\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 58, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/58\hfil Existence of bounded positive solutions]
{Existence of bounded positive solutions of a nonlinear differential
system}

\author[S. Gontara \hfil EJDE-2012/58\hfilneg]
{Sabrine Gontara} 

\address{Sabrine Gontara \newline
D\'{e}patement de math\'{e}matiques, 
Facult\'{e} des sciences de Tunis, Campus Universitaire,
 2092 Tunis, Tunisia}
\email{sabrine-28@hotmail.fr}

\thanks{Submitted January 20, 2012. Published April 10, 2012.}
\subjclass[2000]{35J56, 31B10, 34B16, 34B27}
\keywords{Nonlinear equation; Green's function; asymptotic behavior;
\hfill\break\indent singular operator; positive solution}

\begin{abstract}
 In this article, we study the existence and nonexistence of solutions for the
 system
 \begin{gather*}
 \frac{1}{A}(Au')'=pu^{\alpha }v^{s}\quad \text{on }(0,\infty ), \\
 \frac{1}{B}(Bu')'=qu^{r}v^{\beta }\quad \text{on }(0,\infty ), \\
 Au'(0)=0,\quad u(\infty )=a>0, \\
 Bv'(0)=0,\quad v(\infty )=b>0,
 \end{gather*}
 where $\alpha ,\beta \geq 1$, $s,r\geq 0$, $p,q$ are two nonnegative
 functions on $(0,\infty )$ and $A$, $B$ satisfy appropriate conditions.
 Using potential theory tools, we show the  existence of a positive 
 continuous solution. This allows us to prove the existence of entire 
 positive radial solutions for some elliptic systems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

Existence and nonexistence of solutions of the elliptic system
\begin{equation}
\begin{gathered}
\Delta u=p(|x|)f(v),\quad x\in\mathbb{R}^n \\
\Delta v=q(|x|)g(u),\quad x\in\mathbb{R}^n
\end{gathered}  \label{1.1}
\end{equation}
have been intensively studied in the previous years; see for example
\cite{CR,D,Abd,GR,lair,PS,WW} and the references therein.

Lair and Wood \cite{lair} considered the existence of entire positive
radial solutions to the system \eqref{1.1} when $f(v)=v^{s}$ and
$g(u)=u^{r}$. More precisely, for the sublinear case where
$r,s\in (0,1)$, they proved that if $p$ and $q$ satisfy the decay
conditions
\begin{equation}
\int_0^{\infty }tp(t)dt<\infty ,\quad \int_0^{\infty }tq(t)dt<\infty
\label{1.2}
\end{equation}
then  \eqref{1.1} has bounded solutions, and if
\begin{equation*}
\int_0^{\infty }tp(t)dt=\infty ,\quad \int_0^{\infty }tq(t)dt=\infty
\end{equation*}
then  \eqref{1.1} has large solutions. For the superlinear case, where
$r,s\in (1,\infty )$, the authors proved the existence of an
entire large positive solution of \eqref{1.1}, provided that $p$ and $q$
satisfy \eqref{1.2}.

Later, their results were extended by C\^{\i}rstea and\ Radulescu \cite{CR}
which considered  \eqref{1.1} under the following conditions on $f$
and $g$:
\begin{equation*}
\lim_{t\to \infty } \frac{f(cg(t))}{t}=0,\quad \text{for all }c>0.
\end{equation*}
To study  \eqref{1.1}, Ghanmi et al in \cite{Abd} considered  the  system
\begin{gather*}
\frac{1}{A}(Au')'=p(t)g(v)\quad t\in (0,\infty ), \\
\frac{1}{B}(Bv')'=q(t)f(u)\quad t\in (0,\infty ), \\
u(0)=\alpha >0,\quad v(0)=\beta >0, \\
Au'(0)=0,\quad Bv'(0)=0,
\end{gather*}
where $A,B$ are continuous functions on $(0,\infty)$,
and  $p$, $q$, $f$ and $g$ are nonnegative and
continuous functions on $[0,\infty )$. They proved that if $f$ and $g$ are lipschitz
continuous functions on each interval $[\epsilon ,\infty )$, $\epsilon >0$,
system \eqref{1.1} has a unique bounded positive solution $(u,v)$ satisfying
$u$, $v\in C([0,\infty ))\cap C^1((0,\infty))$.

In this article, we are interested in the study of positive radial
solutions to the semilinear elliptic system
\begin{equation}
\begin{gathered}
\Delta u=p(|x|)u^{\alpha }v^{s},\quad x\in \mathbb{R}^n \\
\Delta v=q(|x|)u^{r}v^{\beta },\quad x\in \mathbb{R}^n
\end{gathered}   \label{1.3}
\end{equation}
where $\alpha ,\beta \geq 1$, $r,s\geq 0$ and $p,q:(0,\infty )
\to [ 0,\infty )$ satisfying \eqref{1.2}. To this end, we
undertake  a study of the system of semilinear
differential equations
\begin{equation}
\begin{gathered}
\frac{1}{A}(Au')'=pu^{\alpha }v^{s}\quad \text{on } (0,\infty ), \\
\frac{1}{B}(Bv')'=qu^{r}v^{\beta }\quad\text{on }(0,\infty ), \\
Au'(0)=0,\quad u(\infty )=a, \\
Bv'(0)=0,\quad v(\infty )=b,
\end{gathered}   \label{s}
\end{equation}
where $a,b>0$ and the functions $A$ and $B$ satisfy  condition
(H0) below. In this paper, we denote
$u(\infty ):=\lim_{x\to \infty } u(x)$ and
 $Au'(0):=\lim_{x\to 0}A(x)u'(x)$.

To simplify our statement, we denote by $B^{+}((0,\infty ))$ the set of nonnegative measurable functions on $(0,\infty )$. Also we refer to
 $C([0,\infty ])$ the collection of all continuous functions
 $u$ in $[0,\infty )$ such that $\lim_{x\to \infty}u(x)$ exists and
$C_0([0,\infty ))$ the subclass of $C([0,\infty] )$ consisting of
functions which vanish continuously at $ \infty $.

Before presenting our main result, we would like to make some assumptions and
recall some properties of the operator $Lu=\frac{1}{A}(Au')'$,
while referring the reader to \cite{HM,MM} for
furthers details. Throughout this paper, we say that a function $A$
satisfies condition (H0) if
\begin{itemize}
\item[(H0)]  $A$ is a continuous function on $[0,\infty)$,
differentiable and positive on $(0,\infty )$ such that
\begin{equation*}
\int_{1}^{\infty }\frac{dt}{A(t)}<\infty \quad \text{and}\quad
\int_0^1\frac{1}{A(t)}\Big(\int_0^{t}A(s)ds\Big)dt<\infty .
\end{equation*}
\end{itemize}

For a function  $A$ satisfying (H0), we denote
by $G$ the Green's function of the operator $Lu=\frac{1}{A}(Au')'$
on $( 0,\infty ) $ with Dirichlet conditions
$Au'(0)=0$, $u(\infty )=0$; that is,
\begin{equation*}
G(x,t)=A(t)\int_{x\vee t}^{\infty }\frac{dr}{A(r)},\quad
\text{for }(x,t)\in ((0,\infty ))^2,
\end{equation*}
where $x\vee t:=$ $\max (x,t)$ and we refer to the potential of a function
 $f $ in $B^{+}((0,\infty ))$ by
\begin{equation*}
Vf(x)=\int_0^{\infty }G(x,t)f(t)dt.
\end{equation*}
We point out that for each $f\in B^{+}((0,\infty ))$
such that $Vf(0)<\infty $, the function 
$Vf\in C_0([0,\infty ))\cap C^1((0,\infty ))$ and satisfies
\begin{gather*}
L(Vf)=-f\quad \text{a.e. on }(0,\infty ), \\
A(Vf)'(0)=0,\quad Vf(\infty )=0.
\end{gather*}
Let us introduce the conditions imposed to the functions $p$ and $q$:
\begin{itemize}
\item[(H1)] $p,q:(0,\infty )\to[ 0,\infty )$ are two measurable functions such that
\begin{equation*}
Vp(0)<\infty \quad\text{and}\quad Wq(0)<\infty .
\end{equation*}
\end{itemize}
Here for $f\in B^{+}((0,\infty ))$, we denote
\begin{equation*}
Wf(x)=\int_0^{\infty }H(x,t)f(t)dt,
\end{equation*}
where
\begin{equation*}
H(x,t)=B(t)\int_{x\vee t}^{\infty }\frac{dr}{B(r)}.
\end{equation*}
Using a fixed point argument, we prove our main result.

\begin{theorem} \label{thm1}
Let $A$ and $B$ be two functions satisfying {\rm (H0)} and let 
$ p,q$ be two functions satisfying {\rm (H1)}. 
Then for each $a,b>0$, system \eqref{s} has a positive solution $(u,v)$
satisfying $u,v\in C([0,\infty ] )\cap C^1((0,\infty ))$.
 Moreover, there exist $c_{1},c_{2}>0$
such that for each $x\in [ 0,\infty )$, we have
\begin{gather*}
a\exp (-c_{1}Vp(0))\leq u(x)\leq a,\\
b\exp (-c_{2}Wq(0))\leq v(x)\leq b.
\end{gather*}
\end{theorem}

\begin{remark} \label{rmk1} \rm
If $A(t)=B(t)=t^{n-1}$, the condition (H1) is equivalent
to \eqref{1.2}. It follows by Theorem \ref{thm1} that if $p,q$ satisfy \eqref{1.2}
then for each $a,b>0$, the elliptic system \eqref{1.3} has a positive radial
solution $(u,v)$ continuous in $\mathbb{R}^n$ such that 
$\lim_{|x|\to \infty}u(x)=a$ and $\lim_{|x|\to \infty}v(x)=b$.
\end{remark}

The outline of this article is as follows. In Section 2, we lay out some
properties pertaining with potential theory and we give some useful results
related to the operator $Lu=\frac{1}{A}(Au')'$.
 In particular, we establish an existence and a uniqueness result to the
problem
\begin{equation}
\begin{gathered}
Lu=p(x)u^{\alpha },\quad x\in (0,\infty )\\
Au'(0)=0,\quad u(\infty )=a>0,
\end{gathered} \label{p'}
\end{equation}
where $\alpha \geq 1$ and $p\in B^{+}((0,\infty ))$
such that $Vp(0)<\infty $. This allows us to prove 
Theorem \ref{thm1} in Section 3 by
using a technical method that requires a potential theory approach.

\section{Preliminary results}

Let $A$ be a function satisfying (H0). The objective of
this section is to give some technical results concerning the operator
 $Lu= \frac{1}{A}(Au')'$ and to recall some
potential theory tools which are crucial to prove our main result.

\begin{proposition} \label{prop1}
Let $q\in B^{+}((0,\infty ))$ such that $Vq(0)<\infty $. 
Then the family of functions
\begin{equation*}
F_{q}=\big\{ x\to Vf(x)=\int_0^{\infty}G(x,t)f(t)dt;|f|\leq q\big\}
\end{equation*}
is uniformly bounded and equicontinuous in $[0,\infty ]$. Consequently 
$F_{q}$ is relatively compact in $C_0([0,\infty ))$.
\end{proposition}

\begin{proof}
By writing
\begin{equation*}
Vf(x)=\int_{x}^{\infty }\frac{1}{A(t)}(\int_0^{t}A(r)f(r)dr)
dt,
\end{equation*}
we deduce that for $x,x'\in [ 0,\infty )$, we have
\begin{equation*}
|Vf(x)-Vf(x')|\leq \int_{x}^{x'}\frac{1}{A(t)}\Big(
\int_0^{t}A(r)q(r)dr\Big)dt.
\end{equation*}
Since $Vq(0)=\int_0^{\infty }\frac{1}{A(t)}(\int_0^{t}A(r)q(r)dr)dt<\infty $, 
it follows by the dominated convergence theorem the equicontinuity of $F_{q}$ 
in $[0,\infty)$.
 Moreover, since
\begin{equation*}
| Vf(x)| \leq \int_{x}^{\infty }\frac{1}{A(t)}\Big(
\int_0^{t}A(r)q(r)dr\Big)dt,
\end{equation*}
we deduce that $\lim_{x\to \infty}Vf(x)=0$,
uniformly in $f$. Which proves that $F_{q}$ is uniformly bounded
in $[0,\infty ] $. Then by Ascoli's theorem, we deduce that
 $F_{q}$ is relatively compact in $C_0([0,\infty ))$.
\end{proof}

In what follows, we need the following lemma and we refer to
 \cite{HM,MM} for more details.

\begin{lemma} \label{lem1}
Let $q\in B^{+}((0,\infty ))$ such that $Vq(0)<\infty $. Then the problem
\begin{equation}
\begin{gathered}
\frac{1}{A}(Au')'-qu=0\quad \text{a.e. on }(0,\infty ), \\
Au'(0)=0,\quad u(0)=1,
\end{gathered}  \label{123}
\end{equation}
has a unique solution 
$\psi \in C([0,\infty ))\cap C^1((0,\infty ))$ satisfying for each 
$t\in [ 0,\infty)$,
\begin{equation*}
1\leq \psi (t)\leq \exp \Big(\int_0^{t}\frac{1}{A(s)}\Big(
\int_0^{s}A(r)q(r)dr\Big)ds\Big).
\end{equation*}
\end{lemma}

\begin{proof}
Let $K$ be the operator defined on $C([0,\infty ))$
by
\begin{equation*}
Kf(t)=\int_0^{t}\frac{1}{A(s)}\Big(\int_0^{s}A(r)q(r)f(r)dr\Big)ds,
\quad t\in [0,\infty ).
\end{equation*}
One can see that
\begin{equation*}
0\leq K^n1(t)\leq \frac{(K1(t))^n}{n!},\quad \text{for }t\in[0,\infty )\text{ and }
n\in\mathbb{N}.
\end{equation*}
Then, the series $\sum_{n\geq 0}K^n1$ converges uniformly to a
function $\psi \in C([0,\infty ))$ satisfying
\begin{equation*}
\psi (t)=1+\int_0^{t}\frac{1}{A(s)}\Big(\int_0^{s}A(r)q(r)\psi
(r)dr\Big)ds,\quad \text{for }t\in [0,\infty ).
\end{equation*}
This implies that $\psi \in C^1((0,\infty ))$ is a solution
of problem \eqref{123}.
Moreover, we have
\begin{equation*}
1\leq \psi (t)\leq \sum_{n\geq 0}\frac{(K1(t))^n}{
n!}=\exp (K1(t)),\quad \text{for }t\in [ 0,\infty ).
\end{equation*}

 Now, let $u$, $v$ be two solutions in $C([0,\infty))\cap C^1((0,\infty ))$ 
of \eqref{123} and $\omega =|u-v|$, then 
\begin{equation*}
0\leq \omega (t)\leq K\omega (t),\quad \text{for }t\in [ 0,\infty ).
\end{equation*}
It follows that for $t\in [ 0,\infty )$ and $n\in\mathbb{N}$
\begin{equation*}
0\leq \omega (t)\leq K^n\omega (t)\leq \|\omega \|_{\infty }K^n1(t)\leq
\|\omega \|_{\infty }\frac{(K1(t))^n}{n!}.
\end{equation*}
By letting $n\to \infty $, we deduce that $\omega (t)=0$, for 
$t\in [ 0,\infty )$ and so $u=v$ on $[0,\infty )$.
\end{proof}

We denote by $G_{q}$ the Green's function of the operator
\begin{equation*}
u\mapsto \frac{1}{A}(Au')'-qu
\end{equation*}
on $(0,\infty )$ with Dirichlet conditions $Au'(0)=0$,
$u(\infty )=0$. Then
\begin{equation*}
G_{q}(x,t)=A(t)\psi (x)\psi (t)\int_{x\vee
t}^{\infty }\frac{dr}{A(r)\psi ^2(r)},\quad \text{for }x,t\in (0,\infty).
\end{equation*}
So we define the potential kernel $V_{q}$ in $B^{+}((0,\infty))$ by
\begin{equation*}
V_{q}f(x)=\int_0^{\infty }G_{q}(x,t)f(t)dt.
\end{equation*}
 Note that $V_{q}$ is the unique kernel which satisfies the
 resolvent equation
\begin{equation}
V=V_{q}+V_{q}(qV)=V_{q}+V(qV_{q}).  \label{11}
\end{equation}
So if $u\in B^{+}((0,\infty ))$ such that $V(qu)(0)<\infty $, we have
\begin{equation}
(I-V_{q}(q.))(I+V(q.))u=(I+V(q.))
(I-V_{q}(q.))u=u.  \label{12}
\end{equation}

To study problem \eqref{p'}, we recall an existence result given in \cite{BS}
for the nonlinear problem
\begin{equation}
\begin{gathered}
Lu=\frac{1}{A}(Au')'=u\varphi (.,u)\quad \text{in }(0,\infty ), \\
Au'(0)=0,\quad u(\infty )=a>0.
\end{gathered}  \label{p}
\end{equation}
 Here the nonlinear term $\varphi $ satisfies the following
hypotheses:
\begin{itemize}
\item[(A1)] $\varphi $ is nonnegative measurable
function in $[0,\infty )\times (0,\infty )$.

\item[(A2)] For each $c > 0$, there exists $q_{c} \in  B^{+}((0,\infty ))$ such that 
$Vq_{c}(0)  <  \infty $ and for each $x\in (0,\infty )$, the
function $t\to t(q_{c}(x)-\varphi (x,t))$ is continuous
and nondecreasing on $[0,c]$.
\end{itemize}

\begin{proposition}[see \cite{BS}] \label{prop2}
For each $a>0$, problem \eqref{p} has a positive bounded
solution $u\in C([0,\infty ])\cap C^1((0,\infty))$ satisfying for
 each $x\in [ 0,\infty )$,
\begin{equation*}
e^{-Vq(0)}a\leq u(x)\leq a,
\end{equation*}
where $q:=q_{a}$ is the function given in {\rm (A2)}.
\end{proposition}

\begin{lemma} \label{lem2}
Let $a>0$ and $\varphi $ be a function satisfying {\rm (A1), (A2)}. 
Let $u$ be a positive function in $C([0,\infty ] )\cap C^1((0,\infty ))$.
Then $u$ is a solution of  \eqref{p} if and only if $u$ satisfies
\begin{equation}
u+V(u\varphi (.,u))=a\quad \text{on }[0,\infty ).  \label{o}
\end{equation}
\end{lemma}

\begin{proof}
Let $u$ be a positive function in $C([0,\infty ] )\cap C^1((0,\infty ))$
 satisfying \eqref{o},
then $u\leq a$. Let $q:=q_{a}$ be the function given by (A2), then we have
\begin{equation*}
u\varphi (.,u)\leq qu\leq aq.
\end{equation*}
Since $Vq(0)<\infty $, it follows by Proposition \ref{prop1} that the function 
$v:=V(u\varphi (.,u))$ is in $C_0([0,\infty))$ and so $v$ satisfies
\begin{equation}
\begin{gathered}
Lv=-u\varphi (.,u)\quad \text{a.e. on }(0,\infty ), \\
Av'(0)=0,\quad v(\infty )=0.
\end{gathered}  \label{ty}
\end{equation}
This together with \eqref{o} proves that $u$ is a solution of \eqref{p}.

 Now, let $u$ be a positive function in $C([0,\infty] )\cap C^1((0,\infty ))$
satisfying \eqref{p}. Since $Au'(0)=0$, then $Au'(x)\geq 0$
 for $x\in (0,\infty )$. It follows by $u(\infty )=a$ that $u\leq a$. 
So, by hypothesis (A2), we have
\begin{equation*}
u\varphi (.,u)\leq aq.
\end{equation*}
Then using again Proposition \ref{prop1}, the function $v:=V(u\varphi (.,u))$ satisfies \eqref{ty}.
 Put $w=u+V(u\varphi (.,u))$. Hence the function $w$
is a solution of
\begin{gather*}
Lw=0\quad \text{a.e. on }(0,\infty ), \\
Aw'(0)=0,\quad w(\infty )=a.
\end{gather*}
It follows that $w=a$ and so $u$ satisfies \eqref{o}.
\end{proof}

\begin{proposition} \label{prop3}
Let $\alpha >1$ and $p\in B^{+}((0,\infty ))$ such
that $Vp(0)<\infty $. Then for each $a>0$, problem \eqref{p'} has a unique
solution $u\in C([0,\infty ] )\cap C^1((0,\infty ))$ satisfying
\begin{equation}
a\exp (-\alpha a^{\alpha -1}Vp(0))\leq u(x)\leq a.  \label{c}
\end{equation}
\end{proposition}

\begin{proof}
Let $\varphi (x,t)=p(x)t^{\alpha -1}$, then it is obvious to see that 
$\varphi $ satisfies (A1) and (A2) where $q_{a}$ is
 explicitly given by $q_{a}(x)=\alpha a^{\alpha -1}p(x)$ for 
$x\in (0,\infty )$. So using Proposition \ref{prop2}, problem \eqref{p'} has a
solution $u$ in $C([0,\infty ])\cap C^1((0,\infty))$ satisfying \eqref{c}.

 Let us prove uniqueness. Let $u$, $v\in C([0,\infty ] )\cap C^1((0,\infty ))$
be two solutions of \eqref{p'} and put $w=u-v$. Then using Lemma \ref{lem2}, the
function $w$ satisfies
\begin{equation}
w+V(hw)=0\text{ on }(0,\infty ),  \label{w}
\end{equation}
where the function $h\in B^{+}((0,\infty ))$ is
defined by
\begin{equation*}
h(x):=\begin{cases}
p(x)\frac{u^{\alpha }(x)-v^{\alpha }(x)}{u(x)-v(x)} &\text{if }u(x)\neq v(x),\\
0 &\text{if }u(x)=v(x).
\end{cases}
\end{equation*}
Now, since $Vh(0)\leq \alpha a^{\alpha -1}Vp(0)<\infty $, we apply the
operator $(I-V_{h}(h.))$ on both sides of \eqref{w}, we obtain
 by \eqref{12} that $w=0$ on $(0,\infty )$. So the
uniqueness is proved.
\end{proof}

\section{Proof of Theorem \ref{thm1}}

Let $E=C([0,\infty ])\times C([0,\infty ])$
endowed with the norm $\|(u,v)\|=\|u\|_{\infty }+\|v\|_{\infty }$. 
Then $(E,\|.\|)$ is a Banach space.
 Now let $a,b>0$, to apply a fixed-point argument, we consider the
set
\begin{equation*}
\Lambda =\big\{ (u,v)\in E:ae^{-V\tilde{p}(0)}\leq u\leq a
\text{ and }be^{-W\tilde{q}(0)}\leq v\leq b\big\} ,
\end{equation*}
where $\tilde{p}:=\alpha a^{\alpha -1}b^{s}p$ and 
$\tilde{q}:=\beta b^{\beta -1}a^{r}q$.
 Then $\Lambda $ is a convex closed subset of $E$.

 We define the operator $T$ on $\Lambda $ by $T(u,v)=(y,z)$ 
where $(y,z)$ is the unique solution of the  problem
\begin{gather*} 
\frac{1}{A}(Ay')'(x)=p(x)v^{s}(x)y^{\alpha}(x),\quad x\in (0,\infty ), \\
\frac{1}{B}(Bz')'(x)=q(x)u^{r}(x)z^{\beta}(x),\quad x\in (0,\infty ), \\
Ay'(0)=0,\quad y(\infty )=a, \\
Bz'(0)=0,\quad z(\infty )=b.
\end{gather*}
Note that if $T(u,v)=(u,v)$ then $(u,v)$ is a solution of \eqref{s}. 
So we will use the Schauder's
fixed point theorem to prove that $T$ has a fixed point in $\Lambda $.

 First, we point out that $T$ is well defined and 
$T\Lambda \subset \Lambda $. Indeed, if $v\leq b$ then using 
Proposition \ref{prop3}, 
the problem
\begin{gather*}
\frac{1}{A}(Ay')'(x)=p(x)v^{s}(x)y^{\alpha}(x),\quad x\in (0,\infty ), \\
Ay'(0)=0,\quad y(\infty )=a,
\end{gather*}
has a unique solution $y$ in $C([0,\infty ] )$
satisfying
\begin{equation*}
a\exp (-V\tilde{p}(0))\leq y\leq a.
\end{equation*}
A similar result holds for the problem
\begin{gather*}
\frac{1}{B}(Bz')'(x)=q(x)u^{r}(x)z^{\beta}(x),\quad x\in (0,\infty ), \\
Bz'(0)=0,\quad z(\infty )=b,
\end{gather*}
if the function $u$ satisfies $u\leq a$.

 Next, we prove that $T\Lambda $ is relatively compact in
 $C([0,\infty ] \times [0,\infty ] )$. Let $(u,v)\in \Lambda $ and put 
$(y,z)=T(u,v)$. Using Lemma \ref{lem2}, the functions $y$ and $z$ satisfy
\begin{gather}
y+V(pv^{s}y^{\alpha })=a\quad \text{on }[0,\infty ),  \label{num1}\\
z+W(qu^{r}z^{\beta })=b\quad \text{on }[0,\infty ).  \label{num2}
\end{gather}
Then for $(x,t)$, $(x',t')\in ([0,\infty ] )^2$, we have
\begin{align*}
&\|T(u,v)(x,t)-T(u,v)(x',t')\|\\
&=|y(x)-y(x')|+|z(t)-z(t')|\\
&=|V(pv^{s}y^{\alpha })(x)-V(pv^{s}y^{\alpha
})(x')|+|W(qu^{r}z^{\beta })(t)
-W(qu^{r}z^{\beta })(t')|.
\end{align*}
 Now, using that $(u,v)$ and $(y,z)$ are
in $\Lambda $, it follows that
$V(pv^{s}y^{\alpha })\in F_{\frac{a}{\alpha }\tilde{p}}$ and 
$W(qu^{r}z^{\beta })\in F_{\frac{b}{\beta }\tilde{q}}$.
This implies, by Proposition \ref{prop1}, that $T\Lambda $ is equicontinuous
in $[0,\infty ] \times [0,\infty ] $. Now, since 
$\{ T(u,v)(x,t);\text{ }(u,v)\in \Lambda \} $ is
uniformly bounded in $[0,\infty ] \times [0,\infty ]$,
 we deduce by Ascoli's Theorem that $T\Lambda $ is relatively compact in 
$C([0,\infty ] \times [0,\infty ] )$.

 Let us prove the continuity of $T$ in $\Lambda $. 
Let $(u_n,v_n)$ be a sequence in $\Lambda $ converging to
$(u,v)\in \Lambda $ with respect to $\|.\|$. 
Put $(y_n,z_n)=T(u_n,v_n)$ and $(y,z)=T(u,v)$. Then
\begin{equation*}
|T(u_n,v_n)(x,t)-T(u,v)(x,t)|=|y_n(x)-y(x)|+|z_n(t)-z(t)|.
\end{equation*}
We denote by $Y_n=y_n-y$ and $Z_n=z_n-z$.
We start by evaluating $Y_n$. By \eqref{num1}, we have for
$x\in[0,\infty ] $
\begin{align*}
Y_n(x)
&=V(pv^{s}y^{\alpha })(x)-V(pv_n^{s}y_n^{\alpha })(x) \\
&= V(py^{\alpha }(v^{s}-v_n^{s}))(x)-V(hY_n)(x),
\end{align*}
where $h\in B^{+}((0,\infty ))$ and defined by
\begin{equation*}
h(x):=\begin{cases}
p(x)v_n^{s}(x)\frac{y_n^{\alpha }(x)-y^{\alpha
}(x)}{y_n(x)-y(x)}&\text{if }y_n(x)\neq y(x), \\
0 &\text{if }y_n(x)=y(x).
\end{cases}
\end{equation*}
Since $Vh(0)<\infty $,  applying the operator $(I-V_{h}(h.))$ on both side of
\begin{equation*}
Y_n+V(hY_n)=V(py^{\alpha }(v^{s}-v_n^{s})),
\end{equation*}
we obtain by \eqref{11} and \eqref{12} that
\begin{equation*}
Y_n=V_{h}(py^{\alpha }(v^{s}-v_n^{s})).
\end{equation*}
So,
\begin{equation*}
|Y_n|\leq V(py^{\alpha }|v^{s}-v_n^{s}|).
\end{equation*}
Now, since $py^{\alpha }|v^{s}-v_n^{s}|\leq 2a^{\alpha }b^{s}p$ and 
$Vp(0)<\infty $, we deduce by the dominated convergence theorem, that
\begin{equation*}
V(y^{\alpha }(v^{s}-v_n^{s})p)(x)\to 0\quad \text{as }n\to \infty .
\end{equation*}
It follows that $Y_n(x)$ converge to $0$ as $n\to \infty $.

Analogously, we have $Z_n(x)$ converge to $0$ as $n\to \infty $.
This proves that for each $(x,t)\in [ 0,\infty)\times [ 0,\infty )$,
\begin{equation*}
T(u_n,v_n)(x,t)\to T(u,v)(x,t)\quad \text{as }n\to \infty .
\end{equation*}
Now, since $T\Lambda $ is relatively compact in 
$C([0,\infty] \times [0,\infty ] )$, we deduce that
\begin{equation*}
\|T(u_n,v_n)-T(u,v)\|\to 0\quad \text{as }n\to \infty .
\end{equation*}
 Hence, $T$ is a compact mapping from $\Lambda $ to itself. Then by
the Schauder's fixed point theorem there exists 
$(u,v)\in \Lambda $ such that $T(u,v)=(u,v)$. 
So $(u,v)$ is the desired solution. This completes the proof.

\subsection*{Acknowledgments}
The author would like to thank Professor Habib Ma\^aagli 
for his guidance and useful discussions, also the editors 
and reviewers for their valuable comments and suggestions which 
contributed to the improvement this article.


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\end{document}