Monotonicity based shape reconstruction in electrical impedance tomography

TL;DR

This paper introduces a monotonicity-based method for shape reconstruction in electrical impedance tomography, enabling the detection of inclusions by comparing measurements with test regions, based on a novel converse relation.

Contribution

It presents a new converse monotonicity relation in EIT that simplifies shape reconstruction by direct comparison of measurements with test regions.

Findings

Effective inclusion detection using monotonicity comparisons.

The method provides a straightforward approach to shape reconstruction.

The approach is based on a novel monotonicity relation in the NtD operators.

Abstract

Current-voltage measurements in electrical impedance tomography can be partially ordered with respect to definiteness of the associated self-adjoint Neumann-to-Dirichlet operators (NtD). With this ordering, a point-wise larger conductivity leads to smaller current-voltage measurements, and smaller conductivities lead to larger measurements. We present a converse of this simple monotonicity relation and use it to solve the shape reconstruction (aka inclusion detection) problem in EIT. The outer shape of a region where the conductivity differs from a known background conductivity can be found by simply comparing the measurements to that of smaller or larger test regions.

Equations

$$\Lambda (\sigma ):g\mapsto u_{\sigma }^g{|}_{\partial \Omega },$$

$$\nabla \cdot \sigma \nabla u_{\sigma }^g=0\text{ in }\Omega ,\sigma {\partial }_{\nu }{u}_{\sigma }^g{|}_{\partial \Omega }=g\text{ on }\partial \Omega ,$$

$$\sigma \le \tau \text{mbox}implies\Lambda (\sigma )\ge \Lambda (\tau ).$$

$$B\subseteq D$$

$$\text{mbox}ifandonlyif\Lambda (1)+\frac{1}{2}{\Lambda }^{\prime }(1){\chi }_B\ge \Lambda (\sigma ).$$

$$D^{+}\cup D^{-}\subseteq C$$

$$\text{mbox}ifandonlyif\Lambda (1)+{\Lambda }^{\prime }(1){\chi }_C\le \Lambda (\sigma )\le \Lambda (1)-{\Lambda }^{\prime }(1){\chi }_C.$$

$$A\ge B$$

$$⟨g,(A-B)g⟩\ge 0,\text{mbox}forallg\in L_{\diamond }^2(\partial \Omega ).$$

$$\Lambda (\sigma ):L_{\diamond }^2(\partial \Omega )\to L_{\diamond }^2(\partial \Omega ),g\mapsto u_{\sigma }^g{|}_{\partial \Omega },$$

$$\nabla \cdot \sigma \nabla u_{\sigma }^g=0\text{ in }\Omega ,\sigma {\partial }_{\nu }{u}_{\sigma }^g{|}_{\partial \Omega }=g\text{ on }\partial \Omega$$

$${\int }_{\Omega }\sigma \nabla u_{\sigma }^g\cdot \nabla v\mathrm{d}x={\int }_{\partial \Omega }gv{|}_{\partial \Omega }\mathrm{d}s\text{mbox}forallv\in H_{\diamond }^1(\Omega ).$$

$$⟨g,\Lambda (\sigma )h⟩={\int }_{\Omega }\sigma \nabla u_{\sigma }^g\cdot \nabla u_{\sigma }^h\mathrm{d}x.$$

$${\Lambda }^{\prime }(\sigma )\kappa :L_{\diamond }^2(\partial \Omega )\to L_{\diamond }^2(\partial \Omega )$$

$$⟨\left({\Lambda }^{\prime }(\sigma )\kappa \right)g,h⟩=-{\int }_{\Omega }\kappa \nabla u_{\sigma }^g\cdot \nabla u_{\sigma }^h\mathrm{d}x.$$

$${\kappa }_1\le {\kappa }_2\text{mbox}implies{\Lambda }^{\prime }(\sigma ){\kappa }_1\ge {\Lambda }^{\prime }(\sigma ){\kappa }_2.$$

$$\Gamma =\partial O\cap B_{ϵ}(z)$$

$$O\cap B_{ϵ}(z)$$

$$\displaystyle\kappa=0\mbox{ (a.e.) on $U$}$$

$$\displaystyle\quad\mbox{ if and only if }\quad\chi_{\mathop{\operator@font supp}\nolimits\kappa}=0\mbox{ (a.e.) on $U$.}$$

$$x\notin \overline{\mathrm{inn}\mathrm{supp}\kappa }\text{mbox}impliesx\notin \text{operator@font}supp\kappa .$$

$$\overline{\mathrm{inn}\mathrm{supp}\kappa }=\text{operator@font}supp\kappa ={\mathrm{out}}_{\partial \Omega }\mathrm{supp}\kappa .$$

$${\int }_{\Omega }({\sigma }_1-{\sigma }_2)|\nabla u_2{|}^2\mathrm{d}x\ge ⟨g,\left(\Lambda ({\sigma }_2)-\Lambda ({\sigma }_1)\right)g⟩\ge {\int }_{\Omega }\frac{{\sigma }_2}{{\sigma }_1}({\sigma }_1-{\sigma }_2)|\nabla u_2{|}^2\mathrm{d}x.$$

$${\int }_{\Omega }{\sigma }_1\nabla u_1\cdot \nabla u_2\mathrm{d}x=⟨g,\Lambda ({\sigma }_2)g⟩={\int }_{\Omega }{\sigma }_2\nabla u_2\cdot \nabla u_2\mathrm{d}x$$

$${\int }_{\Omega }{\sigma }_1|\nabla (u_1-u_2){|}^2\mathrm{d}x$$

$$=⟨g,\Lambda ({\sigma }_1)g⟩-⟨g,\Lambda ({\sigma }_2)g⟩+{\int }_{\Omega }({\sigma }_1-{\sigma }_2)|\nabla u_2{|}^2\mathrm{d}x.$$

$$⟨g,\left(\Lambda ({\sigma }_2)-\Lambda ({\sigma }_1)\right)g⟩$$

$$={\int }_{\Omega }({\sigma }_1-{\sigma }_2)|\nabla u_1{|}^2\mathrm{d}x+{\int }_{\Omega }{\sigma }_2|\nabla (u_2-u_1){|}^2\mathrm{d}x$$

$$={\int }_{\Omega }\left({\sigma }_1|\nabla u_1{|}^2+{\sigma }_2|\nabla u_2{|}^2-2{\sigma }_2\nabla u_1\cdot \nabla u_2\right)\mathrm{d}x$$

$$={\int }_{\Omega }{\sigma }_1{\left|\nabla u_1-\frac{{\sigma }_2}{{\sigma }_1}\nabla u_2\right|}^2\mathrm{d}x+{\int }_{\Omega }\left({\sigma }_2-\frac{{{\sigma }_2}^2}{{\sigma }_1}\right)|\nabla u_2{|}^2\mathrm{d}x.$$

$${\sigma }_1\le {\sigma }_2\text{mbox}implies\Lambda ({\sigma }_1)\ge \Lambda ({\sigma }_2).$$

$$B\subseteq D\text{mbox}implies\Lambda (1+{\chi }_B)\ge \Lambda (\sigma ),$$

$$R:=\bigcup \left\{B\subseteq \Omega :B\text{mbox}measurable,and\Lambda (1+{\chi }_B)\ge \Lambda (\sigma )\right\}$$

$${\Lambda }^{\prime }(1)(1-\sigma )\ge \Lambda (1)-\Lambda (\sigma )\ge {\Lambda }^{\prime }(1)\left(\frac{1}{\sigma }(1-\sigma )\right).$$

Full text

We present a converse of this simple monotonicity relation and use it to solve the shape reconstruction (aka inclusion detection) problem in EIT. The outer shape of a region where the conductivity differs from a known background conductivity can be found by simply comparing the measurements to that of smaller or larger test regions.

1 Introduction

We consider the shape reconstruction (aka inclusion detection) problem in electrical impedance tomography (EIT). Let $\Omega$ describe an electrically conducting object which contains inclusions in which the conductivity $\sigma(x)$ differs from an otherwise known background conductivity. Our aim is to detect these inclusions from current/voltage-measurements on the boundary $\partial\Omega$.

We assume that $\Omega\subset\mathbbm{R}^{n}$, $n\geq 2$ is a domain with smooth boundary $\partial\Omega$ and outer normal vector $\nu$. For ease of presentation we also assume that $\Omega$ is bounded, the background conductivity is equal to $1$ and that we are given measurements on the complete boundary $\partial\Omega$. Our results easily extend to inhomogeneous (but known) backgrounds and partial boundary measurements, cf. section 4.3.

With these assumptions, our goal is to determine the inclusions shape, i.e., the set $\mathop{\operator@font supp}\nolimits(\sigma-1)$, from knowledge of the Neumann-to-Dirichlet (NtD) operator

[TABLE]

where $u^{g}_{\sigma}$ is the solution of

[TABLE]

cf. section 2.1 for the precise mathematical setting.

In this work, we show that $\mathop{\operator@font supp}\nolimits(\sigma-1)$ can be reconstructed by so-called monotonicity tests, which simply compare $\Lambda(\sigma)$ (in the sense of quadratic forms) to NtD-operators $\Lambda(\tau)$ of test conductivities $\tau$. To be more precise, the support of $\sigma-1$ can be reconstructed under the assumption that $\mathop{\operator@font supp}\nolimits(\sigma-1)\subset\Omega$ has connected complement. Otherwise, what we can reconstruct is essentially the support together with all holes that have no connection to the boundary $\partial\Omega$.

Moreover, we show that the test NtDs $\Lambda(\tau)$ can be replaced (without losing any information) by their linear approximations using the Fréchet derivative $\Lambda^{\prime}(1)$ of $\Lambda(\sigma)$ around the background conductivity. Let us stress that the linearized tests still exactly recover the inclusion which is in accordance with the general principle that the linearized EIT problem still contains the exact shape information, cf. [16].

The term monotonicity tests is used because our test criteria are motivated and partly follow from the simple and well-known monotonicity relation

[TABLE]

It seems quite natural and intuitive to probe the domain with test inclusions using the implication (1), and this idea has been worked out and numerically tested in the works of Tamburrino and Rubinacci [47, 46]. The main new part of this work is to rigorously justify this natural idea by proving a non-trivial converse of the implication (1). Our proofs are based on the theory of localized potentials [6].

For a quick impression of our result let us state it for two frequently considered special cases (see examples 15, 17, 21 and 23). (Note that throughout the paper we use the relation symbol ”$\subset$” instead of ”$\subseteq$”, if non-equality of the two related sets is obvious.)

1. (a)

Let $\sigma=1+\chi_{D}$ where $D$ is open, and $\overline{D}\subset\Omega$ has a connected complement. Then for every open ball $B\subseteq\Omega$

[TABLE] 2. (b)

Let $\sigma=1+\chi_{D^{+}}-\frac{1}{2}\chi_{D^{-}}$ where $D^{+},D^{-}\subseteq\Omega$ are open, $\overline{D^{+}}\cap\overline{D^{-}}=\emptyset$, and $\overline{D^{+}}\cup\overline{D^{-}}\subset\Omega$ has a connected complement. Then for every closed $C\subset\Omega$ with connected complement

[TABLE]

(a) is a special case of the definite case in which either all inclusions have a higher conductivity, or all inclusions have a lower conductivity, than the background. (a) shows how to test whether a small ball $B$ lies inside the inclusion or not. The inclusion can thus be obtained as the union of all balls that fulfill the test.

(b) is a special case of the more general indefinite case in which the conductivity may differ in both directions from the background. Using the result in (b) we can test whether a large set $C$ contains the inclusions or not. The inclusion can thus be obtained as the intersection of all these large sets.

Our results show that (under quite general assumptions) monotonicity tests determine $\mathop{\operator@font supp}\nolimits(\sigma-1)$ up to holes that have no connection to the boundary $\partial\Omega$.

Non-iterative methods for shape reconstruction problems have been studied intensively in the last 25 years, cf., e.g., the overview of Potthast [43]. In the context of EIT, the inclusion detection problem was first considered by Friedmann and Isakov [3, 4]. For the following brief overview, we restrict ourselves to the two most prominent and elaborated methods for detecting inclusions of unknown conductivity from the full Neumann-to-Dirichlet (or Dirichlet-to-Neumann) operator on all or part of the boundary: the Factorization Method and the Enclosure Method.

The Factorization Method (FM) was introduced by Kirsch [33, 34] for inverse scattering problems and extended to impedance tomography by Brühl and Hanke [2, 1]. For its further developments in the context of EIT see [35, 10, 5, 11, 20, 40, 42, 7, 36, 9, 15, 44, 17, 45] and the recent review [14]. The Factorization Method reconstructs the shape of inclusions (up to holes that have no connection to the boundary), but two major problems have not been solved so far. First of all, the method relies on a range test (or infinity test) for which there is no known convergent implementation (see, however, Lechleiter [39] for a first step in this direction). Second, the method has only been justified for the definite case (or that the domain can be split into two a-priori known regions with the definiteness property, cf. Schmitt [44] and the review [14]).

The Enclosure Method was introduced by Ikehata [25, 26]. Further extensions including the use of the Sylvester-Uhlmann complex geometrical optics solutions have been worked out in [2, 30, 27, 31, 22, 48, 21]. The method yields a stable testing criterion and it does not require the definiteness assumption (see [22]). However, it does require the construction of special, strongly oscillating probe functions and only reconstructs the convex hull of the inclusions (plus some non-convex features depending on the probe functions).

The herein presented monotonicity tests seem to be a particularly simple and intuitively appealing solution to the long-studied inclusion detection problem. They characterize the outer shape of the inclusions and not just the convex hull. They work for the general indefinite case (though the implementation is simpler in the definite case). Also, they allow a stable implementation (see remark 10), and their linearized versions do not require solving inhomogeneous forward problems.

The paper is organized as follows. Section 2 introduces the mathematical setting and the concept of inner and outer support. In section 3 we derive the main theoretical tools for our proofs: monotonicity estimates and localized potentials. Section 4 then contains our main results: the characterization of inclusion by simple and stable monotonicity tests.

2 Basic notations and support definitions

2.1 Basic notations and the mathematical setting

Let $\Omega\subset\mathbbm{R}^{n}$, $n\geq 2$ be a bounded domain with smooth boundary $\partial\Omega$ and outer normal vector $\nu$. $L_{+}^{\infty}(\Omega)$ denotes the subspace of $L^{\infty}(\Omega)$-functions with positive essential infima. $H_{\diamond}^{1}(\Omega)$ and $L_{\diamond}^{2}(\partial\Omega)$ denote the spaces of $H^{1}$- and $L^{2}$-functions with vanishing integral mean on $\partial\Omega$.

The $L^{2}_{\diamond}(\partial\Omega)$-inner product is denoted by $\langle\cdot,\cdot\rangle$. For two bounded selfadjoint operators $A,B:\ L_{\diamond}^{2}(\partial\Omega)\to L_{\diamond}^{2}(\partial\Omega)$ we write

[TABLE]

if it holds in the sense of quadratic forms, i.e.,

[TABLE]

For $\sigma_{1},\sigma_{2}\in L^{\infty}(\Omega)$ we write $\sigma_{1}\geq\sigma_{2}$ if it holds pointwise (a.e.) on $\Omega$.

For $\sigma\in L^{\infty}_{+}(\Omega)$, the Neumann-to-Dirichlet (NtD) operator $\Lambda(\sigma)$ is defined by

[TABLE]

where $u^{g}_{\sigma}\in H^{1}_{\diamond}(\Omega)$ is the unique solution of

[TABLE]

which is equivalent to

[TABLE]

It is well known $\Lambda(\sigma)$ is a selfadjoint compact linear operator, and that the associated bilinear form is given by

[TABLE]

$\Lambda$ is Fréchet-differentiable, cf., e.g. Lechleiter and Rieder [41] for a recent proof that uses only the abstract variational formulation (see also [23] for similar results). Given some direction $\kappa\in L^{\infty}(\Omega)$ the derivative

[TABLE]

is the selfadjoint compact linear operator associated to the bilinear form

[TABLE]

Note that for $\kappa_{1},\kappa_{2}\in L^{\infty}(\Omega)$ we obviously have that

[TABLE]

The terms piecewise continuous and piecewise analytic are understood in the following sense.

Definition 1**.**

1. (a)

A subset $\Gamma\subseteq\partial O$ of the boundary of an open set $O\subseteq\mathbbm{R}^{n}$ is called a smooth boundary piece if it is a $C^{\infty}$-surface and $O$ lies on one side of it, i.e., if for each $z\in\Gamma$ there exists a ball $B_{\epsilon}(z)$ and a function $\gamma\in C^{\infty}(\mathbbm{R}^{n-1},\mathbbm{R})$ such that upon relabeling and reorienting

[TABLE] 2. (b)

$O$* is said to have smooth boundary if $\partial O$ is a union of smooth boundary pieces. $O$ is said to have piecewise smooth boundary if $\partial O$ is a countable union of the closures of smooth boundary pieces.* 3. (c)

A function $\kappa\in L^{\infty}(\Omega)$ is called piecewise analytic if there exist finitely many pairwise disjoint subdomains $O_{1},\ldots,O_{M}\subset\Omega$ with piecewise smooth boundaries, such that $\overline{\Omega}=\overline{O_{1}\cup\ldots\cup O_{M}}$, and $\kappa|_{O_{m}}$ has an extension which is (real-)analytic in a neighborhood of $\overline{O_{m}}$, $m=1,\ldots,M$. 4. (d)

A function $\kappa\in L^{\infty}(\Omega)$ is called piecewise continuous if $\kappa$ is continuous on an open set $O\subset\Omega$ and $\Omega\setminus O$ is a set of zero measure.

2.2 Inner and outer support

We will show that our method reconstructs $\mathop{\operator@font supp}\nolimits(\sigma-1)$ (the inclusion) up to holes that cannot be connected to the boundary $\partial\Omega$ without crossing the support. For the precise formulation, we will now introduce the concept of the inner and the outer support of a measurable function. For the frequently considered case that the inclusion has a connected complement and the conductivity is piecewise continuous, the inner and the outer support only differ by the boundary of the support, cf. corollary 5. The following has been inspired by the use of the infinity support of Kusiak and Sylvester [38], cf. also [8, 16].

Definition 2**.**

A relatively open set $U\subseteq\overline{\Omega}$ is called connected to $\partial\Omega$ if $U\cap\Omega$ is connected and $U\cap\partial\Omega\neq\emptyset$.

Definition 3**.**

For a measurable function $\kappa:\ \Omega\to\mathbbm{R}$ we define:

1. (a)

the support $\mathrm{supp}\,\kappa$ as the complement (in $\overline{\Omega}$) of the union of those relatively open $U\subseteq\overline{\Omega}$, for which $\kappa|_{U}\equiv 0$, 2. (b)

the inner support $\mathrm{inn\,supp}\,\kappa$ as the union of those open sets $U\subseteq\Omega$, for which $\mathrm{ess\,inf}_{x\in U}|\kappa(x)|>0$. 3. (c)

the outer support $\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,\kappa$ as the complement (in $\overline{\Omega}$) of the union of those relatively open $U\subseteq\overline{\Omega}$ that are connected to $\partial\Omega$ and for which $\kappa|_{U}\equiv 0$.

The interior of a set $M\subseteq\Omega$ is denoted by $\mathrm{int}\,M$ and its closure (with respect to $\mathbbm{R}^{n}$) by $\overline{M}$. If $M$ is measurable we also define

1. (d)

$\mathrm{out}_{\partial\Omega}\,M=\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,\chi_{M},$**

where $\chi_{M}$ is the characteristic function of $M$.

Lemma 4**.**

For every measurable function $\kappa:\Omega\to\mathbbm{R}$ and every measurable set $M$ the following properties hold.

1. (a)

$\mathop{\operator@font supp}\nolimits\kappa,\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,\kappa,\mathrm{out}_{\partial\Omega}\,M\subseteq\overline{\Omega}$* are closed.* 2. (b)

$\mathrm{inn\,supp}\,\kappa\subseteq\Omega$* is open.* 3. (c)

$\mathrm{inn\,supp}\,\kappa\subseteq\mathop{\operator@font supp}\nolimits\kappa\subseteq\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,\kappa$. 4. (d)

$\mathrm{out}_{\partial\Omega}\,(\mathop{\operator@font supp}\nolimits\kappa)=\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,\kappa$** 5. (e)

If $\mathop{\operator@font supp}\nolimits\kappa\subseteq\Omega$ and $\Omega\setminus\mathop{\operator@font supp}\nolimits\kappa$ is connected then $\mathop{\operator@font supp}\nolimits\kappa=\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,\kappa$. 6. (f)

If $\kappa$ is piecewise continuous then $\mathop{\operator@font supp}\nolimits\kappa=\overline{\mathrm{inn\,supp}\,\kappa}$.

Proof.

1. (a)

and (b) immediately follow from definition 3. 2. (c)

If $\kappa=0$ (a.e.) on a relatively open set $U\subset\overline{\Omega}$, then $\kappa=0$ (a.e.) on the open set $U\cap\mathrm{inn\,supp}\,\kappa$. From the definition of the inner support, it follows that $U\cap\mathrm{inn\,supp}\,\kappa=\emptyset$. This shows the first inclusion in (c). The second inclusion is obvious. 3. (d)

follows from the fact that for every relatively open set $U\subseteq\overline{\Omega}$ we have

[TABLE] 4. (e)

Since $\mathop{\operator@font supp}\nolimits\kappa\subseteq\Omega$ implies that $\overline{\Omega}\setminus\mathop{\operator@font supp}\nolimits\kappa$ contains $\partial\Omega$, (e) immediately follows from (c) and definition 3. 5. (f)

Let $\kappa$ be continuous on an open set $O\subset\Omega$ where $\Omega\setminus O$ has zero measure. The assertion follows from (a) and (c) if we can show that for every $x\in\overline{\Omega}$

[TABLE]

Let $x\notin\overline{\mathrm{inn\,supp}\,\kappa}$. Then there exists a relatively open set $B\subset\overline{\Omega}$ with $x\in B$ and $B\cap\overline{\mathrm{inn\,supp}\,\kappa}=\emptyset$. Obviously, $\{\xi\in O:\kappa(\xi)\neq 0\}\subseteq\mathrm{inn\,supp}\,\kappa$, so that $\kappa=0$ on $O\cap B$. Since $\Omega\setminus O$ has zero measure, we have that $\kappa=0$ (a.e.) on $B$ and thus $B\cap\mathop{\operator@font supp}\nolimits\kappa=\emptyset$ which shows the assertion.

∎

As a consequence of Lemma 4(e) and (f) we have

Corollary 5**.**

If $\kappa$ is piecewise continuous, $\mathop{\operator@font supp}\nolimits\kappa\subseteq\Omega$ and $\Omega\setminus\mathop{\operator@font supp}\nolimits\kappa$ is connected then

[TABLE]

3 Monotonicity and localized potentials

3.1 A monotonicity principle

Our main theoretical tools are a monotonicity estimate and the theory of localized potentials. The following estimate goes back to Ikehata, Kang, Seo, and Sheen [32, 24], cf., also the similar results in Ide et al. [22], Kirsch [35], and in [15, 16]. For the convenience of the reader we state the estimate together with a short proof that we copy from [16, lemma 2.1].

Lemma 6**.**

Let $\sigma_{1},\sigma_{2}\in L^{\infty}_{+}(\Omega)$ be two conductivities, $g\in L^{2}_{\diamond}(\Omega)$ be an applied boundary current and $u_{2}:=u^{g}_{\sigma_{2}}\in H^{1}_{\diamond}(\Omega)$. Then

[TABLE]

Proof.

Let $u_{1}:=u^{g}_{\sigma_{1}}\in H^{1}_{\diamond}(\Omega)$. From (3) we deduce

[TABLE]

and thus

[TABLE]

Since the left hand side is non-negative, the first asserted inequality follows.

Interchanging $\sigma_{1}$ and $\sigma_{2}$ we obtain

[TABLE]

Since the first integral on the right hand-side is non-negative, the second asserted inequality follows. ∎

We call lemma 6 a monotonicity estimate because of the following corollary.

Corollary 7**.**

For two conductivities $\sigma_{1},\sigma_{2}\in L^{\infty}_{+}(\Omega)$

[TABLE]

Remark 8**.**

Corollary 7 already yields a simple monotonicity based reconstruction algorithm. Assume that the conductivity in the investigated object is $\sigma=1+\chi_{D}$, where the measurable set $D\subseteq\Omega$ describes the unknown inclusion. Then for all other measurable sets $B\subseteq\Omega$

[TABLE]

so that the set

[TABLE]

is an upper bound of $D$.

A numerical approximation of (this upper bound of) $D$ can be calculated by choosing a number of small balls $B=B_{\epsilon}(z)\subseteq\Omega$ (with center $z\in\Omega$ and radius $\epsilon>0$) and marking all balls where the monotonicity test $\Lambda(1+\chi_{B})\geq\Lambda(\sigma)$ holds true. Algorithms based on this idea have been worked out and numerically tested in the works of Tamburrino and Rubinacci [47, 46].

Also, Lemma 6 gives an estimate for the Fréchet derivative of $\Lambda$ that will be the basis for linearizing our monotonicity tests without losing shape information (cf. [16] for the origin of this idea).

Corollary 9**.**

Let $\sigma\in L^{\infty}_{+}(\Omega)$. Let $\Lambda(1)$ be the NtD-Operator corresponding to the background conductivity $1$ and $\Lambda^{\prime}(1)$ be its Fréchet derivative (see subsection 2.1). Then

[TABLE]

Of course, in practical EIT applications, it is not possible to measure boundary data with infinite precision. Moreover, with a limited number of electrodes on the boundary of an imaging subject (and limited accuracy), we can only obtain a finite-dimensional approximation to the true NtD. Also, we can only calculate finite-dimensional approximations of the NtD for test conductivities (and their linearized counterparts). Hence, let us comment on the stability of monotonicity tests with respect to such errors.

Remark 10**.**

Monotonicity/definiteness tests can be stably implemented in the following sense. Let $A\in\mathcal{L}(H)$ be a selfadjoint compact operator on a Hilbert space $H$, and let $(A^{\delta})_{\delta>0}\subseteq\mathcal{L}(H)$ be a family of compact (e.g. finite dimensional) approximations with

[TABLE]

Possibly replacing $A^{\delta}$ by its symmetric part, we can assume that $A^{\delta}$ is selfadjoint.

For $\alpha>0$, we define the regularized definiteness test

[TABLE]

which is equivalent to checking whether the smallest eigenvalue of $A^{\delta}$ is not below $-\alpha$.

If $A\geq 0$ then $\langle A^{\delta}g,g\rangle\geq-\delta\hskip 0.86108pt\|g\|\hskip 0.86108pt^{2}$ for all $g\in H$. If $A\not\geq 0$ then $A$ has a negative eigenvalue $\lambda<0$ so that $\langle A^{\delta}g,g\rangle\geq-\delta\hskip 0.86108pt\|g\|\hskip 0.86108pt^{2}$ cannot hold for all $g\in H$ for $\delta<|\lambda|/2$. Hence,

[TABLE]

3.2 Localized potentials

We will show that a certain converse of the monotonicity relation (6), resp., (7) holds true. The main theoretical tool for this result is to use the theory of localized potentials by one of the authors [6] to control the energy terms $|\nabla u|^{2}$ in the monotonicity estimate in lemma 6.

Roughly speaking, [6] shows that there exist electric potentials which have arbitrarily large energy $|\nabla u|^{2}$ in some region and arbitrarily small energy in another region, as long as the high-energy region can be reached from the boundary without crossing the low-energy region.

We will make use of the following variant of the result in [6].

Theorem 11**.**

Let $D_{1},D_{2}\subseteq\overline{\Omega}$ be two measurable sets with

[TABLE]

Furthermore let $\sigma\in L^{\infty}_{+}(\Omega)$ be piecewise analytic.

Then there exists $(g_{m})_{m\in\mathbb{N}}\subset L^{2}_{\diamond}(\partial\Omega)$ such that the solutions $(u_{m})_{m\in\mathbb{N}}\subset H_{\diamond}^{1}(\Omega)$ of

[TABLE]

fulfill

[TABLE]

Proof.

The proof is a slight adaptation of the one in [6, Sect. 2.2], see also [13] for the general approach.

1. (a)

Reformulation as range (non-)inclusion

We define the virtual measurement operators $L_{j}$ ($j=1,2$) by

[TABLE]

where $u\in H_{\diamond}^{1}(\Omega)$ solves

[TABLE]

Note that this implies $\nabla\cdot\sigma\nabla u=0$ in $\Omega\setminus\overline{D}_{j}$ and if $\overline{D}_{j}\subseteq\Omega$ then it also implies the homogeneous Neumann boundary condition $\sigma\partial_{\nu}u|_{\partial\Omega}=0$.

It is easily checked that the dual operators

[TABLE]

are given by $L_{j}^{\prime}g=\nabla v|_{D_{j}}$ where $v\in H_{\diamond}^{1}(B)$ solves

[TABLE]

Now the assertion is equivalent to the statement

[TABLE]

which is (see e.g. [6, Lemma 2.5]) equivalent to the range (non-)inclusion

[TABLE] 2. (b)

Proof of the range (non-)inclusion (8)

Since $\mathrm{int}\,D_{1}\nsubseteq\mathrm{out}_{\partial\Omega}\,D_{2}$, the set $\mathrm{int}\,D_{1}$ must intersect one of the sets $U$ in the definition of the outer support of $\chi_{D_{2}}$. Hence, there exists a set $U\subset\overline{\Omega}\setminus\mathrm{out}_{\partial\Omega}\,D_{2}$ with $U$ (relatively) open in $\overline{\Omega}$, $U$ connected to $\partial\Omega$, and $U\cap D_{1}$ contains an open ball $B$. Possibly shrinking the ball we can assume that $\overline{B}\subset\Omega$ and that $(U\cap\Omega)\setminus\overline{B}$ is connected.

Let $L_{B}$ denote the virtual measurement operator corresponding to the ball $B$. Obviously, $B\subseteq D_{1}$ implies $\mathcal{R}(L_{B})\subseteq\mathcal{R}(L_{1})$, so that it suffices to prove that

[TABLE]

To that end let $\varphi\in\mathcal{R}(L_{B})\cap\mathcal{R}(L_{2})$. Then there exist $u_{B},u_{2}\in H_{\diamond}^{1}(\Omega)$ with $u_{B}|_{\partial\Omega}=\varphi=u_{2}|_{\partial\Omega}$, $\sigma\partial_{\nu}u_{B}|_{U\cap\partial\Omega}=0=\sigma\partial_{\nu}u_{2}|_{U\cap\partial\Omega}$, and

[TABLE]

By unique continuation $u_{B}=u_{2}$ in $U\setminus\overline{B}$. Hence

[TABLE]

defines a function $u\in H_{\diamond}^{1}(\Omega)$ with $\nabla\cdot\sigma\nabla u=0$ in $\Omega$ and homogeneous Neumann boundary data $\sigma\partial_{\nu}u|_{\partial\Omega}=0$. It follows that $\varphi=u|_{\partial\Omega}=0$ and thus we have shown that $\mathcal{R}(L_{B})\cap\mathcal{R}(L_{2})=\{0\}$.

Finally, using unique continuation again, we obtain that $L_{B}^{\prime}$ is injective, so that $\mathcal{R}(L_{B})$ is dense in $L_{\diamond}^{2}(\partial\Omega)$. A fortiori, $\mathcal{R}(L_{B})\neq\{0\}$, which, together with $\mathcal{R}(L_{B})\cap\mathcal{R}(L_{2})=\{0\}$, proves (9) and thus the assertion.

∎

Note that theorem 11 also holds for less regular conductivities as long as a unique continuation property is fulfilled, and that localized potentials can be constructed by solving regularized operator equations, cf. [6].

We now show that (regardless of regularity) the properties of the localized potentials do not depend on the conductivity in the low energy region:

Lemma 12**.**

Let $D_{1},D_{2}\subseteq\overline{\Omega}$ be two measurable sets. Let $\sigma,\tau\in L^{\infty}_{+}(\Omega)$ and $u_{m}^{\sigma},u_{m}^{\tau}\in H_{\diamond}^{1}(\Omega)$ denote the corresponing solutions of

[TABLE]

for a sequence of boundary currents $(g_{m})_{m\in\mathbb{N}}\subset L^{2}_{\diamond}(\partial\Omega)$.

If $\mathop{\operator@font supp}\nolimits(\sigma-\tau)\subseteq D_{2}$ then

[TABLE]

holds if and only if

[TABLE]

Proof.

For both conductivities, $\sigma$ and $\tau$, we define the virtual measurement operators

[TABLE]

as in the proof of theorem 11. If $u^{\sigma}|_{\partial\Omega}=L_{2}^{\sigma}F$ with $F\in L^{2}(D_{2})$ and a solution $u^{\sigma}\in H_{\diamond}^{1}(\Omega)$ of

[TABLE]

then $u^{\sigma}$ also solves

[TABLE]

This shows that $\mathcal{R}(L_{2}^{\sigma})\subseteq\mathcal{R}(L_{2}^{\tau})$. As in the proof of theorem 11, this implies that

[TABLE]

By interchanging $\sigma$ and $\tau$, we obtain that

[TABLE]

Using the same argument on $D_{1}\cup D_{2}$ it follows that also

[TABLE]

so that the assertion follows. ∎

Remark 13**.**

Localized potentials can be numerically constructed by solving regularized operator equations (see [6]), and they can be used to probe for an unknown inclusion in the spirit of the probe or needle method, cf. e.g. [28, 29]. We briefly sketch the idea on a simple test example. Assume that the conductivity is $\sigma=1+\chi_{D}$ and that $(g_{m})_{m\in\mathbbm{N}}$ is a sequence such that the solutions $(u_{m})_{m\in\mathbb{N}}\subset H_{\diamond}^{1}(\Omega)$ of $\Delta u_{m}=0$ and $\partial_{\nu}u_{m}|_{\partial\Omega}=g_{m}$ fulfill

[TABLE]

Then the monotonicity estimate in lemma 6 yields that

[TABLE]

Choosing $D_{2}$ to cover most of $\Omega$ and $D_{1}$ to be, e.g., a small ball inside $\Omega\setminus D_{2}$, one may thus estimate the shape of $D$ by slowly shrinking $D_{2}$.

Such an algorithm would however suffer from high computational cost (to construct a high number of localized potentials) and it is not clear how to check the limit of $\langle g_{m},(\Lambda(1)-\Lambda(\sigma))g_{m}\rangle$ in a numerically stable way. Furthermore, the choice of the sets $D_{1}$ and $D_{2}$ would certainly impose some geometrical restrictions on the shapes of inclusions that can be recovered.

In the following, we take a different approach. The monotonicity methods derived in the next section do not require the numerical construction of localized potentials. We will only require the above abstract existence results for localized potentials in order to show that simple monotonicity tests recover the true (outer) shape of an inclusion.

4 Monotonicity based shape reconstruction

4.1 The definite case

We will now show how the shape reconstruction problem can be solved via simple monotonicity tests. We start with the definite case, in which the inclusions conductivity is everywhere higher or everywhere lower than the background. We treat this case separately since it allows a particularly simple reconstruction strategy. Given a small ball the following theorems show how to check whether the ball belongs to the inclusion or not. The proofs of the theorems are postponed until the end of this subsection. The main idea of this subsection has previously been summarized in the extended conference abstract [18].

Theorem 14**.**

Let $\sigma\in L^{\infty}_{+}(\Omega)$ and $\sigma\geq 1$.

For every open ball $B:=B_{\epsilon}(z)$ and every $\alpha>0$,

[TABLE]

Hence, the set

[TABLE]

fulfills

[TABLE]

Example 15**.**

Let $\sigma=1+\chi_{D}$ where the inclusion $D$ is open, and $\overline{D}\subset\Omega$ has a connected complement. Then for every open ball $B\subseteq\Omega$

[TABLE]

Note that implementing the monotonicity tests in theorem 14 or example 15 would be computationally expensive since for each ball $B$ (and possibly also for each test level $\alpha$) we would have to solve the EIT equation with a new inhomogeneous conductivity in order to calculate $\Lambda(1+\alpha\chi_{B})$. The following theorem shows that we can replace the tests by linearized versions, that do not require such inhomogeneous forward solutions. Since this is a bit counterintuitive, let us stress that the following result is not affected by the linearization error, no matter how large that may be. The linearized inverse problem in EIT still contains the exact shape information, cf. [16].

Theorem 16**.**

Let $\sigma\in L^{\infty}_{+}(\Omega)$ and $\sigma\geq 1$.

For every open ball $B:=B_{\epsilon}(z)$ and every $\alpha>0$,

[TABLE]

Hence, the set

[TABLE]

fulfills

[TABLE]

Example 17**.**

Let $\sigma=1+\chi_{D}$ where the inclusion $D$ is open, and $\overline{D}\subset\Omega$ has a connected complement. Then for every ball $B=B_{\epsilon}(z)$

[TABLE]

Proof of theorem 14. Let $\sigma\in L^{\infty}_{+}(\Omega)$, $\sigma\geq 1$. Let $B=B_{\epsilon}(z)$ and $\alpha>0$. Corollary 7 yields that

[TABLE]

Let $B\nsubseteq\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,(\sigma-1)$. Corollary 7 yields that shrinking the ball $B$ only makes $\Lambda(1+\alpha\chi_{B})$ larger, so that we can assume w.l.o.g. that

[TABLE]

We have that $1+\alpha\chi_{B}$ is piecewise analytic,

[TABLE]

(see lemma 4(d)). Hence, we can apply theorem 11 and obtain a sequence of currents $(g_{m})_{m\in\mathbb{N}}\subset L^{2}_{\diamond}(\partial\Omega)$ so that the solutions $(u_{m})_{m\in\mathbb{N}}\subset H_{\diamond}^{1}(\Omega)$ of

[TABLE]

fulfill

[TABLE]

From lemma 6 it follows that

[TABLE]

and hence $\Lambda(1+\alpha\chi_{B})\ngeq\Lambda(\sigma)$. $\Box$

Proof of Theorem 16. Let $\sigma\in L^{\infty}_{+}(\Omega)$, $\sigma\geq 1$. Let $B=B_{\epsilon}(z)$ and $\alpha>0$.

For every $g\in L^{2}_{\diamond}(\partial\Omega)$ and solution $u\in H_{\diamond}^{1}(\Omega)$ of

[TABLE]

we obtain from lemma 6

[TABLE]

This shows that

[TABLE]

To show this let $B\nsubseteq\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,(\sigma-1)$. The linearized monotonicity relation (4) yields that shrinking the ball $B$ only makes $\Lambda(1)+\alpha\Lambda^{\prime}(1)\chi_{B}$ larger, so that we can assume w.l.o.g. that $B\subseteq\Omega\setminus\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,(\sigma-1)$. Then,

[TABLE]

so that the assertion follows using localized potentials for the background conductivity $1$ and the same sets as in theorem 11. $\Box$

Remark 18**.**

If $\sigma\in L^{\infty}_{+}(\Omega)$ and $\sigma\leq 1$ then we obtain with the same arguments that for every open ball $B\subseteq\Omega$ and every $0<\alpha<1$,

[TABLE]

Remark 19**.**

An inspection of the proofs shows that the balls can be replaced by arbitrary measurable sets $B$ with non-empty interior in theorem 16 (and the second part of remark 18). For theorem 14 (and the first part of remark 18) the sets $B$ must additionally possess a piecewise smooth boundary (so that $1+\alpha\chi_{B}$ remains piecewise analytic). We comment on further generalizations in section 4.3.

4.2 The indefinite case

We now consider the general indefinite case where $\sigma$ is no longer required to be everywhere larger or everywhere smaller than the background conductivity $1$. Instead of testing whether a small test region is part of the unknown inclusion, we will now test whether a large test region contains the unknown inclusions.

The main idea is the following. Consider a large test region $C$ with connected complement. If $C$ overlaps the inclusions then a large enough, resp., small enough test conductivity on $C$ will make the corresponding test NtD smaller, resp., larger then the measured NtD. Hence if $C$ overlaps the inclusions then two monotonicity tests (one with a large and one with a small test level on $C$) hold true. On the other hand, if $C$ does not overlap the inclusions then we can connect the non-overlapped part with the boundary, and construct a localized potential with large energy in the non-overlapped part and small energy in $C$. Depending on whether the conductivity is larger, resp., smaller than the background in the non-overlapped part, this localized potential shows that one of the monotonicity tests cannot hold true.

However, for this argument we need a local definiteness property. If a conductivity differs from the background then there must either be a neighborhood of the boundary where it differs from the background in the positive direction, or a neighborhood where it differs in the negative direction. Note that even $C^{\infty}$-conductivities might oscillate infinitely and thus violate this property. This property holds, however, if the conductivity is either piecewise analytic or if the higher-conductivity and lower-conductivity parts have some distance from each other, and the inner support does not deviate too much from the true support (which already holds, e.g., for piecewise continuous functions, see corollary 5).

More precisely, we assume that $\sigma\in L^{\infty}_{+}(\Omega)$ is either piecewise-analytic, or

[TABLE]

where $(\sigma-1)^{+}:=\max\{\sigma-1,0\},\quad(\sigma-1)^{-}:=\min\{\sigma-1,0\}$.

Theorem 20**.**

Let $\sigma\in L^{\infty}_{+}(\Omega)$ either be piecewise-analytic or fufill (10).

Then, for every set $C\subseteq\overline{\Omega}$ with $C=\mathrm{out}_{\partial\Omega}\,C$ and every $\alpha>1$

[TABLE]

and

[TABLE]

Hence,

[TABLE]

fulfills $R=\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,(\sigma-1).$

We postpone the proof until the end of this subsection and first give an example and formulate the linearized version.

Example 21**.**

Let $\sigma=1+\chi_{D^{+}}-\frac{1}{2}\chi_{D^{-}}$ where $D^{+},D^{-}\subseteq\Omega$ are open sets with $\overline{D^{+}}\cap\overline{D^{-}}=\emptyset$, and $\overline{D^{+}}\cup\overline{D^{-}}\subset\Omega$ has a connected complement.

Then for every closed set $C\subset\Omega$ with connected complement $\Omega\setminus C$

[TABLE]

Theorem 22**.**

Under the assumptions of theorem 20 we have that for every set $C\subseteq\overline{\Omega}$ with $C=\mathrm{out}_{\partial\Omega}\,C$ and every $\alpha>0$

[TABLE]

and

[TABLE]

Hence,

[TABLE]

fulfills $R=\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,(\sigma-1).$

Example 23**.**

Let $\sigma=1+\chi_{D^{+}}-\frac{1}{2}\chi_{D^{-}}$ where $D^{+},D^{-}\subseteq\Omega$ are open sets with $\overline{D^{+}}\cap\overline{D^{-}}=\emptyset$, and $\overline{D^{+}}\cup\overline{D^{-}}\subset\Omega$ has a connected complement.

Then for every closed set $C\subset\Omega$ with connected complement $\Omega\setminus C$

[TABLE]

Proof of Theorem 20. Let $\alpha>1$ and $C=\mathrm{out}_{\partial\Omega}\,C\subseteq\overline{\Omega}$. Then $C$ is closed and thus measurable, so that $1-{\textstyle\frac{1}{\alpha}}\chi_{C},1+\alpha\chi_{C}\in L^{\infty}_{+}(\Omega)$.

Corollary 7 yields the first two assertions

[TABLE]

It remains to show that $\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,(\sigma-1)\nsubseteq C$ implies that either

[TABLE]

Let $\mathrm{out}_{\partial\Omega}\,\mathrm{supp}\,(\sigma-1)\nsubseteq C=\mathrm{out}_{\partial\Omega}\,C$. Then there exists a relatively open set $U\subseteq\overline{\Omega}$ that is connected to $\partial\Omega$ where $\sigma|_{U}\not\equiv 1$ and $C\cap U=\emptyset$.

We first prove the assertion for the case that $\sigma$ is piecewise analytic. Using the local definiteness property derived in corollary 25) in the appendix, we can choose (note that $\Omega\setminus D_{2}\subseteq U$ implies $C\subseteq D_{2}$)

[TABLE]

so that either

1. (a)

$\displaystyle\sigma\geq 1\mbox{ on }\Omega\setminus D_{2},\quad\sigma-1\in L^{\infty}_{+}(D_{1})$, or 2. (b)

$\displaystyle\sigma\leq 1\mbox{ on }\Omega\setminus D_{2},\quad 1-\sigma\in L^{\infty}_{+}(D_{1})$.

Replacing $D_{1}$ with $D_{1}\setminus\mathrm{out}_{\partial\Omega}\,D_{2}$, we can also assume that $D_{1}\cap D_{2}=\emptyset$.

We then use the localized potentials theorem 11 for the homogeneous conductivity $\tau=1$ and obtain a sequence $(g_{m})_{m\in\mathbb{N}}\subset L^{2}_{\diamond}(\partial\Omega)$ so that the solutions $(u^{\tau}_{m})_{m\in\mathbb{N}}\subseteq H_{\diamond}^{1}(\Omega)$ of

[TABLE]

fulfill

[TABLE]

Since $C\subseteq D_{2}$ it follows from lemma 12 that the solutions $u^{\tau}_{m}$ for the conductivities $\tau=1-{\textstyle\frac{1}{\alpha}}\chi_{C}$ and $\tau=1+\alpha\chi_{C}$ have the same property.

Hence, in case (a), we apply lemma 6 with $\tau=1+\alpha\chi_{C}$ and obtain (using that $\sigma\geq 1$ on $\Omega\setminus(D_{1}\cup D_{2})$, and that $C\subseteq D_{2}$)

[TABLE]

which proves the assertion for piecewise analytic conductivities.

Now we prove that the assertion also holds for (not necessary piecewise analytic) conductivities fulfilling (10). It suffices to show that also in this case, there exist

[TABLE]

such that either (a) or (b) from above holds.

First note that if $\mathop{\operator@font supp}\nolimits(\sigma-1)^{+}$ and $\mathop{\operator@font supp}\nolimits(\sigma-1)^{-}$ are disjoint compact sets, then

[TABLE]

$\sigma|_{U}\not\equiv 1$ implies that there exists a point $y\in U\cap\mathop{\operator@font supp}\nolimits(\sigma-1)$. Let $x\in\partial\Omega\cap U$. Since $\partial\Omega$ is a smooth boundary and $U\cap\Omega$ is open and connected, we can connect $x$ and $y$ with a continuous path

[TABLE]

Using that $U$ is relatively open, there exists, for each $t\in[0,1]$, a ball $B_{t}:=B_{\epsilon(t)}(\gamma(t))$ with radius $\epsilon(t)<\delta/2$ and $B_{t}\cap\overline{\Omega}\subseteq U$.

By compactness of $\gamma([0,1])$ we can choose a finite number $0\leq t_{1}<\ldots<t_{N}\leq 1$, so that

[TABLE]

Since $\gamma(1)=y\in\mathop{\operator@font supp}\nolimits(\sigma-1)$, there exists a smallest index $J$ for which

[TABLE]

so that there exists an open set $D_{1}\subseteq B_{t_{J}}$ with $|\sigma-1|\in L^{\infty}_{+}(D_{1})$.

We define $D_{2}:=\Omega\setminus\left(B_{t_{1}}\cup\ldots\cup B_{t_{J}}\right)$. Then

[TABLE]

Furthermore, since $B_{t_{J}}$ has diameter less than $\delta$, it can not intersect both $\mathop{\operator@font supp}\nolimits(\sigma-1)^{+}$ and $\mathop{\operator@font supp}\nolimits(\sigma-1)^{-}$, so that either

1. (a)

$\displaystyle\sigma\geq 1\mbox{ on }\Omega\setminus D_{2},\quad\sigma-1\in L^{\infty}_{+}(D_{1})$, or 2. (b)

$\displaystyle\sigma\leq 1\mbox{ on }\Omega\setminus D_{2},\quad 1-\sigma\in L^{\infty}_{+}(D_{1})$,

which finishes the proof. $\Box$

Proof of Theorem 22.

If $1-\alpha\chi_{C}\leq 2-\frac{1}{\sigma}$, then $\alpha\chi_{C}\geq\frac{1}{\sigma}(1-\sigma)$, so that (4) and corollary 9 imply that

[TABLE]

Likewise, if $1+\alpha\chi_{C}\geq\sigma$ then (4) and corollary 9 imply that

[TABLE]

This shows the first two assertions.

Moreover, lemma 6 yields that for all $\alpha\in\mathbbm{R}$

[TABLE]

and

[TABLE]

where $u_{g}\in H_{\diamond}^{1}(\Omega)$ solves $\Delta u_{g}=0$ and $\partial_{\nu}u_{g}|_{\partial\Omega}=g$. Hence, the third assertion follows by using localized potentials for the homogeneous conductivity and the same sets $D_{1}$, $D_{2}$ as in theorem 20. $\Box$

4.3 Remarks and extensions

Let us comment on some extensions and generalizations of our results. Our assumption that the background conductivity is equal to $1$ and that we are given measurements on the complete boundary $\partial\Omega$ have been merely for the ease of presentation. All our results and proofs remain valid if $\partial\Omega$ is replaced by an arbitrarily small open piece $S\subset\partial\Omega$ and we are given the partial Neumann-to-Dirichlet operator

[TABLE]

where $u^{g}_{\sigma}\in H^{1}_{\diamond}(\Omega)$ is the unique solution of

[TABLE]

Also, all the results still hold when the background conductivity $1$ is replaced by a known piecewise analytic function.

Let us also note that our results require piecewise analyticity for only two purposes: the existence of localized potentials and the local definiteness property. Localized potentials exist for less regular conductivities, it only requires that the solutions of the corresponding elliptic EIT equations satisfy a unique continuation property, cf. [6]. Local definiteness can hold for quite general functions, if additional assumption are made (e.g., that positive and negative part are separated as in (10)). However, the authors are not aware of any natural function classes beyond piecewise-analytic functions in which a property in the spirit of theorem 24 holds without further assumptions.

Appendix A Local definiteness of piecewise analytic functions

In this appendix, we show that piecewise analytic functions have a local definiteness property. If they do not vanish identically then there is either a neighborhood of the boundary where they differ from zero in the positive direction, or a neighborhood where they differ in the negative direction.

The property follows from the arguments used in the proofs of [12, Theorem 4.2] and [16, Lemma 3.7]. However, some subtle and not entirely trivial topological details were omitted in [12, 16], which is why we give the proof here in full detail.

Theorem 24**.**

Let $\Omega\subset\mathbbm{R}^{n}$, $n\geq 2$ be a smoothly bounded domain, and let $\sigma\in L_{+}^{\infty}(\Omega)$ be piecewise analytic. Let $U\subseteq\overline{\Omega}$ be relatively open and connected to $\partial\Omega$, and let $\sigma|_{U}\not\equiv 0$.

Then we can find a subset $V\subseteq U$ with the same properties, on which $\sigma$ does not change sign, i.e.

1. (a)

$V\subseteq\overline{\Omega}$* is relatively open, $V$ is connected to $\partial\Omega$, $V\subseteq U$,* 2. (b)

$\sigma|_{V}\not\equiv 0$, and either $\sigma|_{V}\geq 0$ or $\sigma|_{V}\leq 0$.

Obviously, if a piecewise analytic function is not identically zero, we can find a neighborhood where it is bounded away from zero. Hence, choosing $D_{2}:=\Omega\setminus V$, we obtain the following corollary.

Corollary 25**.**

Under the assumptions of theorem 24 we can choose

[TABLE]

and either

[TABLE]

In the following, let $\Omega\subset\mathbbm{R}^{n}$, $n\geq 2$ be a smoothly bounded domain, and $\sigma\in L_{+}^{\infty}(\Omega)$ be piecewise analytic with respect to

[TABLE]

where, w.l.o.g, we assume that every $\partial O_{m}$ consists of infinitely many pieces. Furthemore, let $U\subseteq\overline{\Omega}$ be relatively open and connected to $\partial\Omega$.

Lemma 26**.**

There exists an open ball $B\subseteq\mathbbm{R}^{n}$ such that

[TABLE]

and for one of the $O_{m}$ and one of its smooth boundary pieces $\Gamma_{m}^{k}\subseteq\partial O_{m}$,

[TABLE]

Proof.

Since $U$ is relatively open and $U\cap\partial\Omega\neq\emptyset$ , there exists an open ball $B$ with $\emptyset\neq B\cap\partial\Omega$, $B\cap\overline{\Omega}\subseteq U$, and by shrinking $B$ we can assume that

[TABLE]

$\overline{\Omega}=\overline{O_{1}\cup\ldots\cup O_{M}}$ implies that

[TABLE]

By Baire’s theorem, one of the countably many closed sets $\overline{\Gamma_{m}^{k}}\cap S$ must have non-empty interior in $S$. Hence, for one of the $\overline{\Gamma_{m}^{k}}$, there exists an open ball $B$ with

[TABLE]

Moreover, $B\cap\partial\Omega$ must intersect $\Gamma_{m}^{k}$ because of the following dimension theoretical argument, cf., e.g., the classical book of Hurewicz and Wallman [19, Ch. IV, §4]. $\Omega\cap B$ is an open (neither empty nor dense) subset of the $(n-1)$-dimensional ball $B$. As a subset of $B$, the boundary of $\Omega\cap B$ is $\partial\Omega\cap B$, which shows that $\partial\Omega\cap B$ is $(n-1)$-dimensional (and not of lesser dimension). $\Gamma_{m}^{k}$ is a (neither empty nor dense) open subset of a set that is homeomorphic to $\mathbbm{R}^{n-1}$. Hence, $\Gamma_{m}^{k}$ is $(n-1)$-dimensional and $\overline{\Gamma_{m}^{k}}\setminus\Gamma_{m}^{k}$ is $(n-2)$-dimensional. This shows that $B\cap\partial\Omega\not\subseteq\overline{\Gamma_{m}^{k}}\setminus\Gamma_{m}^{k}$, so that by shrinking $B$ we can assume that

[TABLE]

Finally, we can shrink $B$ so that $B\cap\Omega=B\cap O_{m}$ and that $B\cap\Omega$ is connected. ∎

Lemma 27**.**

Every open ball $B\subseteq\mathbbm{R}^{n}$ that intersects a smooth boundary piece $\Gamma_{m}^{k}$ contains an open subball $B^{\prime}\subseteq B$ intersecting $\Gamma_{m}^{k}$, where either

[TABLE]

Proof.

We use an argument of Kohn and Vogelius [37]. If $\sigma|_{B\cap O_{m}}\equiv 0$ then the assumption is trivial. Otherwise, by analyticity, there must be a smallest $k\in\mathbbm{N}$, so that the normal derivative $\partial^{k}_{\nu(z)}\sigma(z)$ is not identically zero for all $z\in\Gamma_{m}^{k}\cap B$. Hence there is a neighbourhood of a point $z\in\Gamma_{m}^{k}\cap B$ on which either $\sigma\geq 1$ or $\sigma\leq 1$. ∎

Now we are ready to prove the local definiteness property.

Proof of theorem 24. From lemma 26 we obtain an open ball $B\subseteq\mathbbm{R}^{n}$ with

[TABLE]

and (w.l.o.g.)

[TABLE]

If $\sigma$ is not identically zero on $O_{1}$ then the assertion follows from lemma 27.

Otherwise, $M>1$, and the set $V:=B\cap\overline{\Omega}$ has the following properties:

1. (i)

$V$ is a relatively open subset of $\overline{\Omega}$ that is connected to $\partial\Omega$, 2. (ii)

$V$ fulfills $B\cap\overline{\Omega}\subseteq V\subseteq U$, 3. (iii)

$\sigma|_{V}=0$.

Obviously these properties are closed under union, so that we can choose $V$ to be the maximal set fulfilling (i)–(iii).

Now we show that

[TABLE]

Since $V$ is relatively open in $\Omega$ and $V\subseteq U$ it follows that $V\cap\Omega$ is relatively open in $U\cap\Omega$. If $\partial V$ has no intersection with $U\cap\Omega$, then $V\cap\Omega$ is relatively closed, so that $U\cap\Omega=V\cap\Omega$, which contradicts $\sigma|_{U}\not\equiv 0$. Hence, the $\partial V\cap U\cap\Omega\neq\emptyset$. To show the second assertion in (11), assume that there exists $z\in\partial V\cap\Omega\cap U$ with $z\in O_{m}$ for one $O_{m}$. Then we can choose an open ball $B\subseteq O_{m}\cap U$ containing $z$. Since $\sigma$ is analytic on $B$ and $B\cap V$ has non-empty interior, it follows that $\sigma|_{B}\equiv 0$, and hence $V\cup B$ has the properties (i)–(iii). This contradicts the maximality of $V$, so that also the second assertion in (11) must hold.

Because of (11) we can choose an open ball $B$ with $\overline{B}\subseteq U\cap\Omega$ and

[TABLE]

Using the same arguments as in the proof of lemma 26 it follows that by shrinking $B$ we can assume that

[TABLE]

with a smooth boundary piece $\Gamma_{m}^{k}$ of one $O_{m}$.

Since $\partial O_{m}\subseteq\bigcup_{m^{\prime}\neq m}\partial O_{m^{\prime}}\cup\partial\Omega$, equation (12) still holds if we restrict the union to all $m^{\prime}\in\{1,\ldots,M\}\setminus\{m\}$. By repeating the above argument (and possibly shrinking $B$ again) we obtain $m^{\prime}\neq m$ with

[TABLE]

From the definition of smooth boundary pieces it follows that (if we choose $B$ small enough)

[TABLE]

so that either $B\cap O_{m}$ or $B\cap O_{m^{\prime}}$, but not both, intersects $V$. W.l.o.g, let $B\cap O_{m}$ intersect $V$. Then $\sigma|_{O_{m}}\equiv 0$, and using lemma 27 we can shrink $B$ so that $\sigma|_{B\cap O_{m^{\prime}}}$ is either non-negative or non-positive. Hence $B\cup V$ fulfills the above properties (i) and (ii) and it is a proper superset of $V$. Hence, $\sigma$ cannot identically vanish on $B\cup V$, which shows that $B\cup V$ fulfills the assertion of theorem 24. $\Box$