\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 121, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/121\hfil Multiple solutions for a elliptic system] {Multiple solutions for a elliptic system in exterior domain} \author[H. Gu, J. Yang, X. Yu\hfil EJDE-2008/121\hfilneg] {Huijuan Gu, Jianfu Yang, Xiaohui Yu} % in alphabetical order \address{Huijuan Gu \newline Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China} \email{ruobing411@yahoo.com.cn} \address{Jianfu Yang \newline Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China} \email{jfyang\_2000@yahoo.com} \address{Xiaohui Yu \newline China Institute for Advanced Study, Central University of Finance and Economics, Beijing 100081, China} \email{yuxiao\_211@163.com} \thanks{Submitted June 28, 2008. Published August 28, 2008.} \subjclass[2000]{35J50, 35B32} \keywords{Exterior domain; nonlinear elliptic system; existence result} \begin{abstract} In this paper, we study the existence of solutions for the nonlinear elliptic system \begin{gather*} -\Delta u+u=|u|^{p-1}u+\lambda v \quad \text{in } \Omega, \\ -\Delta v+v=|v|^{p-1}v+\lambda u \quad \text{in } \Omega, \\ u=v=0 \quad \text{on } \partial\Omega, \end{gather*} where $\Omega$ is a exterior domain in $\mathbb{R}^N$, $N\geq 3$. We show that the system possesses at least one nontrivial positive solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \section{Introduction} This article concerns the existence of solutions to the semilinear elliptic problem \begin{equation}\label{1.1} \begin{gathered} -\Delta u+u=|u|^{p-1}u+\lambda v \quad \text{in } \Omega, \\ -\Delta v+v=|v|^{p-1}v+\lambda u \quad \text{in } \Omega, \\ u=v=0 \quad \text{on } \partial\Omega, \end{gathered} \end{equation} where $\Omega\subset \mathbb{R}^N, N\geq 3$, is an exterior domain, $0<\lambda<1$ is a real parameter, $\partial \Omega\neq \emptyset$ and $10$ and a constant $\bar\rho=\bar\rho(\lambda)$ such that if $\lambda\in(0,\delta)$ and $$ \mathbb{R}^N\setminus \Omega\subset B_{\bar\rho}(x_0)=\{x\in \mathbb{R}^N: |x-x_0|\leq \bar\rho\}, $$ problem \eqref{1.1} has at least three pairs of nontrivial solutions. \end{theorem} Theorem \ref{thm1.1} will be proved by finding critical points of the corresponding functional of problem \eqref{1.1} \begin{equation}\label{1.6} \begin{aligned} I(u,v) &=\frac 12 \int_\Omega |\nabla u|^2+u^2\,dx +\frac 12 \int_\Omega |\nabla v|^2+v^2\,dx \\ &\quad -\frac 1{p+1}\int_\Omega |u|^{p+1}+|v|^{p+1}\,dx-\lambda\int_\Omega uv\,dx, \end{aligned} \end{equation} where $(u,v)\in E= H_0^1(\Omega)\times H_0^1(\Omega)$. In section 2, we show that ground state solutions are exponentially decaying at infinity and that problem \eqref{1.1} has no ground state solution. In final section, we prove Theorem \ref{thm1.1}. \section{Preliminaries} It was proved in \cite{ACR} that problem \eqref{1.5} has a ground state solution $(u_\lambda,v_\lambda)$ for $0<\lambda<1$, which is positive and radially symmetric. \begin{lemma}\label{thm2.1} There exist $\delta=\delta(\lambda)>0$ and $C>0$ such that \begin{equation}\label{2.1} |D^\alpha u_\lambda(x)|\leq Ce^{-\delta |x|},\quad |D^\alpha v_\lambda(x)|\leq Ce^{-\delta |x|} \quad \forall x\in\mathbb{R}^N \end{equation} for $|\alpha|\leq 2$. \end{lemma} \begin{proof} Let $w_\lambda=u_\lambda+v_\lambda$, then $w_\lambda$ satisfies \begin{equation}\label{2.2} -\Delta w_\lambda+w_\lambda=(u_\lambda^p+v_\lambda^p)+\lambda w_\lambda, \quad {\rm in } \mathbb{R}^N. \end{equation} Since $w=w(r)$ is radially symmetric, let $\phi(r)=r^{\frac{N-1}{2}}w_\lambda$, then $\phi$ satisfies \begin{equation}\label{2.3} \phi_{rr}=[q(r)+\frac {b}{r^2}]\phi \end{equation} with $q(r)=\frac{(1-\lambda)w_\lambda-(u_\lambda^p+v_\lambda^p)}{w_\lambda}$ and $b=\frac {(N-1)(N-3)}{4}$. Since $u_\lambda$ and $v_\lambda$ are radially symmetric, $u_\lambda(r),v_\lambda(r)\to 0$ as $|x|\to \infty$. There is $r_0>0$ such that $q(r)\geq \frac {1-\lambda} 2$ if $r\geq r_0$. Set $\psi=\phi^2$, then $\psi$ satisfies \begin{equation}\label{2.4} \frac 12 \psi_{rr}=\phi_r^2+(q(r)+\frac b {r^2})\psi, \end{equation} this implies that $\psi_{rr}\geq (1-\lambda)\psi$ for $r\geq r_0$. Let $z=e^{-\sqrt{1-\lambda}r}[\psi_r+\sqrt{1-\lambda}\psi]$, we have \begin{equation}\label{2.5} z_r=e^{-\sqrt{1-\lambda}r}[\psi_{rr}-(1-\lambda)\psi]\geq 0 \end{equation} for $r\geq r_0$. So $z$ is nondecreasing on $(r_0,+\infty)$. If there exists $r_1>r_0$ such that $z(r_1)>0$, then $z(r)\geq z(r_1)>0$ for $r\geq r_1$, that is \begin{equation}\label{2.6} \psi_{r}+\sqrt{1-\lambda}\psi\geq (z(r_1))e^{\sqrt{1-\lambda}r}, \end{equation} implying that $\psi_{r}+\sqrt{1-\lambda}\psi$ is not integrable, a contradiction to the fact that both $\psi$ and $\psi_r$ are integrable. Hence, there holds \begin{equation}\label{2.7} (e^{\sqrt{1-\lambda}r}\psi)_r =e^{\sqrt{1-\lambda}r}\psi_r+\sqrt{1-\lambda}e^{\sqrt{1-\lambda}r}\psi =e^{2\sqrt{1-\lambda}r}z\leq 0 \end{equation} for $r\geq r_0$. This implies \begin{equation}\label{2.8} \psi(r)\leq Ce^{-\sqrt{1-\lambda}r}; \end{equation} i.e., \begin{equation}\label{2.9} \phi(r)\leq Ce^{-\frac {\sqrt{1-\lambda}}2r}. \end{equation} By the definition of $\phi,w$ and the fact that $u_\lambda,v_\lambda>0$ we have \begin{equation}\label{2.10} u_\lambda,v_\lambda\leq C r^{-\frac{N-1}2}e^{-\frac {\sqrt{1-\lambda}}2r}. \end{equation} This proves (\ref{2.1}) with $\alpha=0$. Next we estimate the derivatives of $u_\lambda,v_\lambda$. Since \begin{equation}\label{2.11} (r^{N-1}(u_\lambda)_r)_r=-r^{N-1}[-u_\lambda+u_\lambda^p+\lambda v_\lambda], \end{equation} we have \begin{equation}\label{2.12} \begin{split} \int_s^R|(r^{N-1}(u_\lambda)_r)_r|\,dr &=\int_s^Rr^{N-1}[-u_\lambda+u_\lambda^p+\lambda v_\lambda]\,dr\\ &\leq C\int_s^\infty r^{\frac {N-1}2}e^{-\frac {\sqrt{1-\lambda}}{2}r}\,dr\\&\leq Ce^{-\frac {\sqrt{1-\lambda}}{4}s}, \end{split} \end{equation} this means that $r^{N-1}u_r$ has a limit as $r\to \infty$ and this limit can only be $0$ by (\ref{2.12}). Integrating (\ref{2.11}) on $(r,\infty)$ we get \begin{equation}\label{2.13} -r^{N-1}(u_\lambda)_r\leq Ce^{-\frac {\sqrt{1-\lambda}}{4}r}. \end{equation} Similarly, $-r^{N-1}(v_\lambda)_r\leq Ce^{-\frac {\sqrt{1-\lambda}}{4}r}$. Finally the exponential decay of $(u_\lambda)_{rr}$ and $(v_\lambda)_{rr}$ follows from equation \eqref{1.5}. This completes the proof. \end{proof} Now we consider the variational problem \begin{equation}\label{2.14} m_\lambda=\inf_{(u,v)\in \mathcal{N}} I(u,v), \end{equation} where \begin{equation}\label{2.15} \mathcal{N}=\{(u,v)\in E\setminus\{(0,0)\}:\langle I'(u,v),(u,v)\rangle=0 \} \end{equation} is the Nehari manifold related to $I$. Minimizers of $m_\lambda$ are ground state solutions of \eqref{1.1}. By a ground state solution of \eqref{1.1} we mean a nontrivial solution of \eqref{1.1} with the least energy among all nontrivial solutions of \eqref{1.1}. Correspondingly, for the limiting problem \eqref{1.5}, the associated functional \begin{equation}\label{2.16} \begin{aligned} I_\infty(u,v) &=\frac 12 \int_{\mathbb{R}^N} |\nabla u|^2+u^2\,dx +\frac 12 \int_{\mathbb{R}^N} |\nabla v|^2+v^2\,dx\\ &\quad -\frac 1{p+1}\int_{\mathbb{R}^N} |u|^{p+1}+|v|^{p+1}\,dx-\lambda\int_{\mathbb{R}^N} uv\,dx \end{aligned} \end{equation} is well defined in $H^1(\mathbb{R}^N)\times H^1(\mathbb{R}^N)$. We define \begin{equation}\label{2.17} m_\infty^\lambda=\inf_{(u,v)\in \mathcal{N}_\infty} I_\infty(u,v), \end{equation} where \begin{equation}\label{2.18} \mathcal{N}_\infty=\{(u,v)\in H^1(\mathbb{R}^N) \times H^1(\mathbb{R}^N)\setminus\{(0,0)\}:\langle I_\infty'(u,v),(u,v) \rangle=0 \} \end{equation} is the Nehari manifold for $I_\infty$. \begin{lemma}\label{lem2.2} Problem \eqref{1.1} has no ground state solution. \end{lemma} \begin{proof} First we show that $m_\lambda=m_\infty^\lambda$. The fact $H_0^1(\Omega)\subset H^1(\mathbb{R}^N)$ implies $m_\lambda\geq m_\infty^\lambda$. Let $\bar\xi$ be a cutoff function such that $0\leq \bar\xi(t)\leq 1$, $\bar\xi(t)=0$ for $t\leq 1$, $\bar\xi(t)=1$ for $t\geq 2$ and $|\bar\xi'(t)|\leq 2$. Set $\xi(x)=\bar\xi(\frac {|x|}{\rho})$, where $\rho$ is the smallest positive number such that $\mathbb{R}^N\setminus \Omega\subset B_\rho(0)$. Consider the sequence $\{(\phi_n,\psi_n)\}\subset E$ defined by \begin{equation}\label{2.19} (\phi_n,\psi_n)=(\xi(x)u_\lambda(x-y_n),\xi(x)v_\lambda(x-y_n)), \end{equation} where $\{y_n\}\subset \Omega$ is a sequence of points such that $|y_n|\to \infty$. We may verify that there exists a sequence $\{t_n\}\in \mathbb{R}^+$ such that $t_n(\xi(x)u_\lambda(x-y_n),\xi(x)v_\lambda(x-y_n))\in \mathcal{N}$. In fact, we may choose $t_n$ so that \begin{equation}\label{2.20} t_n^{p-1}=\frac{\int_\Omega |\nabla \phi_n|^2+\phi_n^2+|\nabla \psi_n|^2+\psi_n^2-\lambda \phi_n\psi_n\,dx}{\int_\Omega |\phi_n|^{p+1}+|\psi_n|^{p+1}\,dx}. \end{equation} Hence, for $2\leq q<\frac {2N}{N-2}$, \begin{gather*} \|\phi_n(x)-u_\lambda(x-y_n)\|_{L^{q}}^q\leq 2\int_{B_\rho}|u_\lambda(x-y_n)|^{q}\,dx \to 0, \\ \|\psi_n(x)-v_\lambda(x-y_n)\|_{L^{q}}^q\leq 2\int_{B_\rho}|v_\lambda(x-y_n)|^{q}\,dx \to 0, \\ \|\nabla \phi_n(x)-\nabla u_\lambda(x-y_n)\|_{L^{2}}^2\leq C\int_{B_\rho}|\nabla u_\lambda(x-y_n)|^{2}\,dx \to 0, \\ \|\nabla \psi_n(x)-\nabla v_\lambda(x-y_n)\|_{L^{2}}^2\leq C\int_{B_\rho}|\nabla v_\lambda(x-y_n)|^{2}\,dx \to 0 \end{gather*} and $$ \int_{\mathbb{R}^N}\phi(x)\psi(x)-u_\lambda(x-y_n)v_\lambda(x-y_n)\,dx\to 0 $$ as $n\to \infty$. It follows that $t_n\to 1$ as $n\to \infty$ since $(u_\lambda, v_\lambda)\in \mathcal{N}$. By the definition of $m_\lambda$, we have \begin{equation}\label{2.21} m_\lambda\leq I(t_n(\phi_n,\psi_n))=m_\infty^\lambda +o(1) \end{equation} as $n\to\infty$, which implies $m_\lambda=m_\infty^\lambda$. Suppose now that $m_\lambda$ is achieved by $(\bar u, \bar v)$. Extending $(\bar u, \bar v)$ to $\mathbb{R}^N$ by setting $(\bar u, \bar v)=(0,0)$ outside $\Omega$, we see that $(\bar u, \bar v)$ is a minimizer of $m_\infty$. Since we may assume that $\bar u\geq 0, \bar v\geq 0$, we obtain a contradiction by the strong maximum principle. This completes the proof. \end{proof} \section{Proof of Theorem \ref{thm1.1}} Problem \eqref{1.1} is setting in a unbounded, in general, $(PS)$ condition does not hold for $I$. In spirit of \cite[Lemma 3.1]{BC} and \cite[Lemma 4.1]{ACR}, we have the following global compact result. \begin{lemma}\label{thm3.1} Let $\{(u_n,v_n)\}\subset E$ be a sequence such that $I(u_n,v_n)\to c$ and $I'(u_n,v_n)\to 0$ as $n\to \infty$. Then there are a number $K\in \mathbb N$, $K$ sequences of points $\{y_n^j\}$ such that $|y_n^j|\to \infty$ as $n\to \infty$, $1\leq j\leq K$, $K+1$ sequences of functions $(u_n^j,v_n^j)\subset H^1(\mathbb{R}^N)\times H^1(\mathbb{R}^N),0\leq j\leq K$ such that up to a subsequence, \begin{itemize} \item[(i)] $u_n(x)=u_n^0(x)+\sum_{j=1}^{K}u_n^j(x-y_n^j),v_n(x)=v_n^0(x)+\sum_{j=1}^{K}v_n^j(x-y_n^j)$. \item[(ii)] $u_n^0(x)\to u^0(x),v_n^0(x)\to v^0(x)$ as $n\to \infty$ strongly in $H_0^1(\Omega)$. \item[(iii)] $u_n^j(x)\to u^j(x),v_n^j(x)\to v^j(x)$ as $n\to \infty$ strongly in $H^1(\mathbb{R}^N)$, where $1\leq j\leq K$. \item[(iv)] $(u^0,v^0)$ is a solution of \eqref{1.1} and $(u^j,v^j)$ is a solution of \eqref{1.5} for $1\leq j\leq K$. Moreover, when $n\to \infty$ \begin{gather}\label{3.1} \|u_n\|^2\to\|u^0\|^2+\sum_{j=1}^K\|u^j\|^2,\|v_n\|^2\to\|v^0\|^2 +\sum_{j=1}^K\|v^j\|^2, \\ \label{3.2} I(u_n,v_n)\to I(u_0,v_0)+\sum_{j=1}^K I_\infty(u^j,v^j). \end{gather} \end{itemize} \end{lemma} \begin{proof} We sketch the proof for reader's convenience. We may verify that $(u_n,v_n)$ is bounded. Suppose that $u_n\rightharpoonup u^0,v_n\rightharpoonup v^0 $ in $H_0^1(\Omega)$ and $u_n\to u^0,v_n\to v^0$ a.e in $\Omega$. Then, $(u^0,v^0)$ solves \eqref{1.1}. If $(u_n,v_n)\to (u^0,v^0)$, then we are done. Otherwise, let $$ z_n^1(x)= \begin{cases} u_n-u^0(x), &x\in\Omega, \\ 0, &x\in \mathbb{R}^N\setminus \Omega, \end{cases} \quad w_n^1(x)= \begin{cases} v_n-v^0(x), &x\in\Omega, \\ 0, &x\in \mathbb{R}^N\setminus \Omega, \end{cases} $$ then $$ \|u_n\|^2=\|u^0\|^2+\|z_n^1\|^2+o(1), \quad \|v_n\|^2=\|v^0\|^2+\|w_n^1\|^2+o(1). $$ By Brezis-Lieb's Lemma \cite{W}, we deduce $$ \|u_n\|^{p+1}_{L^{p+1}}=\|u^0\|^{p+1}_{L^{p+1}}+\|z_n^1\|^{p+1}_{L^{p+1}}+o(1), \quad \|v_n\|^{p+1}_{L^{p+1}}=\|v^0\|^{p+1}_{L^{p+1}}+\|w_n^1\|^{p+1}_{L^{p+1}}+o(1). $$ Thus, \begin{gather*} I(z_n^1,w_n^1)=I(u_n,v_n)-I(u^0,v^0)+o(1), \\ I'(z_n^1,w_n^1)=I'(u_n,v_n)-I'(u^0,v^0)+o(1)=o(1). \end{gather*} Suppose now that $(z_n^1,w_n^1)\not\to (0,0)$ in $H^1(\mathbb{R}^N)\times H^1(\mathbb{R}^N)$, we define $$ \delta_z=\limsup_{n\to \infty}\sup_{y\in \mathbb{R}^N} \int_{B_1(y)}|z_n^1|^{p+1}\,dx,\quad \delta_w=\limsup_{n\to \infty}\sup_{y\in \mathbb{R}^N} \int_{B_1(y)}|w_n^1|^{p+1}\,dx. $$ We may verify that $\delta_z+\delta_w>0$ since $(z_n^1,w_n^1)\not\to (0,0)$. We may suppose $\delta_z>0$, then there is a sequence $\{y_n^1\}\subset \mathbb{R}^N$ such that $\int_{B_1(y_n^1)}|z_n^1|^{p+1}\geq \frac {\delta_z}2$. Let us consider now the sequence $(z_n^1(x+y_n^1),w_n^1(x+y_n^1))$. We assume that $(z_n^1(x+y_n^1),w_n^1(x+y_n^1))\rightharpoonup (u^1,v^1)$, then $(u^1,v^1)$ is a nontrivial solution of \eqref{1.5}. By the fact that $z_n^1\rightharpoonup 0$ we see that $|y_n^1|\to \infty$. Set $$ z_n^2(x)=z_n^1(x)-u^1(x-y_n^1), w_n^2(x)=w_n^1(x)-v^1(x-y_n^1), $$ and repeat above procedure, it will stop at finite steps. The lemma follows. \end{proof} By \cite[Lemmas 7.8 and 7.9]{ACR}, there exist $0<\lambda_1\leq\lambda_2<1$ such that $m_\infty$ is an isolated critical value\ of $I_\infty$ for $\lambda\in (0,\lambda_1)\cup(\lambda_2,1)$. Denote $m_0=\inf\{\alpha>m_\infty^\lambda: \alpha\ {\rm\ is\ a\ critical \ value\ of}\ I_\infty\}$ and $\bar m=\min\{m_0,2m_\lambda\}$, then we have the following result. \begin{corollary}\label{coro3.2} The functional $I$ satisfies the $(PS)_c$ condition for $c\in (m_\lambda,\bar m)$. \end{corollary} \begin{proof} Let $\{(u_n,v_n)\}\subset E$ be such that $I(u_n,v_n)\to c$ and $I'(u_n,v_n)\to 0$ with $c\in (m_\lambda,\bar m)$. Since $\{(u_n,v_n)\}$ is bounded, we may assume that $u_n\rightharpoonup u$ and $v_n\rightharpoonup v$. By Lemma \ref{thm3.1}, $$ (u_n,v_n)-\sum_{j=1}^K (u^j(x-y_n^j),v^j(x-y_n^j))\to (u,v), $$ where $(u,v)$ is a solution of \eqref{1.1} and $(u^j,v^j)$ is a solution of \eqref{1.5}, $\{y_n^j\}(1\leq j\leq K)$ are $K$ sequences of points in $\mathbb{R}^N$. Moreover, $$ I(u_n,v_n)=I(u,v)+\sum_{j=1}^KI_\infty(u^j,v^j)+o(1). $$ To prove that $u_n\to u,v_n\to v$ in $H_0^1(\Omega)$, we need only to show $K=0$. Since $c<2m_\lambda$, we have $K<2$. We claim that $K=0$. Indeed, if $K=1$, we have either $(u,v)\neq (0,0)$ or $(u,v)= (0,0)$. If $(u,v)\neq (0,0)$, then $I(u_n,v_n)\geq 2m_\lambda+o(1)$, which contradicts to the fact that $c<2m_\lambda$; if $(u,v)= (0,0)$, then $I_\infty(u^1,v^1)=c$, which contradicts the definition of $\bar m$. The assertion follows. \end{proof} Now we introduce a function $\Phi_\rho:\mathbb{R}^N\to H^1(\mathbb{R}^N)\times H^1(\mathbb{R}^N)$ defined by \begin{equation}\label{4.1} \Phi_\rho(y)=t_\rho(\xi(\frac {|x|}{\rho})u_\lambda(x-y), \xi(\frac{|x|}{\rho})v_\lambda(x-y)), \end{equation} where $(u_\lambda,v_\lambda)$ is a ground state solution of equation \eqref{1.5}, $t_\rho$ is chosen such that $t_\rho(\xi(\frac {|x|}{\rho})u_\lambda(x-y), \xi(\frac {|x|}{\rho})v_\lambda(x-y))\in \mathcal{N}$. \begin{lemma}\label{lem4.1} \begin{itemize} \item[(i)] $\Phi_\rho(y)$ is continuous in $y$ for every $\rho>0$. \item[(ii)] $\Phi_\rho(y)\to (u_\lambda(x-y),v_\lambda(x-y))$ strongly in $H^1(\mathbb{R}^N)\times H^1(\mathbb{R}^N)$ uniformly in $y$ as $\rho\to 0$. \item[(iii)] $I(\Phi_\rho(y))\to m_\lambda$ as $|y|\to \infty$ uniformly for every $\rho$. \end{itemize} \end{lemma} \begin{proof} (i) is obvious since $\Phi_\rho(\cdot)$ is the composition of continuous functions. (iii) follows from the same argument of Lemma \ref{lem2.2}. It remains to prove (ii). We claim that \begin{gather*} \|\xi(\frac {|x|}{\rho})u_\lambda(x-y)\|_{L^{p+1}}\to \|u_\lambda(x)\|_{L^{p+1}},\quad \|\xi(\frac {|x|}{\rho})v_\lambda(x-y)\|_{L^{p+1}}\to \|v_\lambda(x)\|_{L^{p+1}},\\ \|\xi(\frac {|x|}{\rho})u_\lambda(x-y)\|\to \|u_\lambda(x)\|,\quad \|\xi(\frac{|x|}{\rho})v_\lambda(x-y)\|\to \|u_\lambda(x)\|,\\ \int_{\mathbb{R}^N}\xi(\frac {|x|}{\rho})u_\lambda(x-y)\xi(\frac {|x|}{\rho})v_\lambda(x-y)\,dx \to \int_{\mathbb{R}^N}u_\lambda(x-y)v_\lambda(x-y)\,dx. \end{gather*} Indeed, \begin{equation}\label{4.2} \begin{split} \|\xi(\frac {|x|}{\rho})u_\lambda(x-y)-u_\lambda(x-y)\|_{L^{p+1}}^{p+1} &\leq 2^{p+1}\int_{B_{2\rho}}|u_\lambda(x-y)|^{p+1}\,dx\\ &\leq 2^{p+1}|\max u_\lambda|^{p+1}\mathop{\rm meas} (B_{2\rho})\to 0. \end{split} \end{equation} Similarly, we have \begin{equation}\label{4.3} \|\xi(\frac {|x|}{\rho})v_\lambda(x-y)-v_\lambda(x-y)\|_{L^{p+1}}\to 0 \end{equation} and \begin{equation}\label{4.4} \begin{split} &\|\xi(\frac{|x|}{\rho})u_\lambda(x-y)-u_\lambda(x-y)\|^2\\ &=\int_{\mathbb{R}^N}|\frac 1{\rho}\nabla \xi(\frac {|x|}{\rho})u_\lambda(x-y)-\xi(\frac {|x|}{\rho})\nabla u_\lambda(x-y)-\nabla u_\lambda(x-y) |^2\,dx+k_2 \mathop{\rm meas}(B_{2\rho})\\ &\leq 2\int_{\rho\leq |x|\leq 2\rho}|\nabla \xi(\frac {|x|}{\rho})u_\lambda(x-y)|^2\,dx\\ &\quad +2\int_{\rho\leq |x|\leq 2\rho}|\xi(\frac {|x|}{\rho})\nabla u_\lambda(x-y)-\nabla u_\lambda(x-y)|^2\,dx +k_2 \mathop{\rm meas}(B_{2\rho})\\&\leq k_3 \rho^{N-2}+k_4 \rho^N\to 0 \end{split} \end{equation} as well as \begin{equation}\label{4.5} \begin{split} &|\int_{\mathbb{R}^N}\xi(\frac {|x|}{\rho})u_\lambda(x-y)\xi(\frac {|x|}{\rho})v_\lambda(x-y)-u_\lambda(x-y)v_\lambda(x-y)\,dx|\\ &\leq \int_{\mathbb{R}^N}|\xi(\frac {|x|}{\rho})u_\lambda(x-y)\xi(\frac {|x|}{\rho})v_\lambda(x-y)-u_\lambda(x-y)v_\lambda(x-y)|\,dx\\ &\leq k_5 \rho^{N}\to 0. \end{split} \end{equation} This proves the claim. The definition of $t_\rho$ and the claim yield that $t_\rho\to 1$ as $\rho\to 0$. This together with equation (\ref{4.4}) imply (ii). \end{proof} Since $I_\infty^\lambda(u_\lambda(x-y),v_\lambda(x-y))=m_\lambda$, the following result is a consequence of (ii) in Lemma \ref{lem4.1}. \begin{corollary}\label{coro4.2} For $0<\lambda<\lambda_1$ or $\lambda_2<\lambda<1$, there exists a $\bar \rho=\bar\rho(\lambda)$ such that for $\rho\leq \bar \rho$, there holds \begin{equation}\label{4.6} \sup_{y\in \mathbb{R}^N}I(\Phi_\rho(y))<\bar m. \end{equation} \end{corollary} From now on we will suppose that $\Omega$ is fixed in such a way that $\rho<\bar \rho$. Now we define a function $\beta: H^1(\mathbb{R}^N)\to R^N$ as follows $$ \beta (u)=\int_{\mathbb{R}^N}u(x)\chi(|x|)x\,dx, $$ where $$ \chi(t)= \begin{cases} 1 & \text{if } 0\leq t\leq R, \\ R/tt & \text{if } t>R \end{cases} $$ and $R$ is chosen such that $\mathbb{R}^N\setminus \Omega \subset B_R$. Let $\mathcal{B}_0:=\{(u,v)\in \mathcal{N}: \beta (u)=0 \text{ or } \beta (v)=0\}$ and let $c_0=\inf_{(u,v)\in \mathcal B_0}I(u,v)$. \begin{lemma}\label{lem4.3} There holds $c_0>m_\lambda$, and there is an $R_0>\rho$ such that \begin{itemize} \item[(a)] if $|y|\geq R_0$, then $I(\Phi_\rho(y))\in (m_\lambda,\frac {m_\lambda+c_0} 2)$; \item[(b)] if $|y|=R_0$, then $\langle \beta \circ P_1\circ\Phi_\rho(y),y\rangle>0$ or $\langle \beta \circ P_2\circ\Phi_\rho(y),y\rangle>0$, where $P_i(u,v)$ is the projection of $(u,v)$ on the $i\rm th$ coordinate. \end{itemize} \end{lemma} \begin{proof} It is obvious that $c_0\geq m_\lambda$. Now suppose that $c_0=m_\lambda$, then there exists a sequence $(u_n,v_n)\in \mathcal{N}$ with $\beta(u_n)=0$ or $\beta(v_n)=0$ such that $I(u_n,v_n)\to m_\lambda$. We may assume that $\beta(u_n)=0$. By Lemma \ref{thm3.1}, $ u_n(x)=u_0(x-y_n)+o(1), v_n=v_0(x-y_n)+o(1) $ with $|y_n|\to \infty$. Denote $(\mathbb{R}^N)_n^+=\{x\in \mathbb{R}^N: \langle x,y_n\rangle >0\}, (\mathbb{R}^N)_n^-=\mathbb{R}^N\setminus (\mathbb{R}^N)_n^+$, then for $n$ large we have $B_{\hat r}(y_n):=\{x:|x-y_n|<\hat r\}\subset (\mathbb{R}^N)_n^+$ for some fixed $\hat r>0$ and $u_0(x-y_n)\geq\delta_0>0,v_0(x-y_n)\geq\delta_0>0$ for $x\in B_{\hat r}(y_n)$ and some $\delta_0>0$. Lemma \ref{thm2.1} implies $$ u_0(x-y_n)\leq \frac{K}{e^{\delta |x-y_n|}|x-y_n|^{\frac {N-1}2}},\quad v_0(x-y_n)\leq \frac{K}{e^{\delta |x-y_n|}|x-y_n|^{\frac {N-1}2}} $$ for $x\in B_{\hat r}(y_n)$. So we have \begin{equation}\label{4.7} \begin{split} &\langle \beta(u_0(x-y_n)),y_n\rangle\\ &=\int_{(\mathbb{R}^N)_n^+}u_0(x-y_n)\chi(|x|)\langle x,y_n\rangle\,dx +\int_{(\mathbb{R}^N)_n^-}u_0(x-y_n)\chi(|x|)\langle x,y_n\rangle\,dx\\ &\geq \int_{B_{\hat r}(y_n)}\delta_0\chi(|x|)\langle x,y_n\rangle\,dx -\int_{(\mathbb{R}^N)_n^-}\frac{KR|y_n|}{e^{\delta |x-y_n|}|x-y_n|^{\frac {N-1}2}}\,dx\\ &\geq \alpha-o(\frac 1 {|y_n|})>0, \end{split} \end{equation} where $\alpha>0$ is a constant. Since $\beta$ is continuous, we have $\beta(u_n)\neq 0$. This contradicts to the fact that $\beta(u_n)=0$. (a) can be proved in the same way as the proof of Lemma \ref{lem2.2} and (b) can be proved as (\ref{4.7}). \end{proof} Now let us consider the set $\Sigma$ given by $$ \Sigma:=\{t_\rho \Phi_\rho (y): |y|\leq R_0\}, $$ where $t_\rho$ is chosen such that $t_\rho \Phi_\rho (y)\in \mathcal{N}$. We define $$ H=\{h\in C(\mathcal{N},\mathcal{N}): h(u,v)=(u,v)\ {\rm for}\ \forall (u,v)\in \mathcal{N}\ {\rm with}\ I(u,v) \leq \frac{c_0+m}{2}\} $$ and $\Gamma=\{A\subset \mathcal{N}, A=h(\Sigma)\}$. \begin{lemma}\label{lem4.4} If $A\in \Gamma$, then $A\cap \mathcal B_0\neq \emptyset$. \end{lemma} \begin{proof} The proof of the lemma is equivalent to prove that for $\forall h\in H$, there is $\bar y\in \mathbb{R}^N$ with $|\bar y|\leq R_0$ such that $\beta\circ h\circ P_1\circ \Phi_\rho (y)=0$ or $\beta\circ h\circ P_2\circ \Phi_\rho (y)=0$. By Lemma \ref{lem4.3}, we have $\langle \beta\circ P_1\circ \Phi_\rho (y),y\rangle>0$ or $\langle \beta\circ P_2\circ \Phi_\rho (y),y\rangle>0$ for $|y|=R_0$. Assume that $\langle \beta\circ P_1\circ \Phi_\rho (y),y\rangle>0$ without of loss generality and define \begin{gather*} f(y)=\beta\circ h\circ P_1\circ \Phi_\rho (y),\\ F(t,y)=tf(y)+(1-t)id. \end{gather*} (b) of Lemma \ref{lem4.3} implies $0\not\in F(t,\partial B_{R_0}$), hence, $deg(F,B_{R_0},0)=\deg (id,B_{R_0},0)=1$. This yields that there exists $\bar y\in B_{R_0}$ such that $\beta\circ h\circ P_1\circ \Phi_\rho (y)=0$. If $\langle \beta\circ P_2\circ \Phi_\rho (y),y\rangle>0$, we may show that there exists a $\bar y\in B_{R_0}$ such that $\beta\circ h\circ P_2\circ \Phi_\rho (y)=0$ in the same way. This proves the Lemma. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.1}] For $\lambda\in (0,\delta)$, obviously, problem \eqref{1.1} has two pair of positive solutions $(U_{1-\lambda}, U_{1-\lambda})$ and $(\pm U_{1+\lambda}, \mp U_{1+\lambda})$, where $U_{1-\lambda}$ and $U_{1+\lambda}$ are positive solutions of \begin{equation}\label{3.10} \begin{gathered} -\Delta u+(1-\lambda)u=|u|^{p-1}u \quad \text{in } \Omega, \\ u=0 \quad \text{on } \partial\Omega, \end{gathered} \end{equation} and \begin{equation}\label{3.11} \begin{gathered} -\Delta u+(1+\lambda)u=|u|^{p-1}u \quad \text{in } \Omega, \\ u=0 \quad \text{on } \partial\Omega, \end{gathered} \end{equation} respectively. It is proved in \cite{BC} that problem (\ref{3.10}) and problem (\ref{3.11}) have nontrivial solutions. Define \begin{equation} c_\lambda=\inf_{A\in \Gamma}\sup_{(u,v)\in A}I(u,v), \end{equation} then we have $\bar m>c_\lambda\geq c_0>m_\lambda$ since $id\in H$ and $A\cap \mathcal B_0\neq \emptyset$. A standard deformation argument implies that $c_\lambda$ is a critical value of $I$. Now, we claim that $c_\lambda0$ small, there exists $\bar\rho=\bar\rho(\lambda_0)$ such that $c_{\lambda_0}