\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 85, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}
\begin{document}
\title[\hfilneg EJDE-2017/85\hfil Positive solutions]
{Positive solutions for a nonlocal problem with singularity}
\author[C.-Y. Lei, C.-M. Chu, H.-M. Suo \hfil EJDE-2017/85\hfilneg]
{Chun-Yu Lei, Chang-Mu Chu, Hong-Min Suo}
\address{Chun-Yu Lei (corresponding author) \newline
School of Sciences,
GuiZhou Minzu University,
Guiyang 550025, China}
\email{leichygzu@sina.cn}
\address{Chang-Mu Chu \newline
School of Sciences,
GuiZhou Minzu University,
Guiyang 550025, China}
\email{372382190@qq.com}
\address{Hong-Min Suo \newline
School of Sciences,
GuiZhou Minzu University,
Guiyang 550025, China}
\email{11394861@qq.com}
\dedicatory{Communicated by Paul H. Rabinowitz}
\thanks{Submitted February 23, 2017. Published March 27, 2017.}
\subjclass[2010]{35G20, 35J60, 35J75}
\keywords{Nonlocal problem; singularity; positive solution}
\begin{abstract}
In this article we study a nonlocal problem involving singular nonlinearity.
Based on the variational and perturbation methods, we obtain the existence
of two positive solutions for this problem.
\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 recent years, the problem
\begin{gather*}
-\Big(a+b\int_\Omega|\nabla u|^2dx\Big)\Delta u=h(x,u), \quad \text{in } \Omega, \\
u=0, \quad \text{on } \partial\Omega,
\end{gather*}
has received considerable attention, we refer to \cite{AM}--\cite{LG}.
In particular, if $h(x,u)=\lambda u^3+\mu u^{-\gamma}$
($0<\gamma<1$), in \cite{LLT}, the existence and multiplicity of solutions
for problem have been considered for this case by using the variational
method and the Nehari manifold. When $h(x,u)=f(x)u^{-\gamma}-\lambda u^p$,
in \cite{LKT}, we have studied the uniqueness of positive solution via
the minima method. In addition, in \cite{LT}, the existence and multiplicity
of positive solutions have been obtained in the cases when
$h(x,u)=\lambda u^{-\gamma}+u^5$.
In particular, Yin and Liu \cite{YL} considered the nonlocal problem
\begin{gather*}
-\Big(a-b\int_\Omega|\nabla u|^2dx\Big)\Delta u=|u|^{p-2}u, \quad \text{in } \Omega, \\
u=0, \quad \text{on } \partial\Omega,
\end{gather*}
where $2
0$, and
$\lambda$ is positive parameter. Now we state our main result.
\begin{theorem} \label{thm1.1}
Assume $a, b>0$, $0<\gamma<1$, there exists $\lambda_*>0$ such that
$0<\lambda<\lambda_*$, then \eqref{1.1} has at least two positive solutions.
\end{theorem}
\section{Proof of main theorem}
Let $H_0^1(\Omega)$ be the usual Sobolev space equipped with the norm
$\|u\|^2=\int_{\Omega}|\nabla u|^2dx$, denote by $B_r$
(respectively, $\partial B_r$) the closed ball (respectively, the sphere)
of center zero and radius $r$, i.e. $B_r=\{u\in H_0^{1}(\Omega): \|u\|\leq r\}$,
$\partial B_r=\{u\in H_0^{1}(\Omega): \|u\|=r\}$ and $C$ be various positive
constant. Let $S$ be the best Sobolev constant, i.e.,
$$
S=\inf\big\{\|u\|^2: u\in H_0^1(\Omega), \int_{\Omega}|u|^6dx=1\big\}.
$$
Consider the energy functional $I_0: H_0^1(\Omega)\to\mathbb{R}$ given by
$$
I_0(u)=\frac{a}{2}\|u\|^2-\frac{b}{4}\|u\|^4
-\frac{\lambda}{1-\gamma}\int_{\Omega}|u|^{1-\gamma}dx.
$$
It is well known that the singular term leads to the non-differentiability
of the functional $I_0$ on $H_0^{1}(\Omega)$, therefore problem \eqref{1.1}
cannot be considered by using critical point theory directly.
Now, we consider the perturbed equation
\begin{equation}\label{1}
\begin{gathered}
-\Big(a-b\int_\Omega|\nabla u|^2dx\Big)\Delta u
=\frac{\lambda}{(|u|+\alpha)^\gamma}, \quad \text{in } \Omega, \\
u=0, \quad \text{on } \partial\Omega,
\end{gathered}
\end{equation}
where $\alpha>0$, the functional associated with \eqref{1} is
$$
I_\alpha=\frac{a}{2}\|u\|^2-\frac{b}{4}\|u\|^4
-\frac{\lambda}{1-\gamma}\int_\Omega[(|u|+\alpha)^{1-\gamma}-\alpha^{1-\gamma}]dx.
$$
\begin{lemma} \label{lem2.1}
Assume $a, b>0$, $0<\gamma<1$, then $I_\alpha$ satisfies the $(PS)_c$ condition
with $c<\frac{a^2}{4b}-D\lambda$, where
$D=\frac{1}{1-\gamma}S^{-\frac{1-\gamma}{2}}|\Omega|^{\frac{5+\gamma}{6}}
\left(\frac{a+1}{b}\right)^{\frac{1-\gamma}{2}}$.
\end{lemma}
\begin{proof}
Let $\{u_n\}\subset H_0^1(\Omega)$ be a nonnegative
($I_\alpha(u_n)=I_\alpha(|u_n|)$) $(PS)_c$ sequence for $I_\alpha$, i. e.,
\begin{equation}\label{2}
I_\alpha(u_n)\to c,\quad I'_\alpha(u_n)\to 0,\quad \text{as } n\to\infty.
\end{equation}
It follows from \eqref{2} that
\begin{equation*}
b\|u_n\|^4= a\|u_n\|^2-\int_\Omega\frac{u_n}{(u_n+\alpha)^\gamma}dx+o(1)
\leq a\|u_n\|^2+o(1),
\end{equation*}
so that
$$
\|u_n\|^2\leq\frac{a+1}{b},
$$
which implies that $\{u_n\}$ is bounded in $H_0^1(\Omega)$.
Therefore, there exist a subsequence (still denoted by $\{u_n\}$) and
$u_*\in H_0^1(\Omega)$ such that
$u_n\rightharpoonup u_*$ weakly in $H_0^{1}(\Omega)$ as $n\to\infty$.
It follows easily from the Vitali Convergence Theorem that
$$
\lim_{n\to\infty}\int_{\Omega}\frac{u_n}{(u_n+\alpha)^\gamma}dx
=\int_{\Omega}\frac{u_*}{(u_*+\alpha)^\gamma}dx.
$$
Set $w_n=u_n-u_*$, then $\|w_n\|\to0$. Otherwise, there exists a subsequence
(still denoted by $w_n$) such that $\lim_{n\to\infty}\|w_n\|=l>0$.
From \eqref{2}, letting $n\to\infty$, it holds
\begin{equation}\label{3}
(a-bl^2-b\|u_*\|^2)\int_{\Omega}(\nabla u_*, \nabla\phi)dx
-\lambda\int_{\Omega}\frac{\phi}{(u_*+\alpha)^\gamma}dx=0,
\quad \forall\phi\in H_0^1(\Omega).
\end{equation}
Taking the test function $\phi=u_*$ in \eqref{3}, it follows
\begin{equation}\label{4}
(a-bl^2-b\|u_*\|^2)\|u_*\|^2-\lambda\int_{\Omega}\frac{u_*}{(u_*+\alpha)^\gamma}dx=0.
\end{equation}
Note that $\langle I'_\alpha(u_n),u_n\rangle\to0$ as $n\to\infty$, it holds
\begin{equation*}
a\|w_n\|^2+a\|u_*\|^2-b\|w_n\|^4-2b\|w_n\|^2\|u_*\|^2-b\|u_*\|^4
-\lambda\int_{\Omega}\frac{u_*}{(u_*+\alpha)^\gamma}dx=o(1).
\end{equation*}
From this and \eqref{4}, it follows
\begin{equation}\label{5}
a\|w_n\|^2-b\|w_n\|^4-b\|w_n\|^2\|u_*\|^2=o(1).
\end{equation}
Consequently,
$$
l^2=\frac{a}{b}-\|u_*\|^2, \quad l>0.
$$
Note that the subadditivity of $t^{1-\gamma}$, namely
\begin{equation}\label{6}
(|v|+\alpha)^{1-\gamma}-\alpha^{1-\gamma}\leq|v|^{1-\gamma}.
\end{equation}
On one hand, recall that $\|u_n\|^2\leq\frac{a}{b}$, then using \eqref{4}
and \eqref{6}, it follows
\begin{align*}
I_{\alpha}(u_*)
&= \frac{a}{2}\|u_*\|^2-\frac{b}{4}\|u_*\|^4
-\frac{\lambda}{1-\gamma}\int_\Omega[(u_*+\alpha)^{1-\gamma}-\alpha^{1-\gamma}]dx\\
&\geq \frac{a}{4}\|u_*\|^2+\frac{b}{4}l^2\|u_*\|^2
-\frac{\lambda}{1-\gamma}S^{-\frac{1-\gamma}{2}}|\Omega|^{\frac{5+\gamma}{6}}
\Big(\frac{a+1}{b}\Big)^{\frac{1-\gamma}{2}}\\
&= \frac{a}{4}\|u_*\|^2+\frac{b}{4}l^2\|u_*\|^2-D\lambda,
\end{align*}
where
\[
D=\frac{1}{1-\gamma}S^{-\frac{1-\gamma}{2}}|\Omega|^{\frac{5+\gamma}{6}}
\Big(\frac{a+1}{b}\Big)^{\frac{1-\gamma}{2}}.
\]
On the other hand, from \eqref{2} and \eqref{5}, it holds
\begin{align*}
I_\alpha(u_*)&= I_\alpha(u_n)-\frac{a}{2}\|w_n\|^2+\frac{b}{4}\|w_n\|^4
+\frac{b}{2}\|w_n\|^2\|u_*\|^2+o(1)\\
&< \frac{a^2}{4b}-D\lambda-\frac{a}{4}\left(\frac{a}{b}-\|u_*\|^2\right)
+\frac{b}{4}l^2\|u_*\|^2\\
&= \frac{a}{4}\|u_*\|^2+\frac{b}{4}l^2\|u_*\|^2-D\lambda.
\end{align*}
This is a contradiction. Therefore, $l=0$, it implies that $u_n\to u_*$
in $H_0^1(\Omega)$. The proof is complete.
\end{proof}
\begin{lemma} \label{lem2.2}
Assume $a, b>0$, there exist $\Lambda_{0}>0$ and $\rho>0$ such that for any
$\lambda\in(0, \Lambda_0)$, it holds
\begin{equation*}
I_\alpha\big|_{u\in \overline{\partial B_\rho}}>0, \quad
\inf_{u\in \overline{B_\rho}}I_\alpha(u)<0.
\end{equation*}
\end{lemma}
\begin{proof}
By H\"older's inequality and \eqref{6}, one has
\begin{align*}
I_\alpha(u)
&= \frac{a}{2}\|u\|^2-\frac{b}{4}\|u\|^4
-\frac{\lambda}{1-\gamma}\int_{\Omega}[(|u|+\alpha)^{1-\gamma}-\alpha^{1-\gamma}]dx\\
&\geq \|u\|^{1-\gamma}\Big(\frac{a}{2}\|u\|^{1+\gamma}-\frac{b}{4}\|u\|^{3+\gamma}
-\frac{\lambda}{1-\gamma}|\Omega|^{\frac{5+\gamma}{6}}S^{-\frac{1-\gamma}{2}}\Big),
\end{align*}
set $h(t)=\frac{a}{2}t^{1+\gamma}-\frac{b}{4}t^{3+\gamma}$, we see that there
exists a constant $\rho=\sqrt{\frac{2a(1+\gamma)}{b(3+\gamma)}}$
such that $\max_{t>0}h(t)=h(\rho)>0$.
Let
\[
\Lambda_0=\frac{(1-\gamma)S^{\frac{1-\gamma}{2}}}{2|\Omega|^{\frac{5+\gamma}{6}}}
h(\rho).
\]
Consequently, $I_\alpha|_{\|u\|=\rho}\geq\frac{h(\rho)}{2}\rho^{1-\gamma}$ for any
$\lambda\in(0, \Lambda_0)$. Moreover, for $u\in H_0^1(\Omega)\backslash\{0\}$
it holds
\begin{align*}
\lim_{t\to0^+}\frac{I_{\alpha}(tu)}{t}
&= -\frac{\lambda}{1-\gamma}\lim_{t\to0^+}\frac{1}{t}
\int_\Omega[(t|u|+\alpha)^{1-\gamma}-\alpha^{1-\gamma}]dx \\
&= -\frac{\lambda}{1-\gamma}\lim_{t\to0^+}\int_\Omega
\frac{(1-\gamma)\xi^{-\gamma}t|u|}{t}dx \quad (\alpha<\xi0$, $0<\lambda<\Lambda_{0}$. Then problem \eqref{1} has a positive
solution $u_\alpha\in H_0^1(\Omega)$, enjoying $I_\alpha(u_\alpha)<0$.
\end{lemma}
\begin{proof}
By Lemmas \ref{lem2.1} and \ref{lem2.2}, similarly to the paper \cite{LG},
we can prove that problem \eqref{1} has a nonzero nonnegative solution
$u_\alpha\in\overline{B_\rho}\subset H_0^1(\Omega)$ such that
$I_\alpha(u_\alpha)=m<0$. Note that $u_\alpha\in\overline{B_\rho}$, it holds
$$
\|u_\alpha\|^2\leq\frac{2a(1+\gamma)}{b(3+\gamma)}<\frac{a}{b},
$$
which implies that $a-b\|u_\alpha\|^2>0$. Therefore, by using the strong
maximum principle, we obtain $u_\alpha>0$ in $\Omega$.
The proof is complete.
\end{proof}
\begin{remark} \label{rmk2.4} \rm
Assume $(U_{1/n})$ is a positive solution of \eqref{1}, then
for every $K\Subset \Omega$, there are $n_0\in\mathbb{N}$ and
$\delta>0$ such that
$$
U_{1/n}(x)\geq \delta,\quad \forall x \in K\text{ and }n\geq n_0.
$$
Indeed, consider $\Psi_n\in H_0^1(\Omega)$ a weak solution of the problem
\begin{gather*}
-\Delta\Psi_n=\frac{\lambda}{a(|\Psi_n|+1)^\gamma}, \quad \text{in } \Omega, \\
\Psi_n=0, \quad \text{on } \partial\Omega.
\end{gather*}
It is easy to prove that $(\Psi_n)$ is a bounded sequence in $H_0^1(\Omega)$,
thus there is $\Psi\in H_0^1(\Omega)$ such that for some subsequence, still
denoted with the same symbol,
\begin{gather*}
\Psi_n\rightharpoonup \Psi \quad \text{in }H_0^1(\Omega), \\
\Psi_n(x)\to \Psi(x)\quad \text{a.e. in } \Omega.
\end{gather*}
Setting
$$
h_n(x)=\frac{\lambda}{a(|\Psi_n(x)|+1)^\gamma},
$$
we see that $(h_n)$ is bounded in $L^{\infty}(\Omega)$, and so, it
is bounded in $L^2(\Omega)$. Then, for some subsequence, we also have
\begin{gather*}
h_n(x)\to h(x)=\frac{\lambda}{a(|\Psi(x)|+1)^\gamma}\quad
\text{a.e. in }\Omega, \\
h_n\rightharpoonup h \quad \text{in } L^2(\Omega).
\end{gather*}
The above information yield
\begin{gather*}
-\Delta\Psi=\frac{\lambda}{a(|\Psi|+1)^\gamma}, \quad \text{in } \Omega, \\
\Psi=0, \quad \text{on } \partial\Omega,
\end{gather*}
from where it follows that $\Psi\in C(\overline{\Omega})$ and
$\Psi(x)>0$ for all $x\in\Omega$. Moreover, the elliptic regularity gives
$$
\Psi_n\to\Psi\quad \text{in } C(\overline{\Omega}).
$$
Thereby, fixed a compact set $K\subset\Omega$, there are $n_0\in\mathbb{N}$
and $\delta>0$ such that
$$
\Psi_n(x)\geq\delta,\quad \forall x\in K\text{ and } n\geq n_0.
$$
On the other hand, let $U_{1/n}$ be a positive solution of \eqref{1}, we know that
\begin{gather*}
-\Delta U_{1/n}\geq=-\Delta\Psi_n, \quad \text{in } \Omega, \\
U_{1/n}=\Psi_n=0, \quad \text{on } \partial\Omega,
\end{gather*}
and so, by maximum principle,
$$
U_{1/n}(x)\geq\Psi_n(x),\quad \forall x\in\Omega \text{ and all } n\in\mathbb{N}.
$$
As a byproduct of above arguments, for each compact set $K\subset\Omega$,
there are $n_0\in\mathbb{N}$ and $\delta>0$ such that
$$
U_{1/n}(x)\geq\delta,\quad \forall x\in K \text{and all } n\geq n_0.
$$
\end{remark}
Now, we show that the functional $I_\alpha$ satisfies the mountain-pass lemma.
\begin{lemma} \label{lem2.5}
The functional $I_\alpha$ satisfies the following conditions for any
$\lambda\in(0, \Lambda_0)$
\begin{itemize}
\item[(i)] $I_\alpha(u)>0$ if $\|u\|=\rho$;
\item[(ii)] There exists $\zeta\in H_0^1(\Omega)$ such that
$I_\alpha(\zeta)<0$.
\end{itemize}
\end{lemma}
\begin{proof}
Conclusion (i) follows from Lemma \ref{lem2.2}.
To prove (ii), let $u\in H_0^1(\Omega)\backslash\{0\}$ and $t>0$, it follows that
\begin{align*}
I_\alpha(tu)
&\leq \frac{at^2}{2}\|u\|^2-\frac{bt^4}{4}\|u\|^4
-\frac{\lambda t^{1-\gamma}}{1-\gamma}\int_{\Omega}[(|u|+\alpha)^{1-\gamma}
-\alpha^{1-\gamma}]dx\\
&\to -\infty
\end{align*}
as $t\to+\infty$. Therefore we can easily find $\zeta\in H_0^1(\Omega)$
with $\|\zeta\|>\rho$, such that $I_\alpha(\zeta)<0$.
The proof is complete.
\end{proof}
Now, it is well known that the function
\begin{equation*}
w_\varepsilon(x)=\frac{(3\varepsilon^2)^{\frac{1}{4}}}
{(\varepsilon^2+|x|^2)^{1/2}},\quad x\in \mathbb{R}^3, \; \varepsilon>0
\end{equation*}
satisfies
\begin{equation*}
-\Delta w_\varepsilon=w_\varepsilon^{5}\quad ]text{in }\mathbb{R}^3.
\end{equation*}
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 $|x|2R$, we set $u_\varepsilon(x)=\eta(x)w_\varepsilon(x)$. Then
\begin{equation*}
\|u_\varepsilon\|^2=S^{\frac{3}{2}}+O(\varepsilon),\quad
|u_\varepsilon|_6^6=S^{\frac{3}{2}}+O(\varepsilon^3).
\end{equation*}
\begin{lemma} \label{lem2.6}
Assume $a, b>0$ and $0<\gamma<1$. Then
\[
\sup_{t\geq0}I_\alpha(u_\alpha+tu_\varepsilon)<\frac{a^2}{4b}-D\lambda.
\]
\end{lemma}
\begin{proof}
As $u_\alpha$ is a positive solution of \eqref{1}, for each
$\varphi\in H_0^1(\Omega)$, it holds
$$
(a-b\|u_\alpha\|^2)\int_{\Omega}(\nabla u_\alpha,\nabla\varphi)dx
=\lambda\int_{\Omega}\frac{\varphi}{(u_\alpha+\alpha)^\gamma}dx.
$$
In particular, it holds
$$
(a-b\|u_\alpha\|^2)\int_{\Omega}(\nabla u_\alpha,\nabla u_\varepsilon)dx
=\lambda\int_{\Omega}\frac{u_\varepsilon}{(u_\alpha+\alpha)^\gamma}dx.
$$
Recalling that $a-b\|u_\alpha\|^2>0$, we have
$$
\int_{\Omega}(\nabla u_\alpha,\nabla u_\varepsilon)dx\geq0.
$$
As $I_\alpha(u_\alpha)<0$, by Remark \ref{rmk2.4}, we have
\begin{align*}
I_\alpha(u_\alpha+tu_\varepsilon)
&= \frac{a}{2}\|u_\alpha\|^2+at\int_{\Omega}(\nabla u_\alpha,
\nabla u_\varepsilon)dx+\frac{at^2}{2}\|u_\varepsilon\|^2-\frac{b}{4}\|u_\alpha\|^4\\
&\quad -\frac{bt^4}{4}\|u_\varepsilon\|^4-bt\|u_\alpha\|^2
\int_{\Omega}(\nabla u_\alpha,\nabla u_\varepsilon)dx
-\frac{bt^2}{2}\|u_\alpha\|^2\|u_\varepsilon\|^2\\
&\quad - bt^2\Big(\int_{\Omega}(\nabla u_\alpha,\nabla u_\varepsilon)dx\big)^2
-bt^3\|u_\varepsilon\|^2\int_{\Omega}(\nabla u_\alpha,\nabla u_\varepsilon)dx\\
&\quad -\frac{\lambda}{1-\gamma}\int_{\Omega}[(u_\alpha
+tu_\varepsilon+\alpha)^{1-\gamma}-\alpha^{1-\gamma}]dx\\
&\leq I_\alpha(u_\alpha)+\frac{at^2}{2}\|u_\varepsilon\|^2
-\frac{bt^4}{4}\|u_\varepsilon\|^4-\frac{bt^2}{2}\|u_\alpha\|^2\|u_\varepsilon\|^2\\
&\quad -\frac{\lambda}{1-\gamma}\int_{\Omega}[(u_\alpha+tu_\varepsilon
+\alpha)^{1-\gamma}-(u_\alpha+\alpha)^{1-\gamma}]dx\\
&\quad +\lambda t\int_{\Omega}\frac{u_\varepsilon}{(u_\alpha+\alpha)^\gamma}dx\\
&\leq \frac{at^2}{2}\|u_\varepsilon\|^2-\frac{bt^4}{4}\|u_\varepsilon\|^4
-\frac{bt^2}{2}\|u_\alpha\|^2\|u_\varepsilon\|^2
+\delta\lambda t\int_{\Omega}u_\varepsilon dx.
\end{align*}
Set
$$
g(t)=\frac{at^2}{2}\|u_\varepsilon\|^2
-\frac{bt^4}{4}\|u_\varepsilon\|^4-\frac{bt^2}{2}\|u_\alpha\|^2
\|u_\varepsilon\|^2+\delta\lambda t\int_{\Omega}u_\varepsilon dx.
$$
It is similar to the paper \cite{LG} that there exist $t_\varepsilon>0$
and positive constants $t_1, t_2$ independent of $\varepsilon, \lambda$,
such that $\sup_{t\geq0}g(t)=g(t_\varepsilon)$ and
$00$ (independent of $\lambda$)
such that $\|u_\alpha\|^2\geq c$. Then, it holds
\begin{align*}
\sup_{t\geq0}I_\alpha(u_\alpha+tu_\varepsilon)
&\leq \sup_{t\geq0}g(t)\\
&\leq \sup_{t\geq0}\big\{\frac{at^2}{2}\|u_\varepsilon\|^2
-\frac{bt^4}{4}\|u_\varepsilon\|^4\big\}-c\|u_\varepsilon\|^2
+\lambda O(\varepsilon^{1/2})\\
&\leq \frac{a^2}{4b}+c_1\varepsilon^{1/2}-c_2S^{\frac{3}{2}},\quad (0<\lambda<1)
\end{align*}
where $c_1, c_2>0$. Let $\varepsilon=\lambda^2$, when
$0<\lambda<\Lambda_1\triangleq\frac{c_2S^{\frac{3}{2}}}{c_1+D}$, it holds
\begin{equation*}
c_1\lambda-c_2S^{\frac{3}{2}}0$ and $\lambda>0$ is sufficiently small, problem \eqref{1}
admits a solution $v_\alpha$ with $I_\alpha(v_\alpha)>0$.
\end{lemma}
\begin{proof}
Set $\lambda^*=\min\{\Lambda_0, \Lambda_1, \frac{a^2}{4bD}, 1\}$.
Then applying the mountain-pass lemma \cite{AR}, there exists a sequence
$\{v_n\}\subset H_0^1(\Omega)$, such that
\begin{equation}\label{8}
I_\alpha(v_n)\to c>\frac{h(\rho)}{2}\rho^{1-\gamma},\quad \text{and}\quad
I_\alpha'(v_n)\to0,
\end{equation}
where
\begin{gather*}
c=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I_\alpha(\gamma(t)), \\
\Gamma=\big\{\gamma\in C([0,1], H_{0}^{1}(\Omega)):
\gamma(0)=u_\alpha, \gamma(1)=\zeta\big\}.
\end{gather*}
By Lemmas \ref{lem2.1} and \ref{lem2.6}, $\{v_n\}\subset H_0^1(\Omega)$ has a convergent subsequence,
say $\{v_n\}$, we may assume that $v_n\to v_\alpha$ in $H_0^1(\Omega)$ as
$n\to\infty$. Hence, from \eqref{8}, it holds
\begin{equation*}
I_\alpha(v_\alpha)=\lim_{n\to\infty}I_\alpha(v_n)
=c>\frac{h(\rho)}{2}\rho^{1-\gamma}>0,
\end{equation*}
this implies that $v_\alpha\not\equiv0$. Furthermore, from the continuity
of $I_\alpha'$, we obtain that $v_\alpha$ is a nonzero nonnegative solution
of \eqref{1}. The proof is complete.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm1.1}]
Let $(U_{1/n})$ be a solution of \eqref{1}, then we can prove that $(U_{1/n})$
is bounded in $H_0^1(\Omega)$, then up to a subsequence, there exists
$u\in H_0^1(\Omega)$ such that
\begin{equation*}
U_{1/n}\rightharpoonup u \text{ weakly in }H_0^{1}(\Omega),\quad
U_{1/n}(x)\to u(x) \text{ a.e. in $\Omega$ as }n\to\infty.
\end{equation*}
By Remark \ref{rmk2.4}, and similar to \cite{LT}, for each
$\phi\in H_0^1(\Omega)$, it holds
\begin{equation}\label{9}
(a-b\lim_{n\to\infty}\|U_{1/n}\|^2)\int_{\Omega}(\nabla u, \nabla\phi)dx
-\lambda\int_{\Omega}\frac{\phi}{u^\gamma}dx=0.
\end{equation}
If $U_{1/n}=u_\alpha$, by Lemma \ref{lem2.1}, Lemma \ref{lem2.3}
and \eqref{9}, we conclude
that $U_{1/n}\to u$ in $H_0^{1}(\Omega)$, and $u$ is a positive solution
of \eqref{1.1} with $I_0(u)=\lim_{n\to\infty}I_{1/n}(U_{1/n})<0$.
If $U_{1/n}=v_a$, combining Lemma \ref{lem2.1}, Lemma \ref{lem2.6} and \eqref{9}, we also deduce
that $U_{1/n}\to u$ in $H_0^{1}(\Omega)$, and $u$ is a positive solution
of \eqref{1.1} with $I_0(u)=\lim_{n\to\infty}I_{1/n}(U_{1/n})>0$.
Therefore problem \eqref{1.1} has at least two different positive solutions.
The proof is complete.
\end{proof}
\subsection*{Acknowledgments}
Thsi research was supported by the National Natural Science Foundation of China
(No. 11661021), and by the Science and Technology Foundation of Guizhou Province
(No. LH[2015]7207, No. KY[2016] 163, No. KY[2016]029).
The authors want to thank the anonymous referee for the careful reading of
the original manuscript, and for valuable comments and
suggestions, which led to an improvement of this article.
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