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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 50, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/50\hfil Existence of positive solutions]
{Existence of positive solutions to Kirchhoff type problems
 involving singular and critical nonlinearities}

\author[C.-Y. Lei, G.-S. Liu \hfil EJDE-2018/50\hfilneg]
{Chun-Yu Lei, Gao-Sheng Liu}

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

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

\dedicatory{Communicated by Paul H. Rabinowitz}

\thanks{Submitted September 1, 2017. Published February 16, 2018.}
\subjclass[2010]{35A15, 35D30, 35J60}
\keywords{Kirchhoff type equation; critical exponents; variational methods}

\begin{abstract}
 In this study, we study a Kirchhoff type problem involving singular
 and critical nonlinearities. With aid of variational methods and
 concentration compactness principle, we prove that the problem admits
 a weak solution.
\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}

We are interested in the  Kirchhoff type problem
\begin{equation}\label{1.1}
\begin{gathered}
-\Big(a+b\int_{\Omega}|\nabla u|^2\,dx\Big)\Delta u=f(x,u), \quad
\text{in }   \Omega,\\
u=0, \quad \text{on }    \partial\Omega,
\end{gathered}
\end{equation}
where $\Omega$ is a bounded smooth domain in $\mathbb{R}^3$, 
$0\in\Omega$, $a>0, b\geq0$.

Existence and multiplicity of solutions to \eqref{1.1} have been studied 
intensively by many researchers. There are lots
of works in the literature not only on the subcritical cases such as 
\cite{AM,CW,HZ,YS,MT,PZ,ST,TC,ZP},
 but also on the critical cases like 
\cite{CF,FV,FJ,GF,LG,LS,LL,LLZ,LKLT,ND,OA,WZ,XZ}. 
In particular, Naimen  \cite{ND1} investigated the  kirchhoff type equation
\begin{equation}\label{A}
\begin{gathered}
-\Big(1+b\int_\Omega|\nabla u|^2\Big)\Delta u=\beta u+u^5, \quad \text{in }  \Omega, \\
u=0,  \quad \text{on }   \partial\Omega,
\end{gathered}
\end{equation}
here $\Omega$ is a 3 dimensional open ball. For the reader's convenience, 
we report here one of the main results of \cite{ND1}.

\begin{theorem}[{\cite[Theorem 1.1]{ND1}}]  \label{thmA} 
Let $\beta\in\mathbb{R}$ be a given constant. Then the following assertions hold.
\begin{itemize}
\item[(i)] If $\beta<\beta_1/4$ ($\beta_1$ is the principal eigenvalue of 
$-\Delta$ on the open ball), problem \eqref{A}
has no solution for all $b\geq0$.

\item[(ii)] If $\beta_1/4<\beta<\beta_1$, there exists a constant 
$A_1=A_1(\beta)>0$ such that \eqref{A} has a
solution for all $0<b<A_1$.

\item[(iii)] If $\beta=\beta_1$, there exists a constant $A_2=A_2(\beta)>0$ 
such that \eqref{A} has a solution
for all $0<b<A_2$ and \eqref{A} has no solution for $b=0$.
\end{itemize}
\end{theorem}

In \eqref{A}, if $b=0$, Brezis-Nirenberg \cite{BN} found a solution provided 
$\beta_1/4<\beta<\beta_1$, thereby, Theorem \ref{thmA} (ii) extends one of the main 
results of Brezis-Nirenberg \cite{BN} to the Kirchhoff type problem. 
When $N=3$, we see that it is not easy to establish a solution in the case
 of $0<\beta<\beta_1$, the reason is that, it is difficult to estimate
 the critical value level for this case. However, for 4-dimensional case, 
Brezis and Nirenberg \cite{BN} obtained a positive solution provided 
$0<\beta<\beta_1$. Therefore, we also see that dimensions of space make an effect on parameter $\beta$.

Recently, Perera et al.\ \cite{PS} considered the  problem
\begin{gather*}
-\Delta u=\beta u+u^{2^*-1}-\mu, \quad \text{in }    \Omega, \\
u=0,   \quad \text{on }   \partial\Omega,
\end{gather*}
where $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$ ($N\geq4$). 
They obtained a ground state solution when $0<\beta<\beta_1$ and $\mu>0$ 
enough small. It remains open to extend this study for the case $N=3$ 
(see \cite[Remark 1.4]{PS}).

Based on the above work, in this article  we consider the case that problem 
has a combination of a critical Sobolev exponent term and a singular term. 
More precisely, we study the  Kirchhoff type equation of the form
\begin{equation}\label{1.2}
\begin{gathered}
-\Big(a+b\int_{\Omega}|\nabla u|^2\,dx\Big)\Delta u
 =\lambda\frac{u}{|x|^{2-s}}+u^5-\mu,\quad  \text{in }    \Omega,\\
u=0, \quad \text{on }   \partial\Omega,
\end{gathered}
\end{equation}
where $0<s<1$, $\lambda, \mu$ are two positive real numbers, and 
$0<\lambda<a\lambda_1$, here $\lambda_1$ is the first eigenvalue for 
eigenvalue problem
\begin{equation}\label{1.3}
\begin{gathered}
-\Delta u=\lambda|x|^{s-2}u \quad \text{in }   \Omega, \\
u=0,   \quad \text{on }  \partial\Omega,
\end{gathered}
\end{equation}
where $0<s<2$,  Chaudhuri et al.\ in \cite{CR} proved  that
problem \eqref{1.3} has a sequence of eigenvalues
$$
0<\lambda_1<\lambda_2\leq\dots\leq\lambda_k\leq\dots\to+\infty.
$$
Moreover, the first eigenvalue is characterized by
\begin{equation*}
  \lambda_1:=\inf_{u\in H_0^1(\Omega)\backslash \{0\}}
\frac{\int_{\Omega}|\nabla u|^2}{\int_{\Omega}|u|^2|x|^{s-2}}.
\end{equation*}
Our main result reads as follows.

\begin{theorem} \label{thm1.1}
 Assume $a>0, b\geq0$ and $0<\lambda<a\lambda_1$. Then there exists $\mu_*>0$ 
such that   \eqref{1.2} has at least a nontrivial solution for every 
$\mu\in(0, \mu_*)$. Moreover, if $\mu=0$, then  \eqref{1.2} admits a 
positive solution. 
\end{theorem}

\begin{remark} \label{rmk1.2} \rm
 On the one hand, compared with Theorem \ref{thmA}, we see that the coefficient
 $b$ is restrained in (ii) and (iii). Moreover, in Theorem \ref{thm1.1}, 
we also see that the singular term $1/|x|^{2-s}$ can release the
restriction on $\beta_1/4<\beta<\beta_1$. 
On the other hand,  the problem mentioned in \cite[Remark 1.4]{PS} 
is hard to tackle, however, if we add a singular term, the problem can 
be solved. So our results can be regarded as partial solution to that problem.
\end{remark}

In the next section we present some lemmas and the proof of Theorem \ref{thm1.1}.

\section{Proof of main results}

Let us give the following some notation:
\begin{itemize}
\item The space $H_0^1(\Omega)$ is equipped with the norm 
$\|u\|^2=\int_{\Omega}|\nabla u|^2\,dx$, the norm in
$L^p(\Omega)$ is denoted by $|\cdot|_p$;

\item $u_n^{+}(x)=\max\{u_n(x),0\}$, $u_n^{-}(x)=\max\{-u_n(x),0\}$; 
$C, C_1, C_2,\dots,$ denote various positive constants, which may vary 
from line to line;

\item Let $S$ be the best Sobolev constant, namely
  \begin{equation}\label{1}
S:=\inf_{u\in H_0^{1}(\Omega)\backslash \{0\}}
\frac{\int_{\Omega}|\nabla u|^2\,dx}{\big(\int_{\Omega}|u|^{6}\,dx\big)^{1/3}}.
\end{equation}
\end{itemize}

\subsection*{Existence of a positive solution}
Consider the energy functional $I_\mu: H_0^{1}(\Omega)\to\mathbb{R}$ given by
\begin{equation*}
I_\mu(u)=\frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4
-\frac{\lambda}{2}\int_{\Omega}\frac{(u^+)^{2}}{|x|^{2-s}}\,dx
-\frac{1}{6}\int_{\Omega}(u^+)^6\,dx+\mu\int_{\Omega}u\,dx.
\end{equation*}

\begin{lemma} \label{lem2.1}
 There exist $\alpha, \rho, \Lambda_0>0$ such that the functional $I_\mu$ 
satisfies the following conditions for each $\mu\in[0, \Lambda_0)$:
\begin{itemize}
\item[(i)]  $I_\mu(u)>\alpha$ if $\|u\|=\rho$;
\item[(ii)] There exists $e\in H_0^1(\Omega)$ such that $I_\mu(e)<0$.
\end{itemize}
\end{lemma}

\begin{proof} 
 (i) For $u\in H_0^1(\Omega)$, by Sobolev and Young inequalities, it holds that
$$
\mu\int_{\Omega}u^-\,dx\leq\frac{5}{6}|\Omega|\mu^{6/5}+\frac{1}{6S^3}\|u\|^6.
$$
Then
\begin{align*}
I_\mu(u)&\geq \frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4
 -\frac{\lambda}{2}\int_{\Omega}\frac{(u^+)^2}{|x|^{2-s}}\,dx
 -\frac{1}{6}\int_{\Omega}(u^+)^6\,dx-\mu\int_{\Omega}u^-\,dx\\
&\geq \frac{a\lambda_1-\lambda}{2\lambda_1}\|u\|^2
 -\frac{1}{3S^3}\|u\|^6-\frac{5}{6}|\Omega|\mu^{6/5}.
\end{align*} 
Set $\rho=[\frac{(a\lambda_1-\lambda)S^3}{2\lambda_1}]^{1/4}$,
$\Lambda_0=[\frac{2}{5}(\frac{S(a\lambda_1-\lambda)}{2\lambda_1})^{3/2}
 |\Omega|^{-1}]^{5/6}$, we have
$$
I_\mu|_{\|u\|=\rho}\geq\frac{1}{3}
\Big[\frac{(a\lambda_1-\lambda)S}{2\lambda_1}\Big]^{3/2} =: \alpha
$$
provided $\mu\in[0, \Lambda_0)$.

(ii) For $u\in H_0^1(\Omega)\backslash\{0\}$, $t>0$, it holds that
\[
I_\mu(tu)\leq \frac{at^2}{2}\|u\|^2+\frac{bt^4}{4}\|u\|^4
 -\frac{t^6}{6}\int_{\Omega}(u^+)^6\,dx+\mu t\int_{\Omega}u\,dx
\to-\infty
\]
as $t\to\infty$. So we can easily find $e\in H_0^1(\Omega)$ with 
$\|e\|>\rho$, such that $I_\mu(e)<0$. The proof is complete.
\end{proof} 

To use variational methods, we firstly derive some results related to 
the Palais-Smale compactness condition.
We say that $I_\mu$ satisfies the $(PS)$ condition at the level 
$c\in\mathbb{R}$ ($(PS)_c$ condition for short) if any sequence 
$\{u_n\}\subset H_0^1(\Omega)$ along with
$$
I_\mu(u_n)\to c,\quad I_\mu'(u_n)\to0 \quad \text{in } (H_0^1(\Omega))^*
$$
as $n\to\infty$ possesses a convergent subsequence. 
If $I_\mu$ satisfies $(PS)_c$ condition for each $c\in\mathbb{R}$, 
then we say that $I_\mu$ satisfies the $(PS)$ condition.
Define
$$
\Lambda=\frac{abS^3}{4}+\frac{b^3S^6}{24}+\frac{(b^2S^4+4aS)^{3/2}}{24}.
$$

\begin{lemma} \label{lem2.2}
 Assume $0<\lambda<a\lambda_1$, then $I_\mu$ satisfies the $(PS)_c$ condition 
for $c<\Lambda-D\mu^{6/5}$, where 
$D=\frac{5}{6}(9|\Omega|^{5/6}6^{-\frac{1}{6}})^{6/5}$.
\end{lemma}

\begin{proof} 
Let $\{u_n\}\subset H_0^{1}(\Omega)$ be a sequence satisfying
\begin{equation}\label{2}
I_\mu(u_n)\to c,\quad I'_\mu(u_n)\to0,\quad\text{as } n\to\infty.
\end{equation}
on the contrary assume $\{u_n\}$ is unbounded,  then
\begin{align*}
1+c+o(1)\|u_n\|&\geq I_\mu(u_n)-\frac{1}{6}\langle I'_\mu(u_n),u_n\rangle\\
&\geq \frac{a}{3}\|u_n\|^2-\frac{\lambda}{3}\int_{\Omega}
 \frac{(u^+)^2}{|x|^{2-s}}\,dx-\frac{5\mu}{6}\int_{\Omega}u_n^-\,dx\\
&\geq \frac{a\lambda_1-\lambda}{3\lambda_1}\|u_n\|^2-C\|u_n\| ,
\end{align*} 
which implies that the last inequality is an absurd. 
So $\{u_n\}$ is bounded in $H_0^1(\Omega)$. Based on the concentration 
compactness principle (see \cite{PL}), there exist a subsequence, 
still denoted by  $\{u_n\}$, such that
\begin{gather*}
|\nabla u_n|^2\rightharpoonup d\eta \geq|\nabla u|^2+\sum_{j\in J}\eta_j\delta_j,\\
|u_n|^{6}\rightharpoonup d\gamma=|u|^{6}+\sum_{j\in J}\gamma_j\delta_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 
$\eta, \gamma$. Moreover, it holds
\begin{equation}\label{3}
\eta_j\geq S\gamma_{j}^{1/3}\quad text{for all } j\in J.
\end{equation}
For $\varepsilon>0$, let $\phi_{\varepsilon,j}(x)$ be a smooth cut-off
function centered at $x_j$ such that $0\leq\phi_{\varepsilon,j}(x)\leq1$, and
\[
\phi_{\varepsilon,j}(x)=
\begin{cases} 1 & \text{in }B(x_j, \\
=0 & \text{in }\Omega\backslash B(x_j,2\varepsilon),
\end{cases}\quad 
|\nabla\phi_{\varepsilon,j}(x)|\leq\frac{2}{\varepsilon}.
\]
By H\"older's inequality and \eqref{1}, 
\begin{align*}
\big|\int_{\Omega}\frac{u_n^+}{|x|^{2-s}}
 \phi_{\varepsilon,j}u_n\,dx\big|
&\leq \Big(\int_{B(x_j,2\varepsilon)}|u_n|^{6}\,dx\Big)^{1/3}
 \Big(\int_{B(x_j,2\varepsilon)}\frac{\,dx}{|x|^{\frac{3(2-s)}{2}}}\Big)^{2/3}\\
&\leq  C\|u_n\|^2\varepsilon^{s}.
\end{align*} 
Note that $\{u_n\}$ is bounded in $H_0^1(\Omega)$, then
\begin{equation*}
\lim_{\varepsilon\to0}\lim_{n\to\infty}\int_{\Omega}\frac{u_n^+}{|x|^{2-s}}
\phi_{\varepsilon,j}u_n\,dx=0.
\end{equation*}
Similarly, we have
\begin{equation*}
\lim_{\varepsilon\to0}\lim_{n\to\infty}\int_{\Omega}\phi_{\varepsilon,j}u_n\,dx=0.
\end{equation*}
As $\phi_{\varepsilon,j}u_n$ is bounded in $H_0^1(\Omega)$, taking the test 
function $\varphi=\phi_{\varepsilon,j}u_n$ in \eqref{2}, it holds
\begin{align*}
0&= \lim_{\varepsilon\to0}\lim_{n\to\infty}
 \langle I_\mu'(u_n),\phi_{\varepsilon,j}u_n\rangle\\
&= \lim_{\varepsilon\to0}\lim_{n\to\infty}
 \Big\{(a+b\|u_n\|^2)\int_{\Omega}(\nabla u_n,\nabla(\phi_{\varepsilon,j}u_n))\,dx\\
&\quad -\lambda\int_{\Omega}\frac{(u^+_n)^2}{|x|^{2-s}}\phi_{\varepsilon,j}\,dx
 -\int_{\Omega}(u_n^+)^{6}\phi_{\varepsilon,j}\,dx
 +\mu\int_{\Omega}u_n\phi_{\varepsilon,j} \,dx\Big\}\\
&= \lim_{\varepsilon\to0}\lim_{n\to\infty}\Big\{(a+b\|u_n\|^2)
 \int_{\Omega}\left(|\nabla u_n|^2\phi_{\varepsilon,j}
 +u_n\nabla u_n\nabla\phi_{\varepsilon,j}\right)\,dx 
  -\int_{\Omega}|u_n|^{6}\phi_{\varepsilon,j}\,dx\Big\}\\
&\geq (a+b\eta_j)\eta_j-\gamma_j,
\end{align*}
so that
$\gamma_j\geq(a+b\eta_j)\eta_j$.
Applying \eqref{3}, we deduce that
\begin{equation}\label{4}
\gamma_j\geq aS\gamma_j^{1/3}+bS^2\gamma_j^{2/3},\quad\text{or}\quad \gamma_j=0.
\end{equation}
Set $X=\nu_j^{1/3}$, it follows from \eqref{4} that
$X^2\geq aS+bS^2X$;
that is,
$$
X\geq\frac{bS^2+\sqrt{b^2S^4+4aS}}{2},
$$
using \eqref{3} again, consequently
$$
\eta_j\geq S X\geq\frac{bS^3+\sqrt{b^2S^6+4aS^3}}{2} =: K.
$$

Next we show that 
\[
\eta_j\geq\frac{bS^3+\sqrt{b^2S^6+4aS^3}}{2}
\]
is impossible. To obtain a contradiction assume that there exists 
 $j_0\in J$ such that 
$\eta_{j_0}\geq\frac{bS^3+\sqrt{b^2S^6+4aS^3}}{2}$. 
By \eqref{2} and Young inequality, 
\begin{align*}
c&= \lim_{n\to\infty}\big\{I_\mu(u_n)-\frac{1}{4}\langle I'_\mu(u_n),u_n\rangle\big\}\\
&= \lim_{n\to\infty}\Big\{(\frac{1}{2}-\frac{1}{4})
 a\|u_n\|^2+b(\frac{1}{4}-\frac{1}{4})\| u_n\|^4
 +(\frac{1}{4}-\frac{1}{6})\int_{\Omega}|u_n|^{6}\,dx \\
&\quad +\frac{3\mu}{4}\int_{\Omega}u_n\,dx
 -\frac{\lambda}{4}\int_{\Omega}|u_n|^{2}|x|^{s-2}\,dx\Big\}\\
&\geq \frac{a}{4}\Big(\|u\|^2+\sum_{j\in J}\eta_j\Big)
 +\frac{1}{12}\Big(\int_{\Omega}|u|^{6}\,dx+\sum_{j\in J}\gamma_j\Big)\\
&\quad -\frac{\lambda}{4}\int_{\Omega}|u|^{2}|x|^{s-2}\,dx-\frac{3\mu}{4}\int_{\Omega}u^-\,dx\\
&\geq \frac{a}{4}\eta_{j_0}+\frac{1}{12}\gamma_{j_0}
 +\frac{a}{4}\|u\|^2-\frac{\lambda}{4}\int_{\Omega}|u|^{2}|x|^{s-2}\,dx\\
&\quad +\frac{1}{12}\int_{\Omega}|u|^6\,dx-\frac{3\mu}{4}\int_{\Omega}|u|\,dx\\
&\geq \frac{aK}{2}+\frac{b}{4}K^2-\frac{K^3}{6S^3}
 -\frac{1}{4}\Big(aK+bK^2-\frac{K^3}{S^3}\Big)-D\mu^{6/5},
\end{align*} 
where $D=\frac{5}{6}(9|\Omega|^{5/6}6^{-\frac{1}{6}})^{6/5}$. 
Easy computations show that
\begin{gather*}
\frac{aK}{2}+\frac{b}{4}K^2-\frac{K^3}{6S^3}=\Lambda,\\
 aK+bK^2-K^3S^{-3}=0.
\end{gather*}
Applying the result, we get $\Lambda-D\mu^{6/5}\leq c<\Lambda-D\mu^{6/5}$.
 This is a contradiction. It indicates that $J$ is empty, which implies that
 $$
\int_{\Omega}(u_n^+)^6\,dx\to\int_{\Omega}(u^+)^6\,dx.
$$
Now, set $\lim_{n\to\infty}\|u_n\|=l$, by \eqref{2}, we have
\begin{equation}\label{5}
(a+b\|u_n\|^2)\|u_n\|^2-\lambda\int_{\Omega}(u_n^+)^2|x|^{s-2}\,dx
-\int_{\Omega}(u_n^+)^6\,dx+\mu\int_{\Omega}u_n\,dx=o(1),
\end{equation}
and
\begin{equation}\label{6}
\begin{aligned}
&(a+b\|u_n\|^2)\int_{\Omega}(\nabla u_n,\nabla\varphi)\,dx\\
&= \lambda\int_{\Omega}u_n^+\varphi|x|^{s-2}\,dx
 +\int_{\Omega}(u_n^+)^5\varphi \,dx-\mu\int_{\Omega}\varphi \,dx+o(1)
\end{aligned}
\end{equation}
for any $\varphi\in H_0^1(\Omega)$. Let $n\to\infty$, then from \eqref{5}, one gets
$$
(a+bl^2)l^2-\lambda\int_{\Omega}(u^+)^2|x|^{s-2}\,dx
-\int_{\Omega}(u^+)^6\,dx+\mu\int_{\Omega}u\,dx=0.
$$
Similarly, from \eqref{6}, 
\begin{equation}\label{7}
\begin{aligned}
&(a+bl^2)\int_{\Omega}(\nabla u,\nabla\varphi)\,dx \\
&= \lambda\int_{\Omega}u^+\varphi|x|^{s-2}\,dx
 +\int_{\Omega}(u^+)^5\varphi \,dx-\mu\int_{\Omega}\varphi \,dx.
\end{aligned}
\end{equation}
Taking the test function $\varphi=u$ in \eqref{7}, we have
\begin{equation*}
(a+bl^2)\|u\|^2-\lambda\int_{\Omega}(u^+)^2|x|^{s-2}\,dx
-\int_{\Omega}(u^+)^6\,dx+\mu\int_{\Omega}u\,dx=0.
\end{equation*}
So we obtain $l=\|u\|$, consequently $u_n\to u$ in $H_0^1(\Omega)$. 
The proof is complete.
\end{proof}

From \cite{BN}, it is well 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{gather*}
-\Delta U_\varepsilon=U_\varepsilon^{5}\quad \text{in }\mathbb{R}^3, \\
\int_{\mathbb{R}^3}|U_\varepsilon|^6
=\int_{\mathbb{R}^3}|\nabla U_\varepsilon|^2=S^{3/2}.
\end{gather*}
Let $\eta\in C_0^\infty(\Omega)$ be a cut-off function such that 
$0\leq\eta\leq1$, $|\nabla\eta|\leq C$ and $\eta(x)=1$ for $|x|<R_0$ and 
$\eta(x)=0$ for $|x|>2R_0$, we set $u_\varepsilon(x)=\eta(x)U_\varepsilon(x)$. 
Then it holds
\begin{gather*}
\|u_\varepsilon\|^2=S^{3/2}+O(\varepsilon^{1/2}),\\
|u_\varepsilon|_6^6=S^{3/2}+O(\varepsilon^{3/2}).
\end{gather*}

\begin{lemma} \label{lem2.3}
 Assume $0<s<1$, then
$\sup_{t\geq0}I_\mu(tu_\varepsilon)<\Lambda-D\mu^{6/5}$
for some $\varepsilon=\varepsilon(\mu)>0$ small enough.
\end{lemma}

\begin{proof} 
Since $\lim_{t\to\infty}I_\mu(tu_\varepsilon)=-\infty$,  which suggests that 
$\sup_{t\geq0}I_\mu(tu_\varepsilon)$
attained at $t_\varepsilon>0$, i.e.,
$$
at_\varepsilon\|u_\varepsilon\|^2+bt_\varepsilon^3\|u_\varepsilon\|^4
-\lambda t_\varepsilon\int_{\Omega}\frac{u_\varepsilon^2}{|x|^{2-s}}\,dx
-t_\varepsilon^5\int_{\Omega} u_\varepsilon^6\,dx
+\mu\int_{\Omega}u_\varepsilon \,dx=0,$$
so that
\begin{equation}\label{8}
t_\varepsilon^4\int_{\Omega} u_\varepsilon^6\,dx
\geq a\|u_\varepsilon\|^2+bt_\varepsilon^2\|u_\varepsilon\|^4
-\lambda\int_{\Omega}\frac{u_\varepsilon^2}{|x|^{2-s}}\,dx.
\end{equation}
It follows from \eqref{8} that $t_\varepsilon$ is bounded below, i.e., 
there exists a positive constant $t_0>0$ (independently of $\varepsilon$) 
such that $0<t_0\leq t_\varepsilon$. Besides, it holds
$$
t_\varepsilon^{2}\int_{\Omega} u_\varepsilon^6\,dx
=\frac{a\|u_\varepsilon\|^2-\lambda\int_{\Omega}u_\varepsilon^2|x|^{s-2}\,dx}
{t_\varepsilon^{2}}+b\|u_\varepsilon\|^4+\frac{\mu}{t_\varepsilon^3}
\int_{\Omega}u_\varepsilon \,dx,
$$
which implies that $t_\varepsilon$ is bounded above for all $\varepsilon>0$; 
that is, there exists a positive real number $t_1>0$ 
(independently of $\varepsilon$), such that $t_\varepsilon\leq t_1<+\infty$.
Set
$$
J(t)=\frac{at^2}{2}\|u_\varepsilon\|^2
 +\frac{bt^4}{4}\|u_\varepsilon\|^4-\frac{t^6}{6}\int_{\Omega} u_\varepsilon^6\,dx.
$$
As in \cite{LT} we have 
$$
\sup_{t\geq0}J(t)\leq\Lambda+O(\varepsilon^{1/2}).
$$
According to the definition $u_\varepsilon$, for $0<\alpha<1$, it holds
\begin{align*}
\int_{\Omega}u_\varepsilon \,dx
&\leq C\varepsilon^{1/4}\int_{|x|\leq R_0} \frac{1}{(\varepsilon+|x|^2)^{1/2}}\,dx\\
&=  C\varepsilon^{1/4}\int_{0}^{R_0}\frac{r^{2}}{(\varepsilon+r^2)^{1/2}}\,dr\\
&\leq C\varepsilon^{1/4}\int_{0}^{R_0}r\,dr
=  C\varepsilon^{1/4}.
\end{align*}
From \cite[Proposition 2.4]{CJR}, for some $K>0$, we have
\begin{equation*}
\int_{\Omega}u_\varepsilon^2|x|^{s-2}\,dx
=K\varepsilon^{\frac{s}{2}}+O(\varepsilon^{1/2}).
\end{equation*}
Consequently,
\begin{align*}
\sup_{t\geq0}I_\mu(tu_\varepsilon)
&\leq \sup_{t\geq0}J(t)-\frac{t_0^2\lambda}{2}
 \int_{\Omega}\frac{u_\varepsilon^2}{|x|^{2-s}}\,dx
 +t_1\mu\int_{\Omega}u_\varepsilon \,dx\\
&\leq \Lambda+C_1\varepsilon^{1/2}-C_2\varepsilon^{\frac{s}{2}}
 +C_3\mu\varepsilon^{1/4},
\end{align*} 
here $C_i$ ($i=1,2,3$) (independently of $\varepsilon, \mu$) are there 
positive constants. Since $0<s<1$, let 
$\varepsilon=\mu^{\frac{12}{5}}$, 
$\mu<\Lambda_1 =\big[\frac{C_2}{C_1+C_3+D}\big]^{\frac{5}{6(1-s)}}$, then
\begin{align*}
 C_1\varepsilon^{1/2}-C_2\varepsilon^{\frac{s}{2}}
+C_3\mu\varepsilon^{1/4}
&=  C_1\mu^{6/5}-C_2\mu^{6s/5} +C_3\mu^{8/5}\\
&\leq (C_1+C_3)\mu^{6/5}-C_2\mu^{6s/5}\\
&<-D\mu^{6/5},
\end{align*} 
so that
\begin{equation*}
\sup_{t\geq0}I_\mu(tu_\varepsilon)\leq\Lambda-D\mu^{6/5},
\end{equation*}
provided $\mu<\Lambda_1$ sufficiently small. The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Let $\mu_*=\min\{\Lambda_0, \Lambda_1\}$, then Lemmas \ref{lem2.1}--\ref{lem2.3},
for all $0\leq\mu<\mu_*$. Assume $\mu\neq0$. Then applying the mountain-pass 
lemma \cite{AR}, there exists a sequence $\{v_n\}\subset H_0^1(\Omega)$ such that
\begin{equation}\label{2.9}
I_\mu(v_n)\to c_\mu>0,\quad \text{and}\quad I_\mu'(v_n)\to0,
\end{equation}
where
\begin{gather*}
c_\mu=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I_\mu(\gamma(t)), \\
\Gamma=\big\{\gamma\in C([0,1], H_{0}^{1}(\Omega)): \gamma(0)=0, 
 \gamma(1)=e\big\}.
\end{gather*}
By Lemmas \ref{lem2.2} and  \ref{lem2.3}, $\{v_n\}\subset H_0^1(\Omega)$ has a convergent 
subsequence, say $\{v_n\}$, we may assume that $v_n\to v_\mu$ in $H_0^1(\Omega)$ 
as $n\to\infty$. Hence, from
\eqref{2.9}, it holds
\begin{equation*}
I_\mu(v_\mu)=\lim_{n\to\infty}I_\mu(v_n)=c_\mu>0,
\end{equation*}
which implies that $v_\mu\not\equiv0$. Furthermore, from the continuity 
of $I_\mu'$, we obtain that
$v_\mu$ is a nontrivial solution of  \eqref{1.1}.

If $\mu=0$, applying the mountain-pass lemma,  there is a sequence 
$\{u_n\}\subset H_0^1(\Omega)$ such that
\begin{equation*}
I_0(u_n)\to c_0\in(0, \Lambda),\quad\text{and}\quad I_0'(u_n)\to 0.
\end{equation*}
Arguing as in the previous proof,  $\{u_n\}$ has a subsequence strongly
convergent in $H_0^1(\Omega)$ to a critical point $v_0$ of $I_0$.
 Moreover, for every $\phi\in H_0^1(\Omega)$, we have
\begin{equation}\label{2.10}
(a+b\|v_0\|^2)\int_{\Omega}(\nabla v_0,\nabla\phi)
 -\lambda\int_{\Omega}v_0^+\phi|x|^{s-2}\,dx
-\int_{\Omega}(v_0^+)^5\phi \,dx=0.
\end{equation}
Taking the test $\phi=v_0^-$ in \eqref{2.10}, it follows that
\begin{equation*}
\|v_0^-\|=0,
\end{equation*}
which implies that $v_0\geq0$ in $\Omega$ and $-(a+b\|v_0\|^2)\Delta v_0\geq 0$.
 Note that $I_0(v_0)=\lim_{n\to\infty}I_0(v_n)=c_0>0$, which means that 
$v_0\not\equiv0$ in $\Omega$. Therefore, by the strong maximum principle, 
we have $v_0>0$ in $\Omega$. The proof is complete.
\end{proof}

\subsection*{\bf Acknowledgments}
This research was supported by the Science and Technology Foundation of
Guizhou Province (No. LH[2015] 7207; No. KY[2016] 163; No. KY[2016] 029),
by the Graduate Innovation Fund Grants of Shanghai University of Finance
and Economics (No. CXJJ-2017-425).

The authors would like to thank the anonymous referees for their very
helpful suggestions and comments which lead to the improvement of this article.



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\end{document}