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

\AtBeginDocument{{\noindent\small
Seventh Mississippi State - UAB Conference on Differential Equations and
Computational Simulations,
{\em Electronic Journal of Differential Equations},
Conf. 17 (2009),  pp.123--131.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document} \setcounter{page}{123}
\title[\hfilneg EJDE-2009/Conf/17\hfil positone to infinite semipositone problems] 
{Subsolutions: A journey from positone to infinite semipositone problems}

\author[E. K. Lee, R. Shivaji, J. Ye\hfil EJDE/Conf/17 \hfilneg]
{Eun Kyoung Lee, Ratnasingham Shivaji, Jinglong Ye}  % in alphabetical order

\address{Eun Kyoung Lee \newline
Department of Mathematics  and Statistics, Center for
Computational Sciences, Mississippi State University, Mississippi
State, MS 39762, USA}
\email{el165@msstate.edu}

\address{Ratnasingham Shivaji \newline
Department of Mathematics  and Statistics, Center
for Computational Sciences, Mississippi State University,
Mississippi State, MS 39762, USA}
\email{shivaji@ra.msstate.edu}

\address{Jinglong Ye \newline
Department of Mathematics and Statistics,
Mississippi State University, Mississippi State, MS 39762, USA}
\email{jy79@msstate.edu}

\thanks{Published April 15, 2009.}
\subjclass[2000]{35J25} 
\keywords{Positone; semipositone; infinite semipositone; sub-super solutions}

\begin{abstract}
 We discuss the existence of positive solutions to
 $-\Delta u=\lambda f(u)$ in $\Omega$, with $u=0$ on the boundary, where
 $\lambda$ is a positive parameter, $\Omega$ is a bounded domain with
 smooth boundary $\Delta $ is the Laplacian operator, and
 $f:(0,\infty)\to R$ is a continuous function. We first
 discuss the cases when $f(0)>0$ (positone), $f(0)=0$ and $f(0)<0$
 (semipositone). In particular, we will review the existence of
 non-negative strict subsolutions. Along with these subsolutions and
 appropriate assumptions on $f(s)$ for $s\gg 1$ (which will lead to
 large supersolutions) we discuss the existence of positive
 solutions. Finally, we obtain new results on the case of infinite
 semipositone problems ($\lim_{s\to 0^{+}}f(s)=-\infty$).
\end{abstract}

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

\section{Introduction}

Nonlinear eigenvalue problems of the form:
\begin{equation}\label{eP}
\begin{gathered}
-\Delta u = \lambda f(u) \quad \text{in }  \Omega  \\
 \quad u = 0  \quad \text{on }   \partial \Omega,
\end{gathered}
\end{equation}
where $\lambda$ is a positive parameter, $\Omega$ is a bounded
domain with smooth boundary $\partial\Omega, \Delta $ is a
Laplacian operator, and $f:(0,\infty)\to R$ is a
continuous function, arise in the study of steady state reaction
diffusion processes, in particular, nonlinear heat generation,
combustion theory, chemical reactor theory and population dynamics
(see Parks \cite{Pk}, Sattinger \cite{Sa}, Parter \cite{Pt}, Tam
\cite{Ta}, Aris \cite{Ar} and Selgrade \cite{SR}). In the case
when $f(0)>0$ (positone problems) there is a very rich history
(spanning over 50 years) on the study of positive solutions (see
Amann \cite{Am}, Brown \cite{BIS}, Cohen \cite{CK}, Grandall
\cite{CR}, de Figueiredo \cite{FLN}, Gidas \cite{GNN}, Joseph
\cite{JL}, Kazdan \cite{KW}, Laetsch \cite{La}, and Rabinowitz
\cite{Ra}). The case when $f(0)<0$ (semipositone problems) is
mathematically more challenging as pointed out by P. L. Lions
\cite{Li}. See also Castro \cite{CMS}. However, in the past 20
years, there has been considerable progress on the study of
semipositone problems (see
\cite{AAB,ACS,AHS,AZ,BCS,BS,CG,CGS1,CGS2,CGS3,CGSh,CHS,CS1,CS2,CS3,Te}).
One of the main
tools used in these studies is the method of sub-super solutions.
By a subsolution of \eqref{eP} we mean a function $\psi \in
C^{2}(\Omega) \cap C(\overline{\Omega})$ that satisfies
\begin{gather*}
-\Delta \psi \leq  \lambda f(\psi)\quad \text{in }  \Omega  \\
\quad \psi \leq 0 \quad \text{on }  \partial \Omega,
\end{gather*}
and by a supersolution of \eqref{eP} we mean a function
$Z \in C^{2}(\Omega) \cap C(\overline{\Omega})$ that satisfies
\begin{gather*}
-\Delta Z \geq  \lambda f(Z)\quad \text{in }   \Omega  \\
\quad Z \geq 0 \quad \text{on }  \partial \Omega.
\end{gather*}
Then it is well known (see Amman \cite{Am}) that if there exists a
subsolution $\psi$ and a supersolution $Z $ such that $\psi \leq Z$
in $\Omega$ then \eqref{eP} has a solution $u$ such that $\psi \leq u
\leq Z$. In applying this method to obtain positive solutions, it is
essential that we are able to construct non-negative strict
subsolutions. By a strict subsolution we mean a subsolution that is
not a solution. In the case when $f(0)>0$, it is trivial to see that
$\psi=0$ is a strict subsolution for every $\lambda >0$. More on
positone problems will be discussed in Section 2.  But the real
challenge occurs in the case when $f(0)<0$ (semipositone problems).
Here our test functions for positive subsolutions must come from
positive functions $\psi$ such that $-\Delta \psi <0$ near $\partial
\Omega$ while $-\Delta \psi >0$ in a large part of the interior of
$\Omega$. We will discuss more on semipositone problems in Section
3. The case when $f(0)=0$ does cause considerable problems in the
construction of positive subsolutions, specially in the case when we
have no other information at the origin. We will address this later
in Section 4.  Finally in Section 5, we will establish new results
for the case when $lim_{s\to 0^{+}}f(s)=-\infty$ (infinite
semipositone problems). We note that once strict non-negative
subsolutions $\psi$ are constructed, with appropriate assumptions on
$f(s)$ for $s\gg 1$ one can obtain large positive supsolutions $Z$
such that $Z\geq \psi$ on $\Omega$, hence establishing positive
solutions to \eqref{eP}.

\section {$f(0)>0$: Positone Problems}

In this section, we consider the problem \eqref{eP} under the following
assumption
\begin{itemize}
\item [(F0)] $f(0)>0$.
\item [(F1)] $lim_{s \to \infty}\frac{f(s)}{s}= 0$.
\end{itemize}
We have the following result

\begin{theorem}\label{Thm2.1}
Assume {\rm (F0), (F1)}. Then \eqref{eP} has a positive solution for all
$\lambda>0$.
\end{theorem}

\begin{proof}
It is clear that $ \psi=0$ is a strict subsolution since $ f(0)>0$.
Let $\widetilde{f}(s):= \max_{t \in [0,s]} f(t)$. Then $f(s)\leq
\widetilde{f}(s)$, $\widetilde{f}$ is nondecreasing and $lim_{s
\to \infty}\frac{\widetilde{f}(s)}{s}= 0$. Hence we can
choose $m(\lambda)\gg 1$ such that
\[
\frac{1}{\| e\|_{\infty}\lambda}\geq
\frac{\widetilde{f}(m(\lambda)\| e\|_{\infty})}{m(\lambda)\|
e\|_{\infty}}
\]
where $e$ is the solution of $-\Delta e = 1$ in
$\Omega$, $ e = 0$ on $
\partial\Omega$. Let $Z:=m(\lambda)e$.
 Then
$$
 -\Delta Z=m(\lambda) \geq \lambda \widetilde{f}(m(\lambda)\|
e\|_{\infty}) \geq \lambda \widetilde{f}(m(\lambda)e) \geq \lambda
f(m(\lambda)e).
$$
Thus $Z$ is a supersolution. Hence \eqref{eP} has a positive solution.
\end{proof}

\section{$f(0)<0$: Semipositone Problems}

In this section, we discuss two results for the problem $(P)$.
First, we assume that
\begin{itemize}
\item [(F2)] There exists $K_0>0$ such that $f(s) \geq -K_0$ for all $s>0$.
\item [(F3)] $\lim_{s \to \infty} f(s) = \infty$.
\end{itemize}

Note that (F2) includes the case $f(0)<0$.

\begin{theorem}\label{Thm3.1}
Assume {\rm (F1), (F2), (F3)}. Then \eqref{eP} has a positive solution
for $ \lambda \gg 1$.
\end{theorem}

\begin{proof}
Let $\lambda_1 >0$ be the  first eigenvalue of the operator
$-\Delta$ with Dirichlet boundary condition and $\phi$ be the
corresponding eigenfunction satisfying $\phi> 0$ in $\Omega$
and $\frac{\partial \phi}{\partial \nu} <0$ on
$\partial \Omega$, where $\nu$ is outward normal vector on $\partial
\Omega$ and $\| \phi \|_\infty = 1$. Note that $\lambda_1$ and
$\phi$ satisfy:
\begin{gather*}
 -\Delta \phi = \lambda_1 \phi \quad  \text{in }  \Omega \\
 \quad \phi = 0  \quad \text{on }  \partial \Omega.
\end{gather*}
Let $ \delta >0$, $\mu>0$, $m>0$ be such that
$| \nabla \phi |^{2} - \lambda_1 \phi^2 \geq m$ in
$\overline{\Omega}_{\delta}$ and $ \mu \leq \phi \leq 1$
 in $\Omega \setminus \overline{\Omega}_{\delta}$ where
$\overline{\Omega}_{\delta}:=\{x \in \Omega : d(x,\partial
\Omega)\leq \delta\}$. This is possible since $| \nabla \phi |
\neq 0$ on $\partial \Omega$. Let $\psi :=\frac{K_0 \lambda}{2m}
\phi^{2}$. Then
$$
\nabla \psi=\frac{K_0 \lambda}{m} \phi \nabla \phi
$$
and
\[
- \Delta \psi = - div( \nabla \psi ) =- \frac{K_0 \lambda}{m} \{
\phi \Delta \phi + | \nabla \phi |^2 \}
=  -\frac{K_0 \lambda}{m} \{ | \nabla \phi |^2 - \lambda_1 \phi^2
\}
\]
Then in $\overline{\Omega}_{\delta}$,
\begin{equation} \label{3.1}
- \Delta \psi \leq  - K_0 \lambda \leq \lambda f( \psi ).
\end{equation}
 From (F3), we know that for $ \lambda \gg 1$
$$
\frac{K_0 \lambda_1}{m} \leq f( \frac{K_0\lambda }{2m} \mu^2).
$$
Thus in $\Omega \setminus \overline{\Omega}_{\delta}$,
\begin{align}\label{3.2}
-\Delta \psi \leq  \frac{K_0 \lambda \lambda_1}{m} \leq \lambda f(
\frac{K_0 \lambda }{2m} \phi^2 )= \lambda f( \psi).
\end{align}
Combining (\ref{3.1}) and (\ref{3.2}), if $\lambda \gg 1$ we see
that
\begin{align*}
-\Delta \psi \leq \lambda f (\psi) \quad \text{ in } \  \Omega.
\end{align*}
Thus $\psi$ is a positive subsolution of \eqref{eP}. Next constructing a
supersolution Z as in the proof of Theorem \ref{Thm2.1}, we can also
choose $m(\lambda)$ large enough so that $Z\geq \psi$ in
$\overline{\Omega}$. This is possible since $e>0$ in $\Omega$ and
$\frac{\partial e}{\partial \nu} <0$ on $\partial\Omega$, where
$\nu$ is outward normal vector on $\partial \Omega$. Thus we know
that \eqref{eP} has a positive solution $u \in [\psi, Z]$ for
$\lambda \gg 1$.
\end{proof}

We next discuss a semipositone problem where $f(u)<0$ for $u \gg
1$. (Hence in this case $(F3)$ will not be satisfied.) In
particular, we recall the example
\begin{equation}\label{ePE}
\begin{gathered}
 -\Delta u = au -bu^2 -c \quad \text{in }  \Omega \\
  \quad  u = 0     \quad  \text{on }  \partial \Omega,
\end{gathered}
\end{equation}
studied in \cite{Shob}. This equation arises in the study of
population dynamics of one species with $u$ representing the
concentration of the species and $c$ representing the rate of
harvesting. To get a positive subsolution, in \cite{Shob} the
authors use the anti-maximum principle by Clement and Peletier
\cite{Clem}, and establish the following result:

\begin{theorem}\label{Thm3.2}
Suppose that $a> \lambda_1 $ and $b>0$. Then there exists $c_1=c_1
(a,b)$ such that for $0<c<c_1$, \eqref{ePE} has a positive solution.
\end{theorem}

\begin{proof}
Consider the boundary-value problem
\begin{equation}\label{3.3}
\begin{gathered}
 -\Delta z - \lambda z = -1 \quad  \text{in }  \Omega \\
\quad z = 0   \quad  \text{on } \partial \Omega.
\end{gathered}
\end{equation}
By the anti-maximum principle in \cite{Clem}, there exist a
$\delta_1 = \delta_1 (\Omega) >0$ such that if $\lambda \in
(\lambda_1, \lambda_1+\delta_1 )$ then (\ref{3.3}) has a solution
$z=z_\lambda$ which is positive in $\Omega$ and $\frac{\partial
z_\lambda}{\partial \nu } < 0 $ on $\partial \Omega$. Fix
$\lambda^* \in (\lambda_1, \min \{ a, \lambda_1 + \delta_1 \})$. Let
$z_{\lambda^* }$ be the solution of (\ref{3.3}) when $\lambda =
\lambda^*$ and $\alpha := \|z_{\lambda^* }\|_\infty $. Define $\psi
= Kc z_{\lambda^* }$, where $K \geq 1$ is to be determined later. We
will choose $K \geq 1$ and $c>0$ properly so that $\psi$ is a
subsolution. We know
$$
- \Delta \psi = Kc ( -\Delta  z_{\lambda^* } ) =Kc(\lambda^*
z_{\lambda^* }) - Kc.
$$
Thus if we prove
\begin{equation}\label{3.5}
(a-\lambda^* ) Kz_{\lambda^* } - bc (Kz_{\lambda^* } )^2 +K-1 \geq
0,
\end{equation}
then
\[
- \Delta \psi =Kc(\lambda^* z_{\lambda^* }) - Kc
 \leq a (Kc z_{\lambda^* }) -b (Kc z_{\lambda^* })^2 -c.
\]
Thus  $\psi = Kc z_{\lambda^* }$ can be a subsolution of \eqref{ePE}. To
show (\ref{3.5}), define $H(y):= (a- \lambda^* ) y -bc y^2 + (K-1)
$. If $H(y) \geq 0$ for all $y \in [0,K\alpha ]$, then (\ref{3.5})
is true. Since $a > \lambda^*$, if $K\geq1$, then it suffice to show
that $H(K\alpha)=(a-\lambda^* ) K\alpha - bc (K\alpha )^2 +(K-1)
\geq 0$, which is equivalent to
$$
c\leq \frac{(a-\lambda^* ) K\alpha  +(K-1)}{ b (K\alpha )^2 }
$$
Thus if we define
$$
c_1 = c_1 (a,b) := \sup_{K\geq 1}\frac{(a-\lambda^* ) K\alpha
+(K-1)}{ b (K\alpha )^2 },
$$
then we know that when $0<c<c_1$, there exist ${\tilde K} \geq 1$
such that $\psi ={\tilde K}c z_{\lambda^* }$ is a subsolution. It is
obvious that $Z=M$ where M is sufficiently large constant is a
supersolution of \eqref{ePE} with $Z \geq \psi$. Thus Theorem
\ref{Thm3.2} is proven.
\end{proof}

\section{$f(0)=0$}

In this section, we consider the problem \eqref{eP} for the case
$f(0)=0$.
 Since $(F2)$ includes the case $f(0)=0$, we've already
discussed this case in Theorem \ref{Thm3.1}. We now discuss a more
precise result under the additional assumption
\begin{itemize}
\item [(F4)]  $f(0)=0$ and $f'(0)>0$.
\end{itemize}


\begin{theorem}\label{Thm4.1}
Assume {\rm (F1), (F4)}. Then \eqref{eP} has a positive solution for
$\lambda > \lambda_1/f'(0)$.
\end{theorem}

\begin{proof}
Since $f'(0)>\lambda_1/\lambda$ we know that there exist
$m=m(\lambda)>0$ such that
\begin{equation}\label{4.1}
f(s) > \frac{\lambda_1}{\lambda}s \quad \text{for all }  s\in(0,m).
\end{equation}
Let $\psi:= m \phi$ where $\phi$ is as defined in the proof of
Theorem \ref{Thm3.1}. Then
$$
-\Delta \psi = \lambda_1 m \phi \leq \lambda f(m\phi )=\lambda
f(\psi).
$$
Thus $\psi$ is a positive subsolution of \eqref{eP}. By the same argument
as in the proof of Theorem \ref{Thm3.1}, we can find a supersolution
$Z$ of \eqref{eP} with $Z \geq \psi$. Thus we know that \eqref{eP} has a
positive solution $u \in [\psi, Z]$ for $\lambda >
\frac{\lambda_1}{f'(0)}$.
\end{proof}

For the case $f(0)=0$, we can also study \eqref{eP} when $f$ does not
satisfy (F1) but satisfies
\begin{itemize}
\item [(F5)] There exists $r_0>0$ such that $f(s)>0$ for
$s \in (0, r_0 )$ and $f(r_0 )=0$.
\end{itemize}

\begin{theorem}\label{Thm4.2}
Assume  {\rm (F5)}. Then \eqref{eP} has a positive solution for
$ \lambda \gg 1$.
\end{theorem}

\begin{proof}
Fix $\sigma \in (0, r_0 )$ and  let $\psi :=\frac{\sigma}{2}\phi^{2}$
where $\phi$ is as defined in the proof of Theorem
\ref{Thm3.1}. Then
$$
-\Delta \psi =-\sigma \{ | \nabla \phi |^{2}- \lambda_1 \phi^2\}.
$$
 Let $ \delta >0$, $ \mu>0$, $m>0$ and $\Omega_{\delta}$ be as before
(see the proof of Theorem \ref{Thm3.1}). We can
choose $\lambda \gg 1$ such that
$$
\sigma \lambda_1 < \lambda \min_{s\in [\frac{\sigma}{2}\mu^2 ,
\sigma] } f(s).
$$
Thus in $\Omega \setminus \overline{\Omega}_{\delta}$ for
$\lambda \gg 1$,
\begin{equation}\label{4.1b}
-\Delta \psi \leq \sigma \lambda_1 < \lambda \min_{s\in
[\frac{\sigma}{2}\mu^2 , \sigma] } f(s) \leq \lambda f(\psi).
\end{equation}
On the other hand, in $\overline{\Omega}_{\delta}$,
\begin{align}\label{4.2}
-\Delta \psi < -\sigma m \leq \lambda f( \psi),
\end{align}
since $\lambda f( \psi) \geq 0$. Combining (\ref{4.1b}) and
(\ref{4.2}), if $\lambda \gg 1$ we see that $\psi$  is a positive
subsolution of \eqref{eP}. Next, it is easy to check that constant
function $Z:= r_0$ is a supersolution of \eqref{eP} with $Z \geq \psi$.
Hence for $\lambda \gg 1$, \eqref{eP} has a positive solution and Theorem
\ref{Thm4.2} is proven.
\end{proof}

\section{$\lim_{s\to 0^{+}}f(s)=-\infty$: Infinite Semipositone Problems}

We discuss the existence of positive solutions to the following
infinite semipositone problem:
\begin{equation}\label{ePI}
\begin{gathered}
-\Delta u = \lambda \frac{g(u)}{u^{\alpha}} \quad \text{in }  \Omega  \\
\quad u = 0 \quad  \quad \text{on } \partial \Omega,
\end{gathered}
\end{equation}
where $\lambda >0$, $\alpha \in (0,1)$, $ g(0)<0$ and  $g $ is
continuous. We introduce the following hypotheses:
\begin{itemize}
\item [(G1)] There exists $\gamma >0$ and $B>0$ such that
 $\alpha \leq \gamma < \alpha + 1$ and $g(s) \leq Bs^\gamma$ for
 $ s \geq 0$.
\item [(G2)] There exists $\beta > 0$ and $A>0$ such that
 $ g(s) \geq As^{\beta}$ for $ s \gg 1$.
\end{itemize}
We establish the following result.

\begin{theorem}
Assume {\rm (G1), (G2)}. Then \eqref{ePI} has a positive solution for
$\lambda \gg 1$.
\end{theorem}

We prove our result by finding sub-super solutions to our singular
equation. By a subsolution of  \eqref{ePI} we mean a function $\psi:
\overline{\Omega}\to R$ such that $\psi \in C^{2}(\Omega)
\cap C(\overline{\Omega})$ and satisfies:
\begin{gather*}
-\Delta \psi \leq  \lambda \frac{g(\psi)}{\psi^{\alpha}}\quad \text{in } \Omega  \\
\quad \psi > 0 \quad \text{in } \Omega  \\
\quad\psi = 0 \quad \text{on }  \partial \Omega
\end{gather*}
and by a supersolution of  \eqref{ePI} we mean a function
$Z: \overline{\Omega}\to R$ such that
$Z \in C^{2}(\Omega) \cap C(\overline{\Omega})$ and satisfies:
\begin{gather*}
-\Delta Z \geq  \lambda \frac{g(Z)}{Z^{\alpha}}\quad \text{in } \Omega  \\
\quad Z > 0\quad \text{in }  \Omega  \\
\quad Z = 0\quad \text{on }   \partial \Omega.
\end{gather*}
Then we have the following Lemma.

\begin{lemma}[\cite{Cui}]
If there exist a subsolution $\psi$ and a
supersolution $Z$ of  \eqref{ePI} such that $\psi \leq Z$ on
$\overline{\Omega}$, then (5.1) has at least one solution $u \in
C^{2}(\Omega) \cap C(\overline{\Omega})$ satisfying $\psi \leq u
\leq Z$ on $\overline{\Omega}$.
\end{lemma}

\begin{proof}
Let $\phi$ be the eigenfunction as defined in the proof of Theorem
\ref{Thm3.1}. Let $\psi:=\lambda^{r}\phi^{\frac{2}{1+\alpha}},r \in
(\frac{1}{1+\alpha},\frac{1}{1+\alpha -\beta})$. Then
\[
\nabla \psi
=\lambda^{r}(\frac{2}{1+\alpha})\phi^{\frac{1-\alpha}{1+\alpha}}\nabla \phi
\]
and
\begin{align*}
- \Delta \psi
&=- \lambda^{r}(\frac{2}{1+\alpha})\{\phi^{\frac{1-\alpha}{1+\alpha}}
  \Delta \phi+\big(\frac{1-\alpha}{1+\alpha}\big)
  \phi^{-\frac{2\alpha}{1+\alpha}}|\nabla \phi |^{2}\}\\
&=\lambda^{r}(\frac{2}{1+\alpha})\frac{1}{(\phi^{\frac{2}{1+\alpha}})
  ^\alpha} \{\lambda_{1}\phi^{2}-\big(\frac{1-\alpha}{1+\alpha}\big)|
  \nabla \phi |^{2}\}.
\end{align*}
Let $\delta>0$, $\mu >0$, $m>0$ be such that
$$
(\frac{2}{1+\alpha})\{(\frac{1-\alpha}{1+\alpha})| \nabla \phi
|^{2}-\lambda_1 \phi^2 \} \geq m\quad \text{in }\overline{\Omega}_{\delta},
$$
and $ \phi \in [\mu,1]$ in $\Omega \setminus
\overline{\Omega}_{\delta}$, where
$\Omega_\delta:=\{x\in \Omega :d(x,\partial \Omega) \leq \delta\}$. Then in
$\overline{\Omega}_{\delta}$, if $\lambda \gg 1$ then
$$
(\frac{2}{1+\alpha})\{\lambda_1
\phi^2-(\frac{1-\alpha}{1+\alpha})| \nabla \phi |^{2}\} \leq
-m \leq \frac{\lambda g(0)}{\lambda^r \lambda^{r
\alpha}}=\lambda^{[1-r-r \alpha]}g(0)
$$
since $g(0)<0$ and $1-r-r \alpha <0$.
Hence in $\overline{\Omega}_{\delta}$, if $\lambda \gg 1$ then
\begin{equation}\label{5.1}
-\Delta \psi
=\lambda^{r}(\frac{2}{1+\alpha})\frac{1}{(\phi^{\frac{2}{1+\alpha}})^\alpha
} \{\lambda_{1}\phi^{2}-\big(\frac{1-\alpha}{1+\alpha}\big)|
\nabla \phi |^{2}\}\leq \lambda \frac{g(\lambda^r \phi
^{\frac{2}{1+ \alpha}})}{(\lambda^r
\phi^{\frac{2}{1+\alpha}})^{\alpha}}.
\end{equation}
Next in $\Omega \setminus \overline{\Omega}_{\delta}$, since $\phi
\geq \mu$, from $(G2)$,
$$
g(\lambda^{r}\phi^{\frac{2}{1+\alpha}})\geq
A(\lambda^{r}\phi^{\frac{2}{1+\alpha}})^{\beta},
$$
for $\lambda \gg 1$. Also since $0< \mu \leq \phi<1$ and
$1+r(\beta-\alpha)-r>0$,
$$
(\frac{2\lambda_1}{1+\alpha})[\lambda^{r}\phi^{\frac{2}{1+\alpha}}]\leq
\lambda A
(\lambda^{r}\phi^{\frac{2}{1+\alpha}})^{\beta-\alpha}=\lambda
\frac{A(\lambda^{r}\phi^{\frac{2}{1+\alpha}})^\beta}
{(\lambda^{r}\phi^{\frac{2}{1+\alpha}})^\alpha},
$$
for $\lambda \gg 1$.
 Hence in $\Omega \setminus \overline{\Omega}_{\delta}$,
for $\lambda \gg 1$,
\begin{equation}\label{5.2}
-\Delta\psi \leq
\lambda^{r}(\frac{2}{1+\alpha})\lambda_{1}\phi^{\frac{2}{1+\alpha}}\leq
\lambda
\frac{A(\lambda^{r}\phi^{\frac{2}{1+\alpha}})^\beta}{(\lambda^{r}\phi^{\frac{2}{1+\alpha}})^\alpha}\leq
\lambda \frac{g(\lambda^r \phi ^{\frac{2}{1+ \alpha}})}{(\lambda^r
\phi^{\frac{2}{1+\alpha}})^{\alpha}}.
\end{equation}
Combining (\ref{5.1}) and (\ref{5.2}), we see that
$$
-\Delta \psi
\leq \lambda \frac{g(\psi)}{\psi^{\alpha}} \quad \text{in }\Omega
$$
for $\lambda \gg 1$. Thus $\psi$ is a positive subsolution.

Now we construct a supersolution $Z\geq \psi$. Since
$1+\alpha-\gamma >0$ and $\gamma - \alpha \geq 0$, we can choose
$m(\lambda)\gg 1$ such that
\[
m(\lambda)^{1+\alpha-\gamma}\geq \lambda B e^{\gamma-\alpha}
\]
where $e$ is the unique positive solution of $-\Delta e = 1$ in
$\Omega, e = 0 $ on $\partial\Omega$. Hence for $m(\lambda)\gg 1$
$$
m(\lambda)\geq \frac{\lambda
B(m(\lambda)e)^{\gamma}}{(m(\lambda)e)^{\alpha}}.
$$
Let
$Z:=m(\lambda)e$. Then by (G1) we have
$$
-\Delta Z =m(\lambda)\geq \frac{\lambda
g(m(\lambda)e)}{(m(\lambda)e)^{\alpha}}.
$$
Thus $Z$ is a supersolution. Further $m(\lambda)$ can be chosen
large enough so that $Z \geq \psi$ on $\overline{\Omega}$. This is
possible since $e>0$ in $ \Omega$ and $ \frac{\partial e}{\partial
\nu}<0$ on $\partial \Omega$, where $\nu$ is outward normal vector
on $\partial \Omega$. Hence for $\lambda \gg 1$, \eqref{ePI} has a
positive solution and the proof is complete.
\end{proof}


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\end{document}
