\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 132, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/132\hfil Perturbation of the free boundary]
{Perturbation of the free boundary in elliptic problem with discontinuities}

\author[S. Bensid \hfil EJDE-2016/132\hfilneg]
{Sabri Bensid}

\address{Sabri Bensid \newline
Department of Mathematics,
Faculty of Sciences,
University of Tlemcen, B.P. 119, Tlemcen 13000,
Algeria}
\email{edp\_sabri@yahoo.fr}

\thanks{Submitted April 14, 2016. Published June 7, 2016.}
\subjclass[2010]{34R35, 35J25}
\keywords{Perturbation; implicit function theorem; free boundary problem}

\begin{abstract}
 We study the  discontinuous elliptic problem
 \begin{gather*}
 -\Delta u =  \lambda H(u-\mu ) \quad \text{in } \Omega,\\
 u =h \quad \text{on }\partial \Omega,
 \end{gather*}
 where $\Omega$ is a regular bounded domain of $\mathbb{R}^n$,
 $H$ is the Heaviside function, $\lambda, \mu$ are a positive real parameters
 and $h$ is a given function.  We prove the existence of  solutions, 
 and characterize the free boundaries  $\{x\in \Omega: u(x)=\mu\}$ 
 using the perturbation of the boundary condition and smooth 
 boundary of the domain.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

Partial differential equations with discontinuous nonlinearities arise
in models from many concrete problems in mathematical physics like those
of combustion theory, porous media, plasma physics. In this article,
we study the existence  of solutions for the problem
\begin{equation} \label{eP}
\begin{gathered}
 -\Delta u =  \lambda H(u-\mu ) \quad \text{in } \Omega,\\
u =h \quad \text{on }\partial \Omega,
\end{gathered}
\end{equation}
 where $\Omega$ is a smooth bounded domain of $\mathbb{R}^n$,
$H$ is the Heaviside function
\[
H(t)=\begin{cases}
1 & \text{if } t \geq 0\\
 0 &\text{if } t<0,
\end{cases}
\]
  $h$ is a given function, and $\lambda, \mu$ are a positive real parameters.

This problem can be reformulated as an equivalent free boundary problem:
Find $u\in C^2(\Omega\setminus \partial w)\cap C^1(\overline{\Omega})$ such that
\begin{equation} \label{eP1}
\begin{gathered}
-\Delta u =  \lambda  \quad \text{in } w,\\
-\Delta u =0  \quad \text{in }\Omega\setminus w,\\
u =h \quad \text{on }\partial \Omega,
\end{gathered}
\end{equation}
where $w=\{x\in \Omega: u(x)>\mu\}$ and $\partial w $ is the free
boundary to be determined. The set $\{x\in \Omega: u(x)=\mu\}$,
dividing the domain $\Omega$ into two (or more) regions where $-\Delta u=\lambda$
or $\Delta u=0$ is satisfied in the classical sense.

To the best of our knowledge, no investigation has been devoted to establishing
the existence of solutions to such  problem when $\Omega$ is a general domain
using perturbation methods. Here, we give a positive answer to  this study.
The problem \eqref{eP} was investigated in the variational context when
$h=0$ by Ambrosetti and Badial \cite{Ambro} and the extension to the
results for p-Laplacian operator $(p>1)$ has been studied by Arcoya and
Calahorrano in \cite{Arcoya}. Further works with the Neumann boundary
conditions can be found in \cite{Candito,Marano}.

When $h$ does not vanish identically, and the domain $\Omega$ is the unit ball
of $\mathbb{R}^n$, the existence of positive  solutions and the behavior
of the corresponding free boundaries with respect to $h$ has been discussed
by Alexander in \cite{Alexander}. In the case $n=2$, Alexander and
Fleishman \cite{Alexander2} studied the problem \eqref{eP} when $\Omega$
is the unit square. Recently,  the authors in \cite{Bensid1} and
\cite{Bensid2} studied the more general following problem when $\Omega$
is the unit ball of $\mathbb{R}^n$,
\begin{gather*}
-\Delta u =  f(u) H(u-\mu ) \quad \text{in } \Omega,\\
u =h \quad \text{on }\partial \Omega,
\end{gather*}
where $f$  is a  given function.

The  methods used hinge on the parametrization of the free boundary which
is the unknown of our problem. This technique  reduces the study to solve
a nonlinear integral equation and  allows us to obtain positive result
regarding the solvability  of the equation of the free boundary under
 perturbation of the boundary without the additional requirement of the
regularity of the boundary of the initial domain. This will be the approach
of this paper.

 In general, the natural way to deal with the discontinuous elliptic problems
are the variational methods, see for instance \cite{Bonano1,Bonano2}.
But  one soon realizes that the study entails serious difficulties mainly
to characterize the variation of the free boundary . Our goal is different,
having a solution of the given problem, we study the effect on the solutions
under perturbations of the boundary conditions and a smooth boundary of the
domain $\Omega$. We point out also that in this work, we can not use the
symmetry of solutions as in \cite{Alexander,Bensid1,Bensid2}.
Hence, we have uses the perturbation techniques to overcome  the encountered
difficulties.
For more works on the discontinuous elliptic problems, we invite the reader
to consult \cite{Alexander,Ambro,Arcoya,Bensid1,Bensid2,Stuart}
and the references given here.

Because the nonlinearity has a discontinuity  at $u=\mu$, so,  a suitable
concept of solution is needed. We say that a function $u\in W^{2,p}(\Omega),(p>1)$
is a solution of problem \eqref{eP} if
$-\Delta u=\lambda H(u-\mu)$ a.e in $\Omega$
and the trace of $u$ on $\partial \Omega$ is equal to $h$.

We need the following assumptions in this work.
\begin{itemize}
\item[(H1)] Assuming that the boundary  $\partial\Omega$ of the domain
$\Omega$ can be parameterized as $R+\beta(\theta)$ where
 $\beta\in C^2(S),\theta\in S$ and $R>0$ where $S$ is the unit sphere.

\item[(H2)] For $n=2$,
$$
\frac{\lambda}{\mu}>\frac{4e}{R^2}.
$$
and for $n\geq 3$,
$$
\frac{\lambda}{\mu}>M_n(R):=\frac{1}{R^2}
\frac{n(n-2)}{(\frac{2}{n})^{\frac{2}{n-2}}-(\frac{2}{n})^{\frac{n}{n-2}}}
$$

\item[(H3)] The function  $h$ is small enough with
$0\leq \|h\|_{\infty}<\mu$ where $\|h\|_{\infty}=\max_{x\in \partial
\Omega}|h(x)|$.
\end{itemize}
This article is organized as follows:
Section 2 collects some  known results for the problem in a ball giving only
slight information on the different methods of proof.
Section 3 contains the statement of the essential result.
Section 4  provides an approach for studying the problem \eqref{eP}.
In Section 5, we treat the regularity of free boundary and finally,
we give an appendix which contains some useful results.

\section{Existence results in a ball}

In this section, we consider the  problem
\begin{equation} \label{e1}
\begin{gathered}
-\Delta u =  \lambda H(u-\mu ) \quad \text{in } B(0,R),\\
 u =h_0 \quad \text{on }\partial B(0,R),
\end{gathered}
\end{equation}
We  assume that the function $h_0$ satisfies  (H3).

\begin{theorem}[\cite{Bensid2}] \label{thm2.1}
 Suppose that there exists $\mu>0$ such that
\begin{gather*}
\frac{\lambda}{\mu}>\frac{4e}{R^2},\quad \text{for }n=2, \\
\frac{\lambda}{\mu}>M_n(R),\quad \text{for } n\geq 3,
\end{gather*}
then  the problem \eqref{e1} has at least two positive solutions
and the free boundaries are  analytic hypersurfaces.
\end{theorem}

\begin{remark} \label{rmk1} \rm
In \cite{Bensid2}, we have treat the case $n\geq 3$ and the more general problem
\begin{gather*} 
-\Delta u =  f(u) H(u-\mu ) \quad \text{in } B(0,1),\\ 
u =h_0 \quad \text{on }\partial B(0,1),
\end{gather*}
There the assumptions are
\begin{itemize}
\item[(H4)] The function $f$ is $k$-Lipstchitzian, non-decreasing, positive and 
there exist two strictly positive constants $k,\beta> 0$ such that 
$f(s)\leq ks+\beta $ with
$k < \min\{\lambda_1,1\}$, where $\lambda_1$ is the first eigenvalue of 
$-\Delta$ under homogeneous Dirichlet boundary conditions.

\item[(H5)] The function $f$ is differentiable and constant on the interval 
of the form $[0,c]$ where $c >\frac{\beta}{2n-k}$ and the function $h_0$  is
small enough, $\|h\|_{\infty}<\mu$ where 
$\|h\|_{\infty}=\max_{x\in \partial B(0,1)}|h(x)|$.

\item[(H6)] There exists $\mu>0$ such that 
$$
\frac{f(\mu)}{\mu}>M_n=\frac{n(n-2)}{(\frac{2}{n})^{\frac{2}{n-2}}
-(\frac{2}{n})^{\frac{n}{n-2}}}, \quad \text{for}\quad n\geq 3 .
$$
\end{itemize}
In this paper, we put $f(u)=\lambda$ to clarify the obtained the results.
For $n=2$, the  treatment is similar with the adequate modification. 
For the convenience of the reader, we give some calculus related to the 
case $\Omega=B(0,R)$ in the appendix.
\end{remark}

The proof of Theorem \ref{thm2.1} is based on the transformation of our problem 
into  an equivalent nonlinear integral equation which is solved with 
respect to the unknown  free boundary $\{x\in \Omega: u(x)=\mu\}$ 
by  the application of the implicit function theorem.
 We can find the complete proof in \cite{Bensid2}.
 Here, we give some ideas of demonstration that will be used throughout this paper. 
 We will denoted the solution of \eqref{e1} by $u_0$ and we consider the 
 problem called the reduced problem
\begin{equation} \label{e2}
\begin{gathered}
-\Delta u_0 =  \lambda H(u_0-\mu ) \quad \text{in } B(0,R),\\
u_0 =0 \quad \text{on }\partial B(0,R),
\end{gathered}
\end{equation}
We remark that since the Heaviside step function is monotone and not decreasing,
the result of Gidas, Ni and  Nirenberg \cite{Gidas} shows that all positive
solutions of \eqref{e2} are radial.
Hence, we look for the free boundary in the form
$\{(r_0,\theta), \theta\in S\}$ for some $r_0\in (0,R)$
and obtain all radial solutions
of \eqref{e2} by finding $u_0,\lambda$ and $r_0$ so that the differential equation
\begin{equation} \label{e3}
\begin{gathered}
r^{1-n}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u_0}{\partial r})
=  \lambda  \quad 0<r<r_0\\
r^{1-n}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u_0}{\partial r}) =  0
 \quad r_0<r<R,\\
u_0(R)=0, \quad \frac{\partial u_0}{\partial r}(0)=0
\end{gathered}
\end{equation}
is satisfied with the following transmission conditions on the free boundary
$$
u_0(r_0)=\mu\quad  \text{and} \quad
\frac{\partial u_0}{\partial r}(r_0-0)
=\frac{\partial u_0}{\partial r}(r_0+0),
$$
where $\frac{\partial u_0}{\partial r}(r_0-0)$ denotes the left derivative
 of $u$ and $\frac{\partial u_0}{\partial r}(r_0+0)$ denotes the right derivative
at the value $r = r_0$.

Using  assumption (H2), the resolution of \eqref{e3} with the previous conditions
  gives two $r_0\in (0,R)$ says $r_1,r_2$. Hence, a simple calculus shows 
that the free boundaries  are spheres with radii $r_1,r_2\in (0,R)$.

\begin{remark} \label{rmk2} \rm
 Let $r_0$ denote one of the values $r_1$ or $r_2$. 
Using  hypothesis (H2), 
 $r_0\neq Re^{-1/2}\quad \text{for } n=2 $
and 
$$
r_0\neq \Big(\frac{2}{nR^{2-n}}\Big)^{\frac{1}{n-2}}\quad \text{for } n\geq 3.
$$
See the Appendix.
\end{remark}

Now, to study the problem
\begin{equation} \label{e4}
\begin{gathered}
 -\Delta u_0 =  \lambda H(u_0-\mu ) \quad \text{in } B(0,R),\\
u_0 =h_0 \quad \text{on }\partial B(0,R),
\end{gathered}
\end{equation}
we use the effect of perturbation  on the solution  in the boundary values.
 More precisely, when $h_0\neq 0$, we look for the free boundary in the
form $r_0+b(\theta), \theta\in S$, where $b(\theta)$
is the perturbation caused by $h_0$.
Let
$$
w=\{(r,\theta)\in (0,R)\times S, 0\leq r < r_0+b(\theta), \theta\in S\}
$$
Now, we denote by $\chi_w$ the characteristic function of $w$.
In the following result, we formulate a nonlinear equation for the unknown
function $b$ and we prove that by solving it, we can
solve the problem \eqref{e4}.

\begin{proposition}[\cite{Bensid2}]
Under assumption {\rm (H2)}, the
problem 
\begin{equation} \label{e5}
\begin{gathered}
-\Delta u
=  \lambda\chi_w(r,\theta) \quad \text{in } B(0,R),\\
u =h_0 \quad \text{on } \partial B(0,R)
\end{gathered}
\end{equation}
 has a unique solution $u_0 \in C^{1,\alpha}(\overline{\Omega},\mathbb{R})$
 with $\alpha =1-\frac{n}{p}$.
Moreover if $u_0(r_0+b(\theta),\theta)=\mu$ with
$0\leq\|h\|_{\infty}<\mu$, then $u_0$ is a solution of \eqref{e4}.
\end{proposition}

As in \cite{Bensid2}, we conclude with  problem \eqref{e4} by showing the 
existence of the function $b$ via the implicit function theorem. 
We can represent the solution of \eqref{e5} in integral form, and we check 
the hypothesis of previous theorem. For more details, we refer the reader 
to  \cite{Bensid2}. In this article, we denote  
$r_0+b(\theta):=b_0(\theta)$, $\theta\in S $.

\section{Perturbation of the free boundary}

In this section, we are concerned with problem \eqref{eP}. 
We introduce some sets necessary for the study using the perturbation of boundary.
We assume that $\Omega$ is diffeomorphic to a ball. 
Then, we can construct a  curvilinear  parametrisation $(r,\theta)$ in a 
neighborhood of any set in $\Omega$, $r\in (0,R]$ and $\theta \in  S$.

Let $\Omega_{\beta}$ be a perturbation of $B(0,R)$ in the sense that the boundary 
$\partial \Omega_{\beta}$ of every smooth open set $\Omega_{\beta}$ 
close to the ball $B(0,R)$ can be described  by 
$R+\beta(\theta)$, $\theta\in S$ and $\beta\in C^2(S)$. 
Hence, from now, 
$\Omega=\Omega_{\beta}$, and 
$\Omega_0=B(0,R)$. We denote also by  $\Omega_{\beta}$ the admissible 
perturbation of the ball $B(0,R)$ and we define the set of admissible 
surfaces in $\Omega_{\beta}$ by
$$
S_{\beta}=\{f\in C(S):(f(\theta),\theta)\in \Omega_{\beta} \text{ for }
 \theta\in S \}.
$$
Now, for   a function $\psi\in S_{\beta}$, we define the set
$$
\Omega_{\beta,\psi}=\{(r,\theta)\in \Omega_{\beta},\quad r<\psi(\theta)\}.
$$
We seek a solution in $W^{2,p}(\Omega_{\beta}), p > 1$, then the boundary 
value function $h$ which is
a trace of $W^{2,p}(\Omega_{\beta})$ function will be taken in the set
$$
H=\{h\in W^{2-\frac{1}{p},p}(\partial \Omega_{\beta},\mathbb{R}),p>n\}.
$$
We denote by $\chi_{\Omega_{\beta,\psi}}$  the characteristic function of 
$\Omega_{\beta,\psi}$, then, we have the following result.

\begin{proposition} \label{prop3.1}
Assume {\rm (H1), (H2)} and that  
\begin{gather*}
\frac{\lambda}{\mu}>\frac{4e}{R^2}\quad \text{for }n=2,\\
\frac{\lambda}{\mu}>M_n(R),\quad \text{for }n\geq 3.
\end{gather*}
Then the problem
\begin{equation} \label{e6}
\begin{gathered}
-\Delta u =  \lambda\chi_{\Omega_{\beta,\psi}}  \quad \text{in } \Omega_{\beta},\\
u =h \quad \text{on } \partial \Omega_{\beta}
\end{gathered}
\end{equation}
 has a unique solution $u \in C^{1,\alpha}(\overline{\Omega_{\beta}},\mathbb{R})$
with $\alpha =1-\frac{n}{p}$.
Moreover, if $u(\psi(\theta),\theta)=\mu$ with
$0\leq\|h\|_{\infty}<\mu$ then $u$ is a solution of \eqref{eP}.
\end{proposition}

\begin{proof}
First, we see that 
$\lambda \chi_{\Omega_{\beta,\psi}}\in L^p(\Omega_{\beta})$, $p>1$.
From \cite[Theorem 9.15]{Gilbarg}, there exists a unique solution of
 \eqref{e6} in $W^{2,p}(\Omega_{\beta})$.
For $p>n$, $W^{2,p}(\Omega_{\beta})\subset C^{1,\alpha}(\Omega_{\beta},\mathbb{R})$ 
with $\alpha=1-\frac{n}{p}$.
Now, we remark that $u$ satisfies
\begin{gather*}
 -\Delta u =  \lambda   \quad \text{in } \Omega_{\beta,\psi},\\ 
-\Delta u =  0 \quad \text{in }  \Omega_{\beta}\setminus \Omega_{\beta,\psi}, \\ 
u =h \quad \text{on } \partial \Omega_{\beta}.
\end{gather*}
If  we prove the existence of a function $\psi$ such that 
$u(\psi(\theta),\theta)=\mu$, then $u$ will be a solution of
\begin{gather*} 
-\Delta u =  \lambda   \quad \text{in } \Omega_{\beta,\psi},\\ 
u =\mu \quad \text{on } \partial \Omega_{\beta,\psi}\\
-\Delta u =  0 \quad \text{in }  \Omega_{\beta}\setminus \Omega_{\beta,\psi},  \\ 
u =h \quad \text{on } \partial \Omega_{\beta}.
\end{gather*}
In $\Omega_{\beta,b}$, the function $u$ satisfies
\begin{gather*}
 -\Delta u =  \lambda  \quad \text{in } \Omega_{\beta,\psi},\\
 u =\mu \quad \text{on } \partial \Omega_{\beta,\psi}
\end{gather*}
The maximum principle implies
$$
\min_{\Omega_{\beta,\psi}}u=\min_{\partial \Omega_{\beta,\psi} }u=\mu.
$$
Hence, 
$u>\mu$ in $\Omega_{\beta,\psi}$.
In $\Omega_{\beta}\setminus \overline{\Omega_{\beta,b}}$, we have
\begin{gather*} 
-\Delta u =  0   \quad \text{in }
 \Omega_{\beta}\setminus \overline{\Omega_{\beta,\psi}},\\
u=\mu \quad \text{on }  \partial \Omega_{\beta,\psi}, \\ 
u =h \quad \text{on } \partial \Omega_{\beta}.
\end{gather*}
As $0\leq \|h\|_{\infty}<\mu$, then
$$
\max_{\Omega_{\beta}\setminus \overline{\Omega_{\beta,\psi}}}u
=\max_{\partial  \overline{\Omega_{\beta,\psi}}}u=\mu
$$
and consequently, $u<\mu$ in $\Omega_{\beta}\setminus \Omega_{\beta,\psi}$. 
Therefore, the function $u$ satisfies
\begin{gather*} 
-\Delta u =  \lambda H(u-\mu ) \quad \text{in } \Omega_{\beta},\\ 
u =h \quad \text{on }\partial \Omega_{\beta}.
\end{gather*}
To conclude with the existence of solutions of problem \eqref{eP}, we need 
only to show  the existence of the function $\psi$ satisfying the equation
 $$
u(\psi(\theta),\theta)=\mu.
$$
Now, since the solution $u$ depend on the domain $\Omega_{\beta}$, then we can 
not use the local methods directly to prove the existence of $\psi$. 
The variation of the domain $\Omega_{\beta}$ suggests to use an adequate
transformation  which maps the changing domain into a fixed domain and
solves the governing equations in the mapped domain.
To exclude this difficulties, we proceed as the following. 
To each admissible perturbation $(\Omega_0,\Omega_{\beta})$ of the domain
 $\Omega_0$ correspond a transformation $T_{\beta}$ of the domain 
$\Omega_{\beta}$ onto the initial domain $\Omega_0$.
$T_{\beta}:\Omega_{\beta}\to\Omega_0$,
$$
(r,\theta)\to(\overline{r},\theta)=(r+r\frac{\beta}{R},\theta)
$$
where $(r,\theta)$ is the coordinates in $\Omega_{\beta}$ and 
$(\overline{r},\theta)$ the coordinates in $\Omega_0$. 
For a small $\beta$, the transformation $T_{\beta}$ is a diffeomorphism of
 class $C^2$ of the domain $\Omega_{\beta}$ into $\Omega_0$. 
The mapping $T_{\beta}$ maps the class of admissible surfaces 
$S_{\beta}$ into $S_0$. Hence,
$$
T_{\beta}(\theta,\psi(\theta))=(\theta, f(\theta))\quad \text{where }\
 f(\theta) \in S_0.
$$
Using the relation 
$$
\overline{u}(T_{\beta}(r,\theta))=u(r,\theta),
$$ 
problem \eqref{e6} with $u(\psi(\theta),\theta)=\mu$ is equivalent to 
the  problem
\begin{equation} \label{e7}
\begin{gathered}
-L_{\beta} \overline{u}
=  \lambda\chi_{\Omega_{\beta,f}}  \quad \text{in } \Omega_0,\\
\overline{u} =h \quad \text{on } \partial \Omega_0
\end{gathered}
\end{equation}
with the equation
$$
\overline{u}(f(\theta),\theta)=\mu,
$$
where $L_{\beta}$ is a linear operator with continuous coefficients depending
 on $\beta$. The following lemma gives the exact expression of $L_{\beta}$.
\end{proof}

\begin{lemma} \label{lem3.1}
The linear operator $L_{\beta}$ is given by $L_{\beta}=\Delta+\delta_{\beta}$
where  $\delta_{\beta}$ has the  form
\begin{align*}
\delta_{\beta}
&=\frac{\beta}{R}(2+\frac{\beta}{R})\frac{\partial^2}{\partial \overline{r}^2}
+\frac{(n-1)}{\overline{r}}\frac{\beta}{R}\frac{\partial }{\partial \overline{r}}\\
&\quad +\frac{1}{\overline{r}^2}\Big[a_{ij}(\theta)
\Big[\frac{r}{R}\frac{\partial \beta}{\partial  \theta_j}
\Big(\frac{\partial^2}{\partial \overline{r}\partial \theta_i}
+\frac{1}{R(1+\frac{\beta}{R})}\frac{\partial \beta}{\partial \theta_i}
\frac{\partial}{\partial \overline{r}}
+\frac{r}{R}\frac{\partial \beta}{\partial \theta_i}
\frac{\partial^2}{\partial \overline{r}^2}\Big)\\
&\quad +\frac{r}{R}\frac{\partial^2 \beta}{\partial \theta_j
\partial \theta_i}\frac{\partial}{\partial \overline{r}}
+\frac{r}{R}\frac{\partial \beta}{\partial \theta_i}
\frac{\partial^2}{\partial \theta_j\partial \overline{r}}\Big]+b_i(\theta)
\big[\frac{r}{R}\frac{\partial \beta}{\partial \theta_i}
\frac{\partial }{\partial \overline{r}}\big]\Big]
\end{align*}
and 
$$
\Delta=\frac{\partial^2 }{\partial \overline{r}^2}
+\frac{(n-1)}{\overline{r}}\frac{\partial}{\partial \overline{r}}
+\frac{1}{\overline{r}^2}\Delta_{\theta}
$$
when $\Delta_{\theta}$ is the Beltrami-Laplace operator.
\end{lemma}

\begin{remark} \label{rmk3} \rm
This formulas are obtained from the  Laplacian in polar coordiantes. 
More precisely, from
\begin{equation} \label{eF}
\Delta=\frac{\partial^2}{\partial r^2}+\frac{(n-1)}{r}
\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_{\theta}
\end{equation}
where $\Delta_{\theta}=a_{ij}(\theta)
\frac{\partial^2}{\partial \theta_i\partial \theta_j}+b_i(\theta)
\frac{\partial}{\partial \theta_i}$.
\end{remark}

\begin{proof}[Proof of Lemma \ref{lem3.1}]
First, we have
\begin{gather*}
\overline{u}(\overline{r},\theta)=u(r,\theta)\quad  \text{where }
\overline{r}=r+r\frac{\beta}{R}\,,\\
\frac{\partial u}{\partial r}
=\frac{\partial \overline{u}}{\partial \overline{r}}
\frac{\partial \overline{r}}{\partial r}
+\frac{\partial \overline{u}}{\partial \theta}
\frac{\partial \theta}{\partial r}
=\frac{\partial \overline{u}}{\partial \overline{r}}
+\frac{\partial \overline{u}}{\partial \overline{r}}\frac{\beta}{R}:=Q\,,
\\
\begin{aligned}
\frac{\partial^2 u}{\partial r}
&=\frac{\partial Q}{\partial \overline{r}}\frac{\partial \overline{r}}{\partial r}
 +\frac{\partial Q}{\partial \theta}\frac{\partial \theta}{\partial r}\\
&=\frac{\partial^2 \overline{u}}{\partial \overline{r}^2}
 \big(1+\frac{\beta}{R}\big)
 +\frac{\partial^2 \overline{u}}{\partial \overline{r}^2}\frac{\beta}{R}
 \big(1+\frac{\beta}{R}\big)\\
&=\frac{\partial^2 \overline{u}}{\partial \overline{r}^2}
 +\frac{\beta}{R}\big(2+\frac{\beta}{R}\big)
 \frac{\partial^2 \overline{u}}{\partial \overline{r}^2}.
\end{aligned}
\\
\frac{\partial u}{\partial \theta_i}
=\frac{\partial \overline{u}}{\partial \overline{r}}
\frac{\partial \overline{r}}{\partial \theta_i}
+\frac{\partial \overline{u}}{\partial \theta_i}
\frac{\partial \theta_i}{\partial \theta_i}
=\frac{r}{R}\frac{\partial \beta}{\partial \theta_i}
\frac{\partial \overline{u}}{\partial \overline{r}}
+\frac{\partial \overline{u}}{\partial \theta_i}:=Z
\\
\begin{aligned}
\frac{\partial^2 u}{\partial \theta_j \partial \theta_i}
&=\frac{\partial Z}{\partial \theta_j}
 =\frac{\partial Z}{\partial \overline{r}}
 \frac{\partial \overline{r}}{\partial \theta_j}
 +\frac{\partial Z}{\partial \theta_j}
 \frac{\partial \theta_i}{\partial \theta_j}\\
&=\frac{r}{R}\frac{\partial \beta}{\partial \theta_j}
 \frac{\partial}{\partial \overline{r}}
 \Big(\frac{\partial \overline{u}}{\partial \theta_i}
 +\frac{r}{R}\frac{\partial \beta}{\partial \theta_i}
 \frac{\partial \overline{u}}{\partial \overline{r}}\Big)
 +\frac{\partial Z}{\partial \theta_i}\\
&=\frac{r}{R}\frac{\partial \beta}{\partial \theta_j}
 \Big(\frac{\partial^2 \overline{u}}
 {\partial \overline{r}\partial \theta_i}
 +\frac{1}{R(1+\frac{\beta}{R})}
 \frac{\partial \beta}{\partial \theta_i}
 \frac{\partial \overline{u}}{\partial \overline{r}}
 +\frac{r}{R}\frac{\partial \beta}{\partial \theta_i}
 \frac{\partial^2 \overline{u}}{\partial \overline{r}^2}\Big)\\
&\quad +\frac{\partial^2 \overline{u}}{\partial \theta_i \partial \theta_j}
 +\frac{r}{R}\frac{\partial^2 \beta}{\partial \theta_j \partial \theta_i}
 \frac{\partial \overline{u}}{\partial \overline{r}}+\frac{r}{R}
 \frac{\partial \beta}{\partial \theta_i}
 \frac{\partial^2 \overline{u}}{\partial \theta_j\partial \overline{r}}.
\end{aligned}
\end{gather*}
If we replace the previous terms in the expression \eqref{eF}, we find the 
formulas of $L_{\beta}$. 
\end{proof}

Thus, to solve problem \eqref{eP}, it is sufficient to prove the existence 
of function $f$ such that
$$
\overline{u}(f(\theta),\theta)=\mu.
$$
In fact, using the implicit function theorem, we prove that the
equation
 $$
\overline{u}(f(\theta),\theta)-\mu=0
$$ 
is uniquely solvable in a given small neighborhood. 
This is the subject of the following section. 
(See Theorem \ref{thm4.1} below).

\section{Solvability in the neighborhood of smooth free boundary}

To each function $f\in S_0$, we associate the solution $\overline{u}$ of 
 problem \eqref{e7}. We have
\begin{equation} \label{e8}
\begin{gathered}
-\Delta \overline{u}-\delta_{\beta}\overline{u}
=  \lambda\chi_{\Omega_{\beta,f}}  \quad \text{in } \Omega_0,\\
\overline{u} =h \quad \text{on } \partial \Omega_0
\end{gathered}
\end{equation}
The solution $\overline{u}$ corresponding to \eqref{e8} has an integral
 representation which is well defined  \cite[Theorem 4.1]{Bensid1},
 given by
$$
\overline{u}(x)=\int_S P(x,y)h(y)dS
-\lambda\int_{\Omega_0}G(x,y)\chi_{\Omega_{\beta,f}}dy
-\int_{\Omega_0}G(x,y)\delta_{\beta}(\overline{u})(y)dy.
$$
Now, we consider  polar coordinates and we define the  operator
$J:H\times S_0\times D\to C(S,\mathbb{R})$ by
$$
J(h,f,\beta)(\theta)=\overline{u}(f(\theta),\theta)-\mu,
$$
where $D$ is the  neighborhood of zero in $C(S)$; i.e.,
\begin{align*}
J(h,f,\beta)(\theta)
&=\int_{\partial \Omega_0}P(f(\theta),\theta,\theta')h(\theta')d\theta'
-\lambda\int_{\Omega_0}G(f(\theta),\theta,r',\theta')
\chi_{\Omega_{\beta,f}}(r',\theta')dr'd\theta'\\
&\quad -\int_{\Omega_0}\delta_{\beta}(\overline{u})(r',\theta')
G(f(\theta),\theta,r',\theta')dr'd\theta'-\mu.
\end{align*}
Then we have the following result.

\begin{lemma} \label{lem4.1}
The operator $J$ is continuously differentiable with respect to the second variable.
\end{lemma}

\begin{proof}
Let $D_j J$ denote the Frechet derivative of $J$, with respect to the variable 
of order $j$  $(j = 1, 2, 3)$.
Let $\varphi(\theta)$ be a small perturbation of $f(\theta)$, then
\begin{align*}
&J(h,f+\varphi,\beta)(\theta)-J(h,f,\beta)(\theta)-D_2J\varphi(\theta)\\
&=\int_{\partial \Omega_0}P(f(\theta)+\varphi(\theta),\theta,\theta')
 h(\theta')d\theta'\\
&\quad -\lambda\int_{\Omega_0}G(f(\theta)
 +\varphi(\theta),\theta,r',\theta')\chi_{\Omega_{f+\varphi}}(r,\theta')dr'd\theta'\\
&\quad -\int_{\Omega_0}l_{\beta}(\overline{u})(r',\theta')
 G(f(\theta)+\varphi(\theta),\theta,r',\theta')dr'd\theta'
 -\int_{\partial \Omega_0}P(f(\theta),\theta,\theta')h(\theta')d\theta'\\
&\quad +\lambda\int_{\Omega_0}G(f(\theta),\theta,r',\theta')
 \chi_{\Omega_{f}}(r',\theta')dr'd\theta'
 +\int_{\Omega_0}l_{\beta}(\overline{u})(r',\theta')
 G(f(\theta),\theta,r',\theta')dr'd\theta'\\
&\quad +\lambda\int_{\Omega_0}G(f(\theta)+\varphi(\theta),\theta,r',\theta')
 \chi_{\Omega_{f}}dr'd\theta'\\
&\quad -\lambda\int_{\Omega_0}G(f(\theta)
 +\varphi(\theta),\theta,r',\theta')\chi_{\Omega_{f}}dr'd\theta'\\
&\quad -\int_{\partial \Omega_0}\frac{\partial P}{\partial r}(f(\theta'),
 \theta,\theta')h(\theta')\varphi(\theta)d\theta' \\
&\quad +\lambda\int_S \int_0^f (r')^{n-1}\frac{\partial G}{\partial r}
 (f(\theta),\theta,r',\theta')\varphi(\theta)dr'd\theta'\\
&\quad +\lambda\int_S f^{n-1}(\theta)G(f(\theta),\theta,f(\theta'),\theta')
 \varphi(\theta')d\theta'\\
&\quad +\int_{\Omega_0}l_{\beta}
 (\overline{u})(r',\theta')\frac{\partial G}{\partial r}
 ((f\theta),\theta,r',\theta')\varphi(\theta)d\theta'
\end{align*}
Hence,
$$
J(h,f+\varphi,\beta)(\theta)-J(h,f,\beta)(\theta)-D_2J\varphi(\theta)
=I_1+I_2+I_3+I_4
$$
where
\begin{gather*}
\begin{aligned}
I_1&=\int_{\partial \Omega_0}P(f(\theta)+\varphi(\theta),\theta,\theta')
 h(\theta')d\theta'-\int_{\partial \Omega_0}
 P(f(\theta),\theta,\theta')h(\theta')d\theta'\\
&\quad -\int_{\partial \Omega_0}\frac{\partial P}{\partial r}
 (f(\theta'),\theta,\theta')h(\theta')\varphi(\theta)d\theta'
\end{aligned} \\
\begin{aligned}
I_2&=-\lambda\int_{\Omega_0}G(f(\theta)+\varphi(\theta),\theta,r',\theta')
 \chi_{\Omega_{f}}dr'd\theta'\\
&\quad +\lambda\int_{\Omega_0}G(f(\theta),\theta,r',\theta')
 \chi_{\Omega_{f}}(r',\theta')dr'd\theta'\\
&\quad +\lambda\int_S \int_0^f (r')^{n-1}\frac{\partial G}{\partial r}(f(\theta),
 \theta,r',\theta')\varphi(\theta)dr'd\theta'
\end{aligned} \\
\begin{aligned}
I_3&=-\int_{\Omega_0}l_{\beta}(\overline{u})(r',\theta')G(f(\theta)
 +\varphi(\theta),\theta,r',\theta')dr'd\theta'\\
&\quad +\int_{\Omega_0}l_{\beta}(\overline{u})(r',\theta')G(f(\theta),\theta,r',
 \theta')dr'd\theta'\\
&\quad +\int_{\Omega_0}l_{\beta}(\overline{u})(r',\theta')
 \frac{\partial G}{\partial r}((f\theta),\theta,r',\theta')\varphi(\theta)d\theta'
\end{aligned} \\
\begin{aligned}
I_4&=\lambda\int_{\Omega_0}G(f(\theta)+\varphi(\theta),\theta,r',\theta')
 [\chi_{\Omega_f}-\chi_{\Omega_{f+\varphi}}]dr'd\theta' \\
&\quad +\lambda\int_S f^{n-1}(\theta)G(f(\theta),\theta,f(\theta'),\theta')
 \varphi(\theta')d\theta'\\
&=-\lambda\int_S d\theta'\Big[\int_f^{f+\varphi}(r')^{n-1}G(f(\theta)
 +\varphi(\theta),\theta,r',\theta')dr'\\
&\quad -f^{n-1}(\theta)G(f(\theta),\theta,f(\theta'),\theta')\varphi(\theta')\Big]
\end{aligned}
\end{gather*}
Using  Taylor's theorem, we have 
$I_1,I_2,I_3,I_4=o(\|\varphi\|_{\infty})$ when 
$\|\varphi\|_{\infty}\to+\infty$.
Hence,
$$
D_2J(h,f,\beta)\varphi(\theta)
=\frac{\partial \overline{u}}{\partial r}(f(\theta),\theta)
\varphi(\theta)-\lambda\int_S f^{n-1}(\theta')
G(f(\theta),\theta,f(\theta'),\theta')\varphi(\theta')d\theta'.
$$
Now, to solve the equation $J(h,f,\beta)(\theta)=0$, for $\theta\in S$, 
in the neighborhood of $(h_0,b_0,0)$. we need to know the invertibility of $J$. 
We recall the reader that we consider here a general domain and that causes 
some difficulties to build clear  conditions so that
the operator $D_2J(h_0,b_0,0)$ either invertible of $S_0$ into $C(S)$ or not.
 We refer to \cite{Bensid1} to see explicit conditions such that the proper 
operator is invertible when the domain is a ball.  \\Let the operator
$$
D_2J(h_0,b_0,0)\varphi(\theta)
=\frac{\partial u}{\partial r}(b_0(\theta),\theta)\varphi(\theta)
-\lambda\int_S b_0^{n-1}(\theta')G(b_0(\theta),\theta,b_0(\theta'),\theta')
\varphi(\theta')d\theta'.
$$
First,  note that $u$ satisfies the equation 
\begin{gather*}
-\Delta u =  \lambda H(u-\mu ) \quad \text{in } B(0,R), \\
u =h_0 \quad \text{on }\partial B(0,R).
\end{gather*}
Then, when $h_0$ is small, the solution $u$ is close in $C^{1,\alpha}$ 
to the solution of
\begin{gather*}
-\Delta u =  \lambda H(u-\mu ) \quad \text{in } B(0,R), \\
u =0 \quad \text{on }\partial B(0,R).
\end{gather*}
Since, $\frac{\partial u}{\partial r}(r_0,\theta)<0$ for 
$r_0\in (0,R),\theta\in S$, then 
$\frac{\partial u}{\partial r}(b_0(\theta),\theta)<0$ for $b_0$ caused 
by the small perturbation $h$.
For the other part, let
$$
K\varphi(\theta)=\lambda\int_S b_0^{n-1}(\theta')G(b_0(\theta),
\theta,b_0(\theta'),\theta')\varphi(\theta')d\theta'.
$$
Let $L_{b_0}^2(S)$ the space of functions belonging to the space $L^2(S)$ 
with the inner product
\begin{equation} \label{eip}
\langle u,v\rangle =\int_Sb_0^{n-1}u(\theta)v(\theta)d\theta.
\end{equation}
We remark that the operator $K$ is negative definite in the space $L_{b_0}^2(S)$,
(The Green function $G$ is negative).
 Since, the function $b_0$ is bounded in $S$, the inner product \eqref{eip}
 is equivalent to the standard product in the space $L^2(S)$.

In \cite{Bensid1}, we have proved that the operator $D_2J(0,r_0,0)$ 
is invertible
(note that $b_0=r_0$ when $h_0=0$ on $\partial \Omega$). 
In this case, we use the explicit eigenvalue of the operator $K$ to 
conclude the nondegeneracy of the operator $D_2J(0,r_0,0)$. 
For more details, see \cite{Bensid1}.

 Finally, using the fact that $D_2J(0,r_0,0)$ is invertible in $L_{r_0}^2(S)$, 
we can be sure that $D_2J(h_0,b_0,0)$ is invertible in $L_{b_0}^2(S)$
 with the previous inner product. The preceding argument shows that 
$D_2J(h_0,b_0,0)$ is invertible in $L^2(S)$. 
Hence, using the implicit function theorem, we have the following result.
\end{proof}

\begin{theorem} \label{thm4.1}
Under  assumptions {\rm (H1)--(H3)}, there exist a neighborhood $V$ 
of $(h_0,0)$ in $H\times C^2(S)$ and a continuous mapping $B:V\to C(S)$ such that
\begin{itemize}
\item[(i)] $B(h_0,0)=b_0$,
\item[(ii)] $J(h,B(h,\beta),\beta)=0$.
\end{itemize}
\end{theorem}

We recall that $h_0$ is a given function satisfying (H3) and 
$b_0(\theta):=r_0+b(\theta)$, where $r_0\in (0,1)$,
 $b\in C^{1,\alpha}(S), \alpha=1-\frac{n}{p}$ and $\theta\in S$.

\begin{proof}[Proof of Theorem \ref{thm4.1}]
Because the operator $J$ is invertible in the neighborhood of $(h_0,b_0,0)$ 
and $J(h_0,b_0,0)(\theta)=0$ for $\theta\in S$, then the implicit function 
theorem implies the existence of function $B$ depending on $h$ and 
$\beta$ such that $B(h,\beta)$ satisfies $J(h,B(h,\beta),\beta)=0$.
\end{proof}

\section{Regularity of the free boundary}

\begin{theorem} \label{thm5.1}
If $\|h\|_{\infty}$ and $\|\beta\|_{\infty}$ are small enough, then 
the free boundary $\{x\in \Omega/u(x)=\mu\}$ is an analytic hypersurface.
\end{theorem}

\begin{proof}
First, let $\overline{u}$ be a solution of \eqref{e8} and let
\begin{align*}
 \Gamma:&=\{x\in \Omega/\overline{u}(x)=\mu\}\\
&=\{(r,\theta)\in (0,R)\times S, \overline{u}(r,\theta)=\mu\}\\
&=\{(f(\theta),\theta), \text{ for } \theta\in S\}.
\end{align*}
 when $\beta$ is sufficiently close to $0$ in $C^2(S)$ and $f$ is close to
 $b_0$ in $C(S)$, then the  solution $\overline{u}$ is close to $u$ 
in $C^{1,\alpha}(\Omega), \alpha\in (0,1)$. 
Since 
$$
\frac{\partial u}{\partial r}(b_0(\theta),\theta)\neq 0,
$$
then 
$$
\frac{\partial \overline{u}}{\partial r}(f(\theta),\theta)\neq 0\quad 
\text{for}\quad \theta\in S.
$$
The implicit function theorem gives that  $f\in C^{1,\alpha}(S)$.
 Hence, the free boundary $\Gamma$ is an hypersurface of class $C^{1,\alpha}$.
Now, we conclude that the free boundary $\Gamma$ is analytic by the application 
of Hodograph transformation. This method was using by the author in \cite{Bensid2}.
\end{proof}

\subsection*{Final remarks}
(1) So far, the problem \eqref{eP} has been  studied for a few class of domain, 
in particular the ball for uses the notion of symmetry. Hence, the main 
results of this paper remain true when $\Omega$ is a ring shaped domain.

(2) The advantage of the perturbation procedure described in this paper is 
the the explicit formula of behavior of the free boundary with respect to $h$. 
For example, in dimension 2, taking 
$\Omega:=\{(r,\theta)/ r<2+\sin(2\theta), \theta\in S\}$ and 
$h(x)=\cos(x)$, we can give an explicit formulation of the solution of 
problem \eqref{eP} and the shape of free boundary.

(3) When $n=1$, the problem \eqref{eP} becomes the following second order 
differential equation. For example, we can see easily that under a suitable 
conditions,  the problem
\begin{equation} \label{eP2}
\begin{gathered}
-u'' =  \lambda H(u-\mu ) \quad \text{for } |x|\leq a, \\
u(\pm a)=0,\quad (a\in \mathbb{R}^+).
\end{gathered}
\end{equation}
 admits multiple solutions. See \cite{Stuart}. In the other part, the
 problem
\begin{equation} \label{eP3}
\begin{gathered}
-u'' =  \lambda H(\mu-u ) \quad \text{for } |x|\leq a,\\
u(\pm a)=0,\quad (a \in \mathbb{R}^+).
\end{gathered}
\end{equation}
has a unique positive solution when the set $\{x\in \Omega, u(x)=\mu\}$
is a given segment included in $|x|\leq a$. In fact, the form of the
discontinuity in the second member has a surprising effects on the set
of solutions and their free boundaries.
In higher dimensions, the result stays true also,we can see \cite{Bensid}
for more details.

(4) The regularity of the free boundary is preserved after perturbations. 
One of the important questions is when will the free boundary develop singularities.

(5) It is also interesting to studied the bifurcation phenomenon. 
In the case when $\Omega$ is the unit ball, the study is in \cite{Bensid2}.

 

\section{Appendix}

 In this section, we  give the proof that the condition in assumption  
(H6) in Remark \ref{rmk1} becomes
\begin{gather*}
\frac{\lambda}{\mu}>\frac{4e}{R^2},\quad \text{for}\quad n=2, \\
\frac{\lambda}{\mu}>M_n(R):=\frac{1}{R^2}\frac{n(n-2)}{(\frac{2}{n})
 ^{\frac{2}{n-2}}-(\frac{2}{n})^{\frac{n}{n-2}}},\quad \text{for } n\geq 3,
\end{gather*}
as in Theorem \ref{thm2.1} when  $\Omega=B(0,R)$ and $f(u)=\lambda$.
 Now, define a function $u$ that satisfies
\begin{equation} \label{eA}
\begin{gathered}
-\Delta  u =  \lambda H(u-\mu ) \quad \text{in } \Omega,\\
u =0 \quad \text{on }\partial \Omega.
\end{gathered}
\end{equation}
Since $H(u-\mu)$ is monotone and not decreasing, the result of
\cite{Gidas} shows that all positive solutions are radial.
Hence, by the maximum principle, a positive radial solution $u$ take
on the value $\mu$ at only one value say $r_0$.
 Then we obtain all radial solutions of \eqref{e2} by finding $\lambda, u$
and $r_0$ so that the  two problems
\begin{equation} \label{eA1}
\begin{gathered}
-r^{1-n}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u}{\partial r})
=  \lambda  \quad \text{for } 0<r<r_0,\\
u'(0)=0, \quad u(r_0)=\mu
\end{gathered}
\end{equation}
and
\begin{equation} \label{eA2}
\begin{gathered}
 -r^{1-n}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u}{\partial r}) =  0
 \quad \text{for } r_0<r<R,\\
u(R)=0, \quad u(r_0)=\mu
\end{gathered}
\end{equation}
are satisfied.
When $n=2$ and from the problems \eqref{eA1}, \eqref{eA2}, we have
$$
u(r)=\begin{cases}
\frac{\lambda}{4}(r_0^2-r^2)+\mu &  0<r\leq r_0,\\
\mu \frac{\ln(R/r)}{\ln(R/r_0)} &  r_0\leq r<R.
\end{cases}
$$
Since $u\in C^{1,\alpha}(\overline{\Omega})$, we have
$$
\frac{\partial u}{\partial r}(r_0^-)=\frac{\partial u}{\partial r}(r_0^+).
$$
We obtain
$$
\frac{\lambda}{2}r_0=\frac{\mu}{r_0(\ln(R/r_0))}
$$
which implies
\begin{equation} \label{eE1}
\frac{2\mu}{\lambda}=r_0^2\ln(R/r_0):=g(r_0).
\end{equation}
Now, by considering the function $g(\rho)$ in $(0,R)$, we have that $f$
has  a maximum value $\frac{R^2}{2e}$ at $\rho=Re^{-\frac{1}{2}}$.
Hence, the equation \eqref{eE1} has  two roots when
$\frac{\lambda}{\mu}>\frac{4e}{R^2}$.

Similarly, we treat the case $n\geq3$. The solution $u$ is given by
$$
u(r)=\begin{cases} 
 \frac{\lambda}{2n}(r_0^2-r^2)+\mu &  0<r\leq r_0,\\[4pt] 
\frac{\mu r^{2-n}}{r_0^{2-n}-R^{2-n}}-\frac{\mu R^{2-n}}{r_0^{2-n}-R^{2-n}} 
&  r_0\leq r<R.
\end{cases}
$$
The transmission conditions imply that
$$
\frac{\partial u}{\partial r}(r_0^+)=\frac{\partial u}{\partial r}(r_0^-),
$$
so
$$
\frac{(2-n)\mu r_0^{1-n}}{r_0^{2-n}-R^{2-n}}=-\frac{\lambda r_0}{n}.
$$
Hence,
\[
\frac{\lambda}{\mu}=\frac{n(n-2)}{r_0^2-r_0^nR^{2-n}}:=g(r_0
\]
It follows that the function $g$ has a minimum value 
\begin{equation} \label{eE2}
M_n(R)=\frac{1}{R^2}\frac{n(n-2)}{(\frac{2}{n})^{\frac{2}{n-2}}
-(\frac{2}{n})^{\frac{n}{n-2}}}
\end{equation}
reached at the point $(\frac{2}{nR^{2-n}})^{\frac{1}{n-2}}$ which implies
that the equation \eqref{eE2} has two roots when
$\frac{\lambda}{\mu}>M_n(R)$.

\subsection*{Acknowledgments}
The author would like to thank the professor B. Abdellaoui  for the discussion
and the suggestions  about the problem in this paper.
I want to thank also the anonymous referee for valuable
suggestions and comments. 


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\end{document}