\documentclass[reqno]{amsart} \usepackage{hyperref} \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 $$\int_D h\,dx=c\int_{\partial D}h\,d\sigma, \label{integral}$$ 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., $$\begin{gathered} -\Delta u=1 \quad \mbox{in D} \\ u=0 \quad \mbox{on \partial D} \end{gathered}\label{Saint}$$ 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 $$\begin{gathered} \Delta u' =0 \quad \mbox{in D} \\ u' =-\frac{\partial u}{\partial\nu} V\cdot\nu \quad \mbox{on \partial D}, \end{gathered}\label{derivative}$$ 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 $$\int_D u'\,dx=c\int_{\partial D}u' \,d\sigma, \label{integrall}$$ 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 $$\int_{\partial D}f V\cdot\nu\,d\sigma=0, \label{vector}$$ 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 $$\int_\gamma f(x)\phi(x)(x-x^*)\cdot \nu (x)\,d\sigma =0. \label{ali}$$ 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 $$\int_Du'\,dx=\int_{\partial D}\big(\frac{\partial u}{\partial\nu}\big)^2 V\cdot\nu \,d\sigma. \label{identity}$$ 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 $$\int_Du'\,dx=-\int_{\partial D}u'\;\frac{\partial u}{\partial\nu}\,d\sigma. \label{identityy}$$ 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 $$\begin{gathered}-\Delta u =1 \quad \mbox{in D} \\ u=0 \quad \mbox{on \partial D} \\ \frac{\partial u}{\partial\nu}=-c \quad \mbox{on \partial D} \end{gathered}\label{overdetermined}$$ 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., $$\begin{gathered} \Delta w=0 \quad \mbox{in D}. \\ \frac{\partial w}{\partial\nu}=pw \quad \mbox{on \partial D} \end{gathered}\label{Stekloff}$$ In (\ref{Stekloff}), $p$ denotes the eigenvalue. It is well known that there are infinitely many eigenvalues \$0=p_1