\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 154, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/154\hfil A characterization of balls]
{A characterization of balls using the domain derivative}

\author[A. Didenko, B. Emamizadeh\hfil EJDE-2006/154\hfilneg]
{Andriy Didenko, Behrouz Emamizadeh}  % in alphabetical order

\address{Andriy Didenko \newline
Department of Mathematics\\
The Petroleum Institute \\
P.O. Box 2533\\
Abu Dhabi, UAE}
\email{adidenko@pi.ac.ae}

\address{Behrouz Emamizadeh\newline
Department of Mathematics\\
The Petroleum Institute \\
P.O. Box 2533\\
Abu Dhabi, UAE}
\email{bemamizadeh@pi.ac.ae}

\thanks{Submitted September 28, 2006. Published December 14, 2006.}
\subjclass[2000]{35J25, 35P99}
\keywords{Domain derivative; overdetermined problems; Stekloff problem}

\begin{abstract}
 In this note we give a characterization of balls in $\mathbb{R}^N$
 using the domain derivative. As a byproduct we will show
 that an overdetermined Stekloff eigenvalue problem is
 solvable if and only if the domain of interest is a ball.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

In this note we give a characterization of balls in $\mathbb{R}^N$ using
the domain derivative. As an application we prove that an
overdetermined Stekloff eigenvalue problem is solvable if the
domain of interest is a ball. This work is motivated by the
following result.

\begin{theorem} \label{thm1}
A domain $D\subset\mathbb{R}^N$ is a ball if and only if
there exists a constant $c$ such that the following integral
equality is valid
\begin{equation}
\int_D h\,dx=c\int_{\partial D}h\,d\sigma,
\label{integral}
\end{equation}
for every harmonic function $h$.
\end{theorem}

For the proof of the above theorem, the reader is referred to
\cite{b1,p2}.

Our characterization replaces (\ref{integral}) by another integral
equation which involves the domain derivative of the solution of
the Saint-Venant equation in $D$. This result will enable us to
show that an overdetermined Stekloff eigenvalue problem is
solvable if and only if the domain of the problem is a ball.

\section{Main result}

To state the main result we need some preparation. Henceforth $D$
is a smooth simply connected bounded domain in $\mathbb{R}^N$. By $u$ we
denote the unique solution of the Saint-Venant problem in $D$;
i.e.,
\begin{equation}
\begin{gathered}
-\Delta u=1 \quad \mbox{in $D$} \\
u=0 \quad \mbox{on $\partial D$}
\end{gathered}\label{Saint}
 \end{equation}
Given a vector field $V\in C^2(\mathbb{R}^N;\mathbb{R}^N)$, we denote by $u'$,
the domain derivative of $u$ at $D$ in direction of $V$; the reader
is referred to \cite{s2} for a thorough treatment of the concept
of domain derivatives.
Using  \cite[Theorems 3.1 and 3.2]{s2}, it follows that
\begin{equation}
\begin{gathered}
\Delta u' =0 \quad \mbox{in $D$} \\
u' =-\frac{\partial u}{\partial\nu} V\cdot\nu \quad
\mbox{on $\partial D$},
\end{gathered}\label{derivative}
\end{equation}
where $\nu$ stands for the unit
outward normal vector on $\partial D$. Now we state our main
result.

\begin{theorem} \label{thm2}
The domain $D$ is a ball if and only if there exists a constant
$c$ such that the following integral equation is valid
 \begin{equation}
\int_D u'\,dx=c\int_{\partial D}u' \,d\sigma, \label{integrall}
\end{equation}
for every vector field $V\in C^2(\mathbb{R}^N;\mathbb{R}^N)$.
\end{theorem}

We need the following result.

\begin{lemma} \label{lem1}
Suppose $f\in C(\partial D)$ and the following equation holds
\begin{equation}
\int_{\partial D}f V\cdot\nu\,d\sigma=0, \label{vector}
\end{equation}
for every $V\in C^2(\mathbb{R}^N,\mathbb{R}^N)$. Then $f$ vanishes on $\partial D$.
\end{lemma}

\begin{proof}
 To derive a contradiction suppose $f(x_0)\neq 0$, for
some $x_0\in \partial D$. Let us assume that in fact $f(x_0)>0$;
the case $f(x_0)<0$ can be addressed similarly. Since $f$ is
continuous, we readily infer existence of an open component of
$\partial D$, denoted $\gamma$, where
\[
f(x)\geq\frac{1}{k},\quad \forall x\in\gamma,
\]
for some integer $k$. Thanks to smoothness of $\partial D$
we can make the following observation; namely, $\partial D$ is locally
star-shaped. This means: For every $\xi\in\partial D$, there
exists a ball $B_{\xi}$ centered at $\xi$, and a point $x_{\xi}\in
D$, such that
\[
(x-x_{\xi})\cdot\nu (x)>0,\quad \forall x\in
B_{\xi}\cap\partial D.
\]
Without loss of generality we may
assume there exists $x^*\in D$ such that
\[
(x-x^*)\cdot\nu (x)>0,\quad \forall x\in\gamma.
 \]
Let us now consider a non-negative test function
$\phi\in C^\infty _0(\mathbb{R}^N)$, where the
intersection of the support of $\phi$ with $\partial D$ is a
proper subset of $\gamma$ and has positive measure. Now we choose
$V=\phi (x) (x-x^*)$ in (\ref{vector}); note that $V$ is
admissible since it belongs to $C^2(\mathbb{R}^N,\mathbb{R}^N)$. Thus
\begin{equation}
\int_\gamma f(x)\phi(x)(x-x^*)\cdot \nu (x)\,d\sigma =0.
\label{ali}
\end{equation}
However
\[
\int_\gamma f(x)\phi(x)\;(x-x^*)\cdot
\nu (x)\,d\sigma\geq\frac{1}{k}\int_{{\rm support}
(\phi)\cap\gamma}\phi (x)(x-x^*)\cdot\nu (x)\,d\sigma>0,
\]
which contradicts (\ref{ali}). Thus $f$ must vanish on $\partial D$, as
desired.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 Assume that \eqref{integrall} is
satisfied. Let us fix $V\in C^2(\mathbb{R}^N;\mathbb{R}^N)$. We claim
\begin{equation}
\int_Du'\,dx=\int_{\partial D}\big(\frac{\partial
u}{\partial\nu}\big)^2 V\cdot\nu \,d\sigma. \label{identity}
\end{equation}
To prove \eqref{identity} we observe that from the differential
equation in (\ref{Saint}) we have
$\int_Du'\,dx=-\int_Du'\Delta u\,dx$. Since $u'$
is harmonic in $D$ it then follows that
\[
\int_Du'\,dx=\int_D(u\Delta u' -u'\Delta u)\,dx.
\]
Now an application of the Green identity to the right
hand side of the above equation yields
\[
\int_Du'\,dx=\int_{\partial D}
\big(u\frac{\partial u'}{\partial\nu}-u'\frac{\partial
u}{\partial\nu}\big)\,d\sigma.
\]
Since $u$ vanishes on $\partial
D$, the above equation implies
\begin{equation}
\int_Du'\,dx=-\int_{\partial D}u'\;\frac{\partial
u}{\partial\nu}\,d\sigma. \label{identityy}
\end{equation}
 From \eqref{identityy} and the boundary condition in (\ref{derivative})
we derive \eqref{identity}. From the hypothesis and
\eqref{identity} we obtain $c\int_{\partial D}u'\,d\sigma
=\int_{\partial D}\left(\frac{\partial
u}{\partial\nu}\right)^2V\cdot\nu \,d\sigma$. So again using the
boundary condition in (\ref{derivative}) we derive
\[
-c\int_{\partial D}\partial u/\partial\nu V\cdot\nu\,d\sigma
=\int_{\partial D}\big(\frac{\partial
u}{\partial\nu}\big)^2V\cdot\nu\,d\sigma.
\]
So
\[
\int_{\partial D}\Big(\big(\frac{\partial u}{\partial
\nu}\big)^2+c\frac{\partial u}{\partial\nu}\Big)V\cdot\nu
\,d\sigma=0.
\]
Since $V\in C^2(\mathbb{R}^N;\mathbb{R}^N)$ is arbitrary Lemma \ref{lem1},
applied to the above equation, guarantees that
\[ \frac{\partial u}{\partial\nu}\big(\frac{\partial
u}{\partial\nu}+c\big)=0\quad \mbox{on }\partial D.
\]
By the Hopf boundary point lemma applied to (\ref{Saint}) we infer
that $\partial u/\partial\nu$ is negative on $\partial D$. So the
last equation implies $\partial u/\partial\nu =-c$ on
$\partial D$. This result added to (\ref{Saint}) yields the following
overdetermined boundary value problem
\begin{equation}
\begin{gathered}-\Delta u =1 \quad \mbox{in $D$} \\
u=0 \quad \mbox{on $\partial D$} \\
\frac{\partial u}{\partial\nu}=-c \quad \mbox{on $\partial D$}
\end{gathered}\label{overdetermined}
\end{equation}
It is classical, see \cite{s1,w1}, that (\ref{overdetermined}) is
solvable if and only if $D$ is a ball.

Conversely, let us assume that $D$ is a ball. Without loss of
generality we may assume that $D$ is the ball with radius $R$
centered at the origin. Note that in this case the solution of
(\ref{Saint}) is
\[
u(x)=\frac{1}{2N}(R^2-|x|^2).
\]
Therefore
$\partial u/\partial\nu$ will be equal to $-R/N$ on $\partial D$.
So if we apply (\ref{identityy}) we find that
\[
\int_Du'\,dx=-\frac{R}{N}\int_{\partial D}u'\,d\sigma,
\]
which coincides with the integral equation \eqref{integrall},
with $c=-R/N$. This completes the proof.
\end{proof}

Note  that $c=-R/N$, as in the above argument, could
also be written as
$c=-\frac{\omega_NR^N}{N\omega_NR^{N-1}}=-\frac{V(D)}{S(D)}$,
where $\omega_N$ stands for the volume of the unit $N$-dimensional
ball, and $V(D)$, $S(D)$ denote the volume and the surface area of
$D$, respectively.

In the remaining of this section we focus on the Stekloff
eigenvalue problem; i.e.,
\begin{equation}
\begin{gathered}
\Delta w=0 \quad \mbox{in $D$}. \\
\frac{\partial w}{\partial\nu}=pw \quad
\mbox{on $\partial D$}
\end{gathered}\label{Stekloff}
\end{equation}
In (\ref{Stekloff}), $p$ denotes the eigenvalue.
It is well known that there are
infinitely many eigenvalues $0=p_1<p_2\leq p_3\leq\dots$
for which (\ref{Stekloff}) has non trivial solutions. These
solutions are the corresponding eigenfunctions denoted by
$w_1,w_2,\dots$, where $w_1$ is clearly constant. We now
prove the following result.

\begin{theorem} \label{thm3}
The overdetermined boundary-value problem
\begin{equation}
\begin{gathered}
\Delta w=0 \quad \mbox{in $D$} \\
\frac{\partial w}{\partial\nu}=pw \quad \mbox{on $\partial D$} \\
\int_D w_k\,dx=0 \quad \forall k\geq 2.
\end{gathered}\label{Steklofff}
\end{equation}
is solvable if and only if $D$ is a ball.
\end{theorem}

\begin{proof}
 Let us assume $D$ is a ball. Let $w_k$ be an
eigenfunction corresponding to $p_k,\;k=2, 3, \dots$.
Since $w_k$ is harmonic it follows from the mean value property
that
\[
\int_Dw_k\,dx=d\int_{\partial D}w_k\,d\sigma,
\]
for some constant $d$. Thus using the boundary condition in
(\ref{Stekloff}) in conjunction with the Divergence Theorem we
infer
\[
\int_Dw_k\,dx=\frac{d}{p_k}\int_D\Delta w_k\,dx.
\]
Since $w_k$ is harmonic in $D$ we obtain $\int_Dw_k\,dx=0$, as
desired.

To prove the converse we proceed along the same lines as
in \cite[Theorem 2]{p1} to prove the converse. To this end, let $u$ be the
solution of the Saint-Venant problem in $D$,
$V\in C^2(\mathbb{R}^N;\mathbb{R}^N)$, and $u'$ the domain derivative of $u$ in
direction of $V$. Since $D$ is smooth it follows from
\eqref{derivative} that $u'\in C^2(\overline D)$. Hence $u'$
can be represented in terms of the eigenfunctions $w_k$ as follows
\[
u' (x)=\sum_{i=1}^{\infty}\gamma _i\;w_i(x),
\]
where
\[
 \gamma_i=\int_{\partial D}w_iu'\,d\sigma.
\]
Integrating the equation before the last, over $D$, and taking
into account that $\int_D w_i\,dx=0$, for $i=2,3,\dots$
yields
\[
\int_Du'\,dx=\gamma_1\int_Dw_1\,dx=k\int_{\partial D}u'\,d\sigma,
\]
where $k$ is a constant independent of the
vector field $V$. Since $V$ is arbitrary we can apply
Theorem \ref{thm2} to conclude that $D$ must be a ball, as desired.
\end{proof}

\subsection*{Acknowledgements}
This research is part of a project entitled ``Applications of the
domain derivative''. The authors would like to thank the
Petroleum Institute for its financial support.


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\end{document}