\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 75, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2016/75\hfil Semi-classical states] {Semi-classical states for Schr\"odinger-Poisson systems on $\mathbb{R}^3$} \author[H. Zhu \hfil EJDE-2016/75\hfilneg] {Hongbo Zhu} \address{Hongbo Zhu \newline School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China. \newline School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, China} \email{zhbxw@126.com} \thanks{Submitted August 14, 2015. Published March 17, 2016.} \subjclass[2010]{35B38, 35J20, 35J50} \keywords{Schr\"odinger-Poisson system; semi-classical states; variational method} \begin{abstract} In this article, we study the nonlinear Schr\"{o}dinger-Poisson equation \begin{gather*} -\epsilon^{2}\Delta u+V(x) u+\phi(x)u=f(u), \quad x\in{\mathbb{R}^{3}}, \\ -\epsilon^{2}\Delta\phi=u^{2},\quad \lim_{|x|\to\infty}\phi(x)=0\,. \end{gather*} Under suitable assumptions on $V(x)$ and $f(s)$, we prove the existence of ground state solution around local minima of the potential $V(x)$ as $\epsilon\to0$. Also, we show the exponential decay of ground state solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Consider the nonlinear Schr\"{o}dinger equation: \begin{equation} i\epsilon\frac{\partial\psi}{\partial t}=-\epsilon^{2}\Delta\psi+\tilde{V}\psi-f(\psi)\label{1.1} \end{equation} coupled with the Poisson equation \begin{equation} -\epsilon^{2}\Delta\phi=|\psi|^{2},\label{1.2} \end{equation} where $\epsilon$ is the planck constant, $i$ is the imaginary unit and $\tilde{ V},\psi$ are real functions on $\mathbb{R}^3$ and represent the effective potential and electric potential respectively. $\psi(x,t)\to \mathbb{C}$ and $f$ is supposed to satisfy $f(\alpha e^{i\theta})=f(\alpha)e^{i\theta}$ for all $\theta,\alpha\in\mathbb{R}$. Problem \eqref{1.1}, \eqref{1.2} arose from semiconductor theory; see e.g, \cite{apa,bf,oj} and the references therein for more physical background. We are interested in standing wave solutions, namely solutions of form $\psi(x,t)=u(x)\exp(i\omega t/\epsilon)$ with $u(x)>0$ in $\mathbb{R}^{3}$ and $\omega>0$ (the frequency), then it is not difficult to see that $u(x)$ must satisfy \begin{equation} \begin{gathered} -\epsilon^{2}\Delta u+V(x) u+\phi(x)u=f(u), \quad x\in{\mathbb{R}^{3}}, \\ -\epsilon^{2}\Delta\phi=u^{2},\lim_{|x|\to\infty}\phi(x)=0. \end{gathered}\label{1.3} \end{equation} An interesting class of solutions of \eqref{1.3}, sometimes called semi-classical states, are families solutions $u_{\epsilon}(x)$ which concentrate and develop a spike shape around one, or more, special points in $\mathbb{R}^{3}$, which vanishing elsewhere as $\epsilon\to0$. Similar equations have been studied extensively by many authors concerning existence, non-existence, multiplicity when the nonlinearity $f(u)=u^{p}$, $10$ for all $x\in\mathbb{R}^3$. \item[(A2)] $f(s)=o(s^{3})$ as $s\to0$. \item[(A3)] There exists $q\in(3,5)$ such that $\lim_{s\to+\infty} f(s)/s^{q}=0$. \item[(A4)] There exists some $4<\theta0. $$ \item[(A5)] For all $x\in \mathbb{R}^{3}$, $f(x, s)/s $ is nondecreasing in $s\geq0$. \end{itemize} The main result of this paper reads as follows. \begin{theorem} \label{thm1.1} Assume that {\rm (A1)--(A5)} hold, and that there is a bounded and compact domain $\Lambda$ in $\mathbb{R}^{3}$ such that $$ \inf_{{x\in\Lambda}}V(x)<\min_{{x\in\partial\Lambda}}V(x). $$ Then there exists $\epsilon_0>0$ such that for any $\epsilon\in(0,\epsilon_0)$, problem \eqref{1.3} has a positive solution $u_{\epsilon}$. Moreover, $u_{\epsilon}$ has at most one local (hence global) maximum $x_{\epsilon}\in\Lambda$ such that $$ \lim_{\epsilon\to0}V(x_{\epsilon})=V_0. $$ Also, there are constants $C,c>0$ such that \begin{equation} u_{\epsilon}(x)\leq C\exp(-c\frac{x-x_{\epsilon}}{\epsilon}).\label{**} \end{equation} \end{theorem} \begin{remark} \label{rmk2} \rm (i) We point out that no restriction on the global behavior of $V(x)$ is required other than condition (A1). This is an improvement on some previous works, see, e.g.,\cite{bc} \cite{hw} and references therein. (ii) Condition (A5) holds if $f(s)/s^{3}$ is an increasing function of $s>0$. In fact, that $f(s)/s^{3}$ is increasing is required in \cite{hw}. \end{remark} This article is organized as follows: In section2, influenced by the work of del Pino and Felmer \cite{d}, we introduce a modified functional for any $\epsilon>0$ and show it has a ground state solution $u_{\epsilon}(x)$. In Section3, we give the uniform boundedness of $\max_{x\in\partial\Lambda}u_{\epsilon}(x)$ and the critical value $ c_{\epsilon}$ when $\epsilon$ goes to zero. In section4, we show the critical point of the modified functional which satisfies the original problem \eqref{1.3}, and investigate its concentration and exponential decay behavior, which completes the proof Theorem \ref{thm1.1}. Hereafter we use the following notation: \noindent$\bullet$ $H^{1}(\mathbb{R}^{3})$ is usual Sobolev space endowed with the standard scalar product and norm $$ (u,v)=\int_{\mathbb{R}^{3}}(\nabla u\nabla v+uv)dx;\|u\|^{2} =\int_{\mathbb{R}^{3}} (|\nabla u|^{2}+u^{2})dx. $$ \noindent$\bullet$ $D^{1,2}(\mathbb{R}^{3})$ is the completion of $C_0^{\infty}(\mathbb{R}^{3})$ with respect to the norm $$ \|u\|^{2}_{D^{1,2}(\mathbb{R}^{3})}=\int_{\mathbb{R}^{3}}|\nabla u|^{2}dx. $$ $\bullet$ $H^{-1}$ denotes the dual space of $H^{1}(\mathbb{R}^{3})$. \noindent$\bullet$ $L^{q}(\Omega),1\leq q \leq+\infty,\Omega\subseteq\mathbb{R}^{3}$, denotes a Lebesgue space, the norm in $L^{q}(\Omega)$ is denoted by $|u|_{q,\Omega}$. \noindent$\bullet$ For any $R>0$ and for any $y\in\mathbb{R}^{3}, B_{R}(y)$ denotes the ball of radius $R$ centered at $y$. \noindent$\bullet$ $C,c$ are various positive constants. \noindent$\bullet$ $o(1)$ denotes the quantity which tends to zero as $n\to\infty$. It is well known that for every $u\in H^{1}(\mathbb{R}^{3})$, the Lax-Milgram theorem implies that there exists a unique $\phi_{u}\in D^{1,2}(\mathbb{R}^{3})$ such that $$ \int_{\mathbb{R}^{3}}\nabla\phi_{u}\nabla vdx =\int_{\mathbb{R}^{3}}u^{2}dx,\quad \forall v\in D^{1,2}(\mathbb{R}^{3}), $$ where $\phi_{u}$ is a weak solution of $-\Delta \phi=u^{2}$ with $$ \phi_{u}(x)=\int_{\mathbb{R}^{3}}\frac{u^{2}(y)}{|x-y|}dy. $$ Substituting $\phi_{u}$ in \eqref{1.3}, we can rewrite \eqref{1.3} as the equivalent equation \begin{equation} -\epsilon^{2}u+V(x)u+\epsilon^{-2}\phi_{u}u=f(u). \label{2.1}% \end{equation} Let $$ H=\{u\in H^{1}(\mathbb{R}^{3}):\int_{\mathbb{R}^{3}}V(x)u^{2}dx<+\infty\} $$ be the Sobolev space endowed with the norm $$ \|u\|_{\epsilon}^{2}=\int_{\mathbb{R}^{3}}(\epsilon^{2}|\nabla u|^{2}+V(x)u^{2})dx. $$ We see that \eqref{2.1} is variational and its solutions are the critical points of the functional \begin{equation} I_{\epsilon}(u)=\frac{1}{2}\int_{\mathbb{R}^{3}}(\epsilon^{2} |\nabla u|^{2}+V(x)u^{2})dx + \frac{1}{4\epsilon^{2}}\int_{\mathbb{R}^{3}}\phi_{u}(x)u^{2}dx -\int_{\mathbb{R}^{3}}F(u)dx.\label{2.2} \end{equation} Clearly, under the hypotheses (A2)--(A5), we see that $I_{\epsilon}$ is well-defined $C^{1}$ functional. In the following proposition, we summarize some properties of $\phi_{u}$, which are useful to study our problem. \begin{proposition}[\cite{cg}] \label{pro} For any $u\in H^{1}(\mathbb{R}^{3})$, we have \begin{itemize} \item[(i)] $\phi_{u}: H^{1}(\mathbb{R}^{3})\to D^{1,2}(\mathbb{R}^{3})$ is continuous, and maps bounded sets into bounded sets; \item[(ii)] if $u_{n}\rightharpoonup u$ in $H^1({\mathbb{R}^{3}})$, then $\phi_{u_{n}}\rightharpoonup\phi_{u}$ in $D^{1,2}(\mathbb{R}^{3});$ \item[(iii)] $\phi_{u}\geq0, \|\phi_{u}\|_{D^{1,2}(\mathbb{R}^{3})}$, and $\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx\leq C\|u\|^{4}$; \item[(iv)] $\phi_{tu}(x)=t^{2}\phi_{u}$ for all $t\in \mathbb{R}$. \end{itemize} \end{proposition} \section{Solution of the modified equation} In this section, we find a solution $u_{\epsilon}$ of problem \eqref{1.3} concentrating on a given set $\Lambda$, we modify the nonlinearity $f(s)$. Here we follow an approach used by del Pino and Felmer \cite{d}. Let $k>\frac{\theta}{\theta-4}$, $a>0$ be such that $\frac{f(a)}{a}=\frac{V_0}{k}$, and set \begin{equation} \tilde{f}(s)= \begin{cases} f(s), &\text{if }s\leq a, \\ \frac{V_0}{k}s, & \text{if } s>a, \end{cases} \label{3.1} \end{equation} and define \begin{equation} g(.,s)=\chi_{\Lambda}f(s)+(1-\chi_{\Lambda})\tilde{f}(s),\label{3.2} \end{equation} where $\Lambda$ is the bounded domain as in the assumptions of Theorem\ref{thm1.1}, and $\chi_{\Lambda}$ denotes its characteristic function. It is easy to check that $g(x,s)$ satisfies the following assumptions: \begin{itemize} \item[(A6)] $g(x,s)=o(s^{3})$ as $s\to0$. \item[(A7)] There exists $q\in(3,5)$ such that $\lim_{s\to+\infty}\frac{g(x,s)}{s^{q}}=0$. \item[(A8)] There exists a bounded subset $K$ of $\mathbb{R}^{3}$, int$(K)\neq\emptyset$ such that \begin{gather*} 0<\theta G(x,s)\leq g(x,s)s\quad \text{for all } x\in K,s>0, \\ 0\leq2G(x,s)\leq g(x,s)s\leq\frac{1}{k}V(x)s^{2} \quad\text{for all } s>0,x\in K^{c}. \end{gather*} \item[(A9)] The function $\frac{g(x,s)}{s}$ is increasing for $s>0$. \end{itemize} Now, we consider the modified equation \begin{equation} \begin{gathered} -\epsilon^{2}\Delta u+V(x)u+\phi(x)u=g(x,s), \quad x\in\mathbb{R}^{3}, \\ -\epsilon^{2}\Delta\phi=u^{2}, \end{gathered}\label{3.3} \end{equation} where $V$ satisfies condition (A1), and $g$ satisfies (A6)--(A9). Here we have set $\epsilon=1$ for notational simplicity. The functional associated with \eqref{3.3} is \begin{equation} J(u)=\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+V(x)u^{2})dx +\frac{1}{4}\int_{\mathbb{R}^{3}}\phi_{u}(x)u^{2}dx -\int_{\mathbb{R}^{3}}G(x,u)dx, \label{3.4} \end{equation} which is of class $C^{1}$ in $H$ with associated norm $\|\cdot\|_{H}$. In the rest of this section, we show some lemmas related to the functional $J$. First, we show the functional $J$ satisfying the mountain pass geometry. \begin{lemma}\label{lem3.1} The functional $J$ satisfies the following conditions: \begin{itemize} \item[(i)] There exist $\alpha,\rho>0$ such that $J(u)\geq\alpha$ for all $\|u\|_{H}=\rho$. \item[(ii)] There exists $e\in B_{\rho}^{c}(0)$ with $J(e)<0$. \end{itemize} \end{lemma} \begin{proof} (i) For any $u\in H\backslash\{0\}$ and $\varepsilon>0$, by (A2) and (A3) there exists $C(\varepsilon)>0$ such that $$ |f(s)|\leq\varepsilon|s|+C_{\varepsilon}|s|^{q}, \quad \forall s\in \mathbb{R}. $$ By the Sobolev embedding $H\hookrightarrow L^{p}(\mathbb{R}^{3})$, with $p\in[2,6]$, we have \begin{align*} J(u) &\geq \frac{1}{2}\|u\|_{H}^{2}-\int_{\mathbb{R}^{3}}[\chi_{\Lambda}(x)F(u) +(1-\chi_{\Lambda}(x))\tilde{F}(u)]dx\\ &\geq \frac{1}{2}\|u\|_{H}^{2}-\int_{\mathbb{R}^{3}}F(u)dx\\ &\geq \frac{1}{2}\|u\|_{H}^{2}-\frac{\varepsilon}{2}\int_{\mathbb{R}^{3}}|u|^{2}dx -\frac{C_{\varepsilon}}{q+1}|u|^{p+1}dx\\ &\geq \frac{1}{2}\|u\|_{H}^{2}-C_{1}\varepsilon\|u\|_{H}^{2} -C_{2}C_{\varepsilon}\|u\|_{H}^{p+1}. \end{align*} Since $\varepsilon$ is arbitrarily small, we can choose constants $\alpha,\rho$ such that $J(u)\geq\rho>0$ for all $\|u\|_{H}=\rho$. (ii) By (A4), we have $F(s)\geq Cs^{\theta}-C$ for all $t>0$, Choosing $u\in H\backslash\{0\}$ not negative, with its support contained in the set $K$, we see that \begin{align*} J(tu)&=\frac{t^{2}}{2}{\|u\|_{H}^{2}} +\frac{t^{4}}{4}\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx -\int_{\mathbb{R}^{3}}G(x,tu)dx\\ &\leq \frac{t^{2}}{2}{\|u\|_{H}^{2}} +\frac{t^{4}}{4}\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx-Ct^{\theta} \int_{K}u^{\theta}dx+C|K| <0 \end{align*} for some $t>$ large enough. So, we can choose $e=t^{*}u$ for some $t^{*}>0$, and (ii) follows. By lemma \ref{lem3.1} and the mountain pass theorem, there is a $(PS)_{c}$ sequence $\{u_{n}\}\subset H$ such that $J(u_{n})\to c$ in $H^{-1}$ with the minmax value \begin{equation} c=\inf_{\gamma\in\Gamma}\max_{0\leq t\leq1}J(\gamma(t)), \label{3.5} \end{equation} where $$ \Gamma:=\{\gamma\in C([0,1],H):\gamma(0)=0, J(\gamma(1))<0\}. $$ \end{proof} \begin{lemma}\label{lem3.2} Let $\{u_{n}\}\subset H$ be a $(PS)_{c}$ sequence for $c>0$. Then $u_{n}$ has a convergent subsequence. \end{lemma} \begin{proof} First, we show that $\{u_{n}\}$ is bounded in $H$. In fact, using $(A8)$ we easily see that \begin{gather} \int_{\mathbb{R}^{3}}(|\nabla u_{n}|^{2}+V(x)u_{n}^{2})dx+\int_{\mathbb{R}^{3}}\phi_{u_{n}}u_{n}^{2}dx \geq\int_{K}g(x,u_{n})u_{n}dx+o(\|u_{n}\|_{H}). \label{3.6} \\ \begin{aligned} &\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u_{n}|^{2}+V(x)u_{n}^{2})dx +\frac{1}{4}\int_{\mathbb{R}^{3}}\phi_{u_{n}}u_{n}^{2}dx\\ &=\int_{\mathbb{R}^{3}}G(x,u_{n})dx+O(1) \\ &\leq \int_{K}G(x,u_{n})dx +\frac{1}{2k}\int_{K^{c}}V(x)u_{n}^{2}dx+O(1) \end{aligned}\label{3.7} \end{gather} Thus, from \eqref{3.6} \eqref{3.7} and (A8) we find \begin{equation} \frac{2}{k}\int_{K^{c}}V(x)u_{n}^{2}dx+o(\|u_{n}\|_{H})+O(1) \geq(1-\frac{2}{k})\int_{\mathbb{R}^{3}}(|\nabla u_{n}|^{2}+V(x)u_{n}^{2})dx .\label{3.8} \end{equation} Then, it follows from the choice of $k$ in $(A8)$ that $\{u_{n}\}$ is bounded in $H$. Then there is a subsequence, still denoted by $\{u_{n}\}$ such that $u_{n}\rightharpoonup u$ weakly in $H$. We now prove this convergence is actually strong. In deed, it suffices to show that, given $\delta>0$, there is an $R>0$ such that \begin{equation} \limsup_{n\to\infty}\int_{B_{R}^{c}}(|\nabla u_{n}|^{2}+V(x)u_{n}^{2})dx\leq\delta.\label{3.9} \end{equation} Let $\xi_{R}(x)\in C^{\infty}(\mathbb{R}^{3},\mathbb{R})$ be a cut-off function such that $0\leq\xi_{R}\leq1$ and $$ \xi_{R}(x)=\begin{cases} 0 &\text{for } |x|\leq\frac{R}{2}, \\ 1, &\text{for } |x|\geq R \end{cases} $$ and $|\nabla\xi_{R}(x)|\leq\frac{C}{R}$ for all $x\in\mathbb{R}^{3}$. Moreover we may assume that $R$ is chosen so that $K\subset B_{\frac{R}{2}}$. Since $\{u_{n}\}$ is a bounded $(PS)_{c}$ sequence, we have that \begin{equation} \langle J'(u_{n}),\xi_{R}u_{n}\rangle=o(1),\label{3.10} \end{equation} so that \begin{equation} \begin{aligned} &\int_{\mathbb{R}^{3}}(|\nabla u_{n}|^{2}+V(x)u_{n}^{2})dx+\int_{\mathbb{R}^{3}}u_{n}\nabla u_{n}\nabla\xi_{R}dx+\int_{\mathbb{R}^{3}}\phi_{u_{n}}u_{n}\xi_{R}dx \\ &=\int_{\mathbb{R}^{3}}g(x,u_{n})u_{n}\xi_{R}dx+o(1) \leq\frac{1}{k}\int_{\mathbb{R}^{3}}V(x)u_{n}^{2}\xi_{R}dx+o(1). \end{aligned} \label{3.11} \end{equation} We conclude that \begin{equation} \int_{B_{R}^{c}}V(x)u_{n}^{2}dx \leq\frac{C}{R}\|u_{n}\|_{L^{2}(\mathbb{R}^{3})}\|\nabla u_{n}\|_{L^{2}(\mathbb{R}^{3})},\label{3.12} \end{equation} from where \eqref{3.9} follows. \end{proof} Lemma \ref{lem3.1} implies that $c$ defined in \eqref{3.5} is a critical value of $J$. \begin{remark} \label{rmk3} \rm Similar to the proof of lemma \ref{lem3.2}, it is not difficult to see that $c$ can be characterized as $$ c=\inf_{u\in H\backslash\{0\}}\sup_{t\geq0}J(tu). $$ \end{remark} Since the modified function $g$ satisfies assumptions (A6)--(A9), the results of the above yield the following lemma. \begin{lemma}\label{lem3.4} For any $\epsilon>0$, there exists a critical point for $J_{\epsilon}$ at level \begin{equation} J_{\epsilon}(u_{\epsilon})=c_{\epsilon} =\inf_{\gamma_{\epsilon}\in\Gamma}\max_{0\leq t\leq1}J_{\epsilon}(\gamma(t)),\label{3.13} \end{equation} where \begin{equation} \begin{gathered} \Gamma_{\epsilon}:=\{\gamma\in C([0,1],H):\gamma(0)=0, J_{\epsilon}(\gamma(1))<0\},\label{3.14} \\ J_{\epsilon}(u)=\frac{1}{2}\int_{\mathbb{R}^{3}} (\epsilon^{2}|\nabla u|^{2}+V(x)u^{2})dx +\frac{1}{4\epsilon^{2}}\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx -\int_{\mathbb{R}^{3}}G(x,u)dx. \end{gathered} \end{equation} \end{lemma} \section{Some estimates} To show that the solution $u_{\epsilon}$ found in lemma \ref{lem3.4} satisfies the original problem and concentrates at some point in $\Lambda$, we need to study the behavior of $u_{\epsilon}$ as $\epsilon\to0$. We begin with an energy estimate. \begin{proposition}[Upper estimate of the critical value] \label{pro3.5} For $\epsilon$ small enough, the critical value $c_{\epsilon}$ defined \eqref{3.13} satisfies \begin{equation} c_{\epsilon}=J_{\epsilon}(u_{\epsilon})\leq\epsilon^{3}(c_0+o(1)) \quad \text{as } \epsilon\to 0.\label{3.15} \end{equation} Moreover, there exists $C>0$ such that \begin{equation} \int_{\mathbb{R}^{3}}(\epsilon^{2}|\nabla u_{\epsilon}|^{2}+V(x)u_{\epsilon}^{2})dx \leq C\epsilon^{3}.\label{3.16} \end{equation} \end{proposition} \begin{proof} Set $V_0=\min_{\Lambda}V=V(x_0)$, and let \begin{equation} c_0:=\inf_{\gamma\in\Gamma}\max_{0\leq t\leq1}I_0(\gamma(t)), \label{3.17} \end{equation} where \begin{gather*} I_0(u)=\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+V_0u^{2})dx + \frac{1}{4}\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx -\int_{\mathbb{R}^{3}}F(u)dx,\\ \Gamma:=\{\gamma\in C([0,1],H^{1}(\mathbb{R}^{3})): \gamma(0)=0, I_0(\gamma(1))<0\}. \end{gather*} From \eqref{3.17}, for any $\delta>0$, there exists a continuous path $\gamma_{\delta}:[0,1]\to H^1({\mathbb{R}^{3}})$ such that $\gamma_{\delta}(0)=0, I_0(\gamma_{\delta}(1))<0$ and $$ c_0\leq\max_{0\leq t\leq1}I_0(\gamma_{\delta}(t))\leq c_0+\delta. $$ Let $\eta$ be a smooth cut-off function with support in $\Lambda$ such that $0\leq\eta\leq1,\eta=1$ in a neighborhood of $x_0$ and $|\nabla\eta|\leq C$. We consider the path $$ \bar{\gamma_{\delta}}(t)(x)=\eta(x)\gamma_{\delta}(t)(\frac{x-x_0}{\epsilon}). $$ Setting $$ \bar{\gamma_{\delta}}(t)(x):=v_{t}(\frac{x-x_0}{\epsilon}), $$ we compute, by a charge of variable \begin{equation} \begin{aligned} &\frac{1}{2}\int_{\mathbb{R}^{3}}[\epsilon^{2}|\nabla \bar{\gamma_{\delta}}(t)|^{2}+V(x)\bar{\gamma_{\delta}}(t)^{2}]dx -\int_{\mathbb{R}^{3}}G(x,\bar{\gamma_{\delta}}(t))dx \\ &=\epsilon^{3}\frac{1}{2}[|\nabla v_{t}(x)|^{2}+V(x_0+\epsilon x)v_{t}^{2}(x)]dx-\epsilon^{3}\int_{\mathbb{R}^{3}}G(x_0+\epsilon x, v_{t}(x))dx. \end{aligned} \end{equation} The Hardy-Littlewood Sobolev inequality leads to \begin{align*} \int_{\mathbb{R}^{3}} \phi_{\bar{\gamma_{\delta}}(t)(x)}\bar{\gamma_{\delta}}(t)(x)^{2}dx &=\int_{\mathbb{R}^{3}}\Big[\int_{\mathbb{R}^{3}} \frac{\bar{\gamma_{\delta}}(t)(y)^{2}}{|x-y|}\bar{\gamma_{\delta}}(t)dy\Big]dx\\ &=\epsilon^{5}\int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}}\frac{v_{t}^{2}(x)v_{t}^{2}(y)}{|x-y|}\,dx\,dy\\ &=\epsilon^{5} \int_{\mathbb{R}^{3}}\phi_{v_{t}}v_{t}^{2}dx. \end{align*} For $\epsilon$ small enough, we obtain $$ \epsilon^{-3}J_{\epsilon}(\bar{\gamma_{\delta}}(t))\to I_0(\gamma_{\delta}(t))+o(1). $$ It follows that $\epsilon$ small enough, $\bar{\gamma_{\delta}}$ belongs to the class of paths $\Gamma_{\epsilon}$ defined by \eqref{3.14}. We deduce that $$ \epsilon^{-3}c_{\epsilon}\leq \epsilon^{-3} \max_{0\leq t\leq1}J_{\epsilon}(\bar{\gamma_{\delta}}(t)) \to\max_{0\leq t\leq1}I_0(\bar{\gamma_{\delta}}(t))+o(1)\leq(c_0+\delta)+o(1). $$ Since $\delta>0$ is arbitrary, \eqref{3.15} is proved. \begin{equation} \begin{aligned} J_{\epsilon}(u_{\epsilon}) &=\frac{1}{2}\int_{\mathbb{R}^{3}}(\epsilon^{2}|\nabla u_{\epsilon}|^{2}+V(x)u_{\epsilon}^{2})dx +\frac{1}{4\epsilon^{2}}\int_{\mathbb{R}^{3}}\phi_{u_{\epsilon}}u_{\epsilon}^{2}dx -\int_{\mathbb{R}^{3}}G(x,u_{\epsilon})dx \\ &= \frac{1}{2}\int_{\mathbb{R}^{3}}(\epsilon^{2}|\nabla u_{\epsilon}|^{2}+V(x)u_{\epsilon}^{2})dx +\frac{1}{4\epsilon^{2}}\int_{\mathbb{R}^{3}}\phi_{u_{\epsilon}}u_{\epsilon}^{2}dx -\int_{K}G(x,u_{\epsilon})dx \\ &\quad -\int_{\mathbb{R}^{3}\backslash\{K\}}G(x,u_{\epsilon})dx \\ &\geq \frac{1}{4}\int_{\mathbb{R}^{3}}(\epsilon^{2}|\nabla u_{\epsilon}|^{2}+V(x)u_{\epsilon}^{2})dx -\frac{1}{2k}\int_{\mathbb{R}^{3}}V(x)u_{\epsilon}^{2}dx \\ &\geq (\frac{1}{4}-\frac{1}{2k})\int_{\mathbb{R}^{3}}(\epsilon^{2}|\nabla u_{\epsilon}|^{2}+V(x)u_{\epsilon}^{2})dx, \end{aligned} \label{3.19} \end{equation} where $C=\frac{1}{4}-\frac{1}{2k}>0$ thanks to the choice of $k$. Combining \eqref{3.15} and \eqref{3.19}, it is easy to obtain \eqref{3.16}. \end{proof} Next, we give a proposition that is the crucial step in the proof of Theorem \ref{thm1.1}. \begin{proposition} \label{pro3.6} \begin{equation} \lim_{\epsilon\to0}\max_{\partial\Lambda}u_{\epsilon}(x)=0.\label{3.20} \end{equation} Moreover, for all $\epsilon$ sufficiently small enough, $u_{\epsilon}$ possesses one local maximum $x_{\epsilon}\in\Lambda$ and we must have \begin{equation} \lim_{\epsilon\to0}V(x_{\epsilon})=V_0=\min_{x\in\Lambda}V(x).\label{3.21}% \end{equation} \end{proposition} \begin{proof} To prove this proposition we establish that If $\epsilon_{n}\to0$ and $x_{n}\in\Lambda$ are such that $u_{\epsilon_{n}}\geq b>0$, then \begin{equation} \lim_{n\to\infty}V(x_{n})=V_0.\label{3.22} \end{equation} We take three steps to prove this claim. \smallskip \noindent\textbf{Step1:} We argue by contradiction. Thus we assume, passing to a subsequence, that $x_{n}\to x^{*}\in\bar{\Lambda}$ and \begin{equation} V(x^{*})>V_0.\label{3.23} \end{equation} We consider the sequence $v_{n}(x)=u_{\epsilon_{n}}(x_{n}+\epsilon_{n}x)$. A simple computation shows $$ \epsilon^{2}\phi_{v_{n}}(x)=\phi_{u_{\epsilon_{n}}}(x_{n}+\epsilon_{n}x). $$ The function $v_{n}$ satisfies the equation \begin{equation} -\Delta v_{n}+V(x_{n}+\epsilon_{n}x)v_{n}+\phi_{v_{n}}v_{n} =g(x_{n}+\epsilon_{n}x,v_{n}), x\in \Omega_{n},\label{3.23b} \end{equation} where $\Omega_{n}=\epsilon_{n}^{-1}\{H-x_{n}\}$. As a consequence of \eqref{3.16}, we see that $v_{n}$ is bounded in $H^1({\mathbb{R}^{3}})$, and from elliptic estimates, we deduce that there exists $v\in H^1({\mathbb{R}^{3}})$ such that $$ v_{n}\to v \quad\text{in } C^{2}_{\rm loc}(\mathbb{R}^{3}). $$ Let $\chi_{n}(x)=\chi_{\Lambda}(x_{n}+\epsilon_{n}x)$, then $\chi_{n}(x)\rightharpoonup \chi$ in any $L^{p}(\mathbb{R}^{3})$ over compacts with $0\leq \chi\leq1$. Now, we claim that $$ \int_{\mathbb{R}^{3}}\phi_{v_{n}}v_{n}\varphi dx \to\int_{\mathbb{R}^{3}}\phi_{v}v\varphi dx\quad \text{for any }\varphi\in C _0^{\infty}(\mathbb{R}^{3}). $$ In fact, we can assume $\operatorname{support}\varphi\subset\Omega$, where $\Omega$ is a bounded domain. Then \begin{align*} &|\int_{\mathbb{R}^{3}}\phi_{v_{n}}v_{n}\varphi dx-\int_{\mathbb{R}^{3}}\phi_{v}v\varphi dx|\\ &= |\int_{\mathbb{R}^{3}}\phi_{v_{n}}(v_{n}-v)\varphi dx+\int_{\mathbb{R}^{3}}(\phi_{v_{n}}-\phi_{v})v\varphi dx|\\ &\leq |\int_{\mathbb{R}^{3}}\phi_{v_{n}}(v_{n}-v)\varphi dx|+|\int_{\mathbb{R}^{3}}(\phi_{v_{n}}-\phi_{v})v\varphi dx|\\ &\leq |\phi_{v_{n}}|_{L_{6}(\Omega)}|v_{n}-v|_{L_{2}(\Omega)} |\varphi|_{L_{3}(\Omega)}+o(1)\to o(1). \end{align*} Therefore, $v$ satisfies the limiting equation \begin{equation} -\Delta v+V(x^{*})v+\phi_{v}v=\bar{g}(x,v),\label{3.25}% \end{equation} where $$ \bar{g}(x,s)=\chi(x)f(s)+(1-\chi(x))\tilde{f}(s). $$ Associated with \eqref{3.25} we have functional $\bar{J}:H^1({\mathbb{R}^{3}})\to\mathbb{R}$ defined as \begin{equation} \begin{aligned} \bar{J}(u) &=\frac{1}{2}\int_{\mathbb{R}^{3}}[|\nabla u|^{2}+V(x^{*})u^{2}]dx +\frac{1}{4} \int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx\\ &\quad -\int_{\mathbb{R}^{3}}\bar{G}(x,u)dx, u\in H^1({\mathbb{R}^{3}}), \end{aligned}\label{3.26} \end{equation} where $\bar{G}(x,s)=\int_0^{s}\bar{g}(x,t)dt$. Then $v$ is a critical point of $\bar{J}$. Set \begin{align*} J_{n}(u) &=\frac{1}{2}\int_{\Omega_{n}}[|\nabla u|^{2}+V(x_{n}+\epsilon_{n}x)u^{2}]dx +\frac{1}{4}\int_{\Omega_{n}}\phi_{u}u^{2}dx\\ &\quad -\int_{\Omega_{n}}G(x_{n}+\epsilon_{n}x,u)dx, u\in H_0^{1}(\Omega_{n}). \end{align*} Then $J_{n}(v_{n})=\epsilon_{n}^{-3}J_{\epsilon_{n}}(u_{\epsilon_{n}})$. So the key step in the proof of proposition is the following step. \smallskip \noindent\textbf{Step2:} $\liminf_{n\to\infty}J_{n}(v_{n})\geq\bar{J}(v)$. In particular, $\bar{J}(v)\leq c_0$, where $c_0$ is given by \eqref{3.17}. Proof: Write $$ h_{n}=\frac{1}{2}[|\nabla v_{n}|^{2}+V(x_{n}+\epsilon_{n}x)v_{n}^{2}] +\frac{1}{4}\phi_{v_{n}}v_{n}^{2}-\bar{G}(x_{n}+\epsilon_{n}x,u). $$ Then, choose $R>0$, since $v_{n}$ converges in the $C^{1}$ sense over compacts to $v$, we have $$ \lim_{n\to\infty}\int_{B_{R}}h_{n}dx =\frac{1}{2}\int_{B_{R}}[|\nabla v|^{2}+V(x^{*})v^{2}]dx +\frac{1}{4}\int_{B_{R}}\phi_{v}v^{2}dx-\int_{B_{R}}\bar{G}(x,v)dx. $$ Since $v\in H^1({\mathbb{R}^{3}})$, for each $\delta>0$, we have \begin{equation} \lim_{n\to\infty}\int_{B_{R}}h_{n}dx\geq\bar{J}(v)-\delta,\label{3.27*} \end{equation} provided that $R$ was chosen sufficiently large. Then it only suffices to check that for large enough $R$ \begin{equation} \lim_{n\to\infty}\int_{\Omega_{n}\backslash B_{R}}h_{n}dx \geq-\delta.\label{3.27} \end{equation} For any fixed $R>0$, let $\xi_{R}(x)\in C_0^{\infty}(\mathbb{R}^{3},\mathbb{R})$ be a cut-off function such that $$ \xi_{R}(x)=\begin{cases} 0 &\text{for } |x|\leq R-1, \\ 1, &\text{for } |x|\geq R, \end{cases} $$ and $|\nabla \xi_{R}(x)|\leq\frac{C}{R}$ for all $x\in\mathbb{R}^{3}$ and $C>0$ is a constant. We use $w_{n}=\xi_{R}v_{n}\in H^{1}(\Omega_{n})$ as a test function for $J'_{n}(v_{n})=0$ to obtain \begin{equation} \begin{aligned} 0=J'_{n}(v_{n})w_{n} &= E_{n}+\int_{\Omega_{n}\backslash B_{R}}(2h_{n}+g_{n})dx+\int_{\Omega_{n}\backslash B_{R}}\phi_{v_{n}}v_{n}^{2}dx \\ &\leq E_{n}+\int_{\Omega_{n}\backslash B_{R}}2h_{n}dx+\int_{\Omega_{n}\backslash B_{R}}\phi_{v_{n}}v_{n}^{2}dx,\label{3.28} \end{aligned} \end{equation} where $g_{n}=2G(x_{n}+\epsilon_{n}x,v_{n})-g(x_{n}+\epsilon_{n}x,v_{n})v_{n}$, and $E_{n}$ is given by \begin{equation} \begin{aligned} E_{n} &=\int_{B_{R}\backslash B_{R-1}}[\nabla v_{n}\nabla(\xi_{R}v_{n}) +V(x_{n}+\epsilon_{n}x)\xi_{R}v_{n}^{2} +\phi_{v_{n}}v_{n}^{2}\xi_{R}]dx \\ &= \int_{B_{R}\backslash B_{R-1}}g(x_{n}+\epsilon_{n}x,v_{n})\xi_{R}v_{n}dx. \end{aligned} \label{3.29} \end{equation} Since $v_{n}$ is bounded in $H^1({\mathbb{R}^{3}})$, it follows that $\int_{\mathbb{R}^{3}}\phi_{v_{n}}v_{n}^{2}dx\leq C\|u_{n}\|^{4}$. The fact that $v\in H^1({\mathbb{R}^{3}})$ implies that for given $\delta>0$, there exists $R>0$ sufficiently large such that $$ \lim_{n\to\infty}|E_{n}|\leq\delta, \quad \int_{\Omega_{n}\backslash B_{R}}\phi_{v_{n}}v_{n}^{2}dx\leq\delta. $$ On the other hand, the definition of $g$ together with the properties of $f$ give that $g_{n}\leq0$. Using this in \eqref{3.28}, \eqref{3.27} follows, and the proof of step2 is complete. \smallskip \noindent\textbf{Step3:} Now, we are ready to obtain a contradiction with \eqref{3.22}. Since $v$ is a critical point of $\bar{J}$, and $\bar{g}$ satisfies $(A9)$, we have that \begin{equation} \bar{J}(v)=\max_{t>0}\bar{J}(tv).\label{3.30}% \end{equation} Then since $f(s)\geq \tilde{f}(s)$ for all $s>0$ we have \begin{equation} \bar{J}(v)\geq\inf_{u\in H^1({\mathbb{R}^{3}})\backslash\{0\}} \sup_{\tau>0}I^{*}(\tau u)\Delta q c^{*},\label{3.31} \end{equation} where \begin{equation} I^{*}(u)=\frac{1}{2}\int_{\mathbb{R}^{3}}[|\nabla u|^{2}+V(x^{*})u^{2}]dx + \frac{1}{4}\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx -\int_{\mathbb{R}^{3}}F(u)dx.\label{3.32} \end{equation} But, since $V(x^{*})>V_0$, we have $c^{*}>c_0$; hence $\bar{J}(v)>c^{*}$, which contradicts step 2, and the proof of the claim, i.e. \eqref{3.22} is follows. To conclude the proof of proposition \ref{pro3.6}, we show that $u_{\epsilon}$ has at most one maximum point in $\Lambda$. The proofs rely on the the arguments carried out in step2 and so we sketch it. By contradiction, assume that, the existence of sequence $\epsilon_{n}\to0$ such that $u_{\epsilon_{n}}$ has two distinct maxima $x_{n}^{1}$ and $x_{n}^{2}$ in $\Lambda$. Set $v_{n}(x)=u_{\epsilon_{n}}(x_{n}^{1}+\epsilon_{n}x)$, and it is easy to check that $\epsilon_{n}^{-1}(x_{n}^{2}-x_{n}^{1})$ is a maximum point of $v_{n}(x)$, two cases occur. \smallskip \noindent\textbf{Case 1:} $\epsilon_{n}^{-1}(x_{n}^{2}-x_{n}^{1})$ is bounded. From \eqref{3.16} and elliptic estimates, up to a subsequence, $v_{n}\to v$ uniformly over compacts, where $v\in H^{1}(\mathbb{R}^{3})$ maximizes at zero and solves $-\Delta v+V(x^{1})v+\phi_{v}v=f(v)$, here $x^{1}=\lim_{n\to\infty}x_{n}^{1}$. Since $\epsilon_{n}^{-1}(x_{n}^{2}-x_{n}^{1})$ is bounded and hence, up to a subsequence, it converges to some $p\in\mathbb{R}^{3}$. So we conclude that $p=0$; therefore $\epsilon_{n}^{-1}(x_{n}^{2}-x_{n}^{1})\in B_{r}$ for $n$ large enough, which is impossible since $0$ is the only critical point of $v$ in $B_{r}$. \smallskip \noindent\textbf{Case2:} $\epsilon_{n}^{-1}(x_{n}^{2}-x_{n}^{1})$ is unbounded. Let $\tilde{v_{n}}(x)=u_{\epsilon_{n}}(\epsilon_{n}x+x_{n}^{2})$, then there exists $\tilde{v}$ such that $\tilde{v}$ is the solution of $-\Delta v+V(x^{2})v+\phi_{v}v=f(v)$, here $x^{2}=\lim_{n\to\infty}x_{n}^{2}$. Note that $|\epsilon_{n}^{-1}(x_{n}^{2}-x_{n}^{1})|\to+\infty$, then for any $R>0$ the balls $\tilde{B}_{R}\cap\bar{B}^{\epsilon}=\emptyset$, where $\bar{B}^{\epsilon}=\tilde{B}_{R}(\epsilon_{n}^{-1}(x_{n}^{2}-x_{n}^{1}))$, repeat the arguments of step2, we find that for any $\nu>0$ it is possible to choose that $R>0$ large enough such that \begin{equation} \lim_{n\to\infty}\int_{\bar{B}^{\epsilon}}h_{n}dx \geq\tilde{J}(\tilde{v})-\nu,\label{3.30*} \end{equation} where $$ \tilde{J}(u)=\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+V(x^{2})u^{2})dx +\frac{1}{4}\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx -\int_{\mathbb{R}^{3}}F(u)dx, $$ and \begin{equation} \lim_{n\to\infty}\int_{\mathbb{R}^{3}\backslash (B_{R}\cup B^{\epsilon})}h_{n}dx \geq-\nu.\label{3.30**} \end{equation} Similar to the argument in \eqref{3.27*}, we obtain \begin{equation} \lim_{n\to\infty}\int_{B_{R}}h_{n}dx\geq J_{1}(v)-\delta,\label{3.27**} \end{equation} where $$ J_{1}(u)=\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+V(x^{1})u^{2})dx+\frac{1}{4}\int_{\mathbb{R}^{3}}\phi_{u}u^{2}dx -\int_{\mathbb{R}^{3}}F(u)dx. $$ From \eqref{3.27**},\eqref{3.30*} and \eqref{3.30**} we conclude that \begin{equation} \lim_{n\to\infty}\int_{\mathbb{R}^{3}}h_{n}dx \geq J_{1}(v)+\tilde{J}(\tilde{v})-3\nu.\label{3.30"} \end{equation} Since $\nu$ is arbitrary we find that $$ \epsilon_{n}^{-3}J_{\epsilon_{n}}(u_{\epsilon_{n}}) =\lim_{n\to\infty}J_{n}(v_{n}) \geq J_{1}(v)+\tilde{J}(\tilde{v})\geq2c_0, $$ which contradicts \eqref{3.15}. The proof of proposition \ref{pro3.6} is now complete. \end{proof} \section{Proof of Theorem \ref{thm1.1}} In this section, we shall prove the existence, concentration, and exponential decay of ground state solution of \eqref{1.3} for small $\epsilon$. \begin{proof}[Proof of Theorem \ref{thm1.1}] By proposition \ref{pro3.6}, there exists $\epsilon_0$ such that for $0<\epsilon<\epsilon_0$, \begin{equation} u_{\epsilon}(x)0$ in $\mathbb{R}^{3}\backslash\{\Lambda\}$, hence all terms in \eqref{3.35} are zero. We conclude in particular $$ u_{\epsilon}(x)\leq a\quad\text{for all } \mathbb{R}^{3}\backslash\{\Lambda\}. $$ Consequently, $u_{\epsilon}$ is a solution to equation \eqref{1.3}, and by proposition \ref{pro3.6}, we know that the maximum value of $u_{\epsilon}$ is achieved at a point $x_{\epsilon}\in\Lambda$ and it is away from zero. To obtain \eqref{**}, we need the following proposition, which is a very particular version of \cite[Theorem 8.17]{tru}. \end{proof} \begin{proposition}[\cite{tru}] \label{pro4.1} Suppose that $t>3$, $h\in L^{t/2}(\Omega)$ and $u\in H^{1}(\Omega)$ satisfies in the weak sense $$ -\Delta u\leq h(x)\quad\text{in }\Omega, $$ where $\Omega$ is an open subset of $\mathbb{R}^{3}$. Then, for any ball $B_{2R}(y)\subset\Omega$, $$ \sup_{x\in B_{R}(y)}u(x)\leq C(\|u^{+}\|_{L^{2}(B_{2R}(y))} +\|h\|_{L^{t/2}(B_{2R}(y))}), $$ where $C$ depends on $t$ and $R$. \end{proposition} \begin{lemma}\label{lem4.2} Let $v_{\epsilon}(x)=u_{\epsilon}(x_{\epsilon}+\epsilon x)$, where $x_{\epsilon}$ is the unique maximum of $u_{\epsilon}$, then there exists $\epsilon^{*}>0$ such that $\lim_{|x|\to\infty}v_{\epsilon}(x)=0$ uniformly on $\epsilon\in(0,\epsilon^{*})$. \end{lemma} \begin{proof} Since $u_{\epsilon}(x)$ is the solution of \eqref{1.3}, by \eqref{3.16} then \begin{equation} \|v_{\epsilon}\|_{H}\leq C,\label{4.1} \end{equation} and also $v_{\epsilon}(x)$ satisfies $$ -\Delta v_{\epsilon}+V(x_{\epsilon}+\epsilon x)v_{\epsilon}(x) +\phi_{v_{\epsilon}}v_{\epsilon}=f(v_{\epsilon}). $$ Now, for any sequence $\epsilon_{n}\to0$, there is a subsequence such that $$ x_{\epsilon_{n}}\to\bar{x}; V(\bar{x})=V_0. $$ From \eqref{4.1} and elliptic estimates, we know that this subsequence can be chosen in such a way that $v_{\epsilon_{n}}\to v$ uniformly over compacts, where $v\in H^1({\mathbb{R}^{3}})$ solves \begin{equation} -\Delta v+V_0v+\phi_{v}=f(v).\label{4.2}% \end{equation} Next, we prove that $v_{\epsilon_{n}}\to v \in \ H^1({\mathbb{R}^{3}})$. Since $\tilde{f}(s)\leq f(s)$ for all $s\geq0$, by \eqref{3.15} we have $$ I_{n}(v_{\epsilon_{n}})\leq\epsilon_{n}^{-3}J_{\epsilon_{n}}(u_{\epsilon_{n}}) \leq c_0, $$ where \begin{equation} \begin{aligned} I_{n}(u) &=\frac{1}{2}\int_{\Omega_{n}}[|\nabla u|^{2} +V(x_{\epsilon_{n}}+\epsilon_{n}x)u^{2}]dx +\frac{1}{4}\int_{\Omega_{n}}\phi_{u}u^{2}dx \\ &\quad -\int_{\Omega_{n}}F(x_{\epsilon_{n}}+\epsilon_{n}x,u)dx, \Omega_{n}\\ &=\epsilon_{n}^{-1}\{\mathbb{R}^{3}-x_{\epsilon_{n}}\}. \end{aligned} \end{equation} On the other hand, using Fatou's lemma and the weak limit of $v_{\epsilon_{n}}$, \begin{align*} I_{n}(v_{\epsilon_{n}}) &=I_{n}(v_{\epsilon_{n}})-\frac{1}{4}\langle I'_{n}(v_{\epsilon_{n}}),v_{\epsilon_{n}}\rangle \\ &=\frac{1}{4}\int_{\Omega_{n}}[|\nabla v_{\epsilon_{n}}|^{2}+V(x_{\epsilon_{n}}+\epsilon_{n}x)v_{\epsilon_{n}}^{2}]dx +\frac{1}{4}\int_{\Omega_{n}}[f(v_{\epsilon_{n}})v_{\epsilon_{n}}-4F(v_{\epsilon_{n}})]dx \\ &\geq \frac{1}{4}\int_{\Omega_{n}}[|\nabla v_{\epsilon_{n}}|^{2}+V_0v_{\epsilon_{n}}^{2}]dx +\frac{1}{4}\int_{\Omega_{n}}[f(v_{\epsilon_{n}})v_{\epsilon_{n}}-4F(v_{\epsilon_{n}})]dx \\ &\geq \frac{1}{4}\int_{\mathbb{R}^{3}}[|\nabla v|^{2}+V_0v^{2}]dx +\frac{1}{4}\int_{\mathbb{R}^{3}}[f(v)v-4F(v)]dx \\ &=I_0(v)-\frac{1}{4}\langle I'_0(v),v\rangle\geq c_0. \end{align*} So, $I_{n}(v_{\epsilon_{n}})\to c_0$ as $n\to\infty$, and it is easy to verify from the above inequalities, $$ \lim_{n\to\infty}\int_{\mathbb{R}^{3}}(|\nabla v_{\epsilon_{n}}|^{2} +V_0v_{\epsilon_{n}}^{2})dx =\int_{\mathbb{R}^{3}}(|\nabla v|^{2}+V_0v^{2})dx. $$ Therefore, using that $v_{\epsilon_{n}}\rightharpoonup v$ weakly in $ H^1({\mathbb{R}^{3}})$, we conclude $v_{\epsilon_{n}}\to v$ in $H^1({\mathbb{R}^{3}})$. As a consequence of the above limit, we have \begin{equation} \lim_{R\to\infty}\int_{|x|\geq R}v_{\epsilon_{n}}^{2}dx=0.\label{4.3}% \end{equation} Applying proposition \ref{pro4.1} in the inequality $$ -\Delta v_{\epsilon_{n}} \leq -\Delta v_{\epsilon_{n}}+V(\epsilon_{n}x +x_{\epsilon_{n}})v_{\epsilon_{n}} +\phi_{v_{\epsilon_{n}}}v_{\epsilon_{n}}=h_{n}(x)\Delta q f(v_{\epsilon_{n}})\quad\text{in } \mathbb{R}^{3}, $$ we have that for some $t>3, \|h_{n}\|_{\frac{t}{2}}\leq C$ for all $n$. Moreover, $$ \sup_{x\in B_{R}(y)}v_{\epsilon_{n}}(x) \leq C(\|v_{\epsilon_{n}}\|_{L^{2}(B_{2R}(y))} +\|h_{n}\|_{L^{t/2}(B_{2R}(y))})\quad\text{for all } y\in\mathbb{R}^{3}, $$ which implies that $\|v_{\epsilon_{n}}\|_{L^{\infty}(\mathbb{R}^{3})}$ is uniformly bounded. Then by \eqref{4.3}, we have $$ \lim_{|x|\to\infty}v_{\epsilon_{n}}(x)=0\quad\text{uniformly on } n\in\mathbb{N}. $$ Consequently, there exists $\epsilon^{*}>0$ such that $$ \lim_{|x|\to\infty}v_{\epsilon}(x)=0\quad\text{uniformly on } \epsilon\in(0,\epsilon^{*})\,. $$ \end{proof} To show the exponential decay of $u_{\epsilon}$, we only need the following result involving of $v_{\epsilon}$. \begin{lemma}\label{lem4.5} There exist constants $C>0$ and $c>0$ such that $$ v_{\epsilon}(x)\leq Ce^{-c|x|}\quad\text{for all } x\in\mathbb{R}^{3}. $$ \end{lemma} \begin{proof} By lemma \ref{lem4.2} and (A2), there exists $R_{1}>0$ such that \begin{equation} \frac{f(v_{\epsilon}(x))}{v_{\epsilon}(x)} \leq\frac{V_0}{2} \quad\text{for all } |x|\geq R_{1},\; \epsilon\in(0,\epsilon^{*}).\label{4.4} \end{equation} Fix $\omega(x)=Ce^{-c|x|}$ with $c^{2}R_{1}.\label{4.6} \end{equation} Define $\omega_{\epsilon}=\omega-v_{\epsilon}$. Using \eqref{4.6} and \eqref{4.5}, we obtain \[ -\Delta \omega_{\epsilon} +\frac{V_0}{2}\omega_{\epsilon}\geq0,\quad \text{in } |x|\leq R_{1}, \\ \omega_{\epsilon}\geq 0,\quad\text{on } |x|= R_{1},\\ \lim_{|x|\to\infty}\omega_{\epsilon}(x)=0. \] The classical maximum principle implies that $\omega_{\epsilon}\geq0$ in $|x|\geq R_{1}$ and by the work in \cite{gnn}, we conclude that $$ v_{\epsilon}(x)\leq Ce^{-c|x|}\quad\text{for all } |x|\geq R_{1}, \epsilon\in(0,\epsilon^{*}). $$ By the definition of $v_{\epsilon}$ and lemma \ref{lem4.5}, we have $$ u_{\epsilon}(x)=v_{\epsilon}(\frac{x-x_{\epsilon}}{\epsilon}) \leq Cexp(-c\frac{|x-x_{\epsilon}|}{\epsilon}) $$ for all $x\in\mathbb{R}^{3}, \epsilon\in(0,\epsilon^{*})$. The proof of Theorem \ref{thm1.1} is complete. \end{proof} \subsection*{Acknowledgments} This work was supported by Natural Science Foundation of China (11201083) and China Postdoctoral Science Foundation (2013M542038). 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