\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \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 $$\label{3} \begin{gathered} -\Delta u= \lambda u-buv\quad \text{in }\Omega,\\ -\Delta v=\mu v\Big(1-\xi \frac{v}{u}\Big)\quad \text{in }\Omega,\\ \frac{\partial u}{\partial \nu}=\frac{\partial v}{\partial \nu}=0 \quad \text{on }\partial\Omega, \end{gathered}$$ 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 ($\beta\to +\infty$): $$\label{2} \begin{gathered} -\Delta u=\lambda u-a(x)u^2- \beta u v \quad \text{in }\Omega,\\ -\Delta v=\mu v\Big(1-\frac{v}{u}\Big) \quad \text{in }\Omega,\\ \frac{\partial u}{\partial \nu}=\frac{\partial v}{\partial \nu}=0 \quad\text{on }\partial\Omega, \end{gathered}$$ 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 $$\label{199} u\leqslant\frac{\xi\mu\lambda}{b(\mu-\lambda)}\quad\text{on }\overline\Omega.$$ \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 $$\label{0} (U, V)=\Big(\frac{b}{\xi}u, b v\Big).$$ Then $(U, V)$ satisfies $$\label{300} \begin{gathered} -\Delta U= U(\lambda-V)\quad \text{in }\Omega,\\ -\Delta V=\mu V\Big(1-\frac{V}{U}\Big)\quad \text{in }\Omega,\\ \frac{\partial U}{\partial \nu}=\frac{\partial V}{\partial \nu}=0 \quad \text{on }\partial\Omega. \end{gathered}$$ Clearly, estimate \eqref{199} is equivalent to $$\label{109} U\leqslant\frac{\mu\lambda}{\mu-\lambda}\quad\text{on }\overline\Omega.$$ 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, $$\label{1911} -\Delta \varphi-\frac{2}{U}\nabla U\cdot\nabla\varphi=-\frac{1}{U}\Delta V +\frac{\varphi}{U}\Delta U\quad\text{in }\Omega.$$ From \eqref{300}, we obtain $$\label{5} \begin{gathered} -\Delta U=U(\lambda-\varphi U)\quad\text{in }\Omega,\\ \frac{\partial U}{\partial \nu}=0 \quad\text{on }\partial\Omega,\\ -\Delta V=\mu\varphi U(1-\varphi)\quad\text{in }\Omega,\\ \frac{\partial V}{\partial \nu}=0 \quad\text{on }\partial\Omega. \end{gathered}$$ Substituting them into \eqref{1911}, we have $$\label{119} -\Delta \varphi-\frac{2}{U}\nabla U\cdot\nabla\varphi=\varphi(\mu-\lambda-\mu\varphi+\varphi U)\quad\text{in }\Omega,$$ 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}): $$\label{55} \int_{\Omega}(U-2\lambda)\frac{|\nabla U|^2}{U^3}dx-\frac{\lambda}{\mu}\int_{\Omega}\frac{|\nabla V|^2}{V^2}dx=\int_{\Omega}\frac{(\lambda-V)^2}{U}dx.$$ 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). \begin{thebibliography}{00} \bibitem{DH} Y. H. Du, S. B. Hsu; \emph{A diffusive predator-prey model in heterogeneous environment}, J. Differential Equations 203 (2004), 331-364. \bibitem {DW} Y. H. Du, M. X. Wang; \emph{Asymptotic behaviour of positive steady states to a predator-prey model}, Proc. Roy. Soc. Edin. 136A(2006), 759-778. \bibitem {LP} J. Lopez-Gomez, R. M. 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