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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 40, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/40\hfil Existence of positive solutions]
{Existence of positive solutions for superlinear $p$-Laplacian equations}

\author[T.-M. Gao, C.-L. Tang \hfil EJDE-2015/40\hfilneg]
{Ting-Mei Gao, Chun-Lei Tang}

\address{Ting-Mei Gao \newline
School of Mathematics and Statistics, Southwest University,
Chongqing 400715,  China.\newline
School of Mathematics and Computer Science,
Shaanxi University of Technology, \newline
Hanzhong 723000, Shaanxi,  China}
\email{gtmgtmgtm@126.com}

\address{Chun-Lei Tang (corresponding author)\newline
School of Mathematics and Statistics, Southwest University,
Chongqing 400715, China}
\email{tangcl@swu.edu.cn, tangcl8888@sina.com}

\thanks{Submitted May 30, 2014. Published February 12, 2015.}
\subjclass[2000]{35J20, 35J92, 35J25, 35B09}
\keywords{Mountain pass theorem; positive solution;
 $p$-Laplacian equation; \hfill\break\indent Dirichlet boundary condition}

\begin{abstract}
 We obtain a positive solution for a superlinear
 $p$-Laplacian equations with the Dirichlet boundary-value
 conditions. Our main tool is a variation of the mountain pass theorem.
\end{abstract}

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


\section{Introduction and statement of the main result}

We consider the  nonlinear elliptic equation of $p$-Laplacian
type
\begin{equation} \label{e1.1}
\begin{gathered}
-\Delta_pu=f(x,u),\quad  x\in\Omega,\\
u=0,\quad x\in\partial\Omega,
\end{gathered}
\end{equation}
 where $\Delta_pu=\operatorname{div}(|\nabla u|^{p-2}\nabla u)$ is the $p$-Laplacian
operator with $p>1$, $\Omega$ is a bounded domain in $R^N(N\geq1)$
with smooth boundary $\partial\Omega$.
The function $f\in C(\overline{\Omega}\times R, R)$ satisfies the
following conditions:
\begin{itemize}

\item[(F1)] $f$ is subcritical in $t$, that is, there is a $q\in(p,
Np/(N-p))$ when $N>p; q\in(p,+\infty)$ when $N\leq p$ such that
\[
\lim_{t\to +\infty} \frac{f(x,t)}{t^{q-1}}=0\quad \text{uniformly in a.e. }
x\in \Omega.
\]

\item[(F2)]
\[
b_0\leq{\liminf_{t\to  0^+}\frac{f(x,t)}{t^{p-1}}}
\leq{\limsup_{t\to  0^+}\frac{f(x,t)}{t^{p-1}}}\leq a(x)
\]
 uniformly in a.e. $x\in \Omega$, where $b_0$ is a constant,
$a\in L^\infty(\Omega)$ satisfies $a(x)\leq \lambda_1$ for all
$x\in\overline{\Omega}$ and
$a(x)<\lambda_1$ on some $\Omega_1\subset \Omega$ with
$|\Omega_1|>0$, $\lambda_1$ is the first eigenvalue of
$-\triangle_p$  and $|\Omega_1|$ is the measure of $\Omega_1$.

\item[(F3)]
 $\lim_{t\to +\infty} f(x,t)/t^{p-1}=+\infty$ uniformly in a.e. $x\in
\Omega$.
\end{itemize}

In this article, we study the existence of a positive solution to
 \eqref{e1.1} under the above assumptions. Since (F3) hods, problem
\eqref{e1.1} is called superlinear in $t$ at $+\infty$. In many studies
involving this superlinear problem, to get a nontrivial solution of
 \eqref{e1.1}, a very famous theorem - Mountain pass theorem is a
common tool, but in applying this theorem, usually, we have to
suppose another condition, that is, for some $\mu>p, M>0$
\begin{equation} \label{AR}
0<\mu F (x, t)\leq f(x, t)t\quad\text{for a.e. }
x\in \overline{\Omega}\text{ and for all }|t|\geq M.
\end{equation}
The condition \eqref{AR} is convenient, but it is very restrictive, in
particular, it implies (F3). To overcome this difficulty,  many efforts
have been made.   Wang and Tang \cite{w1}
proved the following existence theorem without condition \eqref{AR}.

\begin{theorem} \label{thmA}
 Suppose $f$ is subcritical in $t$ and assume that {\rm (F3)} and the following
conditions hold:
\begin{itemize}
\item[(F2')] $\limsup_{t\to  0^+}\frac{f(x,t)}{t^{p-1}}
= c(x)$ uniformly in a.e. $x\in \Omega$, where
$c \in L^\infty(\Omega)$ satisfies $c(x)\leq
\lambda_1$ for all $x\in\overline{\Omega}$ and $c(x)<\lambda_1$ on
some $\Omega'\subset \Omega$ with $|\Omega'|>0$.

\item[(F4')] There exists $\theta\geq 1$ such that $\theta H(x,t)\geq
H(x,\xi t)$ for all $x\in \overline{\Omega}, t\in R$ and $\xi\in [0,
1]$, where $H(x,t)=f(x,t)t-pF(x,t)$ and $F(x,t)=\int_0^tf(x,s)ds$.

\item[(H1)] $f(x,t)\geq 0$ for all $t\geq 0, x\in\overline{\Omega}$.
\end{itemize}
Then \eqref{e1.1} has at least one positive solution.
\end{theorem}

Assumptions (F4') was first introduced by Jeanjean in \cite{j1} for
$p=2$,  Liu and Li in \cite{l2} extended it for general $p>1$, it is
helpful for proving that the functional corresponding to problem
\eqref{e1.1} satisfies Cerami condition (C).

Recently by a monotony condition instead of \eqref{AR},
Iturriage and  Lorca \cite{i1} obtained the following result.

\begin{theorem} \label{thmB}
Suppose  $f(x, t)$ is subcritical in $t$ and satisfies the following:
\begin{itemize}
\item[(G1)] $f: \overline{\Omega}\times [0, +\infty)\to [0,
+\infty)$ is a Carath\'eodory function.

\item[(G2)] ${\limsup_{t\to  0^+}\frac{f(x,t)}{t^{p-1}}}= 0$
 uniformly in a.e. $x\in \Omega$.

\item[(G3)] for any $M>0$, there exists $c_0$ such that
\[
  f(x, t)\geq M|t|^{p-1}-c_0,\quad\text{for all  $t\geq 0$ and $x\in \overline{\Omega}$}.
\]

\item[(G4)] there exists $R>0$ such that the map $t\mapsto f(x, t)t^{1-p}$
 is non-decreasing if $t> R$ for a.e. $x\in\overline{\Omega}$.
\end{itemize}
Then \eqref{e1.1} has a positive solution.
\end{theorem}

Assumptions (G1)--(G3) and the subcritical condition ensure that the energy
functional associated with \eqref{e1.1} has mountain pass geometry.
Assumption (G4) allow us to show the boundedness of the
$(PS)$-sequence at the mountain pass level.
Motivated by Theorem \ref{thmA} and \ref{thmB}, we assume a new condition
and obtain the following result.

\begin{theorem} \label{thm1}
Suppose  {\rm (F1)--(F3)} hold and assume
\begin{itemize}
\item[(F4)] There exist two constants $\theta\geq1, \theta_0>0$ such that
$\theta H(x,s)\geq H(x,t)-\theta_0$ for all $x\in \overline{\Omega},
0\leq t \leq s$, where $H(x,t)=f(x,t)t-pF(x,t)$ and
$F(x,t)=\int_0^tf(x,s)ds$.
\end{itemize}
Then  \eqref{e1.1} has at least one positive solution.
\end{theorem}

\begin{remark} \label{rmk1} \rm
Theorem \ref{thm1} unifies and generalizes
Theorems \ref{thmA} and \ref{thmB} mainly in four aspects:

Firstly, it is obvious that the conditions (F2') in
Theorem \ref{thmA}, and (G2) in Theorem \ref{thmB} are stronger than our condition
(F2).

Secondly, the assumption (G3) can imply  our condition (F3).

Thirdly, it is easily proved that (F4') implies (F4) and  we also can
show (G4) implies (F4) (see Remark \ref{rmk2}), that is to say, (F4)
is weaker than (F4') and (G4).

In addition, as many studies, to prove that a nontrivial
solution of problem \eqref{e1.1} is positive, Theorems \ref{thmA} and 
\ref{thmB} require
$f(x,t)\geq 0$ for all $t\geq 0, x\in\overline{\Omega}$, in this
article, we do not need it any more, in fact, the condition
$b_0\leq{\liminf_{t\to  0^+}\frac{f(x,t)}{t^{p-1}}}$ in (F2)
together with (F3) are sufficient for showing that a nontrivial solution
of \eqref{e1.1} is positive.

Moreover, we can
find some functions that satisfy the conditions of Theorem \ref{thm1}, but
not the conditions of Theorems \ref{thmA} and \ref{thmB}. For instance,
\[
f(x,t)=\begin{cases}
a(x)t^{q-1}-b(x)t^{p-1}, &t\geq 0,\\
 0, &t<0,
\end{cases}
\]
with $a,b\in C(\overline{\Omega})$, $a>0$, $b>0$ and $q\in (p,Np/(N-p)$.
It is obvious that $f$ satisfies (F1)--(F3).
Let $\theta=1,\theta_0=0$, then we also can check that $f$ satisfies (F4),
 and so $f$ satisfies all the conditions
 of Theorem \ref{thm1}, but it does not satisfy the assumptions of 
Theorems \ref{thmA} and \ref{thmB} (in fact, if $t>0$ is small enough, $f(x, t)<0$).
\end{remark}

\begin{remark} \label{rmk2} \rm
Condition (G4) implies condition (F4).
\end{remark}

\begin{proof}
(1) Claim: $H(x, t)$ is nondecreasing on $(R, +\infty)$ for a.e.
$x\in\overline{\Omega}$.
In fact, if we assume $0<R<t<s$, then
\begin{align*}
&H(x, s)-H(x, t)\\
&=p\Big[\frac{1}{p}(f(x, s)s-f(x, t)t)-(F(x, s)-F(x, t))\Big]\\
&= p\Big[\int_R^s\frac{f(x,s)}{s^{p-1}}\tau^{p-1}d\tau
 -\int_R^t\frac{f(x,t)}{t^{p-1}}\tau^{p-1}d\tau
 -\int_t^s\frac{f(x,\tau)}{\tau^{p-1}}\tau^{p-1}d\tau\\
&\quad +\frac{f(x,s)}{ps^{p-1}}R^p-\frac{f(x,t)}{pt^{p-1}}R^p\Big]\\
&= p\Big[\int_t^s\Big(\frac{f(x, s)}{s^{p-1}}-\frac{f(x,
\tau)}{\tau^{p-1}}\Big)\tau^{p-1}d\tau
 +\int_R^t\Big(\frac{f(x,s)}{s^{p-1}}-\frac{f(x, t)}{t^{p-1}}\Big)
 \tau^{p-1}d\tau \\
&\quad +\frac{R^p}{p}\Big(\frac{f(x, s)}{s^{p-1}}
 -\frac{f(x, t)}{t^{p-1}}\Big)\Big]
\geq 0,
\end{align*}
which indicates that
\[
H(x, s)\geq H(x, t)
\]
for $s>t>R$. So $H(x, t)$ is nondecreasing on $(R, +\infty)$ for
a.e. $x\in\overline{\Omega}$.

(2) Now, we prove (F4) holds.
Suppose $\theta=1, \theta_0=2{\max_{\overline{\Omega}\times
[0, R]}}|H(x, s)|$. Since $H(x, t)$ is nondecreasing on
$(R, +\infty)$ for a.e. $x\in\overline{\Omega}$, then
\begin{itemize}
\item[(i)] for all $R\leq t\leq s, H(x, t)-\theta H(x, s)=H(x, t)-H(x,
s)\leq0<\theta_0$.

\item[(ii)]  for all $0\leq t\leq s\leq R$, 
\[
H(x, t)-\theta H(x, s)=H(x,t)-H(x, s)\leq2{\max_{\overline{\Omega}
\times [0,R]}}|H(x, s)|=\theta_0.
\]
\item[(iii)] for all $0\leq t\leq R\leq s$, 
\[
H(x, t)-\theta H(x, s)=
\left(H(x, t)-H(x, R)\right)+\left(H(x, R)-H(x,s)\right)\leq\theta_0+0.
\]
\end{itemize}
So for all $x\in\overline{\Omega}$ and $0\leq t\leq s$, 
$\theta H(x,s)\geq H(x,t)-\theta_0$.
\end{proof}

\section{Preliminaries}

In this section, we give some preliminary knowledge and some
important lemma which will be used to prove our theorem.

Let $\phi$ be a $C^1 $-functional defined on a Banach space $X$, we
say that $\phi$ satisfies the Cerami condition (C), if a
sequence $\{u_n\}\subset X$ is such that $\{\phi(u_n)\}$ is bounded and
$(1+\|u_n\|)\|I'(u_n)\|\to 0$ has a convergent subsequence;
such a sequence is then called a Cerami sequence.

To obtain a positive solution of problem \eqref{e1.1}, we introduced the
$C^1 $-functional $I$, defined by
\[
I(u)=\frac{1}{p}\int_\Omega |\nabla u|^pdx-\int_\Omega F(x,u^+)dx,
\quad \forall u\in W_0^{1,p}(\Omega),
\]
where $F(x,t)=\int_0^tf(x,s)ds$, and $t^+$ denotes the positive part
of $t$. In the proof of our theorem, we shall use the following
lemma.

\begin{lemma} \label{lem2.1}
 If hypothesis {\rm (F1)--(F4)} hold, then the functional
$I$ satisfies the Cerami condition.
\end{lemma}

\begin{proof}
Let $\{u_n\}\subset W_0^{1,p}(\Omega)$ be a sequence such that
\begin{equation} \label{e2.1}
\begin{gathered}
I(u_n)=\frac{1}{p}\int_\Omega |\nabla u_n|^pdx-\int_\Omega
F(x,u_n^+)dx \to c \quad\text{as } n\to \infty, \\
(1+\|u_n\|)\|I'(u_n)\|\to 0\quad \text{as } n\to \infty,
\end{gathered}
\end{equation}
 then
\begin{equation} \label{e2.2}
\int_\Omega\Big(\frac{1}{p}f(x,u_n^+)u_n^+-F(x,u_n^+)\Big) dx=c+o(1).
\end{equation}

 Next, we show that the sequence $\{u_n\}$ is bounded.
Otherwise, there is a subsequence of $\{u_n\}$ (still denoted by
$\{u_n\}$) satisfying $\|u_n\|\to\infty$ as $n\to \infty$.
Set $w_n=u_n/\|u_n\|$, then $\|w_n\|=1$. Up to a subsequence, we  assume that
\begin{equation} \label{e2.3}
\begin{gathered}
w_n\rightharpoonup w\quad \text{in }  W_0^{1,p}(\Omega); \quad
w_n\to w\quad \text{in } L^r(\Omega)\; (1\leq r< p^*);\\
w_n(x)\to w(x)\quad  \text{a.e.} x\in\Omega
\end{gathered}
\end{equation}
for some $w\in W_0^{1,p}(\Omega)$ as $n\to\infty$. It is easily that $w^+$
and $w^-$ have the same convergence which is similar to \eqref{e2.3},
where $u^\pm=\max\{\pm u, 0\}$ for $u\in W_0^{1,p}(\Omega)$.
 We claim that $w^+\equiv 0$.
 Let $\Omega_0=\{x\in\Omega: w^+(x)=0\}$,
$\Omega^+=\{x\in\Omega: w^+(x)>0\}$.
Since $\|u_n\|\to +\infty$, then $u_n^+\to+\infty$ as $n\to+\infty$
for a.e. $x\in\Omega^+$. Since
${\lim_{t\to+\infty}\frac{f(x,t)}{t^{p-1}}}=+\infty$
by (F3), one has
\[
\lim_{n\to\infty}\frac{f(x,u_n^+)}{(u_n^+)^{p-1}}=+\infty
\quad\text{a.e. }x\in\Omega^+.
\]
 From \eqref{e2.1}, we obtain
\begin{equation} \label{e2.4}
|\langle   I'(u_n),u\rangle| \leq\varepsilon_n,
\end{equation}
 where
$\varepsilon_n=(1+\|u_n\|)\|I'(u_n)\|\to 0$ as
$n\to\infty$. It follows from \eqref{e2.4} that
\[
|\|u_n^+\|^p-\int_\Omega f(x,u_n^+)u_n^+dx|\leq\varepsilon_n,
\]
 which implies
\begin{equation} \label{e2.5}
 \int_{\Omega^+} \frac{f(x,u_n^+)}{(u_n^+)^{p-1}}(w_n^+)^pdx
\leq 1+\frac{\varepsilon_n}{\|u_n^+\|^p}.
\end{equation}
 If $|\Omega^+|>0$, since $\|w_n^+\|=1$, from \eqref{e2.5},
one obtains
$$
+\infty\leftarrow\int_{\Omega^+}
\frac{f(x,u_n^+)}{(u_n^+)^{p-1}}(w_n^+)^pdx\leq
1+\frac{\varepsilon_n}{\|u_n^+\|^p}\to 1,
$$
which is a contradiction, so $|\Omega^+|=0$ and $w^+\equiv 0$.

 By  (F1) and (F2), we have
 \[
f(x,t)\leq\left(a(x)+\varepsilon\right)|t|^{p-1}+A |t|^{q-1},\quad
\forall (x,t)\in\overline{\Omega}\times R,
\]
where $A>0$ is a constant, thus
\begin{equation} \label{e2.6}
F(x,t^+)\leq \frac{1}{p}\left(a(x)+\varepsilon\right)|t|^p+A|t|^q,\quad
\forall (x,t)\in\overline{\Omega}\times R.
\end{equation}

Now set a sequence $\{t_n\}$ of real numbers such that
$I(t_n u_n^+)={\max_{t\in[0,1]}I(tu_n^+)}$. For any
integer $m>0$, since $w^+\equiv 0$, then by (F2), \eqref{e2.6},  and the
convergence of $w_n^+$, one has
\begin{align*}
&\limsup_{n\to\infty}\int_\Omega F\left(x,
(2pm)^\frac{1}{p}w_n^+\right)dx \\
&\leq \limsup_{n\to\infty}\Big(\int_\Omega
2m(\lambda_1+\varepsilon)(w_n^+)^pdx
+\int_\Omega A (2pm)^{\frac{q}{p}}(w_n^+)^qdx\Big)\\
&=\lim_{n\to\infty}(C_1\|w_n^+\|^p_p+C_2\|w_n^+\|^q_q)\\
&= C_1\|w^+\|^p_p+C_2\|w^+\|^q_q=0,
\end{align*}
where $C_1, C_2>0$ are constants. Since $\|u_n\|\to+\infty$
as $n\to\infty$, one has 
$0\leq (2pm)^{1/p}/\|u_n\|\leq 1$ when $n$ is big enough.
By the definition of $t_n$, we obtain
\begin{align*}
I(t_nu_n^+)\geq I\left((2pm)^\frac{1}{p}w_n^+\right)\geq
2m-\int_\Omega F\Big(x,(2pm)^\frac{1}{p}w_n^+\Big)\geq m,
\end{align*}
which implies
\begin{equation} \label{e2.7}
I(t_nu_n^+)\to +\infty\quad \text{as }n\to\infty.
\end{equation}
 Noting that $I(0)=0$ and $I(u_n)\to c$, so
$0<t_n<1$ when $n$ is big enough. It follows that
\begin{equation} \label{e2.8}
\begin{aligned}
&\int_\Omega |\nabla (t_nu_n^+)|^pdx-\int_\Omega
f(x,t_nu_n^+)t_nu_n^+dx\\
&=\langle I'(t_nu_n^+),t_nu_n^+\rangle
=t_n\frac{dI (tu_n^+)}{dt}|_{t=t_n}=0.
\end{aligned}
\end{equation}
 But for $0\leq t_n\leq 1$,  $|t_nu_n|\leq|u_n|$,
then (F4), \eqref{e2.7} and \eqref{e2.8} imply
\begin{align*}
&\int_\Omega\left(\frac{1}{p}f(x,u_n^+)u_n^+-F(x,u_n^+)\right)dx\\
&= \frac{1}{p}\int_\Omega
H(x,u_n^+)dx\geq\frac{1}{p\theta}\int_\Omega
\left(H(x,t_nu_n^+)-\theta_0\right)dx\\
&= \frac{1}{\theta}\int_\Omega
\Big(\frac{1}{p}f(x,t_nu_n^+)t_nu_n-F(x,t_nu_n^+)\Big)dx
 -\frac{\theta_0}{p\theta}|\Omega|\\
&= \frac{1}{\theta}\int_\Omega\Big(\frac{1}{p}|\nabla
  t_nu_n^+|^p-F(x,t_nu_n^+)\Big)dx
-\frac{\theta_0}{p\theta}|\Omega|\\
&=\frac{1}{\theta}I(t_nu_n^+)-\frac{\theta_0}{p\theta}|\Omega|\to+\infty
\quad (n\to\infty),
\end{align*}
which contradicts to \eqref{e2.2}, so $\{u_n\}$ is bounded.

By the compactness of Sobolev embedding and the standard procedures, we
know that $\{u_n\}$ has a convergence subsequence. That is to say,
the functional $I$ satisfies the $(C)$ condition.
\end{proof}

\section{Proof of the main result}


\begin{proof}[Proof of Theorem \ref{thm1}]
By  critical point theory, for
finding a positive solution of problem \eqref{e1.1}, we only need to find a
nonzero critical point of the following $C^1 $-functional
\[
I(u)=\frac{1}{p}\int_\Omega |\nabla u|^pdx-\int_\Omega F(x,u^+)dx,
\quad \forall u\in W_0^{1,p}(\Omega).
\]
Since (F2) holds, there exists a positive constant $\alpha<1$ such
that $\int_\Omega a(x)|u|^pdx<\alpha\int_\Omega|\nabla u|^pdx$ for
all $u\in W_0^{1,p}(\Omega)$ (see \cite[Lemma 2.2]{w1}). Let
$\varepsilon>0$ be small enough such that
$\alpha+\varepsilon/\lambda_1<1$. By \eqref{e2.6}, together with the
Poincare inequality and Sobolev inequality, one obtains
\begin{align*}
I(u)&\geq \frac{1}{p}\|u\|^p-\frac{1}{p}\int_\Omega
 \left(a(x)+\varepsilon\right)|u|^pdx-A\int_\Omega |u|^qdx  \\
&\geq \frac{1}{p}\|u\|^p-\frac{1}{p}\int_\Omega
 \big(\alpha+\frac{\varepsilon}{\lambda_1}\big)|\nabla u|^pdx-C\|u\|^q\\
&= \frac{1}{p}\big(1-\alpha-\frac{\varepsilon}{\lambda_1}\big)\|u\|^p-C\|u\|^q,
\end{align*}
 where $C>0$ is a constant. Since
$1-\alpha-\frac{\varepsilon}{\lambda_1}>0$ and $p<q$, when $\rho>0$
be small enough such that
\[
\beta=\frac{1}{p}\Big(1-\alpha-\frac{\varepsilon}{\lambda_1}\Big)\rho^p
-C\rho^q>0,
\]
we have
\begin{equation} \label{e3.1}
I|_{\partial B_\rho}\geq \beta>0.
\end{equation}
Since (F3) holds, then given $\varepsilon>0$, we can find
$c(\varepsilon)> 0$ such that
\[
f(x,t)\geq\frac{t^{p-1}}{\varepsilon}-c(\varepsilon),\quad
\forall(x,t)\in\overline{\Omega}\times R^+,
\]
 which implies
 \[
F(x,t^+)\geq\frac{1}{p\varepsilon}(t^+)^p-c(\varepsilon)t^+,\quad
 \forall(x,t)\in\overline{\Omega}\times R.
 \]
 So, if  $v_1>0$ is the  eigenfunction of
$(-\triangle_p , W_0^{1, p}(\Omega))$ corresponding
  to the first eigenvalue $\lambda_1$ with $\|v_1\|=1$,
 then
\begin{equation} \label{e3.2}
\int_\Omega\frac{F\left(x,(tv_1)^+\right)}{t^p}dx
\geq\int_\Omega\Big(\frac{1}{p\varepsilon}(v_1^+)^p
-\frac{c(\varepsilon){v_1}^+}{t^{p-1}}\Big)dx.
\end{equation}
 Letting $t\to+\infty$ in \eqref{e3.2}, it follows that
\[
\liminf_{t\to+\infty}
\int_\Omega\frac{F\left(x, (tv_1)^+\right)}{t^p}dx
\geq\int_\Omega\frac{1}{p\varepsilon}(v_1^+)^pdx,
\]
for all $\varepsilon>0$. For $\varepsilon>0$ is arbitrary, letting
$\varepsilon\to 0$, we infer that
\[
\lim_{t\to+\infty}\int_\Omega\frac{F\left(x,(tv_1)^+\right)}{t^p}dx=+\infty.
\]
Consequently, one  obtains
\[
\frac{I(tv_1)}{t^p}=\frac{1}{p}\|v_1\|^p-
\int_\Omega\frac{F\left(x,(tv_1)^+\right)}{t^p}dx =
\frac{1}{p}-\int_\Omega\frac{F\left(x,(tv_1)^+\right)}{t^p}dx\to
-\infty
\]
as $t\to+\infty$.
Hence,  when $t_0$  is big enough, there exists $e=t_0v_1\in W_0^{1,
p}(\Omega)\setminus \overline{B_\rho(0)}$ such that
\begin{equation} \label{e3.3}
I(e)\leq 0.
\end{equation}
 Thus,  Lemma \ref{lem2.1} and \eqref{e3.1}, \eqref{e3.3} permit the application of
a variant of mountain pass theorem (see \cite[p. 648]{g1}), so we get a
critical point $u$ of the functional $I$ with $I(u)\geq \beta$. But
from (F2), $f(x, 0)=0$, then $I(0)=0$, that is $u\neq 0$. Since
\[
0=\langle I'(u),u^-\rangle=\|u^-\|^p-\int_\Omega
f(x,u^+)u^-dx=\|u^-\|^p\geq 0,
 \]
which implies that $\|u^-\|=0$, so $u\geq 0$. By the regularity
results (see \cite{l1}), $u\in L^\infty(\Omega)$, and hence
$u\in C^1(\Omega)$( see \cite{t1}). Since $u\in L^\infty(\Omega)$, it is easy to
see that $\Delta_pu=-f(x,u)\in L_{loc}^2(\Omega)$. From
$b_0\leq{\liminf_{t\to  0^+}\frac{f(x,t)}{t^{p-1}}}$
 by (F2), there exist a constant  $\delta>0$ such that
\[
f(x,t)\geq (b_0-1) t^{p-1},\quad  \forall   0\leq t\leq \delta.
\]
By (F3), we can find a positive constant $M$ such that
\[
f(x,t)\geq 0,\quad   \forall   t\geq M.
\]
Because $f\in C(\overline{\Omega}\times R, R)$, then
\[
|f(x,t)|\leq B=B\delta^{-(p-1)}\delta^{p-1}\leq
B\delta^{-(p-1)}t^{p-1},\quad  \forall  \delta\leq t\leq M,
\]
where $B>0$ is a constant, hence
\[
f(x,t)\geq\left(- |b_0-1|-B\delta^{-(p-1)}\right)t^{p-1},\quad
\forall  t\geq 0.
\]
Since $u\geq 0$, it follows that
 $f(x,u)\geq\left(-|b_0-1|-B\delta^{-(p-1)}\right)u^{p-1}=-Du^{p-1}$, where
$D=|b_0-1|+B\delta^{-(p-1)}>0$. Therefore,
$\Delta_pu=-f(x, u)\leq Du^{p-1}$. Hence by the Strong maximum principle
for $p$-Laplacian in \cite{v1} with $\beta(u)=D|u|^{p-1}$, one has $u>0$ a.e. on $\Omega$. That
is, $u$ is a positive solution of problem \eqref{e1.1}. The proof is
complete.
\end{proof}

\subsection*{Acknowledgements}  The authors would like to thank the
anonymous referees and the editors for valuable comments.

This research was supported by National
Natural Science Foundation of China (No. 11471267).

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\end{document}