\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 280, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/280\hfil Kirchhoff problems]
{Existence of positive solutions for \\ Kirchhoff problems}

\author[J.-F. Liao, P. Zhang, X.-P. Wu \hfil EJDE-2015/280\hfilneg]
{Jia-Feng Liao, Peng Zhang, Xing-Ping Wu}

\address{Jia-Feng Liao \newline
School  of  Mathematics  and  Statistics,
Southwest University, Chongqing 400715, China. \newline
School of Mathematics and Computational Science,
Zunyi Normal College, Zunyi, China}
\email{liaojiafeng@163.com}

\address{Peng Zhang \newline
School of Mathematics and Computational Science,
Zunyi Normal College, Zunyi, China}
\email{gzzypd@sina.com}

\address{Xing-Ping Wu (corresponding author)\newline
School  of  Mathematics  and  Statistics,
Southwest University,  Chongqing 400715, China}
\email{wuxp@swu.edu.cn}


\thanks{Submitted June 25, 2015. Published November 10, 2015.}
\subjclass[2010]{35B09, 35B33, 35J20}
\keywords{Kirchhoff type equation; resonance; positive solution; 
\hfill\break\indent mountain pass lemma}

\begin{abstract}
 We study problems for the Kirchhoff equation
 \begin{gather*}
 -\Big(a+b\int_{\Omega}|\nabla u|^2dx\Big)\Delta u
 =\nu u^3+ Q(x)u^{q},\quad \text{in }\Omega, \\
 u=0,  \quad \text{on }\partial\Omega,
 \end{gather*}
 where $\Omega\subset \mathbb{R}^3$ is a bounded domain, $a,b\geq0$
 and $a+b>0$, $\nu>0$, $3<q\leq5$ and $Q(x)>0$ in $\Omega$.
 By the mountain pass lemma, the existence of positive solutions is 
 obtained.  Particularly, we give a condition of $Q$ to ensure the
 existence of solutions for the case of $q=5$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction and main results}

 In this article, we consider the  Kirchhoff type problem
\begin{equation}\label{1.1}
\begin{gathered}
-\Big(a+b\int_{\Omega}|\nabla u|^2dx\Big)\Delta u
 =\nu u^3+ Q(x)u^{q},\quad \text{in }\Omega, \\
u=0,     \quad \text{on }\partial\Omega,
\end{gathered}
\end{equation}
where $\Omega\subset \mathbb{R}^3$ is a bounded domain, $a,b\geq0$ and $a+b>0$,
$\nu>0,3<q\leq5$ are four parameters.
The coefficient function $Q$ is a positive function in $\Omega$.
When $a=0,b>0$, problem \eqref{1.1} is called degenerate, and the case of $a,b>0$
is called non-degenerate.

When $a\geq0$ and $b>0$, problem \eqref{1.1} is called the Kirchhoff type problem.
Kirchhoff type problems are often referred to as being nonlocal because
of the presence of the term $(\int_{\Omega}|\nabla u|^2dx)\Delta u$
which implies that the equation in \eqref{1.1} is no longer a pointwise
equation. The existence and multiplicity of solutions for the  problem
\begin{equation}\label{1.01}
\begin{gathered}
-\Big(a+b\int_{\Omega}|\nabla u|^2dx\Big)\Delta u=f(x,u),\quad \text{in }\Omega, \\
u=0,      \quad \text{on }\partial\Omega,
\end{gathered}
\end{equation}
on a smooth bounded domain $\Omega\subset \mathbb{R}^3$ and
$f:\overline{\Omega}\times \mathbb{R}\to \mathbb{R}$ a continuous function,
has been extensively studied
(see \cite{ACF,GA},\cite{CKW}-\cite{ZP1},
\cite{ST1}-\cite{WTXZ},\cite{XWT,ZP}).

Particularly, Sun and Tang \cite{ST} considered the  problem
\begin{equation}\label{1.02}
\begin{gathered}
-\Big(a+b\int_{\Omega}|\nabla u|^2dx\Big)\Delta u
=\lambda u^3+ g(u)-h(x),\quad \text{in }\Omega, \\
u=0,     \quad \text{on }\partial\Omega,
\end{gathered}
\end{equation}
where $h\in L^2(\Omega)$ and $g\in C(\mathbb{R},\mathbb{R})$ satisfies
\begin{equation}\label{1.4}
\lim_{|t|\to\infty}\frac{g(t)}{t^3}=0.
\end{equation}
Under a Landesman-Lazer type condition, by the minimax methods,
they obtained the existence of solutions for problem \eqref{1.02}.


When $a=1$ and $b=0$, problem \eqref{1.1} reduces to the  semilinear elliptic problem
\begin{equation}\label{1.2}
\begin{gathered}
-\Delta u=\nu u^3+\lambda u^{q},\quad \text{in }\Omega, \\
u=0,     \quad \text{on }\partial\Omega.
\end{gathered}
\end{equation}
Obviously when $3<q<5$, problem \eqref{1.2} has a positive solution
for all $\nu,\lambda>0$.
While for $q=5,\lambda=1$,  Br\'ezis and Nirenberg \cite{HL} studied
problem \eqref{1.2}. By the variant of the mountain pass theorem
of Ambrosetti and Rabinowitz without the (PS) condition, they  obtained
that there exists $\nu_0>0$ such that
problem \eqref{1.2} has a positive solution for each $\nu\geq\nu_0$.


For $u\in H_0^1(\Omega)$, we define
\begin{equation*}
I(u)=\frac{a}{2}\int_{\Omega}|\nabla u|^2dx+\frac{b}{4}
\Big(\int_{\Omega}|\nabla u|^2dx\Big)^2-\frac{\nu}{4}\int_{\Omega}|u|^4dx
-\frac{1}{q+1}\int_{\Omega}Q(x)|u|^{q+1}dx,
\end{equation*}
where $H_0^1(\Omega)$ is a Sobolev space equipped with the norm
$\|u\|=\big(\int_{\Omega}|\nabla u|^2dx\big)^{1/2}$.
Note that a function $u$ is called a weak solution of \eqref{1.1}
if $u\in H_0^1(\Omega)$ such that
\begin{equation}\label{1.00}
\Big(a+b\int_{\Omega}|\nabla u|^2dx\Big)
\int_{\Omega}(\nabla u,\nabla\varphi)dx
-\nu\int_{\Omega}u^3\varphi dx-\int_{\Omega}Q(x)u^{q}\varphi dx=0,
\end{equation}
for all $\varphi\in H_0^1(\Omega)$.

We denote by $\nu_{1}$ is the first eigenfunction of the  eigenvalue problem
\begin{gather*}
-\Big(\int_{\Omega}|\nabla u|^2dx\Big)\Delta u=\nu u^3,  \quad    x\in\Omega, \\
u=0,            \quad   x\in\partial\Omega.
\end{gather*}
From \cite{ZP1}, we know that $\nu_{1}>0$.
Let $S$ be the best Sobolev constant, namely
\begin{equation}\label{1.3}
S:=\inf_{u\in H_0^1(\Omega)\backslash\{0\}}
\frac{\int_{\Omega}|\nabla u|^2dx}{\left(\int_{\Omega}|u|^{6}dx\right)^{1/3}}.
\end{equation}
As well known, the function
\begin{equation}\label{1.5}
 U(x)=\frac{(3\varepsilon^2)^{1/4}}{\left(\varepsilon^2+|x|^2\right)^{1/2}},
\quad x\in \mathbb{R}^3,
\end{equation}
is an extremal function for the minimum problem \eqref{1.3}; that is,
it is a positive solution of the problem
\begin{equation*}
-\Delta u=u^{5},\quad \forall x\in \mathbb{R}^{N}.
\end{equation*}

Problem \eqref{1.1} with $3<q\leq 5$ does not satisfy
condition \eqref{1.4}. It is natural to ask whether \eqref{1.1} has
a positive solution. Using the mountain pass theorem, we study \eqref{1.1}
and give a positive answer. It is worth pointing out that the result of the case
of $q=5$ is more meaningful. Because of $q=5$ is critical case and the
coefficient of critical term is no longer constant. We give the suitable
condition (A1) in the Theorem \ref{thm1.2} below to ensure the existence of solutions
to problem \eqref{1.1}.

Our main results are described as follows.

\begin{theorem} \label{thm1.1}
 Assume $a,b>0$, $3<q<5$ and $Q\in L^{\frac{6}{5-q}}(\Omega)$ is a positive
function, then  \eqref{1.1} possesses a positive solution $u^{*}$ for
all $\nu>0$, and $I(u^{*})>0$.
\end{theorem}

\begin{remark} \label{rmk1}\rm
 Obviously, Theorem \ref{thm1.1} does not apply to \eqref{1.4}. 
For all $\nu>0$, we obtain the existence of positive solutions
for problem \eqref{1.1}. For the degenerate case, that is $a=0$, $b>0$,
we can also obtain that problem \eqref{1.1} possesses a positive solution
for all $0<\nu<b\nu_{1}$.
\end{remark}

\begin{theorem} \label{thm1.2}
 Assume $a,b>0$, $q=5$, $Q\in C(\overline{\Omega})$ is a positive function
and satisfies the assumption
\begin{itemize}
\item[(A1)]  There exists $x_0\in\Omega$ such that
 $Q(x_0)=Q_{M}=\max_{x\in\overline{\Omega}}Q(x)$ and
 $$
Q(x)-Q(x_0)=o(|x-x_0|^{}),\quad\text{as }x\to x_0.
$$
\end{itemize}
Then there exists $\nu^{*}>0$ such that \eqref{1.1} possesses a
 positive solution $u^{*}$ for all $\nu>\nu^{*}$, and $I(u^{*})>0$.
\end{theorem}

\begin{remark} \label{rmk2} \rm
 This case is the critical exponent problem, and Theorem \ref{thm1.2} does not
apply to \eqref{1.4}. When $Q(x)\equiv1$, the Kirchhoff
type problems with critical exponent have been considered by several papers,
such as \cite{ACF,HZ,LLT} \cite{ND1}-\cite{DN3},
\cite{SL2,WTXZ,XWT}.
Particularly, problem \eqref{1.1} with $Q(x)\equiv1$ was been considered
in \cite{DN2}. However, there exists a flaw in the proof of 
\cite[Theorem 1.3]{DN2} with the case $\theta=4$.
\end{remark}

To our best knowledge, problem \eqref{1.1} with $Q(x)$ not constant
has not been considered yet. When $Q(x)$ is not constant, the analysis
of the compactness becomes complicated, which results in much difficulty.
It is worth pointing out that (A1) ensures the existence of solutions.
Obviously, Theorem \ref{thm1.2} extends the corresponding result
of \cite{DN2}.

This article is organized as follows.
In Section 2, we consider the case of $3<q<5$ and prove Theorem \ref{thm1.1}
by the variational methods. We study the critical case of problem \eqref{1.1}
with $q=5$ and give the proof of Theorem \ref{thm1.2} in Section 3.

\section{The case $3<q<5$}

 In this section, suppose that $Q\in L^{\frac{6}{5-q}}(\Omega)$ is a positive
function and $3<q<5$. We will prove Theorem \ref{thm1.1} by the mountain pass theorem.
Before proving Theorem \ref{thm1.1}, we give the following lemma.

\begin{lemma} \label{lem2.1}
Assume $a,b>0$, $3<q<5$ and $Q\in L^{\frac{6}{5-q}}(\Omega)$ is a positive
function, then the functional $I$
 satisfies the (PS)c condition for all $\nu>0$.
\end{lemma}

\begin{proof}
Suppose that $\{u_{n}\}$ is a (PS)c sequence of $I$, that is,
\begin{equation}\label{2.0}
I(u_{n})\to c,\quad I'(u_{n})\to0,
\end{equation}
as $n\to+\infty$. We claim that $\{u_{n}\}$ is bounded in $H_0^1(\Omega)$.
In fact, from \eqref{2.0} one has
\begin{align*}
1+c+o(1)\|u_{n}\|
&\geq  I(u_{n})-\frac{1}{4}\langle I'(u_{n}),u_{n}\rangle\\
&=  \frac{a}{4}\|u_{n}\|^2+\big(\frac{1}{4}-\frac{1}{q+1}\big)
\int_{\Omega}Q(x)|u_{n}|^{q+1}dx\\
&\geq \frac{a}{4}\|u_{n}\|^2.
\end{align*}
Hence, we conclude that $\{u_{n}\}$ is bounded in $H_0^1(\Omega)$.
Going if necessary to a subsequence, still denoted by $\{u_n\}$,
there exists $u\in H_0^1(\Omega)$ such that
\begin{equation}\label{2.1}
\begin{gathered}
u_n\rightharpoonup u,\quad \text{weakly in } H_0^1(\Omega),\\
u_n\to u,\quad \text{strongly in } L^{s}(\Omega),\; 1\leq s<6,\\
u_n(x)\to u(x),\quad \text{a.e. in } \Omega,
\end{gathered}
\end{equation}
as $n\to\infty$. Now, we only need to prove that $u_n\to u$ as
 $n\to\infty$ in $H_0^1(\Omega)$. As usually, letting $w_{n}=u_{n}-u$,
 we need prove that $\|w_{n}\|\to0$ as $n\to\infty$.
By the Vitali theorem (see \cite[p.133]{RW}), we claim that
\begin{equation}\label{2.2}
\lim_{n\to\infty}\int_{\Omega}Q(x)|u_n|^{q+1}dx
=\int_{\Omega}Q(x)|u|^{q+1}dx.
\end{equation}
Indeed, we only need to prove that $\{\int_{\Omega}Q(x)|u_n|^{p+1}dx,n\in N\}$ is
equi-absolutely-continuous. Note that $\{u_{n}\}$ is bounded in $H_0^1(\Omega)$,
by the Sobolev embedding theorem, then exists a constant $C>0$ such that
$|u_{n}|_{6}\leq C<\infty$. From the H$\ddot{\mathrm{o}}$lder inequality, for every
$\varepsilon>0$, setting
$\delta>0$, when $E\subset\Omega$ with
$\operatorname{meas}E<\delta$, we have
\begin{equation*}
\int_{E} Q(x)|u_{n}|^{q+1}dx
\leq |u_n|_{6}^{q+1}\Big(\int_{E}Q^{\frac{6}{5-q}}(x)dx\Big)^{\frac{5-q}{6}}
<\varepsilon,
\end{equation*}
where the last inequality is from the absolutely-continuity
of $\int_{\Omega}Q^{\frac{6}{5-q}}(x)dx$.
Thus, our claim is proved. Moreover, one also has
\begin{gather}\label{2.3}
\int_{\Omega}|\nabla u_{n}|^2dx
=\int_{\Omega}|\nabla w_{n}|^2dx+\int_{\Omega}|\nabla u|^2dx+o(1), \\
\label{2.4}
\Big(\int_{\Omega}|\nabla u_n|^2dx\Big)^2
=\|w_n\|^4+\|u\|^4+2\|w_n\|^2\|u\|^2+o(1).
\end{gather}
Since $I'(u_{n})\to0$, one obtains
$$
a\|u_{n}\|^2+b\|u_{n}\|^4-\nu\int_{\Omega}|u_n|^4dx
-\int_{\Omega}Q(x)|u_n|^{q+1}dx=o(1),
$$
consequently,  from \eqref{2.1}-\eqref{2.4}, we deduce that
\begin{equation}\label{2.5}
a\|w_{n}\|^2+a\|u\|^2+b\|w_{n}\|^4+2b\|w_{n}\|^2\|u\|^2+b\|u\|^4
-\nu|u|_{4}^4-\int_{\Omega}Q(x)|u|^{q+1}dx=o(1).
\end{equation}
From \eqref{2.0} it follows that
\begin{equation}\label{2.6}
\lim_{n\to\infty}\langle I'(u_{n}), u\rangle= a\|u\|^2+bl^2\|u\|^2+b\|u\|^4
-\nu|u|_{4}^4-\int_{\Omega}Q(x)|u|^{q+1}dx=0,
\end{equation}
where $l=\lim_{n\to\infty}\|w_{n}\|$.
According to \eqref{2.5} and \eqref{2.6}, we have
$$
a\|w_{n}\|^2+b\|w_{n}\|^4+b\|w_{n}\|^2\|u\|^2=o(1),
$$
consequently, one has $al^2+bl^4+bl^2\|u\|^2=0$.
Thus $l=0$; that is, $u_n\to u$ as $n\to\infty$ in $H_0^1(\Omega)$.
This completes the proof.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
The main idea is to construct a suitable geometry of mountain pass lemma
(see\cite{AR}). Then obtain a critical point
of $I$ in $H_0^1(\Omega)$. We claim that $I$ has the geometry of mountain
pass lemma in $H_0^1(\Omega)$. Indeed, since
\begin{equation*}
I(u)=\frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4
-\frac{\nu}{4}\int_{\Omega}|u|^4dx-\frac{1}{q+1}\int_{\Omega}Q(x)|u|^{q+1}dx,
\end{equation*}
then $I(0)=0$, and for every $u\in H_0^1(\Omega)\backslash\{0\}$ one has
$$
\lim_{t\to0+}\frac{I(tu)}{t^2}=\frac{a}{2}\|u\|^2,\quad
\lim_{t\to+\infty}\frac{I(tu)}{t^{q+1}}=-\frac{1}{q+1}\int_{\Omega}Q(x)|u|^{q+1}dx.
$$
Since $a>0$ and $\int_{\Omega}Q(x)|u|^{q+1}dx>0$, then there exist $R,\alpha>0$
and $e\in H_0^1(\Omega)$ with $\|e\|>R$ such that
$I|_{\partial B_{R}}\geq \alpha$ and $I(e)<0$, where
$\partial B_{R}=\{u\in H_0^1(\Omega)~|~\|u\|=R\}$. Thus, $I$ satisfies the
geometry of the mountain-pass lemma.

Let
\begin{equation*}
c=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I(\gamma(t)),
\end{equation*}
where $\Gamma=\{\gamma\in C([0,1], H_0^1(\Omega)): \gamma(0)=0, \gamma(1)=e\}$.
Then $c\geq\alpha$. According to Lemma \ref{lem2.1},
$I$ satisfies the conditions of the mountain pass lemma.
Applying the mountain-pass lema, there exists a sequence
$\{u_n\}\subset H_0^1(\Omega)$, such that $I(u_{n})\to c$ and $I'(u_{n})\to0$
as $n\to\infty$. Then $c$ is a critical value of
$I$ and $c>\alpha>0$. Moreover, $\{u_n\}\subset H_0^1(\Omega)$
has a convergent subsequence, still denoted by $\{u_n\}$, we may assume that
$u_n\to u^{*}$ in $H_0^1(\Omega)$ as $n\to\infty$. Thus $I(u^{*})=c>0$ and
$u^{*}$ is a nonzero solution of  \eqref{1.1}. Since
$I(|u|)=I(u)$, by a result due to Br\'ezis and Nirenberg
\cite[Theorem 10]{BCN}, we conclude that $u^{*}\geq0$. By the strong maximum
principle, one has $u^{*}>0$ in $\Omega$. Therefore, $u^{*}$ is a positive
solution of problem \eqref{1.1} with $I(u^{*})>0$.
This completes the proof.
\end{proof}

\section{The case q=5}

 In this part, assume that $Q\in C(\overline{\Omega})$ is a positive
function and satisfies (A1). We study the case of $q=5$.
This case is more delicate, because of the Sobolev embedding
$H_0^1(\Omega)\hookrightarrow L^{6}(\Omega)$ is not compact.
Thus the functional $I$ does not satisfy the $(PS)_{c}$ condition.
When $Q(x)$ is not constant, the analysis of $(PS)$ sequences becomes
complicated, which results in much difficulty. We will complete the proof
of Theorem \ref{thm1.2} by the mountain pass lemma. Now, we prove that $I$ satisfies
the local $(PS)_{c}$ condition.

\begin{lemma} \label{lem3.1}
 Assume $a,b>0$ and the positive function $Q\in C(\overline{\Omega})$
satisfies (A1), then $I$ satisfies the $(PS)_{c}$ condition, where
$c\in(0,\Lambda)$ with
$$
\Lambda=\frac{abS^3}{4Q_{M}}+\frac{b^3S^{6}}{24Q_{M}^2}
+\frac{aS\sqrt{b^2S^4+4aSQ_{M}}}{6Q_{M}}
+\frac{b^2S^4\sqrt{b^2S^4+4aSQ_{M}}}{24Q_{M}^2}.
$$
\end{lemma}

\begin{proof}
Suppose that $\{u_{n}\}$ is a $(PS)_{c}$ sequence for $c\in (0,\Lambda)$; that is,
\begin{equation}\label{3.1}
I(u_{n})\to c,\quad I'(u_{n})\to0,
\end{equation}
as $n\to+\infty$. According to Lemma \ref{lem2.1}, we can easy obtain that
 $\{u_{n}\}$ is bounded in $H_0^1(\Omega)$. Going if necessary to a subsequence,
there exists $u\in H_0^1(\Omega)$ such that \eqref{2.1} holds.
As usually, letting $w_{n}=u_{n}-u$, we need prove that $\|w_{n}\|\to0$ as
$n\to\infty$. We denote $\lim_{n\to\infty}\|w_{n}\|=l$.
As in Lemma \ref{lem2.1}, we  have \eqref{2.3} and \eqref{2.4}.
By Br\'ezis-Lieb's Lemma  \cite{BL}, one has
\begin{equation}\label{3.2}
\int_{\Omega}Q(x)|u_{n}|^{6}dx
=\int_{\Omega}Q(x)|w_{n}|^{6}dx+\int_{\Omega}Q(x)|u|^{6}dx+o(1).
\end{equation}
From \eqref{3.1} and \eqref{2.1}, one obtains
\begin{equation*}
a\|u_{n}\|^2+b\|u_{n}\|^4-\nu\int_{\Omega}|u|^4dx
-\int_{\Omega}Q(x)|u_{n}|^{6}dx=o(1),
\end{equation*}
consequently,  from \eqref{2.3}-\eqref{2.4} and \eqref{3.2} it follows that
\begin{equation}\label{3.3}
\begin{aligned}
& a\|u\|^2+a\|w_n\|^2+b\|u\|^4+b\|w_n\|^4+2b\|w_n\|^2\|u\|^2\\
&-\int_{\Omega}Q(x)|w_n|^6dx-\int_{\Omega}Q(x)|u|^6dx
 -\nu\int_{\Omega}|u|^4dx=o(1).
\end{aligned}
\end{equation}
From \eqref{3.1} it follows that
\begin{equation}\label{3.4}
\lim_{n\to\infty}\langle I'(u_{n}),u\rangle=a\|u\|^2+b\|u\|^4+bl^2\|u\|^2
-\int_{\Omega}Q(x)|u|^6dx-\nu\int_{\Omega}|u|^4dx=0.
\end{equation}
On the one hand, from \eqref{3.4}, we have
\begin{equation}\label{3.0}
\begin{aligned}
 I(u)
&= \frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4-\frac{\nu}{4}\int_{\Omega}|u|^4dx
 -\frac{1}{6}\int_{\Omega}Q(x)|u|^{6}dx\\
&= \frac{a}{4}\|u\|^2+\frac{1}{12}\int_{\Omega}Q(x)|u|^{6}dx
 -\frac{bl^2}{4}\|u\|^2\\
&\geq -\frac{bl^2}{4}\|u\|^2.
\end{aligned}
\end{equation}
On the other hand,  from \eqref{3.3} and \eqref{3.4} it follows that
\begin{equation}\label{3.5}
a\|w_n\|^2+b\|w_n\|^4+b\|w_n\|^2\|u\|^2-\int_{\Omega}Q(x)|w_n|^{6}dx=o(1),
\end{equation}
and
\begin{equation}\label{3.6}
I(u_{n})=I(u)+\frac{a}{2}\|w_n\|^2+\frac{b}{4}\|w_n\|^4
 +\frac{b}{2}\|w_n\|^2\|u\|^2
 -\frac{1}{6}\int_{\Omega}Q(x)|w_n|^{6}dx+o(1).
\end{equation}
From (A1) and \eqref{1.3}, one has
$$
\int_{\Omega}Q(x)|w_n|^{6}dx\leq Q_{M}\int_{\Omega}|w_n|^{6}dx
\leq Q_{M}\frac{\|w_{n}\|^{6}}{S^3},
$$
consequently, it follows from \eqref{3.5} that
$$
al^2+bl^4+bl^2\|u\|^2\leq Q_{M}\frac{l^{6}}{S^3},
$$
which implies that
\begin{equation}\label{3.7}
l^2\geq\frac{1}{2}\Big[\frac{bS^3}{Q_{M}}+\frac{\sqrt{b^2S^{6}
+4S^3Q_{M}(a+b\|u\|^2)}}{Q_{M}}\Big].
\end{equation}
Thus, from \eqref{3.5}-\eqref{3.7}, we obtain
\begin{align*}
 I(u)
&= \lim_{n\to\infty}\Big[I(u_{n})-\frac{a}{2}\|w_n\|^2-\frac{b}{4}\|w_n\|^4
-\frac{b}{2}\|w_n\|^2\|u\|^2
+\frac{1}{6}\int_{\Omega}Q(x)|w_n|^{6}dx\Big]\\
&=  c-\Big(\frac{a}{3}l^2+\frac{b}{12}l^4+\frac{b}{3}l^2\|u\|^2\Big)\\
&\leq c-\Big[\frac{a}{6}\Big(\frac{bS^3}{Q_{M}}+\frac{\sqrt{b^2S^{6}
 +4S^3Q_{M}(a+b\|u\|^2)}}{Q_{M}}\Big)\\
&\quad +\frac{b}{48}\Big(\frac{bS^3}{Q_{M}}+\frac{\sqrt{b^2S^{6}
 +4S^3Q_{M}(a+b\|u\|^2)}}{Q_{M}}\Big)^2\\
&\quad +\frac{b\|u\|^2}{24}\Big(\frac{bS^3}{Q_{M}}+\frac{\sqrt{b^2S^{6}
 +4S^3Q_{M}(a+b\|u\|^2)}}{Q_{M}}\Big)\Big]
-\frac{bl^2}{4}\|u\|^2\\
&\leq  c-\Big(\frac{abS^3}{4Q_{M}}+\frac{b^3S^{6}}{24Q_{M}^2}
 +\frac{aS\sqrt{b^2S^4+4aSQ_{M}}}{6Q_{M}}
+\frac{b^2S^4\sqrt{b^2S^4+4aSQ_{M}}}{24Q_{M}^2}\Big)\\
&\quad  -\frac{bl^2}{4}\|u\|^2\\
&<-\frac{bl^2}{4}\|u\|^2,
\end{align*}
which contradicts \eqref{3.0}. Hence, $l\equiv0$; that is,
$u_{n}\to u$ in $H_0^1(\Omega)$ as $n\to\infty$. Therefore,
$I$ satisfies the $(PS)_{c}$ condition for all $c<\Lambda$.
This completes the proof.
\end{proof}

Next, we estimate the level value of functional $I$ and obtain the following lemma.

\begin{lemma} \label{lem3.2}
 Assume that $a,b>0$ and the positive function $Q\in C(\overline{\Omega})$ 
satisfies {\rm (A1)}. Then there exists $u_0\in H_0^1(\Omega)$, such that 
$\sup_{t\geq0}I(tu_0)<\Lambda$ for all $\nu>\nu^{*}$, where $\Lambda$ 
is defined by Lemma \ref{lem3.1} and $\nu^{*}$ independent of $u_0$ is a positive constant.
\end{lemma}

\begin{proof} 
Define a cut-off function $\eta\in C_0^{\infty}(\Omega)$ such that $0\leq\eta\leq1$,
 $|\nabla \eta|\leq C_{1}$. For some $\tilde{\delta}>0$, we define
\begin{equation*}
\eta(x)=\begin{cases}
1, &|x-x_0|\leq\frac{\tilde{\delta}}{2},\\
0, &|x-x_0|\geq \tilde{\delta},
\end{cases}
\end{equation*}
where $x_0$ is defined by (A1). Set $u_{\varepsilon}=\eta(x)U(x-x_0)$. 
As well known(see \cite{HL,MW}), one has
\begin{gather}\label{3.8}
\|u_{\varepsilon}\|^2=\|U_{\varepsilon}\|^2
+O(\varepsilon)=S^{3/2}+O(\varepsilon), \\
\label{3.9}
|u_{\varepsilon}|_{6}^{6}=|U_{\varepsilon}|_{6}^{6}+O(\varepsilon^3)
=S^{3/2}+O(\varepsilon^3),
\end{gather}
and
\begin{equation}\label{4.9}
\begin{gathered}
C_{2}\varepsilon^{\frac{s}{2}}\leq\int_\Omega u_\varepsilon^sdx
\leq C_{3}\varepsilon^{\frac{s}{2}}, \quad 1\leq s<3,\\
C_{4}\varepsilon^{\frac{s}{2}}|\ln\varepsilon|
 \leq\int_\Omega u_\varepsilon^sdx
 \leq C_{5}\varepsilon^{\frac{s}{2}}|\ln\varepsilon|, \quad s=3,\\
C_{6}\varepsilon^{\frac{6-s}{2}}
 \leq\int_\Omega u_\varepsilon^sdx
 \leq C_{7}\varepsilon^{\frac{6-s}{2}}, \quad 3<s<6.
\end{gathered}
\end{equation}
Moreover, from \cite{XWT}, we  have
\begin{equation}\label{3.10}
\begin{gathered}
\|u_{\varepsilon}\|^4=S^3+O(\varepsilon), \quad
\|u_{\varepsilon}\|^{6}=S^\frac{9}{2}+O(\varepsilon), \\
\|u_{\varepsilon}\|^{8}=S^6+O(\varepsilon), \quad
\|u_{\varepsilon}\|^{12}=S^9+O(\varepsilon).
\end{gathered}
\end{equation}
For all $t\geq0$, we define $I(tu_{\varepsilon})$ by
$$
I(tu_{\varepsilon})=\frac{a}{2}t^2\|u_{\varepsilon}\|^2
+\frac{b}{4}t^4\|u_{\varepsilon}\|^4
-\frac{\nu}{4}t^4\int_{\Omega}|u_{\varepsilon}|^4dx
-\frac{t^{6}}{6}\int_{\Omega}Q(x)|u_{\varepsilon}|^{6}dx,
$$
then we have
\begin{gather*}
\lim_{t\to+0}I(tu_\varepsilon)=0,\quad \text{uniformly for all } 
 0<\varepsilon<\varepsilon_0,\\
\lim_{t\to+\infty}I(tu_\varepsilon)=-\infty,\quad \text{uniformly for all } 
 0<\varepsilon<\varepsilon_0,
\end{gather*}
where $\varepsilon_0>0$ is a small constant. Thus 
$\sup_{t\geq0}I(tu_\varepsilon)$ attains for some $t_\varepsilon>0$. 
Moreover, we can claim that there exist two constants $t_0,T_0>0$, 
which independent of $\varepsilon$, such that
$t_0<t_\varepsilon<T_0$. In fact, from $\lim_{t\to+0}I(tu_\varepsilon)=0$ 
uniformly for all $\varepsilon$, we choose 
$\epsilon=\frac{I(t_\varepsilon u_\varepsilon)}{4}>0$, then there exists 
$t_0>0$ such that $|I(t_0u_\varepsilon)|=|I(t_0u_\varepsilon)-I(0)|<\epsilon$. 
Then according to the monotonicity of $I(tu_\varepsilon)$ near $t=0$, 
we have $t_\varepsilon>t_0$. Similarly, we can obtain that $t_\varepsilon<T_0$. 
Therefore, our claim is proved.

Set $I(tu_\varepsilon)=I_{\varepsilon,1}(t)
-\nu I_{\varepsilon,2}(t)$, where
$$
I_{\varepsilon,1}(t)=\frac{a}{2}t^2\|u_{\varepsilon}\|^2
+\frac{b}{4}t^4\|u_{\varepsilon}\|^4
-\frac{t^6}{6}\int_{\Omega}Q(x)u_{\varepsilon}^{6}dx,
$$
and
$$
I_{\varepsilon,2}(t)=\frac{t^4}{4}\int_{\Omega}u_{\varepsilon}^4dx.
$$
First, we estimate the value $I_{\varepsilon,1}$. Since 
$I_{\varepsilon,1}'(t)=at\|u_{\varepsilon}\|^2+bt^3\|u_{\varepsilon}\|^4
-t^5\int_{\Omega}Q(x)u_{\varepsilon}^{6}dx$,  letting $I_{\varepsilon,1}'(t)=0$; 
that is,
\begin{equation}\label{4.10}
a\|u_{\varepsilon}\|^2+bt^2\|u_{\varepsilon}\|^4-t^4
\int_{\Omega}Q(x)u_{\varepsilon}^{6}dx=0,
\end{equation}
one obtains
$$
T_{\varepsilon}^2=\frac{b\|u_\varepsilon\|^4
+\sqrt{b^2\|u_\varepsilon\|^8+4a\|u_\varepsilon\|^2
\int_{\Omega}Q(x)u_\varepsilon^6dx}}
{2\int_{\Omega}Q(x)u_\varepsilon^6dx}.
$$
Then $I_{\varepsilon,1}'(t)>0$ for all $0<t<T_{\varepsilon}$ and 
$I_{\varepsilon,1}'(t)<0$ for all $t>T_{\varepsilon}$, so 
$I_{\varepsilon,1}(t)$ attains its maximum at $T_{\varepsilon}$.
From (A1), let $\varepsilon\to 0^{+}$, we claim that
\begin{equation}\label{3.11}
\left(\int_\Omega Q(x)u_{\varepsilon}^{6}dx\right)^{1/3}
=Q_{M}^{1/3}|u_\varepsilon|_{6}^2+o(\varepsilon).
\end{equation}
In fact, for all $\varepsilon>0$, it follows that
\begin{equation}\label{3.12}
\begin{aligned}
\left|\int_{\Omega}Q(x)u_{\varepsilon}^{6}dx
-\int_{\Omega}Q_{M}u_{\varepsilon}^{6}dx\right|
&\leq \int_{\Omega}|Q(x)-Q(x_0)|u_{\varepsilon}^{6}dx\\
&\leq \int_{\{x\in\Omega:~ |x-x_0|\leq \tilde{\delta}\}}
 |Q(x)-Q(x_0)|u_{\varepsilon}^{6}dx.
\end{aligned}
\end{equation}
From (A1), for all $\eta>0$, there exists $\delta>0$ such that
$$
|Q(x)-Q(x_0)|<\eta|x-x_0|,\quad \mathrm{for~all}~0<|x-x_0|<\delta.
$$
When $\varepsilon>0$ small enough, for $\delta>\varepsilon^{1/2}$, 
it follows from \eqref{3.12} and (A1) that
\begin{align*}
&\Big|\int_{\Omega}Q(x)u_\varepsilon^{6}dx
 -\int_{\Omega}Q_{M}u_\varepsilon^{6}dx\Big|\\
&\leq \int_{\{x\in\Omega:|x-x_0|\leq \tilde{\delta}\}}
 |Q(x)-Q(x_0)|u_\varepsilon^{6}dx\\
&< \int_{\{x\in\Omega:|x-x_0|\leq\delta\}}\eta|x-x_0|
\frac{(3\varepsilon^2)^{3/2}}{[\varepsilon^2+|x-x_0|^2]^3}dx\\
&\quad +\int_{\{x\in\Omega:~\delta<|x-x_0|\leq \tilde{\delta}\}}
\frac{(3\varepsilon^2)^{3/2}}{[\varepsilon^2+|x-x_0|^2]^3}dx\\
&=  \sqrt{27}\eta\int_0^{\delta}r^3\frac{\varepsilon^3}{(\varepsilon^2+r^2)^3}dr
 + \sqrt{27}\int_{\delta}^{\tilde{\delta}}\frac{\varepsilon^3r^2}{(\varepsilon^2+r^2)^3}dr\\
&=  \sqrt{27}\eta\varepsilon\int_0^{\frac{\delta}{\varepsilon}}
  \frac{r^3}{(1+r^2)^3}dr
 + \sqrt{27}\int_{\frac{\delta}{\varepsilon}}^{\frac{\tilde{\delta}}{\varepsilon}}\frac{r^2}{(1+r^2)^3}dr\\
&\leq C_{8}\eta\varepsilon+C_{9}\varepsilon^3.
\end{align*} 
Consequently, one has
$$
\frac{\left|\int_{\Omega}Q(x)u_\varepsilon^{6}dx
-\int_{\Omega}Q_{M}u_\varepsilon^{6}dx\right|}{\varepsilon}
\leq C_{8}\eta+C_{9}\varepsilon^2,
$$
which implies 
$$
\limsup_{\varepsilon\to 0^{+}}\frac{\big|\int_{\Omega}Q(x)u_\varepsilon^{6}dx
-\int_{\Omega}Q_{M}u_\varepsilon^{6}dx\big|}{\varepsilon}
\leq C_{8}\eta.
$$
Then from the arbitrariness of $\eta$, we obtain \eqref{3.11}. 
Thus, from \eqref{3.9} and \eqref{3.11}, one gets
\begin{equation}\label{3.13}
\int_\Omega Q(x)u_{\varepsilon}^{6}dx
=Q_{M}|u_\varepsilon|_{6}^{6}+o(\varepsilon)=Q_{M}S^{3/2}+o(\varepsilon).
\end{equation}
Thus from \eqref{3.8},\eqref{3.10},\eqref{4.10} and \eqref{3.13}, we have
\begin{align}
I_{\varepsilon,1}(t)
&\leq I_{\varepsilon,1}(T_{\varepsilon}) \nonumber\\
&=  T_{\varepsilon}^2\Big(\frac{a}{2}\|u_{\varepsilon}\|^2
 +\frac{b}{4}T_{\varepsilon}^2\|u_{\varepsilon}\|^4
-\frac{T_{\varepsilon}^4}{6}\int_{\Omega}Q(x)u_{\varepsilon}^{6}dx\Big)  \nonumber\\
&=  T_{\varepsilon}^2\Big(\frac{a}{3}\|u_{\varepsilon}\|^2
 +\frac{b}{12}T_{\varepsilon}^2\|u_{\varepsilon}\|^4\Big)  \nonumber\\
&=  \frac{ab\|u_\varepsilon\|^6+a\|u_\varepsilon\|^2
 \sqrt{b^2\|u_\varepsilon\|^8+4a\|u_\varepsilon\|^2
 \int_{\Omega}Q(x)u_\varepsilon^6dx} }{6\int_{\Omega}Q(x)u_\varepsilon^6dx}  \nonumber\\
&\quad + \frac{b^3\|u_\varepsilon\|^{12}+2ab\|u_\varepsilon\|^{6}
 \int_{\Omega}Q(x)u_\varepsilon^6dx}
 {24 \big(\int_{\Omega}Q(x)u_\varepsilon^6dx\big)^2}  \nonumber\\
&\quad + \frac{b^2\|u_\varepsilon\|^4\sqrt{b^2\|u_\varepsilon\|^8
 +4a\|u_\varepsilon\|^2\int_{\Omega}Q(x)u_\varepsilon^6dx} }
 {24\big(\int_{\Omega}Q(x)u_\varepsilon^6dx\big)^2} \nonumber\\
&= \frac{ab\|u_\varepsilon\|^6}{4\int_{\Omega}Q(x)u_\varepsilon^6dx}
 +\frac{b^3\|u_\varepsilon\|^{12}}{24(\int_{\Omega}Q(x)u_\varepsilon^6dx)^2} \nonumber\\
&\quad +\frac{a\|u_\varepsilon\|^2
\sqrt{b^2\|u_\varepsilon\|^8+4a\|u_\varepsilon\|^2
 \int_{\Omega}Q(x)u_\varepsilon^6dx} }{6\int_{\Omega}Q(x)u_\varepsilon^6dx} \nonumber\\
&\quad + \frac{b^2\|u_\varepsilon\|^4\sqrt{b^2\|u_\varepsilon\|^8
  +4a\|u_\varepsilon\|^2\int_{\Omega}Q(x)u_\varepsilon^6dx} }
 {24\big(\int_{\Omega}Q(x)u_\varepsilon^6dx\big)^2}  \nonumber\\
&= \frac{ab(S^{\frac{9}{2}}+O(\varepsilon))}{4(Q_{M}S^{3/2}+o(\varepsilon))}+
 \frac{b^3(S^9+O(\varepsilon))}{24(Q_{M}S^{3/2}+o(\varepsilon))^2} \nonumber\\
&\quad +\frac{a(S^{3/2}+O(\varepsilon))
 \sqrt{b^2S^{6}+4aS^3+O(\varepsilon)}}{6(Q_{M}S^{3/2}+o(\varepsilon))}  \nonumber\\
&\quad +\frac{b^2(S^6+O(\varepsilon))\sqrt{b^2S^{6}+4aS^3+O(\varepsilon)}}
 {24(Q_{M}S^{3/2}+o(\varepsilon))^2}  \nonumber\\
&= \frac{abS^3}{4Q_{M}}+\frac{b^3S^6}{24Q_{M}^2}
 +\frac{aS\sqrt{b^2S^4+4aSQ_{M}}}{6Q_{M}}  \nonumber\\
&\quad +\frac{b^2S^4\sqrt{b^2S^4+4aSQ_{M}}}{24Q_{M}^2}+O(\varepsilon) \nonumber \\
&= \Lambda+O(\varepsilon).   \label{3.14}
\end{align}

Second, we estimate the value of $I_{\varepsilon,2}$. From \eqref{4.9}, 
since $0<t_0<t_{\varepsilon}<T_0$, one has
\begin{equation}\label{3.15}
I_{\varepsilon,2}(t_{\varepsilon})
= \frac{t_{\varepsilon}^4}{4}\int_{\Omega}u_{\varepsilon}^4dx
\geq \frac{t_0^4}{4}\int_{\Omega}u_{\varepsilon}^4dx
\geq  C_{10}\varepsilon.
\end{equation}
Thus, from \eqref{3.14} and \eqref{3.15}, one gets
\begin{align*}
I(tu_\varepsilon)
&=  I_{\varepsilon,1}(t) -\nu I_{\varepsilon,2}(t)\\
&\leq I_{\varepsilon,1}(t_{\varepsilon})
 -\nu I_{\varepsilon,2}(t_{\varepsilon})\\
&\leq I_{\varepsilon,1}(T_{\varepsilon})
 -\frac{t_0^4}{4}\nu\int_{\Omega}u_{\varepsilon}^4dx\\
&\leq \Lambda+O(\varepsilon)-C_{10}\nu\varepsilon
< \Lambda,
\end{align*}
provided that $\nu$ is large enough. Thus there exists $\nu^{*}>0$ 
such that $I(tu_\varepsilon)<\Lambda$ for all $\nu>\nu^{*}$.
This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
 As in the proof of Theorem \ref{thm1.1}, we can obtain that $I$ has the geometry 
of mountain pass lemma in $H_0^1(\Omega)$. According to Lemmas \ref{lem3.1} and  
\ref{lem3.2},  it follows that $I$ satisfies the conditions of the mountain pass lemma.
Then as in the proof of Theorem \ref{thm1.1}, we obtain that  \eqref{1.1} has a positive 
solution $u^{*}$ with $I(u^{*})>0$ as long as $\nu>\nu^{*}$.
The proof is complete.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by the
National Science Foundation of China (No. 11471267),
by the  Science and Technology Foundation of Guizhou Province (No. LKZS[2014]30,
No. LH[2015]7001, No. LH[2015]7049).
The authors would like to thank the anonymous referees
and Professor G. Molica Bisci for their valuable suggestions.


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\end{document}
