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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 158, 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-2012/158\hfil Positive solutions for boundary-value problems]
{Positive solutions for boundary-value problems with integral
 boundary conditions on infinite intervals}

\author[C. Yu, J. Wang, Y. Guo, H. Wei\hfil EJDE-2012/158\hfilneg]
{Changlong Yu, Jufang Wang, Yanping Guo, Huixian Wei}  

\address{Changlong Yu \newline
College of Sciences, Hebei University of Science and Technology, Shijiazhuang,
 050018, China}
\email{changlongyu@126.com}

\address{Jufang Wang \newline
College of Sciences, Hebei University of Science and Technology, Shijiazhuang,
 050018,  China}
\email{wangjufang1981@126.com}

\address{Yanping Guo  \newline
College of Sciences, Hebei University of Science and Technology, 
Shijiazhuang, 050018, China}

\address{Huixian Wei \newline
Department of Baise, Shijiazhuang Institute of
Railway Technology, Shijiazhuang, 050041, China}


\thanks{Submitted July 5, 2012. Published September 18, 2012.}
\subjclass[2000]{34B18, 34B15}
\keywords{Cone; Avery-Peterson fixed point theorem; positive solution; 
\hfill\break\indent 
integral boundary conditions; infinite interval}

\begin{abstract}
 In this article, we consider the existence of positive solutions
 for a class of boundary value problems with integral boundary conditions
 on infinite intervals
 \begin{gather*}
 (\varphi_{p}(x'(t)))'+\phi(t)f(t,x(t),x'(t))=0, \quad 0<t<+\infty,\\
 x(0)=\int_0^{+\infty}g(s)x'(s)ds,\quad \lim_{t\to+\infty}x'(t)=0,
 \end{gather*}
 where $\varphi_{p}(s)=|s|^{p-2}s$, $p>1$. By applying the Avery-Peterson
 fixed point theorem in a cone, we obtain the existence of three positive
 solutions to the above boundary value problem and give an example at last.
\end{abstract}

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

\def\a{\sum _{i=1}^{m-2}k_{i}u(\xi _{i})}
\def\s3{\sum _{i=1}^{m-2}k_{i}}
\def\w{\sum _{i=1}^{m-2}k_{i}w(\xi _{i})}
\def\aa{p(t)\psi _n(\alpha ^{'}(t) )}
\def\c{\overline{P(\gamma ,c)}}

\section{Introduction}

Boundary value problems (BVPs) on
infinite intervals often appear in applied mathematics and physics
and so on. The existence and multiplicity of positive solutions for
such problems have become an important area of investigation in
recent years. There are many papers concerning the existence of
solutions on the half-line for the boundary value problems, see
\cite{a1,a2,b1,b2,g2,j1,l1,l2,m1,y1,z1}
and the references therein.

At the same time, we notice that a class of BVPs with integral
boundary conditions appeared in heat conduction, chemical
engineering, underground water flow, thermo-elasticity, and plasma
physics. Such problems include two-, three-, multi-point and nonlocal
BVPs as special cases and attract more attention see 
\cite{g1,k1,l4} and the
references therein. For more information about the general theory of
integral equations and their relation with BVPs, we refer to the
book of Corduneanu \cite{c1} and Agarwal and O'Regan \cite{a2}.


Recently,  Lian et al \cite{l2}, studied the existence of positive
solutions for the  boundary-value problem with a
p-Laplacian operator
\begin{gather*}
(\varphi_{p}(x'(t)))'+\phi(t)f(t,x(t),x'(t))=0,\quad  0<t<+\infty,\\
\alpha x(0)-\beta x'(0)=0,x'(\infty)=0.
\end{gather*}

Guo and Yu  \cite{g2} establish the existence of three positive solutions for
$m$-point BVPs on infinite intervals
\begin{gather*}
(\varphi_{p}(x'(t)))'+\phi(t)f(t,x(t),x'(t))=0, \quad 0<t<+\infty,\\
x(0)=\sum_{i=1}^{m-2} \alpha_ix'(\eta_i), \quad
\lim_{t\to +\infty}x'(t)=0,
\end{gather*}
using the Avery-Peterson fixed point theorem in a cone.

Due to the fact that an infinite interval is noncompact, the
discussion about BVPs on the half-line is more complicated, in
particular, for the BVPs with integral boundary conditions on
infinite intervals, few works were done.

Motivated by the results \cite{g2,l2},  we will study the
following BVPs with integral conditions:
\begin{equation} \label{e1.1}
\begin{gathered}
(\varphi_{p}(x'(t)))'+\phi(t)f(t,x(t),x'(t))=0, \quad 0<t<+\infty, \\
x(0)=\int_0^{+\infty}g(s)x'(s)ds,\quad
\lim_{t\to +\infty}x'(t)=0,
\end{gathered}
\end{equation}
where $\varphi_{p}(s)=|s|^{p-2}s$, $p>1$,
$\phi: \mathbb{R}_{+}\to \mathbb{R}_{+}$,
$f(t,u,v):\mathbb{R}_{+}^{3}\to \mathbb{R}_{+}$ is a continuous function,
$\mathbb{R}_{+}=[0,+\infty)$, $g\in L^{1}[0,+\infty)$ is nonnegative.

In this article, we use the following conditions:
\begin{itemize}

\item[(H1)] $\phi\in C(\mathbb{R}_{+},\mathbb{R}_{+})$,
$\phi \not\equiv 0$ on any interval of the form $(t_0,+\infty)$
and
\begin{align*}
\int_0^{+\infty}\phi(s)ds<+\infty,\quad \int_0^{+\infty}
\varphi^{-1}\Big(\int_{\tau}^{+\infty}\phi(s)ds\Big)d\tau<+\infty;
\end{align*}

\item[(H2)] $f(t,(1+t)u,v)\in C(\mathbb{R}_{+}^{3},\mathbb{R}_{+}),f(t,0,0)\neq 0$ on
any subinterval of $(0,+\infty)$ and when $u,v$ are bounded
$f(t,(1+t)u,v)$ is bounded on $[0,+\infty)$.
\end{itemize}

\section{Preliminary results}

In this section, we present some definitions, theorems
and lemmas, which will be needed in the proof of the main results.
We first give the Avery-Peterson fixed point theorem.

\begin{definition} \label{def2.1}\rm
 Let $E$ be a real Banach space. A nonempty
closed convex set $P\subset E$ is called a cone if it satisfies the
following two conditions:
\begin{itemize}
\item[(i)] $x\in P$ and $\lambda\geq 0$ imply $\lambda x\in P$;

\item[(ii)] $x\in P$ and $-x\in P$ imply $x=0$.
\end{itemize}
\end{definition}

\begin{definition} \label{def2.2}\rm
 Given a cone $P$ in a real Banach  space $E$.
 A continuous map $\psi$ is called a concave
(resp. convex) functional on $P$ if  for all
 $x,y\in P$ and $0\leq \lambda\leq 1$, it holds
$\psi(\lambda x+(1-\lambda)y)\geq
\lambda\psi(x)+(1-\lambda)\psi(y)$,
(resp. $\psi(\lambda x+(1-\lambda)y)\leq
\lambda\psi(x)+(1-\lambda)\psi(y)$).
\end{definition}

Let $\alpha,\gamma,\theta,\psi$ be nonnegative continuous maps on
$P$ with $\alpha$ concave, $\gamma,\theta$ convex. Then for positive
numbers $a,b,c,d$, we define the following subsets of $P$:
\begin{gather*}
P(\gamma^{d})=\{x\in P:\gamma(x)<d\}; \\
P(\alpha_{b},\gamma^{d})=\{x\in\overline{P(\gamma^{d})}:b\leq\alpha(x)\};\\
P(\alpha_{b},\theta^{c},\gamma^{d})
=\{x\in\overline{P(\gamma^{d})}:b\leq\alpha(x),\theta(x)\leq c\};\\
R(\psi_{a},\gamma^{d})=\{x\in\overline{P(\gamma^{d})}:a\leq\psi(x)\}.
\end{gather*}
It is obvious that
 $P(\gamma^{d})$, $P(\alpha_{b}$, $\gamma^{d})$, $P(\alpha_{b},\theta^{c},\gamma^{d})$
are convex and $R(\psi_{a},\gamma^{d})$ is closed.

\begin{theorem}[\cite{a3}] \label{thm2.1}
Let $P$ be a cone of a real Banach space $E$. Let $\gamma,\theta$ be nonnegative
convex functional on $P$ satisfying
\begin{gather*}
\psi(\lambda x)\leq \lambda\psi(x), \quad \forall 0\leq\lambda\leq 1,\\
\alpha(x)\leq \psi(x),\quad \|x\|\leq M\gamma(x)\quad \forall
x\in\overline{P(\gamma^{d})}
\end{gather*}
with $M,d$ some positive numbers. Suppose that
$T:\overline{P(\gamma^{d})}\to\overline{P(\gamma^{d})}$ is
completely continuous and there exist positive numbers $a,b,c$ with
$a<b$ such that
\begin{itemize}
\item[(1)] $\{x\in P(\alpha_{b},\theta^{c},\gamma^{d}):\alpha(x)>b\}\neq\emptyset$ and
$\alpha(Tx)>b$ for $x\in P(\alpha_{b},\theta^{c},\gamma^{d})$;

\item[(2)] $\alpha(Tx)>b$ for $x\in P(\alpha_{b},\gamma^{d})$ with
$\theta(Tx)>c$;

\item[(3)] $0\not\in R(\psi_{a},\gamma^{d})$ and $\psi(Tx)<a$ for
$x\in R(\psi_{a},\gamma^{d})$ with $\psi(x)=a$.
\end{itemize}
Then $T$ has at least three fixed points
$x_1,x_{2},x_{3}\in\overline{P(\gamma^{d})}$ such that
$\gamma(x_{i})\leq d$, $i=1,2,3$;
$\psi(x_1)<a$; $\psi(x_{2})>a$ with
$\alpha(x_{2})<b$;  $\alpha(x_{3})>b$.
\end{theorem}

Consider the space
 \begin{equation} \label{e2.1}
 X=\big\{x\in C^{1}[0,+\infty),\sup_{0\leq t<+\infty}
 \frac{|x(t)|}{1+t}<+\infty,\lim_{t\to  +\infty}x'(t)=0\big\}
 \end{equation}
 with the norm $\|x\|=\max\{\|x\|_1,\|x'\|_{\infty}\}$, where
$\|x\|_1=\sup_{0\leq t<+\infty}|x(t)|/(1+t)$,
$\|x'\|_{\infty}=\sup_{0\leq t<+\infty} |x'(t)|$.
By using the standard arguments, we can obtain that
 $(X,\|\cdot\|)$ is a Banach space.
Define the $P\subset X$ by
\begin{equation} \label{e2.2}
\begin{split}
P=\Big\{& x\in X:x(t)\geq0,t\in[0,+\infty),\\
&x(0)=\int_0^{+\infty}g(s)x'(s)ds, x
\text{ is concave on }[0,+\infty)\Big\}.
\end{split}
\end{equation}

\begin{remark} \label{rmk2.1}\rm
 If $x$ satisfies \eqref{e1.1}, then
$(\varphi_{p}(x'(t)))'=-\phi(t)f(t,x(t),x'(t))\leq 0$ on
$[0,+\infty)$, which implies that $x$ is concave on $[0,+\infty)$.
Moreover, if $\lim_{t\to +\infty}x'(t)=0$,  then
$x'(t)\geq 0,t\in[0,+\infty)$ and so $x$ is monotone increasing on
$[0,+\infty)$.
\end{remark}

Let $k>1$ be a constant.  For $x\in P$, define the nonnegative continuous
functionals:
\begin{equation} \label{e2.3}
\begin{gathered}
\alpha(x)=\frac{k}{k+1}\min_{\frac{1}{k}\leq t<+\infty}x(t), \quad
\gamma(x)=\sup_{0\leq t<+\infty}x'(t),\\
\psi(x)=\theta(x)=\sup_{0\leq t<+\infty}\frac{x(t)}{1+t},\quad
A=\int_0^{+\infty}g(s)ds,\\
C=\varphi_{p}^{-1}\Big(\int_0^{+\infty}\phi(s)ds\Big),\quad
C_1(t)=\int_0^t\varphi_{p}^{-1}
\Big(\int_{\tau}^{+\infty}\phi(s)ds\Big)d\tau.
\end{gathered}
\end{equation}

Since the Arzela-Ascoli theorem does not apply in the space $X$, we
need a modified compactness criterion to prove $T$ is compact. In
the following, we present an explicit one. For more general cases,
we refer the readers to \cite{a2,z1} and the reference therein.

\begin{definition} \label{def2.3} \rm
For $l>0$, let $V=\{x\in X:\|x\|<l\}$, and
$$
V_1:=\{\frac{x(t)}{1+t},x\in V\}\cup\{x'(t),x\in V\}
$$
which is called equiconvergent at infinity if  for all
$\varepsilon>0$
there exists $T=T(\varepsilon)>0$ such that for all
$x\in V_1$,
\[
\big|\frac{x(t_1)}{1+t_1}-\frac{x(t_{2})}{1+t_{2}}\big|<\varepsilon,\quad
|x'(t_1)-x'(t_{2})|<\varepsilon, \quad \forall t_1, t_{2}\geq T.
\]
\end{definition}

\begin{lemma}[\cite{l3}] \label{lem2.1}
If $\{\frac{x(t)}{1+t},x\in V\}$ and
$\{x'(t),x\in V\}$ are both equicontinuous on any compact interval
of $[0,+\infty)$ and equiconvergent at infinity . Then $V$ is
relatively compact on $X$.
\end{lemma}

\begin{lemma} \label{lem2.2}
Let $g\in L^{1}[0,+\infty)$ and $g$ is nonnegative, if $v(t)$ is nonnegative
and continuous on
$[0,+\infty)$ and $\lim_{t\to +\infty}v(t)$ exists.
Then there exists at least one $\eta$, $0\leq\eta\leq+\infty$ such
that
\begin{equation} \label{e2.4}
\int_0^{+\infty}g(s)v(s)ds=v(\eta)\int_0^{+\infty}g(s)ds.
\end{equation}
\end{lemma}

\begin{proof}
 It is obvious that the function $v(t)$ exists and has
maxima and minima which are nonnegative and noted by
$M^{*},m^{*}$ on $[0,+\infty)$, then for all $t\in[0,+\infty)$,
we have $m^{*}\leq v(t)\leq M^{*}$, so
\[
 m^{*}\int_0^{+\infty}g(s)ds\leq \int_0^{+\infty}g(s)v(s)ds \leq
M^{*}\int_0^{+\infty}g(s)ds.
\]
If $\int_0^{+\infty}g(s)ds=0$, the result is clear;
If $\int_0^{+\infty}g(s)ds>0$, there is
\[
m^{*}\leq\frac{\int_0^{+\infty}g(s)v(s)ds}
{\int_0^{+\infty}g(s)ds} \leq M^{*}.
\]
Therefore, there exists at least one $\eta$, $0\leq\eta\leq+\infty$ such that
\begin{equation} \label{e2.5}
\int_0^{+\infty}g(s)v(s)ds=v(\eta)\int_0^{+\infty}g(s)ds.
\end{equation}
\end{proof}

\begin{lemma} \label{lem2.3}
Let $y\in C[\mathbb{R}_{+},\mathbb{R}_{+}]$, and
$\int_0^{+\infty}y(t)dt<\infty$, then the boundary-value problem
\begin{equation} \label{e2.6}
\begin{gathered}
 (\varphi_{p}(x'(t)))'+y(t)=0, \quad 0<t<+\infty,\\
x(0)=\int_0^{+\infty}g(s)x'(s)ds,\quad  \lim_{t\to +\infty}x'(t)=0,
\end{gathered}
\end{equation}
has a unique solution
\[
x(t)=\int_0^{+\infty}g(s)\varphi_{p}^{-1}
\Big(\int_{s}^{+\infty}y(\tau)d\tau\Big)ds+\int_0^t\varphi_{p}^{-1}
\Big(\int_{s}^{+\infty}y(\tau)d\tau\Big)ds.
\]
\end{lemma}

Define the operator $T:P\to C^{1}[0,+\infty)$ by
\begin{equation} \label{e2.7}
\begin{split}
(Tx)(t)&=\int_0^{+\infty}g(s)\varphi_{p}^{-1}
\Big(\int_{s}^{+\infty}\phi(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds\\
&\quad +\int_0^t\varphi_{p}^{-1}\Big(\int_{s}^{+\infty}\phi(\tau)f(\tau,x(\tau),
x'(\tau))d\tau\Big)ds,\quad t\in[0,+\infty).
\end{split}
\end{equation}

\begin{lemma} \label{lem2.4}
For $x\in P$, $\|x\|_1\leq M\|x'\|_{\infty}$,
where $M=\max\big\{\int_0^{+\infty}g(s)ds,1\big\}$.
\end{lemma}

\begin{proof}
Since $x\in P$, by Lemma \ref{lem2.2},
\begin{align*}
\frac{x(t)}{1+t}
&=\frac{1}{1+t}(\int_0^tx'(s)ds+\int_0^{+\infty}g(s)x'(s)ds)\\
&\leq\frac{t+\int_0^{+\infty}g(s)ds}{1+t}\|x'\|_{\infty}\leq
M\|x'\|_{\infty}.
\end{align*}
The result follows.
\end{proof}

\begin{lemma}[\cite{g2}] \label{lem2.5}
 For $x\in P$, $\alpha(x)\geq\frac{1}{k+1}\theta(x)$.
\end{lemma}


\begin{lemma} \label{lem2.6}
Let {\rm (H1)--(H2)} hold. Then
$T:P\to P$ is completely continuous.
\end{lemma}

\begin{proof}
It is easy to see that $T:P\to P$ is well defined.
 Now we prove that $T$ is continuous and compact
respectively. Let $x_{n}\to x$ as $n\to +\infty$ in
$P$, then there exists $r_0$ such that
 $\sup_{n\in N\backslash\{0\}}\|x_{n}\|<r_0$. Set
$B_{r_0}=\sup\{f(t,(1+t)u,v),(t,u,v)\in
[0,+\infty)\times[0,r_0]^{2}\}$. Then we have
\[
\int_0^{+\infty}\phi(s)|f(s,x_{n},x_{n}')-f(s,x,x')|ds\leq
2B_{r_0}\int_0^{+\infty}\phi(s)ds.
\]
Therefore, by the Lebesgue dominated convergence theorem,
\begin{align*}
|\varphi_{p}((Tx_{n})'(t))-\varphi_{p}((Tx)'(t))|
&=\big|\int_{t}^{+\infty}\phi(s)(f(s,x_{n},x_{n}^{'})-f(s,x,x'))ds\big|\\
&\leq\int_0^{+\infty}\phi(s)|f(s,x_{n},x_{n}^{'})-f(s,x,x')|ds\\
&\to 0,\quad\text{as }n\to +\infty.
\end{align*}
Furthermore,
$\|Tx_{n}-Tx\|\leq M\|(Tx_{n})'-(Tx)'\|_{\infty}\to 0$, as
$n\to +\infty$. Hence, $T$ is continuous.

$T$ is compact provided that it maps bounded sets into relatively
compact sets. Let $\Omega$ be any bounded subset of $P$. Then there
exists $r>0$ such that $\|x\|\leq r$ for all $x\in\Omega$.
Obviously,
\[
\|(Tx)'\|_{\infty}=\varphi_{p}^{-1}(\int_0^{+\infty}\phi(s)f(s,x(s),x'(s))ds)\leq
C\varphi_{p}^{-1}(B_{r})
\]
for all $x\in\Omega$. Hence,
 $\|T\Omega\|\leq MC\varphi_{p}^{-1}(B_{r})$. So $T\Omega$ is bounded.

Moreover, for any $L\in(0,+\infty)$ and $t_1,t_{2}\in[0,L]$,
\begin{align*}
&\Big|\frac{(Tx)(t_1)}{1+t_1}-\frac{(Tx)(t_{2})}{1+t_{2}}\Big|\\
&\leq \int_0^{+\infty}g(s)\varphi_{p}^{-1}
 \Big(\int_{s}^{+\infty}\phi(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds
\Big|\frac{1}{1+t_1}-\frac{1}{1+t_{2}}\Big|\\
&\quad +\int_0^{t_{2}}\varphi_{p}^{-1}
\Big(\int_{s}^{+\infty}\phi(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds
\big|\frac{1}{1+t_1}-\frac{1}{1+t_{2}}\big|\\
&\quad +\frac{1}{1+t_1}|\int_{t_1}^{t_{2}}\varphi_{p}^{-1}
\Big(\int_{s}^{+\infty}\phi(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds|
\\
&\leq \varphi_{p}^{-1}(B_{r})(AC+C_1(L))|t_1-t_{2}|+|C_1(t_1)-C_1(t_{2})|)\\
&\to 0,\quad\text{uniformly as }t_1\to t_{2},
\end{align*}
and
\begin{align*}
|\varphi_{p}((Tx)'(t_1))-\varphi_{p}((Tx)'(t_{2}))|
&=\big|\int_{t_1}^{t_{2}}\phi(s)(f(s,x,x')ds\big|\\
&\leq B_{r}|\int_{t_1}^{t_{2}}\phi(s)ds|
\to 0, \quad \text{uniformly as }t_1\to t_{2},
\end{align*}
for all $x\in \Omega$. So $T\Omega$ is equicontinuous on any compact
interval of $[0,+\infty)$.

Finally, for any $x\in \Omega$,
\begin{align*}
\lim_{t\to  +\infty}|\frac{(Tx)(t)}{1+t}|
&=\lim_{t\to  +\infty}\frac{1}{1+t}\int_0^t\varphi_{p}^{-1}
\Big(\int_{s}^{+\infty}\phi(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds\\
&\leq M\varphi_{p}^{-1}(B_{r})\lim_{t\to  +\infty}\varphi_{p}^{-1}
\Big(\int_{t}^{+\infty}\phi(s)ds\Big)=0,
 \\
\lim_{t\to +\infty}|(Tx)'(t)|
&=\lim_{t\to  +\infty}\varphi_{p}^{-1}(\int_{t}^{+\infty}\phi(s)f(s,x(s),x'(s))ds)\\
&\leq \varphi_{p}^{-1}(B_{r})\lim_{t\to  +\infty}\varphi_{p}^{-1}
\Big(\int_{t}^{+\infty}\phi(s)ds\Big)=0.
\end{align*}
So $T\Omega$ is equiconvergent at infinity. By using Lemma \ref{lem2.1}, we
obtain that $T\Omega$ is relatively compact, that is, $T$ is a
compact operator.
Hence, $T:P\to P$ is completely continuous. The proof is
complete.
\end{proof}


\section{Main results}

For the main result of this article we sue the hypothesis
\begin{itemize}
\item[(H3)] $f(t,(1+t)u,v)\leq \varphi_{p}(d/C)$, for 
$(t,u,v)\in [0,+\infty)\times[0,Md]\times[0,d]$;

\item[(H4)] $f(t,(1+t)u,v)>\varphi_{p}(b/N)$, for 
$(t,u,v)\in [\frac{1}{k},k]\times[\frac{b}{k},\frac{(k+1)^{2}b}{km}]\times[0,d]$;

\item[(H5)] $f(t,(1+t)u,v)<\varphi_{p}(a/MC)$, for 
$(t,u,v)\in [0,+\infty)\times[0,a]\times[0,d]$;
where
\[
m=\min\{A,1\},\quad 
N=\frac{1}{(k+1)^{2}}\int_{\frac{1}{k}}^{k}(g(s)+1)\varphi_{p}^{-1}
\Big(\int_{s}^{k}\phi(\tau)d\tau\Big)ds.
\]
\end{itemize}

\begin{theorem} \label{thm3.1} 
Let $A>0$. Suppose {\rm (H1)--(H5)} hold.
Suppose further that there exist numbers $a,b,d$ such that
$0<ka<b\leq M m kd/(k+1)^{2}$.
Then \eqref{e1.1} has at least three positive solutions $x_1,x_{2},x_{3}$
such that
\begin{equation} \label{e3.1}
\begin{gathered}
\sup_{0\leq t<+\infty}x'_{i}(t)\leq d,\quad i=1,2,3;\\
\sup_{0\leq t<+\infty}\frac{x_1(t)}{1+t}<a, \quad
a<\sup_{0\leq t<+\infty}\frac{x_{2}(t)}{1+t}<\frac{(k+1)^{2}b}{km}, \quad
\min_{\frac{1}{k}<t<k}x_{2}(t)<\frac{(k+1)}{k}b;
\\
\sup_{0\leq t<+\infty}\frac{x_{3}(t)}{1+t}< Md, \quad
\min_{\frac{1}{k}\leq t<k}x_{3}(t)>\frac{(k+1)}{k}b.
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof} 
Let $X,P,\alpha,\gamma,\theta,\psi$ and $T$ be
defined as \eqref{e2.1}-\eqref{e2.3} and \eqref{e2.7} respectively. 
It is easy to prove that the fixed points of $T$ coincide with the solution of BVP
\eqref{e1.1}. So it is enough to show that $T$ has three positive fixed
points.

In fact, for any 
$x\in\overline{P(\gamma^{d})}$,
$\sup_{0\leq t<+\infty}x'(t)\leq d$ and so
$\sup_{0\leq t<+\infty}\frac{x(t)}{1+t}\leq Md$. 
Condition (H3) implies that
$f(t,x(t),x'(t))\leq \varphi_{p}(d/C)$ for all $t\in [0,+\infty)$.
Therefore,
\begin{align*}
\gamma(Tx)
&=\sup_{0\leq t<+\infty}(Tx)'(t)=(Tx)'(0)\\
&=\varphi_{p}^{-1}\Big(\int_0^{+\infty}\phi(s)f(s,x(s),x'(s))ds\Big)\\
&\leq\frac{d}{C}\varphi_{p}^{-1}\Big(\int_0^{+\infty}\phi(s)ds\Big)=d.
\end{align*}
Hence
$T:\overline{P(\gamma^{d})}\to\overline{P(\gamma^{d})}$ is
completely continuous.

Obviously, $\alpha,\gamma,\theta,\psi$ satisfy the assumptions in
Theorem \ref{thm2.1}. Next we show that conditions (1)-(3) in
 Theorem \ref{thm2.1} hold.

Firstly, choose the function
$x(t)=(1-\frac{1}{k+1}e^{-\frac{k}{A}t})\frac{(k+1)^{2}}{k}b,0\leq
t<+\infty$. It can be checked that 
$x\in P(\alpha_{b},\theta^{c},\gamma^{d})$ with $\alpha(x)>b$, where
$c=\frac{(k+1)^{2}}{km}b$. 
So $\{x\in P(\alpha_{b},\theta^{c},\gamma^{d})|\alpha(x)>b\}\neq\emptyset$. For
any $x\in P(\alpha_{b},\theta^{c},\gamma^{d})$, we obtain
\[
\frac{b}{k}\leq\frac{1}{1+k}\min_{\frac{1}{k}\leq t\leq
k}x(t)\leq\frac{x(t)}{1+k}\leq\frac{x(t)}{1+t}\leq\frac{(k+1)^{2}}{km}b,
\quad t\in\Big[\frac{1}{k},k\Big],
\]
and $0\leq x'(t)\leq d,t\in[0,+\infty)$. In view of assumption
(H4) together with Lemma \ref{lem2.5}, we obtain 
\begin{align*}
\alpha(Tx)
&\geq\frac{1}{k+1}\theta(Tx)
=\frac{1}{k+1}\sup_{0\leq t<+\infty}\frac{(Tx)(t)}{1+t} \\
&=\frac{1}{(k+1)}\sup_{0\leq t<+\infty}\frac{1}{1+t}
\Big[\int_0^{+\infty}g(s)\varphi_{p}^{-1}
\Big(\int_{s}^{+\infty}\phi(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds \\
&\quad +\int_0^t\varphi_{p}^{-1}\Big(\int_{s}^{+\infty}\phi(\tau)
 f(\tau,x(\tau),x'(\tau))d\tau\Big)ds\Big] \\
&\geq\frac{1}{(k+1)^{2}}\Big[\int_{\frac{1}{k}}^{k}g(s)\varphi_{p}^{-1}
\Big(\int_{s}^{k}\phi(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds \\
&\quad +\int_{\frac{1}{k}}^{k}\varphi_{p}^{-1}\Big(\int_{s}^{k}\phi(\tau)
f(\tau,x(\tau),x'(\tau))d\tau\Big)ds\Big] \\
&>\frac{b}{N}\frac{1}{(k+1)^{2}}\int_{\frac{1}{k}}^{k}(g(s)+1)\varphi_{p}^{-1}
\Big(\int_{s}^{k}\phi(\tau)d\tau\Big)ds=b.
\end{align*}
Hence, $\alpha(Tx)>b$ for $x\in P(\alpha_{b},\theta^{c},\gamma^{d})$.

Next we will verify that the condition (2) of Theorem \ref{thm2.1} is
satisfied. Let $x\in P(\alpha_{b},\gamma^{d})$ with $\theta(Tx)>c$,
it follows from Lemma \ref{lem2.5} that
\[
\alpha(Tx)\geq\frac{1}{k+1}\theta(Tx)>\frac{1}{k+1}c
=\frac{1}{k+1}\frac{(k+1)^{2}}{km}b=\frac{(k+1)}{km}b>b,
\]
Thus $\alpha(Tx)>b$ for all $x\in P(\alpha_{b},\gamma^{d})$ with
$\theta(Tx)>c$.

Finally, we show that condition (3) of theorem \ref{thm2.1} is satisfied.
It is clear that $0\in R(\psi_{a},\gamma^{d})$. Suppose that 
$x\in R(\psi_{a},\gamma^{d})$ with $\psi(x)=a$, then by condition
(H5) and Lemma \ref{lem2.4}, we obtain
\begin{align*}
\psi(Tx)&\leq M\gamma(Tx)=M(Tx)'(0)\\
&=M\varphi_{p}^{-1}(\int_0^{+\infty}\phi(s)f(s,x(s),x'(s))ds) \\
&\leq M\cdot\frac{a}{MC}\varphi_{p}^{-1}(\int_0^{+\infty}\phi(s)ds=a.
\end{align*}
Therefore, $T$ has at least three fixed points 
$x_1,x_{2},x_{3}\in \overline{P(\gamma^{d})}$ such that
\[
\psi(x_1)<a,\quad \psi(x_{2})>a \quad\text{with }
\alpha(x_{2})<b,\;\alpha(x_{3})>b.
\]
In addition, condition (H2) guarantees that those fixed points
are positive. So \eqref{e1.1} has at least three positive solutions
$x_1,x_{2},x_{3}$ satisfying \eqref{e3.1} and the proof is complete.
\end{proof}

\section{Example}

Consider the boundary-value problem with integral boundary
value conditions
\begin{equation} \label{e4.1}
\begin{gathered}
(|x'|x')'+e^{-t}f(t,x(t),x'(t))=0, \\
x(0)=\int_0^{+\infty}e^{-2s}x'(s)ds,\quad
\lim_{t\to +\infty}x'(t)=0,
\end{gathered}
\end{equation}
where
\[
f(t,u,v)= \begin{cases}
\frac{|\sin t|}{100}+10^{4}\big(\frac{u}{1+t}\big)^{10}
+\frac{1}{100}\big(\frac{v}{200}\big),\quad u\leq1,\\
\frac{|\sin t|}{100}+10^{4}\big(\frac{1}{1+t}\big)^{10}
+\frac{1}{100}\big(\frac{v}{200}\big),\quad u\geq1.
\end{cases}
\]
Set $\phi(t)=e^{-t}$ and it is easy to verify that (H1) and
(H2) hold. Choose $k=4$, $a=\frac{1}{4}$, $b=2$, $d=200$. Then by simple
calculations, we can obtain
$M=1$, $m=\frac{1}{2}$,
\[
C=1,N=\frac{1}{25}\int_{\frac{1}{4}}^{4}(e^{-2s}+1)
\sqrt{e^{-s}-e^{-4}}ds\geq\frac{1}{25}\int_{\frac{1}{4}}^{4}\sqrt{e^{-s}-e^{-4}}ds>
\frac{1}{48}.
\]
So the nonlinear term $f$ satisfies
\begin{itemize}
\item[(1)] $f(t,(1+t)u,v)\leq 0.01+10^{4}+0.01<4\times
10^{4}=\varphi_{3}(d/C)$, for $(t,u,v)\in
[0,+\infty)\times[0,200]^{2}$;

\item[(2)] $f(t,(1+t)u,v)\geq 10^{4}>96^{2}=\varphi_{3}(b/N)$,
for $(t,u,v)\in [\frac{1}{4},4]\times[\frac{1}{2},25]\times[0,200]$;

\item[(3)] $f(t,(1+t)u,v)\leq 0.01
+10^{4}\times\big(\frac{1}{4}\big)^{10}+0.01<\frac{1}{16}=
\varphi_{3}(a/MC)$, for $(t,u,v)\in
[0,+\infty)\times[0,\frac{1}{4}]\times[0,200]$.
\end{itemize}

Therefore, the conditions in Theorem \ref{thm3.1} are all satisfied. So 
\eqref{e4.1} has at least three positive solutions $x_1,x_{2},x_{3}$ such
that
\begin{gather*}
\sup_{0\leq t<+\infty}x'_{i}(t)\leq 200,\quad i=1,2,3;\\
\sup_{0\leq t<+\infty}\frac{x_1(t)}{1+t}\leq\frac{1}{4}, \quad 
\frac{1}{2}<\sup_{0\leq t<+\infty}\frac{x_{2}(t)}{1+t}< 25, \quad
\min_{\frac{1}{k}\leq t \leq k}x_{2}(t)\leq \frac{5}{2};\\
\sup_{0\leq t<+\infty}\frac{x_{3}(t)}{1+t}\leq 200\quad
\min_{\frac{1}{k}\leq t \leq k}x_{3}(t)>\frac{5}{2}.
\end{gather*}

\subsection*{Acknowledgements}
This research was supported by grants: 10901045 from
the Natural Science Foundation of China,
A2009000664 and A2011208012 from the Natural Science Foundation of
Hebei Province,  and XL201047 from the
Foundation of Hebei University of Science and Technology.


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\end{document}