\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 89, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/89\hfil Singular elliptic boundary-value problems]
{Existence, boundary behavior and asymptotic behavior of solutions
to singular elliptic boundary-value problems}

\author[G. Gao, B. Yan \hfil EJDE-2016/89\hfilneg]
{Ge Gao, Baoqiang Yan}

\address{Ge Gao \newline
School of Mathematical Sciences,
Shandong Normal University,
Jinan 250014, China}
\email{gaoge\_jianxin@163.com}

\address{Baoqiang Yan \newline
School of Mathematical Sciences,
Shandong Normal University,
Jinan 250014, China}
\email{yanbqcn@aliyun.com}

\thanks{Submitted November 6, 2015. Published March 31, 2016.}
\subjclass[2010]{35J25, 35J60, 35J75}
\keywords{Elliptic boundary value problem; existence; 
boundary behavior; \hfill\break\indent  asymptotic behavior}

\begin{abstract}
In this article, we consider the singular elliptic boundary-value problem
$$
-\Delta u+f(u)-u^{-\gamma} =\lambda u \text{ in } \Omega,\quad u>0\text{ in }
 \Omega,\quad u=0 \text{ on } \partial\Omega.
$$
Using the upper-lower solution method, we show the existence and uniqueness of
the solution. Also we study the boundary behavior and asymptotic behavior
of the positive solutions.
\end{abstract}

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

\section{Introduction}

In this article, we consider the singular elliptic boundary-value problem
\begin{equation}
\begin{gathered}
-\Delta u+f(u)-u^{-\gamma} =\lambda u \quad\text{in } \Omega,\\
 u>0 \quad\text{in } \Omega,\\
 u=0 \quad\text{on } \partial\Omega,
\end{gathered} \label{e1.1}
\end{equation}
where $\Omega$ is a bounded domain in $\mathbb{R}^{N}$, $N\geq2$, with $C^{2,\beta}$
boundary $\partial\Omega$, $\gamma>0$, $\beta\in(0,1)$ and $\lambda>0$
is a real parameter. We use the following assumptions in this article.
\begin{itemize}
\item[(A1)] $f: \mathbb{R}^{+}\to \mathbb{R}$ is a continuous function.

\item[(A2)] $s^{-1}{f(s)}$ is strictly increasing for $s>0$.

\item[(A3)] $f: \mathbb{R}^{+}\to \mathbb{R}$ is strictly increasing.
\end{itemize}
Existence, boundary behavior and asymptotic behavior of solutions for 
nonlinear elliptic boundary value problems have been intensively 
studied in the previous decades. 
 Berestycki [1] considered the problem
\begin{equation}
\begin{gathered}
Lu+f(x,u)=\lambda au ,\quad\text{in } \Omega,\\
u=0,\quad \text{on } \partial\Omega,
\end{gathered} \label{e1.2}
\end{equation}
where $L$ is a second order uniformly elliptic operator,
$a\in C^{0}(\bar{\Omega})$, $a>0$ in $\bar{\Omega}$ and under the conditions
that $f(x,0)=0$, $s \to (f(x,s)/s)$ is strictly increasing for $s>0$ and
$ \lim_{s\to\infty}(f(x,s)/s)=+\infty$, it is proved that \eqref{e1.2}
has one and only one positive solution $u_{\lambda}\in W^{2,p}(\Omega)$
for $\lambda>\lambda_1$, where $\lambda_1$ denotes the principle eigenvalue of
$$
L\phi=\lambda a\phi \text{ in } \Omega, \quad
\phi=0 \text{ on } \partial\Omega.
$$
As a special case for \eqref{e1.2},  Fraile, L\'{o}pez-G\'{o}mez and Delis \cite{f1}
studied
\begin{equation}
\begin{gathered}
-\Delta u=\lambda u+f(u)-u^{p+1} \quad\text{in } \Omega,\\
 u>0\quad \text{in } \Omega,\\
 u=0\quad \text{on } \partial\Omega,
\end{gathered} \label{e1.3}
\end{equation}
and described the structure of the positive solutions of \eqref{e1.3}
in detail. Some other asymptotic behavior studies have been shown in
\cite{s3,s4,s5}.

Notice that in the above $f$ needs to be continuous or differential at $x=0$. 
On the other hand, singular elliptic boundary value problems in various 
forms have been studied extensively by many authors,
see \cite{c1,c2,c3,d1,f2,g1,g3,h1,l1,s2,s6,z1,z2}.
 For instance, for the problem
\begin{equation}
\begin{gathered}
-\Delta u-u^{-\alpha} =(\lambda u)^{p} \quad\text{in } \Omega,\\
 u>0\quad \text{in } \Omega,\\
 u=0 \quad \text{on } \partial\Omega,
\end{gathered} \label{e1.4}
\end{equation}
by results in \cite{c3,f2,g1,s6}, it follows that if $\lambda=0$, \eqref{e1.4}
 possesses a unique classical solution and at least one classical solution
for $\lambda\neq0$ and $p<1$.  Coclite and Palmieri \cite{c2}  showed that
if $0<p<1$, \eqref{e1.4} has at least one solution $u_{\lambda}$ for all
$\lambda\geq0$; and if $p\geq1$, there exist $\tilde{\lambda}\in(0,+\infty]$
such that \eqref{e1.4} has at least one classical solution for all
$\lambda\in[0,\tilde{\lambda})$ and has no solution for
$\tilde{\lambda}<\lambda$. See also results in \cite{c1,l1,z2}.
A more generalized work in \cite{s2} is about for the singular problem
\begin{equation}
\begin{gathered}
-\Delta u+K(x)u^{-\alpha}=\lambda u^{p} \quad\text{in } \Omega,\\
 u>0 \quad\text{in } \Omega,\\
 u=0 \quad\text{on } \partial\Omega\,.
\end{gathered}\label{e1.5}
\end{equation}
If $K(x)<0$ for all $x\in\bar{\Omega}$, then \eqref{e1.5} has one and only
 one solution $u_{\lambda}$ for any $\lambda\in\mathbb{R}$,
$c_1d(x)\leq u_{\lambda}(x)\leq c_{2}d(x)$ for any
 $x\in\bar{\Omega}$ and some $c_1, c_{2}>0$ independent of $x$;
if $K(x)>0$ for all $x\in\bar{\Omega}$, there exist a $\tilde{\lambda}>0$
such that \eqref{e1.5} has at least one solution $u_{\lambda}$ for
$\lambda>\tilde{\lambda}$, and \eqref{e1.5} has no solution for
$\lambda<\tilde{\lambda}$. Other results can be found in
\cite{c3,d1,g3,l1,z1}.

Up to now, there are only a few results on existence, boundary behavior and 
asymptotic behavior of positive solutions  for \eqref{e1.1}.
Our goal in this paper is to show existence, boundary behavior and 
asymptotic behavior of the solutions for singular elliptic boundary-value 
problem \eqref{e1.1}. Using the upper-lower solution method, we obtain
that \eqref{e1.1} has at least one solution, and if $0<\gamma<1$, \eqref{e1.1}
has one and only one solution. In the meanwhile, the boundary behavior of the 
solution is established for $0<\gamma<1$. Finally, we obtain the asymptotic 
behavior of solutions under a special form of $f(u)$.

\section{Existence and uniqueness of a solution for \eqref{e1.1}}

First, we introduce notation and present some lemmas. 
In the next lemma, $W^{k,q}(\Omega)$ denotes the usual Sobolev space.


\begin{lemma}[\cite{c3}] \label{lem2.1}
Let $\vartheta_0$, $\vartheta$ be bounded open domains in $\mathbb{R}^{n}$ with
$\bar{\vartheta_0}\subset \vartheta$. Suppose $L$ is a second order
uniformly elliptic operator with coefficients continuous in 
$\bar{\vartheta}$ and $q>n$. Then there is a constant $K$ such that
$$
\|w\|_{W^{2,q}(\vartheta_0)}\leq K(\|Lw\|_{L^{q}(\vartheta)}
+\|w\|_{L^{q}(\vartheta)})
$$
for all $w\in W^{2,q}(\vartheta)$. The constant $K$ depends on $n,q$, 
the diameter of $\vartheta$, the distance from $\vartheta_0$ to
$\partial\vartheta$, the ellipticity constant of $L$, and bounds 
for the coefficients of $L$ (in $L^{\infty}(\vartheta)$) and the 
moduli of continuity of the coefficients.
\end{lemma}

We consider the nonlinear elliptic boundary-value problem
\begin{equation}
\begin{gathered}
Lu+f(x,u)=0 \quad\text{in } \Omega,\\
Bu=g \quad \text{on }\partial\Omega,
\end{gathered}
\end{equation}
where $L$ is a second order uniformly elliptic operator
$$
L=\sum_{i,j=1}^{n}a_{ij}\frac{\partial^{2}u}{\partial x^{i}\partial x^{j}}
+\sum_{i=1}^{n}b_{i}\frac{\partial u}{\partial x^{i}}, \quad x=(x^{1},\dots,x^{n}),
$$
and $B$ is one of the boundary operators
\[
Bu=u,
\]
or
\[
Bu=\frac{\partial u}{\partial v}+\beta(x)u, \quad x\in\partial \Omega.
\]
Here $\partial/\partial v$ denotes the outward conormal derivative, and we
assume $\beta\geq0$ everywhere on the boundary, $\partial \Omega$.

\begin{lemma}[\cite{s1}] \label{lem2.2}
Let there exist two smooth functions $u_0(x)\geq v_0(x)$ such that
\begin{gather*}
Lu_0+f(x,u_0)\leq0,\quad Bu_0\geq g, \\
Lv_0+f(x,v_0)\geq0,\quad Bv_0\leq g.
\end{gather*}
Assume $f$ is a smooth function on $\min v_0 \leq u\leq \max u_0 $.
Then there exists a regular solution $w$ of
$$
Lw+f(x,w)=0,\quad Bw=g,
$$
such that $v_0\leq w \leq u_0$.
\end{lemma}

Let $\phi_1$ denote an eigenfunction corresponding to the first eigenvalue
$\lambda_1$ of
\begin{gather*}
-\Delta u=\lambda u \quad\text{in } \Omega,\\
u=0 \quad\text{on }\partial\Omega.
\end{gather*}
As is known, $\phi_1$ belongs to $C^{2,\beta}(\bar{\Omega})$, $\phi_1>0$ 
in $\Omega$ and $\lambda_1>0$.

\begin{lemma}[\cite{l1}] \label{lem2.3}
 $\int_{\Omega}(\frac{1}{\phi_1})^{s}dx<\infty$ if and only if $s<1$.
\end{lemma}

Assume that \eqref{e1.1} has a positive solution $u_{\lambda}$ and let
$x_0\in\Omega$ be the point where $u_{\lambda}$ reaches its maximum. 
Thus, $-\Delta u_{\lambda}(x_0)\geq 0$, which concludes that
$$
\lambda\geq -u_{\lambda}^{-(\gamma+1)}(x_0)
+\frac{f(u_{\lambda}(x_0))}{u_{\lambda}(x_0)}\,.
$$
Define
$$
G(s)=-s^{-(\gamma+1)}+\frac{f(s)}{s}, s>0.
$$

\begin{theorem} \label{thm2.1}
 If $f(u)$ satisfies {\rm (A1)} and {\rm (A2)}, then \eqref{e1.1}
has at least one solution $u_{\lambda}\in C(\bar{\Omega})\cap C^{2,\beta}(\Omega)$ 
for $\lambda>\lambda_1$, and there exist a constant $C>0$ such that 
$ u_{\lambda}\geq C\phi_1$. Moreover, if $0<\gamma<1$,
\eqref{e1.1} has one and only one solution for $\lambda>\lambda_1$.
\end{theorem}

\begin{proof} 
 (i) (Existence) First we consider the solution of the  nonlinear elliptic 
bound\-ary-value problem
\begin{equation}
\begin{gathered}
 -\Delta u+f(u)-u^{-\gamma}=\lambda u \quad\text{in } \Omega,\\
 u>0 \quad\text{in } \Omega,\\
 u=1/k\quad \text{on }\partial\Omega,
 \end{gathered}\label{Plk}
\end{equation}
where $1/k<G^{-1}(\lambda)$, $k\in N_{+}$. Set
$$
\overline{u_{\lambda}^{(k)}}(x)=G^{-1}(\lambda),
$$
then
\begin{align*}
-\Delta\overline{u_{\lambda}^{(k)}}+f(\overline{u_{\lambda}^{(k)}})
-\overline{u_{\lambda}^{(k)}}^{-\gamma}
&=f(G^{-1}(\lambda))-(G^{-1}(\lambda))^{-\gamma}\\
&=G^{-1}(\lambda)[\frac{f(G^{-1}(\lambda))}{G^{-1}(\lambda)}
 -(G^{-1}(\lambda))^{-(\gamma+1)}]\\
&=G^{-1}(\lambda)G(G^{-1}(\lambda))\\
&=\lambda G^{-1}(\lambda)=\lambda\overline{u_{\lambda}^{(k)}}.
\end{align*}
So
\begin{gather*}
-\Delta\overline{u_{\lambda}^{(k)}}+f(\overline{u_{\lambda}^{(k)}})
 -\overline{u_{\lambda}^{(k)}}^{-\gamma}
\geq\lambda\overline{u_{\lambda}^{(k)}} \quad \text{in }\Omega,\\
\overline{u_{\lambda}^{(k)}}=G^{-1}(\lambda) \quad \text{on }\partial\Omega.
\end{gather*}
Therefore, $\overline{u_{\lambda}^{(k)}}(x)$ is a super-solution of
 \eqref{Plk}. Put
$$
\underline{u_{\lambda}^{(k)}}(x)=h_{k}G^{-1}(\lambda-\lambda_1)
(\phi_1(x)+\delta_{k}),
$$
where $\delta_{k}=1/k$,
$h_{k}=1/(\delta_{k}+1)$, $h_{k}G^{-1}(\lambda-\lambda_1)\delta_{k}\leq1/k$. Since
$$
\underline{u_{\lambda}^{(k)}}\leq 1/k \quad \text{on } \partial\Omega,
$$
and
\begin{align*}
& -\Delta\underline{u_{\lambda}^{(k)}}+f(\underline{u_{\lambda}^{(k)}})
-\underline{u_{\lambda}^{(k)}}^{-\gamma} \\
&=\lambda_1h_{k}G^{-1}(\lambda-\lambda_1)\phi_1(x)
 +f(h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k}))\\
&\quad -[h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k})]^{-\gamma}\\
&\leq h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k})
\big\{\lambda_1+\frac{f(h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)
 +\delta_{k}))}{h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k})}\\
&\quad -[h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k})]^{-(\gamma+1)}\big\} \\
&\leq h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k})
 \big\{\lambda_1+\frac{f(G^{-1}(\lambda-\lambda_1))}{G^{-1}
 (\lambda-\lambda_1)}-[G^{-1}(\lambda-\lambda_1)]^{-(\gamma+1)}\big\}\\
&=  h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k})
 (\lambda_1+\lambda-\lambda_1)\\
&=\lambda h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k}) \\
&=\lambda\underline{u_{\lambda}^{(k)}},
\end{align*}
we obtain that $\underline{u_{\lambda}^{(k)}}(x)$ is a sub-solution of \eqref{Plk}.

When $k=1$, standard elliptic arguments imply that there exist a solution
 $u_{\lambda}^{(1)}\in C(\bar{\Omega})\cap C^{2,\beta}(\Omega)$ such that 
$$
\underline{u_{\lambda}^{(1)}}\leq u_{\lambda}^{(1)}
\leq\overline{ u_{\lambda}^{(1)}}, \quad\text{in}\; \bar{\Omega}.
$$
Now taking $u_{\lambda}^{(1)}(x)$ and $\underline{u_{\lambda}^{(2)}}(x)$ 
as a pair of super and sub-solutions for \eqref{Plk}, we obtain 
a solution $u_{\lambda}^{(2)}\in C(\bar{\Omega})\cap C^{2,\beta}(\Omega)$ such that
$$
\underline{u_{\lambda}^{(2)}}\leq u_{\lambda}^{(2)}\leq u_{\lambda}^{(1)}
\quad\text{in}\; \bar{\Omega}.
$$
In this manner we find a sequence $\{u_{\lambda}^{(k)}\}$, such that
$$
\underline{u_{\lambda}^{(k)}}\leq u_{\lambda}^{(k)} 
\leq u_{\lambda}^{(k-1)}\leq\overline{ u_{\lambda}^{(1)}} \quad \text{in }
 \bar{\Omega}.
$$
Therefore $\lim_{k\to\infty}u_{\lambda}^{(k)}(x)=u_{\lambda}(x)$ 
exists everywhere in $\bar{\Omega}$, where $u_{\lambda}>0$ in $\Omega$ 
and $u_{\lambda}=0$ on $\partial\Omega$. It remains to see that 
$u_{\lambda}\in C^{2}(\Omega)$ and
$$
-\Delta u_{\lambda}+f(u_{\lambda})-u_{\lambda}^{-\gamma}
=\lambda u_{\lambda},\quad \text{in } \Omega.
$$
Choose open subsets $\vartheta_1,\vartheta_{2}$ of $\Omega$ so that 
$\bar{\vartheta_{2}}\subset\vartheta_1\subset\bar{\vartheta_1}\subset\Omega$ 
and $q>N$. By Lemma 2.1 there is a constant 
$K=K(N,q,\vartheta_1,\vartheta_{2},\Delta)$ such that
\begin{align*}
\|u_{\lambda}^{(k)}\|_{W^{2,q}(\vartheta_{2})}
&\leq K(\|\Delta u_{\lambda}^{(k)}\|_{L^{q}(\vartheta_1)}
 +\|u_{\lambda}^{(k)}\|_{L^{q}(\vartheta_1)})\\
&= K(\|f(u_{\lambda}^{(k)})-(u_{\lambda}^{(k)})^{-\gamma}
 -\lambda u_{\lambda}^{(k)}\|_{L^{q}(\vartheta_1)}
 +\|u_{\lambda}^{(k)}\|_{L^{q}(\vartheta_1)}),
\end{align*}
which implies that $\{u_{\lambda}^{(k)}|\,k\in N_{+}\}$ is bounded 
in $W^{2,q}_{\rm loc}(\Omega)$. Therefore $u_{\lambda}^{(k)}\to u_{\lambda}$ 
weakly in $W^{2,q}_{\rm loc}(\Omega)$. Choose $\alpha\in(0,1)$ and 
$q>N(1-\alpha)^{-1}$. If follows from the Sobolev embedding theorems that 
$\{u_{\lambda}^{(k)}\}$ is compact in $C^{1,\alpha}_{\rm loc}(\Omega)$. 
Thus we have $u_{\lambda}\in W^{2,p}_{\rm loc}(\Omega)$. 
The $L^{q}$ regularity theory for $\Delta$ now implies 
$u_{\lambda}\in W^{3,q}_{\rm loc}(\Omega)$ and hence 
$u_{\lambda}\in C^{2,\alpha}_{\rm loc}(\Omega)\subset C^{2}(\Omega)$. 
Notice that Lemma 2.2 ensures that $u_{\lambda}^{(k)}(x)$ is a solution 
of \eqref{Plk}, so
$$
-\Delta u_{\lambda}^{(k)}+f(u_{\lambda}^{(k)})-(u_{\lambda}^{(k)})^{-\gamma}
=\lambda u_{\lambda}^{(k)}\quad \text{in } \Omega.
$$
In conjunction with the results above, let $k\to\infty$, therefore
$$
-\Delta u_{\lambda}+f(u_{\lambda})-u_{\lambda}^{-\gamma}
=\lambda u_{\lambda} \quad \text{in } \Omega.
$$
Consequently, \eqref{e1.1} has at least one solution
$u_{\lambda}\in C(\bar{\Omega})\cap C^{2,\beta}(\Omega)$ for  $\lambda>\lambda_1$.

Furthermore, since $\lim_{k\to\infty}u_{\lambda}^{(k)}(x)=u_{\lambda}(x)$ and
$$
h_{k}G^{-1}(\lambda-\lambda_1)(\phi_1(x)+\delta_{k})
=\underline{u_{\lambda}^{(k)}}(x)\leq u_{\lambda}^{(k)}(x),\quad \text{in }\Omega,
$$
$G^{-1}(\lambda-\lambda_1)>0$, $\delta_{k}=1/k\to0,h_{k}=1/(\delta_{k}+1)\to1$ 
as $k\to\infty$, so there exist a constant $C>0$ such that 
$C\phi_1\leq u_{\lambda}$ in $\Omega$.
\smallskip

(ii) (Uniqueness)  If $0<\gamma<1$, we put
$$
h(s)=\lambda s-f(s)+s^{-\gamma},\quad s>0,
$$
hence $s^{-1}h(s)$ is strictly decreasing for $s>0$.

Assume that $v(x)$ is another solution of \eqref{e1.1}.
 Then we argue by contradiction. Notice that
$$
\Delta u+h(u)=0,\quad  \Delta v+h(v)=0,
$$
and $u, v>0$ in $\Omega$, $u=v$ on $\partial\Omega$. 
Since $C\phi_1(x)\leq u_{\lambda}(x)$, it is easy to see that 
$\Delta u\in L^{1}(\Omega)$ by Lemma 2.3. 
If $u(x)=v(x)$ is not true, we can assume that $u(x)>v(x)$, then there 
exist $\varepsilon_0,\delta_0>0$, and a ball $B\Subset\Omega$ such that
\begin{gather*}
u(x)-v(x)>\varepsilon_0, \quad x\in B, \\
\int_{B}uv(\frac{h(v)}{v}-\frac{h(u)}{u})dx>\delta_0.
\end{gather*}
Let
$$
M=\max\{1,\|\Delta u\|_{L^{1}(\Omega)}\},\quad
\varepsilon =\min\{1,\varepsilon_0,\frac{\delta_0}{4M}\},
$$
and $\theta$ be a smooth function on $R$, such that $\theta(t)=0$ 
if $t\leq \frac{1}{2}$; $\theta(t)=1$ if $t\geq1$; 
$\theta(t)\in(0,1)$ if $t\in(\frac{1}{2},1)$, $\theta'(t)\geq0$ for $t\in\mathbb{R}$.
Then for $\varepsilon>0$, define the function $\theta_{\varepsilon}(t)$ by
$$
\theta_{\varepsilon}(t)=\theta(\frac{t}{\varepsilon}), \quad t\in\mathbb{R}.
$$
It follows from  $\theta_{\varepsilon}(t)\geq0$ for $t\in\mathbb{R}$ 
that
$$
(v\Delta u-u\Delta v)\theta_{\varepsilon}(u-v)
=uv(\frac{h(v)}{v}-\frac{h(u)}{u})\theta_{\varepsilon}(u-v) \quad
\text{in}\; \Omega.
$$
On the other hand, by the continuity of $u,v$ and $\theta_{\varepsilon}$, 
and the fact that $u=v$ on $\partial\Omega$. It is easy to see that there 
exist a subdomain $\hat{\Omega}$ such that 
$B\subset\hat{\Omega}\Subset \Omega$ satisfying that 
$u(x)-v(x)<\frac{\varepsilon}{2}$ for all $x\in\Omega\backslash \hat{\Omega}$. 
Then 
$$
\int_{\hat{\Omega}}(v\Delta u-u\Delta v)\theta_{\varepsilon}(u-v)dx
=\int_{\hat{\Omega}}uv(\frac{h(v)}{v}-\frac{h(u)}{u})\theta_{\varepsilon}(u-v)dx.
$$
Denote
$$
\Theta_{\varepsilon}(t)=\int_0^{t}s \theta'_{\varepsilon}(s)ds,\quad
t\in\mathbb{R}\,.
$$
It is easy to verify that
$0\leq\Theta_{\varepsilon}(t)\leq2\varepsilon$ for $t\in\mathbb{R}$,
and $\Theta_{\varepsilon}(t)=0$ for $t<\frac{\varepsilon}{2}$.
Therefore
\begin{align*}
&\int_{\hat{\Omega}}(v\Delta u-u\Delta v)\theta_{\varepsilon}(u-v)dx \\
&=\int_{\partial\hat\Omega}v\theta_{\varepsilon}(u-v)
 \frac{\partial u}{\partial n}ds
 -\int_{\hat{\Omega}}(\nabla u\cdot\nabla v)\theta_{\varepsilon}(u-v)dx \\
&\quad -\int_{\hat{\Omega}}v\nabla u\theta'_{\varepsilon}(u-v)
 (\nabla u-\nabla v)dx-\int_{\partial\hat\Omega}u\theta_{\varepsilon}(u-v)
 \frac{\partial v}{\partial n}ds\\
&\quad +\int_{\hat{\Omega}}(\nabla v\cdot\nabla u)\theta_{\varepsilon}(u-v)dx
+\int_{\hat{\Omega}}u\nabla v\theta'_{\varepsilon}(u-v)(\nabla u-\nabla v)dx\\
&=\int_{\hat{\Omega}}u\theta'_{\varepsilon}(u-v)(\nabla v-\nabla u)
 (\nabla u-\nabla v)dx
+\int_{\hat{\Omega}}(u-v)\theta'_{\varepsilon}(u-v)
 \nabla u(\nabla u-\nabla v)dx\\
&\leq\int_{\hat{\Omega}}\nabla u\nabla(\Theta'_{\varepsilon}(u-v))dx \\
&=\int_{\partial\hat{\Omega}}\Theta_{\varepsilon}(u-v)
 \frac{\partial u}{\partial n}ds-\int_{\hat{\Omega}}
 \Delta u\Theta_{\varepsilon}(u-v)dx\\
&\leq 2\varepsilon\int_{\hat{\Omega}}|\Delta u|dx \\
&\leq2\varepsilon M<\frac{\delta_0}{2}.
\end{align*}
However,
\begin{align*}
\int_{\hat{\Omega}}uv(\frac{h(v)}{v}-\frac{h(u)}{u})
\theta_{\varepsilon}(u-v)dx
&\geq\int_{B}uv(\frac{h(v)}{v}
-\frac{h(u)}{u})\theta_{\varepsilon}(u-v)dx\\
&=\int_{B}uv(\frac{h(v)}{v}-\frac{h(u)}{u})dx
>\delta_0,
\end{align*}
which is a contradiction. Thus the uniqueness is proved.
\end{proof}

Our method to prove the uniqueness of the solution is similar 
to and motivated by the proof of  Shi and Yao \cite[Lemma 2.3]{s2}.

\section{The boundary behavior of the solution to \eqref{e1.1}}

\begin{theorem} \label{thm3.1}
 If $f(u)$ satisfies {\rm (A1)--(A3)} and $0<\gamma<1$, then the solution
 $u_{\lambda}$ of \eqref{e1.1} is strictly increasing with respect to $\lambda$.
Furthermore, there exist two positive constants $c_1, c_{2}>0$ depending 
on $\lambda$ such that $c_1d(x)\leq u_{\lambda}(x)\leq c_{2}d(x)$ in $\Omega$.
\end{theorem}

\begin{proof}  (i) (Dependence on $\lambda$) 
We assume $0<\lambda_1<\lambda_{2}$, and $u_{\lambda_1}$, $u_{\lambda_{2}}$ 
are corresponding unique solution to $\eqref{e1.1}$. Since
$C\phi_1(x)\leq u_{\lambda}(x)$, it is easy to see that 
$\Delta u_{\lambda}\in L^{1}(\Omega)$ by Lemma \ref{lem2.3}. Thus,
\begin{align*}
0&=\Delta u_{\lambda_{2}}-f(u_{\lambda_{2}})
 +u_{\lambda_{2}}^{-\gamma}+\lambda_{2}u_{\lambda_{2}}\\
 &=\Delta u_{\lambda_1}-f(u_{\lambda_1})+u_{\lambda_1}^{-\gamma}
 +\lambda_1u_{\lambda_1}\\
 &<\Delta u_{\lambda_1}-f(u_{\lambda_1})+u_{\lambda_1}^{-\gamma}
 +\lambda_{2}u_{\lambda_1}
\end{align*}
for $x\in\Omega$ and $u_{\lambda_1}(x)=u_{\lambda_{2}}(x)$ 
on $\partial\Omega$. Therefore, similar to the proof of Theorem \ref{thm2.1} (ii),
$$
u_{\lambda_1}(x)\leq u_{\lambda_{2}}(x),\quad x\in\bar{\Omega}.
$$
Moreover, by the maximum principle,
$$
u_{\lambda_1}(x)< u_{\lambda_{2}}(x),\quad x\in\Omega.
$$
So $u_{\lambda}$ is increasing with respect to $\lambda$.
\smallskip

(ii) (Bounds for the solution) 
Fix $\lambda>0$, let $u_{\lambda}$ be the unique solution of \eqref{e1.1}.
There exists a unique nonnegative solution $\xi\in C^{2,\beta}(\bar{\Omega})$ of
\begin{gather*}
-\Delta\xi=1 \quad \text{in } \Omega,\\
\xi=0 \quad \text{on }\partial\Omega.
\end{gather*}
By the weak maximum principle (see \cite{g2}), $\xi>0$ in $\Omega$. Put $z(x)=c\xi(x)$. 
Consider that we can find $\check{c}>0$ small enough such that
$$
\lambda z(x)+(z(x))^{-\gamma}
\geq \lambda\check{ c}\|\xi\|_{\infty}+(\check{c}\|\xi\|_{\infty})^{-\gamma},
$$
and $\hat{c}>0$ small enough such that
$$
\hat{c}[\frac{-f(\hat{c}\|\xi\|_{\infty})
 +(\hat{c}\|\xi\|_{\infty})^{-\gamma}}{2\hat{c}}-1]
 + \hat{c}\|\xi\|_{\infty}[\frac{-f(\hat{c}\|\xi\|_{\infty})
 +(\hat{c}\|\xi\|_{\infty})^{-\gamma}}{2\hat{c}\|\xi\|_{\infty}}+\lambda]\geq0\,.
$$
Select $0<c<\text{min}\{\check{c},\hat{c}\}$. Then
\begin{align*}
&\Delta z(x)+\lambda z(x)-f(z(x))+(z(x))^{-\gamma}\\ 
&\geq -c+\lambda c\|\xi\|_{\infty}-f(c\|\xi\|_{\infty})+(c\|\xi\|_{\infty})^{-\gamma}\\ \\
&= c[\frac{-f(c\|\xi\|_{\infty})+(c\|\xi\|_{\infty})^{-\gamma}}{2c}-1]\\
&\quad + c\|\xi\|_{\infty}[\frac{-f(c\|\xi\|_{\infty})
 +(c\|\xi\|_{\infty})^{-\gamma}}{2c\|\xi\|_{\infty}}+\lambda]\\
&\geq 0
\end{align*}
for each $x\in\Omega$. Therefore, $z(x)$ is a sub-solution of \eqref{e1.1}
for $c>0$ small enough. Since $\xi\in C^{2,\beta}(\bar{\Omega})$, $\xi>0$ in 
$\Omega$, and $\xi=0$ on $\partial\Omega$, by 
Gilbarg and Trudinger \cite[Lemma 3.4]{g2},
$$
 \frac{\partial\xi}{\partial\nu}(y)<0, \quad\forall y\in\partial\Omega.
$$
Therefore, there exist a positive constant $c_0$ such that
$$
 \frac{\partial\xi}{\partial\nu}(y)
=\lim_{x\in\Omega,x\to y}\frac{\xi(y)-\xi(x)}{|x-y|}\leq -c_0,\quad 
\forall y\in\partial\Omega.
$$
So for each $y\in\Omega$, there exist $r_{y}>0$, such that
\begin{equation}
\frac{\xi(x)}{|x-y|}\geq\frac{c_0}{2}, \quad\forall x\in B_{r_{y}}(y)\cap\Omega.
\label{e3.1}
\end{equation}
Using the compactness of $\partial\Omega$, we can find a finite number $k$
of balls $B_{ry_{i}}(y_{i})$ such that
$$
\partial\Omega\subset\cup_{i=1}^{k}B_{ry_{i}}(y_{i}).
$$
Moreover, assume that for small $d_0>0$,
$$
\{x\in\Omega:d(x)<d_0\}\subset\cup_{i=1}^{k}B_{ry_{i}}(y_{i}).
$$
By \eqref{e3.1} we obtain
$$
\xi(x)\geq\frac{c_0}{2} d(x),\quad \forall x\in\Omega \text{ with } d(x)<d_0.
$$
This fact, combined with $\xi>0$ in $\Omega$, shows that for some
constant $\tilde{c}>0$,
$$
\xi(x)\geq \tilde{c}d(x),\quad \forall x\in \Omega.
$$
Thus, $z(x)\geq c_1d(x)$ in $\Omega$ for some constant $c_1>0$
follows by the definition of $z$. Since
$$
\Delta u_{\lambda}-f(u_{\lambda})+u_{\lambda}^{-\gamma}
+\lambda u_{\lambda}=0\leq \Delta z-f(z)+z^{-\gamma}+\lambda z,
$$
and $u_{\lambda},z>0$ in $\Omega$, $u_{\lambda}=z$ on $\partial\Omega$,
$\Delta z\in L^{1}(\Omega)$.
It follows that $u_{\lambda}\geq z$ in $\bar{\Omega}$. Therefore,
from the above proof, $c_1d(x)\leq u_{\lambda}(x)$ for all $x\in\Omega$,
where $c_1$ is a positive constant.

Next, we prove that $u_{\lambda}(x)\leq c_{2}d(x)$ for some constant $c_{2}>0$.
Our method is similar to that by  Gui and Lin \cite{g3}. 
Using the smoothness of $\partial\Omega$, we can find $\delta\in(0,1)$ 
such that for all
$$
x_0\in \Omega_{\delta}:=\{x\in\Omega; d(x)\leq\delta\},
$$
there exists a $y\in\mathbb{R}^{N}\setminus\bar{\Omega}$, with 
$d(y,\partial\Omega)=\delta$, and
$d(x_0)=|x_0-y|-\delta$.
Let $K>1$ be such that $\text{diam}(\Omega)<(K-1)\delta$ and let $\omega$ 
be the unique solution of the Dirichlet problem
\begin{gather*}
-\Delta\omega=\lambda\omega-f(\omega)+\omega^{-\gamma} \quad \text{in }
 B_{K}(0)\setminus \overline{B_1}(0),\\
\omega>0 \quad \text{in } B_{K}(0)\setminus \overline{B_1}(0),\\
\omega=0 \quad \text{on }\partial(B_{K}(0)\setminus \overline{B_1}(0)),
\end{gather*}
where $B_{r}(0)$ is the open ball in $\mathbb{R}^{N}$ of radius $r$ and centered
at the origin. By uniqueness, $\omega$ is radially symmetric. Hence, 
$\omega(x)=\tilde{\omega}(|x|)$ and
\begin{equation}
\begin{gathered}
\tilde{\omega}''+\frac{N-1}{r}\tilde{\omega}'
+\lambda\tilde{\omega}-f(\tilde{\omega})+\tilde{\omega}^{-\gamma} =0 \quad
\text{for } r\in(1,K),\\
\tilde{\omega}>0 \quad\text{in } (1,K),\\
\tilde{\omega}(1)=\tilde{\omega}(K)=0.
\end{gathered} \label{e3.2}
\end{equation}
Integrating in \eqref{e3.2} yields
\begin{align*}
\tilde{\omega}'(t)
&=\tilde{\omega}'(a)a^{N-1}t^{1-N}-t^{1-N}\int_{a}^{t}r^{N-1}
[\lambda\tilde{\omega}(r)
-f(\tilde{\omega}(r))+(\tilde{\omega}(r))^{-\gamma}]dr\\
&=\tilde{\omega}'(b)b^{N-1}t^{1-N}+t^{1-N}\int_{t}^{b}r^{N-1}
[\lambda\tilde{\omega}(r)-f(\tilde{\omega}(r))
+(\tilde{\omega}(r))^{-\gamma}]dr,
\end{align*}
where $1<a<t<b<K$. Since $-f(\tilde{\omega})+\tilde{\omega}^{-\gamma}\in L^{1}(1,K)$,
 we deduce that both $\tilde{\omega}'(1)$ and $\tilde{\omega}'(K)$ are finite.
Then
$\tilde{\omega}\in C^{2}(1,K)\cap C^{1}[1,K]$ and
\begin{equation}
\omega(x)\leq C\min \{ K-|x|,|x|-1\}, \quad \text{for any }
x\in B_{K}(0)\setminus B_1(0).\label{e3.3}
\end{equation}
Let us fix $x_0\in\Omega_{\delta}$. Then we can find
$y_0\in\mathbb{R}^{N}\setminus\bar{\Omega}$ with
$$
d(y_0,\partial\Omega)=\delta, \quad
d(x_0)=|x_0-y|-\delta.
$$
Thus, $\Omega\subset B_{K\delta}(y_0)\setminus B_{\delta}(y_0)$. Define
$$
v(x)=c\omega(\frac{x-y_0}{\delta}), x\in\bar{\Omega}.
$$
We show that $v$ is a super-solution of \eqref{e1.1} provided that
$c$ is large enough. Indeed, if $c>\max\{ 1,\delta^{2}\}$, then
\begin{align*}
& \Delta v+\lambda v-f(v)+v^{-\gamma}\\
&\leq \frac{c}{\delta^{2}}(\tilde{\omega}''(r)
 +\frac{N-1}{r}\tilde{\omega}'(r))+\lambda(c\tilde{\omega}(r))
-f(c\tilde{\omega}(r))+(c\tilde{\omega}(r))^{-\gamma},
\end{align*}
where
$$
r=\frac{|x-y_0|}{\delta}\in(1,K).
$$
Using  assumption (A2) we obtain
$$
\lambda(c\tilde{\omega})-f(c\tilde{\omega})+(c\tilde{\omega})^{-\gamma}\\
\leq c(\lambda\tilde{\omega}-f(\tilde{\omega})+(\tilde{\omega})^{-\gamma})
\quad \text{in }(1,K).
$$
The above relation leads us to
\begin{align*}
&\Delta v+\lambda v-f(v)+v^{-\gamma}\\
&\leq \frac{c}{\delta^{2}}(\tilde{\omega}''
 + \frac{N-1}{r}\tilde{\omega}'+\lambda\tilde{\omega}-f(\tilde{\omega})
+\tilde{\omega}^{-\gamma})=0.
\end{align*}
Since $\Delta u_{\lambda}\in L^{1}(\Omega)$, then $u_{\lambda}\leq v$ in $\Omega$.
 This combined with \eqref{e3.3} yields
$$
u_{\lambda}(x_0)\leq v(x_0)
\leq C\min \{ K-\frac{|x_0-y_0|}{\delta},\frac{|x_0-y_0|}{\delta}-1\}
\leq\frac{C}{\delta}d(x_0).
$$
Hence $u_{\lambda}\leq(C\setminus \delta)d(x)$ in $\Omega_{\delta}$
and the last inequality follows.
\end{proof}

\section{Asymptotic behavior of the solution}

In this section, we consider the asymptotic behavior of the positive 
solution of \eqref{e1.1} under the assumption that $f(u)=u^{p+1},p>0$,
which satisfy (A1)--(A3). Thus,
\begin{equation}
\begin{gathered}
-\Delta u+u^{p+1}-u^{-\gamma} =\lambda u \quad \text{in } \Omega,\\
u>0 \quad\text{in } \Omega,\\
u=0 \quad \text{on } \partial\Omega.
\end{gathered}\label{e4.1}
\end{equation}
Notice that the function $g(u)$ defined by
\begin{equation}
g(u)=u^{-(\gamma+1)}-u^{p},\quad u>0, \label{e4.2}
\end{equation}
is continuous. Thus $ \lim_{u\to+\infty}g(u)=-\infty$.
In terms of $g(u)$ problem \eqref{e4.1} can be written as
$$
-\Delta u=(\lambda+g(u))u.
$$
In the next two theorems we collect some general features and estimate
the positive solutions of \eqref{e4.1} for $\lambda$ large.

\begin{theorem} \label{thm4.1}
 The following assertions hold
\begin{itemize}
\item[(i)] $u_{\lambda}\leq c_{\lambda}$ for any positive solution
  $u_{\lambda}$ of \eqref{e4.1}, where $c_{\lambda}>0$ is the largest
real number such that
\begin{equation}
\lambda+g(c_{\lambda})=0,\label{e4.3}
\end{equation}
and $g(u)$ is the function defined by \eqref{e4.2}. Moreover,
\begin{equation}
\lim_{\lambda\to\infty}\frac{c_{\lambda}}{\lambda^{1/p}}=1.\label{e4.4}
\end{equation}

\item[(ii)] Given $\varepsilon>0$ arbitrary, there exists 
$\lambda(\varepsilon)>\sigma_1^{\Omega}$ such that
\begin{equation}
|\frac{1}{\lambda}\frac{1}{u_{\lambda}^{\gamma+1}}|\leq\varepsilon\label{e4.5}
\end{equation}
for all $\lambda\geq\lambda(\varepsilon)$ on any compact subsets of $\Omega$.
\end{itemize}
\end{theorem}

\begin{proof} 
(i) Assume that \eqref{e4.1} has a positive solution and let $x_0\in\Omega$
be the point where $u_{\lambda}$ reaches its maximum. Obviously, 
$-\Delta u_{\lambda}(x_0)\geq0$, and
$$
\lambda+g(u_{\lambda}(x_0))\geq0.
$$
Consider that $\lambda+g(u)$  is strictly decreasing for $u>0$ and
 $\lambda+g(c_{\lambda})=0$. Then,
any positive solution $u_{\lambda}$ of \eqref{e4.1} satisfies
$u_{\lambda}(x_0)\leq c_{\lambda}$, where $x_0$ is the point at which 
$u_{\lambda}$ reaches its maximum, hence $u_{\lambda}\leq c_{\lambda}$.
As $\lim_{u\to+\infty}g(u)=-\infty$,
we have
$\lim_{\lambda\to+\infty}c_{\lambda}=+\infty$.
From \eqref{e4.2} and \eqref{e4.3} it follows easily that
$$
\frac{\lambda^{1/p}}{c_{\lambda}}=[1-\frac{1}{c_{\lambda}^{\gamma+p+1}}]^{1/p},
$$
and since $c_{\lambda}\to\infty$ as $\lambda\to\infty$, one has
$$
\lim_{\lambda\to\infty}\frac{1}{c_{\lambda}^{\gamma+p+1}}=0
\quad\text{and}\quad 
\lim_{\lambda\to\infty}\frac{\lambda^{1/p}}{c_{\lambda}}=1.
$$
Hence \eqref{e4.4} holds.
\smallskip

(ii)  Since $G^{-1}(\lambda-\lambda_1)\phi_1(x)<u_{\lambda}(x)$ from the 
proof of Theorem \ref{thm2.1} (i), it is easy to see that 
$u_{\lambda}\to\infty$ as $\lambda\to\infty$
on any compact subsets of $\Omega$. Let $\varepsilon>0$. 
There exists $\lambda_1(\varepsilon)$, if $\lambda\geq\lambda_1(\varepsilon)$ 
such that
$$
|\frac{1}{u^{\gamma+p+1}}|\leq\frac{\varepsilon}{3}.
$$
As $\lim_{\lambda\to\infty}\frac{c_{\lambda}^{p}}{\lambda}=1$,
 there exists $\lambda_{2}(\varepsilon)>0$, if 
$\lambda\geq\lambda_{2}(\varepsilon)$ such that
$$
\frac{c_{\lambda}^{p}}{\lambda}\leq\frac{3}{2}.
$$
Set
$$
\lambda(\varepsilon)=\max\{\lambda_1(\varepsilon),\lambda_{2}(\varepsilon)\}.
$$
Let $\lambda\geq\lambda(\varepsilon)$ and assume that \eqref{e4.1}
has a positive solution $u_{\lambda}$. Then
$$
|\frac{1}{\lambda}\frac{1}{u_{\lambda}(x)^{\gamma+1}}|
=\Big\{\frac{u_{\lambda}(x)}{c_{\lambda}}\Big\}^{p}
\frac{c_{\lambda}^{p}}{\lambda}|\frac{1}{u_{\lambda}(x)^{\gamma+p+1}}|
\leq\frac{3}{2}\frac{\varepsilon}{3}=\frac{\varepsilon}{2}
$$
because $u_{\lambda}\leq c_{\lambda}$. Thus, \eqref{e4.5} holds.
\end{proof}

Note that $\theta_{\lambda, m}$ is the unique positive solution of 
the  problem
\begin{gather*}
-\frac{1}{\lambda}\Delta u=mu-u^{p+1} \quad \text{in } \Omega,\\
u=0 \quad \text{on }\partial\Omega.
\end{gather*}

\begin{theorem} \label{thm4.2}
For each $\varepsilon>0$ arbitrary, there exists 
$\lambda(\varepsilon)>\sigma_1^{\Omega}$ such that
\begin{equation}
\lambda^{1/p}\theta_{\lambda,1-\varepsilon}\leq u_{\lambda}
\leq \lambda^{1/p}\theta_{\lambda,1+\varepsilon}\label{e4.6}
\end{equation}
for all $\lambda\geq\lambda(\varepsilon)$ on any compact subsets
of $\Omega$. In particular,
\begin{equation}
\lim_{\lambda\to\infty}\frac{u_{\lambda}}{\lambda^{1/p}}=1\label{e4.7}
\end{equation}
uniformly on any compact subsets of $\Omega$.
\end{theorem}

\begin{proof}
 The charge of variable $u=\lambda^{1/p}v$ transforms \eqref{e4.1} into
\begin{equation}
\begin{gathered}
 -\frac{1}{\lambda}\Delta v=v+\frac{1}{\lambda^{(\gamma+p+1)/p}v^{\gamma}}-v^{p+1}
\quad \text{in } \Omega,\\
v>0  \quad\text{in }\Omega,\\
v=0\quad\text{on }\partial\Omega,
\end{gathered}\label{e4.8}
\end{equation}
where $p>0$, $\gamma>0$ and $\lambda>0$ is a real parameter.

By Theorem \ref{thm2.1} (see \cite{f1}), to prove this theorem it is 
sufficient to show that
$$
\theta_{\lambda,1-\varepsilon}\leq v_{\lambda}\leq\theta_{\lambda,1+\varepsilon}
$$
for all $\lambda\geq\lambda(\varepsilon)$ on any compact subsets of $\Omega$.

Fixed $\varepsilon>0$, we first show that there exists 
$\lambda_1(\varepsilon)>\sigma_1^{\Omega}$ such that 
$\theta_{\lambda,1-\varepsilon}\leq v_{\lambda}$ for all 
$\lambda\geq\lambda_1(\varepsilon)$ on any compact subsets of $\Omega$.
 Let $v_{\lambda}$ be a positive solution of \eqref{e4.8}.
If $\theta_{\lambda,1-\varepsilon}=v_{\lambda}$ for $\lambda$ large we 
have concluded. If $\theta_{\lambda,1-\varepsilon}\neq v_{\lambda}$, then
$$
\frac{v_{\lambda}^{p+1}-\theta_{\lambda,1-\varepsilon}^{p+1}}{v_{\lambda}
-\theta_{\lambda,1-\varepsilon}}
=\theta_{\lambda,1-\varepsilon}^{p}+Q(x),
$$
with $Q(x)>0$ for some $x\in\Omega$ and hence
$$
\sigma_1[-\frac{1}{\lambda}\Delta-1+\varepsilon
+\frac{v_{\lambda}^{p+1}-\theta_{\lambda,1-\varepsilon}^{p+1}}{v_{\lambda}-
\theta_{\lambda,1-\varepsilon}}]>\sigma_1[-\frac{1}{\lambda}
\Delta-1+\varepsilon+\theta_{\lambda,1-\varepsilon}^{p}].
$$
By the definition of $\theta_{\lambda,1-\varepsilon}$, we find that
$$
\sigma_1[-\frac{1}{\lambda}\Delta-1+\varepsilon
+\theta_{\lambda,1-\varepsilon}^{p}]=0.
$$
Thus,
$$
\sigma_1[-\frac{1}{\lambda}\Delta-1+\varepsilon+\frac{v_{\lambda}^{p+1}-
\theta_{\lambda,1-\varepsilon}^{p+1}}{v_{\lambda}-\theta_{\lambda,1-\varepsilon}}]>0.
$$
On the other hand, after some straightforward manipulations 
it follows from \eqref{e4.8} that
\begin{equation}
\big[-\frac{1}{\lambda}\Delta-1+\varepsilon+\frac{v_{\lambda}^{p+1}-
\theta_{\lambda,1-\varepsilon}^{p+1}}{v_{\lambda}-\theta_{\lambda,1-\varepsilon}}
 \big](v_{\lambda}-\theta_{\lambda,1-\varepsilon})
=\big[\varepsilon+\frac{1}{\lambda}\frac{1}{(\lambda^{1/p}v_{\lambda})^{\gamma+1}}
\big]v_{\lambda}.\label{e4.9}
\end{equation}
Moreover, it follows from \eqref{e4.5} that there exists
$\lambda_1(\varepsilon)>\sigma_1^{\Omega}$ such that
$$
\varepsilon+\frac{1}{\lambda}\frac{1}{(\lambda^{1/p}v_{\lambda})^{\gamma+1}}>0
$$
for all $\lambda\geq\lambda_1(\varepsilon)$ on any compact subsets of
$\Omega$. Therefore, applying the maximum
principle to \eqref{e4.9} we find that
 $\theta_{\lambda,1-\varepsilon}\leq v_{\lambda}$ for all
$\lambda\geq\lambda_1(\varepsilon)$ and any positive solution
$v_{\lambda}$ of \eqref{e4.8} on any compact subsets of $\Omega$
(see \cite{p1,w1}).

We now prove that there exists $\lambda_{2}(\varepsilon)>\sigma_1^{\Omega}$ 
such that $v_{\lambda}\leq \theta_{\lambda,1+\varepsilon}$ for all 
$\lambda\geq\lambda_{2}(\varepsilon)$ on any compact subsets of $\Omega$. 
Let $v_{\lambda}$ be a positive solution of \eqref{e4.8}.
If $\theta_{\lambda,1+\varepsilon}=v_{\lambda}$ for $\lambda$ large 
we have concluded. If $\theta_{\lambda,1+\varepsilon}\neq v_{\lambda}$ 
for $\lambda$, then
$$
\frac{\theta_{\lambda,1+\varepsilon}^{p+1}
-v_{\lambda}^{p+1}}{\theta_{\lambda,1+\varepsilon}-v_{\lambda}}
=\theta_{\lambda,1+\varepsilon}^{p}+\hat{Q}(x),
$$
with $\hat{Q}(x)>0$ for some $x\in \Omega$. Thus, arguing as above we find that
$$
\sigma_1\big[-\frac{1}{\lambda}\Delta-1-\varepsilon
+\frac{\theta_{\lambda,1+\varepsilon}^{p+1}
-v_{\lambda}^{p+1}}{\theta_{\lambda,1+\varepsilon}-v_{\lambda}}\big]
>\sigma_1\big[-\frac{1}{\lambda}\Delta-1-\varepsilon
+\theta_{\lambda,1+\varepsilon}^{p}\big]=0.
$$
On the other hand, after some straightforward manipulations it follows from 
\eqref{e4.8} that
\begin{equation}
\big[-\frac{1}{\lambda}\Delta-1-\varepsilon
+\frac{\theta_{\lambda,1+\varepsilon}^{p+1}
-v_{\lambda}^{p+1}}{\theta_{\lambda,1+\varepsilon}-v_{\lambda}}\big]
(\theta_{\lambda,1+\varepsilon}-v_{\lambda})
=\big[\varepsilon-\frac{1}{\lambda}\frac{1}{(\lambda^{1/p}v_{\lambda})^{\gamma+1}}
\big]v_{\lambda}.\label{e4.10}
\end{equation}
Moreover, it follows from \eqref{e4.5} that there exists
$\lambda_{2}(\varepsilon)>\sigma_1^{\Omega}$ such that
$$
\varepsilon-\frac{1}{\lambda}\frac{1}{(\lambda^{1/p}v_{\lambda})^{\gamma+1}}>0
$$
for all $\lambda\geq\lambda_{2}(\varepsilon)$ on any compact subsets
of $\Omega$. Therefore, applying the maximum principle to \eqref{e4.10}
we find that $v_{\lambda}\leq \theta_{\lambda,1+\varepsilon}$ for all
$\lambda\geq\lambda_{2}(\varepsilon)$ and any positive solution
$v_{\lambda}$ of \eqref{e4.8} on any compact subsets of $\Omega$
(see \cite{p1,w1}).

Taking $\lambda(\varepsilon)=\max\{\lambda_1(\varepsilon),
\lambda_{2}(\varepsilon)\}$,
 the proof of \eqref{e4.6} is completed. The rest of the proof follows from
\cite[Theorem 2.1]{f1}.
\end{proof}

\subsection*{Acknowledgements}
This research is supported by Young Award of Shandong Province
(ZR2013AQ008) and the Fund of Science and Technology Plan
of Shandong Province (2014GGH201010).

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