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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 09, pp. 1--8.\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-2017/09\hfil Multiple positive solutions]
{Multiple positive solutions for nonlocal problems involving
a sign-changing potential}

\author[C.-Y. Lei,  J.-F. Liao, H.-M. Suo \hfil EJDE-2017/09\hfilneg]
{Chun-Yu Lei, Jia-Feng Liao, Hong-Min Suo}

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

\address{Jia-Feng Liao \newline
 School of Mathematics and Information,
China West Normal University, Nanchong 637002, China}
\email{liaojiafeng@163.com}

\address{Hong-Min Suo  \newline
School of Sciences,
Gui Zhou Minzu University,
Guiyang 550025, China}
\email{11394861@qq.com}

\dedicatory{Communicated by Paul Rabinowitz}

\thanks{Submitted October 28, 2016. Published January 10, 2017.}
\subjclass[2010]{35J20, 35J30, 35J66}
\keywords{Nonlocal problem; variational method; sign-changing potential;
\hfill\break\indent positive solution}

\begin{abstract}
 In this article we show the existence and multiplicity of positive
 solutions  for the nonlocal problem with a sign-changing weight function,
 \begin{gather*}
 -(a-b\int_{\Omega}|\nabla u|^2dx)\Delta u= f_\lambda(x)|u|^{q-2}u,
 \quad\text{in }\Omega, \\
 u=0, \quad\text{on }\partial\Omega,
 \end{gather*}
 where $\Omega$ is a smooth bounded domain in $\mathbb{R}^{3}$, $a, b>0$, $1<q<2$.
 Our technical approach is based on the variational method.
\end{abstract}

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


\section{Introduction and statement of main results}

 In this article, we are interested in finding the existence of positive solutions 
to the  nonlocal problem
\begin{equation}\label{1.1}
\begin{gathered}
-M(\int_{\Omega}|\nabla u|^2dx)\Delta u=\lambda f(x,u) \quad\text{in }\Omega, \\
u=0, \quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
where $\Omega$ is a bounded domain in $\mathbb{R}^N$ with $N\geq3$. 
When the continuous function $M: \mathbb{R}^+\to\mathbb{R}^+$ satisfies certain 
conditions, \eqref{1.1} has been investigated by many researchers by 
imposing different types of hypotheses on $f(x,u)$; see for example
 \cite{CB,FV,HT,FJ,FG,LG,ML,XJ,ZN,ND,ZP}.
However, observing the all above studies, we see that the function $M$ 
is assumed to be bounded from below. Recently, Yin and Liu \cite{YL} 
investigated the existence and multiplicity of nontrivial solutions for the 
 a  nonlocal problem
\begin{equation}\label{1.10}
\begin{gathered}
-(a-b\int_{\Omega}|\nabla u|^2dx)\Delta u= |u|^{p-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 $\mathbb{R}^N$ and $N\geq1, a, b>0$ 
are constants and $2<p<p^*$. They obtained a nontrivial non-negative solution 
and a nontrivial non-positive solution by using the mountain-pass lemma. 
Motivated by their work, we consider the equation
\begin{equation}\label{1.3}
\begin{gathered}
-(a-b\int_{\Omega}|\nabla u|^2dx)\Delta u= f_\lambda(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 $\mathbb{R}^3$, $a, b>0, 1<q<2$, 
the weight function $f_\lambda\in L^{\infty}(\Omega)$, defined by 
$f_\lambda=\lambda f_++f_-$, $\lambda>0$, with 
$f_{\pm}=\pm\max\{\pm f, 0\}\not\equiv0$. An interesting question is 
whether multiplicity of positive solutions can
be established for \eqref{1.3}.
 We shall give a positive answer to this question.

Our main existence and multiplicity results for \eqref{1.3} can be stated 
as follows.

\begin{theorem} \label{thm1.1}
Assume that $a, b>0$, $1<q<2$ and $f\in L^\infty(\Omega)$. 
Then there exists $\lambda_{*}>0$, such that for any $\lambda\in(0,\lambda_{*})$,
problem \eqref{1.3} has at least two positive solutions.
\end{theorem}

In this article,  we use the following notation:
 The space $H_0^1(\Omega)$ is equipped with the norm 
$\|u\|^2=\int_{\Omega}|\nabla u|^2dx$, the norm in $L^p(\Omega)$ is denoted by 
$\|\cdot\|_{p}$.
$C, C_1, C_2,\dots$ denote various positive constants, which may vary 
from line to line.
We denote by $B_r$ (respectively, $\partial B_r$) the closed ball 
(respectively, the sphere) of center zero and radius $r$, i.e. 
$B_r=\{u\in H_0^1(\Omega): \|u\|\leq r\}$,
$\partial B_r=\{u\in H_0^1(\Omega): \|u\|=r\}$.
Let $S$ be the best Sobolev embedding constant for the embedding 
$H_0^1(\Omega)\hookrightarrow L^{6}(\Omega)$, namely
\begin{equation*}
S=\inf_{u\in H_0^1(\Omega)\backslash\{0\}}\frac{\int_{\Omega}|\nabla u|^2dx}
{(\int_{\Omega}|u|^{6}dx)^{\frac{2}{3}}}.
\end{equation*}

\section{Proof of main theorem}

\subsection{Existence of a first positive solution of \eqref{1.3}}
We define the functional
\begin{equation*}
I_\lambda(u)=\frac{a}{2}\|u\|^2-\frac{b}{4}\|u\|^4
-\frac{1}{q}\int_{\Omega}f_\lambda(x)|u|^{q}dx.
\end{equation*}
A function $u$ is called a solution of \eqref{1.3} if
$u\in H_0^1(\Omega)$ and for all $v\in H_0^1(\Omega)$ it
holds
\begin{equation*}
(a-b\|u\|^2)\int_{\Omega}(\nabla u,\nabla v)dx
-\int_{\Omega}f_\lambda(x)|u|^{q-2}uv\,dx=0.
\end{equation*}

To prove our main theorem, some preliminary results are needed.
 We first recall the following lemma from \cite{CT}.

\begin{lemma} \label{lem2.1}
 Let $r, s>1$, $\psi\in  L^s(\Omega)$ and $\psi^+=\max\{\psi, 0\}\neq0$. 
Then there exists $\varphi_0\in C_0^{\infty}(\Omega)$ such that 
$\int_{\Omega}\psi(x)|\varphi_0|^rdx>0$.
\end{lemma}

\begin{lemma} \label{lem2.2}
Assume $a, b>0, 1<q<2$ and $f\in L^{\infty}(\Omega)$, then $I_\lambda$ 
satisfies the $(PS)_c$ condition with $c<\frac{a^2}{4b}-D\lambda$, 
where 
\[
D=\big(\frac{1}{q}-\frac{1}{4}\big)|f_+|S^{-q/2}|\Omega|^{(6-q)/6}m^{q}.
\]
\end{lemma}

\begin{proof} 
Let $\{u_n\}\subset H_0^1(\Omega)$ be a $(PS)_c$ sequence for $I_\lambda$, i. e.,
\begin{equation}\label{2.1}
I_\lambda(u_n)\to c,\quad I'_\lambda(u_n)\to 0, \quad\text{as } n\to\infty.
\end{equation}
From \eqref{2.1} it follows that
\begin{align*}
b\|u_n\|^4
&= a\|u_n\|^2-\int_\Omega f_\lambda(x)|u_n|^q dx\\
&\leq a\|u_n\|^2-\int_\Omega f_-(x)|u_n|^q dx\\
&\leq a\|u_n\|^2+|f_-|S^{-q/2}|\Omega|^{(6-q)/6}\|u_n\|^{q},
\end{align*} 
so that
$$
b\|u_n\|^{4-q}\leq a\|u_n\|^2+|f_-|S^{-q/2}|\Omega|^{(6-q)/6},
$$
which implies that $\{u_n\}$ is bounded in $H_0^1(\Omega)$; that is, 
there is $m>0$ (independent of $\lambda$) such that $\|u_n\|\leq m$ for every $n$. 
Moreover, 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*}
as $n\to\infty$. It follows easily from Vitali Convergence Theorem that
$$
\lim_{n\to\infty}\int_{\Omega}f_\lambda(x)|u_n|^q dx
=\int_{\Omega}f_\lambda(x)|u_*|^q dx.
$$
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{2.1}, for every $\phi\in H_0^1(\Omega)$, it holds
\begin{equation*}
(a-b\|u_n\|^2)\int_{\Omega}(\nabla u_n, \nabla\phi)dx
-\int_{\Omega}f_\lambda(x)|u_n|^{q-2}u_n\phi dx=o(1).
\end{equation*}
Letting $n\to\infty$, by using the Br\'ezis-Lieb's lemma (see \cite{BL}), it holds
\begin{equation}\label{2.2}
(a-bl^2-b\|u_*\|^2)\int_{\Omega}(\nabla u_*, \nabla\phi)dx
-\int_{\Omega}f_\lambda(x)|u_*|^{q-2}u_*\phi dx=0.
\end{equation}
Taking the test function $\phi=u_*$ in \eqref{2.2}, it holds
\begin{equation}\label{2.3}
(a-bl^2-b\|u_*\|^2)\|u_*\|^2-\int_{\Omega}f_\lambda(x)|u_*|^qdx=0.
\end{equation}
Note that $\langle I'_\lambda(u_n),u_n\rangle\to0$ as $n\to\infty$, it holds
\begin{equation}\label{2.4}
a\|w_n\|^2+a\|u_*\|^2-b\|w_n\|^4-2b\|w_n\|^2\|u_*\|^2-b\|u_*\|^4
-\int_{\Omega}f_\lambda|u_*|^qdx=o(1).
\end{equation}
It follows from \eqref{2.3} and \eqref{2.4}  that
\begin{equation}\label{2.5}
a\|w_n\|^2-b\|w_n\|^4-b\|w_n\|^2\|u_*\|^2=o(1).
\end{equation}
Consequently
$l^2(a-b\|u_*\|^2-bl^2)=0$, $l>0$,
so that
$$
l^2=\frac{a}{b}-\|u_*\|^2.
$$
On the one hand, recalling that $\|u_n\|\leq m$ and using \eqref{2.3}, it follows
\begin{align*}
I_{\lambda}(u_*)
&= \frac{a}{2}\|u_*\|^2-\frac{b}{4}\|u_*\|^4
 -\frac{1}{q}\int_\Omega f_\lambda(x)|u_*|^{q}dx\\
&= \frac{a}{4}\|u_*\|^2+\frac{b}{4}l^2\|u_*\|^2
 -\big(\frac{1}{q}-\frac{1}{4}\big)\int_\Omega f_\lambda(x)|u_*|^{q}dx\\
&\geq \frac{a}{4}\|u_*\|^2+\frac{b}{4}l^2\|u_*\|^2
 -\lambda\big(\frac{1}{q}-\frac{1}{4}\big)|f_+|S^{-q/2}|\Omega|^{(6-q)/6}\|u_*\|^{q}\\
&\geq \frac{a}{4}\|u_*\|^2+\frac{b}{4}l^2\|u_*\|^2
 -\lambda\big(\frac{1}{q}-\frac{1}{4}\big)|f_+|S^{-q/2}|\Omega|^{(6-q)/6}m^{q}\\
&= \frac{a}{4}\|u_*\|^2+\frac{b}{4}l^2\|u_*\|^2-D\lambda,
\end{align*}
where $D=\big(\frac{1}{q}-\frac{1}{4}\big)|f_+|S^{-q/2}|\Omega|^{(6-q)/6}m^{q}$.

On the other hand, by \eqref{2.1} and \eqref{2.5}, it holds
\begin{align*}
&I_{\lambda}(u_*) \\
&=  I_{\lambda}(u_n)-\frac{a}{2}\|w_n\|^2+\frac{b}{4}\|w_n\|^4
 +\frac{b}{2}\|w_n\|^2\|u_*\|^2+o(1)\\
&\leq I_{\lambda}(u_n)-\frac{a}{2}\|w_n\|^2
 +\frac{1}{4}\left(a\|w_n\|^2-b\|w_n\|^2\|u_*\|^2\right)
 +\frac{b}{2}\|w_n\|^2\|u_*\|^2+o(1)\\
&=  c-\frac{a}{4}\|w_n\|^2+\frac{b}{4}\|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{lem2.3}
There exist $R, \rho, \Lambda_0>0$, such that for each $\lambda\in(0, \Lambda_0)$, 
we have
\begin{equation*}
\inf_{u\in \overline{B_R(0)}}I_{\lambda}(u)<0
\quad\text{and}\quad 
I_{\lambda}|_{u\in \overline{\partial B_R(0)}}>\rho.
\end{equation*}
\end{lemma}

\begin{proof} For $u\in H_0^1(\Omega)$, it holds
\begin{align*}
I_{\lambda}(u)
&= \frac{a}{2}\|u\|^2-\frac{b}{4}\|u\|^4
 -\frac{1}{q}\int_\Omega f_\lambda(x)|u|^qdx\\
&\geq \frac{a}{2}\|u\|^2-\frac{b}{4}\|u\|^4
 -\frac{\lambda}{q}\int_\Omega f_+|u|^{q}dx\\
&\geq \frac{a}{2}\|u\|^2-\frac{b}{4}\|u\|^4
 -\frac{\lambda}{q}|f_+|S^{-q/2}|\Omega|^{(6-q)/6}\|u\|^{q}\\
&= \|u\|^q\big\{\frac{a}{2}\|u\|^{2-q}-\frac{b}{4}\|u\|^{4-q}
 -\frac{\lambda}{q}|f_+|S^{-q/2}|\Omega|^{(6-q)/6}\big\}.
\end{align*}
Set $h(t)=\frac{a}{2}t^{2-q}-\frac{b}{4}t^{4-q}$, we see that there 
exists a constant $R=\left(\frac{2a(2-q)}{b(4-q)}\right)^{1/2}>0$ 
such that $\max_{t>0}h(t)=h(R)>0$. Letting 
$\Lambda_0=\frac{qS^{q/2}h(R)}{2|f_+||\Omega|^{(6-q)/6}}$, 
it follows that $I_{\lambda}|_{\|u\|=R}>0$ for each $\lambda\in(0, \Lambda_0)$.

By Lemma \ref{lem2.1}, there exists $\varphi_0\in C_0^{\infty}(\Omega)\subset H_0^1(\Omega)$ such that
$$\int_{\Omega}f_\lambda(x)|\varphi_0|^qdx>0.$$
Applying the result, it holds
\begin{equation*}
\lim_{t\to0^+}\frac{I_{\lambda}(t\varphi_0)}{t^q}
=-\frac{1}{q}\int_{\Omega}f_\lambda(x)|\varphi_0|^qdx<0.
\end{equation*}
therefore, when $t$ is enough small, we have $I_\lambda(t\varphi_0)<0$. 
Thus there exists $u$ small enough such that $I_\lambda(u)<0$. 
Then we deduce that
\begin{equation}\label{2.6}
d=\inf_{u\in\overline{B_R(0)}}I_\lambda(u)
<0<\inf_{u\in\overline{\partial B_R(0)}}I_\lambda(u).
\end{equation}
\end{proof}

\begin{theorem} \label{thm2.4}
 Assume $a, b>0$, $1<q<2$ and $f\in L^{\infty}(\Omega)$,  problem \eqref{1.3} 
has a positive solution $u_\lambda$ with $I_\lambda(u_\lambda)<0$.
\end{theorem}

\begin{proof} 
From \eqref{2.6}, by applying Ekeland's variational principle in
 $\overline{B_R(0)}$, there exists a minimizing sequence 
$\{u_n\}\subset\overline{B_R(0)}$ such that
$$
I_\lambda(u_n)\leq\inf_{u\in\overline{B_R(0)}}I_\lambda(u)+\frac{1}{n},\quad
I_\lambda(v)\geq I_\lambda(u_n)-\frac{1}{n}\|v-u_n\|, \quad
v\in\overline{B_R(0)}.
$$
Therefore, 
\begin{equation*}
I'_\lambda(u_n)\to0\quad\text{and}\quad I_\lambda(u_n)\to d.
\end{equation*}
Since $\{u_n\}$ is bounded and $\overline{B_R(0)}$ is a closed convex set, 
there exist $u_\lambda\in\overline{B_R(0)}\subset H_0^1(\Omega)$ and a 
subsequence still denoted by $\{u_n\}$, such that $u_n\rightharpoonup u_\lambda$ 
in $H_0^1(\Omega)$ as $n\to\infty$.

Note that $I_\lambda(|u_n|)=I_\lambda(u_n)$, by Lemma \ref{lem2.2}, we can obtain 
$u_n\to u_\lambda$ in $H_0^1(\Omega)$ and 
$d=\lim_{n\to\infty}I_\lambda(u_n)=I_\lambda(u_\lambda)<0$, which suggests 
that $u_\lambda\geq0$ and $u_\lambda\not\equiv0$. Since $u_\lambda\in H_0^1(\Omega)$, 
by the embedding theorem we get $u_\lambda\in L^6(\Omega)$. Besides, as $f_\lambda\in L^\infty(\infty)$, by the regularity of weak solutions,
it holds $u_\lambda\in W^{2, \frac{6}{q}}(\Omega)$. By the embedding theorem again, 
it holds that $u_\lambda\in C^{1,\alpha}(\Omega)$. Therefore, by the Harnack 
inequality, we obtain $u_\lambda>0$ a.e. in $\Omega$. The proof is complete.
\end{proof} 

\subsection{Existence of a second positive solution of \eqref{1.3}}

\begin{lemma} \label{lem2.5}
 Assume that $\lambda\in(0, \Lambda_0)$, for given $R$, the functional 
$I_\lambda$ satisfies the following conditions:
\begin{itemize}
\item[(i)]  $I_\lambda(u)>0$ if $u\in S_R$,
\item[(ii)]  there exists $e\in H_0^1(\Omega)$ such that 
$I_\lambda(e)<0$ when $\|e\|>R$.
\end{itemize}
\end{lemma}

\begin{proof} 
(i) The conclusion follows from Lemma \ref{lem2.3} when $\lambda<\Lambda_0$.

(ii) For $u\in H_0^1(\Omega)\backslash\{0\}$, it holds
\begin{align*}
I_{\lambda}(tu)
&= \frac{at^2}{2}\|u\|^2-\frac{bt^4}{4}\|u\|^4
 -\frac{t^q}{q}\int_\Omega f_\lambda(x)|u|^{q}dx\\
&\leq \frac{at^2}{2}\|u\|^2-\frac{bt^4}{4}\|u\|^4
 +\frac{t^q}{q}\int_\Omega| f_\lambda(x)||u|^{q}dx
\to-\infty
\end{align*}
as $t\to+\infty$. Therefore we can find $e\in H_0^1(\Omega)$ such that 
$I_\lambda(e)<0$ when $\|e\|>R$. The proof is complete.
\end{proof}

It is well known that the function
\begin{equation*}
U_\varepsilon(x)=\frac{(3\varepsilon^2)^{1/4}}{(\varepsilon^2+|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),\\
|u_\varepsilon|_6^6=S^{3/2}+O(\varepsilon^3).
\end{gather*}


\begin{lemma} \label{lem2.6}
 Assume $a, b>0$, $1<q<2$ and $f\in L^{\infty}(\Omega)$, it holds
\begin{equation*}
\sup_{t\geq0}I_\lambda(u_\lambda+tu_\varepsilon)<\frac{a^2}{4b}-D\lambda.
\end{equation*}
\end{lemma}

\begin{proof} 
Since $u_\lambda$ is a positive solution of \eqref{1.3} and 
$I_\lambda(u_\lambda)<0$, it holds
\begin{align*}
&I_{\lambda}(u_\lambda+tu_\varepsilon)\\
&= \frac{a}{2}\|u_\lambda+tu_\varepsilon\|^2-\frac{b}{4}\|u_\lambda
 +tu_\varepsilon\|^4-\frac{1}{q}\int_{\Omega}f_\lambda(x)|u_\lambda
 +tu_\varepsilon|^qdx\\
&= \frac{a}{2}\|u_\lambda\|^2+at\int_{\Omega}(\nabla u_\lambda,\nabla u_\varepsilon)dx
 +\frac{at^2}{2}\|u_\varepsilon\|^2-\frac{b}{4}\|u_\lambda\|^4
 -\frac{bt^4}{4}\|u_\varepsilon\|^4\\
&\quad - bt\|u_\lambda\|^2\int_{\Omega}(\nabla u_\lambda,\nabla u_\varepsilon)dx
 -bt^2\Big(\int_{\Omega}(\nabla u_\lambda,\nabla u_\varepsilon)dx\Big)^2\\
&\quad -\frac{bt^2}{2}\|u_\lambda\|^2\|u_\varepsilon\|^2
 -bt^3\|u_\varepsilon\|^2\int_{\Omega}(\nabla u_\lambda,\nabla u_\varepsilon)dx
 -\frac{1}{q}\int_{\Omega}f_\lambda|u_\lambda+tu_\varepsilon|^qdx\\
&\leq I_\lambda(u_\lambda)+\frac{at^2}{2}\|u_\varepsilon\|^2
 -\frac{bt^4}{4}\|u_\varepsilon\|^4
 -\frac{bt^2}{2}\|u_\lambda\|^2\|u_\varepsilon\|^2\\
&\quad +\int_{\Omega}|f_-|\Big\{\int_{0}^{tu_\varepsilon}[(u_\lambda+\eta)^{q-1}
 -u_\lambda^{q-1}]d\eta\Big\}dx\\
&\leq \frac{at^2}{2}\|u_\varepsilon\|^2-\frac{bt^4}{4}\|u_\varepsilon\|^4
 -\frac{bt^2}{2}\|u_\lambda\|^2\|u_\varepsilon\|^2+\int_{\Omega}|f_-|
 \Big\{\int_{0}^{tu_\varepsilon}\eta^{q-1}d\eta\Big\}dx\\
&\leq \frac{at^2}{2}\|u_\varepsilon\|^2-\frac{bt^4}{4}\|u_\varepsilon\|^4
 -\frac{bt^2}{2}\|u_\lambda\|^2\|u_\varepsilon\|^2
 +Ct^q\int_{\Omega}u_\varepsilon^qdx.
\end{align*} 
Set
$$
g(t)=\frac{at^2}{2}\|u_\varepsilon\|^2
-\frac{bt^4}{4}\|u_\varepsilon\|^4
-\frac{bt^2}{2}\|u_\lambda\|^2\|u_\varepsilon\|^2
+Ct^q\int_{\Omega}u_\varepsilon^qdx.
$$
We prove 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
\begin{equation}\label{2.7}
0<t_1\leq t_\varepsilon \leq t_2<\infty.
\end{equation}
In deed, since $\lim_{t\to0^+}g(t)>0$, $\lim_{t\to+\infty}g(t)=-\infty$, 
there exists $t_\varepsilon>0$ such that
$$
g(t_\varepsilon)=\sup_{t\geq0}g(t)\quad\text{and}\quad 
\frac{dg}{dt}|_{t=t_\varepsilon}=0.
$$
As in \cite{LG} it follows that \eqref{2.7} holds.
 Note that $\int_{\Omega}u_\varepsilon^qdx\leq c\varepsilon^{q/2}$, then it holds
that
\begin{align*}
\sup_{t\geq0}I_{\lambda}(u_\lambda+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_1\|u_\varepsilon\|^2+C_2\varepsilon^{q/2}\\
&\leq \frac{a^2}{4b}+C_3\varepsilon-C_1S^{3/2}+C_2\varepsilon^{q/2}\\
&\leq \frac{a^2}{4b}+(C_2+C_3)\varepsilon^{q/2}-C_1S^{3/2},
\end{align*}
where $C_i>0, i=1,2,3$. Let $\varepsilon=\lambda^{\frac{2}{q}}$,
 when $0<\lambda<\Lambda_1:=\frac{C_1S^{3/2}}{C_2+C_3+D}$, it holds
\begin{equation*}
(C_2+C_3)\lambda-C_1S^{3/2} < (C_2+C_3)\lambda-(C_2+C_3+D)\lambda
=  -D\lambda.
\end{equation*}
Consequently, 
$\sup_{t\geq0}I_{\lambda}(u_\lambda+tu_\varepsilon)<\frac{a^2}{4b}-D\lambda$. 
The proof is complete.
\end{proof}


\begin{theorem} \label{thm2.7}
Assume that $b>0, 1<q<2$ and $f\in L^{\infty}(\Omega)$, there exists 
$\lambda_*>0$ such that for each $\lambda\in(0, \lambda_*)$, problem \eqref{1.3} 
admits a positive solution $v_\lambda$ with $I_\lambda(v_\lambda)>0$.
\end{theorem}

\begin{proof} 
Let $\lambda_*=\min\{\Lambda_0, \Lambda_1, \frac{a^2}{4bD}\}$, then
 Lemmas \ref{lem2.1}, \ref{lem2.2}, \ref{lem2.3}  \ref{lem2.5}, and \ref{lem2.6}
 hold for $\lambda<\lambda_*$. 
Applying the mountain-pass lemma \cite{AR}, there is a sequence 
$\{u_n\}\subset H_0^1(\Omega)$ such that
$$
I_\lambda(u_n)\to c>0\quad\text{and}\quad I'_\lambda(u_n)\to0,
$$
where
\begin{gather*}
c=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I_\lambda(\gamma(t)),\\
\Gamma=\big\{\gamma\in C([0, 1], H_0^1(\Omega)): 
\gamma(0)=u_\lambda, \gamma(1)=e\big\}.
\end{gather*}
From Lemma \ref{lem2.2}, $\{u_n\}$ has a convergent subsequence 
(still denoted by $\{u_n\}$) and there exists $v_\lambda\in H_0^1(\Omega)$ 
such that $u_n\to v_\lambda$ in $H_0^1(\Omega)$. Moreover, we can obtain 
$v_\lambda$ is a non-negative weak solution of \eqref{1.3} and
$$
I_\lambda(v_\lambda)=\lim_{n\to\infty}I_\lambda(u_n)=c>0.
$$
Therefore, we infer that $v_\lambda\not\equiv0$. It is similar to
 Theorem \ref{thm2.4} that $v_\lambda>0$ a.e. in $\Omega$. The proof is complete.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by the National Natural Science Foundation of China
(No. 11661021), Science and Technology Foundation of Guizhou Province
(No. LH[2015]7207), Natural Science Foundation of Education
of Guizhou Province (No. KY[2016]046), and
Science and Technology Foundation of Guizhou Province (No. LH[2016]7033).

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