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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 232, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/232\hfil
changing-sign potential and critical nonlinearities]
{Combined effects of changing-sign potential and critical
 nonlinearities in Kirchhoff type problems}

\author[G.-S. Liu, L.-T. Guo, C.-Y. Lei \hfil EJDE-2016/232\hfilneg]
{Gao-Sheng Liu, Liu-Tao Guo, Chun-Yu Lei}

\address{Gao-Sheng Liu \newline
Shanghai University of Finance and Economics,
Shanghai 200433, China}
\email{772936104@qq.com}

\address{Liu-Tao Guo \newline
Guizhou Minzu University,
Guiyang 550025, China}
\email{350630542@qq.com}

\address{Chun-Yu Lei (corresponding author) \newline
School of Science, 
Guizhou Minzu University, 
Guiyang 550025, China}
\email{leichygzu@sina.cn}

\thanks{Submitted June 8, 2016. Published August 24, 2016.}
\subjclass[2010]{35D05, 35J60, 58J32}
\keywords{Kirchhoff type equation; critical exponents; changing-sign potential;
\hfill\break\indent  Nehari manifold}

\begin{abstract}
 In this article, we study the existence and multiplicity of positive
 solutions for a class of Kirchhoff type problems involving changing-sign
 potential and critical growth terms. Using the concentration
 compactness principle and Nehari manifold, we obtain the existence
 and multiplicity  of nonzero non-negative solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction and statement of main result}

 In this article, we consider the multiplicity of non-negative solutions
of the  Kirchhoff type equation
\begin{equation}\label{1.1}
\begin{gathered}
-\Big(a+ b\int_\Omega|\nabla u|^2dx\Big)\Delta u
 =|u|^4u+\mu|x|^{\alpha-2}u+\lambda f(x)|u|^{q-2}u \quad\text{in }\Omega, \\
u=0,\quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
where $\Omega$ is a smooth bounded domain in $\mathbb{R}^3$, $a, b>0$,
$0<\alpha<1$, $1<q<2$, $\lambda>0$ is a positive real number,
and $0<\mu<a\mu_1$ ($\mu_1$ is the first eigenvalue of
$-\Delta u=\mu|x|^{\alpha-2}u$, under Dirichlet boundary condition).
The weight functions $f\in C(\overline{\Omega})$ is changing-sign potential,
satisfying $f^+=\max\{f, 0\}\neq0$.

In recent years, the existence and multiplicity of solutions to the nonlocal
Kirchhoff type problem
\begin{equation}
\begin{gathered}
-\Big(a+b\int_\Omega|\nabla u|^2dx\Big)\Delta u=g(x,u) \quad\text{in }\Omega, \\
u=0,      \quad\text{on }\partial\Omega
\end{gathered}
\end{equation}
has been the focus of a great deal of research  and some results can be found.
For instance, in  \cite{CF,GA, JX,LY,YJ,TM,WH}.
In particular, when $g(x,u)$ is involved in critical nonlinearities terms,
readers can be referred to \cite{GF,HT,LS,DN,XT} for details.
The authors in \cite{CK,Bi,FJ,LC} have investigated Kirchhoff type equation
 with concave and convex nonlinear. In addition, there are some results for
$g(x,u)$ being changing-sign potential,  see for example
\cite{LL,MZ,ZK}.
Especially, Chen et al.\ \cite{CK} considered the following nonlocal Kirchhoff
type problem
\begin{equation} \label{e1.3}
\begin{gathered}
-\Big(a+b\int_\Omega|\nabla u|^2dx\Big)\Delta u
=f(x)u^{p-2}u+\lambda g(x)|u|^{q-2}u \quad\text{in }\Omega, \\
u=0,       \quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
the authors assumed that $1<q<2<p<6$, the sign-changing weight functions
 $f, g\in C(\overline{\Omega})$ and $f^+=\max\{f, 0\}\neq0$ and
 $g^+=\max\{g, 0\}\neq0$ hold. Then there exists a positive constant
 $\lambda_0(a)>0$ such that for each $a>0$ and
$\lambda\in(0,\lambda_0(a))$, problem \eqref{e1.3} has at least two positive
 solutions. In equation \eqref{e1.3}, assume $1<q<2$, $p=6$, $f(x)\equiv1$
 and add a term of $\mu|x|^{\alpha-2}u$, then an interesting question
is put forward if the existence and multiplicity of solutions can be
established for Kirchhoff type problems with critical and changing-sign terms.

Throughout this paper, we use the following notation:
\begin{itemize}
  \item The space $H_0^{1}(\Omega)$ is equipped with the norm
$\|u\|=(\int_{\Omega}|\nabla u|^2dx)^{1/2}$, the norm in
$L^p(\Omega)$ is represented by $|u|_{p}=(\int_{\Omega}|u|^pdx)^{\frac{1}{p}}$;

\item Let $S$ be the best Sobolev constant, namely
\begin{equation}\label{1.4}
S:=\inf_{u\in D^{1,2}(\mathbb{R}^3)\backslash\{0\}}
\frac{\int_{\mathbb{R}^{3}}|\nabla u|^2dx}
{\big(\int_{\mathbb{R}^{3}}|u|^6dx\big)^{1/3}}.
\end{equation}
\end{itemize}
The energy functional $I_\lambda(u)$: $H_0^{1}(\Omega)\to \mathbb{R}$
corresponding to \eqref{1.1} is defined by
\begin{equation*}
I_\lambda(u)=\frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4
-\frac{\mu}{2}\int_{\Omega}|x|^{\alpha-2}|u|^2dx
-\frac{1}{6}\int_{\Omega}|u|^6dx-\frac{\lambda}{q}\int_{\Omega}f|u|^{q}dx.
\end{equation*}
Generally speaking, a function $u$ is called a weak solution of
\eqref{1.1} if $u\in H_0^{1}(\Omega)$ and for all $\varphi\in
H_0^{1}(\Omega)$ it holds
\begin{equation*}
(a+b\|u\|^2)\int_{\Omega}(\nabla u\cdot\nabla\varphi)dx
=\mu\int_{\Omega}|x|^{\alpha-2}u\varphi dx
+\int_{\Omega}|u|^4u\varphi dx+\lambda\int_{\Omega}f|u|^{q-2}u\varphi dx.
\end{equation*}

Our main result is as follows:

\begin{theorem} \label{thm1.1}
 Assume that $1<q<2$, $0<\alpha<1$, and $f\in L^\infty(\Omega)$ changes sign,
then there exists $\lambda_{*}>0$ such that for every $\lambda\in(0,\lambda_{*})$,
problem \eqref{1.1} has at least two nonzero non-negative solutions,
and one of the solutions is a ground state solution.
\end{theorem}

\begin{remark} \label{rmk1}\rm
 It is well known that the difficulty lies in the lack of compactness of the
embedding: $H_0^{1}\hookrightarrow L^6(\Omega)$, then we overcome
the difficulty by the concentration compactness principle.
The nonlocal Kirchhoff problem becomes difficult when $b>0$ for estimating
 the critical value level, however, by adding a particular term
$\mu|x|^{\alpha-2}u$, we could get over the trouble.
\end{remark}

In section 2 we present some preliminary results,
 while in section 3 we present the proof of Theorem \ref{thm1.1}.

\section{Preliminary results}

 Since $I_\lambda$ is not bounded below on $H_0^1(\Omega)$, we will work on
the Nehari manifold
$$
\mathcal{N}_\lambda=\{u\in H_0^1(\Omega)\backslash\{0\}:
 \langle I'_\lambda(u),u\rangle=0\},
$$
which implies that $\mathcal{N}_\lambda$ holds all nonzero solutions of \eqref{1.1}.
 In addition, $u\in\mathcal{N}_\lambda$ if and only if
$$
a\|u\|^2+b\|u\|^4-\int_{\Omega}|u|^6dx
-\mu\int_{\Omega}|x|^{\alpha-2}|u|^2dx-\lambda\int_{\Omega}f|u|^{q}dx=0.
$$
Let
$$
\psi(u)=a\|u\|^2+b\|u\|^4-\int_{\Omega}|u|^6dx
-\mu\int_{\Omega}|x|^{\alpha-2}|u|^2dx-\lambda\int_{\Omega}f|u|^{q}dx,
$$
and then we obtain
$$
\langle\psi'(u),u\rangle=2a\|u\|^2+4b\|u\|^4
-6\int_{\Omega}|u|^6dx-2\mu\int_{\Omega}|x|^{\alpha-2}|u|^2dx
-q\lambda\int_{\Omega}f|u|^{q}dx.
$$
We split $\mathcal{N}_\lambda$ into three parts:
\begin{gather*}
\mathcal{N}^+_\lambda=\{u\in\mathcal{N}_\lambda: \langle\psi'(u),u\rangle>0\},\\
\mathcal{N}^0_\lambda=\{u\in\mathcal{N}_\lambda: \langle\psi'(u),u\rangle=0\},\\
\mathcal{N}^-_\lambda=\{u\in\mathcal{N}_\lambda: \langle\psi'(u),u\rangle<0\}.
\end{gather*}

\begin{lemma} \label{lem2.1}
 Suppose $\lambda\in(0, T_1)$ with
\[
T_1= \big\{\frac{4-q}{6-q}b[\frac{2b}{6-q}]^{\frac{2}{4-q}}
|\Omega|^{-\frac{6-q}{3(4-q)}}(|f|_{\infty})^{-\frac{2}{4-q}}
S^{\frac{2(6-q)}{4-q}}\big\}^{\frac{4-q}{2}}.
\]
Then
(i) $\mathcal{N}_\lambda^{\pm}\neq\emptyset$, and
(ii) $\mathcal{N}^0_\lambda=\emptyset$.
\end{lemma}

\begin{proof}
 (i) For a given $u\in H_0^1(\Omega)\backslash\{0\}, u\neq0$, as $0<\mu<a\mu_1$,
one has
\begin{align*}
\eta(t):=&
t^{-4}a\|u\|^2+bt^{-2}\|u\|^4-\mu t^{-4}
\int_{\Omega}\frac{|u|^2}{|x|^{2-\alpha}}dx-\lambda t^{q-6}\int_{\Omega}f|u|^qdx\\
\geq& t^{-4}\big(a-\frac{\mu}{\mu_1}\big)\|u\|^2+bt^{-2}\|u\|^4
 -\lambda t^{q-6}\int_{\Omega}f|u|^qdx\\
\geq& t^{-4}\big(a-\frac{\mu}{\mu_1}\big)\|u\|^2
 +bt^{-2}\|u\|^4-\lambda t^{q-6}|f|_{\infty}\int_{\Omega}|u|^qdx.
\end{align*}
We define two functions $\Phi,\Phi_1\in C(\mathbb{R}^+, \mathbb{R})$ by
\begin{gather*}
\Phi(t)=t^{-4}\big(a-\frac{\mu}{\mu_1}\big)\|u\|^2
 +bt^{-2}\|u\|^4-\lambda t^{q-6}|f|_{\infty}\int_{\Omega}|u|^qdx, \\
\Phi_1(t)=bt^{-2}\|u\|^4-\lambda t^{q-6}|f|_{\infty}\int_{\Omega}|u|^qdx.
\end{gather*}
Thus
$$
\Phi_{1}'(t)=-2bt^{-3}\|u\|^4-\lambda(q-6)t^{q-7}|f|_{\infty}\int_{\Omega}|u|^qdx.
$$
Let $\Phi_{1}'(t)=0$, it is simple to show that
$$
t_{\rm max}=\Big[\frac{\lambda(6-q)|f|_{\infty}\int_{\Omega}|u|^qdx}
{2b\|u\|^4}\Big]^{\frac{1}{4-q}}.
$$
Easy computations show that $\Phi_{1}'(t)>0$ for all $0<t<t_{\rm max}$ and
$\Phi_{1}'(t)<0$ for all $t>t_{\rm max}$. Therefore, $\Phi_{1}(t)$ achieves
its maximum at $t_{\rm max}$; that is,
\begin{equation*}
\Phi_{1}(t_{\rm max})=\frac{4-q}{6-q}b\big[\frac{2b}{6-q}\big]^{\frac{2}{4-q}}
\frac{\|u\|^{\frac{4(6-q)}{4-q}}}{\big(\lambda|f|_{\infty}\int_{\Omega}|u|^qdx
\big)^{\frac{2}{4-q}}}\,.
\end{equation*}
Then it follows from \eqref{1.4} that
\begin{align*}
&\eta(t_{\rm max})-\int_{\Omega}|u|^6dx\\
&\geq \Phi(t_{\rm max})-\int_{\Omega}|u|^6dx\\
&\geq \Phi_{1}(t_{\rm max})-\int_{\Omega}|u|^6dx\\
&> \frac{4-q}{6-q}b\big[\frac{2b}{6-q}\big]^{\frac{2}{4-q}}\frac{\|u\|
^{\frac{4(6-q)}{4-q}}}{\big(\lambda|f|_{\infty}\int_{\Omega}|u|^qdx
 \big)^{\frac{2}{4-q}}}-\int_{\Omega}|u|^6dx\\
&> \frac{4-q}{6-q}b\big[\frac{2b}{6-q}\big]^{\frac{2}{4-q}}
 |\Omega|^{-\frac{6-q}{3(4-q)}}(\lambda|f|_{\infty})^{-\frac{2}{4-q}}\frac{\|u\|
^{\frac{4(6-q)}{4-q}}}{|u|_6^{\frac{2q}{4-q}}}-\int_{\Omega}|u|^6dx\\
&= \Big\{\frac{4-q}{6-q}b\big[\frac{2b}{6-q}\big]^{\frac{2}{4-q}}
 |\Omega|^{-\frac{6-q}{3(4-q)}}(\lambda|f|_{\infty})^{-\frac{2}{4-q}}
\Big(\frac{\|u\|^2}{|u|_6^2}\Big)^{\frac{2(6-q)}{4-q}}
-1\Big\}|u|_6^6\\
&\geq \Big\{\frac{4-q}{6-q}b\big[\frac{2b}{6-q}\big]^{\frac{2}{4-q}}
 |\Omega|^{-\frac{6-q}{3(4-q)}}(\lambda|f|_{\infty})^{-\frac{2}{4-q}}
S^{\frac{2(6-q)}{4-q}}
-1\Big\}|u|_6^6
> 0
\end{align*}
when $0<\lambda<T_1$, where we can choose
\begin{equation*}
T_1= \Big\{\frac{4-q}{6-q}b\big[\frac{2b}{6-q}\big]^{\frac{2}{4-q}}
 |\Omega|^{-\frac{6-q}{3(4-q)}}(|f|_{\infty})^{-\frac{2}{4-q}}
S^{\frac{2(6-q)}{4-q}}\Big\}^{\frac{4-q}{2}}.
\end{equation*}
Consequently, there exist constants $t^\pm$ such that
$0<t^+=t^+(u)<t_{\rm max}<t^-=t^-(u)$,
$t^+u\in\mathcal{N}_\lambda^+$ and $t^-u\in\mathcal{N}_\lambda^-$.

(ii) Now we show that $\mathcal{N}^0_\lambda=\emptyset$ for all
$\lambda\in(0, T_1)$. By contradiction, assume there exists $u_0\neq0$
such that $u_0\in\mathcal{N}_\lambda^0$, one obtains
\begin{gather}\label{2.1}
a\|u_0\|^2+b\|u_0\|^4
=\mu\int_{\Omega}\frac{|u_0|^2}{|x|^{2-\alpha}}dx+\int_{\Omega}|u_0|^6dx
+\lambda\int_{\Omega}f|u_0|^{q}dx, \\
\label{2.2}
4a\|u_0\|^2+2b\|u_0\|^4=4\mu\int_{\Omega}\frac{|u_0|^2}
{|x|^{2-\alpha}}dx+\lambda(6-q)\int_{\Omega}f|u_0|^{q}dx.
\end{gather}
It follows from \eqref{2.1} and \eqref{2.2} that
\begin{equation}\label{2.3}
\begin{aligned}
\lambda\int_{\Omega}f|u_0|^{q}dx
&= \frac{4}{6-q}\Big(a\|u_0\|^2
 -\mu\int_{\Omega}\frac{|u_0|^2}{|x|^{2-\alpha}}dx\Big)
 +\frac{2}{6-q}b\|u_0\|^4\\
&\geq \frac{4}{6-q}\big(a-\frac{\mu}{\mu_1}\big)\|u_0\|^2
 +\frac{2}{6-q}b\|u_0\|^4\\
&> \frac{2}{6-q}b\|u_0\|^4.
\end{aligned}
\end{equation}
On the one hand, since the strict inequality $\|u_0\|^2>S|u_0|_6^2$
holds for $u_0\in\mathcal{N}^0_\lambda\backslash\{0\}$, we use a parameter
$\Theta$ by
\begin{equation}\label{2.4}
\begin{aligned}
\Theta&=  |\Omega|^{\frac{6-q}{3(4-q)}}
 S^{-\frac{2(6-q)}{4-q}}|f|_{\infty}^{\frac{2}{4-q}}
\frac{\|u_0\|^{\frac{4(6-q)}{4-q}}}{\big(\int_{\Omega}f(u_0^+)^{q}dx
 \big)^{\frac{2}{4-q}}}-\int_{\Omega}|u_0|^6dx\\
&>  |\Omega|^{\frac{6-q}{3(4-q)}}S^{-\frac{2(6-q)}{4-q}}
 |f|_{\infty}^{\frac{2}{4-q}}
\frac{(S|u_0|_6^2)^{\frac{2(6-q)}{4-q}}}
 {|f|_{\infty}^{\frac{2}{4-q}}|\Omega|^{\frac{6-q}{3(4-q)}}
 |u_0|_6^{\frac{2q}{4-q}}}-\int_{\Omega}|u_0|^6dx\\
&= \int_{\Omega}|u_0|^6dx-\int_{\Omega}|u_0|^6dx
= 0.
\end{aligned}
\end{equation}
On the other hand, by \eqref{2.3}, one deduces that
\begin{align*}
\Theta&=  |\Omega|^{\frac{6-q}{3(4-q)}}S^{-\frac{2(6-q)}{4-q}}
\lambda^{\frac{2}{4-q}}|f|_{\infty}^{\frac{2}{4-q}}
\frac{\|u_0\|^{\frac{4(6-q)}{4-q}}}
{\big(\lambda\int_{\Omega}f(u_0^+)^{q}dx\big)^{\frac{2}{4-q}}}
 -\int_{\Omega}|u_0|^6dx\\
&\leq |\Omega|^{\frac{26-q}{3(4-q)}}S^{-\frac{2(6-q)}{4-q}}
 \lambda^{\frac{2}{4-q}}|f|_{\infty}^{\frac{2}{4-q}}
\frac{\|u_0\|^{\frac{4(6-q)}{4-q}}}
 {\big(\frac{2}{6-q}b\|u_0\|^4\big)^{\frac{2}{4-q}}}\\
&\quad -\frac{2-q}{6-q}\Big(a\|u_0\|^2-\mu\int_{\Omega}
 \frac{|u_0|^2}{|x|^{2-\alpha}}dx\Big)-\frac{b(4-q)}{6-q}\|u_0\|^4\\
&=  |\Omega|^{\frac{6-q}{3(4-q)}}S^{-\frac{2(6-q)}{4-q}}
 \lambda^{\frac{2}{4-q}}|f|_{\infty}^{\frac{2}{4-q}}
\big(\frac{6-q}{2b}\big)^{\frac{2}{4-q}}\|u_0\|^4\\
&\quad -\frac{2-q}{6-q}\big(a\|u_0\|^2-\mu\int_{\Omega}
 \frac{|u_0|^2}{|x|^{2-\alpha}}dx\big)-\frac{b(4-q)}{6-q}\|u_0\|^4\\
&\leq |\Omega|^{\frac{6-q}{3(4-q)}}S^{-\frac{2(6-q)}{4-q}}
 \lambda^{\frac{2}{4-q}}|f|_{\infty}^{\frac{2}{4-q}}
\big(\frac{6-q}{2b}\big)^{\frac{2}{4-q}}
\|u_0\|^4\\
&\quad -\frac{2-q}{6-q}\big(a-\frac{\mu}{\mu_1}\big)\|u_0\|^2
 -\frac{b(4-q)}{6-q}\|u_0\|^4\\
&< \|u_0\|^4
\Big[ |\Omega|^{\frac{6-q}{3(4-q)}}S^{-\frac{2(6-q)}{4-q}}
 \lambda^{\frac{2}{4-q}}|f|_{\infty}^{\frac{2}{4-q}}
 \big(\frac{6-q}{2b}\big)^{\frac{2}{4-q}}-\frac{b(4-q)}{6-q}\Big]
< 0,
\end{align*}
which  contradicts \eqref{2.4}, where the least inequality holds when
$\lambda<T_1$.
The proof  is complete.
\end{proof}


\begin{lemma} \label{lem2.2}
$I_\lambda$ is coercive and bounded below on $\mathcal{N}_\lambda$.
\end{lemma}

\begin{proof}
 Assume $u\in\mathcal{N}_\lambda$, then by \eqref{1.4} we obtain
\begin{align*}
I_\lambda(u)
&= \frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4
 -\frac{\mu}{2}\int_{\Omega}\frac{|u|^2}{|x|^{2-\alpha}}dx
 -\frac{1}{6}\int_{\Omega}|u|^6dx-\frac{\lambda}{q}\int_{\Omega}f|u|^{q}dx\\
&\geq \frac{a}{3}\|u\|^2+\frac{b}{12}\|u\|^4
 -\frac{\mu}{3}\int_{\Omega}\frac{|u|^2}{|x|^{2-\alpha}}dx
 -\lambda\big(\frac{1}{q}-\frac{1}{6}\big)
 \big|\int_{\Omega}f|u|^{q}dx\big|\\
&\geq \frac{a\mu_1-\mu}{3\mu_1}\|u\|^2+\frac{b}{12}\|u\|^4
 -\lambda\big(\frac{1}{q}-\frac{1}{6}\big)|f|_{\infty}|\Omega|^{\frac{6-q}{6}}
S^{-q/2}\|u\|^q.
\end{align*}
Since $1<q<2, 0<\mu<a\mu_1$, it follows that $I_\lambda$ is coercive
and bounded below on $\mathcal{N}_\lambda$. The proof is complete.
\end{proof}

According to Lemma \ref{lem2.1}, we have
 $\mathcal{N}_\lambda=\mathcal{N}^+_\lambda\cup\mathcal{N}^-_\lambda$ for all
$\lambda\in(0, T_1)$. Moreover, we know that $\mathcal{N}^+_\lambda$ and
$\mathcal{N}^-_\lambda$ are nonempty, and by Lemma \ref{lem2.2} we may define
$$
\alpha_\lambda=\inf_{u\in\mathcal{N}_\lambda}I_\lambda(u),\quad
\alpha^+_\lambda=\inf_{u\in\mathcal{N}^+_\lambda}I_\lambda(u), \quad
\alpha^-_\lambda=\inf_{u\in\mathcal{N}^-_\lambda}I_\lambda(u).
$$

\begin{lemma} \label{lem2.3}
 $\alpha_\lambda\leq\alpha^+_\lambda<0$.
\end{lemma}

\begin{proof}
Assume $u\in\mathcal{N}_\lambda^+$, then we have
\begin{equation}\label{2.5}
\int_{\Omega}|u|^6dx<\frac{2-q}{6-q}a\|u\|^2+\frac{4-q}{6-q}b\|u\|^4.
\end{equation}
It follows from \eqref{2.5} that
\begin{align*}
I_\lambda(u)
&= \frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4
 -\frac{\mu}{2}\int_{\Omega}\frac{|u|^2}{|x|^{2-\alpha}}dx
 -\frac{1}{6}\int_{\Omega}|u|^6dx-\frac{\lambda}{q}\int_{\Omega}f|u|^{q}dx\\
&=\big(\frac{1}{2}-\frac{1}{q}\big)(a-\frac{\mu}{\mu_{1}})\|u\|^2
 +\big(\frac{1}{4}-\frac{1}{q}\big)b\|u\|^4
 +\big(\frac{1}{q}-\frac{1}{6}\big)\int_{\Omega}|u|^6dx\\
&< \big(\frac{a}{2}-\frac{1}{q}\big)
 (a-\frac{\mu}{\mu_{1}})\|u\|^2+\big(\frac{1}{4}-\frac{1}{q}\big)b\|u\|^4\\
&\quad +\big(\frac{1}{q}-\frac{1}{6}\big)
 \Big(\frac{2-q}{6-q}a\|u\|^2+\frac{4-q}{6-q}b\|u\|^4\Big)\\
&= \frac{q-2}{6q}\big(4a-\frac{3\mu}{\mu_1}\big)a\|u\|^2
 +\frac{1}{3}\big(\frac{1}{4}-\frac{1}{q}\big)b\|u\|^4
<0.
\end{align*}
By the definitions of $\alpha_\lambda$ and $\alpha_\lambda^+$, we obtain
that $\alpha_\lambda\leq\alpha_\lambda^+<0$.
This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.4}
 For each $u\in\mathcal{N}_\lambda$, there exist $\varepsilon>0$ and a
continuously differentiable function $\hat{f}=\hat{f}(w)>0, w\in H_0^{1}(\Omega)$,
$\|w\|<\varepsilon$ satisfying
\begin{equation*}
\hat{f}(0)=1,\quad \hat{f}(w)(u+w)\in\mathcal{N}_\lambda,\quad
\forall w\in H_0^{1}(\Omega),\; \|w\|<\varepsilon.
\end{equation*}
\end{lemma}

\begin{proof}
For $u\in\mathcal{N}_\lambda$, define
$\hat{F}: \mathbb{R}\times H_0^{1}(\Omega)\to \mathbb{R}$ by
\begin{align*}
\hat{F}(t,w)
&=  t^{2-q}a\int_{\Omega}|\nabla(u+w)|^2dx+t^{4-q}b
 \Big(\int_{\Omega}|\nabla(u+w)|^2dx\Big)^2\\
&\quad - t^{2-q}\mu \int_{\Omega}\frac{|u+w|^2}{|x|^{2-\alpha}}dx
 -t^{6-q}\int_{\Omega}|u+w|^6dx-\lambda\int_{\Omega}f|u+w|^{q}dx.
\end{align*}
As $u\in\mathcal{N}_\lambda$, it is easy to get that $\hat{F}(1,0)=0$ and
$$
\hat{F}_t(1,0)=(2-q)a\|u\|^2+(4-q)b\|u\|^4
-(2-q)\mu\int_{\Omega}\frac{|u|^2}{|x|^{2-\alpha}}dx-(6-q)\int_{\Omega}|u|^6dx.
$$
Since $u\neq0$, by Lemma \ref{lem2.1}, we deduce that $\hat{F}_t(1,0)\neq0$.
Then, applying the implicit function theorem at the point $(0,1)$,
we obtain that $\varepsilon>0$ and a continuously differentiable function
 $\hat{f}: B(0,\varepsilon)\subset H_0^{1}(\Omega)\to \mathbb{R}^+$
satisfying that
\begin{equation*}
\hat{f}(0)=1,\quad \hat{f}(w)>0,\quad
f(w)(u+w)\in\mathcal{N}_\lambda,\quad
\forall w\in H_0^{1}(\Omega)\text{ with }\|w\|<\varepsilon.
\end{equation*}
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.5}
 For each $u\in\mathcal{N}_\lambda^-$, there exist $\varepsilon>0$ and
a continuously differentiable function
$\tilde{f}=\tilde{f}(v)>0, v\in H_0^{1}(\Omega)$,
$\|v\|<\varepsilon$ satisfying that
\begin{equation*}
\tilde{f}(0)=1,\quad \tilde{f}(v)(u+v)\in\mathcal{N}_\lambda^-,
\quad \forall v\in H_0^{1}(\Omega),\; \|v\|<\varepsilon.
\end{equation*}
\end{lemma}

\begin{proof} Similar to the process in Lemma \ref{lem2.4},
for $u\in\mathcal{N}_\lambda^-$, define
$\tilde{F}: \mathbb{R}\times H_0^{1}(\Omega)\to \mathbb{R}$ by
\begin{align*}
\tilde{F}(t,v)
&=  t^{2-q}a\int_{\Omega}|\nabla(u+v)|^2dx+t^{4-q}b
\Big(\int_{\Omega}|\nabla(u+v)|^2dx\Big)^2\\
&\quad - t^{2-q}\mu \int_{\Omega}\frac{|u+v|^2}{|x|^{2-\alpha}}dx
-t^{6-q}\int_{\Omega}|u+v|^6dx-\lambda\int_{\Omega}f|u+v|^{q}dx.
\end{align*}
As $u\in\mathcal{N}_\lambda^-$, we obtain $\tilde{F}(1,0)=0$ and
$\tilde{F}_t(1,0)<0$. Thus, we can apply the implicit function theorem
at the point $(0,1)$ to get the result. This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.6}
If $\{u_n\}\subset\mathcal{N}_\lambda$ is a minimizing sequence of
 $I_\lambda$, for any $\varphi\in H_0^{1}(\Omega)$, then
\begin{equation}\label{2.6}
-\frac{|f'_n(0)|\|u_n\|+\|\varphi\|}{n}
\leq\langle I'_\lambda(u_n),\varphi\rangle
\leq\frac{|f'_n(0)|\|u_n\|+\|\varphi\|}{n}.
\end{equation}
\end{lemma}

\begin{proof}
By Lemma \ref{lem2.2}, we obtain $I_\lambda$ is coercive on $\mathcal{N}_\lambda$.
Then, applying the Ekeland variational principle \cite{IE}, there exists
a minimizing sequence $\{u_n\}\subset\mathcal{N}_\lambda$ of $I_\lambda$ such that
\begin{equation}\label{2.7}
I_\lambda(u_n)<\alpha_\lambda+\frac{1}{n},\quad
I_\lambda(v)-I_\lambda(u_n)\geq-\frac{1}{n}\|v-u_n\|,\quad
\forall v\in\mathcal{N}_\lambda.
\end{equation}
Note that $I_\lambda(|u_n|)=I_\lambda(u_n)$, then we obtain that $u_{n}\geq0$.
Lemma \ref{lem2.2} suggests that $\{u_n\}$ is bounded in $H_0^1(\Omega)$.
Thus there exist a subsequence (still denoted by $\{u_n\}$) and $u_*$ in
 $H_0^1(\Omega)$ such that
\begin{gather*}
u_n\rightharpoonup u_*\quad \text{weakly in } H_0^{1}(\Omega),\\
u_n\to u_*\quad \text{strongly in }L^p(\Omega)\; (1\leq p<6),\\
u_n(x)\to u_*(x)\quad \text{a.e. in }\Omega.
\end{gather*}
Pick $t>0$ sufficiently small, $\varphi\in H_0^{1}(\Omega)$, and let
$u=u_n$, $w=t\varphi\in H_0^{1}(\Omega)$ in Lemma \ref{lem2.4}, then one obtains
$f_n(t)=f_n(t\varphi)$ and
$f_n(0)=1, f_n(t)(u_n+t\varphi)\in\mathcal{N}_\lambda$. Note that
\begin{equation}\label{2.8}
a\|u_n\|^2+b\|u_n\|^4-\int_{\Omega}u_n^6dx
-\mu\int_{\Omega}|x|^{\alpha-2}u_n^2dx
-\lambda\int_{\Omega}fu_n^{q}dx=0.
\end{equation}
From \eqref{2.7}, one has
\begin{equation}\label{2.9}
\begin{aligned}
\frac{1}{n}[|f_n(t)-1|\cdot\|u_n\|+tf_n(t)\|\varphi\|]
&\geq \frac{1}{n}\|f_n(t)(u_n+t\varphi)-u_n\|\\
&\geq I_\lambda(u_n)-I_\lambda[f_n(t)(u_n+t\varphi)],\\
\end{aligned}
\end{equation}
and
\begin{align*}
&I_\lambda(u_n)-I_\lambda[f_n(t)(u_n+t\varphi)]\\
&= \frac{1-f_{n}^2(t)}{2}a\|u_n\|^2+\frac{1-f_{n}^4(t)}{4}b\|u_n\|^4
 +\mu\frac{f_{n}^2(t)-1}{2}\int_{\Omega}(u_n+t\varphi)^2|x|^{\alpha-2}dx\\
&\quad +\frac{f_{n}^6(t)-1}{6}\int_{\Omega}(u_n+t\varphi)^6dx
 +\lambda\frac{f_{n}^{q}(t)-1}{q}\int_{\Omega}f(u_n+t\varphi)^{q}dx\\
&\quad +\frac{f_{n}^2(t)}{2}\Big(a+\frac{f_{n}^2(t)}{2}b(\|u_n\|^2
 +\|u_n+t\varphi\|^2)\Big)\big(\|u_n\|^2-\|u_n+t\varphi\|^2\big)\\
&\quad +\frac{1}{6}\Big(\int_{\Omega}(u_n+t\varphi)^6dx
 -\int_{\Omega}u_n^6dx\Big)
 +\frac{\lambda}{q}\int_{\Omega} f((u_n+t\varphi)^{q}-u_n^{q})dx\\
&\quad +\frac{\mu}{2}\int_{\Omega}|x|^{\alpha-2}((u_n+t\varphi)^2-u_n^2)dx,
\end{align*}
then, by \eqref{2.8} and \eqref{2.9}, dividing by $t$ and letting $t\to0$, we obtain
\begin{align*}
&\frac{|f'_n(0)|\|u_n\|+\|\varphi\|}{n} \\
&\geq -f'_{n}(0)a\|u_n\|^2+f'_{n}(0)b\|u_n\|^4
 +f'_{n}(0)\int_{\Omega}u_n^6dx
 +\lambda f'_{n}(0)\int_{\Omega}fu_n^{q}dx\\
&\quad +\mu f'_{n}(0)\int_{\Omega}|x|^{\alpha-2}u_n^2dx
 -(a+b\|u_n\|^2)\int_{\Omega}(\nabla u_n\cdot\nabla \varphi)dx\\
&\quad +\int_{\Omega}u_n^5\varphi dx
 +\lambda\int_{\Omega}fu_n^{q-1}\varphi dx+\mu\int_{\Omega}|x|^{\alpha-2}u_n\varphi dx\\
&= -f'_{n}(0)(a\|u_n\|^2+b\|u_n\|^4-\int_{\Omega}u_n^6dx
 -\lambda\int_{\Omega}fu_n^{q}dx-\mu\int_{\Omega}|x|^{\alpha-2}u_n^2dx)\\
&\quad -(a+b\|u_n\|^2)\int_{\Omega}(\nabla u_n\cdot\nabla \varphi)dx
 +\int_{\Omega}u_n^5\varphi dx\\
&\quad +\lambda\int_{\Omega}fu_n^{q-1}\varphi dx
 +\mu\int_{\Omega}|x|^{\alpha-2}u_n\varphi dx\\
&= -(a+b\|u_n\|^2)\int_{\Omega}(\nabla u_n\cdot\nabla \varphi)dx
 +\int_{\Omega}u_n^5\varphi dx
 +\lambda\int_{\Omega}fu_n^{q-1}\varphi dx \\
&\quad +\mu\int_{\Omega}|x|^{\alpha-2}u_n\varphi dx.
\end{align*}
Thus, it follows that
\begin{equation}\label{2.10}
\begin{aligned}
-\frac{|f'_n(0)|\|u_n\|+\|\varphi\|}{n}
&\leq(a+b\|u_n\|^2)\int_{\Omega}(\nabla u_n\cdot\nabla\varphi)dx
 -\int_{\Omega}u_n^5\varphi dx\\
&\quad -\lambda\int_{\Omega}fu_n^{q-1}\varphi dx
 -\mu\int_{\Omega}|x|^{\alpha-2}u_n\varphi dx\\
&= \langle I_\lambda'(u_n),\varphi\rangle
\end{aligned}
\end{equation}
for any $\varphi\in H_0^1(\Omega)$. As \eqref{2.10} also holds for $-\varphi$,
one sees that \eqref{2.6} holds.
Moreover,  Lemma \ref{lem2.4} implies that there exists a constant $C>0$,
such that $|f'_n(0)|\leq C$ for all $n\in N$. So, passing to the limit as
 $n\to\infty$ in \eqref{2.6}, we have
\begin{equation}\label{2.11}
\begin{aligned}
&(a+b\lim_{n\to\infty}\|u_n\|^2)\int_{\Omega}(\nabla u_*\cdot\nabla\varphi)dx\\
&-\int_{\Omega}u_*^5\varphi dx
-\lambda\int_{\Omega}fu_*^{q-1}\varphi dx
-\mu\int_{\Omega}|x|^{\alpha-2}u_*\varphi dx=0
\end{aligned}
\end{equation}
for all $\varphi\in H_0^1(\Omega)$. The proof  is complete.
\end{proof}

We define
$$
\Lambda=\frac{abS^3}{4}+\frac{b^3S^6}{24}+\frac{(b^2S^4+4aS)^{3/2}}{24}.
$$

\begin{lemma} \label{lem2.7}
 Suppose $1<q<2$, $0\leq\beta<2$, and let $\{u_n\}\subset\mathcal{N}_\lambda^-$
be a minimizing sequence of $I_\lambda$ with
$\alpha_\lambda^-<\Lambda-D\lambda^{\frac{2}{2-q}}$
where
\[
D=\Big(\frac{(4-q)}{4q}|\Omega|^{\frac{6-q}{6}}|f|_{\infty}S^{-q/2}
\Big)^{\frac{2}{2-q}}\big(\frac{2q}{a}\big)^{\frac{q}{2-q}},
\]
then there exists $u\in H_0^1(\Omega)$ such that $u_n\to u$ in $L^6(\Omega)$.
\end{lemma}

\begin{proof}
Let $\{u_n\}\subset H_0^1(\Omega)$ be a $(PS)_c$ sequence for $I_\lambda$, namely
\begin{equation}\label{2.12}
I_\lambda(u_n)\to c,\quad I'_\lambda(u_n)\to0,\quad \text{as } n\to+\infty.
\end{equation}
We see that $\{u_n\}$ is bounded in $H_0^1(\Omega)$. Indeed, by \eqref{2.12}
and \eqref{1.4}, one has
\begin{align*}
1+c+o(\|u_n\|)
&\geq  I_\lambda(u_n)-\frac{1}{6}\langle I'_\lambda(u_n),u_n\rangle\\
&\geq \frac{1}{3}\big(a-\frac{\mu}{\mu_1}\big)\|u_n\|^2
 +\frac{b}{12}\|u_n\|^4-\lambda\big(\frac{1}{q}-\frac{1}{6}\big)
 |f|_\infty\int_{\Omega}|u_n|^qdx\\
&\geq \frac{1}{3}\big(a-\frac{\mu}{\mu_1}\big)\|u_n\|^2
 -\lambda\big(\frac{1}{q}-\frac{1}{6}\big)|f|_\infty S^{-q/2}
 |\Omega|^{\frac{6-q}{6}}\|u_n\|^q.
\end{align*}
Since $0<\mu<a\mu_1, 1<q<2$, it implies that $\{u_n\}$ is bounded in $H_0^1(\Omega)$.
So there exist a subsequence (still denoted by $\{u_n\}$)
and $u\in H_0^{1}(\Omega)$ such that
\begin{gather*}
u_n\rightharpoonup u,\quad \text{weakly in } H_0^{1}(\Omega),\\
u_n\to u,\quad \text{strongly in } L^p(\Omega)~(1\leq p<6),\\
u_n(x)\to u(x),\quad\text{a.e. in }\Omega.
\end{gather*}
Note that $I_\lambda(|u_n|)=I_\lambda(u_n)$, then we obtain that $u_{n}\geq0$.
According to the concentration compactness principle (see \cite{PL}), there
exists a subsequence, say $\{u_n\}$, such that
\begin{gather*}
|\nabla u_n|^2_{2}\rightharpoonup
d\eta\geq|\nabla u|_{2}^2+\sum_{j\in J}\eta_j\delta_{x_j}, \\
|u_n|^6_6\to d\nu=|u|_6^6+\sum_{j\in J}\nu_j\delta_{x_j},
\end{gather*}
where $J$ is an at most countable index set, $\delta_{x_j}$ is the
Dirac mass at $x_j$, and let $x_j\in\Omega$ in the support of
$\eta, \nu$. we have
\begin{equation}\label{2.13}
\eta_j, \nu_j\geq0,\quad \eta_j\geq S\nu_{j}^{1/3}.
\end{equation}
For any $\varepsilon>0$ sufficiently small, let $\psi_{\varepsilon,j}(x)$
be a smooth cut-off function centered at $x_j$ such that
$0\leq\psi_{\varepsilon,j}(x)\leq1$,
\begin{gather*}
\psi_{\varepsilon,j}(x)=1\quad \text{in }B(x_j,\frac{\varepsilon}{2}),\quad
\psi_{\varepsilon,j}(x)=0\quad \text{in }B(x_j,\varepsilon),\\
|\nabla\psi_{\varepsilon,j}(x)|\leq\frac{4}{\varepsilon}.
\end{gather*}
By \eqref{1.4}, we obtain
\begin{align*}
\big|\int_{\Omega}f|u_n|^{q}\psi_{\varepsilon,j}dx\big|
&\leq |f|_{\infty}\int_{B(x_j,\varepsilon)}|u_n|^{q}dx\\
&\leq |f|_{\infty}\Big(\int_{B(x_j,\varepsilon)}
 |u_n|^{q\cdot\frac{6}{q}}dx\Big)^{q/6}
 \Big(\int_{B(x_j,\varepsilon)}1dx\Big)^{\frac{6-q}{6}}\\
&\leq |f|_{\infty} S^{-q/2}\|u_n\|^q\varepsilon^{\frac{6-q}{2}}.
\end{align*}
Notice that $\{u_n\}$ is bounded in $H_0^1(\Omega)$, and $u_n\rightharpoonup u$
weakly in $L^6(\Omega)$, it implies that
\begin{equation*}
\lim_{\varepsilon\to0}\lim_{n\to\infty}
\int_{\Omega}f|u_n|^{q}\psi_{\varepsilon,j}dx=0.
\end{equation*}
Similarly, we obtain
\begin{equation*}
\lim_{\varepsilon\to0}\lim_{n\to\infty}\mu\int_{\Omega}|x|^{\alpha-2}
|u_n|^2\psi_{\varepsilon,j}dx=0.
\end{equation*}

Since $\{\psi_{\varepsilon,j}u_n\}$ is bounded in $H_0^1(\Omega)$,
taking the test function $\psi_{\varepsilon,j}u_n$ in $I'_\lambda(u_n)\to0$,
one deduces that
\begin{align*}
0&= \lim_{\varepsilon\to0}\lim_{n\to\infty}\langle I'_\lambda(u_n),
  \psi_{\varepsilon,j}u_n\rangle\\
&= \lim_{\varepsilon\to0}\lim_{n\to\infty}
 \Big\{(a+b\|u_n\|^2)\int_{\Omega}(\nabla u_n\cdot\nabla(\psi_{\varepsilon,j}u_n))dx\\
&\quad -\mu\int_{\Omega}|x|^{\alpha-2}u_n^2\psi_{\varepsilon,j}dx\\
&\quad -\int_{\Omega}u_n^6\psi_{\varepsilon,j}dx
 -\lambda\int_{\Omega}fu_n^{q}\psi_{\varepsilon,j}dx\Big\}\\
&\geq \lim_{\varepsilon\to0}\lim_{n\to\infty}
 \Big\{(a+b\|u_n\|^2)\int_{\Omega}(|\nabla u_n|^2\psi_{\varepsilon,j}
 +u_n\nabla u_n\nabla\psi_{\varepsilon,j})dx\\
&\quad -\mu\int_{\Omega}|x|^{\alpha-2}u_n^2\psi_{\varepsilon,j}dx
 -\int_{\Omega}u_n^6\psi_{\varepsilon,j}dx
 -\lambda\int_{\Omega}fu_n^{q}\psi_{\varepsilon,j}dx\Big\}\\
&\geq \lim_{\varepsilon\to0}\lim_{n\to\infty}
 \Big\{(a+b\|u_n\|^2)\int_{\Omega}(|\nabla u_n|^2\psi_{\varepsilon,j}
 +u_n\nabla u_n\nabla\psi_{\varepsilon,j})dx\\
&\quad -\int_{\Omega}u_n^6\psi_{\varepsilon,j}dx\Big\}\\
&\geq  (a+b\eta_j)\eta_j-\nu_j.
\end{align*}
Thus
$\nu_j\geq (a+b\eta_j)\eta_j$.
By \eqref{2.13} we obtain
\begin{equation}\label{2.14}
\nu_j^{2/3}\geq aS+bS^2\nu_j^{1/3},\quad \text{or}\quad \eta_j=\nu_j=0.
\end{equation}
Set $X=\nu_j^{1/3}$, it follows from \eqref{2.14} that
$$
X^2\geq aS+bS^2X,
$$
then
$$
X\geq\frac{bS^2+\sqrt{b^2S^4+4aS}}{2};
$$
therefore
$$
\eta_j\geq SX\geq\frac{bS^3+\sqrt{b^2S^6+4aS^3}}{2}=: K.
$$

Next we show that $\eta_j\geq\sqrt{aS^3}$ is impossible.
So the set $J$ is empty. Assume the contrary, there exists some
$j_0\in J$ such that $\eta_{j_0}\geq\sqrt{aS^3}$. By \eqref{2.12}, \eqref{1.4}
and Young inequality, we obtain
\begin{align*}
c&= \lim_{n\to\infty}I_\lambda(u_n)\\
&= \lim_{n\to\infty}\Big\{\frac{a}{2}\|u_n\|^2+\frac{b}{4}\|u_n\|^4
 -\frac{\mu}{2}\int_{\Omega}|x|^{\alpha-2}|u_n|^2dx
 -\frac{1}{6}\int_{\Omega}|u_n|^6dx\\
&\quad -\frac{\lambda}{q}\int_{\Omega}f|u_n|^{q}dx
-\frac{1}{4}\Big(a\|u_n\|^2+b\|u_n\|^4-\mu\int_{\Omega}|x|^{\alpha-2}|u_n|^2dx\\
&\quad -\int_{\Omega}|u_n|^6dx-\lambda\int_{\Omega}f|u_n|^{q}dx\Big)\Big\}\\
&= \lim_{n\to\infty}\Big\{\big(\frac{1}{2}-\frac{1}{4}\big)a\|u_n\|^2
+b\big(\frac{1}{4}-\frac{1}{4}\big)\| u_n\|^4\\
&\quad +\big(\frac{1}{4}-\frac{1}{6}\big)\int_{\Omega}|u_n|^6dx
-\lambda\big(\frac{1}{q}-\frac{1}{4}\big)\int_{\Omega}f|u_n|^{q}dx\Big\}\\
&\geq \Big\{\big(\frac{1}{2}-\frac{1}{4}\big)
 a\Big(\|u\|^2 +\sum_{j\in J}\mu_j\Big)+\big(\frac{1}{4}-\frac{1}{6}\big)
 \Big(\int_{\Omega}|u|^6dx+\sum_{j\in J}\nu_j\Big)\\
&\quad +\big(\frac{1}{4}-\frac{1}{4}\big)
b\Big(\|u\|^2  +\sum_{j\in J}\mu_j\Big)^2-\lambda\big(\frac{1}{q}-\frac{1}{4}\big)\int_{\Omega}|f||u|^{q}dx\Big\}\\
&\geq \big(\frac{1}{2}-\frac{1}{4}\big)a\eta_{j_0}
 +\big(\frac{1}{4}-\frac{1}{4}\big)b\eta_{j_0}^2
 +\big(\frac{1}{4}-\frac{1}{6}\big)\nu_{j_0}+\frac{1}{4}a\|u\|^2\\
&\quad -\lambda\big(\frac{1}{q}-\frac{1}{4}\big)
 |f|_\infty|\Omega|^{\frac{6-q}{6}}S^{-q/2}\|u\|^{q}\\
&\geq \frac{a}{2}K+\frac{b}{4}K^2-\frac{K^3}{6S^3}
 -\frac{1}{4}\Big(aK+bK^2-\frac{K^3}{S^3}\Big)-D\lambda^{\frac{2}{2-q}},
\end{align*}
where 
\begin{gather*}
D=\Big(\frac{(4-q)}{4q}|\Omega|^{\frac{6-q}{6}}
|f|_{\infty}S^{-q/2}\Big)^{\frac{2}{2-q}}
\Big(\frac{2q}{a}\Big)^{\frac{q}{2-q}},\\
\frac{aK}{2}+\frac{b}{4}K^2-\frac{K^3}{6S^3}=\Lambda,\quad
 K\Big(a+bK-\frac{K^2}{S^3}\Big)=0.
\end{gather*}
Indeed,
\begin{align*}
&\frac{aK}{2}+\frac{b}{4}K^2-\frac{K^3}{6S^3}\\
&=  K\big(\frac{a}{2}+\frac{bK}{4}-\frac{K^2}{6S^3}\big)\\
&=  K\big[\frac{a}{2}+\frac{b}{4}\cdot\frac{bS^3+\sqrt{b^2S^6+4aS^3}}{2}
 -\frac{2b^2S^6+4aS^3+2bS^3\sqrt{b^2S^6+4aS^3}}{24S^3}\big]\\
&=  K\big[\frac{a}{2}+\frac{b^2S^3+b\sqrt{b^2S^6+4aS^3}}{8}
 -\frac{b^2S^3+2+b\sqrt{b^2S^6+4aS^3}}{12}\big]\\
&=  K\big[\frac{8a+b^2S^3+b\sqrt{b^2S^6+4aS^3}}{24}\big]\\
&= \frac{12abS^3+2b^3S^6+(2b^2S^3+8a)\sqrt{b^2S^6+4aS^3}}{48}\\
&= \frac{abS^3}{4}+\frac{b^3S^6}{24}+\frac{(b^2S^3+4a)\sqrt{b^2S^6+4aS^3}}{24}
= \Lambda.
\end{align*}
and
\begin{align*}
& a+bK-\frac{K^2}{S^3}\\
&=  a+b\frac{bS^3+\sqrt{b^2S^6+4aS^3}}{2}
 -\frac{1}{S^3}\frac{2b^2S^6+4aS^3+2bS^3\sqrt{b^2S^6+4aS^3}}{4}\\
&=  a+\frac{b^2S^3+b\sqrt{b^2S^6+4aS^3}}{2}
 -\frac{2S^3(b^2S^3+2a+b\sqrt{b^2S^6+4aS^3})}{4S^3}\\
&= \frac{2a+b^2S^3+b\sqrt{b^2S^6+4aS^3}-b^2S^3-2a-b\sqrt{b^2S^6+4aS^3}}{2}
= 0.
\end{align*}
Therefore, we obtain
 $\Lambda-D\lambda^{\frac{2}{2-q}}\leq c<\Lambda-D\lambda^{\frac{2}{2-q}}$,
which is a contradiction. It suggests that $J$ is empty,
which means that $\int_{\Omega}|u_n|^6dx\to\int_{\Omega}|u|^6dx$ as
 $n\to\infty$. This completes the proof.
\end{proof}

It is known that the function
\begin{equation*}
U_\varepsilon(x)=\frac{(3\varepsilon)^{1/4}}
{(\varepsilon+|x|^2)^{1/2}},\quad x\in \mathbb{R}^3,\; \varepsilon>0
\end{equation*}
satisfies
\begin{equation*}
-\Delta U_\varepsilon=U_\varepsilon^{5}~~~~\text{in}~\mathbb{R}^3.
\end{equation*}
Set
$$
C_\varepsilon=(3\varepsilon)^{1/4},\quad
y_\varepsilon(x)=\frac{U_\varepsilon(x)}{C_\varepsilon}.
$$
We select a cut-off function $\varphi\in C_0^\infty(\Omega)$ such that
$\varphi(x)=1$ for $|x|<R_0$, and $\varphi(x)=0$ for $|x|>2R_0$,
 $0\leq\varphi(x)\leq1$.
Let $u_\varepsilon(x)=\varphi(x)y_\varepsilon(x)$,
$v_\varepsilon(x)=\frac{u_\varepsilon(x)}{|u_\varepsilon|_6}$,
then $|v_\varepsilon|_6^6=1$.
Therefore we obtain the following results (see details in \cite{XT})
\begin{equation}\label{2.15}
\begin{gathered}
\|v_\varepsilon\|^2\leq S+C\varepsilon^{1/2},\\
\|v_\varepsilon\|^6\leq S^3+C\varepsilon^{1/2},\\
\|v_\varepsilon\|^{12}\leq S^6+C\varepsilon^{1/2},\\
\|v_\varepsilon\|^{18}\leq S^9+C\varepsilon^{1/2},\\
\|v_\varepsilon\|^{24}\leq S^{12}+C\varepsilon^{1/2}.
\end{gathered}
\end{equation}
and
\begin{equation}\label{2.16}
\begin{gathered}
O(\varepsilon^{q/4})\leq\int_{\Omega}|u_\varepsilon|^qdx
 \leq O(\varepsilon^{q/4}),\\
\int_{\Omega}|x|^{\alpha-2}|u_\varepsilon|^qdx=O(\varepsilon^{\alpha/2}).
\end{gathered}
\end{equation}

\begin{lemma} \label{lem2.8}
 Assume $1<q<2$, and $0<\alpha<1$. Then
\begin{equation*}
\sup_{t\geq0}I_\lambda(tu_{\varepsilon})<\Lambda-D\lambda^{\frac{2}{2-q}}.
\end{equation*}
\end{lemma}

\begin{proof}
We claim that there exist $t_\varepsilon>0$ and positive constants $t_0, t_1$
which are independent of $\varepsilon, \lambda$, such that
$\sup_{t\geq0}I_\lambda(tu_\varepsilon)=I_\lambda(t_\varepsilon u_\varepsilon)$ and
\begin{equation}\label{2.17}
0<t_0\leq t_\varepsilon\leq t_1<\infty.
\end{equation}
As $\lim_{t\to+\infty}I_\lambda(tu_\varepsilon)=-\infty$,
there exists $t_\varepsilon>0$, such that
\begin{equation*}
I_\lambda(t_\varepsilon u_\varepsilon)
=\sup_{t\geq0}I_\lambda(t_\varepsilon u_\varepsilon),\quad \text{and}\quad
\frac{dI_\lambda(t_\varepsilon u_\varepsilon)}{dt}\big|_{t=t_\varepsilon}=0.
\end{equation*}
Then
\begin{equation}\label{2.18}
t_\varepsilon a\|u_\varepsilon\|^2+t_\varepsilon^3 b\|u_\varepsilon\|^4
-\mu t_\varepsilon\int_{\Omega}|x|^{\alpha-2}u_\varepsilon^2 dx
-t_\varepsilon^5\int_{\Omega}u_\varepsilon^6dx
-\lambda t_\varepsilon^{q-1}\int_{\Omega}fu_\varepsilon^qdx=0,
\end{equation}
and
\begin{equation*}
a\|u_\varepsilon\|^2+3t_\varepsilon^2 b\|u_\varepsilon\|^4
-\mu\int_{\Omega}|x|^{\alpha-2}u_\varepsilon^2dx
-5t_\varepsilon^4\int_{\Omega}u_\varepsilon^6dx
-\lambda(q-1) t_\varepsilon^{q-2}\int_{\Omega}fu_\varepsilon^qdx<0.
\end{equation*}
Therefore,
\begin{equation}\label{2.19}
\begin{aligned}
&(2-q)t_\varepsilon a\|u_\varepsilon\|^2+(4-q)t_\varepsilon^3 b\|u_\varepsilon\|^4
-(2-q)\mu t_\varepsilon\int_{\Omega}\frac{u_\varepsilon^2}{|x|^{2-\alpha}}dx\\
&<(6-q)t_\varepsilon^5\int_{\Omega}u_\varepsilon^6dx.
\end{aligned}
\end{equation}
On the one hand, we can get easily from \eqref{2.19} that $t_\varepsilon$
is bounded below, so, there exists a positive constant $t_0>0$
(independent of $\varepsilon, \lambda$), such that $0<t_0\leq t_\varepsilon$.
On the other hand, it follows from \eqref{2.18} that
$$
\frac{a\|u_\varepsilon\|^2}{t_\varepsilon^2}+b\|u_\varepsilon\|^2
=t_\varepsilon^2\int_{\Omega}u_\varepsilon^6dx
+\frac{\lambda}{t_\varepsilon^{4-q}}\int_{\Omega}fu_\varepsilon^qdx
+\frac{\mu}{t_\varepsilon^2}\int_{\Omega}|x|^{\alpha-2}u_\varepsilon^2dx,
$$
thus, $t_\varepsilon$ is bounded above for all $\varepsilon>0$ sufficiently small.
 Then \eqref{2.17} holds.
Set
$$
h(t_\varepsilon u_\varepsilon)=\frac{a}{2}t_\varepsilon^2\|u_\varepsilon\|^2
+\frac{b}{4}t_\varepsilon^4\|u_\varepsilon\|^4
-\frac{t_\varepsilon^6}{6}\int_{\Omega}u_\varepsilon^6dx.
$$
We claim that there exists a positive constant $c_7$
(independent of $\varepsilon, \lambda$), such that
\begin{equation}\label{2.20}
h(t_\varepsilon u_\varepsilon)\leq\Lambda+c_7\varepsilon^{1/2}.
\end{equation}
Indeed, set
$$
g(t)=\frac{a}{2}t^2\|u_\varepsilon\|^2+\frac{b}{4}t^4\|u_\varepsilon\|^4
-\frac{t^6}{6}\int_{\Omega}u_\varepsilon^6dx.
$$
Since $\lim_{t\to\infty}g(t)=-\infty, g(0)=0$, and
$\lim_{t\to0^+}g(t)>0$, it follows that $\sup_{t\geq0}g(t)$
attained at $T_\varepsilon>0$, namely,
$$
g'(t)|_{T_\varepsilon}=aT_\varepsilon\|u_\varepsilon\|^2
+bT_\varepsilon^3\|u_\varepsilon\|^4-T_\varepsilon^5\int_{\Omega}u_\varepsilon^6dx=0.
$$
Then
$$
T_\varepsilon^4\int_{\Omega}u_\varepsilon^6dx
 -a\|u_\varepsilon\|^2-bT_\varepsilon^2\|u_\varepsilon\|^4=0;
$$
therefore
$$
T_\varepsilon=\bigg(\frac{b\|u_\varepsilon\|^4+\sqrt{b^2\|u_\varepsilon\|^8
+4a\|u_\varepsilon\|^2\int_{\Omega}u_\varepsilon^6dx}}
{2\int_{\Omega}u_\varepsilon^6dx}\bigg)^{1/2}.
$$
Notice that $g(t)$ is increasing in $[0, T_\varepsilon]$, then by \eqref{2.15},
 one has
\begin{align*}
h(t_\varepsilon u_\varepsilon)
&\leq  g(T_\varepsilon)\\
&= \frac{a}{2}T_\varepsilon^2\|u_\varepsilon\|^2+\frac{b}{4}T_\varepsilon^4\|u_\varepsilon\|^4-\frac{T_\varepsilon^6}{6}\int_{\Omega}u_\varepsilon^6dx\\
&=  T_\varepsilon^2\big(\frac{a}{3}\|u_\varepsilon\|^2
 +\frac{b}{12}T_\varepsilon^2\|u_\varepsilon\|^4\big)\\
&=  T_\varepsilon^2\Big(\frac{a}{3}\|u_\varepsilon\|^2
 +\frac{b^2\|u_\varepsilon\|^8+b\|u_\varepsilon\|^4\sqrt{b^2\|u_\varepsilon\|^8
 +4a\|u_\varepsilon\|^2\int_{\Omega}u_\varepsilon^6dx}}
 {24\int_{\Omega}u_\varepsilon^6dx}\Big)\\
&= \frac{ab\|u_\varepsilon\|^6}{6\int_{\Omega}u_\varepsilon^6dx}
 +\frac{b\|u_\varepsilon\|^6}{12\int_{\Omega}u_\varepsilon^6dx}
 +\frac{\|u_\varepsilon\|^2\sqrt{b^2\|u_\varepsilon\|^8
 +4a\|u_\varepsilon\|^2\int_{\Omega}u_\varepsilon^6dx}}
 {6\int_{\Omega}u_\varepsilon^6dx}\\
&\quad +\frac{b^3\|u_\varepsilon\|^{12}}{24(\int_{\Omega}u_\varepsilon^6dx)^2}
 +\frac{b^2\|u_\varepsilon\|^8\sqrt{b^2\|u_\varepsilon\|^8
 +4a\|u_\varepsilon\|^2\int_{\Omega}u_\varepsilon^6dx}}
 {24(\int_{\Omega}u_\varepsilon^6dx)^2}\\
&= \frac{ab\|u_\varepsilon\|^6}{4\int_{\Omega}u_\varepsilon^6dx}
 +\frac{b^3\|u_\varepsilon\|^{12}}{24(\int_{\Omega}u_\varepsilon^6dx)^2}
 +\frac{(b^2\|u_\varepsilon\|^8
 +4a\|u_\varepsilon\|^2\int_{\Omega}u_\varepsilon^6dx)^{3/2}}
 {24(\int_{\Omega}u_\varepsilon^6dx)^2}\\
&\leq \frac{ab(S^{\frac{9}{2}}+c_4\varepsilon^{1/2})}
 {4(S^{3/2}+c_2\varepsilon^{3/2})}+
\frac{b^3(S^9+c_6\varepsilon^{1/2})}{24(S^{3/2}+c_2\varepsilon^{3/2})^2}\\
&\quad +\frac{[b^2(S^6+c_5\varepsilon^{1/2})+4a(S^{3/2}
 +c_1\varepsilon^{1/2})(S^{3/2}+c_2\varepsilon^{3/2})]^{3/2}}
 {24(S^{3/2}+c_2\varepsilon^{3/2})^2}\\
&\leq \frac{abS^3}{4}+\frac{b^3S^6}{24}+\frac{(b^2S^6+4aS^3)^{3/2}}{24S^3}
 +c_7\varepsilon^{1/2}\\
&= \Lambda+c_7\varepsilon^{1/2}.
\end{align*}
Consequently, there exists $c_7>0$ (independent of $\varepsilon,\lambda$)
such that \eqref{2.20} holds.

Since $0<\alpha<1$, from \cite{AJ}, there exists a positive constant $c_8$
(independent of $\varepsilon,\lambda$) such that
\begin{equation}\label{2.21}
\int_{\Omega}|x|^{\alpha-2}u_\varepsilon^2dx=c_8\varepsilon^{\alpha/2}.
\end{equation}
Therefore, from \eqref{2.16}, \eqref{2.20} and \eqref{2.21}, it holds
\begin{equation}\label{2.22}
\begin{aligned}
I_\lambda(t_\varepsilon u_\varepsilon)
&=  h(t_\varepsilon u_\varepsilon)
 -\frac{\mu t_\varepsilon^2}{2}\int_{\Omega}|x|^{\alpha-2}u_\varepsilon^2dx
 -\lambda\frac{t_\varepsilon^q}{q}\int_{\Omega}fu_\varepsilon^qdx\\
&\leq \Lambda+c_7\varepsilon^{1/2}
 -\frac{\mu}{2}t_0^2c_8\varepsilon^{\alpha/2}
 +\lambda\frac{T_1^q|f|_\infty}{q}\int_{\Omega}u_\varepsilon^qdx\\
&= \Lambda+c_7\varepsilon^{1/2}-c_9\varepsilon^{\alpha/2}
 +\lambda c_{10}\varepsilon^{q/4}
\end{aligned}
\end{equation}
where $c_9=\frac{\mu}{2}t_0^2c_8$, $c_{10}=\frac{T_1^q|f|_\infty}{q}$.
Notice that $0<\alpha<1$.
Let
\[
\varepsilon=\lambda^{\frac{4}{2-q}}, \quad
\lambda<\lambda_0=\Big(\frac{c_9}{c_7+c_{10}+D}\Big)^{\frac{2-q}{2(1-\alpha)}}.
\]
Then
\begin{align*}
 c_7\varepsilon^{1/2}-c_9\varepsilon^{\alpha/2}
+c_{10}\lambda\varepsilon^{q/4}
&=  c_7\lambda^{\frac{2}{2-q}}+c_{10}\lambda\lambda^{\frac{q}{2-q}}
 -c_9\lambda^{\frac{2\alpha}{2-q}}\\
&= \lambda^{\frac{2}{2-q}}\Big(c_7+c_{10}
 -c_9\lambda^{-\frac{2(1-\alpha)}{2-q}}\Big)\\
&< -D\lambda^{\frac{2}{2-q}}.
\end{align*}
From \eqref{2.22} it follows that
\begin{align*}
I_\lambda(t_\varepsilon u_\varepsilon)
&=  h(t_\varepsilon u_\varepsilon)
-\frac{\mu t_\varepsilon^2}{2}\int_{\Omega}|x|^{\alpha-2}u_\varepsilon^2dx
-\lambda\frac{t_\varepsilon^q}{q}\int_{\Omega}fu_\varepsilon^qdx\\
&\leq \Lambda-D\lambda^{\frac{2}{2-q}}.
\end{align*}
This completes the proof.
\end{proof}

\section{Proof of main results}

 There exists a constant $\delta>0$ such that
$\Lambda-D\lambda^{\frac{2}{2-q}}>0$ for $\lambda<\delta$.
we set $\lambda_*=\min\{T_1, \delta\}$, thus
 Lemmas \ref{lem2.1}--\ref{lem2.4}, \ref{lem2.6}, \ref{lem2.7} hold for all
$0<\lambda<\lambda_*$.
We shall prove Theorem \ref{thm1.1} in two steps.
\smallskip

\noindent\textbf{Step1}
By Lemma \ref{lem2.6}, there exists a bounded minimizing sequence
$\{u_n\}\subset\mathcal{N}_\lambda$ of $I_\lambda$. Perhaps for a
subsequence, still denoted by $\{u_n\}$, there exists $u_\lambda\in H_0^1(\Omega)$
such that
\begin{gather*}
u_n\rightharpoonup u_\lambda,\quad \text{weakly in }H_0^{1}(\Omega),\\
u_n\to u_\lambda,\quad \text{strongly in } L^{s}(\Omega),\; 1\leq s<6,\\
u_n(x)\to u_\lambda(x),\quad \text{a.e. in } \Omega,
\end{gather*}
as $n\to\infty$. Now we shall prove that $u_\lambda$ is a nonzero
non-negative ground state solution of problem \eqref{1.1}.

At first, we prove that $u_\lambda$ is a non-negative solution of \eqref{1.1}.
Indeed, by \eqref{2.11} in Lemma \ref{lem2.6}, for all $\varphi\in H_0^1(\Omega)$,
 we obtain
\begin{align*}
&(a+b\lim_{n\to\infty}\|u_n\|^2)\int_{\Omega}(\nabla u_\lambda\cdot\nabla\varphi)dx\\
&-\int_{\Omega}u_\lambda^5\varphi dx
-\lambda\int_{\Omega}fu_\lambda^{q-1}\varphi dx
-\mu\int_{\Omega}|x|^{2-\alpha}u_\lambda\varphi dx=0.
\end{align*}
Setting $\lim_{n\to\infty}\|u_n\|=l$, one has
\begin{equation}\label{3.1}
\begin{aligned}
&(a+bl^2)\int_{\Omega}(\nabla u_\lambda\cdot\nabla\varphi)dx
-\int_{\Omega}u_\lambda^5\varphi dx\\
&-\lambda\int_{\Omega}fu_\lambda^{q-1}\varphi dx
-\mu\int_{\Omega}|x|^{2-\alpha}u_\lambda\varphi dx=0.
\end{aligned}
\end{equation}
Taking the test function $\varphi=u_\lambda$ in \eqref{3.1}, we have
\begin{equation}\label{3.2}
(a+bl^2)\|u_\lambda\|^2-\int_{\Omega}u_\lambda^6dx
-\lambda\int_{\Omega}fu_\lambda^{q} dx
-\mu\int_{\Omega}|x|^{2-\alpha}u_\lambda^2 dx=0.
\end{equation}
The fact $u_n\in\mathcal{N}_\lambda$ implies
\begin{equation*}
(a+b\|u_n\|^2)\|u_n\|^2-\int_{\Omega}u_n^6dx
-\lambda\int_{\Omega}fu_n^{q} dx-\mu\int_{\Omega}|x|^{2-\alpha}u_n^2 dx=0.
\end{equation*}
Since $\alpha_\lambda<0<\Lambda-D\lambda^{\frac{2}{2-q}}$,
by Lemma \ref{lem2.7} we obtain
\begin{equation}\label{3.3}
(a+bl^2)l^2-\int_{\Omega}u_\lambda^6dx-\lambda\int_{\Omega}fu_\lambda^{q} dx
-\mu\int_{\Omega}|x|^{2-\alpha}u_\lambda^2 dx=0.
\end{equation}
It follows from \eqref{3.2} and \eqref{3.3} that $\|u_\lambda\|=l$,
 which suggests that $u_n\to u_\lambda$ in $H_0^1(\Omega)$ and $u_\lambda$
is a solution of \eqref{1.1}, namely,
\begin{equation}\label{3.4}
\begin{aligned}
&(a+b\|u_\lambda\|^2)\int_{\Omega}(\nabla u_\lambda\cdot\nabla\varphi)dx
 -\int_{\Omega}u_\lambda^5\varphi dx\\
&-\lambda\int_{\Omega}fu_\lambda^{q-1}\varphi dx
 -\mu\int_{\Omega}|x|^{2-\alpha}u_\lambda\varphi dx=0.
\end{aligned}
\end{equation}
for all $\varphi\in H_0^1(\Omega)$. Recall that $u_\lambda\geq0$.
In addition, note that $u_\lambda\in\mathcal{N}_\lambda$
($u_\lambda$ is a nontrivial solution of  \eqref{1.1}) and $\alpha_\lambda<0$
(by Lemma \ref{lem2.3}), then one obtains
\begin{align*}
&\lambda\big(\frac{1}{q}-\frac{1}{6}\big)\int_{\Omega}f|u_\lambda|^{q}dx\\
&=  \frac{a}{3}\|u_\lambda\|^2
 -\frac{\mu}{3}\int_{\Omega}|x|^{\alpha-2}|u_\lambda|^2dx
 +\frac{b}{12}\|u_\lambda\|^4-I_\lambda(u_\lambda)\\
&\geq \frac{1}{3}(a-\frac{\mu}{\mu_1})\|u_\lambda\|^2+\frac{b}{12}\|u_\lambda\|^4
 -\alpha_\lambda
> 0,
\end{align*}
 which implies that $u_\lambda\not\equiv0$.
By Lemma \ref{lem2.7} we obtain
\begin{equation}\label{3.5}
\alpha_\lambda=\lim_{n\to\infty}I_\lambda(u_n)=I_\lambda(u_\lambda).
\end{equation}
Next, we shall show that $u_\lambda\in\mathcal{N}_\lambda^+$ and
$I_\lambda(u_\lambda)=\alpha_\lambda^+$. We claim that
$u_\lambda\in\mathcal{N}_\lambda^+$. On the contrary, suppose that
$u_\lambda\in\mathcal{N}_\lambda^-$, by Lemma \ref{lem2.1},
 there exist positive numbers $t^+<t_{\rm max}<t^-=1$
such that $t^+u\in\mathcal{N}_\lambda^+$, $t^-u\in\mathcal{N}_\lambda^-$ and
$$
\alpha_\lambda<I_\lambda(t^+u_\lambda)<I_\lambda(t^-u_\lambda)
=I_\lambda(u_\lambda)=\alpha_\lambda,
$$
which is a contradiction. Thus,
 $u_\lambda\in\mathcal{N}_\lambda^+$. From the definition of
$\alpha_\lambda^+$, we obtain $\alpha_\lambda^+\leq I_\lambda(u_\lambda)$.
It follows from Lemma \ref{lem2.3} and \eqref{3.5} that
$$
I_\lambda(u_\lambda)=\alpha_\lambda^+=\alpha_\lambda<0.
$$
From the above discussion, we obtain that $u_\lambda$ is a nonzero
non-negative ground state solution of problem \eqref{1.1}.
\smallskip

\noindent\textbf{Step 2.}
 We shall verify that problem \eqref{1.1} has a second solution
$v_\lambda$ with $v_\lambda\in\mathcal{N}_\lambda^-$.

As $I_\lambda$ is also coercive on $\mathcal{N}_\lambda^-$,
then we apply the Ekeland's variational principle to the
minimization problem
$\alpha_\lambda^{-}=\inf_{v\in\mathcal{N}_\lambda^-}I_\lambda(v)$
to obtain a minimizing sequence
$\{v_{n}\}\subset\mathcal{N}_\lambda^-$ of $I_\lambda$ with the following properties:
\begin{itemize}
\item[(i)] $ I_\lambda(v_{n})<\alpha_\lambda^{-}+\frac{1}{n}$;
\item[(ii)] $ I_\lambda(u)\geq I_\lambda(v_{n})-\frac{1}{n}\|u-v_{n}\|$
for all $u\in\mathcal{N}_\lambda^-$.
\end{itemize}
Since $\{v_{n}\}$ is
bounded in $H_0^{1}(\Omega)$, passing to a subsequence if necessary,
there exists $v_\lambda\in H_0^{1}(\Omega)$ such that
\begin{gather*}
v_n\rightharpoonup v_\lambda,\quad \text{weakly in }H_0^{1}(\Omega),\\
v_n\to v_\lambda,\quad \text{strongly in } L^{s}(\Omega),~1\leq s\leq6,\\
v_n(x)\to v_\lambda(x),\quad \text{a.e. in }\Omega,
\end{gather*}
as $n\to\infty$. Now we shall prove that $v_\lambda$ is a non-negative
solution of  \eqref{1.1}.
Similar to the proof of Theorem \ref{thm1.1}, we obtain $v_n\to v_\lambda$
in $H_0^1(\Omega)$ and $v_\lambda$ is a non-negative solution of \eqref{1.1}.

Now, we prove that $v_\lambda\not\equiv0$ in $\Omega$.
From $v_{n}\in\mathcal{N}_\lambda^-$, we have
\begin{align*}
a(2-q)\|v_{n}\|^2
&\leq (6-q)\int_{\Omega}v_n^6dx+(2-q)\mu\int_{\Omega}|x|^{\alpha-2}v_n^2dx
 -b(4-q)\|v_{n}\|^4\\
&\leq (6-q)\int_{\Omega}v_{n}^6dx+(2-q)\mu\int_{\Omega}|x|^{\alpha-2}v_n^2dx\\
&< (6-q)S^{-3}\|v_n\|^6+(2-q)\frac{\mu}{\mu_1}\|v_n\|^2,
\end{align*}
so that
\begin{equation}\label{3.6}
\|v_n\|>\Big(\frac{(a-\frac{\mu}{\mu_1})(2-q)S^{3}}{(6-q)}\Big)^{1/4},
\quad \forall v_{n}\in\mathcal{N}_\lambda^-.
\end{equation}
Note that $v_n\to v_\lambda$ in $H_0^1(\Omega)$, \eqref{3.6}
implies that $v_\lambda\not\equiv0$.

Next, we  prove that $v_\lambda\in\mathcal{N}_\lambda^-$. It suffices to prove
that $\mathcal{N}_\lambda^-$ is closed.
Indeed, by Lemmas \ref{lem2.7} and \ref{lem2.8},
for $\{v_n\}\subset\mathcal{N}_\lambda^-$,
we obtain
\begin{equation*}
 \lim_{n\to\infty}\int_{\Omega}v_n^6dx=\int_{\Omega}v_\lambda^6dx.
\end{equation*}
From the definition of $\mathcal{N}_\lambda^-$, it holds
\begin{equation*}
(2-q)a\|v_n\|^2+(4-q)b\|v_n\|^4-(6-q)\int_{\Omega}v_n^6dx
-(2-q)\mu\int_{\Omega}|x|^{\alpha-2}v_n^2 dx<0.
\end{equation*}
Then
\begin{equation*}
(2-q)a\|v_\lambda\|^2+(4-q)b\|v_\lambda\|^4-(6-q)\int_{\Omega}v_\lambda^6dx
-(2-q)\mu\int_{\Omega}|x|^{\alpha-2}v_\lambda^2 dx\leq0,
\end{equation*}
which implies that $v_\lambda\in\mathcal{N}_\lambda^0\cup\mathcal{N}_\lambda^-$.
If $\mathcal{N}_\lambda^-$ is not closed, then one obtains
$v_\lambda\in\mathcal{N}_\lambda^0$.
By Lemma \ref{lem2.1}, it follows that $v_\lambda=0$, which contradicts
$v_\lambda\not\equiv0$.
Therefore, $v_\lambda\in\mathcal{N}_\lambda^-$.
Since $\mathcal{N}_\lambda^{+}\cap\mathcal{N}_\lambda^{-}=\emptyset$,
$u_{\lambda}$ and $v_{\lambda}$ are different.


\subsection*{Acknowledgments}
This research was supported by the Science and Technology Foundation
of Guizhou Province (No. LH[2015]7207), and by the
 Natural Science Foundation of Guizhou Provincial Department of
 Education [2013]405.


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\end{document}