On the Inverse Conductivity Problem · Chapter 1. Introduction 1 x1.1. The inverse conductivity problem 2 x1.2. Notes 5 Chapter 2. Preliminary reductions and results 7 x2.1. Reduction

On the Inverse ConductivityProblem

Kim Knudsen

Ph.D. Thesis

2002

DEPARTMENT OF MATHEMATICAL SCIENCESAALBORG UNIVERSITY

Preface

This thesis presents the results of my Ph.D. studies at the Department ofMathematical Sciences, Aalborg University, Denmark, in the period August1 1999 – August 1 2002.

The aim of the thesis is two-fold. First, the thesis is a presentation ofthe scientific results obtained. These new results concern three different as-pects of the inverse conductivity problem and include the results obtainedjointly in the papers [KT01, CK02, CKS02]. Second, the thesis is intendedas a self-contained and convenient introduction to the inverse conductivityproblem. To accomplish this the following exposition contains both a num-ber of well-known results, which are either explained briefly or proved rig-orously, and the new results, which are then given in a suitable context.Furthermore, at the end of each chapter a section with notes can be found.Here further results, references, and comments are given.

The mathematical notation used in the thesis is standard; a review isgiven in Appendix A. References are displayed as [Knu02] and refers tothe bibliography pp. 97.

Part of the work was done in the fall 2001 while I was visiting the Math-ematical Sciences Research Institute in Berkeley, California, as a Ph.D. stu-dent associate in the Inverse Problem program. I thank Professor GuntherUhlmann for making this visit possible and hereby giving me a wonderfulexperience.

I am deeply indebted to my collaborators Horia Cornean, Samuli Sil-tanen, and Alexandru Tamasan. Thanks to Horia for always being openfor discussions on all kinds of subjects, thanks to Samu for having a great

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iv Preface

spirit in our joint work and for the generosity in sharing with me the mat-lab code, and thanks to Alex for many good discussions in Berkeley. I amgrateful also to my fellow students Lasse and Steen for both our mathemat-ical discussions and a good climate in the office.

I would also like to thank my advisor, Professor Arne Jensen, for sug-gesting this interesting problem to me in the first place, for encourage-ment through the studies, and for giving me constructive comments onthe manu-script.

I am grateful to Professor Russell Brown, Professor Gunther Uhlmann,and Associate Professor Svend Berntsen for being in the assessment com-mittee for this thesis. Thanks also to Svend for sharing with us the ideasthat lead to the joint project with Horia.

Finally, I wish to thank my wife, Janne, for coming with me to Berkeley,for understanding (at least some of the mathematics), encouraging support,and love!

Aalborg, Denmark, September 2002 Kim Knudsen

Revised, November 2002

Summary

This thesis concerns the inverse conductivity problem. The physical moti-vation for the study of this mathematical problem is a new technology formedical imaging called electrical impedance tomography. In this methodone looks to find the interior conductivity in body from static electric mea-surements on the boundary of the body. The underlying mathematicalproblem is then to determine and reconstruct a coefficient in an elliptic par-tial differential equation on a smooth and bounded domain from full orpartial knowledge of the associated Dirichlet-to-Neumann on the bound-ary, i.e. the map that takes a voltage potential on the boundary (Dirichletdata) into the resulting current flux through the boundary (Neumann data).

In the literature many of the important questions concerning unique-ness, reconstruction, and stability have been answered affirmatively forsufficiently regular conductivities, but for less regular conductivities in-cluding merely bounded conductivities the questions still remain open.This thesis surveys the available methods and results. The emphasis is puton the following three aspects, where also new results are obtained:

Reconstruction of conductivities in the plane (Chapter 3): focusing onthe interplay between the two-dimensional inverse conductivity problemand complex analysis, a review of the uniqueness proof and reconstruc-tion algorithm due to Nachman for conductivities having essentially twoderivatives is given. Moreover, the uniqueness proof due to Brown andUhlmann for conductivities having essentially one derivative is expounded.Then by making each step constructive in the last uniqueness proof, a newreconstruction algorithm for less regular conductivities is proposed, andthis algorithm is compared to the method of Nachman’s. Furthermore, the

v

vi Summary

algorithm is implemented numerically and tested on synthetic, noiselessdata.

The inverse conductivity problem in higher dimensions (Chapter 4):the uniqueness proof for the higher dimensional problem is reviewed aswell as the existing reconstruction method. Then a result concerning ab-sence of exceptional points near zero for merely bounded conductivities isproved. As a byproduct of this result, a new algorithm for the reconstruc-tion of a sufficiently regular conductivity close to constant is proposed.

Inverse boundary value problem with a spherical potential (Chapter 5):here an inverse problem for the related Schrodinger equation in three di-mensions is considered. The problem is from a finite number of boundarymeasurements to reconstruct the potential. We prove that a square inte-grable and sufficiently small potential, which is assumed to be indepen-dent of the radial coordinate, can be reconstructed in a stable way by justtwo boundary measurements.

Dansk resume (summaryin Danish)

Denne afhandling omhandler det inverse ledningsproblem. Den fysiskeproblemstilling, der motiverer studierne af dette matematiske problem, eren nyere teknologi for billeddannelse til eksempelvis medicinsk diagnostikkaldet Elektrisk Impedans Tomografi. Ideen bag denne metode er udfrastatiske elektriske malinger pa overfladen af et objekt at bestemme den in-dre ledningsevne i objektet. Det bagvedliggende matematiske problem erda at bestemme og rekonstruere en koefficient i en elliptisk partiel differ-entialligning pa et glat og begrænset omrade udfra helt eller delvist kend-skab til den tilhørende Dirichlet-til-Neumann afbildning. Denne afbildninger defineret som afbildningen, der tager et spændingsfelt pa overfladenaf omradet (Dirichlet data) til den resulterende strøm gennem overfladen(Neumann data).

I litteraturen er mange af de vigtige spørgsmal omkring entydighed,rekonstruktion og stabilitet besvaret for tilstrækkeligt glatte ledningsevner,men for mindre regulære ledningsevner er flere af spørgsmalene stadigabne. Denne afhandling giver et overblik over metoder og resultater. Ho-vedvægten er især lagt pa tre aspekter af problemet, indenfor hvilke derogsa gives en række nye resultater:

Rekonstruktion af ledningsevner i planen (Kapitel 3): Fokus er her pasamspillet mellem det todimensionale inverse problem og kompleks ana-lyse. Der redegøres for Nachmans metode til rekonstruktion af ledning-sevner, der er to gange differentiable, og for et nyere entydighedsbevis afBrown og Uhlmann for ledningsevner, der kun er en gang differentiable.

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viii Dansk resume (summary in Danish)

Ved at lave hvert skridt i denne senere metode konstruktivt, gives en nyrekonstruktionsalgoritme for en klasse af mindre regulære ledningsevner.Endvidere gives en foreløbig numerisk implementation af denne algoritme.

Det inverse ledningsproblem i flere dimensioner (Kapitel 4): Der re-degøres for løsning af spørgsmalene omkring entydighed og rekonstruk-tion for det inverse problem i flere dimensioner. Endvidere gives et nyt re-sultat, som udelukker eksistensen af exceptionelle punkter nær origo. Somen konsekvens af dette resultat foreslas en ny rekonstruktionsalgoritme fortilpas glatte ledningsevner, som er næsten konstante.

Et inverst randværdiproblem med et sfærisk potentiale (Kapitel 5): Herbehandles et inverst problem for den beslægtede Schrodinger ligning medet sfærisk potentiale. Problemet er udfra et begrænset sæt af malinger paranden at rekonstruere potentialet. Det vises, at et kvadratisk integrabeltpotentiale, som ikke afhænger af den radiale variabel, og som desuden hartilpas lille norm, kan rekonstrueres udfra kun to malinger pa randen.

Contents

Preface iii

Summary v

Dansk resume (summary in Danish) vii

Chapter 1. Introduction 1

§1.1. The inverse conductivity problem 2

§1.2. Notes 5

Chapter 2. Preliminary reductions and results 7

§2.1. Reduction to the case γ = 1 near ∂Ω 7

§2.2. Reduction to a Schrodinger equation 10

§2.3. An integral identity 12

§2.4. Exponentially growing solutions 13

§2.5. Notes 24

Chapter 3. Uniqueness and reconstruction in two dimensions 29

§3.1. A few results from complex analysis 30

§3.2. Uniqueness and reconstruction for γ ∈ W2,p(Ω) 34

§3.3. Uniqueness for γ ∈ W1,p(Ω) 40

§3.4. Reconstruction of γ ∈ C1+ε(Ω) 44

§3.5. Comparison of the two reconstruction algorithms 50

§3.6. Numerical implementation of the algorithm 53

§3.7. Notes 66

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x Contents

Chapter 4. Uniqueness and reconstruction in higher dimensions 69§4.1. Uniqueness for the inverse conductivity problem 69§4.2. Absence of exceptional points near zero for γ ∈ L∞

+ (Ω) 71§4.3. Reconstruction of potentials and conductivities 76§4.4. Notes 81

Chapter 5. Local uniqueness, stability, and reconstruction from twoboundary measurements 85

§5.1. Outline of the method and results 86§5.2. Regularity of solutions 88§5.3. An equation on the boundary 91§5.4. Solving the boundary integral equation 94§5.5. Stability 96§5.6. Application to the conductivity problem 96§5.7. Notes 97

Appendix A. Notation 99

Bibliography 103

Chapter 1

Introduction

The inverse conductivity problem is the mathematical problem behind anew method for medical imaging called Electrical Impedance Tomogra-phy (EIT). In EIT one has a conductive body with unknown conductivity,and from static electric measurements on the boundary, i.e. by applying avoltage potential on the boundary and measuring the current flux throughthe boundary, one would like to monitor the interior conductivity. Sincemuscle tissue, fat tissue, bones, inner organs, lungs etc. have different con-ductive properties (see [BB90]), an image of the conductivity distributioninside a body may be used for medical diagnostics.

The expected advantages of this technology are many. First of all EITis inexpensive compared to other imaging technologies such as Comput-erized Tomography and Magnetic Resonance Imaging. A second perspec-tive is that images from EIT may complement the images obtained by theother technologies, i.e. a combination of images obtained from differenttechnologies may give a better basis for medical diagnostics. In other situ-ations where other technologies do not distinguish the features one wantsto observe, EIT may be the only alternative for medical imaging. A thirdperspective is that EIT is considered to be less harmful and more comfort-able to the patient. Finally, since the data collection in EIT can be done veryfast, it may be possible with EIT to monitor certain dynamic features inreal-time. The limitation of the technology is, however, that the expectedresolution of images obtained by EIT is very low when compared to im-ages obtained by the other imaging methods. This is in part due to theill-posedness of the underlying mathematical problem.

In the wide-spread literature on EIT, many concrete applications havebeen proposed, for example in non-invasive monitoring of blood-flow, of

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2 1. Introduction

heart and lung function, in the detection of pulmonary emboli and breastcancer, and in imaging of the brain function. Besides medical imaging EITmay also be useful in other imaging problems. We mention only geoelec-trical exploration of ground water flow [CS98], an application in the diag-noses of trees, where EIT is proposed as a method for estimating the extentof interior damage to trees, and hence the danger for the surroundings dueto falling, and a recent application in the determination of the fat marblingand hence the quality in beef. Even though there are many possible appli-cations, the technology is still considered to be at an early stage, and thereare many practical and theoretical problems to be solved before successfulcommercial applications of EIT can be seen. These problems include ques-tions concerning the design of the measurement device, the design of theexperiments, construction of the inversion algorithm. We refer to [CIN99]for a thorough review of the state of the art of EIT.

In this thesis the mathematical issues of EIT are considered. In the nextsection we will in greater detail introduce these mathematical problemsand give an overview of the contents of the thesis.

1.1. The inverse conductivity problem

Let Ω ⊂ Rn be an open and bounded subset with smooth boundary ∂Ω.Define the open subset

L∞+ (Ω) = γ ∈ L∞(Ω) : γ > 0 and γ−1 ∈ L∞(Ω)

of L∞(Ω) and assume henceforth that the conductivity γ ∈ L∞+ (Ω). Then

γ satisfies a uniform ellipticity condition, i.e. there is a θ > 0 such thatγ(x) ≥ θ > 0, a.e. x ∈ Ω. Assuming that there are no sources or sinksof current in Ω, the application of a voltage potential f on ∂Ω induces avoltage potential u inside Ω defined as the unique solution to the boundaryvalue problem

∇ · (γ∇u) = 0 in Ω, u = f on ∂Ω. (1.1)

The solvability of this elliptic boundary value problems is well-known.When f ∈ H1/2(∂Ω) there is a unique solution u ∈ H1(Ω) which can beestimated by

‖u‖H1(Ω) ≤ C(‖γ‖L∞(Ω)θ−1 + 1)‖ f ‖H1/2(∂Ω), (1.2)

where the constant C depends only on Ω. A proof of this statement can befound in any of the monographs [Eva98, GT83, Gri85]. The estimate (1.2) isobtained by keeping track of constants in the specialization of the generaltheory to the conductivity problem. The solution to (1.1) is slightly more

1.1. The inverse conductivity problem 3

regular in that the normal derivative on the boundary can be defined in acoherent way as an element of H−1/2(∂Ω) by

〈γ(∂νu)|∂Ω, g〉 =

∫

Ω

γ(x)∇u(x) · ∇v(x)dx, (1.3)

where v ∈ H1(Ω) satisfies v|∂Ω = g ∈ H1/2(∂Ω). Here 〈·, ·〉 denotes thedual pairing between Hs(∂Ω) and its dual space H−s(∂Ω). Using that u isa weak solution to (1.1), it is straightforward to see that the definition (1.3)is independent of the extension v of g. Note that when u, v ∈ C2(Ω) andγ ∈ C1(Ω) integration by parts in (1.3) gives

〈γ(∂νu)|∂Ω, g〉 =

∫

∂Ω

γ(x)∂νu(x)v(x)dσ(x),

where dσ is the usual Euclidean surface measure on ∂Ω. Hence the general-ized definition (1.3) of a normal derivative at the boundary coincides withthe classical definition, when this makes sense.

The distribution γ(∂νu)|∂Ω describes the current flux through the bound-ary, and it is the natural Neumann data for the equation (1.1). Thus we candefine the Dirichlet-to-Neumann map Λγ : H1/2(∂Ω) → H−1/2(∂Ω) by

Λγ f = γ(∂νu)|∂Ω. (1.4)

This map encodes all possible boundary measurements which can in prin-ciple be considered for EIT.

Consider the map

Λ : γ 7→ Λγ

defined on L∞+ (Ω) with values in B(H1/2(∂Ω), H−1/2(∂Ω)). This map is

continuous with respect to the norm topologies on the Banach spaces L∞(Ω)

and B(H1/2(∂Ω), H−1/2(∂Ω)) (see section 2.3 below), and hence the prob-lem of computing Λγ from γ is well-posed in the sense of Hadamard [Had53],i.e. for any γ ∈ L∞

+ (Ω) there is a well-defined Dirichlet-to-Neumann mapwhich depends continuously on γ. This problem is our forward problem.Note that Λ is a non-linear map since the integrand in the right hand sideof (1.3) depends on γ in a non-linear way.

The inverse conductivity problem concerns the inverse of the map Λ.There are several important questions:

• The injectivity of Λ (uniqueness).

• The continuity of Λ−1 (stability).• Reconstruction of γ from Λγ.• Numerical realization of a reconstruction algorithm.• Determination of the range of Λ (existence).

4 1. Introduction

• Finite number of measurements, i.e. the action of Λγ on a finitedimensional subspace of H1/2(∂Ω) is known.

The uniqueness question is extremely relevant, since an affirmative answertells that two different conductivities can in principle be distinguished fromboundary measurements, and hence EIT is in principle possible. Stabilityis important for the analysis of the propagation of errors in the data andalso for numerical algorithms. The reconstruction question is relevant bothfrom a theoretical and applied point of view, and a numerical realizationof an algorithm may on one hand be a practical tool in applications and onthe other hand give new insight into the theoretical problem. Concerningthe existence question little seems to be known, even though it is a veryinteresting mathematical question. Moreover, in applications it may be ad-vantageous to project noisy data onto the range of Λ before an inversionis carried out. Finally the question of having only a finite number of mea-surements is in some sense more realistic and a greater understanding ofthis problem and the limitations of finite data is definitely useful. Note thatthe issues listed are by no means independent. Indeed, as we shall see, auniqueness proof can provide a stable reconstruction algorithm, which canbe realized numerically.

An interesting fact about the inverse conductivity problem is that theamount of data given in Λγ is depending on the dimension of the un-derlying space. The data Λγ is a map from a space with n − 1 variablesinto a space with the same number of variables, so the Schwartz kernel ofthat operator is a function of 2 ∗ (n − 1) variables. On the other hand thefunction γ, we are looking to recover, is depending on n variables. So, indimension n = 2 the problem is formally determined, whereas in higherdimensions it is over determined. This motivates somewhat that to solvethe two-dimensional problem we have to invoke a different method thanthe one used for the higher dimensional problem.

In the literature many of the inverse problems have been treated, andfor sufficiently regular conductivities affirmative answers have been given.However, when the conductivity is merely bounded, only little seems tobe know; this is the starting point of this project. In this thesis we willboth review available results and give a number of new results. The em-phasis will be put on the uniqueness and the reconstruction questions forthe inverse conductivity problem in both two and higher dimensions, andon the reconstruction problem for a related Schrodinger equation from twoboundary measurements.

The thesis is divided into 4 chapters. In Chapter 2 we describe twobasic reductions of the inverse conductivity problem, which simplifies theproblem. Then we will introduce the fundamental concept of exponentially

1.2. Notes 5

growing solutions and the notion of exceptional points, and we will see thatthese concepts are equivalent, and moreover characterized by a boundaryintegral equation.

In Chapter 3 we draw the attention to the two-dimensional problem.Focusing on the interplay between the two-dimensional inverse conduc-tivity problem and complex analysis we will review the uniqueness proofand reconstruction algorithm due to Nachman for conductivities having es-sentially two derivatives. Moreover, we will review the uniqueness proofdue to Brown and Uhlmann for conductivities having essentially one de-rivative, and by making each step constructive in the last uniqueness proofwe propose a reconstruction algorithm for a class of less regular conduc-tivities. This reconstruction algorithm is then compared to the method ofNachman’s. Furthermore, the algorithm is implemented numerically andtested on synthetic, noiseless data. The theoretical results in this chapterare in part joint with Alexandru Tamasan, University of Washington, andthe implementation is joint with Samuli Siltanen, Instrumentarium Corp.,Finland.

In Chapter 4 we focus on the three-dimensional problem and consideralso here the uniqueness and reconstruction issues. The uniqueness prooffor the higher dimensional problem is reviewed as well as the reconstruc-tion method due to Nachman. Moreover, a new result concerning ab-sence of exceptional points near zero for merely bounded conductivitiesis proved. As a byproduct of this result, a new reconstruction algorithm forsufficiently small and regular conductivities is proposed. These results areobtained jointly with Horia Cornean, Aalborg University, Denmark, andSamuli Siltanen, Instrumentarium Corp., Finland.

Finally in Chapter 5 we study a related inverse problem for a Schro-dinger equation in three dimensions with a spherical potential. We provethat if the the potential is square integrable, sufficiently small, and indepen-dent of the radial coordinate, then it can be reconstructed in a stable way byjust two boundary measurements. Moreover, we see how this result relatesto the inverse conductivity problem. This result is joint work with HoriaCornean, Aalborg University, Denmark.

1.2. Notes

The mathematical formulation of the inverse conductivity problem as de-scribed here was given by Calderon in [Cal80], a paper that put the di-rection for an intensive research in a new area in applied PDE. Calderonwas motivated by a problem in geophysics he worked on as an engineerin the 1940’s. He considered instead of the Dirichlet-to-Neumann map theunbounded quadratic form Qγ in L2(∂Ω) given on the domain H1/2(∂Ω)

6 1. Introduction

by

Qγ( f ) =

∫

Ω

γ(x)|∇u(x)|2dx, (1.5)

where u solves (1.1). This form can be interpreted as the power needed tomaintain the voltage potential f on ∂Ω. We note that since Qγ is positive,symmetric and closed, it defines a unique selfadjoint operator in L2(∂Ω),which is exactly Λγ. We will return to the results of Calderon in section 2.5.Although Calderon was the first author to state the problem as we do here,the problem has roots far back. In fact Langer considered a related problemin 1933 [Lan33]. For a thorough review on the history and developmentsof the inverse conductivity problem we refer to the review papers [SU90,IN99, Uhl99].

We note that the same inverse problems can be stated in case the con-ductivity is anisotropic, i.e. when the coefficient γ = (γij)ij entering (1.1) isa strictly positive definite n × n matrix. In fact in medical imaging the con-ductive properties of certain tissue does depend on directions; for instancemuscle tissue has different conductivity in the longitudinal and transversaldirections. For the anisotropic inverse conductivity problem uniqueness isknown not to hold. Let Φ : Ω 7→ Ω be a diffeomorphism with Φ|∂Ω = I.Then for

Φ∗γ(x) =(DΦ)Tγ(x)(DΦ)

|DΦ| Φ−1(x),

where DΦ is the Jacobian of Φ, it can be seen [KV84b] that Λγ = ΛΦ∗γ.Hence this kind of diffeomorphic transformations violates uniqueness. Intwo dimensions these transformations are known to be the only obstructionto uniqueness [Syl90, Nac96], but in higher dimensions this question isopen. We refer to [Uhl00] and the references given there for anisotropicinverse problems.

The assumptions that ∂Ω is smooth is taken mainly for simplicity. Mostresults to be presented are valid for more general domains, for instancewhen ∂Ω is assumed only to be Lipschitz. However, this assumption is, aswe shall see in Chapter 2, in general not a restriction.

Chapter 2

Preliminary reductionsand results

In this chapter we describe two reductions of the inverse conductivity prob-lem to problems, which are in some sense simpler. Furthermore, we definethe important concepts of exponentially growing solutions and exceptionalpoints, and give a number of results concerning these objects.

The outline of the chapter is the following: in section 2.1 the generalinverse problem is reduced to a special problem, where the conductivity isconstant near the boundary. In section 2.2 the conductivity equation is thenreduced to a Schrodinger equation, and for that equation a related inverseproblem is posed. In section 2.3 we derive a simple integral identity, whichlinks a difference of two Dirichlet-to-Neumann maps to the difference ofthe conductivities. This identity motivates the interest in finding solutionsto the equations with exponential growth at infinity, the so-called exponen-tially growing solutions. The construction and properties of such solutionsare the contents of section 2.4.

2.1. Reduction to the case γ = 1 near ∂Ω

In this section the inverse conductivity problem is reduced to a problem,where the conductivity is constant near the boundary of the domain. Themain idea behind this reduction is based on three steps. First the bound-ary value of the unknown conductivity and its derivatives are calculatedfrom the Dirichlet-to-Neumann map. Then from the boundary values theconductivity is extended outside the domain of interest in a way such thatthe regularity is preserved. Finally, the Dirichlet-to-Neumann map on the

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8 2. Preliminary reductions and results

boundary of some larger domain is computed from the known extensionand the original Dirichlet-to-Neumann map. We emphasize that this re-duction is constructive.

The first step concerns reconstruction of the conductivity at the bound-ary. Results of this kind rely both on the regularity of the conductivity andthe regularity of the boundary of the domain. A formulation suitable forour purpose is the following.

Lemma 2.1.1. Assume that γ ∈ W1,r(Ω), r > n. Then the boundary value γ|∂Ω

can be reconstructed from Λγ. If further γ ∈ W2,p(Ω)∪ C1+ε(Ω), p > n/2, ε >

0, then (∂νγ)|∂Ω can be computed from Λγ.

Proof. A proof of the result concerning reconstruction of conductivities inSobolev spaces can be found in [Nac96], and a proof of the result concern-ing Holder continuous conductivities can be found in [KY01].

Note that by the Sobolev embedding theorem the conductivities in Lem-ma 2.1.1 are continuous functions on Ω.

The next results concerns the extension to Rn of a function given onthe boundary of a domain. We require in this context that the procedure isconstructive, and that the extension relies only on the boundary values ofthe function and its derivatives.

Lemma 2.1.2. Let Ω be a smooth bounded domain and let u ∈ Cl+ε(Ω), ε > 0.Then from the trace (u|∂Ω, (∂νu)|∂Ω, . . . , ((∂ν)

lu)|∂Ω) ∈ ∏lj=0 Cl−j+ε(∂Ω), we

can construct function u′ ∈ Cl+ε(R2) such that u′ = u on Ω.Furthermore, let u ∈ Ws,p(Ω) and let l ∈ Z+ satisfy s − l − 1/p > 0.

Then from (u|∂Ω, (∂νu)|∂Ω), . . . , ((∂ν)lu)|∂Ω) ∈ ∏

lj=0 Ws−j−1/p,p(∂Ω) we can

construct a function u′ ∈ Ws,p(Rn) such that u′ = u on Ω.

Proof. The first claim follows from the construction of the extension oper-ator of Whitney type [Ste70, pp. 176], which takes into account only theboundary values of the function and its derivative.

To prove the second claim we use the fact that the trace operator

u 7→ (γ|∂Ω, (∂νu)|∂Ω, . . . , ((∂ν)lu)|∂Ω)

is bounded from Ws,p(Rn) into ∏lj=0 Ws−j−1/p,p(∂Ω) and has a continuous

right inverse E (see for instance [Gri85, Theorem 1.5.1.2]). Then we definethe extension u′ by

u′ =

u on Ω,E(γ|∂Ω, (∂νu)|∂Ω, . . . , ((∂ν)

lu)|∂Ω) on Rn \ Ω.

This function is easily seen to be in Ws,p(Rn).

2.1. Reduction to the case γ = 1 near ∂Ω 9

When an extension has been constructed, it is straightforward to mod-ify this function by multiplying and adding smooth functions supportedaway from Ω and thereby construct another extension u, with the propertythat u − 1 has compact support.

We will finally prove that if the Dirichlet-to-Neumann map is knownat the boundary of some domain and the conductivity is known in the ex-terior, then the Dirichlet-to-Neumann map at the boundary of a larger en-capsulating domain can be computed.

Lemma 2.1.3. Let Ωe, Ω ⊂ Rn be smooth and bounded domains such that Ω ⊂Ωe. Let γ ∈ L∞

+ (Ωe), let Λγ be the Dirichlet-to-Neumann map on ∂Ω correspond-ing to γ|Ω and let Λγe be the Dirichlet-to-Neumann map on ∂Ωe corresponding toγ. Then Λγe can be reconstructed from Λγ and γ|Ωe\Ω.

Proof. From the definition of the Dirichlet-to-Neumann map we have forany f , g ∈ H1/2(∂Ωe), that

〈Λγe f , g〉 =

∫

Ωe

γ∇u · ∇v

=

∫

Ωe\Ω

γ∇u · ∇v + 〈Λγ(u|∂Ω), v|∂Ω〉,

where u ∈ H1(Ωe) denotes the unique solution to

∇ · γ∇u = 0, in Ωe,

u = f , on ∂Ωe,(2.1)

and v ∈ H1(Ωe) is any function with v|∂Ωe = g. Hence we see that Λγe

can be found from γ|Ωe\Ω and Λγ without explicit knowledge of γ in Ω

provided that the solution u to (2.1) can be found in Ωe \ Ω.

We claim that u in Ωe \Ω can be found as the unique solution to bound-ary value problem

∇ · γ∇u = 0 in Ωe \ Ω,

u = f on ∂Ωe,

γ(∂νu)|∂Ω = Λγ(u|∂Ω) on ∂Ω.

(2.2)

Here γ(∂νu)|∂Ω ∈ H−1/2(∂Ω) is the normal derivative on ∂Ω defined in theweak sense for g ∈ H1/2(∂Ω) as

〈γ(∂νu)|∂Ω, g〉 =

∫

Ω\Ω

γ∇u · ∇v, (2.3)

with v ∈ H1(Ω \ Ω), v|∂Ω = g, v|∂Ωe = 0.


That u|Ωe\Ω solves (2.2) is trivial. For the uniqueness we assume thatu0 ∈ H1(Ωe \ Ω) solves (2.2) with f = 0. Extend u0 into Ω as the solution to

∇ · γ∇u = 0 in Ω, u = u0 on ∂Ω.

Since the extended u0 is in H1(Ωe) and solves (2.1) with f = 0, we concludethat u0 = 0 in Ωe.

Note that the solution to (2.2) can be constructed explicitly using forinstance pseudodifferential calculus, see [Gru96].

The conclusion of this section is now that if the domain Ω is smooth andthe conductivity is known a priori to be in either W1,p(Ω), p > n, W2,p(Ω),p > n/2, or C1+ε(Ω), ε > 0, then the inverse problem can be transformedinto an inverse problem on a larger domain (a ball for instance) for an ex-tended conductivity γ, which has exactly the same regularity as before, andthe property that γ = 1 near the boundary of the larger domain. Hence ifwe can solve the reduced problem on the larger domain, we will implicitlysolve the original problem.

2.2. Reduction to a Schrodinger equation

In this section we review the well-known transformation of the conductiv-ity problem (1.1) to a Dirichlet problem for a Schrodinger operator, and forthis operator we pose an inverse problem. This reduction is central in thesolution of the inverse conductivity problem, and moreover it motivatesthe studies in Chapter 5.

Since the reduction requires a certain amount of smoothness, we willassume that γ ∈ W2,∞(Ω). Let u be a solution to ∇ · γ∇u = 0 and introducev = γ1/2u. Then

0 = ∇ · γ(v∇γ−1/2 + γ−1/2∇v)

= v∇ · γ∇γ−1/2 + (γ∇γ−1/2 + ∇γ1/2) · ∇v + γ1/2∆v

= −v∆γ1/2 + γ1/2∆v,

which implies that

(−∆ + q)v = 0 (2.4)

for

q = γ−1/2∆γ1/2. (2.5)

Conversely we could take any solution v to (2.4) and show that u = γ−1/2vsolves the conductivity equation. Hence there is a one to one correspon-dence between solutions to the conductivity equation and solutions to the

2.2. Reduction to a Schrodinger equation 11

Schrodinger equation. This shows that the problem

(−∆ + q)v = 0 in Ω, v = f on ∂Ω (2.6)

has a unique solution v ∈ H1(Ω) for any f ∈ H1/2(∂Ω), i.e. zero is not aDirichlet eigenvalue for (−∆ + q).

Consider now (2.6) with a general potential q ∈ L∞(Ω), and assumethat zero is not a Dirichlet eigenvalue for (−∆ + q). Define the Dirichlet-to-Neumann map Λq : H1/2(Ω) → H−1/2(Ω) associated with q by

Λq f = (∂νv)|∂Ω, (2.7)

where v ∈ H1(Ω) is the unique solution to (2.6) with trace f ∈ H1/2(∂Ω).Here the normal derivative on the boundary is defined in the weak senseby

〈(∂νv)|∂Ω, g〉 =

∫

Ω

∇v · ∇w + qvw, (2.8)

where w ∈ H1(Ω) has trace g ∈ H1/2(∂Ω) (cf. (1.3)).A useful fact is that when q is defined from (2.5), then Λq can be recov-

ered from Λγ. Let u ∈ H1(Ω) denote the unique solutions to ∇ · γ∇u = 0with boundary value γ−1/2 f ∈ H1/2(∂Ω). Then γ1/2u solves (2.6), and forw ∈ H1(Ω) with w|∂Ω = g ∈ H1/2(∂Ω) we have that

〈Λq f , g〉 =

∫

Ω

∇(γ1/2u) · ∇w +

∫

Ω

∆γ1/2uw

=

∫

Ω

(γ∇u · ∇(γ−1/2w) + ∇γ1/2 · ∇(uw))

+

∫

∂Ω

(∂νγ1/2)uw −∫

Ω

∇γ1/2 · ∇(uw)

= 〈Λγ(γ−1/2 f ), γ−1/2g〉 + 〈(∂νγ1/2)(γ−1/2 f ), g〉.Therefore the Dirichlet-to-Neumann map Λq associated to the Schrodingerequation can be obtained from the Dirichlet-to-Neumann map Λγ associ-ated to the conductivity equation by first calculating the boundary valuesof γ and ∂νγ by Lemma 2.1.1 and then Λq by

Λq = γ−1/2(

12

∂νγ + Λγ

)

(γ−1/2·). (2.9)

Note that when γ = 1 near ∂Ω then (∂νγ1/2)|∂Ω = 0 and hence Λq = Λγ.We can now pose the inverse problems concerning injectivity, stability,

invertibility etc. of the map

Λ : L∞ → B(H1/2(∂Ω), H−1/2(∂Ω))

q 7→ Λq.


As we shall see below in section 4.1, an answer to this inverse problem willgive at least a partial answer to the inverse conductivity problem.

We note that for a general potential q ∈ L∞(Ω), zero may be a Dirich-let eigenvalue of (−∆ + q), and then the Dirichlet-to-Neumann map is notwell-defined. The relevant data for the inverse problem is then the set ofCauchy data given by

Cq = (u|∂Ω, (∂νu)|∂Ω)∣

∣ u ∈ H1(Ω), (−∆ + q)u = 0 in Ω. (2.10)

For a potential with a well-defined Dirichlet-to-Neumann the set of Cauchydata is just the graph of this map, i.e.

Cq = ( f , Λq f ) ∈ H1/2(∂Ω) × H−1/2(∂Ω).

We will return to this in section 4.1.

2.3. An integral identity

The inverse conductivity problem is a problem where one from cleverlychosen Dirichlet data and the observation of the resulting Neumann datalooks for a function inside a domain. The next result is an integral iden-tity, which turns the focus from the boundary values of solutions to thesolutions inside the domain.

Lemma 2.3.1. Let γ1, γ2 ∈ L∞(Ω). Then for any solutions u1, u2 ∈ H1(Ω) to(1.1) with boundary values f1, f2 ∈ H1/2(∂Ω) we have

〈(Λγ1 − Λγ2) f1, f2〉 =

∫

Ω

(γ1 − γ2)∇u1 · ∇u2. (2.11)

Proof. This is a direct consequence of the symmetry of the Dirichlet-to-Neumann map, i.e.

〈(Λγ1 − Λγ2) f1, f2〉 = 〈Λγ1 f1, f2〉 − 〈Λγ2 f2, f1〉 =

∫

Ω

(γ1 − γ2)∇u1 · ∇u2.

Note that (2.11) implies that the forward map Λ : γ 7→ Λγ is continu-ous. In fact the estimate (1.2) gives that

‖Λγ1 − Λγ2‖B(H1/2(∂Ω),H−1/2(∂Ω)) ≤ C‖γ1 − γ2‖L∞(Ω), (2.12)

where the constant C = C1(‖γ1‖L∞ /θ1 + 1)(‖γ2‖L∞ /θ2 + 1), C1 dependsonly on Ω, and θi is the ellipticity constant for γi, i = 1, 2.

For the Dirichlet-to-Neumann maps associated the Schrodinger equa-tion we find a similar equation. In this case the relation is

〈(Λq1 − Λq2) f1, f2〉 =

∫

Ω

(q1 − q2)u1u2, (2.13)

2.4. Exponentially growing solutions 13

where ui ∈ H1(Ω) solve (−∆ + qi)ui = 0 with fi = ui|∂Ω, i = 1, 2.The identity (2.11) is especially useful in relation to the uniqueness

question. If two Dirichlet-to-Neumann maps agree then the difference ofthe underlying coefficients multiplied by a product of gradients of solu-tions to the relevant equations will integrate to zero. Hence uniquenesswill follow, if products of gradients of solutions to the conductivity prob-lem are dense in L1(Ω).

2.4. Exponentially growing solutions

In the following we will consider the Schrodinger equation and the con-ductivity equation in the entire space Rn. Tacitly we assume that the givenpotential and conductivity is extended by zero and one respectively outsidethe domain Ω of interest.

Let V = Cn \ 0 | ζ · ζ = 0, where ζ · ζ = ∑nj=1 ζ2

j is the real innerproduct. With this definition eix·ζ is harmonic if and only if ζ ∈ V . The ideadue to Sylvester and Uhlmann [SU87] is then to look for a family of specialsolutions ψ(x, ζ), ζ ∈ V , to the Schrodinger equation, which are asymptoti-cally exponential, i.e.

(−∆ + q)ψ(x, ζ) = 0 in Rn,

ψ(x, ζ) ∼ eix·ζ when |x| → ∞.(2.14)

The asymptotic property means that the function ω defined by

ω(x, ζ) = µ(x, ζ) − 1, (2.15)

with

µ(x, ζ) = e−ix·ζψ(x, ζ), (2.16)

decays to zero when x tends to infinity. Note that by plugging (2.15) into(2.14) we get the equation

(−∆ − 2iζ · ∇)ω(x, ζ) + q(x)ω(x, ζ) = −q(x) in Rn. (2.17)

The function ψ is called an exponentially growing solution or a complexgeometrical optics solution.

In this section we outline the construction and analyze the propertiesof exponentially growing solutions. First we will define Faddeev’s Green’sfunctions and review the properties of these. This leads then to the con-struction of exponentially growing solutions and the definition of the so-called exceptional points. We then characterize the exceptional points interms of the solvability of a certain boundary integral equation, an equa-tion that enables a coherent definition of exponentially growing solutions


and exceptional points in relation to the conductivity equation with a non-smooth conductivity.

2.4.1. Faddeev’s Green’s functions. Consider the equation

(−∆ − 2iζ · ∇)u = f

and note that on S ′(Rn)

(−∆ − 2iζ · ∇)u = F−1((|ξ|2 + 2ξ · ζ)u(ξ)).

Hence formally

(−∆ − 2iζ · ∇)−1 f = F−1((|ξ|2 + 2ξ · ζ)−1u(ξ)) = gζ ∗ f ,

where

gζ(x) =1

(2π)n

∫

Rn

eix·ξ

|ξ|2 + 2ξ · ζdξ, Gζ(x) = eix·ζ gζ(x) (2.18)

are the Faddeev’s Green’s functions for (−∆ − 2iζ · ∇) and (−∆) respec-tively. Note that for fixed ζ ∈ V , the zeroes of |ξ|2 + 2ξ · ζ are located inthe hyper plane ξ · Im ζ = 0 on the circle |ξ + Re ζ| = |Re ζ|. Moreover,(|ξ|2 + 2ξ · ζ)−1 ∈ L1

loc(Rn), and hence the integral in the definition of gζ

makes sense when interpreted as the inverse Fourier transform defined onthe space of tempered distributions.

We now collect a few useful facts about gζ. First note that ζ = Re(ζ) +

i Im(ζ) ∈ V satisfies ζ · ζ = Re(ζ)2 − Im(ζ)2 + 2i Re(ζ) · Im(ζ) = 0, whichimplies that

|Re(ζ)| = | Im(ζ)|, Re(ζ) · Im(ζ) = 0.

Hence ζ has the form

ζ = κ(k⊥ + ik), (2.19)

where k⊥, k ∈ Rn, |k⊥| = |k| = 1/√

2, k · k⊥ = 0, and |ζ| = κ. We have nowthe following result, which is useful in the analysis of gζ for growing κ.

Proposition 2.4.1. Let ζ ∈ V be decomposed as (2.19). Then

gζ(x) = κn−2gk⊥+ik(κx). (2.20)

Let further R be a real orthogonal n × n matrix with det(R) = 1. Then

gζ(x) = gRζ(Rx). (2.21)

Proof. We prove the results by manipulating the formal integral (2.18) (arigorous but tedious proof by distribution theory can also be given). The


substitution ξ = ξ/κ implies

gζ(x) =1

(2π)n

∫

Rn

eix·ξ

|ξ|2 + 2ζ · ξdξ

=1

(2π)n

∫

Rn

eiκx·ξ

κ2(|ξ|2 + 2 ζκ · ξ)

κndξ

= κn−2gk⊥+ik(κx),

which shows (2.20).To prove (2.21), write ξ′ = RTξ and dξ′ = dξ to get

gζ(Rx) =1

(2π)n

∫

Rn

ei(Rx)·ξ

|ξ|2 + 2ζ · ξdξ

=1

(2π)n

∫

Rn

eix·(RTξ)

|ξ|2 + 2ζ · ξdξ

=1

(2π)n

∫

Rn

eix·ξ′

|(RT)−1ξ′|2 + 2ζ · ((RT)−1ξ′)dξ′

=1

(2π)n

∫

Rn

eix·ξ′

|ξ′|2 + 2(R−1ζ) · ξ′dξ′

= gR−1ζ(x).

The next result concerns the mapping properties of the convolutionoperator with kernel gζ. Consult the appendix for the definition of theweighted Sobolev spaces Hs

δ(Rn).

Proposition 2.4.2. Let n ≥ 2, ζ ∈ V with |ζ| > ε > 0 and take −1 < δ < 0.Then there is a constant C > 0 depending on δ, ε, n such that

‖gζ ∗ f ‖Hsδ(Rn) ≤

C|ζ| ‖ f ‖Hs

δ+1(Rn), (2.22)

‖gζ ∗ f ‖Hsδ(Rn) ≤ C‖ f ‖Hs−1

δ+1(Rn), (2.23)

for any s ≥ 0.

Proof. The key estimate (2.22) is proved in [SU87]. By a careful analy-sis in the Fourier domain it is proved that locally gζ∗ can be seen as theFourier multiplier given by the function (ξ1 + iξ2)

−1. An estimate for thistwo-dimensional operator due to Nirenberg and Walker [NW73] gives theresult.

The estimate (2.23) is proved using (2.22), see [Bro96].


The estimates (2.22) and (2.23) are sufficient for our purpose. Consultthe notes for further results along this line.

2.4.2. Existence of exponentially growing solutions and definition of ex-ceptional points. The question considered in this section concerns the ex-istence and uniqueness of exponentially growing solutions. To proceed weconvolve with gζ in (2.17) and obtain the so-called Lippmann-Schwinger-Faddeev integral equation

(I + gζ ∗ (q·))ω(·, ζ) = gζ ∗ q, (2.24)

which is essentially equivalent to the equation for ψ

ψ(x, ζ) +

∫

RnGζ(x − y)q(y)ψ(y, ζ)dy = eix·ζ. (2.25)

We have then the following fundamental result concerning the existence ofexponentially growing solutions for large |ζ| :

Theorem 2.4.3. Let q ∈ L∞(Ω) and let −1 < δ < 0. Then there is a constantC > 0 depending only on Ω and δ such that for |ζ| > C‖q‖L∞(Ω), there is a uniquesolution ω(·, ζ) ∈ H1

δ (Rn) to (2.17). Furthermore, we have the estimate

‖ω(·, ζ)‖L2δ(Rn) ≤

C|ζ| ‖q‖L∞(Ω). (2.26)

Proof. We intend to solve (2.24) in L2δ(Rn) Let |ζ| > ε > 0. Since multiplica-

tion by the compactly supported q ∈ L∞(Rn) maps L2δ(Rn) into L2

δ+1(Rn),the operator gζ ∗ (q·) is bounded on L2

δ(Rn) with norm

‖gζ ∗ (q·)‖B(L2δ(Rn)) ≤

C|ζ| ‖q‖L∞(Rn).

Hence the operator (I + gζ ∗ (q·)) can be inverted in L2δ(Rn) by a Neumann

series whenever |ζ| > C‖q‖L∞(Ω). Note that q ∈ L∞(Ω) extended by zerointo Rn \ Ω is in L2

δ+1(Rn). This implies by (2.22) that gζ ∗ q ∈ H1δ (Rn) and

we can then define

ω(·, ζ) = (I + gζ ∗ (q·))−1(gζ ∗ q),

and derive the norm estimate of ω(·, ζ) from

‖gζ ∗ q‖L2δ(Rn) ≤

C|ζ| ‖q‖L∞(Ω).

Since gζ ∗ q ∈ H1δ (Rn) and gζ ∗ (qω) ∈ H1

δ (Rn) it follows that ω ∈ H1δ (Rn).


We note that a simple bootstrapping argument shows that in fact ψ ∈H2

loc(Rn) and smooth wherever q = 0.Having established the existence and uniqueness of exponentially grow-

ing solutions for sufficiently large |ζ|, we now consider the same questionsfor general ζ ∈ V . It turns out that this question is highly related to theexistence of a non-vanishing solution to the homogeneous equation

(−∆ − 2iζ · ∇)ω0(x, ζ) + q(x)ω0(x, ζ) = 0 in Rn. (2.27)

Lemma 2.4.4. Let ζ ∈ V . Then there is a unique exponentially growing solutionto (2.14) with e−ix·ζψ(x, ζ) − 1 ∈ H1

δ (Rn) if and only if (2.27) have in H1δ (Rn)

only the trivial solution.

Proof. Decompose gζ ∗ (q·) on H1δ (Rn) as

H1δ (R

n)r→ H1(Ω)

j→ L2(Ω)

q·→ L2(Ω)e→ L2

δ+1(Rn)

gζ∗·→ H1δ (R

n),

where r is the restriction operator from Rn to Ω, j is the identity operator,e is the extension outside Ω by zero. Since j : H1(Ω) 7→ L2(Ω) is compactby Rellich’s theorem (see for instance [Eva98]), the compound operator iscompact on H1

δ (Rn). Moreover, the right hand side gζ ∗ q ∈ H1δ (Rn) by

(2.23). Hence (2.24) is a Fredholm equation of the second kind in H1δ (Rn), so

if the homogeneous equation (2.27) has only the trivial solution then I + gζ ∗(q·) is invertible by the Fredholm alternative. This proves the claim.

The idea of converting the question of existence of exponentially grow-ing solutions into a uniqueness question for a Fredholm equation of thesecond kind is in particular useful for the two-dimensional inverse prob-lem for potentials q coming from a conductivity, since ideas from complexanalysis (see Chapter 3) provide us with a uniqueness proof. In dimen-sion n ≥ 3 the uniqueness question seems harder to settle and hence thisapproach has not yet been useful.

Those points ζ ∈ V , for which there is no unique exponentially growingsolution, are called exceptional. By Lemma 2.4.4 these are the points where(2.27) has a non trivial solution ω0 ∈ H1

δ (Rn). Since ψ0 = eix·ζω0 solves theSchrodinger equation we define:

Definition 2.4.5. A point ζ ∈ V is said to be exceptional for q, if there is a non-vanishing solution ψ0(·, ζ) to (2.14) with e−ix·ζψ0(x, ζ) ∈ H1

δ (Rn).

The content of Theorem 2.4.3 can be formulated in terms of absence ofexceptional points:

Corollary 2.4.6. Let q ∈ L∞(Ω). Then there is a constant C depending on q suchthat there are no exceptional points ζ for q with |ζ| > C.


We will later (Theorem 4.2.4) the possibility of having exceptional pointsnear ζ = 0.

2.4.3. A boundary integral equation. Let Sζ be the single layer potentialcorresponding to Faddeev’s Green’s function defined initially on C(∂Ω) by

Sζφ(x) =

∫

∂Ω

Gζ(x − y)φ(y)dσ(y), x ∈ Rn \ ∂Ω, (2.28)

and let

Sζφ(x) =

∫

∂Ω

Gζ(x − y)φ(y)dσ(y), x ∈ ∂Ω

be the boundary single layer potential. Denote by G0 the standard Green’sfunction for (−∆) defined by

G0(x) = − 12π

log |x|, n = 2, (2.29)

G0(x) =1

n(n − 2)α(n)|x|n−2 , n ≥ 3, (2.30)

where α(n), n ≥ 3, is the volume of the unit ball in Rn. Since Gζ − G0 isharmonic and hence smooth in Rn, the operators Sζ, Sζ inherit many usefulproperties from the standard single layer potentials

S0φ(x) =

∫

∂Ω

G0(x − y)φ(y)dσ(y), x ∈ Rn \ ∂Ω, (2.31)

S0φ(x) =

∫

∂Ω

G0(x − y)φ(y)dσ(y), x ∈ ∂Ω.

We collect a few facts in the following proposition. To fix notation we definethe boundary operator K′

ζ by

K′ζφ(x) = p.v.

∫

∂Ω

∂Gζ(x − y)

∂ν(x)φ(y)dσ(y). (2.32)

Moreover, for u ∈ C1(Rn \ ∂Ω) we write ∂ν±u = limε→0 ν · ∇u(x ± εν)|∂Ω

for the interior and exterior normal derivatives, and use the weak extensionfor less regular functions (cf. (2.3)). Then we have

Proposition 2.4.7. Let −1 ≤ s ≤ 0. Then Sζ ∈ B(Hs(∂Ω), Hs+1(∂Ω)) andSζ ∈ B(Hs(∂Ω), Hs+3/2(Ω)). For φ ∈ Hs(∂Ω) and u = Sζφ we have

(1) ∆u = 0 in Rn \ ∂Ω.

(2) The trace of u on ∂Ω is given by

u|∂Ω = Sζφ. (2.33)


(3) The operator K′ζ ∈ B(Hs(∂Ω)), and the normal derivative has across ∂Ω

the jump

∂ν±u(x) = (∓12

+ K′ζ)φ(x). (2.34)

In particular

(∂ν− − ∂ν+)u(x) = φ(x), x ∈ ∂Ω. (2.35)

Proof. Since Gζ − G0 is harmonic it is sufficient to validate the propertiesfor S0. It is a classical result that S0 ∈ B(L2(∂Ω), H1(∂Ω)), and that S0φis harmonic in Rn \ ∂Ω and has trace S0φ|∂Ω = S0φ for φ ∈ L2(∂Ω) (see[CK98] for the case of a smooth boundary and [Ver84] for the case of aLipschitz boundary). This shows that S0 ∈ B(L2(∂Ω), H3/2(Ω)).

Since the adjoint S′0 ∈ B(L2(∂Ω), H1(∂Ω)), a duality argument shows

that S0 ∈ B(H−1(∂Ω), L2(∂Ω)). The claim S0 ∈ B(Hs(∂Ω), Hs+1(∂Ω)),−1 < s < 0, then follows by interpolation (see for instance [BL76]). Now(1) follows by an approximation argument and (2) follows by elliptic regu-larity.

It is also classical that K′0 ∈ B(L2(∂Ω)) (defined by (2.32) with Gζ re-

placed by G0) and that the L2(∂Ω) adjoint is bounded on H1(∂Ω). By du-ality we then see that K′

0 ∈ B(H−1(∂Ω)), and by interpolation that K′0 ∈

B(Hs(∂Ω)) for −1 ≤ s ≤ 0. Hence (2.34) and (2.35) follows by a densityargument.

As a consequence of (2.33) we will abuse notation and let Sζ denoteboth the single layer potential and the boundary single layer potential cor-responding to Gζ.

An exponentially growing solution is inside Ω a solution to a bound-ary value problem for the Schrodinger equation and outside Ω a solutionto Laplace’s equation with a certain exponential growth condition near in-finity. We can think of the boundary of Ω as the manifold, where theseproperties meet. In fact the boundary value of the exponentially growingsolution is easily seen to satisfy a single equation, which takes into accountboth behaviors. Assume zero is not a Dirichlet eigenvalue of (−∆ + q) andlet Λq be the associated Dirichlet-to-Neumann map. Assume further that ζis not exceptional and let ψ be the unique exponentially growing solution.Then for x ∈ Rn \ Ω we find from (2.13) and (2.25) that

(Sζ(Λq − Λ0)ψ)(x) = 〈Gζ(x − ·), (Λq − Λ0)ψ(·, ζ)〉

=

∫

RnGζ(x − y)q(y)ψ(y, ζ)dy

= eix·ζ − ψ(x, ζ),


and by the continuity of the trace operator (2.33) we see that fζ = ψ(·, ζ)|∂Ω

solves

fζ = eix·ζ − Sζ(Λq − Λ0) fζ. (2.36)

The following theorem relates the set of exceptional points and hence theexistence of exponentially growing solutions to the solvability of this equa-tion.

Theorem 2.4.8. A point ζ ∈ V is not exceptional if and only if the equation (2.36)has a unique solution fζ ∈ H1/2(∂Ω).

Proof. We will first see that any solution fζ ∈ H1/2(∂Ω) to (2.36) can beextended to an exponentially growing solution. Define ψ(·, ζ) on Ω to bethe solution to (−∆ + q)ψ = 0 with boundary value ψ|∂Ω = fζ, and defineψ on Rn \ Ω by

ψ(x, ζ) = eix·ζ − Sζ(Λq − Λ0) fζ

= eix·ζ −∫

Ω

Gζ(x − y)q(y)ψ(y, ζ)dy, x in Rn \ Ω. (2.37)

By assumption on fζ

ψ|∂Ω = fζ,

where the trace is taken from either side of ∂Ω. Furthermore, since eix·ζ −Sζ(Λq − Λ0) fζ is harmonic in Ω and equals fζ on ∂Ω, we see that

∂ν−(eix·ζ − Sζ(Λq − Λ0) fζ) = Λ0 fζ.

The jump relation (2.35) now shows that

(∂ν+ − ∂ν−)(eix·ζ − Sζ(Λq − Λ0) fζ) = (Λq − Λ0) fζ,

and hence we conclude that

∂ν+ ψ(·, ζ) = Λq fζ.

This shows that ψ ∈ H1loc(Rn) is in fact a weak solution to the Schrodinger

equation across ∂Ω. Moreover, Proposition 2.4.2 applied to (2.37) showsthat ψ is exponentially growing.

To prove the theorem assume ζ is not exceptional and let ψ be theunique exponentially growing solution. Then ψ(·, ζ)|∂Ω solves (2.36). Iffζ is another solution, then it has by the above argument an extension toan exponentially growing solution, which differs from Ψ; this violates theuniqueness of the exponentially growing solution and hence the assump-tion on ζ.


Conversely, assume (2.36) has a unique solution fζ, which extends to anexponentially growing solution ψ. Let ψ be a different exponentially grow-ing solution. Then ψ(·, ζ)|∂Ω solves (2.36) and thus ψ(·, ζ)|∂Ω = ψ(·, ζ)|∂Ω =fζ. By uniqueness for the Dirichlet problem we see that ψ(·, ζ) = ψ(·, ζ)in Ω and then by (2.25) that ψ(·, ζ) = ψ(·, ζ) everywhere. This shows theuniqueness of exponentially growing solutions, and by Lemma 2.4.4 we seethat ζ is not exceptional.

2.4.4. Exceptional points for the conductivity equation. When the con-ductivity γ ∈ W2,∞(Ω) and γ = 1 near ∂Ω the Dirichlet-to-Neumann mapsΛγ and Λq coincide. We can then write the boundary integral equation(2.36) as

ψ(x, ζ) = eix·ζ − Sζ(Λγ − Λ1)ψ(·, ζ). (2.38)

By Theorem 2.4.8 the solvability of this equation is equivalent to ζ not beingexceptional, but since (2.38) makes sense for γ ∈ L∞

+ (Ω) we can generalizethe definition of exceptional points in a coherent way.

Definition 2.4.9. Let γ ∈ L∞+ (Ω) and assume that γ = 1 near ∂Ω. Then a point

ζ ∈ V is said not to be exceptional for γ if the boundary integral equation (2.38)has a unique solution in H1/2(∂Ω).

The definition of exceptional points for a potential (Definition 2.4.5)was motivated by the existence of exponentially growing solutions to theSchrodinger equation. The following theorem shows that a exceptionalpoint in the sense of Definition 2.4.9 has an analogue property.

Theorem 2.4.10. Let γ ∈ L∞+ (Ω) and assume γ = 1 near ∂Ω and in Rn \ Ω.

Then for ζ not exceptional, there is a solution φ(·, ζ) to

∇ · γ∇φ = 0 ∈ Rn (2.39)

with

(φ(·, ζ)e−ix·ζ − 1) ∈ L2δ(R

n) (2.40)

for −1 < δ < 0.

Proof. Let fζ ∈ H1/2(∂Ω) be the unique solution to (2.38), and define φ(·, ζ)in Ω as the solution to the conductivity equation with φ|∂Ω = fζ and inRn \ Ω by

φ(x, ζ) = eix·ζ − Sζ(Λγ − Λ1) fζ.

This defines a solution φ ∈ H1loc(Rn) to (2.39) (cf. the proof of Theorem 2.4.8).


To prove (2.40) we see by (2.11) that

eix·ζ − φ(x, ζ) = Sζ(Λγ − Λ1)φ(·, ζ)

= 〈Gζ(x − ·), (Λγ − Λ1)φ(·, ζ)〉

=

∫

Ω

(γ(y) − 1)∇yGζ(x − y) · ∇φ(y)dy

for x ∈ Rn \ Ω. Multiplying this equation by e−ix·ζ and replacing Gζ by gζ

gives

1 − φ(x, ζ)e−ix·ζ =

∫

Ω

(γ(y) − 1)∇y(e−iy·ζgζ(x − y)) · ∇φ(y)dy

=

∫

Ω

(γ(y) − 1)(∇y − iζ)gζ(x − y) · e−iy·ζ∇φ(y)dy

= −(∇x + iζ) ·∫

Ω

gζ(x − y)(γ(y) − 1)e−iy·ζ∇φ(y)dy.

Proposition 2.4.2 now implies for −1 < δ < 0 that

‖φ(x, ζ)e−ix·ζ − 1‖L2δ(Rn) ≤ C‖γ − 1‖L∞(Ω)‖e−iy·ζ∇φ‖L2

δ+1(Ω), (2.41)

which shows the claim.

We note that it seems impossible to obtain a decay estimate like (2.26)for (φ(x, ζ)e−ix·ζ − 1) when ζ grows. In the estimate (2.41) above, the con-stant C is independent of ζ, but ‖e−iy·ζ∇φ‖L2

δ+1(Rn) seems to be growing.This estimate may, however, not be optimal. We note further that the inte-gral equation for ω(x, ζ) = φ(x, ζ)e−ix·ζ − 1 is

(I + (∇x + iζ) · gζ ∗ ((γ − 1)(∇ + iζ)·))ω = −iζ · ∇gζ ∗ (γ − 1),

but the estimates in Proposition 2.4.2 seem insufficient for solving this equa-tion.

To actually solve the boundary integral equation the following result isuseful. Here it is shown that Sζ(Λγ − Λ1) is compact, and thus (2.38) is aFredholm equation of the second kind. When γ is sufficiently regular thisfact can be proved by transforming the equation into the Fredholm equa-tion (2.24), but when γ is only bounded we have to argue differently. Mostprobably the result is well-known, but in the lack of a proper reference, wegive the proof.

Lemma 2.4.11. Assume γ ∈ L∞+ (Ω) satisfies γ = 1 near ∂Ω. Then (Λγ −Λ1) ∈

B(H1/2(∂Ω), Hs(∂Ω)) for any s ∈ R, and we have the estimate

‖Λγ − Λ1‖B(H1/2(∂Ω),Hs(∂Ω)) ≤ C‖γ − 1‖L∞(Ω), (2.42)

where C depends on s, Ω, the ellipticity constant for γ, and the support of (γ− 1).

Moreover, the composite operator Sζ(Λγ − Λ1) is compact on H1/2(∂Ω).


Proof. We will prove that Λγ − Λ1 ∈ B(H1/2(∂Ω), Hm+1/2(∂Ω)) for anym ∈ Z+. Take an open domain Ω′ ⊂ Ω such that γ = 1 on Ω0 = Ω \ Ω′.Take further a set of smooth cut-off functions φkm

k=0 supported near ∂Ω

such that

(1) 0 ≤ φk ≤ 1, k = 0, . . . , m(2) φm = 1 near ∂Ω

(3) φ0 = 0 near ∂Ω′

(4) φk = 1 on Ωk+1 := supp(φk+1) for k = 0, . . . , m − 1

Then since φk(u − v) ∈ H10(Ωk) solves

∆(φk(u − v)) = ∆φk(u − v) + ∇φk · ∇(u − v) in Ωk,

we can estimate

‖φk(u − v)‖Hk+2(Ω) ≤ C‖∆φk(u − v) + ∇φk · ∇(u − v)‖Hk(Ωk)

≤ C‖u − v‖Hk+1(Ωk)

≤ C‖φk−1(u − v)‖Hk+1(Ω), k = 1, 2, . . . , m, (2.43)

where C depends on the cut-off function φk. A similar argument shows that

‖φ0(u − v)‖H2(Ω) ≤ C‖u − v‖H1(Ω). (2.44)

Now since u − v ∈ H10(Ω) satisfies

∇ · γ∇(u − v) = −∇ · (γ − 1)∇v in Ω,

standard elliptic estimates show that

‖u − v‖H1(Ω) ≤ C‖∇ · (γ − 1)∇v‖H−1(Ω)

≤ C‖γ − 1‖L∞(Ω)‖v‖H1(Ω)

≤ C‖γ − 1‖L∞(Ω)‖ f ‖H1/2(∂Ω), (2.45)

where C now depends on the ellipticity constant for γ and Ω. Combining(2.43), (2.44) and (2.45) gives

‖(Λγ − Λ1) f ‖Hm+1/2(∂Ω) ≤ ‖∂ν(φm(u − v))‖Hm+1/2(∂Ω)

≤ C‖φm(u − v)‖Hm+2(Ω)

≤ C‖φ0(u − v)‖H2(Ω)

≤ C‖u − v‖H1(Ω)

≤ C‖γ − 1‖L∞(Ω)‖ f ‖H1/2(∂Ω),

which proves the claim.

The compactness of Sζ(Λγ −Λ1) in H1/2(∂Ω) follows from the first partof the lemma, since Sζ ∈ B(L2(∂Ω), H1(∂Ω)) (see Proposition 2.4.7) and theinclusion Hs(∂Ω) ⊂ Hs−ε(∂Ω) is compact for s ∈ R, ε > 0.


Since Sζ(Λγ − Λ1) is compact on H1/2(∂Ω), (I + Sζ(Λγ − Λ1)) is invert-ible when ζ is not exceptional. This gives a formula for the computation ofψ(·, ζ)|∂Ω, i.e.

ψ(·, ζ)|∂Ω = (I + Sζ(Λγ − Λ1))−1(eix·ζ),

which will later be useful in the reconstruction procedures.

2.5. Notes

Concerning the determination and reconstruction of the conductivity γ andits derivatives on the boundary from Λγ, there are many results available.The first result was due to Kohn and Vogelius [KV84a], who showed that ifγ is smooth on a smooth domain Ω, then Λγ uniquely determines ∂m

ν γ|∂Ω

for any m ≥ 0. Hereby they also solved the uniqueness question for theclass of real-analytic conductivities, a result that was later generalized topiecewise real-analytic conductivities [KV85]. The result on the boundarywas improved by Sylvester and Uhlmann [SU88], who proved by pseudo-differential techniques that if ∂Ω is smooth and γ ∈ Ck(Ω) then Λγ de-termines ∂m

ν γ|∂Ω for m ≤ k. Moreover, a stability estimate for the bound-ary values was given. For Lipschitz domains Alessandrini [Ale88, Ale90]proved uniqueness at the boundary of the conductivity γ ∈ W1,∞(Ω) andfor the derivatives ∂m

ν γ|∂Ω m ≤ k, provided that γ is Ck near the boundary.More recent results include [Bro01],[NT01], and [KY01], where the last tworeferences treat the determination of anisotropic conductivities as well.

The results in Lemma 2.1.1 concerning γ ∈ W1,p(Ω), p > n and γ ∈W2,p(Ω), p > n/2 are valid also when ∂Ω is only Lipschitz. Also Lemma2.1.2 is valid for less regular domains. The result for functions in the Holderspaces holds for any Lipschitz domain, while the result for functions inSobolev spaces seems to require the boundary to be Ck,1, k + 1 > s. Hencethe method outlined in section 2.1 may be used to reduce an inverse con-ductivity problem on a non-smooth domain into a problem on a smoothdomain.

The reduction to the case γ = 1 near ∂Ω was first given by Nachman[Nac96]. He considered the case γ ∈ W2,p(Ω), p > n, and gave an ex-plicit formula for the Dirichlet-to-Neumann map on the outer boundary:let Ω1 := Ω and let Ω2 := Ωe. Define the operators Λij : H1/2(∂Ωi) →H−1/2(∂Ωj), i, j ∈ 1, 2, by

〈Λij f , g〉 = (−1)i∫

Ω1\Ω2

γ∇ui · ∇vj,

where ∇ · γ∇ui = 0, ui|∂Ωi = f , u1|∂Ω2 = 0 = u1|∂Ω1 and vj|∂Ωj = g, v1|∂Ω2 =

0 = v2|∂Ω1 . This defines Λ11 and Λ22 as reduced Dirichlet-to-Neumannmaps on ∂Ω1 and ∂Ω2 respectively, and Λ12, Λ21 as transition Dirichlet-to-Neumann maps from one boundary to the other. Then it is proved that

2.5. Notes 25

(1) (Λγ − Λ11) is invertible,

(2) Λγe = Λ22 + Λ21(Λγ − Λ11)−1Λ12,

which gives the formula for Λγe . The proof of these statements (see thepreprint [Nac93]) relies on the assumption that γ ∈ W2,p(Ω), p > 1. How-ever, in a recent paper by Ikehata [Ike02], (1) was proved in a differentcontext using only the assumption that γ ∈ L∞

+ (Ω). Moreover, by usingIkehata’s result, (2) is easily seen to be valid in the more general case. Thisconstitutes a different proof of Lemma 2.1.3; the proof given in section 2.1is based on [KT01].

An interesting question related to Lemma 2.1.3 is, whether it is pos-sible to propagate the Dirichlet-to-Neumann inwards through a knownmedium, i.e. whether Λγ is determined by Λγe and γ|Ωe\Ω. Ikehata [Ike02]gave an affirmative answer assuming γ ∈ C0,1(Ωe) when n ≥ 3 and γ ∈L∞(Ωe) when n = 2. The method of proof relies on the Runge approxima-tion theorem and is thus in essence non-constructive.

The study of the exponentially growing solutions is motivated by Calde-ron’s fundamental paper [Cal80]. Consider the bilinear form associated thequadratic form (1.5)

Qγ( f , g) =

∫

Ω

γ∇u · ∇v,

where u, v solves (1.1) with f , g on the boundary. Calderon showed in hispaper that the Frechet derivative of Qγ in direction h ∈ L∞(Ω) is

(dQγ(h))( f , g) = limε→0

1ε(Qγ+εh( f , g) − Qγ( f , g)) =

∫

Ω

h∇u · ∇v, (2.46)

where again u, v solves (1.1) with f , g on the boundary. Hence the deriv-ative is injective if products of gradients to the conductivity equation aredense in L1(Ω). For γ = 1 and ζ ∈ V the harmonic functions eix·ζ, e−ix·ζ

are solutions to the conductivity equation, and by using these solutionsin (2.46) Calderon proved that dQ1(h) = 0 if and only if h = 0. However,since, as it was pointed out by Calderon, Ran(dQ1) is not closed (see [IN99]for a proof), the inverse mapping theorem cannot be applied to give localuniqueness around constant conductivities. Calderon [Cal80] gave also us-ing this differential an approximate reconstruction method for γ close to aconstant.

In [Isa91] Isakov proved by a different method the existence of expo-nentially growing solutions by taking into account the fact that the poten-tial is compactly supported. The idea is to show the existence of a Green’soperator Eζ : L2(Ω) → H2(Ω) with

(∆ + 2iζ · ∇)(Eζ f ) = f ,


such that

‖Eζ f ‖L2(Ω) ≤C|ζ| ‖ f ‖L2(Ω),

‖Eζ f ‖H1(Ω) ≤ C‖ f ‖L2(Ω),

where the generic C is independent of ζ. The ideas in the proof goes backto the work of Ehrenpreis and Malgrange in the 1950’s and is based on ageneral method for constructing Green’s functions. This method works infact for more general types of equations, and hence exponentially growingsolutions for these equations can be constructed. In [Hah96] Hahner gavean elegant and elementary proof of a similar result using Fourier seriestechniques.

Another result concerning the convolution operator gζ∗ in dimensionn ≥ 3 is the following due to Jerison and Kenig (as found in [Cha90]): letn ≥ 3 and let Ω ⊂ Rn be an open bounded domain. Assume that q ∈ Lr(Ω)for some r with n/2 ≤ r ≤ (n + 1)/2. Then for ζ ∈ V and p = 2r/(r − 1)

‖gζ ∗ (q f )‖Lp(Rn) ≤C

|ζ|2−n/r ‖q‖Lr(Ω)‖ f ‖Lp(Rn).

This result follows by applying to gζ/|ζ| the uniform estimates in [KRS87]for second order differential operators with constant coefficients. As a con-sequence we obtain for

C|ζ|−(2−n/r)‖q‖Lr(Ω) < 1,

the existence of a unique exponentially growing solution with

‖ω(·, ζ)‖Lp(Rn) ≤C

|ζ|2−n/r ‖q‖Lr(Ω). (2.47)

We refer to Chapter 3 for an improvement of this Lp-estimate in two dimen-sions.

We mention also the following extension of Proposition 2.4.2 due toBrown [Bro96, Theorem 0.3]. By interpolating (2.22) and (2.23) we get


C|ζ|1−s ‖ f ‖L2

δ+1(Rn), 0 ≤ s ≤ 1, −1 < δ < 0,

which by duality implies that

‖gζ ∗ f ‖L2δ(Rn) ≤

C|ζ|1−s ‖ f ‖H−s

δ+1(Rn), 0 ≤ s ≤ 1, −1 < δ < 0.

Interpolating these two estimates with interpolation parameter θ = 1/2gives


C|ζ|1−2s ‖ f ‖H−s

δ+1(Rn), 0 ≤ s ≤ 12

, −1 < δ < 0.

2.5. Notes 27

The study of the special Green’s function goes back to Faddeev [Fad65],who considered the inverse scattering problem for the Schrodinger equa-tion. Faddeev suggested a generalization of the method for the one-dimen-sional inverse scattering problem due to Gelfand-Levitan-Marchenko, buta major obstacle was the presence of real exceptional points. For a com-prehensive survey on the inverse scattering problem in dimension n = 3we refer to the monograph [New89]. In [NA84, BC85, SU87] Faddeev’sGreen’s functions were rediscovered independently.

The boundary integral equation was derived in Nachman [Nac88] andNovikov [NK87] independently. Nachman derived the slightly differentboundary integral equation

ψ(x, ζ) = eix·ζ − (SζΛq − Kζ −12

I)ψ(·, ζ) (2.48)

based on a combined single and double layer potential representation of theexponentially growing solution. Here Kζ is the boundary double layer po-tential based the Faddeev Green’s function. Since an easy argument showsthat (1/2I + Kζ) = SζΛ0 we see that (2.48) is equivalent to (2.36).

Chapter 3

Uniqueness andreconstruction in twodimensions

In this chapter we will turn our attention to the two-dimensional inverseconductivity problem. As stated in Chapter 1 this problem seems at firstglance harder than the higher dimensional problem, since it is only for-mally determined. One great advantage in two dimensions is, however,the applicability of the well developed tools from complex analysis.

The problem of uniqueness and reconstruction was solved by Nachmanin [Nac96] for conductivities having essentially two derivatives. Shortly af-ter Brown and Uhlmann relaxed the assumption and gave an affirmativeanswer to the uniqueness question for conductivities having only one de-rivative [BU97]. The technique of proof in the two papers are very sim-ilar in that both methods exploit results from complex analysis and arebased on the so-called ∂-method of inverse scattering. Based on Brown andUhlmann’s approach in a joint work with Alexandru Tamasan we gave areconstruction algorithm for the less regular class of conductivities [KT01].

In this chapter we will first review a few basic and useful results fromcomplex analysis. Next we review the uniqueness proof and reconstructionmethod for conductivities having two derivatives. We will then focus onthe uniqueness proof for less regular conductivities and then show howthis method can be turned into a reconstruction algorithm. In section 3.5the two reconstruction methods are compared, and we will see that in somesense the new method is a direct generalization of the previous one. In the

29

30 3. Uniqueness and reconstruction in two dimensions

last section we describe a numerical implementation of the algorithm andcomment on the results.

As seen in section 2.1 the general inverse conductivity problem can bereduced to the case, where the conductivity γ = 1 near the boundary of thedomain Ω. We will throughout this chapter take this assumption as well asthe ellipticity assumption γ ∈ L∞

+ (Ω).

3.1. A few results from complex analysis

Let z = x1 + ix2 ∈ C and define the complex differential operators ∂ = ∂/∂zand ∂ = ∂/∂z by

∂ =12

(

∂

∂x1− i

∂

∂x2

)

, ∂ =12

(

∂

∂x1+ i

∂

∂x2

)

.

To define the inverse of these operators we introduce

g(z) = F−1(

2i(ξ1 + iξ2)

)

=1

πz, (3.1)

which is a Green’s function for ∂. In the definition of g and henceforth weidentify the complex number z = x1 + ix2 ∈ C and (x1, x2) ∈ R2. Then we

define on S(R2) the weakly singular integral operators ∂−1, ∂−1

by

∂−1 f (z) = g ∗ f (z) =1π

∫

C

f (z′)z − z′

dµ(z′),

∂−1

f (z) = g ∗ f (z) =1π

∫

C

f (z′)z − z′

dµ(z′),

where dµ(z) = dz1dz2 is the usual Lebesgue measure in the plane. Themapping properties of these operators in conventional spaces are well an-alyzed in the literature. We denote below by p, p′ the parameters definedby

1p

=1p− 1

2for 1 ≤ p ≤ 2

1p′ = 1 − 1

pfor 1 ≤ p ≤ ∞.

Then we can state the following useful facts.

Proposition 3.1.1. Let f ∈ S(R2) and let T be either ∂−1 or ∂−1

. Define u = T fin R2. Then for 1 < p < 2

‖u‖L p(R2) ≤ C‖ f ‖Lp(R2), (3.2)

and hence T ∈ B(Lp(R2), L p(R2)).

3.1. A few results from complex analysis 31

Moreover, if f ∈ Lp1(R2) ∩ Lp2(R2) for some 1 < p1 < 2 < p2 < ∞ then

u ∈ Cα(R2), for α = 1 − 2

p2. (3.3)

Finally, if f ∈ Cα(R2) is compactly supported then u ∈ C1+αloc (R2).

Proof. The first result is a consequence of the Hardy-Littlewood-Sobolevtheorem of fractional integration for the Riesz transform, see [Ste70, The-orem 1, p. 119]. The results concerning Holder regularity are from theclassical monograph of Vekua, see [Vek62, Theorem 1.21 and 1.32].

Note that since T commutes with differential operators, more regularityin f implies more regularity in T f .

We will later need the following extension of Proposition 3.1.1. Definethe unimodular function

e(z, k) = exp(i(zk + zk)) = exp(2i Re(zk)), z, k ∈ C.

Then

(∂ + ik)u = e(z,−k)∂(e(z, k)u)

motivates the definition

(∂ + ik)−1 f = e(z,−k)∂−1(e(z, k) f ).

For this operator we have the following result.

Proposition 3.1.2. Let 1 < p < 2 and assume that v ∈ L p(R2) has ∂v ∈Lp(R2). Then for k ∈ C \ 0, the function u = (∂ + ik)−1v ∈ W1,p(R2) and wehave the estimates

‖(∂ + ik)−1v‖L p(R2) ≤C|k| (‖v‖L p(R2) + ‖∂v‖Lp(R2)), (3.4)

‖(∂ + ik)−1v‖W1,p(R2) ≤ C(

1 +1|k|

)

(‖v‖L p(R2) + ‖∂v‖Lp(R2)), (3.5)

where C is independent of k.

Proof. We refer to [Nac96].

A fundamental result in the theory of complex functions is Liouville’stheorem, which states that any bounded function u satisfying ∂u = 0 inC must be constant. A consequence is that the only entire function inLp(R2), 1 ≤ p < ∞, is the zero function. This fundamental result can begeneralized to a class of functions satisfying a more complicated equation.


In the following we will say that a function u is pseudoanalytic in C withcoefficients a, b if and only if it satisfies the equation

∂u = au + bu in C. (3.6)

The generalized Liouville theorem for pseudoanalytic functions is then

Theorem 3.1.3. If u ∈ Lp(R2), 1 ≤ p < ∞ is pseudoanalytic in C with thecoefficients a, b ∈ L2(R2), then u = 0.

Proof. We will give the proof when a, b ∈ Lp1(R2) ∩ Lp2(R2) for 1 ≤ p1 <

2 < p2 < ∞. Let

v = auu

+ b ∈ Lp1(R2) ∩ Lp2(R

2).

Note that ∂−1

v ∈ L∞(R2) by (3.3). Let v = u exp(−∂−1

v) and note thatv ∈ Lp(R2) by assumption on u. Since ∂v = 0 everywhere, v is an entirefunction in Lp(R2), and by the Liouville theorem we conclude that v = 0.This implies u = 0.

When p1 = 2 = p2 the proof is more complicated, since in general ∂−1

fis only in BMO(R2) and not bounded for f ∈ L2(R2). For a proof of thetheorem in this case we refer to [BU97].

In the following we will several times have to solve a certain pseudo-analytic equation with a decay condition at infinity. The next result givesan exact condition for this to be possible.

Corollary 3.1.4. Let 1 < p < 2 and assume that a ∈ Lp(R2) ∩ L2(R2). Thenthe equation

∂m = am (3.7)

has a unique solution m with m − 1 ∈ L p(R2).

Proof. The equation for m − 1 is

∂(m − 1) = a(m − 1) + a

or equivalently

(I − ∂−1

(a ·))(m − 1) = ∂−1

a. (3.8)

Since a ∈ Lp(R2) implies ∂−1

a ∈ L p(R2) by (3.2), it makes sense to considerthis integral equation in L p(R2).

From [Nac93] it is known that when a ∈ L2(R2), the operator ∂−1

(a ·)is real-linear and compact in the Lr(R2), r > 2. Hence invertibility of

(I − ∂−1

(a ·)) in L p(R2) is by the Fredholm alternative a consequence of

3.1. A few results from complex analysis 33

uniqueness of a solution to the homogeneous equation. However, a solu-tion h ∈ L p(R2) to the homogeneous equation would satisfy ∂h = ah, andhence h = 0 by the generalized Liouville theorem, Theorem 3.1.3. Thisproves the result.

We note that when a is only in L2(R2), then the generalized Liouvilletheorem does give uniqueness of a solution to (3.7). But for a ∈ L2(R2),

∂−1

a is generally only of bounded mean oscillation, and hence it seems notpossible to actually solve (3.8) in a conventional Lebesgue space.

In applications of the above result we will need a result concerning thedependence of the solution on the parameter a. We have the following con-tinuity result.

Lemma 3.1.5. Let 1 < p1 < 2 < p2 < ∞ and assume that ai ∈ Lp1(R2) ∩Lp2(R2), i = 1, 2. Let mi, mi − 1 ∈ L p1(R2) be the unique solution to (3.7) withcoefficient ai. Then

‖m1 − m2‖L p1 (R2) ≤ C2(‖a1 − a2‖Lp1 (R2) + C1‖a1 − a2‖L2(R2)‖a1‖Lp1 (R2)),

where

Ci = C exp(C(‖ai‖Lp1 (R2) + ‖ai‖Lp2 (R2))).

Proof. From [BBR01, Lemma 2.6] we have for the solution to ∂u = au + bthe stability estimate

‖u‖L p(R2) ≤ C‖b‖Lp(R2) exp(C(‖a‖Lp1 (R2) + ‖a‖Lp1 (R2))) (3.9)

for 1 < p < 2 and 1 < p1 < 2 < p2 < ∞. This estimate applied to (3.7) gives

‖mi − 1‖L p1 (R2) ≤ C‖ai‖Lp1 (R2) exp(C(‖ai‖Lp1 (R2) + ‖ai‖Lp2 (R2)))

= Ci‖ai‖Lp1 (R2). (3.10)

Since

∂(m1 − m2) = a1(m1 − m2) + (a1 − a2)m2,

the estimate (3.9) applied to m1 − m2 then gives

‖m1 − m2‖L p1 (R2) ≤ C1‖(a1 − a2)m2‖Lp1 (R2)

≤ C1(‖a1 − a2‖Lp1 (R2) + ‖a1 − a2‖L2(R2)‖m2 − 1‖L p1 (R2))

by Holders inequality. The claim now follows by combining this with(3.10).

The last result we need from complex analysis is a result concerning thesingle layer operator associated with the Green’s functions for ∂ and ∂. Let


D ⊂ C be an open, bounded and smooth domain and let f ∈ Cα(∂D), 0 <

α < 1. Define the Cauchy integral operator

S f (z) =1

2πi

∫

∂D

f (z′)z′ − z

dz′, z ∈ R2 \ ∂D,

where the integral is understood as a path integral along a positively ori-ented path describing the boundary ∂D. This integral defines a functionwhich is holomorphic away from ∂D. When z ∈ ∂D the integral kernel isnot integrable, so to define an operator on the boundary we have to workwith principal values of the integral. Thus we define the singular integraloperator

K f (z) =1

2πip.v.

∫

∂D

f (z′)z′ − z

dz′, z ∈ ∂D.

A few facts about K and the relation to S is collected in the next result.

Proposition 3.1.6. The singular integral operator K ∈ B(Cm+α(∂D)) for m ∈Z+ and 0 < α < 1.

Let ν be the outpointing normal to ∂D. Then on the boundary we have thejump condition

limε→0

S f (z ± νε) = ∓12

f (z) + K f (z), z ∈ ∂D. (3.11)

Proof. For a proof we refer to the classical monograph of Muskhelishvilion singular integral equations [Mus53, p. 42].

3.2. Uniqueness and reconstruction for γ ∈ W2,p(Ω)

In this section we briefly describe Nachman’s [Nac96] reconstruction me-thod for the two-dimensional inverse conductivity problem. The main re-sult is

Theorem 3.2.1. Let Ω ⊂ R2 be open, bounded and smooth, and let γ ∈ W2,p(Ω),p > 1. Then γ can be reconstructed uniquely from Λγ.

The method of proof is based on the reduction of the conductivity equa-tion to the Schrodinger equation as described in section 2.2, and the con-struction of exponentially growing solutions to this equation. In two di-mensions the variety V = ζ ∈ C2 \ 0 | ζ · ζ = 0 = ζ ∈ C2 \ 0 | ζ =(k,±ik), k ∈ C, and since an exponentially growing solution ψ to theSchrodinger equation has the symmetry ψ(x, (k, ik)) = ψ(x, (−k, ik)) whenq = ∆γ1/2/γ1/2 is real, it suffices to consider only ζ = (k, ik), k ∈ C \ 0. Inthis case we note that x · ζ = x1k + ix2k = zk for z = x1 + ix2. Hence we willin the following abuse notation and write ψ(z, k) := ψ(x, (k, ik)) indicatingthat this is an exponentially growing solution with growth eizk.

3.2. Uniqueness and reconstruction for γ ∈ W2,p(Ω) 35

From Theorem 2.4.6 it is known, that there are no exceptional pointsoutside a sufficiently large ball. In two dimensions it can be shown (seeLemma 3.2.3 below) that when the potential comes from a conductivity,then there are no exceptional points at all, i.e. the exponentially growingsolutions ψ(·, k) are defined for any k ∈ C \ 0. This allows the introduc-tion of the function

t(k) =

∫

Ω

eikzq(z)ψ(z, k)dµ(z), k ∈ C \ 0, (3.12)

the so-called non-physical scattering transform of the potential q. We notethat t is a nonlinear transform of q, since in the integrand both q itself and ψ

depend on the potential. Note further that since ψ(z, k) ∼ eizk, the integrandhas the asymptotics ei(kz+zk)q(x) ∼ e2i(x1k1−x2k2)q(x). This implies t(k) ∼2πq(−2k1, 2k2) for large k. Not only is t asymptotically close to the Fouriertransform of the potential, but it has also many similar properties. It canfor instance be shown that smoothness in q corresponds to decay in t, andsymmetries in q implies symmetries in t, see [Sil99].

The non-physical scattering transform is the key intermediate object inthe reconstruction algorithm consisting of the two steps:

(1) Reconstruct t from Λγ.

(2) Reconstruct γ from t.

In the following subsections we will first review the result concerningabsence of exceptional points. Then we will be more explicit about thedetails in the two steps in the reconstruction.

3.2.1. Absence of exceptional points. Before giving the result we state thefollowing extension of Proposition 2.4.2. Let gζ be Faddeev’s Green’s func-tion defined in (2.18) and denote

gk(z) := g(k,ik)(x), k ∈ C \ 0.

Proposition 3.2.2. Let 1 < p < 2 and k ∈ C \ 0. Then the operator gk∗ is abounded operator from Lp(R2) into W1,p(R2) and satisfies

‖gk ∗ f ‖Ws,p(R2) ≤ C|k|s−1‖ f ‖Lp(R2), (3.13)

for 0 ≤ s ≤ 1. Here the constant C is uniform for |k| ≥ ε > 0.

Proof. Since −∆ − 2(k, ik) · ∇ = −4(∂(∂ + ik)), we have

gk ∗ f = −14(∂ + ik)−1∂−1 f .

The result is then obtained by invoking Proposition 3.1.2 and interpolating(3.4) and (3.5).


The definition of exceptional points (Definition 2.4.5) were based on thesolvability of the Lippmann-Schwinger-Faddeev equation (2.24) in H1

δ (Rn).Due to the improved estimate (3.13), we will in the following equivalentlyconsider (2.24) in a usual Sobolev space and in this context adopt the notionof exceptional points.

The absence of exceptional points is the result of the next lemma. Wewill give the proof of the result, since it is illustrative in relation to thereconstruction for less regular conductivities to follow later.

Lemma 3.2.3. Let q = ∆γ1/2/γ1/2 for γ ∈ W2,p(Ω), 1 < p < 2. Then for anyk ∈ C \ 0 there exists a unique solution ψ(·, k) to

(−∆ + q)ψ = 0 in R2 (3.14)

with ω(·, k) = ψ(·, k)e−ikz − 1 ∈ W1,p(R2). For k sufficiently large we have theestimate

‖ω(·, k)‖Ws,p(R2) ≤ C|k|s−1‖q‖Lp(R2) (3.15)

for any 0 ≤ s ≤ 1.

Proof. The integral equation for ω is (cf. (2.24))

ω(z, k) + gk ∗ (qω(·, k))(z) = (gk ∗ q)(z). (3.16)

We consider this integral equation in W1,p(R2). Since the inclusion W1,p(R2)⊂ L∞(R2) is compact due to Rellich’s theorem, multiplication by q ∈ Lp(R2)is bounded from L∞(R2) into Lp(R2), and the operator gk∗ is boundedfrom Lp(R2) into W1,p(R2), we deduce that gk ∗ (q·) is compact in W1,p(R2).Hence (3.16) is a Fredholm equation of the second kind in W1,p(R2). Thusuniqueness of a solution to (3.16) implies existence.

To prove uniqueness assume ω0 ∈ W1,p(R2) solves the homogeneousequation and let ψ0 = eizkω0. Then ψ0 solves (3.14), and by the reality of q,ψ0 is another solution with the same asymptotic behavior. Hence we canwithout loss of generality assume that ψ0 is real. Introduce

v = γ1/2∂(γ−1/2ψ0)e−izk,

which is in L p(R2). Now since γ−1/2ψ0 solves the conductivity equation, asimple calculation using (3.14) and the reality of ψ0 shows that

∂v =∂(γ1/2)

γ1/2 e(z,−k)v. (3.17)

Theorem 3.1.3 then implies that v = 0, since γ−1/2∂(γ1/2) ∈ L2(R2). Thisgives the equation

∂(γ−1/2ψ0)e−izk = 0,


which implies

0 = ∂(γ−1/2ψ0)e−izke(z, k)

= ∂(γ−1/2ψ0eizk)

= ∂(γ−1/2ψ0e−izk)

for the the function γ−1/2ψ0e−izk ∈ L p(R2). By the usual Liouville theoremwe conclude that ψ0 = 0 and hence ω0 = 0.

The estimate (3.15) is a direct consequence of (3.16) and (3.13).

Note that the existence of exponentially growing solutions to the Schro-dinger equation relies on the particular form of the potential q, i.e. the ex-istence of a positive function γ1/2 such that q = ∆γ1/2/γ1/2. When this isnot the case, there will be exceptional points even for small potentials, see[Nac96, Tsa93].

3.2.2. From Λq to t. Since eikz is harmonic the identity (2.13) implies that

t(k) =

∫

∂Ω

eikz(Λq − Λ0)ψ(·, k)dσ(z). (3.18)

From Theorem 2.4.8 and Lemma 3.2.3 it follows that for any k ∈ C \ 0the exponentially growing solution ψ(·, k)|∂Ω can be constructed from Λqby solving the boundary integral equation (2.36). This enables the recon-struction of the scattering transform from the Dirichlet-to-Neumann mapand hence we have carried out the first step in the reconstruction.

From the boundary integral equation we have for small k the followinguseful estimate.

Proposition 3.2.4. The solution ψ(z, k), z ∈ ∂Ω to (2.36) satisfies for k suffi-ciently small the estimate

‖ψ(·, k) − 1‖H3/2(∂Ω) ≤ C|k|. (3.19)

Proof. In [SMI00] the estimate ‖ψ(·, k) − 1‖H1/2(∂Ω) ≤ C|k| was obtainedby decomposing Sζ = S0 + Hζ, where S0 is the usual single layer operatorand Hζ is a convolution operator with a harmonic kernel, and then analyz-ing the dependency of k in Hζ. From this result we get

‖ψ(·, k) − 1‖H3/2(∂Ω)

≤ ‖eizk − 1‖H3/2(∂Ω) + ‖Sζ(Λγ − Λ1)(ψ(·, k) − 1)‖H3/2(∂Ω)

≤ C|k|,since ‖Sζ(Λγ − Λ1)‖B(H1/2(∂Ω),H3/2(∂Ω)) is uniformly bounded for small ζ.


We refer to section 4.2.6 for a generalization of this result to higher di-mensions, where we also give a more detailed proof. Note that this resultshows that ψ(·, 0)|∂Ω is well-defined.

3.2.3. From t to γ. This second step is divided into two steps. First theexponentially growing solutions ψ are computed inside Ω from t, next theconductivity is obtained from ψ.From t to ψ. The following lemma gives the key relation between the scat-tering transform and the exponentially growing solution through a differ-ential equation in the parameter variable k. We let µ(z, k) = ψ(z, k)e−izk anddefine ∂k = ∂

∂k.

Lemma 3.2.5. The mapping k 7→ µ(·, k) is differentiable as a mapping from C \0 into W1,p

−β(R2), β p > 2, and it satisfies the equation

∂kµ(z, k) =t(k)4πk

e(z,−k)µ(z, k). (3.20)

Proof. See [Nac96].

Since W1,p−β(R2) ⊂ Cloc(R2), we can consider (3.20) pointwise for z ∈ Ω

fixed.The equation (3.20) is the equation from which ψ is to be found. The

equation is a pseudoanalytic equation, and to solve it we will need thefollowing decay property of t :

Proposition 3.2.6. For any s ∈ ( p′, p)

t(k)/k ∈ Ls(R2).

Proof. A proof can be found in [Nac96].

As a consequence of Corollary 3.1.4 and Proposition 3.2.6, (3.20) has aunique solution v with v − 1 ∈ Lr(R2) for r > ( p′ ) = p′. To relate thissolution to µ we need a decay condition of ω(z, ·) = µ(z, ·) − 1 when |k|grows.

Proposition 3.2.7. For ω(z, ·) we have the estimate

supz

‖ω(z, ·)‖Lr(R2) < C

for r > p′.

Proof. Note that only the large k behavior has to be considered; (3.19) dealswith the small k behavior and for k in a compact set away from k = 0, ω(z, ·)


is clearly uniformly bounded in k. Choose now 2/ p < s < 1. Then for ksufficiently large, (3.15) and the Sobolev embedding theorem implies that

supz

|ω(z, ·)| ≤ C|k|s−1.

Hence for r(1 − s) > 2 and z ∈ R2 fixed it follows that (µ(z, ·) − 1) ∈Lr(R2). The result then follows by noting that r(1 − 2/ p) = r(2 − 2/p) > 2implies r > p/(p − 1) = p′.

We can now conclude that for fixed z ∈ Ω, µ(z, ·) is the unique solutionto (3.20) with µ(z, ·) − 1 ∈ Lr(Ω), r > p′.

Moreover, Proposition 3.2.7 shows that this solution must coincide withµ(z, ·). Hence µ(z, k) and then ψ(z, k) can be obtained from t by solving(3.20).From ψ to γ. From the exponentially growing solution ψ(z, k) = eizkµ(z, k)for z ∈ Ω, the potential q can be obtained by inserting ψ into (3.14) andthen the conductivity γ can be obtained by solving (2.5). However, we canobtain γ directly from ψ as the following result shows:

Corollary 3.2.8. γ can be obtained from ψ by

limk→0

ψ(z, k) = γ1/2(z).

Proof. Since ψ(·, k) − γ1/2(·) ∈ W2,p(Ω) solves the equation

(−∆ + q)(ψ(z, k) − γ1/2(z, k)) = 0 in Ω,

it can be estimated by

‖ψ(·, k) − γ1/2‖W2,p(Ω) ≤ C‖ψ(·, k) − γ1/2‖W2−1/p,p(∂Ω).

Now the result follows by combining (3.19) with the embedding H3/2(∂Ω) ⊂W2−1/p,p(∂Ω), 1 < p < 2, i.e.

‖ψ(·, k) − γ1/2‖W2−1/p,p(∂Ω) ≤ C‖ψ(·, k) − γ1/2‖H3/2(∂Ω) ≤ C|k|.

The reconstruction algorithm for conductivities γ ∈ W2,p(Ω) is now

(1) Solve (2.36) for ψ(·, k)|∂Ω for any k ∈ C \ 0.

(2) Compute t by (3.18).

(3) Solve the ∂k-equation for µ(z, ·).

(4) Reconstruct γ by the formula

limk→0

ψ(z, k) = γ1/2(z).


Since each step is governed by uniqueness, this method also solves theuniqueness question for the inverse problem.

3.3. Uniqueness for γ ∈ W1,p(Ω)

In this section we review the uniqueness proof by Brown and Uhlmann forless regular conductivities. We will describe the basic reductions, notions,and ideas and hence give the basis for the reconstruction in the next section.The main result is the following:

Theorem 3.3.1. Let Ω ⊂ R2 be open, bounded and smooth, and let γ ∈ W1,p(Ω),p > 2. Then γ is uniquely determined by Λγ.

We note that the Sobolev embedding W2,p(Ω) ⊂ W1,p(Ω), 1 < p < 2,shows that the above theorem extends Theorem 3.2.1.

The idea of the proof of Theorem 3.3.1 is the following. Instead of re-ducing the conductivity equation to a Schrodinger equation it is reducedto a first order 2 × 2 matrix equation. Let u ∈ H1(Ω) be a solution to theconductivity equation (1.1) and define the vector valued function (v, w) by

(

vw

)

= γ1/2(

∂u∂u

)

. (3.21)

Since

∇ · γ∇u = 2∂γ∂u + 2∂γ∂u + 4γ∂∂u = 0,

one easily checks that

(D − Q)

(

vw

)

= 0, (3.22)

where the matrix operator D and the matrix potential Q are defined by

D =

(

∂ 00 ∂

)

, Q =

(

0 qq 0

)

(3.23)

with

q = −γ−1/2∂γ1/2. (3.24)

By assumption q ∈ Lp(Ω), p > 2.In analogy with the scattering theory for the two-dimensional Schro-

dinger equation described above, Beals and Coifman [BC88] developed ascattering theory for (3.22). For the direct scattering problem we look fora family of functions, which solve the equation (3.22) in the whole plane

3.3. Uniqueness for γ ∈ W1,p(Ω) 41

(with Q extended by zero outside Ω) and have a certain exponential be-havior at infinity. More precisely, for each k ∈ C, we seek solutions Ψ(·, k)of the form

Ψ(z, k) = m(z, k)(

eizk 00 e−izk

)

, (3.25)

where m is a 2 × 2 matrix valued function, which approaches the identitymatrix I as |z| → ∞. The solution Ψ is called an exponentially growingsolution.

A simple calculation shows that m solves the equation

(Dk − Q)m = 0, (3.26)

where Dk is the matrix operator

Dk A =

(

∂A11 (∂ − ik)A12(∂ + ik)A21 ∂A22

)

and Q denotes multiplication by the matrix potential Q.Given our assumptions on Q we have the following result from [BU97].

We will give the proof since it shows nicely the use of complex analysis andsince the intermediate calculations are useful for the numerical considera-tions later.

Lemma 3.3.2. Let Q ∈ Lp(R2;M2), p > 2, be an off-diagonal, hermitian,compactly supported matrix. Then for any k ∈ C there is a unique solutionm(·, k) with m(·, k) − I ∈ Lr(R2;M2) for any any r > 2. Moreover, the mapQ 7→ m − I is continuous with respect to the norm topologies on Lp(R2;M2) andLr(R2;M2).

Furthermore, for any r > 2p/(p − 2) we have that

supz∈C

‖m(z, ·) − I‖Lr(R2;M2) < C (3.27)

with C depending on p, r, Q.

Proof. The equation for m is

Dkm =

(

∂m11 (∂ − ik)m12(∂ + ik)m21 ∂m22

)

=

(

qm21 qm22qm11 qm12

)

.

Now to find the first column of m, (m11, m21)T, note that

m± = m11 ± m21e(z,−k) (3.28)

solves (cf. (3.17))

∂m± = ±qe(z,−k)m±. (3.29)

Since q ∈ Lp(R2) is compactly supported q ∈ Ls(R2), 1 ≤ s ≤ p and henceCorollary 3.1.4 gives the existence of a unique m± with m± − 1 ∈ Lr(R2)


for any r > 2. The second column of m is now found from (3.30). That mdepends continuously on Q is a consequence of Lemma 3.1.5.

The estimate (3.27) is due to Li-Yeng Sung; a proof can be found in[BU97]. We note that this fact relies on q ∈ Lp(Ω), p > 2.

Note that unlike the result of Lemma 3.2.3, the proof of existence of ex-ponentially growing solutions to (3.22) is also valid when Q is not definedfrom a conductivity. Note further that (3.3) implies that m ∈ Cα(R2;M2),α < 1 − 1/p.

Due to the symmetry in Dk and Q, the uniqueness in Lemma 3.3.2 im-plies that

m11(z, k) = m22(z, k), m21(z, k) = m12(z, k), (3.30)

so it suffices to consider only one column of m. We will occasionally do so.Define now in analogy with (3.12) the scattering transform of the po-

tential Q by

S(k) =iπ

∫

C

(

0 e−izkq(z)Ψ22(z, k)−eizkq(z)Ψ11(z, k) 0

)

dµ(z). (3.31)

S relates for large k to the Fourier transform of q

S(k) ∼ i2(

0 q(2k1, 2k2)

−q(−2k1, k2) 0

)

as k → ∞. (3.32)

Since Ψ solves (3.22), Q is supported in Ω, and ∂eizk = ∂e−izk = 0, integra-tion by parts in (3.31) shows that

S(k) =i

2π

∫

∂Ω

(

0 e−izkν(z)Ψ12(z, k)−eizkν(z)Ψ21(z, k) 0

)

dσ(z). (3.33)

In (3.33) ν is the complex normal at the boundary, i.e. if (ν1(z), ν2(z)) de-notes the outer unit normal at z ∈ ∂Ω, then ν(z) = ν1(z) + iν2(z) andν = ν1(z) − iν2(z).

The scattering transform S is the key intermediate object in the solutionof the inverse problem. Indeed, the uniqueness proof can decomposed intothree steps. The first step is to see that Λγ determines S, the second step isto determine Q from S, and the third step is to determine γ from Q.

The first step is treated by the following lemma. Let γi ∈ W1,p(Ω), i =

1, 2, define Qi by (3.23), and let Ψ(i) be the exponentially growing solutionto (3.22). Then

Lemma 3.3.3. If Λγ1 = Λγ2 then Ψ(1)(·, k) = Ψ(2)(·, k) for any in R2 \ Ω, k ∈C.

3.3. Uniqueness for γ ∈ W1,p(Ω) 43

Proof. See [BU97].

This lemma says by (3.33) in particular that Λγ determines S.The next step in the uniqueness proof is to prove injectivity of the non-

linear map Q 7→ S. Consider m(z, ·) as a function of the dual variable k ∈ C.Then we have the following useful relation between m and S.

Lemma 3.3.4. The scattering transform S ∈ L2(R2;M2). Furthermore, for fixedz ∈ R2, the map k 7→ m(z, k) is a differentiable map from C into Lr

−β(R2;M2),and we have the equation

∂km(z, k) = m(z, k)Λk(z)S(k), (3.34)

where

Λk(z) =

(

e(z, k) 00 e(z,−k)

)

.

Proof. This is a combination of results from [BC88] and [BU97].

The equation (3.34) is not a pseudoanalytic equation in the sense of(3.6), but it can be rewritten in a suitable way.

Lemma 3.3.5. Let z ∈ C be fixed and let

m±(z, k) = m11(z, k) ± m12(z, k). (3.35)

Then we have the equation

∂km±(z, k) = ±S21(k)e(z,−k)m±(z, k). (3.36)

Proof. This is a consequence of (3.30) and (3.34).

Now since S ∈ L2(R2;M2), Liouville’s theorem for pseudoanalyticfunctions states that (3.36) has at most one solution subject to the condi-tion m±(z, ·) − 1 ∈ Lr(R2) (cf. (3.27)). Hence m(z, k), z, k ∈ C is uniquelydetermined by S. Now to obtain Q from m we can either use the equation(3.22) directly or use the formula

Q(z) = limk0→∞

1π

∫

k:|k−k0|<1Dkm(z, k)dµ(k) (3.37)

from [BU97]. This shows that S determines Q.The last step is to determine γ from Q. Once again we invoke the gen-

eralized Liouville theorem to conclude that γ1/2 is the unique solution to(3.24) with γ1/2 ∼ 1. This ends our sketch of the uniqueness proof for con-ductivities γ ∈ W1,p(Ω), p > 2.


3.4. Reconstruction of γ ∈ C1+ε(Ω)

In this section we will turn the uniqueness proof in section 3.3 into a recon-struction algorithm. The main result is

Theorem 3.4.1. Let Ω be open, bounded and smooth, and let γ ∈ C1+ε(Ω) forsome ε > 0. Then γ can be reconstructed uniquely from Λγ.

We note that compared to the Theorem 3.3.1 we assume in Theorem 3.4.1slightly more regularity. We need this extra smoothness only when solvingthe pseudoanalytic equation (3.36) constructively. We will return to thislater.

In the proof of Theorem 3.4.1 we follow closely the method described inthe previous section, and we will see how each step can be made construc-tive. In particular we give a method for reconstruction of the exponentiallygrowing solutions on the boundary of Ω. This gives a constructive wayof obtaining the scattering transform S(k) from Λγ. Next we will considerthe solvability of the ∂k-equation (3.36). To be precise, we consider the ∂k-equation for the function m given by Lemma 3.3.2 as the unique solution toDkm = −QTm with m ∼ I. It turns out that m+ (defined in (3.35)) satisfiesthe pseudoanalytic equation

∂km+(z, k) = S21(−k)e(z,−k)m+(z, k) (3.38)

with the asymptotic condition m+ ∼ 1. From this equation we can recon-struct m+ from S. Moreover, we will see that the conductivity γ can bereconstructed from m+ by

γ1/2(z) = Re(m+(z, 0)). (3.39)

The outline of the reconstruction method is then:

(1) Determine Ψ on ∂Ω (see Theorem 3.4.4 below).

(2) Define the scattering transform S by (3.33).

(3) Solve (3.38) for m+(z, ·), z ∈ Ω.

(4) Recover γ from (3.39).

3.4.1. A boundary relation. In this section we give an explicit characteri-zation of the Cauchy data for the first order system (3.22), in case the po-tential Q comes from a conductivity. This characterization will enable us toconstruct Ψ on ∂Ω.

To fix notation let s : [0, |∂Ω|] → ∂Ω be an arclength parameterizationof ∂Ω and let for any f ∈ C1(∂Ω), ∂τ f = d/dt( f (s(t))) be the derivativealong the boundary. Let further C0(∂Ω) = f ∈ C(∂Ω) :

∫

∂Ωf dσ = 0,

3.4. Reconstruction of γ ∈ C1+ε(Ω) 45

and define on C0(∂Ω)

(∂−1τ f )(s(t)) =

∫ t

0f (s(t′))dt′.

Note that (∂−1τ ∂τ f )(s(t)) = f (s(t)) − f (s(0)) for f ∈ C1(∂Ω).

When γ ∈ C1+ε(Ω) it is natural to measure the smoothness of solutionsto the conductivity equation in Holder spaces. Assume u ∈ C2+ε(Ω) solves(1.1) for some f ∈ C2+ε(∂Ω). Then (v, w) = γ1/2(∂u, ∂u) solves (3.22).Decomposing ∂ and ∂ at the boundary in the normal direction ν = (ν1, ν2)and the tangential direction τ = (−ν2, ν1) gives

2v = 2∂u = (1,−i) · ∇u = ((ν1 − iν2)ν + (−ν2 − iν1)τ) · ∇u

= (v1 − iv2)Λγ − (ν1 − iν2)i∂τ,

2w = 2∂u = (ν1 + iν2)Λγ + (ν1 + iν2)i∂τ.

Equivalently(

vw

)∣

∣

∣

∣

∂Ω

=12

(

ν −iνν iν

)(

Λγ f∂τ f

)

, on ∂Ω. (3.40)

Inverting (3.40) gives(

Λγ f∂τ f

)

=

(

ν νiν −iν

)(

vw

)∣

∣

∣

∣

∂Ω

. (3.41)

Now since f ∈ C2+ε(∂Ω), we find by the fundamental theorem of calculusthat

0 =

∫

∂Ω

∂τ f dσ = i∫

∂Ω

(νv − νw)dσ.

Hence (νv − νw)|∂Ω ∈ C1+ε0 (∂Ω) and from (3.41) we arrive at the relation

iHγ(νv|∂Ω − νw|∂Ω) = (νv|∂Ω + νw|∂Ω),

where the operator Hγ = Λγ∂−1τ is defined on C1+ε

0 (∂Ω). This relation mo-tivates the definition of the space

BR = (h1, h2) ∈ C1+ε(∂Ω) × C1+ε(∂Ω) : (νh1 − νh2) ∈ C1+ε0 (∂Ω)

iHγ(νh1 − νh2) = νh1 + νh2.(3.42)

A pair of functions (h1, h2) ∈ BR is said to satisfy the boundary relation.For Q ∈ Cε(R2) it follows as a consequence of Proposition 3.1.1 that the

natural Cauchy data for the system (3.22) is the set

CQ = (v|∂Ω, w|∂Ω) : (v, w) ∈ C1+ε(Ω) × C1+ε(Ω), (D − Q)(v, w)T = 0.

The boundary value of any solution to (3.22) is in CQ, and we just saw that ifthe solution is defined through (3.21), then on the boundary the boundary


relation is satisfied. However, as the following theorem shows, (3.42) is infact a complete characterization of CQ :

Lemma 3.4.2. If Q ∈ Cε(Ω) is given by (3.24), then CQ = BR

Proof. First we show that BR ⊂ CQ. Let (h1, h2) ∈ BR and let u ∈ C2+ε(Ω)be the unique solution to the Dirichlet problem

∇ · γ∇u = 0, in Ω,u = i∂−1

τ (νh1 − νh2), on ∂Ω.

Define a solution (v, w) to (3.22) by the relation (3.21) with u from above.Then (v, w)|∂Ω = (h1, h2) by (3.40) and (3.42), i.e. (h1, h2) ∈ CQ.

Next we see CQ ⊂ BR : let (h1, h2) ∈ CQ and let (v, w) ∈ C1+ε(Ω) ×C1+ε(Ω) be a solution to (3.22) with Cauchy data (h1, h2). Since Q is of theform (3.24), we have the compatibility relation

∂(γ−1/2v) = ∂(γ−1/2w),

which ensures the existence of a u ∈ C2+ε(Ω) such that

γ−1/2(

vw

)

=

(

∂u∂u

)

.

It is easy to check that u is a solution of the conductivity equation in theform 2∂u∂γ + 2∂u∂γ + 4γ∂∂u = 0. Now relation (3.40) with f = u|∂Ω showsthat (h1, h2) ∈ BR.

3.4.2. Construction of Ψ(·, k)|∂Ω. In this section we show how to recon-struct the trace on ∂Ω of the exponentially growing solutions Ψ definedby (3.25). Due to (3.30) it suffices to consider only the first column of Ψ.Note that Ψ11 is analytic outside Ω while Ψ21 is anti-analytic outside Ω.Moreover, they have a prescribed behavior at infinity and their traces on∂Ω satisfy the boundary relation (3.42). We will prove that these relationscharacterize the trace of (Ψ11, Ψ21) uniquely.

Introduce

gk(z) =1π

e−ikz

z,

a Green’s kernel for ∂ which takes into account exponential growth at in-finity. Define the single layer potentials Sk and Sk as boundary integraloperators by

Sk f (z) = p.v.∫

∂Ω

f (ζ)gk(ζ − z)dζ, Sk f (z) = p.v.∫

∂Ω

f (ζ)gk(ζ − z)dζ.

(3.43)


Since the integral kernels have the same singularity as the one treated inProposition 3.1.6, Sk, Sk ∈ B(Cm+α(∂Ω)) for 0 < α < 1 and m ∈ Z+.

The following result gives an equation on the boundary ∂Ω for the ex-ponentially growing solutions. This equation encodes in some sense theexponential growth.

Lemma 3.4.3. Let r > 2 and assume that the functions v, w satisfy (e−izkv −1), e−izkw ∈ Lr(R2), and that ∂v, ∂w ∈ Cα(R2), 0 < α < 1, are compactlysupported in Ω. Then on ∂Ω

(I − iSk)v = 2eizk,

(I + iSk)w = 0.(3.44)

Proof. Let z ∈ C \ Ω. Then since ∂v ∈ Cα(R2) is compactly supported in Ω,Proposition 3.1.1 implies v ∈ C1+α

loc (R2). Moreover, v(z)e−izk − 1 ∈ Lr(R2)and (3.2) shows that

v(z)e−izk − 1 =1π

∫

R2

∂(v(ζ)e−iζk − 1)

z − ζdµ(ζ)

=1π

∫

Ω

∂(v(ζ)e−iζk)

z − ζdµ(ζ)

=1

2πi

∫

∂Ω

v(ζ)e−iζk

z − ζdζ

where the last equality follows by integration by parts. This is equivalentto

v(z) +1

2πi

∫

∂Ω

v(ζ)e−i(ζ−z)k

ζ − zdζ = eizk, (3.45)

and the equation for v now follows by letting z ∈ C \ Ω approach theboundary and then applying (3.11).

To prove the result for w, we note that we−izke(z, k) = weizk ∈ Lr(R2).As above we then get

w(z) − 12πi

∫

∂Ω

w(ζ)e−i(ζ−z)k

ζ − zdζ = 0, z ∈ C \ Ω, (3.46)

and again the result follows by invoking (3.11).

The next result shows that the conditions (3.42) and (3.44) are a com-plete characterization of the traces of the exponentially growing solutions.


Theorem 3.4.4. The system of equations on ∂Ω

(I − iSk) 00 (I + iSk)

(iHγ − 1)ν −(iHγ + 1)ν

Φ =

2eizk

00

, (3.47)

has in Φ ∈ C1+ε(∂Ω) × C1+ε(∂Ω) : (νΦ1 − νΦ2) ∈ C1+ε0 (∂Ω) the unique

solution Φ = (Ψ11, Ψ21)|∂Ω.

Proof. Since (Ψ11, Ψ21)|∂Ω ∈ CQ and satisfies the assumptions of Lemma3.4.3, it is a solution to (3.47).

Let h ∈ (h1, h2) ∈ C1+ε(∂Ω) × C1+ε(∂Ω) : (νh1 − νh2) ∈ C1+ε0 (∂Ω) be

another solution to (3.47). Then

(iHγ − 1)νh1 − (iHγ + 1)νh2 = 0

shows that h ∈ BR. Then by Lemma 3.4.2 we may extend h inside Ω toa solution (v, w) to the system (3.22), and guided by (3.45) and (3.46) weextend h in C \ Ω to (v, w) by

v(z) = − 12i

∫

∂Ω

h1(ζ)gk(ζ − z)dζ + eizk,

w(z) =12i

∫

∂Ω

h2(ζ)gk(ζ − z)dζ.

We will prove that defined this way, (v, w) is a solution to (3.22) and thatve−izk − 1, we−izk ∈ Lr(R2), r > 2. Then the result follows from the unique-ness in Lemma 3.3.2.

By construction (v, w) solves (3.22) in Ω and in C \ Ω, so we have toshow that the equation holds as we cross ∂Ω. Since the support of Q isstrictly inside Ω, v is analytic and w is anti-analytic in a neighborhood ofthe boundary. Hence for the equation to hold it suffices to have continuityacross the boundary and then invoke Morera’s theorem (see for instance[Ahl53, p. 98]). We shall show the continuity of v, for w similar reasoningworks. Let z approach some point z0 ∈ ∂Ω from outside. Then using (3.11)we get

limz→z+

0

v(z) = −(

−h1(z0)

2+

12iSkh1(z0)

)

+ eiz0k.

Now use (3.47) to conclude limz→z+0

v(z) = h1(z0). The continuity of v frominside follows by the construction.

That ve−izk − 1 ∈ Lr(R2) follows since

v(z)e−izk − 1 = − 12πi

∫

∂Ω

h1(ζ)e−ikζ

ζ − zdζ = O

(

1|z|

)

, as |z| → ∞.

This proves the theorem.


This characterization gives a method for finding (Ψ11, Ψ21) on ∂Ω andhence by (3.33) a method for finding the scattering transform.

3.4.3. From S(k) to γ. In this section we will see how γ can be recon-structed from S(k). More precisely we will solve the equation (3.38) for m,and then finally find γ by the formula (3.39).

First the pseudoanalytic equation (3.38) for m+ will be derived.

Lemma 3.4.5. Let for some p > 2, Q ∈ Lp(R2;M2) be an off-diagonal, her-mitian, compactly supported matrix and let Q = −QT. Denote by S and S theassociated scattering transforms. Then

S(k) = S(−k). (3.48)

Proof. In [BC88] it was noted that

S(k) = S(−k)T;

a proof can be found in [BBR01]. Furthermore, by (3.30) it follows that

S12(k) =iπ

∫

q(z)e(z,−k)m22(z, k)dµ(z)

=−iπ

∫

q(z)e(z, k)m11(z, k)dµ(z)

= S21(k),

which shows the claim.

The pseudoanalytic equation for m is (3.36) with S21(k) = S21(−k) in-stead of S21. This substitution shows that (3.38) holds.

When γ ∈ C1+ε the special solution m and the scattering transform Sare slightly more regular than what is stated in Lemma 3.3.2. This extraregularity is what we need for (3.38) to be uniquely solvable.

Lemma 3.4.6. Assume Q ∈ Cε(R2;M2), 0 < ε < 1, is off-diagonal, hermitianand compactly supported. Then the scattering transform of S ∈ Ls(R2;M2) forany s > 4/(2 + ε). Furthermore, the equation (3.38) has the unique solutionm+(z, ·) − 1 in Lr(R2) ∩ Cα(R2) for r > 4/ε and α < (1 + ε)/2.

Proof. The Ls(R2) property of S can be found in [BU96, BBR01]. Then theunique solvability of (3.38) in Lr(R2) follows from Corollary 3.1.4, sinceS12 ∈ Ls(R2), s > 4/(2 + ε), ε > 0, and m+(z, ·) − 1 ∈ Lr(R2) for any r > 2by (3.27). The Holder continuity of m+ follows from (3.3).

This lemma is where the assumption γ ∈ C1+ε(R2) is used. If q isonly in Lp(Ω) for some p > 2, then a result from [BU97] shows that S ∈L2(R2;M2). This does imply uniqueness of a solution to (3.38) (cf. Theorem


3.1.3), but seems insufficient for solvability of the equation in a suitableLebesgue space.

By the previous lemma we find m+(z, k) for fixed z, k ∈ C. To recon-struct γ we use the formula [BBR01, Proposition 4.2]

γ1/2(z) = m11(z, 0) + m21(z, 0),

which follows from Corollary 3.1.4 since

∂(m+(z, 0) − γ1/2(z)) = −q(m+(z, 0) − γ1/2(z)),

(m+(z, 0) − γ1/2(z)) ∈ Lr(R2).

Since γ is real, we rewrite the formula using (3.30) and (3.35)

γ1/2(z) = m11(z, 0) + m21(z, 0)

= Re m11(z, 0) + Re m21(z, 0)

= Re m11(z, 0) + Re m12(z, 0)

= Re m+(z, 0).

This ends the reconstruction of conductivities γ ∈ C1+ε(Ω).

3.5. Comparison of the two reconstruction algorithms

We have now two algorithms available for the reconstruction of a conduc-tivity γ ∈ C1+ε(Ω) ∩W2,p(Ω), ε > 0, p > 1. Both methods are based on theexistence of exponentially growing solutions, the construction of a scatter-ing transform from the boundary data, and the solution of a ∂k-equation.Moreover, the conductivity is in both cases obtained as a low frequencylimit. In this section we will investigate the similarities of the two methodsfurther. We will see that the method presented in section 3.4 in some sensecan be seen as a direct generalization of the method presented in section 3.2.

As noted in the proof of Lemma 3.4.2 it is possible to go back and forthbetween solutions to the conductivity equation and solutions to (3.22). Us-ing this idea Barcelo, Barcelo and Ruiz proved the existence of exponen-tially growing solutions to the conductivity equation [BBR01]. Moreover,they showed that the scattering transform S can be found using these ex-ponentially growing solutions. Their result is

Lemma 3.5.1. Let γ ∈ W1,p(Ω), p > 2. Then for any k ∈ C \ 0 there is aunique solution

u(z, k) = eizk(1ik

+ ω(z, k)) (3.49)

3.5. Comparison of the two reconstruction algorithms 51

to ∇ · γ∇u = 0 in R2 such that

ω(·, k) ∈ W1,r(R2), 2 < r < ∞.

The norm is estimated by

‖ω(·, k)‖W1,r(R2)

≤ C(1 +1k)(‖γ−1/2m11 − 1‖Lr(R2) + ‖m11∂γ−1/2 + γ−1/2qm21‖Lr(R2)),

(3.50)

where C is independent of k.

Assume further that γ ∈ C1+ε(Ω) is such that γ = 1 near ∂Ω. Then

1k(S(1)

21 (k) − S(2)21 (k)) =

12

∫

∂Ω

u1(z,−k)(Λγ1 − Λγ2)u2(·, k)dσ(z), (3.51)

where S(i)21 and ui are the scattering transform and special solution corresponding

to γi, i = 1, 2.

Proof. We will review the construction of the solutions. For a proof of (3.51)we refer to [BBR01, Proposition 3.1].

Assume there is a solution u to the conductivity equation of the form(3.49) and let (v, w) = γ1/2(∂u, ∂u) be the corresponding solutions to (3.22).Then

e−izkv − 1 = (γ1/2 − 1) + γ1/2(∂ + ik)ω ∈ Lr(R2),

and by the uniqueness in Theorem 3.3.1 we see that e−izkv = m11. Thiscalculation motivates the definition

ω = (∂ + ik)−1(γ−1/2m11 − 1).

The W1,r(R2) property and (3.50) now follow by Proposition 3.1.2, since

∂(γ−1/2m11 − 1) = ∂(γ−1/2)m11 + γ−1/2qm21 ∈ Lr(R2)

for any 1 ≤ r ≤ 2.

Specializing (3.51) to the case γ1 = 1 in Ω we have that

S21(k) =−i2

∫

∂Ω

eizk(Λγ − Λ1)u(·, k)dσ(z), (3.52)

since for this particular conductivity u1(z, k) = 1ik eizk and S(1) = 0. In this

formula u(z, k) is the special solution of Lemma 3.5.1 corresponding to theconductivity γ. This formula for the scattering transform S21 is similar tothe formula (3.18) for t, and in fact it enables us to relate the two quantitiesin case γ is sufficiently smooth. We denote as usual by ψ the exponentiallygrowing solution to the Schrodinger equation.


Theorem 3.5.2. Let γ ∈ C1+ε(Ω) ∩ W2,p(Ω) for p > 1, ε > 0, and assumeγ = 1 near ∂Ω. Then

iku(z, k) = γ−1/2(z)ψ(z, k)

and the scattering transforms satisfy

t(k) = −2kS21(k). (3.53)

Proof. Since the function

iku(z, k) = eizk(1 + ikω(z, k))

= eizkγ−1/2(γ1/2 + γ1/2ikω(z, k))

= eizkγ−1/2(1 + (γ1/2 − 1) + γ1/2ikω(z, k))

is a solution to the conductivity equation with (γ1/2 − 1) + γ1/2ikω(·, k) ∈W1,r(Ω), and

γ−1/2(z)ψ(z, k) = eizkγ−1/2(1 + ω(z, k)),

is another solution to the conductivity equation with ω(·, k) ∈ W1,p(R2),by the uniqueness of exponentially growing solutions (Lemma 3.2.3 andLemma 3.5.1), we see that they must agree.

Furthermore, (3.18) and (3.52) implies by the assumption that γ = 1 on∂Ω that

S21(k) =−i2

∫

∂Ω

eizk(Λγ − Λ1)u(·, k)dσ(z)

=−12k

∫

∂Ω

eizk(Λγ − Λ1)ψ(·, k)dσ(z)

=−12k

t(k).

Equation (3.53) relates the coefficients in the pseudoanalytic equations(3.20) and (3.38) for γ ∈ C1+ε(Ω) ∩ W2,p(Ω), p > 1. This shows that the∂k-equations solved as the third step in the two reconstruction methods areessentially identical.

An interesting question is now whether the exponentially growing so-lutions u from Lemma 3.5.1 satisfy a boundary integral equation similar to(2.36). An affirmative answer is given in the following result.

Theorem 3.5.3. Let γ ∈ W1,p(Ω) for a p > 2 and assume that γ = 1 near ∂Ω.Then for any k ∈ C \ 0 the trace of the exponentially growing solution u(·, k)|∂Ω

3.6. Numerical implementation of the algorithm 53

is the unique solution to

u(z, k) =1ik

eizk − Sk(Λγ − Λ1)u(·, k), z ∈ ∂Ω. (3.54)

Moreover, the operator (I + Sk(Λγ − Λ1)) is invertible in H1/2(∂Ω).

Proof. Let γnn∈N ⊂ W2,r(Ω), 1/r = 1/p + 1/2 be a sequence converg-ing to γ in W1,p(Ω). Denote by ψn the exponentially growing solution tothe Schrodinger equation with potential γ−1/2

n ∆γ1/2n and by u, un the expo-

nentially growing solutions to the conductivity equation with conductivityγ, γn respectively. Since m depends continuously on γ (Lemma 3.3.2), (3.50)implies that so does ω. Hence it follows that for k 6= 0

γ−1/2nik

ψn(·, k) = un(·, k) → u(·, k) in H1/2(∂Ω).

This result and the continuity of the Dirichlet-to-Neumann map with re-spect to the conductivity (cf. (2.12)) now implies that

Sk(Λγn − Λ1)un(·, k) → Sk(Λγ − Λ1)u(·, k) in H1/2(∂Ω),

which shows that u(·, ζ)|∂Ω satisfies (3.54). Uniqueness follows from theuniqueness in Lemma 3.5.1, since by Theorem 2.4.10 any solution to (3.54)can be extended to an exponentially growing solution to the conductivityequation.

Invertibility of (I + Sk(Λγ − Λ1)) in H1/2(∂Ω) follows from the com-pactness of Sk(Λγ − Λ1) (see Lemma 2.4.11) and the uniqueness above.

This theorem validates that the computation of the scattering transformS can be done by first computing the function u(·, k)|∂Ω and then S21 by(3.52). With this modification the algorithm is

(1) Solve (3.54) for u(·, k)|∂Ω for any k ∈ C \ 0.

(2) Compute S21 by (3.52).

(3) Solve (3.38) for m+(z, ·) for any z ∈ R2.

(4) Reconstruct γ by

Re(m(z, 0)) = γ1/2(z).

We see now that this new procedure is a direct generalization of Nachman’smethod to the class of less regular conductivities.

3.6. Numerical implementation of the algorithm

In this section we describe a preliminary implementation of the algorithmdescribed in section 3.4. We will test the algorithm on a radial conductivity


defined on the ball Ω = B(0, 1). We choose the conductivity to be con-stant 1 near the boundary. This example is chosen because we then have afast and reliable algorithm available for the computation of the Dirichlet-to-Neumann map from the conductivity. Furthermore, as we will see, sym-metry in γ implies symmetry in S, and this will be extremely useful from acomputational point of view.

We will first describe our test example and see how the Dirichlet-to-Neumann map can be computed. Second we will describe an approximatemethod for computing the scattering transform. Next we will focus on thenumerical solution method for pseudoanalytic equations. Then we see howsymmetry in the conductivity gives symmetry in the scattering transform,and finally we will show some numerical results.

3.6.1. Test example and computation of the Dirichlet-to-Neumann map.Our test example is the radial conductivity

γ(z) =

(1 + 10F(|z|)), −ρ < |z| < ρ,1, |z| ≥ ρ,

where

F(t) = |t2 − ρ2|1.1|t2 − ρ2/4|1.1 cos(π

ρt),

and ρ = 3/4. This defines γ ∈ C1.1(B(0, 1)), with γ − 1 supported onB(0, 3/4). We note that γ is smooth away from the C1.1 singularities ρ, ρ/2.A plot of the test conductivity can be seen in Figure 1.

−1 −0.5 0 0.5 1

0.6

0.8

1

1.2

1.4

1.6

Test conductivity

x

γ(x)

Figure 1. Profile of the test conductivity.


For this conductivity we will compute the Dirichlet-to-Neumann map.As noted by [Syl92] in the case of a radial conductivity on B(0, 1), theDirichlet-to-Neumann map has eigenfunctions einθ , n ∈ Z, i.e.

Λγeinθ = λneinθ .

In particular

Λ1einθ = |n|einθ , n ∈ Z.

Thus the Dirichlet-to-Neumann map can be represented by the eigenvaluesλn and the difference Λγ − Λ1 can be represented by λn − |n|. In [SMI00] adirect method for the computation of the eigenvalues is proposed. The ideais to approximate a radial conductivity from below and above by piecewiseconstant conductivities and then compute by explicit formulas the eigen-values for the Dirichlet-to-Neumann associated the approximate conduc-tivities. This gives upper and lower bounds for the eigenvalues λn, whichare then approximated by taking the average of the bounds. We have usedthis method in the computation of the eigenvalues for our specific example.We refer to [SMI00] for further details concerning this algorithm.

Figure 2 is a logarithmic plot of the eigenvalues λn − |n| for the specifictest example. It shows that the the difference is exponentially decreasing.

0 10 20 30 40 50 60 70 80−45

−40

−35

−30

−25

−20

−15

−10

−5

0

Eigenvalues for Λγ−Λ

n, logarithmic plot

|n|

ln(λ

n −|n

|)

Figure 2. The eigenvalues of Λγ − Λ1.

3.6.2. Computing the scattering transform from boundary data. The equa-tion (3.47) is a complete characterization of Ψ|∂Ω. To compute Ψ|∂Ω fromthis equation numerically, a suggestion could be to discretize the boundaryoperators involved as a matrix A and the right hand side in the equation


as a vector b, and then consider the linear system Ax = b. Then a candi-date for a solution could be found using a numerical linear solver. Havingfound the boundary values Ψ|∂Ω the scattering transform can be computedfrom (3.33) by numerical integration. This idea has in preliminary testscaused numerical difficulties, due to the fact that when the equation (3.47)is solved numerically, the exponentially growing diagonal part will domi-nate the solution. However, the computation of S by (3.33) involves onlythe off-diagonal entries of Ψ|∂Ω, which are small and dominated by com-putational errors. In some sense this makes the computation of S by (3.33)based on information obtained from (3.47) hard, and the practical value of(3.47) needs further investigation.

To overcome this difficulty we suggest a different method for the com-putation of an approximation of S from Λγ based on the formula (3.52). Inthe implementation of Nachman’s method [SMI00] the non-physical scat-tering transform is not computed by solving the boundary integral equa-tion (2.36) and then inserting the boundary values into (3.18). Instead a firstorder approximation of t is computed by inserting the approximation eizk

for ψ in (3.18). Following this idea we suggest the approximation

Sexp21 (k) =

−12k

∫

∂Ω

eizk(Λγ − Λ1)eizkdσ(z). (3.55)

We will use this formula for our computation of the scattering transform.We note that (3.19) and

‖eizk − 1‖H3/2(∂Ω) < C|k|

easily imply that for |k| sufficiently small

‖ψ(·, k) − eizk‖H3/2(∂Ω) < C|k|.

Since Ran(Λγ − Λ1) ⊂ H1/20 (∂Ω) this shows that for k > 0

|S21(k) − Sexp21 (k)| ≤

∣

∣

∣

∣

Ck

∫

∂Ω

(eizk − 1)(Λγ − Λ1)(ψ(z, k) − eizk)dσ(z)∣

∣

∣

∣

≤ C|k|.

Hence we can expect the approximation to be accurate for small values ofk.

Sexp21 is implemented by decomposing eizk in a Fourier basis on einθ ,

which makes the application of Λγ − Λ1 trivial. The formula is then (cf.[SMI00, formula (41)]

Sexp21 (k) =

−12k

∞

∑n=0

(−1)n|k|2n

(n!)2 (λn − n). (3.56)


In the implementation this series is simply cut off at a sufficiently largen (= 75).

3.6.3. Solving pseudoanalytic equations numerically. In the algorithm wehave proposed, a pseudoanalytic equation appears both in the forwardproblem (3.29) and in the inverse problem (3.38). In our numerical testsolving (3.29) will be the crucial step in the computation of the exact S21from the known potential, while solving (3.38) is an essential step in thealgorithm for reconstruction of the conductivity. Both equations can betransformed into Fredholm equations of the second kind having the genericform

(I − ∂−1

(T(·, p)·))m(v, p) = 1, (3.57)

where v is the variable (z, k, respectively), p is the parameter (k, z, respec-tively), T : C × C 7→ C is the multiplier in the pseudoanalytic equation(±q(z)e(z,−k), S21(−k)e(z,−k), respectively) and

∂−1

(T(·, p) ·))m(·, p) =1π

∫

R2

T(v′, p)

v − v′ m(v′, p)dµ(v′).

In the following we will suppress the dependency on the parameter p.The first problem in the numerical realization of (3.57) is that the equa-

tion is an equation in an unbounded domain, so to make it practicable wewill have to truncate in a suitable way. The first truncation is done on thedomain of integration for the convolution operator. In the forward problem(3.29) this is not a issue since q is compactly supported, but in the inverseproblem this cut-off will introduce a systematic error. Lemma 3.1.5 ensures,however, that if the domain of integration is sufficiently large then the er-ror can be neglected. We note that choosing a smooth cut-off instead mayimprove the numerical results.

Assuming that the domain of integration in (3.57) is bounded we willnow make a second truncation by considering the equation as an integralequation in a bounded domain Ω, i.e.

(I − RΩ∂−1

(T ·))m = 1 in Ω, (3.58)

where RΩ is the restriction operator to Ω. The following lemma states thatthe restricted equation is uniquely solvable, and that the solution coincideswith the solution to (3.57) on Ω.

Lemma 3.6.1. Assume that T ∈ Lp(R2), p > 2, is compactly supported in thebounded domain Ω. Let m be the unique solution to (3.57) with m − 1 ∈ Lr(R2),r > 2. Then (3.58) has the unique solution m|Ω in Lr(Ω).


Proof. It is clear that m|Ω is a solution to (3.58) in Lr(Ω). Uniqueness fol-lows from the uniqueness of the solution to (3.57), since any solution m0 to(3.58) can be extended beyond Ω by

m0(v) =1π

∫

Ω

T(v′)v − v′ m0(v′)dµ(v′), v ∈ R

2 \ Ω

as a solution to (3.57).

Note that the uniqueness of a solution to (3.58) gives invertibility in

Lr(Ω) of (I − RΩ∂−1

(T ·)), since RΩ∂−1

(T ·) is compact.We now need a fast and reliable implementation of (3.58). To serve

this purpose we adapt a method developed by Vainikko [Vai93, Vai00] forsolving Lippmann-Schwinger type equations. This method has been usedsuccessfully in the numerical computation of t from a known q in the im-plementation of Nachman’s algorithm [MS01].

Assume that Ω = [−1, 1]2. Let n be a positive integer and let

Z2n = (z1, z2) ∈ Z

2 | − 2n−1 ≤ zj ≤ 2n−1, j = 1, 2.

Fix the discretization level h = 21−n. Then hZ2n will be a uniform grid on

Ω consisting of N2 points with N = (2n + 1). The grid approximation of afunction f on Ω is defined by

f j,h = f (jh), j ∈ Z2n.

Discretizing (3.58) by the piecewise constant collocation method (see [Vai93,section 5.3]) yields

mj,h + (Thm)j,h = 1, j ∈ Z2n

where

(Thφ)j,h = ∑k∈Z2

nj 6=k

gj−k,h(Tk,hφk,h),

and g(z) = 1/(πz). The convolution operator can be implemented numer-ically as a matrix directly, which would give the complexity O(N4) for oneapplication of Th. Instead of doing so we implement the convolution oper-ator using fast Fourier transform, i.e. by computing as matrices

(Thφ)·,h = iFFT(FFT(g·,h). ∗ FFT(Thφ·,h))

where .∗ denotes pointwise multiplication of matrices. This method givesa complexity O(N2 log N) for one application of the convolution operator.Having implemented the convolution operator, the equation can be solvednumerically using an iterative algorithm like GMRES [SS86].

From the general convergence analysis of algorithms for solving weaklysingular Fredholm equations of the second kind [Vai93, Theorem 5.1] we


can deduce that the convergence rate for the particular implementation de-scribed here is O(h). In our particular implementation we have the coeffi-cient N = 7. This choice gives a good accuracy with reasonable computa-tion time.

3.6.4. Symmetry in the scattering transform. The following result showshow radial symmetry in the conductivity implies symmetry in the scatter-ing transform. This result is an analogue of a result for the non-physicalscattering transform t, see [SMI00, Theorem 3.3].

Lemma 3.6.2. Let γ ∈ W1,p(Ω), p > 2 and assume γ(z) = γ(eiθz) for someangle θ. Then eiθS21(eiθk) = S21(k) and e−iθS12(eiθk) = S12(k).

Proof. First note that γ(z) = γ(eiθz) implies

q(z) = −∂γ1/2(z)γ1/2(z)

= −∂(γ1/2(eiθz))γ1/2(eiθz)

= eiθq(eiθz).

Next note that since the transformation z 7→ eiθz implies the change z 7→e−iθz, the exponentially growing solution (Ψ11(eiθz, k), e−2iθΨ21(eiθz, k)) sa-tisfies

∂Ψ11(eiθz, k) = e−iθ(∂Ψ11)(eiθz, k)

= e−iθq(eiθz)Ψ21(eiθz, k)

= q(z)e−2iθΨ21(eiθz, k)

(3.59)

∂(e−2iθΨ21(eiθz, k)) = e−iθ(∂Ψ21)(eiθz, k)

= e−iθq(eiθz)Ψ11(eiθz, k)

= q(z)Ψ11(eiθz, k).

(3.60)

Now since the vector (Ψ11(z, eiθk), Ψ21(z, eiθk)) also solves (3.59) and (3.60)and has the same asymptotic behavior when |z| → ∞, we deduce by theuniqueness in Lemma 3.3.2 that

Ψ11(eiθz, k) = Ψ11(z, eiθk)

Ψ21(eiθz, k) = e2iθΨ21(z, eiθk)

which implies

m11(eiθz, k) = m11(z, eiθk)

m21(eiθz, k) = e2iθm21(z, eiθk).


Hence

eiθS21(eiθk) =−iπ

∫

R2e−iθq(z)e(z, eiθk)m11(z, eiθk)dµ(z)

=−iπ

∫

R2q(eiθz)e(eiθz, k)m11(eiθz, k)dµ(z)

=−iπ

∫

R2q(z′)e(z′, k)m11(z′, k)dµ(z′)

= S21(k).

For S12 a similar calculation can be carried out.

This result is useful when dealing with radial conductivities since oneonly has to compute the scattering transform for real values of the para-meter k and then obtain

S21(k) = e−i arg(k)S21(|k|), k ∈ C. (3.61)

A similar identity can be found for Sexp21 .

3.6.5. Numerical results.Computing the scattering transform. We compute the scattering transformboth by solving the forward problem Q 7→ S21 and by the inverse prob-lem Λγ 7→ S21. In the first method we start by solving the pseudoanalyticequation (3.29) for m±, then solving the linear systems (3.28), and finallyintegrating in (3.31). In the implementation of this method we use themethod for solving pseudoanalytic equations numerically outlined above.The second method is the more realistic method of computing the scatter-ing transform from the Dirichlet-to-Neumann map. This method relies onthe approximation (3.55) computed by (3.56).

In Figure 3 the scattering transform S21 is displayed, and in Figure 4it is displayed together with the approximation Sexp

21 . Moreover, the errorS21(k) − Sexp

21 is displayed. Both functions are computed for a range of realpositive k. In Figure 4 the axis is truncated at k = 30, since Sexp

21 becomeshighly inaccurate beyond k = 25. We note that below this value the ap-proximation seems reasonable accurate.Reconstructing the conductivity. The reconstruction of the conductivity fromthe scattering transform is done by solving (3.38) for m+(z, k) for a suitablerange of real z and then obtaining γ(z) = (Re m+(z, 0))2. The first recon-struction is done from the true scattering transform S21 (Figure 3), which isknown for real k and extended to the complex plane by (3.61). The resultof this inversion is displayed in Figure 5 together with the true conductiv-ity. Moreover, we have plotted the relative error |γ(x) − γrec(x)|/|γ(x)|.We see that the reconstruction is very accurate where γ is smooth, but the


0 10 20 30 40 50−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

Scattering transform S21

k

S21

(k)

Figure 3. The true scattering transform.

0 10 20 30−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

Scattering transforms S21

, S21exp

k

S21

(k)

0 10 20 30−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1Error

k

Figure 4. The scattering transforms S21 (red) and Sexp21 (green).

singularities are smoothed out. This is no surprise, since we from (3.32) doexpect the singularities of γ to be represented in the large k behavior of S21,but this behavior is discarded in the numerical implementation.

The second reconstruction is based on the approximation Sexp21 . As can

be seen from Figure 4, Sexp21 (k) is unreliable for k > 25. Hence we will have

to truncate this function before doing the inversion. To illustrate how deli-cate the choice of cut-off parameter is, we have chosen to truncate at the


different values k = 10, 15, 20, 25, 28 and then reconstruct from the trun-cated Sexp

21 . In Figure 6 – 11 the reconstructions are displayed together withthe true conductivity. Furthermore, we have for comparison displayed thereconstructed conductivity based on the same truncation of the true S21. Wesee from the figures that when the cut-off is small, then the inversion givesa low frequency approximation of the conductivity. The quality of theselow frequency approximations obtained from Sexp

21 are comparable to thoseobtained from S21. When the cut-off parameter increases more detailed re-constructions appears. These reconstructions both capture some features inthe original conductivity and introduce certain artifacts. When the cut-offis taken beyond k = 25 the error in Sexp

21 dominates.

0 0.2 0.4 0.6 0.8 1

0.6

0.8

1

1.2

1.4

1.6

Reconstructed conductivity

|x|

γ(x)

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5Relative error

|x|

Figure 5. The reconstructed conductivities. Blue curve is the true con-ductivity and red curve is the conductivity reconstructed from the trueS21(k), k < 50.

3.6.6. Conclusion. In this section we have seen a numerical implementa-tion of the reconstruction algorithm for conductivities in C1+ε(Ω). We havetested the algorithm on synthetic, noiseless data. The conclusions are asfollows:

Concerning the computation of the scattering transform from the Dirich-let-to-Neumann map, the implementation based on solving (3.47) to obtainΨ21 and computing S21 by (3.33) seems to be unstable. Hence we have sug-gested a different method, which computes the approximation Sexp

21 . Thisapproximation shows good accuracy for small k < 25 and gives fair lowfrequency reconstructions, however, if more detailed images are needed,we need a better method for the computation of an approximate scattering


0 0.2 0.4 0.6 0.8 1

0.6

0.8

1

1.2

1.4

1.6

Reconstructed conductivity, truncation |k|<5

|x|

γ(x)

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5Relative error, truncation |k|<5

|x|

Figure 6. The reconstructed conductivities. Green curve is the conduc-tivity reconstructed from Sexp

21 (k), k < 5, red curve is the conductivityreconstructed from the true S21, k < 5, and blue curve is the true con-ductivity.

0 0.2 0.4 0.6 0.8 1

0.6

0.8

1

1.2

1.4

1.6


|x|

γ(x)

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4


|x|



transform. An important issue in this implementation concerns the choiceof cut-off parameter for truncating the scattering transform before the in-version is done. We have seen that this choice has a great impact on the


0 0.2 0.4 0.6 0.8 1

0.6

0.8

1

1.2

1.4

1.6


|x|

γ(x)

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4


|x|



0 0.2 0.4 0.6 0.8 1

0.6

0.8

1

1.2

1.4

1.6


|x|

γ(x)

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4


|x|



reconstructions. In presence of noise this method of cutting of the scatter-ing transform may be seen as a non-linear regularization strategy [MS01].


0 0.2 0.4 0.6 0.8 1

0.6

0.8

1

1.2

1.4

1.6


|x|

γ(x)

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4


|x|



0 0.2 0.4 0.6 0.8 1

0.6

0.8

1

1.2

1.4

1.6


|x|

γ(x)

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4


|x|



The implementation has shown that the direct and inverse scatteringtransform (the numerical solver for pseudoanalytic equations) are numeri-cally stable operations, even when performed by the fast method based onFFT. The fast implementation of a solver for (3.57) may be useful in other


applications as well, for instance in the numerical solution of the non-linearDavey-Stewartson equations. This is a project for further studies. Concern-ing computation speed and accuracy we note that the method in [Vai00] isactually a multi-grid method, which can give more accuracy at a low cost.We do not exploit this feature in the current implementation.

3.7. Notes

The results in section 3.1 are classical. A good reference for pseudoanalyticequations is [Vek62]. The formula (3.11) goes back to Plemelj in 1908.

Both Nachman’s method and Brown-Uhlmann’s method rely on the∂-method in inverse scattering. This method was initially introduced inthe study of some one-dimensional non-linear evolution equations, whichcould be linearized by the scattering transform [BC81, BC82]. The methodwas later developed and extended to higher dimensional problems by anumber of people. For an general introduction to the ∂-method we refer tothe review papers [BC85, BC89].

For the two-dimensional inverse conductivity problem local unique-ness for small γ ∈ W3,∞(Ω) was proved in [SU86]; the general questionremained open until 1996, when Nachman [Nac96] gave the uniquenessproof and a reconstruction algorithm outlined in section 3.2. Based on thismethod Liu [Liu97] proved a conditional stability estimate of the form

‖γ1 − γ2‖L∞(Ω) ≤ Cλ(‖Λγ1 − Λγ2‖B(H1/2(∂Ω),H−1/2(∂Ω))).

Here the constant C depends on an a priori bound of ‖γi‖W2,p(Ω), 1 < p < 2,and λ is a real function with

λ(t) ≤ | log(t)|−δ

for 0 < δ < 2/p′ and t sufficiently small. Furthermore, the algorithm hasbeen tried out numerically (see [SMI00, SMI01a, SMI01b]). We note that ascattering theory for the two-dimensional Schrodinger problem similar tothe theory outlined in section 3.2 was considered by Boite, Leon, Mannaand Pempinelli in [BLMP87] and Tsai in [Tsa93].

The inverse problem for the Schrodinger equation (when the poten-tial is not defined from a conductivity) concerning injectivity of the mapq 7→ Λq is open. There are partial results due to Sun and Uhlmann concern-ing generic uniqueness [SU91] and the recovery of the singularities of thepotential [SU93].

In 1997 Brown and Uhlmann ([BU97]) improved the uniqueness resultfor the class of less regular conductivities γ ∈ W1,p(Ω), p > 2. Basedon their method Barcelo, Barcelo and Ruiz proved conditional stability

3.7. Notes 67

[BBR01]. Indeed, if γi ∈ C1+ε(Ω) and ‖γi‖C1+ε(Ω) < M, then

‖γ1 − γ2‖L∞(Ω) ≤ Cλ(‖Λγ1 − Λγ2‖B(H1/2(∂Ω),H−1/2(∂Ω))).

Here the constant C depends on M and λ is a real function with

λ(t) ≤ | log(t)|−δ

for some δ > 0 and t sufficiently small. Following the same method Franciniproved uniqueness for conductivities γ = γR + iγI , with γR, γI real and γIis small [Fra00].

The motivation of Beals and Coifman in [BC88] for constructing a scat-tering theory for (3.22) was the applicability of this theory in the analysisof certain non-linear evolution equations. The crucial observation is thatthe transformation q 7→ S linearizes the non-linear equations, a fact thatmakes the evolution trivial. Among the results in [BC88] is that q ∈ S(R2)implies S ∈ S(R2) with norm equality ‖q‖L2(R2) = ‖S‖L2(R2). This showsthat the map q 7→ S is isometric on S(R2) equipped with the L2(R2) norm.Due to the non-linearity of the map a density argument does not directlyextend the result to L2(R2). Further results concerning the scattering theoryfor (3.22) can be found in [Sun94a],[Sun94b],[Sun94c]. The motivation isagain the solvability of a non-linear evolution equation. There the analysisis given for Q ∈ L1(R2) ∩ L∞(R2).

Chapter 4

Uniqueness andreconstruction in higherdimensions

In this chapter we will focus on the questions of uniqueness and reconstruc-tion for the inverse conductivity problem in dimensions n ≥ 3. We will firstreview the proof of uniqueness for sufficiently regular conductivities. Thenwe will present a new result concerning absence of small exceptional pointsfor a merely bounded conductivity. Next we will review the two known re-construction algorithms, including one based on the so-called ∂-method ofinverse scattering, and finally we will propose a new method for the recon-struction of almost constant conductivities.

4.1. Uniqueness for the inverse conductivity problem

The uniqueness results in this section are all well-known. We give theproofs anyway, since they show the importance of having exponentiallygrowing solutions.

The following theorem due to Sylvester and Uhlmann [SU87] answersthe uniqueness question for the inverse problem for the Schrodinger op-erator. Recall from (2.10) the definition of the set of Cauchy data Cq for(−∆ + q).

Theorem 4.1.1. Let n ≥ 3 and assume qi ∈ L∞(Ω). Then Cq1 = Cq2 impliesq1 = q2 in Ω.

69

70 4. Uniqueness and reconstruction in higher dimensions

Proof. Let ui satisfy (−∆ + qi)ui = 0 in Ω. By assumption (u1, ∂νu1)|∂Ω ∈Cq1 = Cq2 and then integration by parts shows that

∫

Ω

(q1 − q2)(x)u1(x)u2(x)dx = = 〈u1, ∂νu2〉 − 〈u2, ∂νu1〉

=

∫

Ω

(q2 − q2)(x)u1(x)u2(x)dx

= 0,

where u1 solves (−∆ + q2)u1 = 0 in Ω with (u1, ∂νu1)|∂Ω = (u1, ∂νu1)|∂Ω.Take now two parameters ζ1, ζ2 ∈ V defined by

ζ1 =12(ξ + κk⊥ + κ′ik),

ζ2 =12(ξ − κk⊥ + κ′ik),

where ξ ∈ Rn is arbitrary, k⊥, k ∈ Rn are chosen such that ξ · k⊥ = ξ · k =k⊥ · k = 0, and κ, κ′ ∈ R satisfy κ′2 = κ2 + |ξ|2 (cf. (2.19)). This choice ispossible since n ≥ 3. Then for κ sufficiently large, Theorem 2.4.3 guaranteesthe existence of exponentially growing solutions

ui(x, ζi) = eix·ζ(1 + ωi(x, ζi))

corresponding to the potential qi and parameter ζi, i = 1, 2.Since (2.26) implies that

‖eix·ξ − u1(·, ζ1)u2(·, ζ2)‖L1(Ω)

≤ C(‖ω1(·, ζ1)‖L2δ(Ω)‖ω2(·, ζ2)‖L2

δ(Ω) + ‖ω1(·, ζ1)‖L2δ(Ω) + ‖ω2(·, ζ2)‖L2

δ(Ω))

≤ C|ζ|

≤ Cκ

it follows by taking κ → ∞ that

0 = limκ→∞

∫

Ω

(q1 − q2)(x)u1(x, ζ1)u2(x, ζ2)dx

=

∫

Ω

(q1 − q2)(x)eix·ξdx

= (q1 − q2)(ξ).

Since ξ is arbitrary the result now follows by inverting the Fourier trans-form.

4.2. Absence of exceptional points near zero for γ ∈ L∞+ (Ω) 71

Note that in two dimensions it is impossible to choose the two para-meters ζ1 and ζ2 as above. This is essentially why the uniqueness proof inthis case has to be based on different methods.

Now we can apply the previous result to settle the uniqueness questionfor the inverse conductivity problem.

Corollary 4.1.2. Let γi ∈ W2,∞(Ω). Then Λγ1 = Λγ2 implies γ1 = γ2.

Proof. From Lemma 2.1.1 it follows that γ = γ1 = γ2 and ∂νγ1 = ∂νγ2

on ∂Ω. Hence with qi = ∆γ1/2i /γ1/2

i we have Λq1 = Λq2 by (2.9), whichimplies Cq1 = Cq2 . We conclude by Theorem 4.1.1 that q = q1 = q2 in Ω. Theconductivities γi thus satisfy the boundary value problem

(−∆ + q)γ1/2i = 0 in Ω, γ1/2

i = γ1/2|∂Ω on ∂Ω,

which has a unique solution γ1/2 = γ1/21 = γ1/2

2 , since the potential isdefined from a conductivity.

4.2. Absence of exceptional points near zero for γ ∈ L∞+ (Ω)

From Corollary 2.4.6 we know that for any potential there are no excep-tional points outside a sufficiently large ball. In this section we considerthe other extreme, i.e. the possibility of having small exceptional points.We assume without loss of generality that Ω = Ba, γ ∈ L∞

+ (Ba) and γ = 1near ∂Ba. Then we will show that the boundary integral equation (2.36) issolvable in H1/2(∂Ba) for any ζ ∈ V sufficiently small, and hence that sucha ζ is not exceptional for γ. Moreover, we prove that for γ sufficiently closeto 1, there are no exceptional points satisfying |ζ| ≤ 1. We emphasize herethat the main achievement is that we only assume that γ ∈ L∞

+ (Ω), i.e. nofurther regularity is assumed for the conductivity.

Recall from (2.29) and (2.18) the Green’s functions G0 and Gζ for theLaplacian and define the harmonic function

Hζ(x) = Gζ(x) − G0(x). (4.1)

The single layer potential Sζ defined in (2.28) can then be decomposed as

Sζ = S0 + Hζ,

where S0 is the usual single layer potential and

Hζφ(x) =

∫

∂Ba

Hζ(x − y)φ(y)dσ(y), x ∈ ∂Ba. (4.2)

This decomposition motivates a splitting of the operator

(I + Sζ(Λγ − Λ1)) = (I + S0(Λγ − Λ1)) + Hζ(Λγ − Λ1) (4.3)


into a free part independent of ζ and a perturbation. For the free part wehave the result.

Lemma 4.2.1. Assume γ ∈ L∞+ (Ba) satisfies γ = 1 near ∂Ba. Then the boundary

integral operator I + S0(Λγ − Λ1) is invertible in H1/2(∂Ba).

Proof. The operator S0(Λγ −Λ1) is compact in H1/2(∂Ba) by Lemma 2.4.11,and hence injectivity of I + S0(Λγ −Λ1) implies invertibility by Fredholm’salternative.

Assume now that h ∈ H1/2(∂Ba) and

(I + S0(Λγ − Λ1))h = 0. (4.4)

Let v be the unique harmonic function in Ω with v|∂Ω = h and let w =S0(Λγ − Λ1)h ∈ H1(Ba), where S0 denotes the standard single layer poten-tial. Then v + w is harmonic in Ba and has trace h + S0(Λγ − Λ1)h = 0. Itfollows that v + w = 0 in Ba. Hence from the jump in the normal derivativeof the single layer potential (see (2.34)) it follows from taking the trace of∂ν(v + w) from inside Ω that

0 = Λ1h +12(Λγ − Λ1)h + K′

0(Λγ − Λ1)h

=12(Λγ + Λ1)h + K′

0(Λγ − Λ1)h.(4.5)

Since on ∂Ba

∂νx G0(x − y) =x|x| · ∇x

1nωn|x − y|n−2

=x|x| ·

−(n − 2)(x − y)

nωn|x − y|n

=−(n − 2)

2a|x|2 + |y|2 − 2x · y

nωn|x − y|n

=−(n − 2)

2aG0(x − y)

we get by (4.4)

K′0(Λγ − Λ1)h =

−(n − 2)

2aS0(Λγ − Λ1)h =

(n − 2)

2ah. (4.6)

Inserting (4.6) into (4.5) gives

0 = (Λγ + Λ1)h +(n − 2)

ah

which implies h = 0, since

0 = 〈(Λγ + Λ1)h +(n − 2)

ah, h〉 ≥ (n − 2)

a‖h‖2

L2(∂Ba).


Next we will investigate the dependence of the perturbation in (4.3)on ζ. We need the following lemma concerning properties of the harmonicfunction Hζ:

Lemma 4.2.2. Let ζ = κ(k⊥ + ik) ∈ V with κ ∈ R, k⊥, k ∈ Rn, k⊥ · k = 0 (cf.(2.19)), and let R be a real orthogonal n × n matrix with det(R) = 1. Then theharmonic function Hζ = Gζ − G0 satisfies

Hζ(x) = κn−2Hk⊥+ik(κx), (4.7)

Hζ(x) = HRTζ(RTx). (4.8)

Proof. From (2.20) it follows that

Hζ(x) = Gζ(x) − G0(x)

= eix·ζ gζ(x) − G0(x)

= eiκx·(k⊥+ik)κn−2g(k⊥+ik)(κx) − κn−2G0(κx)

= κn−2H(k⊥+ik)(κx).

Further,

Hζ(Rx) = Gζ(Rx) − G0(Rx)

= ei(Rx)·ζ gζ(Rx) − G0(|Rx|)= eix·(RTζ)gR−1ζ(x) − G0(|x|)= HRTζ(x).

The result now follows since det(R) = 1 together with the reality of Rimplies RT = R−1.

From the properties of Hζ it is now straightforward to derive estimatesfor Hζ defined by (4.2).

Lemma 4.2.3. The integral operator Hζ satisfies for any ζ ∈ V and 0 < |ζ| ≤ 1the estimates

‖Hζ‖B(H1/2(∂Ba))≤ C|ζ|n−2, (4.9)

‖Hζ‖B(H1/20 (∂Ba),H1/2(∂Ba))

≤ C|ζ|n−1, (4.10)

where the constant C depends only on the radius a.

Proof. From (4.8) we can without loss of generality assume that ζ = κ(e1 +ie2) with e1 = (1, 0, · · · , 0), e2 = (0, 1, 0, · · · , 0). Then an application of (4.7)


gives

Hκ(e1+ie2) f (x) =

∫

∂Ba

Hκ(e1+ie2)(x − y) f (y)dσ(y)

= κn−2∫

∂Ba

He1+ie2(κ(x − y)) f (y)dσ(y). (4.11)

Define the smooth function H(x) = He1+ie2(x) − He1+ie2(0) and set

Kκ f (x) =

∫

∂Ba

H(κ(x − y)) f (y)dσ(y), x ∈ ∂Ba. (4.12)

Then

Hκ(e1+ie2) f (x) = κn−2Kκ f (x) + κn−2He1+ie2(0)

∫

∂Ba

f (y)dσ(y).

In particular

Hκ(e1+ie2) f (x) = κn−2Kκ f (x)

for f ∈ H1/20 (∂Ba).

Since H(0) = 0 and H is smooth, there is a constant C depending onlyon a such that supp|x|≤2a |H(x)| ≤ C|x|. Then for 0 ≤ κ ≤ 1

∫

∂Ba

∫

∂Ba

|H(κ(x − y))|2dσ(y)dσ(x) ≤ Cκ2,

which implies‖Kκ‖B(L2(∂Ba)) ≤ Cκ. (4.13)

Furthermore, since H is a smooth function, supp|x|≤2a |∇H(x)| ≤ C forsome constant C > 0. Hence by taking derivatives in (4.12) we find that

‖Kκ‖B(H1(∂Ba)) ≤ Cκ. (4.14)

Thus the estimate‖Kκ‖B(H1/2(∂Ba))

≤ Cκ (4.15)

can be obtained for 0 ≤ κ ≤ 1 from (4.13) and (4.14) by interpolation.

The solvability of the boundary integral equation for small ζ can nowbe proved using a standard perturbation argument.

Theorem 4.2.4. Assume γ ∈ L∞+ (Ba) satisfies γ = 1 near ∂Ba. Then the bound-

ary integral equation (2.36) is uniquely solvable in H1/2(∂Ba) for ζ ∈ V suffi-ciently small. Furthermore, if ‖γ − 1‖L∞(Rn) is sufficiently small, then solvabilityholds for all |ζ| ≤ 1.


Proof. Let

Aγ,0 = S0(Λγ − Λ1), Aγ,ζ = Sζ(Λγ − Λ1).

Then by Lemma 4.2.1 the operator I + Aγ,0 is invertible in H1/2(∂Ω). Thus

I + Aγ,ζ = I + Aγ,0 + (Aγ,ζ − Aγ,0)

= (I + Aγ,0)(I + (I + Aγ,0)−1(Aγ,ζ − Aγ,0))

is invertible if

‖(I + Aγ,0)−1(Aγ,ζ − Aγ,0)‖B(H1/2(∂Ba))

< 1.

From the decomposition Sζ = S0 + Hζ it now follows from (4.9) that for afixed γ ∈ L∞

+ (Ba)

‖(I + Aγ,0)−1(Aγ,ζ − Aγ,0)‖B(H1/2(∂Ba))

= ‖(I + Aγ,0)−1Hζ(Λγ − Λ1)‖B(H1/2(∂Ba))

≤ C|ζ|n−2‖(I + Aγ,0)−1‖B(H1/2(∂Ba))

‖Λγ − Λ1‖B(H1/2(∂Ba))

< 1

(4.16)

for ζ sufficiently small.If γ − 1 is sufficiently small then by Lemma 2.4.11, I + Aγ,0 can be in-

verted by a Neumann series which gives the estimate

‖(I + Aγ,0)−1‖B(H1/2(∂Ba))

≤ 11 − ‖S0(Λγ − Λ1)‖B(H1/2(∂Ba))

.

Hence from (4.16) and Lemma 2.4.11 we conclude that if ζ and γ satisfy theestimate

0 < |ζ|n−2 C‖γ − 1‖L∞(Ba)

1 − C‖S0‖B(H1/2(∂Ba))‖γ − 1‖L∞(Ba)

< 1,

with C being the constant from Lemma 2.4.11, then ζ is not exceptional forγ. Furthermore, we conclude that if ‖γ − 1‖L∞(Ba) is sufficiently small, thenthere are no exceptional points ζ ∈ V with |ζ| ≤ 1.

From the invertibility of the boundary integral equation we get the fol-lowing estimate of the exponentially growing solutions for small ζ.

Lemma 4.2.5. Assume γ ∈ L∞+ (Ba) satisfies γ = 1 near ∂Ba. Then for ζ ∈ V

sufficiently small

‖ψ(·, ζ) − 1‖H3/2(∂Ba)≤ C|ζ|. (4.17)

Proof. Let Aγ,ζ = Sζ(Λγ − Λ1). Then by Theorem 4.2.4 the operator I +Aγ,ζ is invertible for ζ near zero. Since (Λγ − Λ1)1 = 0 we have

ψ(x, ζ) − 1 = (I + Aγ,ζ)−1(eix·ζ − 1). (4.18)


Now ‖(I + Aγ,ζ)−1‖B(H1/2(∂Ba))

is uniformly bounded for small |ζ| and thus

‖ψ(·, ζ) − 1‖H1/2(∂Ba)≤ C‖eix·ζ − 1‖H1/2(∂Ba)

≤ C|ζ|,where the last estimate can be proved by writing the Taylor expansion foreiζ·x around zero and then using interpolation similarly to the proof ofLemma 4.2.3. To get (4.17) we note that Aγ,ζ is uniformly bounded fromH1/2(∂Ba) to H3/2(∂Ba) for |ζ| < 1. This implies that

‖ψ(·, ζ) − 1‖H3/2(∂Ba)

= ‖eix·ζ − 1‖H3/2(∂Ba)+ ‖Aγ,ζ‖B(H1/2(∂Ba),H3/2(∂Ba))

‖ψ(·, ζ) − 1‖H1/2(∂Ba),

which gives the result.

The next step would now be to prove solvability of (2.36) for a generalγ ∈ L∞

+ (Ba) and ζ ∈ V without the smallness assumptions. This wouldrequire a decay estimate for Sζ, when |ζ| → ∞, which does not seem to beavailable. However, when the conductivity is sufficiently regular, we knowfrom Corollary 2.4.6 that there are no exceptional points for large ζ. On theother hand Theorem 4.2.4 shows the absence of exceptional points for smallζ. Balancing these two statements gives the following result.

Corollary 4.2.6. Let γ ∈ W2,∞(Ba) with γ = 1 near ∂Ba. Then there is a δ > 0depending only on the radius a, such that if with ‖γ − 1‖W2,∞(Ba) < δ, then thereare no exceptional points for γ.

Proof. From Theorem 4.2.4 it follows that there is a δ1, such that when‖γ − 1‖L∞(Ba) < δ1 there are no exceptional points ζ ∈ V with |ζ| ≤ 1. Fur-thermore, from Theorem 2.4.3 there is a constant δ2 such that for ‖q‖L∞(Ba) ≤‖γ − 1‖W2,∞(Ba) < δ2 there are no exceptional points with ζ ∈ V with |ζ| ≥ 1.The result now follows by choosing δ = min(δ1, δ2).

4.3. Reconstruction of potentials and conductivities

In this section we will focus on the reconstruction issue of the inverse con-ductivity problem in higher dimensions. The methods we will outlineall rely on the reduction of the conductivity equation to a Schrodingerequation, and hence we will generally assume that the conductivity γ ∈W2,∞(Ω). First we will define the higher dimensional non-physical scat-tering transform and recall the ∂ζ-equation satisfied by the exponentiallygrowing solutions. Next we will review the two reconstruction algorithmsavailable in the literature. Finally we turn our attention to the case when theconductivity is smooth and close to a constant. The absence of exceptional

4.3. Reconstruction of potentials and conductivities 77

points for such conductivities enables us to give a new reconstruction algo-rithm, which can be seen as a direct analogue of the algorithm presented insection 3.2.

4.3.1. Scattering transform and the ∂ζ-equation. Let ζ ∈ V and assume itis not exceptional. Define for ζ ∈ V , which is not exceptional, and ξ ∈ Rn

the non-physical scattering transform of the potential q by

t(ξ, ζ) =

∫

Ω

e−ix·(ξ+ζ)q(x)ψ(x, ζ)dx, (4.19)

where ψ(x, ζ) is the unique solution to (2.25). To turn the integral into anintegral on the boundary we define for any ξ ∈ Rn

Vξ = ζ ∈ V : (ξ + ζ)2 = |ξ|2 + 2ξ · ζ = 0.

Then e−ix·(ξ+ζ) is harmonic if and only if ζ ∈ Vξ, and when ζ ∈ Vξ is notexceptional we get by (2.13) the formula

t(ξ, ζ) =

∫

∂Ω

e−ix·(ξ+ζ)(Λq − Λ0)ψ(·, ζ)dσ(x) ξ ∈ Rn, ζ ∈ Vξ. (4.20)

We note that for fixed ξ ∈ Rn, the set Vξ consists of those ζ ∈ V withreal part in the hyper plane |Re(ζ) + ξ| = |Re(ζ)| and imaginary part inthe hyper plane Im ζ · ξ = 0. Hence Vξ is a variety of complex dimensionsn − 2. Note that in two dimensions the choice ξ = (−2k1, 2k2), k1, k2 ∈ R,implies that Vξ = (k, ik), (k,−ik), k = k1 + ik2 ∈ C, and that the scatteringtransform evaluated at this ξ and ζ = (k, ik) has exactly the form (3.12).

The usefulness of the scattering transform in relation to the inverseboundary value problem is that it can be computed from the Dirichlet-to-Neumann map as just like in two dimensions. For ξ ∈ Rn and ζ ∈ Vξ notexceptional, we can first solve (2.36) to find ψ(·, ζ)|∂Ω, and then computet(ξ, ζ) from Λq by (4.20).

Consider now for fixed x ∈ Rn, µ(x, ·) as a function on V . In two di-mensions V is parameterized by one complex variable, and the absenceof exceptional points implies that µ(x, ·) can be seen as a function of onecomplex variable defined everywhere in the plane. This brings into playthe classical tools from complex analysis. In higher dimensions the issueis more complicated, since V is a variety of complex dimension n − 1, andthere there may be exceptional points. Therefore µ(x, ·) is locally a func-tion of n − 1 complex variables and has singularities at exceptional points.However, away from the singularities µ is differentiable:

Let ζ ∈ V and let Wζ = w ∈ Cn : w · ζ = 0. Using a local coordinatechart on V it can be seen that Wζ is the tangent space to V in the point ζ.


Then for f ∈ C1(V) we define the ∂ζ-derivative in the point ζ and directionw ∈ Wζ by

w · ∂ζ f (ζ) =n

∑j=1

wj∂ζ j f (ζ).

Similarly, for ξ ∈ Rn, ζ ∈ Vξ we denote by Wξ,ζ = w ∈ Cn : w · ζ =w · ξ = 0 the tangent space to Vξ in ζ, and then we define for ξ ∈ Rn andf ∈ C1(Vξ) the derivative in the point ζ ∈ Vξ in direction w ∈ Wξ,ζ by

w · ∂ζ f (ζ) =n

∑j=1

wj∂ζ j f (ζ).

For µ and t it is now possible to state the following ∂ζ-equations.

Lemma 4.3.1. Assume that ζ ∈ V is not exceptional. Then for w ∈ Vζ we have

w · ∂ζµ(x, ζ) =−1

(2π)n−1

∫

Bζ

eix·ξt(ξ, ζ)µ(x, ζ + ξ)(w · ξ)dσ(ξ), (4.21)

where Bζ = ξ ∈ Rn : (ξ + ζ)2 = 0 is the ball in the plane ξ · Im ζ = 0 centeredat c = −Re ζ with radius r = |Re ζ|.

Furthermore, for ξ ∈ Rn, ζ ∈ Vξ and w ∈ Wξ,ζ we have

w · ∂ζt(ξ, ζ) =−1

(2π)n−1

∫

Bζ

t(ξ − η, ζ + η)t(η, ζ)(w · η)dσ(η). (4.22)

for any ξ ∈ Rn, ζ ∈ Vξ.

Proof. See any of the references [NA84, BC85, NK87, Nac88] for the higherdimensional ∂ζ-equations; the formulation here is taken from [Nac88].

Note that integrating by parts in (2.18) shows that Gζ+ξ(x) = Gζ(x) forξ ∈ Rn with (ξ + ζ)2 = 0, so from (2.25) we have that ζ is exceptional ifand only if ζ + ξ ∈ V is exceptional for ξ ∈ Rn. Hence µ(x, ζ + ξ) in theright hand side of (4.21) is well-defined on the domain of integration. Thesame argument shows that the scattering transform in the right hand sideof (4.21) and (4.22) is well-defined and can be evaluated from boundarymeasurements by (4.20).

The ∂-equation (4.22) for t is in fact an implicit characterization of theadmissible set of scattering data, i.e. t is the scattering transform of some qif and only if it solves (4.22) [BC85]. In presence of noisy data it could beuseful to project the scattering transform onto the admissible set defined bythat equation before solving the inverse scattering problem of computing qfrom t.

4.3. Reconstruction of potentials and conductivities 79

4.3.2. Reconstruction of potentials and conductivities. There are two meth-ods available for reconstruction of the potential q from its scattering trans-form t. Since (4.19) and (2.26) implies that

|q(ξ) − t(ξ, ζ)| ≤ C|ζ| , (4.23)

we can directly reconstruct

q(ξ) = lim|ζ|→∞ζ∈Vξ

t(ξ, ζ), (4.24)

and then q by inverting the Fourier transform. This method is alreadyimplicitly in the proof of Theorem 4.1.1 in [SU87] and given explicitly in[NSU88, Nov88]. This reconstruction procedure seems, however, imprac-tical since it relies only on the large (complex) frequency information inΛγ. We note again that in two dimensions it is impossible to use (4.24) forreconstruction, since the set Vξ consists of two isolated points only .

The second reconstruction method is based on the ∂ζ-equation (4.22).This idea was suggested in a different context in [NA84, LN87]; a rigor-ous treatment was given by Nachman in [Nac88]. The method is based ona higher dimensional version of the generalized Cauchy formula, whichgives a formula for q in terms of t, ∂ζt and the asymptotic value (4.24) oft. A suitable version due to Hatziafratis [Hat86] of this so-called Bochner-Martinelli formula gives the existence of a differential form K on Vξ suchthat for ξ ∈ Rn, R > 0, f ∈ C1(Vξ \ B(0, R)) and R′

> R

f (ζ) =

∫

z∈Vξ ,|z|=R′f (z)K(z, ζ) −

∫

z∈Vξ ,|z|=Rf (z)K(z, ζ)

+

∫

z∈Vξ ,|z|>R∂z f (z) ∧ K(z, ζ).

(4.25)

For the exact definition of the form K(z, ζ) we refer to [Nac88].To avoid the appearance of exceptional points, the reconstruction for-

mula is based only on a range of large ζ ∈ Vξ, where by Theorem 2.4.6it is known that there are no exceptional points. Choose R > 0 such thatthere are no exceptional points with |ζ| ≥ R and consider this formula withf (ζ) = t(ξ, ζ). For R′ → ∞ the estimate (4.23) can be seen to imply that

limR→∞

∫

z∈Vξ ,|z|=R′t(ξ, z)K(z, ζ) = q(ξ).


The resulting formula is then

q(ξ) = t(ξ, ζ) +

∫

z∈Vξ ,|z|=Rt(ξ, z)K(z, ζ) −

∫

z∈Vξ ,|z|>R∂zt(ξ, z) ∧ K(z, ζ),

(4.26)

and since the right hand side is known from the boundary measurements,this formula can be interpreted as a reconstruction formula for q. We notethat contrary to the method described in section 3.2 for the two-dimensionalproblem, the method (4.26) is not based on solving a ∂ζ-equation.

The algorithm for reconstructing the conductivity γ ∈ W2,∞(Ω) nowconsists of the steps

(1) Solve the boundary integral equation (2.36) for ψ(·, ζ)|∂Ω for ζ ∈ Vsufficiently large.

(2) Calculate the scattering transform t by (4.20) and ∂ζt by (4.22) forξ ∈ Rn, ζ ∈ Vξ.

(3) Calculate q by (4.24) or (4.22) and then q by inverting the Fouriertransform.

(4) Solve (2.5) with γ1/2|∂Ω = 1 for γ1/2.

4.3.3. Reconstruction of a conductivity close to constant. In this sectionwe will show how to reconstruct a conductivity γ ∈ C∞(Ω) close to a con-stant from boundary data. This is a well-known result, however, the in-teresting point here is that the algorithm to be given is a direct analogue ofthe reconstruction algorithm for the two-dimensional problem given in sec-tion 3.2. The idea is first to compute the scattering transform from bound-ary measurements, then to solve the ∂-equation (4.21) for µ, and finally toreconstruct the conductivity from µ by taking ζ → 0. A difficulty in dimen-sion n ≥ 3 is the possibility of having exceptional points. In order to avoidthis we will assume that γ is sufficiently close to a constant such that thereare no exceptional points by Corollary 4.2.6. The smallness assumption willalso together with the smoothness enable us to solve (4.21). We will in thissection without loss of generality assume that Ω = Ba and γ = 1 near ∂Ω.

The next result concerns the solvability of (4.21).

Lemma 4.3.2. For q ∈ C∞(Ω) sufficiently small and x ∈ Ω fixed, there is aunique solution µ(x, ·) ∈ C(V) to (4.21) with

lim|ζ|→∞

µ(x, ζ) = 1. (4.27)

Proof. We refer to [BC85, Theorem 4].

4.4. Notes 81

The proof of this theorem is based on the fact that smoothness in qis equivalent to decay of t and that smallness in q implies smallness of t.Beals and Coifman do actually give a method for obtaining the unique so-lution µ to (4.21) subject to the asymptotic condition (4.27). Hence from thescattering transform t(ξ, ζ), the exponentially growing solution ψ(x, ζ) =

eix·ζµ(x, ζ) can be computed.Next we will show that a conductivity can be reconstructed from the

exponentially growing solutions by taking the parameter ζ to zero. Thisis a consequence of the estimate on ∂Ba for the solution to the boundaryintegral equation.

Theorem 4.3.3. Let ζ ∈ V be sufficiently small and let ψ(x, ζ) be the uniqueexponentially growing solution. Then

‖ψ(·, ζ) − γ1/2‖H2(Ba) ≤ C|ζ|.

Proof. Since γ = 1 near ∂Ω we have by Lemma 4.2.5 that for ζ near zero

‖ψ(x, ζ) − γ1/2‖H3/2(∂Ba)= ‖ψ(x, ζ) − 1‖H3/2(∂Ba)

≤ C|ζ|.

Since ψ(x, ζ)− γ1/2(x) is the unique solution to (−∆ + q)(ψ(x, ζ)− γ1/2) =0 in Ba elliptic regularity implies that

‖ψ(·, ζ) − γ1/2(·)‖H2(Ba) ≤ C‖ψ(·, ζ) − γ1/2(·)‖H3/2(∂Ba)≤ C|ζ|.

To sum up the proposed reconstruction algorithm for conductivitiesclose to constant we outline the main steps:

(1) Solve (2.36) for ψ(·, ζ), ζ ∈ V .

(2) Calculate the scattering transform t by (4.20) for ξ ∈ Rn, ζ ∈ Vξ.

(3) Solve (4.21) for µ(x, ·), x ∈ Ω.

(4) Reconstruct γ from

γ1/2(x) = lim|ζ|→0

ψ(x, ζ).

We note that a generalization of this algorithm to conductivities, which arenot close to one, requires the solution of two problems. First we would haveto show absence of exceptional points for this class conductivities, next wewould have to show solvability of (4.21).

4.4. Notes

In higher dimensions a major breakthrough in the theory of inverse bound-ary value problems was the paper [SU87]. There Sylvester and Uhlmann


proved the existence of exponentially growing solutions and that Λγ de-termines γ ∈ C∞(Ω) when Ω is a smooth domain. The proof of Theo-rem 4.1.1 using exponentially growing solutions in (2.13) as presented herewas given by Nachman, Sylvester, and Uhlmann [NSU88] for smooth do-mains. The smoothness of the boundary was then relaxed by Nachman[Nac88] to cover C1,1 domains and Alessandrini [Ale88] to Lipschitz do-mains. We note that by using (2.47), Theorem 4.1.1 can be seen to be validfor q ∈ Lp(Ω), p > n/2 when Ω is a Lipschitz domain.

The a priori assumption on γ for uniqueness to hold has since been re-laxed beyond W2,∞(Ω) by a number of authors. The method of proof seemsin all of the works to be based on the construction of exponentially grow-ing solutions. Brown [Bro96] showed existence of exponentially grow-ing solutions to the Schrodinger equation with a more singular potentialcoming from a conductivity. This gave uniqueness for conductivities inγ ∈ C3/2+ε(Ω). In a recent paper by Paivarinta, Panchenko, and Uhlmann[PPU] uniqueness was obtained for γ ∈ C3/2(Ω). In their approach theconductivity equation is reduced to a Schrodinger equation with a mag-netic potential instead of an electric potential, again the uniqueness proofis based on the construction of exponentially growing solutions. Finally ina recent preprint Brown and Torres [BT02] proved uniqueness for conduc-tivities γ ∈ W3/2,p(Ω), p > 2n improving the uniqueness result slightly.However, there seems to be no reason to believe that this result is optimal.

The solvability of the boundary integral equation for small ζ and γ ∈L∞(Ω) is based on the joint work [CKS02]. The technique of proof is analo-gous to the two-dimensional case analyzed by Siltanen, Mueller, and Isaac-son [SMI00]. Concerning the absence of exceptional points for small poten-tials, we note that Corollary 4.2.6 can also be deduced from (2.47) by takingr = n/2. A similar result was obtained by Beals and Coifman [BC85] in di-mension n = 3 assuming that q ∈ L3/2−ε(R3) ∩ L3/2+ε(R3). For the inverseconductivity problem it seems natural to conjecture that if γ ∈ W2,∞(Ω)

and q = ∆γ1/2/γ1/2, then there are no exceptional points for q. This wouldextend the result in two dimensions (Theorem 3.2.3) to higher dimensions.

The reconstruction method outlined in section 4.3.2 is due to Nachman[Nac88] and Novikov [Nov88] independently. The method is based on theso-called ∂-approach to inverse scattering, which was introduced by Bealsand Coifman [BC81], [BC82] in the study of integrable evolution equations.The theory was developed further by a number of authors, see [ABYF83,NA84, BC85, Nac88, NK87, New89].

The idea of solving (4.21) for µ(x, ζ) is not new. In [BC85] a smallpotential q is reconstructed by first solving (4.21) and then by pluggingω = µ − 1 into the equation (2.17). A similar approach is suggested in

4.4. Notes 83

[NA84, NK87, Nov88]. However, the idea of reconstruction the conduc-tivity by taking the parameter to zero seems to be new. The estimates forsmall ζ (Lemma 4.2.5) and the direct reconstruction of γ close to constantare based on [CKS02].

Concerning stability for the higher dimensional problem we mentionthat Alessandrini [Ale88, Ale90] proved a conditional stability result usingexponentially growing solutions. He proved that if γi ∈ W2,∞(Ω) with

0 < θ < γi, ‖γi‖W2,∞(Ω) < M,

then

‖γ1 − γ2‖L∞(Ω) ≤ Cλ(‖Λγ1 − Λγ2‖B(H1/2(∂Ω),H−1/2(∂Ω)),

where λ(t) = | log t|−δ for 0 < t < 1, δ < 1 depends on n, s, and C dependson M. Mandache showed by example recently in [Man01] that this stabilityresult is optimal.

Chapter 5

Local uniqueness,stability, andreconstruction from twoboundary measurements

This chapter concerns the question of inferring knowledge of a conductiv-ity or a potential from a finite number of boundary measurements. Whenthe boundary data is limited we cannot hope to recover an arbitrary coeffi-cient, so we will have to restrict the set of admissible coefficients.

In the classical paper [CDJ63], Cannon, Douglas and Jones consideredthe inverse conductivity problem for a class of conductivities homogeneousin one direction. More precisely let Ω = D × [0, a] ⊂ Rn for a bounded andsmooth D ⊂ Rn−1 and a > 0, and assume that the smooth conductivity γ isindependent of the cylindrical variable, i.e. γ(x′, z) = γ(x′, 0), x′ ∈ D, 0 <

z ≤ a. Then the result is that γ can be recovered from only one boundarymeasurement.

The fundamental result of this chapter can in some sense be seen as ageneralization of the result of Cannon, Douglas and Jones to the Schrodingerequation and to a different geometry. Let Ba = B(0, a) ⊂ R3 be the ballcentered at zero with radius 0 < a < ∞, and let the potential q be homo-geneous in the radial direction, i.e. q ∈ L2(∂Ba) is extended inside Ba by

85

86 5. . . . two boundary measurements

q(x) = q(ax), where x = x/|x|. We call such a potential a spherical poten-tial. Consider the boundary value problem

(−∆ + q)u = 0 in Ba, u = f on ∂Ba. (5.1)

Assuming that zero is not a Dirichlet eigenvalue of (−∆ + q), the ellipticboundary value problem (5.1) has a unique weak solution u ∈ H1(Ba) forany f ∈ H1/2(∂Ba) (see section 5.2 for further details on the existence andregularity of solutions). Moreover, the unique solution admits a normalderivative at the boundary ∂νu as defined in section 2.2.

The inverse problem can then be stated: assume that zero is not a Dirich-let eigenvalue for (−∆ + q) in Ba. Let I be a finite index set and let

(ui, ∂νui)|∂Ba : (−∆ + q)ui = 0i∈I ⊂ H1/2(∂Ba) × H−1/2(∂Ba)

be the data. Now, is q uniquely determined from this set of data, and inthat case, how can it be reconstructed?

The main result to be presented in this chapter is that a sufficientlysmall spherical potential can be found from only two boundary measure-ments, i.e. by first applying f1 = 1 on ∂Ba and measuring Λq(1), then ap-plying f2 = Λq1 and measuring Λ2

q(1). Note that for the class of sphericalpotentials reconstructing q on ∂Ba is equivalent to reconstructing q every-where. Hence this inverse problem is fundamentally different from theinverse problems studied in Chapter 3 and Chapter 4.

In the following section we state the exact results and outline the methodof proof. Then in the following sections the proofs of the theorems will begiven. In section 5.2 we consider the regularity properties of solutions to(5.1). These results are well-known but included here for completeness.Then in section 5.3 we derive a fundamental equation relating the data andthe potential, and in section 5.4 we see how this equation can be solved.This gives existence and uniqueness. Stability is then proved in section 5.5,and finally in section 5.6 we give an application to the inverse conductivityproblem. The results presented in this chapter are based on a joint workwith Horia Cornean [CK02].

5.1. Outline of the method and results

Our first result is an equation relating the data and the potential q. Let−∆D be the Friedrichs extension of (−∆)|C∞

0 (Ba), which is selfadjoint onthe domain D(−∆D) = H1

0(Ba) ∩ H2(Ba). For q ∈ L2(Ba) we can define(−∆D + q) as a selfadjoint operator on the same domain, and when zerois not an eigenvalue of this operator, Rq = (−∆D + q)−1 exists and is asmoothing operator of degree two, see section 5.2 below. Let now forq ∈ L2(Ba), Mq be the multiplication operator Mq : φ 7→ qφ defined in

5.1. Outline of the method and results 87

L2(Ba) on the domain D(Mq) = φ ∈ L2(Ba) | qφ ∈ L2(Ba). Let furtherρ0 : Hs(Ba) → Hs−1/2(∂Ba), s > 1/2, be the usual trace operator. A funda-mental result relating Λq and q is then

Proposition 5.1.1. Let q ∈ L2(Ba) be spherical and assume that zero is not aneigenvalue of (−∆D + q). Then in L2(∂Ba) we have

q = Λ2q(1) +

3a

Λq(1) +2a

ρ0∂rRq MqRq(q). (5.2)

The equation (5.2) is the fundamental equation in the solution of theinverse problem. Define the non-linear operator

F(p) =2a

ρ0∂rRp MpRp(p) (5.3)

for any spherical potential q for which zero is not an eigenvalue of (−∆D +q). Then the forward problem is defined by the operator

I − F : q 7→ q0 = Λ2q(1) +

3a

Λq(1)

and the inverse problem concerns the inversion of this map. We will showthat when q is sufficiently small then I − F can be inverted. To controlthe size of the potentials, we introduce the set Wλ of spherical potentialsq, which satisfy ‖q‖L2(∂Ba) < λ. This bound implies the norm bound on thewhole domain ‖q‖L2(Ba) < (1/3a3)1/2λ. In Lemma 5.2.4 below we show thatwhen λ1 is sufficiently small and q ∈ Wλ1 , then zero is not an eigenvalue of(−∆D + q), and therefore (5.2) holds.

In the inversion procedure the equation (5.2) is the equation to be solved.Assume that q0 = Λ2

q(1) + 3a Λq(1) is given. Define the non-linear operator

T : Wλ1 → L2(∂Ba) by

T(p) = q0 + F(p). (5.4)

We see from (5.2) that q is a fixed point for T. Now, to solve the inverseproblem we will prove that for a λ0 ≤ λ1 sufficiently small, T has a uniquefixed point in the set Wλ0 , and moreover, that this fixed point can be foundby iteration. The following theorem states the result:

Theorem 5.1.2. There is a constant λ0 depending only on the radius a, such thatif q ∈ Wλ0 and q0 = Λ2

q(1) + 3a Λq(1), then (5.4) has a unique solution. More-

over, this solution can be found as the L2(∂Ba) limit of the convergent sequenceTn(0)n∈N ⊂ L2(∂Ba), i.e.

q = limn→∞

Tn(0).

Not only can q be reconstructed uniquely but we can also easily provethe following stability result:


Theorem 5.1.3. Let λ0 be given from Theorem 5.1.2 and assume that q1, q2 ∈Wλ0 . Let q(i)

0 = Λ2qi(1) + 3

a Λqi(1) for i = 1, 2 be the given data. Then the stabilityestimate holds

‖q1 − q2‖L2(∂Ba) ≤ 2‖q(1)0 − q(2)

0 ‖L2(∂Ba).

The simplicity in the proofs given below relies heavily on the assump-tion that the potential is spherical. A very interesting open problem alongthis line is to determine a layered potential from boundary measurements.In the two-layered case the potential is assumed to have the form q(x) =q1(x) + q2(x)χBa1

(x) in Ba, where 0 < a1 < a. The problem is then to deter-mine q1, q2 and a1 from only a finite number of measurements. The solu-tion of this problem and its generalization to a finite number of layers maylead to a layer stripping approach for solving the general problem withq ∈ L2(Ba). This is a project for further studies.

5.2. Regularity of solutions

In this section we collect some results concerning perturbations of −∆Dand inversion of such perturbation operators. Then we use these results toprove regularity properties of solutions to boundary value problems. Theresults in this section do not require q to be spherical and the domain to bethe sphere Ba.

We start out by defining −∆D + q rigorously as a selfadjoint operator:

Lemma 5.2.1. For q ∈ L2(Ba) the operator (−∆D + q) in L2(Ba) is selfadjointon the domain D(−∆D + q) = D(−∆D) = H1

0(Ba) ∩ H2(Ba).Furthermore, (−∆D + q) has discrete spectrum, and if zero is not an eigen-

value of (−∆D + q), then Rq = (−∆D + q)−1 is in B(L2(Ba), H2(Ba)) and inB(H−1(Ba), H1(Ba)).

Proof. We will prove that the multiplication operator Mq : φ 7→ qφ is oper-ator bounded by (−∆D) with bound less than 1, and then apply the Kato-Rellich theorem (see [Kat66, Theorem 4.3]).

Since for f ∈ L∞(Ba)

‖q f ‖L2(Ba) ≤ ‖q‖L2(Ba)‖ f ‖L∞(Ba)

it follows for φ ∈ D(−∆D) and µ > 0 that with Rµ = (−∆D + µ)−1

‖qφ‖L2(Ba)

≤ ‖qRµ(−∆D + µ)φ‖L2(Ba)

≤ ‖q‖L2(Ba)‖Rµ‖B(L2(Ba),L∞(Ba))‖(−∆D + µ)φ‖L2(Ba)

≤ ‖q‖L2(Ba)‖Rµ‖B(L2(Ba),L∞(Ba))(‖ − ∆Dφ‖L2(Ba) + µ‖φ‖L2(Ba)). (5.5)

5.2. Regularity of solutions 89

Let Gµ(x, y) denote the Dirichlet Green’s function for (−∆D + µ) definedfor fixed x ∈ Ba as the unique solution to

(−∆D + µ)Gµ(x, y) = δ(x − y) in Ba, Gµ(x, y) = 0 on ∂Ba.

Let further Φµ(x, y) = e−√

µ|x−y|(4π|x − y|)−1 be the standard fundamentalsolution. Since Φµ(x, y) − Gµ(x, y) is harmonic on Ba and hence bounded,Gµ(x, y) is positive near the singularity x. By the maximum principle ap-plied to Gµ(x, y) in Ba \ B(x, ε), x ∈ Ba, ε > 0 then shows, that Gµ(x, y)is positive everywhere. Another application of the maximum principle toΦµ(x, y) − Gµ(x, y) gives for x, y ∈ Ba, x 6= y the estimate

Gµ(x, y) <e−

√µ|x−y|

4π|x − y| .

This shows that

‖Rµ‖B(L2(Ba),L∞(Ba)) = supf∈L2(Ba), ‖ f ‖L2(Ba)

=1

∥

∥

∥

∥

∫

Ba

Gµ(x, y) f (y)dy∥

∥

∥

∥

L∞(Ba)

≤ supx∈Ba

(∫

Ba

|Gµ(x, y)|2dy)1/2

≤ supx∈Ba

14π

(

∫

Ba

e−2√µ|x−y|

|x − y|2 dy

)1/2

≤(∫ 2a

0e−2√µrdr

)1/2

≤ Cµ−1/4,

and by (5.5) we find that φ 7→ qφ is (−∆D)-bounded, i.e. for φ ∈ D(−∆D)

‖qφ‖L2(Ba) ≤ b‖(−∆D)φ‖L2(Ba) + c‖φ‖L2(Ba), (5.6)

where b = Cµ−1/4‖q‖L2(Ba), c = Cµ3/4‖q‖L2(Ba). The choice µ > C‖q‖4L2(Ba)

implies that b < 1, and then it follows from the Kato-Rellich theorem that−∆D + q is well-defined in L2(Ba) and selfadjoint on D(−∆D).

It is well-known that (−∆D + q) has a purely discrete spectrum. Whenzero is not an eigenvalue the inverse Rq exists as a bounded operator onL2(Ba). The smoothing property of Rq is an easy consequence of the map-ping properties of (−∆D)−1.

To solve the problem (5.1) we apply the standard procedure of trans-forming the problem into a problem with a source term and zero boundarycondition. This reduction relies on well-known properties of solutions tothe Laplace equation:


Proposition 5.2.2. For any f ∈ H1/2+s(∂Ba), s ≥ 0, there is a unique solutionv ∈ H1+s(Ba) to

−∆v = 0 in Ba, v = f on ∂Ba. (5.7)

Proof. See for instance [GT83] for a proof.

We can now prove the result

Proposition 5.2.3. Assume zero is not an eigenvalue of (−∆D + q). Then fors = 0, 1 there is for any f ∈ H1/2+s(∂Ba) a unique solution u ∈ H1+s(Ba) to(5.1).

Proof. Let v ∈ H1+s(Ba) solve (5.7) and introduce w = u − v. Then w solves(−∆ + q)w = −(qv) in Ba and vanishes on ∂Ba, and formally we definew = u − v = −Rq(qv). For f ∈ H1/2+s(∂Ba) Proposition 5.2.2 gives thatv ∈ H1+s(Ba). Hence the result follows from the mapping properties ofRq given in Lemma 5.2.1 provided that qv ∈ H−1+s(Ba). For s = 1 theSobolev embedding H2(Ba) ⊂ L∞(Ba) gives that qv ∈ L2(Ba). For s = 0 theSobolev embedding H1(Ba) ⊂ L6(Ba) and Holder’s inequality implies thatqv ∈ L3/2(Ba). By the Sobolev embedding qv ∈ H−1/2(Ba) ⊂ H−1(Ba) weget the result.

Uniqueness follows from the injectivity of Rq.

It is well-known that the spectrum of (−∆D) is positive. When q issmall this property is inherited by (−∆D + q). In particular zero cannot bean eigenvalue. We will give the easy proof of this claim.

Lemma 5.2.4. There is a constant Ca depending only on a such that if λ1 =(Ca‖(−∆D)−1‖B(L2(Ba),H2(Ba)))

−1 and q ∈ Wλ1 , then zero is not an eigenvalue of(−∆D + q).

Proof. To prove the result, we write in D(−∆D)

(−∆D + q) = (1 + q(−∆D)−1)(−∆D), (5.8)

and hence formally

Rq = (−∆D + q)−1 = (−∆D)−1(1 + q(−∆D)−1)−1.

Since (−∆D)−1 is a smoothing operator of degree two, the result follows if(1 + q(−∆D)−1) is invertible in L2(Ba).

The inverse (1 + q(−∆D)−1) is given by a convergent Neumann seriesif

‖q(−∆D)−1‖B(L2(Ba),L2(Ba)) < 1.

5.3. An equation on the boundary 91

By the estimates

‖q(−∆D)−1‖B(L2(Ba),L2(Ba))) ≤ ‖q‖L2(Ba)‖(−∆D)−1‖B(L2(Ba),L∞(Ba))

≤ a‖q‖L2(∂Ba)‖(−∆D)−1‖B(L2(Ba),L∞(Ba))

≤ Ca‖q‖L2(∂Ba)‖(−∆D)−1‖B(L2(Ba),H2(Ba)),

where the last estimate follows from the Sobolev embedding theorem andCa is a constant depending on a, the result is obtained when ‖q‖L2(∂Ba) <

λ1 = (Ca‖(−∆D)−1‖B(L2(Ba),H2(Ba)))−1.

Note that from the convergent Neumann series we can get the estimate

‖Rq‖B(L2(Ba),H2(Ba)) ≤ (‖(−∆D)−1‖−1B(L2(Ba),H2(Ba))

− Ca‖q‖L2(Ba))−1, (5.9)

provided ‖q‖L2(Ba) < (Ca‖(−∆D)−1‖B(L2(Ba),H2(Ba)))−1.

5.3. An equation on the boundary

In this section we derive the equation (5.2) for q. The idea is to establish arelation between Λ2

q( f ) and the second order derivative ∂2r u|∂Ba of the solu-

tion to (5.1), and then use the partial differential equation for u to express∂2

r u|∂Ba as a sum of lower order terms in the special case f = 1.The details will be given in two lemmas:

Lemma 5.3.1. Assume zero is not an eigenvalue of (−∆D + q). Then for anyf ∈ H3/2(Ba) we have the relation

Λ2q( f ) − ρ0∂2

r u =2a

Λ0( f ) − 1a

Λq f − 2a

ρ0∂rRq MqRq(qv), (5.10)

where u, v are solutions to (5.1) and (5.7) respectively.

Note that when f ∈ H3/2(∂Ba), then the right hand side in (5.10) isin H1/2(∂Ba) and hence the H−1/2(∂Ba) singularity of the left hand sidecancels.

Proof. Introduce the operator A = ( xa · ∇) and note that ρ0 A = ρ0∂r on the

boundary ∂Ba. Hence ρ0 Au = Λq f ∈ H1/2(∂Ba), where u ∈ H2(Ba) is thesolution to (5.1) according to Proposition 5.2.3. To get an equation for Auwe apply A to (−∆ + q)u in the distributional sense. By the definition of Awe get for φ ∈ D(Ba) that

〈A(qu), φ〉 = −3a〈qu, φ〉 − 〈qu, Aφ〉.


Now since q is spherical, an integration in spherical coordinates gives that

〈qu, Aφ〉 =

∫

Ba

quAφ

=

∫ 2π

0

∫ π/2

0q(x)

∫ a

0u(x)

r3

a∂rφ(x)dr sin ψdψdθ

= −∫ 2π

0

∫ π/2

0q(x)

∫ a

0

(

r3

a∂ru(x) + u(x)

3r2

a

)

φ(x)dr sin ψdψdθ

= −〈qAu, φ〉 − 3a〈qu, φ〉,

which shows that

Aqu = qAu.

Moreover, an easy calculation shows that

[A,−∆] =2a

∆.

Hence Au solves the boundary value problem

(−∆ + q)Au = −2a

∆u = −2a

qu in Ba, Au = Λq f on ∂Ba.

Let u ∈ H1(Ba) be the unique solution to (−∆ + q)u = 0 in Ba withu|∂Ba = Λq f . The function w = u − Au ∈ H1

0(Ba) is then a solution to(−∆ + q)w = 2

a qu, and since q ∈ L2(Ba) and u ∈ H2(Ba) ⊂ L∞(Ba) bythe Sobolev embedding theorem, the product qu ∈ L2(Ba). Hence we caninvert the equation to find

u − Au =2a

Rq(qu) ∈ H2(Ba). (5.11)

Apply now ρ0∂r to this equation and note that

ρ0∂ru = Λ2q( f ),

ρ0∂r Au = ρ0∂rra

∂ru = ρ0∂2r u +

1a

Λq f .(5.12)

Furthermore, recall from the proof of Proposition 5.2.3 that u can be ex-pressed in terms of the solution to the free problem by u = −Rq(qv) + v.Hence

Rq(qu) = Rq(Mqv − MqRq(qv))

= v − u − Rq MqRq(qv).

5.3. An equation on the boundary 93

By applying ρ0∂r to this expression and combining this with (5.11) and(5.12) we find

Λ2q( f ) − ρ0∂2

r u − 1a

Λq f =2a(Λ0( f ) − Λq f ) − 2

aρ0∂rRq MqRq(qv)

from which the result follows.

For the special choice of boundary field f = 1 we are able to removethe ρ0∂2

r u term from (5.10).

Lemma 5.3.2. Let u1 ∈ H2(Ba) be the solution to (5.1) with f = 1. Then

ρ0∂2r u1 = q − 2

aΛq1. (5.13)

Note that ρ0∂2r u1 a priori is no better than H−1/2(∂Ba), but the lemma

implies that ρ0∂2r u1 ∈ L2(∂Ba).

Proof. Write the Laplace operator in spherical coordinates

∆ = ∂2r +

2r

∂r +1r2 Lθ,φ

where

Lθ,φ =1

sin θ

∂

∂θ

(

sin θ∂

∂θ

)

+1

sin θ

∂2

(∂φ)2 .

Hence

0 = ρ0(−∆ + q)u1

= ρ0(−∂2r −

2r

∂r − 1r2 Lθ,φ + q)u1

= −ρ0∂2r u1 −

2a

Λq1 + q,

since ρ0Lθ,φu1 = Lθ,φρ0u1 = Lθ,φ1 = 0.

Proof of Proposition 5.1.1. The result follows from (5.10) and (5.13) sincev = 1 is the unique harmonic function with v|∂Ba = 1 and therefore Λ0(1) =0.

We note that when q is sufficiently small, it follows from Lemma 5.2.4that (5.2) holds.


5.4. Solving the boundary integral equation

In this section the proof of Theorem 5.1.2 will be given. The first step is thefollowing lemma concerning the contraction properties of the operator Fdefined in (5.3):

Lemma 5.4.1. There is a λ2 > 0 depending only on the radius a such that ifp, r ∈ Wλ2 , then

‖F(p) − F(r)‖L2(∂Ba) ≤12‖p − r‖L2(∂Ba).

Proof. From Lemma 5.2.4 we can find λ1 such that if q ∈ Wλ1 , then Rq =

(−∆D + q)−1 ∈ B(L2(Ba), H2(Ba)). Then for λ ≤ λ1 and p, r ∈ Wλ theoperator F is well-defined, and we have

F(p) − F(r) =2a

ρ0∂r(Rp MpRp p − Rr MrRrr)

=2a

ρ0∂r(Rp MpRp(p − r) + Rp Mp(Rp − Rr)r

+ Rp Mp−rRrr + (Rp − Rr)MrRrr).

Since the trace theorem for this particular geometry has the explicit form

‖u‖L2(∂B(0,a)) ≤ (2a

+ 1)‖u‖H1(B(0,a)),

we see by the triangle inequality that

‖F(p) − F(r)‖L2(∂Ba) ≤ ‖2a

ρ0∂r(Rp MpRp p − Rr MrRrr)‖L2(∂Ba)

≤ Ca‖Rp MpRp p − Rr MrRrr‖H2(Ba)

≤ Ca(‖Rp MpRp(p − r)‖H2(Ba) + ‖Rp Mp(Rp − Rr)r‖H2(Ba)

+ ‖Rp Mp−rRrr‖H2(Ba) + ‖(Rp − Rr)MrRrr‖H2(Ba))

≤ Ca

(

‖Rp − Rr‖B(L2(Ba),H2(Ba)) supz∈Wλ

(

‖Rz‖B(L2(Ba),H2(Ba))

)

· supz∈Wλ

(

‖z‖2L2(Ba)

)

+ ‖p − r‖L2(Ba) supz∈Wλ

(

‖Rz‖2B(L2(Ba),H2(Ba))

)

supz∈Wλ

(

‖z‖L2(Ba)

)

)

.

The resolvent identity Rp − Rr = −Rp Mp−rRr now implies

‖Rp − Rr‖B(L2(Ba),H2(Ba)) ≤ ‖r − p‖L2(Ba) supz∈Wλ

(

‖Rz‖2B(L2(Ba),H2(Ba))

)

.

Hence

‖F(p) − F(r)‖L2(Ba) ≤ CaC(λ)‖p − r‖L2(Ba),

5.4. Solving the boundary integral equation 95

where

C(λ) = maxk=0,1

supz∈Wλ

(

‖Rz‖2+kB(L2(Ba),H2(Ba))

)

supz∈Wλ

(

‖z‖1+kL2(Ba)

)

.

Since by (5.9) implies that ‖Rz‖B(L2(Ba),H2(Ba)) is uniformly bounded for z ∈Wλ, λ < λ1, the result now follows by taking λ2 < λ1 such that C(λ2) ≤C−1

a /2.

Proof of Theorem 5.1.2. Let λ2 be given from Lemma 5.4.1. For the unique-ness, assume q1, q2 ∈ Wλ2 are fixed points for T. Then from Lemma 5.4.1 wehave

‖q1 − q2‖L2(∂Ba) = ‖T(q1) − T(q2)‖L2(∂Ba)

= ‖F(q1) − F(q2)‖L2(∂Ba)

≤ 12‖q1 − q2‖L2(∂Ba)

which implies ‖q1 − q2‖L2(∂Ba) = 0.

Concerning the convergence of Tn(0)n∈N we note that if q0 = 0, thenthe existence of a fixed point follows by Schauder’s fixed point theorem[Eva98, section 9.2.2], since F (and hence T) is a contraction in Wλ2 byLemma 5.4.1. In case q0 6= 0 we cannot immediately use a general fixedpoint theorem, so we will have to rely on the smallness assumption. Forq ∈ Wλ1 it follows from (5.2) that

‖Λ2q(1) +

3a

Λq1‖L2(∂Ba)

≤ Ca(1 + ‖Rq‖B(L2(Ba),H2(Ba))‖2‖q‖L2(∂Ba))‖q‖L2(∂Ba),

where Ca is a constant depending only on a. Hence there is a constant λ0 ≤λ2 ≤ λ1 such that if q ∈ Wλ0 , then q0 = T(0) = Λ2

q(1) + 3a Λq1 ∈ Wλ2/2. An

induction argument now shows that if

‖Tm−1(0)‖L2(∂Ba) ≤m−1

∑k=1

(

12

)k

λ2,

which in particular implies that Tm−1(0) ∈ Wλ2 , then by Lemma 5.4.1

‖Tm(0)‖L2(∂Ba) ≤ ‖F(Tm−1(0)) − F(Tm−2(0))‖L2(∂Ba) + ‖Tm−1(0)‖L2(∂Ba)

=m

∑k=1

(

12

)m

λ2.


By the same type of argument we find that

‖Tm+1(0) − Tm(0)‖L2(∂Ba) ≤12‖Tm(0) − Tm−1(0)‖L2(∂Ba)

≤ 12m+1 ‖T(0)‖L2(∂Ba)

and hence

‖Tm+k(0) − Tm(0)‖L2(∂Ba) ≤k

∑j=1

‖Tm+j(0) − Tm+(j−1)(0)‖L2(∂Ba)

≤k

∑j=1

12m+j+1 ‖T(0)‖L2(∂Ba).

From this expression it follows that Tn(0)n∈N is a Cauchy sequence inL2(∂Ba), and hence the sequence converges to a p ∈ Wλ2 . Since

‖p − T(p)‖L2(∂Ba) ≤ limn→∞

(‖p − Tn(0)‖L2(∂Ba) + ‖Tn(0) − T(p)‖L2(∂Ba))

≤ limn→∞

32‖p − Tn(0)‖L2(∂Ba)

= 0,

we see that p is indeed a fixed point for T.

5.5. Stability

Proof of Theorem 5.1.3. In the inversion procedure we find q1 and q2 asunique fixed points for the operators T(1), T(2) respectively, where

T(i)(p) = q(i)0 + F(p), i = 1, 2,

for F(p) = 2a ρ0∂rRp MpRp p. Hence

‖q1 − q2‖L2(∂Ba) = ‖T(1)(q1) − T(2)(q2)‖L2(∂Ba)

≤ ‖q(1)0 − q(2)

0 ‖L2(∂Ba) + ‖F(q1) − F(q2)‖L2(∂Ba)

≤ ‖q(1)0 − q(2)

0 ‖L2(∂Ba) +12‖q1 − q2‖L2(∂Ba),

since q1, q2 ∈ Wλ. From this inequality the stability result follows.

5.6. Application to the conductivity problem

We would like to apply the results obtained concerning the spherical po-tentials to the inverse conductivity problem. To fix notation we denote bySλ the set

Sλ = γ ∈ H2(Ba) ∩ L∞+ (Ba) : γ−1/2∆γ1/2 ∈ Wλ.

5.7. Notes 97

Note that since q is assumed to be spherical in the definition of Sλ, thenfrom the equation (−∆ + q)γ1/2 = 0 it can be that γ is actually dependingon both the radial and spherical variables. The following lemma showsthat Sλ is not empty.

Lemma 5.6.1. For λ < λ0 sufficiently small the set Sλ is not empty.

Proof. For any harmonic function v the function

u = −(−∆D + q)−1(vq) + v

satisfies (−∆ + q)u = 0. Let v = c > 0 and note that

u ≥ −c‖Rq‖B(L2(Ba),H2(Ba))‖q‖L2(Ba) + c.

Now q ∈ Wλ implies that M = ‖Rq‖B(L2(Ba),H2(Ba))‖q‖L2(Ba) < 1, and hence

u ≥ c(1 − M) > 0,

which shows that u ∈ Sλ.

Now the the following corollary to Theorem 5.1.2 can be obtained:

Corollary 5.6.2. Let γ ∈ Sλ0 be a conductivity and let γ, ∂νγ be given at theboundary ∂Ba. Then γ can be reconstructed from the two boundary measurementsΛγ(γ−1/2) and Λγ(1/2γ−3/2∂νγ + γ−1Λγ(γ−1/2)).

Proof. Since γ ∈ Sλ0 we have the potential q = ∆γ1/2

γ1/2 ∈ Wλ0 . Moreover,since γ, ∂νγ are known on ∂Ba we know from the boundary measurementsby (2.9) the quantities

Λq1 = γ−1/2(

12

∂νγ + Λγ

)

(γ−1/2)

Λ2q(1) = γ−1/2

(

12

∂νγ + Λγ

)(

12

γ−3/2∂νγ + γ−1Λγ(γ−1/2)

)

.

Hence q can be reconstructed according to Theorem 5.1.2 and then γ can befound by solving the equation (−∆ + q)γ1/2 = 0 with the known boundarydata γ1/2|∂Ba .

5.7. Notes

Another way of restricting the class of potentials in the inverse conductiv-ity problem with a finite number of measurements is by assuming that theconductivity has the form γ(x) = 1 + χD(x) in the domain Ω, where χDis the characteristic function on the inclusion D ⊂ D ⊂ Ω. The problem isthen to determine the inclusion D. There are in the literature uniquenessresults available for certain classes of domains including polygons in theplane, convex polyhedra in three dimensions, and balls in two and three


dimensions. We refer to [KS00] and [Ike00] for this problem. For the in-verse problem for the Schrodinger operator a similar structure of the poten-tial is considered in [KKY01], where uniqueness for balls in two and threedimensions is shown.

We note also that in the recent paper [Ber02], Berntsen considered therelated inverse scattering problem for the Schrodinger equation having apotential of the form q(x) = 1 + ∑

Nj=1 k jχDj(x), where the location of the N

disjoint scatterers Dj is known and the constants k j are unknown. The resultis then that the constants k j are uniquely determined from one scatteringexperiment.

Appendix A

Notation

In this appendix we will review the notation used in the thesis. The defini-tions are standard but included here for completeness and reference.Spaces. We will denote by S(Rn) the Schwartz space and by S ′(Rn) thespace of tempered distributions. Furthermore for any measurable set Ω ⊂Rn we denote by D(Ω) the space consisting of smooth functions supportedinside Ω and by D′(Ω) the space of distributions. The spaces of distribu-tions are in particular relevant when taken derivatives, and we will alwaysunderstand derivatives as defined in the weak sense on S ′(Rn) or D′(Ω).

Let Ω ⊂ Rn be a measurable set and let the Lebesgue space Lp(Ω), 1 ≤p < ∞ be the space of (equivalence classes of) measurable functions in Ω

with

‖u‖Lp(Ω) =

(∫

Ω

|u(x)|pdx)1/p

< ∞,

and let L∞(Ω) be the space of essentially bounded measurable functionsequipped with

‖u‖L∞(Ω) = ess sup(u) < ∞.

With the norm ‖ · ‖Lp(Ω), Lp(Ω), 1 ≤ p ≤ ∞ is a Banach space.For m ∈ Z+ and 1 ≤ p ≤ ∞ we denote by Wm,p(Ω) the Lp-based

Sobolev space of order m consisting of functions u ∈ Lp(Ω) with

Dαu ∈ Lp(Ω)

99

100 A. Notation

for any multi-index α with |α| < m. Wm,p(Ω) is a Banach space with norm

‖u‖Wm,p(Ω) =

(

∑|α|≤m

‖Dαu‖pLp(Ω)

)1/p

< ∞.

We write Hm(Ω) = Wm,2(Ω). Note that Wm,p(Rn) coincides with the Sobolevspace Hm

p (Rn) defined by the Fourier transform. The completion of D(Ω)

in H1(Ω) is denoted H10(Ω) and consists of the elements in H1(Ω) that van-

ishes on the boundary ∂Ω.Denote by M2 the vector space of complex 2 × 2 matrices. Then we

denote by Lp(Rn;M2) the Lp-space consisting of functions defined on Rn

with values in M2 having each matrix-element in Lp(Rn). The natural normon this space is

‖u‖Lp(Rn;M2) =2

∑i,j=1

‖uij‖Lp(Rn).

Let Lpδ (Rn) be the weighted Lp-space

Lpδ (R

n) = u ∈ S′(Rn) | 〈x〉δu ∈ Lp(R

n)with norm

‖u‖Lpδ (Rn) = ‖〈x〉δu‖Lp(Rn).

Denote by Ws,pδ (Rn) the natural weighted Sobolev space. When p = 2 we

write Hsδ(Rn) = Ws,2

δ (Rn).In the context of Lp-spaces we use the notation p′ for the conjugate ex-

ponent of p defined by

1p

+1p′ = 1,

and the notation p defined for 1 < p < 2 by

1p

=1p− 1

2.

For m ∈ Z+ we denote by Cm(Ω) the space of bounded continuousfunctions on Ω with continuous and bounded derivatives up to order m.This space we equip with the norm

‖u‖Cm(Ω) = ∑|α|≤m

‖Dαu‖L∞(Ω).

We write C(Ω) = C0(Ω). Furthermore, for 0 < α < 1 we denote by Cα(Ω)the Holder space with exponent α. This space consists of functions u ∈

A. Notation 101

C(Ω) for which there is a constant C such that

|u(x) − u(y)| ≤ C|x − y|α

for all x, y ∈ Ω. This space is normed by

‖u‖Cα(Ω) = infx,y∈Ω

|u(x) − u(y)||x − y|α .

By C0,1(Ω) we denote the space of Lipschitz continuous functions consist-ing of functions u ∈ C(Ω) satisfying

|u(x) − u(y)| ≤ C|x − y|.The natural norm is

‖u‖C0,1(Ω) = infx,y∈Ω

|u(x) − u(y)||x − y| .

Boundaries. Concerning regularity of the boundary let Ω ⊂ Rn, n ≥ 2, bea bounded and open subset. Then we say that ∂Ω is globally Ck providedthat for any x ∈ ∂Ω there is an open neighborhood U, which has a localcoordinate patch such that

Ω ∩ U = x = (x′, xn) ∈ U | xn > σ(x′),

∂Ω ∩ U = x = (x′, xn) ∈ U | xn = σ(x′),(A.1)

where the real function σ ∈ Ck(Rn−1). When σ is only Lipschitz continuouswe shall say that Ω has Lipschitz boundary. We note that when ∂Ω is C1

then a normal vector field to ∂Ω is well-defined. We will by ν = ν(x) denotethe out pointing unit normal at x ∈ ∂Ω.

To define Sobolev spaces on the boundary, let UjNj=1 be an open cover

of ∂Ω consisting of such neighborhoods in (A.1). Due to the compactnessof ∂Ω we can assume N < ∞. Let φjN

j=1 be a smooth partition of unitysubordinate to this covering and define the map Φ : Rn → Rn by

Φj(x1, · · · , xn) = (x1, · · · , xn − σj(x′)).

This map straightens out the boundary, i.e. if y = Φ j(x) for some point x ∈∂Ω∩Uj, then yn = 0. Then we will say that f ∈ Ws,p(∂Ω), s > 0, 1 ≤ p ≤ ∞

if and only ifN

∑j=1

f (φjΦ−1j (y′, 0)) ∈ Ws,p(R

n−1).

The norm on Hs(∂Ω) is defined as

‖ f ‖Ws,p(∂Ω) = ‖N

∑j=1

f (φjΦ−1j (y′, 0))‖Ws,p(Rn−1).

102 A. Notation

Again we write Hs(∂Ω) = W2,p(∂Ω). A fundamental result states that thetrace operator initially defined on C(Ω) extends to a bounded operator onWs,p(Ω), s > 1/p with range Ws−1/p,p(∂Ω). We note that the definitionsof the spaces on the boundary are independent of the choice of the coverUjN

j=1. For s < 0, we define Hs(∂Ω) = (H−s(∂Ω))∗, the dual space ofH−s(∂Ω) taken with respect to the norm topology.

A useful space when dealing with boundary value problems is the sub-space Hs

0(∂Ω) of Hs(∂Ω) consisting of functions, which integrate to zeroalong the boundary, i.e.

Hs0(∂Ω) = g ∈ Hs(∂Ω) | 〈g, 1〉 = 0. (A.2)

Here we use the fact that 1 ∈ H−s(∂Ω) for any s ∈ R. We will in the sameway denote by Cα

0 (∂Ω) ⊂ Cα(∂Ω) the subset of functions, which integrateto zero along the boundary. The space H1

0(∂Ω) should not be confused withthe space H1

0(Ω) defined above.

Concerning integration along C1 curves in the plane we will use thenotation

∫

Γ

f (z)dz =

∫ b

af (γ(t))γ′(t)dt

where γ : [a, b] 7→ Γ is a parameterization of Γ. We note that if dσ is theusual surface measure on Γ, then

∫

Γ

f (z)dz =

∫

Γ

f (x)i(ν1 + iν2)(x)dσ(x),

where x1 + ix2 = z and (ν1, ν2) is the unit normal vector to Γ in x pointingto the right relative to the direction of γ.Fourier transform. The n-dimensional Fourier transform is defined for φ ∈S(Rn) by

φ(ξ) = Fφ(ξ) =1

(2π)n/2

∫

Rne−ix·ξφ(x)dx

and extended in the usual way to L1(Rn), L2(Rn) and S ′(Rn). The inverseFourier transform is denoted by F−1.Miscellaneous. All constants are generically denoted by C. Occasionally weelaborate the dependency of a constant on certain parameters. Furthermorewe will use the notation f ∼ g to indicate that f − g is asymptotically zero.In the concrete applications we are more exact about in which sense thisholds.

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Documents

On the Inverse Conductivity Problem · Chapter 1. Introduction 1 x1.1. The inverse conductivity problem 2 x1.2. Notes 5 Chapter 2. Preliminary reductions and results 7 x2.1. Reduction