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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 154, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/154\hfil Existence of positive solutions]
{Existence of positive solutions  for nonlinear systems
of second-order differential equations with integral boundary
conditions on an infinite interval in  Banach spaces}

\author[X. Chen, X. Zhang \hfil EJDE-2011/154\hfilneg]
{Xu Chen, Xingqiu Zhang} % in alphabetical order

\address{Xu Chen \newline
School of Mathematics, Liaocheng University,
Liaocheng, Shandong, China}
\email{woshchxu@163.com}

\address{Xingqiu Zhang \newline
School of Mathematics, Liaocheng University,
Liaocheng, Shandong, China}
\email{zhxq197508@163.com}

\thanks{Submitted August 22, 2011. Published November 10, 2011.}
\subjclass[2000]{34B15, 34B16, 34B40}
\keywords{Singular differential equations;
integral boundary; cone; ordering; \hfill\break\indent
positive solution; M\"onch  fixed point theorem}

\begin{abstract}
 The article shows the existence of positive solutions for
 systems of nonlinear singular differential equations with integral
 boundary conditions on an infinite interval in Banach spaces.
 Our main tool is the M\"onch fixed point theorem combined with
 a monotone iterative technique.  In addition, an explicit
 iterative approximation of the solution is provided.
\end{abstract}

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

\section{Introduction}

The theory of boundary-value problems with integral boundary
conditions for ordinary differential equations arises in different
areas of applied mathematics and physics. For example, heat
conduction, chemical engineering,underground water flow,
thermo-elasticity, and plasma physics can be reduced to the
nonlocal problems with integral boundary conditions. There are
many excellent results about the existence of positive solutions
for integral boundary value problems in scalar case (see, for
instance, \cite{l2,y1,y2,z1}
and references therein). Very recently, by using
Schauder fixed point theorem, Guo \cite{g1} obtained the existence of
positive solutions for a class of nth-order nonlinear impulsive
singular integro-differential equations in a Banach space.

Recently, Zhang et al \cite{z2}, using the cone theory and monotone
iterative technique, investigated the existence of minimal
nonnegative solution of the following nonlocal boundary value
problems for second-order nonlinear impulsive differential
equations on an infinite interval with an infinite number of
impulsive times
\begin{gather*}
-x''(t)=f(t,x(t),x'(t)),\quad t\in J,t\neq t_k,\\
\Delta x|_{t=t_k}=I_{k}(x(t_k)),\quad k=1,2,\dots,\\
\Delta x'|_{t=t_k}=\overline{I}_{k}(x(t_k)),\quad k=1,2,\dots,\\
x(0)=\int_0^\infty g(t)x(t)\,\mathrm{d}t,\quad
x'(\infty)=0,
\end{gather*}
where $J=[0,+\infty)$,
$f\in C(J\times \mathbb{R}^+\times \mathbb{R}^+,\mathbb{R}^+)$,
$\mathbb{R}^+=[0,+\infty]$, $0<t_1<t_2<\dots<t_k<\dots$,
$t_k\to \infty$,
$I_k\in C[\mathbb{R}^+,\mathbb{R}^+],\overline{I}_k\in C[\mathbb{R}^+,
\mathbb{R}^+]$,
 $g(t)\in C[\mathbb{R}^+,\mathbb{R}^+)$, with
$\int_0^\infty g(t)\,\mathrm{d}t<1$.

To the best of
our knowledge, only a few authors  have studied integral
boundary value problems in Banach spaces and results for systems of
second-order differential equation are rarely seen.
Motivated by Zhang and Guo's work, in this paper,
we consider the following singular integral boundary value problem
on an infinite interval in a Banach space $E$:
\begin{equation}
\begin{gathered}
x''(t)+f(t,x(t),x'(t),y(t),y'(t))=0,\\
y''(t)+g(t,x(t),x'(t),y(t),y'(t))=0,\quad t\in J_{+},\\
x(0)=\int_0^\infty q(t)x(t)\,\mathrm{d}t,\quad x'(\infty)=x_{\infty},\\
y(0)=\int_0^\infty h(t)y(t)\,\mathrm{d}t,\quad
y'(\infty)=y_{\infty},
\end{gathered} \label{e1}
\end{equation}
where
$J=[0,\infty)$, $J_+=(0,\infty)$, and the functions
$q(t),h(t)$ are in $L[0,\infty)$ with
$\int_0^\infty q(t)\,\mathrm{d}t<1$,
$\int_0^\infty h(t)\,\mathrm{d}t<1$ and
$\int_0^\infty tq(t)\,\mathrm{d}t<\infty$,
$\int_0^\infty th(t)\,\mathrm{d}t<\infty$.
$x'(\infty)=\lim_{t\to\infty}x'(t)$,
$y'(\infty)=\lim_{t\to\infty}y'(t)$.
The nonlinear terms $f(t,x_0,x_1,y_0,y_1)$ and $g(t,x_0,x_1,y_0,y_1)$
permit singularities at $t=0$ and $x_i, y_i=\theta$ $(i=0,1)$,
where $\theta$ denotes the zero element of Banach space $E$.
By singularity, we mean that $\|f(t,x_0,x_1,y_0,y_1)\|\to\infty$
as $t\to 0^+$ or $x_{i},y_i\to\theta$.

 The main features of this article are as follows:
Firstly, compared with \cite{z2}, the systems of integral
boundary value problem we discussed here is in Banach spaces
and nonlinear term permits singularity not
only at $t=0$ but also at $x_i, y_i =\theta$ $(i=0,1)$.
Secondly, compared with \cite{g1}, the problem
we discussed here is systems of integral boundary value problem,
since the problem we discuss is the integral boundary value
problems, the construction of bounded convex closed set is different
from that in \cite{g1}.
Furthermore, the relative compact conditions we
used are weaker. Finally, an iterative sequence for the solution
under some normal type conditions is established which makes it
 convenient in applications.

\section{Preliminaries}
Let
\begin{gather*}
FC[J,E]=\{x\in C[J,E]:\sup_{t\in J}\frac{\|x(t)\|}{t+1}<\infty\},\\
DC^1[J,E]=\{x\in C^1[J,E]:\sup_{t\in J}\frac{\|x(t)\|}{t+1}
<\infty\text{ and }\sup_{t\in J}\|x'(t)\|<\infty\}.
\end{gather*}
Evidently,
$C^1[J,E]\subset C[J,E]$ and $DC^1[J,E]\subset FC[J,E]$.
It is easy to see that $FC[J,E]$ is a Banach space with norm
$$
\|x\|_F=\sup_{t\in J}\frac{\|x(t)\|}{t+1},
$$
and $DC^1[J,E]$ is also  a Banach space with norm
$$
\|x\|_D=\max\{\|x\|_F,\|x'\|_1\},
$$
where
$$
\|x'\|_1=\sup_{t\in J}\|x'(t)\|.
$$
Let $X=DC^1[J,E]\times DC^1[J,E]$ with norm
$\|(x,y)\|_X=\max\{\|x\|_D,\|y\|_D\}$, for $(x,y)\in X$. Then
$(X,\|\cdot,\cdot\|_X)$ is also a Banach space. The basic space
using in this paper is $(X,\|\cdot,\cdot\|_X)$.

Let $P$ be a normal cone in $E$ with normal constant $N$ which
defines a partial ordering in $E$ by $x\leq y$. If $x\leq y$ and
$x\neq y$, we write $x<y$. Let $P_+=P\backslash \{\theta\}$. So,
$x\in P_+$ if and only if $x>\theta$. For details on cone theory,
see \cite{g3}.

In what follows, we always assume that
$x_{\infty}\geq x_0^*$,
$y_{\infty}\geq y_0^*$, $x_0^*, y_0^*\in P_+$.
Let $P_{0\lambda}=\{x\in P:x\geq \lambda x_0^*\}$,
$P_{1\lambda}=\{y\in P:y\geq \lambda y_0^*\}(\lambda>0)$.
Obviously, $P_{0\lambda},P_{1\lambda}\subset P_+$ for
any $\lambda>0$.
When $\lambda=1$, we write $P_0=P_{01},P_1=P_{11}$;
i.e., $P_0=\{x\in P:x\geq x_0^*\}$,
$P_1=\{y\in P:y\geq y_0^*\}$. Let
$P(F)=\{x\in FC[J,E]:x(t)\geq\theta, \forall\ t\in J\}$, and
$P(D)=\{x\in DC^1[J,E]:x(t)\geq\theta, x'(t)\geq\theta, \forall\
t\in J\}$.
It is clear, $P(F)$ and $P(D)$ are cones in $FC[J,E]$ and
$DC^1[J,E]$, respectively.
A map $(x,y) \in DC^1[J, E]\cap C^2[J_+,E]$ is called a
positive solution of \eqref{e1} if
$(x,y)\in P(D)\times P(D)$ and $(x(t),y(t))$ satisfies
\eqref{e1}.

Let $\alpha, \alpha_F, \alpha_D,\alpha_X$  denote the Kuratowski
measure of non-compactness in the sets
$E, FC[J,E],DC^1[J, E]$ and $X$,
respectively. For details on the definition and properties of the
measure of non-compactness, the reader is referred to references
\cite{d1,g2,g3,l1}.
Let
\begin{equation}
D_0=\Big(1+\frac{\int_0^\infty q(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t}\Big),\quad
D_1=\Big(1+\frac{\int_0^\infty h(t)\,\mathrm{d}t}
{1-\int_0^\infty h(t)\,\mathrm{d}t}\Big),
\label{e2}
\end{equation}
$D^*=\max\{D_0,D_1\}$.
Denote
$$
\lambda_0^*= \min\{\frac{\int_0^\infty tq(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t},1\},\quad
\lambda_1^*= \min\{\frac{\int_0^\infty th(t)\,\mathrm{d}t}
{1-\int_0^\infty h(t)\,\mathrm{d}t},1\}
$$
Let us list some conditions for convenience.

\begin{itemize}
\item[(H1)] $f,g\in C[J_+\times P_{0\lambda}\times
P_{0\lambda}\times P_{1\lambda}\times P_{1\lambda}, P]$ for any
$\lambda>0$ and there exist $a_i,b_i,c_i\in L[J_+,J]$ and $z_i\in
C[J_+\times J_+\times J_+\times J_+, J]\ (i=0,1)$ such that
$$
\|f(t,x_0,x_1,y_0,y_1)\|\leq a_0(t)+b_0(t)z_0(\|x_0\|,\|x_1\|,\|y_0\|,
\|y_1\|),
$$
for all $t\in J_+,x_i\in P_{0\lambda_0^*},y_i\in P_{1\lambda_1^*}$;
$$
\|g(t,x_0,x_1,y_0,y_1)\|\leq a_1(t)+b_1(t)z_1(\|x_0\|,\|x_1\|,\|y_0\|,
 \|y_1\|),
$$
for all $t\in J_+,x_i\in P_{0\lambda_0^*},y_i\in P_{1\lambda_1^*}$;
and
\begin{gather*}
\frac{\|f(t,x_0,x_1,y_0,y_1)\|}{c_0(t)(\|x_0\|+\|x_1\|+\|y_0\|
+\|y_1\|)}\to 0,\\
\frac{\|g(t,x_0,x_1,y_0,y_1)\|}{c_1(t)(\|x_0\|+\|x_1\|+\|y_0\|
+\|y_1\|)}\to 0,
\end{gather*}
as $x_i\in P_{0\lambda_0^*}$, $y_i\in P_{1\lambda_1^*}$ ($i=0,1$),
$\|x_0\|+\|x_1\|+\|y_0\|+\|y_1\|\to\infty$,
uniformly for $t\in J_+$; and for $i=0,1$:
$$
\int_0^\infty a_i(t)\mathrm{d}t=a_i^*<\infty,\quad
\int_0^\infty b_i(t)\mathrm{d}t=b_i^*<\infty, \quad
\int_0^\infty c_i(t)(1+t)\mathrm{d}t=c_i^*<\infty.
$$

\item[(H2)] For any $t\in J_{+}$ and countable bounded
set $V_i\subset DC^1[J,P_{0\lambda_0^*}]$,
$W_i\subset DC^1[J,P_{1\lambda_1^*}]$  ($i=0,1$), there exist
$L_i(t),K_i(t)\in L[J, J]$ ($i=0,1$) such that
\begin{gather*}
\alpha(f(t,V_0(t),V_1(t),W_0(t),W_1(t)))
 \leq \sum_{i=0}^1L_{0i}(t)\alpha(V_i(t))+K_{0i}(t)\alpha(W_i(t)),\\
\alpha(g(t,V_0(t),V_1(t),W_0(t),W_1(t)))
 \leq \sum_{i=0}^1L_{1i}(t)\alpha(V_i(t))+K_{1i}(t)\alpha(W_i(t)),
\end{gather*}
with
\begin{gather*}
G_i^*=\int_0^{+\infty}[(L_{i0}(s)+K_{i0}(s))(1+s)+L_{i1}(s)
+K_{i1}(s)]\mathrm{d}s<\infty\quad (i=0,1),\\
G^*=\max\{G_0^*,G_1^*\}.
\end{gather*}

\item[(H3)]  $t\in J_{+}$,
$\lambda_0^*x_0^*\leq x_i\leq\overline{x}_i$,
$\lambda_1^*y_0^*\leq y_i\leq\overline{y}_i$ ($i=0,1$) imply
$$
f(t,x_0,x_1,y_0,y_1)\leq f(t,\overline{x}_0,\overline{x}_1,
\overline{y}_0,\overline{y}_1),\quad
g(t,x_0,x_1,y_0,y_1)\leq
g(t,\overline{x}_0,\overline{x}_1,\overline{y}_0,\overline{y}_1).
$$
\end{itemize}

In what follows, we write
\begin{gather*}
Q_1=\{x\in DC^1[J,P]:x^{(i)}(t)\geq
\lambda_0^*x_0^*, \forall\ t \in J,i=0,1\},\\
Q_2=\{y\in
DC^1[J,P]:y^{(i)}(t)\geq \lambda_1^*y_0^*, \forall\ t \in
J,i=0,1\},
\end{gather*}
and $Q=Q_1\times Q_2$. Evidently, $Q_1,Q_2$ and $Q$ are
closed convex set in $DC^1[J ,E]$ and $X$, respectively.

We shall reduce BVP \eqref{e1} to a system of integral equations in
$E$. To this end, we first consider operator $A$ defined by
\begin{equation}
A(x,y)(t)=(A_1(x,y)(t), A_2(x,y)(t)),\label{e3}
\end{equation}
where
\begin{equation}
\begin{split}
 A_1(x,y)(t)
&= \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{x_{\infty}\int_0^{\infty}tq(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)f(s,x(s),x'(s),y(s),y'(s))\,\mathrm{d}s\,\mathrm{d}t\Big\} \\
&\quad +tx_{\infty}+\int_0^{\infty}G(t,s)f(s,x(s),x'(s),y(s),
y'(s))\,\mathrm{d}s,\\
\end{split}\label{e4}
\end{equation}
and
\begin{equation}
\begin{split}
 A_2(x,y)(t)&= \frac{1}{1-\int_0^{\infty}h(t)\,\mathrm{d}t}
\Big\{y_{\infty}\int_0^{\infty}th(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}h(t)G(t,s)g(s,x(s),x'(s),y(s),y'(s))\,\mathrm{d}s\,\mathrm{d}t\Big\} \\
&\quad +ty_{\infty}+\int_0^{\infty}G(t,s)g(s,x(s),x'(s),y(s),y'(s))\,\mathrm{d}s,\\
\end{split}\label{e5}
\end{equation}
where
$$
G(t,s)=\begin{cases}
t,& 0\leq t\leq s<+\infty,\\
s,& 0\leq s\leq t<+\infty.
\end{cases}
 $$

\begin{lemma}\label{lem1}
  If {\rm(H1)} is satisfied, then the operator $A$ defined
 by \eqref{e3} is a continuous operator  from $Q$ to $Q$.
\end{lemma}

\begin{proof} Let
\begin{gather}
\varepsilon_0=\min\Big\{\frac{1}
{8c_0^*\big(1+\frac{\int_0^\infty q(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t}\big)},
\frac{1}{8c_1^*\big(1+\frac{\int_0^\infty h(t)\,\mathrm{d}t}
{1-\int_0^\infty h(t)\,\mathrm{d}t}\big)}\Big\},\label{e6}
\\
r=\min\Big\{\frac{\lambda_0^*\|x_0^*\|}{N},
\frac{\lambda_1^*\|y_0^*\|}{N}\Big\}>0.\label{e7}
\end{gather}
By (H1), there exists a $R>r$ such that
\[
\|f(t,x_0,x_1,y_0,y_1)\|\leq
 \varepsilon_0c_0(t)(\|x_0\|+\|x_1\|+\|y_0\|+\|y_1\|),
\]
for all $t\in J_+$, $x_i\in P_{0\lambda_0^*}$,
$y_i\in P_{1\lambda_1^*}$ ($i=0,1$),
$\|x_0\|+\|x_1\|+\|y_0\|+\|y_1\|>R$;
 and
\[
\|f(t,x_0,x_1,y_0,y_1)\| \leq  a_0(t)+M_0b_0(t),
\]
for all $t\in J_+$, $x_i\in P_{0\lambda_0^*}$,
$y_i\in P_{1\lambda_1^*}$ ($i=0,1$),
$\|x_0\|+\|x_1\|+\|y_0\|+\|y_1\| \leq R$,
 where
$$
M_0=\max\{z_0(u_0,u_1,v_0,v_1):r\leq u_i,v_i\leq R\; (i=0,1)\}.
$$
Hence
\begin{equation} \label{e8}
\|f(t,x_0,x_1,y_0,y_1)\| \leq
 \varepsilon_0c_0(t)(\|x_0\|+\|x_1\|+\|y_0\|+\|y_1\|)+a_0(t)+M_0b_0(t),
\end{equation}
for all $t\in J_+$, $x_i\in P_{0\lambda_0^*}$,
$y_i\in P_{1\lambda_1^*}$  ($i=0,1$).
 Let $(x,y)\in Q$.
 By \eqref{e8} we have
 \begin{equation}
\begin{split}
&\|f(t,x(t),x'(t),y(t),y'(t))\|\\
&\leq \varepsilon_0c_0(t)(1+t)\Big(\frac{\|x(t)\|}{t+1}
+\frac{\|x'(t)\|}{t+1}+\frac{\|y(t)\|}{t+1}+\frac{\|y'(t)\|}{t+1}\Big)
 +a_0(t)+M_0b_0(t)\\
&\leq \varepsilon_0c_0(t)(1+t)(\|x\|_F+\|x'\|_1+\|y\|_F
 +\|y'\|_1)+a_0(t)+M_0b_0(t)\\
&\leq 2\varepsilon_0c_0(t)(1+t)(\|x\|_D+\|y\|_D)+a_0(t)+M_0b_0(t),\\
&\leq 4\varepsilon_0c_0(t)(1+t)\|(x,y)\|_X+a_0(t)+M_0b_0(t),\quad
 \forall\ t\in J_+,
\end{split}\label{e9}
\end{equation}
which together with  (H1) implies the
convergence of the infinite integral
\begin{equation}
\int_0^\infty \|f(s,x(s),x'(s),y(s),y'(s))\|\mathrm{d}s,\label{e10}
\end{equation}
which together with \eqref{e4}  and (H1) implies
\begin{align*}
&\| A_1(x,y)(t)\| \\
&\leq \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)\|f(s,x(s),x'(s),y(s),y'(s))\|\,\mathrm{d}s\,\mathrm{d}t \\
&\quad +\|x_{\infty}\|\int_0^{\infty}tq(t)\,\mathrm{d}t\Big\}+t\|x_{\infty}\|+\int_0^{\infty}G(t,s)\|f(s,x(s),x'(s),y(s),y'(s))\|\,\mathrm{d}s.
%\label{11}
\end{align*}
Therefore,
\begin{equation}
\begin{split}
&\frac{\|A_1(x,y)(t)\|}{1+t}\\
&\leq \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{\int_0^{\infty}\int_0^{\infty}q(t)\|f(s,x(s),x'(s),y(s),y'(s))
\|\,\mathrm{d}s\,\mathrm{d}t \\
&\quad +\|x_{\infty}\|\int_0^{\infty}tq(t)\,\mathrm{d}t\Big\}
+\|x_{\infty}\|+\int_0^{\infty}\|f(s,x(s),x'(s),y(s),y'(s))\|
 \,\mathrm{d}s\\
&\leq \Big(1+\frac{\int_0^\infty q(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t}\Big)
[4\varepsilon_0c_0^*\|(x,y)\|_X+a_0^*+M_0b_0^*]\\
&\quad +\Big(1+\frac{\int_0^\infty tq(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t}\Big)\|x_{\infty}\|\\
&\leq
\frac{1}{2}\|(x,y)\|_X+\Big(1+\frac{\int_0^\infty q(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t}\Big)(a_0^*+M_0b_0^*)\\
&\quad +\Big(1+\frac{\int_0^\infty tq(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t}\Big)\|x_{\infty}\|.
\end{split}\label{e12}
\end{equation}
Differentiating \eqref{e4}, we obtain
 \begin{equation}
A_1'(x,y)(t)=\int_t^{\infty}f(s,x(s),x'(s),y(s),y'(s))\mathrm{d}s
+x_{\infty}.\label{e13}
\end{equation}
Hence,
\begin{equation}
\begin{split} \|A'_1(x,y)(t)\|
&\leq \int_0^{+\infty}\|f(s,x(s),x'(s),y(s),y'(s))\|\mathrm{d}s+\|x_{\infty}\|\\
&\leq  4\varepsilon_0c_0^*\|(x,y)\|_X+a_0^*+M_0b_0^*+\|x_{\infty}\|\\
&\leq \frac{1}{2}\|(x,y)\|_X+a_0^*+M_0b_0^*+\|x_{\infty}\|,\quad
 \forall t\in J.
\end{split}\label{e14}
\end{equation}
It follows from \eqref{e12} and \eqref{e14} that
\begin{equation}
\begin{split}
\|A_1(x,y)\|_D
&\leq \frac{1}{2}\|(x,y)\|_X+\Big(1+\frac{\int_0^\infty
q(t)\,\mathrm{d}t}{1-\int_0^\infty
q(t)\,\mathrm{d}t}\Big)(a_0^*+M_0b_0^*)\\
&\quad +\Big(1+\frac{\int_0^\infty tq(t)\,\mathrm{d}t}{1-\int_0^\infty
q(t)\,\mathrm{d}t}\Big)\|x_{\infty}\|.
\end{split}\label{e15}
\end{equation}
So, $A_1(x,y)\in DC^1[J,E]$. On the other hand,
it can be easily seen that
\begin{gather*}
 A_1(x,y)(t)\geq\Big(\frac{\int_0^\infty
q(t)\,\mathrm{d}t}{1-\int_0^\infty
q(t)\,\mathrm{d}t}\Big)x_{\infty}\geq\lambda_0^*x_{\infty}\geq\lambda_0^*x_0^*,
\quad \forall\ t\in J,
\\
A'_1(x,y)(t)\geq x_{\infty}\geq\lambda_0^*x_{\infty}
\geq\lambda_0^*x_0^*, \quad \forall\ t\in J.
\end{gather*}
so, $A_1(x,y)\in Q_1$. In the same way, we
can easily obtain
\begin{equation}
\begin{split}
\|A_2(x,y)\|_D& \leq\frac{1}{2}\|(x,y)\|_X+\Big(1+\frac{\int_0^\infty
h(t)\,\mathrm{d}t}{1-\int_0^\infty
h(t)\,\mathrm{d}t}\Big)(a_0^*+M_0b_0^*)\\
&\quad +\Big(1+\frac{\int_0^\infty th(t)\,\mathrm{d}t}{1-\int_0^\infty
h(t)\,\mathrm{d}t}\Big)\|y_{\infty}\|,
\end{split} \label{16}
\end{equation}
 and
\begin{gather*}
A_2(x,y)(t)\geq\Big(\frac{\int_0^\infty
h(t)\,\mathrm{d}t}{1-\int_0^\infty
h(t)\,\mathrm{d}t}\Big)y_{\infty}\geq
 \lambda_1^*y_{\infty}\geq\lambda_1^*y_0^*,
\quad \forall\ t\in J,\\
 A'_2(x,y)(t)\geq
y_{\infty}\geq\lambda_1^*y_{\infty}\geq\lambda_1^*y_0^*, \quad
\forall t\in J.
\end{gather*}
where $M_1=\max\{z_1(u_0,u_1,v_0,v_1):r\leq
u_i,v_i\leq R\; (i=0,1)\}$.
Thus, we have proved that $A$ maps $Q$
to $Q$ and we have
\begin{equation}
\|A(x,y)\|_X\leq \frac{1}{2}\|(x,y)\|_X+\gamma,\label{e17}
\end{equation}
where
\begin{equation}
\begin{split}
\gamma&= \max\Big\{\Big(1+\frac{\int_0^\infty
q(t)\,\mathrm{d}t}{1-\int_0^\infty
q(t)\,\mathrm{d}t}\Big)(a_0^*+M_0b_0^*)
+\Big(1+\frac{\int_0^\infty tq(t)\,\mathrm{d}t}
{1-\int_0^\infty q(t)\,\mathrm{d}t}\Big)\|x_{\infty}\|,\\
&\Big(\Big(1+\frac{\int_0^\infty h(t)\,\mathrm{d}t}{1-\int_0^\infty h(t)\,\mathrm{d}t}\Big)(a_0^*+M_0b_0^*)
+\Big(1+\frac{\int_0^\infty th(t)\,\mathrm{d}t}
{1-\int_0^\infty h(t)\,\mathrm{d}t}\Big)\|y_{\infty}\|\Big\}.
\end{split} \label{e18}
\end{equation}

Finally, we show that $A$ is continuous. Let $(x_m,y_m),
(\overline{x},\overline{y})\in Q$,
$\|(x_m,y_m)-(\overline{x},\overline{y})\|_X\to 0\ (m\to\infty)$.
Then $\{(x_m,y_m)\}$ is a bounded subset of $Q$. Thus, there
exists $r>0$ such that $\sup_m\|(x_m,y_m)\|_X<r$ for $m\geq 1$ and
$\|(\overline{x},\overline{y})\|_X\leq r+1$. Similar to
\eqref{e12} and \eqref{e14}, it is easy to show that
\begin{equation}
\begin{split}
&\|A_1(x_m,y_m)-A_1(\overline{x},\overline{y})\|_X\\
&\leq\int_0^{\infty}\|f(s,x_m(s),x'_m(s),y_m(s),y'_m(s))
-f(s,\overline{x}(s),\overline{x'}(s),\overline{y}(s),\overline{y'}(s))
 \|\mathrm{d}s\\
&\quad +\frac{\int_0^\infty q(t)\,\mathrm{d}t}
 {1-\int_0^\infty q(t)\,\mathrm{d}t}
 \int_0^{\infty}\|f(s,x_m(s),x'_m(s),y_m(s),y'_m(s))\\
&\quad -f(s,\overline{x}(s),\overline{x}'(s),
\overline{y}(s),\overline{y}'(s))\|\mathrm{d}s.
\end{split}\label{e19}
\end{equation}
It is clear that
\begin{equation}
 f(t,x_m(t),x'_m(t),y_m(t),y'_m(t))\to
f(t,\overline{x}(t),\overline{x}'(t),\overline{y}(t),\overline{y}'(t))
\label{e20}
\end{equation}
as $m\to\infty$, for all $t\in J_{+}$.
By \eqref{e9}, we obtain
\begin{equation}
\begin{split}
&\|f(t,x_m(t),x'_m(t),y_m(t),y'_m(t))-
f(t,\overline{x}(t),\overline{x}'(t),\overline{y}(t),\overline{y}'(t))\|\\
&\leq 8\epsilon_0c_0(t)(1+t)r+2a_0(t)+2M_0b_0(t)\\
&=\sigma_0(t), \quad \sigma_0(t)\in L[J,J],\quad m=1,2,3,\dots,\;
 \forall t\in J_+.
\end{split}\label{e21}
\end{equation}
It follows from \eqref{e20}, \eqref{e21}, and the dominated
convergence theorem that
\[
\lim_{m\to\infty}\int_0^\infty\|f(s,x_m(s),x_m'(s),y_m(s),y_m'(s))
- f(s,\overline{x}(s),\overline{x}'(s),\overline{y}(s),
\overline{y}'(s))\|\mathrm{d}s=0. %\label{e22}
\]
It follows from the above inequality and \eqref{e19} that
$\|A_1(x_m,y_m)-A_1(\overline{x},\overline{y})\|_D\to 0$ as
$m\to \infty$. By the same method, we have
$\|A_2(x_m,y_m)-A_2(\overline{x},\overline{y})\|_D\to 0$ as
$m\to \infty$. Therefore, the continuity of $A$ is proved.
\end{proof}

\begin{lemma}\label{lem2}
 Under assumption {\rm (H1)},  $(x,y)\in
Q\cap (C^2[J_{+},E]\times  C^2[J_{+},E])$ is a solution of
\eqref{e1} if and only if $(x,y)\in Q$ is a fixed point of $A$.
\end{lemma}

\begin{proof}
Suppose that $(x,y)\in Q\cap (C^2[J_{+},E]\times
C^2[J_{+},E])$ is a solution of \eqref{e1}. For $t\in J$,
integrating \eqref{e1} from $0$ to $t$, we have
\begin{equation}
\begin{gathered}
-x'(t)+x'(0)=\int_0^tf(s,x(s),x'(s))\,\mathrm{d}s, \\
-y'(t)+y'(0)=\int_0^tg(s,x(s),x'(s))\,\mathrm{d}s.
\end{gathered}\label{e23}
\end{equation}
Taking the limit as $t\to\infty$, we obtain
\begin{equation}
\begin{gathered}
-x_{\infty}+x'(0)=\int_0^{\infty}f(s,x(s),x'(s))\,\mathrm{d}s,\\
-y_{\infty}+y'(0)=\int_0^{\infty}g(s,x(s),x'(s))\,\mathrm{d}s,
\end{gathered} \label{e24}
\end{equation}
Thus,
\begin{equation}
\begin{gathered}
x'(0)=x_{\infty}+\int_0^{\infty}f(s,x(s),x'(s))\,\mathrm{d}s,\\
y'(0)=y_{\infty}+\int_0^{\infty}g(s,x(s),x'(s))\,\mathrm{d}s.
\label{e25}
\end{gathered}
\end{equation}
We obtain
\begin{gather}
\begin{gathered}
 x'(t)= x_{\infty}+\int_0^{\infty}f(s,x(s),x'(s))\,\mathrm{d}s
 -\int_0^tf(s,x(s),x'(s))\,\mathrm{d}s,\\
y'(t)= y_{\infty}+\int_0^{\infty}g(s,x(s),x'(s))\,\mathrm{d}s
 -\int_0^tg(s,x(s),x'(s))\,\mathrm{d}s;
\end{gathered}\label{e26}\\
x'(t)=x_{\infty}+\int_t^{\infty}f(s,x(s),x'(s))\,\mathrm{d}s,
\label{e27}\\
y'(t)=y_{\infty}+\int_t^{\infty}g(s,x(s),x'(s))\,\mathrm{d}s.
\label{e28}
\end{gather}
Integrating \eqref{e27} and \eqref{e28} from $0$ to $t$, we obtain
\begin{gather}
x(t)=x(0)+tx_{\infty}+\int_0^{\infty}G(t,s)f(s,x(s),x'(s))
\,\mathrm{d}s,
\label{e29}\\
y(t)=y(0)+ty_{\infty}+\int_0^{\infty}G(t,s)g(s,x(s),x'(s))
 \,\mathrm{d}s,\label{e30}
\end{gather}
which together with the boundary-value condition implies that
\begin{equation}
\begin{split}
x(0)&= \int_0^{\infty}q(t)x(t)\,\mathrm{d}t=x(0)
 \int_0^{\infty}q(t)\,\mathrm{d}t+x_{\infty}
 \int_0^{\infty}tq(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)f(s,x(s),x'(s))
 \,\mathrm{d}s \,\mathrm{d}t,
\\
y(0)&= \int_0^{\infty}h(t)x(t)\,\mathrm{d}t=y(0)
 \int_0^{\infty}h(t)\,\mathrm{d}t+y_{\infty}
 \int_0^{\infty}th(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}h(t)G(t,s)g(s,x(s),x'(s))
 \,\mathrm{d}s \,\mathrm{d}t.
\end{split}\label{e31}
\end{equation}
Thus,
\begin{equation}
\begin{split}
 x(0)&= \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{x_{\infty}\int_0^{\infty}tq(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)f(s,x(s),x'(s))\,\mathrm{d}s
\,\mathrm{d}t\Big\},
\end{split} \label{e32}
\end{equation}
\begin{equation}
\begin{split}
y(0)&= \frac{1}{1-\int_0^{\infty}h(t)\,\mathrm{d}t}
 \Big\{y_{\infty}\int_0^{\infty}th(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}h(t)G(t,s)g(s,x(s),x'(s))\,\mathrm{d}s
\,\mathrm{d}t\Big\}.
\end{split} \label{e33}
\end{equation}
Substituting \eqref{e32} and \eqref{e33} in \eqref{e29}
and \eqref{e30},
\begin{equation}
\begin{aligned}
x(t)
&= \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{x_{\infty}\int_0^{\infty}tq(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)f(s,x(s),x'(s))\,\mathrm{d}s
\,\mathrm{d}t\Big\}\\
&\quad +tx_{\infty} +\int_0^{\infty}G(t,s)f(s,x(s),x'(s))
 \,\mathrm{d}s, \\
y(t)
&= \frac{1}{1-\int_0^{\infty}h(t)\,\mathrm{d}t}
\Big\{y_{\infty}\int_0^{\infty}th(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}h(t)G(t,s)g(s,x(s),x'(s))\,\mathrm{d}s
\,\mathrm{d}t\Big\}\\
&\quad +ty_{\infty} +\int_0^{\infty}G(t,s)g(s,x(s),x'(s))
 \,\mathrm{d}s.
\end{aligned} \label{e34}
\end{equation}
Obviously, the integral
$\int_0^t\int_s^{\infty}f(\tau,x(\tau),x'(\tau),y(\tau),y'(\tau))
\mathrm{d}\tau
\mathrm{d}s$ and the integral\\
$\int_0^t\int_s^{\infty}g(\tau,x(\tau),x'(\tau),y(\tau),y'(\tau))\mathrm{d}\tau
\mathrm{d}s$ are convergent. Therefore, $(x,y)$ is a fixed point of
operator $A$.
Conversely, if $(x,y)$ is fixed point of operator $A$, then direct
differentiation gives the proof.
\end{proof}

\begin{lemma}\label{lem3}
 Let {\rm (H1)} be satisfied, and $V$  be a bounded set of $Q$.
Then $\frac{(A_iV)(t)}{1+t}$ and $(A'_iV)(t)$ are
equicontinuous on any finite subinterval of $J$; and for any
$\epsilon>0$, there exists $N=\max\{N_1,N_2\}>0$ such that
$$
\|\frac{A_i(x,y)(t_1)}{1+t_1}-\frac{A_i(x,y)(t_2)}{1+t_2}\|
<\epsilon, \quad \|A'_i(x,y)(t_1)-A'_i(x,y)(t_2)\|<\epsilon
$$
uniformly with respect to $(x,y)\in V$ as $t_1, t_2\geq N$.
\end{lemma}

\begin{proof}
 We only give the proof for operator $A_1$.
For $(x,y)\in V$, $t_2>t_1$, we have
\begin{equation}
\begin{split}
&\|\frac{A_1(x,y)(t_1)}{1+t_1}-\frac{A_1(x,y)(t_2)}{1+t_2}\|\\
&\leq\big|\frac{1}{1+t_1}-\frac{1}{1+t_2}\big|
 \Big(\frac{\int_0^\infty tq(t)\,\mathrm{d}t}{1-\int_0^\infty q(t)
 \,\mathrm{d}t}\Big)\|x_{\infty}\|\\
&\quad +\big|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}\big|\|x_{\infty}\|
 +\Big(1+\frac{\int_0^\infty q(t)\,\mathrm{d}t}{1-\int_0^\infty q(t)
 \,\mathrm{d}t}\Big)\\
&\quad\times \Big\{\|\frac{t_1}{1+t_1}\int_{t_1}^{\infty}
f(s,x(s),x'(s),y(s),y'(s))\mathrm{d}s\\
&\quad -\frac{t_2}{1+t_2}\int_{t_2}^{\infty}
f(s,x(s),x'(s),y(s),y'(s)\mathrm{d}s\|\\
&\quad +\|\int_0^{t_1}\frac{s}{1+t_1}f(s,x(s),x'(s),y(s),y'(s)
 \mathrm{d}s\\
&\quad -\int_0^{t_2}
 \frac{s}{1+t_2}f(s,x(s),x'(s),y(s),y'(s)\mathrm{d}s\|\Big\}\\
&\leq \big|t_1-t_2\big|\Big(1+\frac{\int_0^\infty tq(t)\,\mathrm{d}t}
 {1-\int_0^\infty q(t)\,\mathrm{d}t}\Big)
 \|x_{\infty}\|+\Big(1+\frac{\int_0^\infty q(t)\,\mathrm{d}t}
 {1-\int_0^\infty q(t)\,\mathrm{d}t}\Big)\\
&\quad\times \Big\{\big|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}\big|\,
\|\int_0^{\infty}f(s,x(s),x'(s),y(s),y'(s)\mathrm{d}s\|\\
&\quad +\|\int_{t_1}^{t_2}sf(s,x(s),x'(s),y(s),y'(s)\mathrm{d}s\|\\
&\quad +\frac{t_2}{1+t_2}\|\int_{t_1}^{t_2}f(s,x(s),x'(s),y(s),y'(s)
 \mathrm{d}s\|\\
&\quad +\big|\frac{1}{1+t_1}-\frac{1}{1+t_2}\big|\,\|\int_0^{t_1}s
f(s,x(s),x'(s),y(s),y'(s)\mathrm{d}s\|\\
&\quad +\big|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}\big|\,\|\int_0^{t_1}
f(s,x(s),x'(s),y(s),y'(s)\mathrm{d}s\|\Big\}.
\end{split}\label{e35}
\end{equation}
Then, it is easy to see that by the above inequality and (H1),
$\{\frac{A_1V(t)}{1+t}\} $ is equicontinuous on any finite
subinterval of $J$.

Since $V\subset Q\ $ is bounded, there exists $r>0$ such that for
any $(x,y)$ in $V$, $\|(x,y)\|_X\leq r$. By \eqref{e13}, we obtain
\begin{equation}
\begin{split}
&\|A'_1(x,y)(t_1)-A'_1(x,y)(t_2)\|\\
&= \|\int_{t_1}^{t_2}f(s,x(s),x'(s),y(s),y'(s))\mathrm{d}s\|\\
&\leq \int_{t_1}^{t_2}[4\epsilon_0rc_0(s)(1+s)+a_0(s)+M_0b_0(s)]
 \mathrm{d}s.
\end{split}\label{e36}
\end{equation}
It follows from \eqref{e36}, (H1), and the absolute continuity of
Lebesgue integral that $\{A'_1V(t)\} $ is equicontinuous on any
finite subinterval of $J$.

We are in position to show that for any
$\epsilon>0$, there exists $N_1>0$ such that
$$
\|\frac{A_1(x,y)(t_1)}{1+t_1}-\frac{A_1(x,y)(t_2)}{1+t_2}\|<\epsilon,
\quad \|A'_1(x,y)(t_1)-A'_1(x,y)(t_2)\|<\epsilon,
$$
uniformly with respect to $x\in V$ as $t_1, t_2\geq N_1$.
Combining this with \eqref{e35}, we need only to show that for any
$\epsilon>0$, there exists  sufficiently large $N_1>0$ such that
\begin{align*}
&\|\int_0^{t_1}\frac{s}{1+t_1}f(s,x(s),x'(s),y(s),y'(s))\mathrm{d}s\\
&-\int_0^{t_2}\frac{s}{1+t_2}
f(s,x(s),x'(s),y(s),y'(s))\mathrm{d}s\|<\epsilon,
\end{align*}
for all $x\in V$ as $t_1, t_2\geq N_1$. The rest part of the proof
is very similar to \cite[Lemma 2.3]{l3}, we omit the details.

The proof for operator $A_2$ can be given in a similar way.
Then the proof is complete.
\end{proof}

\begin{lemma}\label{lem4}
 Let {\rm (H1)} be satisfied, $V$ be a bounded
set in $DC^1[J,E]\times DC^1[J,E]$. Then
$$
\alpha_D(A_iV)=\max\big\{ \sup_{t\in J}\alpha
\Big(\frac{(A_iV)(t)}{1+t}\Big),\;
\sup_{t\in J}\alpha((A_iV)'(t))\big\}\quad (i=0,1).
$$
\end{lemma}

The proof of the above lemma is similar to that of
\cite[Lemma 2.4]{l3}, we omit it.

\begin{lemma}[M\"onch Fixed-Point Theorem \cite{d1,g2}] \label{lem5}
Let $Q$ be a closed convex set of $E$ and $u\in Q$.
Assume that the continuous operator $F : Q \to Q $ has the
following property: $V \subset Q $
countable, $V\subset \overline{\rm co}(\{u\}\cup F(V))\Rightarrow V$
is relatively compact. Then $F$ has a fixed point in $Q$.
\end{lemma}

\begin{lemma}\label{lem6}
 If {\rm (H3)} is satisfied, then for $x,y\in Q$,
$x^{(i)}\leq y^{(i)}$, $t\in J$ $(i=0,1)$ imply
 $(Ax)^{(i)}\leq (Ay)^{(i)}$, $t\in J$ $(i=0,1)$.
\end{lemma}

It is easy to see that the above lemma follows from
\eqref{e4} \eqref{e5} \eqref{e13} and condition (H3).


\begin{lemma}[\cite{g4}] \label{lem7}
Let $D$ and $F$ be bounded sets in $E$, then
$$
\widetilde{\alpha}(D\times F)=\max\{\alpha(D), \alpha(F)\},
$$
where $\widetilde{\alpha}$ and $\alpha$ denote the Kuratowski
measure of non-compactness in $E\times E$ and $E$, respectively.
\end{lemma}

\begin{lemma}[\cite{g4}] \label{lem8}
  Let $P$ be normal (fully regular) in $E$, $\widetilde{P}=P\times
P$, then $\widetilde{P}$ is  normal  (fully regular) in $E\times E$.
\end{lemma}

\section{Main results}

\begin{theorem}\label{thm1}
Assume {\rm (H1), (H2)} and that
 $2D^*\cdot G^*<1$. Then  \eqref{e1} has a positive solution
$(\overline{x},\overline{y})\in (DC^1[J, E]\cap C^2[J_+,E])\times
(DC^1[J, E]\cap C^2[J_+,E])$ satisfying
$(\overline{x})^{(i)}(t)\geq \lambda_0^*x_0^*,
(\overline{y})^{(i)}(t)\geq\lambda_1^* y_0^*$ for $t\in J\ (i=0,1)$.
\end{theorem}

\begin{proof}
By Lemma \ref{lem1}, the operator $A$ defined by \eqref{e3} is
a continuous operator from $Q$ to $Q$. By Lemma \ref{lem2},
we need only to show that $A$ has a fixed point
$(\overline{x},\overline{y})$ in $Q$.
Choose $R>2\gamma$ and let
$Q^*=\{(x,y)\in Q:\|(x,y)\|_X\leq R\}$. Obviously, $Q^*$ is a
bounded closed convex set in space $DC^1[J, E]\times DC^1[J, E]$.
It is easy to see that $Q^*$ is not empty since
$((1+t)x_{\infty},(1+t)y_{\infty})\in Q^*$. It follows from
\eqref{e17} \eqref{e18} that $(x,y)\in Q^*$ implies that
$A(x,y)\in Q^*$; i.e., $A$ maps $Q^*$ to $Q^*$.  Let
$V=\{(x_m,y_m):m=1,2,\dots\}\subset Q^*$ satisfying
$V\subset\overline{co}\{\{(u_0,v_0)\}\cup AV\}$ for some
$(u_0,v_0)\in Q^*$. Then $\|(x_m,y_m)\|_X\leq R$. By
\eqref{e4} and \eqref{e13}, we have
\begin{equation}
\begin{split}
&A_1(x_m,y_m)(t)\\
&= \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{\int_0^{\infty}\int_0^{\infty}q(t)
G(t,s)f(s,x_m(s),x'_m(s),y_m(s),y'_m(s))\,\mathrm{d}s\,\mathrm{d}t \\
&\quad +x_{\infty}\int_0^{\infty}tq(t)\,\mathrm{d}t\Big\}
+tx_{\infty}+\int_0^{\infty}G(t,s)f(s,x_m(s),x'_m(s),y_m(s),y'_m(s))
\,\mathrm{d}s,\\
\end{split}\label{e37}
\end{equation}
and
\begin{equation}
A'_1(x_m,y_m)(t)=\int_t^{\infty}f(s,x_m(s),x'_m(s),y_m(s),y'_m(s))
\mathrm{d}s+x_{\infty}.\label{e38}
\end{equation}
By Lemma \ref{lem4}, we have
\begin{equation}
\alpha_D(A_1V)=\max\Big\{\sup_{t\in J}\alpha((A_1V)'(t)),\;
\sup_{t\in J}\alpha
\Big(\frac{(A_1V)(t)}{1+t}\Big)\Big\},\label{e39}
\end{equation}
where
$(A_1V)(t)=\{A_1(x_m,y_m)(t):m=1,2,\dots\}$,
$(A_1V)'(t)=\{A'_1(x_m,y_m)(t):m=1,2,\dots\}$.

By \eqref{e10}, we know that the infinite integral
$\int_0^{\infty}\|f(t,x(t),x'(t),y(t),y'(t))\|\mathrm{d}t$ is
convergent uniformly for for $m=1,2,3,\dots$.  So, for any
$\epsilon>0$, we can choose a sufficiently large
$T > \xi_i$ $(i=1,2,\dots, m-2)>0$ such that
\begin{equation}
\int_T^{\infty}\|f(t,x(t),x'(t),y(t),y'(t))\|\mathrm{d}t<\epsilon.
\label{e40}
\end{equation}
Then, by Guo et al. \cite[Theorem 1.2.3]{y1},  \eqref{e37}, \eqref{e38},
\eqref{e40},  (H2), and Lemma \ref{lem7}, we obtain
\begin{equation}
\begin{split}
&\alpha\Big(\frac{(A_1V)(t)}{1+t}\Big)\\
&\leq \frac{D_0}{1+t} \Big\{2\int_0^T\alpha(f(t,x_m(t),x'_m(t),y_m(t),
 y'_m(t))\,\mathrm{d}t+2\varepsilon\Big\} \\
&\leq 2D_0\int_0^{\infty}\alpha(f(t,x_m(t),x'_m(t),y_m(t),y'_m(t))
 \,\mathrm{d}t+2\varepsilon\\
&\leq 2D_0\alpha_X(V)\int_0^\infty (L_{00}(s)+K_{00}(s))(1+s)
 +(L_{01}(s)+K_{01}(s))\,\mathrm{d}t +2\varepsilon.\\
&\leq 2D_0G_0^*\alpha_X(V)+2\varepsilon,
\end{split}\label{e41}
\end{equation}
and
\[
\alpha({(A_1V)'(t)})\leq2\int_0^{\infty}
\alpha(f(t,x_m(t),x'_m(t),y_m(t),y'_m(t))\,\mathrm{d}s+2\varepsilon
\leq 2G_0^*\alpha_X(V) +2\varepsilon.
\] %\label{42}
From this inequality, \eqref{e39}  and\eqref{e41}, it follows
that
\begin{equation}
\alpha_D(A_1V)\leq2D_0\alpha_X(V)G_0^*.\label{e43}
\end{equation}
In the same way, we obtain
\begin{equation}
\alpha_D(A_2V)\leq2D_1\alpha_X(V)G_1^*.\label{e44}
\end{equation}
On the other hand, $\alpha_X(V )
 \leq \alpha_X\{\overline{\rm co}(\{u\}\cup(AV))\}
= \alpha_X(AV )$. Then, \eqref{e43}, \eqref{e44}, (H2),
and Lemma \ref{lem7} imply $\alpha_X(V) = 0$; i.e.,
$V$ is relatively compact in $DC^1[J,E]\times DC^1[J,E]$. Hence, the
M\"onch fixed point theorem guarantees that $A$ has a fixed point
$(\overline{x},\overline{y})$ in $Q_1$.
\end{proof}

\begin{theorem}\label{thm2}
Let cone $P$ be normal and conditions
{\rm (H1)--(H3)} be satisfied.  Then  \eqref{e1}
has a positive solution $(\overline{x},\overline{y}) \in Q\cap
(C^2[J_+ ,E]\times C^2[J_+ ,E])$ which is minimal in the sense
that $u^{(i)}(t)\geq \overline{x}^{(i)}(t),v^{(i)}(t)\geq
\overline{y}^{(i)}(t)$, $t \in J$ $(i = 0, 1)$ for any positive
solution $(u,v) \in Q\cap(C^2[J_+ ,E]\times C^2[J_+ ,E])$ of
\eqref{e1}. Moreover, $\|(\overline{x},\overline{y})\|_X \leq
2\gamma+\|(u_0,v_0)\|_X$, and there exists a monotone iterative
sequence $\{(u_m(t),v_m(t))\}$ such that $u^{(i)}_ m (t)\to
\overline{x}^{(i)}(t), v^{(i)}_ m (t)\to \overline{y}^{(i)}(t)$ as
$m\to\infty (i = 0, 1)$ uniformly on $J$ and $u_m''(t)\to
\overline{x}''(t), v_m''(t)\to \overline{y}''(t)$ as
$m\to\infty$ for any $t \in  J_+$, where
\begin{equation}
\begin{split}
&u_0(t)\\
&= \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{x_{\infty}\int_0^{\infty}tq(t)\,\mathrm{d}t
+\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)f(s,\lambda_0^*x_0^*,
\lambda_0^*x_0^*,\lambda_1^*y_0^*,\\
&\quad \lambda_1^*y_0^*)\,\mathrm{d}s\,\mathrm{d}t\Big\}
+tx_{\infty}+\int_0^{\infty}G(t,s)f(s,\lambda_0^*x_0^*,
\lambda_0^*x_0^*,\lambda_1^*y_0^*,\lambda_1^*y_0^*)\,\mathrm{d}s,
\end{split}\label{e45}
\end{equation}
\begin{equation}
\begin{split}
&v_0(t)\\
&= \frac{1}{1-\int_0^{\infty}h(t)\,\mathrm{d}t}
\Big\{y_{\infty}\int_0^{\infty}th(t)\,\mathrm{d}t
 +\int_0^{\infty}\int_0^{\infty}h(t)G(t,s)
 g(s,\lambda_0^*x_0^*,\lambda_0^*x_0^*,\lambda_1^*y_0^*,\\
&\quad \lambda_1^*y_0^*)\,\mathrm{d}s\,\mathrm{d}t\Big\}
 +ty_{\infty}+\int_0^{\infty}G(t,s)g(s,\lambda_0^*x_0^*,
 \lambda_0^*x_0^*,\lambda_1^*y_0^*,\lambda_1^*y_0^*)
 \,\mathrm{d}s,
\end{split}\label{e46}
\end{equation} and
\begin{equation}
\begin{split}
&u_m(t)\\
&= \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{x_{\infty}\int_0^{\infty}tq(t)\,\mathrm{d}t
 +\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)f(s,u_{m-1}(s),u'_{m-1}(s), \\
&\quad v_{m-1}(s),v'_{m-1}(s))\,\mathrm{d}s\,\mathrm{d}t\Big\}
 +tx_{\infty}+\int_0^{\infty}G(t,s)f(s,u_{m-1}(s),u'_{m-1}(s),
 v_{m-1}(s),\\
&\quad v'_{m-1}(s))\,\mathrm{d}s,\quad \forall t\in J (m=1,2,3,\dots),
\end{split} \label{e47}
\end{equation}
\begin{equation}\begin{split}
&v_m(t)\\
&= \frac{1}{1-\int_0^{\infty}h(t)\,\mathrm{d}t}
\Big\{y_{\infty}\int_0^{\infty}th(t)\,\mathrm{d}t
+\int_0^{\infty}\int_0^{\infty}h(t)G(t,s)g(s,u_{m-1}(s),u'_{m-1}(s), \\
&\quad v_{m-1}(s),v'_{m-1}(s))\,\mathrm{d}s\,\mathrm{d}t\Big\}
+ty_{\infty}+\int_0^{\infty}G(t,s)g(s,u_{m-1}(s),u'_{m-1}(s),
 v_{m-1}(s),\\
&\quad v'_{m-1}(s))\,\mathrm{d}s,\quad \forall
 t\in J (m=1,2,3,\dots).
\end{split}\label{e48}
\end{equation}
\end{theorem}

\begin{proof}
From \eqref{e45},\eqref{e46} one sees that
$(u_0,v_0)\in C[J,E]\times  C[J,E]$ and
\begin{equation}
u_0'(t)=\int_t^{+\infty}f(s,\lambda_0^*x_0^*,\lambda_0^*x_0^*,
\lambda_0^*y_0^*,\lambda_0^*y_0^*)\mathrm{d}s+x_{\infty}.
\label{e49}
\end{equation}
By \eqref{e45} and \eqref{e49}, we have
$u_0^{(i)}\geq \lambda_0^*x_{\infty}\geq \lambda_0^*x_0^*\ (i=0,1)$
and
\begin{align*}
&\frac{\|u_0(t)\|}{1+t}\\
&\leq \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)\|f(s,
\lambda_0^*x_0^*,\lambda_0^*x_0^*,\lambda_1^*y_0^*,
\lambda_1^*y_0^*)\|\,\mathrm{d}s\,\mathrm{d}t \\
&\quad +\|x_{\infty}\|\int_0^{\infty}tq(t)\,\mathrm{d}t\Big\}
 +t\|x_{\infty}\|+\int_0^{\infty}G(t,s)\|f(s,\lambda_0^*x_0^*,
 \lambda_0^*x_0^*,\lambda_1^*y_0^*,\lambda_1^*y_0^*)\|\,\mathrm{d}s,\\
&\leq\Big(1+\frac{\int_0^\infty g(t)\,\mathrm{d}t}
 {1-\int_0^\infty g(t)\,\mathrm{d}t}\Big)
\int_0^{\infty}a_0(s)+b_0(s)z_0(\|\lambda_0^*x_0^*\|,\|
\lambda_0^*x_0^*\|,\|\lambda_0^*y_0^*\|,\|\lambda_0^*y_0^*\|
\mathrm{d}s\\
&\quad +\Big(1+\frac{\int_0^\infty tg(t)\,\mathrm{d}t}
 {1-\int_0^\infty g(t)\,\mathrm{d}t}\Big)\|x_{\infty}\|,
\end{align*} %\label{50}
\begin{align*}
\|u_0'(t)\|
&\leq \int_t^{\infty}\|f(s,\lambda_0^*x_0^*,\lambda_0^*x_0^*,
 \lambda_1^*y_0^*,\lambda_1^*y_0^*)\|\mathrm{d}\tau+\|x_{\infty}\|\\
&\leq \int_0^{\infty}a_0(s)+b_0(s)h_0(\|\lambda_0^*x_0^*\|,
 \|\lambda_0^*x_0^*\|,\|\lambda_0^*y_0^*\|,\|\lambda_0^*y_0^*\|)
 \mathrm{d}s+\|x_{\infty}\|,
\end{align*}
which implies  $\|u_0\|_D<\infty$. Similarly, we have
$\|v_0\|_D<\infty$. Thus,
 $(u_0,v_0)\in DC^1[J,E]\times DC^1[J,E]$.
It follows from \eqref{e4} and \eqref{e47} that
\begin{equation}
(u_m,v_m)(t)=A(u_{m-1},v_{m-1})(t), \quad \forall t\in J,\;
m=1,2,3,\dots.\label{e50}
\end{equation}
By Lemma \ref{lem1}, we obtain
$(u_m,v_m)\in Q$ and
\begin{equation}
\|(u_m,v_m)\|_X=\|A(u_{m-1},v_{m_1})\|_X\leq
\frac{1}{2}\|(u_{m-1},v_{m-1})\|_X+\gamma.\label{e51}
\end{equation}
By (H3) and \eqref{e50}, we have
\begin{equation}
u_1(t)=A_1(u_0(t),v_0(t))\geq
A_1(\lambda_0^*x_0^*,\lambda_1^*y_0^*)=u_0(t),\quad
 \forall\ t\in J,\label{52}
\end{equation}
and
\begin{equation}
v_1(t)=A_2(u_0(t),v_0(t))\geq
A_2(\lambda_0^*x_0^*,\lambda_1^*y_0^*)=v_0(t),\quad \forall t\in
J.\label{53}
\end{equation}
By induction, we obtain
 \begin{equation}
 (\lambda_0^*x_0^*,\lambda_1^*y_0^*)\leq
(u_0(t),v_0(t))\leq (u_1(t),v_1(t))\leq \dots\leq
(u_m(t),v_m(t))\leq\dots, \label{e54}
\end{equation}
for all $t\in J$.
By induction and Lemma \ref{lem6} and \eqref{e50}, we have
\begin{equation}
(\lambda_0^*x_0^*,\lambda_1^*y_0^*)\leq
(u_0^{(i)}(t),v_0^{(i)}(t))\leq (u_1^{(i)}(t),v_1^{(i)}(t))\leq
\dots\leq (u_m^{(i)}(t),v_m^{(i)}(t))\leq\dots, \label{e55}
\end{equation}
for all $t\in J$.
It follows from \eqref{e51}, by induction, that
\begin{equation}
\begin{split}
\|(u_m,v_m)\|_X &\leq \gamma+\frac{1}{2}\gamma+\dots
 +\big(\frac{1}{2}\big)^{m-1}\gamma
+\big(\frac{1}{2}\big)^m\|(u_0,v_0)\|_X\\
&\leq  \frac{\gamma(1-(\frac{1}{2})^m)}{1-\frac{1}{2}}+\|(u_0,v_0)\|_X\\
&\leq   2\gamma+\|(u_0,v_0)\|_X\quad
(m=1,2,3,\dots).
\end{split}\label{e56}
\end{equation}
Let $K = \{(x,y)\in Q:\|(x,y)\|_X\leq 2\gamma+\|(u_0,v_0)\|_X\}$.
Then $K$ is a bounded closed convex set in space
$DC^1[J,E]\times DC^1[J,E]$ and operator $A$ maps $K$ into $K$.
 Clearly, $K$ is not empty since $(u_0,v_0)\in K$.
Let $W = \{(u_m,v_m):m=0,1,2,\dots\},AW =
\{A(u_m,v_m) : m = 0, 1, 2,\dots\}$. Obviously, $W\subset K$ and
$W = \{(u_0,v_0)\}\cup A(W)$. As in the proof of Theorem \ref{thm1},
we obtain $\alpha_X(AW) = 0$; i.e., $W$ is relatively
compact in $DC^1[J,E]\times DC^1[J,E]$. So, there exists a
$(\overline{x},\overline{y})\in DC^1[J,E]\times DC^1[J,E]$ and a
subsequence $\{(u_{m_j},v_{m_j}) : j = 1, 2, 3,\dots\} \subset W$
such that $\{(u_{m_j},v_{m_j})(t) : j = 1, 2, 3, \dots\}$
converges to $(\overline{x}^{(i)}(t),\overline{y}^{(i)}(t))$
uniformly on $J\ (i = 0, 1)$. Since that $P$ is normal and
$\{(u^{(i)}_ m (t),v^{(i)}_ m (t)) : m = 1, 2, 3, \dots\}$ is
nondecreasing, by Lemma \ref{lem8} it is easy to see that the
entire sequence $\{(u^{(i)}_ m (t),v^{(i)}_ m (t)) : m = 1, 2, 3,
\dots\}$ converges to
$(\overline{x}^{(i)}(t),\overline{y}^{(i)}(t))$ uniformly on
$J$ $(i = 0, 1)$. Since $(u_m,v_m)\in K$ and $K$ is a closed
convex set in space $DC^1[J,E]\times DC^1[J,E]$, we have
$(\overline{x},\overline{y})\in K$. It is clear,
\begin{equation}
f(s,u_m(s),u_m'(s),v_m(s),v_m'(s))\to
f(s,\overline{x}(s),\overline{x}'(s),\overline{y}(s),
\overline{y}'(s)),\label{e57}
\end{equation}
as $m\to\infty$, for all $s\in J_+$. By (H1)
and \eqref{e56}, we have
\begin{equation}
\begin{split}
&\|f(s,u_m(s),u_m'(s),v_m(s),v_m'(s))-
f(s,\overline{x}(s),\overline{x}'(s),\overline{y}(s),
 \overline{y}'(s))\|\\
& \leq 8\epsilon_0c_0(s)(1+s)\|(u_m,v_m)\|_X+2a_0(s)+2M_0b_0(s)\\
&\leq 8\epsilon_0c_0(s)(1+s)(2\gamma+\|(u_0,v_0)\|_X)+2a_0(s)
 +2M_0b_0(s)
\end{split} \label{e58}.
\end{equation}
Noticing \eqref{e57} and \eqref{e58}
and taking limit as $m\to  \infty$ in \eqref{e47}, we obtain
\begin{equation}
\begin{split}
\overline{x}(t)
&= \frac{1}{1-\int_0^{\infty}q(t)\,\mathrm{d}t}
\Big\{x_{\infty}\int_0^{\infty}tq(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}q(t)G(t,s)f(s,\overline{x}(s),
 \overline{x}'(s),\overline{y}(s),\overline{y}'(s))\,\mathrm{d}s
 \,\mathrm{d}t\Big\} \\
&\quad +tx_{\infty}+\int_0^{\infty}G(t,s)f(s,\overline{x}(s),
 \overline{x}'(s),\overline{y}(s),\overline{y}'(s))\,\mathrm{d}s,
\end{split}\label{e59}
\end{equation}
In the same way, taking limit as $m\to  \infty$ in \eqref{e48},
we obtain
\begin{equation}
\begin{split}
\overline{y}(t)
&= \frac{1}{1-\int_0^{\infty}h(t)\,\mathrm{d}t}
\Big\{y_{\infty}\int_0^{\infty}th(t)\,\mathrm{d}t\\
&\quad +\int_0^{\infty}\int_0^{\infty}h(t)G(t,s)g(s,\overline{x}(s),
 \overline{x}'(s),\overline{y}(s),\overline{y}'(s))\,\mathrm{d}s\,
 \mathrm{d}t\Big\} \\
&\quad +ty_{\infty}+\int_0^{\infty}G(t,s)g(s,\overline{x}(s),
 \overline{x}'(s),\overline{y}(s),\overline{y}'(s))\,\mathrm{d}s,
\end{split}\label{e60}
\end{equation}
which together with \eqref{e59} and Lemma \ref{lem2}
implies  that
$(\overline{x},\overline{y})\in K\cap C^2[J_+,E]\times C^2[J_+,E]$
and $(\overline{x}(t),\overline{y}(t))$ is a positive solution of
\eqref{e1}. Differentiating \eqref{e47} twice, we obtain
\[
u''_m(t)=-f(t,u_{m-1}(t),u'_{m-1}(t),v_{m-1}(t),v'_{m-1}(t)),\quad
\forall t\in J_+',\; m=1,2,3,\dots.
\]
Hence, by \eqref{e57}, we obtain
\[
\lim_{m\to\infty}u_m''(t)=-f(t,\overline{x}(t),
\overline{x}'(t),\overline{y}(t),\overline{y}'(t))
=\overline{x}''(t),\quad \forall t\in J_+'.
\]
 Similarly, we have
\[
\lim_{m\to\infty}v_m''(t)
=-g(t,\overline{x}(t),\overline{x}'(t),\overline{y}(t),\overline{y}'(t))
=\overline{y}''(t),\quad \forall\ t\in J_+'.
\]
Let $(m(t),n(t))$ be any positive
solution of \eqref{e1}. By Lemma \ref{lem2}, we have
$(m,n)\in Q$ and $(m(t),n(t)) = A(m,n)(t)$, for $t \in J$.
It is clear that $m^{(i)}(t)\geq\lambda_0^* x_0^*>\theta$,
$n^{(i)}(t)\geq \lambda_1^*y_0^*>\theta$ for any
$t\in J$ $(i = 0, 1)$.
So, by Lemma \ref{lem6}, we have
$m^{(i)}(t)\geq u_0^{(i)}(t)$,
$n^{(i)}(t)\geq v_0^{(i)}(t)$ for any $t \in J$ $(i = 0, 1)$.
Assume that $m^{(i)}(t)\geq u^{(i)}_{ m-1}(t)$,
$n^{(i)}(t)\geq v^{(i)}_{m-1}(t)$ for $ t \in J$,
$m \geq 1$  $(i = 0, 1)$.
From Lemma \ref{lem6} it follows that
\[
(A_1^{(i)}(m,n)(t),A_2^{(i)}(m,n)(t))\geq(A_1^{(i)}(u_{m-1},
v_{m-1}))(t), A_2^{(i)}(u_{m-1},v_{m-1}))(t))\]
 for $t \in J$ $(i = 0, 1)$; i.e.,
$(m^{(i)}(t),n^{(i)}(t))\geq (u^{(i)}_m (t),v^{(i)}_m (t))$ for
$t \in J$ $(i = 0, 1)$. Hence, by induction, we obtain
\begin{equation}
m^{(i)}(t)\geq \overline{x}^{(i)}_m (t), n^{(i)}(t)\geq
\overline{y}^{(i)}_m (t) \quad \forall t \in J\;
 (i = 0,1;m = 0, 1, 2, \dots).
\label{e61}
\end{equation}
Now, taking limits in \eqref{e61}, we obtain
$m^{(i)}(t)\geq \overline{x}^{(i)}(t),n^{(i)}(t)\geq
\overline{y}^{(i)}(t)$ for $t \in J\ (i = 0, 1)$. The proof is
complete.
\end{proof}

\begin{theorem}\label{thm3}
Let cone $P$ be fully regular and conditions
{\rm (H1)} and {\rm (H3)} be satisfied. Then the
conclusion of Theorem \ref{thm2} holds.
\end{theorem}

\begin{proof}
The proof is almost the same as that of Theorem
\ref{thm2}. The only difference is that, instead of using
condition (H2), the conclusion $\alpha_X(W) = 0 $ is
implied directly by \eqref{e55} and \eqref{e56}, the full
regularity of $P$ and Lemma \ref{lem8}.
\end{proof}

\section{An example}

Consider the infinite system of scalar singular second-order
three-point boundary value problems:
\begin{equation}
\begin{gathered}
\begin{aligned}
-x_n''(t)&=\frac{1}{9n^3\sqrt[3]{e^{2t}}(2+5t)^9}\Big(5+y_n(t)
 +x_{2n}'(t)+y_{3n}'(t)\\
&\quad +\frac{2}{3n^2x_n(t)}
 +\frac{4}{7n^5x'_{2n}(t)}\Big)^{1/2}
 +\frac{1}{9\sqrt[6]{t}(1+3t)^2}\ln\big[(1+3t)x_n(t)\big],
\end{aligned}\\
\begin{aligned}
-y_n''(t)&=\frac{1}{7n^4\sqrt[3]{e^{2t}}(4+5t)^8}\Big(6+x_{3n}(t)
  +x_{4n}'(t)+\frac{1}{8n^3y_{2n}(t)}\\
 &\quad +\frac{5}{16n^4y'_{4n}(t)}\Big)^{1/3}
  +\frac{1}{7\sqrt[6]{t}(3+4t)^3}\ln\big[(3+4t)y'_{4n}(t)\big],
\end{aligned}\\
x_n(0)=\int_0^\infty e^{-t^2}x_n(t)\,\mathrm{d}t,\quad
x_n'(\infty)=\frac{1}{n},\\
 y_n(0)=\int_0^\infty\frac{1}{2} e^{\frac{-t^2}{2}}y_n(t)
\,\mathrm{d}t,\quad
y_n'(\infty)=\frac{1}{2n},
\quad ( n=1,2,\dots).
\end{gathered}\label{e62}
\end{equation}

\begin{proposition} \label{prop1}
System \eqref{e62} has a
minimal positive solution $(x_n(t),y_n(t))$ satisfying
$x_n(t),x'_n(t)\geq 1/n$, $y_n(t), y'_n(t)\geq 1/(2n)$
for $0\leq t <+\infty$ $(n= 1, 2, 3, \dots)$.
\end{proposition}

\begin{proof}
Let $E=c_0=\{x=(x_1,\dots,x_n,\dots):x_n\to 0 \}$
with the norm $\|x\|=\sup_n|x_n|$. Obviously, $(E, \|\cdot\|)$ is
a real Banach space. Choose
$P=\{x=(x_n)\in c_0:x_n\geq 0, n=1,2,3,\dots\}$. It is easy
to verify that $P$ is a normal cone in $E$ with normal constant 1.
Now we consider \eqref{e62} which can be regarded as a
boundary-value problem of form \eqref{e1} in $E$ with
$x_{\infty}=(1,\frac{1}{2},\frac{1}{3},\dots)$,
$y_{\infty}=(\frac{1}{2},\frac{1}{4},\frac{1}{6},\dots)$.
In this situation, $x=(x_1,\dots,x_n,\dots)$,
$u=(u_1,\dots,u_n,\dots)$,
$y=(y_1,\dots,y_n,\dots)$,
$v=(v_1,\dots,v_n,\dots)$,
$f=(f_1,\dots,f_n,\dots)$, in which
\begin{equation}
\begin{aligned}
&f_n(t,x,u,y,v)\\
&= \frac{1}{9n^3\sqrt[3]{e^{2t}}(2+5t)^9}\Big(5+y_n+u_{2n}+v_{3n}
+\frac{2}{3n^2x_n} +\frac{4}{7n^5u_{2n}}\Big)^{1/2}\\
&\quad +\frac{1}{9\sqrt[6]{t}(1+3t)^2}\ln[(1+3t)x_n],
\end{aligned}\label{e63}
\end{equation}
and
\begin{equation}
\begin{aligned}
& g_n(t,x,u,y,v))\\
&= \frac{1}{7n^4\sqrt[3]{e^{2t}}(4+5t)^8}\Big(6+x_{3n}+u_{4n}
+\frac{1}{8n^3y_{2n}}
 + \frac{5}{16n^4v_{4n}}\Big)^{1/3}\\
&\quad +\frac{1}{7\sqrt[6]{t}(3+4t)^3}\ln[(3+4t)v_{4n}].
\end{aligned}\label{e64}
\end{equation}
Let $x_0^*=x_{\infty}=(1,\frac{1}{2},\frac{1}{3},\dots)$,
$y_0^*=y_{\infty}=(\frac{1}{2},\frac{1}{4},\frac{1}{6},\dots)$.
Then $P_{0\lambda}=\{x=(x_1,x_2,\dots,x_n,\dots):
x_n\geq\frac{\lambda}{n},\
n=1,2,3,\dots\},P_{1\lambda}=\{y=(y_1,y_2,\dots,y_n,\dots):
y_n\geq\frac{\lambda}{2n},\ n=1,2,3,\dots\}$, for $\lambda>0$.
By a simple computation, we have
$D_0=8.7912$, $D_1=2.6787$,
$\int_0^\infty e^{-t^2}\,\mathrm{d}t\approx0.8863<1$,
$\int_0^\infty te^{-t^2}\,\mathrm{d}t=0.5$,
$\int_0^\infty\frac{1}{2} e^{\frac{-t^2}{2}}\,\mathrm{d}t\approx0.6267<1,$
$\int_0^\infty \frac{1}{2}te^{\frac{-t^2}{2}}\,\mathrm{d}t=0.5$,
$\lambda_0^*=\lambda_1^*=1$. It is
clear, $f,g\in C[J_+\times P_{0\lambda}\times P_{0\lambda}\times
P_{1\lambda}\times P_{1\lambda},P]$ for any $\lambda>0$. Note that
 $\sqrt[3]{e^{2t}}>\sqrt[6]{t}$ for $t>0$ for $t>0$, by \eqref{e63} and
\eqref{e64}, we obtain
\begin{equation}
\begin{split}
\|f(t,x,u,y,v)\|&\leq\frac{1}{9\sqrt[6]{t}(1+3t)^2}\Big\{\Big(
\frac{167}{21}+\|y\|+\|u\|+\|v\|\Big)^{1/2}\\
&\quad +\ln\big[(1+3t)\|x\|\big]\Big\},
\end{split}\label{e65}
\end{equation}
and
\begin{equation}
\|g(t,x,u,y,v)\|\leq\frac{1}{7\sqrt[6]{t}(3+4t)^2}\Big\{\big(
9+\|x\|+\|u\|\big)^{1/3}+\ln\big[(3+4t)\|v\|\big]\Big\},
\label{e66}
\end{equation}
which imply (H1) is satisfied for $a_0(t)=0$,
$b_0(t)=c_0(t)=\frac{1}{9\sqrt[6]{t}(1+3t)^2}$, $a_1(t)=0$,
$b_1(t)=c_1(t)=\frac{1}{7\sqrt[6]{t}(3+4t)^2}$ and
\begin{gather*}
z_0(u_0,u_1,u_2,u_3)=\big(\frac{167}{21}+u_1+u_2+u_3\big)^{1/2}
+\ln[(1+3t)u_0],\\
z_1(u_0,u_1,u_2,u_3)=\big(9+u_0+u_1
\big)^{1/3}+\ln[(3+4t)u_3].
\end{gather*}
Let $f^1=\{f^1_1, f^1_2,\dots, f^1_n,\dots\}$,
$f^2=\{f^2_1, f^2_2,\dots,f^2_n,\dots\}$,
$g^1=\{g^1_1, g^1_2,\dots, g^1_n,\dots\}$,
$g^2=\{g^2_1, g^2_2,\dots, g^2_n,\dots\}$, where
\begin{gather}
f^1_{n}(t,x,u,y,v)=\frac{1}{9n^3\sqrt[3]{e^{2t}}(2+5t)^9}
\Big(5+y_n+u_{2n}+v_{3n}+\frac{2}{3n^2x_n}+
\frac{4}{7n^5u_{2n}}\Big)^{1/2},\label{e67}
\\
f^2_{n}(t,x,u,y,v)=\frac{1}{9\sqrt[6]{t}(1+3t)^2}
\ln[(1+3t)x_n],\label{e68}
\\
g^1_{n}(t,x,u,y,v)=\frac{1}{7n^4\sqrt[3]{e^{2t}}(4+5t)^8}
 \Big(6+x_{3n}+u_{4n}+\frac{1}{8n^3y_{2n}}+
\frac{5}{16n^4v_{4n}}\Big)^{1/3},\label{e69}
\\
g^2_{n}(t,x,u,y,v)=\frac{1}{7\sqrt[6]{t}(3+4t)^3}
\ln[(3+4t)v_{4n}].\label{e70}
\end{gather}
Let $t\in J_+$, and $R>0$, and $\{z^{(m)}\}$ be any
sequence in $f^1(t, P_{0R}^*,P_{0R}^*,P_{1R}^*,P_{1R}^*)$, where
$z^{(m)}=(z^{(m)}_1,\dots,z^{(m)}_n,\dots)$. By \eqref{e67}, we
have
\begin{equation}
0\leq z^{(m)}_n\leq\frac{1}{9n^3\sqrt[3]{e^{2t}}(2+5t)^9}
\big(\frac{167}{21}+ 3R\big)^{1/2}\quad (n,m=1,2,3,\dots).
\label{e71}
\end{equation}
So, $\{z^{(m)}_n\}$ is bounded, and by the diagonal method
we can choose a subsequence $\{m_i\} \subset \{m\}$ such that
\begin{equation}
\{z^{(m)}_n\}\to\overline{z}_n\quad \text{as $i\to\infty$
$(n=1,2,3,\dots)$},\label{e72}
\end{equation}
which  by \eqref{e71} implies
\begin{equation}
0\leq \overline{z}_n\leq\frac{1}{9n^3\sqrt[3]{e^{2t}}(2+5t)^9}
\big(\frac{167}{21}+ 3R\big)^{1/2}\quad (n=1,2,3,\dots).\label{e73}
\end{equation}
Hence $\overline{z}=(\overline{z}_1,\dots,\overline{z}_n,\dots)\in c_0$.
It is easy to see from \eqref{e71}-\eqref{e73} that
\[
\|z^{(m_i)} -\overline{z}\| = \sup_n |z^{(m_i)}_n-\overline{z}_n|
\to 0 \quad \text{as } i \to\infty.
\]
Thus, we have proved that
$f^1(t, P_{0R}^*,P_{0R}^*,P_{1R}^*,P_{1R}^*)$ is relatively
compact in $c_0$.

For any $t\in J_+,  R>0,x,y,\overline{x}, \overline{y}\in D\subset
P_{0R}^*$,  by \eqref{e68} we have
\begin{equation}
\begin{split}
 |f^2_n(t,x,y,u,v)-f^2_n(t,\overline{x},\overline{y},\overline{u},
\overline{v})|
&= \frac{1}{9\sqrt[6]{t}(1+3t)^2}|\ln[(1+3t)x_n]-
\ln[(1+3t)\overline{x}_n]|\\
&\leq \frac{1}{9\sqrt[6]{t}(1+3t)}\frac{|x_n-\overline{x}_n|}{(1+3t)
\xi_n},
\end{split}\label{e74}
\end{equation}
where $\xi_n$ is between $x_n$ and $\overline{x}_n$. By
\eqref{e74}, we obtain
\begin{equation}
\|f^2(t,x,y,u,v)-f^2(t,\overline{x},\overline{y},\overline{u},
\overline{v})\|
\leq\frac{1}{9\sqrt[6]{t}(1+3t)}\|x-\overline{x}\|,\quad
 x, y, \overline{x}, \overline{y}\in D.\label{e75}
\end{equation}
In the same way, we can prove that
$g^1(t, P_{0R}^*,P_{0R}^*,P_{1R}^*,P_{1R}^*)$ is
relatively compact in $c_0$. Also we can obtain
\[
\|g^2(t,x,u,y,v)-g^2(t,\overline{x},\overline{u},\overline{y},
\overline{v})\|
\leq\frac{1}{7\sqrt[6]{t}(3+4t)^2}\|v-\overline{v}\|,\quad
x,y,\overline{x}, \overline{y}\in D. %\label{e76}
\]
From this inequality and \eqref{e75}, it is easy to see that
(H2) holds for
$L_{00}(t)=1/(9\sqrt[6]{t}(1+3t))$,
$L_{10}(t)=1/(7\sqrt[6]{t}(3+4t)^2)$. By a simple computation,
we have $G^*_0\approx0.044$, $G^*_1\approx0.013$,
$2G^*\cdot D^*\approx0.7736<1$. Our conclusion follows from Theorem
\ref{thm1}.
\end{proof}

\subsection*{Acknowledgements}
Supported  by grants 20110491154 from the China
Postdoctoral Science Foundation,  BS2010SF004 from the Foundation
for Outstanding Middle-Aged and Young Scientists of Shandong
Province,  and J10LA53 from  a Project of Shandong Province Higher
Educational Science and Technology Program.
The authors want to thank the anonymous referee for his/her valuable 
comments.


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\end{document}