\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 $$\begin{gathered} \Delta u=p(|x|)f(v),\quad x\in\mathbb{R}^n \\ \Delta v=q(|x|)g(u),\quad x\in\mathbb{R}^n \end{gathered} \label{1.1}$$ 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 $$\int_0^{\infty }tp(t)dt<\infty ,\quad \int_0^{\infty }tq(t)dt<\infty \label{1.2}$$ 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 $$\begin{gathered} \Delta u=p(|x|)u^{\alpha }v^{s},\quad x\in \mathbb{R}^n \\ \Delta v=q(|x|)u^{r}v^{\beta },\quad x\in \mathbb{R}^n \end{gathered} \label{1.3}$$ 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 $$\begin{gathered} \frac{1}{A}(Au')'=pu^{\alpha }v^{s}\quad \text{on } (0,\infty ), \\ \frac{1}{B}(Bv')'=qu^{r}v^{\beta }\quad\text{on }(0,\infty ), \\ Au'(0)=0,\quad u(\infty )=a, \\ Bv'(0)=0,\quad v(\infty )=b, \end{gathered} \label{s}$$ 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 $$\begin{gathered} Lu=p(x)u^{\alpha },\quad x\in (0,\infty )\\ Au'(0)=0,\quad u(\infty )=a>0, \end{gathered} \label{p'}$$ 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 $$\begin{gathered} \frac{1}{A}(Au')'-qu=0\quad \text{a.e. on }(0,\infty ), \\ Au'(0)=0,\quad u(0)=1, \end{gathered} \label{123}$$ 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 $$V=V_{q}+V_{q}(qV)=V_{q}+V(qV_{q}). \label{11}$$ So if $u\in B^{+}((0,\infty ))$ such that $V(qu)(0)<\infty$, we have $$(I-V_{q}(q.))(I+V(q.))u=(I+V(q.)) (I-V_{q}(q.))u=u. \label{12}$$ To study problem \eqref{p'}, we recall an existence result given in \cite{BS} for the nonlinear problem $$\begin{gathered} Lu=\frac{1}{A}(Au')'=u\varphi (.,u)\quad \text{in }(0,\infty ), \\ Au'(0)=0,\quad u(\infty )=a>0. \end{gathered} \label{p}$$ 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 $$u+V(u\varphi (.,u))=a\quad \text{on }[0,\infty ). \label{o}$$ \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 $$\begin{gathered} Lv=-u\varphi (.,u)\quad \text{a.e. on }(0,\infty ), \\ Av'(0)=0,\quad v(\infty )=0. \end{gathered} \label{ty}$$ 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 $$a\exp (-\alpha a^{\alpha -1}Vp(0))\leq u(x)\leq a. \label{c}$$ \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 $$w+V(hw)=0\text{ on }(0,\infty ), \label{w}$$ 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. \begin{thebibliography}{99} \bibitem{BS} S. Ben Othman, H. M\^{a}agli, N. 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