\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 $\mu \geq \mu_0$. 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