\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 178, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{7mm}}

\begin{document}
\title[\hfilneg EJDE-2016/178\hfil Kirchhoff type problems]
{Kirchhoff type problems with potential well and indefinite potential}

\author[Y. Wu, Y. Huang, Z. Liu \hfil EJDE-2016/178\hfilneg]
{Yuanze Wu, Yisheng Huang, Zeng Liu}

\address{Yuanze Wu (corresponding author) \newline
 College of Sciences,
China University of Mining and Technology,
Xuzhou 221116,  China}
\email{wuyz850306@cumt.edu.cn}

\address{Yisheng Huang \newline
Department of Mathematics,
Soochow University, Suzhou 215006, China}
\email{yishengh@suda.edu.cn}

\address{Zeng Liu \newline
Department of Mathematics,
Suzhou University of Science and Technology,
Suzhou 215009,  China}
\email{luckliuz@163.com}

\thanks{Submitted February 19, 2016. Published July 6, 2016.}
\subjclass[2010]{35B38, 35B40, 35J10, 35J20}
\keywords{Kirchhoff type problem; indefinite potential; potential well; 
\hfill\break\indent variational method}

\begin{abstract}
 In this article, we study the  Kirchhoff type problem
 \begin{gather*}
 -\Big(\alpha\int_{\mathbb{R}^3}|\nabla u|^2dx+1\Big)\Delta u
 +(\lambda a(x)+a_0)u=|u|^{p-2}u \quad\text{in }\mathbb{R}^3,\\
 u\in  H^1(\mathbb{R}^3),
 \end{gather*}
  where $4<p<6$, $\alpha$ and $\lambda$ are two positive parameters,
 $a_0\in\mathbb{R}$ is a (possibly negative) constant and $a(x)\geq0$ is
 the potential well.  Using the variational method, we show the existence 
 of nontrivial solutions. We also obtain the concentration behavior 
 of the solutions as $\lambda\to+\infty$.
\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 will study the  Kirchhoff type problem
\begin{equation}
\begin{gathered}
-\Big(\alpha\int_{\mathbb{R}^3}|\nabla u|^2dx+1\Big)\Delta u
+(\lambda a(x)+a_0)u=|u|^{p-2}u \quad \text{in }\mathbb{R}^3,\\
 u\in H^1(\mathbb{R}^3),
\end{gathered} \label{ePal}
\end{equation}
where $4<p<6$, $\alpha$ and $\lambda$ are two positive parameters,
$a_0\in\mathbb{R}$ is a constant and $a(x)$ is a potential satisfying some
conditions to be specified later.

The Kirchhoff type problems in bounded domains is one of most popular 
nonlocal problems in the study areas of elliptic equations 
(cf. \cite{CKW11,CWL12,LLS14,LLT15,N14,N141,ZP06} and the references therein).  
One motivation comes from the very important application to such problems 
in physics. Indeed, The Kirchhoff type problem in bounded domains is related 
to the stationary analogue of the model
\begin{equation}\label{eq991}
\begin{gathered}
 u_{tt}-\Big(\alpha\int_{\Omega}|\nabla u|^2dx+\beta\Big)\Delta u=h(x,u)\quad
\text{in }\Omega\times(0, T),\\
u=0\quad\text{on }\partial\Omega\times(0, T),\\
u(x,0)=u_0(x),\quad u_t(x,0)=u^*(x),
\end{gathered}
\end{equation}
where $T>0$ is a constant, $u_0, u^*$ are continuous functions.  
Such model was first proposed by Kirchhoff in 1883 as an extension of the 
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{eq991}, $u$ denotes the displacement, 
$h(x,u)$ the external force and $\beta$ the initial tension while $\alpha$ 
is related to the intrinsic properties of the string (such as Young¡¯s modulus).  
For more details on the physical background of Kirchhoff type problems,
we refer the readers to \cite{A12,K83}.

The Kirchhoff type nonlocal term was introduced to the elliptic equations 
in $\mathbb{R}^3$ by He and Zou in \cite{HZ12}, where, by using the 
variational method, some existence results of the nontrivial solutions 
were obtained.  Since then, many papers have been devoted to such topic,
 see for example \cite{AF12,HLP14,HLW15,LLS12,LY131,SW14,WHL15} 
and the references therein.  In particular, in a  recent article \cite{SW14}, 
Sun and Wu have studied the  Kirchhoff type problem
\begin{gather*}
-\Big(\mu\int_{\mathbb{R}^3}|\nabla u|^2dx+\nu\Big)\Delta u
 +\lambda a(x)u=f(x,u) \quad\text{in }\mathbb{R}^3,\\
u\in H^1(\mathbb{R}^N),
\end{gather*}
where $\mu,\nu,\lambda>0$ are parameters and $a(x)$ satisfies the 
following conditions:
\begin{itemize}
\item[(A1)] $a(x)\in C(\mathbb{R}^3)$ and $a(x)\geq0$ on $\mathbb{R}^3$.

\item[(A2)] There exists $a_\infty>0$ such that 
$|\mathcal{A}_\infty|<+\infty$, where 
$\mathcal{A}_\infty=\{x\in\mathbb{R}^3: a(x)<a_\infty\}$ and 
$|\mathcal{A}_\infty|$ is the Lebesgue measure of the set $\mathcal{A}_\infty$.

\item[(A3)] $\Omega=\text{int} a^{-1}(0)$ is a bounded domain and has smooth 
boundaries with $\overline{\Omega}=a^{-1}(0)$.
\end{itemize}
Using the variational method, they obtain some existence and non-existence 
results of the nontrivial solutions when $f(x,u)$ is $1$-asymptotically linear, 
$3$-asymptotically linear or $4$-asymptotically linear at infinity.

Under the conditions (A1)--(A3), $\lambda a(x)$ is called as the steep 
potential well for $\lambda$ sufficiently large and the depth of the well 
is controlled by the parameter $\lambda$.  Such potentials were first 
introduced by Bartsch and Wang in \cite{BW95} for the scalar 
Schr\"odinger equations.  An interesting phenomenon for this kind 
of Schr\"odinger equations is that, one can expect to find the solutions 
which are concentrated at the bottom of the wells as the depth goes to infinity.  
Because this interesting property, such topic for the scalar Schr\"odinger 
equations was studied extensively in the past decade.  
We refer the readers to \cite{BT13,DS07,LHL11,ST09,WZ09} and 
the references therein.  Recently, the steep potential well was also 
considered for some other elliptic equations and systems, see for 
example \cite{FSX10,GT121,JZ11,YT14,ZLZ13} and the references therein.  
To our best knowledge, most of the literatures on this topic are devoted 
to the definite case while the indefinite case was only considered 
in \cite{BT13,DS07} for the the scalar Schr\"odinger equations and 
in \cite{ZLZ13} for the Schr\"odinger-Poisson systems.

Inspired by the above facts, we wonder what will happen for the Kirchhoff 
type problem with steep potential wells in the indefinite case of $a<0$?  
To our best knowledge, this kind of  problems has not been studied yet 
in the literatures.  Thus, the purpose of this paper is to explore the 
preceding problems.

Before stating our results, we shall introduce some notation.  
By  condition (A3), it is well known that in the case of $a_0\neq0$, all the 
eigenvalues $\{\gamma_i\}$ of the  problem
\begin{equation}\label{eq001}
-\Delta u=\gamma |a_0|u\quad u\in H_0^1(\Omega)
\end{equation}
satisfy $\gamma_1<\gamma_2<\gamma_3<\dots<\gamma_i<\dots$ with 
$\gamma_i\to+\infty$ as $i\to\infty$ and the multiplicity of $\gamma_i$ 
is finite for every $i\in\mathbb{N}$.  In particular, $\gamma_1$ is simple.  
For each $i\in\mathbb{N}$, denote the corresponding eigenfunctions and the
 eigenspace of $\gamma_i$ by $\{\varphi_{i,j}\}_{j=1,2,\dots,k_i}$ and 
$\mathcal{N}_i=$span$\{\varphi_{i,j}\}_{j=1,2,\dots,k_i}$ respectively, 
where $k_i$ are the multiplicity of $\gamma_i$, then $\varphi_{i,j}$ 
can be chosen so that $\|\varphi_{i,j}\|_{L^2(\Omega)}=\frac{1}{|a_0|^2}$ 
and $\{\varphi_{i,j}\}$ can form a basis of $H_0^1(\Omega)$.  Let
\begin{equation}\label{eq9998}
k_0^*=\inf\{k:\gamma_k>1\},
\end{equation}
then our main result in this paper can be stated as follows.

\begin{theorem}\label{thm001}
Suppose that  {\rm (A1)--(A3)} hold. 
 If either $a_0\geq0$ or $a_0<0$ with $\gamma_{k_0^*-1}<1$ then there 
exist positive constants $\alpha_*$ and $\Lambda_*$ such that 
$(\mathcal{P}_{\alpha, \lambda})$ has a nontrivial solution 
$u_{\alpha, \lambda}$ for all $\lambda>\Lambda_*$ and $\alpha\in(0 ,\alpha_*)$.  
Moreover, $u_{\alpha,\lambda}\to u_{\alpha}$ strongly in $H^1(\mathbb{R}^3)$ 
as $\lambda\to+\infty$ up to a subsequence and $u_\alpha$ is a nontrivial 
solution of the following Kirchhoff type problem:
\begin{equation} \label{ePa*}
\begin{gathered}
-\Big(\alpha\int_{\Omega}|\nabla u|^2dx+1\Big)\Delta u
+a_0 u =|u|^{p-2}u \quad\text{in }\Omega,\\
u=0\quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
\end{theorem}

\begin{remark} \em
(a)  If $a_0<0$ with $|a_0|$ large enough then it is easy to see that $k_0^*>1$.  
It follows that \eqref{ePal} is indefinite in a suitable Hilbert space 
(see Lemma \ref{lem005} for more details).  To out best knowledge, 
Theorem \ref{thm001} is the first result for the Kirchhoff type problem in 
$\mathbb{R}^3$ for the indefinite case.

(b)  Theorem~\ref{thm001} also gives the existence of nontrivial solutions 
to \eqref{ePa*}.  Note that \eqref{ePa*} is also indefinite if $a_0<0$ with 
$|a_0|$ large enough.  Thus, to our best knowledge, it is also the 
first result for the Kirchhoff type problem on bounded domains in the 
indefinite case.
\end{remark}

Through this paper, $C$ and $C_i$ $(i=1,2,\dots)$ will be indiscriminately 
used to denote various positive constants.  
$o_n(1)$ and $o_\lambda(1)$ will always denote the quantities tending 
towards zero as $n\to\infty$ and $\lambda\to+\infty$ respectively.

\section{Variational setting}

By  condition (A1), we see that for every $a_0\in\mathbb{R}$ and 
$\lambda>\max\{0, \frac{-a_0}{a_\infty}\}$,
\[
E=\{u\in D^{1,2}(\mathbb{R}^3):\int_{\mathbb{R}^3}a(x)u^2dx<+\infty\}
\]
equipped with the inner product
\[
\langle u,v\rangle_{\lambda}=\int_{\mathbb{R}^3}
(\nabla u\nabla v+(\lambda a(x)+a_0)^+uv)dx
\]
is a Hilbert space, which we will denote by $E_{\lambda}$,
where $(\lambda a(x)+a_0)^+=\max\{\lambda a(x)+a_0, 0\}$.
 The corresponding norm on $E_\lambda$ is
\[
\|u\|_{\lambda}=\Big(\int_{\mathbb{R}^3}(|\nabla u|^2
+(\lambda a(x)+a_0)^+u^2)dx\Big)^{1/2}.
\]
It follows from the H\"older inequality, the Sobolev inequality and
the conditions (A1)--(A2) that for every $u\in E_\lambda$ with
$\lambda>\max\{0, -a_0/a_\infty\}$,
\begin{align*}
\int_{\mathbb{R}^3}u^2dx
&= \int_{\mathcal{A}_\infty}u^2dx
 +\int_{\mathbb{R}^3\backslash\mathcal{A}_\infty}u^2dx \\
&\leq |\mathcal{A}_\infty|^{2/3}S^{-1}\int_{\mathbb{R}^3}|\nabla u|^2dx+\frac{1}{a_0+a_\infty\lambda}\int_{\mathbb{R}^3}(\lambda a(x)+a_0)^+u^2dx \\
&\leq \max\big\{|\mathcal{A}_\infty|^{2/3}S^{-1},
\frac{1}{a_0+a_\infty\lambda}\big\}
\int_{\mathbb{R}^3}(|\nabla u|^2+(\lambda a(x)+a_0)^+u^2)dx
\end{align*}
and
\begin{align*}
\Big(\int_{\mathbb{R}^3}|u|^pdx\Big)^{1/p} 
&\leq  S_p^{-1/2}\Big(\int_{\mathbb{R}^3}(|\nabla u|^2+u^2)dx\Big)^{1/2} \\
&\leq S_p^{-1/2}\sqrt{1+\max\{|\mathcal{A}_\infty|^{2/3}S^{-1},
\frac{1}{a_0+a_\infty\lambda}\}} \\
&\quad\times \Big(\int_{\mathbb{R}^3}(|\nabla u|^2+(\lambda a(x)+a_0)^+u^2)dx\Big)^{1/2},
\end{align*}
where $S$ and $S_p$ are the best Sobolev embedding constant from
$D^{1,2}(\mathbb{R}^3)$ to $L^6(\mathbb{R}^3)$ and $H^1(\mathbb{R}^3)$
to $L^{p}(\mathbb{R}^3)$ respectively; that is,
\[
S=\inf\{\|\nabla u\|_{L^2(\mathbb{R}^3)}^2 : u\in D^{1,2}(\mathbb{R}^3),
\|u\|_{L^{6}(\mathbb{R}^3)}^2=1\}
\]
and
\[
S_p=\inf\{\|\nabla u\|_{L^2(\mathbb{R}^3)}^2+\|u\|_{L^2(\mathbb{R}^3)}^2 :
 u\in H^{1}(\mathbb{R}^3), \|u\|_{L^{p}(\mathbb{R}^3)}^2=1\},
\]
where $\|\cdot\|_{L^p(\mathbb{R}^3)}$ is the usual norm in $L^p(\mathbb{R}^3)$ 
for all $p\geq1$.

Let $d_\lambda=\sqrt{\max\{|\mathcal{A}_\infty|^{2/3}S^{-1}, 
\frac{1}{a_0+a_\infty\lambda}\}}$. Then we have
\begin{equation}\label{eq0001}
\|u\|_{L^2(\mathbb{R}^N)}\leq d_\lambda\|u\|_{\lambda}\quad\text{and}\quad 
\|u\|_{L^p(\mathbb{R}^3)}\leq S_p^{-1/2}\sqrt{1+d_\lambda^2}\|u\|_{\lambda},
\end{equation}
which yields that $E_\lambda$ is embedded continuously into $ H^1(\mathbb{R}^3)$ for 
$\lambda>\max\{0, \frac{-a_0}{a_\infty}\}$.
Moreover, by using \eqref{eq0001}, the conditions (A1)--(A2) and by 
following a standard argument, we can show that corresponding energy 
functional $J_{\alpha, \lambda}(u)$ to the Problem \eqref{ePal}, given by
\[ %\label{eq0131}
J_{\alpha,\lambda}(u)=\frac\alpha4\|\nabla u\|_{L^2(\mathbb{R}^3)}^4
+\frac12\int_{\mathbb{R}^3}(|\nabla u|^2+(\lambda a(x)
+a_0)u^2)dx-\frac{1}{p}\|u\|_{L^{p}(\mathbb{R}^3)}^{p},
\]
is $C^2$ in $E_\lambda$  for $\lambda>\max\{0, \frac{-a_0}{a_\infty}\}$. 
 For the sake of convenience, we re-write the energy functional $J_{\lambda}(u)$ by
\[
J_{\alpha,\lambda}(u)=\frac\alpha4\|\nabla u\|_{L^2(\mathbb{R}^3)}^4
+\frac12\|u\|_\lambda^2-\frac12\mathcal{D}_{\lambda}(u,u)
-\frac{1}{p}\|u\|_{L^{p}(\mathbb{R}^3)}^{p},
\]
where $\mathcal{D}_{\lambda}(u,v)=\int_{\mathbb{R}^3}(\lambda a(x)+a_0)^-uvdx$ 
and $(\lambda a(x)+a_0)^-=\max\{-(\lambda a(x)+a_0), 0\}$.  
In what follows, inspired by \cite{DS07,ZLZ13}, we shall make some
 further observations on the functional $\mathcal{D}_{\lambda}(u,u)$.

By  condition (A1), $\int_{\mathbb{R}^3}(\lambda a(x)+a_0)u^2dx\geq0$ 
for all $u\in E_\lambda$ with $\lambda>0$ in the case of $a_0\geq0$.  
It follows that $\mathcal{D}_{\lambda}(u,u)$ is definite on 
$E_\lambda$ with $\lambda>0$ in the case of $a_0\geq0$.  
Let us consider the case of $a_0<0$ in what follows.  Let
\[
\mathcal{A}_\lambda:=\{x\in\mathbb{R}^3:  \lambda a(x)+a_0<0\},
\]
then by the condition (A3), we have $\Omega\subset\mathcal{A}_\lambda$, 
which means that $\mathcal{A}_\lambda\neq \emptyset$ for every
 $\lambda>0$, and moreover, by the conditions (A1)--(A2), the real number
\[
\Lambda_{0}:=\inf\{\lambda>0: |\mathcal{A}_\lambda|<+\infty\}.
\]
satisfies $0<\Lambda_{0}\leq\frac{-a_0}{a_\infty}$.  
For $\lambda>\Lambda_{0}$, we define
\[
\mathcal{F}_{\lambda}:=\{u\in E_{\lambda}:\operatorname{supp}u
\subset\mathbb{R}^3\backslash\mathcal{A}_\lambda\}.
\]
It follows from the conditions (A1)--(A3) that $\mathcal{F}_{\lambda}$ 
is nonempty, closed and convex with $\mathcal{F}_{\lambda}\neq E_{\lambda}$.
 Hence, $E_{\lambda}=\mathcal{F}_{\lambda}\oplus\mathcal{F}_{\lambda}^\perp$ 
and $\mathcal{F}_{\lambda}^\perp\neq \emptyset$ for $\lambda>\Lambda_{0}$
in the case of $a_0<0$, where $\mathcal{F}_{\lambda}^\perp$ is the orthogonal 
complement of $\mathcal{F}_{\lambda}$ in $E_{\lambda}$.

\begin{lemma}\label{lem001}
Let
\[
\beta(\lambda):=\inf_{u\in\mathcal{F}_{\lambda}^\perp\cap\mathbb{D}_\lambda}
\|u\|_\lambda^2,
\]
where $\mathbb{D}_\lambda:=\{u\in E_\lambda:  \mathcal{D}_\lambda(u,u)=1\}$.
If the conditions (A1)--(A3) hold, then $\beta(\lambda)$ is nondecreasing 
as the function of $\lambda$ on $(\Lambda_0, +\infty)$ and $\beta(\lambda)$
 can be attained by some $e(\lambda)\in\mathcal{F}_{\lambda}^\perp$.  
Furthermore, $(e(\lambda), \beta(\lambda))\to (\varphi_1,\gamma_1)$ strongly 
in $ H^1(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence.
\end{lemma}

\begin{proof}
First, thanks to the definition of $\Lambda_0$, we see that 
$\mathcal{D}_\lambda(u,u)$ and $\|u\|_\lambda^2$ are weakly continuous and 
weakly low semi-continuous on $\mathcal{F}_{\lambda}^\perp$ respectively.  
Thus, we can use a standard argument to show that $\beta(\lambda)$ can be 
attained by some $e(\lambda)\in\mathcal{F}_{\lambda}^\perp\cap\mathbb{D}_\lambda$ 
for all $\lambda>\Lambda_0$.

 Next, we  show that $\beta(\lambda)$ is nondecreasing as the function of 
$\lambda$ on $(\Lambda_0, +\infty)$.  Indeed, let $\lambda_1\geq\lambda_2$, 
then by the definition of $E_\lambda$, we have $E_{\lambda_1}=E_{\lambda_2}$ 
in the sense of sets.  It follows that 
$\mathcal{F}_{\lambda_2}\subset\mathcal{F}_{\lambda_1}$,
 which implies 
$\mathcal{F}_{\lambda_1}^{\perp}\subset\mathcal{F}_{\lambda_2}^{\perp}$.  
Note that $\|u\|_{\lambda_1}^2\geq\|u\|_{\lambda_2}^2$ and 
$\mathcal{D}_{\lambda_1}(u,u)\leq\mathcal{D}_{\lambda_2}(u,u)$ for all 
$u\in E_{\lambda_1}$ by the condition (A1).  Thus, due to the definition of 
$\beta(\lambda_1)$ and $\beta(\lambda_2)$, we can see that 
$\beta(\lambda_2)\leq\beta(\lambda_1)$; that is, $\beta(\lambda)$ 
is nondecreasing as a functional of $\lambda$ on $(\Lambda_0, +\infty)$.

 Finally, we shall prove that $(e(\lambda), \beta(\lambda))\to (\varphi_1,\gamma_1)$ 
strongly in $ H^1(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence.
In fact, since $\int_{\mathbb{R}^3}(\lambda a(x)+a_0)^-[e(\lambda)]^2dx=1$, 
it implies that
\begin{eqnarray}\label{eq9999}
\lim_{\lambda\to+\infty}\int_{\mathbb{R}^3}(a(x)+\frac{a_0}{\lambda})^+
[e(\lambda)]^2dx=0.
\end{eqnarray}
Note that  $H_0^1(\Omega)\subset\mathcal{F}_{\lambda}^\perp$ for all
 $\lambda>\Lambda_0$ due to the condition (A3), we can easily show that
 $0<\beta(\lambda)\leq\gamma_{1}$ for all $\lambda>\Lambda_0$.  
It follows from \eqref{eq0001} that $\{e(\lambda)\}$ is bounded in $ H^1(\mathbb{R}^3)$ 
for $\lambda$.  Without loss of generality, we assume that 
$e(\lambda)\rightharpoonup e^*$ weakly in $ H^1(\mathbb{R}^3)$ and $\beta(\lambda)\to\beta^*$  
as $\lambda\to+\infty$.
By the Sobolev embedding theorem, the condition (A2) and \eqref{eq9999}, 
we must have $(e^*,\beta^*)\in H^1_0(\Omega)\times\mathbb{R}^+$  
satisfying $e^*\equiv0$ outside $\Omega$ and $|a_0|^2\int_{\Omega}|e^*|^2dx=1$ 
and $(e(\lambda), \beta(\lambda))\to (e^*,\beta^*)$ strongly in 
$L^2(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence, 
which gives 
\begin{equation}
\begin{aligned}
\gamma_1&\geq \int_{\mathbb{R}^3}(|\nabla e(\lambda)|^2
 +(\lambda a(x)+a_0)^+[e(\lambda)]^2)dx \\
&\geq \int_{\Omega}|\nabla e^*|^2dx+o_\lambda(1) \\
&\geq \gamma_1+o_\lambda(1).
\end{aligned} \label{eq1002}
\end{equation}
Hence, $(e(\lambda), \beta(\lambda))\to (e^*,\gamma_1)$ strongly in
$ H^1(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence and
$(e^*,\gamma_1)$ satisfies
\[
\gamma_1=\int_{\Omega}|\nabla e^*|^2dx
=\inf_{u\in H^1_0(\Omega)\backslash\{0\}}
\frac{\int_{\Omega}|\nabla u|^2dx}{|a_0|^2\int_{\Omega}|u|^2dx}.
\]
Thus, $e_\alpha^*=\varphi_{1}$ and the proof is complete.
\end{proof}

We re-denote the above 
 $(e(\lambda), \beta(\lambda))$ by $(e_{1}(\lambda), \beta_{1}(\lambda))$ and define
\[
\mathcal{F}_{\lambda,1}^\perp:=\big\{u\in\mathcal{F}_{\lambda}^\perp: 
\frac{\|u\|_\lambda^2}{\mathcal{D}_\lambda(u,u)}=\beta_{1}(\lambda)\big\}.
\]
Since $\gamma_{2}>\gamma_{1}\geq\beta_{1}(\lambda)$ for $\lambda>\Lambda_0$, 
it is easy to see that 
$\mathcal{F}_{\lambda,1}^\perp\neq \mathcal{F}_{\lambda}^\perp$.
Thus, we have $\mathcal{F}_{\lambda}^\perp=\mathcal{F}_{\lambda,1}^\perp
\oplus\mathcal{F}_{\lambda,1}^{\perp,*}$, where $\mathcal{F}_{\lambda,1}^{\perp,*}$ 
is the orthogonal complement of $\mathcal{F}_{\lambda,1}^\perp$ in 
$\mathcal{F}_{\lambda}^\perp$.

\begin{lemma}\label{lem002}
Suppose that  {\rm (A1)--(A3)} hold. 
 Then there exists $\Lambda_{1}\geq\Lambda_0$ such that 
$\mathcal{F}_{\lambda,1}^{\perp}=\operatorname{span}\{e_{1}(\lambda)\}$ and
$\beta_{2}(\lambda)$ can be attained by some 
$e_{2}(\lambda)\in \mathcal{F}_{\lambda,1}^{\perp,*}$ for $\lambda>\Lambda_{1}$, 
where
\[
\beta_{2}(\lambda):=\inf_{\mathcal{F}_{\lambda,1}^{\perp,*}
\cap\mathbb{D}_\lambda}\|u\|_\lambda^2.
\]
Furthermore, $(e_{2}(\lambda),\beta_{2}(\lambda))\to(\varphi_{2,j},\gamma_2)$ 
strongly in $ H^1(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence for some 
$j\in\mathbb{N}$ with $1\leq j\leq k_2$.
\end{lemma}

\begin{proof}
Since $\mathcal{D}_\lambda(u,u)$ and $\|u\|_\lambda^2$ are weakly continuous 
and weakly low semi-con\-tinuous on $\mathcal{F}_{\lambda,1}^{\perp,*}$ respectively, 
by the fact that $\mathcal{F}_{\lambda,1}^{\perp,*}$ is weakly closed 
for $\lambda>\Lambda_0$, we can also use a standard argument to show that 
$\beta_{2}(\lambda)$ can be attained by some 
$e_{2}(\lambda)\in \mathcal{F}_{\lambda,1}^{\perp,*}$ for $\lambda>\Lambda_0$.  
For the sake of clarity, the remaining proof will be performed through the 
following steps.
\smallskip

\noindent\textbf{Step 1.}
 We prove that there exists $\Lambda_{1}\geq\Lambda_0$ 
such that $\mathcal{F}_{\lambda,1}^{\perp}=\operatorname{span}\{e_{1}(\lambda)\}$ 
for $\lambda>\Lambda_{1}$.
Indeed, suppose on the contrary that there exist 
$e_{1}^*(\lambda_n),e_{1}^0(\lambda_n)\in\mathcal{F}_{\alpha,\lambda,1}^{\perp}$ 
with
\begin{gather*}
\langle e_{1}^*(\lambda_n), e_{1}^0(\lambda_n)\rangle_{E_{\lambda_n},
 E_{\lambda_n}}=0, \\
\int_{\mathbb{R}^3}(\lambda_n a(x)+a_0)^-[e_{1}^*(\lambda_n)]^2dx
 =\int_{\mathbb{R}^3}(\lambda_n a(x)+a_0)^-[e_{1}^0(\lambda_n)]^2dx=1
\end{gather*}
for $\{\lambda_n\}$ satisfying $\lambda_n\to+\infty$ as $n\to\infty$.  
By Lemma \ref{lem001}, we can see that 
$e_{1}^*(\lambda_n)\to\varphi_1$ and $e_{1}^0(\lambda_n)\to\varphi_1$
 strongly in $ H^1(\mathbb{R}^3)$ as $n\to\infty$ up to a subsequence.  
It follows from \eqref{eq1002} and Lemma \ref{lem001} once more that
\begin{equation}
\begin{aligned}
2\gamma_1
&= 2\beta_{1}(\lambda_n)+o_n(1) \\
&= \|e_{1}^*(\lambda_n)\|_{\lambda_n}^2+\|e_{1}^0(\lambda_n)\|_{\lambda_n}^2+o_n(1) \\
&= \|\nabla (e_{1}^*(\lambda_n)-e_{1}^0(\lambda_n))\|_{L^2(\mathbb{R}^3)}^2+o_n(1)
= o_n(1),
\end{aligned}\label{eq100}
\end{equation}
which is a contradiction.
\smallskip

\noindent\textbf{Step 2.}
 We show that $\limsup_{\lambda\to+\infty}\beta_{2}(\lambda)\leq\gamma_2$.
In fact, by Step 1, we have
 $\varphi_{2,1}=d_{\lambda}^* e_{1}(\lambda)+\varphi_{2,1,\lambda}^*$, 
where $d_{\lambda}^*$ is a constant and $\varphi_{2,1,\lambda}^*$ is the 
projection of $\varphi_{2,1}$ in $\mathcal{F}_{\lambda,1}^{\perp,*}$.  
Thus, $\langle e_{1}(\lambda),\varphi_{2,1}\rangle_{E_\lambda,E_\lambda}
=d_{\lambda}^*\|e_{1}(\lambda)\|_\lambda^2$.  It follows from the condition 
(A3) and Lemma \ref{lem001} that $d_{\lambda}^*\to0$ as $\lambda\to+\infty$ 
up to a subsequence.  Now, by the definition of $\beta_{2}(\lambda)$, 
we can see from Lemma \ref{lem001}, \eqref{eq1002} and a variant of the 
Lebesgue dominated convergence theorem (cf. \cite[Theorem~2.2]{PK74}) that
\begin{align*}
\limsup_{\lambda\to+\infty}\beta_{2}(\lambda)
&\leq \limsup_{\lambda\to+\infty}\frac{\|\varphi_{2,1,\lambda}^*\|_\lambda^2}
{\mathcal{D}_\lambda(\varphi_{2,1,\lambda}^*,\varphi_{2,1,\lambda}^*)}\\
&= \limsup_{\lambda\to+\infty}\frac{\|\varphi_{2,1}-d_{\lambda}^* 
 e_{1}(\lambda)\|_\lambda^2}{\mathcal{D}_\lambda(\varphi_{2,1}
 -d_{\lambda}^* e_{1}(\lambda),\varphi_{2,1}-d_{\lambda}^* e_{1}(\lambda))}\\
&= \frac{\|\nabla\varphi_{2,1}\|_{L^2(\mathbb{R}^3)}^2}{|a_0|^2
 \|\varphi_{2,1}\|_{L^2(\mathbb{R}^3)}^2}
= \gamma_2.
\end{align*}
\smallskip

\noindent\textbf{Step 3.}
 We prove that $\limsup_{\lambda\to+\infty}\beta_{2}(\lambda)\geq\gamma_2$ 
and $(e_{2}(\lambda),\beta_{2}(\lambda))\to(\varphi_{2,j},\gamma_2)$ 
strongly in $ H^1(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence for 
some $j\in\mathbb{N}$ with $1\leq j\leq k_2$.
Actually, by Step 2, we know that $\{e_{2}(\lambda)\}$ is bounded in
 $D^{1,2}(\mathbb{R}^3)$.  Similarly as in the proof of Lemma \ref{lem001}, 
we can see that $(e_{2}(\lambda),\beta_{2}(\lambda))\to(e_{2}^*,\beta_{2}^*)$ 
strongly in $L^2(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a 
subsequence with $e_{2}^*\in H^1_0(\Omega)$ and $e_{2}^*\equiv0$ outside $\Omega$.
  Since $\mathcal{D}_\lambda(u,u)$ is weakly continuous on 
$\mathcal{F}_{\lambda}^\perp$, we also have $|a_0|^2\int_{\Omega}|e_2^*|^2dx=1$. 
 Furthermore, by the theory of Lagrange multipliers, we can also see that 
$(e_{2}^*,\beta_{2}^*)$ satisfies \eqref{eq001}.  
It follows from a variant of the Lebesgue dominated convergence theorem 
(cf. \cite[Theorem~2.2]{PK74}) that
\begin{align*}
\|\nabla e_{2}^*\|_{L^2(\mathbb{R}^3)}^2
&= \beta_{2}^*|a_0|\|e_{2}^*\|_{L^2(\mathbb{R}^3)}^2\\
&= \beta_{2}(\lambda)\mathcal{D}_\lambda(e_{2}(\lambda),
 e_{2}^*(\lambda))+o_\lambda(1)\\
&= \int_{\mathbb{R}^3}|\nabla e_{2}(\lambda)|^2dx+o_\lambda(1)\\
&\geq \|\nabla e_{2}^*\|_{L^2(\mathbb{R}^3)}^2.
\end{align*}
Thus, $(e_{2}(\lambda),\beta_{2}(\lambda))\to(e_{2}^*,\beta_{2}^*)$
strongly in $ H^1(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence. 
By Step~2, we must have $\beta_{2}^*=\gamma_1$ or $\beta_{2}^*=\gamma_2$.  
If $\limsup_{\lambda\to+\infty}\beta_{2}(\lambda)<\gamma_2$ then there exists 
$\{\lambda_n\}$ such that 
$(e_{2}(\lambda_n),\beta_{2}(\lambda_n))\to(\varphi_1,\gamma_1)$ 
strongly in $L^2(\mathbb{R}^3)\times\mathbb{R}$ as $n\to\infty$ up to a subsequence.  
It follows from Lemma~\ref{lem001}, \eqref{eq1002} and Step 2 that
\[
0=\langle e_{1}(\lambda_n), e_{2}(\lambda_n)\rangle_{\lambda_n, \lambda_n}
=\|\nabla\varphi_1\|_{L^2(\mathbb{R}^3)}^2+o_n(1),
\]
which is a contradiction.
\end{proof}

Let
\[
\mathcal{F}_{\lambda,2}^\perp:=\big\{u\in\mathcal{F}_{\lambda}^\perp: 
\frac{\|u\|_\lambda^2}{\mathcal{D}_\lambda(u,u)}=\beta_{2}(\lambda)\big\}.
\]
Since $\gamma_{3}>\gamma_{2}$, it yields from  Lemma~\ref{lem002} and 
condition (A3) that 
$\mathcal{F}_{\lambda,1}^\perp\oplus\mathcal{F}_{\lambda,2}^\perp\neq 
\mathcal{F}_{\lambda}^\perp$.

\begin{lemma}\label{lem003}
Suppose that {\rm (A1)--(A3)} hold.  Then there exists $\Lambda_{2}\geq\Lambda_{1}$ 
such that dim$(\mathcal{F}_{\lambda,2}^\perp)\leq k_2$ for $\lambda>\Lambda_{2}$.
\end{lemma}

\begin{proof}
Let $e_{2}(\lambda),e_{2}'(\lambda)\in\mathcal{F}_{\lambda,2}^\perp$.  
By Lemma \ref{lem002}, $e_{2}(\lambda)\to\varphi_{2,j}$ and
 $e_{2}'(\lambda)\to\varphi_{2,j'}$ strongly in $ H^1(\mathbb{R}^3)\times\mathbb{R}$ as 
$\lambda\to+\infty$ up to a subsequence for some $j,j'\in\mathbb{N}$ with 
$1\leq j,j'\leq k_2$.  Clearly, two cases may occur:
\begin{itemize}
\item[(1)] $\varphi_{2,j}=\varphi_{2,j'}$;
\item[(2)] $\varphi_{2,j}\neq \varphi_{2,j'}$ and
$\int_{\Omega}\varphi_{2,j}\varphi_{2,j'}dx=0$.
\end{itemize}
If case $(1)$ happens then by a similar argument used in the proof
of \eqref{eq100}, we can get that $\gamma_{2}=0$, which is a contradiction.
Thus, we must have case $(2)$.  It follows that there exists
$\Lambda_{2}\geq\Lambda_{1}$ such that dim$(\mathcal{F}_{\lambda,2}^\perp)\leq k_2$
for $\lambda>\Lambda_{2}$.
\end{proof}

Now, by iterating, for $m=3,4,\dots$, we can define $\beta_{m}(\lambda)$
 as follows:
\[
\beta_{m}(\lambda):=\inf_{\mathcal{F}_{\lambda,m}^{\perp,*}
\cap\mathbb{D}_\lambda}\|u\|_\lambda^2,
\]
where
\begin{gather*}
\mathcal{F}_{\lambda,m}^{\perp,*}:=\{u\in\mathcal{F}_{\lambda}^\perp:
 \langle u,v\rangle_\lambda=0,\text{ for all }v\in\oplus_{i=1}^{m-1}
\mathcal{F}_{\lambda,i}^\perp\}, \\
\mathcal{F}_{\lambda,i}^\perp:=\Big\{u\in\mathcal{F}_{\lambda}^\perp:
 \frac{\|u\|_\lambda^2}{\mathcal{D}_\lambda(u,u)}=\beta_{i}(\lambda)\Big\}.
\end{gather*}
Similarly as Lemmas \ref{lem002} and \ref{lem003}, we can obtain the
following result.

\begin{lemma}\label{lem004}
Suppose that {\rm (A1)--(A3)} hold.  Then there exists $\Lambda_{m}\geq\Lambda_{m-1}$
such that $\beta_{m}(\lambda)$ can be attained by some
$e_{m}(\lambda)\in \mathcal{F}_{\lambda,m}^{\perp,*}$ for $\lambda>\Lambda_{m}$.
Furthermore, $(e_{m}(\lambda),\beta_{m}(\lambda))\to(\varphi_{m,j},\gamma_m)$
strongly in $ H^1(\mathbb{R}^3)\times\mathbb{R}$ as $\lambda\to+\infty$ up to a subsequence for some
$j\in\mathbb{N}$ with $1\leq j\leq k_m$ and dim$(\mathcal{F}_{\lambda,m}^\perp)\leq k_m$
for $\lambda>\Lambda_{m}$, where
\[
\mathcal{F}_{\lambda,m}^\perp:=\Big\{u\in\mathcal{F}_{\lambda}^\perp:
\frac{\|u\|_\lambda^2}{\mathcal{D}_\lambda(u,u)}=\beta_{m}(\lambda)\Big\}.
\]
\end{lemma}

Let $k_0^*$ be given by \eqref{eq9998}, then by Lemmas \ref{lem001}, \ref{lem002}
and \ref{lem004}, $\oplus_{i=1}^{k_0^*-1}\mathcal{F}_{\lambda,i}^\perp$
 and $\mathcal{F}_{\lambda,k_0^*}^{\perp,*}$ are well defined for
$\lambda>\Lambda_{k_0^*}$.

\begin{lemma}\label{lem005}
Suppose that {\rm (A1)--(A3)} hold.  If $\gamma_{k_0^*-1}<1$
then there exists $\widetilde{\Lambda}_{k_0^*}\geq\Lambda_{k_0^*}$
such that for $\lambda>\widetilde{\Lambda}_{k_0^*}$, it holds that
\begin{itemize}
\item[(1)] $\|u\|_{\lambda}^2-\mathcal{D}_\lambda(u,u)
 \leq\frac12(1-\frac{1}{\gamma_{k_0^*-1}})\|u\|_{\lambda}^2$ in
 $\oplus_{i=1}^{k_0^*-1}\mathcal{F}_{\lambda,i}^\perp$;
\item[(2)] $\|u\|_{\lambda}^2-\mathcal{D}_\lambda(u,u)
 \geq\frac12(1-\frac{1}{\gamma_{k_0^*}})\|u\|_{\lambda}^2$ in
$\mathcal{F}_{\lambda,k_0^*}^{\perp,*}$.
\end{itemize}
\end{lemma}

The proof of the above lemma follows immediately from Lemmas \ref{lem001},
\ref{lem002} and \ref{lem004}.


\begin{remark}\label{rmk001} \rm
By Lemmas~\ref{lem002}--\ref{lem004}, we also have
 $\oplus_{i=1}^{k_0^*-1}\mathcal{F}_{\lambda,i}^\perp=\emptyset$
in the case of $\gamma_1>1$ while
$\oplus_{i=1}^{k_0^*-1}\mathcal{F}_{\lambda,i}^\perp\neq \emptyset$
and $\dim (\oplus_{i=1}^{k_0^*-1}\mathcal{F}_{\lambda,i}^\perp)
\leq\sum_{i=1}^{k_0^*-1}k_i$ in the case of $\gamma_1<1$.
\end{remark}

\section{Nontrivial solution}

We first consider the case of $a_0<0$.  By the decomposition of $E_\lambda$,
 we will find the nontrivial solution by the linking theorem.
Let us first verify that $J_{\alpha,\lambda}(u)$ has a linking structure
in $E_\lambda$ in the case of $a_0<0$.

\begin{lemma}\label{lem006}
Suppose that {\rm  (A1)--(A3)} hold and $a_0<0$.  For every $\alpha>0$,
if $\beta_{k_0^*-1}<1$ then there exists $\rho>0$ independent of $\lambda$
such that
\begin{eqnarray}
\inf_{\mathcal{F}_{\lambda,k_0^*}^{\perp,*}\cap\mathbb{S}_{\lambda,\rho}}
J_{\alpha,\lambda}(u)\geq d_0\label{eq102}
\end{eqnarray}
for all $\lambda>\widetilde{\Lambda}_{k_0^*}$, where
$\mathbb{S}_{\lambda,\rho}:=\{u\in E_\lambda: \|u\|_\lambda=\rho\}$ and
$d_0$ is a constant independent of $\lambda$ and $\alpha$.
\end{lemma}

\begin{proof}
By \eqref{eq0001} and Lemma~\ref{lem005}, for every
 $u\in \mathcal{F}_{\lambda,k_0^*}^{\perp,*}$, we have
\begin{equation}
\begin{aligned}
J_{\alpha,\lambda}(u)
&= \frac\alpha4\|\nabla u\|_{L^2(\mathbb{R}^3)}^4
 +\frac12\|u\|_\lambda^2-\frac12\mathcal{D}_{\lambda}(u,u)
 -\frac{1}{p}\|u\|_{L^{p}(\mathbb{R}^3)}^{p} \\
&\geq \frac14(1-\frac{1}{\gamma_{k_0^*}})\|u\|_{\lambda}^2-S_p^{-\frac{p}{2}}
 (1+d_\lambda^2)^{\frac p2}\|u\|_{\lambda}^p \\
&\geq \|u\|_{\lambda}^2\Big(\frac14(1-\frac{1}{\gamma_{k_0^*}})-S_p^{-\frac{p}{2}}
 (1+d_\lambda^2)^{\frac p2}\|u\|_{\lambda}^{p-2}\Big).
\end{aligned}\label{eq1001}
\end{equation}
Note that 
$d_\lambda=\sqrt{\max\{|\mathcal{A}_\infty|^{2/3}S^{-1},
\frac{1}{a_0+a_\infty\lambda}\}}$, so that
\[
 d_\lambda\leq\sqrt{\max\{|\mathcal{A}_\infty|^{2/3}S^{-1},
\frac{1}{a_0+a_\infty\widetilde{\Lambda}_{k_0^*}}\}}
\]
 for
$\lambda>\widetilde{\Lambda}_{k_0^*}$.  It follows from \eqref{eq1001}
that there exists $\rho>0$ independent on $\lambda$ such that \eqref{eq102}
holds for all $\lambda>\widetilde{\Lambda}_{k_0^*}$.
\end{proof}

Let
\[
\mathcal{Q}_{\lambda,k_0^*}:=\{u=v+te_{k_0^*}(\lambda):
t\geq0\text{ and }v\in\oplus_{i=1}^{k_0^*-1}\mathcal{F}_{\lambda,i}^\perp\}.
\]

\begin{lemma}\label{lem007}
Suppose that {\rm  (A1)--(A3)} hold and $a_0<0$.
If $\gamma_{k_0^*-1}<1$ then there exist $\alpha_0>0$ and $R_0>\rho$
independent of $\lambda$ such that
\begin{eqnarray*}
\sup_{\partial\mathcal{Q}_{\lambda,k_0^*}^{R_0}}J_{\alpha,\lambda}(u)\leq\frac12d_0
\end{eqnarray*}
for all $\lambda>\widetilde{\Lambda}_{k_0^*}$  in the case of
$\alpha\in(0, \alpha_0)$, where $d_0$ is given in lemma \ref{lem006},
$\mathcal{Q}_{\lambda,k_0^*}^{R_0}:=\mathcal{Q}_{\lambda,k_0^*}
\cap \mathbb{B}_{\lambda,R_0}$ and
$\mathbb{B}_{\lambda,R_0}:=\{u\in E_\lambda:  \|u\|_\lambda\leq R_0\}$.
\end{lemma}

\begin{proof}
Let $u_{\lambda}\in\partial\mathcal{Q}_{\lambda,k_0^*}^{R}$.
Then one of the following two cases must happen:
\begin{itemize}
\item[(a)] $u_{\lambda}=R\widetilde{u}_{\lambda}$ with 
 $\widetilde{u}_{\lambda}\in \oplus_{i=1}^{k_0^*-1}\mathcal{F}_{\lambda,i}^\perp$ 
 and $\|\widetilde{u}_{\lambda}\|_\lambda\leq1$.
\item[(b)] $u_{\lambda}=R\widetilde{u}_{\lambda}$ with 
 $\widetilde{u}_{\lambda}\in 
 \mathcal{Q}_{\lambda,k_0^*}^{1}\backslash\oplus_{i=1}^{k_0^*-1}
 \mathcal{F}_{\lambda,i}^\perp$ and $\|\widetilde{u}_{\lambda}\|_\lambda=1$.
\end{itemize}
If the case (b) happens then by Lemma \ref{lem005}, we deduce that
\begin{equation} \label{eq103}
J_{\alpha,\lambda}(u_{\lambda})=J_{\alpha,\lambda}(R\widetilde{u}_{\lambda})
\leq\frac{\alpha}{4}R^4+\frac12(1-\frac{1}{\gamma_{k_0^*}})R^2
 -\frac1p\|R\widetilde{u}_{\lambda}\|_{L^p(\mathbb{R}^3)}^p.
\end{equation}
Since $\widetilde{u}_{\lambda}\in \oplus_{i=1}^{k_0^*}\mathcal{F}_{\lambda,i}^\perp$,
by Lemmas~\ref{lem001}, \ref{lem002} and \ref{lem004},
$\widetilde{u}_{\lambda}=\widetilde{u}+o_\lambda(1)$ strongly in $ H^1(\mathbb{R}^3)$ for some
$\widetilde{u}\in\operatorname{span}\{\varphi_{i,j}\}
^{i=1,2,\dots,k_0^*}_{j=1,2,\dots,k_i}$
and $\int_{\Omega}|\nabla \widetilde{u}|^2dx=1$.
Thus, $\|\widetilde{u}_{\lambda}\|_{L^p(\mathbb{R}^3)}^p
=\|\widetilde{u}\|_{L^p(\mathbb{R}^3)}^p+o_\lambda(1)$ by the Sobolev
embedding theorem.  

Note that $\dim\operatorname{span}\{\varphi_{i,j}\}^{i=1,2,\dots,k_0^*}_{j=1,2,\dots,k_i})
\leq\sum_{i=1}^{k_0^*-1}k_i+1$ for all $\lambda>\widetilde{\Lambda}_{k_0^*}$
by Remark \ref{rmk001}.  Therefore, there exists a constant $M>0$ such that
$\|u\|_{L^p(\mathbb{R}^3)}\geq M$ for all
$u\in$span$\{\varphi_{i,j}\}^{i=1,2,\dots,k_0^*}_{j=1,2,\dots,k_i}$ with
$\int_{\Omega}|\nabla u|^2dx=1$.  In particular,
$\|\widetilde{u}\|_{L^p(\mathbb{R}^3)}\geq M$.  It follows from $4<p<6$ and
\eqref{eq103} that there exists a constant $R_0(>\rho)$ such that
$J_{\alpha,\lambda}(R_0\widetilde{u}_{\lambda})\leq0$ for all
$\lambda>\widetilde{\Lambda}_{k_0^*}$.
Now, we consider the case of $(a)$.  By Lemma~\ref{lem005} once more, we know that
\begin{eqnarray*}
J_{\alpha,\lambda}(u_{\lambda})
=J_{\alpha,\lambda}(R\widetilde{u}_{\lambda})\leq\frac{\alpha}{4}R_0^4.
\end{eqnarray*}
Thus, there exists $\alpha_0>0$ such that
$J_{\alpha,\lambda}(u_{\lambda})\leq \frac12 d_0$ for
$\lambda>\widetilde{\Lambda}_{k_0^*}$ and $\alpha\in(0, \alpha_0)$.
\end{proof}

From Lemmas \ref{lem006} and \ref{lem007}, we can see that 
$J_{\alpha,\lambda}(u)$ has a linking structure in $E_\lambda$ with 
$\lambda>\widetilde{\Lambda}_{k_0^*}$ and $\alpha\in(0, \alpha_0)$ 
in the case of $a_0<0$.
By the linking theorem, there exists $\{u_n\}\subset E_\lambda$ such that
 $(1+\|u_n\|_\lambda)J_{\alpha,\lambda}'(u_n)=o_n(1)$ strongly in $E_\lambda^*$ and
$J_{\alpha,\lambda}(u_n)=c_{\alpha,\lambda}+o_n(1)$, where $E_\lambda^*$ is the 
dual space of $E_\lambda$.  Furthermore, 
$c_{\alpha,\lambda}\in[d_0, \frac{\alpha}{4}R_0^4+\frac12(1-\frac{1}{\gamma_{k_0^*}}) 
R_0^2]$.  Note that in the special case $\gamma_1>1$, the linking structure is 
actually the mountain pass geometry.  Thus, the linking theorem can be replaced 
by the mountain pass theorem and we can also obtain a sequence 
$\{u_n\}\subset E_\lambda$ such that 
$(1+\|u_n\|_\lambda)J_{\alpha,\lambda}'(u_n)=o_n(1)$ strongly in $E_\lambda^*$ and
$J_{\alpha,\lambda}(u_n)=c_{\alpha,\lambda}+o_n(1)$.  In the case $a_0\geq0$, 
since $4<p<6$ and the fact that $\mathcal{D}_\lambda(u,u)=0$ in 
$E_\lambda$, by using a standard argument, we can verify that 
$J_{\alpha, \lambda}(u)$ has a mountain pass geometry in $E_\lambda$ for 
$\lambda>0$; that is,
\begin{itemize}
\item[(a)] $\inf_{\mathbb{S}_{\lambda,\overline{\rho}}}J_{\alpha,\lambda}(u)\geq C$ 
 for some $\overline{\rho}>0$;
\item[(b)] $J_{\alpha,\lambda}(\overline{R}_0\phi)\leq0$ for some 
 $\overline{R}_0>\overline{\rho}$ and $\phi\in H_0^1(\Omega)$.
\end{itemize}

This also gives the existence of a sequence $\{u_n\}\subset E_\lambda$ such that
\[
(1+\|u_n\|_\lambda)J_{\alpha,\lambda}'(u_n)=o_n(1)
\]
 strongly in $E_\lambda^*$ and
$J_{\alpha,\lambda}(u_n)=c_{\alpha,\lambda}+o_n(1)$ with 
$c_{\alpha,\lambda}\in [C_\alpha, C_\alpha']$, where $C_\alpha,C'_\alpha$ 
are two positive constants independent of $\lambda$.  
In a word, in both cases of $a_0<0$ and $a_0\geq 0$, for
 $\lambda>\widetilde{\Lambda}_{k_0^*}$, there exists $\{u_n\}\subset E_\lambda$ 
such that $(1+\|u_n\|_\lambda)J_{\alpha,\lambda}'(u_n)=o_n(1)$ strongly in
 $E_\lambda^*$ and
$J_{\alpha,\lambda}(u_n)=c_{\alpha,\lambda}+o_n(1)$ with 
$c_{\alpha, \lambda}\in[C_\alpha, C'_\alpha]$.

\begin{lemma}\label{lem008}
Suppose that  {\rm (A1)--(A3)} hold.  For every $\alpha>0$, if either 
$a_0\geq0$ or $a_0<0$ with $\beta_{k_0^*-1}<1$ then $\{\|u_n\|_\lambda\}$ is bounded.
\end{lemma}

\begin{proof}
Since $\lambda>\widetilde{\Lambda}_{k_0^*}$, by the condition (A2) and 
the H\"older and the Sobolev inequalities, we obtain that
\begin{eqnarray*}
\mathcal{D}_\lambda(u_n,u_n)\leq|a_0|\int_{\mathcal{A}_\infty}|u_n|^2dx
\leq|a_0||\mathcal{A}_\infty|^{\frac{2}{3}}S^{-1}
\|\nabla u_n\|_{L^2(\mathbb{R}^3)}^2.
\end{eqnarray*}
Note that $(1+\|u_n\|_\lambda)J_{\alpha,\lambda}'(u_n)=o_n(1)$ strongly in 
$E_\lambda^*$ and
$J_{\alpha,\lambda}(u_n)=c_{\alpha,\lambda}+o_n(1)$, by the Young inequality 
and the fact that $4<p<6$, we deduce that
\begin{align*}
&c_{\alpha,\lambda}+o_n(1)\\
&= J_{\alpha,\lambda}(u_n)-\frac1p\langle J_{\alpha,\lambda}'(u_n), 
 u_n\rangle_{E_\lambda^*, E_\lambda}\\
&= \alpha(\frac{1}{4}-\frac1p)\|\nabla u_n\|_{L^2(\mathbb{R}^3)}^4
 +(\frac12-\frac1p)\|u_n\|_\lambda^2-(\frac12-\frac1p)\mathcal{D}_\lambda(u_n,u_n)\\
&\geq \frac{p-4}{4p}(\alpha\|\nabla u_n\|_{L^2(\mathbb{R}^3)}^4
 +\|u_n\|_\lambda^2)-\frac{p-2}{2p}|a_0||\mathcal{A}_\infty|^{\frac{2}{3}}S^{-1}
 \|\nabla u_n\|_{L^2(\mathbb{R}^3)}^2\\
&\geq \frac{p-4}{8p}(\alpha\|\nabla u_n\|_{L^2(\mathbb{R}^3)}^4
 +\|u_n\|_\lambda^2)-\frac{2(p-2)^2}{\alpha(p-4)p}|a_0|^2
 |\mathcal{A}_\infty|^{\frac{4}{3}}S^{-2},
\end{align*}
where $\langle \cdot,\cdot\rangle_{E_{\lambda}^*, E_{\lambda}}$ is the duality 
pairing of $E_{\lambda}^*$ and $E_{\lambda}$.
The preceding inequality, together with $c_{\alpha,\lambda}\in[C_\alpha, C'_\alpha]$ 
and $4<p<6$, implies $\{\|u_n\|_\lambda\}$ is bounded.
\end{proof}

By Lemma~\ref{lem008}, we can see that $u_n=u_{\alpha,\lambda}+o_n(1)$ weakly 
in $E_\lambda$ for some $u_{\alpha,\lambda}\in E_\lambda$ up to a subsequence.  
Without loss of generality, we may assume that $u_n=u_{\alpha,\lambda}+o_n(1)$ 
weakly in $E_\lambda$.

\begin{lemma}\label{lem009}
Suppose that {\rm (A1)--(A3)} hold.  For every $\alpha>0$, if either
 $a_0\geq0$ or $a_0<0$ with $\beta_{k_0^*-1}<1$ then there exists 
$\overline{\Lambda}_{k_0^*}>\widetilde{\Lambda}_{k_0^*}$ such that
 $u_{\alpha,\lambda}$ is a nontrivial solution of \eqref{ePal} for 
$\lambda>\overline{\Lambda}_{k_0^*}$.
\end{lemma}

\begin{proof}
We first prove that $u_{\alpha,\lambda}\neq 0$ in $E_\lambda$.  Indeed, 
suppose on the contrary, then by the Sobolev embedding theorem, we can see 
that $u_n=o_n(1)$ strongly in $L^2_{loc}(\mathbb{R}^3)$, which, together 
with  (A2), implies $u_n=o_n(1)$ strongly in $L^2(\mathcal{A}_\infty)$. 
 It follows from Lemma~\ref{lem008},  conditions (A1)--(A2) and the H\"older 
and the Sobolev inequality that
\begin{equation}
\begin{aligned}
\int_{\mathbb{R}^3}|u_n|^pdx
&\leq \Big(\int_{\mathbb{R}^3}|u_n|^2dx\Big)^{\frac{6-p}{4}}
\Big(\int_{\mathbb{R}^3}|u_n|^6dx\Big)^{\frac{p-2}{4}} \\
&\leq S^{-\frac{3(p-2)}{4}}\|\nabla u_n\|_{L^2(\mathbb{R}^3)}^{\frac{3(p-2)}{2}}
\Big(\int_{\mathbb{R}^3\backslash\mathcal{A}_\infty}|u_n|^2dx+o_n(1)
\Big)^{\frac{6-p}{4}} \\
&\leq S^{-\frac{3(p-2)}{4}}(C_{1}+o_n(1))^{\frac{5p-10}{4}}
\Big(\frac{1}{a_0+a_\infty\lambda}\Big)^{\frac{6-p}{4}}
\|u_n\|_\lambda^2+o_n(1).
\end{aligned}\label{eq8888}
\end{equation}
On the other hand, by  conditions (A1)--(A2) once more, we have
\begin{equation}
\mathcal{D}_\lambda(u_n, u_n)\leq|a_0|\int_{\mathcal{A}_\infty}|u_n|^2dx=o_n(1).
\label{eq8889}
\end{equation}
Therefore, we deduce from the fact that
$(1+\|u_n\|_\lambda)J_{\alpha,\lambda}'(u_n)=o_n(1)$ strongly in
$E_\lambda^*$ that
\begin{align*}
&\alpha\|\nabla u_n\|_{L^2(\mathbb{R}^2)}^4+\|u_n\|_\lambda^2\\
&\leq S^{-3(p-2)}(C_{1}+o_n(1))^{\frac{5p-10}{4}}
\Big(\frac{1}{a_0+a_\infty\lambda}\Big)^{\frac{6-p}{4}}\|u_n\|_\lambda^2+o_n(1),
\end{align*}
which yields that there exists
$\overline{\Lambda}_{k_0^*}>\widetilde{\Lambda}_{k_0^*}$ dependent of
$\alpha$ such that $u_n=o_n(1)$ strongly in $E_\lambda$ with
$\lambda>\overline{\Lambda}_{k_0^*}$.  It is impossible since
$c_{\alpha,\lambda}\geq C_\alpha>0$ for all
$\lambda>\widetilde{\Lambda}_{k_0^*}$.
Therefore $u_{\alpha,\lambda}\neq 0$ in $E_\lambda$. It remains to show that
$J_{\alpha,\beta}'(u_{\alpha,\beta})=0$ in $E_\lambda^*$.
In fact, without loss of generality, we may assume that
$\|u_n\|_{L^2(\mathbb{R}^3)}^2=A+o_n(1)$ and consider the following energy functional
\[
I_{\alpha, \lambda}(u)=\frac{\alpha A}{2}\|u\|_{L^2(\mathbb{R}^3)}^2
+\frac12\|u\|_\lambda^2-\frac12 \mathcal{D}_\lambda(u, u)
-\frac1p\|u\|_{L^p(\mathbb{R}^3)}^p.
\]
Clearly, by \eqref{eq0001}, $I_{\alpha, \lambda}(u)$ is of $C^2$ in $E_\lambda$
for $\lambda>\overline{\Lambda}_{k_0^*}$.
Since $(1+\|u_n\|_\lambda)J_{\alpha,\lambda}'(u_n)=o_n(1)$ strongly in
$E_\lambda^*$, it is easy to see from $\|u_n\|_\lambda^2=A+o_n(1)$ and
$u_n=u_{\alpha,\lambda}+o_n(1)$ weakly in $E_\lambda$ that
 $\langle I_{\alpha, \lambda}'(u_n), u_n-u_{\alpha,\beta}\rangle_{E_\lambda^*,
 E_\lambda}=o_n(1)$ and $I_{\alpha, \lambda}'(u_n)=o_n(1)$ strongly in
$E_\lambda^*$, so that $I_{\alpha, \lambda}'(u_{\alpha, \lambda})=0$ in
$E_\lambda^*$.  In particular, $\langle I_{\alpha, \lambda}'
(u_{\alpha, \lambda}), u_n-u_{\alpha,\beta}\rangle_{E_\lambda^*, E_\lambda}=0$.
Now, we  obtain
\begin{align*}
o_n(1)&= \langle I_{\alpha, \lambda}'(u_n)-I_{\alpha, \lambda}'(u_{\alpha, \lambda}),
 u_n-u_{\alpha,\beta}\rangle_{E_\lambda^*, E_\lambda}\\
&= \alpha A\|u_n-u_{\alpha,\beta}\|_{L^2(\mathbb{R}^3)}^2
 +\|u_n-u_{\alpha,\beta}\|_\lambda^2\\
&\quad -\mathcal{D}_\lambda(u_n-u_{\alpha,\beta}, u_n-u_{\alpha,\beta})
 -\|u_n-u_{\alpha,\beta}\|_{L^p(\mathbb{R}^3)}^p.
\end{align*}
Since $u_n-u_{\alpha,\beta}=o_n(1)$ weakly in $E_\lambda$, by using similar
 arguments in the proofs of \eqref{eq8888} and \eqref{eq8889}, we can see that
$u_n-u_{\alpha,\beta}=o_n(1)$ strongly in $E_\lambda$ for $\lambda$
sufficiently large, say $\lambda>\overline{\Lambda}_{k_0^*}$.
Thus, we must have that $J_{\alpha,\beta}'(u_{\alpha,\beta})=0$ in $E_\lambda^*$
for $\lambda>\overline{\Lambda}_{k_0^*}$.
\end{proof}

The following lemma will give a description on the concentration behavior 
of the nontrivial solutions $u_{\alpha,\lambda}$ as $\lambda\to+\infty$.
\begin{lemma}\label{lem010}
Suppose that  (A1)--(A3) hold.  For every $\alpha>0$, if either $a_0\geq0$ 
or $a_0<0$ with $\beta_{k_0^*-1}<1$ then we have  
$u_{\alpha,\lambda}\to u_{\alpha}$ strongly in $H^1(\mathbb{R}^3)$ as 
$\lambda\to+\infty$ up to a subsequence.  Furthermore, $u_\alpha$ 
is a nontrivial solution of \eqref{ePa*}.
\end{lemma}

\begin{proof}
Let $u_{\alpha,\lambda_n}$ be the nontrivial solution obtained in 
Lemma~\ref{lem009} with $\lambda_n\to+\infty$ as $n\to\infty$.  
By Lemma~\ref{lem008}, we can see that
\[
\int_{\mathbb{R}^3}(|\nabla u_{\alpha, \lambda_n}|^2
+(\lambda_n a(x)+a_0)^+|u_{\alpha,\lambda_n}|^2)dx
\leq C_1\quad\text{for all }n\in\mathbb{N}.
\]
It follows that $\{u_{\alpha, \lambda_n}\}$ is bounded in 
$D^{1,2}(\mathbb{R}^3)$ for $n$ and
\begin{eqnarray*}
\int_{\mathbb{R}^3}(a(x)+\frac{a_0}{\lambda_n})^+|u_{\alpha,\lambda_n}|^2dx=o_n(1).
\end{eqnarray*}
Without loss of generality, we may assume that 
$u_{\alpha, \lambda_n}=u_\alpha+o_n(1)$ weakly in $D^{1,2}(\mathbb{R}^3)$. 
 Thanks to the Sobolev embedding theorem and  conditions (A1)--(A3), we can see 
that $u_{\alpha, \lambda_n}=u_\alpha+o_n(1)$ strongly in 
$L^2(\mathbb{R}^3)$ and $u_\alpha\in H_0^1(\Omega)$ with 
$u_\alpha\equiv0$ on $\mathbb{R}^3\backslash\Omega$.  
Therefore, by the H\"older and the Sobolev inequality, we obtain
\begin{align*}
&\|u_{\alpha, \lambda_n}-u_\alpha\|_{L^p(\mathbb{R}^3)}\\
&\leq\|u_{\alpha, \lambda_n}-u_\alpha\|_{L^2(\mathbb{R}^3)}^{\frac{6-p}{2p}}
(\|u_{\alpha, \lambda_n}\|_{L^6(\mathbb{R}^3)}
+\|u_{\alpha}\|_{L^6(\mathbb{R}^3)})^{\frac{3p-6}{2p}}=o_n(1).
\end{align*}
On the other hand, by a variant of the Lebesgue dominated convergence 
theorem (cf. \cite[Theorem~2.2]{PK74}) and the condition (A1), 
we also have $\mathcal{D}_{\lambda_n}(u_{\alpha, \lambda_n}-u_\alpha,
 u_{\alpha, \lambda_n}-u_\alpha)=o_n(1)$.  Therefore,
\begin{align*}
\int_{\Omega}|u_\alpha|^pdx
&= \|u_{\alpha, \lambda_n}\|_{L^p(\mathbb{R}^3)}^p+o_n(1)\\
&= \mathcal{D}_{\lambda_n}(u_{\alpha, \lambda_n}, u_{\alpha, \lambda_n})
 +\|u_{\alpha, \lambda_n}\|_{\lambda_n}^2
 +\alpha\|\nabla u_{\alpha, \lambda_n}\|_{L^2(\mathbb{R}^3)}^4\\
&\geq \int_{\Omega}\alpha|\nabla u_\alpha|^4+|\nabla u_\alpha|^2
 +a_0|u_\alpha|^2dx+o_n(1).
\end{align*}
Note that $u_\alpha\in H_0^1(\Omega)\subset H^1(\mathbb{R}^3)$, it is easy to see from 
$J_{\alpha,\lambda_n}'(u_{\alpha, \lambda_n})=0$ in 
$E_{\lambda_n}^*$ that $u_\alpha$ is a solution of \eqref{ePa*}.  
In particular,
\[
\int_{\Omega}\alpha|\nabla u_\alpha|^4+|\nabla u_\alpha|^2
+a_0|u_\alpha|^2dx=\int_{\Omega}|u_\alpha|^pdx.
\]
Thus, $u_{\alpha, \lambda_n}=u_\alpha+o_n(1)$ strongly in $D^{1,2}(\mathbb{R}^3)$ and
\[
\int_{\mathbb{R}^3}\lambda_na(x)u_{\alpha,\lambda_n}^2dx=o_n(1).
\]
It follows that $u_{\alpha, \lambda_n}=u_\alpha+o_n(1)$ strongly in $ H^1(\mathbb{R}^3)$.  
Thanks to $c_{\alpha, \lambda}\geq C_\alpha>0$, $u_{\alpha}$ must be nonzero. 
 Hence, $u_\alpha$ is a nontrivial solution of \eqref{ePa*}.
\end{proof}


\begin{proof}[Proof of Theorem~\ref{thm001}]
The statement of the theorem follows immediately from 
Lemmas~\ref{lem009} and \ref{lem010}.
\end{proof}

\subsection*{Acknowledgements}
Y. Wu is supported by the Fundamental Research Funds for the Central
 Universities (2014QNA67).


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\end{document}