Calderón's inverse conductivity problem in two dimensions: Nachman's approach and the case for complex conductivity

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Universidade de Aveiro Departamento de Matem´atica 2019

Ivan Mira

Pombo

O problema de conductividade inverso de Calder´

on

a duas dimens˜

oes: a abordagem de Nachman e o

caso de conductividade complexa

Calder´

on’s inverse conductivity problem in two

dimensions: Nachman’s approach and the case for

complex conductivity

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Universidade de Aveiro Departamento de Matem´atica 2019

Ivan Mira

Pombo

O problema de conductividade inverso de Calder´

on

a duas dimens˜

oes: a abordagem de Nachman e o

caso de conductividade complexa

Calder´

on’s inverse conductivity problem in two

dimensions: Nachman’s approach and the case for

complex conductivity

Disserta¸cão apresentada à Universidade de Aveiro para cumprimento dos requisitos necessários à obten¸cão do grau de Mestre em Matemática e Aplica¸cões, realizada sob a orienta¸cão cient´ıfica do Doutor Uwe Kahler, Professor Associado com Agrega¸cão do Departamento de Matemática da Universidade de Aveiro.

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Ao eu do passado e ao eu do futuro.

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o j´uri / the jury

presidente / president Professor Eug´enio Alexandre Miguel Rocha

Professor Auxiliar do Departamento de Matem´atica da Universidade de Aveiro

Vogal-arguente principal Professor Jens Wirth

Professor Associado do Departamento de Matem´atica da Universidade de Estu-garda

Vogal-orientador Professor Uwe Kahler

Professor Associado com Agrega¸c˜ao do Departamento de Matem´atica da Univer-sidade de Aveiro (orientador)

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agradecimentos / acknowledgements

Gostaria de agradecer ao Professor Uwe Kahler e à professora Paula Cere-jeiras, por me darem a oportunidade de trabalhar com eles desde o inicio da minha Licenciatura, por me darem a conhecer o mundo da matemática e por me darem as condi¸cões para atingir o meu potentical enquanto matemático. Professor Jens Wirth für alle Antworten auf meine verschiedenen Fragen und für das Abendessen während meines Aufenthalts in Stuttgart.

Ao Don¸ca, por ser o meu a¸coriano favorito para ir relaxar e beber uns canecos.

`

A Liliana, por me aturar nos bons e maus momentos, por me fortalecer todos os dias e ainda que n˜ao perceba quando lhe apresento as coisas em inglˆes, tem sempre uma palavra a dizer para fazer de mim uma pessoa melhor e um melhor profissional. Agrade¸co-te do mais profundo do meu ser.

Finalmente, agrade¸co aos meus pais, porque literalmente sem eles não es-tava aqui e por me darem sempre abertura e apoio a que eu me dedique a novas oportunidades e que dê sempre o meu máximo.

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palavras-Chave problema de conductividade inverso, conductividade complexa, tomografia de impedˆancia el´ectrica, operador Dirichlet-to-Neumann

resumo O problema de Calderón consiste na determina¸cão da conductividade (real ou complexa) de um corpo baseado em medidas na fronteira. Este forma o modelo matemático para métodos de imagem médica não invasiva, como por exemplo tomografia de impedância eléctrica. Neste trabalho, apresen-tamos como introdu¸cão ao problema, algumas ferramentas necessárias que são usadas na abordagem de resolu¸cão do problema a duas dimensões, para conductividades reais, publicado por Nachman, no qual obteve o primeiro método de reconstru¸cão. No entanto, o problema só esta completamente resolvido nestas condi¸cões. Assim sendo, é importante olhar para o caso em que a conductividade têm valores complexos e para o caso de dimensões maiores que dois. Com este intuito, mostramos também um novo conceito para abordar o problema de conductividades complexas e mencionamos os problemas que têm que ser ultrapassados para obter resultados para di-mensões maiores.

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

abstract The Calder´on problem consists in the determination of the (real- or complex-valued) conductivity of a body based on measurements at its boundary. It forms the mathematical model for methods of non-invasive medical imaging, like electric impedance tomography. In two dimensions the problem was fully solved for real conductivities with tools from complex analysis. In this work, we present as an introduction to the problem, the necessary tools that are used in the approach to solve the problem in two dimensions, for real conductivities, which was published by Nachman as first reconstruction method. However, the problem is only fully solved on this conditions. In this 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 also show a new concept to approach the problem for complex conductivities and mention the problemas that need to be overcomed to obter results for higher dimensions.

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List of Figures

1.1 Domain Ω ⊂ R2, with different tissues. The colors represent different intensities of the conductivity γ. At black we have the heart, the lighter gray is the lungs and the rest is the background. . . 2 2.1 The diagram shows in a simple manner Nachman’s idea to recover γ from

knowledge of the Dirichlet-to-Neumann map. . . 12 3.1 Simple representation on a type of domain we can consider for the transmission

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Chapter 1

Introduction

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

The field of Inverse Problems is the process of obtaining from a set of observations the casual 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 in 1911, where he describes the asymptotic behavior of the eigenvalues of the Laplace-Beltrami Operator. Today this well-known Weyl’s law is trivialized in the question: ”Is it possible to hear the shape of a drum?”. Weyl conjectured that the eigenfrequencies of a drum would be related 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 in trying 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 computational applications, 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 Calder´on problem. In the 1950’s Alberto Calder´on 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 conductivity of a body from measurements at its surface.

Currently, the determination of the electrical conductivity of a body is a key instrument in 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

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explanations.

Let Ω be a domain in Rn, with n ≥ 2. This domain represents an inhomogeneous, electric conductive isotropic body, (i.e. the conductivity property is independent of directions), with electrical conductivity γ : Ω → [0, ∞). In this application we consider two things: first, the conductivity is a bounded function, (given that the domain is bounded); second, it has a positive 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 of the conductivity γ. At black we have the heart, the lighter gray is the lungs and the rest is the background.

The plan here is to determine the conductivity inside Ω, without previous information directly 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 human body magnetic permeability can be neglected, (i.e. in response to a magnetic field the human body 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 voltage potential f set at the boundary.

The boundary value problem (1.1.1) is elliptic due to the positive lower bound of γ, if γ as first derivatives. In praxis, we want to find an approach for the worst case scenario, which for us means that we require the conductivity to be in L∞(Ω). This assumption is supported on one hand, by the fact that, physically, it does not make sense to have infinite values on the conductivity. 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 the boundary 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, which is 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-normal derivative 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

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Dirichlet-to-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 weak formulation by: hΛγf, gi = Z ∂Ω gγ∂u ∂νdS for all g ∈ H 1/2_(∂Ω)

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

hΛγf, gi = Z Ω ∇(γv) · ∇u + vγ∆u dV = Z Ω

γ∇v · ∇u + v(∇γ · ∇u + γ∆u) dV =

Z Ω

γ∇v · ∇u dV (1.1.3)

as the weak formulation of the DtN map.

In this framework, Calderón published a paper (Calderón [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, Calderón 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) depends on γ we can observe from the weak formulation (1.1.3) that the product between γ and ∇u makes the DtN a non-linear mapping. This makes the Calder´on problem a very ill-posed problem.

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 only partial boundary data is available. Even in the ideal case, the solution fails in the continuous dependence of the data.

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

kΛγ1− Λγ2kH1/2_(∂Ω)→H−1/2_(∂Ω) < δ, but still kγ₁− γ₂k_L∞ > ,

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 ≤ R 1, R < r ≤ 1

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where A is a positive value.

By explicitly calculations the Dirichlet-to-Neumann maps for both conductivities can be shown to satisfy

kΛγ1 − Λγ2kH1/2_(∂Ω)→H−1/2_(∂Ω) ≤ |1 − A|R → 0, R → 0. while

kγ₁− γ₂k_L∞_(Ω) = |1 − A|.

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

kγ₁− γ₂k_L∞_(Ω) ≤ f (kΛ_γ₁ − Λ_γ₂k_H1/2_(∂Ω)→H−1/2_(∂Ω)).

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

The extension of Calder´on work in two dimensions was based on the conversion of the conductivity 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 Calder´on 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 quadratic form 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-analytic on Ω, then it is uniquely determinedthe DtN map;

Sylvester and Uhlmann [1986]: In this paper it is present uniqueness in R2 _{for γ} close to 1 in W3,∞(Ω) (as a local result);

Sylvester and Uhlmann [1987]: They presented a first global result, which extends the result of Kohn and Vogelius to the whole domain, i.e. in Rn, n ≥ 3. Namely, they showed that smooth conductivities are uniquely determined from the quadratic form in the 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 in dimension 2 was still an open question.

Nachman [1996]: First global result in two dimensions through a reconstruction method for conductivities W2,p, p > 1, by converting (1.1.1) to the Schr¨odinger equa-tion. We will take a closer look into this later on;

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Brown and Uhlmann [1997]: They reduce the need requirement for uniqueness to only conductivities which are in W1,p. For this they use the Dirac scattering equation, which had already been studied in the literature of inverse scattering problems. In the paper by Knudsen and Tamasan [2005], they obtain a reconstruction method to this uniqueness result ;

Päivärinta et al. [2003] This paper yields the current best result for three dimensions. It provides a uniqueness proof for γ ∈ W3/2,∞(Ω), i.e., for Lipschitz conductivities; Astala and Päivärinta [2006]: In this paper finally a positive answer was given

to the Calder´on problem, i.e. the Dirichlet to Neumann map defines uniquely real-valued conductivities in L∞, which have a positive lower bound. This was shown by converting (1.1.1) into the Beltrami equation and applying results from the theory of quasi-conformal mappings.

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

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

γ(x) = σ(x) + iω(x), (1.1.4)

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

Francini [2000]: This was one of the first explicit treatments of this case. It adapts the 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 Schr¨odinger equation, to the case of any complex conductivity with two derivatives ;

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

Lakshtanov and Vainberg [2016]: This result uses the transformation to the Dirac equation 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 three

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This thesis is structured in three chapters. In chapter 2, we present Nachman’s approach in two dimensions. Here we follow closely the paper (Nachman [1996]). We show every step of his reconstruction in detail. Nachman’s paper was the first to present a reconstruction method, so it is ideal to make an introduction to the problem, as it contains all necessary tools and 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 practical reconstruction method in several works (Mueller and Siltanen [2012]).

In the third chapter we present an approach to the study of the inverse conductivity problem 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 treating them.

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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 tomography viable. Many methods have been used and tested on real data, like linearization and iteratively non-linear approaches, although each of them has its fragilities and from the point of view of applications 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-Neumann map. We present the way Nachman used to solved this problem with of tools from complex and spectral analysis.

In Nachman [1996] the first reconstruction work for two-dimensions was presented. Under some conditions Nachman proves uniqueness from the DtN map in a constructive manner for twice-differentiable conductivities. Later, many numerical implementations of his approach have been realized and each step seems to be have been viably implemented. Even though there are always some questions on how to implement them properly, in a direct approach we only 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 of the following theorem.

Theorem 2.1.1. Let Ω be a bounded Lipschitz domain in R2. Let γ1 and γ2 be in W2,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 Schr¨odinger equation. In parallel with Nachman’s paper, there have been many works on the Inverse Scattering problem, but there has always been some growth conditions on the potential. With Nachman’s approach we give a partial answer to Calder´on problem.

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

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obtain: ∇ · γ∇u = 0 ⇔∇γ · ∇γ−1/2u˜+ γ∆γ−1/2u˜= 0 ⇔∇γ · −1 2γ −3/2_{u∇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 + γ ˜u ∆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 going to multiply what remains by −γ1/2 to obtain:

− ∆˜u + 1 2γ −2_{(∇γ · ∇γ) − γ}1/2_∆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 + q ˜u = 0, in Ω. (2.1.1)

Similar as before, we can define the Dirichlet-to-Neumann map for the Schr¨odinger equa-tion. By the conditions on γ and recalling that the solution u of (1.1.1) is in H1(Ω), it follows that ˜u ∈ H1_{(Ω) is a weak solution to the above Schr¨}_{odinger equation. Moreover, we also have} that ˜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 Schr¨odinger 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 analogously through Green identities via the bilinear form:

(2.1.2) DΛqf , ˜˜g E = Z Ω ∇˜v · ∇˜u + q˜v ˜u dV, for all ˜v

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

D Λqf , ˜˜g E = Z ∂Ω ˜ g∂(γ 1/2_u) ∂ν dS = Z ∂Ω ˜ g∇ γ1/2u · ν dS = Z ∂Ω ˜ g u1 2γ −1/2∂γ ∂ν + γ 1/2∂u ∂ν dS = Z ∂Ω ˜ gγ−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 on the Schr¨odinger equation this reduction removes the unknown coefficient in (1.1.1) from the higher order terms which will simplify the problem.

Given that our method will use the Schr¨odinger equation and, therefore, the DtN map Λq in some steps of our reconstruction, we will need to determine first γ|∂Ω and ∂γ_∂ν|∂Ω since in 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 Λqin this way. The proof could follow this path but it is simpler to use the boundary values of the conductivity to reduce the problem to 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 ≡ Λγfor this new conductivity. Moreover, one of the important steps for the proof is the knowledge we can obtain from the forward problem, i.e. given a conductivity and Dirichlet data we ask the question: What type of solutions to (1.1.1) and (2.1.1) do we obtain?;

Rather than looking at the Schr¨odinger 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 solutions ψ(x, ζ) (when they exist) of

(2.1.4) (−∆ + q)ψ(x, ζ) = 0 in R2,

where e−ix·ζ− 1 ∈ W1, ˜p_(R2_{), with} 1 ˜ p =

1 p−

1

2 is the Sobolev conjugate and ζ ∈ V =: {(ζ1, ζ2) ∈ C2\{0} : ζ2:= ζ12+ζ22 = 0}. This type of solutions are what we call the exponentially growing solutions. They were first introduced in (Faddeev [1966]) for the inverse scattering problem

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and later rediscovered in the ICP by Sylvester and Uhlmann, but in some sense already used by Calder´on.

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

−∆φ(x, χ) = − ∂ 2 ∂x2 1 + ∂ 2 ∂x2 2 ei(x1χ1+x2χ2)_{= (χ}2 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 ∈ W1, ˜p(R2), as mentioned previously. The idea to pick this space, comes from 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 two and for complex conductivities, the concept of exceptional points. In the last chapter we explain 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 ∈ W1, ˜p(R2), then we say that ˜ζ is an exceptional point. This is equivalent to say that exists h 6= 0, with he−ix·˜ζ ∈ W1, ˜p_(R2_{) is a solution of (2.1.4), since we can pick h = ψ}

1− ψ2. During the section 2, we will show that for each k ∈ C \ {0} there is a unique solution on the desired conditions to (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 directly measurable 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) =

Z 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-exceptional points, see section 2. This will be the case in two dimensions for potentials of the following type:

Definition 2.1. Let q ∈ Lp(R2), 1 < p < 2. We say that q is of conductivity type, if exists ψ0 ∈ 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 reconstruction of γ from it. This step will in fact only require the values of t(ξ, ζ) with ξ = −2<(ζ) and ζ ∈ V. In this sense we can make a change of notation. The set V can be parameterized as

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follows: {(k, ik) : k ∈ C\{0}}∪{(k, −ik) : k ∈ C\{0}}. Henceforth, we denote ψ(x, (k, ik)) by ψ(x, k) and by observing that the assumption on q being real-valued, we have that uniqueness for (2.1.4) yields ψ(x, (−¯k, i¯k)) = ψ(x, (k, ik)) = ψ(x, k), it follows that for reconstruction purposes 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, for convenience. In this way, we use z = x1+ix2to 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)) =

Z R2

e−i(x·(−2(k1,−k2)+(k1+ik2,−k2+ik1)_{q(x)ψ(x, k) dx} = Z R2 e−ix·(−¯k,i¯k)q(x)ψ(x, k) dx = Z R2 ei(x1−ix2)¯k_{q(x)ψ(x, k) dx} = Z R2 ei¯z¯kq(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 tools from complex analysis and the ¯∂ establish the foundational approach for the two-dimensional inverse conductivity problem, being presented in the works: (Nachman [1996]), (Brown and Uhlmann [1997]) and (Astala and P¨aiv¨arinta [2006]).

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

∂¯kµ(x, k) = 1

4π¯kt(k)e−k(x)µ(x, k), (2.1.8)

in the weighted Sobolev space W_β1, ˜p(R2) = {f : hxi−βf ∈ W1, ˜p(R2)}, where hxi = (1 + |x|2₎1/2_{, for β > 2/˜}_{p; moreover we also define e}

k as:

ek(x) := ek(z) = ei(kz+¯k¯z). (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 the usual large k asymptotics of µ, which need more decay on t than what is available in our case. On this sense, we can develop the idea that γ1/2(x) = limk→0µ(x, k), and we will show that 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 we have, 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 Λqand the solutions of (1.1.1) at the boundary ∂Ω.

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

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map has a role to play, but is not as straightforward, since do not know what Dirichlet data gives 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 and the 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 what an implementation of his method should look like:

(i) Determine γ|∂Ωand ∂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 8π2_i Z C t(k0) (k0− k)¯k0e−x(k 0 )µ(x, k0) dk0∧ d¯k0; (2.1.10)

(v) Obtain γ(x) from the (absolutely convergent) integral formula γ1/2(x)1 + 1 8π2_i Z C t(k0) |k0_|2e−x(k 0_{)µ(x, k}0_{) dk}0_{∧ d¯}_k0_. (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 on what 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 the equation (2.1.4) in the desired spaces for the case that the potential is of conductivity type. For this we need to present and show some results and estimates essential for simplifying further proofs as well as showing that potentials of conductivity type do not possess exceptional points.

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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 ∈ W1, ˜p and (2.2.1) e −izk ψ(·, k) − 1 Ws, ˜p ≤ c|k| s−1_kqk Lp

for 0 ≤ s ≤ 1 and k sufficiently large.

Notation: We denote ∂ and ¯∂, to be the differential operators, corresponding to the Wirtinger 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 _πz1 and 1

π ¯z, and, hence, we can define the inverse operator (left inverse modulo analytic functions) to be given by convolution with the fundamental solutions as

¯ ∂−1f (z) = 1 2πi Z C f (w) w − zdw ∧ d ¯w,

where dw ∧ d ¯w = −2idx ∧ dy, and which we call Teodorescu transform. Similarly, we can define ∂−1.

Finally, we just introduce a function that will be useful in simplifying the notation by ek(z) = ei(kz+¯k¯z)= 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) k(∂ + ik)−1f k_Lp˜≤ ckf k_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 solution w in Lp˜of (∂ + ik)w = v. Moreover, w ∈ W1, ˜p and

(2.2.3) kwk_Lp˜ ≤ c1

|k|(kvkLp˜+ k∂vkLp) and k∇wkLp˜ ≤ c2k∂vkLp

Proof. i. Straightforward computation gives that the fundamental solution to (∂ + ik) is given by e−k(z)

π ¯z .Therefore, we have that u(z) = (∂ + ik)−1f (z) = 1 2πi Z C e−k(z − w)f (w) ¯ w − ¯z dw ∧ d ¯w = e−k(z) 1 2πi Z C ek(w)f (w) ¯ w − ¯z dw ∧ d ¯w = e−k(z)∂−1(ekf )(z) (2.2.4)

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With this in mind, the desired estimate (2.2.2) follows by a simple modification of the Hardy-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 them satisfies:

∂(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) is an entire function. By Liouville’s theorem for Lp spaces it follows that this function equals zero and since |ek(z)| = 1, ∀ z ∈ C, we obtain u1− u2 = 0 and, consequently, the uniqueness result holds.

ii. We start by looking at the desired equation and changing it a little bit to get a case similar 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 h (∂ + ik)v − ∂v i . 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 k(∂ + ik)−1∂vk_Lp˜ ≤ ck∂vk_Lp. Consequently, from v ∈ Lp˜, we derive:

kwk_Lp˜ = 1 ik(I − (∂ + ik) −1 ∂)v/ik Lp˜ ≤ 1 |k| kvkLp˜+ (∂ + ik)−1∂v Lp˜ ≤ 1 |k|(kvkLp˜+ ck∂vkLp) , which shows the first desired estimate.

For the second estimate, we can first bound the gradient of w in terms of the Wirtinger Derivatives. 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

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

k ¯∂wk_Lp˜ = k ¯∂∂−1(∂w)k_Lp˜ ≤ ˜ck∂wk_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 exponentially growing solutions, we can modify the equation in question to be given only in terms of only of µ. We start with ψ(z, k) = eikz_{µ(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) − e¯ ikz4 ¯∂∂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 the differential operator on the left-hand side. In this case, we pick c = 1 in order to satisfy our assumptions.

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

(|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 Z R2 eix·ξ |ξ|2_{+ 2k(ξ} 1+ iξ2) dξ1dξ2, is the fundamental solution to (2.2.7).

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

Gk(x) = ei(x1+ix2)kgk(x). Thus follows: −∆Gk(x) = −∆ eizkgk(x) = −4∂ ¯∂eizkgk(x) = −4∂eizk∂g¯ k(x) = −4ikeizk∂g¯ k(x) − eizk∆gk(x) = eizk(−∆ − 2iζ · ∇)gk(x) = δ(x).

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

Skf (x) = Z

∂Ω

Gk(x − y)f (y) dσ(y),

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

The essence of the above lemma consists in allowing us to prove some new estimates on the convolution operator defined via gk. The idea now is to use smoothing properties of this operator 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 ∈ W1, ˜p and (2.2.10) kuk_Ws, ˜p ≤

c

|ζ|1−skf kLp for |ζ| ≥ const. > 0, 0 ≤ s ≤ 1.

Proof. As already explained we can assume without loss of generality that ζ = (k, ik) for some k ∈ C \ {0}. To verify uniqueness we use the properties of the Fourier transform and note that any tempered distribution solution u0 of (2.2.9) with f ≡ 0 must be of the form u0 = 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 and use 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, in the sense of obtaining a factor of 1/|k| in the estimates, we derive another formula for the inversion,namely, u = gk∗ f = (∂ + ik)−1v = 1 ik v − v + ik(∂ + ik) −1 v = 1 ik v − (∂ + ik) −1 ((∂ + ik)v − ikv) = − 1 4ik ¯ ∂−1− (∂ + ik)−1∂ ¯∂−1f. (2.2.11)

Finally, we just need to look over the estimates. For this we use Sobolev interpolation theorems and Estimate (2.2.3). First, we look at u in Lp˜. By the first inequality of (2.2.3) and the definition of v we obtain kuk_Lp˜ ≤ c |k| k ¯∂ −1_{f k} Lp˜+ k∂ ¯∂−1f k_Lp

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

(2.2.12) kuk_Lp˜ ≤

c1 |k|kf kLp

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

(2.2.13) k∇uk_Lp˜≤ c₂kf k_Lp

Joining (2.2.12) and (2.2.13) together and paying attention to the definition of the norm of ∇u, we get kuk_W1, ˜p ≤ C 1 |k| + 1 kf k_Lp

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

kuk_Ws, ˜p ≤ kukσ_p_˜kuk1−σ_W1, ˜p ≤ C 1 |k|1−s 1 |k|+ 1 s kf kLp.

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

(2.2.14) kwk_Ws, ˜p ≤ C1

|ζ|1−skf kLp, |ζ| ≥ const. > 0

At several points in the paper we will recall the following lemma for which we have to introduce the weighted spaces Lpρ(R2) with norm kf kLpρ = k(1 + | · |

2₎ρ/2_{f k} Lp. Lemma 2.2.3. If f ∈ Lp1_(R2_{) ∩ L}p2_(R2_{) with 1 < p}

1 < 2 < p2, then the function u = ¯∂−1f satisfies lim|z|→∞u(z) = 0,

kukL∞ ≤ c_p 1,p2(kf kLp1 + kf kLp2), and (2.2.15) |u(z1) − u(z2)| ≤ cp2|z1− z2| 1−2/p2_{kf k} 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 that lim|z|→∞u(z) = 0 we use the fact that the space Lp11 ∩ L

p2

1 is dense in Lp1 ∩ Lp2, given that Schwartz functions are also a dense subset of the first intersection. So let f ∈ Lp1

1 ∩ L p2

1 , then we get f ∈ L1 and the identity

(2.2.17) ∂¯−1f (z) = 1 z h − 1 2πi Z C f (w)dw ∧ d ¯w + ¯∂−1(wf (w))(z)i for z 6= 0

holds. Since wf (w) ∈ Lp1 _{∩ L}p2_{, 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 by a density argument.

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With this lemma, we are now ready to show the absence of exceptional points in the case where 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 ∈ W1, ˜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 under th 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 ∈ W1,p which follows by

(2.2.18) +∞ > kqkp_Lp = Z R2 ∆ψ0 ψ0 p dx ≥ 1 ||ψ₀||p_L∞ Z R2 |∇ · ∇ψ₀|pdx

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 to deduce the space to which v belongs.

We have

v1= ψ0∂he−ikz = ψ0(∂(he−ikz) − ik(he−ikz)).

By reason of q being of conductivity type ψ0 ∈ L∞ and he−ikz ∈ W1, ˜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 ∈ W1,p, (ψ0 real-valued), and he−ikz ∈ W1, ˜p _{we get that their product is also in W}1,p_{, and its norm is} estimated by the product of the norms of both functions in the spaces where we know they exist. So v2 ∈ W1,p ⊂ Lp˜. Further, we have v ∈ Lp˜ and by h and ψ0 being real-valued, we deduce (2.2.19) (−∆ + q)h = 0 ⇔ (−4 ¯∂∂ +4 ¯∂∂ψ0) ψ0 )h = 0 ⇔ ψ0∂∂h = h ¯¯ ∂∂ψ0. This we use on ¯ ∂v =∂ψ¯ 0∂h + ψ0∂∂h − h ¯¯ ∂∂ψ0− ∂ψ0∂he¯ −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∂ψ0e−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 that A := −( ¯∂ψ0)/ψ0∈ Lp∩ Lp˜(R2) and B := (e−k∂ψ0)/ψ0 ∈ Lp∩ Lp˜(R2). Thus we can construct an auxiliary function w = ¯∂−1(( ¯∂ψ0)/ψ0 − ∂ψ0e−kv/(ψ¯ 0v)) since the set of zeros of ¯v is of measure zero. Then we are applying ¯∂−1to a function in Lp∩ Lp˜_{and, hence, by Lemma 2.2.3} we have that w is bounded and continuous.

Given that w is bounded and continuous, we have that ve−w∈ Lp˜_{. We want to show that} v 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

¯

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Thus ve−w is entire and belongs to Lp˜. By Liouville theorem for Lp functions it follows that this function is zero everywhere, hence, due to the boundedness of w this implies that v is 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)ei¯z¯kis 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 just need to prove existence. For this we only need to be able to solve the integral equation, which we already derived before,

µ = 1 − gk∗ (qµ) We define two operators:

Ck: Lp(R2) → W1, ˜p(R2), defined by Ck(f ) = gk∗ f ; Mq : W1, ˜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 {qn}n∈N be a sequence of smooth compactly supported functions which converges to q in Lp(R2) (possible due to density of the first functions in the second space).

First, we show that the operator Mqn : W

1, ˜p_(R2_{) → L}p_(R2₎ is a compact, for each n ∈ N.

Let {ψk}k∈N be a bounded sequence in W1, ˜p(R2). The theorem of Rellich-Kondrachov states that the embedding W1, ˜p(R2) ,→ CB(supp(qn)) is compact. Hence, there exists a subsequence {ψki}ki∈N which converges to a ψ in the norm CB(supp(qn)).

Therefore,

kM_q_nψki− Mqnψk_Lp_(R2₎= kMqnψki− Mqnψk_Lp_(supp(q n))

≤ kψ_k_i− ψk_C_B_(supp(q_n₎₎kqnkLp_(R2₎ → 0, as k_i → +∞

Consequently, {Mqnψk}k∈N has a convergent subsequence and thus Mqn is a compact operator.

To conclude that Mqis compact we show that Mqn converges to it. By Sobolev embedding theorem we have W1, ˜p(R2) ,→ CB(R2), thus we obtain

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kM_q− M_q_nk_L(W1, ˜p_(R2_),Lp_(R2₎₎≤ sup kφk_{W 1, ˜}_p(R2)≤1 kM_qφ − MqnφkLp_(R2₎ ≤ sup kφk_{W 1, ˜}_p(R2)≤1 kφk_C B(R2)kqn− qkLp(R2) ≤ C sup kφk_{W 1, ˜}_p(R2)≤1 kφk_W1, ˜p_(R2₎kq_n− qk_Lp_(R2₎ ≤ Ckq_n− qk_Lp_(R2₎→ 0, as n → +∞. Hence, Mq is indeed a compact operator between these spaces.

Consequently, also CkMqis, and by Fredholm theory, we have that I +CkMqis a Fredholm operator of index zero on the space W1, ˜p. Since uniqueness is already shown, then the operator is 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

Ws0, ˜p_(R2_{) ,→ L}∞_(R2_). Therefore, together with (2.2.10) we have:

(2.2.21) kg_k∗ (qf )k_W_s0,˜p ≤ c₁|k|s0−1kqk_Lpkf k_W_s0,˜p Further, for |k| sufficiently large, we have

kC_kMqf k ≤ 1/2kf | s0, ˜p. So by Neumann series, we have

[I + CkMq]−1= ∞ X

l=0

(−1)l(CkMq)l,

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

kµ − 1k_Ws0,˜p = k[I + C_kM_q]−1(g_k∗ q)k_W_s0,˜p ≤ 2kg_k∗ qk_W_s0,˜p ≤ 2c₁|k|s0−1_kqk

Lp ≤ 1 (2.2.22)

Finally, using (2.2.8), we obtain

kµ − 1k_W_s0,˜p ≤ c₁|k|s0−1kqk_Lpkµ − 1k

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by applying both (2.2.21) and (2.2.22).

Now, we take a look at s1 ∈ [0, 2/˜p]. We have that kµ − 1k_Ws1,˜p = kg_k∗ (qµ)k_W_s1,˜p ≤ c1 |k|1−s1kqµkL p ≤ c1 |k|1−s1 (kq(µ − 1)kL p+ kqk_Lp) ≤ c1 |k|1−s1 (c2kqkL pkµ − 1k_W_s0,˜p+ kqk_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||L p, 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 unique solution to the Schr¨odinger equation from which we started. Therefore, there exists a unique solution depending on both parameters x and k. Up until now, we only have been worried about the behavior with respect to the first parameter. In this section we will take a look at the behavior of µ in terms of the other parameter and derive an operator, which will simplify our 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 + C_kMq] ∂ ∂¯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 Z R2 ∂ ∂¯k eix·ξ (ξ1+ iξ2)((ξ1− iξ2) + 2k) ! dξ1dξ2 = 1 4π2 Z R2 eix·ξ 2(ξ1+ iξ2) ! ∂ ∂¯k 1 (ξ1− iξ2)/2 + (k1+ ik2) ! dξ1dξ2 = 1 4π Z R2 eix·ξ 2(ξ1+ iξ2) ! δ(ξ1/2 + k1) + i(k2− ξ2/2) dξ1dξ2 = − 1 4π Z R2 ei2x·( ˜ξ1,− ˜ξ2) ( ˜ξ1− i ˜ξ2) ! δ(( ˜ξ1+ i ˜ξ2) + k) d ˜ξ1d ˜ξ2 = − 1 4π e−2i(x1k1−x2k2) k1− ik2 = − 1 4π e−k(x) ¯ k (2.3.2)

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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 4π¯k Z R2 e−k(x − y)q(y)µ(y, k) dy

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

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

(2.3.4) t(k) =

Z R2

ek(y)q(y)µ(y, k) dy.

Remark. If, in fact, the potential is of conductivity type we eill not have exceptional points, i.e., the solutions for each k ∈ C \ {0} will be unique. This implies that the operator t 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

4π¯ke−k(x)t(k) Combining Equations (2.3.1) and (2.3.5) we obtain

(2.3.6) ∂ ∂¯kµ(x, k) = 1 4π¯kt(k)[I + CkMq] −1_(e −k)(x)

One aspect to take into account is that the inverse operator applied to e−k does not make sense analytically since the latter function is not in W1, ˜p. So we proceed formally for now and 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 + g¯ k∗ (e−k(q ¯µ))

= e−k[¯µ + ¯gk∗ (q ¯µ)] = 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

4π¯kt(k)e−k(x)µ(x, k)

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

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

gk∗ (e−kf )(x) = Z

R2

gk(x − y)e−k(y)f (y)dy = Z R2 Z R2 ei(x1ξ1+x2ξ2) |ξ|2_{+ 2k(ξ} 1+ iξ2)

e−i(y1(ξ1+2k1)+y2(ξ2−2k2))_{f (y) dξ}

1dξ2dy

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) = Z R2 Z R2 ei(x1(ξ1−2k1)+x2(ξ2+2k2)) |ξ|2_{− 2k(ξ} 1+ iξ2) e−i(y1ξ1+y2ξ2)_{f (y) dξ} 1dξ2dy = e−k(x) Z R2 Z R2 ei(x−y)·ξ |ξ|2_{− 2k(ξ} 1+ 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−β1, ˜p topology, for β > 2/˜p.

It will be helpful to use the notation (similar to Fourier transform): F f (k) = i/2

Z C

ek(z)f (z)dz ∧ d¯z. (2.3.10)

Lemma 2.3.1. Let α > 2/p0, β > 2/˜p (where p0 denotes the exponent dual to p). The map k 7→ (∂ + ik)−1 is differentiable on C in the strong operator topology: Lpα → Lp−β˜ and

(2.3.11) ∂

∂¯k(∂ + ik)

−1_{f (z) = −}i

πF f (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 Z C ek(w − z) ¯ w − ¯z f (w) dw ∧ d ¯w. 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 Z C yj− xj ¯ w − ¯z ek(w − z)f (w) dw ∧ d ¯w, w = y1+ iy2.

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A first step is to show that _∂k∂ j(∂ + ik) −1 _{= D} j(k). Clearly, kDj(k)f kL∞ ≤ 2 πkf kL1 since yj−xj ¯ w−¯z

≤ 1 and |ek(w − z)| = 1. Furthermore, via Proposition B.2.1 we have for α > 2/p 0

that Lpα ⊂ L1 and, for β > 2/˜p, we have that hxi−β ∈ Lp˜. Hence, using the representation of inverse operators via convolution with fundamental solutions we get

lim h→0 (∂ + i((k1+ h) + ik2))−1− (∂ + ik)−1 h − D1(k) f ˜ p Lp_−β˜ = lim h→0 Z z∈R2 hzi−β 2πi Z ek(w − z) ¯ w − ¯z eh(w − z) − 1 h − 2i(y1− x1) f (w) dw ∧ d ¯w ˜ p . Moreover, we can apply the dominated convergence theorem twice, and use the limit

lim h→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 to show _∂k∂

2(∂ + ik)

−1_{= D} 2(k). Joining both together, we get

∂ ∂¯k(∂ + ik) −1 f = 1 2(D1(k) + iD2(k))f (z) = 1 2π Z C ek(w − z)f (w) dw ∧ d ¯w = −i πe−k(z) i 2 Z C ek(w)f (w) dw ∧ d ¯w = −i πe−k(z)F f (k) (2.3.13)

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

Lemma 2.3.2. If 2/p0 < α < 1 and α + 1 − 2/p < δ < 2/p0, then (2.3.14) ∂ ¯∂−1Lp_α ⊂ Lp α+ {u ∈ L p δ: ∂u ∈ L p α} Proof of Lemma 3.2. Let f ∈ Lpα. We first prove the identity (2.3.15) ∂ ¯∂−1f = hzi−α∂ ¯∂−1hziαf −α 2 hzi −2 ¯ z ¯∂−1f +α 2hzi −α ∂ ¯∂−1hziα−2z ¯∂−1f To do so we take into consideration that ∂ hziα= α₂ hziα−2z, and start with¯ (2.3.16) ∂ hzi¯ α∂¯−1f = α

2 hzi

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

¯

∂−1f ∈ Lp_δ−1 ⇒ hziδ−1∂¯−1f ∈ Lp ⇒ hziδ−αhziα−1∂¯−1f ∈ Lp ⇒ hziα−2z ¯∂−1f ∈ Lp_δ−α,

and for the second term it follows, similarly,

f ∈ Lp_δ ⇒ hziαhziδ−αf ∈ Lp ⇒ hziαf ∈ Lp_δ−α.

Furthermore, we have 1 − 2/p < δ − α < 2/p0 so we can apply ∂ ¯∂−1 to (2.3.16) and deduce ∂ ¯∂−1 ∂ hzi¯ α∂¯−1f = α 2∂ ¯∂ −1_hziα−2_{z ¯}_∂−1_{f + ∂ ¯}_∂−1_hziα_f ⇔ α 2hzi α−2_{z + hzi}_¯ α_{∂ ¯}_∂−1_{f =} α 2∂ ¯∂ −1_hziα−2_{z ¯}_∂−1_{f + ∂ ¯}_∂−1_hziα_f. Dividing both sides by hziα gives the expected equation (2.3.15).

The first term on the right hand side of (2.3.15) is in Lpαbecause we can use Lemma B.3.1 in the particular case where δ = 0. The other two terms were in Lp_δ−α before the division by hziα so by definition they are now in Lp_δ. 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) ∂hzi−2z ¯¯∂−1f= − hzi−4¯z2∂¯−1f + hzi−2z∂ ¯¯ ∂−1f

Since f ∈ Lp_δ, by Lemma B.3.1 we have the right-hand side in Lp_δ+1, and therefore, Lp_δ+1⊂ Lpα, because α < α + 2/p0= α + 2 − 2/p < δ + 1.

For the remaining ter, we start by computing ∂hzi−α∂ ¯∂−1hziα−2z ¯∂−1f (2.3.18) = −α 2 hzi −α−2 ¯ z∂ ¯∂−1hziα−2z ¯∂−1f + α − 2 2 hzi −α ∂ ¯∂−1hziα−4|z|2∂¯−1f + hzi−α∂ ¯∂−1hziα−2∂¯−1f + hzi−α∂ ¯∂−1hziα−2z∂ ¯∂−1f =: a + b + c + d

(2.3.19)

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

Case a: Similar as before we have hziα−2z ¯∂−1f is in Lp_δ−α and ∂ ¯∂−1 is bounded on Lp_δ−α, so that a ∈ Lp_δ+1 ⊂ Lpα

Case b: hziα−4∂¯−1f ∈ Lp_δ−α+3 ⇒ hziα−4|z|2_∂_¯−1_{f ∈ L}p

δ−α+1. We can now choose δ 0 _∈ (α, α + 2/p0). Then, we have δ0 < δ + 1, so that hziα−4|z|2_∂¯−1_{f ∈ L}p

δ0_−α. Furthermore, 0 < δ0− α < 2/p0 so that we can apply ∂ ¯∂−1 and we get that b ∈ Lp_δ0 ⊂ L

p α.

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Case c: hziα−2∂¯−1f ∈ Lp_δ−α+1 ⊂ Lp_δ0_−α with δ0 defined as above. Therefore, we have ∂ ¯∂−1(hziα−2∂¯−1f ) ∈ L_δp0_−α, this implies c ∈ Lp_δ0 ⊂ Lpα.

Case d: Finally, from ∂ ¯∂−1 ∈ Lp_δ, we have hziα−2z∂ ¯∂−1f ∈ Lp_δ−α+1 ⊂ Lp_δ0_−α. Hence, d ∈ Lpα.

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

Lemma 2.3.3. Let 2/p0 < α < 1, β > 2/˜p. The map k 7→ gk∗ · is differentiable on C \ {0} in the strong operator topology: Lpα→ W−β1, ˜p and

(2.3.20) ∂

∂¯k(gk∗ f ) = − 1

4π¯kF f (k)e−k

Proof. Let f ∈ Lpα then by Lemma 2.3.2 we can write ∂ ¯∂−1f = f1+ f2 with f1∈ Lpα, f2 ∈ Lp_δ, for some δ ∈ (α + 1 − 2/p, 2/p0) and ∂f2 ∈ Lpα. Then we get

(∂ + ik)−1(∂ ¯∂−1f ) = (∂ + ik)−1f1+ ik ik(∂ + ik) −1 f2+ 1 ikf2− 1 ik(∂ + ik) −1 (∂ + ik)f2 = (∂ + ik)−1f1+ 1 ikf2− 1 ik(∂ + ik) −1_∂f 2 (2.3.21)

(small remark: f2 ∈ W1,p ⊂ Lp˜). Therefore, by Lemma 2.3.1., we have that the right-hand side of (2.3.21) is differentiable in Lp_−β˜ for k 6= 0 and

(2.3.22) ∂ ∂¯k(∂ + ik) −1_{∂ ¯}_∂−1_{f = −}i πe−kF (∂ ¯∂ −1_{f )(k) = −}i π k ¯ ke−kF f (k),

where the last equality follows using the Fourier multiplier symbol of ∂ ¯∂−1. Now remembering formula (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 second we can use the above equation (2.3.21). With this, the differentiability of gk∗ f in Lp−β˜ follows as well as the desired formula (2.3.20). To prove the differentiability in W_−β1, ˜p by (2.2.11) and Lemma 2.3.1 we also observe that

¯ ∂(gk∗ f ) = − 1 ik ¯ ∂∂ − (∂ + ik)¯ −1∂ ¯∂−1 f = − f 4ik + ¯ ∂ 4ike−k∂ −1 (ek∂ ¯∂−1) f = − f 4ik + ¯ ∂ 4ike−k∂ −1 _{∂ e} k∂¯−1f − ikek∂¯−1f = −∂¯ 4(∂ + ik) −1_∂_¯−1_{f = −(∂ + ik)}−1_{f /4,}

so we can use Lemma 3.1. again. For the derivative with respect to z we use the above equation 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 in Lp_−β˜ .

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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 the coefficients, i.e., we consider only 2/˜p < β < ρ − 2/p0, which is not a problem since the norm || · ||_W1, ˜p

−β is decreasing in β. Furthermore, we choose α such that 2/p

0 _{< α < min(1, ρ − β).} We rewrite Equation (2.2.8), to take into account the desired growth

hxi−βµ = hxi−β− K(k)(hxi−βµ), with the operator K(k)f = hxi−βgk∗ (h·i qf ) .

By our choice of α and β, we have

(2.3.24) q ∈ Lp_ρ⇒ hxiβ ∈ Lp_ρ−β ⇒ hxiβ ∈ Lp_α,

where the last implication follows because of α < min(1, ρ−β) < ρ−β. Therefore, analogously as before, we have that multiplication by hxiβq is a compact operator: W1, ˜p → Lpα. From this and Lemma 2.3.3 it follows that k → K(k) is differentiable on C \ {0} in the uniform operator topology W1, ˜p → W1, ˜p_.

Let us suppose that ˜h is a solution in W1, ˜p of ˜h = −K(k)˜h. Then hxiβ˜h satisfies hxiβ˜h = −gk∗ (q h·iβh) so that hxi˜ βh is in W˜ 1, ˜p and, hence, vanishes since k is assumed to be a non-exceptional point. Thus I + K(k) is invertible on W1, ˜p _and

∂ ∂¯k hxi−βµ(x, k)= − ∂ ∂¯kK(k) hxi−βµ(x, k)− K(k) hxi −β ∂ ∂¯kµ(x, k) (2.3.25) ⇔ (I + K(k)) ∂ ∂¯k hxi−βµ(x, k)= − ∂ ∂¯kK(k) hxi−βµ(x, k) (2.3.26) ∂ ∂¯k hxi−βµ(x, k) = −(I + K(k))−1 ∂ ∂¯kK(k) hxi−βµ(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) hxi−β . We use (2.3.7) and given that hxi is real-valued we get (2.3.28) (I + K(k))−1e−k(x) hxi−β

= e−k(I + K(k))−1hxi−β Thus (2.3.9) holds in the W_−β1, ˜p.

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) =pγ(z)

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2.4 Behavior near k = 0

In this section the assumptions on the potential q are more general, i.e., the results will apply 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 hxi−aψ0 ∈ L∞(R2_{) for some a < min(1, ρ − 1) and ∆ψ}

0 = 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) = −_2π1 log |x|, and the following are equivalent: (a) ψ0 ∈ L∞(R2); (b) lim R→∞ 1 πR2 Z |x|<R

ψ0(x)dx exists (and equals c∞); (c) R

R2qψ0 = 0;

(d) ψ0− c∞∈ W1, ˜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 the Laplacian G0. Lemma 2.4.2. If f ∈ Lpρ(R2), 1 < p < 2, ρ > 1, then: (i) kG₀∗ f + 1 2π(log |x|) Z R2 f k_Lp˜ ≤ ckf k_Lp ρ; (ii) k∇G0∗ f kLp˜ ≤ ckf k_Lp.

Proof. To the first inequality, we will use some properties of the logarithm, to split one of the integrals into two parts.

G0∗ f + 1 2π(log |x|) Z R2 f Lp˜ = " Z R2 Z R2 1

2π(log |x| − ln |x − y|)f (y) dy ˜ p dx #1/ ˜p (2.4.2) ≤ 1 2π Z R2 Z 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) := " Z |x|≤|y| ln|x + y| |x| ˜ p dx #1/ ˜p and, I2(y) := " Z |x|>|y| ln|x + y| |x| ˜ p dx #1/ ˜p ,

Calderón's inverse conductivity problem in two dimensions: Nachman's approach and the case for complex conductivity

Ivan Mira

Pombo

O problema de conductividade inverso de Calder´

on

a duas dimens˜

oes: a abordagem de Nachman e o

caso de conductividade complexa

Calder´

on’s inverse conductivity problem in two

dimensions: Nachman’s approach and the case for

complex conductivity

Ivan Mira

Pombo

O problema de conductividade inverso de Calder´

on

a duas dimens˜

oes: a abordagem de Nachman e o

caso de conductividade complexa

Calder´

on’s inverse conductivity problem in two

dimensions: Nachman’s approach and the case for

complex conductivity

Contents

List of Figures

Chapter 1

Introduction

1.1

Inverse Conductivity Problem (ICP)

Chapter 2

Nachman’s Approach

2.1

The ways of Nachman

2.2

Exponentially Growing Solutions

2.3

The ¯

∂ equation

2.4

Behavior near k = 0