\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 126, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/126\hfil Uniqueness of positive solutions]
{Uniqueness of positive solutions for an elliptic system}

\author[W. Zhou, X. Wei\hfil EJDE-2011/126\hfilneg]
{Wenshu Zhou, Xiaodan Wei} 

\address{Wenshu Zhou \newline
Department of Mathematics,
Dalian Nationalities University, Dalian 116600, China}
\email{pdezhou@126.com, wolfzws@163.com}

\address{Xiaodan Wei \newline
School of Computer Science, Dalian Nationalities University, 
Dalian 116600, China}
\email{weixiaodancat@126.com}


\thanks{Submitted April 19, 2011. Published September 29, 2011.}
\thanks{Supported by grants 10901030 and 11071100 from the
 National Natural Science Foundation \hfill\break\indent
 of China, 2009A152
 from the Department of Education of Liaoning Province.}
\subjclass[2000]{35J57, 92D25}
\keywords{Predator-prey model; strong-predator;
  positive solution; uniqueness}

\begin{abstract}
 We prove the uniqueness of positive solutions  for an elliptic
 system that appears in the study of solutions for a degenerate
 predator-prey model in the strong-predator case.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

 This article is devoted to showing the uniqueness of positive
solutions for the  elliptic system
\begin{equation}\label{3}
\begin{gathered}
 -\Delta  u= \lambda u-buv\quad \text{in }\Omega,\\
 -\Delta v=\mu v\big(1-\xi \frac{v}{u}\big)\quad\text{in }\Omega,\\
 \partial_\nu u=\partial_\nu v=0\quad \text{on }\partial\Omega,
 \end{gathered}
\end{equation}
where $\Omega \subset \mathbb{R}^N$ is a smooth bounded domain,
$\nu$ is the outward unit normal vector on $\partial\Omega$,
$\partial_\nu=\frac{\partial}{\partial \nu}$, $\lambda, b,\mu$  and
$\xi$ are positive constants.

Problem \eqref{3} appears in the study of positive
solutions of the degenerate preda\-tor-prey model in the
strong-predator case
\begin{equation}\label{2}
\begin{gathered}
-\Delta u=\lambda u-a(x)u^2- \beta u v\quad  \text{in }\Omega,\\
-\Delta v=\mu v\big(1-\frac{v}{u}\big)\quad \text{in }\Omega,\\
\partial_\nu u=\partial_\nu v=0\quad \text{on }\partial\Omega,
\end{gathered}
\end{equation}
where  $\beta$ is a positive constant, and $a(x)$ is a
continuous function satisfying   $a(x)=0$ on
$\overline \Omega_0$ and $a(x)>0$ in
$\overline\Omega \setminus \overline \Omega_0$, where $\Omega_0$
is a  smooth domain with $\overline\Omega_0 \subset \Omega$.
Recently, problem \eqref{2} has been studied in
 \cite{DH, DW}. Under the condition
$\mu>\lambda\geq\lambda_1$, where
$\lambda_1$ denotes the first eigenvalue of the Laplace equation on
$\Omega_0$ with homogenous Dirichlet boundary condition,
Du and Wang \cite{DW} described  spatial patterns of positive
solutions  of problem \eqref{2} by studying asymptotic behavior
of positive solutions as  $\beta \to 0^+$ (weak-predator),
$\beta \to +\infty$ (strong-predator) and
$\mu \to +\infty$ (small-predator diffusion), respectively.
For related work on problem \eqref{2}, please refer to \cite{WP}.

  Clearly, problem \eqref{3} has a  positive
solution $(u, v)=(\frac{\xi\lambda}{b}, \frac{\lambda}{b})$. In
\cite[Remark 3.2]{DW}, the authors pointed out that when the spatial
dimension $N = 1$, the positive solution of problem \eqref{3} is
unique for any $\mu>0$ by a simple variation of the arguments in
\cite{LP}.  In the present paper, we prove the uniqueness for all
sufficiently large $\mu$ in the high dimensional case, which can be
stated as follows

\begin{theorem} \label{thm1.1}
Let $N \geq 2$. Then there exists a positive constant $\mu_0$
depending only on $\lambda$ and $\Omega$ such that problem \eqref{3}
admits a unique positive solution  for any  $\mu \geq \mu_0$.
\end{theorem}

\begin{remark} \label{rmk1.1} \rm
The proof to Theorem \ref{thm1.1} is based on the fact that
$ (\hat{u},\hat{v})$ is a positive solution of problem \eqref{3}
if and only if $ (\frac{b}{\xi}\hat{u}, b\hat{v})$ is a positive
solution of
 \begin{equation}\label{300}
\begin{gathered}
-\Delta u= u(\lambda-v)\quad\text{in }\Omega,\\
-\Delta v=\mu v\big(1-\frac{v}{u}\big)\quad\text{in }\Omega,\\
\partial_\nu u=\partial_\nu v=0\quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
\end{remark}

\begin{remark} \label{rmk1.2} \rm
As a result of Theorem \ref{thm1.1} and \cite[Remarks 3.1-3.2]{DW},
 one can prove that if $(u_\beta, v_\beta)$ is a solution of problem
\eqref{2}, then for any $\mu \geq \mu_0$, we have, as $\beta
\to +\infty$,
 \begin{gather*}
\Big(\frac{u_\beta}{\|u_\beta\|_\infty},
\frac{v_\beta}{\|v_\beta\|_\infty}\Big) \rightharpoonup
(1,1)\quad \text{in }[H^1(\Omega)]^2,\\
\Big(\frac{u_\beta}{\|u_\beta\|_\infty},
\frac{v_\beta}{\|v_\beta\|_\infty}\Big)\to (1, 1)\quad
\text{in } [L^p(\Omega)]^2, \forall p>1.
\end{gather*}
\end{remark}

\section{Proof of Theorem \ref{thm1.1}}

First recall several preliminary results.

\begin{lemma}[Harnack Inequality \cite{LNT}] \label{lem2.1}
 Let $w \in C^2(\Omega)\cap C^1(\overline\Omega)$ be
a positive solution to $\Delta w(x)+c(x)w(x)=0$, where $c \in
C(\overline\Omega)$, satisfying the homogeneous Neumann boundary
condition. Then there exists a positive constant $C$ which depends
only on $B$ where $\|c\|_\infty \leq B$ such that
$\max_{\overline\Omega}w \leq C
\min_{\overline\Omega}w$.
\end{lemma}

\begin{lemma}[Maximum Principle \cite{LN}] \label{lem2.2}
Suppose that $g \in C^1(\Omega\times\mathbb{R}^1)$. Then
\begin{itemize}
\item[(i)] if $w \in C^2(\Omega)\cap C^1(\overline\Omega)$ satisfies
$\Delta w(x)+g(x, w) \geq 0$ in $\Omega$,
$\partial_\nu w \leq 0$ on $\partial\Omega$,
and $w(x_0)=\max_{\overline\Omega}w$,
then $g(x_0,w(x_0))\geq 0$.


\item[(ii)] if $w \in C^2(\Omega)\cap C^1(\overline\Omega)$ satisfies
$\Delta w(x)+g(x, w) \leq 0$ in $\Omega$,
$\partial_\nu w \geq 0$ on $\partial\Omega$,
and $w(x_0) = \min_{\overline\Omega}w$, then
$g(x_0,w(x_0))\leq 0$.
\end{itemize}
\end{lemma}

The following lemma can be inferred from \cite[Lemma 3.7]{DH} (see
also \cite{WP}).

\begin{lemma} \label{lem2.3}
Let $\{u_n\} \subset H^1(\Omega)$ satisfy, in the weak sense,
$$
-\Delta u_n \leq A u_n,\quad u_n \geq 0,\quad
\partial_\nu u_n|_{\partial\Omega}=0,\quad\|u_n\|_\infty
\leq B,~~\forall n \geq 1,
 $$
where $A$ and $B$ are positive constants. Then  there
exists a subsequence of $\{u_n\}$, still denoted by $\{u_n\}$,
and a nonnegative function $u\in H^1(\Omega) \cap L^p(\Omega)$
for all $p> 1$, such that
$$
u_n \rightharpoonup u\quad\text{in } H^1(\Omega),
\quad u_n \to u\quad\text{in } L^p\Omega).
$$
If we further assume that $\|u_n\|_\infty \geq \delta>0$
for all  $n\geq  1$, then $u\neq 0$.
\end{lemma}

The following lemma gives the uniform bounds of the positive
solutions for problem \eqref{300}.

\begin{lemma} \label{lem2.4}
Let $(u_\mu, v_\mu)$ be a positive solution of
problem \eqref{300}. Then there exist a positive constant
$\mu_0=\mu_0(\lambda,\Omega)$ and two positive constants
$C_2, C_1$ independent of $\mu$ such that for all $\mu \geq
\mu_0$,
\begin{equation}\label{199}
\begin{aligned}
C_1\leq u_\mu,\quad v_\mu\leq C_2\quad \text{on }
\overline\Omega.
\end{aligned}
\end{equation}
Moreover, as $\mu \to  +\infty$,
\begin{equation}\label{200}
u_\mu\to  \lambda  \quad \text{in }C^1(\overline\Omega).
\end{equation}
\end{lemma}


\begin{proof}
 By Lemma \ref{lem2.2}  and the definition of
$v_\mu$,  it follows that
\begin{equation}\label{19}
\max_{\overline\Omega}u_{\mu}\geq
\max_{\overline\Omega}v_{\mu},\quad
\min_{\overline\Omega}v_{\mu}\geq
\min_{\overline\Omega}u_{\mu}.
\end{equation}
Hence, to prove \eqref{199}, it suffices to show that there exist
a positive constant $\mu_0=\mu_0(\lambda,\Omega)$ and two positive
constants $C_2, C_1$ independent of $\mu$ such that
\begin{equation}\label{44}
C_1\leq \min_{\overline\Omega}u_{\mu},\quad
\max_{\overline\Omega}u_{\mu}\leq C_2,\quad \forall
\mu \geq \mu_0.
\end{equation}

We first prove the second inequality of \eqref{44}.
Assume on the contrary that there exist  a sequence $\{\mu_n\}$
converging to $+\infty$ and the
corresponding solution $(u_{\mu_n}, v_{\mu_n})$, such that
$$
 \|u_{\mu_n}\|_\infty\to+\infty\quad\text{as } n\to+\infty.
$$
Denote
\begin{equation*}
\hat{u}_{\mu_n}=\frac{u_{\mu_n}}{\|u_{\mu_n}\|_\infty
+\|v_{\mu_n}\|_\infty},\quad
\hat{v}_{\mu_n}=\frac{v_{\mu_n}}{\|u_{\mu_n}\|_\infty
+\|v_{\mu_n}\|_\infty}.
\end{equation*}
Then $\hat{u}_{\mu_n}$ and $\hat{v}_{\mu_n}$ satisfy
$\|\hat{u}_{\mu_n}\|_\infty+\|\hat{v}_{\mu_n}\|_\infty=1$,
$\|\hat{u}_{\mu_n}\|_\infty \geq \frac12$ by \eqref{19}, and
\begin{equation}\label{5}
\begin{gathered}
-\Delta \hat{u}_{\mu_n}= \hat{u}_{\mu_n}(\lambda-v_{\mu_n})\quad
\text{in }\Omega,\\
-\Delta  \hat{v}_{\mu_n}=\mu_n\hat{v}_{\mu_n}
 \big(1-\frac{\hat{v}_{\mu_n}}{\hat{u}_{\mu_n}}\big)\quad
 \text{in }\Omega,\\
\partial_\nu \hat{u}_{\mu_n}=\partial_\nu \hat{v}_{\mu_n}=0\quad
\text{on }\partial\Omega.
\end{gathered}
\end{equation}
In particular, we have
\begin{equation}\label{6}
-\Delta \hat{u}_{\mu_n}\leq\lambda
 \hat{u}_{\mu_n}\quad\text{in }\Omega,\quad
\partial_\nu \hat{u}_{\mu_n}=0\quad \text{on }\partial\Omega.
\end{equation}
By Lemma \ref{lem2.3} and $\|\hat{v}_{\mu_n}\|_\infty \leq 1$,
there exist a subsequence of $\{(\hat{u}_{\mu_n}, \hat{v}_{\mu_n})\}$,
still denoted by itself, and a pair of non-negative functions
$ (\hat{u}, \hat{v}) \in
\big(H^1(\Omega)\cap L^p(\Omega)\big)\times L^\infty(\Omega)$
for all $p>1$, $\hat{u} \neq 0$, such that
\begin{equation*}
\hat{u}_{\mu_n} \rightharpoonup\hat{u}\quad\text{in }H^1(\Omega),\quad
\hat{u}_{\mu_n} \to\hat{u}\quad \text{in } L^p(\Omega),\quad
\hat{v}_{\mu_n} \rightharpoonup\hat{v}\quad \text{in }L^2(\Omega).
\end{equation*}
 Integrating the first equation of \eqref{5} over $\Omega$ yields
\begin{equation*}
 \lambda\int_{\Omega}\hat{u}_{\mu_n}dx
=\int_{\Omega}v_{\mu_n}\hat{u}_{\mu_n}dx
= (\|u_{\mu_n}\|_\infty+\|v_{\mu_n}\|_\infty)
 \int_\Omega\hat{u}_{\mu_n}\hat{v}_{\mu_n}dx.
\end{equation*}
From $\|u_{\mu_n}\|_\infty \to+\infty~~ (n\to +\infty)$,
we have
\begin{equation}\label{16}
\int_\Omega \hat{u}\hat{v}dx=\lim_{n\to+\infty}
\int_\Omega\hat{u}_{\mu_n}\hat{v}_{\mu_n}dx=\lim_{n\to+\infty}
\frac{\lambda}{\|u_{\mu_n}\|_\infty+\|v_{\mu_n}\|_\infty}
\int_\Omega\hat{u}_{\mu_n}dx=0.
\end{equation}
By the second equation in \eqref{5},
$\hat{v}_{\mu_n}$ is a positive solution of
\begin{equation}\label{7}
 -\Delta  w+\mu_n\frac{\hat{v}_{\mu_n}}{\hat{u}_{\mu_n}}w
=\mu_n w\quad \text{in }\Omega,\quad
\partial_\nu w=0\quad \text{on }\partial\Omega.
\end{equation}
From  the variational characterization of the first eigenvalue it
follows that
\begin{equation*}
\int_\Omega |\nabla \phi|^2dx +\mu_n\int_\Omega
\frac{\hat{v}_{\mu_n}}{\hat{u}_{\mu_n}}\phi^2dx \geq \mu_n
\int_{\Omega}\phi^2dx
\end{equation*}
for any $\phi\in \{w\in H^2(\Omega); \partial_\nu w=0
\text{ on }\partial\Omega \}$ (cf. \cite{BL}). Taking
$\phi=\hat{u}_{\mu_n}$ yields
\begin{equation*}
\frac{1}{\mu_n}\int_\Omega |\nabla \hat{u}_{\mu_n}|^2dx +\int_\Omega
 \hat{v}_{\mu_n} \hat{u}_{\mu_n} dx \geq
\int_{\Omega}\hat{u}_{\mu_n}^2dx.
\end{equation*}
Passing to the limit and using \eqref{16}, we obtain
$ \int_\Omega \hat{u}^2 dx=0$,
so $\hat{u}=0$, which is a contradiction.
Thus there exist a positive constant $\mu_0=\mu_0(\lambda,\Omega)$
and a positive constant $C_2$ independent of $\mu$ such that
\begin{equation}\label{444}
\begin{aligned}
 \max_{\overline\Omega}u_{\mu}\leq C_2, \quad \forall
\mu \geq \mu_0.
 \end{aligned}
\end{equation}

Next we prove the first inequality in \eqref{44}.
Suppose that this is not so. Then there exist $\{\mu_n\}$
converging to $+\infty$ and the
corresponding solution $(u_{\mu_n}, v_{\mu_n})$ such that
\begin{equation}\label{17}
\begin{aligned}
\lim_{n\to+\infty} \min_{\overline\Omega}u_{\mu_n}=0.
\end{aligned}
\end{equation}
Now rewrite the equation of $u_{\mu_n}$ as
\begin{equation*}
 \Delta  u_{\mu_n}+f(x)u_{\mu_n}=0  \quad\text{in }\Omega,\quad
\partial_\nu u_{\mu_n}=0\quad \text{on }\partial\Omega,
\end{equation*}
where $f(x)=\lambda-v_{\mu_n}$. By the first estimate of
\eqref{19} and \eqref{444}, we have, for all sufficiently large $n$,
$$
\|f\|_\infty \leq \lambda+\|v_{\mu_n}\|_\infty
\leq\lambda+C_2,
$$
 by Lemma \ref{lem2.1}, there exists a positive constant $C_3$
independent of $n$ such that for all sufficiently large $n$,
$$
\max_{\overline\Omega}u_{\mu_n} \leq
C_3\min_{\overline\Omega}u_{\mu_n}.
$$
Therefore, it follows from \eqref{17} and the first estimate
of \eqref{19} that
\begin{equation}\label{18}
\lim_{n\to+\infty} \max_{\overline\Omega} u_{\mu_n} =0,\quad
\lim_{n\to+\infty} \max_{\overline\Omega} v_{\mu_n} =0.
\end{equation}
Denote $\tilde{u}_{\mu_n}=u_{\mu_n}/\|u_{\mu_n}\|_\infty$. Then
$\tilde{u}_{\mu_n}$ satisfies $\|\tilde{u}_{\mu_n}\|_\infty=1$, and
\begin{equation*}
-\Delta  \tilde{u}_{\mu_n}=\tilde{u}_{\mu_n}(\lambda-v_{\mu_n})
\quad \text{in }\Omega,\quad
\partial_\nu \tilde{u}_{\mu_n}=0\quad \text{on }\partial\Omega.
\end{equation*}
By \eqref{19}, \eqref{444} and the definition of $\tilde{u}_{\mu_n}$,
both
$\{-\Delta \tilde{u}_{\mu_n}\}$ and $ \{\tilde{u}_{\mu_n}\}$ are
bounded sets in $L^\infty(\Omega)$. By the standard elliptic theory
(cf. \cite[Theorem 9.9]{GT}),  $\{\tilde{u}_{\mu_n}\}$ is bounded in
$W^{2,p}(\Omega)$ for any $p>1$. Therefore, there exist a
subsequence of $\{\tilde{u}_{\mu_n}\}$, still denoted by
itself, and a nonnegative function $\tilde{u} \in
C^1(\overline\Omega)$ with $\|\tilde{u}\|_\infty=1$, such that
\begin{equation*}
\tilde{u}_{\mu_n} \to \tilde{u} \quad\text{in }
C^1(\overline\Omega),
\end{equation*}
by \eqref{18} and the definition of $\tilde{u}_{\mu_n}$,
we derive that
\begin{equation*}
 -\Delta  \tilde{u} =\lambda\tilde{u}\quad \text{in }\Omega,\quad
\partial_\nu \tilde{u}=0\quad \text{on }\partial\Omega.
\end{equation*}
This implies $\tilde{u}=0$, which is a contradiction.
This proves \eqref{199}.

Next we show \eqref{200}.  By \eqref{199} and the equation
of $u_{\mu}$,  $\{-\Delta u_\mu\}_{\mu \geq\mu_0}$,
$\{u_{\mu}\}_{\mu \geq\mu_0}$ and $\{v_{\mu}\}_{\mu \geq\mu_0}$
are bounded sets in $L^\infty(\Omega)$.
By the standard elliptic theory,
there exist a sequence $\{\mu_n\}$  converging to $+\infty$,
the corresponding solution $(u_{\mu_n}, v_{\mu_n})$ of
problem \eqref{3} and a pair of functions
$(u, v) \in C^1(\overline\Omega) \times L^\infty(\Omega)$ with
$C_1 \leq u, v \leq C_2$, such that
\begin{equation*}
u_{\mu_n} \to u\quad \text{in }C^1(\overline\Omega),\quad
 v_{\mu_n} \rightharpoonup v\quad \text{in } L^2(\Omega).
\end{equation*}
Clearly, $(u, v)$ satisfies, in the weak sense,
$$
-\Delta u=u(\lambda-v)\quad \text{in }\Omega,\quad
\partial_\nu u=0\quad \text{on }\partial\Omega.
$$
 Multiplying the  equation of  $v_{\mu_n}$ by
$\phi \in C_0^\infty(\Omega)$ and integrating
over $\Omega$, we get
$$
-\frac{1}{\mu_n}\int_\Omega v_{\mu_n}\Delta \phi dx
=\int_\Omega v_{\mu_n}\Big(1-\frac{v_{\mu_n}}{u_{\mu_n}}\Big)\phi dx.
$$
Passing to the limit yields
$$
\int_\Omega v \big(1-\frac{v}{u}\big)\phi dx=0,
$$
which implies that $v \big(1-\frac{v}{u}\big)=0$. Since $v \neq 0$,
we must have $v=u$. By the regularity theory of elliptic equation,
$u \in C^2(\overline\Omega)$ and satisfies
$$
-\Delta u=u(\lambda-u)\quad\text{in }\Omega,\quad
\partial_\nu u=0\quad \text{on }\partial\Omega.
$$
Then $u=\lambda$. The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 Let $(u_\mu,
v_\mu)$ be a positive solution of problem \eqref{300}. By
\eqref{200},  there exists a constant
$\mu_0=\mu_0(\lambda, \Omega)$ such that for all $\mu \geq
\mu_0$,
\begin{equation}\label{8}
\begin{aligned} u_\mu \leq 2\lambda\quad \text{on }
\overline\Omega.
 \end{aligned}
\end{equation}
Multiplying the equations of $u_\mu$ and $v_\mu$ by
$\frac{\lambda-u_\mu}{u_\mu^2}$ and
$\frac{1}{\mu}\frac{\lambda-v_\mu}{v_\mu}$, respectively, we obtain
\begin{equation*}
 -2\lambda\int_{\Omega}\frac{|\nabla
u_\mu|^2}{u_\mu^3}dx+\int_{\Omega}\frac{|\nabla
u_\mu|^2}{u_\mu^2}dx=\int_{\Omega}
\frac{(\lambda-u_\mu)(\lambda-v_\mu)}{u_\mu}dx,
\end{equation*}
and
\begin{align*}
-\frac{\lambda}{\mu}\int_{\Omega}\frac{|\nabla
v_\mu|^2}{v_\mu^2}dx
&=\int_{\Omega}\frac{(u_\mu-v_\mu)(\lambda-v_\mu)}{u_\mu}dx\\
&=\int_{\Omega}\frac{(u_\mu-\lambda)(\lambda-v_\mu)}{u_\mu}dx
+\int_{\Omega}\frac{(\lambda-v_\mu)^2}{u_\mu}dx.
\end{align*}
Adding these two equalities yields
\begin{equation}\label{55}
-2\lambda\int_{\Omega}\frac{|\nabla
u_\mu|^2}{u_\mu^3}dx+\int_{\Omega}\frac{|\nabla
u_\mu|^2}{u_\mu^2}dx-\frac{\lambda}{\mu}\int_{\Omega}\frac{|\nabla
v_\mu|^2}{v_\mu^2}dx=\int_{\Omega}\frac{(\lambda-v_\mu)^2}{u_\mu}dx.
\end{equation}
Noting \eqref{8},  for all $\mu \geq \mu_0$, we obtain
\begin{equation*}
 -2\lambda\int_{\Omega}\frac{|\nabla
u_\mu|^2}{u_\mu^3}dx+\int_{\Omega}\frac{|\nabla
u_\mu|^2}{u_\mu^2}dx=\int_{\Omega}(u_\mu-2\lambda)\frac{|\nabla
u_\mu|^2}{u_\mu^3}dx\leq 0,
\end{equation*}
 which and \eqref{55} implies that
$ \int_{\Omega}\frac{(\lambda-v_\mu)^2}{u_\mu}dx \leq 0$, hence
$v_\mu=\lambda$ for all $\mu \geq \mu_0$, so
$u_\mu=\lambda$ for all $\mu \geq \mu_0$. Combining this and
Remark \ref{rmk1.1} completes the proof.
\end{proof}


\subsection*{Acknowledgments} The authors thank  the
anonymous referee for his/her insightful comments.

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\end{document}
