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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 141, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/141\hfil Existence of positive solutions]
{Existence of positive solutions for boundary value problems
of second-order nonlinear differential equations on the half line}

\author[X. Zhang\hfil EJDE-2009/141\hfilneg]
{Xingqiu Zhang}

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


\thanks{Submitted October 22, 2008. Published October 30, 2009.}
\thanks{Supported by grant 10671167 from the National Natural Science
Foundation of China, \hfill\break\indent
and 31805 from the the Natural Science
Foundation of Liaocheng University}
\subjclass[2000]{34B10, 34B15, 34B18, 34B40}
\keywords{Successive iteration; singular differential
equation; half line; \hfill\break\indent
positive solution}

\begin{abstract}
 In this article, we study the existence of positive solutions
 for Sturm-Liouville boundary-value problems of a second-order
 nonlinear differential equation on the half line.
 Our approach is based on the fixed point theorem and the
 monotone iterative technique. Without  assuming the existence
 of lower and upper solution, we obtain the existence of
 positive solutions, and establish iterative schemes for
 approximating the solutions.
\end{abstract}

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

\section{Introduction}

 In this article, we prove the existence positive solutions,
and establish a an iterative scheme for their approximation, for
the following Sturm-Liouville boundary value problem of second-order
differential equation  on the half line
\begin{equation} \begin{gathered}
x''(t)+q(t)f(t,x(t),x'(t))=0,\quad t\in J_{+},\\
\alpha x(0)-\beta x'(0)=0,\quad x'(\infty)=x_{\infty}\geq0,
\end{gathered} \label{e1}
\end{equation}
 where $J=[0,+\infty)$, $J_+=(0,+\infty)$,
$\alpha>0$, $\beta\geq 0$, $x'(\infty)=\lim_{t\to+\infty}x'(t)$ and
$f(t,u,v):J\times J\times J\to J$ is continuous. Throughout this
paper, we assume the following conditions.
\begin{itemize}
\item[(H1)]   $f(t, u, v)\in C(J\times J\times J,J), f(t,0,0)\not\equiv0$
on any subinterval of $J$ and, when $u, v$ are bounded,
$f(t,(1+t)u,v)$ is bounded  on $J$;

\item[(H2)]  $q(t)$ is a nonnegative measurable function defined in
$J_+$ and $q(t)$ does not identically
vanish on any subinterval of $J_+$ and
$$
0<\int_0^{+\infty}q(t)\mathrm{d}t<+\infty,\quad
0<\int_0^{+\infty}tq(t)\mathrm{d}t<+\infty.
$$
\end{itemize}

Boundary value problems on half-line arise quite naturally in the
study of radially symmetric solutions of nonlinear elliptic
equations and models of gas pressure in a semi-infinite porous
medium; see for example \cite{a1,l4,l5,o1,z1}.
In the past few years, the existence
and multiplicity of bounded or unbounded positive solutions to
nonlinear differential equations on the half line have been studied
by different type of techniques, we refer the reader to
\cite{a1,l1,l3,l4,l5,l6,l7,o1,y1,z1}
and references therein. Most of papers only considered the existence
of positive solutions of various boundary value problems. Seeing
such a fact, we cannot but ask ``How can we find the solutions when
they are known to exist?" More recently, Ma, Du and Ge \cite{m1}, Sun and
Ge \cite{s1,s2} proved the existence of positive solutions for some
second-order $p$-Laplacian boundary value problems which are defined
on finite intervals by virtue of the iterative technique.

To the best of our knowledge, up to now, it seems that no results in
the literature are available for the computation of positive
solutions for boundary value problems on the half line. Motivated by
above papers, the purpose of this paper is to fill this gap. As we
know, it is very important to check the compactness of the
corresponding operator when we use the monotone iterative technique
and Ascoli-Arzela theorem plays a very important role. However,
Ascoli-Arzela theorem is not suitable for operators on the half
line. So, it is needed to list some new conditions to meet the
requirement of compactness.

\section{Preliminaries}

First, we give some  definitions.

\begin{definition} \label{def1} \rm
Let $E$ be a real Banach space. A
nonempty closed set $P\subset E$ is said to be a cone provided that
\begin{itemize}
\item[(1)] $au+bv\in  P$ for all $u, v\in P$ and all $a\geq 0, b\geq 0$;
\item[(2)] $u, -u \in P$ implies $u=0$.
\end{itemize}

A map $\alpha:P\to [0,+\infty)$ is said to be concave  on $P$, if
$$
\alpha(tu + (1-t)v) \geq t\alpha(u)+(1-t)\alpha(v)
$$
for all $u, v \in P$ and $t\in [0,1]$.
\end{definition}

We will use the following space to study \eqref{e1}
$$
E=\big\{x\in C^1[0,+\infty):\sup_{t\in J}
\frac{|x(t)|}{t+1}<\infty,\; \lim_{t\to+\infty}x'(t)\mbox{
exists}\big\}.
$$
 Then $E$ is a Banach space equipped with the
norm $\|x\|=\max\{\|x\|_1, \|x'\|_{\infty}\}$, where
$\|x\|_1=\sup_{t\in J}\frac{|x(t)|}{t+1}$,
$\|x'\|_{\infty}=\sup_{t\in J}|x'(t)|$.
Let $E_+=\{x\in E: x(t)\geq0\}$. Define the cone $P\subset E$
 by
\begin{align*}
P=\big\{& x\in E: x(t)\geq 0, t\in [0,+\infty),
x\  \mbox{is concave on } [0,+\infty),\\
&\alpha x(0)-\beta x'(0)=0, \mbox{ and }
\lim_{t\to+\infty}x'(t)\mbox{ exists}\big\}.
\end{align*}

\begin{remark} \label{rmk1} \rm
If $x$ satisfies \eqref{e1}, then $x''(t)=-q(t)f(t,x(t),x'(t))\leq 0$
on $[0,+\infty)$, which implies that $x$ is concave on
$[0,+\infty)$. Moreover if $x'(\infty)=x_{\infty}\geq 0$, then
$x'(t)\geq 0, t\in [0,+\infty)$ and so $x$ is monotone increasing on
$[0,+\infty)$.
\end{remark}

\begin{lemma}\label{lem1}
 Assume that {\rm (H1)-(H2)} hold. Then
  $x\in E_+\cap C^2[J_+,J]$ is a solution of
\eqref{e1} if and only if $x\in C[J,E]$ is a solution of the
integral equation
\begin{equation}
\begin{aligned}
x(t)&=\frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)
 f(s,x(s),x'(s))\mathrm{d}s+x_{\infty}\Big)\\
&\quad+\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau
\mathrm{d}s+tx_{\infty}.
\end{aligned}\label{e2}
\end{equation}
\end{lemma}

\begin{proof}
 Suppose that $x\in E_+\cap C^2[J_+,J]$ is a solution of
 \eqref{e1}. For $t\in J$, integrating \eqref{e1} from $t$ to
$+\infty$, we have
\begin{equation}
x'(t)=x_{\infty}+\int_t^{+\infty}f(s,x(s),x'(s))\mathrm{d}s.\label{e3}
\end{equation}
Integrating  from 0 to $t$,
\begin{equation}
x(t)=x(0)+tx_{\infty}+\int_0^t\int_s^{+\infty}q(\tau)
f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau
\mathrm{d}s.\label{e4}
\end{equation}
Thus, by \eqref{e3} we obtain
\[
x'(0)=x_{\infty}+\int_0^{+\infty}f(s,x(s),x'(s))\mathrm{d}s,
\]
which together with the boundary value condition implies
\begin{equation}
x(0)=\frac{\beta}{\alpha}\Big(x_{\infty}+\int_0^{+\infty}f(s,x(s),x'(s))
\mathrm{d}s\Big).\label{e5}
\end{equation}
Substituting the above expression into \eqref{e4}, we have
\begin{align*}
x(t)&=\frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)
f(s,x(s),x'(s))\mathrm{d}s+x_{\infty}\Big)\\
&\quad +\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))
\mathrm{d}\tau \mathrm{d}s+tx_{\infty}.
\end{align*}

Next, we show that the integral
$\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau
\mathrm{d}s$ are convergent.
Since $x\in E_+$,  then there exists $r_0$ such that $\|x\|<r_0$.
Set $B_{r_0}=\sup\{f(t,(1+t)u,v)|(t,u,v)\in J\times
[0,r_0]\times[0,r_0]\}$, and we have by exchanging the integral
order
\begin{equation}
\begin{aligned}
\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau
\mathrm{d}s
&\leq\int_0^{+\infty}\int_s^{+\infty}q(\tau)f(\tau,x(\tau),
 x'(\tau))\mathrm{d}\tau\mathrm{d}s\\
&\leq\int_0^{+\infty}sq(s)\mathrm{d}s\cdot B_{r_0}.
\end{aligned} \label{e13}
\end{equation}
 By (H2), we know that
$\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau
\mathrm{d}s$ is convergent. Thus, we have proved that the right term
in \eqref{e2} is well defined.
Conversely, if $x$ is a solution of integral equation, then direct
differentiation gives the proof.
\end{proof}

Now, we define an operator $A:P\to C^1[0,+\infty)$ by
\begin{equation}
\begin{aligned}
(Ax)(t)&=\frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)
f(s,x(s),x'(s))\mathrm{d}s +x_{\infty}\Big)\\
&\quad +\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau
\mathrm{d}s+tx_{\infty}.
\end{aligned}\label{e6}
\end{equation}
 To obtain the complete continuity of $A$, the following lemma is
 needed.

\begin{lemma}[\cite{c1,l6}] \label{l2}
Let $W$ be a bounded subset of $P$. Then
$W$ is relatively compact in $E$ if $\{W(t)/(1+t)\}$ and $\{W'(t)\}$
are both equicontinuous on any finite subinterval of $[0,+\infty)$
and for any $\varepsilon>0$, there exists $N>0$ such that
$$
\Big|\frac{x(t_1)}{1+t_1}-\frac{x(t_2)}{1+t_2}\Big|<\varepsilon,\quad
|x'(t_1)-x'(t_2)|<\varepsilon,\quad \forall\ t_1, t_2\geq N,
$$
uniformly with respect to $x\in W$ as
$t_1,t_2\geq N$, where $W(t)=\{x(t)|x\in W\}$,
$W'(t)=\{x'(t)|x\in W\}$, $t\in [0,+\infty)$.
\end{lemma}

\begin{lemma}\label{l3}
Assume that {\rm (H1)-(H2)} hold. Then $A:P\to P$ is completely
continuous.
\end{lemma}

\begin{proof} It is clear that $(Ax)(t)\geq 0$ for any $x\in P$,
$t\in J$. By \eqref{e6}, we have
\begin{gather}
(Ax)'(t)=\int_t^{+\infty}q(s)f(s,x(s),x'(s))\mathrm{d}s+x_{\infty}\geq
0,\label{e7}\\
(Ax)''(t)=-q(t)f(t,x(t))\leq 0.\label{e8}
\end{gather}
These two inequalities  imply  that $(TP)\subset P$. Now, we prove
that $A$ is continuous and compact respectively. Let $x_n\to x$ as
$n\to\infty$ in $P$, then there exists $r_0$ such that $\sup_{n\in
N\backslash \{0\}}\|x_n\|<r_0$. Let
$B_{r_0}=\sup\{f(t,(1+t)u,v)|(t,u,v)\in J\times [0,r_0]\times
[0,r_0]\}$. By (H2), we have
\begin{equation}
\int_0^{+\infty}q(\tau)|f(\tau,x_n(\tau),x_n'(\tau))
-f(\tau,x(\tau),x'(\tau))|d \tau
\leq 2B_{r_0}\cdot\int_0^{+\infty}q(s)\mathrm{d}s<+\infty,\label{e9}
\end{equation}
and
\begin{equation} \begin{aligned}
&\int_0^t\int_s^{+\infty}q(\tau)|f(\tau,x_n(\tau),x_n'(\tau))
-f(\tau,x(\tau),x'(\tau))|\mathrm{d}\tau \mathrm{d}s\\
&\leq\int_0^{+\infty}\int_s^{+\infty}q(\tau)|f(\tau,x_n(\tau))
-f(\tau,x(\tau),x'(\tau))|\mathrm{d}\tau\mathrm{d}s\\
&\leq 2B_{r_0}\int_0^{+\infty}sq(s)\mathrm{d}s<+\infty.
\end{aligned}\label{e10}
\end{equation}
It follows from \eqref{e6}, \eqref{e9}, \eqref{e10}, and the Lebesgue
dominated convergence theorem that
\begin{align*}
\|Ax_n-Ax\|_1
&= \sup_{t\in J}\Big\{\frac{1}{1+t}\Big|\frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)(f(s,x_n(s),x_n'(s))-f(s,x(s),x'(s)))\mathrm{d}s
\Big)\\
&\quad +\int_0^t\int_s^{+\infty}q(\tau)(f(\tau,x_n(\tau),x_n'(\tau))
-f(\tau,x(\tau),x'(\tau)))\mathrm{d}\tau \mathrm{d}s\Big|\Big\}\\
&\leq \sup_{t\in J}\Big\{\frac{1}{1+t}\frac{\beta}{\alpha}
\Big(\int_0^{+\infty}q(s)|f(s,x_n(s),x_n'(s))-f(s,x(s),x'(s))|\mathrm{d}s
\Big)\Big\}\\
&\quad +\sup_{t\in
J}\Big\{\int_0^t\int_s^{+\infty}q(\tau)|f(\tau,x_n(\tau))
-f(\tau,x(\tau),x'(\tau))|\mathrm{d}\tau \mathrm{d}s\Big\}\to0,
\end{align*}
as $n\to\infty$.
Also
\begin{align*}
\|(Ax_n)'-(Ax)'\|_{\infty}
&= \sup_{t\in J}\Big\{\int_t^{+\infty}q(s)|f(s,x_n(s),x_n'(s))
 -f(s,x(s),x'(s))|\mathrm{d}s\Big\}\\
&\leq 2B_{r_0}\int_0^{+\infty}q(s)\mathrm{d}s<+\infty.
\end{align*}
Therefore, $A$ is continuous.

Let $\Omega$ be any bounded subset of $P$. Then, there exists $r>0$
such that $\|x\|\leq r$ for any $x\in \Omega$. Therefore, we have
\begin{align*}
&\|Ax\|_1\\
&= \sup_{t\in J}\Big\{\frac{1}{1+t}\Big|
\frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)f(s,x(s),x'(s))\mathrm{d}s
+x_{\infty}\Big)\\
&\quad +\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),
x'(\tau))\mathrm{d}\tau \mathrm{d}s+tx_{\infty}\Big|\Big\}\\
&\leq \frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)\mathrm{d}s\cdot
B_r+x_{\infty}\Big)+\sup_{t\in
J}\Big\{\frac{t}{1+t}\int_0^{+\infty}q(s)f(s,x(s),x'(s))\mathrm{d}s\Big\}
 +x_{\infty}\\
&\leq \Big(\frac{\beta}{\alpha}+1\Big)
 \Big(\int_0^{+\infty}q(s)\mathrm{d}s\cdot
B_r+x_{\infty}\Big),
\end{align*}
 and
\[
\|(Ax)'\|_{\infty}=\sup_{t\in
J}\Big\{\Big|\int_t^{+\infty}q(s)f(s,x(s),x'(s))\mathrm{d}s
+x_{\infty}\Big|\Big\}
\leq \int_0^{+\infty}q(s)\mathrm{d}s\cdot B_r+x_{\infty}.
\]
So, $T\Omega$ is bounded. Remembering that the integral
$\int_0^{+\infty}\int_s^{+\infty}q(\tau)\mathrm{d}\tau \mathrm{d}s$
is convergent, so for any $T\in J_+$ and $t_1,t_2\in [0,T]$, by the
absolute continuity of the integral, we have
\begin{align*}
&\Big|\frac{(Ax)(t_1)}{1+t_1}-\frac{(Ax)(t_2)}{1+t_2}\Big|\\
&\leq \frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)
f(s,x(s),x'(s))\mathrm{d}s
+x_{\infty}\Big)\cdot\Big|\frac{1}{1+t_1} -\frac{1}{1+t_2}\Big|\\
&\quad +\Big|\frac{1}{1+t_1}\int_0^{t_1}\int_s^{+\infty}
 q(\tau)f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau \mathrm{d}s\\
&\quad -\frac{1}{1+t_2}\int_0^{t_2}\int_s^{+\infty}q(\tau)
 f(\tau,x(\tau),x'(\tau))\mathrm{d}\tau \mathrm{d}s\Big|\\
&\quad +\Big|\frac{t_1}{1+t_1}-\frac{t_1}{1+t_1}\Big|x_{\infty}\\
&\leq \frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)\mathrm{d}s
 \cdot B_r+x_{\infty}\Big)\Big|\frac{1}{1+t_1}-\frac{1}{1+t_2}\Big|
 +\frac{1}{1+t_1}\Big|\int_{t_1}^{t_2}\int_s^{+\infty}q(\tau)
 \mathrm{d}\tau \mathrm{d}s\cdot B_r\Big|\\
&\quad +\int_0^{t_2}\int_s^{+\infty}q(\tau)\mathrm{d}\tau
\mathrm{d}s\cdot B_r\Big|\frac{1}{1+t_1}-\frac{1}{1+t_2}\Big|
+\Big|\frac{t_1}{1+t_1}-\frac{t_1}{1+t_1}\Big|x_{\infty}
\end{align*}
which approaches zero, uniformly as $t_1\to t_2$.
Also
\[
|(Ax)'(t_1)-(Ax)'(t_2)|
= \Big|\int_{t_1}^{t_2}q(s)f(s,x(s),x'(s))\mathrm{d}s\Big|
\leq B_r\cdot\Big|\int_{t_1}^{t_2}q(s)\mathrm{d}s\Big|
\]
which approaches zero, uniformly as $t_1\to t_2$.
Thus, we have proved that $T\Omega$ is equicontinuous on any finite
subinterval of $[0,+\infty)$.

Next, we prove that for any  $\varepsilon>0$, there exits
sufficiently large $N>0$ such that
\begin{equation}
\Big|\frac{(Ax)(t_1)}{1+t_1}-\frac{(Ax)(t_2)}{1+t_2}\Big|<\varepsilon,\quad
|(Ax)'(t_1)-(Ax)'(t_2)|<\varepsilon, \label{e11}
\end{equation}
for all $t_1, t_2\geq N$ and all $x\in\Omega$.
 For any $x\in\Omega$, we have
\begin{equation}
\lim_{t\to\infty}\Big|\frac{(Ax)(t)}{1+t}\Big|
=\lim_{t\to\infty}\frac{1}{1+t}
\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))\mathrm{d}
\tau\mathrm{d}s+x_{\infty}.\label{e12}
\end{equation}
Similar to \eqref{e13}, we get
$$
\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))
\mathrm{d}\tau\mathrm{d}s
\leq B_r\int_0^{+\infty}\tau q(\tau)\mathrm{d}\tau<+\infty,
$$
which shows that
\begin{equation}
\lim_{t\to\infty}\frac{1}{1+t}
\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))
\mathrm{d}\tau\mathrm{d}s=0.\label{e14}
\end{equation}
On the other hand, we arrive at
\begin{equation}
\begin{aligned}
\lim_{t\to\infty}|(Ax)'(t)|
&= \lim_{t\to\infty}\int_t^{+\infty}q(s)f(s,x(s),x'(s))\mathrm{d}s
 +x_{\infty}\\
&\leq  B_r\cdot \lim_{t\to\infty}\int_t^{+\infty}q(s)\mathrm{d}s
 +x_{\infty}=x_{\infty}.\label{e20}
\end{aligned}
\end{equation}
It follows from \eqref{e12},  \eqref{e14}, and \eqref{e20} that
$\big|\frac{(Ax)(t)}{1+t}\big|$ and $|(Ax)'(t)|$ tend to
$x_{\infty}$ uniformly as $t\to\infty$. So, for any $\varepsilon>0$,
there exists $N_1>0$ such that
$$
\Big|\frac{(Ax)(t)}{1+t}-x_{\infty}\Big|
<\frac{\varepsilon}{2},\ \ \forall\ t\geq N_1.
$$
Consequently, for any $t_1,t_2\geq N_1$, we have
\begin{equation}
\Big|\frac{(Ax)(t_1)}{1+t_1}-x_{\infty}\Big|<\frac{\varepsilon}{2},\quad
\Big|\frac{(Ax)(t_2)}{1+t_2}-x_{\infty}\Big|<\frac{\varepsilon}{2}.
\label{e15}
\end{equation}
Similarly, we can prove that there exists $N_2>0$ such that
\begin{equation}
\Big|(Ax)'(t_1)-x_{\infty}\Big|<\frac{\varepsilon}{2},\quad
\Big|(Ax)'(t_2)-x_{\infty}\Big|<\frac{\varepsilon}{2}, \quad
\forall\ t_1, t_2\geq N_2.\label{e21}
\end{equation}
 Choose $N=\max\{N_1, N_2\}$, then
\eqref{e11} can be easily seen by \eqref{e15} and \eqref{e21}.
By Lemma \ref{l2}, we know that $A:P\to P$ is completely continuous.
\end{proof}

\section{Main results}

For notational convenience, we denote
$$
m=\Big(\frac{\beta}{\alpha}+1\Big)x_{\infty}, \quad
n=\frac{\beta}{\alpha}\int_0^{+\infty}q(\tau)\mathrm{d}\tau
+\max\Big\{\int_0^{+\infty}q(\tau)\mathrm{d}\tau,\int_0^{+\infty}\tau
q(\tau)\mathrm{d}\tau\Big\}.
$$
We will prove the following existence results.

\begin{theorem}\label{thm1}
 Assume that {\rm (H1)-(H2)} hold, and there exists $a>2m$ such that
\begin{itemize}
\item[(S1)] $f(t,x_1,y_1)\leq f(t,x_2,y_2)$ for any $0\leq
t<+\infty$, $0\leq x_1\leq x_2$, $0\leq y_1\leq y_2$;
\item[(S2)] $f(t,(1+t)u,v)\leq \frac{a}{2n}$,
$(t,u,v)\in [0,+\infty)\times [0,a]\times [0,a]$.
\end{itemize}
Then the boundary value problem \eqref{e1} has two positive
nondecreasing on $[0,+\infty)$ and concave solutions $w^*$ and
$v^*$, such that $0<\|w^*\|\leq a$, and
$\lim_{n\to\infty}w_n=\lim_{n\to\infty}A^nw_0=w^*$,
where
$$
w_0(t)=\frac{a}{2}(t+1)+\frac{\beta}{\alpha}x_{\infty}
+tx_{\infty},\ \ t\in J,
$$
and $0<\|v^*\|\leq a$,
$\lim_{n\to\infty}v_n=\lim_{n\to\infty}A^nv_0=v^*$,
where  $v_0(t)=0$, $t\in J$.
\end{theorem}

\begin{proof}
By Lemma \ref{l3}, we know that $A:P\to P$ is
completely continuous. For any $x_1, x_2\in P$ with $x_1\leq x_2$,
$x'_1\leq x'_2$, from the definition of $A$ and (S1), we
can easily get that $Ax_1\leq Ax_2$. We denote
$$
\overline{P}_a=\{x\in P:\|x\|\leq a\}.
$$
First, we prove that $A:\overline{P}_a\to
\overline{P}_a$. If $x\in\overline{P}_a$, then $\|x\|\leq a$. By
\eqref{e2}, (S1) and $(\mathrm{S_2})$, we get
\begin{align*}
\|Ax\|_1&= \sup_{t\in J}\Big\{\frac{1}{1+t}\Big|
\frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)f(s,x(s),x'(s))\mathrm{d}s
+x_{\infty}\Big)\\
&\quad +\int_0^t\int_s^{+\infty}q(\tau)f(\tau,x(\tau),x'(\tau))
\mathrm{d}\tau \mathrm{d}s+tx_{\infty}\Big|\Big\}\\
&\leq \Big(\frac{\beta}{\alpha}+1\Big)x_{\infty}
 +\Big(\frac{\beta}{\alpha}+1\Big)\int_0^{+\infty}q(\tau)\mathrm{d}
\tau\cdot \frac{a}{2n}\\
&\leq  m +n\cdot \frac{a}{2n}\leq a,
\end{align*}
and
\begin{align*}
\|(Ax)'\|_{\infty}
&=\sup_{t\in J}\Big\{\Big|\int_t^{+\infty}q(s)f(s,x(s),x'(s))\mathrm{d}s
+x_{\infty}\Big|\Big\}\\
&\leq \int_0^{+\infty}q(s)\mathrm{d}s\cdot  \frac{a}{2n}
+x_{\infty}\leq a.
\end{align*}
Hence, we have proved that $A:\overline{P}_a\to \overline{P}_a$.
Let $w_0(t)=\frac{a}{2}(t+1)+\frac{\beta}{\alpha}x_{\infty}
+tx_{\infty}, 0\leq t<+\infty$, then $w_0(t)\in\overline{P}_a$. Let
$w_1=Aw_0, w_2=A^2w_0$, then by Lemma \ref{l3}, we have that
$w_1\in \overline{P}_a$ and $w_2\in \overline{P}_a$. We denote
$w_{n+1}=Aw_n=A^nw_0, n=0,1,2,\dots$. Since $A:\overline{P}_a\to
\overline{P}_a$, we have $w_n\in A(\overline{P}_a)\subset
\overline{P}_a, n=1,2,3,\dots$. It follows from the complete
continuity of $A$ that $\{w_n\}_{n=1}^{\infty}$ is a sequentially
compact set.

By \eqref{e2} and (S2), we get
\begin{equation}
\begin{aligned}
w_1(t)
&= \frac{\beta}{\alpha}\Big(\int_0^{+\infty}q(s)
 f(s,\omega_0(s),\omega_0'(s))\mathrm{d}s +x_{\infty}\Big)\\
&\quad +\int_0^t\int_s^{+\infty}q(\tau)
 f(\tau,\omega_0(\tau),\omega_0'(\tau))\mathrm{d}\tau
 \mathrm{d}s+tx_{\infty}\\
&\leq \frac{\beta}{\alpha}\int_0^{+\infty}q(s)\mathrm{d}s
 \cdot\frac{a}{2n} +\frac{\beta}{\alpha}x_{\infty}
 +\int_0^{+\infty}\tau q(\tau)\mathrm{d}\tau
 \cdot\frac{a}{2n}+tx_{\infty}\\
&\leq \frac{a}{2}+\frac{\beta}{\alpha}x_{\infty}
+tx_{\infty}\leq w_0(t),\ 0\leq t<+\infty,
\end{aligned} \label{e16}
\end{equation}
and
 \begin{equation}
\begin{aligned}
\omega_1'(t)&= (A\omega_0)'(t)
 =\int_t^{+\infty}q(s)f(s,\omega_0(s),\omega_0'(s))\mathrm{d}s+x_{\infty}\\
&\leq \int_0^{+\infty}q(s)\mathrm{d}s\cdot\frac{a}{2n}+x_{\infty}\\
&\leq \frac{a}{2}+x_{\infty}=\omega_0'(t),\quad
 0\leq t<+\infty.
\end{aligned} \label{e22}
\end{equation}
 So, by \eqref{e16}, \eqref{e22} and (S1) we have
\begin{gather*}
w_2(t)=(Aw_1)(t)\leq (Aw_0)(t)=w_1(t), \quad 0\leq t<+\infty,\\
w_2'(t)=(Aw_1)'(t)\leq (Aw_0)'(t)=(w_1)'(t), \ 0\leq t<+\infty.
\end{gather*}
By induction, we get
$$
w_{n+1}(t)\leq w_n(t),\quad
w'_{n+1}(t)\leq w'_n(t), \quad 0\leq t<+\infty,\; n=0,1,2,\dots.
$$
Thus, there exists $w^*\in\overline{P}_a$ such that $w_n\to w^*$ as
$n\to\infty$. Applying the continuity of $A$ and $w_{n+1}=Aw_n$, we
get that $Aw^*=w^*$.

Let $v_0(t)=0, 0\leq t<+\infty$, then $v_0(t)\in\overline{P}_a$. Let
$v_1=Av_0, v_2=A^2v_0$, then by Lemma \ref{l3}, we have that $v_1\in
\overline{P}_a$ and $v_2\in \overline{P}_a$. We denote
$v_{n+1}=Av_n=A^nv_0, n=0,1,2,\dots$. Since $A:\overline{P}_a\to
\overline{P}_a$, we have $v_n\in A(\overline{P}_a)\subset
\overline{P}_a, n=1,2,3,\dots$. It follows from the complete
continuity of $A$ that $\{v_n\}_{n=1}^{\infty}$ is a sequentially
compact set.

Since $v_1=Av_0\in\overline{P}_a$, we have
\begin{gather*}
v_1(t)=(Av_0)(t) =(A0)(t)\geq 0,\quad 0\leq t<+\infty,\\
v'_1(t)=(Av_0)'(t) =(A0)'(t)=v_0'(t)\geq 0,\quad 0\leq t<+\infty.
\end{gather*}
So, that we have
\begin{gather*}
v_2(t)=(Av_1)(t)\geq (A0)(t)=v_1(t), \quad 0\leq t<+\infty,\\
v'_2(t)=(Av_1)'(t)\geq (A0)'(t)=v_1'(t), \quad 0\leq t<+\infty.
\end{gather*}
By induction, we get
$$
v_{n+1}(t)\geq v_n(t),\quad v'_{n+1}(t)\geq v'_n(t), \quad
 0\leq t<+\infty,\; n=0,1,2,\dots.
$$
Thus, there exists $v^*\in\overline{P}_a$ such that $v_n\to v^*$ as
$n\to\infty$. Applying the continuity of $A$ and $v_{n+1}=Av_n$, we
get that $Av^*=v^*$.

If $f(t,0,0)\not\equiv 0$, $0\leq t<\infty$, then the zero function
is not the solution of  \eqref{e1}. Thus, $v^*$ is a positive
solution of  \eqref{e1}.
It is well known that each fixed point of $A$ in $P$ is a solution
of  \eqref{e1}. Hence, we assert that $w^*$ and $v^*$ are two
positive, nondecreasing on $[0,+\infty)$ and concave solutions of
 \eqref{e1}.
\end{proof}

\begin{remark}  \label{rmk2} \rm
The iterative schemes in Theorem \ref{thm1} are
 $w_0(t)=\frac{a}{2}(t+1)+\frac{\beta}{\alpha}x_{\infty}
+tx_{\infty}, w_{n+1}=Aw_n=A^nw_0, n=0,1,2,\dots$ and $v_0(t)=0,
v_{n+1}=Av_n=A^nv_0, n=0,1,2,\dots$. They start off with a known
simple linear function and the zero function respectively. It is
convenient in application. We can easily get that $w^*$ and $v^*$
are the maximal and minimal solutions of the boundary value problem
\eqref{e1}. Of course $w^*$ and $v^*$ may coincide and then the
boundary value problem \eqref{e1} has only one solution in $P$.
\end{remark}

The following theorem can be obtained directly from Theorem
\ref{thm1}.

\begin{theorem}\label{thm2}
Assume that {\rm (H1)-(H2)} hold and there exists
$2m<a_1<a_2<\dots<a_n$ such that
\begin{itemize}
\item[(S1)] $f(t,x_1,y_1)\leq f(t,x_2,y_2)$ for any $0\leq
t<+\infty$, $0\leq x_1\leq x_2$,  $0\leq y_1\leq y_2$;

\item[(S2)] $f(t,(1+t)u,v)\leq \frac{a_k}{2n}$, $(t,u,v)\in
[0,+\infty)\times [0,a_k]\times [0,a_k]$, $k=1,2,\dots,n$.
\end{itemize}
Then the boundary value problem \eqref{e1} has $2n$ positive
nondecreasing on $[0,+\infty)$ and concave solutions $w_k^*$ and
$v_k^*$, such that $0<\|w_k^*\|\leq a_k$, and
$\lim_{n\to\infty}w_{kn}=\lim_{n\to\infty}A^nw_{k0}=w_k^*$, where
$$
w_{k0}(t)=\frac{a_k}{2}(t+1)+\frac{\beta}{\alpha}x_{\infty}
+tx_{\infty},\quad  t\in J,
$$
and $0<\|v_k^*\|\leq a_k$,
$\lim_{n\to\infty}v_{kn}=\lim_{n\to\infty}A^nv_0=v_k^*$,
where  $v_{k0}(t)=0,\ t\in J$.
\end{theorem}


\section{Example}
Consider the boundary-value problem
 \begin{equation}
\begin{gathered}
 x''(t)+\frac{1}{\sqrt{t}(1+t)^2}f(t,x(t),x'(t))=0,\quad t\in J_{+},\\
2x(0)-3x'(0)=0,\quad x'(\infty)=0,
\end{gathered} \label{e17}
\end{equation}
 where
\[
 f(t,u,v)=\begin{cases}
10^{-2}|\cos(2t+1)|+10^{-2}\Big(\frac{u}{1+t}\Big)^4
  +\frac{1}{10}\Big(\frac{v}{400}\Big),& u\leq3,\\
10^{-2}|\cos(2t+1)|+10^{-2}\Big(\frac{3}{1+t}\Big)^4
  +\frac{1}{10}\Big(\frac{v}{400}\Big),& u\geq3.
\end{cases}
\]
 Set $q(t)=\frac{1}{\sqrt{t}(1+t)^2}$. It is clear that
(H1) and (S2) hold. Let $\alpha=2$,
$\beta=3$, $x_{\infty}=0$. By direct computation, we can obtain that
\begin{gather}
\int_0^{+\infty}q(t)\mathrm{d}t=\int_0^{+\infty}
 \frac{1}{\sqrt{t}(1+t)^2}\mathrm{d}t
<\int_0^1\frac{1}{\sqrt{t}}\mathrm{d}t
 +\int_1^{+\infty}\frac{1}{\sqrt{t}\cdot
t^2}\mathrm{d}t=\frac{8}{3},\label{e18}
\\
\int_0^{+\infty}tq(t)\mathrm{d}t
 =\int_0^{+\infty}\frac{t}{\sqrt{t}(1+t)^2}\mathrm{d}t
<\int_0^1\sqrt{t}\mathrm{d}t+\int_1^{+\infty}\frac{\sqrt{t}}{
t^2}\mathrm{d}t=\frac{8}{3} .\label{e19}
\end{gather}
By \eqref{e18} and \eqref{e19}, we have $m=0$, $n<\frac{20}{3}$.
Choose $a=300$ and check check (S2). Since nonlinear term $f$
satisfies
$$
f(t,(1+t)u,v)\leq \frac{1}{10^2}+\frac{81}{100}+\frac{3}{40}
 =\frac{179}{200}<\frac{300}{2\cdot \frac{20}{3}}<\frac{300}{2n},
$$
for $t\in [0,+\infty)$, $u, v\in [0,300]$; then all  the conditions
in Theorem \ref{thm1} are  satisfied. Therefore, the conclusion of
Theorem \ref{thm1} holds.


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