\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 05, pp. 1--18.\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-2016/05\hfil Ground state solutions]
{Existence and concentration of ground state solutions
for a Kirchhoff type problem}
\author[H. Fan \hfil EJDE-2016/05\hfilneg]
{Haining Fan}
\address{Haining Fan \newline
School of Sciences, China University of Mining and Technology,
Xuzhou 221116, China}
\email{fanhaining888@163.com}
\thanks{Submitted July 6, 2015. Published January 4, 2016.}
\subjclass[2010]{35A15, 35B33, 35J62}
\keywords{Nehari-Pohozaev manifold; nonlocal problem; positive solution;
\hfill\break\indent concentration property}
\begin{abstract}
This article concerns the Kirchhoff type problem
\begin{gather*}
-\Big(\varepsilon^2a+\varepsilon b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u
+V(x)u= K(x)|u|^{p-1}u,\quad x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gather*}
where $a,b$ are positive constants, $2< p<5$, $\varepsilon>0$ is a small parameter,
and $V(x),K(x)\in C^1(\mathbb{R}^3)$. Under certain assumptions on the non-constant
potentials $V(x)$ and $K(x)$, we prove the existence and concentration properties of
a positive ground state solution as $\varepsilon\to 0$.
Our main tool is a Nehari-Pohozaev manifold.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks
\section{Introduction}
In this article we study the Kirchhoff type problem
\begin{equation} \label{Pe}
\begin{gathered}
-\Big(\varepsilon^2a+\varepsilon b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u
+V(x)u= K(x)|u|^{p-1}u,\quad x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gathered}
\end{equation}
where $a,b$ are positive constants, $2< p<5$, $\varepsilon>0$ is a small parameter,
$V(x),K(x)\in C^1(\mathbb{R}^3)$. Such problems are often referred as being
nonlocal because of the presence of the term
$\big(\int_{\mathbb{R}^3}|\nabla u|^2dx\big)\Delta u$ which implies that
\eqref{Pe} is no longer a point-wise equation. Problem \eqref{Pe}
is related to the stationary analogue of the equation
\begin{equation} \label{e1.1}
u_{tt}-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u=f(x,u),
\end{equation}
presented by Kirchhoff in \cite{k1} as an extension of classical
D'Alembert's wave equations for free vibration of elastic strings.
Kirchhoff's model takes into account the changes in length of the
string produced by transverse vibrations. In \eqref{e1.1}, $u$ denotes the
displacement, $f(x,u)$ the external force and $b$ the initial tension while
$a$ is related to the intrinsic properties of the string
(such as Young's modulus). We have to point out that nonlocal problems
also appear in other fields such as biological systems, where $u$ describes
a process which depends on the average of itself (for example, population
density, see \cite{a1,c1}).
In recent years, there have been many works concerned with the existence
of solutions to the problems similar to \eqref{Pe} via variational methods,
see e.g. \cite{a2,f1,f2,l1,l4,n1,w5}.
Also, there are some recent works considered
the concentration property of solutions as $\varepsilon\to 0$, see for
instance \cite{h1,l5,w1,w2,w3}
and the references therein. Indeed, a typical way
to deal with \eqref{Pe} is to use the mountain pass theorem. For this purpose,
the most of the above results focused on the nonlinear model $|u|^{p-1}u$ with
$3
0$ such that
\[
\overline{c}=\inf_{u\in H^1(\mathbb{R}^3)\backslash\{0\} }
\frac{\int_{\mathbb{R}^3}(|\nabla u|^2+V(x)u^2)dx}{\int_{\mathbb{R}^3}u^2dx}>0.
\]
\end{itemize}
By using a monotonicity trick and constructing a new version of global compactness
Lemma, they proved that \eqref{Pe} has a positive ground state solution.
More recently, Ye \cite{y1} studied \eqref{Pe} under different conditions.
On one hand, if $1
0$.
\item[(A2'')] $\nabla V(x)\cdot x\leq0$ for all $x\in \mathbb{R}^3$ and
the inequality is strict in a subset of positive Lebesgue measure.
\item[(A3'')] $V(x)+\frac{\nabla V(x)\cdot x}{4}\geq V_\infty$ for all
$x\in\mathbb{R}^3$.
\item[(A4'')] $\nabla V(x)\cdot x+\frac{xH(x)x}{4}\leq0$ for all
$x\in\mathbb{R}^3$, where $H$ denotes the Hessian matrix of $V$.
\item[(A5'')] there exists a constant $T>1$ which is defined in
\cite{y1} such that
\[
\sup_{x\in\mathbb{R}^3}V(x)\leq V_\infty+T.
\]
\end{itemize}
Ye \cite{y1} proved that \eqref{Pe} has a high energy solution.
However, to the best of our knowledge, for the case $2
0$ such that
$\nabla V(x)\equiv0$ for all $|x|\geq R_V$.
\item[(A5)] $K(x)\in C^1(\mathbb{R}^3,\mathbb{R})$ and
$00$ such that $\nabla K(x)\equiv0$ for all
$|x|\geq R_K$.
\end{itemize}
\begin{remark} \label{rmk1.1} \rm
There are many examples of $V$ and $K$ that satisfy the hypotheses above.
For example, define $\eta\in C^\infty(\mathbb{R}^3)$ by
\[
\eta(x):=\begin{cases}
C \exp(\frac{1}{|x|^2-1}), & \text{if } |x|<1,\\
0, & \text{if } |x|>1,
\end{cases}
\]
where $C>0$ is a constant. Then $V(x)=C-\eta(x)$ satisfies $(V_1)-(V_4)$
and $K(x)=\frac{C}{2}+\eta(x)$ satisfies $(K_1)-(K_4)$.
\end{remark}
Clearly, the above assumptions imply that there exists an $\overline{x}\in\Omega_1$
such that $K(\overline{x})\geq K(x)$ for all $|x|\geq R$ and some $R>0$.
Here, we denote
\begin{gather*}
\Omega_1:=\{x\in\mathbb{R}^3;V(x)=V_{\rm min}\},
\Omega_2:=\{x\in\mathbb{R}^3;K(x)=K_{\rm max}:=\max_{x\in\mathbb{R}^3}K(x)\},\\
\mathcal{H}:=\{x\in \Omega_1;K(x)=K(\overline{x})\}\cup\{x\not\in \Omega_1;K(x)>K(\overline{x})\}.
\end{gather*}
\begin{remark} \label{rmk1.2} \rm
Obviously, $\mathcal{H}\neq\emptyset$ because $\overline{x}\in \mathcal{H}$.
It is clear that $ \mathcal{H}=\Omega_1\cap\Omega_2$ when
$\Omega_1\cap\Omega_2\neq\emptyset$. For example, let $V(x)=C-\eta(x)$ and
$K(x)=\frac{C}{2}+\eta(x)$ as in Remark \ref{rmk1.1},
then $\Omega_1=\{0\}, \Omega_2=\{0\}$ and $\mathcal{H}=\{0\}$.
If we set $V(x)=C-\eta(x-x_0)$ and $K(x)=\frac{C}{2}+\eta(x)$ and $x_0\neq0$,
we can easily see that $\Omega_1=\{x_0\}, \Omega_2=\{0\}$ and
$\Omega_1\cap\Omega_2=\emptyset$. We obtain that $\mathcal{H}=\{x;|x|\leq|x_0|\}$.
\end{remark}
The main result of this article reads as follows.
\begin{theorem} \label{thm1.1}
\begin{itemize}
\item[(I)] Assume {\rm (A1)--(A3), (A5)--(A7} hold. Then \eqref{Pe} possesses
a positive ground state solution $u_\varepsilon$ for all $\varepsilon>0$.
\item[(II)] Suppose {\rm (A1), (A3), (A4), (A5), (A7), (A8)} are satisfied. Then
\begin{enumerate}
\item $u_\varepsilon$ possesses one maximum point $x_\varepsilon$ such that,
up to a subsequence, $x_\varepsilon\to x_0$ as $\varepsilon\to 0$,
$\lim_{\varepsilon\to 0}\operatorname{dist}(x_\varepsilon, \mathcal{H})=0$,
$\omega_\varepsilon(x):=u_\varepsilon(\varepsilon x+x_\varepsilon)$
converges in $H^1(\mathbb{R}^3)$ to a positive ground state solution of
\[
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u+V(x_0)u= K(x_0)|u|^{p-1}u,
\quad x\in \mathbb{R}^3.
\]
In particular, if $\Omega_1\cap\Omega_2\neq\emptyset$, then
$\lim_{\varepsilon\to 0}\operatorname{dist}(x_\varepsilon, \Omega_1\cap\Omega_2)=0$ and
$\omega_\varepsilon$ converges in $H^1(\mathbb{R}^3)$ to a positive ground
state solution of
\[
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u+V_{\rm min}u
= K_{\rm max}|u|^{p-1}u,~ x\in \mathbb{R}^3.
\]
\item There exist $C_1,C_2>0$ such that
\[
u_\varepsilon(x)\leq C_1e^{-C_2|\frac{x-x_\varepsilon}{\varepsilon}|}.
\]
\end{enumerate}
\end{itemize}
\end{theorem}
\begin{remark} \label{rmk} \rm
Note that (A1) and (A4) imply (A2). Also (A5) and (A8) imply (A6).
\end{remark}
This article is organized as follows. In Section 2, we establish some
preliminary results. Section 3 is to prove the existence of ground states.
Section 4 is devoted to the proof of Theorem \ref{thm1.1}. Throughout this paper
we denote by $\to $ (resp. $\rightharpoonup$) the strong (resp. weak) convergence.
The letters $C,C_1,C_2,\dots $ will be repeatedly used to denote various
positive constants whose exact values are irrelevant.
\section{Preliminaries}
Throughout this article by $|\cdot|_r$ we denote the $L^r$-norm.
On the space $H^1(\mathbb{R}^3)$ we consider the norm
\[
\|u\|=\Big(\int_{\mathbb{R}^3}(|\nabla u|^2+u^2)dx\Big)^{1/2}.
\]
Without loss of generality, we may assume that $\varepsilon=1$,
then \eqref{Pe} becomes
\begin{equation} \label{P1}
\begin{gathered}
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u+V(x)u
= K(x)|u|^{p-1}u,~ x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gathered}
\end{equation}
At this step, we see that \eqref{P1} is variational and its weak solutions
are the critical points of the functional given by
\begin{align*}
J(u)&=\frac{a}{2}\int_{\mathbb{R}^3}|\nabla u|^2dx
+\frac{1}{2}\int_{\mathbb{R}^3}V(x)u^2dx
+\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2\\
&\quad -\frac{1}{p+1}\int_{\mathbb{R}^3}K(x)|u|^{p+1}dx.
\end{align*}
For $20$ such that $f'_u(t)=0$, $f_u(\cdot)$ is increasing for $(0,t_u)$
and decreasing for $(t_u,\infty)$. That is, there is a unique $t_u$
such that $u_{t_u}\in M$.
\end{lemma}
\begin{proof}
By making the change of variable $s=t^{4+p}$, we obtain
\begin{align*}
f_u(s)
&=\frac{a}{2}s^\frac{3}{4+p}\int_{\mathbb{R}^3}|\nabla u|^2dx
+\frac{s^\frac{5}{4+p}}{2}\int_{\mathbb{R}^3}V(s^\frac{1}{4+p}x)u^2dx\\
&\quad +\frac{bs^\frac{6}{4+p}}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2
-\frac{s}{p+1}\int_{\mathbb{R}^3}K(s^\frac{1}{4+p}x)|u|^{p+1}dx.
\end{align*}
By (A3) and (A7), $f_u(s)$ is a concave function. We already know that attains
its maximum. Let $t_u$ be the unique point at which this maximum is achieved.
Then $t_u$ is the unique critical point of $f_u$ and $f_u(t_u)$ is positive and
$f_u(\cdot)$ is increasing for $0t_u$.
In particular, for any $u\in H^1(\mathbb{R}^3)\backslash\{0\}$,
$t_u\in\mathbb{R}$ is the unique value such that $u_{t_u}$ belongs to $M$,
and $J(u_t)$ reaches global maximum for $t=t_u$. This completes the proof.
\end{proof}
Set
\[
m:=\inf_{u\in M}J(u), \quad
m^*:=\inf_{u\in H^1(\mathbb{R}^3)\backslash\{0\}}\max_{t>0}J(u_t).
\]
By Lemma \ref{lem2.2}, we have $m=m^*\geq0$.
\begin{lemma} \label{lem2.3}
There holds $m>0$.
\end{lemma}
\begin{proof}
Let us define
\begin{align*}
\overline{J}(u)
&=\frac{a}{2}\int_{\mathbb{R}^3}|\nabla u|^2dx
+\frac{1}{2}\int_{\mathbb{R}^3}V_{\rm min}u^2dx
+\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2\\
&\quad -\frac{1}{p+1}\int_{\mathbb{R}^3}K_{\rm max}|u|^{p+1}dx.
\end{align*}
Obviously, $\overline{J}(u)\leq J(u)$, and this implies that
\[
\overline{m}:=\inf_{u\in H^1(\mathbb{R}^3)\backslash\{0\}}
\max_{t>0}\overline{J}(u_t)\leq \inf_{u\in H^1(\mathbb{R}^3)
\backslash\{0\}}\max_{t>0}J(u_t)=m.
\]
It suffices to show that $\overline{m}>0$. Define
\[
\overline{M}:=\{u\in H^1(\mathbb{R}^3)\backslash\{0\};g'_u(1)=0\},
\]
where $g_u(t)=\overline{J}(u_t)$. For any $u\in \overline{M}$,
\[
C\|u\|_{H^1}^2\leq\frac{3a}{2}\int_{\mathbb{R}^3}|\nabla u|^2dx
+\frac{5}{2}\int_{\mathbb{R}^3}V_{\rm min}u^2dx
\leq\frac{4+p}{p+1}\int_{\mathbb{R}^3}K_{\rm max}|u|^{p+1}dx
\leq C\|u\|_{H^1}^{p+1}.
\]
Thus we obtain
$C\leq\|u\|_{H^1}^{p-1}$.
Consequently,
\begin{align*}
\overline{J}(u)
&=\overline{J}(u)-\frac{1}{p+4}g'_u(1)\\
&=\frac{(p+1)a}{2(p+4)}\int_{\mathbb{R}^3}|\nabla u|^2dx
+\frac{p-1}{2(p+4)}\int_{\mathbb{R}^3}V_{\rm min}u^2dx
+\frac{(p-2)b}{4(p+4)}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2\\
&\geq C\|u\|_{H^1}^2\geq C>0.
\end{align*}
\end{proof}
\begin{lemma} \label{lem2.4}
There exists $C>0$ such that for any $u\in M$,
\[
J(u)\geq C\|u\|_{H^1}^2.
\]
\end{lemma}
\begin{proof}
Fix $t\in(0,1)$. Then there exist $\delta,\gamma>0$ such that
\begin{gather*}
V(tx)\geq V_{\rm min}\geq\delta V_\infty\geq\delta V(x), \\
K(tx)\leq K_{\rm max}\leq\gamma K_\infty\leq\gamma K(x)
\end{gather*}
for all $x\in\mathbb{R}^3$.
For $u\in M$, we compute
\begin{align*}
&J(u_t)-t^{\lambda+4}J(u)\\
&=\Big(\frac{t^3}{2}-\frac{t^{\lambda+4}}{2}\Big)a\int_{\mathbb{R}^3}|\nabla u|^2dx
+\Big(\frac{t^6}{4}-\frac{t^{\lambda+4}}{4}\Big)
b\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2\\
&\quad +\int_{\mathbb{R}^3}
\Big(\frac{t^5}{2}V(tx)-\frac{t^{\lambda+4}}{2}V(x)\Big)u^2dx
+\int_{\mathbb{R}^3}\Big(\frac{t^{\lambda+4}}{p+1}K(x)-\frac{t^{p+4}}{p+1}
\Big)|u|^{p+1}dx,
\end{align*}
where $2<\lambda0$ such that
\begin{gather*}
\frac{t^5}{2}V(tx)-\frac{t^{\lambda+4}}{2}V(x)
\geq\Big(\delta\frac{t^5}{2}-\frac{t^{\lambda+4}}{2}\Big)V(x)\geq\varepsilon_0,
\\
\frac{t^{\lambda+4}}{p+1}K(x)-\frac{t^{p+4}}{p+1}K(tx)
\geq\Big(t^{\lambda+4}-\gamma t^{p+4}\Big)\frac{K(x)}{p+1}\geq 0.
\end{gather*}
From these two inequalities and Lemma \ref{lem2.2}, taking a smaller $\varepsilon_0>0$
if necessary, we obtain
\[
(1-t^{\lambda+4})J(u)\geq J(u_t)-t^{\lambda+4}J(u)\geq\varepsilon_0\|u\|_{H^1}^2.
\]
Taking $C=\varepsilon_0/(1-t^{\lambda+4})$, we complete the proof.
\end{proof}
\section{Existence result}
In this section, we combine the Nehari-Pohozaev manifold with the concentration
compactness principle to prove the existence of a ground state solution
for \eqref{P1}. Initially, we give the following concentration-compactness
principle.
\begin{lemma}[{\cite[Lemma 1.1]{b1}}] \label{lem3.1}
Let $\{\rho_n\}$ be a sequence of nonnegative $L^1$ functions on
$\mathbb{R}^N$ satisfying $\lim_{n\to \infty}\int_{\mathbb{R}^N}\rho_ndx=c_0>0$.
There exists a subsequence, still denoted by $\{\rho_n\}$ satisfying one of
the following three possibilities:
\begin{itemize}
\item[(i)] (Vanishing) for all $R>0$,
\[
\lim_{n\to \infty}\sup_{y\in\mathbb{R}^N}\int_{B_R(y_n)}\rho_ndx=0;
\]
\item[(ii)] (compactness) there exists $\{y_n\}\subset \mathbb{R}^N$ such that,
for any $\varepsilon>0$, there exists an $R>0$ satisfying
\[
\lim_{n\to \infty}\inf\int_{B_R(y_n)}\rho_ndx\geq c_0-\varepsilon;
\]
\item[(iii)] (Dichotomy) there exists an $\alpha\in(0,c_0)$ and
$\{y_n\}\subset \mathbb{R}^N$ such that, for any $\varepsilon>0$,
there exists an $R>0$, for all $r\geq R$ and $r'\geq R$,
\[
\lim_{n\to \infty}\sup\Big(\big|\alpha-\int_{B_r{y_n}}\rho_n\,dx\big|
+\big|(c_0-\alpha)-\int_{\mathbb{R}^N\backslash B_{r'}({y_n})}\rho_n\,dx\big|
\Big)<\varepsilon;
\]
\end{itemize}
\end{lemma}
\begin{lemma}[{\cite[Lemma 1.21]{w4}}] \label{lem3.2}
Let $r>0$ and $2\leq q<2^*$. If $\{u_n\}$ is bounded in $H^1(\mathbb{R}^N)$ and
\[
\sup_{y\in\mathbb{R}^N}\int_{B_r(y)}|u_n|^qdx\to 0,~\text{as}~n\to +\infty,
\]
then $u_n\to 0$ in $L^s(\mathbb{R}^N)$ for $20$, there exists
an $R>0$ satisfying
\[
\int_{\mathbb{R}^3\backslash B_R(y_n)}(|\nabla u_n|^2+|u_n|^2)dx\leq\varepsilon.
\]
\end{lemma}
\begin{proof}
First, we claim that $\int_{\mathbb{R}^3}|u_n|^{p+1}dx\nrightarrow0$, as
$n\to \infty$. Indeed, since $m>0$, it is easy to obtain that
$\|u_n\|_{H^1}\nrightarrow0$ by the Sobolev embedding theorem.
By Lemma \ref{lem2.2}, for any $t>1$,
\begin{align}
m\leftarrow &J(u_n)\geq J((u_n)_t)
\\
&=\frac{at^3}{2}\int_{\mathbb{R}^3}|\nabla u_n|^2dx
+\frac{t^5}{2}\int_{\mathbb{R}^3}V(x)u_n^2dx
+\frac{bt^6}{4}\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2\\
&\quad -\frac{t^{4+p}}{p+1}\int_{\mathbb{R}^3}K(tx)|u_n|^{p+1}dx \\
&\geq\frac{t^3}{2}\int_{\mathbb{R}^3}(a|\nabla u_n|^2
+V_{\rm min}u_n^2)dx-\frac{t^{p+4}}{p+1}K_{\rm max}\int_{\mathbb{R}^3}|u_n|^{p+1}dx\\
&\geq\frac{t^3}{2}\sigma-\frac{t^{p+4}}{p+1}K_{\rm max}
\int_{\mathbb{R}^3}|u_n|^{p+1}dx,
\end{align}
where $\sigma$ is a fixed constant. It suffices to take $t>1$ so that
$\frac{t^3\sigma}{2}>2m$ to get a lower bound for
$\int_{\mathbb{R}^3}|u_n|^{p+1}dx$.
Let us assume that
\begin{equation} \label{e3.1}
\lim_{n\to \infty}\int_{\mathbb{R}^3}|u_n|^{p+1}dx\to A\in(0,+\infty).
\end{equation}
By Lemma \ref{lem3.2}, we obtain that there exist $\delta>0$ and
$\{x_n\}\subset\mathbb{R}^3$ such that
\[
\int_{B(x_n)}|u_n|^{p+1}dx>\delta>0.
\]
Take $R>\max\{1,\varepsilon^{-1}\}$, $\phi_R(t)$ a smooth function such that
\begin{itemize}
\item $\phi_R(t)=1$ for $0\leq t\leq R$.
\item $\phi_R(t)=0$ for $t\geq 2R$.
\item $\phi'_R(t)\leq 2/R$.
\end{itemize}
Write
\[
u_n(x)=\phi_R(|x-x_n|)u_n(x)+(1-\phi_R(|x-x_n|))u_n(x):=v_n+\omega_n.
\]
Then
\begin{equation} \label{e3.2}
\lim_{n\to \infty}\int_{B_R(x_n)}|v_n|^{p+1}dx\geq\delta.
\end{equation}
To complete the proof, we only need to prove that there exist constants
$C>0$ independent of $\varepsilon$ and $n_0=n_0(\varepsilon)$ such that
$\|\omega_n\|_{H^1}\leq C\varepsilon$ for all $n\geq n_0$.
Define $z_n=u_n(\cdot+x_n)$, and then $z_n\rightharpoonup z$ weakly in
$H^1(\mathbb{R}^3)$. By taking a larger $R$, if necessary, we can assume
that $\int_{A_0(R,2R)}|z|^{p+1}dx<\varepsilon$, where $A_0(R,2R)$
denotes the annulus centered in $0$ with radii $R$ and $2R$. Then, for $n$ large
enough, we have
\begin{equation} \label{e3.3}
\Big|\int_{\mathbb{R}^3}K(tx)(|u_n|^{p+1}-|v_n|^{p+1}
-|\omega_n|^{p+1})dx\Big|\leq C\varepsilon.
\end{equation}
Since $|\nabla z_n|^2$ is uniformly bounded in $L^1(\mathbb{R}^3)$,
up to a subsequence, $|\nabla z_n|^2$ converges (in the sense of measure)
to a certain positive measure $\mu$ with $\mu(\mathbb{R}^3)<+\infty$.
By enlarging $R$ necessary, we can assume that $\mu(A_0(R,2R))<\varepsilon$.
Then, for $n$ large enough,
\[
\int_{\mathbb{R}^3}|\nabla u_n|^2\phi_R(|x-x_n|)(1-\phi_R(|x-x_n|))dx<\varepsilon.
\]
Taking this into account, direct calculations show that for $n$ large enough,
\begin{equation}
\Big|\int_{\mathbb{R}^3}|\nabla u_n|^2dx
-\int_{\mathbb{R}^3}|\nabla v_n|^2dx
-\int_{\mathbb{R}^3}|\nabla \omega_n|^2dx\Big|
=\Big|2\int_{\mathbb{R}^3}\nabla v_n\nabla \omega_ndx\Big|\leq C\varepsilon,
\end{equation}
and thus
\begin{align*}
&\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2 \\
&=\Big(\int_{\mathbb{R}^3}|\nabla v_n|^2dx
+\int_{\mathbb{R}^3}|\nabla \omega_n|^2dx
+C\varepsilon\Big)^2 \\
&=\Big(\int_{\mathbb{R}^3}|\nabla v_n|^2dx\Big)^2
+\Big(\int_{\mathbb{R}^3}|\nabla \omega_n|^2dx\Big)^2
+2\int_{\mathbb{R}^3}|\nabla v_n|^2dx
\int_{\mathbb{R}^3}|\nabla \omega_n|^2dx+C\varepsilon\\
&\geq\Big(\int_{\mathbb{R}^3}|\nabla v_n|^2dx\Big)^2
+\Big(\int_{\mathbb{R}^3}|\nabla \omega_n|^2dx\Big)^2+C\varepsilon.
\end{align*}
Arguing as before, for $R$ large enough, we obtain
\begin{equation} \label{e3.6}
\Big|\int_{\mathbb{R}^3}V(tx) u_n^2dx-\int_{\mathbb{R}^3}V(tx) v_n^2dx
-\int_{\mathbb{R}^3}V(tx) \omega_n^2dx\Big|\leq C\varepsilon.
\end{equation}
Putting together \eqref{e3.3}-\eqref{e3.6} we obtain that for $n$ sufficient
large and $t>0$,
\begin{equation} \label{e3.7}
J((u_n)_t)\geq J((v_n)_t)+J((\omega_n)_t)-C\varepsilon.
\end{equation}
Now let us denote with $t_{v_n}$ and $t_{\omega_n}$ the positive values
which maximize $f_{v_n}(t)$ and $f_{\omega_n}(t)$ respectively, namely,
\[
J((v_n)_{t_{v_n}})=\max_{t>0}J((v_n)_t)~\text{and}~J((\omega_n)_{t_{\omega_n}})
=\max_{t>0}J((\omega_n)_t).
\]
Let us assume that $t_{v_n}\leq t_{\omega_n}$(the other case will be treated later).
Then
\[
J((\omega_n)_t)\geq0~\text{for}~t\leq t_{v_n}.
\]
We claim that there exist $0<\widetilde{t}<1<\overline{t}$ independent
of $\varepsilon$ such that $t_{v_n}\in(\widetilde{t},\overline{t})$.
Indeed, take $\overline{t}=(2(p+1)(K_{\rm max}A)^{-1}B)^\frac{1}{p-2}$,
where $A$ comes from \eqref{e3.1} and $B$ is large enough such that
$\overline{t}>1$ and moreover,
\[
B\geq a\int_{\mathbb{R}^3}|\nabla u_n|^2dx
+\int_{\mathbb{R}^3}V_\infty|u_n|^2dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2.
\]
Then
\begin{align*}
J((u_n)_{\overline{t}})
&\leq\frac{\overline{t}^6}{2}\bigg(a\int_{\mathbb{R}^3}|\nabla u_n|^2dx
+\int_{\mathbb{R}^3}V_\infty|u_n|^2dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2\\
&\quad - \frac{\overline{t}^{p-2}}{p+1}\int_{\mathbb{R}^3}K_{\rm max}|u_n|^{p+1}dx\bigg)
\\
&\leq-B\frac{\overline{t}^6}{2}<0.
\end{align*}
Taking a smaller $\varepsilon$ in \eqref{e3.7}, we obtain
\[
J((v_n)_{\overline{t}})+J((\omega_n)_{\overline{t}})<0.
\]
Then $J((v_n)_{\overline{t}})<0$ or $J((\omega_n)_{\overline{t}})<0$.
In any case, Lemma \ref{lem2.2} implies that $t_{v_n}<\overline{t}$
(recall that we are assuming $t_{v_n}\leq t_{\omega_n}$).
For the lower bound, take $\widetilde{t}=\left(\frac{m}{B}\right)^\frac{1}{3}$.
Let us point out that $\widetilde{t}<1$. For any $t\leq\widetilde{t}$,
\[
J((u_n)_t)\leq \frac{\widetilde{t}^3}{2}
\Big(a\int_{\mathbb{R}^3}|\nabla u_n|^2dx
+\int_{\mathbb{R}^3}V_\infty|u_n|^2dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2\Big)\leq\frac{m}{2}.
\]
Since
\begin{equation} \label{e3.8}
m\leftarrow J(u_n)\geq J((u_n)_{t_{v_n}})\geq J((v_n)_{t_{v_n}})
+J((\omega_n)_{t_{v_n}})-c\varepsilon\geq m-C\varepsilon
\end{equation}
and the right hand side can be made greater than $\frac{m}{2}$ by choosing
a small $\varepsilon$, we conclude that $t_{v_n}>\widetilde{t}$ and the
claim is proved.
Using \eqref{e3.8} we deduce, for $n$ large, $J((\omega_n)_t)\leq 2C\varepsilon$
for all $t\in(0,t_{v_n})$. Moreover, for any $t\in(0,\widetilde{t})$, we have
\begin{align*}
2C\varepsilon
&\geq J((\omega_n)_t)\\
&\geq\frac{t^6}{4}\Big(a\int_{\mathbb{R}^3}|\nabla \omega_n|^2dx
+\int_{\mathbb{R}^3}V_{\rm min}\omega_n^2dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2\Big)\\
&\quad -\frac{t^{p+4}}{p+1}\int_{\mathbb{R}^3}K_{\rm max}|\omega_n|^{p+1}dx\\
&\geq\frac{t^6}{4}q_n-Dt^{p+4},
\end{align*}
where
\[
q_n=a\int_{\mathbb{R}^3}|\nabla \omega_n|^2dx
+\int_{\mathbb{R}^3}V_{\rm min}\omega_n^2dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2
\]
and $D>A$. Observe that
\[
\frac{t^6}{4}q_n-Dt^{p+4}=\frac{(p+2)D}{2}
\Big(\frac{q_n}{2(p+4)D}\Big)^{p+4}
\quad \text{for } t=\Big(\frac{q_n}{2(p+4)D}\Big)^\frac{1}{p-2}.
\]
By taking a large $D$ we can assume that
$\big(\frac{q_n}{2(p+4)D}\big)^\frac{1}{p-2}\leq\widetilde{t}$.
With this choice of $t$, we obtain
\[
2C\varepsilon\geq J((\omega_n)_t)
\geq\frac{(p+2)D}{2}\Big(\frac{q_n}{2(p+4)D}\Big)^{p+4}\geq Cq_n^{p+4}.
\]
Thus we have
\begin{equation} \label{e3.9}
\|\omega_n\|_{H^1}\leq C\varepsilon\quad \text{for some } C>0.
\end{equation}
In the case $t_{v_n}>t_{\omega_n}$, we can assume analogously to conclude
that $\|v_n\|_{H^1}\leq C\varepsilon$ for some $C>0$. But, choosing small
$\varepsilon$, this contradicts \eqref{e3.2}, so \eqref{e3.9} holds.
This completes the proof.
\end{proof}
\begin{lemma} \label{lem3.4}
The value $m$ is achieved at some $u\in M$.
\end{lemma}
\begin{proof}
Recall that $z_n\rightharpoonup z$ in $H^1(\mathbb{R}^3)$, we have
$z_n\to z$ in $L_{loc}^q(\mathbb{R}^3)$ for $10.
\]
Therefore, $\max_{t\geq0}J(u_t)=m$ and $u_n\to u$ in $H^1(\mathbb{R}^3)$.
In particular, $u\in M$ is a minimizer of $J|_M$.
\smallskip
\noindent\textbf{Case 2:} $\{x_n\}$ is unbounded.
In this case, by Lebesgue convergence Theorem and (A1), we have
\begin{align*}
\lim_{n\to \infty} \int_{\mathbb{R}^3}V(tx)(u_n(x))_t^2dx
&=\lim_{n\to \infty} \int_{\mathbb{R}^3}V(t(x+x_n))(z_n(x))_t^2dx\\
&=V_\infty\int_{\mathbb{R}^3}z_t^2dx\geq\int_{\mathbb{R}^3}V(tx)z_t^2dx\\
&=\lim_{n\to \infty} \int_{\mathbb{R}^3}V(tx))(z_n(x))_t^2dx
\end{align*}
for any $t>0$ fixed. Moreover,
\begin{align*}
\lim_{n\to \infty} \int_{\mathbb{R}^3}K(tx)|u_n(x))_t|^{p+1}dx
&=\lim_{n\to \infty} \int_{\mathbb{R}^3}K(t(x+x_n))|(z_n(x))_t|^{p+1}dx\\
&=K_\infty\int_{\mathbb{R}^3}|z_t|^{p+1}dx\\
&\leq\lim_{n\to \infty} \int_{\mathbb{R}^3}K(tx))|(z_n(x))_t|^{p+1}dx
\end{align*}
for any $t>0$ fixed. Therefore,
\[
m=\lim_{n\to \infty}J(u_n)\geq\liminf_{n\to \infty}J((z_n)_t)
\geq J(z_t),\quad \text{for any }t>0.
\]
So, taking $t_z$ so that $f_z(t)=J(z_t)$ reaches its maximum,
we obtain that $z_{t_z}\in M$ and is a minimizer for $J|_M$.
\end{proof}
\begin{theorem} \label{thm3.1}
The minimizer $u$ of $J|_M$ is a positive ground state solution of \eqref{P1}.
\end{theorem}
\begin{proof}
Let $u\in M$ be a minimizer of the functional $J|_M$. We will prove that
$u$ is a positive ground state solution of $(P)$ in the following.
Recall that, by Lemma \ref{lem2.2},
\[
J(u)=\inf_{u\in H^1(\mathbb{R}^3)\backslash\{0\}}\max_{t>0}J(u_t)=m.
\]
We argue by contradiction. Suppose that $u$ is not a weak solution of \eqref{P1}.
Then we can choose $\phi\in C_0^\infty(\mathbb{R}^3)$ such that
\begin{align*}
\langle J'(u),\phi\rangle
&=a\int_{\mathbb{R}^3}\nabla u\nabla\phi dx
+\int_{\mathbb{R}^3}V(x)u\phi dx
+b\int_{\mathbb{R}^3}|\nabla u|^2dx\int_{\mathbb{R}^3}\nabla u\nabla\phi dx\\
&\quad -\int_{\mathbb{R}^3}K(x)|u|^{p-1}u\phi dx<-1.
\end{align*}
We fix $\varepsilon>0$ sufficiently small such that
\[
\langle J'(u_t+\sigma\phi),\phi\rangle\leq-\frac{1}{2},\quad
\forall |t-1|,|\sigma|\leq\varepsilon.
\]
and introduce a cutoff function $0\leq\eta\leq1$ such that $\eta(t)=1$
for $|t-1|\leq\frac{\varepsilon}{2}$ and $\eta(t)=0$ for
$|t-1|\geq\varepsilon$. Set
\[
\gamma(t)=\begin{cases}
u_t, & \text{if } |t-1|\geq\varepsilon,\\
u_t+\varepsilon\eta(t)\phi, &\text{if } |t-1|<\varepsilon.
\end{cases}
\]
Note that $\gamma(t)$ is a continuous curve in $H^1(\mathbb{R}^3)$ and,
eventually choosing a smaller $\varepsilon$, we obtain that
$\|\gamma(t)\|_{H^1}>0$ for $|t-1|<\varepsilon$.
We claim $\sup_{t\geq0}J(\gamma(t))0$ and $G(\gamma(1+\varepsilon))<0$,
there exists $t_0\in (1-\varepsilon,1+\varepsilon)$ such that
$G(\gamma(t_0))=0$, i.e., $\gamma(t_0)=u_{t_0}+\varepsilon\eta(t_0)\phi\in M$
and $J(\gamma(t_0))0$.
\end{proof}
\section{Concentration behavior}
In this section, we study the concentration behavior of the ground state
solutions $u_\varepsilon$ as $\varepsilon\to 0$. From now on, we assume
(A1), (A3), (A4), (A5), (A7), (A8) are satisfied. Introducing the
re-scaled transformation $x\mapsto\varepsilon x$ we can rewrite \eqref{Pe} as
\begin{equation} \label{PPe}
\begin{gathered}
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u
+V(\varepsilon x)u= K(\varepsilon x)|u|^{p-1}u,~ x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gathered}
\end{equation}
Let
\begin{align*}
J_\varepsilon(u)
&=\frac{a}{2}\int_{\mathbb{R}^3}|\nabla u|^2dx
+\frac{1}{2}\int_{\mathbb{R}^3}V(\varepsilon x)u^2dx
+\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2\\
&\quad -\frac{1}{p+1}\int_{\mathbb{R}^3}K(\varepsilon x)|u|^{p+1}dx
\end{align*}
be the associated energy functional, $P_\varepsilon(u)$,
\[
M_\varepsilon:=\{u\in H^1(\mathbb{R}^3);G_\varepsilon(u)
=P_\varepsilon(u)+\langle J'_\varepsilon(u),u\rangle=0\}
\]
and $m_\varepsilon=\inf_{u\in M_\varepsilon}J_\varepsilon(u)$
be the corresponding Pohozaev identity, the Nehari-Pohozaev manifold
and the least energy, respectively. We need the following constant
coefficients problem
\begin{equation} \label{Plam}
\begin{gathered}
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u+\lambda u
= \mu|u|^{p-1}u,\quad x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gathered}
\end{equation}
where $\lambda,\mu>0$. In the same way, we use the notations
$J_{\lambda\mu},P_{\lambda\mu},M_{\lambda\mu}, G_{\lambda\mu}$ and
$m_{\lambda\mu}$. In a similar way to Section 3, there exists
some $u\in M_{\lambda\mu}$ such that $J_{\lambda\mu}(u)=m_{\lambda\mu}$.
\begin{lemma} \label{lem4.1}
Suppose $\lambda_1\geq\lambda_2$ and $\mu_2\geq\mu_1$. Then
$m_{\lambda_1\mu_1}\geq m_{\lambda_2\mu_2}$ is achieved at some $u\in M$.
\end{lemma}
\begin{proof}
Let $u\in M_{\lambda_1\mu_1}$ be such that
$m_{\lambda_1\mu_1}=J_{\lambda_1\mu_1}(u)=\max_{t>0}J_{\lambda_1\mu_1}(u_t)$.
Then there exists a unique $t_{\lambda_2\mu_2}$ such that
$u_{t_{\lambda_2\mu_2}}\in M_{\lambda_2\mu_2}$, and hence
\begin{align*}
m_{\lambda_1\mu_1}
&=J_{\lambda_1\mu_1}(u)\\
&\geq J_{\lambda_1\mu_1}(u_{t_{\lambda_2\mu_2}})\\
&=J_{\lambda_2\mu_2}(u_{t_{\lambda_2\mu_2}})
+\frac{(\lambda_1-\lambda_2)(t_{\lambda_2\mu_2})^5}{2}
\int_{\mathbb{R}^3}|u_{t_{\lambda_2\mu_2}}|^2dx\\
&\quad +\frac{(\mu_1-\mu_2)(t_{\lambda_2\mu_2})^{p+4}}{p+1}
\int_{\mathbb{R}^3}|u_{t_{\lambda_2\mu_2}}|^{p+1}dx\\
&\geq m_{\lambda_1\mu_1}.
\end{align*}
\end{proof}
Without loss of generality, up to translation, we assume that
\[
K(\overline{x})=\max_{x\in\Omega_1}K(x)\quad \text{and}\quad
\overline{x}=0\in\Omega_1.
\]
Thus
\[
V(0)=V_{\rm min}\quad \text{and}\quad
k:=K(0)\geq K(x)\quad \text{for all } |x|\geq R.
\]
\begin{lemma} \label{lem4.2}
There exists $C>0$ independent of $\varepsilon$ such that $m_\varepsilon\geq C$.
On the other hand, $\limsup_{\varepsilon\to 0}m_\varepsilon\leq m_{V_{\rm min}k}$.
\end{lemma}
\begin{proof}
Since $m_\varepsilon\geq m_{V_{\rm min}K_{\rm max}}>0$, we only need to
prove the second part. Take $u\in M_{V_{\rm min}k}$ satisfying
$J_{V_{\rm min}k}(u)=m_{V_{\rm min}k}$.
By Lemma \ref{lem2.2}, we know that there is a unique $t_\varepsilon>0$
such that $u_{t_\varepsilon}\in M_\varepsilon$ and
\begin{equation} \label{e4.1}
\begin{aligned}
m_\varepsilon
&\leq \max_{t>0}J_\varepsilon(u_t)\\
&=\frac{at_\varepsilon^3}{2}\int_{\mathbb{R}^3}|\nabla u|^2dx
+\frac{t_\varepsilon^5}{2}\int_{\mathbb{R}^3}V(t_\varepsilon\varepsilon x)u^2dx\\
&\quad +\frac{bt_\varepsilon^6}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2
-\frac{t_\varepsilon^{4+p}}{p+1}\int_{\mathbb{R}^3}
K(t_\varepsilon\varepsilon x)|u|^{p+1}dx.
\end{aligned}
\end{equation}
This combining with $m_\varepsilon>0$, we have $\{t_\varepsilon\}$ is bounded
with respect to $\varepsilon$. For each $\varepsilon>0$, there exists an $R>0$
such that
\begin{equation}
\big|\int_{|x|>R}(V(t_\varepsilon\varepsilon x)-V_{\rm min})u^2dx\big|<\varepsilon.
\end{equation}
Since $0\in\Omega_1$, we obtain
\begin{equation}
\lim_{\varepsilon\to 0}\big|\int_{|x|\leq R}(V(t_\varepsilon\varepsilon x)
-V_{\rm min})u^2dx\big|=0.
\end{equation}
Similarly, there holds
\begin{equation} \label{e4.4}
\lim_{\varepsilon\to 0}\int_{\mathbb{R}^3}(K(t_\varepsilon\varepsilon x)-k)|u|^{p+1}dx=0.
\end{equation}
From \eqref{e4.1}-\eqref{e4.4}, we can draw the conclusion that
\[
m_\varepsilon\leq J_\varepsilon(u_{t_\varepsilon})=J_{V_{\rm min}k}(u_{t_\varepsilon})+o(1)\leq J_{V_{\rm min}k}(u)+o(1)=m_{V_{\rm min}k}+o(1).
\]
Thus
\[
\limsup_{\varepsilon\to 0}m_\varepsilon\leq m_{V_{\rm min}k}.
\]
\end{proof}
Let $v_\varepsilon$ be the ground state solution of \eqref{PPe}.
\begin{lemma} \label{lem4.3}
There exists $\varepsilon^*>0$ such that, for all $\varepsilon\in(0,\varepsilon^*)$,
there exist $y_\varepsilon\in\mathbb{R}^3$ and $R,C>0$ such that
\[
\int_{B_R(y_\varepsilon)}v_\varepsilon^2dx>C.
\]
\end{lemma}
\begin{proof}
Suppose by contradiction that there is a sequence $\varepsilon_n\to 0$ as
$n\to \infty$ such that for all $R>0$,
\[
\lim_{\varepsilon\to 0}\sup_{y\in\mathbb{R}^3}\int_{B_R(y_\varepsilon)}
v_\varepsilon^2dx=0.
\]
From Lemma \ref{lem3.2}, we can deduce that $v_{\varepsilon_n}\to 0$ in
$L^q(\mathbb{R}^3)$ for $q\in(2,6)$. Since
\begin{align*}
m_{\varepsilon_n}
&=J_{\varepsilon_n}(v_{\varepsilon_n})
-\frac{1}{2}\langle J'_{\varepsilon_n}(v_{\varepsilon_n}),v_{\varepsilon_n}\rangle
\\
&=-\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla v_{\varepsilon_n}|^2dx\Big)^2
+\Big(\frac{1}{2}-\frac{1}{p+1}\int_{\mathbb{R}^3}K(\varepsilon_nx)
|v_{\varepsilon_n}|^{p+1}dx\Big).
\end{align*}
Letting $n\to \infty$, we have
\[
0<\liminf_{\varepsilon\to 0}m_{\varepsilon_n}
=-\liminf_{\varepsilon\to 0}\frac{b}{4}
\Big(\int_{\mathbb{R}^3}|\nabla v_{\varepsilon_n}|^2dx\Big)^2\leq0.
\]
Which is absurd.
\end{proof}
We denote
\[
\omega_\varepsilon(x):=v_\varepsilon(x+y_\varepsilon)
=u_\varepsilon(\varepsilon x+\varepsilon y_\varepsilon).
\]
So $\omega_\varepsilon$ is a positive ground state solution to
\begin{equation} \label{PPPe}
\begin{gathered}
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u
+V(\varepsilon x+\varepsilon y_\varepsilon)u
= K(\varepsilon x+\varepsilon y_\varepsilon)|u|^{p-1}u,\quad x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gathered}
\end{equation}
Denote the corresponding energy functional by $\Phi_\varepsilon$.
Set $\phi(\omega_\varepsilon)=\Phi'_\varepsilon((\omega_\varepsilon)_t)|_{t=1}$.
Thus
\begin{align*}
\phi(\omega_\varepsilon)
&=\frac{3a}{2}\int_{\mathbb{R}^3}|\nabla \omega_\varepsilon|^2dx
+\frac{5}{2}\int_{\mathbb{R}^3}V(\varepsilon x
+\varepsilon y_\varepsilon)\omega_\varepsilon^2dx
+\frac{1}{2}\int_{\mathbb{R}^3}\nabla V(\varepsilon x+\varepsilon y_\varepsilon)\varepsilon x\omega_\varepsilon^2dx
\\
&\quad +\frac{3b}{2}\Big(\int_{\mathbb{R}^3}|\nabla \omega_\varepsilon|^2dx\Big)^2
-\frac{4+p}{p+1}\int_{\mathbb{R}^3}K(\varepsilon x+\varepsilon y_\varepsilon)
|\omega_\varepsilon|^{p+1}dx\\
&\quad -\frac{1}{p+1}\int_{\mathbb{R}^3}
\nabla K(\varepsilon x+\varepsilon y_\varepsilon)\varepsilon x
|\omega_\varepsilon|^{p+1}dx=0.
\end{align*}
\begin{lemma} \label{lem4.4}
The sequence $\{\varepsilon y_\varepsilon\}$ is bounded.
\end{lemma}
\begin{proof}
It is easy to know that $\{\omega_\varepsilon\}$ is bounded in
$H^1(\mathbb{R}^3)$. We may assume that
\[
\omega_\varepsilon\rightharpoonup\omega_0\geq0\quad \text{in } H^1(\mathbb{R}^3).
\]
It follows from Lemma \ref{lem4.3} that $\omega_0\not\equiv0$.
Suppose to the contrary that, after passing to a subsequence,
\[
|\varepsilon y_\varepsilon|\to \infty.
\]
Clearly, we have $V(\varepsilon y_\varepsilon)\to V_\infty$ and
$K(\varepsilon y_\varepsilon)\to K_\infty$ as $\varepsilon\to 0$.
Thus $\omega_0$ is a solution of
\begin{equation} \label{e4.5}
-(a+bA)\Delta u+V_\infty u= K_\infty|u|^{p-1}u,\quad x\in \mathbb{R}^3,
\end{equation}
where $A=\lim_{\varepsilon\to 0}\int_{\mathbb{R}^3}|\nabla \omega_\varepsilon|^2dx$.
Similarly as Lemma \ref{lem2.1}, we have the Pohozaev identity
\[
P_{A,\infty}(\omega_0):=\frac{a+bA}{2}\int_{\mathbb{R}^3}
|\nabla \omega_0|^2dx-\frac{3K_\infty}{p+1}\int_{\mathbb{R}^3}
|\omega_0|^{p+1}dx+\frac{3V_\infty}{2}\int_{\mathbb{R}^3}|\omega_0|^2dx=0.
\]
Let us define
\begin{align*}
g_{\omega_0}(t):&=I_\infty((\omega_0)_t) \\
&=\frac{a+bA}{2}t^3\int_{\mathbb{R}^3}|\nabla \omega_0|^2dx
+\frac{t^5}{2}\int_{\mathbb{R}^3}V_\infty\omega_0^2dx
-\frac{t^{4+p}}{p+1}\int_{\mathbb{R}^3}\nabla K_\infty|\omega_0|^{p+1}dx\\
&=0,
\end{align*}
where $I_\infty$ is the energy functional associated to \eqref{e4.5}.
Obviously, $g_{\omega_0}(t)$ attains its unique maximum since $2K(\overline{x})\}.
\]
We see that $x_0\in(\Omega_1\backslash \mathcal{A})\cup(\Omega_1^c\backslash
\mathcal{B})$. As mentioned early, we may assume $\overline{x}=0$ and
$K(0)=\max_{x\in\Omega_1}K(x)=k$. When $x_0\in\Omega_1\backslash \mathcal{A}$,
then $V(x_0)=V_{\rm min}$ and $K(x_0)0$ such that
\[
\omega_\varepsilon(x)\leq C_1e^{-C_2|x|}.
\]
for all $x\in\mathbb{R}^3$.
\end{lemma}
\begin{proof}[Proof of Theorem \ref{thm1.1}]
Let $\delta_\varepsilon$ be the global maximum of $\omega_\varepsilon$.
By Lemma \ref{lem4.7}, we see that $\delta_\varepsilon\in B_R(0)$ for some $R>0$.
Thus the global maximum of $v_\varepsilon$, given by
$z_\varepsilon=y_\varepsilon+\delta_\varepsilon$, satisfies
$\varepsilon z_\varepsilon=\varepsilon y_\varepsilon+\varepsilon \delta_\varepsilon$.
Note that $u_\varepsilon(x)=(x/\varepsilon)$, then we see that
$u_\varepsilon(x)$ is positive ground state solution to \eqref{Pe}
with $\varepsilon>0$ and has a global maximum point
$x_\varepsilon=\varepsilon z_\varepsilon$. Since $\{\delta_\varepsilon\}$
is bounded, it follows from \eqref{PPPe} and Lemma \ref{lem4.5} that
$\varepsilon z_\varepsilon\to x_0$ and
$\lim_{\varepsilon\to 0}\operatorname{dist}(\varepsilon z_\varepsilon,\mathcal{H})=0$.
In particular, if $\Omega_1\cap\Omega_2\neq\emptyset$, then
$\lim_{\varepsilon\to 0}\operatorname{dist}(\varepsilon z_\varepsilon,
\Omega_1\cap\Omega_2)=0$. Moreover, since $\omega_\varepsilon$ is a
$(PS)_{m_{V(x_0)K(x_0)}}$ sequence for $J_{m_{V(x_0)K(x_0)}}$ and
$\omega_\varepsilon\to \omega_0$ in $H^1(\mathbb{R}^3)$, we deduce that
$\omega_0$ is a positive ground state solution of
\begin{gather*}
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u+V(x_0)u
= K(x_0)|u|^{p-1}u,\quad x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gather*}
In particular, if $\Omega_1\cap\Omega_2\neq\emptyset$, we have
$V(x_0)=V_{\rm min}$, $K(x_0)=K_{\rm max}$ and $\omega_0$ is a positive
ground state solution of
\begin{gather*}
-\Big(a+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u
+V_{\rm min}u= K_{\rm max}|u|^{p-1}u,\quad x\in \mathbb{R}^3,\\
u\in H^1(\mathbb{R}^3),
\end{gather*}
In view of the definition of $v_\varepsilon$, from Lemma \ref{lem4.7} we obtain
\[
u_\varepsilon(x)=v_\varepsilon(\frac{x}{\varepsilon})
=\omega_\varepsilon(\varepsilon^{-1}x-y_\varepsilon)
=\omega_\varepsilon(\varepsilon^{-1}x-\varepsilon^{-1}x_\varepsilon
+\delta_\varepsilon)
\leq C_1e^{-C_2|\frac{x-x_\varepsilon}{\varepsilon}|}.
\]
The proof is complete.
\end{proof}
\subsection*{Acknowledgments}
This research was supported by the Fundamental Research Funds for the
Central Universities (Grant No. 2015QNA45).
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