\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 $f(u)=u^{p}$, $1
0$ 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}