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\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 49, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/49\hfil Existence of positive solutions]
{Existence of positive solutions for Dirichlet problems of some singular
elliptic  equations}

\author[Zhiren Jin\hfil EJDE--2003/49\hfilneg]
{Zhiren Jin}

\address{Zhiren Jin\newline
   Department of Mathematics and Statistics\\
   Wichita State University\\
   Wichita, Kansas, 67260-0033, USA}
\email{zhiren@math.twsu.edu}

\date{}
\thanks{Submitted December 11, 2002. Published April 29, 2003.}
\subjclass[2000]{35J25, 35J60, 35J65}
\keywords{Elliptic boundary value problems,
positive solutions, \hfill\break\indent
singular semilinear equations, unbounded domains,
Perron's method, super solutions.}


\begin{abstract}
  When an unbounded domain is inside a slab, existence of a positive
  solution is proved for the Dirichlet problem of a class of semilinear
  elliptic equations similar to the singular Emden-Fowler equation.
  The proof is based on a super and sub-solution method.
  A super solution is constructed by Perron's method together with a
  family of auxiliary functions.
\end{abstract}

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


\section{Introduction and Main Results}

Let $\Omega $ be an unbounded domain in $\mathbb{R}^n$ ($n\geq 3$) with
$C^{2,\alpha }$ ($0<\alpha <1$) boundary. We assume that $\Omega $ is
inside a slab of width $2M$:
$$
\Omega \subset  S_M=\{ (\mathbf{x} , y )\in \mathbb{R}^n : |y|<M \}
$$
where $\mathbf{x}=(x_1, x_2, \dots ,x_{n-1})$ and throughout the paper,
$y$ will be identified with $x_n$.

We consider the
existence of positive solutions for the Dirichlet problem
\begin{equation}
-\sum_{i,j=1}^n a_{ij} (\mathbf{x},y)D_{ij} u = p(\mathbf{x},y)u^{-\gamma }
\quad\mbox{on } \Omega;
\quad u=0 \quad \mbox{on } \partial \Omega ;
\label{eq:theproblem}
\end{equation}
where $(a_{ij})$ is a positive definite matrix in which
each entry is a local H{\"o}lder continuous function on ${\overline{\Omega}}$,
$p(\mathbf{x},y)$ is a also local H{\"o}lder continuous on $\overline{\Omega}$,
$\gamma > 0$ is a constant.

The main result of the paper is as follows.

\begin{theorem} \label{thm1}Assume
\begin{enumerate}
\item $p(\mathbf{x}_0,y_0)>0$ for
some $(\mathbf{x}_0,y_0)\in \Omega$;
\item there is a positive constant $C$ such that
\begin{equation}
0\leq p(\mathbf{x},y) \leq C(|\mathbf{x}|+1)^{\gamma}
\quad \rm{for} \quad (\mathbf{x},y)\in \Omega ;
\label{eq:boundofp}
\end{equation}
\item $\mathop{\rm Trace}(a_{ij})=1$ and there is a constant $c_1>0$, such that
\begin{equation}
a_{nn}(\mathbf{x},y)\geq c_1 \quad \mbox{on } \overline{\Omega }.
\label{eq:boundofann}
\end{equation}
\end{enumerate}
Then (\ref{eq:theproblem}) has a positive solution
$u\in C^{2}(\Omega )\cap C^{0}(\overline{\Omega } )$.
\end{theorem}

When the principal part in (\ref{eq:theproblem}) is the Laplace
operator, (\ref{eq:theproblem}) becomes a boundary value problem for the
singular  Emden-Fowler equation
\begin{equation}
-\Delta  u = p(\mathbf{x},y)u^{-\gamma } \quad\mbox{on } \Omega;
\quad u=0 \quad\mbox{on }  \partial \Omega .
\label{eq:theproblem1}
\end{equation}
The singular Emden-Fowler is related to the theory of heat
conduction in electrical conduction materials and in the studies
of boundary layer phenomena for viscous fluids \cite{Callegari1,Wong}.
The existence of positive solutions of the equation on
exterior domains (including $\mathbb{R}^n$) has been considered by
quite a number of authors (for example, see
\cite{Dalmasso,Edelson, Zjin,Kusano,Lair,Shaker}, and references
therein). The main approach used to prove existence is to
construct super and sub- solutions. To construct super solutions,
one needs to assume that $p(\mathbf{x},y)$ decays near infinity in
an appropriate rate. A super solution is usually found in the
class of radial symmetric functions. If $\Omega $ is an exterior
domain (not inside a slab), $\gamma >0$ and there is $C$ such that
$p(\mathbf{x},y)\geq \frac{C}{(1+|\mathbf{x}|^{2}+y^{2})}$ for
$|\mathbf{x}|^{2}+y^{2}$ large, then (\ref{eq:theproblem1}) has no
positive solutions (\cite{Kusano}). On the other hand, if  there
are constants $\sigma >1$ and $C$, such that $0\leq
p(\mathbf{x},y)\leq \frac{C}{(1+|\mathbf{x}|^{2}+y^{2})^{\sigma
}}$ for $|\mathbf{x}|^{2}+y^{2}$ large, (\ref{eq:theproblem1}) has
a positive solution (\cite{Zjin}). When $\Omega$ is an unbounded
domain inside a slab, the situation is quite different. The
traditional way to construct a super solution by finding an
appropriate radial symmetric function is no longer valid since the
domain now is inside a slab (the generality of the coefficient
matrix $(a_{ij})$ also makes finding a radial symmetric super
solution impossible). In this paper, we combine an idea from
\cite{Lopez} and a family of auxiliary functions constructed in
\cite{JL4} to construct a super solution which is then used to
prove the existence of a positive solution of
(\ref{eq:theproblem}).


Actually the procedure in the paper can be applied to prove the existence
of a positive solution for the Dirichlet problem of more general elliptic
equations. A statement for the general case will be given in the last
section of the paper. Here we just state a special case of the general result.

\begin{theorem} \label{thm2} Assume
\begin{enumerate}
\item $p(\mathbf{x}_0,y_0)>0$ for
some $(\mathbf{x}_0,y_0)\in \Omega$;

\item there is a positive constant $C$ such that
\begin{equation}
0\leq p(\mathbf{x},y) \leq Ce^{|\mathbf{x}|}
\quad \rm{for} \quad (\mathbf{x},y)\in \Omega ,
\label{eq:boundofp1}
\end{equation}

\item $\mathop{\rm Trace}(a_{ij})=1$, and there is a constant $c_1>0$, such that
\begin{equation}
a_{nn}(\mathbf{x},y)\geq c_1 \quad {\rm{on}} \quad
\overline{\Omega }.
\label{eq:boundofann2}
\end{equation}
\end{enumerate}
Then the problem
\begin{equation}
-\sum_{i,j=1}^n a_{ij} (\mathbf{x},y)D_{ij} u = p(\mathbf{x},y)e^{-u}
\quad\mbox{on } \Omega;
\quad u=0 \quad\mbox{on }  \partial \Omega
\label{eq:theproblem3}
\end{equation}
has a positive solution $u\in C^{2}(\Omega )\cap C^{0}(\overline{\Omega } )$.
\end{theorem}

This paper is organized as follows. In Section 2, we construct a family of auxiliary functions that are defined on
a family of subdomains of $\Omega$. In Section 3,
we combine the family of auxiliary functions constructed in Section 2 and an idea
from \cite{Lopez} to prove that (\ref{eq:theproblem}) has a positive supper solution.
In Section 4, we prove that (\ref{eq:theproblem}) has a positive solution by the procedure used in
\cite{Zjin}. In Section 5, we discuss the general case.

\section{A Family of Auxiliary Functions}

In this section, we will construct families of sub-domains $\Omega_{\mathbf{x}_0}$ of $\Omega $ and functions
$T_{\mathbf{x}_0} +z$ (see definitions below) so that
\begin{equation}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x}, y)D_{ij}(T_{\mathbf{x}_0} +z)
\geq p(\mathbf{x},y) (T_{\mathbf{x}_0} +z)^{-\gamma}
\quad \mbox{on } \Omega_{\mathbf{x}_0}
\label{eq:auxiliary}
\end{equation}
and the graphs of the functions $T_{\mathbf{x}_0} +z$ have special relative
positions (see below).

Our construction is based on the construction of a family of auxiliary functions
used in \cite{JL4}
(the construction in \cite{JL4} was adapted from
\cite{JL1} which in turn was inspired from \cite{Finn} and \cite{Serrin}).
We consider the operator
$$
Qu =\sum_{i,j=1}^n a_{ij}(\mathbf{x}, y)D_{ij}u .
$$
We first extend $a_{ij}$ ($1\leq i,j \leq n$) to be continuous functions on $\overline{S_M}$ in such a way that we still have
$\mathop{\rm Trace}(a_{ij})=1$ and
\begin{equation}
a_{nn}({\bf x},y) \ge c_1  \quad \mbox{on } S_M.
\label{eq:coeffbound}
\end{equation}
In the rest of the paper, we will use
$c_{m}$ (for some integer $m\geq 2$) to denote a constant depending only on $c_1$
and $M$. Once a constant $c_{m}$ is
used in a formula, it will represent the same constant if the same notation
appears again in the paper.

It was proved in \cite{JL4} (also see Appendix I) that
there are positive decreasing functions  $\chi (t)$,
$h_a(t)$  and a positive increasing function $A(t)$ ($\chi (t)$
depending on $c_1$ only, $h_a(t)$ and $A(t)$ depending on $c_1$ and $M$ only),
such that for any number $K$,
there is a number $H_0$, depending only on $K$, $M$ and $c_1$,
such that for $H\geq H_0$, we have (for $0<t<2M$)
\begin{gather}
 A(H)\leq h_a^{-1}(t) \leq A(H)e^{\chi (H)}, \quad
22MH\leq c_1A(H)e^{\chi (H)} \leq 66MH, \label{eq:boundofae} \\
 8K \leq A(H)e^{\chi (H)}, \quad 0<\chi (H) <1,
\label{eq:boundofae1}
\end{gather}
and the non-negative function
\begin{equation}
z=z_{\mathbf{x}_0}=A(H)e^{\chi (H)} -\{ (h_a^{-1}(y+M))^2-|{\bf x}-\mathbf{x}_0|^2 \}^{1/2}
\label{eq:solution111}
\end{equation}
satisfies
\begin{gather}
Qz  \le  \frac{-3c_1}{22eMH}  \quad \mbox{in }\Omega_{\mathbf{x}_0,H,K} ,
\label{eq:supersolution1} \\
z \ge   K  \quad \mbox{on } \partial\Omega_{\mathbf{x}_0,H,K}\cap \{ |y|<M \},
\quad z(\mathbf{x}_0,y) \le \frac{2M}{H} \quad \mbox{for }  |y|\le M ,
\label{eq:supersolution2}
\end{gather}
where
\begin{equation}
\Omega_{\mathbf{x}_0,H,K}
=\{(\mathbf{x},y):|y|< M,|\mathbf{x}-\mathbf{x}_0|<
\sqrt{\frac{2K}{A(H)e^{\chi (H)}}}h_a^{-1}(y+M) \} .
\label{eq:domain1}
\end{equation}
(For verifications of (\ref{eq:boundofae})-(\ref{eq:boundofae1}) and
(\ref{eq:supersolution1})-(\ref{eq:supersolution2}), see Appendix I.)

Now we set
\begin{equation}
K=100, \quad H=H_0+4M, \quad
\Omega_{\mathbf{x}_0}=\Omega_{\mathbf{x}_0,H,K} . \label{eq:specialvalue}
\end{equation}
Then (\ref{eq:supersolution1})-(\ref{eq:supersolution2}) becomes
\begin{gather}
Qz  \le  -c_2  \quad\mbox{in }\Omega_{\mathbf{x}_0}  , \label{eq:supersolution3}\\
z \ge   100  \quad \mbox{on } \partial\Omega_{\mathbf{x}_0}\cap \{ |y|<M \}, \quad
z(\mathbf{x}_0,y) \le 1 \quad \mbox{for }  |y|\le M.
\label{eq:supersolution4}
\end{gather}
Now we construct a family of auxiliary functions as follows.

If $(\mathbf{x},y)\in \Omega_{\mathbf{x}_0}$, from (\ref{eq:boundofae}) and (\ref{eq:domain1}), we have
$$
|\mathbf{x}-\mathbf{x}_0|< \sqrt{200A(H)e^{\chi (H)}}
\leq \sqrt{13200MH/c_1}=c_{4}.
$$
For $C$ defined in (\ref{eq:boundofp}), we set
\begin{equation}
T_{\mathbf{x}_0}= (\frac{C}{c_2})^{1/\gamma} (|\mathbf{x}_0|+c_{4}+1) .
\label{eq:definitionoft}
\end{equation}
Then we have that on $\Omega_{\mathbf{x}_0}$,
$$
p(\mathbf{x},y) (T_{\mathbf{x}_0}+z)^{-\gamma} \leq
C(|\mathbf{x}|+1)^{\gamma } T_{\mathbf{x}_0}^{-\gamma }
\leq
\frac{C(|\mathbf{x}_0|+c_{4}+1)^{\gamma}}{T_{\mathbf{x}_0}^{\gamma }}
= c_2.
$$
Thus
\begin{equation}
-Q(T_{\mathbf{x}_0} +z) \geq c_2
\geq p(\mathbf{x},y) (T_{\mathbf{x}_0} +z)^{-\gamma}
\quad \mbox{on } \Omega_{\mathbf{x}_0}.
\label{eq:supersolution}
\end{equation}
When $\mathbf{x}_0$ changes, we obtain families of auxiliary functions
$T_{\mathbf{x}_0} +z$ and domains $\Omega_{\mathbf{x}_0}$ satisfying
(\ref{eq:auxiliary}).

To be able to use the family of auxiliary functions, we need to investigate
relative positions of the graphs of these auxiliary functions.

For two points ${\mathbf{x}_0}$ and ${\mathbf{x}_1}$ in $R^{n-1}$,
when $\Omega_{\mathbf{x}_1}$ either covers the whole segment of the set
$\{ ({\mathbf{x}_0},y)||y|\leq M\}$ or does not intersect with the set, from (\ref{eq:boundofae}) and
(\ref{eq:domain1}), we have either
\begin{equation}
|\mathbf{x}_1-\mathbf{x}_0|\leq \sqrt{200A(H)e^{-\chi (H)}} \quad\mbox{or}\quad
|\mathbf{x}_1-\mathbf{x}_0|\geq \sqrt{200A(H)e^{\chi (H)}}.
\label{eq:overlapping0}
\end{equation}
Then when
$\Omega_{\mathbf{x}_1}$ covers part of some neighborhood of
$\{ ({\mathbf{x}_0},y): |y|\leq M\}$,
we have
\begin{equation}
\sqrt{195A(H) e^{-\chi (H)}}
\leq |\mathbf{x}_1-\mathbf{x}_0|\leq \sqrt{205A(H)e^{\chi (H)}} .
\label{eq:overlapping}
\end{equation}
Let
$\mathbf{x}_1$ and $\mathbf{x}_0$ satisfy
(\ref{eq:overlapping}) and
$\delta_0$ be a small positive
number such that $2\delta_0<\sqrt{195A(H)e^{-\chi (H)}}$.
If $(\mathbf{x},y)\in \Omega_{\mathbf{x}_1}$ for some $y$ and
$|\mathbf{x}-\mathbf{x}_0|\leq \delta_0$,
by (\ref{eq:boundofae}), (\ref{eq:solution111}) and (\ref{eq:overlapping}), we have
\begin{align*}
&T_{\mathbf{x}_1} + z_{\mathbf{x}_1}(\mathbf{x}, y) \\
&\geq T_{\mathbf{x}_1} +A(H)e^{\chi (H)} -\{ A(H)^{2} e^{2\chi (H)}
- |\mathbf{x}-\mathbf{x}_1| \}^{1/2} \\
&\geq T_{\mathbf{x}_1} +A(H)e^{\chi (H)} -\big\{ A(H)^{2} e^{2\chi (H)}
-(\sqrt{195A(H)e^{-\chi (H)}} -\delta_0)^{2} \big\}^{1/2} \\
& \geq T_{\mathbf{x}_1} +A(H)e^{\chi (H)} \\
&\quad -\big\{ A(H)^{2} e^{2\chi (H)}
-195A(H)e^{-\chi (H)} +2\delta_0 \sqrt{195A(H)e^{-\chi (H)}} \big\}^{1/2} \\
&\geq T_{\mathbf{x}_1} + A(H)e^{\chi (H) } \Big(1-\Big(1-\frac{195}{A(H)e^{3\chi (H)}}+
\frac{2\delta_0\sqrt{195A(H)e^{-\chi (H)}}}{A(H)^{2}e^{2\chi (H)}} \Big)^{1/2}\Big) \\
&\mbox{(by the inequality $\sqrt{1-t}\leq 1-\frac{1}{2}t$ for $0<t<1$ and
(\ref{eq:boundofae1}))}\\
& \geq T_{\mathbf{x}_1} +A(H)e^{\chi (H) } (\frac{195}{2A(H)e^{3\chi (H)}}
-\frac{2\delta_0\sqrt{195A(H)e^{-\chi (H)}}}{2A(H)^{2}e^{2\chi(H)}} ) \\
&= T_{\mathbf{x}_1}+\frac{195}{2e^{2\chi (H)}}
-\frac{\delta_0\sqrt{195A(H)e^{-\chi (H)}}}{A(H)e^{\chi(H)}}
 >T_{\mathbf{x}_1} +10-\frac{\delta_0\sqrt{195A(H)e^{-\chi (H)}}}{A(H)e^{\chi(H)}}.
\end{align*}
Thus there is a $\delta_0$ small such that for all
$|\mathbf{x}-\mathbf{x}_0|\leq \delta_0$ with
$(\mathbf{x},y)\in \Omega_{\mathbf{x}_1}$,
if $\mathbf{x}_1$ and $\mathbf{x}_0$ satisfy
(\ref{eq:overlapping}),
we have
\begin{equation}
T_{\mathbf{x}_1} + z_{\mathbf{x}_1}(\mathbf{x}, y) \geq
T_{\mathbf{x}_1} +8. \label{eq:test1}
\end{equation}
Further for all $\mathbf{x}_0$ and $\mathbf{x}_1$ satisfying
(\ref{eq:overlapping}),
\begin{align*}
T_{\mathbf{x}_0} +2 &\leq T_{\mathbf{x}_1} + T_{\mathbf{x}_0}
- T_{\mathbf{x}_1} +2 \\
&\leq T_{\mathbf{x}_1} + (\frac{C}{c_2})^{\frac{1}{\gamma}} (|\mathbf{x}_0|
- |\mathbf{x}_1|)+2 \\
&\leq T_{\mathbf{x}_1} + (\frac{C}{c_2})^{\frac{1}{\gamma}}
|\mathbf{x}_1-\mathbf{x}_0|+2 \\
&\leq T_{\mathbf{x}_1} + (\frac{C}{c_2})^{\frac{1}{\gamma}}
\sqrt{205A(H)e^{\chi (H)}} +2\\
&\leq T_{\mathbf{x}_1} + (\frac{C}{c_2})^{\frac{1}{\gamma}}c_{5}+2
\end{align*}
where $c_{5} = \sqrt{205A(H)e^{\chi (H)}}$.
Thus if we assume that $C$ in (\ref{eq:boundofp}) satisfies
\begin{equation}
C\leq 6^{\gamma}c_{5}^{-\gamma }c_2 ,
\label{eq:smallofp}
\end{equation}
we have that for all $\mathbf{x}_0$ and $\mathbf{x}_1$ satisfying
(\ref{eq:overlapping}),
\begin{equation}
T_{\mathbf{x}_0} +2 \leq T_{\mathbf{x}_1} +8\,. \label{eq:test2}
\end{equation}
 From (\ref{eq:domain1}) and (\ref{eq:supersolution4}), we can choose a number $\delta_2({\mathbf{x}_0})>0$ such that
for all $\mathbf{x}\in R^{n-1}$ with
$|\mathbf{x}_0 - \mathbf{x}|\leq \delta_2({\mathbf{x}_0})$, we have
$(\mathbf{x},y)\in \Omega_{\mathbf{x}_0}$ for all $|y|< M$, and
\begin{equation}
T_{\mathbf{x}_0} + z_{\mathbf{x}_0}(\mathbf{x}, y)
\leq T_{\mathbf{x}_0} +2\,. \label{eq:test3}
\end{equation}
Now if we set $\delta_{\mathbf{x}_0} =\min \{ \delta_0, \delta_2({\mathbf{x}_0})\}$,
from (\ref{eq:test1}), (\ref{eq:test2}) and (\ref{eq:test3}), we have
\begin{equation}
T_{\mathbf{x}_0}+z_{\mathbf{x}_0}(\mathbf{x},y)
\leq T_{\mathbf{x}_1} +z_{\mathbf{x}_1}(\mathbf{x},y)
\label{eq:definitionofradius}
\end{equation}
for all $\mathbf{x}_0$ and $\mathbf{x}_1$ satisfying (\ref{eq:overlapping}),
$|\mathbf{x}_0 - \mathbf{x}|\leq \delta_{\mathbf{x}_0}$ and
$(\mathbf{x},y)\in \Omega_{\mathbf{x}_1}$.

Finally we define a family of open subsets of $\Omega $ that will be needed in
next section.

For each point $(\mathbf{x}_0,y_0)\in \overline{\Omega} $, we define an open set
$O(\mathbf{x}_0,y_0)$ as follows:
\begin{enumerate}
\item If $(\mathbf{x}_0,y_0)\in \Omega$, we choose a ball $B$ with
center $(\mathbf{x}_0,y_0)$ and a radius less than $\delta_{\mathbf{x}_0}$ so that $B\subset \Omega$.
We then set $O(\mathbf{x}_0,y_0)=B$;

\item If $(\mathbf{x}_0,y_0)\in \partial \Omega$, since $\Omega $ has $C^{2,\alpha }$ boundary, there is a
ball $B$ with center $(\mathbf{x}_0,y_0)$ and a radius less than $\delta_{\mathbf{x}_0}$, such that there is a
$C^{2,\alpha }$ diffeomorphism $\Phi $ satisfying
$$
\Phi (B\cap \Omega ) \subset \mathbb{R}^n_{+}, \quad
\Phi (B\cap \partial \Omega ) \subset \partial \mathbb{R}^n_{+}; \quad
\Phi (\mathbf{x}_0,y_0)={\bf{0}}.
$$
\end{enumerate}
Now we choose a domain $J$ with $C^{3}$ boundary with following properties:
(a) $J\subset \Phi (B\cap \Omega )$; (b) $\partial J\cap \partial \mathbb{R}^n_{+}$ is a neighborhood
of ${\bf{0}}$ in $\partial \mathbb{R}^n_{+}$. Certainly there are many different $J$'s having those properties.
One example is given in the Appendix II at the end of paper to illustrate
how to construct such a domain $J$.

Now we set $O(\mathbf{x}_0,y_0)=\Phi^{-1}(J)$. It is easy to see that
$O(\mathbf{x}_0,y_0)\subset B\cap \Omega $, $O(\mathbf{x}_0,y_0)$
has a $C^{2,\alpha }$ boundary and
$\partial O(\mathbf{x}_0,y_0)\cap \partial \Omega$ is a neighborhood of
$(\mathbf{x}_0,y_0)$ in $\partial \Omega$.
Let $\Pi$ be the collection of all such open sets $O(\mathbf{x}_0,y_0)$
defined in (1) and (2).


\section{A Super Solution of (\ref{eq:theproblem})}

In this section, using the family of auxiliary functions
$T_{\mathbf{x}_0}+z$ constructed in Section 2 and an idea
from \cite{Lopez} (that basically says that the Perron's method still works if we can find a family of appropriate
auxiliary functions that works like a super solution), we will show that there is a positive
function $u\in C^{2}(\Omega )\cap C^{0}(\overline{\Omega})$, satisfies
$$
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}u = p(\mathbf{x},y)u^{-\gamma }
\quad \mbox{on } \Omega ,
\quad u=\tau  \quad \mbox{on } \partial \Omega  .
$$
for some constant $\tau >0$. Then $u$ will be a super solution of (\ref{eq:theproblem}).

If $u=c_0v$ for some constant $c_0$, $v$ will satisfy
$$
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}v = c_0^{-\gamma -1}
p(\mathbf{x},y)v^{-\gamma } \quad\mbox{on } \Omega ,
\quad v=\tau /c_0 \quad \mbox{on } \partial \Omega  \,.
$$
Thus without loss of generality, we may assume $C$ in (\ref{eq:boundofp})
satisfying (\ref{eq:smallofp}).
Then all constructions in Section 2 are valid.

Let $v>0$ be a function on $\overline{\Omega}$, for a point $(\mathbf{x}_0,y_0)\in \overline{\Omega } $,
we define a new function $M_{(\mathbf{x}_0,y_0)}(v)$, called the
lift of $v$ over $O(\mathbf{x}_0,y_0)$ as follows:
\begin{gather*}
M_{(\mathbf{x}_0,y_0)}(v)(\mathbf{x},y)=v(\mathbf{x},y) \quad {\rm{if}}
\quad (\mathbf{x},y)\in \Omega\setminus O(\mathbf{x}_0,y_0) \\
M_{(\mathbf{x}_0,y_0)}(v)(\mathbf{x},y)=w(\mathbf{x},y) \quad {\rm{if}}
\quad (\mathbf{x},y)\in O(\mathbf{x}_0,y_0)
\end{gather*}
where $w(\mathbf{x},y)$ is the positive solution of the boundary-value problem
\begin{equation}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w = p(\mathbf{x},y) w^{-\gamma }
\quad  \mbox{in } O(\mathbf{x}_0,y_0), \quad
w=v \quad \mbox{on } \partial O(\mathbf{x}_0,y_0) \,. \label{eq:lift}
\end{equation}
It is easy to see (\ref{eq:lift}) has a unique positive solution in
$C^{2}(O(\mathbf{x}_0,y_0))\cap
C^{0}(\overline{O(\mathbf{x}_0,y_0)})$. Indeed
$m_1=\min \{ v (\mathbf{x},y): (\mathbf{x},y)\in \partial O(\mathbf{x}_0,y_0)\}$ is a sub-solution since
$p(\mathbf{x},y)$ is non-negative,
$m_2+T_{\mathbf{x}_0} + z_{\mathbf{x}_0}$ is a super solution  by
(\ref{eq:auxiliary}), where
$m_2=\max \{ v (\mathbf{x},y): (\mathbf{x},y)\in \partial O(\mathbf{x}_0,y_0)\}$.
Then we can conclude the existence of a desired solution (for example, see \cite{Amann} or \cite{Crandall}).
Uniqueness of positive
solutions of (\ref{eq:lift}) follows from a standard argument.

Set $\tau = (C/c_2)^{1/\gamma} c_{4}$ (see (\ref{eq:definitionoft}) for
the source of the constants).

We define a class $\Xi $ of functions as follows: a function $v$ is in $\Xi $ if
\begin{enumerate}
\item $v\in C^{0}(\overline{\Omega })$, $v>0$ on $\overline{\Omega}$ and
$v\leq \tau $ on $\partial \Omega$;

\item For any $(\mathbf{x}_0,y_0)\in \overline{\Omega}$,
$v\leq M_{(\mathbf{x}_0,y_0)}(v)$;

\item $v\leq T_{\mathbf{x}_0} +z_{\mathbf{x}_0}$ on $\Omega_{\mathbf{x}_0} \cap \Omega$
for any $(\mathbf{x}_0,y_0)\in \overline{\Omega}$.
\end{enumerate}
By the following well-known lemma, it is easy to check the function $v=\tau $
is in $\Xi$. Thus $\Xi$ is not empty.

\begin{lemma} \label{lm1}
Let $D$ be a bounded domain, $f(\mathbf{x},y,t)$ be a $C^{1}$ function that is
decreasing in $t$. If $w_1$, $w_2$ are in $C^{2}(D)\cap C^{0}(\overline{D})$,
$w_1\leq w_2$  on $\partial D$, and
\begin{gather*}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w_1 \leq f(\mathbf{x},y,w_1)
\quad  \mbox{in } D, \\
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w_2 \geq  f(\mathbf{x},y, w_2)
\quad\mbox{in }D
\end{gather*}
then $w_1\leq w_2$ on $D$.
\end{lemma}

Now we set
$$
u(\mathbf{x},y)=\sup_{v\in {\Xi }} v(\mathbf{x},y), \quad
 (\mathbf{x},y)\in \overline{\Omega} \,.
$$
We will show that $u$ is in $C^{2}(\Omega )\cap C^{0}(\overline{\Omega } )$
and satisfies
$$
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}u = p(\mathbf{x},y) u^{-\gamma }
 \quad  \mbox{on } \Omega ;\quad
u=\tau  \quad \mbox{on } \partial \Omega .
$$

First we need some lemmas.

\begin{lemma} \label{lm2}
If $0<v_1\leq v_2$, then
$M_{(\mathbf{x}_0,y_0)}(v_1)\leq M_{(\mathbf{x}_0,y_0)}(v_2)$
for any $(\mathbf{x}_0,y_0)\in \overline{\Omega} $.
\end{lemma}

\begin{proof} Let $w_1$, $w_2$ be the positive solutions for the following problems
\begin{gather*}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w_{k} = p(\mathbf{x},y)
w_{k}^{-\gamma } \quad\mbox{in }O(\mathbf{x}_0,y_0), \\
w_{k}=v_{k}\quad \mbox{on }  \partial O(\mathbf{x}_0,y_0), \quad k=1,2.
\end{gather*}
Since $w_1=v_1\leq v_2=w_2$ on $\partial O(\mathbf{x}_0,y_0)$,
 $p(\mathbf{x},y) t^{-\gamma }$ is decreasing on $t$, from lemma 1, we see
$w_1\leq w_2$ on $O(\mathbf{x}_0,y_0)$. On $\Omega \setminus  O(\mathbf{x}_0,y_0)$,
$M_{(\mathbf{x}_0,y_0)}(v_1)=v_1$, $M_{(\mathbf{x}_0,y_0)}(v_2)=v_2$.
Thus $M_{(\mathbf{x}_0,y_0)}(v_1)\leq M_{(\mathbf{x}_0,y_0)}(v_2)$.
\end{proof}

\begin{lemma} \label{lm3}
If $v_1\in \Xi$, $v_2\in \Xi$, then $\max \{ v_1, v_2 \} \in \Xi$.
\end{lemma}

\begin{proof} If $v_1\in \Xi$, $v_2\in \Xi$, it is clear that
$\max \{ v_1, v_2 \}\in C^{0}(\overline{\Omega })$,
$\max \{ v_1, v_2 \}>0$ on $\overline{\Omega}$ and $\max \{ v_1, v_2 \}\leq \tau $ on $\partial \Omega$.
It is also clear that
$\max \{ v_1, v_2 \}\leq T_{\mathbf{x}_0} +z_{\mathbf{x}_0}$ on $\Omega_{\mathbf{x}_0} \cap \Omega$
for any $(\mathbf{x}_0,y_0)\in \overline{\Omega}$.
Since
$$
v_1\leq \max \{ v_1, v_2 \}, \quad v_2 \leq \max \{ v_1, v_2\}
$$
we have (by lemma 2) that for any $(\mathbf{x}_0,y_0)\in \overline{\Omega} $,
$$
M_{(\mathbf{x}_0,y_0)}(v_1) \leq
M_{(\mathbf{x}_0,y_0)}(\max \{ v_1, v_2\} ),\quad
M_{(\mathbf{x}_0,y_0)}(v_2) \leq
M_{(\mathbf{x}_0,y_0)}(\max \{ v_1, v_2\} ).
$$
Since $v_1\in \Xi $ and $v_2\in \Xi $ imply
$$
v_1 \leq M_{(\mathbf{x}_0,y_0)}(v_1),\quad
v_2 \leq M_{(\mathbf{x}_0,y_0)}(v_2),
$$
we have
$$
\max \{ v_1, v_2 \}\leq M_{(\mathbf{x}_0,y_0)}(\max \{ v_1, v_2\}) .
$$
Thus $\max \{ v_1, v_2 \}\in \Xi$.
\end{proof}

\begin{lemma} \label{lm4}
If $v\in \Xi $, then $M_{(\mathbf{x}_0,y_0)}(v)\in \Xi$
for any $(\mathbf{x}_0,y_0)\in \overline{\Omega} $.
\end{lemma}

\begin{proof} By the definition of $M_{(\mathbf{x}_0,y_0)}( v)$, it is clear
that $M_{(\mathbf{x}_0,y_0)}( v)>0$ on $\overline{\Omega }$,
$M_{(\mathbf{x}_0,y_0)}( v)\in C^{0}(\overline{\Omega} )$ and
$M_{(\mathbf{x}_0,y_0)}(v) \leq \tau $ on $\partial \Omega $.

For any $(\mathbf{x}^{*},y^{*})\in \overline{\Omega} $,
we first show that
\begin{equation}
M_{(\mathbf{x}_0,y_0)}( v)(\mathbf{x},y)\leq
M_{(\mathbf{x}^{*},y^{*})}(M_{(\mathbf{x}_0,y_0)}( v))(\mathbf{x},y).
\label{eq:mv}
\end{equation}
We only need to prove that (\ref{eq:mv}) is true for
$(\mathbf{x},y)\in O(\mathbf{x}^{*},y^{*})$.
Since
$$
v\leq M_{(\mathbf{x}_0,y_0)}( v),
$$
we have (by lemma 2)
$$
M_{(\mathbf{x}^{*},y^{*})}( v)\leq M_{(\mathbf{x}^{*},y^{*})}(M_{(\mathbf{x}_0,y_0)}( v)).
$$
Then from $v\leq M_{(\mathbf{x}^{*},y^{*})}( v)$ (by lemma 2 again), we have
$$
v\leq M_{(\mathbf{x}^{*},y^{*})}(M_{(\mathbf{x}_0,y_0)}( v)).
$$
Thus for $(\mathbf{x},y)\in
O(\mathbf{x}^{*},y^{*}) \setminus O(\mathbf{x}_0,y_0)$,
\begin{equation}
M_{(\mathbf{x}_0,y_0)}( v)(\mathbf{x},y)=v(\mathbf{x},y)
\leq M_{(\mathbf{x}^{*},y^{*})}(M_{(\mathbf{x}_0,y_0)}( v))(\mathbf{x},y).
\label{eq:boundary1}
\end{equation}
That is, (\ref{eq:mv}) is true on $O(\mathbf{x}^{*},y^{*}) \setminus O(\mathbf{x}_0,y_0)$,
Now for $\Omega_1=O(\mathbf{x}^{*},y^{*}) \cap O(\mathbf{x}_0,y_0)$,
if we set
$$
M_{(\mathbf{x}_0,y_0)}( v)=w_1,
\quad
M_{(\mathbf{x}^{*},y^{*})}(M_{(\mathbf{x}_0,y_0)}( v))=w_2
$$
we have
\begin{gather*}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w_1 = p(\mathbf{x},y) w_1^{-\gamma }
 \quad\mbox{on }  \Omega_1, \\
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w_2 = p(\mathbf{x},y) w_2^{-\gamma }
\quad \mbox{on }\Omega_1 .
\end{gather*}
On $\partial \Omega_1$,
$w_1\leq w_2$ on $O(\mathbf{x}^{*},y^{*})\cap \partial O(\mathbf{x}_0,y_0)$
by (\ref{eq:boundary1}) and
$w_1\leq w_2$ on $\partial O(\mathbf{x}^{*},y^{*})\cap O(\mathbf{x}_0,y_0)$  since
(\ref{eq:mv}) is true on $\Omega\setminus O(\mathbf{x}^{*},y^{*})$.
Then
lemma 1 implies $w_1\leq w_2$ on $\Omega_1$.
Thus (\ref{eq:mv}) is true on $O(\mathbf{x}^{*},y^{*}) \cap O(\mathbf{x}_0,y_0)$ and on
$O(\mathbf{x}^{*},y^{*})$.
\end{proof}

Now we prove that $M_{(\mathbf{x}_0,y_0)}(v)\leq T_{\mathbf{x}_1}+z_{\mathbf{x}_1}$
on $\Omega_{\mathbf{x}_1}\cap \Omega$
for all $(\mathbf{x}_1,y_1)\in \overline{\Omega }$.

By the definition of $M_{(\mathbf{x}_0,y_0)}(v)$, we only need to consider the graph of the function
$M_{(\mathbf{x}_0,y_0)}(v)$ over $O(\mathbf{x}_0,y_0)$.
If $O(\mathbf{x}_0,y_0)$ is covered completely by $\Omega_{\mathbf{x}_1}$, since
$v\leq T_{\mathbf{x}_1}+z_{\mathbf{x}_1}$ and $T_{\mathbf{x}_1}+z_{\mathbf{x}_1}$ satisfies (\ref{eq:auxiliary}),
$T_{\mathbf{x}_1}+z_{\mathbf{x}_1}$ is a super solution of (\ref{eq:lift}) on $O(\mathbf{x}_0,y_0)$. Then
Lemma \ref{lm1} implies
$M_{(\mathbf{x}_0,y_0)}(v)\leq T_{\mathbf{x}_1}+z_{\mathbf{x}_1}$ on $O(\mathbf{x}_0,y_0)$.
In the case that
$O(\mathbf{x}_0,y_0)$ does not intersect with $\Omega_{\mathbf{x}_1}$, the conclusion is trivial.
Now we consider the case that
$O(\mathbf{x}_0,y_0)$ is partially covered by $\Omega_{\mathbf{x}_1}$.
Since $O(\mathbf{x}_0,y_0)$ is covered by 
$\Omega_{\mathbf{x}_0}$, we always have
\begin{equation}
M_{(\mathbf{x}_0,y_0)}(v)\leq T_{\mathbf{x}_0}+z_{\mathbf{x}_0}
\quad \mbox{on } O(\mathbf{x}_0,y_0).
\label{eq:control}
\end{equation}
Then by the choice of $\delta_{\mathbf{x}_0}$, $O(\mathbf{x}_0,y_0)$,
and the fact that $O(\mathbf{x}_0,y_0) \cap T_{\mathbf{x}_1}$ is not empty,  
we have that $\mathbf{x}_0$ and $\mathbf{x}_1$ satisfy (\ref{eq:overlapping}), and for 
all $(\mathbf{x},y)\in O(\mathbf{x}_0,y_0)\cap \Omega_{\mathbf{x}_1}$,  
$|\mathbf{x}_0 - \mathbf{x}|\leq \delta_{\mathbf{x}_0}$.
Then by (\ref{eq:definitionofradius}),
the graph of
$T_{\mathbf{x}_0}+z_{\mathbf{x}_0}$  over $O(\mathbf{x}_0,y_0)\cap \Omega_{\mathbf{x}_1}$ is under the graph
of  $T_{\mathbf{x}_1}+z_{\mathbf{x}_1}$. Thus the conclusion follows from (\ref{eq:control}).

Now we are ready to prove that $u$ has the desired properties.

Let $(\mathbf{x}_0,y_0)\in \overline{\Omega }$.
By the definition of $u(\mathbf{x}_0,y_0)$, there is a sequence of functions
$v_{k}$ in $\Xi $ such that
$$
u(\mathbf{x}_0,y_0)=\lim_{k\to \infty } v_{k}(\mathbf{x}_0,y_0).
$$
By lemma 3 and the fact that $v=\tau $ is in $\Xi$,
replacing $v_{k}$ by $\max \{ v_{k}, \tau \}$ if it is necessary, we may assume
that $v_{k}\geq \tau $ on $\Omega$.
We replace $v_{k}$ by $M_{(\mathbf{x}_0,y_0)}(v_{k})$. Then we have a sequence of
functions $w_{k}$ satisfying
\begin{gather*}
u(\mathbf{x}_0,y_0)=\lim_{k\to \infty } w_{k}(\mathbf{x}_0,y_0) ,\\
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w_{k} =p(\mathbf{x},y) w_{k}^{-\gamma } \quad on \quad
O(\mathbf{x}_0,y_0), \\
w_{k}=v_{k} \quad on \quad \partial O(\mathbf{x}_0,y_0).
\end{gather*}
Since for all $k$,
$$
\tau \leq v_{k}\leq w_{k} \leq T_{\mathbf{x}_0} +z_{\mathbf{x}_0}
\quad \mbox{on } O(\mathbf{x}_0,y_0).
$$
By \cite[Theorem 9.11]{Trudinger} and an approximation of the boundary value by smooth functions,
we see that there is a subsequence of $w_{k}$, for convenience still denoted by $w_{k}$, converges to a
$C^{2}(O(\mathbf{x}_0,y_0))\cap C^{0}(\overline{O(\mathbf{x}_0,y_0)})$
function $w(x)$ in $C^{2}(O(\mathbf{x}_0,y_0)) \cap C^{0}(\overline{O(\mathbf{x}_0,y_0)})$. Thus $w(x)$ satisfies
$$
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w = p(\mathbf{x},y)w^{-\gamma} \quad on \quad
O(\mathbf{x}_0,y_0)
$$
and $u(\mathbf{x}_0,y_0)=w(\mathbf{x}_0,y_0)$.
We claim that $u=w$ on $O(\mathbf{x}_0,y_0)$. Indeed, if there is another point
$(\mathbf{x}_2,y_2) \in O(\mathbf{x}_0,y_0)$ such that
$u(\mathbf{x}_2,y_2)$ is not equal to $w(\mathbf{x}_2,y_2)$, then
$u(\mathbf{x}_2,y_2)>w(\mathbf{x}_2,y_2)$. Then there is a function $u_0\in \Xi $, such
that
$$
w(\mathbf{x}_2,y_2)< u_0(\mathbf{x}_2,y_2)\leq
u(\mathbf{x}_2,y_2).
$$
Now the sequence $\max \{ u_0, M_{(\mathbf{x}_0,y_0)}(v_{k}) \}$ satisfying
$$
v_{k} \leq \max \{ u_0, M_{(\mathbf{x}_0,y_0)}(v_{k}) \} \leq u .
$$
Then similar to the way we obtain $w$, $M_{(\mathbf{x}_0,y_0)}(\max \{ u_0, M_{(\mathbf{x}_0,y_0)}(v_{k}) \})$
will produce a function
$w_1$ satisfying
\begin{gather*}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w_1 =p(\mathbf{x},y) w_1^{-\gamma }
\quad\mbox{on } O(\mathbf{x}_0,y_0), \\
w\leq w_1 \quad on \quad O(\mathbf{x}_0,y_0), \quad
w(\mathbf{x}_2,y_2)<u_0(\mathbf{x}_2,y_2)\leq  w_1(\mathbf{x}_2,y_2), \\
w(\mathbf{x}_0,y_0)=w_1(\mathbf{x}_0,y_0)=u(\mathbf{x}_0,y_0) .
\end{gather*}
That is,  $w_1(\mathbf{x},y)-w(\mathbf{x},y)$ is non-negative, not identically zero on
$O(\mathbf{x}_0,y_0)$
and
achieves its minimum value zero inside $O(\mathbf{x}_0,y_0)$.
However, from the equations satisfied by $w$ and $w_1$, we have
that on $O(\mathbf{x}_0,y_0)$,
$$
-\sum_{i,j=1}^n a_{i,j}(\mathbf{x},y) D_{ij}(w_1-w)+
\gamma p(\mathbf{x},y) (w+\theta (w_1-w))^{-\gamma -1 } (w_1-w)=0
$$
for some continuous function $\theta$.
Then by the standard maximum principle (for example, see
\cite[Theorem 3.5]{Trudinger}), we get a contradiction.
Thus $u=w$ on $O(\mathbf{x}_0,y_0)$. Therefore $u\in C^{2}(\Omega )$ and
$$
-\sum_{i,j=1}^n a_{i,j}(\mathbf{x},y) D_{ij}u =p(\mathbf{x},y) u^{-\gamma }
\quad\mbox{on } \Omega .
$$
When $(\mathbf{x}_0,y_0)\in \partial \Omega$,  $\partial O(\mathbf{x}_0,y_0) \cap \partial \Omega $
is a neighborhood of $(\mathbf{x}_0,y_0)$ in $\partial \Omega$. Since
$\max\{ \tau ,v_{k}\}=\tau $ on $\partial \Omega$, $u=\tau $ on $\partial \Omega$ and
$w=\tau $
on $\partial O(\mathbf{x}_0,y_0) \cap \partial \Omega$. Since $w$ is continuous up to the boundary of
$O(\mathbf{x}_0,y_0)$, $u$ is continuous on $\partial O(\mathbf{x}_0,y_0) \cap \partial \Omega$ from inside
$O(\mathbf{x}_0,y_0)$. Thus $u\in C^{0}(\overline{\Omega})$ and
$u=\tau $ on $\partial \Omega $.


\section{Proof of Existence}

Using the super solution $u$ constructed in Section 3, we can prove
the existence of a positive solution of (\ref{eq:theproblem}) exactly in the
same way as that in \cite{Zjin}
(the generality of the principal term of the elliptic operator will not cause any extra
difficulty). We just sketch the proof here.

Since $\Omega $ is an unbounded domain with $C^{2, \alpha}$ boundary,
we can choose a sequence of subdomains of $\Omega$, denoted by
${\Omega }_{m}$, $m=1,2,3, \dots$, such that
\begin{enumerate}
\item ${\Omega}_{m} \subset {\Omega}_{m+1} \subset \Omega$ for all
$m$;

\item $\cup {\Omega}_{m} = \Omega$;

\item each ${\Omega }_{m}$ is a
bounded domain with $C^{2,\alpha}$ boundary;

\item $dist(0, \partial \Omega \setminus \partial {\Omega}_{m})
\to \infty $ as $m\to \infty $.
\end{enumerate}
We can find a number $\mu$, such that
for each large $m$, the eigenvalue problem
$$
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w =\lambda (\mu p(\mathbf{x},y)) w
\quad\mbox{on }\Omega_{m}, \quad
w=0 \quad\mbox{on } \partial \Omega_{m}
$$
has a first eigenvalue ${\lambda_1}<1$ with its first eigenfunction $\phi_{m}$.
We can assume $\max \phi_{m}=1$. Choose $\delta_{m}$ such that
$\delta_{m}\leq \frac{1}{2} \tau $ and
$$
\mu p(\mathbf{x},y) t \leq p(\mathbf{x},y) t^{-\gamma }\quad\mbox{for }(\mathbf{x},y)
\in \Omega_{m}, \quad
0<t<\delta_{m} .
$$
Then
\begin{equation}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w = p(\mathbf{x},y)w^{-\gamma } \quad\mbox{on }\Omega_{m}, \quad
w=0 \quad\mbox{on }\partial \Omega_{m}
\label{eq:app}
\end{equation}
has a pair of super and sub solutions $u(\mathbf{x},y)$, $\delta_{m} \phi_{m}$. Thus (\ref{eq:app})
has a solution
$w_{m}$ that can be proved to satisfy
\begin{gather*}
0<w_{m}<u \quad\mbox{on }\Omega_{m}, \\
\frac{1}{2}\delta_{s} \phi_{s} \leq w_{m} \quad\mbox{on }\Omega_{m}
\end{gather*}
for all $m>s$. Finally we take limit of $w_{m}$ to get a desired solution.


\section{The General Case}

Now we consider the boundary-value problem
\begin{equation}
- \sum_{i,j=1}^n a_{ij}(\mathbf{x},y)D_{ij}u =g(\mathbf{x},y,u) \quad\mbox{on }\Omega ,
\quad u=0 \quad\mbox{on }\partial \Omega .
\label{eq:general}
\end{equation}
In addition to the assumptions on $(a_{ij})$ and $\Omega$ given at the beginning of the
paper, we assume the following conditions.
\begin{enumerate}
\item $\mathop{\rm Trace}a_{ij})=1$;

\item There is a constant $c_1>0$ such that
$a_{nn} \geq c_1$ on $\overline{\Omega}$;

\item There is a family of increasing positive functions $T=T(t)$ satisfying
(with $T_{\mathbf{x}}=T(|\mathbf{x}|)$)
\begin{itemize}
\item[(a)] $|T_{\mathbf{x}_0}-T_{\mathbf{x}}|\leq |\mathbf{x}_0-\mathbf{x}|/c_{5}$;

\item[(b)] $g(\mathbf{x},y,T_{\mathbf{x}_0}+z_{\mathbf{x}_0} )\leq c_2 $ on $\Omega_{\mathbf{x}_0}$
($\Omega_{\mathbf{x}_0}$, $z_{\mathbf{x}_0}$ and $c_2$ are defined in Section 2);
\end{itemize}


\item $g(\mathbf{x},y,t)$ is non-negative, in $C^{1}(\overline{\Omega}\times \mathbb{R}^n_{+})$
and decreasing on $t$.

\item ${\lim }_{t\longrightarrow 0^{+}}
\frac{g(\mathbf{x},y,t)}{t} \geq v_0(\mathbf{x},y)$ uniformly
for $(\mathbf{x},y)$ in any bounded subset on $\overline{\Omega}$,
where $v_0(\mathbf{x},y)$ is a non-negative function satisfying that when $m$ is
large, the eigenvalue problem
$$
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w = \lambda v_0(\mathbf{x},y) w \ \  on \ \
{\Omega }_{m}, \quad  u =0 \ \  on \ \ \partial {\Omega }_{m} .
$$
has a first eigenvalue $\lambda_1<1$.

\end{enumerate}
Then we have the following conclusion.

\begin{theorem}  \label{thm3}
Under the assumptions (1)-(5), (\ref{eq:general}) has a
positive solution.
\end{theorem}


\begin{proof} We just sketch the proof here.
Assumptions (1)--(3) assure that $T_{\mathbf{x}_0}+z_{\mathbf{x}_0} $
is a family of auxiliary functions satisfying (\ref{eq:auxiliary})
on $\Omega_{\mathbf{x}_0}$ and the graphs of these function have the
desired relative positions as discussed in Section 2.

Assumption (4) assures that lemma 1 can be applied and the boundary value problem
\begin{equation}
-\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w = g(\mathbf{x},y, w)
\quad \mbox{in } O(\mathbf{x}_0,y_0),
\quad
w=v \mbox{on } \partial O(\mathbf{x}_0,y_0)
\label{eq:lift2}
\end{equation}
has a unique positive solution for each positive function $v$ on
$\overline{\Omega }$.
Thus the lift $M_{(\mathbf{x}_0, y_0)}$ and the class $\Xi$ of functions
are well defined. The proofs of lemmas 2-4 and the existence of the super solution
$u$ are the same.

Finally the assumption (5) assures that the proof in Section 4 still works
out like that in \cite{Zjin}.
\end{proof}

Now we apply theorem 3 to the case that $g(\mathbf{x},y,u)= p(\mathbf{x},y)e^{-u}$.
We consider a modified problem:
\begin{equation}
- \sum_{i,j=1}^n a_{ij}(\mathbf{x},y)D_{ij}u
=\frac{p(\mathbf{x},y)e^{-c_{5}u}}{c_{5}} \quad\mbox{on }\Omega ,\quad
u=0 \quad\mbox{on }\partial \Omega .
\label{eq:general2}
\end{equation}
If we can find a positive solution $u$ of (\ref{eq:general2}),
then $c_{5}u$ is a positive solution of (\ref{eq:theproblem3}).

For (\ref{eq:general2}), we set
$$
T(t) =\frac{1}{c_{5}} (t+c_{4}) + \frac{1}{c_{5}} \ln \frac{C}{c_2c_{5}} +A
$$
where $A$ is a positive
constant such that $\frac{1}{c_{5}} \ln \frac{C}{c_{5}} +A>1$,
$C$ is defined in (\ref{eq:boundofp1}) and $c_2$, $c_{4}$, $c_{5}$ are defined in Section 2.
Then $T(t)$ is increasing and the assumption (3)(a) is obviously satisfied for $T_{\mathbf{x}}=T(|\mathbf{x}|)$.
For (3)(b), on $\Omega_{\mathbf{x}_0}$,
\begin{align*}
\frac{1}{c_{5}}p(\mathbf{x},y)e^{-c_{5}(T_{\mathbf{x}_0} +z_{\mathbf{x}_0})}
&\leq \frac{C}{c_{5}} e^{|\mathbf{x}|}e^{-c_{5}T_{\mathbf{x}_0}} \\
&\leq \frac{C}{c_{5}} e^{|\mathbf{x}_0|+c_{4}}e^{-c_{5}T_{\mathbf{x}_0}}\\
&=\frac{C}{c_{5}} e^{|\mathbf{x}_0|+c_{4}}e^{-|\mathbf{x}_0| -c_{4}
- \ln \frac{C}{c_2c_{5}} -c_{5}A}\\
&= c_2e^{-c_{5}A} <c_2 \,.
\end{align*}

Assumption (4) is obvious. For assumption (5), let $\lambda_1$ be the first eigenvalue of
the eigenvalue problem ($\Omega_1$ is defined in Section 4)
$$
- \sum_{i,j=1}^n a_{ij}(\mathbf{x},y)D_{ij}w =\lambda p(\mathbf{x},y)w \quad\mbox{on }
\Omega_1, \quad w=0 \quad\mbox{on }\partial \Omega_1.
$$
Set $v_0=2\lambda_1 p(\mathbf{x},y)$, then it is easy to see that
$$
\lim_{t\to 0^{+}} \frac{p(\mathbf{x},y)e^{-t}}{t} \geq v_0 (\mathbf{x},y)
\quad\mbox{uniformly  on } \overline{\Omega} .
$$
It is also easy to see that $v_0$ has the desired property.
Thus assumption (5) is satisfied. Therefore we can conclude that Theorem \ref{thm2}2
is true.


\section{Appendix I: Verifications of (\ref{eq:boundofae}),
(\ref{eq:boundofae1}), (\ref{eq:supersolution1}),(\ref{eq:supersolution2}) }

In this appendix, we verify (\ref{eq:boundofae})-(\ref{eq:boundofae1}) and
(\ref{eq:supersolution1})-(\ref{eq:supersolution2}) used in
Section 2. All the computations here are copied from \cite{JL4}.

Set $\Phi_1(\rho)=\rho^{-2}$  if $0<\rho<1$  and
$\Phi_1(\rho)=\frac{11}{c_1}$
if $\rho \geq 1$, and define a function $\chi$  by
$$
\chi(\alpha)=\int_{\alpha}^{\infty}\ \frac{d\rho}{\rho^{3}\Phi_1(\rho)}
\quad\mbox{for } \alpha>0.
$$
It is clear that $\chi (\alpha )$ is a decreasing function with
range $(0,\infty).$  Let $\eta$  be the inverse of $\chi.$
Then $\eta$  is a positive, decreasing function with range $(0,\infty)$.
Let $c^{*}
=11/c_1$.  For $\alpha >1$,  we have
\begin{equation}
\chi(\alpha)=\int_{\alpha}^{\infty}\ \frac{d\rho}{\rho^{3}\Phi_1(\rho)}
=\int_{\alpha}^{\infty}\ \frac{d\rho}{c^{*}\rho^{3}}
= \frac{1}{2c^{*}} \alpha^{-2} .
\label{eq:chi2}
\end{equation}
Thus
\begin{equation}
\eta (\beta ) = (2c^{*}\beta)^{-\frac{1}{2}}\quad for \quad 0<\beta <(2c^{*})^{-1}.
\label{eq:beta1}
\end{equation}
Let $H\ge 2$. Since $\eta(\chi(H))=H$ and $\eta $ is decreasing, we have
$\eta(\beta)> H$ for $0<\beta< \chi(H)$. We define a function $A(H)$ by
\begin{equation}
A(H) = 2M (\int_1^{e^{\chi(H)}}\ \eta (\ln t)  dt)^{-1} .
\label{eq:ah}
\end{equation}
 For the rest of this article, we set $a=A(H)$ and define
\begin{equation}
h_a(r)=\int_{r}^{ae^{\chi(H)}}
\eta (\ln \frac{t}{a} )\ dt \quad \mbox{for } a\le r\le ae^{\chi(H)}.
\end{equation}
Then
\begin{equation}
h_a(ae^{\chi (H)})=0, \quad h_a(a) =h_{A(H)}(A(H))= 2M.
\label{eq:chi3}
\end{equation}
For $a<r\le ae^{\chi(H)}$,
\begin{equation}
h_a'(r)=-\eta(\ln \frac{r}{a} )<0,\  |h_a'(r)|>H, \quad
h_a''(r)=\frac{1}{r}(\eta(\ln \frac{r}{a} ))^{3}\Phi_1
(\eta(\ln \frac{r}{a} )).
\label{eq:derivativelarge}
\end{equation}
Thus for $a<r\le ae^{\chi(H)}$,
\begin{equation}
\frac{h_a''(r)}{(h_a'(r))^2}=-\frac{h_a'(r)}{r}\Phi_1(-h_a'(r)) .
\label{eq:hequ}
\end{equation}
Let $h_a^{-1}$ be the inverse of $h_a.$  Then $h_a^{-1}$ is decreasing and
\begin{equation}
h_a^{-1}(0)=A(H)e^{\chi (H)}, \quad
h_a^{-1}(2M) = A(H) .
\label{eq:valueofinverseh}
\end{equation}
Thus we have the first half of (\ref{eq:boundofae}). Further for $-M\leq y\leq M$,
$$
(h_a^{-1})'(y+M) =\frac{1}{h'_a(h_a^{-1}(y+M))}
$$
\begin{align*}
(h_a^{-1})''(y+M) &=(\frac{1}{h'_a(h_a^{-1}(y+M))})' \\
&= - \frac{h''_a(h_a^{-1}(y+M))(h_a^{-1})'(y+M)}{ (h'_a(h_a^{-1}(y+M)))^{2}} \\
&= - \frac{h''_a(h_a^{-1}(y+M))}{ (h'_a(h_a^{-1}(y+M)))^{3}}\\
&= \frac{1}{h_a^{-1}(y+M)} \Phi_1 (-h'_a(h_a^{-1}(y+M))) .
\end{align*}
Thus
\begin{equation}
(h_a^{-1})''(y+M)h_a^{-1}(y+M) =\Phi_1 (-h'_a(h_a^{-1}(y+M))) .
\label{eq:equationofinverse}
\end{equation}
Now we choose an $H_0>2$ such that for $H\geq H_0$,
\begin{equation}
H_0>\frac{1}{\sqrt{2c^{*}}} +3M+4+\frac{24nc_1K}{M},
\quad
\sqrt{\frac{4K}{A(H)e^{\chi (H)}}}\leq \frac{1}{\sqrt{2}}.
\label{eq:Hlarge}
\end{equation}
Then we have (\ref{eq:boundofae1}).  For $H>H_0$,
by (\ref{eq:chi2}), (\ref{eq:beta1}), we have
\begin{align*}
A(H)^{-1} &= (2M)^{-1} \int_1^{e^{\chi(H)}}\ \eta (\ln t)  dt \\
&=(2M)^{-1} \int_0^{\chi(H)}\ \eta (m)e^{m}  dm \\
&=(2M)^{-1} \int_0^{\chi(H)}\ \frac{e^{m}}{\sqrt{2c^{*}m}}   dm\,.
\end{align*}
From
$$
\frac{1}{\sqrt{2c^{*}}}\int_0^{\chi(H)}\ \frac{1}{\sqrt{m}}   dm
\leq
\int_0^{\chi(H)}\ \frac{e^{m}}{\sqrt{2c^{*}m}}   dm
\leq
\frac{e^{\chi (H)}}{\sqrt{2c^{*}}}
\int_0^{\chi(H)}\ \frac{1}{\sqrt{m}}  dm\,,
$$
we have
$$
\frac{1}{c^{*}H} =
\frac{2\sqrt{\chi (H)}}{\sqrt{2c^{*}}} \leq
\int_0^{\chi(H)}\ \frac{e^{m}}{\sqrt{2c^{*}m}}   dm
\leq \frac{2e^{\chi (H)}\sqrt{\chi (H)}}{\sqrt{2c^{*}}}
=\frac{e^{\frac{1}{2c^{*}H^{2}}}}{c^{*} H} .
$$
Thus
\begin{equation}
2Mc^{*}H \geq A(H)\geq 2Mc^{*}He^{-\chi (H)} = 2Mc^{*}H e^{-\frac{1}{2c^{*}H^{2}}} .
\label{eq:ah2}
\end{equation}
Thus we have the second half of (\ref{eq:boundofae}) since $c^{*}=11/c_1$.

For $\mathbf{x}_0\in \mathbb{R}^{n-1}$, and a fixed constant $K$,
we define a domain $\Omega_{\mathbf{x}_0,H,K}$
in $(\mathbf{x},y)$ space by (\ref{eq:domain1}) and
define a function $z=z(\mathbf{x}, y)$
by (\ref{eq:solution111}).
Since $h_a^{-1}(y+M)\geq 0$ for $|y|\leq M$,
$(\mathbf{x}_0, y)\in \Omega_{\mathbf{x}_0,H,K}$ for $|y|< M$.
Further it is clear that the function
$z=z(\mathbf{x},y)$
is well defined on $\Omega_{\mathbf{x}_0,H,K}$.

Now we verify the first half of (\ref{eq:supersolution2}),
on $\partial \Omega_{\mathbf{x}_0,H,K} \cap \{ (\mathbf{x},y): |y|<M \}$,
$$
|\mathbf{x}-\mathbf{x}_0|=
\sqrt{\frac{2K}{A(H)e^{\chi (H)}}}h^{-1}(y+M) ;
$$
then from (\ref{eq:valueofinverseh}), we have
\begin{align*}
z&=A(H)e^{\chi (H)} -\{ (h_a^{-1}(y+M))^2-|\mathbf{x}-\mathbf{x}_0|^2 \}^{1/2}\\
&=  A(H)e^{\chi (H)} - h_a^{-1}(y+M) (1- \frac{2K}{A(H)e^{\chi (H)} } )^{1/2}\\
&\geq  A(H)e^{\chi (H)} -
A(H)e^{\chi (H)} (1- \frac{2K}{A(H)e^{\chi (H)} } )^{1/2} \\
&\geq  A(H)e^{\chi (H)}(1 - (1- \frac{2K}{2A(H)e^{\chi (H)} } ))
=K.
\end{align*}
Here we have used (\ref{eq:Hlarge}) and
the fact that $\sqrt{1-t}\leq 1-\frac{1}{2}t $ for $0<t<1$.
For the second half of (\ref{eq:supersolution2}),
since $h_a^{-1}(r)$ and $\eta $ are decreasing functions, we have
\begin{equation}
\begin{aligned}
\frac{-1}{h_a'(h_a^{-1}(y+M))}
&=\frac{1}{ \eta(\ln(\frac{1}{a}h_a^{-1}(y+M))) }\\
&\leq \frac{1}{\eta (\ln e^{\chi (H)})}\\
&=\frac{1}{\eta(\chi(H))}= \frac{1}{H},
\quad \mbox{for } |y|\le -M.
\end{aligned}
\label{eq:neededagain}
\end{equation}
Then by (\ref{eq:solution111}), we have
$$
\frac{\partial z}{\partial y}(\mathbf{x}_0,y)
=\frac{-1}{h_a'(h_a^{-1}(y+M))}\leq  \frac{1}{H},
\quad \mbox{for }  |y|\le -M.
$$
Now the second half of (\ref{eq:supersolution2}) follows from this and
$$
z(\mathbf{x}_0,-M)=A(H)e^{\chi (H)}  -h_a^{-1}(0) =A(H)e^{\chi (H)}
 -A(H)e^{\chi (H)} =0.
$$
For (\ref{eq:supersolution1}), we
set $S= \{(h_a^{-1}(y+M))^2-|\mathbf{x}-\mathbf{x}_0|^2 \}^{1/2}$. Then
we have that for $1\le i\le n-1 $,
$$
\frac{\partial z}{\partial x_i}=\frac{1}{S}(x_i-x_{0i}),\quad
\frac{\partial z}{\partial y}=- \frac{1}{S}h_a^{-1} (h_a^{-1})'.
$$
By (\ref{eq:Hlarge}) and (\ref{eq:ah2}), on $\Omega_{\mathbf{x}_0, H, K}$, we have
$$
\frac{1}{2} h_a^{-1}(y+M) \leq S \leq h_a^{-1}(y+M),
$$
and
$$
\frac{|\mathbf{x}-\mathbf{x}_0|}{S}
\leq 2(\frac{2K}{A(H)e^{\chi (H)}})^{1/2}
\leq 2(\frac{2K}{2Mc^{*}H} )^{1/2} .
$$
Thus, by (\ref{eq:neededagain}), we have
\begin{equation}
|\frac{\partial z}{\partial x_{i}}| \leq 2(\frac{c_1K}{MH})^{1/2},
\quad |\frac{\partial z}{\partial y}|
\leq \frac{h_a^{-1}(y+M)}{S |h'_a(h_a^{-1}(y+M)|}
\leq \frac{2}{H} .
\label{eq:gradient2}
\end{equation}
Hence from (\ref{eq:Hlarge}),
and the assumption that $\mathop{\rm Trace}a_{ij})=1$
(hence all eigenvalues of
$(a_{ij})$ are less than or equal to 1), we have
\begin{equation}
|\sum_{i,j=1}^n a_{ij} \frac{\partial z}{\partial x_{i}}
\frac{\partial z}{\partial x_{j}}| \leq |Dz|^2 \leq 1 .
\label{eq:normofgradient}
\end{equation}
Now we have
\begin{align*}
Qz&=\sum_{i,j=1}^n\ a_{ij}(\mathbf{x},y)D_{ij}z\\
&=\frac{1}{S}\sum_{i=1}^{n-1} a_{ii}
+\frac{1}{S^{3}}\sum_{i,j=1}^{n-1} a_{ij}(x_{i}-x_{i}^{0})(x_{j}-x_{j}^{0})
-\frac{1}{S^{3}}\sum_{i=1}^{n-1} a_{in}(x_{i}-x_{i}^{0})h_a^{-1} (h_a^{-1})'\\
&\quad -\frac{1}{S}a_{nn}((h_a^{-1})^{2} + h_a^{-1} (h_a^{-1})'')
+\frac{1}{S^{3}}a_{nn}(h_a^{-1})^{2} ((h_a^{-1})')^{2}\\
&= \frac{1}{S} \big\{1-a_{nn}+\sum_{i,j=1}^n a_{ij} \frac{\partial z}{\partial x_{i}} \frac{\partial z}{\partial x_{j}}
-a_{nn} ((h_a^{-1})^{2} + h_a^{-1} (h_a^{-1})'') \big\}\quad
\mbox{(since $a_{nn}>0$)}\\
&\leq \frac{1}{S}\big\{ 1+\sum_{i,j=1}^n a_{ij} \frac{\partial z}{\partial x_{i}}
\frac{\partial z}{\partial x_{j}}
-a_{nn} h_a^{-1} (h_a^{-1})'' \big\}\,.
\end{align*}
By (\ref{eq:coeffbound}), (\ref{eq:equationofinverse})), (\ref{eq:ah2}) and
(\ref{eq:normofgradient})) the above expression is bounded by
$$
\frac{-9}{S} \leq \frac{-9} {h_a^{-1}(y+M)} \leq  \frac{-9}
{A(H)e^{\chi (H)}} \leq \frac{-9}{2Mc^{*}He^{\frac{1}{2c^{*}H^{2}}}}
\leq \frac{-3c_1}{22eMH}.
$$
This shows (\ref{eq:supersolution1}).

\section{Appendix II: A Construction of the Domain $J$}

In this part, we give a construction of the domain $J$ used
at the end of Section 2 in the definition of $\Pi$.
Let
\begin{gather*}
\mathbb{R}^n_{+}=\{ (y_1, y_2, \dots , y_n)|y_n>0\}\,,\\
J_1=\big\{ (y_1, y_n) : y_1=\pm 1, \; |y_n|\leq 1 \;
{\rm{or}} \; y_n=\pm 1, \; |y_1|\leq 1 \big\}
\end{gather*}
That is, $J_1$ is a square with side length 2 and center $(0,0)$ in $(y_1,y_n)$
plane. In polar coordinate we can write $\partial J_1$ as
$$
(y_1, y_n) =(k(\theta )\cos \theta , k(\theta ) \sin \theta ), \quad
0\leq \theta \leq 2\pi,
$$
where $k(\theta )$ is a positive, continuous, periodic function of period $2\pi$,
$k(\theta )$ is $C^{\infty }$ except at
$\theta =\pm \frac{\pi}{4}$, $\pm \frac{3\pi}{4}$. Then we can smooth out
$k(\theta )$ near those points to get a function $k_1(\theta )$ such that
$k_1(\theta )$ is a positive, $C^{\infty }$, periodic function of period $2\pi$,
$k_1(\theta )=k(\theta )$ except in some small neighborhoods of
$\theta =\pm \frac{\pi}{4}$, $\pm \frac{3\pi}{4}$, and
$k_1(\theta )\leq k(\theta )$ for all $\theta $.
Indeed we can modify $k(\theta )$ as follows:

Let $s(t)$ be a $C^{\infty }$ function satisfying
\begin{enumerate}
\item $s(t)=0$ if $t\leq 1$;

\item $0<s(t)\leq \frac{1}{8}$ if $1<t\leq 2$;

\item $s(t)\geq 0$ for all $t$;

\item $s(t)=1$ if $t\geq 4$.
\end{enumerate}
Fixed a positive constant $\epsilon <\frac{\pi}{100}$. Near $\theta =\frac{\pi}{4}$,
we define
$$
k_1(\theta ) = k(\theta ) s\big(\frac{1}{\epsilon} |\theta -\frac{\pi}{4}|\big) +
\frac{1}{8} \big(1-s(\frac{2}{\epsilon}|\theta -\frac{\pi}{4}|)\big).
$$
Then using the fact that $\max k(\theta )=\sqrt{2}$, $\min k(\theta )=1$, we can verify that
$k_1(\theta )$ is positive, smooth and
$$
k_1(\theta )=k(\theta ) \ \ {\rm{if}} \ \ |\theta -\frac{\pi}{4}|\geq 4\epsilon ;
\quad 0<k_1(\theta )\leq k(\theta ).
$$
In a similar way, we can modify $k(\theta )$ near other points $- \pi/4$ and
$\pm 3\pi/4$. Now let
$J_2$ be the domain in $(y_1,y_n)$ plane bounded by the curve
$$
(y_1, y_n) =(k_1(\theta )\cos \theta , k_1(\theta ) \sin \theta ), \quad 0\leq \theta \leq 2\pi .
$$
We then rotate the set
$$
\big\{ (y_1, 0, \cdot ,\cdot ,\cdot , 0, y_n ):(y_1,y_n)\in J_2 \big\}
$$
with respect to $y_n$ axis to get a domain $J_{3}$. Finally, $J$ is obtained from
$J_{3}$ by appropriate translation and scaling.

\subsection*{Acknowledgement}
 The author thanks the anonymous referee for his/her comments and suggestions
that make the paper more readable.


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\end{document}