Universidade de Aveiro Departamento de Matematica2019

Ivan MiraPombo

O problema de conductividade inverso de Calderona duas dimensoes: a abordagem de Nachman e ocaso de conductividade complexa

Calderon’s inverse conductivity problem in twodimensions: Nachman’s approach and the case forcomplex conductivity

Universidade de Aveiro Departamento de Matematica2019

Ivan MiraPombo

O problema de conductividade inverso de Calderona duas dimensoes: a abordagem de Nachman e ocaso de conductividade complexa

Calderon’s inverse conductivity problem in twodimensions: Nachman’s approach and the case forcomplex conductivity

Dissertacao apresentada a Universidade de Aveiro para cumprimento dosrequisitos necessarios a obtencao do grau de Mestre em Matematica eAplicacoes, realizada sob a orientacao cientıfica do Doutor Uwe Kahler,Professor Associado com Agregacao do Departamento de Matematica daUniversidade de Aveiro.

Ao eu do passado e ao eu dofuturo.

o juri / the jury

presidente / president Professor Eugenio Alexandre Miguel RochaProfessor Auxiliar do Departamento de Matematica da Universidade de Aveiro

Vogal-arguente principal Professor Jens WirthProfessor Associado do Departamento de Matematica da Universidade de Estu-

garda

Vogal-orientador Professor Uwe KahlerProfessor Associado com Agregacao do Departamento de Matematica da Univer-

sidade de Aveiro (orientador)

agradecimentos /acknowledgements

Gostaria de agradecer ao Professor Uwe Kahler e a professora Paula Cere-jeiras, por me darem a oportunidade de trabalhar com eles desde o inicio daminha Licenciatura, por me darem a conhecer o mundo da matematica e porme darem as condicoes para atingir o meu potentical enquanto matematico.

Professor Jens Wirth fur alle Antworten auf meine verschiedenen Fragenund fur das Abendessen wahrend meines Aufenthalts in Stuttgart.

Ao Donca, por ser o meu acoriano favorito para ir relaxar e beber unscanecos.

A Liliana, por me aturar nos bons e maus momentos, por me fortalecer todosos dias e ainda que nao perceba quando lhe apresento as coisas em ingles,tem sempre uma palavra a dizer para fazer de mim uma pessoa melhor eum melhor profissional. Agradeco-te do mais profundo do meu ser.

Finalmente, agradeco aos meus pais, porque literalmente sem eles nao es-tava aqui e por me darem sempre abertura e apoio a que eu me dedique anovas oportunidades e que de sempre o meu maximo.

palavras-Chave problema de conductividade inverso, conductividade complexa, tomografiade impedancia electrica, operador Dirichlet-to-Neumann

resumo O problema de Calderon consiste na determinacao da conductividade (realou complexa) de um corpo baseado em medidas na fronteira. Este formao modelo matematico para metodos de imagem medica nao invasiva, comopor exemplo tomografia de impedancia electrica. Neste trabalho, apresen-tamos como introducao ao problema, algumas ferramentas necessarias quesao usadas na abordagem de resolucao do problema a duas dimensoes, paraconductividades reais, publicado por Nachman, no qual obteve o primeirometodo de reconstrucao. No entanto, o problema so esta completamenteresolvido nestas condicoes. Assim sendo, e importante olhar para o casoem que a conductividade tem valores complexos e para o caso de dimensoesmaiores que dois. Com este intuito, mostramos tambem um novo conceitopara abordar o problema de conductividades complexas e mencionamos osproblemas que tem que ser ultrapassados para obter resultados para di-mensoes maiores.

keywords inverse conductivity problem, complex conductivity, electrical impendancetomography, Dirichlet-to-Neumann operator

abstract The Calderon problem consists in the determination of the (real- or complex-valued) conductivity of a body based on measurements at its boundary. Itforms the mathematical model for methods of non-invasive medical imaging,like electric impedance tomography. In two dimensions the problem wasfully solved for real conductivities with tools from complex analysis. In thiswork, we present as an introduction to the problem, the necessary toolsthat are used in the approach to solve the problem in two dimensions, forreal conductivities, which was published by Nachman as first reconstructionmethod. However, the problem is only fully solved on this conditions. Inthis way, is of great importance to look at the case where we have complex-conductivites and the case of higher dimensions. In this sense, we alsoshow a new concept to approach the problem for complex conductivitiesand mention the problemas that need to be overcomed to obter results forhigher dimensions.

Contents

Contents i

List of Figures iii

1 Introduction 11.1 Inverse Conductivity Problem (ICP) . . . . . . . . . . . . . . . . . . . . . . . 1

2 Nachman’s Approach 72.1 The ways of Nachman . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Exponentially Growing Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 The ∂ equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Behavior near k = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Reconstruction of the conductivity at the boundary . . . . . . . . . . . . . . . 462.6 Reduction to the case γ ≡ 1 close to the boundary . . . . . . . . . . . . . . . 512.7 From Λ to t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542.8 From t to γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3 Complex Conductivities 633.1 Our ways . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.2 Main construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.2.1 Transmission condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.2.2 The Lippmann-Schwinger equation for CGO-Faddeev solutions . . . . 67

3.3 Technical details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.3.1 The choice of the functional space . . . . . . . . . . . . . . . . . . . . 683.3.2 Analysis of the Lippmann-Schwinger equation . . . . . . . . . . . . . . 693.3.3 Enrichment of the set of CGO incident waves . . . . . . . . . . . . . . 703.3.4 Scattering data and reconstruction of the potential in admissible points 713.3.5 Necessary results for the proof of Theorem 3.3.1 . . . . . . . . . . . . 723.3.6 Proof of Theorem 3.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 77

3.4 Scattering data for Dirac equation via the Dirichlet to Neumann map . . . . 78

4 From the current problems to future works 81

A 83A.1 Laplace Transform analogue of the Haussdorf-Young inequality . . . . . . . . 83A.2 Proof of Lemma 3.3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

i

B 86B.1 Domains and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86B.2 Lp and Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87B.3 Complex function Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 92B.4 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93B.5 Other Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

Bibliography 97

ii

List of Figures

1.1 Domain Ω ⊂ R2, with different tissues. The colors represent different intensitiesof the conductivity γ. At black we have the heart, the lighter gray is the lungsand the rest is the background. . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.1 The diagram shows in a simple manner Nachman’s idea to recover γ fromknowledge of the Dirichlet-to-Neumann map. . . . . . . . . . . . . . . . . . . 12

3.1 Simple representation on a type of domain we can consider for the transmissionproblem. The inside boundary Γ represents the contour where γ is not smooth. 63

iii

Chapter 1

Introduction

In this day and age, we are able to obtain data from almost every aspect of human life. Oncontemporary mathematics, one of the areas that has been taking advantage of this and hasbeen growing steadily for the past 30 years is the field of Inverse Problems. Modern digitalsensors produce vast amounts of measurements related to various areas including engineering,geophysics, medicine, biology, physics, chemistry and finance. At this stage, we need to pointout that these this measurements are indirect and most of the cases incomplete, i.e., the dataobtained in applications does not give full information about the physical property that weare trying to measure. Inverse problems is the established mathematical approach to interpretthis data.

The field of Inverse Problems is the process of obtaining from a set of observations thecasual factors that produced them. The area took its first steps through a successful combi-nation between physics and pure mathematics.

One of the first solved inverse problems was due to Hermann Weyl and published in1911, where he describes the asymptotic behavior of the eigenvalues of the Laplace-BeltramiOperator. Today this well-known Weyl’s law is trivialized in the question: ”Is it possible tohear the shape of a drum?”. Weyl conjectured that the eigenfrequencies of a drum would berelated to the area and perimeter of the drum by a particular equation.

During the period of emergence of this field, Gelfand and Levitan put their efforts intrying to solve the inverse scattering problem in a direct manner, i.e., to discover an ana-lytic constructive method of the solution. Their success was translated into computationalapplications, but it was rapidly discovered that the method is very unstable.

1.1 Inverse Conductivity Problem (ICP)

On this thesis we are concerned with the Calderon problem. In the 1950’s AlbertoCalderon worked as an engineer for a company which used electromagnetic methods in geo-physical prospecting. One of the challenges that he faced was to find the electrical conductivityof a body from measurements at its surface.

Currently, the determination of the electrical conductivity of a body is a key instrumentin Medical imaging, where the problem is known as Electrical Impedance Tomography (EIT).It has been proposed as a valuable diagnostic tool especially for detecting pulmonary emboli,for example. To properly present the problem we use this specific application to simplify

1

explanations.

Let Ω be a domain in Rn, with n ≥ 2. This domain represents an inhomogeneous, electricconductive isotropic body, (i.e. the conductivity property is independent of directions), withelectrical conductivity γ : Ω → [0,∞). In this application we consider two things: first, theconductivity is a bounded function, (given that the domain is bounded); second, it has apositive lower bound γ ≥ c0 > 0, (as we do not want the existence of annulus inside the body,with conductivity zero).

Figure 1.1: Domain Ω ⊂ R2, with different tis-sues. The colors represent different intensities ofthe conductivity γ. At black we have the heart,the lighter gray is the lungs and the rest is thebackground.

The plan here is to determine the conductivity inside Ω, without previous informationdirectly on inner body. We do this by setting a voltage potential at the boundary of the do-main ∂Ω and measuring the current-flux which comes across it. Therefore, the observations(in our case) are electrical measurements at the boundary.

The mathematical model is deduced from the Maxwell equations, given that in the humanbody magnetic permeability can be neglected, (i.e. in response to a magnetic field the humanbody can not be magnetized). Hence, we obtain:

∇ · (γ∇u) = 0 in Ω(1.1.1)

u|∂Ω = f

where u is the electrical potential inside our particular body which corresponds to the voltagepotential f set at the boundary.

The boundary value problem (1.1.1) is elliptic due to the positive lower bound of γ, if γas first derivatives. In praxis, we want to find an approach for the worst case scenario, whichfor us means that we require the conductivity to be in L∞(Ω). This assumption is supportedon one hand, by the fact that, physically, it does not make sense to have infinite values on theconductivity. On the other hand, we can have several tissues with very different conductivity(e.g. look at the heart compared with the background). This condition, together with theboundary value f ∈ H1/2(∂Ω), implies the existence of a unique (weak) solution u ∈ H1(Ω)of (1.1.1).

Another realistic measurement we can perform at the boundary is the current-flux, whichis the co-normal derivative associated to the elliptic operator. In the case we have the conduc-tivity and we can set a voltage f at the boundary, then we obtain u and hence the co-normalderivative on the boundary will be given by γ ∂u∂ν

∣∣∂Ω

, (see Grisvard [2011]).Under the above explanation, we can convert the Dirichlet condition into Neumann con-

ditions at the boundary, by formally defining a map, which is designated by Dirichlet-to-

2

Neumann map (DtN):

Λγ : H1/2(∂Ω)→ H−1/2(∂Ω)(1.1.2)

f 7→ γ∂u

∂ν

∣∣∣∂Ω

For reconstruction, it is easier to establish the Dirichlet-to-Neumann map in a weak sense.Let u be a solution to (1.1.1), with Dirichlet data f ∈ H1/2(∂Ω). Then, we define the weakformulation by:

〈Λγf, g〉 =

∫∂Ωgγ∂u

∂νdS for all g ∈ H1/2(∂Ω)

where dS is the surface measure associated with Ω. Let v ∈ H1(Ω) be such that its trace isg. By using Green’s first identity we obtain

〈Λγf, g〉 =

∫Ω∇(γv) · ∇u+ vγ∆u dV

=

∫Ωγ∇v · ∇u+ v(∇γ · ∇u+ γ∆u) dV

=

∫Ωγ∇v · ∇u dV(1.1.3)

as the weak formulation of the DtN map.In this framework, Calderon published a paper (Calderon [1980]) where the following prob-

lem was posed: to decide on whether a real and positive lower bounded function γ ∈ L∞(Ω)is uniquely determined by the Λγ map and, if so, how to reconstruct it. In that same paper,Calderon proved that the linearized problem is in fact injective.

Even though the DtN map is a linear operator in terms of the boundary data, it is non-linear in terms of the conductivity. This is easy to see since the solution u of (1.1.1) dependson γ we can observe from the weak formulation (1.1.3) that the product between γ and ∇umakes the DtN a non-linear mapping. This makes the Calderon problem a very ill-posedproblem.

This becomes even worse when we pass from the theoretical framework to the applied one,where the measurements are discrete and of finite precision (hence noisy) and most likely onlypartial boundary data is available. Even in the ideal case, the solution fails in the continuousdependence of the data.

An example is given by Alessandrini in his paper (c.f. Alessandrini [1988]). Discontinuousdependence is established by showing that given any ε > 0 and any δ > 0, there existconductivities γ1, γ2 such that:

‖Λγ1 − Λγ2‖H1/2(∂Ω)→H−1/2(∂Ω) < δ, but still ‖γ1 − γ2‖L∞ > ε,

that is to say arbitrarily close DtN maps may correspond to too different conductivities.Indeed, set Ω to be the unit disk in R2 and consider the radial conductivities:

γ1(r, θ) = 1, γ2(r, θ) =

A, 0 ≤ r ≤ R1, R < r ≤ 1

3

where A is a positive value.By explicitly calculations the Dirichlet-to-Neumann maps for both conductivities can be

shown to satisfy

‖Λγ1 − Λγ2‖H1/2(∂Ω)→H−1/2(∂Ω) ≤ |1−A|R→ 0, R→ 0.

while

‖γ1 − γ2‖L∞(Ω) = |1−A|.

Since this is independent of R, the result holds. Therefore, it follows that there cannot exista continuous real function with f(0) = 0 and satisfying an inequality of the type:

‖γ1 − γ2‖L∞(Ω) ≤ f(‖Λγ1 − Λγ2‖H1/2(∂Ω)→H−1/2(∂Ω)).

In terms of applications, this might seem devastating, but with some a priori knowledgeof the conductivity it is possible to obtain estimates like the above (Novikov and Santacesaria[2010]).

The extension of Calderon work in two dimensions was based on the conversion of theconductivity equation into another equation, and to study the inverse problem in that case.For the case of three dimensions the results are in general based on the relation of the bound-ary data to the Fourier transform of the conductivity.

After Calderon posed the problem, various authors investigated it and extended his results.Here we give a short list of the larger steps:

Kohn and Vogelius [1984]: They show that for smooth conductivities, the quadraticform of the DtN map defines uniquely all derivatives of the conductivities at the bound-ary of the domain. Hence, by analytic continuation, it follows that if γ is real-analyticon Ω, then it is uniquely determinedthe DtN map;

Sylvester and Uhlmann [1986]: In this paper it is present uniqueness in R2 for γclose to 1 in W 3,∞(Ω) (as a local result);

Sylvester and Uhlmann [1987]: They presented a first global result, which extendsthe result of Kohn and Vogelius to the whole domain, i.e. in Rn, n ≥ 3. Namely, theyshowed that smooth conductivities are uniquely determined from the quadratic form inthe whole domain;

Nachman [1988]: This was the first reconstruction procedure given in the literature.It holds for dimensions higher than 3 and gives uniqueness for C1,1- conductivities.

Nevertheless, up to this time global uniqueness for smooth conductivities indimension 2 was still an open question.

Nachman [1996]: First global result in two dimensions through a reconstructionmethod for conductivities W 2,p, p > 1, by converting (1.1.1) to the Schrodinger equa-tion. We will take a closer look into this later on;

4

Brown and Uhlmann [1997]: They reduce the need requirement for uniqueness toonly conductivities which are in W 1,p. For this they use the Dirac scattering equation,which had already been studied in the literature of inverse scattering problems. In thepaper by Knudsen and Tamasan [2005], they obtain a reconstruction method to thisuniqueness result ;

Paivarinta et al. [2003] This paper yields the current best result for three dimensions.It provides a uniqueness proof for γ ∈W 3/2,∞(Ω), i.e., for Lipschitz conductivities;

Astala and Paivarinta [2006]: In this paper finally a positive answer was givento the Calderon problem, i.e. the Dirichlet to Neumann map defines uniquely real-valued conductivities in L∞, which have a positive lower bound. This was shown byconverting (1.1.1) into the Beltrami equation and applying results from the theory ofquasi-conformal mappings.

The Calderon problem for three and higher dimensions is still open. It is interesting tonotice that the problem was always thought to be tougher in two dimensions, since in thiscase the problem is not overdetermined. A naive reason for this is that in n dimensions theSchwartz kernel of Λγ depends on 2(n− 1) variables while γ is a function of n variables; thismeans that in a sense there is no over-determinancy in two dimensions and all the data hasto be used for the reconstruction.

Naturally, there is an extension of the initial question of Calderon to the case in which γis complex-valued. In this case we designate γ as admittivity which is defined as:

γ(x) = σ(x) + iωε(x),(1.1.4)

where σ is the conductivity, ε is the electric permittivity (physically meaning a substancecapacity to resist an electrical field from an induced charge), and ω is the electro-magneticfrequency of the waves. In this scenario, there are much more problems to study, one ofthem we will introduce and study later, and there is still much to be done. As a short list ofprevious works we mention a selected few:

Francini [2000]: This was one of the first explicit treatments of this case. It adaptsthe method of Brown and Uhlmann, needing only one derivative on the conductivity,but it only works for small frequencies;

Bukhgeim [2008]: In his approach, Bukhgeim used the change to the Schrodingerequation, to the case of any complex conductivity with two derivatives ;

Many works are done for twice-differentiable conductivities, treating theproblem in the sense of the inverse scattering problem;

Lakshtanov and Vainberg [2016]: This result uses the transformation to the Diracequation in two dimensions and gives a reconstruction method for complex conductivi-ties that are one time differentiable, but have some higher regularity than just that;

Lakshtanov et al. [2017]: This paper gives a similar result to the current best on threedimensions and shows uniqueness for complex Lipschitz conductivities in the plane;

5

This thesis is structured in three chapters. In chapter 2, we present Nachman’s approachin two dimensions. Here we follow closely the paper (Nachman [1996]). We show everystep of his reconstruction in detail. Nachman’s paper was the first to present a reconstructionmethod, so it is ideal to make an introduction to the problem, as it contains all necessary toolsand results used in later developments of this problem (including some in later dimensions).We would like to point out that indeed Nachman’s technique was implemented as a practicalreconstruction method in several works (Mueller and Siltanen [2012]).

In the third chapter we present an approach to the study of the inverse conductivityproblem with discontinuous complex conductivities. For this work we follow the paper (Pombo[2019]).

In the last chapter we present an overview of current problems and some ideas for treatingthem.

6

Chapter 2

Nachman’s Approach

2.1 The ways of Nachman

In the last decades there has been a great effort to make electrical impedance tomographyviable. Many methods have been used and tested on real data, like linearization and iterativelynon-linear approaches, although each of them has its fragilities and from the point of view ofapplications is not fully viable.

The purpose of this chapter is to present the inverse conductivity problem directly, i.e.,finding an analytic approach to determine the conductivity from the Dirichlet-to-Neumannmap. We present the way Nachman used to solved this problem with of tools from complexand spectral analysis.

In Nachman [1996] the first reconstruction work for two-dimensions was presented. Undersome conditions Nachman proves uniqueness from the DtN map in a constructive manner fortwice-differentiable conductivities. Later, many numerical implementations of his approachhave been realized and each step seems to be have been viably implemented. Even thoughthere are always some questions on how to implement them properly, in a direct approach weonly need to worry with the errors present in the numerical implementation.

By the end of this chapter Nachman’s reconstruction method will lead us to the proof ofthe following theorem.

Theorem 2.1.1. Let Ω be a bounded Lipschitz domain in R2. Let γ1 and γ2 be in W 2,p(Ω)for some p > 1, and having positive lower bounds. If

Λγ1 = Λγ2 then γ1 = γ2.

The need for two derivatives in the conductivity comes from the fact of the conversion of(1.1.1) to a Schrodinger equation. In parallel with Nachman’s paper, there have been manyworks on the Inverse Scattering problem, but there has always been some growth conditionson the potential. With Nachman’s approach we give a partial answer to Calderon problem.

During this chapter, unless explicitly mentioned otherwise, we assume that γ is a real andbounded function, and there exists c0 ∈ R+ such that γ(x) ≥ c0 for any x ∈ Ω. Let u be asolution of (1.1.1) and set u = γ1/2u. We substitute it in the conductivity equation and we

7

obtain:

∇ · γ∇u = 0

⇔∇γ · ∇(γ−1/2u

)+ γ∆

(γ−1/2u

)= 0

⇔∇γ ·(−1

2γ−3/2u∇γ + γ−1/2∇u

)+ γ∇ ·

(u∇γ−1/2 + γ−1/2∇u

)= 0

⇔− 1

2γ−3/2 (∇γ · ∇γ) + γ−1/2∇γ · ∇u+ γu

(∆γ−1/2

)+ 2γ∇γ−1/2 · ∇u+ γ1/2∆u = 0

Even though the last formula seems awfully complicated, by using ∇γ−1/2 = −12γ−3/2∇γ,

as we have done to obtain the second equation, the second and fourth terms cancel out.Given that we working under the assumption that γ as a positive lower bound, we are goingto multiply what remains by −γ1/2 to obtain:

−∆u+

(1

2γ−2 (∇γ · ∇γ)− γ1/2∆γ−1/2

)= 0

⇔−∆u+

(1

2γ−2 (∇γ · ∇γ)− γ1/2

[−1

2γ−3/2∆γ +

3

4γ−5/2 (∇γ · ∇γ)

])u = 0

⇔−∆u+

(1

2γ−1∆γ − 1

4γ−2∇γ · ∇γ

)u = 0.

Thus if we set

q = γ−1/2∆γ1/2 = γ−1/2∇ ·(∇γ1/2

)= γ−1/2

(1

2γ−1/2∆γ − 1

4γ−3/2∇γ · ∇γ

)we get that u solves:

−∆u+ qu = 0, in Ω.(2.1.1)

Similar as before, we can define the Dirichlet-to-Neumann map for the Schrodinger equa-tion. By the conditions on γ and recalling that the solution u of (1.1.1) is in H1(Ω), it followsthat u ∈ H1(Ω) is a weak solution to the above Schrodinger equation. Moreover, we also havethat u fulfills the Dirichlet condition on the boundary, i.e. u|∂Ω = γ1/2f = f ∈ H1/2(∂Ω).

Hence, we define the formal DtN map for the Schrodinger equation as the map:

Λq : H1/2(∂Ω)→ H−1/2(∂Ω)

f 7→ ∂u

∂ν

∣∣∣∣∂Ω

By considering v to be any function in H1(Ω) with v|∂Ω = g, we can define it analogouslythrough Green identities via the bilinear form:

(2.1.2)⟨

Λqf , g⟩

=

∫Ω∇v · ∇u+ qvu dV, for all v

8

The good thing about this transformation is that we can still relate Dirichlet-to-Neumannmaps of both equations. Let the functions be defined as above. We get⟨

Λqf , g⟩

=

∫∂Ωg∂(γ1/2u)

∂νdS =

∫∂Ωg∇(γ1/2u · ν

)dS

=

∫∂Ωg

(u

1

2γ−1/2∂γ

∂ν+ γ1/2∂u

∂ν

)dS

=

∫∂Ωgγ−1/2

(1

2

∂γ

∂ν

(γ−1/2u

)+ Λγ

(γ−1/2u

))dS

=

⟨γ−1/2

(Λγ +

1

2

∂γ

∂ν

)γ−1/2f , g

⟩Given that this holds for all g ∈ H1/2(∂Ω), we have the operators identity

(2.1.3) Λq = γ−1/2(

Λγ +1

2

∂γ

∂ν

)γ−1/2

Therefore, to compute Λq we just need Λγ and γ|∂Ω, ∂γ∂ν |∂Ω. With this we can reduce

the inverse problem to the recovery of q from Λq. Besides the knowledge we already have onthe Schrodinger equation this reduction removes the unknown coefficient in (1.1.1) from thehigher order terms which will simplify the problem.

Given that our method will use the Schrodinger equation and, therefore, the DtN mapΛq in some steps of our reconstruction, we will need to determine first γ|∂Ω and ∂γ

∂ν |∂Ω sincein theory we only have information about Λγ . A method for this will be presented in Sec-tion 2.5. In spite of this equality, we won’t compute Λq in this way. The proof could follow thispath but it is simpler to use the boundary values of the conductivity to reduce the problemto one with γ equal to 1 near the boundary, which follows on section 6. By this construc-tion and the equality between DtN maps, it will follow that Λq ≡ Λγ for this new conductivity.

Moreover, one of the important steps for the proof is the knowledge we can obtain fromthe forward problem, i.e. given a conductivity and Dirichlet data we ask the question: Whattype of solutions to (1.1.1) and (2.1.1) do we obtain?;

Rather than looking at the Schrodinger equation just on Ω, we extend it to the whole R2,by setting:

q =

γ−1/2(x)∆γ1/2 , for x ∈ Ω;0 , for x ∈ R2 \ Ω.

Following this, we can assume more generally that q ∈ Lp(R2) and we look at the solutionsψ(x, ζ) (when they exist) of

(2.1.4) (−∆ + q)ψ(x, ζ) = 0 in R2,

where e−ix·ζ−1 ∈W 1,p(R2), with 1p = 1

p−12 is the Sobolev conjugate and ζ ∈ V =: (ζ1, ζ2) ∈

C2\0 : ζ2 := ζ21 +ζ2

2 = 0. This type of solutions are what we call the exponentially growingsolutions. They were first introduced in (Faddeev [1966]) for the inverse scattering problem

9

and later rediscovered in the ICP by Sylvester and Uhlmann, but in some sense already usedby Calderon.

In the case that the potential comes from the conductivity, then we have that in R2\Ω thistype of solutions fulfill the Laplace equation. Hence the idea here is to look to eigenfunctions ofthe Laplace operator, with especially focus in exponential growing solutions. Let χ ∈ C2 \0and set for now φ(x, χ) to which we call non-physical eigenfunctions. Then we obtain:

−∆φ(x, χ) = −(∂2

∂x21

+∂2

∂x22

)ei(x1χ1+x2χ2) = (χ2

1 + χ22)φ(x, χ).(2.1.5)

Therefore, φ(·, χ) are eigenfunctions with respect to the eigenvalue χ2. If we pick, χ ∈ V,then φ(·, χ) is a solution of Laplace equation.

This puts some interest in solutions to (2.1.4) that asymptotically behave like this eigen-functions. In this manner, we set our solutions to be of the type:

ψ(x, ζ) = eix·ζµ(x, ζ),(2.1.6)

and µ→ 1 as x tends to infinity, since we want eigenfunction asymptotic behavior.With this type of solutions in mind, and since we always desire some regularity, we will

assume that µ − 1 ∈ W 1,p(R2), as mentioned previously. The idea to pick this space, comesfrom the fact that during the proofs we use a lot of Sobolev embeddings and inequalities.

Now we are ready to introduce one of the main headaches on dimensions higher than twoand for complex conductivities, the concept of exceptional points. In the last chapter weexplain a little of what happens in this cases. Lets fix ζ ∈ V. If for (2.1.4) there exist two

solutions ψ1(·, ζ), ψ2(·, ζ), of the type e−ix·ζψ1(x, ζ) − 1, e−ix·ζψ2(x, ζ) − 1 ∈ W 1,p(R2), thenwe say that ζ is an exceptional point. This is equivalent to say that exists h 6= 0, withhe−ix·ζ ∈W 1,p(R2) is a solution of (2.1.4), since we can pick h = ψ1−ψ2. During the section2, we will show that for each k ∈ C \ 0 there is a unique solution on the desired conditionsto (2.1.4) and prove a asymptotic estimate.

The importance of this set of points lies on the proper definition of the scattering trans-form, which relates the solutions and the potential, and even though it can not be directlymeasurable in experiments, is a key intermediate object in our reconstruction.

The Scattering transform is defined for ζ ∈ V non-exceptional and ξ ∈ R2 by:

t(ξ, ζ) =

∫R2

e−ix·(ξ+ζ)q(x)ψ(x, ζ) dx.(2.1.7)

So to have the most information possible, we would like that all ζ ∈ V are non-exceptionalpoints, see section 2. This will be the case in two dimensions for potentials of the followingtype:

Definition 2.1. Let q ∈ Lp(R2), 1 < p < 2. We say that q is of conductivity type, if existsψ0 ∈ L∞(R2), such that q = (∆ψ0)/ψ0, ψ0(x) ≥ c > 0 and ∇ψ0 ∈ Lp(R2).

Remark: On the conductivity scenario, we have that ψ0 = γ1/2.

To be more precise, the importance of the scattering transform lies on the reconstructionof γ from it. This step will in fact only require the values of t(ξ, ζ) with ξ = −2<(ζ) andζ ∈ V. In this sense we can make a change of notation. The set V can be parameterized as

10

follows: (k, ik) : k ∈ C\0∪(k,−ik) : k ∈ C\0. Henceforth, we denote ψ(x, (k, ik)) byψ(x, k) and by observing that the assumption on q being real-valued, we have that uniquenessfor (2.1.4) yields ψ(x, (−k, ik)) = ψ(x, (k, ik)) = ψ(x, k), it follows that for reconstructionpurposes it suffices to work on the sheet ζ = (k, ik).

For the rest of the chapter, we will interchange between with C and R2 back and forth, forconvenience. In this way, we use z = x1+ix2 to represent the complex number of x = (x1, x2).Therefore, for reconstruction purposes, by k = k1 + ik2 with k1, k2 ∈ R it follows:

t(k) =: t(−2<(k, ik), (k, ik)) = t(−2(k1,−k2), (k, ik))

=

∫R2

e−i(x·(−2(k1,−k2)+(k1+ik2,−k2+ik1)q(x)ψ(x, k) dx

=

∫R2

e−ix·(−k,ik)q(x)ψ(x, k) dx

=

∫R2

ei(x1−ix2)kq(x)ψ(x, k) dx

=

∫R2

eizkq(x)ψ(x, k) dx

The conductivity reconstruction from the scattering transform lies on using the ∂ method.This method was initially introduced in inverse scattering on (Beals and Coifman [1981])for the case of one dimension and later extended to two dimensions. The usage of toolsfrom complex analysis and the ∂ establish the foundational approach for the two-dimensionalinverse conductivity problem, being presented in the works: (Nachman [1996]), (Brown andUhlmann [1997]) and (Astala and Paivarinta [2006]).

We will show on section 3, that for k 6= 0 it holds:

∂

∂kµ(x, k) =

1

4πkt(k)e−k(x)µ(x, k),(2.1.8)

in the weighted Sobolev space W 1,pβ (R2) = f : 〈x〉−β f ∈ W 1,p(R2), where 〈x〉 = (1 +

|x|2)1/2, for β > 2/p; moreover we also define ek as:

ek(x) := ek(z) = ei(kz+kz).(2.1.9)

From previous works to his, Nachman also takes a different path on determining the con-ductivity. He uses the behavior of µ(·, k) for k near zero, studied in section 4, instead of theusual large k asymptotics of µ, which need more decay on t than what is available in ourcase. On this sense, we can develop the idea that γ1/2(x) = limk→0 µ(x, k), and we will showthat this is in fact true on section 5.

By our explanation up to now, it is enough to solve (2.1.8) to obtain the conductivity.Although, currently we don’t have the scattering transform from the only information wehave, the DtN map. Here lies the usefulness of the ∂ method, because through this approach,as long as q is known outside ∂Ω, then we can determine t from Λq and the solutions of (1.1.1)at the boundary ∂Ω.

So now a problem seems to appear. How can we determine the solutions of ourequation at the boundary? We can immediately expect that the Dirichlet-to-Neumann

11

map has a role to play, but is not as straightforward, since do not know what Dirichlet datagives rise to each ψ(·, k) to construct a proper relation with t(k).

In section 7, we explain how to obtain them through a integral boundary equation andthe relation between the scattering transform and the Dirichlet-to-Neumman map.

With the purpose of clarification of Nachman’s method, we give a quick overview of whatan implementation of his method should look like:

(i) Determine γ|∂Ω and ∂γ∂ν |∂Ω. Constructively reduce the problem to one with conductivity

γ ≡ 1 at ∂Ω;

(ii) Solve an integral boundary equation to obtain ψ(·, k) at ∂Ω;

(iii) Compute t(k) with help of the Dirichlet-to-Neumann map and the solutions ψ(·, k);

(iv) Solve the ∂ equation through the integral equation:

µ(x, k) = 1 +1

8π2i

∫C

t(k′)

(k′ − k)k′e−x(k′)µ(x, k′) dk′ ∧ dk′;(2.1.10)

(v) Obtain γ(x) from the (absolutely convergent) integral formula

γ1/2(x)1 +1

8π2i

∫C

t(k′)

|k′|2e−x(k′)µ(x, k′) dk′ ∧ dk′.(2.1.11)

We can establish the following relation between the direct and inverse problem.

Λq

q

γ

µ

t

Figure 2.1: The diagram shows in a simple manner Nachman’s idea to recover γ from knowl-edge of the Dirichlet-to-Neumann map.

In this manner, we study the direct problem in the following first three sections, and onwhat is left of the chapter we prove each step of the algorithm.

2.2 Exponentially Growing Solutions

We start by introducing the main result of this section which concerns the injectivity of theequation (2.1.4) in the desired spaces for the case that the potential is of conductivity type. Forthis we need to present and show some results and estimates essential for simplifying furtherproofs as well as showing that potentials of conductivity type do not possess exceptionalpoints.

12

Theorem 2.2.1. Let q ∈ Lp(R2), 1 < p < 2, be of conductivity type. Then for any k ∈ C\0,there is a unique solution ψ(x, k) of (2.1.4) with e−izkψ(·, k) − 1 ∈ Lp ∩ L∞. Furthermore,e−izkψ(·, k)− 1 ∈W 1,p and

(2.2.1)∥∥∥e−izkψ(·, k)− 1

∥∥∥W s,p

≤ c|k|s−1 ‖q‖Lp

for 0 ≤ s ≤ 1 and k sufficiently large.

Notation: We denote ∂ and ∂, to be the differential operators, corresponding to theWirtinger derivatives, defined as

∂ =1

2(∂x1 − i∂x2) and ∂ =

1

2(∂x1 + i∂x2)

.In addition, we have that the fundamental solutions to ∂ and ∂ are the functions 1

πz and1πz , and, hence, we can define the inverse operator (left inverse modulo analytic functions) tobe given by convolution with the fundamental solutions as

∂−1f(z) =1

2πi

∫C

f(w)

w − zdw ∧ dw,

where dw ∧ dw = −2idx ∧ dy, and which we call Teodorescu transform. Similarly, we candefine ∂−1.

Finally, we just introduce a function that will be useful in simplifying the notation byek(z) = ei(kz+kz) = e2i(x1k1−x2k2), considering k = k1 + ik2.

Lemma 2.2.1. i. For any f ∈ Lp(R2) and any k in C there is a unique u ∈ Lp(R2)satisfying (∂ + ik)u = f . Further, u = (∂ + ik)−1f =: e−k∂

−1(ekf) and

(2.2.2) ‖(∂ + ik)−1f‖Lp ≤ c‖f‖Lp ,

with c independent of k.

ii. If v is a function in Lp with ∂v ∈ Lp, then for any k ∈ C \ 0 there is a unique solutionw in Lp of (∂ + ik)w = v. Moreover, w ∈W 1,p and

(2.2.3) ‖w‖Lp ≤c1

|k|(‖v‖Lp + ‖∂v‖Lp) and ‖∇w‖Lp ≤ c2‖∂v‖Lp

Proof. i. Straightforward computation gives that the fundamental solution to (∂ + ik) is

given bye−k(z)πz .Therefore, we have that

u(z) = (∂ + ik)−1f(z) =1

2πi

∫C

e−k(z − w)f(w)

w − zdw ∧ dw

= e−k(z)1

2πi

∫C

ek(w)f(w)

w − zdw ∧ dw

= e−k(z)∂−1(ekf)(z)(2.2.4)

13

With this in mind, the desired estimate (2.2.2) follows by a simple modification of theHardy-Littlewood-Sobolev Inequality [Stein, 1970], given that the functions ek has mod-ulus 1. Thus the existence of a solution and the estimate follows.

For uniqueness, we consider a similar conversion of the equation:

∂(eku) = ek(∂ + ik)u.(2.2.5)

Thus, suppose u1, u2 ∈ Lp(R2) are solutions to the equation (∂ + ik)uj = f , for f ∈Lp(R2) and j = 1, 2. Hence, it immediately follows that the difference between themsatisfies:

∂(ek(u1 − u2)) = ek(∂ + ik)(u1 − u2) = 0 ⇒ ∂(ek(u1 − u2)) = 0

Thus, ek(u1 − u2) is an anti-holomorphic function in Lp. Consequently, ek(u1 − u2) isan entire function. By Liouville’s theorem for Lp spaces it follows that this functionequals zero and since |ek(z)| = 1, ∀ z ∈ C, we obtain u1 − u2 = 0 and, consequently, theuniqueness result holds.

ii. We start by looking at the desired equation and changing it a little bit to get a casesimilar to (i). Suppose that for v under the desired conditions there exist w fulfilling

(∂ + ik)w = v =ikv

ik+∂v

ik− ∂v

ik=

1

ik

[(∂ + ik)v − ∂v

].

Therefore, we can consider the following function

(2.2.6) w = (I − (∂ + ik)−1∂)v/ik,

which by above fulfills the desired equation. Moreover, due to ∂v ∈ Lp it is well-defined,and below we show that in fact belongs to Lp.

In view of ∂v being in Lp by part (i), it follows the estimate

‖(∂ + ik)−1∂v‖Lp ≤ c‖∂v‖Lp .

Consequently, from v ∈ Lp, we derive:

‖w‖Lp =

∥∥∥∥ 1

ik

[(I − (∂ + ik)−1∂)v/ik

]∥∥∥∥Lp

≤ 1

|k|(‖v‖Lp +

∥∥(∂ + ik)−1∂v∥∥Lp

)≤ 1

|k|(‖v‖Lp + c‖∂v‖Lp) ,

which shows the first desired estimate.

For the second estimate, we can first bound the gradient of w in terms of the WirtingerDerivatives. For this reason we only need to look at the bounds of ∂w and ∂w in Lp.In the first case, we have ∂w = (∂ + ik)−1∂v and the estimate itself follows by part

14

(i). For the second case, we rely on the boundedness of the Beurling transform ∂∂−1 inLr, 1 < r <∞, as well as on the estimate for ∂w as deduced above. Then, we have

‖∂w‖Lp = ‖∂∂−1(∂w)‖Lp ≤ c‖∂w‖Lp

So the second estimate follows, and the proof is complete.

In view of our assumption that the solutions to the Equation (2.1.4) are exponentiallygrowing solutions, we can modify the equation in question to be given only in terms of onlyof µ. We start with ψ(z, k) = eikzµ(z, k) and remembering that ∆ = 4∂∂ we obtain

0 = (−∆ + q)ψ(z, k) = (−4∂∂ + q)(eikzµ(z, k)

)= −4∂

(ikeikzµ(z, k) + eikz∂µ(z, k)

)+ qeikzµ(z, k)

= −4ikeikz∂µ(z, k)− eikz4∂∂µ(z, k) + qeikzµ(z, k)

So we can divide by eikz on both sides to achieve the equation:(−∆− 4ik∂

)µ(z, k) = −qµ(z, k)

(−∆)− 2ik(∂x1 + i∂x2)µ(z, k) = −qµ(z, k)

(−∆− 2i(k, ik) · (∂x1 , ∂x2))µ(z, k) = −qµ(z, k)

(−∆− 2iζ · ∇)µ(z, k) = −qµ(z, k),

(−∆− 2iζ · ∇) (µ(z, k)− 1) = −qµ(z, k),(2.2.7)

with ζ = (k, ik). The last equation follows since constants c belong to the kernel of thedifferential operator on the left-hand side. In this case, we pick c = 1 in order to satisfy ourassumptions.

In a tempered distribution sense, we can apply the Fourier transform to both sides andget

(|ξ|2 + 2k(ξ1 + iξ2))F(µ− 1)(ξ) = −F(qµ)(ξ).

Through division and the convolution theorem follows

(2.2.8) µ = 1− gk ∗ (qµ),

where

gk(x) =1

4π2

∫R2

eix·ξ

|ξ|2 + 2k(ξ1 + iξ2)dξ1dξ2,

is the fundamental solution to (2.2.7).

This function enables us to compute a new fundamental solution for the Laplacian whichis better known as Fadeev Green function:

Gk(x) = ei(x1+ix2)kgk(x).

Thus follows:

−∆Gk(x) = −∆(eizkgk(x)

)= −4∂∂

(eizkgk(x)

)= −4∂

(eizk∂gk(x)

)= −4ikeizk∂gk(x)− eizk∆gk(x) = eizk(−∆− 2iζ · ∇)gk(x) = δ(x).

15

For this new Fadeev Green function we can define a corresponding single layer operatorover Ω for k 6= 0 by:

Skf(x) =

∫∂ΩGk(x− y)f(y) dσ(y),

where dσ is the surface measure which exists when Ω is a Lipschitz domain, our domain ofinterest for Nachman approach.

The essence of the above lemma consists in allowing us to prove some new estimates onthe convolution operator defined via gk. The idea now is to use smoothing properties of thisoperator to compensate for the roughness of the potentials.

We start by showing a lemma for the most general case of a right-hand side in (2.2.7).

Lemma 2.2.2. For any f ∈ Lp(R2) and any ζ ∈ V there is a unique u ∈ Lp(R2) satisfying

(2.2.9) (−∆− 2iζ · ∇)u = f in R2

Furthermore, u ∈W 1,p and

(2.2.10) ‖u‖W s,p ≤c

|ζ|1−s‖f‖Lp for |ζ| ≥ const. > 0, 0 ≤ s ≤ 1.

Proof. As already explained we can assume without loss of generality that ζ = (k, ik) forsome k ∈ C \ 0. To verify uniqueness we use the properties of the Fourier transform andnote that any tempered distribution solution u0 of (2.2.9) with f ≡ 0 must be of the formu0 = p(z) + e−kq(z). Thus, for u0 ∈ Lr for some 1 ≤ r <∞, u0 ≡ 0.

For existence we start by considering our equation in terms of Wirtinger Derivatives anduse the Teodorescu transform:

−∂(∂ + ik)u = −1

4f

(∂ + ik)u = −1

4∂−1f

We now pick v = −14 ∂−1f and notice that (in a similar way to Lemma 2.2.1) v ∈ Lp and

∂v ∈ Lp, since by assumption f ∈ Lp. So we are in conditions to apply Lemma 2.2.1(ii), which guarantees the existence of a u ∈ Lp, which solves our problem. Moreover, inthe sense of obtaining a factor of 1/|k| in the estimates, we derive another formula for theinversion,namely,

u = gk ∗ f = (∂ + ik)−1v =1

ik

(v − v + ik(∂ + ik)−1v

)=

1

ik

(v − (∂ + ik)−1((∂ + ik)v − ikv)

)= − 1

4ik

(∂−1 − (∂ + ik)−1∂∂−1

)f.(2.2.11)

Finally, we just need to look over the estimates. For this we use Sobolev interpolation theoremsand Estimate (2.2.3). First, we look at u in Lp. By the first inequality of (2.2.3) and thedefinition of v we obtain

‖u‖Lp ≤c

|k|(‖∂−1f‖Lp + ‖∂∂−1f‖Lp

)16

Further, by the boundedness of the Beurling transform and using (2.2.2) we get

(2.2.12) ‖u‖Lp ≤c1

|k|‖f‖Lp

Now, instead of looking at the normal of each partial derivative of first order, we can justlook at the norm of ∇u in Lp. We use the second estimate of (2.2.3) and once again theBeurling transform, to derive

(2.2.13) ‖∇u‖Lp ≤ c2‖f‖Lp

Joining (2.2.12) and (2.2.13) together and paying attention to the definition of the normof ∇u, we get

‖u‖W 1,p ≤ C(

1

|k|+ 1

)‖f‖Lp

Therefore, using Sobolev interpolation theorem with s ∈ [0, 1], t0 = 1, t1 = 0, and σ = 1−swe get

‖u‖W s,p ≤ ‖u‖σp‖u‖1−σW 1,p

≤ C 1

|k|1−s

(1

|k|+ 1

)s‖f‖Lp .

So for |k| ≥ const., we have 1 + 1/|k| ≤ 1 + 1/const. and we obtain what we aimed for witha C not depending on k,i.e. we have

(2.2.14) ‖w‖W s,p ≤C1

|ζ|1−s‖f‖Lp , |ζ| ≥ const. > 0

At several points in the paper we will recall the following lemma for which we have tointroduce the weighted spaces Lpρ(R2) with norm ‖f‖Lpρ = ‖(1 + | · |2)ρ/2f‖Lp .

Lemma 2.2.3. If f ∈ Lp1(R2) ∩ Lp2(R2) with 1 < p1 < 2 < p2, then the function u = ∂−1fsatisfies lim|z|→∞u(z) = 0,

‖u‖L∞ ≤ cp1,p2(‖f‖Lp1 + ‖f‖Lp2 ), and(2.2.15)

|u(z1)− u(z2)| ≤ cp2 |z1 − z2|1−2/p2‖f‖Lp2 ,(2.2.16)

where cp1 , cp2 are constants depending only on the respective indexes.

Proof. The estimates above are proved in Vekua [2014] [Thm 1.21,pg. 42]. To prove thatlim|z|→∞ u(z) = 0 we use the fact that the space Lp1

1 ∩ Lp21 is dense in Lp1 ∩ Lp2 , given that

Schwartz functions are also a dense subset of the first intersection. So let f ∈ Lp11 ∩L

p21 , then

we get f ∈ L1 and the identity

(2.2.17) ∂−1f(z) =1

z

[− 1

2πi

∫Cf(w)dw ∧ dw + ∂−1(wf(w))(z)

]for z 6= 0

holds. Since wf(w) ∈ Lp1 ∩ Lp2 , then we have that the second term is bounded by (2.2.15)and, therefore. u = O(1/|z|). This gives the desired decay at infinity and the rest follows bya density argument.

17

With this lemma, we are now ready to show the absence of exceptional points in the casewhere our potential is of is of conductivity type.

Lemma 2.2.4. Let q ∈ Lp(R2) be of conductivity type. If h satisfies to (−∆ + q)h = 0 in R2

and he−izk ∈W 1,p(R2) for some k ∈ C then h ≡ 0.

Proof. We can assume, without loss of generality, that h is real-valued since q is real-valued.Moreover, from q being of conductivity type there exists ψ0 ∈ L∞, which generates q underth conditions of Definition 2.1 .We start by defining a new function v = (ψ0e

−ikz∂h − he−ikz∂ψ0) = v1 − v2. Furthermore,we claim that ∇ψ0 ∈W 1,p which follows by

(2.2.18) +∞ > ‖q‖pLp =

∫R2

∣∣∣∆ψ0

ψ0

∣∣∣p dx ≥ 1

||ψ0||pL∞

∫R2

|∇ · ∇ψ0|p dx

and given that (∇×∇)ψ0 = 0 we can get a bound on the mix derivatives, showing the claim.Furthermore, the Sobolev embedding theorem implies that ∇ψ0 ∈ Lp. So this enable us todeduce the space to which v belongs.

We havev1 = ψ0∂he

−ikz = ψ0(∂(he−ikz)− ik(he−ikz)).

By reason of q being of conductivity type ψ0 ∈ L∞ and he−ikz ∈W 1,p it follows that v1 ∈ Lp.To bound v2, we use a property of multiplication operators in Sobolev spaces Valent [1985]

with r = p and taking into account that R2 has the cone property. Since ∂ψ0 ∈ W 1,p, (ψ0

real-valued), and he−ikz ∈ W 1,p we get that their product is also in W 1,p, and its norm isestimated by the product of the norms of both functions in the spaces where we know theyexist. So v2 ∈ W 1,p ⊂ Lp. Further, we have v ∈ Lp and by h and ψ0 being real-valued, wededuce

(2.2.19) (−∆ + q)h = 0⇔ (−4∂∂ +4∂∂ψ0)

ψ0)h = 0⇔ ψ0∂∂h = h∂∂ψ0.

This we use on

∂v =[∂ψ0∂h+ ψ0∂∂h− h∂∂ψ0 − ∂ψ0∂h

]e−ikz

=( ∂ψ0

ψ0

)(ψ0∂he

−ikz)− (∂ψ0

ψ0

)(ψ0∂he

−ikz)=( ∂ψ0

ψ0

)v + he−ikz

∂ψ0∂ψ0

ψ0−(e−k

∂ψ0

ψ0

)(ψ0∂h− h∂ψ0

)e−ikz − he−ikz ∂ψ0∂ψ0

ψ0

= (∂ψ0/ψ0)v −(e−k∂ψ0/ψ0

)v.(2.2.20)

Consequently, v is a generalized analytic function. Moreover, ∂ψ0, ∂ψ0 ∈ Lp∩Lp implies thatA := −(∂ψ0)/ψ0 ∈ Lp∩Lp(R2) and B := (e−k∂ψ0)/ψ0 ∈ Lp∩Lp(R2). Thus we can constructan auxiliary function w = ∂−1((∂ψ0)/ψ0 − ∂ψ0e−kv/(ψ0v)) since the set of zeros of v is ofmeasure zero. Then we are applying ∂−1 to a function in Lp∩Lp and, hence, by Lemma 2.2.3we have that w is bounded and continuous.

Given that w is bounded and continuous, we have that ve−w ∈ Lp. We want to show thatv vanishes, because this will lead to h also vanishing. To do so we start by showing that ve−w

is an entire function. By definition of w, we get

∂(ve−w) = ∂ve−w − ∂w + ve−w = (∂v − ∂v)e−w = 0

18

Thus ve−w is entire and belongs to Lp. By Liouville theorem for Lp functions it followsthat this function is zero everywhere, hence, due to the boundedness of w this implies that vis also 0. Therefore, through the definition of v we obtain:

(ψ0∂h− ∂ψ0h)e−ikz = 0⇔ ψ0∂h = h∂ψ0

which helps us prove that (h/ψ0) is antiholomorphic:

∂(h/ψ0) = (ψ0∂h− h∂ψ0)/(ψ0)2 = 0.

Finally, we have that (h/ψ0)eizk is also antiholomorphic in C and, moreover, it is in Lp, hence,it to must vanish. Thus h ≡ 0.

Now we are ready, to prove Theorem 2.2.1 which is the main result of this section.

Proof of 2.2.1. Lemma 2.4 gives the uniqueness of solutions to Equation (2.1.4). We justneed to prove existence. For this we only need to be able to solve the integral equation, whichwe already derived before,

µ = 1− gk ∗ (qµ)

We define two operators:

Ck : Lp(R2)→W 1,p(R2), defined by Ck(f) = gk ∗ f ;

Mq : W 1,p(R2)→ Lp(R2), defined by Mq(f) = qf.

By Lemma 2.2.2. we have that the convolution operator is bounded between this spaces.Now, we show that the multiplication operator is a compact operator. Let qnn∈N be asequence of smooth compactly supported functions which converges to q in Lp(R2) (possibledue to density of the first functions in the second space).

First, we show that the operator

Mqn : W 1,p(R2)→ Lp(R2)

is a compact, for each n ∈ N.Let ψkk∈N be a bounded sequence in W 1,p(R2). The theorem of Rellich-Kondrachov

states that the embedding W 1,p(R2) → CB(supp(qn)) is compact. Hence, there exists asubsequence ψkiki∈N which converges to a ψ in the norm CB(supp(qn)).

Therefore,

‖Mqnψki −Mqnψ‖Lp(R2) = ‖Mqnψki −Mqnψ‖Lp(supp(qn))

≤ ‖ψki − ψ‖CB(supp(qn))‖qn‖Lp(R2) → 0, as ki → +∞

Consequently, Mqnψkk∈N has a convergent subsequence and thus Mqn is a compactoperator.

To conclude that Mq is compact we show that Mqn converges to it. By Sobolev embeddingtheorem we have W 1,p(R2) → CB(R2), thus we obtain

19

‖Mq −Mqn‖L(W 1,p(R2),Lp(R2)) ≤ sup‖φ‖

W1,p(R2)≤1‖Mqφ−Mqnφ‖Lp(R2)

≤ sup‖φ‖

W1,p(R2)≤1‖φ‖CB(R2)‖qn − q‖Lp(R2)

≤ C sup‖φ‖

W1,p(R2)≤1‖φ‖W 1,p(R2)‖qn − q‖Lp(R2)

≤ C‖qn − q‖Lp(R2) → 0, as n→ +∞.

Hence, Mq is indeed a compact operator between these spaces.Consequently, also CkMq is, and by Fredholm theory, we have that I+CkMq is a Fredholm

operator of index zero on the space W 1,p. Since uniqueness is already shown, then the operatoris in fact invertible.

Now, we take into account the following, purely formal, deduction:

[I + CkMq]µ = 1 and [I + CkMq]1 = 1 + gk ∗ q,subtracting them [I + CkMq](µ− 1) = −gk ∗ q

This allows to consider µ = 1− [I +CkMq]−1(gk ∗ q), and notice it is well-defined by Lemma

2.2.2, because q ∈ Lp(R2).For the desired estimate we use the fractional Sobolev embedding theorem, which in our

specific case says that for fixed s0 ∈ (2/p, 1), we have

W s0,p(R2) → L∞(R2).

Therefore, together with (2.2.10) we have:

(2.2.21) ‖gk ∗ (qf)‖W s0,p ≤ c1|k|s0−1‖q‖Lp‖f‖W s0,p

Further, for |k| sufficiently large, we have

‖CkMqf‖ ≤ 1/2‖f | s0, p.

So by Neumann series, we have

[I + CkMq]−1 =

∞∑l=0

(−1)l(CkMq)l,

with norm less or equal to 2.For |k| sufficiently large it holds

‖µ− 1‖W s0,p = ‖[I + CkMq]−1(gk ∗ q)‖W s0,p ≤ 2‖gk ∗ q‖W s0,p

≤ 2c1|k|s0−1‖q‖Lp ≤ 1(2.2.22)

Finally, using (2.2.8), we obtain

‖µ− 1‖W s0,p ≤ c1|k|s0−1‖q‖Lp‖µ− 1‖W s0,p + ‖gk ∗ q‖W s0,p ≤ c2|k|s0−1‖q‖Lp

20

by applying both (2.2.21) and (2.2.22).Now, we take a look at s1 ∈ [0, 2/p]. We have that

‖µ− 1‖W s1,p = ‖gk ∗ (qµ)‖W s1,p ≤c1

|k|1−s1‖qµ‖Lp

≤ c1

|k|1−s1(‖q(µ− 1)‖Lp + ‖q‖Lp)

≤ c1

|k|1−s1(c2‖q‖Lp‖µ− 1‖W s0,p + ‖q‖Lp)

Thus, by (2.2.22) for sufficiently large |k| it follows

(2.2.23) ||µ− 1||W s1,p ≤c3

|k|1−s1||q||Lp ,

and, therefore, the inequality (2.2.1), follows for all s ∈ [0, 1].

2.3 The ∂ equation

From the previous chapter, we observe that at least for each k ∈ C \ 0 there is a uniquesolution to the Schrodinger equation from which we started. Therefore, there exists a uniquesolution depending on both parameters x and k. Up until now, we only have been worriedabout the behavior with respect to the first parameter. In this section we will take a look atthe behavior of µ in terms of the other parameter and derive an operator, which will simplifyour recovery of q from the measurements at the boundary.

The first step is to use (2.2.8) to obtain a differential equation in terms of ∂∂k

.Straightforward computation leads us to

∂

∂kµ(x, k) = − ∂

∂k(gk ∗ (qµ(·, k))) (x)

= −(∂

∂kgk

)∗ (qµ(·, k)) (x)−

(gk ∗

(q∂

∂kµ(·, k)

))(x)

⇔ [I + CkMq]

(∂

∂kµ(x, k)

)= −

(∂

∂kgk

)∗ (qµ(·, k)) (x)(2.3.1)

We simplify the right-hand side by computing the derivative in a distributional sense:

∂

∂kgk(x) =

1

4π2

∫R2

∂

∂k

(eix·ξ

(ξ1 + iξ2)((ξ1 − iξ2) + 2k)

)dξ1dξ2

=1

4π2

∫R2

(eix·ξ

2(ξ1 + iξ2)

)∂

∂k

(1

(ξ1 − iξ2)/2 + (k1 + ik2)

)dξ1dξ2

=1

4π

∫R2

(eix·ξ

2(ξ1 + iξ2)

)δ(

(ξ1/2 + k1) + i(k2 − ξ2/2))dξ1dξ2

= − 1

4π

∫R2

(ei2x·(ξ1,−ξ2)

(ξ1 − iξ2)

)δ((ξ1 + iξ2) + k) dξ1dξ2

= − 1

4π

e−2i(x1k1−x2k2)

k1 − ik2= − 1

4π

e−k(x)

k(2.3.2)

21

where we use the substitution ξ1 = ξ1/2, ξ2 = ξ2/2.This calculations can be directly substituted on the right-hand side of (2.3.1) and we

obtain

(2.3.3) −(∂

∂kgk

)∗ (qµ(·, k))(x) =

1

4πk

∫R2

e−k(x− y)q(y)µ(y, k) dy

Hence, we can define the scattering transform, sometimes also called non-linear Fouriertransform, by:

Definition 2.2. Let q ∈ Lp be a fixed potential and ψ(x, k) = eizkµ(x, k) the solutions toSchrodinger’s equation (2.1.4). We define the scattering transform as:

(2.3.4) t(k) =

∫R2

ek(y)q(y)µ(y, k) dy.

Remark. If, in fact, the potential is of conductivity type we eill not have exceptionalpoints, i.e., the solutions for each k ∈ C \ 0 will be unique. This implies that the operatort will be well-defined. Further ahead we will look at other properties of this operator.

Using our scattering transform (2.3.3) can be formulated as:

(2.3.5) −(∂

∂kgk

)∗ (qµ(·, k))(x) =

1

4πke−k(x)t(k)

Combining Equations (2.3.1) and (2.3.5) we obtain

(2.3.6)∂

∂kµ(x, k) =

1

4πkt(k)[I + CkMq]

−1(e−k)(x)

One aspect to take into account is that the inverse operator applied to e−k does not makesense analytically since the latter function is not in W 1,p. So we proceed formally for nowand we will look at the proper spaces later on.

The idea now is to show that [I + CkMq](e−kµ) = e−k.We claim that, for some complex-valued function f , we have

(2.3.7) gk ∗ (e−kf) = e−k(gk ∗ f).

Assuming this and not forgetting (2.2.8) then it follows that

[I + CkMq](e−kµ) = e−kµ+ gk ∗ (e−k(qµ))

= e−k[µ+ gk ∗ (qµ)] = e−k[µ+ gk ∗ (qµ)] = e−k,(2.3.8)

and, therefore, from (2.3.6) it follows that we have (at least formally)

(2.3.9)∂

∂kµ(x, k) =

1

4πkt(k)e−k(x)µ(x, k)

Now, the following work on this chapter is all focused on showing that this equation holdsin a certain Sobolev space topology.

22

Before we give the main theorem we present the necessary computations which show thatthe above claim, that (2.3.7) holds for some complex-valued function f , is true. First of all,we have

gk ∗ (e−kf)(x) =

∫R2

gk(x− y)e−k(y)f(y)dy

=

∫R2

∫R2

ei(x1ξ1+x2ξ2)

|ξ|2 + 2k(ξ1 + iξ2)e−i(y1(ξ1+2k1)+y2(ξ2−2k2))f(y) dξ1dξ2dy

We do a linear change of variables ξ1 → ξ1 + 2k1, ξ2 → ξ2 + 2k2, and use

(ξ1 − 2k1)2 + (x2 + 2k2)2 + 2k(ξ1 + iξ2)− 4|k|2 = |ξ|2 − 4(k1ξ1 − k2ξ2) + 4|k|2+

+ 2(k1ξ1 − k2ξ2) + 2i(ξ1k2 + k1ξ2)− 4|k|2

= |ξ|2 − 2k(ξ1 + iξ2)

to obtain

gk ∗ (e−kf)(x) =

∫R2

∫R2

ei(x1(ξ1−2k1)+x2(ξ2+2k2))

|ξ|2 − 2k(ξ1 + iξ2)e−i(y1ξ1+y2ξ2)f(y) dξ1dξ2dy

= e−k(x)

∫R2

(∫R2

ei(x−y)·ξ

|ξ|2 − 2k(ξ1 + iξ2)dξ

)f(y) dy.

By making the substitution ξ → −ξ, the claim (2.3.7) follows.

Theorem 2.3.1. Let q be real-valued and in Lpρ(R2), 1 < p < 2, ρ > 1. Then for any

k ∈ C \ 0 which is not an exceptional point, equation (2.3.9) holds in the W 1,p−β topology, for

β > 2/p.

It will be helpful to use the notation (similar to Fourier transform):

Ff(k) = i/2

∫Cek(z)f(z)dz ∧ dz.(2.3.10)

Lemma 2.3.1. Let α > 2/p′, β > 2/p (where p′ denotes the exponent dual to p). The mapk 7→ (∂ + ik)−1 is differentiable on C in the strong operator topology: Lpα → Lp−β and

(2.3.11)∂

∂k(∂ + ik)−1f(z) = − i

πFf(k)e−k(z)

Proof. We start by noticing that the fundamental solution to (∂+ ik) is e−k(z)/z. Therefore,we can define the inverse operator by convolution:

(∂ + ik)−1f(z) =1

2πi

∫C

ek(w − z)w − z

f(w) dw ∧ dw.

We will write k = k1 + ik2 and define Dj(k), j = 1, 2, to be the operators

(2.3.12) Dj(k)f(z) =(−1)j−1

π

∫C

yj − xjw − z

ek(w − z)f(w) dw ∧ dw, w = y1 + iy2.

23

A first step is to show that ∂∂kj

(∂ + ik)−1 = Dj(k). Clearly, ‖Dj(k)f‖L∞ ≤ 2π‖f‖L1 since∣∣∣yj−xjw−z

∣∣∣ ≤ 1 and |ek(w − z)| = 1. Furthermore, via Proposition B.2.1 we have for α > 2/p′

that Lpα ⊂ L1 and, for β > 2/p, we have that 〈x〉−β ∈ Lp. Hence, using the representation ofinverse operators via convolution with fundamental solutions we get

limh→0

∥∥∥∥((∂ + i((k1 + h) + ik2))−1 − (∂ + ik)−1

h−D1(k)

)f

∥∥∥∥pLp−β

= limh→0

∫z∈R2

∣∣∣∣∣〈z〉−β2πi

∫ek(w − z)w − z

(eh(w − z)− 1

h− 2i(y1 − x1)

)f(w) dw ∧ dw

∣∣∣∣∣p

.

Moreover, we can apply the dominated convergence theorem twice, and use the limit

limh→0

eh(w − z)− 1

h= 2i(y1 − x1),

we get that the limit above is equal to zero. Analogously, we can apply the same process toshow ∂

∂k2(∂ + ik)−1 = D2(k).

Joining both together, we get

∂

∂k(∂ + ik)−1f =

1

2(D1(k) + iD2(k))f(z)

=1

2π

∫Cek(w − z)f(w) dw ∧ dw

= − iπe−k(z)

i

2

∫Cek(w)f(w) dw ∧ dw

= − iπe−k(z)Ff(k)(2.3.13)

Remark. Further ahead we will need to differentiate gk ∗ f . While the previous proofsappear to require f ∈ L1, the function ∂∂−1f (see formula (2.2.11) need not be in L1 evenwhen f has compact support. The idea is to work in a smaller space Lpα for which we showthat any function in ∂∂−1Lpα can be written as the sum of an L1 function and another whichis less decaying but has a derivative in Lpα. This will enable us to apply (∂+ik)−1 to the latter.

Lemma 2.3.2. If 2/p′ < α < 1 and α+ 1− 2/p < δ < 2/p′, then

(2.3.14) ∂∂−1Lpα ⊂ Lpα + u ∈ Lpδ : ∂u ∈ Lpα

Proof of Lemma 3.2. Let f ∈ Lpα. We first prove the identity

(2.3.15) ∂∂−1f = 〈z〉−α ∂∂−1 〈z〉α f − α

2〈z〉−2 z∂−1f +

α

2〈z〉−α ∂∂−1 〈z〉α−2 z∂−1f

To do so we take into consideration that ∂ 〈z〉α = α2 〈z〉

α−2 z, and start with

(2.3.16) ∂ 〈z〉α ∂−1f =α

2〈z〉α−2 z∂−1f + 〈z〉α f

24

Given that δ < α by definition of the weighted space we also have f ∈ Lpδ and so ∂−1f ∈ Lpδ−1.The right hand side of (2.3.16) is in Lpδ−α, since for the first term we have

∂−1f ∈ Lpδ−1 ⇒ 〈z〉δ−1 ∂−1f ∈ Lp ⇒ 〈z〉δ−α 〈z〉α−1 ∂−1f ∈ Lp

⇒ 〈z〉α−2 z∂−1f ∈ Lpδ−α,

and for the second term it follows, similarly,

f ∈ Lpδ ⇒ 〈z〉α 〈z〉δ−α f ∈ Lp ⇒ 〈z〉α f ∈ Lpδ−α.

Furthermore, we have 1−2/p < δ−α < 2/p′ so we can apply ∂∂−1 to (2.3.16) and deduce

∂∂−1(∂ 〈z〉α ∂−1f

)=α

2∂∂−1 〈z〉α−2 z∂−1f + ∂∂−1 〈z〉α f

⇔ α

2〈z〉α−2 z + 〈z〉α ∂∂−1f =

α

2∂∂−1 〈z〉α−2 z∂−1f + ∂∂−1 〈z〉α f.

Dividing both sides by 〈z〉α gives the expected equation (2.3.15).The first term on the right hand side of (2.3.15) is in Lpα because we can use Lemma B.3.1

in the particular case where δ = 0. The other two terms were in Lpδ−α before the division by〈z〉α so by definition they are now in Lpδ . To finish the proof we only need to show that their∂-derivatives are in Lpα.

We start by looking at what happens with the second term in (2.3.15) (minus the con-stants):

(2.3.17) ∂(〈z〉−2 z∂−1f

)= −〈z〉−4 z2∂−1f + 〈z〉−2 z∂∂−1f

Since f ∈ Lpδ , by Lemma B.3.1 we have the right-hand side in Lpδ+1, and therefore, Lpδ+1 ⊂ Lpα,

because α < α+ 2/p′ = α+ 2− 2/p < δ + 1.For the remaining ter, we start by computing

∂(〈z〉−α ∂∂−1 〈z〉α−2 z∂−1f

)(2.3.18)

= −α2〈z〉−α−2 z∂∂−1 〈z〉α−2 z∂−1f +

α− 2

2〈z〉−α ∂∂−1 〈z〉α−4 |z|2∂−1f

+ 〈z〉−α ∂∂−1 〈z〉α−2 ∂−1f + 〈z〉−α ∂∂−1 〈z〉α−2 z∂∂−1f

=: a+ b+ c+ d(2.3.19)

where we are using ∂∂−1 = ∂−1∂, which holds in these weighted spaces. Now, let us lookat the four terms above:

Case a: Similar as before we have 〈z〉α−2 z∂−1f is in Lpδ−α and ∂∂−1 is bounded onLpδ−α, so that a ∈ Lpδ+1 ⊂ L

pα

Case b: 〈z〉α−4 ∂−1f ∈ Lpδ−α+3 ⇒ 〈z〉α−4 |z|2∂−1f ∈ Lpδ−α+1. We can now choose δ′ ∈

(α, α + 2/p′). Then, we have δ′ < δ + 1, so that 〈z〉α−4 |z|2∂−1f ∈ Lpδ′−α.Furthermore, 0 < δ′ − α < 2/p′ so that we can apply ∂∂−1 and we get thatb ∈ Lpδ′ ⊂ L

pα.

25

Case c: 〈z〉α−2 ∂−1f ∈ Lpδ−α+1 ⊂ Lpδ′−α with δ′ defined as above. Therefore, we have

∂∂−1(〈z〉α−2 ∂−1f) ∈ Lpδ′−α, this implies c ∈ Lpδ′ ⊂ Lpα.

Case d: Finally, from ∂∂−1 ∈ Lpδ , we have 〈z〉α−2 z∂∂−1f ∈ Lpδ−α+1 ⊂ Lpδ′−α. Hence,d ∈ Lpα.

Therefore, the derivative ∂ is in Lpα and the result follows.

Lemma 2.3.3. Let 2/p′ < α < 1, β > 2/p. The map k 7→ gk ∗ · is differentiable on C \ 0in the strong operator topology: Lpα →W 1,p

−β and

(2.3.20)∂

∂k(gk ∗ f) = − 1

4πkFf(k)e−k

Proof. Let f ∈ Lpα then by Lemma 2.3.2 we can write ∂∂−1f = f1 + f2 with f1 ∈ Lpα, f2 ∈ Lpδ ,for some δ ∈ (α+ 1− 2/p, 2/p′) and ∂f2 ∈ Lpα. Then we get

(∂ + ik)−1(∂∂−1f) = (∂ + ik)−1f1 +ik

ik(∂ + ik)−1f2 +

1

ikf2 −

1

ik(∂ + ik)−1(∂ + ik)f2

= (∂ + ik)−1f1 +1

ikf2 −

1

ik(∂ + ik)−1∂f2(2.3.21)

(small remark: f2 ∈ W 1,p ⊂ Lp). Therefore, by Lemma 2.3.1., we have that the right-handside of (2.3.21) is differentiable in Lp−β for k 6= 0 and

(2.3.22)∂

∂k(∂ + ik)−1∂∂−1f = − i

πe−kF(∂∂−1f)(k) = − i

π

k

ke−kFf(k),

where the last equality follows using the Fourier multiplier symbol of ∂∂−1. Now rememberingformula (2.2.11),

gk ∗ f = − 1

4ik

[∂−1f − (∂ + ik)−1∂∂−1f

],

we notice that the first term on the right-hand side, does not depend on k and for the secondwe can use the above equation (2.3.21). With this, the differentiability of gk ∗f in Lp−βfollows

as well as the desired formula (2.3.20). To prove the differentiability in W 1,p−β by (2.2.11) and

Lemma 2.3.1 we also observe that

∂(gk ∗ f) = − 1

ik∂[∂ − (∂ + ik)−1∂∂−1

]f = − f

4ik+

∂

4ik

[e−k∂

−1(ek∂∂−1)]f

= − f

4ik+

∂

4ik

[e−k∂

−1(∂(ek∂−1f

)− ikek∂−1f

)]= − ∂

4(∂ + ik)−1∂−1f = −(∂ + ik)−1f/4,

so we can use Lemma 3.1. again. For the derivative with respect to z we use the aboveequation to get the identity

∂(gk ∗ f) = (−ik)(gk ∗ f) + (∂ + ik)(gk ∗ f)

= (−ik)(gk ∗ f) + (∂ + ik)∂−1(∂gk ∗ f)

= (−ik)(gk ∗ f)− ∂−1f/4.(2.3.23)

With this only the first term depends on k and we already know that this term holds inLp−β.

26

Now we are finally ready to show the main theorem of this section.

Proof of Theorem 3.1. We may start by assuming some restrictions on the intervals of thecoefficients, i.e., we consider only 2/p < β < ρ− 2/p′, which is not a problem since the norm|| · ||

W 1,p−β

is decreasing in β. Furthermore, we choose α such that 2/p′ < α < min(1, ρ− β).

We rewrite Equation (2.2.8), to take into account the desired growth

〈x〉−β µ = 〈x〉−β −K(k)(〈x〉−β µ),

with the operator K(k)f = 〈x〉−β gk ∗ (〈·〉 qf) .

By our choice of α and β, we have

(2.3.24) q ∈ Lpρ ⇒ 〈x〉β ∈ Lpρ−β ⇒ 〈x〉

β ∈ Lpα,

where the last implication follows because of α < min(1, ρ−β) < ρ−β. Therefore, analogouslyas before, we have that multiplication by 〈x〉β q is a compact operator: W 1,p → Lpα. Fromthis and Lemma 2.3.3 it follows that k → K(k) is differentiable on C \ 0 in the uniformoperator topology W 1,p →W 1,p.

Let us suppose that h is a solution in W 1,p of h = −K(k)h. Then 〈x〉β h satisfies〈x〉β h = −gk ∗ (q 〈·〉β h) so that 〈x〉β h is in W 1,p and, hence, vanishes since k is assumed tobe a non-exceptional point. Thus I +K(k) is invertible on W 1,p and

∂

∂k

(〈x〉−β µ(x, k)

)= −

(∂

∂kK(k)

)(〈x〉−β µ(x, k)

)−K(k)

(〈x〉−β ∂

∂kµ(x, k)

)(2.3.25)

⇔ (I +K(k))∂

∂k

(〈x〉−β µ(x, k)

)= −

(∂

∂kK(k)

)(〈x〉−β µ(x, k)

)(2.3.26)

∂

∂k

(〈x〉−β µ(x, k)

)= −(I +K(k))−1

(∂

∂kK(k)

)(〈x〉−β µ(x, k)

)(2.3.27)

By Lemma 2.3.3, we have that ∂∂kK(k) is an operator of rank 1 and, therefore, we only

need to compute

(I +K(k))−1(e−k(x) 〈x〉−β

).

We use (2.3.7) and given that 〈x〉 is real-valued we get

(2.3.28) (I +K(k))−1(e−k(x) 〈x〉−β

)= e−k(I +K(k))−1 〈x〉−β

Thus (2.3.9) holds in the W 1,p−β .

Observation: The importance of understanding the behavior in k of the solutions µ(x, k)comes from the fact that at the limit k → 0 we have

(2.3.29) (−∆ + q)µ(·, 0) = 0⇒(−∆ +

∆√γ

γ

)µ(·, 0)⇒ µ(z, 0) =

√γ(z)

27

2.4 Behavior near k = 0

In this section the assumptions on the potential q are more general, i.e., the results willapply not only to a potential of conductivity type.

Lemma 2.4.1. Let q be a real-valued function in Lpρ(R2) , 1 < p < 2, ρ > 1. If 〈x〉−a ψ0 ∈L∞(R2) for some a < min(1, ρ− 1) and ∆ψ0 = qψ0 (as distributions), then ψ0 is continuous,there is a constant c∞ such that ψ0 satisfies

(2.4.1) ψ0 = c∞ −G0 ∗ (qψ0),

with G0(x) = − 12π log |x|, and the following are equivalent:

(a) ψ0 ∈ L∞(R2);

(b) limR→∞

1

πR2

∫|x|<R

ψ0(x)dx exists (and equals c∞);

(c)∫R2 qψ0 = 0;

(d) ψ0 − c∞ ∈W 1,p(R2);

(e) ∇ψ0 ∈ Lp1(R2);

(f) ∇ψ0 ∈ Lp(R2).

To prove the above lemma, we will use an estimate on the fundamental solution of theLaplacian G0.

Lemma 2.4.2. If f ∈ Lpρ(R2), 1 < p < 2, ρ > 1, then:

(i) ‖G0 ∗ f +1

2π(log |x|)

∫R2

f‖Lp ≤ c‖f‖Lpρ ;

(ii) ‖∇G0 ∗ f‖Lp ≤ c‖f‖Lp .

Proof. To the first inequality, we will use some properties of the logarithm, to split one of theintegrals into two parts.∥∥∥∥G0 ∗ f +

1

2π(log |x|)

∫R2

f

∥∥∥∥Lp

=

[∫R2

∣∣∣∣∫R2

1

2π(log |x| − ln |x− y|)f(y) dy

∣∣∣∣p dx]1/p

(2.4.2)

≤ 1

2π

∫R2

[∫R2

|ln |x+ y| − ln|x||p dx]1/p

|f(y)| dy.

Here we applied Minkowski integral inequality and made a change of variable on x.Furthermore, by defining

I1(y) :=

[∫|x|≤|y|

∣∣∣∣ln |x+ y||x|

∣∣∣∣p dx]1/p

and,

I2(y) :=

[∫|x|>|y|

∣∣∣∣ln |x+ y||x|

∣∣∣∣p dx]1/p

,

28

we can treat the singularities at 0 and infinity separately.In the first integral since we are working with a bounded domain even though it depends

on |y| we can estimate I1 by taking a limit to 0 and Sobolev Inequality:

I1(y) ≤ limε→0

[∫ε≤|x|≤|y|

∣∣∣∣∇ ln

(1 +|y||x|

)∣∣∣∣p dx]1/p

= limε→0

[∫ε≤|x|≤|y|

∣∣∣∣ |y||x|(|y|+ |x|)

∣∣∣∣p dx]1/p

= limε→0

[∫ 2π

0

∫ |y|ε

∣∣∣∣ |y|r(|y|+ r)

∣∣∣∣p r dr dθ]1/p

≤ limε→0

(1

2π

)1/p[∫

ε≤|x|≤|y|r1−p dr

]1/p

= limε→0

(1

2π

)1/p [ r2−p

2− p

]1/p∣∣∣∣∣r=|y|

r=ε

= Cp|y|(2−p)/p = Cp|y|2/p.

The convergence of the integral and the limit follows from p ∈ (1, 2) and the Sobolev conju-gate gives the last equality.

For the second integral, we work with the series approximation of the logarithm since thedomain of integration gives |y|/|x| < 1, i.e., log(1 + |y|/|x|) ≤ |y|/|x|. Hence,

I2(y) ≤

[∫|x|>|y|

|y|p

|x|pdx

]1/p

= |y|

[∫ 2π

0

∫ ∞|y|

r1−p dr dθ

]1/p

= Cp|y|1+(2−p)/p = Cp|y|2/p

Therefore, substituting in (2.4.2) we obtain:

‖G0 ∗ f +1

2π(log |x|)

∫R2

f‖Lp ≤∫R2

(Cp + Cp

)|y|2/p|f(y)| dy

≤ cp

∥∥∥∥∥ |y|2/p〈y〉ρ

∥∥∥∥∥Lp′‖f‖Lpρ

The first norm converges given that p′/p = p′/2− 1 implies∣∣∣∣∣ |y|2/p〈y〉ρ

∣∣∣∣∣p′

≤ 1

(1 + |y|2)

〈y〉p′

〈y〉ρp′=

1

〈y〉21

〈y〉(ρ−1)p′=

1

〈y〉2ρ,

which is in L1(R2).This concludes the proof of inequality (i). The second equality follows by looking at

the gradient of the fundamental solution of the laplacian and the use of Hardy-LittlewoodInequality on fractional integration [Stein, 1970].

This lemma will allow us to demonstrate that a certain constructed harmonic function isconstant.

Proof of Lemma 2.4.1. In order to obtain Formula (2.4.1) we define the functionh = ψ0 + G0 ∗ (qψ0). By hypothesis it follows that, in distribution sense, we have

29

∆h = ∆ψ0 + ∆(G0 ∗ qψ0) = ∆ψ0− qψ0 = 0, hence, h is harmonic. Moreover, qψ0 ∈ Lpρ−a andρ− a > 1, so by Lemma 2.4.2 it follows

h(x) = − 1

2π

(∫R2

qψ0

)log |x|+ ψ0(x) +

((G0 ∗ (qψ0)) (x) +

1

2π

(∫R2

qψ0

)log |x|

)(2.4.3)

= − 1

2π

(∫R2

qψ0

)log |x| mod (Lp + L∞−a).

To obtain that h ≡ c∞, we will be using the mean value theorem over disks and Liouville’stheorem for harmonic functions. Let y ∈ R2, with |y| > 1. Since the mean value theoremholds for any disk of radius R we fix R = |y|2. Then we obtain

|h(y)− h(0)| = 1

4πR

∣∣∣∣∣− 1

2π

(∫R2

qψ0

)∫|x|=R

(log |x+ y| − log |x|) dx +

+

(∫|x|=R

ψ0(x+ y)− ψ0(x) dx

)+

(∫|x−y|=R

φ(x) dx−∫|x|=R

φ(x) dx

)∣∣∣∣∣,where φ is the last term on the left-hand side of (2.4.3). This term is of order O(R1/p) whichfollows by Lemma 2.4.2 and the integral of the indicator functions over the surface of thesphere of radius R. Hence, we obtain

|h(y)− h(0)| ≤ 1

4πR

1

2π

∣∣∣∣∣(∫

R2

qψ0

)∫|x|=R

log|x+ y||x|

dx

∣∣∣∣∣ +

+1

4πR

∣∣∣∣∣∫|x|=R

ψ0(x+ y)− ψ0(x) dx

∣∣∣∣∣+O(R−1/p),

For the first integral we obtain∫|x|=R

log|x+ y||x|

dx ≤∫|x|=R

log(1 + |y|/|x|) dx ≤ |y|R

2π =1

R1/2

and because qψ0 ∈ Lpρ−a and ρ − a > 1, it follows by Holder inequality that qψ0 ∈ L1.Substituting we obtain

|h(y)− h(0)| = O(R−3/2 +R−1/p

)+

1

4πR

∣∣∣∣∣∫|x|=R

ψ0(x+ y)− ψ0(x) dx

∣∣∣∣∣≤ O

(R−1/p

)+‖ψ0‖L∞−a

4πR

∣∣∣∣∣∫|x|=R

((1 + |x+ y|2

)a/2 − (1 + |x|2)a/2)

dx

∣∣∣∣∣≤ O

(R−1/p

)+‖ψ0‖L∞−a

4πR

∣∣∣∣∣∫|x|=R

((1 + |x|+ |y|2

)a/2 − (1 + |x|2)a/2)

dx

∣∣∣∣∣= O

(R−1/p

)+‖ψ0‖L∞−a

4πR

∣∣∣∣∣∫|x|=R

(1 + |x|2

)a/2((1 +

|y|2

1 + |x|2

)a/2− 1

)dx

∣∣∣∣∣≤ O

(R−1/p

)+‖ψ0‖L∞−a

4πR

(a/2)(1 +R2)a/2|y|2

(1 +R2)

30

= O(R−1/p

)+‖ψ0‖L∞−a

2R

(a/2)R(1 +R2)a/2

(1 +R2)

= O(R−1/p +R−(2−a)

)= O

(R−1/p

),

where for to evaluation of the last integral we used the binomial series. Therefore, since foreach y we can always pick an R big enough and we have that R−1/p < 1 there exists a constantC > 0 such that

|h(y)− h(0)| ≤ C, ∀y ∈ R2 such that |y| > 1.

Hence by continuity of harmonic functions we have that h is a bounded harmonic functionwhich by Liouville’s theorem implies that there exists a constant c∞ > 0 with h ≡ c∞ and theFormula (2.4.1) follows. The second statement of Lemma 2.4.2 gives∇ψ0 ∈ Lp. Consequently,by the inclusion of Lp spaces it follows that ψ0 ∈ W 1,p

loc (R2). Hence, by Sobolev embeddingtheorem it follows that ψ0 is continuous.

To prove the equivalence of the statements we integrate h on a disk of radius R to have∫|x|<R

h(x) dx =

∫|x|<R

ψ0(x) dx− 1

2π

(∫R2

qψ0

)∫|x|<R

log |x| dx±∫|x|<R

|φ(x)| dx

c∞πR2 ≤

∫|x|<R

ψ0(x) dx− 1

2π

(∫R2

qψ0

)[∫ 2π

0

∫ R

0r log r dr dθ

]+ ‖φ‖Lp

(∫|x|<R

1 dx

)1/p′

c∞πR2 =

∫|x|<R

ψ0(x) dx−(

1

2R2 logR− 1

4R2

)∫R2

qψ0 +O(R2/p′

),

where the last equation follows by integration by parts on the logarithmic integral and thevolume of the ball of radius R.

Given that the formula holds for all R > 0, we can take the limit to have

c∞ = limR→∞

[1

πR2

∫|x|<R

ψ0(x) dx−(

1

2πlogR− 1

4π

)∫R2

qψ0 +O(R−2/p

)]

= limR→∞

[1

πR2

∫|x|<R

ψ0(x) dx−(

1

2πlogR− 1

4π

)∫R2

qψ0

]

This allows to prove now the equivalences.Equivalences:

[(b)⇔(c)]: By the properties of the limits we immediately obtain the equivalencebetween (b) and (c) from the above identity;

[(a)⇒(b),(c)]: If we have (a), i.e., ψ0 ∈ L∞(R2), then (b) follows by straightforwardcomputation of the absolute limit and, hence, also (c);

[(c)⇒(d)⇒(a)⇒(c)]: If (c) holds then ψ0 − c∞ = −G0 ∗ (qψ0) = −G0 ∗ (qψ0) −1

2π (log |x|)∫R2 qψ0 ∈ Lp(R2) via Lemma 2.4.2 (i). By part (ii) it follows ψ0 − c∞ ∈

W 1,p(R2), hence (d). By Sobolev embedding theorem we obtain (a). Up to now weshowed that the first four statements are equivalent;

31

[(c)⇒(e)]: By ∆ψ0 = qψ0 ⇒ ∂(∂ψ0) = qψ0/4 and ∂ψ0 ∈ Lp. If (c) holds then we have

∂ψ0(z) = ∂−1(∂∂ψ0

)(z) = ∂−1(qψ0(z))/4 =

1

8πi

∫C

q(w)ψ0(w)

w − zdw ∧ dw

=1

8πi

∫C

(z − w + w)q(w)ψ0(w)

z(w − z)dw ∧ dw

=1

8πi

∫Cq(w)ψ0(w) dw ∧ dw +

1

z8πi

∫C

wq(w)ψ0(w)

w − zdw ∧ dw

=∂−1(wq(w)ψ0(w))(z)

4z.

Analogously, we can obtain the a similar formula for ∂ψ0. By Lemma 2.2.1 and thefact that qψ0 ∈ Lp1 it follows that ∂ψ0, ∂ψ0 ∈ Lp1 and, hence, the statement (e) follows

∇ψ0 ∈ Lp1;

[(c)⇒(f)]: To show this implication we use Lemma B.3.1. We have wqψ0 ∈ Lp. There-fore, ∂−1(wqψ0) ∈ Lp−1. By the above proof we obtain that ∂ψ0, ∂ψ0 ∈ Lp and (f)follows;

[(e)⇒(c)]: Since (e) holds we pick p1 with p < p1 < 2 and let p2 = (1/p1− 1/p)−1 > 2.Hence, by Holder inequality ‖∇ψ0‖Lp1 ≤ ‖ 〈x〉∇ψ0‖Lp‖ 〈x〉

−1 ‖Lp2 < ∞. It followsthat the Fourier transform (∂ψ0)∧(ξ) is in Lp

′1 . Furthermore, qψ0 is in L1. Hence,

it’s Fourier transform is continuous. From (∂ψ0)∧(ξ) = −i(qψ0)∧(ξ)/(2(ξ1 + iξ2)), andp′1 > 2 it follows that in a ball around zero it does not have a singularity of order 1.Therefore, we deduce (qψ0)∧(0) = 0;

[(f)⇒(c)]: We can use the argument of the previous implication to show the finalimplication.

All the implications were shown, consequently the result follows.

The main result of this section is the following:

Theorem 2.4.1. Suppose q ∈Lpρ(R2), with 1 < p < 2 and ρ > 1, is of the form q = (∆ψ0)/ψ0

with ψ0 ∈ L∞(R2) a real-valued function and ψ0(x) ≥ c0 > 0. Then

|t(k)| ≤ c|k|ε(2.4.4)

for k close to zero, and (with c∞ as in Lemma 2.4.1)

‖µ(·, k)− ψ0/c∞‖W 1,p−β≤ c|k|ε(2.4.5)

for all ε, β satisfying

0 < ε < min((ρ− 1)/2, 2/p′) and β > 2/p+ ε.(2.4.6)

For the proof of this theorem we will use the well known-device of separating a rank 1piece from the operator of convolution by gk which is responsible for the log k singularity at kequal zero. In this sense we define the modified Faddeev Green function gk(x) = gk(x) + `(k),with `(k) given by

`(k) = (log |k|+ γ0)/2π for |k| ≤ 1 and `(k) = γ0/2π for |k| > 1(2.4.7)

where γ0 denotes not a conductivity but the Euler constant!

32

Lemma 2.4.3. For any ε, 0 < ε < 1. there is a constant Cε such that for all 0 < |k| ≤ 1/2there is an inequality |gk(x)−G0(x)| ≤ Cε|k|ε 〈x〉ε.

To prove this inequality we will use the following identities:

Proposition 2.4.1. For any non-zero x ∈ R2, we have

1

2π

[∫|ξ|<2/|x|

eix·ξ − 1

|ξ|2dξ +

∫|ξ|>2/|x|

eix·ξ

|ξ|2

]= −γ0.(2.4.8)

For any k ∈ C \ 0 a contour integration yields

1

2π

∫|ξ|<2

1

|ξ|2 + 2k(ξ1 + iξ2)dξ = − log |k| if |k| ≤ 1, and = 0 if |k| > 1.(2.4.9)

For any non-zero x ∈ R2 we have

G0(x) =1

4π2

∫2<|ξ|<2/|x|

1

|ξ|2dξ.(2.4.10)

Proof. To determine an integral formula for the Euler-Mascheroni constant let x ∈ R2 benon-zero. Hence, we have by the change to the complex plane and then by the change ofvariables: ξ′ = xξ, ξ′ = xξ′; we obtain

I1 :=1

2π

(∫|ξ|<2/|x|

eix·ξ − 1

|ξ|2dξ +

∫|ξ|>2/|x|

eix·ξ

|ξ|2dξ

)

=i

4π

(∫|ξ|<2/|x|

ei12

(xξ+ξx) − 1

|ξ|2dξ ∧ dξ +

∫|ξ|>2/|x|

ei12

(xξ+ξx)

|ξ|2dξ ∧ dξ

)

=i

4π

(∫|ξ|<2

ei12

(ξ+ξ) − 1

|ξ|2dξ ∧ dξ +

∫|ξ|>2

ei12

(ξ+ξ)

|ξ|2dξ ∧ dξ

)

=1

2π

(∫|ξ|<2

eiξ1 − 1

|ξ|2dξ +

∫|ξ|>2

eiξ1

|ξ|2dξ

)

=1

2π

(∫ 2

0

∫ 2π

0

eir cos(θ) − 1

rdθdr +

∫ +∞

2

∫ 2π

0

eir cos(θ)

rdθdr

)

=

∫ 2

0

J0(r)− 1

rdr +

∫ +∞

2

J0(r)

rdr.

We estimate the first integral by using the series representation of the Bessel function J0,

J0(r) =

∞∑m=0

(−1)m

(m!)2

(r2

)2m,

and the second integral equals −Ji0(2) (where Ji0 denotes the integral Bessel function oforder zero) which is defined as

Ji0(2) = γ0 +∞∑m=1

(−1)m

2m(m!)2.

33

Thus follows the identity

I1 =

∫ 2

0

+∞∑m=1

(−1)m

(m!)2

(1

2

)2m

r2m−1 dr − Ji0(2)

=+∞∑m=1

(−1)m

(m!)2

(1

2

)2m 22m

2m− γ0 −

∞∑m=1

(−1)m

2m(m!)2= −γ0.

For the second identity the result follows by the following changes of coordinates: first wetake the k from the denominator, by scaling, then we apply a polar coordinates change andthen from substituting z = eiθ, which implies dz

iz = dθ.Then we can easily use the residuetheorem to obtain the final result.

I2 : =1

2π

∫|ξ|<2

1

|ξ|2 + 2k(ξ1 + iξ2)dξ =

1

2π

∫|ξ|<1/|k|

1

|ξ|2 + (ξ1 + iξ2)dξ

=1

2π

∫ 1/|k|

0

∫ 2π

0

1

r + eiθdrdθ =

1

2πi

∫ 1/|k|

0

∫|z|=1

1

r(r + z)dzdr

Now if |k| > 1 then we have that r ∈ (0, 1/|k|) will always be a pole for f(r, z) = 1z(r+z) .

I2 =

∫ 1/|k|

0

(1

2πi

∫|z|=1

f(r, z) dz

)dr =

∫ 1/|k|

0(Res(f(r, z),−r)) dr

=

∫ 1/|k|

0

(limz→0

f(r, z)z + limz→−r

f(r, z)(z + r)

)dr

=

∫ 1/|k|

0

(1

r− 1

r

)dr = 0

The computation for 0 < |k| ≤ 1 has a small difference because now for r ∈ (0, 1/|k|) thefunction g(r, z) = 1

z(r+z) can have two poles or one depending on the value of r. In this caseit follows

I2 = limε→0

[∫ 1−ε

0

(1

2πi

∫|z|=1

g(r, z) dz

)dr +

∫ 1/|k|

1+ε

(1

2πi

∫|z|=1

g(r, r) dz

)dr

]The first integration can be treated as in case of |k| > 1 and thus equals 0. In the second

integration only 0 will be a pole and the residue theorem gives the result:

I2 =

∫ 1/|k|

1

(1

2πi

∫|z|=1

g(z) dz

)dr =

∫ 1/|k|

1Res (g(r, z), 0) dr =

∫ 1/|k|

1

1

rdr

= ln(r)|1/|k|1 = ln(1/|k|) = − ln |k|.

Thus the second identity follows.Now for the third statement a simple computation is all that we need:

1

4π2

∫2<|ξ|<2/|x|

1

|ξ|2dξ =

1

2π

∫ 2/|x|

2

1

rdr =

1

2πln(r)|2/|x|2 = − 1

2πln |x| = G0(x)

34

Thus we are ready to use the identities to prove the desired lemma.

Proof of Lemma 2.4.3. By the identities above we compute the following formula for thedifference of the fundamental solutions

gk(x)−G0(x) = gk(x) + `(k)−G0(x)k(x) +1

2πln |k|+ 1

2πγ0 −G0(x)

=1

4π2

(∫R2

eix·ξ

|ξ2|+ 2k(ξ1 + iξ2)dξ −

∫|ξ|<2

1

|ξ2|+ 2k(ξ1 + iξ2)dξ −

∫|ξ|<2/|x|

eix·ξ

|ξ|2dξ

−∫|ξ|>2/|x|

eix·ξ

|ξ|2dξ −

∫2<|ξ|<2/|x|

1

|ξ|2dξ

)

=1

4π2

(∫|ξ|<2

eix·ξ − 1

|ξ2|+ 2k(ξ1 + iξ2)dξ +

∫|ξ|>2

eix·ξ

|ξ2|+ 2k(ξ1 + iξ2)dξ −

∫|ξ|<2

eix·ξ − 1

|ξ|2dξ

−∫

2<|ξ|<2/|x|

eix·ξ

|ξ|2dξ −

∫|ξ|>2/|x|

eix·ξ

|ξ|2dξ

)

=1

4π2

(∫|ξ|<2

(eix·ξ − 1)

[1

|ξ2|+ 2k(ξ1 + iξ2)− 1

|ξ|2

]dξ

+

∫|ξ|>2

eix·ξ[

1

|ξ|2 + 2k(ξ1 + iξ2)− 1

|ξ|2

]dξ

)

To obtain the desired inequality, we look at two cases, when |x| > 1/2 and when |x| < 1/2.Hence, in the following we consider that |x| > 1/2. Under this condition we can bound

(eix·ξ − 1) by |x||ξ| for |ξ| < 1/|x| through series expansion and by 2 otherwise. Keeping thisin mind it follows:

|gk(x)−G0(x)| = 1

4π2

∣∣∣∣∣∫|ξ|<2

−2k(eix·ξ − 1)

|ξ|2(ξ1 − iξ2 + 2k)dξ +

∫|ξ|>2

−2keix·ξ

|ξ|2(ξ1 − iξ2 + 2k)dξ

∣∣∣∣∣≤ 1

2π2

(∫|ξ|<1/|x|

|x||k||ξ| ||ξ|2 + 2k(ξ1 + iξ2)|

dξ+

+

∫1/|x|<|ξ|<2

2|k||ξ|2|ξ − iξ2 + 2k|

dξ +

∫|ξ|>2

|k||ξ|2|ξ − iξ2 + 2k|

dξ

)

≤ 1

2π2

(∫|ξ|<1/|x|

|x||k||ξ| ||ξ|2 + 2k(ξ1 + iξ2)|

dξ +

∫|ξ|>1/|x|

2|k||ξ|2|ξ − iξ2 + 2k|

dξ

)

Now by changing the integration to the complex plane and making the change of variablesξ = kξ′, ξ = kξ′, and keeping in the end ξ notation, it follows for ε ∈ (0, 1)

|gk(x)−G0(x)| ≤ 1

4π2

(∫|ξ|< 1

|k||x|

|x||k||ξ||ξ + 2| dξ ∧ dξ

+ 2

∫|ξ|> 1

|x||k|

1

|ξ|2|ξ + 2|dξ ∧ dξ

)

35

=1

4π2

(|x||k|

∫|ξ|< 1

|k||x|

|ξ|1−ε

|ξ|2−ε|ξ + 2|dξ ∧ dξ +

∫|ξ|> 1

|k||x|

1

|ξ|ε1

|ξ|2−ε|ξ + 2|dξ ∧ dξ

)

≤ 3

4π2|x|ε|k|ε

∫C

1

|ξ|2−ε|ξ + 2|dξ ∧ dξ

=:3

4π2|x|ε|k|εI1 ≤

3

4π2

(1 + |x|2

)ε/2 |k|εI1

Hence, to conclude this first part we need to show that I1 is a constant depending just onε. To do so we split the integral into three parts isolating each of the singularities and usethe following inequalities:

For |ξ| < 1 we have |ξ + 2| > ||ξ| − 2| = 2− |ξ| > 1⇒ 1|ξ+2| < 1;

For |ξ| > 1, |ξ + 2| > 1 and ξ1 > −1 we have |ξ + 2| > |ξ|;

For |ξ| > 1, |ξ + 2| > 1 and ξ1 < −1 we have |ξ + 2| < |ξ|.

Thus we obtain

I1 =

∫R2

1

|ξ|2−ε|ξ + 2|dξ ∧ dξ

=

∫|ξ|<1

1

|ξ|2−ε|ξ + 2|dξ ∧ dξ +

∫|ξ|>1,|ξ+2|<1

1

|ξ|2−ε|ξ + 2|dξ ∧ dξ+

+

∫|ξ|>1,|ξ+2|>1,ξ1>−1

1

|ξ|2−ε|ξ + 2|dξ ∧ dξ +

∫|ξ|>1,|ξ+2|>1,ξ1<−1

1

|ξ|2−ε|ξ + 2|dξ ∧ dξ

≤∫|ξ|<1

1

|ξ|2−εdξ ∧ dξ +

∫|ξ+2|<1

1

|ξ + 2|dξ ∧ dξ

+

∫|ξ|>1, |ξ+2|>1,ξ1>−1

1

|ξ|3−εdξ ∧ dξ +

∫|ξ|>1, |ξ+2|>1,ξ1<−1

1

|ξ + 2|3−εdξ ∧ dξ

≤∫|ξ|<1

1

|ξ|2−εdξ ∧ dξ +

∫|ξ+2|<1

1

|ξ + 2|dξ ∧ dξ + +2

∫|ξ|>1

1

|ξ|3−εdξ ∧ dξ

=2π

ε+ 2π +

4π

1− ε=: C1,ε,

where the last equality follows by by changing to polar coordinates.Hence, we have the result for |x| > 1/2. Now we conclude the proof by calculating the

inequality for the case |x| < 1/2. For this case we can bound (eix·ξ − 1) by |x||ξ| for any|ξ| < 2. Hence, by the same change of variables as we did in the previous computation itfollows

|gk(x)−G0(x)| ≤ 1

4π2

(∫|ξ|<2/|k|

|x||k||ξ||ξ + 2|

dξ ∧ dξ +

∫|ξ|>2/|k|

1

|ξ|2|ξ + 2|dξ ∧ dξ

).

Now, since |k| ≤ 1/2, we can make the splitting of domains |ξ| < 4 and 4 < |ξ| < 2/|k|and use the inequality |ξ + 2| > |ξ|/2, which follows for 4 < |ξ| < 2/|k| since |ξ + 2| >

36

|ξ|(

1− 2|ξ|

)> |ξ|/2. From this we obtain:

|gk(x)−G0(x)| ≤ 1

4π2

(|k|2

∫|ξ|<4

1

|ξ||ξ + 2|dξ ∧ dξ +

∫4<|ξ|<2/|k|

|xk||ξ||ξ + 2|

dξ ∧ dξ

+

∫|ξ|>2/|k|

1

|ξ|2|ξ + 2|dξ ∧ dξ

)

≤ 1

4π2

(|k|2

∫|ξ|<4

1

|ξ||ξ + 2|dξ ∧ dξ + |k|

∫4<|ξ|<2/|k|

1

|ξ|2, dξ ∧ dξ

+ 2

∫|ξ|> 2

|k|

1

|ξ|3dξ ∧ dξ

)

≤ 1

4π2

(|k|2I2 + 2π|k|

∫ 2/|k|

4

1

rdr + 4π

∫ +∞

2/|k|

1

r2dr

)

=1

4π2

(|k|2I2 + 2π|k|

(ln

(2

|k|

)− ln(4)

)+ 4π

|k|2

)

=1

4π2

(|k|2I2 + 2π|k| |ln |k| − ln(1/2)|+ 4π

|k|2

)≤ C1,ε|k|| ln |k||≤ C1,ε|k|| ln |k|| 〈x〉ε

In fact, this constant does not depend on ε. It is just a matter of notation. Although, westill need to show that the constant is, in fact, finite. For this we compute the integral sinceits the only thing which it will depend on.

I2 =

∫|ξ|<4

1

|ξ||ξ + 2|dξ ∧ dξ

=

∫|ξ|<4,|ξ+2|<1

1

|ξ||ξ + 2|dξ ∧ dξ +

∫|ξ|<4,|ξ+2|>1

1

|ξ||ξ + 2|dξ ∧ dξ

≤∫|ξ+2|<1

1

|ξ + 2|dξ ∧ dξ +

∫|ξ|<4

1

|ξ|dξ ∧ dξ

= 2π + 8π <∞,

where the last equality follows by a polar coordinates.Therefore, it follows for any x ∈ R2 \ 0

|gk(x)−G0(x)| ≤ C1,ε|k|ε 〈x〉ε + C2,ε|k|| ln |k|| 〈x〉ε ≤ C1,ε|k|ε 〈x〉ε + C2,ε|k|ε 〈x〉ε

≤ (C1,ε + C2,ε)|k|ε 〈x〉ε ,

given that |k| < |k|ε and | ln |k|| < 1 for |k| ≤ 1/2.

Now, let K(k)f = 〈x〉−β gk ∗ (〈·〉β qf) if k 6= 0, and K(0)f = 〈x〉−β G0 ∗ (〈·〉β qf).

37

Lemma 2.4.4. Let q, β, ε satisfy the hypothesis of Theorem 2.4.1 and in addition β <min(1, ρ− ε− 2/p′). Then K(k) is bounded on W 1,p for all k,

1. ‖K(k)− K(0)‖W 1,p 7→W 1,p ≤ c|k|ε for|k| < 1/2, and

2. I + K(k) is invertible on W 1,p for k sufficiently small.

Proof. (i) By hypothesis on β and ε we have by Holder Inequality 〈x〉β+ε q ∈ L1 ∩ Lp. Forthe boundedness, we first look at the case when k = 0. Let f ∈W 1,p, we start by lookingat the norm in Lp:

‖K(0)f‖Lp =1

2π

[∫R2

∣∣∣∣〈x〉−β ∫R2

log |x− y| 〈y〉β q(y)f(y) dy

∣∣∣∣p dx]1/p

≤ 1

2π

∫R2

[∫R2

∣∣∣〈x〉−β log |x− y| 〈y〉β q(y)f(y)∣∣∣p dx]1/p

dy

=1

2π

∫R2

[∫|x−y|≥R

∣∣∣〈x〉−β log |x− y| 〈y〉β q(y)f(y)∣∣∣p dx]1/p

dy

+1

2π

∫R2

[∫B(y,R)

∣∣∣〈x〉−β log |x− y| 〈y〉β q(y)f(y)∣∣∣p dx]1/p

dy

=: I1 + I2.

Above we used Minkowski integral inequality and we split the integrals into two domains,picking R big enough, such that log |x− y| ≤ 〈x− y〉ε . Similar as before, in this way wecan treat both singularities separately.

Therefore, by applying the properties of 〈x〉 we have

I1 ≤1

2π

∫R2

[∫|x−y|≥R

∣∣∣〈x〉−β 〈x− y〉ε 〈y〉β q(y)f(y)∣∣∣p dx]1/p

dy

≤ 1

2π

∫R2

[∫|x−y|≥R

∣∣∣〈x〉−β 〈x〉ε 〈y〉β 〈y〉ε q(y)f(y)∣∣∣p dx]1/p

dy

≤ 1

2π

∫R2

[∫R2

∣∣∣〈x〉−β+ε∣∣∣p dx]1/p

| 〈y〉β+ε q(y)f(y)| dy

=

[∫R2

∣∣∣〈x〉−β+ε∣∣∣p dx]1/p ∫

R2

| 〈y〉β+ε q(y)f(y)| dy

The first integral is finite due to β − ε > 2/p ⇒ p(β − ε) > 2. For the second, wehave the Sobolev embedding W 1,p(R2) → CB(R2). Hence, there exists C > 0 such thatI1 ≤ C‖f‖W 1,p .

38

Furthermore, by also using the properties on 〈x〉 we have

I1 =1

2π

∫R2

[∫B(0,R)

∣∣∣〈x+ y〉−β log |x| 〈x+ y − x〉β q(y)f(y)∣∣∣p dx]1/p

dy

≤ 1

2π

∫R2

[∫B(0,R)

∣∣∣log |x| 〈x〉β∣∣∣p dx]1/p

|q(y)f(y)| dy

≤ 1

2π

[∫B(0,R)

∣∣∣log |x| 〈x〉β∣∣∣p dx]1/p ∫

R2

|q(y)f(y)| dy

The first integral converges because the term 〈x〉β goes faster to zero and the convergenceof the second integral follows by the same reasons as above. Hence, there exists C ′ > 0such that I2 ≤ C ′‖f‖W 1,p .

Joining both integrals we obtain:

‖K(0)f‖Lp ≤ (C + C ′)‖f‖W 1,p

.

Now we need to look at the derivatives of K(0)f to estimate the Sobolev norm. To useLemma 2.4.2 we look at the gradient.

∇(K(0)f

)(x) = ∇

(〈x〉−β

(G0 ∗ (〈·〉β qf

)(x))

=(∇〈x〉−β

)(G0 ∗

(〈·〉β qf

)(x))

+ 〈x〉−β∇(G0 ∗

(〈·〉β qf(x)

))= −(x1, x2)β 〈x〉−(β+2)

(G0 ∗

(〈·〉β qf

)(x))

+ 〈x〉−β∇(G0 ∗

(〈·〉β qf

)(x))

=: a0 + a1

To estimate a0 in the Lp norm we can redo the previous process with some slight changeson the coefficient we use to estimate the logarithm. Given that 〈x〉−β ≤ 1 it follows thata1 is estimated by Lemma 2.4.2 (ii) and similar as before we use the Sobolev embeddingto obtain the estimate.

During the proof of Theorem 2.3.1 we showed that K(k) is bounded in W 1,p for k 6= 0and the operator K(k) can be given in terms of it. We have for k 6= 0:

K(k)f = 〈x〉−β (gk + `(k)) ∗ (〈·〉β qf) = K(k)f + `(k) 〈x〉−β∫R2

〈y〉β q(y)f(y) dy

(2.4.11)

Therefore, by Minkowski integral inequality and knowing that 〈y〉β q(y)f(y) is in L1

we just need to make sure that ‖ 〈x〉−β ‖W 1,p is finite, but that immediately follows byβ > 2/p+ ε. Hence, the boundedness of K(k) in W 1,p follows.

39

Now, we start by showing the statement (i). Similarly to above, we start by estimatingthe norm in Lp. Let f ∈ W 1,p and |k| < 1/2 then by Lemma 2.4.3 and similarly usingthe properties of 〈x〉 it follows

∥∥∥(K(k)− K(0))f∥∥∥Lp

=

[∫R2

∣∣∣∣〈x〉−β ∫R2

(gk −G0)(x− y) 〈y〉β q(y)f(y) dy

∣∣∣∣p dx]1/p

(2.4.12)

≤∫R2

[∫R2

∣∣∣〈x〉−β Cε|k|ε 〈x− y〉ε 〈y〉β q(y)f(y)∣∣∣p dx]1/p

dy

Cε|k|ε∫R2

| 〈y〉β+ε q(y)f(y)| dy[∫

R2

| 〈x〉−β+ε |p dx]1/p

≤ C ′ε|k|ε ‖f‖W 1,p for |k| < 1/2,

and both integrals converge by the above presented arguments.

Furthermore, we have that

−4(∂ + ik)∂(gk ∗ f) = f

−4∂∂(G0 ∗ f) = f

Therefore, we obtain the expression

∂((gk −G0) ∗ (·)) = −1

4

[(∂ + ik)−1 − ∂−1

](2.4.13)

It follows that for f ∈W 1,p

‖∂((K(k)− K(0))f‖Lp ≤∥∥∥(∂ 〈x〉−β)((gk −G0) ∗

(〈·〉β qf

))∥∥∥Lp

+∥∥∥〈x〉β (∂ (gk −G0) ∗

(〈·〉β qf

))∥∥∥Lp

By using the Lemma 2.4.3 we can estimate the first norm similarly to above.

For the second norm by using 2.4.13 we obtain∥∥∥〈x〉β (∂ (gk −G0) ∗(〈·〉β qf

))∥∥∥Lp

=

∥∥∥∥1

4

[(∂ + ik)−1 − ∂−1

] (〈·〉β qf

)∥∥∥∥Lp−β

(2.4.14)

Now, Lemma 2.3.1 with α = 2/p′ + ε implies that the map (∂ + ik)−1 is stronglydifferentiable Lp−β and hence continuous which implies by 2.4.14 that∥∥∥〈x〉β (∂ (gk −G0) ∗

(〈·〉β qf

))∥∥∥Lp≤ c|k|

∥∥∥〈·〉β qf∥∥∥Lpα.

By our upper restriction on β it follows β +α < ρ and by Sobolev embedding it followsfor |k| < 1/2 ∥∥∥∂ (K(k)− K(0)

)f∥∥∥Lp≤(C ′′ε |k|ε + C ′|k|

)‖f‖W 1,p ≤ c|k|ε ‖f‖W 1,p(2.4.15)

Joining the estimates (2.4.12) and (2.4.15), we obtain the desired statement (i).

40

(ii) Due to the set of invertible operators being open and due to the (i) we only need toshow that I + K(0) is invertible in W 1,p. So let h ∈ W 1,p be a real-valued functionsuch that it satisfies h = −〈x〉−β G0 ∗ (q(〈·〉β h)). Now, we set h = 〈x〉β h which fulfillsh = −G0 ∗ (qh). Differentiating we have ∇h = −∇(G0 ∗ (qh)) and with qh ∈ Lp byLemma 2.4.2 (ii) we obtain that ∇h ∈ Lp. To conclude our result we proceed as in theproof of Lemma 2.2.4. Remembering that q = (∆ψ0)/ψ0, with ψ0 ∈ L∞, we have thatby equivalence with (a) from Lemma 2.4.1 it follows ∂ψ0 ∈ Lp1, (e), and in Lp, (f).

Then we set v = (ψ0∂h−h∂ψ0). By Sobolev embedding theorem h ∈ L∞, which impliesthat h ∈ L∞−β ⊂ L∞−1, since β < 1. Therefore, we have ψ0∂h ∈ Lp and h∂ψ0 ∈ Lp, hence

v ∈ Lp.Now we look at ∂v. First, we have that h satisfies

−∆h = −qh

, given that G0 is the fundamental solution of the Laplacian. This implies the followingrelation between h and ψ0:

(−∆ + q)h = 0⇔(−∆ +

∆ψ0

ψ0

)h = 0⇒ ψ0∂∂h = h∂∂ψ0

With this it follows that:

∂v = ∂ψ0∂h+ ψ0∂∂h− ∂h∂ψ0 − h∂∂ψ0 = ∂ψ0∂h−−∂h∂ψ0

=∂ψ0

ψ0(ψ0∂h)− ∂ψ0

ψ0(ψ0∂h) =

∂ψ0

ψ0(ψ0∂h− h∂ψ0)− ∂ψ0

ψ0(ψ0∂h− h∂ψ0)

=∂ψ0

ψ0v − ∂ψ0

ψ0v =: Av +Bv

Here, given that ψ0 is bounded from below, we have that A,B ∈ Lp ∩ Lp and, hence,v is a generalized analytic function. Consequently, we can construct a bounded andcontinuous function w such that ve−w is entire and in Lp. Therefore, it must vanish byLiouville theorem for Lp-functions. This in turn implies that v must vanish since w isbounded.

This gives that v = 0⇔ (ψ0∂h−h∂ψ0) = 0. Now this means that h/ψ0 is antiholomor-phic:

∂(hψ0) =(∂h)ψ0 − (∂ψ0)h

(ψ0)2= 0.

Given that h/ψ0 ∈ L∞−β and β < 1 the generalized Liouville’s theorem implies thath/ψ0 = c, because it has less growth than |z|. Thus by 2.4.1 it follows that cψ0 =−cG0 ∗ (qψ0) = c(ψ0 − c∞). Since ψ0 is assumed to be bounded away from zero c∞cannot be zero so c must be. Hence, h = cψ0 vanishes.

Similarly to the proof of Theorem 2.2.1 and by the estimates already computed weobtain that K(0) is a compact operator in W 1,p. Hence, I + K(0) is Fredholm of indexzero and by the injectivity shown above the invertibility follows. This concludes theproof of (ii).

41

With these statements proven, the proof of Theorem 2.4.1 follows in a quick succession.

Proof of Theorem 2.4.1. In addition to the established conditions on β and ε in (2.4.6) weassume, without loss of generality, that

β < min(1, ρ− ε− 2/p′).(2.4.16)

Now we fix these coefficients to fulfill (2.4.6) and (2.4.16) and choose ε′ such that

ε < ε′ < min((ρ− 1)/2, β − 2/p, ρ− β − 2/p′).(2.4.17)

Under these conditions Lemma 2.4.4 holds for β and ε′. Hence, for k close to zero we definea family of functions

µ(x, k) = 〈x〉β(I + K(k)

)−1(〈·〉−β)(x).(2.4.18)

In the same way as the solution of 2.2.8 fulfills also 2.2.7, we have that µ(x, k) is also a solutionof 2.2.7, but in W 1,p

−β .

For k = 0, we claim that µ(x, 0) = ψ0

c∞. Due to Lemma 2.4.1 and 〈x〉−β ∈ Lp we have that

ψ0 ∈W 1,p−β . Moreover, we get[

I + K(0)] (〈x〉−β ψ0/c∞

)= 〈x〉−βψ0

c∞+ 〈x〉−β G0 ∗

(〈·〉β q

(〈·〉−β ψ0

c∞

))= 〈x〉−β 1

c∞[ψ0 +G0 ∗ (qψ0)] =

(〈x〉−β /c∞

)c∞ = 〈x〉−β ,

where the last inequality follows by (2.4.1) in Lemma 2.4.1. Given that the operator isinvertible the solutions to the equation (2.4.18) are unique. Hence, the claim follows.

Now we use this and (2.4.18) to obtain (2.4.5):

‖µ(·, k)− ψ0/c∞‖W 1,p−β

=∥∥∥〈x〉β [((I + K(k))−1 − (I + K(0))−1

)〈·〉β]∥∥∥

W 1,p−β

=∥∥∥((I + K(k))−1 − (I + K(0))−1

)〈·〉β∥∥∥W 1,p

≤ C∥∥∥(I + K(k))−1

[(I + K(0)− (I + K(k))

](I + K(0))−1 〈·〉β

∥∥∥W 1,p

≤ c′∥∥∥(I + K(0)− (I + K(k))

∥∥∥W 1,p

≤ c′∥∥∥K(0)− K(k)

∥∥∥W 1,p

≤ c|k|ε,

where the last inequality holds for k small and follows from the first statement in Lemma2.4.4.

Furthermore, we have

(I +K(k))−1 =(I + K(k)

)−1 (I + K(k)

)(I +K(k))−1

=(I + K(k)

)−1 (I + K(k) +K(k)−K(k)

)(I +K(k))−1

=(I + K(k)

)−1 ((I +K(k)) +

(K(k)−K(k)

))(I +K(k))−1

=(I + K(k)

)−1+(I + K(k)

)−1 [K(k)−K(k)

](I +K(k))−1(2.4.19)

42

We have that the difference in (2.4.19) is of the form:[K(k)−K(k)

]f = 〈x〉−β `(k)

∫R2

〈y〉β q(y)f(y) dy.

Therefore, it is an operator of rank 1. We apply (2.4.19) to 〈x〉−β and we obtain

(I +K(k))−1 〈·〉β =(I + K(k)

)−1+(I + K(k)

)−1 [K(k)−K(k)

](I +K(k))−1 〈·〉β

⇔ 〈x〉−β µ(x, k) = 〈x〉−β µ(x, k) +(I + K(k)

)−1 [K(k)−K(k)

]〈·〉−β µ(·, k)

⇔ 〈x〉−β µ(x, k) = 〈x〉−β µ(x, k) +(I + K(k)

)−1〈·〉−β `(k)

∫R2

q(y)µ(y, k) dy

⇔ µ(x, k) = µ(x, k)

[1 + `(k)

∫R2

q(y)µ(y, k) dy

]By defining

τ(k) =:

∫R2

q(y)µ(y, k) dy,(2.4.20)

we obtain the following identity

µ(x, k) = (1 + `(k)τ(k))µ(x, k)(2.4.21)

Furthermore, we also define τ similar to τ , but involving µ instead of µ. By integrating(2.4.21) against q yields the following equation:

τ(k) = (1 + `(k)τ(k)) τ(k);

Thus, we can write τ in terms of τ :

τ(k) = (1 + `(k)τ(k)) τ(k) = τ(k) + `(k)τ(k)τ(k) + τ(k)− τ(k)

= (1 + `(k)τ(k)) τ(k)− τ(k) + τ(k)

⇒2τ(k)− (1 + `(k)τ(k)) τ(k) = τ(k)⇒ τ(k) = (1− `(k)τ(k))τ(k)

τ(k) =τ(k)

1− `(k)τ(k)

Now, we can estimate τ(k) in terms of τ(k). Using the estimate for µ and Lemma 2.4.1(c) we have for k small enough

|τ(k)| =∣∣∣∣∫

R2

q(x)µ(x, k) dx

∣∣∣∣ ≤ ∣∣∣∣∫R2

q(x) (µ(x, k)− ψ0/c∞)

∣∣∣∣≤ ‖ 〈x〉β q‖L1‖µ(x, k)− ψ0/c∞‖W 1,p

−β≤ C|k|ε′

where we used the Sobolev embedding W 1,p → CB.Hence, we have

|τ(k)| =∣∣∣∣ τ(k)

1− `(k)τ(k)

∣∣∣∣ ≤ c|k|ε′ ∣∣∣∣ 1

1− `(k)τ(k)

∣∣∣∣ ≤ C ′|k|ε′ ,43

where we used the fact that for k small enough we have that |k|ε′`(k) = |k|ε′(log |k| +γ0)/(2π) → 0 as |k| → 0. Hence, it is bounded and, therefore, |1 − |`(k)τ(k)|| is boundedfrom below and, consequently, the fraction is also bounded.

Under this condition on τ(k) and using (2.4.21) we will obtain Estimate (2.4.5) for µ

‖µ(x, k)− ψ0/c∞‖W 1,p−β

= ‖(1 + `(k)τ(k))µ(·, k)− ψ0/c∞‖W 1,p−β

≤ ‖µ(x, k)− ψ0/c∞‖W 1,p−β

+ |`(k)τ(k)| ‖µ(·, k)‖W 1,p−β

≤ c|k|ε′ +O(log |k||k|ε′)(‖µ(·, k)− ψ0/c∞‖W 1,p

−β+ ‖ψ0/c∞‖W 1,p

−β

)≤ c|k|ε′ +O(log |k||k|ε′)

(c|k|ε′ +

∥∥∥〈x〉−β ψ0/c∞

∥∥∥W 1,p

).

The last norm converges due to Lemma 2.4.1 and β > 2/+ε.Due to ε′ > ε for k small enough we have log |k||k|ε′ ≤ c′|k|ε and |k|s ≤ |k|t, for s > t and

|k| < 1. Consequently, it follows that

‖µ(x, k)− ψ0/c∞‖W 1,p−β≤ c|k|ε(2.4.22)

To finally conclude Estimate (2.4.4) we will come back to the formula that defines t anduse the above estimate (2.4.22). First by Lemma 2.4.1 (c) we have

∫R2 q(x)ψ0(x) dx = 0. This

implies

|t(k)| =∣∣∣∣∫

R2

ek(x)q(x)µ(x, k) dx

∣∣∣∣≤∫R2

|q(x)µ(x, k)| dx−∫R2

q(x)ψ0(x)/c∞ dx

=

∫R2

q(x) [sgn(q)(x)|µ(x, k)| − ψ0(x)/c∞] dx

≤∫R2

q(x) [|µ(x, k)− ψ0(x)/c∞|] dx

≤∫R2

|q(x)||µ(x, k)− ψ0(x)/c∞| dx

≤ c‖µ(·, k)− ψ0/c∞‖W 1,p−β‖q‖L1

β

≤ C|k|ε,

where the norm of q in L1β follows by Sobolev embedding theorem.

With the above results we obtained enough information from the direct problem to lookat a property of the scattering transform.

Theorem 2.4.2. Suppose q ∈ Lpρ(R2), with 1 < p < 2, ρ > 1, is a real-valued function of theform q = (∆ψ0)/ψ0 with ψ0 ∈ L∞(R2) and ψ0(x) ≥ c0 > 0. Then t(k)/k is in Lr(R2) for allr ∈ (p′, r2), where

r2 = p if ρ > 5− 4/p, r2 = 4/(3− ρ) if 1 < ρ < 5− 4/p.(2.4.23)

44

Proof. We start by writing t(k) = Fq(k) + F(q(µ(·, k) − 1)) where we use the notation in(2.3.10). Then the proof will be done in three steps, which correspond to the division of theintegral over R2 into three pieces:

t(k)/k ∈ Lr(|k| ≥ k0) for r > p′ and k0 large enough:

To use the formula (2.3.10) we look first at the behavior of Fq(k):

Fq(k) =i

2

∫Cek(z)q(z) dz ∧ dz =

i

2|k|2

∫Ce1(z)q(kz) dz ∧ dz.

From Theorem 2.2.1, with s > 2/p, we have ‖µ(·, k)− 1‖W s,p ≤ c|k|s−1 ‖q‖Lp for |k| ≥k0, with k0 large enough. Hence, we obtain

|F(q(µ(·, k)− 1))(k)| ≤∫R2

|q(x)(µ(x, k)− 1)| dx ≤ ‖µ(·, k)− 1‖W s,p‖q‖L1

≤ c|k|s−1‖q‖Lp‖q‖L1

Therefore, given that q ∈ L1∩Lp and by the above formula and estimate we can obtainthe desired result by the following computations:∫

|k|≥k0

∣∣∣∣ t(k)

k

∣∣∣∣r dk ≤ ∫|k|≥k0

∣∣∣∣Fq(k)

k

∣∣∣∣r dk +

∫|k|≥k0

∣∣∣∣F(q(µ(·, k)− 1))(k)

k

∣∣∣∣r dk≤ ‖q‖L1

∫|k|≥k0

1

|k|3rdk + ‖q‖L1‖q‖Lp

∫|k|≥k0

1

|k|r(2−s)dk

For the second integral to converge we need to have r(2− s) > 2. Given that

s > 2/p⇒ 2− s < 2− 2/p⇒ (2− s)/2 < 1/p′ ⇒ 2/(2− s) > p′,(2.4.24)

we have the condition r > p′ > 1. Moreover, under this condition the first integralconverges. Therefore, we conclude the statement.

t(k)/k ∈ Lr(|k| ≤ k1) for r < 2(1 − ε) and k1 small enough such that (2.4.4)holds:

We have that for such k1, |t(k)| < c|k|ε, hence it follows:∫|k|≤k1

∣∣∣∣t(k)

k

∣∣∣∣r dk ≤ c∫|k|≤k1

|k|(ε−1)r dk

For the last integral to converge, we need (1 − ε)r < 2 and, hence, it follows thatr < 2(1 − ε). Now, by condition (2.4.6), we either have ε < (ρ − 1)/2 < 2/p′ orε < 2/p′ < (ρ − 1)/2 depending on which is the minimum. Now, suppose the firstinequality holds we have:

(ρ− 1)/2 < 2/p′ ⇒ (ρ− 1) < 4(1− 1/p)⇒ ρ < 5− 4/p.

Otherwise, if the second inequality holds we obtain

ρ > 5− 4/p.

45

In the first case it follows that

ε < (ρ− 1)/2⇒ 1− ε > 1− (ρ− 1)/2⇒ 1/(1− ε) < 2/(3− ρ)

⇒ r < 2/(1− ε) < 4/(3− ρ)

For the second we get

ε < 2/p′ ⇒ (1− ε) > 1− 2/p′ ⇒ 2/(1− ε) < 2p′/(p′ − 2) = p⇒ r < p

With this we obtain the condition (2.4.23).

Finally t(k)/k ∈ Lr(k1 < |k| < k0) :

This follows because µ(·, k) is differentiable in W 1,p−β and, therefore, is continuous in

k 6= 0. So in the compact domain k ∈ C : k1 < |k| < k0, we have uniform continuity.Hence, also t(k)/k is continuous because for k, k′ ∈ k : k1 < |k| < k0, it follows that

|t(k)− t(k′)||k − k′|

≤ 1

|k − k′|

∫R2

|q(x)||(µ(x, k)− 1)− (µ(x, k′)− 1)| dx

≤‖(µ(·, k)− 1)− (µ(·, k′)− 1)‖

W 1,p−β

|k − k′|

∫R2

| 〈x〉β q(x)| dx

≤ c|k − k′||k − k′|

‖q‖L1β≤ C,

where we have q ∈ L1β by Holder inequality. Now, the result follows immediately.

2.5 Reconstruction of the conductivity at the boundary

In this section, we will work with a domain Ω ∈ Rn, for n ≥ 2, which is of Lipschitz type.Hence the domain can be partitioned into N ≥ 1 connected components Ω1, ...,ΩN .

We defined the Neumann-to-Dirichlet map (NtD), R, on the space

H−1/2(∂Ω) = h ∈ H−1/2(∂Ω) : 〈h, 1〉∂Ωj = 0, j = 1, ..., N(2.5.1)

by Rh = w|∂Ω, with w ∈ H1(Ω) the weak solution (unique modulo functions constant on eachΩj) of ∆w = 0, ∂w∂ν = h.

For the proofs we will need to define a new function. Let x0 ∈ ∂Ωj0 , let U = B × I ⊂ Rnbe a cylindrical neighborhood with coordinates chosen so as to have Ω ∩ U = Ωj0 ∩ U =(x′, xn) ∈ U : xn < φ(x′) with φ a Lipschitz function. On this conditions we define forh ∈ L2(∂Ω) that as support in ∂Ωj0 ∩ U and for any η ∈ Rn−1 × 0, the function hη to beidentically zero on ∂Ωj0 for j 6= j0, and

hη(x) = h(x)e−ix·η − 1

|∂Ωj0

∫∂Ω∩U

h(y)e−iy·η dσ(y) for x ∈ ∂Ωj0 .(2.5.2)

46

Lemma 2.5.1. Let Ω be a bounded Lipschitz domain in Rn, n ≥ 2. Assume γ is in W 1,r for

some r > n and has a positive lower bound. For any f ∈ H1/2(∂Ω) and h ∈H −1/2(∂Ω) there

exists the identity:

〈h, (γ −RΛγ)f〉 =

∫Ωu∇w · ∇γ,(2.5.3)

where u is the H1(Ω) solution of ∇ · (γ∇u) = 0 in Ω with u|∂Ω = f, w ∈ H1(Ω) is a weaksolution of ∆w = 0 in Ω with ∂w

∂ν

∣∣∂Ω

= h, and R denotes the Neumann-to-Dirichlet map forthe Laplacian, and ν is the normal unit vector to the boundary ∂Ω.

Proof. Since w is a weak solution of the Neumann problem by the weak Green identity wehave

〈h, v|∂Ω〉 =

∫∂Ωhv|∂Ω dS =

∫∂Ω

∂w

∂ν

∣∣∣∣∂Ω

v|∂Ω dS =

∫Ωv∆w +∇v · ∇w dV

=

∫Ω∇v · ∇w dV(2.5.4)

Moreover, by definition we have Rh = w|∂Ω (modulo constant functions on each ∂Ωj), soby the weak definition of the Dirichlet-to-Neumann map (1.1.3) it follows

〈Rh,Λγf〉 =

∫Ωγ∇w · ∇u,(2.5.5)

where the constant mentioned above does not affect the computations because by the sameweak formulation we have 〈1,Λγf〉∂Ωj = 0.

Multiplication by γ is a bounded operator on H1(Ω) due to the Sobolev embedding ofW 1,r(Ω) ⊂ CB(Ω). Therefore, we can apply (2.5.4) to v = γu

〈h, γf〉 =

∫Ω∇w · ∇(γu) =

∫Ωγ∇w · ∇u+ u∇w · ∇γ.(2.5.6)

Now, we subtract the equation (2.5.5) from the above one (2.5.6). Hereby, we use the Rsymmetry property, which is obtained through application of the Green identities. Thus, wehave

〈h, γf〉 − 〈h,RΛγf〉 = 〈h, γf〉 − 〈Rh,Λγf〉 =

∫Ωu∇w · ∇γ

Note, that the new identity (2.5.3) unlike the others in the proof does not involve deriva-tives of u.

Furthermore, to prove the main theorem of this section we need another auxiliary Lemma.

Lemma 2.5.2. Let Ω be a bounded Lipschitz domain in Rn, n ≥ 2. Assume γ is in W 2,p(Ω)for some p > n

2 and has a positive lower bound. For any f, g ∈ H1/2(∂Ω) there exists theidentity ⟨

g,

(2Λγ − Λ1γ − γΛ1 +

∂γ

∂ν

)f

⟩=

∫Ω

2v∇(u− u0) · ∇γ + v(2u− u0)∆γ,(2.5.7)

where u , u0, v are respectively the H1(Ω) solutions of ∇ · (γ∇u) = 0,∆u0 = 0 and ∆v = 0 inΩ, with u|∂Ω = u0|∂Ω = f and v|∂Ω = g.

47

Proof. First by Sobolev embedding theorem it follows that γv ∈ H1(Ω), and by (1.1.3) wehave

〈g, γΛ1f〉 = 〈γg,Λ1f〉 =

∫Ω∇(γv) · ∇u.(2.5.8)

Moreover, by noting that γ(2u − u0)|∂Ω = γf and γ(2u − u0) ∈ H1(Ω) and that Λ1 is theDirichlet-to-Neumann map of the Laplacian we obtain

〈g,Λ1γf〉 =

∫Ω∇v · ∇(γ(2u− u0)).(2.5.9)

By hypothesis we have γ ∈ W 2,p(Ω), p > n2 , hence, by Sobolev embedding theorem it

follows that ∂γ∂ν defines a bounded operator from H1/2(∂Ω) to H−1/2(∂Ω) defined via through:⟨

g,∂γ

∂νf

⟩=

∫∂Ωfg∂γ

∂νdS;

which also satisfies (obtained through Green identities)⟨g,∂γ

∂νf

⟩=

∫Ω∇(uv) · ∇γ + uv∆γ(2.5.10)

for any u, v ∈ H1(Ω) with traces f , g on ∂Ω, respectively.Now, if we use the above equation with 2u− u0 instead of just u, we obtain

⟨g,∂γ

∂νf

⟩=

∫Ω∇((2u− u0)v) · ∇γ + (2u− u0)v∆γ

=

∫Ω

(2u− u0)∇v · ∇γ + v∇(2u− u0) · ∇γ + (2u− u0)v∆γ.

Hence, by using this formula, the (1.1.3), the above weak formulations (2.5.8), and (2.5.9),we have

⟨g,

(2Λγ − Λ1γ − γΛ1 +

∂γ

∂ν

)f

⟩= 2 〈g,Λγf〉 − 〈g,Λ1γf〉 − 〈g, γΛ1f〉+

⟨g,∂γ

∂νf

⟩=

∫Ω

2γ∇v · ∇u−∫

Ω∇v · ∇(γ(2u− u0))

−∫

Ωγ∇v · ∇u0 + v∇γ · ∇u0 + (2u− u0)∇v · ∇γ

+

∫Ωv∇(2u− u0) · ∇γ + (2u− u0)v∆γ

=

∫Ω

2v∇(u− u0) · ∇γ + v(2u− u0)∆γ.

The last equality follows by expansion of all terms and eliminating the terms that cut out.Consequently, the result follows.

48

Theorem 2.5.1. Let Ω be a bounded Lipschitz domain in R2, n ≥ 2. Suppose γ is in W 1,r(Ω)for some r > n and has a positive lower bound c0. Let x0 ∈ ∂Ω and let U be a cylindricalneighborhood of x0 as described above. Then

(i) γ|∂Ω∩U can be recovered from Λγ by the formula:

〈h, γf〉 = lim|η|→∞

η∈Rn−1×0

⟨hη, RΛγe

−i〈·,η〉f⟩,(2.5.11)

where f ∈ H1/2(∂Ω)∩C(∂Ω) and h ∈ L2(∂Ω) are assumed supported in U ∩ ∂Ω and hηis defined as zero outside ∂Ω ∩ U and:

hη(x) = h(x)e−ix·η − 1

|∂Ω ∩ U |

∫∂Ω∩U

h(y)e−iy·ηdσ(y) for x ∈ ∂Ω ∩ U(2.5.12)

(ii) If, moreover, γ ∈W 2,p(Ω) for some p > n/2, then for any continuous functions f, g inH1/2(∂Ω) with support in U ∩ ∂Ω,⟨

g,∂γ

∂νf

⟩= lim

|η|→∞η∈Rn−1×0

⟨g, e−i〈·,η〉(γΛ1 + Λ1γ − 2Λγ)ei〈·,η〉f

⟩,(2.5.13)

with Λ1 the Dirichlet-to-Neumann ma corresponding to γ ≡ 1.

Proof. (i) We start by defining fη = ei〈·,η〉f and apply Lemma 2.5.1 to this function and

hη. The condition hη ∈H −1/2(∂Ω) follows by duality L2(∂Ω) ⊂ H−1/2(∂Ω) and due

to computation in ∂Ω ∩ U and the supported being in this space for the other parts ofthe boundary.

Hence, it follows

〈hη, (γ −RΛγ)fη〉 =

∫Ωuη∇wη · ∇γ,

where ∇ · (γ∇uη) = 0, uη|∂Ω = fη and ∆wη = 0,∂wη∂ν = hη.

From this we obtain the inequality

| 〈hη, (γ −RΛγ)fη〉 | ≤ ‖uη‖L∞(Ω)‖∇wη‖L2(Ω)‖∇γ‖L2(Ω).(2.5.14)

The above estimate is finite because the weak solutions uη satisfy‖uη‖L∞(Ω) ≤ c‖f‖L∞(∂Ω) (B.5.1) and the operator taking hη into the solution of the

Neumann problem wη is bounded from L2(∂Ω) to H3/2(Ω).

The functions hη converge weakly in L2(∂Ω) to zero by Riemann-Lebesgue lemma.Moreover, compact operators map weakly convergent sequences to strongly convergentsequences (B.5.3), hence, due to the embedding H3/2(Ω) in H1(Ω) being compact andthe composition with a compact operator still being compact, follows that wη convergesto 0 in the H1 norm.

49

With this convergence in mind and with (2.5.14) it follows:

lim|η|→∞

η∈Rn−1×0

〈hη, RΛγfη〉 = lim|η|→∞

η∈Rn−1×0

〈hη, (γ −RΛγ)fη〉+ lim|η|→∞

η∈Rn−1×0

〈hη, γfη〉

= lim|η|→∞

η∈Rn−1×0

〈hη, γfη〉

= lim|η|→∞

η∈Rn−1×0

∫∂Ωhη(x)γ(x)eix·ηf(x) dσ(x)

=

∫∂Ωh(x)γ(x)f(x) dσ(x)

− lim|η|→∞

η∈Rn−1×0

1

|∂Ω ∩ U |

∫∂Ω∩U

h(y)e−iy·η dσ(y)

∫∂Ωfη(x) dσ(x)

= 〈h, γf〉

(ii) Now, consider gη and let u0η, vη ∈ H1(Ω) be the solution to ∆u0

η = 0, u0η|∂Ω = 0 and

∆vη = 0, vη|∂Ω = gη, respectively. As in part (i) we have ‖uη‖L∞(Ω), ‖u0η‖L∞(Ω) and

‖vη‖L∞(Ω) are bounded uniformly in η (B.5.1).

The function uη − u0η satisfies

∆(uη − u0η) = −(∇γ/γ) · ∇uη, and (uη − u0

η)|∂Ω = 0.(2.5.15)

Moreover, the Laplacian ∆ has a bounded inverse as an operator from H10 (Ω) to H−1(Ω)

(Lax-Milgram). By hypothesis γ ∈ W 2,p(Ω), p > n/2, therefore, both γ and it’s firstderivative are bounded on the closure of Ω. Then it follows by embeddings that

∥∥uη − u0η

∥∥H1(Ω)

≤ c∥∥∥∥1

γ∇γ · ∇uη

∥∥∥∥H−1(Ω)

≤c‖γ‖L∞(Ω)‖γ′‖L∞(Ω)

c0‖∇uη‖H−1(Ω)

≤ c′‖uη‖L∞(Ω) ≤ c′′‖f‖L∞(Ω)(2.5.16)

Now Lemma 2.5.2, together with the Holder inequality for 1/2 = 1/pn + (1/p − 1/n)and the Sobolev embedding W 1,p(Ω)→ Lnp/(n−p)(Ω) gives∣∣∣∣⟨gη,(2Λγ − Λ1γ − γΛ1 +

∂γ

∂ν

)⟩∣∣∣∣ ≤ 2

∫Ω

∣∣vη∇(uη − u0η) · ∇γ

∣∣+

∫Ω

∣∣vη(2uη − u0η)∆γ

∣∣≤ 2

∥∥∇(uη − u0η)∥∥L2(Ω)

‖vη∇γ‖L2(Ω)

+ c‖f‖L∞(Ω)‖γ‖W 2,p(Ω)‖vη‖Lp′ (Ω)

≤ 2∥∥uη − u0

η

∥∥H1(Ω)

‖vη‖Lpn (Ω) ‖∇γ‖Lnp/(p−n)

+ c‖f‖L∞(Ω)‖γ‖W 2,p(Ω)‖vη‖Lp′ (Ω)

≤ c′′‖f‖L∞(Ω)‖vη‖Lpn (Ω)‖∇γ‖W 1,p(Ω)

+ c‖f‖L∞(Ω)‖γ‖W 2,p(Ω)‖vη‖Lp′ (Ω)

≤ C‖f‖L∞(Ω)‖γ‖W 2,p(Ω)

(‖vη‖Lpn (Ω) + ‖vη‖Lp′ (Ω)

)50

The solution operator for the Dirichlet problem is bounded from L2(∂Ω) to H1/2(Ω).Similarly as above, by Riemann-Lebesgue lemma gη converges weakly in L2(∂Ω) so bycompactness of the embedding H1/2(Ω) → L2(Ω) it follows that vη converges to zero inthe L2-norm. Now convergence in the Lpn(Ω) immediately follows by Lp inclusions ifpn ≤ 2, and by interpolation if 2 < pn <∞, since in this case

‖vη‖Lpn (Ω) ≤ ‖vη‖1−2/pnL∞(Ω) ‖vη‖

2/pnL2(Ω)

≤ c‖g‖1−2/pnL∞(∂Ω)‖vη‖

2/pnL2(Ω)

.

Due to Lp inclusions, it follows that Lpn(Ω) ⊂ Lp′(Ω), hence, we get convergence to zeroin the Lp

′-norm.

The desired formulation follows immediately by re-arranging the terms:

lim|η|→∞

η∈Rn−1×0

⟨e−i〈·,η〉g,

(2Λγ − Λ1γ − γΛ1 +

∂γ

∂ν

)f

⟩

2.6 Reduction to the case γ ≡ 1 close to the boundary

The basic idea in this section comes from the fact that if γ ≡ 1 near the boundary thenthe Dirichlet-to-Neumann map for the conductivity equation with this conductivity is equalto the Dirichlet-to-Neumann map for the Schrodinger equation.

Here, we extend the domain Ω into a new domain Ω2. To make a proper expansion, weneed γ|∂Ω and ∂γ

∂ν |∂Ω, as we have done in the previous section. Hence, we pick Ω2 ⊃ Ω andextend γ to a conductivity in W 2,p(Ω2) with γ ≡ 1 near ∂Ω2 and infΩ2

γ > 0. We introducethe notation Λ2 for the Dirichlet-to-Neumann map on ∂Ω2. Moreover, we set Ω1 = Ω andΛγ = Λ1, for simpler explanation.

The idea is to give a constructive way to determine Λ2 from knowledge of Λ1 on ∂Ω.

What we want is the Dirichlet-to-Neumann map of the equation:

∇ · (γ∇u) = 0, in Ω2(2.6.1)

u|∂Ω2 = f2,

where f2 ∈ H1/2(∂Ω2). Given that the (weak) solution in H1(Ω1) to (1.1.1) is unique thenwe have by setting f1 = u|∂Ω1 that u also solves the equation (1.1.1) with the Dirichlet data f1.

In this sense we split the equation (2.6.1) into two over different domains to obtain twosolutions u1 ∈ H1(Ω1) and u2 ∈ H1(Ω2 \ Ω1) which fulfill

∇ · (γ∇u1) = 0, in Ω1(2.6.2)

u1|∂Ω1 = f1

∇ · (γ∇u2) = 0, in Ω2 \ Ω1(2.6.3)

u2|∂Ω1 = f1

u2|∂Ω2 = f2

Moreover, we can split the equation (2.6.3) again, and instead look at the solutions u12, u

22 ∈

H1(Ω2 \ Ω1) which satisfy

51

∇ · (γ∇u12) = 0, in Ω2 \ Ω1(2.6.4)

u12|∂Ω1 = f1

u12|∂Ω2 = 0

∇ · (γ∇u22) = 0, in Ω2 \ Ω1(2.6.5)

u22|∂Ω1 = 0

u22|∂Ω2 = f2

With these equations in mind, we have that u12 + u2

2 = u2, by uniqueness of the weakproblem in H1 and in an analogous manner we have u = u1 + u2 (where we make a properextension of the solutions by zero to the rest of the domain).

In this setting the Dirichlet-to-Neumann map with respect to γ in the domain Ω2 \ Ω1 canbe viewed as a 2 × 2 matrix of operators Λij : H1/2(∂Ωj) −→ H−1/2(∂Ωi), i, j ∈ 1, 2. Letfj ∈ H1/2(∂Ωj) for j = 1, 2 and gi ∈ H1/2(∂Ωi), i = 1, , 2. Thus we define each entry of thematrix, formally, as follows

Λijfj = γ∂uj2∂ν

∣∣∣∣∣∂Ωi

and thus its weak formulation⟨gi,Λ

ijfj⟩∂Ωi

=: (−1)i∫

Ω2\Ω1

γ∇vi · ∇uj2,(2.6.6)

where uj is the solution to one of the problems above, (2.6.4) and (2.6.5), corresponding tofj . Furthermore, vi are functions in H1(Ω2 \ Ω1) with traces gi in ∂Ωi and zero on the otherboundary.

Hence, hereby, we state the main result of this section concerns the reconstruction of Λ2.

Proposition 2.6.1. Let Ω1 ⊂ Ω2 be bounded Lipschitz domains in R2 and let γ ∈W 2,p(Ω2),p > 1, with γ(x) ≥ c0 > 0. Then Λ1 − Λ11 : H1/2(∂Ω1) −→ H1/2(∂Ω1) is an invertibleoperator and

Λ2 = Λ22 + Λ21(Λ1 − Λ11)−1Λ12.(2.6.7)

Proof. For the proof we will use the notation explained above and further let v∗ ∈ H1(Ω1),v∗|∂Ω1 = g1. To prove the first statement we will use the Lax-Milgram theorem applied tothe sesquilinear form

b : H1(Ω)×H1(Ω)→ C with

b(f1, g1) =

∫Ω1

γ(x)∇u1(x) · ∇v∗(x) dx+

∫Ω2\Ω1

γ(x)∇u12(x) · ∇v1(x) dx

To use Lax-Milgram theorem we need that this sesquilinear form is bounded and coercive.For boundedness, we do a simple computation, using Holder inequality and the definition

of the H1-norm. In particular, we pick v∗, v1 as solutions of −∆v∗ = 0 in Ω1, v∗|∂Ω1 = g1

and −∆v1 = 0 in Ω2 \ Ω1, v1|∂Ω1 = g1, v1|∂Ω2 = 0, respectively. We get

|b(f, g)| ≤ ‖γ‖L∞(Ω2)

(‖∇u1‖L2(Ω1)‖∇v∗‖L2(Ω1) + ‖∇u1

2‖L2(Ω2\Ω1)‖∇v1‖L2(Ω2\Ω1)

)≤ ‖γ‖L∞(Ω2)

(‖u1‖H1(Ω1)‖v∗‖H1(Ω1) + ‖u1

2‖H1(Ω2\Ω1)‖v1‖Ω2\Ω1

)≤ 2(c+ c′)‖γ‖L∞(Ω2)‖f1‖H1/2(∂Ω1)‖g1‖H1/2(∂Ω1),

52

where the last inequality follows by the boundedness of Dirichlet elliptic problems.Now let u∗ = u1 +u1

2 ∈ H1(Ω2), hence u∗|∂Ω2 = 0 and u∗|∂Ω1 = f1 ∈ H1/2(∂Ω1). Thus byPoincare inequality and the boundedness of the trace operator, we have

b(f1, f1) =

∫Ω1

γ(x)|∇u1(x)|2 dx+

∫Ω2\Ω1

γ(x)|∇u(x)|2 dx

≥ c0

(∫Ω2

|∇(u1 + u12)(x)|2 dx

)≥ c0 ‖∇u∗‖L2(Ω2)

≥ Cc0/2(‖∇u∗‖L2(Ω2) + ‖u∗‖L2(Ω2))

= Cc0/2(‖u∗‖H1(Ω2)) = Cc0/2(‖u1‖H1(Ω1) + ‖u1

2‖H1(Ω2)

)≥ C ′‖f‖H1/2(∂Ω1)

In this sense, we can conclude by Lax-Milgram that exists a uniquely determined isomor-phism B : H1/2(∂Ω1)→ H−1/2(∂Ω1), such that 〈Bf1, g1〉H1/2(∂Ω1) = b(f1, g1).

Furthermore, our operator in question is a bilinear form which is uniquely determined byb. We have

〈Bf1, g1〉H1/2(∂Ω1) = b(f1, g1) =⟨(Λ1 − Λ11)f1, g1

⟩(2.6.8)

Hence, given that Λ1 −Λ11 is well-defined from H1/2(∂Ω1) to H−1/2(∂Ω) an that B is anisomorphism, then our operator in question is also invertible.

Since we have two non-overlapping domains (Quarteroni and Valli [1999]) and we splitour main equation into two with respect to this domains at ∂Ω1 we have

γ∂u1

∂ν

∣∣∣∣∂Ω1

= γ∂u2

∂ν

∣∣∣∣∂Ω1

γ∂u1

∂ν

∣∣∣∣∂Ω1

= γ(∂u1

2 + u22)

∂ν

∣∣∣∣∂Ω1

.

By the invertibility of the operator we obtain:

Λ1f1 = Λ11f1 + Λ12f2

f1 = (Λ1 − Λ11)−1Λ12f2.

To finally deduce the desired formula (2.6.7) we will use this fact on f1, the decompositionpreviously done on the solution of the conductivity problem in Ω2 for u and by pickingv ∈ H1(Ω2), such that in Ω1 is zero and v|∂Ω2 = g2 ∈ H1/2(∂Ω2). Thus, it follows:

〈g2,Λ2f2〉 =

∫Ω2

γ∇u · ∇v =

∫Ω1

γ∇u1 · ∇v +

∫Ω2\Ω1

γ∇(u22 + u1

2) · ∇v

=

∫Ω2\Ω1

γ∇u22 · ∇v +

∫Ω2\Ω1

γ∇u12 · ∇v

=⟨g2,Λ

22f2

⟩+⟨g2,Λ

21f1

⟩=⟨g2,Λ

22f2

⟩+⟨g2,Λ

21(Λ1 − Λ11)−1Λ12f2

⟩=⟨g2,(Λ22 + Λ21(Λ1 − Λ11)−1Λ12

)f2

⟩Hence, we hereby conclude the proof of the proposition.

53

2.7 From Λ to t

Given that our reconstruction of γ is based on solving an equation in terms of the spectralparameter and t we will need to show is how we can obtain the scattering transform from thedata we in fact have, that is, the Dirichlet-to-Neumann map.

We start by showing a useful result for the proof.

Lemma 2.7.1. Assume Ω is a bounded domain with a Lipschitz boundary. The single layeroperator Sk corresponding to Faddeev’s Green function is bounded: Hs(∂Ω)→ Hs+1(∂Ω), for−1 ≤ s ≤ 0.

Proof. Gk(x) differs from the classical fundamental solution G0 by a harmonic function soit suffices to prove the lemma for the standard single layer operator S0. For s = 0 this is aconsequence of the theorem of Coifman, McIntosh and Meyer (Coifman et al. [1982], see alsothe appendix in Nachman [1988]). The statement for s = −1 is obtained by duality, and thatfor −1 < s < 0 by interpolation.

Theorem 2.7.1. Let Ω be a bounded Lipschitz domain in R2 and let q ∈ Lp(Ω) for somep > 1. Assume 0 is not a Dirichlet eigenvalue of −∆ + q in Ω. Then, for any k ∈ C \ 0:

(i) If ψ(·, k)|∂Ω has been determined, then t(k) can be recovered from the formula

t(k) =

∫∂Ωeizk(Λq − Λ0)ψ(·, k) dσ.(2.7.1)

(ii) The trace on ∂Ω of the function ψ(·, k) satisfies the integral equation

ψ(·, k)|∂Ω = eizk − Sk(Λq − Λ0)ψ(·, k);(2.7.2)

(iii) Sk(Λq − Λ0) is a compact operator from H1/2(∂Ω) to H1/2(∂Ω);

(iv) I +Sk(Λq −Λ0) is invertible on H1/2(∂Ω) if and only if k is not an exceptional point ofq when we extend it to zero outside Ω;

Proof. We extend q to be 0 outside Ω. Let u, u0 ∈ H1(Ω) be (weak) solutions of (−∆+q)u = 0and ∆u0 in Ω, respectively. Then, for q and 0 it follows from (2.1.2)

〈u0|∂Ω, (Λq − Λ0)u|∂Ω〉 =

∫Ωqu0u.(2.7.3)

To verify (2.7.1) we apply identity (2.7.3) to u = ψ(·, k) and u0 = eizk where we obtain

∫∂Ωeizk(Λq − Λ0)ψ(·, k) dσ =

⟨eizk, (Λq − Λ0)ψ(·, k)

⟩=

∫Ωeizkqψ(·, k) = t(k)(2.7.4)

Under the condition that we have ψ(·, k) statement (i) follows.

54

Let x /∈ Ω. Set u0(y) = Gk(x − y) and u(y) = ψ(y, k) which fulfill the Laplace andSchrodinger equation, respectively, in Ω. Hence, by applying (2.7.3), it follows

(Sk(Λq − Λ0)ψ(·, k))(x) =

∫∂ΩGk(x− y)(Λq − Λ0)ψ(y, k) dσ(y)

(2.7.5)

= 〈Gk(x− ·), (Λq − Λ0)ψ(·, k)〉 =

∫ΩGk(x− y)q(y)ψ(y, k) dy

= eizk − ψ(x, k).

Taking (2.7.5) the trace to ∂Ω from the exterior yields Formula (2.7.2) and, therefore, itfollows (ii).

More generally, we define Pq : f 7→ u ∈ H1(Ω) solving (−∆+q)u = 0 in Ω, with u|∂Ω = f .Hence, we obtain

Sk(Λq − Λ0)f(x) =

∫ΩGk(x− y)q(y)u(y) dσ(y)(2.7.6)

Choose p1 such that 1 < p1 < p and let r be defined by 1/r = 1/p1−1/p. Then 2 < r <∞and by (2.7.6) we have the factorization

Sk(Λq − Λ0) = RGkqIPq,(2.7.7)

where the operators are defined by

R : H1(Ω)→ H1/2(∂Ω), u 7→ u|∂Ω;(2.7.8)

Gk : Lp1(Ω)→W 1,p1(Ω), u 7→ Gk ∗ u;(2.7.9)

q : Lr(Ω)→ Lp1(Ω), u 7→ qu;(2.7.10)

I : H1(Ω)→ Lr(Ω), (Sobolev compact Embedding);(2.7.11)

Pq : H1/2(∂Ω)→ H1(Ω), f 7→ Pqf = u.(2.7.12)

Hence due to the compact embedding and the decomposition (2.7.7) it follows thatSk(Λq − Λ0) is compact in H1/2(∂Ω). Thus we obtain (iii).

Now, to finally conclude, we prove (iv).Suppose k is not an exceptional point of the Schrodinger operator with potential q. By

definition, there exists a function h, not zero everywhere, such that he−izk ∈ W 1,p(R2) and(−∆ + q)h = 0 in R2. Therefore, by Lemma 2.2.2 it follows that h = −Gk ∗ (qh).

Applying identity (2.7.3) to h(y) and Gk(x− y) and taking the limit from the exterior of∂Ω, as previously, we get

Sk(Λq − Λ0)h(x)|∂Ω = 〈Gk(x− ·)|∂Ω, (Λq − Λ0)h|∂Ω〉

=

∫ΩGk(x− y)h(y) dy = −h|∂Ω

⇒ [I + Sk(Λq − Λ0)]h|∂Ω = 0.(2.7.13)

55

Furthermore, h will not vanish identically on ∂Ω. If it would, h would be an interiorDirichlet eigenfunction, but by hypothesis, since 0 is not an eigenvalue, this implies that h iszero in Ω. By our zero extension of q outside Ω it follows that h is harmonic with h|∂Ω = 0and ∂h

∂νh|∂Ω = 0. Through the identity principle for harmonic functions it follows that h ≡ 0everywhere.

Conversely, suppose a nonzero h ∈ H1/2(∂Ω) satisfies (2.7.13), i.e., I +Sk(Λq −Λ0) is notinvertible in this space. We set v(x) = −(Sk(Λq − Λ0)h)(x) for x throughout R2. Moreover,we denote v−, v+ as the boundary values (as given by the trace operators) of v from theinside and outside of Ω, respectively. Given that h is a solution of (2.7.13) it follows thatv− = v+ = h on ∂Ω. Moreover, by the jump relations for the single-layer operator we obtain∂v+

∂ν −∂v−

∂ν = (Λq − Λ0)h (Theorem B.5.4).We show that v is harmonic inside Ω. Let x ∈ Ω and w, u ∈ H1(Ω) satisfying −∆w =

0 in Ω, w|∂Ω = h and (−∆ + q)u = 0 in Ω, u|∂Ω = h. By using the Green identities we have

(Sk(Λq − Λ0)h)(x) =

∫∂ΩGk(x− y)

[∂u

∂ν(y)− ∂w

∂ν(y)

]dσ(y)

=

∫Ω

∆u(y)Gk(x− y) +∇yGk(x− y) · ∇(u− w)(y)−∆w(y)Gk(x− y)dy

=

∫Ω∇yGk(x− y) · ∇(u− w)(y) + q(y)u(y)Gk(x− y) dy

=

∫Ω

(∆yGk(x− y))(u− w)(y) +∇yGk(x− y) · ∇(u− w)(y) dy

−∫

Ω∆yGk(x− y))(u− w)(y) dy +Gk ∗ (qu)(x)

=

∫∂Ω

(h− h)∂Gk∂ν

(x− y) dσ(y) +

∫Ω

∆x(Gk(x− y))(u− w)(y) dy

+Gk ∗ (qu)(x)

= (w − u)(x) +Gk ∗ (qu)(x) = w(x)

where we used the fact that Gk is the Faddeev Green function of the Laplacian. Thus v isindeed harmonic inside Ω.

Given this, it follows that ∂v−

∂ν = Λ0h, henceforth, we also have ∂v+

∂ν = Λqh.Now we define a function u by

u =

Pqh, in Ω

v, in R2 \ Ω(2.7.14)

Then, u− = h = v+ = u+ on ∂Ω and ∂u−

∂ν = Λqh = ∂v+

∂ν = ∂u+

∂ν . Hence, u solves(−∆ + q)u = 0 in R2, due to q = 0 outside Ω.

We claim that ue−izk ∈W 1,p(R2). Then there exists a solution to our Schrodinger equationin the above sense, which implies that k is an exceptional point, henceforth, (iv) follows.

Thus to conclude the proof we show this claim. Let u = −Gk ∗ (qu). We know that u|Ω isin H1(Ω) by the boundedness of Pq, as previously defined. Further, by the cutoff of q outsideof Ω we have that qu ∈ Lp1(R2) by (2.7.11) and (2.7.10), for 1 < p1 < p. Analogously to theproof of Theorem 2.2.1 we obtain e−izku ∈W 1,p1(R2).

56

Moreover, for x /∈ Ω, we have by applying (2.7.3) once again, −(Sk(Λq−Λ0)h)(x) = u(x).The left-hand side, by definition, is v and, therefore, corresponds to u outside Ω. Hence, itfollows u− u ≡ 0 in R2 \ Ω. Given that u− u is harmonic in R2, it follows from the identityprinciple that u ≡ u everywhere, and thus e−izku ∈ W 1,p1(R2). Due to e−izku ∈ W 1,p1(Ω)and p1 > p, then it follows that e−izku ∈W 1,p(Ω). Similar to the proof of Theorem 2.2.1 andu = −Gk ∗ (qu) it follows that e−izku ∈W 1,p(R2).

Therefore, it follows that k is an exceptional point, and thus we conclude the proof of(iv).

2.8 From t to γ

Looking at things from the perspective of our problem at hand, we have the Schrodingerequation with a potential q ∈ Lp(R), for p ∈ (1, 2), given by q = (∆γ1/2)/γ1/2 in case γ isin W 2,p(Ω) where Ω is the bounded domain for the conductivity equation. By construction,q = 0 in R2 \ Ω, hence q ∈ Lpρ for any ρ > 1.

So we can set ψ0 = γ1/2 in Ω and ψ0 = 1 outside of it. Hence, the conditions in Lemma2.4.1 are fulfilled and by straightforward computations we have the constant c∞ = 1. Sincewe also have ψ0 ∈ L∞ the equivalences follow. In this setting, all the results in previouschapters hold. From Theorem 2.4.2 and the ∂-equation (2.3.9) we have that for each x ∈ R2

the function µ(x, k) is a pseudo-analytic function of k.Accordingly, the setup we are creating holds for our current problem.

In this chapter we want to show that this function also fulfils the following properties:

(i) µ(x, ·)− 1 ∈ Lr(C), p′ < r ≤ ∞,

(ii) limk→0 µ(x, k) = γ1/2(x).

Still keeping the more generic setting we will show that given t(k) as in Theorem 2.4.2,there is for every x a unique solution of (2.3.9) satisfying (i). If the statement (ii) holds thenit follows that if two different conductivities γ1, γ2 have the same scattering transform t,thenγ1 = γ2.

Theorem 2.8.1. Let t(k) be such that t(k)/k ∈ Lr(C) for all r ∈ (p′, r2), r2 > 2.Fix r0 > max(p′, 2r2/(r2 − 2)). For all x ∈ R2 there is a unique solution µ(x, ·) of (2.3.9)

with µ(x, ·)− 1 ∈ Lr0 ∩ L∞(C). Moreover, infx,k |µ(x, k)| > 0,

supx‖µ(x, ·)− 1‖Lr <∞ for all r ∈ (p′,∞], and(2.8.1)

supx|µ(x, k1)− µ(x, k2)| ≤ c|k1 − k2|ε for 0 < ε < 1− 2/r2.(2.8.2)

If r2 is as given in Theorem 2.4.2 then it follows that 1 − 2/r2 = min((ρ − 1)/2, 2/p′).Hence, in (2.8.2) we have the same range for ε as in (2.4.5).

Before giving the proof of the Theorem above we present an auxiliary lemma which fulfilsthe same conditions and will be used during the proof.

57

Lemma 2.8.1. Let t(k) be such that t(k)/k ∈ Lr(C) for all r ∈ (p′, r2), r2 > 2 and fix

r0 > max(p′, 2r2/(r2 − 2)). Then we have that the operator I − ∂−1k

(e−x(k) t(k)

4πk·)

is a

Fredholm operator of index zero on Lr0(C). Moreover, the operator is in fact invertible in thisspace.

Proof. To facilitate notation, we define t#x = t(k)

4πke−x(k) and set K = ∂−1

k

(t#x ·)

.

The proof is based on the well-known result that if K is a compact operator in a Banachspace, then I −K is a Fredholm Operator of index zero. Therefore, we will show first thatK is in fact a compact operator. Moreover, since the composition of compact operators iscompact, during the compactness proof we assume K is defined without the conjugation.

Due to Schauder’s Theorem, an operator K is compact if and only if its dual operator iscompact. We use this, because the dual of K has the multiplication operator as the secondoperator in the composition which makes it easier to use the well-known result of Rellich-Kondrachov.

First, the dual space of Lr0(C), with r0 > max(p′, 2r2/(r2−2)), is the space Lr∗0 (C), where

1/r∗0 + 1/r0 = 1, and we have r∗0 < min(p, 2r2/(r2 + 2)) < 2. Now, we quickly determine thedual operator. Let g ∈ Lr∗0 (C) and f ∈ Lr0 then by Fubini we have:

〈Kf, g〉 =

∫C

(∂−1k

t#x f)

(k) g(k) dk ∧ dk =

∫C

(∫C

t#x (k′)f(k′)

k′ − kdk′ ∧ dk′

)g(k) dk ∧ dk

= −∫Cf(k′)t#

x (k′)

∫C

g(k)

k − k′dk ∧ dk dk′ ∧ dk′

= −∫Cf(k′)t#

x (k′)∂−1kg(k′) dk ∧ dk

= −⟨f, t#

x ∂−1k g

⟩.

Therefore, we have the dual operator being defined by:

−t#x ∂−1k

: Lr∗0 (C) −→ Lr

∗0 (C).(2.8.3)

By Lemma 2.2.1 and Holder inequality, with 1/r∗0 = 1/2 + 1/r∗0, the boundedness of thisoperator follows

‖t#x ∂−1kf‖

Lr∗0≤ ‖t#

x ‖L2‖∂−1kf‖

Lr∗0≤ c‖t#

x ‖L2‖f‖Lr∗0.(2.8.4)

To simplify the proof we will work with the operators

−t#x,n∂

−1k

: Lr∗0 → Lr

∗0 , for n ∈ N

where we take a sequence t#x,n ∈ C∞c (C) which converges to t#

x in L2 since the space of smoothand compactly supported functions is dense in Lp-spaces. Furthermore, for f ∈ Lr∗0 we havethat ∂k(∂

−1k

)f ∈ Lr∗0 and ∂k(∂−1k

)f ∈ Lr∗0 . This enables us to estimate ∇t#x,n∂

−1kf by ∂k and

∂k, separately, via

‖∂kt#x,n∂

−1kf‖

Lr∗0≤ ‖∂kt#

x,n‖L2‖∂−1kf‖

Lr∗0

+ ‖t#x,n‖L∞‖f‖Lr∗0 ≤ C‖f‖Lr∗0 .

58

Consequently, we have ‖t#x,n∂

−1kf‖

W 1,r∗0 ≤ C ′‖f‖Lr∗0

which implies t#x,n∂

−1kf is in W 1,r∗0 and

has compact support. By the use of Rellich-Kondrachov embedding theorem, we have theabove Sobolev space compactly embedded in Lq. This immediately gives compactness of theoperator t#

x,n∂−1k

in Lr∗0 .

Since the set of compact operators defined on a Banach space is closed in the algebra ofbounded linear operators we just need to show that the operator with t#

x,n converges to theoperator in (2.8.3). This actually follows by an application of the above inequality (2.8.4)and the density of C∞c -functions

‖t#x ∂−1kf − t#

x,n∂−1kf‖

Lr∗0c ≤ ‖t#

x − t#x,n‖L2‖f‖

Lr∗0

n→∞−−−→ 0.

Thus it follows that the dual operator is compact which then implies that the operator in

question K is also compact. This gives the first part of the result that I − ∂−1k

(e−x(k) t(k)

4πk·)

is Fredholm of index zero in Lr0 .To show invertibility is enough to show that the operator is injective. For that let ν ∈ Lr0

be in its null-space then ν = ∂−1k

(t#x ν). By Holder inequality it follows that t#

x ν ∈ Lrr0/(r+r0)

for all r ∈ (p′, r2). Taking a further look it follows

rr0/(r + r0) ∈ (s, r2),

where s ∈ (2/3, 2), because 1/p′ + 1/p′ ∈ (1/2, 3/2). Hence, by Lemma 2.2.3, we deduce thatν ∈ L∞(C) and vanishes at infinity.

Moreover, it is a pseudo-analytic function since it fulfils ∂kν = t#x . Hence, as before,

we can construct a continuous and bounded function w, such that v exp(−w) is holomorphic.Therefore, by Liouville’s theorem it follows that it is a constant c. Given that w, v are boundedand v vanishes at infinity then c = 0 which implies v = 0. Hence, the operator is invertiblein Lr0(C).

Proof of Theorem 2.8.1. We fix x ∈ R2 and r0 > max(p′, 2r2/(r2 − 2)) and define t#x (k) =

t(k)e−x(k)/4πk.To show uniqueness we use Liouville’s theorem in the same line of reasoning as in the

previous proofs. Suppose ν is a solution of (2.3.9) and it belongs to Lr0 ∩ L∞(C). By

t#x ∈ Lr(C), r ∈ (p′, r2), we have that t#

xνν ∈ L

p1 ∩ Lp2 , for 1 < p1 < 2 < p2. So by Lemma

2.2.3 it follows that w = ∂−1k

(t#xνν

)is a continuous and bounded function. Thus it follows

that νe−w ∈ Lr0 is holomorphic in k

∂k(νe−w

)= (∂kν) e−w − (∂kw)νe−w = t#

x (k)νe−w − t#x (k)

ν

ννe−w = 0.

Hence Liouville’s theorem gives that ve−w = 0 which, therefore, implies that v ≡ 0 anduniqueness follows.

For existence we consider the integral equation equivalent to (2.3.9) that is its solutionalso solves the differential equation. The integral equation is given by convolution with thefundamental solution of ∂k:

µ(x, k) = 1 +1

8π2i

∫C

t(k′)

(k′ − k)k′e−x(k′)barµ(x, k′) dk′ ∧ dk′.(2.8.5)

59

The integral equation translates simply to

µ(x, ·)− 1 = ∂−1k

(t#x µ).

For this equation we can obtain the solution in terms of the operator defined in Lemma 2.8.1.

µ(x, ·)− 1 = ∂−1k

(t#x µ)

⇔µ(x, ·)− 1 = ∂−1k

(t#x µ) + ∂−1

kt#x − ∂−1

kt#x

⇔(µ(x, ·)− 1)−[∂−1k

(t#x µ)− ∂−1

k(t#x )]

= ∂−1k

t#x

⇔[I − ∂−1

k(t#x · )

](µ(x, ·)− 1) = ∂−1

kt#x

⇔µ(x, ·)− 1 =[I − ∂−1

k(t#x · )

]−1 (∂−1k

t#x

).

The last equation holds because t#x ∈ Lr0 , where 1

r0= 1

r0− 1

2 and r0 being the Sobolev

conjugate of r0. Hence, we can use Lemma 2.2.1 to show that ∂−1k

t#x ∈ Lr0 .

[Note: r0 ∈ (p′, 2) because p′ < r0 ≤ ∞ ⇒ 1/2 ≤ 1/r0 < 1/2 + 1/p′ ⇒ 1/2 ≤ 1/r0 <1/p′ ⇒ p′ < r0 ≤ 2.]

Therefore, there exists a solution to the integral equation and, consequently, to the desireddifferential equation.

Now, we define a new function

v(x, ·) = ∂−1k

[t#x (µ(x, ·)− 1)/(µ(x, ·)− 1)

].

By using (2.2.3) we obtain that ‖v(x, ·)‖L∞ is bounded independently of x via (2.2.15)

and∥∥∥t#x (µ(x, ·)− 1)/(µ(x, ·)− 1)

∥∥∥Lr

=∥∥∥ t(·)· ∥∥∥Lr . Furthermore, we have

∂k(µ(x, k)e−v(x,k)) = (∂k(µ(x, k)− 1))e−v(x,k) − (µ(x, k)− 1)(∂kv(x, k))e−v(x,k) =

= t#x (k)(µ(x, k))e−v(x,k) − (µ(x, k)− 1)t#

x (k)µ(x, k)− 1

µ(x, k)− 1e−v(x,k)

= t#x (k)e−v(x,k).

Hence, we also have that µ(x, ·)−1 = ev(x,·)∂−1k

(t#x e−v(x,·)). By the boundedness of v(x, ·)

and by ∂k(t#x (k)e−v(x,k)) ∈ Lr0 for r0 > p′, through the same reasons as in the note above,

follows (2.8.1).

Due to e−v(x,·)t#x ∈ Lr(C), r ∈ (p′, r2), the statement (2.8.2) with p2 = r2 − ε > 2

and ε > 0, follows by using (2.2.16), the boundedness of v(x, ·) for any x, and the uniformboundedness of µ in x.

To conclude the proof we only need to show that |µ| is bounded below. To achieve thiswe define a new function

v(x, ·) = ∂−1k

(t#x µ(x, ·)/µ(x, k)).

Similarly to v, by Lemma 2.2.3, we have that v(x, ·) is bounded independently of x andlim|k|→0 v(x, k) = 0. Moreover, we have that for k 6= 0

∂k(µe−v) = t#

x µe−v − t#

x

µ

µµe−v = 0,

60

and, hence, µe−v is holomorphic for k 6= 0 and bounded on C by what we have done above.Hence, it is entire. Therefore, by Liouville’s theorem it must be a constant c. By (2.8.1) wehave that lim|k|→0 µ(x, k) = 1. Hence, by the limit of v, it follows that

lim|k|→0

µ(x, k)e−v(x,k) = 1⇒ c = 1⇒ µ(x, k) = ev(x,k).(2.8.6)

Furthermore, we have by Lemma 2.2.3: ‖v(x, ·)‖L∞ ≤ cr1,r3

(‖t#x ‖Lr1 + ‖t#x ‖Lr3

)with p′ <

r1 < 2 < r3 < r2. This implies −|v(x, k)| ≥ cr1,r3(‖t#x ‖Lr1 + ‖t#x ‖Lr3

)and thus it follows

|µ(x, k)| ≥ exp[−cr1,r3

(‖t#x ‖Lr1 + ‖t#x ‖Lr3

)].(2.8.7)

Remark:With this theorem the equation µ(x, k) = ev(x,k) gives a formula for γ1/2(x) = µ(x, 0) in

terms of t and the phase µ/µ of µ.

In the sense of reconstruction we are able to show a result which shows that if the scat-tering data fulfils an estimate similar to (2.4.4) then the potential which generates it is ofconductivity type.

Theorem 2.8.2. Let q be a real-valued function in Lpρ(R2), 1 < p < 2, ρ > 1.The following statements are equivalent:

(a) q = (∆ψ0)/ψ0 for some ψ0 ∈ L∞(R)2 with ψ0 ≥ c0 > 0 a.e.

(b) There are no exceptional points ζ ∈ C2 with ζ2 = 0, and the scattering transform satisfies

|t(−2ζR, ζ)| ≤ c|ζ|ε(2.8.8)

for some ε > 0 and all sufficiently small ζ = ζR + iζI in V.

Proof.

(a)⇒(b): Suppose (a) is true then by our hypothesis we have that q ∈ Lp(R2) and byLemma 2.4.1 (f) it follows that ∇ψ0 ∈ Lp. Therefore, we can use Lemma 2.2.4 which showsthat under these conditions there are no exceptional points. Moreover, Theorem 2.4.1 givesthe desired estimate (2.8.8) with ε < min((ρ− 1)/2, 2/p′).

(b)⇒(a): Assume q ∈ Lpρ is real-valued and has no exceptional points k in C. Followingthe proof of Theorem 2.2.1 we can construct µ(·, k) for all k ∈ C, satisfying (2.2.1) and(∆ + 4ik∂)µ = qµ. Furthermore, Theorem 2.3.1 holds, hence, we have that these solutionsfulfill the ∂-equation (2.3.9) in C\0 and, therefore, it follows that t(k) is continuous in C\0.

By using the above estimate for large k, the t-continuity away from zero and assumption(2.8.8) it follows, analogously to the proof of Theorem 2.4.2, that t(k)/k ∈ Lr(C) for allr ∈ (p′, rε), where rε = 2/(1− ε).

61

Now, we are in the conditions of Theorem 2.8.1. Thus we are able to define

µ0(x) = 1 +[∂−1k

(t#x µ(x, ·)

)] ∣∣∣k=0

.

Moreover, by estimate (2.8.1) it follows that µ0 ∈ L∞(R2), infx |µ0(x)| > 0 andsupx |µ0(x) − µ(x, k)| ≤ c|k|ε′ for any ε′ ≤ ε. The latter implies that in distributional senseit follows that (∆ + 4ik∂)µ(·, k) and qµ(·, k) converges to ∆µ0 and qµ0 as k goes to 0, re-spectively. Thus, also in a distributional sense ∆µ0 = qµ0. Henceforth, conditions (b)-(f) ofLemma 2.4.1 are satisfied by µ0 and given that it is lower bounded it follows by (b), c∞ 6= 0.In this sense, and to have normalization, we define ψ0(x) = µ0(x)/c∞.

To conclude, we just want to show that the obtain function ψ0 is real-valued. Indeed, letφ0 be its imaginary part. Then φ0 is a bounded solution of (−∆ + q)φ0 = 0 and thus satisfies(b)-(f) of Lemma (3.1). Since the limit in (b) for ψ0 equals 1, by the above normalization, iffollows that for φ0 the limit is 0. Thus by (d), we obtain φ0 ∈ W 1,p. It follows by definitionof exceptional points and due to k = 0 not being one that φ0 ≡ 0. Hence, ψ0 is real-valuedand, therefore, it has a positive lower bound.

62

Chapter 3

Complex Conductivities

3.1 Our ways

After the discussion of real conductivities in the previous chapter we are going to considerthe inverse conductivity problem for complex conductivities.

The principal question is still the same: Does the Dirichlet-to-Neumann map de-termine the conductivity uniquely?

Most of the results in this chapter were published in Pombo [2019]. Hereby, our principalassumption is that the conductivity function γ is somehow smooth except on a closed contourΓ ⊂ O. Let γ+ be the trace of γ from the exterior part of the Γ and γ− be the trace fromthe interior part. By D we denote the interior part of Γ.

Under our assumption on γ we look at solutions of the problem (1.1.1) which are smoothin each domain, u− ∈ D and u+ ∈ O \ D, and satisfy the following transmission condition atΓ,

(3.1.1)

u−(z)− u+(z) = 0,

γ− ∂u−

∂ν (z)− ∂u+

∂ν (z)γ+ = 0, z ∈ Γ.

Problem (1.1.1) together with the condition (3.1.1) is called an (interior) transmissionproblem.

Figure 3.1: Simple representation on a type ofdomain we can consider for the transmissionproblem. The inside boundary Γ represents thecontour where γ is not smooth.

The purpose of this approach is to establish a new method to overcome the limitation ofLipschitz conductivities in the current literature. In particular, we have in mind the handlingof cases where separation of tissues is an important issue, like in detection of nodules throughmedical imaging.

63

The reconstruction procedure of γ follows a different approach than the one presented inthe previous chapter. Instead of conversion of the conductivity equation to a Schrodingerequation we follow the observation made in Brown and Uhlmann [1997] and Francini [2000].Let u be a solution of (1.1.1) but on the domain Ω \ Γ satisfying the transmission conditionabove (3.1.1). Then, the pair

(3.1.2) φ = γ1/2(∂u, ∂u)t = γ1/2

(∂u∂u

)satisfies the Dirac equation

(3.1.3)

(∂ 00 ∂

)φ = qφ, z = x+ iy ∈ C \ Γ,

with

q(z) =

(0 q12(z)

q21(z) 0

), q12 = −1

2∂ log γ, q21 = −1

2∂ log γ,(3.1.4)

where we extend γ to the outside of O by setting γ = 1. On Γ, the pair φ satisfies atransmission condition for the Dirac equation which is derived from the previous one.

Thus, it is enough to solve the inverse Dirac scattering problem instead of the ICP. If itis solvable and q can be found then the conductivity γ is immediately obtained from (3.1.4),up to a constant.

In order to complete the reduction of the ICP to the inverse Dirac problem one needsto show that the scattering data for the Dirac equation can be found via the Cauchy data(u|∂O , ∂u∂ν |∂O

). In fact, the scattering data for the Dirac equation can be obtained by simple

integration of its Dirichlet data, see formula (3.3.13).We assume that log γ is well defined in the whole complex plane, that is, there exists a

ray that does not intersect the range of γ or the real part of the conductivity is positive lowerbounded.

We have to remark that, in fact, we are going to present only a partial result, given thatwe cannot yet reconstruct, and show uniqueness of, the potential q in the whole of C \ Γ.Our proof is based on a new concept, that is based on a specific set of points, which we nowdefine:

Definition 3.1. We say that a point w is an admissible point if there is a number λO suchthat

A := supz∈O<[λO(z − w)2] < 1/2

B := supz∈D<[λO(z − w)2] < −1/2,

Moreover, if w is an admissible point and the constants A and B fulfillsA = 1/2−ε1, B = −1/2−ε2, with ε2−ε1 > 0, we further say that w is a proper admissible

point.

64

The main theorem of this chapter will be obtained using this novel idea. Even though theproof follows by reconstruction, we give here the uniqueness theorem without introducing thescattering data first.

Theorem 3.1.1. Let Ω be a bounded Lipschitz domain in the plane, and let γ ∈W 2,∞(D) ∩W 2,∞(O \ D). If

√γ−

γ+ − 1 is small enough on Γ (in the pointwise sense), we have that the

Dirichlet-to-Neumann map Λγ determines the conductivity γ uniquely in any proper admis-sible point.

Hereby, we want to point out that the Dirichlet-to-Neumann map determines the scatter-ing data uniquely can be proven similarly to Lakshtanov and Vainberg [2016] (see Section 3.4).

As mentioned, Theorem 3.3.1 will provide a reconstruction formula for the potential qin so-called proper admissible points. This is an improvement of previous existent methodsinsofar as a convenient enlargement of the set of CGO incident waves allows to highlight thedesirable areas around such points. This approach provides a 2D reconstruction result which,although being apparently a rather weak result, cannot possibly be obtained by any previoustechnique and represents a first step in this direction. As the methods for 2D and 3D arequite different even at the level of Faddeev Green function analysis, we focus this analysis onthe 2D case only.

The main tools developed in the paper to facilitate further study include: the right choiceof the function space, a set of admissible points (essential to the reconstruction), and theenrichment of the set of CGO incident waves (i.e. we use solutions like |λ|f(z) which highlightdesirable areas). The latter solutions are unlimited even after the CGO-Faddeev normaliza-tion and we are required to obtain many-dimensional Laplace Transform analogues of theHausdorff-Young inequality.

This chapter has a simple organization: In Section 2 we recall the necessary facts on thetransmission condition and the construction of the Lippmann-Schwinger equation for CGO-Faddeev solutions in our case. In Section 3 we introduce the necessary function spaces aswell as related lemmas. We present the novel concept of admissible points (see Definition3.1) based on a convenient enrichment of the set of CGO incident waves and we study thescattering data and reconstruction of the potential in these type of points. We finalize thissection with two subsections containing some more necessary results and the proof of ourmain theorem. For the sake of readability we placed some additional results together withtheir proofs in an appendix.

3.2 Main construction

3.2.1 Transmission condition

We denote by ν(z) = (nx, ny) = nx + iny the mean unit normal vector in Γ.

Lemma 3.2.1. The transmission condition (3.1.1) is equivalent to

(3.2.1)

(φ+

1 − φ−1

φ+2 − φ

−2

)=

1

2

(α+ 1

α − 2 (α− 1α)ν2

(α− 1α)ν2 α+ 1

α − 2

)(φ−1φ−2

)

65

where α =√

γ−

γ+ .

Proof. From the first equation of (3.1.1) follows for the tangential derivative that ∂∂l (u

+(z)−u−(z)) = 0 and, therefore, √

γ+u+l −

√γ−u−l = u−l

√γ−(

1

α− 1),

where ul = ∂u∂l . From the second equation of (3.1.1) we get u+

n = γ−

γ+u−n , where u±n denotes

the normal derivative of u±, so that√γ+u+

n −√γ−u−n =

√γ−u−n (α− 1).

Note that we have now

∂u =1

2(νun − iνul),(3.2.2)

∂u =1

2(νun + iνul),(3.2.3)

φ+1 − φ

−1 =

√γ+∂u+ −

√γ−∂u− =

(1

α− 1

)u−l√γ−

1

2(−iν) + (α− 1)u−n

√γ−

1

2ν,

φ+2 − φ

−2 =

√γ+∂u+ −

√γ−∂u− =

(1

α− 1

)u−l√γ−

1

2(iν) + (α− 1)u−n

√γ−

1

2ν.

These relations take the matricial form(φ+

1 − φ−1

φ+2 − φ

−2

)=

1

2

((α− 1)ν ( 1

α − 1)(−iν)(α− 1)ν ( 1

α − 1)(iν)

)(u−n√γ−

u−l√γ−

).

Using (3.2.2) and (3.2.3), together with the definition of φ, we obtain the relation(u−n√γ−

u−l√γ−

)=

(ν νiν −iν

)(φ−1φ−2

)we allows to derive the result.

Lemma 3.2.2. Consider any functions u+ ∈ Hp(O \ D) and u− ∈ Hp(D), with p > 3/2,and let vector-functions, (φ+

1 , φ+2 ) and (φ−1 , φ

−2 ), be defined by formula (3.1.2). Suppose that

the boundary relation (3.2.1) holds at Γ. Then the following relations hold.∫Γ+

∂u+dz +

∫Γ+

∂u+dz = 0.

Proof. Let us show thatν2dz = −dz

. If (lx, ly) is a tangent vector, then n = σ(−ly, lx), σ2 = 1, so that σν = −ly+ilx = i(lx+ily).Now, remark that dz = (lx − ily)dt and

ν2dz = i2(lx + ily)|(lx + ily)|2dt = −dz.

Hence, the proof follows straightforward from (3.2.1).

66

3.2.2 The Lippmann-Schwinger equation for CGO-Faddeev solutions

Consider the vector φ which satisfies (3.1.3) and the following asymptotic

(3.2.4) φ1 = eλ(z−w)2/4U(z) + eλ(z−w)2/4o(1), φ2 = eλ(z−w)2/4o(1), z →∞,

where U(z) is entire and can depend on the parameter λ.We denote

(3.2.5) µ1 = φ1e−λ(z−w)2/4, µ2 = φ2e

−λ(z−w)2/4

so that we get the following integral equation for φ:

(3.2.6) (I + PAλ −DQλ)µ =

(U0

),

where D =

(∂−1 0

0 ∂−1

), the matrices Aλ and Qλ have the following form

Aλ =1

2

(α+ 1

α − 2 (α− 1α)ν2e−i=[λ(z−w)2/2]

(α− 1α)ν2ei=[λ(z−w)2/2] α+ 1

α − 2

),

Qλ =1

2

(0 Q12e

−i=[λ(z−w)2/2]

Q21ei=[λ(z−w)2/2] 0

),

where Q12, Q21 are L∞ extensions of q12, q21 to Γ. Moreover, P is a projector

(3.2.7) P =

(P+ 00 P−

),

where P+, P− are the Cauchy projector and its complex adjoint, respectively:

P+f =1

2πi

∫Γ

f(z)dz

z − w, P−f =

1

2πi

∫Γ

f(z) dz

z − w, w ∈ C.

Hereby, f represents the trace values of f taken from the interior of Γ.

Proof of (3.2.6). We use the same approach as in Lakshtanov et al. [2016]. The followingCauchy-Green formulas hold for each f ∈ C1(Ω) and an arbitrary bounded domain Ω withsmooth boundary:

f(z) = − 1

π

∫Ω

∂f(ς)

∂ς

dςRdςIς − z

+1

2πi

∫∂Ω

f(ς)

ς − zdς, z ∈ Ω,(3.2.8)

0 = − 1

π

∫Ω

∂f(ς)

∂ς

dςRdςIς − z

+1

2πi

∫∂Ω

f(ς)

ς − zdς, z 6∈ Ω.(3.2.9)

Denote by DR a disk of radius R and centered at z, and take D−R = DR \D. We recall, Dis the interior part of Γ. Assume that z ∈ D, f = φ1 in both formulas, and Ω = D in (3.2.8)and Ω = D−R in (3.2.9). We add the left- and right-hand sides in formulas (3.2.8) and (3.2.9).If we take (3.2.1) into account, we obtain that

(3.2.10) φ1(z, λ) = − 1

π

∫C

(Qλφ)1(z, ς)dςRdςIς − z

+1

2πi

∫Γ−

[φ](ς)

ς − zdς +

1

2πi

∫∂DR

φ1(ς)

ς − zdς,

where [φ] = φ− − φ+. It remains to note that the last term on the right converges to U asR→∞ due to (3.2.4). A simple reordering gives now (3.2.6).

67

3.3 Technical details

3.3.1 The choice of the functional space

Let 1 < p <∞ and define the space Hp1 as the intersection of L∞z (Lpλ(|λ| > R)) with thespace of continuous functions L∞z (L∞λ (|λ| > R)). Also, define the space Hp2 of all functionsPf such that (Pf)|Γ− = f, where P is the projector (3.2.7), endowed with the norm

‖Pf‖pHp2 =

∫|λ|>R

∫Γ|f(z, λ)|p d|z|dλ.

The space Hp = Hp1 +Hp2 is endowed with the norm

(3.3.1) ‖t‖Hp = infu∈Hp1,v∈H

p2,u+v=t

max(‖u‖Hp1 , ‖v‖Hp2).

Let us remind that the operations of intersection and union of two Banach spaces are correctlydefined if all terms can be continuously embedded into a common locally convex space. Inour situation this common locally convex space will be a space endowed with the semi-norms∫

|λ|>R

∫z∈O

1

|λ|2|f(z, λ)| dσzdλ.

If f ∈ Hp1 the embedding is evident. For f ∈ Hp2 we have

‖Pf‖Lp(O) ≤ C‖f‖Lp(Γ),

so that[‖Pf‖Lp(O)

]p ≤ [‖f‖Lp(Γ)

]pand∫ (∫

O1

|Pf |pdσz)dλ ≤

∫[‖f‖Lp(Γ)]

pdλ =

∫|λ|>R

∫Γ|f(z, λ)|pdλd|z|.

The boundedness of each semi-norm follows from the continuity of the embedding of Lp(O)into L1(O).

Lemma 3.3.1. The operators P± : f → (Pf)|Γ± are bounded in the space with norm∫|λ|>R

∫Γ|f(z, λ)|p dzdλ

.

Proof. During the proof the sign ± in the projectors will be omitted. From the continuity ofCauchy projectors in Lp(Γ) follows

‖P f‖Lp(Γ) ≤ C‖f‖Lp(Γ).

and therefore (‖P f‖Lp(Γ)

)p≤ Cp

(‖f‖Lp(Γ)

)p.

Finally

‖PPf‖Hp2 =

∫|λ|>R

(‖P f‖Lp(Γ)

)pdλ ≤ Cp

∫|λ|>R

(‖f‖Lp(Γ)

)pdλ = Cp

∫|λ|>R

∫Γ

|f(z, λ)|pdλd|z|.

68

Lemma 3.3.2. Let u ∈ Hp1. Then P (u|Γ) ∈ Hp2.

Proof. From the definition of Hp1, combined with the fact that u is a continuous function, weget

‖u‖Lpλ ∈ L∞z (Γ).

Since Γ is a bounded set, the Lp norm does not exceed (up to a constant) the L∞ norm and,therefore

‖‖u‖Lpλ‖Lpz(Γ) ≤ C‖u‖Hp1 .

Now we just note the left-hand side of the above inequality is the norm Hp2 norm.

.

3.3.2 Analysis of the Lippmann-Schwinger equation

Multiplying equation (3.2.6) by I +DQλ we get

(3.3.2) (I +M)µ = (I +DQλ)

(U0

)where

(3.3.3) M = PAλ +DQλPAλ −DQλDQλ.

Lemma 3.3.3. If the jump α−1 is small, then the operators DQλPAλ, DQλDQλ are boundedin Hp, p > 1. Moreover, if R > 0 is large enough they are contractions.

Proof. In order to estimate ‖(DQλPAλ)t‖Hp and ‖(DQλDQλ)t‖Hp (recall to Definition 3.3.1)we consider the representation t = u + v where the infimum is (almost) achieved. It is easyto see that the desirable estimate follows from the fact that these operators are a contractionin each of the spaces, H1 and H2. This fact can be shown as follows.

In Lakshtanov et al. [2017] it was proved that the operator DQλDQλ is bounded in Hp1.

The statement for the DQλPAλ can be proved in a similar way:

DQλPu =

∫ΓA(z, z2, λ, w)u(z2) dσz2

where

(3.3.4) A(z, z2, λ, w) = π−2

∫O

e−i=[λ(z−w)2]/2

z − z1Q12(z1)

1

z1 − z2dσz1 ,

andsupz‖A(z, ·, λ, w)‖Lqz2 (Γ) = o(1), λ→∞, 1/p+ 1/q = 1

so that|DQλPu|(z) ≤ ‖A(z, ·, λ, w)‖Lqz2 (Γ)‖u‖Lpz2 (Γ).

Then, we have‖DQλPu‖Lpλ ≤ ‖A(z, ·, λ, w)‖Lqz2 (Γ)‖u‖Hp2

where we used the fact that u ∈ Hp2 is the same as ‖u‖Lpz2 (Γ) ∈ Lpλ.

69

3.3.3 Enrichment of the set of CGO incident waves

Let w ∈ O be a fixed point. We are going to consider functions of the type

(3.3.5) U = eln |λ|λO(z−w)2, z ∈ R2,

where λO is a parameter.These functions lead us to the concept of admissible points. We recall here their definition:

Definition 3.2. We say that a point w is an admissible point, if there is a number λO suchthat

A := supz∈O<[λO(z − w)2] < 1/2

B := supz∈D<[λO(z − w)2] < −1/2,

Moreover, if w is an admissible point and A and B fulfills A = 1/2− ε1, B = −1/2− ε2, withε2 − ε1 > 0, we further say that w is a proper admissible point.

Note: The set of admissible points is not empty. In order to see this we consider aboundary point w0 ∈ ∂O which belongs also to the convex hull of O. It is easy to see thatall interior points w ∈ O near the w0 would be admissible.

We will not try to give a general geometric description of admissible points. Instead, weare only aiming to show the viability of the concept.

Denote

(3.3.6) f = µ− (I +DQλ)

(U0

),

where µ is defined in (3.2.5).The vector f satisfies the equation

(3.3.7) (I +M)f = −M(I +DQλ)

(U0

).

We know already that for R > 0 large enough the operator in the left-hand side of thisequation is a contraction in Hp, p > 1 and below we show that in fact we have for theright-hand side:

(3.3.8)1

|λ|AM(I +DQλ)

(U0

)∈ Hp, p > 2.

Therefore, we get the following statement

Lemma 3.3.4. For any p > 2 and R large enough such that U is given in terms of a properadmissible point w we have

(3.3.9)1

|λ|A

[µ− (I +DQλ)

(U0

)]∈ Hp.

70

Proof. We start by showing that

(3.3.10)1

|λ|A

(U0

)∈ Hp2

and1

|λ|AMDQλ

(U0

)∈ Hp1.

Since M is a contraction for R > 0 big enough we are going to obtain (3.3.8) and the resultwill immediately follows for p > 2.

To show (3.3.10) we refer to the following simple estimate[∫|λ|>R

∫Γ

∣∣∣∣∣ 1

|λ|Aeln |λ|λs(z−w)2

∣∣∣∣∣ d|z| dσλ]1/p

≤

[∫|λ|>R

∫Γ

∣∣∣∣∣ 1

|λ|A|λ|B

∣∣∣∣∣p

d|z| dσλ

]1/p

=

[∫|λ|>R

∫Γ

∣∣∣∣∣ 1

|λ|1+(ε2−ε1)

∣∣∣∣∣p

d|z| dσλ

]1/p

<∞

For the second statement we need to dismantle M into its various parts and show thatthe statement holds for each one of them. The trick is always the same, so we will only show

one of the computations, namely the one corresponding to the term 1|λ|A (DQλ)3

(U0

). By

Lemma 3.3.5 we get

supz∈O

[∫|λ|>R

∣∣∣∣∣ 1

|λ|A

∫O

ei=(λ(z1−w)2)

z1 − zQ21(z1)

∫O

e−i=(λ(z2−w)2)

z2 − z1Q12(z2)·

·∫O

eρ(z3)

z3 − z2Q21(z3) dσz3 dσz2 dσz1

∣∣∣∣∣p

dσλ

]1/p

≤ supz∈O

[C||Q||3L∞

∫O

∫O

1

|z1 − z|1

|z2 − z1|1

|z2 − w|1−δdσz2 dσz1

]< C ′

Thus, the result (3.3.8) follows, and in consequence also (3.3.9) holds from (3.3.6) and(3.3.7).

3.3.4 Scattering data and reconstruction of the potential in admissiblepoints

We consider the function

(3.3.11) eln |λ|λs(z−w)2 ,

where the number λs is chosen such that

(3.3.12) supz∈O<[λs(z − w)2] < 1/2, sup

z∈D<[λs(z − w)2] < −1/2,

71

We could also choose λs equal to λO here but this is not necessary. Nevertheless, in all theproofs ahead we assume λO = λs.

Consider now our scattering data

(3.3.13) h(λ,w) =

∫∂Oeln |λ|λs(z−w)2µ2(z)dz.

Using Green’s formula ∫∂Of dz = −2i

∫O∂f dσz

we can see that(3.3.14)

h(λ,w) =

∫Γ+

eln |λ|λs(z−w)2µ2(z)dz +

∫O\D

eln |λ|λs(z−w)2e−i=[λ(z−w)2]/2q21(z)µ1(z)dσz.

This formula gives raise to an operator that we denote by T and it is defined by

T [G](λ) =

∫O\D

eln |λ|λs(z−w)2e−i=[λ(z−w)2]/2q21(z)G(z)dσz.

From our representation for the solution µ (3.3.6) and the fact that the matrix Q isoff-diagonal we get

T

[((I +DQλ)

(U0

))1

]= T [U ].

This allows us to state our main theorem.

Theorem 3.3.1. Let the potential q be in W 1,∞(D) ∩W 1,∞(O \ D). If the jump α − 1 issmall enough and w is a proper admissible point for a spectral parameter λs, then

(3.3.15)λs

4π2 ln 2limR→∞

∫R<|λ|<2R

|λ|−1 h(λ,w) dσλ = q21(w).

The proof of this theorem requires some additional results concerning the behavior ofh(λ,w)/|λ|. These results will be given in the form of three lemmas which we establish in thenext section.

3.3.5 Necessary results for the proof of Theorem 3.3.1

We start by presenting a result which we need afterwards. For its proof we refer toAppendix A. Consider an arbitrary number λ0 ∈ C, denote ρ(z) = −i=[λ(z − w)2]/2 +ln |λ|λ0(z − w)2, and let A0 = supz∈O <[λ0(z − w)2].

Lemma 3.3.5. Let z1, w ∈ C, p > 2, and ϕ ∈ L∞comp. Then∥∥∥∥∥ 1

|λ|A0

∫Cϕ(z)

eρ(z)

z − z1dσz

∥∥∥∥∥Lpλ(C)

≤ C ‖ϕ‖L∞|z1 − w|1−δ

,

where the constant C depends only on the support of ϕ and on δ = δ(p) > 0.

72

To study the main term in (3.3.15), we have the following lemma.

Lemma 3.3.6. Let ϕ ∈W 1,∞(O), and supp(ϕ) ⊂ O, where O is a domain in R2 such that

(3.3.16) supz∈O<(z − w)2 < 1.

Then the following asymptotic holds

(3.3.17)

∫e−i=(λ(z−w)2)+ln |λ|(z−w)2

ϕ(z)dσz =2π

|λ|ϕ(w) +Rw(λ),

where |λ|−1Rw ∈ L1(λ : |λ| > R).

Proof. Consider two domains

I1 = z ∈ O : |z − w| < 3ε, I2 = z ∈ O : |z − w| > ε,

where ε > 0 is an a priori chosen arbitrarily small but fixed number. Furthermore, we picktwo functions δ1 and δ2 with supports I1 and I2, respectively, such that δ1 + δ2 ≡ 1 in O.Moreover, we assume that δ1(z − w) is represented as a product of δ1(x)δ1(y) and that thefunction δ1(x) decreases monotonically as |x| grows.

The integrand is multiplied by (δ1 + δ2) and this naturally splits the integral into twoterms. The term corresponding to I2 can be integrated by parts once and then the requiredestimate follows from the Hausdorff-Young inequality (A.1.1) for p = q = 2. We also use herethe fact that the estimate (3.3.16) is sharp.

Now, we consider the term corresponding to integration in I1. This term will be dividedinto two parts as well correspondent to the representation

δ1(z)ϕ(z) = δ1(z)ϕ(w) + δ1(z)(ϕ(z)− ϕ(w))

The second part can be treated as in Lemma 3.3.5, i.e., one has to realize the change ofvariables and this leads to an expression similar to (A.2.1) wich, after a change of variablesgenerates a singularity in |u| of total order |u|1/2 - since φ ∈W 1,∞ and has compact support,so that |φ(z) − φ(w)| is of order |u|1/2. Then, one has to integrate by parts again and thesingularity will be of the order |u|3/2 so that one can apply Hausdorff-Young Lemma for theLaplace transform for p = 4/3 and get the required estimate.

Consider the change of variables y =√|λ|(z − w). Due to the separation of variables in

δ1 the asymptotic of

(3.3.18)ϕ(w)

|λ|

∫e−i=y2+

ln |λ||λ| y

2

δ1

(∣∣∣∣∣w +y√|λ|

∣∣∣∣∣)dσy

follows from the formula

(3.3.19)

∫ λ1/2δ

0e−ix2+

ln |λ||λ| x

2

δ1

(|x|√|λ|

)dx =

1√2π

(1 + o(1)), λ→∞.

This can be proven in the following way: consider the change of variables

x2 = t, g(t) := δ1

(|x(t)|√|λ|

)

73

then, we have ∫ λ1δ2

0e−it+ ln |λ|

|λ| t1√tg(t)dt =

∫ 1

0e−it+ ln |λ|

|λ| t1√tg(t)dt+

1

−i+ ln |λ||λ|

∫ λδ2

1

(e−it+ ln |λ|

|λ| t)′ 1√

tg(t)dt.

For the second term we obtain

1

−i+ ln |λ||λ|

∫ λδ2

1

(e−it+ ln |λ|

|λ| t)′ 1√

tg(t)dt =

1

−i+ ln |λ||λ|

(e−it+ ln |λ|

|λ| t)g(t)√t

∣∣∣∣∣∣λδ2

1

+

1

−i+ ln |λ||λ|

∫ λδ2

1e−it+ ln |λ|

|λ| t1

2t3/2g(t)dt =

−1

−i(e−i) g(1)√

1+

1

−i

∫ λδ2

1e−it

(g(t)

2t1/2

)′dt+ o(1), λ→∞.

We used here the fact that the last integral is absolutely convergent (g has a finite support)and

(3.3.20) supz∈I1

∣∣∣∣e ln |λ||λ| y

2

− 1

∣∣∣∣ = o(1), λ→∞.

Therefore, we get∫ λδ2

0e−ix2+

ln |λ||λ| x

2

f(x)dx =

∫ λδ2

0e−it

1√tg(t)dt =

∫ ∞0

e−it1√tg(t)dt+ o(1).

Now the result of our lemma is an immediate consequence of this formula.

The following two lemmas assure that the remaining terms in (3.3.15) is integrable and,therefore, their impact vanishes.

Lemma 3.3.7. For some p < 2, with R large enough and f defined as in (3.3.6), we get

(3.3.21)1

|λ|T

[M(I +DQλ)

(U0

)]∈ Lp(λ : |λ| > R),

and

(3.3.22)1

|λ|T [Mf ] ∈ Lp(λ : |λ| > R).

Proof. Given the structure of M = PAλ+DQλ−DQλDQλ and that 1|λ|T is a linear operator,

it is enough to show that each term applied to both,

(U0

)and DQλ

(U0

), belongs to Lp(λ :

|λ| > R).We look directly at the computations of each term. By using Fubini’s Theorem, Minkowski

integral inequality, Holder inequality, and Lemma 3.3.5 we can show that all of these terms

74

are in fact in Lp(λ : |λ| > R). Since the computations for each term follow roughly the samelines, and for the convenience of the reader, we present just the computation in one of thesecases, the computations of the remaining terms being analogous, with special attention to theconvergence of the integrals.

We look at the term

1

|λ|T

[DQλDQλ

(U0

)]∈ Lp(λ : |λ| > R).

Let us denote ρ(z) = i=[λ(z−w)2]/2+ln |λ|λs(z−w)2 and A = S = supz∈O <[λ(z−w)2] <1/2.

∥∥∥∥∥ 1

|λ|T

[DQλDQλ

(U0

)]∥∥∥∥∥Lp(λ:|λ|>R)

=

=

[∫|λ|>R

∣∣∣∣∣ 1

4π2|λ|

∫O\D

eρ(z)q21(z)

∫O

e−i=[λ(z1−w)2]/2

z1 − zQ12(z1)

∫O

eρ(z2)

z2 − z1Q21(z2) dσz2 dσz1 dσz

∣∣∣∣∣p

dσλ

]1/p

=

[∫|λ|>R

∣∣∣∣∣ 1

4π2|λ|

∫O

(∫O\D

eρ(z)

z1 − zq21(z) dσz

)(∫O

eρ(z2)

z2 − z1Q21(z2) dσz2

)·

·Q12(z1)e−i=[λ(z1−w)2]/2 dσz1

∣∣∣∣∣p

dσλ

]1/p

≤∫O

[∫|λ|>R

∣∣∣∣∣ |λ|A+S

|λ|

(1

|λ|A

∫O\D

eρ(z)

z1 − zq21(z) dσz

)(1

|λ|S

∫O

eρ(z2)

z2 − z1Q21(z2) dσz2

)∣∣∣∣∣p

dσλ

]1/p

|Q12(z1)| dσz1

≤ ‖Q‖L∞

∫O

∣∣∣∣∣∣∣∣∣∣ 1

|λ|A

∫O\D

eρ(z)

z1 − zq21(z) dσz

∣∣∣∣∣∣∣∣∣∣L2p(λ:|λ|>R)

∣∣∣∣∣∣∣∣∣∣ 1

|λ|S

∫O

eρ(z2)

z2 − z1Q21(z2) dσz2

∣∣∣∣∣∣∣∣∣∣L2p(λ:|λ|>R)

dσz1

≤ C‖Q‖L∞

∫O

1

|z1 − w|1−δ1

|z1 − w|1−δdσz1 <∞.

With these calculations we obtain (3.3.21). To show (3.3.22) we have that 1|λ|A f ∈ H

p, for

p > 2, by Lemma 3.3.4. We consider T applied to each term of M . Again, we present only thecomputations for the case 1

|λ|T [DQλDQλf ], since the other computations are analogous, with

special attention to the behavior of 1|λ|A f . In the same spirit, we only present the calculation

for the first term of the vector.

[∫|λ|>R

∣∣∣∣∣ 1

|λ|

∫O\D

eρ(z)Q21(z)

∫O

e−i=(λ(z1−w)2)

z1 − zQ12(z1)·

·∫O

ei=(λ(z−w)2)

z2 − z1Q21(z2)f1(z2) dσz2 dσz1 dσz

∣∣∣∣∣p

dσλ

]1/p

≤ C‖Q‖3L∞∫O

∫O

1

|z2 − z1|1

|z1 − w|1−δ

∥∥∥∥ 1

|λ|Af1(z2)

∥∥∥∥L2pλ

dσz2 dσz1 <∞

The boundedness of the last integral follows from the fact that 1|λ|A f ∈ H

p implies its

boundedness with respect to the z variable.

75

Lemma 3.3.8. For R large enough, and w being a proper admissible point, we have

1

|λ|

∫Γ+

eln |λ|λs(z−w)2µ2(z)dz ∈ L1(|λ| > R).

Proof. We divide the integral

(3.3.23)1

|λ|

∫Γ+

eln |λ|λs(z−w)2µ2(z)dz,

into two pieces, according to the decomposition of µ2 given by formula (3.3.6), that is

(3.3.24) µ2 =

[DQλ

(U0

)]2

+ f2.

By Lemma 3.3.4 we have that 1|λ|A f ∈ H

p, for any p > 2. Therefore, we apply (3.3.24) to

(3.3.23) and we split the integral into I1 and I2, according to the order in (3.3.24).Since, by assumption, w is an admissible point there exists a λs fulfilling the inequality

supz∈D <[λs(z − w)2] < −1/2.So, for z ∈ Γ+ we get∣∣∣|λ|Aeln |λ|λs(z−w)2

∣∣∣ = |λ|A|eln |λ|<[λs(z−w)2]| < |λ|Ae−1/2 ln |λ| = |λ|A−1/2∣∣∣|λ|Aeln |λ|λs(z−w)2∣∣∣ < |λ|−δ,(3.3.25)

where we choose −δ = A− 1/2 < 0 (recall, A < 1/2). Hence, we obtain

|I2| ≤1

|λ|

∫Γ+

∣∣∣∣|λ|Aeln |λ|λs(z−w)2

(1

|λ|Af2

)∣∣∣∣ d|z|<

1

|λ|1+δ

∫Γ+

∣∣∣∣ 1

|λ|Af2

∣∣∣∣ d|z|.Integrating with respect to the spectral parameter, we have for R > 0 large enough∫

|λ|>R|I2|dσλ ≤

∫|λ|>R

1

|λ|1+δ

∫Γ+

∣∣∣∣ 1

|λ|Af2

∣∣∣∣ d|z|dσλ≤∥∥∥∥ 1

|λ|1+δ

∥∥∥∥Lqλ

∥∥∥∥∫Γ+

∣∣∣∣ 1

|λ|Af2

∣∣∣∣ d|z|∥∥∥∥Lpλ

.

Therefore, by Lemma 3.3.4 the second norm is finite for p > 2. We now pick q such thatq(1 + δ) > 2, which is always possible given that δ > 0. Hence, I2 is in L1(λ : |λ| > R).

Now, we look at I1. By definition we have

(3.3.26) I1 =1

2|λ|

∫Γ+

eln |λ|λs(z−w)2

∫O

eln |λ|λs(z1−w)2+i=(λ(z1−w)2)/2

z − z1Q21(z1) dσz1 dz.

Again, integrating against the spectral parameter we get:∫|λ|>R

|I1|dσλ ≤∫|λ|>R

1

2|λ|

∫Γ+

|λ|−δ∣∣∣∣∣ 1

|λ|A

∫O

eln |λ|λs(z1−w)2+i=(λ(z1−w)2)/2

z − z1Q21(z1)dσz1

∣∣∣∣∣ d|z| dσλ=

∫Γ+

∫|λ|>R

1

2|λ|1+δ

∣∣∣∣∣ 1

|λ|A

∫O

eln |λ|λs(z1−w)2+i=(λ(z1−w)2)/2

z − z1Q21(z1)dσz1

∣∣∣∣∣ dσλ d|z≤

∣∣∣∣∣∣∣∣∣∣ 1

2|λ|1+δ

∣∣∣∣∣∣∣∣∣∣Lqλ

∫Γ+

∣∣∣∣∣∣∣∣∣∣ 1

|λ|A

∫O

eln |λ|λs(z1−w)2+i=(λ(z1−w)2)/2

z − z1Q21(z1)dσz1

∣∣∣∣∣∣∣∣∣∣Lpλ

d|z|,

76

where we use Fubini’s theorem and Holder inequality, with p > 2 small enough so that thefirst norm is finite as in the computation of I2.

Now, we can use Lemma 3.3.5, given that we assume that our potential Q has support inO and it is in L∞z , to obtain a constant C > 0 depending only on the support of the potentialand on a certain δ > 0:∫

|λ|>R|I1|dσλ ≤ C

∥∥∥∥ 1

2|λ|1+δ

∥∥∥∥Lqλ

‖Q21‖L∞z∫

Γ+

1

|z − w|1−δd|z|.

Given that the last integral is finite, we have I1 ∈ L1(λ : |λ| > R) and the desired resultfollows.

3.3.6 Proof of Theorem 3.3.1

Now we can present the proof of our main theorem, using the lemmas of the previoussection while paying close attention to how µ and f are defined.

Proof. Let us start by taking a look at the following term

h(λ,w)

|λ|=

1

|λ|

[∫Γ+

eln |λ|λs(z−w)2µ2(z)dz

+

∫O\D

eln|λ|λs(z−w)2e−i=(λ(z−w)2)/2q21(z)µ1(z)dσz

].(3.3.27)

From (3.3.6) we have

µ = f + (I +DQλ)

(U0

),

whereby f is a solution of

f = −

(Mf +M(I +DQλ)

(U0

)).

This leads to µ1 = −[Mf +M(I +DQλ)

(U0

)]1

+ U. Therefore, by (3.3.27) and the

definition of the operator T , we get

h(λ,w)

|λ|=

1

|λ|

∫Γ+

eln |λ|λs(z−w)2µ2(z)dz − 1

|λ|T

([Mf

]1

)− 1

|λ|T

([M(I +DQλ)

(U0

)]1

)+

1

|λ|T [U ] =: A+B + C +D.(3.3.28)

We need to study the terms A, B, C, D. By Lemma 3.3.7, we have for p < 2 and R largeenough that:

B,C ∈ Lp(λ : |λ| > R).

From Lemma 3.3.8, we obtain

A ∈ L1(λ : |λ| > R).

77

Hence, we just need to analyze the behavior of the last term.

T [U ] =

∫O\D

eln|λ|λs(z−w)2e−i=(λ(z−w)2)/2q21(z)eln |λ|λs(z−w)2

dσz

=

∫O\D

eln|λ|(√λsz−

√λsw)2

e−i=(λ(z−w)2)/2q21(z)eln |λ|(√λsz−

√λsw)2

dσz

=1

λs

∫O\D

eln|λ|(z−√λsw)2

e−i=(λ/√λs(z−

√λsw)2)/2q21(z)

eln |λ|(√λsz−

√λsw)2

e−i=(λ(z−√λsw)2)ei=(λ(z−

√λsw)2)dσz,

where we did a simple change of variables. We define

φ(z) = e−i=(λ/λs(z−√λsw)2)/2ei=(λ(z−

√λsw)2)eln|λ|(z−

√λsw)2

q21(z/√λs).

Given that the conditions of Lemma 3.3.6 are fulfilled, we obtain:

T [U ] =1

λs

[2π

|λ|φ(√λsw) +R√λsw(λ)

],(3.3.29)

which by substitution implies:

1

|λ|T [U ] =

1

λs

2π

|λ|q21(w) +

1

λs|λ|−1R√λsw(λ) =: D1 +D2

By Lemma 3.3.6, we have D2 ∈ L1(λ : |λ| > R).So finally we are ready to evaluate the left-hand side of (3.3.15):

limR→∞

∫R<|λ|<2R

|λ|−1h(λ,w)dσλ = limR→∞

∫R<|λ|<2R

2π

λs|λ|−2q21(w)dσλ

= q21(w)4π2

λslimR→∞

∫ 2R

Rr−1dr

= q21(w)4π2

λslimR→∞

ln r∣∣∣2RR

= q21(w)4π2 ln 2

λs.

From this we get the desired asymptotic:

q21(w) =λs

4π2 ln 2limR→∞

∫R<|λ|<2R

|λ|−1h(λ,w)dσλ.

3.4 Scattering data for Dirac equation via the Dirichlet toNeumann map

In this section, we consider that are working with the conductivity γ in W 1,∞(Ω). Thegeneralization to our conductivity comes from this construction with some small changes.

78

Our next goal is to establish a relation between the Dirichlet-to-Neumann map for equation(1.1.1) and the traces of the solutions of (3.1.3) on ∂O. Let

Tq :=

φ|∂O : φ =

(φ1

φ2

)is a solution of (3.1.3), φ1, φ2 ∈ H1(O)

.

Let u ∈ H2(O) be a solution of (1.1.1) with u|∂O = f ∈ H3/2(∂O). Consider φ =γ1/2(∂u, ∂u) ∈ H1(O). Then, formally

(3.4.1) φ|∂O =1

2

(ν −iνν iν

)(Λγf∂sf

),

where Λγ is the co-normal D-t-N map and ∂s is the operator of the tangential derivative.Inverting we get

(3.4.2)

(Λγf∂sf

)=

(ν νiν −iν

)φ|∂O.

We normalize ∂−1s in such a way that ∫

∂O∂−1s fds = 0.

Then (3.4.2) could be rewritten as a boundary relation

(3.4.3) (I − iΛγ∂−1s )(νφ1|∂O) = (I + iΛγ∂

−1s )(νφ2|∂O)

Let us show the generalization of [Knudsen and Tamasan, 2005, Thm 3.2] for the case ofnon-continuous γ.

Theorem 3.4.1.

Tq = (h1, h2) ∈ H1/2(∂O)×H1/2(∂O) : (I − iΛγ∂−1s )(νh1) = (I + iΛγ∂

−1s )(νh2)

Proof. First we show that any pair (h1, h2)t ∈ H1/2(∂O)×H1/2(∂O) that satisfies the bound-ary relation above is in Tq. Consider a solution u ∈ H2(O) of (1.1.1) with the boundarycondition

u|∂O = i∂−1s (νh1 − νh2) ∈ H3/2(∂O).

Since γ ∈W 1,∞(O) and γ is separated from zero, it follows that γ1/2 ∈W 1,∞(O). Then, bothcomponents of the vector φ = γ1/2(∂u, ∂u)t belong to H1(O) and φ satisfies (3.1.3). The factφ|∂O = (h1, h2)t follows from (3.4.1) and (3.4.3).

Conversely, we start with a solution φ ∈ H1(O) of (3.1.3). From (3.1.3) and (3.1.4) wehave a compatibility condition

∂(γ−1/2φ1) = ∂(γ−1/2φ2).

The Poincare lemma and Lemma 3.2.2 ensure the existence of a function u such that(φ1

φ2

)= γ1/2

(∂u∂u

).

It is easy to check that u is a solution to (1.1.1) and belongs to H2(O). Then, (3.4.1)-(3.4.3)prove that h = φ|∂O satisfies the boundary relation stated in the theorem.

79

Denote Sk : H1/2(∂O)→ H1/2(∂O)

Sλ,wf(z) =1

iπ

∫∂Of(ς)

e−λ(z−w)2+λ(ς−w)2

ς − zdς

This integral is understood in the sense of principal value.The next theorem gives conditions to find the trace of ψ(z, k) at ∂O. (a proof can be

found in [Knudsen and Tamasan, 2005, Th.4.3] for γ ∈ C1+ε(R2)).

Theorem 3.4.2. The only pair (h1, h2) ∈ H1/2(∂O)×H1/2(∂O) which satisfies

(I − Sλ,w)h1 = 2eλ(z−w)2,(3.4.4)

(I − Sλ,w)h2 = 0,(3.4.5)

(I − iΛγ∂−1s )(νh1) = (I + iΛγ∂

−1s )(νh2)(3.4.6)

is solution of (3.2.4).

80

Chapter 4

From the current problems tofuture works

For proper application on electrical impedance tomography the Calderon is very impor-tant. In our world, the objects of interest are obviously three dimensional. Hence, not havingyet a positive answer to Calderon problem for higher dimensions is a big constraint to makingthis medical imaging technique viable.

One thing to be noticed is the different approaches taken on dimension two and on higherdimensions. In the first case, proofs of uniqueness and reconstruction are based on the useof ∂-equation, while for the second approaches are based on using the large asymptotics ofthe scattering transform to determined the Fourier transform of the potential. Although,this limit makes a numerical implementation difficult. A similar problem is found for whenthe conductivity is complex-valued. Both of this problems are related to the fact that inthis conditions we are not sure if exceptional points exist or not. The only thing makingthis approaches useful on establishing uniqueness is the fact that outside a ball of radius bigenough exceptional points do not exist.

The solution to higher dimensions may pass through an adaptation of the ∂-methodand of the approach of Astala and Paivarinta [2006]. This could be done with the tools ofquaternionic analysis to obtain a Beltrami-equation analogue to higher dimensions.

For the case of complex conductivities the established framework in Chapter 3 needsfurther work. We can have more discontinuous conductivities by looking at the case of morediscontinuity curves. The case to three dimensions keeps the ideas intact, but needs toestablish each step without complex analysis.

Recently, there as been some considerations on applications of the inverse conductivityproblem which turns the tables around. If up to now mathematicians have been worried intrying to give a positive answer to Calderon problem some recent works have been trying toestablish the possibility of making an object ”invisible”. This concept of invisibility is basedon recent articles (Leonhardt [2006], Pendry et al. [2006] and others) and tries to establishtheoretical ”cloaking”. In principle there would exist a device that makes so that a region oran object is shielded or cloaked from detection via electromagnetic waves.

The theoretical framework is based on giving counterexamples to an extension of theCalderon inverse problem for the conductivity equation. Most of the work is based on creatingconductivities which are not bounded below, hence creating an annulus which possibilitateswhat is inside to be invisible. This is of extreme interest in terms of military applications and

81

great efforts are been put into this look on the inverse conductivity problem.In this sense each problem is of separate interest but they can complement each other to

finally obtain a positive answer to Calderon problem in higher dimensions and to cases whereother conductivities are present.

82

Appendix A

The first appendix is a follow up of the last chapter. Here we present some proofs thatare used to prove the main theorem on Chapter 3, but have their one research interest withinthemselves. We show the proof of Lemma 3.3.5, which corresponds to the application of theLaplace Transform analogue of the Hausdorff-Young inequality.

A.1 Laplace Transform analogue of the Haussdorf-Young in-equality

We need to recall some statements on the Laplace Transform.The following results hold (see Merzon and Sadov [2011]): consider the map

Lγf(s) =

∫ ∞0

e−z(s)tf(t)dt

where s > 0 is a natural parameter of a contour γ : z(s), s > 0,<(z(s)) > 0.Theorem 7 from Merzon and Sadov [2011] claims that Lγ is a bounded operator from

Lq(R+) to Lp(γ), where 1 ≤ q ≤ 2 and 1/p + 1/q = 1. Moreover, the norm of this map isbounded uniformly in the class of convex contours.

Now we consider only contours such that |(<z(s))′| < 1/2 for s >> 1. This means thatthe spaces Lp, p > 1 for the variable s > 0 and for variable =z(s) are equivalent. We nowprove that the result of the Hausdorff-Young inequality is valid for the following map 2D:

Lf(λ1, λ2) =

∫ ∞0

∫ ∞1

e−iλ1xe−iλ2y−ln |λ2|yf(x, y)dxdy, λ1, λ2 > 0,

namely we prove that for some fixed domain D and constant C = C(D) > 0, we have

(A.1.1) ‖Lf‖Lpλ1,λ2≤ C‖f‖Lqx,y , Suppf ⊂ D.

Proof Consider the function A(y, λ1):

(A.1.2) A(y, λ1) =

∫ ∞0

e−iλ1xf(x, y)dx, y > 1

and note that by Hausdorff-Young inequality we get

(A.1.3) ‖A(y, ·)‖Lpλ1=

(∫|A(y, λ1)|pdλ1

)1/p

≤(∫|f(x, y)|qdx

)1/q

.

83

For the sake of simplicity we omit all positive constants here and in further inequalities. Weclaim that A(·, λ1) ∈ Lqy and we prove this fact later. Accepting this claim and using theabove mentioned theorem from Merzon and Sadov [2011] we get

(A.1.4)

∫ ∣∣∣∣∫ e−iλ2y−ln |λ2|yA(y, λ1)dy

∣∣∣∣p dλ2 ≤ (‖A(·, λ1)‖Lqy)p.

Further we use the notation ‖ · ‖ for ‖ · ‖Lpλ1,λ2. Now we are ready to estimate ‖Lf‖:

(A.1.5) ‖Lf‖p =

∫ ∫ ∣∣∣∣∫ e−iλ2y−ln |λ2|yA(y, λ1)dy

∣∣∣∣p dλ1dλ2 ≤∫

(‖A(·, λ1)‖Lqy)pdλ1.

First we apply the integral form of the Minkowski inequality, and then (A.1.3). Hence, weget:

(A.1.6)

(∫‖A(·, λ1)‖p

Lqydλ1

)q/p=

(∫ ∣∣∣∣∫ |A(y, λ1)|qdy∣∣∣∣p/q dλ1

)q/p≤

∫ (∫|A(y, λ1)|pdλ1

)q/pdy ≤

∫ (∫|f(x, y)|qdx

)dy.

This proves (A.1.1). Now let us show that A(·, λ1) ∈ Lqy. From Minkowski inequality we get

‖A(·, λ1)‖Lqy =

(∫ ∣∣∣∣∫ ∞0

e−iλ1xf(x, y)dx

∣∣∣∣q dy)1/q

≤∫ (∫

|f(x, y)|qdy)1/q

dx.

Since f has finite support, then the function∫|f(x, y)|qdy has finite support too. Let us

denote by C1 the length of its support. Therefore∫ (∫|f(x, y)|qdy

)1/q

dx ≤∫ (∫

|f(x, y)|qdy)dx+ C1.

A.2 Proof of Lemma 3.3.5

The following lemma represents a generalization of Lemma 3.2 from Lakshtanov et al.[2017]. Consider λ0 ∈ C, denote by ρ(z) = −i=[λ(z − w)2]/2 + ln |λ|λ0(z − w)2, and letA0 = supz∈O <[λ0(z − w)2]. For convenience we recall Lemma 3.3.5:

Lemma Let z1, w ∈ C, p > 2 and ϕ ∈ L∞comp. Then∥∥∥∥∥ 1

|λ|A0

∫Cϕ(z)

eρ(z)

z − z1dσz

∥∥∥∥∥Lpλ(C)

≤ C ‖ϕ‖L∞|z1 − w|1−δ

,

where the constant C depends only on the support of ϕ and on δ = δ(p) > 0.

Proof. Denote by F = F (λ,w, z1) the integral on the left-hand side of the inequality above.In order to have non-positiveness of the real part of the phase we make a change of variablesu = (z − w)2 in F and take into account that dσu = 4|z − w|2dσz. Then

(A.2.1) F =1

4

∑±

∫Cϕ(w ±

√u)

ei=(λu)/2+λ0 ln |λ|u

|u|(±√u− (z1 − w))

dσu.

84

Now, we consider a new change of variable u = u− u0, where

u0 = argmaxw±√

u∈suppϕ<(λ0u)

and apply the Hausdorff-Young inequality for the Laplace transform on a contour (A.1.1).The result on Lemma 3.3.5 follows immediately from Lakshtanov et al. [2017], Lemma 3.1which we recall here for the reader’s convenience

[Lakshtanov et al., 2017, Lemma 3.1] Let 1 ≤ p < 2. Then the following estimate is validfor an arbitrary 0 6= a ∈ C and some constants C = C(p,R) and δ = δ(p) > 0:∥∥∥∥ 1

u(√u− a)

∥∥∥∥Lp(u∈C:|u|<R)

≤ C(1 + |a|−1+δ).

85

Appendix B

In this appendix we present the necessary definitions and results which we use duringthe thesis. Some of them are well-known to mathematicians in the area of analysis but theirdeep importance for our results justify their inclusion here. Others are less well-known andtherefore have to be mentioned for a better understanding of this work. Given that thisthesis concerns the inverse conductivity problem, of which there exists already an extensiveliterature, we will not present all the proofs in detail. Instead we will refer to the appropriateliterature in the relevant cases. Moreover, sometimes we present only partial results in orderto avoid a too long appendix.

B.1 Domains and Properties

Let Ω be a bounded and open set of Rn and define ∂Ω = Ω∩ (Rn \Ω) to be its boundary.For most results on this thesis concerning Sobolev spaces we need our domain Ω to satisfysome properties described below, as well as a certain regularity on its boundary. All thedefinitions below are present in Adams and Fournier [2003].

Definition B.1 (Lipschitz Domain). We say that Ω is a Lipschitz domain if its boundarycan be locally represented by Lipschitz continuous function. Namely, for any x ∈ ∂Ω thereexists a neighborhood of x, Vx ⊂ Rn, such that Vx∩∂Ω is the graph of a Lipschitz continuousfunction under a proper local coordinate system.

In this sense, for some N ≥ 1, there exists Ω1, ...,ΩN connected components of the Lips-chitz domain, such that Ω =

⋃Nj=1 Ωj and for each j ∈ 1, ..., N the boundary ∂Ωj is given by

the graph of a Lipschitz function φj : Rn−1 → R through ∂Ωj = x ∈ Rn : xn = φ(x′)∀x′ ∈Rn−1.

Moreover, for any Lipschitz domain Ω there exists a surface measure dσ and we can definethe integration over the boundary also through:∫

∂Ω

f(x) dσ(x) =

N∑j=1

∫∂Ωj

f(x1, x2, ..., xn−1, φj(x1, x2, ..., xn−1))√

1 + |∇φj(x1, ..., xn−1)|2 dx1...dxn−1;

By Rademacher theorem and the fact that is Lipschitz, φj is differentiable almost everywhereand bounded for any j ∈ 1, ..., N.

Definition B.2 (Higher regularity domains). We say that Ω is of class Ck,α, 0 ≤ α ≤ 1, if ateach point x0 ∈ ∂Ω there is a ball B = B(x0) and a one-to-one mapping ψ of B onto D ⊂ Rnsuch that:

86

(i) ψ(B ∩ Ω) ⊂ Rn+;

(ii) ψ(B ∩ ∂Ω) ⊂ ∂Rn+;

(iii) ψ ∈ Ck,α(B), ψ−1 ∈ Ck,α(D).

Definition B.3 (Cone Condition). Let v be a nonzero vector in Rn and let ∠(x, v) be theangle between the position vector x 6= 0 and v. For such given v, and ρ > 0, and κ satisfyingto 0 < κ ≤ π, the set

C = x ∈ Rn : x = 0 or 0 < |x| ≤ ρ,∠(x, v) ≤ π/2

is called a finite cone of height ρ, axis direction v, and aperture angle κ with vertex at theorigin.

We say that a domain Ω satisfies the cone condition if there exists a finite cone C suchthat each x ∈ Ω is the vertex of a finite cone Cx contained in Ω and congruent to C. Notethat Cx need not be obtained from C by a parallel translation, but simply by rigid motions(rotation, reflection and/or translation).

B.2 Lp and Sobolev Spaces

We start with the definition of Lp spaces and immediately proceed with the Sobolevspaces.

Definition B.4 (Adams and Fournier [2003]). Let Ω be a domain in Rn and let p ∈ [1,∞].We denote by Lp(Ω) the class of all measurable functions u defined on Ω for which∫

Ω|u(x)|p dx <∞, if 1 ≤ p <∞(B.2.1)

supx∈Ω|u(x)| <∞(B.2.2)

Two functions are said equivalent if they are equal almost everywhere in Ω. In this sense,an element of Lp(Ω) is just a representative of an equivalence class of measurable functionssatisfying (B.2.1 or B.2.2) , and we will work with these equivalence classes of functionswithout making further comments.

The space is a Banach Space with the norm

‖u‖Lp(Ω) =

[∫Ω|u(x)|p dx

]1/p

(B.2.3)

‖u‖L∞(Ω) = supx∈Ω|u(x)|(B.2.4)

Moreover, for p = 2, the space is in fact a Hilbert Space, with the inner product:

〈f, g〉 =

∫Ωf(x)g(x) dx(B.2.5)

In our thesis, the measure dx will always represent the Lebesgue measure in Rn, although,the above statement work for any measure.

87

The concept of Sobolev Spaces comes from the definition of weak derivative, which we willnot present here, and from the above definition ofLp space.

Definition B.5 (Adams and Fournier [2003]). Let Ω be a domain in Rn, p ∈ [1,∞] andk be a positive integer. We denote by W k,p(Ω) the space of functions such that all the weak(or distributional) partial derivatives Dα, for 0 ≤ |α| ≤ k, are in Lp(Ω), that is:

W k,p(Ω) = u ∈ Lp(Ω) : Dαu ∈ Lp(Ω) for 0 ≤ |α| ≤ k(B.2.6)

This space is a Banach space under the norm:

‖u‖Wk,p(Ω) =

∑0≤|α|≤k

‖Dαu‖pLp(Ω)

1/p

, if 1 ≤ p <∞,(B.2.7)

‖u‖Wk,∞ = max0≤|α|≤m

‖Dαu‖L∞(Ω)(B.2.8)

Similarly to above, when p = 2 it is an Hilbert space, under the sum of the inner productsin L2 of all the derivatives. Hence, we change notations in this case: W k,2(Ω) = Hk(Ω).

Theorem B.2.1 (Adams and Fournier [2003]). Suppose that vol(Ω) =∫

Ω 1 dx <∞ and1 ≤ p ≤ q ≤ ∞. If u ∈ Lq(Ω) then u ∈ Lp(Ω), and

‖u‖Lp(Ω) ≤ (vol(Ω))1/p−1/q ‖u‖Lq(Ω)(B.2.9)

Hence,Lq(Ω) → Lp(Ω).

Proposition B.2.1 (Properties of Lpρ spaces). Let p ≥ 1 and α > β > 1. Then, under thedefinition that 〈x〉 = (1 + |x|2)1/2 and setting Lpα(R2) = f ∈ Lp(R2) : 〈·〉α f ∈ Lp(R2), wehave that:

(i) Lpα(R2) ⊂ Lpβ(R2)

(ii) Lp−β(R2) ⊂ Lp−α(R2)

(iii) If q ∈ Lpρ(R2) for p > 1 and ρ > 2/p′, where p′ is the Holder conjugate, then q ∈L1(R2) ∩ Lp(R2). In particular, if 1 < p < 2, then it holds for ρ > 1;

Proof. (i) Due to β − α < 0 we have 〈x〉β−α ≤ 1. Hence, it follows:

‖f‖Lpβ = ‖ 〈x〉β−α 〈x〉α f‖Lp ≤ ‖ 〈x〉α f‖Lp = ‖f‖Lpα

(ii) For the same reason as above it follows:

‖f‖Lp−α = ‖ 〈x〉−α f‖Lp = ‖ 〈x〉β−α 〈x〉−β f‖Lp ≤ ‖ 〈x〉−β f‖Lp = ‖f‖Lp−β

88

(iii) We show the most general case. Let ρ > 2/p′. We have:

‖q‖L1 = ‖ 〈x〉−ρ 〈x〉ρ f‖L1 ≤ ‖ 〈x〉−ρ ‖Lp′‖f‖LpρThus, it is enough to evaluate when the first norm is finite. By hypothesis we haveρ p′ > 2, so that:

‖ 〈x〉−ρ ‖p′

Lp′=

∫R2

1

(1 + |x|2)p′ρ/2dx

≤∫|x|≤1

1

(1 + |x|2)ρp′/2dx+

∫|x|>1

1

|x|ρp′dx <∞.

The first integral converges because the integrand, as well as the domain, are bounded.The second integral converges by the hypothesis (simple change to polar coordinates).

Theorem B.2.2 (Adams and Fournier [2003]). Let 1 ≤ p < q < r be such that

1

q=θ

p+

1− θr

for some θ ∈ (0, 1). If u ∈ Lp(Ω) ∩ Lr(Ω), then u ∈ Lq(Ω) and

‖u‖Lq ≤ ‖u‖θLp‖u‖1−θLr .

Theorem B.2.3 (Adams and Fournier [2003]). Let 1 ≤ p < ∞. Suppose that f ismeasurable on Rm × Rn, that f(·, y) ∈ Lp(Rm) for almost all y ∈ Rn, and that the functiony → ‖f(·, y)‖Lp(Rm) belongs to L1(Rn). Then the function x →

∫Rn f(x, y) dy belongs to

Lp(Rm) and (∫Rm

∣∣∣∣∫Rnf(x, y) dy

∣∣∣∣p dx)1/p

≤∫Rn

(∫Rm|f(x, y)|p dx

)1/p

dy.

Theorem B.2.4 (Adams and Fournier [2003], Sobolev Embedding Theorem). Let Ω bea domain in Rn. Let j ≥ 0 and m ≥ 1 be integers and let 1 ≤ p <∞.

PART I Suppose that Ω satisfies the cone condition.Case A If either mp > n or m = n and p = 1, then

W j+m,p(Ω)→ CjB(Ω).(B.2.10)

For p ≤ q ≤ ∞, we have

Wm,p → Lq(Ω)(B.2.11)

Case B If mp < n then, for p ≤ q ≤ p∗ = np/(n−mp), we have

Wm,p(Ω)→ Lq(Ω)(B.2.12)

PART II Suppose that Ω is a Lipschitz domain. Then, we have the following refinementholds: if mp > n > (m− 1)p then

W j+m,p(Ω)→ Cj,λ(Ω) for 0 < λ ≤ m− (n/p).(B.2.13)

89

Theorem B.2.5 (Adams and Fournier [2003], Sobolev Inequality). When mp < n thenthere exists a finite constant K such that:

‖φ‖Lp∗ (Rn) ≤ K

∑|α|=m

‖Dαφ‖pLp(Rn)

1/p

for p∗ = np/(n−mp) and φ ∈ C∞c (Rn).

(B.2.14)

In particular, for n ≥ 2 and 1 ≤ p < 2 we get

‖φ‖Lp∗ (Rn) ≤ K‖∇φ‖Lp(Rn)(B.2.15)

Theorem B.2.6 (Valent [1985]). Assume that Ω is a domain in Rn that has the coneproperty. Let p, q ≥ r and let m ≥ 1 be a positive integer such that

m

n>

1

p+

1

q− 1

r.

If the volume of Ω is infinite, assume further that mp ≤ n when q 6= r, that mq ≤ n whenp 6= r, and that

m− 1

n≤ 1

p+

1

q− 1

r, when p 6= r, q 6= r.

Then, if u ∈ Wm,p(Ω) and v ∈ Wm,q(Ω), we have uv ∈ Wm,r(Ω) and it exists a positivenumber c independent of u and v such that ‖uv‖Wm,r ≤ c‖u‖Wm,p‖v‖Wm,q .

Theorem B.2.7 (Adams and Fournier [2003], Rellich-Kondrachov Theorem). Let Ω bea domain in Rn and Ω0 a bounded subdomain of Ω. Let j ≥ 0 and m ≥ 1 be integers and letp ∈ [1,∞).

PART I If Ω satisfies the cone condition and mp ≤ n, then the following embedding iscompact:

Wm,p(Ω)→ Lq(Ω0), if 1 ≤ q ≤ np/(n−mp).(B.2.16)

PART II If Ω satisfies the cone condition and mp > n, then the following embedding iscompact:

Wm,p(Ω)→ CB(Ω0).(B.2.17)

Definition B.6 (Demengel et al. [2012], Fractional Sobolev spaces). Let Ω be a open setin Rn. We fix s ∈ (0, 1) and for any p ∈ [1,∞) we define the fractional Sobolev space asfollows:

W s,p(Ω) := u ∈ Lp(Ω) :|u(x)− u(y)||x− y|

np

+s∈ Lp(Ω× Ω),(B.2.18)

90

i.e., W s,p(Ω) is an intermediary Banach space between Lp(Ω) and W 1,p(Ω), endowed with thenatural normal

‖u‖W s,p(Ω) =

(∫Ω|u|p dx+

∫Ω

∫Ω

|u(x)− u(y)||x− y|

np

+sdxdy

)1/p

.(B.2.19)

Also, we define for p = 2 the space Hs ≡W s,p.

The definition of fractional Sobolev space can be given in a more abstract way by inter-polation of Lp and W 1,p.

Theorem B.2.8 (Adams and Fournier [2003], Sobolev interpolation inequality). Wehave the following inequality for s ∈ (0, 1) and p > 1:

‖u‖W s,p(Rn) ≤ K‖u‖1−sLp(Rn)‖u‖sW 1,p(Rn),(B.2.20)

where K is a constant depending only on n and p.

Theorem B.2.9 (Demengel et al. [2012], Fractional Sobolev embeddings). Let Ω be eithera Lipschitz open domain or Ω = Rn. Moreover, let s ∈ (0, 1) and p ≥ 1. Then:

If sp < n, then W s,p(Ω) → Lq(Ω) for every q ≤ np/(n− sp).

If n = sp, then W s,p(Ω) → Lq(Ω) for every q <∞.

If sp > n, then W s,p(Ω) → L∞(Ω).

Theorem B.2.10 (McLean [2000], Trace Operator). Let Ω be a domain in Rn. We definethe trace operator in this domain by:

tr : D(Ω)→ D(Γ)

u 7→ tr(u) = u|Γ.

(i) If Ω is a Ck−1,1 and 12 < s ≤ k, then tr has a unique extension to a bounded linear

operator

tr : Hs(Ω)→ Hs−1/2(Γ).

(ii) If Ω be a Lipschitz domain and 12 < s < 3

2 , then tr has a unique extension to a boundedlinear operator

tr : Hs(Ω)→ Hs−1/2(Γ).

Both extensions have a continuous right inverse.

Theorem B.2.11. Let 1 ≤ p < ∞ and let Ω be a bounded domain. Then, there exists aconstant C, depending only on Ω and p, so that for every u in the Sobolev space W 1,p(Ω) withzero-trace we have

‖u‖Lp(Ω) ≤ C‖∇u‖Lp(Ω).

91

B.3 Complex function Properties

In this section we consider the following spaces:

We denote by A(C) the set of the entire functions, i.e., v ∈ A(C) if and only if ∂v(z) = 0for all z ∈ C;

Moreover, A∗(C) is the set of holomorphic functions on C, with possible exception of adiscrete set of isolated singular points;

We denote by Dz(C) the set of locally integrable functions in C which have a weakderivative with respect to z. The concept is the same as in the weak Sobolev derivative,whereas the derivative is replaced by ∂;

Let A,B and F be Lp1 ∩ Lp2 functions with 1 ≤ p1 < 2 < p2. We define the followingoperator CA,B(w) = ∂w +Aw +Bw. We denote by:

A∗(A,B, F ) = w ∈ Dz(C) : CA,Bw = F,

the set of generalized solutions of the above equation. We call generalized analyticfunctions to the elements of the set

A∗ =⋃A∗(A,B, 0).

Theorem B.3.1 (Remmert [2012], Liouville Theorem). Every bounded function f in A(C)is constant.

Theorem B.3.2 (Liouville theorem for Lp-functions). Let f ∈ A(C) ∪ Lp, p ≥ 1. Then wehave f ≡ 0.

Proof. Let f be an entire function in Lp. Hence, f is holomorphic inside any contour of thetype:

γz0,R = z ∈ C : |z − z0| = R

for any z0 ∈ C and R > 0. By the Cauchy’s integral formula we have:

f(z0) =1

2πi

∫γz0,R

f(z)

z − z0dz

By Holder inequality we get

|f(z0)| ≤

(∫γz0,R

|f(z)|p |dz|

)1/p(∫γz0,R

1

|z − z0|q|dz|

)1/q

=

(∫γz0,R

|f(z)|p |dz|

)1/p(∫ 2π

0R1−q dθ

)1/q

= (2π)−1/pR−1/p

(∫γz0,R

|f(z)|p |dz|

)1/p

92

Thus, we have∫ +∞

0|f(z0)R1/p(2π)1/p|p dR ≤

∫ +∞

0

∫γz0,R

|f(z)|1/p |dz|dR = ‖f‖pLp <∞.

Taking the limit R → ∞ implies f(z0) = 0. The arbitrariness of z0 holds now the resultf ≡0.

Theorem B.3.3 (Vekua [2014], Beurling Transform). We define the Beurling transform off ∈ Lp(C), 1 < p <∞, as

Bf(z) =1

2πi

∫C

f(w)

(w − z)2dw ∧ dw.

This operator is bounded in Lp(C). Moreover, it can be formally defined as the derivative ofTeodorescu transform B = ∂∂−1.

Furthermore, ∂∂−1 is also a bounded operator on Lp(C).

Lemma B.3.1. If 1− 2/p < δ < 2/p′, then:

1. ∂−1 is a bounded operator: Lpδ → Lpδ−1;

2. ∂∂−1 is a bounded operator on Lpδ .

The proof of this lemma is a special case of Theorem 2.1 of Nirenberg and Walker [1973].These ideas are also present in Lemma A Sylvester and Uhlmann [1986].

Proposition B.3.1 (Vekua [2014]). If ∂f ∈ Lp(C), p > 1, then ∂f exists and also belongsto Lp(C).

B.4 Harmonic Functions

Let u be a C2(Rn) function. We say that u is harmonic if ∆u =∑∞

i=1 ∂2xiu = 0.

Proposition B.4.1 (Evans [2010], Fundamental Solution of Laplace of Laplace equation).The function

G0(x) =

− 1

2π log |x|, n = 21

n(n−2)α(n)1

|x|n−2 , n ≥ 3,

where α(n) = πn/2Γ(n2 + 1) and Γ denotes the gamma function, is a fundamental solution ofthe Laplace operator.

Theorem B.4.1 (Evans [2010], Analyticity). If u is harmonic in a domain Ω, then u isanalytic in Ω.

93

Theorem B.4.2 (Evans [2010], Mean Value formulas). If u ∈ C2(Ω) is harmonic, then

u(x) =1

nωnrn−1

∫∂B(x,r)

u(y) dS =1

ωnrn

∫B(x,r)

u(y) dy,(B.4.1)

where dS is the surface measure, ωn is the area of the n-dimensional unit ball in Euclideanspace and B(x, r) is a ball in Ω centered in x and with positive radius r.

Theorem B.4.3 (Evans [2010], Maximum principle). Assume u ∈ C(Ω) ∩ C(Ω) to beharmonic within Ω.

(i) Thenmax

Ωu = max

∂Ωu.

(ii) Furthermore, if Ω is connected and there exists a point x0 ∈ Ω such that

u(x0) = maxΩ

u,

thenu is constant within Ω.

Substituting u by −u, we obtain similar assertions with ”max” replaced by ”min”.

Theorem B.4.4 (Evans [2010], Liouville theorem for harmonic functions). Let u : Rn → Rbe harmonic and bounded. Then u is constant.

Theorem B.4.5 (Identity theorem for harmonic functions). Let u be harmonic on a regionΩ ⊂ C. If u = 0 at every point of a nonempty open subset U ⊂ Ω then u ≡ 0 at every pointof Ω.

Proof. Construct the holomorphic function f = ux− iuy on Ω. Since f = 0 on U , then f = 0on all of Ω by the Identity Theorem for holomorphic functions. Hence, u is a constant.

B.5 Other Results

Proposition B.5.1. Let x, y ∈ R2 ∼ C. Then, we have the following property:

〈x− y〉 ≤ 〈x〉 〈y〉 .

Proof. We do this by straight computation:

〈x− y〉2 = (1 + |x− y|2) ≤ (1 + |x|2 + |y|2) ≤ (1 + |x|2|y|2 + |x|2 + |y|2)

= (1 + |x|2)(1 + |y|2) = 〈x〉2 〈y〉2

Hence, the result immediately follows.

Proposition B.5.2. There exists an R > 0 big enough such that for x, y ∈ R2 where |x−y| >R, and all ε > 0 it holds:

log |x− y| ≤ 〈x− y〉ε

94

Proof. The proof follows by computing the limit with L’Hopital’s rule:

limr→+∞

log r

(1 + r2)ε/2= lim

r→+∞

1/r

εr/(1 + r2)ε/2−1= lim

r→+∞

(1 + r2)1−ε/2

r2= 0,

because 1− ε/2 < 1.

Theorem B.5.1 (Gilbarg and Trudinger [2015], Maximum and Minimum principles forelliptic operators). We consider the differential operator L defined as:

Lu =

n∑i,j=1

Di(aij(x)Dju(x)),

whose coefficients aij (i, j = 1, ..., n) are assumed to be measurable functions on a domainΩ ∈ Rn (we consider this operator in a weak sense).

Let the operator L satisfy the conditions:

1. The operator L is strictly elliptic: ∃λ > 0 :∑n

i,j=1 aij(x)ξiξj ≥ λ|ξ|2, ∀x ∈ Ω, ξ ∈ Rn

2. L has bounded coefficients:∑n

i,j=1 |aij(x)|2 ≤ Λ2

Then, if u ∈ H1(Ω) is a solution of Lu = 0, with u|∂Ω = f ∈ H1/2(∂Ω) we have

supΩu ≤ sup

∂Ωf+ inf

Ωu ≤ sup

∂Ωf−,

where f = f+ + f−, f+ = max f(x), 0, f− = min f(x), 0.

Theorem B.5.2 (Stein [1970], Hardy-Littlewood-Sobolev inequality). Let f ∈ Lp(Rn). Wedefined the Riesz potential of f by:

(Iαf) (x) =1

γ(α)

∫Rn

f(y)

|x− y|n−αdy, for 0 < α < n(B.5.1)

with γ(α) = πn/22αΓ(α/2)/

Γ ((n− α)/2). Moreover, for 1 < p, q <∞, such that 1q = 1

p −αn ,

it holds

‖Iα(f)‖Lq(Rn) ≤ Ap,q‖f‖Lp(Rn).

In particular, for 1 < p < 2, we have:

‖I1(f)‖Lp(R2) ≤ Ap‖f‖Lp(R2),

where p is the Sobolev conjugate 1p = 1

p −12 .

Theorem B.5.3 (Reed et al. [1972], Compact Operator properties). Let X1, X2, X3, Xand Y be Banach spaces. Then, the following properties follow:

1. Let A ∈ L(X1, X2), B ∈ L(X2, X3) and C ∈ L(X3, X1). Moreover, let B be a compactoperator. Then BA ∈ L(X1, X3) and CB ∈ L(X2, X1) are compact operators;

2. A compact operator K : X → Y maps weakly convergent sequences into norm convergentsequences.

95

Theorem B.5.4 (McLean [2000], Jump Relations of Single Layer Operator). Let Sk be thesingle layer operator associated with the Fadeev Green function defined on the boundary of aLipschitz domain Ω. Given f ∈ H−1/2(∂Ω), we set v(x) = −Skf(x) and denote by v+, v− its

traces from exterior and interior, respectively. Then we have: ∂v+

∂ν (z), ∂v−

∂ν ∈ H−1/2(∂Ω) and

they satisfy the following jump relations:

∂v+

∂ν(z)− ∂v−

∂ν(z) = f(z), for almost every z ∈ ∂Ω.

Theorem B.5.5 (Bochner et al. [1949], Riemann-Lebesgue Lemma). Let f be a functionin L1(Rn). Then its Fourier transform is defined as:

Ff(ξ) :=

∫Rnf(x)e−ix·ξ dx,

and fulfillsFf(ξ)→ 0, as |ξ| → +∞.

The result holds also by extension for L2 functions.

Moreover, for any f ∈ L2(∂Ω) where Ω is a Lipschitz domain, we have by the aboveintegration over the surface that the Riemann-Lebesgue lemma also holds for this functions,since

√1 + |∇φj(x′)|2f(x′) is in L2(Rn−1).

96

Bibliography

Adams, R. A. and Fournier, J. J. (2003). Sobolev spaces, volume 140. Elsevier.

Alessandrini, G. (1988). Stable determination of conductivity by boundary measurements.Applicable Analysis: An International Journal, 27:153–172.

Astala, K., Faraco, D., and Rogers, K. M. (2016). Unbounded potential recovery in the plane.Ann. Sci. Ec. Norm. Super.(4), 49:1023–1047.

Astala, K. and Paivarinta, L. (2006). Calderon’s inverse conductivity problem in the plane.Annals of Mathematics, pages 265–299.

Beals, R. and Coifman, R. (1981). Scattering, transformations spectrales et equationsd’evolution non lineaires. Seminaire Equations aux derivees partielles (Polytechnique),pages 1–9.

Blasten, E., Imanuvilov, O. Y., and Yamamoto, M. (2015). Stability and uniqueness for a two-dimensional inverse boundary value problem for less regular potentials. Inverse Problemsand Imaging, 9(3):709–723.

Bochner, S., Chandrasekharan, K., et al. (1949). Fourier transforms. Princeton UniversityPress.

Borcea, L. (2002). Electrical impedance tomography. Inverse problems, 18(6):R99.

Brown, R. M. and Uhlmann, G. A. (1997). Uniqueness in the inverse conductivity problemfor nonsmooth conductivities in two dimensions. Communications in partial differentialequations, 22(5-6):1009–1027.

Bukhgeim, A. L. (2008). Recovering a potential from cauchy data in the two-dimensionalcase. Journal of Inverse and Ill-posed Problems jiip, 16(1):19–33.

Calderon, A. P. (1980). On an inverse boundary value problem. Seminar on NumericalAnalysis and its Applications to Continuum Physics, Soc. Brasileira de Matematica, Riode Janeiro, pages 65–73.

Coifman, R. R., McIntosh, A., and Meyer, Y. (1982). L’integrale de cauchy definit unoperateur borne sur l2 pour les courbes lipschitziennes. Annals of Mathematics, pages361–387.

Demengel, F., Demengel, G., and Erne, R. (2012). Functional spaces for the theory of ellipticpartial differential equations. Springer.

97

Evans, L. (2010). Partial Differential Equations. Graduate studies in mathematics. AmericanMathematical Society.

Faddeev, L. D. (1966). Increasing solutions of the schrodinger equation. In Soviet PhysicsDoklady, volume 10, page 1033.

Francini, E. (2000). Recovering a complex coefficient in a planar domain from the dirichlet-to-neumann map. Inverse Problems, 16(1):107.

Gerken, T. and Lechleiter, A. (2017). Reconstruction of a time-dependent potential fromwave measurements. Inverse Problems, 33(9):094001.

Gilbarg, D. and Trudinger, N. S. (2015). Elliptic partial differential equations of second order.springer.

Greenleaf, A., Lassas, M., Santacesaria, M., Siltanen, S., and Uhlmann, G. (2018). Propa-gation and recovery of singularities in the inverse conductivity problem. Analysis & PDE,11(8):1901–1943.

Grisvard, P. (2011). Elliptic problems in nonsmooth domains. SIAM.

Haberman, B., Tataru, D., et al. (2013). Uniqueness in calderon’s problem with lipschitzconductivities. Duke Mathematical Journal, 162(3):497–516.

Knudsen, K. and Tamasan, A. (2005). Reconstruction of less regular conductivities in theplane. Communications in Partial Differential Equations, 29(3-4):361–381.

Kohn, R. and Vogelius, M. (1984). Determining conductivity by boundary measurements.Communications on Pure and Applied Mathematics, 37(3):289–298.

Lakshtanov, E., Tejero, J., and Vainberg, B. (2017). Uniqueness in the inverse conductiv-ity problem for complex-valued lipschitz conductivities in the plane. SIAM Journal onMathematical Analysis, 49(5):3766–3775.

Lakshtanov, E. and Vainberg, B. (2016). On reconstruction of complex-valued once differen-tiable conductivities. J. Spectr. Theory, 6(4):881–902.

Lakshtanov, E. L., Novikov, R. G., and Vainberg, B. R. (2016). A global riemann-hilbertproblem for two-dimensional inverse scattering at fixed energy. Rend. Istit. Mat. Univ.Trieste, 48:1–26.

Leonhardt, U. (2006). Optical conformal mapping. science, 312(5781):1777–1780.

McLean, W. (2000). Strongly elliptic systems and boundary integral equations. Cambridgeuniversity press.

Merzon, A. and Sadov, S. (2011). Hausdorff-young type theorems for the laplace transformrestricted to a ray or to a curve in the complex plane. arXiv preprint arXiv:1109.6085.

Mueller, J. L. and Siltanen, S. (2012). Linear and nonlinear inverse problems with practicalapplications, volume 10. Siam.

98

Nachman, A. I. (1988). Reconstructions from boundary measurements. Annals of Mathemat-ics, 128(3):531–576.

Nachman, A. I. (1996). Global uniqueness for a two-dimensional inverse boundary valueproblem. Annals of Mathematics, pages 71–96.

Nirenberg, L. and Walker, H. F. (1973). The null spaces of elliptic partial differential operatorsin rn. Journal of Mathematical Analysis and Applications, 42(2):271–301.

Novikov, R. (1992). The inverse scattering problem on a fixed energy level for the two-dimensional schrodinger operator. Journal of functional analysis, 103(2):409–463.

Novikov, R. and Santacesaria, M. (2010). A global stability estimate for the gel’fand–calderoninverse problem in two dimensions. Journal of Inverse and Ill-Posed Problems, 18(7):765–785.

Novikov, R. G. and Santacesaria, M. (2013). Monochromatic reconstruction algorithms fortwo-dimensional multi-channel inverse problems. International Mathematics Research No-tices, 2013(6):1205–1229.

Paivarinta, L., Panchenko, A., Uhlmann, G., et al. (2003). Complex geometrical opticssolutions for lipschitz conductivities. Revista Matematica Iberoamericana, 19(1):57–72.

Pendry, J. B., Schurig, D., and Smith, D. R. (2006). Controlling electromagnetic fields.science, 312(5781):1780–1782.

Pombo, I. (2019). Cgo-faddeev approach for complex conductivities with regular jumps. arXivpreprint arXiv:1903.03485.

Quarteroni, A. and Valli, A. (1999). Domain decomposition methods for partial differentialequations. Oxford University Press.

Reed, M., Simon, B., Simon, B., and Simon, B. (1972). Methods of modern mathematicalphysics, volume 1. Elsevier.

Remmert, R. (2012). Theory of complex functions, volume 122. Springer Science & BusinessMedia.

Serov, V. and Paivarinta, L. (2006). Inverse scattering problem for two-dimensionalschrodinger operator. Journal of Inverse and Ill-posed Problems jiip, 14(3):295–305.

Stein, E. M. (1970). Singular integrals and differentiability properties of functions. PrincetonMathematics Series, Princeton.

Stein, E. M. and Weiss, G. (1970). Introduction to Fourier analysis on Euclidean spaces.Princeton Mathematics Series, Princeton.

Sylvester, J. and Uhlmann, G. (1986). A uniqueness theorem for an inverse boundary valueproblem in electrical prospection. Communications on Pure and Applied Mathematics,39(1):91–112.

Sylvester, J. and Uhlmann, G. (1987). A global uniqueness theorem for an inverse boundaryvalue problem. Annals of mathematics, pages 153–169.

99

Tejero, J. (2017). Reconstruction and stability for piecewise smooth potentials in the plane.SIAM Journal on Mathematical Analysis, 49(1):398–420.

Valent, T. (1985). A property of multiplication in sobolev spaces. some applications. Rendi-conti del Seminario Matematico della Universita di Padova, 74:63–73.

Vekua, I. N. (2014). Generalized analytic functions, volume 25. Elsevier.

100