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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 34, pp. 1--4.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/34\hfil Uniqueness of positive solutions]
{Uniqueness of positive solutions for an elliptic system arising in
  a diffusive predator-prey model}

\author[X. Wei,  W. Zhou\hfil EJDE-2013/34\hfilneg]
{Xiaodan Wei, Wenshu Zhou}  % in alphabetical order

\address{Xiaodan Wei \newline
School of Computer Science, Dalian Nationalities
University, Dalian 116600, China}
\email{weixiaodancat@126.com}

\address{Wenshu Zhou \newline
 Department of Mathematics, Dalian Nationalities
University, Dalian 116600, China}
\email{pdezhou@126.com}

\thanks{Submitted September 3, 2012. Published January 30, 2013.}
\subjclass[2000]{35J57, 92D25}
\keywords{Predator-prey model; strong-predator;  positive solution;
 uniqueness}

\begin{abstract}
 In this note, we study the uniqueness of positive solutions
 for an elliptic system which arises in a  diffusive predator-prey
 model in the strong-predator case.
 The  main result extends an earlier results by the same authors.
\end{abstract}

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

\section{Introduction}

 In this note, we study the uniqueness of positive solutions
 for the 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,\\
\frac{\partial u}{\partial \nu}=\frac{\partial v}{\partial \nu}=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 $\partial\Omega$,
$\lambda, b,\mu$  and $\xi$ are positive constants, which arises in
the  diffusive predator-prey model in the strong-predator case
($\beta\to +\infty$):
\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,\\
\frac{\partial u}{\partial \nu}=\frac{\partial v}{\partial \nu}=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 $\overline\Omega \setminus \overline
\Omega_0$ for some smooth domain $\Omega_0$   with $\overline\Omega_0
\subset \Omega$. We refer the reader to
 \cite{DH, DW, WP} for some related studies on \eqref{2}.

  It is easy to see that  $(u, v)=(\frac{\xi\lambda}{b}, \frac{\lambda}{b})$ 
is a  positive solution for problem \eqref{3}. 
In \cite[Remark 3.2]{DW}, the
authors pointed out that when  $N = 1$, the positive solution of
 \eqref{3} is unique for any $\mu>0$ by a simple variation of
the arguments in \cite{LP}.  When $N \geqslant 2$, it was proved in
\cite{ZW}  that the uniqueness holds for all sufficiently large
$\mu$. In the present paper, we prove the uniqueness for $\mu
\geqslant 2\lambda$. We point out that a key step of the proof is to
establish a new a priori estimate on $u$ for the solution $(u, v)$
of problem \eqref{3}, which is stated as follows.


\begin{theorem} \label{thm1.1}
Let $(u, v)$ be a positive solution of  \eqref{3}. If $\mu >\lambda$, then
\begin{equation}\label{199}
u\leqslant\frac{\xi\mu\lambda}{b(\mu-\lambda)}\quad\text{on }\overline\Omega.
\end{equation}
\end{theorem}

Based on this  estimate  and the identity in \cite[(2.13)]{ZW}, we have

\begin{theorem} \label{thm1.2}
Let $N \geqslant 2$. If $\mu \geqslant 2\lambda$, then there is a
unique positive solution for \eqref{3}.
\end{theorem}


\section{Proofs of main theorems}

To prove Theorem \ref{thm1.1}, we need the following maximal principle due to
Lou and Ni \cite[Lemma 2.1]{LN2}.

\begin{lemma}  \label{lem2.1}
Suppose that $g \in C^1(\overline\Omega\times\mathbb{R}^1)$, 
$b_j \in C(\overline\Omega)$ for  $j=1,2,\dots,N$.

(i) If $w \in C^2(\Omega)\cap C^1(\overline\Omega)$ satisfies
\begin{gather*}
\Delta w(x)+\sum_{j=1}^{N}b_j(x)w_{x_j}+g(x, w(x)) \geqslant 0
\quad\text{in }\Omega,\\
\partial_\nu w \leqslant 0\quad\text{on }\partial\Omega,
\end{gather*}
and $w(x_0)=\max_{\overline\Omega}w$, then $g(x_0,w(x_0))\geqslant 0$. 

(ii) If $w \in C^2(\Omega)\cap C^1(\overline\Omega)$ satisfies
\begin{gather*}
\Delta w(x)+\sum_{j=1}^{N}b_j(x)w_{x_j}+g(x, w(x)) \leqslant 0
\quad\text{in }\Omega,\\
\partial_\nu w \geqslant 0\quad\text{on }\partial\Omega,
\end{gather*}
and $w(x_0) = \min_{\overline\Omega}w$, then $g(x_0,w(x_0))\leqslant 0$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm1.1}] 
 Denote us denote
   \begin{equation}\label{0}
(U, V)=\Big(\frac{b}{\xi}u, b v\Big).
\end{equation}
 Then $(U, V)$ satisfies
 \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,\\
 \frac{\partial U}{\partial \nu}=\frac{\partial V}{\partial \nu}=0
\quad \text{on }\partial\Omega.
 \end{gathered}
\end{equation}
Clearly,  estimate \eqref{199} is equivalent to 
\begin{equation}\label{109}
U\leqslant\frac{\mu\lambda}{\mu-\lambda}\quad\text{on }\overline\Omega.
\end{equation}

 Let $\varphi=V/U$. Then $V=\varphi U$, and differentiating it twice yields
\begin{equation*}\label{19}
\Delta V=\varphi\Delta U+2\nabla U\cdot\nabla\varphi+U\Delta \varphi
\quad\text{in }\Omega;
\end{equation*}
therefore,
\begin{equation}\label{1911}
-\Delta \varphi-\frac{2}{U}\nabla U\cdot\nabla\varphi=-\frac{1}{U}\Delta V
+\frac{\varphi}{U}\Delta U\quad\text{in }\Omega.
\end{equation}
From  \eqref{300}, we obtain
\begin{equation}\label{5}
 \begin{gathered}
 -\Delta U=U(\lambda-\varphi U)\quad\text{in }\Omega,\\
  \frac{\partial U}{\partial \nu}=0
 \quad\text{on }\partial\Omega,\\
 -\Delta V=\mu\varphi U(1-\varphi)\quad\text{in }\Omega,\\
\frac{\partial V}{\partial \nu}=0
 \quad\text{on }\partial\Omega.
 \end{gathered}
\end{equation}
Substituting them into \eqref{1911}, we have
\begin{equation}\label{119}
-\Delta \varphi-\frac{2}{U}\nabla
U\cdot\nabla\varphi=\varphi(\mu-\lambda-\mu\varphi+\varphi
U)\quad\text{in }\Omega,
\end{equation}
and hence
\begin{equation*}%\label{44}
-\Delta \varphi-\frac{2}{U}\nabla
U\cdot\nabla\varphi\geqslant\varphi(\mu-\lambda-\mu\varphi)
\quad\text{in }\Omega.
\end{equation*}
Using Lemma \ref{lem2.1} (ii) and noticing that 
$ \frac{\partial \varphi}{\partial \nu}=0$ on $\partial\Omega$, we obtain
\begin{equation*}
 \varphi \geqslant\frac{\mu-\lambda}{\mu}\quad\text{on }\overline\Omega.
\end{equation*}
From the estimate and the first equation of \eqref{5} it follows
that
\begin{gather*}
 -\Delta U  \leqslant U\Big(\lambda- \frac{\mu-\lambda}{\mu} U\Big)
\quad\text{in }\Omega,\\
\frac{\partial U}{\partial \nu}=0\quad\text{on }\partial\Omega.
 \end{gather*}
By Lemma \ref{lem2.1} (i), we obtain \eqref{109}.  The proof is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.2}]
 It suffices to show that
 $(u, v)=(\frac{\xi\lambda}{b}, \frac{\lambda}{b})$
 for any positive solution $(u, v)$   of\eqref{3}. 
Let $(U,V)$ be  the same as that in \eqref{0}. Then $(U, V)$ satisfies 
\eqref{300}.

 On the other hand, one can show the following identity 
(i.e. \cite[(2.13)]{ZW}):
 \begin{equation}\label{55}
\int_{\Omega}(U-2\lambda)\frac{|\nabla
U|^2}{U^3}dx-\frac{\lambda}{\mu}\int_{\Omega}\frac{|\nabla
V|^2}{V^2}dx=\int_{\Omega}\frac{(\lambda-V)^2}{U}dx.
\end{equation}
Indeed, multiplying the equations of $U$ and $V$ by
$\frac{\lambda-U}{U^2}$ and
$\frac{1}{\mu}\frac{\lambda-V}{V}$, respectively, we obtain
\begin{equation*}
-2\lambda\int_{\Omega}\frac{|\nabla
U|^2}{U^3}dx+\int_{\Omega}\frac{|\nabla
U|^2}{U^2}dx=\int_{\Omega}\frac{(\lambda-U)(\lambda-V)}{U}dx,
\end{equation*}
and
\begin{align*} 
-\frac{\lambda}{\mu}\int_{\Omega}\frac{|\nabla V|^2}{V^2}dx
&=\int_{\Omega}\frac{(U-V)(\lambda-V)}{U}dx\\
&=\int_{\Omega}\frac{(U-\lambda)(\lambda-V)}{U}dx
+\int_{\Omega}\frac{(\lambda-V)^2}{U}dx.
\end{align*}
Adding the two identities yields \eqref{55}.

Noticing $\mu \geqslant 2\lambda$, we deduce from \eqref{109} that
$U\leqslant 2\lambda$, so the first integral of left hand side of
\eqref{55} is non-positive, hence
\begin{equation*}
 \int_{\Omega}\frac{(\lambda-V)^2}{U}dx \leqslant 0,
\end{equation*}
 which  implies that  $V=\lambda$, so
 $U=\lambda$. Recalling \eqref{0}, we complete the proof.
\end{proof}

\subsection*{Acknowledgments}
This research was Supported by   National Natural Science Foundation of China
 (grants 10901030, 11071100),
and by Dalian Nationalities University (grants DC110109, DC120101064).


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\end{thebibliography}

\end{document}