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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 78, pp. 1--11.\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-2015/78\hfil Multiple positive solutions]
{Multiple positive solutions for elliptic problem with concave
 and convex nonlinearities}

\author[J. Liu, L. Zhao, P. Zhao \hfil EJDE-2015/78\hfilneg]
{Jiayin Liu, Lin Zhao, Peihao Zhao}

\address{Jiayin Liu (corresponding author) \newline
School of Mathematics and Statistics, Lanzhou University,
Lanzhou, Gansu 730000, China}
\email{xecd@163.com}

\address{Lin Zhao \newline
Department of Mathematics, China University of Mining and Technology,
Xuzhou, Jiangsu 221116, China}
\email{zhaolinmath@gmail.com}

\address{Peihao Zhao \newline
School of Mathematics and Statistics, Lanzhou University,
Lanzhou, Gansu 730000, China}
\email{zhaoph@lzu.edu.cn}

\thanks{Submitted November 27, 2014. Published March 31, 2015.}
\subjclass[2000]{35J20, 35J25, 35J60}
\keywords{Variational methods; supercritical exponent;
 mountain pass theorem, \hfill\break\indent Moser iteration technique}

\begin{abstract}
 In this article, we consider the existence of multiple solutions to
 the elliptic problem
 \begin{gather*}
 -\Delta u=\lambda u^q+u^s+\mu u^p\quad \text{in } \Omega,\\
 u>0\quad \text{in } \Omega,\\
 u=0\quad \text{on }  \partial\Omega,
 \end{gather*}
 where $\Omega\subseteq \mathbb{R}^N$ $(N\geq3)$ is a bounded domain with
 smooth boundary $\partial\Omega$, $0<q<1<s<2^*-1\leq p$, $2^*:=\frac{2N}{N-2}$,
 $\lambda$ and $\mu$ are nonnegative parameters. By using variational methods,
 truncation and Moser iteration techniques, we show that if the parameters
 $\lambda$ and $\mu$ are small enough, then the problem has at least two
 positive solutions.
\end{abstract}

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


\section{Introduction and main results}

In this article we study the existence of nontrivial solutions for
the  elliptic problem
\begin{equation} \label{eP}
\begin{gathered}
-\Delta u=\lambda u^q+u^s+\mu u^p\quad \text{in }  \Omega,\\
u>0\quad \text{in }  \Omega,\\
u=0\quad \text{on } \partial\Omega,
\end{gathered}
\end{equation}
where $\Omega\subset \mathbb{R}^N (N\geq3)$ is a bounded smooth domain,
$\lambda$ and $\mu$ are nonnegative parameters, $0<q<1$,
$1<s<2^*-1$, $p\geq 2^*-1$, $ 2^*=\frac{2N}{N-2}$,
i.e.  the nonlinearity is a combination of a sublinear term, a
subcritical term and a critical or supercritical term. From the
perspective of the concavity and convexity of a function,  problem \eqref{eP}
 has one concave term, two convex terms.

 We want to remark that if the subcritical term $u^s(1<s<2^*-1)$ does not
appear in our problem \eqref{eP}, i.e.
 \begin{equation}\label{1.1}
\begin{gathered}
-\Delta u=\lambda u^q+\mu u^p\quad\text{in }\Omega,\\
u>0\quad\text{in }\Omega,\\
u=0\quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
 by the linear transformation $v=\mu^{\frac{1}{p-1}}u$, problem \eqref{1.1}
is equivalent to
  \begin{equation}\label{1.2}
\begin{gathered}
-\Delta v=\widetilde{\lambda} v^q+v^p\quad\text{in }\Omega,\\
v>0\quad\text{in }\Omega,\\
v=0\quad\text{on }\partial\Omega
\end{gathered}
\end{equation}
 with $\widetilde{\lambda}=\lambda\mu^{\frac{1-q}{p-1}}$.
This concave-convex problem was first considered by Ambrosetti, Brezis and
Cerami \cite{ABC},  they discover that there exists $\Lambda >0$ such that
for $0<\widetilde{\lambda}<\Lambda$, problem \eqref{1.2} has a solution if $p>1$,
and has a second solution if $1<p\leq (N+2)/(N-2)$.
For supercritical case, i.e.  $p > (N +2)/(N -2)$, the authors poses an open
problem: When $\Omega$ is a ball in $\mathbb{R}^N$, does problem \eqref{1.2} have
two solutions for $\widetilde{\lambda}>0$ small enough? After this seminal work,
many works have been devoted to problems with concave-convex nonlinearities,
see for example \cite{Alves,Brown,Gar,WTF,ZL,ZC}.
Especially in the literature \cite{ZC}, using a concept of radial singular solution,
Zhao and Zhong prove that if $\widetilde{\lambda} > 0$ is small enough and
$p > 2^*-1$, then problem \eqref{1.2} has exactly one solution.
In particular, this means that problem  \eqref{1.1} cannot have a second solution
if $p > (N +2)/(N -2)$, and gives a negative answer to that open problem.
In other words, \eqref{1.1} has exactly one solution for $\lambda$ and $\mu$
small enough. Now, we are interested in  what will happen with adding a subcritical
term $u^s$ in \eqref{1.1}.
 In this paper, we  show that the appearance of the subcritical term $u^s$
in \eqref{1.1}  destroys the uniqueness result.
 More precisely, we prove the following main results.

 \begin{theorem}\label{thm1.1}
Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain.
Assume $0<q<1<s<2^*-1\leq p$, then  \eqref{eP} has at least
two positive solutions if $\lambda$ and $\mu$ are sufficiently small.
\end{theorem}

Our approach is variational, based on the critical point theory and we
use truncation methods and Moser iteration technique to  deal with
the critical case and supercritical case in a unified approach.

Before we proceed, we  recall that to use the Mountain Pass
Theorem \cite{AR,PH,Willem} the Palais-Smale (PS) condition is needed.
A $C^1$ functional $J$ on a Banach space $X$ is said to satisfy the
 (PS) condition at $c\in \mathbb{R}$ if every sequence $u_n\subset X$ satisfying
\[
\text{$J(u_n) \to c$ and $\|J'(u_n)\|_{X'}\to 0$   as  $n\to \infty$}
\]
admits a strongly convergent subsequence. We say that $J$ satisfies the
(PS) condition if $J$ satisfies the (PS) condition at any $c\in \mathbb{R}$.
This compactness type condition, which compensates for the lack of local compactness
in the underlying space $X$ being in general infinite dimensional, leads
to the following well known Mountain Pass Theorem.

\begin{lemma} \label{lem1.2}
Let $X$ be a Banach space and $J\in C^1(X,\mathbb{R})$ satisfying the (PS) condition.
Suppose $J(0)=0$ and
\begin{itemize}
\item[(J1)] there are constants $\rho,\alpha>0$ such that
$J|_{\partial B_\rho}\geq \alpha$, and
\item[(J2)] there is an $e\in X\setminus B_\rho$ such that $J(e)\leq0$.
\end{itemize}
Then $J$ possesses a critical value $c\geq\alpha$. Moreover, $c$ can
be characterized as
\begin{equation*}
c=\inf_{g\in\Gamma}\max_{u\in g([0,1])}J(u),
\end{equation*}
where $\Gamma=\{g\in C([0,1],X): g(0)=0,\; g(1)=e\}$.
\end{lemma}

In this article, the norm in  $L^r(\Omega)$ $(1<r<\infty)$ 
is $\|u\|_r=\big(\int_\Omega|u|^r{\rm d}x\big)^{{1}/{r}}$,
and the norm in $H_0^1(\Omega)$ is
$\|u\|=\big(\int_\Omega|\nabla u|^2{\rm d}x\big)^{1/2}$.
Here $X'$ denotes the dual space of $X$.
 $S$ is the best Sobolev embedding constant
\begin{equation}\label{1.3}
S=\inf_{u\in H_0^1(\Omega)\setminus\{0\}}
\frac{\int_\Omega|\nabla u|^2{\rm d}x}{\big(\int_\Omega|u|^{2^*}{\rm d}x\big)^{2/2^*}}.
\end{equation}

This article is organized as follows.
In section 2, we consider a truncated problem \eqref{ePK} and obtain two
solutions by using variational methods.
In section 3, we finish the proof of Theorem \ref{thm1.1}
by demonstrating that solutions of \eqref{ePK} are actually solutions of
the original problem \eqref{eP}, this reduces to an $L^{\infty}$  estimate.

\section{Truncated problem}

One of the main difficulty to prove the existence solutions of problem
\eqref{eP} by using variational methods is that $J(u)$ does not satisfying
the (PS) condition for large energy level for $p = \frac{N+2}{N-2}$ and
 $J(u)$ is not well defined on $H_0^1(\Omega)$ for $p > \frac{N+2}{N-2}$.
Following the idea in \cite{FZ,MMT,PH,ZZ}, we first investigate the
 truncated problem
\begin{equation} \label{ePK}
\begin{gathered}
-\Delta u=\lambda u^q+u^s+\mu g_K(u)\quad\text{in }\Omega,\\
u>0\quad\text{in }\Omega,\\
u=0\quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
where $K>0$ is a real number, whose value will be fixed later,
$g_K(u)$ is given by
\begin{equation}\label{2.1}
g_K(u)= \begin{cases}
u^p, &|u|\leq K,\\
K^{p-r+1}u^{r-1}, &|u|\geq K,
\end{cases}
\end{equation}
where $p\geq 2^*-1$, $2<r:=s+1<2^*$, then
\begin{equation}\label{2.2}
G_K(u):=\int_0^ug_K(t){\rm d}t=
\begin{cases}
\frac{1}{p+1}u^{p+1}, &|u|\leq K,\\[4pt]
\big(\frac{1}{p+1}-\frac{1}{r}\big)K^{p+1}+\frac{1}{r}K^{p-r+1}u^r,
 &|u|\geq K,
\end{cases}
\end{equation}
and
\begin{equation}\label{2.3}
|g_K(u)|\leq K^{p-r+1}u^{r-1}, \ |G_K(u)|\leq\frac{1}{r}K^{p-r+1}u^r.
\end{equation}
The associated functional in $H_0^1(\Omega)$ is
\begin{equation} \label{2.4}
\begin{aligned}
J_K(u)&=\frac{1}{2}\int_\Omega|\nabla u|^2{\rm d}x\\
&\quad -\frac{\lambda}{q+1}\int_\Omega u^{q+1}{\rm d}x
 -\frac{1}{s+1}\int_\Omega u^{s+1}{\rm d}x-\mu\int_\Omega G_K(u){\rm d}x.
\end{aligned}
\end{equation}

\begin{remark} \rm
The original problem \eqref{eP} is critical and supercritical, after
truncation, it becomes subcritical and the functional $J_K(u)\in C^1$
 is well defined, this fact allows us to use the usual minimax methods.
\end{remark}

We have the following multiplicity theorem for problem \eqref{ePK}.

\begin{theorem}\label{thm2.2}
There exist two positive constants $\lambda_0$ and $\mu_0$ such that for all
$\lambda,\mu$ with $0<\lambda<\lambda_0$ and
$0<\mu<\mu_0$, problem \eqref{ePK} has at least two positive solutions.
\end{theorem}

\begin{proof}
Let $e$ denote the solution of
\begin{gather*}
-\Delta e=1\quad\text{in }\Omega,\\
e=0\quad\text{on }\partial\Omega,
\end{gather*}
then $e\in C_0^{\infty}(\Omega)$ is nonnegative, and
$\|e\|_\infty\leq C$ for some positive constant $C>0$.
Since $0<q<1<s<2^*-1$, and $2<r=s+1<2^*$, we can find $\lambda_0>0$
and $\mu_0>0$ such that
for all $0<\lambda<\lambda_0$ and  $0<\mu<\mu_0$, there exits
$M=M(\lambda, \mu)>0$ satisfying
\[
M\geq \lambda M^q\|e\|^q_\infty + M^s\|e\|^s_\infty
+ \mu K^{p-r+1}M^{r-1}\|e\|^{r-1}_\infty.
\]
As a consequence, the function $Me$ satisfies
\begin{align*}
-\Delta(Me)=(-\Delta e)M=M
& \geq \lambda M^q\|e\|^q_\infty + M^s\|e\|^s_\infty
 + \mu K^{p-r+1}M^{r-1}\|e\|^{r-1}_\infty\\
& \geq \lambda(Me)^q+(Me)^s+\mu g_K(Me),
\end{align*}
and hence it is a supersolution of \eqref{ePK}. Moreover, any
$\varepsilon\varphi_1$ is a subsolution of \eqref{ePK}, provided
\[
-\Delta(\varepsilon\varphi_1)=\lambda_1\varepsilon\varphi_1
\leq \lambda(\varepsilon\varphi_1)^q+(\varepsilon\varphi_1)^s
+\mu g_K(\varepsilon\varphi_1),
\]
which is satisfied for all $\varepsilon>0$ small enough and all $\lambda>0, \mu>0$.
Taking $\varepsilon$ possibly smaller, we also have
\[
\varepsilon\varphi_1<Me
\]
If follows that \eqref{ePK} has a solution
$\varepsilon\varphi_1\leq u_1\leq Me$ whenever $\lambda\leq \lambda_0$
and $\mu\leq\mu_0$. Actually, $u_1$ is a local minimum of $J_K$ in the
$C^1$-topology, hence a local minimum for $J_K$ in the $H_0^1(\Omega)$-topology,
see \cite{ABC} for details.

Next, we look for a second solution of \eqref{ePK} by Mountain Pass Theorem,
since $u_1$ is a local minimum in the $H_0^1(\Omega)$-topology, we only need
to show that the $(PS) $ condition is satisfied and $J_K(tu)\to -\infty$,
as $t\to +\infty$.
\smallskip

\noindent\textbf{Claim 1.} The functional $J_K(u)$ satisfies $(PS)_c$ for
any $c\in\mathbb{R}$.
To see this, take $c\in\mathbb{R}$ and assume that $\{u_n\}$ is a Palais-Smale
sequence at level $c$, namely such that
\[
J_K(u_n)\to c \text{ and } J_K'(u_n)\to0 (\text{in }H_0^1(\Omega)'),
\]
Consequently we obtain, by Sobolev embedding theorem, together with \eqref{2.1}
and \eqref{2.2},
\begin{equation} \label{2.5}
\begin{aligned}
c(1+\|u_n\|)
&\geq J_K(u_n)-\frac{1}{s+1}J_K'(u_n)u_n  \\
&=\Big(\frac{1}{2}-\frac{1}{s+1}\Big)
\int_\Omega|\nabla u_n|^2{\rm d}x
-\Big(\frac{\lambda}{q+1}-\frac{\lambda}{s+1}\Big)\int_\Omega
u_n^{q+1}{\rm d}x\\
&\quad +\mu\int_\Omega\Big[\frac{1}{s+1}g_K(u_n)u_n-G_K(u_n)\Big]{\rm
d}x                       \\
&\geq \Big(\frac{1}{2}-\frac{1}{s+1}\Big)\int_\Omega|\nabla u_n|^2{\rm d}x
-\Big(\frac{\lambda}{q+1}-\frac{\lambda}{s+1}\Big)\int_\Omega u_n^{q+1}{\rm d}x                                  \\
&\geq \Big(\frac{1}{2}-\frac{1}{s+1}\Big)\|u_n\|^2
-\Big(\frac{\lambda}{q+1}-\frac{\lambda}{s+1}\Big)S_b^{q+1}\|u_n\|^{q+1},
\end{aligned}
\end{equation}
where $S_b$ is the Sobolev constant, and we have also used the fact
 that \eqref{2.1} and \eqref{2.2} imply
$g_K(t)t\geq(s+1)G_K(t)$ for all $t\in\mathbb{R}$.
It follows from \eqref{2.5} (note $1<q+1<2$),
$\{u_n\}$ is bounded in $H_0^1(\Omega)$.
Then we can assume that, up to a subsequence,
there exists $u\in H_0^1(\Omega)$ such that
\begin{gather*}
u_n\rightharpoonup u\quad \text{in } H_0^1(\Omega),\\
u_n(x)\to u(x)\quad \text{for almost every } x\in\Omega,\\
u_n\to u\quad \text{in } L^s(\Omega).
\end{gather*}
As a consequence,
\begin{gather*}
\int_\Omega(u_n^q-u^q)(u_n-u){\rm d}x\to0,\quad
\int_\Omega\left(u_n^s-u^s\right)(u_n-u){\rm d}x\to0,\\
\int_\Omega\left[g_K(u_n)-g_K(u)\right](u_n-u){\rm
d}x\to0\quad \text{as } n\to\infty.
\end{gather*}
We conclude by computing
\begin{align*}
o(1)&=\left(J_K'(u_n)-J_K'(u)\right)(u_n-u)\\
&=\int_\Omega|\nabla(u_n-u)|^2{\rm d}x
 -\lambda\int_\Omega(u_n^q-u^q)(u_n-u){\rm d}x\\
&\quad -\int_\Omega\left(u_n^s-u^s\right)(u_n-u){\rm
d}x-\mu\int_\Omega\left[g_K(u_n)-g_K(u)\right](u_n-u){\rm d}x\\
&=\|u_n-u\|^2+o(1),
\end{align*}
which shows that $u_n\to u$ in $H_0^1(\Omega)$. This proves Claim 1.
\smallskip

\noindent\textbf{Claim 2.} $J_K(tu)\to -\infty$, as $t\to+\infty$.
For every $u\in H_0^1(\Omega)\setminus\{0\}$ and $t>0$ we have
\begin{align*}
J_K(tu)
&=\frac{t^2}{2}\|u\|^2-\frac{\lambda t^{q+1}}{q+1}\int_\Omega
u^{q+1}{\rm d}x-\frac{ t^{s+1}}{s+1}\int_\Omega
u^{s+1}{\rm d}x-\mu\int_\Omega G_K(tu){\rm d}x\\
&=\frac{t^2}{2}\|u\|^2-\frac{\lambda t^{q+1}}{q+1}\int_\Omega
u^{q+1}{\rm d}x-\frac{ t^{s+1}}{s+1}\int_\Omega
u^{s+1}{\rm d}x\\
&\quad -\frac{\mu t^{p+1}}{p+1}\int_{\{|tu|\leq
K\}}u^{p+1}{\rm d}x-\frac{\mu t^r
K^{p-r+1}}{r}\int_{\{|tu|\geq K\}}u^r{\rm d}x.
\end{align*}
Since
\[
\int_{\{|tu|\leq K\}}u^{p+1}{\rm d}x\to0\quad \text{as }t\to+\infty,
\]
and  $1<q+1<2<s+1=r<2^*$, it follows that $J(tu)\to-\infty$ as
$t\to+\infty$. This proves Claim 2.

 Since Claims 1 and  2 hold, by the mountain pass theorem there exists a
 $u_2\in H_0^1(\Omega)$ such
 that $J_K(u_2)=c_M$, where
 \[
 c_M=\inf_{\omega\in W}\max_{t\in [0,1]} J_K(\omega(t))
\quad\text{and}\quad
W=\{\omega\in C([0,1]): \omega(0)=u_1, J_K(\omega(1))<0\}.
 \]
We may assume that $u_2$ is positive. Indeed, we can extend the nonlinearity
to zero if $u<0$, with this extension, the maximum principle implies
that every nontrivial solutions of \eqref{ePK} is positive.
\end{proof}

\begin{lemma}\label{lem2.3}
The solutions for problem \eqref{ePK} obtained by Theorem \ref{thm2.2}
are bounded in $H_0^1(\Omega)$, i.e.
\[
\|u_i\|\leq\gamma,\quad i=1,2.
\]
where $\gamma>0$ is independent of $\mu$.
\end{lemma}

\begin{proof}
Let $c_M$ be the mountain pass level for $J_K$ obtained in previous section,
\begin{align*}
c_M
&\geq J_K(u_i)=J_K(u_i)-\frac{1}{s+1}J_K'(u_i)u_i                   \\
&=\Big(\frac{1}{2}-\frac{1}{s+1}\Big)\int_\Omega|\nabla u_i|^2{\rm
d}x-\Big(\frac{\lambda}{q+1}-\frac{\lambda}{s+1}\Big)\int_\Omega
u_i^{q+1}{\rm d}x\\
&\quad +\mu\int_\Omega\Big[\frac{1}{s+1}g_K(u_i)u_i-G_K(u_i)\Big]{\rm
d}x                      \\
&\geq \Big(\frac{1}{2}-\frac{1}{s+1}\Big)\int_\Omega|\nabla u_n|^2{\rm
d}x-\Big(\frac{\lambda}{q+1}-\frac{\lambda}{s+1}\Big)\int_\Omega
u_i^{q+1}{\rm d}x                                  \\
&\geq \Big(\frac{1}{2}-\frac{1}{s+1}\Big)\|u_i\|^2
-\Big(\frac{\lambda}{q+1}-\frac{\lambda}{s+1}\Big)S_b^{q+1}\|u_i\|^{q+1}.
\end{align*}
Since $1<q+1<2$, we infer that $\|u_i\|\leq \gamma$ which is independent of $\mu$.
\end{proof}

\begin{remark}\label{rmk2.8} \rm
Actually, $u_1$ and $u_2$ also solve problem \eqref{eP}, to show this,
 we only need to prove $\|u_i\|_{L^{\infty}(\Omega)}\leq K$, $i=1, 2$.
One should note that $c_M$ is decreasing with respect to $K$, so, $\gamma$
is also decreasing with respect to $K$, this fact is important in the
following $L^{\infty}(\Omega)$ estimate (see inequality \eqref{3.14}
in next section).
\end{remark}

\section{Proof of main result}

To prove Theorem \ref{thm1.1}, we only need to show that solutions
of \eqref{ePK} are actually bounded by some $K$. Our approach is a variant
 of Moser iteration technique inspired by \cite{Corr,FZ,MMT,ZZ}.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
For convenience, set $u:=u_i$, $i=1,2$. Let $u$ be a weak solution of \eqref{ePK}.
Hence, for any $\varphi\in H_0^1(\Omega)$,
\begin{equation}\label{3.1}
\int_\Omega\nabla u\nabla\varphi{\rm d}x=\lambda\int_\Omega
u^q\varphi{\rm d}x+\int_\Omega
u^s\varphi{\rm d}x+\mu\int_\Omega g_K(u)\varphi{\rm d}x.
\end{equation}
For each $L>0$, let us define the following functions
\begin{equation*}
u_L(x)= \begin{cases}
u(x),&\text{if }u(x)\leq L,\\
L, &\text{if }u(x)>L,
\end{cases}
\end{equation*}
$z_L=u_L^{2(\beta-1)}u$ and $w_L=u_L^{\beta-1}u$, where
$\beta>1$ will be fixed later.
Taking $z_L$ as a test function in  \eqref{3.1}, we obtain
\begin{equation}\label{3.2}
\int_\Omega\nabla u\nabla z_L{\rm d}x=\lambda\int_\Omega
u^q z_L{\rm d}x+\int_\Omega
u^sz_L{\rm d}x+\mu\int_\Omega g_K(u)z_L{\rm d}x.
\end{equation}
The left hand side of the above equality is
\begin{align*}
\int_\Omega\nabla u\nabla z_L{\rm d}x
&=\int_\Omega\nabla u\nabla (u_L^{2(\beta-1)}u){\rm d}x\\
&=\int_\Omega|\nabla u|^2u_L^{2(\beta-1)}{\rm d}x
 +2(\beta-1)\int_\Omega uu_L^{2(\beta-1)-1}\nabla u \nabla u_L {\rm d}x\\
&=\int_\Omega|\nabla u|^2u_L^{2(\beta-1)}{\rm d}x
 +2(\beta-1)\int_{\{0\leq u\leq L\}}|\nabla u|^2u_L^{2(\beta-1)}{\rm d}x
\end{align*}
Since $2(\beta-1)\int_{\{0\leq u\leq L\}}|\nabla u|^2u_L^{2(\beta-1)}{\rm d}x\geq 0$,
 it follows that
\begin{equation} \label{3.3}
\begin{aligned}
&\int_\Omega|\nabla u|^2u_L^{2(\beta-1)}{\rm d}x\\
&\leq \int_\Omega\nabla u\nabla z_L{\rm d}x \\
&=\lambda\int_\Omega u^q z_L{\rm d}x+\int_\Omega
u^s z_L{\rm d}x+\mu\int_\Omega g_K(u)z_L{\rm d}x \\
&=\lambda\int_\Omega u^q u_L^{2(\beta-1)}u{\rm d}x+\int_\Omega
u^s u_L^{2(\beta-1)}u{\rm d}x
 +\mu\int_\Omega g_K(u)u_L^{2(\beta-1)}u{\rm d}x \\
&\leq \lambda\int_\Omega u^{q+1} u_L^{2(\beta-1)}{\rm d}x+\int_\Omega
u^{s+1}u_L^{2(\beta-1)}{\rm d}x
 +\mu K^{p-r+1}\int_\Omega u^r u_L^{2(\beta-1)}{\rm d}x
\end{aligned}
\end{equation}
where we have used \eqref{2.3}, \eqref{3.1} and \eqref{3.2}.
By \eqref{1.3}, we obtain
\begin{equation} \label{3.4}
\begin{aligned}
&\Big(\int_\Omega |w_L|^{2^*}\Big)^{2/2^*}{\rm d}x\\
&\leq S^{-1}\int_\Omega |\nabla w_L|^2{\rm d}x
= S^{-1}\int_\Omega |\nabla (u_L^{\beta-1}u)|^2{\rm d}x \\
&= S^{-1}\int_\Omega |(\beta-1)u u_L^{\beta-2}\nabla u_L
 +u_L^{\beta-1}\nabla u|^2{\rm d}x \\
&\leq 2S^{-1}\int_\Omega |(\beta-1)u u_L^{\beta-2}\nabla u_L|^2{\rm d}x
 +\int_\Omega |u_L^{\beta-1}\nabla u|^2{\rm d}x \\
&=2S^{-1}\int_{\{0\leq u\leq L\}}(\beta-1)^2 u_L^{2(\beta-1)}|\nabla u|^2{\rm d}x
+\int_\Omega u_L^{2(\beta-1)}|\nabla u|^2{\rm d}x \\
&\leq 2S^{-1}[(\beta-1)^2+1]\int_\Omega u_L^{2(\beta-1)}|\nabla u|^2{\rm d}x \\
&=2S^{-1}\beta^2\big[\big(\frac{\beta-1}{\beta}\big)^2+\frac{1}{\beta^2}\big]
\int_\Omega u_L^{2(\beta-1)}|\nabla u|^2{\rm d}x \\
&\leq 4S^{-1}\beta^2\int_\Omega u_L^{2(\beta-1)}|\nabla u|^2{\rm d}x.
\end{aligned}
\end{equation}
Since $u_L\leq u, 0<q<1$, we can use \eqref{3.3} and \eqref{3.4} to obtain
\begin{equation} \label{3.5}
\begin{aligned}
\Big(\int_\Omega |w_L|^{2^*}\Big)^{2/2^*}{\rm d}x
&\leq 4S^{-1}\beta^2\Big[\lambda\int_\Omega u^{q+1} u_L^{2(\beta-1)}{\rm d}x
 +\int_\Omega  u^{s+1}u_L^{2(\beta-1)}{\rm d}x \\
&\quad +\mu K^{p-r+1}\int_\Omega u^r u_L^{2(\beta-1)}{\rm d}x\Big] \\
&\leq 4S^{-1}\beta^2\Big[\lambda|\Omega|
 +\lambda\int_\Omega u^2 u_L^{2(\beta-1)}{\rm d}x+\int_\Omega
  u^{s+1}u_L^{2(\beta-1)}{\rm d}x \\
&\quad +\mu K^{p-r+1}\int_\Omega u^r u_L^{2(\beta-1)}{\rm d}x\Big]
\end{aligned}
\end{equation}
Considering the Sobolev embedding $H_0^1(\Omega))\hookrightarrow L^{2^*}(\Omega)$,
and $\|u\|\leq\gamma$ (see Lemma \ref{lem2.3}), we have
\[
S^{1/2}\Big(\int_\Omega|u|^{2^*}{\rm d}x\Big)^{1/{2^*}}
\leq\Big(\int_\Omega|\nabla u|^2{\rm d}x\Big)^{1/2}\leq\gamma,
\]
then
\begin{equation}\label{3.6}
\|u\|_{2^*}\leq\gamma S^{-1/2}.
\end{equation}
Let $\alpha^*=\frac{2^*\cdot2}{2^*-r+2}$. Since
\[
u^r u_L^{2(\beta-1)}=u^{r-2}w_L^2,~u^{s+1} u_L^{2(\beta-1)}=u^{s-1}w_L^2
\]
and $u^2 u_L^{2(\beta-1)}=w_L^2$, we now use the H\"{o}lder inequality,
\eqref{3.4}, \eqref{3.5} and \eqref{3.6}  to
conclude that, whenever $w_L\in L^{\alpha^*}(\Omega)$, it holds
\begin{equation} \label{3.7}
\begin{aligned}
\|w_L\|^2_{2^*}
&\leq 4S^{-1}\beta^2\Big[\lambda|\Omega|+\lambda\int_\Omega w_L^2{\rm d}x
+\int_\Omega u^{s-1} w_L^2{\rm d}x\\
&\quad +\mu K^{p-r+1}\int_\Omega u^{r-2} w_L^2{\rm d}x\Big] \\
&\leq  4S^{-1}\beta^2\Big[\lambda|\Omega|
 +\lambda|\Omega|^{\frac{\alpha^*-2}{\alpha^*}} \|w_L\|^2_{\alpha^*}
 +\|u\|_{2^*}^{s-1}\|w_L\|^2_{\alpha^*}\\
&\quad  +\mu K^{p-r+1}\|u\|^{2^*(1-\frac{2}{\alpha^*})}_{2^*}\|w_L\|^2_{\alpha^*}\Big] \\
&\leq  4S^{-1}\beta^2\Big[\lambda|\Omega|
 +\Big(\lambda|\Omega|^{\frac{\alpha^*-2}{\alpha^*}}+(\gamma S^{-1/2})^{s-1}\\
&\quad +\mu K^{p-r+1}(\gamma S^{-1/2})^{2^*(1-\frac{2}{\alpha^*})}\Big)
 \|w_L\|^2_{\alpha^*}\Big] \\
&\leq  4S^{-1}\beta^2\Big[2\lambda(1+|\Omega|)+\gamma^{s-1} S^{-\frac{s-1}{2}}\\
&\quad +\mu K^{p-r+1}(\gamma S^{-1/2}+1)^{2^*}\Big]
\max\{1,\|w_L\|^2_{\alpha^*}\}
\end{aligned}
\end{equation}
Set $\beta:=2^*/\alpha^*$, then $w_L\in L^{\alpha^*}(\Omega)$.
From \eqref{3.7} we have
\begin{equation}\label{3.8}
\|w_L\|^2_{2^*}\leq \beta^2 C_{\lambda,\mu,K}
\max\{1,\|w_L\|^2_{\alpha^*}\}
\end{equation}
where $C_{\lambda,\mu,K} = 4S^{-1}
\left[2\lambda(1+|\Omega|)+\gamma^{s-1} S^{-\frac{s-1}{2}}
+\mu K^{p-r+1}(\gamma S^{-1/2}+1)^{2^*}\right]$,
which is independent of $u,~\beta,~\alpha^*$ and $L$.
From \eqref{3.8} and the definition of $w_L$, we obtain
\begin{align*}
\Big(\int_\Omega u_L^{(\beta-1)2^*} u^{2^*}{\rm d}x\Big)^{2/2^*}
&\leq \beta^2 C_{\lambda,\mu,K}
 \max\Big\{1,\Big(\int_\Omega u_L^{(\beta-1)\alpha^*} u^{\alpha^*}{\rm d}x
 \Big)^{2/\alpha^*}\Big\}\\
&\leq \beta^2 C_{\lambda,\mu,K}
 \max\Big\{1,\Big(\int_\Omega u^{\beta\alpha^*}{\rm d}x
 \Big)^{2/\alpha^*}\Big\}
\end{align*}
By Fatou's Lemma,
\[
\Big(\int_\Omega u^{\beta2^*}{\rm d}x\Big)^{2/2^*}
\leq \beta^2 C_{\lambda,\mu,K}
\max\Big\{1,\Big(\int_\Omega u^{\beta\alpha^*}{\rm d}x\Big)^{2/\alpha^*}\Big\},
\]
which is equivalent to
\begin{equation}\label{3.9}
\|u\|_{\beta2^*}\leq\beta^{1/\beta}
 C_{\lambda,\mu,K}^{\frac{1}{2\beta}}\max\{1,\|u\|_{\beta\alpha^*}\}
\end{equation}
Since $\beta=\frac{2^*}{\alpha^*}>1$ and $u\in L^{2^*}(\Omega)$, the
inequality \eqref{3.9} holds for this choice of $\beta$. Now, let us
choose a sequence of positive numbers $\{\beta_m\}_m$ in the
following way:
\begin{equation}\label{3.10}
\beta_0=\beta,\quad
\beta_m=\beta^m.
\end{equation}
Noting that $\beta^2\alpha^*=\beta2^*$, we have
\begin{equation}\label{3.11}
\beta_{m+1}\alpha^*=\beta^{m+1}\alpha^*=\beta^{m-1}
\left(\beta^2\alpha^*\right)=\beta^{m-1}\cdot
\beta2^*=\beta^m2^*=\beta_m2^*.
\end{equation}
In view of \eqref{3.10} and \eqref{3.11}, we can restate  \eqref{3.9}
as
\[
\|u\|_{\beta_m\alpha^*}\leq\beta_{m-1}^{\frac{1}{\beta_{m-1}}}
C_{\lambda,\mu,K}^{\frac{1}{2\beta_{m-1}}}
\max\{1,\|u\|_{\beta_{m-1}\alpha^*}\}.
\]
Define $b_m=\max\{1,\|u\|_{\beta_m\alpha^*}\}$, then
\begin{equation} \label{3.12}
\begin{aligned}
\log b_m
&\leq \frac{1}{\beta_{m-1}}\log\beta_{m-1}+\frac{1}{2\beta_{m-1}}\log
C_{\lambda,\mu,K}+\log b_{m-1} \\
&\leq \sum_{i=1}^{m-1}\frac{\log\beta_i}{\beta_i}+\frac{\log
C_{\lambda,\mu,K}}{2}\sum_{i=1}^{m-1}\frac{1}{\beta_i}+\log b_0\\
&=\sum_{i=1}^{m-1}\frac{\log\beta^i}{\beta^i}+\frac{\log
C_{\lambda,\mu,K}}{2}\sum_{i=1}^{m-1}\frac{1}{\beta^i}
+\log\max\{1,\|u\|_{2^*}\}.
\end{aligned}
\end{equation}
Notice that
\[
\sum_{i=1}^{m-1}\frac{\log\beta^i}{\beta^i}+\frac{\log
C_{\lambda,\mu,K}}{2}\sum_{i=1}^{m-1}\frac{1}{\beta^i}\to
C_\beta+C'_\beta\log C_{\lambda,\mu,K}:=C_0
\]
as $m\to\infty$, with
\[
C_\beta=\sum_{i=1}^{\infty}\frac{\log\beta^i}{\beta^i},\quad
2C'_\beta=\sum_{i=1}^{\infty}\frac{1}{\beta^i},~\beta>1.
\]
Taking the limit as $m\to\infty$ in \eqref{3.12}, and using \eqref{3.6},
we deduce that
\[
\|u\|_\infty\leq e^{C_0}\max\{1,\|u\|_{2^*}\}
\leq e^{C_0}\max\{1,\gamma S^{-1/2}\}.
\]
We should pay attention that $C_0$ depends on $\lambda,\mu,K, |\Omega|, S, \gamma$
and control the dependence of $C_0$ on $|\Omega|,S$ and $\gamma$.
Now, to prove our theorem, we need choose suitable value of
$\lambda,\mu,K$ carefully, such that
\begin{equation}\label{3.13}
e^{C_0}\max\big\{1,\gamma S^{-1/2}\big\}
=e^{C_\beta+C'_\beta\log C_{\lambda,\mu,K}}\max\{1,\gamma S^{-1/2}\}\leq K.
\end{equation}
this is equivalent to
\[
C^{C'_\beta}_{\lambda,\mu,K}e^{C_\beta}\max\{1,\gamma S^{-1/2}\}\leq K.
\]
That is,
\begin{align*}
&\Big[4S^{-1}(1+|\Omega|)(2\lambda+\gamma^{s-1} S^{-\frac{s-1}{2}}) \\
&+\mu K^{p-r+1}(\gamma S^{-1/2}+1)^{2^*}\Big]^{C'_\beta}
e^{C_\beta}\max\{1,\gamma S^{-1/2}\}\leq K.
\end{align*}
Choose $K>0$ to satisfy the inequality (note that $\lambda\leq\lambda_0$)
\begin{equation}\label{3.14}
\Big(\frac{K}{e^{C_\beta}\max\{1,\gamma S^{-1/2}\}}\Big)^{1/C'_\beta}
-4S^{-1}(1+|\Omega|)(2\lambda+\gamma^{s-1} S^{-\frac{s-1}{2}})>0,
\end{equation}
and then fix $\mu_K$ such that
\begin{align*}
\mu_K
&:=\frac{1}{K^{p-r+1}(\gamma S^{-1/2}+1)^{2^*}}
\Big[\Big(\frac{K}{e^{C_\beta}\max\{1,\gamma S^{-1/2}\}}\Big)^{1/C'_\beta}\\
&\quad- 4S^{-1}(1+|\Omega|)(2\lambda+\gamma^{s-1} S^{-\frac{s-1}{2}})\Big].
\end{align*}
Let $\mu^*:=\min\{\mu_0,\mu_K\}$, we obtain \eqref{3.13} for $\mu\in[0,\mu^*]$
and some $K$ satisfying \eqref{3.14}.
This completes the proof.
\end{proof}

Since $u_i\in L^{\infty}(\Omega),~i=1, 2$, using bootstrap technique, 
we obtain $u_i\in C^{2,\alpha}(\Omega)$, $i=1, 2$ for some constant $0<\alpha<1$.

\begin{corollary} \label{coro3.1}
The solutions obtained in Theorem \ref{thm2.2} are smooth; i.e., 
$u_i$ belongs to $ C^{2,\alpha}(\overline{\Omega})$, $i=1, 2$ for some 
constant $0<\alpha<1$.
\end{corollary}

\begin{remark} \rm
Our method could be generalized to obtain analogous results for
equations with more general perturbation $h(x,u)$, i.e.
\begin{equation}\label{3.15}
\begin{gathered}
-\Delta u=\lambda u^q+u^s+\mu h(x,u)\quad\text{in }\Omega,\\
u>0\quad\text{in }\Omega,\\
u=0\quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
where $0<q<1<s<2^*-1$, $h(x,t)\geq 0 $ for $t\geq 0$ and satisfies the 
growth condition $|h(x,t)|\leq C_0\left(1+|t|^{p-1}\right)$, 
$p\geq 2^*$ and $C_0>0$ is a constant.
\end{remark}

We have the following result similar to Theorem \ref{thm1.1}.

\begin{theorem}
Problem \eqref{3.15} has at least two positive solutions for
$\lambda$ and $\mu$ small enough.
\end{theorem}

\begin{proof}
In fact, the truncation of $h(x,t)$ can be given by
\begin{equation}\label{3.16}
h_K(x,t)=\begin{cases}
h(x,t),&|t|\leq K,\\
\min\{h(x,t),C_0(1+K^{p-r}t^{r-1})\},&|t|>K,
\end{cases}
\end{equation}
where $r\in(2,2^*)$. Then $h_K$ satisfies
\begin{equation}\label{3.17}
|h_K(x,t)|\leq C_0 (1+K^{p-r}|t|^{r-1}).
\end{equation}
The truncated problem associated to problem \eqref{3.15} becomes
\begin{equation}\label{3.18}
\begin{gathered}
-\Delta u=\lambda u^q+u^s+\mu h_K(x,u)\quad\text{in }\Omega,\\
u>0\quad\text{in }\Omega,\\
u=0\quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
By \eqref{3.16}--\eqref{3.18} and a technique similar to the one in
Theorem \ref{thm1.1}, we can prove that the two solutions (one is a local
minimum, the other is of Mountain Pass type) for truncated problem
\eqref{3.18} satisfy $\|u_i\|\leq K$, $i=1,2$. In view of the
definition of $h_K$, we know that $u_1$ and  $u_2$ are also solutions of the
original problem \eqref{3.15}.
\end{proof}


\subsection*{Acknowledgements}
This work is supported by the NSF of China (No. 11471147 and 11201206),
and by the Fundamental Research
Funds for the Central Universities (No. lzujbky-2014-25).


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\end{document}