\documentclass[twoside]{article} \usepackage{amssymb, amsmath} \pagestyle{myheadings} \markboth{\hfil Positive solutions of nonlinear elliptic equations \hfil EJDE--2002/41} {EJDE--2002/41\hfil Imed Bachar, Habib M\^aagli, \& Lamia M\^aatoug \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 41, pp. 1--24. \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 nonlinear elliptic equations in a half space in $\mathbb{R}^2$ % \thanks{ {\em Mathematics Subject Classifications:} 31A25, 31A35, 34B15, 34B27, 35J65. \hfil\break\indent {\em Key words:} Singular elliptic equation, superharmonic function, Green function, \hfil\break\indent Schauder fixed point theorem, maximun principle. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Submitted December 14, 2001. Published May 16, 2002.} } \date{} % \author{Imed Bachar, Habib M\^aagli, \& Lamia M\^aatoug} \maketitle \begin{abstract} We study the existence and the asymptotic behaviour of positive solutions of the nonlinear equation $\Delta u+f(.,u)=0$, in the domain $D=\{(x_1,x_2)\in \mathbb{R}^2:x_2>0\}$, with $u=0$ on the boundary. The aim is to prove some existence results for the above equation in a general setting by using a fixed-point argument. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \allowdisplaybreaks \numberwithin{equation}{section} \section{Introduction} In [12], Zeddini considered the nonlinear elliptic problem $$\label{P} \begin{gathered} \Delta u\ +\ f(.,u)=0 \quad \text{in }D \\ u>0 \quad \text{in }D \\ u=0 \quad \text{on }\partial D, \end{gathered}$$ in the sense of distributions, where $D$ is the outside of the unit disk in $\mathbb{R}_{}^2$ and $f$ is a nonnegative function in $D\times (0,\infty )$ non-increasing with respect to the second variable. Then, when $f$ is in a certain Kato class, he proved the existence of infinitely many positive continuous solutions on $\overline{D}$. More precisely, he showed that for each $b>0$, there exists a positive continuous solution $u$ satisfying \begin{equation*} \underset{| x| \to \infty }{\lim }\frac{u(x)}{Log|x| }=b. \end{equation*} Note that the existence results of problem (\ref{P}) have been extensively studied for the special nonlinearity $f(x,t)=p(x)q(t)$, for both bounded and unbounded domain $D$ in $\mathbb{R}^n(n\geq 1)$, with smooth compact boundary (see for example [3, 4, 5, 6] and the references therein). On the other hand, in [7, 9, 10, 11], the authors considered the problem $$\label{Q} \begin{gathered} \Delta u\ +\ g(.,u)=0 \quad \text{in }D \\ u>0 \quad \text{in }D \\ u=0 \quad \text{on }\partial D, \end{gathered}$$ where there is no restriction on the sign of $g$, and $D$ is an unbounded domain in $\mathbb{R}^n$ ($n\geq 1$) with a compact Lipschitz boundary. Then they proved the existence of infinitely many solutions provided that $g$ is in a certain Kato class. Namely, they showed that there exists a number $b_0>0$ such that for each $b\in (0,b_0]$, there exists a positive continuous solution $u$ in $\overline{D}$ satisfying \begin{equation*} \lim_{| x| \to \infty }\frac{u(x)}{h(x)}=b, \end{equation*} where $h$ is a positive solution of the homogeneous Dirichlet problem $\triangle u=0 \mbox{ in } D\,,u=0 \mbox{ on }{\partial D}$.\\ In this paper, we consider the domain \begin{equation*} D=\mathbb{R}_{+}^2=\{(x_1,x_2)\in \mathbb{R}^2:x_2>0\}, \end{equation*} which has a non-compact boundary. The purpose of this paper is two-folded. One is to introduce a new Kato class $K$ of functions on $D$ and to study the properties of this class. The other is to investigate the existence of positive continuous solutions on (\ref{P}) and (\ref{Q}). Indeed, we shall establish some existence theorems for problems (\ref{P}) and (\ref{Q}), when $f$ and $g$ are required to satisfy suitable assumptions related to the class $K$. Note that solutions of these problems are understood as distributional solutions in $D$. The outline of the paper is as follows. In section 2, we prove some inequalities on the Green's function $G(x,y)=\frac 1{4\pi }Log(1+\frac{% 4x_2y_2}{| x-y| ^2})$ of the Laplacian in $D$. In particular, we establish the fundamental inequality \begin{equation*} \frac{G(x,y)G(y,z)}{G(x,z)}\leq C_0\big[\frac{y_2}{x_2}G(x,y) +\frac{y_2}{z_2}G(y,z)\big] \end{equation*} which is called the 3G-Theorem. This enable us to define and study, in section 3, a new Kato class $K$ on $D$. \paragraph{Definition} A Borel measurable function $\varphi$ in $D$ belongs to the class $K$ if $\varphi$ satisfies \begin{gather} \label{e1.1} \lim_{\alpha \to 0} \sup_{x\in D} \int_{(| x-y| \leq \alpha )\cap D}\,\frac{y_2}{x_2}G(x,y)| \varphi (y)| dy=0\,,\\ \label{e1.2} \lim_{M\to \infty }\sup_{x\in D}\int_{(| y| \geq M)\cap D}\, \frac{y_2}{x_2}G(x,y)| \varphi(y)| \,dy=0\,. \end{gather} To study Problem (\ref{P}) in section 4, we assume that $f$ satisfies: \begin{enumerate} \item[(H1)] $f:D\times (0,\infty )\to [0,\infty )$ is measurable, continuous and non-increasing with respect to the second variable. \item[(H2)] For all $c>0$, $f(.,c)\in K$. \item[(H3)] For all $c>0$, $V(f(.,c))>0$, where $V=(-\Delta )^{-1}$ is the potential kernel associated to $\Delta$. \end{enumerate} As usual, we denote by $\mathcal{B}(D)$ the set of Borel measurable functions in $D$ and $\mathcal{B}^{+}(D)$ the set of nonnegative functions. $C(D)$ will denote the set of continuous functions in $D$ and \begin{equation*} C_0(D)=\{v\in C(D):\lim_{x\to \partial D} v(x)=\lim_{| x| \to \infty } v(x)=0\}. \end{equation*} Throughout this paper, the letter $C$ will denote a generic positive constant which may vary from line to line. \begin{theorem} \label{thm1.1} Assume (H1)-(H3). Then for each $b>0$, the problem (\ref{P}) has at least one positive solution $u$ continuous on $\overline{D}$ and satisfying \begin{equation*} \lim_{x_2\to \infty }\frac{u(x)}{x_2}=b. \end{equation*} Moreover, we have for $x$ in $D$, \begin{equation*} bx_2\leq u(x)\leq bx_2+\min \big(\delta ,\int_DG(x,y)f(y,by_2)dy\big), \end{equation*} where $\delta =\inf_{\alpha >0}(\alpha +\| Vf(.,\alpha)\|_\infty)$. \end{theorem} \begin{theorem} \label{thm1.2} Assume (H1)-(H3). Then the problem (\ref{P}) has a unique solution $u\in C_0(D)$, satisfying \begin{equation*} \frac{x_2}{C(| x| +1)^2}\leq u(x)\leq \min (\delta ,\int_DG(x,y)f(y,\frac{y_2}{C(| y| +1)^2})dy),\quad \forall x\in D. \end{equation*} \end{theorem} We point out, that for some functions $f$ of the type $f(x,t)=p(x)t^{-\sigma }$, with $\sigma \geq 0$, we get better estimates on the solution. Namely for each $x\in D$, we have \begin{equation*} u(x)\leq C\frac{x_2^{\frac 1{1+\sigma }}}{(| x| +1)^{\frac 2{1+\sigma }}}, \end{equation*} for some positive constant $C$. In section 5, we consider Problem (\ref{Q}) under the following hypotheses: \begin{enumerate} \item[(A1)] The function $g$ is measurable on $D\times (0,\infty )$, continuous with respect to the second variable and satisfies \begin{equation*} | g(x,t| \leq t\psi (x,t)\quad\text{for } (x,t)\in D\times (0,\infty ), \end{equation*} where $\psi$ is a nonnegative measurable function on $D\times (0,\infty )$ such that the function $t\to \psi (x,t)$ is nondecreasing on $(0,\infty )$ and $\lim_{t\to 0} \psi (x,t)=0$. \item[(A2)] The function defined as $x\to \psi (x,x_2)$ on $D$ belongs to the class $K$. \end{enumerate} \begin{theorem} \label{thm1.3} Assume (A1)-(A2). Then (\ref{Q}) has infinitely many solutions. More precisely, there exists $b_0>0$ such that for each $b\in (0,b_0]$, there exists a solution $u$ of (\ref{Q}) continuous on $D$ and satisfying \begin{equation*} \frac b2 x_2\leq u(x)\leq \frac{3b}2x_2\quad\text{and}\quad \lim_{x_2\to \infty } \frac{u(x)}{x_2}=b. \end{equation*} \end{theorem} \section{Properties of Green's function} \begin{lemma} \label{lm2.1} For $x$ and $y$ in $D$, we have the following properties: \begin{enumerate} \item[(i)] If $x_2y_2\leq | x-y| ^2$, then $\max(x_2,y_2)\leq \frac{\sqrt{5}+1}2| x-y|$. \item[(ii)] If $| x-y| ^2\leq x_2y_2$, then $\frac{3-\sqrt{5}}2x_2\leq y_2\leq \frac{3+\sqrt{5}}2x_2$. \end{enumerate} \end{lemma} \paragraph{Proof} (i) If $x_2y_2\leq | x-y| ^2$ then $| y-\widetilde{x}| \geq \frac{\sqrt{5}}2x_2$, where $\widetilde{x}=(x_1,\frac 32x_2)$. It follows that \begin{equation*} | y-x| \geq | y-\widetilde{x}| -| x-\widetilde{x} | \geq \frac{\sqrt{5}-1}2x_2. \end{equation*} i.e., $x_2\leq \frac{\sqrt{5}+1}2| x-y|$. Thus, interchange the role of $x$ and $y$, we obtain (i). \\ (ii) If $|x-y| ^2\leq x_2y_2$ then $| x_2-y_2| ^2\leq x_2y_2$. Hence \begin{equation*} \big[ y_2-\frac{3+\sqrt{5}}2x_2\big] \big[ y_2-\frac{3-\sqrt{5}} 2x_2\big] \leq 0. \end{equation*} \begin{proposition} \label{prop2.2} There exists $C>0$ such that, for all $x$ and $y$ in $D$ \begin{gather} \frac{x_2y_2}{C(| x| +1)^2(| y| +1)^2}\leq G(x,y)\leq \frac 1\pi \frac{x_2y_2}{| x-y| ^2}. \label{e2.1}\\ \frac 1\pi \frac{y_2^2}{| x-y| ^2+4x_2y_2}\leq \frac{y_2}{x_2}% G(x,y)\leq C(1+G(x,y)). \label{e2.2} \end{gather} \end{proposition} \paragraph{Proof} Recall that the Green's function $G$ of $\Delta$ in $D$ is $$\label{e2.3} G(x,y)=\frac 1{4\pi }Log(1+\frac{4x_2y_2}{| x-y| ^2}).$$ To prove (\ref{e2.1}) and the first inequality in (\ref{e2.2}), we use that \begin{equation*} \frac t{1+t}\leq Log(1+t)\leq t,\forall t\geq 0,\quad\text{and} \quad |x-y| \leq (| x| +1)(| y| +1),\quad \forall x,y\in D. \end{equation*} The second inequality in (\ref{e2.2}) follows from Lemma \ref{lm2.1}. Indeed, if $x_2y_2\leq | x-y| ^2$ then \begin{equation*} \frac{y_2}{x_2}G(x,y)\leq C\frac{y_2^2}{| x-y| ^2}\leq C \end{equation*} and if $| x-y| ^2\leq x_2y_2$ then \begin{equation*} \frac{y_2}{x_2}G(x,y)\leq CG(x,y).\quad \diamondsuit \end{equation*} \begin{theorem}[3G-Theorem] \label{thm2.3} There exists a constant $C_0>0$ such that for all $x$, $y$ and $z$ in $D$, we have $$\label{e2.4} \frac{G(x,z)\,G(z,y)}{G(x,y)}\leq C_0 \big[ \frac{z_2}{x_2}G(x,z)+\frac{z_2}{y_2}G(y,z)\big] .$$ \end{theorem} \paragraph{Proof.} Let $N(x,y)=\frac{x_2y_2}{G(x,y)}$, for $x$ and $y$ in $D$. Then (\ref{e2.4}) is equivalent to $$\label{e2.5} N(x,y)\leq C_0(N(y,z)+N(z,x)).$$ Using the inequalities $\frac t{1+t}\leq Log(1+t)\leq t,\forall t\geq 0$, we deduce by (\ref{e2.1}) and (\ref{e2.2}) that for all $x$ and $y$ in $D$, $$\label{e2.6} \pi | x-y| ^2\leq N(x,y)\leq \pi (| x-y| ^2+4x_2y_2).$$ Then to prove (\ref{e2.5}), we need to consider two cases:\\ Case i: $x$ and $y$ in $D$ with $x_2y_2\leq | x-y| ^2$. Then by (\ref{e2.6}), for all $z$ in $D$, $$N(x,y) \leq 5\pi | x-y| ^2 \leq 10\pi (| x-z| ^2+| z-y| ^2) \leq 10(N(x,z)+N(z,y)).$$ Case ii: $x$ and $y$ in $D$ with $| x-y| ^2\leq x_2y_2$. Then by Lemma \ref{lm2.1}, \begin{equation*} \frac{3-\sqrt{5}}2x_2\leq y_2\leq \frac{3+\sqrt{5}}2x_2. \end{equation*} If $| x-z| ^2\leq x_2z_2$ or $| y-z| ^2\leq y_2z_2$, then by Lemma \ref{lm2.1} \begin{equation*} \frac{3-\sqrt{5}}2x_2\leq z_2\leq \frac{3+\sqrt{5}}2x_2, \quad\text{or}\quad \frac{3-\sqrt{5}}2y_2\leq z_2\leq \frac{3+\sqrt{5}}2y_2. \end{equation*} Recall that for all $a$ and $b$ in $(0,\infty )$, \begin{equation*} \frac{ab}{a+b}\leq \min (a,b)\leq 2\frac{ab}{a+b}, \end{equation*} and for all $x,y$ and $z$ in $D$, $| x-y| ^2\leq 4\max (| x-z| ^2,| z-y| ^2)$, then in this case we have \begin{equation*} Log(1+\frac{4x_2y_2}{| x-y| ^2})\geq C\min \big[Log(1+\frac{4z_2y_2 }{| z-y| ^2}),Log(1+\frac{4x_2z_2}{| x-z| ^2})\big]. \end{equation*} Which is equivalent to (\ref{e2.5}).\\ If $| x-z| ^2\geq x_2z_2$ and $| y-z| ^2\geq y_2z_2$, then using (\ref{e2.6}) and Lemma \ref{lm2.1}, we obtain \begin{eqnarray*} N(x,y) &\leq &5\pi x_2y_2\leq C| x-z| | y-z| \\ &\leq& C(| x-z| ^2+| y-z| ^2) \leq C(N(x,z)+N(y,z)). \quad \diamondsuit \end{eqnarray*} Now we are ready to study the properties of the functional class $K$. \section{The class K.} \begin{proposition} \label{prop3.1} Let $\varphi$ \thinspace be a\ function in $K$. Then the function $y\to y_2^2\varphi (y)$ is in $L_{{\rm loc}}^1(\overline{D})$. \end{proposition} \paragraph{Proof} Since $\varphi \in K$, then by (\ref{e1.1}) there exists $\alpha >0$ such that \begin{equation*} \sup_{x\in D} \int_{(| x-y| \leq \alpha )\cap D} \frac{y_2}{x_2}G(x,y)| \varphi (y)| \,dy\leq 1. \end{equation*} Let $R>0$ and $a_1,\dots ,a_{n}$ in $B(0,R)\cap D$ with $B(0,R)\cap D\subset \cup_{1\leq i\leq n} B(a_i,\alpha )$. Then by (\ref{e2.2}), there exists $C>0$ such that for all $i\in \{1,\dots ,n\}$ and $y\in B(a_i,\alpha )\cap D$ \begin{equation*} y_2^2\leq C\frac{y_2}{(a_i)_2}G(a_i,y). \end{equation*} Hence, we have \begin{eqnarray*} \int_{B(0,R)\cap D}y_2^2| \varphi (y)| dy &\leq &C \sum_{1\leq i\leq n} \int_{(|x_i-y| \leq \alpha )\cap D} \frac{y_2}{(a_i)_2}\,G(a_i,y)| \varphi (y)| dy \\ &\leq & Cn\sup_{x\in D} \int_{(|x-y| \leq \alpha )\cap D} \frac{y_2}{x_2}G(x,y)| \varphi (y)| dy \\ &\leq &Cn <\infty \hspace{5cm} \diamondsuit \end{eqnarray*} In the sequel, we use the notation $$\label{e3.1} \| \varphi \| = \sup_{x\in D} \int_D \frac{y_2}{x_2}G(x,y)| \varphi (y)| \,dy\,.$$ \begin{proposition} \label{prop3.2} If $\varphi \in K$, then $\| \varphi \| <+\infty$. \end{proposition} \paragraph{Proof} Let $\alpha >0$ and $M>0$. Then we have \begin{eqnarray*} \int_D\frac{y_2}{x_2}G(x,y)\,| \varphi (y)| \,dy &\leq& \int_{(| x-y| \leq \alpha )\cap D}\frac{y_2}{x_2}G(x,y)| \varphi (y)| dy \\ &&+\int_{(| y| \geq M)\cap D}\frac{y_2}{x_2}G(x,y)| \varphi (y)| dy \\ &&+\int_{(| x-y| \geq \alpha )\cap (| y|\leq M)\cap D} \frac{y_2}{x_2}G(x,y)| \varphi (y)| dy. \end{eqnarray*} By (\ref{e2.3}), we have $$\int_{(| x-y| \geq \alpha )\cap (| y| \leq M)\cap D} \frac{y_2}{x_2} G(x,y)| \varphi (y)| dy \leq C\int_{B(0,M)\cap D}y_2^2| \varphi (y)| dy.$$ Thus the result follows immediately from (\ref{e1.1})), (\ref{e1.2}) and Proposition \ref{prop3.1}. \hfill$\diamondsuit$ \begin{proposition} \label{prop3.3} Let $\varphi$ be a function in $K$ and $h$ be a positive superharmonic function in $D$.\\ a) For $x_0\in \overline{D}$, \begin{gather} \label{e3.2} \lim_{r\to 0}\sup_{x\in D}\frac1{h(x)} \int_{B(x_{_{0\,}},\,r)\cap D} G(x,y)\,h(y)\,| \varphi (y)| \,dy\,=0 \\ \lim_{M\to +\infty }\sup_{x\in D}\frac1{h(x)} \int_{D\cap (| y| \geq M)}G(x,y)\,h(y)\,| \varphi(y)| \,dy=0. \label{e3.3} \end{gather} b) For all $x\in D$ and $C_0$ as in Theorem \ref{thm2.3}, $$\int_DG(x,y)\,h(y)\,| \varphi (y)| \,dy\leq 2C_0\| \varphi \| \,h(x). \label{e3.4}$$ \end{proposition} \paragraph{Proof} Let $h$ be a positive superharmonic function in $D$. Then by [8;Theorem 2.1, p.164], there exists a sequence $(f_n)_n$ of positive measurable functions in $D$ such that \begin{equation*} h(y)=\sup_{n}\int_D G(y,z)f_n(z)\,dz\,. \end{equation*} Hence, we need only to verify (\ref{e3.2}), (\ref{e3.3}) and (\ref{e3.4}) for $h(y)=G(y,z)$, uniformly for $z\in D$.\\ a) Let $r>0$. By using Theorem \ref{thm2.3}, we obtain \begin{eqnarray*} \lefteqn{ \frac 1{G(x,z)}\int_{B(x_{_{0\,}},\,r)\cap D}G(x,y)\,G(y,z)\,| \varphi (y)| \,dy\, }\\ &\leq & 2C_0\sup_{\xi \in D} \int_{B(x_{_{0\,}},\,r)\cap D} \frac{y_2}{\xi _2}G(\xi ,y)\,| \varphi (y)| \,dy\,. \end{eqnarray*} Let $\alpha >0$ and $M>0$. Then by (\ref{e2.1}), we have \begin{align*} \int_{B(x_{_{0\,}},\,r)\cap D}\frac{y_2}{x_2}G(x,y)\,| \varphi (y)| \,dy \leq &\int_{B(x_{_{0\,}},\,r)\cap D\cap (| x-y| \leq \alpha )}% \frac{y_2}{x_2}G(x,y)\,| \varphi (y)| \,dy \\ &+C\int_{B(x_{_{0\,}},\,r)\cap D\cap (| x-y| \geq \alpha )\cap (| y| \leq M)}\,y_2^2| \varphi (y)| \,dy \\ &+\int_{B(x_{_{0\,}},\,r)\cap D\cap (| y| \geq M)}\frac{y_2}{x_2}% G(x,y)\,| \varphi (y)| \,dy. \end{align*} Then (\ref{e3.2}) follows from (\ref{e1.1}), (\ref{e1.2}) and Proposition \ref{prop3.1}. On the other hand, we have \begin{eqnarray*} \lefteqn{\frac 1{G(x,z)}\int_{(| y| \geq M)\cap D}G(x,y)\,G(y,z)\,| \varphi (y)| \,dy}\\ &\leq& 2C_0\sup_{\xi \in D} \int_{(| y| \geq M)\cap D} \frac{y_2}{\xi _2}G(\xi ,y)\,| \varphi (y)| \,dy \end{eqnarray*} which converges to zero as $M\to \infty$. This gives (\ref{e3.3}).\\ b) By using Theorem \ref{thm2.3}, we obtain $$\frac 1{G(x,z)}\int_DG(x,y)\,G(y,z)\,| \varphi (y)| \,dy \leq 2C_0\| \varphi \| .$$ \begin{corollary} \label{coro3.4} Let $\varphi$ be a function in $K$. Then we have \begin{gather} \sup_{x\in D} \int_D G(x,y)\,| \varphi (y)| \,dy<\infty , \label{e3.5}\\ \int_D\frac{y_2}{(| y| +1)^2}|\varphi (y)|<\infty ,\label{e3.6}\\ \int_{D\cap (| y| \leq M)}y_2|\varphi (y)|dy<\infty ,\quad \forall M>0. \label{e3.7} \end{gather} \end{corollary} \paragraph{Proof} Inequality (\ref{e3.5}) follows from (\ref{e3.4}) with $h=1$ in $D$ and Proposition \ref{prop3.2}. Let $x_0\in D$. Then by (\ref{e2.1}) and (\ref{e3.5}), we have \begin{equation*} \int_D\frac{y_2}{(| y| +1)^2}| \varphi (y)| dy\leq C% \frac{(| x_0| +1)^2}{| x_0| }(\sup_{x\in D} \int_DG(x,y)| \varphi (y)| dy) <\infty , \end{equation*} which gives (\ref{e3.6}). Inequality (\ref{e3.7}) follows immediately from (\ref{e3.6}). \begin{proposition} \label{prop3.5} Let $\varphi \in K$. Then the function \begin{equation*} V\varphi (x)=\int_DG(x,y)\varphi (y)dy \end{equation*} is defined in $D$ and is in $C_0(D)$. \end{proposition} \paragraph{Proof} Let $x_0\in D$ and $r>0$. Let $x,x'\in B(x_0,\frac r2)\cap D$. Then for $M>0$ \begin{eqnarray*} \lefteqn{| V\varphi (x)-V\varphi (x')| }\\ &\leq &\int_D| G(x,y)-G(x',y)| | \varphi (y)|dy \\ &\leq &2\sup_{\xi \in D} \int_{B(x_0,r)\cap D}G(\xi ,y)| \varphi (y)| dy+2\sup_{\xi \in D} \int_{(| y| \geq M)\cap D}G(\xi ,y)| \varphi (y)| dy \\ &&+\int_{D\cap (| y-x_0| \geq r)\cap (| y| \leq M)}| G(x,y)-G(x',y)| | \varphi (y)| dy. \end{eqnarray*} By (\ref{e2.1}), there exists $C>0$ such that for all $x\in B(x_0,\frac r2)\cap D$, for all $y\in B(0,M)\cap (D\backslash B(x_0,r))$, \begin{equation*} G(x,y)\leq Cy_2. \end{equation*} Moreover, $G(x,y)$ is continuous on $(x,y)\in (B(x_0,\frac r2)\cap D)\times (D\backslash B(x_0,r))$. Then by (\ref{e3.7}) and Lebesgue's theorem, we have that \begin{equation*} \int_{D\cap (| y-x_0| \geq r)\cap (| y| \leq M)}| G(x,y)-G(x',y)| | \varphi (y)| dy\to 0 \quad \text{as }| x-x'| \to 0. \end{equation*} Hence, we obtain by (\ref{e3.2}) and (\ref{e3.3}) with $h=1$ that $V\varphi$ is continuous in $D$. Now, we will show that \begin{equation*} \lim_{x\to \partial D} V\varphi (x)=\lim_{|x| \to +\infty } V\varphi (x)=0. \end{equation*} Let $x_0\in \partial D$ and $r>0$. Let $x\in B(x_0,\frac r2)\cap D$. Then for $M>0$, \begin{eqnarray*} | V\varphi (x)| &\leq &\int_DG(x,y)\,| \varphi (y)|\,dy \\ &\leq &\sup_{\xi \in D} \int_{B(x_0,r)\cap D}G(\xi ,y)| \varphi (y)| dy+\sup_{\xi \in D} \int_{(| y| \geq M)\cap D}G(\xi ,y)| \varphi (y)| dy \\ && +\int_{D\cap (| y-x_0| \geq r)\cap (| y| \leq M)}G(x,y)| \varphi (y)| dy. \end{eqnarray*} Since \begin{equation*} \int_{D\cap (| y-x_0| \geq r)\cap (| y| \leq M)}G(x,y)| \varphi (y)| dy\leq Cx_2\int_{D\cap (| y| \leq M)}y_2| \varphi (y)| dy, \end{equation*} then we obtain by (\ref{e3.7}), (\ref{e3.2}) and (\ref{e3.3}) with $h=1$ that \begin{equation*} \lim_{x\to \partial D} V\varphi (x)=0. \end{equation*} Let $M>0$ and $x$ in $D$ such that $| x| \geq M+1$, then we have \begin{eqnarray*} | V\varphi (x)| &\leq & \int_DG(x,y)| \varphi (y)| \,dy\\ &\leq & \int_{(| y| \leq M)\cap D}G(x,y)\,| \varphi (y)| \,dy +\int_{(| y| \geq M)\cap D}G(x,y)| \varphi(y)| \,dy\,. \end{eqnarray*} Since $G(x,y)\leq C\frac{x_2y_2}{(| x| -M)^2}$, for $|y| \leq M$, then from (\ref{e3.7}) and (\ref{e3.3}) with $h=1$, we deduce that \begin{equation*} \lim_{| x| \to +\infty } V\varphi (x)=0 \end{equation*} \begin{proposition} \label{prop3.6} Let $\lambda$, $\mu$ be in $\mathbb{R}$ and $\theta$ be the function defined on $D$ by \begin{equation*} \theta (y)=\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }. \end{equation*} Then $\theta \in K$ if and only if $\lambda <2<\mu$. \end{proposition} \paragraph{Proof} Let $\lambda <2<\mu$ and $\alpha >0$. Then we have \begin{eqnarray*} I &=&\int_{(| x-y| \leq \alpha )\cap D}\frac{y_2}{x_2}Log(1+\frac{% 4x_2y_2}{| x-y| ^2})\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }dy \\ &\leq &\int_{(| x-y| \leq \alpha )\cap D_1}\frac{y_2}{x_2}Log(1+% \frac{4x_2y_2}{| x-y| ^2})\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }dy \\ &&+\int_{(| x-y| \leq \alpha )\cap D_2}\frac{y_2}{x_2}Log(1+\frac{% 4x_2y_2}{| x-y| ^2})\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }dy \\ &=&I_1+I_2, \end{eqnarray*} where $$D_1=\{y\in D:x_2y_2\leq | x-y| ^2\}\quad\text{and}\quad D_2=\{y\in D:| x-y| ^2\leq x_2y_2\}.$$ So, using $Log(1+t)\leq t$, for $t>0$ and Lemma \ref{lm2.1}, we obtain \begin{eqnarray*} I_1 &\leq &\int_{(| x-y| \leq \alpha )\cap D_1}\frac{% y_2^{2-\lambda }}{| x-y| ^2}dy \\ &\leq &C\int_{(| x-y| \leq \alpha )\cap D_1}\frac{_1}{| x-y| ^\lambda }dy\leq C\int_0^\alpha t^{1-\lambda }dt, \end{eqnarray*} which converges to zero as $\alpha \to 0.$\\On the other hand, we have from Lemma \ref{lm2.1}, that there is $C>0$ such that if $y\in D_2$, \begin{equation*} \frac 1C(| x| +1)\leq | y| +1\leq C(| x|+1). \end{equation*} Hence \begin{equation*} I_2\leq C\frac 1{x_2^\lambda (| x| +1)^{\mu -\lambda }}\int_{(| x-y| \leq \alpha )\cap D_2}Log(1+\frac{(cx_2)^2}{% | x-y| ^2})dy, \end{equation*} where $c=1+\sqrt{5}$. Let $\gamma \in \, ]\max(0,\lambda ),2[$. Since $Log(1+t^2)\leq Ct^\gamma ,\forall t\geq 0$, then \begin{equation*} I_2\leq C\frac{x_2^{\gamma -\lambda }}{(| x| +1)^{\mu -\lambda }}% \int_0^{\inf (\alpha ,cx_2)}t^{1-\gamma }dt\leq C\max (\alpha ^{2-\lambda },\alpha ^{2-\gamma }), \end{equation*} which converges to zero as $\alpha \to 0$. Now, we will show that \begin{equation*} \lim_{M\to \infty } \Big(\sup_{x\in D} \int_{(| y| \geq M)} \frac{y_2}{x_2}Log(1+\frac{4x_2y_2}{| x-y| ^2}) \frac 1{(| y| +1)^{\mu -\lambda}y_2^\lambda }\,dy\Big)=0. \end{equation*} By the above argument, for $\varepsilon >0$, there exists $\alpha >0$ such that \begin{equation*} \sup_{x\in D} \int_{(| y| \geq M)\cap D\cap (| x-y| \leq \alpha )}\,\frac{y_2}{x_2}Log(1+\frac{4x_2y_2}{| x-y| ^2})\,\,\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }\,dy\leq \varepsilon . \end{equation*} Fixing this $\alpha$ and letting $M>1$, we have \begin{eqnarray*} \lefteqn{ \sup_{x\in D} \int_{(| y| \geq M)\cap D\cap (| x-y| \geq \alpha )}\frac{y_2}{x_2}Log(1+\frac{4x_2y_2}{| x-y| ^2})\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }dy }\\ &\leq &\sup_{x\in D} \int_{(| y| \geq M)\cap D\cap (| x-y| \geq \alpha )}\frac{y_2^{2-\lambda }}{| x-y| ^2| y| ^{\mu -\lambda }}dy \\ &\leq & \sup_{| x| \leq M/2} \int_{(| y| \geq M)\cap D}\frac{dy}{| x-y| ^2| y| ^{\mu -2}} \\ &&+\sup_{| x| \geq M/2} \Big[\int_{(M\vee \frac{| x| }2\leq | y| \leq 2| x| )\cap D\cap (| x-y| \geq \alpha )}\frac{dy}{| x-y| ^2| y| ^{\mu-2}}\\ &&+\int_{(| y| \geq 2| x| )\cap D}\frac{dy}{| x-y| ^2| y| ^{\mu -2}}\Big] +\sup_{| x| \geq 2 M}\int_{(M\leq | y| \leq \frac{| x| }2)\cap D}\frac{dy}{| x-y| ^2|y| ^{\mu -2}} \\ &\leq &C(\int_{(| y| \geq M)\cap D}\frac 1{| y| ^\mu}dy +\sup_{| x| \geq M/2} \frac{Log\frac{3|x| }\alpha }{| x| ^{\mu -2}})\\ &\leq& C(\frac 1{M^{\mu -2}}+\sup_{| x| \geq M/2} \frac{Log\frac{3| x| }\alpha }{| x| ^{\mu -2}}), \end{eqnarray*} which converges to zero as $M\to \infty$. Conversely, if $\theta \in K$ then we have by Proposition \ref{prop3.5} that \begin{equation*} \lim_{x_2\to 0} V\theta (x)=\underset{x_2\to +\infty }{\lim }V\theta (x)=0,\quad \text{for } x=( 0,x_2). \end{equation*} On the other hand, it follows from Lemma \ref{lm2.1} that \begin{eqnarray*} V\theta (x) &=&\frac 1{4\pi }\int_DLog(1+\frac{4x_2y_2}{| x-y| ^2}% )\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }dy \\ &\geq &C\int_{D\cap (| x-y| ^2\leq x_2y_2)}\frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda }dy \\ &\geq &C\frac 1{x_2^\lambda (| x| +1)^{\mu -\lambda }}\int_{| \widetilde{x}-y| \leq \frac{\sqrt{5}}2x_2}dy\geq C\frac{% x_2^{2-\lambda }}{(x_2+1)^{\mu -\lambda }}, \end{eqnarray*} where $\widetilde{x}=(0,\frac 32x_2)$. Hence, it is necessary that $\lambda <2<\mu .\quad \diamondsuit$ \\ Moreover, we have the following estimates. \begin{proposition} \label{prop3.7} There exists $C>0$ such that for all $x$ in $D$, we have \begin{gather} V\theta (x)\leq C\frac{x_2^{\mu -2}}{(| x| +1)^{2\mu -4}}, \quad\text{if } 2<\mu <\min (3,4-\lambda) \label{e3.8} \\ V\theta (x)\leq C\frac{x_2}{(| x| +1)^2},\quad \text{if } \lambda<1 \text{ and }\mu >3 \label{e3.9} \\ V\theta (x)\leq C\frac{x_2}{(| x| +1)^2}Log(\frac{(| x| +1)^2}{x_2}),\quad \text{if } \left\{ \begin{array}{c} \lambda <1 \\ \mu =3 \end{array} \right. \text{ or } \left\{\begin{array}{c} \lambda =1 \\ \mu \geq 3 \end{array} \right. \label{e3.10} \\ V\theta (x)\leq C\frac{x_2^{2-\lambda }}{(| x| +1)^{4-2\lambda }} ,\quad\text{if } 1<\lambda <2\text{ and }\mu \geq 4-\lambda. \label{e3.11} \end{gather} \end{proposition} For the proof, we need the following lemma. \begin{lemma} \label{lm3.8} Let $\lambda <2$, $B:=\{x\in \mathbb{R}^2,| x| <1\}$, and \begin{equation*} w(x)=\int_BG_B(x,y)\frac 1{(1-| y| )^\lambda }dy,\quad \text{for } x\in B, \end{equation*} where $G_B$ is the Green's function of $\Delta$ in $B$. Then for each $x\in B$, \begin{enumerate} \item[1)] $w(x) \leq C(1-| x| )$, if $\lambda<1$ \item[2)] $w(x) \leq C(1-| x| )Log(\frac 2{1-| x| })$, if $\lambda =1$ \item[3)] $w(x) \leq C(1-| x| )^{2-\lambda }$, if $1<\lambda <2$. \end{enumerate} \end{lemma} \paragraph{Proof} Since \begin{equation*} G_B(x,y)=\frac 1{4\pi }Log(1+\frac{(1-| x| ^2)(1-| y| ^2)}{| x-y| ^2}), \end{equation*} and the function $w$ is radial, then by elementary calculus we have \begin{equation*} w(x)=C\int_0^1Log(\frac 1{r\vee | x| })\frac r{(1-r)^\lambda }dr, \end{equation*} where $r\vee | x| =\max (r,| x| )$. Since $tLog(\frac 1t)\leq 1-t,\forall t\in [0,1]$, then we have \begin{equation*} w(x)\leq C\int_0^1\frac{1-(r\vee | x| )}{(1-r)^\lambda }dr. \end{equation*} Hence, if $| x| \leq \frac 12$ then \begin{equation*} w(x)\leq C\int_0^1(1-r)^{1-\lambda }dr<\infty , \end{equation*} and if $| x| \geq \frac 12$ then \begin{eqnarray*} w(x) &\leq &C[(1-| x| )(\int_0^{\frac 12}\frac 1{(1-r)^\lambda }dr+\int_{\frac 12}^{| x| }\frac 1{(1-r)^\lambda }dr)+\int_{| x| }^1(1-r)^{1-\lambda }dr] \\ &\leq &C[(1-| x| )+(1-| x| )\int_{\frac 12}^{| x| }\frac 1{(1-r)^\lambda }dr+(1-| x| )^{2-\lambda }]. \end{eqnarray*} Which implies the result. \hfill$\diamondsuit$ \paragraph{Proof of Proposition \ref{prop3.7}} Let $\gamma :D\to B$ be the M\"obius transformation defined by $\gamma (x)=x^{*}=e-\frac{2(x+e)}{% | x+e| ^2}$, where $e=(0,1)$. Then for $x,y\in D$, \begin{equation*} G(x,y)=G_B(x^{*},y^{*}). \end{equation*} On the other hand, it is easy to see that $$\label{e3.12} \frac 1{\sqrt{2}}(| x| +1)\leq | x+e| \leq (| x| +1),\forall x\in D.$$ Since for $x\in D$, we have $1-| x^{*}| ^2=\frac{4x_2}{| x+e| ^2}$, then by $(\ref{e3.12})$ we obtain that $$\label{e3.13} \frac{2x_2}{(| x| +1)^2}\leq \delta _B(x^{*})=1-| x^{*}| \leq \frac{8x_2}{(| x| +1)^2}.$$ It follows that \begin{eqnarray*} V\theta (x) &\leq &C\int_DG_B(x^{*},y^{*})\frac 1{(| y| +1)^{\mu +\lambda }(\delta _B(y^{*}))^\lambda }dy \\ \ &\leq &C\int_BG_B(x^{*},\xi )\frac 1{| \xi -e| ^{4-\mu -\lambda }}\frac 1{(\delta _B(\xi ))^\lambda }d\xi . \end{eqnarray*} Since $1-| \xi | \leq | \xi -e| \leq 2,\forall \xi \in B$, we have \begin{equation*} V\theta (x)\leq C\int_BG_B(x^{*},\xi )\frac 1{(\delta _B(\xi ))}d\xi ,\text{ if }4-\mu -\lambda \leq 0 \end{equation*} and \begin{equation*} V\theta (x)\leq C\int_BG_B(x^{*},\xi )\frac 1{(\delta _B(\xi ))^{4-\mu }}d\xi ,\text{ if }4-\mu -\lambda >0. \end{equation*} Thus the required inequalities follow from Lemma \ref{lm3.8} and (\ref{e3.13}). \section{Proofs of Theorems \ref{thm1.1} and \ref{thm1.2} } For this section, we need some preliminary results. Recall that the potential kernel $V$ is defined on $B^{+}(D)$ by \begin{equation*} V\phi (x)=\int_DG(x,y)\phi (y)dy,\text{\thinspace }x\in D. \end{equation*} Hence, for $\phi$ $\in B^{+}(D)$ such that \ $\phi \in L_{{\rm loc}}^1(D)$ and $V\phi \in L_{{\rm loc}}^1(D)$, we have in the distributional sense that $\Delta(V\phi )=-\phi$, in $D$. We point out if $V\phi \neq \infty$, we have $V\phi \in L_{{\rm loc}}^1(D)$, (see [1], p.51). Let us recall that $V$ satisfies the complete maximum principle, i.e for each $\phi \in \mathcal{B} ^{+}(D)$ and \ $v$ a nonnegative superharmonic function on $D$ such that $V\phi \leq v$ in $\{\phi >0\}$ we have $V\phi \leq v$ in $D$, (cf. [8], Theorem 3.6, p.175]). \begin{lemma} \label{lm4.1} Let $h\in \mathcal{B}^{+}(D)$ 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{lemma} \paragraph{Proof} We denote by $w^{+}=\max (w,0)$ and $w^{-}=\max(-w,0)$. Since V(h$| w| )<\infty$, then we have \begin{equation*} w^{+}+V(hw^{+})=v+w^{-}+V(hw^{-}). \end{equation*} Hence \begin{equation*} V(hw^{+})\leq v+V(hw^{-})\quad \text{in } \{w^{+}>0\}. \end{equation*} Since $v+V(hw^{-})$ is a nonnegative superharmonic function in $D$, then we have as consequence of the complete maximum principle that \begin{equation*} V(hw^{+})\leq v+V(hw^{-})\quad \text{in }D, \end{equation*} that is $V(hw)\leq v=w+V(hw)$. This implies that $0\leq w\leq v$. \begin{theorem} \label{thm4.2} Assume (H1)-(H3). Let $\alpha >0$ and $b>0$. Then the problem \begin{equation*} (P_{\alpha})\quad \quad \quad \begin{gathered} \Delta u+f(.,u)=0 \quad\text{in } D \\ u>0 \quad\text{in } D \\ u=\alpha \quad\text{on }\partial D \end{gathered} \end{equation*} has at least one positive solution $u_\alpha \in C(\overline{D})$ satisfying \begin{equation*} \underset{x_2\to \infty }{\lim }\frac{u_\alpha (x)}{x_2}=b. \end{equation*} \end{theorem} \paragraph{Proof} Let $\alpha >0$. It follows from (H2) and Proposition \ref{prop3.5} that $V(f(.,\alpha ))\in C_0(D)$. So, in the sequel, we denote \begin{equation*} \beta =\alpha +\| V(f(.,\alpha ))\| _\infty . \end{equation*} To apply a fixed-point argument, we consider the convex set \begin{equation*} F=\{w\in C(\overline{D}\cup \{\infty \}):\alpha \leq w(x)\leq \beta ,\,\,\forall x\in D\}. \end{equation*} and on this set we define the integral operator \begin{equation*} Tw(x)=\alpha +\frac \alpha {\alpha +bx_2}\int_DG(x,y)f(y, \frac{(\alpha +by_2)}\alpha w(y))dy,\quad x\in D. \end{equation*} By (H1), we have $$\label{e4.1} f(y,\frac{(\alpha +by_2)}\alpha w(y))\leq f(y,\alpha ),\forall w\in F.$$ Then for $w\in F$ \begin{equation*} \alpha \leq Tw(x)\leq \beta \quad \forall x\in D. \end{equation*} As in the proof of Proposition \ref{prop3.5} we show that the family $TF$ is equicontinuous in $\overline{D}\cup \{\infty \}$. In particular, for all $v\in F,$ $Tw\in C(\overline{D}\cup \{\infty \})$ and so $TF\subset F$. Moreover, the family $\{Tw(x),w\in F\}$ 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 $Y$. We consider a sequence $(w_n)$ in $F$ which converges uniformly to a function $w$ in $F$. Then we have \begin{eqnarray*} \lefteqn{| Tw_n(x)-Tw(x)| }\\ &\leq &\frac \alpha {\alpha +bx_2}\int_DG(x,y)| f(y,\frac{(\alpha +by_2)% }\alpha w_n(y))-f(y,\frac{(\alpha +by_2)}\alpha w(y))| dy. \end{eqnarray*} Since $f$ is continuous with respect to the second variable, we deduce by (\ref{e4.1}), (H2), (\ref{e3.5}) and the Lebesgue's theorem that for each $x\in \overline{D}\cup \{\infty \}$ \begin{equation*} Tw_n(x)\to Tw(x) \quad\text{as }n\to \infty . \end{equation*} Since $TY$ is a relatively compact family in $C(\overline{D}\cup \{\infty \})$, we have the uniform convergence, namely \begin{equation*} \| Tw_n-Tw\| _\infty \to 0\quad\text{as } n\to \infty. \end{equation*} Thus we have proved that $T$ is a compact mapping from $F$ to itself. Hence, by the Schauder's fixed point-theorem, there exists $w_\alpha \in F$ such that \begin{equation*} w_\alpha (x)=\alpha +\frac \alpha {\alpha +bx_2}\int_DG(x,y)f(y,\frac{% (\alpha +by_2)}\alpha w_\alpha (y))dy,\forall x\in D. \end{equation*} Put $u_\alpha (x)=\frac{(\alpha +bx_2)}\alpha w_\alpha (x)$, for $x\in D$. Then we have $$\label{e4.2} u_\alpha (x)=\alpha +bx_2+\int_DG(x,y)f(y,u_\alpha (y))dy,\forall x\in D.$$ By (H1), we have for each $y\in D$, $$\label{e4.3} f(y,u_\alpha (y))\leq f(y,\alpha ).$$ Then we deduce by (H2) and Proposition \ref{prop3.1} that the map $y\to f(y,u_\alpha (y))\in L_{{\rm loc}}^1(D)$, and by Proposition \ref{prop3.5}, that $V(f(.,u_\alpha ))\in C_0(D)\subset L_{{\rm loc}}^1(D)$. Apply $\Delta$ on both sides of equality (\ref{e4.2}), we obtain that \begin{equation*} \Delta u_\alpha +f(.,u_\alpha )=0\quad\text{in $D$ (in the sense of distributions).} \end{equation*} Furthermore, it follows from (\ref{e4.2}) that $$\label{e4.4} \alpha +bx_2\leq u_\alpha (x)\leq \beta +bx_2,\,\forall x\in D.$$ Hence \begin{equation*} \underset{x_2\to \infty }{\lim }\frac{u_\alpha (x)}{x_2}=b. \end{equation*} Now, using (\ref{e4.3}), (H2), Proposition \ref{prop3.5} and (\ref{e4.2}), we obtain $\lim_{x\to \partial D} u_{\alpha}(x)=\alpha$. Then, $u_\alpha$ is a positive continuous solution of the problem $(P_\alpha )$. \begin{proposition} \label{prop4.3} Let $f:D\times (0,\infty )\to [0, \infty )$ be a measurable function satisfying (H1) and $\alpha _1,\alpha _2,b_1,b_2$ be real numbers such that $0\leq \alpha _1\leq \alpha _2$ and $0\leq b_1\leq b_2$. If $u_1$ and $u_2$ are two positive functions continuous on D satisfying for each $x$ in $D$ \begin{gather*} u_1(x)=\alpha _1+b_1x_2+V(f(.,u_1))(x),\\ u_2(x)=\alpha _2+b_2x_2+V(f(.,u_2))(x). \end{gather*} Then \begin{equation*} 0\leq u_2(x)-u_1(x)\leq \alpha _2-\alpha _1+(b_2-b_1)x_2,\quad \forall x\in D. \end{equation*} \end{proposition} \paragraph{Proof} Let $h$ be the function defined on $D$ as \begin{equation*} h(x)=\begin{cases} \frac{f(x,u_1(x))-f(x,u_2(x))}{u_2(x)-u_1(x)} &\text{if } u_1(x)\neq u_2(x) \\ 0 & \text{if }u_1(x)=u_2(x). \end{cases} \end{equation*} Then $h\in B^{+}(D)$ and \begin{equation*} u_2(x)-u_1(x)+V(h(u_2-u_1))(x)=\alpha _2-\alpha _1+(b_2-b_1)x_2. \end{equation*} Now, since $$V(h| u_2-u_1| ) \leq V(f(.,u_1))+V(f(.,u_2)) \leq u_1+u_2<\infty ,$$ we deduce the result from Lemma \ref{lm4.1}. \paragraph{Proof of Theorem \ref{thm1.1}} Let $(\alpha _n)$ be a sequence of positive real numbers, non-increasing to zero. For each $n\in \mathbb{N}$, we denote by $u_n$ the continuous solution of the problem given by the integral equation (\ref{e4.2}) with $\alpha =\alpha _n$. Then, by Proposition \ref{prop4.3}, the sequence $(u_n)$ decreases to a function $u$. Since $$\label{e4.5} u_n(x)-\alpha _n=b x_2+\int_DG(x,y)f(y,u_n(y))dy\geq bx_2>0.$$ Then the sequence $(u_n-\alpha _n)$ increases to $u$ and so $u>0$ in $D$. Hence, \begin{equation*} u=\inf_{n} u_n= \sup_{n}(u_n-\alpha _n) \end{equation*} is a positive continuous function in $D$. Using (H1) and applying the monotone convergence theorem, we get $$\label{e4.6} u(x)=bx_2+\int_DG(x,y)f(y,u(y))dy \,,\quad \forall x\in D.$$ Then, it follows from (\ref{e4.6}) that $V(f(.,u))\in L_{{\rm loc}}^1(D)$. On the other hand, since $u$ is positive in $D$, then by (H2) and Proposition \ref{prop3.1}, the function $y\to f(y,u(y))\in L_{{\rm loc}}^1(D)$. Applying $\Delta$ on both sides of equality (\ref{e4.6}), we conclude that $u$ satisfies \begin{equation*} \Delta u+f(.,u)=0\quad \text{in }D. \end{equation*} Since for $x$ in $D$ and $n$ in $\mathbb{N}$, \begin{equation*} 0\leq u_n(x)-\alpha _n\leq u(x)\leq u_n(x)\quad \text{and}\quad \lim_{x_2\to \infty } \frac{u_n(x)}{x_2}=b, \end{equation*} we deduce that \begin{equation*} \underset{x\to \partial D}{\lim }u(x)=0\,\,\text{\thinspace \thinspace \thinspace and\thinspace \thinspace }\,\,\,\underset{% x_2\to \infty }{\lim }\frac{u(x)}{x_2}=b. \end{equation*} Thus, $u\in C(\overline{D})$ and u is a positive solution of the problem(\ref{P}). Now, let \begin{equation*} \delta =\inf_{\alpha >0} (\alpha +\| Vf(.,\alpha \|_\infty ). \end{equation*} Then by (H3) and (H1)$, \delta >0$. By (\ref{e4.4}) we have that \begin{equation*} bx_2\leq u(x)\leq bx_2+\delta . \end{equation*} By (H1) and (\ref{e4.6}), \begin{equation*} bx_2\leq u(x)\leq bx_2+\int_DG(x,y)f(y,by_2)dy. \end{equation*} Which implies that \begin{equation*} bx_2\leq u(x)\leq bx_2+\min (\delta ,\int_DG(x,y)f(y,by_2)dy). \end{equation*} \begin{corollary} \label{coro4.4} Let $00$, $\lambda <1-\sigma$ and $\mu >\max (2,3-\sigma )$. Suppose that the function $f$ satisfies (H1), (H3) and such that \begin{equation*} f(y,t)\leq \frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda t^\sigma}. \end{equation*} Then for each $b>0$, there exists $C>0$ such that the problem \begin{gather*} \Delta u+f(.,u)=0 \quad \text{in }D \\ u>0\quad \text{in }D \\ u=0 \quad\text{on }\partial D \\ \lim_{x_2\to \infty} \frac{u(x)}{x_2}=b, \end{gather*} has a continuous solution $u$ in $D$ satisfying \begin{equation*} bx_2\leq u(x)\leq Cx_2\,, \quad \forall x\in D. \end{equation*} \paragraph{Proof of Theorem \ref{thm1.2}} Let $\alpha >0$ and $(b_n)$ be a sequence of positive real numbers, non-increasing to zero. If $u_{\alpha ,n}$ denotes the positive continuous solution of the problem $(P_\alpha )$ given by (\ref{e4.2}) for $b=b_n$, then for each $x$ in $D$ $$\label{e4.7} u_{\alpha ,n}(x)=\alpha +b_nx_2+\int_DG(x,y)f(y,u_{\alpha ,n}(y))dy,$$ and if $u_n$ denotes the positive continuous solution of the problem (\ref{P}) given by (\ref{e4.6}) for $b=b_n$, then $$\label{e4.8} u_n(x)=b_nx_2+\int_DG(x,y)f(y,u_n(y))dy,\forall x\in D.$$ By Proposition \ref{prop4.3}, the sequence $(u_n)$ decreases to a function $u$ and by (H1) the sequence $(u_n-b_nx_2)$ increases to $u$. Then $u$ is a positive continuous function in $D$. Using the monotone convergence theorem, we deduce that $u$ satisfies $$\label{e4.9} u(x)=\int_DG(x,y)f(y,u(y))dy, \quad \forall x\in D.$$ Moreover, from Proposition \ref{prop4.3} and (\ref{e4.3}), we have $$\label{e4.10} u(x)\leq u_{\alpha ,n}(x)\leq \alpha +Vf(.,\alpha )(x),\quad \forall x\in D.$$ Then it follows from Proposition \ref{prop3.5} that \begin{equation*} \lim_{x\to \partial D}u(x)=\lim_{| x|\to \infty }u(x)=0\,. \end{equation*} Now, by (\ref{e4.10}) and (H1), we have \begin{equation*} \int_DG(x,y)f(y,\delta )dy\leq u(x)\leq \delta \quad \forall x\in D, \end{equation*} where $\delta =\underset{\alpha >0}{\inf }(\alpha +\| Vf(.,\alpha\| _\infty )$. Then, we get from (\ref{e2.1}) that \begin{equation*} \frac{x_2}{C(| x| +1)^2}\int_D\frac{y_2}{(| y| +1)^2} f(y,\delta )dy\leq u(x)\,,\quad \forall x\in D. \end{equation*} Hence we deduce from (H2) and (\ref{e3.6}) that $$\label{e4.11} \frac{x_2}{C(| x| +1)^2}\leq u(x).$$ Since f is non-increasing with respect to the second variable, then we have \begin{equation*} u(x)\leq \min (\delta ,\int_DG(x,y)f(y,\frac{y_2}{C(| y| +1)^2})dy). \end{equation*} Finally, we intend to show the uniqueness of the solution. Let u and v be two solutions of (\ref{P}) in $C_0(D)$. Suppose that there exists $x_0\in D$ such that $u(x_0)0\}$. Then $\Omega$ is an open nonempty set in $D$ and by (H3) we deduce that $\Delta w\geq 0$, in $\Omega$ with $w=0$ on $\partial \Omega$. Hence, by the maximum principle ([2], p.465-466), we get $w\leq 0$ in $\Omega$. Which is in contradiction with the definition of $\Omega$. \hfill$\diamondsuit$ We close this section by giving another comparison result for the solutions $u$ of the problem (\ref{P}), in the case of the special nonlinearity $f(x,t)=p(x)q(t)$. The following hypotheses on $p$ and $q$ are adopted. \begin{enumerate} \item[i)] The function $p$ is nontrivial nonnegative and is in $K\cap C_{{\rm loc}}^\gamma (D)$, $0<\gamma <1$. \item[ii)] The function $q:(0, \infty )\to (0,\infty )$ is a continuously differentiable and non-increasing. \end{enumerate} In the sequel, we define the function $Q$ in $[0,\infty )$ by \begin{equation*} Q(t)=\int_0^t\frac 1{q(s)}ds. \end{equation*} From the hypothesis adopted on $q$, we note that the function $Q$ is a bijection from $[0,\infty )$ to itself. Then we have the following theorem. \begin{theorem} \label{thm4.5} Let $u$ be the positive solution of $$\label{e*} \Delta u(x)+p(x)q(u(x))=0\quad x\in D, \;u\in C_0(D).$$ Then $q(\delta )Vp\leq u\leq Q^{-1}(Vp)$ in $D$. \end{theorem} \paragraph{Proof} Since $u\leq \delta$ in $D$ and $q$ is non-increasing, we deduce from (\ref{e4.9}) that \begin{equation*} q(\delta )Vp(x)\leq u(x)=\int_DG(x,y)p(y)q(u(y))dy,\quad \forall x\in D. \end{equation*} To show the upper estimate, we consider the function $v$ defined in $D$ by \begin{equation*} v=Q(u)-Vp. \end{equation*} Then $v\in C^2(D)$ and \begin{equation*} \Delta v=\frac 1{q(u)}\Delta u+p-\frac{q'(u)}{q^2(u)}| \nabla u| ^2\geq 0. \end{equation*} In addition, since $Vp\in C_0(D)$, we deduce that $v\in C_0(D)$. Thus, the maximum principle implies that $v\leq 0$. \begin{corollary} \label{coro4.6} Let $\lambda <2<\mu$. Suppose further that the function $p$ satisfies \begin{equation*} p(y)\leq \theta (y)\quad \forall y\in D, \end{equation*} where $\theta (y)=1/((| y| +1)^{\mu -\lambda }y_2^\lambda)$. Let $u$ be the positive solution of (\ref{e*}). Then there exists $C>0$ such that for each $x\in D$, \begin{equation*} \frac 1C\frac{x_2}{\ (| x| +1)^2}\leq u(x)\leq Q^{-1}(r_{\lambda ,\mu }(x)),\forall x\in D, \end{equation*} where $r_{\lambda ,\mu }$ is the right hand function in the inequalities of Proposition \ref{prop3.7}. \end{corollary} \paragraph{Proof} The lower estimate is obtained from (\ref{e4.11}). Using Theorem \ref{thm4.5}, the upper estimate follows from the monotonicity of $Q^{-1}$ and Proposition \ref{prop3.7}. \paragraph{Example} Let $\lambda <2$, $\mu \geq 4-\lambda$ and $\sigma \geq 0$. Suppose further that the function $p$ satisfies \begin{equation*} p(y)\leq \frac 1{(| y| +1)^{\mu -\lambda }y_2^\lambda },\text{ \ for }y\in D. \end{equation*} Then the equation \begin{equation*} \Delta u+pu^{-\sigma }=0 \quad \text{in D},\; u\in C_0(D) \end{equation*} has a unique positive solution $u\in C^{2+\gamma }(D)$ which for each $x\in D$ it satisfies: \begin{enumerate} \item[i)] $\frac 1C\frac{x_2}{\ (| x| +1)^2} \leq u(x) \leq C \frac{x_2^{\frac{2-\lambda }{1+\sigma }}}{(| x| +1)^{\frac{ 4-2\lambda }{1+\sigma }}}$, if $1<\lambda <2$. \item[ii)] $\frac 1C\frac{x_2}{\ (| x| +1)^2} \leq u(x) \leq C\frac{x_2^{\frac 1{1+\sigma }}}{(| x| +1)^{\frac 2{1+\sigma }}} \big[ Log(\frac{2(| x| +1)^2}{\ x_2})\big] ^{\frac 1{1+\sigma }}$, if $\lambda =1$. \item[iii)] $\frac 1C\frac{x_2}{\ (| x| +1)^2} \leq u(x) \leq C\frac{x_2^{\frac 1{1+\sigma }}}{(| x| +1)^{\frac 2{1+\sigma }}}$, if $\lambda <1$. \end{enumerate} \section{Proof of Theorem \ref{thm1.3}} Let \begin{equation*} C_0(\overline{D}):=\{w\in C(\overline{D}):\underset{| x| \to \infty }{\lim }w(x)=0\}. \end{equation*} Then $C_0(\overline{D})$ is a Banach space with the uniform norm $\| w\| _\infty \,=\sup_{x\in D} | w(x)| .$\\Let $% \varphi _0$ be a positive function belonging to K and let \begin{equation*} F_0:=\{\varphi \in K:\,\,| \varphi (x)| \leq \varphi _0(x),\,\,\forall x\in D\}. \end{equation*} \begin{lemma} \label{lm5.1} The family of the functions \begin{equation*} \Big\{\int_D\frac{y_2}{x_2}G(.,y)\varphi (y)dy,\varphi \in F_0\Big\} \end{equation*} is uniformly bounded and equicontinuous on $\overline{D}\cup \{\infty \}$. Consequently it is relatively compact in $C_0(\overline{D})$. \end{lemma} \paragraph{Proof} Let $T$ be the operator defined on $F_0$ as \begin{equation*} T\varphi (x)=\,\int_D\frac{y_2}{x_2}G(x,y)\varphi (y)\,dy\,. \end{equation*} Then for all $\varphi \in F_0$, \begin{equation*} | T\varphi (x)| \leq \int_D\frac{y_2}{x_2}G(x,y)\varphi _0(y)\,dy\,. \end{equation*} Since $\varphi _0\in K$, from Proposition \ref{prop3.2}, $\| T\varphi \| _\infty \leq \| \varphi _0\|$ for all $\varphi \in F_0$. Thus the family $T(F_0)=\{T\varphi ,\varphi \in F_0\}$ is uniformly bounded. Now, we prove the equicontinuity of $T(F)$ on $\overline{D}\cup\{\infty \}$. Let $x_0\in \overline{D}$ and $r>0$. Let $x$, $x'\in B(x_0,\frac r2)\cap D$ and $\varphi \in F_0$, then for $M>0$, \begin{eqnarray*} | T\varphi (x)-T\varphi (x')| &\leq &2\,\sup_{x\in D} \int_{B(x_0,r)\cap D}\frac{y_2}{x_2} G(x,y)\varphi _0(y)\,dy \\ &&+2\,\sup_{x\in D} \int_{(| y| \geq M)\cap D}\frac{% y_2}{x_2}G(x,y)\varphi _0(y)dy \\ &&+\int_{(| x_0-y| \geq r)\cap (| y| \leq M)\cap D}| \frac{G(x,y)}{x_2}-\frac{G(x',y)}{x_2'}| y_2\varphi _0(y)dy. \end{eqnarray*} By (\ref{e2.1}), there exists $C>0$ such that for all $x\in B(x_0,\frac r2)\cap D$, for all $y\in B(0,M)\cap (D\backslash B(x_0,r))$, \begin{equation*} \frac{y_2}{x_2}G(x,y)\varphi _0(y)\leq Cy_2^2\varphi _0(y). \end{equation*} Moreover, $\frac{G(x,y)}{x_2}$ is continuous on $(x,y)\in (B(x_0,\frac r2)\cap D)\times (D\backslash B(x_0,r))$. Then by Proposition \ref{prop3.1} and Lebesgue's theorem, we have \begin{equation*} \int_{(| x_0-y| \geq r)\cap (| y| \leq M)\cap D}| \frac{G(x,y)}{x_2}-\frac{G(x',y)}{x_2'} |y_2\varphi _0(y)\,dy\to 0, \end{equation*} as $| x-x'| \to 0$. Then it follows from (\ref{e3.2}) that \begin{equation*} | T\varphi (x)-T\varphi (x')| \to 0\quad\text{as }| x-x'| \to 0 \end{equation*} uniformly for all $\varphi \in F_0$. On the other hand, to establish compactness we need to show that \begin{equation*} \lim_{| x| \to +\infty } T\varphi (x)=0, \quad\text{uniformly for } \varphi \in F_0. \end{equation*} Let $M>0$ and $x$ in $D$ such that $| x| \geq M+1$, then \begin{eqnarray*} | T\varphi (x)| &\leq &\int_D\frac{y_2}{x_2}G(x,y)\varphi_0(y)\,dy \\ \ &\leq &\,\int_{(| y| \leq M)\cap D}\frac{y_2}{x_2} G(x,y)\varphi _0(y)\,dy +\int_{(| y| \geq M)\cap D}\frac{y_2}{x_2}G(x,y)\varphi_0(y)\,dy\,. \end{eqnarray*} Since $\lim_{| x| \to +\infty } y_2\frac{G(x,y)}{x_2}=0$ uniformly for $| y| \leq M$, and $\frac{y_2}{x_2}G(x,y)\leq \frac 1\pi y_2^2$, for $| x-y| \geq 1$, then from Proposition \ref{prop3.1}, Lebesgue's theorem and (\ref{e3.3}) with $h=1$, we deduce that \begin{equation*} \lim_{| x| \to +\infty }T\varphi (x)=0 \end{equation*} uniformly for all $\varphi \in F_0$. Finally, by Ascoli's theorem, the family $T(F_0)$ is relatively compact in $C_0(\overline{D})$. \paragraph{Proof of Theorem \ref{thm1.3}} Let $\beta \in (0,1)$. Then, by (A1),(A2) and Lemma \ref{lm5.1}, the function \begin{equation*} T_\beta (x)=\int_D\frac{y_2}{x_2}G(x,y)\,\,\psi (y,\beta y_2)\,dy \end{equation*} is continuous on $\overline{D}$ satisfying \begin{equation*} \lim_{| x| \to +\infty} T_\beta (x)=0\quad\text{and}\quad \lim_{\beta\to 0} T_\beta (x)=0 \; \forall x\in \overline{D}. \end{equation*} Moreover, the function $\beta \to T_\beta (x)$ is nondecreasing on $(0,1)$. Then, by Dini Lemma, we have \begin{equation*} \lim_{\beta \to 0}\sup_{x\in D} \int_D\frac{y_2}{x_2}G(x,y)\psi (y,\beta y_2) \,dy=0. \end{equation*} Thus, there exists $\beta \in (0,1)$ such that for each $x\in D$, \begin{equation*} \int_D\frac{y_2}{x_2}G(x,y)\psi (y,\beta y_2)\,dy\leq \frac 13. \end{equation*} Let $b_0=\frac 23\beta$ and $b\in (0,\,b_0]$. In order to apply a fixed-point argument, set \begin{equation*} S=\big\{w\in C(\overline{D}\cup \{\infty \}):\frac b2\leq w(x)\leq \frac{3b}2,\,x\in D\}. \end{equation*} Then, $S$ is a nonempty closed bounded and convex set in $C(\overline{D}\cup \{\infty \})$. Define the operator $\Gamma$ on $S$ as \begin{equation*} \Gamma w(x)=b+\frac 1{x_2}\,\int_DG(x,y) g(\,y,y_2 w(y))\,dy\,,\quad x\in D. \end{equation*} First, we shall prove that the operator $\Gamma$ maps $S$ into itself. Let $v\in S$, then for any $x\in D$, we have by (A1) that \begin{equation*} | \Gamma w(x)-b| \leq \frac{3b}2\int_D\frac{y_2}{x_2}G(x,y)\,\psi (y,\beta y_2)\,dy \leq \frac b2. \end{equation*} It follows that $\frac b2\leq \Gamma w\leq \frac{3b}2$ and by Lemma \ref{lm5.1}, $\Gamma (S$) is included in $C(\overline{D}\cup \{\infty \})$. So $% \Gamma S\subset S$. Next, we shall prove the continuity of $\Gamma$ in the supremum norm. Let $(w_k)_k$ be a sequence in $S$ which converges uniformly to $w\in S$. It follows from (A1) and Lebesgue's theorem that \begin{equation*} \forall \,x\in D, \quad \Gamma w_k(x)\to\Gamma w(x)\quad\text{as }k\to +\infty . \end{equation*} Since $\Gamma (S)$ is a relatively compact family in $C(\overline{D}\cup \{\infty \})$, then the pointwise convergence implies the uniform convergence. Thus we have proved that $\Gamma$ is a compact mapping from $S$ to itself. Now the Schauder fixed-point theorem implies the existence of $% w\in S$ such that $\Gamma w=w$. For $x\in D$, put $u(x)=x_2 w(x)$. Therefore we have \begin{equation*} u(x)=b x_2+\int_DG_D(x,y) g(y,u(y))\,dy. \end{equation*} Since $g(y,u(y))\leq y_2\psi (y,y_2)$, then we have by $(A_2)$ and (\ref{e3.7}) that $y\to g(y,u(y)$ is in L$_{{\rm loc}}^1(D)$. Applying $\Delta$ in both sides of the above equation, we get \begin{equation*} \Delta u+g(.,u)=0,\quad\text{in} D. \end{equation*} It is clear that $u$ is a solution of (\ref{Q}), continuous on $D$, \begin{equation*} \frac b2x_2\leq u(x)\leq \frac{3b}2x_2\quad\text{and}\quad \lim_{x_2\to +\infty }\frac{u(x)}{x_2}=b. \end{equation*} \paragraph{Example} Let $\sigma >0$ and $\lambda <2<\mu$. Let $p$ be a measurable function in $D$ such that \begin{equation*} | p(x)| \leq \frac C{(| x| +1)^{\mu -\lambda }x_2^{\lambda +\sigma }},\quad \forall x\in D. \end{equation*} Then there exists $b_0>0$ such that for each $b\in (0,b_0]$, the problem \begin{gather*} \Delta u(x)+p(x)u^{\sigma +1}(x)=0,\quad x\in D \\ u(x)>0,\quad x\in D \\ u\big|_{\partial D}=0 \end{gather*} has a solution $u$ continuous on $D$ and satisfying \begin{equation*} \frac b2x_2\leq u(x)\leq \frac{3b}2x_2\quad \text{and}\quad \lim_{x_2\to \infty }\frac{u(x)}{x_2}=b. \end{equation*} \begin{thebibliography}{99} \bibitem{ch} K. 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No. 53, 1-20 (2001). \end{thebibliography} \noindent\textsc{Imed Bachar } (e-mail: Imed.Bachar@ipeigb.rnu.tn)\\ \textsc{Habib M\^aagli } (e-mail: habib.maagli@fst.rnu.tn) \\ \textsc{Lamia M\^aatoug } (e-mail: Lamia.Maatoug@ipeit.rnu.tn ) \\[2pt] D\'epartement de Math\'ematiques, Facult\'e des Sciences de Tunis,\\ Campus universitaire 1060 Tunis, Tunisia. \end{document}