\pdfoutput=1\relax\pdfpagewidth=8.26in\pdfpageheight=11.69in\pdfcompresslevel=9 \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 $2M$: $$\Omega \subset S_M=\{ (\mathbf{x} , y )\in \mathbb{R}^n : |y| 0 is a constant. The main result of the paper is as follows. \begin{theorem} \label{thm1}Assume \begin{enumerate} \item p(\mathbf{x}_0,y_0)>0 for some (\mathbf{x}_0,y_0)\in \Omega; \item there is a positive constant C such that $$0\leq p(\mathbf{x},y) \leq C(|\mathbf{x}|+1)^{\gamma} \quad \rm{for} \quad (\mathbf{x},y)\in \Omega ; \label{eq:boundofp}$$ \item \mathop{\rm Trace}(a_{ij})=1 and there is a constant c_1>0, such that $$a_{nn}(\mathbf{x},y)\geq c_1 \quad \mbox{on } \overline{\Omega }. \label{eq:boundofann}$$ \end{enumerate} Then (\ref{eq:theproblem}) has a positive solution u\in C^{2}(\Omega )\cap C^{0}(\overline{\Omega } ). \end{theorem} When the principal part in (\ref{eq:theproblem}) is the Laplace operator, (\ref{eq:theproblem}) becomes a boundary value problem for the singular Emden-Fowler equation $$-\Delta u = p(\mathbf{x},y)u^{-\gamma } \quad\mbox{on } \Omega; \quad u=0 \quad\mbox{on } \partial \Omega . \label{eq:theproblem1}$$ 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 $$0\leq p(\mathbf{x},y) \leq Ce^{|\mathbf{x}|} \quad \rm{for} \quad (\mathbf{x},y)\in \Omega , \label{eq:boundofp1}$$ \item \mathop{\rm Trace}(a_{ij})=1, and there is a constant c_1>0, such that $$a_{nn}(\mathbf{x},y)\geq c_1 \quad {\rm{on}} \quad \overline{\Omega }. \label{eq:boundofann2}$$ \end{enumerate} Then the problem $$-\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}$$ has a positive solution u\in C^{2}(\Omega )\cap C^{0}(\overline{\Omega } ). \end{theorem} This paper is organized as follows. In Section 2, we construct a family of auxiliary functions that are defined on a family of subdomains of \Omega. In Section 3, we combine the family of auxiliary functions constructed in Section 2 and an idea from \cite{Lopez} to prove that (\ref{eq:theproblem}) has a positive supper solution. In Section 4, we prove that (\ref{eq:theproblem}) has a positive solution by the procedure used in \cite{Zjin}. In Section 5, we discuss the general case. \section{A Family of Auxiliary Functions} In this section, we will construct families of sub-domains \Omega_{\mathbf{x}_0} of \Omega  and functions T_{\mathbf{x}_0} +z (see definitions below) so that $$-\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}$$ and the graphs of the functions T_{\mathbf{x}_0} +z have special relative positions (see below). Our construction is based on the construction of a family of auxiliary functions used in \cite{JL4} (the construction in \cite{JL4} was adapted from \cite{JL1} which in turn was inspired from \cite{Finn} and \cite{Serrin}). We consider the operator$$ 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 $$a_{nn}({\bf x},y) \ge c_1 \quad \mbox{on } S_M. \label{eq:coeffbound}$$ 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 0T_{\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 $$T_{\mathbf{x}_1} + z_{\mathbf{x}_1}(\mathbf{x}, y) \geq T_{\mathbf{x}_1} +8. \label{eq:test1}$$ 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 $$C\leq 6^{\gamma}c_{5}^{-\gamma }c_2 , \label{eq:smallofp}$$ we have that for all \mathbf{x}_0 and \mathbf{x}_1 satisfying (\ref{eq:overlapping}), $$T_{\mathbf{x}_0} +2 \leq T_{\mathbf{x}_1} +8\,. \label{eq:test2}$$ 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 $$T_{\mathbf{x}_0} + z_{\mathbf{x}_0}(\mathbf{x}, y) \leq T_{\mathbf{x}_0} +2\,. \label{eq:test3}$$ 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 $$T_{\mathbf{x}_0}+z_{\mathbf{x}_0}(\mathbf{x},y) \leq T_{\mathbf{x}_1} +z_{\mathbf{x}_1}(\mathbf{x},y) \label{eq:definitionofradius}$$ 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 \,. $$Thus without loss of generality, we may assume C in (\ref{eq:boundofp}) satisfying (\ref{eq:smallofp}). Then all constructions in Section 2 are valid. Let v>0 be a function on \overline{\Omega}, for a point (\mathbf{x}_0,y_0)\in \overline{\Omega } , we define a new function M_{(\mathbf{x}_0,y_0)}(v), called the lift of v over O(\mathbf{x}_0,y_0) as follows: \begin{gather*} M_{(\mathbf{x}_0,y_0)}(v)(\mathbf{x},y)=v(\mathbf{x},y) \quad {\rm{if}} \quad (\mathbf{x},y)\in \Omega\setminus O(\mathbf{x}_0,y_0) \\ M_{(\mathbf{x}_0,y_0)}(v)(\mathbf{x},y)=w(\mathbf{x},y) \quad {\rm{if}} \quad (\mathbf{x},y)\in O(\mathbf{x}_0,y_0) \end{gather*} where w(\mathbf{x},y) is the positive solution of the boundary-value problem $$-\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}$$ It is easy to see (\ref{eq:lift}) has a unique positive solution in C^{2}(O(\mathbf{x}_0,y_0))\cap C^{0}(\overline{O(\mathbf{x}_0,y_0)}). Indeed m_1=\min \{ v (\mathbf{x},y): (\mathbf{x},y)\in \partial O(\mathbf{x}_0,y_0)\} is a sub-solution since p(\mathbf{x},y) is non-negative, m_2+T_{\mathbf{x}_0} + z_{\mathbf{x}_0} is a super solution by (\ref{eq:auxiliary}), where m_2=\max \{ v (\mathbf{x},y): (\mathbf{x},y)\in \partial O(\mathbf{x}_0,y_0)\}. Then we can conclude the existence of a desired solution (for example, see \cite{Amann} or \cite{Crandall}). Uniqueness of positive solutions of (\ref{eq:lift}) follows from a standard argument. Set \tau = (C/c_2)^{1/\gamma} c_{4} (see (\ref{eq:definitionoft}) for the source of the constants). We define a class \Xi  of functions as follows: a function v is in \Xi  if \begin{enumerate} \item v\in C^{0}(\overline{\Omega }), v>0 on \overline{\Omega} and v\leq \tau  on \partial \Omega; \item For any (\mathbf{x}_0,y_0)\in \overline{\Omega}, v\leq M_{(\mathbf{x}_0,y_0)}(v); \item v\leq T_{\mathbf{x}_0} +z_{\mathbf{x}_0} on \Omega_{\mathbf{x}_0} \cap \Omega for any (\mathbf{x}_0,y_0)\in \overline{\Omega}. \end{enumerate} By the following well-known lemma, it is easy to check the function v=\tau  is in \Xi. Thus \Xi is not empty. \begin{lemma} \label{lm1} Let D be a bounded domain, f(\mathbf{x},y,t) be a C^{1} function that is decreasing in t. If w_1, w_2 are in C^{2}(D)\cap C^{0}(\overline{D}), w_1\leq w_2 on \partial D, and \begin{gather*} -\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w_1 \leq f(\mathbf{x},y,w_1) \quad \mbox{in } D, \\ -\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij}w_2 \geq f(\mathbf{x},y, w_2) \quad\mbox{in }D \end{gather*} then w_1\leq w_2 on D. \end{lemma} Now we set$$ 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 . $$First we need some lemmas. \begin{lemma} \label{lm2} If 00 on \overline{\Omega} and \max \{ v_1, v_2 \}\leq \tau  on \partial \Omega. It is also clear that \max \{ v_1, v_2 \}\leq T_{\mathbf{x}_0} +z_{\mathbf{x}_0} on \Omega_{\mathbf{x}_0} \cap \Omega for any (\mathbf{x}_0,y_0)\in \overline{\Omega}. Since$$ 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\}) . $$Thus \max \{ v_1, v_2 \}\in \Xi. \end{proof} \begin{lemma} \label{lm4} If v\in \Xi , then M_{(\mathbf{x}_0,y_0)}(v)\in \Xi for any (\mathbf{x}_0,y_0)\in \overline{\Omega} . \end{lemma} \begin{proof} By the definition of M_{(\mathbf{x}_0,y_0)}( v), it is clear that M_{(\mathbf{x}_0,y_0)}( v)>0 on \overline{\Omega }, M_{(\mathbf{x}_0,y_0)}( v)\in C^{0}(\overline{\Omega} ) and M_{(\mathbf{x}_0,y_0)}(v) \leq \tau  on \partial \Omega . For any (\mathbf{x}^{*},y^{*})\in \overline{\Omega} , we first show that $$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}$$ We only need to prove that (\ref{eq:mv}) is true for (\mathbf{x},y)\in O(\mathbf{x}^{*},y^{*}). Since$$ 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)). $$Thus for (\mathbf{x},y)\in O(\mathbf{x}^{*},y^{*}) \setminus O(\mathbf{x}_0,y_0), $$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}$$ That is, (\ref{eq:mv}) is true on O(\mathbf{x}^{*},y^{*}) \setminus O(\mathbf{x}_0,y_0), Now for \Omega_1=O(\mathbf{x}^{*},y^{*}) \cap O(\mathbf{x}_0,y_0), if we set$$ M_{(\mathbf{x}_0,y_0)}( v)=w_1, \quad M_{(\mathbf{x}^{*},y^{*})}(M_{(\mathbf{x}_0,y_0)}( v))=w_2 $$we have \begin{gather*} -\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w_1 = p(\mathbf{x},y) w_1^{-\gamma } \quad\mbox{on } \Omega_1, \\ -\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w_2 = p(\mathbf{x},y) w_2^{-\gamma } \quad \mbox{on }\Omega_1 . \end{gather*} On \partial \Omega_1, w_1\leq w_2 on O(\mathbf{x}^{*},y^{*})\cap \partial O(\mathbf{x}_0,y_0) by (\ref{eq:boundary1}) and w_1\leq w_2 on \partial O(\mathbf{x}^{*},y^{*})\cap O(\mathbf{x}_0,y_0) since (\ref{eq:mv}) is true on \Omega\setminus O(\mathbf{x}^{*},y^{*}). Then lemma 1 implies w_1\leq w_2 on \Omega_1. Thus (\ref{eq:mv}) is true on O(\mathbf{x}^{*},y^{*}) \cap O(\mathbf{x}_0,y_0) and on O(\mathbf{x}^{*},y^{*}). \end{proof} Now we prove that M_{(\mathbf{x}_0,y_0)}(v)\leq T_{\mathbf{x}_1}+z_{\mathbf{x}_1} on \Omega_{\mathbf{x}_1}\cap \Omega for all (\mathbf{x}_1,y_1)\in \overline{\Omega }. By the definition of M_{(\mathbf{x}_0,y_0)}(v), we only need to consider the graph of the function M_{(\mathbf{x}_0,y_0)}(v) over O(\mathbf{x}_0,y_0). If O(\mathbf{x}_0,y_0) is covered completely by \Omega_{\mathbf{x}_1}, since v\leq T_{\mathbf{x}_1}+z_{\mathbf{x}_1} and T_{\mathbf{x}_1}+z_{\mathbf{x}_1} satisfies (\ref{eq:auxiliary}), T_{\mathbf{x}_1}+z_{\mathbf{x}_1} is a super solution of (\ref{eq:lift}) on O(\mathbf{x}_0,y_0). Then Lemma \ref{lm1} implies M_{(\mathbf{x}_0,y_0)}(v)\leq T_{\mathbf{x}_1}+z_{\mathbf{x}_1} on O(\mathbf{x}_0,y_0). In the case that O(\mathbf{x}_0,y_0) does not intersect with \Omega_{\mathbf{x}_1}, the conclusion is trivial. Now we consider the case that O(\mathbf{x}_0,y_0) is partially covered by \Omega_{\mathbf{x}_1}. Since O(\mathbf{x}_0,y_0) is covered by \Omega_{\mathbf{x}_0}, we always have $$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}$$ Then by the choice of \delta_{\mathbf{x}_0}, O(\mathbf{x}_0,y_0), and the fact that O(\mathbf{x}_0,y_0) \cap T_{\mathbf{x}_1} is not empty, we have that \mathbf{x}_0 and \mathbf{x}_1 satisfy (\ref{eq:overlapping}), and for all (\mathbf{x},y)\in O(\mathbf{x}_0,y_0)\cap \Omega_{\mathbf{x}_1}, |\mathbf{x}_0 - \mathbf{x}|\leq \delta_{\mathbf{x}_0}. Then by (\ref{eq:definitionofradius}), the graph of T_{\mathbf{x}_0}+z_{\mathbf{x}_0} over O(\mathbf{x}_0,y_0)\cap \Omega_{\mathbf{x}_1} is under the graph of T_{\mathbf{x}_1}+z_{\mathbf{x}_1}. Thus the conclusion follows from (\ref{eq:control}). Now we are ready to prove that u has the desired properties. Let (\mathbf{x}_0,y_0)\in \overline{\Omega }. By the definition of u(\mathbf{x}_0,y_0), there is a sequence of functions v_{k} in \Xi  such that$$ 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 . $$Then similar to the way we obtain w, M_{(\mathbf{x}_0,y_0)}(\max \{ u_0, M_{(\mathbf{x}_0,y_0)}(v_{k}) \}) will produce a function w_1 satisfying \begin{gather*} -\sum_{i,j=1}^n a_{ij}(\mathbf{x},y) D_{ij} w_1 =p(\mathbf{x},y) w_1^{-\gamma } \quad\mbox{on } O(\mathbf{x}_0,y_0), \\ w\leq w_1 \quad on \quad O(\mathbf{x}_0,y_0), \quad w(\mathbf{x}_2,y_2)s. Finally we take limit of w_{m} to get a desired solution. \section{The General Case} Now we consider the boundary-value problem $$- \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}$$ In addition to the assumptions on (a_{ij}) and \Omega given at the beginning of the paper, we assume the following conditions. \begin{enumerate} \item \mathop{\rm Trace}a_{ij})=1; \item There is a constant c_1>0 such that a_{nn} \geq c_1 on \overline{\Omega}; \item There is a family of increasing positive functions T=T(t) satisfying (with T_{\mathbf{x}}=T(|\mathbf{x}|)) \begin{itemize} \item[(a)] |T_{\mathbf{x}_0}-T_{\mathbf{x}}|\leq |\mathbf{x}_0-\mathbf{x}|/c_{5}; \item[(b)] g(\mathbf{x},y,T_{\mathbf{x}_0}+z_{\mathbf{x}_0} )\leq c_2  on \Omega_{\mathbf{x}_0} (\Omega_{\mathbf{x}_0}, z_{\mathbf{x}_0} and c_2 are defined in Section 2); \end{itemize} \item g(\mathbf{x},y,t) is non-negative, in C^{1}(\overline{\Omega}\times \mathbb{R}^n_{+}) and decreasing on t. \item {\lim }_{t\longrightarrow 0^{+}} \frac{g(\mathbf{x},y,t)}{t} \geq v_0(\mathbf{x},y) uniformly for (\mathbf{x},y) in any bounded subset on \overline{\Omega}, where v_0(\mathbf{x},y) is a non-negative function satisfying that when m is large, the eigenvalue problem$$ -\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} . $$has a first eigenvalue \lambda_1<1. \end{enumerate} Then we have the following conclusion. \begin{theorem} \label{thm3} Under the assumptions (1)-(5), (\ref{eq:general}) has a positive solution. \end{theorem} \begin{proof} We just sketch the proof here. Assumptions (1)--(3) assure that T_{\mathbf{x}_0}+z_{\mathbf{x}_0}  is a family of auxiliary functions satisfying (\ref{eq:auxiliary}) on \Omega_{\mathbf{x}_0} and the graphs of these function have the desired relative positions as discussed in Section 2. Assumption (4) assures that lemma 1 can be applied and the boundary value problem $$-\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}$$ has a unique positive solution for each positive function v on \overline{\Omega }. Thus the lift M_{(\mathbf{x}_0, y_0)} and the class \Xi of functions are well defined. The proofs of lemmas 2-4 and the existence of the super solution u are the same. Finally the assumption (5) assures that the proof in Section 4 still works out like that in \cite{Zjin}. \end{proof} Now we apply theorem 3 to the case that g(\mathbf{x},y,u)= p(\mathbf{x},y)e^{-u}. We consider a modified problem: $$- \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}$$ If we can find a positive solution u of (\ref{eq:general2}), then c_{5}u is a positive solution of (\ref{eq:theproblem3}). For (\ref{eq:general2}), we set$$ 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} 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 $$\chi(\alpha)=\int_{\alpha}^{\infty}\ \frac{d\rho}{\rho^{3}\Phi_1(\rho)} =\int_{\alpha}^{\infty}\ \frac{d\rho}{c^{*}\rho^{3}} = \frac{1}{2c^{*}} \alpha^{-2} . \label{eq:chi2}$$ Thus $$\eta (\beta ) = (2c^{*}\beta)^{-\frac{1}{2}}\quad for \quad 0<\beta <(2c^{*})^{-1}. \label{eq:beta1}$$ 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 $$A(H) = 2M (\int_1^{e^{\chi(H)}}\ \eta (\ln t) dt)^{-1} . \label{eq:ah}$$ For the rest of this article, we set $a=A(H)$ and define $$h_a(r)=\int_{r}^{ae^{\chi(H)}} \eta (\ln \frac{t}{a} )\ dt \quad \mbox{for } a\le r\le ae^{\chi(H)}.$$ Then $$h_a(ae^{\chi (H)})=0, \quad h_a(a) =h_{A(H)}(A(H))= 2M. \label{eq:chi3}$$ For $aH, \quad h_a''(r)=\frac{1}{r}(\eta(\ln \frac{r}{a} ))^{3}\Phi_1 (\eta(\ln \frac{r}{a} )). \label{eq:derivativelarge} Thus for$a2$such that for$H\geq H_0$, $$H_0>\frac{1}{\sqrt{2c^{*}}} +3M+4+\frac{24nc_1K}{M}, \quad \sqrt{\frac{4K}{A(H)e^{\chi (H)}}}\leq \frac{1}{\sqrt{2}}. \label{eq:Hlarge}$$ 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 $$2Mc^{*}H \geq A(H)\geq 2Mc^{*}He^{-\chi (H)} = 2Mc^{*}H e^{-\frac{1}{2c^{*}H^{2}}} . \label{eq:ah2}$$ Thus we have the second half of (\ref{eq:boundofae}) sincec^{*}=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|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 DomainJ$} In this part, we give a construction of the domain$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