

(r), (9.6.3)

exp(ikθ · x) = 4π

∞∑

`=0

∑

m=−ì`j`(k|x|)Y`m(x0)Y`m(θ), x0 :=

x

|x| . (9.6.4)

Substitute (9.6.2) into (9.6.1) and get

− k2∞∑

`,`′=0

∑

|m|≤`,|m′ |≤`′

∫v(z)j`(k|z|)Y`m(z0)j`′(k|z|)Y`′m′ (z0)dzY`m(x0)Y`′m′(y0) ·

· h(1)` (k|x|)h(1)

`′ (k|y|) =

∞∑

`,`′=0

∑

|m|≤`,|m′|≤`′f`m,`′m′Y`m(x0)Y`′m′ (y0).

(9.6.5)

Thus

v``′mm′ : =

∫v(z)j`(k|z|)j`′(k|z|)Y`m(z0)Y`′m′(z0)dz

= −k−2f`m,`′m′h(1)` (kR)h

(1)`′ (kR)−1.

(9.6.6)

Thus v``′mm′ are expressed in terms of the data analytically. Multiply v``′mm′ by 4πY`m(θ)Y`′m′ (−θ′)i`+`′ ,sum over `m, `′m′ and use (9.6.4) to get

−4πk2∞∑

`,`′=0

∑

|m|≤`,|m′ |≤`′f`m,`′m′h(1)

` (kR)h(1)`′ (kR)−1i`+`

′

Y`m(θ)Y`′m′(−θ′) =

∫expik(θ − θ′) · zv(z)dz.

(9.6.7)

Choose an arbitrary λ ∈ R3, |λ| ≤ 2k, and θ, θ′ ∈ S2 such that k(θ − θ′) = −λ. The left-hand side of(9.6.7) is a known function, which we denote v(λ) if k(θ − θ′) = −λ, and (9.6.7) gives v(λ) for |λ| ≤ 2k.Since v(z) is compactly supported, we reduced the problem of solving (9.6.1) to inversion of the Fouriertransform from the compact |λ| ≤ 2k. This problem is solved analytically with arbitrary accuracy forexact data in [R51] and the solution was used in [R83], [R139] and [RKa], see also Theorem 5.8.4. Wehave proved that equation (9.6.1) can be solved analytically if k > 0. If k = 0 then the above argumentgives v(0), and this is not sufficient for recovery of v(z). However, equation (9.6.1) at k = 0 has at mostone solution. The data at k = 0 determine uniquely the numbers

v0``′mm′ =

∫v(z)|z|`+`′Y`m(z0)Y`′m′ (z0)dz (9.6.8)

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 givenwith x, y ∈ CR = x : ρ = R, ρ = (x2

1 +x22)

1/2, D ⊂ CR. The idea of inversion is the same as in section9.6. We start with the formula

g(x, z, k) =

∞∑

h=−∞

expin(φz − φx)4π2

∫ ∞

−∞expiλ(z3 − x3)

iπ

2Jn(αρz)H

(1)n (αρx)dλ, (9.7.1)

where ρz < ρx, α = (k2 − λ2)1/2, ρx = ρy = R, ρx := (x21 + x2

2)1/2, and

expi[λx3 + αρ cos(φ− φ′)] = exp(iλx3)∞∑

n=−∞in expin(φ − φ′)Jn(αρ). (9.7.2)

Equation (9.6.1) can be written as

− 1

64π2

∞∑

n,n′=−∞

∫∫ ∞

−∞dλdλ′ exp−i(nφx + n′φy) − i(λx3 + λ′y3)H(1)

n (αR)H(1)n (α′R)

×∫v(z) expi(n+ n′)φz + i(λ + λ′)z3Jn(αρz)Jn′(α′ρz)dz

=

∞∑

n,n′=−∞

∫∫ ∞

−∞dλdλ′fλnλ′n′ exp−i(nφx + n′φy) − i(λx3 + λ′y3),

(9.7.3)

where α′ = (k2 − λ′2)1/2 and

fλnλ′n′ =1

(2π)4

∫ 2π

0

∫ 2π

0

∫ ∞

−∞

∫ ∞

−∞expi(nφx + n′φy) + i(λx3 + λ′y3)f(φx, x3, φy, y3)dφxdφydx3dy3.

(9.7.4)Here f(φx, x3, φy, y3) is the datum on CR. It follows from (9.7.3) that

∫v(z) expi(n+ n′)φz + i(λ + λ′)z3Jn(αρz)Jn′(α′ρz)dz = − 64π2fλnλ′n′

H(1)n (αR)H

(1)n (α′R)

. (9.7.5)

Multiply (9.7.5) by in+n′

exp−i(nφ+ n′φ′), sum over n, n′ and use (9.7.2) to get

∫v(z) expik(ν + ν ′) · zdz = F (k(ν + ν ′)) (9.7.6)

where

F (k(ν + ν ′)) := −64π2∞∑

n,n′=−∞

fλnλ′n′in+n′

exp−i(nφ+ n′φ′)H

(1)n (αR)H

(1)n (α′R)

, (9.7.7)

λ = k cos θ, α = k sin θ, λ′ = k cos θ′, α′ = k sin θ′, ν = (sin θ cos φ, sin θ sinφ, cos θ), ν ′ = (sin θ′ cosφ′, sin θ′ sinφ′, cos θ′).Let k(ν + ν ′) = µ. When ν, ν ′ run through S2 vector µ runs through the ball |µ| ≤ 2k in R3. Thus,equation (9.7.6) reduces to finding v(z), a compactly supported function, from its Fourier transform v(µ)known for |µ| ≤ 2k.

If k = 0 then equation (9.6.1) with x, y ∈ CR has at most one solution. We leave further details forthe reader.

9.8. TWO-DIMENSIONAL INVERSE PROBLEMS 271

9.8 Two-dimensional inverse problems

In this section we consider two-dimensional problems. The difficulty in two-dimensional problems comes

from the fact that Green’s function of the Helmholtz equation in R2 is i4H

(1)0 (k|x − y|) and it has no

finite limit as k → 0 :

g :=i

4H

(1)0 (kr) = α(k) + g0 + O

(k2r2 log

1

kr

)as k → 0 (9.8.1)

where g0 := − 12π ln r, r := |x− y|, and we denote in this section α := α(k) := − 1

2π ln k2 − γ

2π + i4 , where

γ = 0.5572 . . . is Euler’s constant. Consider equation (9.1.6) in R2 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

usc := u− g = k2α2

∫v(z)dz + α

∫[g0(x, z) + g0(y, z)]v(z)dz

+

∫g0(x, z)v(z)g0(z, y)dz + o(k2) as k → 0.

(9.8.2)

Therefore

limk→0

usck2α2

:= f0 =

∫v(z)dz, (9.8.3)

limk→0

usc − α2k2f0αk2

:= f1(x, y) =

∫[g0(x, z) + g0(y, z)]v(z)dz, (9.8.4)

and

limk→0

usc − α2k2f0 − αk2f1(x, y)

k2:= f2(x, y) =

∫g0(x, z)v(z)g0(z, y)dz. (9.8.5)

The quantities f0, f1(x, y) and f2(x, y) are calculated explicitly given the low-frequency data. Thenumber f0 is the intensity of the inhomogeneity. Equation (9.8.5) is analogous to (9.1.6). It follows from(9.8.4) that ∫

g0(x, z)v(z)dz =1

2f1(x, x). (9.8.6)

This is an integral equation arising in inverse problems of potential theory. In general, it is not uniquelysolvable. However, if v(z1, z2) = v(z2) for a ≤ z1 ≤ b, −R < z2 < 0 and is zero otherwise, then takex = (x1, 0) in (9.8.6), and get

∫ 0

−Rdz2v(z2)

∫ b

a

ln[(x1 − z1)2 + z2

2]1/2dz1 = −πf1(x1, 0;x1, 0) := h(x1). (9.8.7)

Differentiate (9.8.7) with respect to x1, put z2 = η, z1 = ξ, x1 = ζ, h′(ζ) := φ(ζ), and get:

∫ 0

−Rdηv(η)

∫ b

a

ζ − ξ

(ζ − ξ)1 + η2dξ = φ(ζ), −∞ < ζ <∞. (9.8.8)

Take the Fourier transform of (9.8.8) in ζ, φ(λ) :=∫∞−∞ exp(−iλζ)φ(ζ)dξ, to get

∫ 0

−Rdyv(y)

∫ b

a

dξ exp(iλξ)

∫ ∞

−∞

s exp(iλs)ds

s2 + y2= φ(λ). (9.8.9)


One has ([Er]) ∫ ∞

−∞

s exp(iλs)ds

s2 + η2= iπsgnλ exp(−|λ||η|),

so ∫ ∞

−∞

ζ − ξ

(ζ − ξ) + η2exp(iλζ)dλ = exp(iλξ)iπsgnλ exp(−|λ||η|). (9.8.10)

Thus (9.8.9) becomes

∫ 0

−Rdηv(η)

∫ b

a

dξ exp(iλξ)iπsgnλ exp(−|λ||η|) = φ(λ) (9.8.11)

or ∫ 0

−Rdηv(η) exp(−|λ||η|) = φ4(λ) :=

λφ(λ)sgnλ

π[exp(iλb) − exp(iλa)]. (9.8.12)

Therefore the Laplace transform of v(η) is known and v(η) can be recovered analytically by the methodgiven in [R139]. Thus, under the assumption v = v(z2) an analytical inversion is given of the zero offsetdata (i.e. x = y) f1(x, x) known at the line L := x : x2 = 0.

Consider equation (9.8.5). Let x, y ∈ L. Differentiate (9.8.5) with respect to x1 := ζ and y1 := η and

let 4π2 ∂2f(x,y)∂x1∂y1

|x2=y2=0 := h(ζ, η). Then

∫dzv(z)

ζ − z1[(ζ − z1)2 + z2

2 ]

η − z1[(η − z1)2 + z2

2]= h(ζ, η). (9.8.13)

Take the Fourier transform in ζ and η of (9.8.13) and use (9.8.10) to get

−π2

∫dzv(z) expi(λ+ µ)z1 − (|λ|+ |µ|)|z2| = h(λ, µ). (9.8.14)

Let λ + µ = ν, |λ| + |µ| = t. Then (9.8.14) can be written as

∫dzv(z) exp(iνz1 − t|z2|) = w(ν, t), (9.8.15)

where w(ν, t) is h(λ, µ) calculated in the new coordinates. The Jacobian of the transformation λ, µ →ν, t is J =

∂(ν,t)∂(λ,µ) = sgnµ−sgnλ 6= 0 if λµ < 0. Take µ > 0 and λ < 0 and invert the Fourier transform

in ν 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 v(x), x ∈ R1, given the values of u(x, y, k)for x ≥ 0, y = 0, k ∈ (0, k0), k0 > 0 is a fixed (small) number. This section is of methodological nature.The basic integral equation is

u(x, k) = g(x, k) + k2

∫g(x, z, k)v(z)u(z, k)dz,

∫=

∫

R1

, (9.9.1)

g(x, y, k) =i

2kexp(ik|x− y|), g′′ + k2y = −δ(x− y) (9.9.2)

g =i

2k− |x− y|

2+ O(k|x− y|2) as k → 0 (9.9.3)

9.10. INVERSION OF THE BACKSCATTERING DATA 273

Let u− g = φ. Then (9.9.1) can be written as

φ = φ0 + k2

∫gvφdz, φ0 := k2

∫gvgdz = −1

4

∫exp(ik(|x− z| + |z|)vdz (9.9.4)

Equation (9.9.4), as k → 0, is uniquely solvable by iterations. One has

limk→0

φ = −1

4

∫vdz := −v0

4. (9.9.5)

Let

φ =

∞∑

j=0

Cj(x)(ik)j , φ0 = −1

4

∞∑

j=0

φ0j(x)(ik)j , (9.9.6)

φ0j :=1

j!

∫(|x− z| + |z|)jv(z)dz. (9.9.7)

Substitute (9.9.6) into (9.9.4) and equate coefficients in front of similar powers of (ik)j . This givessome recurrence formulas for Cj, C0 = −v0

4. From these formulas one can find the moments vn :=∫ 0

−R v(z)zndz, and, therefore, v(z).

9.10 Inversion of the backscattering data and a problem of in-

tegral geometry

In this section we invert analytically the backscattering data in the Born approximatisn. Sut x = y in(9.6.1) and assume that 16π2f(x, x, k) := F (x, k) is known for all x ∈ P := x : x3 = 0 and all k > 0.The problem is to find v(z) from equation (9.6.1) with the above data:

∫

D

dzv(z)exp(2ik|x− z|)

|x− z|2 = F (x, k) ∀x ∈ P, ∀k > 0. (9.10.1)

Differentiate (9.15.1) in k, put 2k = λ, 12i∂F∂k |k=λ

2:= φ(λ) and get

∫

D

dzv(z)exp(iλ|x− z|)

|x− z| = φ(x, λ), ∀x ∈ P, ∀λ > 0. (9.10.2)

Take the Fourier transform of (9.10.2) in x ∈ P using the formula

∫

P

dxexp(iλ|x − z|)

|x− z| exp(−iν · x) =2πi exp−iν · z + i|z3|(λ2 − ν2)1/2

(λ2 − ν2)1/2, λ > |ν|, (9.10.3)

where for λ < |ν| one has (λ2 − ν2)1/1 = i(|ν|2 − λ2)1/2, and get

∫

D

dzv(z) exp−iν · z + i|z3|(λ2 − ν2)1/2 =φ(ν, λ)(λ2 − ν2)1/2

2πi, λ > |ν|. (9.10.4)

Define α ∈ R3, α = (ν1, ν2, (λ2 − ν2)1/2) for |ν| < λ. Since |z3| = −z3, (9.10.4) can be written as

∫

D

dzv(z) exp(−iα · z) = A(α) |α| ≤ λ (9.10.5)

where A(α) is the right-hand side of (9.10.5) expressed in the variable α. Thus v(z) can be founduniquely from (9.10.5) by inversion of the Fourier transform. Note that A(α) is an entire function of α


of exponential type, that is |A(α)| ≤ c1 exp(c2|α|) ∀α ∈ C3, since v(z) is compactly supported. Thisnecessary condition on the data is also sufficient for v(z) to have compact support, as follows from thePaley-Wiener theorem.

Let us give a simple example:

Let v(z) = δ(z), where δ(z) is Dirac’s function. Then φ(x, λ) = exp(iλ|x|)|x| , φ(ν, λ) = 2πi

(λ2−|ν2|)1/2 ,

A(α) = 1. Thus A(α) is an entire function and its inverse Fourier transform is v(z) = δ(z).Let us summarize the result.

Proposition 9.10.1. Equation (9.10.2) has at most one solution in the class of compactly supporteddistributions. For equation (9.10.2) to have a compactly supported solution it is necessary and sufficientthat the function φ(ν, λ)(λ2 − ν2)1/2 be an entire function of α ∈ C3 of exponential type, where α :=(ν1, ν3, (λ

2 − |ν|2)1/2).Equation (9.10.2) is equivalent to anIntegral geometry problem of finding v(z) from its integrals over the spheres centered

at x ∈ P and of radii t ∈ (0,∞).Indeed, define φ(x,−λ) = φ(x, λ) where the bar stands for complex conjugate. Multiply (9.10.2) by

(4π)−1 exp(−iλt) and integrate in λ over (−∞,∞) to get

∫

D

dzv(z)δ(|x− z| − t)

|x− z| = h(x, t), t > 0, x ∈ P (9.10.6)

where δ(|x− z| − t) is the delta function and h(x, t) := (2π)−1∫∞−∞ φ(x, λ) exp(−iλt)dλ, h = 0 for t < 0.

Equation (9.10.6) can be written as∫

|x−z|=tv(z)ds = th(x, t) (9.10.7)

Thus the integrals of v(z) over the family of spheres S(x, k) = z : |x − z| = t, z ∈ R3 are known,and v(z) is recovered from these data via equation (9.10.5). Necessary and sufficient conditions on thefunction th(x, t) for this function to be of the form (9.10.7) follow from the given above necessary andsufficient conditions on φ(x, λ) 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∫

D

g(x, z, k)v(z)g(z, y, k)dz = f(x, y, k) x ∈ `−a, y ∈ à (9.11.1)

where g(x, y, k) = i4H

(1)0 (k|x − y|), à = x : x1 = a,−∞ < x3 < ∞, z = (z1, z3), y = (y1, y3), and

k > 0 is fixed. Equation (9.11.1) models the well-to-well imaging problem in the Born approximationin the case when the inhomogeneity v(z) does not depend on one of the coordinates, say z2, and theacoustic geophysical data are collected along two boreholes, lines `−a and à, at a fixed k > 0. We wishto solve equation (9.11.1) for v(z). Take the Fourier transform of (9.11.1) with respect to x3 and y3 andget

f (λ, µ) = (2π)−2

∫∫ ∞

−∞exp(−iλx3 − iµy3)f(x3, y3)dx3dy3 = − 1

16π2

∫v(z) exp−i(λ + µ)z3·

· expi(a+ z1)(k2 − λ2)1/2 + i(a− z1)(k

2 − µ2)1/2(k2 − λ2)1/2(k2 − µ2)1/2

dz, k2 > λ2, k2 > µ2.

(9.11.2)

9.11. INVERSION OF THE WELL-TO-WELL DATA 275

If k2 > λ2 then (k2 − λ2)1/2 = i(λ2 − k2)1/2. The radical is defined on the complex plane λ with the cut(−k, k) so that (λ2 − k2)1/2 > 0 for λ2 > k2 and in the calculations k is taken to be k + i0.

In particular, for k2 < µ2 and k2 < λ2 one obtains:

f (λ, µ) =1

16π2

∫

D

v(z) exp−i(λ+ µ)z3exp−(a + z1)(λ

2 − k2)1/2 − (a − z1)(µ2 − k2)1/2

(λ2 − k2)1/2(µ2 − k2)1/2dz,

k2 < λ2, k2 < µ2.

(9.11.3)

We have used the formula

(2π)−1

∫ ∞

−∞exp(−iλx3)g(x, z, k)dx3 =

iπ exp−iλz3+i(a+z1)(k2−λ2)1/2(2π)2(k2−λ2)1/2 if k2 > λ2

π exp−iλz3−(a+z1)(λ2−k2)1/2

(2π)2(λ2−k2)1/2 if k2 < λ2

∞ if k2 = λ2.

(9.11.4)

Consider the case when the data are given for (λ, µ) ∈ S := λ, µ : −k < λ < k and−k < µ < k. Let p := λ + µ, q := (k2 − µ2)1/2 − (k2 − λ2)1/2,

F (p, q) := −16π2f (λ, µ)(k2 − λ2)1/2(k2 − µ2)1/2 exp−ia[(k2 − λ2)1/2 + (k2 − µ2)1/2] (9.11.5)

and the right-hand side of (9.11.5) should be expressed in the variables p, q. The Jacobian

J :=

∣∣∣∣∂(p, q)

∂(λ, µ)

∣∣∣∣ = − µ

(k2 − µ2)1/2− λ

(k2 − λ2)1/26= 0 if λ 6= −µ (9.11.6)

Equation (9.11.2) can be written as

∫

D

dzv(z) exp(−ipz3 − iqz1) = F (p, q) (9.11.7)

Consider the image S of S under the transformation (λ, µ) → (p, q). When (λ, µ) run through theboundary of S then (p, q) run through the boundary of S. Take λ = −k and let µ run through thesegment (−k, k). On (p, q) plane one gets the curve (p + k)2 + q2 = k2, q > 0, that is an upperhalf of the circle centered at (−k, 0) with radius k. A similar argument shows that the boundary ofS consists of the circles C± : (p + k)2 + q2 = k2 and (p − k)2 + q2 = k2. The diagonal λ = −µof S is mapped into the origin of the (p, q) plane. Therefore, given the data f (λ, µ) in S one knowsF (p, q) in the disks D± with the boundaries C±. Equation (9.11.7) shows that the Fourier transform ofv(z) is known in D+ ∪D−. Since v(z) is compactly supported, the Fourier transform can be uniquelyinverted from D+ ∪ D− for v(z) by the method of [R139] (see also [RKa]). Consider the case whenλ, µ ∈ S′ = λ, µ : |λ| > k and |µ| > k, λ, µ are real. In this case imaging with evanescent wavesis studied. Write (9.11.3) as

∫dzv(z) exp(−ipz3 + qz1) = φ(p, q), |λ| > k, |µ| > k, (9.11.8)

where p := λ + µ, q := (µ2 − k2)1/2 − (λ2 − k2)1/2,

φ(p, q) := 16π2f(λ, µ)(λ2 − k2)1/2(µ2 − k2)1/2 expa(λ2 − k2)1/2 + a(µ2 − k2)1/2, (9.11.9)

where the right-hand side of (9.11.9) should be expressed in the variables p, q. As above, the Jacobian

of the transformation (λ, µ) → (p, q) does not vanish for λ 6= −µ. The image S′ of the region S′

under this transformation has the boundary which consists of the hyperbolas (p + k)2 − q2 = k2 and


(p − k)2 − q2 = k2. The knowlesge of φ(p, q) in S′ allows one to uniquely recover v(z) by inverting theFourier and Laplace transforms in (9.11.6). One can consider the case when k = 0, so that (9.11.8) holds

with p = λ+ µ, q = |µ| − |λ|, and again v(z) can be uniquely recovered from the data given in S′.Finally, let us briefly discuss the low frequency exact inversion theory for the problem at hand. One

starts with the equation

u =

∫

D

g(x, z, k)v(z)g(z, y, k)dz +O(k4) as k → 0, (9.11.10)

takes the Fourier transform of (9.11.10) with respect to x3 and y3, and let k → 0 to get equation (9.11.8),where p = λ+ µ and q = |µ|− |λ|. Therefore the inversion problem is solved as above. In this argumentwe avoided the direct study of the asymptotic behavior of the solution to (9.11.10) as k → 0. If onetakes the Fourier transform of (9.11.10), then the singular term α(k)δ(λ)δ(µ), with α(k) defined belowformula (9.8.1), vanishes if one considers the region λ2µ2 > 0 as we do in (9.11.8). This makes it possibleto pass to the limit k → 0 in the Fourier transformed equation (9.10.2) while it is not possible to passto the limit k → 0 in (9.11.9) directly since g(x, y, k) does not have a finite limit as k → 0 (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 v = v(z3) for|z3| ≤ R and r ≤ R, v = 0 otherwise. Here r = (z2

1 + z22)1/2. Let x = (0, 0, x), y = (0, 0, y), y = x − d,

d = const, z3 = ζ. Then one has:

∫ ∞

−∞dζv(ζ)h(x − ζ) = φ(x), −∞ < x <∞ (9.12.1)

where

φ(x) := (2π)−1f(x, x− d), h(x− ζ) =

∫ R

0

drr[(ζ − x)2 + r2][(ζ − x+ d)2 + r2]−1/2 (9.12.2)

Equation (9.12.1) is of convolution type and can be solved by the Fourier transform. If φδ is knownin place of the exact data φ, |φδ − φ| < δ, δ > 0 is a known small number, the noise level, then aregularization is needed for a stable solution of (9.7.7), that is, for finding vδ such that |vδ(ζ)−v(ζ)| → 0as δ → 0.

If the measurements are taken strictly inside the borehole, then equation (9.12.1) is to be replacedby the equation ∫ ∞

0

dζv(ζ)h(x − ζ) = φ(x), x > 0 (9.12.3)

This is a Wiener-Hopf equation of the first kind which can be solved by the known factorization methodsor 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 h(x − ζ, k) =∫ R0drr exp[ik(α1 + α2)](α1α2)

−1, where

α1 := [(ζ − x)2 + r2]1/2, α2 := [(ζ − x+ d)2 + r2]1/2.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 x3-axis. The acoustic field is measured at thepoint x which is also on the x3-axis. The distance between x and y is d. One wishes to find theinhomogeneity v from the logging measurements, that is, from φ(x).

Consider now the two-dimensional case in which v = v(r, ζ), r = (z21 +z2

2 )1/2, ζ = z3, v = 0 for r ≥ Ror |ζ| > R, f(x, y, k) is measured for all x, y on the x3-axis L that is on the axis of the borehole, k > 0.

9.12. INDUCTION LOGGING PROBLEMS 277

The basic equation is∫

D

expik(|x− z| + |z − y|)|x− z||y − z| v(z)dz = f(x, y, k), x, y ∈ L (9.12.4)

Let us use the spheroidal coordinates:

z3 = `st +x3 + y3

2, z2 = `(s2 − 1)1/2(1 − t2)1/2 cos θ +

x2 + y22

(9.12.5)

z1 = `(s2 − 1)1/2(1 − t2)1/2 sin θ +x1 + y1

2, ` = |x− y|/2. (9.12.6)

The Jacobian J = |∂(z1, z2, z3)/∂(s, t, θ)| = `3(s2−t2), |x−z|+|z−y| = 2`s, 1 ≤ s <∞, |x−z|−|z−y| =2`t, −1 ≤ t ≤ 1, θ is the angle betwean the plane containing x and y and the plane containing x− z andy − z. In our case x and y are directed along L and we take as the plane containing x and y the planez2 = 0. In the variables (s, t, θ) equation (9.12.4) takes the form:

2π`

∫ ∞

1

ds exp(2ik`s)

∫ 1

−1

v

(`(s2 − 1)1/2(1 − t2)1/2, `st +

x+ y

2

)dt = f(x, y, k). (9.12.7)

It is clear from (9.12.7) that the data f should depend on ` = |x−y|2 and m := x+y

2 .Assume that f(x, y, k) is given for all k > 0, and denote

p(s, `,m) :=

∫ 1

−1

v(`(s2 − 1)1/2(1 − t2)1/2, `st +m)dt, s ≥ 1, ` ≥ 0 (9.12.8)

The function p(s, `,m) can be found from (9.12.7) by taking the inverse Fourier transform in the variablek, and equation (9.12.8) can be solved for v(r, ζ) given p(s, `,m). Letξ := `(s2 − 1)1/2(1− t)1/2, η := `st+m. Denote by E(s, `) the ellipse ξ2(s2 − 1)−1 + (η−m)2s−2 = `2.Equation (9.12.8) can be written as

∫

E(s,`)

w(ξ, η)dσ = p(s, `,m), s ≥ 1, ` ≥ 0. (9.12.9)

Here dσ is the element of the arc length of the ellipse, dσ = `(s2−t2)1/2(1−t2)−1/2dt, w := v(ξ, η)a(ξ, η),where a(ξ, η) := dt

dσexpressed in the variables (ξ, η). 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 similarproblem was studied in the case when the family of ellipses was different: one focus was at the originand the other one ran through a straight line.

Suppose that v(r, ζ) = v(ζ). Then (9.12.8) reduces to

∫ 1

−1

v(`st +m)dt = p(s, `,m), s ≥ 1, ` ≥ 0. (9.12.10)

Let s = m` , m(t+ 1) = τ , and write (9.12.10) as

∫ 2m

0

v(τ )dτ = mp(m

`, `,m) := h(m, `). (9.12.11)

Differentiate (9.12.11) to get

v(2m) =1

2

dh(m, `)

dm, v(ζ) =

1

2

dh

dm

∣∣∣m= ζ2

(9.12.12)

Several other cases of analytic solvability of equation (9.12.4) are considered in [R139].


9.13 Examples of non-uniqueness of the solution to an inverseproblem 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 arecollected at the point at which the source is, that is, at the zero-offset. The uniqueness problem reducesto the homogeneous equation ∫

D

v(z)dz

|x− z|2 = 0 ∀x ∈ P, D ⊂ R3− (9.13.1)

A similar equation corresponding to the Born inversion is (k > 0 is fixed):

∫

D

exp2ik|x− z||x− z|2 v(z)dz = 0 ∀x ∈ P, D ⊂ R3

− (9.13.2)

Proposition 9.13.1. Equation (9.13.1) has infinitely many non-trivial solutions.

Remark 9.13.2. Equation (9.13.1) ∀x ∈ B, where B is an open set in R3 \ D, has only the trivialsolution.

Remark 9.13.3. Equation ∫

D

v(z)dz

|x− z| = 0 in D′ = R3 \D (9.13.3)

has infinitely many non-trivial solutions.

Proof of Proposition 9.13.1. Define

v(z) = w1(z1)φ1(ζ) − w2(z

1)φ2(ζ), z1 = (z1, z2), ζ = z3. (9.13.4)

Take an arbitrary w(z1) ∈ C∞0 (R2), w = 0 if |z1| > R, and φ(ζ) ∈ C∞

0 (−R, 0), put w1 = w, w2 = ∆w,

∆ = ∂2

∂z21+ ∂2

∂z22, φ2 = −ζφ(ζ), φ1 = (ζφ′)′. Then v defined by (9.13.4) is a non-trivial solution to (9.13.1).

Indeed, take the Fourier transform in x1 of (9.13.1) to get

∫dζv(µ, ζ)K0(|µ||ζ|) = 0 ∀µ = (µ1, µ2) ∈ R2 (9.13.5)

Here v(µ, ζ) =∫∫∞

−∞ exp(−iµ · z1)v(z1, ζ)dz1, and ([GR], formula 6.65.4):

K0(|µ||ζ|) = (2π)−1

∫∫ ∞

−∞(ζ2 + |z1|2)−1 exp(iµ · z1)dz1, (9.13.6)

where K0 is the modified Bessel function. Substitute (9.13.4) into (9.13.5) to get

w(µ)

∫dζ(ζφ′)′K0(|µ||ζ|) = µ2w(µ)

∫dζζφ(ζ)K0(|µ||ζ|). (9.13.7)

After an integration by parts on the left, and taking into account that [ζK ′0(|µ||ζ|)]′−µ2ζK0(|µ||ζ|) =

0, one obtains an identity which implies that the function (9.13.4): v(z) = w(z1)[ζφ′(ζ)]′ +∆w(z1)ζφ(ζ)is a non-trivial solution to (9.13.1) for any w ∈ C∞

0 (R2), w = 0 if |z1| > R, and any φ ∈ C∞0 (−R, 0),

where the cylinder C := z : |z1| ≤ R,−R < ζ < 0 ⊂ D. The proof is completed. 2

Proof of Remark 9.13.2. Suppose (9.13.1) holds for all x ∈ B where B ⊂ R3 \ D is an open set. Thefunction

h(x) :=

∫

D

v(z)dz

|x− z|2 (9.13.8)

9.14. SCATTERING IN ABSORPTIVE MEDIUM 279

is an analytic function of x1, x2 and x3 in a neighborhood in C3 of the region R3 \D. Therefore, by theunique continuation property, the condition h(x) = 0 in B ⊂ R3 \D implies h(x) = 0 in R3 \D. Takethe Fourier transform of (9.13.8) in x ∈ R3 to get

cv(λ)|λ|−1 = h(λ), c =3

2π3/2. (9.13.9)

Since v(z) and h(x) are compactly supported (they vanish outside D), it follows that v and h are entirefunctions of λ ∈ C3. Thus (9.13.9) yields a contradiction since |λ|−1 is not a meromorphic function whileh(λ)/v(λ) is. This contradiction proves the Remark. 2

Exercise 9.13.4. Does there exist a non-trivial solution to (9.13.1) in the case when D ⊂ CR := z :z21 + z2

2 ≤ R2, (9.13.1) holds for all x ∈ ∂CR and v = 0 for |z3| > R1, R1 < R?

Proof of Remark 9.13.3. Take any ball B1 ⊂ D let B2 ⊂ B1 be a concentric ball and take v(z) = c2 inB2, v(z) = c1 in B1 \ B2, c1 and c2 are constants, chosen so that c2|B2| + c1(|B1| − |B2|) = 0, where

|B| is the volume of B. Then the function φ(x) :=∫ v(z)dz

|x−z| is a potential with charge density v(z). By

symmetry it is clear that φ(x) = φ(|x|) and φ(x) = const |x|−1 for x 6∈ B1, where const is proportionalto the total charge

∫v(z)dz which is zero by the construction. So φ(x) = 0 in R3 \B but v 6≡ 0. 2

9.14 Scattering in absorptive medium

In this section we consider inverse scattering in an absorptive medium. Let

∇2 + k2 + ika(x)u+ k2v(x)u = −δ(x − y) in R3 (9.14.1)

u satisfies the radiation condition, v(x) is the same as in Section 9.1 and a(x) ∈ L2(D), Ima = 0,a(x) = 0 outside D. We are interested in

IP : given u(x, y, k) ∀x, y ∈ P and ∀k ∈ (0, k0), find v(x) and a(x).The problem with the data given for ∀x ∈ P1, ∀y ∈ P2 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

u = g + ik

∫gaudz + k2

∫

D

gvudz, g :=exp(ik|x− y|

4π|x− y| . (9.14.2)

This equation can be solved uniquely by iterations if k is small enough and

f0(x, y) := 16π2 limk→0

u− g

ik=

∫

D

a(z)dz

|x− z||y − z| . (9.14.3)

The function f0(x, y) 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:

∫

D

v(z)dz

|x− z||y − z| = f(x, y) :=

∫

D

a(z)f0(z, y)dz

4π|x− z| + 16π2 limk→0

u− g − ik∫gagdz

k2. (9.14.4)

Again, f(x, y) is computable from the data and therefore can be considered known, and (9.14.4) canbe 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 normalcomponent of its velocity are known on Γ0, and one wants to find the surface Γ1 on which the normalcomponent 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 Γ1 and Γ2, and D′ is a region bounded by Γ′ := Γ1 ∪ Γ2 or by acomponent of Γ′. Then u solves the homogeneous Dirichlet problem for the Laplace equation, so u = 0in D′. By the unique continuation property, u = 0 outside the region D0 which lies inside Γ0. This is acontradiction since |u0| + |u1| 6≡ 0. Proposition 9.15.1 is proved. 2

Let us give a method for solving IP. First, without loss of generality assume that Γ0 is the unit circle.If not, one first maps D0 conformally onto the unit disk. Then Γ0 is mapped onto the unit circle. So,assume that D0 is the unit disk. Any harmonic function outside D0 can be written as

u(r, φ) = a0 + b0 ln r +

∞∑

n=1

[rn(an cos nφ+ bn sinnφ) + r−n(cn cos nφ+ dn sinnφ)

](9.15.1)

The constants an, bn, cn, dn are uniquely and constructively determined, given u0 and u1, from theequations

a0 = A0, an + cn = An, bn + dn = Bn, n ≥ 1 (9.15.2)

b0 = A′0, n(an − cn) = A′

n, n(bn − dn) = B′n, n ≥ 1 (9.15.3)

where

u0 = A0 +

∞∑

h=1

(An cosnφ+ Bn sinnφ) (9.15.4)

u1 = A′0 +

∞∑

h=1

(A′n cosnφ+ B′

n sinnφ) (9.15.5)

Let r = f(φ) be the equation of Γ1. Then the unknown function f(φ) can be found from the equationu = 0 on Γ1:

u(f(φ), φ) = 0 (9.15.6)

The same idea can be easily used for other boundary conditions on Γ1. For example, if ∂u∂N |Γ1 = 0

then, using the formulas

∇u =∂u

∂rer +

1

r

∂u

∂φeφ, N = er −

1

rf ′(φ)eφ (9.15.7)

where er and eφ are the coordinate unit vectors in polar coordinates, N is the outer normal to Γ1, andu(r, φ) is given by formula (9.15.1) with the coefficients found from (9.15.4) and (9.15.5). The equationfor f(φ) in the case of the boundary condition ∂u

∂N|r=f(φ) = 0 is

∂u

∂r− 1

r2∂u

∂φf ′(φ)|r=f(φ) = 0 (9.15.8)

where u(r, φ) is given by (9.15.1) as explained above.

9.16. AN INVERSE PROBLEM FOR A BIHARMONIC EQUATION 281

Example 9.15.2. Let u0 = 1, u1 = −1. Then An = Bn = A′n = B′

n = 0, n ≥ 1, A0 = 1, A′0 = −1.

One finds an = bn = cn = dn = 0, n ≥ 1, a0 = 1, b0 = 1. Thus u = 1− ln r and (9.15.6) is 1− ln r = 0,r = e. Therefore Γ2 is the circle |x| = e.

Exercise 9.15.3. Does the problem in the above Example have a solution if ∂u∂N

|Γ1 = 0?

Example 9.15.4. Let u0 = cos φ, u1 = cos φ. Then, one finds, u = r cosφ and (9.15.8) is cos φ +f(φ) sin φf2(φ) f ′(φ) = 0, f ′

f = − (sin φ)′

(sin φ) , f = c(sinφ)−1, c = const. Therefore, there is no bounded f(φ),

0 ≤ φ < 2π, which solves the problem.

Example 9.15.5. Let u0 = 1, u1 = 0. Then u(r, φ) = 1 and any closed curve Γ1 solves the problem

∆u = 0 in D, u|Γ0 = 1,∂u

∂r|Γ0 = 0,

∂u

∂N|Γ1 = 0 (9.15.9)

where D is the region bounded by Γ0 ∪ Γ1.

Remark 9.15.6. The last example shows that the inverse problem analogous to IP with the Neumannboundary condition on Γ1 is not uniquely solvable. Example 2 shows that the inverse problem may haveno solutions.

9.16 An inverse problem for a biharmonic equation

Consider the problem

∆2u = 0 in R2− := x, y : y < 0 (9.16.1)

u(x, 0) = u0(x), uy(x, 0) = u1(x) (9.16.2)

∆u(x, 0) = u2(x),∂∆u(x, y)

∂y|y=0 = u3(x) (9.16.3)

The inverse problem, IP, is:

Given the data (9.16.2) - (9.16.3), find a curve Γ such that

u = 0 on Γ. (9.16.4)

One can use other boundary conditions on Γ, such as uN = 0 on Γ, for example, where N is the outernormal to Γ.

The basic idea is to measure the Cauchy data for elliptic equation, to determine from these datathe solution, and to find the zero set of the solution from (9.16.4). Of course, the Cauchy problem forelliptic equations is ill-posed, but its solution may still be of practical use. The problems such as IP canbe of interest in elasticity theory as a way to determine cracks given boundary measurements. Note thatuniqueness of the solution to IP is not claimed. What we outline is a way to find some of the solutionsto IP.

Let us look for a solution to (9.16.1) of the form

u(x, y) =

∫ ∞

−∞dα [A0(α) +A1(α)x+A2(α)y] exp(iαx+ αy)

+ [B0(α) + B1(α)x+ B2(α)y] exp(iαx− αy) ,(9.16.5)


where Aj(α), Bj(α) should be chosen so that the boundary conditions (9.16.2), (9.16.3) are satisfied.This requirement is equivalent to the equations

u0 = A0 + B0 + iA′1 + iB′

1 (9.16.6)

u1 = A2 + B2 + αA0 − αB0 + i(αA1)′ − i(αB1)

′ (9.16.7)

u2 = 2iα(A1 + B1) + 2α(A2 − B2) (9.16.8)

u3 = 2iα2A1 + 2α2A2 − 2iα2B1 + 2α2B2 (9.16.9)

Here A′ = dAdα , u := (2π)−1

∫∞−∞ exp(−iαx)u(x)dx, and the Fourier transform is taken in the sense of

distributions. We have six functions Aj, Bj , 0 ≤ j ≤ 2, and four equations (9.16.6) - (9.16.9). Twoadditional equations can be imposed. Let us choose these additional conditions as:

A1 = 0, B1 = 0 (9.16.10)

Then (9.16.6) - (9.16.9) can be solved uniquely and explicitly

A0 = [u0 + α−1u1 − (2α2)−1u3]/2, B0 = [u0 − α−1u1 + (2α2)−1u3]/2

A2 = [(2α)−1u2 + (2α2)−1u3]/2, B2 = [(2α2)−1u3 − (2α)−1u2]/2(9.16.11)

Formulas ((9.16.5) , (9.16.10) and (9.16.11) solve problem (9.16.1) - (9.16.3). The solution is smoothif the data are such that the integral (9.16.5) with Aj and Bj given by (9.16.10) and (9.16.11) convergesrapidly.

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),

u(x, y(x)) = 0 (9.16.12)

where y = y(x) is the equation of Γ.

Remark 9.16.1. The IP may have no solutions.Example:

Let A0 = 1 if −1 ≤ α ≤ 1 and zero otherwise, A1 = B1 = A2 = B2 = 0, then u(x, y) =∫ 1

−1exp(iαx+

αy)dα = (exp(ix+ y) − exp(−ix − y))/(ix+ y). Equation (9.16.4) is exp[2(ix+ y)] = 1, or 2(ix+ y) =2inπ, y = 0, x = nπ, n = 0,±1, . . . . Thus, no closed curve in R2 solves IP.

9.17 Inverse scattering when the background is variable

Let[∇2 + k2n0(x) + k2v(x)]u = −δ(x − y) in R3 (9.17.1)

then, with [∇2 + k2n0(x)]G = −δ(x− y) in R3, one has:

u(x, y, k) = G(x, y, k) + k2

∫

D

G(x, z, k)v(z)u(z, y, k)dz. (9.17.2)

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 n0(x) one can prove that G(x, y, k) has a limit as k → 0

G(x, y, k) → g(x, y) as k → 0, x 6= y, g = (4π|x− y|)−1. (9.17.3)

If (9.17.3) holds, then one derives from (9.17.2) that∫

D

v(z)dz

|x− z||y − z| = f(x, y) (9.17.4)

9.17. INVERSE SCATTERING WHEN THE BACKGROUND IS VARIABLE 283

where

f(x, y) = 16π2 limk→0

k−2[u(x, y, k)−G(x, y, k)]. (9.17.5)

Therefore, the basic equation (9.17.4) is the same as (9.1.6). The crucial difference is in the low-frequencydata f(x, y): the data (9.17.5) differ from the function on the left-hand side of (9.1.5). Indeed, formally,

16π2 limk→0

k−2[u− g(x, y, k)] =

∫

D

[n0(z) − 1 + v(z)]

|x− z||y − z| dz, g(x, y, k) :=exp(ik|x− y|)

4π|x− y| . (9.17.6)

The practical conclusion is: in order to use the knowledge of the Variable background n0(x), one shouldcalculate accurately the Green function G(x, y, k) in (9.17.1) and use G(x, y, k) in (9.17.5) for calculatingf(x, y) from the data u(x, y, k), x, y ∈ P . Although (9.17.3) holds, one cannot use g(x, y, k) in place ofG(x, y, k) in (9.17.5).

A simple sufficient condition for (9.17.3) to hold is: n0(z) = n1(z)+ν(z), where |ν(z)| ≤ c(1+ |z|)−β,β > 3, c = const > 0, and n1(z) is such that for G1, the Green function corresponding to n1(z), equation(9.17.3) holds. For example, n1 = const has this property, and many other examples can be given.

In some cases G(x, y, k) can be calculated analytically. For example, let

n0(z) =

1 if z3 ≥ 0

1 − 1−c0d z3 if − d ≤ z3 ≤ 0, c0 < 1

c0 if z3 < −d.

(9.17.7)

This refraction coefficient is of interest in the study of layered structures. One can calculate G(x, y, k)for n0(z) defined in (9.17.7) by taking the Fourier transform in x1 and x2 of (9.17.1). Let

G(x, y, k) = (2π)−2

∫∫ ∞

−∞expiλ · (x1 − y1)G(x3, y3, λ, k)dλ, x1 = (x1, x2), λ = (λ1, λ2) (9.17.8)

Then

G′′ − λ2G+ k2n0(x3)G = −δ(x3 − y3), (9.17.9)

where G′ = ∂G∂x3

. If n0(x3) is given by (9.17.7) then G, the solution to (9.17.9), can be calculatedanalytically:

G =h+(x3)h−(y3)

−2ik1, x3 ≥ y3, k1 := (k2c0 − λ2)1/2, (9.17.10)

G for x3 ≤ y3 is defined by symmetry,

h±(x3) = exp(±ik0x3) for x3 > 0; k0 := (k2 − λ2)1/2, (9.17.11)

h±(x3) = A±ξ1/2J1/3

(2β

3ξ3/2

)+B±ξ

1/2J−1/3

(2β

3ξ3/2

)for − d < x3 < 0, (9.17.12)

ξ := k20 + βx3, β := k2(1 − c0)

−1, (9.17.13)

h±(x3) = C± exp(ik1x3) +D± exp(−ik1x3) for x3 < −d. (9.17.14)

The constants A±, B±, C±, and D±, are uniquely determined by the requirement that h±(x3) andh′±(x3) are continuous at x3 = 0 and x3 = −d.


9.18 Remarks concerning the basic equation

We have:

w(x, y) :=

∫

D

v(z)dz

|x− z||y − z| = 0 ∀x ∈ P1, ∀y ∈ L. (9.18.1)

Here P1 ⊂ P is an open set on P := x : x3 = 0 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 y 6∈ D, in particular, in a neighborhood of P .

Therefore, the set of zeros of this function is an analytic set whose intersection with an analytic setx3 = 0 has to be piecewise-analytic. Therefore, if it is not, one concludes that an open set P2 on P ,containing L, has to be the set of zeros of w(x, y). If w(x, y) = 0 for all x ∈ P1 and all y ∈ P2, whereP1 and P2 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 L = L1 ∪ L2, and their angle of intersection isω, where ω

π is an irrational number, then (9.18.1) implies that v(z) = 0. Indeed, the line L3, which isthe reflection of L1 with respect to L2, has to be the line of zeros of a harmonic function if L1 is. Thisfollows from the fact that −u(x1,−x2, x3) is harmonic if u(x1, x2, x3) is and u(x1, 0, x3) = 0. Thus, theset of zeros of the harmonic function w(x, y) of y will be dense on P if L1 ∪L2 belongs to the set of itszeros. Thus w(x, y) = 0 ∀y ∈ P if w(x, y) = 0 for y ∈ L. If w(x, y) = 0 for all x ∈ P1 and all y ∈ P ,then v(z) = 0.

Chapter 10

Wave scattering by small bodies ofarbitrary 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 beenpublished on the subject because of its importance in applications. From a theoretical point of viewthere are two directions of investigation:

(i) to prove that the scattering amplitude can be expanded in powers of ka, where k = 2π/λ and a is acharacteristic 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, Kleinmanand 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 scatteringby small bodies of arbitrary shapes with the emphasis on practical applicability of the formulas obtainedand on the mathematical rigor of the theory. For the scalar wave scattering by a single body, our mainresults 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 smallbody of an arbitrary shape is given. Applications of these results are outlined (calculation of theproperties of a rarefied medium; inverse radio measurement problem; formulas for the polarizationtensors and capacitances).

(3) The multi-particle scattering problem is analyzed and interaction of the scattered waves is taken intoaccount. For the self-consistent field in a medium consisting of many particles (∼ 1023) integral-differential equations are found. The equations depend on the boundary conditions on the particlesurfaces. These equations offer a possibility of solving the inverse problem of finding the medium

285

286 CHAPTER 10. WAVE SCATTERING BY SMALL BODIES OF ARBITRARY SHAPES

properties from the scattering data. For about 5 to 10 bodies the fundamental integral equationsof 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 10.1.3the electromagnetic scattering is studied and the solution of the inverse problem of radio measurementsis outlined. In Section 10.2 the many-body problem is examined. We use results from [R65] and [R215].

10.1.2 Scalar wave scattering by a single body

In this Section we denote by Γ the boundary of the body D, by S the surface area of Γ, by f thescattering amplitude, by n the unit vector in the direction of scattering, and by ν the unit vector in thedirection of the incident plane wave.

Consider the problem

(∇2 + k2

)v = 0 in Ω;

(∂v

∂N− hv

) ∣∣∣∣Γ = −(∂u0

∂N− hu0

)∣∣∣∣Γ

, (10.1.1)

v ∼ exp (ik|x|)|x| f(n, k) as |x| → ∞,

x

|x| = n, (10.1.2)

where Ω = R3\D, D is a bounded domain with a smooth boundary Γ, N is the outer normal to Γ, u0

is the initial field which is usually taken in the form u0 = exp ik(ν, x). We look for a solution of theproblem (10.1.1)-(10.1.2) of the form

v =

∫

Γ

exp (ikrxt)

4πrxtσ(t)dt, rxt = |x− t|, (10.1.3)

and for the scattering amplitude f we have the formula

f =1

4π

∫

Γ

exp−ik(n, t)σ(t, k)dt =1

4π

∫

Γ

σ0(t)dt +O(ka), (10.1.4)

where

σ(t, k) = σ0(t) + ikσ1(t) +(ik)2

2σ2(t) + . . . (10.1.5)

Putting (10.1.3) in the boundary conditions (10.1.1) we get the integral equation for σ:

σ = A(k)σ − hT (k)σ − 2hu0 + 2∂u0

∂N, (10.1.6)

where

A(k)σ =

∫

Γ

∂

∂Ns

exp (ikrst)

2πrstσ(t)dt, T (k)σ =

∫

Γ

exp (ikrst)

2πrstσ(t)dt. (10.1.7)

Expanding σ, 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:

σ0 = A0σ0, (10.1.8)

σ1 = A0σ1 + A1σ0 + 2∂u01

∂N, (10.1.9)

σ2 = A0σ2 + 2A1σ1 +A2σ0 + 2∂u02

∂N, (10.1.10)

σ3 = . . . (10.1.11)

10.1. WAVE SCATTERING BY SMALL BODIES 287

etc, where

A(k) = A0 + ikA1 +(ik)2

2A2 + . . . ; u0 = u00 + iku01 +

(ik)2

2u02 + . . . . (10.1.12)

Expanding f in formula (10.1.4) we obtain, up to the terms of the second order:

f = 14π

∫Γσ0dt+

ik4π

∫Γσ1dt+

(n,∫Γσ0(t)tdt

)

+ (ik)2

8π

∫Γσ2dt+ 2

(n,∫Γσ1tdt

)+∫Γσ0(n, t)

2dt.

(10.1.13)

From (10.1.9) it follows that σ0 = 0 and from (10.1.10) it follows that∫Γσ1dt = 0. Some calculations

lead to the following final result (see [R65]):

f =ikV

4πβpqnp

∂u0

∂xq

∣∣∣∣x=0 +V

4π∆u0

∣∣∣∣x=0

. (10.1.14)

Usually u0 = expik(ν, x), and in this case formula (10.1.15) can be written as:

f = −k2V

4π(βpqνqnp + 1) , (10.1.15)

where over the repeated indices the summation is understood, V is the volume of the body D and βpq isthe magnetic polarizability tensor of D. Note that f ∼ k2a3, the scattering is anisotropic and is definedby the tensor βpq . Formula (10.1.27) below allows one to calculate this tensor.

For h = ∞ (the Dirichet boundary condition) integral equation (10.1.7) takes the form:

T (k)σ = −2u0. (10.1.16)

Hence ∫

Γ

σ0dt

4πrst= −u0

∣∣∣∣Γ

.

Since ka << 1 the field u0|Γ = u0(x, k)|x=0, where the origin is assumed to be inside the body D. Fromthe above equation it follows that

∫Γσ0dt = −Cu0, and

f = −Cu0

4π, (10.1.17)

where C is the electrical capacitance of a conductor with the shape D. Hence, for the Dirichet boundarycondition, f ∼ a, where a is a characteristic radius of D, and the scattering is isotropic.

For h 6= 0, using the same line of arguments, it is possible to obtain the following approximateformula for the scattering amplitude:

f ≈ − hS

4π (1 + hSC−1)u00, (10.1.18)

where S = meas (Γ), i.e., the surface area of Γ, and C is the electrical capacitance of the conductorD. If h is very small (h ∼ k2a3), then the formula for f should be changed and the terms analogous to(10.1.15) should be taken into account.

10.1.3 Electromagnetic wave scattering by a single body

If a homogeneous bodyD with parameters ε, µ, σ, is placed into a homogeneous medium with parametersε0, µ0, σ0, then the following formula for the scattering matrix S was established by the author (see[R65]):

S =k2V

4π

[µ0β11 + α22 cos θ − α32 sin θ, α21 cos θ − α31 sin θ − µ0β12

α12 − µ0β21 cos θ + µ0β31 sin θ, α11 + µ0β22 cos θ − µ0β32 sin θ

]

,

(10.1.19)


where S is defined by the formula

(f2f1

)= S

(E2

E1

)=

(S2S3

S4S1

)(E2

E1

)

,

(10.1.20)

θ is the angle of scattering, E1, E2 are the components of the initial field, f1, f2 are the componentsof the scattered field in the far-field region multiplied by |x|−1 exp(ik|x|), the plane YOZ is the planeof scattering, αij = αij(γ), is the polarizability tensor, γ = (ε − ε0)/(ε + ε0), and βij = αij(−1) is themagnetic polarizability tensor. Analytic formulas for the polarizability tensors αij(γ) are given below.

If one knows S one can find all the values of interest to the physicists for electromagnetic wavepropagation in a rarefied medium consisting of small bodies. The tensor of refraction coefficient can becalculated by the formula nij = δij + 2πNk−2Sij(0), where N is the number of bodies per unit volume,and Sij(0) are the elements of the S-matrix corresponding to the forward scattering, that is for θ = 0(see e.g. [N]). The tensor αij(γ) can be calculated analytically by the formula (see [R65]):

∣∣∣αij(γ) − α(n)ij (γ)

∣∣∣ ≤ Aqn, 0 < q < 1, γ :=ε − ε0ε + ε0

, (10.1.21)

where A, and q are some constants depending only on the geometry of the surface, and

α(n)ij (γ) :=

2

V

n∑

m=0

(−1)m

(2π)mγn+2 − γm+1

γ − 1b(m)ij , n ≥ 1. (10.1.22)

In (10.1.22)

b(0)ij = V δij, b

(1)ij =

∫

Γ

∫

Γ

Ni(s)Nj (t)dsdt

rst, (10.1.23)

b(m)ij =

∫Γ

∫ΓdsdtNi(s)Nj(t)

∫

Γ

. . .

∫

Γ︸︷︷︸m−1

1rstψ (t1, t) . . .ψ (tm−1, tm−2) dt1 . . . dtm−1,

ψ(t, s) ≡ ∂∂Nt

1rst.

(10.1.24)

In particular

α(1)ij (γ) = 2

(γ + γ2

)δij −

γ2b(1)ij

πV, β

(1)ij = −

b(1)ij

πV. (10.1.25)

For the particles with µ = µ0 and ε not very large, so that the depth δ of the skin layer is considerablylarger than a, one can neglect the magnetic dipole radiation and in formula (10.1.19) for the scatteringmatrix one can omit the terms with the multipliers βij .

The vectors of the electric P and magnetic M polarizations can be found by the following formulas,respectively,

Pi = αij(γ)V ε0Ej, γ :=ε− ε0ε+ ε0

, (10.1.26)

where Ej is the initial field, over the repeated indices one sums up, and

Mi = αij(γ)V µ0Hj + βijV µ0Hj, γ :=µ − µ0

µ + µ0, βij := αij(−1), (10.1.27)

where Hj is the initial field, and the second term on the right hand side of equality (10.1.27) should beomitted if the skin-layer depth δ >> a.

The scattering amplitudes can be found from the formulas:

fE =k2

4πε0[n, [P, n]]+

k2

4π

√µ0

ε0[M,n], (10.1.28)

10.1. WAVE SCATTERING BY SMALL BODIES 289

fH =

√ε0µ0

[n, fE], (10.1.29)

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 δ >> a one can neglect the second term on the right-handside of (10.1.28).

It is possible to give a simple solution to the following inverse problem which can be called the inverseproblem of radiomeasurements.

Suppose an initial electromagnetic field is scattered by a small probe. Assume that the scatteredfield E′, H ′ can be measured in the far field region. The problem is: calculate the initial field at thepoint where the small probe detects E′, H ′. This problem is of interest, for example, when one wants todetermine the electromagnetic field distribution in an antenna’s aperture. Let us assume for simplicitythat for the probe δ >> a, so that

E′ =exp(ikr)

r

k2

4πε0[n[P, n]] . (10.1.30)

From (10.1.30) one can find P − n1(P, n1) = E′(n1)b where b = exp(ikr)r

k2

4πε0. A measurement in

the n2 direction, where (n1, n2) = 0, results in P − n2(P, n2) = E′(n2)b. Hence (n1, P ) = b(E′(n2), n1).Thus P = bE′(n1) + n1(E

′(n2), n1). But (cf (10.1.26)):

Pi = αij(γ)V ε0Ej, 1 ≤ i ≤ 3. (10.1.31)

Since V and ε0 are known and αij(γ) can be calculated by formulas (10.1.21), (10.1.22) and the matrixαij is positive definite (because 1

2αijV ε0EjEi is the energy) it follows that system (10.1.31) is uniquely

solvable for Ej. 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 tobe very useful in practice ([R65]):

C(n) = 4πε0S2

(−1)n

(2π)n

∫

Γ

∫

Γ

dsdt

rst

∫

Γ

. . .

∫

Γ︸︷︷︸n times

ψ(t, t1) . . . ψ(tn−1, tn)dt1 . . .dtn

−1

,

(10.1.32)

C(0) =4πε0S

2

J≤ C, J ≡

∫

Γ

∫

Γ

dsdt

rst, S = measΓ. (10.1.33)

It can be proved ([R65]) that

∣∣∣C − C(n)∣∣∣ ≤ Aqn, 0 < q < 1, (10.1.34)

where A and q are constants which depend only on the geometry of Γ.

It is shown in [R65] that formula (10.1.32) with n = 0 gives a relative error not more than 0.03, andwith n = 1 not more than 0.01, for a wide variety of shapes. Therefore this formula can be used fordeveloping a universal computer code for calculation of electrical capacitances for conductors of arbitraryshapes. 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 αij 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 derivean integral-differential equation for the self-consistent field in a medium consisting of many (r ∼ 1023)small bodies. We look for a solution of the scalar wave scattering problem of the form

u = u0 +

r∑

j=1

∫

Γj

exp(ikrst)

4πrxtσj(t)dt. (10.1.35)

Applying the boundary condition,

u

∣∣∣∣Γj

= 0, 1 ≤ j ≤ r, (10.1.36)

we obtain the system of r integral equations for the r unknown functions σj. In general this systemcan be solved numerically. When d << λ, where d = mini6=jdij, and dij is the distance between i − thand j − th body, the system of the integral equations has dominant diagonal terms and it can be easilysolved by an iterative process, the zero approximation being the initial field u0.

If ka >> 1, d >> a, but not necessarily d >> λ, the average (self-consistent) field in the mediumconsisting of small particles can be found from the integral equation ([R144]):

u(x, k) = u0(x) −∫

R3

exp(ikrxy)

4πrxyq(y)u(y, k)dy. (10.1.37)

Here q(y) is the average value of hjSj(1 + hjSjC−1j )−1 over the volume dy in a neighborhood of y

for bodies with impedance boundary conditions. For hj = ∞ (the Dirichlet boundary condition) andidentical bodies, one has q(y) = N (y)C, where N (y) is the number of the bodies per unit volume and Cis the capacitance of a body. For the Neumann boundary condition the corresponding equation is theintegral-differential equation (cf (10.1.14)):

u(x, k) = u0(x, k) +

∫

R3

exp(ikrxy)

4πrxy

Bpq(y)

∂u(y, k)

∂yq

xp − yprxy

+ b(y)∆u(y, k)

dy, (10.1.38)

whereb(y) = N (y)V, Bpq(y) = ikV βpqN (y), (10.1.39)

V is the volume of a body, and βpq is its magnetic polarizability tensor (see formula (10.1.27)). Thesolution to equations (10.1.35) and (10.1.36) can be considered as the self-consistent (effective) fieldacting in the medium.

Equations (10.1.35) and (10.1.36) allow one to solve the inverse problems of the determination of themedium properties from the scattering data. For example, from (10.1.35) it follows that the scatteringamplitude has the form

f = − 1

4π

∫exp−ik(n, y)q(y)u(y, k)dy. (10.1.40)

For a rarefied medium it is reasonable to replace u by u0 (the Born approximation) and to obtain

f ≈ − 1

4π

∫

R3

exp−ik(n, y)q(y)u0(y, k)dy. (10.1.41)

If u0 = expik(ν, x) formula (10.1.39) is valid for k >> 1 with the error O(k−1) if

|q(x)| + |∇q(x)| ≤ c

1 + |x|3+ε, ε > 0. (10.1.42)

Hence if f is known for all 0 < k < ∞, and all (n − ν) ∈ S2, where S2 is the unit sphere in R3, theFourier transform of q(y) is known, and q can be uniquely determined. If q(y) is compactly supported,

10.2. EQUATIONS FOR THE SELF-CONSISTENT FIELD. 291

i.e. q(y) = 0 outside some bounded domain, then f is an entire function of k, and knowing f in anyinterval [k0, k1], 0 < k0 < k1, for all (n− ν) ∈ S2 one can find f for all 0 < k < ∞ uniquely by analyticcontinuation, 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 Dirichletboundary condition holds. Let us look for a solution of the form

u(x) = u0 +

r∑

j=1

∫

Γj

exp(ik|x− t|)4π|x− t| σj(t)dt. (10.1.43)

The scattering amplitude is equal to

f(n, k) =1

4π

r∑

j=1

exp −ik(n, tj)∫

Γj

exp −k(n, t− tj)σj(t)dt, (10.1.44)

where tj is some point inside the j-th body. Since ka << 1 this formula can be rewritten as:

f(n, k) =1

4π

r∑

j=1

exp −ik(n, tj)Qj , (10.1.45)

where

Qj =

∫

Γj

σj0dt+O(ka), σj0 = σj

∣∣∣∣k=0

. (10.1.46)

This is the same line of arguments as in Section 10.1.2. Using the Dirichlet boundary condition one gets:

r∑

j 6=m,j=1

∫

Γj

exp(ik|xm − t|)4π|xm − t| σj(t)dt+

∫

Γm

exp(ik|xm − t|)4π|xm − t| σmdt = −u0(xm). (10.1.47)

With the accuracy of O(ka), this can be written as:∫

Γm

σm(t)dt

4π|xm − t| +

r∑

j=1,j 6=m

exp(ikdmj)

4πdmjQj = −u0m, 1 ≤ m ≤ r, (10.1.48)

where dmj = |xm − tj|. If Cm is the capacitance of the m-th body one can rewrite (10.1.48) as:

Qm = −Cmu0m −r∑

j=1,j 6=mCm

exp(ikdmj)

4πdmjQj, 1 ≤ m ≤ r. (10.1.49)

This is a linear system for Qj, 1 ≤ m ≤ r. If dmjC−1m >> 1, this system can be easily solved by an

iterative process. If Qj 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 mediaconsisting of many small bodies randomly distributed in some region. Acoustic and electromagneticfields are considered in such a medium. Each body has a characteristic dimension a λ, where λ is thewavelength in the free space.

The minimal distance d between any of the two bodies satisfies the condition d a, but it may alsosatisfy the condition d λ in acoustic scattering. In electromagnetic scattering our assumptions area λ and λ d.

Let us derive an integral-differential equation for the self-consistent acoustic or electromagnetic fieldsin the above media.


10.2.1 Introduction

We propose a general method for derivation of equations for the self-consistent (effective) field in amedium consisting of many small particles and illustrate it by the derivation of such equations foracoustic and electromagnetic waves.

Equation (10.2.17) (see below) is of the type obtained in [R65], and earlier in [MK] by a differentargument. 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, acousticallysoft body is isotropic, and the scattered field in the far-field zone is described by one scalar, namely bythe electrical capacitance of the perfect conductor of the same shape as the small body, and the scatteredfield is of order O(a), where a is the characteristic size of the small body. If a small body is acousticallyhard, that is, condition (10.2.21) holds on its boundary, then the scattering is anisotropic, the scatteredfield in the far-field zone is described by a tensor, and it is of order k2a3. Therefore, this field is muchsmaller (by a factor of order O((ka)2)) than the one for an acoustically soft body of the same size a andof 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 uniformdielectric to the density of the distribution of molecules and the polarizability of these molecules. Inthis 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 thepolarizability tensor is proportional to the unit matrix, and the coefficient of proportionality is the cubeof 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, andthe number of the molecules per unit volume. The derivation of this formula in [BoW] is based on theequation of electrostatics.

Our basic physical assumptions, (10.2.1), allow for rarefied medium, when d >> λ, but also, inacoustic wave scattering theory, for medium which is dense, when a << d << λ. Equations (10.2.25)and (10.2.33)-(10.2.36), that we will derive, have an unusual feature: the integrand depends on thedirection from y to x. This happens because of the anisotropy of the scattered field in the case ofnon-spherical homogeneous small bodies.

A possible application of equation (10.2.17) is a method for finding the density of the distribution ofsmall bodies from the scattering data. Namely, the function C(y) in (10.2.17)-(10.2.18) determines thisdistribution. On the other hand, this function can be determined from the field scattered by the regionR. The uniqueness results and computational methods for solving this inverse scattering problem aredeveloped in [R139].

Below we study the dynamic fields, so that the wavenumber is positive.Consider a random medium consisting of many small bodies Dj , 1 ≤ j ≤ J , J ∼ 1023, located in a

region R. Let aj be the radius of the body Dj, defined as aj := 12

maxx,y∈Dj

|x− y|, and a := max1≤j≤J

aj . We

assume a λ, where λ is the wavelength of the field in the free space (or in a homogeneous space inwhich the small bodies are embedded). Let d := min

x∈Dj ,y∈Dj ,1≤i,j≤J,i6=j|x− y|. Assume

a λ, a d. (10.2.1)

We do not assume that d λ in the acoustic wave scattering, but assume this in electromagnetic wavescattering. 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 p eikr

r in theelectromagnetic theory, where p is a vector independent of r and r is the distance from a small body tothe observation point.

We consider acoustic field in the above medium, and derive in Section 10.2.2 an integral-differentialequation for the self-consistent field in this medium. The notion of the self-consistent field is defined inSection 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 themedium.

Each small body may have an arbitrary shape. The key results from [R65], that we use, are theformulas for the S-matrix for acoustic and electromagnetic wave scattering by a single homogeneoussmall body of an arbitrary shape. These formulas are given in Section 10.2.2 and Section 10.2.3. Wavepropagation 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 u |Sj= 0, where Sj is the boundary ofDj . The governing equation for the acoustic pressure u is

(∇2 + k2)u = 0 in D′ := R3\D, D := UJj=1Dj , (10.2.2)

u = 0 on S := UJj=1Sj , (10.2.3)

u = u0 + v, u0 := eikν·x, ν ∈ S2, (10.2.4)

r

(∂v

∂r− ikv

)→ 0, r := |x|, (10.2.5)

where the direction ν ∈ S2 of the incident field is given, k = 2πλ = const > 0 is the wave number, S2 is

the unit sphere in R3.Let us look for v of the form:

v(x) =

J∑

j=1

∫

Sj

g(x, s)σj(s)ds, g(x, y) :=eik|x−y|

4π|x− y| . (10.2.6)

If (10.2.1) holds, and min1≤i≤J

|x− sj | a, then (10.2.6) can be written as

v(x) =J∑

j=1

g(x, sj)Qj [1 + O(ka)], Qj :=

∫

Sj

σjds, (10.2.7)

Define the self-consistent field ue at the point sm ∈ Sm by the formula

ue(sm) = u0(sm) +

J∑

j=1,j 6=mg(sm , sj)Qj , (10.2.8)

and at any point x, such thatinfj|x− sj | a, (10.2.9)

by the formula:

ue(x) = u0(x) +

J∑

j=1

g(x, sj)Qj. (10.2.10)

If (10.2.9) holds, then ∣∣∣∣J∑

j=1

g(x, sj)Qj −∑

j=1,j 6=mg(x, sj)Qj

∣∣∣∣ |ue(x)|,

that is, removal of one small body does not change the self-consistent field in the region which containsno immediate neighborhood of this body. On the surface S of the body the total field u = 0, so the


self-consistent field on Sm differs from u, while at a point x such that (10.2.9) holds, u(x) = ue(x)+o(1),if (10.2.1) holds.

Let us derive a formula for Qj. By (10.2.3) one gets:

0 = ue(sm) +

∫

Sm

σm(s)ds

4π|sm − s| + εm, (10.2.11)

where

εm :=

∫

Sm

[g(sm, s) − g0(sm, s)]σmds, g0 :=1

4π|sm − s| , (10.2.12)

so thatmaxm

|εm| = o(1) as ka→ 0. (10.2.13)

Thus, one may neglect εm in (10.2.11) and consider the resulting equation∫

Sm

σmds

4π|sm − s| = −ue(sm) (10.2.14)

as an equation for the charge distribution σm on the surface Sm of a perfect conductor charged to thepotential −ue(sm). Then, by (10.2.7),

Qm :=

∫

Sm

σmds = −Cmue(sm), (10.2.15)

where Cm is the electrostatic capacitance of the conductor Dm. From (10.2.10) and (10.2.15) one gets

ue(x) = u0(x) −J∑

j=1

g(x, sj)Cjue(sj) + o(1), (10.2.16)

as ka→ 0.Let us emphasize the physical assumptions we have used in the derivation of (10.2.16). First, the

assumption ka << 1 allows one to claim that, uniformly with respect to all small bodies, the term o(1)in (10.2.16) tends to zero as ka → 0. Secondly, the assumption d >> a allows one to claim that them−th small body is in the far zone with respect to the j−th body for any j 6= m. The expressionunder the sum in (10.2.16) is the field, scattered by j−th body and calculated at the point x, suchthat |x − sj | >> a, that is, in the far zone from the j−th body. So, physically, the equations for theself-consistent field in the medium, derived in this Section, are valid not only for the rarified medium(that is, when a << λ and d >> λ, but also for not too dense medium, that is, when d >> a anda << λ, but, possibly, d << λ.

The limiting equation for ue(x) is:

ue(x) = u0(x) −∫

R

g(x, y)C(y)ue(y)dy, (10.2.17)

whereC(y)dy =

∑

sj∈dyCj, (10.2.18)

and the summation is taken over all small bodies located in the volume dy around point y. If one assumesthat the capacitances Cj are the same for all these bodies around point y, and are equal to c(y), thenC(y) = c(y)N (y), where N (y) is the number of small bodies in the volume dy.

Equation (10.2.17) is the integral equation for the self-consistent field in the medium in the regionR. This field satisfies the Schrodinger equation:

[∇2 + k2 − q(x)]ue = 0, q(x) := C(x). (10.2.19)


Since Cj ∼ a in (10.2.18), and the number N of the terms in the sum (10.2.18) is N = O( 1d3 ), provided

that dy is a unit cube, one concludes that ad3

= O(1), so

d = O(a

13

). (10.2.20)

If one had Na3 = O(1), i.e., small bodies have nonzero limit of volume density, then the assumptiond a would be violated.

Let us now assume that the small bodies are acoustically hard, i.e., the Neumann boundary condition

∂u

∂N= 0 on S, (10.2.21)

replaces (10.2.3), N is the outer normal to S. In this case the derivation of the equation for ue(x) ismore complicated, because the formula for Qj is less simple. It is proved in [R65] that, for the boundarycondition (10.2.21), one gets

Qj = ikVjβ(j)pq np

∂ue∂xq

+ Vj∆ue. (10.2.22)

Here and below, one sums up over the repeated indices, Vj is the volume of Dj , ∆ue is the Laplacean,

np :=xp−yp

|x−y| , the small body Dj is located around point y, the scattered field is calculated at point x,

β(j)pq is the magnetic polarizability tensor of Dj which is defined by the formula [R65]

βpq := αpq(−1), (10.2.23)

and αpq(γ), γ := ε−ε0ε+ε0

, is the electric polarizability tensor, defined by the formula:

Pp = αpq(γ)Vqε0Eq. (10.2.24)

Here P is the dipole moment induced on the dielectric body Dj , with the dielectric constant ε, placedin the electrostatic field E in the homogeneous medium with the dielectric constant ε0.

Analytical formulas for calculation of αpq(γ) with an arbitrary accuracy, in terms of the geometry ofSj , are derived in [R65] and given in Section 10.1.

From (10.2.10) and (10.2.22) one gets

ue(x) = u0(x) +

∫

R

g(x, y)

[ikβpq(y)

xp − yp|x− y|

∂ue(y)

∂yq+ ∆ue(y)

]V (y)dy, (10.2.25)

where V (y) and βpq(y) are defined by the formulas

V (y)dy =∑

sj∈dyVj, V (y)βpq (y)dy =

∑

sj∈dyVjβpq,j , (10.2.26)

where βpq,j is the magnetic polarizability tensor of the j−th small body, and the summation is overall small bodies in the volume dy around point y, so that V (y) is the density of the distribution of thevolumes 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 directionx− y. This one can understand, if one knows that the acoustic wave scattering by a small soft body Dj

is isotropic and depends on one scalar, the electrostatic capacitance of the conductor Dj , while acoustic

wave scattering by a small hard obstacle is unisotropic and depends on the tensor β(j)pq .

Finally, if the third boundary condition holds:

∂u

∂N+ hu = 0 on S (10.2.27)


then (see [R65]), if h is not too small, one has:

Qj = − h|Sj|1 + h|Sj|C−1

j

ue, (10.2.28)

where |Sj | is the area of Sj , so that

ue(x) = u0(x) −∫

R

g(x, y)b(y)ue(y)dy, (10.2.29)

where

b(y)dy =∑

sj∈dy

h|Sj|1 + h|Sj|C−1

. (10.2.30)

10.2.3 Electromagnetic waves in random media

In this section our basic assumptions are:

a << λ d >> λ. (10.2.31)

The reason for the change in the assumption compared with (10.2.2), where d is not necessarily greaterthan λ, is the following: formula (10.2.35) below is valid if a << λ and d >> λ, where d is the distancefrom a small body to the point of observation. This comes from applying twice the operation of ∇× tothe vector potential. In acoustic scattering the formula for the scattered field (see e.g. (10.2.8)) is validif d >> a, while formula (10.2.35) (see below) is valid for d >> λ and a << λ.

Let U :=(EH

). Denote by S the 6x6 matrix which sends U into gUsc, where gUsc is the scattered

field, g := eikr

r , r is the distance from a small body located at a point y to the observation point x,r = |x− y|.

If S is known, then the equation for the self-consistent field Ue in the random medium situated ina region R, and consisting of many small bodies, satisfying conditions (10.2.32), is

Ue = U0 +

∫

R

g(x, y)S

(y,x− y

r

)Ue(y)dy, (10.2.32)

where

S

(y,x− y

r

)dy =

∑

sj∈dyS (j)

(sj ,

x− sj|x− sj |

)(10.2.33)

as follows from the argument given for the derivation of (10.2.25).Let us give a formula for S (j), assuming without loss of generality, that the origin is situated inside

a single body Dj , which has parameters εj, µj, and σj, (dielectric permittivity, magnetic permeability,and conductivity, respectively), and drop index j in the formula for S (j).

In [R65] one can find the formulas

Esc(ν′) =

k2

4π

(1

ε0[ν ′, [P, ν ′]] +

√µ0

ε0[M,ν ′]

), (10.2.34)

Hsc(ν′) =

√ε0µ0

[ν ′, Esc(ν′)] =

k2

4π

(1√ε0µ0

[ν ′, P ] + [ν ′, [M,ν ′]]

), (10.2.35)

where ν ′ is the unit vector in the direction of the scattered wave.Formulas (10.2.34), (10.2.35) can be written as

Usc = S U =k2V

4π

(α− ν ′(ν ′, α·), −(µ3

0ε−10 )

12 [ν ′, β·]

−(ε30µ0)− 1

2 [ν ′, α·], β − ν ′(ν ′, β·)

)(E

H

)(10.2.36)

10.3. FINDING SMALL SUBSURFACE INHOMOGENEITIES FROM SCATTERING DATA 297

where the matrix S is expressed in terms of the tensors α := αpq and β := βpq , because P is calculatedin (10.2.24),

Mi = αij(γ)V µ0Hj + βijV µ0Hj := βijV µ0Hj, γ :=µ − µ0

µ + µ0, βij := αij(γ) + βij . (10.2.37)

βij is defined in (10.2.23), and (ν, α·)E := (ν, αE), [ν, α·] := [ν, αE].

10.3 Finding small subsurface inhomogeneities from scattering

data

A new method for finding small inhomogeneities from surface scattering data is proposed and mathe-matically justified in this Section. The method allows one to find small holes and cracks in metallic andother objects from the observation of the acoustic field scattered by the objects.

10.3.1 Introduction

In many applications one is interested in finding small inhomogeneities in a medium from the observationof 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 materialscience and technology. Suppose that a piece of metal or other material is given and one wants toexamine if it has small cavities (holes or cracks) inside. One irradiates the metal by acoustic waves andobserves 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 theprobability for a woman to get a new cancer cell in her breast as a result of an x-ray mammographytest is rather high. Therefore it is quite important to introduce ultrasound mammography tests. Thisis being done currently. A new cancer cells can be considered as small inhomogeneities in the healthybreast tissue. The problem is to localize them from the observation on the surface of the breast of thescattered acoustic field.

The purpose of this Section is to describe a new idea of solving the problem of finding inhomo-geneities, small in comparison with the wavelength, from the observation of the scattered acoustic orelectromagnetic waves on the surface of the medium.

For simplicity we present the basic ideas in the case of acoustic wave scattering. These ideas arebased 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 findingsmall inhomogeneities from surface scattering data are:

1) to develop a computationally simple and stable method for a partial solution of the above inversescattering problem. The exact inversion procedures (see Chapter 5, [R139] and references therein) arecomputationally difficult and unstable. In practice it is often quite important, and sometimes sufficientfor 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 inhomogeneitiesinside the body? If yes, where are these inhomogeneities located? What are their intensities? We definethe notion of intensity vm 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 justificationis presented. Some theoretical and numerical results based on a version of the proposed approach onecan find in [R193], [GR1].


10.3.2 Basic equations

Let the governing equation be

[∇2 + k2 + k2v(x)]u = −δ(x− y) in R3, (10.3.1)

where u satisfies the radiation condition, k = const > 0, and v(x) is the inhomogeneity in the velocityprofile.

Assume that supx∈R3 |v(x)| ≤ c0, supp v = UMm=1Bm(zm, ρm) ⊂ R3− = x | x(3) < 0, where x(3)

denotes the third component of vector x in Cartesian coordinates, Bm(zm, ρm) is a ball, centered at zmwith radius ρm, kρm 1.

Denote

vm :=

∫

Bm

v(x)dx.

Problem 4 (Inverse Problem (IP):). Given u(x, y, k) for all x, y ∈ P , P = x | x(3) = 0 and afixed k > 0, find zm, vm, 1 ≤ m ≤M .

In this Section we propose a numerical method for solving the (IP).To describe this method let us introduce the following notations:

P := x | x(3) = 0 (10.3.2)

xj, yj := ξj , 1 ≤ j ≤ J, xj, yj ∈ P

are the points at which the data u(xj, yj, k) are collected(10.3.3)

k > 0 is fixed (10.3.4)

g(x, y, k) :=exp(ik|x− y|)

4π|x− y| (10.3.5)

Gj(z) := G(ξj , z) := g(xj , z, k)g(yj, z, k) (10.3.6)

fj :=u(xj, yj, k) − g(xj , yj, k)

k2(10.3.7)

Φ(z1, . . . , zM , v1, . . . , vM ) :=

J∑

j=1

∣∣∣∣fj −M∑

m=1

Gj(zm)vm

∣∣∣∣2

. (10.3.8)

The proposed method for solving the (IP) consists in finding the global minimizer of function (10.3.8).This minimizer (z1, . . . , zM , v1, . . . , vM) gives the estimates of the positions zm of the small inhomo-geneities and their intensities vm. This is explained in more detail below formula (10.3.14). Numericalrealization of the proposed method, including a numerical procedure for estimating the number M ofsmall inhomogeneities from the surface scattering data is described in [GR1].

Our approach with a suitable modification is valid in the situation when the Born approximationfails, for example, in the case of scattering by delta-type inhomogeneities [AlbS].

In this case the basic condition Mk2c0ρ2 1 (∗), which guarantees the applicability of the Born

approximation, is violated. Here ρ := max1≤m≤M ρm and c0 was defined below formula (10.3.1). Weassume throughout that M is not very large, between 1 and 15.

In the scattering by a delta-type inhomogeneity the assumption is c0ρ3 = const := V as ρ → 0, so

that for any fixed k > 0 one has k2c0ρ2 = k2V ρ−1 → ∞ as ρ → 0, and clearly condition (∗) is violated.

In our notations this delta-type inhomogeneity is of the form k2v(x) = k2∑M

m=1 vmδ(x − zm).

10.3. FINDING SMALL SUBSURFACE INHOMOGENEITIES FROM SCATTERING DATA 299

The scattering theory by the delta-type potentials (see [AlbS]) requires some facts from the theoryof selfadjoint extensions of symmetric operators in Hilbert spaces and in this Section we will not go intodetail (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:

u(x, y, k) = g(x, y, k) + k2M∑

m=1

∫

Bm

g(x, z, k)v(z)u(z, y, k)dz. (10.3.9)

For small inhomogeneities the integral on the right-hand side of (10.3.9) can be approximately writtenas

k2

∫

Bm

g(x, z, k)v(z)u(z, y, k)dz : = k2

∫

Bm

g(x, z, k)v(z)g(z, y, k)dz + ε2

= k2G(x, y, zm)

∫

Bm

vdz + ε2 = k2G(ξ, zm)vm + ε2, 1 ≤ m ≤ M

(10.3.10)where ε2 is defined by the first equation in formula (10.3.10), it is the error due to replacing u under thesign of integral in (10.3.9) by g, and zm is a point close to zm.

One has |u− g| = O(Mk2c0ρ3/d2) if x, y ∈ P, and |u− g| = O(Mk2c0ρ

2/d) if x ∈ D,y ∈ P . Thus,the error term ε2 in (10.3.10) equals to O(M 2k4c20ρ

5/d2) if x, y ∈ P .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 ε2

|g| , where y ∈ P and z ∈ D, provided that:

c20M2k2 ρ

5

d 1, x, y ∈ P, (10.3.11)

where ρ = max1≤m≤M ρm, c0 := maxx∈R3 |v(x)|, M is the number of inhomogeneities, d is the minimaldistance from Bm, m = 1, 2, . . . ,M to the surface P .

A sufficient condition for the validity of the Born approximation, that is, the approximationu(x, y, k) ∼g(x, y, k) for x, y ∈ D, is the smallness of the relative error |u(x,y,k)−g(x,y,k)|

|g(x,y,k)| for x ∈ D, y ∈ P , which

holds if:Mk2c0ρ

2 := δ 1. (10.3.12)

One has:

ε2 = O(M2k4c20ρ

5

d2

)= O

(δ2ρd2

) 1,

if ρ d and if δ is not small, so that the Born approximation may be not applicable. Note that uin (10.3.9) has dimension L−1, where L is the length, v(z) is dimensionless, and ε2 has dimension L−1.In many applications it is natural to assume ρ d.

If the Born approximation is not valid, for example, if c0ρ3 = V 6= 0 as ρ → 0, which is the case

of scattering by delta-type inhomogeneities, then the error term ε2 in formula (10.3.10) can still benegligible: ε2 = O(M 2k4c0V ρ

2/d2), so ε2 1 if M 2k4V ρ2/d2 1.If one understands a sufficient condition for the validity of the Born approximation as the condition

which guarantees the smallness of ε2 for all x, y ∈ R3 then condition (10.3.12) is such a condition.However, if one understands a sufficient condition for the validity of the Born approximation as thecondition which guarantees the smallness of ε2 for x, y running only through the region where thescattered field is measured, in our case when x, y ∈ P , then a much weaker condition (10.3.11) willsuffice.


In the limit ρ → 0 and c0ρ3 = V 6= 0 formula (10.3.10) takes the form (10.3.13), (see [GRa]). It is

shown in [GRa] (see also [AlbS]) that the resolvent kernel of the Schrodinger operator with the delta-typepotential supported on a finite set of points (in our case on the set of points z1, . . . , zM) has the form

u(x, y, k) = g(x, y, k) + k2M∑

m=1

cmm′g(x, zm)g(y, zm′ ) (10.3.13)

where cmm′ are some constants. These constants are determined by a selfadjoint realization of thecorresponding Schrodinger operator with delta-type potential. There is an M 2-parametric family ofsuch realizations (see [GRa] more details).

Although in general the matrix cmm′ is not diagonal, under a practically reasonable assumption (10.3.11)one can neglect the off-diagonal terms of the matrix cmm′ and then formula (10.3.13) reduces practicallyto the form (10.3.10) with the term ε2 neglected.

We have assumed in (10.3.10) that the point zm exists such that

∫

Bm

g(x, z, k)v(z)g(z, y, k)dz = G(x, y, zm)vm.

This is an equation of the type of mean-value theorem. However, such a theorem does not hold, ingeneral, for complex-valued functions. Therefore, if one wishes to have a rigorous derivation, onehas to add to the error term ε2 in (10.3.10) the error which comes from replacing of the integral∫Bm

g(x, z, k)v(z)g(z, y, k)dz in (10.3.10) by the term G(x, y, zm)vm. The error of such an approxi-mation can be easily estimated. We do not give such an estimate, because the basic conclusion that theerror term is negligible compared with the main term k2G(x, y, zm)vm remains valid under our basicassumption kρ 1. From (10.3.10) and (10.3.7) it follows that

fj ≈M∑

m=1

Gj(zm)vm, Gj(zm) := G(ξj, zm, k). (10.3.14)

Therefore, parameters zm and vm can be estimated by the least-squares method if one finds the globalminimum of the function (10.3.8):

Φ(z1, . . . , zM , v1, . . . , vM ) = min . (10.3.15)

Indeed, if one neglects the error of the approximation (10.3.10), then the function (10.3.8) is a smoothfunction of several variables, namely, of z1, z2, . . . , zM , v1, v2, . . . , vM , and the global minimum of thisfunction is zero and is attained at the actual intensities v1, v2, . . . , vM and at the values zi = zi, i =1, 2, . . . ,M .

This follows from the simple argument: if the error of approximation is neglected, then the ap-proximate equality in (10.3.14) becomes an exact one. Therefore fj −

∑Mm=1Gj(zm)vm = 0, so that

function (10.3.8) equals to zero. Since this function is non-negative by definition, it follows that the val-ues zm and vm are global minimizers of the function (10.3.8). Therefore we take the global minimizersof 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 ρm 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 electromagneticscattering problems. An important ingredient of our approach from the numerical point of view is thesolution of the global minimization problem (10.3.14). The theory of global minimization is developedextensively and the literature of this subject is quite large ( see, e.g., [BarJ] and [BarJ1] and referencestherein).

10.4. INVERSE PROBLEM OF RADIOMEASUREMENTS 301

10.4 Inverse problem of radiomeasurements

1. Suppose that we are interested in measuring the electromagnetic field in the aperture of a mirrorantenna. A possible method for making such measurements is as follows. Let us assume that thewavelength range is λ ∼ 3 cm and let us place at some point x0 in the aperture of the antenna a smallprobe of dimension a, ka << 1, k = 2πλ−1. Let E0, H0 denote the electromagnetic field at the point x0

and E′ = gE, H ′ = gH, where g := eikr

r , 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 ka2r−1 << 1,is in fact close to the probe. For example, if λ = 3 cm, a = 0.3 cm then ka2 = 0.19 cm. Therefore ifr = 0.2 cm then ka2r−1 ∼ 0.1 << 1. Let us assume for simplicity that the probe material is such thatthe magnetic dipole radiation from the probe is negligible. In this case the electric field scattered by theprobe in the direction n can be calculated from formula (10.2.34) and assume M = 0. Then

E =k2

4πε0[ν ′, [P, ν ′]], (10.4.1)

where

Pi = αij(γ)ε0V E0j, γ =ε′ − ε0ε′ + ε0

. (10.4.2)

Here V is the volume of the probe, ε0 is its dielectric constant, αij(γ) is the electric polarizability tensor,k is the wave number of the field in the aperture, E0 is the dielectric field at the point x0 where theprobe was placed, and ν ′ is the unit vector. Let n1 and n2 be two noncollinear unit vectors, and Ej,j = 1, 2, be the scattered fields corresponding to nj . We will solve the following

Problem 5. Find E0, H0 from the measured Ej, j = 1, 2

We assume that the tensor αij(γ) is known. In Section 10.1.3 explicit analytical approximate formulasfor αij(γ) are given, see (10.1.21)-(10.1.25). From (10.4.1) it follows that

Ej = bP − nj(P, nj), b =k2

4πε0j = 1, 2. (10.4.3)

ThereforebP = E1 + P − n1(P, n1) = E2 + P − n2(P, n2). (10.4.4)

Let us choose for simplicity n1 perpendicular to n0. Then it follows from (10.4.4) that

b(P, n2) = (E1, n2) (10.4.5)

b(P, n1) = (E2, n1). (10.4.6)

ThereforeP = b−1E1 + b−1n1(E2, n1) = b−1E2 + b−1n2(E1, n2), (10.4.7)

Thus one can find vector P from the knowledge of E1 and E2. If P is known then E0 can be found fromthe linear system

1

2αijε0V E0j = Pi, 1 ≤ i ≤ 3. (10.4.8)

The matrix of this system is positive definite because the tensor αij has this property, as follows fromthe fact that 1

2αijE0V E0jE0i is the energy of the dipole P in the field E0. Therefore the system (10.4.8)can be uniquely solved for E0j, 1 ≤ j ≤ 3. We have proved that the above problem has a unique solutionand gave a simple algorithm for the solution of this problem. The key point in the above argument isthe fact that the matrix αij(γ) 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 |ε′−ε0||ε0| << 1 is


satisfied. We assume that the medium is rarefied, i.e., d >> a, where a is the characteristic dimension ofthe 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 thatφ(r) is density of the distribution of the particles according to their sizes, so that φ(r)dr is the numberof the particles per unit volume with the radius in the interval (r, r+ dr). Then the total scattered fieldin the direction n can be calculated by the formula

F (n, k) =

∫ ∞

0

φ(r)f(n, k, r)dr. (10.4.9)

Let us assume that we can measure F (n, k) for a fixed k and all directions n. Then (10.4.9) can beconsidered as an integral equation of the first kind for the unknown function φ(r).

3. Suppose that we can measure the electric field scattered by a small particle (ka << 1) of anunknown shape. The initial field we denote by E0j, the scattered field by fj . Let us assume that themagnetic 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 thescattered field is the dipole scattering. We have seen in Section 10.4.1 that the knowledge of the scatteredfield allows one to find the dipole moment P and equation (10.4.2) holds. This equation allows one tofind the αij(γ) corresponding to the particle. This tensor is determined if one knows its diagonal form.Let α1, α2, α3 be the eigenvalues of tensor αij(γ). Then an ellipsoid with the semiaxes proportional toαj scatters as the above body. Therefore one can identify the shape of the small scatterer by giving thethree numbers (α1, α2, α3). These numbers are eigenvalues of the tensor αij(γ). They can be calculatedfrom the knowledge of the initial field E0j and the measured scattered field fj . For example, one cantake E0j = δij . Then Pi = αij(γ)V ε0. We assume that the particle is homogeneous and its dielectricconstant ε is known, so that γ in (10.4.2) is known. For an ellipoid the polarizability tensor in thediagonal form is αij = αjδij , where αj = (ε′ − ε0)(ε0 + (ε′ − ε0)n

(j))−1, ε′ is the dielectric constant ofthe ellipsoid and n(j) are the depolarization coefficients. These coefficients are calculated explicitly withthe help of the elliptic integrals and they are tabulated ([LaL]).

Chapter 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 Pompeiuproblem is given. In particular, an author’s result is proved. This result says: if D1 has Pompeiu’sproperty (P -property) then D2 has it, provided D2 is sufficiently close to D1 in the following sense:meas(D12 \D12) is sufficiently small. Here D12 := D1 ∪D2, D

12 := D1 ∩D2.

11.1.1 Introduction

The Pompeiu problem (P -problem) can be stated as follows:

Let f ∈ L1loc(R

n) ∩ S , where S is the Schwartz class of distributions, and

∫

σ(D)

f(x)dx = 0 ∀σ ∈ G, (11.1.1)

where D ⊂ Rn is a bounded domain with Lipschitz boundary and G is the group of all rigid motion ofRn, (G consists of all translations and rotations). Does (11.1.1) imply that f = 0? If yes, one says thatD has Pompeiu’s property, P -property. Otherwise, one says that D has P−property. Equation (11.1.1)can be written as ∫

D

f(y + gx)dx = 0 ∀y ∈ Rn, ∀g ∈ SO(n), (11.1.2)

where SO(n) is the group of rotations, so g is an arbitrary orthogonal matrix with det g = 1. Since 1929the P -problem has been open. Large literature exists on this problem. It was studied in non-euclideanspaces, 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 leastone 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 transformχB(ξ) of the characteristic function of a ball B will satisfy (11.1.1) (see formula (11.1.6) below). ThisFourier transform is (2πa)

n2 |ξ|−n

2 Jn2(a|ξ|). Here Jm(a|ξ|) is the Bessel function of the argument a|ξ|,

where a is the radius of the ball centered at the origin, and ξ is the Fourier transform variable. The setof zeros of Jn

2(t) is a discrete set µj, j = 1, 2, 3, . . ., where the dependence on n is suppressed, and as a

compact subset of zeros of χB(ξ) one can take a sphere |ξ| =µj

a := bj for some fixed positive integer j.

Thus, f (ξ) = A(ξ)δ(|ξ| − bj), where δ(|ξ| − bj) is the delta function supported on the sphere of radius bjand A(ξ) is a certain function. If one takes A(ξ) = 1, then one gets f(x) = cn,j|x|1−n

2 Jn−22

(bj|x|) which

is a solution to (11.1.1). Here cn,j is a positive constant which can be written down explicitly. If one

303

304 CHAPTER 11. THE POMPEIU PROBLEM

takes f (ξ) = δ(ξ)[δ(ξ1 + bj) − δ(ξ1 − bj)], where ξ := (ξ2, . . . , ξn), then f(x) = Cn sin(bjx1) is a solution

to (11.1.1). Here Cn is another constant which could be written explicitly, and ξ := (ξ2, . . . , ξn).

A group of σ smaller than G was considered (see [Z]). In this paper we assume D to be strictlyconvex, homeomorphic to a ball, and its boundary S to be piecewise-smooth. We consider P -problemin Rn. The following results are proved in section 11.1.2 in a self-contained way:

1) a bounded domain D ⊂ Rn, n ≥ 2, does not have P -property iff χD(kα) = 0 ∀α ∈ Sn−1 andsome k > 0; here χD(x) is the characteristic function of D, χ(ξ) is its Fourier transform, ξ = |ξ|α,α ∈ Sn−1, |ξ| = k;

2) a connected bounded domain D ⊂ Rn does not have P -property iff the problem

(∆ + k2)u = −1 in D, u = uN = 0 on S := ∂D, k > 0, (11.1.3)

where N is the unit exterior normal to S, has a solution for some k > 0, or, equivalently, the problem

(∆ + k2)V = 0 in D, VN = 0, V =1

k2on S, k > 0, (11.1.4)

has a solution; here V := u+ 1k2 . Note that the two boundary conditions (11.1.4) imply OV = 0 on S.

2a) If (11.1.3) (or (11.1.4)) holds and S ∈ C1 (or Lipschitz), then k2 is necessarily simultaneously aNeumann and a Dirichlet eigenfunction of the Laplacian, and

2b) S is a real analytic hypersurface.

3) if a bounded, strictly convex domain D with smooth boundary Γ is not a ball, then all the surfacesof zeros of the function χD(ξ) in Rn for sufficiently large |ξ| are not spheres,

or, equivalently,3a) if D ⊂ Rn is a bounded strictly convex domain and χD(tmα) = 0 for all α ∈ Sn−1 and for a

sequence tm → +∞, then D is a ball.

4) assume that Dj ⊂ D , j = 1, 2, where D is a class of smooth strictly convex domains withuniformly bounded C3− norm of the functions representing locally the boundaries of D ⊂ D andGaussian curvatures, uniformly bounded from below by a positive constant.

The author’s main result, Theorem 11.1.6, is:

if a bounded domain D1 has P -property and meas(D12 \ D12) is sufficiently small, then D2 hasP -property.

Here D12 := D1 ∪D2, D12 := D1 ∩D2.

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 X be the space of theFourier transformed elements of X. Let us assume that the only element of X supported on the setkj : χ(kjα) = 0 ∀α ∈ Sn−1 is the zero element. Then any bounded domain D ⊂ Rn, homeomorphicto a ball, has P−property. Therefore we assume in this Section that f(x) ∈ L1

loc.

Examples of functional spaces X with the above property are spaces Lp(Rn) with 1 ≤ p ≤ 2.

Results 1) and 2) can be found in [BST], [W1],[W2], and result 3) is from [Ber1], and result 4) isfrom [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 N , defined below formula (11.1.6), and its rotational invariance, andb) 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:

11.1. THE POMPEIU PROBLEM 305

Theorem 11.1.1. Equation (11.1.1) holds for some f 6≡ 0, iff (11.1.3) is solvable. Moreover:i) (11.1.3) is solvable iff there exists a k > 0 such that χ(ξ) = 0, |ξ| = k, ξ ∈ Rn;ii) if (11.1.3) is solvable then S is an analytic hypersurface;andiii) If (11.1.3) is solvable then k2 > 0 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

F f := f (ξ) :=

∫

Rn

f(x) exp(iξ · x)dx.

It follows from (11.1.2) that

0 =

∫

Rn

dξf(ξ)

∫

D

dx exp(−iξ · gx) exp(−iξ · y) ∀y ∈ Rn, ∀g ∈ SO(n). (11.1.5)

In fact (11.1.2) and (11.1.5) are equivalent, and they are equivalent to

f (ξ)χ(g−1ξ) = 0 ∀g ∈ SO(n), (11.1.6)

where the bar stands for complex conjugate and χ(x) is the indicator of D :

χ(x) =

1 in D

0 in D′, D′ := Rn \D.

Let N :=⋂g Ng, where Ng := ξ : ξ ∈ Rn, χ(gξ) = 0. Thus, N is rotation invariant. Since D is

compact, χ(ξ) is an entire function of exponential type, so Ng is an analytic set. If a point ξ, |ξ| = a > 0,belongs to N , then all the points of the sphere Sa := ξ : |ξ| = a belong to N due to the rotationinvariance of N . It follows from (11.1.6) that iff N is not empty there exists an f 6≡ 0, suppf (ξ) = N ,which satisfies (11.1.1).

It follows that any bounded domain D has P -property if one restricts the class of admissible f(x) informula (11.1.1) to the set of functions f(x) with the following property:

(P): if f (ξ) vanishes on the complement to N then f = 0.The set N is rotationally invariant and can be identified with the discrete set S consisting of those

k > 0 for which N contains the spheres Sk := ξ : ξ ∈ Rn, |ξ| = k. This set is discrete since thefunction χ(ξ) is an entire function of ξ.

For example, if we restrict the set of f(x) in (11.1.1) to be Lp(Rn), p = 1, 2, then the above property(P) holds, and any bounded domain D, including balls, has Pompeiu property.

If f 6≡ 0, then f (ξ) 6≡ 0, and, since N is nonempty and rotation invariant, it either contains a sphereSa, a > 0, or N = 0. The last case cannot occur since χ(0) = measD > 0. Consider the first case. IfN contains Sa, a > 0, then χ(ξ) = 0 for |ξ| = a. This implies χ(ξ) = (ξ2 − a2)u(ξ), where u(ξ) is anentire function. Taking the inverse Fourier transform, one gets equation (11.1.3) for u(x) := F −1u. Thisequation is satisfied in Rn, k2 = a2, suppu = D, so u = 0 in D′. Since u ∈ H2

loc by elliptic regularityand u = 0 in D′, 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 D′ and let

U (x) =

u(x) in D

0 in D′.

Then, because of the boundary conditions (11.1.3), the function U (x) solves the equation

(∆ + k2)U (x) = −χ(x) in Rn,


and its Fourier transform yields (−ξ2 + k2)U = −χ(ξ), where U is an entire function of exponentialtype. Thus, χ(ξ) vanishes on the sphere |ξ|2 = k2, N contains Sk, and there exists an f 6≡ 0, 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 ψ of the homogeneous

equation (11.1.3). Since u0 := exp(ikα · x), ∀α ∈ Sn−1, are such solutions, it follows that

0 =

∫

D

exp(ikα · x)dx := χ(kα) ∀α ∈ Sn−1. (11.1.7)

This yields an independent proof of the implication: existence of a solution to (3) ⇒ χ(ξ) = 0 for all ξwith |ξ| = k.

Claim ii) follows from the results on regularity of free boundary [KN]. Namely, it is proved in [KN]that if S is C1, u ∈ C2, Imu = 0, and (11.1.3) holds, then S is a real-analytic hypersurface. In [W2]this result is proved for Lipschitz S.

Let us prove claim iii). If equation (11.1.3) is solvable, then so is (11.1.4). Thus k2 is a Neumanneigenvalue of the Laplacian in D. Moreover, the solution of (11.1.4) satisfies the condition OV = 0 onS. Therefore, for each j = 1, 2, 3, the function wj := ∂V

∂xjis a solution to the equation

(∇2 + k2)wj = 0, (11.1.8)

which satisfies the Dirichlet boundary condition

wj = 0 on S. (11.1.9)

Claim iii) is proved.Thus, Theorem 11.1.1 is proved. 2

Let α ∈ Sn−1 and denote x+(x−) the point on S at which the tangent to S plane is orthogonal to αand α is directed along inner (outer) normal to S at x+(x−). Denote by

d = d(α) := α · (x− − x+) > 0

the width of D in the direction α, and by K± = K±(α) the Gaussian curvature at the points x±.

Theorem 11.1.2. If there is an α ∈ Sn−1 such that K+(α) 6= K−(α), or there are αj, j = 1, 2, suchthat d(α1) 6= d(α2), then the set N is compact.

Proof. By formula (4.7.1) in [RKa] we have

χ(tα) = t−n+1

2

[eitα·x+a+ + eitα·x−a− +O(

1

t)

], t→ +∞ (11.1.10)

wherea± = (2π)

n−12 e±i

π4 (n+1)K

−1/2± .

Thus, the zeros of χ(tα) in Rn, when t → ∞, can be found asymptotically from the equation

eitd = −a+

a−= −

(K−K+

)1/2

eiπ2 (n+1),

or

t =1

id

[1

2ln

K−K+

+ iπ +iπ

2(n+ 1) + 2iπν

], (11.1.11)

where ν is an integer. One sees that t > 0 cannot satisfy (11.1.11) unless K+ = K−. Since the set Nis rotation invariant, it is empty for sufficiently large t = |ξ| provided that there is an α ∈ Sn−1 such

11.1. THE POMPEIU PROBLEM 307

that K+(α) 6= K−(α). For equation (11.1.11) to be satisfied by t independent of α, it is necessary and

sufficient that d(α) = const and K+(α)K−(α)

= 1. 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 assumptionsof Theorem 11.1.2 imply that, for sufficiently large t, the surface of real zeros of χ(tα) is not a sphere.Thus, the set N is compact. 2

It is well-known that there are convex smooth (and not smooth) bodies of constant width which arenot balls. However, the following lemma holds:

Lemma 11.1.3. If D ⊂ R2 is a strictly convex connected domain with a smooth boundary S, whosewidth is constant: α · (x−(α) − x+(α)) = d = const, and K+(α) = K−(α), then D is a disk.

Proof. Let s be the length of S considered as a natural parameter on S: each point on S is uniquelydefined by the value of this parameter. Since D is convex, each point of S is also uniquely defined by aunit vector α. Namely, given α, one defines x+(α) as the (unique) point of S at which α is the interiorunit normal to S. Thus, α can be considered as a function of s, and, by Frenet’s formulas,

dα

ds= −K+(α)τ,

dx+

ds= τ, τ ·α = 0,

dτ

ds= K+(α)α, (11.1.12)

where τ is the unit vector tangent to S at the point x+, K+ = 1R+

is the curvature of S at the point

x+, and R+ is the radius of curvature, K+ > 0. Differentiating the equation α · (x−(α) − x+(α)) = dwith respect to s yields

−K+(α)τ · (x−(α) − x+(α)) − α · (−τ − τ ) = 0.

Since K+(α) 6= 0, and α · τ = 0, one gets

τ · (x−(α) − x+(α)) = 0. (11.1.13)

In this calculation we have used the assumption K−(α) = K+(α), which allowed us to conclude thatdx− = −τds, where ds is the same as in the formula dx+ = τds. Differentiating (11.1.13) and using(11.1.12), one gets

K+α · (x− − x+) − 2τ · τ = 0,

or d = 2R+, or

R+ =d

2= const. (11.1.14)

Equation (11.1.14) implies that S is a circle of radius d2. 2

Lemma 11.1.4 ([Al, p.304]). If D ⊂ Rn, n > 2, is a strictly convex connected domain with a smoothboundary S, whose width is constant and K+(α) = K−(α), then D is a ball.

In fact, it is proved in [Al] that any C2-surface S in Rn, n ≥ 2, homeomorphic to a sphere and suchthat there exists a C1-function φ(k1, k2, . . . ., kn) with the properties φ(k1, k2, . . . , kn) = const, φkj >0, j = 1, 2, . . . , n, must be a sphere. Here kj, j = 1, 2, . . . , n, are the principal curvatures of S.

Corollary 11.1.5. Assume that D ⊂ Rn is a bounded strictly convex domain with a smooth boundaryS. If χ(tmα) = 0 for all α ∈ Sn−1 and for a sequence tm → +∞, then D is a ball.

Proof. It follows from the assumption and formula (11.1.10) that α · (x+ − x−) = const, K+(α) =K−(α) ∀α ∈ Sn−1. This implies that D is a ball. For n = 2 this is proved in Lemma 11.1.3. For n > 2this is a result from [Al] as stated in Lemma 11.1.4. 2


Let D12 := D1 ∪D2, D12 := D1 ∩D2, and assume for simplicity that D1 and D2 belong to a class D

of strictly convex smooth domains with Gaussian curvature, uniformly bounded from below by a positiveconstant and the C3− norm of the functions representing locally the boundary of D ⊂ D is uniformlybounded by an arbitrary large positive constant C, which characterizes the class D together with thelower bound on the Gaussian curvature.

Theorem 11.1.6. If D1 has P -property, D2 ⊂ D , and meas(D12 \D12) is sufficiently small, then D2

has P -property.

Proof. If D1 has P -property then there is no k > 0 such that χ1(kα) = 0 ∀α ∈ Sn−1. Let∑

be aconnected component of the set of real zeros of χ1(ξ). Note that |χ2 − χ1| ≤ meas(D12 \ D12) := δ.Therefore if

inf0<k<k0

supα∈Sn−1

|χ1(kα)| ≥ ε > 0,

then

inf0<k<k0

supα∈Sn−1

|χ2(kα)| ≥ ε− δ.

Here k0 > 0 is an arbitrary large fixed number. The above argument shows that there are no sphericalsurfaces of zeros of χ2(ξ) in the ball of radius k0 if there are no such surfaces for χ1(ξ) and if D2 differsfrom D1 sufficiently little (precisely, if δ < ε). Outside this ball, for sufficiently large k0, the set of zerosof χ(ξ), ξ = tα, is given asymptotically by the equation

eitα·x+a+ + eitα·x−a− = 0. (11.1.15)

For t sufficiently large this equation yields equation (11.1.11) as the asymptotic equation for∑

. It isclear from (11.1.11) and from Lemma 11.1.4 that various components

∑are, as t → ∞, different from

spheres if D1 is not a ball. Since the sets of zeros of χ2(ξ) is in a δ−neighborhood locally of the set ofzeros of χ1(ξ) for all k > k0, and the sets of zeros of χ1(ξ) in this region are not in a δ−neighborhoodof any sphere centered at the origin, if δ > 0 is sufficiently small, it follows that χ2(ξ) does not havespherical surfaces of zeros in the region k > k0, if δ > 0 is sufficiently small, and, as we proved above,χ2(ξ) does not have spherical surfaces of zeros in the region 0 < k < k0. Clearly, χ2(0) > 0. Thus, χ2(ξ)does not have spherical surfaces of zeros, and therefore D2 has P -property. Thus, we have proved thatthe set of the domains having P -property is open if the distance between D1 and D2 is defined to bemeas(D12 \D12).

Let us now discuss in more detail the choice of the number k0 above. If D1 has P -property, then it isnot a ball. If it is not a ball, then any domain D2, sufficiently close to D1 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 α, such that K−(α) isdifferent from K+(α). This implies that χD2(ξ) does not have spherical surfaces of zeros on any a priorifixed compact K. If this domain is smooth and strictly convex, and if one assumes a uniform positivebound on the Gaussian curvature of all these domains from below and a uniform bound in C3− norm ofthe functions representing locally the boundaries of the domains, then outside K there are no sphericalsurfaces of zeros of χD2 (ξ) either, and the existence of the k0, which can be chosen simultaneously for allsuch domains in the proof of Theorem 11.1.6 is clear. Namely, choose t0 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 onthe C3− norm of the functions representing the boundaries of the domains. Note that c can be chosenuniformly 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 α (depending possibly on the choice of the domain). For this t0 findε > 0 as mint maxα |χD1(ξ)|, where max is taken over α ∈ S2, and min is taken over t in the interval[0, t0]. Now choose 0 < δ < ε and let meas (D12 \D12) < δ. Then for any domain D2 in this set χD2 (ξ)does not have spherical surface of zeros. Thus D2 has P -property. This argument does not requirethat the boundaries of the domains should be close in some Sobolev norm. It does require an a prioriuniform positive lower bound on Gaussian curvatures and a uniform upper bound on the C3− norm of

11.2. NECESSARY AND SUFFICIENT CONDITION 309

the functions representing locally the boundaries of the domains in D , and the convexity of the domains.The last requirement is not crucial (see [RKa], ch.4), however a detailed discussion in the absence ofconvexity is longer, and one has to exclude various pathologies, e.g., existence of countably many criticalpoints of the functions representing the surfaces locally.


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

u(x) =

∫

D

G(x, y, k)dy, G =exp(ik|x− y|)

4π|x− y| if n = 3. (11.1.16)

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 D′ = Rn \D. This follows from (11.1.3), Green’s formula, and the assumption that thesolution to (11.1.3) has compact support. Thus, existence of the solution to (11.1.3), which has compactsupport, implies that the inverse problem for metaharmonic potentials, which consists of finding D,given the values of the potential near infinity, does not have a unique solution.

2

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, whichfails to have Pompeiu property, to be a ball (cf [R186]).

11.2.1 Introduction

Let D ⊂ Rn be a bounded domain with Lipschitz boundary S, SO(n) be the rotational group.Suppose that f ∈ L1

loc(Rn) ∩ S , where S is the Schwartz class of distributions, condition (11.1.2)

holds, and f 6≡ 0, so D has P−property.If f(x) belongs to some functional space the elements of which decay at infinity sufficiently fast, for

example f ∈ Lp(Rn), p = 1, 2, then any bounded domain D, including balls, has P−property. Indeed,arguing as in Section 11.1, let

F f := f (ξ) :=

∫

Rn

f(x) exp(iξ · x)dx.

Let χ(x) denote the characteristic function ofD. One can write (11.1.2) as a convolution of a distributionf , which is a locally integrable function, and a compactly supported function χ(−g−1x). Taking theFourier transform, one gets:

f (ξ)χ(g−1ξ) = 0 ∀g ∈ SO(n), (11.2.1)

where the bar stands for complex conjugate and

χ(x) =

1 in D

0 in D′, D′ := Rn \D.

Since g in (11.2.1) is arbitrary, one concludes that the support of f (ξ) is the discrete set of spheres ofradii kj > 0, kj → ∞, such that χ(ξ) = 0 for |ξ| = kj. This set is discrete since χ(ξ) is an entire functionof ξ if the domain D is bounded. If X is any functional space such that the only element f(x) ∈ X withf (ξ) 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 mentionedtwo 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.In this Section we give a result related to this conjecture.We write D ⊂ P if D has P-property, and D ⊂ P if it fails to have P-property.We need the following result proved in Section 1.1: D ⊂ P iff the following problem is solvable for a

positive number k2 > 0:

(∇2 + k2

)u = −1 in D, u = uN = 0 on S. (11.2.2)

Here, uN is the normal derivative of u, Ns is the exterior unit normal to S at the point s ∈ S.It follows immediately from (11.2.2) that k2 is both a Dirichlet and Neumann eigenvalue of the

Laplacian in D . Indeed, define w := u+ k−2. Then (∆ + k2)w = 0 in D, and wN = 0 on S,w = k−2 =const on S. Thus k2 is a Neumann eigenvalue of the Laplacian. Moreover, since w = const on S, andwN = 0 on S, it follows that grad w = 0 on S, so wxj := v (for any j = 1, 2, 3, . . .) solves the problem(∆ + k2)v = 0 in D, v = 0 on S. Thus k2 is a Neumann and a Dirichlet eigenvalue simultaneously.

Denote by N the eigenspace of the Dirichlet Laplacian corresponding to the eigenvalue k2, byuj(x)1≤j≤J an orthonormal basis of N , by L , the linear span of the functions ujN (s)1≤j≤J , s ∈ S,and by M the orthogonal complement of L in L2(S).

Let α ∈ Sn−1 be an arbitrary unit vector, Sn−1 is the unit sphere in Rn. Let v be the velocitycorresponding to the rotation of Rn about the axis directed along α and passing through the gravitycenter of D.

If n = 3 then v = [α, x] where [α, x] is the vector product. For simplicity let us take n = 3 in whatfollows, but the argument and the result hold for n ≥ 2 if one writes Gx in place of [α, x]. By (a, b) theinner product in Rn 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,D ⊂ P . Then D is a ball if and only if for any α ∈ S2 one has:

([α, s], Ns) ∈ M . (11.2.3)

.

The above result says thatthe conjecture (C) is true provided that (11.2.3) holds.In other words, equation (11.2.3) means that the function ([α, s], Ns), for any α ∈ S2, does not have

a non-zero projection onto the finite-dimensional subspace L in L2(S).If D is a ball, then

([α, s], Ns) = (α, [s,Ns]) ≡ 0,

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.

Lemma 11.2.2. Iff v ·N ≡ 0 on S for all α ∈ S2, then S is a sphere. Iff [s,Ns] ≡ 0 on S, then S is asphere.

In Section 11.2.2 a proof of Theorem 11.2.1 is given.

11.2.2 Proof

Proof. 1) Necessity: Suppose D ⊂ P only if D = B. Then, if D ⊂ P , one concludes that ([α, s], Ns) ≡0 ∀α ∈ S2, so (11.2.3) is trivially satisfied.


2) Sufficiency: Suppose (11.2.3) holds and D ⊂ P . We want to prove that D = B. Let H2(D) bethe Sobolev space, and B0 be a ball containing D and centered at the gravity center of D. If D ⊂ Pthen (11.2.2) holds. Therefore, for any

h ∈ N0 := h : (∇2 + k2)h = 0 in B0, h ∈ H2(B0) (11.2.4)

one has ∫

D

h(x) dx = 0 ∀h ∈ N0. (11.2.5)

This is verified by multiplying (11.2.2) by h, integrating by parts and taking into account the zeroCauchy data for u in (11.2.2).

If h ∈ N0, then for any g ∈ SO(3), one has h(gx) ∈ N0.Fix an arbitrary α ∈ S2 and let the x3-axis of the coordinate system be directed along α. Choose as

g a rotation about α, which is given by the matrix

g =

cosϕ − sinϕ 0

sinϕ cosϕ 0

0 0 1

(11.2.6)

Then

dg

dϕ

∣∣∣∣ϕ=0

=

0 −1 0

1 0 0

0 0 0

:= G. (11.2.7)

Therefore the velocity field in R3, corresponding to this rotation, is

v = Gx = [α, x]. (11.2.8)

Taking h(gx) in place of h(x) in (11.2.5), differentiating with respect to ϕ and taking ϕ = 0 afterwards,one gets: ∫

D

∇h · [α, x] dx= 0 ∀h ∈ N0, ∀α ∈ S2. (11.2.9)

Using Gauss formula, one obtains from (11.2.9) the following equation:

∫

S

h(s) ([α, s], Ns) ds = 0 ∀h ∈ N0, ∀α ∈ S2. (11.2.10)

Denote

([α, s], Ns) := f(s, α), (11.2.11)

and write (11.2.10) as ∫

S

h(s)f(s, α) ds = 0 ∀h ∈ N0, ∀α ∈ S2. (11.2.12)

If, for any α ∈ S2, one could choose h ∈ N0 such that

||h(s) − f(s, α)|| < ε, (11.2.13)

where ε > 0 is arbitrarily small and the norm in (11.2.13) is L2(S)-norm, then (11.2.12) would imply

f(s, α) = 0 ∀α ∈ S2, (11.2.14)


or (use (11.2.11) and take into account that α is arbitrary) the following equation:

[s,Ns] ≡ 0 on S. (11.2.15)

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 f(s, α) ∈ M .We drop in what follows the α-dependence for brevity.Let us first prove that the boundary-value problem (11.2.21) (see below) has a solution iff f(s, α) ∈

M . To prove this, pick an arbitrary F (x) ∈ H2(D) such that

F = f(s) on S, (11.2.16)

and define w by the formula:h = w + F. (11.2.17)

Then (∇2 + k2

)w = −

(∇2 + k2

)F in D, w = 0 on S. (11.2.18)

It is known that (11.2.18) is solvable iff the following orthogonality conditions hold:

∫

D

uj(x)(∇2 + k2

)F dx = 0, 1 ≤ j ≤ J, (11.2.19)

where uj is a basis of N , the eigenspace of the Dirichlet Laplacian corresponding to the eigenvalue k2.Integrating by parts in (11.2.19) yields

∫

S

ujN f(s) ds = 0, 1 ≤ j ≤ J. (11.2.20)

This means that f ∈ M . Conditions (11.2.20) are necessary and sufficient for the solvability of theproblem

Lh :=(∇2 + k2

)h = 0 in D, h = f on S. (11.2.21)

To complete the proof of Theorem 11.2.1, it is sufficient to verify the following Lemma 11.2.3, whichimplies inequality (11.2.13) for f(s, α) ∈ M .

Lemma 11.2.3. The set of restrictions of the elements of N0 to S is complete in M .

Let us prove Lemma 11.2.3.

Proof. We start the proof by noting that the set g(x, y)y/∈B0where

g :=exp(ik|x− y|)

4π|x− y| ,

is total (complete) in N0 in L2(B0). Indeed, if h ∈ N0 and

w(y) :=

∫

B0

g(x, y)h(x)dx = 0 ∀y /∈ B0,

thenLw = −h ∈ B0, w = wN = 0on ∂B0.

Multiply the last equation by h, integrate over B0 and then by parts, use equation (11.2.21) for h andthe zero Cauchy data for w and get ∫

B0

|h|2dx = 0,


so h = 0 as we wanted to show.Let us now finish the proof of Lemma 11.2.3.Assume that

v(x) :=

∫

S

g(x, s)f(s)ds = 0 ∀x /∈ B0. (11.2.22)

We want to prove that (11.2.19) holds iff f ∈ L . Equation (11.2.22) implies v(x) = 0 in D′ := R3 \Dand v solves the homogeneous problem (11.2.21) in D. By the jump formula for the normal derivativeof the single layer potential v(x), one has f = ∂v

∂N. Thus, (11.2.22) implies that f ∈ L . The converse is

easy to prove also. Therefore Lemma 11.2.3 is proved. 2

The proof of Theorem 11.2.1 is complete. 2

Bibliographical notes

In Chapter 1 statements of many inverse problems are given. In Chapter 2, Sections 2.1-2.3 containmaterial that can be found in [IVT], [Mor], [MG], [Gro], [VV], [TLY], [R121], [EHN]. There are somenew results regarding regularization of unbounded nonlinear ill-posed operator equations, and a newversion of the discrepancy principle. In the presentation of the Backus-Gilbert method the rate ofconvergence of the proposed version of this method is estimated. We did not discuss the optimality ofthe methods for solving ill-posed problems (see, e.g., [IVT], [EHN]), but argue that a stable solution ofan 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 problemsearlier. 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 numericaldifferentiation 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 CuboMathematical Journal for permission to use paper [R221]. Chapter 3 contains many new results, andthe presentation of the classical results contains many novel points, especially in the presentation ofGel’fand-Levitan’s (GL) theory and Marchenko’s theory. Krein’s inversion theory is presented withcomplete proofs for the first time. Its presentation in Section 3.9, is based on [R197]. The analysisof the invertibility of the inversion steps in the GL’s theory and in Marchenko’s theory is discussed indetail, which was not done earlier. A detailed analysis of the Newton-Sabatier (NS) theory for invertingfixed-energy phase shifts is given in Section 3.6. This theory was presented in [N], [CS]. Our analysisconcludes that the NS inversion theory is fundamentally wrong in the sense that its foundations arewrong. In [Sab] P.Sabatier made an attempt to disagree with the above conclusion but, in fact, failedto address the main point from [R206], namely, that there is no proof that the basic equation in the NStheory (equation (3.6.65) in Section 3.6.4) has a solution for all r > 0, and that this solution is unique. Itis shown in [R206] (and in Section 3.6) that equation (3.6.65) generically does not have a unique solutionfor some r > 0 and in this case the NS inversion procedure breaks down. In [R207] a counterexampleis 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 problemof finding a confining potential (a quarkonium system) from a few measurements is given. This problemwas considered in [TQR], but the approach in [TQR] was not correct by the reasons explained in Section3.8 (and in [R183]). Section 3.10 contains a solution of some new Inverse problems for the heat and waveequations. In Section 3.11 the result from [R191] is presented. In Section 3.12 an inverse problem ofocean acoustics is solved ([R199]), and it is shown that the method used for the study of a related problemin [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 uniquenesstheorem for I−function and its numerous applications. The I−function for the class of potentials, usedin scattering theory, is equal to the Weyl’s m−function, which is in one-to-one correspondence with thespectral function. In [GS] and [GS1] some properties of m−functions are given and their applicationsto inverse problems are discussed. In [R181], [R184] a uniqueness theorem and a convergent iterative

314


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 numeri-cal methods for solving inverse obstacle scattering problems are discussed. These methods and othernumerical methods are analyzed in Section 4.5-4.6. The novel points in this Chapter include (but notlimited 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 dataand with high-frequency data, and the analysis of the numerical methods for solving obstacle inversescattering 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 propersubclass.

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 Birkhauser for permission to use this paper. Thenotion 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 beenstudied 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 6.4is based on [Ro1]. The example of non-uniqueness in Section 6.1 shows that the usage of numericalparameter-fitting as a method for solving inverse problems may be meaningless. This is important tohave 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 theuniqueness 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 onantenna 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 prob-lems are obtained for overdetermined problems. We formulate several basic non-overdetermined inverseproblems 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 themonographs [R83]. Our presentation is based on these papers and on the book [R139]. Almost all of theresults 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 arbitraryshapes. This is a classical area of research, originated by Lord Rayleigh in 1871, who understood thatthe main term in the acoustic field, scattered by a small in comparison with the wavelength body isgiven by a dipole radiation. Lord Rayleigh had published many papers on this subject before his deathin 1919. There are hundreds of papers, mostly dealing with applied sciences, which use wave scatteringtheory by small bodies. However, there were no analytic formulas for calculation of this dipole radiationfor bodies of arbitrary shapes. Such formulas were found in [R22], [R25], [R32], and the theory waspresented in [R50], [R51] and summarized in [R65]. Our basic results include analytic formulas, whichallow one to calculate the capacitances of conductors of arbitrary shapes and the polarizability tensorsof homogeneous dielectric bodies of arbitrary shapes with arbitrary accuracy in terms of the geometry ofthese bodies and their dielectric constant. These formulas allow one to derive an analytic formula for thescattering matrix for electromagnetic wave scattering by small bodies of arbitrary shapes. This formulaallows one to solve an inverse problem of radiomeasurements ([R65], [R33]). Section 10.2 deals withwaves 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 a large bodyof literature on wave scattering in random media [Ish]. Section 10.3 is based on [R193], and numericalresults 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 manypapers 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 problemshave not been discussed. A short and incomplete list of these include Carleman estimates and theirapplications, inverse problems for systems, in particular, for elasticity and Maxwell’s equations, controltheory methods, integral geometry problems, to say nothing about many concrete inverse problems.

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[R51] Ramm, A.G., Theory and applications of some new classes of integral equations,Springer-Verlag, New York, 1980.

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[R57] Ramm, A.G., Existence uniqueness and stability of solutions to some nonlinear problems.Appl. Analysis, 11, (1981), 223-232.

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[R66] Ramm, A.G., Mathematical foundations of the singularity and eigenmode expansion meth-ods. J. Math. Anal. Appl., 86, (1982), 562-591.

[R67] Ramm, A.G., Perturbations preserving asymptotics of spectrum with a remainder. Proc.Amer. Math.Soc., 85, N2, (1982), 209-212.

[R68] Ramm, A.G., Inverse scattering for geophysical problems. Phys. Letters, 99A, (1983), 258-260.

[R69] Ramm, A.G., An inversion formula in scattering theory. Phys. Lett., 99A, (1983), 201-204.

[R70] Ramm, A.G., On a property of the set of radiation patterns. J. Math. Anal. Appl. 98,(1984), 92-98.

[R71] Ramm, A.G., Scattering by a penetrable body. J. Math. Phys. 25, N3, (1984), 469-471.

[R72] Ramm, A.G., Representations of solutions to Helmholtz’s equation. J.Math.Phys., 25, N4,(1984), 807-809.

[R73] Ramm, A.G., Description of the degree of nonuniqueness in inverse source problem. J.Math. Phys., 25, N6, (1984), 1791-1793.

[R74] Ramm, A.G., Remarks about inverse diffraction problem. J. Math. Phys., 25, N11, (1984),2672-2674.

[R75] Ramm, A.G., On inverse diffraction problem. J. Math. Anal. Appl., 103, (1984), 139-147.

[R76] Ramm, A.G., Inverse diffraction problem. Inverse methods in electromagnetic imaging, Ed.Boerner, W., Reidel, Dordrecht, 1985, pp. 231-249.

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[R79] Ramm, A.G., Offset measurements on a sphere at a fixed frequency do not determine theinhomogeneity uniquely, Inverse problems, 1, (1985), L35-L37.

[R80] Ramm, A.G., Inversion of the back scattering data and a problem of integral geometry.Phys. Lett. 113A, (1985), 172-176.

[R81] Ramm, A.G., Inverse scattering in an absorptive medium. Inverse problems, 2, (1986),L5-L7.

[R82] Ramm, A.G., Singularities of the inverses of Fredholm operators. Proc. of Roy. Soc. Edin-burgh, 102A, (1986), 117-121.

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[R87] Ramm, A.G., On completeness of the products of harmonic functions, Proc. A.M.S., 99,(1986), 253-256.

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[R93] Ramm, A.G., Optimal estimation from limited noisy data, J. Math. Anal. Appl., 125 (1987),258-266.

[R94] Ramm, A.G., Signal estimation from incomplete data, J. Math. Anal. Appl., 125 (1987),267-271.

[R95] Ramm, A.G., Recovery of the potential from I-function, Math. Reports of the Acad. of Sci.,Canada, 9, (1987), 177-182.

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[R108] Ramm, A.G., A criterion for completeness of the set of scattering amplitudes, Phys. LettA.129, (1988), 191-194.

[R109] Ramm, A.G., Recovery of the potential from fixed energy scattering data, Inverse Problems,4, (1988), 877-886; 5, (1989) 255.

[R110] Ramm, A.G., A simple proof of uniqueness theorem in impedance tomography, Appl. Math.Lett., 1, N3, (1988), 287-290.

[R111] Ramm, A.G., Numerical method for solving 3D inverse scattering problems, Appl. Math.Lett., 1, N4, (1988), 381-384.

[R112] Ramm, A.G., Uniqueness theorems for multidimensional inverse problems with unboundedcoefficients. J. Math. Anal. Appl. 136, (1988), 568-574.

[R113] Ramm, A.G., Numerical method for solving 3D inverse problems of geophysics, J. Math.Anal. Appl. 136, (1988), 352-356.

[R114] Ramm, A.G., Multidimensional inverse problems: Uniqueness theorems, Appl. Math. Lett.1, N4, (1988), 377-380.

[R115] Ramm, A.G., Multidimensional inverse scattering problems and completeness of the prod-ucts of solutions to homogeneous PDE. Zeitschr. f. angew. Math. u. Mech., 69, (1989) N4,T13-T22.

[R116] Ramm, A.G., Numerical method for solving 3D inverse problems with complete and in-complete data, In the book: “Wave Phenomena”, Springer-Verlag, New York 1989, (eds.L. Lam and H. Morris), 34- 43.

[R117] Ramm, A.G., Numerical recovery of the 3D potential from fixed energy incomplete scatter-ing data, Appl. Math.Lett., 2, N1, (1989), 101-104.

[R118] Ramm, A.G., Necessary and sufficient condition for a scattering amplitude to correspondto a spherically symmetric scatterer, Appl.Math.Let. 2, (1989), 263-265.

[R119] Ramm, A.G., An inverse problem for Maxwell’s equations, Phys.Let. A 138(1989), 459-462.

[R120] Ramm, A.G., Property C and uniqueness theorems for multidimensional inverse spectralproblem, Appl.Math.Lett., 3, (1990), 57-60.

[R121] Ramm, A.G., Random fields estimation theory, Longman Scientific and Wiley,New York, 1990.

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[R122] Ramm, A.G., Stability of the numerical method for solving the 3D inverse scattering prob-lem with fixed energy data, Inverse problems 6, (1990), L7-12.

[R123] Ramm, A.G., Algorithmically verifiable characterization of the class of scattering ampli-tudes for small potentials, Appl. Math. Lett, 3, (1990), 61-65.

[R124] Ramm, A.G., Is the Born approximation good for solving the inverse problem when thepotential is small? J. Math. Anal. Appl., 147, (1990), 480-485.

[R125] Ramm, A.G., Completeness of the products of solutions of PDE and inverse problems,Inverse Probl.6, (1990), 643-664.

[R126] Ramm, A.G., Uniqueness result for inverse problem of geophysics I,Inverse Probl.6, (1990),635-642.

[R127] Ramm, A.G., Uniqueness theorems for geophysical problems with incomplete surface data,Appl. Math. Lett.3, (1990), N4, 41-44.

[R128] Ramm, A.G., Symmetry properties for scattering amplitudes and applications to inverseproblems, J. Math. Anal. Appl., 156, (1991), 333-340.

[R129] Ramm, A.G., Necessary and sufficient condition for a PDE to have property C, J. Math.Anal. Appl.156, (1991), 505-509.

[R130] Ramm, A.G., Stability of the numerical method for solving 3D inverse scattering problemwith fixed energy data, J.f.die reine und angew. Math, 414, (1991), 1-21.

[R131] Ramm, A.G., Finding conductivity from boundary measurements, Comp.& Math.withAppl., 21, N8, (1991), 85-91

[R132] Ramm, A.G., Exact inversion of fixed-energy data, in the book Mathematical and NumericalAspects of Wave Propagation Phenomena, SIAM, Philadelphia, (1991), pp.481-486

[R133] Ramm, A.G., Property C and inverse problems, ICM-90 Satellite Conference Proceedings,Inverse Problems in Engineering Sciences, Proc. of a conference held in Osaka, Japan, Aug.1990, Springer Verlag, New York, 1991, pp. 139-144.

[R134] Ramm, A.G., Can a constant be a scattering amplitude? Phys.Lett., 154A, (1991), 35-37

[R135] Ramm, A.G., Inversion of limited-angle tomographic data, Comp. and Math. with Applic.,22, 4/5, (1991), 101-112.

[R136] Ramm, A.G., On a problem of integral geometry, Comp. and Math. with Appl., 22, 4/5,(1991), 113-118.

[R137] Ramm, A.G., Solution of some integral equations arising in integral geometry,Appl.Math.Lett., 4, (1991), 177-181

[R138] Ramm, A.G., Inversion of the Radon transform with incomplete data, Math.Methods inthe Appl.Sci., 15, N3, (1992), 159-166.

[R139] Ramm, A.G., Multidimensional inverse scattering problems, Longman/Wiley,New York, 1992, pp.1-385.

[R140] Ramm, A.G., Stability of the solution to inverse scattering problem with exact data,Appl.Math.Lett., 5, 1, (1992), 91-94

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[R141] Ramm, A.G., Inversion of limited-angle tomographic data II, Appl.Math.Lett., 5, N2,(1992), 47-49.

[R142] Ramm, A.G., Inverse scattering problem with fixed-energy data, Appl.Math.Lett., 5, N4,(1992), 63-67.

[R143] Ramm, A.G., Stability estimates in inverse scattering, Acta Appl. Math., 28, N1, (1992),1-42.

[R144] Ramm, A.G., Numerical solution of 3D inverse scattering problems with noisy discretefixed-energy data, Appl. Math. Lett., 5, N6, (1992), 15- 18.

[R145] Ramm, A.G., Stability of the inversion of 3D fixed-frequency data, J.Math.Anal.Appl., 169,N2(1992), 329-349.

[R146] Ramm, A.G., Stability of the solution to 3D fixed-energy inverse scattering problem,J.Math.Anal.Appl., 170, N1 (1992), 1-15.

[R147] Ramm, A.G., Uniqueness and inversion of cone-beam data, Appl.Math.Lett., 6, N1 (1993),35-38.

[R148] Ramm, A.G., An inverse problem for the heat equation, Proc.Roy.Soc. Edinburgh, 123, N6,(1993), 973-976.

[R149] Ramm, A.G., Approximation by the scattering solutions and applications to inverse scat-tering, Math.Comp.Modelling, 18, N1, (1993), 47-56.

[R150] Ramm, A.G., Comments on the letter of P.Sabatier ”An erroneous statement”,http://www.arxiv.org, paper math-ph/0308025,

[R151] Ramm, A.G., Property C with constraints and inverse problems, J.of Inverse and Ill-PosedProblems, 1, N3 (1993), 227-230.

[R152] Ramm, A.G., Algorithm for solving 3D inverse scattering problems with noisy discretefixed-energy data, Proceedings of ICES-92 conference on inverse problems. In the book:Inverse Problems, Atlanta Technology Publications, Atlanta, Georgia, (1993), pp.70-74.

[R153] Ramm, A.G., Scattering amplitude is not a finite rank kernel in the basis of sphericalharmonics, Appl.Math.Lett., 6, N5, (1993), 89-92.

[R154] Ramm, A.G., New method for proving uniqueness theorems for obstacle inverse scatteringproblems, Appl.Math.Lett., 6, N6, (1993), 19-22.

[R155] Ramm, A.G., Scattering amplitude as a function of the obstacle, Appl.Math.Lett., 6, N5,(1993), 85-87.

[R156] Ramm, A.G., Simplified optimal differentiators, Radiotech.i Electron.17, (1972), 1325-1328.

[R157] Ramm, A.G., Multidimensional inverse scattering problems, Mir Publishers,Moscow, 1994, pp.1-496. (Russian translation of the expanded monograph 278).

[R158] Ramm, A.G., A numerical approach to 3D inverse scattering problems, Appl.Math.Lett.7,N2, (1994), 57-61

[R159] Ramm, A.G., Numerical method for solving inverse scattering problems, Doklady of RussianAcad. of Sci., 337, N1, (1994), 20-22

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[R160] Ramm, A.G., Multidimensional inverse scattering: solved and unsolved problems,Proc.Intern.Confer. on Dynamical Syst.and Applic., Vol.1, Atlanta, (1994), pp.287-296.(Eds. G.Ladde and M.Sambandham)

[R161] Ramm, A.G., Inversion of fixed-frequency surface data for layered medium, J. of Inverseand Ill-Posed Problems, 2, N3, (1994), 263-268

[R162] Ramm, A.G., Stability of the solution to inverse obstacle scattering problem, J.Inverse andIll-Posed Problems, 2, N3, (1994), 269-275.

[R163] Ramm, A.G., Stability of the solution to 3D inverse scattering problems with fixed-energydata. Proc. ASME Nov.6-11, 1994, meeting. Inverse problems in mechanics, AMD-Vol 186,pp.99-102.

[R164] Ramm, A.G., Stability estimates for obstacle scattering, J.Math.Anal.Appl. 188, N3, (1994),743-751.

[R165] Ramm, A.G., Examples of nonuniqueness for an inverse problems of geophysics, Appl.Math. Lett., 8, N4, (1995), 87-90.

[R166] Ramm, A.G., Property C with constraints, Compt.Rendus, Paris, ser 1,321, N 11, (1995),1413-1417.

[R167] Ramm, A.G., Continuous dependence of the scattering amplitude on the surface of anobstacle, Math. Methods in the Appl. Sci., 18, (1995), 121-126.

[R168] Ramm, A.G., Estimates of the derivatives of random functions, J. Math. Anal. Appl., 102,(1984), 244-250.

[R169] Ramm, A.G., A formula for inversion of boundary data, J. of Inverse and Ill- Posed Prob-lems, 3, N5, (1995), 411-415.

[R170] Ramm, A.G., Radon transform on distributions, Proc.Japan Acad., ser A, 71, N9, (1995),202-206.

[R171] Ramm, A.G., Uniqueness theorems for inverse obstacle scattering problems in Lipschitzdomains, Applic. Analysis, 59, (1995), 377-383.

[R172] Ramm, A.G., Finding potential from the fixed-energy scattering data via D-N map, J. ofInverse and Ill-Posed Problems, 4, N2,(1996), 145-152.

[R173] Ramm, A.G., Necessary and sufficient conditions for a PDO to be a local tomographyoperator, Comptes Rend Acad Sci, Paris, 332, N7, (1996), 613-618.

[R174] Ramm, A.G., Minimization of the total radiation from an obstacle by a control function ona part of the boundary, Jour. of Inverse and Ill-posed Prob., 4, N6, (1996), 531-534.

[R175] Ramm, A.G., Random fields estimation theory, MIR, Moscow, 1996, pp.1-352

[R176] Ramm, A.G., New methods for finding discontinuities of functions from local tomographicdata, Jour. of Inverse and Ill-Posed Problems, 5, N2, (1997), 165-174.

[R177] Ramm, A.G., The Pompeiu problem, Applicable Analysis, 64, N1-2, (1997), 19-26.

[R178] Ramm, A.G., Multidimensional inverse scattering with fixed-energy data, in the book”Quantum Inversion”, Lecture Notes in physics vol.488, pp. 373-384, Springer Verlag,Berlin, 1997 (ed. B.Apagyi).

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[R179] Ramm, A.G., A method for finding small inhomogeneities from surface data, Math.Sci.Research Hot-Line, 1, N10, (1997), 40-42.

[R180] Ramm, A.G., Fundamental solutions to elliptic equations with discontinuous senior coeffi-cients and an inequality for these solutions. Math. Ineq. and Applic., 1, N1, (1998), 99-104.

[R181] Ramm, A.G., Recovery of compactly supported spherically symmetric potentials from thephase shift of s-wave, In the book: Spectral and scattering theory, Plenum publishers, NewYork, 1998 (ed. A.G.R), pp.111-130.

[R182] Ramm, A.G., Inequalities for norms of some integral operators Math. Ineq. and Applic. 1,N2, (1998), 259-265.

[R183] Ramm, A.G., Recovery of a quarkonium system from experimental data, Jour. of Phys. A,31, N15, (1998), L295-L299.

[R184] Ramm, A.G., Compactly supported spherically symmetric potentials are uniquely deter-mined by the phase shift of s-wave, Phys. Lett. A, 242, N4-5, (1998), 215-219.

[R185] Ramm, A.G., Theory of ground-penetrating radars II, Jour of Inverse and Ill- PosedProbl.,6, N6, (1998), 619-624.

[R186] Ramm, A.G., Necessary and sufficient condition for a domain, which fails to have Pompeiuproperty, to be a ball, Jour of Inverse and Ill-Posed Probl., 6, N2, (1998), 165-171.

[R187] Ramm, A.G., A new approach to the inverse scattering and spectral problems for theSturm-Liouville equation, Ann. der Phys., 7, N4, (1998), 321- 338.

[R188] Ramm, A.G., Property C for ODE and applications to inverse scattering, Zeit. fuer Angew.Analysis, 18, N2, (1999), 331-348.

[R189] Ramm, A.G., Analytical solution of a new class of integral equations, Diff. Integral Eqs,16, N2, (2003), 231-240.

[R190] Ramm, A.G., A numerical method for some nonlinear problems, Math. Models and Meth.in Appl.Sci., 9, N2, (1999), 325-335.

[R191] Ramm, A.G., Inverse problem for an inhomogeneous Schrodinger equation, Jour. Math.Phys, 40, N8, (1999),3876-3880.

[R192] Ramm, A.G., Inverse scattering problem with part of the fixed-energy phase shifts, Comm.Math. Phys. 207, N1, (1999), 231-247.

[R193] Ramm, A.G., Finding small inhomogeneities from surface scattering data, Jour. of Inverseand Ill-Posed Problems, 8, N2, (2000), 205-210.

[R194] Ramm, A.G., Inequalities for the derivatives, Math. Ineq. and Appl., 3, N1, (2000), 129-132.

[R195] Ramm, A.G., The ground-penetrating radar problem III Jour. of Inverse and Ill- PosedProblems, 8, N1, (2000), 23-31.

[R196] Ramm, A.G., Property C for ODE and applications to inverse problems, in the book ”Op-erator Theory and Its Applications”, Amer. Math. Soc., Fields Institute Communicationsvol. 25, (2000), pp.15-75, Providence, RI. (editors A.G.R, P.N.Shivakumar, A.V.Strauss).

[R197] Ramm, A.G., Krein’s method in inverse scattering, in the book ”Operator Theory andIts Applications”, Amer. Math. Soc., Fields Institute Communications vol.25, pp.441-456,Providence, RI, 2000. (editors A.G.R, P.N.Shivakumar, A.V.Strauss).

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[R198] Ramm, A.G., A non-overdetermined inverse problem of finding the potential from thespectral function, IJDEA (Intern. J. of Diff. Eq. and Appl.), 3, N1, (2001), 15-29.

[R199] Ramm, A.G., An inverse problem of ocean acoustics, Jour. of Inverse and Ill- Posed Probl.,9, N1, (2001), 95-102.

[R200] Ramm, A.G., Linear ill-posed problems and dynamical systems, Jour. Math. Anal. Appl.,258, N1, (2001), 448-456.

[R201] Ramm, A.G., An inverse problem for the heat equation, Jour. of Math. Anal. Appl., 264,N2, (2001), 691-697.

[R202] Ramm, A.G., Reconstruction of the potential from I-function, Jour. of Inverse and Ill-PosedProbl., 10, N4, (2002), 385-395.

[R203] Ramm, A.G., Stability of solutions to inverse scattering problems with fixed- energy data,Milan Journ of Math., 70, (2002), 97-161.

[R204] Ramm, A.G., Regularization of ill-posed problems with unbounded operators, J. Math.Anal. Appl., 271, (2002), 547-550.

[R205] Ramm, A.G., Modified Rayleigh conjecture and applications, Jour. Phys. A, 35, (2002),L357-361.

[R206] Ramm, A.G., Analysis of the Newton-Sabatier scheme for inverting fixed-energy phaseshifts, Applic. Analysis, 81, N4, (2002), 965-975.

[R207] Ramm, A.G., A counterexample to a uniqueness result, Applic. Analysis, 81, N4, (2002),833-836.

[R208] Ramm, A.G., Exterior boundary value problems as limits of interface problems, J. Math.Anal. Appl. 84, (1981), 256-263.

[R209] Ramm, A.G., An inverse problem for the heat equation II, Applic. Analysis, 81, N4, (2002),929-937.

[R210] Ramm, A.G., Acceleration of convergence of a continuous analog of the Newton method,Applic. Analysis, 81, N4, (2002), 1001-1004.

[R211] Ramm, A.G., Injectivity of the spherical means operator, Compt. Rend. Acad Sci., Paris,Ser I, 335, N12, (2002), 1033-1038.

[R212] Ramm, A.G., Dynamical systems method for solving equations with non-smooth monotoneoperators

[R213] Ramm, A.G., On a new notion of regularizer, J.Phys A, 36 (2003), 2191-2195.

[R214] Inequalities for the transformation operators and applications, JIPAM (Jour. of Inequalitiesin Pure and Appl. Math.) 4, N4, (2003), pp.1-9, (paper69).

[R215] Ramm, A.G., Equations for the self-consistent field in random medium, Phys.Lett. A, 312,N3-4, (2003), 256-261.

[R216] Ramm, A.G., Dynamical systems method for solving operator equations, Communic. inNonlinear Sci. and Numer. Simulation, 9, N2, (2004), 383-402.

[R217] Ramm, A.G., Global convergence for ill-posed equations with monotone operators: thedynamical systems method, J. Phys A, 36, (2003), L249-L254.

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[R218] Ramm, A.G., On the discrepancy principle, Nonlinear Functional Anal. and Applic., 8, N2,(2003), 307-312.

[R219] Ramm, A.G., DSM for ill-posed equations with monotone operators, Communic. in Non-linear Sci. and Numer. Simulation, 10, N8, (2005), 935-940.

[R220] Ramm, A.G., Discrepancy principle for the dynamical systems method, Communic. inNonlinear Sci. and Numer. Simulation, 10, N1, (2005), 95-101.

[R221] Ramm, A.G., One-dimensional inverse scattering and spectral problems Revista Cubo, 6,N1, (2004), 313-426.

[R222] Ramm, A.G., Dynamical systems method and surjectivity of nonlinear maps, Commun. inNonlin. Sci. and Numer. Simul., 10, N8, (2005), 931-934.

[RAI] Ramm, A.G., Arredondo, J., Izquierdo, B., Formula for the radius of the support of thepotential in terms of the scattering data, Jour. of Phys. A, 31, N1, (1998), L39-L44.

[RF] Ramm, A.G., Frolov, L., Calculation of the magnetization of thin films, Microelectronics 6,(1971), 65-68.

[RG] Ramm, A.G., Galstian, A., On deconvolution methods, Internat. Jour. of Engin. Sci., 41,N1, (2003), 31-43.

[RGo] Ramm, A.G., Ghosh Roy, D, Inverse geophysical problems for some non-compactly sup-ported inhomogeneities, Appl. Math. Lett., 6, N6, (1993), 15-18.

[RGu1] Ramm, A.G., Gutman, S., Modified Rayleigh conjecture for scattering by periodic struc-tures, Internat. Jour. of Appl. Math. Sci., 1, N1, (2004), 55-66.

[RGu2] Ramm, A.G., Gutman, S., Analysis of a linear sampling method for identification of obsta-cles, Acta Appl. Math. Sinica, 21, N3, (2005), 1-6.

[RGu3] Ramm, A.G., Gutman, S., Numerical solution of obstacle scattering problems, Internat.Jour. of Appl. Math. and Mech., 1, (2005), 1-32.

[RGu4] Optimization methods in direct and inverse scattering, in the book: Continuous Opti-mization: Current Trends and Modern Applications, Springer, New York, 2005, pp.51-110.(Editors: V.jeyakumar and A.Rubinov).

[RH] Ramm, A.G., Harris, J., Inversion of the acoustic well to well data, Appl. Math. Letters, 1,(1988), 127-131.

[RKa] Ramm, A.G., Katsevich, A.I., The Radon Transform and Local Tomography,CRC Press, Boca Raton 1996.

[RKo] Ramm, A.G., Koshkin, S., An inverse problem for an abstract evolution equation, Applic.Analysis, 79, N3-4, (2001), 475-482.

[RLi] Ramm, A.G., Li. P., Numerical recovery of the layered medium, J.of Comput.andAppl.Math., 25, N3, (1989), 267-276.

[RPY] Ramm, A.G., Pang, P., Yan, G., A uniqueness result for the inverse transmission problem,Internat. Jour. of Appl. Math., 2, N5, (2000), 625-634.

[RRa] Ramm, A.G., Rakesh, Property C and an inverse problem for a hyperbolic equation, J.Math. Anal. Appl., 156, (1991), 209-219.

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[RRu] Ramm, A.G., Ruiz, A., Existence and uniqueness of scattering solutions in non-smoothdomains, J. Math. Anal. Appl., 201, (1996), 329-338.

[RSa] Ramm, A.G., Sammartino, M., Existence and uniqueness of the scattering solutions in theexterior of rough domains, in the book ”Operator Theory and Its Applications”, Amer.Math. Soc., Fields Institute Communications vol.25, pp.457-472, Providence, RI, 2000.(editors A.G.Ramm, P.N.Shivakumar, A.V.Strauss).

[RSch] Ramm, A.G., Scheid, W., An approximate method for solving inverse scattering problemwith fixed-energy data, Jour. of Inverse and Ill-Posed Problems, 7, N6, (1999), 561-571.

[RSh] Ramm, A.G., Shcheprov, A., Theory of ground-penetrating radars, Jour. of Inverse andIll-Posed Problems, 5, N4, (1997), 377-384.

[RSi] Ramm, A.G., Simon, B., A new approach to inverse spectral theory III, Short range po-tentials, J. d’Analyse Math., 80, (2000), 319-334.

[RSj] Ramm, A.G., Sjostrand,J., An inverse problem for the wave equation, Math. Zeitschr., 206,(1991) 119-130.

[RSm1] Ramm, A.G., Smirnova, A. B., A numerical method for solving nonlinear ill-posed problems,Nonlinear Funct. Anal. and Optimiz., 20, N3, (1999), 317-332.

[RSm2] Ramm, A.G., Smirnova, A. B., A numerical method for solving the inverse scatteringproblem with fixed-energy phase shifts, Jour. of Inverse and Ill-Posed Problems, 8, N3,(2000), 307-322.

[RSm3] Ramm, A.G., Smirnova, A. B., On stable numerical differentiation, Mathem. of Computa-tion, 70, (2001), 1131-1153.

[RSm4] Ramm, A.G., Smirnova, A. B., Continuous regularized Gauss-Newton-type algorithm fornonlinear ill-posed equations with simultaneous updates of inverse derivative, Intern. Jour.of Pure and Appl Math., 2, N1, (2002), 23-34.

[RSm5] Ramm, A.G., Smirnova, A. B., Stable numerical differentiation: when is it possible? Jour.Korean SIAM, 7, N1, (2003), 47-61.

[RSm6] Ramm, A.G., Smirnova, A. B., On deconvolution problems: numerical aspects, Jour. ofComp. Appl. Math., 176, N2, (2005), 445-460.

[RSF] Ramm, A.G., Smirnova, A. B., Favini, A., Continuous modified Newton’s-type method fornonlinear operator equations, Ann. di Mat. Pure Appl, 182, N1, (2003), 37-52.

[RSo] Ramm, A.G., Somersalso, E., Electromagnetic inverse problem with surface measurementsat low frequencies, Inverse Probl., 5, (1989), 1107-1116.

[RSt1] Ramm, A.G., Stefanov, P., A multidimensional Ambartsumian’s theorem, Appl.Math.Lett,5, N5, (1992), 87-88.

[RSt2] Ramm, A.G., Stefanov, P., Fixed-energy inverse scattering for non-compactly supportedpotentials, Math.Comp.Modelling, 18, N1, (1993), 57-64

[RSZ] Ramm, A.G., Steinberg, A., Zaslavsky, A., Stable calculation of the Legendre transform ofnoisy data, J.Math.Anal.Appl.178, N2, (1993), 592-602.

[RUG] Ramm, A.G., Usoskin, V., Golubkova, M., Calculation of the capacitance of a paral-lelepiped, Electricity, 5, (1972), 90-91

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[RW2] Ramm, A.G., Weaver, O., A characterization of the scattering data in 3D inverse scatteringproblem, Inverse problems, 3, (1987), L49-52.

[RWWW] Ramm, A.G., Weaver, O., Weck, N, Witsch, K., Dissipative Maxwell’s equations at lowfrequencies, Math. Meth. in the Appl. Sci. 13, (1990), 305-322.

[RWeg] Ramm, A.G., Weglein, A., Inverse scattering for geophysical problems II, Inversion of acous-tical data, J. Math. Phys., 25, N11, (1984), 3231-3234.

[RX] Ramm, A.G., Xie, G., Uniqueness result for inverse problem of geophysics II, Appl. Math.Lett., 3, (1990), 103-105.

[RZ] Ramm, A.G., Zou, Q., Numerical solution of some inverse problems of geophysics, Com-puters and Mathematics with Applications, 21, (1991), 75-80

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Index

Antenna synthesis, 18, 254antenna synthesis, 3, 255, 315

Backus-Gilbert, 31, 32backwards heat equation, 22biharmonic equation, 281Born inversion, 2, 15, 233–235, 278

Deconvolution, 73deconvolution, 19, 21, 22Dynamical systems method, 314dynamical systems method, 2Dynamical systems method (DSM), 39

Finding ODE from a trajectory, 20

Gel’fand-Levitan (GL) method, 143geometrical inverse problem, 16, 280ground-penetrating radars, 166ground-penetrating radars is given, 75

I-function, 84, 87, 89, 90Ill-posed problems, 20, 54ill-posed problems, 2, 20, 23, 25, 28, 30, 42, 45,

50, 52, 60, 63, 67, 74, 314image processing, 19Induction logging, 276Integral geometry, 274, 277integral geometry, 3, 13, 14, 273Inverse, 11inverse, 80, 236Inverse geophysical problems, 256Inverse gravimetry problem, 18Inverse obstacle scattering, 12, 195, 256inverse obstacle scattering, 2, 12, 315inverse problem of ocean acoustics, 75, 160Inverse problems for the heat and wave equa-

tions, 12, 155, 314inverse problems for the heat and wave equa-

tions, 75Inverse problems of potential theory, 10inverse problems of potential theory, 3, 252, 271inverse radiomeasurements problem, 3

Inverse scattering problems, 148, 195, 266inverse scattering problems, 11, 19, 315Inverse source problem, 18inverse source problem, 17, 255inverse source problems, 17, 252

Krein’s method, 144, 147

Low-frequency inversion, 263

Marchenko inversion procedure, 75Monotone operators, 25monotone operators, 28, 45, 50, 51, 55

non-overdetermined problems, 19Non-uniqueness, 240, 252non-uniqueness, 3, 242, 278

Pompeiu problem, 3, 14, 15, 303, 304, 310, 316Projection method, 74projection method, 74Property C, 2, 75, 76, 80, 129, 226, 266, 315Pseudoinverse, 38pseudoinverse, 23

Quasiinversion, 32Quasisolution, 31, 32, 74

random media, 293, 296Random medium, 296random medium, 292Regularization, 23, 26, 27, 63, 74, 169regularization, 23, 24, 63, 73, 276

S-matrix, 288singular value decomposition, 23Small bodies, 290, 292, 294–297small bodies, 3, 13, 285, 288, 291–293, 315spectral assumption, 55Spectral problems, 75spectral problems, 11, 80, 236Stability estimates, 195, 211stability estimates, 2, 15, 315stable differentiation, 63, 65, 70

340

INDEX 341

Stable summation, 21, 71

Uniqueness theorems, 83, 154, 256, 266uniqueness theorems, 3, 83, 232, 236, 243, 257

Variable background, 283Variational, 23, 74variational, 63Volterra equations, 72, 77

Wave scattering, 285, 290, 292, 293, 295, 297wave scattering, 3, 140, 285–287, 290, 292, 315well-to-well data, 274

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