\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Positive solutions \hfil EJDE--2001/53} {EJDE--2001/53\hfil Noureddine Zeddini \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2001}(2001), No. 53, pp. 1--20. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Positive solutions of singular elliptic equations outside the unit disk % \thanks{ {\em Mathematics Subject Classifications:} 34B16, 34B27, 35J65. \hfil\break\indent {\em Key words:} Singular elliptic equation, Green function, Schauder fixed point theorem, \hfil\break\indent maximum principle, superharmonic function. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Submitted February 13, 2001. Published July 24, 2001.} } \date{} % \author{Noureddine Zeddini} \maketitle \begin{abstract} We study the existence and the asymptotic behaviour of positive solutions for the nonlinear singular elliptic equation $\Delta u +\varphi(.,u)=0$ in the outside of the unit disk in $\mathbb{R}^2$, with homogeneous Dirichlet boundary condition. The aim is to prove some existence results for the above equation in a general setting by using a potential theory approach. \end{abstract} \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \newtheorem{theo}{Theorem}[section] \newtheorem{prop}[theo]{Proposition} \newtheorem{lem}[theo]{Lemma} \newtheorem{coro}[theo]{Corollary} \newtheorem{rem}{Remark}[section] \newtheorem{exa} {Example}[section] \section{Introduction} The singular semi-linear elliptic equation $$\Delta u + q(x) u^{-\gamma} = 0,\quad x \in \Omega \subset \mathbb{R}^n, \quad \gamma > 0, \label{e1.1}$$ has been extensively studied for both bounded and unbounded domain $\Omega$ (see for example \cite{e1,k1,l1,l2} and the references therein). For $0<\gamma <1$, Edelson \cite{e1} proved the existence of an entire positive solution $u \in C_{{\rm loc}}^{2+\alpha}(\mathbb{R}^2)$ of (\ref{e1.1}), having logarithmic growth as $|x|\to \infty$, provided that $q \in C_{{\rm loc}}^{\alpha}(\mathbb{R}^2)$, $0<\alpha<1$, $q(x)>0$ for $|x|>0$ and $$\int_e^{\infty}t(Log t)^{-\gamma}(\displaystyle\max_{|x|=t}q(x)) dt <\infty.$$ Lazer and Mckenna \cite{l2} considered (\ref{e1.1}) in the case where $\Omega \subset \mathbb{R}^n$ ($n \geq 1$) is a bounded domain with smooth boundary. They proved the existence and the uniqueness of a positive solution $u \in C^{2+\alpha}_{{\rm loc}} (\Omega) \cap C(\overline{\Omega})$ with homogeneous Dirichlet boundary condition, provided that $q \in C^\alpha(\overline{\Omega})$ and $q(x) > 0$ for all $x \in \overline{\Omega}$. Kusano and Swanson \cite{k1} considered the generalized equation $$\Delta u + f(x, u) = 0,\; x \in \Omega, \label{e1.2}$$ where $\Omega$ is an exterior domain of $\mathbb{R}^n$, $n\geq 2$. For $n=2$, they proved the existence of an exterior domain $\Omega_T = \{x \in \mathbb{R}^2 : |x| > T >1 \}$ and a positive solution $u$ on $\Omega_T$ such that $u(x)/Log |x|$ is bounded and bounded away from zero provided that the following conditions are satisfied \begin{enumerate} \item[C1)] $f \in C^\alpha_{{\rm loc}}(\Omega \times (0, \infty))$. \item[C2)] There exist two functions $\psi$ and $\phi : (0, \infty) \times (0, \infty) \to (0, \infty)$ of class $C^\alpha_{{\rm loc}}((0,\infty) \times (0, \infty))$, such that $\psi(t, u)$ and $\phi(t,u)$ are non-increasing functions of $u$ for each fixed $t > 0$, and $$\psi(|x|, u) \leq f(x, u) \leq \phi(|x|,u), \mbox{ for all } (x,u) \in \Omega \times (0, \infty).$$ \item[C3)] $\int^\infty \phi(t, c Log t ) dt < \infty$, for some positive constant $c$. \end{enumerate} Kusano and Swanson showed also for $n=2$, the existence of a bounded positive solution of (\ref{e1.2}) in some exterior domain $\Omega_T$, $T$ sufficiently large, provided that $\phi$ satisfies C1 and C2, and $\int^\infty t \phi(t,c )Log t dt < \infty$, for some constant $c>0$. In this article, we improve the results of \cite{e1} by letting the exponent $\gamma$ be unbounded. More precisely, we are concerned with the following problem \begin{eqnarray} &\Delta u + \varphi(x, u) = 0, \quad\mbox{in $D$, (in the weak sense)}& \label{e1.3}\\ & u \mid _{\partial D} = 0,\quad \quad \quad \quad & \nonumber \end{eqnarray} where $D = \{x \in \mathbb{R}^2 : |x| > 1\}$ and $\varphi$ is a nonnegative Borel measurable function in $D \times (0, \infty)$ that belongs to a convex cone which contains, in particular, all functions $$\varphi(x,t)= q(x) t^{-\gamma},\quad\gamma>0$$ with nonnegative Borel function $q$. Under appropriate conditions on $\varphi$, we show that (\ref{e1.3}) has infinitely many positive solutions continuous on $\overline{D}$. More precisely, for each $\mu > 0$, there exists a positive solution $u \in C(\overline{D})$ such that $\lim_{|x|\to\infty} u(x)/Log|x| = \mu$. Under additional conditions on $\varphi$, we prove that (\ref{e1.3}) has a bounded positive solution continuous on $\overline{D}$. This paper is organized as follows. In section 2, we recall and establish some properties of functions belonging to the Kato class introduced in \cite{m2}. In section 3, we prove the existence of many positive solutions of (\ref{e1.3}) which are continuous on $\overline{D}$. In the last section, we give some estimates on the solutions of (\ref{e1.3}). We point out that for some functions $\varphi$, we get better estimates on the solutions; namely for each $x\in \overline{D}$, we have $$\mu Log|x| \leq u(x) \leq C Log |x| ,\quad \mbox{if } \lim_{|x|\to \infty} \frac{u(x)}{Log|x|} = \mu > 0$$ and $$\frac{1}{C} \big(1-\frac{1}{|x|} \big) \leq u(x) \leq C \big(1-\frac{1}{|x|} \big) , \quad \mbox{if u is bounded},$$ where $C$ is a positive constant. As usual let $B(D)$ be the set of Borel measurable functions in $D$ and let $B^+(D)$ be the subset of the nonnegative functions. We recall that the potential kernel $V$ associated to $\Delta$ is defined on $B^+(D)$ by $$V \Psi(x) = \int_D G(x, y) \Psi(y)dy,\quad \mbox{for } x \in D,$$ where $G$ is the Green's function of the Laplacian in $D$. Hence, for any\linebreak $\Psi \in B^+(D)$ such that $\Psi \in L^1_{{\rm loc}}(D)$ and $V \Psi \in L^1_{{\rm loc}}(D)$, we have (in the distributional sense) $$\Delta (V \Psi) = - \Psi \mbox{ in } D. \label{e1.4}$$ We note that for any $\Psi \in B^+(D)$ such that $V\Psi \neq \infty$, we have $V \Psi \in L^1_{{\rm loc}}(D)$ (see \cite[p.51]{c1}). Let us recall that $V$ satisfies the complete maximum principle \cite[p.175]{p1}, i.e for each $f \in B^+(D)$ and $v$ a nonnegative superharmonic function on $D$ such that $Vf \leq v$ in $\{f > 0\}$ we have $Vf \leq v$ in $D$. Throughout this paper, the function $\varphi$ is required to satisfy combinations of the following hypotheses \begin{enumerate} \item[H1)] $\varphi$ is continuous and non-increasing with respect to the second variable. \item[H2)] $\varphi (.,c) \in K^{\infty }(D)$ for every $c > 0$. \item[H3)] $V \varphi (.,c) > 0$ for every $c > 0$. \end{enumerate} Finally we mention that the letter $C$ will denote a generic positive constant which may vary from line to line. \section{The Kato class $K^{\infty}(D)$} Throughout this paper, let $D=\left\{x \in \mathbb{R}^{2},|x|>1\right\}$, $\overline{D} =\left\{x \in \mathbb{R}^{2},|x| \geq1\right\}$, and $G(x,y)=\frac{1}{2\pi} Log (1+\displaystyle\frac{(|x|^2-1)(|y|^2-1)}{|x-y|^2})$ be the Green's function of $\Delta$ in $D$. \paragraph{Definition} A Borel measurable function $q$ in $D$ belongs to the Kato class $K^{\infty}(D)$ if $q$ satisfies the following conditions \begin{eqnarray} &\lim_{\alpha\to 0} \sup_{x\in D}\int _{(|x-y|\leq \alpha) \cap D} \frac{|y|-1}{|y|} \frac{|x|}{|x|-1}G(x,y)|q(y)| dy = 0 & \label{e2.1}\\ &\lim_{M\to\infty} \sup_{x\in D} \int_{(|y| \geq M)} \frac{|y|-1}{|y|} \frac{|x|}{|x|-1} G(x,y) |q(y)| dy = 0. & \label{e2.2} \end{eqnarray} \begin{lem} \label{lm1} For each x,y in $D$, $$\frac{1}{2\pi} (1-\frac{1}{|x|})(1- \frac{1}{|y|}) \leq G(x,y).$$ \end{lem} \paragraph{Proof} By the definition of $G$, we have \begin{eqnarray*} G(x,y) &=& \frac{1}{2\pi} Log \left( 1+\frac{(|x|^2-1)(|y|^2-1)}{|x-y|^2} \right) \\ & =& \frac{1}{2\pi}\frac{(|x|-1)}{|x|} \frac{(|y|-1)}{|y|} \int_{0}^{1} \frac{|x| |y| (1+|x|)(1+|y|)}{|x-y|^2+t(|x|^2-1)(|y|^2-1)} dt.\end{eqnarray*} For every $t\in \left[0,1\right]$ and $x,y$ in $D$, we have \begin{eqnarray*} \frac{|x-y|^2+t(|x|^2-1)(|y|^2-1)}{|x| |y|(1+|x|)(1+|y|)} &\leq& \frac{(|x|+|y|)^2+(|x|^2-1)(|y|^2-1)}{|x\|y|(1+|x|)(1+|y|)}\\ &=&\frac{(|x\|y|+1)^2}{|x| |y| (1+|x|)(1+|y|)} \leq 1.\end{eqnarray*} Hence $G(x,y) \geq \frac{1}{2 \pi}(1 - \displaystyle\frac{ 1 }{ |x| })(1- \displaystyle\frac{ 1 }{ |y| })$. \begin{prop} \label{prop1} Let $q$ be a function in the class $K^{\infty}(D)$.Then the function $y\to (1- \displaystyle\frac{1}{|y|})^2 q(y)$ is in $L^1(D)$. In particular $q\in L_{{\rm loc}}^1(D)$. \end{prop} \paragraph{Proof.} Let $q \in K^{\infty}(D)$. Then by (\ref{e2.1}) and (\ref{e2.2}), there exist $\alpha > 0$ and $M > 1$ such that $$\sup_{x\in D}\int_{(|x-y|\leq \alpha)} \frac{|y|-1}{|y|} \frac{|x|}{|x|-1} G(x,y) |q(y)| dy \leq 1$$ and $$\sup_{x\in D} \int_{(|y| \geq M)} \frac{|y|-1}{|y|} \frac{|x|}{|x|-1} G(x,y) |q(y)| dy \leq 1.$$ Let $x_1,x_2,\ldots ,x_n$ in $D$ such that $\overline{D} \cap \overline{B}(0,M) \subset \bigcup_{1\leq i \leq n} B(x_i,\alpha)$. By using Lemma \ref{lm1}, we get \begin{eqnarray*} \lefteqn{\int_{D} (1- \frac{1}{|y|})^2 |q(y)| dy }\\ & \leq & \int_{( |y| \geq M )}(1- \frac{1}{|y|})^2 |q(y)| dy + \int_{( 1\leq |y| \leq M ) \cap D} (1- \frac{1}{|y|})^2 |q(y)| dy \\ & \leq & 2\pi \sup_{x \in D} \int_{( |y| \geq M)} \frac{|y|-1}{|y|} \frac{|x|}{|x|-1} G(x,y) |q(y)| dy \\ &&+ \sum_{i = 1}^{n} \int_{B(x_i,\alpha ) \cap D} (1- \frac{1}{|y|})^2 |q(y)| dy \\ & \leq & 2\pi + 2\pi \sum_{i = 1}^{n} \int_{B(x_i,\alpha ) \cap D} \frac{|y| - 1}{|y|} \frac{|x_i|}{|x_i| - 1} G(x_i,y) |q(y)| dy \\ & \leq & 2\pi +2\pi n \sup_{x \in D} \int_{B(x,\alpha ) \cap D} \frac{|y| -1}{|y|} \frac{|x|}{|x| -1} G(x,y)|q(y)| dy \\ & \leq & 2\pi (1+n) <\infty . \end {eqnarray*} \begin{lem} \label{lm2} Let $M > 1$ and $r > 0$. Then there exists a constant $C > 0$ such that for each $x \in D$ and $y \in D$ satisfying $|x - y | \geq r$ and $|y| \leq M$, $$G(x,y) \leq C (1 - \frac{1}{|x|}) (1 - \frac{1}{|y|} ) .$$ \end{lem} \paragraph{Proof.} We have for $|x -y| \geq r$ and $|y| \leq M$, \begin{eqnarray*} \frac{|x|}{|x|-1} \frac{|y|-1}{|y|} G(x,y) & \leq & \frac{1}{2 \pi } \frac{|x| (|y| -1)}{(|x| -1)|y|} \frac{(|x|^2 - 1)(|y|^2 - 1)}{|x -y|^2} \\ & = & \frac{1}{2 \pi } \frac{(|y| -1)^2 ( |y| + 1)}{|y|} \frac{ |x| (|x| + 1)}{|x - y|^2} \\ & \leq & \frac{1}{2 \pi } (1 - \frac{1}{|y|})^2 M (M + 1) \frac{|x|(|x| + 1)}{( (|x| - M) \vee r)^2 } , \end{eqnarray*} where $( |x| - M ) \vee r = \max ( |x| -M , r )$. Since the function $t \to \displaystyle\frac{ t(t+1)}{((t-M)\vee r )^2}$ is continuous and positive on $\left [1,\infty \right )$ and $\lim_{t\to +\infty} \displaystyle\frac{t(t+1)}{((t-M)\vee r )^2} = 1$, then there exists $C > 0$ such that $$\frac{|x|}{|x| -1} \frac{|y| -1}{|y|} G(x,y) \leq C (1 - \frac{1}{|y|} )^2 .\quad \diamondsuit$$ In the sequel , we use the notation $$\|q\|_{D} = \sup_{x \in D} \int_{D} \frac{|y| - 1}{|y|} \frac{|x|}{|x| -1}G(x,y)|q(y)| dy .$$ \begin{prop} \label{prop2} If $q \in K^{\infty}(D)$, then $\|q\|_{D} < \infty$. \end{prop} \paragraph{Proof.} Let $\alpha > 0$ and $M > 1$. Then \begin{eqnarray*} \lefteqn{\int_{D} \frac{|y| - 1}{|y|} \frac{|x|}{|x| - 1} G(x,y) |q(y)| dy }\\ & \leq & \int_{(|x-y| \leq \alpha) \cap D} \frac{|y| - 1}{|y|} \frac{|x|}{|x| - 1} G(x,y) |q(y)| dy \\ & & + \int_{(|y| \geq M)} \frac{|y| - 1}{|y|} \frac{|x|}{|x| - 1} G(x,y) |q(y)| dy \\ & & + \int_{(|x-y| \geq \alpha ) \cap (|y| \leq M) \cap D} \frac{|y| - 1}{|y|} \frac{|x|}{|x| - 1} G(x,y) |q(y)| dy . \end{eqnarray*} By Lemma \ref{lm2}, $$\int_{(|x-y| \geq \alpha ) \cap (|y| \leq M) \cap D} \frac{|y| - 1}{|y|} \frac{|x|}{|x| - 1} G(x,y) |q(y)| dy \leq C \int_{D} ( 1 - \frac{1}{|y|})^2 |q(y)| dy .$$ Thus, the result follows from (\ref{e2.1}) , (\ref{e2.2}) and Proposition \ref{prop1}. \quad$\diamondsuit$ The following result of Selmi \cite{s1}, will be needed in the sequel. \begin{theo} \label{3g} There exists a constant $C_0 > 0$ depending only on $D$ such that for all $x,y$ and $z$ in $D$, we have $$\frac{G(x,z) G(z,y)}{G(x,y)} \leq C_0 \left[ \frac{|z| - 1}{|z|} \frac{|x|}{|x| - 1}G(x,z) +\frac{|z| -1}{|z|} \frac{|y|}{|y| -1} G(z,y) \right]. \label{e2.3}$$ \end{theo} By using the above theorem we have the following \begin{prop} \label{prop3} There exists a constant $C_D > 0$ depending only on $D$ such that for any function $q$ belonging to $K^{\infty}(D)$, any nonnegative superharmonic function h in $D$ and all $x \in D$ $$\int_D G(x,y) h(y) |q(y)| dy \leq C_D \| q\|_D h(x). \label{e2.4}$$ \end{prop} \paragraph{Proof.} Let $h$ be a nonnegative superharmonic function in $D$, then there exists a sequence ${( f_n )}_n$ of nonnegative measurable functions in $D$ such that $$h(y) = \sup_{n} \int_D G(y,z)f_n(z) dz \;,\; \forall y \in D.$$ Hence, we need only to verify (\ref{e2.4}) for $h(y) = G(y,z)$ for all $z \in D$. By using (\ref{e2.3}), we obtain \begin{eqnarray*} \frac{1}{G(x,z)} \int_D G(x,y) G(y,z) |q(y)| dy \leq 2 C_0 \| q\|_D.\quad \diamondsuit \end{eqnarray*} If we take $h=1$ in Proposition \ref{prop3}, we obtain the following statement. \begin{coro} \label{coro1} Let q be a function in $K^{\infty} (D)$ . Then $$\sup_{x \in D} \int_D G(x,y) |q(y)| dy < \infty . \label{e2.5}$$ \end{coro} \begin{coro} \label{coro2} Let $q$ be a function in the class $K^{\infty} (D)$. Then the function $y \to ( 1 - \displaystyle\frac{1}{|y|} )q(y)$ is in $L^1(D)$. \end{coro} \paragraph{Proof.} For each $x$, $y$ in $D$, by Lemma \ref{lm1} we have $$\frac{1}{2\pi} (1-\frac{1}{|x|})(1- \frac{1}{|y|}) \leq G(x,y).$$ Hence $\int_D (1 - \displaystyle\frac{1}{|y|}) |q(y)| dy \leq 2\pi \displaystyle\frac{|x|}{|x|-1} \int_D G(x,y)|q(y)| dy$. The result follows from Corollary \ref{coro1}. \quad$\diamondsuit$ In the next proposition we prove that for $q$ radial , $q \in K^{\infty}(D)$ if and only if (\ref{e2.5}) is satisfied. \begin{prop} \label{prop4} Let $q$ be a radial function in $D$. Then $q \in K^{\infty}(D)$ if and only if $$\int_1^{+ \infty } r Log ( r ) |q(r)| dr < \infty . \label{e2.6}$$ \end{prop} \paragraph{Proof.} By elementary calculus, we have $$\int_D G(x,y) |q(y)| dy = \int_1^{+ \infty} r Log ( r \wedge R ) |q( r )| dr \, ,$$ where $R = |x|$ and $r \wedge R = \min (r,R)$. Hence by (\ref{e2.5}), we deduce that if $q \in K^{\infty}(D)$ then (\ref{e2.6}) is satisfied. The proof of the converse is found in \cite[Prop.2]{m2}. \quad$\diamondsuit$ Using the same argument as in the proof of Proposition \ref{prop3}, we establish the following lemma (see also \cite{m2}). \begin{lem} \label{lm3} Let $x_0 \in \overline{D}$. Then for any function $q$ belonging to $K^{\infty}(D)$ and any positive superharmonic function $h$ in $D$, we have $$\lim_{r \to 0} \sup_{x \in D} \frac{1}{h(x)} \int_{B(x_0,r) \cap D} G(x,y) h(y) |q(y)| dy = 0$$ and $$\lim_{M \to + \infty } \sup_{x \in D} \frac{1}{h(x)} \int_{( |y| \geq M)} G(x,y) h(y) |q(y)| dy = 0.$$ \end{lem} \begin{prop} \label{prop5} Let $q$ be a function in $K^{\infty}(D)$. Then $Vq \in C(D)$ and $\lim_{x \to \partial D} Vq(x)= 0$. \end{prop} \paragraph{Proof.} Without loss of generality, assume that $q$ is nonnegative. Let $x_0 \in D$ and $\varepsilon > 0$. By Lemma \ref{lm3}, there exist $r > 0$ and $M > 1$ such that $$\sup_{z \in D} \int_{B(x_0,2r) \cap D} G(z,y)q(y) dy \leq \frac{ \varepsilon }{4}$$ and $$\sup_{z \in D} \int_{(|y| \geq M)} G(z,y)q(y) dy \leq \frac{ \varepsilon }{4}.$$ Let $x, x' \in B(x_0,r) \cap D$, then we have \begin{eqnarray*} \lefteqn{| Vq(x) - Vq(x' ) | }\\ & \leq & 2 \sup_{z \in D} \int_{B(x_0,2r) \cap D} G(z,y)q(y) dy + 2 \sup_{z \in D} \int_{( |y| \geq M)} G(z,y)q(y) dy \\ & & + \int_{( |x_0 - y | \geq 2r ) \cap (1< |y| \leq M)} |G(x,y) - G(x',y)| q(y) dy \\ & \leq & \varepsilon + \int_{( |x_0 - y | \geq 2r ) \cap (1< |y| \leq M)} |G(x,y) - G(x',y)| q(y) dy . \end{eqnarray*} For every $y \in ( |x_0 -y| \geq 2r ) \cap ( 1 < |y| \leq M )$ and $x,x' \in B(x_0,r) \cap D$, using Lemma \ref{lm2} we obtain $$|G(x,y) - G(x',y)| \leq G(x,y) + G(x',y) \leq C \big( 1 - \frac{1}{|y|} \big).$$ Now since $G$ is continuous out the diagonal, we deduce by Corollary \ref{coro2} and the Lebesgue's theorem that $$\int_{(|x_0 - y| \geq 2r ) \cap (1 < |y| \leq M)} | G(x,y) -G(x',y) |q(y) dy \to 0 \; as \; |x - x'| \to 0.$$ Hence $Vq \in C(D)$. Next, we consider $x_0 \in \partial D$ and $\varepsilon> 0$. By Lemma \ref{lm3}, there exist $r > 0$ and $M > 1$ such that $$\sup_{z \in D} \int_{B(x_0, 2r) \cap D} G(z, y) q(y)dy \leq \frac{\varepsilon}{4}$$ and $$\sup_{z \in D} \int_{(|y| \geq M)}G(z, y) q(y) dy \leq \frac{\varepsilon}{4}.$$ Let $x \in B(x_0, r) \cap D$, then we have \begin{eqnarray*} Vq(x)&=& \int_D G(x,y) q(y)dy \\ &=& \int_{B(x_0,2r) \cap D} G(x,y) q(y)dy + \int_{(|y| \geq M)} G(x,y) q(y)dy\\ & & + \int_{B^c(x_0, 2r) \cap (1 \leq |y| \leq M)} G(x,y) q(y)dy\\ &\leq& \frac{\varepsilon}{2} + \int_{B^c(x_0, 2r) \cap (1 < |y| \leq M)} G(x,y) q(y)dy. \end{eqnarray*} For every $y \in B^c(x_0, 2r) \cap D \cap \overline{B}(0, M)$ and $x \in B(x_0, r)$, we get by using Lemma \ref{lm2} $$G(x, y) q(y) \leq C(1 - \frac{1}{|y|}) q(y).$$ Now, since for all $y \in D, \lim_{x \to \partial D}G(x,y) = 0$, then as in the above argument, we get $\lim_{x \to x_0}Vq(x) = 0$. This achieves the proof of the proposition. \section{Positive solutions of $\Delta u + \varphi ( . , u) = 0$} In this section, we study the existence of positive solutions for the nonlinear singular elliptic boundary value problem (\ref{e1.3}). \begin{lem} \label{lm4} Let $h \in B^+ (D)$ and $v$ be a nonnegative superharmonic function on $D$. Then for all $w \in B(D)$ such that $V(h |w|) < \infty$ and $w + V(hw) = v$, we have $0 \leq w \leq v$. \end{lem} \paragraph{Proof.} Let $w^+ = \max(w,0)$ and $w^- =\max(-w,0)$. Since $V(h|w|) < \infty$, then $$w^+ + V(hw^+) = v + w^- + V(hw^-).$$ Hence $$V(hw^+) \leq v + V(hw^-) \quad \mbox{in } \left\{ w^+ > 0\right\}.$$ Since $v + V(hw^-)$ is a nonnegative superharmonic function in $D$, we have as consequence of the complete maximum principle $$V(hw^+) \leq v + V(hw^-) \quad\mbox{in } D,$$ that is $V(hw) \leq v = w + V(hw)$. This implies that $0 \leq w \leq v$. \begin{prop} \label{prop6} Let $\varphi : D \times (0, \infty) \to [0, \infty)$ be a measurable function satisfying H1 and $\lambda_1, \lambda_2, \mu_1, \mu_2$ be real numbers such that $0 \leq \lambda_1 \leq \lambda_2$ and $0 \leq \mu_1 \leq \mu_2$. If $u_1$ and $u_2$ are two positive functions continuous on $\overline{D}$ satisfying for each $x \in D$ $$u_1(x)=\lambda_1 +\mu_1 Log|x| + V(\varphi(\cdot,u_1))(x)$$ and $$u_2(x)=\lambda_2 +\mu_2 Log|x| + V(\varphi(\cdot,u_2))(x).$$ Then we have $$0 \leq u_2(x) - u_1(x) \leq \lambda_2 - \lambda_1 +(\mu_2 - \mu_1)Log|x| \;,\; \forall x \in \overline{D} .$$ \end{prop} \paragraph{Proof.} Let $h$ be the function defined on $D$ by $$h(x) = \left\{ \begin{array}{ll} \displaystyle \frac{\varphi(x, u_1(x)) - \varphi(x, u_2(x))} {u_2(x) - u_1(x)} &\mbox{if } u_2(x) \neq u_1(x) \\[3pt] 0, &\mbox{if } u_2(x) = u_1(x). \end{array}\right.$$ Then $h \in B^+(D)$ and we have $$u_2 - u_1 + V(h(u_2 - u_1)) = \lambda_2 - \lambda_1+(\mu_2-\mu_1)Log|\cdot|.$$ Furthermore, we have $$V(h|u_2 - u_1|) \leq V(\varphi(\cdot, u_2)) + V(\varphi(\cdot, u_1)) \leq u_2 + u_1 < \infty.$$ Hence we deduce the result from Lemma \ref{lm4}. \begin{theo} \label{thm1} Let $\lambda > 0$, $\mu > 0$ and $\varphi : D \times (0, \infty) \to [ 0, \infty )$ be a Borel measurable function satisfying H1 and H2. Then the problem \begin{eqnarray} &\Delta u + \varphi ( . , u) = 0,\quad \mbox{in } D \mbox{ (in the weak sense}), & \label{e3.1} \\ &u \mid_{\partial D} = \lambda, \quad \lim_{|x| \to \infty} {u(x)\over Log |x| } = \mu, \quad &\nonumber \end{eqnarray} has a unique positive solution $u_{\lambda } \in C( \overline{D})$. \end{theo} \paragraph{Proof.} Let $\lambda > 0$. Then by hypothesis H2, the function $\varphi (.,\lambda) \in K^{\infty }(D)$ and by Corollary \ref{coro1}, we deduce that $\| V \varphi (.,\lambda)\|_{\infty} < \infty$. To apply a fixed point argument, we consider the convex set $$F = \Big\{\omega \in C(\overline{D} \cup \{\infty\}) : \lambda \leq \omega (x) \leq \lambda + \frac{\lambda\|V \varphi(\cdot,\lambda)\|_\infty} {\lambda + \mu Log |x|}, \, \forall x \in \overline{D}\Big\}$$ and define the operator $T$ on $F$ by $$T \omega (x) = \lambda + \frac{\lambda}{\lambda + \mu Log |x|} \int_D G(x,y) \varphi \Big(y, \omega(y) (1 + \frac{\mu}{\lambda} Log |y|)\Big) dy \,,\; x \in \overline{D}.$$ Since for all $\omega \in F$ and $y \in D, \varphi \left(y, \omega (y) (1 + \frac{\mu}{\lambda} Log |y|) \right) \leq \varphi (y, \lambda)$, then for each $\omega \in F, \lambda \leq T \omega \leq \lambda + \displaystyle\frac{\lambda\|V\varphi(\cdot, \lambda)\|_\infty}{\lambda + \mu Log |x|}$ and as in the proof of Proposition \ref{prop5}, we show that the family $TF$ is equicontinuous in $\overline{D} \cup \{\infty\}$. In particular, for all $\omega \in F, T \omega \in F$. Moreover, the family $\{T \omega (x), \omega \in F\}$ is uniformly bounded in $\overline{D} \cup \{\infty\}$. It follows by Ascoli's theorem that $TF$ is relatively compact in $C(\overline{D} \cup \{\infty\})$. Next, we prove the continuity of $T$ in $F$. Consider a sequence $(\omega_k)_{k \in I\!\!N}$ in $F$ which converges uniformly to a function $\omega \in F$. Then \begin{eqnarray*} |T \omega_k(x) - T \omega(x)| &\leq& \frac{\lambda}{\lambda + \mu Log |x|} \int_D G(x,y) \Big|\varphi\Big(y, \omega_k(y) (1 + \frac{\mu}{\lambda} Log|y|)\Big) \\ &&- \varphi\Big(y, \omega (y)(1 + \frac{\mu}{\lambda} Log |y|)\Big)\Big|dy. \end{eqnarray*} Now by the monotonocity of $\varphi$, we have $$\Big|\varphi\Big(y, \omega_k(y) (1 + \frac{\mu}{\lambda} Log |y|)\Big) - \varphi\Big(y, \omega (y) (1 + \frac{\mu}{\lambda} Log |y|)\Big)\Big| \leq 2 \varphi(y, \lambda),$$ and since $\varphi$ is continuous with respect to the second variable, we deduce by the dominated convergence theorem and Corollary \ref{coro1}, that $$\forall\; x \in \overline{D},\; T \omega_k(x) \to T \omega(x) \quad\mbox{as } k \to \infty.$$ Since $TF$ is relatively compact in $C(\overline{D} \cup \{\infty\})$, then $T \omega_k$ converges uniformly to $T\omega$ as $k \to \infty$. Thus we have proved that $T$ is a compact mapping from $F$ to itself. Hence by the Schauder's fixed point theorem, there exists $\omega_\lambda \in F$ such that for each $x \in D$, $$\omega_\lambda(x) = \lambda + \frac{\lambda}{\lambda + \mu Log |x|} \int_D G(x,y) \varphi\Big(y, \omega_\lambda(y) (1 + \frac{\mu}{\lambda} Log |y|)\Big)dy.$$ Put $u_\lambda(x) = \omega_\lambda(x) (1 + \frac{\mu}{\lambda} Log |x|)$, for $x \in \overline{D}$. Then we have $$u_\lambda(x) = \lambda + \mu Log |x| + \int_D G(x, y) \varphi(y, u_\lambda(y))dy. \label{e3.2}$$ In addition, since for each $y \in D, \;\varphi(y, u_\lambda(y)) \leq \varphi (y, \lambda)$, we deduce by hypothesis H2 and Proposition \ref{prop1} that the map $y \to \varphi(y, u_\lambda(y)) \in L^1_{{\rm loc}}(D)$. On the other hand, using Proposition \ref{prop5}, it follows that $V(\varphi(\cdot, u_\lambda)) \in C(\overline{D})$ and $\lim_{x \to \partial D} V(\varphi(\cdot, u_\lambda))(x)=0$. So we can apply $\Delta$ to the equation (\ref{e3.2}) to obtain $\Delta u_\lambda + \varphi(\cdot, u_\lambda) = 0$ (in the weak sense). Furthermore, for every $x \in D$, we have $$\mu + \frac{\lambda}{Log|x|} \leq \frac{u_\lambda(x)}{Log|x|} \leq \mu + \frac{\lambda + \|V\varphi(\cdot, \lambda)\|_\infty}{Log |x|}.$$ Thus $\lim_{|x|\to \infty} \displaystyle\frac{u_\lambda(x)}{Log |x|}=\mu$, and by (\ref{e3.2}), we have $u_\lambda /_{\partial D} =\lambda$. This shows that $u_\lambda$ is a positive continuous solution of (\ref{e3.1}). Finally, we show the uniqueness of the solution. Let $u$ be a positive continuous solution of the problem in Theorem \ref{thm1}. Clearly $u$ is a superharmonic function with boundary value $\lambda$ and $\lim_{|x|\to \infty} (u(x) - \lambda) \geq 0$. So, we have by the maximum principle \cite[p.465]{d1} that $u \geq \lambda$ on $D$. Which together with the monotonicity of $\varphi$ imply that $\varphi(\cdot, \lambda) \geq \varphi(\cdot, u) \in K^\infty(D)$. So, we deduce by Proposition \ref{prop1} and Proposition \ref{prop5} that the functions $\varphi(\cdot, u)$ and $V \varphi(\cdot, u)$ are in $L^1_{{\rm loc}}(D)$ and $C(\overline{D})$ respectively with $\lim_{x \to \partial D} V \varphi(\cdot, u) (x) = 0$. Hence $u$ satisfies $\Delta(u - V \varphi(\cdot, u)) = 0$ (in the weak sense). It follows that the function $h = u - V\varphi(., u) - \mu Log |x| - \lambda$ is harmonic in $D$ satisfying $h /_ {\partial D} = 0$ and $\lim_{|x|\to \infty} \displaystyle\frac{h(x)}{Log |x|} = 0$. Thus by \cite[p.419]{d1}, we have $h = 0$. So $u$ satisfies (\ref{e3.2}), which yields with Proposition \ref{prop6} to the uniqueness of $u_\lambda$. \quad$\diamondsuit$ \begin{lem} \label{lm5} If $u \in C(\overline{D})$ is a nonnegative solution of the problem \begin{eqnarray} &\Delta u + \varphi ( ., u) = 0,\quad \mbox{ in $D$ (in the weak sense)}& \label{e3.3}\\ &u\mid_{\partial D} = 0 , \quad \lim_{|x| \to \infty} {u(x)\over Log |x| } = \mu \geq 0,& \nonumber \end{eqnarray} then for each $x \in D$, $$\mu Log |x| \leq u(x) \leq \mu Log|x|+V\left(\varphi(\cdot,u)\right)(x). \label{e3.4}$$ \end{lem} \paragraph{Proof.} We assume that $V\varphi\left(\cdot,u\right)\neq \infty$, otherwise the upper inequality is satisfied. Let $\varepsilon > 0$. Since $\lim_{|x|\to \infty} \displaystyle\frac{u(x)}{Log|x|} = \mu$, there exists $M > 1$ such that $$(\mu- \varepsilon) Log |x| \leq u(x) \leq (\mu + \varepsilon) Log |x|,\quad \mbox{for } |x| \geq M.$$ The functions defined on $D$ by $v_\varepsilon(x) = u(x) + (\varepsilon - \mu) Log |x|$ and \linebreak $w_\varepsilon(x) =V\varphi(., u)(x) - u(x) +(\mu+\varepsilon)Log|x|$ satisfy the following properties: $$\displaylines{ v_\varepsilon \in C(\overline{D}) , \quad \Delta v_\varepsilon = \Delta u \leq 0 \quad \mbox{in } D , \cr v_\varepsilon = 0 \quad \mbox{in }{\partial D}, \quad \liminf_{|x| \to \infty} v_\varepsilon (x) \geq 0 }$$ The function $w_\varepsilon$ is lower semi-continuous on $D$, $$\displaylines{ \Delta w_\varepsilon = -\varphi(.,u)- \Delta u = 0 \quad \mbox{in }D ,\cr w_\varepsilon \geq 0 \quad \mbox{in }{\partial D}, \quad \liminf_{|x| \to \infty} w_\varepsilon (x) \geq 0. }$$ Hence by \cite[p.465]{d1}, we get $$(\mu -\varepsilon) Log |x|\leq u(x) \leq (\mu +\varepsilon)Log|x|+V\varphi(.,u)\; \mbox{ in }D.$$ Since $\varepsilon$ is arbitrary, we obtain (\ref{e3.4}). \quad$\diamondsuit$ Now we are ready to prove one of the main results of this section. \begin{theo} \label{thm2} Let $\varphi : D \times (0, \infty) \to [0, \infty)$ be a measurable function satisfying H1 and H2, and $\mu > 0$. Then the problem (\ref{e3.3}) has a unique positive solution $u \in C(\overline{D})$ satisfying $V \varphi(\cdot, u) \neq \infty$. If we suppose further that \linebreak$\varphi \in C^\alpha_{{\rm loc}}(D \times (0, \infty))$, $(0 < \alpha < 1)$, then the solution $u \in C^{2+\alpha}_ {loc}(D) \cap C(\overline{D})$. \end{theo} \paragraph{Proof.} Let $(\lambda_n)_{n \geq 0}$ be a sequence of real numbers that decreases to zero. For each $n \in \mathbb{N}$, we denote by $u_n$ the unique positive solution of problem (\ref{e3.1}) given in Theorem \ref{thm1} for $\lambda=\lambda_n$. Then by Proposition \ref{prop6}, the sequence $(u_n)_{n \geq 0}$ decreases to a function $u$ and so by (\ref{e3.2}), the sequence $(u_n - \lambda_n)_{n \geq 0}$ increases to $u$. Due to the monotonicity of $\varphi$, we have for each $x \in D$ \begin{eqnarray*} u(x) &\geq& u_n(x) - \lambda_n = \mu Log |x| + \int_D G(x,y) \varphi(y, u_n(y)) dy\\ &\geq& \mu Log |x| > 0. \end{eqnarray*} Hence, applying the monotone convergence theorem, we get $$u(x) = \mu Log |x| + \int_D G(x,y) \varphi(y, u(y)) dy \;, \;\forall x \in D. \label{e3.5}$$ Since $u =\sup_n(u_n-\lambda_n) = \inf_n(u_n)$ and for each $n \in\mathbb{N}$ the function $u_n$ is continuous on $D$, then $u$ is a positive continuous function on $D$. Which together with (\ref{e3.5}) imply that $V(\varphi(\cdot, u)) \in L^1_{{\rm loc}}(D)$. So using hypothesis H2 and Proposition \ref{prop1}, we deduce that the map $y \to \varphi(y, u(y)) \in L^1_{{\rm loc}}(D)$. Applying $\Delta$ on both sides of equality (\ref{e3.5}) we obtain $$\Delta u + \varphi(\cdot, u) = 0, \quad\mbox{ in D (in the weak sense)}.$$ Now, since for each $x \in D$ and $n \in \mathbb{N}$, we have $0 \leq u_n(x) - \lambda_n \leq u(x) \leq u_n(x)$ and $\lim_{|x| \to \infty} \displaystyle\frac{u_n(x)}{Log|x|} = \mu$, then $$\lim_{x \to \partial D} u(x) = 0\quad\mbox{and}\quad \lim_{|x|\to \infty} \frac{u(x)}{Log|x|} = \mu.$$ Thus $u \in C(\overline{D})$ and $u$ is a positive solution of the problem (\ref{e3.3}). Finally, we intend to show the uniqueness of the solution. Let $u$ be a positive continuous solution of the problem (\ref{e3.3}) such that $V(\varphi(\cdot, u)) \neq \infty$. Then the functions $\varphi(\cdot, u)$ and $V \varphi(\cdot , u)$ are in $L^1_{{\rm loc}}(D)$. We deduce by \cite[p.52]{c1} that $\Delta (V \varphi(\cdot, u)) = - \varphi(\cdot, u)$, in $D$ (in the weak sense) and consequently $\Delta \left(V \varphi(\cdot, u) + \mu Log |\cdot| - u\right) = 0$ in $D$ (in the weak sense). Hence there exists a harmonic function $h$ in $D$ such that $$h(x) +u(x) - \mu Log |x| = V \varphi(\cdot, u) (x)\quad a.e \mbox{ on }D.$$ Since $u$ and $V\varphi(.,u)$ are superharmonic functions in $D$, we get by \cite[p.134]{p1} that $$h(x) + u(x) - \mu Log |x| = V \varphi (\cdot, u)(x) \mbox{ on } D.$$ Now using (\ref{e3.4}), we get $0 \leq h \leq V \varphi(\cdot, u) < \infty$. Hence by \cite[p.158]{p1}, we deduce that $h = 0$. The function $u$ satisfies (\ref{e3.5}) and the uniqueness follows by Proposition \ref{prop6}. \quad$\diamondsuit$ \begin{coro}\label{coro3} Let $\varphi_1, \varphi_2$ be nonnegative measurable functions in $D \times (0, \infty)$ satisfying the hypotheses H1 and H2, and $\mu_1, \mu_2 \in \mathbb{R}_+$ such that $0 \leq \varphi_1 \leq \varphi_2$ and $0 < \mu_1 \leq \mu_2$. If we denote by $u_j \in C(\overline{D})$ the unique positive solution of the problem (\ref{e3.3}) with $\varphi = \varphi_j$ and $\mu=\mu_j$, $j \in \{1, 2\}$, given in Theorem \ref{thm2}, then we have $$0 \leq u_2 - u_1 \leq (\mu_2 - \mu_1) Log |\cdot| + V\left(\varphi_2(\cdot, u_2) - \varphi_1(\cdot, u_2)\right)\;\mbox{in}\; D.$$ \end{coro} \paragraph{Proof.} It follows by Theorem \ref{thm2}, that $$u_1 = \mu_1 Log |\cdot| + V \varphi_1(\cdot, u_1)\; \mbox{ and }\; u_2 = \mu_2 Log |\cdot| + V \varphi_2(\cdot, u_2).$$ Let $h$ be the nonnegative measurable function defined on $D$ by $$h(x) = \left\{ \begin{array}{ll} \displaystyle \frac{\varphi_1(x, u_2(x)) - \varphi_1(x, u_1(x))}{u_1(x) - u_2(x)} , &\mbox{ if } u_1(x) \neq u_2(x)\\ 0, &\mbox{ if } u_1(x) = u_2(x). \end{array}\right.$$ Then $h \in B^+(D)$ and we have $$u_2 - u_1 + V(h(u_2 -u_1)) = (\mu_2 - \mu_1) Log |\cdot| + V \big(\varphi_2 (\cdot, u_2) - \varphi_1 (\cdot , u_2)\big).$$ Now, since $$V(h|u_2 - u_1|) \leq V \varphi_1(\cdot, u_2) + V \varphi_1(\cdot, u_1) \leq V \varphi_2 (\cdot, u_2) + V \varphi_1 (\cdot,u_1) \leq u_1 + u_2 < \infty$$ and $(\mu_2 - \mu_1) Log |\cdot| + V (\varphi_2 (\cdot, u_2) - \varphi_1(\cdot, u_2))$ is a nonnegative superharmonic function on $D$, we deduce the result from Lemma \ref{lm4}. \begin{theo} \label{thm3} Let $\; \varphi : D \times (0, \infty) \to [0, \infty)$ be a measurable function satisfying H1-H3. Then the problem (\ref{e1.3}) has a unique positive bounded solution $u \in C(\overline{D})$ satisfying $V \varphi(\cdot, u) \neq \infty$. \end{theo} \paragraph{Proof.} Let $\lambda > 0$ and $(\mu_n)$ be a sequence of real numbers that decreases to zero. For each $n \in \mathbb{N}$, we denote by $u_{\lambda, \mu_n}$ the unique positive continuous solution of the problem (\ref{e3.1}) and by $v_n$ the unique positive continuous solution of the problem (\ref{e3.3}), given in Theorem \ref{thm2} for $\mu=\mu_n$. Then by Corollary \ref{coro3}, the sequence $(v_n)_{n\in \mathbb{N}}$ decreases to a function $u$ and so by (\ref{e3.5}), the sequence $(v_n - \mu_n Log|\cdot|)_n$ increases to $u$. Due to the monotonicity of $\varphi$ and by (\ref{e3.2}), we have for each $x \in D$ \begin{eqnarray*} \lefteqn{\lambda +\|V\varphi(\cdot,\lambda)\|_\infty+ \mu_nLog|x|}\\ &\geq& u_{\lambda, \mu_n}(x) \geq v_n(x) \\ & \geq & u(x) \geq v_n(x) - \mu_n Log|x|\\ & \geq &\int_D G(x,y) \varphi\left(y, \mu_n Log |y| + \lambda + \|V\varphi (\cdot, \lambda)\|_\infty\right)dy. \end{eqnarray*} Letting $n$ tends to infinity, we get $$\lambda + \|V\varphi (\cdot, \lambda)\|_\infty \geq u(x) \geq V\varphi\left(\cdot, \lambda + \|V \varphi(\cdot, \lambda)\|_\infty\right)(x) \; ,\;\; \forall\; x \in D.$$ By H2, H3 and Corollary \ref{coro1}, $u$ is a positive bounded function in $D$. Since, for each $n\in\mathbb{N}$ and $x \in D$ $$v_n -\mu_n Log|x| =\int_D G(x,y)\varphi(y,v_n(y))\;dy,$$ we obtain, as $n\to\infty$, that $$u(x) = \int_DG(x, y) \varphi(y, u(y))dy \;,\; \forall\; x \in D. \label{e3.6}$$ As in the proof of Theorem \ref{thm2}, we show that $u \in C(\overline{D})$ and $u$ is a positive bounded solution of (\ref{e1.3}). Using (\ref{e3.4}), we establish the uniqueness of such a solution. \quad$\diamondsuit$ \begin{coro} \label{coro4} Let $\varphi : D \times (0, \infty) \to [0, \infty)$ be a measurable function satisfying H1 and H2. Then for each $\mu> 0$ and $f$ a nonnegative continuous function on $\partial D$, the following nonlinear problem \begin{eqnarray} &\Delta u + \varphi(\cdot, u) = 0, \quad\mbox{in $D$ (in the weak sense)}& \label{e3.7}\\ &u_{/ \partial D } = f,\quad \lim_{|x|\to \infty} \displaystyle\frac{u(x)}{Log|x|} = \mu,\quad \quad & \nonumber \end{eqnarray} has a unique positive solution $u \in C(\overline{D})$ satisfying $V\varphi (\cdot,u) \neq \infty$. \end{coro} \paragraph{Proof.} Let $H^D_f$ denotes the unique bounded solution of the following Dirichlet problem $$\displaylines{ \Delta \omega = 0, \quad\mbox{in } D,\cr \omega_{/ \partial D} = f.\quad }$$ We note that if $u$ is a continuous solution of (\ref{e3.7}), then as $\varphi$ is a nonnegative function, we deduce that $u - H^D_f$ is superharmonic such that $u - H^D_f = 0$ on $\partial D$ and $\lim_{|x| \to \infty} \Big(u(x) - H^D_f(x)\Big) = + \infty$. We conclude by the maximum principle, that $$u \geq H^D_f \quad \mbox{in } D.$$ Let $\Psi$ be the function defined on $D \times(0, \infty)$ by $\Psi(x,t) = \varphi(x, t + H^D_f(x))$. It is clear to verify that $\Psi$ satisfies the same hypotheses H1-H2 as $\varphi$. Hence by Theorem \ref{thm2}, the problem $$\displaylines{ \Delta v + \Psi(\cdot, v) = 0, \quad\mbox{in } D,\cr v_{/\partial D} = 0,\quad \lim_{|x|\to \infty} \frac{v(x)}{Log |x|} = \mu, }$$ has a unique positive solution $v \in C(\overline{D})$ satisfying $V \Psi(\cdot, v) \neq \infty$. Moreover, $u$ is a solution of (\ref{e3.7}) if and only if $u = v + H^D_f$. This completes the proof. \quad$\diamondsuit$ \begin{coro} \label{coro5} Let $\varphi : D \times (0, \infty) \to [0, \infty)$ be a measurable function satisfying H1-H3 and $f$ be a nonnegative continuous function on $\partial D$. Then the nonlinear Dirichlet problem \begin{eqnarray} &\Delta u + \varphi ( . , u) = 0,\quad\mbox{in $D$ (in the weak sense)} & \label{e3.8}\\ &u_{/\partial D}= f,\quad \quad \quad \quad \quad \quad \quad \quad &\nonumber \end{eqnarray} has a unique positive bounded solution $u \in C(\overline{D})$ satisfying $V\varphi(\cdot, u) \neq \infty$. \end{coro} \section{Estimates on solutions} In this section, we give some estimates on the solutions of (\ref{e1.3}) given by (\ref{e3.5}) and (\ref{e3.6}). We denote by $\beta =\displaystyle\inf_{\lambda > 0} \{\lambda + \|V \varphi(\cdot, \lambda) \|_\infty\}$ and we remark that if H3 is satisfied then $\beta > 0$. \begin{theo} \label{thm4} Let $\mu > 0$ and $u$ be the solution of problem (\ref{e3.3}) given by (\ref{e3.5}). Then for each $x \in \overline{D}$, we have $$\mu Log |x| \leq u(x) \leq \mu Log |x| + \min\Big(\beta, V \varphi\left(. , \mu Log |\cdot|\right) (x)\Big).$$ \end{theo} \paragraph{Proof.} For each $\lambda > 0$ and each $x \in \overline{D}$, we have $$\mu Log |x| \leq u(x) \leq u_\lambda(x) \leq \mu Log |x| + \lambda + \|V \varphi(\cdot, \lambda)\|_\infty,$$ where $u_\lambda$ is the solution of the problem (\ref{e3.3}). Thus $$\mu Log |x| \leq u(x) \leq \mu Log|x| + \beta\;,\; \forall\; x \in \overline{D}.$$ Since $\varphi$ is non-increasing with respect to the second variable, from (\ref{e3.5}) we obtain $$\mu Log |x| \leq u(x) \leq \mu Log |x| + \int_D G(x,y)\varphi(y, \mu Log |y|) dy \,,\; \forall x \in \overline{D}.$$ Which completes the proof. \quad$\diamondsuit$ \begin{rem} \rm Let $\varepsilon > 0$, sufficiently small, $D_\varepsilon = \{x \in \mathbb{R}^2, 1 < |x| \leq 1 + \varepsilon\}$ and $u$ be the solution of (\ref{e3.3}) given by (\ref{e3.5}). If $\varphi$ satisfies $$\sup_{x \in D_\varepsilon} \int_D \frac{1}{|x-y|}\varphi(y, \mu Log |y|) dy < \infty,$$ then there exists a constant $c > 0$ such that for each $x \in \overline{D}$, $$\mu Log |x| \leq u(x) \leq (\mu + c) Log |x|.$$ Indeed, there exists $C > 0$ such that for every $x, y$ in $D$, we have $$G(x,y) \leq C \frac{(|x| - 1) \wedge (|y| -1)}{|x-y|}.$$ Hence, for each $x \in D_\varepsilon$ \begin{eqnarray*} u(x) &\leq& \mu Log |x| + V (\varphi(\cdot, \mu Log |\cdot|))(x)\\ &\leq& \mu Log |x| + C\Big(\int_D \frac{\varphi(y, \mu Log |y|)}{|x-y|} dy\Big)(|x| -1)\\ &\leq& \mu Log |x| + C\Big(\sup_{z \in D_\varepsilon} \int_D \frac{\varphi(y, \mu Log |y|)}{|z - y|}\Big) (1 + \varepsilon) Log |x|\\ &=& \mu Log |x| + C_1 Log |x|. \end{eqnarray*} Moreover, $$\forall\; x \in \overline{D} \backslash D_\varepsilon,\;\; u(x) \leq \mu Log |x| + \beta \leq \mu Log |x| + \frac{\beta}{Log (1 + \varepsilon)} Log |x|.$$ Consequently, for each $x \in \overline{D}$ $$\mu Log |x| \leq u(x) \leq \mu Log |x| + \max (C_1, \frac{\beta} {Log(1 + \varepsilon)}) Log |x| = (\mu + c) Log |x|.$$ \end{rem} \begin{exa} \rm Let $u$ be the positive solution of (\ref{e3.3}) given by (\ref{e3.5}). For\linebreak $r \in [1, \infty)$, we denote by $\phi(r, \cdot) = \displaystyle{\sup_{|x| = r}} \varphi(x, \cdot)$. If $$\int^\infty_1 r \phi (r, \mu Log r)dr < \infty,$$ then there exists $c > 0$ such that for every $x \in \overline{D}$, $$\mu Log |x| \leq u(x) \leq (\mu + c) Log |x|.$$ Indeed, by Theorem \ref{thm4}, we have for each $x \in \overline{D}$, \begin{eqnarray*} \mu Log |x| \leq u(x)&\leq& \mu Log |x| + \int_D G(x,y) \varphi(y, \mu Log|y|)dy\\ &\leq& \mu Log |x| + \int^\infty_1 r Log (|x| \wedge r) \phi(r, \mu Log r)dr\\ &\leq& \mu Log |x| + (\int^\infty_1 r \phi(r, \mu Log r)dr) Log |x|\\ &=& (\mu + c) Log |x|. \end{eqnarray*} \end{exa} \begin{theo} \label{thm5} Let $u \in C(\overline{D})$ be the unique positive bounded solution of (\ref{e1.3}). Then there exists $c > 0$ such that $$c(1 - \frac{1}{|x|}) \leq u(x) \leq \min \Big(\beta, V\varphi \Big(. , c (1 - \frac{1}{|\cdot|})\Big)(x)\Big)\;,\; \forall\; x \in \overline{D}.$$ \end{theo} \paragraph{Proof.} As it can be seen in the proof of Theorem \ref{thm3}, we have $$V\varphi(\cdot, \beta)(x) \leq u(x) \leq \beta\;,\; \forall\; x \in D.$$ On the other hand, from Lemma \ref{lm1}, we have $$\frac{1}{2\pi} (1 - \frac{1}{|x|}) \Big(\int_D (1 - \frac{1}{|y|}) \varphi(y, \beta) dy \Big) \leq V\varphi(\cdot, \beta)(x) \;,\; \forall\; x\in D.$$ Hence, the lower bound inequality follows from H2 and Corollary \ref{coro2}, with $$c = \frac{1}{2\pi} \int_D(1 -\frac{1}{|y|}) \varphi(y, \beta)dy.$$ Now, since $\varphi$ is non-increasing with respect to the second variable, we get by using (\ref{e3.6}) that $$u(x) \leq \int_DG(x,y) \varphi\Big(y, c(1 - \frac{1}{|y|})\Big)\,dy.$$ This completes the proof. \begin{rem} \rm Let $\varepsilon > 0$, sufficiently small, $D_\varepsilon = \{x \in \mathbb{R}^2, 1 < |x| \leq 1 + \varepsilon\}$ and $u$ be the unique positive bounded solution of (\ref{e1.3}). If $\varphi$ satisfies $$\sup_{z \in D_\varepsilon} \int_D \frac{1}{|z-y|}\varphi \Big(y, c(1-\frac{1}{|y|})\Big) dy < \infty \;,\;\forall \,c>0,$$ then there exists a constant $C > 0$ such that for each $x \in \overline{D}$, $$\frac{1}{C}\big(1-\frac{1}{|x|} \big) \leq u(x) \leq C \big(1-\frac{1}{|x|}\big).$$ \end{rem} \begin{exa} \rm Let $u$ be the positive bounded solution of (\ref{e1.3}) given by (\ref{e3.6}) and $\phi$ defined as in Example 4.1. If $$\forall c> 0,\; \; \int_1^{\infty} r \phi \Big(r,c(1-\frac{1}{r}) \Big) \;dr < \infty ,$$ then there exists $C > 0$ such that $$\frac{1}{C} \big(1- \frac{1}{|x|} \big) \leq u(x) \leq C \big(1- \frac{1}{|x|} \big) \;,\; \forall x \in \overline{D}.$$ Indeed, for $1\leq |x| \leq 2$, we have \begin{eqnarray*} c \big(1-\frac{1}{|x|} \big) \leq u(x) & \leq & \int_D G(x,y) \varphi \Big(y,c(1-\frac{1}{|y|}) \Big)\;dy\\ & \leq & \int_1^{\infty} rLog(|x|\wedge r)\phi \Big(r,c(1-\frac{1}{r})\Big) \;dr \\ & \leq & Log|x| \int_1^{\infty} r\phi \Big(r,c(1-\frac{1}{r})\Big) \;dr \\ & \leq & \big(1-\frac{1}{|x|}\big) \Big( 2 \int_1^{\infty} r\phi \big(r,c(1-\frac{1}{r})\big)\;dr \Big) \end{eqnarray*} Moreover, for $|x|>2$, \begin{eqnarray*} c \big(1-\frac{1}{|x|}\big) \leq u(x) \leq \beta \leq 2\beta \big(1-\frac{1}{|x|}\big). \end{eqnarray*} This gives the desired estimates. \end{exa} We close this paper by giving an other comparison result for the solutions of the problem (\ref{e1.3}), in the case of the special nonlinearity $\varphi(x,t) = q(x) f(t)$. The following hypotheses on $q$ and $f$ are adopted. \begin{itemize} \item $f : (0, \infty) \to (0, \infty)$ is a continuously differentiable non-increasing function. \item $q \in C^\alpha_{{\rm loc}}(D) \cap K^\infty(D)$, $0<\alpha <1$, is a nontrivial nonnegative function in $D$. \end{itemize} We define the function $F$ in $[0, \infty)$ by $F(t) = \int_0^t 1/f(s)ds$. From the hypotheses on $f$, we note that the function $F$ is a bijection from $[0, \infty)$ to itself. Then, we have the following statement. \begin{theo} \label{thm6} Let $u \in C(\overline{D})$ be the positive solution of the problem \begin{eqnarray} &\Delta u + q(x) f(u) = 0 \quad \mbox{in $D$ (in the weak sense)}& \label{e4.1}\\ &u\mid _{\partial D} = 0, \quad \lim_{|x|\to \infty}\frac{u(x)}{Log|x|} =\mu >0,&\nonumber \end{eqnarray} such that $V\big(qf(u)\big) \neq \infty$. Then $$V\big(qf(\beta + \mu Log |\cdot|)\big)(x) + \mu Log |x| \leq u(x) \leq F^{-1} (Vq(x)) + \mu Log |x|\;, \;\forall\; x \in \overline{D}.$$ \end{theo} \paragraph{Proof.} Since $u \leq \beta + \mu Log |\cdot|$ in $D$ and $f$ is nonincreasing with respect to the second variable, we deduce that for each $x$ in $D$, $$V(qf(\beta + \mu Log|\cdot|))(x) + \mu Log |x| \leq u(x) = \mu Log |x| + \int_D G(x,y) q(y) f(u(y)) dy.$$ To show the upper estimate, we consider $\varepsilon >0$ and define the function $v_{\varepsilon}$ in $D$ by $v_{\varepsilon}(x) = F \big(u(x) - \mu Log |x|\big) - Vq(x) -\varepsilon Log|x|$. Then, $v_{\varepsilon} \in C^2(D)$ and \begin{eqnarray*} \Delta v_{\varepsilon}(x) &=& \frac{1}{f\big(u(x) -\mu Log |x|\big)} \Delta u(x) + q(x)\\ &&- \frac{f'\big(u(x) - \mu Log |x|\big)}{f^2\big(u(x) - \mu Log |x|\big)} \|\nabla (u - \mu Log |\cdot|) (x)\|^2. \end{eqnarray*} Thus, $\Delta v_{\varepsilon} \geq 0$. Moreover, since $0 \leq u - \mu Log|.|\leq \beta$, $Vq$ is bounded in $D$ and $\lim_{x \to \partial D} Vq(x) = 0$, we get ${v_{\varepsilon}}_{/\partial D}=0$ and $\lim_{|x|\to \infty}v_{\varepsilon}(x) \leq 0$. Hence, by \cite[p.465]{d1} we deduce that $v_{\varepsilon} \leq 0$ in $D$. Since $\varepsilon$ is arbitrary, we get the upper inequality. \quad$\diamondsuit$ Using the same arguments as in the proof above, we can prove the following theorem. \begin{theo} \label{thm7} Let $u \in C(\overline{D})$ be the positive bounded solution of the problem(\ref{e1.3}) with $\varphi(.,u)=q(x)f(u)$, such that $V\big(qf(u)\big) \neq \infty$. Then we have $$f(\beta) Vq(x) \leq u(x) \leq F^{-1}\big(Vq(x)\big)\,, \; \forall x \in \overline{D}.$$ \end{theo} \begin{coro} \label{coro6} Let $\varphi : D \times (0, \infty) \to [0, \infty)$ be a measurable function satisfying H1. Further, assume that $\varphi$ satisfies $$\varphi(x, t) \leq q(x) f(t) \;,\; \forall\; (x, t) \in D \times (0, \infty).$$ Let $\mu >0$ and $u$ be the solution of the problem (\ref{e3.3}). Then $u$ satisfies $$\mu Log |x| \leq u(x) \leq \mu Log |x| + F^{-1} (Vq(x)) \;,\; \forall\; x \in \overline{D}.\label{e4.2}$$ \end{coro} \paragraph{Proof.} Let $v$ be the solution of the problem (\ref{e4.1}). Then, by Corollary \ref{coro4}, we deduce that $u \leq v$ in $D$. Which together with Theorems \ref{thm4} and \ref{thm6}, give (\ref{e4.2}). \begin{exa} \rm Let $\gamma > 0$. Then the problem \begin{eqnarray*} & \Delta u + q(x) u^{-\gamma} = 0, \quad \mbox{ in } D&\\ &u\mid_{\partial D} = 0\,,\quad \lim_{|x|\to \infty} \frac{u(x)}{Log |x|} = \mu \geq 0& \end{eqnarray*} has a positive solution $u \in C(\overline{D})$ satisfying $$V\big((\beta + \mu Log|\cdot|)^{-\gamma}q\big)(x) \leq u(x) - \mu Log |x| \leq \big[(\gamma + 1)Vq(x)\big]^{\frac{1}{1+\gamma}} \;,\; \forall x \in \overline{D}.$$ \end{exa} \paragraph{Acknowledgements.} I want to thank Professor Habib Ma\^agli, my adviser, for his valuable help. I also want to thank the anonymous referee for his/her suggestions. \begin{thebibliography}{00} \bibitem{b1} J. Bliedtner, W. Hansen, {\em Potential theory. An analytic and probabilistic approach to balayage}, Springer Verlag (1986). \bibitem{c1} K.L. Chung, Z. Zhao, {\em From Brownian motion to Schr\"odinger's equation}, Springer Verlag (1995). \bibitem{d1} R. Dautray, J.L. 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