On reconstruction from a partial knowledge of the Neumann-to-Dirichlet operator

TL;DR

This paper presents formulae to determine the location of an unknown polygonal inclusion with unknown conductivity inside a known material, using partial data from the Neumann-to-Dirichlet operator.

Contribution

It introduces new formulae that utilize partial Neumann-to-Dirichlet data to locate polygonal inclusions with unknown conductivities.

Findings

Formulae successfully identify inclusion location

Method works with partial Neumann-to-Dirichlet data

Applicable to polygonal inclusions with unknown conductivities

Abstract

We give formulae that yield an information about the location of an unknown polygonal inclusion having unknown constant conductivity inside a known conductive material having known constant conductivity from a partial knowledge of the Neumann -to-Dirichlet operator.

Equations

$$\nabla \cdot \gamma \nabla u=0\text{mbox}in\Omega ,\frac{\partial u}{\partial \nu }=g\text{mbox}on\partial \Omega .$$

$${\Lambda }_{\gamma }(P,Q):g⟼u(P)-u(Q)$$

$${\Lambda }_{\gamma }:g⟼u{|}_{\partial \Omega }.$$

$$\text{mbox}diamD<\text{mbox}dist(D,\partial \Omega ).$$

$$h_D(\omega )=\underset{x\in D}{\sup }x\cdot \omega ,\omega \in S^1.$$

$$\{x\in {\mathbf{R}}^2|x\cdot \omega =h_D(\omega )\}\cap \partial D$$

$$v_{\omega }=v_{\omega }(x;\tau )=e^{\tau x\cdot (\omega +i{\omega }^{\perp })},\tau >0$$

$$\omega \cdot {\omega }^{\perp }=0,\text{mbox}det(\omega {\omega }^{\perp })<0.$$

$$g=g_{\omega }(y;\tau )=\frac{\partial v_{\omega }}{\partial \nu }{|}_{\partial \Omega }.$$

$$I_{\omega }(\tau ,t)=e^{-\tau t}\{{\Lambda }_{\gamma }(P,Q)-{\Lambda }_1(P,Q)\}{g}_{\omega }(\cdot ;\tau ),\tau >0,t\in \mathbf{R}.$$

$$I_{\omega }(\tau ,t)=\{{\Lambda }_{\gamma }(P,Q)-{\Lambda }_1(P,Q)\}\left(\frac{\partial (e^{-\tau t}{v}_{\omega })}{\partial \nu }{|}_{\partial \Omega }\right).$$

$$e^{-\tau t}{v}_{\omega }=e^{\tau (x\cdot \omega -t)}{e}^{i\tau x\cdot {\omega }^{\perp }},$$

$$\begin{array}[]{c}\displaystyle\{t\in{\rm\bf R}\,|\,\lim_{\tau\longrightarrow\infty}I_{\omega}(\tau,t)=0\}=[h_{D}(\omega),\,\infty[\\ \\ \displaystyle\lim_{\tau\longrightarrow\infty}\frac{\log|I_{\omega}(\tau,t)|}{\tau}=h_{D}(\omega)-t,\,\,\forall t\in{\rm\bf R},\end{array}$$

$$h_D(\omega )\tau \approx \log |\{{\Lambda }_{\gamma }(P,Q)-{\Lambda }_1(P,Q)\}{g}_{\omega }(\cdot ;\tau )|$$

$$\displaystyle\nabla\cdot\gamma\nabla\mbox{$\cal D$}=0\,\,\mbox{in}\,\Omega\,\,\,\,\frac{\partial\mbox{$\cal D$}}{\partial\nu}=\delta_{P}-\delta_{Q}\,\,\mbox{on}\,\partial\Omega$$

$$\displaystyle I_{\omega}(\tau,t)=e^{-\tau t}\sum_{j}(k_{j}-1)\int_{\partial D_{j}}\mbox{$\cal D$}(P,Q;\,\cdot\,)\frac{\partial v_{\omega}(\,\cdot\,;\tau)}{\partial\nu}.$$

$$I_{\omega }(\tau ,t)=e^{\tau (h_D(\omega )-t)}{I}_{\omega }(\tau ,h_D(\omega )),$$

$$e^{-\tau t}\sum _j(k_j-1){\int }_{\partial D_j}u\frac{\partial v_{\omega }(\cdot ;\tau )}{\partial \nu }$$

$$\begin{array}[]{c}\displaystyle V=V(P,Q;x)=-\frac{1}{\pi}(\log|P-x|-\log|Q-x|),\,\,x\in{\rm\bf R}^{2},\\ \\ \displaystyle\Psi=\Psi(P,Q;x)=-\frac{1}{\pi}\left(\frac{y-P}{|y-P|^{2}}-\frac{y-Q}{|y-Q|^{2}}\right)\cdot\nu(y),\,\,y\in\partial\Omega\setminus\{P,Q\}.\end{array}$$

$${\int }_{\partial \Omega }\Psi (P,Q;y)d\sigma (y)=0.$$

$$(\gamma -1)\nabla V\in L^{\infty }(\Omega ).$$

$$\begin{array}[]{c}\displaystyle\nabla\cdot\gamma\nabla\mbox{$\cal E$}=-\nabla\cdot(\gamma-1)\nabla V\,\,\mbox{in}\,\Omega,\\ \\ \displaystyle\frac{\partial}{\partial\nu}\mbox{$\cal E$}=-\Psi\,\,\mbox{on}\,\partial\Omega,\\ \\ \displaystyle\int_{\partial\Omega}\mbox{$\cal E$}=0.\end{array}$$

$$\displaystyle\mbox{$\cal D$}=\mbox{$\cal D$}(P,Q;x)=V(P,Q;x)+\mbox{$\cal E$}(P,Q;x),\,\,x\in\Omega.$$

$$\displaystyle\int_{\Omega}\gamma\nabla\mbox{$\cal D$}(P,Q;x)\cdot\nabla\varphi dx=\varphi(P)-\varphi(Q),$$

$$\displaystyle\int_{\Omega}\gamma\nabla\mbox{$\cal D$}\cdot\nabla\varphi dx=\int_{\Omega}\gamma\nabla V\cdot\nabla\varphi dx+\int_{\Omega}\gamma\nabla\mbox{$\cal E$}\cdot\nabla\varphi dx.$$

$$\displaystyle\int_{\Omega}\gamma\nabla\mbox{$\cal E$}\cdot\nabla\varphi dx=-\int_{\Omega}(\gamma-1)\nabla V\cdot\nabla\varphi dx-\int_{\partial\Omega}\Psi\varphi.$$

$${\int }_{\Omega }\gamma \nabla V\cdot \nabla \phi dx={\int }_{\partial \Omega }\Psi \phi +\phi (P)-\phi (Q).$$

$$\nabla \cdot \gamma \nabla w=-\nabla \cdot F\text{mbox}in\Omega ,\frac{\partial }{\partial \nu }w=0\text{mbox}on\partial \Omega$$

$$\text{mbox}suppF\subset \overline{D}.$$

$$\displaystyle w(P)-w(Q)=-\int_{D}F(x)\cdot\nabla\mbox{$\cal D$}(P,Q;x)dx,$$

Full text

1 Statement of the result

We give another application of the method developed in [5] and [6] to a special, however, important version of the inverse boundary value problem formulated by Calderón [2]. It is a mathematical formulation of electrical impedance tomography.

Let $\Omega$ be a two-dimensional bounded domain with smooth boundary. We consider $\Omega$ an isotropic, electrically conductive material. Let $\gamma$ be the conductivity of $\Omega$, and $D$ an open set of $\Omega$ such that $\overline{D}\subset\Omega$ and $\partial D$ is Lipschitz. Assume that $\gamma$ takes a positive constant $k_{j}$ on each connected component $D_{j}$ of $D$ with $k_{j}$, $j=1,\cdots,m\leq\infty$ and is equal to $1$ on $\Omega\setminus D$. In this paper, we always assume that $m<\infty$ and that $\overline{D_{j}}\cap\overline{D_{j^{\prime}}}=\emptyset$ if $j\not=j^{\prime}$. We call $D$ an inclusion and $\gamma$ the corresponding conductivity.

Let $\nu$ denote the unit outward normal vector field to $\partial(\Omega\setminus\overline{D})$. We prescribe the electric current distribution $g\in L^{\infty}(\partial\Omega)$ satisfying $\displaystyle\int_{\partial\Omega}g=0$. Consider the elliptic problem

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The problem (1.1) has a $H^{1}(\Omega)$ solution $u$ and any solution has the form $u+c$ where $c$ is a constant. It is well known that $u$ is Hölder continuous on $\overline{\Omega}$ [8].

We consider the following problem.

Inverse Problem

Let $P$, $Q$ be two arbitrary distinct points on $\partial\Omega$. We fix $P$ and $Q$. Then the map

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is well defined where $u$ is a solution to the problem (1.1). Assume that $D$ and $\gamma$ on $D$ are unknown. The problem is to find a formula that yields an information about the location of $D$ from $\Lambda_{\gamma}(P,Q)$. This map is a partial knowledge of the Neumann-to-Dirichlet operator:

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$\Lambda_{\gamma}$ uniquely determines $D$ and $\gamma$. This is a corollary of Isakov’s uniqueness theorem [7] which covers a more general case. Under suitable regularity assumption on $\gamma$, Nachman [9] established a reconstruction formula of $\gamma$ itself from the full knowledge of $\Lambda_{\gamma}$. See the survey paper [10] for several other results.

In [6], using the method developed in [5], we gave formulae that yield the information about the convex hull of $D$ from $\{u(y)-u(Q)\,|\,y\in\partial\Omega\}$ for fixed $Q$ and $g\not\equiv 0$ provided each connected component of $D$ is a polygon and $D$ satisfies the condition

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This result gives a constructive proof of Friedman-Isakov’s uniqueness theorem [3]. However, the formulae involve an integral of $g(y)$ and of the measured data $u(y)-u(Q)$.

The strong motivation of our study is to seek formulae that avoid any integration of the measured data on $\partial\Omega$. In this paper we give two formulae that yield the information about the convex hull of $D$ from $\Lambda_{\gamma}(P,Q)$ for fixed $P$ and $Q$ provided that each connected component of $D$ is a polygon and $D$ satisfies (1.2).

Now we describe the result more precisely. Let $S^{1}$ denote the set of all unit vectors of ${\rm\bf R}^{2}$. Define

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This function is called the support function of $D$ and the convex hull of $D$ can be reconstructed from this function.

We say that $\omega\in S^{1}$ is regular with respect to $D$ if the set

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consists of only one point.

Note that the set of all unit vectors that are not regular with respect to $D$ is a finite set. Since $h_{D}(\,\cdot\,)$ is continuous on $S^{1}$, the convex hull of $D$ can be reconstructed from the restriction of $h_{D}(\,\cdot\,)$ to the set of all unit vectors that are regular with respect to $D$.

The following special harmonic functions play the central role:

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where $\omega,\omega\in S^{1}$ and satisfy

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Define

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Note that $g\in C^{\infty}(\partial\Omega)$ and $\displaystyle\int_{\partial\Omega}g=0$.

Definition 1.1(Indicator function). Define

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Note that one can rewrite

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Since

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the function $e^{-\tau t}v_{\omega}$ has a special property as $\tau\longrightarrow\infty$: $|e^{-\tau t}v_{\omega}|$ is exponentially growing in the half space $x\cdot\omega>t$ and exponentially decaying in $x\cdot\omega<t$; $e^{-\tau t}v_{\omega}$ is oscillating on the line $x\cdot\omega=t$.

The result is the following two formulae.

Theorem 1.1.* Assume that each connected component of $D$ is a polygon and $D$ satisfies (1.2). Let $\omega$ be regular with respect to $D$. The formulae*

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are valid.

In particular, this theorem tells us that

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for $\tau>>1$. This will be usefull to calculate the approximate value of $h_{D}(\omega)$ from $\Lambda_{\gamma}(P,Q)-\Lambda_{1}(P,Q)$. We will do the test in the future.

A brief outline of the proof of Theorem 1.1 is as follows. In Section 2 we construct a solution \mbox{\cal D}=\mbox{\cal D}(P,Q;x) to the elliptic problem (see Proposition 2.1)

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where for $x\in\partial\Omega$, $\delta_{x}$ denotes the Dirac measure on $\partial\Omega$ concentrated at $x$. Using this solution, in Section 3 we establish the representation formula of the indicator function:

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From this and the trivial identity

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we know that it suffices to study the asymptotic behaviour as $\tau\longrightarrow\infty$ of the oscillatory integral

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for u=\mbox{\cal D}(P,Q;x). However, we have already encountered this type of integral [5, 6]. we see that this integral decays algebraically as $\tau\longrightarrow\infty$ provided (1.2) (see Lemma 3.1). Then from (1.5) we obtain the two formulae in Theorem 1.1.

2 Preliminaries

First we construct a solution to the problem (1.4). The construction is similar to that of a solution to the crack problem described in Appendix D, D.1 of [1]. Set

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Since we assumed that $\partial\Omega$ is smooth, $\Psi(P,Q;\,\cdot\,)\in L^{\infty}(\partial\Omega)$. Moreover, we know that

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These are well known facts in the potential theory. From the assumption $\mbox{supp}\,(\gamma-1)\subset\overline{D}$, we see that

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Therefore we have the unique $H^{1}(\Omega)$ solution \mbox{\cal E}=\mbox{\cal E}(P,Q;x) to the elliptic problem

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Define

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We prove that this satisfies (1.4) in the following sense.

Proposition 2.1.* For any $\varphi\in H^{1}(\Omega)$ that is smooth in a neighbourhood of $\partial\Omega$ the formula*

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is valid.

Proof. Note that $\nabla V\cdot\nabla\varphi$ is absolutely integrable in the whole domain because of the regularity assumption of $\varphi$. Therefore, we have

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From (2.1) we have

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It is well known that

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A combination of (2.3)-(2.5) gives (2.2).

$\Box$

Using the function \mbox{\cal D}(P,Q;x), we obtain a representation formula of $w(P)-w(Q)$ for any $H^{1}(\Omega)$ solution $w$ to the problem

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where $F\in L^{2}(\Omega)$ and satisfy

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From this and $\gamma=1$ in a neighbourhood of $\partial\Omega$, $w$ is smooth in a neighbourhood of $\partial\Omega$.

Proposition 2.2.* The formula*

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is valid.

Proof. From (2.2) we have

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For $\epsilon>0$ define

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where

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From the regularity of $w$, Lebesgue’s dominated convergence theorem and (2.6) we obtain

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This together with (2.8) gives (2.7).

$\Box$

3 Proof of Theorem 1.1

First we give the representation formula of the indicator function.

Proposition 3.1.* The formula*

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is valid.

Proof. Let $u$ be a $H^{1}(\Omega)$ solution to the problem (1.1). Given $g\in L^{\infty}(\partial\Omega)$ satisfying $\displaystyle\int_{\partial\Omega}g=0$ let $v\in H^{1}(\Omega)$ be a harmonic function satisfying $\displaystyle\frac{\partial v}{\partial\nu}=g$ on $\partial\Omega$. Then $w=u-v$ satisfies

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Since $v\in C^{\infty}(\Omega)$ and $\mbox{supp}\,(\gamma-1)\subset\overline{D}$, we have $(\gamma-1)\nabla v\in L^{\infty}(\Omega)$. Then one obtains from (2.7) that

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Since \mbox{\cal D}(P,Q;x) is in $H^{1}(D_{j})$, $v$ is in $H^{2}(D_{j})$ and $\partial D_{j}$ is Lipschitz, from Lemma 1.5.3.7 in [4] one has

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Note that $v$ is outward to $\partial(\Omega\setminus\overline{D})$. From (3.2), (3.3) and the definition of $\Lambda_{\gamma}(P,Q)$ we obtain

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Now (3.1) is clear.

$\Box$

Equation (3.4) is the representation formula of $\Lambda_{\gamma}(P,Q)-\Lambda_{1}(P,Q)$.

Now we describe a lemma which is the key for the proof of Theorem 1.1.

Lemma 3.1.* Assume that each connected component of $D$ is a polygon and $D$ satisfies (1,2); $u\in H^{1}_{\mbox{loc}}(\Omega)$ satisfies*

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for all $\varphi\in C^{\infty}_{0}(\Omega)$ and is not a constant function. Let $\omega$ be regular with respect to $D$ and $v=v_{\omega}(x;\tau)$. There exist positive constants $L$ and $\mu$ such that

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The proof of this lemma is essentially the same as that of the key lemma in [6]. We describe only the outline.

Outline of the proof. We have

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for any constant $c$ since $v$ is harmonic in $\Omega$.

From the regularity of $\omega$ we know that the line $x\cdot\omega=h_{D}(\omega)$ meets $\partial D$ at a vertex $x_{0}$ of $D$. Moreover, there exist $j_{0}$ and $\delta>0$ such that $x_{0}$ is a vertex of $D_{j_{0}}$, $h_{D_{j_{0}}}(\omega)=h_{D}(\omega)$ and $\cup_{j\not=j_{0}}D_{j}$ is located in the half-space $x\cdot\omega<h_{D}(\omega)-\delta$. Therefore, we have

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as $\tau\longrightarrow\infty$.

Using a well known expansion of $u$ about $x_{0}$ (see, for instance, Proposition 2.1 in [6]) and (3.6) for $c=u(x_{0})$, we obtain the asymptotic exapnsion as $\tau\longrightarrow\infty$:

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where $0<\mu_{1}<\mu_{2}<\cdots$ (Proposition 3.2 in [6]). Then the problem is to show that $L_{j}\not=0$ for some $j$. We see that if $L_{j}=0$ for all $j$, then $u$ has a harmonic continuation in a neighbourhood of $x_{0}$ having a rotation invariance property with respect to some angle $0<\theta\leq\pi$ (Lemma 4.1 in [6]). Note that $x_{0}$ is also a vertex of the convex hull of $D$ which is a polygon. Then applying Friedman-Isakov’s extension argument [3] to $u$ outside the convex hull of $D$, we see that $u$ has a harmonic continuation in the whole domain. Then it is easy to see that $u$ has to be a constant: a contradiction.

$\Box$

From (2.2) we see that \mbox{\cal D}=\mbox{\cal D}(P,Q;\,\cdot\,)\in H^{1}_{\mbox{loc}}(\Omega) satisfies

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for all $\varphi\in C^{\infty}_{0}(\Omega)$; \mbox{\cal D}(P,Q;\,\cdot\,) is not a constant function. Therefore, from Lemma 3.1 for u=\mbox{\cal D}(P,Q;\,\cdot\,), (1.5) and (3.1) we obtain the two formulae in Theorem 1.1.

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Acknowledgments

The author thanks the referees for several suggestions for the improvement of the manuscript. This research was partially supported by Grant-in-Aid for Scientific Research (C)(No. 11640151) of Japan Society for the Promotion of Science.

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