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\begin{document}



 \title[A sampling method]
{A sampling method for an inverse boundary value problem
for harmonic vector fields}

\author{Rainer Kress}



\address{\hskip-\parindent
        Rainer  Kress\\
        Institut f\"ur Numerische und Angewandte Mathematik\\
        Universit\"at G\"ottingen\\
        37083 G\"ottingen\\
        Germany}
\email{kress@math.uni-goettingen.de}


\thanks{This research was
carried out at MSRI and  supported in part by NSF grant DMS-9701755.
The hospitality and the support 
are gratefully acknowledged.
}




\begin{abstract}
Two variants of 
 sampling methods have been recently suggested
and developed by Colton and Kirsch for the approximate
solution of inverse obstacle scattering problems for acoustic waves,
i.e., for inverse boundary value problems for the Helmholtz
equation. We present an analogue of the second variant of
these methods for the solution
of an inverse boundary value problem from electrostatics, i.e.,
for
 an inverse Dirichlet problem
for harmonic vector fields.
\end{abstract}

\maketitle

\section{Introduction}


Inverse boundary value problems for partial differential
equations, in principle, consist of 
 determining the shape of an unknown object $D$
 from a knowledge of the type of the
 boundary condition on its boundary $\pd$ and of one or several
 solutions to the differential equation
at locations away from the boundary. 
They  are
difficult to solve since they are  both nonlinear and
improperly posed. 
Recently  new  solution methods  have  been developed 
which circumvent the issue of nonlinearity
by introducing a parameter point 
$z$ and then solving a linear operator equation
 to determine whether or not
\(z \in D\).  These methods are called   sampling
methods and two different variants
were introduced first for inverse obstacle scattering problems,
i.e., for inverse boundary value problems for the Helmholtz equation,
by Colton and Kirsch~\cite{CoKi} and by Kirsch~\cite{Ki}
(see also the recent survey~\cite{CCM}).
The second variant  was extended
to an inverse Dirichlet problem for the Laplace equation
by H\"ahner~\cite{Haehnerhabil} in three dimensions and 
by Kress~\cite{Krbook,Cubo} in two dimensions.
For the Laplace equation, the two-dimensional case is slightly
more involved due to the behavior of the logarithmic fundamental
solution at infinity. 
The inverse Neumann problem for the Laplace equation equation
in two dimensions was considered by Kress and K\"uhn~\cite{KrKuehn}.
A corresponding approach 
for the inverse impedance tomography problem, i.e.,
for transmission problems for the Laplace equation, was initiated
by Br\"uhl and Hanke~\cite{Br,BrHa}.



The sampling method of Kirsch has also been extended
to inverse obstacle scattering problems for elastic
waves by Alves and Kress~\cite{AlvKr} 
and by Arens~\cite{arens}. However, at the
time this is being written, a corresponding analysis
for the inverse obstacle scattering problem for electromagnetic
waves has no yet been completed, although with factorization
theorems some of the ingredients are already
 available (see~\cite{PikeSab}).
Since the theoretical foundation of the sampling method
in potential theory is conceptionally  simpler than in scattering
theory, in this paper we will consider the sampling
method for an inverse obstacle scattering problem in
electrostatics. We anticipate that investigating this
inverse boundary value problem for harmonic vector fields, i.e.,
for solutions of 
\begin{equation}\label{harm}
\divv v=0,\quad \curl v=0
\end{equation}
might help to finally extend Kirsch's sampling method
also to the case of the time-harmonic Maxwell equations.


If $D$ is a bounded simply connected
domain in $\real^3$ with a connected $C^2$ boundary
$\partial D$ and outward unit normal $\nu$
and $y$ is a point in $\Dex$ and $p$ a vector in $\real^3$,
 we consider the exterior boundary value 
problem to find a harmonic vector field 
$v(\cdot,\,y,p)\in C^1(\Dex)\cap C(\Dexab)$ in $\Dexab$
satisfying the Dirichlet boundary condition
\begin{equation}\label{ein1}    
\nu\times v(\cdot,\,y,p) =-
\nu\times \grad\divv \left\{\Phi(\cdot\,,y)p\right\}
\quad\mbox{on }\partial D,
\end{equation}
where $\Phi$ denotes the fundamental solution to the Laplace equation.
Furthermore, we require that
\begin{equation}\label{ein1*}
\int_{\partial D}\nu\cdot v\,ds=0
\end{equation}
and 
$v(x)\to 0$ as $|x|\to \infty $ 
uniformly for all directions.
We may interpret $v+\grad\divv\{ \Phi p\}$ as the total electrostatic
 field of a dipole 
at the point $y$ with polarization $p$
 in the exterior of a  perfect conductor $D$.
The condition \eref{ein1*} implies that the total charge of
$D$ vanishes.
Since  $w$  depends linearly on the polarization vector
$p$ we may write
\[
v(\cdot\,,\cdot\,,p)=wp,
\]
where $w$ is a $3\times3$--matrix depending on $x,y\in \Dex$.



Let $B$ be an additional bounded
 simply connected domain with a connected $C^2$ boundary
$\pb$ that contains $\bar D$. Then the inverse problem
we are interested in is to determine the shape of $D$, i.e., the boundary
$\pd$, from a knowledge of $w(x,y)$ for all $x,y$ in $\pb$.
To solve this inverse problem,
we will characterize $D$ in terms of spectral data of the integral
operator  
$W$
defined by
\begin{equation}\label{ein2}
(Wg)(x):=\int_{\partial B}w(x,y)g(y)\,ds(y),
\quad x\in \partial B.
\end{equation}
We will consider the two cases where the density
$g$ is either  normal or tangential to the boundary
$\pb$. In the case, where $g=\tilde g \nu$ with some
scalar function $\tilde g$ and
$\nu$ denoting the outward normal to $\pb$, we replace \eref{ein2}
by 
the integral
operator  
$W_\nu:L^2(\partial B)\to L^2(\partial B)$
defined by
\begin{equation}\label{ein2normal}
(W_\nu\tilde g)(x):=\int_{\partial B}w_\nu(x,y)\tilde g(y)\,ds(y),
\quad x\in \partial B,
\end{equation}
with the scalar kernel
$$w_\nu(x,y)=\nu(x)\cdot w(x,y)\nu(y),\quad x,y\in\pb,
$$
i.e., related to the direction of the dipoles
we  only use the normal components of the
field $v$ as data.
In the second case, where $g$ is a tangential field, we view \eref{ein2}
as 
an integral
operator  
$ W:L^2_t(\partial B)\to L^2_t(\partial B)$
mapping the space $L^2_t(\partial B)$ of tangential fields on
$\pb$ into itself,
i.e., we use only the tangential components of the
field $v$ as data.
To distinguish between the two cases,
in the sequel, we use a subscript $\nu$ 
for those operators that correspond  to
the first case.


In particular, we will show that $W_\nu$ and $W$ are compact, self-adjoint, and
positive semi-definite operators.
 Therefore their square roots $W_\nu^{1/2}$ and
$W^{1/2}$ are well defined. If for some
 constant unit vector $e$ and a parameter point $z\in\real^3$
we define the dipole field
$
H(\cdot\,,z):=\grad\divv \{\Phi(\cdot\,,z)e\}$
in $\real^3\setminus\{z\}$,
then our main result is the characterization of the domain $D$ by
the property that the improperly posed linear operator equations
\begin{equation}\label{ein3n}
W^{1/2}_\nu\tilde g=\nu\cdot H(\cdot\,,z)|_{\pb}
\end{equation}
and 
\begin{equation}\label{ein3t}
W^{1/2}g=\left\{\nu\times H(\cdot\,,z)|_{\pb}\right\}\times \nu
\end{equation}
have  solutions $\tilde g\in L^2(\partial B)$
and $g\in L^2_t(\partial B)$, respectively, if and only if $z\in D$.
With the aid of  Picard's theorem this can be used numerically for
a visualization of the unknown domain.

The plan of this paper is as follows. Since the study of an inverse
problem always requires a solid foundation of the corresponding
direct problem, in Section \ref{sec2} we summarize the
classical existence and uniqueness results  for the exterior
Dirichlet problem for harmonic vector fields
 extended by an investigation of the solution operator
as needed in the following analysis. Then, in Section \ref{sec3}
we introduce the inverse problem
with dipoles in normal direction.
Our main result will be 
the theoretical foundation of the sampling method
through Theorem \ref{haehner} and its two corollaries.
The final Section \ref{sec4} 
is devoted to the inverse problem with dipoles in tangential
directions and is kept shorter, since the analysis is analogous
to that in Section \ref{sec3}. 

For simplicity, we have assumed that the domain $D$ is
simply connected. In principle, the analysis can be extended
to multiply connected domains. However, technical difficulties
arise through nonuniqueness issues in connection with the Neumann problem
for harmonic vector fields and the integral operator $I+M$ as used
in the existence analysis. The case when $D$ has a finite
number of components can also be considered. Here, the
flux condition \eref{ein1*} has to be imposed for each
component of $D$ to ensure uniqueness for the
exterior Dirichlet problem.




\section{Exterior Dirichlet problem}\label{sec2}
\setcounter{equation}{0}

Recall that $D$ is a bounded simply connected
domain in $\real^3$ with a connected $C^2$ boundary
$\partial D$ and exterior unit normal $\nu$.
By $C^{0,\alpha}_{t,div}(\pd)$ we denote the
space of all 
$C^{0,\alpha}(\pd)$ H\"older continuous tangential fields $a$ that possess
a weak surface divergence $\Divv a \in C^{0,\alpha}(\pd)$
equipped with the norm
$$
\|a\|_{C^{0,\alpha}_{t,div}}:=
\|a\|_{C^{0,\alpha}} +  \|\Divv a\|_{C^{0,\alpha}}
$$
(see~\cite{CoKr2},~p.~169).
Now we consider the exterior
Dirichlet problem for harmonic vector fields:
Given a tangential field $c\in C^{0,\alpha}_{t,div}(\pd)$
with $\Divv c=0$,
find a  vector field 
$v\in C^1(\Dex)\cap C(\Dexab)$ satisfying the differential
equations
\begin{equation}\label{d1}
\divv v=0,\quad \curl v=0\quad\mbox{in }\Dex,
\end{equation}
the boundary condition
\begin{equation}\label{d2}
\nu\times v=c\quad\mbox{on }\pd,
\end{equation}
and the flux condition
\begin{equation}\label{d3}
\int_{\partial D}\nu\cdot v\,ds=0.
\end{equation}
At infinity it is required that $v(x)\to 0$
for $|x|\to\infty$ uniformly for all directions.
In order to introduce notations
for the subsequent analysis we briefly recall the
classical results on existence and uniqueness for \eref{d1}--\eref{d3}.
For a comprehensive study of boundary value problems for
harmonic vector fields we refer to Martensen~\cite{Martensen}.

Since we assume that $D$ and, consequently, $\Dex$ are
simply connected, from \eref{d1} we have that
$v=\grad u$ for some harmonic function $u$.
Then the homogeneous form $\nu\times v=0$ on $\pd$ of
the boundary condition \eref{d2}
implies that $u=\mbox{const}$ on $\pd$.
Now, using Green's integral theorem
and the behavior of  harmonic vector fields at infinity,
from \eref{d3} it can be deduced that $\grad u=0$
and consequently $v=0$ for any solution to
the homogeneous boundary value problem \eref{d1}--\eref{d3}.
From the vector formula
\begin{equation}\label{vecform}
\Divv \{\nu\times v\}=-\nu\cdot \curl v
\end{equation}
(see ~\cite{CoKr2},~p.~169) we observe that the
condition $\Divv c=0$ is necessary for the
solvability of \eref{d1}--\eref{d3}.

If we denote by
\[
\Phi (x,y):=  {1 \over 4 \pi} \;  {1 \over |x-y|}\; , 
\quad x\neq y,
\]
 the
 fundamental solution
of Laplace's equation in $\real^3$, then the vector field
\begin{equation}\label{1.5}
v(x)= \curl \int_{\partial D}  \Phi  (x,y) a (y)\,ds(y),
\quad x \in \Dex, 
\end{equation}
with   density $a\in C^{0,\alpha}_{t,div}(\pd)$
 is a solution of \eref{d1}--\eref{d3} if $a$ solves the
 integral equation
\begin{equation}\label{1.6}
a  + Ma=2c, 
\end{equation}
where the 
 integral operator
$M:C^{0,\alpha}_{t,div}(\pd)\to C^{0,\alpha}_{t,div}(\pd)$ is given by
\begin{equation}\label{1.6*}
(Ma )(x):=2
\int_{\pd} \nu(x)\times \curl_x\{ \Phi (x,y) a(y)\}
 ds(y),\quad x \in 
\pd.
\end{equation}
Clearly, $v$ as defined by \eref{1.5} satisfies $\divv v=0$
in $\Dex$ 
and, by Stokes' theorem, the flux condition \eref{d3}.
Using the relation
\begin{equation}\label{mdiv}
\Divv Ma = -K^*\Divv a
\end{equation}
(see ~\cite{CoKr2},~p.~169), 
where $K^*:C(\pd)\to C(\pd)$  denotes the normal derivative of the
single-layer potential operator, given by
$$
(K^*\varphi)(x):=
2 \int_{\pd}  {\partial \Phi (x,y) \over 
\partial \nu(x)}\; \varphi (y) \,ds(y),\quad x \in 
\pd,
$$
from \eref{1.6} 
and the solvability condition $\Divv c=0$ we deduce
that 
$$
\Divv a - K^* \Divv a=0.
$$
This implies that $\Divv a=0$, since $I-K^*$
is injective (see Theorem 6.20 in~\cite{Krbook}).
Because of the vector identity
$
\curl\curl=-\Delta +\grad\divv
$
we have that
\begin{equation}\label{transf0}
\curl \curl \int_{\partial D}  \Phi  (x,y) a (y)\,ds(y)
=\grad \divv \int_{\partial D}  \Phi  (x,y) a (y)\,ds(y)
\end{equation}
for integrable densities $a$ and  $x\in
\real^3\setminus \pd$.
From this,
with the aid of
the Gauss surface divergence theorem, 
we can deduce
that
\begin{equation}\label{transf}
\curl \curl \int_{\partial D}  \Phi  (x,y) a (y)\,ds(y)
=\grad \int_{\partial D}  \Phi  (x,y)\Divv a (y)\,ds(y)
\end{equation}
for $a\in C^{0,\alpha}_{t,div}(\pd)$ and  $x\in
\real^3\setminus \pd$ (see~\cite{CoKr2},~p.~170).
Therefore, $\Divv a=0$ now implies that 
$\curl v=0$
in $\Dex$. 
Finally,
by the potential theoretic jump
relations (see Theorem 6.11 in~\cite{CoKr2}), the integral equation \eref{1.6} 
implies that the boundary condition \eref{d2} is fulfilled.

The integral operator 
$M:C^{0,\alpha}_{t,div}(\pd)\to C^{0,\alpha}_{t,div}(\pd)$
is compact (see Theorem 6.16 in~\cite{CoKr2}).
Since for a simply connected domain $D$ the operator
$I+M$  can be shown to be injective (see Theorem 5.4 in~\cite{CoKr1}),
by the Riesz theory, it
has a bounded inverse 
$(I+M)^{-1}:C^{0,\alpha}_{t,div}(\pd)\to C^{0,\alpha}_{t,div}(\pd)$.
 This concludes the classical uniqueness and
existence analysis for the exterior
Dirichlet problem \eref{d1}--\eref{d3}.

Recall that $B\subset \real^2$ is an additional bounded, simply
connected domain with
connected $C^2$ boundary such that $\bar D\subset B$.
We introduce an operator 
\[
A_\nu :\left\{\nu\times v|_{\pd}\right\}\times \nu\mapsto \nu\cdot v|_{\pb}
\]
that, for solutions
to \eref{d1}--\eref{d3}, maps the  tangential component on $\pd$ 
onto the normal component on $\pb$. 
From the above existence analysis we have that
\begin{equation}\label{a}
A_\nu =2U_\nu(I+M)^{-1}R,
\end{equation}
where  $R:L^2_t(\pd)\to L^2_t(\pd)$ is defined by
\[Ra:=\nu\times a\]
and 
$U_\nu:L^2_t(\pd)\to L^2(\pb)$  by
\[
(U_\nu a)(x):=\nu(x)\cdot\curl
 \int_{\pd}   \Phi (x,y) a (y)\, ds(y),\quad x \in 
\pb.
\]
For the subsequent analysis we need to consider
the operator $A_\nu $ in appropriate $L^2$ spaces, i.e.,
Hilbert spaces.
 By
$L^2_t(\pd)$ we denote the space of square integrable
tangential fields on   $\partial D$ and,
further, we denote
\[
L^2_{t,div}(\partial D):
=\{a\in L^2_t(\partial D):\Divv a\in L^2(\partial D)\}
\]
equipped with the norm
\[
\|a\|_{L^2_{t,div}}^2
:=\|a\|_{L^2_{t}}^2+
\|\Divv a\|_{L^2}^2.
\]
Because its  kernel is weakly singular,
 $M$ is a
compact operator from 
$L^2_t(\pd)$
into itself.
Using the Fredholm alternative it can be shown that
the nullspace
of $I+M$  with respect to $L^2_t(\pd)$ coincides with
the nullspace with respect to $C^{0,\alpha}_{t,div}(\pd)$ 
(see also~\cite{CoKr2},~p.~59). Therefore,
 the inverse 
operator $(I+M)^{-1}:L^2_t(\pd)\to L^2_t(\pd)$
also exists and is bounded. 
Since $U_\nu$ is an integral operator with continuous kernel,
it is compact from  $L^2_t(\pd)$ into $L^2(\pb)$
and, consequently, 
the operator $A_\nu $ as given by \eref{a} is a compact operator from
$L^2_t(\pd)$ into  $L^2(\pb)$. 
We note that due to the solvability condition $\Divv c=0$
the operator $A_\nu $ given by \eref{a} corresponds to the
mapping of the tangential component of harmonic vector fields on $\pd$
onto the normal component
on $\pb$
only on the subspace $R(L^{2,0}_{t,div}(\pd))$, where
$$
L^{2,0}_{t,div}(\pd):=\left\{
a\in L^{2}_{t,div}(\pd):
\Divv  a=0\right\}.
$$
Since $M$ 
is also compact from $L^2_{t,div}(\pd)$
into itself (see~\cite{Haehner}), using \eref{mdiv}
it can be seen that 
\[
(I+M)^{-1}(L^{2,0}_{t,div}(\pd))\subset (L^{2,0}_{t,div}(\pd)).
\]





\begin{theorem}\label{satz3.1}
The compact operator
$A_\nu :L^2_t(\pd)\to L^2(\pb)$
is injective on the subspace
$R(L^{2,0}_{t,div}(\pd))$.
\end{theorem}
\proof
From above we already know that 
$(I+M)^{-1}:L^{2,0}_{t,div}(\pd)\to L^{2,0}_{t,div}(\pd)$
is injective and $R$ trivially is injective. Therefore,
to establish injectivity of $A_\nu $ it remains
to show that $U_\nu:L^{2,0}_{t,div}(\pd)\to L^2(\pb)$ is injective.
The property $\Divv a$ for $a\in L^{2,0}_{t,div}(\pd)$ implies
that the  vector field $v$ defined by
\eref{1.5} is harmonic (see \eref{transf}).
From
$U_\nu a=0$ we have   $\nu\cdot v=0=0$ on $\pb$,
that is, $v$ solves the homogeneous Neumann problem
for harmonic vector fields in $\real^3\setminus B$.
Since for harmonic vector fields in simply connected
domains we can write $v=\grad u$ for some harmonic
function $u$, from the uniqueness results for
the Neumann problem for harmonic functions we can
deduce that the homogeneous Neumann problem
in simply connected domains only allows the trivial solution.
Therefore we have $v=0$ in $\real^3\setminus B$, whence
by 
 analyticity   $v=0$ in $\Dex$ follows.
From this, using 
 the potential theoretic jump-relations in the
$L^2$ sense due to Kersten~\cite{Kersten} 
(see~\cite{Krbook}~p.~172)
we obtain that $a+Ma=0$. Hence $a=0$, since
$I+M$ is injective.
\beweisende



Note that by the Gauss divergence theorem 
and the condition \eref{ein1*} we have that
\begin{equation}\label{bilda}
A_\nu R(L^{2,0}_{t,div}(\pd))\subset L^2_0(\pb),
\end{equation}
where
\[
 L^2_0(\pb):=\left\{
g \in L^2(\pd):\int_{\pb}g\,ds=0\right\}.
\]
The $L^2$ adjoint $A^*_\nu: L^2(\pb)\to L_t^2(\pd)$
is given by 
\begin{equation}\label{a*}
 A^*_\nu=2R^*(I+M^*)^{-1}U_\nu^*.
\end{equation}
Here, 
as can be seen by interchanging orders of integration,
the adjoint \linebreak $U_\nu^*: L^2(\pb)\to L_t^2(\pd)$
of $U_\nu$ is described by
\[
(RU_\nu^*g)(x)=\nu(x)\times\curl
 \int_{\pb}   \Phi (x,y) g (y)\nu(y)\, ds(y),\quad x \in 
\pd,
\]
the adjoint $M^*:L^2_{t}(\pd)\to L^2_{t}(\pd)$
of $M$  by $M^*=RMR$ (see~\cite{CoKr2},~p.~171), and the adjoint
of $R$  by $R^*=-R$. This can be used
to transform \eref{a*} into
\[
 A^*_\nu=2(I-M)^{-1}R^*U_\nu^*,
\]
i.e., for $g\in L^2(\pb)$ we have that $a:= A^*_\nu g$
solves
\[
(I-M)a=2R^*U_\nu^*g.
\]
Hence, since $M$ is  compact from $L^2_{t,div}(\pd)$
into itself, by the Fredholm alternative 
and $R^*U_\nu^*g\in L^2_{t,div}(\pd)$
we can conclude
 that 
$A_\nu^*$ maps $L^2(\pb)$ boundedly into $L^2_{t,div}(\pd)$.



\section{Inverse problem with normal dipoles}\label{sec3}
\setcounter{equation}{0}

We now 
consider the special case of the exterior Neumann problem
\eref{d1}--\eref{d3} with the boundary data
 given by a  dipole with polarization
vector $p$ located in some point $y\in\Dex$, i.e.,
the field
 $v=v(\cdot\,,y,p)$
is harmonic
 in $\Dex$, satisfies the
 boundary condition \eref{ein1}, has zero flux
and vanishes at infinity.
We note that in view of \eref{vecform} the solvability 
condition $\Divv c$  is satisfied for the particular boundary
data \eref{ein1}.
The
 inverse problem we want to consider
in this section  is, given 
 $\nu(x)\cdot v(x,y,\nu(y))$  for all $x,y\in \partial B$,
determine the shape of  $D$.
We will
 develop an explicit characterization of the
unknown domain $D$ in terms of spectral data of the
integral operator 
$W_\nu:L^2(\partial B)\to L^2(\partial B)$ with kernel $w_\nu$ as
defined by \eref{ein2normal}.

In the sequel, by $(\cdot\,,\cdot)$ we denote
the inner product in $L^2(\pb)$ and $L^2(\pd)$, respectively.
We define a bounded
 operator
$N:L^2_{t,div}(\partial D)\to L^2_{t}(\partial D)$ by
\begin{equation}\label{defn}
N=\Grad S \Divv,
\end{equation}
where $S:L^2(\partial D)\to L^2(\partial D)$
denotes the single-layer  potential operator 
$$
(S\varphi)(x)
:=  \int_{\partial D}
\Phi(x,y)\,\varphi(y)\, ds(y),\quad
x\in\partial D.
$$
The operator $N$ is well defined, since $S$ maps
$L^2(\pd)$ boundedly into the Sobolev space $H^1(\pd)$
(see Theorem 3.6 in~\cite{CoKr2}). 

\begin{theorem}\label{s1}
The operator $N$ is negative semi-definite with
respect to the $L^2$ inner product. The nullspace of
$N$ is given by $L^{2,0}_{t,div}(\partial D)$.
\end{theorem}
\proof
This follows from the Gauss surface divergence
theorem, i.e.,
\begin{equation}\label{ndefinit}
(Na,b)=(\Grad S \Divv a, b)
=- ( S \Divv a,\Divv b),
\end{equation}
and the positive definiteness of the single-layer operator $S$.
The latter is a consequence of the relation
$$
(S\varphi,\varphi)
=\int_{\real^3}|\grad u|^2dx
$$
for the single-layer potential $u$ with density $\varphi$
which follows by
applying  Green's theorem and using  the jump relations. 
\beweisende










For the following theorem we note that
$A_\nu NA_\nu^*$ is well defined on $L^2(\pb)$,
since $A_\nu^*$ is bounded from $L^2(\pb)$
into $L^2_{t,div}(\pd)$ and $N$ is bounded from
$L^2_{t,div}(\pd)$ into $R(L^{2,0}_{t,div}(\pd))$
as consequence of \eref{vecform}.


\begin{theorem}\label{fact}
The operators $W_\nu$, $A_\nu $, and $N$ are related through
\begin{equation}\label{18n.3}
W_\nu=-A_\nu  N A_\nu^*.
\end{equation}
\end{theorem}
\proof
Define operators $P_\nu :L^2_{t}(\pd)\to L^2(\pb)$
and $P_\nu^*:L^2(\pb)\to L^2_{t}(\pd)$  by
\[(P_\nu a)(x):=\nu(x)\cdot \curl\curl
\int_{\partial D} \Phi(x,y)
 a(y)\,ds(y),
\quad x\in \partial B,\]
and
\[(P_\nu^*g)(x):=-\Grad 
\int_{\partial B}\frac{\partial \Phi(x,y)}{\partial \nu(y)}\;g(y)\,ds(y),
\quad x\in \partial D.
\]
Then, using \eref{transf}, for $a\in L^2_{t,div}(\pd)$ we can write
\[(P_\nu a)(x)=
\int_{\partial D} \frac{\partial \Phi(x,y)}{\partial \nu(x)}\,\Divv
 a(y)\,ds(y),
\quad x\in \partial B.\]
Now, we can interchange orders of integration
and use the Gauss surface divergence theorem
to see that $P_\nu$ and $P_\nu^*$
 are adjoint with respect to the $L^2$ inner product, since
$L^2_{t,div}(\pd)$ is dense in $L^2_{t}(\pd)$.
Noting that 
\[
\divv_x \left\{\Phi(\cdot\,,y)\,g(y)\nu(y)\right\}=
-\frac{\partial \Phi(\cdot\,,y)}{\partial \nu(y)}\;g(y)
\]
for $g\in L^2(\pb)$ from the boundary condition \eref{ein1}
we have that
\[
\nu(x)\cdot v(x,y,g(y)\nu(y))
= \left(A_\nu  \Grad\frac{\partial \Phi(\cdot\,,y)}{\partial \nu(y)}
\;g(y)\right)(x),\quad x,y\in \pb.
\]
Integrating this over $\pb$ and using
the boundedness of $A_\nu $ we deduce that
\begin{equation}\label{10.1}
W_\nu= -A_\nu  P_\nu^*.
\end{equation}
With the aid of  \eref{transf},
by the definition of the operators, we  have
$P_\nu =A_\nu N$ on $L^2_{t,div}(\pd)$,
 whence
\begin{equation}\label{10.1*}
P_\nu^*=NA_\nu^*
\end{equation}
on $L^2(\pb)$ follows, since
$L^2_{t,div}(\pd)$ is dense in $L^2_{t}(\pd)$.
Inserting this
into \eref{10.1} completes the proof.
\beweisende

As a first consequence of the factorization \eref{18n.3}
we note that the operator 
$W:L^2(\pb)\to L^2(\pb)$ is compact,
since $A:L^2(\pd)\to L^2(\pb)$ is compact
and  $N A_\nu^*:L^2(\pb)\to L^2(\pd)$ is bounded.


\begin{theorem}\label{fak2}
The compact operator $W_\nu:L^2(\pb)\to L^2(\pb)$ is 
positive
semi-definite. 
The nullspace of $W_\nu$ is given by
$\spa\{1\}$ and 
there exists a complete orthonormal system
$g_n$, $n=1,2,\ldots,$ of $ L^2_0(\pb)$ of eigenelements of $W_\nu$ with
positive eigenvalues $\lambda_n$, i.e.,
\begin{equation}\label{eig}
W_\nu g_n=\lambda_ng_n,\quad n=1,2,\ldots.
\end{equation}
\end{theorem}
\proof 
From \eref{18n.3} and the self-adjointness of 
$N$ we observe that $W_\nu$ is a self-adjoint operator.
For $g\in L^2(\partial B)$ we have 
$
(W_\nu g,g)=-(N A_\nu^*g, A_\nu^*g)
$
and the negative semi-definiteness of $N$ 
implies that $W_\nu$ is  positive semi-definite.
From \eref{ndefinit} and the positive definiteness
of $S$ it follows that
equality $(W_\nu g,g)=0$ holds if and only if 
$NA_\nu^*g=0$.

To characterize the nullspace of $W_\nu$, as consequence of \eref{10.1*}
we note that the nullspaces of 
$NA_\nu^*$ and $P_\nu^*$ coincide.
If $P_\nu^*g=0$ then the double-layer potential
\begin{equation}\label{defu}
u(x):=
\int_{\partial B}\frac{\partial \Phi(x,y)}{\partial \nu(y)}\;g(y)\,ds(y),
\quad x\in \real^3\setminus\partial B,
\end{equation}
satisfies
$\Grad u=0$ on $\pd$, that is, $u=\mbox{const}$
on $\pd$. Consequently,
by the maximum-minimum principle for harmonic functions and
analyticity we have
 $u=\mbox{const}$ in $B$.
The jump-relations for the double-layer potential
in the $L^2$ sense (see~\cite{CoKr2} p.~45) imply that
$g-K_Bg=0$
where
$$
(K_Bg)(x):=
2 \int_{\pb}  {\partial \Phi (x,y) \over 
\partial \nu(y)}\; g (y) \,ds(y),\quad x \in 
\pb.
$$
This implies that $g$ is continuous,
since with the help of the Fredholm alternative it can be seen that the
nullspaces of $I-K_B$ with respect to $L^2(\pb)$ and
$C(\pb)$ coincide (see~\cite{CoKr2}~p.~59). Hence, because the
normal derivative of the double layer potential with continuous
density
is continuous across $\pb$, from the uniqueness
for the exterior Neumann problem for Laplace's equation
 we have $u=0$ in
$\real^3\setminus B$. Therefore, again the jump relations
finally imply that $g=\mbox{const}$.
Conversely, if $g=\mbox{const}$ then, by
Green's theorem, the double-layer
potential $u$ defined by \eref{defu} is constant
in $B$, whence $P_\nu^*g=-\Grad u|_{\pd} =0$ follows.

Because in view of \eref{bilda}
 we may consider  $W_\nu$ as a compact and positive definite
operator from $L^2_0(\pb)$ into itself, the statement
on the eigenvalues and eigenfunctions are straightforward consequences
of the spectral theory for such operators 
(see Theorem 15.12
in~\cite{Krbook}).
\beweisende

Since the single-layer potential operator $S:L^2(\pd)\to L^2(\pd)$
is compact and positive definite, there exists
a well defined compact and positive definite 
operator $S^{1/2}:L^2(\pd)\to L^2(\pd)$ that satisfies
$S^{1/2}S^{1/2}=S$ (see \eref{wurzel} below).
Because $S$ is a bounded operator 
from the Sobolev space $H^{-1/2}(\pd)$ into $H^{1/2}(\pd)$ (see~\cite{Mclean})
we have that
\[
\sup_{\|\psi\|_{H^{-1/2}}=1}
\|S^{1/2}\psi\|^2_{L^{2}}=\sup_{\|\psi\|_{H^{-1/2}}=1}
(S\psi,\psi)\leq c^2
\]
for some constant $c>0$. Therefore
\[
\|S^{1/2}\varphi\|_{H^{1/2}}
=\sup_{\|\psi\|_{H^{-1/2}}=1}|(S^{1/2}\varphi,\psi)|
=\sup_{\|\psi\|_{H^{-1/2}}=1}|(\varphi,S^{1/2}
\psi)|
\leq c\|\varphi\|_{L^{2}}
\]
for all $\varphi\in L^{2}(\pd)$, that is,
$S^{1/2}$ is a bounded operator from $L^2(\pd)$ 
into $H^{1/2}(\pd)$.






Using the eigenfunctions $g_n$
of the operator $W_\nu$ we are now in a position to define
 the functions
\begin{equation}\label{18n.5}
\varphi_n:=-\frac{1}{\sqrt{\lambda_n}}\;
S^{1/2}\Divv A_\nu^*g_n,\quad n=1,2,\ldots.
\end{equation}
Since $A_\nu^*g_n\in L^2_{t,div}(\pd)$ 
we  have $\varphi_n\in H^{1/2}(\partial D)$.
Using the Gauss surface divergence theorem, \eref{18n.3}, 
\eref{eig}, and the orthonormality of the $g_n$   we find that
\[
(\varphi_n,\varphi_m)
=-\frac{1}{\sqrt{\lambda_n\lambda_m}}\;
(g_n,A_\nu \Grad S\Divv A_\nu^*g_n)=(g_n,g_m),
\]
i.e.,
the $\varphi_n$ form
 an orthonormal system in $L^2(\partial D)$.

For a constant unit vector $e$ and $z\in B$ we define
\[
\Psi(x,z):=\nu(x)\cdot \grad_x\divv_x\{\Phi(x,z)e\}
,\quad   x\in\partial B.\]
In view of 
\begin{equation}\label{333}
\grad\divv\{\Phi(\cdot\,,z)e\}=
\curl\curl\{\Phi(\cdot\,,z)e\},
\end{equation}
by Stokes' theorem, we have $\Psi(\cdot\,,z)\in L^2_0(\pb)$.
Hence
 the Fourier series
\begin{equation}\label{ent}
\Psi(\cdot\,,z)=\sum_{n=1}^\infty(\Psi(\cdot\,,z),g_n)g_n
\end{equation}
converges in $L^2(\pb)$,
because the $g_n$ are complete in  $L^2_0(\pb)$.




\begin{theorem}\label{haehner}
The point $z\in B$  belongs to $D$ if and only if 
\begin{equation}\label{18n.6}
\sum_{n=1}^\infty
\frac{1}{\lambda_n}\;|(\Psi(\cdot\,,z),g_n)|^2<\infty.
\end{equation}
\end{theorem}
\proof
Let $z\in D$. Then as consequence of  of \eref{333} the field 
\[
H(\cdot,z):= \grad\divv\{\Phi(\cdot\,,z)e\}
\]
is harmonic in $\real^3\setminus\{z\}$, vanishes at infinity,
and, by Stokes' theorem, has vanishing flux
through $\pd$. Therefore, setting
\[b(\cdot\,,z):=\Grad\divv\{\Phi(\cdot\,,z)e\}|_{\pd}
\]
 we have 
$A_\nu b=\Psi$. 
Define $f\in H^1(\pd)$ by
 $f(\cdot\,,z):=\divv\{\Phi(\cdot\,,z)e\}|_{\pd}$.
Then there exists a function $\psi\in L^2(\partial D)$
such that $S\psi=f$, since $S$
is an isomorphism from $L^2(\pd)$ into $H^1(\pd)$ (see~\cite{Mclean}).
Hence, we have \[\Psi=A_\nu b=A_\nu \Grad f=A_\nu \Grad S\psi\] and consequently 
$\Psi=A_\nu \Grad S^{1/2}\varphi$, where 
$\varphi=S^{1/2}\psi$ is in $H^{1/2}(\partial D)$.
Now, from \eref{18n.3} and \eref{18n.5} we find that
$$
\sqrt{\lambda_n}\,(\varphi,\varphi_n)
=
-(\varphi,S^{1/2}\Divv A_\nu^*g_n)
=
(A_\nu  \Grad S^{1/2}\varphi,g_n)=(\Psi,g_n)
$$
for $n=1,2,\ldots$ 
and inserting this into 
Bessel's inequality for $\varphi$ in terms of 
the orthonormal system \eref{18n.5}
 yields the convergence
of the series \eref{18n.6}.

 Conversely, assume that the series \eref{18n.6} converges. Then,
$$
\varphi:=
\sum_{n=1}^\infty \frac{1}{\sqrt{\lambda_n}}\;(\Psi(\cdot\,,z),g_n)
\varphi_n
$$
defines a function $\varphi\in L^2(\partial D)$.
Using \eref{18n.3}, 
\eref{eig}, and \eref{18n.5} we obtain
 the relation
\begin{equation}\label{444}
 A_\nu \Grad S^{1/2}\varphi_n= \sqrt{\lambda_n}\,g_n,\quad n=1,2,\ldots.
\end{equation}
From the well-posedness of the exterior Dirichlet problem
for  Laplace's equation with $H^{1/2}(\pd)$
boundary data it can be seen that the
operator $A_\nu $ maps $\Grad (H^{1/2}(\pd))$
boundedly on $L^2(\pb)$. Therefore, with
the aid of  the
boundedness of the operator
$S^{1/2}:L^2(\pd)\to H^{1/2}(\pd)$, \eref{ent},
and \eref{444}
we deduce that $\varphi$ satisfies
$
A_\nu \Grad S^{1/2}\varphi=\Psi(\cdot\,,z).
$
By the uniqueness for the Neumann problem 
for harmonic vector fields in the exterior
of $B$ and analyticity, this implies that for the weak solution 
$v \in L^2_{loc}(\Dex)$
to the  Dirichlet problem for harmonic vector fields
in the exterior of $D$ with tangential
component
$\Grad S^{1/2}\varphi$
we have $v=H(\cdot\,,z)$.
Therefore the point
 $z$ cannot belong to $B\setminus D$, since 
 $H(\cdot\,,z)\not\in L^2_{ loc}(\Dex)$ if $z\in B\setminus D$.
\beweisende



Since the operator $W_\nu$ is compact, self-adjoint, and positive semi-definite
the square root operator
$W_\nu^{1/2}:L^2(\pb)\to L^2(\pb)$ is well defined
 by
\begin{equation}\label{wurzel}
W^{1/2}_\nu g=\sum_{n=1}^\infty \sqrt{\lambda_n}\,(g,g_n)g_n.\end{equation}
In terms of this operator, by using Picard's theorem
(see Theorem 15.18 in~\cite{Krbook}), we can reformulate
Theorem \ref{haehner} as the following corollary.


\begin{kor}
The point $z\in B$  belongs to $D$ if and only if
the operator equation
\[
W^{1/2}_\nu g=\Psi(\cdot\,,z)
\]
has a solution $g\in L^2(\pb)$.
\end{kor}



\begin{kor}
The domain $D$ is uniquely determined by a knowledge
of \linebreak $\nu(x)\cdot v(x,y,\nu(y))$ for  $x,y\in \pb$.
\end{kor}
\proof This is an immediate consequence of Theorem \ref{haehner}.
\beweis




\section{Inverse problem with tangential dipoles}\label{sec4}
\setcounter{equation}{0}

We now turn to the  case of the inverse problem
using tangential dipole fields $g\in L^2_t(\pd)$
 and the tangential
components of $v(x,y,g(y))$ as data.
Again we will
 develop an explicit characterization of the
unknown domain $D$ in terms of spectral data of the
integral operator 
$W:L^2_t(\partial B)\to L^2_t(\partial B)$ with kernel $w$ as
defined by \eref{ein2}.
For doing so, we need to replace the
operator $A_\nu$ mapping the tangential components
of harmonic vector fields on $\pd$ onto the normal components
on $\pb$ by the operator $A$ that maps the
tangential components
 on $\pd$ onto the tangential components
on $\pb$, i.e.,
\[
A:\left\{\nu\times v|_{\pd}\right \}\times \nu\mapsto 
\left\{\nu\times v|_{\pb}\right \}\times \nu.
\]
Then \eref{a} has to be replaced by
\begin{equation}\label{aneu}
A=2U(I+M)^{-1}R,
\end{equation}
where $U:L^2_t(\pd)\to L^2_t(\pb)$ now is given by
\[
(Ua)(x):=\left[\nu(x)\times \curl
 \int_{\pd}   \Phi (x,y) a (y)\, ds(y)\right]
\times \nu(x)
,\quad x \in 
\pb.
\]
For the analogue of Theorem \ref{satz3.1}  we only need
to be concerned with the injectivity of $U$. However, the
injectivity of $U:L^{2,0}_{t,div}(\pd)\to L^2_t(\pb)$ is obtained
as in the proof of Theorem \ref{satz3.1} by using the
uniqueness for the  Dirichlet problem for harmonic vector fields with
zero flux through $\pb$ in the exterior of $B$ rather than
the uniqueness for the Neumann problem. 
Hence we can state the following theorem.

\begin{theorem}\label{satz3.1neu}
The compact operator
$A:L^2_t(\pd)\to L^2_t(\pb)$
is injective on the subspace
$R(L^{2,0}_{t,div}(\pd))$.
\end{theorem}

As in Section \ref{sec2} we note 
that due to the solvability condition $\Divv c=0$
the operator $A$ given by \eref{aneu} corresponds to the
mapping of the tangential components of harmonic vector fields on $\pd$
onto the tangential components
on $\pb$ only on the subspace $R(L^{2,0}_{t,div}(\pd))$.





The $L^2$ adjoint $A^*: L^2_t(\pb)\to L_t^2(\pd)$
is given by 
\begin{equation}\label{a*neu}
 A^*=2R^*(I+M^*)^{-1}U^*,
\end{equation}
where 
\[
(U^*g)(x)=\left[\nu(x)\times\curl
 \int_{\pb}   \Phi (x,y) g (y)\, ds(y)\right]\times
\nu(x),\quad x \in 
\pd,
\]
is the adjoint $U^*: L^2_t(\pb)\to L_t^2(\pd)$
of $U$.
 This, as in Section \ref{sec2}, can be used
to show that $A^*$ maps $L^2_t(\pb)$ boundedly into $L^2_{t,div}(\pd)$.


For the sequel, the spaces 
$L^2_{t,div}(\pb)$ 
and $L^{2,0}_{t,div}(\pb)$
are defined analogously to the corresponding spaces
on $\pd$.


\begin{theorem}\label{factneu}
The operators $W$, $A$, and $N$ are related through
\begin{equation}\label{18n.3neu}
W=-A N A^*.
\end{equation}
\end{theorem}
\proof
Define operators $P:L^2_{t}(\pd)\to L^2_{t}(\pb)$
and $P^*:L^2_{t}(\pb)\to L^2_{t}(\pd)$ by
\[(Pa)(x):=\Grad \divv \int_{\partial D} \Phi(x,y)
 a(y)\,ds(y),
\quad x\in \partial B,\]
and \[(P^*g)(x):=\Grad \divv
\int_{\partial B} \Phi(x,y) g(y)\,ds(y),
\quad x\in \partial D.
\]
In view of \eref{transf0} and \eref{transf},
for $a\in L^2_{t,div}(\pd)$ and $g\in L^2_{t,div}(\pb)$,
we can write
\[(Pa)(x):=\Grad  \int_{\partial D} \Phi(x,y)
\Divv  a(y)\,ds(y),
\quad x\in \partial B,\]
and \[(P^*g)(x):=\Grad 
\int_{\partial B} \Phi(x,y) \Divv g(y)\,ds(y),
\quad x\in \partial D,
\]
and from this interchanging integrations and employing
a denseness argument it can be seen that
$P$ and $P^*$
are adjoint with respect to the $L^2$ inner product.
Integrating 
\[
 v(x,y,g(y))
= -\left(A \Grad \divv\{\Phi(\cdot\,,y)g(y)\}\right)(x)
\]
 over $\pb$ and using the boundedness of $A$
we can deduce that
\begin{equation}\label{10.1neu}
W= -A P^*.
\end{equation}
Using \eref{transf},
by the definition of the operators, we  have
$P=AN$,
 whence
\begin{equation}\label{10.1*neu}
P^*=NA^*
\end{equation}
 follows. Inserting this
into \eref{10.1neu} completes the proof.
\beweis

\begin{theorem}\label{fak2neu}
The compact operator $W:L^2_{t}(\pb)\to L^2_{t}(\pb)$ is 
positive semi-definite. 
The nullspace of $W$ is given by
$L^{2,0}_{t,div}(\pb)$ and 
there exists an  orthonormal system
$g_n\in L^2_t(\pb)$ of eigenelements of $W$ with
positive eigenvalues $\lambda_n$, i.e.,
\begin{equation}\label{eigneu}
W_\nu g_n=\lambda_ng_n,\quad n=1,2,\ldots.
\end{equation}
The eigenelements are complete in the orthogonal complement
of $L^{2,0}_{t,div}(\pb)$.
\end{theorem}
\proof 
As in the proof of Theorem \ref{fak2},
from \eref{18n.3neu} we deduce that 
 $W$ is   self-adjoint and that  equality holds in
$
(W g,g)\geq 0
$
 if and only if
 $P^* g=0$. 
To characterize the nullspace of $P^*$ let $P^*g=0$. Then 
\begin{equation}\label{defuneu}
u(x):= \divv
\int_{\partial B} \Phi(x,y)  g(y)\,ds(y),
\quad x\in \real^3\setminus\partial B,
\end{equation}
satisfies $\Grad u=0$ on $\pd$.
As in the proof of Theorem \ref{fak2} this implies
that $u=\mbox{const}$ in $B$. Using the jump relations
in the $L^2$ sense for the gradient of a single-layer
potential (see~\cite{CoKr2}~p.~45), the transformation
\eref{transf0} and Stokes' theorem, arguing as in
Section \ref{sec2} for the uniqueness of the solution to the exterior Dirichlet
problem we can conclude that $u=0$ in $\real^3\setminus B$.
In view of the jump relations this implies that $u=0$ everywhere.
Now let $f\in H^1(\pb)$ and denote by $S_B$ the single-layer
operator for the domain $B$. Since $S_B$ is an isomorphism
from $L^2(\pb)$ onto $H^1(\pb)$, there exists $\varphi\in
L^2(\pb)$ such that $f=S_B\varphi$. Then we can interchange
orders of integration to obtain
that
\[
(\Grad f,g)=(\Grad S_B\varphi ,g)=-(\varphi,u)=0
\]
for the inner product in $L^2(\pb)$. Hence $g\in L^2_{t,div}(\pb)$
with $\Divv g=0$. 
Conversely from \eref{transf0}, \eref{transf},
and the definition of $P^*$ it is
obvious that for $g\in L^2_{t,div}(\pb)$
with $\Divv g=0$ we have that $P^*g=0$.
Now as in the proof of Theorem \ref{fak2}
the statement on the eigenelements and eigenvalues follows
from the spectral theorem.
\beweisende

 
Using the eigenelements  $g_n$
of the operator $W$ we can define 
 the functions
\begin{equation}\label{18n.5neu}
\varphi_n:=-\frac{1}{\sqrt{\lambda_n}}\;
S^{1/2}\Divv A^*g_n,\quad n=1,2,\ldots.
\end{equation}
As in Section \ref{sec3} it can be seen that
 $\varphi_n\in H^{1/2}(\partial D)$ and that 
the $\varphi_n$ form
 an orthonormal system in $L^2(\partial D)$.

For a constant unit vector $e$ and $z\in B$ we define
\[
\chi(x,z):=[ \nu(x)\times \grad_x\divv_x\{\Phi(x,z)e\}]
\times \nu(x)
,\quad   x\in\partial B.\]
With the aid of the Gauss surface divergence theorem
it can be seen that
$\chi(\cdot\,,z)$ is orthogonal to the
nullspace $L^{2,0}_{t,div}(\pb)$ of $W$. Therefore,
 we have the Fourier series
\begin{equation}\label{entneu}
\chi(\cdot\,,z)=\sum_{n=1}^\infty(\Psi(\cdot\,,z),g_n)g_n.
\end{equation}

\begin{theorem}\label{haehnerneu}
The point $z\in B$  belongs to $D$ if and only if 
\begin{equation}\label{18n.6neu}
\sum_{n=1}^\infty
\frac{1}{\lambda_n}\;|(\chi(\cdot\,,z),g_n)|^2<\infty.
\end{equation}
\end{theorem}
\proof
If $z\in D$ then, obviously,  
we can write $\chi=A\Grad S^{1/2} \varphi$ with
$\varphi$ given as in the proof of Theorem \ref{haehner}
and from this convergence of the series \eref{18n.6neu} follows.
Conversely, if the series \eref{18n.6neu} converges
then $$
\varphi:=
\sum_{n=1}^\infty \frac{1}{\sqrt{\lambda_n}}\;(\chi(\cdot\,,z),g_n)
\varphi_n
$$
defines a function $\varphi\in L^2(\partial D)$.
Again
proceeding as in the proof of Theorem \ref{haehner},
with the aid of  the expansion \eref{entneu} it can be seen
that $\varphi$ satisfies
$
A \Grad S^{1/2}\varphi=\chi(\cdot\,,z).
$
Now, using the uniqueness for the exterior Dirichlet problem
for harmonic vector fields 
instead of the uniqueness for the exterior Neumann problem
continuing as in  the proof of Theorem \ref{haehner}, it
can be shown that $z$ cannot belong to 
 $B\setminus D$.
\beweis


\begin{kor}
The point $z\in B$  belongs to $D$ if and only if
the operator equation
\[
W^{1/2} g=\chi(\cdot\,,z)
\]
has a solution $g\in L^2(\pb)$.
\end{kor}



\begin{kor}
The domain $D$ is uniquely determined by a knowledge
of \linebreak $\nu(x) \times v(x,y,p)$ for  all $x,y\in \pb$ and
$p\in\real^3$ with $p\cdot \nu(y)=0$.
\end{kor}











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\end{thebibliography}

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