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\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}

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\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
\begin{equation}
\Delta u + q(x) u^{-\gamma} = 0,\quad x \in \Omega \subset \mathbb{R}^n,
\quad \gamma > 0, \label{e1.1}
\end{equation}
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
\begin{equation}
\Delta u + f(x, u) = 0,\; x \in \Omega, \label{e1.2}
\end{equation}
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)
\begin{equation}
\Delta (V \Psi) = - \Psi \mbox{ in } D. \label{e1.4}
\end{equation}
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
\begin{equation}
 \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{equation}
 \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 $
\begin{equation}
 \int_D G(x,y) h(y) |q(y)|   dy \leq C_D   \| q\|_D    h(x). \label{e2.4}
 \end{equation}
\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
\begin{equation}
 \sup_{x \in D} \int_D G(x,y) |q(y)|  dy <  \infty .
\label{e2.5} \end{equation}
\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 \begin{equation}
 \int_1^{+ \infty } r Log ( r ) |q(r)| dr < \infty . \label{e2.6}
\end{equation}
\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
\begin{equation}
u_\lambda(x) = \lambda + \mu Log |x| + \int_D G(x, y) \varphi(y,
u_\lambda(y))dy. \label{e3.2}
\end{equation}
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 $x \in D$,
\begin{equation}
\mu Log |x| \leq u(x) \leq \mu
Log|x|+V\left(\varphi(\cdot,u)\right)(x). \label{e3.4}
\end{equation}
\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
\begin{equation}
u(x) = \mu Log |x| + \int_D G(x,y) \varphi(y, u(y)) dy \;,
\;\forall x \in D. \label{e3.5}
\end{equation}
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
\begin{equation}
u(x) = \int_DG(x, y) \varphi(y, u(y))dy \;,\; \forall\; x \in D.
\label{e3.6}
\end{equation}
 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
\begin{equation}
\mu Log |x| \leq u(x) \leq \mu Log |x| + F^{-1} (Vq(x)) \;,\;
\forall\; x \in  \overline{D}.\label{e4.2}
\end{equation}
\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.

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\noindent\textsc{Noureddine  Zeddini}\\
 D\'epartement de Math\'ematiques, Facult\'e des Sciences de Tunis\\
Campus Universitaire, 1060 Tunis, Tunisia.\\
e-mail: Noureddine.Zeddini@ipein.rnu.tn

\end{document}