Calder\'on's problem for some classes of conductivities in circularly symmetric domains

TL;DR

This paper addresses Calderón's inverse conductivity problem in circularly symmetric domains, providing explicit reconstruction formulas and demonstrating Lipschitz stability of the solution.

Contribution

The paper introduces explicit formulas for conductivity reconstruction in symmetric domains and proves Lipschitz stability, advancing understanding of inverse problems in these settings.

Findings

Explicit reconstruction formulas derived for symmetric domains

Lipschitz stability of the conductivity reconstruction established

Enhanced understanding of Calderón's problem in specific geometries

Abstract

In this note, we study Calder\'on's problem for certain classes of conductivities in domains with circular symmetry in two and three dimensions. Explicit formulas are obtained for the reconstruction of the conductivity from the Dirichlet-to-Neumann map. As a consequence, we show that the reconstruction is Lipschitz stable.

Equations

$$\{\begin{array}{ll}\nabla \cdot (\gamma \nabla u)=0& \text{mbox}in\Omega ,\\ u=f& \text{mbox}on\partial \Omega .\end{array}$$

$$||{\Lambda }_1-{\Lambda }_2|{|}_{\star }=\underset{\genfrac{}{}{0ex}{}{f\in H^{\frac{1}{2}}(\partial \Omega )}{f\ne 0}}{\sup }\frac{||({\Lambda }_1-{\Lambda }_2)f|{|}_{H^{-\frac{1}{2}}(\partial \Omega )}}{||f|{|}_{H^{\frac{1}{2}}(\partial \Omega )}}.$$

$$||\gamma_{1}-\gamma_{2}||_{L^{\infty}(\Omega)}\leq C\big{(}\log(1+||\Lambda_{1}-\Lambda_{2}||_{\star}^{-1})\big{)}^{-\sigma}$$

$${\gamma }_{\alpha }(x)=\{\begin{array}{ll}{\alpha }_1+{\alpha }_2(a-r)& \text{mbox}if0\le r<a,\\ {\alpha }_0& \text{mbox}ifa\le r<1,\end{array}$$

$${\gamma }_{\alpha }(z)=\{\begin{array}{ll}1+{\alpha }_1& \text{mbox}ifh\le z<\infty ,\\ 1+{\alpha }_2& \text{mbox}if0\le z<h,\end{array}$$

$${\Vert {\Lambda }_{\alpha }-{\Lambda }_{\beta }\Vert }_{\star }\ge C\left(\left|{\alpha }_0-{\beta }_0\right|+\left|{\alpha }_1-{\beta }_1\right|+\left|{\alpha }_2-{\beta }_2\right|\right),\forall {\gamma }_{\alpha },{\gamma }_{\beta }\in \mu (a,{\epsilon }_0,M,N).$$

$${\Vert {\Lambda }_{\alpha }-{\Lambda }_{\beta }\Vert }_{H_{rad}^{\frac{1}{2}}(B)\to H_{rad}^{-\frac{1}{2}}(B)}\ge C\left(\left|{\alpha }_2-{\beta }_2\right|+\left|{\alpha }_1-{\beta }_1\right|\right),\forall {\gamma }_{\alpha },{\gamma }_{\beta }\in \mu (h,M).$$

$$\{\begin{array}{ll}\nabla \cdot ({\gamma }_{\alpha }\nabla u)=0& \text{mbox}inB,\\ u=f& \text{mbox}on\partial B,\end{array}$$

$$\{\begin{array}{rl}({\gamma }_{\alpha }{u}_n^{\prime }{)}^{\prime }+\frac{{\gamma }_{\alpha }}{r}{u}_n^{\prime }-\frac{n^2{\gamma }_{\alpha }}{r^2}{u}_n& =0,\forall n\in \mathbb{Z},\\ \underset{r\to a^{-}}{\lim }{u}_n(r)& =\underset{r\to a^{+}}{\lim }{u}_n(r),\\ \underset{r\to a^{-}}{\lim }\left(\gamma u_n^{\prime }\right)(r)& =\underset{r\to a^{+}}{\lim }\left(\gamma u_n^{\prime }\right)(r).\end{array}$$

$$u_0(r)=c_0,0\le r<1,$$

$$u_n(r)=\{\begin{array}{ll}{b}_n{r}^{|n|}+c_n{r}^{-|n|}& \text{mbox}ifa\le r<1,\\ \sum _{k\ge |n|}{a}_k{r}^k,& \text{mbox}if0<r<a,\end{array}$$

$$a_{|n|+m}$$

$$={\left(\frac{{\alpha }_2}{{\alpha }_1+a{\alpha }_2}\right)}^m\prod _{j=1}^m\frac{(2j-1)\left|n\right|+j(j-1)}{2j\left|n\right|+j^2}{a}_{|n|},m=1,2,\cdots .$$

$$\frac{c_n}{b_n}=\frac{a^{2|n|}\left[\left|n\right|u_n(a^{-})-a\frac{{\alpha }_1}{{\alpha }_0}{u}_n^{\prime }(a^{-})\right]}{\left|n\right|u_n(a^{-})+a\frac{{\alpha }_1}{{\alpha }_0}{u}_n^{\prime }(a^{-})}.$$

$${\Lambda }_{\alpha }f(\theta )=\sum _{n\in \mathbb{Z}}\widehat{{\Lambda }_{\alpha }f}(n)e^{in\theta }$$

$$\widehat{{\Lambda }_{\alpha }f}(n)$$

$$={\alpha }_0\left|n\right|\hat{f}(n)\frac{1-a^{2|n|}+(1+a^{2|n|})\frac{{\alpha }_1}{{\alpha }_0}{B}_n(b)}{1+a^{2|n|}+(1-a^{2|n|})\frac{{\alpha }_1}{{\alpha }_0}{B}_n(b)},\forall n\in \mathbb{Z},$$

$$B_n(b)=1+\frac{b}{2\left|n\right|+1}\times \frac{1+\sum _{m=2}^{\infty }m{b}^{m-1}{h}_{m,n}}{1+\frac{|n|}{2|n|+1}b\left(1+\sum _{m=2}^{\infty }{b}^{m-1}{h}_{m,n}\right)},$$

$$h_{m,n}=\prod _{j=2}^m\frac{(2j-1)\left|n\right|+j(j-1)}{2j\left|n\right|+j^2},b=\frac{a{\alpha }_2}{{\alpha }_1+a{\alpha }_2}.$$

$$B_n(b)=1+\frac{b}{2\left|n\right|+1}\times \frac{1+\sum _{m=2}^{\infty }m{b}^{m-1}{h}_{m,n}}{1+\frac{|n|}{2|n|+1}b\left(1+\sum _{m=2}^{\infty }{b}^{m-1}{h}_{m,n}\right)}.$$

$$B_n(b)$$

$$\le 1+b_0+\sum _{m=2}^{\infty }m{b}_0^m\prod _{j=2}^m\frac{2j-1}{2j}=d_0.$$

$$\left(2\left|n\right|+1\right)\left(B_n(b)-1\right)=\frac{b\left(1+\sum _{m=2}^{\infty }m{b}^{m-1}{h}_{m,n}\right)}{1+\frac{|n|}{2|n|+1}b\left(1+\sum _{m=2}^{\infty }{b}^{m-1}{h}_{m,n}\right)}.$$

$$\underset{n\to \infty }{\lim }\left(2\left|n\right|+1\right)(B_n(b)-1)=\frac{b{(1-b)}^{-\frac{3}{2}}}{{(1-b)}^{-\frac{1}{2}}}=\frac{b}{1-b}.$$

$$\underset{n\to \infty }{\lim }\frac{\frac{{\alpha }_1}{{\alpha }_0}{B}_n(b)-1}{\frac{{\alpha }_1}{{\alpha }_0}{B}_{n+1}(b)-1}=\frac{\frac{{\alpha }_1}{{\alpha }_0}-1}{\frac{{\alpha }_1}{{\alpha }_0}-1}=1.$$

$$\underset{n\to \infty }{\lim }\frac{B_n(b)-1}{B_{n+1}(b)-1}=\underset{n\to \infty }{\lim }\frac{(2\left|n\right|+1)\left(B_n(b)-1\right)}{(2\left|n\right|+3)\left(B_{n+1}(b)-1\right)}=1.$$

$$M_n(b)=\frac{1}{2\left|n\right|+1}+\sum _{m=2}^{\infty }\frac{\left|n\right|{(m-1)}^2}{{(2\left|n\right|+1)}^2}{b}^m{h}_{m,n}+\sum _{m=2}^{\infty }\frac{m^2}{2\left|n\right|+1}{b}^{m-1}{h}_{m,n}+I_n(b),$$

$$I_n(b)=\left(\sum _{l=2}^{\infty }\frac{l^2{b}^{l-1}}{2\left|n\right|+1}{h}_{l,n}\right)\left(\sum _{k=2}^{\infty }\frac{\left|n\right|b^k}{2\left|n\right|+1}{h}_{k,n}\right)-\left(\sum _{l=2}^{\infty }\frac{l\left|n\right|b^{l-1}}{2\left|n\right|+1}{h}_{l,n}\right)\left(\sum _{k=2}^{\infty }\frac{k{b}^k}{2\left|n\right|+1}{h}_{k,n}\right).$$

$$\sum _{k+l-1=m}\left(\frac{l^2\left|n\right|}{{\left(2\left|n\right|+1\right)}^2}-\frac{l\left|n\right|k}{{\left(2\left|n\right|+1\right)}^2}\right)h_{l,n}{h}_{k,n}$$

$$=\frac{1}{2}\sum _{k+l-1=m}\frac{\left|n\right|{(k-l)}^2}{{\left(2\left|n\right|+1\right)}^2}{h}_{l,n}{h}_{k,n}.$$

$$M_n(b)\ge \frac{1}{2\left|n\right|+1}.$$

$$\sum _{m=2}^{\infty }\frac{\left|n\right|{(m-1)}^2}{{(2\left|n\right|+1)}^2}{b}^m{h}_{m,n}$$

Full text

Part of this work was done when the second author visited the Vietnam Institute for Advanced Study in Mathematics (VIASM) whom we thank for support and hospitality. We thank N.A.Tu for useful conversation and the referee for helpfull comments.

1. Introduction

Consider a conductor in a domain $\Omega\subset\mathbb{R}^{n}$ with conductivity $\gamma(x).$ When a voltage potential $f\in{H^{\frac{1}{2}}}(\partial\Omega)$ is applied at the boundary $\partial\Omega$, the induced potential $u$ in $\Omega$ is the unique weak solution in $H^{1}(\Omega)$ of

[TABLE]

The Dirichlet-to-Neumann map is given by $\Lambda_{\gamma}(f)=\gamma{\partial_{\nu}}u|_{\partial\Omega}\in{H^{-\frac{1}{2}}}(\partial\Omega).$ Here $\nu$ denotes the exterior unit normal to $\partial\Omega.$ The problem studied by Calderón in [8] is to determine the conductivity $\gamma$ from $\Lambda_{\gamma}$.

For $n\geq 3$ and $\gamma\in C^{2}$, that $\Lambda_{\gamma}$ uniquely determine $\gamma$ was proved by Sylvester and Uhlmann in [16]. Recently, based on the breakthrough work by Haberman and Tataru [13], Caro and Rogers [9] proved uniqueness for Lipschitz conductivities. There is also related work by Haberman [12].

In two dimension, and $C^{2}$ conductivities, the uniqueness was proved by Nachman [15]. Later, Astala and Päivärinta [4] proved uniqueness for bounded measurable conductivities.

After the uniqueness has been established, it is natural to study the stability of the reconstruction, i.e., we would like to estimate $\gamma_{1}-\gamma_{2}$ in certain norm by

[TABLE]

In [1], Alessandrini proved that the following log-stability estimate hold

[TABLE]

where $C,\sigma$ are positive constants and $\gamma_{j}\in H^{s+2}(\Omega),s>n/2$. Later, Mandache [14] showed that such estimate is optimal.

To improve the stability estimate, Alessandrini and Vessella [3] consider the special classes of piecewise constant conductivities, for $n\geq 3$. The Lipschitz stability obtained therein has been generalized to other classes of conductivities in [2], [7] and [11].

The analog of the result of [3] was proved for the two dimensional case in [5]. Subsequent generalizations of this result are obtained in [6] and [10].

In this paper, we proved Lipschitz stability estimate for two special cases of domains with circular symmetry. In the first case, we consider $\Omega=B(0,1)\subset\mathbb{R}^{2}$ with conductivities of the form

[TABLE]

where $r=|x|$ and $\varepsilon_{0}\leq\alpha_{0},\alpha_{1}\leq M,$ $0\leq\alpha_{2}\leq N.$ We denote this set of conductivities $\mu(a,\varepsilon_{0},M,N).$

In the second case, we consider $\Omega=B(0,1)\times(0,+\infty)\subset\mathbb{R}^{3}$ with conductivities of the form

[TABLE]

where ${\alpha_{j}}\in\left[{0,M}\right],j=1,2,M>0,h>0$. We denote this set of conductivities $\mu(h,M).$

We give a formula for the Dirichlet-to-Neumann map in each case, together with a formula to recover the conductivity from the Dirichlet-to-Neumann map. As a consequence, we show that the map $\Lambda_{\gamma}\mapsto\gamma$ is Lipschitz. More precisely our main results are as follows.

Theorem 1.1**.**

Let $\Omega=B(0,1)$ and $a\in(0,1),\varepsilon_{0},M>0,N\geq 0$. There exists a positive constant $C=C(a,\varepsilon_{0},M,N)$ such that

[TABLE]

Theorem 1.2**.**

Let $\Omega=B(0,1)\times(0,\infty)$ and $h\in(0,\infty),M>0$. There exists a positive constant $C=C(h,M)$ such that

[TABLE]

2. Proof of Theorem 1.1

Consider the Dirichlet problem in the unit disc $B=B(0,1)$ on the plane

[TABLE]

where the conductivity $\gamma_{\alpha}\in\mu(a,\varepsilon_{0},M,N)$.

In the polar coordinate, if $u(x)=\sum\limits_{n\in\mathbb{Z}}{u_{n}(r)e^{in\theta}}\in H^{1}(B)$ then the equation in (2.1) is

[TABLE]

Solving these systems, we obtain

[TABLE]

and for $n\not=0$,

[TABLE]

where

[TABLE]

Note that $\alpha_{2}\geq 0,\alpha_{1}\geq\epsilon_{0}>0,$ the power series $\sum_{k\geq|n|}a_{k}r^{k}$ is uniformly convergent on $[0,a].$

From that we get

[TABLE]

The Dirichlet-to-Neumann map $\Lambda_{\alpha}:{H^{\frac{1}{2}}}(\partial B)\to{H^{-\frac{1}{2}}}(\partial B)$ is determined by

[TABLE]

where $f(\theta)=\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{in\theta}\in H^{\frac{1}{2}}(\partial B)$ and

[TABLE]

[TABLE]

[TABLE]

Note that $0\leq b\leq b_{0}=\dfrac{{aN}}{{{\varepsilon_{0}}+aN}}<1.$

To obtain some properties of $B_{n}(b)$ we need the following technical lemma.

Lemma 2.1**.**

*(i) $\mathop{\lim}\limits_{n\to\infty}\sum\limits_{m=2}^{\infty}{m{b^{m-1}}{h_{m,n}}}=\sum\limits_{m=2}^{\infty}{m{b^{m-1}}\prod\limits_{j=2}^{m}{\dfrac{{2j-1}}{{2j}}}}={(1-b)^{-\frac{3}{2}}}-1.$

(ii) $\mathop{\lim}\limits_{n\to\infty}\sum\limits_{m=2}^{\infty}{{b^{m-1}}{h_{m,n}}}=\sum\limits_{m=2}^{\infty}{{b^{m-1}}\prod\limits_{j=2}^{m}{\dfrac{{2j-1}}{{2j}}}}=\dfrac{2}{{1-b+\sqrt{1-b}}}-1.$*

We have the following proposition:

Proposition 2.2**.**

$B_{n}$’s satisfy:

1. (i)

$1\leq B_{n}(b)\leq d_{0},$ where $d_{0}=1+\frac{{{b_{0}}}}{{{{(1-{b_{0}})}^{\frac{3}{2}}}}}.$ 2. (ii)

$\mathop{\lim}\limits_{n\to\infty}{B_{n}}(b)=1.$ 3. (iii)

$\mathop{\lim}\limits_{n\to\infty}(2\left|n\right|+1)({B_{n}}(b)-1)=\dfrac{b}{{1-b}}.$ 4. (iv)

$\mathop{\lim}\limits_{n\to\infty}\dfrac{{\dfrac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)-1}}{{\dfrac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n+1}}(b)-1}}=1,b\not=0.$ 5. (v)

$\dfrac{{1-{b_{0}}}}{{2\left|n\right|+1}}\leq{B^{\prime}_{n}}(b)\leq\dfrac{A}{{2\left|n\right|+1}}\mbox{ where }A=A(a,\varepsilon_{0},N)\mbox{ is a constant}.$

Proof.

We rewrite $B_{n}(b)$ as follows

[TABLE]

(i) From (2.2) it is easy to see that $B_{n}(b)\geq 1.$

We now show that $B_{n}(b)\leq d_{0}.$ Indeed, using (i) in Lemma 2.2 we have

[TABLE]

(ii) From (2.2) it is not difficult to get $\mathop{\lim}\limits_{n\to\infty}{B_{n}}(b)=1$.

(iii) We have

[TABLE]

Hence, from Lemma 2.1 we obtain

[TABLE]

(iv) We consider two cases:

$\star$ Case $1$: $\alpha_{0}\neq\alpha_{1}$.

From (ii) we have

[TABLE]

$\star$ Case $2$: $\alpha_{0}=\alpha_{1}.$

We need to prove $\mathop{\lim}\limits_{n\to\infty}\dfrac{{{B_{n}}(b)-1}}{{{B_{n+1}}\left(b\right)-1}}=1$. From (iii) we get

[TABLE]

(v) We denote by $M_{n}(b)$ and $N_{n}(b)$ the numerator and denominator of $B^{\prime}_{n}(b)$, respectively. Direct computation gives

[TABLE]

where

[TABLE]

The coefficient of $b^{m}$ in $I_{n}(b)$ is:

[TABLE]

From this we obtain

[TABLE]

Moreover, we have

[TABLE]

Next, we have

[TABLE]

We see that

[TABLE]

It follows that

[TABLE]

From (2.4), (2),(2) and (2) we deduce that

[TABLE]

On the other hand we have

[TABLE]

From (2.8) and (2.9) we have

[TABLE]

where $A$ is a constant depending on $a,\varepsilon_{0},N.$ ∎

We now give an explicit formula to reconstruct the parameters $a$ and $\alpha$ from the Dirichlet-to-Neumann map. We define

[TABLE]

If there is a strictly increasing sequence of positive integers $\{n_{k}\}_{k=1}^{\infty}$ such that $C_{n_{k}}=\alpha_{0}$, it is easy to obtain $\alpha_{0}=\alpha_{1},\alpha_{2}=0,$ i.e. the conductor is homogeneous. Otherwise we have the following proposition.

Proposition 2.3**.**

The following formulas hold

1. (i)

$\alpha_{0}=\mathop{\lim}\limits_{n\to\infty}{C_{n}}.$ 2. (ii)

$a^{-2}=\mathop{\lim}\limits_{n\to\infty}\frac{{{C_{n}}-{\alpha_{0}}}}{{{C_{n+1}}-{\alpha_{0}}}}.$ 3. (iii)

${\alpha_{1}}=\alpha_{0}D,$ where

[TABLE] 4. (iv)

$a\alpha_{2}=\alpha_{1}E,$ where

[TABLE]

Proof.

(i) From (ii) in Proposition 2.2

[TABLE]

(ii) Next we have

[TABLE]

Using (ii) and (iv) in Proposition 2.2 we obtain

[TABLE]

(iii) Using (ii) in Proposition 2.2 we have

[TABLE]

This leads to

[TABLE]

(iv) We now calculate $\alpha_{2}$. From

[TABLE]

we calculate

[TABLE]

From that and (iii) in Proposition 2.2 we get

[TABLE]

From $b={\dfrac{{a{\alpha_{2}}}}{{{\alpha_{1}}+a{\alpha_{2}}}}}$ we obtain $a\alpha_{2}=\alpha_{1}E.$ ∎

We now prove Theorem 1.1.

Proof.

For $\gamma_{\alpha},\gamma_{\beta}\in\mu(a,\varepsilon_{0},M,N),f\in H^{\frac{1}{2}}(\partial B)$ we have

[TABLE]

where $b=a{\alpha_{2}}/(\alpha_{1}+a{\alpha_{2}}),c=a{\beta_{2}}/(\beta_{1}+a{\beta_{2}}),$ and

[TABLE]

By direct computation, we obtain

[TABLE]

We denote by $K_{n}$ and $H_{n}$ the numerator and denominator of $A_{n}-B_{n}$, respectively. We have ${H_{n}}\leq{\left({2+\frac{M}{{{\varepsilon_{0}}}}{d_{0}}}\right)^{2}}$ and

[TABLE]

[TABLE]

When $\alpha_{0}\neq\beta_{0}$, for $n$ big enough, we obtain

[TABLE]

Hence

[TABLE]

For $\alpha_{0}=\beta_{0}$ we also have (2.10).

Next, we have

[TABLE]

From (2.10) we have

[TABLE]

where $C_{1}=C_{1}(a,\varepsilon_{0},M)$ is a constant. We now consider

[TABLE]

So from (2.10) and (2.11) we have

[TABLE]

where $C_{2}=C_{2}(a,\varepsilon_{0},M)$ and

[TABLE]

Using (i) and (v) in Proposition 2.2 we get

[TABLE]

[TABLE]

There exists an $n_{0}=n_{0}(a,\varepsilon_{0},N)$ such that for every $n\geq n_{0}$ then

[TABLE]

We now show that

[TABLE]

We consider three cases.

$\star$ Case 1: $({\alpha_{1}}-{\beta_{1}})({\alpha_{2}}-{\beta_{2}})\geq 0.$

We have

[TABLE]

From (2.12) and (2) we obtain (2.15).

$\star$ Case 2: ${\left({{\alpha_{1}}-\beta_{1}}\right)}\left({{\alpha_{2}}-{\beta_{2}}}\right)<0$ and

[TABLE]

From that we have

[TABLE]

Then there exists an $n_{1}=n_{1}(a,\varepsilon_{0},M,N)>n_{0}$ such that

[TABLE]

We get

[TABLE]

From (2.12) and (2) we have (2.15)

$\star$ Case 3: ${\left({{\alpha_{1}}-\beta_{1}}\right)}\left({{\alpha_{2}}-{\beta_{2}}}\right)<0$ and

[TABLE]

There exists an $n_{2}=n_{2}(a,\varepsilon_{0},M,N)>n_{0}$ such that

[TABLE]

If

[TABLE]

we return to Case 2. Otherwise,

[TABLE]

From (2.18) and (2.19) we obtain

[TABLE]

Moreover, we have

[TABLE]

From (2.12) and (2) we have (2.15). From (2.10) and (2.15) the conclusion follows. ∎

3. Proof of Theorem 1.2

Consider the Dirichlet problem

[TABLE]

where the conductivity $\gamma_{\alpha}\in\mu(h,M).$

Definition 3.1**.**

(i) We denote

[TABLE]

[TABLE]

(ii) Let

[TABLE]

where

[TABLE]

$J_{0}(\lambda_{n}r)$ is Bessel function of order zero, $\lambda_{n}$ is positive zero of function $J_{0}$,

[TABLE]

${J_{1}}({\lambda_{n}})$ is Bessel function of order one and

[TABLE]

with ${J_{1}}({\lambda_{n}})\sim\sqrt{\dfrac{2}{{\pi{\lambda_{n}}}}}\cos\left({{\lambda_{n}}-\dfrac{{3\pi}}{4}}\right)+{\rm O}\left({\dfrac{1}{{\lambda_{n}^{3/2}}}}\right)$ when $n\to\infty.$

The norm of $f\in H_{rad}^{\frac{1}{2}}(B)$ is given by

[TABLE]

(iii) The dual space of $H_{rad}^{\frac{1}{2}}(B)$ is defined by

[TABLE]

with norm

[TABLE]

(iv) We denote

[TABLE]

In the cylindrical coordinates, if $u(r,z)=\sum\limits_{n=1}^{\infty}u_{n}(z)J_{0}(\lambda_{n}r)$ we have

[TABLE]

For $f\in{H_{rad}^{\frac{1}{2}}(B)}$, the Dirichlet problem (3.1) in cylindrical coordinates is

[TABLE]

have unique solution $u\in H^{1}_{rad}\left({B\times(0,\infty)}\right)$.

We expand $u=\sum\limits_{n=1}^{\infty}{{u_{n}}(z){J_{0}}({\lambda_{n}}r)}$. By direct computation we have

[TABLE]

At $z=h$ we have

[TABLE]

It follows that

[TABLE]

The Dirichlet-to-Neumann map $\Lambda_{\alpha}:{H_{rad}^{\frac{1}{2}}(B)}\to{H_{rad}^{-\frac{1}{2}}(B)}$ is determined by

[TABLE]

We now give an explicit formula to reconstruct the parameters $h$, $\alpha$ from the Dirichlet-to-Nemann map. Define

[TABLE]

If $A_{1}=1+\alpha_{2}$ then $\alpha_{1}=\alpha_{2},$ i.e. the conductor is homogeneous. Otherwise $A_{n}\not=1+\alpha_{2}\;\forall n\in\mathbb{N}$ and we have the following proposition.

Proposition 3.2**.**

We recontruct $h,\alpha_{j}$ as follows

1. (i)

${\alpha_{2}}=\mathop{\lim}\limits_{n\to\infty}{A_{n}}-1,$ 2. (ii)

$h=\dfrac{1}{2\pi}\ln\left({\mathop{\lim}\limits_{n\to\infty}\dfrac{{{A_{n}}-1-\alpha_{2}}}{{{A_{n+1}}-1-\alpha_{2}}}}\right).$ 3. (iii)

${\alpha_{1}}=\dfrac{{2A+(A+2){\alpha_{2}}}}{{2-A}},$ where

[TABLE]

Proof.

(i) It is easy to show that ${\alpha_{2}}=\mathop{\lim}\limits_{n\to\infty}{A_{n}}-1.$

(ii) We have

[TABLE]

Note that ${\lambda_{n}}\sim\left({n-\frac{1}{4}}\right)\pi,\mbox{ when }n\to\infty$. We obtain

[TABLE]

Hence

[TABLE]

(iii) Since

[TABLE]

so ${\alpha_{1}}=(2A+(A+2)\alpha_{2})/(2-A).$ ∎

Remark 3.3*.*

We can reconstruct $h,\alpha_{1}$ from $\alpha_{2},A_{1},A_{2}$ as follows

[TABLE]

We now prove Theorem 1.2.

Proof.

Firstly, for each $\gamma_{\alpha},\gamma_{\beta}\in\mu(h,M),f\in H^{\frac{1}{2}}_{rad}(B)$ we have

[TABLE]

where

[TABLE]

[TABLE]

By direct computation we obtain

[TABLE]

where

[TABLE]

We denote by $K_{n}$ and $H_{n}$ the numerator and denominator of $(A_{n}-B_{n}),$ respectively. We have $H_{n}\leq(2+3M)^{2}$ and

[TABLE]

Hence

[TABLE]

For $\alpha_{2}\neq\beta_{2}$, we choose $n$ big enough so that

[TABLE]

From (3.4), (3) becomes

[TABLE]

For $\alpha_{2}=\beta_{2}$ we also have (3.5).

It is easy to get

[TABLE]

Therefore, from (3) and (3.5) we have

[TABLE]

From (3.5) and (3.6) we are done. ∎